IndexTAG: 0 TitleTAG: A prize for Myrimit I want to propose that MITx or edX give a prize or certificate to Myrimit due to her great job helping fellows to understand all topics on the course, even if the rest of the CTA helped, Myrimit was amazing. So if you agree with my propose, please vote here. Greetings from Venezuela Merry Xmas and happy new year UserIdTAG: 296419 UserNameTAG: oscman CreateTimeTAG: 2012-12-24T22:26:11Z VoteTAG: 94 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 33 FirstChildTAG: I do agree FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-12-24T22:31:18Z FirstChildTAG: Count me in ! FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-12-24T22:43:56Z FirstChildTAG: Absolutely! FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-12-24T22:48:19Z FirstChildTAG: I agree. FirstChildUserIdTAG: 364126 FirstChildUserNameTAG: shunyi FirstChildCreateTimeTAG: 2012-12-25T10:38:25Z FirstChildTAG: Absolutely! I think she'd be happiest if that certificate was an admit card to MIT! :) But I'm sure she would still be happy with anything! Borrowing Myriam's term, haha! :) FirstChildUserIdTAG: 232157 FirstChildUserNameTAG: GayatriTR FirstChildCreateTimeTAG: 2012-12-25T03:36:33Z FirstChildTAG: Vote me in! FirstChildUserIdTAG: 303762 FirstChildUserNameTAG: leocarrasco FirstChildCreateTimeTAG: 2012-12-24T23:45:51Z FirstChildTAG: I'm agree. That is a GREAT IDEA! :) FirstChildUserIdTAG: 110802 FirstChildUserNameTAG: leoblack FirstChildCreateTimeTAG: 2012-12-24T23:51:35Z SecondChildTAG: Agree SecondChildUserIdTAG: 368981 SecondChildUserNameTAG: asimakhtar SecondChildCreateTimeTAG: 2012-12-25T02:14:03Z FirstChildTAG: Excelent idea oscman. Of course, I am agree with this idea!!! Merry Christmas and Happy New Year too. Luis Excelente idea oscman. Por supuesto, estoy de acuerdo con esta idea!!! Feliz Navidad y Feliz Año Nuevo. Luis FirstChildUserIdTAG: 227223 FirstChildUserNameTAG: luisbonelli FirstChildCreateTimeTAG: 2012-12-25T04:42:06Z FirstChildTAG: Great Idea :)... FirstChildUserIdTAG: 122513 FirstChildUserNameTAG: Mona77 FirstChildCreateTimeTAG: 2012-12-25T06:32:33Z FirstChildTAG: I fully agree with oscman. FirstChildUserIdTAG: 279687 FirstChildUserNameTAG: TanujDhawan FirstChildCreateTimeTAG: 2012-12-25T06:33:14Z FirstChildTAG: I agree with oscman 1000%. Myrimit is in her own league of helping out other students. God bless Myrimit. FirstChildUserIdTAG: 183507 FirstChildUserNameTAG: obiradaniel FirstChildCreateTimeTAG: 2012-12-25T07:43:16Z FirstChildTAG: AGREE..!!!!!!!!! FirstChildUserIdTAG: 316761 FirstChildUserNameTAG: gk_goel FirstChildCreateTimeTAG: 2012-12-25T08:38:01Z FirstChildTAG: Agree :-) FirstChildUserIdTAG: 398594 FirstChildUserNameTAG: DaveyJC FirstChildCreateTimeTAG: 2012-12-24T22:56:15Z FirstChildTAG: I 'm with OSCMAN Greetings from Yemen FirstChildUserIdTAG: 318037 FirstChildUserNameTAG: Maher-84 FirstChildCreateTimeTAG: 2012-12-25T11:07:05Z FirstChildTAG: I agree :-) FirstChildUserIdTAG: 222098 FirstChildUserNameTAG: Avneendra FirstChildCreateTimeTAG: 2012-12-25T12:27:14Z FirstChildTAG: I do agree :) FirstChildUserIdTAG: 244706 FirstChildUserNameTAG: Miguel_Angel FirstChildCreateTimeTAG: 2012-12-25T12:27:26Z FirstChildTAG: I do agree!! Henry FirstChildUserIdTAG: 175734 FirstChildUserNameTAG: hestrada FirstChildCreateTimeTAG: 2012-12-25T12:57:02Z FirstChildTAG: Wonderfull initiative! I agree FirstChildUserIdTAG: 220837 FirstChildUserNameTAG: Calsomus FirstChildCreateTimeTAG: 2012-12-25T15:32:15Z FirstChildTAG: I do agree dudes FirstChildUserIdTAG: 145544 FirstChildUserNameTAG: pandiya FirstChildCreateTimeTAG: 2012-12-25T07:09:21Z FirstChildTAG: i agree. FirstChildUserIdTAG: 357747 FirstChildUserNameTAG: kishores FirstChildCreateTimeTAG: 2012-12-25T14:31:47Z FirstChildTAG: I agree as well! Besides being a Community TA, I was also a student this semester, and I've utilized **Myrimit**'s resources and study guides on more than one occasion when I didn't have time to watch a week's lecture. Congratulations **Myrimit**! Jersey Mark FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-12-25T18:26:33Z FirstChildTAG: Hi oscman, I am blushing haha. I am really happy for all this nice words from all my Classmates. Thank you very much. I have to say, that I am already, as I always say, happy. All this for me is a reward: all this kind gesture from my classmates, this nice words really have touched my heart, this is really valuable for me; the opportunity that edX gave me, no mattering my last Term grade - I got a happy B, to be a CTA this Fall; also all the support that edX gave us in order to run the Contest again, thank you. Also, some students have opened their doors to their Countries if I were near there someday haha, so I am welcome to Spain, Brazil, India and China, so what else can I ask ? I feel like the happiest student of all the world haha ! Thank you all my Classmates for all this nice words, I don´t want to cry -I am sensitive most of the time-:p. I hope to have reach to all of you, so that you could be inspired too to contribute, no mattering being or not CTA, with the new students in the future Courses of edX :), that would be so cool! Thank you edX for all this amazing experience of learning! All of you are awesome! Keep doing this wonderful work always! Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-25T23:58:40Z SecondChildTAG: Myriam is a pleasure for me to ask for a reward for your big contribution to this community, again thanks for all your help and you are welcome to Venezuela when you want to come here. Best Regards Oscar SecondChildUserIdTAG: 296419 SecondChildUserNameTAG: oscman SecondChildCreateTimeTAG: 2012-12-26T17:16:14Z SecondChildTAG: Myrimit, thanks for everything... And as you said if you want to come here in Brazil, we are going to be waiting for you with arms wide open... Success in your life... =) SecondChildUserIdTAG: 128634 SecondChildUserNameTAG: BrunoMoraes SecondChildCreateTimeTAG: 2013-01-05T00:33:16Z FirstChildTAG: agree :) FirstChildUserIdTAG: 108863 FirstChildUserNameTAG: shiviz FirstChildCreateTimeTAG: 2012-12-26T04:58:17Z FirstChildTAG: Since Myriam is everyone's princess from Argentina, I looked this up as a present for her. Maybe this Argentine music, Adios Nonino can make you cry from happiness, like our princess Maxima (also from Argentina) at her wedding 10 yrs ago. A beautiful interpretation. http://www.youtube.com/watch?v=bJWD_789-OE FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-12-26T17:48:35Z SecondChildTAG: I prefer this ;) http://www.youtube.com/watch?v=4r8sz23qvHA SecondChildUserIdTAG: 244706 SecondChildUserNameTAG: Miguel_Angel SecondChildCreateTimeTAG: 2012-12-26T18:17:07Z SecondChildTAG: At least everyone applauded! SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-12-26T18:23:57Z SecondChildTAG: Thank you salsero, I really like Adios Nonino :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-26T23:43:46Z FirstChildTAG: I agree!! FirstChildUserIdTAG: 355773 FirstChildUserNameTAG: Albright4edx FirstChildCreateTimeTAG: 2012-12-29T15:42:09Z FirstChildTAG: Yes,i will agree. FirstChildUserIdTAG: 371498 FirstChildUserNameTAG: Engineer001 FirstChildCreateTimeTAG: 2012-12-29T02:46:40Z FirstChildTAG: Helping others is not an opportunity given to everyone, May God bless you more...:) FirstChildUserIdTAG: 96011 FirstChildUserNameTAG: amanwalia92 FirstChildCreateTimeTAG: 2012-12-30T14:56:12Z FirstChildTAG: Yes! FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-12-25T03:39:23Z FirstChildTAG: Myrimit, thanks for everything... And as you said if you want to come here in Brazil, we are going to be waiting for you with arms wide open... Success in your life... =) FirstChildUserIdTAG: 128634 FirstChildUserNameTAG: BrunoMoraes FirstChildCreateTimeTAG: 2013-01-05T00:31:41Z SecondChildTAG: You are welcome, I wish you all the best to you too! =) Take care, Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2013-01-05T15:10:16Z FirstChildTAG: Absolutley agree! FirstChildUserIdTAG: 145540 FirstChildUserNameTAG: Anatoly FirstChildCreateTimeTAG: 2013-01-05T21:55:20Z FirstChildTAG: Me too! FirstChildUserIdTAG: 136489 FirstChildUserNameTAG: ShahabSafa FirstChildCreateTimeTAG: 2012-12-29T09:58:54Z FirstChildTAG: i agree FirstChildUserIdTAG: 533341 FirstChildUserNameTAG: adjmhatre FirstChildCreateTimeTAG: 2013-01-22T07:27:31Z FirstChildTAG: **I am strongly Agree , even any body make a small fund AC I am ready to transfer my contribution** MK.prasanth FirstChildUserIdTAG: 156694 FirstChildUserNameTAG: mkprasanth FirstChildCreateTimeTAG: 2013-03-11T11:38:25Z IndexTAG: 1 TitleTAG: Welcome to 6.002x! We hope you enjoy using our platform. :-) Dave (edX developer) UserIdTAG: 11 UserNameTAG: dormsbee CreateTimeTAG: 2012-09-05T03:27:34Z VoteTAG: 94 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 21 FirstChildTAG: Did I loose anything? I thought, about noon, there would be some online and live video or something alike. Is it noon only the time the course would be open with all the material available? FirstChildUserIdTAG: 175378 FirstChildUserNameTAG: Ivanildo FirstChildCreateTimeTAG: 2012-09-05T12:28:16Z SecondChildTAG: Noon is just when we opened. :-) SecondChildUserIdTAG: 11 SecondChildUserNameTAG: dormsbee SecondChildCreateTimeTAG: 2012-09-05T12:59:36Z FirstChildTAG: Thanks. It's pretty good, but I think, older platform was better. It was more clear and easy in use. Here in the discussion panel I can't find my discussion profil page :( Please add **"left panel minimize option"** in the courseware section. In all, thanks for your effort :) FirstChildUserIdTAG: 821 FirstChildUserNameTAG: mpaluch FirstChildCreateTimeTAG: 2012-09-05T12:29:58Z SecondChildTAG: We're actually working on a fairly significant overhaul of the discussion forums, of which this is only the first step. We've got a version in testing now and hope to have it out for you folks soon. SecondChildUserIdTAG: 11 SecondChildUserNameTAG: dormsbee SecondChildCreateTimeTAG: 2012-09-05T12:58:29Z FirstChildTAG: thanks so much MIT folks, you are doing a great cause. it should have began several months ago. im so thrilled. FirstChildUserIdTAG: 359504 FirstChildUserNameTAG: Sisbboy FirstChildCreateTimeTAG: 2012-09-05T12:34:54Z FirstChildTAG: I will move over a bunch of prior wiki stuff, probably some pages or a topic at a time (my mental 'down time'). Over the next few days. FirstChildUserIdTAG: 9147 FirstChildUserNameTAG: SueP FirstChildCreateTimeTAG: 2012-09-05T16:20:28Z FirstChildTAG: Thank you and to the staff for your effort. I'm so excited to take this course! FirstChildUserIdTAG: 147593 FirstChildUserNameTAG: knack FirstChildCreateTimeTAG: 2012-09-05T17:13:06Z FirstChildTAG: Thank you Dave! Thank you all the edX Staff! ;) FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-06T00:37:47Z FirstChildTAG: Thank You! FirstChildUserIdTAG: 88550 FirstChildUserNameTAG: Neil_S_Berry FirstChildCreateTimeTAG: 2012-09-06T01:45:13Z SecondChildTAG: u welcome SecondChildUserIdTAG: 474881 SecondChildUserNameTAG: linklist SecondChildCreateTimeTAG: 2012-09-21T18:38:13Z FirstChildTAG: Thanks very much for this opportunity! FirstChildUserIdTAG: 281552 FirstChildUserNameTAG: MartinRamos FirstChildCreateTimeTAG: 2012-09-06T03:46:47Z SecondChildTAG: is my pleasure SecondChildUserIdTAG: 474881 SecondChildUserNameTAG: linklist SecondChildCreateTimeTAG: 2012-09-21T18:38:21Z FirstChildTAG: Thank you so much. I am looking fordward to take this course! FirstChildUserIdTAG: 387521 FirstChildUserNameTAG: Horas FirstChildCreateTimeTAG: 2012-09-06T12:21:53Z FirstChildTAG: Thank you for an awesome product. Can I beg for one improvement? When we choose to make a video fullscreen, the video navigation menu sits across the middle of the video and obscures whatever is happening behind it. Can you move it to the top above the video? (or at least make it dragable?) Thanks, and keep up the good work. FirstChildUserIdTAG: 256569 FirstChildUserNameTAG: stridera FirstChildCreateTimeTAG: 2012-09-07T09:48:11Z FirstChildTAG: Yeah!wonderful experience indeed. It's really feeling great to learn whenever and however you want!But I can not find Where to talk with TA? I posted my query with the tag "staff help" but I am not able to see that if TA can see my question or did answer???? How can I present myself and get answer from Faculties? Please guide.Thank you :) FirstChildUserIdTAG: 281159 FirstChildUserNameTAG: ManasiS FirstChildCreateTimeTAG: 2012-09-10T18:39:24Z FirstChildTAG: Thank you so much. I was so much searching for a teacher to teach me electronics. I got it here for free FirstChildUserIdTAG: 211715 FirstChildUserNameTAG: pitankar FirstChildCreateTimeTAG: 2012-09-13T08:52:03Z FirstChildTAG: Dear Sir , Amazing never ever got chance Just like this Thanks a lot for edX Team and Dr. Anant Agarwal, Prof. Gerald Sussman, Dr. Piotr Mitros, Prof. Chris Terman, Prof. Khurram Afridi. Thanks FirstChildUserIdTAG: 156694 FirstChildUserNameTAG: mkprasanth FirstChildCreateTimeTAG: 2012-09-13T14:38:40Z FirstChildTAG: Thank you EDX FirstChildUserIdTAG: 251792 FirstChildUserNameTAG: Ahmux FirstChildCreateTimeTAG: 2012-09-15T14:27:33Z FirstChildTAG: thank you dave and edX team for making this great course available for free!! Cheers! Looking forward to have fun. I extend my Heartfelt Gratitude all the way from Bhutan!! FirstChildUserIdTAG: 12144 FirstChildUserNameTAG: Sonam FirstChildCreateTimeTAG: 2012-09-05T12:19:17Z SecondChildTAG: Hey hav u heard about Nirma University? SecondChildUserIdTAG: 64566 SecondChildUserNameTAG: tvl SecondChildCreateTimeTAG: 2012-09-05T16:20:40Z FirstChildTAG: Hi, I agree Sonam, I just can be grateful of having this opportunity to do this course.... for free, and from MITx. Thank you so much!! FirstChildUserIdTAG: 342966 FirstChildUserNameTAG: SandraNavarro FirstChildCreateTimeTAG: 2012-09-05T12:34:40Z FirstChildTAG: Dear Sir , aim not able to do wire connecting in my lab I am using Google chrome OS : win XP 5.1.2600 Kindly help is possible My lap top DELL E5410 I have another stimulator it is Yanka its work well and i am very much experience to work with power point to make electronic circuit I hove some issue with this application please help FirstChildUserIdTAG: 156694 FirstChildUserNameTAG: mkprasanth FirstChildCreateTimeTAG: 2012-09-09T17:30:35Z FirstChildTAG: hi dave i have a difficulty in home work week 1 on syntax of algebraic total resistance,,please help FirstChildUserIdTAG: 324463 FirstChildUserNameTAG: Junbacs FirstChildCreateTimeTAG: 2012-09-15T17:09:29Z FirstChildTAG: How many classmates do we have? :) FirstChildUserIdTAG: 455902 FirstChildUserNameTAG: emersonmalca FirstChildCreateTimeTAG: 2012-09-26T00:09:23Z FirstChildTAG: Thank you Dave. Hola y muchas gracias FirstChildUserIdTAG: 184827 FirstChildUserNameTAG: DiegoT FirstChildCreateTimeTAG: 2012-09-28T15:50:45Z FirstChildTAG: Okay I voted to make it 100. Thanks for the opportunity. The platform is fantastic. Thanks very much. FirstChildUserIdTAG: 623210 FirstChildUserNameTAG: preveen FirstChildCreateTimeTAG: 2012-10-29T13:42:34Z IndexTAG: 2 TitleTAG: Copy video I think this video is just the same as the video before; I think it doesn't really matter, but I thought to point out that. UserIdTAG: 353589 UserNameTAG: klaramun CreateTimeTAG: 2012-09-11T21:40:07Z VoteTAG: 73 CoursewareTAG: Week 3 / Graphical Analysis of Nonlinear Circuits CommentableIdTAG: 6002x_graphical_analysis_of_nonlinear_circuits NumberOfReplyTAG: 3 FirstChildTAG: Thank you for pointing this out, we'll fix this. FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-12T03:11:38Z SecondChildTAG: I agree with you, It is basically the same video. SecondChildUserIdTAG: 143268 SecondChildUserNameTAG: jborrego SecondChildCreateTimeTAG: 2012-09-16T14:43:29Z SecondChildTAG: Sorry for bothering you. The variable "a" must have "Apms" as a measure unit. It is not critically important, but in context that Prof. Agarwal writes down other units, correction might be justified =) SecondChildUserIdTAG: 187024 SecondChildUserNameTAG: gkcalat SecondChildCreateTimeTAG: 2012-09-23T10:11:04Z SecondChildTAG: should it be an another video here or it's just dup? SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-25T18:49:03Z SecondChildTAG: Yeah, it should be cut at the point he paused for a few seconds for us to think. SecondChildUserIdTAG: 98259 SecondChildUserNameTAG: rogerloh0 SecondChildCreateTimeTAG: 2012-09-26T00:38:23Z FirstChildTAG: There is a pause in the video(asking the slope of the load line), maybe it used to be two videos, but merged with another one FirstChildUserIdTAG: 653 FirstChildUserNameTAG: ziyou FirstChildCreateTimeTAG: 2012-09-27T05:30:14Z FirstChildTAG: I like it, imma watch it twice. FirstChildUserIdTAG: 280220 FirstChildUserNameTAG: pilgrimCycle FirstChildCreateTimeTAG: 2012-09-25T18:00:38Z SecondChildTAG: agree ;) SecondChildUserIdTAG: 343633 SecondChildUserNameTAG: lexder SecondChildCreateTimeTAG: 2012-09-26T17:39:04Z IndexTAG: 3 TitleTAG: Hints for those struggling with H6P1 Q2, Q3 and Q4!! ![enter image description here][1] ![enter image description here][2] ![enter image description here][3] ![enter image description here][4] Here is the link to a tutorial on how to calculate partial derivatives on Wolfram Alpha https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/507d2b565ae43d1f00000140 ![enter image description here][5] ![enter image description here][6] ![enter image description here][7] ![enter image description here][8] ![enter image description here][14] ![enter image description here][9] ![enter image description here][10] ![enter image description here][11] ![enter image description here][12] ![enter image description here][13] And Voila! Hazel. [1]: https://edxuploads.s3.amazonaws.com/13506638211343619.jpg [2]: https://edxuploads.s3.amazonaws.com/13506712271343656.jpg [3]: https://edxuploads.s3.amazonaws.com/13506638431343663.jpg [4]: https://edxuploads.s3.amazonaws.com/13506638551343638.bmp [5]: https://edxuploads.s3.amazonaws.com/13506639971343607.bmp [6]: https://edxuploads.s3.amazonaws.com/13506640141343679.bmp [7]: https://edxuploads.s3.amazonaws.com/1350664710134368.bmp [8]: https://edxuploads.s3.amazonaws.com/13506653921343664.bmp [9]: https://edxuploads.s3.amazonaws.com/13506654101343685.bmp [10]: https://edxuploads.s3.amazonaws.com/13506654411343679.bmp [11]: https://edxuploads.s3.amazonaws.com/13506654711343668.bmp [12]: https://edxuploads.s3.amazonaws.com/13506654851343681.jpg [13]: https://edxuploads.s3.amazonaws.com/13506655033000377.bmp [14]: https://edxuploads.s3.amazonaws.com/13506793071343676.jpg UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-10-19T16:38:40Z VoteTAG: 64 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 12 FirstChildTAG: this is what i calculated for H6P1 but it said that cannot parse ... where i am wrong.? FirstChildUserIdTAG: 136490 FirstChildUserNameTAG: Ali_PU FirstChildCreateTimeTAG: 2012-10-19T16:41:38Z SecondChildTAG: Ali_PU, do not post your answer again. I responded to your other post. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-19T16:46:25Z SecondChildTAG: ok i will not do again IN SHA ALLAH :) SecondChildUserIdTAG: 136490 SecondChildUserNameTAG: Ali_PU SecondChildCreateTimeTAG: 2012-10-19T17:16:00Z SecondChildTAG: in the calculation of gm i defrentiate ids wrt to VGS and got answer in terms of X^2.X but it said x^2 not allowed where i am wrong.. derivative of vgs wrt to vgs is 1 and derivative of constant is zero m i right.? SecondChildUserIdTAG: 136490 SecondChildUserNameTAG: Ali_PU SecondChildCreateTimeTAG: 2012-10-19T17:41:17Z FirstChildTAG: Wolfram can also use subscripts, if that's helpful, with a letter and a subscript number, like this: x_1, x_2, y_3, etc, etc. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-19T17:34:42Z SecondChildTAG: Awesome! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-19T17:44:07Z SecondChildTAG: Thanks! I did not know this! SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-19T17:48:09Z FirstChildTAG: Great job, hazel1919! I have one question. You put this: ![enter image description here][1] and then the next line is this: ![enter image description here][2] So shouldn't it be iDS=K*(VGS-VT)*VDS^2? Or am I thinking of it incorrectly? [1]: https://edxuploads.s3.amazonaws.com/13506701251343635.png [2]: https://edxuploads.s3.amazonaws.com/13506702094543858.png FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-19T18:11:36Z SecondChildTAG: Ah! I should put 'and VDS=VOUT'. Thanks for pointing this out, and taking the time to review! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-19T18:22:09Z SecondChildTAG: No worries. Great job, again. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-19T18:23:03Z SecondChildTAG: Done! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-19T18:29:58Z FirstChildTAG: I am amazed by how much time people are willing to put in preparing hints for their fellow students. Very nice write up. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-19T18:55:07Z SecondChildTAG: Like I said, I do this for myself. The concepts of calculus are extremely shaky in my head, so I have to do a complete step by step analysis. Then I just replace the answers with X's and post! :) (It is a time consuming process to upload 11+ images mind!) SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-19T19:10:07Z SecondChildTAG: thanks hazel, and good luck with your calculus ;) SecondChildUserIdTAG: 47372 SecondChildUserNameTAG: Digius SecondChildCreateTimeTAG: 2012-10-20T12:22:52Z SecondChildTAG: Thanks! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-20T12:32:17Z SecondChildTAG: Thanks Hazel SecondChildUserIdTAG: 362299 SecondChildUserNameTAG: anandbaskaran SecondChildCreateTimeTAG: 2012-10-20T17:18:37Z SecondChildTAG: GREAT WORK THANX A LOT HAZEL SecondChildUserIdTAG: 285222 SecondChildUserNameTAG: dhaval24 SecondChildCreateTimeTAG: 2012-10-20T17:47:22Z FirstChildTAG: ![enter image description here][1] [1]: http://www.mindguru-consultants.com/wp-content/uploads/png-vs-jpeg.jpg FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-19T19:49:14Z SecondChildTAG: lol SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-19T20:28:24Z SecondChildTAG: saved as .jpeg SecondChildUserIdTAG: 241232 SecondChildUserNameTAG: hypnotic SecondChildCreateTimeTAG: 2012-10-21T13:16:33Z SecondChildTAG: Actually that is one of the best suggestions! Thank you YakovO SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-22T11:26:23Z FirstChildTAG: Hazel1919, Thank you so very much...the Wolfram is a wonderful find and your explanation is even better. Much appreciated Sam FirstChildUserIdTAG: 57012 FirstChildUserNameTAG: sam747 FirstChildCreateTimeTAG: 2012-10-19T23:17:00Z FirstChildTAG: awesome thanks a ton.:)) FirstChildUserIdTAG: 266739 FirstChildUserNameTAG: satvikchugh FirstChildCreateTimeTAG: 2012-10-20T07:43:41Z FirstChildTAG: wxMaxima is also good and free)) FirstChildUserIdTAG: 406023 FirstChildUserNameTAG: neitrino FirstChildCreateTimeTAG: 2012-10-20T20:19:55Z FirstChildTAG: Thank you Hazel1919 for the thorough explanation. It must have taken forever to create that and paste all of it into this forum. The part I didn't realize or catch is that when calculating gm and ro, each were dependent on different variables, thereby having to take the partial derivative two different times. I would have never caught that without your help. Now I know to look for situations like these. Calculus can be a life saver sometimes, especially L'Hopital's rule, it has saved me many times. Thanks again FirstChildUserIdTAG: 132828 FirstChildUserNameTAG: rharris FirstChildCreateTimeTAG: 2012-10-20T22:18:13Z SecondChildTAG: I don't think I would have understood that too if I did not watch week 6 tutorial. That is a pretty good tutorial (although the similarities could have been better explained). Thanks for the nice comments! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-21T08:30:57Z FirstChildTAG: r0=1/(2*k*VDS*(VGS-VT)) whats wrong in dis?? FirstChildUserIdTAG: 224318 FirstChildUserNameTAG: ashwin12312 FirstChildCreateTimeTAG: 2012-10-21T03:46:21Z SecondChildTAG: May be K insted k? SecondChildUserIdTAG: 133015 SecondChildUserNameTAG: GeksogeN SecondChildCreateTimeTAG: 2012-10-21T05:41:42Z SecondChildTAG: instead* SecondChildUserIdTAG: 133015 SecondChildUserNameTAG: GeksogeN SecondChildCreateTimeTAG: 2012-10-21T05:42:17Z SecondChildTAG: Vds vgs not permitted in answer SecondChildUserIdTAG: 319286 SecondChildUserNameTAG: thouheed SecondChildCreateTimeTAG: 2012-10-21T09:36:54Z FirstChildTAG: LOVE U HAZEL.. FirstChildUserIdTAG: 259970 FirstChildUserNameTAG: anilgovindaraj FirstChildCreateTimeTAG: 2012-10-21T05:41:31Z FirstChildTAG: thanx alot it was great help....:) FirstChildUserIdTAG: 117319 FirstChildUserNameTAG: iqramalik FirstChildCreateTimeTAG: 2012-10-21T19:40:49Z IndexTAG: 4 TitleTAG: Got this from old MITx ![enter image description here][1] [1]: https://mitx_askbot_stage.s3.amazonaws.com/13312756901518897.jpg UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-07T13:18:15Z VoteTAG: 58 CoursewareTAG: Week 1 / Battery model CommentableIdTAG: 6002x_S1E9_Battery_Model NumberOfReplyTAG: 6 FirstChildTAG: thanks :) FirstChildUserIdTAG: 122513 FirstChildUserNameTAG: Mona77 FirstChildCreateTimeTAG: 2012-09-08T20:04:22Z FirstChildTAG: Can anyone kindly explain why is i2*0.25 not included in the KVL Loop I calculation w.r.t S1E9 ? Thanks in advance !! FirstChildUserIdTAG: 170349 FirstChildUserNameTAG: amitGuliya FirstChildCreateTimeTAG: 2012-09-09T09:30:24Z SecondChildTAG: Loop2 has 2 elements : V2 and R2 (0.32 Ohm) (As you can see the Loop II in the picture) 0.25 is in Loop I (which has i1, not i2) Is it OK for you? SecondChildUserIdTAG: 229745 SecondChildUserNameTAG: LeQuang SecondChildCreateTimeTAG: 2012-09-10T13:07:05Z FirstChildTAG: thanks for teaching the methodical way to solve :) FirstChildUserIdTAG: 400239 FirstChildUserNameTAG: LincyG FirstChildCreateTimeTAG: 2012-09-09T17:22:19Z SecondChildTAG: Another way is to use KCL and KVL. **KVL**: You have V1-v1=0V where v1 is the voltage across R1 (v1=R1*i1; Ohm's law). So you find V1=R1*i1 and i1=V1/R1=6A. Same for the branch of V2 -> i2=V2/R2=4.6875A. Via **KCL**: i1+i2=I=10.6875A. SecondChildUserIdTAG: 396446 SecondChildUserNameTAG: RousseauxS SecondChildCreateTimeTAG: 2012-09-10T13:13:35Z FirstChildTAG: S1E9 the second question symbolic answer is V/R1+V/R2 FirstChildUserIdTAG: 345464 FirstChildUserNameTAG: Constantine_ru FirstChildCreateTimeTAG: 2012-09-11T19:03:58Z FirstChildTAG: You used Ohm's law for solving that, I really though that there were another way to do that, without Ohm, cause they haven't taught it yet. FirstChildUserIdTAG: 265133 FirstChildUserNameTAG: Elienai FirstChildCreateTimeTAG: 2012-09-09T01:05:58Z FirstChildTAG: Hi. What software did you use when writing this? I have bamboo table and using OneNote but your software seems better. FirstChildUserIdTAG: 89084 FirstChildUserNameTAG: anakluhur FirstChildCreateTimeTAG: 2012-09-15T03:39:39Z SecondChildTAG: On the last question you neglected i_2. Is this because the circuit is not short-circuit anymore? SecondChildUserIdTAG: 89084 SecondChildUserNameTAG: anakluhur SecondChildCreateTimeTAG: 2012-09-15T04:06:09Z IndexTAG: 5 TitleTAG: [PETITION] More MITx courses! Please! =] Hello to everyone! I've been enjoying this course a lot and I would like to continue enrolling myself to new MITx courses, if possible, already next spring. I'd like to ask to everyone who share this same feeling to up vote this post **in hope to Staff keep up this great work**. Please Staff understand that at least all of us are interested in continue learning with you! Of course we know that OCW offers many other courses, but it is much more exciting study through this platform. Thanks to all! UserIdTAG: 288637 UserNameTAG: gotchi CreateTimeTAG: 2012-11-06T00:43:04Z VoteTAG: 53 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Seconded! Preferably with Professor Anant Agarwal as lecturer! FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-11-06T10:08:39Z FirstChildTAG: No need for a petition... Please continue to visit edX as we do plan on continually adding more courses each semester! - Rohan FirstChildUserIdTAG: 391929 FirstChildUserNameTAG: RohanNagarkar FirstChildCreateTimeTAG: 2012-11-06T14:56:07Z SecondChildTAG: Wonderful news! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-06T15:54:16Z SecondChildTAG: Great! =D SecondChildUserIdTAG: 288637 SecondChildUserNameTAG: gotchi SecondChildCreateTimeTAG: 2012-11-06T17:02:48Z SecondChildTAG: good to hear any more with professor agarwal? SecondChildUserIdTAG: 160242 SecondChildUserNameTAG: Mlevins35 SecondChildCreateTimeTAG: 2012-11-06T20:21:28Z SecondChildTAG: Sir please add some more excellent courses on electronics(digital integrated circuits etc) it would be a great help for us SecondChildUserIdTAG: 321556 SecondChildUserNameTAG: sj31867 SecondChildCreateTimeTAG: 2012-11-07T11:23:09Z SecondChildTAG: Lectures on Automatic Control Theory will be very useful...I think... SecondChildUserIdTAG: 104522 SecondChildUserNameTAG: IgorNovice SecondChildCreateTimeTAG: 2012-12-02T19:37:09Z SecondChildTAG: I would also find an automatic control theory course useful SecondChildUserIdTAG: 77891 SecondChildUserNameTAG: morecowbell89 SecondChildCreateTimeTAG: 2012-12-11T22:29:39Z SecondChildTAG: advanced electronics courses will be great and courses in electromagnetic waves and antenna design SecondChildUserIdTAG: 322638 SecondChildUserNameTAG: khalid1988 SecondChildCreateTimeTAG: 2012-12-16T11:05:40Z SecondChildTAG: Voting for a Digital Electronics course :) and preferably taught by Professor Agarwal :) SecondChildUserIdTAG: 298775 SecondChildUserNameTAG: surja SecondChildCreateTimeTAG: 2012-12-18T07:10:41Z SecondChildTAG: I would love to see a practical digital electronics course based around projects on CPUs microcontrollers, FPGAs interfaces with analogue circuitry like oscillators etc. SecondChildUserIdTAG: 61469 SecondChildUserNameTAG: Spacedog SecondChildCreateTimeTAG: 2012-12-24T02:03:43Z SecondChildTAG: I would love to have a course on automatic control systems and designing of electronic circuits. SecondChildUserIdTAG: 449211 SecondChildUserNameTAG: Deveshmonga SecondChildCreateTimeTAG: 2012-12-27T16:31:21Z SecondChildTAG: any course pleaaassseeeee. i love the way it is taught on edx SecondChildUserIdTAG: 386372 SecondChildUserNameTAG: akashrao1991 SecondChildCreateTimeTAG: 2012-12-29T17:04:52Z IndexTAG: 6 TitleTAG: Answer Apply KCL: i1+i2+i3=0 ---> (e-V1)/R1+(e-V2)/R2+e/R3 =0 Rearrange: e(1/R1 + 1/R2 + 1/R3) = V1/R1 + V2/R2 e[(R1*R2+R2*R3+R1*R3)/(R1*R2*R3)]= V1/R1 + V2/R2 Replacing S= R1*R2+R2*R3+R1*R3 e[S/(R1*R2*R3)] = V1/R1 + V2/R2 e = R2*R3*V1/S + R1*R3*V2/S ...........(1) How do you get a1, a2, b1, b2, c1, c2? Easy!!!! v3=e ---> v3 = (R2*R3/S)*V1 + (R1*R3/S)*V2 so.... a1 = R2*R3/S ..... a2 = R1*R3/S i1=(e-V1)/R1 .....Let's play with eq (1): e - V1 = (R2*R3-S)*V1/S + R1*R3*V2/S = (-R1*R2-R1*R3)*V1/S + R1*R3*V2/S (e-V1)/R1 = i1 = (-R2-R3)*V1/S + R3*V2/S so b1 = (-R2-R3)/S .....b2 = R3/S i2 = (e-V2)/R2 .... Let's play with eq (1) again: e- V2 = R2*R3*V1/S + (R1*R3-S)*V2/S = R2*R3*V1/S + (-R1*R2-R2*R3)*V2/S (e-V2)/R2 = R3*V1/S + (-R1-R3)*V2/S so....c1 = R3/S ........ c2 = (-R1-R3)/S That's all !!!!!!!! UserIdTAG: 268104 UserNameTAG: SaulCardenasG CreateTimeTAG: 2012-09-18T03:17:41Z VoteTAG: 52 CoursewareTAG: Week 2 / Linear combinations of source strenghts CommentableIdTAG: 6002x_S3E2_Linear_Combinations_of_Source_Strengths NumberOfReplyTAG: 5 FirstChildTAG: You are my hero. FirstChildUserIdTAG: 364798 FirstChildUserNameTAG: atari1994 FirstChildCreateTimeTAG: 2012-09-18T03:45:51Z SecondChildTAG: nice explanation SecondChildUserIdTAG: 131565 SecondChildUserNameTAG: vdprasad22 SecondChildCreateTimeTAG: 2012-09-18T13:09:25Z SecondChildTAG: thx.. SecondChildUserIdTAG: 303485 SecondChildUserNameTAG: iparitosh SecondChildCreateTimeTAG: 2012-09-18T18:01:36Z SecondChildTAG: save my life really.... thanks so much SecondChildUserIdTAG: 225436 SecondChildUserNameTAG: ElmerTsai SecondChildCreateTimeTAG: 2012-09-21T00:18:14Z SecondChildTAG: nice explanation.. i was tryin this prob from an hour but got stuck.. SecondChildUserIdTAG: 337430 SecondChildUserNameTAG: aparna_b SecondChildCreateTimeTAG: 2012-09-21T15:20:22Z SecondChildTAG: thanks my head is aching and my eyes are burning with this... great, a lot of thanks... i wasn't getting this... SecondChildUserIdTAG: 317836 SecondChildUserNameTAG: VitorRodrigues SecondChildCreateTimeTAG: 2012-09-21T21:38:35Z SecondChildTAG: cooooooool jaja SecondChildUserIdTAG: 127021 SecondChildUserNameTAG: victorag SecondChildCreateTimeTAG: 2012-09-22T16:12:39Z SecondChildTAG: Wonderfully explained. Thank you! SecondChildUserIdTAG: 218159 SecondChildUserNameTAG: Miss SecondChildCreateTimeTAG: 2012-09-23T16:12:52Z SecondChildTAG: thankkyou SecondChildUserIdTAG: 188643 SecondChildUserNameTAG: ANNADIVS SecondChildCreateTimeTAG: 2012-11-14T10:08:32Z FirstChildTAG: Thank you for showing the work. Here is my attempt at converting your post into "math mode". $i_1+i_2+i_3=0$ ---> $\frac{e-V_1}{R_1}+\frac{e-V_2}{R_2}+\frac{e}{R_3} =0$ Rearrange: $e\cdot(\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}) = \frac{V_1}{R_1} + \frac{V_2}{R_2}$ $e\cdot\frac{R_1 \cdot R_2+R_2 \cdot R_3+R_1 \cdot R_3}{R_1 \cdot R_2 \cdot R_3}= \frac{V_1}{R_1} + \frac{V_2}{R_2}$ Replacing $S= R_1 \cdot R_2+R_2 \cdot R_3+R_1 \cdot R_3$ $e \cdot \frac{S}{R_1 \cdot R_2 \cdot R_3} = \frac{V_1}{R_1} + \frac{V_2}{R_2}$ $\boxed {e = \frac{R_2 \cdot R_3 \cdot V_1}{S} + \frac{R_1 \cdot R_3 \cdot V_2}{S}} \Longleftarrow \boxed {eq. 1}$ How do you get $a_1, a_2, b_1, b_2, c_1, c_2$? Easy!!!! $v_3=e \Longrightarrow v_3 = \frac{R_2 \cdot R_3}{S} \cdot V_1 + \frac{R_1 \cdot R_3}{S} \cdot V_2$ $so \dots \boxed{a_1 = \frac{R_2 \cdot R_3}{S}} \dots \boxed{a_2 = \frac{R_1 \cdot R_3}{S}}$ $i_1=\frac{e-V_1}{R_1} \dots \text{Let's play with eq (1):}$ $e - V_1 = (R_2 \cdot R_3-S) \cdot \frac{V_1}{S} + R_1 \cdot R_3 \cdot \frac{V_2}{S} = (-R_1 \cdot R_2-R_1 \cdot R_3) \cdot \frac{V_1}{S} + R_1 \cdot R_3 \cdot \frac{V_2}{S}$ $\frac{e-V_1}{R_1} = i_1 = (-R_2-R_3) \cdot \frac{V_1}{S} + R_3 \cdot \frac{V_2}{S}$ so $\dots \boxed{b_1 = \frac{-R_2-R_3}{S}} \dots \boxed{b_2 = \frac{R_3}{S}}$ $i_2 = \frac{e-V_2}{R_2} \dots$ Let's play with eq (1) again: $e-V_2 = R_2 \cdot R_3 \cdot \frac{V_1}{S} + (R_1 \cdot R_3-S) \cdot \frac{V_2}{S} = R_2 \cdot R_3 \cdot \frac{V_1}{S} + (-R_1 \cdot R_2-R_2 \cdot R_3) \cdot \frac{V_2}{S}$ $\frac{e-V_2}{R_2} = R_3 \cdot \frac{V_1}{S} + (-R_1-R_3) \cdot \frac{V_2}{S}$ so$\dots \boxed{c_1 = \frac{R_3}{S}} \dots \boxed{c_2 = \frac{-R_1-R_3}{S}}$ FirstChildUserIdTAG: 332090 FirstChildUserNameTAG: WPurin FirstChildCreateTimeTAG: 2012-09-18T07:54:39Z SecondChildTAG: nice solution mind SecondChildUserIdTAG: 146995 SecondChildUserNameTAG: ZYJ SecondChildCreateTimeTAG: 2012-09-18T14:20:19Z SecondChildTAG: Fantastic! SecondChildUserIdTAG: 309952 SecondChildUserNameTAG: alexjose SecondChildCreateTimeTAG: 2012-09-18T16:37:05Z SecondChildTAG: I do not get this part, could someone kindly eplain the algebra behind this? i1=e−V1R1…Let's play with eq (1): e−V1=(R2⋅R3−S)⋅V1S+R1⋅R3⋅V2S SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-09-18T19:15:07Z SecondChildTAG: make e=i1R1+V1,then take i1 itself on LHS do usual maths operations like subtraction,division,taking LCM.u will get the answer! SecondChildUserIdTAG: 353405 SecondChildUserNameTAG: karthickks SecondChildCreateTimeTAG: 2012-09-18T19:36:01Z SecondChildTAG: thanks!!! specially to WPurin for her(his) "math mode".... SecondChildUserIdTAG: 268104 SecondChildUserNameTAG: SaulCardenasG SecondChildCreateTimeTAG: 2012-09-19T04:11:56Z SecondChildTAG: Ok, take a look: e1-V1 = (R2*R3/S)*V1 - V1 + ..... e1-V1 = (R2*R3/S)*V1 -(S/S)*V1 ....homogeneous equations e1-V1 = [(R2*R3-S)/S]*V1 e1-V1 = [-R1*R3-R1*R2)/S]/V1 I hope i was clear..... SecondChildUserIdTAG: 268104 SecondChildUserNameTAG: SaulCardenasG SecondChildCreateTimeTAG: 2012-09-19T04:19:43Z SecondChildTAG: Thanks!!! Nice solution. SecondChildUserIdTAG: 312043 SecondChildUserNameTAG: Osmar17 SecondChildCreateTimeTAG: 2012-09-21T03:18:57Z SecondChildTAG: How did you come to the conclusion that R1*R3 = -R1*R2 -R2*R3 ....(1) and R2*R3 = -R1*R2 - R1*R3 ....(2)? SecondChildUserIdTAG: 32244 SecondChildUserNameTAG: cyrildex SecondChildCreateTimeTAG: 2012-09-21T20:49:12Z SecondChildTAG: In eq.(1) we replaced R1*R2+R2*R3+R1*R3 =S, so V1*R1*R2*R3/R1*S = R2*R3/(V1*S)....that's your question? SecondChildUserIdTAG: 268104 SecondChildUserNameTAG: SaulCardenasG SecondChildCreateTimeTAG: 2012-09-25T03:52:17Z SecondChildTAG: Thank you so much! SecondChildUserIdTAG: 795417 SecondChildUserNameTAG: rnepxk SecondChildCreateTimeTAG: 2012-12-29T06:26:49Z FirstChildTAG: This helped me a lot, thanks. FirstChildUserIdTAG: 302713 FirstChildUserNameTAG: EVega FirstChildCreateTimeTAG: 2012-09-18T08:38:00Z FirstChildTAG: What with Honor Code?! "By enrolling in an edX course, I agree that I will: Complete all mid-terms and final exams with my own work and only my own work. I will not submit the work of any other person. Maintain only one user account and not let anyone else use my username and/or password. **Not engage in any activity that would dishonestly improve my results, or improve or hurt the results of others.** **Not post answers to problems that are being used to assess student performance.**" What's wrong with you? FirstChildUserIdTAG: 394836 FirstChildUserNameTAG: v2g6ch4 FirstChildCreateTimeTAG: 2012-09-18T09:47:07Z SecondChildTAG: "Complete all mid term and final exams with my work" .. this is isn't either :-) What's wrong with you ? ;-) SecondChildUserIdTAG: 160683 SecondChildUserNameTAG: Jinish SecondChildCreateTimeTAG: 2012-09-18T12:18:48Z SecondChildTAG: I'm not a staff member, but I don't see any violations of the honor code. The problem referenced is an example problem, not a graded homework, lab, or exam. Example problems are fair game. I don't see that he or anyone else in the thread is using more than one account. So I think he's all good. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-09-18T14:38:58Z SecondChildTAG: ![enter image description here][1] [1]: http://t0.gstatic.com/images?q=tbn:ANd9GcQP3kLW-LVhzWBpKdfEYZg4McuCxGSvtQAiS_tzN0AHA5MB72Mk SecondChildUserIdTAG: 394836 SecondChildUserNameTAG: v2g6ch4 SecondChildCreateTimeTAG: 2012-09-19T04:28:19Z SecondChildTAG: Look here: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_S1E3_AC_power/threads/504c53e607d5d11f00000004 https://www.edx.org/courses/MITx/6.002x/2012_Fall/info They were congratulated by staff members, i guess i didn't violate the honor code....anyway, any staff member can say if i did it, please?!! if i did it i will delete this post. SecondChildUserIdTAG: 268104 SecondChildUserNameTAG: SaulCardenasG SecondChildCreateTimeTAG: 2012-09-19T04:35:16Z SecondChildTAG: Don't worry, thanks for your comment anyway!!!! SecondChildUserIdTAG: 268104 SecondChildUserNameTAG: SaulCardenasG SecondChildCreateTimeTAG: 2012-09-19T06:09:33Z SecondChildTAG: Well, I can appreciate the information, and do not think that this violated the honor code because, as you say it is a simple task. In my case, I have knowledge of electronics but due to my limitations with the language could not understand what was being asked and this post has helped me tremendously. I do not want the answers, just know that you are asking because so far the exercises are very basic and do not present any complications except the language. Thanks SecondChildUserIdTAG: 244706 SecondChildUserNameTAG: Miguel_Angel SecondChildCreateTimeTAG: 2012-09-20T16:10:50Z SecondChildTAG: So in case it was still in question, posting worked solutions for example problems is not a violation of the honor code. If this had been about a HW or exam problem before it was due, then it would have been a violation. SecondChildUserIdTAG: 11 SecondChildUserNameTAG: dormsbee SecondChildCreateTimeTAG: 2012-09-21T13:22:57Z SecondChildTAG: Thanks!!!! dormsbee....i was very worried!!!! SecondChildUserIdTAG: 268104 SecondChildUserNameTAG: SaulCardenasG SecondChildCreateTimeTAG: 2012-09-24T04:26:16Z FirstChildTAG: Please help me in H2P1 FirstChildUserIdTAG: 293719 FirstChildUserNameTAG: jaiprakash1 FirstChildCreateTimeTAG: 2012-09-19T14:01:05Z IndexTAG: 7 TitleTAG: Suggestion: sharing models from circuit sandbox I wonder if the staff can implement the sandbox circuit model sharing, so we can share our circuits in this forum easily. Sure I can upload a print screen but then how many people would want to recreate it by hand so they can play with it and perhaps see if there's something wrong. Thx UserIdTAG: 5325 UserNameTAG: vkz CreateTimeTAG: 2012-09-09T09:30:04Z VoteTAG: 49 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: This is a really good idea FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-09T12:08:14Z SecondChildTAG: Why don't you up vote it then. Use the arrows on the left of the post. Let the staff know. SecondChildUserIdTAG: 5325 SecondChildUserNameTAG: vkz SecondChildCreateTimeTAG: 2012-09-09T12:27:56Z FirstChildTAG: Yes, vkz is ryt! If u want this, then up vote and let the staff know FirstChildUserIdTAG: 314624 FirstChildUserNameTAG: Owais001 FirstChildCreateTimeTAG: 2012-09-10T10:16:11Z IndexTAG: 8 TitleTAG: Question 2! My work for the Q2: ![enter image description here][1] [1]: http://puu.sh/12InC http://puu.sh/12InC for bigger image :x UserIdTAG: 255404 UserNameTAG: zepp CreateTimeTAG: 2012-09-07T01:28:40Z VoteTAG: 46 CoursewareTAG: Week 1 / AC power CommentableIdTAG: 6002x_S1E3_AC_power NumberOfReplyTAG: 13 FirstChildTAG: Very nice. Do this as much as you can. It will help many people on this course. This is fun. trig identities are always nice when things fall into place. But many will not be thinking that! ha ha! Nicely written. Well done FirstChildUserIdTAG: 15344 FirstChildUserNameTAG: kob FirstChildCreateTimeTAG: 2012-09-07T01:57:43Z SecondChildTAG: Thanks, will try! SecondChildUserIdTAG: 255404 SecondChildUserNameTAG: zepp SecondChildCreateTimeTAG: 2012-09-07T02:12:49Z FirstChildTAG: bracket is misplaced,its cos^2(120*pi*t)....not cos(120*pi*t)^2 FirstChildUserIdTAG: 153869 FirstChildUserNameTAG: rabindra FirstChildCreateTimeTAG: 2012-09-07T06:54:48Z SecondChildTAG: Well spotted, I am sure he'll update it. (But in above context is pretty clear what it is) SecondChildUserIdTAG: 15344 SecondChildUserNameTAG: kob SecondChildCreateTimeTAG: 2012-09-07T07:36:23Z FirstChildTAG: THATS NICE HIGH SCHOOL INTEGRATION! FirstChildUserIdTAG: 102370 FirstChildUserNameTAG: aninda FirstChildCreateTimeTAG: 2012-09-07T14:55:30Z SecondChildTAG: So why say its high school integration? Have you started doing it differently since you left high school? ;-) That reads a little bit like a sneery comment. Ha ha! Probably was not meant that way. SecondChildUserIdTAG: 15344 SecondChildUserNameTAG: kob SecondChildCreateTimeTAG: 2012-09-07T15:56:56Z SecondChildTAG: Maybe, but the last time I used this stuff is about 15 years ago. To freshen up your knowledge of integral calculus (if needed) you can have a look at [Khan Academy][1] [1]: http://www.khanacademy.org/math/calculus/integral-calculus SecondChildUserIdTAG: 114881 SecondChildUserNameTAG: xvink SecondChildCreateTimeTAG: 2012-09-08T18:30:13Z FirstChildTAG: buddy thanks for your help ,it was really heilful.africa FirstChildUserIdTAG: 273039 FirstChildUserNameTAG: SOCRATE23 FirstChildCreateTimeTAG: 2012-09-07T12:10:16Z FirstChildTAG: may i know why you choose the upper limit as 1/60 FirstChildUserIdTAG: 118611 FirstChildUserNameTAG: mitianhari FirstChildCreateTimeTAG: 2012-09-07T14:51:55Z SecondChildTAG: HE DENOTES 1/60 AS THE TIME PERIOD OF THE WAVE..WHICH IS FOR ONE CYCLE..SINCE FREQUENCY IS 60 HZ AND T=1/F SecondChildUserIdTAG: 102370 SecondChildUserNameTAG: aninda SecondChildCreateTimeTAG: 2012-09-07T14:57:07Z FirstChildTAG: Thanks Zepp. One question (please pardon my simplicity) - where does the 1/T term before the integral come from? The rest of the proof is wicked! FirstChildUserIdTAG: 166031 FirstChildUserNameTAG: krebryna FirstChildCreateTimeTAG: 2012-09-07T21:44:48Z SecondChildTAG: I found this formula in the textbook, page 42; the rest of the thing is just calculus :) I also used this video: http://www.youtube.com/watch?v=9ogLqX5uFfM&feature=related Hope it would be helpful! ![enter image description here][1] http://puu.sh/12ZmW [1]: http://puu.sh/12ZmW SecondChildUserIdTAG: 255404 SecondChildUserNameTAG: zepp SecondChildCreateTimeTAG: 2012-09-08T00:38:19Z FirstChildTAG: Every one must "think simple" FirstChildUserIdTAG: 272523 FirstChildUserNameTAG: Jivraj FirstChildCreateTimeTAG: 2012-09-07T17:56:34Z FirstChildTAG: I ran out of patience after the first four steps FirstChildUserIdTAG: 395953 FirstChildUserNameTAG: dustinge FirstChildCreateTimeTAG: 2012-09-08T20:16:03Z FirstChildTAG: Good job mate. Found it very useful. FirstChildUserIdTAG: 284058 FirstChildUserNameTAG: GowthamAR FirstChildCreateTimeTAG: 2012-09-08T19:15:44Z FirstChildTAG: . ![][1] [1]: http://bayimg.com/cACHpAAeC FirstChildUserIdTAG: 142857 FirstChildUserNameTAG: viking2 FirstChildCreateTimeTAG: 2012-09-12T12:37:06Z FirstChildTAG: why 1/T at the start???????? start of intigral? FirstChildUserIdTAG: 388747 FirstChildUserNameTAG: Azeem1721 FirstChildCreateTimeTAG: 2012-09-09T17:08:33Z SecondChildTAG: Because 1/T is a constant and could be factored out of the integral; and it's also part of the formula. SecondChildUserIdTAG: 255404 SecondChildUserNameTAG: zepp SecondChildCreateTimeTAG: 2012-09-09T18:54:05Z SecondChildTAG: Because by integrating from 0 to 1/60 you're calculating the total power dissipated over one cycle. The definition of average power is total power divided by the total time (total time being T = 1/60 in this case) SecondChildUserIdTAG: 248902 SecondChildUserNameTAG: Chanute SecondChildCreateTimeTAG: 2012-09-24T03:02:43Z FirstChildTAG: Thanks so much zepp, my knowledge of integrals has always been a bit shaky at best. This helped me get through it! FirstChildUserIdTAG: 208175 FirstChildUserNameTAG: DavidSteele FirstChildCreateTimeTAG: 2012-09-11T02:14:05Z SecondChildTAG: You are very welcome :) SecondChildUserIdTAG: 255404 SecondChildUserNameTAG: zepp SecondChildCreateTimeTAG: 2012-09-12T01:44:58Z FirstChildTAG: Integrating ![enter image description here][1] [1]: http://puu.sh/14o5v FirstChildUserIdTAG: 142857 FirstChildUserNameTAG: viking2 FirstChildCreateTimeTAG: 2012-09-12T12:44:26Z IndexTAG: 9 TitleTAG: 'Building Midterm Statistics' Please take out this survey. I wish to do some statistical analysis. Actually lets computer do this. Check it out [here][1] PS: Please don't forget to +1 it. As it will help it in becoming popular & giving better results...' People here must have fallen in love with the deadlines! So, deadline of this survey is 4 Nov 2012 (Timings midnight Boston time). And results will be published after that. [1]: http://www.surveymonkey.com/s/LG7KBHP UserIdTAG: 374590 UserNameTAG: ahope CreateTimeTAG: 2012-10-29T04:38:01Z VoteTAG: 44 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 23 FirstChildTAG: Midterm: 100% Overall: 56% (With HW and LAB 7, 8, 9) FirstChildUserIdTAG: 357453 FirstChildUserNameTAG: BrunoCanoso FirstChildCreateTimeTAG: 2012-10-30T18:59:17Z FirstChildTAG: MidTerm 83% Overall 37% FirstChildUserIdTAG: 378096 FirstChildUserNameTAG: Jamshaid271 FirstChildCreateTimeTAG: 2012-10-29T08:05:54Z SecondChildTAG: MT 97%, Overall 50% SecondChildUserIdTAG: 61207 SecondChildUserNameTAG: cvayalas SecondChildCreateTimeTAG: 2012-10-29T18:21:02Z FirstChildTAG: *Midterm*: 66% **Overall**: 23% FirstChildUserIdTAG: 374590 FirstChildUserNameTAG: ahope FirstChildCreateTimeTAG: 2012-10-29T04:40:29Z SecondChildTAG: Midterm: 100% Overall: 54% SecondChildUserIdTAG: 270011 SecondChildUserNameTAG: SnowmanZA SecondChildCreateTimeTAG: 2012-10-29T05:18:02Z SecondChildTAG: MidTerm 76% Overall 44% SecondChildUserIdTAG: 413002 SecondChildUserNameTAG: Gauravjain88 SecondChildCreateTimeTAG: 2012-10-29T05:22:29Z SecondChildTAG: Joined in week 5 only. MT 100% OA 38% SecondChildUserIdTAG: 623210 SecondChildUserNameTAG: preveen SecondChildCreateTimeTAG: 2012-10-29T07:00:06Z SecondChildTAG: MT:76% overall:32% :'( SecondChildUserIdTAG: 158367 SecondChildUserNameTAG: brainyash1990 SecondChildCreateTimeTAG: 2012-10-29T07:18:35Z SecondChildTAG: MT 93% overall 49% SecondChildUserIdTAG: 297655 SecondChildUserNameTAG: Aljoska SecondChildCreateTimeTAG: 2012-10-29T08:01:27Z SecondChildTAG: MT 93% OA 49% SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-29T08:07:19Z SecondChildTAG: MT 93% OA 49% SecondChildUserIdTAG: 489065 SecondChildUserNameTAG: amiths SecondChildCreateTimeTAG: 2012-10-29T08:09:53Z SecondChildTAG: Midterm 100% Overall 45% SecondChildUserIdTAG: 405936 SecondChildUserNameTAG: Whible SecondChildCreateTimeTAG: 2012-10-29T08:13:42Z SecondChildTAG: Midterm 83% Overall 44% SecondChildUserIdTAG: 114913 SecondChildUserNameTAG: ngoctuan SecondChildCreateTimeTAG: 2012-10-29T08:42:32Z SecondChildTAG: MT : 97 % , OA : 41 % SecondChildUserIdTAG: 319109 SecondChildUserNameTAG: Ali__H SecondChildCreateTimeTAG: 2012-10-29T08:58:18Z SecondChildTAG: MT=90 , OA=48 SecondChildUserIdTAG: 358539 SecondChildUserNameTAG: syed_abdullah SecondChildCreateTimeTAG: 2012-10-29T10:06:35Z SecondChildTAG: Midterm 93% Overall 41% SecondChildUserIdTAG: 364126 SecondChildUserNameTAG: shunyi SecondChildCreateTimeTAG: 2012-10-29T10:15:21Z SecondChildTAG: MT: 100%, OA: 48% SecondChildUserIdTAG: 357924 SecondChildUserNameTAG: DanI20 SecondChildCreateTimeTAG: 2012-10-29T10:41:02Z SecondChildTAG: MT: 62%, OA: 30% ... :( SecondChildUserIdTAG: 342966 SecondChildUserNameTAG: SandraNavarro SecondChildCreateTimeTAG: 2012-10-29T10:56:57Z SecondChildTAG: MT : 97%, OA: 47% SecondChildUserIdTAG: 329254 SecondChildUserNameTAG: KGabor SecondChildCreateTimeTAG: 2012-10-29T11:23:25Z SecondChildTAG: MT 100% overall 53% SecondChildUserIdTAG: 99628 SecondChildUserNameTAG: Pranjal16 SecondChildCreateTimeTAG: 2012-10-29T12:31:26Z SecondChildTAG: MT 93% OA 47% SecondChildUserIdTAG: 358764 SecondChildUserNameTAG: BrianKolb SecondChildCreateTimeTAG: 2012-10-29T12:57:04Z SecondChildTAG: MT 93%, OA 28% joined 2 days before exam :) SecondChildUserIdTAG: 714242 SecondChildUserNameTAG: AntM SecondChildCreateTimeTAG: 2012-10-29T13:42:15Z SecondChildTAG: MT 93%, OA 49% SecondChildUserIdTAG: 301962 SecondChildUserNameTAG: labrinim SecondChildCreateTimeTAG: 2012-10-29T13:59:20Z SecondChildTAG: Hi, MT 100 %, OA 54% SecondChildUserIdTAG: 220837 SecondChildUserNameTAG: Calsomus SecondChildCreateTimeTAG: 2012-10-29T13:59:42Z SecondChildTAG: Midterm: 100% Overall: 45% 6th week done SecondChildUserIdTAG: 148389 SecondChildUserNameTAG: chento SecondChildCreateTimeTAG: 2012-10-29T15:16:17Z SecondChildTAG: Mid-term 90%... Overall 35% SecondChildUserIdTAG: 434904 SecondChildUserNameTAG: mtrav SecondChildCreateTimeTAG: 2012-10-30T02:09:57Z SecondChildTAG: ..Up to week 6. SecondChildUserIdTAG: 434904 SecondChildUserNameTAG: mtrav SecondChildCreateTimeTAG: 2012-10-30T02:11:11Z SecondChildTAG: midterm--90%...overall--45% SecondChildUserIdTAG: 278792 SecondChildUserNameTAG: sali SecondChildCreateTimeTAG: 2012-10-30T04:28:30Z SecondChildTAG: midterm-100%...overall-48% upto 6 week SecondChildUserIdTAG: 321556 SecondChildUserNameTAG: sj31867 SecondChildCreateTimeTAG: 2012-10-30T14:15:22Z SecondChildTAG: MT 96, overall 46 SecondChildUserIdTAG: 12905 SecondChildUserNameTAG: Baer SecondChildCreateTimeTAG: 2012-10-30T20:07:02Z SecondChildTAG: sorry typo in last post. MT 93, overall 46 SecondChildUserIdTAG: 12905 SecondChildUserNameTAG: Baer SecondChildCreateTimeTAG: 2012-10-30T20:08:23Z SecondChildTAG: Midterm 76%, Overall 35% SecondChildUserIdTAG: 70519 SecondChildUserNameTAG: Fipe SecondChildCreateTimeTAG: 2012-10-31T03:10:44Z SecondChildTAG: MT 97%, OA 42% SecondChildUserIdTAG: 362900 SecondChildUserNameTAG: Jacey123 SecondChildCreateTimeTAG: 2012-11-02T01:22:04Z SecondChildTAG: Midterm 97%, Overall 50% SecondChildUserIdTAG: 111780 SecondChildUserNameTAG: Ranand SecondChildCreateTimeTAG: 2012-11-03T05:56:10Z FirstChildTAG: midterm 100%, overall 100% FirstChildUserIdTAG: 97581 FirstChildUserNameTAG: Asimakis FirstChildCreateTimeTAG: 2012-10-29T07:35:01Z SecondChildTAG: are you chuck norris? :P SecondChildUserIdTAG: 329254 SecondChildUserNameTAG: KGabor SecondChildCreateTimeTAG: 2012-10-29T11:23:58Z SecondChildTAG: probably he meant so far he has scored full marks in all hw, lab and exam;-) SecondChildUserIdTAG: 623210 SecondChildUserNameTAG: preveen SecondChildCreateTimeTAG: 2012-10-29T12:01:38Z SecondChildTAG: he is not chuck norris, maybe "pontios"! SecondChildUserIdTAG: 301962 SecondChildUserNameTAG: labrinim SecondChildCreateTimeTAG: 2012-10-29T14:00:28Z SecondChildTAG: lol Don't use Chuck's name in vain..its dangerous SecondChildUserIdTAG: 97581 SecondChildUserNameTAG: Asimakis SecondChildCreateTimeTAG: 2012-10-29T15:52:34Z SecondChildTAG: hell no.... it can't ever be overall 100% without giving final exam.... SecondChildUserIdTAG: 134922 SecondChildUserNameTAG: spriha2010 SecondChildCreateTimeTAG: 2012-10-30T04:48:10Z SecondChildTAG: 100% Midterm and all labs and HW 100% SecondChildUserIdTAG: 99441 SecondChildUserNameTAG: coyarce SecondChildCreateTimeTAG: 2012-11-01T05:54:33Z FirstChildTAG: **Asimakis:** How did you get an "Overall percentage" of 100%? Did you take the course last semester? You cannot get a 100% overall without taking the Final Exam! **ahope:** Your statistics on "Overall percentage" will be a little off because not everyone is in the same place in the course. Some students have only finished Homework and Lab Week 6, others have done Homework 7 but not Lab 7, and yet others have completed both Homework 8 and Lab 8! So to put everyone on "even ground", you should have asked everyone for their "Overall percentage" with respect to a particular week; and adjusted it from there; as each successfully completed Lab and Homework set adds 3% to your "Overall percentage." FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-10-29T09:05:19Z SecondChildTAG: My mistake,I meant overall up to this point.So, to be exact 100% midterm and 54% overall (up to week 8). SecondChildUserIdTAG: 97581 SecondChildUserNameTAG: Asimakis SecondChildCreateTimeTAG: 2012-10-29T09:33:41Z SecondChildTAG: congrats. that is a great score;-) SecondChildUserIdTAG: 623210 SecondChildUserNameTAG: preveen SecondChildCreateTimeTAG: 2012-10-29T12:04:35Z SecondChildTAG: just 6% away from scoring a C already! SecondChildUserIdTAG: 623210 SecondChildUserNameTAG: preveen SecondChildCreateTimeTAG: 2012-10-29T12:05:04Z SecondChildTAG: TA's are really smart... @JereseyMark SecondChildUserIdTAG: 374590 SecondChildUserNameTAG: ahope SecondChildCreateTimeTAG: 2012-10-29T13:18:45Z SecondChildTAG: Hi Jersey Mark, can we have real statistics? in order to get an idea "where" exactly we are in the course, over/ under the average.... Thanks!! Regards, Sandra SecondChildUserIdTAG: 342966 SecondChildUserNameTAG: SandraNavarro SecondChildCreateTimeTAG: 2012-10-29T13:24:12Z SecondChildTAG: @Jersey Mark can you upload the result analysis or percentile so that we can get an idea regarding our position. SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-10-29T14:10:33Z SecondChildTAG: follow the link in the question above. SecondChildUserIdTAG: 374590 SecondChildUserNameTAG: ahope SecondChildCreateTimeTAG: 2012-10-29T15:52:54Z SecondChildTAG: **SandraNavarro** and **Vikaash:** Please understand that I do *not* have access to the "official" statistics on the percentage of students that passed, the mean grade, the standard deviation, and all of that good stuff. You are thinking of **staff**. Community TAs like me are just like the rest of you guys; MIT contacts us via email and lets us moderate the Discussion board. That's it. Oh, and we get a cool green tag next to our name. Personally, I have never even been to MIT's campus, but my university was in the general area. **All:** I just try to help out when I can; from what I hear the edX platform is supposed to be open-source, so edX / MITx may (not exactly sure) release the numbers that are used to come up with the statistics, so that others can analyze them for themselves. I don't know when / if they will release the source code to things like the circuit sandbox, so others can contribute code and make it truly great. One thing I do know is that last year when edX did the statistics, they found that most of the approximately 100k students that signed up for the course **and** did not complete the course - that it was the first homework set that was the drop-off point, not the mid-term; this is opposite from where "traditional" university students usually drop a course. That is interesting in that many people signed up just to sign up, on a whim, to evaluate the course's difficulty, because the course is "free", etc. If you sign up for a traditional university course, you never drop after the first homework set. You cannot really "evaluate" a course without paying the registrar to enroll. Most of the students that made it through the first homework set, suprisingly, stayed on until finish: that's a very good sign of both the determination of the students sticking with the course, and the quality and attractiveness of 6.002x itself. Jersey Mark SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-10-29T17:11:24Z SecondChildTAG: Thanks Jersey Mark :)) Good that Community TA existsssssss Cheers :)) SecondChildUserIdTAG: 342966 SecondChildUserNameTAG: SandraNavarro SecondChildCreateTimeTAG: 2012-10-29T20:15:45Z FirstChildTAG: MIDTERM 93% overall 45% FirstChildUserIdTAG: 219204 FirstChildUserNameTAG: vsriram FirstChildCreateTimeTAG: 2012-10-29T13:57:41Z SecondChildTAG: mdtrm 79% ovrall 44% SecondChildUserIdTAG: 95280 SecondChildUserNameTAG: Fortyq SecondChildCreateTimeTAG: 2012-10-29T14:22:21Z FirstChildTAG: MIDTERM 100% OVERALL 53% FirstChildUserIdTAG: 309359 FirstChildUserNameTAG: abarea10 FirstChildCreateTimeTAG: 2012-10-29T15:22:27Z FirstChildTAG: Midterm 100 % Overall 48% FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-10-29T15:37:24Z FirstChildTAG: Midterm 100% Overall 48% FirstChildUserIdTAG: 110802 FirstChildUserNameTAG: leoblack FirstChildCreateTimeTAG: 2012-10-30T01:12:25Z FirstChildTAG: Midterm 100% Overall 53% (Labs done through week 8, homework through week 7). :-) FirstChildUserIdTAG: 428560 FirstChildUserNameTAG: MikeDayton FirstChildCreateTimeTAG: 2012-10-30T02:32:37Z FirstChildTAG: Midterm 100% Overall 40% FirstChildUserIdTAG: 187466 FirstChildUserNameTAG: a9verma FirstChildCreateTimeTAG: 2012-10-30T07:13:11Z FirstChildTAG: Midterm 100% Overall 40% FirstChildUserIdTAG: 216684 FirstChildUserNameTAG: Taimoor1017 FirstChildCreateTimeTAG: 2012-10-30T12:31:00Z FirstChildTAG: Midterm 93%, Overall 45% FirstChildUserIdTAG: 114881 FirstChildUserNameTAG: xvink FirstChildCreateTimeTAG: 2012-10-30T20:03:01Z FirstChildTAG: Midterm 100%, Overall 45% (missed HW&Lab 06) FirstChildUserIdTAG: 383496 FirstChildUserNameTAG: winzard111 FirstChildCreateTimeTAG: 2012-10-30T23:37:33Z FirstChildTAG: This is a good example of a poorly constructed survey that yield meaningless data. The sample space will clearly be biased toward "eager beavers" with high scores. FirstChildUserIdTAG: 142857 FirstChildUserNameTAG: viking2 FirstChildCreateTimeTAG: 2012-11-01T16:19:47Z SecondChildTAG: That is unfortunately a good point. SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-06T18:20:16Z FirstChildTAG: mid term 100% with 1 hw and 1 lab not submitted...overall 47% FirstChildUserIdTAG: 146770 FirstChildUserNameTAG: solelonerajm FirstChildCreateTimeTAG: 2012-11-02T16:22:33Z FirstChildTAG: mid term 100% with 1 hw and 1 lab not submitted...overall 54% I am registration, after deadline 1 week. FirstChildUserIdTAG: 464744 FirstChildUserNameTAG: attache FirstChildCreateTimeTAG: 2012-11-02T17:22:34Z FirstChildTAG: MT: 100% , OA 54 % FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-11-03T17:54:43Z FirstChildTAG: M done with HW and Lab7... Midterm 100% and OA 51% FirstChildUserIdTAG: 399003 FirstChildUserNameTAG: SNEHALY FirstChildCreateTimeTAG: 2012-11-04T05:38:51Z FirstChildTAG: MT 97% OA 50% with HW & lab 7 submitted FirstChildUserIdTAG: 159850 FirstChildUserNameTAG: reticularformat FirstChildCreateTimeTAG: 2012-11-03T21:14:55Z FirstChildTAG: **Dissapointed MIDTERM** **MIDTERM:** 45% FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-11-04T07:25:22Z FirstChildTAG: Mid Term : 97.. Over all : 50 (till Week 7) FirstChildUserIdTAG: 332360 FirstChildUserNameTAG: srinivasav FirstChildCreateTimeTAG: 2012-11-04T09:57:02Z FirstChildTAG: Hey! Where is results? FirstChildUserIdTAG: 394836 FirstChildUserNameTAG: v2g6ch4 FirstChildCreateTimeTAG: 2012-11-06T15:28:33Z SecondChildTAG: survey results would be nice. SecondChildUserIdTAG: 12905 SecondChildUserNameTAG: Baer SecondChildCreateTimeTAG: 2012-11-12T19:34:20Z IndexTAG: 10 TitleTAG: H5P 1 : Zero offset Amplifier!! Actually in finding valid Operating ranges V0 = [-1 + \sqrt{1+2*K*RL*Vs}] /KRL In the above formula Vs= applied voltage across RL and Drain to Source( i.e Vds).. Vs=ids∗ RL +Vds (or) Vs=ids∗RL+V0 But in given problem , by applying KVL to output loop we get Vds+idsRL=Vs+(−Vs−) Vds+ids∗RL=1−(−1)=2 So in finding V0 substitute Vs=2 (NOT 1) And we know Valid maximum input range is Vgs = V0 + Vthershold Applying KVL to input loop Vin=Vgs+Vs- Substitue the value ...You will get it !!!!!! Cheers!!!! Hope it helps you .. UserIdTAG: 147459 UserNameTAG: Khanth CreateTimeTAG: 2012-10-11T15:19:11Z VoteTAG: 44 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 14 FirstChildTAG: thanks a lot... FirstChildUserIdTAG: 178840 FirstChildUserNameTAG: mehtanayanv FirstChildCreateTimeTAG: 2012-10-13T22:44:37Z SecondChildTAG: For part 3, do you have another way to show me why Vs = 2 and not 1? I used Vs = 1 for the second part and it worked, but maybe this was because it was for the node between the drain and RL, which didn't incorporate VS-. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-14T21:12:40Z FirstChildTAG: thanks. it helped alot... FirstChildUserIdTAG: 535777 FirstChildUserNameTAG: amnaliaquat10 FirstChildCreateTimeTAG: 2012-10-11T20:21:01Z SecondChildTAG: BRILLIANT way of explanation...Thank You :) SecondChildUserIdTAG: 122513 SecondChildUserNameTAG: Mona77 SecondChildCreateTimeTAG: 2012-10-12T06:29:41Z SecondChildTAG: but how to calculate RL for the above equation? need some help SecondChildUserIdTAG: 558276 SecondChildUserNameTAG: noormm SecondChildCreateTimeTAG: 2012-10-13T08:02:59Z FirstChildTAG: Thanks a lot you are the best :) FirstChildUserIdTAG: 309933 FirstChildUserNameTAG: miramar FirstChildCreateTimeTAG: 2012-10-14T11:17:56Z FirstChildTAG: Thanks for stopping many sleepless nights FirstChildUserIdTAG: 296419 FirstChildUserNameTAG: oscman FirstChildCreateTimeTAG: 2012-10-12T19:26:23Z FirstChildTAG: thank u soo much :) FirstChildUserIdTAG: 294497 FirstChildUserNameTAG: Saurabhkotian FirstChildCreateTimeTAG: 2012-10-14T11:38:38Z FirstChildTAG: great FirstChildUserIdTAG: 558276 FirstChildUserNameTAG: noormm FirstChildCreateTimeTAG: 2012-10-12T22:11:16Z SecondChildTAG: but how to calculate RL for the above equation? need some help SecondChildUserIdTAG: 558276 SecondChildUserNameTAG: noormm SecondChildCreateTimeTAG: 2012-10-13T08:20:41Z FirstChildTAG: thanks a lot. :) FirstChildUserIdTAG: 222911 FirstChildUserNameTAG: bhaswardg FirstChildCreateTimeTAG: 2012-10-13T04:59:09Z FirstChildTAG: Thanks a lot,,, FirstChildUserIdTAG: 368981 FirstChildUserNameTAG: asimakhtar FirstChildCreateTimeTAG: 2012-10-13T15:06:09Z FirstChildTAG: I feel you are very brilliant.It was till now the most troubling one for me.THANK YOU VERY MUCH. FirstChildUserIdTAG: 99628 FirstChildUserNameTAG: Pranjal16 FirstChildCreateTimeTAG: 2012-10-13T18:25:31Z FirstChildTAG: I have used your equations, but i can't find the right answer for the last question for the minimun value of Vin FirstChildUserIdTAG: 308861 FirstChildUserNameTAG: Ignaas FirstChildCreateTimeTAG: 2012-10-13T18:43:14Z FirstChildTAG: Khanth. I do not have words to thank you.. You helped me solve the problem of my life :p FirstChildUserIdTAG: 131726 FirstChildUserNameTAG: chemiboy11 FirstChildCreateTimeTAG: 2012-10-13T19:23:27Z FirstChildTAG: awesome.:)) FirstChildUserIdTAG: 266739 FirstChildUserNameTAG: satvikchugh FirstChildCreateTimeTAG: 2012-10-13T04:21:35Z FirstChildTAG: So the voltage from gate to source -- Vgs, actually creates a loop that KVL can be used on? I thought that it was supposed to be effectively an open circuit from gate to source... FirstChildUserIdTAG: 393689 FirstChildUserNameTAG: etindell FirstChildCreateTimeTAG: 2012-10-14T13:06:03Z FirstChildTAG: Thanks for the help. I kept trying it over and over and could not get the right answer even after reading all the suggestions on here. Turns out I didn't read the units of K very well. I had the right formulas all along but was always a factor of 1000 off. Check all your units guys! It could save you hours of frustration haha! FirstChildUserIdTAG: 417864 FirstChildUserNameTAG: beauclark FirstChildCreateTimeTAG: 2012-10-16T18:18:30Z IndexTAG: 11 TitleTAG: Have you seen this - AWESSSUMM ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13573619448853366.jpg Needless to say the Photo is ONE-O-ONE Percent Awesummm I am so happy to see that Mr Anant Agarwal Still Uses DEll, In a world of MAC BOOKS. I guess a Part Of him is Still Indian. If you are an indian YOu will know what i mean. REst of y'll :- Atleast Smile UserIdTAG: 378150 UserNameTAG: GladIDoThis CreateTimeTAG: 2013-01-05T05:01:52Z VoteTAG: 43 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: Like it! FirstChildUserIdTAG: 420339 FirstChildUserNameTAG: AliJenabi FirstChildCreateTimeTAG: 2013-01-05T06:03:26Z FirstChildTAG: And Yes i admit i did break the Forum Guidelines .. For repetition FirstChildUserIdTAG: 378150 FirstChildUserNameTAG: GladIDoThis FirstChildCreateTimeTAG: 2013-01-05T05:02:51Z FirstChildTAG: east or west INDIA is d best...INDIANS rock.. FirstChildUserIdTAG: 169416 FirstChildUserNameTAG: subramanya26shin FirstChildCreateTimeTAG: 2013-01-05T07:34:54Z FirstChildTAG: You don't have to be indian to use normal computers, and NOT overpriced garbage with a microkernel as OS. I am indian that way and always be. FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2013-01-06T15:51:47Z SecondChildTAG: I think he's referring to the fact that Prof. Agarwal has **Dell** monitors. It's the most popular PC brand here (I doubt this is the case only in India however). Virtually every laptop that people own are Dells. Both my monitors are Dell too. My parents use Dell laptops. At work I had Dell Optiplex desktops. I think HP used to be really popular before the Dell Inspiron and Studio series started catching up (the latter was unfortunately discontinued). SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2013-01-07T09:01:02Z SecondChildTAG: Ashwith i use dell too Inspiron one 2320 Yes a bit costly (60K ) But Equivalent to the I mac worth 2-4 Lakh Rupees So The Best deal in the market for Last 1.3 year SecondChildUserIdTAG: 378150 SecondChildUserNameTAG: GladIDoThis SecondChildCreateTimeTAG: 2013-01-07T10:44:44Z SecondChildTAG: Well only my monitors are dell. I use an assembled PC so almost each component comes from a different place. :-) SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2013-01-07T13:39:12Z SecondChildTAG: I was referring to macs.lol. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2013-02-11T08:24:58Z IndexTAG: 12 TitleTAG: Appreciation to MIT for this course Let me use this time to extend my thanks and appreciation to MIT for this course. UserIdTAG: 8419 UserNameTAG: Darlington CreateTimeTAG: 2012-09-05T22:57:00Z VoteTAG: 43 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 3 FirstChildTAG: I agree. FirstChildUserIdTAG: 280220 FirstChildUserNameTAG: pilgrimCycle FirstChildCreateTimeTAG: 2012-09-05T23:54:29Z SecondChildTAG: Ditto. SecondChildUserIdTAG: 376507 SecondChildUserNameTAG: kioko SecondChildCreateTimeTAG: 2012-09-06T00:16:47Z FirstChildTAG: I'd like to add that, although I'd read about some people having slight issues, the interface is the best I've seen. I think their consideration is amazing to provide some beautifully created tools. FirstChildUserIdTAG: 280220 FirstChildUserNameTAG: pilgrimCycle FirstChildCreateTimeTAG: 2012-09-07T11:21:17Z FirstChildTAG: I'd like to extend my thanks and appreciation to MIT for this course too. FirstChildUserIdTAG: 285612 FirstChildUserNameTAG: abdallah2012 FirstChildCreateTimeTAG: 2012-09-23T10:31:49Z IndexTAG: 13 TitleTAG: H8P1,H8P2, H8P3 Hints Requested by Maheenjd Hi Maheenjd, It was suppoused to Post this yesterday night but my internet service in my building was cut, luckily backed today. Lets see this so that later you can solve it by your own. ----- **H8P1.** The impulse response of a circuit is its response to a unit impulse, δ(t). Knowing the impulse response of a linear circuit is extremely valuable as we can figure out the circuit's response to an arbitrary input from it. In this problem you will find the response of a circuit when it is driven by a unit impulse. ![1][1] Consider the circuit shown above in which vIN(t)=1δ(t) volt-seconds. That is, vIN is a unit impulse at time t = 0. In the circuit, L=5mH, R1=10Ω, C=33nF and R2=10kΩ. Note that since this circuit is driven by an impulse and there is no other source of energy, the capacitor voltage and inductor current at t = 0 will be zero. **(a) What is the value of vR1(t) at t=0− in Volts (V)?** Part a. Hint. Well here the t=0- means a time a little bit before the time 0 , take a look at the orange zero-minus of the down image. ![IM1][2] Ok, now, they say that in the time zero , the green zero of the above image, you will have an impulse that will excite the circuit. So, What is an impulse? Take a look at page 485 , figure 9.44 [read here][3] You will see that you will have your impulse only in t=0. So, the Circuit will be excited thereafter to that time. So, the question is, will you have a supply before t=0? Hmmm, so, what would be your voltage in the resistor 1? ;) ---------- **(b) What is the value of vR1(t) at t=0+ in volts (V)? Hint: Recall that that ∫0+0−δ(t)dt=1** Hint. How can you find the voltage of R1 in a certain time? Well, that will be your task. But, lets see some hints that might be helpful to you : - Recall KVL. How can you relate the voltages in a loop? Take a look at the green LOOP. Can you relate the red voltage, the light blue voltage and the violet voltage? - What current do you have in R1? - Is R1 in series with L? So what current will you have? Isn´t it the same as the current in the inductor L? and if the current in the L vary in the time, so, the current in R1 will vary in the time too! - If you have the current that vary in the time of the element R1. Can you find the voltage that vary in the time of R1? Recall Ohm´s Law. - So, hmmm, if you have some expression of vR1 in terms of the time and elements, and they are requesting you to find it for a time 0+ , you can take t=0.000000001 or any close time of t=0 can you find the voltage for that time? ![IM2][4] - You should take a look at page 518. [read here][5] Note. You can solve this by Laplace too. If not, use the mechanism of homogeneus and particular solution, take a look at page 518... ---------- **(c) What is the value for vR1(t) at t=1ms in volts(V)? (Hint: The L-R1 and C-R2 branches of the circuit are decoupled.)** - You have from the previous part, part b, the voltage of vR1 in terms of the time, so... Can you find the voltage in a certain time ? yes :) ---- (**d) What is the value of vC(t) at t=0+ in volts (V)?** Hint. This is similar to part a. But, now you have to find the voltage in the capacitor and not in R2. ![IM3][6] - You should take a look at page 515 [read here][7] ---- **(e) What is the value for vC(t) at t=1ms in volts (V)?** - From your previous part d. , evaluate that expression for that time. ----- **H8P2** The electrical circuit language is a powerful language for modeling other kinds of systems, such as mechanical, chemical, thermal, and fluidic systems. Here is an example from physiology: drug delivery. We can model the amount of the drug by charge and the concentration of the drug by voltage. Thus, we get a very simple model: ![im4][8] I models the rate of injection of the drug into the body. vC models the concentration of the drug in the body. 1/R models the rate of elimination of the drug. C models the size of the body. So, let's do some numbers. Consider the consequences of injecting Q=400.0mg of cipro into a C=70kg person. Assume The person is entirely made of water, which weighs 1kg/L. **Part 1**. **What is the initial concentration after the injection vC(0+), of the antibiotic in the body, in mg/L?** - Hint. Recall that in a circuit a capacitance C multiplied by a voltage V will be equal to the charge Q. So, what will be the capacitance? and what will be the charge ? Can you find your v ? :). ---- **Part 2**. The half-life of the antibiotic in the person is about th=4.0Hours. Thus, the concentration will drop to half of its initial value in that time. **What is the resistance, in seconds/liter, in our model circuit, that will produce this rate of loss of concentration?** **A visual Hint.** You know that the drug will behave inside the body like a discharging capacitor in a circuit. Also, they tell you that your concentration [voltage] will be the half, so if your inicial value is Vinitial, by that time you will have Vinitial/2. So, do you know how a Capacitor dischages in a Circuit? Do you know your C ?Can you find your R? yes. Be careful with Units. ![im4][9] ---- **Part 3.** If the doctor wants to keep the concentration between 1/4 and 2/3 of the initial concentration, **what is longest time, in hours, that may be allowed to elapse before a new injection is required?** **A visual Hint.** ![im7][10] **P.D. I corrected the above image, I mistake in marking the T haha. Thank you Sergtronix for the correction in the post response ;).** ---- **Part 4.** **What dose, in milligrams, will be required at that time to get the concentration back to 2/3 of the initial concentration?** ![im4][11] --- **H8P3.** One possible implementation of one bit of a digital memory might be something like the circuit below. The memory element here is the gate-to-source capacitance of transistor Q2. Since the capacitance is leaky, this can hold a value only for a time determined by the time constant of the capacitance and the leakage resistance. Thus, such a memory has to be "refreshed" periodically, by reading out its value and rewriting that value. ![IM][12] We assume an "ancient" 1μm technology satisfying the static discipline: **VS=5.0V, VOH=3.5V, VIH=3.0V, VIL=0.9V, VOL=0.5V** The gate-source capacitance CGS=3.5fF and VT=1.0V. **RON=1850.0Ω, ROFF=95.0MΩ and RPU=18.0kΩ.** **Part 1.** The specifications for this logic family say that we have to refresh this memory every 64ms. So, if we charge the gate capacitance to VOH it will take more than 64ms to decay to VIH. If it takes 0.1s to decay that far, **what must be the parasitic resistance, in TerraOhms from the gate to the source of Q2?** Note: (1 TΩ=1012Ω). Hint . You know that it is charged to VOH. So, your initial value will be VOH. Do you now how to get the voltage when you discharge a Capacitor? Is that in terms of t, your R and your C? Can you find your R ? yes ;) ---- Part 2.**When DIN is low Q1 turns off and its drain goes high. What is the maximum value of voltage on the drain of Q1, in Volts?** In this part, you have to be careful with the resistences that you have to consider. RON or ROFF ?? They tell you that when DIN is low Q1 turns off. So what R , take a look at the red R in the down image, will you have, the ROFF or the RON? Hint. You know R, Rpul and Vs, Can you find the voltage on the drain of Q1, green V in the image? Another Hint. Recall voltage dividers. ---- Part 3. Now, suppose the drain of Q1 is high, as above, and the store line is held at the same voltage as the drain of Q1. **What is the maximum voltage, in Volts, that the gate of Q2 can be charged to?** Note, this value must be larger than VOH to satisfy the static discipline Visual Hint. ![im4][13] - Hint. Do you know your green V from the previous part ? - Hint. What was the condition that has to verify VGS3? more or equal than...., Recall Previous Week assingments :) - Hint. Recall KVL , light blue loop. ---- **Part 4.** **The drain of Q1 is still high and the gate of Q2 is at VOL. A STORE pulse comes in and turns on Q3. How long must the STORE pulse be on, in picoseconds, to charge the gate capacitance of Q2 to VOH?** Visual Hint. ![im5][14] - Hint. Consider all the R invlved, Rthevenin... - Your Orange R will be RON or ROFF? - Your Red R will be RON or ROFF? ---------- **Part 5.** Now, suppose DIN is high, the drain of Q1 is low, and the gate of Q2 is at VOH. A STORE pulse comes in and turns on Q3. **How long must the STORE pulse be on, in picoseconds, to discharge the gate capacitance of Q2 to VIL?** Hint. Really similar of Part 4. But here, you have to be careful with the red R, it is not the same as the part 4.... Hint. Consider a Capacitor discharging... Hint. Your final value will tend to VOL. ---- I hope this can help you. See you, Myriam. [1]: https://www.edx.org/static/content-mit-6002x/images/circuits/H8P1Impulse.63e08049f83e.png [2]: https://edxuploads.s3.amazonaws.com/13525502101818754.png [3]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/509 [4]: https://edxuploads.s3.amazonaws.com/1352555442983032.png [5]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/542 [6]: https://edxuploads.s3.amazonaws.com/13525578541343662.png [7]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/539 [8]: https://www.edx.org/static/content-mit-6002x/images/circuits/norton-capacitor.af0dd9613fc1.gif [9]: https://edxuploads.s3.amazonaws.com/13525601784250427.png [10]: https://edxuploads.s3.amazonaws.com/13525709446971753.png [11]: https://edxuploads.s3.amazonaws.com/13525611501343682.png [12]: https://www.edx.org/static/content-mit-6002x/images/circuits/memory.3e1311910007.gif [13]: https://edxuploads.s3.amazonaws.com/13525647931343686.png [14]: https://edxuploads.s3.amazonaws.com/13525660581343641.png UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2012-11-10T14:44:38Z VoteTAG: 42 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 21 FirstChildTAG: hey i am not able to solve h8p2 last part....pls help FirstChildUserIdTAG: 346960 FirstChildUserNameTAG: priya91 FirstChildCreateTimeTAG: 2012-11-11T12:10:25Z FirstChildTAG: Thanks Myrimit for the part you've illustrated and also in advance. FirstChildUserIdTAG: 183507 FirstChildUserNameTAG: obiradaniel FirstChildCreateTimeTAG: 2012-11-10T14:50:52Z SecondChildTAG: You are welcome obiradaniel. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-10T18:12:39Z SecondChildTAG: hey for the 3rd question 4th part, i have got my resistance values right i guess..but the graph shows voltage B as maximum..according to the question,we have to find the time taken to reach the value VOH..but maximum value is calculated from the part 3 right? can you just tell me if my eqn is right.. VOH=VOL+(Vmax-VOL)*(1-e^(-t/(Rth*C))..is this equation right? SecondChildUserIdTAG: 156585 SecondChildUserNameTAG: AdiRanga SecondChildCreateTimeTAG: 2012-11-10T18:51:27Z FirstChildTAG: Miriam, allow me to ask.. Your visual hint for H8P2 part3 "what is longest time, in hours, that may be allowed to elapse before a new injection is required" I think that is the time for 1/4 concentration.Am I wrong? FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-11-10T16:08:11Z SecondChildTAG: No, you are right :) SecondChildUserIdTAG: 137087 SecondChildUserNameTAG: daxiel22 SecondChildCreateTimeTAG: 2012-11-10T16:22:29Z SecondChildTAG: Thank you Sergtronix. I was testing you of you were sleepy (nah joke) haha. I corrected the image, I marked incorreclty the T when I draw the image. Thank you thank you. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-10T18:14:07Z SecondChildTAG: .if SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-10T18:15:53Z SecondChildTAG: No, Miriam.. Thanks to you! Your understanding this course is amazing. I do really admire you! My best wishes, Serge SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-11-10T18:22:53Z FirstChildTAG: Hi Miriam Regarding H8P3, part 1, I can't seem to find the right combination of variables to plug into the equation for calculating the voltage when discharging the capacitor. First, I'm using the equation Vc=Vfinal+(Vinital-Vfinal)*e^(-t/(R*C)), and I'm assuming we solve for R, which in this case is the parasitic resistance we're looking for. My problem is, I seem to be missing a variable. I set Vinitial = VOH, Vfinal = VIH, and of course we're given t and C, but what do I use for Vc? I would think it would be Vfinal, but then R = 0. What am I missing? Thanks FirstChildUserIdTAG: 132828 FirstChildUserNameTAG: rharris FirstChildCreateTimeTAG: 2012-11-10T18:03:54Z SecondChildTAG: You do have two voltages Vcharged and Vdefined, where is Vdefined is VIH. Next, your net is only Rpar and Cgs.What is eqation should be used here? I think that VIH=VOH*e(T/tau) SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-11-10T18:56:37Z SecondChildTAG: I had the correct answer all along. No wonder I was so confused, the question is so simple. I have converted so many milliseconds and microseconds to seconds that in this case I was moving the decimal point to the left, where I should have been moving it to the right. My answer from my calculator was similar to .yyy X 10^14, so my answer should be in the form yy.y X 10^12. I guess it doesn't pay to work until 1:30 AM in the morning. Thanks for putting up with my convoluted brain. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-11-10T20:16:35Z SecondChildTAG: Hi rharris, Can I help you ? Lets suppouse that your answer is 5*10^14 Ohms. They ask you, the answer in Tera Ohms, a Tera Ohm is 1*10^12 Ohms. So, you have to use a 3 simple rule like this, ![im][1] [1]: https://edxuploads.s3.amazonaws.com/1352590823806461.png SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-10T23:40:34Z SecondChildTAG: I still don't understand what Vc is supposed to be. It does seem like it should be VIH, but, if you assign it that, then R=0 as rharris mentioned. SecondChildUserIdTAG: 346777 SecondChildUserNameTAG: fishmael SecondChildCreateTimeTAG: 2012-11-11T03:51:17Z SecondChildTAG: Never mind. I managed to figure it out. Thanks for all the hints. SecondChildUserIdTAG: 346777 SecondChildUserNameTAG: fishmael SecondChildCreateTimeTAG: 2012-11-11T04:24:00Z SecondChildTAG: Well done fishmael :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-11T05:09:53Z FirstChildTAG: I'm trying to do H8P3 part 5. We're told that dIN is high, so that the drain of Q1 must be low. To calculate this voltage, I simply focus on the Q1 portion of the circuit and do a simple voltage divider relationship with VS, Rpu, and Ron. ...but Ron is a quite high resistance, and I end up with a 'low' Q1 drain voltage of about 2.69V. This is well over VOL, and breaks the static discipline, so I think I've done something wrong... but my whole analysis can't proceed any further because that value really makes no sense. Would anyone like to set me straight on this point? FirstChildUserIdTAG: 287805 FirstChildUserNameTAG: Beneficial FirstChildCreateTimeTAG: 2012-11-10T18:24:13Z SecondChildTAG: for Q1 Vd is 2.69V Ron should be comparable with Rpu. I think somewhere your calculations are incorrect. Really you should get Vd(Q1 is ON) =<0.9V....Tryy to check your calculations again. PS While an initial values differs , you may give us values from your homework version.We may try to help you SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-11-10T18:38:47Z SecondChildTAG: Hi Benefical, Hint. On state and Off state image, ![state][1] Another Hint. ![ima][2] [1]: https://edxuploads.s3.amazonaws.com/13525917727110422.png [2]: https://edxuploads.s3.amazonaws.com/1352591823220929.png SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-10T23:58:06Z SecondChildTAG: I'm really still struggling with this initial step in the analysis. We're told Q1 is a logical low, which means q1 mosfet is on, so the resistor in red that you've drawn is RON. We're also told the source has turned Q3 on...so the Orange branch you've drawn is also RON. I tried a bunch of methods for solving for the drain voltage of Q1, but all the numbers I get are ridiculous. I eventually resorted to a good old nodal analysis, but that didn't get me anywhere either. I ended up with a value of 8V for Q1, but as far as I understand the question, it should have to at least be a value below 0.9V. My nodal analysis, in case I've done something boneheaded, you can correct it: Let Q1 drain be unknown voltage e1. Taking all currents to be leaving the node and to be positive: 5-e1/RPU + e1/RON + 3.5-e1/RON = 0 a little algebra later, e1 = 8V I'm still confused. SecondChildUserIdTAG: 287805 SecondChildUserNameTAG: Beneficial SecondChildCreateTimeTAG: 2012-11-11T06:25:00Z SecondChildTAG: For greater clarity: (5-e1)/RPU + e1/RON + (3.5-e1)/RON = 0 SecondChildUserIdTAG: 287805 SecondChildUserNameTAG: Beneficial SecondChildCreateTimeTAG: 2012-11-11T06:26:31Z SecondChildTAG: can you xplain H8P3 part 5 and 4 clearly!! thanks a lot.. SecondChildUserIdTAG: 424518 SecondChildUserNameTAG: sathvika SecondChildCreateTimeTAG: 2012-11-11T06:31:49Z SecondChildTAG: **Beneficial** As example.Let Rpu is 15 kOhm, Ron is 1 kOhm and Roff is 100 MOhm. For Q1ON Vdrain is: Vdrain=Vs*(Ron)/(Ron+Rpu)=0.3125V. Note that Rdischarge will be Rthev(Q1)+Ron(Q3)=[Rpu*Ron/(Rpu+Ron)]+Ron. So you have Vth=0.3125V and Rdis is 1937.5Ohm; For Q1(off) Vdrain is Vd(Q1)=Vs*Roff)/(Roff+Rpu)=4.9999V I hope it will help you. SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-11-11T10:13:39Z SecondChildTAG: Ugh... oh boy. I feel a little silly. I think the mistake I've been making for two straight days on this issue, is that I've been using RON as 1800, rather than 18000 as given. Sigh... SecondChildUserIdTAG: 287805 SecondChildUserNameTAG: Beneficial SecondChildCreateTimeTAG: 2012-11-11T16:04:12Z SecondChildTAG: yep, as I wrote you above :) Good luck! SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-11-11T16:38:30Z SecondChildTAG: Thanks for all the help Sergtronix and Myrimit. SecondChildUserIdTAG: 287805 SecondChildUserNameTAG: Beneficial SecondChildCreateTimeTAG: 2012-11-11T16:39:03Z SecondChildTAG: You are welcome Beneficial. Well done :)! Haha, yes sometimes we miss some zeros by accident haha - I am on the list haha-. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-11T19:20:00Z FirstChildTAG: By the meantime, while I do afraid any doctors and of course any injections :) I cant understand H8P2 last question.. Miriam, could you give more hints for part4 , please ? Ok, I got green sign..but it is crazy.Im disappointed.. FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-11-10T18:25:53Z SecondChildTAG: please give me the hint about the solution. i am simply taking it as the subtraction between the *2/3.V - 1/4.V*. But it marks my answer with a red cross. SecondChildUserIdTAG: 64618 SecondChildUserNameTAG: ashfaq2419 SecondChildCreateTimeTAG: 2012-11-10T22:54:00Z SecondChildTAG: done it!!! thanks a lot dear SecondChildUserIdTAG: 64618 SecondChildUserNameTAG: ashfaq2419 SecondChildCreateTimeTAG: 2012-11-10T22:58:03Z SecondChildTAG: Well done ashfaq2419 ! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-11T00:01:22Z SecondChildTAG: Can I help you Sergtronix? SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-11T00:02:01Z SecondChildTAG: plz help me..h8p1 last part.. SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-11-11T09:04:32Z SecondChildTAG: its h8p2 last part SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-11-11T09:24:24Z SecondChildTAG: > Can I help you Sergtronix? –posted about 10 hours ago by > MyrimitCOMMUNITY TA Miriam, thanks! I coulndt imagine that solution is so crazy simple for this part.. It is done already SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-11-11T10:18:17Z SecondChildTAG: i need help with h8p2 last part too . plz SecondChildUserIdTAG: 285675 SecondChildUserNameTAG: Ascot SecondChildCreateTimeTAG: 2012-11-11T15:34:08Z SecondChildTAG: **Ascot** Youre asked about How many milligrams ... Could you calculate amount of drug for 1/4 and 2/3 from an initial value? SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-11-11T16:48:16Z SecondChildTAG: Well done Sergtronix! ---- Hi Ascot. Can I help you? Hint. Take a look at the **blue line** with the symbol - **?**- Ok, now, lets try to think, How much is that quantity? e g., if you have for example 10 Volts and then 3 Volts, How much will be that quantity? 10V-3V=7 V. Ok, but is 7 V the concentration? Remember that the voltage was analogue to units mg/L and they are requesting you the mg ;). So..., you are missing something... Take a look again to Part 1 Hint, you will see How V is related with others parameters ;) I hope this can help you, Myriam. ![image][1] [1]: https://edxuploads.s3.amazonaws.com/13525611501343682.png SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-11T19:34:16Z FirstChildTAG: About H8P3 Q4. I can't undestand why Rth not eq to Ron and charge by voltage divider FirstChildUserIdTAG: 406023 FirstChildUserNameTAG: neitrino FirstChildCreateTimeTAG: 2012-11-10T20:58:15Z SecondChildTAG: Hi neitrino, Hint. What resistance do you have between the gate-source of the Transistor Q2? ![IM9][1] [1]: https://edxuploads.s3.amazonaws.com/13525922729145258.png SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-11T00:04:45Z SecondChildTAG: I have the same question. It would be an series RC circuit with a supply voltage of 4.9994 and a resistance of Ron... But the result is wrong.... Time goes by... SecondChildUserIdTAG: 239608 SecondChildUserNameTAG: guillegf84 SecondChildCreateTimeTAG: 2012-11-11T17:05:54Z FirstChildTAG: Hello everyone. I have some problems to solve H8P3 Q3. I think it has not to be very difficult, but I cannot really see how to solve it. I know the condition that has to satisfy the VGS3 and although I have written the KVL ecuation I have two unknowns, VGS3 and VGS2, and I do not know how to use the condition for VGS3. Can anyone help me? Thank you a lot. FirstChildUserIdTAG: 384763 FirstChildUserNameTAG: DanielRR FirstChildCreateTimeTAG: 2012-11-11T05:21:26Z SecondChildTAG: VGS3 is the max. value of the input voltage for which it take as logic '0'. SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-11-11T09:38:38Z SecondChildTAG: OH, I didn't understand the question until now that you have written that to me and I have read the question again. Thank you so much Vikaash for your help ^^. SecondChildUserIdTAG: 384763 SecondChildUserNameTAG: DanielRR SecondChildCreateTimeTAG: 2012-11-11T12:11:43Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-11T19:36:19Z FirstChildTAG: HI Myrimit!! I sovle the h8p1 question B and D ,but I cant work out the question C and E! As you say, I just put the certain time 1ms in my equation V-V*e^(-tR/L) for question C or V*e^(-t/RC) for question E, It doesn't work! I just think about that should I change the value of V to 1000 when the time is 1ms??? Give me a hand!! FirstChildUserIdTAG: 296511 FirstChildUserNameTAG: sky0917 FirstChildCreateTimeTAG: 2012-11-11T08:21:55Z SecondChildTAG: i am facing the same problem here.. SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-11-11T09:40:32Z SecondChildTAG: wow~ if you work it out ~ please let me know! SecondChildUserIdTAG: 296511 SecondChildUserNameTAG: sky0917 SecondChildCreateTimeTAG: 2012-11-11T10:23:24Z SecondChildTAG: @Vikaash for RL Vl(1ms) = (R/L)*e^(-R*t/L) for RC VC(1ms)= 1/(RC)*e^(-t/RC) SecondChildUserIdTAG: 296511 SecondChildUserNameTAG: sky0917 SecondChildCreateTimeTAG: 2012-11-11T11:43:59Z SecondChildTAG: @sky0917 thanks a lot... SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-11-11T12:01:42Z SecondChildTAG: just differentiate the normal formula you work on , thats where the formulas Vikaash wrote , come from SecondChildUserIdTAG: 285675 SecondChildUserNameTAG: Ascot SecondChildCreateTimeTAG: 2012-11-11T15:35:49Z FirstChildTAG: Dear friends, In H8P1, when we deal with t=0+, what value of t should be inserted in the equation to find the voltage/current e.g. across the inductor where VL=V.e^(-Rt/L)? FirstChildUserIdTAG: 64618 FirstChildUserNameTAG: ashfaq2419 FirstChildCreateTimeTAG: 2012-11-11T08:51:44Z FirstChildTAG: Awesome Myrimit! Very helpfull, many thanks! FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-11-11T09:27:49Z SecondChildTAG: You are welcome hazel1919. Nice to know that they are been helpul :). By the way, How are your projects? SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-11T19:39:23Z FirstChildTAG: Thanks Myriam :) You are really doing a great job !!!!! A big thanks to you :) May God bless you !!! FirstChildUserIdTAG: 145239 FirstChildUserNameTAG: Maheenjd FirstChildCreateTimeTAG: 2012-11-11T09:32:16Z SecondChildTAG: You are welcome Maheenjd :). I hope that this Hints could help you. My best wish to you! Myriam P.D. If you still have any doubt please tell me, I would like to help you. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-11T19:41:22Z FirstChildTAG: In H8P1, we know that the area of the pulse is 1.So , without a lot of headaches look at the page 553 from the book https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/577 FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-11-11T11:54:59Z FirstChildTAG: PLEASE GIVE SOME EASY HINT ABOUT H8P3 (LAST PART), WHERE DIN IS LOW AND WE HAVE TO FIND THE TIME SO THAT THE CAPACITANCE OF Q2 SET THE VOLTAGE TO VIL. I HAVE TRIED EVERYTHING BUT TO NO AVAIL. I HAVE ALSO SEEN THE ABOVE EXPLANATIONS BUT STILL CONFUSED. NEED SOME URGENT HELP. THANKS IN ADVANCE. FirstChildUserIdTAG: 64618 FirstChildUserNameTAG: ashfaq2419 FirstChildCreateTimeTAG: 2012-11-11T12:47:17Z SecondChildTAG: Please please read carefuly. All necessary infos are here thanks SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-11-11T16:35:46Z FirstChildTAG: Thank you very much, Miriam ! You do really ahuge job for all of us. Good luck in everything!!! FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-11-11T10:28:37Z SecondChildTAG: You are welcome Sergtronix ;)! I really enjoy doing this ! Thank you, good luck to you too! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-11T19:42:56Z FirstChildTAG: in H8P2 part 4 , shouldn't i calcualte on the rising curve as i want to calculate from 1/4 to 2/3 or it doesn't matter ? FirstChildUserIdTAG: 285675 FirstChildUserNameTAG: Ascot FirstChildCreateTimeTAG: 2012-11-11T15:39:40Z FirstChildTAG: I have a question, I just got H8P3, part 4 correct using the voltage from part 2 as VIN, however the first time I calculated t, I got it wrong. I used VIN from part 3, as it was the maximum voltage the gate of Q2 could be charged. If using the answer from part 3 is the highest the gate of Q2 can be charged, why is the answer from part 2 used for VIN, which is higher than the answer from part 3? FirstChildUserIdTAG: 132828 FirstChildUserNameTAG: rharris FirstChildCreateTimeTAG: 2012-11-11T20:48:24Z FirstChildTAG: @Myriam, Thanks for your thorough insight and thoughtfulness in explaining these problems. I attempted H8P1 and the last parts of H8P3 for the last two weeks without achieving the coveted green checks! Your hints and the discussions that they generated allowed me to work through these problems. I can't thank you enough. Ron FirstChildUserIdTAG: 11730 FirstChildUserNameTAG: RonFlip FirstChildCreateTimeTAG: 2012-11-12T01:02:28Z FirstChildTAG: Thanks a lot! brilliant work FirstChildUserIdTAG: 374228 FirstChildUserNameTAG: geechu FirstChildCreateTimeTAG: 2012-11-12T00:14:43Z FirstChildTAG: Myrimit, your hints rock! Thank you!! FirstChildUserIdTAG: 393045 FirstChildUserNameTAG: SrChasJC FirstChildCreateTimeTAG: 2012-11-12T04:24:48Z FirstChildTAG: Hey everyone, I need help on H8 P1 part (b).I know that VL = V.e^(-R/L).t.But as t tends to 0, VL becomes just V. Isn't VL hence 1 since V = 1(length of impulse) and thus by KVL, VR1 = 0? But thats not the ans. Why? FirstChildUserIdTAG: 536922 FirstChildUserNameTAG: arjshar FirstChildCreateTimeTAG: 2012-11-12T06:22:09Z IndexTAG: 14 TitleTAG: H9P1, H9P2, H9P3 Hints requested by arjshar H9P1 -- Note: In this problem we have chosen numbers for the part parameters to make it easier to compute an answer :-). By the way, it is also hard to arrange zero resistance, except with superconducting materials at very low temperatures. In the circuit shown below L=125.0H and C=3.24mF. ![im1](https://www.edx.org/static/content-mit-6002x/images/circuits/ILC.2ef75c14167d.gif) The current source puts out an impulse of area A=2/π=0.64 Coulombs at time t=5.0s. At t=0 the state is: vC(0)=0.0 and iL(0)=1.0. The equation governing the evolution of the inductor current in this circuit is Hints. **Part 1.** - Take a look at the page 629 [read here][1]. - Remember that $2*\pi*f = \omega_0$ as this is in Hz. **Part 2.** - Read page 634 equation 12.28. [read here][2] - Do you have your initial conditions? Can you find the energy ? ;) **Part 3.** - Do you have disipation, resistive components?, So, what will be the energy? Will it change? ;). **Visual Hints - Part 4. and 5.** ![image][3] **Part 6.** At the time just after the impulse happens what is the current iL(5.0+), in Amperes, through the inductor? - The current can change that quickly, so, what it would be your value? ;) **Part 7.** At the time just after the impulse happens what is the voltage vC(5.0+), in Volts, across the capacitor? - What is your value of your vC(5.0-) and what happens in t=5? vC(5.0+) = vC(5.0-) + some_voltage(pink line of previous image) ;) **Part 8** At the time just after the impulse happens what is the total energy, in Joules, stored in the circuit? - Do you have the current of the inductor and the voltage of the capacitor in the part 6 and 7 ? So, recall page 634. Can you find the energy after that time? ;). ----- H9P2 -- Consider the second order low pass filter circuit shown below. ![im2][4] The differential equation associated with the input voltage vi(t) and the output voltage vo(t) of this system can be written in the following form: Hints. $A \cdot \frac{d^2v_o}{dt^2} + B \cdot \frac{dv_o}{dt} + C \cdot v_o = v_i$ **General Hints:** First of all, lets try to find an equation that relates iL with vO. ![circ 2][5] - KCL - Can you relate the iC in terms of vO and C? Yes! - Can you find your iR in terms of vO and R? Yes! - So, can you find your iL in terms of Vo? Yes! ;) Now take a look at the light blue loop ;). ![loop][6] - KVL. - Can you express your vL in terms of iL? ;) - Do you have your iL in terms of vO?Yes! - Can you obtain your Vi in function of your Vo, R,L and C with lapace? ;) **Hints Part a, b and c**. By comparing your Vi in terms of Vo, R, L and C, can you obtain your requested coefficients? :). Yes! **Part d)** The circuit would ring when excited with a step response vi=u(t). What is the natural frequency of this circuit, in terms of the component values? *Hint part d)* Now, with your expression of Vi in terms of Vo, R, L nad C , find its characteristic equation. Remember that the coefficient of s^2 has to be 1 for this equation. Take a look at page 641 (12.43) . By comparing, what is your wo value? ;) *Another part d) Hint.* Remember that $2*\pi* f = \omega_0$ as the uhnit is in Hz. **Part e)**. The ringing will be damped by the factor e−αt. What is the expression for α in terms of the component values? *Hint part e).* Compare the characteristic equation with the $\alpha$ given in 641 page (12.43) :) **Part f).** What is the expression for the "Quality Factor" Q of this circuit, in terms of the component values? *Hint Part f).* Remember that $Q = \frac{\omega_0}{2* \alpha}$ **Part g).** Suppose we need to suppress the ringing. We could change the value of R to make this circuit critically damped (Hint: make the Q = 0.5). What is the expression for this critical R, in terms of L and/or C? *Hint Part g).* Remember that the condition when is critically dumped is when $\alpha = \omega_0$ . Do you have $\alpha $ ? Do you have $\omega_0$ from your previous part? Can you find your R? ;) ----- H9P3 -- Hints H9P3. One way to model a vehicle is as a massive object that is connected by springs and shock absorbers to the wheels. Here is a simple model: ![im][7] $m \frac{d^2 y(t)}{dt^2} + b \frac{d y(t)}{dt} + k (y(t) - Y_{road}(t)) = 0$ Now this equation looks very much like the equation governing a driven RLC circuit: ![IM][8] $L C \frac{d^2 v_C(t)}{dt^2} + R C \frac{d v_C(t)}{dt} + v_C(t) = V(t)$ The also tell you that Since this is second order the characteristic polynomial is of the form: $ s^2 + 2 \alpha s + \omega_0^2 = 0$ **Hints Part 1 and 2:** *Hint .* Now, try to find your characteristic equation of your model ;), that is to say, the coefficient of s^2 have to be 1 ;). *Hint .* If you have done Hint 1, by comparing with the given formula in the statement of the characteristic polynomial. Can you find your $\alpha$, $\omega_0$ and ? **Part 3.** Suppose that the shock absorber is dead, so b=0. Write an algebraic expression for the frequency, in Hertz, of the bouncing of our vehicle. Your expression should be in terms of m and k. *Hint. Part 3* Make b=0, take a look at your characteristic equation. ;). What will be know your wo? *Another Hint Part 3.* $2*\pi*f$ = wo in Hz *Another Hint Part 3 too.* You can write pi instead of $\pi$ in the Text box. **Part 4.** We can make the frequency of the bouncing go to zero by picking an appropriate value of the viscous-friction coefficient b. (This is called critical damping.) Write an algebraic expression for the value of b that will make this vehicle critically damped. *Hint part 4.* Critically dumped condition page 644 [read here]( https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/668) ;). Take a look at that condition, you will reach to your answer! --- Oh my, I had a lot of trouble posting this haha, two times my hints were accidentally erased :P- Now I am going to review for 3.091x Exam, I haven’t opened yet Ooops. See you at night or tomorrow. If you have doubts post it , I will try to help you later :p See you, Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/653 [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/658 [3]: https://edxuploads.s3.amazonaws.com/13531693699488369.bmp [4]: https://www.edx.org/static/content-mit-6002x/images/circuits/H9P2DSOC.549931dd0d35.png [5]: https://edxuploads.s3.amazonaws.com/13531731801343676.bmp [6]: https://edxuploads.s3.amazonaws.com/13531733914685996.bmp [7]: https://www.edx.org/static/content-mit-6002x/images/circuits/Vehicle.6a336fd3cd85.gif [8]: https://www.edx.org/static/content-mit-6002x/images/circuits/VRLC.ac3871e39d10.gif UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2012-11-17T14:43:21Z VoteTAG: 40 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 12 FirstChildTAG: i don't know how get h9p2(e). I tried to compare (b) solution with 2*alpha but i don't get right solution FirstChildUserIdTAG: 314294 FirstChildUserNameTAG: victormp FirstChildCreateTimeTAG: 2012-11-18T20:03:43Z SecondChildTAG: Hi victormp, Can I help you? Have you obtained your characteristic equation with the components of the Circuit? Hint. Try to analyze the coefficients of your characteristic equation of the Circuit - If you take a look at H9P3 in the statement they tell you about how is a characteristic equation. So, by comparing, what is your alpha? Hint 2. an arbitrary example, lets say that in your circuit you have the value 8 in the coefficient of s, also you know that in the characteristic equation, the coefficient of s is $2\alpha$ , so $2\alpha$*s =8*s , your circuit alpha will be 4 ;). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-18T21:14:35Z FirstChildTAG: @H9P1 part 4 Graph makes it clear, but how do I get the result without solving DE and getting the equation for iL(t) ? FirstChildUserIdTAG: 270284 FirstChildUserNameTAG: nkukushkin FirstChildCreateTimeTAG: 2012-11-17T16:59:46Z SecondChildTAG: no need of calculation...you just need to "observe" and think :) SecondChildUserIdTAG: 122513 SecondChildUserNameTAG: Mona77 SecondChildCreateTimeTAG: 2012-11-17T17:35:14Z SecondChildTAG: could you be a bit more precise? :P SecondChildUserIdTAG: 270284 SecondChildUserNameTAG: nkukushkin SecondChildCreateTimeTAG: 2012-11-17T17:44:33Z SecondChildTAG: is current changing? SecondChildUserIdTAG: 122513 SecondChildUserNameTAG: Mona77 SecondChildCreateTimeTAG: 2012-11-17T19:10:21Z SecondChildTAG: inductor current goes up and down, yes but how do I know it's position at a certain time? SecondChildUserIdTAG: 270284 SecondChildUserNameTAG: nkukushkin SecondChildCreateTimeTAG: 2012-11-17T19:16:35Z SecondChildTAG: The initial condition tells you that at t=0 the inductor has 1A. In part one you calculated the frequency. With that information you know the position at any given time SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-17T21:27:47Z FirstChildTAG: Myrimit, u rock! U know that right? :) FirstChildUserIdTAG: 499671 FirstChildUserNameTAG: maliha266 FirstChildCreateTimeTAG: 2012-11-17T16:43:32Z SecondChildTAG: Hahaha Thank you :)! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-17T18:48:50Z FirstChildTAG: Thanks a lot Myrimit for the hints! FirstChildUserIdTAG: 536922 FirstChildUserNameTAG: arjshar FirstChildCreateTimeTAG: 2012-11-17T18:10:02Z SecondChildTAG: You are welcome. :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-17T18:48:22Z FirstChildTAG: Hi) Thank's for your hints) Can you tell more information about H9P1 task5?) FirstChildUserIdTAG: 192114 FirstChildUserNameTAG: chuba FirstChildCreateTimeTAG: 2012-11-17T18:42:23Z SecondChildTAG: All problem was in sine) SecondChildUserIdTAG: 192114 SecondChildUserNameTAG: chuba SecondChildCreateTimeTAG: 2012-11-17T19:40:13Z FirstChildTAG: I can not understand why the question 4 H9P1 current is zero, and the question number 5 maximum voltage. Please explain this with the help of equations. FirstChildUserIdTAG: 196404 FirstChildUserNameTAG: Yel1owstone FirstChildCreateTimeTAG: 2012-11-17T21:13:10Z SecondChildTAG: In part 4 I think that you know that current is zero because the initial condition and frequency that you calculated on part one. Part 5, you need to remember how to calculate the energy for inductor and capacitor (S17V17). On part 2 you calculated the total energy of the system. So the total energy of the system is the energy on the inductor + the energy on the capacitor. From there you can calculate the capacitor voltage (you already know the inductor current) SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-17T21:45:41Z SecondChildTAG: Matias was subtle.The natural frequency is 0.25 , so the period of one oscillation is 4 seconds.So in second 4, iL is 1 again since there is no loss, and in second 5- it drops to 0.Just make a draw man. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-11-17T22:40:06Z SecondChildTAG: Thank you. I decided this way. But when I'm trying to solve analytically, it does not work. (In my opinion) i(t)=V*C*w0*sin(w0*t); where w0=1/(sqrt(L*C))=1.572; t=1 s. And sin(1.572*1)=0.999999. Not zero. that is, the current is not zero. Where is mistake? SecondChildUserIdTAG: 196404 SecondChildUserNameTAG: Yel1owstone SecondChildCreateTimeTAG: 2012-11-18T11:52:16Z FirstChildTAG: Thanks Myriam ! It is always a pleasure to read your post and explanations, founded on logic and book examples. It is uncommon to see so much dedication. FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-11-17T22:47:08Z SecondChildTAG: You are welcome @AlexAlexandrescu. Thank you for the nice words haha. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-18T21:17:30Z FirstChildTAG: H9P1, parts 3, 4 and 5 with impulse at 9s. Ok, It's my turn. I took a different approach and solved the differential equation and came up with Vp = A and Vh = A1*e^.618j +A2*e^-.618j, with s = +/-.618j. I then took the leap of faith and considered that since the left side of the equation was real, the right side had to be real also. So, using the concept of complex conjugates and Euler's theorem, I came up with Vh = A + A1*Cos(.618t) + A2*Sin(.618t). I then added Vp to Vh and got .64 - .64*Cos(.618t) + A2*Sin(.618t). I then found the current equation by using C*dv/dt and using the initial conditions I got A2 = 360 (Ok, it thought this was high, but I continued). I(t) turned out to be = Cos(.618t) - .00178*Sin(.618t). I then calculated i(9) and came up with the current, and then calculated the energy with E = (L/2)*i^2 and got the **correct** answer. The funny thing is, I used degrees, because radians didn't work (rare chance this is a coincidence, but I got the correct answer). Ok, I thought that was great and then used the current I got to calculate the energy for question 4 and it was wrong. Could it be the answer I got for the energy was somehow a coincidence? If is was, that was bazaar. Can someone point me in the right direction? I'll now try it graphically like Myriam did, however it may take a while as Dr. Agarwal's example was for a series RLC circuit. For the heck of it I used the value of current at 9s the same as the current from the graph that Myriam displayed in the hints and it worked. Now I'll have to reevaluate my equation I calculated and see what went wrong. I should get the same answer whether It came from a graph or an equation. FirstChildUserIdTAG: 132828 FirstChildUserNameTAG: rharris FirstChildCreateTimeTAG: 2012-11-18T00:28:58Z SecondChildTAG: Hi Myriam First, to get the waveforms in your visual hints, you calculated V(t) = Vp + Vh and obtained the constants, correct? Otherwise, we were never shown an intuitive method to figure out the zero crossings of the current, which would give the current after 9s. - Thanks Also, I tried to determine by intuition which way the current and voltage went, up or down, after time t = 0. The professor was clear on how to do it with a series circuit, but not with a parallel circuit. How did you do that? ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/1353210617134367.jpg SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-11-18T04:10:13Z SecondChildTAG: Hi, rharris. Look at videos S17V16 and S17V17 (Undriven LC Network Response). Note that before the impulse the system behaves just like the circuit in those videos (the initial conditions are different but the mechanics is the same). Like a pendulum. Also take a look above in this same thread to the question asked by Yel1owstone. Good luck SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-18T04:32:18Z SecondChildTAG: Thanks matiasgrodriguez. I neglected to go back to part 1 and use the frequency I calculated, and therefore the period. I did that and it worked. However, the difference between this problem and the one in S17V16 and S17V17 is that i(0) = 1, and therefore j in the equation for the current doesn't cancel out. As it turns out this is a moot point, since there is no need to find the values of the constants (A1 and A2) because the answer is found graphically. Same goes for the voltage, as they said above this value can be obtained using energy. I'll do that tomorrow. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-11-18T07:50:23Z FirstChildTAG: Myriam, thanks for your hints, but can you give more particularly hint for problem 3 last part of value b? FirstChildUserIdTAG: 96873 FirstChildUserNameTAG: Santyaga FirstChildCreateTimeTAG: 2012-11-18T12:13:40Z SecondChildTAG: Hi Santyaga, see if this helps: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/undefined/threads/50a7bd5824371a2800000022 SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-18T13:52:58Z SecondChildTAG: Hi, matiasgrodriguez! As i understand we have dumped system. Also we have ωo and α. according ecuation for factor Q Q=ωo/2*α but how get from this value for b unfortunatly can`t see... SecondChildUserIdTAG: 96873 SecondChildUserNameTAG: Santyaga SecondChildCreateTimeTAG: 2012-11-18T15:28:04Z SecondChildTAG: The problem states: *Write an algebraic expression for the value of b that will make this vehicle critically damped*. From the previous tips you know that this means that the quality factor should be 0.5. Then 1/2=$\omega$o/2*$\alpha$ just find the b that is inside your $\alpha$ SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-18T16:28:29Z SecondChildTAG: GOT IT!!! Thanks, matiasgrodriguez!!!!! :-) SecondChildUserIdTAG: 96873 SecondChildUserNameTAG: Santyaga SecondChildCreateTimeTAG: 2012-11-18T19:07:46Z SecondChildTAG: you're welcome! SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-18T20:34:18Z SecondChildTAG: Well done Santyaga! :) , Thank you matiasgrodriguez for helping to Santyaga :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-18T21:16:40Z SecondChildTAG: Thank you Myrimit for helping everybody. If you have time I need some help on H10P3 https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/undefined/threads/50a825c23dae702300000056 lol :) SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-18T21:24:17Z SecondChildTAG: I will take a look at it :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-19T13:19:24Z FirstChildTAG: yu did a good job man.. thamx a lot FirstChildUserIdTAG: 509517 FirstChildUserNameTAG: randima FirstChildCreateTimeTAG: 2012-11-18T18:55:20Z SecondChildTAG: You are welcome @randima :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-18T21:15:29Z FirstChildTAG: @Myrimit, as always, you make this course make sense. Thanks for all the time you put into this. FirstChildUserIdTAG: 248807 FirstChildUserNameTAG: 1kingsman FirstChildCreateTimeTAG: 2012-11-19T04:40:35Z SecondChildTAG: :). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-19T13:18:44Z FirstChildTAG: Thank you. FirstChildUserIdTAG: 411504 FirstChildUserNameTAG: taubrafi FirstChildCreateTimeTAG: 2012-11-19T05:12:37Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-19T13:18:55Z IndexTAG: 15 TitleTAG: Divide y conquer ![S5E1][1] [1]: https://edxuploads.s3.amazonaws.com/13485441005063727.jpg UserIdTAG: 268104 UserNameTAG: SaulCardenasG CreateTimeTAG: 2012-09-25T03:36:10Z VoteTAG: 39 CoursewareTAG: Week 3 / Switch Model Exercise CommentableIdTAG: 6002x_switch_model_exercise NumberOfReplyTAG: 5 FirstChildTAG: It is this. Very good! FirstChildUserIdTAG: 338518 FirstChildUserNameTAG: PHLMenezes FirstChildCreateTimeTAG: 2012-09-25T12:05:12Z FirstChildTAG: +1 for this. Very good way of solving the problem. FirstChildUserIdTAG: 305491 FirstChildUserNameTAG: lefam FirstChildCreateTimeTAG: 2012-09-25T21:15:37Z FirstChildTAG: Thanks FirstChildUserIdTAG: 158740 FirstChildUserNameTAG: Ili FirstChildCreateTimeTAG: 2012-09-28T05:55:56Z FirstChildTAG: why is it y and not y bar? FirstChildUserIdTAG: 226266 FirstChildUserNameTAG: anjanasgf FirstChildCreateTimeTAG: 2012-09-29T16:23:03Z SecondChildTAG: Follow the secuence: 1. Red Circuit: input = X output = X bar 2. Sky Blue Circuit: input = X bar and Y output = bar(X bar + Y) 3. Blue Circuit: input = bar(X bar + Y) output = bar(bar(X bar + Y) = X bar + Y I hope I was clear. SecondChildUserIdTAG: 268104 SecondChildUserNameTAG: SaulCardenasG SecondChildCreateTimeTAG: 2012-09-30T02:03:46Z SecondChildTAG: Third switch from left is the first input for y SecondChildUserIdTAG: 320661 SecondChildUserNameTAG: AbbaHanna SecondChildCreateTimeTAG: 2012-09-30T08:19:24Z SecondChildTAG: Thanks for the drawing! 1. The Red Circuit just as labled 'a', it is a inverter gate so output=X bar 2. The Sky Blue Circuit as labled 'b', it is a NOR gate so output=(Xbar + y)bar 3. The Blue Circuit as labled 'a', it is a inverter gate so output=((Xbar+y)bar)bar=Xbar+y SecondChildUserIdTAG: 458098 SecondChildUserNameTAG: BobDuan SecondChildCreateTimeTAG: 2012-10-02T19:58:26Z FirstChildTAG: nice! thanks. FirstChildUserIdTAG: 623210 FirstChildUserNameTAG: preveen FirstChildCreateTimeTAG: 2012-10-29T13:45:37Z IndexTAG: 16 TitleTAG: S4E2 Distinct Boolean-Valued Functions Can someone please explain what is meant by "distinct boolean-valued functions"? UserIdTAG: 248902 UserNameTAG: Chanute CreateTimeTAG: 2012-09-07T09:26:44Z VoteTAG: 37 CoursewareTAG: Week 2 / Boolean functions CommentableIdTAG: 6002x_S4E2_Boolean_Functions NumberOfReplyTAG: 2 FirstChildTAG: It refers to how many truth tables there can be. I believe that ends up being 2 raised to the number of rows in the truth table. Also the number of rows in the truth table is 2^n, where n is the number of boolean-valued signals. So the number of boolean-valued functions is 2^(2^n). I'm not 100% sure of this so I'd appreciate feedback. I had trouble with the question as well and was looking for an answer in the discussion board. The Wikipedia article on boolean algebra helped me out. FirstChildUserIdTAG: 174100 FirstChildUserNameTAG: markpolak FirstChildCreateTimeTAG: 2012-09-08T02:23:01Z SecondChildTAG: the number of boolean-valued functions is not 2^(2^n).the answer is 4^n SecondChildUserIdTAG: 118611 SecondChildUserNameTAG: mitianhari SecondChildCreateTimeTAG: 2012-09-09T09:38:12Z SecondChildTAG: "I believe that ends up being 2 raised to the number of rows in the truth table." Not quite sure I understand this. Can you please explain it in more details? SecondChildUserIdTAG: 288018 SecondChildUserNameTAG: njsss SecondChildCreateTimeTAG: 2012-09-10T02:12:55Z SecondChildTAG: Thanks! That makes a lot of sense. I was simply confused as to what the definition of a "Boolean-Valued Function" was, and your explanation was excellent. It would seem the general form would be F#v^(S#v^S#), where F#v = number of values a function output can take on, S#v = number of values the signal can take on, and S# is the number of signals. The answer for the first half of the exercise is unique and can be expressed by 4^S# since (2^2)^2 = 2^(2^2), but that is not generally the case, since (x^y)^z != x^(y^z) SecondChildUserIdTAG: 248902 SecondChildUserNameTAG: Chanute SecondChildCreateTimeTAG: 2012-09-10T22:20:46Z SecondChildTAG: Thanks a lot. SecondChildUserIdTAG: 129288 SecondChildUserNameTAG: Tinchito SecondChildCreateTimeTAG: 2012-09-14T18:31:23Z SecondChildTAG: ![answer][1] [1]: https://edxuploads.s3.amazonaws.com/13479724689171592.jpg SecondChildUserIdTAG: 378967 SecondChildUserNameTAG: krish2012 SecondChildCreateTimeTAG: 2012-09-18T12:48:18Z SecondChildTAG: This is the intuitive way of answering this question. SecondChildUserIdTAG: 378967 SecondChildUserNameTAG: krish2012 SecondChildCreateTimeTAG: 2012-09-18T12:48:54Z SecondChildTAG: f(0)=00 here i wants to mention 00 as a function zero,not directly equals to it SecondChildUserIdTAG: 378967 SecondChildUserNameTAG: krish2012 SecondChildCreateTimeTAG: 2012-09-18T12:51:28Z FirstChildTAG: Boolean valued functions doesn't indicate no of boolean operations performed on a signal???? FirstChildUserIdTAG: 340640 FirstChildUserNameTAG: Hariprathin FirstChildCreateTimeTAG: 2012-09-14T13:54:39Z SecondChildTAG: As we are just talking about combinational logic, a function is just the relation between the inputs and the output. Let's say we have n=2 inputs. a b | f -------- 0 0 | 0 0 1 | 0 1 0 | 0 1 1 | 1 In this case the function f=0001, which is the AND gate f=a·b, but we could think in any other result for the function (f=0111,f=1000, ...) ¿How many? Well, we can just simply calculate the permutations with repetition of n=2 elements (two output values 0 or 1) in sets of r=4 (maximum number of differen inputs). So the maximum number of functions can be calculated as n^r, in this case 2⁴=16. If we have 3 boolean inputs, we can have up to 2³=8 different inputs a b c | f --------- 0 0 0 | 0 0 0 1 | 0 0 1 0 | 0 0 1 1 | 0 1 0 0 | 0 1 0 1 | 0 1 1 0 | 0 1 1 1 | 1 and functions will be like f=00000001 (AND), f=011111111 (OR), ... so the maximum number of functions can be calculated again as the permutation with repetition of n=2 elements in sets of r=8, 2⁸=256. I think that's the idea, isn't it? SecondChildUserIdTAG: 370084 SecondChildUserNameTAG: jmiguelsalgado SecondChildCreateTimeTAG: 2012-09-14T21:30:02Z SecondChildTAG: A boolean-valued function is a mapping of type f:X→B, where X is an arbitrary set and where B is a boolean domain（http://beta.wikiversity.org/wiki/Boolean-valued_function） in the second situation,the number of element in set X is 8,and the boolean domain is fixed {0,1},so the total number of function is 2⁸=256 ---------- - SecondChildUserIdTAG: 331687 SecondChildUserNameTAG: haijohn SecondChildCreateTimeTAG: 2012-09-16T12:36:47Z IndexTAG: 17 TitleTAG: Evidences that proves that Prof. Agarwal is awesome, cool and a big hearted person. Thank you for being like this! :) **This are the pics of Textbooks signed presonally for Prof. Agarwal for the winners of the CECC 1, last semester. The students uploaded it in the Forum Discussion.** **So, if you have doubts if you suspected that *Prof. Agarwal is awesome,cool and a big hearted person*, well here is the evidence that prove it :). He kindly donated the Textbooks for the Contest organized by students to the students of the last semester - CECC 1 .** *Thank you for being like this! :)* **1st. Place.** komisz. ![im1][1] **2nd. Place.** ruudoleo. ![im3][2] **3rd. Place.** Danik. ![im][3] You can read here if you want to participate this term in the CECC 2 [here][4]. Also you can check Course Info update about this Contest :). CECC Team. [1]: https://mitx_askbot_stage.s3.amazonaws.com/1349831706763785.jpg [2]: https://mitx_askbot_stage.s3.amazonaws.com/1349778393134368.jpg [3]: https://mitx_askbot_stage.s3.amazonaws.com/13497016987110511.jpg [4]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50b2892629b7b1270000003b UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2012-12-06T12:07:43Z VoteTAG: 36 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 8 FirstChildTAG: I want one! FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-12-06T13:40:02Z FirstChildTAG: Thank you ! I really believe that without him , this course was not so cool, and simple, and student oriented . I already have an engineer education , but this helped me to fill some big holes in my knowledge, meet wonderful people all over the world, and to study at the biggest engineering university of the world. To finish this course is my biggest accomplishment in a very long time. So again, THANK YOU ! FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-12-06T14:51:29Z FirstChildTAG: Wow, so cool! FirstChildUserIdTAG: 361137 FirstChildUserNameTAG: Ericson FirstChildCreateTimeTAG: 2012-12-06T16:03:53Z FirstChildTAG: very big Aha! moment FirstChildUserIdTAG: 352647 FirstChildUserNameTAG: praveenjugge FirstChildCreateTimeTAG: 2012-12-07T08:06:09Z FirstChildTAG: Thanks again Professor AA ! I hope this course also has showed as many AHA moments for the students as the previous one had given to me. The Final Exam is going to be challenging but not Mission Impossible if you followed 6002.x mysterious ways through the Math labyrinth. Keep up your good job. FirstChildUserIdTAG: 35103 FirstChildUserNameTAG: komisz FirstChildCreateTimeTAG: 2012-12-15T20:05:39Z FirstChildTAG: To be honest I have no any doubt that Prof. Agarwal is awesome, cool and a big heart person. you can see that from his way of teaching in the videos. FirstChildUserIdTAG: 226085 FirstChildUserNameTAG: Teto FirstChildCreateTimeTAG: 2012-12-17T12:15:05Z SecondChildTAG: Recall the video of the chainsaw. When he looked like an MIB agent. That was awesome ! SecondChildUserIdTAG: 418182 SecondChildUserNameTAG: euldji2005 SecondChildCreateTimeTAG: 2012-12-22T17:14:01Z SecondChildTAG: yeah :) SecondChildUserIdTAG: 226085 SecondChildUserNameTAG: Teto SecondChildCreateTimeTAG: 2012-12-25T04:08:55Z FirstChildTAG: an "AHA" moment! :) FirstChildUserIdTAG: 175205 FirstChildUserNameTAG: KT_MM FirstChildCreateTimeTAG: 2012-12-18T02:13:09Z FirstChildTAG: Prof, U are owesome! FirstChildUserIdTAG: 384515 FirstChildUserNameTAG: KEDI FirstChildCreateTimeTAG: 2013-01-11T13:44:45Z IndexTAG: 18 TitleTAG: How to write the algebraic equation This is a problem in **S6E0: THEVENIN ISOLATES NONLINEAR ELEMENT** The Thevinin equivalent voltage and Resistance is in terms of Rp, Rs and Vs. But when I submit the results and hit the CHECK it prints the following. **Invalid input: Vs Rp Rp Rs not permitted in answer** So, how is the expression to be written? Which variables to be used? There are no numeric values I can see. I use latest version of Chrome on Windows Xp SP2. UserIdTAG: 211715 UserNameTAG: pitankar CreateTimeTAG: 2012-09-13T08:29:33Z VoteTAG: 36 CoursewareTAG: Week 3 / Isolation Via Thevenin CommentableIdTAG: 6002x_isolation_via_thevenin NumberOfReplyTAG: 4 FirstChildTAG: I am unable to check my answer. FirstChildUserIdTAG: 246991 FirstChildUserNameTAG: spatra FirstChildCreateTimeTAG: 2012-09-21T22:49:46Z SecondChildTAG: TRY PUTTING IN nothing, hit check then show answer. You can then sse the accepatble format. SecondChildUserIdTAG: 440714 SecondChildUserNameTAG: mcktim SecondChildCreateTimeTAG: 2012-09-23T15:26:48Z FirstChildTAG: I think we need handout or tutorial on acceptable formats for algebraic expressions...It seems much less than clear what is acceptable. FirstChildUserIdTAG: 211150 FirstChildUserNameTAG: Bernie1961 FirstChildCreateTimeTAG: 2012-09-23T20:54:11Z FirstChildTAG: Try VS RP RS etc instead. FirstChildUserIdTAG: 312035 FirstChildUserNameTAG: philhiggins FirstChildCreateTimeTAG: 2012-09-13T08:34:20Z SecondChildTAG: Thank you. that worked ^_^ SecondChildUserIdTAG: 211715 SecondChildUserNameTAG: pitankar SecondChildCreateTimeTAG: 2012-09-13T09:15:38Z SecondChildTAG: It is a bug, I think. It is clearly shown on the picture Vs, and not VS. SecondChildUserIdTAG: 208296 SecondChildUserNameTAG: OZ1 SecondChildCreateTimeTAG: 2012-09-13T20:12:07Z SecondChildTAG: The symbol shown is in the pic is V, subscript capital S. V_S or $V_S$. You can enter into the answer box as simply "VS" but not vs or vS or Vs. SecondChildUserIdTAG: 151427 SecondChildUserNameTAG: RichardZ SecondChildCreateTimeTAG: 2012-09-18T19:43:31Z SecondChildTAG: Thanks for the tip! SecondChildUserIdTAG: 202518 SecondChildUserNameTAG: MCN SecondChildCreateTimeTAG: 2012-09-20T22:50:22Z SecondChildTAG: Ya... writing VS for Vs really matters for this answer. SecondChildUserIdTAG: 374185 SecondChildUserNameTAG: shiva05 SecondChildCreateTimeTAG: 2012-09-26T16:09:26Z SecondChildTAG: It is because is case sensitive. SecondChildUserIdTAG: 155605 SecondChildUserNameTAG: JaimeLopezCerezo SecondChildCreateTimeTAG: 2012-09-28T06:17:03Z SecondChildTAG: Thanks! SecondChildUserIdTAG: 230787 SecondChildUserNameTAG: Gabriel007 SecondChildCreateTimeTAG: 2012-09-30T21:22:48Z FirstChildTAG: Why is the expression format different in every one of these challenges????????? The picture clearly indicates Vs, Rp and Rs. Nowhere in description does it say it's all capital. I tried to enter values as they were written in conditions and got Invalid input: Vs Rp Rp Rs not permitted in answer. I knew I didn't mess up so I opened this discussion and found that everything had to be capitalized! What's going on? In one the earlier challenges it wouldn't accept * as multiplication symbol, turned out I had to use "x" instead? How do we know these things in advance??? Seems like conditions differ from challenge to challenge. After all, opening the discussion BEFORE submitting an answer is really ... cheating. I'm using Firefox on x64 platform. Are these lectures geared towards IE? FirstChildUserIdTAG: 559256 FirstChildUserNameTAG: BrianLind FirstChildCreateTimeTAG: 2012-10-06T03:51:10Z IndexTAG: 19 TitleTAG: Questions explained This is the first video, and this explains question 1: http://www.youtube.com/watch?v=wmTbg-Qv-5U (Sorry about the bad audio, I have a crappy mic) Question 2: http://www.youtube.com/watch?v=-pwyVAZB86s (no audio, but everything should be explained) Screen Shots: ![enter image description here][1] ![enter image description here][2] Question 2 is on the way. I want to Zepp for his working of question 2, and the video he pointed to, I am going to give a detailed explanation of the question, to the best of my abilities. In the mean time, just tell me how the vid for question 1 was. If anyone wants to help me make future vids, I would be more than happy to work with them, but please make sure that you have good bandwidth. [1]: http://83.imagebam.com/download/fyaqR4VJ9SjQFtE0uEoLZg/20993/209923047/vlcsnap-2012-09-09-14h32m11s254.png [2]: http://86.imagebam.com/download/S2YQfTRwPLJ7d34oIydm0g/20993/209923062/vlcsnap-2012-09-09-14h32m24s131.png UserIdTAG: 193033 UserNameTAG: NIslam CreateTimeTAG: 2012-09-09T08:31:34Z VoteTAG: 35 CoursewareTAG: Week 1 / AC power CommentableIdTAG: 6002x_S1E3_AC_power NumberOfReplyTAG: 15 FirstChildTAG: I think you forget about the negative sign in your equation. That's why you end up with a minimum value. While if you put the negative sign you will have a maximum value. FirstChildUserIdTAG: 164513 FirstChildUserNameTAG: MZeeshanSheikh FirstChildCreateTimeTAG: 2012-09-09T09:16:33Z SecondChildTAG: Nope, f(1/120) is negative, you can see for yourself. Later on, you just use the equation for power to calculate max power. SecondChildUserIdTAG: 193033 SecondChildUserNameTAG: NIslam SecondChildCreateTimeTAG: 2012-09-09T09:27:20Z SecondChildTAG: it is helpful to finding peak and least voltage. SecondChildUserIdTAG: 456614 SecondChildUserNameTAG: atulnarayan786 SecondChildCreateTimeTAG: 2012-09-18T10:32:50Z FirstChildTAG: Thank You :) FirstChildUserIdTAG: 62332 FirstChildUserNameTAG: aymanyounis FirstChildCreateTimeTAG: 2012-09-10T21:58:04Z FirstChildTAG: Question 2 posted. FirstChildUserIdTAG: 193033 FirstChildUserNameTAG: NIslam FirstChildCreateTimeTAG: 2012-09-09T15:47:19Z FirstChildTAG: Thank you! this actually made me realize my reasoning was ok, I didn't resolve the integral right though. FirstChildUserIdTAG: 339382 FirstChildUserNameTAG: gaastonsr FirstChildCreateTimeTAG: 2012-09-09T23:44:43Z SecondChildTAG: We all get stuck at some point, I just hope I was able to make things a bit simpler. SecondChildUserIdTAG: 193033 SecondChildUserNameTAG: NIslam SecondChildCreateTimeTAG: 2012-09-10T09:11:50Z FirstChildTAG: > f(t)=120*sqrt(2)*cos(**120***pi*t) It should be 2 not 120? But great tutorial any way. FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-09T11:52:08Z SecondChildTAG: $f\left( t \right) =120\cdot \sqrt { 2 } \cdot \cos { (60\quad\cdot\quad 2\pi\quad\cdot\quad t) } $ $f\left( t \right) \quad =\quad 120\sqrt { 2 } \cdot \cos { (120\pi t) } $ SecondChildUserIdTAG: 193033 SecondChildUserNameTAG: NIslam SecondChildCreateTimeTAG: 2012-09-09T13:32:18Z FirstChildTAG: Thank you for your trouble ! FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-09-10T07:29:18Z SecondChildTAG: No trouble at all my friend! :) SecondChildUserIdTAG: 193033 SecondChildUserNameTAG: NIslam SecondChildCreateTimeTAG: 2012-09-10T09:11:11Z FirstChildTAG: Or simpler, for sin wave we know that $ V_\mathrm{peak}=\sqrt{2}\ V_\mathrm{rms} $ FirstChildUserIdTAG: 207456 FirstChildUserNameTAG: Octomos FirstChildCreateTimeTAG: 2012-09-09T10:18:54Z SecondChildTAG: Thats good, but this is how you understand the formula you just gave us. Thanks though, I really appreciate it.! :) Would you be kind enough to show the working behind that in your post? SecondChildUserIdTAG: 193033 SecondChildUserNameTAG: NIslam SecondChildCreateTimeTAG: 2012-09-09T13:50:55Z SecondChildTAG: Shouldn't the equation be: $V_{rms}=\sqrt{\cfrac{1}{T}\int_0^T(V_{peak}\cdot\sin(2\pi\cdot f\cdot t))^2 dt}$ ? SecondChildUserIdTAG: 276409 SecondChildUserNameTAG: IgnacioUY SecondChildCreateTimeTAG: 2012-09-19T00:50:59Z FirstChildTAG: My dear friend, there is a lot easier way: v(t)=Vm cos(2*pi*60*t) Vm=Vrms/sqrt(2):here rms voltage=120V so,v(t)=120*sqrt(2) cos(2*pi*60*t) Pm=peak power=Vm*Im=Vm*Vm*(1/R) OHMS LAW Pm=120*120*2/110=261.739W FirstChildUserIdTAG: 372535 FirstChildUserNameTAG: _Infinity FirstChildCreateTimeTAG: 2012-09-10T11:46:45Z SecondChildTAG: Thats simply applying a formula. If you understand what you are doing in the formula, then please be my guest. But the way I have done it, makes sense. Your method is faster, but I prefer a method that I understand, even if its longer. SecondChildUserIdTAG: 193033 SecondChildUserNameTAG: NIslam SecondChildCreateTimeTAG: 2012-09-10T12:44:33Z FirstChildTAG: don't you think your calculus approach is an overkill for question 1.. i mean the question asks for 'peak power'.. that is simply the value of power when it is at its maximum.. **power = V*I** => power = V*(V/R) = (V^2)/R R the resistance..it is a constant.. => power will be max when V^2 is max..i.e you will get power at its peak..when V^2 is at its peak.. V at any time t = 120*sqrt(2)*cos(2π⋅60⋅t) = A*cos(theta) say, V^2 = A^2* (cos(theta))^2 its max value will be A^2 , when (cos(theta))^2 will be at its max.. max value of (cos(theta))^2 can only be 1.. so, V will be max when cos part is 1.. hence, power at its peak when cos part =1, and at that time..power = A^2/R = 261.8181 by the way.. your drawings are beautiful ! FirstChildUserIdTAG: 15237 FirstChildUserNameTAG: anubhavsinha FirstChildCreateTimeTAG: 2012-09-10T19:27:17Z FirstChildTAG: Why is this here. This is not right and a waste of space .... FirstChildUserIdTAG: 359302 FirstChildUserNameTAG: GaryG FirstChildCreateTimeTAG: 2012-09-11T23:04:23Z SecondChildTAG: I apologize completely. I tried to view this in closed caption and it says all kinds of strange things. So I thought it was some kind of joke. Must be your bad audio too. SecondChildUserIdTAG: 359302 SecondChildUserNameTAG: GaryG SecondChildCreateTimeTAG: 2012-09-14T22:24:36Z FirstChildTAG: Thanks for the solution. But hooold on, the question is asking for peak power, not peak voltage. FirstChildUserIdTAG: 208105 FirstChildUserNameTAG: jmunya FirstChildCreateTimeTAG: 2012-09-14T22:42:20Z FirstChildTAG: Great work Thanks very much for this very helpful Post. One small tiny tweak at the second video, at 1:40 it is written: "The total power generated over one cycle" referring to the integration over one cycle .. it should be "The Total Energy generated over one cycle" as E=(integration of) P(t)dt and average power then equal: P = Energy/time Thanks again for the effort done here FirstChildUserIdTAG: 398394 FirstChildUserNameTAG: Sakr85 FirstChildCreateTimeTAG: 2012-09-16T04:26:09Z FirstChildTAG: it is good. FirstChildUserIdTAG: 453055 FirstChildUserNameTAG: DAWIT999 FirstChildCreateTimeTAG: 2012-09-17T15:25:25Z FirstChildTAG: Great work! FirstChildUserIdTAG: 175734 FirstChildUserNameTAG: hestrada FirstChildCreateTimeTAG: 2012-09-19T01:11:01Z FirstChildTAG: Really helpful..:) FirstChildUserIdTAG: 314460 FirstChildUserNameTAG: PriyaPrabhakar FirstChildCreateTimeTAG: 2012-09-22T15:53:41Z IndexTAG: 20 TitleTAG: H11P1, H11P2, H11P3 Hints Requested by who need it :). ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- > **First of all, a little bit of Marketing haha, take a look at this Post [CECC 2 - Contest][1] . I hope that you can Participate :). You > can Win one of the 3 Textbooks signed personally by Prof. Agarwal!** ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- (For spanish translation read in the end) Ok, now we will go to the Hints :). Here I write some clues/hints to the Homework 11. So that later you can solve it by your own. I hope you find it helpful. H11P1 Hints: --- LC Tank In this first part, we have a model of a radio equipment (see Fig 1.) ![image description](https://mitx_askbot_stage.s3.amazonaws.com/1337742779730269.png) **Part 1:** They ask us to determine the bandwith (delta w). But, how do we calculate the bandwith of the given circuit? Ok, let's see. To understand the concept of bandwith, you should read the excercise in the page 798 of the Textbook [read here][2] In this page of the Textbook, they tell us about the way to find the canonic equation (see Textbook page 798, here I will paste a part, so that the explanation is more easy, I do not have copyright infringement intended). If you see the transference, you have to compare the denominator (red circle) with the canonic equation (red rectangle). ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13377439036059747.png) Hint part 1: So, if we know Z (I will not explain how to obtain Z because it is supoused that you already can handle it to find it). We will get a transference like the previous image. Ok, now, leaving our numerator, we will have our canonic equation (but with the values of L,RL,RC,C). Now, for the billion question, which is the value of te bandwith?? Clue: remember that the relation of wo/Q is the bandwith. So, if we have the canonic equation, we can find wo, alpha. Again, What is the value (in function of the parameters) of the bandwith? **Part 2:** Hint: if they give us a fo, and we know that the value of wo=2pifo, and also, we have wo in function of RL,L,RC,C ; and we know as a data the values of RL,L and RC. So, which is the value of C for that given fo? Part 3: Hint: If you have the expression of the part 1 (bandwith) and also you have the value of C that you have calculated of the part 2. So, what is the bandwith? Another Hint: remember that the equation of part 1 it is expressed in radians per second (delta w), remember to convert that value to Hz, that is to say, divide the result by 2pi. **Part 4:** It is the same as part 2 but with other value of fo. **Part 5:** It is the same as part 3 but with other value of fo. ---------- H11P2 Hints: --- Scope Probe This example is very simple. It show us a model of a scope probe of an oscilloscope. ![image description](https://mitx_askbot_stage.s3.amazonaws.com/1337746295726337.png) ![image description](https://mitx_askbot_stage.s3.amazonaws.com/1337746455223348.jpg) **Part 1:** they tell you that you have a x10 probe (that means that reduces 10 times what you measure). So, if you have in the input vM, what do you have in the output Vs? **Part 2:** Ok., know, if you have a signal with an amplitude with a peak value of 0.06 V ; and in the output you have 0.006 V. So, you have Z2, Vm, Vo, but not all the data of Z1. Can you use a resistive divider? Yes! You know, how would be the expression of an impedance (C parallel R), so, if you know Z1, R, what is the value of C?? ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13377471234544686.png) **Part 3:** Same as part 2. But, wait a minute. The relationship of Vo/Vin is equal, so, does it modify something? **Part 4:** Independent of the signal, we have seen a correspond from the attenuation with the Vo/Vin. Again, if we have vin=0.1, we now Vo/Vi, what is the value of Vo?? ------ H11P3 Hints: --- Branch Voltages **Part 1,2,3:** you should read page 640 of the Textbook [read here][3] and compare the curves (fig.12.16, etc...). Also you can simulate in the sandbox to arrive to conclusions about it* **Part 4:** Hint: How many Q cycles of the signal do we have till the amplitude be equal or less that d... ?You can read the page 648 of the Textbook (explanation).[read here][4] ![read here](https://mitx_askbot_stage.s3.amazonaws.com/1337748236309465.png) **Part 5:** OMG! Those mischevous engineers! Hahaha! Why they complicate the life of the 6.002x Students haha ;). By the way, I will become one of them in the future and I will do the same haha - nah, joke. ![image][5] You get the following Frequency-domain plot of the magnitude of the impedance: ![image][6] **Part 6.** Based off of the graph of the magnitude of the impedance, which element was removed? Enter L or C in the space provided. This part is conceptual. Try to think how was an RL and RC plot was. If you don´t remember this, try to watch again the explanations of Prof. Agarwal... **Part 7.**What is the value of the resistor in Ohms (Ω)? Hint. Once you have known How is your Circuit: R,L? or R,C?. - Do you know How is the module of the impedance Z of that circuit? ;). - Can you have graphically, for a value of w, the value of the Z module? - Hmmm, what happens for a high value of w? What is your Z? Can you find your R? ;). **Part 8:** Do you have your Z module expression in function of R,w and (L or C)? yES! Do you have your R, from your previous part? yes! can you pick a value of w? Can you find graphically for that w the value of the module of Z? Yes!! So, can you find your (L or C?) ? YES!!! **Part 9:** *A big hint:* Take a look at the page 647 of the Textbook :) [read here][7] That is all! :). I hope this can help you and don´t forget to participate in the Contest :). That will be awesome! We will be waiting for your video! Myriam. ------- > **Now in Spanish.** H11P1 --- LC Tank En la primera parte, tenemos un modelo de un equipo de radio (ver Fig. 1). ![image description](https://mitx_askbot_stage.s3.amazonaws.com/1337742779730269.png) **Part 1:** Se nos solicita el ancho de banda (delta w). Pero, cómo calculamos el ancho de banda de aquél circuito? Bien, para comprender el concepto del ancho de banda, es recomendable echarle un vistazo al ejercicio de la página 798 del Textbook [leer aquí][8] En esta página del libro, se nos habla el procedimiento para hallar la ecuación canónica de la forma (ver Textbook pág. 798, aquí pego una parte para que sea más cómoda la explicación, no tengo intenciones de infracción del copyright). Si se ve la transferencia, tenemos que comparar el denominador (círculo rojo) con la ecuación canónica (rectangulo rojo). ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13377439036059747.png) Pista part 1: Entonces, si tenemos Z (no voy a explicar cómo obtener Z porque eso se supone que ya lo deben manejar bien). Obtendremos una transferencia algo así como la transferencia de la imagen anterior. Bien, ahora, independientemente de nuestro numerador, tendremos nuestra ecuación canónica (pero con los valores de L, RL, RC, C). Ahora, la pregunta del millón, cuál es el ancho de banda?? Pista: recordar que la relación wo/Q es el ancho de banda. Entonces, si tenemos de nuestra ecuación canónica wo, alfa, cuál es el valor del ancho de banda? **Part 2:** Pista: si nos dan el valor de una fo, y sabemos que el valor de wo=2pifo , y además tenemos wo en función de RL,L,RC,C ; y conocemos como dato RL,L y RC. Entonces, cuál es el valor de C para dicha fo dada? **Part 3:** Pista: Si ya tienes la expresión de la parte 1 (ancho de banda) como así también el valor de C calculado para la parte 2. Entonces, cuál es el ancho de banda? Otra pista: recuerda que la ecuación de la parte 1 está expresada en radianes por segundos (delta w), recuerda pasar el dato a Hz, es decir, dividir el resultado por 2pi. **Part 4:** Es lo mismo que la parte 2 pero con otra fo como dato. **Part 5:** Es lo mismo que la parte 3 pero con otra fo como dato. ---------- H11P2 Hints: --- Scope Probe Este ejercicio es bastante simple. Nos enseña un modelo de una punta de prueba de un osciloscopio. ![image description](https://mitx_askbot_stage.s3.amazonaws.com/1337746295726337.png) ![image description](https://mitx_askbot_stage.s3.amazonaws.com/1337746455223348.jpg) **Part 1:** Se nos da el dato que tenemos una punta de prueba x10 (esto significa que atenuamos 10 veces lo que estamos midiendo a la entrada). Entonces, si tenemos la entrada vM, qué valor tenemos a la salida Vs ? **Part 2:** Si se tiene una señal cualquiera de la cual sabemos su valor pico de 0.06V; y si a la salida tenemos 0.006V. Entonces, si además tenemos Z2, vM, Vo, pero no todo el dato de Z1. Podemos utilizar un dividor resistivo? Sí! Ahora, si sabes la expresión de la impedancia (C paralelo R), entonces, si conoces Z1 y R, cuál es el valor de C?? ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13377471234544686.png) **Part 3:** Idem parte 2. Pero, un minuto! La relación de Vo/Vin sigue siendo igual, entonces, aquello modifica algo? **Part 4:** Independientemente de qué señal se trate, hemos visto una correspondencia entre la atenuación de Vo/Vin. Otra vez, si tenemos Vin=0.1, sabemos Vo/Vi, Cuál es el valor de Vo?? ---- H11P3 Hints: --- H11P3: Branch Voltages **Part 1,2,3:** Es recomendable leer la página 640 del Textbook [leer aqí][9]y comparar las curvas (fig 12.16, etc..). También puedes simular en el sandbox para arribar a conclusiones. **Part 4:** Hint: Cuántos Q ciclos de la señal tenemos hasta que la amplitud sea menor o igual a d...? Puedes leer la página 648 del Textbook, allí se explica. [leer aquí][10] ![read here](https://mitx_askbot_stage.s3.amazonaws.com/1337748236309465.png) **Part 5:** Por favor! Estos pícaros Ingenieros! jaja! Por qué complican tanto la existencia de los alumnos del 6.002x jaja ;). Por cierto, me convertiré en el futuro en uno de ellos y haré lo mismo - no, en broma jaja. ![image][5] El problema dice: You get the following Frequency-domain plot of the magnitude of the impedance: ![image][6] **Part 6.** El problema dice: Based off of the graph of the magnitude of the impedance, which element was removed? Enter L or C in the space provided. Esta parte es muy conceptual. Trata de pensar cómo era la curva de impedancia de un circuito RL y RC. Si no lo recuerdas, trata de mirar nuevamente los videos explicativos del Prof. Agarwal... **Part 7.** El problema dice: What is the value of the resistor in Ohms (Ω)? Hint. Una vez que ya has averiguado de qué circuito se trata, es decir RL o RC? - Puedes saber cómo es el módulo de la impedancia Z de este Circuito? ;). - Puedes entonces,para un dado valor de w, obtener gráficamente el valor del módulo de Z ? - mmm, Qué sucede para un valor alto de w? Cuál es tu valor de Z? Puedes entonces hallar tu R ? ;). **Part 8:** Tienes la expresión del módulo de la impedancia Z de tu Circuito en función de R, w y (L o C)? Sí! Tienes tu R de la parte previa? Sí! Entonces, puedes elegir un valor de w y hallar graficamente, para dicho valor, su correspondiente valor de impedancia - en módulo-? Sí!! Entonces, puedes hallar tu valor de L o C? sí!!! **Part 9:** *Una gran Hint:* Leer la página 647 del Textbook :) [Aquí][7] Eso es todo! :). Espero que esto pueda servirles de ayuda y no se olviden de participar en el Concurso :). Eso sería grandioso! Estamos ansiosos por ver tu video! Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50b2892629b7b1270000003b [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/822 [3]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/664 [4]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/672 [5]: https://www.edx.org/static/content-mit-6002x/images/circuits/H11P3_Impedance.f0fb3ea5925a.png [6]: https://www.edx.org/static/content-mit-6002x/images/circuits/H11P3_part2.965ea6adfd70.jpg [7]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/671 [8]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/822 [9]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/664 [10]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/672 UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2012-11-27T23:42:06Z VoteTAG: 34 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 9 FirstChildTAG: hello Myrimit thanku for the hints, quite helpfull. but i'm having a problem finding Cp in P2... to find it we need frequency to substitut ine 's=jw'.... how cud we get that... Vm gives a value... are we supposed to use that...?? FirstChildUserIdTAG: 117319 FirstChildUserNameTAG: iqramalik FirstChildCreateTimeTAG: 2012-11-28T21:59:37Z SecondChildTAG: Hint: $V_M = A \cos({\omega t})$ ---------- So, What is you w ? ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-29T12:27:56Z SecondChildTAG: *your SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-29T12:28:07Z SecondChildTAG: Thank SecondChildUserIdTAG: 99628 SecondChildUserNameTAG: Pranjal16 SecondChildCreateTimeTAG: 2012-12-26T09:03:05Z FirstChildTAG: Hi Myriam, replacing values of components into P1 to find the bandwith is straightforward or should be. However I'm wasting an enormous amount of time and not getting the checkmark. I enter the values using the short notation like 0.65m 56.56845p 490k and divide the result by 2pi. Also I multiply by 10^3 since it's kHz to no avail. Any hint you can provide? FirstChildUserIdTAG: 11075 FirstChildUserNameTAG: roncada FirstChildCreateTimeTAG: 2012-11-28T03:06:27Z SecondChildTAG: Hi roncada, Can I help you? Be careful with the Units... Hints. Do you have correctly your expression of $\Delta w$ -Part 1- ? Isn´t it? If yes, Can you obtain your Capacitor C that vary with the frecuency? Hint. Recall the characteristic equation, term $wo^2$ :) Warning Moment: Be careful with the units. When they ask you the value in picoFarads, means, eg. if your capacitance is 10pF you must write only 10. Also, in the bandwith, remember that to express the bandwith in Hz, you should divide $\Delta w / 2\pi$ . This will give you the result in Hz. But here is another warning moment too, AHA! your answer has to be in kHz units, So, eg., if you have $\Delta w / 2\pi$ = 15000 Hz , your answer in kHz should it be 15kHz, so you must write 15. I hope this can help you :) , Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-28T04:12:39Z SecondChildTAG: Thanks Myriam. This part is what i needed "if you have Δw/2π = 15000 Hz , your answer in kHz should it be 15kHz, so you must write 15." I was multiplying by 10^3 when it should be by 10^-3. aiaiai! SecondChildUserIdTAG: 11075 SecondChildUserNameTAG: roncada SecondChildCreateTimeTAG: 2012-11-28T04:42:05Z SecondChildTAG: ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-28T04:51:10Z FirstChildTAG: Hi, Miriam! Thank you for explanation. I have question. For the part 1, L is connected to the $R_L$.So we have $\frac {1}{SerialImpedance}$. Next, my equation seems to be like this: H=$\frac{1}{\frac{1}{SerialImpedance}+Admitance1+Admitance2}$ FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-11-28T10:35:21Z SecondChildTAG: Hi Serge , I have edited your Post ;). You are really close! Try to work more with the expression in order to get in the denominator the characteristic equation form: $s^2+2 \alpha s+ wo^2$ Myriam SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-28T11:58:56Z SecondChildTAG: Im feeling that Im so close :) but Im stuck at this point.I have tried, of course..Result is awful SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-11-28T12:06:58Z SecondChildTAG: ok, if youre talking that Im close - youre right, as always. Part 1 is done... Thank you, Miriam! hmm...I do expect I'll ask more questions soon SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-11-28T13:57:47Z SecondChildTAG: Well done Serge :)! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-28T18:25:03Z SecondChildTAG: Thank you, Miriam :) But Im stuck again, with value of Capasitor/Inductor in the H11P3.. Nevermind..Wy equation for the sequentally connected resistor and capasitor is wrong..Any hints? SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-11-28T20:08:28Z SecondChildTAG: Hmm, nevermind..3 hours are lost!I have similar problem as: "MEng_IIT I have also the same problem. My answer is 1.00503781526, but only 1 works." So, answer is 1..strange SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-11-28T20:19:51Z FirstChildTAG: I give up: P. The last question will go blank. Says the exercise "Based off of the value of Q Determined you earlier", **when we calculate Q? ¿?**. ------------------------ I have not Q, What I do with the formula on page 647?. How it calculated the coil?. I leave, I still have Lab 12. :) FirstChildUserIdTAG: 244706 FirstChildUserNameTAG: Miguel_Angel FirstChildCreateTimeTAG: 2012-11-28T19:20:08Z SecondChildTAG: Hi Miguel_Angel, Can I help you? Try to take a look at the part 4 of the H11P3. They ask you to find the Q graphically ;). The last part, says, that based on the value of Q you determined earlier (part 4)...Enter the capacitance in nF or the inductance in mH, depending on which element you determined was removed from the circuit. Hint: Recall part 4 and take a look at page 647 ;). Do you have your Q, R ? yes! Can you calculate L or C ? Yes! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-29T12:21:35Z SecondChildTAG: Dear Myrimit, I still do not get this. Looking at equ 12.66 on pg647, L and C remains unknown. Given that one is removed, I am not able to derive what happens with the formula of Q. I have tried making LorC = 1, but that gives me a wrong answer. So what should we do with the removed parameter? SecondChildUserIdTAG: 329254 SecondChildUserNameTAG: KGabor SecondChildCreateTimeTAG: 2012-12-01T13:30:18Z SecondChildTAG: Hi KGabor. Let me try to help. Please take a look at video S22V4. There we worked on the transfer function for an LCR circuit like in this problem. There you will understand the formula $Q=\frac{\omega o}{2\alpha}$ in terms of L, C and R. Based on the previuous answers you already have the value to Q, R, and (C or L). Using the formula above you can calculate the remaining unknown. SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-12-01T17:08:52Z SecondChildTAG: KGabor if your doubt is about how to discover which element was removed. Then my hints are: - Write the transfer function for the LR circuit (S, R and L) and other for the CR circuit (S, R and C). - Look at both the graph and the two formulas... what happens when $\omega $ tends to infinity? (you will be able to match the graph with one of the equations. You just found which element should have been removed and which one still is in the circuit. - Looking at the graph you should be able find R. - Using the graph and the magnitude formula you should be able to find C or L. To solve the last question see my previous post. Good luck SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-12-01T17:23:11Z SecondChildTAG: Hello, I am trying to cope with H11P3 Q7. It seems that something is wrong with my answer. abs(Z)=sqrt(R^2+(wL)^2), I have abs(Z) value from the graph, w and R values and I need to find L. Seems to be quite easy, but checker is not accepting my answer. SecondChildUserIdTAG: 190265 SecondChildUserNameTAG: Kostiantyn SecondChildCreateTimeTAG: 2012-12-01T17:30:15Z SecondChildTAG: Hi Kostiantyn, just checking... if you are trying to calculate L is because the removed element was C (the capacitor)? SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-12-01T17:38:46Z SecondChildTAG: Kostiantyn, if the removed element was the capacitor... mmm... I might be wrong... but the magnitude function for the LR circuit does not look correct to me.. SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-12-01T17:47:42Z SecondChildTAG: Hi KGabor, Can I help you? You know Q,R, and also you know your L or C - I can not tell you which you know as this value is the one that you must answer in part H11P3 part 6, so you will only will have one incognit L or C. So, can you get the L or C value? yes ;) Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-01T19:43:49Z SecondChildTAG: Just had to read carefully the question(( SecondChildUserIdTAG: 190265 SecondChildUserNameTAG: Kostiantyn SecondChildCreateTimeTAG: 2012-12-02T13:59:45Z FirstChildTAG: Thanks a lot, hint to P1 was VERY usefull ) FirstChildUserIdTAG: 394836 FirstChildUserNameTAG: v2g6ch4 FirstChildCreateTimeTAG: 2012-11-29T06:50:00Z SecondChildTAG: You are welcome v2g6ch4 :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-29T12:21:56Z SecondChildTAG: I'm stuck with P1 still. Although I've tried to solve it as many possible ways as I know. I've used the hints as well but no luck. when i take Rc in parallel with cap in parallel RL series with L. I can't get the denominator in the form of S^2+2aS+w0^2 why is that? any help please SecondChildUserIdTAG: 8897 SecondChildUserNameTAG: sam1202 SecondChildCreateTimeTAG: 2012-12-02T00:43:35Z SecondChildTAG: Hi sam1202, Can I help you? Try to work with your expression, eg., an arbitrary example that might help you a little bit. Lets see that you have something like this, Ztotal = $\frac{D}{\frac{1}{A}+B+C+\frac{1}{E}}$ Ztotal = $\frac{D}{\frac{E+B.A.E+C.A.E+A}{A.E}}$ Ztotal = $\frac{D.A.E}{E+B.A.E+C.A.E+A}$ Lets suppouse that A=B=s and that C=2, D=1 and E=3 Ztotal = $\frac{1.S.3}{3+s.s.3+2.s.3+s}$ Ztotal = $\frac{3.S}{3+ 3.s^2 +7.s}$ We need something like s^2 +number1.s+number2 Ztotal = $\frac{3.S}{3.s^2 +7.s+3}$ Divide numerator and denominator by the coefficient of s^2, that is 3. Ztotal = $\frac{(3/3).S}{(3/3).s^2 +(7/3).s+(3/3)}$ Ztotal = $\frac{S}{s^2 +(7/3).s+1}$ So there you have, your characteristic equation will be the denominator s^2 +(7/3).s+1 SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-02T01:41:35Z SecondChildTAG: thanks Myrimit..it helped me..and solved it..an aha moment!! SecondChildUserIdTAG: 8897 SecondChildUserNameTAG: sam1202 SecondChildCreateTimeTAG: 2012-12-02T11:24:34Z SecondChildTAG: Well done sam1202 ! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-02T15:08:32Z FirstChildTAG: I've read the hints for Problem 2 Part 2...but I'm still having a lot of trouble ...I just keep getting stuck working with very complicated and long algebraic equations. SO complicated and time consuming that I'm pretty certain I must be missing something. I wish everything were truly as easy as the magic words "voltage divider", but I'm going to need a bit more help on this one. Here are some, but not all of the difficulties I'm having. I'll refer to Z1 and Z2 as labelled in the diagram above. The value of the transfer function |Vo/Vi| is easy enough to determine because Vo and Vi are given. But if |Vo/Vi| = |Z2/(Z1 + Z2)| (by the voltage divider) and I want to solve for Cp (which is contained in Z1), then my expression becomes very very complicated complex algebra which is just impractical to solve. So I think I'm doing something wrong. Would appreciate a few more suggestions. FirstChildUserIdTAG: 287805 FirstChildUserNameTAG: Beneficial FirstChildCreateTimeTAG: 2012-12-01T08:24:22Z SecondChildTAG: |Vo/Vi|=|Z1/(Z1+Z2)| Try this..... SecondChildUserIdTAG: 477198 SecondChildUserNameTAG: Rajesh1993 SecondChildCreateTimeTAG: 2012-12-01T09:54:49Z SecondChildTAG: Woah...hang on though. I thought the output we were given was the output which appeared at Vs (as labelled in the problem) ie. at Z2 (in the diagram above). SecondChildUserIdTAG: 287805 SecondChildUserNameTAG: Beneficial SecondChildCreateTimeTAG: 2012-12-01T15:38:25Z SecondChildTAG: try think of this way , Vo is equal to 0.1 of the Vin , so the impedence which Vo see ( Z1) equals to the impedence which Vin see ( Z1 + Z2 ) , all in magnitude ofcorse i did it that way SecondChildUserIdTAG: 285675 SecondChildUserNameTAG: Ascot SecondChildCreateTimeTAG: 2012-12-01T16:42:47Z SecondChildTAG: Hi Benefical, Can I help you? Another way to think this, Hint. You know |Vo/Vi|. So, you will have that Z1 will be some proportional number to Z2. So, Z1 = some.number Z2 . So, there must to be a relation of proportionality between the components in Z1 with the components of Z2. And what is that some.number ? ;) Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-01T19:37:46Z SecondChildTAG: Hi Myriam. I was thinking something along those lines. Let's say I get a relation wheren Z1 = aZ2 (where a is some constant). If I work out Z2, which is simply ZCs || ZRs, I end up with an expression which contains the complex j. So I can an "expression" for Z2, but i can't get a "value" per se, with which I could calculate Z1 or Cp. SecondChildUserIdTAG: 287805 SecondChildUserNameTAG: Beneficial SecondChildCreateTimeTAG: 2012-12-01T22:28:51Z SecondChildTAG: Maybe a slightly different approach will work. Compute the complex impedance and determine the condition that makes it pure real. I ended up with a real numerator and complex denominator. I set the complex part of the denominator to zero, and there was the condition that I needed. The solution took about five lines and ten minutes. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-12-01T23:22:06Z SecondChildTAG: Hi Beneficial, Yes, you are close. Some Hints. You know the Z2, as you have said. Also, your Z1 will be similar of your Z2 but with Cp and Rp. So, try to work with the expression of Z2 and try to reach to something like this Z2 = A + j B . Also with, Z1= C + j D . As you said, Z1 = a*Z2, Also can you compare your B with your a*D ? yes. So, if B is in function of Cp and a*D is in function of Cp, and you know Cs. Can you get your Cp? Yep :) I hope this can help you. I can not tell more ;) Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-02T00:55:38Z SecondChildTAG: I've found these comments helpful but I'm still running into trouble. My value for 'a' keeps coming out as a negative value, such that my calculation for Cp in the end turns out negative. I'm not sure where I'm going wrong. SecondChildUserIdTAG: 287805 SecondChildUserNameTAG: Beneficial SecondChildCreateTimeTAG: 2012-12-02T02:29:24Z SecondChildTAG: Here is more about my approach. I first calculate the Z for each resistor-capacitor pair. I simplify the Z's to have real numerators and a complex denominators. I then form a voltage divider, and simplify again to get a real numerator with a complex denominator. In the denominator there is an expression that is the ratio of two complex terms that I can force to equal one by the proper choice of Cp. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-12-02T02:59:30Z SecondChildTAG: Hi Beneficial, are you sure that a is a negative value. **Here is a math Hint.** Lets suppose that you have the following ratio. In order to solve this equation, you can apply the following rule, that is to say, multiply the green numerator with the green denominator, and make it equal to the orange expression, that is to say multiply the orange denominator with the orange numerator. ![IM][1] Now it is your turn. Try to do it again. Are you sure that your a is negative ?;) [1]: https://edxuploads.s3.amazonaws.com/13544362598474447.png SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-02T08:23:11Z SecondChildTAG: Here is more detail for an alternate approach: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50bbabcf4df6b51f0000004d SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-12-02T19:35:42Z SecondChildTAG: thank you, Myrimit SecondChildUserIdTAG: 413784 SecondChildUserNameTAG: Dorotka SecondChildCreateTimeTAG: 2012-12-02T21:43:41Z FirstChildTAG: Thank you Myrimit, Your hints are always a very useful reference. I had a lot of trouble finding the values of L/C in the last exercice just because I took the numerical values of L/C as I found in the calculator, the same thing happened to me in many previous homeworks. I appreciate your hints very much. FirstChildUserIdTAG: 418182 FirstChildUserNameTAG: euldji2005 FirstChildCreateTimeTAG: 2012-12-01T18:49:29Z SecondChildTAG: You are welcome euldji2005 ;). Nice to know that I am being hepful here haha. Take care, Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-01T19:31:02Z FirstChildTAG: Hi Myriam that is a really good job that you did here . I did not start solving the home work yet but it seems that it,s easy. I just want to say Hi and am sorry that I did not response fast on the past post.(related to "log"). all the best to you. :) FirstChildUserIdTAG: 226085 FirstChildUserNameTAG: Teto FirstChildCreateTimeTAG: 2012-12-02T03:53:33Z SecondChildTAG: Thank you Teto. Don´t worry :). I will try to make that video explanation as soon as possible. Take care, my best wish to you too. Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-02T08:08:42Z SecondChildTAG: I bull back my word it is not easy. I don't know what is wrong I tried so many ways and did the calculation step by step but it didn't work. Myriam please some help here. Q1 part 3 I read all what you said about this part and I did it exactly I don't know what is wrong. SecondChildUserIdTAG: 226085 SecondChildUserNameTAG: Teto SecondChildCreateTimeTAG: 2012-12-02T10:39:03Z SecondChildTAG: I solve the fifth part by one of the ways it's work when I use it to part three it doesn't work..... help (S.O.S) :) SecondChildUserIdTAG: 226085 SecondChildUserNameTAG: Teto SecondChildCreateTimeTAG: 2012-12-02T10:46:52Z SecondChildTAG: I think the best solution for "mischievous engineers!" is to shock him before solving the problem. that will keep him away for a while or may take him to his grave. SecondChildUserIdTAG: 226085 SecondChildUserNameTAG: Teto SecondChildCreateTimeTAG: 2012-12-02T13:02:43Z FirstChildTAG: I'm really struggling with H11P1. I understand the concept of bandwidth etc, but apparently I'm lacking some math skills here. I know what the formula for Z is, but whatever I try, I'm not able to get to something that even looks similar to the canonical equation. What would you suggest? Is there a particular on-line course that will enable me to solve this stuff? FirstChildUserIdTAG: 114881 FirstChildUserNameTAG: xvink FirstChildCreateTimeTAG: 2012-12-02T12:26:21Z SecondChildTAG: Hi xvink, Try to take a look at my comment to Hi sam1202. You have to work with the expression a little bit more in order to get the expression of the characteristic equation... SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-02T15:12:51Z SecondChildTAG: The more I work with the expression, the more confused I get :( I'm skipping the 3 questions where I need this and see if the answer will bring some enlightenment. This really frustrates me; I know what to do, but not how to get there. Thanks for all the hints. SecondChildUserIdTAG: 114881 SecondChildUserNameTAG: xvink SecondChildCreateTimeTAG: 2012-12-02T19:40:38Z SecondChildTAG: Be careful with your subscript Ls and Cs-- I just spent an hour scratching my head because when I was writing everything out on paper by the end I was confusing RL and RC. SecondChildUserIdTAG: 393689 SecondChildUserNameTAG: etindell SecondChildCreateTimeTAG: 2012-12-03T02:04:03Z SecondChildTAG: I've read the answer and still don't get this: $(jwR_cL+R_cR_l) \cdot$ $\displaystyle\frac{1}{LCR_c}$ $= jw \cfrac{R_cL+R_cR_l}{LCR_C}$ ???? Shouldn't this be: $(jwR_cL+R_cR_l) \cdot$ $\displaystyle\frac{1}{LCR_c}$ $= \cfrac{jwR_cL+R_cR_l}{LCR_C}$ Can someone please explain this to me? SecondChildUserIdTAG: 114881 SecondChildUserNameTAG: xvink SecondChildCreateTimeTAG: 2012-12-04T16:57:32Z IndexTAG: 21 TitleTAG: Separate Letter with distinction Thank you edX & MITx Team for this wonderful experience and learning opportunity. I'm hoping that MITx 6.002x team consider issuing a separate letter for those who achieved distinction in the course. Similiar with Berkeleyx CS188.1x Artificial Inteligence course signed by professors. Here's the sample Letter ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/1357281230134360.png UserIdTAG: 185303 UserNameTAG: NormanDP CreateTimeTAG: 2013-01-04T06:35:02Z VoteTAG: 31 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I am afraid we will not be listing grades on certificates or issuing letters of distinction. Although it is not the same thing, you are always welcome to print out a screenshot of your grades page and use that to demonstrate that you did well on 6.002x. FirstChildUserIdTAG: 8 FirstChildUserNameTAG: aa FirstChildCreateTimeTAG: 2013-01-05T22:35:03Z SecondChildTAG: I absolutely agree. IMHO, there's much less rigor in an online course like 6.002x, for example, 24 hours were allowed for both the midterm and the final; doing exercises and problems from the textbook was voluntary; there was no real laboratory experience, no real projects. Therefore, I would say the course is an outstanding complementary tool. To me it doesn't have the foundation for granting a distinction, and even mentioning the grade on the certificate. I hope more exercises and problems from the textbook will be interactive in the future. And thank you for this outstanding experience! SecondChildUserIdTAG: 252722 SecondChildUserNameTAG: vaboro SecondChildCreateTimeTAG: 2013-01-06T03:16:20Z FirstChildTAG: Nice !!! That course was way too tough for me . Great job ! FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2013-01-06T15:57:59Z IndexTAG: 22 TitleTAG: What about @MITx.edu Maybe everyone passing these classes could get an MITx.edu email forwading address. That would make it possible to get software and books and other materials as a student. It could also be used to keep in touch with classmates. Vote up if you support this. The domain is available. UserIdTAG: 142857 UserNameTAG: viking2 CreateTimeTAG: 2012-12-30T23:39:54Z VoteTAG: 31 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: Good idea, but maybe an @edx.edu would be better, so we can buy books used at other universities too. Maybe interesting for edx too, if they can get a small percentage of the resellers? Think off all the hundreds of thousends of students in the future, could be nice to cover some of the costs. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-12-30T23:57:39Z FirstChildTAG: That's totally wonderful. It will help alot. Thanks MITx.edu. FirstChildUserIdTAG: 341850 FirstChildUserNameTAG: Juhi FirstChildCreateTimeTAG: 2012-12-31T05:02:24Z FirstChildTAG: I too agree FirstChildUserIdTAG: 146930 FirstChildUserNameTAG: fayzan_007 FirstChildCreateTimeTAG: 2012-12-31T08:55:17Z FirstChildTAG: Why to waste storage space and resources on this ? Have gmail went out of space ? Wouldn't it be better if those resources would be spent on other online courses ? And why not to all ? If anybody wants the whole world to know that he passed this course, i have an idea : just make an avatar with your progress.A lot of sites will accept large avatar formats, like 256 x 256. FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-12-31T13:24:57Z SecondChildTAG: I think viking2 is talking about a forwarding address that would redirect to someone's Gmail account, thereby using no long-term space. SecondChildUserIdTAG: 45933 SecondChildUserNameTAG: Qubit SecondChildCreateTimeTAG: 2012-12-31T15:39:29Z SecondChildTAG: The comment was edited. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-12-31T18:56:33Z FirstChildTAG: Thumbs Up :-) FirstChildUserIdTAG: 199839 FirstChildUserNameTAG: vijaykrishnankk FirstChildCreateTimeTAG: 2012-12-31T17:01:59Z IndexTAG: 23 TitleTAG: [Tutorial] Write mathematical formula ---------- **Intro** ----- Hi, You might have noticed that we can write equations in a nice style in the forum. In fact, the forum follows the "LaTeX language" via MathJax. You must enclose your commands between two dollars `$$` symbols to access the Math mode. You will find here some useful commands. Don't hesitate to ask for other ones ; ). I'll give some commands twice with the "\displaystyle" variation. I find it nicer but it takes more place. You can find more commands anywhere on the web typing "LaTeX commands". For example: http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Symbols If you want try some formula : http://www.onemathematicalcat.org/MathJaxDocumentation/TeXSyntax.htm EDIT : Thanks Chanute ; ) ---------- **Table** ----- The most useful commands are : - **Constants**: - `$\pi$` gives $\pi$ - *Lowercase Greek alphabet* : `$\alpha, \beta, \omega, \dots$` gives $\alpha, \beta, \omega, \dots$ - *Uppercase Greek alphabet* : `$\Delta, \Sigma, \Omega, \dots$` gives $\Delta, \Sigma, \Omega, \dots$ - `$\sin, \cos, \tan$` gives $\sin, \cos, \tan$ - `$\arcsin, \arccos, \arctan$` gives $\arcsin, \arccos, \arctan$ - `$\ln, \exp$` gives $\ln, \exp$ ---------- - **Division / Fraction**: - `$\frac{A}{B}$` gives $\frac{A}{B}$ - or `$\displaystyle\frac{A}{B}$` gives $\displaystyle\frac{A}{B}$ - *Recurring fractions* : `$\cfrac {V^2}{R_1+\cfrac{R_2\cdot R_3}{R_2+R_3}}$` gives $\cfrac {V^2}{R_1+\cfrac{R_2\cdot R_3}{R_2+R_3}}$ ---------- - **Subscript:** - `$A_B$` gives $A_B$ ---------- - **Superscript / Exponent**: - `$A^B$` gives $A^B$ ---------- - **Multiplication**: - `$A \cdot B$` gives $A \cdot B$ - or `$A * B$` gives $A * B$ ---------- - **Square root \ root**: - `$\sqrt{AB}$` gives $\sqrt{AB}$ - `$\sqrt[3]{AB}$` gives $\sqrt[3]{AB}$ ---------- - **Integration**: - `$\int_0^t f(t)\cdot dt$` gives $\int_0^t f(t)\cdot dt$ - or `$\displaystyle\int_0^t f(t)\cdot dt$` gives $\displaystyle\int_0^t f(t)\cdot dt$ - or `$\displaystyle\int\limits_1^2 f(x)~dx$` gives $\displaystyle\int\limits_1^2 f(x)~dx$ ---------- - **Time derivative**: - `$\dot v$` gives $\dot v$ - `$\ddot v$` gives $\ddot v$ ---------- - **Sum**: - `$\sum_{n=0}^N C_n$` gives $\sum_{n=0}^N C_n$ - or `$\displaystyle\sum_{n=0}^N C_n$` gives $\displaystyle\sum_{n=0}^N C_n$ ---------- - **Condition**: - `$I|_{v=0}$` gives $I|_{v=0}$ ---------- - **Parallel**: - `$R_1 \parallel R_2$` gives $R_1 \parallel R_2$ ---------- - **Checkmark**: - `$\checkmark$` gives $\checkmark$ ---------- - **Fractions and power**: - `$(\frac{A}{B})^2$` gives $(\frac{A}{B})^2$ - or nicer `$\left(\frac{A}{B}\right)^2$` gives $\left(\frac{A}{B}\right)^2$ Note that if you want to use more letters in a subscript you can do it using {}, e.g. `$A_{BC}$` gives $A_{BC}$. You can play with this to do things like $A^{B_C}$. UserIdTAG: 396446 UserNameTAG: RousseauxS CreateTimeTAG: 2012-09-16T17:52:24Z VoteTAG: 30 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: Yes, thank you RousseauxS. This is very helpful. FirstChildUserIdTAG: 419029 FirstChildUserNameTAG: resonate FirstChildCreateTimeTAG: 2012-09-17T00:29:08Z SecondChildTAG: Thanks SecondChildUserIdTAG: 530744 SecondChildUserNameTAG: Sholleybobo SecondChildCreateTimeTAG: 2012-10-01T10:05:06Z FirstChildTAG: That is a sticky if I ever saw one. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-16T18:15:52Z SecondChildTAG: Thanks! This is a really good list. It'll be nice if you add this to the wiki as it will be easily accessible over there. SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-09-16T18:51:08Z SecondChildTAG: Thanks again, for the expanded list, **RousseauxS**! LaTeX makes it a lot easier to ask and/or help someone with the math in the discussion forums. It also looks cool and is easy to use: $P_1 = I_1^2 \cdot R_1$ SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-09-16T19:23:38Z SecondChildTAG: It is in the Wiki now ! https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/mathjax-tutorial-write-formula-forum/ SecondChildUserIdTAG: 396446 SecondChildUserNameTAG: RousseauxS SecondChildCreateTimeTAG: 2012-09-17T17:42:24Z FirstChildTAG: Thank you RousseauxS! Good work! ;) FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-16T19:35:43Z FirstChildTAG: That's awesome RousseauxS! I just want to add to your lovely collection, demonstrating a few things I thought were going to be useful. So that I can follow your post and have quick access to all of the things! And a way-overkill link to a spot with every command under the sun: http://www.onemathematicalcat.org/MathJaxDocumentation/TeXSyntax.htm For recurring fractions, or simply as a shortcut to display style fractions use \cfrac. `$ \cfrac {V^2}{R_1 + \cfrac{R_2\cdot R_3}{R_2+R_3}} $` gives $ \cfrac {V^2}{R_1 + \cfrac{R_2\cdot R_3}{R_2+R_3}} $ Using the \limits_1^2 tag with the int tag in \displaystyle mode makes the limits appear above and below. When using the \right and \left tags, if you want one to not have an item, use a . to denote it should be empty. Other tags demo'd here are \dot and \ddot denoting single and double derivatives, ~ which is a non-linebreaking space, and \checkmark! `$ \displaystyle\int\limits_1^2\ddot f(x)~dx = \left . \dot f(x) \right |_{x=1}^{x=2}~~~\checkmark $` gives $ \displaystyle\int\limits_1^2\ddot f(x)~dx = \left . \dot f(x) \right |_{x=1}^{x=2}~~~\checkmark $ And, and example constructing a matrix! `$ \left [ \matrix { a & b & c \cr d & e & f \cr g & h & i} \right ] $` gives $ \left [ \matrix { a & b & c \cr d & e & f \cr g & h & i} \right ] $ FirstChildUserIdTAG: 248902 FirstChildUserNameTAG: Chanute FirstChildCreateTimeTAG: 2012-09-17T00:51:52Z SecondChildTAG: Thanks Chanute ! I've updated my thread with most of your formula. I didn't put the matrix because the "&" sign is replaced by "amp;". Do you also get that ? I didn't put multi-lines equation for the same reason. It could be related to the MathJax version of the forum : [Link][1] [1]: https://groups.google.com/forum/?fromgroups#!searchin/mathjax-users/&/mathjax-users/QLpeqFQdk88/pMBKR1NYI-QJ SecondChildUserIdTAG: 396446 SecondChildUserNameTAG: RousseauxS SecondChildCreateTimeTAG: 2012-09-17T17:29:48Z SecondChildTAG: Oh wow! They must have changed the forum software since I posted that, because it worked alright back then! They must have applied a filter that replaced all ampersands with & and now you can see that the ampersands are doing their job in the matrix...but the amp;'s are left as an artifact. I've edited it back and it should be showing up alright now! SecondChildUserIdTAG: 248902 SecondChildUserNameTAG: Chanute SecondChildCreateTimeTAG: 2012-09-21T23:40:24Z SecondChildTAG: Ok, nevermind. I think it's a bug in the software, the preview turns out alright, but the filter must be applied post-submittal to ampersands within "code chunks" and Math chunks. This seems like a bug for the techs to look at, as I'm pretty sure we'll be dealing with matrices later in the course, no? SecondChildUserIdTAG: 248902 SecondChildUserNameTAG: Chanute SecondChildCreateTimeTAG: 2012-09-21T23:44:46Z IndexTAG: 24 TitleTAG: Statistics... Hi there, Since this course has ended for 3 weeks already :( can we have some stats about it? I've read the stats from the Spring 2012; it would be interesting to have our statistics as well.. Thanks UserIdTAG: 277787 UserNameTAG: kirilaska CreateTimeTAG: 2013-01-15T14:41:07Z VoteTAG: 29 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 25 TitleTAG: My experience as CTA in edX! Hi everyone! I am writting this Post because I would like to share with you my experience as CTA in edX :). **Some brief 6.002x Historial:** I have to confess that I am really curious and when I first got noticed about the Prototype Course of 6.002x of MITx (I am a big Fan of MIT since my childhood) I suscribed inmediately to the first online Course! I took 6.002x last semester and it was awesome! I met a lot of people and I learnt a lot of interesting things. We keep in touch via that old Forum Discussion. It was my first Online Course :). **What is Next?** Well, I felt sad when I got noticed that the 6.002x Spring 2012 Course was ended. I guess that most of you are asking this by this time haha. So, I can tell you that this is the begining of new things and creativity ;). Here some examples: - *Ammuhave* started with the idea of making the 6.003z Course (Signals and Systems) with the OCW of MIT and run it last semester,in the Gap before we were waiting to start this. Also, students like *ashwith, dantyrant, amyc and juancho* colaborated a lot with that group (juancho posting simulations im maple, amyc colaborating with the Homework explanations, dantyrant and ashwith made video explanations). - *Siliconbronze* offer himself to teach the Course by email that he was teaching since a lot of years ME3521x - Mechanical & Structural Vibration. - *Barrabas* proposed MycroftsMaze Project : build a practice engine in the form of a game. This would be in the form of a text adventure, where the puzzles are practice exercises in the form of electronics questions. (In the manner of Zork or Adventure or Legacy.). With the colaboration of students like *logi540,tags07,* ... - and so many many things happened :). This is amazing! I hope this can happen after the end of this Course too! ;). **What I discovered as CTA? :** I registered again in this Course, mainly, because I wanted to help the new students, more to the Hispanic Community as English is not my native language, I speak Spanish. I didn´t knew that of the CTA, the edX offered me that and I accepted ;). I am really thankful that they could give me that opportunity. First, I thought I could not handle it as I couldn´t downvote any of my Classmates last semester (in the last semester students could downvote others students , increasing and decreasing a scoring system), so I were afraid to be weak as I am really sensitive haha. But, I guess that I did well as CTA haha. I really enjoy doing this and I hope to be inspiring for those students that want to help, no mattering being or not CTA, you can do it next semester to the new students, this experience is really a feedback for me: I learnt a lot, I never being a TA, so this opportunity was amazing, thank you edX! - I discovered that there many people involved behind edX (you don´t have any idea of how many they are!). They are excellent people and all are authentic, I will repeat this comment always haha - don´t think that they are paying me to say it haha-this is the totally true. So, all this, is the work of all that Team that is always present behind the scene and you don´t see them but there they are :). Lyla is one of them, she is always there, working at the speed of light! - Also, I discovered that the Staff members are academically prepared. In 3.091x, eg., most of them are studying a PhD or have a PhD, wow! I always feel so tiny, like a mosquito student haha! They are gorgeous! I like that because I can learn a lot of them and I can feel inspired to keep studying more and more :P. Now, I am more confident to do a PhD in the future - In my work, Engineering Student Fellowship, they always tells me that I should apply to a PhD Fellowship once I get the engineer degree haha - anyway, I am still a student, but I am motivated to do it! I like Research so I would love to do it - I don´t know if my brain will follow me, it is sometimes in idle state,needs holidays -nah joke, haha :P! - I discovered that Prof. Agarwal is really cool haha! I guess that you already know that ;). Well that is all! I just wanted to share with you all this beautiful things of edX and that the quality of the Courses are excellent not only because for the material also because all the people involved here, they are really kind and this makes me feel comfortable like being at my own home :). I am so happy for that! Ah! Good luck in the Final Exam! Review all the concepts! ;) Take care, Merry Christmas and Happy New Year! Myriam. P.D. Don´t forget to participate in the Contest, you can read the rules in the Wiki ;). UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2012-12-18T06:37:16Z VoteTAG: 29 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 10 FirstChildTAG: Thank you Myriam! FirstChildUserIdTAG: 175734 FirstChildUserNameTAG: hestrada FirstChildCreateTimeTAG: 2012-12-18T12:52:23Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-18T23:20:22Z SecondChildTAG: thanx a lot Myriam. you had been very helpful. Hope to see you in spring courses as well. SecondChildUserIdTAG: 381797 SecondChildUserNameTAG: Saira180 SecondChildCreateTimeTAG: 2012-12-20T05:18:03Z SecondChildTAG: myriam exams got over for me...i have scored 92% overall other then MR. AGRAWAL ur support make my path so much easy...the journey i have started with this first online course i haven't publish so much neither i attracted with so many people, the reason was my collage also upto the edge i need to keep that in mind too.... i didn't even get so much time for this course...so all ur suggestion for homeworks makes it easy to me.....one more time thanks a lot dude and no one paying me for this either :) n haapy chrismas n new year to u too ;...i hope if i could add u on facebook.... :) SecondChildUserIdTAG: 108863 SecondChildUserNameTAG: shiviz SecondChildCreateTimeTAG: 2012-12-22T17:39:17Z SecondChildTAG: You are welcome Saira180. Me too, I hope to see yoy in spring courses as well !:) Congratulations shiviz!:) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-23T07:57:18Z FirstChildTAG: Your hints are really helpful,thank you! FirstChildUserIdTAG: 136959 FirstChildUserNameTAG: christerpher FirstChildCreateTimeTAG: 2012-12-19T01:11:12Z SecondChildTAG: You are welcome christerpher! :) I am happy that the hints could help you. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-19T01:44:30Z FirstChildTAG: Im sure, you will be excellent engineer. Thank you , Myriam! Good luck FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-12-18T15:34:53Z SecondChildTAG: Thank you Serge! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-18T23:20:43Z FirstChildTAG: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50d1a2cf361ec12500000005 FirstChildUserIdTAG: 368712 FirstChildUserNameTAG: jmen FirstChildCreateTimeTAG: 2012-12-19T11:20:34Z FirstChildTAG: You are a great assistant, and also a great person. Thank you ! FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-12-19T17:36:17Z SecondChildTAG: Thank you AlexAlexandrescu! I am so happy that I could be hepful this Fall. I really like doing this, helping the new students. Thank you for your kind words haha. Good luck in the Final Exam! :) Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-19T21:46:30Z FirstChildTAG: **Myrimit:** If we could get Staff to enact **6.003z** (Signals and Systems) in edX / MITx. that would be great, as that course is usually the next, follow-up course to **6.002x**. I remember at University of Connecticut, our first Electrical Engineering course was EE201 (Fundamentals of Circuit Analysis) which is almost exactly like 6.002x. It was followed by EE202 (Signals and Systems), which I bet is the same as 6.003z! A lot of students are asking "What's next?", and the obvious logical next step is for MITx to implement Signals and Systems. After retaking a course in circuit analysis, it would great (for me personally, to review) and for others to follow up with a signals analysis course. Jersey Mark FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-12-19T18:27:04Z SecondChildTAG: Totally agree with you :). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-19T21:43:56Z FirstChildTAG: Good Luck to everyone and i'll really miss this next week :( FirstChildUserIdTAG: 51259 FirstChildUserNameTAG: Wardah FirstChildCreateTimeTAG: 2012-12-20T18:45:13Z SecondChildTAG: Don´t feel sad Wardah. Is not the end, is the beginning of new things, Good luck to you too :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-20T20:08:36Z FirstChildTAG: Many thanks for your help. You are great. Hope for your succeses in life. FirstChildUserIdTAG: 455950 FirstChildUserNameTAG: Wahabbaluch FirstChildCreateTimeTAG: 2012-12-19T05:51:52Z SecondChildTAG: You are welcome Wahabbaluch . Thank you for your nice words :). I am so happy to helping here. I really enjoy that. I wish you the best to you to, make all that you want and have a lot of fun always! Take care and good luck in the Final Exam! Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-19T21:42:41Z FirstChildTAG: myriam exams got over for me...i have scored 92% overall other then MR. AGRAWAL ur support make my path so much easy...the journey i have started with this first online course i haven't publish so much neither i attracted with so many people, the reason was my collage also upto the edge i need to keep that in mind too.... i didn't even get so much time for this course...so all ur suggestion for homeworks makes it easy to me.....one more time thanks a lot dude and no one paying me for this either :) n haapy chrismas n new year to u too ;...i hope if i could add u on facebook.... :) FirstChildUserIdTAG: 108863 FirstChildUserNameTAG: shiviz FirstChildCreateTimeTAG: 2012-12-22T17:40:05Z SecondChildTAG: Hi shiviz!:) Congratulations! that is an excellent score !:) I had once facebook, but I have closed it years ago as I didn´t used it that much haha. Also I don´t have linkedin profile - anyway my curriculum vitae is so short that better don´t published it haha :p. But I made an email account for my username, you can contactme if anything you need in myrimit at gmail dot com See you, My best wish to you , mYRIAM. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-22T20:52:25Z SecondChildTAG: Thanks myriam hey I have read about student video competition but there are many of things I’m not certain about…… 1) Is it necessary that the project should contain 6.002x relating data or relevant to 6.002x(as the previous year student did), can’t we go to beyond that thing, obviously that will include electronics ; but can we use microcontroller or else… 2) Do we need to show it’s working and idea only or to show the whole process of making. 3) Expected video length…? Other then this when we will get our 6.002x certificate, n when will 6.003 (signal n system) starts… Hey sorry to bother you about that but this was my first course n u could imagine how much excited I am for this….^_^ SecondChildUserIdTAG: 108863 SecondChildUserNameTAG: shiviz SecondChildCreateTimeTAG: 2012-12-23T06:54:18Z SecondChildTAG: Hi shiviz, 1) You can choose a topic of 6.002x, it can be anything related of what we have seen. We encourage to use real components. Yes, you can use microcontroller if you want. 2) We haven´t done tight rules, so we leave to you the creativity, if you want to show the device or components working, you can do it. If you want to show the whole process of making you can do it too. 3) The CECC1 was about 15 min. But, this term we didn´t restrict the time. So, you can manage the time of your video as you like. Nice to see all that enthusiasm!:). As far as I have read, the Certificates will be available before New Year. Based on my experience, last term appeares a button in the profile Progress where you could Download it :). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-23T07:55:09Z SecondChildTAG: thanks....:) SecondChildUserIdTAG: 108863 SecondChildUserNameTAG: shiviz SecondChildCreateTimeTAG: 2012-12-23T10:58:13Z FirstChildTAG: MyriMIT ! You are a very cool person. Thank you so much FirstChildUserIdTAG: 418182 FirstChildUserNameTAG: euldji2005 FirstChildCreateTimeTAG: 2012-12-23T12:15:23Z SecondChildTAG: You are welcome euldji2005 :). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-23T16:38:34Z IndexTAG: 26 TitleTAG: Applause to the best teacher! Thank you very muck for your efforts! UserIdTAG: 97529 UserNameTAG: ivanesses CreateTimeTAG: 2012-12-11T19:40:54Z VoteTAG: 29 CoursewareTAG: Week 14 / S28V13_Voltage_Drop_across_the_Parasitic_Inductor CommentableIdTAG: 6002x_S28V13_Voltage_Drop_across_the_Parasitic_Inductor NumberOfReplyTAG: 10 FirstChildTAG: Thanks professor.... FirstChildUserIdTAG: 613206 FirstChildUserNameTAG: grubio FirstChildCreateTimeTAG: 2012-12-11T21:51:46Z FirstChildTAG: He definitely is the best teacher I've ever seen! FirstChildUserIdTAG: 136959 FirstChildUserNameTAG: christerpher FirstChildCreateTimeTAG: 2012-12-12T00:43:37Z SecondChildTAG: His teaching methods and attitude can be used as a guidance in all Teachers Training courses, right from Pre-school/Kindergarten to Advanced Degrees. SecondChildUserIdTAG: 232157 SecondChildUserNameTAG: GayatriTR SecondChildCreateTimeTAG: 2012-12-12T01:59:27Z SecondChildTAG: I agree. SecondChildUserIdTAG: 40209 SecondChildUserNameTAG: sadaf78 SecondChildCreateTimeTAG: 2012-12-12T11:37:33Z FirstChildTAG: Fantastic!! With professors like you is easy to learn the most difficult things. FirstChildUserIdTAG: 250199 FirstChildUserNameTAG: Expo FirstChildCreateTimeTAG: 2012-12-12T17:07:16Z FirstChildTAG: thanks a lot professor, the best online course I've had so far. greetings from Colombia FirstChildUserIdTAG: 385046 FirstChildUserNameTAG: SergioRod FirstChildCreateTimeTAG: 2012-12-12T02:07:17Z FirstChildTAG: Thanks to the teachers an all the Staff, i really learnt a lot. Applause from Ecuador. FirstChildUserIdTAG: 289949 FirstChildUserNameTAG: manolito FirstChildCreateTimeTAG: 2012-12-15T18:49:16Z FirstChildTAG: Thanks a lot to our dear professor,all staffs and community TAs for your efforts,learing this course while I'm still in college is an unforgettable and great journey for me.Prof. Anant Agarwal makes me much more interested in my major.Best greetings from China. FirstChildUserIdTAG: 361137 FirstChildUserNameTAG: Ericson FirstChildCreateTimeTAG: 2012-12-14T09:40:33Z FirstChildTAG: professor I love this course and material because of you .....thanks a lot FirstChildUserIdTAG: 428658 FirstChildUserNameTAG: Moza FirstChildCreateTimeTAG: 2012-12-15T19:40:52Z FirstChildTAG: I wish also to express my gratitude for having this course with the quality of teaching as well as humor and pleasant atmosphere distilled in the videos. Thank you very much ... FirstChildUserIdTAG: 311074 FirstChildUserNameTAG: BIS001 FirstChildCreateTimeTAG: 2012-12-15T22:05:22Z FirstChildTAG: Thank you professor for all your efforts. Greetings form Spain! FirstChildUserIdTAG: 414462 FirstChildUserNameTAG: Pablo_C FirstChildCreateTimeTAG: 2012-12-19T11:23:25Z FirstChildTAG: Thanks a lot, greetings from poland FirstChildUserIdTAG: 158754 FirstChildUserNameTAG: hyro FirstChildCreateTimeTAG: 2012-12-18T19:51:39Z IndexTAG: 27 TitleTAG: Lab3 Hints requested by ManasiS/mkprasanth/nitesh2703 ;) Hi ManasiS/mkprasanth ! **Your goal for this lab is to design a circuit that implements a 3-input logic gate that implements Z=¬(C(A+B)) where the ¬ symbol stands for logical negation.** Remember that the following colors means: **A**, it is the place/output where it is the **red** voltage source’s probe. **B**, it is the place/output where it is the **green** voltage source’s probe. **C**, it is the place/output where it is the **blue** voltage source’s probe. **Z**, if you see it is the place/output where it is the **cyan** voltage source’s probe ;). ![enter image description here][1] I will give you some hints/aditional data, in order that you can solve this by yourself. Ok, the statements says, that you have to find a circuit (look at the violet box with ???? ), ![4][2] So that Circuit that you have inside the violet box (the output) behaves as the Z (cyan wave) with the different inputs (A,B,C): ![3][3] **Let’s review the basic configurations: True Tables:** Go to the Textbook, page 294 The Circuit for a NAND Figure. (6.20) [HERE][4] ![AND][5] Go to the Textbook, page 294 The Circuit for a multiple NOR Figure. (6.22a) [HERE][6] ![OR][7] Go to the Textbook, page 291 The Circuit for a INVERTER Figure. (6.14b)[HERE][8] ![INVERTER][9] ---------- **Quick Reference Symbols meanings:** ---------- Output=(expression1)**+**(expression2) = do the (expresión1) OR (expression2) Output=(expression1)**.**(expression2) = do the (expresión1) AND(expression2) Output= **¬** (expression) = INVERTER the (expression) Output= **¬** (expression **AND** expression) = NAND Output=**¬** (expression **OR** expression) = NOR ---------- ---------- **Hint 1:** So, if you have your expresssions A,B and C (like we have seen in expression 1 and 2 examples) and the output Z, can we find our circuit combining NAND, NOR, AND, OR, INVERTER, etc?? Yes! Can we find our network? yes!! Hint: take the differents inputs near each voltage source probe. The input A, will be the in the place where it is the red voltage probe ;). Take a look to the exercise of the Textbook Example 6.2 of page 294 [HERE][10] ;) ---------- Ok, Once we have our Circuit (that is your task, to find it), lets modify the W/L parameter: First of all, let’s see how do we change in Sandbox that famous L/W. Let’s see an arbitrary example, ---------- Suppouse that Ron=10kOhm; - Rn=20kOhm; Having: - RON=Rn(L/W) So, - (L/W)=RON/Rn - (W/L)=Rn/RON=20kOhm/10kOhm=2. ![UNO][11] *Remember that we have to write W/L value, only value not unit, in the box:* ![DOS][12] Now, we have our Rn given in the statement so we have to find our RON and then we can have the parameter of the MOSFER W/L. Once you have your Circuit, remember that you have to take a look to the voltaje Output. The statemens say **Please add the appropriate pulldown network of mosfet switches connected to node Z to implement the truth table above, with RON of the mosfets chosen so that Vol of the logic gate is less than 0.25V** **Hint:** We need a voltaje output less of Vol… So, when do we have our highest voltaje in the output? That is tos say, in which combination of ON and OFF the switches do we have the highest voltaje, our worst case? Isn’t it when we have the maximun RPD(Rpulldown)? ;). [Tutorials][13] And if we have the value of RPD that is in function of RON, can we find RON??? Yes!! And if we have RON can we find W/L??? Yes!!! If you replace that value in your Circuit, in each mosfet and you then clic on TRAN, do you have all done? yes!! ;). See you! Myriam. ---------- Now in Spanish: Hola ManasiS/mkprasanth! **El objetivo de este Lab es el diseño de un Circuito que implemente 3 entradas de compuertas lógicas de tal forma que satisfagan la siguiente expresión Z=¬(C(A+B)), en donde el símbolo ¬ , significa negación lógica.** Recuerda que los colores que tú ves en la gráfica que se te da en el enunciado provienen de: En el caso de la **curva roja**, del lugar en donde se encuentra la punta de prueba de tensión referida a **A** en el Circuito. En el caso de la **curva verde**, del lugar en donde se encuentra la punta de prueba de tensión referida a **B** en el Circuito. En el caso de la **curva azul**, del lugar en donde se encuentra la punta de prueba de tensión referida a **C** en el Circuito. Finalmente, la curva **color cyan** es la salida, es decir, del lugar en donde se encuentra la punta de prueba de tensión referida a **Z** en el Circuito. ![enter image description here][14] Aquí te escribiré algunas pistas e información adicional a modo de clarificación, para que tú puedas luego resolverlo solo. Bien, el enunciado nos dice que debemos hallar un circuito. Si ves en la imagen allí verás un recuadro violeta con signos de interrogación, ese recuadro será el circuito que tú deberás hallar, ![4][15] Entonces, ese circuito incógnito dentro de la caja violeta combinará tus entradas (A,B,C) de tal forma que a la salida Z (curva color cyan), se comporte de acuerdo a la gráfica que se solicita: ![3][16] **Manos a la obra! Repasemos las configuraciones básica, comenzando primero por las Tablas de verdad:** Lee la página 294 del Textbook , aquí encontrarás la disposición respectiva de los MOSFETS que te darán una compuerta lógica NAND (ver Figura 6.20) . [Aquí][17] ![AND][18] Lee la página 294 del Textbook , aquí encontrarás la disposición respectiva de los MOSFETS que te darán una compuerta lógica NOR (ver Figura 6.22a) . [Aquí][19] ![OR][20] Lee la página 291 del Textbook , aquí encontrarás la disposición respectiva de los MOSFETS que te darán una compuerta lógica INVERSORA (ver Figura 6.14b) . [Aquí][21] ![INVERTER][22] ---------- **Ayuda Rápida de los significados de la simbología lógica combinacional:** ---------- Output=(expresión1)**+**(expresión2) = do the (expresión1) OR (expresión2) Output=(expresión1)**.**(expresión2) = do the (expresión1) AND(expresión2) Output= **¬** (expresión) = INVERTER the (expresión) Output= **¬** (expresión **AND** expresión) = NAND Output=**¬** (expresión **OR** expresión) = NOR ---------- ---------- **Hint 1:** Si tenemos nuestras expresiones A,B,C (que, como vimos arriba, podrían bien reemplazarse por las expresiones 1 y 2 que se han ejemplificado, es decir, A podría ser la expresión 2, por ejemplo) , y si además, se tiene la salida-output Z, podemos hallar nuestro circuito combinando las distintas compuertas lógicas NAND, NOR, AND, OR e INVERSORA?? Sí! Entonces, podemos hallar nuestro circuito? vaya que sí! Hint: Tomar las distintas entradas cerca de cada punta de prueba de tensión. La entrada A, será por ejemplo, la que se encuentra cerca de la punta de prueba de tensión roja. Si también te es de ayuda, puedes ver el ejercicio del Texbook, Example 6.2 de la página 294 [HERE][23] ;). ---------- Ok, una vez que hallamos el circuito (eso te lo dejo a ti), veamos cómo modificar ese confuso parámetro W/L que se menciona en el enunciado: Vayamos al Sandbox, Intentemos con un ejemplo arbitrario, ---------- Supongamos que tenemos un Ron=10kOhm; - Rn=20kOhm; De la ecuación que relaciona Rn, Ron y (L/W): - RON=Rn(L/W) Podemos reemplazar los valores, y obtener (L/W) - (L/W)=RON/Rn Ahora si lo invertimos, tendremos (W/L) - (W/L)=Rn/RON=20kOhm/10kOhm=2. ![UNO][24] **Recuerda que se debe escribir el valor de W/L sin unidades en el box, y no L/W.** ![DOS][25] Ahora, ya que en el enunciado se nos da la Rn tenemos que hallar nuestro RON para luego así obtener nuestro W/L del MOSFET. Una vez que tienes el circuito, recuerda que tienes que prestar atención al voltaje de salida Vol que se requiere. El enunciado dice: **Por favor, agregue una red pulldown de compuertas mosfets que se conectan al nodo Z para implementar la tabla de verdad de arriba, con RON de los mosfets elegidas de tal manera que el Vol de la compuerta lógica sea menor a 0.25V** **Hint:** se necesita que la salida sea menor a Vol... Entonces, cuando es que tenemos nuestro mayor voltaje posible a la salida? Es decir, en que combinación posible, de encendido y apagado de las compuertas lógicas, tenemos nuestro peor caso, o sea el mayor voltaje? No es acaso cuando tenemos nuestro máximo valor de RPD (Rpulldown) ? ;)[ ver Tutoriales de la Week3][26] Y en el caso que tuvieramos la RPD que a su vez se encuentra en función de RON, podemos hallar nuestro RON??? Sí!! Y si tenemos nuestro RON podemos hallar nuestro W/L?? Sí!!! Entonces, si remplazas el valor de (W/L) en casa mosfet de tu circuito y luego haces clic en el botón TRAN, tienes resuelto el Circuito??? Sí!!! Hasta Pronto! Myriam. ---- Edit: I forgot to include nitesh2703 ;). He posted [here][27] [1]: https://www.edx.org/static/content-mit-6002x/images/circuits/Lab3_1.e51463be4f12.png [2]: https://edxuploads.s3.amazonaws.com/13489414283670985.png [3]: https://edxuploads.s3.amazonaws.com/13489415553347277.png [4]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/318 [5]: https://edxuploads.s3.amazonaws.com/1348941853134361.png [6]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/318 [7]: https://edxuploads.s3.amazonaws.com/1348942027391101.png [8]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/315 [9]: https://edxuploads.s3.amazonaws.com/13489421793034176.png [10]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/315 [11]: https://edxuploads.s3.amazonaws.com/13489424608529173.png [12]: https://edxuploads.s3.amazonaws.com/1348942494453794.png [13]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_3/week3_gjs/ [14]: https://www.edx.org/static/content-mit-6002x/images/circuits/Lab3_1.e51463be4f12.png [15]: https://edxuploads.s3.amazonaws.com/13489414283670985.png [16]: https://edxuploads.s3.amazonaws.com/13489415553347277.png [17]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/318 [18]: https://edxuploads.s3.amazonaws.com/1348941853134361.png [19]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/318 [20]: https://edxuploads.s3.amazonaws.com/1348942027391101.png [21]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/315 [22]: https://edxuploads.s3.amazonaws.com/13489421793034176.png [23]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/315 [24]: https://edxuploads.s3.amazonaws.com/13489424608529173.png [25]: https://edxuploads.s3.amazonaws.com/1348942494453794.png [26]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_3/week3_gjs/ [27]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50646730c65fb72300000008 UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2012-09-29T18:30:36Z VoteTAG: 29 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: Gracias Myrimit =D! me sirvió de mucho tus posts! FirstChildUserIdTAG: 113110 FirstChildUserNameTAG: Menphys FirstChildCreateTimeTAG: 2012-09-30T01:19:57Z SecondChildTAG: Por nada Menphys! Me alegro mucho que te hayan sido de ayuda! ;) Saludos!!! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-30T01:28:07Z SecondChildTAG: Dear Myrimit, Thanks a lot I completed the same Thanks for Helping us I completed the lab and HW now Need to start look on the week 4 Class on Monday Thanks MK.Prasanth SecondChildUserIdTAG: 156694 SecondChildUserNameTAG: mkprasanth SecondChildCreateTimeTAG: 2012-09-30T15:11:37Z SecondChildTAG: Congratulations mkprasanth! Well done! ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-30T16:21:04Z FirstChildTAG: Myrimit, thank you for this post! I am sure many students will find this very useful FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-29T19:46:32Z SecondChildTAG: You are welcome jelizon ;). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-29T20:38:48Z FirstChildTAG: Thank you so much Myrimit! Your explanation is wonderful! I was stuck with the MOSFET design and I think I still have to work on that.Logic gates have been so easy for me. Never though NAND and NOR with MOSFET will be difficult for me! Now I got the circuit searching for w/l. hehe Thanks again! :) FirstChildUserIdTAG: 281159 FirstChildUserNameTAG: ManasiS FirstChildCreateTimeTAG: 2012-09-29T20:47:53Z SecondChildTAG: You are welcome ManasiS ;). I am here to help. If you have any doubt please ask, I will try to do my best! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-29T21:01:23Z SecondChildTAG: Got it! Got it! Got it! Got it! Got it! Got it! Got it! Got it! Got it! :D So happy. SecondChildUserIdTAG: 281159 SecondChildUserNameTAG: ManasiS SecondChildCreateTimeTAG: 2012-09-29T21:38:18Z SecondChildTAG: Congratulations! Well done ;)! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-29T22:04:19Z SecondChildTAG: ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13490132141343619.png I got this result when I did the trans analysis, which is just like the output desired, except for the glitch at 3 micro seconds. can you help me out of this??? SecondChildUserIdTAG: 171378 SecondChildUserNameTAG: ABHISHEKFROMINDIA SecondChildCreateTimeTAG: 2012-09-30T13:56:04Z SecondChildTAG: Hi ABHISHEKFROMINDIA! Are you sure that you are getting the correct output? Remember that your output Z has to be the one that you see in the cyan color. In the statement they give you this image and comparing with the image above it is different (take a look to the last wave, it is a line in your image, and it sould be similar to a step )...: ![image][1] 1- Are you sure that you have the Correct Circuit? 2- Are you sure that you are getting the correct W/L? [1]: https://www.edx.org/static/content-mit-6002x/images/circuits/Lab3_1.e51463be4f12.png SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-30T16:28:27Z FirstChildTAG: Great stuff Myriam! You should become a teacher :-) Or are you one already? FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-09-30T13:33:43Z SecondChildTAG: Hi ashwith! Thank you! ;). I am not a Teacher haha, just a Student. I have to confess that I always wanted to be a Teaching Assistant in my University haha , but I am not feeling confident yet, because it is a big responsability with the Students in a Class. The only thing that I am contributing for the moment, is to help free as a Student in a Investigation Group that analyze the dropping out of studies (University), it is really interesting knowing all the difficulties of the Students, and also to know that I am helping to the University and to the Students, both ;). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-30T15:40:42Z FirstChildTAG: Hello everyone. I got the same output as desired one but still it says on checking that its wrong. What should I do? ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13494983676365312.png FirstChildUserIdTAG: 177968 FirstChildUserNameTAG: manmeet143 FirstChildCreateTimeTAG: 2012-10-06T04:39:41Z SecondChildTAG: Hi manmeet143, The Lab3 deadline have already passed, might it is because of that (deadline september 30th)... But, can I help you? Have you checked your W/L ratio with the solution provided in the Homework by the Staff?. Have you calculated correctly your RON? Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T04:41:21Z IndexTAG: 29 TitleTAG: Add your marker on the Google map In the Spring course of 2012 a google map was created where everyone who followed the course could place a mark with his location on the map. [googlemap spring course 6002x][1] Please do not change this map!!!!! I propose we create a new map for this course. You need a gmailaccount (free) to be able to add your marker. Click on Edit Button and place your mark on the map at the location you reside. Don't change anyone's else's mark. ![markermenu][2] Click on the blue flag so you can create a red marker on the map Label your marker with your Edx alias name. Save the changes. This is the link to the new map for the fall course 6002x: http://goo.gl/maps/kR6AK [1]: https://maps.google.com/maps/ms?ll=26.74561,17.578125&spn=133.365185,343.125&t=m&z=2&msa=0&msid=216538751025155157244.0004c280194f8253493c4&source=embed [2]: https://edxuploads.s3.amazonaws.com/13564036961343669.jpg UserIdTAG: 79885 UserNameTAG: ruudoleo CreateTimeTAG: 2012-12-25T02:51:39Z VoteTAG: 27 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 10 FirstChildTAG: Done, thanks! FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-12-25T14:24:39Z FirstChildTAG: I solved the problem. Thanks. FirstChildUserIdTAG: 260994 FirstChildUserNameTAG: capvl FirstChildCreateTimeTAG: 2012-12-25T13:46:13Z FirstChildTAG: Nice idea! Done FirstChildUserIdTAG: 325197 FirstChildUserNameTAG: Vitali_Jerin FirstChildCreateTimeTAG: 2012-12-25T08:53:11Z FirstChildTAG: Nice!! But...Am i the only Greek who completed 6.002?? FirstChildUserIdTAG: 304587 FirstChildUserNameTAG: Vasso FirstChildCreateTimeTAG: 2012-12-25T12:39:42Z FirstChildTAG: How do I do to clean mark on the map ? FirstChildUserIdTAG: 260994 FirstChildUserNameTAG: capvl FirstChildCreateTimeTAG: 2012-12-25T13:09:39Z FirstChildTAG: Give even more data to google? No thanks. FirstChildUserIdTAG: 406420 FirstChildUserNameTAG: Picolo FirstChildCreateTimeTAG: 2012-12-25T16:34:32Z FirstChildTAG: Its nice to see that I had my fellow mates from every corner of the world. FirstChildUserIdTAG: 154016 FirstChildUserNameTAG: Albatross FirstChildCreateTimeTAG: 2012-12-25T14:50:01Z FirstChildTAG: A great idea! FirstChildUserIdTAG: 361137 FirstChildUserNameTAG: Ericson FirstChildCreateTimeTAG: 2012-12-30T17:42:33Z FirstChildTAG: Another one =) FirstChildUserIdTAG: 379741 FirstChildUserNameTAG: Tolcheev_Art FirstChildCreateTimeTAG: 2012-12-27T11:11:21Z FirstChildTAG: Great ! FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2013-01-03T21:03:11Z IndexTAG: 30 TitleTAG: Stats about 6.002x Fall 2012 ? Hi, By curiosity: - how many people signed up to the class ? - from how many countries ? (it would be cool to have pie charts with the number of people by country) - what are the average results for the HW and Labs ? I think I read somewhere that more than 150.000 people registered last time and found that really huge. Could some stats be available on the platform ? : ) Thanks! UserIdTAG: 396446 UserNameTAG: RousseauxS CreateTimeTAG: 2012-09-24T19:43:36Z VoteTAG: 27 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Well you can have some idea. The old registrations are still active as I saw I was member 79885 the same as in the mitx course. I have seen member 400207. So at least 400000-150000= 250000 new members but that would be for the whole Edx. 400207 is not the last. I only scanned a few members. You can see your number if you click on your name in the discussion forum- it's at the end of the browserlink. So members with higher numbers post here your membernumber. It takes some time before they release the actual subscriber numbers in a press conference probably before or after the midterm exam. Ofcourse they could do it earlier to contradict me :<). FirstChildUserIdTAG: 79885 FirstChildUserNameTAG: ruudoleo FirstChildCreateTimeTAG: 2012-09-28T20:41:25Z SecondChildTAG: I didn't know the trick of the number ; ). If the staff comes here, it could be great to release some stats or at least say when it will be the case. SecondChildUserIdTAG: 396446 SecondChildUserNameTAG: RousseauxS SecondChildCreateTimeTAG: 2012-09-30T14:56:15Z SecondChildTAG: I saw member 619805 who logged in 3 days ago so we have 619805-154763=465042 new members. SecondChildUserIdTAG: 79885 SecondChildUserNameTAG: ruudoleo SecondChildCreateTimeTAG: 2012-10-14T11:15:52Z IndexTAG: 31 TitleTAG: complete solutions/answers to Midterm For those who have asked for solutions, here is my work. Excuse my writing. feel free to ask for any clarifications.. Up vote this if it helps. OOPS soory that it came jumbled, look for page number in comment sectiopn.![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13515197921343671.jpg UserIdTAG: 148523 UserNameTAG: span993 CreateTimeTAG: 2012-10-29T14:10:26Z VoteTAG: 26 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 11 FirstChildTAG: ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13515200071343663.jpg FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-10-29T14:14:18Z SecondChildTAG: page 3 SecondChildUserIdTAG: 148523 SecondChildUserNameTAG: span993 SecondChildCreateTimeTAG: 2012-10-29T14:42:47Z FirstChildTAG: ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13515205261343658.jpg FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-10-29T14:22:44Z SecondChildTAG: page 6 SecondChildUserIdTAG: 148523 SecondChildUserNameTAG: span993 SecondChildCreateTimeTAG: 2012-10-29T14:43:41Z SecondChildTAG: the only i didn't get =( SecondChildUserIdTAG: 308853 SecondChildUserNameTAG: fmorato SecondChildCreateTimeTAG: 2012-10-29T23:54:32Z FirstChildTAG: ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13515206301343669.jpg FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-10-29T14:24:46Z SecondChildTAG: page 7 SecondChildUserIdTAG: 148523 SecondChildUserNameTAG: span993 SecondChildCreateTimeTAG: 2012-10-29T14:43:54Z FirstChildTAG: ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13515210171343632.jpg FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-10-29T14:31:31Z SecondChildTAG: page 8 SecondChildUserIdTAG: 148523 SecondChildUserNameTAG: span993 SecondChildCreateTimeTAG: 2012-10-29T14:44:07Z FirstChildTAG: ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/1351520226134362.jpg FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-10-29T14:17:20Z SecondChildTAG: page 5 SecondChildUserIdTAG: 148523 SecondChildUserNameTAG: span993 SecondChildCreateTimeTAG: 2012-10-29T14:44:23Z FirstChildTAG: ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13515224261343627.jpg FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-10-29T14:54:34Z SecondChildTAG: page 9 SecondChildUserIdTAG: 148523 SecondChildUserNameTAG: span993 SecondChildCreateTimeTAG: 2012-10-29T14:54:59Z FirstChildTAG: ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13515225351343674.jpg FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-10-29T14:56:15Z SecondChildTAG: page 10 SecondChildUserIdTAG: 148523 SecondChildUserNameTAG: span993 SecondChildCreateTimeTAG: 2012-10-29T14:56:28Z FirstChildTAG: **All:** Please note that **span993**'s answers may be different than your answers because initial conditions such as V1, R1, etc. vary and can be different between two students taking the same exam. I guess this is in place to hinder cheating, as copying a list of answers will not work. However, the theory behind, and the format of, the questions are the same for all students. Therefore, look at the equations that **span993** presents, and the logic he/she uses to solve the problems. If you want to check, use the numerical values that you were given in your particular exam, and plug those into the equations in the photos / pictures shown. Note that the "official" answers will come out tomorrow. FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-10-29T15:23:22Z FirstChildTAG: ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13515198811343639.jpg FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-10-29T14:11:33Z SecondChildTAG: page 2 SecondChildUserIdTAG: 148523 SecondChildUserNameTAG: span993 SecondChildCreateTimeTAG: 2012-10-29T14:42:27Z FirstChildTAG: ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13515201063145627.jpg FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-10-29T14:15:52Z SecondChildTAG: page 4 SecondChildUserIdTAG: 148523 SecondChildUserNameTAG: span993 SecondChildCreateTimeTAG: 2012-10-29T14:43:04Z FirstChildTAG: Nice Job FirstChildUserIdTAG: 455950 FirstChildUserNameTAG: Wahabbaluch FirstChildCreateTimeTAG: 2012-11-01T04:52:15Z SecondChildTAG: good nice effort SecondChildUserIdTAG: 550401 SecondChildUserNameTAG: revathisingh SecondChildCreateTimeTAG: 2012-11-01T15:35:54Z SecondChildTAG: Thank you! Wahabbaluch and revathisingh. SecondChildUserIdTAG: 148523 SecondChildUserNameTAG: span993 SecondChildCreateTimeTAG: 2012-11-01T16:38:38Z IndexTAG: 32 TitleTAG: Lab6 Hints requested by dandradet (Read down for the Spanish translation) Hi dandradet, sorry for the delay. You asked me [here][1] if I could guide you in Lab6 as you were having difficulties with the English language. Here I have translated to you the statement of Lab6 in [Spanish][2] :). **Lab6:** Lets see this together dandradet, so later you can solve it by yourself, ---------- In the First figure they give you a circuit of an inverter. The goal is that you can click on TRAN and obtain by graphical measurements the RON, VTH and RTH and $t_{pd,0 \rightarrow 1}$ and the $t_{pd,1 \rightarrow 0}$ . But in the second part, they give you inverters one ahead of the other (9 inverters in total), there they explain you about the propagation delay and they request you to find it graphically. ---------- Ok, that is the general idea of this Lab, but lets go to the start: ![circuit][3] The statement says: Run a 2ns transient analysis on this circuit and use measurements from the plot to answer the following questions. **Part 1:** Measure final output voltage of the inverter when the input is high and use that to estimate RON for the mosfet switch. So, I am sure that you are asking yourself what is the final output of the inverter when the input is high? Hint : Read page 526 of the Textbook [read here][4] and focus yourself in the Figure 10.16 , you will observe 3 curves (input curve, ideal output curve and actual output curve). See the input curve, see when it goes from the 0 to the 1 (low and high) and then, see the third curve... Hmmm, so, What it is on your plot when you click on TRAN? Isn’t it your output voltage? Can you tell me, to which value it reachs when your input reachs to the high? take a look again to the Figure 10.16, you will find the answer :). Hint : Lets back to the inverter, read the page 306 [read here][5] of the Textbook. Do you have your required vOUT from the previous parragraph? yes! so, can you find your RON? Yes! ---- **Part 2:** Estimate VTH and RTH for the Thevenin equivalent circuit. Hint: Take a look at Figure 10.22 [see here][6] the Figure 10.22 a) is the circuit model and the 10.22 b) it is its equivalent thevenin. So, what is your VTH? Can you calculate the voltage supply that you observe in the equivalent thevenin? yes! :p Hint: The same goes for your RTH :). Can you calculate your resistance that you observe in the equivalent thevenin? yes! ---- **Part 3:** Hint: In they statement they are telling you to read the equation 10.66 [see here][7]Be careful with the value of vC(t), in the Textbook they are giving an specific value in order to reach to 10.68 equation result :). So what would be your vC(t) in your case? ;) ----- **Part 4:** Hint: Lets see again the page 526 [read here][8], the figure 10.16, third curve :). Hint: Read definition of tpd 0->1 of page 527 [read here][9]… is the time interval pair for a low to high instantaneous transition at the input, more precisely, is the time interval from the moment that the input changes from 0 to 1 to the moment that the output reaches a corresponding valid output voltage level… So, what is the valid output voltage level that we reached it? Take a look again to the third graph, where valid voltage does it reach in the output when passing 0 to 1 in the input? ---- **Part 5:** Ok, here the statement request you to change interchange the initial and plateau voltages. To do that, double click on Vin and interchange the values :)and just do what the statement says: and measure the value of tpd,1→0, the time it takes for the inverter's output to reach VOH. ![change][10] Hint: you should change your TRAN to a upper time like 5 nano seconds or 5n :) ---- **Part 6:** Some Hints: The problem says: **Run a 50ns transient simulation on the ring oscillator and measure the period of oscillation. Be patient! It can take a moment for the simulation to complete. Divide the result by 9 to get an estimate for the time it takes one inverter to make a 0→1 transition followed by a 1→0 transition.** *So, what is the period of oscillator?* To calculate the period of oscillator you have to undestand: what is a period of a signal? Let's play with the sandbox: - make the next circuit: Voltage Source signal type: sin. ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13350501687740329.png) - Click on TRAN and watch the graph ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13350504641639809.png) - Now, to measure the period of a signal you have to choose two points of the curve **a** and **b**: ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13350508142068495.png) So, the **period** of this signal is **T =b-a** Now, it is your turn. Click on TRAN of the problem circuit. ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13350498368201185.png) Now, measure the T period. Remember to wait to the signal to be stable. ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13350511497002498.png) Once you have the period T , you have to find just like the problem says **"'Divide the result by 9 to get an estimate for the time it takes one inverter to make a 0→1 transition followed by a 1→0 transition", that is to say = T/9**. That is all, I hope this can help you :). Myriam. ----- Now in Spanish: **Lab6:** dandradet, vamos a ver esto juntos, así luego tú lo puedes resolver por ti mismo. ---------- En la primera figura, se te da un circuito de un inversor. El objetivo es que al hacer clic sobre TRAN tú puedas obtener por métodos gráficos,mediante mediciones que tú harás, los valores que te permitirán obtener RON, VTH, RTH, $t_{pd,0 \rightarrow 1}$ y $t_{pd,1 \rightarrow 0}$ . Pero en la segunda parte, tendrás 9 inversores concatenados, por eso dicen que se encuentran en forma de anillo, uno atrás del otro. Allí, en el enunciado se te explica más sobre el delay de propagación y se te pide que lo halles gráficamente (puedes ver la traducción del enunciado [aquí][11]). ---------- Bien, ahora que ya tienes una idea general de lo que se te solicitará, comencemos desde el principio: ![circuit][12] El enunciado dice: Compile a 2ns el análisis transitorio de este circuito y utilice las mediciones del plot para responder a las siguientes preguntas. **Parte 1:** Medir la tensión final del inversor cuando la entrada se encuentra en estado alto y utilícela para estimar el RON del mosfet switch. Entonces, de seguro que ahora te estás preguntando: Cuál es el valor final del inversor cuando la entrada se encuentra en estado alto? Hint : Leer la página 526 del Textbook [leer aquí][13] y concéntrate en la Figura 10.16 , verás que habrá tres gráficas (una correspondiente a la entrada, otra a la salida pero considerándola en un contexto ideal y finalmente, una tercera que será la salida pero en un contexto real). Si miras la gráfica correspondiente a la entrada, y observas cuando pasa del 0 al 1 (es decir, del estado bajo al alto) y luego, observas paralelamente cómo se comporta la tercera curva, la salida.... Entonces, mmmm, puedes decirme qué es lo que estás viendo en tu plot cuando realizas clic en TRAN? No es acaso que estás viendo tu tensión de salida? Entonces, puedes decirme, a qué valor llega la salida cuando nuestra entrada llega al valor alto? Mira nuevamente la Figura 10.16, allí encontrarás la respuesta :). Hint : Volvamos al inversor, leer la página 306 del Textbook [leer aquí][14]. Tienes tu vOUT requerida del párrafo posterior ? sí! entonces, puedes hallar tus RON? Sí! ---- **Parte 2:** Estimar VTH y RTH para el circuito equivalente Thevenin. Hint: Mirar la Figura 10.22 [ver aquí][15]. La Figura 10.22 a) es el modelo circuital del inversor y la Figura 10.22 b) es el equivalente Thevenin. Entonces, cuál será tu VTH? Puedes calcular la fuente de alimentación que ves en el equivalente Thevenin? Sí! :p Hint: Lo mismo aplica para la RTH :). Puedes calcular ;a resistencia que observas en el equivalente Thevenin ? Sí! ---- **Parte 3:** Hint: En el enunciado se te dice que tienes que leer la ecuación 10.66 [ver aquí][16]. Ten cuidado con el valor de vC(t), en el Textbook se te está dando un valor específico para el cálculo que puedes observar en la ecuación 10.68 :). Entonces, en tu caso, qué vC(t) deberás elegir? ;) ----- **Parte 4:** Hint: Ver nuevamente la página 526 [leer aquí][17], y observar la tercera gráfica de ;a figura 10.16 :). Hint: Leer la definición de tpd 0->1 en la página 527 [leer aquí][18]… , básicamente, es el intervalo de tiempo para el cuál la entrada cambia de 0 a 1 al momento que la salida llega al correspondiente valor válido de salida.... Entonces, cuál es valor válido de nivel de salida al cuál llegamos? Mirar nuevamente el tercer gráfico, a qué valor se corresponde cuando la salida llega al valor válido cuando la entrada pasa de 0 a 1 ? ---- **Parte 5:** Bien, aquí el enunciado te solicita intercambiar el valor initial y plateu de las tensiones. Para realizarlo, debes hacer doble clic sobre Vun e intercambiar ambos valores :) y hacer, lo que se te solicita: medir el valor de tpd,1→0, el tiempo que tarda la salida del inversor en llegar a VOH. ![change][19] Hint: cambia tu TRAN a un tiempo un poco más grande, como 5 nano seconds o 5n :) ---- **Part 6:** Algunas Hints: El problema dice: **Compila la simulación transitoria a 50 ns y mide el período de oscilación. Sé paciente! Puede ser que tarde un momento hasta que se complete la simulación. Divide el resultado por nueve y obtén un tiempo estimativo para el cual el inversor tarde en realizar una transición de 0->1 seguida de una transición de 1->0.** *Entonces, cuál es el periodo del oscilador?* En orden de calcular el periodo de oscilación tienes que entender: qué es el periodo de una señal? Juguemos con el sandbox: - Hacer el siguiente circuito: Voltage Source signal type: sin. ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13350501687740329.png) - Hacer clic en TRAN y ver el gráfico ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13350504641639809.png) - Ahora, medir el periodo de la señal. Deberás elegir dos puntos de la curva **a** y **b**: ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13350508142068495.png) Entonces, el **periodo** de esta señal es **T =b-a** Ahora es tu turno! Hacer Clic en TRAN en el circuito del Lab. ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13350498368201185.png) Ahora, mide el periodo T. Recuerda que tienes que esperar a que la señal sea estable. ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13350511497002498.png) Una vez que tienes tu periodo T, tienes que realizar lo que dice el enunciado **Divide el resultado por nueve y obtén un tiempo estimativo para el cual el inversor tarde en realizar una transición de 0->1 seguida de una transición de 1->0**, es decir, T/9. Eso es todo, espero que esto te haya servido de ayuda :). Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/507ef87ef4d89e1f00000137 [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5081c45132aa871f00000002 [3]: https://edxuploads.s3.amazonaws.com/13507412791343667.png [4]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/550 [5]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/330 [6]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/555 [7]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/557 [8]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/550 [9]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/551 [10]: https://edxuploads.s3.amazonaws.com/13507443211343617.png [11]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5081c45132aa871f00000002 [12]: https://edxuploads.s3.amazonaws.com/13507412791343667.png [13]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/550 [14]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/330 [15]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/555 [16]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/557 [17]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/550 [18]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/551 [19]: https://edxuploads.s3.amazonaws.com/13507443211343617.png UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2012-10-20T15:40:51Z VoteTAG: 26 CoursewareTAG: Week 6 / Ring Oscillator CommentableIdTAG: 6002x_Ring_Oscillator NumberOfReplyTAG: 10 FirstChildTAG: Eres una crack!! muy util FirstChildUserIdTAG: 250473 FirstChildUserNameTAG: pajaropica FirstChildCreateTimeTAG: 2012-10-20T16:06:12Z SecondChildTAG: Gracias pajaropica! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-20T17:02:21Z FirstChildTAG: Awsome work Myriam!!! This post helped me a lot.. Thank You FirstChildUserIdTAG: 362299 FirstChildUserNameTAG: anandbaskaran FirstChildCreateTimeTAG: 2012-10-20T16:59:33Z SecondChildTAG: You are welcome anandbaskaran! Nice to know that it was helpful:) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-20T17:03:33Z FirstChildTAG: Very nice, Myrimit. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-20T17:47:54Z SecondChildTAG: Thank you JSChambers :). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-20T17:52:29Z FirstChildTAG: thanks alot Myrimit...ur hints always work :) FirstChildUserIdTAG: 278792 FirstChildUserNameTAG: sali FirstChildCreateTimeTAG: 2012-10-21T05:10:48Z SecondChildTAG: You are welcome sali :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-21T05:58:13Z SecondChildTAG: Can't understand, where I must measure vout to find Ron( In my measurement vout=Vs when input is high. What I do wrong? SecondChildUserIdTAG: 133015 SecondChildUserNameTAG: GeksogeN SecondChildCreateTimeTAG: 2012-10-21T09:56:51Z FirstChildTAG: You are very very very Clever ....... Are you GRADUATE or underdrgaduate ? FirstChildUserIdTAG: 382505 FirstChildUserNameTAG: AhmedGalal2 FirstChildCreateTimeTAG: 2012-10-21T14:53:56Z SecondChildTAG: Hi AhmedGalal2, I am undergraduate student of Engineering and you? Thank you for the compliment, haha! :). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-21T21:24:22Z SecondChildTAG: Myrimit ..ur younger than me , still ur such a genius...i completed my engineering (undergrad) 2 yrs back n now i cant remember anything.i have a 9 month baby with me so its really difficult studying now..:( but this course is sure a refresher...i hope i get through it SecondChildUserIdTAG: 278792 SecondChildUserNameTAG: sali SecondChildCreateTimeTAG: 2012-10-22T03:25:57Z SecondChildTAG: Hi sali! Congratulations for the baby! I admire you sali, taking care of a baby it is a really hard work and simultaneusly doing courses and all, it should be for you a lot of effort, and I have to say that I admire you :). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-22T03:40:02Z FirstChildTAG: Can't understand, where I must measure vout to find Ron( In my measurement vout=Vs when input is high. What I do wrong? FirstChildUserIdTAG: 133015 FirstChildUserNameTAG: GeksogeN FirstChildCreateTimeTAG: 2012-10-21T09:57:07Z SecondChildTAG: Hi GeksogeN! Are you sure that your output voltage it is VS when your input is high? To which value it tends to? - Take a look at the 1st curve (input voltage), when it reachs to 1 logical, - What would be your output voltage (third curve)? And what are you observing in your plot? isn't it your output voltage in this condition? So, to which value it reachs vout? ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13508247331343602.png SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-21T21:19:39Z SecondChildTAG: Myrimit can you please tell which app you are using to make these illustrations. Thank you very much for putting them together. They are indeed helpful. SecondChildUserIdTAG: 623210 SecondChildUserNameTAG: preveen SecondChildCreateTimeTAG: 2012-10-29T13:49:26Z SecondChildTAG: @preveen, just paint hahaha! ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-29T15:09:02Z FirstChildTAG: Hola Myriam.. muchas gracias Rohit FirstChildUserIdTAG: 35584 FirstChildUserNameTAG: rohit7gupta FirstChildCreateTimeTAG: 2012-10-21T18:00:44Z SecondChildTAG: Por nada Rohit :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-21T21:24:37Z FirstChildTAG: hello Myriam "my savior :) " I would like to ask you about the value of vCin part3 how to find it? thank you very much FirstChildUserIdTAG: 226085 FirstChildUserNameTAG: Teto FirstChildCreateTimeTAG: 2012-10-21T21:26:13Z SecondChildTAG: You are welcome Teto! :) **Hint:** Take a look at page 532 [read here][1], in the part "computing tpd,0->1", 2nd parragraph . What value are they taking for vC? Isn't it something related to a valid output ? ;). [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/556 SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-21T23:14:17Z SecondChildTAG: aaaaa!!! actually I still can not get it. sorry can you explain more please SecondChildUserIdTAG: 226085 SecondChildUserNameTAG: Teto SecondChildCreateTimeTAG: 2012-10-21T23:28:00Z SecondChildTAG: this is a really aha...! moment. finally I found it. thank you very much. SecondChildUserIdTAG: 226085 SecondChildUserNameTAG: Teto SecondChildCreateTimeTAG: 2012-10-21T23:50:06Z SecondChildTAG: Aha! Well done! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-22T00:16:03Z SecondChildTAG: Hi Myrimit. In Part 1 of Lab 6, vOUT = 250m, isn't? I can't get the green mark! SecondChildUserIdTAG: 288637 SecondChildUserNameTAG: gotchi SecondChildCreateTimeTAG: 2012-10-22T00:55:32Z SecondChildTAG: Hi gotchi! Hint: Are you sure that that is the final value that reachs the output? To which value does it tends to? ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-22T01:26:35Z SecondChildTAG: Myrimit, thank you so much! You do saved my life this time!! =D SecondChildUserIdTAG: 288637 SecondChildUserNameTAG: gotchi SecondChildCreateTimeTAG: 2012-10-22T02:18:28Z SecondChildTAG: Hahaha! Well done gotchi XD SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-22T03:31:24Z FirstChildTAG: i devided by 7 because the my plotter is at the 7th inverter , seems it works with you when u devided by 9 , but works for me when devided by 7 , and thanks for the great work FirstChildUserIdTAG: 285675 FirstChildUserNameTAG: Ascot FirstChildCreateTimeTAG: 2012-10-21T21:41:04Z SecondChildTAG: You are welcome! I have 9 inverters :). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-21T23:15:37Z FirstChildTAG: hi why did in the question he said the period of oscillation being determined by the time it takes for a signal to propagate twice around the loop can't understand ? FirstChildUserIdTAG: 221617 FirstChildUserNameTAG: konan FirstChildCreateTimeTAG: 2012-10-22T06:01:09Z IndexTAG: 33 TitleTAG: Difference between "Distinct value"and "Distinct boolean valued function"..... Firstly by **Distinct value** means here that how many combination would be created by the value..so its simple by **2^n..** Secondly by **Distinct boolean valued function** means here how many truth table can be created by the value... so this by **2^(2^n)** It is the correct way to calculate..... Thank you UserIdTAG: 282892 UserNameTAG: wiky CreateTimeTAG: 2012-09-16T19:21:07Z VoteTAG: 26 CoursewareTAG: Week 2 / Boolean functions CommentableIdTAG: 6002x_S4E2_Boolean_Functions NumberOfReplyTAG: 4 FirstChildTAG: thank you!! FirstChildUserIdTAG: 51696 FirstChildUserNameTAG: blasco23 FirstChildCreateTimeTAG: 2012-09-17T21:09:16Z SecondChildTAG: Thanks But I supppose the n is the number of inputs ie n=number of inputs SecondChildUserIdTAG: 382532 SecondChildUserNameTAG: emmanuelpeace SecondChildCreateTimeTAG: 2012-09-23T08:16:05Z SecondChildTAG: so appreciate SecondChildUserIdTAG: 176234 SecondChildUserNameTAG: hejinjie SecondChildCreateTimeTAG: 2012-09-29T08:23:17Z SecondChildTAG: thank you very very much SecondChildUserIdTAG: 226085 SecondChildUserNameTAG: Teto SecondChildCreateTimeTAG: 2012-10-04T15:19:34Z FirstChildTAG: Yet, it doesn't explain what a 'Boolean-valued function' is. I also figured out a way of calculating the number of 'Boolean-valued functions'; however, it's not quite clear to me what it really is. FirstChildUserIdTAG: 252722 FirstChildUserNameTAG: vaboro FirstChildCreateTimeTAG: 2012-09-18T20:22:59Z SecondChildTAG: It's the number of truth-tables you could compose. So if you have n inputs, each of those can take on 2 different values, for a total of 2^n possible tuples of input values. And for each input tuple, the function can return 0 or 1. So you want to count all the possible ways to assign 0 or 1 to each input tuple, because that's exactly what a Boolean-valued (output is 0 or 1) function does. SecondChildUserIdTAG: 325730 SecondChildUserNameTAG: gadzin1203 SecondChildCreateTimeTAG: 2012-09-18T22:15:31Z SecondChildTAG: I would appreciate further explanation. SecondChildUserIdTAG: 164898 SecondChildUserNameTAG: jbparkes SecondChildCreateTimeTAG: 2012-09-18T23:30:22Z SecondChildTAG: It seems that for each tuple, you can assign either a 0 or a 1, which is two options per tuple. That means 2*2^n, not 2^2^n. I don't understand how it becomes exponential again. SecondChildUserIdTAG: 164898 SecondChildUserNameTAG: jbparkes SecondChildCreateTimeTAG: 2012-09-18T23:31:47Z SecondChildTAG: it suppose to be 2^(2^n) since each function also has two possible inputs either 1 or zero Thanks SecondChildUserIdTAG: 382532 SecondChildUserNameTAG: emmanuelpeace SecondChildCreateTimeTAG: 2012-09-23T08:29:52Z FirstChildTAG: The link below http://mathworld.wolfram.com/BooleanFunction.html gives an example for two values. It might help :) FirstChildUserIdTAG: 319299 FirstChildUserNameTAG: sumanthy FirstChildCreateTimeTAG: 2012-09-20T00:37:39Z SecondChildTAG: really thank u i pleased to see Ur link SecondChildUserIdTAG: 295983 SecondChildUserNameTAG: qassam SecondChildCreateTimeTAG: 2012-09-20T04:13:05Z SecondChildTAG: Your link really explains it. Thanks. The number of functions are 2^(number of distinct values represented), i can see it clearly in your link SecondChildUserIdTAG: 55392 SecondChildUserNameTAG: eff SecondChildCreateTimeTAG: 2012-09-20T17:16:35Z FirstChildTAG: Thank you! I stuck at those "boolean-valued funtions" questions. FirstChildUserIdTAG: 201508 FirstChildUserNameTAG: datle FirstChildCreateTimeTAG: 2012-09-21T09:15:43Z IndexTAG: 34 TitleTAG: Inclusion of 2D barcode for verification link in the certificate I have a humble suggestion that these certificates should also have a 2D barcode encoding of the verification link. Please see a sample certificate in the link below http://1.bp.blogspot.com/-PYTkDqIDsQU/T9l2IfdQUnI/AAAAAAAABvU/z2OMvC9BN8M/s1600/mitx_certificate.png The verification link is difficult to type and a small typing error would show the certificate as invalid. As most smartphones come with a barcode reader application these days, verifying the certificate would be very convenient this way. UserIdTAG: 297370 UserNameTAG: ayush3504 CreateTimeTAG: 2013-01-01T16:36:38Z VoteTAG: 25 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 3 FirstChildTAG: Excellent Idea ! Enjoy this new year FirstChildUserIdTAG: 183712 FirstChildUserNameTAG: colegiocientifico FirstChildCreateTimeTAG: 2013-01-01T16:58:38Z SecondChildTAG: Thank you. You too :) SecondChildUserIdTAG: 297370 SecondChildUserNameTAG: ayush3504 SecondChildCreateTimeTAG: 2013-01-01T18:46:49Z FirstChildTAG: Great Idea! FirstChildUserIdTAG: 454609 FirstChildUserNameTAG: Alwahsh FirstChildCreateTimeTAG: 2013-01-01T19:13:49Z FirstChildTAG: Yes very good idea!!! FirstChildUserIdTAG: 108929 FirstChildUserNameTAG: namit FirstChildCreateTimeTAG: 2013-01-02T09:42:10Z IndexTAG: 35 TitleTAG: Welcome to the CECC 2 ! Do you want to win a Textbook signed personally by Prof. Agarwal? Read here :) > **UPDATED. submition deadline of Videos due January 14th 2013.** **Welcome to the CECC 2!** *(Circuits and Electronics Classmates Contest 2 )* ---------- Take a look at this video introduction to CECC 2 [YouTube - Watch here][1] - For those that are facing issues with YouTube - banned - you can download the video introduction to the Contest [4Shared - here][2] Read the .pdf rules of the Contest at the Wiki [here][3] ---------- > We expect to receive many creative and educational videos! We > encourage you to use real electronic components in your video. You > should send the link of your video uploaded on YouTube or the video > file to the email box ceccmitx@gmail.com. Entries may be submitted > anytime until January 5th, 2013 (two weeks after the final exam). Any > question about the CECC 2 can be requested from ceccmitx@gmail.com. > > The selection of the winning videos will be made by the jury > consisting of **dantyrant, ChaunceyGardiner, JSChambers, komisz, Danik > and Barrabas**. The results of the CECC2 will be published in the forum > few weeks after the date of the Final Exam. We will also contact the > winners via e-mail. > > The organizers are **ashwith, juancho and Myrimit**. > > All students involved in this Contest as the CECC 2 Team do not > receive any kind of remuneration and their participation is voluntary. CECC Team [1]: http://www.youtube.com/watch?v=OvtyCzxEOCE&feature=youtu.be [2]: http://www.4shared.com/video/Oct7xKcz/C12F.html [3]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/ UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2012-11-25T21:09:58Z VoteTAG: 25 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 9 FirstChildTAG: Sounds like a great idea and I wish all the entrants the best of luck. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-11-25T22:22:01Z SecondChildTAG: Thanks for the new deadline!!! SecondChildUserIdTAG: 99441 SecondChildUserNameTAG: coyarce SecondChildCreateTimeTAG: 2013-01-03T18:05:43Z FirstChildTAG: It has to be in english? Best regards! FirstChildUserIdTAG: 298547 FirstChildUserNameTAG: AnthonyRF FirstChildCreateTimeTAG: 2012-11-25T23:11:39Z SecondChildTAG: Hi AnthonyRF, We prefer to be in English as the Jury, that will evaluate the videos, speaks English... But if you want to make the video in your native language you can, but I has to have yes or yes the English subtitles so that the Jury can understand it and also your Classmates :). I hope that you can participate and win one of the three Textbooks signed personally by Prof. Agarwal that will be delivered to your home by post. Once we have the Winners we will put in contact with Dianna of edX and she will send the Textbooks. If you decide to participate, have a lot of fun and make the best video ever haha! :) I wish you the best, Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-25T23:29:42Z SecondChildTAG: It doesn't have to be perfect English, though. If you can make yourself understood, pronunciation and grammar doesn't matter. SecondChildUserIdTAG: 90796 SecondChildUserNameTAG: ChaunceyGardiner SecondChildCreateTimeTAG: 2012-11-26T00:02:23Z SecondChildTAG: Yep, I am totally agree with ChaunceyGardiner ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-26T00:08:38Z SecondChildTAG: Is Klingon accepted? SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-12-03T23:38:22Z SecondChildTAG: Hahaha @salsero! While subtitled in english, you can use it XD. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-04T11:13:36Z FirstChildTAG: Thanks for the announcement Myriam! Staff, is it possible to put this on the Course info page? It'll be really helpful if this is done. This thread will disappear in a few days. FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-11-26T15:30:43Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-26T17:08:44Z FirstChildTAG: As participant of CECC1 I want to give some tips to people who want to participate. 1)A lot of people will be afraid their submission will not be to MIT standards. Don't be a simple experiment can be awesome (blow up a resistor or open up a power transistor 2N3055: just be carefull some components can be harmfull if breathed in), and once you start making the video you will find quickly you will be short in time (youtube video uploads are max 15 minutes). 2)You can use a videocamera but a webcam works also fine. If you want to make a screencast microsoft expression has a good free version to grab the screenformat in high quality. http://www.microsoft.com/expression/products/Encoder4_Overview.aspx 3)Don't worry about compression of the video's youtube will convert to a high compressed mp4 once you have uploaded it. 4) if you make the video in english and you have the spoken text written out you can automatically synchronize your text in youtube after uploading your transcript. 5) most people don't have hightech equipment to experiment. I used a digilent's module which comes at the price of a big book. A worldwide distributor http://be.farnell.com/digilent/410-244/analog-discovery-design-kit/dp/2143587 , just change your country selection to see the offer in your continent. US students best order at Digilent as they give discounts. Just make sure you have a small breadboard to connect it (ebay). ![breadboard][1] Also don't forget to order some components; you can make already awesome simulations with a few resistors and capacitors. 6) ofcourse you can use EDX lab or another free simulator like LTspice to make your experiments but seeing the physical thing real makes it a lot more interesting. 7)Don't wait for the last 14 days to start thinking about it, if you are prepared you'll make the video in a few hours. Test your screengrab software , adjustment of the soundlevel etc. before you start the live recording so you don't have to search for that during the recording. 8) a good tip try to smile while you are recording, your voice automatically changes to a more agreable tone. I am a grumpy old bagger so that didn't helped me much. 9) make sure you have applied for a patent before uploading your inventions to CECC2 :<) 10)it's a great opportunity and a kind gesture from EDx so JOIN IN [1]: https://edxuploads.s3.amazonaws.com/1354010579134361.jpg FirstChildUserIdTAG: 79885 FirstChildUserNameTAG: ruudoleo FirstChildCreateTimeTAG: 2012-11-27T10:35:42Z SecondChildTAG: This is great advice. Go for it! SecondChildUserIdTAG: 90796 SecondChildUserNameTAG: ChaunceyGardiner SecondChildCreateTimeTAG: 2012-11-27T11:02:04Z SecondChildTAG: I love these tips! I'm adding these to the wiki. :) SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-11-28T15:49:09Z SecondChildTAG: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/tips-ruudoleo/ SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-11-28T15:57:22Z FirstChildTAG: Myrimit or any organizer, I've just found this contest and I need to say I find it terrific!!! I've read the rules pdf file but I don't SEE any specific rule! Can anyone tell me which ones are the rules? Looks like this: - a video of no more than 15 mins - a video that is related with the subjects of 6.002x - it should be in English of with English subtitles Is there anything else I need to know? BTW: I'd love to win a book since I find it a good material and even more if it is signed personally by Prof. Agarwal! FirstChildUserIdTAG: 99441 FirstChildUserNameTAG: coyarce FirstChildCreateTimeTAG: 2012-11-28T20:28:47Z SecondChildTAG: Just pick a topic from 6002x that interests you and do your best to make it interesting for your fellow students. That really is all there is to it. (We do encourage the use of actual physical electronic components.) SecondChildUserIdTAG: 90796 SecondChildUserNameTAG: ChaunceyGardiner SecondChildCreateTimeTAG: 2012-11-28T23:26:43Z SecondChildTAG: Hi coyarce, As ChaunceyGardiner said just pick a topic from 6002x that interests you :) -We do encourage the use of actual physical electronic components- . You are free to use your creativity :)- the video can be of 15 min or more or less, we didn´t write tight rules, so you can manage the time as you wish. We prefer the video in english, as Chauncey said previously: It doesn't have to be perfect English, though. If you can make yourself understood, pronunciation and grammar doesn't matter. But, if you want to speak in your native language you must yes or yes subtitle the video. That is all, Have fun! :) Myriam. P.D: Oh yes! A signed Textbook from Prof. Agarwal is awesome! He is so cool, coolest ever Professor haha! :). Thank you Prof. for donating the Textbooks to this Contest :). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-29T12:48:16Z FirstChildTAG: That's sounds great. I will try to arrange some time for it. FirstChildUserIdTAG: 226085 FirstChildUserNameTAG: Teto FirstChildCreateTimeTAG: 2012-12-02T04:50:15Z SecondChildTAG: Excellent Teto! ;) That would be awesome! Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-03T23:16:21Z SecondChildTAG: I tried but my mechanical courses restrict me. SecondChildUserIdTAG: 226085 SecondChildUserNameTAG: Teto SecondChildCreateTimeTAG: 2012-12-17T12:18:20Z SecondChildTAG: I am sorry miryam SecondChildUserIdTAG: 226085 SecondChildUserNameTAG: Teto SecondChildCreateTimeTAG: 2012-12-17T12:19:14Z SecondChildTAG: This is my face now ---> T_T Haha, nah joke. You don´t need to apologies Teto, because I totally understand you , it happened to me with the Exam2 of 3.091x, I was planning to do it on sunday but I couldn´t due other obligations like my Exams at University and finish a Project :P. But remember that you still have time to submit your video for the Contest, the important is that you can participate and have a lot of fun, no mattering the award - of course that all wants the signed Textbook haha- but the point is to participate, that would be awesome Teto ;). See you! Myriam. P.D. I owe you the video explanation of log :P. I will try to do it now that I am more free ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-18T03:06:31Z SecondChildTAG: at least I saw some of my name letter in your face :) this is realy honer for me. ;) I really wish if I can make a video to win the textbook with the signature on it ( it is all about the sign from the great professor mr. Anant Agarwal) I'll try my best maybe I can do it. and about the log video if you made it please but the link here so I can find it easily see you :) SecondChildUserIdTAG: 226085 SecondChildUserNameTAG: Teto SecondChildCreateTimeTAG: 2012-12-25T03:53:21Z FirstChildTAG: Is it possible to watch the winning videos of CECC 1? You know it provides some kind of great idea about application of logic. FirstChildUserIdTAG: 370089 FirstChildUserNameTAG: Quarck FirstChildCreateTimeTAG: 2012-12-04T13:17:40Z SecondChildTAG: There are links in the [Wiki][1]. Scroll to the bottom of the page. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/ SecondChildUserIdTAG: 90796 SecondChildUserNameTAG: ChaunceyGardiner SecondChildCreateTimeTAG: 2012-12-04T17:48:35Z FirstChildTAG: Is it okay to join from other country? and by group? FirstChildUserIdTAG: 340272 FirstChildUserNameTAG: ezekielbrizuela FirstChildCreateTimeTAG: 2012-12-08T13:35:35Z SecondChildTAG: Yes, this contest is open to all the students of 6002x this fall. It doesn't matter where you live. There is nothing in the rules that prevents students from forming teams to make videos. You'll have to figure out who gets the book if you win, though. SecondChildUserIdTAG: 90796 SecondChildUserNameTAG: ChaunceyGardiner SecondChildCreateTimeTAG: 2012-12-08T22:07:17Z SecondChildTAG: What ChaunceyGardiner said it is true, you can join no matter where you live :) . Also, you can submit the video by Team if you want - we haven´t wrote tight rules, we encourage the creativity in your videos, so you are free to make the video as you wish-, as Chauncey also said, the problem will be who will get the book if your Team win :p Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-08T23:17:44Z FirstChildTAG: to ashwith rego(CCEC) my name is aditya jayant mhatre this is the name which i use on this edx forum my email address-adjmhatre@gmail.com FirstChildUserIdTAG: 533341 FirstChildUserNameTAG: adjmhatre FirstChildCreateTimeTAG: 2013-01-08T11:49:11Z IndexTAG: 36 TitleTAG: Hints Lab4 requested by maraivette **LAB 4 Hints** Hi! I hope this Post can be helpful for those that have difficulties in this Lab. (read down for Spanish translation). Ok, the goal of this lab, is that you can plot this Graph: ![imagen][1] But, What does it mean all those colorful curves? The answer is that they represent the variation of the current iDS (Drain-Source)of the MOSFET in function of voltage values of vDS (Drain - Source). But, why there are so many curves? What is the difference between the yellow, magenta, black, cyan, etc.. color wave? Your Aha moment! : Because there are the representation of the iDS vs vDS but with different constant value of vGS (Gate Source)! So, the magenta will be the same circuit of the yellow one but with a different constant value of vGS. Now, I am sure that you are asking yourself: How is it the circuit that I need? The fact is that I can not tell you, but, let´s see the following Hints so that you can do it by yourself :). **Circuit Hints:** Hint 1: You can take a look to this page of the Textbook [page 336][2]and pay attention to the Figure 7.7 - Setup for observing MOSFET Characteristics. Hint 2: Remember that they are asking to plot different curves. We have seen that they were the same circuit but with a difference... So, if we have two voltages sources, which will be the constant voltage source vDS or vGS? and which will be the variable voltage source vDS or vGS? Aha moment again! ;) Hint 3: If we need n different curves, How many circuits do we need before clicking on TRAN? Hint 4: see the Figure 1. that they are giving you in the statement. They are using a triangle voltage source, triangle(0,3,1000). What this will mean? this will mean: - 0 : the wave will start at 0 Volt. - 3: the wave will have its maximun value at 3 Volt. - 1000: means the frecuency in Hertz of the wave. So, it will have a period of T=1/f = 1ms , that is to say, it will repeat it shape each 1ms. ![one][3] Hint 4: If they are asking you plotting the current iDS, where will you locate the current probe? ;)- you can see again the figure 7.7. Hint 5: If they are asking you plotting the voltage vDS with different constant values of vGS, where will you locate the voltage probe? Hmmm, isn't it the same place you have located iDS? Ok, I shouldn't talk that much. I think that now it is your turn! with all this Hints you can design your circuit :). ---- Once you have your Circuit, They say: **Please answer the following questions using measurements taken from the v-i plots you created.** PART 1: **Compute the effective RON in the triode region of the v-i characteritic when vGS=3V, i.e., measure the current through the mosfet switch when, say, vDS=1V and report vDS/iDS as your answer below. Remember this should be for a device whose W/L is 1.** Hint PART 1: What is the color of the curve that has the vGS=3V? Can you graphically read by one value of voltage vDS and it correspondant value of iDS? Can you find that ratio? Can you find so RON? Yes!!! Where can I know if my vDS it is in the Triode Region? Read page 337 of the Textbook [here][4] (I will post the image, but please read that page of the Textbook). ![snap][5] Also, you can see more info about current probe [here][6] (in the wiki I wrote about how to meassure a current in DC), is not the case that we are needing but might help you if you have doubts of how we can move this element, etc.. PART 2: ![enun][7] Hint Part 2: What does it mean vGS1 or vGS2? Hmmm..., Didn't you have different curves with different values of vGS? Which color of curve has the vGS1? which color of curve has the vGS2? If know my two curves... can you find vDS1 and iDS1 of the vGS1 curve? Yes! can you find vDS2 and iDS2 of the vGS2 curve? yes! So, can you find the value of VT by replacing in the given formula? Yes!!! Another Hint Part 2: Here a mathematical property that might help you :). ![propiedad][8] ---- Part 3: **Finally, using your measurements and calculated value for VT, compute the value for the constant K in equation 7.8 (see above) and report the results.** Hi maraivette! Sorry for the delay in this [here][9] you asked me this: "Good afternoon Myrimit, I almost have finished lab4 but I have an issue with the last question... what values of ids and vGS we have to take in order to find the K?..." Hint: If they give us the formula iDS in function of vGS and VT in the saturation region, and if we already have the value VT of the part 2, can we **choose** a curve of **any vGS** and find graphically it correspondant iDS? yes! and if you have iDS, VT and vGS, can you find your K? yes! So, maraivette, the answer will be that you can choose any value of iDS while it be **in the saturated region** [more info in the Textbook here][10], it is the plain region ;). Greetings! Myriam. ---- Now in Spanish! **LAB 4 Hints** Hola! Espero que este Post les pueda ser de ayuda para aquellos que se encuentran con dificultades en la resolución de este Lab. Muy bien, el objetivo de este Lab, es que ustedes puedan lograr obtener de alguna forma las siguientes características dadas por el enunciado: ![imagen][1] Bien, pero qué significan todas esas curvas coloridas? La respuesta es que ellas representan la variación de la corriente drenaje fuente iDS del MOSFET en función de la tensión drenaje fuente vDS. Pero, un momento, por qué hay tantas curvas? cuál es la diferencia entre la curva amarilla, negra, cyan, magenta, etc..? Presta atención, porque aquí debería venir tu Aha moment! : Porque si bien representan la relación corriente (iDS) tensión (vDS), la diferencia radica en que cada una de ellas corresponde a un valor constante de compuerta fuente vGS! Esto quiere decir, que por ejemplo, la curva de color magenta será el mismo circuito que el de la curva amarilla pero con la diferencia que poseerá un valor constante distinto de vGS. Ahora, estoy más que segura que te estás preguntando: Cómo es el circuito que necesito para resolver este Lab? Bueno, no quiero ser aguafiestas, pero esta pregunta no se las puedo decir porque lo deben hacer ustedes, pero, veamos algunas Hints que les podrán ser de ayuda para que luego ustedes mismos lo puedan resolver :). **Hints del Circuito a diseñar:** Hint 1: Es muy importante que le den una mirada a la página 336 del Textbook[leer aquí][2] y que presten pero extrema atención a la Figura 7.7 de dicha página - Seteo para observar las características del MOSFET. Hint 2: Recuerden que ellos te estan pidiendo que realices el gráfico de las distintas curvas con el simulador. Hemos visto, que eran los mismos circuitos pero con una esencial diferencia... Entonces, si tenemos dos fuentes de tensión, cuál será la fuente que deberá ser la de valor constante, la que esta entre el drenaje-fuente o la que está entre la compuerta-fuente? y cuál será la que deberá ser de valor variable, la que esta entre el drenaje-fuente o la que está entre la compuerta-fuente? Otra vez creo que este debe haber sido también tu Aha moment! ;) Hint 3: Si se necesitan n diferentes curvas, cuantos circuitos necesitamos antes de hacer clic en TRAN? Hint 4: Si vemos la Figura 1 (Figure 1)que ellos nos brindan en el enunciado. Se ve que se utiliza una fuente de tensión triangular, triangulo(0,3,1000). Qué significará esto? Esto significará lo siguiente: - 0 : la onda comenzará en 0 Volt. - 3: la onda tendrá su máximo en 3 Volt. - 1000: significa la frecuencia de la onda. Entonces, tendrá un periodo de T=1/f = 1ms , es decir, que su forma se repetirá cada 1ms. ![one][3] Hint 4: Si ellos te solicitan que graficar la corriente iDS, en donde localizarás la punta de prueba de corriente? ;) - mira nuevamente la figura 7.7. Hint 5: Si ellos te solicitan que grafiques la vDS pero con diferentes valores constantes de vGS, en dónde deberás localizar la punta de prueba de tensión? Hmmm, acaso no es el mismo lugar que iDS? Creo que ahora es tu turno! Con todas estas Hints creo que ya estás listo/lista para el diseño de tu circuito :). ---- Una vez que hayas hallado tu circuito, Te dicen en el enunciado: ** Por Favor responda las siguientes preguntas utilizando las mediciones de los plots v-i que has creado** PARTE 1: **Cacular la RON efectiva en la región del triodo de las curvas características de v-i cuando vGS=3V, es decir, medir la corriente a través del mostfet cuando cambia, o sea, cuando vDS=1V y anota como respuesta tu vDS/iDS. Recuerda que estoo debe ser para tu dispositivo para cuando W/L=1** Hint PARTE 1: Cuál es el color de la curva para la cuál vGS=3V? Gráficamente, es posible la lectura de una iDS que se corresponde con un valor dado de vDS? Puedes hallar el cociente? Entonces, puedes calcular RON? Sí!!! Cómo puedo saber si mi vDS se encuentra en la región del triodo? Leer página 337 del Textbook[aquí][4] ( aquí pondré la imagen por comodidad, pero por favor lee esta página del Textbook ya que es muy informativa y te puede ser de ayuda para la comprensión de este Lab). ![snap][5] También, puedes leer más información sobre la punta de prueba de corriente [aquí][6] ( en la wiki, escribí hace un tiempo, cómo poder medir la corriente en DC con una punta de prueba de corriente), no es el caso que necesitas aquí, pero puede servirte por si tienes duda de cómo mover el elemento, etc... PARTE 2: ![enun][7] (Aquí no voy a traducir lo de arriba) Hint Parte 2: Qué significa vGS1 o vGS2? Hmmm..., No era que tenías diferentes curvas que se correspondían con diferentes valores de vGS? Qué color de curva corresponde a la vGS1 solicitada? Qué color de curva corresponde a la vGS2 solicitada? Si sé cuales son las ds curvas... Puedes hallar la vDS1 y la iDS1 de la correspondiente curva correspondiente a vGS1? Sí! Entonces, puedes hallar el valor de VT reemplazando en la fórmula dada por el enunciado? Sí!!! Otra Hint de la Parte 2: Aquí te daré una propiedad matemática que quizás pueda serte de ayuda :). ![propiedad][8] ---- Parte 3: **Finalmente, utilizando tus mediciones y el valor calculado para VT, calcula el valor de la constante K en la ecuación 7.8 (mira arriba)y escribe el resultado** Hola maraivette! Disculpa por la tardanza en mi respuesta a tu pregunta que me has hecho[aquí][9] : "buenas tardes Myrimit ya casi termino el lab 4 pero tengo un problema conla ultima pregunta que valores de ids y cgs hay que tomar para el calculo de k los ores o la diferencia de ellos ? me puedes orientar al respecto ..." Hint: Si nos dan la fórmula de iDS en función de vGS y VT en la zona de saturación, y si ya tenemos el valor de VT que hemos hallado en la parte 2, podemos con cualquier curva, es decir, cualquier curva con cualquier vGS, hallar gráficamente nuestra iDS? Sí! y, si tienes iDS, VT y vGS, puedes hallar tu K? Sí! Entonces, maraivette, la respuesta a tu inquietud, es que **puedes elegir cualquier valor de iDS mientras se encuentre en la zona de saturación** [lee aquí sobre la zona de saturación][10], verás que es la zona plana de la gráfica ;). Saludos!! Myriam. [1]: https://www.edx.org/static/content-mit-6002x/images/circuits/Lab4_1.ee850d29dfef.png [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/360 [3]: https://edxuploads.s3.amazonaws.com/13495680481343692.png [4]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/361 [5]: https://edxuploads.s3.amazonaws.com/1349568996732725.png [6]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/how-measure-current-sandbox/ [7]: https://edxuploads.s3.amazonaws.com/1349569137134368.png [8]: https://edxuploads.s3.amazonaws.com/13495701465917771.png [9]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_graph_interp_inc_method/threads/506b7c614f5ec3270000007d [10]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/361 UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2012-10-07T01:58:15Z VoteTAG: 25 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 16 FirstChildTAG: Thanks **Myrimit**...ur doing a great job..keep it up.. from **Aninda,India** FirstChildUserIdTAG: 102370 FirstChildUserNameTAG: aninda FirstChildCreateTimeTAG: 2012-10-07T06:17:08Z SecondChildTAG: Thank you aninda! :). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T06:18:57Z FirstChildTAG: Myrimit i have a problem regarding hint no2&4. plss explain these two hints FirstChildUserIdTAG: 395966 FirstChildUserNameTAG: sambo007 FirstChildCreateTimeTAG: 2012-10-07T06:34:34Z SecondChildTAG: Yes sure, sambo007. What are your difficulties? In which part are you lost? Ok, no problem ;). in Hint 2: if you alredy take a look to Figure 7.7 of the Textbook, you will notice two voltage sources. You know that you have n curves by observing the graph. Also, that each curve corresponds to a circuit. That all the circuits are similar, but with one different constant voltage source value (that is why you see n curves). Ok, here your goal it is discover which of those 2 voltages sources will be the one that is constant and have some of the value that they ask you in the statement {0 V; 0.5V; 1V; 1.5V, etc...}. So, where will you put that voltage souce? in the gate or in the source of the MOSFET? Try to re-read the Lab statement again: **The behavior of the mosfet is determined by two voltages: vDS, the potential difference between the drain and source terminals and vGS, the potential difference between the gate and source terminals. As the test device use a mosfet with a W/L parameter of 1. Use a DC voltage source to supply vGS: you'll want to try a range of values -- 0V, 0.5V, 1.0V, 1.5V, 2.0V, 2.5V and 3V -- varying from ground up to the power supply voltage.** Ok, now lets go to hint 4: Hint 4: If they are asking you plotting the current iDS, where will you locate the current probe? ;)- you can see again the figure 7.7. lets see a MOSFET, you will have 3 terminals: Gate, Source and Drain. Take a look at figure 7.7. The Gate it is in the midle of the MOSFET; the Drain it is up and the Source it is down. So, where does the Ids current goes trough? isn't it from Drain to Source? So, where do you have to put your current Probe? up or midle? ;). See you! Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T08:02:34Z SecondChildTAG: I think I said already everthing.... it's amazing the support you are providing for students. I didn't study electronics before, so I just was a bit "autodidacta" before the course and now and with your help I'm more or less doing it. I admire scientists and people interested in science with such a good will like yours!!!! regards, Sandra SecondChildUserIdTAG: 342966 SecondChildUserNameTAG: SandraNavarro SecondChildCreateTimeTAG: 2012-10-07T23:30:28Z SecondChildTAG: Thank you Sandra! My best wish to you! I have to say that I admire you, because you didn't study electronics before and you are doing great ;)! See you! Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T23:48:21Z FirstChildTAG: Myrimit, you're awesome. Although I've done that lab by myself, but I'm highly amazed by your work. I can only guess how much time you spend to write that useful posts for every task... FirstChildUserIdTAG: 326787 FirstChildUserNameTAG: Angstrem FirstChildCreateTimeTAG: 2012-10-07T08:34:30Z SecondChildTAG: Thank you Angstrem! :). How are you? I am really enjoying doing this, I really like it. I hope to be helpful for the New Students:). This was one of my objectives when I registered again to 6.002x (I did the Prototype course in the spring) and I really wanted to help here haha (trying to help in Spanish and English, both). My original objective when I first registered to 6.002x Prototype Course was because when I was a Child I got really amazed with the MIT research in a Documental (I think that was in Discovery Channel) and I always wanted to meet the scientist of MIT and the Laboratories haha, so this is a kind of placebo haha, so helping here, makes me feel, in someway, part of this huge iniciative that shares MIT with Harvard and Berkeley. So, I am really happy to help you. So, If you have any difficulty here I will be :) See you! Now I am going to sleep haha! Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T09:34:14Z SecondChildTAG: I'm fine, thanks! And you? It's the first time I'm taking this course. The reason I'm here is because I love physics. Although I prefer studying the world rather than engineering, I think, this is a great experience to take part in this course. Is your field of occupation connected to electronics somehow? Anyway, I think, it's very nobly to spend your time and efforts in order to help others. Maybe, I'll follow your example in the Cryptography course, the next iteration of which starts soon on coursera. I've being inspired by your activity here :) SecondChildUserIdTAG: 326787 SecondChildUserNameTAG: Angstrem SecondChildCreateTimeTAG: 2012-10-07T09:56:00Z SecondChildTAG: Yes, I am luckily fine haha. Cool Angstrem! That will be nice! You should share your knowdlege ;). Nice that you were motivated! Yes I am a Student of engineering in Argentina ;). I also have a fellowship in research (Solid State Laser) and helping free since two years ago in an investigation group of Education in my University (I really like that, we are analyzing the drop out of the students in order to find the difficulties and causes). Cryptography on coursera sounds interesting! See you! Myriam. P.D: now I am going to sleep, haha. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T10:21:12Z FirstChildTAG: Dear Myrimit, Can you please guide that what values do i have to use for Vgs and Vds ,i also mean the type of signal, Dc ,triangular etc. Also the Simulation time for Transient response. Thanks FirstChildUserIdTAG: 232499 FirstChildUserNameTAG: Asim09 FirstChildCreateTimeTAG: 2012-10-07T08:43:10Z SecondChildTAG: Hi Asim09, Remember that one of those two, have to be constant and , as the statement says, you have to try with a range of values given in the {.....}, take a look at the statement values ;). The other one, it should vary , hint: you can use one as the one that they say in the statement (take a look at Figure 1 - statement). TRAN, you can use the TRAN time given in the statement: *The Vtest voltage source generates a triangle waveform that ramps from 0V to 3V then back again over a period of 1ms. The stop time of the transient analysis is chosen to be **0.5ms** so that the voltage on node vtest will be just the rising ramp from 0V to 3V.* See you! Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T09:46:54Z FirstChildTAG: I am not getting the exact graph in lab 4...please help me![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13496005806047951.jpg FirstChildUserIdTAG: 410204 FirstChildUserNameTAG: hkaushalya FirstChildCreateTimeTAG: 2012-10-07T09:03:36Z SecondChildTAG: Hi hkaushalya, Hmm... Take a look in your Plot.. there are 3 Axis! two of voltages and one of current. Are you sure that you need the cyan voltage probe? you already have a voltage probe (transparent, x-axis)... Remember that each curve has a different constant value. So, are you changing the values of the voltages sources for each circuit? See you! :) Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T09:40:46Z FirstChildTAG: Really need help!! Myrimit i really need help please responds as the dead line is closing in ,in my time zone. The problem is that when i set up the ckt as shown in the text book page 336. I get two parallel lines and the current and voltage both are on the Y axis and time being the X axis. I am running a simulation of 1 sec. with all voltages as DC. Vgs = 3V and Vds = 2V. need help really!! image is here: http://www.image-share.com/ijpg-1780-221.html FirstChildUserIdTAG: 232499 FirstChildUserNameTAG: Asim09 FirstChildCreateTimeTAG: 2012-10-07T09:49:19Z SecondChildTAG: Hi Asim09! - Try with TRAN = 0.5m . - You only need a plot where you can meassure current vs voltage, that is to say, that you need a **one current probe** and a **one voltage probe**. You should delete one of the 2 voltages probes (remember to set x-axis, so that you can see voltage in the x axis). - The thing here, is that you will need one voltage source constant (but remember that each circuit has to have a different constant value, for example, 8, 10, etc..., in the statement they give you a {..,...,...,} values to use). - Are you sure that vDS = 2V? ;). Hint: remember that one should vary like the example of the source given in the Figure 1 of the statement. I hope this can be helpful. I wish to help you more but I don't want to cross the line... See you! :) Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T10:03:27Z SecondChildTAG: more help: are you sure that the voltage probe goes in the gate? Try to remember to double click on the voltage probe and set it to x-axis as the statement says. This is in order that the plot will be in function of voltage and not time... Try to think again about the voltage of drain source... Is it right? Do you need there an DC voltage source? Are you sure? I think I can not say more... SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T10:11:11Z SecondChildTAG: Thanks a lot for the reply bu the problem i am experiencing is that I cant set the voltage or current probe axis. That is perhaps i am not able to see the correct wave form. I believe its a bug of the sandbox.Please see below and advise http://www.image-share.com/ijpg-1780-227.html Thanks SecondChildUserIdTAG: 232499 SecondChildUserNameTAG: Asim09 SecondChildCreateTimeTAG: 2012-10-07T10:30:02Z SecondChildTAG: stupid me for being so blind!!! sorry there is x axis in the plot color!! hurry makes a bad curry!! lolz SecondChildUserIdTAG: 232499 SecondChildUserNameTAG: Asim09 SecondChildCreateTimeTAG: 2012-10-07T10:49:07Z SecondChildTAG: Hi Asim09! Could you solve your difficulties? ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T17:37:28Z FirstChildTAG: @Myrimit you are doing a great job....... thanks a lot FirstChildUserIdTAG: 114864 FirstChildUserNameTAG: Vikaash FirstChildCreateTimeTAG: 2012-10-07T14:09:21Z SecondChildTAG: Thank you Vikaash! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T17:37:42Z FirstChildTAG: i have completed the circuit and got the correct mark but my graph doesn't look like the given graph. so i'm unable to find iDS. since for triode region, vDS < vGS - VT, and given vGS=3 volt, vDS= 1 volt i have vGS, vDS value. from here i found VT<2 now how to proceed?? plz help thanks. FirstChildUserIdTAG: 204213 FirstChildUserNameTAG: ratneshray FirstChildCreateTimeTAG: 2012-10-07T14:36:24Z SecondChildTAG: Hi ratneshray, Can I help you? - How does your graph look like? Are you seeing the different curves with the respective axis? SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T17:42:00Z FirstChildTAG: Hi Myrimit.. I need a study partner . Could you please be my partner? FirstChildUserIdTAG: 329525 FirstChildUserNameTAG: udhayaraj12 FirstChildCreateTimeTAG: 2012-10-07T17:14:01Z SecondChildTAG: Hi udhayaraj12! Yes sure, here I will be here if you need to review, and if you have doubts, :). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T17:48:32Z FirstChildTAG: Hello guys, I still have problem with plot axis, I can't put the current on y and voltage on x??? Could somebody help me ?Thanks FirstChildUserIdTAG: 443442 FirstChildUserNameTAG: Nelutu FirstChildCreateTimeTAG: 2012-10-07T17:32:55Z SecondChildTAG: Hi Nelutu, Can I help you? You can plot the voltage in the x-axis by double click on the element (voltage source) and by choosing the x-axis option :). What did you mean whit that you couldn't put the current on? If you see in the right you will see a dash with a magenta arrow (current probe), just dragg it on the place you want to meassure your current. See you! Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T17:46:31Z FirstChildTAG: ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13496336351043384.png Hi Myrimit .. I am not abt to get correct answer for second part of Lab 2. Here is my ckt please help ........ dead line is about to finish. FirstChildUserIdTAG: 413002 FirstChildUserNameTAG: Gauravjain88 FirstChildCreateTimeTAG: 2012-10-07T18:15:37Z SecondChildTAG: Try to be careful with the voltages sources. Are you sure that the one that is is in the gate should be variable ? Remember that you need one voltage probe and one current probe... So that, you can see in the plot current vs voltage... SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T18:44:38Z FirstChildTAG: Wops , it was so simple, now I get it . I know how to put the current probe , but I didn't know how to put the voltage on x-axis, because I had the time:(. Thanks a lot. FirstChildUserIdTAG: 443442 FirstChildUserNameTAG: Nelutu FirstChildCreateTimeTAG: 2012-10-07T18:25:45Z SecondChildTAG: You are welcome! Cool Nelutu! Nice to see that you got it :)! Yes, sometimes missing a little detail it is like the butterfly effect haha! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T18:46:54Z FirstChildTAG: Hola Miryam. Sigue así de juiciosa y colaboradora. Que chevere que haya gente tan dedicada como tu. Eres un excelente complemente para este curso. Chau FirstChildUserIdTAG: 184827 FirstChildUserNameTAG: DiegoT FirstChildCreateTimeTAG: 2012-10-07T18:33:45Z SecondChildTAG: Muchas Gracias DiegoT! Me alegra mucho saber que estoy siendo de ayuda en este curso. Cualquier duda sabes que estaré aquí tratando de ayudar a la comunidad hispanohablante :). Saludos!! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T18:48:23Z FirstChildTAG: I'm getting what looks like a "good" graph - curves look like the desired output - but they're all magenta, despite my setting different probe colors! (And it looks like I have two voltage axes [left and bottom] and one current [right]. Blah!) Not sure what I've done. ;-/ Any advice would be appreciated! FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-10-07T19:38:35Z SecondChildTAG: Hi planetscape! Can I help you? Hmm... Are you sure that you are changing the colors of the current probes (the one with a dash and a magenta color)? Might you are changing the voltage probe and not the current probe ;). If you are seeing two voltages axes it is might because you are using an extra voltage source probe... remember that the voltage source probe has to be set with the x-axis option and in the correct place of the MOSFET... SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T19:59:01Z SecondChildTAG: I'm not sure what I've done. I seem to be getting good numbers from the graph, in that answers 1 and 2 of Lab 4 are now correct, though I can't seem to get 3. And I do have a 75% score ATM, which is 100% more than it was an hour ago. ;-) I do not see a magenta probe with a dash. I see no dashes, actually... SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-07T20:11:14Z SecondChildTAG: Hi planetscape: The current probe it is like this: ![it is a dash with a magenta arrow][1] You can find it here: ![other][2] ---- Hmm... Remember that in the last part of the lab, the part3, where you have to find VT you have to be careful, because the 1st formula that they give you, it is for the saturated region, so, you have to choose your values for a certain part of the curve, it will not be valid for the others parts ;). [1]: https://edxuploads.s3.amazonaws.com/13481128761343615.png [2]: https://edxuploads.s3.amazonaws.com/1348113222234286.png SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T20:17:33Z SecondChildTAG: Ah, that explains the probe confusion. I have numerous little ones that look like a cyan hypodermic. Maybe I should have paid more attention to the lab instruction back in the overview? ;-) Thanks, I shall have a bite of food and some coffee and look into your hints in more detail! Thanks! You're a lifesaver! SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-07T20:36:28Z SecondChildTAG: Ah yes, now lines are multi-colored! Thanks, Myrimit! Those little suckers are hard to manipulate! Now to the math. ;-) SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-07T20:59:23Z SecondChildTAG: haha! well done! ;). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T21:04:47Z FirstChildTAG: I'm having problems to figure out Vt... I tried to use all your hint for this part, but I can't get the right answer for Vt, hence I can't solve K neither, because I need to use Vt. Also I tried first Vgs1=3 and Vgs2=2.5 and its corresponding currents, and then I tried Vgs1=2.5 and Vgs2=3, and they give me similar values for Vt but any of them is right. Could you help me? FirstChildUserIdTAG: 45307 FirstChildUserNameTAG: josejimenez2 FirstChildCreateTimeTAG: 2012-10-07T20:51:42Z SecondChildTAG: Do you have a check for question 1? Under Progress, do you have at least 50%? That would be a way to verify your circuit is correct. Garbage in, garbage out... SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-07T21:00:51Z SecondChildTAG: I checked my circuit and it's right, then I calculated Ron, and it's right too, but when I calculate Vt and check it, it's wrong. I don't know what I'm doing wrong SecondChildUserIdTAG: 45307 SecondChildUserNameTAG: josejimenez2 SecondChildCreateTimeTAG: 2012-10-07T21:08:41Z SecondChildTAG: Hi josejimenez2! Can I help you? They give you two values of vGS (vGS1 and vGS2), you will identify in the graph (if you have changing the colors to each circuit - current probe) the correspondant color for the vGS1 and vGS2 ;). Ok, now your goal is to try to find the correspondant iD for the vDS given... for the two conditions of vGS ... Then as you have the formula that they give you , you could find VT :) [1]: https://edxuploads.s3.amazonaws.com/1349568996732725.png SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T21:16:13Z SecondChildTAG: I got it Myrimit =) Actually my problem was that I wasn't working on the **saturation region**, I was using values for Ids1 and Ids2 from the triode region of the MOSFET. When I read the book, and then your hints again I realized it. Again, muchas gracias por toda tu ayuda, tus hints son realmente muy útiles y me han ayudado muchísimo a progresar y entender muchas cosas de 6.002x. *You're amazing* SecondChildUserIdTAG: 45307 SecondChildUserNameTAG: josejimenez2 SecondChildCreateTimeTAG: 2012-10-07T21:24:58Z SecondChildTAG: Super cool josejimenez2! Well done! I am really happy for you! ;). You are welcome. P.D: josejimenez2, cuenta conmigo cuando tengas dificultades, desde aquí haré mi mejor esfuerzo para ayudarte! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T21:29:50Z SecondChildTAG: Thanks a lot :) Muchas gracias... You're such a great partner! SecondChildUserIdTAG: 45307 SecondChildUserNameTAG: josejimenez2 SecondChildCreateTimeTAG: 2012-10-07T21:43:35Z FirstChildTAG: Part 3 is still bothering me - checker says VT is correct, I'm sure I've taken IDS values from the saturated region (I've tried two samples, both give similar values for K) - but still can't get a green check by #3... FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-10-07T21:54:09Z SecondChildTAG: Hi planetscape! Can I help you? Ok, if you have ok, your VT value. Now, try to use the other formula that they give you , the one that has the K. Ok, You should choose a curve and try to obtain the data of that curve in the plain region ;) (saturated). Be careful with units, are you sure that you are writting correctly? hint: it is a really small value be careful to not missing a zero in your answer... SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T22:08:01Z SecondChildTAG: Units bit me for the second time (::blush::), *and* dropped a zero on the floor somewhere. So embarrassed. Thank you for your tireless help! I'm sure I'm not the only one here who would love to buy you a drink! SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-07T23:02:58Z SecondChildTAG: You are welcome! ;). haha! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T23:07:20Z IndexTAG: 37 TitleTAG: More delay in certificates Hi All, We were hoping to be able to coordinate signing the certificate before the new year, but due to holidays and complications with travel, we have to postpone the certificate distribution for another couple of days. We appreciate your patience. UserIdTAG: 96452 UserNameTAG: Lyla CreateTimeTAG: 2012-12-31T15:56:06Z VoteTAG: 24 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 8 FirstChildTAG: Thanks Lyla! ** ** Happy New Year!!! ================= ** ----------------- ** FirstChildUserIdTAG: 402193 FirstChildUserNameTAG: Feramico FirstChildCreateTimeTAG: 2012-12-31T18:50:32Z FirstChildTAG: Thanks Lyla. & Happy New Year!!! FirstChildUserIdTAG: 146930 FirstChildUserNameTAG: fayzan_007 FirstChildCreateTimeTAG: 2012-12-31T16:37:28Z FirstChildTAG: Thank you Lyla! Why don't you update the *course info* page so everybody will know as soon as they enter the site. **Happy New Year** to you and all other behind the screen people. BTW you guys should find some time to put together a video about your experiences regarding the fall 2012 course. FirstChildUserIdTAG: 623210 FirstChildUserNameTAG: preveen FirstChildCreateTimeTAG: 2012-12-31T16:02:23Z SecondChildTAG: Okay. SecondChildUserIdTAG: 96452 SecondChildUserNameTAG: Lyla SecondChildCreateTimeTAG: 2012-12-31T16:30:33Z FirstChildTAG: Lyla, thanks for your post, I appreciate the courtesy of keeping us informed about it; happy new year! FirstChildUserIdTAG: 61723 FirstChildUserNameTAG: lvence FirstChildCreateTimeTAG: 2012-12-31T16:03:29Z FirstChildTAG: i hope you announce a specific day for the certificate the next time because i open my home page every hour to check . thanx FirstChildUserIdTAG: 464827 FirstChildUserNameTAG: princeemad FirstChildCreateTimeTAG: 2012-12-31T16:45:24Z SecondChildTAG: Ohhh seems like some people have lost patience :D SecondChildUserIdTAG: 365201 SecondChildUserNameTAG: sirajmuhammad SecondChildCreateTimeTAG: 2012-12-31T17:07:16Z FirstChildTAG: Hi all, Will we get a hard copy of the Certificate? I asked so, because Lyla mentioned about signing the certificates.. FirstChildUserIdTAG: 199839 FirstChildUserNameTAG: vijaykrishnankk FirstChildCreateTimeTAG: 2012-12-31T17:00:38Z FirstChildTAG: @Lyla Just let us know by email when the certificates are put up for downloading. It would be so helpful or we need to check every now and then. FirstChildUserIdTAG: 390093 FirstChildUserNameTAG: 5riram FirstChildCreateTimeTAG: 2012-12-31T17:42:28Z FirstChildTAG: ![enter image description here][1] [1]: http://www.sotojohnson.net/2013/1.jpg FirstChildUserIdTAG: 183712 FirstChildUserNameTAG: colegiocientifico FirstChildCreateTimeTAG: 2012-12-31T17:19:33Z IndexTAG: 38 TitleTAG: Thanks a lot COMMUNITY TA Especially to Myrimit and Skyhawk for their extraordinary hints and advices. You have done a great,great work.Congratulations. By the way...Will the Final Exam cover topics from weeks 1-12, or 13 and 14 must be also included? UserIdTAG: 329444 UserNameTAG: albmartin CreateTimeTAG: 2012-12-11T19:07:14Z VoteTAG: 24 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: Thank you ! FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-12-12T14:41:51Z FirstChildTAG: They are simply the best! FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-12-11T20:47:25Z FirstChildTAG: The final covers weeks 1-12 only. FirstChildUserIdTAG: 90796 FirstChildUserNameTAG: ChaunceyGardiner FirstChildCreateTimeTAG: 2012-12-12T03:15:21Z FirstChildTAG: Hi albmartin, You are welcome :). It is nice to know that I were helpful this Fall in 6.002x haha. I really enjoy helping here and I hope that many of you, can back next term and help the new students, no mattering of your grade, I got a happy grade B last term - so anything is impossible haha, or being or not CTA next spring. That would be awesome! Believe me that it is a nice experience! I like it! I am also in 3.091x, I like it too, but I am a disaster haha, I have to study more, I guess that I will do it better next spring ;). I wish to help more there. Prof. Cima is great and also all the people of there like eg. John Rogosic are excellent! So, you can take it next spring if you hadn´t take it. I will review the material in my holidays and help more next spring in 3.091x. Also I will make some experiments with cristaline structures by my own in my holidays, I am planning one haha. Also I would like to mix the knowdlege of EE of 6.002x with 3.091x, another project for my holidays too haha! Take care, Myriam. P.D. I am with a lot of exams and projects at my university and work, might I will be a little be offline these days, but I will be here haha. Good luck in the Final Exam!!!!and don´t forget to participate in the Contest!!! XD FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-12T12:24:51Z SecondChildTAG: Hay Myriam I'm thinking of volunteering next semester for this class, if they will have me, and I've noticed how excellent your text and graphics are. What tools do you use, and is there anything you can recommend that will make it easy to post material on this discussion forum? I'm also going to try and find out what Dr. Agarwal uses for his video's. Thanks. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-12-12T21:23:24Z SecondChildTAG: Hi rharris :) Cool! I am really happy for your iniciative! That would be great! I will share the secret with you haha XD: - In the videos I have used Windows Movie Maker. - The images of the graphics are in paint. - For formulas, in some Posts, I have used Mathjax. - Then, I have used the icons that you can see when you post a response in order to make the post more friendly: like inserting images, changing from **this** to *this* haha, but anything else. For the Wiki, I have to confess that I didn´t know how to use it haha. Then I started to try . If you take a look, when you click on edit, you can see the code, etc..., I learned by reading the edits haha. If you need help with that I will be pleasured to guide you and help you in order that you can learn how to use it. Prof. Agarwal uses camtasia - I couldn´t find the link of one post that was posted by Ashwith some days ago about this-. > I guess that is all haha, I lied to you, I have forgotten to tell you > the main secret, is your enthusiam :). Take care and good luck in the Final Exam! See you, Myriam. PD. I am planning to prepare materials in my summer holidays -here in my Country is spring- for tmarch, like eg. some help in math , etc., but I have to throw the ideas in a paper , I have them in some part of my few remaining neurons in my brain - I hope they don´t lost in synapses- haha. I haven´t started yet. But if you are interested, might we can work together after the course ends. What do you think? This was an idea that suddendly came up right now haha, I don´t know, might it would be interesting to make a list of difficulties of the students of this term and see what we can do for the next term students, might other students are interested too - I haven´t told them but if they are reading this you are invited haha- . If yes, I have created an email myrimit at gmail dot com , is not my personal email, but you can contact me there. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-13T01:13:35Z SecondChildTAG: Hi Miriam, We cant thank you enough for the awesome work you've been doing! We greatly admire your infinite enthusiasm, energy, and kindness to help others! Just on a lighter note :-), like the way we enjoy Aha! moments in the course, I think you enjoy many 'haha' moments when you write! haha! :) SecondChildUserIdTAG: 232157 SecondChildUserNameTAG: GayatriTR SecondChildCreateTimeTAG: 2012-12-13T03:57:39Z SecondChildTAG: MyriMIT ! You are the best ! along with the Skyhawk. Thank you so much. SecondChildUserIdTAG: 418182 SecondChildUserNameTAG: euldji2005 SecondChildCreateTimeTAG: 2012-12-18T01:02:45Z SecondChildTAG: Hi @GayatriTR and @ euldji2005, Thank you for your nice words :). I really like doing this and helping as far as I can. I am really happy to know this. I want that students can be inspired and do this, that they feel confident to do things, projects, to work in Team with your classmates, that you can make the CECC in the future 6.002x, that is why we left the open door to everyone who would like to do that, and feel that anything is impossible ;). Yep, haha is funny haha! XD I wish you to you that have a Happy New Year ;)! Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-18T02:51:23Z SecondChildTAG: @rharris I have received your email ;). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-18T02:51:55Z IndexTAG: 39 TitleTAG: [STAFF] Problems entering long algebraic fomulæ Please vote up if you have the same issue. Having problems with parser in H10 (and others). If I have anything more than a simple algebraic fomula, Chrome crashes with the "Aw-Snap!" message, and I have to enter the entire formula again. How I solve this problem is to 1. enter a symbol or two at a time, 2. check if it is correct (i know it is not), just so I don't get a full reset. 3. Repeat (1) until done. I have been successful so far, but it can take 100 clicks on the 'check' button especially if multiple formlæ must be entered. I worry that I will not be able to succeed at the final exam. Please a) Dont have long algebraic fomulæ on final exam and/or b) Have 'check' buttons after each algebraic formula (i have Win 7 64 and Chrome with all updates) UserIdTAG: 142857 UserNameTAG: viking2 CreateTimeTAG: 2012-11-08T00:12:43Z VoteTAG: 24 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: I have the same problem but instead of clicking the check button you can copy what you have entered on your clipboard. That is what i do. I thought that this happens because of my internet connection, but i guess i have to change my browser to mozzila or something else. FirstChildUserIdTAG: 304587 FirstChildUserNameTAG: Vasso FirstChildCreateTimeTAG: 2012-11-08T00:36:34Z SecondChildTAG: That is one way to address it. Have found that to be error prone as well since it is easy to miss when you can not see the real formula. Please vote post up to get some visibility. Thanks. SecondChildUserIdTAG: 142857 SecondChildUserNameTAG: viking2 SecondChildCreateTimeTAG: 2012-11-08T01:07:23Z SecondChildTAG: Definitely understand why you need to see how your formula is being interpreted when using [MathJax][1] to insure fractions, parens, all are being taken the way you intend. A suggestion might be to use another version of MathJax so you can fine-tune your formula before entering. It doesn't guarantee a correct answer, of course, but at least you can make sure it looks the way you want: Use something like [TiddlyWiki][2] with the [MathJax Plugin][3]. Both are very easy to set up and learn to use. Not only will it permit you to preview your formulas, but now you also have a really handy place to keep your course notes. ;-) [1]: http://www.mathjax.org/ [2]: http://tiddlywiki.com/ [3]: http://math-template.tiddlyspace.com/#%5B%5BPluginMathJax%20v1.3%5D%5D SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-11-08T03:17:27Z SecondChildTAG: same problem here(win7 chrome with all updates) SecondChildUserIdTAG: 124534 SecondChildUserNameTAG: srihari46 SecondChildCreateTimeTAG: 2012-11-08T14:28:17Z SecondChildTAG: Use Notepad++ or similar text editor to enter the formula, then cut&paste to check it. The browser will not crash as often and even if it do, the formulas won´t be lost. SecondChildUserIdTAG: 296965 SecondChildUserNameTAG: LGMailhos SecondChildCreateTimeTAG: 2012-11-08T14:52:43Z FirstChildTAG: During the Spring 2012 pilot 6.002x course, there was no display of equations at all, we entered them in the boxes that were too small and spent hours troubleshooting missing parentheses, etc. Very frustrating, and a major complaint by students. I have been very pleased with the MathJax display in the Fall 6.002x course assignments, and have not yet experienced problems with it (I am using Portable Firefox 10.0.2 under WinXP Pro). In the Spring 6.002x course, I used the equation display of ASCII Math Editor (see link below) to fine tune my answer and ensure it represented what I intended, then cut-and-pasted it into the answer box. Perhaps this will help you work around the Chrome problem. http://www1.chapman.edu/~jipsen/mathml/asciimatheditor/ When using ASCII Math Editor, make sure you enclose your equation between left-quotes (accent grave) or dollar signs, but don't include them in your answer when you paste it in the 6.002x answer box. FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-11-08T03:42:20Z FirstChildTAG: Thank you for reporting the bug, we are looking into this. FirstChildUserIdTAG: 151427 FirstChildUserNameTAG: RichardZ FirstChildCreateTimeTAG: 2012-11-08T05:45:50Z SecondChildTAG: Thanks to staff for troubleshooting! SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-11-08T07:13:47Z SecondChildTAG: still happening to me SecondChildUserIdTAG: 18073 SecondChildUserNameTAG: gburkhart SecondChildCreateTimeTAG: 2012-11-11T05:04:30Z IndexTAG: 40 TitleTAG: Graphing Calculator The online graphing calculator at [www.desmos.com][1] is pretty good for solving this problem by the graphical method. [1]: http://www.desmos.com UserIdTAG: 366083 UserNameTAG: smath CreateTimeTAG: 2012-09-16T21:00:46Z VoteTAG: 24 CoursewareTAG: Week 3 / Graphical Analysis of Nonlinear Circuits Exercise CommentableIdTAG: 6002x_graphical_analysis_of_nonlinear_circuits_exercise NumberOfReplyTAG: 2 FirstChildTAG: Thank you.Nice tool ! FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-09-19T10:20:27Z SecondChildTAG: nice one thnx.... it SecondChildUserIdTAG: 282892 SecondChildUserNameTAG: wiky SecondChildCreateTimeTAG: 2012-09-20T18:31:51Z FirstChildTAG: operating point is the point where the load line will intersect the exiting curve...use desmos graphic calculator ...write two of the equation there. and finally you will get the all highlighted points.. FirstChildUserIdTAG: 282892 FirstChildUserNameTAG: wiky FirstChildCreateTimeTAG: 2012-09-20T18:33:23Z IndexTAG: 41 TitleTAG: S6E1: Why resistance outside of the range 0-1mA is 2? According to chart it should be 3/2. Do I miss something? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-13T04:06:34Z VoteTAG: 24 CoursewareTAG: Week 3 / Analyzing Nonlinear Circuits Exercise CommentableIdTAG: 6002x_analyzing_nonlinear_circuits_exercise NumberOfReplyTAG: 6 FirstChildTAG: Try doing the slope calculation again. Make sure you're using a section of line with constant slope. FirstChildUserIdTAG: 248902 FirstChildUserNameTAG: Chanute FirstChildCreateTimeTAG: 2012-09-13T04:44:45Z SecondChildTAG: (Y2-Y1)/(X2-X1)=SLOPE =RESISTANCE=2KOHMS;WHEREY2=3,Y1=1,X2=2,X1=1 SecondChildUserIdTAG: 133068 SecondChildUserNameTAG: swarn99 SecondChildCreateTimeTAG: 2012-09-26T05:23:02Z SecondChildTAG: See resistance is graphically the slope of line depicting V-I relationship. Initially between voltage range 0V to 1V (which will be our X - coordinates, X1, X2 for our calculation of resistance), the current range is 0mA to 1mA (which will be our Y - coordinates, Y1, Y2 for our calculation of resistance). So for that part it is Y2 - Y1/X2 - X1 i.e. 1-0/1-0 = 1kohm. Next, for voltage range 1V to 2V (which will be our X - coordinates, X1, X2 for our calculation of resistance for that range), the current range is 0mA to 1mA (which will be our Y - coordinates, Y1, Y2 for our calculation of resistance). So for that part it is Y2 - Y1/X2 - X1 i.e. 3-1/2-1 = 2kohm. SecondChildUserIdTAG: 408765 SecondChildUserNameTAG: NeevGhodasara SecondChildCreateTimeTAG: 2012-09-28T15:29:10Z FirstChildTAG: Try doing the slope calculation again FirstChildUserIdTAG: 360956 FirstChildUserNameTAG: prabhavathi234 FirstChildCreateTimeTAG: 2012-09-13T09:37:41Z FirstChildTAG: Check that in the chart that pendent starts at 1 in the x axes and in 1 in the y axes. FirstChildUserIdTAG: 401611 FirstChildUserNameTAG: FedericojjjkQkkuevedo FirstChildCreateTimeTAG: 2012-09-14T03:49:00Z FirstChildTAG: Seems to me that resistance should be from 1k to 1k5 when current is outside 1ma range (and less than 2ma). FirstChildUserIdTAG: 345655 FirstChildUserNameTAG: roginvs FirstChildCreateTimeTAG: 2012-09-15T18:28:34Z FirstChildTAG: Do the slope calculation for both linear sections of slope. 0 to 1 volt and for the section 1 to 3 volts. Also bear in mind the V/I values calculated for milliamps not amps. The first part slope is 1V/1mA or 1V/0.001A giving 1000 ohms. FirstChildUserIdTAG: 345958 FirstChildUserNameTAG: PWilson123 FirstChildCreateTimeTAG: 2012-09-15T04:29:57Z SecondChildTAG: To add - make sure you calculate the slope from beginning to end, not just the values at the end - V is at 3 and I is at 2ma, but slope is the rise over the run, so (3-startV)/(2ma-startI) - plug in for the start values from when the slope changes. SecondChildUserIdTAG: 323230 SecondChildUserNameTAG: Xango5346 SecondChildCreateTimeTAG: 2012-09-15T20:16:24Z SecondChildTAG: WELL Sir here remember in such cases the resistance is the Slop... SecondChildUserIdTAG: 94924 SecondChildUserNameTAG: ahmedlamine SecondChildCreateTimeTAG: 2012-09-18T15:10:11Z SecondChildTAG: Why slope? The resistance should be something related to voltage. When measuring res, instruments doesn't care what the slope is, it just calculate the V/I, so the answer is ---- V/2+0.5. It's a nonlinear device when I>1mA SecondChildUserIdTAG: 143150 SecondChildUserNameTAG: dafshi SecondChildCreateTimeTAG: 2012-09-18T22:57:57Z SecondChildTAG: dafshi is right, if there is a resistance jump at 1mA, the Voltage should jump up as well, otherwise Ohms Law would be violated ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13480517491343659.png SecondChildUserIdTAG: 295197 SecondChildUserNameTAG: Ratio SecondChildCreateTimeTAG: 2012-09-19T10:51:43Z SecondChildTAG: ups, i forgot the negative part... SecondChildUserIdTAG: 295197 SecondChildUserNameTAG: Ratio SecondChildCreateTimeTAG: 2012-09-19T10:52:37Z SecondChildTAG: I think the correct answer should be *Rk=2-1/ik* where *ik* is in **mA** and *ik>1ma* SecondChildUserIdTAG: 316984 SecondChildUserNameTAG: voffka SecondChildCreateTimeTAG: 2012-09-19T18:24:50Z SecondChildTAG: What is the resistance (in kOhms) when the current is **outside that range**? R=V/I , V=3-1 , I=2*(10^-3) - 1*(10^-3) R=2 / 10^-3 =2kOhm SecondChildUserIdTAG: 194717 SecondChildUserNameTAG: kaa SecondChildCreateTimeTAG: 2012-09-21T14:47:04Z SecondChildTAG: I agree with dafshi and voffka. According to the Ohm's Law, resistance of element changes from 1k on 1mA, to 1.5k on 2mA. Why resistance is a slope, why not V/I? Why I should assume that element like a resistor in series with a voltage source? SecondChildUserIdTAG: 257498 SecondChildUserNameTAG: vtk SecondChildCreateTimeTAG: 2012-09-25T06:41:04Z SecondChildTAG: The slope of the curve is more correctly called "differential resistance", which is dV/dI, meaning the V-variation to I-variation ratio. You infer that we're talking about this differential resistance in the question about the resistance "outside that range" because otherwise you couldn't provide a unique value. By the way, if you want to consider the resistance as V/I, you can think of it as a variable resistance RD=2vD/(vD+1), so that RD=1k when vD=1V and RD=1.5k when vD=3V (or RD=2-1/iD in terms of iD). SecondChildUserIdTAG: 84801 SecondChildUserNameTAG: marcuspag SecondChildCreateTimeTAG: 2012-09-25T09:33:06Z FirstChildTAG: We've been "lucky" because the solution is within the "small resistance" range, so if you try to substitute the nonlinear element with a linear resistor R=1k, you find the correct solution. If you try to solve the same problem with VS=10V, you end up in the range i>1mA, v>1V, where the differential resistance is 2k. Unfortunately, you cannot use this 2k resistance and simply solve a voltage divider. You need to use the relation vD=2iD-1 (with iD in mA) provided by AndBre instead (or iD=(vD+1)/2). If you solve by using a 2k resistance, you get a solution which falls in the "correct range" (where the larger resistance holds), but still is incorrect. Please try and let me know what you get! FirstChildUserIdTAG: 84801 FirstChildUserNameTAG: marcuspag FirstChildCreateTimeTAG: 2012-09-25T09:44:25Z IndexTAG: 42 TitleTAG: Hints H12P1, H12P2, H12P3 requested by jmen Hi jmen, sorry for the delay. You ask me in a Post if I could post some hints on H12P2. Here they are as I have promised. I hope is not late... Myriam. ------ Before starting this Homework, you should read Chapter 15 of Textbook which explains the issue of Operational Amplifiers, page 837. ---------- ---------- ---------- But for those who want to know quickly the "modus operandi" to solve exercises that involve the use of Operational Amplifiers, do not despair, here I'll give you some key concepts:* ---------- ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13384202612905768.png) ---------- ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13384203152699093.png) ---------- In the case of an Operational Amplifier (see Fig. 1), in ideal condition we have that (we only mention 3): - *The Gain A, is a very large value that tends to infinity (A → ∞).* - *The impedance Zin, if we see the entrance to the two terminals (+, -), is a very high value: (Zin → ∞).* - *The impedance Zout, we see Operational amplifier output is of a very small value: (Zin → 0).* OK, but how important are these considerations? Can I solve the exercises involving the use of Operational Amplifiers with this concepts ? Yes ! ---------- Now, think, see Formula 1, if A is very large, what happens to the voltage difference between the positive terminal V + and V-negative terminal? Vo = A * (V + - V-)? ![image description](https://mitx_askbot_stage.s3.amazonaws.com/1338421654680479.png) That is, ideally, we see the entrance, between the two terminals, we have a short: ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13384214381028736.png) With this concept, we can solve a lot of exercises! ---------- *Enough Theory! We will stop here and we will go to the Homework. The point is to give you a general idea so that later you can solve it by yourself. Lets go! ---------- **H12P1: CURRENT SOURCE** **Part 1:** In this exercise they said to us the importance of having a controlled current source. For these applications, it is very common to use Operational Amplifiers. Consider the circuit of Figure 3. Since we are told that the amplifier operates with a gain A infinite, then we will be under ideal conditions and (see red circle in Figure 3) the voltage drop at the entrance of AO (Operational Amplifier, hereinafter we refer to these acronyms) will be zero. ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13384224815652255.png) Now, they tell us that the power supply can no longer tolerate a certain amount of current (the statement give us this amount), suppose we call this current as i_KB (i KA BOOM, because if we overcome the circuit may burn). From Figure 3, considering ideal conditions, we would have the circuit like this (see Fig 4): ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13384227524981501.png) Now, it is very simple, see the node that is formed above the red line in Figure 4. There is this current i_KB, current i, and the resistor current, call it i_R. This exercise, which at first seems complicated, in fact, is just a very simple exercise! Of course, here's the Hint: if we do not want to burn our circuit, we must ensure that the value of i_KB be very small, almost zero ... Then, if we remember the Kirchhoff Law tells us that the sum of currents into a node is equal to zero and having the value of i (given by the statement), also having the value of v, and looking up the circuit (mesh that involves v and R), so then, what value of R do we need?? It's easy! Let's try! ---------- **Part 2:** In this part, we are told that in fact the AO are imperfect and we are given a more realistic version of the circuit (Figure 5) that has been seen previously (Figure 3). ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13387460369832142.png) Now let's analyze the V-(in blue), the Vo (in red) and Vs (in green). Is there any way to relate these data in such a way to get the load resistance (Load)? Yes. Here comes an Illustrated Hint: ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13384235503264897.png) Now knowing R and V, I think you are able to calculate the load resistance (Load). I will not give you more clues! Out fast! ---------- **Part 3:** Here we say that we must assume that we are not working in ideal conditions, and therefore, we must consider our gain A ... Therefore, we can not do what we did in Figure 4 (look at the red line). :( Also, They give as the input load resistance (Load), as well as they say, we have to take the R calculated in Part 1 ... Well, here I recommend reading it (not excuses for not reading it, yes or yes) page 842 of Textbook, Section 15.3.1. Now, if we gain A, we have R (from part 1), we have V ^ +, we have load resistance (Load), then can we calculate the current i? Yes. Hint: Check the topology of this connection. Is it looks like the one you have on page 842 of Textbook? So, can we calculate the output voltage? (see Equation 15.7 of this page). In turn, if we look at Figure 15.7, we note that Vo is equal to the current (i) that we need to find multiplied it by the sum of two resistances (Do we know what those resistances are? Yes). No more hints for this part. ---------- H12P2 --- The Bipolar Junction Transistor (BJT) is a device that can be modeled with a current controlled dependent sourve and a diode; the gain K depends on the specific BJT being used. For our analysis we will use that model, and therefore the circuit is: ![im1][1] The Zener diode, as we studied before, has the following IV characteristic: ![im2][2] As part of your job as an engineer, your company has asked you to design a new regulator. Before actually doing the design, let's do some calculations. For your results you may consider the BJT, diode and the amplifier to be ideal (A→∞, Zin→∞, Zout=0). **Part a.** Write an expression for the output voltage (vout) when the input voltage is below the Zener voltage (vinvz and, for now, assume Rz=0. Write an expression for the output voltage as a function of vz and resistances. Hint. You know that the equivalent circuit of the zener is the following: a source in series with a resistance. So, try to make Rz=0 :) ![im][5] Another Hint. Now, what do you have in the v+? Can you find your output voltage as a function of vz and resistances? ![im][6] **Now we are ready for the actual design. The goal of your project is to select the appropriate resistances so that the output voltage is 5V±0.1V when the input voltage varies between 10V and 20V. Assume that the Zener diode you are using has a vz=2.5V and Rz=1Ω (Notice Rz is not zero anymore!).** **Part c.** What combination of R0, R1 and R2 satisfies the output voltage requirement? Well here you have to write all the equations that you will need and adopt some criteria. Lets see this together in order that you can solve it later by your own. Hint1. You know that your zener will be a source vz in serial with a resistence Rz. So, what will you have in v+? Isn´t it a source plus some votage related to the Rz? What current goes through your Rz? Isn´t it the same that goes through your Ro? Hint2. A visual Hint. OA is ideal so you will not have current in the terminals, see the red i that is equal to zero. Take a look at the orange current iD, How can you calculate this current? Remember that a current is Vt/Rt What is the total voltage in this loop? Dont´t forget to consider the source of the zener :p. What is the total resistence in this loop? ![im4][7] Hint3. Try to find your delta v+, that is to say v+ evaluated in one extreme condition minus v+ evaluated in the other extreme condition. Isn´t it a variation? Hmmmm... - So, what variation are they asking you in the satatement? Isn´t it less than a delta value of voltage? How do you find that delta? - eg., if the statement tell you 12V plus 0.5 or minus 0.5, your delta will be 1. - e.g, if the statement tell you 19V plus 0.35 or minus 0.35, your delta will be 0.70. - So, what will be your delta? Can you find your Ro? ;) Suggestion. adopt a value large enough to verify the condition... Hint4. Ok, once you have your v+, can you related it with your vout? Recall voltage dividers, this last will be the gain. Hint5. Now try to find your gain, you can try to take vout equal 5 for a average between the range given in the statement. Try to check if given some vin verifies the range of tolerance given in the statement. I can not tell more. warning. values must be in ohms, eg., if you have 1k you should enter 1000 and not 1. --- **Part d.** Hint1. The efficiency is the following ratio : **efficiency = OutputVoltage*OutputCurrent/InputVoltage*InputCurrent** Hint2. Visual hint. Recall KCL. ![im][8] Hint3. Try to think when you will have the minimun efficiency. What pair of vout and vin will give you the smaller value? voutmax and vinmin, voutmin and vinmax, voutmax and vinmax, etc..? Hint4. The values are not in porcentual. eg., they are like 0.75 and not 75 or like 0.10 and not 10. ---- H12P3 --- **H12P3:OPAMPS AND FILTER DESIGN** In this last part, we are asked to design a filter that meets certain requirements. They show us 4 circuits (A, B, C, D) and tell us thattwo of them are low-pass filters and the others, are High Pass Filters. ![image description](https://mitx_askbot_stage.s3.amazonaws.com/1338427123331308.png) Here give us a filter and requested we find the values ​​of its components. ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13384272825857654.png) This exercise is not difficult. We have to ask ourselves: _ What kind of circuit is the one that we see in the green box? Does not resemble the circuit B of Figure 10? So what type of filter is it? ................. _What Kind of circuit is the one that we see in the blue box? Does not resemble the circuit A of Figure 10? So what type of filter is it? .................... So, If A1=Vo1/VI A2= Vo/VI2 A = VI2/Vo1 We know our total transfer of voltages will be the product (since the three circuits are cascaded) of the 3 gains: H=Vo/VI=AT=A1*A*A2= (Vo1/VI)*( VI2/Vo1)* (Vo/VI2) ![image description](https://mitx_askbot_stage.s3.amazonaws.com/133842762245142.png) Well, all that is new here is the calculation of the gain of the red square in Figure 11 (A) ... I will give you a Illustrated Hint: ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13384277723339659.png) Ok, ince that we have the total gain AT, we will have: H=AT= (numerador)/(denominador). Lets see a general example: Supouse that you have a expression like this:* H= (s*M)/((s+N)*(S+P)) ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13384279351181185.png) We know that: 2pi*f1=N 2pi*f2=P **Part 1,2,4,5:** Now it's your turn! A search for the expression of AT Total Income and pay attention in its denominator. If you like data f1, f2, (C1 and C2 suitable choice, the statement gives us two values, but does not tell which is which), taking the expression N and P (remember that N and P may be a function of R1, C1 , R2, C2, etc ...), we can find the values ​​that we requested. ---------- **Part 3:** With C1, C2, R1, R2, RS, and knowing that the AT module is equal to a value given by the statement at a frequency, you can find the one missing data, ie, RF. Hint: Remember that s = jw and w = 2pi * f.* ---------- Next, we will have a circuit like this: ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13384282571365309.png) This exercise is very simple. Forget the Operational Amplifier and analyze the currents in Z1 and Z2. Hint1: The current in Z1 and Z2 is the same. So, knowing that the node between Z1 and Z2 acts as virtual ground, we can raise a VI equation in terms of Z1 i, and also get another function depending on Z2 vo e i. Now we have to raise the H = Vo / Vi and see the denominator and analyze the performance of that is very similar to Fig 14 ... With this, we obtain the values ​​of R4, R3 (conveniently, the two capacitors are available, assign conveniently C4 and C3, remember that if we're wrong we realize C as the resistance R will be much greater range of resistance that we have available to use in the design). If we calculated the values ​​of R3, R4, C4, C3, the last part (midband gain), we can simply get in the Sandbox. ---------- What is the gain value for midband gain ? *Hint: Consider the mid frequencies to any value in where the transfer is flat. Then, the midband gain value will be where the transfer is flat (value in the y-axis). Remember that the value is given in dB and the results need to be at times ... 20 * log (readed_value) = average profit. So average profit = 10 ^ (readed_value/20).* *Hint: Watch out for the sign within the plot read the module! Is negative or positive??* ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13384286775242575.png) ---------- ---------- ---------- > Now in Spanish Antes de comenzar con este Homework, es recomendable leer el Capítulo 15 del Textbook que trata el tema de Amplificadores Operacionales. ---------- ---------- ---------- Pero para aquellos que deseen saber rápidamente el “modus operandi” para resolver ejercicios que involucren el uso de Amplificadores Operacionales, no desesperen, que aquí les daré algunos conceptos claves: ---------- ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13384202612905768.png) ---------- ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13384203152699093.png) ---------- Para el caso de un Amplificador Operacional (ver **Fig. 1**), en condiciones ideales tenemos que (solo mencionamos 3): - La Ganancia A, es un valor muy grande, tiende a infinito (A→∞). - La Impedancia Zin, que vemos a la entrada de los dos terminales (+,-), es de un valor muy elevado: (Zin →∞). - La Impedancia Zout, que vemos a la salida del Amplificador Operacional, es de un valor muy pequeño: (Zin →0). Bien, pero qué tienen estas consideraciones de importante? Pueden facilitarme la resolución que impliquen el uso de Amplificadores Operacionales? **Sí!** ---------- Ahora, pensemos, veamos la Fórmula 1, si A es muy grande, que sucede con la diferencia de tensiones entre el terminal positivo V+y el terminal negativo V-? Vo= A*(V+ - V-) ? ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13384213999289919.png) Es decir, en condiciones ideales, vemos que a la entrada, entre los dos terminales, tenemos un corto: ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13384214381028736.png) Con este concepto, podemos resolver un montón de ejercicios!! ---------- Ahora, basta de explicaciones y vayamos a los ejercicios. La idea es darles una idea general para que luego puedan resolverlos ustedes mismos. Manos a la Obra! **EN:** *Enough Theory! We will stop here and we will go to the Homework. The point is to give you a general idea so that later you can solve it by yourself.* *Let's go!* ---------- > Now in Spanish **H12P1: CURRENT SOURCE** **Part 1:** En este ejercicio nos explican la importancia de tener una Fuente de Corriente Controlada. Para ello, nos dicen que para este tipo de aplicaciones, es muy común el uso de Amplificadores Operacionales. Veamos el Circuito de la Figura 3. Ya que se nos dice que el amplificador trabaja con una ganancia A infinita, entonces estaremos en condiciones ideales y tendremos (ver el círculo rojo en la Figura 3) una caída de tensión nula en la entrada del A.O (Amplificador Operacional, de aquí en adelante nos referiremos con estas siglas ). ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13384224815652255.png) Ahora, nos dicen que la fuente de tensión no puede tolerar más de una cierta cantidad de corriente (nos lo dan en el enunciado), supongamos que llamamos dicha corriente como i_KB(i de KA BUM!, porque si la superamos el circuito se nos puede quemar ). De la figura 3, con las condiciones ideales, tendríamos un circuito así (ver la Fig. 4): ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13384227524981501.png) Ahora, es muy simple, veamos en el nodo que se forma arriba de la línea roja de la Figura 4. Allí estará esta corriente i_KB , la corriente i, y la corriente del resistor, llamémosla i_R. Este ejercicio que a simple vista parece complicado,en realidad, no es más que un ejercicio muy simple! Eso sí, aquí va la Pista: si no queremos quemar nuestro circuito, debemos asegurarnos que el valor de i_KB sea muy pequeño, casi nulo… Entonces, teniendo la consideración, recordando las Ley de Kirchhoff que nos dice que la suma de Corrientes en un nodo es igual a cero y teniendo el valor de i (dado por el enunciado), teniendo el valor de v, y teniendo nuestra malla que involucra v y R, entonces, la pregunta es , qué valor de R necesitamos?? Es fácil! A trabajar! ---------- **Part 2:** SP: En esta parte, se nos explica que en realidad los A.O son imperfectos y se nos da una versión más real del circuito (Figura 5) que se ha visto con anterioridad (Figura 3). ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13387460369832142.png) Ahora, tratemos de analizar la V- (en color azul), la Vo (en color rojo), y la Vs (en color verde). Hay alguna forma de relacionar estos datos de tal forma de obtener la Resistencia de carga (Load)? Sí. Aquí viene una pista gráfica: ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13384235503264897.png) Ahora sabiendo R y Vs, creo que ya están en condiciones de calcular la resistencia de carga (Load). No les doy más pistas! Sale rápido! ---------- **Part 3:** Aquí nos dicen que debemos suponer que no estamos trabajando en condiciones ideales, y que por ello, debemos considerar nuestra ganancia A… Por lo tanto, ya no podemos hacer lo que habíamos hecho en la figura 4 (línea roja debido al corto). ;( También, nos dan como dato la resistencia de carga (Load), como así también nos dicen que tomemos la R que calculamos de la parte 1… Bien, aquí les recomiendo leer sí o sí la página 842 del Textbook, Sección 15.3.1. Ahora, si tenemos la ganancia A, tenemos la R (de la part 1), tenemos la V^+, tenemos la resistencia de carga (Load), entonces, podemos calcular la corriente i ? Sí. Pista: Ver la topología de esta conexión. No se parece a la que está en la página 842 del Textbook? Entonces, podemos calcular la tensión de salida? (ver Ecuación 15.7 de dicha página). A su vez, si miramos la Figura 15.7, notaremos que la Vo es igual a la corriente (i) que buscamos multiplicada por la suma de dos resistencias (sabemos cuáles son dichas resistencias? sí). No más pistas para esta parte. ---- H12P2 --- H12P2 --- El enunciado dice: The Bipolar Junction Transistor (BJT) is a device that can be modeled with a current controlled dependent source and a diode; the gain K depends on the specific BJT being used. For our analysis we will use that model, and therefore the circuit is: ![im1][1] También se da la curva del diodo zener. The Zener diode, as we studied before, has the following IV characteristic: ![im2][2] Como parte de tu trabajo como ingeniero, tu compañía te ha solicitado diseñar un nuevo regulador. Antes de realizar el diseño, hagamos algunas cuentas. Para tus resultados debes considerar ideales al BJT, diodo y al amplificador (A→∞, Zin→∞, Zout=0). **Part a.** El enunciado dice: Write an expression for the output voltage (vout) when the input voltage is below the Zener voltage (vinvz and, for now, assume Rz=0. Write an expression for the output voltage as a function of vz and resistances. Hint. Se sabe que el circuito equivalente que representa a un diodo zener es el siguiente: una fuente en serie con una resistencia. Ahora que sabes este equivalente, trata de hacer Rz=0 :) ![im][5] Otra Hint. Ahora, puedes decirme qué tensión hay en el terminal v+? Puedes de alguna forma hallar el voltaje de salida en función de vz y las resistencias? ![im][6] El enunciado dice: **Now we are ready for the actual design. The goal of your project is to select the appropriate resistances so that the output voltage is 5V±0.1V when the input voltage varies between 10V and 20V. Assume that the Zener diode you are using has a vz=2.5V and Rz=1Ω (Notice Rz is not zero anymore!).** **Part c.** Eñ enunciado dice: What combination of R0, R1 and R2 satisfies the output voltage requirement? Bueno, aquí tienes que tratar de escribir todas las ecuaciones que necesitarás para hallar tus incógnitas y cabe destacar, que deberás utilizar algún criterio para la resolución. Veamos esto juntos así luego lo puedes resolver por ti mismo. Hint1. Sabes qie tu zener será una fuente vz en serie con una resistencia Rz. Entonces, qué tensión tendrán enganchada en v+? No es acaso una fuente de tensión debida al zener más otra tensión debido a la caída de tensión de la Rz? Qué corriente circula a través de tu Rz? Acaso no es la misma que la de la Ro? Hint2. Hint Visual. El AO es ideal, por lo cual, no tienes corriente en los terminales del operacional. Mira la corriente iD que se encuentra en color naranja, Cómo puedes calcular esta corriente? Recuerda que una corriente es Vt/Rt. Cuál esel voltaje total en este loop ? No te olvides de considerar la fuente del zener en tus cálculos :p. Cuál es la resistencia total en este loop? ![im4][7] Hint3. Intenta hallar tu delta v+, esto es, la diferencia entre el v+ evaluado en una condición extrema menos el v+ evaluado en la otra condición extrema. Qué es esta diferencia, no es acaso una variación? Mmmmm... - Entonces, cuál es la variación que se te solicita satisfacer en el enunciado? Cómo hallas este delta? - ej., si el enunciado te da 12V más o menos 0.5 V, tu delta será 1V. - ej., si el enunciado te da 19V más o menos 0.35 V, tu delta será 0.70V. - Entonces, cuál es tu delta? Puedes hallar ahora tu Ro? ;) Sugerencia. adoptar un valor bastante grande de tal forma que verifique la condición del delta menor que ... Hint4. Una vez que tienes tu v+, puedes relacionarla de alguna forma con tu vout? Recuerda el concepto de divisores resistivos de tensión, este último será la ganancia. Hint5. Ahora intenta hallar tu ganancia, puedes tomar un valor de vout igual a 5 para un cierto valor promedio del rango dado en el enunciado. Intenta verificar si tus valores elegidos concuerdan con la tolerancia permitida en el diseño. No puedo decir más. Warning. Los valores deben estar expresados en ohms, por ejemplo, si tienes 1k debes escribir 1000 y no 1. --- **Part d.** Hint1. La eficiencia se define como el siguiente cociente: **efficiency = OutputVoltage*OutputCurrent/InputVoltage*InputCurrent** Hint2. Visual hint. Recordar KCL. ![im][8] Hint3. Ahora intenta pensar cuál es la mínima eficiencia. Qué par de vout y vin te arrojarán el menor valor? voutmax and vinmin, voutmin and vinmax, voutmax and vinmax, etc..? Hint4. Los valores no se encuentran en porcentual. ej, es correcto escribir 0.75 y no 75 , lo mismo si tienes 0.10, no será correcto escribir 10. ---- H12P3 --- **H12P3:OPAMPS AND FILTER DESIGN** En esta última parte, nos piden diseñar un Filtro que cumpla con ciertos requisitos. Nos muestran 4 circuitos (A,B,C,D) y nos dicen que dos son Filtros Pasa Bajos y otros, Filtros Pasa Alto. ![image description](https://mitx_askbot_stage.s3.amazonaws.com/1338427123331308.png) A continuación, nos dan un Filtro y se nos solicita hallar los valores de sus componentes. ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13384272825857654.png) Este ejercicio, no es difícil. Tenemos que preguntarnos: _ Qué clase de Circuito es el que vemos en el recuadro verde? No se parece al circuito B de la Figura 10? Entonces, qué tipo de Filtro es? …………….. _Qué clase de circuito es el que vemos en el recuadro azul? No se parece al circuito A de la Figura 10? Entonces, qué tipo de Filtro es? ……………….. Entonces, Si A1=Vo1/VI A2= Vo/VI2 A = VI2/Vo1 SP:Sabemos que nuestra Transferencia Total de Tensiones será el Producto (ya que los tres circuitos se encuentran en cascada), de las 3 ganancias: H=Vo/VI=AT=A1*A*A2= (Vo1/VI)*( VI2/Vo1)* (Vo/VI2) ![image description](https://mitx_askbot_stage.s3.amazonaws.com/133842762245142.png) Bien, lo único nuevo que hay aquí es el cálculo de la ganancia del cuadrado rojo de la figura 11 (A)… Les daré una pista gráfica: ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13384277723339659.png) Bien, una vez que tenemos la ganancia Total, tendremos H=AT= (numerador)/(denominador). Veamos un ejemplo genérico cualquiera: Supongamos tener una expresión de la forma: H= (s*M)/((s+N)*(S+P)) ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13384279351181185.png) Sabemos que : 2pi*f1=N 2pi*f2=P **Part 1,2,4,5:** Ahora, es tu turno! A buscar la expresión de la Ganancia Total AT y prestar atención en su denominador. Si tienes como dato f1, f2, (eligiendo convenientemente C1 y C2 , el enunciado nos da dos valores, pero no nos dice cuál es cual ), teniendo la Expresión N y P (recordar que N y P pueden ser función de R1, C1, R2,C2, etc…), podemos hallar los valores que se nos solicita. ---------- **Part 3:** Teniendo C1,C2,R1,R2,RS, y sabiendo que el módulo de AT es igual a un valor dado por el enunciado a una frecuencia , se puede hallar el único dato que falta, es decir, RF. Pista: Recuerda que s=jw y w=2pi*f . ---------- A continuación, tendremos un circuito como sigue: ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13384282571365309.png) Este ejercicio es muy sencillo. Dejemos de lado el Amplificador Operacional y analicemos las corrientes en Z1 y Z2. Pista1: La corriente en Z1 y Z2 es igual. Entonces, sabiendo que el nodo entre Z1 y Z2 se comporta como tierra virtual, podemos plantear una ecuación de VI en función de Z1 e i ; y, otra función de vo en función de Z2 e i. Ahora, tenemos que plantear la H=Vo/Vi y ver el denominador y analizar el ejercicio de forma muy similar a la Fig. 14… Con esto, podremos obtener los valores de R4,R3 (convenientemente, de los dos capacitores disponibles, asignamos convenientemente C4 y C3, recordar que si nos equivocamos de C nos daremos cuenta ya que la resistencia R será mucho mayor del rango de resistencias que tenemos disponibles para utilizar en el diseño). Si tenemos calculados los valores de R3,R4,C4,C3, la última parte (midband gain), la podemos obtener sencillamente en el Sandbox. ---------- Cuál es el valor de ganancia para frecuencias medias? Pista: Considera frecuencia media a cualquier valor en donde la transferencia es plana. Luego, el valor de ganancia media será en donde la Transferencia es plana (valor en el eje de las ordenadas). Recuerda que el valor está dado en dB y el resultado que necesitas debe estar en veces… 20 *log (valor_leido)=ganancia media. Entonces, ganancia media=10^(valor_leido/20). Pista: Cuidado con el signo, en el plot leemos el módulo! Es negativa o positiva??? ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13384286775242575.png) [1]: https://www.edx.org/static/content-mit-6002x/images/circuits/H12P2Schematic2.d054b9d07ac0.png [2]: https://www.edx.org/static/content-mit-6002x/images/circuits/ZenerCharacteristic.36c39c4c3b80.png [3]: https://edxuploads.s3.amazonaws.com/13550165861343631.png [4]: https://edxuploads.s3.amazonaws.com/13550170181343683.png [5]: https://edxuploads.s3.amazonaws.com/13550173911343685.png [6]: https://edxuploads.s3.amazonaws.com/13550177211343655.png [7]: https://edxuploads.s3.amazonaws.com/13550209071343618.png [8]: https://edxuploads.s3.amazonaws.com/13550234101343648.png UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2012-12-08T22:37:29Z VoteTAG: 23 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 9 FirstChildTAG: Thank you Myriam. Where would we be without you? I'm anxiously awaiting the explanation for H12P2. I can't seem to get started, as I can't figure out which region the zener diode is in, whether the diode is an open circuit, or has a negative or positive source for Vz. FirstChildUserIdTAG: 132828 FirstChildUserNameTAG: rharris FirstChildCreateTimeTAG: 2012-12-08T23:33:43Z SecondChildTAG: Haha, I guess it will be the same without me :p I hope this hints can help you. Take care, Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-09T04:11:52Z SecondChildTAG: The hints helped and now a and b are correct, thank you. I am still having difficulty understanding how you determined what region the zener diode was in for Vin < Vz and Vin > Vz. How did you do that? Once I saw your hint that told me what regions the diode was in, calculating Vout was easy. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-12-09T06:45:28Z SecondChildTAG: For Vin < Vz: Iz=0; for Vin > Vz Iz>0. SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-09T10:57:01Z SecondChildTAG: I'm confused why the Zener has i=0. If the Vz > Vin, doesn't backwards flow happen which would provide |i|>0? I thought that was the reason for using zeners, to allow flow in both directions at certain thresholds. SecondChildUserIdTAG: 355773 SecondChildUserNameTAG: Albright4edx SecondChildCreateTimeTAG: 2012-12-09T16:12:28Z SecondChildTAG: Please read [here][1]. Note that OUR Zener is simplified, ie no current at $V_{reverse} SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-09T16:29:14Z SecondChildTAG: Corrected :$V_{reverse}$<$V_Z$ SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-09T16:30:06Z SecondChildTAG: What I don't understand is, how do you determine Iz = 0 when Vin < Vz or Iz > 0 when Vin > Vz? If the problem said VD < Vz or VD > Vf or somewhere in between it would be clear what region the diode is in. Do you see how this is confusing? One has to take into consideration R0 and how it affects the voltage level and polarity on the diode, and the polarity on the zener diode and how it is affected by where VD is on the VD axis of the ID vs VD curve. Also, Vin is dropped across both R0 and the diode, so how do you come to the conclusion that VD is between Vz and Vf, as in the case with part a? SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-12-09T18:07:30Z SecondChildTAG: R0 will not affect polarity or voltage level on the diode.It is impossible.Zener isnt independent voltage source. For solving this task you should assume that no current when $v_z$<$v_{IN}$, in this case zener is simply removed from the consideration. (Real Zener has some leakage current on the reverse branch) SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-09T18:44:12Z SecondChildTAG: OOPS Sorry, typo again.. CORRECT HERE: WHEN $v_{IN}$ less then $V_Z$ SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-09T18:46:25Z FirstChildTAG: I'm done with the rest of it.... thnx to these hints.... but i am having trouble with finding the resistnce values in H12P2 finding the delta V+ value. the thing i find confusing is that how to find delta V+ from the given deltas in Vin and Vout. i got an expression for V+ in terms of R0 and other. where R0 is still unkonown. so how do we get the delta V+ ..... could some one please help ... thanx in advance:) FirstChildUserIdTAG: 117319 FirstChildUserNameTAG: iqramalik FirstChildCreateTimeTAG: 2012-12-09T18:05:46Z SecondChildTAG: You can find few ways discussed already. My way was to define more strong restriction for $v_O$ deviation (for example +-0.001V) and from this point to calculate $v^+$..By the other words I do involve 5.001V value to my equation toghether with $v_O$=5.000V. While we know $v_Z$=2.5V we can estimate gain $G$=~2.All next from this point should be clear. SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-09T18:14:06Z SecondChildTAG: Hi iqramalik, Can I help you? Try to think how to find a V+. You will see that the V+ is the voltage that you have in the + terminal of the OA. You know that you have the zener there. So, how was the equvalent circuit of the zener? ![im][1] Now, all that green box will be connected to the V+ terminal. So, you will have that your voltage there will be the sum of the vz plus a current multiplied by Rz. But, what, is that current? Try to take a look to the orange current. How can you calculate that current, what voltage do you have in that two serial resistences? Isn´t it a voltage minus another voltage ? and what resistances are involved? So, isn´t it V divided by R a current? ![im][2] Might, this is confusing. But lets see this together so that later you can solve it by your own, Your V+ will be in terms of Ro,Rz, vin and vz. They give you a range of vin, like, eg., from 40 to 50. So, your vin min will be 40 and your vin máx will be 50. If you see, you have your V+ equation. For vin máx you can have an equation, the only value missing will be Ro. For vin min you can have an equation, the only value missing will be Ro too. Also, your tolerance will be defined as the variation in the output of a certain value. You know that your Delta Vout = Delta V+ and your Delta V+ will be the V+max minus V+min.... They give you your Delta Vout, so can you choose a value of Ro. Try to pick one that verifies your condition, suggestion choose a large one, in order to verify the restriction.... I hope this can help you. I am tempted to say more but I can´t because I can´t post a step by step solution.... But tell me if you are still having problems to find V+, I will like to help you :) Myriam. [1]: https://edxuploads.s3.amazonaws.com/13550173911343685.png [2]: https://edxuploads.s3.amazonaws.com/13550209071343618.png SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-09T19:28:43Z SecondChildTAG: Am I misreading here? Delta Vout is going to be Delta V+ **times the gain of the op amp circuit as determined by R1 and R2**?? SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-12-09T19:46:50Z SecondChildTAG: Skyhawk - youre right :) SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-09T20:22:01Z SecondChildTAG: Sorry, yes, that is correct. Delta Vout = Gain* Delta V+ SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-09T23:56:50Z FirstChildTAG: Oh Myriam, I wish I could see that this was easy, but I'm stuck with this for the total gain for P3 and I can't do anything with it: $ \frac {sC_1R_1(R_S + R_F)}{(sC_1C_2R_1 + C_2)(sC_1C_2R_2 + C_1)(sC_1C_2R_S)} $ It's doing my head in and I have no other approach I can come up with. Thanks so much for all your hard work, but is there something simple that I'm missing here, because "easy" is not how I could describe this problem? FirstChildUserIdTAG: 166031 FirstChildUserNameTAG: krebryna FirstChildCreateTimeTAG: 2012-12-09T04:38:12Z SecondChildTAG: Do you need Gain on the breakpoint? Or fo you need Gain at the middle of bandwidth? For the last case you can completely ignore both capasitors. SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-09T09:37:06Z SecondChildTAG: Hi krebryna, Can I help you? ![ima][1] Another alternative, for part 3, it is to think it in this way, that A will contribute to the midband gain. So, how is the gain of A - the red box-. Do you have A in terms of RF and Rs? and what says the statement about the value of the midband gain? I hope this can help you. Myriam. [1]: https://mitx_askbot_stage.s3.amazonaws.com/133842762245142.png SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-09T18:41:29Z SecondChildTAG: Krebryna Have a look here: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50c493e4468bdc230000003b It was meant to focus on the second part of the problem, but it has information on transfer functions for low and high pass filters, which you can combine with Myriam's hints. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-12-09T19:54:04Z SecondChildTAG: Thank you Myriam and Skyhawk! I went to bed feeling really quite stupid, but with your guidance, I've managed to knock the edges (just the edges, mind you) off that edifice. However, I'm not completely sure why I'm right. Are we taking advantage of the fact that at break frequencies, the real and imaginary parts of the transfer functions are equal? And that therefore in the total gain equation that I (sort of) have above, if I can keep track of which parts of that equation are responsible for the low- and high-pass parts of the circuit, I can individually figure out the resistance/capacitance values for each filter component? i.e. in the most general case, I can derive the expression for $ \omega $ (as a function of L, C, whatever) that will equal 1. My fears centre around these cascaded circuits. I'm scared to be extracting small parts of much larger equations to figure out the behaviour of interest. That's a bit of a ramble. But, nonetheless, you guys are stars. Thank you. SecondChildUserIdTAG: 166031 SecondChildUserNameTAG: krebryna SecondChildCreateTimeTAG: 2012-12-10T03:43:30Z SecondChildTAG: Good to hear that you you got it. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-12-10T12:06:00Z FirstChildTAG: I am completly stuck with h12p2 c and d For find R0 i think Vout2/Vout1=V+2/V+1 next I suppose R1+R2=100 R1=100/(k+1) R2=100*k/(k+1) Vin work against Vz Iin = current in branch (R0+Rz)+ K (Itotal) Itotal=Ivoltage divider+ IL, Pout=IL^2*RL, Pin=Vin*Iin, eff=Pout/Pin but i have 4 red cross FirstChildUserIdTAG: 406023 FirstChildUserNameTAG: neitrino FirstChildCreateTimeTAG: 2012-12-09T20:28:56Z SecondChildTAG: Yure wrong, and do hmm..calculationsby the not so correct way SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-09T20:47:02Z SecondChildTAG: where SecondChildUserIdTAG: 406023 SecondChildUserNameTAG: neitrino SecondChildCreateTimeTAG: 2012-12-09T20:51:09Z FirstChildTAG: no way....i am leaving...only 15 mins more FirstChildUserIdTAG: 156835 FirstChildUserNameTAG: kphariprakash1968 FirstChildCreateTimeTAG: 2012-12-09T20:38:41Z SecondChildTAG: no way for what? SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-09T20:48:27Z SecondChildTAG: H12P3,,,,,R0 ,R1.R2 efficiency not getting.... SecondChildUserIdTAG: 156835 SecondChildUserNameTAG: kphariprakash1968 SecondChildCreateTimeTAG: 2012-12-09T20:54:33Z SecondChildTAG: You may get some values by the "hand" manner.I cant write equations here, sorry.You may find them easily..I posted some.. Spreadsheet will help you. Note: although it is absolutelly equal will you use 10Ohm for R1(R2) or 10MOhm in this homework, try to use "real" values like 10kOhm, for example. You may play with received values :) 1)try to get $v^+$; 2)with this $v^+$ you will find G:apply equation for non-inverting OA 3)Check $v_O$ at the corners $v_I$ 4)Increase R0 if need it. Ok, you do have R0..R3 now 5)Calculate $P_{load}$ 6)calculate $P_{total}$=$V_{in20V}$* $I_{total}$(all currents in this voltage regulator).What they are?It is load, zener net, BJT net, R1?R2 divider net.. 7) Relation between 5 and 6 will be efficiency. SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-09T22:08:05Z SecondChildTAG: Hi kphariprakash1968, Can I help you? SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-09T23:55:12Z FirstChildTAG: I think i got H12P2 c) d) right, i even simulated the circuit with circuitlabs and everything works as expected (vin from 10V to 20V leads to a vout from 4.9 to 5.1) but still i receive red marks four the last four answers. I am a little desperate. Any hints on what stupid mistake i stumbled upon? The answers are R0, R1 and R2 in this order right? I know its difficult to judge since i am not allowed to post my solutions here. But maybe someone still can help. Maybe i can send the circuitlab design to one of the TAs? FirstChildUserIdTAG: 388514 FirstChildUserNameTAG: hdbam FirstChildCreateTimeTAG: 2012-12-09T22:02:29Z SecondChildTAG: Do you have calculated efficiency? SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-09T22:15:35Z SecondChildTAG: Yeah i did. That even fits to the calculated value in the simulation as well. Still no luck (the efficiency is low btw. I think i dont tell too much if i say that it is below 10% right? SecondChildUserIdTAG: 388514 SecondChildUserNameTAG: hdbam SecondChildCreateTimeTAG: 2012-12-09T22:30:19Z SecondChildTAG: it is possible..You should not use %, correct value (example) 0.1, not 10% SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-09T22:53:25Z SecondChildTAG: Yes you are right. I didnt use the percentage value but 0.0... Still i am stuck which drives me crazy a little :D Am i allowed to verify my R0 in someway with someone who got it right? It should be an Integer and its kinda low resistance right (no kiloohms). SecondChildUserIdTAG: 388514 SecondChildUserNameTAG: hdbam SecondChildCreateTimeTAG: 2012-12-09T23:23:42Z SecondChildTAG: Hi hdbam, Can I help you? Might your 3 resistences values are correct and you have incorrect your eficciency. So that is why the check shows you incorrect the 4 ones, instead of only the efficiency....It says that in the statement, that you have to have the 4 values correcly in order to recieve the green check.... Are you sure that you are calculating correctly the efficiency? Are you calculating correclty the currents? Are you calculating the minimun efficiency? what vout and vin are you taking? Are you sure that that pair of values gives the minimun result? Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-09T23:53:48Z SecondChildTAG: There is a limit on how small R0 can be, but the grader will accept unrealistically large values. You do realize, don't you, that the voltage doesn't have to swing from 4.9V to 5.1V. A swing from 4.99V to 5.01V is acceptable or even 4.999V to 5.001V. The latter cases make for an easier design. In fact, the voltage swing does not have to be symmetric about 5V. A swing from 5.01V to 5.02V is OK. In the real world there is a minimum current for which the Zener will give good regulation. That corresponds to a minimum Rz, but we are given a constant Rz so any current works here! SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-12-09T23:57:00Z SecondChildTAG: @Myriam: Well i think my efficiency is correct. I took the larger vin which gives me the larger vout which should be the lower efficiency. Btw I think your hint for the efficiency might be confusing since you marked the current through the Diode as output current but it should be the current through RL or am i wrong? SecondChildUserIdTAG: 388514 SecondChildUserNameTAG: hdbam SecondChildCreateTimeTAG: 2012-12-10T00:27:48Z SecondChildTAG: Ok, I got it now. My calculations were correct but the grader didnt accept my values for R0,R1,R2 and efficiency when i calculated them such that i receive an output from 4.9V to 5.1V. It worked when i changed the output swing to 4.99V to 5.01V (which of course changes R0 and the dependency of R1 to R2). @Skyhawk: Thx for the hint! SecondChildUserIdTAG: 388514 SecondChildUserNameTAG: hdbam SecondChildCreateTimeTAG: 2012-12-10T00:37:47Z SecondChildTAG: My pleasure! SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-12-10T01:05:29Z SecondChildTAG: @Myriam: Btw i forgot to thank you for your effort as well and your great explanation which helped me to get through H12! Keep on the good work and good luck for the final :-) SecondChildUserIdTAG: 388514 SecondChildUserNameTAG: hdbam SecondChildCreateTimeTAG: 2012-12-10T16:47:42Z FirstChildTAG: Hello. I don't understand how to apply KVL correctly in H12P2. I keep calculating the following: v0 (voltage drop across R0) - vD - vin = 0, but this does not result in the correct answers for parts a & b. I know what the answers to parts a & b are supposed to be but I cannot derive them using KVL. Could anyone point out the error in my thinking? FirstChildUserIdTAG: 345398 FirstChildUserNameTAG: xost33 FirstChildCreateTimeTAG: 2012-12-10T01:42:04Z SecondChildTAG: You need V+ to solve the problem. For parts a and b where Rz is assumed zero. V+ = vin, vin < vz and V+ = vz, vin >= vz. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-12-10T02:05:02Z FirstChildTAG: @ myriam: i understud the xplanation u gav above for al 3 probs..:)) now come to h12p3 second half..i hav eqns for H and A(=-R4/R3)...u easily told tat R4 n R3 can b found...explain it pls :).. FirstChildUserIdTAG: 356056 FirstChildUserNameTAG: Ganesh2810 FirstChildCreateTimeTAG: 2012-12-10T07:25:26Z FirstChildTAG: thanks myrmit, got all the answers right,couldn't clear the 60% bar but I AM PRETTY CLOSE!! thanks a ton again! good luck for end term!! FirstChildUserIdTAG: 368712 FirstChildUserNameTAG: jmen FirstChildCreateTimeTAG: 2012-12-10T11:42:59Z SecondChildTAG: Well done! I am so happy for that! Good luck to you too!:) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-10T11:58:47Z IndexTAG: 43 TitleTAG: Sound pitch messed up on the 1.25x version On the 1.25x speed version of this video, the pitch of the sound is messed up. The professor sounds like a smurf :-). UserIdTAG: 341293 UserNameTAG: Pietr CreateTimeTAG: 2012-11-18T21:21:11Z VoteTAG: 23 CoursewareTAG: Week 10 / Series RLC Deriving The Frequency Response CommentableIdTAG: 6002x_Series_RLC_deriving_the_frequency_response NumberOfReplyTAG: 1 FirstChildTAG: Glad I am not the only one. I just assumed that he got cold or something and that made it sound funny. FirstChildUserIdTAG: 460122 FirstChildUserNameTAG: sandred FirstChildCreateTimeTAG: 2012-11-19T03:30:56Z SecondChildTAG: It's even funnier on 1.50x ! SecondChildUserIdTAG: 351871 SecondChildUserNameTAG: Lamarque SecondChildCreateTimeTAG: 2012-11-20T22:46:04Z SecondChildTAG: try 0.75x :) SecondChildUserIdTAG: 447663 SecondChildUserNameTAG: brigadakuznetsov SecondChildCreateTimeTAG: 2012-11-23T16:32:13Z SecondChildTAG: 0.75x and 1.5x commendable!! SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-12-18T13:33:00Z SecondChildTAG: Needs pitch correction! SecondChildUserIdTAG: 248902 SecondChildUserNameTAG: Chanute SecondChildCreateTimeTAG: 2012-12-21T21:52:18Z IndexTAG: 44 TitleTAG: CONGRATULATIONS THIS IS TO ANANTH AGRAWAL SIR FOR BEING ENTERING IN FORBES 15 MOST REVOLUTIONARY PERSONS HATS OFF TO U SIR!!! UserIdTAG: 285222 UserNameTAG: dhaval24 CreateTimeTAG: 2012-11-11T04:57:28Z VoteTAG: 23 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 12 FirstChildTAG: Congratulations! FirstChildUserIdTAG: 329361 FirstChildUserNameTAG: ikrukov FirstChildCreateTimeTAG: 2012-11-13T19:10:46Z FirstChildTAG: congrats:-)) FirstChildUserIdTAG: 114864 FirstChildUserNameTAG: Vikaash FirstChildCreateTimeTAG: 2012-11-11T05:41:56Z FirstChildTAG: Congratulations! FirstChildUserIdTAG: 278792 FirstChildUserNameTAG: sali FirstChildCreateTimeTAG: 2012-11-11T05:12:05Z FirstChildTAG: Congratulations Prof. Anant Agarwal. You deserve it. FirstChildUserIdTAG: 183507 FirstChildUserNameTAG: obiradaniel FirstChildCreateTimeTAG: 2012-11-11T07:53:17Z FirstChildTAG: Nothing can defeat nature and hardwork..... Congrats....!! FirstChildUserIdTAG: 477198 FirstChildUserNameTAG: Rajesh1993 FirstChildCreateTimeTAG: 2012-11-11T09:04:50Z FirstChildTAG: like like like FirstChildUserIdTAG: 499268 FirstChildUserNameTAG: ruinoah FirstChildCreateTimeTAG: 2012-11-11T05:05:08Z FirstChildTAG: Congratulations!:) FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-11T05:07:46Z FirstChildTAG: congos.;) FirstChildUserIdTAG: 266739 FirstChildUserNameTAG: satvikchugh FirstChildCreateTimeTAG: 2012-11-11T08:14:13Z FirstChildTAG: Many Many congrats sir. I have never seen such a nice teacher of electronics like professor agarwal :D FirstChildUserIdTAG: 64618 FirstChildUserNameTAG: ashfaq2419 FirstChildCreateTimeTAG: 2012-11-11T16:25:04Z FirstChildTAG: Wow! Congratulations! FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-11-12T01:24:11Z FirstChildTAG: Excellent Teacher!!!!! FirstChildUserIdTAG: 369339 FirstChildUserNameTAG: vargaslen FirstChildCreateTimeTAG: 2012-11-12T10:32:13Z FirstChildTAG: Congratulations FirstChildUserIdTAG: 244225 FirstChildUserNameTAG: lerimock FirstChildCreateTimeTAG: 2012-11-13T12:41:02Z IndexTAG: 45 TitleTAG: Midterm Review Packet From 6.002x Spring 2012 [Without Solutions][1] [With Solutions][2] [1]: https://docs.google.com/open?id=0BzOcj2-0MDJeMDFnUFNjRGxraWs [2]: https://docs.google.com/open?id=0BzOcj2-0MDJeZDJqZVB3MGdpZ3M UserIdTAG: 127195 UserNameTAG: princeofsudan CreateTimeTAG: 2012-10-23T14:21:36Z VoteTAG: 23 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: thanks a lot.. FirstChildUserIdTAG: 114864 FirstChildUserNameTAG: Vikaash FirstChildCreateTimeTAG: 2012-10-24T09:04:02Z FirstChildTAG: This looks almost ***do-able***! FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-10-23T18:43:12Z SecondChildTAG: Thanks!! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-23T18:43:38Z FirstChildTAG: Thanks, **princeofsudan!** FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-10-25T16:23:52Z IndexTAG: 46 TitleTAG: Here's a picture of my lab! ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13509137641343666.jpg UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-10-22T13:49:29Z VoteTAG: 23 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 15 FirstChildTAG: woww..:) FirstChildUserIdTAG: 122513 FirstChildUserNameTAG: Mona77 FirstChildCreateTimeTAG: 2012-10-23T05:15:50Z SecondChildTAG: nice .. SecondChildUserIdTAG: 381619 SecondChildUserNameTAG: siddhantmishra007 SecondChildCreateTimeTAG: 2012-10-24T11:13:47Z SecondChildTAG: Hazel- wat u do? SecondChildUserIdTAG: 130980 SecondChildUserNameTAG: madhumohan23 SecondChildCreateTimeTAG: 2012-10-24T17:29:28Z SecondChildTAG: I do 6002x! :) I am a student SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-24T20:24:20Z FirstChildTAG: Wow! So neat! Wonderful! Note to self: Get your parts into bins then you can find them when you need them.. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-22T14:13:13Z SecondChildTAG: Thanks skyhawk! But believe me, this morning was cleanup morning, and I take pictures after cleanup mornings! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-22T14:19:13Z SecondChildTAG: Very wise. ;-) SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-23T03:33:26Z FirstChildTAG: I like the exhaust duct you have for soldering, good idea. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-22T14:07:05Z SecondChildTAG: I'm in preparations for moving, so my stuff is kinda scattered all around, otherwise I would post a picture. Thanks for sharing. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-22T14:07:57Z SecondChildTAG: Do so as soon as you can! We all love lab shots! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-22T14:20:44Z SecondChildTAG: BTW thanks! There is a little computer fan in there, it's surprisingly effective! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-22T14:29:20Z SecondChildTAG: Brilliant! SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-23T03:34:16Z SecondChildTAG: Nice.. Lab Hazel.. SecondChildUserIdTAG: 111917 SecondChildUserNameTAG: ashish_mit SecondChildCreateTimeTAG: 2012-10-24T02:54:56Z FirstChildTAG: Nice hazel1919! Than you for sharing this! :) Just for curious, what do you have in that boxes? FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-22T14:00:53Z SecondChildTAG: Which boxes? SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-22T14:17:45Z SecondChildTAG: The one that looks like it's being worked on is a PLL SSTC Driver. Here is the whole circuit...![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/1350916023182495.png SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-22T14:27:10Z SecondChildTAG: ![imag][1] That boxes :) [1]: https://edxuploads.s3.amazonaws.com/13509172091343675.png SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-22T14:47:26Z SecondChildTAG: Components! Thousands upon thousands of resistor, capacitors, inductors, diodes, LEDs, microchips and transistors. + connectors, heatsinks and switches of all sorts! I don't need any component until I bin it though (if you know what I mean)! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-22T14:58:35Z SecondChildTAG: Oh, and I think one of those compartments is full of dead MOS-FETs. They are useless now, but they cost too much to throw away! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-22T15:00:37Z SecondChildTAG: Your lab is so clean ! I renounce to post a photo of mine ! SecondChildUserIdTAG: 373752 SecondChildUserNameTAG: Croak SecondChildCreateTimeTAG: 2012-10-22T15:38:47Z SecondChildTAG: @hazel1919 :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-22T22:33:24Z SecondChildTAG: ::drool:: SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-23T03:35:10Z FirstChildTAG: Hazel1919, what scope do you have there? FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-22T16:05:31Z SecondChildTAG: HAMEG 20MHZ 203-7, not tektronix by any stretch, but functional. I used to have two Telequipment scopes. How about you? SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-22T16:16:02Z SecondChildTAG: Mine's a Tektronix 464. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-23T14:41:57Z SecondChildTAG: http://www.testbuyer.com/pdf/specs.cfm?pdf_id=550E5C0F6B SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-23T14:52:18Z SecondChildTAG: Awesome! I am jealous, I don't know why I always get jealous when I see an Oscilloscope. SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-23T19:31:09Z SecondChildTAG: Hazel Sir , You can make your PC as a oscilloscope I buy one for my official use it is cheap Thanks , SecondChildUserIdTAG: 156694 SecondChildUserNameTAG: mkprasanth SecondChildCreateTimeTAG: 2012-10-30T10:53:46Z SecondChildTAG: Yes good suggestion, o-scopes are pretty dear! There are also ones you can download. But as far as I'm aware, it is limited frequency wise by your sound card (20khz). SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-08T17:25:02Z FirstChildTAG: Cool! FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-10-22T16:35:56Z SecondChildTAG: Gracias SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-08T17:25:16Z FirstChildTAG: On the right of picture High voltage design? FirstChildUserIdTAG: 464744 FirstChildUserNameTAG: attache FirstChildCreateTimeTAG: 2012-10-22T13:59:53Z SecondChildTAG: Hydrogen generator ;) All Acrylic and Stainless! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-22T14:20:00Z SecondChildTAG: What for do you use hydrogen? SecondChildUserIdTAG: 208296 SecondChildUserNameTAG: OZ1 SecondChildCreateTimeTAG: 2012-10-22T15:29:51Z SecondChildTAG: Just experimentation, the odd demo, (BOOM). SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-22T15:50:54Z SecondChildTAG: ::cackles evilly:: SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-23T03:35:47Z SecondChildTAG: lol! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-23T08:11:22Z FirstChildTAG: Where do you live, Hazel? (Approx. answer is fine. :-)) FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-10-23T03:38:04Z SecondChildTAG: South of France! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-23T08:12:09Z SecondChildTAG: Darn! I can't just pop over for a visit, then! ;-) SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-24T13:21:28Z SecondChildTAG: Now, why the heck not? SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-08T17:25:38Z FirstChildTAG: **Nice Hazel!** Looks like *my* lab with all those parts bins: Those are a must for any electrical lab. Nice fume remover / exhaust for soldering smoke. Looks like you have a Weller iron, like me, a necessity. For equipment, I use BK Precision because I knew a rep who lived in N.J. (locally) and gave me a good deal on a BK digital signal source (with LCD readout), and a BK 5A 30V dual power supply. I also have a Fluke 179 multimeter, an old scope, a logic probe, and a signal tracer and injector that I bought during my university years; plus a Fisher scientific hot-plate with digital readout for chem/heating uses, and a oxy-MAPP gas torch when I need to heat things really hot! (Oh, and all sorts of mechanics' tools, but that's in the garage, I work on cars, too!) *Let me know your wish list!* My **wish list** is a for (1) a digital phosphor scope (Tektronix TDS1000-EDU is my first choice but expensive; also thinking of a BK Precision or Chinese knockoff equivalent depending on my funds), (2) a data recorder to store lab data on my PC, and (3) a microcontroller programmer and software platform (for a PIC or Arduino, maybe even a BASIC Stamp but that's less advanced - right now I'm limited to analog and basic logic circuits with discrete ICs). (I really appreciate those who build their own gear, but building a proper oscilloscope from parts would probably cost more than buying a finished model!) Thanks to all for sharing! Mark in N.J. FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-10-23T03:37:13Z SecondChildTAG: Love this community and the sharing! :-D SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-23T03:40:30Z SecondChildTAG: My wish list: - A CNC router - A Tektronix digital storage oscilloscope - A selection of High voltage probes - Full version of Wolfram Mathematica - Metcal soldering station - TI85 - Fluke multimeter - A garage SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-23T08:19:28Z SecondChildTAG: You can't beat a good old Weller TCP! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-23T20:26:52Z FirstChildTAG: great laboratory for a student to have FirstChildUserIdTAG: 352647 FirstChildUserNameTAG: praveenjugge FirstChildCreateTimeTAG: 2012-10-29T08:06:24Z SecondChildTAG: I agree ;) SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-08T17:25:55Z FirstChildTAG: Thanks for sharing everything, knowledge, lab photos! Very, very cool! FirstChildUserIdTAG: 175205 FirstChildUserNameTAG: KT_MM FirstChildCreateTimeTAG: 2012-11-02T01:06:27Z FirstChildTAG: \* drool * One day I will have some of those... One day! Really interesting setup :-) Would love to see your projects. Do you have a website/blog where you put these up? Right now I have a small table so things are packed away and taken out only when I need them. Not very comfortable but that's all I have space for at the moment. FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-11-02T18:08:22Z SecondChildTAG: Hye, there is no shame in that! I don't have a blog though, i'll probably post a few projects here though. I also frequent 4HV.org. SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-08T17:27:20Z SecondChildTAG: Thanks! hazel1919 :) The first thing I want to buy in March next year is a decent oscilloscope. I made a mistake of buying the DSO quad :-/ Thanks for that link! That forum looks really cool! SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-11-17T17:08:53Z FirstChildTAG: Bonjour, hazel1919, I'm en edX staff member, and we here at edX loved seeing the photo of your lab. We often get requests from the media for information about edX learners. Would you be interested in sharing your 6.002X story with us (e.g., why you are taking the course, how you have found it, what impact it is having on your life/education), for possible sharing with the press? I you are, please email me at margaret@edx.org. FirstChildUserIdTAG: 455109 FirstChildUserNameTAG: ElementsofStyle FirstChildCreateTimeTAG: 2012-11-07T15:13:01Z SecondChildTAG: Hi, Please excuse my skepticism, but how do I know you are staff? SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-08T09:26:02Z FirstChildTAG: good one FirstChildUserIdTAG: 336466 FirstChildUserNameTAG: kenstan FirstChildCreateTimeTAG: 2012-12-09T19:57:51Z FirstChildTAG: Awesome, hazel1919! I too am drooling like planetscape and ashwith! Would be great if you or anyone in this group can give some advice on how to go about in setting up a small to medium-sized home-laboratory for learning and experimenting. FirstChildUserIdTAG: 232157 FirstChildUserNameTAG: GayatriTR FirstChildCreateTimeTAG: 2012-12-31T09:04:22Z IndexTAG: 47 TitleTAG: Thoughts on Week 2 Work **Hello fellow students:** (And wow, a new discussion editor, a lot better than when I was in discussion last week) Did anyone think Week 2's Homework was WAY easier than Week 1 but Week 2's Lab was tougher? If yes, **give this post a plus (+)**. I want to see if anyone felt the same! In the middle of the week, after watching the lectures, Homework 2 was done in ONE hour for me (5 minutes for the LOGIC stuff). So I thought that Lab 2 would be just as easy, and I waited until the last minute. I waited until 4 hours before 12:00 midnight, thinking I could quickly finish it. **I was wrong.** It took me two hours of getting the equations using superposition correct, with a few algebra mistakes I had to fix. Then I went into the simulator. The waveform was **correct** and came very close to 0.667 and -0.167 when I ran the TRANSIENT analysis. But I kept getting **red X's!** I was wondering what was wrong! I knew the values were related by 0.5 and 0.3, so kept tweaking my answers, changing 0.5 to 0.66666, etc. hoping it would work, but it would mess up my waveform so it didn't match. With 40 minutes left, I finally figured out the problem: I was **closing** my transient analysis WINDOW before pressing CHECK. This time, I went back to my original, correct, values. I left the transient analysis window (the window that shows the waveform) **open**. Then I also pressed **check**. Finally, I got the *green mark* I desperately needed! Another 40 minutes and I would have gotten 0 instead of 100! Another week, **another mistake that has nothing to do with my work, but with the protocol**, that wastes an hour or two: That is 6.002x! I just had to let that frustration vent out! Mark in New Jersey USA UserIdTAG: 292546 UserNameTAG: JerseyMark CreateTimeTAG: 2012-09-24T03:32:15Z VoteTAG: 23 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Well to be honest, I started the second set of lectures for week 2 (Static Discipline and Boolean Logic) at about 10:40 PM, because I thought it would be as easy as last week. After I got through the lectures (at about 11:05), I quickly did the homework in about 20 minutes or so. Naturally, I thought the lab would be just as simple. Of course, I was wrong. The lab was devastatingly difficult this week - at first, I had absolutely no clue how to do it. At 11:45, after trying various approaches, I started to panic. Fortunately for me, I finally tried something and submitted the correct answer at 11:54. Originally, when I submitted, I didn't expect the answer to be correct at all - I thought my answer was just a little BS. But, in the time after I submitted my answer (coincidentally, after midnight passed), I finally understood why my circuit worked. And hey - just in time, too! So, I'm not really sure why I said all that. Just a storytelling habit of mine, I suppose. But, I do agree with you - this week's homework was pretty easy, but the lab was freakishly difficult. Fortunately, I left my transient analysis window open the first time around. Who knows what would have happened if I didn't. But I don't know. A lot of things can happen in 6 minutes, I guess. FirstChildUserIdTAG: 224087 FirstChildUserNameTAG: epark FirstChildCreateTimeTAG: 2012-09-24T04:12:32Z SecondChildTAG: I've got the same habit as you...I tend to wait until the last minute. Often work comes first, and the only time I have when I'm not busy is Saturday and Sunday. Next week that will change. I will watch a bit of the lecture each night, and look at the lab and homework by Wednesday at the latest. I can always solve EECS problems while waiting at the doctor's office or when in line (queue) buying groceries. **Panic** at 11:45PM causes too much stress in life, and needs to be avoided :) SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-09-24T04:29:04Z SecondChildTAG: I had the same issue as you guys ;) I've learned from last weeks Lab, not to leave it till last, but to try and understand the problem at the beginning of the week and work through the lecture sequences with the problem in mind! That way you don't get caught off guard like me! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-09-24T19:49:09Z IndexTAG: 48 TitleTAG: Fullscreen video problem When I click the "Fill browser" button during a video, the video fills the browser screen, but the annotations and video selection bar (the bar on the top) stay where they are, causing them to be in the middle of the video. Browser used: Firefox 15.0 Flash version: 11.4 r402 UserIdTAG: 106837 UserNameTAG: rgiesler CreateTimeTAG: 2012-09-05T12:34:44Z VoteTAG: 23 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 3 FirstChildTAG: I also have the same bug. Browser: Rockmelt 0.16.91.483 FirstChildUserIdTAG: 310007 FirstChildUserNameTAG: ErinF FirstChildCreateTimeTAG: 2012-09-06T20:08:28Z FirstChildTAG: I can confirm this problem. I have the same bug. - Browser: Google Chrome 21.0.1180.89 - OS: Ubuntu 12.04 - Flash version: 11.2 r202 **Possible fix:** Change "z-index" property of the ".sequence-list-wrapper" CSS class from "9999" to "999". FirstChildUserIdTAG: 190466 FirstChildUserNameTAG: kop FirstChildCreateTimeTAG: 2012-09-05T12:47:48Z FirstChildTAG: I have the same problem no matter what OS (tried Windoze and GNU/Linux) or browser (Chrome, Firefox, IE, Epiphany) I use. FirstChildUserIdTAG: 93403 FirstChildUserNameTAG: tdsamardzhiev FirstChildCreateTimeTAG: 2012-09-06T21:49:43Z IndexTAG: 49 TitleTAG: edX model and 50 plus students First of all, a big thank you Prof Agarwal and your team for teaching a wonderful course 6.002X. I got introduced to edx during a discussion with NMEICT (National Mission for Education through ICT, India) friends. I looked up the web and when I saw 6.002X listed, I registered. I am a 57, going 58, years old (wonder how many of us on the wrong side of fifty were there in this course) who did his B Tech in Electronics and Communications in 1975. I thought I will just have a look at some course material to get a feel of the online courses and then move on. But I was totally hooked. The course is a pedagogic revelation. We do have nptel (National Programme for Technology Enhanced Learning) videos on NMEICT portal ( www.sakshat.ac.in). They are by, mostly, IIT Professors and are good no doubt, but are (from what I saw) monolithic, not broken down into small capsules, last upwards of 40 minutes and are not backed by any homework or lab or finger exercises. This does not make for a product which is pedagogically sound. But happily,the things are changing. I just visited their website. They have announced online courses, with tests, exams, assignments beginning Jan 2013. Good effect of edX model, I am sure. After being in the field for these many years (though not doing much technical work for years now), I want to confess that your course gave me many insights and aha moments. Wish I had teachers like you when I was an undergrad myself. With best wishes. A great 2013 to you, your family and your edX team. (BTW, I will treasure an A from MITx, though I could not get 100% in the final). UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-12-27T04:07:56Z VoteTAG: 22 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: You are on the wiser side of the 50;-) and electrical engineering courses now look rather like applied math courses. FirstChildUserIdTAG: 623210 FirstChildUserNameTAG: preveen FirstChildCreateTimeTAG: 2012-12-27T10:05:57Z FirstChildTAG: I'm 57 and I tried a few years ago to follow some of the nptel courses. They are very interesting, and they offered a lot of courses, that's not the point. What I have to say is not ment disrespectfully, but the teachers spoke often so slow! I've add/adhd and I wasn't able to hangon for 50 minutes, and I need a book or some other materials. I was taking notes, but I gave up after a few months, because every time I had to go back a few seconds in the video, but it jumped back halve a minute or more, so I had to listen over and over again. And it was sometimes very difficult for me to understand what the teachers said, because some of them had a very heavvy accent. Who doesn't? But some accents are easy to decode and others are not. And there was no homework, so I couldn't check if I comprehended the material. On the other hand, some teachers on mit/edx speak sometimes very fast, too fast for me, so I've to playback again. In fact, I skipped most videos, because I prefer to read and make notes in my book. I also had that problem of long videos with the mit/ocw material, but it was much better, because you could download some pdfs. But I also need the books at hand. For me video in chunks of 5 minutes is doable. I like this system from edx/mit very much, but I have problems with the deadlines, so now I try Udacity too. But I think all the course teams are doing a great job and I understand that it's almost impossible to create a course in such a way that the whole world can be satisfied. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-12-27T14:37:27Z SecondChildTAG: I strongly recommend to watch the videos of prof anant agarval. they will give you insight into the material 10x times reading and in fact enhance your reading very much. Did you try the speed adjustment and the closed captions to help you with comprehension? SecondChildUserIdTAG: 623210 SecondChildUserNameTAG: preveen SecondChildCreateTimeTAG: 2012-12-27T16:43:37Z SecondChildTAG: Glad to hear from another 50+ young. Glad that you actually tried nptel videos. I hope they emulate edX and make the material more learning friendly. There are some early signs that they are bringing in changes of format, making it more like edX. Will have to wait to see how it rolls out. SecondChildUserIdTAG: 190670 SecondChildUserNameTAG: chauhan1955 SecondChildCreateTimeTAG: 2012-12-27T17:07:25Z SecondChildTAG: To preveen: Yes, I tried in the spring course, but somehow I can't memorize more than a few seconds of what I see on the video's. Has something to do with my short-term memory (I had 10 ect treatments the year before and it still has negative effects on my memory). And in combination with my attention problem and bad eyesight, it's a problem to make notes at the sametime, because I've to switch glasses everytime. Reading text from the online book was also not a success, so I bought the book and that made life much easier. It's a personal problem that edx cannot solve for me. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-12-27T17:41:28Z SecondChildTAG: To chauhan: nptel offered a lot of interesting courses and I'll certainly take a look again in the near future, to see what changes they have made. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-12-27T17:46:55Z SecondChildTAG: @salsero I am sure that your attempts to keep learning will improve your memory, attention, and eyesight. You brain will figure out what you want and will rewire itself to help you with your endeavors. Keep learning. thank you for sharing your experiences SecondChildUserIdTAG: 623210 SecondChildUserNameTAG: preveen SecondChildCreateTimeTAG: 2012-12-27T18:10:40Z SecondChildTAG: preveen is right, salsero, and this is one reason we all ought to challenge our brains to keep Alzheimer's away. Try to learn a new trick every day:) & watch the brain re-wire. Best wishes to you all. SecondChildUserIdTAG: 285616 SecondChildUserNameTAG: AmiyaX SecondChildCreateTimeTAG: 2012-12-28T08:29:23Z SecondChildTAG: Glade to read comments by contemporary 50+, Just 59 and half. SecondChildUserIdTAG: 455950 SecondChildUserNameTAG: Wahabbaluch SecondChildCreateTimeTAG: 2012-12-28T12:42:55Z FirstChildTAG: I am the other side of 50 and have a degree in Computing and Electronics from 1982! I did this course as a refresher in electronics since I've worked on the computing side since I graduated. The course 'attacked' the subject matter completely differently from what I'd done before. It was much more challenging to me than it should have been (not a criticism of the course, more of my memory!) The course builds a solid foundation for more advanced topics in electronics. Given that the field of electronics is so broad, it cannot cover everything, but rather gives you tools to analyze devices that aren't covered. Rather than saying here is a capacitor or inductor and this is the equation that describes it, capacitance and inductance are introduced through the real problems of crosstalk and ringing. I still find it amazing that we got all the way through impedance without 'reactance' being mentioned! The analysis of the MOSFET and circuits containing it gives a good foundation in analyzing non-linear devices. The messy quadratics aren't fun, but give essential practice in circuit analysis. The fake devices we analyzed built on this foundation nicely. One has the tools to analyze a circuit containing a new device. Some problems were surprising. The antenna tuner - so that's how an antenna tuner works! The 'scope probe. Real rather than abstract problems. Even though I already knew the basics of op-amp circuits, going through it again was instructive. I finally understood the integrator, differentiator and logarithmic amplifier. Again, the course progression made it easy to analyze these circuits and derive my own answer rather than being presented with a transfer function and wondering where did that come from. I understand that this is just a foundation. I hope that the more advanced electronics courses will be available in this format and look forward to doing them if the do. FirstChildUserIdTAG: 141000 FirstChildUserNameTAG: OrinE FirstChildCreateTimeTAG: 2012-12-27T20:21:39Z SecondChildTAG: Hi OrinE. Thank you for taking the time to help us on the forum with the homework. I appreciated your posts a lot. SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-12-30T00:42:43Z SecondChildTAG: Glad that I was of help. Good luck with your endeavors in Electronics! SecondChildUserIdTAG: 141000 SecondChildUserNameTAG: OrinE SecondChildCreateTimeTAG: 2012-12-30T05:04:02Z IndexTAG: 50 TitleTAG: H5P2 Hints requested by maraivette (for Spanish translation see down). Hi maraivette, Sorry for the delay, you ask me some doubts in spanish in this [Post][1]: *In your comments you told me that you were a little bit confused on how to replace one equation into other and reach to the mentioned quadratic equation.* As I have promised to you, I will help you so later you can solve it by your own. Ok, lets start :)! ---------- **H5P2 SOURCE FOLLOWER LARGE SIGNAL** One of the most valuable of the amplifier configurations is the source follower, shown in the figure below: ![imagen][2] As usual, the MOSFET **is set up to** operate in the **saturated region**. **Part 1:** Write an algebraic expression for iDS in terms of **K**, **vIN**, **vOUT**, and **VT**. Remember, algebraic expressions are case sensitive. *Warning! : be careful while entering the variables:* - **vIN** it is not VIN or vin - **vOUT** it is not VOUT or vout - **K** it is not writing k - **VT** is not vt **Hint Part1:** Take a look at this [page 341][3] of the Textbook, you will find how is the drain source current (iDS) in the saturated region. Your question ask you to find that current in terms of iDS in terms of K, vIN, vOUT, and VT , but the expression that you have found reading that page of the Textbook it is not expressed by the requested terms, so, hmmm..., Can you find a way to express them in function of K, vIN, vOUT, and VT? Remember this: vDS is the Drain Source voltage, it is the blue one in the following image. vGS is the Gate Source voltage, it is the green one in the following image. ![image][4] Now, take a look at the light blue loop. ![imagen2][5] So, if you use KVL, can you relate vIN with vOUT and vGS? Hmmm.... wait a minute, if I know now vIN in function of vGS... And I also have iDS in function of vGS... Can I reach in someway to the answer of this Part 1? Yes! :). ---------- **Part 2:** Write an algebraic expression for vOUT in terms of iDS and RS. **Hint Part 2:** Where is vOUT in this circuit? What current pass through there? Can I use Ohm's Law? yes! ---------- **Part 3**:Solve the previous two results for iDS. You will get a quadratic equation, and you will have to choose the correct root: make sure that iDS=0 when vIN=VT. What is your solution for iDS? Hi maraivette, here was your difficultie, ok, let me help you here. As the statement says, you will have your iDS from Part 1 and also you will have your vOUT expressed in terms of iD from part 2. **Hint Part 3:** From Part 1, you will have iDS in function of K, vOUT,vIN and VT; in Part 2 you will have your vOUT expressed in function of iDS. So, can you "mix" both expressions in order to get iDS not in function of vOUT? yes! Here you should review quadratics equations, lets see some concepts that might can help you: Suppouse that you have this quadratic equation: ![quadratica][6] And also suppouse that A, B and C coefficients are constants, that is to say their value is a number, like, 1, 56, 3.4, etc... You have your variable "x", so, if you have this equation, what will the value of x that verifies the equation, verifies means that if you replace your quadratic with the x value that you got from solving the quadratic the result in the equation it will be zero. What are the possibles x values for that quadratic equation? One value is: $x1 = \frac{ -B + \sqrt{B^2 - 4*A*C}}{ 2*A}$ The Other value is: $x2 = \frac{ -B - \sqrt{B^2 - 4*A*C}}{ 2*A}$ Lets see an arbitrary numerical example: 2x^2 +4x-6=0 A=2 B=4 C=-6 $x1 = \frac{ -4 + \sqrt{4^2 - 4*2*(-6)}}{ 2*2} $ x1= 1 $x2 = \frac{ -4 - \sqrt{4^2 - 4*2*(-6)}}{ 2*2} $ x2 = -3 That is to say with x1 value: 2*X1^2 +4*x1-6=0 2*1^2+4*1-6=0 verifies! That is to say with x2 value: 2*X1^2 +4*x1-6=0 2*(-3)^2+4*(-3)-6=0 verifies! Now, they ask you to choose the correct iDS, that is to say, as iDS will have an iDS1 an iDS2 , like you previously saw with x1 and x2. You will have to find the expression when iDS=0 when vIN=VT. If you have you quadratic, Can you tell which will be the correct iDS? The value of iDS1 or the value of iDS2? Yes! **Part 4,5and 6 :** You can get it if you have done the previous step ;). **Part 7:** In order to allow this much swing on the output voltage, we must supply a sufficiently high supply voltage. What is the minimum value of VDD (in Volts) that we must supply to keep the transistor in the region of saturated operation? Hint Part 7: Take a look at this page of the Textbook 341, where is the saturated region in function vDS in function of VT and vGS? Can you relate with KVL vDS(vIN,vOUT)? Yes! Can you relate with KVL vGS(vIN,vOUT)? Yes! So, which will the value of vIN that you should consider? Is it the small one or the big one? So, once you have your vIN can you find the minimun value of VDD? Yes! I hope this could be helpful for you. If you have any doubt please ask me. :) Myriam. ----- Now in Spanish, Hola maraivette, Discúlpame por la tardanza de esta respuesta a tu duda, que tú me has planteado aquí [Post][1]: *“hola espero estés bien aquí solicitando de tu ayuda estoy atorada en la tarea 5 problema 2 en la 3er pregunta en donde hay me tiene que salir una ecuación cuadrática estoy confundida con el procedimiento me podrías ayudar gracias”* Tal y como te he prometido, te voy a ayudar para que luego puedas resolverlo tú misma. Bien, empecemos : ) ! ---------- **H5P2 SOURCE FOLLOWER LARGE SIGNAL** Una de las configuraciones que un amplificador tiene es el de seguidor de fuente, que se muestra en la figura de abajo: ![imagen][2] Por lo general, el MOSFET se encuentra diseñado para que opere en su región de saturación. **Parte 1:** Escriba una expresión algebraica de iDS en términos de **K**, **vIN**, **vOUT**, and **VT**. Recuerde que, las expresiones algebraicas son sensible a las mayúsculas y minúsculas. *Warning! : Ten cuidado cuando escribeslas variables* - **vIN** no es lo mismo que VIN or vin - **vOUT** no es lo mismo que VOUT or vout - **K** no el lo mismo que k - **VT** no es lo mismo que vt **Hint Parte1:** Mira la página 341 del Textbook [leer aquí][3], aquí encontrarás cómo se define la corriente drenaje fuente (iDS) en su region saturada. En este enunciado se te solicita que tú puedas hallar la corriente iDS en términos de K,vIN,VOUT y VT. Sin embargo, si has visto la expresión que te he indicado en el párrafo anterior, versa que no se encuentra expresada en los términos que se solicitan, mmmm…, entonces, Puedes hallar alguna forma de expresarlas en función de K, vIN, vOUT, y VT? Recuerda esto: vDS es la tension drenaje fuente, es el que te he marcado con azul en la imagen de abajo. vGS es la tension fuente compuerta, que te he marcado en la imagen de abajo. ![image][4] Ahora, te recomiendo que mires el loop celeste. ![imagen2][5] Entonces, si utilizas KVL, Puedes reacionar vIN con vOUT y vGS? Mmm… Un momento, si tú sabes vIN en function de Vgs… Y si también tienes la iDS en function de vGS… Puedes resolver de alguna forma esta Parte1? Sí! : ) ---------- **Parte 2:** Escribir una función algebraica de vOUT en términos de iDS y RS. **Hint Parte 2:** En dónde se encuentra vOUT? En dónde está “colgado”? Qué corriente atraviesa por allí? Puedes usar la Ley de Ohm? Sí! ---------- **Parte 3**: Resuelve lo anterior. Obtendrás una ecuación cuadrática, y deberás elegir el resultado correcto de raíz: asegúrate que iDS=0 cuando vIN=VT. Cuál es la solución para iDS? Maraivette, aquí estaba tu dificultad, no? Déjame ayudarte en esta parte. El enunciado dice que, tu tendrás tu iDS de la Parte 1 y que también tendrás tu vOUT expresada en terminus de iD de la part 2. **Hint Parte 3:** De la parte 1, tendrás iDS en función de K, vOUT,vIN and VT; en la Part 2 tendrás tu vOUT expresada en función de iDS. Entonces, puedes de alguna forma mezclar esas dos expresiones para obtener una iDS pero que no esté en función de vOUT? sí! Aquí deberías revisar el concepto de ecuaciones cuadráticas, veamos algunos conceptos que quizás puedan ayudarte: Supongamos que tienes la siguiente ecuación cuadrática: ![quadratica][6] También, que tenemos A,B y C como coeficientes. Recuerda que estos son constants y tienen un valor numérico, tal y como 1,56,3.4, etc… También tienes una variable “x”. Entonces, si tienes esta ecuación cuadrática, cuál es valor de x que verifica la ecuación?, verificar la ecuación significa que al reemplazar el valor que has obtenido de x en esa ecuación cuadrática en cada sumando te deberá dar como resultado cero. Entonces, cuáles son los posibles valores de x para dicha ecuación cuadrática? Un valor posible es: $x1 = \frac{ -B + \sqrt{B^2 - 4*A*C}}{ 2*A}$ El otro es: $x2 = \frac{ -B - \sqrt{B^2 - 4*A*C}}{ 2*A}$ Veamos un ejemplo numérico arbitrario: 2x^2 +4x-6=0 A=2 B=4 C=-6 $x1 = \frac{ -4 + \sqrt{4^2 - 4*2*(-6)}}{ 2*2} $ x1=1 $x2 = \frac{ -4 - \sqrt{4^2 - 4*2*(-6)}}{ 2*2}$ x2=-3 Esto quiere decir que con el valor de x1 en nuestra cuadrática nos dará: 2*X1^2 +4*x1-6=0 2*1^2+4*1-6=0 verifica que sea cero! Esto quiere decir que con el valor de x2 en nuestra cuadrática nos dará: 2*X1^2 +4*x1-6=0 2*(-3)^2+4*(-3)-6=0 verifica que sea cero! Ahora, se te pide hallar el valor correcto de iDS. Tal y como ellos te dicen, la expresión de iDS será una cuadrática y, como toda cuadrática, tendrá dos posibles valores, es decir, iDS tendrá una iDS1 y una iDS2 , tal y como hemos visto con el ejemplo de x1 y x2. Entonces, deberás hallar la expresión tal que tu iDS sea nula cuando tu vIN=VT. Entonces, si tienes tu cuadrática, puedes decir cuál será la expresión de iDS que cumpla con lo solicitado? Será la expresión de iDS1 o la de iDS2? Sí! **Parte 4,5 y 6 :** Lo puedes obtener si has hecho los pasos previos ;). **Parte 7:** Para poder permitir esta fluctuación en la tensión de salida, se debe tener una tensión de alimentación suficiente. Cuál es el mínimo valor de VDD (en volts) que se debe proveer para mantener al transistor en la zona de saturación? Hint Parte 7: Mira nuevamente la página 341, En dónde se encuentra la región de saturación expresada en vDS en función de VT y vGS? Puedes relacionar con KVL tu vDS(vIN,vOUT)? Sí! Puedes relacionarla con KVL vGS(vIN,vOUT)? Sí! Entonces, cuál será el valor de vIN que deberás considerar? Es el valor más grande o el más chico que te han dado? Entonces, una vez que tienes vIN puedes hallar el valor mínimo de VDD? Sí ! Espero que te haya sido de ayuda. Cualquier duda pregúntame.:) Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5076d96c4dd0a9270000008f [2]: https://www.edx.org/static/content-mit-6002x/images/circuits/source-follower.7552cff69ae2.gif [3]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/365 [4]: https://edxuploads.s3.amazonaws.com/13501487967705749.png [5]: https://edxuploads.s3.amazonaws.com/13501489291343616.png [6]: https://edxuploads.s3.amazonaws.com/1350150223134369.png UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2012-10-13T19:40:03Z VoteTAG: 22 CoursewareTAG: Week 5 / Large Signal Analysis CommentableIdTAG: 6002x_large_sig NumberOfReplyTAG: 10 FirstChildTAG: thanx myrimit for awesome explanation. FirstChildUserIdTAG: 269641 FirstChildUserNameTAG: BAUWA FirstChildCreateTimeTAG: 2012-10-13T19:48:20Z SecondChildTAG: You are welcome BAUWA! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-13T19:56:33Z FirstChildTAG: Thank u Myrimit u flash my memory FirstChildUserIdTAG: 318037 FirstChildUserNameTAG: Maher-84 FirstChildCreateTimeTAG: 2012-10-13T21:46:20Z SecondChildTAG: Hahaha! Thank you Maher-84! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T00:16:06Z FirstChildTAG: thanks Myrimit , i got my expression of iDS it is (1+(K*RS*(vIN-VT))-sqrt(1+2*K*RS*(vIN-VT)^2))/(K*RS^2) , is it wrong , i solved it samee way you explained and i chosed the root that will make ids =0 when Vin = VT FirstChildUserIdTAG: 285675 FirstChildUserNameTAG: Ascot FirstChildCreateTimeTAG: 2012-10-13T22:14:20Z SecondChildTAG: Hi Ascot, There something wrong the, take a look when vIN=VT in your expression: iDS differnt from 0... Remember when vIN=VT it is the same as (vIN-VT)=0 ... and 1/K*RS^2 is not zero...iDS is not zero for vIN=VT, so you are missing something... Your expression is close, but you are missing something in one of the terms and try to reduce it at the minimun expression as posible, if you can factorize terms like (vIN-VT) , etc... I hope this can help you. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T00:42:24Z SecondChildTAG: my equation is iDS=[{RS(vIN-VT)+2/K}-sqrt{(4RS/K)(vIN-VT)+(2/K)^2}]RS^2 and it satisfies vIN-VT=0 but it says invalid input. why? SecondChildUserIdTAG: 259693 SecondChildUserNameTAG: MehrozKhan SecondChildCreateTimeTAG: 2012-10-14T07:33:51Z FirstChildTAG: Hola myri, muchas gracias por tu explicación en español, la necesitaba. He podido avanzar con el problema 2, pero en la tercera pregunta he quedado pillado. Hago el siguiente procedimiento, reemplazo el valor de vout que corresponde, y me queda una sóla ecuación en términos de ids, una ecuación cuadrática tal como dice el problema. Sin embargo, cuando la resuelvo para vin=vt, no obtengo el resultado correcto. Hay algo que no estoy entendiendo bien del problema, y debe estar relacionado a la condición de que ids = 0. No me llevo muy bien con el inglés, te agradecería si me pudieras ayudar a comprender bien que pide esa pregunta. Saludos FirstChildUserIdTAG: 87950 FirstChildUserNameTAG: neoacademic FirstChildCreateTimeTAG: 2012-10-13T22:27:18Z SecondChildTAG: Hola neoacademic! Sí, no hay problema. Pregunta, estoy aquí para ayudarlos lo más que pueda. Bien, si ya tienes la ecuación cuadrática de iDS, verás, que como toda equación cuadrática tendrá dos raíces iDS1 e iDS2 (estas dos raíces son análogas al ejemplo que di de la cuadrática, x1 y x2). Bien, aquí es lo que por ahí no se entiende mucho, y más se complica al hablar en otro idioma, te comprendo... La idea es que ellos te dan una condición que tu ecuación cuadrática debe satisfacer, es decir, si bien las dos ,iDS1 e iDS2, funcionan "matemáticamente" ellos te solicitan aquella raíz que verifique su comportamiento "físico", es decir para cuando vIN=VT o bien es lo mismo que decir (vIN-VT)=0 su corriente de drenaje fuente sea igual a cero... Entonces, ahora, básicamente, tu tarea sería hallar cuál de esas dos expresiones es la correcta. Como, pista, verás que sólo una verificará esa condición... entonces, la pregunta es: cuál es la ecuación que satisface las condiciones al reemplazar vIN=VT? la que surge de ...+...sqrt(...) o la que ...-...sqrt(...)? Espero que te haya sido de ayuda, Saludos! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T00:53:54Z SecondChildTAG: Myri, muchas gracias por tu respuesta. Al final pude dar con la repuesta, sin embargo era otro el error que estaba cometiendo, al considerar la ecuación para Ids. De todos modos, tus respuestas en este apartado y otros me han ayudado demasiado, y aunque tardé. creo que nunca es tarde para agradecer . Ahora iré por la semana 6, cualquier duda que tenga, ya sé de alguien que habla español , jaja :p. Saludos SecondChildUserIdTAG: 87950 SecondChildUserNameTAG: neoacademic SecondChildCreateTimeTAG: 2012-10-15T15:50:54Z FirstChildTAG: Myrimit, thank you so much! Your post helped me to distinguish Drain and Source of that mosfet... FirstChildUserIdTAG: 326787 FirstChildUserNameTAG: Angstrem FirstChildCreateTimeTAG: 2012-10-13T21:07:40Z SecondChildTAG: :) You are welcome Angstrem ! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T00:15:47Z FirstChildTAG: GRACIAS POR LA AYUDA Y SI NO ESTOY MAL PARA LA ULTIMA PARTE TENGO QUE DERIVAR ESTA MISMA FUNCION DE VOUT * VIN FirstChildUserIdTAG: 320715 FirstChildUserNameTAG: maraivette FirstChildCreateTimeTAG: 2012-10-14T04:50:54Z SecondChildTAG: Por nada maraivette :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T14:42:20Z FirstChildTAG: Hi Myrimit, lots of thanks for your support. I erased my post I wrote before because I thought that, writting my possible answer could have been against the honor code. But I'm sure I was far away from the correct answer. I amended my k and K and still wrong :( I'm sure I might having problems with my maths, anyway, will keep on trying and tomorow will see the correct answer. Thankssssss!!!! Sandra FirstChildUserIdTAG: 342966 FirstChildUserNameTAG: SandraNavarro FirstChildCreateTimeTAG: 2012-10-14T16:03:04Z SecondChildTAG: Hi SandraNavarro, You are welcome. Please If you have any doubts I would like to help you :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T19:00:36Z SecondChildTAG: You are already helping a lot!! I write in english just in case other students have the same questions, although we have the same mother tongue :)) Question: for the grading, homework 15% and labs 15%. Are video lectures and tutorials exercises also included? I hope no :-P Gracias!! Sandra SecondChildUserIdTAG: 342966 SecondChildUserNameTAG: SandraNavarro SecondChildCreateTimeTAG: 2012-10-14T21:39:20Z SecondChildTAG: Hi Sandra! No, remember that in this course the Final Score will be based only on Homeworks, Labs and Exams (Midterm and Final). *Letter grades will be based on the following weighting: homework 15%, labs 15%, midterm 30%, and final exam 40%. Each of the homework and labs carries equal weight. You will need to get a total mark of 60% for a C, 70% for a B, and 87% for an A.* [read here - syllabus][1] ---- Hola Sandra! Te respondo en Español. No se considerarán los ejercicios o exercises (son los que aparecen en la secuencia de los videos de las lecturas) para la calificación-puntaje final. Sí lo harán los Homeworks, Labs y Examenes (Midterm y Final). Puedes leer en syllabus, allí dice que la ponderación para la puntuación final será 15% para los homeworks, 15% para los labs, 30% para el midterm, y 40% para el final. Necesitarás sumar un total de 60% para un C, 70% para un B y un 87% para un A. [leer aquí - syllabus][1] [1]: https://www.edx.org/static/content-mit-6002x/handouts/syllabus.a477535058a1.pdf SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T22:54:55Z SecondChildTAG: Gracias Myriam :)))) Thanks a lot!!!!!!!!!!!!!!!! SecondChildUserIdTAG: 342966 SecondChildUserNameTAG: SandraNavarro SecondChildCreateTimeTAG: 2012-10-15T06:09:07Z FirstChildTAG: Hi Myrimit I solve all the part in this home work except part 7 can you help me FirstChildUserIdTAG: 226085 FirstChildUserNameTAG: Teto FirstChildCreateTimeTAG: 2012-10-14T23:25:50Z SecondChildTAG: all the three value of vIN give me same answer SecondChildUserIdTAG: 226085 SecondChildUserNameTAG: Teto SecondChildCreateTimeTAG: 2012-10-14T23:26:26Z SecondChildTAG: I used VDD=vIN-vOUT-VT is that correct SecondChildUserIdTAG: 226085 SecondChildUserNameTAG: Teto SecondChildCreateTimeTAG: 2012-10-14T23:27:10Z SecondChildTAG: Hi Teto, I can not tell you if your formula it is correct or not... Hint 1: Are you sure that you are getting correctly the KVL equations of vDS(vIN,vOUT) and the vGS(vIN,vOUT)? Hint 2: If in someway you can relate both equation that involves vDS(vIN,vOUT) and vGS(vIN,vOUT), and both have the vOUT but in different sides ... What happened with the vOUT? ;). I hope this can help you. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T23:42:40Z SecondChildTAG: give me few minute I'll try something SecondChildUserIdTAG: 226085 SecondChildUserNameTAG: Teto SecondChildCreateTimeTAG: 2012-10-14T23:47:23Z SecondChildTAG: I still can not find the answer ok look at my steps please. first, I take the loop in the right and apply KVL I get VDD=vIN+vOUT-VT then I just substitute by the valure that we have in part's 4,5,6 and see which one is correct SecondChildUserIdTAG: 226085 SecondChildUserNameTAG: Teto SecondChildCreateTimeTAG: 2012-10-14T23:54:33Z SecondChildTAG: O.K O.K I found it thank you very much for your help :) to be honest I tried to many ways I would like to ask you about it if it may be in some way correct. SecondChildUserIdTAG: 226085 SecondChildUserNameTAG: Teto SecondChildCreateTimeTAG: 2012-10-15T00:08:24Z SecondChildTAG: Hi Teto, I can not tell you if your procedure it is ok. Remember that you should also evaluate the saturation condition too... Anothers Hints: Hint: Take a look at (7.10) of the Textbook [here][1]. Hint: Ok, now read again the two hints that I have wrote you in the previous comment;) Hint: vDS and vGS are: ![image][2] Hint: Can you find vDS and vGS? So, can you find vIN with the saturation condition? yes! [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/365 [2]: https://edxuploads.s3.amazonaws.com/13501487967705749.png SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-15T00:16:07Z SecondChildTAG: Well done Teto! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-15T00:18:09Z SecondChildTAG: for example can we use the current equation of the mosfet to solve this problem?? SecondChildUserIdTAG: 226085 SecondChildUserNameTAG: Teto SecondChildCreateTimeTAG: 2012-10-15T00:18:38Z SecondChildTAG: thanks a lot Myriam. SecondChildUserIdTAG: 226085 SecondChildUserNameTAG: Teto SecondChildCreateTimeTAG: 2012-10-15T11:29:00Z FirstChildTAG: Hay Myriam You sure are a wealth of information - Thanks I have come up with the quadratic equation (checked it 3 times now), and I want to make sure I'm calculating it properly when they say, "make sure that iDS=0 when vIN=VT". Everywhere in the quadratic equation where vIN-VT (vIN-vIN) occurred, I set it = 0, which greatly reduced the equation. Is this the correct way to handle this? Also, I'm not sure how to handle setting iDS =0. In the quadratic equation, when iDS = (-b+sqrt(b^2-4*a*c))/2*a do I set iDS=0 so I have 0 = (-b+sqrt(b^2-4*a*c))/2*a, then simplify? Somehow I don't believe this is correct because I get 1 +/- 1 = 0. FirstChildUserIdTAG: 132828 FirstChildUserNameTAG: rharris FirstChildCreateTimeTAG: 2012-10-15T04:08:16Z SecondChildTAG: By the way, here is my quadratic equation: iDS = [(K*RS*(vIN-VT)+1)+sqrt([K*RS*(vIN-VT)+1]^2-K^2*RS^2* (vIN-VT)^2)]/(K*RS^2)) Do you see anything wrong? SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-15T04:24:09Z SecondChildTAG: Hi rharris, The equation is reduced doing vIN=VT, this is only in order to verify the statement condition only and to choose the correct iDS equation...Remember that iDS is a quadratic equation and it has 2 possible results. So, the statement ask you which of those two is the one that when you evaluate vIN=VT is iDS=0... SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-15T11:39:56Z FirstChildTAG: this is my first time using the discussion in this course. It was soooo helpful. Big thanks to you Myriam. FirstChildUserIdTAG: 226085 FirstChildUserNameTAG: Teto FirstChildCreateTimeTAG: 2012-10-15T11:33:23Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-15T11:35:15Z IndexTAG: 51 TitleTAG: S8E2 Closed Form Solution Here's my scanned solution. Maybe someone would find it useful. It nicely shows, that $R_{TH}$ is independent from value of the external excitation. It's not neccessary to assume any exact value, as I've seen in other posts. ![Closed Form Solution S8E2][1] [1]: https://edxuploads.s3.amazonaws.com/13495771101343674.png UserIdTAG: 4076 UserNameTAG: damians CreateTimeTAG: 2012-10-07T02:33:29Z VoteTAG: 22 CoursewareTAG: Week 4 / Dependent Sources Exercise 1 CommentableIdTAG: 6002x_dep_src_exsz_1 NumberOfReplyTAG: 7 FirstChildTAG: Great job! Thanks! FirstChildUserIdTAG: 301141 FirstChildUserNameTAG: Smaragda FirstChildCreateTimeTAG: 2012-10-07T07:56:55Z FirstChildTAG: Thanks a lot for this. Very helpful indeed! FirstChildUserIdTAG: 194509 FirstChildUserNameTAG: blackcompe FirstChildCreateTimeTAG: 2012-10-07T09:11:04Z FirstChildTAG: How did you solve the second part?? I got the relation between Vth and U. But i'm unable to find U.. FirstChildUserIdTAG: 309722 FirstChildUserNameTAG: kaushikraghavan1992 FirstChildCreateTimeTAG: 2012-10-07T12:03:15Z SecondChildTAG: Just do another node analysis for circuit b), compute the potential at $e_2$ and divide by $i_{TH}$. SecondChildUserIdTAG: 4076 SecondChildUserNameTAG: damians SecondChildCreateTimeTAG: 2012-10-07T15:37:04Z FirstChildTAG: For RTH I didn't take ccvs into consideration :) and RTH=R1*R2/(R1+R2)=398.4375 is correct answer. U're right, RTH=(R1+alp)*R2/(alp+R1+R2), But u could solve it easier. Cause RTH is just R1 + alp parallel with R2 FirstChildUserIdTAG: 215271 FirstChildUserNameTAG: JustStudent FirstChildCreateTimeTAG: 2012-10-07T09:22:24Z SecondChildTAG: Yes, but I wanted to show how to do it in a direct way and that $i_{TH}$ really doesn't matter. The easy solution would be to calculate the resistance of our dependent voltage source A, which is voltage across $V_A=\alpha i$, divided by the current $i$. We get $R_A=\frac{\alpha i}{i}=\alpha$. From that is follows that $R_{TH}=(R_1+R_A)||R2=(R_1+\alpha)||R2$, where $||$ means parallel connection. SecondChildUserIdTAG: 4076 SecondChildUserNameTAG: damians SecondChildCreateTimeTAG: 2012-10-07T15:48:41Z FirstChildTAG: Whi I=e1/r1? FirstChildUserIdTAG: 214083 FirstChildUserNameTAG: rosvld FirstChildCreateTimeTAG: 2012-10-13T09:15:25Z FirstChildTAG: good one . thaks FirstChildUserIdTAG: 296303 FirstChildUserNameTAG: DEEPatXUniv FirstChildCreateTimeTAG: 2012-10-17T10:47:13Z FirstChildTAG: Thax dude FirstChildUserIdTAG: 303485 FirstChildUserNameTAG: iparitosh FirstChildCreateTimeTAG: 2012-10-27T14:41:47Z IndexTAG: 52 TitleTAG: Thevenin/Norton Equivalent with a Dependent Source: A Worked Example I noticed that there was a distinct lack of worked examples of finding Thevenin/Norton equivalents when dependent sources are present. I feel like I really could have used one that didn't trivially eliminate the dependent source, so I'm going to do my best to present one here. If anybody catches a mistake or something please let me know so I can fix it! Let's examine the following circuit, and assume we are asked to find the Thevenin or Norton Equivalents. ![Thevenin Voltage Schematic][1] **Thevenin Voltage:** To first find $V_{TH}$, leave the terminal open circuited and solve for $v_p$ (which is $V_{TH}$ under these conditions since $v_p = V_{TH}+I_{terminal} \cdot R_{TH}$ and leaving the terminal open-circuited forces $I_{terminal}=0$). KCL equations are: $ \cfrac {v_0-(-100)}{10k} + \cfrac {v_0-v_p}{22k}+ \cfrac {v_0}{40k} = 0 $ and $ \cfrac {v_p-v_0}{22k} + \cfrac {v_p}{30k} - (0.003 \cdot v_0) = 0 $ with some algebra we can solve the first equation such that $v_0 = \cfrac {4}{15} \cdot v_p - \cfrac {176} {3} ~~~~\boxed {eq1}$ substituting into the second eqn and solving for $v_p$ gives $ \boxed {v_p = V_{TH} = \cfrac {2680}{11} = 243.64 V}$ **Norton Current:** To find $I_N$, short circuit the terminal and figure out what the current across the short should be. ![Norton Current Schematic][2] The KCL equations look similar to last time, with an additional current through the short circuit (and no current through the 30kΩ resistor): $ \cfrac {v_0-(-100)}{10k} + \cfrac {v_0-v_p}{22k}+ \cfrac {v_0}{40k} = 0 $ and $ \cfrac {v_p-v_0}{22k} - (0.003 \cdot v_0) - I_N = 0 \boxed {eq2}$ And we have an extra special piece of information because of the short we've implied: $v_p =0$(!) From the rearranged $ \boxed {eq1} $ above this tells us that $v_0 = - \frac {176}{3} $, and substituting those two pieces of information into $ \boxed {eq2}$ and solving for $I_N$ gives: $ \boxed {I_N = \cfrac {67}{375} = 0.1787 A} $ **Thevenin/Norton Resistance:** Here's the clever part! Now we do some clever manipulation of: $v_p = V_{TH} + I_{terminal} \cdot R_{TH}$ First we turn the independent sources to zero (in this case that means short-circuiting the 100V source). That sets $V_{TH}=0$ in our new circuit. Now let's add a 1A independent current source, making $I_{terminal}=1A$. This means if we figure out $v_p$ we'll know $R_{TH}$! ![Thevenin/Norton Resistance Schematic][3] Again, our KCL equations look very similar with only 2 minor changes: $ \cfrac {v_0}{10k} + \cfrac {v_0-v_p}{22k}+ \cfrac {v_0}{40k} = 0 $ and $ \cfrac {v_p-v_0}{22k} + \cfrac {v_p}{30k} - (0.003 \cdot v_0) + 1 = 0 $ Rearranging the first equation gives: $ v_0 = \cfrac {4}{15} \cdot v_p$ Substituting the results into the second equation gives: $ \boxed {v_p = R_{TH} = \cfrac {15k}{11} = 1.36 k \Omega} $ **Sanity Check:** We've been told that the Norton current, Thevenin voltage, and corresponding resistance should obey Ohm's law, so let's check that: $ R_{TH} = \cfrac {V_{TH}}{I_N} = \cfrac {\cfrac {2680}{11}}{\cfrac {67}{375}} = \cfrac {15000}{11} = 1363 \Omega ~~~~~\checkmark$ QED! [1]: https://edxuploads.s3.amazonaws.com/1348530250520921.jpg [2]: https://edxuploads.s3.amazonaws.com/1348530176425192.jpg [3]: https://edxuploads.s3.amazonaws.com/1348530192847448.jpg UserIdTAG: 248902 UserNameTAG: Chanute CreateTimeTAG: 2012-09-25T00:02:48Z VoteTAG: 22 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 6 FirstChildTAG: Really nice explanation Chanute! Thank you :-) Your solution becomes really important in networks where there are no independent sources. The sanity check trick won't work because if we try to calculate $V_{TH}$ and $I_N$, we'll get 0 for both values. Therefore the only way out is to connect a test source to the terminals and proceed as in your example. FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-09-25T03:18:32Z FirstChildTAG: indeed very nice !!! FirstChildUserIdTAG: 331483 FirstChildUserNameTAG: Doru FirstChildCreateTimeTAG: 2012-09-25T08:10:28Z FirstChildTAG: @ ashwith: Hmmm, interesting point. You **would** calculate $V_{TH}=I_{TH}=0$ with no independent sources present, because you'd have a purely resistive network! It is interesting to note that the way I presented above might not actually be ideal for really complicated circuits; if you have lots of independent sources then you have to solve 2 very different circuits to come up with the answers. Perhaps a better way would be simply to leave the circuit exactly as is and apply 2 different test currents (say... 1 mA and 2 mA). Then you'd have two independent equations: $v_{p1} = V_{TH}+ 1mA \cdot R_{TH}$ and $v_{p2} = V_{TH}+ 2mA \cdot R_{TH}$ and with only two unknowns you could calculate everything. That way you're only solving the circuit once, and doing some matrix algebra. As an added bonus that method should work on real world live systems where you wouldn't even have to solve the circuit, but rather you could just measure $v_{p1}$ and $v_{p2}$ as you apply the different test currents! Could be a useful trick when reverse engineering equipment.... FirstChildUserIdTAG: 248902 FirstChildUserNameTAG: Chanute FirstChildCreateTimeTAG: 2012-09-25T23:48:42Z SecondChildTAG: Actually in the case where there are independent sources you do not need a test source. The easiest numbers to work with are zero and infinity - things cancel out. So when there are independent source, we just calculate $V_{TH}$ and $I_N$ and divide the two - this is what you did for the sanity check. You could have used that directly instead. We get the Thevenin voltage buy open circuiting the terminals and the Norton current by short circuiting the terminals. So it may be easy to not use test currents/voltages at all in some cases. SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-09-26T05:49:06Z SecondChildTAG: I think that's probably true, if the circuit is easy enough to analyze. But if it's a tough nut to crack (say has lots of branches, some maybe with only sources and whatnot), I feel like it would be easier to just solve it once without worrying about re-solving it for a short-circuited situation. Maybe I'm wrong? The other thing is when reverse engineering equipment, I think you'd probably want to avoid short-circuiting random terminals, no? SecondChildUserIdTAG: 248902 SecondChildUserNameTAG: Chanute SecondChildCreateTimeTAG: 2012-09-29T04:58:44Z SecondChildTAG: Oh that's right. In a lab you should use a reference source value :-) We have Prof. Agarawal, Prof. Gerry and Piotr to tell us what happens when we short terminals of a component. Stay tuned for that :-). SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-09-29T10:21:44Z FirstChildTAG: Explicitly Explained !!! Indeed helpful!!!! FirstChildUserIdTAG: 147459 FirstChildUserNameTAG: Khanth FirstChildCreateTimeTAG: 2012-10-04T18:26:46Z FirstChildTAG: Thanks a lot! It really helps! FirstChildUserIdTAG: 653 FirstChildUserNameTAG: ziyou FirstChildCreateTimeTAG: 2012-10-07T09:57:41Z FirstChildTAG: grateful to you,thanks FirstChildUserIdTAG: 86976 FirstChildUserNameTAG: kavita FirstChildCreateTimeTAG: 2012-10-07T08:06:19Z IndexTAG: 53 TitleTAG: COMPLIMENTS FROM INDIA Respected edx community, Respected Prof Anant Agarwal, Respected Siteadmin, Dear fellow students, Firstly I would wish you all Merry christmas and an exciting wonderful life ahead. To introduce myself, I am a soldier who discontinued my studies four years before. This site I accidentally found while browsing. Registered for the course and did my level best. During one homework I was actually travelling in Train in a two day leave trip to my home. Great was the experience. I was simulating one of the world's best virtual electronic lab and attending a class virtually in one of the best institutes in the world. I really feel proud to be student......I am a MIT edx Student........of MIT. A special thanks to Myrimit. Whoever out there organizing this wonderful thing I bid a high hat salute!!!!!! With Regards and compliments, Pandiya Rajan V. UserIdTAG: 145544 UserNameTAG: pandiya CreateTimeTAG: 2012-12-25T07:14:00Z VoteTAG: 21 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: great job man........ FirstChildUserIdTAG: 108863 FirstChildUserNameTAG: shiviz FirstChildCreateTimeTAG: 2012-12-25T07:40:37Z FirstChildTAG: Merry Christmas everyone ! Gratefulness at its epitome here from New Delhi. Great work done relentlessly by the whole team. Kudos. God bless ! FirstChildUserIdTAG: 97340 FirstChildUserNameTAG: AnkitRana FirstChildCreateTimeTAG: 2012-12-25T09:07:31Z FirstChildTAG: Great and amazing... I'm a engineering student. This course helped me too. And I feel proud to be MITx student. -Anand from India FirstChildUserIdTAG: 362299 FirstChildUserNameTAG: anandbaskaran FirstChildCreateTimeTAG: 2012-12-26T02:55:52Z IndexTAG: 54 TitleTAG: Tran When clicking the TRAN button and put in the information that is required, no graph is showing up. UserIdTAG: 271670 UserNameTAG: moncapitane CreateTimeTAG: 2012-09-05T13:13:05Z VoteTAG: 21 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 8 FirstChildTAG: I am also facing the same problem i have movied to chrome for transient but still no resultant graph! FirstChildUserIdTAG: 375082 FirstChildUserNameTAG: satya1889 FirstChildCreateTimeTAG: 2012-09-05T13:33:20Z SecondChildTAG: place the probes at the node points and not at A,B and C SecondChildUserIdTAG: 124534 SecondChildUserNameTAG: srihari46 SecondChildCreateTimeTAG: 2012-09-05T13:56:20Z FirstChildTAG: Crashes Safari as well. Make sure you click the "check" button to save work or you will lose it. No way to answer. FirstChildUserIdTAG: 243189 FirstChildUserNameTAG: NickNy516 FirstChildCreateTimeTAG: 2012-09-05T13:39:14Z FirstChildTAG: place the probes at the node points and not at A,B and C FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-05T13:55:50Z SecondChildTAG: I've placed it at the node points but still no graph, in Firefox or Chrome. SecondChildUserIdTAG: 298775 SecondChildUserNameTAG: surja SecondChildCreateTimeTAG: 2012-09-05T14:06:59Z FirstChildTAG: Try to model the circuit in the Circuit Sandbox, then just copy the results. For some reason it works there and doesn't here. Also make sure to input the time correctly, if you try too big a number it will crash. FirstChildUserIdTAG: 267993 FirstChildUserNameTAG: mavlijas FirstChildCreateTimeTAG: 2012-09-05T17:08:02Z SecondChildTAG: Yup, I did the same as mavlijas and it's the only way I could get the answer. SecondChildUserIdTAG: 256249 SecondChildUserNameTAG: And90 SecondChildCreateTimeTAG: 2012-09-06T00:51:12Z SecondChildTAG: Thanks, this worked! SecondChildUserIdTAG: 298775 SecondChildUserNameTAG: surja SecondChildCreateTimeTAG: 2012-09-06T01:42:13Z SecondChildTAG: I am still not getting graph in IE 9 or FF 15.0 both in the sandbox and here. Any ideas? SecondChildUserIdTAG: 271670 SecondChildUserNameTAG: moncapitane SecondChildCreateTimeTAG: 2012-09-06T06:35:22Z SecondChildTAG: I'm unfortunately experiencing the same issue. Both in Firefox 15 and Chrome 21.0.1180.89 m - AC and DC work but TRANS doesn't show any graphs. Running Windows 7 here; will try another system soon (Ubuntu 12.04) to see if it makes any difference. SecondChildUserIdTAG: 152722 SecondChildUserNameTAG: Adad SecondChildCreateTimeTAG: 2012-09-06T10:54:12Z SecondChildTAG: I got a graph but the upper peak is not changing only lower value is coming as -167mv. Can you tell me how we can change the upper peak value.? SecondChildUserIdTAG: 182470 SecondChildUserNameTAG: nitesh2703 SecondChildCreateTimeTAG: 2012-09-16T15:34:38Z FirstChildTAG: Same problem here, DC works but TRAN doesn't. Tried Firefox 15.0 and Chrome 21.0.1180.89, OS: Windows 8. FirstChildUserIdTAG: 311022 FirstChildUserNameTAG: GordanS FirstChildCreateTimeTAG: 2012-09-05T19:00:14Z FirstChildTAG: I have the same problem. I'm using FF 15.0, Windows 7. FirstChildUserIdTAG: 364822 FirstChildUserNameTAG: Strus FirstChildCreateTimeTAG: 2012-09-05T13:54:12Z SecondChildTAG: Have you installed java virtual environment for windows? SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-09-24T18:25:35Z FirstChildTAG: Hey u might have grounded any resistors across which u have connected the output probe . Check it FirstChildUserIdTAG: 45699 FirstChildUserNameTAG: Praveenkrish FirstChildCreateTimeTAG: 2012-09-09T01:04:42Z FirstChildTAG: In H2L2 the tran response is showing lowest peak i.e, -167mv. But can any one tell me how to change the highest peak value, because for any resistance value its 999mv. I guess R(2) = 5R(1) FirstChildUserIdTAG: 182470 FirstChildUserNameTAG: nitesh2703 FirstChildCreateTimeTAG: 2012-09-16T15:39:34Z IndexTAG: 55 TitleTAG: H6P2: Phase Inverter Q5 Q6 Hints To find Thevenin equivalent of small signal circuit, (vin) think about the name 'Phase Inverter' ( What will I get If I inverted vin?). No Calculations!! To find the thevenin resistance looking in through the output port, ratio vout/i(current) will help you. UserIdTAG: 268444 UserNameTAG: Marlonabeykoon CreateTimeTAG: 2012-10-20T14:50:34Z VoteTAG: 20 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: Thanks, suffered almost an hour because a missing signXD FirstChildUserIdTAG: 70211 FirstChildUserNameTAG: Emberfej FirstChildCreateTimeTAG: 2012-10-20T15:48:43Z SecondChildTAG: welcome :) SecondChildUserIdTAG: 268444 SecondChildUserNameTAG: Marlonabeykoon SecondChildCreateTimeTAG: 2012-10-21T13:40:55Z SecondChildTAG: icant understand what r u saying plzzz elaborate it SecondChildUserIdTAG: 164689 SecondChildUserNameTAG: muhammadfaizan SecondChildCreateTimeTAG: 2012-10-21T16:35:56Z FirstChildTAG: Thankssssssssssss, I can rest. FirstChildUserIdTAG: 296419 FirstChildUserNameTAG: oscman FirstChildCreateTimeTAG: 2012-10-21T02:16:43Z SecondChildTAG: welcome :) SecondChildUserIdTAG: 268444 SecondChildUserNameTAG: Marlonabeykoon SecondChildCreateTimeTAG: 2012-10-21T13:41:04Z FirstChildTAG: This is very correct-- they name the circuits these things for a reason! Although note it specifically refers to an inversion of phase, not sign-- if you shift a sinusoid 180 degrees that inverts its "phase" and will also make it become the negative of itself. FirstChildUserIdTAG: 390034 FirstChildUserNameTAG: kahlil FirstChildCreateTimeTAG: 2012-10-20T20:09:50Z FirstChildTAG: tell me what pages from text should i read to find the answers or help me in h6p2 plzzz FirstChildUserIdTAG: 164689 FirstChildUserNameTAG: muhammadfaizan FirstChildCreateTimeTAG: 2012-10-21T14:52:40Z IndexTAG: 56 TitleTAG: Write algebraic expression Can somebody explain what does it mean? Invalid input: Rs Rp Rs Rp not permitted in answer UserIdTAG: 188219 UserNameTAG: GenadiOs CreateTimeTAG: 2012-09-23T23:12:53Z VoteTAG: 20 CoursewareTAG: Week 3 / Isolation Via Thevenin CommentableIdTAG: 6002x_isolation_via_thevenin NumberOfReplyTAG: 2 FirstChildTAG: Hi I believe this has already been answered...try RS, RP etc FirstChildUserIdTAG: 377602 FirstChildUserNameTAG: Goby FirstChildCreateTimeTAG: 2012-09-23T23:16:06Z SecondChildTAG: IF ONLY WORKS WITH CAPITALS :P SecondChildUserIdTAG: 244706 SecondChildUserNameTAG: Miguel_Angel SecondChildCreateTimeTAG: 2012-09-24T16:57:23Z SecondChildTAG: Hint and help : VTH=RP*VS/(RS+RP) will be checked correct.....try RTH now.... SecondChildUserIdTAG: 164471 SecondChildUserNameTAG: BillNieves SecondChildCreateTimeTAG: 2012-09-27T00:24:54Z FirstChildTAG: We have to be careful because the course answers are case sensitive. FirstChildUserIdTAG: 152106 FirstChildUserNameTAG: JoaoBR FirstChildCreateTimeTAG: 2012-09-24T22:52:57Z SecondChildTAG: nin tale SecondChildUserIdTAG: 414370 SecondChildUserNameTAG: prathi SecondChildCreateTimeTAG: 2012-09-28T11:30:10Z SecondChildTAG: sorry i meant in taking we have to be careful SecondChildUserIdTAG: 414370 SecondChildUserNameTAG: prathi SecondChildCreateTimeTAG: 2012-09-28T11:31:16Z IndexTAG: 57 TitleTAG: Solution 1.Total power dissipated by the load is 1000 watts(I had initially thought that totall power transmitted by joe's house is 1000watts Pd = Vd^2 /R = 240^2/1000 = 57.6ohms 2. Another way of find the voltage drop on the transmission line 1. i = V/(Rt + RL) = 240/(0.358+57.6) = 4.14A 2. Vt = i * Rt = 4.14 * 0.358 = 1.482volts Regards Hari UserIdTAG: 123484 UserNameTAG: hudkmr CreateTimeTAG: 2012-09-10T03:53:00Z VoteTAG: 20 CoursewareTAG: Week 1 / Power delivery modeling CommentableIdTAG: 6002x_S2E6_Modeling NumberOfReplyTAG: 4 FirstChildTAG: dear sir what it means D FirstChildUserIdTAG: 156694 FirstChildUserNameTAG: mkprasanth FirstChildCreateTimeTAG: 2012-09-13T17:44:44Z SecondChildTAG: Thanks! It was really helpful. SecondChildUserIdTAG: 291362 SecondChildUserNameTAG: Gudson SecondChildCreateTimeTAG: 2012-09-19T13:32:59Z SecondChildTAG: I think d for dissipation SecondChildUserIdTAG: 428658 SecondChildUserNameTAG: Moza SecondChildCreateTimeTAG: 2012-10-24T13:43:11Z FirstChildTAG: That is wrong FirstChildUserIdTAG: 217427 FirstChildUserNameTAG: Riyadh_144 FirstChildCreateTimeTAG: 2012-09-12T00:21:59Z FirstChildTAG: I used the first method you found, but before I saw your post. So, kudos. Up-vote from me. FirstChildUserIdTAG: 248807 FirstChildUserNameTAG: 1kingsman FirstChildCreateTimeTAG: 2012-09-12T04:07:47Z FirstChildTAG: r u sure tat it is power to be dissipated & not to be transmitted FirstChildUserIdTAG: 372535 FirstChildUserNameTAG: _Infinity FirstChildCreateTimeTAG: 2012-09-13T18:23:14Z IndexTAG: 58 TitleTAG: case sensitivity This diagram is misleading. v0 is really V0. UserIdTAG: 279379 UserNameTAG: RajaSrinivasan CreateTimeTAG: 2012-09-08T11:34:03Z VoteTAG: 20 CoursewareTAG: Week 1 / Node analysis practice, part 1 CommentableIdTAG: 6002x_L2Node0 NumberOfReplyTAG: 4 FirstChildTAG: "Note, this is case sensitive." FirstChildUserIdTAG: 394836 FirstChildUserNameTAG: v2g6ch4 FirstChildCreateTimeTAG: 2012-09-08T11:35:55Z SecondChildTAG: yeah... it is case sensitive...:) i got it wrong initially... but when i checked the case, i got it right...:D SecondChildUserIdTAG: 497378 SecondChildUserNameTAG: zaherva SecondChildCreateTimeTAG: 2012-09-30T01:58:34Z FirstChildTAG: I was confused too. 0 for V and 1 for I should be a subscript. FirstChildUserIdTAG: 208296 FirstChildUserNameTAG: OZ1 FirstChildCreateTimeTAG: 2012-09-08T17:34:07Z FirstChildTAG: well, v0 is the same as (a-e1)/R1 + e1/R2 if you feel that v0 is misleading FirstChildUserIdTAG: 254607 FirstChildUserNameTAG: werehenry FirstChildCreateTimeTAG: 2012-09-11T12:23:13Z FirstChildTAG: In which terms should be answer? should consider single loop or all loops? FirstChildUserIdTAG: 67648 FirstChildUserNameTAG: mgm FirstChildCreateTimeTAG: 2012-09-18T13:41:40Z IndexTAG: 59 TitleTAG: The "Virtual Short" method RAISES to an aha moment (Vote if you believe) The professor said if you think the short method is at a level of an aha moment, vote for it. This is just for fun. UserIdTAG: 259238 UserNameTAG: omidsadeghi CreateTimeTAG: 2012-11-26T13:58:42Z VoteTAG: 19 CoursewareTAG: Week 12 / S23V21_Inverting_amplifier_input_resistance_-_2 CommentableIdTAG: 6002x_S23V21_Inverting_amplifier_input_resistance_-_2 NumberOfReplyTAG: 4 FirstChildTAG: Yes it should be FirstChildUserIdTAG: 219780 FirstChildUserNameTAG: ReconIII FirstChildCreateTimeTAG: 2012-12-18T16:09:56Z FirstChildTAG: Yes FirstChildUserIdTAG: 231764 FirstChildUserNameTAG: PaulP4881 FirstChildCreateTimeTAG: 2012-12-03T05:17:39Z FirstChildTAG: Nice trick to assume the end result(input potential difference to be zero) and work backwards to obtain the output voltage. Deserves the status of an AHA moment:) FirstChildUserIdTAG: 154016 FirstChildUserNameTAG: Albatross FirstChildCreateTimeTAG: 2012-12-04T16:30:04Z FirstChildTAG: indeed AHA!! moment it makes life simpler. FirstChildUserIdTAG: 431908 FirstChildUserNameTAG: navneet_ipu FirstChildCreateTimeTAG: 2012-12-08T07:38:32Z IndexTAG: 60 TitleTAG: Lab7 Hints **Task 1:** Energy storage in capacitors The circuit below contains a 1A sinusoidal current source driving a 1mF capacitor. We've added both a voltage probe and a current probe so we can see what's going on. $v_C(t) = \frac{1}{C} \int^{t}_{-\infty} i(t)dt$ In this example, C=.001 and i(t)=sin(1000∗2π∗t) **PART 1:** Formula for vC(t): ---------- **Hint:** *I will try to explain how to use a method (sustitution) so that later you can solve this problem part:* - For example (i will invent an arbitrary example) , supoused that you have a function, such as **cos (56*x)** , where: - **in the argument** of the **function** you have: - a **constant** value (56) and - a **variable** value named x - Now, supoused that you have to find the **integral** of that function **evaluated** between two range values (**a** and **b**): ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13356696649808311.png) - Now, it would be easy if the function where something like this **cos y**: Because, you can use a calculus table like this one: ![image description](https://mitx_askbot_stage.s3.amazonaws.com/1335668695348142.png) and then the result will be like : ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13356689701539048.png) - **But the problem** is that we have **cos (56*x)** and not **cos y** or **cos x**.... ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13356693133003414.png) - Luckily exists the **method of sustitution** to solve that integral! So, This method consists in repace your argument of your function as it would be lika another variable. Is something like this: ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13356695168474432.png) Now, choose a variable , for example we will choose the variable y. ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13356697728064014.png) So, you know that the diferential ( increment) of y will be something like: ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13356698903966668.png) Replacing, ![image description](https://mitx_askbot_stage.s3.amazonaws.com/1335670140495442.png) Solving by table the integral of **cos y** ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13356704819020793.png) Now, supoused that b = ((1/112)*PI) so, sin (56*b) =1 , your expression will be reduced as (1/56).(1- sin(56*a) ). - **So, now is your turn! Try to use the sustitution method to solve the problem** :) ---- **Part 2:** **Hint Part 2:** Do you have your expression from part 1? are they giving you a certain t? Can you find what the statement is aking you? :) be careful with the digits 6 digits of precision! ---- **Part 3:** **Hint Part 3:** What is the color of the voltage probe of the Circuit? Can we find what they are requesting you for a certain time? Yes! ---- **Part 4:** Estimated value for electrical energy stored at time 0.0005, in joules: Hints: Do you have your value from part 3? ;)[Equation 9.18 given in the statement][1] ---- **Task 2: Using energy storage in a circuit** **Part 5:** Power delivered to the load in watts: **Hint:** What is the color of the curve that belongs to the current that is passing through the resistor? Can you find it in the plot? Yes! And, what voltage is between the resistor? Isn't what are you measuring with the voltage probe? and How do you calculate a power in terms of a voltage an a current? ;) ---- **Part 6:** Average current supplied by bridge rectifier, in amps: **Visual Hint 1:** Average an arbitrary signal. ![im][2] **Hint 2:** warning, wait till the signal is stable before meassuring :). ---- I hope this can help you. Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/489 [2]: https://edxuploads.s3.amazonaws.com/13517361921343617.png UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2012-11-01T02:24:29Z VoteTAG: 19 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: Excellent, very well explained. Wahab FirstChildUserIdTAG: 455950 FirstChildUserNameTAG: Wahabbaluch FirstChildCreateTimeTAG: 2012-11-01T03:08:02Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-01T14:01:31Z SecondChildTAG: Really awesome explanation.....a school student can also understand easily....thank you... SecondChildUserIdTAG: 316761 SecondChildUserNameTAG: gk_goel SecondChildCreateTimeTAG: 2012-11-03T13:30:51Z SecondChildTAG: Excellent!! SecondChildUserIdTAG: 114913 SecondChildUserNameTAG: ngoctuan SecondChildCreateTimeTAG: 2012-11-04T18:19:36Z SecondChildTAG: @ ngoctuan and @ gk_goel :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-04T18:55:23Z SecondChildTAG: Really Good explanation Myrimit..:) SecondChildUserIdTAG: 111917 SecondChildUserNameTAG: ashish_mit SecondChildCreateTimeTAG: 2012-11-04T20:25:24Z SecondChildTAG: Thank you ashish_mit ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-04T20:33:31Z FirstChildTAG: Hi, i tried to follow your great explanation and i got (-cos(2000*PI*t))/(2*PI) for the first part. I don't understand why i don't get right solution. Saludos y muchas gracias de antemano FirstChildUserIdTAG: 314294 FirstChildUserNameTAG: victormp FirstChildCreateTimeTAG: 2012-11-02T17:45:08Z SecondChildTAG: Are you sure that you are replacing corretly the result of the integral? Hint: Remember that you have to evaluate two extremes: A and B. What value is your A and what value is your B? SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-02T18:00:31Z SecondChildTAG: :) Thanks. Muchas gracias SecondChildUserIdTAG: 314294 SecondChildUserNameTAG: victormp SecondChildCreateTimeTAG: 2012-11-02T18:15:22Z SecondChildTAG: Hi **Myrimit** I was solve as you say and get this -(1/.001*2000*pi) * (cos( 2000 * pi * t ) - cos( 2000 * pi * (- infinity) )) then -(1/.001*2000*pi) * (cos( 2000 * pi * t ) - 1) but they give me wrong ans. can you give me Hint SecondChildUserIdTAG: 318037 SecondChildUserNameTAG: Maher-84 SecondChildCreateTimeTAG: 2012-11-02T20:47:24Z SecondChildTAG: Ok. I find it. thank you SecondChildUserIdTAG: 318037 SecondChildUserNameTAG: Maher-84 SecondChildCreateTimeTAG: 2012-11-02T21:52:50Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-02T21:54:55Z SecondChildTAG: Really awesome explanation.....a school student can also understand easily....thank you... SecondChildUserIdTAG: 316761 SecondChildUserNameTAG: gk_goel SecondChildCreateTimeTAG: 2012-11-03T13:30:30Z SecondChildTAG: You are welcome gk_goel! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-03T15:02:42Z FirstChildTAG: The integral is a definite integral. There are contributions from both the upper and **lower** limit. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-02T18:04:07Z FirstChildTAG: hey Myrimit, I got my formula right when i checked. but the value of voltage at particular time 't' when substituted in the formula, the result is marked as wrong by the checker. please help. Thanks. FirstChildUserIdTAG: 204213 FirstChildUserNameTAG: ratneshray FirstChildCreateTimeTAG: 2012-11-03T19:46:11Z SecondChildTAG: Hi ratneshray! Can I help you? Be careful with the angle, it is in radians and not in sexagesimal degrees ;). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-04T00:47:47Z SecondChildTAG: i got this term containing "t" in my formula "cos(2000*PI*t)". and i'm substituting t=0.0005 in the formula. and for "PI" i'm substituting "3.14". please help where i'm doing wrong. thanks. SecondChildUserIdTAG: 204213 SecondChildUserNameTAG: ratneshray SecondChildCreateTimeTAG: 2012-11-04T13:40:20Z SecondChildTAG: (0.159) *(1- cos( 1000*2 * PI * t ) ) the system is not accepting my answer SecondChildUserIdTAG: 338685 SecondChildUserNameTAG: varshaD SecondChildCreateTimeTAG: 2012-11-04T14:55:50Z SecondChildTAG: Hi @ratneshray, Try to not round the $\pi$ value, use more digits 3.141592654 ... SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-04T16:54:09Z FirstChildTAG: Hi Myrimit, I am having difficulty getting the correct answer for Lab 7 problem 4. I have tried in both Matlab and Python. **Matlab:** eq = (C*v*(t)^2)/2 C = .001;v = {correct answer to problem 3};t = .0005; eval(eq) **Python:** (.001*{correct answer to problem 3}*.0005**2)/2; There is very slight delta between the answers, but basically the same. Please advise. Thanks, Ryan PS - Thanks so much for your contribution to the Discussions, you have really helped me in the past 7 weeks. FirstChildUserIdTAG: 355773 FirstChildUserNameTAG: Albright4edx FirstChildCreateTimeTAG: 2012-11-04T23:55:16Z SecondChildTAG: Hi Ryan, Be careful with your equation v(t) **is not** v multiplied by t **v(t) means the value of a voltage in the time t.** Hint: our v(t)= some_voltage_value and what have you calculated in the previous question isn't it the value of the voltage in a certain t? ;) I hope this can help you. You are welcome Ryan. I am really happy, nice to see that I am been helpul here ;). Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-05T00:08:55Z SecondChildTAG: Perfect! I should have recognized that, thanks! SecondChildUserIdTAG: 355773 SecondChildUserNameTAG: Albright4edx SecondChildCreateTimeTAG: 2012-11-05T00:27:16Z IndexTAG: 61 TitleTAG: These are my Babies. Hello people,i present to you my passions and motivations which Electronica study. I hope to one day work together on a robotics project. A Tricycle Robot with ultrasonic sensor and a Spider Robot, both programmed with Arduino Mini Pro Mini from ATMEGA168. ![][1] This Biped Robot was programmed with Pic 18f4620 from Microchip. ![][2] Well, I'm from Argentina and as you can see this is not a limitation for me to access to technology, since I am a hobbyst and when I need something i buy it on ebay in china. Today is not a excuse geographic location. Thank you very much and congratulations to MIT for this beautiful course. Konredus. [1]: http://i49.tinypic.com/281ymhw.jpg [2]: http://i47.tinypic.com/ivaey0.jpg UserIdTAG: 274475 UserNameTAG: Konredus CreateTimeTAG: 2012-10-28T05:53:53Z VoteTAG: 19 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 10 FirstChildTAG: Nice Robot, Konrad! Good work! I am from Argentina too:) Que lindo Robot! Te felicito por tu trabajo! Yo también soy de Argentina! Saludos! Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-28T06:09:45Z SecondChildTAG: Thanks Myrimit ... is a pleasure to share my humble hobby. Gracias Myrimit... es todo un placer poder compartir mi humilde hobby. Soy de Quilmes - Buenos Aires, vos? SecondChildUserIdTAG: 274475 SecondChildUserNameTAG: Konredus SecondChildCreateTimeTAG: 2012-10-28T07:02:55Z SecondChildTAG: :). CABA! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-28T15:27:50Z SecondChildTAG: Dear Konredus Sir , Can you please help to make a simple USB programer for Mico chip 16f84 I want to design a Electronic note pad for my son ( he is just 18 month old ) i have project in my mind I am planing to make the note pad with touch screen and LCD screen Thanks a lot Thanks SecondChildUserIdTAG: 156694 SecondChildUserNameTAG: mkprasanth SecondChildCreateTimeTAG: 2012-10-28T17:02:31Z SecondChildTAG: Hi mkprasanth, i'm happy to help you. The problem with USB programmers is that you need a microcontroller to be programmed initially that will manage the communication. So you need to contact a friend to initially program the programmer. Here is the link of a Pickit2-Clone made ​​in Argentina, download it and see if it helps to you: https://skydrive.live.com/?cid=adcff3854d40b985&id=ADCFF3854D40B985!193 Any other questions I'll be here... :) SecondChildUserIdTAG: 274475 SecondChildUserNameTAG: Konredus SecondChildCreateTimeTAG: 2012-10-28T17:57:02Z SecondChildTAG: $Gracias$ $Konredus$. (sorry I can't speak any further spanish) Can you help me learn robotics. Kindly give some pointers for a complete newbie. $Gracias$ SecondChildUserIdTAG: 623210 SecondChildUserNameTAG: preveen SecondChildCreateTimeTAG: 2012-10-29T06:01:51Z SecondChildTAG: Konredus sir , Thanks a lot for your support , I think Better I buy the pick kit from microchip , If any support I need i will come to you Thanks MK.Prasanth SecondChildUserIdTAG: 156694 SecondChildUserNameTAG: mkprasanth SecondChildCreateTimeTAG: 2012-10-29T10:45:40Z FirstChildTAG: Fantastic! By any chance have you open sourced your projects? If you have would it be possible to get the schematics and the code? Great to hear that everybody can enjoy robotics. FirstChildUserIdTAG: 85490 FirstChildUserNameTAG: yashsavani FirstChildCreateTimeTAG: 2012-10-28T07:24:07Z FirstChildTAG: Awesome work :) I will be waiting for the details. Mean while if you are on any social media. please share your profile, that would help us have a conversation FirstChildUserIdTAG: 156974 FirstChildUserNameTAG: ManojKumar FirstChildCreateTimeTAG: 2012-10-28T08:14:21Z FirstChildTAG: This is brilliant! Is it possible for you to share the details of your work? Thank you so much for sharing! :-) FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-10-28T07:19:29Z SecondChildTAG: Thanks for your interest. I have documents, but only in Spanish. Sorry. SecondChildUserIdTAG: 274475 SecondChildUserNameTAG: Konredus SecondChildCreateTimeTAG: 2012-10-28T07:21:27Z SecondChildTAG: Well you could post them anyway if it's not a problem with you. Maybe Google translate will be accurate enough :-) Otherwise I hope I can bug one of you for help in translating parts I can't understand. :-) SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-10-28T07:23:57Z SecondChildTAG: I agree. I would love to get the schematics. I can get it translated from a Spanish professor at my school and repost it here if you like. SecondChildUserIdTAG: 85490 SecondChildUserNameTAG: yashsavani SecondChildCreateTimeTAG: 2012-10-28T07:27:13Z SecondChildTAG: I must be honest guys, I have many codes for these robots and how they are for different microcontrollers i dont nou in what language i will send to you. The other issue is the PCB that are made ​​on a temporary basis. Be patient until they are finished. Thank you. SecondChildUserIdTAG: 274475 SecondChildUserNameTAG: Konredus SecondChildCreateTimeTAG: 2012-10-28T07:51:31Z SecondChildTAG: Alright. I'll be waiting then. Thanks again :-) SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-10-28T07:56:25Z SecondChildTAG: intrested in all the details...!! GR8 WoRK!! SecondChildUserIdTAG: 127800 SecondChildUserNameTAG: MNB SecondChildCreateTimeTAG: 2012-10-28T09:51:31Z SecondChildTAG: Dear Ashwith sir , Please visit the Basicstamp /Ardunio web site it may help you I don't know if I paste the URL it may be problem it is out of subject so please Google it Thanks MK.Prasaanth SecondChildUserIdTAG: 156694 SecondChildUserNameTAG: mkprasanth SecondChildCreateTimeTAG: 2012-10-29T10:45:33Z FirstChildTAG: That's great! I want to build something similar... What do I need to know to do something like that? FirstChildUserIdTAG: 326787 FirstChildUserNameTAG: Angstrem FirstChildCreateTimeTAG: 2012-10-28T10:33:03Z SecondChildTAG: I think you refer to the Biped Robot. First, you need know something of robotic move algebra, then a Programing language how "C" and the last knowledge what you need is put this "C" or another High Lenguage to one Microcontroller.... :) I Hope this was usefull. SecondChildUserIdTAG: 274475 SecondChildUserNameTAG: Konredus SecondChildCreateTimeTAG: 2012-10-28T12:59:51Z SecondChildTAG: Please tell more about robotics algebra. Gracias SecondChildUserIdTAG: 623210 SecondChildUserNameTAG: preveen SecondChildCreateTimeTAG: 2012-10-29T06:11:24Z FirstChildTAG: Very nice. FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-10-29T00:43:54Z FirstChildTAG: Great job! FirstChildUserIdTAG: 216684 FirstChildUserNameTAG: Taimoor1017 FirstChildCreateTimeTAG: 2012-10-28T18:37:56Z FirstChildTAG: Wonderful work! I am sure any details that you are willing to share will be greatly appreciated. I like all three but would especially be interested in hearing about the biped. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-28T18:55:28Z FirstChildTAG: Cool! FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-29T04:57:50Z SecondChildTAG: Thanks Pennypacker.. :P SecondChildUserIdTAG: 274475 SecondChildUserNameTAG: Konredus SecondChildCreateTimeTAG: 2012-10-29T04:58:55Z FirstChildTAG: **Thoughts on Micro-controllers, and Assembly vs. High-Level** **Konredus:** Do you also use Lego (or similar blocks) for the mechanics in the two robots in the first picture? I know many people that started with Lego and then moved on to micro-controller programming in high school or college, and re-used the same Lego blocks they used as children to build mechanics for their robots. Your "biped" 'bot in the second photo looks more "professional", but at the expense of construction time. With robotics you have to know more than just electronics; as evidently you had to take time to drill / shape / machine the metal brackets / flanges; and then correctly place the motors for proper operation: It's as much *mechanical engineering* and computer programming as it is electronics. Is the RJ-11 jack and the telephone cord the physical *computer interface* for the micro-controller on the "biped"? The programmer we had to buy for class (expensive!) had the same style "hookup". And you are using PICs? I was programming PICs with their proprietary debugger/programmer, but at the university level we had to use **assembly language**! I only wish we could have used C or Basic, or another high-level language. With assembly, we had to write subroutines for our particular micro-controller PIC16F873 just to perform seemingly simple operations as *multiplication* and *division*! We needed to perform these operations to translate the 16-bit pulse-width-modulated input from a temperature sensor to convert it into Fahrenheit or Celsius degrees and display it on an LCD screen. These lower-level PICs have no built-in multiplication (*) or division (/) instructions! Plus since these lower-level PICs are only 8-bit (ALU and MUX), we had to use 2 registers and do the processing by splitting the input up into the higher 8-bit portion and the lower 8-bit portion! Instead of all this nonsense, in **high-level language** you can just write: Tc = (Tf-32) * 5/9. A few lines of code versus two subroutines and one main routine. Programming in assembly on a uC is not very intuitive, mistakes are easy to make, and coding is tedious: Instead of focusing on making a robot motor move in response to stimulus, or making a sensor read out correctly, you spend more time worrying about syntax, registers, and memory. Of course we also programmed PICs to learn assembly on, and the differences between Harvard / RISC / micro-controller architecture versus x86 / CISC / IBM PC architecture, and we really learned! We would have had more **fun**, though, using C or Basic! Personally, I want to start using the Arduino; I heard they're cool. Jersey Mark FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-10-29T09:46:50Z SecondChildTAG: Dear Mr Jersey Mark Sir , Yes you are right Arduino is easy to learn so many it is open source too. more over you can write the command jut like how I am writing to you but i like to do in Assembly we are very close to the chip and less memory space and more speed But Arduino is good now i have one I am playing with him always SecondChildUserIdTAG: 156694 SecondChildUserNameTAG: mkprasanth SecondChildCreateTimeTAG: 2012-10-29T10:41:02Z SecondChildTAG: Thanks Jersey for your time and for your good explanation of context. The Lego pieces that you see are just the motherboard, the rest of the components are tied or glued to other lego parts, but it's not a lego kit as Mindstrom or anything like that, it's just to get what I had to hand. The RJ45 connector that you see is effectively the connector to the Microchip ICD2, I got it when I studied Masters in PIC programming here in Argentina, and yes, it is a expensive programmer / debugger. With respect to the programming language, I program in Assembler, Basic, C16, C18,CCS and ANSI C. It's clear that they are all C, but with certain variations or modifications by each company. With respect to the Biped Robot, it was something more professional, Why could not use Lego pieces for mechanical parts, and I did it together with a friend who studied Mechanical Engineering, but it was no big deal, we did an shematic on paper and start to cut a casing of a PC power source ... :) Arduino also is C, and it is REALLY EASY to use as it brings many templates and examples. I use Arduino when I do not care much processing speed, or when I'm hurry .. :) As you noted, it's all about taking advantage of what you got on hand and apply a little ingenuity. Konredus SecondChildUserIdTAG: 274475 SecondChildUserNameTAG: Konredus SecondChildCreateTimeTAG: 2012-10-29T15:10:49Z IndexTAG: 62 TitleTAG: Lab5 Hints Hi! I hope that you are doing good! I am preparing a super cool DEMO material for the Mid-Term Exam for you, so that you can practice and go confident to the Exam, haha, I hope to finish it in a few days :). I am also in the 3.091x haha, I distracted a little bit there haha showing the pic of my cat "El Rey Gato" or "The King Cat" haha! I found there some Classmates from here or 6.002x Spring: like dantyrant, ashwith, MobiusTruth, JerseyMark, a lot more too like super-awesome Suyog that asked me if I could guide 3.091x like 6.002x haha (The problem is that I haven't seen pure Chemistry since High School haha, but I will try to help there as far as I can) ! and also I have found to Aumatar! (Aumatar was a referent in the Prototype Course 6.002x, he was in fact, the "The Father of Hints", he started this movement, I hope he can read this, thank you for being inspirational, thank you, I hope you don't mine that I have named you as "The Father of Hints" here haha). --- Ok, lets go to some Hints that can help you to solve this Lab by your own: ---------- **Lab5 Hints** The goal of this lab is to analyze the performance of an inverting mosfet amplifier operating in its linear region, where the output waveform is an undistorted but amplified replica of the input waveform. ![mosfet][1] ---------- **Part 1:** Input voltage at lower end of linear operating range: **Hint Part1** : Watch the video lecture [S9V17 - watch here][2]: RANGES OF OPERATION - ANOTHER WAY Once you have seen the video, what is the valid region, that is to say, the linear region?. Ok, If you know that region and you know that the x-axis is the input voltaje, can you tell the Input voltage at lower end (up extreme - curve)of that region? Yes! ---------- **Part 2:** Input voltage upper end of linear operating range: **Hint Part 2:** Idem Part 1 but with the upper case (down extreme - curve). ![mosfet2][3] ---------- **Part 3:** **Hint Part 3:** Isn´t it the magenta voltage probe where you get the output voltaje? Isn´t it the blue voltage probe where you get the input voltaje? Yes! Can you graphically by the TRAN analysis get the ratio of the Vout and Vin? Yes! Hint: take peak to peak values (from a the up extreme to the down extreme of the wave). ---------- **Part 4:** **Hint Part 4:** Now change the amplitude of Vsignal in Figure 2 from 0.1V to 1V and rerun the TRAN analysis. You should see significant distortion in the output signal, in this case clipping or truncation of the max and min signal values. Experiment with amplitudes of Vbias and Vsignal to find the largest amplitude for Vsignal for which amplifier produces an unclipped output. ![enter image description here][4] This part is the complicated one, ok, lets see a Little hint so that later you can solve it by your own: I will explain you what it is the idea of the problem so that you can later solve it: - Now change the amplitude of Vsignal in Figure 2 from 0.1V to 1V and rerun the TRAN analysis. That means that you have to change the value of the voltage source between 0.1 V (100mV) to 1V. Look at the red circle from the next picture: ![image description](https://mitx_askbot_stage.s3.amazonaws.com/1334345757192777.png) - You should see significant distortion in the output signal, in this case clipping or truncation of the max and min signal values. Experiment with amplitudes of Vbias and Vsignal to find the largest amplitude for Vsignal for which amplifier produces an unclipped output. Here, you have to observe what happens when you start to change the values range. For that, double clic on the source element and change it to the different values of the problem. Look at the magenta and blue wave. One belongs to the output and the other to the input. ![image description](https://mitx_askbot_stage.s3.amazonaws.com/1334346018803450.png) - Playing with the values, watch that for some values happens a distorsion that cuts your wave. You have to find the maximun value of the input (blue wave), that do not cause the distortion (the limit value). Hint: remember that the value is peak to peak (double)... ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13343466638579712.png) :) I hope this can help you. Myriam. ---- Now in Spanish: Hola! Espero que les esté yendo bien! Estuve media descolgada, porque estoy preparando un DEMO super cool como material de estudio para que ustedes puedan practicar para el Examen, jaja. Espero poder terminarlo en estos días :). También les cuento que estoy registrada en el curso 3.091x jaja, y sinceramente me distraje mucho allí, hasta subí una foto de mi gato que se llama "El Rey Gato" jaja! En el 3.091x encontré a muchos compañeros de aquí y del 6.002x (Spring): como ashwith, MobiusTruth, JerseyMark, y muchos más tal y como el genial Suyog que allí me preguntó si planeaba guiar en 3.091x como lo estaba haciendo en 6.002x porque le era de mucha ayuda, jaja (lamentablemente mis conocimientos en química pura están oxidados jaj, creo que no veo química química desde el Secundario, pero desde ya, en todo lo que pueda ayudar lo haré)! También me encontré con Aumatar!, se preguntarán quién es. Muchos no lo conocen porque él ha estado en el 6.002x (Spring); Aumatar, fue un referente en el Curso Prototipo 6.002x, de hecho, él fue "El Padre de los Hints", comenzando con esta especie de movimiento, tal y como lo es ahora este Post. Espero que si él lee esto no se enoje por el apodo que se ha ganado "El Padre de los Hints", espero que él pueda leer esto, para mí fue muy inspirador, gracias! --- Bueno, veamos algunas Hints que pueden serte de ayuda para que luego puedas resolverlo por ti mismo: ---------- **Lab5 Hints** El objetivo de este Lab es analizar cómo actúa el amplificador inversor mosfet en su región lineal, en donde su onda de salida no se distorsiona pero realiza una réplica amplificada de la onda de entrada. ![mosfet][1] ---------- **Parte 1:** Tensión de entrada del extremo bajo de la zona de operación lineal: **Hint Parte 1** : Mirar el video de lectura [S9V17 - watch here][2]: RANGES OF OPERATION - ANOTHER WAY Una vez que hayas visto el video, Cuál es la región válida, es decir, la región lineal? Bien, si sabes la región y también que el eje x representa la tensión de entrada, puedes decir cuál es el valor de tensión de entrada en el extremo bajo (curva- lado arriba) de la región? Sí! ---------- **Parte 2:** Tensión de entrada en el extremo alto de la zona de operación lineal: **Hint Parte 2:** Idem Parte 1 pero con el caso del extremo alto (curva - lado bajo). ![mosfet2][3] ---------- **Parte 3:** **Hint Parte 3:** No es acaso en donde está la punta de prueba color magenta el lugar en donde se mide la tensión de salida? Sí! No es acaso en donde está la punta de prueba color azul el lugar en donde se mide la tensión de entrada? Puedes obtener los datos gráficamente en el análisis TRAN, es decir, obtener el cociente de Vout respecto Vin? Sí. Hint: tomar el valor pico a pico como valor ( desde el extremo superior al inferior de la onda). P.D: Perdón amigos Chilenos, sé que pico es una mala palabra en Chile, pero estoy acostrumbrada a nombrar peak to peak como pico a pico, tal y como lo llamamos aquí en Argentina, mil disculpas, gracias. ---------- **Parte 4:** **Hint Parte 4:** Ahora cambiar la amplitud de la Vsignal de la Figura 2 desde 0.1V a 1V y nuevamente hacer clic en el análisis TRAN. Deberías ver una significativa distorsión en la señal de salida, en este caso cortando el valor máximo y mínimo de los valores de la señal. Experimenta con valores de Vbias y de Vsignal en orden de hallar la mayor amplitud posible en donde el mplificador produce una salida sin recorte. ![enter image description here][4] Esta es la parte complicada, bien, veamos una pequeña Hint que pueda serte de ayuda para que luego puedas experimentar tú y obtener la solución por ti mismo: Te explicaré en forma breve cuál es la idea de este punto del problema para que luego lo puedas resolver tú mismo: - Ahora se debe cambiar la amplitud de Vsignal en la Figura 2 desde 0.1V a 1V Now change the amplitude of Vsignal in Figure 2 from 0.1V to 1V y ejecutar nuevamente el análisis TRAN. Esto significa que tienes que cambiar la fuente de tensión entre .1V (100mV)a 1V. Mira el círculo rojo de la siguiente imagen: ![image description](https://mitx_askbot_stage.s3.amazonaws.com/1334345757192777.png) - Deberías ver una significativa distorsión en la señal de salida, en este caso un recorte del mínimo y del máximo valor de los valores de la señal. Experimenta con amplitudes de Vbias y Vsignal en orden de hallar la amplitud más grande posible para Vsignal que no provoque recorte de la señal de salida. Aquí, debes observar qué sucede al cambiar los valores en un cierto rango. Para ello, realiza un doble clic sobre el elemento (fuente de tensión) y cámbialo a los distintos valores del problema. Recuerda que el magenta es la salida y el azul la entrada. ![image description](https://mitx_askbot_stage.s3.amazonaws.com/1334346018803450.png) - Jugando con los valores, se observará que para algunos valores la onda comienza a distorsionarse, finalmente se recorta en un cierto valor que tú debes hallar. Deberás encontrar el máximo valor de entrada para el cuál la entrada (onda azul), no causa distorsión (el valor límite). Hint: el valor se encuentra dado en pico a pico (doble). ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13343466638579712.png) :) Espero que esto les haya servido de ayuda. Myriam. [1]: https://edxuploads.s3.amazonaws.com/1349916128134369.png [2]: https://www.youtube.com/watch?feature=player_embedded&v=WdugcnssxHA [3]: https://edxuploads.s3.amazonaws.com/13499174511343694.png [4]: https://edxuploads.s3.amazonaws.com/1349917535134360.png UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2012-10-11T02:02:26Z VoteTAG: 19 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 9 FirstChildTAG: Myrimit rocks like socks in a box. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-11T11:49:21Z SecondChildTAG: Haha! Thanks JSChambers! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-11T15:33:23Z SecondChildTAG: Haha! Thanks JSChambers! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-11T15:33:35Z SecondChildTAG: i didn't understand the last part Myrimit. can you tell me something more about it? SecondChildUserIdTAG: 259693 SecondChildUserNameTAG: MehrozKhan SecondChildCreateTimeTAG: 2012-10-13T19:38:54Z SecondChildTAG: Hi MehrozKhan! Yes sure, can I help you? Hint: In other words, the last part they are asking you which voltage of Vsignal (blue)is the one that cut the wave shape of the output(magenta). :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-13T20:58:33Z FirstChildTAG: thanks a lot ..:) FirstChildUserIdTAG: 101902 FirstChildUserNameTAG: DHEERAJK_VITS FirstChildCreateTimeTAG: 2012-10-14T17:53:56Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-15T03:19:54Z FirstChildTAG: Dear Myriam Mit Thanks a lot Prasanth FirstChildUserIdTAG: 156694 FirstChildUserNameTAG: mkprasanth FirstChildCreateTimeTAG: 2012-10-11T03:06:27Z SecondChildTAG: You are welcome mkprasant! I will try to post Hints of Homework 5 :). How are you? How are you doing with this course and the other courses? My best wish to you! Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-11T04:04:13Z FirstChildTAG: hola Myrimit no me da el resultaod correcto en la ultima pregunta no se que estoy hciendo mal espero puedas orientarme gracias FirstChildUserIdTAG: 320715 FirstChildUserNameTAG: maraivette FirstChildCreateTimeTAG: 2012-10-12T16:16:13Z SecondChildTAG: Hola maraivette! Has variado el valor de Vsignal? Hint: recuerda que tienes que hallar hasta qué punto puedes incrementar tu Vsignal (valor de onda azul)hasta antes que la onda color magenta se recorte. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-13T20:41:05Z FirstChildTAG: Hola Myrimit, primeramente las disculpas a todos por escribir en español pero mi ingles es todavía malito. Mi pregunta es con respecto a la ultima interrogante del Lab 5 y solo motivo de comprender mas a fondo el funcionamiento ya que ya lo he resuelto, pero la verdad me intriga saber por que las respuestas pueden ser variadas, es decir que encontré una respuesta favorable para el problema pero por curiosidad decidí escribir otro valor y resulta que también es efectiva. Espero no estar violando ninguna regla con mi comentario solo me gustaría saber que es lo que esta pasando con este problema para ampliar mis conocimientos. FirstChildUserIdTAG: 256189 FirstChildUserNameTAG: MarquezMario FirstChildCreateTimeTAG: 2012-10-12T22:45:01Z SecondChildTAG: Hola MarquezMario, Si quieres lo vemos después de la fecha límite de entrega :). Sinceramente, no sé muy bien porqué en tu segunda opción te dio el check en verde, recuerda que las respuestas tienen cierta tolerancia + - un valor respecto del que debería... también podría ser por un bug, pero la verdad es que no lo sé. Saludos! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-13T20:45:26Z SecondChildTAG: ok Myrimit me parece bien, porque de verdad a mi lo que me interesa es aprender un poco más. Me escribes cuando ya se pueda verlo a mayor profundidad. Bug no creo que sea porque es un rango bastante amplio de valores que me permite. Saludos. SecondChildUserIdTAG: 256189 SecondChildUserNameTAG: MarquezMario SecondChildCreateTimeTAG: 2012-10-15T05:15:09Z FirstChildTAG: hello Myriam!! thanks ! for being so helpful. I have a very bad habit of doing the HW at the 11th hour and need some help. I have problem in only 1 question. which is as follows: for the LAB5 the second question asking for : Input voltage upper end of linear operating range: If i understand the question correctly, it is asking me to tell the voltage of the point from where the curve starts to fall. I have got the first answer correct and all other parts ,but this one is not letting me have the green tick !! its sucking my brain out!! I have tried all nearby values for that curve from the graph please advise. Thanks FirstChildUserIdTAG: 232499 FirstChildUserNameTAG: Asim09 FirstChildCreateTimeTAG: 2012-10-13T18:05:48Z SecondChildTAG: Hi Asim09! How are you? Hint: this question in fact, it is a little bit confusing, it is the inverse of what you could consider upper ;). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-13T20:48:48Z SecondChildTAG: sorry for being so dumb! but i still cant get the hint and the answer!! SecondChildUserIdTAG: 232499 SecondChildUserNameTAG: Asim09 SecondChildCreateTimeTAG: 2012-10-14T09:04:27Z SecondChildTAG: Hi Asim09. Can I help you? SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T19:02:55Z SecondChildTAG: I still have the same problem, i cannot get the second part in lab 5. Any hint ? SecondChildUserIdTAG: 141709 SecondChildUserNameTAG: charbelantonios SecondChildCreateTimeTAG: 2012-10-15T05:47:40Z FirstChildTAG: I change the values of Vbias and Vsignal...and got the unclipped signal..then I calculate the Amplitude by (Vp_max - Vp_min)/2...but not getting the right answer...Am I missing something here...thanks in advance :) FirstChildUserIdTAG: 122513 FirstChildUserNameTAG: Mona77 FirstChildCreateTimeTAG: 2012-10-12T12:45:10Z SecondChildTAG: and my answer is in 0.00XXX form! SecondChildUserIdTAG: 122513 SecondChildUserNameTAG: Mona77 SecondChildCreateTimeTAG: 2012-10-12T12:47:22Z SecondChildTAG: the answer will be in volts something like x/10 volts. I mean like .9 or .4 0r.3 volts. These values are just to make it clear. ol the best!! SecondChildUserIdTAG: 232499 SecondChildUserNameTAG: Asim09 SecondChildCreateTimeTAG: 2012-10-13T17:55:44Z SecondChildTAG: Hi Mona77! You are calculating the Amplitude with the formula that you are writting.... Rember that you have to give the answers in peak to peak. Remember that if you have an arbitrary wave, your peak to peak value would be from the top to the other extreme (red line) ;). ![peak][1] [1]: https://edxuploads.s3.amazonaws.com/13501616911343608.png SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-13T20:55:07Z SecondChildTAG: hi myrimit, the qstn is asking to say the amplitude, not peak to peak SecondChildUserIdTAG: 284628 SecondChildUserNameTAG: jumana_mp SecondChildCreateTimeTAG: 2012-10-14T13:23:21Z SecondChildTAG: Hi jumana_mp, yes, the statement is confusing... but they meant peak to peak... SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T14:56:39Z SecondChildTAG: oooo GOT It :)...you know you are a life saver :)..cheers SecondChildUserIdTAG: 122513 SecondChildUserNameTAG: Mona77 SecondChildCreateTimeTAG: 2012-10-14T15:52:50Z SecondChildTAG: Well done Mona77! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T19:01:36Z SecondChildTAG: AM STILL NOT DONE WITH LAB 5 LAST QSTN, SOMEBODY PLEASE HELP ME SecondChildUserIdTAG: 284628 SecondChildUserNameTAG: jumana_mp SecondChildCreateTimeTAG: 2012-10-14T20:10:34Z SecondChildTAG: HEY MYRIMIT, PLEASE HELP ME, LAB 5 LAST QSTN HASNT CLICKED ME YET SecondChildUserIdTAG: 284628 SecondChildUserNameTAG: jumana_mp SecondChildCreateTimeTAG: 2012-10-14T20:10:59Z FirstChildTAG: Hi Myrimit, I am stucked in Part 3 of lab5. I have run the tran analysis and got the wavefrom. From the wavefrom i got (Vo)peak = 3.148 and (Vi)=900.049m. After taking the Vo/Vi i am not getting the correct gain. Please correct me where i am doing wrong. Thanks in advance ... Gaurav FirstChildUserIdTAG: 413002 FirstChildUserNameTAG: Gauravjain88 FirstChildCreateTimeTAG: 2012-10-14T09:37:14Z SecondChildTAG: Got the answer from your previous explanation ... Thanks SecondChildUserIdTAG: 413002 SecondChildUserNameTAG: Gauravjain88 SecondChildCreateTimeTAG: 2012-10-14T09:43:18Z SecondChildTAG: Well done! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T14:53:31Z FirstChildTAG: Hola Myrimit gracias por ayudar a tantas personas en el curso. :) Por otra parte no termino de entender la parte 1 del lab, estoy enredado con las palabras. Se que la parte lineal es aquella en que la curva decrece despues de la zona de saturacion, pero no consigo dar con los valores. ojala puedas ayudar,me, muchas gracias. FirstChildUserIdTAG: 110802 FirstChildUserNameTAG: leoblack FirstChildCreateTimeTAG: 2012-10-14T20:52:16Z SecondChildTAG: Hola leoblack, Disculpa que no he leído tu Post antes... Hint: Mira la curva 7.21 de la página 346 [leer aquí][1] ![here][2] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/370 [2]: https://edxuploads.s3.amazonaws.com/13502714181343631.png SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-15T03:25:54Z IndexTAG: 63 TitleTAG: H4P1 Hints : VACUUM DIODE **H4P1: VACUUM DIODE** Although vacuum tubes are no longer commonly used in computer or consumer electronics, they still have a substantial niche in high-end audio, high-power radio transmitters, particle accelerators, and microwave ovens. A vacuum diode, the simplest vacuum tube, is an interesting two-terminal device. ![image][1] A vacuum diode's voltage-current characteristic is closely approximated by the Childs-Langmuir Law, with one parameter P called the perveance: ![formula1][2] **Hint Part 1:** They give us our iP in function of vPK (that is to say the red curve that you see down here) : ![Imagen2][3] Ok, you know how does iP behaves as they give you the expression of that red curve. So, what is the value of the current IP of an offset of voltage that we call VPK? Isn't it when our vPK it is that value of voltage? Can we get the expression of IP? Yes!!! ![Imagen3][4] I suggest to watch and pay attention to the Zeng thing haha [watch here (S7V4 video)][5]! **Part 2 Hint:** You should watch this Tutorial Video explanation [watch here (Week 4 Tutorials - An abstract element)][6] :). So, what is, in our case the incremental resistance? isn't it a variation of the vPK respect of the iP? so, do we know how to calculate a small small variation? yes! **Hint Part 3:** idem part 1 but with a different value of vPK. **Hint Part 4:** idem part 2 but with a different value of vPK. ---- Some curious stuff: ![supuestos][7] Let's see in the arbitrary case 1), we will have that the violet line it is less than the violet line of 2). So, the ratio (violet 1/ light-blue 1)= value 1 and (violet 2/ light-blue 2) = value 2... so, we will see that value 1 will be higher than value 2 . **Aha Moment!** So, can you predict which value of incremental resistance will be higher? the one tht you calculated in part 2) or the one that you have calculated in the part 4? ;). That is all! I hope this could be helpful to you so that you can solve it by yourself, be careful with the units :) some are given in mA (10^-3 A). Myriam. PD: I will try to update the Wiki Hints with more hints of Week 4(Wiki -> Myrimit's guide to 6.002x -> [Myrimit's Hints][8]) ---- Now in Spanish: **H4P1: Diodo de Vacío** Aunque los tubos de vacío ya no son comúnmente utilizados en las computadoras o electrodomésticos, ellos aún poseen una sustancial aplicación en audio de alta frecuencia, transmisores de alta potencia, aceleradores de partículas, y microondas. Un diodo de vacío, el más simple tubo de vacío, es un interesante dispositivo de dos terminales. ![image][1] La característica tensión-corriente de un diodo de vacío es aproximadamente dada por la Ley de Childs-Langmuir, con un parámetro P deneminado Perveancia: ![formula1][2] **Hint Parte 1:** Ellos te dan una iP, esta iP estará expresada en función de vPK (es decir, la curva que puedes ver abajo): ![Imagen2][3] Bien, ahora que sabemos que la iP se comporta de una forma tal como la curva roja. En el caso de tener un valor de offset de tensión, VPK, podemos hallar su correspondiente IP? No es acaso cuando nuesta vPK toma aquél valor de tensión? Entonces, podemos hallar nuestra IP? Sí!!! ![Imagen3][4] Te sugiero que mires y prestes mucha atención a la cosa Zeng jaja [mirar aquí el (video S7V4 )][5]! **Hint Parte 2:** Deberías ver este video tutorial explicativo [ver aquí (Week 4 Tutorials - An abstract element)][6] :). Una vez que lo viste, cuál es entonces nuestra resistencia incremental solicitada? No es acaso una variación de nuestra vPK respecto de nuestra iP? Entonces, nosotros sabemos cómo calcular una variación muy pero muy pequeña de algo? entonces sabemos cómo calcular dicha resistencia incremental? Sí! **Hint Parte 3:** Idem parte 1 pero con diferente valor de vPK. **Hint Parte 4:** idem parte 2 pero con diferente valor de vPK. ---- Una cosa curiosa que vale la pena pensar: ![supuestos][7] Veamos que en el caso arbitrario 1), tendremos que la línea violeta de 1) es menor que la línea violeta de 2). Por lo cual, el cociente de (violeta 1/ celeste 1) = valor 1 y el cociente (violeta 2/ celeste 2)= valor 2. Podemos observar que el valor 1 será mayor que el valor 2. **Aha Moment!** Entonces, podemos predecir cuál de las resitencias incrementales será mayor? es decir, previo al cálculo de las resistencias incrementales de la parte 2 y 4, podrías predecir cuál será mayor o cuál menor? Eso es todo! Espero que esto les haya sido de ayuda para que luego puedan plantearlos ustedes solos en casa. Cuidado con las unidades :), algunas estan dadas en mA (10^-3 A). Myriam. PD: Recuerden que voy a tratar de actualizar las Wiki Hints de la semana 4, pueden revisar en Wiki -> Myrimit's guide to 6.002x -> [Myrimit's Hints][8]) [1]: https://www.edx.org/static/content-mit-6002x/images/circuits/VacuumDiode.847843a3f788.gif [2]: https://edxuploads.s3.amazonaws.com/13492159153015487.png [3]: https://edxuploads.s3.amazonaws.com/13492163707448034.png [4]: https://edxuploads.s3.amazonaws.com/13492168501343622.png [5]: https://www.youtube.com/watch?v=ou7LgMGBmkQ&feature=player_embedded#! [6]: https://www.youtube.com/watch?v=M6zrANNaY9Q&feature=player_embedded#! [7]: https://edxuploads.s3.amazonaws.com/13492189331343669.png [8]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/myrimits-hints-links/ UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2012-10-02T23:44:33Z VoteTAG: 19 CoursewareTAG: Week 4 / Graphical Interpretation of Incremental Method CommentableIdTAG: 6002x_graph_interp_inc_method NumberOfReplyTAG: 8 FirstChildTAG: Good explanation though this is the one problem I did not have issue with this week, except maybe the units! Any insight to H4P3 with the zener would be appreciated. I think I'll be working on it for a while... FirstChildUserIdTAG: 244115 FirstChildUserNameTAG: Pedro1969 FirstChildCreateTimeTAG: 2012-10-03T00:08:31Z SecondChildTAG: Thank you Pedro1969! :) Ok, I will post later some hints about H4P3. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-03T00:11:42Z SecondChildTAG: Hi Pedro1969! Could you solve the Zener issue? ;). Here some hints [H4P2 here][1]. Do you still having problems with H4P3 or did you mean H4P2? [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_graph_interp_inc_method/threads/5070be20fabaf62b0000004d SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T17:53:45Z FirstChildTAG: hello I'm still really lost not understand what is the function that devemos try use the exponential id = Is (e ^ vd / vt -1) but not where VPK substitute values ​​of p and also with fincion of P * v ^ 3/2 and no longer saw the video several times and I'm still in the same Myrimit can help me thanks hola yo aun sigo realmente perdida no entiendo cual es la funcion que devemos utilizar lo intente con la exponencial id=Is(e^vd/vt -1) pero no se en donde sustituir los valores de vpk y de p tambien con la fincion de P*v^3/2 y nada ya vi varias veces el video y sigo en las mismas me puedes ayudar Myrimit gracias FirstChildUserIdTAG: 320715 FirstChildUserNameTAG: maraivette FirstChildCreateTimeTAG: 2012-10-03T16:12:18Z SecondChildTAG: Hola maraivette! En el enunciado te dicen que el diodo en vacío se comporta como iP=P*(vPK^3/2) para vPK >0, es decir, para valores de vPK mayores que cero, la corriente iP será o estará dada por P*(vPK^3/2)... en cambio, cuando tienes valores de vPK menores a cero entonces, el valor de iP es igual a cero. Debes utilizar esta expresión que se te da en el enunciado. Recuerda que el valor de P es una constante cuyo valor se da como dato. :) Cualquier duda, avísame. Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-03T20:55:55Z SecondChildTAG: gracias por tu ayuda estaba haciendo mal mis calculos pero ahora para la resistencia tengo que derivar esta funcion cierto de p*vpk^3/2 qie seria 1/df(v) pero al parecer algo estoy haciendo mal porque no me da el resultado deseado segun yo stoy derivando mal me puedes guiar gracias SecondChildUserIdTAG: 321559 SecondChildUserNameTAG: dandradet SecondChildCreateTimeTAG: 2012-10-03T21:18:43Z SecondChildTAG: gracias por tu ayuda estaba haciendo mal mis calculos pero ahora para la resistencia tengo que derivar esta funcion cierto de p*vpk^3/2 qie seria 1/df(v) pero al parecer algo estoy haciendo mal porque no me da el resultado deseado segun yo stoy derivando mal me puedes guiar gracias SecondChildUserIdTAG: 320715 SecondChildUserNameTAG: maraivette SecondChildCreateTimeTAG: 2012-10-03T21:20:27Z SecondChildTAG: ok merezco un sape definitivamente estaba haciendo mal mis calculos ya pude resolver el problema completo y me vas a decir que como doy lata pero aun no me sale el calculo de v0 ya lo intente de diferente modos hasta con un simulador y no obtengo la respuesta solo de V0 en ambos casos me puedes explicar por favor en esta semana se me complico el curso gracias y disculpa la molestia es bueno poder hablar en mi idioma SecondChildUserIdTAG: 320715 SecondChildUserNameTAG: maraivette SecondChildCreateTimeTAG: 2012-10-03T21:38:42Z SecondChildTAG: Hola maraivette! Me alegra mucho que te haya salido el problema. Voy a tratar de seguir posteando más pistas para que puedan serte de ayuda :). No hay inconvenientes, pregúntame cuando tengas dudas, aquí estoy para poder ayudar a la comunidad de habla hispana. Saludos! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-04T00:00:43Z SecondChildTAG: Hola dandratet , te dare una pista, intenta primero expresar vPK en función de iP. Luego te será más fácil hallar la resistencia incremental ;). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-04T00:03:04Z SecondChildTAG: gracias myrimit de verdad espero ya no darte tanta lata es que esta semana se me complico el curso por cierto ya intente hacer el laboratorio pero no me sale el valor de vt creo estoy malen mi despeje si sabes de alguna tecnica para despejar te lo agradecere SecondChildUserIdTAG: 321559 SecondChildUserNameTAG: dandradet SecondChildCreateTimeTAG: 2012-10-04T03:58:10Z SecondChildTAG: Hola dandradet, hoy voy a intentar postear algo del lab4 :). No hay problema, estoy aquí para ayudarles en lo que pueda. Sinceramente, disfruto hacer esto. Saludos! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-04T10:55:04Z SecondChildTAG: HOLA MUCHAS GRACIAS MYRIMIT ESTA SEMANA REALMENTE ESTA COMPLICADA. SALUDOS SecondChildUserIdTAG: 321559 SecondChildUserNameTAG: dandradet SecondChildCreateTimeTAG: 2012-10-04T14:27:32Z FirstChildTAG: I dn't understand how to calculate Incremental Resistance... FirstChildUserIdTAG: 122513 FirstChildUserNameTAG: Mona77 FirstChildCreateTimeTAG: 2012-10-04T18:35:53Z SecondChildTAG: Hi Mona77! Hint: try to express your vPK in function of iP. You know that the incremental resistance it is a variation (part 2 hint). Isn't it a variation of a voltage respect a current? :). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-05T01:40:00Z SecondChildTAG: the ans is again putting in 204.77ohm... which is incorrect by the system SecondChildUserIdTAG: 395966 SecondChildUserNameTAG: sambo007 SecondChildCreateTimeTAG: 2012-10-07T07:22:57Z FirstChildTAG: Also, just in case id like to add that ya'll should be a little careful with the units given in this particular question. Take a good look at the units and then proceed with the calculations. I had the answer in front of me for hours, just had to tweak it a bit. FirstChildUserIdTAG: 204745 FirstChildUserNameTAG: allwynmendes FirstChildCreateTimeTAG: 2012-10-05T22:44:24Z SecondChildTAG: Thank you for the advice allwynmendes! :). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-06T02:00:15Z FirstChildTAG: buenas tardes Myrimit ya casi termino el lab 4 pero tengo un problema conla ultima pregunta que valores de ids y cgs hay que tomar para el calculo de k los ores o la diferencia de ellos ? me puedes orientar al respecto ahh y aunn no logro resolver el problema 2 de la tarea no encuentro valores de v0? gracias FirstChildUserIdTAG: 320715 FirstChildUserNameTAG: maraivette FirstChildCreateTimeTAG: 2012-10-04T17:54:08Z SecondChildTAG: Hola maraivette! Disculpa por la demora en responderte [aquí está la respuesta a tu pregunta][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5070e1b72210d02400000055 SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T01:59:36Z FirstChildTAG: it is not working with the incremental resistance.... by this trick,(0.0293/6)^-1 and it is actually not working.... huh!!!!!!! FirstChildUserIdTAG: 395966 FirstChildUserNameTAG: sambo007 FirstChildCreateTimeTAG: 2012-10-07T07:18:59Z SecondChildTAG: Hi sambo007! Can I help you? SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T08:04:02Z FirstChildTAG: Hi Myrimit, as from the beginning you've been my favorite. Always posting valuable material for helping. Hats off...! FirstChildUserIdTAG: 378096 FirstChildUserNameTAG: Jamshaid271 FirstChildCreateTimeTAG: 2012-10-07T08:26:11Z SecondChildTAG: Thank you Jamshaid271 :)! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T17:49:57Z FirstChildTAG: I am still having a problem calculating the resistance, I have the derivatives at the two voltages but I still get them marked wrong, I have tried plotting the graph and doing them by hand!!! I am lost HELP please! FirstChildUserIdTAG: 238005 FirstChildUserNameTAG: isisbocardo FirstChildCreateTimeTAG: 2012-10-07T19:31:30Z SecondChildTAG: Can I help you? Hint: Try to express your vPK in function of iP ;). If you derivate and obtain a ratio (slope) you will find it ;). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T20:04:31Z SecondChildTAG: So I plotted iP=0.002*vPK^(3/2) then took the tangents at the vPK I was given those formulas should give me resistance but the numbers I get are wrong! I got vPK at 4V and at 10V and the formulas I get from the tangents are: @ 4V iP=0.006vPK-0.008 @ 10V iP = 0.00948vPK-0.32 I am stuck pass this point! SecondChildUserIdTAG: 238005 SecondChildUserNameTAG: isisbocardo SecondChildCreateTimeTAG: 2012-10-07T20:54:38Z SecondChildTAG: Remember that if you plot graphically iP vs vPK, your tangent will be the inverse of the resistance ;). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T21:00:29Z SecondChildTAG: I am sorry but I am still not getting it, I tried the inverse of the two equations and the resistances I got are still wrong, I ended up with (R^-1)=0.006x-.008 1/R=0.006x-0.008 R=500/(-4+3x) plug in my 4 for x and got 62.5 but that marks me wrong for the resistance, I did something similar for the 10 V SecondChildUserIdTAG: 238005 SecondChildUserNameTAG: isisbocardo SecondChildCreateTimeTAG: 2012-10-07T21:42:36Z SecondChildTAG: I found the answer after I saw your post thank you so much for you help, the inverse part made sense and when I plugged the numbers it came clear also reading page 222!!! :D Wonderful help SecondChildUserIdTAG: 238005 SecondChildUserNameTAG: isisbocardo SecondChildCreateTimeTAG: 2012-10-08T05:30:37Z SecondChildTAG: You are welcome isisbocardo! Well done :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-08T12:09:21Z IndexTAG: 64 TitleTAG: D'oh! Ids is in mA Don't enter your answer in amperes like I did. UserIdTAG: 202518 UserNameTAG: MCN CreateTimeTAG: 2012-09-28T02:10:07Z VoteTAG: 19 CoursewareTAG: Week 5 / MOSFET model exercise CommentableIdTAG: 6002x_mosfet_mod_e NumberOfReplyTAG: 2 FirstChildTAG: Friend, 1 mA = 0.001 Ampers or 1x10^-3 Ampers FirstChildUserIdTAG: 459662 FirstChildUserNameTAG: FRANCISCOG FirstChildCreateTimeTAG: 2012-09-28T02:32:32Z FirstChildTAG: too late FirstChildUserIdTAG: 262875 FirstChildUserNameTAG: GrantDennison FirstChildCreateTimeTAG: 2012-10-08T23:21:02Z IndexTAG: 65 TitleTAG: Video Freezing YouTube made a recent change to their Flash player that broke our video embeds. As a temporary workaround while we figure things out, you can use YouTube's HTML5 player. To do so, please go to YouTube's [Join the HTML5 Trial][1] page and click "Join the HTML5 Trial" at the bottom. Then reload our site in your browser. Thank you for your patience as we work on this. UPDATE: We've temporarily had to revert to using just the plain YouTube player without our UI enhancements. This is just a temporary measure to allow people to watch the full video while we work on a more permanent fix that restores the previous functionality. Thank you. [1]: http://www.youtube.com/html5 UserIdTAG: 11 UserNameTAG: dormsbee CreateTimeTAG: 2012-09-13T02:25:43Z VoteTAG: 19 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 3 FirstChildTAG: Thanks Dave! :D FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-13T02:29:39Z FirstChildTAG: **Thanks!!** FirstChildUserIdTAG: 118093 FirstChildUserNameTAG: Oktavious FirstChildCreateTimeTAG: 2012-09-13T05:30:33Z FirstChildTAG: still same problem here .... transcript on the left side disappears, plz help FirstChildUserIdTAG: 96584 FirstChildUserNameTAG: mujahid FirstChildCreateTimeTAG: 2012-09-13T09:15:25Z IndexTAG: 66 TitleTAG: MIT hello sir, I'm 22 year old guy belongs to INDIA ,doing my engineering from electronics and communication. I have recently finished with 6.002,that was awesome; but sir i just wanted to know that if we can have practical work or lab work related to electronics gadgets and some exciting stuff(as MIT is known for)...there is no such facility in my collage we are doing lots of boring practicals in lab,there isn't such new things.... 6.002x provide some demo which was really good, but i and some of my friend who have joined edx wanted a total lab work . so sir can we have some courses like EMBEDDED SYSTEM,ROBOTICS or MECHATRONICS... UserIdTAG: 108863 UserNameTAG: shiviz CreateTimeTAG: 2012-12-26T04:46:46Z VoteTAG: 18 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: same as said by shiviz. FirstChildUserIdTAG: 404188 FirstChildUserNameTAG: jishudasorissa FirstChildCreateTimeTAG: 2012-12-26T14:39:51Z IndexTAG: 67 TitleTAG: Midterm Exam Due Date Appeal Hello 6.002x Staff, Thank you very much for this course! The due date for the midterm exam being on a Thursday could be somewhat difficult. I know that for me in particular the last week of October is one of the busiest weeks of the year at work. Changing the date of the midterm to the following Sunday would give students the opportunity to prepare for and do the midterm during a full week and/or weekend in place of a homework and lab regularly due on that day. That might be more convenient and advantageous for many. I know that it would be for me! It appears (from the Course Calendar and Course-at-Glance sheet) that such a convenient lag is already set to precede the final. Thanks again! Laureen PS: When signing up for edX, I mistakenly listed only my last name, rather than my first and last name, etc., for my full name and have not yet found a way to fix that. Does anyone know how? Thanks in advance! UserIdTAG: 188609 UserNameTAG: Laureen CreateTimeTAG: 2012-10-06T20:34:54Z VoteTAG: 18 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 1 FirstChildTAG: I'm not sure if you are still looking at this, or if you've figured this out by now, but you can change your listed name by clicking on "edit" next to your full name on the edX [Dashboard][1]. Cheers. [1]: https://www.edx.org/dashboard FirstChildUserIdTAG: 359310 FirstChildUserNameTAG: AaronYeoh FirstChildCreateTimeTAG: 2012-11-26T11:50:16Z IndexTAG: 68 TitleTAG: Add a Clock to the website for the Spring? I was wondering if it would be possible to have a clock built into the edX platform. This could either be: 1. A Master Clock which displays Boston time to everyone or, 2. Have a location specific calibrated clock when users are logged in. The first option being a simplistic, reliable and a practical benchmark. The second option is more convenient for logged-in users, but it may still cause confusion as everyone will be talking with respect to their local time in the forum. The first option is simple in theory but may require the clock to be real-time or have the page autorefresh etc. The second option does not have to be a real-time clock, it just has to adjust the start and finish times for the specific location. Anyhow I thought I would put this here to see what others think as well. UserIdTAG: 292543 UserNameTAG: Pennypacker CreateTimeTAG: 2012-12-23T15:35:08Z VoteTAG: 17 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 2 FirstChildTAG: It will be helpful. FirstChildUserIdTAG: 128276 FirstChildUserNameTAG: ChengBin FirstChildCreateTimeTAG: 2012-12-23T16:04:55Z SecondChildTAG: I concur; especially with everyone confused when exams start or end, even more so at the beginning of each semester with all the new students. It would take a lot of commonly-asked questions from taking up Community TA time, so we can concentrate on helping students with Homeworks and Labs. SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-12-23T17:24:35Z FirstChildTAG: In similar situations, I've kept a browser tab open to a site providing [online world time][1], but I believe this suggestion has a great deal of merit. [1]: http://wwp.greenwichmeantime.com/ FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-12-23T18:45:11Z IndexTAG: 69 TitleTAG: New courses about Eletronics Engineering Hi , I am very grateful for this amazing online course platform , it's pretty helpful for everyone who need to emphasize and improve their skills on electronics engineering .Presently , we almost finish the course of 6.002x. then , i have a concern : Will there be some more courses in the future and when are they avialable online ? Thanks UserIdTAG: 232667 UserNameTAG: KyleLiux CreateTimeTAG: 2012-12-17T04:31:28Z VoteTAG: 17 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I also hope to see 6.004x and 6.014x FirstChildUserIdTAG: 136959 FirstChildUserNameTAG: christerpher FirstChildCreateTimeTAG: 2012-12-17T05:05:20Z SecondChildTAG: what is the different between 6.004 and 6.00? SecondChildUserIdTAG: 318037 SecondChildUserNameTAG: Maher-84 SecondChildCreateTimeTAG: 2012-12-17T16:06:44Z SecondChildTAG: you can search it on mit ocw.600 is about programming.but6004 is computation structure SecondChildUserIdTAG: 136959 SecondChildUserNameTAG: christerpher SecondChildCreateTimeTAG: 2012-12-18T00:39:01Z IndexTAG: 70 TitleTAG: Small piece of advice to EDX developers. Hello, I thank you for such a great website. But a small advice, Please let the Boston time Ticking in some corner of this website. That would be better for every one. Thank you Manoj Kumar S UserIdTAG: 156974 UserNameTAG: ManojKumar CreateTimeTAG: 2012-10-26T10:07:57Z VoteTAG: 17 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: I do agree with ManojKumar, but in the mean while we have http://www.timeanddate.com/worldclock/converter.html wiz. the 1st link that you'll get if you google "time converter". Its pretty good and fairly simple to use. FirstChildUserIdTAG: 204745 FirstChildUserNameTAG: allwynmendes FirstChildCreateTimeTAG: 2012-10-26T13:23:26Z SecondChildTAG: Good tip. Here is the specific Boston page they can bookmark from your link in the meantime. http://www.timeanddate.com/worldclock/city.html?n=43 SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-26T14:46:55Z FirstChildTAG: If you want to put a clock with Boston time on it on your computer screen, go knock yourself out. I rather like having a midnight Sunday deadline (my time) every week. You can take the mid-term and final exams anytime over several days. There is almost never a reason to know or care about Boston time. I don't need unnecessary clutter on my screen. Clutter is the least of the problems with your suggestion. If you think about it seriously, you will realize this is not the only course. There are several edx courses, with deadlines in various time-zones. If you add in Coursera and other sources of online classes, the time-zones are around the world. Your plan, if adopted, is just going to create confusion and anguish. Many students are taking several classes simultaneously, and they are very busy trying to keep up with all the deadlines. Personally I keep them straight by converting them to my time-zone. Having multiple class windows open with different times displayed would not help at all. FirstChildUserIdTAG: 9673 FirstChildUserNameTAG: xp42 FirstChildCreateTimeTAG: 2012-10-26T15:18:43Z SecondChildTAG: Thanks for taking pain in explain why there should not be a clock in the sceen!! :P SecondChildUserIdTAG: 156974 SecondChildUserNameTAG: ManojKumar SecondChildCreateTimeTAG: 2012-10-26T15:26:05Z FirstChildTAG: Yes, that's a good idea. But I'd like that clocks to be re removeable. P.S: If you use Linux Ubuntu, than you can add a Boston time ticking to the calendar panel. FirstChildUserIdTAG: 326787 FirstChildUserNameTAG: Angstrem FirstChildCreateTimeTAG: 2012-10-26T17:56:07Z FirstChildTAG: I agree. While it is not necessary, it could be very useful and save some annoyance. I suggest creating an "Exam Remaining Time" countdown too. FirstChildUserIdTAG: 392546 FirstChildUserNameTAG: JohnWayne FirstChildCreateTimeTAG: 2012-10-26T10:14:11Z SecondChildTAG: Just let the clock ticking with time and date! not specially for exam. For everything. SecondChildUserIdTAG: 156974 SecondChildUserNameTAG: ManojKumar SecondChildCreateTimeTAG: 2012-10-26T10:40:30Z IndexTAG: 71 TitleTAG: Staff: A Request for Spring Courses It has been a exciting month all along.... With Edx offering so many courses,many students are enjoying as well as gaining experience like never before... For me, I had already taken the prototype course offered by MITX in this spring..... Realy amazed by the beauty of al-together new concepts and different point of views Offered by Agarwal sir... But here i would like u to add more courses related to Electronics and its application (may be the Course Succeeding 6002x). This Fall majority of the courses offered by EDX are "Computer Courses"...... I am a computer lover too but would love to Gain knowledge relating to Electronic Circuits and their Applications.... Courses like: "Signals And System" "Electro-magnetics And its application" "Digital and analog Communication" "Analog Circuits design"...etc would be really Fun to learn..... Especially Dr Agarwals "Ahaa!!!" moments in them would be great to catch.... Just a little curiosity made me to post this....cant help it!! :P I know its a bit demanding but really many of us are greeding for more...(Electronics!!!) Just want to say that i cant thank you all with all my words for this Extraordinary initiative that u have taken for mankind!!!! Keep it up guys!!! :) :) I invite all members of 6002x to add courses appealing to them....(but all of them should be related to Electronics!!!...) :) :) What say guys??? UserIdTAG: 82597 UserNameTAG: bondrajat CreateTimeTAG: 2012-10-19T20:03:03Z VoteTAG: 17 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 3 FirstChildTAG: i would love to add some courses to the above list like- digital design electronic and electrical instrumentation control systems robotics FirstChildUserIdTAG: 368712 FirstChildUserNameTAG: jmen FirstChildCreateTimeTAG: 2012-10-20T12:14:45Z SecondChildTAG: it would be a great aha moment if control system is offered by MITX SecondChildUserIdTAG: 358539 SecondChildUserNameTAG: syed_abdullah SecondChildCreateTimeTAG: 2012-10-20T12:58:15Z SecondChildTAG: Yup i missed that!!! Instead i invite all the members to Add courses of their choice to the above list.... (but the courses should be relating to electronics applications)........ SecondChildUserIdTAG: 82597 SecondChildUserNameTAG: bondrajat SecondChildCreateTimeTAG: 2012-10-20T12:59:38Z FirstChildTAG: I would like to add few more courses to the above list - Digital IC Design - Analog IC Design - VLSI Design - Embedded Systems FirstChildUserIdTAG: 362299 FirstChildUserNameTAG: anandbaskaran FirstChildCreateTimeTAG: 2012-10-20T17:49:15Z SecondChildTAG: The above courses are amazing.... would be fun to learn.... But hope that some of them will be released soon enough!!!! :) By the way good choices anandbaskaran!!!! SecondChildUserIdTAG: 82597 SecondChildUserNameTAG: bondrajat SecondChildCreateTimeTAG: 2012-10-20T18:21:38Z FirstChildTAG: I hope this too Could any staff give an official reply? Thanks FirstChildUserIdTAG: 136959 FirstChildUserNameTAG: christerpher FirstChildCreateTimeTAG: 2012-10-21T00:49:05Z IndexTAG: 72 TitleTAG: Hint for those struggling with H6P1 Q1!! ![enter image description here][1] ![enter image description here][2] ![enter image description here][3] [1]: https://edxuploads.s3.amazonaws.com/13504899119301294.jpg [2]: https://edxuploads.s3.amazonaws.com/13504899171343651.jpg [3]: https://edxuploads.s3.amazonaws.com/1350489923811414.jpg And **Voila!** Hazel. UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-10-17T15:52:59Z VoteTAG: 17 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 10 FirstChildTAG: Why then it can not get my answer right then? FirstChildUserIdTAG: 286954 FirstChildUserNameTAG: ododo FirstChildCreateTimeTAG: 2012-10-17T17:35:12Z SecondChildTAG: Post what you have done so far. I know one thing that got me was the signs. SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-17T20:20:21Z SecondChildTAG: got exactly same like in your formula. And it don't work SecondChildUserIdTAG: 286954 SecondChildUserNameTAG: ododo SecondChildCreateTimeTAG: 2012-10-17T21:47:16Z SecondChildTAG: Then you are writing it wrong, or you have a sign wrong or have taken the wrong root. Double check! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-18T07:01:31Z SecondChildTAG: Looks like this (1+-sqrt(1+4krVS(...))/(-2kr(...)). IS it right? SecondChildUserIdTAG: 286954 SecondChildUserNameTAG: ododo SecondChildCreateTimeTAG: 2012-10-18T23:07:31Z SecondChildTAG: check the sign before 2 and 1 and R instead of r! SecondChildUserIdTAG: 122513 SecondChildUserNameTAG: Mona77 SecondChildCreateTimeTAG: 2012-10-19T08:13:02Z SecondChildTAG: (­-1+-sqrt(1+4*R*K*VS*(....)))/(2*R*K*(....)) Here, still not working. Is that's a way you wrote. SecondChildUserIdTAG: 286954 SecondChildUserNameTAG: ododo SecondChildCreateTimeTAG: 2012-10-19T13:51:24Z SecondChildTAG: ('-'1'+-'sqrt(1+4*R*K*VS*(....)))/('+'2*R*K*(....)) Everything quoted must be changed, please be extremely careful with your signs! Remember you can only have either + or a minus before the sqrt. SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-19T14:03:25Z SecondChildTAG: Hazel. .plz help me.i am is got stucked in the Homework6 question 1,i have solved the quadratic equation and got the ans like this Vout=+-X/[1+4*X*X*X(X-X)]^1/2 ,But not getting the green tick,plz help me SecondChildUserIdTAG: 97359 SecondChildUserNameTAG: azeem179 SecondChildCreateTimeTAG: 2012-10-19T20:13:02Z SecondChildTAG: Your equation should be **(1-sqrt(1+4*x*x*(x-x)*x))/(-2*x*x*(x-x))** SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-19T20:34:59Z SecondChildTAG: Where are your problems? SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-19T20:35:29Z SecondChildTAG: (1-sqrt(1+4*k*R*VS*(vIN-VT)))/(-2*k*R*(vIN-VT)) Whats wrong here??? Why it does not work??? SecondChildUserIdTAG: 286954 SecondChildUserNameTAG: ododo SecondChildCreateTimeTAG: 2012-10-20T00:34:09Z SecondChildTAG: got it the green tick. .thanks hazel1919 SecondChildUserIdTAG: 97359 SecondChildUserNameTAG: azeem179 SecondChildCreateTimeTAG: 2012-10-20T07:51:51Z SecondChildTAG: i am getting answer after apply L-hospital is vOUT=X/sqrt((1+4*X*X*(X-X)*X)...still getting wrong...please help SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-10-20T09:58:13Z SecondChildTAG: That is the wrong topology of the equation. It should look exactly as follows, if you follow the instructions above. **(1-sqrt(1+4*k*R*VS*(vIN-VT)))/(-2*k*R*(vIN-VT))** SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-20T10:24:21Z SecondChildTAG: Srry I above equation is wrong. I mean: **(1-sqrt(1+4*x*x*(x-x)*x))/(-2*x*x*(x-x))** SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-20T10:25:29Z SecondChildTAG: Finally, Thanks a bunch SecondChildUserIdTAG: 286954 SecondChildUserNameTAG: ododo SecondChildCreateTimeTAG: 2012-10-20T14:05:59Z SecondChildTAG: Still didnt get it..have the same solution as given above..but still no green tick..someone please help.. SecondChildUserIdTAG: 173147 SecondChildUserNameTAG: cruiser_rahit SecondChildCreateTimeTAG: 2012-10-20T20:53:51Z FirstChildTAG: Now this'll help! Thanks! :D FirstChildUserIdTAG: 108454 FirstChildUserNameTAG: Raven7281 FirstChildCreateTimeTAG: 2012-10-17T16:18:46Z SecondChildTAG: I hope so! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-19T14:03:34Z FirstChildTAG: Very kind of you to take the time to prepare a nice write up for others to follow. One reaction, as a long time electronics tinkerer/hobbyist using the node method to get vOUT looks like massive overkill, but your mileage may vary! Best regards, Skyhawk FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-17T16:21:30Z SecondChildTAG: Thanks! How would you suggest an improvement? I do this mainly for my self, the concepts are really shaky in my head! :) SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-19T14:04:27Z FirstChildTAG: Many many thanks FirstChildUserIdTAG: 455950 FirstChildUserNameTAG: Wahabbaluch FirstChildCreateTimeTAG: 2012-10-18T04:29:07Z SecondChildTAG: A pleasure!! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-19T14:04:46Z FirstChildTAG: Thanks hazel1919 I got answer, I was struggling with and L hospital rule Thanks again FirstChildUserIdTAG: 352647 FirstChildUserNameTAG: praveenjugge FirstChildCreateTimeTAG: 2012-10-18T06:14:58Z SecondChildTAG: sings SecondChildUserIdTAG: 352647 SecondChildUserNameTAG: praveenjugge SecondChildCreateTimeTAG: 2012-10-18T06:15:26Z SecondChildTAG: I am too much confused. Do I have to apply L Hospital rule or not, after I get the expression for vOUT? SecondChildUserIdTAG: 381797 SecondChildUserNameTAG: Saira180 SecondChildCreateTimeTAG: 2012-10-18T06:32:20Z SecondChildTAG: No it is no necessary apply L'Hôpital rule to get the correct answer SecondChildUserIdTAG: 362840 SecondChildUserNameTAG: emata SecondChildCreateTimeTAG: 2012-10-18T08:36:02Z SecondChildTAG: Well to choose the correct sign for the radical you do need to take the limit as vIN -> goes to VT and see which sign gives the correct result. Of course you can guess or someone can tell you! SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-18T11:58:42Z SecondChildTAG: How do you apply L'Hospital rule?? SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-19T14:05:09Z FirstChildTAG: Could someone explain to me why iDS is negative? I'm getting confused about the sign conventions for the node method/equation. FirstChildUserIdTAG: 194270 FirstChildUserNameTAG: dphung FirstChildCreateTimeTAG: 2012-10-18T03:54:48Z SecondChildTAG: Is it because the current is going from top to bottom through the resistor? what simple thing am i missing? SecondChildUserIdTAG: 194270 SecondChildUserNameTAG: dphung SecondChildCreateTimeTAG: 2012-10-18T03:57:53Z SecondChildTAG: It is because we moved it from the left hand side of the equation to the right hand side. Algebraically if: **((X-X)/X)+iDS=0** Then... **((X-X)/X)=iDS** SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-18T07:10:43Z SecondChildTAG: Sorry this is important! Ignore above! Algebraically if: **((X-X)/X)+iDS=0** Then... **((X-X)/X)=-iDS** SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-18T07:11:55Z FirstChildTAG: to what extent do we have to simplify the equation ? mines in giving "couldnot parse" FirstChildUserIdTAG: 278792 FirstChildUserNameTAG: sali FirstChildCreateTimeTAG: 2012-10-19T03:26:55Z SecondChildTAG: Post your problem. SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-19T14:05:26Z FirstChildTAG: yay for the tick of niceness +4 FirstChildUserIdTAG: 335803 FirstChildUserNameTAG: bobbanovski FirstChildCreateTimeTAG: 2012-10-22T12:52:12Z FirstChildTAG: These are some legit steps. Got the 1st part right. Thanks Hazel FirstChildUserIdTAG: 204745 FirstChildUserNameTAG: allwynmendes FirstChildCreateTimeTAG: 2012-10-21T12:31:14Z FirstChildTAG: how do we get vOUT^2 in equation 1 of iDS? FirstChildUserIdTAG: 401175 FirstChildUserNameTAG: Raghav12 FirstChildCreateTimeTAG: 2012-10-21T21:40:25Z IndexTAG: 73 TitleTAG: Wolfram Tutorial on Calculating Partial Derivatives (as encoundered in S11E1) Dear all, If you, like me, don't have all the calculus prerequisites, then Wolfram Alpha is going to be an important tool. http://www.wolframalpha.com/ *I recommend creating an account with Wolfram, as this unlocks a lot of extra features (Such as converting the results of your equations into Copyable plain text)!* Unfortunately, the Wolfram Alpha website is also subtle when it comes to it's features. I am going to use Week Six S11V7 as an example, to show you how to use Wolfram Alpha as a tool to take the **partial differential** of **K/2*(VGS-VT)^2** : https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_6/Small-Signal_Circuit_Models/ So the first step is to get to the correct calculator that we want... click on the link to Wolfram: http://www.wolframalpha.com/ Then click 'examples' just below the search bar. Then click on 'Calculus' under the maths subheading. ![enter image description here][1] Then click on the 'derivatives>>' button, which will give more options. ![enter image description here][2] Now we want to calculate a partial derivative with respect to **vGS**. So we select the appropriate calculator for that task. ![enter image description here][3] [1]: https://edxuploads.s3.amazonaws.com/13503782302771469.jpg [2]: https://edxuploads.s3.amazonaws.com/1350378378134369.jpg [3]: https://edxuploads.s3.amazonaws.com/13503786681343603.jpg Now, how do we enter the expression? We want the partial derivative of **K/2*(vGS-VT)^2** with respect to **vGS**. Wolfram doesn't like **vGS's** and **VT's** so we substitute them for alphabetic letters, i.e. **vGS=a**, **VT=b** and **K**. *Just remember what equals what.* Now our expression looks like this **K/2*(a-b)^2**. So how do we parse this expression keeping in mind we want it with respect to **vGS** (which is a)? We simply write: **Differentiate (equation) wrt (variable)** So for our expression we would write: **Differentiate K/2*(a-b)^2 wrt a** This gives us **(d)/(da)(1/2 K (a-b)^2) = K (a-b)** So... **K(a-b)** or substituting **'a'** for our operating point **VGS** and **'b'** for **VT** we get: **K(VGS-VT)** Multiplying by our small signal **vgs** we get the formula for our small signal **ids**: **ids=K(VGS-VT)*vgs** And that's it! Hopefully this will help you like it did me. Hazel, UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-10-16T09:39:34Z VoteTAG: 17 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Thanks!!! FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-10-19T00:51:44Z FirstChildTAG: Yes, thank you very much for explaining how to use this extremely useful tool! FirstChildUserIdTAG: 175205 FirstChildUserNameTAG: KT_MM FirstChildCreateTimeTAG: 2012-10-19T01:24:50Z FirstChildTAG: http://www.youtube.com/watch?v=UFWAu8Ptth0&feature=relmfu FirstChildUserIdTAG: 73699 FirstChildUserNameTAG: FranciscoAguilera FirstChildCreateTimeTAG: 2012-10-22T08:40:45Z IndexTAG: 74 TitleTAG: H5P1 Hints Requested by Ignaas (Read down for Spanish translation ) **H5P1 ZERO-OFFSET AMPLIFIER** In many amplifiers, we use dual power supplies so we can obtain a 0V offset at the output, that is, the DC operating point at the output is 0V. An example amplifier circuit that can achieve this is shown below. ![image][1] In this circuit, VS+=1.0V, VS−=−1.0V, and the MOSFET parameters are K=1 mAV2 and VT=0.5V. - Before to start lets see some clarifications of signs with an arbitrary example, that this was a little bit confusing: Suppouse that we have an arbitrary circuit like this one and try to find VX: ![imagen][2] WAY 1) If you write the KVL equations, you will have: (VS-) + VX - (VS+) = 0 VX = (VS+) -(VS-) Lets also assume that they give us (VS-)= -2 V and (VS+)= 5 V , Ok, so your VX will be VX= 5V - (-2V) = 7V WAY 2) If the way 1) confuses you, you can take the following circuit, it is the same but represented in a different way: ![im2][3] invert the source and take the module: /Vs-/= VM = /-2 V/ = 2 V KVL -VM +Vx-(Vs+)=0 Vx= (Vs+) + VM = 5 V + 2 V = 7 V ---------- Ok, now that the sign convention was clarified, lets start: **Part 1:** What is the minimum value of input voltage VIN in volts for the MOSFET to be operating in saturation region? **Visual Hint Part1:** vDS (Drain Source) , vGS (Gate Source). ![mos][4] **Another Visual Hint Part 1:** Can you write an KVL expression in this loop? ![mos2][5] **Hint Textbook Part 1:** What value it takes vGS in the saturated region? What is that minimun value of vGS? :). Textbook page 338 [read here][6] . If we have that value and our KVL equation can we find our vIN? Yes! ---------- Part 2: What must be the value of RL in Ohms such that it achieves VOUT=0V when VIN=0V? Assume that the MOSFET is operating in saturation. They are telling us that: - VOUT=0V - VIN=0V And we want to know RL = ?? **Hint Part 2:** How does the current of drain source iDS behaves in the saturated region? Read the page 341 [here][7] **Visual Hint Part 2:** Can you write a KVL in the violet loop? ;). If you have the value of iDS, and you have this KLV can you find RL? Yes! Remember that some values given in the statement are set to zero :p ![mos3][8] ---- **Part 3:** As VIN is increased, the output voltage VOUT decreases and the MOSFET goes out of saturation. For the value or RL found above, what is the maximum input voltage VIN in volts such that the MOSFET will remain in the saturation region? **Hint Part 3:** Take a look at the page 358 of the Textbook [here][9] . There it is explaned how to determine the highest value of input voltage for which MOSFET satisfies the saturation discipline. You should take a look at this part of the Textbook. But, here, it is not developed for our specific circuit, so, there will be some differences between that and this. **Another Hint Part 3:** be careful with signs! some value is negative and your voltage shift not :P Remember that you can search in the Forum Discussion there is a lot of data of this last part ;) I hope this can help you. Myriam. ---- (Now in spanish ) **H5P1 ZERO-OFFSET AMPLIFIER** En muchos amplificadores, se utilizan dos fuentes de alimentación para obtener un offset de 0V a la salida, esto es, el punto de operación a la salida es de 0V. Un ejemplo de este circuito amplificador es el que se muestra abajo. ![image][1] En este circuito, VS+=1.0V, VS−=−1.0V, y el parámetro K del MOSFET es K=1 mAV2 y VT=0.5V. - Antes de comenzar veamos algunas cosas a tener en cuenta o a poner en claro sobre los signos. Veamos esto con un ejemplo arbitrario, ya que esto fue de mucha confusión: Supongamos un circuito arbitrario como el que esta abajo y tratemos de hallar VX: ![imagen][2] Forma 1) Si escribes la ecuación utilizando KVL, tendrás: (VS-) + VX - (VS+) = 0 VX = (VS+) -(VS-) También, adoptemos para este ejemplo que se nos dan los siguientes valores (VS-)= -2 V and (VS+)= 5 V , Entonces, nuestra VX será: VX= 5V - (-2V) = 7V FORMA 2) Si la FORMA 1) te es de mucha confusión, puedes tomar el siguiente circuito equivalente, que es lo mismo que el anterior, pero con la diferencia que se encuentra representado en una forma diferente: ![im2][3] Invierte la fuente pero toma su valor en módulo : /Vs-/= VM = /-2 V/ = 2 V KVL -VM +Vx-(Vs+)=0 Vx= (Vs+) + VM = 5 V + 2 V = 7 V ---------- Bien, ahora que hemos entendido la convención de signos, manos a la obra! **Parte 1:** Cuál es valor mínimo de tensión de entrada vIN en volts para el MOSFET que opera en la región de saturación? **Visual Hint Parte1:** vDS (Drain Source) , vGS (Gate Source). ![mos][4] **Otra Visual Hint Parte 1:** Puedes escribir una ecuación utilizando KVL en este loop? ![mos2][5] **Hint Textbook Parte 1:** Qué valor tomo vGS en la región de saturación? Cuál es el mínimo valor de vGS? :). Leer el Texbook en la página 338 [leer aquí][6] . Si tenemos dicho valor como así también nuestra ecuación KVL podemos hallar nuestra vIN? Sí! ---------- Parte 2: Cuál debe ser el valor de RL expresado en Ohms tal que VOUT =0V cuando VIN=0? Se debe asumir que el MOSFET opera en la zona de saturación. Nos están diciendo que: - VOUT=0V - VIN=0V Y se quiere saber el valor de RL = ?? **Hint Parte 2:** Cómo es que la corriente drenaje fuente iDS se comporta en la región de saturación? Leer la página 341 [aquí][7] **Visual Hint Parte 2:** Es posible escribir una ecuación utilizando KVL en el loop violeta? ;). Si tienes el valor de iDS, y también esta ecuación que surge de aplicar el KVL, puedes hallar tu RL? Sí! Recuerda que alguno de los valores que se dan en el enunciado tienen valores nulos :p ![mos3][8] ---- **Parte 3:** A medida que VIN se incrementa, la tensión de salida VOUT se decrementa y el MOSFET sale de la región de saturación. Para el valor de RL del punto anterior, cuál es el voltaje VIN en volts tal que lo haga permanecer en la zona de saturación? **Hint Parte 3:** Mirar la página 358 del Textbook [aquí][9] . Aquí se explica cómo determinar el valor más alto para el cual la tensión de entrada del MOSFET satisface la saturación. Deberías leer con minuciosidad esta página del Textbook. Pero, encontrarás, que el desarrollo hecho en el libro no es para nuestro circuito en particular, entonces, existirán algunas pequeñas diferencias entre aquello y esto, pero sí les servirá como guía. **Otra Hint Parte 3:** Ten cuidado con los signos!Habrá algún valor negativo y otro gap o intervalo de desplazamiento de tensión que no lo es :P Recuerda que puedes buscar cn search en este Foro, hay mucha información sobre esta última parte ;) Espero que esto te haya servido de ayuda. Myriam. [1]: https://www.edx.org/static/content-mit-6002x/images/circuits/H5P1_Zero_Offset_Amplifier.7faf7a368520.png [2]: https://edxuploads.s3.amazonaws.com/13501765601343603.png [3]: https://edxuploads.s3.amazonaws.com/1350177376134367.png [4]: https://edxuploads.s3.amazonaws.com/13501780631343628.png [5]: https://edxuploads.s3.amazonaws.com/1350178267694178.png [6]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/362 [7]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/341 [8]: https://edxuploads.s3.amazonaws.com/13501789641343661.png [9]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/382 UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2012-10-14T02:03:25Z VoteTAG: 17 CoursewareTAG: Week 5 / MOSFET Amplifier CommentableIdTAG: 6002x_mosfet_amp NumberOfReplyTAG: 4 FirstChildTAG: thnx!!!!!!!!!!!!!! FirstChildUserIdTAG: 308861 FirstChildUserNameTAG: Ignaas FirstChildCreateTimeTAG: 2012-10-14T02:16:59Z SecondChildTAG: You are welcome Ignaas! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T02:33:00Z FirstChildTAG: Thanks Myriam, for putting time into this explanation. FirstChildUserIdTAG: 194270 FirstChildUserNameTAG: dphung FirstChildCreateTimeTAG: 2012-10-14T04:51:44Z SecondChildTAG: You are welcome dphung :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T14:29:00Z FirstChildTAG: but isn't the minimum value of input voltage VIN simply the VT? FirstChildUserIdTAG: 260272 FirstChildUserNameTAG: saikat24 FirstChildCreateTimeTAG: 2012-10-14T15:05:31Z SecondChildTAG: No, because VIN is not VGS for this circuit. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-14T18:34:51Z SecondChildTAG: Why not? Or, how can you tell? SecondChildUserIdTAG: 325730 SecondChildUserNameTAG: gadzin1203 SecondChildCreateTimeTAG: 2012-10-14T22:30:54Z SecondChildTAG: Look at what VIN is referenced to, and look at the voltage on the source. SecondChildUserIdTAG: 308617 SecondChildUserNameTAG: jrosenberger SecondChildCreateTimeTAG: 2012-10-14T23:21:47Z FirstChildTAG: Hello, I need your help with part (a). I think the answer specified only ensures conduction, not saturation. What do you think? See the full post here: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/507eda1981bd8ada12000023 Thanks, Ignacio FirstChildUserIdTAG: 276409 FirstChildUserNameTAG: IgnacioUY FirstChildCreateTimeTAG: 2012-10-18T18:35:34Z SecondChildTAG: Hi Ignacio :). Ok, I will take a look. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-19T03:42:28Z IndexTAG: 75 TitleTAG: Recomendation see page 417 of book UserIdTAG: 155008 UserNameTAG: sohailahmed CreateTimeTAG: 2012-10-02T18:39:42Z VoteTAG: 17 CoursewareTAG: Week 5 / Two Terminal Connection Exercise CommentableIdTAG: 6002x_2_term_con_e NumberOfReplyTAG: 0 IndexTAG: 76 TitleTAG: Progress I love how the scores are stored in the "Progress" section, and when you get those green ticks next to your answers. If there were "achievments" to unlock, or "secrets" you could discover by dutifully going through the lectures/excercises/textbook material, it would feel almost like a video game :) UserIdTAG: 364822 UserNameTAG: Strus CreateTimeTAG: 2012-09-08T01:40:18Z VoteTAG: 17 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Way more fun than my highschool's math homework. FirstChildUserIdTAG: 270160 FirstChildUserNameTAG: Sethhhh FirstChildCreateTimeTAG: 2012-09-08T07:43:05Z IndexTAG: 77 TitleTAG: Sir Isaac Newton Today is Sir Isaac Newton 370 years of his birth. How this man is the greatest symbol of the changes of Science for the modern world, we honor him. UserIdTAG: 361823 UserNameTAG: EliasOak CreateTimeTAG: 2013-01-04T11:55:23Z VoteTAG: 16 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 78 TitleTAG: No Grades in certificate No Grades in certificate :( UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2013-01-03T14:07:21Z VoteTAG: 16 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 5 FirstChildTAG: We need grades in certificate and a ranking. It is a great motivation. FirstChildUserIdTAG: 43957 FirstChildUserNameTAG: Rafael_Nunes FirstChildCreateTimeTAG: 2013-01-03T18:15:56Z FirstChildTAG: i second it!! FirstChildUserIdTAG: 76418 FirstChildUserNameTAG: ankitarora FirstChildCreateTimeTAG: 2013-01-03T14:42:36Z FirstChildTAG: I wish they reconsider showing the grades. The course has been tough. FirstChildUserIdTAG: 398594 FirstChildUserNameTAG: DaveyJC FirstChildCreateTimeTAG: 2013-01-03T14:33:18Z SecondChildTAG: i completed it with 100%. So i wish the grades to be mentioned. As last year they provided the choice of downloading the certificate either with grades or without grades. So please do consider that option again. SecondChildUserIdTAG: 149119 SecondChildUserNameTAG: mnasirshahzad SecondChildCreateTimeTAG: 2013-01-03T14:42:30Z SecondChildTAG: without grade means just a piece of memorial paper. haaa haaa SecondChildUserIdTAG: 374393 SecondChildUserNameTAG: rmaleki SecondChildCreateTimeTAG: 2013-01-03T15:19:50Z SecondChildTAG: I just did a google image search for certificate as well as diploma, I literally don't see any certificates or diplomas with grades on them, nor I have I ever seen one in real life with a letter grade. I just check my Drivers License, it doesn't have a grade either! lol Have a great New Year. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2013-01-03T15:30:20Z SecondChildTAG: Penneypacker,you showed us informal but convincing reasons for not showing grades.I'm on your side! lol SecondChildUserIdTAG: 361137 SecondChildUserNameTAG: Ericson SecondChildCreateTimeTAG: 2013-01-03T16:04:59Z SecondChildTAG: I am also can't find any grade on my degree scroll although I got first class. Sigh.... SecondChildUserIdTAG: 529515 SecondChildUserNameTAG: Low SecondChildCreateTimeTAG: 2013-01-04T00:31:11Z FirstChildTAG: I hope edx will make the provision for certificates with grades on them FirstChildUserIdTAG: 449211 FirstChildUserNameTAG: Deveshmonga FirstChildCreateTimeTAG: 2013-01-03T18:36:10Z FirstChildTAG: I think the problem is that you were too generous to us and a lot of people got an excellent grade, so they want more and more now. Make the exam and the homework more real with limited number of answers and the exam with more realistic duration say with 2 additional hours from the beginning not +22, and every answer on the exam with only a try. Then the results will be very close to the real knowledge of the people, and also people won't be so overexcited by themselves... FirstChildUserIdTAG: 415375 FirstChildUserNameTAG: ZWX FirstChildCreateTimeTAG: 2013-01-03T19:38:39Z SecondChildTAG: I cannot but agree with the statement though I got an A grade but understand that this is not a real campus taken course and a lot of liberties were given to the students to earn a higher grade. I never studied electronics before but I was a student some 20 years ago to know that many of those luxuries are not provided in a real life. Those courses are created to give an access to a quality study material worldwide and not to earn any degree credits. I merely prefer it to OCW because it helps me to manage my time better having the deadlines and because it gives my some community sense SecondChildUserIdTAG: 16265 SecondChildUserNameTAG: serge_korolev SecondChildCreateTimeTAG: 2013-01-04T00:19:28Z SecondChildTAG: **serge:** When I took the course at my university we had the "2 hour exam, no internet, no textbook" standard also, so the exam was difficult as well; however there was a lot more one-on-one help with the professor available, and a lot of worked out examples on the blackboard, and although our professor was strict with time limits, his teaching method was very good (He was from Pakistan and I still remember when he said "when I was an engineering undergrad we had to learn welding, machine operation, metal fabrication..."; and his style and accent were very similar to our Prof. Agarwal, although if you weren't paying attention he would get mad!) Anyways...enough memories...when I read stories of **today's** university students (and I am not that old myself) it seems like university is a vacation...luxury hotel-style dormitories, restaurant-style cafeterias, etc., and parents are paying and arm-and-a-leg for tuition (lucky students-I had to pay my own way with jobs) so they expect that their kid gets a 100% even though the kid puts in no effort and drinks beer all semester. I know it's not all true what is reported in the media, but I do know from my experience that each new school year brings students who have more and more electronic devices (first it was programmable calculators, then graphing calculators, then laptops, now ipads and smartphones) to do their "work" for them. You don't even have to perform an integration or a differentiation anymore, much less add and subtract, much less think! With circuit simulation becoming so prevalent, I had to resist the temptation of using MATLAB, Wolfram-Alpha, and PSpice to solve all my 6.002x homework for me! But I guess it's not "cheating" when no real-world engineer uses the pencil-and-paper method anyways... SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2013-01-04T15:46:34Z SecondChildTAG: Huh? Pen?? Paper??? SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2013-01-04T18:09:31Z IndexTAG: 79 TitleTAG: TO ALL FELLOW CLASS MATES I found in this discussion forum that some of our class mates are spearing the community TAs for the release of certificates. Please keep patience. They know when to release it and as an obedient student we have to wait for it. Please give them feedback of the course rather than drilling them for certificates. UserIdTAG: 145544 UserNameTAG: pandiya CreateTimeTAG: 2012-12-30T19:06:59Z VoteTAG: 16 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Thanks! FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-12-30T21:35:27Z FirstChildTAG: I absolutely agree. I think there should be more exercises during the course. There is an option of solving exercises and problems from the textbook but answers are available not for all of them and I know nothing about solutions manual or textbook. I believe the course would benefit very much if most exercises and problems were in electronic form and should be solved for credit, maybe a small credit, but the more exercises the better. The main advantage of the system I see in the possibility of discussion: in the end of every page there is a 'Post New Message' button. So, if most of exercises and problems are in electronic form, I think, students would discuss them. One might think that it's an individual decision whether to do a particular exercise or problem. This thought is absolutely valid. However, obtaining an answer whether right or wrong delivers only a fraction of value, the most part comes from the process of coming to the answer. I suppose, the value of this process is enhanced if students come in groups. It is one thing to make it alone, it is another - to make it together. A student might take the textbook, pick up an exercise or a problem and think of solving it with somebody else, go to the Internet and start searching for an interested community. On the other hand during this course there is already an interested community available, so one doesn't need to search for one if she or he already joined. Therefore, the value of having many more exercises and problems in electronic form is very, very high. Maybe all of the exercises and problems from the textbook should be interactive?... FirstChildUserIdTAG: 252722 FirstChildUserNameTAG: vaboro FirstChildCreateTimeTAG: 2012-12-30T21:39:22Z SecondChildTAG: This is the elsevier companion site with extra exercises/solutions and additions: http://www.elsevierdirect.com/v2/companion.jsp?ISBN=9781558607354 SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-12-31T00:23:14Z SecondChildTAG: Yes, I know this site. It contains the material marked with an asterisk in the main textbook. So, this is just a complementary material to the main textbook. SecondChildUserIdTAG: 252722 SecondChildUserNameTAG: vaboro SecondChildCreateTimeTAG: 2013-01-01T21:45:20Z IndexTAG: 80 TitleTAG: A personal opinion about certificates This course is a free course from a very good University. An opportunity for everybody from anywhere,poor or rich,to learn something more.It is not a competition between members of a "High IQ club". What is the goal, or if you prefer, what is the "big picture" of the course? The Knowledge itself for everybody, or a certificate? For me, the answer is obvious. Knowledge. If we are talking about certificates, there are two of them. The first, and most important,is in our mind, knowing that we took this course, spending hours and hours of effort. The second one, is a piece of paper. Of course it has its own value,it has our name and colors, but it is still a paper. If this certificate has an A or B or C or nothing on it, it is not important. The important is, that you are travelling. Enjoy it. Lets remember Ithaka. You can listen to Sean Connery reciting this poem http://www.youtube.com/watch?v=1n3n2Ox4Yfk and if you are Greek (or you know Greek) http://www.youtube.com/watch?v=2BkB1iEG_PA *"Keep Ithaka always in your mind. Arriving there is what you are destined for. But do not hurry the journey at all. Better if it lasts for years, so you are old by the time you reach the island, wealthy with all you have gained on the way, not expecting Ithaka to make you rich."* http://www.cavafy.com/poems/content.asp?cat=1&id=74 Thank you all, and I wish you the best for 2013 UserIdTAG: 146969 UserNameTAG: Andreas-N CreateTimeTAG: 2012-12-30T09:01:53Z VoteTAG: 16 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: Brilliant words! This perfect world of knowledge was discovered by edX for everyone for free. Be thankful! Enjoy of each moment, keep in touch with your classmates and the staff. Remember that the road will be passed by walking man only. Keep going! Happy New Year! FirstChildUserIdTAG: 325197 FirstChildUserNameTAG: Vitali_Jerin FirstChildCreateTimeTAG: 2012-12-30T09:30:44Z FirstChildTAG: I agree with you Andreas!! This is a great opportunity for anybody who is interested in the offered knowledge. A piece of paper (or link) can not even compare to the quality of knowledge that someone that doesn't have much for college, or refresh what he already knows (like me), can get out of this wonderful opportunity! Although i hold a degree in Electrical Engineering, this has proved to be a brilliant refresher for me(and perhaps a stepping stone for further education). Please do not let a piece of paper ruin this once in a lifetime opportunity for you people! Happy New Year!! (Andrea if you are indeed Greek, it will make me extremely proud that i wasn't the only one taking this course from my Country :) ) FirstChildUserIdTAG: 304587 FirstChildUserNameTAG: Vasso FirstChildCreateTimeTAG: 2012-12-30T11:21:38Z SecondChildTAG: You are not the only Greek!! I am also Greek and I suppose there are a lot of others Greeks that have take this course. Xronia polla! SecondChildUserIdTAG: 383635 SecondChildUserNameTAG: mixxmike SecondChildCreateTimeTAG: 2012-12-30T14:56:38Z SecondChildTAG: Βάσω, yes, I am also Greek. Σας εύχομαι καλή χρονιά. SecondChildUserIdTAG: 146969 SecondChildUserNameTAG: Andreas-N SecondChildCreateTimeTAG: 2012-12-30T18:38:20Z SecondChildTAG: Χρονια πολλα παιδες! SecondChildUserIdTAG: 214275 SecondChildUserNameTAG: TheodoreGr SecondChildCreateTimeTAG: 2012-12-30T20:07:43Z SecondChildTAG: Χρόνια πολλά παιδιά!!! :))) SecondChildUserIdTAG: 304587 SecondChildUserNameTAG: Vasso SecondChildCreateTimeTAG: 2012-12-31T15:00:18Z FirstChildTAG: I agree too. FirstChildUserIdTAG: 273912 FirstChildUserNameTAG: raj2691 FirstChildCreateTimeTAG: 2012-12-30T15:54:10Z FirstChildTAG: I find your comments very wise. Your citation made me think of a poem of the swedish writer Karin Boye http://www.karinboye.se/verk/dikter/dikter-mcduff/in-motion.shtml *IN MOTION* *The sated day is never first. The best day is a day of thirst. Yes, there is goal and meaning in our path - but it's the way that is the labour's worth. The best goal is a night-long rest, fire lit, and bread broken in haste. In places where one sleeps but once, sleep is secure, dreams full of songs. Strike camp, strike camp! The new day shows its light. Our great adventure has no end in sight.* The meaning is the same, its the effort that is rewarding. To stand up and keep going. The world is not "fair" and often your hard work don't pay off so soon. If the certificate is important I can see no other way than some type of controlled environment during the exam. Happy new year, with new adventures! Stig FirstChildUserIdTAG: 151472 FirstChildUserNameTAG: Stensmed FirstChildCreateTimeTAG: 2012-12-31T00:49:37Z SecondChildTAG: Very nice poem Stig, and thank you for your comments. We will find a new destination to continue our journey. Happy new year. SecondChildUserIdTAG: 146969 SecondChildUserNameTAG: Andreas-N SecondChildCreateTimeTAG: 2012-12-31T06:39:08Z IndexTAG: 81 TitleTAG: Thank you! Professor Anant Agarwal from Peru (Machu Picchu) Thank you Professor Anant Agarwal and all course staff: Gerald Sussman and Piotr Mitros, passionate educators. Thanks to our friend Lorenzo. Thanks also to Lyla Fischer, an Edx fellow and entire team of edX. Thanks to Myriam Nonaka (Myrimit) and entire Community TA. Thank you very much to edX, Harvard University and Massachusetts Institute of Technology (MIT). And to my classmates worldwide, thanks! Merry Christmas and Happy New Year to all! ![Anant Agarwal][1] Prof. Anant Agarwal ![Gerald Sussman][2] Prof. Gerald Sussman ![Piotr Mitros][3] Prof. Piotr Mitros ---------- ![Lorenzo][4] Lorenzo ---------- ![Lyla Fischer][5] Lyla Fischer ---------- ![Myriam Nonaka][6] Myriam Nonaka (Myrimit) ---------- ![edX][7] ![Harvard University][8] ![MIT][9] [1]: https://edxuploads.s3.amazonaws.com/13563996011343679.jpg [2]: https://edxuploads.s3.amazonaws.com/13563996551343654.jpg [3]: https://edxuploads.s3.amazonaws.com/13563996961343641.jpg [4]: https://edxuploads.s3.amazonaws.com/13564738351343613.jpg [5]: https://edxuploads.s3.amazonaws.com/13563998128474442.jpg [6]: https://edxuploads.s3.amazonaws.com/13563998591343688.jpg [7]: https://edxuploads.s3.amazonaws.com/13563998971343633.png [8]: https://edxuploads.s3.amazonaws.com/13563999501343622.png [9]: https://edxuploads.s3.amazonaws.com/1356399978176469.gif UserIdTAG: 402193 UserNameTAG: Feramico CreateTimeTAG: 2012-12-25T01:47:17Z VoteTAG: 16 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I think I would like to say as our friend Feramico "Thank you very much" and also big thanks to the people behind this project. FirstChildUserIdTAG: 226085 FirstChildUserNameTAG: Teto FirstChildCreateTimeTAG: 2012-12-25T03:34:29Z SecondChildTAG: I think you forgot Lorenso![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13564097101343681.png SecondChildUserIdTAG: 226085 SecondChildUserNameTAG: Teto SecondChildCreateTimeTAG: 2012-12-25T04:28:55Z SecondChildTAG: I'm sorry. Thank you very much! Professor Lorenso and to all course staff I forgot to mention, and to my classmates worldwide, thanks! SecondChildUserIdTAG: 402193 SecondChildUserNameTAG: Feramico SecondChildCreateTimeTAG: 2012-12-25T05:37:58Z IndexTAG: 82 TitleTAG: edx staff A TREMENDOUS THANKS TO THE (MITx: 6.002x Circuits and Electronics) PROFESSORS UserIdTAG: 462861 UserNameTAG: edxforme CreateTimeTAG: 2012-12-23T23:00:16Z VoteTAG: 16 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 83 TitleTAG: Looking forward to 6.004x Hi, 6.002x was one of the best online courses I have enrolled for; the lectures, homeworks, labs and discussion forum are, by far some of the most exceptional advances in online education I have seen. Congratulations to the entire team for such a stellar effort. 6.004 (Computation Structures) was one of the most exciting course that I have undertaken. It is the true bridge between electronics and programming. I am looking forward to seeing 6.004x on edX. Hopefully by the spring semester. Thank you edX once again for providing us with such an exceptional opportunity. UserIdTAG: 85490 UserNameTAG: yashsavani CreateTimeTAG: 2012-12-15T09:21:31Z VoteTAG: 16 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: 6.004 looks really awesome, more details here: http://6004.mit.edu/ FirstChildUserIdTAG: 189680 FirstChildUserNameTAG: vnd FirstChildCreateTimeTAG: 2012-12-15T10:26:04Z SecondChildTAG: Woah, seriously seconded. SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-12-15T17:25:35Z SecondChildTAG: what is the different between 6.004 and 6.00? SecondChildUserIdTAG: 318037 SecondChildUserNameTAG: Maher-84 SecondChildCreateTimeTAG: 2012-12-17T16:05:53Z IndexTAG: 84 TitleTAG: virtual aha! ground I vote for and 'Aha' momment for virtual short method. UserIdTAG: 250199 UserNameTAG: Expo CreateTimeTAG: 2012-11-27T18:12:28Z VoteTAG: 16 CoursewareTAG: Week 12 / S23V22 Inverting amplifier input resistance using virtual short method CommentableIdTAG: 6002x_S23V22_Inverting_amplifier_input_resistance_using_virtual_short_method NumberOfReplyTAG: 1 FirstChildTAG: Definitely Agree!! Working in electronics for 30 years with a much less rigorous understanding than I now have (Thanks so much Professor), the virtual short was THE way to understand OpAmp circuits. FirstChildUserIdTAG: 369519 FirstChildUserNameTAG: perpstud FirstChildCreateTimeTAG: 2012-12-13T04:45:01Z IndexTAG: 85 TitleTAG: Dr. Agrawal is extraordinary professor His teaching is an artistic piece of GOLD UserIdTAG: 382505 UserNameTAG: AhmedGalal2 CreateTimeTAG: 2012-11-07T11:51:49Z VoteTAG: 16 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: I agree to.....:) His method of teaching is to attractive...:) > **Regards:** asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-11-08T16:15:04Z FirstChildTAG: I agree. Like day and night if you compare him to my local university computer science "professors". Kudos for him and the EDX platform! FirstChildUserIdTAG: 398594 FirstChildUserNameTAG: DaveyJC FirstChildCreateTimeTAG: 2012-11-07T13:12:28Z SecondChildTAG: He has passion for the subject, an astute sense of humor; above all his love for the students. SecondChildUserIdTAG: 285616 SecondChildUserNameTAG: AmiyaX SecondChildCreateTimeTAG: 2012-11-08T09:02:06Z FirstChildTAG: Dr Agarwal is great professor and has good humor aha FirstChildUserIdTAG: 352647 FirstChildUserNameTAG: praveenjugge FirstChildCreateTimeTAG: 2012-11-09T10:37:16Z FirstChildTAG: I will better say, DIAMOND ;) Thank you Prof. Agarwal! You are amazing and I appeciate you infinite and if I can let myself dream for a femto second haha, I hope to meet you someday in the future, might I could appear out of the blue in the real 6.002 Class, who knows haha, so funny! Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-10T17:46:06Z IndexTAG: 86 TitleTAG: Midterm Exam 6.002x (Fall) -2012 edX- Q4 Solution video explanation (Not Official) Hi! How are you? I hope that you did well in the Midterm Exam! :) I made a video explanation of Q4 of Midterm Exam. This video it took me a lot of time, almost two days haha. I wish to have more time and make more videos explanations about the Midterm Exam Problems, because I really really like doing this, I really enjoy it! But I have a lot of FIFO things to do offline haha! The explanation it is a little bit long haha, I hope that you don't get bored. This explanation it is not official. I made it based on what I did in my Midterm Exam. The officials answers as far as I have read will be posted soon. Also I have to say that Rana still being the Star of 6.002x! Hahaha! It was out of the blue that happened, I did not purposely haha! no copyright infringement intended Prof. Agarwal haha! X'D [Watch here - Q4 Explanation video][1] Have fun! Myriam. P.D: Sorry for my english. UPDATED: For those who are facing YouTube Problems I uploaded the video. [Click here - DOWNLOAD][2] - 55.2MB (28 min video) [1]: http://www.youtube.com/watch?v=08IdYWeM2qk&feature=youtu.be [2]: http://www.4shared.com/video/i48zYuCK/MidQ4.html UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2012-10-30T03:04:09Z VoteTAG: 16 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 7 FirstChildTAG: Thanks Myrimit!!! FirstChildUserIdTAG: 175734 FirstChildUserNameTAG: hestrada FirstChildCreateTimeTAG: 2012-10-31T03:01:13Z SecondChildTAG: You are welcome hestrada! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-31T11:00:00Z FirstChildTAG: Awesome! Thank you, I will get some popcorn for this. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-30T03:23:39Z SecondChildTAG: Hahaha! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-30T03:31:04Z FirstChildTAG: Dear Myriam, Many many thanks for helping your course mates in studies. Itried to wath your above video but due ban on youtube, I was unable to do so. Can it be watch otherway? Wahab FirstChildUserIdTAG: 455950 FirstChildUserNameTAG: Wahabbaluch FirstChildCreateTimeTAG: 2012-10-30T03:24:43Z SecondChildTAG: You are welcome Wahabbaluch! :). I will try to upload the file tomorrow so you can download it and watch it. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-30T03:33:48Z SecondChildTAG: I have updated the Post. I uploaded the video explanation of 28 min in 55.2 MB :). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-30T22:31:56Z FirstChildTAG: Awesome!! haha Goooooddddd!! Many thanks :)) FirstChildUserIdTAG: 342966 FirstChildUserNameTAG: SandraNavarro FirstChildCreateTimeTAG: 2012-10-30T08:17:22Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-30T11:39:43Z FirstChildTAG: $gracias$ FirstChildUserIdTAG: 623210 FirstChildUserNameTAG: preveen FirstChildCreateTimeTAG: 2012-10-30T11:06:50Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-30T11:39:52Z FirstChildTAG: That is awesome. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-10-30T15:44:55Z SecondChildTAG: Thank you Lyla! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-30T17:27:04Z FirstChildTAG: Gracias no entendía por qué las corrientes eran iguales :) Thanks i don't unsderstand why i_q1 was equal to i_q2, now is clear :) FirstChildUserIdTAG: 85931 FirstChildUserNameTAG: rafaelmtz FirstChildCreateTimeTAG: 2012-10-30T22:23:25Z SecondChildTAG: :) por nada ! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-30T22:34:48Z SecondChildTAG: thanks SecondChildUserIdTAG: 550401 SecondChildUserNameTAG: revathisingh SecondChildCreateTimeTAG: 2012-11-01T15:12:52Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-02T11:48:44Z IndexTAG: 87 TitleTAG: My two cents for edX! My Video Tutorial of Lab2. Hi! How are you doing with 6.002x? I hope that you are doing fantastic! I made a Video Tutorial about Lab2. There I explain why we can´t use 2 resistors :) and how to reach to the answer. Believe me that I did a lot of effort in the video speaking in English haha – I have to back to my English classes again-, also I tried to not be boring (you will see why I am not good at soccer and discover a super cool hiden message in the calculator while watching the video in the Aha Moment! part haha). The quality of the video were not as good as I will wished haha, and if you hear some back sounds of steps, like ghost steps , don´t scare, it is my keyboard when I edited the video haha! I also make a version in Spanish, it is the same video but speaked in Spanish for the Hispanic Community. The video it is a quite long haha, I added also some music in the middle, so don´t be scared again if you in the middle are concentrated and then the music suddenly appears , don´t say that I didn´t warned you haha (It was purposely, so you can wake up in the middle haha, catch you)! I really enjoyed the 6.002x Prototype Course (Spring)! I had a lot of fun, I met a lot of People there, I miss them so much – we still in contact in the old Forum-! they were excellent (some are here in this fall, so you will dicover who they are with the time, haha)! Also, I have to tell you that Piotr Mitros is a wonderful person, he was really supportive with our ideas in the Prototype Course, he encouraged so many Students in the Spring (at a first look you think that he is distant haha, but then you discover that he is really involved with the students, that he worries about your opinion and trust me that he is amazing and open-minded , I guess that he is reading this and reads all your comments haha, Thank you Piotr! I really appreciate it!) and Prof. Agarwal, he is incredible,I have no words for him, he is a big-hearted person, believe me, it is true! , he is really concerned about the Students, he demostrated it in the Prototype Course, and all the Staff that you don´t know but they are there present, behind (Lauren, Peter, Kimberly, -Dave and Lyla- you already know them :) , all the Staff), they are amazing! Thank you! I hope that this could be useful for those that had difficulties or doubts in this Lab! And also for those that want to see more explanation of this Lab! Have fun! - Video Tutorial – English Version: [Watch Video - here][1] - Video Tutorial – Spanish Version: [Ver video - aquí][2] My best wish to all of you with 6.002x! See you! Myriam. P.D: Remember that I will try to update the Wiki Hints. [1]: http://www.youtube.com/watch?v=qzORok-sXz0&feature=youtu.be [2]: http://www.youtube.com/watch?v=VhY2kzVR9RI&feature=youtu.be UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2012-10-01T23:32:29Z VoteTAG: 16 CoursewareTAG: Week 2 / Linear combinations of source strenghts CommentableIdTAG: 6002x_S3E2_Linear_Combinations_of_Source_Strengths NumberOfReplyTAG: 4 FirstChildTAG: Bravo, Miriam! A wonderfully entertaining video and some fine analysis! FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-10-02T03:25:36Z SecondChildTAG: You are welcome MobiusTruth! :). How are you? SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-02T11:14:40Z SecondChildTAG: thnk...really helpful SecondChildUserIdTAG: 804955 SecondChildUserNameTAG: alchy SecondChildCreateTimeTAG: 2012-12-03T20:25:28Z FirstChildTAG: Thanks! FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-02T23:27:04Z SecondChildTAG: You are welcome Pennypacker :)! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-03T00:32:54Z FirstChildTAG: Xevere fiieeeriiitaaaaaaaa. Sos great! FirstChildUserIdTAG: 138769 FirstChildUserNameTAG: ArturoPrado FirstChildCreateTimeTAG: 2012-10-09T02:14:17Z SecondChildTAG: Thank you ArturoPrado!:) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-09T03:43:27Z FirstChildTAG: thank for you FirstChildUserIdTAG: 345915 FirstChildUserNameTAG: thabuty_rudi FirstChildCreateTimeTAG: 2012-10-15T07:11:57Z SecondChildTAG: You are welcome :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-15T12:31:28Z IndexTAG: 88 TitleTAG: Homework Hints H3P1: A Logic Family **H3P1:** A Logic Family For many purposes of gate design, we can model a MOSFET used as a switch simply as as an ideal switch and an "on-state resistor" RON. This is the SR model. Assuming this model for the MOSFET, consider the inverter in the figure. This inverter is intended to be used as an element in a logic family with NAND and NOR gates. ![image1][1] he static discipline required for this family is: VS=5.0V, VOH=4.5V, VIH=4.0V, VIL=1.5V, VOL=1.0V. ---------- **Hints for Part 1, 2 and 3:** Read the 250 and [251][2] of the Textbook :). Remember that: MN0 = Noise Margin for logical zero. MN1 = Noise Margin for logical one. The forbidden region you will find a clue down the equation (5.6) of that page of the Textbook ;). ---- Part 1: What is the low noise margin (in Volts)? Part 2: What is the high noise margin (in Volts)? Part 3: What is the width of the forbidden region (in Volts)? ---------- Suppose that the threshold voltage for the MOSFET is VT=2.0V and RON=19000.0Ω. **Hints for Part 4, Part 5 and Part 6:** Hint: Watch this Video Tutorial [Here][3] Hint: Once you have seen Hint1 video, remember that pullup resistor it is calculated with the worst case of RPD and with VOL ;). ---- Part 4: What is the minimum value of the pullup resistor RPuI (in Ohms) for which this inverter can obey the required static discipline? Part 5: Now, consider the NAND gate of this family. What is the minimum value of the pullup resistor RPuA (in Ohms) for which this inverter can obey the required static discipline? Part 6: How about the NOR gate of this family. What is the minimum value of the pullup resistor RPuO (in Ohms) for which this inverter can obey the required static discipline? ---------- Assume that we implemented this family with the minimum pullup resistors that you have already calculated. **Hints for Part 7, Part 8, Part 9:** Hint: Watch this video Tutorial of Week 3 [Here][4] Hint: Once you have watched hint1 video, you will see that power will maximized where the resistance is minimized ;). ---- Part 7: What is the maximum power (in Watts) consumed by the inverter? Part 8:What is the maximum power (in Watts) consumed by the NAND? Part 9:What is the maximum power (in Watts) consumed by the NOR? ---- ---- See you! Myriam. P.D: I will be posting others hints of week3 . mkprasanth as you requested me [here][5] I also will posting some hints of lab 3 ;) EDIT: [Lab 3 Hints posted here][6] [1]: https://www.edx.org/static/content-mit-6002x/images/circuits/mos-gates.0f48c3719ec8.gif [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/275 [3]: https://www.youtube.com/watch?v=HrZS106zxI4&feature=player_embedded#! [4]: http://%20https://www.youtube.com/watch?v=qsPg03qsxvA&feature=player_embedded#! [5]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/505e2fd7e903b1230000001f [6]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50673e4c2193ea2300000009 UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2012-09-28T21:28:38Z VoteTAG: 16 CoursewareTAG: Week 3 / Logic Gates CommentableIdTAG: 6002x_logic_gates NumberOfReplyTAG: 7 FirstChildTAG: Muchas gracias, en serio, que amable eres. FirstChildUserIdTAG: 184827 FirstChildUserNameTAG: DiegoT FirstChildCreateTimeTAG: 2012-09-29T02:58:33Z SecondChildTAG: :) Gracias DiegoT! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-29T04:48:19Z SecondChildTAG: Dear Myrimit , I was calculated it is saying wrong I am confused only power calculation not coming correct all other are ok SecondChildUserIdTAG: 156694 SecondChildUserNameTAG: mkprasanth SecondChildCreateTimeTAG: 2012-09-29T08:29:04Z SecondChildTAG: Dear Myriam , Thanks a lot for support Thanks a completed the Home work and lab Thanks MK.Prasanth SecondChildUserIdTAG: 156694 SecondChildUserNameTAG: mkprasanth SecondChildCreateTimeTAG: 2012-09-29T17:43:12Z SecondChildTAG: ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-29T19:30:28Z SecondChildTAG: gracias Myriam por tu apoyo / Thanks Myriam for your support :)) SecondChildUserIdTAG: 342966 SecondChildUserNameTAG: SandraNavarro SecondChildCreateTimeTAG: 2012-09-30T18:13:57Z SecondChildTAG: Por nada SandraNavarro ;). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-30T22:40:01Z FirstChildTAG: Yeah please really do for lab 3 FirstChildUserIdTAG: 281159 FirstChildUserNameTAG: ManasiS FirstChildCreateTimeTAG: 2012-09-29T01:29:18Z SecondChildTAG: Yes I will ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-29T04:48:36Z SecondChildTAG: [Hints of Lab here posted here][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50673e4c2193ea2300000009 SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-29T19:31:00Z FirstChildTAG: Where are you getting these videos from? They don't appear anywhere in courseware that I see. FirstChildUserIdTAG: 264596 FirstChildUserNameTAG: Nuru FirstChildCreateTimeTAG: 2012-09-30T01:43:04Z SecondChildTAG: They are in the Week 3 Tutorials (Courseware->Week3-> Week3 Tutorials ;)) If you click on there, you will find a lot of video Tutorials ;). I just Linked the videos (from YouTube) ;). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-30T02:06:49Z SecondChildTAG: Ah, I had looked there but not clicked on the third example thinking it was something else entirely. Thanks :) SecondChildUserIdTAG: 264596 SecondChildUserNameTAG: Nuru SecondChildCreateTimeTAG: 2012-09-30T02:25:59Z SecondChildTAG: You are welcome Nuru ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-30T02:29:52Z FirstChildTAG: And once you understand a problem, don't do what I did: become too confident. Part of this one took me several tries - just because I figured I could just "run the numbers" without paying enough attention to my reasoning. It was pretty frustrating for a while! FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-09-29T19:37:15Z SecondChildTAG: Hi MobiusTruth! ;). How are you? Yes, I agree with you! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-29T20:56:42Z FirstChildTAG: Thank you Myrimit. It really was a very very useful post. FirstChildUserIdTAG: 378096 FirstChildUserNameTAG: Jamshaid271 FirstChildCreateTimeTAG: 2012-09-30T02:51:18Z SecondChildTAG: You are welcome Jamshaid271 ;). I will be posting updates of Weekly Hints in the Wiki [you can read this Post][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_logic_gates/threads/5067b3ab2193ea2300000022 SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-30T03:15:51Z FirstChildTAG: Kindly plz help me on questions 4-7 as I am unable to view tutorials due to banning of youtube in my country FirstChildUserIdTAG: 414201 FirstChildUserNameTAG: uzaifakram FirstChildCreateTimeTAG: 2012-09-30T11:56:55Z SecondChildTAG: Kindly plz anybody give me the NOR formula for Vout SecondChildUserIdTAG: 414201 SecondChildUserNameTAG: uzaifakram SecondChildCreateTimeTAG: 2012-09-30T14:00:34Z FirstChildTAG: I have got the ans. for nand also by using formula vs2/2Ron+Rpull but still dn't know how to find power for Nor FirstChildUserIdTAG: 414201 FirstChildUserNameTAG: uzaifakram FirstChildCreateTimeTAG: 2012-09-30T17:44:32Z SecondChildTAG: I also got all of the problem except for the power for Nor. I hope to figure it out before you reply, though. ;-) Thanks for all of your generous help to everyone, Myrimit! SecondChildUserIdTAG: 86632 SecondChildUserNameTAG: LCL SecondChildCreateTimeTAG: 2012-10-01T03:22:51Z SecondChildTAG: Yeah. I figured it out. On the NOR the minimum needed is the amount needed to bring the Vs to Vol for any of the cases. So if one is On, then find it there. Since both are the same resistance, that should be the minimum. The maximum is what the tutorial shows.... Please someone bring light if I am wrong. SecondChildUserIdTAG: 230787 SecondChildUserNameTAG: Gabriel007 SecondChildCreateTimeTAG: 2012-10-01T03:28:19Z IndexTAG: 89 TitleTAG: S8E0 exercise It is not clearly stated in S8E0 what parameters should be used to express vO. Besides, it is always a puzzle to choose the right letter and index. For example, the letters such as '1', 'I', 'l' look very similar and it is better not to use them together in formulas and expressions. UserIdTAG: 143593 UserNameTAG: Tsybulkin CreateTimeTAG: 2012-09-20T05:23:57Z VoteTAG: 16 CoursewareTAG: Week 4 / Dependent Sources Exercise 0 CommentableIdTAG: 6002x_dep_src_exsz_0 NumberOfReplyTAG: 2 FirstChildTAG: You have to find: vB=?? and iB=?? in function of VI and/or R1 FirstChildUserIdTAG: 197569 FirstChildUserNameTAG: darksamaro FirstChildCreateTimeTAG: 2012-09-22T21:35:10Z FirstChildTAG: Yes, i agree.It is always a puzzle with the variables.I propose that you should spell them before, as given parameters. FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-09-23T08:01:20Z SecondChildTAG: yes. I do agree. specially for letter "o" in this example. I inserted little "o" instead of capital "O" and it made lots of problem for me. SecondChildUserIdTAG: 374393 SecondChildUserNameTAG: rmaleki SecondChildCreateTimeTAG: 2012-09-27T05:35:26Z SecondChildTAG: Yes. It's our goal to learn the theory of dependent source (voltage or current) from control. Not the goal to learn mat expression. Can someone help or hint. SecondChildUserIdTAG: 229018 SecondChildUserNameTAG: Changming SecondChildCreateTimeTAG: 2012-10-02T23:47:50Z IndexTAG: 90 TitleTAG: H2P1 Hints requested by sandeshacharya. Hi sandeshacharya! How are you? Sorry for the delay. You requested me this 8 days ago: *“hello friends i am new course taking student of fall September and would like to ask a question to you.could you please help me with H2P1.i got really confused with the type of question and the resistors to be chosen!!!”* [posted here][1] ---------- As this is a graded part, I can not give you an implicit result but can colaborate with you with hints in order that you can solve this problem by yourself. ---------- Let’s see this together! **H2P1: VOLTAGE – DIVIDER DESIGN** *The resistances of commercially-available discrete resistors are restricted to particular sets. For example, the available values of resistors with 10% tolerance are selections from the E12 set multiplied by a power of ten from 10^0 through 10^5. The E12 set is: E12={10,12,15,18,22,27,33,39,47,56,68,82}* - Ok, What are that numbers inside the {}? They are standard values of resistors, values that you can buy comercially. Imagine if there were not this values available and any manufacturer could make any arbitrary value depending on their available materials in stock … it would be a caos! Fortunately, exists standards values ;) - What they mean when they say that they have a 10% of a tolerance??? That means that if you buy a resistor of 22 Ohm, if you meassure that resistor it values can be 10% more or less: the resistor can be R= 22+2.2=**24.2 Ohm** or R=22-2.2=**19.8 Ohm**. (*) - What about that range that they give (10^0 to 10^5)? The values inside the {} are: 10 * **10^0** Ohm= 10 Ohm 10 * 10^1 Ohm = 100 Ohm 10* 10^2=1000 Ohm = 1kOhm 10* 10^3=10000Ohm = 10kOhm 10*10^4=100000 Ohm = 100kOhm 10***10^5**=1000000 Ohm = 1 MOhm . . . 47 * 10^0 Ohm= 47 Ohm 47* 10^1 Ohm = 470 Ohm 47* 10^2=4700 Ohm = 4.7kOhm 47* 10^3=47000Ohm = 47kOhm 47*10^4=470000 Ohm = 470kOhm 47*10^5=4700000 Ohm = 4.7 MOhm And so on with all the values inside the {}… The statement says : **In this problem we need to choose 10% resistors to make a voltage divider that meets a given specification.** **We are given an input voltage Vin=70.0V, and we need to provide an open-circuit output voltage of Vout≈24.5V. An additional requirement is that the Thevenin resistance as seen from the output terminals is between 10kΩ and 30kΩ. Assume first that the resistors have their nominal resistance. Come up with resistors R1 and R2 such that the divider ratio Vout/Vin is within 10% of the requirement.** **Of course, the resistances you chose are just nominal. Given that they are only guaranteed to have resistances within 10% of the nominal value, what is the largest and smallest value that Vout may have?** **Here my first Hint to you:** review this part of the Textbook [Read Here][2]. The statement it is a Little confusing. Ok, let’s try to understand what are they asking in order that you can later solve this by your own: **Hint PART 1 and PART 2:** They give you Vin and Vout. You know how to relate Vout in parameters of R1, R2 and Vin (have you read the first hint? - Textbook Chapter). Can you find R1 and R2? … No, we need another equation more... (because we have 2 incognits and to solve the system we need the same number of equations as variables that we have)… ... But they tell you: An additional requirement is that the thevenin resistance as seen from the output terminals is between 10kΩ and 30kΩ. (**Here it is your Aha moment!**). (Read Thevenin in the [Textbook][3]) Can you find R1 and R2 now? Yes!!!! .... Ok, but wait a minute, what does it mean: **“Come up with resistors R1 and R2such that the divider ratio Vout/Vin is within 10% of the requirement”** This is, that once you have your values, with the values that you have chosen, that voutput/vinput has to be less-equal than a 10%. It is to verify your chosen values! If it is less or equal it is correct if not, it is wrong and you have to re-calculate again. ---- **Hint PART 4 and 5:** If you Know that the resistances that you have chosen has an Rmáx value and Rmín value (Remember that this is because they have a 10% tolerance), So, If you have chosen R1 and R2, what combination of the possibles values gives me the V output máx and the Voutput mín of the Circuit? ----- See you! Myriam. EDIT: (*) I have corrected that equation thank you to moutasem. I have put 2 instead of 2.2. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/MITx_6_002x_2012_Fall/threads/50474faf685941270000000d [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/99 [3]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/182 UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2012-09-19T07:19:16Z VoteTAG: 16 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: Hi guys,I am actually stark on how to use Thevenin in this question.Any help. FirstChildUserIdTAG: 429851 FirstChildUserNameTAG: ssembajjwe FirstChildCreateTimeTAG: 2012-09-19T10:18:03Z SecondChildTAG: Hi, Short out the voltage source to calculate the 2 resistors in parallel. This will be equal to your thevenin resistance. SecondChildUserIdTAG: 320661 SecondChildUserNameTAG: AbbaHanna SecondChildCreateTimeTAG: 2012-09-30T15:46:36Z FirstChildTAG: hi, I write from Colombia my name is jonathan, and based on the response of myriam also i suggest you take 20k as RTH, this you gives better degree of accuracy and does not allow you to leave your value of 10k and 30k, I take this value and it worked hope you too. Cordially Jonathan!!!! FirstChildUserIdTAG: 441072 FirstChildUserNameTAG: jonathan726 FirstChildCreateTimeTAG: 2012-09-19T08:04:33Z FirstChildTAG: Hi Myrimit, what do you think about [my problem][1] thanks before and i also like that "AHA moment" [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/505aa3fae87fdd2800000001 FirstChildUserIdTAG: 452559 FirstChildUserNameTAG: kuz1toro FirstChildCreateTimeTAG: 2012-09-20T16:58:11Z SecondChildTAG: Ok, I will take a look at night ;) and try to help you. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-20T20:57:40Z SecondChildTAG: Hi kuz1toro! I saw your Post: I think that a) condition refers about an extreme of vout/vin and not an interval and b) condition refers to the other extreme condition of vout/vin. If you know which corresponds to vomax/vin and which to vomin/vin (it is your task to find which corresponds to which)and you also know the value of Rth have to be from ... to ..., you can obtain the R1 and R2 ;). I hope this can help you. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-21T02:26:58Z FirstChildTAG: Hello Myrimit,I really need your help.I just got the mail about the up coming examination.I am in Africa,Nigeria to be precised.Due to power fluctuation and and unstable internet connection,i have missed lectures and several home works and i really need to meet up before the commencement of the exams.Could you please help me with the links on how to get the past home works?I will guide myself with the online text books.I don't want to lose points.Kindly assist me. Thanks, Joseph FirstChildUserIdTAG: 417489 FirstChildUserNameTAG: Lakeside FirstChildCreateTimeTAG: 2012-10-10T08:00:00Z SecondChildTAG: Yes, sure Joseph. Can I help you? In which part are you lost? SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-10T22:09:22Z FirstChildTAG: "That means that if you buy a resistor of 22 Ohm, if you meassure that resistor it values can be 10% more or less: the resistor can be R= 22+2=24 Ohm or R=22-2=20 Ohm. " 10% from 22 is 2,2 and not 2 as you explained in your aide example , if your explain wrong , how i can complete ??? FirstChildUserIdTAG: 231318 FirstChildUserNameTAG: moutasem FirstChildCreateTimeTAG: 2012-10-16T22:10:04Z SecondChildTAG: Hi moutasem!, yes, you are right, I haven't noticed that before. I don't know why I wrote that .Thank you, I will correct it. It should said 2.2 instead of 2. Thank you! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-17T01:25:06Z IndexTAG: 91 TitleTAG: Thanks edX!! I loved this class and learned a ton! I am going to pursue this further and continue to learn more in electrical engineering. Thank you edX for such an amazing class this class made an impact in my life and it's amazing to see the good these classes are doing around the world. Thanks for everything!! ![enter image description here][1] [1]: http://imgur.com/umCT6.jpg UserIdTAG: 219780 UserNameTAG: ReconIII CreateTimeTAG: 2013-01-04T03:47:34Z VoteTAG: 15 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: Congratulations. FirstChildUserIdTAG: 110802 FirstChildUserNameTAG: leoblack FirstChildCreateTimeTAG: 2013-01-04T12:54:07Z FirstChildTAG: Is this original or copy? FirstChildUserIdTAG: 318037 FirstChildUserNameTAG: Maher-84 FirstChildCreateTimeTAG: 2013-01-04T05:20:33Z FirstChildTAG: Lol I already have mine framed too. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2013-01-04T13:27:35Z FirstChildTAG: That is totally awesome! Congratulations Matthew! Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2013-01-04T22:03:51Z FirstChildTAG: Wow,so cool!I will have mine framed and hang on the wall. Very happy to get the certificate! FirstChildUserIdTAG: 361137 FirstChildUserNameTAG: Ericson FirstChildCreateTimeTAG: 2013-01-04T16:13:31Z IndexTAG: 92 TitleTAG: How to get a certificate after completing the course, for example: Hi Since everyone is asking, it is better to show you. Following the course, the course appears the label: Your certificate generated. It can hang on the course a few hours or days. That will cause even more questions. And you will see a button labeled as below: Download Your PDF certificate. If the position of points were not be scored, there will be another polite label, as shown (below). ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13565184977637822.png UserIdTAG: 402617 UserNameTAG: Vl CreateTimeTAG: 2012-12-26T10:56:48Z VoteTAG: 15 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: Thanks ! FirstChildUserIdTAG: 454609 FirstChildUserNameTAG: Alwahsh FirstChildCreateTimeTAG: 2012-12-26T11:17:36Z FirstChildTAG: Additionally, the certificate looks like this. ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13565221361343648.png FirstChildUserIdTAG: 402617 FirstChildUserNameTAG: Vl FirstChildCreateTimeTAG: 2012-12-26T11:43:57Z SecondChildTAG: when was it available?i mean how long after final? SecondChildUserIdTAG: 342221 SecondChildUserNameTAG: deepkar SecondChildCreateTimeTAG: 2012-12-26T11:48:23Z SecondChildTAG: hey it doesn't shows your grade...........is it?????????? SecondChildUserIdTAG: 108863 SecondChildUserNameTAG: shiviz SecondChildCreateTimeTAG: 2012-12-26T11:53:40Z SecondChildTAG: Two or three days, but during which time there was no holidays. A many of people a lot of time on the generation and preservation of certificates. SecondChildUserIdTAG: 402617 SecondChildUserNameTAG: Vl SecondChildCreateTimeTAG: 2012-12-26T12:02:24Z SecondChildTAG: how to check the validity of this certificate? SecondChildUserIdTAG: 342221 SecondChildUserNameTAG: deepkar SecondChildCreateTimeTAG: 2012-12-26T12:18:30Z SecondChildTAG: At the bottom of the certificate is a link to a resource where you can check it. But I have not tried it, you can try to check yourself. SecondChildUserIdTAG: 402617 SecondChildUserNameTAG: Vl SecondChildCreateTimeTAG: 2012-12-26T13:06:22Z FirstChildTAG: Спасибо Владимир! Красивый сертификат. Thanks a lot to Vladimir! Nice certificate! FirstChildUserIdTAG: 325197 FirstChildUserNameTAG: Vitali_Jerin FirstChildCreateTimeTAG: 2012-12-26T11:51:51Z FirstChildTAG: I hope we'll get the certificate in january, because than it will be new. If we get it before, then in january it will be from last year! I want a new one! FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-12-26T12:19:22Z SecondChildTAG: Yes, most likely they will do after the new year, after the holidays. SecondChildUserIdTAG: 402617 SecondChildUserNameTAG: Vl SecondChildCreateTimeTAG: 2012-12-26T13:08:14Z SecondChildTAG: What means New? New design of some sort? or just '13th dated? SecondChildUserIdTAG: 345502 SecondChildUserNameTAG: ElectroVova SecondChildCreateTimeTAG: 2012-12-26T14:38:45Z SecondChildTAG: 13th dated. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-12-26T14:44:28Z SecondChildTAG: January, 13 2013 :) SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-26T19:43:58Z IndexTAG: 93 TitleTAG: Please in advance stop saying your score Please stop saying your score until the exam is finished, for a few reasons : - the forum will be filled up with unnecessary posts and perhaps hard to search in, for an explanation to a homework, that could help others to solve a question ; - other students with lower scores will feel bad, and it's no need.Besides, it's almost Christmas ; - a person with 100% score it's a hard working and intelligent individual. Saying that, score comes from vanity .It's not an intelligent move to brag. - Someone will make a list with all scores at the final. Good luck and see you after. UserIdTAG: 240487 UserNameTAG: AlexAlexandrescu CreateTimeTAG: 2012-12-19T16:56:55Z VoteTAG: 15 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: i was about to post something on similiar lines!! u did my job well!! thanks and good luck! FirstChildUserIdTAG: 368712 FirstChildUserNameTAG: jmen FirstChildCreateTimeTAG: 2012-12-19T17:02:20Z SecondChildTAG: Godd luck and happy exam ! :) SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-12-19T17:33:43Z FirstChildTAG: You are right. I completely agree. FirstChildUserIdTAG: 418182 FirstChildUserNameTAG: euldji2005 FirstChildCreateTimeTAG: 2012-12-19T22:12:19Z IndexTAG: 94 TitleTAG: Hola Myriam and other Classmates,Folks,Blokes,Guys.. I was a student of last semester 6002.x course.I would like to give a little help to other students on the field of Op Amp. I have made a video with LM358 Op Amp in order to show how the inverting and non-inverting connection work on a breadboard comparing to the calculated data on a sheet of paper. If you are interested to see this video click on this link: http://www.youtube.com/watch?v=nx2R0UcfDYU P.S. Here,on this link you can find the datasheet: http://www.ti.com/lit/ds/symlink/lm158-n.pdf UserIdTAG: 35103 UserNameTAG: komisz CreateTimeTAG: 2012-12-02T16:59:46Z VoteTAG: 15 CoursewareTAG: Week 12 / S23E2 Inverting amplifier analysis CommentableIdTAG: 6002x_S23E2_Inverting_amplifier_analysis NumberOfReplyTAG: 4 FirstChildTAG: Like like :) FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-02T19:28:23Z SecondChildTAG: Thank You SecondChildUserIdTAG: 35103 SecondChildUserNameTAG: komisz SecondChildCreateTimeTAG: 2012-12-02T19:49:02Z FirstChildTAG: good video komisz I would like to ask about the program and the devise that you use to supply and measure the A.C. power. could you please give us information about it ( name, price,...) FirstChildUserIdTAG: 226085 FirstChildUserNameTAG: Teto FirstChildCreateTimeTAG: 2012-12-02T23:08:16Z SecondChildTAG: Yes sure, Here is the link: http://www.parallax.com/go/propscope SecondChildUserIdTAG: 35103 SecondChildUserNameTAG: komisz SecondChildCreateTimeTAG: 2012-12-02T23:23:21Z SecondChildTAG: thank you very much SecondChildUserIdTAG: 226085 SecondChildUserNameTAG: Teto SecondChildCreateTimeTAG: 2012-12-07T10:49:55Z FirstChildTAG: Very nice demonstration. Well done. FirstChildUserIdTAG: 90796 FirstChildUserNameTAG: ChaunceyGardiner FirstChildCreateTimeTAG: 2012-12-03T21:07:23Z SecondChildTAG: Very inspiring too. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-12-03T21:34:13Z FirstChildTAG: Thanks komisz! FirstChildUserIdTAG: 398594 FirstChildUserNameTAG: DaveyJC FirstChildCreateTimeTAG: 2012-12-04T21:11:04Z IndexTAG: 95 TitleTAG: where is the midterm link? Could anybody help me? How to get to the midterm exam link? UserIdTAG: 380287 UserNameTAG: gayetan CreateTimeTAG: 2012-10-25T04:55:00Z VoteTAG: 15 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: The test is not release yet !! FirstChildUserIdTAG: 628745 FirstChildUserNameTAG: albertclint FirstChildCreateTimeTAG: 2012-10-25T05:26:43Z SecondChildTAG: can we have an email notification mentioning the test's link too SecondChildUserIdTAG: 694092 SecondChildUserNameTAG: emanurag SecondChildCreateTimeTAG: 2012-10-25T05:28:26Z FirstChildTAG: Hi. Mid term available at **course ware** just under the **Week6**. > **Regards:** asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-10-25T14:22:56Z IndexTAG: 96 TitleTAG: Hints for those struggling with Lab-6: Q1 and Q2 part a and b! ![enter image description here][1] ![enter image description here][2] ![enter image description here][3] ![enter image description here][4] And Voila. Hazel [1]: https://edxuploads.s3.amazonaws.com/1350843061255076.png [2]: https://edxuploads.s3.amazonaws.com/13508430671343661.png [3]: https://edxuploads.s3.amazonaws.com/13508430821343696.png [4]: https://edxuploads.s3.amazonaws.com/13508430921343615.png UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-10-21T18:11:57Z VoteTAG: 15 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 97 TitleTAG: Neon Relaxation Oscillator without differential equations here is another solution without using differential equations figure A shows what really happens, after an initial charge we have sequential charges and discharges so we divide the problem to 2 sets : charge (B) and discharge (C) ![enter image description here][1] charge (B) we use the general exponential formula for charging a capacitor from zero volts to a target voltage Vsteady: Vc=Vsteady(1-e^(-t/RC)) watch for the 32volts vertical axis displacement A <-> C the solution does not change if we accept that the capacitor tries to charge from 0volts to 92-35=57 and we are interested for the time it takes the voltage to get to 77-35=42volts discharge (C) (***) we use the general exponential formula for discharging a capacitor from an initial voltage Vinit to a target zero voltage : Vc=Vinit*e^(-t/RC)) (we do not need vertical displacement here as that is exactly what happens) Hope this helps some people without diff equations mathematical background. Coupled with thevenin and norton and voltage dividers these generic charging and discharging formulas (with vertical displacement modifications when necessary) can solve many RC problems *** as YakovO correctly pointed out: "for discharge VI will not be zero, it will be VS*RB/(R+RB)=92*10k/(1.5M+10k)=0.613 also R will not be 10k, but 1.5M||10k=9933.77 it's not a big deal for numbers, but for understanding :)" That is why I mentioned voltage dividers at the end of my post [1]: https://edxuploads.s3.amazonaws.com/13498119526515985.jpg UserIdTAG: 319453 UserNameTAG: leonidasGr CreateTimeTAG: 2012-10-09T20:11:51Z VoteTAG: 15 CoursewareTAG: Week 6 / Neon relaxation oscillator exercise CommentableIdTAG: 6002x_neon_relaxation_oscillator_exercise NumberOfReplyTAG: 2 FirstChildTAG: Thank you for posting this - it was very helpful. By the way, I get the right answers for this, but negative. I end up with an equation like this: t = ln(15/57)*1.5 = -2.002 Can you tell me where I am slipping up? FirstChildUserIdTAG: 434869 FirstChildUserNameTAG: patey FirstChildCreateTimeTAG: 2012-10-14T18:33:26Z SecondChildTAG: (-t) from e^(-t/RC) SecondChildUserIdTAG: 237941 SecondChildUserNameTAG: per2x SecondChildCreateTimeTAG: 2012-10-18T03:42:19Z FirstChildTAG: I do not understand where did you get the 10^2 in the "e^-t/10^2". thanks FirstChildUserIdTAG: 166869 FirstChildUserNameTAG: Vasco FirstChildCreateTimeTAG: 2012-10-20T19:17:54Z SecondChildTAG: do you mean e^(-t/10^-2) ? It is the time constant for the discharge: RC = 10kOhm * 1microFarad = 0.01 s SecondChildUserIdTAG: 434869 SecondChildUserNameTAG: patey SecondChildCreateTimeTAG: 2012-10-21T20:22:14Z IndexTAG: 98 TitleTAG: H4P3 Part B (Problem solved) Hey guys, I see that a lot of you have been struggling with the Norton equivalent. Here's where most of you are going wrong. Firstly, you CAN'T take the equivalent Resistance as (R1+R2)||R3, because there's a dependent source lying, whose value depends on the point between R1 and R2. So we assume that the "U" is our first voltage source, and that part's equivalent resistance is R2||R3. Then connect this equivalent circuit with the 5V voltage source and R1. So, we finally get the equivalent Norton/Thevenin resistance as [R1+(R2||R3)], and one can calculate the current following the same steps. It's piecewise-linear, so break it into pieces UserIdTAG: 118266 UserNameTAG: Nisheet CreateTimeTAG: 2012-10-07T14:29:46Z VoteTAG: 15 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: I really don't get the current part. Does Au stands for gain as it doesn't have any units? I got the Rn early but I'm stuck with In until now. FirstChildUserIdTAG: 352373 FirstChildUserNameTAG: keyholder FirstChildCreateTimeTAG: 2012-10-07T14:48:32Z SecondChildTAG: I thought AU was possibly a current controlling the VCVS, but as it turns out it this is not true. As far as the units go, we know u is a voltage, so Au seems to be a voltage increased by constant A. So, now that I think about it, A would be dimensionless since u has the units of volts. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-07T17:29:30Z SecondChildTAG: Hmm, "a current controlling the VCVS", sounds contradicting, therefore Au can't be a current, otherwise the dependent source would be a CCVS. Go figure? SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-07T17:32:33Z FirstChildTAG: Thanks alot Nisheet FirstChildUserIdTAG: 441436 FirstChildUserNameTAG: Ooffy FirstChildCreateTimeTAG: 2012-10-07T17:52:18Z SecondChildTAG: :) SecondChildUserIdTAG: 118266 SecondChildUserNameTAG: Nisheet SecondChildCreateTimeTAG: 2012-10-10T11:51:06Z FirstChildTAG: I don't understand why the "U"'s equivalent resistance is R2||R3. I can't "see" it! FirstChildUserIdTAG: 359310 FirstChildUserNameTAG: AaronYeoh FirstChildCreateTimeTAG: 2012-10-07T23:39:10Z SecondChildTAG: Thanks a lot but I still have some problems to "see it" too. SecondChildUserIdTAG: 298547 SecondChildUserNameTAG: AnthonyRF SecondChildCreateTimeTAG: 2012-10-08T04:06:24Z SecondChildTAG: I guess it's too late to reply, but if you still can't see it, try this:![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/1349863644134366.png SecondChildUserIdTAG: 118266 SecondChildUserNameTAG: Nisheet SecondChildCreateTimeTAG: 2012-10-10T10:07:38Z SecondChildTAG: Sorry for the stupid/annoying arrows, but I wanted to emphasize on this thing: Break it into pieces! The red and the blue are linear individually, thus, one can use Thevinin/Norton tricks on those, but you can't simply use it on the whole circuit as it is piecewise linear, i.e., non-linear (unless you break it into linear pieces). SecondChildUserIdTAG: 118266 SecondChildUserNameTAG: Nisheet SecondChildCreateTimeTAG: 2012-10-10T10:12:01Z IndexTAG: 99 TitleTAG: In attention to those so desperate to obtain the certificate I don't know that any of you realize that this certificate doesn't have a real value in the real tough world , except maybe vanity, self respect , etc. BUT, what many of you should understand it's that the real value is GOOD QUALITY FREE KNOWLEDGE . If we are after a certificate, rather than free knowledge, then i think we are waaay off the point of this initiative. They have spent money, their time and a touch of heart maybe, for us to have access to this wonderful system of education. Our society evolution won't be based on certificates , but on education, respect for other people culture and pace. In my opinion i think that is OUR TURN not to disappoint them . UserIdTAG: 240487 UserNameTAG: AlexAlexandrescu CreateTimeTAG: 2012-12-22T20:10:08Z VoteTAG: 14 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: yes you guy has the point FirstChildUserIdTAG: 475448 FirstChildUserNameTAG: msamwelmollel FirstChildCreateTimeTAG: 2012-12-22T20:21:45Z FirstChildTAG: Scientia potentia est. Francis Bacon FirstChildUserIdTAG: 402617 FirstChildUserNameTAG: Vl FirstChildCreateTimeTAG: 2012-12-22T20:24:37Z FirstChildTAG: Leonardo da Vinci : "The sea with it's waves won't make so much noise, as the desire for knowledge of the human heart" . FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-12-22T20:27:27Z FirstChildTAG: Hmm..Alex, each student has own motivation which may be depended from the different conditions, reasons and etc. I hope that all students do understand that "possible" certificate does not has real weight, though. But it may be dream to have it..Or someone have motivation to show this certificate. So we cant judge their wishes. The EDX team doing Excellent work. And who knows, maybe because of this work many students discovered electronics for itself, and among of us we have future designers of the magnificent and complex devices.. FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-12-22T20:46:47Z SecondChildTAG: I agree Serghei. I was only making an appeal to moderation. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-12-22T21:08:52Z SecondChildTAG: me three, I agree. I was always afraid to talk about electronics. Now, I can't say I became genius in this domain, but it made my fear toward this domain less than I ever expected. Furthermore, having such experience made my soul very thirsty to learn even more about this domain. I'm so eager to learn even more about transmission lines and antennaes and complicated analog and digital circuits. May Allah bless all of your souls. SecondChildUserIdTAG: 418182 SecondChildUserNameTAG: euldji2005 SecondChildCreateTimeTAG: 2012-12-22T23:24:51Z SecondChildTAG: Now you may learn much more if you decide to build something. SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-23T11:19:28Z SecondChildTAG: That's correct. I intend to do so SecondChildUserIdTAG: 418182 SecondChildUserNameTAG: euldji2005 SecondChildCreateTimeTAG: 2012-12-23T11:59:17Z FirstChildTAG: Very wise words, thank you. In my country, universities do not believe in any certificate from other countries (besides few), so it will be a self award to remind me that I may have what it's necessary to start learning even much complicated sciences. and also, to help me learn about my capacity (which is below average :( ) in order to make me push my self even more in order to repelnish my skills. FirstChildUserIdTAG: 418182 FirstChildUserNameTAG: euldji2005 FirstChildCreateTimeTAG: 2012-12-22T23:36:38Z SecondChildTAG: Dear euldji , when i started to learn about electronics i was in highschool. It was very difficult, because there was limited knowledge on the internet and google was at the beginings . So i had to go to the town library and to spend hours reading about monostable and astable circuits, about amplification and other stuff . I went to university of automatic control hoping to learn the right way, but the level seemed to me very low so i have lost interess . I started to work in cg, and i've stumbeld almost accidentaly over edx on youtube .Now i have a wider range of understanding , and i can finally put all the pieces together . You never know when you will find what you are looking for, but if you keep the flame burning, sooner or later you'll find your answers. Allah bless you too ! SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-12-23T00:10:48Z SecondChildTAG: Wow ! cool ! I started to learn about electronics 7 years ago. I had a teacher that kept teching as electronics as physics, so it became difficult to me to learn it in the most appropriate way, since his way of thinking was so different than mine. Anyway, This opportunity prooved to me that pedagogy is the most important piece in order to render learning a very amusing experience. SecondChildUserIdTAG: 418182 SecondChildUserNameTAG: euldji2005 SecondChildCreateTimeTAG: 2012-12-23T12:05:47Z IndexTAG: 100 TitleTAG: Announcement: Extra attempts and time for questions 4 and 6 Due to some confusing statements of questions 4 and 6, we have extended the number of attempts on those problems to a total of 6 attempts for the next 24 hours. If you used all of your attempts and/or all of your time and suffered from those confusing statements, please feel free to go back, re-read the question, and correct your answers. UserIdTAG: 96452 UserNameTAG: Lyla CreateTimeTAG: 2012-12-21T13:40:55Z VoteTAG: 14 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Thanks a lot! FirstChildUserIdTAG: 309359 FirstChildUserNameTAG: abarea10 FirstChildCreateTimeTAG: 2012-12-21T15:51:19Z FirstChildTAG: Thanks a lot! ) FirstChildUserIdTAG: 394836 FirstChildUserNameTAG: v2g6ch4 FirstChildCreateTimeTAG: 2012-12-21T15:55:18Z IndexTAG: 101 TitleTAG: We should have T-Shirts It would be cool if edx offered T-Shirts :D lol what do you think? ![enter image description here][1] ![enter image description here][2] [1]: https://edxuploads.s3.amazonaws.com/13558463503499024.png [2]: https://edxuploads.s3.amazonaws.com/13558463631343694.png UserIdTAG: 219780 UserNameTAG: ReconIII CreateTimeTAG: 2012-12-18T16:04:33Z VoteTAG: 14 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: I believe that edX is going to change the world, but before it can, people have to know that it exists. Word of mouth is going to be important. T-shirts can help with that. FirstChildUserIdTAG: 202032 FirstChildUserNameTAG: KeithD FirstChildCreateTimeTAG: 2012-12-18T19:47:14Z FirstChildTAG: You asked :)... Imho: its a marketing means / promotional method for a world with 3 dimensions. Here we are in a virtual world. So this fits like a nut into a wall. So it might work on a campus, but cant on forum. To find promotional gifts and methods suitable for virtual space one has to be more inventive than that: make an edx screen saver with actual useful formulas, filter circuits with bode plots passing by, small circuits, etc. Ofc from time to time AA saying AHA!. Background pictures, prefferably some either funny or embarassing ones with staff, pref ppls in videos would pwn. :) Also, promotional stuff is used to promote your service, but for the ppls already here it should not be done since we are all already "clients", we already subscribed, so no point. So no "promovability" in this direction. Then its either educative or "funny". If ppls in videos are to wear them, it might work, perhaps promotes some kind of unity. But might need baseball caps with embroidery to go along with the t-shirts. You havent established yet what you want to achieve with it. Ideas are worth nothing by themselves. Hitting and objective is. You just need an objective first, then find the best way to achieve it. Not the other way around. Else its just an expense with no purpose or utility. FirstChildUserIdTAG: 331483 FirstChildUserNameTAG: Doru FirstChildCreateTimeTAG: 2012-12-18T19:12:40Z FirstChildTAG: all this is possible via ebay dropshipping order FirstChildUserIdTAG: 323963 FirstChildUserNameTAG: gtrf FirstChildCreateTimeTAG: 2012-12-19T10:00:19Z FirstChildTAG: it's nice man!! FirstChildUserIdTAG: 509517 FirstChildUserNameTAG: randima FirstChildCreateTimeTAG: 2012-12-20T11:48:05Z FirstChildTAG: cool idea! FirstChildUserIdTAG: 721593 FirstChildUserNameTAG: CircuitsRule FirstChildCreateTimeTAG: 2012-12-20T04:03:19Z IndexTAG: 102 TitleTAG: The course by itself is one big AHA! I've never imagined that learning Analog Electronics would be so enchanting and fascinating. Dr.Agarwal's way of explaining things made me feel like a kindergarten kid listening with wide-eyes to fantastic fairy tales full of adventures narrated by her teacher. I wish the stories never end and his narration goes on and on. In his own terms, the course by itself calls for one big AHA! moment. I yearningly hope that he offers a lot more number of courses. Would be nice if he can indicate something regarding his future plans for new courses, as he sometimes mentions in his lectures as "upcoming attractions"!! :) Dr.Agarwal's teaching methods and attitude should be used as a guidance in all Teachers Training Courses, right from Pre-school/Kindergarten to Advanced Degrees. Thank you, Professor Anant! And a big thanks to all the staff who provided such fabulous support to the course! अन्नदानं परं दानं विद्यादानं अत: परम् | अन्न्ेान क्षणिका तॄप्ति: यावज्जीवं च विद्यया || Giving food to the hungry is a good deed indeed ('anna dAnam'). But more than that, giving knowledge to people ('vidya dAnam') is better because by giving food one's hunger would be satisfied for a short time. But knowledge is helpful for a whole life. Therefore it is considered that giving knowledge ('vidya dAna') is the most important among various other types of giving. (Reference: Samskruta Subhashitani, Sloka 70, http://sa.wiktionary.org) UserIdTAG: 232157 UserNameTAG: GayatriTR CreateTimeTAG: 2012-12-12T03:03:03Z VoteTAG: 14 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Count me in for the aha. FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-12-12T09:01:33Z FirstChildTAG: Well said GayatriTR I have struggled all my life with learning through my EGR BS Degree and working for 25 years, and when I came across Dr. Agarwal's videos I was astounded. I thought, where was he when I was in school? He makes everything so clear on topics I've been stumped on for years. Well, its never too late to learn, so I'm really looking forward to many more classes. FirstChildUserIdTAG: 132828 FirstChildUserNameTAG: rharris FirstChildCreateTimeTAG: 2012-12-12T21:13:22Z FirstChildTAG: I am also very much eagerly waiting for the upcoming analog and digital electronics courses. thanks for those already running. FirstChildUserIdTAG: 149146 FirstChildUserNameTAG: shahudullah FirstChildCreateTimeTAG: 2012-12-12T17:22:31Z IndexTAG: 103 TitleTAG: Little mistake with number ranges The number examples of tau=Rin*C Rin = 1 GOhms = 10^9 Ohms and C= 1 fFarads = 10^-15 Farads So, just correcting the numbers approximation we will get tau = 1us = 10^-6 seconds, and not 1ms. UserIdTAG: 278353 UserNameTAG: vgmariucci CreateTimeTAG: 2012-11-03T18:37:56Z VoteTAG: 14 CoursewareTAG: Week 8 / Building A Better Static Memory Element CommentableIdTAG: 6002x_Building_A_Better_Static_Memory_Element NumberOfReplyTAG: 1 FirstChildTAG: exactly:) FirstChildUserIdTAG: 209164 FirstChildUserNameTAG: KarthikHegde FirstChildCreateTimeTAG: 2012-11-10T15:21:37Z IndexTAG: 104 TitleTAG: [Staff] Will be some deadline extension of Hw and Labs of Week 7 for those who where affected by the Hurricane Sandy? Dear Staff, As a way of solidarity with my Classmates, I was wondering if it will be possible some deadline extension of Asignments of Week 7 for those who where affected by the Hurricane Sandy... It is not my case, because I live in South America and I have already finished my Hw and Lab of Week 7, but I was thinking of my Classmates who were affected with Power cut, Online Connection shortage, water services interruption, etc... and couldn't follow the videos lectures and doing properly the assignments. Thank you very much, Best regards, Myriam. P.D for the 6.002x Classmates: JerseyMark was affected for the Hurricane Sandy, he was worried about the students - he is really responsible with his commitment as Community TA - he requested us (others Community TA's) to tell you about his situation in the case that you asked... So he will be offline some days more... (I hope he don't mind that I have posted this) ... Back soon JerseyMark! My best wish to you! UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2012-11-03T02:17:27Z VoteTAG: 14 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: Yes. I think the staff should have to be considering this. Since it is not about only a few people but the whole humanity. FirstChildUserIdTAG: 131726 FirstChildUserNameTAG: chemiboy11 FirstChildCreateTimeTAG: 2012-11-03T11:26:38Z FirstChildTAG: Staff, Hope the request for extention would be considered sympathatically. Wahab FirstChildUserIdTAG: 455950 FirstChildUserNameTAG: Wahabbaluch FirstChildCreateTimeTAG: 2012-11-03T03:50:37Z SecondChildTAG: Andy-Sandy :((( SecondChildUserIdTAG: 286954 SecondChildUserNameTAG: ododo SecondChildCreateTimeTAG: 2012-11-03T13:09:49Z FirstChildTAG: I guess I will have the best excuse for 6.002x and 3.091x... the hurricane ate my homework. I am one of those affected by Hurricane Sandy. I live and work in midtown Manhattan (New York City), and I spent the entire week at work from Monday until Saturday morning without power, working with little sleep to get generators transported from other areas of the US and connected to the buildings I am responsible for. I am still working, trying to get all building systems, communications systems, and industrial processing equipment working so we can be back in business ASAP. This part of Manhattan lost power on Monday night and didn't get it back until late Friday evening. I spent the week hooking up generators large and small to keep things going, about 3.5 megawatts total, plus an army of porta-potties (no power means no water pumps or sewage ejectors, thus no flush toilets or running water). We had only one freight elevator running on emergency generator power, and it only went up to the third floor in a 10-floor building. I have climbed more stairs in the past week than I have in probably the last year. The buildings total 2 million square feet, so when I wasn't climbing stairs, I was walking my legs off. First, we had to get water out of the electrical switchgear rooms and the lower floors of the buildings so we could safely hook up generator power to the panel boards (4000 amps at 480 volts each). We had an entire city block (avenue to avenue) lined with large generators on tractor trailers, along with trailers full of supporting electrical equipment. We also had many smaller generators operating critical systems. We were just finishing the connections on a 1.8 megawatt generator when Con Ed restored power. That doesn't end the work involved, switchover from generator power back to Con Ed is a pain for these humongous generators. We had to evacuate the areas of the affected buildings, load shed and shutdown electrical distribution panels, undo the generator cabling and reinstall bus bars, then reconnect the switchgear to Con Ed service entrance panels. It's all done manually with lots of trepidation and multiple layers of safety checks because of the massive power involved. Arc flash and arc blast are extremely dangerous hazards, the people who have to manually operate the electrical switchgear are literally taking their lives in their own hands. I missed all the video news coverage of the storm because I was working (and had no power), and only heard some of the descriptions via battery-powered radio. When I came home Saturday and saw some of the pictures, I was floored. On Monday night after we lost power, I watched parts of New Jersey go dark from my office window, and then saw the transformers explode and light up the night sky, but I couldn't really visualize the destruction from the storm surge. I am very fortunate because my apartment was not damaged, and none of my friends, family, and loved ones were injured or killed. Some of my work colleagues have damaged homes and family members with injuries from the storm or severe medical conditions. I am tired and sore, and most likely won't get a day off from work until sometime next week, but I was able to go home yesterday and take a shower and sleep in a bed instead of on my office floor or stretched across a few chairs. Heat is still a problem, but it came back today, so maybe I will have a warm apartment when I finally go home. I hope everyone in the edX community is safe and sound. I'll touch base later in the week when things calm down. FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-11-04T15:17:19Z SecondChildTAG: Bufff g_hopper, my best wishes for you and your family. In Spain this is the first new you can see in TV and newspapers and a lot of people is making efforts to help. Hope at least for YOU and other affected, MITx can postpone the homework due date this week. SecondChildUserIdTAG: 342966 SecondChildUserNameTAG: SandraNavarro SecondChildCreateTimeTAG: 2012-11-04T15:39:38Z SecondChildTAG: I was worried about you g_hopper ... Nice to know that you are fine. If there is anything that I can do for you, please tell me, I will try to help you. I am sure that everything will be getting better soon, All the best wish for you, your family and your friends, Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-04T16:10:51Z SecondChildTAG: @g_hopper...wow. a fascinating first-person account. It's too easy to forget how fortunate we usually are. SecondChildUserIdTAG: 92346 SecondChildUserNameTAG: MobiusTruth SecondChildCreateTimeTAG: 2012-11-05T02:08:26Z FirstChildTAG: Thanks for pointing this out, and our hearts go out to all those effected by this storm. We have extended the deadline by two days for those who have been out of power for the past week. We hope that the recovery from the storm goes smoothly. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-11-04T23:58:32Z SecondChildTAG: Thank you, thank you Lyla! I am sure that they will be really grateful with edX :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-05T00:10:48Z SecondChildTAG: Thanks Lyla! Much appreciated. SecondChildUserIdTAG: 94557 SecondChildUserNameTAG: g_hopper SecondChildCreateTimeTAG: 2012-11-05T01:12:40Z SecondChildTAG: A nice gesture. SecondChildUserIdTAG: 92346 SecondChildUserNameTAG: MobiusTruth SecondChildCreateTimeTAG: 2012-11-05T02:03:45Z SecondChildTAG: Great job. Cosidering the devastating situation, students of the affacted area still need some relaxation while grading. SecondChildUserIdTAG: 455950 SecondChildUserNameTAG: Wahabbaluch SecondChildCreateTimeTAG: 2012-11-05T03:30:19Z IndexTAG: 105 TitleTAG: Future MITx courses Hello I just want to know if there's a plan to offer advanced electrical engineering courses (Power Systems, Power Electronics, Electric Machines) in the future. The 6.002x was offered last semester and it's offered again this semester, I wonder if MITx will offer other electrical engineering courses in the coming semester. I'm an electrical engineer and I'm enjoying the course very much. Thanks to MITx for giving us a chance to get an MIT education :-) UserIdTAG: 94596 UserNameTAG: Alkhdour CreateTimeTAG: 2012-10-30T17:49:29Z VoteTAG: 14 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: It would be much grateful if MIT offers courses in power systems and electric machines in near future. FirstChildUserIdTAG: 219204 FirstChildUserNameTAG: vsriram FirstChildCreateTimeTAG: 2012-10-31T06:23:40Z FirstChildTAG: Every now and then someone reminds us that OCW offers a lot of courses. I wonder if it offers a community to the learner as well. I think OCW Scholar does, doesn't it...is that just for selected topics? FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-10-31T14:59:37Z IndexTAG: 106 TitleTAG: Isn't it past mid-term exam start time? I don't see any link? UserIdTAG: 106776 UserNameTAG: skcvr CreateTimeTAG: 2012-10-25T04:09:09Z VoteTAG: 14 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 7 FirstChildTAG: 17 mins gone now FirstChildUserIdTAG: 266739 FirstChildUserNameTAG: satvikchugh FirstChildCreateTimeTAG: 2012-10-25T04:17:30Z SecondChildTAG: I don't understand. The exam should have been posted by now. I don't see how it might be late. Its supposed to be auto-generated by the system. SecondChildUserIdTAG: 179140 SecondChildUserNameTAG: JoydeepSil SecondChildCreateTimeTAG: 2012-10-25T04:26:02Z FirstChildTAG: already 15 mins... FirstChildUserIdTAG: 200319 FirstChildUserNameTAG: Virviil FirstChildCreateTimeTAG: 2012-10-25T04:14:35Z FirstChildTAG: same here... FirstChildUserIdTAG: 120850 FirstChildUserNameTAG: Nitin1A FirstChildCreateTimeTAG: 2012-10-25T04:10:00Z SecondChildTAG: Same here SecondChildUserIdTAG: 719214 SecondChildUserNameTAG: Anonwok SecondChildCreateTimeTAG: 2012-10-25T04:11:41Z FirstChildTAG: same here FirstChildUserIdTAG: 365309 FirstChildUserNameTAG: chetnasinghaldas FirstChildCreateTimeTAG: 2012-10-25T04:11:38Z SecondChildTAG: Same here. SecondChildUserIdTAG: 464744 SecondChildUserNameTAG: attache SecondChildCreateTimeTAG: 2012-10-25T04:13:34Z FirstChildTAG: It is still not there after ten mitnutes? FirstChildUserIdTAG: 99628 FirstChildUserNameTAG: Pranjal16 FirstChildCreateTimeTAG: 2012-10-25T04:12:06Z SecondChildTAG: waiting waiting waiting!!! SecondChildUserIdTAG: 365309 SecondChildUserNameTAG: chetnasinghaldas SecondChildCreateTimeTAG: 2012-10-25T04:22:01Z FirstChildTAG: D: is'n good... in Chile es 1:33 am i need go to sleep. i hope begin the exam at the morning... im so anxious :( FirstChildUserIdTAG: 70519 FirstChildUserNameTAG: Fipe FirstChildCreateTimeTAG: 2012-10-25T04:37:07Z FirstChildTAG: it's a time warp in Boston :) FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-25T04:42:50Z IndexTAG: 107 TitleTAG: Thank you All thanks to you for your appreciation of us Muslims for(holiday of Eid al-Adha).This is progress in thought .Thank you to all the members of this course and all the officials of edX UserIdTAG: 318037 UserNameTAG: Maher-84 CreateTimeTAG: 2012-10-23T21:04:47Z VoteTAG: 14 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: you cannot imagine my happiness for this news ..... many thanks ... FirstChildUserIdTAG: 428658 FirstChildUserNameTAG: Moza FirstChildCreateTimeTAG: 2012-10-23T21:33:23Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-24T00:31:46Z SecondChildTAG: Me too SecondChildUserIdTAG: 109941 SecondChildUserNameTAG: Alhashemy SecondChildCreateTimeTAG: 2012-10-26T13:20:37Z FirstChildTAG: Thanks EDX management for extending the midterm deadline :-) FirstChildUserIdTAG: 460906 FirstChildUserNameTAG: comsat002 FirstChildCreateTimeTAG: 2012-10-24T05:41:30Z FirstChildTAG: THANK U VERY MUCH FirstChildUserIdTAG: 156535 FirstChildUserNameTAG: badawiIiIi FirstChildCreateTimeTAG: 2012-10-24T09:04:25Z FirstChildTAG: For real, I'm now very very happy thank you very very much And,Kool Sana we ento 6aibeen :) FirstChildUserIdTAG: 395257 FirstChildUserNameTAG: blackguitar FirstChildCreateTimeTAG: 2012-10-24T19:54:36Z SecondChildTAG: I'm so happy SecondChildUserIdTAG: 109941 SecondChildUserNameTAG: Alhashemy SecondChildCreateTimeTAG: 2012-10-26T13:22:46Z FirstChildTAG: Indeed!!..Thank you so much :)) FirstChildUserIdTAG: 122513 FirstChildUserNameTAG: Mona77 FirstChildCreateTimeTAG: 2012-10-24T06:23:45Z IndexTAG: 108 TitleTAG: Midterm submission extended to Sunday In order to accommodate the Muslim holiday of Eid al-Adha, we have extended the deadline to submit the midterm exam to 23:59 (11:59 pm) on Sunday, October 28th, Boston time. The exam will still be released on Thursday, October 25th at 00:01 (12:01 am) Boston time. UserIdTAG: 96452 UserNameTAG: Lyla CreateTimeTAG: 2012-10-23T14:53:59Z VoteTAG: 14 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 10 FirstChildTAG: Thanks! FirstChildUserIdTAG: 397838 FirstChildUserNameTAG: Muhamad_Alaa FirstChildCreateTimeTAG: 2012-10-24T00:31:04Z SecondChildTAG: thanQ SecondChildUserIdTAG: 215974 SecondChildUserNameTAG: Esmail SecondChildCreateTimeTAG: 2012-10-24T08:49:50Z SecondChildTAG: Thanks very match for this news SecondChildUserIdTAG: 109941 SecondChildUserNameTAG: Alhashemy SecondChildCreateTimeTAG: 2012-10-26T13:24:45Z FirstChildTAG: really thanks .. FirstChildUserIdTAG: 428658 FirstChildUserNameTAG: Moza FirstChildCreateTimeTAG: 2012-10-23T21:37:40Z FirstChildTAG: And yes, I am staff and this announcement is real. It should be on Course Info before the midterm is released. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-10-23T14:54:42Z SecondChildTAG: Hi Lyla, will there be any revision help, e.g past exams uploaded soon? And, it is mentioned in the Course Synopsis that the best 10 of 12 Homework Scores will be used. However, we only have 10 homeworks. Any idea on how the homework scores will be tallied. Thanks! SecondChildUserIdTAG: 143823 SecondChildUserNameTAG: NathanNadeson SecondChildCreateTimeTAG: 2012-10-23T15:03:26Z SecondChildTAG: Yes, we are going to post last year's exam, and the same review packet that last year's students got. There will be 12 home works, and only 10 of them will count towards your final grade. SecondChildUserIdTAG: 96452 SecondChildUserNameTAG: Lyla SecondChildCreateTimeTAG: 2012-10-23T15:06:32Z SecondChildTAG: Thanks! SecondChildUserIdTAG: 143823 SecondChildUserNameTAG: NathanNadeson SecondChildCreateTimeTAG: 2012-10-23T15:20:48Z SecondChildTAG: Thank you Lyla! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-23T15:24:04Z SecondChildTAG: Hi Lyla, I was wondering why sometimes the posts are made with the staf/community ta tag and sometimes they're not. Is it something you can toggle? SecondChildUserIdTAG: 154440 SecondChildUserNameTAG: Aahlad SecondChildCreateTimeTAG: 2012-10-23T15:52:57Z SecondChildTAG: thank you lyla.. SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-23T16:43:31Z SecondChildTAG: @Aahlad - Top level posts do not currently have a staff/community ta tag, but comments do. I also have an "incognito" account that only has normal student permissions. Sometimes I will forget that I am using that account, and will post with it, and there will be no flag on those either. SecondChildUserIdTAG: 96452 SecondChildUserNameTAG: Lyla SecondChildCreateTimeTAG: 2012-10-23T16:51:36Z SecondChildTAG: Thank you for your consideration :)) SecondChildUserIdTAG: 122513 SecondChildUserNameTAG: Mona77 SecondChildCreateTimeTAG: 2012-10-23T18:17:42Z SecondChildTAG: Thank you Respected Staff :)) SecondChildUserIdTAG: 271448 SecondChildUserNameTAG: UsmanRashid SecondChildCreateTimeTAG: 2012-10-23T20:15:24Z SecondChildTAG: I'm a Muslim and I want to say thank you Lyla and all the staff of Edx **عيدمبارك وكل عام والجميع بالف خير** SecondChildUserIdTAG: 109941 SecondChildUserNameTAG: Alhashemy SecondChildCreateTimeTAG: 2012-10-26T13:31:21Z FirstChildTAG: WOW! Thanks! I am a Muslim, so I wanted to say thanks for considering us although I will probably do it on Thursday. Thanks again :D FirstChildUserIdTAG: 454609 FirstChildUserNameTAG: Alwahsh FirstChildCreateTimeTAG: 2012-10-23T18:19:41Z SecondChildTAG: i think am going to do so :D lets finish what we started :D SecondChildUserIdTAG: 84213 SecondChildUserNameTAG: mohamed373 SecondChildCreateTimeTAG: 2012-10-24T14:08:40Z FirstChildTAG: awwwwsome.... thanx alot..... FirstChildUserIdTAG: 117319 FirstChildUserNameTAG: iqramalik FirstChildCreateTimeTAG: 2012-10-23T18:52:07Z FirstChildTAG: thanx alot :) FirstChildUserIdTAG: 278792 FirstChildUserNameTAG: sali FirstChildCreateTimeTAG: 2012-10-23T20:45:19Z FirstChildTAG: Best News Ever! FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-10-23T19:48:13Z FirstChildTAG: I plan to do it on Friday as I want to keep my Saturday (as here Eid is on 27th) and Sunday free for friends and Family :))..but those who are celebrating Eid on 26th this is going to be AWESOME :)....Best of Luck for the Mid-term and A very Happy Eid :) FirstChildUserIdTAG: 122513 FirstChildUserNameTAG: Mona77 FirstChildCreateTimeTAG: 2012-10-24T06:29:20Z FirstChildTAG: Thnx a lot to **Lyla** and **edX** for giving inconvinance to Muslims students...:) > **Regards:** asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-10-24T15:00:57Z SecondChildTAG: giving inconvinance??? they have helped them by considering their situation!! god bless them!! GOOD LUCK FOR YOUR MID-TERMS!! SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-24T15:34:15Z FirstChildTAG: I'm a Muslim and I want to say thank you Lyla and all the staff of Edx **عيدمبارك وكل عام والجميع بالف خير** FirstChildUserIdTAG: 109941 FirstChildUserNameTAG: Alhashemy FirstChildCreateTimeTAG: 2012-10-26T13:31:48Z IndexTAG: 109 TitleTAG: model question paper Can you upload some sample question paper for the mid term test? So that it will be useful for us.. UserIdTAG: 359968 UserNameTAG: Anirudh796 CreateTimeTAG: 2012-10-08T15:12:17Z VoteTAG: 14 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: If the other class was anything to judge by, they will post a practice exam, which is one they gave in years past to 6.002 students at MIT. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-08T18:37:56Z IndexTAG: 110 TitleTAG: Graphing on Google 1. Change the equation(s) variable to 'x'. 2. Put the two equations, separated by comma, in the search field. 3. The Graphic is show in result. "4-x, x^3" without quotes on Google... UserIdTAG: 180092 UserNameTAG: MrBondBoca CreateTimeTAG: 2012-09-19T20:41:31Z VoteTAG: 14 CoursewareTAG: Week 3 / Graphical Analysis of Nonlinear Circuits Exercise CommentableIdTAG: 6002x_graphical_analysis_of_nonlinear_circuits_exercise NumberOfReplyTAG: 3 FirstChildTAG: Cool beans! FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-19T21:41:37Z SecondChildTAG: better put it this way in google 4-​x/​8.2, x^​3 The graph of above two lines will give the exact operating point i.e. intersection of the two curves. SecondChildUserIdTAG: 200355 SecondChildUserNameTAG: bhavj SecondChildCreateTimeTAG: 2012-09-24T07:42:35Z FirstChildTAG: Fantastic! Where do they describe syntaxes for keyboard math symbolization? For instance, how does one type a logarithm? FirstChildUserIdTAG: 280220 FirstChildUserNameTAG: pilgrimCycle FirstChildCreateTimeTAG: 2012-09-19T23:48:14Z SecondChildTAG: I suggest you just experiment. ln(x), log(x) work for natural and 10-based logs. Google also tries to guess the input: sine(x) or sin(x) both work sqrt(x) works as well. SecondChildUserIdTAG: 276409 SecondChildUserNameTAG: IgnacioUY SecondChildCreateTimeTAG: 2012-09-20T02:19:46Z FirstChildTAG: you can also solve the qubic equation that results from the two equations of I(d) using wolframaplha to get a value for V(d) for the operating voltage. FirstChildUserIdTAG: 150267 FirstChildUserNameTAG: vwsingh FirstChildCreateTimeTAG: 2012-09-23T08:57:35Z IndexTAG: 111 TitleTAG: Interesting, This is so interesting that why I love electronics UserIdTAG: 286165 UserNameTAG: nchibwe CreateTimeTAG: 2012-09-17T14:50:02Z VoteTAG: 14 CoursewareTAG: Week 3 / Switch Resister Model Exercise CommentableIdTAG: 6002x_switch_resistor_model_exercise NumberOfReplyTAG: 0 IndexTAG: 112 TitleTAG: Staff: Why are people posting answers online? Sept. 16, 2012 **Dear Staff:** I've noticed that many users follow the honor code, and when someone asks for an answer for the homework problems, *they give a hint, but not the actual answer*, which seems O.k. to me. Other users in posting give out the *actual answers for graded problems*, which seems to me like helping others to cheat. It also seems like a blatant violation of any academic standards anywhere. Finally, there's this 'study group' on Facebook, which *purports to give out the answers* to Homework 1 even though it isn't due for everyone yet (Here in N.Y. we still have about 12 hours left). I know that, personally, I work out the problems multiple times, spending **hours upon hours** to catch stupid mistakes such as a wrong sign, or discovering for myself how to set up the equations to solve some of the more complicated problems. Sure, I can use a hint when I am stuck, but I feel that some users are enabling others to just type in answers and get an easy A. (Trust me it won't be so easy on the mid-term or final when there's no one to ask for help, except your past experience in actually going through the problems step-by-step.) Is the staff doing anything to address this besides the honor code? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-16T16:57:19Z VoteTAG: 14 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: I will say that the staff does actively remove answers from the forum. I have noticed on more then one occasion they have deleted threads within minutes. That being said the forum could maybe use a "report post" button. As the OP mentioned, I do not think they will succeed during the exam, unless they are all online together and take the work from one knowledgeable student. I myself would be amazed if I even got 60% on an exam, based on my current confidence level, so I don't know what they are going to do come exam time. At the end of the day, I hear that most of us have different values for the questions, so perhaps it would have the opposite effect for them, one would think. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-16T17:22:29Z SecondChildTAG: Yup, I just saw two more threads being deleted since my last post here. They were requests for me, or anyone to email them. Not very subtle, lol are they? SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-16T17:42:27Z FirstChildTAG: Sharing answers on our forum, facebook, or anywhere else is definitely against the honor code. We do not distribute any honor code certificates to accounts that we believe were violating the honor code. We performed similar actions last term, especially during the exams. We do rely on students to hold themselves to a higher standard than what we are currently able to directly enforce. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-09-16T17:34:19Z SecondChildTAG: Thank you Lyla, I didn't know being in a Facebook group violated the honor code. I just dropped out of the one I was in. There didn't seem to be any honor code violations, but I didn't want to take a chance. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-09-30T08:06:20Z FirstChildTAG: Yeah, and anyone who thinks that they can just post in an EE or circuits forum, and get the answers, let me tell you, they have already wised up about newcomers coming around and looking for these kinds of answers. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-09-16T18:36:56Z FirstChildTAG: Whatever they do in this forum, what would you have them do about the facebook group? Send federal marshals? EDIT: I didn't mean to imply that it was ok. It's not. It's clearly violating the Honor Code. If you are giving or getting the answers on graded material to or from anyone else, you are cheating. Doesn't matter where it happens. The honor code clearly says so. They can and do police the forums to the best of their abilities and people get removed from class for cheating. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-09-16T17:02:55Z SecondChildTAG: How about explicitly stating that visiting the Facebook group breaks the honor code, or letting students know if it's allowed 'by the rules'. I want to check it out, but not if it means violating a code I agreed to. SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-09-16T17:14:56Z SecondChildTAG: It does, in fact, violate the honor code. We are aware of it, and will not distribute any honor code certificates to accounts that we believe were violating the honor code. We performed similar actions last term, especially during the exams. SecondChildUserIdTAG: 96452 SecondChildUserNameTAG: Lyla SecondChildCreateTimeTAG: 2012-09-16T17:20:13Z SecondChildTAG: **Lyla:** If I **help** someone with a homework problem, by posting a *general outline* of how to solve the problem, including the *general equations* to use (i.e. KCL, power dissipated by a resistor), as long as I do **not** provide the *answer*, or a *schematic*, or *worked-out equations*, is that O.k.? SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-09-16T17:42:20Z SecondChildTAG: Yes, and I have to tell you, having taken this course once already, cheating on homework is not going to make the mid-terms and finals easier. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-09-16T18:38:55Z SecondChildTAG: I agree. I am the OP; I posted anonymously because I didn't want to piss off the people that just want answers. I, too, took this course YEARS ago at a major public university, and it was bad then, too; we didn't have the internet, but in the computer lab students would hound those known to be the "brains" for the "answers". Those types of students ended up flunking the mid-term. I can understand the need for answers: This is typically a **weed-out course**, whoever cannot pass it on their own (with only a little help) has no future or business in the electrical engineering field. Similar courses are mechanics (analyzing static forces on three-dimensional structures) and thermodynamics: although we had to take them, I don't think EECS majors have to take them anymore, only civil and mechanical engineering students. People do not want to be told they can't make the cut. You can tell when, after mid-terms, 1/3 of the class is missing because they dropped the course! Another reason people need answers, and this one is not that bad a reason as the other, is that they are lazy and they do not want to re-check their answers. It is often easy enough with the proper answer, to re-engineeer your equations and find out where your mistake was. Just the other day, I made the stupid mistake of $\frac{v_1-4}{3} = -\frac{4}{3}v$ when in reality it is $=\frac{1}{3}v_1-\frac{4}{3}$, stumping me for a good half-hour on HW1P2 until I found the basic algebra (what, grade 5) mistake! After all these years, I still mess up on signs, too. This is one reason I am re-taking this course, is to catch the errors that I didn't when I took the course the first time at age 20. Back then, partying was more on my mind than school work. For those that take this course and do it by themselves, those who do every practice problem and view every lecture if they can't grasp the material, those who spend hours upon hours on the homework, you have a bright future! I just had to get that out there. SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-09-16T19:14:16Z SecondChildTAG: Well said, JerseyMark, well said. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-09-16T19:20:03Z SecondChildTAG: agree, we as engineer or student or maybe just someone that like engineering and science must hold in the high esteem the code of honor. But as a social creature we also has to use our heart to help each other without has to breaking the code. I deeply thank you for the existence of this program and I hope it open more course SecondChildUserIdTAG: 452559 SecondChildUserNameTAG: kuz1toro SecondChildCreateTimeTAG: 2012-09-23T17:28:31Z IndexTAG: 113 TitleTAG: Two ways ![enter image description here][1] For me it's simply unclear. Most people who hear "-7.20V" do the picture that is up.However picture was given and therefore we were supposed to insert -7.2 into it (so minus became plus) and we got second picture. It should be told in a more convienient way I think. [1]: https://dl.dropbox.com/u/61908039/Capture.PNG UserIdTAG: 139775 UserNameTAG: 184052 CreateTimeTAG: 2012-09-11T09:08:11Z VoteTAG: 14 CoursewareTAG: Week 2 / Node equation review CommentableIdTAG: 6002x_S3E1_NodeEquationReview NumberOfReplyTAG: 3 FirstChildTAG: -7.2V is a property of the votage source FirstChildUserIdTAG: 331687 FirstChildUserNameTAG: haijohn FirstChildCreateTimeTAG: 2012-09-13T12:41:33Z SecondChildTAG: i agree with you, it should be given in more convenient way.. SecondChildUserIdTAG: 357440 SecondChildUserNameTAG: agadaria SecondChildCreateTimeTAG: 2012-09-16T13:43:29Z SecondChildTAG: Normally a circuit in the field will not be described like this with a negative dc power supply but the instructors want you to be aware that this can happen. SecondChildUserIdTAG: 256543 SecondChildUserNameTAG: sidney23 SecondChildCreateTimeTAG: 2012-09-17T04:17:49Z FirstChildTAG: The first picture is wrong. When you see that a voltage source has a negative value it means that its positive terminal "+" has a voltage that is lower with respect to the negative terminal "-". This is independent on how the source is oriented, the positive terminal can be up, down, right or left, and the previous statement should be always true. FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-11T13:59:39Z SecondChildTAG: Is this because of DC source?. What about AC source? SecondChildUserIdTAG: 378967 SecondChildUserNameTAG: krish2012 SecondChildCreateTimeTAG: 2012-09-12T06:32:08Z SecondChildTAG: jelizon: is this physically possible? I thought a DC voltage source had its positive side at a higher potential than its negative side by definition. SecondChildUserIdTAG: 393930 SecondChildUserNameTAG: tmciver SecondChildCreateTimeTAG: 2012-09-18T23:46:07Z SecondChildTAG: Not necessarily. The "+" terminal just a voltage "V" (the value of the source) higher than the "-" terminal. However if the value of the source is negative, then "+" is really at a lower voltage than "-". I know it's quite confusing, but that is the way the convention works. SecondChildUserIdTAG: 381746 SecondChildUserNameTAG: jelizon SecondChildCreateTimeTAG: 2012-09-25T01:05:35Z FirstChildTAG: I did the same! And fot 1.69! I agree. How am I suppose to think in the test! FirstChildUserIdTAG: 281159 FirstChildUserNameTAG: ManasiS FirstChildCreateTimeTAG: 2012-09-14T20:01:38Z IndexTAG: 114 TitleTAG: Transient analysis not working I changed the Vsupply to the sinusoid form, placed the scope probes (or voltage probes?), clicked the trans button an set the time to 5m and later to 0.005. However, once I click the ok button nothing happens, there isn't any graph. Anyone knows what's the problem? UserIdTAG: 256249 UserNameTAG: And90 CreateTimeTAG: 2012-09-05T19:58:05Z VoteTAG: 14 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 13 FirstChildTAG: This isn't an ideal solution, but after reading Kob's reply I repeated the exercise in the Circuit Sandbox and got the correct answers. It's probably a bug or something. I just hope this doesn't happen in an exam or anything like that... FirstChildUserIdTAG: 256249 FirstChildUserNameTAG: And90 FirstChildCreateTimeTAG: 2012-09-05T23:27:56Z FirstChildTAG: Hi you have to put a voltage probe for you can doing the transient, that for each node. excuse me for me English I'm speak spanish :D FirstChildUserIdTAG: 97946 FirstChildUserNameTAG: julianezapata FirstChildCreateTimeTAG: 2012-09-05T20:01:06Z SecondChildTAG: Thanks, but I did put the voltage probes in nodes A, B and C. It still doesn't work. También hablo español. SecondChildUserIdTAG: 256249 SecondChildUserNameTAG: And90 SecondChildCreateTimeTAG: 2012-09-05T20:06:02Z FirstChildTAG: i used it and it's work, look again if you have put all the data needed in trans mode like the amp (1V) the offset etc.... if anything is missing it will not work!! FirstChildUserIdTAG: 151444 FirstChildUserNameTAG: mael FirstChildCreateTimeTAG: 2012-09-05T20:12:29Z SecondChildTAG: 1,1,1k,0,0. And i tried IE9, still not working SecondChildUserIdTAG: 214584 SecondChildUserNameTAG: Askeroff SecondChildCreateTimeTAG: 2012-09-05T20:16:14Z FirstChildTAG: Make sure that you draw a connection between the probe and the circuit node FirstChildUserIdTAG: 277787 FirstChildUserNameTAG: kirilaska FirstChildCreateTimeTAG: 2012-09-05T21:53:52Z FirstChildTAG: I have the same problem. No graph or "scope trace" shows up when I do the transient analysis. I am using firefox as my browser. Which browser are you using And90? The browser is the most likely suspect, if it works for some people and not for others. FirstChildUserIdTAG: 347289 FirstChildUserNameTAG: kellrobinson FirstChildCreateTimeTAG: 2012-09-05T20:06:00Z SecondChildTAG: Am...i use firefox too. Maybe this is a problem?) SecondChildUserIdTAG: 214584 SecondChildUserNameTAG: Askeroff SecondChildCreateTimeTAG: 2012-09-05T20:11:28Z SecondChildTAG: I tried it in both Firefox and Chrome. Apparently for other parts of Lab0 Chrome is working better. SecondChildUserIdTAG: 256249 SecondChildUserNameTAG: And90 SecondChildCreateTimeTAG: 2012-09-05T20:51:32Z SecondChildTAG: Probably not a browser issue as sandbox is working for me but lab version not. Unless they are using two different versions of the code for lab and sandbox. Where lab version is failing with certain browsers. If it is that then all they got to do is use sandbox version which seems to be okay for everyone. SecondChildUserIdTAG: 15344 SecondChildUserNameTAG: kob SecondChildCreateTimeTAG: 2012-09-05T20:53:52Z FirstChildTAG: Exact same problem here, everything appears to work except the transient analysis. I am using Portable Firefox v.10.0.2 on WinXP SP3. This combination of browser and OS worked on the pilot 6.002x course, and from that 14 week marathon, I certainly know how to run a transient analysis, so I can only conclude there is a bug or limitation in the latest edX iteration of the circuit simulator. FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-09-06T01:53:49Z SecondChildTAG: Yes this is not due to any browser. Definite bug in edx systems implementation of the simulator. I suspect people have been divided up into sets and some of us are running within a flacky virtual server (I am guessing it is VMs not sure). Which is why some and not all are seeing problems. Anyway agree with you 100% its the system not the browser. SecondChildUserIdTAG: 15344 SecondChildUserNameTAG: kob SecondChildCreateTimeTAG: 2012-09-06T10:04:43Z FirstChildTAG: Another one with the same problem, running on Waterfox (64bits firefox version). sin(1,1,1k,0,0) and the three voltage probes, click on TRAN, put "5m", hit "ok" and nothing happens... ![Circuit][1] [1]: http://i45.tinypic.com/1j3ig1.png FirstChildUserIdTAG: 66241 FirstChildUserNameTAG: Ephexis FirstChildCreateTimeTAG: 2012-09-05T23:55:25Z SecondChildTAG: Exactly the same here - tried on WinXP64 with IE8, latest FF, Safari and Chrome and with Mac OS X Lion with latest Chrome, FF and Safari. Same result as above i every configuration... SecondChildUserIdTAG: 389424 SecondChildUserNameTAG: jmlietaer SecondChildCreateTimeTAG: 2012-09-09T09:22:58Z FirstChildTAG: Yea, i got the same problem. Transient analysis is not working even if i have only one probe FirstChildUserIdTAG: 214584 FirstChildUserNameTAG: Askeroff FirstChildCreateTimeTAG: 2012-09-05T20:08:04Z FirstChildTAG: my sand box is also not working for transient analysis.:( i hope it gets fixed soon as i am loving working with edX:) FirstChildUserIdTAG: 155935 FirstChildUserNameTAG: farhanakausar FirstChildCreateTimeTAG: 2012-09-06T13:17:33Z FirstChildTAG: Got the problem only in sandbox, but in the lab it was working. So quite the opposite of what other people were experiencing with the lab not working, but the sandbox. Using Firefox 15 with Mac OS X 10.7.4 FirstChildUserIdTAG: 388844 FirstChildUserNameTAG: AnAppleADay FirstChildCreateTimeTAG: 2012-09-07T13:08:24Z FirstChildTAG: I think I know what it is. They must allocate different servers to groups of students. To make things easier for them to handle thousands of clients. So most of students are probably seeing the correct simulator. I can not prove it but I would suspect that this is only a small subset of all the students who happen to be unlucky enough to be on some flaky server (guessing of course) that would explain why most do not have this problem. It is not to do with the web browser. If it was none of the simulators would work properly for me. But I do see sandbox okay. So this is a problem to do with the system. I would imagine they are trying to trace it. But would be nice if we can get some notification on progress. But this is all free so I am not holding out much chance of that. Ha ha! **THEY FIXED IT YESTERDAY** BY ADDING A RESET BUTTON. JUST CLICK IT AND YOU"LL SEE ALL COMPONENTS NOW APPEAR FirstChildUserIdTAG: 15344 FirstChildUserNameTAG: kob FirstChildCreateTimeTAG: 2012-09-06T01:56:23Z FirstChildTAG: I can't get this working either - tried updated firefox, chrome and internet explorer - guess I might have to postpone this until another day. FirstChildUserIdTAG: 231247 FirstChildUserNameTAG: Doppi FirstChildCreateTimeTAG: 2012-09-05T22:56:21Z SecondChildTAG: I also can't get a graph. SecondChildUserIdTAG: 266912 SecondChildUserNameTAG: pietvo SecondChildCreateTimeTAG: 2012-09-05T23:28:03Z FirstChildTAG: **!!!!!!!** For transient analysis please use this sandbox! https://6002x.mitx.mit.edu/courseware/6.002_Spring_2012/Overview/Circuit_Sandbox/ FirstChildUserIdTAG: 295103 FirstChildUserNameTAG: Syavick FirstChildCreateTimeTAG: 2012-09-22T23:15:10Z IndexTAG: 115 TitleTAG: Average power I got the first question and why there is the RMS equivalency, but I don't get how to do the average power. I tried integrating the power from 0 to 1/60 (which would be 120^2 * 2 * cos^2(2*pi*60*t) divided by 110) and from 0 to 1/120 but the results don't match up. What am I doing wrong? What exactly should I integrate? UserIdTAG: 308260 UserNameTAG: andreasvitikan CreateTimeTAG: 2012-09-05T18:46:41Z VoteTAG: 14 CoursewareTAG: Week 1 / AC power CommentableIdTAG: 6002x_S1E3_AC_power NumberOfReplyTAG: 3 FirstChildTAG: Look at section 1.8.1 in the textbook. Your integral is correct, but the hint neglects to say that you also need to multiple by 1/T. (T=1/60) FirstChildUserIdTAG: 179027 FirstChildUserNameTAG: Gretchen FirstChildCreateTimeTAG: 2012-09-05T18:57:04Z SecondChildTAG: You can also multiply the RMS value by 1.414 to obtain the peak value. To obtain the average value multiply the peak value by 0.637 SecondChildUserIdTAG: 372866 SecondChildUserNameTAG: jonmark67 SecondChildCreateTimeTAG: 2012-09-05T20:30:09Z FirstChildTAG: You need integrate the P(t)dt from 0 to 1/60. FirstChildUserIdTAG: 370855 FirstChildUserNameTAG: EnocB FirstChildCreateTimeTAG: 2012-09-05T21:43:29Z SecondChildTAG: 1. the figure shows 0.1s of the waveform. T = 1/60s. 2.you integrate the instantaneous power ((120*2*cos^2(2*pi*60*t)/110) on T. 3. to integrate this,use the trigonometric formula: cos^2(x) = (1 + cos(2*x))/2 you shouldget that result. SecondChildUserIdTAG: 65216 SecondChildUserNameTAG: peresbri SecondChildCreateTimeTAG: 2012-09-06T03:37:45Z FirstChildTAG: What you are calculating is overall power. the average power is just the result divided by the time interval. Which is 1/60. So just divide your result by 1/60 and you should get the average power. FirstChildUserIdTAG: 211715 FirstChildUserNameTAG: pitankar FirstChildCreateTimeTAG: 2012-09-13T14:45:30Z IndexTAG: 116 TitleTAG: Teacher's Day In India, on 5th september we celebrate teacher's day....so on the behalf of every student Happy Teacher's day to every teaching staff of MITX............. UserIdTAG: 87035 UserNameTAG: Ritu CreateTimeTAG: 2012-09-05T12:33:11Z VoteTAG: 14 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 9 FirstChildTAG: Yeah Happy teachers day from India FirstChildUserIdTAG: 162567 FirstChildUserNameTAG: pkalein FirstChildCreateTimeTAG: 2012-09-05T12:40:50Z FirstChildTAG: Happy Teachers Day to all the teachers of edX:) FirstChildUserIdTAG: 11538 FirstChildUserNameTAG: trishul FirstChildCreateTimeTAG: 2012-09-05T12:40:52Z FirstChildTAG: Happy Teachers Day to all the respected and honoured teaching staff of MITx.... FirstChildUserIdTAG: 128409 FirstChildUserNameTAG: arijitbme FirstChildCreateTimeTAG: 2012-09-05T12:47:07Z FirstChildTAG: HAPPY TEACHER'S DAY FROM INDIA....! FirstChildUserIdTAG: 357045 FirstChildUserNameTAG: SarathG FirstChildCreateTimeTAG: 2012-09-05T12:47:36Z FirstChildTAG: yeaa Happy teachers day to all the teachers of edx FirstChildUserIdTAG: 374440 FirstChildUserNameTAG: Arpit_Narang FirstChildCreateTimeTAG: 2012-09-05T15:07:30Z FirstChildTAG: Happy Teachers Day to the all the edX Staff! Quoting Rabindranath Tagore: "... Where Knowledge is free..Into that Heaven of freedom, my Father, let my country awake!" Thanks, edX, for providing valuable knowledge for free. FirstChildUserIdTAG: 232157 FirstChildUserNameTAG: GayatriTR FirstChildCreateTimeTAG: 2012-09-05T16:05:58Z FirstChildTAG: Happy Teachers Day to all FirstChildUserIdTAG: 64566 FirstChildUserNameTAG: tvl FirstChildCreateTimeTAG: 2012-09-05T16:13:43Z FirstChildTAG: happy teachers day ! FirstChildUserIdTAG: 357270 FirstChildUserNameTAG: priya_4 FirstChildCreateTimeTAG: 2012-09-05T16:39:35Z FirstChildTAG: Happy teacher's day ^_^ FirstChildUserIdTAG: 211715 FirstChildUserNameTAG: pitankar FirstChildCreateTimeTAG: 2012-09-13T14:48:03Z IndexTAG: 117 TitleTAG: Link to article: Students Rush to Web Classes Hi everyone, I copied this from the old MITx site, PLEASE read NY Times recent article on this subject. I STRONGLY believe it is very INSIGHTFUL. http://www.nytimes.com/2013/01/07/education/massive-open-online-courses-prove-popular-if-not-lucrative-yet.html?pagewanted=all&_r=1& UserIdTAG: 83276 UserNameTAG: salsero CreateTimeTAG: 2013-01-07T17:55:00Z VoteTAG: 13 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 118 TitleTAG: NEW YEAR WISHES TO Prof. ANANT AGARWAL AND edX. HAPPY 2013 and a great year ahead. :-) :-) UserIdTAG: 526154 UserNameTAG: RAJAT09 CreateTimeTAG: 2013-01-01T04:16:42Z VoteTAG: 13 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: :) FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2013-01-01T12:59:23Z FirstChildTAG: HAPPY NEWYEAR............... FirstChildUserIdTAG: 169361 FirstChildUserNameTAG: nimnall FirstChildCreateTimeTAG: 2013-01-01T16:16:43Z IndexTAG: 119 TitleTAG: h12p2 official explanation could be better Since deadline is passed, why is the official explanation so threadbare? Another explanation posted by a user is an improvement but a guide is appropriate as pre-deadline hint since it is in the nature of (useful) hand-waving. Can someone just write the explanation -- step by step -- and equations with numbers? The grading deadline is passed and there seems to be no reason not to. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-12-10T15:21:30Z VoteTAG: 13 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I agree. This H12P2 caused alot of threads and discussions. It will be nice to see detailed explanation. FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-12-10T15:56:58Z SecondChildTAG: I really struggled with this problem and was looking forward to seeing the solution. The solution provided is useless in helping me understand the circuit. SecondChildUserIdTAG: 345671 SecondChildUserNameTAG: cbjerregaard SecondChildCreateTimeTAG: 2012-12-10T22:31:21Z SecondChildTAG: Agreed. SecondChildUserIdTAG: 398594 SecondChildUserNameTAG: DaveyJC SecondChildCreateTimeTAG: 2012-12-10T22:38:40Z FirstChildTAG: Weird. I got all green check marks. But, my ans is different from the official solution. They include the Op Amp power in the expression for Pin, and I didn't. My Pin = vin * i_in, where, i_in = i' + K * i i' = (vin - vz)/(R0 + Rz) the current flowing into the R0, resistor. i = vout/Ro the current flowing through the regular diode, where Ro is the composite output resistance, Ro = RL*(R1+R2)/(R1+R2+RL) so, i_in = (vin-vz)/(R0+Rz) + K * vout * (R1+R2+RL)/((R1+R2)*RL) Hence, Pin = vin*(vin-vz)/(R0+Rz) + K * vin*vout*(R1+R2+RL)/((R1+R2)*RL) But, when I reduce the offical solution Pin' = Pin_1 + Pin_2, their solution yields, Pin' = Pin + (1-K)*vout*vout*((R1+R2+RL)/((R1+R2)*RL) Where Pin is my input power, and Pin' is the official input power. Looking at the circuit, I can see where the exta term is coming from. It's the current flowing into the Op Amp's output, i.e. the difference between K*i current flowing in the BJT's dependent source part and the i flowing in the regular diode of the BJT, the node between these two parts yields, i - K*i, flowing down into the OpAmp. Since that node is effectively at a potential vout relative to ground, the power connected with the OpAmp is vout*(1-K)*i, a part I totally ignored. Somehow, I thought we were supposed to ignore the opamp's input, which comes from it's own power supply, which isn't shown in the diagram. Hmmm...yet, I got it right. Go figure. Yes, it would be nice to see a full official solution, with all detail revealed. Much to learn in this circuit. So, it's well worth it to give students a full explanation of the problem. I vote too, for this. FirstChildUserIdTAG: 4463 FirstChildUserNameTAG: pmj FirstChildCreateTimeTAG: 2012-12-11T02:54:04Z SecondChildTAG: agreed...official solution must be revealed... SecondChildUserIdTAG: 249575 SecondChildUserNameTAG: badimalamadhu SecondChildCreateTimeTAG: 2012-12-11T05:01:20Z IndexTAG: 120 TitleTAG: Thank You TAs!!!! I just wanted to say thank you to all of the community TAs out there as well as all the staff. Not sure if i'd have made it all the way to the end without the gentle hints left in this discussion area!! Thanks!! UserIdTAG: 278602 UserNameTAG: Timothyking2011 CreateTimeTAG: 2012-12-09T14:37:09Z VoteTAG: 13 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: Anyone coming back next year!? FirstChildUserIdTAG: 183507 FirstChildUserNameTAG: obiradaniel FirstChildCreateTimeTAG: 2012-12-09T19:47:24Z SecondChildTAG: I will for the third time in 6.002x haha! The first time was in the Prototype Course of 6.002x spring 2012, the second is in this Fall 2012 and the third will be the next spring 2013 ;) I really enjoy colaborating here as student to the new students, because I can help them and also, at the same time, I can reinforce concepts, so I will be next year too ;). I love doing this so much haha. I have to confess that I am a big Fan of MIT since my childhood haha, I love it, so helping here is like to be in someway part of all this amazing iniciative! So, yes I will come back next year too. Take care, Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-10T00:19:51Z SecondChildTAG: That is very delighting, can't wait. SecondChildUserIdTAG: 183507 SecondChildUserNameTAG: obiradaniel SecondChildCreateTimeTAG: 2012-12-10T10:03:56Z FirstChildTAG: You are welcome Timothyking2011, I am blushing myself ;) haha. I am happy helping here and is really nice to know that I am being helpful here, this is one of the main reasons that I have re-taken this Course, to help the new students. My best wishes from now on, in your projects, studies, family and all! Take care, Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-09T18:21:34Z SecondChildTAG: Miriam, I was really smitten by you. For my opinion it is so rare to meet young girl with excellent understanding pretty hard and specific engineering science like Electronics. I wish you all the best, good luck in the learning and excellent work in the future! Serge PS sorry for my English SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-09T18:30:38Z SecondChildTAG: Haha, thank you Serge! Your english is excellent, mine one is a disaster haha. Take care, I wish you the best too ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-09T19:36:07Z SecondChildTAG: Myriam, skyhawk and others like you, thank you so much for your continued efforts helping others through the course. A special one for you Myriam, you have been the biggest contributor of this current 6.002x and helped me and am sure many others do HWS and labs with much better explanation. Wish the best for u in future, I will offer 6.002x the next time it is offered, hope to "see" you there. SecondChildUserIdTAG: 183507 SecondChildUserNameTAG: obiradaniel SecondChildCreateTimeTAG: 2012-12-09T19:46:17Z SecondChildTAG: Cool obiradaniel! That will be nice. See you there too! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-10T00:00:25Z FirstChildTAG: I'll be there again. I love Prof. Sussman's problems (assuming he designed them this time as well). H12P2 was quite interesting. It wasn't there last time. Looking forward to another mind numbing final exam. :-) FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-12-10T04:54:35Z FirstChildTAG: Definitely a BIG thank you to the TAs without which I would have been overwhelmed from the beginning. Your help had been one of the main sources of learning for me!!! FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-12-10T08:36:04Z SecondChildTAG: You have also been a great help, [hazel1919][1], with your instructions for using Wolfram Alpha with the course... You saved me much frustration and hair-pulling! :-) [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/users/365551 SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-12-13T23:50:00Z IndexTAG: 121 TitleTAG: Lab12 Hints Requested by who need it [for Spanish translation read in the end of the post] **Lab 12: OP AMPS** Sorry for the delay of this Post. This Last Lab, is really easy. But, you have to read without any excuse the Textbook in section 15.6.3 on page 863 [read here][1] and the section 15.6.4 on page 865 [read here][2]. In the statement they give us a design of a filter on (Fig.1): ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13386799779534531.png) Hint: Is Figure 1 very similar to the Figure 15.23 on page 863 [read here][3]? ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13386802793594549.png) They are asking us to design an active filter and our task is to find C1, C2, R1 and R2, provided in the requirements, that is to say that the value of Q> valor_Q and the value of fo is fo = valor_fo more less a certain tolerance (1%). They also give us a picture of how our design should be clicking on the button AC Analysis (remember that the value in the X axis is in logarithmic form, so you need to find this value as 10 ^ readed_value): ![image description](https://mitx_askbot_stage.s3.amazonaws.com/1338680784461514.png) Let's design it! If we have the data fo, can we find wo? Hint: recall that* w = 2pi * f. For this topology, we know how to relate our wo with R1, R2, C1 and C2? Hint: See Equation 15.84 on page 865. Remember that g1 = (1/R1) and g2 = (1/R2). - Since there are many unknowns, we adopt a value. Hint: We could propose a capacitance value such that C1 = C2 = C. Why do this? Why choose a C and not an R? The reason is very simple, because if we adopt a value of resistance, we would have less to choose values ​​of C, whereas if you look C, we have a lot of values ​​of R to choose from! Remember to choose the value between the values ​​in this list [standard Capacitors](https://6002x.mitx.mit.edu/static/circuits/Lab12_4.png): ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13386818527721427.png) with the 15.84 expression, we will obtain an expression 1, R2 = R1 * value_R. Well, let this expression 1 for later. - Now, they give us a Q> value_Q. We know that by definition Q = (wo / 2 * alpha). If we know wo and Q = value_Q (we will use this limit value), can we find the value of alpha?? Yes! - Can we express alpha as a function of R2, C (recall that we did C1 = C2 = C)??. Hint: see Equation 15.83 on page 865. - If we know alpha = value_alpha, plus we have C, and alpha (R2, C), can we find R2??? Yes! - Now, going back to our expression 1, and with our value of R2, can we find R1? Yes! - Finally, our task is to find normalized values ​​of R1 and R2 (no need to choose new values ​​for C1 and C2, since we adopted standard values​​!) ... *Example (arbitrary)* If our value of R2 = 1.19MOHM, from the table we should choose a value of R2 = 1.2MOhm. [Table](https://6002x.mitx.mit.edu/static/circuits/Lab12_3.png) If our value of _ = 23.9kOHM, from the table we should choose a value of _ = 24.3kOHM. [Table](https://6002x.mitx.mit.edu/static/circuits/Lab12_3.png) ![image description](https://mitx_askbot_stage.s3.amazonaws.com/1338682770688074.png) Now it is your turn! Design your own filter! Remember to set the sin voltage source to sin (0,1,10k,0,0) Good luck in the Final Exam and in all your future projects :). And don´t forget to help the New Students of the next spring and also, I hope that you can participate in the CECC 2 [read here][4]. Myriam. ----- > Now in spanish **Lab 12: OP AMPS** Disculpen por la demora de este Post. Este último Lab, es muy simple. Lo único que hay que leer sí o sí del Textbook es la sección 15.6.3 de la página 863 [leer aquí][5] y la sección 15.6.4 de la página 865 [leer aquí][6]. En el enunciado nos dan un diseño de un Filtro activo (Fig.1): ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13386799779534531.png) Pista: La Figura 1 no es muy parecida a la Figura 15.23 de la página 863[read here][7]?? Bien, nos piden diseñar un Filtro activo y nuestra tarea es hallar C1,C2,R1 y R2; tal que cumplan con los requisitos solicitados, es decir, que el valor de Q>valor_Q y que el valor de fo sea fo=valor_fo más menos una cierta tolerancia (al 1%). Ellos mismos nos dan una gráfica de cómo debería resultar nuestro diseño al hacer clic sobre el botón de AC Analysis (recuerda que el valor en el eje X , se encuentra en la forma logarítmica, por lo cual, hay que hallar dicho valor como 10^valor_leído): ![image description](https://mitx_askbot_stage.s3.amazonaws.com/1338680784461514.png) Bien, ahora a diseñar!! - Si tenemos el dato de fo, podemos hallar wo?? Pista: recordar que w=2pi*f. - Para esta topología, sabemos cómo relacionar nuestra wo con R1,R2,C1 y C2?? Pista: Ver Ecuación 15.84 de la página 865. Recordar que g1=(1/R1) y g2=(1/R2). - Ya que hay muchas incógnitas, adoptemos algún valor. Pista: Podríamos proponer un valor de capacitancia tal que C1=C2=C. Por qué hacemos esto?? Por qué elegimos un C y no un R?? La razón es muy simple, ya que si adoptáramos un valor de resistencia, tendríamos menos valores para elegir de C; en cambio, si ya fijamos C, tenemos un montón de valores de R para elegir! Recordar elegir el valor entre los valores de esta lista [Capacitores normalizados ](https://6002x.mitx.mit.edu/static/circuits/Lab12_4.png) : ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13386818527721427.png) - Con la expresión 15.84, obtendremos una expresión 1 ) R1*R2 = valor_R. Bien, dejemos esta expresión 1 para más adelante. - Ahora, ellos nos dan un Q > valor_Q . Sabemos que por definición Q=(wo/2*alpha) . Si tenemos wo y Q=valor_Q (usemos este valor límite), Podemos hallar el valor de alpha???? Sí! - Ahora, podemos expresar alpha en función de R2,C (recordar que habíamos hecho C1=C2=C)???. Pista: ver Ecuación 15.83 de la página 865. - Si tenemos alpha=valor_alpha ; además tenemos C ; y alfa(R2,C), podemos hallar R2??? Sí! - Ahora sí, retomando a nuestra expresión 1, y teniendo nuestro valor de R2, podemos hallar R1?? Sí! - Finalmente, nuestra tarea es hallar valores normalizados de R1 y R2 (no hay necesidad de elegir valores nuevos para C1 y C2, ya que habíamos adoptado valores normalizados!)... Ejemplo (arbitrario), Si nuestro valor de R2= 1.19MOHM, deberíamos escoger de la tabla un valor de R2=1.2MOhm.[Tabla](https://6002x.mitx.mit.edu/static/circuits/Lab12_3.png) Si nuestro valor de R1= 23.9kOHM, deberíamos escoger de la tabla un valor de R1= 24.3kOHM.[Tabla](https://6002x.mitx.mit.edu/static/circuits/Lab12_3.png) ![image description](https://mitx_askbot_stage.s3.amazonaws.com/1338682770688074.png) Ahora es tu turno! A diseñar! Recuerda setear la Fuente a sin (0,1,10k,0,0) Suerte en el Exámen final y en sus futuros proyectos :). No se olviden también de ayudar a los nuevos estudiantes del próximo semestre. Además, espero verlos en el CECC 2 [leer aquí][4]. Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/887 [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/889 [3]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/887 [4]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50b2892629b7b1270000003b [5]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/887 [6]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/889 [7]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/887 UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2012-12-08T21:59:11Z VoteTAG: 13 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 7 FirstChildTAG: Thank you so much for the hints! :D FirstChildUserIdTAG: 310147 FirstChildUserNameTAG: ildomarcarvalho FirstChildCreateTimeTAG: 2012-12-09T06:15:37Z SecondChildTAG: You are welcome ildomarcarvalho :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-10T02:15:22Z FirstChildTAG: sorry for the irrelevance question Myrimit. But could you please upload the hints for homework 12 as well or the links ? I've completed lab 12 thanks to the discussion people have here. I've done the homework 12 all but P3..midband gain. Can you help me with this? thanks . FirstChildUserIdTAG: 8897 FirstChildUserNameTAG: sam1202 FirstChildCreateTimeTAG: 2012-12-08T22:10:39Z SecondChildTAG: Yes, sam1202. Sure. I were a little but delayed, sorry... I have promised to Post Hw12 Hints in a Post some days ago. As I have promised I will post it right now :p Don´t forget to participate in the Contest sam1202 :) Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-08T22:28:22Z SecondChildTAG: Here they are ;) https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50c3c129468bdc270000001c SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-09T04:13:00Z FirstChildTAG: A little bit more of a hint that builds on what has been said already. If you use equations 15.83 and 15.84 from the text and let C1 = C2 and finally form Q by taking the ratio, a very nice simplification occurs. You get an even nicer expression if you calculate Q^2!!! Also, the design requirement is that Q > 5 so, you have the flexibility to choose Q to be a value that will simplify the design so long as it is greater than 5. Added note: If you work through the algebra you can get an expression for w0 in terms of Q, C (the common value for C1 and C2) and either one of the resistors. If you choose C and Q this equation will give an expression for one of the Rs. The result may not be a nice number, and you will have to approximate with a standard value. If, on the other hand, you choose C and the R in the equation you can calculate Q. Any Q is OK as long as it is greater than 5. You are then left with calculating the final R. With this approach you only have to approximate the calculated value of one R with a standard value. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-12-08T23:01:46Z SecondChildTAG: Thank you for the contribution Hint , skyhawk. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-09T04:14:45Z FirstChildTAG: Thanks! Myriam (Myrimit). Of course. FirstChildUserIdTAG: 402193 FirstChildUserNameTAG: Feramico FirstChildCreateTimeTAG: 2012-12-09T00:40:21Z SecondChildTAG: You are welcome Feramico. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-09T04:13:26Z SecondChildTAG: Thank you very much! SecondChildUserIdTAG: 402193 SecondChildUserNameTAG: Feramico SecondChildCreateTimeTAG: 2012-12-10T02:09:46Z FirstChildTAG: i got the green tick but my graph is nowhere close to required one.Rather it resembles a low pass filter. FirstChildUserIdTAG: 357747 FirstChildUserNameTAG: kishores FirstChildCreateTimeTAG: 2012-12-09T17:03:05Z SecondChildTAG: Can I help you kishores? Have you received the green check for the elements of your circuits? Are you sure that you have connected correctly the Capacitors and Resistors? Have you click on the AC button? SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-09T23:44:57Z FirstChildTAG: Thank you so much, Myromit, for your posts. You truly do have a gift for teaching. You are are a godsend. FirstChildUserIdTAG: 77598 FirstChildUserNameTAG: Spacecadet1974 FirstChildCreateTimeTAG: 2012-12-09T21:18:33Z SecondChildTAG: Hahaha, you are welcome Spacecadet1974 ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-09T23:41:33Z FirstChildTAG: Thanks a lot for your hints, Myrimit and Skyhawk. By the way, will The Final Exam cover topics from weeks 11-12, or 13 and 14 must be also included? Thanks again. FirstChildUserIdTAG: 329444 FirstChildUserNameTAG: albmartin FirstChildCreateTimeTAG: 2012-12-10T19:01:02Z IndexTAG: 122 TitleTAG: Lab11 Hints Requested by who need it :). ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- > **First of all, a little bit of Marketing haha, take a look at this Post > [CECC 2 - Contest][1]. I hope that you can Participate :). You can Win > one of the 3 Textbooks signed personally by Prof. Agarwal!** ------ ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- (For the Spanish translation read in the end) Lab11 Hints. ---------- Task1. In this part, we have a RLC series, and they ask us to find the L and C Circuits elements, so that, with that values, we can obtain a graph like the Figure 2. ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13378125164318829.png) The problem says: "Consider the series RLC circuit shown in Figure 1. Leaving the resistance fixed at 10Ω your task is choose values of L and C such that the undamped resonance frequency, ω0, is 10kHz, with Q=10." So, we have the value of R, the value of Q, the value of fo (remember that wo=2pi*fo). Now, for the billion question, How do we calculate the values of L and C in order to have the amplitude graph like the Figure 2??? Hint: In order that you can answer it by your own, you should take a look to the equation 14.60 of the page 801 of the Textbook [read here][2] and the equation 14.62 of the page 802 [read here][3]. Remember that you have the value of R, the value of Q and the value of fo (wo=2pi*fo). Can you calculate the L and C with this equations? Yes! Let's work! Find the values of L and C, and the press the AC buttom! Note: be careful with the units of the answers that they ask us ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13378167447443612.png) ---------- Task 2. In this part, the problem ask us to change the source voltage to a new source voltage (a step voltage source). For those that do not know how to change that voltage source, the first thing that you should do is a double clic on the element, in this case, on the voltage source element. You have to choose: **- Type: step.** **- TRAN: 2ms.** ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13378172663766421.png) In the problem they tell us that we have to obtain a plot like the Figure 3. Ok, here, they ask us to find the Ratio of V2 @ 1025us to V2 @ 25us. But, what is that? It is a relation between the voltage V2 at the time 1025us (VB) respect the voltage V2 for the time 25us (VA). Hint: Isn't it a relation a division?? ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13378182603291137.png) ---------- Task 3. Finally, we arrive to the Task 3! In resume, the statement ask us to exchange the capacitor and the resistor. In addition, we have also to change the voltage source that we have to a sinusoidal one (sin). Then, they show and tell us, that with this changes, we have to obtain a curve like the Figure 4. ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13378188952802809.png) Then, just as they say, we have to press the TRAN buttom and run it to 2ms. Which is the steady-state? Pista: In an infinite time, we will have a steady-state, a balance, so what is the value of the voltage where we can not increase any more? to what value of voltage that do we tend? That is all! I hope this can help you :). And see you in the CECC 2 ! We will be waiting for your video :). Myriam. ----- > (Now in Spanish) **Tak 1.** En esta parte, tenemos un circuito RLC serie, y se nos solicita hallar los componentes de L y C, de tal forma que con estos nuevos valores, obtengamos la gráfica de magnitud que nos dan ( ver Figura 2). ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13378125164318829.png) El enunciado dice: "Consider the series RLC circuit shown in Figure 1. Leaving the resistance fixed at 10Ω your task is choose values of L and C such that the undamped resonance frequency, ω0, is 10kHz, with Q=10." Entonces, tenemos el valor de R, el valor de Q, el valor de fo (recordar que wo=2pi*fo). Ahora, la pregunta del millón es: cómo obtengo la L y C para que el gráfico de la amplitud sea igual a la Figura 2??? Pista: Para que ustedes se puedan responder la pregunta por sí mismos , es recomendable que le echen un vistazo a la ecuación 14.60 de la página 801 del Textbook [leer aquí][4] y la ecuación 14.62 de la página 802[leer aquí][5] . Recuerda que tienes el valor de R, el valor de Q y el valor de fo (wo=2pi*fo). Se pueden calcular con dichas ecuaciones la inductancia L y la capacitancia C? Sí! A trabajar! Calcula los valores de L y C, y luego presiona el botón AC! Nota: ten cuidado con las unidades solicitadas en las respuestas. ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13378167447443612.png) ---------- Task 2. En esta parte, el problema nos solicita cambiar la fuente de alimentación a una fuente de alimentación escalón (step). Para aquellos que no saben cómo modificar dicha fuente, lo que primero deben hacer es un doble clic sobre el elemento, en este caso, sobre la fuente de alimentación. Se debe elegir: **- Type: step.** **- TRAN: 2ms.** ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13378172663766421.png) Tal y como ellos nos muestran en la Figura 3, al realizar los pasos previos, obtendremos una curva idéntica a la mostrada anteriormente. Bien, aquí nos piden que hallemos el "Ratio of V2 @ 1025us to V2 @ 25us". Pero qué es eso? Es la relación entre la tensión V2 para el tiempo 1025us (VB) respecto de la tensión de V2 para el tiempo 25us (VA). Pista: una relación no es una división?? ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13378182603291137.png) ---- Task 3: Finalmente, llegamos al Task 3! En definitiva, el enunciado nos pide que cambiemos de lugar dos componentes: la resistencia en el lugar que ocupa el capacitor y el capacitor, en el lugar que ocupa la resistencia. Además, nos piden que cambiemos la fuente de alimentación a una fuente senoidal. Luego, nos indican que debemos obtener, con estos cambios, una curva igual a la de la Figura 4. ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13378188952802809.png) Luego, tal y como lo dice el enunciado, se debe presionar el botón de TRAN a 2ms. Cuál es el steady-state? Cuál es el estado de equilibrio? Pista: en un tiempo infinito, tendríamos equilibrio, de qué valor de tensión estamos hablando? cuál es el valor en donde ya no podemos crecer más? a qué valor tiende?? Eso es todo! Espero que les haya servido de ayuda! :). Espero verlos en el CECC 2! :). Myriam. ---------- [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50b2892629b7b1270000003b [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/825 [3]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/826 [4]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/825 [5]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/825 UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2012-11-28T01:44:34Z VoteTAG: 13 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: Someone's taking Chauncey's marketing comment very seriously :P Excellent hints Myriam! FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-11-28T17:09:11Z SecondChildTAG: Hahaha :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-28T18:25:31Z FirstChildTAG: Myrian Si tu remplazas los siguientes valores, logras la curva esperada. L=0,10 H C=2500pF R= 620 ohm pero no logro que los valores sean correctos.... a ver si me puedes ayudar... Raúl FirstChildUserIdTAG: 60541 FirstChildUserNameTAG: raulcastdiaz FirstChildCreateTimeTAG: 2012-11-29T02:43:01Z SecondChildTAG: Hola Raúl, Te puedo ayudar? En qué parte estás perdido? Hint: Has leído la página 801? Ahí podrás ver cómo la frecuencia de resonancia se relaciona, de alguna forma, con tu L y C.... Hint: Recuerda que la wo=2pi*fo... No te olvides de hacer clic en el botón de AC... SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-29T12:10:00Z FirstChildTAG: myrimit i caNT UNDERSTAND THE NO 4 QUESTION FirstChildUserIdTAG: 128409 FirstChildUserNameTAG: arijitbme FirstChildCreateTimeTAG: 2012-11-30T10:57:01Z SecondChildTAG: Hi arijitbme, Can I help you? Have you exchanged the C and R of place? Have you click on TRAN (2m)? Hint: what is the value where we can not grow anymore (infinite time)? What value does it tends? SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-30T12:47:55Z FirstChildTAG: myrimit you are great FirstChildUserIdTAG: 352647 FirstChildUserNameTAG: praveenjugge FirstChildCreateTimeTAG: 2012-11-30T08:23:24Z SecondChildTAG: Thank you praveenjugge ;). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-30T12:43:18Z IndexTAG: 123 TitleTAG: [STAFF] Question about new course Is there new course from MIT x for 6.004 ? and it will be offered in the spring ? thanks advance UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-29T05:43:27Z VoteTAG: 13 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I agree 6.004 will be a great fun . 6.004 lecture videos ,tutorials and quizzes are available at http://6004.mit.edu/ .I would definitely love to be a part of future electronics and electrical courses at edx .If a course like 6.004 follows just after 6.002x ends , that will be awesome . Best, bikz05 FirstChildUserIdTAG: 125353 FirstChildUserNameTAG: bikz05 FirstChildCreateTimeTAG: 2012-10-29T10:10:57Z SecondChildTAG: Ya!! All 6.002x students prefer to have courses like 6.004x where we will apply the concepts we learn in 6.002x. SecondChildUserIdTAG: 362299 SecondChildUserNameTAG: anandbaskaran SecondChildCreateTimeTAG: 2012-10-29T16:47:42Z FirstChildTAG: I'm going to ask again - Signals and Systems! :-) FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-10-29T17:35:38Z IndexTAG: 124 TitleTAG: Thanks a lot Anant!!!!! Thanks!!!! You are a great teacher and a great person for this idea and the course! UserIdTAG: 299251 UserNameTAG: TheRedBlackOne CreateTimeTAG: 2012-10-26T05:15:42Z VoteTAG: 13 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I guess you meant course xD FirstChildUserIdTAG: 454609 FirstChildUserNameTAG: Alwahsh FirstChildCreateTimeTAG: 2012-10-26T05:18:53Z SecondChildTAG: yes, I have already edited the word "course" :) SecondChildUserIdTAG: 299251 SecondChildUserNameTAG: TheRedBlackOne SecondChildCreateTimeTAG: 2012-10-26T05:24:58Z SecondChildTAG: Are you referring to Dr. Agarwal? His name is Anant, not Adnan. SecondChildUserIdTAG: 9673 SecondChildUserNameTAG: xp42 SecondChildCreateTimeTAG: 2012-10-26T07:25:59Z SecondChildTAG: updated! SecondChildUserIdTAG: 299251 SecondChildUserNameTAG: TheRedBlackOne SecondChildCreateTimeTAG: 2012-10-26T07:56:11Z IndexTAG: 125 TitleTAG: EXAM DATE(MIDTERM) Can the exam date be moved to sunday so that everyone are free most of the day which will benefit us UserIdTAG: 224318 UserNameTAG: ashwin12312 CreateTimeTAG: 2012-10-08T15:29:46Z VoteTAG: 13 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: yes. I agree. FirstChildUserIdTAG: 355370 FirstChildUserNameTAG: omaransari FirstChildCreateTimeTAG: 2012-10-08T17:51:58Z FirstChildTAG: yes that is true..but better if it is made exam week by keeping open the exams for a week...let 24hr deadline shall be same... FirstChildUserIdTAG: 169416 FirstChildUserNameTAG: subramanya26shin FirstChildCreateTimeTAG: 2012-10-09T13:46:34Z IndexTAG: 126 TitleTAG: H4P2 ZENER REGULATOR I got the right ans for V0 and v0 before inserting zener diode but after inserting zener i m not getting Right vo. UserIdTAG: 346003 UserNameTAG: anupamshakya CreateTimeTAG: 2012-10-04T11:36:33Z VoteTAG: 13 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: u must now that zener will operate in negative side >and u will apply large signal so u need to find equation satsify the negative part of curve >think of it<<< simply ID=VD-5 >now u can apply node method and replace id with last equation>>> when apply small signal can replace non linear device(zener)with resistance and can estimate it by derivative like lecture >and solve by node method.i hope i will help u. ammar samir,egypt FirstChildUserIdTAG: 292299 FirstChildUserNameTAG: ammarsamir FirstChildCreateTimeTAG: 2012-10-05T03:57:26Z SecondChildTAG: i cant understand kindly help me .... i can calculate small vo ... just this part is remaining. SecondChildUserIdTAG: 438395 SecondChildUserNameTAG: ali_PU1 SecondChildCreateTimeTAG: 2012-10-06T12:50:19Z FirstChildTAG: Zener works on breakdown reverse voltage - the IV you given is forward voltage. FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-05T04:49:38Z IndexTAG: 127 TitleTAG: QED......Quite easily Done No need to include another source or to apply superposition.... here it is.....V across R1 and combination of "CCVS and R2" must be same. So,I put them equal and found out 'i'. thereafter calculated Vth as (I0-i)*R2 . And for Rth open the CS and combined 'R1 in series with αi' in parallel with'R2'and .............it worked !!! Apply-> i*R1=α*i+(I0-i)*R2 UserIdTAG: 477713 UserNameTAG: ikm104 CreateTimeTAG: 2012-10-03T07:50:58Z VoteTAG: 13 CoursewareTAG: Week 4 / Dependent Sources Exercise 1 CommentableIdTAG: 6002x_dep_src_exsz_1 NumberOfReplyTAG: 4 FirstChildTAG: dynamite! FirstChildUserIdTAG: 12905 FirstChildUserNameTAG: Baer FirstChildCreateTimeTAG: 2012-10-04T20:30:40Z SecondChildTAG: great one.. SecondChildUserIdTAG: 111540 SecondChildUserNameTAG: tksanthosh SecondChildCreateTimeTAG: 2012-10-24T05:06:51Z FirstChildTAG: Can you spell out a little more what you equated? This problem is doing my head in. FirstChildUserIdTAG: 166031 FirstChildUserNameTAG: krebryna FirstChildCreateTimeTAG: 2012-10-04T21:50:31Z SecondChildTAG: i*R1=α*i+(I0-i)*R2 SecondChildUserIdTAG: 477713 SecondChildUserNameTAG: ikm104 SecondChildCreateTimeTAG: 2012-10-06T08:02:12Z FirstChildTAG: how do u get v across R1 and V across ccvs. FirstChildUserIdTAG: 439427 FirstChildUserNameTAG: dany87 FirstChildCreateTimeTAG: 2012-10-05T17:06:59Z SecondChildTAG: v across R1 will be i times R1. And as "ccvs and R2 are in series" in parallel with R1 so voltage will be shared by both ccvs and R2 in ratio of their impedances.. SecondChildUserIdTAG: 477713 SecondChildUserNameTAG: ikm104 SecondChildCreateTimeTAG: 2012-10-06T08:05:46Z FirstChildTAG: It is wrong. Right answer: i*R1+α*i=(I0-i)*R2 FirstChildUserIdTAG: 407765 FirstChildUserNameTAG: Igor14 FirstChildCreateTimeTAG: 2012-10-06T19:14:41Z SecondChildTAG: This equation is correct one. SecondChildUserIdTAG: 92229 SecondChildUserNameTAG: muralicv SecondChildCreateTimeTAG: 2012-10-06T20:53:27Z SecondChildTAG: Dear from the eqn. quoted earlier,we can get the solution too.You need to look into,again.. SecondChildUserIdTAG: 477713 SecondChildUserNameTAG: ikm104 SecondChildCreateTimeTAG: 2012-10-07T10:38:10Z IndexTAG: 128 TitleTAG: Playlists for blocked Youtube users Hello, I made some playlists using JWPlayer and the provided MP4 files, for those who can't access the Youtube Playlists. Here they are: Week 2 https://dl.dropbox.com/u/24096724/6002xMP4/S3.html https://dl.dropbox.com/u/24096724/6002xMP4/S4.html Week 3 https://dl.dropbox.com/u/24096724/6002xMP4/S5.html If this is useful for you, please vote for this post! UserIdTAG: 276409 UserNameTAG: IgnacioUY CreateTimeTAG: 2012-09-25T17:53:33Z VoteTAG: 13 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Thank you. YouTube is blocked here in my country. Kindly do it for other weeks as wel. Regards FirstChildUserIdTAG: 39896 FirstChildUserNameTAG: garrisonian FirstChildCreateTimeTAG: 2012-09-25T19:19:41Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-26T11:27:53Z FirstChildTAG: This is one of the things that I loved about the Spring 2012 session of 6.002x: the community building on each others' efforts. Thank you Ignacio FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-25T19:32:01Z SecondChildTAG: I totally agree! :) Thank you IgnacioUY! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-26T11:27:46Z FirstChildTAG: Added week 1 and week 4, part 2. You can find it here. Week 1: https://dl.dropbox.com/u/24096724/6002xMP4/S1.html https://dl.dropbox.com/u/24096724/6002xMP4/S2.html Week 3: https://dl.dropbox.com/u/24096724/6002xMP4/S5.html https://dl.dropbox.com/u/24096724/6002xMP4/S6.html FirstChildUserIdTAG: 276409 FirstChildUserNameTAG: IgnacioUY FirstChildCreateTimeTAG: 2012-09-26T20:36:57Z IndexTAG: 129 TitleTAG: 6004x course Hello. in the S4V14: DIGITAL LOGIC CIRCUITS class, prof. agarwal comments about the 6004x mit's course. is there any plan to make this course available to edx in the future? thanks UserIdTAG: 249194 UserNameTAG: heuvaladao CreateTimeTAG: 2012-09-22T19:54:05Z VoteTAG: 13 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: some material for this course is already available on MIT OpenCourseWare: http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-004-computation-structures-spring-2009/ FirstChildUserIdTAG: 372321 FirstChildUserNameTAG: EnricoDona FirstChildCreateTimeTAG: 2012-11-22T10:04:59Z IndexTAG: 130 TitleTAG: To All Asking for Help: A few suggestions. Sept 16, 2012 **To All Asking for Help:** To all those asking for help, just a few suggestions: 1. Please be **very specific** on where you are getting stuck in your problem. Try not to write "I need help with Homework 1 Question 1", because, first of all, each homework question has multiple parts to it, and second of all, it is important to explain the work you have done so far to get to where you are stuck. 2. It is against the honor code to give out answers, or ask for answers. It may be **useless** anyways, because I've seen two different people have the same homework problem, but with different **starting variables**. Understand this when you ask a question, too: * For example, my Homework 1 Problem 3 was worded: "...three $1480.0W$ baseboard heaters to provide a total heating capacity of $4440.0W$...", * But someone else had the same exact question worded "...three $1300.W$ baseboard heaters to provide a total heating capacity of $3900.0W$..." * Of course, my answers will be different than yours, unless we are lucky to get the same starting variables assigned by the system. 3. Use the search function, or just search back through the posts, to see if your question has already been answered. There are a **small** number of people answering questions compared to the **large** amount of people asking for help; so resources are limited. Once one of these helpful posters answers a question, they will probably not want to re-write the *same* answer, just to a different person. 4. Check and re-check your algebra before you ask any questions, because if you are "getting it", 99% of mistakes are simple, overlooked mistakes. Make sure you also follow the conventions in assigning signs (+) and (-), because this determines current flow direction, and voltage orientation with respect to ground; and a mistake in one sign can mess up your answer even after you've spent a half-hour figuring out the node equations. If you are "not getting it", make sure you read and re-read the textbook, work out the practice problems, ask the teaching assistants for help at their designated times, etc. UserIdTAG: 292546 UserNameTAG: JerseyMark CreateTimeTAG: 2012-09-16T22:26:03Z VoteTAG: 13 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Thank you for this helpful post! FirstChildUserIdTAG: 151427 FirstChildUserNameTAG: RichardZ FirstChildCreateTimeTAG: 2012-09-16T23:41:14Z IndexTAG: 131 TitleTAG: Schematic tool Hi, I am curious to know the technology behind the schematic tool. Do you have a spice engine on your servers that works with the browser as the client. Can you plz provide some details. I am quite delighted to see this in action. Regards, Manu UserIdTAG: 403960 UserNameTAG: manubatura CreateTimeTAG: 2012-09-13T10:33:06Z VoteTAG: 13 CoursewareTAG: Week 2 / Superposition experiment CommentableIdTAG: 6002x_Superposition_experiment NumberOfReplyTAG: 2 FirstChildTAG: Hi, the schematic editor is a Javascript program that was written specifically for 6.002x. Thanks for the feedback! FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-15T19:56:37Z SecondChildTAG: I was quite impressed with it too. It fulfills the rule of a simple spice simulator quite well. If expanded on, I feel that it could be a valuable asset to the Open Source Hardware and the beginner/hobbyist communities. Keeping the GUI simple and then perhaps displaying the equivalent SPICE code could prove to be a very valuable asset to those learning circuit simulation tools. SecondChildUserIdTAG: 227432 SecondChildUserNameTAG: JasonTraud SecondChildCreateTimeTAG: 2012-09-16T20:01:04Z FirstChildTAG: http://www.falstad.com/circuit/ ---------- enjoy :) FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-09-18T05:14:33Z IndexTAG: 132 TitleTAG: A Good Tutorial about Parallel batteries connection! http://www.youtube.com/watch?v=w1Mqn6Ewvio UserIdTAG: 153707 UserNameTAG: masoud_np CreateTimeTAG: 2012-09-12T19:53:50Z VoteTAG: 13 CoursewareTAG: Week 1 / Battery model CommentableIdTAG: 6002x_S1E9_Battery_Model NumberOfReplyTAG: 0 IndexTAG: 133 TitleTAG: mighty science behind 7 loops Please some 1 briefly explain me that why there are 7 loops in this system ? thanks UserIdTAG: 415376 UserNameTAG: imab90 CreateTimeTAG: 2012-09-12T15:37:36Z VoteTAG: 13 CoursewareTAG: Week 1 / Circuit topology CommentableIdTAG: 6002x_S2E1_Circuit_Topology NumberOfReplyTAG: 3 FirstChildTAG: I'll quote from what your fellow student jovan said earlier: *Loop does not need to contain voltage source, there are loops: V R1 R2; V R1 R3 R5; V R4 R5; V R4 R3 R2; R4 R3 R1; R5 R2 R3; R4 R5 R2 R1;* FirstChildUserIdTAG: 390034 FirstChildUserNameTAG: kahlil FirstChildCreateTimeTAG: 2012-09-12T16:08:51Z SecondChildTAG: Thanks as well :)) SecondChildUserIdTAG: 342966 SecondChildUserNameTAG: SandraNavarro SecondChildCreateTimeTAG: 2012-09-15T16:25:53Z FirstChildTAG: first pick a node then figure out how many ways you can come back to that node through the circuit. do this for all the nodes in the circuit and then add the results. when figuring out the paths back to that node take care not to cross one branch more than once. thats the hard way. simpler, if u figured out 3 loops then you are half way through. adjacent loops can combined together an considered as a loop. the 3 loops can be combined among themselves in 4 more ways. thus 3 + 4 = 7. FirstChildUserIdTAG: 152516 FirstChildUserNameTAG: shuvo915 FirstChildCreateTimeTAG: 2012-09-12T16:31:12Z FirstChildTAG: I had the same doubt as imab90. Because I attended a course where we only have take into count the loops V R1 R2; R1 R4 R3 and R2 R3 R5. Or three loops, three ecuations. Thanks kahlil and shuvo915. FirstChildUserIdTAG: 265027 FirstChildUserNameTAG: edumm1 FirstChildCreateTimeTAG: 2012-09-14T23:33:22Z IndexTAG: 134 TitleTAG: Lab 2 hints I saw several posts asking for help with Lab 2. Here are a couple of hints: 1. You have two voltage sources and the output voltage is a combination of them. It is simpler to calculate the output voltage one term at a time (i.e. first the term for V1 and then the term for V2). Do you know any technique to do that?... Use it! 2. Is the output voltage affected if you multiply all the resistances by the same number? Try to understand why or why not! 3. Notice that the coefficients of the output voltage doesn't add to 1. Can you achieve that using only two resistors as shown in the simple resistive mixer of Figure 2? I hope this helps! UserIdTAG: 381746 UserNameTAG: jelizon CreateTimeTAG: 2012-09-12T13:40:23Z VoteTAG: 13 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: In the schematic I had integer value and fractional value resistors and could not produce a proper plot. I tried modeling the same circuit with Falstads simulator, boom, looks good. Finally tried changing the fractional resistor value to a 4-place decimal approximation and the transient analysis looked good. I can confirm this, twice. Once using Iceweasel 11.0 on Crunchbang Squeeze, once on Firefox 15 on Win7 x64. FirstChildUserIdTAG: 257171 FirstChildUserNameTAG: arcticSearcher FirstChildCreateTimeTAG: 2012-09-14T20:06:11Z FirstChildTAG: I cant get it right :( the main problem is that the coefficient Of V1 and V2 doesn't add up to 1. How can i solve this ???? FirstChildUserIdTAG: 336001 FirstChildUserNameTAG: syd_buet12 FirstChildCreateTimeTAG: 2012-09-20T15:56:55Z SecondChildTAG: I don't get it. Why is it relevant that the coefficients don't add up to 1? A few people have mentioned this now. I must be missing something SecondChildUserIdTAG: 287805 SecondChildUserNameTAG: Beneficial SecondChildCreateTimeTAG: 2012-09-23T06:19:08Z IndexTAG: 135 TitleTAG: Fullscreen Bug When Using Safari on Mac OSX Lion I use safari on mac OSX and whenever I try to go fulscreen on any lecture video the courseware navigation icons remain on the screen and can't be moved, they look like this : ![Screenshot][1]http://i49.tinypic.com/2l46ya.jpg [1]: http://i49.tinypic.com/2l46ya.jpg UserIdTAG: 7778 UserNameTAG: OscarBarajas CreateTimeTAG: 2012-09-09T00:33:09Z VoteTAG: 13 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 8 FirstChildTAG: yes, me too FirstChildUserIdTAG: 284395 FirstChildUserNameTAG: AhmedAbuZeid FirstChildCreateTimeTAG: 2012-09-09T00:52:05Z FirstChildTAG: Yep, I get the same thing. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-09-09T00:39:09Z FirstChildTAG: It's not your computers, it is a bug that needs to be tweaked on the website. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-09T02:11:15Z FirstChildTAG: It can be fixed if the css value for the `sequence-list-wrapper` z-index: is lower than 1000. Value of 1 works fine. FirstChildUserIdTAG: 207456 FirstChildUserNameTAG: Octomos FirstChildCreateTimeTAG: 2012-09-09T09:25:24Z FirstChildTAG: Apple SUX xD FirstChildUserIdTAG: 394836 FirstChildUserNameTAG: v2g6ch4 FirstChildCreateTimeTAG: 2012-09-09T02:51:20Z FirstChildTAG: still having the same problem. Browser: Firefox 15.0.1 on Windows 7 FirstChildUserIdTAG: 204213 FirstChildUserNameTAG: ratneshray FirstChildCreateTimeTAG: 2012-09-11T13:58:51Z SecondChildTAG: the same for Win7, Google Chrome 21.0.1180.89 m. SecondChildUserIdTAG: 415594 SecondChildUserNameTAG: barniley SecondChildCreateTimeTAG: 2012-09-11T18:13:58Z FirstChildTAG: I also get that, and have the same problem on my iPad 2 running iOS. FirstChildUserIdTAG: 359097 FirstChildUserNameTAG: MaxAbramson FirstChildCreateTimeTAG: 2012-09-09T01:07:24Z FirstChildTAG: Safari has this problem with a lot of websites that use embedded youtube videos like so. It will remain this way until Apple updates Safari's compatibility with embedded videos. FirstChildUserIdTAG: 243871 FirstChildUserNameTAG: Vanilly FirstChildCreateTimeTAG: 2012-09-19T20:57:15Z IndexTAG: 136 TitleTAG: Formula memorizing tip. Twinkle-twinkle little star, Power equals to I square R i.e. P=I^2*R ; P=power, I=current, R=resistance. UserIdTAG: 315157 UserNameTAG: Kachhy CreateTimeTAG: 2012-09-07T11:25:13Z VoteTAG: 13 CoursewareTAG: undefined CommentableIdTAG: 5_Simple_Power NumberOfReplyTAG: 1 FirstChildTAG: That's one way to do it, but there are so many formulas that you will need to remember in EE that it will, in the end, be much easier to just learn to derive what you need. In this case, you really don't even need to find the current. You're given V & R and asked to find P so: V=I*R or I=V/R, P=V*I, substitute I to get P=V*(V/R) and you get P=(V^2)/R. You should become familiar with all the variations of Ohm's law and the power formula and to help with that, there is a very handy chart sometimes called the Ohm's Law pie chart. You should easily be able to derive all the formulas on the chart. ![enter image description here][1] [1]: http://www.electronics-tutorials.ws/dccircuits/dcp4.gif FirstChildUserIdTAG: 208891 FirstChildUserNameTAG: tjwuth FirstChildCreateTimeTAG: 2012-09-07T14:37:32Z SecondChildTAG: I appreciate your concern, thank you. But, I already have it. SecondChildUserIdTAG: 315157 SecondChildUserNameTAG: Kachhy SecondChildCreateTimeTAG: 2012-09-07T17:32:33Z IndexTAG: 137 TitleTAG: RMS: definition RMS means Root Mean Square. You can calculate the RMS value for any waveform: ![enter image description here][1] [1]: http://img266.imageshack.us/img266/7344/daumequation13468970327.png For a sine or cosine wave (centered at zero), the RMS value is the peak value divided by √2. UserIdTAG: 188586 UserNameTAG: FLara CreateTimeTAG: 2012-09-06T02:15:52Z VoteTAG: 13 CoursewareTAG: Week 1 / AC power CommentableIdTAG: 6002x_S1E3_AC_power NumberOfReplyTAG: 0 IndexTAG: 138 TitleTAG: Possible error in toolbox In firefox 15 button DC and TRAN analysis are not displayed. Is this problem only in my browser? UserIdTAG: 325246 UserNameTAG: efys CreateTimeTAG: 2012-09-05T12:51:45Z VoteTAG: 13 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 14 FirstChildTAG: its ok with Chrome..i experienced the same,and shifted things to chrome FirstChildUserIdTAG: 212417 FirstChildUserNameTAG: saiteja25 FirstChildCreateTimeTAG: 2012-09-05T13:30:17Z FirstChildTAG: I see the same thing in firefox. FirstChildUserIdTAG: 271670 FirstChildUserNameTAG: moncapitane FirstChildCreateTimeTAG: 2012-09-05T12:53:42Z FirstChildTAG: Pressing "Check" made the DC analysis work for me. But TRAN is not working. FirstChildUserIdTAG: 127195 FirstChildUserNameTAG: princeofsudan FirstChildCreateTimeTAG: 2012-09-05T12:54:52Z SecondChildTAG: Same problem here, DC works but TRAN doesn't. Tried Firefox 15.0 and Chrome 21.0.1180.89, OS: Windows 8. SecondChildUserIdTAG: 311022 SecondChildUserNameTAG: GordanS SecondChildCreateTimeTAG: 2012-09-05T16:23:03Z FirstChildTAG: I have the same version and it works. What operating system are you using? FirstChildUserIdTAG: 6977 FirstChildUserNameTAG: rocha FirstChildCreateTimeTAG: 2012-09-05T12:55:00Z SecondChildTAG: I am in windows 7 firefox 15.0. SecondChildUserIdTAG: 271670 SecondChildUserNameTAG: moncapitane SecondChildCreateTimeTAG: 2012-09-05T12:55:36Z FirstChildTAG: I have the same thing in Firefox 15 on OS X(Snow Leopard). There is no DC, AC or TRAN button. In Chrome, they are there. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-09-05T13:04:26Z FirstChildTAG: even i tried using firefox, wasnt working so i switched to chrome and it works fine. FirstChildUserIdTAG: 330881 FirstChildUserNameTAG: mmmishra FirstChildCreateTimeTAG: 2012-09-05T13:15:38Z FirstChildTAG: Same for me on a Firefox 12, under WinXP: ![enter image description here][1] [1]: https://lh5.googleusercontent.com/-9KgzFZRCSh4/UEdQS18iCOI/AAAAAAAAJ-M/LwPo7gYSsa0/s800/first.JPG FirstChildUserIdTAG: 144671 FirstChildUserNameTAG: andresdans FirstChildCreateTimeTAG: 2012-09-05T13:17:01Z FirstChildTAG: I use Chrome for this course. FirstChildUserIdTAG: 191549 FirstChildUserNameTAG: Fodin FirstChildCreateTimeTAG: 2012-09-05T13:19:26Z FirstChildTAG: I use chrome and it's all riht. FirstChildUserIdTAG: 192114 FirstChildUserNameTAG: chuba FirstChildCreateTimeTAG: 2012-09-05T13:32:47Z FirstChildTAG: Pressing Check button worked for me. Probably you may need to refresh the page afterwards (ctrl+f5) FirstChildUserIdTAG: 320863 FirstChildUserNameTAG: Extrapolator FirstChildCreateTimeTAG: 2012-09-05T13:58:16Z SecondChildTAG: same prob here.. not solved even changed to chrome from firefox SecondChildUserIdTAG: 168602 SecondChildUserNameTAG: chittiprolu SecondChildCreateTimeTAG: 2012-09-05T14:57:58Z FirstChildTAG: I had the same problem. I use firefox. About an hour the page refreshed itself and DC and TRAN analysis were displayed. It is interesting that in the Circuit Sandbox it worked all the time. I'm so sorry for my english. FirstChildUserIdTAG: 190448 FirstChildUserNameTAG: Axmap FirstChildCreateTimeTAG: 2012-09-05T16:40:40Z FirstChildTAG: Same for me here. How do I run this? FirstChildUserIdTAG: 298775 FirstChildUserNameTAG: surja FirstChildCreateTimeTAG: 2012-09-05T13:28:27Z FirstChildTAG: Well i am using Google Chrome browser in Windows 7, with very low internet speed still its working fine. FirstChildUserIdTAG: 408765 FirstChildUserNameTAG: NeevGhodasara FirstChildCreateTimeTAG: 2012-09-12T15:09:35Z FirstChildTAG: I currently do not have access to a linux machine with a GUI. Will have to test when I get home. FirstChildUserIdTAG: 271670 FirstChildUserNameTAG: moncapitane FirstChildCreateTimeTAG: 2012-09-05T13:00:53Z IndexTAG: 139 TitleTAG: CECC 2 - winners The winners of the Circuits and Electronics Classmates Contest 2 are: **1st place: AlexAlexandrescu**, [PWM motor control](http://youtu.be/Jk6OIjlnBUs) **2nd place: dmgongora**, [On/off temperature controller with hysteresis](http://youtu.be/boA7kC9dRvg) **3rd place : Enrique**, [Ring oscillator](http://youtu.be/EFKJO8G62dQ) The winners are awarded the course textbook *“Foundations of Analog and Digital Electronic Circuits”* signed personally by Professor Anant Agarwal. We received a total of 9 entries and all the entries are very interesting and well made. The jury had a very hard time selecting just three winners and we encourage our fellow students to watch all the videos. The remaining entries are (in alphabetical order) : adjmhatre, [Microphone amplifier](http://youtu.be/yu_6UWFozmY) AndBre, [Music over a light beam](http://youtu.be/vB3hoj1A-QI) AnthonyRF, [Manhattan method](http://youtu.be/8BB7_gcFKTo) Chingun Khasar, [Light following car Part 1](http://youtu.be/P3WLEc4gPyE) [Part 2](http://youtu.be/mK_uyiUBhx0) ReconIII, [ALU in Cedar Logic Simulator and Minecraft](http://youtu.be/xwQMRtPSxYI) Shmatko_Nazar, [Compensated voltage divider](http://youtu.be/gU9hKdsS9-I) **The jury and the organizers congratulate all the contestants with a job well done.** The organizers are ashwith, juancho and Myrimit. The jury members are Barrabas, Danik, dantyrant, JSChambers, komisz and ChaunceyGardiner. UserIdTAG: 90796 UserNameTAG: ChaunceyGardiner CreateTimeTAG: 2013-01-29T12:11:46Z VoteTAG: 12 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: Congratulations to the winners and everyone who participated. Your videos were all really great! FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2013-01-29T15:36:19Z FirstChildTAG: Congratulations! :) Thank you very much for all the Contestants, we really appreciate it. All the entries were really good! Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2013-01-29T12:38:37Z SecondChildTAG: Yeah! Thanks to all the jury members for selecting my video, and thanks to the organizers for making it possible. SecondChildUserIdTAG: 9100 SecondChildUserNameTAG: dmgongora SecondChildCreateTimeTAG: 2013-01-29T14:23:06Z SecondChildTAG: I'm so happy! Thank you very much! There were a lot of good projects, no doubt it has been a difficult decision! Well done people! SecondChildUserIdTAG: 87808 SecondChildUserNameTAG: pumoneon SecondChildCreateTimeTAG: 2013-01-29T14:44:52Z FirstChildTAG: Thank you very much ! I didn't expect to be among the winners and this comes as a BIG surprise for me . The best day in a very long time ! I would like to congratulate all my colleagues who participated and perhaps to encourage future participants.I had big doubts about my entry, and i have submitted in the 11-th hour just to fulfill my new year's resolution :) . I am feeling very honored and i would like to thank you again for this amazing opportunity . FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2013-01-29T13:09:13Z SecondChildTAG: Alex that was great ... You are **Awesome** With a big A in it SecondChildUserIdTAG: 378150 SecondChildUserNameTAG: GladIDoThis SecondChildCreateTimeTAG: 2013-01-29T13:11:56Z SecondChildTAG: Thanks ! SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2013-01-29T17:10:37Z FirstChildTAG: :L oh well lol, congrats to the winners and thanks for the opportunity to participate in this! FirstChildUserIdTAG: 219780 FirstChildUserNameTAG: ReconIII FirstChildCreateTimeTAG: 2013-01-29T17:31:38Z SecondChildTAG: Thank you for participating in this Contest @ReconIII. Your participation was really valuable and important. All the videos for me were really good. I bet the Jury had a difficult debate by chosing only three of them... Again thank you. We really appreciate your involvement and participation. Thank you :). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2013-01-29T18:17:00Z IndexTAG: 140 TitleTAG: Analog Integrated Circuit Design Course Hi, Waiting for new EDX courses I would suggets to watch these interesting lectures on Analog Circuit Design on OpenCourseWare: http://ocw.tudelft.nl/courses/microelectronics/analog-integrated-circuit-design/course-home/ Have fun! Enrico UserIdTAG: 372321 UserNameTAG: EnricoDona CreateTimeTAG: 2013-01-08T15:43:52Z VoteTAG: 12 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: thanks for the link! FirstChildUserIdTAG: 434869 FirstChildUserNameTAG: patey FirstChildCreateTimeTAG: 2013-01-09T23:07:14Z FirstChildTAG: Thanks Enrico!! FirstChildUserIdTAG: 132591 FirstChildUserNameTAG: juantronic FirstChildCreateTimeTAG: 2013-01-10T09:11:21Z IndexTAG: 141 TitleTAG: Thank You! I just wanted to thank Anant Agarwal and the entire edX team for the initiative and the immense dedication put into this project. What MITx (and edX as a whole) is (are) providing for so many people around the world is not only inspiring but also world changing. This project extinguishes the boundaries to knowledge and makes it accessible to millions of people who have the will, but many times lack the opportunity. I know there will be a lot of improvement going forward, especially with the massive participation and feedback in the discussion forums; but the first steps have been taken thanks to you. In terms of the course itself, the material was presented in an extremely professional and intuitive way, which made me to rethink and improve my learning capabilities. Congratulations for the initiative, dedication and commitment to expand knowledge throughout the world! It was an amazing experience... UserIdTAG: 342148 UserNameTAG: gustavopereira CreateTimeTAG: 2013-01-08T14:07:36Z VoteTAG: 12 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 142 TitleTAG: Do you have doubts regarding Proctored Exam? Ask here. ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13573387871343677.png) ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13573388947637853.png) UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2013-01-04T22:45:41Z VoteTAG: 12 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 6 FirstChildTAG: This is great news. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2013-01-04T23:29:19Z FirstChildTAG: Hi!, good to hear about this proctored exam! I have some questions: Is it a paper-based or an internet-based exam? I mean, if it is going to be similar to what we have done with the final and midterm exams, or it is similar to a standard university exam like the handouts that we were given. What is the process that we have to follow? Should I contact one of these centres before a certain date and they have to contact you? Is everything done automatically by Pearson? Regarding the date of the exam, is it possible to choose the day and hours from a wide range, lets say a couple of days, like in the online version? Thanks in advance! FirstChildUserIdTAG: 7057 FirstChildUserNameTAG: raiabril FirstChildCreateTimeTAG: 2013-01-04T23:52:20Z SecondChildTAG: I believe it is computer based at a physical location, but don't quote me on that. Once it becomes online at Pearson Vue, you will be able to sign up to a facility close to you. http://www.pearsonvue.com/vtclocator/ As far as time I'm sure you will be able to choose from a range of testing times. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2013-01-05T01:22:37Z SecondChildTAG: The exam will be internet-based, but there will be restrictions on what resources you can use during the exam. More details will be announced as those details are finalized. The registration procedure will be announced once registration details have been finalized. Please do not expect the same level of timing flexibility that we were able to offer in the online exams. SecondChildUserIdTAG: 96452 SecondChildUserNameTAG: Lyla SecondChildCreateTimeTAG: 2013-01-05T07:00:48Z SecondChildTAG: Thanks for your answers! We will make an effort to adjust our schedules to the testing times SecondChildUserIdTAG: 7057 SecondChildUserNameTAG: raiabril SecondChildCreateTimeTAG: 2013-01-05T10:18:10Z FirstChildTAG: I will start asking my doubts: 1- **How will be the Proctored Exam Format ?** 1.a - Will it be similar to the Format of the Final Exam? That is to say, having 3 attemps in a computer from a Examination Centre? 1.b - Will it be by pencil and paper? If yes, will it be a multiple choice, by writting our own answers by handwritting? - In the case that it will be in pencil and paper, is any specification of the pencil that we have to take. I am asking this because I took a Japanese International Exam and they requested as an specific type of pencil - HB - for the Exam... 2- **Will be at open book like the Final Exam?** - If yes, can we have access to the our Course in 6.002x during the Exam. The Centre will provide the computer with internet? or do we have to take our own notes , textbooks , etc? 3- **How much time we will have to solve our Exam?** 3 hours ? To be sincere I haven´t meassured my time as I did the Exams really by cutting my time doing others activities while doing the Exam - like sleeping, launch, the cat, etc...- I used to distract a lot, :p, it is possible that I could take more than time, I am not sure... - Oops, ignore this point, I have read above that it is 2 hours. 4- **As my native language is not English, can we take to the Exam a Dictionary English-Spanish?** 5- **Can we take to the Exam our calculator?** 6- **What documentation we have to take to the Exam in order to identify ourselves?** - Passport? - Do we have to complete a voucher with our Picture? Most of the International Exams ask you to do this... - DNI? - Here in my country we have a National Document Identity. Do we have to take it to the Exam? 7- **Do we have to pay in dollars or in our local currency ,to the official exchange local currency-dollar, in the more near Pearson VUE Centre ?** I like my Country, it has a lot of good things, but due an Economical-Political Situation, we can not buy dollars here unless you request a permition that must be eccepted by the Federal Public Revenue Administration in order that you can buy it, your request it has to be consistent and you have to specify why you need it, also they verify if you spend it on that... I prefer not give my personal opinion about it :/ ..., by the way, that explains a lot another reason of why I have chosen Science haha. Anyway, **Can we pay it by Credit Card?** , I have to say that, OMG, I don´t have a credit card haha, but I guess that I can use my family contacts, some generous uncle or cool friend will pay with his or her credit card - of course I will pay them haha. I fortunately can afford the cost of the Proctored Exam. But, **if some students can not afford it in one pay, can they pay it in fee plan?** like in two fee plan? 8- **Can you provide the http:// link of the Proctored Exam 6.002x on Pearson VUE?** 9- **The Proctored Exam will be on a Weekend day like sunday or saturday?** I took a International Exam at an University when I was a teen on a sunday, I remember that. 10- **This Proctored Exam will be offered next year too? How many often will be offered?** twice a year? once a year? 11- **The Proctored Certificate will be sent by Post to our home? How much time will delay?** 12- **As this is the last question, I am tempted to ask, as I have seen this question really frecuently asked in the old forum and in this one during the Course, will this Proctored Exam count as a credit?** I guess I have asked a lot of questions :p, sorry about that... I guess that I am curious and I start to ask and so on... Thank you for reading this, Myriam. P.D. I took as a fact that in my Country we can take the Proctored Exam, but How can we check if we can set it in our Country? FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2013-01-05T00:29:05Z SecondChildTAG: 13- Will they play a recording of aa's dog barking encouragement (and hints) in the background? SecondChildUserIdTAG: 9673 SecondChildUserNameTAG: xp42 SecondChildCreateTimeTAG: 2013-01-05T00:41:16Z SecondChildTAG: Haha, that is funny, I vote for it ;), Rana -Prof. Agarwal´s pet- is the most famous dog of edX! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2013-01-05T00:48:31Z SecondChildTAG: Wow, I was going to ask some questions but I think you've covered everything I wanted to ask and more! As a thought though, I expect it will be at least partially open book as all the tests for review I looked at had a note saying you could bring in 3 double sided sheets of notes. If this is at least true, I'll feel a lot more relieved and relaxed about this. I spend half my time looking simple things up that I know but am too nervous to trust myself. :D SecondChildUserIdTAG: 467169 SecondChildUserNameTAG: Eyowzitgoin SecondChildCreateTimeTAG: 2013-01-05T07:00:12Z SecondChildTAG: You bring up a lot of good points, several of which we hadn't thought of before. Thank you so much for posting them! I will bring this up with the appropriate people and get back to you next week. SecondChildUserIdTAG: 96452 SecondChildUserNameTAG: Lyla SecondChildCreateTimeTAG: 2013-01-05T07:05:51Z SecondChildTAG: Thanks Lyla! and enjoy the [Dakar Rally (Peru)][1]. [1]: http://elcomercio.pe/dakar2013/1518774/noticia-peruano-ignacio-flores-hace-historia-al-ganar-primera-etapa-dakar SecondChildUserIdTAG: 402193 SecondChildUserNameTAG: Feramico SecondChildCreateTimeTAG: 2013-01-05T18:21:41Z FirstChildTAG: These certificates don't have a grade either. Will MIT (all the edx partners for that matter) issue transcripts if we need them? FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2013-01-05T18:47:54Z FirstChildTAG: And new question : what is "pilot" means here? For example, what are edX team plans for the future about the proctored exam? And, may we try to pass proctored exam in our choosen period, for example, this Spring if it will be available? And what we can expect about cost of this exam.Now it is $95, will it be changed (logically it may be only increased) in the future? Thank you, and best regards to all. FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2013-01-05T21:55:29Z FirstChildTAG: Considering that 'Circuits and Electronics' is a basic course and not a specialty course, for people in India, $95 maybe on the costly side. We would have to think a bit before registering for such an exam. Also there are no credits associated with this course for us in our engineering colleges. But considering that this is a certificate from the great MIT, this would surely carry some value. My question is- how much? How useful would a proctored exam certificate be? What value will it carry? One of the reasons we like getting certificates is it could be helpful in a job interview. How useful? Will it be qualified enough to be considered as a major certificate? FirstChildUserIdTAG: 131526 FirstChildUserNameTAG: nithin_243 FirstChildCreateTimeTAG: 2013-01-06T10:53:14Z IndexTAG: 143 TitleTAG: Can we have access to this entire courseware after it's over? Maybe I'll want to watch these videos again in the future UserIdTAG: 136959 UserNameTAG: christerpher CreateTimeTAG: 2013-01-01T06:59:24Z VoteTAG: 12 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Yes, I believe you will continue to have access to your past materials. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2013-01-01T12:58:30Z FirstChildTAG: Even if it's not accessible you can see all here: https://6002x.mitx.mit.edu/ FirstChildUserIdTAG: 297370 FirstChildUserNameTAG: ayush3504 FirstChildCreateTimeTAG: 2013-01-01T18:45:20Z IndexTAG: 144 TitleTAG: A Wonderful message of Peace for this 2013 from Antártida-Antarctica. Hi to all of you, I wish you the best and a Happy New Year 2013!:) I have a work mate that is now in the Antártida-Antarctica and he have sent to me this Pic. It was really beautiful so I decided to share it with all my Classmates, as it is really peaceful and wonderful, and in someway is a lovely message of life and nature for all the World, I wish and I hope the World Peace for this 2013 :) ![Ant][1] Myriam. [1]: https://edxuploads.s3.amazonaws.com/13570039581343618.jpg UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2013-01-01T01:48:54Z VoteTAG: 12 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: nice pic......a happy new year to all FirstChildUserIdTAG: 114864 FirstChildUserNameTAG: Vikaash FirstChildCreateTimeTAG: 2013-01-01T08:12:54Z FirstChildTAG: Equally for you Myriam (Myrimit) and for all! FirstChildUserIdTAG: 402193 FirstChildUserNameTAG: Feramico FirstChildCreateTimeTAG: 2013-01-01T01:59:56Z SecondChildTAG: Thank you Feramico, Happy New Year, my best wish to you!:) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2013-01-01T02:02:45Z SecondChildTAG: Thanks! Myriam where are you? SecondChildUserIdTAG: 402193 SecondChildUserNameTAG: Feramico SecondChildCreateTimeTAG: 2013-01-01T02:10:52Z SecondChildTAG: Where are you from? SecondChildUserIdTAG: 402193 SecondChildUserNameTAG: Feramico SecondChildCreateTimeTAG: 2013-01-01T02:12:40Z SecondChildTAG: Are you from Argentina? SecondChildUserIdTAG: 402193 SecondChildUserNameTAG: Feramico SecondChildCreateTimeTAG: 2013-01-01T02:13:43Z SecondChildTAG: Yes, I am from Argentina :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2013-01-01T02:38:02Z SecondChildTAG: where do You live now? SecondChildUserIdTAG: 402193 SecondChildUserNameTAG: Feramico SecondChildCreateTimeTAG: 2013-01-01T03:08:34Z SecondChildTAG: Do you live in the U.S.? SecondChildUserIdTAG: 402193 SecondChildUserNameTAG: Feramico SecondChildCreateTimeTAG: 2013-01-01T03:09:46Z SecondChildTAG: Or in Argentina? SecondChildUserIdTAG: 402193 SecondChildUserNameTAG: Feramico SecondChildCreateTimeTAG: 2013-01-01T03:36:02Z SecondChildTAG: Is this a Survey haha - nah joke. Yep, I live in Argentina ;) and you? SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2013-01-01T03:37:46Z SecondChildTAG: Haha, I live in Peru. SecondChildUserIdTAG: 402193 SecondChildUserNameTAG: Feramico SecondChildCreateTimeTAG: 2013-01-01T03:51:28Z SecondChildTAG: The female penguin is tender as Myriam. SecondChildUserIdTAG: 402193 SecondChildUserNameTAG: Feramico SecondChildCreateTimeTAG: 2013-01-01T04:02:14Z SecondChildTAG: Thanks for being so! SecondChildUserIdTAG: 402193 SecondChildUserNameTAG: Feramico SecondChildCreateTimeTAG: 2013-01-01T04:08:25Z SecondChildTAG: One hour for 2013 in Peru! SecondChildUserIdTAG: 402193 SecondChildUserNameTAG: Feramico SecondChildCreateTimeTAG: 2013-01-01T04:09:06Z SecondChildTAG: Does the image comes from Argentina San Martin Base? SecondChildUserIdTAG: 402193 SecondChildUserNameTAG: Feramico SecondChildCreateTimeTAG: 2013-01-01T04:40:12Z SecondChildTAG: Gracias por toda tu fineza Miriam, que Dios te bendiga y que te de un año lleno de éxitos y triunfos ty que te permita lograr todo lo que te propongas. Aunque no te conozco en persona, un fuerte abrazo a la distancia. SecondChildUserIdTAG: 183712 SecondChildUserNameTAG: colegiocientifico SecondChildCreateTimeTAG: 2013-01-01T15:00:39Z FirstChildTAG: Peru is a member consultative (consultative adherent country) Antarctic Treaty of 1959, with voice and vote in taking all decisions related to the austral area. FirstChildUserIdTAG: 402193 FirstChildUserNameTAG: Feramico FirstChildCreateTimeTAG: 2013-01-01T03:53:31Z SecondChildTAG: The Base Machu Picchu is a polar research station, established in Antarctica by the Peruvian State, consulting member of the Antarctic Treaty. The purpose is the realization of studies geographical, geological, climatic and biological in this area. The base is located in Admiralty Bay, Cove Mc Kellar of King George Island, South Shetland's. SecondChildUserIdTAG: 402193 SecondChildUserNameTAG: Feramico SecondChildCreateTimeTAG: 2013-01-01T04:16:30Z SecondChildTAG: ![Base Machu Picchu of Peru in Antarctica][1] [1]: https://edxuploads.s3.amazonaws.com/13570145841343603.png SecondChildUserIdTAG: 402193 SecondChildUserNameTAG: Feramico SecondChildCreateTimeTAG: 2013-01-01T04:33:53Z IndexTAG: 145 TitleTAG: A word of thanks to MITx and edX Starting from registering for this course, it was both excitement and anxiety. Excitement for joining a course provided by MIT and nervousness to able to complete it. But as the time passed, all the lectures, video series, eBook material and full help from community TA's made every thing so easy.This journey, to learn was so wonderful.And was all about improving knowledge.Its an wonderful initiative from all the participating Universities. I am totally thankful to this system,Circuit and Electronics Course Staff,Community TA's and every person associated with this service. I wholeheartedly thank all. Have a wonderful year ahead. Thanks again and Happy New Year. Cheers :) UserIdTAG: 341850 UserNameTAG: Juhi CreateTimeTAG: 2012-12-31T05:29:47Z VoteTAG: 12 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 146 TitleTAG: EDX and career!!! What other courses are you doing?? Let's start with me, I have just finished Biosystems Engineering in Uganda (East Africa) this June and was **looking for cheap/free courses on computer science and computer engineering around July this year**, despite the fact that **I had and have completely no money, job or way of earning an independent living from then till now, till may be around Jan-Feb 2013.** So I **googled free courses**. **Edx, Udacity and Coursera** popped up instantaneously (around July this year). **Though they have no University Credit, what is most important is to learn not a certificate or transcript.** Because if we were **very unlucky** and **this platforms did not exist** ****we still would look for knowledge elsewhere but definitely would most likely not get the quality and depth we got here especially for free at the comfort of a PC**.** **Edx** struck me immediately because I could not believe it, **MIT and Harvard together**. I mean who doesn't know what MIT stands for in the world of Science and Engineering and Harvard is pretty much in it's own league too. So i signed up for courses related to Computer Science, computer engineering and fundamental Science. Currently I thus offered, **6.002x, 3.091x, CS50x, 6.00x and CS188.1x.** After the first two weeks of **CS188.1x**, **I had to unregister from it**. I just simply **did not have the prerequisites** and the fact that I had four other courses meant they was no way I could catch up with prerequisites and do the course concurrently. I just finished **6.002x**, so **6.00x, 3.091x and CS50x are** left. **The quality, depth of material and tools used are simply fantastic**. In **6.002x** there is a **circuit simulator** (very vital), **6.00x** has an **online python shell environment**, **3.091x** has a **detailed periodic table** and **CS50x** has **whitespaces and CS50 appliance**. **Most importantly 6.00x, 6.002x and 3.091x** have their **books online** and **a good mathematical calculator**. I have not seen this anywhere else. **So here is why am writing this**, we all signed for **different but similar reasons**. **Mine was knowledge and skills enhance in computer science/engineering and general/fundamental science**. On this platform, thus there is a **good chance** that out of the **global student/learner community** there are individuals with **similar career/work interests**. This would be a nice opportunity to meet one another because with the ever increasing **globalism**, there are **high chances** that **careers/work may coincide** at point t in their lifetime most likely for the **better of each coinciding career**. **Udacity mesmerised me**, a kind of **automated University** where you take courses at your pace and an accomplishment statement is automatically availed after finishing the course. **However, Udacity mainly has courses on computer science and computer engineering which are ideal for me.** I have signed for many of them, **I estimate I will take about 1-2 years to finish the ones I have signed up for starting in order of prerequisites. They are really wonderful courses.** **Coursera just silenced my mind**, currently they are 33 universities and ~209 courses. **So I signed up for some courses** starting next year and one is already running, So I also expect to finish some courses by mid 2013. **So in 1-3 years I will have had a significant education/knowledge of my own interst** courtesy of **Edx, Udacity and cousera**. **I don't care if I get a job or anything, I have to study here no matter what. I love the experience.** **But of all Edx has the highest quality but Udacity and Coursera are very good too. (My own opinion and am free to share it)** **I think and believe it is possible to orient your career courtesy of these platforms, even though they don't give University credit what they give is more than enough to get you started. Am personally trying to fulfill this in about 2 years** *** **MOST IMPORTANTLY MERRY CHRISTMAS AND HAPPY HOLIDAYS TO EVERYONE ON EDX** **SEASONS GREETINGS FROM OBIRADANIEL, Kampala Uganda.** ------------------------------------------------------------------------ *** UserIdTAG: 183507 UserNameTAG: obiradaniel CreateTimeTAG: 2012-12-25T09:45:17Z VoteTAG: 12 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: Merry X'mas and Happy New Year! Thanks for taking time to write a detailed post with nice formatting. With these explosion around anybody with free time and and internet connection now has an opportunity to engage their brain in challenging ways, in lots of areas of interest. I would say, 2012 has been a **FANTASTIC** year! It marked a revolution in education. FirstChildUserIdTAG: 623210 FirstChildUserNameTAG: preveen FirstChildCreateTimeTAG: 2012-12-25T10:02:43Z SecondChildTAG: thanks preveen SecondChildUserIdTAG: 183507 SecondChildUserNameTAG: obiradaniel SecondChildCreateTimeTAG: 2012-12-25T21:31:46Z SecondChildTAG: honestly 2012 rekindled my undying love for CS study after Udacity, Edx and coursera. I agree with you and on it being a really big revolution in Education. SecondChildUserIdTAG: 183507 SecondChildUserNameTAG: obiradaniel SecondChildCreateTimeTAG: 2012-12-25T21:38:03Z FirstChildTAG: Hi! Thank you for Udacity reference. I found an interesting course there - CS348. Merry Christmass and Happy New Year! Serge, Ukraine FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-12-25T10:19:33Z SecondChildTAG: **Functional Hardware Verification: How to Verify Chips and Eliminate Bugs** Very nice course, I will offer it in my 2nd or 3rd year of study. Can't do it now. Good luck Serge. SecondChildUserIdTAG: 183507 SecondChildUserNameTAG: obiradaniel SecondChildCreateTimeTAG: 2012-12-25T22:06:34Z FirstChildTAG: Obiradaniel thanks for sharing. I'll probably take some coursera courses. Good luck, Matias FirstChildUserIdTAG: 288174 FirstChildUserNameTAG: matiasgrodriguez FirstChildCreateTimeTAG: 2012-12-25T12:24:42Z FirstChildTAG: Another fantastic repository of video lectures is http://nptel.iitm.ac.in The lectures are based on syllabus for courses taught in different engineering streams in Indian Institute of Technology and Indian Institute of Science. These institutes are renowned for producing some of the best engineers in the world. FirstChildUserIdTAG: 232157 FirstChildUserNameTAG: GayatriTR FirstChildCreateTimeTAG: 2012-12-25T16:09:41Z FirstChildTAG: **obiradaniel**: Good luck with your professional career, just getting started off! I am happy that we can bring quality university-level courses to such places as Uganda. You mentioned Biosystems Engineering. Did you obtain a degree in that major, or are you looking to enroll in a university for a degree and / or credit? As for me, in the Spring I am planning to take **6.00x** and **3.091x** (both from MITx / edX) as well as **VLSI CAD: Logic to Layout** (from University of Illinois at Urbana-Champaign / Coursera) and **Digital Signal Processing** (from École Polytechnique Fédérale de Lausanne / Coursera) depending on how much time I have after obligations to my job, family, friends, and hobbies (automotive restoration on a 1987 Ford Mustang convertible, chemistry - setting up a home lab, and electronics experimentation - getting started with Arduino programming). Sincerely, Jersey Mark, New Jersey, U.S.A. FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-12-25T17:24:53Z SecondChildTAG: You *have* time after all those things you listed? SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-12-25T18:59:27Z SecondChildTAG: ^ lol SecondChildUserIdTAG: 454609 SecondChildUserNameTAG: Alwahsh SecondChildCreateTimeTAG: 2012-12-25T19:20:27Z SecondChildTAG: **JerseyMark,** I just finished this June. it's a four year programme, it's basically a biological approach to agricultural and environmental engineering. So graduation is next month, yeah this is Africa, you graduate after about 8 months of finishing a course. My only interest in it is renewable energy with a strong focus on solar and bioenergy. However my interest is really CS. So that's why I want to do this CS courses. SecondChildUserIdTAG: 183507 SecondChildUserNameTAG: obiradaniel SecondChildCreateTimeTAG: 2012-12-25T21:25:50Z SecondChildTAG: **JerseyMark**, your projects are **advanced Engineering**, I just hope to get there after a good deal of time. SecondChildUserIdTAG: 183507 SecondChildUserNameTAG: obiradaniel SecondChildCreateTimeTAG: 2012-12-25T21:28:39Z IndexTAG: 147 TitleTAG: lecture sequences completed.. big thanks to all associated people finally completed all the lecture sequences..its time for final battle..i will never forget this experience of online teaching and especially "**AHA moments"**...a big thanks to Anant Sir and all associated professors and communityTA... the great thing about this course is that we are learning by doing and the teaching style..hope more courses will be available next time.. **all the best to all for final exam** UserIdTAG: 114864 UserNameTAG: Vikaash CreateTimeTAG: 2012-12-13T08:42:28Z VoteTAG: 12 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You too! FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-12-13T13:45:52Z IndexTAG: 148 TitleTAG: Thanks EDx Thank You So much Edx I cant believe Edx and its domain and Progressed so majestically that it astounds me in every way. Luckily NEw colleges and Gates Foundation too have also joined, So WE can really expect More if not better courses than 6.002x Because although i have taken other courses No one Can Beat 6.002x As a course and no one can beat ,Professor Cima As a Teacher(Myrimit Knows What i mean!!!) Thanks a Lot to Tons Of Rich People Who Took a Great step To help the Less Fortunate ones like us, I cant Pay for Mit Even i sell myself and all my family assets(I will Still apply for anyway!!) but i can take The Course for Free , Thanks a Lot UserIdTAG: 378150 UserNameTAG: GladIDoThis CreateTimeTAG: 2012-11-20T17:00:00Z VoteTAG: 12 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: So nice words GladIDoThis :)! I am really thankful too!! See you! Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-20T20:38:28Z IndexTAG: 149 TitleTAG: Where's the exam I thought I was the only one who's waiting and still waiting. UserIdTAG: 144115 UserNameTAG: mitangel CreateTimeTAG: 2012-10-25T04:33:59Z VoteTAG: 12 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: i double it.. FirstChildUserIdTAG: 266739 FirstChildUserNameTAG: satvikchugh FirstChildCreateTimeTAG: 2012-10-25T04:36:09Z SecondChildTAG: me too!! where is the exam released??? SecondChildUserIdTAG: 346016 SecondChildUserNameTAG: muthukrishna SecondChildCreateTimeTAG: 2012-10-25T04:38:47Z SecondChildTAG: im also asking the same.... its almost an hour late SecondChildUserIdTAG: 371712 SecondChildUserNameTAG: RayBueno SecondChildCreateTimeTAG: 2012-10-25T04:40:01Z SecondChildTAG: when??? SecondChildUserIdTAG: 328920 SecondChildUserNameTAG: TisO SecondChildCreateTimeTAG: 2012-10-25T04:42:05Z SecondChildTAG: I am sure they might have some issue. we are also waiting anxiously here in Pakistan. Hope to get it soon. SecondChildUserIdTAG: 149119 SecondChildUserNameTAG: mnasirshahzad SecondChildCreateTimeTAG: 2012-10-25T04:45:41Z SecondChildTAG: I see just weeks 1-8, and don't see week 9 and MidTerm page at all!! Where is link??? SecondChildUserIdTAG: 86420 SecondChildUserNameTAG: Aleksei_Katkov SecondChildCreateTimeTAG: 2012-10-25T04:46:10Z SecondChildTAG: Week nine will be posted later but midterm must be here till now. SecondChildUserIdTAG: 99628 SecondChildUserNameTAG: Pranjal16 SecondChildCreateTimeTAG: 2012-10-25T04:51:31Z SecondChildTAG: Well...I'm going to bed. I hope I have time to finish before traveling to a wedding on Friday morning (PST), and I hope this does not inconvenience all of you guys too much! And can I say that I am so pleased and honored to have classmates in all different parts of the world! Chile, Pakistan, etc. Amazing! SecondChildUserIdTAG: 311920 SecondChildUserNameTAG: ghowell SecondChildCreateTimeTAG: 2012-10-25T05:08:51Z FirstChildTAG: it's a time warp in Boston :) FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-25T04:42:16Z IndexTAG: 150 TitleTAG: H5P1 HELP I think the people stuck with the 3rd part can get help from this.Most of you are making same mistake.You are taking Vgs=Vi & Vds=Vo which is not true due to present of offset.So take Vi=Vgs+Vs(-) & Vo=Vds+Vs(-) instead.Hope this will help. UserIdTAG: 92895 UserNameTAG: shihab2555 CreateTimeTAG: 2012-10-11T08:22:27Z VoteTAG: 12 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I think they should give to us more examples with this situations... I know we should know how to do these exercises, but after 17 videos showing the most basic example, its easy to get confused... FirstChildUserIdTAG: 310147 FirstChildUserNameTAG: ildomarcarvalho FirstChildCreateTimeTAG: 2012-10-12T21:58:35Z SecondChildTAG: gr8 help dere!!! SecondChildUserIdTAG: 127800 SecondChildUserNameTAG: MNB SecondChildCreateTimeTAG: 2012-10-14T18:22:42Z IndexTAG: 151 TitleTAG: Great Experiment Great experiment. Very interesting and practical and at the same time funny. Thank you UserIdTAG: 242529 UserNameTAG: antpc CreateTimeTAG: 2012-10-02T21:48:33Z VoteTAG: 12 CoursewareTAG: Week 5 / Music Amplifier Demonstration CommentableIdTAG: 6002x_snd_d NumberOfReplyTAG: 0 IndexTAG: 152 TitleTAG: Congratulations wow, it is a very interesting class, thanks for share it with us UserIdTAG: 26082 UserNameTAG: aiaa CreateTimeTAG: 2012-09-16T21:34:58Z VoteTAG: 12 CoursewareTAG: Week 1 / Lumped Element Abstraction CommentableIdTAG: 6002x_lumped_element_abstraction NumberOfReplyTAG: 0 IndexTAG: 153 TitleTAG: Videos in Youtube The videos are, today, host just in Youtube (I'm using Windows 7 and Google Chrome, but I tested it in Mozilla and continued the problem)... Its temporary or permanently? I enjoy the old subtitles shape, and now it doesn't appear =) UserIdTAG: 291362 UserNameTAG: Gudson CreateTimeTAG: 2012-09-13T09:59:52Z VoteTAG: 12 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 3 FirstChildTAG: yes there is problem in converting lecture to words what is the problem FirstChildUserIdTAG: 295983 FirstChildUserNameTAG: qassam FirstChildCreateTimeTAG: 2012-09-13T10:49:49Z FirstChildTAG: I have the same problem. Looks awful. I am watching from Ubuntu (Chrome). FirstChildUserIdTAG: 240170 FirstChildUserNameTAG: seca FirstChildCreateTimeTAG: 2012-09-13T10:37:24Z FirstChildTAG: Please see [Dave's (edX developer) post][1]. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/MITx_6_002x_2012_Fall/threads/5051442739934a2b0000001a FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-13T12:44:57Z SecondChildTAG: Thanks! I saw that the videos at youtube also presents subtitles, and this is very helpful to me, although the speed option was off. I verified now, the old videos shape is back! Thanks a lot, staff! SecondChildUserIdTAG: 291362 SecondChildUserNameTAG: Gudson SecondChildCreateTimeTAG: 2012-09-14T13:18:56Z IndexTAG: 154 TitleTAG: How to find the voltages in Transient Analysis based on the display? Requested by mkprasanth Hi mkprasanth! mkprasanth, you requested me in this [Post][1]: "*Thanks a lot for your support I need to understand the Lab usage how we can find the voltages in Trans analyze based on the display*" Remember that you can use the [Sandbox][2] to Simulate Circuits. This Tool will be really important during this Course: You can meassure, analyze, etc... You ask me How you could meassure voltages in Transient Analysis: Lets First Try to make a Circuit. I will invent an arbitrary Circuit: ![enter image description here][3] I am sure that you are thinking: Nice Circuit, but, - How had you put a Sinusoidal Voltage Source in the Circuit? Easy: select the element (Voltage Source) . If you double click on it, It will appear a down arrow, select sin (sinusoidal). Then it will appear all the Parameters of this sinusoidal voltage source. Try it! ;) ![enter image description here][4] - How did you change the Colour of the Scopes (blue and red Scope) ? This is not for fashion ;). You have to change the colour of the Scopes in order to identify the waves related with each scope. Each wave will have the same color of the scope that you have chosen. Just double click on this element and in the down arrow you will find a lot of colours. ![enter image description here][5] ---- Lets back to our issue mkprasanth ! How do we meassure in Transient analysis? oK, Now change the 1 Ohm Resistor for the values given in the following image: 10k,20k and 10k. Click on TRAN and put 10n (this is 10 nanosecons - what you will see in the maximun time in your window graph). Click on OK. You will see two values. If you move your mouse, you will see that for one time it will appear two values in the up of the window. ![enter image description here][6] So, for time t=6.975ns you will have : In the Node where is the blue scope a voltage of 186.073nV In the Node where is the blue scope a voltage of 157.105nV I hope this can help you! You can do it! See you! Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/MITx_6_002x_2012_Fall/threads/50474faf685941270000000d [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Overview/Circuit_Sandbox/ [3]: http://s17.postimage.org/4x0hfduf3/sand.png [4]: http://s12.postimage.org/yikq0lzct/source.png [5]: http://s8.postimage.org/4e2s9ez6d/scope.png [6]: http://s7.postimage.org/js9pdn7vf/GRAFI.png UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2012-09-12T00:08:58Z VoteTAG: 12 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 5 FirstChildTAG: Hi Myrimit, I would strongly suggest that you create a Wiki page to document all your work, in addition to having them in the discussion forums. FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-12T03:10:02Z SecondChildTAG: Thank you kimt ;). I will do that! Thank you for your suggestion. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-12T11:11:49Z SecondChildTAG: **Gracias** SecondChildUserIdTAG: 298577 SecondChildUserNameTAG: msolis SecondChildCreateTimeTAG: 2012-09-16T12:24:49Z FirstChildTAG: Dear Myrim Sir , Thanks a lot for your support , I am sorry for taking you time , i will come back if any support required understand well Thanks MK.Prasanth FirstChildUserIdTAG: 156694 FirstChildUserNameTAG: mkprasanth FirstChildCreateTimeTAG: 2012-09-12T07:15:11Z SecondChildTAG: You are welcome mkprasanth! ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-12T11:38:29Z FirstChildTAG: Dear Myrim Sir I COULD NOT UNDERSTAND HOW TO FIND THE VOLTAGES AND CURRENT USING KCL ,KVL OR BY NODAL ANALYSIS .I NEED HELP SO THAT I CAN UNDERSTAND AND DO MY HOMEWORK .BECAUSE OF THIS I COULD NOT FINISH MY HOMEWORK IN TIME . IT WOULD BE MORE HELPFUL FOR ME IF I UNDERSTAND THE WORKING OF HOW TO CALCULATE KVL, KCL AND NODAL ANALYSIS WITH AN EXAMPLE OF A CIRCUIT. THE ANOTHER THING IS THAT I HAVE DESIGNED THE CIRCUIT IN LAB AND CHECKED I COULD GET THE VALUES BUT WHEN I CLICK ON CHECK IT SAY IT IS WRONG . I WOULD BE GLAD IF U COULD HELP ME OUT FROM THIS PROBLEM ALSO. IF U NEED ME TO SEND THE SCREEN SHOT OF THE CIRCUIT DESIGNED BY ME I COULD SEND IT ALSO .AS MY LAST DATE TO FINISH MY HOMEWORK AND LAB IS SEPTEMBER 16 ,I BELIEVE I WOULD BE ABLE DO MY HOMEWORK AND LAB IF I GET ANY HELP OR I WOULD NOT ABLE TO SOLVE THIS PROBLEM. FirstChildUserIdTAG: 429168 FirstChildUserNameTAG: arunprakashavm FirstChildCreateTimeTAG: 2012-09-15T06:42:05Z SecondChildTAG: Hi arunprakashavm! I haven't seen this post before... Can I help you? For KCL and KVL I suggest you to watch Prof. Agarwal video explanation: ---------- [KVL,KCL][1] [S2V1: REVIEW KVL, KCL][2] ---------- Remember that: KVL: "The sum of voltages in a loop it is zero". KCL: "The sum of currents ina node it is zero ". What is a loop? what is a node? [Take a look at this page of the Textbook][3] ---------- Convention sign of Currents: You can use: 1) If the current enters to a node, it is positive and sum; if a current get outs from a node, it is negative and rest. ![enter image description here][4] 2) If the current enters to a node, it is negative and rest; if a current get outs from a node, it is positive. ![enter image description here][5] Both conventions are valid. But you have to choose only one when solving a Circuit, you can not mix the two methods. ---------- About the Lab 1 you can see my answer in this [Post][6] there are some hints... ---------- I hope this can help you. Please tell me if I can assist you more... Myriam. [1]: https://www.youtube.com/watch?v=SaieZYN_WR0&feature=player_embedded#! [2]: https://www.youtube.com/watch?v=eLAyO33baQ8&feature=player_embedded#! [3]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/78 [4]: https://edxuploads.s3.amazonaws.com/13477775069540922.png [5]: https://edxuploads.s3.amazonaws.com/1347777652104503.png [6]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/MITx_6_002x_2012_Fall/threads/50555d8278dbb72200000001 SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-16T06:44:09Z SecondChildTAG: **Gracias** SecondChildUserIdTAG: 298577 SecondChildUserNameTAG: msolis SecondChildCreateTimeTAG: 2012-09-16T12:25:11Z SecondChildTAG: Por nada msolis ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-16T23:59:35Z FirstChildTAG: DEAR SIR , I DON'T UNDERSTAND THAT HOW TO ROTATE THE SELECTED RESISTANCE IN SANDBOX TOOL,WILL YOU PLEASE HELP ME? FirstChildUserIdTAG: 561964 FirstChildUserNameTAG: shashishirsat3 FirstChildCreateTimeTAG: 2012-10-05T06:08:57Z SecondChildTAG: Hi shashishirsat3! Yes, sure I want to help you. If you want to rotate an element just click on the element and press the "R" of the keboard :). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-05T11:40:43Z FirstChildTAG: please i do not understand the small signal calculations on diodes and algebraic expressions of capacitors in parallel FirstChildUserIdTAG: 400788 FirstChildUserNameTAG: Patrickn FirstChildCreateTimeTAG: 2012-10-27T11:59:35Z IndexTAG: 155 TitleTAG: Calculator Problem Dear Sir, Ananat Agarwal and Khurram Afridi, I am relay missing the online calculator as it is in Spring 2012 under 6.002x MITx. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-05T13:16:27Z VoteTAG: 12 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: I am also having the same problem. FirstChildUserIdTAG: 259693 FirstChildUserNameTAG: MehrozKhan FirstChildCreateTimeTAG: 2012-09-05T21:57:41Z FirstChildTAG: Yeah, I am missing the calculator too. Is there a calculator? FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-09-05T14:27:00Z IndexTAG: 156 TitleTAG: Current sign Direction of the current seems to be wrong,flowing into the positive terminal instead of out of it. UserIdTAG: 162351 UserNameTAG: ericpts CreateTimeTAG: 2012-09-05T13:15:29Z VoteTAG: 12 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 6 FirstChildTAG: Hi, I am also confused about the first circuit when we did DC analysis. Why its showing -500mA. Why current is orginating from negative terminal and end at Positive terminal. It should be other way around. Please correct me if I am wrong!! FirstChildUserIdTAG: 181368 FirstChildUserNameTAG: Hanumanta FirstChildCreateTimeTAG: 2012-09-05T14:38:01Z FirstChildTAG: You can always presume direction of the current, and if you get a minus it means your presumption was wrong. FirstChildUserIdTAG: 235637 FirstChildUserNameTAG: Mandic FirstChildCreateTimeTAG: 2012-09-05T13:25:11Z FirstChildTAG: This is not a bug. FirstChildUserIdTAG: 259238 FirstChildUserNameTAG: omidsadeghi FirstChildCreateTimeTAG: 2012-09-06T05:53:22Z FirstChildTAG: I think i made a mistake looking at that direction symbol for the current also. FirstChildUserIdTAG: 175033 FirstChildUserNameTAG: cenzin FirstChildCreateTimeTAG: 2012-09-06T02:43:47Z FirstChildTAG: Have you check the sign of the current? Current arrow is pointing into the positive terminal but it has a negative sign, meaning that it is actually going in the other direction. FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-05T13:21:19Z SecondChildTAG: Thanks for that explanation! Was wondering why it seemed negative as well. Is there any way to change the direction of the arrow? SecondChildUserIdTAG: 272193 SecondChildUserNameTAG: jarrelscy SecondChildCreateTimeTAG: 2012-09-05T13:26:36Z SecondChildTAG: wow, did not notice the sign, unforgivable! SecondChildUserIdTAG: 175033 SecondChildUserNameTAG: cenzin SecondChildCreateTimeTAG: 2012-09-06T02:45:52Z FirstChildTAG: According to their notes and I state "And the source current sign is negative if the current is flowing out of the negative terminal of the source, through the circuit, and into the positive terminal of the source." I am also confuse with their answer as why positive instead of negative because the arrow indicates into the positive terminal. can someone please explain? FirstChildUserIdTAG: 132229 FirstChildUserNameTAG: ahlo FirstChildCreateTimeTAG: 2012-09-05T16:46:19Z SecondChildTAG: The arrow is an asumption. You first assume a positive direction for the current. The assumption here is the counterclockwise direction. So if the value for the current is positive it is in the assumed direction and otherwise if the value is negative it is in the reverse direction of the assumed direction. the value of the current is negative which means the direction is reverse of the assumed direction So it is from positive to negative because the picture assumes that negative to positive is positive direction. Now you know the actual direction of current following in the circuit is clockwise. But the question says that we assume again that the positive to negative (clockwise) is positive direction. With this assupmption, the current value becomes positive. SecondChildUserIdTAG: 259238 SecondChildUserNameTAG: omidsadeghi SecondChildCreateTimeTAG: 2012-09-05T20:12:48Z SecondChildTAG: I believe the answer should be "negative" 500e-6 not positive too based on the lecture. My idea is because the simulator or the diagram itself shown the current flowing here is actually negative. That means the simulator is considering the what is called electron flow or negative flow or negative convention flow. Electron flow or negative flow or negative convention flow ( for the source component) is wherein the "current is flowing out of the negative terminal of the source, through the circuit, and into the positive terminal of the source". Positive flow or hole flow or positive convention flow (for the source component) in contrast pertains to the first statement in the course which says "the sign of the source current as positive if the current is flowing out of the positive terminal of the source, through the rest of the circuit, and then back in to the negative terminal." SecondChildUserIdTAG: 309255 SecondChildUserNameTAG: mamba747 SecondChildCreateTimeTAG: 2012-09-06T04:07:00Z IndexTAG: 157 TitleTAG: Virtual Certificate: Good or Bad? Ideas? I think that a virtual certificate is actually **better** than a "mailed" certificate. You can customize it to your needs! - First, you can print the certificate out any way you like. If you want to use **fancy paper**, you know the kind that has a heavy weight, a nice feel to it, and whatever color you want, you can do so. Even *translucent* or *pink* paper! You can then frame it. - If not, you can simply print it on regular copy paper in black and white. You can make multiple cheap black and white copies if you're mailing them out with college / university applications or attaching to your resume / C.V. - Or just keep it stored electronically if you're lazy or don't feel the need. Many employment applications are completed online, and there's almost always an option to attach your credentials via a .PDF or other image file. This saves you from having to scan your certificate into electronic form. - Mailing out paper certificates would waste a lot of paper, a natural resource. For you environmentalists out there, you should appreciate this. Also, postage takes time, money, and items can be lost in the mail. It is better that edX use it's resources to develop additional courses instead of worrying about certificates. Even if you get a "real" paper certificate, the printing and signatures on them are going to be computer-generated anyways! - You can go down to your local print shop, a copy center (In the U.S. stores like Staples, OfficeMax, or Fedex Office), or even one of those kiosks where you can print photos from a USB or other memory stick, and print the certificate with a **quality color laser printer** if you want to. - Some print shops can even customize your certificate for an extra charge. You can get the logos embossed with metal foil so it looks nice, or even add a border, etc. - With a virtual certificate, you can resize it and print it however big or small you want. You can **reduce it to credit-card size**, laminate it with clear plastic, and then you can keep all your edX certificates in your wallet! How cool is that? - The certificate has an **URL**/web address on it, so anyone in the world can verify that it is real, even without physically examining the document, by going online. This deters even the best counterfeiters. Even my home state of New Jersey is eliminating all of the old-fashioned stamps, seals, etc. and adding a verification URL to important government documents such as birth certificate apostilles, certificates of good standing for corporations, etc. If it's good enough for the U.S. government, then that means quite a bit. - Some have suggested, in another post, adding a **bar code** that is smart-phone readable onto the certificate so that the URL can be accessed quickly without having to type that long string into a web browser. This can be easily done yourself! Since the certificate is in electronic form, you just have to find an application that will generate a barcode, and then you can copy-and-paste that barcode into your .PDF file. Or if you're not that good with digital image manipulation, many places will let you use their barcode label printer (you can buy one but they're expensive). You can print a barcode sticker with your specific URL, and just stick it on the certificate. So easy! - I know that some of the more advanced students among us can come up with additional ideas. Being an Electrical and Computer Engineering student, it is possible to even embed an RFID microchip inside of your certificate, just like they do in passports. Resist the temptation on this one, unless you want to be labeled an uber-**nerd**. - ***Note / Caution:** You could even add other things to / digitally manipulate your certificate (i.e. your grade, a picture of your cat as a background) but I am **not** recommending this, as it will not show up if whomever authenticating your certificate goes and checks it on the web. Depending on the use of the certificate and your intent, electronically adding / changing the certificate can range anywhere from harmless (i.e. adding your grade for home display) to criminal fraud (i.e. putting someone else's name on the certificate and trying to pass it off as genuine). Don't do it!* Jersey Mark UserIdTAG: 292546 UserNameTAG: JerseyMark CreateTimeTAG: 2013-01-02T21:29:42Z VoteTAG: 11 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: What I did to give authenticity to my grade was carrying a lawyer who is also a public notary so he could see that it was me who did the final exam and certify from the computer directly that the degree and score I obtained was true, I printed progress and he officially certify that's m MITX grade and score was legally obtained. FirstChildUserIdTAG: 183712 FirstChildUserNameTAG: colegiocientifico FirstChildCreateTimeTAG: 2013-01-02T22:34:36Z SecondChildTAG: Please remember that the certificate has not any kind of legal or academic validity is only a simple object with symbolic value. SecondChildUserIdTAG: 27399 SecondChildUserNameTAG: juancho SecondChildCreateTimeTAG: 2013-01-03T00:19:41Z SecondChildTAG: I strongly agree with juancho. But I like JerseyMark's perspective too. SecondChildUserIdTAG: 297370 SecondChildUserNameTAG: ayush3504 SecondChildCreateTimeTAG: 2013-01-03T00:23:46Z SecondChildTAG: Not always the case, for example many universities in Europe, attached to the respect of the right to education as enshrined in international human rights instruments recognize studies and to nonacademic experiences conducive to credit not a race but have at least the same degree depth and difficulty of the subjects they teach. So, an approved course at MIT with a degree of excellence far superior to any university in the world, will be recognized in lieu of any course of circuits. All you have to prove is that the knowledge gained via a curriculum, which is available on the same platform of MITX and be authenticated by a notary public lawyer. Is a case in England where a friend recognized him earlier during circuit-6.002x 1 in a prestigious British university. The following was taken from http://education-portal.com. An Overview Types of Accredited Life Experience Based College Degrees Some colleges offer associate's and bachelor's life experience based college degrees. They are available in areas such as engineering, public administration, culinary arts, drafting, human resources and project management. Students may pursue accredited life experience based college degrees to help them compete with others who already have degrees but lack practical job experience. Some employers may not consider accredited life experience based college degrees to be as prestigious as traditional degrees. Requirements for an Accredited Life Experience Based College Degree Accredited life experience based college degree programs require applicants to have completed a minimum amount of life experience related to the desired program. Life experience may include employer-sponsored seminars, workshops and training or military education, as well as courses taken at a college, university, vocational or technical school. Volunteer work with relevant organizations and independent writing and reading may also count toward an accredited life experience based college degree. Applicants for an accredited life experience based college degrees must submit credentials such as college course transcripts, awards, certificates, portfolios and job reviews. Some life experience degree programs may accept letters from former employers describing job responsibilities and accomplishments in detail. Accredited life experience based college degree programs may administer exams to assess life experience and previous learning. Accreditation of Life Experience Based College Degree Programs Some life experience based college degree programs are not accredited and are not highly regarded. Students should be especially wary of master's and Ph.D. life experience based degrees because they are typically not accredited by creditable organizations. To check the credibility of organizations that accredit life experience based college degrees, search a database of 6,900 accredited post-secondary institutions provided by the U.S. Department of Education (DOE). Users can search the database by names of accrediting organizations, school names and city and state (http://ope.ed.gov/accreditation). SecondChildUserIdTAG: 183712 SecondChildUserNameTAG: colegiocientifico SecondChildCreateTimeTAG: 2013-01-03T00:57:18Z SecondChildTAG: En español No siempre es así, por ejemplo muchas universidades en Europa, apegados al respeto del derecho a la educación consagrado en los instrumentos internacionales de derechos humanos reconocen estudios y hasta experiencias no académicas no conducentes a créditos en una carrera pero que tengan como mínimo el mismo grado de profundidad y dificultad que las materias que ellos imparten. Así las cosas, un curso aprobado en el MIT, con un grado de excelencia infinitamente superior a cualquier universidad del mundo, va a ser reconocido en sustitución de cualquier curso de circuitos. Lo único que tenés que demostrar es que los conocimientos adquiridos vía un programa de estudios, el cual se puede obtener en la misma plataforma de mitx y ser autenticada por un abogado notario público. Se de un caso en inglaterra a donde un amigo le reconocieron el curso anterior de 6.002x por circuitos 1 en una prestigiosa universidad británica. SecondChildUserIdTAG: 183712 SecondChildUserNameTAG: colegiocientifico SecondChildCreateTimeTAG: 2013-01-03T01:00:50Z SecondChildTAG: Estoy seguro que el curso 6002x no difiere ni en contenido ni en profundidad de cualquier otro curso introductorio de cualquier otra universidad del mundo. Simplemente es una cuestion de "marca", la marca MIT es más deslumbrante pero el curso 6002x no tiene nada que no pueda ser encontrada en cualquier otra universidad. SecondChildUserIdTAG: 27399 SecondChildUserNameTAG: juancho SecondChildCreateTimeTAG: 2013-01-03T01:13:42Z SecondChildTAG: I'm sure that the course 6002x does not differ in content or depth of any introductory course from any other university in the world. It is simply a matter of "label", the label MIT is more dazzling but the course 6002x has not nothing that can not be found in any other university. SecondChildUserIdTAG: 27399 SecondChildUserNameTAG: juancho SecondChildCreateTimeTAG: 2013-01-03T01:18:58Z SecondChildTAG: I disagree, http://www.topuniversities.com/university-rankings/world-university-rankings/2012/subject-rankings/technology/electrical-engineering SecondChildUserIdTAG: 183712 SecondChildUserNameTAG: colegiocientifico SecondChildCreateTimeTAG: 2013-01-03T01:34:03Z SecondChildTAG: Please let me know one topic that is presented in 6002x but it is not presented in any other course of any other university. SecondChildUserIdTAG: 27399 SecondChildUserNameTAG: juancho SecondChildCreateTimeTAG: 2013-01-03T01:44:14Z SecondChildTAG: It is not a topic, is the depth, the quality of teachers, the kind of problems developed, teaching methodology, the technology used and the college experience, and MIT is the best. SecondChildUserIdTAG: 183712 SecondChildUserNameTAG: colegiocientifico SecondChildCreateTimeTAG: 2013-01-03T01:50:00Z SecondChildTAG: Okay but such "depth", "quality of teachers", "kind of problems developed", "teaching methodology", "technology" were not presented in the 6002x. SecondChildUserIdTAG: 27399 SecondChildUserNameTAG: juancho SecondChildCreateTimeTAG: 2013-01-03T01:56:11Z SecondChildTAG: **juancho:** I can speak as to U.S. academic institutions, as I am familiar with their admissions process and their transfer of credit process through personal experience. Each academic institution in the U.S. determines the validity and transferability of student credentials, whether for admissions purposes or for credit, according to that particular institution's policies. For the most part, this determination is made on a case-by-case basis; normally there is no blanket policy in place. By stating that an edX certificate "has not any kind of legal or academic validity" without restriction, you are (A) making a broad, generalized, blanket statement, without offering specifics or sources - in my opinion something that should be avoided, (B) and not qualifying such a statement as a personal opinion, where there is a likely possibility that others may take it as fact, and (C) by doing so, whether intentionally or not, **misleading** everyone on this Discussion Board, and (D) doing so when posting off-topic, as this post concerns student opinion on virtual vs. mailed paper certificates. In effect, you are claiming that it is the policy of every single academic institution (high school, college, university, etc.) in the entire world, that an edX certificate carries no weight, and by stating such, you imply that you are **intimately familiar** with the policies of every such institution. There is no way that this is even remotely possible. I even doubt (please correct me if I am wrong) that you are even affiliated with **any** single institution of higher learning in an official capacity, much less qualified to speak for that institution; but you are egregious enough to go as far as to speak for *every* institution, whether you know their policy or not. In short, please **stop** claiming policy regarding the academic validity of an edX certificate unless you either (A) can specifically cite a verifiable source that states your position, or (B) state that you are authorized to speak on behalf of an educational institution concerning that institution's position on edX certificates. As far as I know, not a **single** educational institution has come out on record that, as a blanket policy, an edX certificate carries no academic validity with that institution (again, if I am wrong, please provide a source and I will stand corrected), much less every one in existence as you claim! Finally, you speak of the "legal...validity" concerning edX certificates. This is somewhat vague, but I will attempt to decipher your assertion. The Eighth Edition of Black's Law Dictionary defines something as "valid" when it is "Legally sufficient; binding." Using this definition, the latter term, "binding", cannot apply because an edX certificate is not a contract nor can it be construed as such. Therefore, I can deduce that an edX certificate is valid *when it is legally sufficient for its purpose*. Since you fail to mention what **specific purpose** an edX certificate would be used for (i.e. as a credential accompanying a job application, as consideration for acceptance at a university), I look at the general purpose of an edX certificate and the authority of edX to issue one. The syllabus for this course states that "*Those who successfully earn enough points will receive an honor code certificate from MITx.*" Thus, according to the instructor(s), the purpose of an 6.002x edX certificate is to recognize those students who successfully earned the 60 points or more in this course. As U.S. law does not specify a particular manner in which students must be recognized - leaving it up to individual institutions - 6.002x certificates issued by edX are legally valid in the U.S. to show that a student has successfully passed **this** course. In conclusion, your statement "Please remember that the certificate has not any kind of legal or academic validity" is **wrong** on both counts. A 6.002x Honor Code certificate issued by edX is legally valid in the U.S. when used to show that a student has successfully completed this course. And as determination of the academic validity of edX certificates (as with any new type of credential) are determined by, and may differ according to, each **individual** institution, there has not been, as of now, any official statement either way from even a single school. While you are free to criticize edX certificates, please make sure that you do not confuse fact with opinion, and please do so on-topic, or simply hit the "New Post" button instead of threadjacking. Jersey Mark *(Note 1: Portions of this post contained topics of a legal nature, and as such I feel that it requires a disclaimer. I am not an attorney and the material in this post is not to be construed as legal advice. Also remember that the internet is a global medium and that while U.S. laws govern MITx and edX, each country has its own specific laws that may or may not apply to you. If you have a specific question concerning the law and its application to an edX certificate, please contact an attorney licensed in your jurisdiction.)* *(Note 2: Portions of this post not relevant to my Community Teaching Assistant functions are my opinion, especially where stated as such, or when qualified by terms such as "As far as I know", "I think", etc. I alone am responsible for such content, and such may not reflect the official positions of edX, MITx, instructors, sponsors, etc. especially when inconsistent with written policy.)* SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2013-01-03T03:08:41Z SecondChildTAG: farigh loug.... SecondChildUserIdTAG: 127800 SecondChildUserNameTAG: MNB SecondChildCreateTimeTAG: 2013-01-03T10:16:19Z IndexTAG: 158 TitleTAG: Thanks Dr. Agarwal and staff. Great course! But DANG! Perfect all the way through the course, then missed one on the final. Kinda like bowling a 299... As it says in the subject, thank you Dr. A and staff for the great course, and the opportunity to take it here on edx. UserIdTAG: 428560 UserNameTAG: MikeDayton CreateTimeTAG: 2012-12-22T20:19:30Z VoteTAG: 11 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 159 TitleTAG: Thanking edX and M.I.T. (Prof. Agrawal)! It was an awesome experience being a part of M.I.T. standard education. Even though I have already completed my B.E. in E&C Engineering, it was really a great time learning with Prof. Anant Agrawal and his team. Well, I could not be the part of some top ranked institutes like M.I.T. for my B.E. but at least, I could get glimpse of experience about their education system. The lessons covered at 6.002 were not new to me but the teaching methodology and the knowledge I gained was totally "Aaha moment" for me. Glad I took this course. P.S. I don't think we are supposed to discuss on the final score so, I don't want to break the rules. . Greetings from Nepal! UserIdTAG: 405473 UserNameTAG: Avisec CreateTimeTAG: 2012-12-22T13:05:11Z VoteTAG: 11 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 160 TitleTAG: Cannot stop praising the quality of this course I've already written praises about this course. But cant stop talking more and more about how wonderful the whole journey was. I very much hope that educationists from the school sector in India take a serious note of not only the quality of content of this course, but also the immense kindness involved in the content and its presentation in all forms! I don't know how to put it in right words, but I hope they get my point. If that is done, then the number of suicides in the schools due to poor performance, unhealthy competition, and pressure from parents and teachers could be brought down drastically. "Infinite patience, infinite purity, infinite perseverance, and above all, infinite love", these are words of Vivekananda. This course has it all. UserIdTAG: 232157 UserNameTAG: GayatriTR CreateTimeTAG: 2012-12-21T14:23:36Z VoteTAG: 11 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Rightly said GayatriTR. FirstChildUserIdTAG: 18877 FirstChildUserNameTAG: johnkhiangte FirstChildCreateTimeTAG: 2012-12-21T14:57:34Z FirstChildTAG: I feel the same things than you mate. FirstChildUserIdTAG: 87808 FirstChildUserNameTAG: pumoneon FirstChildCreateTimeTAG: 2012-12-22T22:05:57Z FirstChildTAG: I agree FirstChildUserIdTAG: 418182 FirstChildUserNameTAG: euldji2005 FirstChildCreateTimeTAG: 2012-12-23T12:11:14Z IndexTAG: 161 TitleTAG: A 'Thank You' isn't enough! Now after I finished the final exam and passed successfully I'd like to thank everyone who participated in this great and fabulous work! A very very very big and enormous thank you to Professor Anant Agarwal, my words become useless when it come to praise you! I hope that there will be another Electronics courses, I can't wait actually :) Thank you again edX Team. UserIdTAG: 365201 UserNameTAG: sirajmuhammad CreateTimeTAG: 2012-12-21T08:47:04Z VoteTAG: 11 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 162 TitleTAG: Hint for H11P2 - A slightly different approach If you are having trouble with H11P2, perhaps this will help you. You have two parallel RC circuits in series. The complex impedance for a parallel RC circuit is given by Zrc = R/D where D is the complex expression 1 + j*w*R*C. To solve the problem you need the transfer function Vo/Vin, which is given by the voltage divider equation Z1/(Z1+Z2) where 1 and 2 refer to the probe and scope. You have to figure out which is which. Substituting for the Z's gives Vo/Vin = (R1/D1)/(R1/D1+R2/D2). Multiplying numerator and denominator by D1 gives Vo/Vin = R1/(R1+(D1/D2)*R2). If D1/D2 can be made equal to one the equation simplifies to that for a resistive voltage divider. The requirement that D1/D2 = 1 gives a condition for Cp that solves the problem, and Vo/Vin = R1/(R1+R2). UserIdTAG: 406202 UserNameTAG: skyhawk CreateTimeTAG: 2012-12-02T19:28:15Z VoteTAG: 11 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: The best hint that could be, not merely paraphrasing the same question! FirstChildUserIdTAG: 285616 FirstChildUserNameTAG: AmiyaX FirstChildCreateTimeTAG: 2012-12-02T20:30:14Z SecondChildTAG: Thanks! SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-12-02T21:42:19Z FirstChildTAG: Thanks! FirstChildUserIdTAG: 199758 FirstChildUserNameTAG: Varinia FirstChildCreateTimeTAG: 2012-12-03T01:04:33Z FirstChildTAG: I'd been reading about 'scope probes before this question came up. The 'trick' to getting a flat frequency response is simply to make the time constants equal. I.e. $R_pC_p = R_sC_s$ Rearranging it, you get $\frac{R_s}{R_p} = \frac{C_p}{C_s}$ You can think of the voltage divider in two ways: the ratio of the two halves ($R_s:R_p$) or the way we've usually seen ($\frac{R_s}{R_s+R_p}$) and here we have two ratios being equal. Look at $\frac{C_p}{C_s}$. It's the impedance of $C_s%$ ($\frac{1}{j\omega C_s}$) divided by the impedance of $C_p$ ($\frac{1}{j\omega C_p}$). The $j\omega$ s cancel out. So another way to look at this (rearranged equation) is to say that the capacitors divide the voltage in exactly the same ratio as the resistors. FirstChildUserIdTAG: 141000 FirstChildUserNameTAG: OrinE FirstChildCreateTimeTAG: 2012-12-03T06:07:52Z FirstChildTAG: In literature they often use the symbol for tau (can't make it here) and it means tau=R*C is a time constant. So for this probe question tau_probe=Rp*Cp must be equal to tau_scope=Rs*Cs. So you'll get tau_probe=tau_scope. So if you know only one of the tau's and know the ratio of the divider, and one R or C of the unknown tau, you can easily obtain that tau to equal it to the known one and than it's easy to find the unknown R or C. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-12-03T16:35:23Z FirstChildTAG: www.wolframalpha.com/input/?i=solve+R_s%2F%281%2Bi*w*R_s*C_s%29%2F%28R_p%2F%281%2Bi*w*R_p*x%29%2BR_s%2F%281%2Bi*w*R_s*C_s%29%29+%3D1%2F10+for+x FirstChildUserIdTAG: 714237 FirstChildUserNameTAG: PaxPolaris FirstChildCreateTimeTAG: 2012-12-03T17:05:23Z IndexTAG: 163 TitleTAG: Lab9 Hints requested by cruiser_rahit / arjshar [For Spanish explanation , read at the end of this Post] Hi cruiser_rahit and arjshar, Here are the Hints that I have promised to you. Sorry for the delay of this Post. ---------- **LAB 9 HINTS** In this lab we'll be exploring the properities of second-order circuits, i.e., circuits with two energy storage elements. ---------- **Part 1.** Hints. - You should take a look at the equation 12.45 of page 641 [Read here][1] - Important one: Don't forget that $\omega_0 = 2 \pi f$ *Note. don´t forget to click on TRAN* ---------- **Part 2.** Hints. Take a look at the page 644 [Read here][2]. Do you have your L and C from your previous part? Can you find the Resistance? ;) ---------- **Part 3.** Hints. I guess that this part of this lab is in order that you can play with the plot and discover how can you reach to the answer by analysis of the curves, varying the values of the components, etc :). Ok, lets see some Hints so that later you can solve this by your own: They request you to design a DC-DC power supply. ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13365638294918287.png) In order to design the Circuit, we must find the values of L -Inductor-,C - Capacitor- and D - duty cycle. ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13365659276761658.png) In the statement they ask you to design your Circuit with a certain tolerance, but what does this mean? Well, that tolerance means that in the output we will have an output voltage which it value will be between the two values **(Voutmín, Voutmáx)**. see the next image: fluctuación (Spanish) means fluctuation. ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13365702767240844.png) Try to click on the TRAN 10m button before changing the values of the components. You will see something like this: ![image description](https://mitx_askbot_stage.s3.amazonaws.com/1336570510301777.png) Evidently this will not verifying the requisits of the Statement, as the minimun voltage output is not 5.9 V , is less, so is not in the tolerance allowed range- might some students will have a different value of tolerance given in the statement-. So, the idea is that you can choose : - L - C - D Hint. Test and error experiment: Try to increase C value. You will see that this will not enogh to reach to the requirements so you will need to change another parameter too, the duty cycle. As the statement says to you, if you know your input voltage VIN and the output voltage (Vout max), you can find your D :) - Take a look at the formula in the statement of the Lab. Where is that D? How do I set that value that you have found? Take a look at the blue box in the next image. ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13365722164754694.png) Another Hint: Once you have play with the D - try to test upper values and down values of the D that you have obtained with the statement formula- you will notice that the blue curve will move up and down. Once you have seen that it verify the requirement (5.9V and 6.1 V). Change the value of L (decrease in .1mH for example) and adjust C again and so on, this is Test and error . Try it , now it is your turn!;) I hope this can hçbe helpful for you. This Lab part is really intuitive and you have to use plot analysis, test and error, etc :). See you, Myriam. ---- Now in Spanish! Hola cruiser_rahit and arjshar, Aquí están las Hints que les había prometido. Mil disculpas por la tardanza de este Post. ---------- **LAB 9 HINTS [Español]** En este Lab, se explorarán las propiedades de los Circuitos de segundo orden, por ejemplo, los Circuitos con dos elementos de almacenamiento de energía. ---------- **Parte 1.** Hints. - Intenta leer la página 641 del Libro de Texto y presta atención a la ecuación 12.45 [Leer aquí][1] - Y esto es muy muy importante: No te olvides esto $\omega_0 = 2 \pi f$ *Nota. Recuerda en hacer clic en el botón TRAN, no te ovides de esto tampoco.* ---------- **Parte 2.** Hints. Lee la página 644 del Libro de Texto [Aquí][2]. Tienes tu L y C de la parte previa? Entonces, puedes hallar la resistencia? ;) ---------- **Parte 3.** Hints. La idea general de la parte 3 de este Lab, es el diseño de una **Fuente de DC-DC** (continua a la entrada - continua a la salida). Es por ello, que nos dan el circuito: ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13365638294918287.png) El objetivo de este problema es **que nosotros diseñemos** dicha fuente, es decir, que modifiquemos los valores del capacitor **C**, el inductor **L** y el ciclo de actividad **D** (ver los círculos en rojo), para que a la salida tengamos una tensión del voltaje de salida solicitado con una cierta tolerancia. ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13365659276761658.png) Dicha **tolerancia**, significa que a la salida tendremos un voltaje de salida Vout cuyo valor de salida estará comprendido entre dos valores **(Voutmín, Voutmáx)**: ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13365702767240844.png) Entonces, si hacemos **TRAN 10ms** obtendremos algo así: ![image description](https://mitx_askbot_stage.s3.amazonaws.com/1336570510301777.png) Evidentemente vemos que **no cumple** con los requisitos (la tensión de salida mínima no es 5.9V, sino un valor menor...) Por ello, debemos **elegir**: - **C** - **L** - **D** (ciclo de actividad) Hint. Experimento de Prueba y error: Trata de incrementar el valor de C. Verás que este no será sufiente para que puedas cumplir con los requerimientos solicitados. Es por ello, que verás que será necesario también modificar otro parámetro, tal y como lo es el Ciclo de Actividad - Duty Cycle. Tal y como se dice en el enunciado, con la tensión de entrada VIN y la de salida (Vout máx) podrás hallar tu D , Duty Cycle :)- mirar nuevamente la fórmula en el enunciado del Lab. Bien ahora resta hallar **L y D**. El mismo enunciado, nos dice que al aumentar **D (el ciclo de actividad)**, **aumenta nuestro voltaje de salida**. Entonces, conocido el voltaje de entrada VIN y la tensión de salida (Voutmáx), podemos hallar el D aproximado... En el recuadro azul de la imagen se señala en donde cambiar el D. ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13365722164754694.png) Pista: Una vez que jugamos con los D (intentar varios valores por arriba o por debajo del D que se obtuvo con la fórmula) veremos que la curva azul se moverá arriba y abajo. Una vez que ya casi cumpla con el requisito (5.9V a 6.1 V). Cambiar L (decrementar en .1mH) y ajustar C. Espero que les haya sido de ayuda. Esta parte del lab es muy intuitiva y a prueba y error. Saludos. Ahora es tu turno, a trabajar! :) Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/665 [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/668 UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2012-11-17T13:20:13Z VoteTAG: 11 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: I haven't really posted before but I think I should let you know how grateful I am for your help Myriam, and the rest of you guys :) FirstChildUserIdTAG: 435193 FirstChildUserNameTAG: ManosP FirstChildCreateTimeTAG: 2012-11-18T23:27:11Z SecondChildTAG: You are welcome @ManosP :). I am really happy to know that, thank you for the nice words :p SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-18T23:34:25Z FirstChildTAG: i got it..thanks @Myrimit...@Myrimit i have solved 3rd part by analyzing the response of graph corresponding to different value of C and L just like trial and error method..is there any mathematical approach to this problem? FirstChildUserIdTAG: 114864 FirstChildUserNameTAG: Vikaash FirstChildCreateTimeTAG: 2012-11-17T14:05:21Z SecondChildTAG: Well done Vikaash! :). Yes, there is a way, but you have to write all the equations :p. If you want we can discuss this after the deadline of Lab9 :). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-17T14:08:01Z SecondChildTAG: i'm also waiting for the mathematical approach discussion!! :) SecondChildUserIdTAG: 301962 SecondChildUserNameTAG: labrinim SecondChildCreateTimeTAG: 2012-11-17T15:42:58Z SecondChildTAG: @Myrimit I am also waiting for mathematical approach...Hope you help me... Cheers:) SecondChildUserIdTAG: 147459 SecondChildUserNameTAG: Khanth SecondChildCreateTimeTAG: 2012-11-17T19:31:38Z SecondChildTAG: i am also waiting..:-) SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-11-18T11:04:36Z FirstChildTAG: Thank you Myrimit.........Trial and error method worked out........... FirstChildUserIdTAG: 156835 FirstChildUserNameTAG: kphariprakash1968 FirstChildCreateTimeTAG: 2012-11-17T16:29:02Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-17T16:37:52Z FirstChildTAG: You are an Angel Myrimit, thanks again. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-11-17T23:04:55Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-18T23:02:12Z FirstChildTAG: I got it...Thanks a lot Myrimit!!!!! :) FirstChildUserIdTAG: 9396 FirstChildUserNameTAG: NathalieCeron FirstChildCreateTimeTAG: 2012-11-18T22:42:59Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-18T23:01:57Z IndexTAG: 164 TitleTAG: Goodbye, adios, au revoir, auf wiedersehen, namaste', وداعا , अलविदा के Well, folks... It looks like time has finally won out. :) I have taken on a few extra obligations and now I cannot complete them all. Sadly, I will not be able to complete this class, this time around. Can't tell you how much fun it's been, though. The enthusiasm is great fun. :) Regards. (and acknowledgement to Google translate...hope I didn't mess up too badly with that title line! :) ) UserIdTAG: 92346 UserNameTAG: MobiusTruth CreateTimeTAG: 2012-11-12T00:01:34Z VoteTAG: 11 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 9 FirstChildTAG: Take care... Best wishes! FirstChildUserIdTAG: 184827 FirstChildUserNameTAG: DiegoT FirstChildCreateTimeTAG: 2012-11-12T00:58:03Z FirstChildTAG: So sorry to hear that! I do hope we will see you again in the Spring! I'm pretty sure I'll be here - this has gone so fast I know I've missed a great deal. Best wishes on the tasks that occupy you now! FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-11-12T00:19:44Z FirstChildTAG: My best Wish to you! We will keep in touch! :) Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-12T01:22:17Z FirstChildTAG: I am sorry to see someone who can do the work not complete the task due to other events. Best wishes! FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-12T01:09:36Z FirstChildTAG: Mobius - Best wishes to you. See you in other classes. Third time is the charm. ;) FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-11-12T06:32:14Z SecondChildTAG: Hello, Grace...I did make it through the first time. This was for some additional depth and reinforcement. I got a bit of that, so all is good. See ya'! :) SecondChildUserIdTAG: 92346 SecondChildUserNameTAG: MobiusTruth SecondChildCreateTimeTAG: 2012-11-13T05:11:55Z FirstChildTAG: see you in the spring! FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-11-12T08:13:01Z FirstChildTAG: :( FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-11-12T03:55:12Z FirstChildTAG: Thanks شكرا FirstChildUserIdTAG: 106816 FirstChildUserNameTAG: Laith FirstChildCreateTimeTAG: 2012-11-12T11:44:32Z FirstChildTAG: :( :( FirstChildUserIdTAG: 285945 FirstChildUserNameTAG: lindalapiso FirstChildCreateTimeTAG: 2012-11-16T09:06:17Z IndexTAG: 165 TitleTAG: For those who would like to see another approach to solving H8P1 I solved this problem three different ways. The following is the way that I found to be the easiest and most intuitive. Of course, opinions may vary. First of all let's deal with part a. For t = 0- the current i1 is just the initial condition given in the problem statement. Therefore, v1(0-) = R1*i1(0-). Now for parts b and c. Write the differential equation for the loop containing the inductor using the component relations: L*(di1/dt) + i1*R1 = Vin where Vin is an impulse at t=0. Now some will object to using a differential equation as too difficult, but when dealing with the dynamics of inductors and/or capacitors either you or someone else must solve a differential equation! The interesting fact is that we can answer part b without solving the DE. We simply integrate each term in the equation from 0- to 0+. Integrationing the first term gives: L*(i1(0+)-i1(0-)). Integrating Vin simply gives S, where S is the strength of the impulse, which in this case is 1, but I will carry the S to be more general. Finally integrating the i1*R1 term gives zero, since it is a finite term integrated over an infinitesimal time. Therefore, the result is: L*(i1(0+)-i1(0-)) = S or solving for i1(0+) we get i1(0+) = i1(0-) + S/L, then multiplying by R1 gives the voltage drop across R1. Since the impulse is zero for t>=0+, the DE is homogeneous for that time interval. Therefore, we need to solve (L/R1)*(di1/dt) + i1 = 0. This is nothing more the the equation for exponential decay with time constant T, where T is the expression multiplying di1/dt. The result is: i1(t) = i1(0+)*exp(-t/T) and VR1(t) = R1*i1(t). The solution for parts d and e follows a similar procedure. Using the node method for the node between R2 and the capacitor, we obtain: (Vin-Vc)/R2 = C*(dVc/dt), which can be rearranged as follows: (R2*C)*(dVc/dt) + Vc = Vin. This has a similar form to the DE that we have already treated! Integrating term by term from t = 0- to 0+ gives: (R2*C)*(Vc(0+)-Vc(0-)) = S, which when rearranged yields: Vc(0+) = Vc(0-) + S/(R2*C), where Vc(0-) comes from the initial cconditions in the problem statement. Finally, since the DE is homogeneous for t >= 0+, we can write the solution for exponential as: Vc(0+)*exp(-t/T), where the time constant for this decay is the term multiplying dVc/dt. Note that the two branches are uncoupled as the hint states because both ends of each branch are held at ground potential except for the instant when the impulse is non-zero. Also note that in the inductor branch the current is flowing downward, which means that the potential of the resistor-inductor node is less the ground potential, whereas the potential of the resistor-capacitor node is greater that the ground potential, and current is flowing upward in that branch Edited to it typos! UserIdTAG: 406202 UserNameTAG: skyhawk CreateTimeTAG: 2012-11-11T19:30:03Z VoteTAG: 11 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Inspiring explanation :). Thanks. FirstChildUserIdTAG: 153707 FirstChildUserNameTAG: masoud_np FirstChildCreateTimeTAG: 2012-11-11T21:17:06Z SecondChildTAG: My pleasure! SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-11-11T23:32:22Z SecondChildTAG: Awesome! SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-11-12T00:54:04Z SecondChildTAG: Thank you! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-12T01:23:30Z SecondChildTAG: hey very elaborated explanation skyhawk.. but I don't understand the "T" part for part c :( SecondChildUserIdTAG: 499671 SecondChildUserNameTAG: maliha266 SecondChildCreateTimeTAG: 2012-11-12T04:25:21Z SecondChildTAG: Got it.. it's weird how stupid one can't be sometimes :) SecondChildUserIdTAG: 499671 SecondChildUserNameTAG: maliha266 SecondChildCreateTimeTAG: 2012-11-12T04:33:34Z SecondChildTAG: *can SecondChildUserIdTAG: 499671 SecondChildUserNameTAG: maliha266 SecondChildCreateTimeTAG: 2012-11-12T04:33:42Z FirstChildTAG: I felt it Easier with Laplace Transformations :) FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-11-12T08:48:05Z IndexTAG: 166 TitleTAG: VS is missed Here VS is considered to be 5V. UserIdTAG: 326787 UserNameTAG: Angstrem CreateTimeTAG: 2012-10-29T17:12:00Z VoteTAG: 11 CoursewareTAG: Week 7 / Fall Time CommentableIdTAG: 6002x_Fall_Time NumberOfReplyTAG: 0 IndexTAG: 167 TitleTAG: Very IMPORTANT Question Regarding MID-TERM (Admins please Answer!) I know on the Mid-Term exam that we get three strikes until we're out, just like in baseball... **BUT** I know that in Homework, each "question" has several "parts" to it. My question: Do we have **three chances per question**, or three chances per ***each "part" of a question*** (i.e. for every box that an answer goes in)? There's a big difference, depending on the form the exam can take. Here are two possible examples: **- - - - - - - - - - - - - -** *EECS 6.002x MID-TERM EXAM VERSION I* Question 1: * Find $V_{IN}$, [CHECK ANSWER FOR QUESTION 1] Question 2: * Find $i_D$, * Find $g_m$, * Find small signal $V_{TH}$, and * Find small signal $R_{TH}$, [CHECK ANSWERS FOR QUESTION 2] **- - - - - - versus - - - - - -** *EECS 6.002x MID-TERM EXAM VERSION II* Question 1: * Find $V_{IN}$, [CHECK ANSWER] Question 2: * Find $i_D$, [CHECK ANSWER] * Find $g_m$, [CHECK ANSWER] * Find small signal $V_{TH}$, [CHECK ANSWER] and * Find small signal $R_{TH}$, [CHECK ANSWER] **- - - - - - - - - - - - - -** Which version will the exam take? If there's a check-box next to every "part", that's O.K., but if there's a check-box only at the end of each Question of two-or-more "parts", then that's **more difficult**! I just want to prepared for what the online exam will be like! Mark in N.J., USA UserIdTAG: 292546 UserNameTAG: JerseyMark CreateTimeTAG: 2012-10-23T04:11:28Z VoteTAG: 11 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I think it would be the set of problem with one check... For detail please visit [MIDTERM Info of MITx Spring FALL-2012][1] [1]: https://6002x.mitx.mit.edu/wiki/view/MidtermInfo > **Regards:** asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-10-23T10:18:30Z SecondChildTAG: yes its for th entire set and not for individual questions....:( SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-23T11:34:50Z FirstChildTAG: Hi JerseyMark, Based on my experience in the Prototype Course 6.002x (Spring), we had **one buttom per problem** and **not per each question**... So, **it will be something like this:** Question 2: - Find iD, - Find gm, - Find small signal VTH, and - Find small signal RTH, [CHECK ANSWERS FOR QUESTION 2] **And NOT** like this: Question 2: - Find iD, [CHECK ANSWER] - Find gm, [CHECK ANSWER] - Find small signal VTH, [CHECK ANSWER] and - Find small signal RTH, [CHECK ANSWER] FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-23T13:05:40Z SecondChildTAG: We will be using the same format this semester as well. SecondChildUserIdTAG: 96452 SecondChildUserNameTAG: Lyla SecondChildCreateTimeTAG: 2012-10-23T14:51:24Z SecondChildTAG: Thanks **Mirimit** and **Lyla** for clarifying this point! SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-10-25T16:18:05Z IndexTAG: 168 TitleTAG: H8P3 magic Q3? For the first question, it would seem that we have to ignore Q3 and look at Q2 in isolation to get the right answer. Certainly, if you were to consider Q3's Roff and Din was low, then Q2s gate wouldn't discharge at all! If Din was high, then the 105Mohm Roff of Q3 would swamp the gate leakage of Q2 and the gate would discharge in the order of a microsecond. But then for the final two questions, we definitely have to use the given value of Ron for Q3... So what is going on with Q3 when the STORE pulse is off? Why doesn't its Roff charge/discharge Q2's gate? UserIdTAG: 141000 UserNameTAG: OrinE CreateTimeTAG: 2012-10-19T04:06:26Z VoteTAG: 11 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 169 TitleTAG: H7P2 TIME CONSTANTS In second question of H7P2, the correct answer has to be (L/(R1*R2))*(R1+R2), which it is not accepting as correct. But it's accepting (L/(R1+R2))*(R1*R2) as correct which is actually wrong. Please check this. UserIdTAG: 409985 UserNameTAG: jveesa CreateTimeTAG: 2012-10-08T12:29:54Z VoteTAG: 11 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: I wonder if the exercise H7P2 B is incorrect answer with the system. I managed to do it but I do not believe that is the correct answer. FirstChildUserIdTAG: 361823 FirstChildUserNameTAG: EliasOak FirstChildCreateTimeTAG: 2012-10-08T13:37:48Z SecondChildTAG: So, are you agreeing with me? SecondChildUserIdTAG: 409985 SecondChildUserNameTAG: jveesa SecondChildCreateTimeTAG: 2012-10-08T13:54:40Z SecondChildTAG: I agree. I think that system is not correct. I reviewed s12v11, and according to that system is wrong SecondChildUserIdTAG: 81344 SecondChildUserNameTAG: DmitriyL SecondChildCreateTimeTAG: 2012-10-08T15:13:43Z SecondChildTAG: YES it's wrong SecondChildUserIdTAG: 310474 SecondChildUserNameTAG: aldaris565 SecondChildCreateTimeTAG: 2012-10-08T15:50:47Z SecondChildTAG: given answer is wrong. correct answer is L*(1/R1+1/R2) SecondChildUserIdTAG: 414078 SecondChildUserNameTAG: rsiddu SecondChildCreateTimeTAG: 2012-10-08T15:54:30Z SecondChildTAG: YUP.DEFINITELY SecondChildUserIdTAG: 357747 SecondChildUserNameTAG: kishores SecondChildCreateTimeTAG: 2012-10-08T16:08:28Z SecondChildTAG: Yes, it definitely accepts the wrong answer. It basically accepts L*R as correct instead of L/R. SecondChildUserIdTAG: 270011 SecondChildUserNameTAG: SnowmanZA SecondChildCreateTimeTAG: 2012-10-08T19:33:15Z SecondChildTAG: Thanks, it really helped me!!! SecondChildUserIdTAG: 396141 SecondChildUserNameTAG: nuidem SecondChildCreateTimeTAG: 2012-11-03T05:39:52Z FirstChildTAG: Thanks for pointing out this error in the system. We are correcting it right now. Anybody who got the same incorrect answer while it was being accepted should still get credit, but any further answer submissions will have to get the actually-correct answer. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-10-08T21:28:20Z FirstChildTAG: STILL NOT GETTING THE ANSWER .THIS (L/(R1+R2))*(R1*R2) ANS IS NOT ACCEPTING TELL ME HOW TO SOLVE . FirstChildUserIdTAG: 694173 FirstChildUserNameTAG: trishna1523 FirstChildCreateTimeTAG: 2012-10-22T10:39:31Z IndexTAG: 170 TitleTAG: Week 4 = hell I just wanted to say that I'm very frustrated after going through week 4. The jump in difficulty from last week was just too big for me. Although I got 100% on the homework, I don't feel I learned a thing. I did it all by trial and error, which took me at least 17 hours, going back and watching the lectures 5 times trying to figure it out. If I had to go through homework again it'd take me the same amount of time as it did the first time. The dependent sources are way too abstracted. They should've been covered by giving real world examples, and showing why the abstraction is beneficial. Right now, I have zero intuition about them. I just hope that I won't hit the same roadblock again further in the course, because I'd probably give up. UserIdTAG: 395349 UserNameTAG: bgr CreateTimeTAG: 2012-10-07T20:40:33Z VoteTAG: 11 CoursewareTAG: Week 4 / Dependent Sources Example 1 CommentableIdTAG: 6002x_dep_src_ex_1 NumberOfReplyTAG: 15 FirstChildTAG: MOSFET is a VCCS in it's saturation regime with iDS=K/2*(vGS-VT)^2 FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-07T20:51:12Z SecondChildTAG: **YakovO**: As the professor likes to say: "An A-ha moment". **bgr:** Personally, I thought this Lab was the easiest so far (Week 2 was difficult), and the Homework easy, but only because I know how Zener diodes work in reality, so I took a few "short-cuts" with H4P2. Otherwise, if you've never dealt with diodes, I feel your pain on P2! And this week is the first time we've needed to use a bit of calculus, in the incremental resistance problems. For those who struggle with, or have never taken, calculus, I see how they would feel it's too "abstract." SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-10-07T21:38:31Z FirstChildTAG: Incremental Analysis was well explained by both the professor and the text. I think the text did a pretty good job of explaining dependent sources. I had trouble with H4P3, but mostly because the Nodal Analysis was a bit involved. Just a word of advice: Don't depend solely on the video lectures to do this work; go through the text and do **most** of the problems. I generally try to have a solid knowledge of the topic before watching the lectures. And don't get discouraged either. I'm sure a lot of people feel the same way. It took me a good two days to finish Wk 4. I'm doing well in the class, but there have been times when I felt like I just didn't get it. You have to fail to learn. FirstChildUserIdTAG: 194509 FirstChildUserNameTAG: blackcompe FirstChildCreateTimeTAG: 2012-10-07T21:05:50Z SecondChildTAG: Thanks for the tip, I'll pay attention to textbook more. SecondChildUserIdTAG: 395349 SecondChildUserNameTAG: bgr SecondChildCreateTimeTAG: 2012-10-08T00:04:59Z FirstChildTAG: hi my name is mohammed boussetta i'm from algeria and i'm student at university 2st years master mechatronics FirstChildUserIdTAG: 587066 FirstChildUserNameTAG: lifewin FirstChildCreateTimeTAG: 2012-10-07T21:17:59Z FirstChildTAG: bgr, I know it's hard. You will develop intuition about circuits eventually, maybe at the end of the course. That's when I developed intuition about them. This class is very hard, and it's not going to get easier. Week 6 is particularly hairy, especially with mid-terms hot on it's heels. Last class I spent about 20 hours a week doing the homework, so don't feel bad about the time you had to put into it. You got the right answers. Take your victory and move on to the next week. Sometimes you have to fake it before you make it. You'll make it, just keep hanging on. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-07T21:31:07Z SecondChildTAG: Totally agree with you JSChambers! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-08T00:03:28Z SecondChildTAG: Yeah, but if this ware a real test and those green checkmarks weren't there, my results on this homework would be hitting 0% :D I'll take your word on gaining the intuition and try to push through ;) SecondChildUserIdTAG: 395349 SecondChildUserNameTAG: bgr SecondChildCreateTimeTAG: 2012-10-08T00:09:27Z FirstChildTAG: Maybe you should try reading the textbook. FirstChildUserIdTAG: 90796 FirstChildUserNameTAG: ChaunceyGardiner FirstChildCreateTimeTAG: 2012-10-07T21:53:18Z SecondChildTAG: Absolutely agree! The textbook is awesome. Everything is explained at the toddlers' level, I mean, same concepts are repeated again, again and again, and then once more:) and many examples are provided. Their might be a feeling of going round in circles but it's worth it:) SecondChildUserIdTAG: 252722 SecondChildUserNameTAG: vaboro SecondChildCreateTimeTAG: 2012-10-07T23:16:29Z SecondChildTAG: I think I know this @super-cool person haha! ;) Hi @ChaunceyGardiner! So much time! How are you? Are you in other edX Course? ;). Just I was thinking of e-mailing all the Team (Dan, Brian ,Juan and you)in order to know how were you :). I promise that I will write ! See you! Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-08T00:00:25Z SecondChildTAG: Maybe I should :) SecondChildUserIdTAG: 395349 SecondChildUserNameTAG: bgr SecondChildCreateTimeTAG: 2012-10-08T00:10:10Z SecondChildTAG: I'm on the other side of the fence - the textbook does nothing for me. I find it really hard to look up things. Nothing is explained, but rather loads of examples and this makes it really hard to use it as a reference book. Might suit some, but unfortunately not me. Shame. SecondChildUserIdTAG: 207204 SecondChildUserNameTAG: hgustavii SecondChildCreateTimeTAG: 2012-10-08T14:08:24Z SecondChildTAG: Consider reviewing material from similar classes online or from different textbooks as you go through this course, along with more general electronics learning sites. Google is your friend; you may find alternative explanations for the concepts that make everything clear to you. All About Circuits http://www.allaboutcircuits.com/ and Electronics Tutorials http://www.electronics-tutorials.ws/ are two good places to start. Also look at the MIT OpenCourseware version of 6.002 http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-electronics-spring-2007/. SecondChildUserIdTAG: 94557 SecondChildUserNameTAG: g_hopper SecondChildCreateTimeTAG: 2012-10-09T16:34:15Z FirstChildTAG: Text book example problems would help, of course its kinda confusing. But worth trying and facing different problems.... FirstChildUserIdTAG: 21687 FirstChildUserNameTAG: Benadicta FirstChildCreateTimeTAG: 2012-10-07T22:20:54Z SecondChildTAG: Hi Benadicta! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-08T00:01:29Z FirstChildTAG: learning curve is off the wall steep this week. my main criticism of the course as a whole is there is insufficient examples relevant to the questions that follow. the tutorials are good, but more examples to re-enforce the ideas would be very helpful this week. FirstChildUserIdTAG: 61469 FirstChildUserNameTAG: Spacedog FirstChildCreateTimeTAG: 2012-10-07T22:33:25Z SecondChildTAG: More examples are in the textbook. And even more examples are online, I mean the online companion of the textbook. Well, again: together the textbook and its online companion have billions and billions of examples:) SecondChildUserIdTAG: 252722 SecondChildUserNameTAG: vaboro SecondChildCreateTimeTAG: 2012-10-07T23:18:50Z FirstChildTAG: Well as a student we must pass the challenges :D it takes me more than two days just for doing homework. FirstChildUserIdTAG: 114913 FirstChildUserNameTAG: ngoctuan FirstChildCreateTimeTAG: 2012-10-07T20:52:11Z FirstChildTAG: "The dependent sources are way too abstracted. They should've been covered by giving real world examples, and showing why the abstraction is beneficial." Right! Excellent observation in my opinion! This is what this course is mostly about: about the lumped-circuit abstractions. They help to get a grasp of reality, or, as it was told in one of the demos: "It's exciting when you predict exactly how nature works!" I don't know if it helps, but try to think about a 'source' (a voltage source or a current source) as anything with a vertical or horizontal line on the v-i graph. If the line is parallel to the v-axis - it's a current source because no matter what the voltage is, the current is the same. If the line is parallel to the i-axis - it's a voltage source, because the voltage is constant no matter what the current is. A source can't work in isolation. If it's a current source, it will not 'produce' any voltage until it's connected to a load. The same is about the voltage source: there will be no current until there is a load (of course because real sources are not ideal there might be some internal resistance and leakage, but the lumped-circuit elements don't have this; if one needs this property, he or she needs to incorporate it into the lumped-circuit model) Well... this is how I understand sources... Hope my understanding is not very much incorrect... Please, correct me, if I'm wrong somewhere:) FirstChildUserIdTAG: 252722 FirstChildUserNameTAG: vaboro FirstChildCreateTimeTAG: 2012-10-07T23:30:20Z FirstChildTAG: 3 12+ hour days... Got a few by guessing. Relieved week 4 is over for me, but I think there is no way in hell I'll ever pass a Midterm if it looks anything like this... FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-10-08T03:20:52Z SecondChildTAG: hang in there, week 4 was a killer alright, but keep putting the time , rack up on the labs, and do as best you can on homework, and study for the exam, we can ask for past papers and practise before (leave lots of time for this on week 6) SecondChildUserIdTAG: 61469 SecondChildUserNameTAG: Spacedog SecondChildCreateTimeTAG: 2012-10-09T02:01:05Z FirstChildTAG: what can i do for calculating Norton current and resistor ? FirstChildUserIdTAG: 369692 FirstChildUserNameTAG: andhale FirstChildCreateTimeTAG: 2012-10-08T05:43:11Z SecondChildTAG: If you are able to calculate the Thevenin resistance, you have the Norton resistance, and can find the Norton Current. Where exactly are you stuck? SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-08T23:39:02Z FirstChildTAG: Hang in there, folks. I'm repeating the class, even though I passed, just because I felt I missed so much the first time. This time around I'm paying more attention to the book. The development there is more gradual than in the videos - or maybe my reading speed is a better "impedance match" to my brain processing speed! :) It's good to see the encouragement of others. FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-10-08T16:17:25Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-08T18:49:02Z SecondChildTAG: I can't thank those who've already taken the course and are hanging around (or have come back) to help newbs like myself, enough! I suspect I'll be making a second pass at this, too. SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-08T23:40:44Z FirstChildTAG: Part of the problem, is that we are used to a different style than MIT's style. Most classes give you examples, show you how to do them, and then tests you on them. That's just not how this class works. They give you the basics, and you have to discover the other stuff for yourself. Nobody will lead you to it by the nose, as the saying goes. That's what makes it so hard and so rewarding. You really have to learn for yourself. It's really meant to be done in study groups, I think. Four or five at a time all around a table trying to plow through it. If you can do it with a friend or two taking the class with you, that's probably the way to do it. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-08T18:32:39Z FirstChildTAG: If it is easy to learn the course,no challenges,no difficulties,we may not find out any fun enjoying the feeling of discovering new things. If you want to finish a job,even to do one thing perfectly,try everything which can help u,make full use of materials the course has privided, like the wiki(very useful notes) ,discussion part(collaboration) and so on. Keep on ! FirstChildUserIdTAG: 361137 FirstChildUserNameTAG: Ericson FirstChildCreateTimeTAG: 2012-10-09T10:16:43Z FirstChildTAG: The most difficult part of HW4 was taking the derivative correctly to obtain the small signal model from the large signal circuit. The professor has done this numerous times in lecture. I found success in H4P3 depended on how good you are at basic algebra. Even HW5 and HW6 are pretty simple. The difficulty in these problem sets arises in solving quadratic equations correctly and once again taking derivatives correctly to get the ss circuit. It is pretty mechanical stuff. FirstChildUserIdTAG: 370247 FirstChildUserNameTAG: Dijkstra FirstChildCreateTimeTAG: 2012-10-09T22:06:53Z IndexTAG: 171 TitleTAG: The lab of week 2 Are you going to solve the lab after the due date ? If yes where can I find the solution and the explanation ? Thanks UserIdTAG: 246143 UserNameTAG: edx90 CreateTimeTAG: 2012-09-23T20:54:23Z VoteTAG: 11 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 172 TitleTAG: Lab 2 help/hints requested by DanielCuartas/sakar_abhijoy/alexha/jumana_mp/werehenry **Lab 2 Hints:** I hope that this is not late for you [DanielCuartas][1]/[sakar_abhijoy/alexha][2]/[juamana_mp][3]/[werehenry][4]. **A circuit that combines two or more signals is called a mixer. In this lab, your goal is to build a mixer that combines the signals generated by two voltage sources, V1 and V2, where:** - **V1 is a 1 kHz square wave that varies between 0V and +1V, and** If we simulate with Sandbox, we will see that this V1 source it will behaves like the following graph: ![squarewave][5] You will also observe, that the values that this wave will take it is from 0 V to 1 V (Voltage Axis) and the period of this wave will be T1=1ms , that is to say f1=1/T1=1kHz. ![pwave][6] - **V2 is a 5 kHz sine wave that varies between -1V and +1V.** If we simulate with Sandbox, we will see that this V2 source it will behaves like the following graph: ![sinwave][7] You will also observe, that the values that this wave will take it is from -1 V to +1 V (Voltage Axis) and the period of this wave will be T2=0.2ms (1ms we will have 5 sin wave with a period of T2=1ms/5=0.2ms ) , that is to say f2=1/T2=5kHz. ![psin][8] - **Please design a circuit that mixes V1 and V2 to produce Vout such that** - Vout≈(1/2)*V1+(1/6)*V2 Ok, what are they telling us with that Vout? Why we have Vout expressed in terms of V1 and V2? Let's see the following visual **HINT 1**: The ? is the resistive network that you have to find :) ![voutsch][9] In order to understand the idea, you shoud review the superposition method. If you go to the Textbook and read the page [here][10] **The Superposition Method:** **1-** For each independent source, from a subcircuit with all other independent sources set to zero. Setting a voltage source to zero implies replacing the voltage source with a short circuit, and setting a current source to zero implies replacing the current source with an open circuit. ![Supercirc][11] **2-** From each subcircuit corresponding to a given independent source, find the response to that independent source acting alone. This step results in a set of individual responses. ![supercir2][12] **3-** Obtain the total response by summing together each of the individual responses. ![vout][13] You can go to the Textbook and see an example of this superposition method [Example 3.14][14] ---- Now, a **HINT 2** is: If you have your Vout = (1/2)*V1 + (1/6)*V2 given by the statement and if with the superposition method you have your Vout=function1(R1,R2,...)*V1+function(R1,R2...)*V2... by **comparing** and **adopting** some value, (1/2)= function1(R1,R2,...) (1/6)= function2(R1,R2...) can you find your resistive values of your chosen network?? Yes!!! Now it is your turn, try to find what kind of a resistive network you need and try to apply the superposition method! :) Also, you can review this part of the textbook [voltage dividers][15] As the statement says: **The resulting output should be similar to that shown in Figure 1.** ![enter image description here][16] Remember that you **must not to** change the voltage sources and wiring parameters. Only you have to find the **? network** that **verifies your output condition** given in the statement ;). I hope this can be useful for you! Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/505e92c6db0aa01f00000033 [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/505e8e1ee89693230000002d [3]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/505def15e58fce2b0000001d [4]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/505e51868db1f3270000002b [5]: https://edxuploads.s3.amazonaws.com/13484160021343622.png [6]: https://edxuploads.s3.amazonaws.com/1348359994906241.png [7]: https://edxuploads.s3.amazonaws.com/13484161197188445.png [8]: https://edxuploads.s3.amazonaws.com/13483596197979652.png [9]: https://edxuploads.s3.amazonaws.com/13484173924318865.png [10]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/173 [11]: https://edxuploads.s3.amazonaws.com/13484182995947425.png [12]: https://edxuploads.s3.amazonaws.com/13484184615367767.png [13]: https://edxuploads.s3.amazonaws.com/13484186226517531.png [14]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/173 [15]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/99 [16]: https://www.edx.org/static/content-mit-6002x/images/circuits/Lab2B_1.55d00ab4a4f5.png [17]: https://www.edx.org/static/content-mit-6002x/images/circuits/Lab2B_1.55d00ab4a4f5.png UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2012-09-23T17:07:10Z VoteTAG: 11 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Great job! FirstChildUserIdTAG: 195218 FirstChildUserNameTAG: Al_Incognito FirstChildCreateTimeTAG: 2012-09-23T17:19:50Z SecondChildTAG: You are welcome Al_Incognito ;)! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-23T17:46:01Z SecondChildTAG: THANKS Myrimit.I can die in peace now. SecondChildUserIdTAG: 357747 SecondChildUserNameTAG: kishores SecondChildCreateTimeTAG: 2012-09-23T18:05:17Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-23T18:23:36Z SecondChildTAG: But how to get max 667 v SecondChildUserIdTAG: 162670 SecondChildUserNameTAG: charlesbabyt SecondChildCreateTimeTAG: 2012-09-23T18:37:31Z SecondChildTAG: Hi charlesbabyt, Hint: are you sure that you need that value while you design in this Lab2? ;). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-23T19:16:26Z SecondChildTAG: do we have to use more then two resistors? SecondChildUserIdTAG: 401175 SecondChildUserNameTAG: Raghav12 SecondChildCreateTimeTAG: 2012-09-23T19:43:28Z SecondChildTAG: Hi Raghav12 , Hint: mean more than 2 ;). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-23T20:34:55Z SecondChildTAG: thank you very much Myriam =) I just solve it =D! SecondChildUserIdTAG: 113110 SecondChildUserNameTAG: Menphys SecondChildCreateTimeTAG: 2012-09-23T20:44:31Z SecondChildTAG: You are welcome Menphys! ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-23T20:49:38Z SecondChildTAG: brilliant. :) SecondChildUserIdTAG: 380743 SecondChildUserNameTAG: MANQ SecondChildCreateTimeTAG: 2012-09-24T00:34:39Z SecondChildTAG: nice post Myriam , quit helpful..... SecondChildUserIdTAG: 306294 SecondChildUserNameTAG: swpnljoshi SecondChildCreateTimeTAG: 2012-09-30T07:17:30Z SecondChildTAG: You are welcome swpnljoshi :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-30T16:41:41Z IndexTAG: 173 TitleTAG: Good for deaf people Good thing in the classroom is an interpreter for deaf people. UserIdTAG: 184827 UserNameTAG: DiegoT CreateTimeTAG: 2012-09-21T19:19:35Z VoteTAG: 11 CoursewareTAG: Week 3 / MOSFET CommentableIdTAG: 6002x_mosfet_model NumberOfReplyTAG: 0 IndexTAG: 174 TitleTAG: speed change not working The S7V2 video does not change speeds. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-17T12:19:38Z VoteTAG: 11 CoursewareTAG: Week 4 / Introduction to Incremental Analysis CommentableIdTAG: 6002x_intro_to_inc_analysis NumberOfReplyTAG: 1 FirstChildTAG: It feels like watching things in slo-mo after watching everything else at 1.5x! Haha FirstChildUserIdTAG: 248902 FirstChildUserNameTAG: Chanute FirstChildCreateTimeTAG: 2012-09-17T20:42:05Z SecondChildTAG: Its working fine with me now SecondChildUserIdTAG: 251792 SecondChildUserNameTAG: Ahmux SecondChildCreateTimeTAG: 2012-10-02T09:21:06Z SecondChildTAG: I have the same problem, can anyone help? SecondChildUserIdTAG: 82183 SecondChildUserNameTAG: 0Bojan SecondChildCreateTimeTAG: 2012-10-06T22:45:37Z SecondChildTAG: not working fine still!! i need a 2x speed!! SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-12-19T17:04:58Z IndexTAG: 175 TitleTAG: Why there is no voltage drop on Rs ? Sorry if this is a trivial question, but how do i calculate Vth in this case? I saw that vth= I*Rp, but why? Thanks in advance ! UserIdTAG: 240487 UserNameTAG: AlexAlexandrescu CreateTimeTAG: 2012-09-14T20:19:17Z VoteTAG: 11 CoursewareTAG: Week 2 / Thevenin model CommentableIdTAG: 6002x_S3E5_Thevenin_Model NumberOfReplyTAG: 1 FirstChildTAG: The Thevenin voltage of a network is calculated considering an open circuit at its output. So, if you have no load on the output, there is no current and no current means no drop. The effect of Rs will be taken into account in Rth. FirstChildUserIdTAG: 396446 FirstChildUserNameTAG: RousseauxS FirstChildCreateTimeTAG: 2012-09-14T20:35:32Z SecondChildTAG: Thanks RousseauxS, but this is what's bothering me.The potential on the node +, near v, isn't the same potential from the top node of the resistor Rp. It's a small thing, but it just keeps bothering me. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-09-14T20:41:30Z SecondChildTAG: Hi Alex, I was a little confused by this myself(perhaps still am). Anyway let me try and explain what I understand about this. Firstly I think you don't realise that there is a difference between an open-circuit voltage and a closed circuit voltage. I suggest you read section 1.7 of the textbook. Now we're considering the open circuit i.e. there is no connection between the positive and negative terminals. So there is no current flowing through either of the Rs'. There is basically nowhere for the current to go. The only current flowing in that situation is through the smaller loop including the current source and the resistor Rp. The voltage across the two nodes of this loop is then the same as that across Rp. The voltage across Rp is simply V=Rp*I. The whole of the current is used, as once again, current is only flowing through the small loop, so there is no splitting of the current at the nodes. (Here my understanding gets a bit fuzzier but I'll try anyway and hopefully some others would correct me if I get something wrong.) Now that you know the voltage across the nodes of the smaller loop, we can think about what the voltages are at the ends of the circuit, i.e. at the points marked + and -. Now, as the circuit is still open, and there is still no current flowing through either of the Rs', the voltage is unaffected. So the voltage at the point marked + is the same as the voltage at the top node. Similarly the voltage at the point marked - is the same as the lower node. I.e. the voltage difference or the voltage across them is the same as the voltage across the nodes i.e. the voltage across Rp. Therefore the voltage across Rp is the required Thevenin voltage. SecondChildUserIdTAG: 434904 SecondChildUserNameTAG: mtrav SecondChildCreateTimeTAG: 2012-09-23T01:01:40Z SecondChildTAG: I understood it.Thank you ! SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-09-23T11:10:09Z IndexTAG: 176 TitleTAG: Number of Loops 1. c-v-a-b-c 2. c-b-d-c 3. b-a-d-b 4. c-v-a-d-c 5. c-b-a-d-c 6. c-v-a-b-d-c 7. c-v-a-d-b-c UserIdTAG: 16093 UserNameTAG: hossamnagy CreateTimeTAG: 2012-09-05T16:41:31Z VoteTAG: 11 CoursewareTAG: Week 1 / Circuit topology CommentableIdTAG: 6002x_S2E1_Circuit_Topology NumberOfReplyTAG: 0 IndexTAG: 177 TitleTAG: 8.02x: Electricity and Magnetism, MITx, edX (New!) Register here: [8.02x: Electricity and Magnetism, MIT, edX][1] [1]: https://www.edx.org/courses/MITx/8.02x/2013_Spring/about UserIdTAG: 402193 UserNameTAG: Feramico CreateTimeTAG: 2013-01-17T23:45:19Z VoteTAG: 10 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 6 FirstChildTAG: I have registered. I love professor Walter Lewin. He inspired me to become a good Physics teacher. FirstChildUserIdTAG: 529515 FirstChildUserNameTAG: Low FirstChildCreateTimeTAG: 2013-01-18T03:19:26Z FirstChildTAG: Like, like ! Thank you for sharing this, I didn´t knew this, but I was waiting for this XD. I will register to 8.02x . I love Electricity and Magnetism XD. I also have read a lot of good comments about Prof. Lewin! Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2013-01-18T00:33:16Z SecondChildTAG: You're welcome. SecondChildUserIdTAG: 402193 SecondChildUserNameTAG: Feramico SecondChildCreateTimeTAG: 2013-01-18T01:17:56Z FirstChildTAG: Awesome! FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2013-01-18T00:53:38Z FirstChildTAG: Dr Lewin is an excellent teacher. I have watched all his lectures in this introductory sequence at least once, some of them a number of times. In this course his, lectures on Faraday's Law and on the displacement current are especially effective at explaining some hard to grasp ideas. This course is a masterpiece of teaching performance. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2013-01-18T10:52:04Z SecondChildTAG: Will you register skyhawk? :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2013-01-18T11:32:32Z SecondChildTAG: I may register in order to get access to course materials, but I don't see any point in doing the course for a grade or certificate. I already to plan to follow four other courses this spring, two at edX and two at Coursera. The robot controller course at Coursera, if it is even half way good, will be the number one priority. That course is only seven weeks long, so I will have time to study other things. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2013-01-18T12:48:04Z SecondChildTAG: Skyhawk, is it correct "Robot controller course" ?I cant find it.. SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2013-01-18T14:00:59Z SecondChildTAG: https://www.coursera.org/#course/conrob SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2013-01-18T14:10:34Z SecondChildTAG: "Control of Mobile Robots"..Ok, thanks! SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2013-01-18T17:32:33Z SecondChildTAG: even i have registered for the same course at coursera..nice to see all of you to another great platform:-) SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2013-01-19T06:15:12Z SecondChildTAG: Coursera has announced "MOS transistors" course;As for me - I have registered to the VLSI CAD:Logic to layout and from the UDACity :Hardware verification.I do curious these two courses SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2013-01-19T09:39:33Z FirstChildTAG: prof walter is one of the inspiring physiscts to me :) FirstChildUserIdTAG: 221617 FirstChildUserNameTAG: konan FirstChildCreateTimeTAG: 2013-01-19T16:11:47Z FirstChildTAG: Thank you very much Feramico. I never saw the electromagnetism's videos of Walter Lewin, but I saw some of his mechanics's videos. Awesome, really awesome. I took several ideas from the videos and shared it with my students. It will be an excellent course!!! I'd like to be there. Luis FirstChildUserIdTAG: 227223 FirstChildUserNameTAG: luisbonelli FirstChildCreateTimeTAG: 2013-01-22T22:28:18Z SecondChildTAG: You're welcome! SecondChildUserIdTAG: 402193 SecondChildUserNameTAG: Feramico SecondChildCreateTimeTAG: 2013-01-24T17:24:59Z IndexTAG: 178 TitleTAG: SUGGESTION TO STAFF -labs in final exam! Dear Staff, I think the labs we do weekly should play a significant role in the final exam too.The final exam should have a problem set on lab too to be completed in a limited number of checks. this would enhance the efficiency of the course and make all the concepts learned through lab a bit more effective. Thanks. UserIdTAG: 368712 UserNameTAG: jmen CreateTimeTAG: 2013-01-02T16:49:22Z VoteTAG: 10 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: Hi jmen ! Almost any numerical problem can be solved in Sandbox.Even the ones from the exams. :)) FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2013-01-02T20:34:15Z SecondChildTAG: PS: Ssssstttt .. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2013-01-02T20:34:30Z SecondChildTAG: Final Exam Question 6 took me only a few minutes because of the Sandbox Circuit Simulator (It ought to have a more "official"-sounding name, shouldn't it?) Of course I know how to solve the problem mathematically, but on an exam you want to (A) get the correct answer by avoiding mistakes, and (B) save time. The pencil-and-paper method, while very instructional, is also prone to mistake-making. It's also slow if you're like me as I like to write neatly, and sketch nice plots, and use different color pencils for different voltage signals, etc. Looking Final Exam Question 6 over, it is immediately apparent that simulation is ideal: The schematic is simple, and the waveform input is a simple square wave. All that is necessary is measurement at the correct $v_{max}$ and $v_{min}$ points SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2013-01-02T23:43:00Z FirstChildTAG: Thanks for your suggestion, **Jmen**! Personally, I think that these "labs" are fun, and practical as well (e.g. "real-world" design versus theory.) Speaking of the "**real-world**": You will likely never have to analyze a *complex* circuit for node voltages and branch currents using the paper-and-pencil method in your career as an engineer or technician (except for maybe on an employment application). But you will need to know how to run a computerized SPICE simulation. Remember that theory is still *very important*, for example if you are involved in research and development, and have to create new models of transistors in the SPICE software, or experiment with parameters to come up with an optimal device, you will need to know things like $V_T$, $g_m$, and how to measure and/or calculate them (using the small-signal model, for example) because these values will not be found in a datasheet; you may be the engineer responsible for creating it! On the other hand, the ability to analyze *simplified* circuits is very useful. If you are an electrician, and the circuit beaker is rated for 15 amperes, you have to be able to quickly calculate the maximum number of electrical appliances you can plug into that circuit safely, given their power consumption in watts. Ohm's law is used all the time like this. Also remember that in the "real-world", wires have a finite DC resistance. This is based on the material (typically the purity of copper used), along with the wire's cross section (specified as wire gauge), and it's length. Of course the laws of physics can be used to determine the resistance per unit length, and then Ohm's law can be used to determine the maximum amount of current safely permitted, or the minimum wire gauge necessary, but to save time, electricians have a table where they can quickly look up such information while minimizing the math (and therefore error) involved. Remember, most electricians are paid by the job (i.e. "Can you give me an estimate of how much you will charge me to add a new circuit in my office?"), so the faster one job is finished, the faster they can move on to a new job, and the more money they can make. Jersey Mark FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2013-01-02T20:31:35Z SecondChildTAG: Aren't the majority of complex circuits designed in a modular fashion out of many smaller circuits hence making the techniques learned on 6002X invaluable? ;-) SmallBlindChris SecondChildUserIdTAG: 259719 SecondChildUserNameTAG: Smallblindchris SecondChildCreateTimeTAG: 2013-01-02T22:04:16Z FirstChildTAG: Hi Mark ! Truth been told, i am not sure that this course is intended for electricians . Perhaps you just gave a very common example , i don't know. But, many of us, i am 100% sure that will use Spice or Eagle or Multisim at some point . I am a hobbist and i am using them pretty often. When i find a new schematic, the first thing than i do is to inspect it into a simulator to see how exactly reacts to my expectations.Of course that real world components have tolerances and so on , but you can also include these into simulators as well. And i bet that i am not the only one by far. And i don't think that the simulators are only for research and development since even the people with less education in electronics use them successfully and it helps a lot. just do a search on youtube, and you will find more simulations that "electrician problems " hehehe. Jmen has a BIG POINT in my opinion, emphasizing the importance of laboratories . My previous comment was that it would be redundant, since almost every numerical problem from an exam can be simulated in Sandbox. Perhaps that's why so many people have problems with the exercises where you have to input a formula. ;) FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2013-01-02T22:28:24Z SecondChildTAG: **AlexAlexandrescu**: You're right, this course is not intended for electricians. You were also right in that I wanted to give a simple example of where the pencil-and-paper method is used more often than a computerized simulation, and I chose electricians because when they go out to a job site, they need to make a quick calculation in the absence of a computer. I also used the example because, if you remember, Homework 1 Problem 3 deals with the same exact type of math that most electricians run into on a daily basis. I will try to edit this post to clarify my point. On a related note, a good Engineer (or any professional for that matter) should broaden his or her outlook and cross-train and overlap other related fields if possible. It makes you less expendable in today's workforce. I am a hobbyist like you as well; and in university we used PSpice (now offered by OrCAD). I started off my career as an electronic tech, doing the "hands-on" work, prototyping, diagnosing, a lot of soldering and wiring, you know. However, the pay is not the greatest; and in this bad economy, electrical engineering is a tough field to compete in, especially at the entry-level as a junior engineer. So many new graduates (I am still relatively young) are competing for only a limited number of spots. Those with much, much more experience are willing to work for less and less. And companies are hesitant to invest in and train new talent when there's a downturn in demand. And there is still this stigma among professionals that blue-collar work (trade work such as Electrician, Automotive Mechanic, or Plumber) is beneath them. But the pay is surprisingly high. In my state, to get an Electrician's license, you have to go to a trade school, graduate, work for at least six years, if not more, under a licensed electrician as an apprentice (a low-paying, tough-to-get, backbreaking job), and then take the National Electric Code (NEC) exam to get that coveted license. There is a **short-cut** in the law that almost no one takes: If you have your Bachelor's degree in Electrical Engineering, pass your NEC exam, and get a licensed electrician to state you are competent, you get one as well. There is a great economic benefit in a License: You get to join a union; and hourly rates are very high compared to technicians, and can easily exceed an engineer's salary depending on your business. You get a chance at a pension, at a time when pensions are non-existent in corporate America. You can also work a regular corporate engineering job during the day, and moonlight as an electrician on the weekends. You can even start your own business - and in the U.S. this means you hire undocumented workers - or they come to you - and because they cannot legally install wiring without a Licensed Electrician's signature, your signature is worth 1000 - 1500 USD per job. Of course you have to supervise your workers and inspect their work, but most of these workers were highly skilled in their home countries, and need little supervision. I've seen, on most jobs, the Licensed Electrician show up just twice. The first time to do the estimate (this takes about an hour), and the second time to do a quick inspection (usually another hour). 1500 dollars is not bad for two hours' work. SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2013-01-03T00:28:50Z SecondChildTAG: Hi Mark ! Now i see your point, and you bring into discussion a very interesting aspect . I am from Romania, and here the things are pretty bad. I cannot find anything in electronics or automated systems control, so electronics is just a hobby for me. I make computer graphics for a living.But, some of my highschool friends are doing pretty well as electricians, wiring houses and installing heat systems . So yes, here is also more profitable to be a Electrician. Thank you for presenting so kindly your point ! SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2013-01-03T08:07:57Z FirstChildTAG: Oh such a healthy discussion!! well what i really meant was that we should have a separate qs pertaining to lab in which we r asked to solve or design a circuit, implement it using sandbox, get values of components nd the waveforms and then solve qs on them.Just like the qs we got a few weeks back in filters and the MIXER circuit. IT is similiar to a PSPICE SIMULATION which i have been frequently using in past! This will not only enhance our knowledge but help us to solve problems which engineers face in professional world..Purely hands on!! so wat say?? r u game for it? FirstChildUserIdTAG: 368712 FirstChildUserNameTAG: jmen FirstChildCreateTimeTAG: 2013-01-03T11:33:00Z IndexTAG: 179 TitleTAG: good news for you Congratulations to everyone who earned a certificate! Due to some complications surrounding Christmas and the New Year, we still need a couple of days to get the certificates signed and distributed. They will be available by Jan 3rd. We appreciate your patience. UserIdTAG: 174229 UserNameTAG: fares27 CreateTimeTAG: 2013-01-01T14:26:20Z VoteTAG: 10 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Thanks, and have a Happy New Year. FirstChildUserIdTAG: 183712 FirstChildUserNameTAG: colegiocientifico FirstChildCreateTimeTAG: 2013-01-01T14:51:23Z IndexTAG: 180 TitleTAG: Honor to whom honor due Honor to whom honor is due. It is very important all that I have learned in this course, and I am very grateful to the Dr.Anant Agarwal and his entire staff, honor to them, and I sincerely must say, this course has a greater depth than the other engineering courses I have taken in the two Costa Rica's most prestigious universities, and not to mention the excellent quality of teachers and assistants who guided us in this course and also the infinite resources of both didactic and quality and content they provided us in a way just only MIT can do it , and for that I am very proud to have passed this course, in my 52 years old, the effort that I personally I had to do was huge, I had to sacrifice my children and family on weekends to study and perform duties that I could not do during the week because of a lot of work,certainly, with other courses wich I had registered I had to put them aside to devote 100% of my effort to pass this course which I feel so proud, so I would feel very honored to hang and display in my office a certificate indicating that I got a B at MIT. I'm grateful for this free opportunity I had to be able to do so and that even with five years of my work would have not been able to accumulate the money to take MIT a course like this. And about a degree certificate, I think I deserve it because I take a course given properly structured for that, and unlike OCW models that allow learning the contents of 6002 at your own pace, 6.002x requirements are very different and must be lead subject to a very demandant schedule and evaluation. En Español Honor a quien Honor Merece Es muy importante lo que he aprendido en este curso, y estoy muy agradecido con el Dr.Anant Agarwal y a todo su staff, honor para ellos, y sinceramente lo digo, este curso tiene una profundidad más grande que los otros cursos de ingeniería que he tomado en las dos universidades más prestigiosas de Costa Rica, y ni que decir de la excelente calidad de los profesores y asistentes que nos orientaron en este curso y de los infinitos recursos tanto didacticos como de calidad y contenido que ellos nos facilitaron en una forma como solo MIT lo puede hacer, y por eso estoy muy orgulloso de haber aprobado este curso, pues a mis 52 años, el esfuerzo que yo en lo personal yo tuve que hacer fué enorme, tuve que sacrificar a mis hijos y familia los fines de semana para poder estudiar y realizar los deberes que no podía hacer entre semana por motivo de trabajo, inclusive otros cursos que había matriculado tuve que dejarlos de lado para dedicar el 100% de mi esfuerzo en aprobar este curso del cual me siento tan orgulloso, por lo cual, me sentiría muy honrado el poder colgar y exhibir en mi oficina un certificado que indique que obtuve una B en el MIT, gracias a una oportunidad gratuita que yo tuve de poder hacerlo y que ni siquiera con cinco años de mi trabajo hubiera podido acumular el dinero para tomar en MIT un curso como este. Y respecto a un grado en el certificado, creo que me lo merezco dado que tome un curso debidamente estructurado para eso, y a diferencia delos OCW que permiten aprender los contenidos del 6.002 a un ritmo que uno quiera, las exigencias del 6.002x son muy diferentes y se debe llevar sujeto a un calendario y evaluación muy exigentes. Gracias por todo y Felices Fiestas Navideñas a todos y un Próspero y Venturoso Año Nuevo UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-12-30T15:37:31Z VoteTAG: 10 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Un cordial saludo, creo que todos los que colocamos un esfuerzo de mas para realizar y aprobar este curso, tuvimos que realizar ciertos sacrificios a nivel personal y laboral, pero el hecho de pasar un curso en circuitos y electrónica del MIT es algo muy gratificante, a pesar de los muchos comentarios que hay en los demás foros donde hay gente disgustada porque en EEUU no tiene validez, como dijiste, es una gran oportunidad ya que es un curso gratuito pero donde se realizan las mismas exigencias que en los cursos pagos y más allá del hecho académico para mi es un hecho personal, gracias por sus comentarios y muchos saludos desde Colombia... A los integrantes que están criticando la respectiva certificación y demás lineamientos del curso, solo queda decirles que son muy desagradecidos y que si necesitan un curso certificado que lo paguen... Feliz Año nuevo. FirstChildUserIdTAG: 459662 FirstChildUserNameTAG: FRANCISCOG FirstChildCreateTimeTAG: 2012-12-31T15:09:17Z SecondChildTAG: FransiscoG un cordial saludo y Feliz Año Nuevo, así es no me importa si tiene vlidez o no, pero para mi será un orgullo tener colgado en mi oficina este diploma, no es cualquiera el que tiene el privilegio de aprobar 6.002x en MIT, las estadísticas del curso inaugural indican que de 155 000 personas que se inscribieron solo 3000 lo aprobaron. SecondChildUserIdTAG: 183712 SecondChildUserNameTAG: colegiocientifico SecondChildCreateTimeTAG: 2013-01-01T01:13:58Z FirstChildTAG: Un gran saludo a todos los compañeros de curso a nivel mundial y en especial a los latinoamericanos. No recuerdo haber estudiado tanto en mis años de universitario como lo hice en este curso 6.002x, pero en realidad valió la pena el esfuerzo, los conocimientos adquiridos son inmensos y que mejor regalo que un certificado del MIT, la verdad me siento bendecido por esto. Le doy muchas gracias a todo el equipo del MITx, también al Dr. Rafael Reif director del MIT por apoyar esta iniciativa, me siento orgulloso de que sea venezolano y latinoamericano. Saludos desde Venezuela FirstChildUserIdTAG: 296419 FirstChildUserNameTAG: oscman FirstChildCreateTimeTAG: 2013-01-02T15:32:47Z IndexTAG: 181 TitleTAG: Rant: An Educational Revolution! As EdX is *not-for-profit*, and the world is just beginning to see the impact of massive, open-enrollment online courses (MOOCs), we are just at the very beginning of an educational revolution! *As for me*, I foresee a day when all students, no matter their personal situation, can get an *economical university education* [***See Note 1***] (universally necessary for economic and social advancement) by watching **world-renowned professors** (Mr. Agarwal and Mr. Cima are great examples of such professors) giving *exciting* lectures online, in an environment that *immerses* the student with choices, full interactivity, and community, such as the EdX platform looks to provide. **[Note 1]** ***The higher-education "bubble" is set to burst any day:** U.S. students are crumbling under the weight of the massive loans currently needed to obtain a decent education, with little possibility for repayment in the current economic situation; Furthermore, many South Asian students feel their school systems are not preparing them properly; and students world-wide are feeling so fed-up with the current state of education that they think little of plagiarism or blatant cheating, especially when they see world and financial leaders engaging in unstoppable corruption and malfeasance.* Students should always have the convenience of submitting Homeworks and Mid-Term exams online, **and perhaps** going to a local community or technical college, or any location with the proper facilities, for the "hands-on" portion of classes, for a small fee (for example, the circuit prototyping necessary for introduction to circuits courses, or laboratory portion for chemistry courses), taught by teaching assistants to leave costs at a minimum. Extra tutoring / assistance could also be provided by graduate students (or even at this level by qualified professionals who are under-employed or in-between jobs; this would surely be positive for the job market.) The Final Exam, being cumulative and the most important - if a student wants to get **college credit** for a course - would be taken at a testing center in the student's city, or also at a local community college, or even any centralized location with an internet connection; proctored of course, to maintain the integrity of the issued credential(s). [*Note:* I believe that some type of transferable credit is necessary for MOOCs such as EdX to succeed in the long-run, especially economically.] Others just wanting to gain knowledge for knowledge's sake, or to improve their professional standing, or to add to their resume (C.V.), could opt to take the Final online and get the Honor Code certificate currently offered. Even mere *participation* in a MOOC, without even competing for a high grade or certificate, allows the student access to a global network of peers and the opportunity to learn how to function within the growing global community! With the world rapidly changing, many students feel that a traditional on-campus education is not for, or has failed, them . On a regular campus, you have to put up with **not enough seats** for a popular class, not being able to **"test drive" a course** before putting up hundreds or even thousands of dollars in tuition, registration, and the ever-increasing fees [***See Note 2***]; to getting **stuck** with a professor / class which you do not like; having to drive vast distances (and paying fuel costs) if you're a commuter student (typically treated as a second-class student) at a far-away university; to taking classes at unpopular hours (7:00AM), or hours that conflict with work / dates that conflict with religious obligations; dealing with inclement weather; not being able to work at your own pace (an impediment to education for students with learning and other disabilities), etc. etc. The list of "black marks" against "brick-and-mortar" universities goes on and on... **[Note 2]** *"Educational" institutions have been moving beyond their core focus and establishing their own fiefdoms: running their own construction companies through infrastructure fees to cover overpriced, unnecessary restaurant-style cafeterias, hotel-like dorms, mall-like stores complete with parking garages, luxury sports stadiums, etc.; and financing their own police and fire personnel; eventually controlling and overtaking whole municipalities and thus getting resentment from the local citizenry.* Plus, in many U.S. universities, the learning environment is rapidly deteriorating *(**except** at the upper-level schools like the Top 25, Ivy League, and flagship state Universities)* because, just to turn a profit or boost numbers, they readily accept unqualified, immature, and/or educationally unprepared students and then thrust them into remedial classes! Such "students" then lose confidence and treat college like an extension of a typical U.S. high school - with all the cliques and social grouping, popularity contests, and juvenile partying that goes along with that. Thus the students that are actually at university to learn, and especially those students from non-Western cultures or from a lower socioeconomic background, feel **excluded**. While traditional, small liberal-arts colleges (for their intimate atmosphere), as well as major universities (dedicated to research and promoting the arts and sciences) will always have a place, I feel that an educational revolution will see the many sub-par, for-profit colleges go out-of-business, and MOOCs can expand to provide a broad curriculum; a *good* thing as: - Finally there is an educational opportunity for the **rest of us**! Jersey Mark *(Please note that my rant and views expressed above are not approved or endorsed by EdX / MITx or their sponsors, members, staff, etc. and are entirely my own. I appreciate any comments.)* UserIdTAG: 292546 UserNameTAG: JerseyMark CreateTimeTAG: 2012-12-24T07:36:35Z VoteTAG: 10 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: Amen!!! I too hope that the day that it will be possible for everyone to educate themselves in a proper manner, is comming soon! The only reason that i hold two higher education degrees, is that in my country education is free (for the moment). Nevertheless, although i feel very lucky about that, the fact that education is so pricy in other couthries, holds me back from further educating myself. Let's hope for a brighter future!! Merry Christmas!! :) FirstChildUserIdTAG: 304587 FirstChildUserNameTAG: Vasso FirstChildCreateTimeTAG: 2012-12-24T08:20:01Z FirstChildTAG: Wow! Was that a thesis statement...hehehehe..anyways, there were good suggestions from you, JerseyMark! FirstChildUserIdTAG: 143823 FirstChildUserNameTAG: NathanNadeson FirstChildCreateTimeTAG: 2012-12-24T09:20:42Z FirstChildTAG: Wow!!!, **AMEN**, deep reasoning and highly valid arguments **JerseyMark**. I agree with you 100% , considering my home country **Uganda** in **East Africa**, the situation is actually alot much more worse. This is one of the best opportunities for students to enrich our **really low quality education** for free. Thanks to the community TAS like Myrimit, Jerseymark and most importantly to MIT, Prof. Agarwal and the entire class. **God bless Edx** FirstChildUserIdTAG: 183507 FirstChildUserNameTAG: obiradaniel FirstChildCreateTimeTAG: 2012-12-24T10:35:20Z SecondChildTAG: My sincere thanks to Skyhawk too! Somehow, I've become a fan of chainsaw too. Hats off! SecondChildUserIdTAG: 285616 SecondChildUserNameTAG: AmiyaX SecondChildCreateTimeTAG: 2012-12-24T15:12:04Z FirstChildTAG: I am an older student who studied engineering to gain a masters at a quality university in the 1980's. I did well in the final score but felt as if it would be better to repeat the course before sitting the final. I felt that I had only a superficial knowledge and wasn't up to it. I was so LUCKY to attend a live university before the costs became prohibitive and these days it would be impossible for me. My feelings are the same as the first poster. This is how the future of higher education will be for humanity. I thank those who make it possible and hold them in such high regard. We cannot afford to waste the talent that otherwise would go to waste. FirstChildUserIdTAG: 345958 FirstChildUserNameTAG: PWilson123 FirstChildCreateTimeTAG: 2012-12-24T10:47:36Z FirstChildTAG: Very relevant and valid points, Mark. Well at the same time I must admit that I was rather disappointed when a Professor misspelled 'Potassium'; frequently consulted his/her handouts even while writing equations; chalked/rubbed-off (on the blackboard) time and again etc. Being a teacher myself, I understand that these do happen occasionally, especially so when you are unprepared, but that is why we need to do 'OUR homework' first. I must say I am very impressed by the 6.002x team, their dedication, passion & knowledge; and Harvard PH 207x (a course I could not complete, due to time constraints) team. Prof. Pagano & Cook are so very amiable and informative. Will take that course again, if offered free. Learning, sans frontiers. By the way, what does EdX stand for? dX=distance unknown (radio dx-ers), e=electronic (as in e-mail); or is it Ed=education, for x population. Don't tell me it's some 'differentiation'(dx):=) Have a very happy 2013 & Merry Christmas to you all. FirstChildUserIdTAG: 285616 FirstChildUserNameTAG: AmiyaX FirstChildCreateTimeTAG: 2012-12-24T11:23:19Z SecondChildTAG: Maybe the X stands for: Xenia (Greek: ξενία, xenía) is the Greek word for the concept of hospitality, or generosity and courtesy shown to those who are far from home. It is often translated as "guest-friendship" (or "ritualized friendship") because the rituals of hospitality created and expressed a reciprocal relationship between guest and host. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-12-24T16:55:47Z SecondChildTAG: That's innovative thinking, salsero! SecondChildUserIdTAG: 285616 SecondChildUserNameTAG: AmiyaX SecondChildCreateTimeTAG: 2012-12-28T08:33:00Z IndexTAG: 182 TitleTAG: [DEADLINE EXTENDED TO JAN 14th] To those who have Finished the exam You do want a book signed by Prof. Agarwal don't you? So stop relaxing and start working on the contest! ;-) Time to put all those assignments, labs and exams to use. Here are the details again: ----- **Evidences that proves that Prof. Agarwal is awesome, cool and a big hearted person. Thank you for being like this! :)** [Read Here][1] ------ **Welcome to the CECC 2 ! Do you want to win a Textbook signed personally by Prof. Agarwal?** [Read here][2] ----- And if you think it's impossible to come up with anything have a look at what last time's contestants did. They were all 6.002x students just like you and they were all amazing! 1. [Komisz][3] 2. [Ruudoleo][4] 3. [Danik][5] Now get to work! (I wish I was eligible for this :-() ![enter image description here][6] --- [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50b2892629b7b1270000003b [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50c08a8f2196ec2300000022 [3]: http://youtu.be/tMGXvU9o3kM [4]: http://youtu.be/swFbTCG05sM [5]: http://youtu.be/eMmb6zgLpqU [6]: https://edxuploads.s3.amazonaws.com/13561862038474432.jpg UserIdTAG: 14386 UserNameTAG: ashwith CreateTimeTAG: 2012-12-22T14:23:53Z VoteTAG: 10 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: hmm..videos should be submitted until Jan, 5...It seems almost impossible FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-12-23T11:40:07Z SecondChildTAG: You have two weeks. I think it should be possible. It doesn't need to be "flashy". Just show off your ideas based on what you've learned here. One of them created a binary adder in mine craft. The adder is a very simple circuit. But the idea of coming up with it inside the game is what was awesome. The chainsaw demo is another nice example of applying simple stuff (I'm not asking you to dance though don't worry :-)) SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-12-23T15:25:03Z SecondChildTAG: I have an idea completely related to few parts of our course, but it seems to be much more nice to show working sample.Unfortunately, it is impossible because holidays and etc SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-23T20:20:51Z SecondChildTAG: Would a week's extension help? (We are considering it.) SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-12-24T03:42:56Z FirstChildTAG: I am working on a motor speed regulator (PWM) with a forward/back option with mosfet H bridge.I am almost done, and i am planing to explain how mosfet works for real, describe 555 timer, and how a cmos 4011 nand and 4013 d latch is working , because i am using them also. Unfortunately i don't have an oscilloscope or any fancy instruments beside a multimeter and two breadboards and a lot of scrap components.Am i eligible ? Thanks ! PS: Maybe if i will have enough time i will do also a PID controller with AO to explain why is an integrator or derivative so important in automatic control, but since i don't have an oscilloscope, it will be time consuming to tune it. FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-12-24T22:17:11Z SecondChildTAG: No answer ? Am i not eligible ? SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-12-28T23:18:51Z SecondChildTAG: Disregard this message.I saw the main thread. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-12-29T19:35:15Z SecondChildTAG: Hi AlexAlexandrescu, I'm really sorry I didn't see your question before. Yes the project is eligible. You can explain how the circuit works and also you should be able to show the motor speed change and that should be more than enough. One of the reasons why Myriam started this context (she did this last time as well) was to encourage future students to join 6.002x by showing what they can do after they are through with this course. Hopefully this time we'll be able to get the winner's videos on the 6.002x homepage for people to see when they sign up (no promises though). So your viewers will appreciate the motor control aspect of it more than the oscilloscope output, which would probably be quite cryptic unless they've seen this before. SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2013-01-03T04:33:43Z IndexTAG: 183 TitleTAG: MIT rocks ......Quality Education ALLAH bless u all :) AMEEN UserIdTAG: 415376 UserNameTAG: imab90 CreateTimeTAG: 2012-12-22T13:45:45Z VoteTAG: 10 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 2 FirstChildTAG: That is correct. I wish I have the opportunity to come to USA to visit Mr. Agarwal and the staff of 6.002x and to express my gratitude to them face to face FirstChildUserIdTAG: 418182 FirstChildUserNameTAG: euldji2005 FirstChildCreateTimeTAG: 2012-12-22T17:25:49Z SecondChildTAG: Cool euldji2005 !:) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-22T20:03:25Z SecondChildTAG: Give my best wishes to Sir Anand Agarwal, when you visit him, euldji2005! Many thanks in advance! SecondChildUserIdTAG: 455950 SecondChildUserNameTAG: Wahabbaluch SecondChildCreateTimeTAG: 2012-12-23T05:17:43Z SecondChildTAG: Thank you Dr. Agarwal for sharing your knowledge and time for us through EDX....u simply Rock!!!! SecondChildUserIdTAG: 375464 SecondChildUserNameTAG: mohdnits SecondChildCreateTimeTAG: 2012-12-23T09:11:21Z FirstChildTAG: I think this little community we have built here could serve as an example to all. People working hand-in-hand, helping each other, encouraging each other, regardless of whether our respective governments are at odds... We'll never get rid of governments, and I'm not advocating that we should. But take this experience as proof we can get along and work side by side as *people with common desires and interests*, not just as citizens of countries that may be in conflict. FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-12-23T11:38:48Z IndexTAG: 184 TitleTAG: Amazing experience with MITx 6.002x:Circuit &Electronics I never enjoy this kind of efforts form technical side(website maintainance & lecture video upload) as well as Helping desk(TA). Thank you very much to MITx for providing me a platform to explore my concepts in electronics community & to improve it. Heartly thankful to #Prof.Anant Agrawal & his team for this collabrative outstanding course. Many many congratulation for themExpecting more & more good courses from team in Future. UserIdTAG: 136634 UserNameTAG: akash96 CreateTimeTAG: 2012-12-21T03:47:53Z VoteTAG: 10 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: SAme here... Prof. Agrawal never let us loose interest in the topic... He presents each and every topic in a very interesting and lively way....It feels as if we are sitting in the class, talking to him directly.. and his INDIAN accent really helps :-) FirstChildUserIdTAG: 126705 FirstChildUserNameTAG: arpitchugh FirstChildCreateTimeTAG: 2012-12-21T07:37:50Z FirstChildTAG: I agree !!! Thank you so much for all, and i hope one day to meet and take a coffee all the classroom and Prof.s together ^__^ FirstChildUserIdTAG: 200496 FirstChildUserNameTAG: Ranieri FirstChildCreateTimeTAG: 2012-12-21T09:26:40Z IndexTAG: 185 TitleTAG: Thank you ALL the guys ..who are part of this team 6.002x... Thank you all of you .... pretty well organized course.. UserIdTAG: 209930 UserNameTAG: saikiraniitr CreateTimeTAG: 2012-12-20T18:17:47Z VoteTAG: 10 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Yes,the best of the best online course! FirstChildUserIdTAG: 230076 FirstChildUserNameTAG: ANANDAM FirstChildCreateTimeTAG: 2012-12-20T18:37:07Z FirstChildTAG: I'll just add on here rather than start another post. Thank you to Prof. Agarwal and the other professors, all of the TA's and staff who kept this site going along, and especially to all of the other students who were also a wealth of information and assistance. I actually really enjoyed this course! FirstChildUserIdTAG: 334855 FirstChildUserNameTAG: JimMonty FirstChildCreateTimeTAG: 2012-12-21T00:03:06Z IndexTAG: 186 TitleTAG: H12P2-MYRMIT'S HELP REQUIRED Hi Myrimit, Well, we may not have communicated earlier in the course. but going by the way you help people out here and due to your highly appreciable efforts to present hints to the problem, I am really impressed and hence look forward to help from you. The course has been running fine, could clear through it but now I am stuck with this problem H12P2. Can you please provide a step by step approach to solve this problem as you have done in previous questions?? Moreover, I have yet not cleared the 60% mark since I had my regular examinations at college but I am pretty close.I would be grateful if you could help me at the earliest. Regards Ur well wisher jmen UserIdTAG: 368712 UserNameTAG: jmen CreateTimeTAG: 2012-11-30T18:40:06Z VoteTAG: 10 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi jmen, Ok, I will try to post hints of Hw12 ;) Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-01T19:25:13Z SecondChildTAG: thank you.. SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-12-03T07:09:00Z SecondChildTAG: Sorry for the delay, here they are https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50c3c129468bdc270000001c SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-09T04:10:11Z IndexTAG: 187 TitleTAG: Lab8 Hints Requested by mkprasanth I will try to translate this to spanish too ... **LAB 8** Here there are some Hints of Lab8, so that later you can solve the problem by yourself: In this lab, they give you a circuit: ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13361373838628283.png) You have two devices: - Device 1 - Device 2 This devices can be : - Device 1: Resistor R ; Device 2: Capacitor C; - Device 1: Capacitor C; Device 2: Inductor L; - Device 1: Inductor L; Device 2: Inductor L; - etc..., try to combine the possibilities. If you notice, you have a current probe and a voltage probe: ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13361382628274182.png) Look at that probes and where are measuring the current (red color) and the voltage (blue color). So, the red color plot will show the variation of the current in the two elements (bothe devices are in serial , so the share the same current). So, the blue color plot will show the variation of the voltage in the second device (look where this blue probe is put). Now, that you understand what current or voltage we see in the plot. Just try to discover wich elements could be in device 1 and 2, to have a plot in each situation: ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13361389585298184.png) You can get a help with the sandbox! Try to make the next circuit and explore! Make the circuit, and try to combine elements and see the plots and compare with the plots that they give you. Use a TRAN time 5n ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13361394431845809.png) Once you have the topology and you know the elements (eg., device 1 = R, device 2= L). You have to solve it as you have been doing till now (using kirchhoff law, charging/descharging capacitor, etc..). But, the question is: how ? how can i find the values???? Well, you have to analyze the values of the plot. I will give you an invented example. So that, you can do by yourself. Look at the green circle and light blue circle. You will have two datas ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13361412252795679.png) Look thee blue plot. The only alternative is to be a resistor , because you have a constant voltage R (V=R*i). Now, what is the value of that resistor?. Use kichhoff R= (Vsource-V)/i if you see Vsource=1V (step 1V). V=900mV (constant value all over the time) i=500mA (constant value all over the time) So, you will get a R= 0.2 Ohm **Now, it is your turn, try to use all the tools** (sandbox, theory [Go to chapter 10 and take a look at the graphs and compare it with your graph ;) ][1], graphical analysis, etc)! ** in order to solve it by your own!** I hope this can help you. Myriam. ---- P.D. The dashed lines means that whe are meassuring in a time t . And where the dashed lines cut the curve is the related voltage of that time. That value of voltage is displayed in the left top of the plot. ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13362467779620418.png) ---- [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/504 UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2012-11-07T01:57:49Z VoteTAG: 10 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: Myrimit, muchas gracias por la explicación FirstChildUserIdTAG: 175734 FirstChildUserNameTAG: hestrada FirstChildCreateTimeTAG: 2012-11-07T12:40:11Z SecondChildTAG: Por nada :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-07T17:33:43Z FirstChildTAG: Myrimit, thanks! it helps a lot!... FirstChildUserIdTAG: 349139 FirstChildUserNameTAG: 1977ROYELMER FirstChildCreateTimeTAG: 2012-11-09T06:30:00Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-09T13:01:05Z FirstChildTAG: Shouldn't one of them be a resistor, so the case L - L is excluded? Thanks! FirstChildUserIdTAG: 456761 FirstChildUserNameTAG: marian1 FirstChildCreateTimeTAG: 2012-11-11T16:22:35Z FirstChildTAG: really....wonderful explanation......thank you for such nice explanation.,.,.,. FirstChildUserIdTAG: 316761 FirstChildUserNameTAG: gk_goel FirstChildCreateTimeTAG: 2012-11-11T10:13:58Z SecondChildTAG: You are welcome gk_goel ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-11T23:19:41Z FirstChildTAG: it was expected to have some tips about C and L. dont know why? you didnt give any explanation about finding the value of C and L what you provided: finding R was the easiest part. it would be appreciated if the work is helpful as a whole. otherwise it was very goood. FirstChildUserIdTAG: 154172 FirstChildUserNameTAG: mofassair FirstChildCreateTimeTAG: 2012-11-12T11:15:16Z IndexTAG: 188 TitleTAG: Definite Integral Calculus with Wolfram Alpha: A guide (as encountered in LAB 7) Wolfram Alpha, for me, is a life saver. Of course, it is imperative that we learn the fundamentals that underpin the operations preformed by Wolfram Alpha but still Wolfram is a powerful tool. As with all tools, we need to know how to use it. The following is a little guide on how to preform a definite integration with wolfram alpha (LAB-7). The first time we are asked to preform an integration in 6.002x is (if my memory serves me correctly) in LAB-7. ![enter image description here][1] This blew my mind. So let's dive right in. First open Wolfram Alpha: http://www.wolframalpha.com/ Go to examples and select 'Calculus' under the mathematics subheading. ![enter image description here][2] When in the calculus menu, choose the correct calculator to calculate a definite integral... ![enter image description here][3] Now how do we parse this equation? Remember that **i(t) = sin(1000*2*PI*t)** Also **t=<0** So we simply write: **1/c integrate sin(1000*2*pi*t) from t=0 to t** So 1/c is not to be integrated (as it is to the left of the integral symbol), therefore we put it **before** we write 'integrate' in Wolfram. Or, generally: **xx integrate xxxx from x=x to x** It really is that simple! I hope it helps you like it helped me. And any questions just ask!! [1]: https://edxuploads.s3.amazonaws.com/13514599001343655.bmp [2]: https://edxuploads.s3.amazonaws.com/13514600815763598.png [3]: https://edxuploads.s3.amazonaws.com/13514601652596587.png UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-10-28T21:49:32Z VoteTAG: 10 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Great post, Hazel1919. Thanks a lot for this. Jordi FirstChildUserIdTAG: 366320 FirstChildUserNameTAG: Galli FirstChildCreateTimeTAG: 2012-10-28T22:22:41Z SecondChildTAG: In Matlab: syms t x v=1000*int(sin(2000*pi*x),x,0,t) SecondChildUserIdTAG: 475577 SecondChildUserNameTAG: kibens SecondChildCreateTimeTAG: 2012-10-29T06:43:23Z SecondChildTAG: but i got this 'Invalid input: integrate from to not permitted in answer' when i wrote in the formula? for lab 7 ofcourse. pls help SecondChildUserIdTAG: 285945 SecondChildUserNameTAG: lindalapiso SecondChildCreateTimeTAG: 2012-10-30T07:13:18Z SecondChildTAG: Thanks. SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-08T09:23:40Z FirstChildTAG: I signed up for Wolfram Alpha and I entered the equation and here is what I got: ![enter image description here][1] which is totally wrong. I did it by hand using u substitution and here is what I got: ![enter image description here][2] Am I doing something wrong here with Wolfram? By looking at there result it doesn't seem so. If it turns out something is wrong with Wolfram, a lot of people will be disappointed. What do you think hazel1919? [1]: https://edxuploads.s3.amazonaws.com/13514800701343654.jpg [2]: https://edxuploads.s3.amazonaws.com/13514806601343634.jpg FirstChildUserIdTAG: 132828 FirstChildUserNameTAG: rharris FirstChildCreateTimeTAG: 2012-10-29T03:27:09Z SecondChildTAG: You confused dx and dt while typing in Wolfram Alpha. SecondChildUserIdTAG: 405936 SecondChildUserNameTAG: Whible SecondChildCreateTimeTAG: 2012-10-29T05:36:48Z SecondChildTAG: Please excuse my bad grammar. SecondChildUserIdTAG: 405936 SecondChildUserNameTAG: Whible SecondChildCreateTimeTAG: 2012-10-29T05:42:18Z SecondChildTAG: Hi Rharris, it's simple where you went wrong. You need to write exactly: **1/c integrate sin(1000*2*pi*t) from t=0 to t** You wrote: **integrate sine(1000(2)(pi)(t))dx from x=0 to pi** First off notice that you forgot to put 1/c before the **word** integrate in the wolfram bar, if you don't do that you will not get a complete equation. Secondly, we want to measure the integral from "t=0" to "t" not x=0 to PI, as this will give a **completely** wrong result. Finally, you don't have to worry about putting in 'dx' as the wolfram software can do this for you! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-29T08:03:44Z SecondChildTAG: Hi hazel1919 I was not confused about using dx, I have been doing math and calculus for so long, x (and also y) is imprinted on my brain, if you know what I mean. I didn't even notice it. Also, if you look at the result, it is correct because sin(1000*2*pi*t) is a constant with respect to x. My view of Wolfram was tainted originally, because I did actually find a bug with a test function I entered, the result of sqrt(3*x^2) = 3*x when it should have been = to x*sqrt(3). I reported it to their technical support team. Regarding inputting the equation into Wolfram, you actually don't need to use "*" between variables when multiplying, with the exception of between the 1000 and the 2 (1000*2). If you notice, Wolfram parsed my equation correctly, even with the parentheses and without "*" for multiplication. I agree, I did forget 1/c which would change the result by a constant, however I was focusing on the integral of sin t = sin t, not realizing I had put dx in the equation. Be careful in leaving dx out of the equation and letting Wolfram determine which variable to integrate with respect to. It would have worked in this case, but what happens when there is more than one variable in the equation (like in partial differentiation or partial differential equations, and I'm pretty sure there is partial integration, but I don't remember doing any), such as what I did (t and dx), you get totally different results. For example, what if I actually wanted to integrate a sin t with respect to x, leaving out the dx would cause problems. One last thing, regarding the limits of the integration, in my manual method I should have put the integral from t=0 to t=pi, since the variable in the integral was u, then after replacing the u with the original value it would be from 0 to pi without specifying t. I haven't looked at week 7 yet, so I arbitrarily picked pi as the upper limit of the integral for sake of the discussion. This was a good discussion, thank you. As always it strengthens our knowledge of the subject, if not refreshes our memories from our school days. I feel bad for someone without a math background who has to use Wolfram, as I can imagine it can get pretty confusing. Hay whible - Don't worry about your grammar, you are doing fine. Mine is not any better. We're just thankful you are available to help. - Thanks. I think MIT should make you both Community TA's, you are good at it. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-29T21:27:13Z SecondChildTAG: Haha! Thanks, I appreciate that! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-08T09:24:02Z FirstChildTAG: Thank you hazel1919! Also you can take a look at an arbitrary example here [Lab7 Hints][1]. You can calculate the integral by the sustitution method ;). Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5091dd5d3fc09f2b0000018a FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-01T02:44:49Z SecondChildTAG: Thanks Myrimit! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-08T09:24:39Z SecondChildTAG: Myriam's explanation was lucid & helpful (intuitive:=) SecondChildUserIdTAG: 285616 SecondChildUserNameTAG: AmiyaX SecondChildCreateTimeTAG: 2012-11-08T17:26:36Z IndexTAG: 189 TitleTAG: 6.002x use in hobbyist electronics Hi, First of all I would like to thank MIT for providing such a fantastic opportunity for students like me to learn so much about circuits and electronics. However as a hobbyist I was wondering whether this degree of complexity is required to make circuits that do cool things. I am not denying that this course is great for providing the jump from physics to electronic engineering, however I would be grateful if somebody could tell me whether this degree of complexity is required when designing hobbyist electronics such as those that may be used in robotics? Thank You in advance UserIdTAG: 85490 UserNameTAG: yashsavani CreateTimeTAG: 2012-10-27T09:30:10Z VoteTAG: 10 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Yes, this course will be is very empowering to the robotic hobbyist. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-27T11:27:00Z SecondChildTAG: As a robotics enthusiast, the important aspects are the protocols, timing, logic and the correct use of ICs to do what you want. Though this course provides the fundamentals behind the discrete electronics how exactly would it help demystify the complexities of such aspects of robotics? SecondChildUserIdTAG: 85490 SecondChildUserNameTAG: yashsavani SecondChildCreateTimeTAG: 2012-10-27T11:49:17Z SecondChildTAG: As you have mentioned, this course helps to understand the *fundamentals* of analog and digital circuits. You will definitely need pieces of pure analog circuitry in robotics while working, for example, with various analog sensors, and that's when it comes to a number of ways of signal amplification, noise filtering and so on. And I think that the role of digital circuitry in robotics is obvious. And it's always easier to understand the specifics of use of, for example, complex IC, when you know various methods of circuit analysis and can apply them freely to the introduced in datasheet sample circuit, then just stare at the datasheet making some assumptions based on the pure intuition. SecondChildUserIdTAG: 192703 SecondChildUserNameTAG: voffch SecondChildCreateTimeTAG: 2012-10-27T12:48:46Z SecondChildTAG: As you know, ICs (Integrated Circuits) are really just devices with multiple discrete transistors, amongst other things. Knowing how to diagnose, analyze and implement discrete devices is the foundation for building more complex devices. So far we have learned about logic, MOSfets, resistors timing and to some degree protocol, depending on your definition. Now with just some of these mentioned things, you should be able to design and build your own Digital IC. It will not be very small. You could even replace the MOSfets with mechanical relays. To control this device you will need the device's VIH and VIL to be compatible with the VOH and VOL that is fed from the device you are installing it in. Now you can do that. In week six we are getting into capacitors, specifically charge times and oscillators. You will also find more uses for capacitors. They are useful in removing unwanted noise from an analog AC signal, say to prevent interference from your local radio station. They can also be used to block DC signals, only passing AC. In either case the math is useful in determining the appropriate size for the capacitor. You may want to communicate with your device using some kind of primitive radio. Well with just a simple LC circuit, (Inductor-Capacitor), you are able to build a useful tank.(Resonator). The math will be useful in having the signal oscillate at the desired frequency. Or maybe this tank will be useful somewhere else? I could go on and on. The main thing is you will be a more knowledgeable hobbyist when you are done. Even if you only use some of the tools. As the Professor mentioned early on in the course, experience will show you where to use specific tools. I can see most of these topics extremely relevant in your specific hobby, especially when you want to move beyond plug and play. Have fun. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-27T12:52:32Z SecondChildTAG: Thank you, I always thought that this course would be extremely helpful in my EECS endeavors. However some of my friends disagreed, saying that a comprehensive understanding of the discrete workings of the electronics in robotics was unnecessary. Now I can better explain the course's relevance and hopefully persuade them to join the Spring version of the course. SecondChildUserIdTAG: 85490 SecondChildUserNameTAG: yashsavani SecondChildCreateTimeTAG: 2012-10-27T13:56:28Z SecondChildTAG: In order to assess as to whether your high level abstractions are reasonable you need to be familiar with the primitives. It is all in the text book. Read and savor yourself. SecondChildUserIdTAG: 623210 SecondChildUserNameTAG: preveen SecondChildCreateTimeTAG: 2012-10-27T16:09:14Z SecondChildTAG: Dear Penny Packer Sir , I am agreeing most of the case I was stuck up with new devices especially when trouble shouting my own design or I want to designs an new circuit most of the time the semiconductors get heat or it will blown of then blindly changing the value of biasing resistors etc. but it consume more time and money , now at lest I can save my time and money any was I like this course I working too.. I want to attend one more time this course to master in this subject Thanks MK.Prasanth SecondChildUserIdTAG: 156694 SecondChildUserNameTAG: mkprasanth SecondChildCreateTimeTAG: 2012-10-29T10:55:28Z IndexTAG: 190 TitleTAG: PLEASE STOP SAYING YOUR SCORES Please, but please stop saying your scores. 1) Congratulations to us all ! 2) Nobody cares. PS: 2) Except your family.Go crazy on your uncle. UserIdTAG: 240487 UserNameTAG: AlexAlexandrescu CreateTimeTAG: 2012-10-25T18:44:43Z VoteTAG: 10 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: LOL xD FirstChildUserIdTAG: 454609 FirstChildUserNameTAG: Alwahsh FirstChildCreateTimeTAG: 2012-10-25T19:40:21Z FirstChildTAG: Per Honor Code and from what I've seen Staff say, there should be no talk of the Mid-Term on the Discussion board **until Monday**, and that includes grades, questions you thought were difficult, etc. FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-10-25T22:15:00Z SecondChildTAG: "Please stop saying your scores" breaks somehow the honor code? I don't get it Mark. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-10-26T07:56:19Z IndexTAG: 191 TitleTAG: Such a great experiment! It's really amazing and unexpected. It is extremely interesting to study on both practical and very interesting objects. Thank you a lot, 6.002x staff! You never cease to surprise again and again. UserIdTAG: 377277 UserNameTAG: Peragian CreateTimeTAG: 2012-10-12T11:35:14Z VoteTAG: 10 CoursewareTAG: Week 5 / Music Amplifier Demonstration CommentableIdTAG: 6002x_snd_d NumberOfReplyTAG: 0 IndexTAG: 192 TitleTAG: Too late Caution in this lecture is somewhat late -- VGS wasn't equal to VI in the previous HW, so by now most of the people probably know this, and some of them learned this in a bit hard way while getting constantly incorrect results. So might make sense to move this lecture earlier in the future. UserIdTAG: 189680 UserNameTAG: vnd CreateTimeTAG: 2012-10-11T07:07:30Z VoteTAG: 10 CoursewareTAG: Week 6 / Perspective on the small-signal circuit CommentableIdTAG: 6002x_small_signal_perspective NumberOfReplyTAG: 2 FirstChildTAG: yeah, spent 3 days last week stuck on the homework, only now seeing that there is a whole section explaining the topic a week after the homework is given. beating my head against a wall. with no relevant tutorials or support, I cancelled a weekend trip to focus on it and still ended up nowhere. My morale of beneath the floor because of questions appearing well ahead of the learning curve in this course :( FirstChildUserIdTAG: 61469 FirstChildUserNameTAG: Spacedog FirstChildCreateTimeTAG: 2012-10-21T09:40:41Z FirstChildTAG: oh yea I've spent so much time figuring this out last hw :S FirstChildUserIdTAG: 270284 FirstChildUserNameTAG: nkukushkin FirstChildCreateTimeTAG: 2012-10-21T15:39:38Z IndexTAG: 193 TitleTAG: Wolfram Alpha There is a wonderful online website, http://www.wolframalpha.com, that can be used to solve equations (and in some cases, show the steps). Type something like "**solve** x + e^x + 1 = 0" and it will spit out the answer with some elegant graphs and info regarding the question! UserIdTAG: 334183 UserNameTAG: stonefruit CreateTimeTAG: 2012-10-07T05:06:54Z VoteTAG: 10 CoursewareTAG: Week 4 / Linearization Exercise 1 CommentableIdTAG: 6002x_linearization_ex_1 NumberOfReplyTAG: 5 FirstChildTAG: Thanks. FirstChildUserIdTAG: 158740 FirstChildUserNameTAG: Ili FirstChildCreateTimeTAG: 2012-10-07T06:00:12Z FirstChildTAG: Otherwise, you could use the convergence method as suggested. FirstChildUserIdTAG: 334183 FirstChildUserNameTAG: stonefruit FirstChildCreateTimeTAG: 2012-10-07T05:07:53Z FirstChildTAG: Yes, that's a brilliant site. Very comfortable and useful. FirstChildUserIdTAG: 326787 FirstChildUserNameTAG: Angstrem FirstChildCreateTimeTAG: 2012-10-07T08:27:02Z FirstChildTAG: I am not getting the exact graph in lab 4![enter image description here][1]..please help me [1]: https://edxuploads.s3.amazonaws.com/13496003901343602.jpg FirstChildUserIdTAG: 410204 FirstChildUserNameTAG: hkaushalya FirstChildCreateTimeTAG: 2012-10-07T09:00:06Z FirstChildTAG: also you can just google it :) http://www.google.com/search?hl=en&q=x%2Be%5Ex%2B1 FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-07T09:00:10Z SecondChildTAG: Great!:) SecondChildUserIdTAG: 252722 SecondChildUserNameTAG: vaboro SecondChildCreateTimeTAG: 2012-10-08T00:16:34Z IndexTAG: 194 TitleTAG: S9E2 - Threshold Voltage not explicitly stated I had a difficult time entering in this answer because I typed in VTH for the threshold voltage variable. It is not stated, but it happens to be VT in this problem. UserIdTAG: 274263 UserNameTAG: CoreyO CreateTimeTAG: 2012-09-26T00:54:22Z VoteTAG: 10 CoursewareTAG: Week 5 / MOSFET Amplifier Exercise 1 CommentableIdTAG: 6002x_mosfet_amp_e1 NumberOfReplyTAG: 1 FirstChildTAG: This is a very easy problem, but the wording made it seem tougher than it was. As you point out VT was not explicitly pointed out as a variable we could use. I thought there might be a clever way to remove it from the equation. The first time I tried to solve the first part, I did it terms of vOUT. I was told I was not supposed to use vOUT. This made me assume that I also was not supposed to use vIN (lower case v being dependent variable, upper case independent variable). So I tried to solve the second part without using vIN, and tried to solve in terms of VS and RL. I don't think this can be done without bringing K into the equation, which is not an allowed variable. Incredibly easy problem, but poses as much tougher one due to ambiguity. FirstChildUserIdTAG: 174100 FirstChildUserNameTAG: markpolak FirstChildCreateTimeTAG: 2012-10-01T04:34:47Z IndexTAG: 195 TitleTAG: This is creative Practical learnig! This is how every student want to learn and enjoy every subject.So it keeps us connected. Thanks Prof. Agarwal and the edX team for giving an AMAZING fun learning experience. UserIdTAG: 281159 UserNameTAG: ManasiS CreateTimeTAG: 2012-09-25T19:19:56Z VoteTAG: 10 CoursewareTAG: Week 2 / Chain saw CommentableIdTAG: 6002x_chain_saw NumberOfReplyTAG: 1 FirstChildTAG: Thank you for that it made my day!!!! God how i wish I had teachers like that. Now I do thanks to Prof. Agarwal. FirstChildUserIdTAG: 396673 FirstChildUserNameTAG: anthonypraveen FirstChildCreateTimeTAG: 2012-09-26T07:54:32Z SecondChildTAG: I agree with you anthonypraveen, Dr. Agarwal is the best. Thank you Dr. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-09-30T04:45:43Z SecondChildTAG: yes guys u r right .. reminded me of proff . walter lewin ... really awesome SecondChildUserIdTAG: 242267 SecondChildUserNameTAG: MOHITVASHISHTA SecondChildCreateTimeTAG: 2012-10-05T18:13:16Z IndexTAG: 196 TitleTAG: Official answers to Homework and Labs after deadline? Hi, After a homework deadline, will 'official' answers with explanations be posted? I believe this would be very beneficial. I have already seen several posts on the discussion forum asking for explanations to various problems and labs. (In particular the Week2 lab) I believe an official answer from the course providers should be the primary source to explain how a particular question or lab should have been approached and solved. It would also minimize the requests for explanations in the forum. Many thanks. M. UserIdTAG: 385677 UserNameTAG: Michaelc1 CreateTimeTAG: 2012-09-24T00:02:58Z VoteTAG: 10 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi In addition to answers to the homework and labs, it would also be very helpful to have the workings to the questions within the lectures on a weekly basis. I know the answers are given, but sometimes it is really not obvious how to get them. The workings were given to all of these in the first edition of the course! Many thanks Melbur FirstChildUserIdTAG: 95673 FirstChildUserNameTAG: melbur FirstChildCreateTimeTAG: 2012-09-27T10:57:42Z IndexTAG: 197 TitleTAG: Software for designing logic circuits For those of you who don't know it, there's an excellent open source software called [Logisim][1] for designing logic circuits. It's a wonderful tool for experimenting with logic gates and boolean logic applied to circuit design. [1]: http://ozark.hendrix.edu/~burch/logisim/ UserIdTAG: 185338 UserNameTAG: mgaldieri CreateTimeTAG: 2012-09-21T17:48:33Z VoteTAG: 10 CoursewareTAG: Week 2 / Boolean functions CommentableIdTAG: 6002x_S4E2_Boolean_Functions NumberOfReplyTAG: 1 FirstChildTAG: any free software that runs verilog? FirstChildUserIdTAG: 173147 FirstChildUserNameTAG: cruiser_rahit FirstChildCreateTimeTAG: 2012-09-21T17:57:34Z SecondChildTAG: You can simulate Verilog with the free Icarus or Verilator simulators (and for a subset also synthesize it). SecondChildUserIdTAG: 326664 SecondChildUserNameTAG: Sebastian1 SecondChildCreateTimeTAG: 2012-09-22T15:57:30Z IndexTAG: 198 TitleTAG: Hints on y1 and y2 Hi everyone, I noticed a lot of confusion in calculating y1 and y2. Here's a repost of some hints I gave earlier as a reply to one of the questions below. You should be taking the following steps: - Set $V_2=0$, leaving only $V_1\neq 0$. - Solve for the current into $R_1$, call this value $y_1$. - Set $V_1=0$, leaving only $V_2 \neq 0V$. - Solve for the current into $R_1$, call this value $y_2$. - Sum $y_1$ and $y_2$ to get $i_1$, the total current through resistor 1 due to both sources. UserIdTAG: 151427 UserNameTAG: RichardZ CreateTimeTAG: 2012-09-18T19:51:31Z VoteTAG: 10 CoursewareTAG: Week 2 / Superposition example CommentableIdTAG: 6002x_S3E3_superposition NumberOfReplyTAG: 3 FirstChildTAG: thanks for the hint. I thought that y2 was the i2 current with only using the V2. FirstChildUserIdTAG: 433574 FirstChildUserNameTAG: endika86 FirstChildCreateTimeTAG: 2012-09-26T15:26:06Z FirstChildTAG: Thanks FirstChildUserIdTAG: 413784 FirstChildUserNameTAG: Dorotka FirstChildCreateTimeTAG: 2012-09-20T13:27:26Z SecondChildTAG: Hi I can't figure out how to solve the current into R1 ? What formula use ? SecondChildUserIdTAG: 413714 SecondChildUserNameTAG: Banszki_Istvan SecondChildCreateTimeTAG: 2012-09-21T17:59:15Z FirstChildTAG: Well, I suppose you all were here many weeks ago but if someone is still around here and around solving y1 and y2 I would like to ask if you can draw a sketch highlighting the differences between solving for x1, x2 and y1,y2. I had no problems to solve x1 and x2 but I have failed to solve for y1,y2. I remember that under superposition if you want to find one voltage you set up the other one to zero and for current you need to open the circuit and here is where I can not visualise this properly. Thank you if anyone is passing by with a hint. FirstChildUserIdTAG: 134637 FirstChildUserNameTAG: runsbarefeet FirstChildCreateTimeTAG: 2012-11-26T00:53:30Z IndexTAG: 199 TitleTAG: Typo @ 6:11? "if VS is 10 volts, then VS divided by 11 is 0.45" Should this be 0.909 because 10/11 = 0.909 or am I missing something here? UserIdTAG: 86986 UserNameTAG: wwluo CreateTimeTAG: 2012-09-18T18:58:07Z VoteTAG: 10 CoursewareTAG: Week 3 / Switch Resistor Model of a MOSFET CommentableIdTAG: 6002x_switch_resistor_model NumberOfReplyTAG: 1 FirstChildTAG: the Vs value is 5 here...by mistake,he said as 10 FirstChildUserIdTAG: 118611 FirstChildUserNameTAG: mitianhari FirstChildCreateTimeTAG: 2012-09-21T16:49:09Z SecondChildTAG: that's true by mistake he said 10 rather than 5 SecondChildUserIdTAG: 341818 SecondChildUserNameTAG: Elshazly SecondChildCreateTimeTAG: 2012-09-23T00:17:14Z SecondChildTAG: aha, there is the lamb's mother SecondChildUserIdTAG: 149549 SecondChildUserNameTAG: Java_Dido SecondChildCreateTimeTAG: 2012-09-25T20:20:48Z SecondChildTAG: it should be (5+1)/11, so it should be 0.55? SecondChildUserIdTAG: 311516 SecondChildUserNameTAG: queiss89 SecondChildCreateTimeTAG: 2012-09-25T23:27:15Z SecondChildTAG: never mind SecondChildUserIdTAG: 311516 SecondChildUserNameTAG: queiss89 SecondChildCreateTimeTAG: 2012-09-25T23:27:58Z SecondChildTAG: it is (5*1)/(10+1), that's why it is 0.45v SecondChildUserIdTAG: 148389 SecondChildUserNameTAG: chento SecondChildCreateTimeTAG: 2012-09-26T16:45:43Z SecondChildTAG: better chento SecondChildUserIdTAG: 336466 SecondChildUserNameTAG: kenstan SecondChildCreateTimeTAG: 2012-09-27T06:02:17Z IndexTAG: 200 TitleTAG: H4P2 zener regulator For some reason I can't seem to get the values of vo for when the Rl is 2k and 4k in the 2nd part of H4P2. This is the one where the zener is in parallel with the load resistance and they give you a characteristic graph. Can anyone give a hint for how to find vo? I figured it would just be 2/3 of the AC voltage and the zener will always act in breakdown since its DC bias will always be above 9.5*(2/3) UserIdTAG: 272853 UserNameTAG: nirvanaguy CreateTimeTAG: 2012-09-18T13:03:50Z VoteTAG: 10 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 8 FirstChildTAG: Have you notice the hint in the problem itself? You should model the zener as a combination of an independent voltage source an a resistor, with values depending on the operation region. Replacing the zener with the appropriate model should allow you to calculate the answer by simple circuit techniques. FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-18T14:22:12Z SecondChildTAG: So I replaced diode with DC source and resistor. Then I counted V0 and v0. For source 12V I got good answer. But when I get v0 with 2 different methods I got similar values but they are wrong. So to get v0 I counted V on this resistor with Vi=12.02 and then I subtract voltage on resistor for source 12.02V and voltage for 12V. What am I doing wrong? SecondChildUserIdTAG: 302188 SecondChildUserNameTAG: Pepek SecondChildCreateTimeTAG: 2012-09-19T09:18:05Z SecondChildTAG: To compute v0 I have substituted the small signal equivalent of the DC source and resistor and got the correct answer. SecondChildUserIdTAG: 375382 SecondChildUserNameTAG: apatriarca SecondChildCreateTimeTAG: 2012-09-22T00:31:30Z SecondChildTAG: Anyone can help I CAN CALCULATE v0 already tried vi RELATIONSHIP WITH RESISTANCE AND ME BUT NO THANKS gives the desired result SecondChildUserIdTAG: 320715 SecondChildUserNameTAG: maraivette SecondChildCreateTimeTAG: 2012-10-01T15:08:16Z FirstChildTAG: I have the same problem. I get wrong values for v0 in the second part of the exercise (with the zener regulator). I have proved to solve the exercise in the three zones of working of the Zener and the result is wrong in everyone. The solution for v0 of the first part is correct, but the second part for v0 (spite of I think the solution should be the same because the Zener is working in the Zone where Id=0) is incorrect. Could someone tell me what I am doing wrong if I am doing the same to calculate V0 and v0? FirstChildUserIdTAG: 370769 FirstChildUserNameTAG: PabloFCid FirstChildCreateTimeTAG: 2012-09-21T10:15:52Z SecondChildTAG: Count me in... SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-09-24T18:28:28Z SecondChildTAG: Hi ! Please use all digits in doing your math.This is outrageous precise , and comparing with the rest of the approximations it's a bit stupid.The thinking is correct. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-09-24T19:58:04Z SecondChildTAG: Thevnin equivalent resistance RTH=1||2KOhm or RTH=1||4Kohm. Whay v0 (AC) equal in both cases? Incorrect aswer? Sorry for my English... SecondChildUserIdTAG: 464744 SecondChildUserNameTAG: attache SecondChildCreateTimeTAG: 2012-09-24T23:06:59Z SecondChildTAG: All OK. SecondChildUserIdTAG: 464744 SecondChildUserNameTAG: attache SecondChildCreateTimeTAG: 2012-09-25T08:21:41Z SecondChildTAG: The same problem. I use vi*RL/(Rin+RL) to calculate vo in second part (spite of I think the solution should be the same because the Zener is working in the Zone where Id=0) but result is incorrect. SecondChildUserIdTAG: 387154 SecondChildUserNameTAG: Junglist SecondChildCreateTimeTAG: 2012-09-30T14:12:30Z SecondChildTAG: i get it SecondChildUserIdTAG: 387154 SecondChildUserNameTAG: Junglist SecondChildCreateTimeTAG: 2012-09-30T14:54:11Z SecondChildTAG: the same problem,i've managed finding V0(for the 13.5V DC source) but i can not calculate v0, i also think that it should be the same as in the first part,because diod is closed in thet region, but it's incorrect,where am i wrong? SecondChildUserIdTAG: 376851 SecondChildUserNameTAG: twoleggedeye SecondChildCreateTimeTAG: 2012-09-30T15:54:19Z SecondChildTAG: after many many time of puzzling all around, vo is solved!!! hint: the zener acts like an resistance in the region where she is closed... :) SecondChildUserIdTAG: 301962 SecondChildUserNameTAG: labrinim SecondChildCreateTimeTAG: 2012-10-01T08:13:29Z SecondChildTAG: (or much much time? i've forgotten my english...) SecondChildUserIdTAG: 301962 SecondChildUserNameTAG: labrinim SecondChildCreateTimeTAG: 2012-10-01T08:14:16Z SecondChildTAG: i dont follow, zener is acting in the region where Id=0, why model it as a resistor? we should rather treat it as an open circuit and use voltage divider equation to find vo right? SecondChildUserIdTAG: 285945 SecondChildUserNameTAG: lindalapiso SecondChildCreateTimeTAG: 2012-10-01T08:17:55Z SecondChildTAG: try it. first find the equivalent thevevin resistance and voltage. and, voila!!! it works! SecondChildUserIdTAG: 301962 SecondChildUserNameTAG: labrinim SecondChildCreateTimeTAG: 2012-10-01T10:59:16Z SecondChildTAG: :( im about to go crazy over this.... :( SecondChildUserIdTAG: 285945 SecondChildUserNameTAG: lindalapiso SecondChildCreateTimeTAG: 2012-10-01T11:51:15Z SecondChildTAG: find first the equivalent resistance and voltage(thevenin) without the zener and then connect the zener... SecondChildUserIdTAG: 301962 SecondChildUserNameTAG: labrinim SecondChildCreateTimeTAG: 2012-10-01T12:12:21Z SecondChildTAG: thank you anyway :( SecondChildUserIdTAG: 285945 SecondChildUserNameTAG: lindalapiso SecondChildCreateTimeTAG: 2012-10-02T08:27:53Z SecondChildTAG: Remember the small signal analysis technique here. Fix the DC offset using the DC voltage source and then do the small signal analysis. Use Thevenin theorem for ease SecondChildUserIdTAG: 321556 SecondChildUserNameTAG: sj31867 SecondChildCreateTimeTAG: 2012-10-02T11:24:05Z SecondChildTAG: I just can't make this, i've gone over and over it trying a thousand methods and I can't get the values for vo with the zener. To start with I really don't get why it isn't just an open circuit. It's driving me crazy SecondChildUserIdTAG: 319905 SecondChildUserNameTAG: andrea490 SecondChildCreateTimeTAG: 2012-10-02T15:21:30Z SecondChildTAG: THANK YOU ALL!!(finally,i got all the answers right). can someone please post the answers(showing the steps and reasoning behind) once the due date is over? SecondChildUserIdTAG: 285945 SecondChildUserNameTAG: lindalapiso SecondChildCreateTimeTAG: 2012-10-04T00:19:35Z SecondChildTAG: How can i find the Zener resistance?? from the graph the resistance comes out to be infinite for small signal.. SecondChildUserIdTAG: 213509 SecondChildUserNameTAG: vishalshaw62 SecondChildCreateTimeTAG: 2012-10-06T06:12:11Z FirstChildTAG: Can anybody give me a hint about the voltage of the independent voltage source and the resistance which should replace the Ziner Diode? FirstChildUserIdTAG: 210954 FirstChildUserNameTAG: Shahrouz FirstChildCreateTimeTAG: 2012-09-22T21:18:27Z SecondChildTAG: Look at the graph. Zener-resistor's zero point is moved by 5 volts. This can be modeled with DC. But maybe I'm wrong. Try it yourself. SecondChildUserIdTAG: 208296 SecondChildUserNameTAG: OZ1 SecondChildCreateTimeTAG: 2012-09-24T00:54:44Z SecondChildTAG: It's in a tutorial from week 2, where the question about a composite circuit is linear or nor. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-09-25T10:50:39Z SecondChildTAG: Lets just take a Zener diode(think without any circuit) now take a DC voltage source and a resistance in series(draw it below zener diode) now write down the KVL equation for it. now compare to the graph given in the question for zener and from that fix the value of DC voltage and resistance which will give to the desired curve SecondChildUserIdTAG: 321556 SecondChildUserNameTAG: sj31867 SecondChildCreateTimeTAG: 2012-10-02T11:28:18Z SecondChildTAG: no! it doesnt work in that way. if the current is zerothen either a 0.6v or a -5v source and a resistance of 1k or 1ohm should be used.but that gives the wrong answer... SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-03T11:58:08Z FirstChildTAG: Anyone can help I CAN CALCULATE v0 already tried vi RELATIONSHIP WITH RESISTANCE AND ME BUT NO THANKS gives the desired result FirstChildUserIdTAG: 320715 FirstChildUserNameTAG: maraivette FirstChildCreateTimeTAG: 2012-10-01T15:38:40Z FirstChildTAG: I dont understand what is wrong when we calculate vo for RL=2Kohm. If I build the thevenin equivalent I can see that the VTH is so low that the zener is in the zone where the iD is ZERO. So I replace it with an open circuit (zero current flowing across it) and I get an easy voltage divider circuit. When I calculate the output at the 2Kohm resistor, it says that it is wrong. What is happening? I've read that I have to replace the zener with a voltage source in series with a resistor, but in this case the current is ZERO, so the real replacement would be just an open circuit. I dont understand what is wrong. Can anyone explain better what is happening when we are calculating vo for RL=2kohm with the zener connected??? Thanks FirstChildUserIdTAG: 181432 FirstChildUserNameTAG: enriqueferreralcala FirstChildCreateTimeTAG: 2012-10-01T23:34:58Z SecondChildTAG: i have the same problem.I think for v0 is also an open circuit. So the solution must be like the original circuit. I dnt know whats going wrong. Some help plz. SecondChildUserIdTAG: 78992 SecondChildUserNameTAG: chuso06pdm SecondChildCreateTimeTAG: 2012-10-02T19:07:58Z SecondChildTAG: while calculating v0 do we find out the region with the mv source or do we first find the region of operation using the dc source? i guess this might be the mistake u r committing. SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-03T12:28:42Z FirstChildTAG: Ok, I think I have the solution. The problem is that with v0 you can't do the same thing than with V0. The Zener is working as a Non-Linear Device (Lesson 7), so you should treat the zener as a Non-Linear Device working with small signals ("converting" the zener in a ...). Remember the zener is working in a zone defined by V0. Then you have a circuit with a "new" component instead of the zener, and you can solve using KVL and KCL to get v0. FirstChildUserIdTAG: 370769 FirstChildUserNameTAG: PabloFCid FirstChildCreateTimeTAG: 2012-10-03T10:09:36Z SecondChildTAG: while calculating v0 do we find out the region with the mv source or do we first find the resion of operation using the dc source? secondly, i somehow managed to calculate V0 but do not know the approach for v0.the diode is replaced by an independent source of 2V and a resistance.is the value of this resistance constant or does it change with vd.My dc source voltage is 9v. while finding v0 do we short the constant dc source and the independent source we found out for V0 and use only the mv source and the resistances?? PLEASE HELP!! SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-03T12:26:46Z SecondChildTAG: Check the slides of the Lesson 7 (slide 25). For a Non-Linear Device in small signals you only need one component instead the Zener diode, and this component is constant. You need to get V0, then working in the same region, you have a expresion: iD=iD(vD). Then doing the same thing than in the slides you get a value for your new component, and you only need to solve the circuit with that new component and vi using KVL and KCL. SecondChildUserIdTAG: 370769 SecondChildUserNameTAG: PabloFCid SecondChildCreateTimeTAG: 2012-10-03T13:28:53Z SecondChildTAG: hey i have checked that slide.actually we require to use a resistance of some value and that value is given to be inverse of slope of curve which is 1A/1V. SO it should be 1 ohm . Or let it be 1kohm. now if we replace the diode by this resistance and use the mv source and other resistance i am not getting an answer. My actual qs is what is the value of resistance?is it constant=1 ohm or kom or does it vary. It should not vary as slope is constant. what do you say? I am tired of solving this question..Please help man! Thanks in advance! SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-03T14:32:45Z SecondChildTAG: Why 1 Kohm?? If you get one 1 ohm then use 1 ohm, not 1 kohm... Use the value which you get for the resistance using the same method than in slide 25? Replace it, and solve the circuit with exactly the same resistances and the small signall. SecondChildUserIdTAG: 370769 SecondChildUserNameTAG: PabloFCid SecondChildCreateTimeTAG: 2012-10-03T15:38:30Z SecondChildTAG: na na.. I have used 1 ohm too. Its not working ...:( SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-03T15:54:05Z SecondChildTAG: lesson 7(slide 25)?! SecondChildUserIdTAG: 285945 SecondChildUserNameTAG: lindalapiso SecondChildCreateTimeTAG: 2012-10-03T16:27:48Z SecondChildTAG: thank you guys...I got the answer...Referred another author's book... SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-03T16:29:28Z SecondChildTAG: I tried like the slide 25 and is still wrong. i take VD which is correct and the Id=0. So the R=1/Id --> Open circuit. I dnt know what im doing wrong.Plz help!! this problem is making me crazy SecondChildUserIdTAG: 78992 SecondChildUserNameTAG: chuso06pdm SecondChildCreateTimeTAG: 2012-10-03T17:24:13Z SecondChildTAG: For v0 use the same formula only turn off all the DC sources including the voltage source you added. We are looking for the small signal response. This was the problem for me. SecondChildUserIdTAG: 392100 SecondChildUserNameTAG: glenng SecondChildCreateTimeTAG: 2012-10-03T21:53:11Z SecondChildTAG: jmen what book did you use? SecondChildUserIdTAG: 261190 SecondChildUserNameTAG: ItachiItachi SecondChildCreateTimeTAG: 2012-10-04T01:32:48Z SecondChildTAG: well,got a simple answer for this...Zener diode acts as a voltage regulator i.e it doesn't allow voltage across load to change much by incorporating the entire change across itself. In this case being parallel to the load resistance,for part of V0 one doesn't require any calculations.Its pretty straightforward.Its just the voltage of knee point of curve but the catch is you must know which knee point to select? and that is decided by the operating point...Good luck guys! Its simpler and hardly requires any calculation. SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-04T02:41:04Z SecondChildTAG: jmen. Can you please guide me on how to get the RL. I am stuck with that problem. Thanks! SecondChildUserIdTAG: 371000 SecondChildUserNameTAG: Joseph090892 SecondChildCreateTimeTAG: 2012-10-04T02:51:27Z SecondChildTAG: once you follow my approach you will notice that the value of V0 comes out to be same in both cases(approximately). Now apply a voltage divider b/w Rin and Rl and the given dc source.The voltage across the load should be less than that of V0(the fixed value of voltage you got for both the parts).so you'll get the range of RL SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-04T03:00:24Z SecondChildTAG: wot to do for v0 da noise one?? CONFUSED??:0 SecondChildUserIdTAG: 127800 SecondChildUserNameTAG: MNB SecondChildCreateTimeTAG: 2012-10-07T14:08:34Z FirstChildTAG: v0 is extemely small.It should be zero, but they wanted perhaps to show us how the noise is rejected by trying to compute the residual voltage. Use 6 or 7 decimals after 0 in your computations and you will get it. FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-10-04T19:45:31Z FirstChildTAG: So to find v0 I substituted all the element in the scheme with their linear equivalents and used node method to find the voltage on Rl. Then I assumed that v0=V0-v0', where v0' is the voltage in the linearised scheme. I was crazy accurate with the numbers, but it doesn't seem to work... FirstChildUserIdTAG: 380703 FirstChildUserNameTAG: vargy FirstChildCreateTimeTAG: 2012-10-07T11:13:11Z IndexTAG: 201 TitleTAG: Thanks To My Government ..... Need Help From Management URGENT Youtube is Blocked in Pakistan and due to this i don't have access to Week 2 or any other Video of this Course ..... I am enable to submit my Assignment due to this issue please need help from Management ... DO Something for Pakistani Student because i hope everyone from Pakistan has no Access to Videos due to Ban on youtube UserIdTAG: 176353 UserNameTAG: amJunaid CreateTimeTAG: 2012-09-17T16:17:30Z VoteTAG: 10 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 6 FirstChildTAG: he is right....... staff should give a solution for this problem........:( FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-09-17T16:28:44Z SecondChildTAG: https://www.facebook.com/groups/483354858342165/ join this group on faebook and saram kumar has the videos SecondChildUserIdTAG: 436749 SecondChildUserNameTAG: waynebrown SecondChildCreateTimeTAG: 2012-09-19T07:58:25Z SecondChildTAG: he only has week 4 videos...what should we do abt week 2 and week 3? SecondChildUserIdTAG: 122513 SecondChildUserNameTAG: Mona77 SecondChildCreateTimeTAG: 2012-09-20T09:54:22Z FirstChildTAG: I believe Youtube has been banned because Google refused to remove a particular video. I really hope edX has plans to not use Youtube in the future as these bans will really stop these courses from being accessible to everyone. FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-09-17T17:56:52Z FirstChildTAG: Dear friends, The challenge you are facing is not palatable to the ears and for this I sympathize with you as well as say: 'it is my pleasure to be in class with people from various countries sharing thoughts and experiences worldwide'. As respect the issue of banned YouTube videos in your country - I must confess, it will be a serious challenge which must be tackled in a timely manner. I will like to suggest that you guys can overcome this stumbling block by using the annotated copies of the lectures, print them out or read directly from your systems in order to meet up with deadline for submissions of assignment before **edx** team find a solution. Sorry for this discomfort as we herald the importance of this course in our various countries. Regards, Ugo. FirstChildUserIdTAG: 329879 FirstChildUserNameTAG: Ugochukwu FirstChildCreateTimeTAG: 2012-09-17T18:31:55Z FirstChildTAG: You can download the MIT OCW version of 6.002, it was taped live in front of actual MIT students. The link is here: http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-electronics-spring-2007/ You can also try to use a proxy server or VPN to bypass country restrictions on Youtube. FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-09-18T02:17:19Z FirstChildTAG: I am also facing the same problem...now I am stuck in the middle of the homework as well as lab...is there any other way we can watch the videos and do our assignments...thanks FirstChildUserIdTAG: 122513 FirstChildUserNameTAG: Mona77 FirstChildCreateTimeTAG: 2012-09-19T05:15:08Z FirstChildTAG: You can see in the Wiki [Lecture Videos Download][1]:). Also see more info [here][2]: **Due to the YouTube outage in certain parts of the world, we have provided direct links to download the lecture videos, and extended the Week 2 homework and lab deadlines to the coming Sunday (September 30). Please note that Week 3 deadlines have not changed.** [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/ [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/info FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-26T11:50:45Z IndexTAG: 202 TitleTAG: Thank You Chainsaw Man! Throughout my life, no one bothered to explain the context of the things that I am learning. It was all just about passing exams. But thanks to Prof. Agarwal, I get the entirety of the stuff that I am learning. I mean, I just went through the first video lecture, and everything was explained in a logical way... What is engineering > electrical engineering > laws are abstractions > equations I just wanted to say thank you for explaining things so simply and delivering this course. I hope that I meet you one day, and talk over a coffee! :) UserIdTAG: 193033 UserNameTAG: NIslam CreateTimeTAG: 2012-09-07T13:10:37Z VoteTAG: 10 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: I kinda miss the "like" button now :) FirstChildUserIdTAG: 151751 FirstChildUserNameTAG: Senade FirstChildCreateTimeTAG: 2012-09-07T13:47:50Z IndexTAG: 203 TitleTAG: Lower case letters make sure you use lower case letters ot it will show that the answer is wrong. UserIdTAG: 287284 UserNameTAG: bikerideraustin CreateTimeTAG: 2012-09-05T16:20:40Z VoteTAG: 10 CoursewareTAG: Week 1 / Various V-I characteristics CommentableIdTAG: 6002x_S1E1_Various_V-I_characteristics NumberOfReplyTAG: 4 FirstChildTAG: Yes, this got me, too. Very tricky as I went back and read the book because I was confused as to why I was wrong. FirstChildUserIdTAG: 265346 FirstChildUserNameTAG: k_candiotti FirstChildCreateTimeTAG: 2012-09-06T03:09:55Z FirstChildTAG: I seem to get only caps on my iPad. I suppose I'll have to use the computer. FirstChildUserIdTAG: 161978 FirstChildUserNameTAG: PriscillaN FirstChildCreateTimeTAG: 2012-09-06T04:36:37Z FirstChildTAG: I agree. Pay attention that it may be a huge difference between two cases of one letter. For example V and v, as a lecturer said in this task FirstChildUserIdTAG: 205332 FirstChildUserNameTAG: Vslav FirstChildCreateTimeTAG: 2012-09-11T20:00:34Z FirstChildTAG: The problem was updated to accept lower and upper case FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-16T06:06:34Z IndexTAG: 204 TitleTAG: Analog circuits course by Prof.Shanthi pavan,Indian Institute of Technology,madras(IITM) hi friends,,those who want to know design of ampilifiers(MOS,BJT) and negative feedback and design of opamps etc,may refer to this course http://www.ee.iitm.ac.in/~nagendra/videolectures/doku.php?id=ec201_2010:start UserIdTAG: 857413 UserNameTAG: Naveenkrishna CreateTimeTAG: 2013-01-25T18:23:00Z VoteTAG: 9 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Very good!! Naveenkrishna. FirstChildUserIdTAG: 361823 FirstChildUserNameTAG: EliasOak FirstChildCreateTimeTAG: 2013-01-25T23:36:43Z IndexTAG: 205 TitleTAG: (A small part of) The 6502 chip explained down to the silicon A real-world example of some of the ideas we learned in this course: http://www.arcfn.com/2013/01/a-small-part-of-6502-chip-explained.html UserIdTAG: 69075 UserNameTAG: ErikR CreateTimeTAG: 2013-01-15T16:37:00Z VoteTAG: 9 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Excelent web site!!! FirstChildUserIdTAG: 61723 FirstChildUserNameTAG: lvence FirstChildCreateTimeTAG: 2013-01-16T03:27:52Z FirstChildTAG: Indeed, nice site! The Rockwell 6502 was my first processor that I used to experiment with (about 36 years ago) with the TIM (Terminal Input Monitor), that contained ROM with a small monitor program and some I/O ports and I was sOO happy that I had 1kbyte (!!!) of ram. It still works! First I had to build a video interface with about 50 TTL chips or so, and keyboards weren't almost impossible to get (pc didn''t exist at that time). And I used The First Book of KIM to try out some programs. I've spend hundreds of happy hours with it. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2013-01-16T16:33:22Z SecondChildTAG: I had a MOS KIM-1 Computer around 1978. I bought a huge surplus power supply to power it. My first effort at hardware design was interfacing a multi-digit seven segment display to one of the output ports. I used interrupts to sequentially enable each digit in turn. The code was written in assembly and hand assembled. Programs were entered on a hex key pad and stored on a cassette recorder. I remember spending much time playing a lunar lander game on the KIm. Then I learned about optimal control and that the optimal solution (least fuel used) was bang-bang control. Ah, such pleasure in the 'Good Ole Days"!! SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2013-01-16T16:58:32Z SecondChildTAG: That does sound like fun. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2013-01-18T00:55:06Z FirstChildTAG: Please keep this posts comming ... Super ! FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2013-01-17T23:43:28Z IndexTAG: 206 TitleTAG: Happy New Year very very Happy New Year to Respected Prof.Anant Agarwal,Prof.Gerald Sussman, r. Piotr Mitros,Prof.Chris Terman,Prof.Khurram Afridi.MIT,All loving and caring TA's, Classmates and Edx team. I love you all. UserIdTAG: 40209 UserNameTAG: sadaf78 CreateTimeTAG: 2012-12-31T10:02:40Z VoteTAG: 9 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 207 TitleTAG: A Possible Solution for Grading on Certificate Problem Hey Sirs and Guys "Grading in Certificate": ------------------------ So many different Minds...so many concepts. I would Like to come up with Mine that can be practiced, like in my University (Renowned one in my Country) there are two things after completion of Degree in Engineering, A Certificate Like the one the kind edx team is going to award us (without Grades just Name,mentioning of Course name and Teachers Signatures, and in my University Vise Chancellor signature). The other Related Page is the Detailed Marks Sheet, "at first" its Headings are merely: University Logo and Name ----------------------- Unofficial mark sheet --------------------- then comes the Students Name and Marks details etc... If a Student need to use that Sheet for a Reason, This Certificate holds no Authenticity or weight so he/she have to get that marks sheet Printed and Signed, attested , stamped from his department. My Point is that Mostly Certificates do not show Grades at this Level and as we are involved in an Online Course which has its own Protocols and its Methods are Still under development ,the Certificate may never show any Grade (no matter how much we miss it) But edx can give another Option "unofficial type" But with Name ,Logo whatever, a marks (Grade Sheet of home works....Final Term with Grades) This Idea may Look Childish to some But it is Like "Something better than Nothing" (in Grading prospect). The Students who worked really hard will feel Acknowledged for their work in the past many Months. The students with Low Grades will have option not to get that additional sheet if they feel like) As certificate will be separate Some may say that it is Same Like getting Printed our "Progress web Page" But my Point is for Something Different ,unofficial no doubt ( not like Certificate with Signatures ) But should Carry something to Praise to Right full Genius Students ( The same as they are in the Whole Educational World) I am Hope full that it will cost not too much for the Respected Team, as you have already done too much for us. It is a just act to be done for the sake of Grades recognition ( Hard Work Acknowledgement ) or some may say Mental satisfaction for Brains Thanks What Community TA and others say........ UserIdTAG: 37464 UserNameTAG: DoubleA CreateTimeTAG: 2012-12-30T00:12:22Z VoteTAG: 9 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: > The Students who worked really hard will feel Acknowledged for their work in the past many Months. Please don't take this very hard , but if you really worked that hard, aren't you already satisfied with your knowledge ? FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-12-31T13:19:28Z SecondChildTAG: Come on my dear I am already in my Electrical Engineering Finals and surely know the weightage of this certificate in real world ,I just happened to saw so many Posts in here complaining about this Grading Problem So i tried to come up with a most possible Suggestion for their Problem, and talking of Knowledge....No doubt You work you get, But in here we got so much compared to our own Effort.. Well Happy New Year SecondChildUserIdTAG: 37464 SecondChildUserNameTAG: DoubleA SecondChildCreateTimeTAG: 2012-12-31T13:58:56Z FirstChildTAG: Thanks for pointing this out, and I can bring this up again if we talk about certificate format again, but at least for this semester, we are only distributing grade-less certificates. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-12-31T15:57:22Z FirstChildTAG: As a Community TA, I assure you that I have no special powers. Your idea sounds reasonable enough to me, essentially you want to be able to print out something with your grades that is similar but different then the progress page. That being said, I myself feel that the certificate, link as a well as continued access to my progress page sufficient. Nonetheless I will give you the email to which you can forward your suggestions and feedback. I assure you they do read it, as well as these forums. mit-6002x@edx.org Congratulations on your completion of the course and have a great New Year. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-12-31T13:10:26Z SecondChildTAG: hahaha , I was not pointing to any special Powers etc ,I just meant a little better Approach that might have to Our Respected Sirs, I guess you are really through a tough duty listening to so many odd demands and Issues ;) Thanks and Many Happy Returns SecondChildUserIdTAG: 37464 SecondChildUserNameTAG: DoubleA SecondChildCreateTimeTAG: 2012-12-31T15:36:41Z IndexTAG: 208 TitleTAG: MITx at google groups This discussion forum will go away. If you are a graduate of this class, you can join mitx@googlegroups.com. The goal is to have a forum of people with similar interests. Maybe we can also be a group that can serve the interests of the students as well as contribute to this amazing transformation of global learning. Vote this post up, and hope to see you there. Visit this group's homepage: https://groups.google.com/d/forum/mitx Email this group: mitx@googlegroups.com UserIdTAG: 142857 UserNameTAG: viking2 CreateTimeTAG: 2012-12-29T00:32:13Z VoteTAG: 9 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Myrimit, hopefully you would also join this group. Thanks FirstChildUserIdTAG: 142857 FirstChildUserNameTAG: viking2 FirstChildCreateTimeTAG: 2012-12-29T12:12:24Z FirstChildTAG: You should be able to sign up, post, and use the group right away. FirstChildUserIdTAG: 142857 FirstChildUserNameTAG: viking2 FirstChildCreateTimeTAG: 2012-12-29T00:36:45Z IndexTAG: 209 TitleTAG: Thanks, and some old reflexions Thank you for the course. I hope that a lot of people around the world get and take the chance of learning this way. To learn and share knowledge is a strong force and give us possibilities to meet the challenge in front of humanity. In the more than 1000 years old Viking poetry Eddan a couple of verses are: *A better burden may no man bear, For wanderings wide than wisdom; It is better than wealth on unknown ways, And in grief a refuge it gives.* *A better burden may no man bear, For wanderings wide than wisdom; Worse food for the journey he brings not afield, Than an over-drinking of ale.* Those words indicate for me the personal gain you get from the knowledge itself. To open up those courses in this way is for me also important in quite another way. As nearly 70 years I hope it helps me get my brain working some more years. At the same time I don't take up place for any young. Hopefully I can pass some of the enthusiasm from Prof Agarwal and his team to young people in my surrounding. A special thank to Myriam you have contributed in a very constructive way. UserIdTAG: 151472 UserNameTAG: Stensmed CreateTimeTAG: 2012-12-28T22:18:01Z VoteTAG: 9 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Amazing to see how people with similar genuine interests from all over the world come together here. Good luck, and hope to see you in other courses. Good Yule from Bergen FirstChildUserIdTAG: 142857 FirstChildUserNameTAG: viking2 FirstChildCreateTimeTAG: 2012-12-28T23:52:01Z FirstChildTAG: Hi Stensmed, You are welcome, I am really happy for being helpful, this Fall, with my Classmates . I have never imagine that I could reach to so many students, I got really amazed at the end of the Course with all the nice words that I have received of students worldwide - [like this Post , haha][1]. I took 6.002x Spring 2012 - Prototype Course, and for me it was amazing, when I ended that Course MITx posted about edX,I suscribed inmediately in order to be e-mailed for new courses. Then I didn´t doubt it and registered again to 6.002x in order to help to the new students, and more if I could, to help ,as support, in my native language, Spanish :p I am totally agree with you: learn and share knowledge is really important. In my point of view, we interpret this world based our knowledge and as more we can learn as more far we can go, because we can see the world in a different perspective and we can open our minds. I am really happy that you had a nice experience in 6.002x and I wish that you can keep learning, sharing, and doing the things that you like :), you are really inspiring. All the best, Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50d8d6835ea03c1f0000002b FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-29T01:04:39Z FirstChildTAG: Hi Stensmed, Great quote from Viking poetry. Indian tradition also puts knowledge and knowledge giver (Guru, Prof Agarwal in this case) at a very high pedestal. In fact there is a couplet by Kabir, a famous saint poet, which says that if I see God and Guru before me, I will first bow to Guru because it was he who showed me the path to God. Amazed at your hunger for knowledge at your age. I am but a couple of years away from sixties and would consider myself fortunate to retain my hunger for knowledge by 70 and beyond. Wow, now we are talking of poetry. One creative pursuit leads to another. FirstChildUserIdTAG: 190670 FirstChildUserNameTAG: chauhan1955 FirstChildCreateTimeTAG: 2012-12-29T04:28:13Z SecondChildTAG: Very inspiring comments from old buddies, a good motivation for young generation. SecondChildUserIdTAG: 455950 SecondChildUserNameTAG: Wahabbaluch SecondChildCreateTimeTAG: 2012-12-29T07:15:40Z IndexTAG: 210 TitleTAG: To professor Agarwal and wonderful Staff I Never thought to learn so much about electronic circuits as I did in these three months. Thanks to the talent and professional level of Dr. Agarwal and his team, this has been possible. I'm 56 years old and at the beginning of the course I wasn't able to tell a coulomb from a ampere.I had to put great effort,since it's been a while I left College (mathematics were a wild ride),but It was worth-ed indeed.Thanks a lot! professor (I bow to you Sr.) and team. Have a very Merry Christmast! Apologies If I did any mistake, my natural language is Spanish! Lenin Vargas Venezuela UserIdTAG: 369339 UserNameTAG: vargaslen CreateTimeTAG: 2012-12-23T22:10:21Z VoteTAG: 9 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 211 TitleTAG: Things to Consider for Future Improvements to The edx Platform First off, I would like to thank Professor Agarwal and the 6.002x staff for a great experience. I would like to share a few thoughts for improvement though. Some context so that you all know where I'm coming from: I'm from Canada, 26 years old, I have a liberal arts undergraduate degree and a law degree. About a year and 1/2 ago, I graduated from law school, and struggled to find work in a tough job market. With a lot of free time on my hands for the first time in a long time, I discovered that I really enjoyed maths and sciences (and even started to seriously doubt whether I had chosen the right career path!). I started working through textbooks on my own to learn these interesting subjects, which I had never studied in school, using the internet as a resource for materials. That's how I eventually heard about edx. Needless to say, I was incredibly excited about this new community and encouraged to see that educators at respected institutions were beginning to recognize that the internet is a tool which could be used to dramatically change the way people learn. I excitedly signed up for 6.002x as well as Intro Chemistry and Intro Programming/Comp Sci, and I eagerly waited for the courses to begin. The disappointing thing is this: as I accustomed myself to the organization of the edx course materials, and dehabituated myself from the study habits I had developed through my self-directed learning, a little of my enthusiasm for science got killed. And while I really did get a lot out of this course, I don't think anything more damning can be said about a course than that it had the effect of suffocating a students inherent excitement for the subject matter. It seems that rather than use the internet as a tool for developing better teaching methods, the edx materials are merely exactly supplanted and reproduced online from their real-life campus equivalents. I have been reminded of everything that I hated about school. Class lectures, generally speaking, are boring! The format of having a professor prattle on at the front of the classroom full of students is tremendously outdated. Students have either not done the textbook readings in advance and so they can barely follow what is being said, OR students have done the textbook readings in advance and the professor's lecture tends to only serve as a less cogent rehashing of the same material. There is, of course, tremendous value in the presence of a knowledgeable person to emphasize and explain difficult topics in a more colloquial and intuitive way. But I would estimate that these little pearls of teaching represent only about 1% of lecture content, while easily 40% is dead air-time and 'uhms' and 'ahs'. Even at 1.50x speed, I found the lecture material in 6.002x slow. There was obviously a structure to the order in which Professor Agarwal chose to present ideas, but then the lectures conformed to that structure only loosely and on the whole gave the impression of a rushed, often repetitive, and generally unpolished discussion. The fact of the matter is that presenting material by talking freely off the top of one's head with only a vague sense of the direction of thought is not the best way of communicating ideas. It was an opportunity squandered not to use the internet and pre-recorded videos to present clear, cogent, and compelling videos. People respond to good story-telling, and the careful progressions of thought which take place in science make for great stories! My excitement for science was muted a little through these lecture materials which lacked sufficient artistry or craft. I enjoyed Professor Agarwal's genuine enthusiasm for the subject matter and it was definitely contagious, but not enough to overcome the shortfalls in structure. Walter Lewin's lectures are a wonderful example of the kinds of very carefully crafted lectures I have in mind. It's perhaps not fair of me to hold everyone up to that gold standard, and yet, isn't that the whole point? This is the internet, and having access to the very best in teaching is as easy as copy/paste, so why should we expect anything less than the best? Summary: Lectures were sloppy and slow. Killed a little of my enthusiasm. Assignment deadlines are oppressive! I think the edx staff have gone to great lengths to give the platform an air of legitimacy. And, certainly, a course wherein a student's progress is measured by some rigorous standard would seem to have a greater sense of authenticity, but, the backwards notion that the measure of a student's success is his ability to keep pace with everyone else is one which is obstructive of the goals of genuine learning. When I was learning on my own, if there was something I didn't quite grasp, I would dwell on the material a little longer, taking the extra time needed to fully grasp it before moving on. But, with regular deadlines in place, there's simply no time. During this course, I often found myself hazy on topics. Even when I was submitting answers and getting green check marks, often times I submitted those answers with very little confidence. Yet, with another deadline fast approaching, there is simply no time to stop and work things out in your mind. As a result, you move on, understanding things well-enough, but not truly understanding them in depth. As such, I found the assignment deadlines encouraged laziness in my habits: the goal becomes learning enough to keep up rather than learning for the sake of understanding or for the sake of exploring a subject in detail. Quite simply, with deadlines, there's no time to think! I think the clearest example that 6.002x falls short of its goals is that, while I achieved a respectable final grade, my understanding of the material at the end of the course leaves much to be desired. During the final exam, there were definitely questions I struggled with tremendously, and when I referred back to previous problem sets for guidance, I found myself unable to even recall what methods I had previously used to solve those problems which I had ostensibly already mastered! Worst of all, the stress and time commitments related to meeting deadlines has, by the end of the course, left me so exhausted that I can't be bothered anymore to think about this material. I need a break. I would suggest that these courses should have no start date and no end date. They should be always open and students should work through them at their own pace, such that at any given time there's a whole continuum of students at various stages of the course material. So in theory the edx platform is great. And, though I haven't said much positive in this forum post, there are lots of great things about it. But, now that my first experience with edx is complete, the bottom line is that I went from having an insatiable desire to learn maths and sciences...to being exhausted, unenthused, and having only a passing knowledge of the materials to show for all of my efforts. I'm left wondering if it's worth the bother signing up for any more courses, or whether I should rather just go back to the relaxed and successful learning habits of learning on my own. All in all, I'm encouraged that an attempt has been made to make these materials available to the world, but it falls somewhat short of the stated goals. UserIdTAG: 287805 UserNameTAG: Beneficial CreateTimeTAG: 2012-12-23T05:43:43Z VoteTAG: 9 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 15 FirstChildTAG: To move with science is to understand the reasons. Many philosophers have tried to elucidate these fundamentals. For me, science depends on persuasion. The more a person makes contact with the themes científicoa, the more she will understand science. This takes a lot of time and experience. It's not a course only that can arrive at an answer. The Course EDX is only one way, but not the only one. Even for those who like the EDX also advise you to see the lectures of Professor Walter Lewin as quoted by Beneficial. Read also about how philosophers define science. Then advise the books of Thomas Kuhn and Popper. Are good examples. FirstChildUserIdTAG: 361823 FirstChildUserNameTAG: EliasOak FirstChildCreateTimeTAG: 2012-12-23T11:12:04Z FirstChildTAG: Consumption is quick. Production takes lots of time. FirstChildUserIdTAG: 4463 FirstChildUserNameTAG: pmj FirstChildCreateTimeTAG: 2012-12-23T07:00:47Z SecondChildTAG: True, and fair enough. SecondChildUserIdTAG: 287805 SecondChildUserNameTAG: Beneficial SecondChildCreateTimeTAG: 2012-12-23T14:13:14Z FirstChildTAG: This is really a beautiful discussion. I don't usually like to read long articles but I read your words, Beneficial and Myrimit and your words both are great. You seem to me as two great persons. Keep it up people. FirstChildUserIdTAG: 454609 FirstChildUserNameTAG: Alwahsh FirstChildCreateTimeTAG: 2012-12-23T09:52:28Z FirstChildTAG: Hi Beneficial, Thank you for sharing your thoughts about your experience on this Course. As I know, for them is really important your feedback, I know this, because once we finished the 6.002x Spring Piotr Mitros was really concern about our opinion. So, your opinion is really valuable. In my point of view Beneficial, please don´t leave your enthusiasm for the Science and Math. I can tell you that this is only the beginning of a hard journey of Science. I have to confess that when I started my career I was anxious to see the Electronics quickly, but I have found that before of that I had to study some other things first, and I have to confess that it was a little bit boring at a first sight haha...Now all that knowlege can be used in Projects, sometimes I get amazed of what I can do myself. I have learnt that you have to encouraged by your own to do things, to experiment, to give the first step no mattering if components explotes haha, because any Subject at the University will teach you how you can handle in the real life. Science for me is the best thing, after working as Intern in different places of Industy, part time, I discovered that my real passion was really Science - Research. I feel really lucky as now I am doing what I really enjoy: I won a Fellowship as student of engineering in Research and now I am also colaborating free at my University since now 3 years ago, wow so much time, in order to help the students and also in a way to return my University all that is giving me, and I discovered a new world in Science ! I really love that and believe me that this is only the start of a long journey. Science is amazing. So, give it another chance Beneficial, you will discover a new world in Science. Remember that during your journey is possible that you can find things that you didn´t like or that it was difficult, but Science is so amazing that it worth to pass trough all that things. It is a challenge. Never abandone your enthusiasm in Science and Math if that is what you really like. Please give it another chance :). Remember that it will not be easy neither impossible, if you are perseverant and have enthusiasm, you can reach to anywhere , you don´t need to be smart, I am the living example of that haha -of course it is better if you are, I bet you are as you already are a Lawyer and you did really well in 6.002x, wow! So, you have all in favour Beneficial! So, you can do it Beneficial! My best wish to you, Take care, Myriam. P.D. It is never late to do what you want. My mother is Lawyer - she studied that career because of my grandfather will, as he was Lawyer too, in fact, my Japanese ancestors had something connected to Law-Politics, so it was a hard burden to her- and she discovered a lot of years later that her vocation was to be Teacher rather than Lawer, because she enjoy to teach. Anyway, I have to say that she implicit did the same with me haha,I don´t blame her as I know all that she had to going through, she didn´t want that I studied engineer as she considered that it was a career for man haha. So, like I am, perseverant of my ideas, I negociated with her in order to do what I like it haha, I told her if my 1st year of engineer go wrong I will quit and I will do what she suggested, of Course, I put all my effort to pass 1st year, studying a lot, haha, so here I am. I am really happy to do what I like. I will never regret. So, if you like Science Beneficial, give it another chance- as I always say , sometimes the students need a little push to be up and to know that they can do it, so this will be, in someway, my push to you Beneficial :) FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-23T07:29:30Z SecondChildTAG: Thanks Myriam for your encouragement. I'll definitely be keeping up with it in one way or another. And thank you for all your help throughout the course. About halfway through 6.002x, whenever I went looking for help in the forums, I learned to look for your posts first :D. You must have inherited the ability to teach just like your mother. SecondChildUserIdTAG: 287805 SecondChildUserNameTAG: Beneficial SecondChildCreateTimeTAG: 2012-12-23T14:18:17Z SecondChildTAG: Cool Beneficial :). I am sure you will succeed in Since and Math. Haha, I will tell that to my mom :p Nice to know that I were helpful here, I am really happy knowing that XD. I hope to be a source of inspiration to others, so that they can encourage to help the New Students as I did this Term but in the next Term :), no mattering being or not being CTA. That will be awesome! My best wish to you Beneficial! Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-23T16:50:53Z FirstChildTAG: I completely agree with your para on assignment deadlines. I could have written it myself, if only I could be so eloquent on the topic of autodidactism. I, too, felt very pressured by the deadlines, in particular because I very much lacked the math prereqs. I found myself with very little time to *think*, instead being too busy *doing*. I recall thinking many times: "I need to come back and think about this point some more;" sadly, I now do not even recall the particular topic I was studying when those moments occurred. I opined elsewhere that I planned to re-take the course in the Spring. I hoped that with this semester as a "preview", the time pressures (in particular wrt math) would be somewhat eased, and I might have the luxury of *thinking*. Already real life commitments have me doubting that premise. Thank you so much for taking the time to write out this cogent and valuable essay! I believe you speak for many here, myself included. Best wishes on your further endeavours! FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-12-23T11:30:57Z FirstChildTAG: In my point of view, deadlines are necessary in order to render the student have commitement to the course itself. If there are no deadlines, I myself will leave the homeworks until the last moments, which is something very incorrect to do (but I'm still doing so ! :( ). As a recommendation, I would like to see more solved problems for the last 6 weeks, I think they were few. Also, I'd like to see more Tutorials, Mr. Piotr and Mr. Gerald are so cool. Their tutorials were very useful to me as well as the sidebars and the worked problems. Thank you. FirstChildUserIdTAG: 418182 FirstChildUserNameTAG: euldji2005 FirstChildCreateTimeTAG: 2012-12-23T12:36:17Z SecondChildTAG: I agree that more Tutorials from Piotr and Gerald would be wonderful. I really enjoyed the tutorial near the end of the course on soldering. The tutorials did a good job of bringing the theory of the course back to the real world, but I would have liked a lot more of that: to see the circuit elements we're studying and hold them in our hands. Also, I enjoyed the dynamic of those two on camera together. I disagree that deadlines are necessary to keep people committed, after all, everyone who comes here to take this course is self-motivated and interested in the material. That should be commitment enough. SecondChildUserIdTAG: 287805 SecondChildUserNameTAG: Beneficial SecondChildCreateTimeTAG: 2012-12-23T14:23:09Z FirstChildTAG: Maybe Dr. Agarwal seemed rushed because you were watching him at 1.5x? I am sure that, should I ever meet him, he will seem sluggish to me in person! FirstChildUserIdTAG: 428560 FirstChildUserNameTAG: MikeDayton FirstChildCreateTimeTAG: 2012-12-23T12:45:13Z SecondChildTAG: Haha, for sure! I hope I didn't come across as ragging too hard on Dr. Agarwal. As lecturers go, I actually think he's quite a good one. It's just that lecturing itself is an inherently poor way of communicating information, and I was disappointed that the course seemed to make little use of modern technology beyond uploading the experience of being trapped in a lecture hall for two or more hours. SecondChildUserIdTAG: 287805 SecondChildUserNameTAG: Beneficial SecondChildCreateTimeTAG: 2012-12-23T14:31:10Z FirstChildTAG: I was going to address specific points in your post, but frankly the whole thing is depressing, subtlety mean spirited and has a sense of entitlement. Your complaints offer little in the way of constructive feedback. I truly hope they don't take your suggestions to heart. While anything can be improved, I like to think the majority of students feel the **opposite** of you and enjoy the casual intimacy that the videos bring into our home. It's authentic and new. You cannot really compare this to your traditional experiences. You have to remember you are here for a **free** education in **Electronic and Circuits**, perhaps you were looking for something different. Best of luck in the future! FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-12-23T12:36:21Z SecondChildTAG: My hope was, at least, to capture how I felt at the end of my first edx experience. I think I captured that honestly, and was able to describe the aspects of the course which mainly caused me to feel the way that I did. There are lots of positive things to say about edx and 6.002x in particular, and they're definitely worth stating, but my post was getting long enough as it is. I wanted my feedback to have a little more substance than a mere "lectures are boring and deadlines are difficult!" lest people take my comments as only an empty whine. I took the course to learn something I found truly interesting, and, by the end, I was tired out and had good reason to question whether the posted grade truly reflected the amount I had learned. That was something worth reporting, I think. And the truth is, I took the time to make my comments because I think this is a great product and can be made an even better one with a little refocusing and by getting rid of some tired old schoolroom attitudes about what constitutes successful learning. As for mean-spiritedness, I definitely didn't want to come across as mean, but I'll admit the tone of my comment is at times biting. Welcome to the internet, where you're going to get some honest feedback. This is something I would have never written on course feedback sheets at the end of a course in school (but I should have!). I judge these materials by the same standard that I would judge ANY youtube video, and as we all know, that can be an exacting (but honest!) standard. Billing something as 'free' shouldn't be an excuse for it to be 'bad'. The edx team doesn't act that way, and I gave my feedback because I genuinely do believe it will be received and considered for what it's worth. And then, in another sense, it's definitely not 'free'. Working full time while I took this course, with only a little window at the end of each day from about 7-11pm, I dedicated almost every evening to studying the material. I would even squeeze in pages of textbook reading on my phone while I waited for motions to be heard in court. I did that, because I genuinely enjoy it, but it's nevertheless a huge time investment, and one has to ask is it WORTH this time investment? Am I getting enough out of it? And those are fair questions to ask, not 'entitled' ones. In fact, I would describe the reverse sentiment as 'entitled': the notion that we should all be so tremendously thankful and privileged to be participating that the materials are beyond criticism. I agree that the videos had a very nice intimate quality. I still think you can keep that quality while putting together a more polished set of videos. SecondChildUserIdTAG: 287805 SecondChildUserNameTAG: Beneficial SecondChildCreateTimeTAG: 2012-12-23T15:06:47Z SecondChildTAG: I admit my post maybe a little biting as well, I am sorry for that. You really got me felling down for a bit! lol I feel that while this is the internet, we should still appreciate the specific *tools* that are given for free here. People can literally be taught electronics while virtually sitting beside others at MIT. (The big deal here is, this is obviously *the best* school in the world for these specific types of courses, in my opinion) I wonder how many real MIT students would have rather done the lectures in their pajamas. It is a given that in time things will becomes a little more polished, as we know this is new and evolutionary. Anyways at the end of the day we are given the tools, it's up to the individual what the build with them. These are not really cookie cutter courses designed to constrain you to some specific job. What we have learned are like tangible tools that can be used in many different fields or trades. Similar to those "Choose your own adventure" books we had as a kids. Thanks for your sentiments, have a Merry Season. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-12-23T16:04:09Z SecondChildTAG: Oh, as though "real" students never show up in their pajamas! What? ;-) SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-12-23T18:52:09Z FirstChildTAG: I appreciate the availability of this course and the fact that it is free. Taking this course is an experience that I am glad that I have had. I also see areas for improvement if an honest evaluation of the course is desired by those in charge. I agree with some of the statements made by beneficial but disagree on others. I do not believe the homework deadlines were a problem for anyone with the specified prerequisites for the course. On more than one occasion I saw students struggling to get some software to do a calculus calculation that could be written on inspection by anyone with the calculus prerequisite. On the other hand, I did waste considerable time on homework problems that were either poorly written or outright buggy. There were also quirky problems where the approximate answer got a check and the exact answer got an x. When I found these problems, I tried to help students wrestling with these quirks to see their way through to the correct answer. I will write more of my impressions in the coming days, I hope they will be accepted as an effort at an objective appraisal. One final comment for now. I found the work to be much easier than expected. I had seen the videos of the live class, so I knew the content was not deep. However, I expected the homework and exams to have some very challenging problems, but they did not. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-12-23T14:05:25Z SecondChildTAG: You are so right . Since that problem with the 0.0000xyz zenner diode current, i was always carring with me 7 or 8 decimals. I think that they wanted to help people with different backgrounds . Harvard has however on their CS50x course an alternative called hacker edition for people who want a challenge so i think that it would be nice to have on 6.002x also. I found some of the tutorials hard to understand and i was expecting some "surprises" like when we have learned about mosfets, but there wasn't any videos with a mosfet turning on/off onto a breadboard. Because if there were, many of us would realize that the gate of the mosfet remains polarized even after you remove the potential, thanks to the gate source capacitor and if you touch the gate with your hand it will discharge.Just a random example of how a video would do more than a thousand words. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-12-23T15:55:47Z SecondChildTAG: I've heard more than once in this forum about "the easiness" of the course. I think you forget that this course is not meant to be a game with puzzles that present challenges and that just a few would actually complete it. The course is an introductory course in electrical engineering. The keyword here is INTRODUCTORY. If you feel it was easy, then congratulations, you have aptitude to learn electronics. I actually think most people found the course hard as it is and I think you can learn a lot from the course as it is. I really don't see more challenging problems improving the content for most people, except for a selected few eager for tough problems. SecondChildUserIdTAG: 318577 SecondChildUserNameTAG: takeuchi SecondChildCreateTimeTAG: 2012-12-24T22:10:31Z FirstChildTAG: I firmly disagree with all of your arguments. You are just being unfair to yourself. Just listen to myrimit's words above "....sometimes I get amazed of what I can do myself...." If you persist you will be amazed too. gracias Myrimit. You are an inspiration. FirstChildUserIdTAG: 623210 FirstChildUserNameTAG: preveen FirstChildCreateTimeTAG: 2012-12-23T17:31:40Z FirstChildTAG: The following is a bit of speculation on my part, but I believe that it captures certain aspects of human nature. I believe if a course with the same content, taught in the same order, by the same professor or one with equivalent teaching skills but was put out by "Podunk U" located in the southern or mid-western United States it would not be viewed by nearly as many students or would there be nearly the reticence to criticize aspects of the course. The fact is, the MIT brand sells. People are inclined to say, "If that's the way MIT does it, then that's the way it should be done. After all MIT is the best engineering school." Of course, this is a logical fallacy called "appeal to authority". The term "best engineering school" is ambiguous. Best at what? The best at advanced research or the best at teaching undergraduates? Departments within a school also vary in quality, so the best engineering school (if there is such a thing!) need not have the best electrical/electronic engineering department. People taking the course have a vested interest in "talking up" the quality of the school and course. They like to be able to say that they passed an engineering course from MIT or that they got an A or that they were 100% and you know that "MIT is the best!" I've looked over at Coursera, and they are offering a course this spring called "Fundamentals of Electrical Engineering". From the description it sounds as though it is the Rice University equivalent of 6.002. From the FAQ it says, "**This course is routinely taken by second-year electrical engineering students at Rice as their first electrical engineering course. Its reputation can be summarized as “the hardest course I have ever taken but I learned a lot.**” Looking at the syllabus it looks like quite a "meaty" course: Elements of signal and system theory Week 1: Digital and analog information; block diagrams: sources, systems, sinks. Simple signals and systems. Complex numbers. Analog Signal Processing Weeks 2-4: Representation of signals by electrical quantities (electric currents and electromagnetic radiation). Elementary circuit theory: resistors and sources, KVL and KCL, power, equivalent circuits. Circuits with memory: impedance, transfer functions, Thévenin and Mayer-Norton equivalent circuits. Formal circuit methods. Conservation of power. Electronics: operational amplifiers and active filters.Frequency Domain Ideas Weeks 5-7: Fourier series and Fourier transforms. Signals in time and frequency domains. Encoding information in the frequency domain. Filtering signals. Modeling the speech signal. Digital Signal Processing Weeks 8-9: Analog-to-digital (A/D) conversion: Sampling Theorem, amplitude quantization, data rate. Discrete-time signals and systems. Discrete-time Fourier transform, discrete Fourier transform and the fast Fourier transform. Digital implementation of analog filtering. Communicating information Weeks 10-11: Fundamentals of communication: channel models, wireline and wireless channels. Analog (AM) communication: modulation and demodulation, noise (signal-to-noise ratio, white noise models), linear filters for noise reduction. Weeks 12-14: Digital communication: binary signal sets, digital channel models. Entropy and Shannon's Source Coding Theorem: lossless and lossy compression; redundancy. Error-correcting codes: Shannon’s Noisy Channel Coding Theorem, channel capacity, Hamming codes. Comparison of analog and digital communication. It's just my opinion and nothing more, but from the description the course at Rice appears to be more demanding than MIT's 6.002. I may take the course, but there are also other courses that I'm interested in, and there is only so much time available for self study for someone working a full time job! FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-12-23T19:32:28Z SecondChildTAG: Covering more topics does not mean covering them in the same depth, so it's not necessarily more demanding. It may be, but you won't really know until you take it. SecondChildUserIdTAG: 304535 SecondChildUserNameTAG: hugo_h SecondChildCreateTimeTAG: 2012-12-23T20:55:24Z SecondChildTAG: I have downloaded the book for the course, and the material looks quite substantial. Maybe Rice University is engaged in false advertising, but I am familiar with Rice, and I think not! I just finished 6.002 and know the depth of coverage, and as I said previously the coverage wasn't that deep. I expected some of the homework and exam problems to be very challenging, but they were not. All of this is just my opinion, and your mileage may vary. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-12-23T21:38:14Z SecondChildTAG: Thank you for sharing your point of you, your feedback is really valuable. That sounds interesting, it looks like more oriented to Communications and Digital ;). I never took a Coursera Course, but I know a lot of people that have taken other Courses at Coursera, some are satisfied and other not, some wants to try to take the edX version of a Course that they have taken in Coursera, but now in edX - New Courses are comming :p- and I think that is perfect that you would like to take it in Coursera. I am curious now haha, I hope that in the future you can feedback us, to your 6.002x Classmates how it was that Course, that would be awesome to know ;). I am sure you will find nice people there as I have found here too, for me edX is like a big family. My best wish to you skyhawk, Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-23T21:49:14Z SecondChildTAG: coursera courses tend to be difficult in many ways and mostly reformatted versions of the real course offered at the university. SecondChildUserIdTAG: 623210 SecondChildUserNameTAG: preveen SecondChildCreateTimeTAG: 2012-12-24T05:53:37Z SecondChildTAG: Hi praveen, have you taken a Course in Coursera? If yes, just for curious, which one it was? :p Sorry for my ignorance haha, but How does they compare the difficult of the Course Campus vs the Online version ? Do you have Students from the Campus taking the same Course at the same time? SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-24T06:09:54Z SecondChildTAG: Oh, my english is terrible haha. I mean difficulty and Do they have Students from the Campus taking... SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-24T06:11:46Z SecondChildTAG: I tried to take some courses at coursea but dropped out mainly because of lack of quality. Lectures were not really intuitive at all. they were introductions to genetic and engineering and computational investing. I don't think students from campus were there, but TAs are from the campus the quiz / homework platform is not as aesthetic as edx's. Only one try was allowed on the courses I took:=( generally a "turn off" for enthusiastic learners. As a result the *noise* on the forum tends to be very less SecondChildUserIdTAG: 623210 SecondChildUserNameTAG: preveen SecondChildCreateTimeTAG: 2012-12-24T06:47:08Z SecondChildTAG: I agree with you skyhawk . I had no ideea about coursera . Thank you ! SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-12-24T22:28:01Z SecondChildTAG: Thank you again for the heads up about coursera . It is amazing . I had no ideea about it . SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-12-24T22:56:23Z FirstChildTAG: Oh, edX hasn't heard the last from me. I have signed up for 6.00x and CS191x (How can one resist a course whose logo is the superposition of a dead cat and a live cat!) in the new year. I've started watching the videos from 6.00x for last semester. I like the professor's lecturing style. It is relaxed but very systematic. As far as 6.002x, I'm a realist. The course content and sequence is wedded to the professor's textbook, so it's not going to change. I just hope that there are some here who will look outside in the larger world and see what it has to offer. There are some good (perhaps even very good) things there too! FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-12-23T22:56:37Z SecondChildTAG: Hello skyhawk. Did you know that the image of cats comes of the allegory of Schrödinger's cat to explain Quantum Physics? SecondChildUserIdTAG: 361823 SecondChildUserNameTAG: EliasOak SecondChildCreateTimeTAG: 2012-12-24T03:50:49Z SecondChildTAG: Yes, I'm a theoretical physicist by original education. I did my research in computational quantum mechanics. Now among my varied interests are the philosophical foundations of quantum mechanics. Such topics as well as electronics are far removed from the way I earn my living. These are the ways I relax. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-12-24T04:46:02Z SecondChildTAG: Wow skyhawk, I didn´t knew that you were theoretical physicist. Many of my work mates are physicists. Are currently you working in Research? Will you take CS191.x in edX? Take care, Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-24T05:14:05Z SecondChildTAG: Haha, I have read your post response again, I am distracted, I have noticed that you have already signed up for CS191.x :p. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-24T05:17:57Z SecondChildTAG: See you in CS191x . Now i know who to ask..., if i will make it further than week 2 . :) SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-12-24T22:30:38Z SecondChildTAG: I am also Theoretical Physicist. I studied quantum physics at the University and currently read about the philosophical aspects of modern physics. SecondChildUserIdTAG: 361823 SecondChildUserNameTAG: EliasOak SecondChildCreateTimeTAG: 2012-12-24T22:56:27Z SecondChildTAG: When I received my PhD in 1972 there were precious few jobs in physics. Like many of my classmates I took a job at the Manned Spacecraft Center/Johnson Space Center and worked on the early development of the space shuttle. I worked on simulation in support of the development of the avionics system (GN&C). I left after the first atmospheric drop test (Approach and Landing Test) and went back to school to study and teach chemical engineering. My research area was in chemical process dynamics. My dissertation topic was "The Wrong Way Effect". Before writing the final draft of my dissertation one of my former students introduced me to his boss at a petroleum consulting firm, and I was offered a position in petroleum reservoir simulation that was too good to refuse. As a result of the crash in the petroleum industry in 1986, I moved to a job with a contractor at a US department of energy National Lab in 1987. I left my old job on a Thursday, drove across country and reported for work at my new job on Monday! I've been in that job nearly 26 years. After the move, for four semesters I also taught a senior/graduate course on math modelling/numerical methods in the chemical and petroleum engineering department of a nearby university. My specialty is modelling production from shale reservoirs. It's a hot topic these days! Currently I am 2 1/2 years past my normal retirement age. With all that background in engineering I still have a love for physics and quantum mechanics. I am currently reading "The Quantum Challenge" by Greenstein and Zajonc. I have enjoyed watching the video sequences on physics by Leonard Susskind of Stanford. And the basic physics sequence by MIT's Walter Lewin is the "gold standard". And the old videos like "The Character of Physical Law" by Feynman are wonderful. Of course, Feynman was **THE** physics teacher to my generation. When a professor from my university taught a course on quantum mechanics out of Feynman volume 3 (the red books) I first got a glimpse of the wonder of quantum mechanics. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-12-25T00:33:34Z SecondChildTAG: His resume is very impressive. I notice that you took the very conditions that his country offers a profession poeuco highlighted, at least in my country. I am Brazilian and here physics is undervalued. I do my studies by passion and altruism. I graduated in 2000 and did Masters in Electrical Engineering in the area of ​​Power Systems. I started my PhD in the same line of research. Richard Feynman consider a major reference in physics, especially for those who like electronics, after all he is one of the pioneers of nanotechnology. You also quotes Professor Walter Lewin. I'm currently doing extensive work on popular science using materials from MIT OpenCourse. It was through him that I found this course of EDX. I love reading about the philosophy of science. I like the bedside livrom Structure of Scientific Revolutions, Thomas Kuhn. Other authors that I really like is Karl Popper, Lakatos and Feyerabend, though the latter is not that I most admire. Some authors you mentioned that I know of, but I'll look for them yourself. Here in Brazil there are many people who are interested in physics, especially in the philosophical aspects. Mitas times we feel like lone wolves, but with the Internet I have connected with many people from other countries. SecondChildUserIdTAG: 361823 SecondChildUserNameTAG: EliasOak SecondChildCreateTimeTAG: 2012-12-25T14:20:07Z SecondChildTAG: His resume is very impressive. I notice that you took the very conditions that his country offers a little detached profession, at least in my country. I am Brazilian and here physics is undervalued. I do my studies by passion and altruism. I graduated in 2000 and did Masters in Electrical Engineering in the area of ​​Power Systems. I started my PhD in the same line of research. Richard Feynman consider a major reference in physics, especially for those who like electronics, after all he is one of the pioneers of nanotechnology. You also quotes Professor Walter Lewin. I'm currently doing extensive work on popular science using materials from MIT OpenCourse. It was through him that I found this course of EDX. I love reading about the philosophy of science. I like the pillow book Structure of Scientific Revolutions, Thomas Kuhn. Other authors that I really like is Karl Popper, Lakatos and Feyerabend, though the latter is not that I most admire. Some authors you mentioned that I know of, but I'll look for them yourself. Here in Brazil there are many people who are interested in physics, especially in the philosophical aspects. Often we feel like lone wolves, but with the Internet I have connected with many people from other countries. SecondChildUserIdTAG: 361823 SecondChildUserNameTAG: EliasOak SecondChildCreateTimeTAG: 2012-12-25T14:30:12Z SecondChildTAG: lol SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-12-26T00:10:26Z SecondChildTAG: Wow skyhawk, you must be intelligent ! That explains a lot :p Would you like to continue teaching in the future? I don´t know if you are interested, but with ashwith and rharris we will prepare some materials for the students of the next term, we would like to collect some topics that were consider difficult and try to make some video explanations for the Wiki, for complementary support of their studies. We haven´t started yet. If you are interested you are welcome, you can e-mail me to myrimit at gmail dot com :). That would be awesome! Take care, Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-26T03:20:34Z FirstChildTAG: Hi,nice to meet you! I really admire you that you take the course in science world even you have a liberal arts undergraduate degree and a law degree. On my part,science courses are mostly not easy to understand in one semester,many basic concepts and tools like math...we may hope we can learn and command the knowledge after taking the edx course immediately,but actually it is kind of impossible as edx platform does not provide real lab situation---experiments are important for science students to deeply understand those difficult concepts. If possible, I suggest you do some real labs,which is of great fun! Just like you, when I returned to the previous homeworks or labs,sometimes I forgot how I solved.So I am considering taking the next spring 6.002x course again and have a deep understanding.Deadline might be a good thing,and I may be lazy enough to forget my course without it,though it truly made me ignore something that should be reconsidered and jump to next week assignment.Anyway, it is a great experience here,with so many helpful friends(ps:I often see your name in the discussion forum) . Best wishes to you! FirstChildUserIdTAG: 361137 FirstChildUserNameTAG: Ericson FirstChildCreateTimeTAG: 2012-12-24T05:55:34Z FirstChildTAG: Heartily Thanks to EDX team for providing such great platform with free education to the world. For 6.002x course : I am a software guy , but I always had keen interest in Electronics and wanted to have such knowledge in this field , so that i can make simple gadgets myself , something near to what Tony shark does in IRON MAN , at least something like my own radio receiver,transmitter . But didnt know where to start from , this course provided me with that direction, and with professor like Mr. Agarawal , it became very easy and interesting , after all we have faculty from best Technical institute of the world. And Video sequences are such that almost no doubt remains after viewing them about concerned topic, one good thing of Mr. Agarwal he always tried to co relate the new things with old, thus making our knowledge as continuous flow of topics each chained to one another. These days I have been very busy in my job , so this time i couldnt make in final exams because I left many lecture sequences in a row in second half of the course , i did well in first half , but instead of getting disappointed , iam taking it as I have learned alot because i didnt enrolled this course hust to get the grade , i took because of my interest in electronics , Iam lookinf forward to again enroll in this course , till then irrespective of the fact that course has been completed i continue to fill gaps in my 6.002x knowledge. I always thought that if I had came across with such a great thing in my student life , i would have made much out of it, but still "There is never too late" One request with 6.002x team is that although you provide us lab session but along with that if with every week you can guide us with little kind of project so as to how we can put out knowledge of that session into real working hardware , MOSFETS etc , just to get help in getting started in putting ideas into real hardware. Again many thanks to 6.002x team for enlightening world . FirstChildUserIdTAG: 187466 FirstChildUserNameTAG: a9verma FirstChildCreateTimeTAG: 2012-12-25T07:05:53Z FirstChildTAG: Well, I do not share the views of Beneficial about traditional teaching and learning. I have been teaching at University for more than ten years by now and I think there are valuable things to keep from class lectures dynamics. On the other hand, I encourage Beneficial to keep on learning about science and maths. There are many resources available in the net, and many people willing to help and share knowledge and ways to expand it. Merry Christmas and Happy New Year to everybody! FirstChildUserIdTAG: 341020 FirstChildUserNameTAG: franjescribano FirstChildCreateTimeTAG: 2012-12-27T10:40:00Z SecondChildTAG: I agree about the traditional teaching and learning, especially when the lecturer is video recorded, enabling one to view it multiple times if necessary. A lot, if not most of the free courses at MIT so far do not have lecture videos, therefore I don't consider them near as valuable as the ones that do. I also take exception with Beneficial's characterization of Dr. Agarwal's abilities. I believe he is an excellent professor and I believe there should be more like him. I've taken classes from many professors over the years and I believe he is one of the best. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-12-30T19:21:54Z IndexTAG: 212 TitleTAG: Thanks MitX Team and TA for your work and help My native language is spanish but i want to say you thank you very much, although my English is not so good. I finish the course and the final exam. I hope to continue learning and help others students too. I am very grateful to Professor Anant Agarwal, his team and the teacher assistant for giving me and give us the opportunity to study again. I am also very grateful to those who have written in the forum, sharing ideas and clarifying ways. As the years pass, when we have family, work and little time, is imposible to continue studying at an university going to it. This kind of courses allow us to continuing studing. I felt like this course as a very good course, and I fill, you did the best work and help for helping us to go ahead. That motivates us to continue studying, to continue learning. Thank you very much, thank you Anant Agarwal!, team! and TA! and go ahead! Luis Bonelli Finalicé el curso y el examen final. Deseo seguir aprendiendo y también en lo que pueda ayudar a otros estudiantes. Estoy muy agradecido al profesor Anant Agarwal, a su equipo y a los teacher assistant por darme y darnos la posibilidad de estudiar nuevamente. Estoy muy agradecido también con quienes han escrito en el foro, clarificando ideas y compartiendo caminos. Cuando los años pasan, hay familia y trabajos es difícil poder seguir estudiando en una universidad en modo presencial. Esta clase de cursos nos permiten continuar estudiando. Lo he sentido como un muy buen curso y sobre todo siento que han hecho el mejor trabajo y ayuda, para ayudarnos a seguir adelante. Eso nos motiva a nosotros a seguir estudiando, a seguir aprendiendo. Muchas gracias, muchas gracias Anant Agarwal, a Ud., a su equipo y a TA, y a seguir adelante !!! Luis Bonelli UserIdTAG: 227223 UserNameTAG: luisbonelli CreateTimeTAG: 2012-12-22T02:09:41Z VoteTAG: 9 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: That is so nice to know Luis :). I am happy that you have enjoyed this Course, 6.002x is amazing ! edX is doing an excellent work, all the Team is great! And as I always say, Prof. Agarwal is really cool haha! Take care and my best wish to you, Myriam. P.D. Que pases unas lindas fiestas y mi mejor deseo para ti para todo lo que emprendas de aquí en adelante!:) FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-22T19:51:00Z SecondChildTAG: Muchas gracias Myriam por tus palabras y deseos. Tu me has ayudado mucho para que en este curso haya podido llegar hasta el final. Muchas gracias, no solamente por los conocimientos que compartistes, también por tus palabras de fuerza, tratando de ayudar al máximo para que compañeros no abandonaran y siguieran con nosotros hasta finalizar el curso. También para tí mis deseos de unas lindas fiestas y no tengo dudas que te irá bien en lo que emprendas, por todo el bien que haces y tu valor como persona. Un saludo con cariño desde Uruguay. Luis SecondChildUserIdTAG: 227223 SecondChildUserNameTAG: luisbonelli SecondChildCreateTimeTAG: 2012-12-22T20:23:59Z SecondChildTAG: Que lindas palabras, muchísimas gracias Luis. Estoy muy feliz de haberles sido de ayuda aquí. Sí, la verdad es que a mi siempre me ha gustado darle el empujón a aquél que lo necesita, a veces, es necesario una palabra de aliento. Estoy muy feliz de poder haberles sido de ayuda no solo en la parte académica sino también en la parte humana, que es muy importante. Un abrazo grande Luis, que la pases muy lindo en estas Fiestas! Saludos a todo Uruguay! Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-24T02:03:39Z IndexTAG: 213 TitleTAG: Certificates? Well first congrats to everyone who passed the class! :D I was just wondering when and how we will get the certificate for completing the class? Will they mail it or will we just be able to print it out or something? UserIdTAG: 219780 UserNameTAG: ReconIII CreateTimeTAG: 2012-12-20T16:56:24Z VoteTAG: 9 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: You will be able to print it, I'm not sure when, check in the "progress" area. Congratulations! FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-12-20T17:09:37Z SecondChildTAG: Thanks! And yeah I don't see it yet :L I suppose it will be up when the final is over for everyone. Or maybe just when my 24 hours is offically over. SecondChildUserIdTAG: 219780 SecondChildUserNameTAG: ReconIII SecondChildCreateTimeTAG: 2012-12-20T18:36:40Z FirstChildTAG: **i completed the cs188.1X and this how we get the certificate** :![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13560363931343612.jpg FirstChildUserIdTAG: 236136 FirstChildUserNameTAG: Hamid-ch FirstChildCreateTimeTAG: 2012-12-20T20:46:50Z SecondChildTAG: Thanks Hamid! SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-12-20T20:57:17Z SecondChildTAG: Cool, and congrats! :D When did that show up for you? Was it after everyone completed the final? SecondChildUserIdTAG: 219780 SecondChildUserNameTAG: ReconIII SecondChildCreateTimeTAG: 2012-12-20T20:59:14Z SecondChildTAG: i dont remember exactely but , it's between 1 week and 10 days SecondChildUserIdTAG: 236136 SecondChildUserNameTAG: Hamid-ch SecondChildCreateTimeTAG: 2012-12-20T21:03:07Z SecondChildTAG: Reconlll i recomend you to change your name (put your real full name), in the certificate they will put the name you subscribed with SecondChildUserIdTAG: 236136 SecondChildUserNameTAG: Hamid-ch SecondChildCreateTimeTAG: 2012-12-20T21:05:51Z SecondChildTAG: Well ReconIII is just my display name/username, when you sign up it asks you for your full name then a display name/username. So won't the certificate have my full name on it? Or are you saying they only put your display name/username on it? SecondChildUserIdTAG: 219780 SecondChildUserNameTAG: ReconIII SecondChildCreateTimeTAG: 2012-12-20T21:24:11Z SecondChildTAG: aha ok , then no problem (they will put the full name, not the display) SecondChildUserIdTAG: 236136 SecondChildUserNameTAG: Hamid-ch SecondChildCreateTimeTAG: 2012-12-20T21:29:00Z SecondChildTAG: I want to express my congratulations Hamid ! That was a tough one in my opinion, since the period was so short . I have also subscribed, but i realized that i'm not good at probabilities and programming.Will take first the programming course at MIT. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-12-20T22:41:36Z SecondChildTAG: thanks Alex , for the first part (cs188.1x)you dont need a strong probabilities (take it easy),but good programming is required SecondChildUserIdTAG: 236136 SecondChildUserNameTAG: Hamid-ch SecondChildCreateTimeTAG: 2012-12-20T23:45:14Z SecondChildTAG: thanks Hamid SecondChildUserIdTAG: 404188 SecondChildUserNameTAG: jishudasorissa SecondChildCreateTimeTAG: 2012-12-26T13:50:56Z SecondChildTAG: u r welcom jishuda SecondChildUserIdTAG: 236136 SecondChildUserNameTAG: Hamid-ch SecondChildCreateTimeTAG: 2012-12-26T15:20:34Z FirstChildTAG: and will it have the grade in it? FirstChildUserIdTAG: 173147 FirstChildUserNameTAG: cruiser_rahit FirstChildCreateTimeTAG: 2012-12-22T14:18:41Z SecondChildTAG: no SecondChildUserIdTAG: 236136 SecondChildUserNameTAG: Hamid-ch SecondChildCreateTimeTAG: 2012-12-23T22:42:24Z FirstChildTAG: We are hoping to distribute certificates before the end of the year to students who passed the course. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-12-21T13:37:26Z SecondChildTAG: i hope so SecondChildUserIdTAG: 236136 SecondChildUserNameTAG: Hamid-ch SecondChildCreateTimeTAG: 2012-12-21T13:53:36Z SecondChildTAG: Just for curiosity... we can download the certificate or have a option to receive the "paper" my regular mail? SecondChildUserIdTAG: 153313 SecondChildUserNameTAG: fernandofei SecondChildCreateTimeTAG: 2012-12-21T17:02:28Z SecondChildTAG: download the certificat (PDF format) SecondChildUserIdTAG: 236136 SecondChildUserNameTAG: Hamid-ch SecondChildCreateTimeTAG: 2012-12-21T19:59:45Z SecondChildTAG: Thanks, **Lyla**! Will there be an option to include the grade as in 6.002x Spring 2012? Jersey Mark SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-12-25T16:54:02Z SecondChildTAG: hey lyla thanks for ur cooperation throught the course ......a big big thanks.... SecondChildUserIdTAG: 358198 SecondChildUserNameTAG: Jay1492 SecondChildCreateTimeTAG: 2012-12-26T17:50:23Z SecondChildTAG: Nope - there will not be an option to include a grade on your certificate. SecondChildUserIdTAG: 96452 SecondChildUserNameTAG: Lyla SecondChildCreateTimeTAG: 2012-12-26T18:15:10Z IndexTAG: 214 TitleTAG: Open-Badges for Course Completion Hello, First of all, I'd like to thank everyone for the great course. Now that I've spent a couple months, and have scored an A, I wanted to know whether it would be possible to receive an [Open-Badge][1] for the completion of the course. Open-Badges is a project by Mozilla, which aims to create a platform for allowing recognition of abilities and achievements. It would be really nice if in addition to certificates one could get an Open-Badge for the course. For more information on Open-Badges, one can see their [site][1], the [wiki][2], and the [GitHub project][3]. Thank you! [1]: http://openbadges.org/en-US/ [2]: https://wiki.mozilla.org/Badges [3]: https://github.com/mozilla/openbadges UserIdTAG: 414306 UserNameTAG: YN300 CreateTimeTAG: 2012-12-20T16:50:02Z VoteTAG: 9 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 215 TitleTAG: MITx is the best ! Don't you agree ? I was always afraid of talking about electronics because of my previous "full-of-failures" experiences in this domain. Now, I am proud to say that I LOVE this course very much because of Mr. Anant. I don't find the words to express my gratutude to him. He helped me to bypass my fears toward electronics. Thank you so much. Without to forget the TA community because of their wonderfully important feedbacks. This course itself prooved to me without a doubt that MIT deserve to be the BEST university at all times. Thank you. I hope I get to see other courses from MITx. I am so eager for them. UserIdTAG: 418182 UserNameTAG: euldji2005 CreateTimeTAG: 2012-12-18T00:58:38Z VoteTAG: 9 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 216 TitleTAG: Congratulations Dr. Agarwal and your staff. Thank you. UserIdTAG: 260994 UserNameTAG: capvl CreateTimeTAG: 2012-12-13T20:07:52Z VoteTAG: 9 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 217 TitleTAG: When starts 6.014? In the week 14 video S28V5 I listen, that 6.014 course will be in the future. When we can see that course on Edx?) UserIdTAG: 104522 UserNameTAG: IgorNovice CreateTimeTAG: 2012-12-11T18:54:30Z VoteTAG: 9 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 3 FirstChildTAG: Hope to see 6004x.. FirstChildUserIdTAG: 136959 FirstChildUserNameTAG: christerpher FirstChildCreateTimeTAG: 2012-12-14T02:56:47Z FirstChildTAG: And will Dr.Agarwal be the one teaching the course? FirstChildUserIdTAG: 232157 FirstChildUserNameTAG: GayatriTR FirstChildCreateTimeTAG: 2012-12-12T01:53:25Z FirstChildTAG: http://web.mit.edu/6.014/www/index.html FirstChildUserIdTAG: 136489 FirstChildUserNameTAG: ShahabSafa FirstChildCreateTimeTAG: 2012-12-12T06:37:18Z IndexTAG: 218 TitleTAG: What we knew or refreshed in our memory during this course In the beggin was Ohm law![enter image description here][1] Then we "knew" about parallel or series connextion of elements![series and parallel connection][2] Do you remember about voltage devider? The next step was our favorit theorems! Then was nonlinear elements![enter image description here][3] [tvin diode][4] And MOSFET with amplifiers ![enter image description here][5] ![enter image description here][6] Of course energy storage elements was in our course like this ![capasitor][7] or this![inductor][8] parallel and series RLC sinusoidal response. Some logic elements;digital amplifiers and filters. Don`t forget about [wolfram Alpha][9] and [Khan Academy][10] . Also was useful to knew about [desmos][11] I hope i didn`t forget anything! ;-) In other way I will be grateful for your additions [1]: https://edxuploads.s3.amazonaws.com/13545493941343671.png [2]: https://edxuploads.s3.amazonaws.com/13545496811343649.gif [3]: https://edxuploads.s3.amazonaws.com/13545502421343651.gif [4]: http://www.r-type.org/pdfs/6al5.pdf [5]: https://edxuploads.s3.amazonaws.com/13545504861343697.gif [6]: https://edxuploads.s3.amazonaws.com/13545505121343645.gif [7]: https://edxuploads.s3.amazonaws.com/13545507821343632.jpg [8]: https://edxuploads.s3.amazonaws.com/13545508731343661.jpg [9]: http://www.wolframalpha.com/ [10]: http://www.khanacademy.org/ [11]: https://www.desmos.com/ UserIdTAG: 96873 UserNameTAG: Santyaga CreateTimeTAG: 2012-12-03T16:22:31Z VoteTAG: 9 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: Don't forget the op-amps! By the way: nice coil, with silver-galvanized wire (I suppose). Think it's part of an antenna-tuner for a short-wave transmitter, because I see so many taps on it. 73 de PA2LTH. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-12-03T17:05:59Z FirstChildTAG: You are right Salero! But usually I prefer to use MFJ-969 antenna tuner. I hope see you on the bands! 73 de RZ3DBN FirstChildUserIdTAG: 96873 FirstChildUserNameTAG: Santyaga FirstChildCreateTimeTAG: 2012-12-03T19:11:52Z FirstChildTAG: Cool Post ! :) FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-03T23:07:58Z SecondChildTAG: Thank you Myrimit! ;-) SecondChildUserIdTAG: 96873 SecondChildUserNameTAG: Santyaga SecondChildCreateTimeTAG: 2012-12-05T17:21:33Z FirstChildTAG: is it an air-gang capacitor? only read about it. FirstChildUserIdTAG: 357747 FirstChildUserNameTAG: kishores FirstChildCreateTimeTAG: 2012-12-07T12:50:55Z FirstChildTAG: Yes Kishores! You are right! :) FirstChildUserIdTAG: 96873 FirstChildUserNameTAG: Santyaga FirstChildCreateTimeTAG: 2012-12-08T11:31:20Z IndexTAG: 219 TitleTAG: More advanced courses in electrical engineering? This course has been awesome. Can you speak about any plans in the works to offer courses further along in the MIT curriculum? Rob UserIdTAG: 468623 UserNameTAG: RobNik CreateTimeTAG: 2012-11-29T00:09:52Z VoteTAG: 9 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Sister Maryum, Can you throw any light on this subject question? Almost day has passed without response. FirstChildUserIdTAG: 455950 FirstChildUserNameTAG: Wahabbaluch FirstChildCreateTimeTAG: 2012-11-30T05:40:25Z IndexTAG: 220 TitleTAG: S22V7 How is the mag(Vc/Vi) more than 1 in critical damping case, where Q (peak value at w0) is 1/2 ? UserIdTAG: 185815 UserNameTAG: adilmeersha CreateTimeTAG: 2012-11-26T08:49:19Z VoteTAG: 9 CoursewareTAG: Week 11 / S22V7: Using_Q_to_compare_TD_and_FD_critically_damped_case CommentableIdTAG: 6002x_S22V7_Using_Q_to_compare_TD_and_FD_critically_damped_case NumberOfReplyTAG: 3 FirstChildTAG: i tested dat ckt in sandbox also,and it seems the plots shown in the lecture should be wrong! @Myrimit help me please..! FirstChildUserIdTAG: 185815 FirstChildUserNameTAG: adilmeersha FirstChildCreateTimeTAG: 2012-11-26T22:22:11Z FirstChildTAG: In fact the magnitude of Vc/Vi has not a peak when Q=1/2 but when Q=1, so the plot in the lecture refers to the case Q=1 (but it's not critical dumping): in that case (Q=1) the peak is greater than one. Besides, and more importantly, the peak is not at w0 but at a little bit lower value. Of course at w0 the magnitude equals Q so it is 1. You can even check the textbook at pag. 812 fig. 14.36 where the magnitudes for different values of Q are plotted. FirstChildUserIdTAG: 376877 FirstChildUserNameTAG: AndBre FirstChildCreateTimeTAG: 2012-11-29T11:12:50Z FirstChildTAG: same question@Myrimit FirstChildUserIdTAG: 136959 FirstChildUserNameTAG: christerpher FirstChildCreateTimeTAG: 2012-11-30T05:42:38Z IndexTAG: 221 TitleTAG: Ohm's Law ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13534556991343605.jpg UserIdTAG: 127195 UserNameTAG: princeofsudan CreateTimeTAG: 2012-11-20T23:55:39Z VoteTAG: 9 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 6 FirstChildTAG: wow !!! FirstChildUserIdTAG: 477713 FirstChildUserNameTAG: ikm104 FirstChildCreateTimeTAG: 2012-11-21T05:54:51Z FirstChildTAG: +1 FirstChildUserIdTAG: 323416 FirstChildUserNameTAG: Alexxkr FirstChildCreateTimeTAG: 2012-11-21T00:47:57Z FirstChildTAG: lol nice that is a cool way to think of it FirstChildUserIdTAG: 219780 FirstChildUserNameTAG: ReconIII FirstChildCreateTimeTAG: 2012-11-21T02:07:59Z FirstChildTAG: rule 34 FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-11-21T06:09:15Z FirstChildTAG: The horror. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-11-21T12:49:08Z FirstChildTAG: Бедный Ампер! Poor Amper! FirstChildUserIdTAG: 208296 FirstChildUserNameTAG: OZ1 FirstChildCreateTimeTAG: 2012-11-28T06:58:52Z IndexTAG: 222 TitleTAG: Hints for H7P3. - Q1)Remember Inductor acts as a short circuit initially when there is no initial current. Now what will be V0..Simple voltage divider pattern. ---------- 2.q2)Time constant =L/R. To find R , draw the thevenin equivalent for the given circuit. ---------- Q3)Il(current through inductor)=iro(current through Ro).We know RHS, since V0 is given. And LHS is simply the Equation 10.59 from the text.So what would be t? ---------- Q4)Simple elementary calculations..SPEED=DISTANCE/TIME. ---------- Hope this helps...... UserIdTAG: 148523 UserNameTAG: span993 CreateTimeTAG: 2012-11-04T12:51:40Z VoteTAG: 9 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: @pedroramus:did u get the answer?? FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-11-04T15:13:03Z SecondChildTAG: PLEAse see that the answer is asked in nanoseconds SecondChildUserIdTAG: 148523 SecondChildUserNameTAG: span993 SecondChildCreateTimeTAG: 2012-11-04T15:13:52Z SecondChildTAG: My L = 15.44nH, so the answer is nanoseconds. SecondChildUserIdTAG: 366669 SecondChildUserNameTAG: pedroramus SecondChildCreateTimeTAG: 2012-11-04T15:16:16Z SecondChildTAG: But the answer is asked in nano seconds so if the answer is mE(-9) then u should enter m in the checkbox. SecondChildUserIdTAG: 148523 SecondChildUserNameTAG: span993 SecondChildCreateTimeTAG: 2012-11-04T15:18:05Z SecondChildTAG: I didn't understand, because my L is in nH, so my answer is implicit that will be in nS. Didn't? SecondChildUserIdTAG: 366669 SecondChildUserNameTAG: pedroramus SecondChildCreateTimeTAG: 2012-11-04T15:25:26Z SecondChildTAG: for say let your answer be 9nanoseconds then you should enter 9 in the check box since the answer they have asked is in nanoseconds. Try doing it.. SecondChildUserIdTAG: 148523 SecondChildUserNameTAG: span993 SecondChildCreateTimeTAG: 2012-11-04T15:27:23Z SecondChildTAG: This or you probably made some mistake in your procedure. SecondChildUserIdTAG: 148523 SecondChildUserNameTAG: span993 SecondChildCreateTimeTAG: 2012-11-04T15:28:25Z SecondChildTAG: In my question Vo = 2.5V, Ro = 50ohm, Rs = 22ohm, L = 15.44nH. Assembling the problem: iL = (V/R)*(1 - e^(-(R/L)*t)) 2.5/50 = (5/77)*(1 - e^(-(77/15.44)*t)) Isn't correct? SecondChildUserIdTAG: 366669 SecondChildUserNameTAG: pedroramus SecondChildCreateTimeTAG: 2012-11-04T15:43:15Z SecondChildTAG: Very correct, except that 50+22 is 77. SecondChildUserIdTAG: 148523 SecondChildUserNameTAG: span993 SecondChildCreateTimeTAG: 2012-11-04T16:02:18Z SecondChildTAG: SORRY 72 SecondChildUserIdTAG: 148523 SecondChildUserNameTAG: span993 SecondChildCreateTimeTAG: 2012-11-04T16:02:30Z SecondChildTAG: LOL!!!!!!!!!!!!!! I don't believe. Thank you span993. Was my mistake. You help me a lot. SecondChildUserIdTAG: 366669 SecondChildUserNameTAG: pedroramus SecondChildCreateTimeTAG: 2012-11-04T16:14:12Z SecondChildTAG: Finally :D SecondChildUserIdTAG: 148523 SecondChildUserNameTAG: span993 SecondChildCreateTimeTAG: 2012-11-04T16:20:05Z SecondChildTAG: Help me... Vo=1.65 Ro = 50ohm, Rs = 22ohm, L = 15.44nH. 1.65/50 = (3.3/72)(1 - e^(-(72/15.44)*t)) I get number but it isn't correct.... SecondChildUserIdTAG: 192114 SecondChildUserNameTAG: chuba SecondChildCreateTimeTAG: 2012-11-04T16:29:38Z SecondChildTAG: http://www.wolframalpha.com/input/?i=1.65%2F50%3D%283.3%2F%2850%2B22%29%29*%281-e%5E%28-%2850%2B22%29%2F15.44%29*t%29%29 I even used this but it isn't also correct... SecondChildUserIdTAG: 192114 SecondChildUserNameTAG: chuba SecondChildCreateTimeTAG: 2012-11-04T16:30:48Z SecondChildTAG: Try again, You definitely made some mistake in calculations. SecondChildUserIdTAG: 148523 SecondChildUserNameTAG: span993 SecondChildCreateTimeTAG: 2012-11-04T16:37:36Z SecondChildTAG: I found) Thanks) SecondChildUserIdTAG: 192114 SecondChildUserNameTAG: chuba SecondChildCreateTimeTAG: 2012-11-04T16:41:53Z SecondChildTAG: still stuck with this problem. SecondChildUserIdTAG: 352553 SecondChildUserNameTAG: AbbyMacalla SecondChildCreateTimeTAG: 2012-11-04T16:48:02Z SecondChildTAG: okay got it...thanks a lot =) SecondChildUserIdTAG: 352553 SecondChildUserNameTAG: AbbyMacalla SecondChildCreateTimeTAG: 2012-11-04T17:09:09Z SecondChildTAG: Jst Cnt Get Q3 SecondChildUserIdTAG: 478827 SecondChildUserNameTAG: Abbas1989 SecondChildCreateTimeTAG: 2012-11-04T17:50:30Z SecondChildTAG: Done...phew! SecondChildUserIdTAG: 478827 SecondChildUserNameTAG: Abbas1989 SecondChildCreateTimeTAG: 2012-11-04T18:06:41Z SecondChildTAG: Why is V0 2.5. I did't get that. can you please explain? SecondChildUserIdTAG: 244670 SecondChildUserNameTAG: ravikiran1201 SecondChildCreateTimeTAG: 2012-11-04T20:19:28Z SecondChildTAG: Thanks Span993 for pointing out error, I was doing the same mistake of taking sum of 50 and 22 as 77. A simple mistake causing 3 days unrest. But for searching the way for solution, have seen many other related topics of interest. Anyhow it is done. Thanks to all for helping your mates, special thanks to sister Myrimit. SecondChildUserIdTAG: 455950 SecondChildUserNameTAG: Wahabbaluch SecondChildCreateTimeTAG: 2012-11-05T04:29:12Z FirstChildTAG: span993, the inductor acts as a short circuit NOT when there is no initial current, but when there is no variation in the current so that di/dt = 0 and consequently v = 0. FirstChildUserIdTAG: 376877 FirstChildUserNameTAG: AndBre FirstChildCreateTimeTAG: 2012-11-04T16:46:03Z SecondChildTAG: Sorry My mistake, voltage across inductor is -ldi/dt becomes zero coz i is constant(for dc source). And also at low frequency the Impedence offered by inductor (2*pi*f*L) is negligilble. Thus we consider short circuit..Thanks Andbre. SecondChildUserIdTAG: 148523 SecondChildUserNameTAG: span993 SecondChildCreateTimeTAG: 2012-11-04T17:19:14Z SecondChildTAG: Exactly! ;-) SecondChildUserIdTAG: 376877 SecondChildUserNameTAG: AndBre SecondChildCreateTimeTAG: 2012-11-04T21:11:47Z SecondChildTAG: Thanks Span993 for pointing out error, I was doing the same mistake of taking sum of 50 and 22 as 77. A simple mistake causing 3 days unrest. But for searching the way for solution, have seen many other related topics of interest. Anyhow it is done. Thanks to all for helping your mates, special thanks to sister Myrimit. SecondChildUserIdTAG: 455950 SecondChildUserNameTAG: Wahabbaluch SecondChildCreateTimeTAG: 2012-11-05T04:28:34Z FirstChildTAG: Thank you for your contribution span993 :)! FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-04T13:25:20Z SecondChildTAG: :) SecondChildUserIdTAG: 148523 SecondChildUserNameTAG: span993 SecondChildCreateTimeTAG: 2012-11-04T14:02:42Z SecondChildTAG: I didn't manager to find the solution. By the equation 10.59 I found Iro = IL = Vo/Ro and V = 5V and R = Rs+Ro. But the solution is not correct. Where's my mistake? Please help me. SecondChildUserIdTAG: 366669 SecondChildUserNameTAG: pedroramus SecondChildCreateTimeTAG: 2012-11-04T15:07:21Z SecondChildTAG: The question above is about H7P3 Q3 SecondChildUserIdTAG: 366669 SecondChildUserNameTAG: pedroramus SecondChildCreateTimeTAG: 2012-11-04T15:12:47Z IndexTAG: 223 TitleTAG: Notation I can't seem to establish the "correct" format for this exercise. What is meant by a "standard ordering convention"? UserIdTAG: 286880 UserNameTAG: herbsteiner CreateTimeTAG: 2012-10-26T22:19:47Z VoteTAG: 9 CoursewareTAG: Week 8 / Review A Step Up CommentableIdTAG: 6002x_Review_A_Step_Up NumberOfReplyTAG: 3 FirstChildTAG: To make sure that you are inputting the order of operations correctly. Here is some additional information. http://en.wikipedia.org/wiki/Order_of_operations FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-26T22:58:38Z SecondChildTAG: I'm having the same problem. I believe my order of operations are correct. [ () first; * or / second; addition, subtraction last. If there is an exponent, it will be evaluated first, following the previous mentioned order of operations], however the system doesn't like my answer. My question is, if the syntax is correct, no matter if the answer is correct or not, will the system only give me a red x, or is there any situation where the answer is incorrect, the syntax is correct but I still get complaints about the syntax? Here is the general form of my answer: Z*(1-e^((x-y)*(A/B))) Any idea what is wrong? SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-11-01T21:54:11Z SecondChildTAG: Never mind, my format worked. I had the correct answer, although in a different order, however in the confusion of trying to get the syntax correct for both answers I must have missed a syntax error in the 2nd question, as the answer was correct. Regarding the first question, I never expected the answer to be in that form ( IS*u(t-t0) ). Since t0 ≠ 0, wouldn't the answer also be: IS*(1-e^((t0-t)*(R/L)))? SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-11-01T22:08:35Z SecondChildTAG: I must agree with rharris about the answer of the first question. I expected: IS*(1-e^(-t*(R/L))) since t0 should be zero. SecondChildUserIdTAG: 296965 SecondChildUserNameTAG: LGMailhos SecondChildCreateTimeTAG: 2012-11-06T04:57:08Z FirstChildTAG: Invalid input: Is not permitted in answer this what i see FirstChildUserIdTAG: 322638 FirstChildUserNameTAG: khalid1988 FirstChildCreateTimeTAG: 2012-11-10T12:49:04Z SecondChildTAG: try IS (capital I and capital S) SecondChildUserIdTAG: 69075 SecondChildUserNameTAG: ErikR SecondChildCreateTimeTAG: 2012-11-10T18:10:29Z FirstChildTAG: I have the same problem: "Invalid Input: u not permitted in answer", any suggestions? FirstChildUserIdTAG: 147561 FirstChildUserNameTAG: nbala FirstChildCreateTimeTAG: 2012-11-11T02:45:52Z IndexTAG: 224 TitleTAG: H6P2 Thevenin equivalent hint look at example 8.I in the book, Q2 can be modeled as a resistor. UserIdTAG: 276058 UserNameTAG: Ofer CreateTimeTAG: 2012-10-20T09:13:58Z VoteTAG: 9 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 225 TitleTAG: in love with electronics Dr. Agarwal u made me fall in love with electronics !! UserIdTAG: 190441 UserNameTAG: JOHN92RAY CreateTimeTAG: 2012-10-19T07:58:32Z VoteTAG: 9 CoursewareTAG: Week 6 / Amplifier operating point CommentableIdTAG: 6002x_amplifier_operating_point NumberOfReplyTAG: 1 FirstChildTAG: I hate statistics but now I feel if professor Agarwal would be my statistics teacher I would have ended as a Statician. FirstChildUserIdTAG: 99628 FirstChildUserNameTAG: Pranjal16 FirstChildCreateTimeTAG: 2012-10-19T14:13:02Z SecondChildTAG: Yeah, I get that same feeling. I would be interested in anything he was teaching. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-19T16:20:31Z IndexTAG: 226 TitleTAG: Volume amplifier error! Although I clearly see the signal is amplified on the graph. But I can't hear the louder sound comparing output signal to input signal. UserIdTAG: 201508 UserNameTAG: datle CreateTimeTAG: 2012-10-13T13:47:17Z VoteTAG: 9 CoursewareTAG: Week 5 / Music Amplifier Demonstration CommentableIdTAG: 6002x_snd_d NumberOfReplyTAG: 1 FirstChildTAG: Same here FirstChildUserIdTAG: 392100 FirstChildUserNameTAG: glenng FirstChildCreateTimeTAG: 2012-10-13T23:30:08Z IndexTAG: 227 TitleTAG: Mid-Term Exam date released! : October 25th at 00:01 am to October 28th at 11:59 pm (almost midnight) Boston time. The mid-term exam will be **released on October 25th at 00:01 am Boston time**. The exam is designed to take 2 hours; however, in order to compensate for any Internet or power outages in your area **you will have 24 hours to finish this exam**. You can start the exam when it is convenient for you, but you must complete this exam by **11:59 pm (almost midnight) Boston time on October 27th**, even if that is less than 24 hours after you started the exam. The exam will cover material through Small-Signal Analysis of Transistor Circuits (end of first lecture video sequence of Week 6). The exam must be completed online. Please understand that we cannot accept submissions in any format past the deadline, or delay the deadline for individual students for to any reason. Please read [here][1] P.D: I hope that they could extend it to monday, October 29th so we could be more relaxed, setting at saturday/sunday :). In the Prototype course I had a power energy cut while I were doing the Mid-Term Exam :P. My best wish to you in the Exam! UPDATED 23/10/12: $\color{blue}{Change\ in\ Midterm\ Timing:}$ In order to accommodate the Muslim holiday of Eid al-Adha, we have extended the deadline to submit the midterm exam to 23:59 (11:59 pm) on Sunday, October 28th, Boston time. The exam will still be released on Thursday, October 25th at 00:01 (12:01 am) Boston time. [read here][2] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/info [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/info UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2012-10-07T22:24:12Z VoteTAG: 9 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 11 FirstChildTAG: Sorry, can you give the exam 27-28 October (Saturday and Sunday) ? Because a lot people, I think, will be busy in this days. FirstChildUserIdTAG: 192114 FirstChildUserNameTAG: chuba FirstChildCreateTimeTAG: 2012-10-07T22:36:53Z SecondChildTAG: That would be nice :). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T22:40:30Z SecondChildTAG: Unfortunately, 27-28 wouldn't be quite convenient for me... I'd prefer the 25th. Yet, I didn't quite get the idea: if its 00:01 on Oct 25, and 24 hours are allowed, why it's 11:59 on Oct 27? SecondChildUserIdTAG: 252722 SecondChildUserNameTAG: vaboro SecondChildCreateTimeTAG: 2012-10-07T23:51:06Z SecondChildTAG: ...I mean since 00:01 am of Oct 25 till 11:59 pm of Oct 27 there are (72 hours - 2 minutes)... SecondChildUserIdTAG: 252722 SecondChildUserNameTAG: vaboro SecondChildCreateTimeTAG: 2012-10-07T23:53:40Z SecondChildTAG: it would be nice if mid term start from after 6pm on every Friday, every one can do it on night of friday and saturday. SecondChildUserIdTAG: 114913 SecondChildUserNameTAG: ngoctuan SecondChildCreateTimeTAG: 2012-10-08T08:07:14Z SecondChildTAG: For me also 27-28 will not be convenient. I.will prefer 26th. SecondChildUserIdTAG: 365309 SecondChildUserNameTAG: chetnasinghaldas SecondChildCreateTimeTAG: 2012-10-08T15:15:42Z FirstChildTAG: Oh my god, does it means that I have to hurry up and complete all the weekly lectures, HWs and Labs by that time? I'm only at the beginning of Week 5 by now. And yes, it would be great if the exam is on the weekends! FirstChildUserIdTAG: 326787 FirstChildUserNameTAG: Angstrem FirstChildCreateTimeTAG: 2012-10-07T23:01:29Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T23:05:42Z FirstChildTAG: Should I just have my stroke now? ;-/ FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-10-07T23:31:14Z SecondChildTAG: Not yet. As far as I understand the min-term will cover material up to chapter 8 of the textbook including ch 8. The fun starts from chapter 9:)) SecondChildUserIdTAG: 252722 SecondChildUserNameTAG: vaboro SecondChildCreateTimeTAG: 2012-10-07T23:42:14Z FirstChildTAG: Please change the title of this topic to read 11:59 pm instead of 1:59 pm, I was quite confused for a moment. (How is that almost midnight!?) FirstChildUserIdTAG: 143420 FirstChildUserNameTAG: ChiggerPepi FirstChildCreateTimeTAG: 2012-10-08T03:20:02Z SecondChildTAG: Hi ChiggerPepi! I have changed it, I was missing a 1 in 1:59pm :p. Thank you for the advice :). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-08T12:03:56Z FirstChildTAG: if i understand it correctly...from week 1 till first lecture of week 6 are included in the course for mid-term..does it include home work for week 6 as well? FirstChildUserIdTAG: 122513 FirstChildUserNameTAG: Mona77 FirstChildCreateTimeTAG: 2012-10-08T05:03:31Z FirstChildTAG: I think it is very important to mention that "Boston time" is actually US Eastern Time,which is GMT-4 (Eastern Daylight Time - EDT). FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-10-08T06:31:49Z FirstChildTAG: In homework, we can submit the answer as many times as we want. But is this the case with midterms as well or there are amendments ? FirstChildUserIdTAG: 211164 FirstChildUserNameTAG: Hamoodi FirstChildCreateTimeTAG: 2012-10-08T14:59:20Z SecondChildTAG: same thing i also want to know SecondChildUserIdTAG: 365309 SecondChildUserNameTAG: chetnasinghaldas SecondChildCreateTimeTAG: 2012-10-08T15:14:57Z SecondChildTAG: Mid-Term = limited times of check (only 3). Homweworks/Labs = Unlimited times of check. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-08T17:43:24Z SecondChildTAG: I think this is fair for an exam, and it's for our own good, if we want our final online certifications to be respected. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-10-08T19:58:10Z SecondChildTAG: In the HW4, I've pressed the "Check" button for hundreds of times. It seems that I'll get a little nervous before the exam :) The worst thing is that in many cases the answers are connected to each other: you have to use data from one answers to compute the following one. So if one answer fails - the following will also fail. SecondChildUserIdTAG: 326787 SecondChildUserNameTAG: Angstrem SecondChildCreateTimeTAG: 2012-10-08T21:38:38Z SecondChildTAG: @Myrim, Is this declared officially that 3 chances would be given this time ? SecondChildUserIdTAG: 211164 SecondChildUserNameTAG: Hamoodi SecondChildCreateTimeTAG: 2012-10-09T01:02:56Z SecondChildTAG: Hi Hamoodi, if they are based on the prototype course, yes. We had only 3 times of check by each buttom in the Midterm and Final Exam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-09T01:51:24Z SecondChildTAG: Okay! Thank you Myrim for your response. By the way, What is meant by prototype course ? SecondChildUserIdTAG: 211164 SecondChildUserNameTAG: Hamoodi SecondChildCreateTimeTAG: 2012-10-09T10:00:57Z SecondChildTAG: Hi Hamoodi! The 6.002x has its Pilot version online (Circuits and Electronics Prototype Course 6.002x)in the spring, I have participated there :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-09T11:27:51Z SecondChildTAG: Thank you :) SecondChildUserIdTAG: 211164 SecondChildUserNameTAG: Hamoodi SecondChildCreateTimeTAG: 2012-10-09T15:19:06Z FirstChildTAG: **One more thing I need to ask is that whether the lectures, homework's and labs would be available ?** FirstChildUserIdTAG: 211164 FirstChildUserNameTAG: Hamoodi FirstChildCreateTimeTAG: 2012-10-09T10:48:32Z SecondChildTAG: If they are based on the Prototype course,all material will be available ;). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-09T11:29:04Z SecondChildTAG: Thanks a lot dear **Myrim** :) ! SecondChildUserIdTAG: 211164 SecondChildUserNameTAG: Hamoodi SecondChildCreateTimeTAG: 2012-10-09T15:18:51Z SecondChildTAG: :) you are welcome Hamoodi! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-15T21:04:29Z FirstChildTAG: I'm not a staff member, just a student from the pilot 6.002x who is taking this class for the second time (there are many of us from the pilot 6.002x doing this). The edX staff are the authoritative source for info, but here are my observations and recommendations from the pilot 6.002x regarding midterm and final exams. 1) Do the all homeworks, labs, and sequence exercises, and review the detailed answers given for the homeworks and labs (click on the 'Show Answer' button after the weekly deadline has passed). The worked problems in the tutorials look like a good resource, they are a new feature to this iteration of 6.002x. 2) In general, one problem on each exam was very similar to a homework or sequence exercise. I found it helpful to review my notes made while working on the homeworks and labs, so I could quickly locate needed formulas and concepts during the exam. If you don't take or retain your notes while working on the weekly assignments, you will waste time and energy searching through the textbook during the exams. 3) Some students made study sheets for themselves before the exam which contained important formulas, concepts, etc. I didn't do that last time, but I will this time. It saves time during the exam, and reinforces the material in your head as you prepare for the exam. 4) Each exam had one problem that contained material not explicitly covered in the lectures. Don't go screaming "Not fair!" The point of these problems was to apply the general methods and techniques to a novel situation. For example, if the covered lecture and homework problems have the circuit in a specific configuration, expect to see an exam problem with a different circuit configuration, or using a different type of component. 5) The exams are open book, and open Internet. That means you can use online calculators and equation solvers such as Wolfram Alpha http://www.wolframalpha.com, Number Empire http://www.numberempire.com/equationsolver.php, etc. You can also use the discussion forum to review posts. What you ***CAN'T*** do is cheat or collaborate with others. You can't ask questions that pertain to the exam material on this forum or elsewhere, or reveal exam material until the exam deadline has closed for everyone. The staff monitor other online venues and use statistical techniques to identify plagiarism and dishonest students. Don't embarrass yourself and get banned from edX. You only cheat yourself if you aren't making honest original effort in these classes. 6) Set aside as much time as possible to take the exam. It's open for 24 hours, and many of us used all or nearly all of that for the final. I can't imagine taking the final as an on-campus MIT student with a two or three-hour limit, and no access to online resources, but they do it. Of course, the typical on-campus MIT student is much younger than me, and probably a lot smarter, so perhaps some of us won't need all 24 hours to complete the exams. 7) Consider reviewing material from similar classes online or from different textbooks as you go through this course, along with more general electronics learning sites. Google is your friend; you may find alternative explanations for the concepts that make everything clear to you. All About Circuits http://www.allaboutcircuits.com/ and Electronics Tutorials http://www.electronics-tutorials.ws/ are two good places to start. Also look at the MIT OpenCourseware version of 6.002 http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-electronics-spring-2007/. Good luck! FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-10-09T16:30:35Z SecondChildTAG: Thank you g_hopper for all this suggestions! I am sure that the new students will be guided by this :). Thank you! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-09T23:44:58Z SecondChildTAG: These pointers will definitely help us. THANK YOU :) SecondChildUserIdTAG: 386004 SecondChildUserNameTAG: ravinarv SecondChildCreateTimeTAG: 2012-10-15T09:08:26Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-15T21:04:57Z SecondChildTAG: thank you man very useful tips SecondChildUserIdTAG: 188662 SecondChildUserNameTAG: MOHAMED22 SecondChildCreateTimeTAG: 2012-10-23T17:54:58Z FirstChildTAG: will there be any lab questions in the mid-term exam? FirstChildUserIdTAG: 401175 FirstChildUserNameTAG: Raghav12 FirstChildCreateTimeTAG: 2012-10-10T19:39:20Z SecondChildTAG: Hi Raghav12! How are you? If they are based on the Prototype Course of 6.002x, the answer is no. We haven't labs in the Mid-Term Exam :). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-10T22:07:25Z SecondChildTAG: thanks for the info, Myrimit! i'm good, thanks :) hope you're fine too. SecondChildUserIdTAG: 401175 SecondChildUserNameTAG: Raghav12 SecondChildCreateTimeTAG: 2012-10-11T18:34:58Z SecondChildTAG: I am fine. Nice to know that you ar ok :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-11T19:36:58Z FirstChildTAG: I really hope that the mid terms (and finals) have new problems. :-( The 24-hours of brain grinding is so much fun! :-) FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-10-16T08:22:28Z IndexTAG: 228 TitleTAG: Recomendation see page 412 of book UserIdTAG: 155008 UserNameTAG: sohailahmed CreateTimeTAG: 2012-10-02T19:20:35Z VoteTAG: 9 CoursewareTAG: Week 5 / Small Signal Amplifier Exercise CommentableIdTAG: 6002x_small_sig_amp_e NumberOfReplyTAG: 0 IndexTAG: 229 TitleTAG: Incremental resistance what does incremental resistance of the device mean UserIdTAG: 343388 UserNameTAG: bijojoseph CreateTimeTAG: 2012-09-23T13:13:26Z VoteTAG: 9 CoursewareTAG: Week 4 / Graphs Exercise CommentableIdTAG: 6002x_graphs_exercise NumberOfReplyTAG: 1 FirstChildTAG: It gives the small change in voltage across a device for a small change in current through the device. Think of it as the effect of the signal on the output in the presence of a bias or the effect of a variation about an operating point. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-09-23T13:35:20Z SecondChildTAG: Thank u. For incremental resistance I tried taking the slope of the curve i=v^3. one pt as the operating pt (1.5617,3.808) and the other pt I assumed a value for current like i=3.7 and calculated v which gives result(1.5466,3.7).Then I calculated inverse of slope and the (ans=1.398) .Is this the actual value of incremental resistance of device.Or is it calculated in any other way. SecondChildUserIdTAG: 343388 SecondChildUserNameTAG: bijojoseph SecondChildCreateTimeTAG: 2012-09-23T13:51:25Z SecondChildTAG: Incremental resistance is dV/di, but if you you have I as a function of V and can't or don't wish to invert the function you calculate di/dV and take the reciprocal. For example, if i = V^3, then V = i^1/3, then dV/di = (1/3)i^(-2/3). Instead you can compute dI/dV = 3*V^2 and take the reciprocal. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-09-23T17:43:45Z SecondChildTAG: page 221 and equation 4-75 of textbook. SecondChildUserIdTAG: 374393 SecondChildUserNameTAG: rmaleki SecondChildCreateTimeTAG: 2012-09-25T16:19:43Z IndexTAG: 230 TitleTAG: my take on this problem https://www.dropbox.com/s/klws8rin30pcie1/S3E2%20Page%201.jpg https://www.dropbox.com/s/mnk6nligilcor4k/S3E2%20Page%202.jpg UserIdTAG: 230787 UserNameTAG: Gabriel007 CreateTimeTAG: 2012-09-21T16:57:06Z VoteTAG: 9 CoursewareTAG: Week 2 / Linear combinations of source strenghts CommentableIdTAG: 6002x_S3E2_Linear_Combinations_of_Source_Strengths NumberOfReplyTAG: 2 FirstChildTAG: it was helpful thanks .. ! i was confused with the coefficients ! FirstChildUserIdTAG: 389831 FirstChildUserNameTAG: saahilaa FirstChildCreateTimeTAG: 2012-09-22T06:48:30Z SecondChildTAG: THANK YOU SecondChildUserIdTAG: 140076 SecondChildUserNameTAG: snarayana SecondChildCreateTimeTAG: 2012-09-22T16:08:09Z SecondChildTAG: Thank you. SecondChildUserIdTAG: 206196 SecondChildUserNameTAG: victorrsoliveira SecondChildCreateTimeTAG: 2012-09-22T23:24:00Z SecondChildTAG: thank u sooo much!!! :) SecondChildUserIdTAG: 400239 SecondChildUserNameTAG: LincyG SecondChildCreateTimeTAG: 2012-09-23T12:01:33Z SecondChildTAG: thank u ssooooooo much i was totally in mess while solving ur method was easy SecondChildUserIdTAG: 397322 SecondChildUserNameTAG: sarvani692 SecondChildCreateTimeTAG: 2012-09-25T13:02:21Z FirstChildTAG: The question should Read : "In each of the following, write algebraic expressions for the coefficients **in terms of R**" FirstChildUserIdTAG: 256356 FirstChildUserNameTAG: WesleyHanes FirstChildCreateTimeTAG: 2012-09-23T17:49:16Z IndexTAG: 231 TitleTAG: S3E2 Go : (1) (e-V1)/R1 + (e-V2)/ R2 + e/R3 = 0 . . (2) e = (R2*R3*V1)/(R2*R3+R1*R3+R1*R2)+(R1*R3*V2)/(R2*R3+R1*R3+R1*R2), e = V3, so a1 and a2 (OK). For i1 and i2 : (3) i1 = (e-V1)/R1 and i2 = (e-V2)/R2 ( to see expression of "e" in (2) above ) All right. UserIdTAG: 260994 UserNameTAG: capvl CreateTimeTAG: 2012-09-18T21:18:17Z VoteTAG: 9 CoursewareTAG: Week 2 / Linear combinations of source strenghts CommentableIdTAG: 6002x_S3E2_Linear_Combinations_of_Source_Strengths NumberOfReplyTAG: 0 IndexTAG: 232 TitleTAG: power entering the source It's explained in textbook in paragraph 1.5.3 UserIdTAG: 177560 UserNameTAG: ss7loginov CreateTimeTAG: 2012-09-08T11:52:38Z VoteTAG: 9 CoursewareTAG: undefined CommentableIdTAG: 5_Simple_Power NumberOfReplyTAG: 0 IndexTAG: 233 TitleTAG: Improve subtitles: suggestions to clarify [word?]s Sometimes in the subtitles there's a term that was not clearly understood by the person writing the transcript. Can we send our suggestions somewhere to help improve the subtitles? For example: [voltage read?] is probably "Voltage V". UserIdTAG: 227945 UserNameTAG: daebwae CreateTimeTAG: 2012-09-05T14:16:52Z VoteTAG: 9 CoursewareTAG: Week 1 / Lumped Element Abstraction CommentableIdTAG: 6002x_lumped_element_abstraction NumberOfReplyTAG: 1 FirstChildTAG: The Subtitle from my observation is not been typed! Rather it is a technical issue with he speech recognision system... K! The computer interprets the spoken words into text. (SPEECH TO TEXT SOFTWARE) FirstChildUserIdTAG: 117661 FirstChildUserNameTAG: BUNDAY FirstChildCreateTimeTAG: 2012-09-07T12:36:30Z IndexTAG: 234 TitleTAG: Fullscreen video problem When you make the videos fullscreen, the lecture selection toolbar floats in the middle of the video. This happens on Firefox 15 and Chrome. UserIdTAG: 21541 UserNameTAG: JSChambers CreateTimeTAG: 2012-09-05T13:09:17Z VoteTAG: 9 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 4 FirstChildTAG: Same here. FirstChildUserIdTAG: 4463 FirstChildUserNameTAG: pmj FirstChildCreateTimeTAG: 2012-09-05T16:32:06Z FirstChildTAG: Same here FirstChildUserIdTAG: 189740 FirstChildUserNameTAG: MoHeY25 FirstChildCreateTimeTAG: 2012-09-05T13:31:55Z FirstChildTAG: Same here on Firefox 15.0 on Windows 7. FirstChildUserIdTAG: 182168 FirstChildUserNameTAG: arthurdent FirstChildCreateTimeTAG: 2012-09-05T18:16:13Z FirstChildTAG: same problem on firefox 15.0 FirstChildUserIdTAG: 301303 FirstChildUserNameTAG: era20 FirstChildCreateTimeTAG: 2012-09-05T13:51:32Z IndexTAG: 235 TitleTAG: Thanks alot MITx-- edX -- this is what humanity is about . History tells us that knowledge is easily missing if not passed on or spread around.We still don't have the clue as to how the pyramids in Egypt where build, and this should serve as a lesson to mankind not repeat the same mistakes the ancient did, but keep on spreading knowledge as MITx - edX and all other Institutions are doing. that is wonderful.Even individual on YouTube sacrificing their time spreading knowledge around, and that is a great thing to do.**This is what humanity is all about**. So thanks to you all- Mitx-edX and all who are helping to make this world a better place. And let keep on doing the good works. Peace UserIdTAG: 11607 UserNameTAG: deyoung36 CreateTimeTAG: 2013-01-07T04:27:25Z VoteTAG: 8 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: naively said......... FirstChildUserIdTAG: 91261 FirstChildUserNameTAG: mr_sophisticated FirstChildCreateTimeTAG: 2013-01-07T08:52:40Z IndexTAG: 236 TitleTAG: Feedback Dear course-mates, I would like to congratulate all the batch-mates who have got completed the Course successfully. I would like to thank Community -TAs for their continuous support during discussions. I thank Miramit for giving such a nice hints during Homework and Lab problems solving. I thank all the course coordinators for preparing such a nice,lucid and informative videos,presentations etc. Last but not the least Anant agarwal ji I wish to thank you for getting this course on edx platform and encouraging the world to learn at free. I want to see many other courses to be launched through MITX like Digital signal processing etc (for which video lectures already available in MIT OCW). Thanks everyone. UserIdTAG: 4342 UserNameTAG: SEENU CreateTimeTAG: 2013-01-03T17:13:08Z VoteTAG: 8 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 237 TitleTAG: Coursera course: Fundamentals of Electrical Engineering starting soon Dear all, For information there is a course called 'Fundamentals of Electrical Engineering' starting on Jan 21st 2013 which runs for 12 weeks. It is based on a course given at Rice University. Link: https://www.coursera.org/course/eefun. It seems to have a strong signal processing component and may be a good follow on from 6.002x. From what I can see there are videos, short quizzes and weekly homeworks but I didn't see any mention of an exam, and there is no certificate. There is a free online book that goes with the course. Quoting the information 'This course is routinely taken by second-year electrical engineering students at Rice as their first electrical engineering course. Its reputation can be summarized as “the hardest course I have ever taken but I learned a lot.”' I'm thinking of doing it. Is anyone else interested? UserIdTAG: 3646 UserNameTAG: freespirit CreateTimeTAG: 2012-12-31T00:28:06Z VoteTAG: 8 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 8 FirstChildTAG: me too. singed up FirstChildUserIdTAG: 267220 FirstChildUserNameTAG: ABHINAVSAXENA318619 FirstChildCreateTimeTAG: 2012-12-31T06:11:09Z FirstChildTAG: I also signed up FirstChildUserIdTAG: 136959 FirstChildUserNameTAG: christerpher FirstChildCreateTimeTAG: 2012-12-31T09:19:37Z FirstChildTAG: Easy course, not so difficult as 6.002x FirstChildUserIdTAG: 183712 FirstChildUserNameTAG: colegiocientifico FirstChildCreateTimeTAG: 2012-12-31T04:33:22Z SecondChildTAG: Do you speak from experience? SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-12-31T19:06:14Z SecondChildTAG: It is not so obvious from looking at the free online book that it is easier. It has a much stronger focus on signal processing, and looks more mathematical. I think depending on your previous experience it may be easier or harder. SecondChildUserIdTAG: 3646 SecondChildUserNameTAG: freespirit SecondChildCreateTimeTAG: 2013-01-01T18:37:16Z FirstChildTAG: I am signed up. FirstChildUserIdTAG: 61923 FirstChildUserNameTAG: MikeJones FirstChildCreateTimeTAG: 2012-12-31T03:28:53Z FirstChildTAG: There is a certificate for participation, without grade. Same as here. I had some courses there and it's really fun and you can learn a lot, too. FirstChildUserIdTAG: 207728 FirstChildUserNameTAG: RossyFromBulgaria FirstChildCreateTimeTAG: 2012-12-31T07:29:57Z FirstChildTAG: Looks interesting, also taught by another MIT graduate, who worked at MIT Lincoln Lab before moving over to Rice. Should be fun. The prof claims the students taking his course routinely describe it as the hardest course they've ever taken. He claims to cover every aspect of EE, but I doubt it, from a brief look at the material. I think you're right, its heavy on the signal processing side. FirstChildUserIdTAG: 4463 FirstChildUserNameTAG: pmj FirstChildCreateTimeTAG: 2012-12-31T01:18:06Z SecondChildTAG: Seems very, very interesting. I will enroll for sure. SecondChildUserIdTAG: 330357 SecondChildUserNameTAG: steryd SecondChildCreateTimeTAG: 2012-12-31T01:26:48Z FirstChildTAG: Hi all I signed up, will see you there. FirstChildUserIdTAG: 296419 FirstChildUserNameTAG: oscman FirstChildCreateTimeTAG: 2013-01-02T20:50:25Z FirstChildTAG: Hi all some of you signed up for the 3.091x? FirstChildUserIdTAG: 296419 FirstChildUserNameTAG: oscman FirstChildCreateTimeTAG: 2013-01-02T20:58:28Z IndexTAG: 238 TitleTAG: Why certificate is late, Human Nature or what? You wonder why the certificate is not here yet? Because staff have to deal with this kind of posts everyday: I copied this from the old mitx discussion: questions about the deadlines for example and reason people have for being late : MITx NOT FAIR! My watch broke not my fault! HAD TO LEAVE AND GOT BAK BUT TIME WAS UP STAFF NEED MORE TIME HELP! There was earthquake and my timezone move!! WHY IS EXAM NOT TAKE MY ANSWER MORE??!! These problem too long for internet connect! STAFF YOU FIX THIS I VERY UPSET A LOT!!! Why do the answers have to be so, like, exact or something, really? HAY I CNT GT DONE CUZ THEY TOL ME GMT TIME N ITS DIFFRNT HERE!! Hi I just found out about this coarse and it looks like a lot of work,how to get a extension? WINDOWS TIME NOT SAME AS STUPID INTERNET TIME STAFF HELP!!! Guys, is there any way I can just get a certificate say if I turn in a few problems or something? I TRY TOO MANY FORMULA FOR BOX REALLY HARD BUT NEVER GET GREEN TICK WHY? You know, if this is really a good university, then they should HELP you with stuff to get a good grade you know an I tried real hard but it just kept giving me the red X! I HAD ALL THE ANSWERS WROTE OUT BUT WHEN THE CABLE GUY CAME AND GOT MY NET WORKIN TIME WAS UP WHAT CAN I DO? Me from parallel universe. In our universe lightspeed lower then in yours, so took my answer longer to reach you ... Pse correct me grade, yes?? UserIdTAG: 83276 UserNameTAG: salsero CreateTimeTAG: 2012-12-30T14:01:24Z VoteTAG: 8 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: this is crazy.. FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-12-30T14:09:04Z FirstChildTAG: Be patient ! Enjoy Christmas Holidays and Live a Happy New Year. http://www.sotojohnson.net/navidad Animation is best view with Google Chrome. FirstChildUserIdTAG: 183712 FirstChildUserNameTAG: colegiocientifico FirstChildCreateTimeTAG: 2012-12-30T14:55:24Z SecondChildTAG: Cool XD SecondChildUserIdTAG: 289949 SecondChildUserNameTAG: manolito SecondChildCreateTimeTAG: 2012-12-30T15:04:49Z SecondChildTAG: WebGL? Too slow... SecondChildUserIdTAG: 394836 SecondChildUserNameTAG: v2g6ch4 SecondChildCreateTimeTAG: 2012-12-30T15:25:25Z FirstChildTAG: Hi salsero, Haha, I almost forgot that from the old Forum!;) In my point of view, I guess that they are having a lot of requests due the Certificates. I remember last Term when students started to ask them to change their names or, in the Spanish cases, adding the accent like this María instead of Maria, etc...and hundreds of requests... Also, I guess that they are analyzing the cases were the students broke the Honor Code Agreement, might they delete that accounts or decide not to give them their Certificates...I am sure that they are working hard. I remember last Term haha, OMG, I went to check a lot of exagerated times if the Certificate was available , haha, so I totally understand the anxiety of the students of this Fall, as I had that same anxiety last Spring ;). And when I finally got it, Oh my, I had a smile from one side to the other side of the cheek haha, I have shown the Certificate to all my family, I got a happy B and I was so happy, I remember that moment haha. So, I am sure that you will have that smile from face to face soon, before New Year as it was posted some days ago in the Forum by a Staff. My best wish to all of you, Have a Happy New Year 2013! Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-30T15:08:54Z SecondChildTAG: I agree, I think they just need time to process, analyze and extract information about the course. The delay is not out of spite, I'm sure. I think it is a reasonable amount of time given the complexity and amount of students involved. In the end, it will still probably be faster then getting a certificate from a "brick and mortar" institution. Plus you don't have to wait for postage, you can print it out and stick it to the fridge within minutes of it becoming available. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-12-30T16:14:37Z SecondChildTAG: **Salsero**...wow..and I thought the current forum is bad with all the repetitive questions! That aggregation of old forum posts seems like it was written either by impatient children or cavemen. And I mean no insult to either children nor cavepersons! **Myrimit** said "*Also, I guess that they are analyzing the cases were the students broke the Honor Code Agreement, might they delete that accounts or decide not to give them their Certificates*" I had no idea that they would wait for the final moment to do all that. But I guess cheating during the Final Exam would be the most egregious violation of the code. I'm glad that they're analyzing the Final Exams for signs of impropriety. ***Just a quick question:*** Do you have any official word (Lyla etc.) that edX is actually doing this, or is your stated "guess" just an educated guess? Also **best wishes for a Happy New Year 2013** from New Jersey, USA to you both!!! I wish that this semester of 6.002x would go on forever, but all good things come to an end; and may we see each other in future edX courses! The arrival of the New Year is so sudden this year, but I am **glad it is over** because for me personally this was the year of Hurricane Sandy and the floods that destroyed my cars and the winds that took electricity out for two weeks. We also had those shootings of schoolchildren in Newtown, Connecticut; near where I grew up and in the same state where I went to university! So close to home. Jersey Mark SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-12-30T18:06:05Z SecondChildTAG: Oh, and **Pennypacker**...best wishes for a Happy New Year 2013 from New Jersey, USA to **you** too!!! (Didn't see you up there!) Jersey Mark SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-12-30T18:08:41Z SecondChildTAG: Hey Admin you take your own time. My hearty new year wishes for you all. SecondChildUserIdTAG: 145544 SecondChildUserNameTAG: pandiya SecondChildCreateTimeTAG: 2012-12-30T19:00:15Z FirstChildTAG: Well, i guess that people cry out for education . And what better way to demonstrate that you are educated, than a certificate ? FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-12-31T09:47:22Z FirstChildTAG: Believe us, we are not punishing annoying posts. It just so happened that this class ended on Christmas Eve, and it is difficult to coordinate things during that holiday week. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-12-31T16:00:59Z SecondChildTAG: Thank you for the response Lyla. Happy New Year 2013!:) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2013-01-01T00:53:51Z IndexTAG: 239 TitleTAG: Without a Grade it is a simple participation certificate Without a grade, people not knows if there were an evaluation or examination to achieve it or if this was a simple participation certificate. In my university the minimun score to pass a course is 70% B, I need to approve Circuits to achieve my BSc degree in civil engineering and I don't think they are going to recognize this certificate if there is not a grade B. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-12-29T15:30:00Z VoteTAG: 8 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 8 FirstChildTAG: If a grade is given, then it doesn't prove that the person with the name on the certificate, did the exam himself. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-12-29T16:08:35Z SecondChildTAG: I suppose that is where the proctored exam (If it becomes available) comes into play, for those who need further verification. Perhaps this weighs heavy in their decision on whether or not to display a grade. I try to be grateful for anything they give us, primarily the education, the certificate is more of a cherished souvenir which would not have meaning without the experience. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-12-29T16:19:12Z SecondChildTAG: Like as all the homeworks, and exams to do in house in all presencial universities. They only accept an honor code of his students, like you and me and all honest people did with edx. SecondChildUserIdTAG: 183712 SecondChildUserNameTAG: colegiocientifico SecondChildCreateTimeTAG: 2012-12-29T16:30:31Z SecondChildTAG: Salsero is 100% right. Hope grade demanding students will appreciate this and prepare for a proctored exam! SecondChildUserIdTAG: 455950 SecondChildUserNameTAG: Wahabbaluch SecondChildCreateTimeTAG: 2012-12-30T07:59:22Z SecondChildTAG: And here in Costa Rica, who is going to take the exam for me a got a 100%, Dr. Frankling Chang Díaz maybe in the moon or in the sideral space?, ja ja SecondChildUserIdTAG: 183712 SecondChildUserNameTAG: colegiocientifico SecondChildCreateTimeTAG: 2012-12-30T15:19:32Z FirstChildTAG: Not necessarily, in general, a certificate would imply that you have successfully completed some sort of predefined course or curriculum. I can understand that you would want to display an "A" if you achieved one, however a "C" might look a little awkward on a certificate. I suppose that is why they had the option for both types last semester. (So I hear). I myself am undecided in the matter. If I had to choose one over the other, I think I would choose the ungraded certificate for the reason I stated above, although it would be nice to always have some sort of access to your grade if needed. I am unaware if the link they provide us has a grade or not. Again I could go either way on this depending on what my grade was in a particular course. :+ FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-12-29T16:06:52Z SecondChildTAG: There is always the option to print or display one's progress bars... ;-) SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-12-29T17:00:15Z SecondChildTAG: @planetscape: Exactly. You can also print them on the backside of your certificate. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-12-29T18:57:02Z SecondChildTAG: Come on, its like if you write something on your university diploma, so not serious... SecondChildUserIdTAG: 207728 SecondChildUserNameTAG: RossyFromBulgaria SecondChildCreateTimeTAG: 2012-12-29T19:17:07Z SecondChildTAG: Yes, but it isn't ... SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-12-29T20:19:05Z SecondChildTAG: Sorry, didn't know that these four months were playing a game, not something serious. SecondChildUserIdTAG: 207728 SecondChildUserNameTAG: RossyFromBulgaria SecondChildCreateTimeTAG: 2012-12-29T20:27:04Z SecondChildTAG: They didn't originally intend to give us an option last semester, that was just a solution to an unforeseen problem. They promised graded certificates more or less by mistake before the course started, so they had to back down from their later decision to not include the grade in the certificates. They also made it clear that this would not apply to later courses. As far as I know, they never bothered explaining *why* they won't include the grade - in spite of many questions about it. SecondChildUserIdTAG: 90796 SecondChildUserNameTAG: ChaunceyGardiner SecondChildCreateTimeTAG: 2012-12-29T23:20:20Z SecondChildTAG: I believe in grades. The purpose of the grade is to provide information: (a) to the student to indicate whether they have or have not mastered the material and (b) to the potential employer or graduate program to indicate whether the student mastered the material or not. However, this course was free and I think everyone should be able to agree that they got a lot more than they paid for. SecondChildUserIdTAG: 126825 SecondChildUserNameTAG: aphoenix SecondChildCreateTimeTAG: 2012-12-30T14:44:47Z FirstChildTAG: Myself, I am just happy that not only did I improve my grade on the final vs the midterm, I came very very close to an "A" overall... A piece of paper is nice; that feeling I got is way better. FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-12-29T17:01:45Z FirstChildTAG: Without a grade it is simply nothing. If I knew the grade won't be available in the certificate, I wouldn't waste time to achieve each next level and it was really hard for me and I would stop at 60%. Then why you need that progress chart at all, when you could only draw one line - 60%? I am very disappointed, that was something edx should told us earlier. Sorry if I hurt anyone's feelings, but I am one of the many people which grade is A. FirstChildUserIdTAG: 207728 FirstChildUserNameTAG: RossyFromBulgaria FirstChildCreateTimeTAG: 2012-12-29T19:14:01Z SecondChildTAG: But you know about the very popular bulgarian adage, that the teethes of a free given horse shouldn't be examined ... And I think the idea of the edx is to give knowledge not certificates, they are completely useless. SecondChildUserIdTAG: 415375 SecondChildUserNameTAG: ZWX SecondChildCreateTimeTAG: 2012-12-30T21:50:45Z SecondChildTAG: @ZWX: Funny, we have almost the same saying in Dutch: Don't look into the mouth of a given horse. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-12-30T22:40:59Z SecondChildTAG: Even in spanish we have the same saying. I appreciate very much what I learned through the course, although I would loved having a grade included in my certificate. Regards! and happy new year Carlos SecondChildUserIdTAG: 220837 SecondChildUserNameTAG: Calsomus SecondChildCreateTimeTAG: 2012-12-31T03:30:27Z SecondChildTAG: So the lies should be accepted just because they are free? At the beginning of the course they said there will be grades. After the final exam they said there will be no grades. Therefore there's a lie to all of the students of edx who gave their time to solve the cases of the homeworks, labs and exams... Thank you for the multicultural journey guys, but not everything that you receive as a gift is usable. Guess your grades wasn't very high. There is one more usable bulgarian adage: The chicken you bought for one coin, is working exactly for one coin. If they were said earlier that 60% average progress is enough for this "certificate" and there won't be grades in it, the things now would be different. SecondChildUserIdTAG: 207728 SecondChildUserNameTAG: RossyFromBulgaria SecondChildCreateTimeTAG: 2012-12-31T07:43:15Z FirstChildTAG: I think one reason NOT to show a grade, could be to attract only people who are really interested in learning the course material. Even if this course wouldn't have offered a certificate, I would have followed it. The growing number of progress bars alone, are giving me a good feeling. Also solving the harder HW, midterm- and final problems made my heart jumps with joy. I think that this kind of courses offer the ability to study for yourself on a very good level, and that people can use the material to compare the level of their own skills with other people and other educational institutions all over the world. And a nice side effect might be that people are improving their language skills, learn how to read a foreign language better and can practice this via this discussion. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-12-29T19:31:41Z FirstChildTAG: I took this course twice now.. Just for the fun of it. And to "perfect" my understanding and mastery of material. You don't need credit for this course, just use the "knowledge" gained from this course, and apply it to get a better grade in your regular university course when you take a similar course. The idea of taking a course once, for credit, is now old fashioned and archaic. We never absorb everything the first time we hear it, or read it, or see it, repetition is required for mastery of any subject. Now, we can repeat a course without repeating the "fee" associated with taking the same or similar course over again. So, for the first time in history of education, we can master the material taught, and not just learn it. When a professor teaches the same subject over and over again, he gains complete mastery of the material. Now the student too, can go over the material again and again to master it like the professor has. Think different. FirstChildUserIdTAG: 4463 FirstChildUserNameTAG: pmj FirstChildCreateTimeTAG: 2012-12-29T19:38:21Z SecondChildTAG: Totally agree. SecondChildUserIdTAG: 361137 SecondChildUserNameTAG: Ericson SecondChildCreateTimeTAG: 2012-12-30T03:04:42Z FirstChildTAG: **RE: Grades and another Rant** (definitely TL;DR) My view is that if edX wants to take itself **seriously**, it ought to offer students official transcripts of courses completed, just like every other brick-and-mortar university (and for-profit, credit-granting online university, for that matter). To me, it doesn't matter if obtaining a transcript requires proctored examinations, fees of some sort, stipulations stating that the grade was obtained under an Honor Code and may not reflect that the actual student participated, etc.; the point is that the system has to **evolve**. More and more universities are coming aboard edX and are conducting lessons under the MOOC model, but without some leadership, direction, and response to feedback - other concepts (i.e. Coursera) - may gain the competitive advantage. Maintaining an edge in massive open-enrollment online course education means offering students something of **value**. While the majority of us simply value learning for learning's sake - and the honor of being able to learn under the worlds' greatest professors is itself very valuable and humbling - I see that many students aren't satisfied as much as they could be. This is because education, by its very nature, is a very competitive field. There are a limited number of slots at the very top, whether it be for financial aid, for admission to a prestigious university, for qualification to practice in a professional field and to obtain those credentials, etc. and not everyone is going to make it. Those that have the ability to do so should be allowed to stand out and above the rest. Thus while we can make everyone feel good by issuing an unlimited number of participatory certificates to whomever signs up (achieving the necessary 60% is actually pretty easy), the reality is that colleauges, evaluators, educators, employers, and others measure success by raw numbers and through verification, hence why two concepts need to be implemented if edX wants to keep up and prevent others from taking it's share of students: 1. ***Verification***. While this is not necessary for the student who wants simply to learn the subject matter, verification of one's student body is is important if a university consortium wants to award any type of certificate and for these certificates to mean anything. Fraud and identity theft abound in today's world. Verification of students can be as simple as having a third-party (such as our announced partner Pearson VUE, or one of the many member universities, even -via a simple email from a .gov address- from a local government official) cross-check a student's identity by locally verifying that student's government-issued ID card for a small fee. Proctored exams can also accomplish this, but at a greater cost. Students will be classified as verified or non-verified; thus ***the market*** can establish the worth of certificates and transcripts that are verified versus those that are not. 2. ***Access to Records.*** This is also not necessary for those students just wanting to learn for it's own sake. However, as edX expands and hopefully becomes prestigious and the premier online education destination , it is going to be difficult for students, employers, university admissions officers, guidance counselors, social workers, military recruiters, governments, etc. to obtain ***transcripts*** of a student's progress because, let's face it, the needs for records are many: Employers extend job offers based on educational achievements and include these in performance reviews and assessments for promotion; universities accept students based on their academic success, part of which includes success in edX courses - an excellent predictor of how a student will fare in a brick-and-mortar university by the way; governments decide benefits such as unemployment compensation on whether a person is enrolled in a course of study; criminal justice agencies judge the moral character of an ex-convict based on if he is reforming himself through an educational course, etc. The list goes on and on for a need for a record to be provided to various parties in any educational endeavor. 3. ***Competitiveness.*** Democratization of, and accessibility to, education is one of the top goals of any MOOC program, but there is always a limited amount of resources; therefore it could be beneficial to ensure that everyone who enrolls in some of the top-tier courses be screened for admission, and might have to show a basic level of competence on a basic entrance examination. Edx is currently experimenting with this in HarvardX's Spring course on Copyright, so this idea may not be as controversial as I first thought. Students having basic competency -and a showing of netiquette- would ease the burden on Staff and TAs sifting through repetitive, basic questions that most students would find an answer to on their own initiative by consulting already-available resources. This resources would be freed up to help students genuinely needing help, not asking "when do I get my certificate" a thousand times over. 4. ***Sunshine and Privacy.*** It would be a great idea if edX meetings concerning decisions on upcoming courses and policies would be ***recorded and open***, and it would be even better with student participation. I often find myself in the dark, and changes to edX often take me by surprise. It does help to be proactive and read all of the news releases, and it would help to have somewhat of an official "student union" considering this is an educational setting. It would also be great if edX could formulate a ***privacy policy***. Right now, we don't know what our names are being used for, how long (if/if not) our records will be kept, how our data is disclosed (of course all Discussion is open record), etc. It would also be great if we could opt out of certain policy we are averse to. I know this is a lot, but this is not too much for the world's largest and greatest educational institutions to implement in a reasonable period of time. Jersey Mark *(Note that the opinions I have expressed above do not necessarily reflect the view of edX, MITx, Staff, sponsors, or anyone affiliated with them; they are solely my own view and I alone am responsible for the content).* FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-12-30T06:39:03Z SecondChildTAG: I don't understand why they don't want to give a grade in the honor code certificates. So be it. Proctored exams seem like an okay solution for those who need the certificate, but it's not perfect. I don't know how well I would have done if I had to take the exam without access to my computer. I am talking about my method for dealing with the problems. I guess my approach differs from that of most students - I don't take notes, and I barely use any paper during the exams. I do most of the calculations in Notepad, and paste the final expressions into the parser input field. I use a couple of other tools when necessary, such as the Windows Calculator and at times Microsoft Math. On the other hand, I don't need the certificate. A screening process would most likely exclude people like me. My math skills were lacking (an understatement), and I learned a lot of math while progressing with 6002x. In spite of my lack of math skills, and not having used my brain for academic stuff since the 1980s, I got a total score close to 100% in the spring 6002x (which IMO had a tougher final than the fall 6002x). I think that proves that screening is a bad idea. I also think that the reason they do select candidates for the Harvard copyright course is that the way the course is set up requires more staff per student than 6002x did. SecondChildUserIdTAG: 90796 SecondChildUserNameTAG: ChaunceyGardiner SecondChildCreateTimeTAG: 2012-12-30T11:02:43Z SecondChildTAG: **Chauncey:** Perhaps you missed a few points (glad you're probably the only one who read my long rant though!): "Screening" would not mean a math test. It would be optional, and it would simply be a ***verification of your identity*** so that when you get your certificate, it would say "Verified", and if you would show it to your employer, he would know it was actually you that took the course, not your rocket-scientist friend. Proctoring would also be optional (and separate from verification). You would still get a certificate, but it would say "Honor Code" instead of "Proctored". Also, FYI: ***Proctored*** testing centers (like our published partner, Pearson VUE) use **computers** to administer exams. You would definitely **not** have to take a "paper-and-pencil" Final for 6.002x! It would be the **same exact format** as the other students take! You would (should?) have full access to the internet and all other resources (Wolfram-alpha, online graphing calculators, etc). You also would (should?) have available to you the full 24-hours should the need arise; you could go out for lunch, come back to the testing center, and finish your proctored exam as often as you wish, within the 24-hour period. SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-12-30T17:47:38Z SecondChildTAG: If that's what you meant, I don't see any harm in what you suggest. SecondChildUserIdTAG: 90796 SecondChildUserNameTAG: ChaunceyGardiner SecondChildCreateTimeTAG: 2012-12-31T02:34:09Z FirstChildTAG: I got your ideas, but for some of us this certificate is not just for fun. We have investing time and work for achieving this grades and for what? For example, I need this certificate for my job, not just for having fun, and not having a grade in it makes me feel lied. I have registered for three more courses here and left them because I needed more time for 6.002x to score better. I have registered for some of the spring 2013 courses, although they will be not free. But if I have to pay for something that wont say what grade did I have, then I'll find an alternative. Why they don't just add our scores like in the previous course? FirstChildUserIdTAG: 207728 FirstChildUserNameTAG: RossyFromBulgaria FirstChildCreateTimeTAG: 2012-12-29T19:51:13Z SecondChildTAG: I would think you could "prove" your scores by logging in and displaying your progress page for your employer. With a login, you "prove" your ownership of the account. OTOH, if I were hiring, I'd want you to show me what you can DO, not just hand me a piece of paper anyway. SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-12-29T20:34:55Z SecondChildTAG: I agree with planetscape. I've had very interesting jobs, but MOSTLY because I could prove myself. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-12-29T21:01:08Z SecondChildTAG: Yeah, I completely agree for the part of showing what you can DO. BUT, first you have to make the HR or your future boss to notice you and invite you to see what really you can do. And here's whet diplomas and certificates come first. Or at least here, in Bulgaria that's the way things work, and it is completely not easy. SecondChildUserIdTAG: 207728 SecondChildUserNameTAG: RossyFromBulgaria SecondChildCreateTimeTAG: 2012-12-29T21:02:05Z SecondChildTAG: I agree with RossyFromBulgaria SecondChildUserIdTAG: 183712 SecondChildUserNameTAG: colegiocientifico SecondChildCreateTimeTAG: 2012-12-30T03:27:42Z IndexTAG: 240 TitleTAG: Many thanks from a 50++ year old student I want to thank all MITX people for this opportunity for knowledge (or ...re-knowledge).The trip was beautiful. Full of adventure, full of discovery. Thank you again. UserIdTAG: 146969 UserNameTAG: Andreas-N CreateTimeTAG: 2012-12-28T18:55:58Z VoteTAG: 8 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Spanish is my native language. Sorry for the mistakes. Andreas-N 50++ is an excelent time for knowledge or re-knowledge !!! Go ahead !!! Is very good to know that a lot of persons with 50 or more be with us sharing ideas and knowledge. I'm agree with you saying thanks to all MitX Team and TA for their work and help. Greetings for you. FirstChildUserIdTAG: 227223 FirstChildUserNameTAG: luisbonelli FirstChildCreateTimeTAG: 2012-12-28T20:49:47Z IndexTAG: 241 TitleTAG: I respectfully disagree about THE expressions for 4b & 4c First let me say I took the first version of this class on mitx. It was a very rewarding experience. I have followed along with this class to see how the edx platform compares. Suffice to say it has had its teething problems across may different courses, but on the whole I will say it is worthwhile to take classes on edx. Hopefully one day soon edx will catch up to the standard that was set by mitx. I am keeping my fingers crossed. But this post is to voice an observation about the grading of the final exam. The accepted answers for 4b &4c are only **approximate** expressions. If that is what was wanted, I believe that is what the question should have explicitly asked for - especially since I started at the last minute so to speak, after all the corrections and clarifications were made, and the extra attempts had already been given and taken away... I undertook this question with the clear understanding that it had been clarified to remove **any** ambiguity. Below are my answers. When they were rejected by the grader, my reaction was 'here we go again'. This looked like the last version of the course where correct answers were rejected if an equation wasn't entered in the "official" order. So that I can complain without sour-grapes I took a screen shot and then used my final attempts to ensure my final submissions were incorrect, so even if this were to be retroactively graded as correct it would not benefit me. So there you have it. I am complaining about **the** 'official' answers. I think after two attempts at phrasing them the questions should be obvious. I do not think I should be using extra attempts to guess the intent of the examiner. Also (and this may be just my perception) I felt this final was not up to par with the final for the last rendition of the course. The previous final was significantly longer and more difficult to finish on time. Considering that final was given to real MIT students who were concurrently taking the same online class, I can only speculate that either that final was too hard (but this has never been stated to my knowledge) or this course's final was a watered down version. I find this second alternative sad and somehow insulting to the students who have spent the best part of 4 months preparing for this event. Hopefully this is not the case. However, upon first looking over the final I felt cheated. I was looking forward to a challenge that never appeared. ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13565953121343634.png UserIdTAG: 9673 UserNameTAG: xp42 CreateTimeTAG: 2012-12-27T08:44:40Z VoteTAG: 8 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 20 FirstChildTAG: er..what exactly is the point that you are trying to make by posting these rejected answers?..Not being rude..but I did not understand.. FirstChildUserIdTAG: 173147 FirstChildUserNameTAG: cruiser_rahit FirstChildCreateTimeTAG: 2012-12-27T08:53:59Z FirstChildTAG: T is large enough to ignore (1-e^(-T/(C.R1||R2))) from appearing in result The question wording was " After a VERY LONG time T , s1 is opened and s2 is simultaneously closed." so you shouldn't have included that term in to your answer. the parser does allow a lot of flexibility. as you can see in the pic![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13566001811343643.png FirstChildUserIdTAG: 623210 FirstChildUserNameTAG: preveen FirstChildCreateTimeTAG: 2012-12-27T09:12:50Z SecondChildTAG: Agree. By the words " after a VERY long time T" is just explained, that capacitor is completely charged to the voltage, produced by R1/R2 voltage divider, and processes of Phase2 of this quiz starting from exactly that voltage of capacitor. SecondChildUserIdTAG: 345502 SecondChildUserNameTAG: ElectroVova SecondChildCreateTimeTAG: 2012-12-27T11:51:08Z FirstChildTAG: I missed the (t-T) part and hence, a perfect score. While one question was a X'mas gift (though I did not benefit much because I got a vague deja' vu feeling only after I had actually solved it), it was a somewhat easy final exam. The next course students will certainly curse us if they get a tough paper as a reaction of this thread! FirstChildUserIdTAG: 190670 FirstChildUserNameTAG: chauhan1955 FirstChildCreateTimeTAG: 2012-12-27T11:17:40Z SecondChildTAG: I missed the time shift too. UGGH! But an A is an A. SecondChildUserIdTAG: 349862 SecondChildUserNameTAG: edsaf8 SecondChildCreateTimeTAG: 2012-12-27T12:53:47Z SecondChildTAG: Probably, I thought subconsciously that when the oscillator was going to oscillate to infinite ('anant' in Hindi, that explains the infinite wisdom of Prof Agarwal) time, what is a measly T in its lifetime :-). SecondChildUserIdTAG: 190670 SecondChildUserNameTAG: chauhan1955 SecondChildCreateTimeTAG: 2012-12-28T03:58:48Z FirstChildTAG: As a review exercise I did the fall 2008 6.002 final exam and, well, I found the difficulty level comparable to fall 2012 6.002x final; the problem could have been the 3 hours time limit (probably I would have scored around 90% within 3 hours) but I didn't find it incredibly hard. It is possible that the difficulty level varies with time (even at MIT, not only at edX): sometimes the final is harder, sometimes it's simpler... Having said that, it's even possible that edX main goal is to increase the base of its students and, as a consequence, they modulate the difficulty level accordingly, making the midterm/final a little bit less hard than that provided to a real class. I would like to see the spring 2012 6.002x final (I don't know if it's available online somewhere) so I can judge it for myself because, to be honest, you xp42 don't seem to be flexible enough to evaluate difficulty levels if even a simple "after a very long time" makes you in troubles. FirstChildUserIdTAG: 376877 FirstChildUserNameTAG: AndBre FirstChildCreateTimeTAG: 2012-12-27T12:23:40Z SecondChildTAG: 'Broader base' is probably the right word, AndBre. Tough or simple, we will always be on either side of the digital divide (and NOT in the forbidden region; please do not confuse this NOT with the inverter gate). What actually matters to me personally is that how we can implement those lessons into practical uses (or career opportunities, if at all). It's still a long way to go, for me, practical-wise. SecondChildUserIdTAG: 285616 SecondChildUserNameTAG: AmiyaX SecondChildCreateTimeTAG: 2012-12-27T15:44:53Z SecondChildTAG: Without a doubt 'broader base' is the right word: English is not my native language and it's easier to realize that when I am in a hurry... Of course there are several reasons for taking a MOOC course: for someone it's important the learning experience (no matter the grade) whilst for others it's important the grade and the competition (comparing own performance with others'); I understand both the positions: competition is fun but at the end it's important what you learned (whatever the use of your learning). SecondChildUserIdTAG: 376877 SecondChildUserNameTAG: AndBre SecondChildCreateTimeTAG: 2012-12-27T16:19:33Z SecondChildTAG: In the end, (if I may borrow the pitch from cs50x) it only matters where you are compared to yourself, when you started the course. SecondChildUserIdTAG: 623210 SecondChildUserNameTAG: preveen SecondChildCreateTimeTAG: 2012-12-27T16:31:44Z SecondChildTAG: well said (or quoted) Preveen. SecondChildUserIdTAG: 190670 SecondChildUserNameTAG: chauhan1955 SecondChildCreateTimeTAG: 2012-12-27T17:00:09Z SecondChildTAG: Very right AndBre. It was difficult to do it in 2 hrs or even 3 hrs but relatively easy if you give it 5-6 hrs. I gave it 5 hrs, 11 pm - 4 am, after a hectic day. Not at my best but there is no excuse really for not beginning well in time. edX gave enough time. Just one thought, what was the greatest motivation to keep going? I think those green ticks, the thrill of it all when you get something right. SecondChildUserIdTAG: 190670 SecondChildUserNameTAG: chauhan1955 SecondChildCreateTimeTAG: 2012-12-28T04:05:03Z FirstChildTAG: In my point of view, an statement like this would be less confusing: ---------- Until time t=0, both switches are open and the circuit is initially at rest, i.e., vC(t=0−)=0 and iL(t=0−)=0. At t=0, switch S1 is closed but S2 stays open. After a time T, S1 is opened and S2 is simultaneously closed. Answer the following questions in terms of the parameters: V, R1 R2, C, L, T and t, as applicable. a *What is the expression for vC(t) during the time interval 0 < t < T?* b *What is the expression for vC(t) during the time interval t>T?* c *What is the expression for iL(t) during the time interval t>T?* In the case that the time T is a VERY LONG time, find the following reduced expressions (Hint: the capacitor is completely charged): d *What is the Reduced expression for vC(t) during the time interval t>T?* e *What is the Reduced expression for iL(t) during the time interval t>T?* ---------- FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-27T12:37:11Z FirstChildTAG: The letter T (uppercase) is a constant. To be accepted as an independent variable has to be written in lowercase. FirstChildUserIdTAG: 361823 FirstChildUserNameTAG: EliasOak FirstChildCreateTimeTAG: 2012-12-27T13:11:01Z SecondChildTAG: But the lightspeed c is a constant too, but I've never seen it written in uppercase. "Has to be written" is just a convention. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-12-29T15:18:53Z FirstChildTAG: I see xp42's point. His answer is in fact closer to the "correct" answer, than the given/required answer. What I discovered at this course, and in the spring one, is that (I don't mean this in a negative way) sometimes a lack of knowledge is an advantage. This question made me also very suspicious about the required answer. In fact I almost would have given his answer too, but I thougth by myself "They don't expect me to fill in this very large formula, don't they?" Let's try the simple one first. With HW12P2 I had a discussion about how to interpret efficiency, and an ideal OA. From a practical point of view, with a lot experience with building circuits, all kinds of questions popped up in my mind. Where did the supply for the OA come from? I mean, if the OA was ideal, then in fact it wasn't even necessary to use the bjt. If you calculate efficiency, then I expect that one include all powersources and so on. So indeed staff at last changed the text a few times, because I was nagging. A very long time is not the same as infinite time, but infinite time is a VERY long time. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-12-27T13:42:30Z SecondChildTAG: You definitely have a good sense of humor, salsero:) SecondChildUserIdTAG: 285616 SecondChildUserNameTAG: AmiyaX SecondChildCreateTimeTAG: 2012-12-27T15:50:40Z SecondChildTAG: Thank you! SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-12-27T16:36:11Z SecondChildTAG: Dear Salsero, having some experience with circuits does not mean that you will be right no matter what. I am sorry to disappoint you, but you'll have some surprises in the future too. > What I discovered at this course, and in the spring one, is that sometimes a lack of knowledge is an advantage. I hope you will realize that it depends if your knowledge is correct or not.There are a bunch of guys that repairs tvs just based on experience.If a transistor will fail, they will find an "equivalent " that will burn up within a few months .My point is that it is best to learn things the right way. I am a bit surprised that you will say something like that, since i always considered you a good student. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-12-27T20:06:56Z SecondChildTAG: To AlexAlexandrescu: I agree with you and I can see your point. I will elaborate more on my interpretation of a lack of knowledge in a new post. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-12-28T12:56:41Z FirstChildTAG: ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13566341891343658.bmp Even i faced similar problems while attempting this question.I certainly dont find it acceptable to reject this answer on grounds of VERY LONG time T. Infact this answer is more accurate than the approx. answer being accepted correctly.I sincerely request the staff to please consider this case as it stands on genuine grounds. FirstChildUserIdTAG: 257681 FirstChildUserNameTAG: Bhumi FirstChildCreateTimeTAG: 2012-12-27T19:03:30Z SecondChildTAG: It IS NOT the first exercise of this kind. If you are not familiar with it, i kindly suggest to go on and to study more. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-12-27T19:50:12Z SecondChildTAG: Hi Bhumi, approximation is very important in engineering science, EECS included. It is just math which mostly requires precison.So in the case, we do not choose the complex,more precise answer with the hint " very long time". SecondChildUserIdTAG: 361137 SecondChildUserNameTAG: Ericson SecondChildCreateTimeTAG: 2012-12-28T06:47:29Z FirstChildTAG: I think that those complaining about the official answer for 4b and 4c miss an essential part of the course. Considerable time was spent in the lectures on intuitive solutions and simplifications. In this case, they expect you to understand that after a "VERY LONG" time, the capacitor is fully charged. If they had wanted the $(1-e^{-\frac{T}{\tau}})$ term to appear, then they would not have said a VERY LONG time. Recognizing this fact was required for a correct answer. In fact, my reaction to seeing "a VERY LONG time" was "Oh good, the capacitor is fully charged". In fact, Q6 covers the territory where you do have both rising and falling exponentials. I found the explanation for Q4 parts B and C rather long winded. At $t=T$, it becomes a simple undamped LC circuit, so $v_C(t)$ will be of the form $Acos(\omega_0 t)$ where A is $v_C(t)$ at $t = 0$ and $\omega_0$ is the resonant frequency, i.e. $\frac{1}{\sqrt{LC}}$. I too had issues with HW12P2. I had no such issues with this problem and would have got it in two tries if I hadn't had a typo with an $R_2$ turning into an $R_1$ (I blew one try missing the minus on the exponential and another because T was missing from the list of paramaters, so I tried once without T in my expression). FirstChildUserIdTAG: 141000 FirstChildUserNameTAG: OrinE FirstChildCreateTimeTAG: 2012-12-27T19:47:07Z FirstChildTAG: I got this wrong on first attempt too. ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13566341891343658.bmp But I was "aware" even as I entered this formula, that the wording of the question might be indicating that the simpler form was wanted here. I did have 3 tries. So, I said, what the heck, let me do it this way first, and if it doesn't work, try the simplification. So, I got it on the second submit. When trying to figure out what is right and what is wrong, we also have to figure out what we're being asked to do. There's some ambiguity here. And given the nuances in the English language, I wouldn't be surprised if alot of people whose native language is not English, got this question wrong. A careful reading of the question, however, reveals what the Prof really wants the student to do. A major part of the question is seeking whether the student understands what part of the formula can be tossed away, and what part must be kept. In other words, the question is really seeking the answer to the question: "What is the correct and appropriate approximation to use here." This was supposed to be indicated by the "VERY LONG" time phrase. But, the phrase doesn't quite do it, because "VERY LONG" relative to what? We have to "guess" what is meant and being inferred to, that is, that the circuit reached some stable and unchanging state relative to our measuring instruments, etc..and we can neglect further changes. Of course, if the circuit was on a spacecraft flying by the earth, that VERY LONG could become a surprisingly short time to observers on earth, according to relativity. Or a student, taking in the Prof's remarks very seriously, might think that VERY LONG indicated that relative to the speed of light, the signals flowing around the circuit take a VERY LONG time, so that the circuit obeys the lumped matter discipline (LMD), and therefore all circuit laws apply. Approximation, is, however, a major part of science. Knowing what approximations to make is often more important than knowing the exact formulas. I always found math easy, and physics hard, and this was one of the main reasons. Mathematicians have exact answers for everything, every problem has a unique solution, and there is such a thing as right and wrong. I decided on Physics partly because it was more of a challenge to understand why I didn't understand things. When I got the graduate school, my adviser, who was also the Dean of the Physics dept at the time, told me that I was suffering from "student's syndrome." This he explained to me in detail. When you're an undergrad, or back in high school, the type of questions and problems that you're challenged with, all have exact answers. You can always just look at the back of the book, to see what the "right" answer to the problem is, to check if you got it right. But when you enter graduate school, you now begin to grapple with real world problems. These problems have no exact answers, there's a large grey area, appropriate approximations have to be made, lots of untested assumptions are introduced, many "intelligent guesses" are required to arrive at a solution that may even only partly work. This is where I had the difficulty with my Physics. Why are these intelligent guesses made and not those, how do I know that I've made the correct assumptions, or made the right approximations? Did I leave anything out? Is the approximation justified? These are the kinds of things that make Physics much harder than Mathematics. The mathematician always knows, at the end of the day, whether he's right or wrong. But, the Physicist, after all the work, is still making "intelligent guesses," and is never sure. There's always some new parameter that if adjusted might throw out the whole calculation. So, what this question was really testing, was "intelligence", rather than knowledge. Can you make the "intelligent guess," here, to figure out what the Professor might have intended you put there in the answer box? That's not a knowledge guestion, it's an intelligence test. I see alot of these "I.Q" type tests over in 6.00x and 3.091x too. Anyway, that's my take on it. Your ability to figure out what the question really want you to put in the box is part of the testing process. FirstChildUserIdTAG: 4463 FirstChildUserNameTAG: pmj FirstChildCreateTimeTAG: 2012-12-27T20:33:28Z SecondChildTAG: > "VERY LONG" relative to what? Well, it's relative to the time constant (of the charging circuit in this case): after 3*tau the exponential term is very small relative to 1; e^-3 is always 0.05 even if we used infinite precision instruments. I learned this during my physics course at University 20 years ago and I never forgot it. SecondChildUserIdTAG: 376877 SecondChildUserNameTAG: AndBre SecondChildCreateTimeTAG: 2012-12-27T22:26:18Z SecondChildTAG: Wouldn't then be a lot clearer to just say T >> 3*tau in the problem specs? That's the thing. If we only learn one thing, then there's only one possible response. But, if we learn many things, then there arises the possibility of alternative answers. At that point, there's ambiguity in the question. So, it all depends on what you know. It's also necessary for the times in the circuit to be VERY LONG compared to the time electromagnetic waves move about the length scales involved, before we can use the LMD approximation. Many types of approximation require VERY LONG times too. SecondChildUserIdTAG: 4463 SecondChildUserNameTAG: pmj SecondChildCreateTimeTAG: 2012-12-27T23:25:44Z SecondChildTAG: > to just say T >> 3*tau in the problem specs And what would have been the fun part of it? It would be like reading a crime novel and knowing who the murderer is before getting to the last page... ;-) SecondChildUserIdTAG: 376877 SecondChildUserNameTAG: AndBre SecondChildCreateTimeTAG: 2012-12-28T11:38:55Z FirstChildTAG: With every exam there will always be a lot of students complaining that they were confused by something or that there was something wrong in the answer.What they omit is that there were also a bunch of students that solved the exam as it was. It is not the first time when you see this type of exercise. Enrolling in this course DOES NOT guarantee that you will take grade A, or that you will pass. It is very accessible from my part and there was also the review packet far more difficult . There was a H8P1 offered for free in the exam since you can just copy the answers.Of course that if someone will say that they haven't noticed that, it means that they didn't invested enough time in the course. It is not a charity for certificates in my opinion. FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-12-27T20:14:13Z SecondChildTAG: I simply reworked the problem that was a duplicate of H8P1. I recognized that it was very similar, but it would probably have taken me more time to go back and find the problem in the homework, then check that the details were the same, than to rework the problem. SecondChildUserIdTAG: 141000 SecondChildUserNameTAG: OrinE SecondChildCreateTimeTAG: 2012-12-27T20:43:58Z SecondChildTAG: Hey, you're right. They did give H8P1 free in exam. Bummer. I didn't notice that. Wow! I wasted time doing what I didn't need to, all over again. Hmm...Gotta pay more attention next time. They did the same thing last time, but the Prof warned us that he would do so. Somehow, I didn't recall any warning this time around. Who knew? Oh well, we did get 24 hrs, which was more than enough time to do things over again. ;) SecondChildUserIdTAG: 4463 SecondChildUserNameTAG: pmj SecondChildCreateTimeTAG: 2012-12-28T02:54:45Z FirstChildTAG: My friends,remember that a long time for this type of circuits is t>=5*TC.It is common to assume for the t greater that five time constants (5*TC), that the function e^(-t/TC) is essentially zero. FirstChildUserIdTAG: 282828 FirstChildUserNameTAG: Chibolator FirstChildCreateTimeTAG: 2012-12-28T00:30:10Z FirstChildTAG: I think the wording was good on this one. It allows you to prove that you understand what you're doing, and not just how to solve explicit math problems. FirstChildUserIdTAG: 90796 FirstChildUserNameTAG: ChaunceyGardiner FirstChildCreateTimeTAG: 2012-12-28T04:39:10Z FirstChildTAG: Tsk, tsk, @Chauncey. If you needed to put pen to paper for question 4 you haven't learnt something fundamental to this class. **Intuition**, dear Chauncey, *intuition*. This problem was so simple that you should have obtained the answer by inspection. There was no need for any calculations or math. Shame on you for implying otherwise! --- What the problem **did not state** was that the voltage on the capacitor reached a steady state or that the system reached equilibrium by time T. **Absent this assertion, T is just a very long time: nothing more!** *You need an infinitely long time to actually reach steady state.* 4b asked for **'the'** expression for the voltage on the capacitor, 4c asked for **'the'** expression for the current in the inductor. In English, the expression **'the' anything** means either: everyone already knows '*which anything*' you are referring to, because it is '*the anything*' we are currently talking about; or else, it is **'the' one and only anything**, and it is unique. In the context of the exam question 4b and 4c, **the expressions** asked for can only mean the *unique* expressions. There are two sets of expressions on the table at this point: the exact equations, and a set of approximations. Neither one is currently being talked about. Only one of these sets can be **the expressions** for the capacitor voltage and the inductor current at time T. So what do we have to help us decide? * T is a very long time, but not infinite. So the actual values at time T are exponentially close to the steady state solutions, but they are clearly different from the steady state solutions. * The exact expressions account for the exponential approach to the steady state values, and at time T they evaluate to values exponentially close to, but different from the steady state values. * The approximation expressions define the steady state values. * Thus the approximate solutions (i.e. the steady state solutions) can not be **the unique expressions** since they do not uniquely express the values at time T. They only estimate them, and clearly many other expressions can also describe approximate solutions (and more accurately as well). * Only the exact expressions for the capacitor voltage and the inductor current can uniquely express the values of these quantities. Thus only they can be **the unique expressions** at time T. ---- Now before you blow a gasket Chauncey, let me say that it became apparent what the intended answers were: obviously, red X's serve some purpose! Unfortunately, the intended answers were not the ones actually asked for, regardless of the intent. --- Considering the international reach of this class, I think the words used should have been chosen more carefully. Had 4b and 4c each asked for **an expression**, or **an approximate expression**, or even **a simplified expression** there would have been no foul. Had **steady state** or **equilibrium** been mentioned there would have been no foul. It's just my opinion, but I think edx failed on this one. Two attempts to get it right, and still they swung and missed. --- FirstChildUserIdTAG: 9673 FirstChildUserNameTAG: xp42 FirstChildCreateTimeTAG: 2012-12-28T08:42:29Z SecondChildTAG: Are you psychic now ? You don't have any reason to assume anything about my approach. I was talking about the wording of the question, based on the complaint you posted here. I don't see any justification for you to blame me for your lack of intuition and practical skills. Most people that reads your first post here can see that you approached this problem with nothing but theoretical math skills. SecondChildUserIdTAG: 90796 SecondChildUserNameTAG: ChaunceyGardiner SecondChildCreateTimeTAG: 2012-12-28T09:03:11Z SecondChildTAG: > you haven't learnt something fundamental to this class. Intuition, dear Chauncey, intuition. This problem was so simple that you should have obtained the answer by inspection. There was no need for any calculations or math I totally agree with you, xp42, but, I'm very sorry to have to say this, if there is someone who hasn't learnt something fundamental in this class, well, that's YOU. If after taking this course two times you are still making these complaints, you missed a very import point of the entire class: getting a fast answer making the right approximations. This is electronics, without approximations is math. And you are talking about math. But this course is called Circuits and Electronics. Have an happy new year and try to relax. SecondChildUserIdTAG: 376877 SecondChildUserNameTAG: AndBre SecondChildCreateTimeTAG: 2012-12-28T11:07:48Z SecondChildTAG: I still agree with xp42 and I absolutely don't have the skills to become a mathematician ever. I think xp42 has an excellent intuition and I will elaborate on that in new post here. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-12-28T13:45:06Z SecondChildTAG: > You need an infinitely long time to actually reach steady state. If we are going to be this pedantic, no you don't. Remember charge is carried by electrons, so at some point, there will be not enough voltage difference to get another electron into or out of the capacitor... Run some numbers - the charge on an electron is $1.6X10^{-19}$ - taking a 1pF capacitor, and $Q=CV$ and discharging from 1V to one electron, I get just under 16 time constants; certainly not infinite. Oh, but you say, that's a tiny capacitor - put in 1F - now I get 43 time constants. How about from 1000V, add 7 more time constants. So discharging a 1F capacitor to 1 electron takes about 50 time constants. Of course, for real circuits, noise will enter the picture and after a VERY LONG time, the capacitor voltage will bounce around the theoretical fully charged voltage and you'll never get so close that one electron will make a difference. SecondChildUserIdTAG: 141000 SecondChildUserNameTAG: OrinE SecondChildCreateTimeTAG: 2012-12-28T22:10:06Z SecondChildTAG: If he had excellent intuition he would have used it, instead of complaining in here. Besides - he brought it up. I stated that I thought the wording was good and I gave a reason why. He responded by attacking me and using his usual psycho techniques to make it look like I am to blame for his problem with the question. Some people are simply unable to handle anybody disagreeing with them. SecondChildUserIdTAG: 90796 SecondChildUserNameTAG: ChaunceyGardiner SecondChildCreateTimeTAG: 2012-12-28T22:10:09Z SecondChildTAG: ... and professor on the videos said "very long time" a lot of times... I thought that the meaning should have been clear for everybody watching the videos. SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-12-30T00:29:05Z FirstChildTAG: I feel the problem was fine. "VERY LONG time T" - with the first two words written in caps was enough to let us know we should assume infinite time. I started working out the exact expression first because I didn't notice those two words. I spotted it mid-way through my calculations. I have a tendency to just see the circuit and then jump to the questions above the textboxes. I made a fool of myself a couple times in the Spring session thanks to this habit. "The expression" = "the only and only expression when T is very long - VERY LONG - enough for you to consider T as infinity" = "Steady state value" If such an approximation is incorrect, we shouldn't be studying steady state analysis at all because infinite time is never possible and we will never reach the steady state value. AFAIK, anything beyond 5 time constants (some books say 4, some 5) is considered as steady state (maybe things are different when you go down to a 10nm gate length, maybe not - $5\tau$ is based on a ratio anyway). Also about the difficulty about the paper, I do agree the paper was easier. However we may not have found it as easy if this was our first time. Piotr did say that the Spring 2012 paper was "on the longer side" (or tougher side. I don't remember exactly). I really feel he was understating the difficulty and I feel the paper was hard even by MIT standards. I didn't find all of the review packet problems as hard as the finals. Some were hard. Not all. I would love a hard paper for a good mental workout but that may not serve the purpose for someone new to the course. FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-12-28T10:56:15Z FirstChildTAG: The term "**after a very long time**" always was used like that. Its pretty clear it refers to a long-term stability point. Hence those questions were not about the short term expression. FirstChildUserIdTAG: 331483 FirstChildUserNameTAG: Doru FirstChildCreateTimeTAG: 2012-12-28T11:00:19Z FirstChildTAG: xp42 says: "Considering the international reach of this class, I think the words used should have been chosen more carefully." And that is why I totally agree with xp42. The so called 'intuition' and 'testing intelligence', used by some people here, is a large part of investigation of psychologists. A lot of psychologists disagree about what intelligence really is, what one is really measuring with a so called intelligence test. There are many types of intelligence and intuition. Every psychologist knows that you can't simply translate an IQ-test from one language into another language. It's a big mistake to think that's that simple. Things like ethnetical background, culture and religion, language, to mention a few, are even of importance within countries themselves. So it's very important when you deal with an international audience, to take into account the way you ask things. Once, in an electronics midterm, I had to name 5 kinds of measuring devices, but we had learned 10 types of them. I gave 5 correct answers, but I only got 3 points out of a possible 10 for that question. So I asked for an explanation. Why 3 points only? And the teacher replied: Because you didn't mention the MOST important ones ... I replied, they didn't ask for the that, but he stated it was logical to expect that he ment that. Hmm! Sorry! I think I deserved extra points, because AT LEAST I showed that I had learned the least important ones too! And my logic, my friends call it LeoLogic ( because of my totally different way to look at things in general and leo is my name), is that if you are 'expected' to give the most important ones, then you (if you are lazy) develop the 'intuition' only to learn what is important in the eyes of someone else. Intuition works in several ways. Because I make so many mistakes with calculating and with manipulating formulas (because of a lack of concentration, I can hardly copy a long formula to an other piece of paper), I developed a kind of a sense for a 'beautiful' or an 'ugly' formula. It often helps me to find mistakes I made somewhere. Teachers and students come in different sizes. Some 'intelligent' teachers are not so intelligent, that they are able to understand why a student doesn't understand something. But a 'less' intelligent teacher understands sometimes perfectly why a student doesn't understand something. Why? Because, for example, the teacher had problems himself with learning the material. Not every student learns in the same way, it's not always intelligence only, that is of importance. It's also the way how you present the material and so on .. In one course you learn how to be precise and exact, like math and on highschool with physics, but in other courses you 'learn' how to be precise, but not exact. Hmm! Sometimes with HW we were expected to be exact and sometimes not. Not only in this course, but also in other courses I did. Go and figure what the prof means ... I'm not a mind reader! That brings me on a lack of knowledge, but that's for another post. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-12-28T15:53:02Z SecondChildTAG: **The so called 'intuition' and 'testing intelligence', used by some people here, is a large part of investigation of psychologists.** You must remember, that MIT is not doing these online courses for naught. They expect to get something in return. Prof Agarwal mentioned the critical thing they are collecting, and its "data on how we learn". So, it's not surprising that some ambiguity will be thrown in to questions now and then. Part of the testing is probably to determine how students from around the world handle various kinds of ambiguity. How does someone in "China" interpret the "VERY LONG" phrase, verses someone in "Brazil" ? All the data they collect is only valuable if the questions are designed in such a way that they reveal something about our minds. ;) SecondChildUserIdTAG: 4463 SecondChildUserNameTAG: pmj SecondChildCreateTimeTAG: 2012-12-29T01:34:26Z SecondChildTAG: Yes, I noticed. And if they want to know how a certified nut learns, then they only have to study me ... SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-12-29T13:22:10Z SecondChildTAG: I resemble those implications, [salsero][1]. ;-) [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/users/83276 SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-12-30T01:22:40Z FirstChildTAG: @AlexAlexandrescu: You wrote: "having some experience with circuits does not mean that you will be right no matter what. I am sorry to disappoint you, but you'll have some surprises in the future too." Yes, you are so right. However, I never mentioned that I was totally right. And I'm sure that I'm in for more surprises in the future too, and even better: I sincerely hope so! My reason for learning at my age, is not to get a job, but to gather knowledge. I realize that there is more that I don't know, than that I know. I've no intention to compete with anyone of the staff, because I accept the fact that the worst of them probably always will outsmart me. But that doesn't mean that I'm not allowed to give my point of view on how a problem is stated, and why I think it's confusing ME, and some other students agree and/or disagree on certain points. I can live with that. And I learn also from the comments of the staff and from other students: if I had not asked or pointed out my point of view, I wouldn't have learned certain things. But I notice, let me speak for myself, that I often "over think" problems, I make them more complicated than they are, or are ment to be (and often the opposite too). And if you know more theory about something, or have practical experience, then it can sometimes be a disadvantage. I remember from one of the homeworks, but there are more examles, that a 0 (zero) was only accepted as correct, while the real answer was in fact very close to it. Many students had problems, probably not to find the answer, but to find the expected "correct" answer that one had to put into the editbox. Such an "error" drives my crazy! If I take H12P2 again: I found the problem itself very simple, but what made it problematic in my case, was for example that the grader did not accept the correct answer in the first place to begin with, because the text in the problemset was incorrect. I don't blame the staff for making such a mistake, that's not the point, but what happens to me is, that I ALWAYS doubt my own answer FIRST, because I've add/adhd and make so many mistakes, you won't believe! And than I'll dive "deeper" into the problem, trying to find out if I overlooked something. So the staff corrected the text, but still no green mark (no red either!!). And why? Because, I discovered that much later, now an answer like 10M or 10000k (10 mega ohm) was not accepted, (nowhere mentioned in the problemset that a suffix was not allowed) while it normally does accept such an anwer. It's both common practice and practical knowledge to use suffix k or M. But the students without that knowledge, or not yet used to that 'practice' or habit, didn't encounter that problem. So also in this case I started to doubt my answer. It was correct, but not accepted. And I couldn't change the input of the first and second editbox of H12P2, (it was not updated) read my post about that over the experiments with the on purpose created errors with the "inf" in the answer,it took me many hours to find out how to work around that. How? I put the right answers in box 1 and 2 and created an error in box 3, by multiplying the resistance of 10M with inf/inf. Now were at least box 1 and 2 accepted (updated), but got the error in box3. (Notice: I could NOT change (update) box 1 and 2 without creating the error in box 3! And I could not change the other boxes, because for one or another reason the update wasn't propagated). And after that I removed the inf/inf, than I was able to remove the suffix. I'm sure it's an error somewhere in the program for the grader, or maybe it was created because staff changed something, while I already had put the values with the suffix in the boxes before the text was changed. I still don't know. So yes, because of these problems I tried out many things, and my practical experience helped me to look at it from different points of view, because changing the editboxes wasn't working properly in my case, and I realized that my experiences could help other students too, especially the problem with the connection to the pws of the OA. So my practical knowledge is by times working against me and sometimes advantageous. You said: "My point is that it is best to learn things the right way." What is the right way? And for who? I think that discussions like this, can be very interesting for students too. I've seen hundreds of questions about the same subject from students on this forum and that of spring course because they lack a certain insight. I've no problem with that, it's the main reason why we are here: to learn. And I personally think that suggestions of practical experiences can be very important too. Because it also creates a certain intuition to look at problems. Maybe not at this very moment, but someday later. "If a transistor will fail, they will find an "equivalent " that will burn up within a few months." Yes, I know! Or that you can replace an inductor in a color tv deflection system by some other "equivalent" inductor, an electrolytic capacitor in switched PS by "any" with the same values for capacitance and voltage. Or, even a stronger example: Any 74LS74 with any other from an other chip maker. Not so! The designers of an Intel SBC80/10 or 80/20, I don't remember anymore, "abused" the fact that the chip produced and based on nor-technology, can have different effects than one with nand-technology. They didn't mention it in their doc that it was necessary to use that specific brand, and they made a kind of one-shot of it in a reset after power up. Your collegue can't solve the problem and gives you the board (with the wrong chip already mounted) and solve that!! I replaced hundreds of chips in my time, but this was a real surprise! I knew the theory, and I solved it, but it took me hours! I always check the specifications if I replace whatever component. My collegues always gave the 'unsolvable' problems to me, because I don't give up easily, that's an advantage of add/adhd: if something is really interesting I don't let go! And I love that kind of problems. And that is a reason why I never ask questions at the discussion about the course material itself, I want to find out myself. And with the answer of xp42: his answer itself is just correct. It's as simple as that. I think it's totally legitimate to mention it, because it wouldn't surprise me, if an other professor or on another exam, or another course, expects you to give that answer. That's also knowledge that I've gathered during my years with studying. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-12-28T20:49:46Z FirstChildTAG: About intuition: Many things in math and physics were only discovered by people with the "wrong" intuition and later they were proven to be right and created totally new fields of sciences. And in this case: I mention Cantor, thanks to him we are able to work with infinities; he discovered there are several kinds of inifinity. The poor man spend a lot of time in mental hospitals and suffered from depressions, partly because the "scientific" society rejected his ideas. Take Boltzman, Einstein, Galileo and so on, all with the "wrong" intuition. My problem with a lot of teachers and academics, is that they always think that what they know is absolute truth. I appreciate teachers telling you that what you learn today, might proven to be untrue tomorrow. People believe in numbers and statistics and use them as an excuse to treat other people in an unpleasant way. If 99.9% disagree with me, then it still doesn't prove that they are right, even if I can't prove that I'm right. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-12-29T14:49:25Z FirstChildTAG: The evololução Science is accompanied by several times. There was a time when a simple look to an apple in freefall enabled the discovery of laws that led a man to the moon All this using intuition and genealidade. However, with the technological developments achieved by the development of Science, reached a status where intuition has become an afterthought. We can build buildings or even reach the Moon with the same Newton's Laws. But to find a Higgs Boson in the middle of a bunch of elementary particles that make up matter becomes more complicated if we do not use large particle accelerators and increasingly sophisticated computers, and billions of dollars to invest in this venture. Today Science is not done with one person, but with groups of scientists and organizations. This evolution seems to biological. A mutation that made genática humans from individualism and become multicelur, even if the mode of constrir knowledge. The EDX platform is an example of this evolution. It is made by a lot of people together that allows knowledge to flow like clouds in the sky. FirstChildUserIdTAG: 361823 FirstChildUserNameTAG: EliasOak FirstChildCreateTimeTAG: 2012-12-30T10:18:56Z IndexTAG: 242 TitleTAG: Again, Popping up some Posts! :) Hi all, I am popping up some Posts that might got lost in the list. If you haven´t read them, you should haha ;) Here they are: ---------- **Evidences that proves that Prof. Agarwal is awesome, cool and a big hearted person. Thank you for being like this! :)** [Read Here][1] ---------- **Welcome to the CECC 2 ! Do you want to win a Textbook signed personally by Prof. Agarwal? Read here :)** > You have time to submit your video till january 14th 2013. [Read Here][2] ---- See you!;) [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50c08a8f2196ec2300000022 [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50b2892629b7b1270000003b UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2012-12-26T14:49:57Z VoteTAG: 8 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi,Myrimit.I can't have access to the introduction video.Would you please help me? FirstChildUserIdTAG: 361137 FirstChildUserNameTAG: Ericson FirstChildCreateTimeTAG: 2012-12-27T07:37:27Z SecondChildTAG: Hi Ericson, I uploaded it on 4shared... Is that link not currently working? Please let me know... SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-27T12:46:22Z SecondChildTAG: Thanks Myrimit,I have downloaded it.Than link works. SecondChildUserIdTAG: 361137 SecondChildUserNameTAG: Ericson SecondChildCreateTimeTAG: 2012-12-28T06:11:11Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-28T11:29:54Z IndexTAG: 243 TitleTAG: Big Thank to Edx Thank you very much professor Anant agrawal and team Edx. It was great learning experience from best.Really after completion of this course i will more confident in dealing with any type of electronics circuit and has developed more interest in learning more.Looking forward for more courses from MITx. THANK YOU UserIdTAG: 431908 UserNameTAG: navneet_ipu CreateTimeTAG: 2012-12-25T17:13:44Z VoteTAG: 8 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 244 TitleTAG: Thanks to edX Staff, fellow Comunnity TAs and especially edX President - Anant Agarwal First of all, thanks to my fellow Students for tolerating me this semester. Thanks to whoever selected and/or approved me as a TA. This is something that I truly cherish and I am humbled to be amongst such fine company. JerseyMark, Myrimit and JSChambers. Thanks to Professors Gerald Sussman, Khurram Afridi, Chris Terman and Piotr Mitros edX Cheif Scientist. Special thanks to Professor Anant Agarwal who is also the President of edX. For this initiative I feel he deserves special acknowledgment. Thanks to the folks who keep this learning machine going - Lyla Fischer, Rohan Nagarkar, Howard Lurie, John Jarvis, Teppo Jouttenus, Victor Shnayder, Mark Chang, Tena Herlihy and Elina Hu. (No particular order) Please forgive me for the people I have missed, I'm sure there are many more out there. So I wish the best to everyone in your future ventures, I hope that many of you will visit the forum in the off season, perhaps we can use it as an impromptu penny-university. I myself am not a Religious man anymore, but may your particular God bless you and your families this Holiday Season and through the New Year. Be safe. UserIdTAG: 292543 UserNameTAG: Pennypacker CreateTimeTAG: 2012-12-25T15:43:54Z VoteTAG: 8 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Thanks, **Pennypacker**, for all your help, as well, throughout this semester of 6.002x! Happy holidays and I wish you success in both your professional and personal life! Jersey Mark FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-12-25T18:16:16Z SecondChildTAG: Thanks JerseyMark! SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-12-25T18:28:45Z SecondChildTAG: Thanks to all! SecondChildUserIdTAG: 402193 SecondChildUserNameTAG: Feramico SecondChildCreateTimeTAG: 2012-12-25T18:52:41Z IndexTAG: 245 TitleTAG: Do you want to win a Textbook signed personally by Prof. Agarwal? Contest extended due January 14th 2013!:) Hi Classmates of 6.002x, I don´t know if you knew, but we are running a Contest. You can win a Textbook Signed Personally by Professor Agarwal delivered at your home. If you will to participate you can check the rules at the Wiki and send us a video of any related topic of 6.002x till January 14th 2013! We haven´t made tight rules, you can handle the time duration of the video and you can use all that you want, we encourage to use real components in your video. Take a look at the original Post [here][1] See you!:) I hope that you can participate. CECC Team. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50b2892629b7b1270000003b UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2012-12-24T15:27:07Z VoteTAG: 8 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: **Cool!** I thought I would *miss* the CECC Contest because of Christmas and New Years celebrations and getting together with my family and friends after a very, very busy Fall, but since my new classes don't start 'till February, I'm sure I can devise a quick project and knock out a quick video in 15 days! I was originally planning on designing my own Control System for an Anti-lock Brake System as my Senior Design project for University (I have a classic 1987 Ford Mustang, as well as the anti-lock actuators from a 1995 Mustang, the first year ABS was implemented in these cars, as well as the 1995 sensors (Hall-effect pickup type for those interested). I went as far as getting the mounting points for the sensors and wiring up a harness, as well as a pin-out to a breadboard. I saw guys do this project the conventional way (with the Ford factory Anti-Lock Controller transplanted into the earlier-model cars; but this does not involve ECE 6.002x theory, only a bit of wire tracing, like hooking the controller to the actuators, power, ground, and the ignition key), but the wiring for the actuators and sensors is actually pretty simple. So: My goal was to use **op-amps**, a **micro-controller** (probably with as simple of a program as possible), probably **A/D converters**, and at least have a functioning front-wheel anti-lock system (the rear is trickier as these are the drive wheels and involve reducing engine output) and demonstrate it at work on an icy road - make a quick video of locking wheels vs. the "pulsating" lock-unlock-lock-unlock cycle. If not, I would document all the signals from the sensors and document where the failure might be. It would be cool for the students to have a "real" anti-lock braking system (ABS) demonstration video for 6.002x, instead of the Prof.'s poor handwriting! After this, I would replace my "project" with the stock Ford controller for **safety reasons** so I could drive the vehicle on public roads. The problem is I never got around to fabricating and plumbing the hydraulic lines to the brakes themselves (not fun for the home / hobbyist mechanic, though pre-made sections are available), and with winter here, that looks like a project for the summer. I am also in desperate need of a digital oscilloscope so I can trace the digital and analog signals from the sensors and through the circuit, and get rid of any glitches since there is much sources of noise in an automobile. It's now a project for another time, and will take months as opposed to days to realize, but I need to think of something "quick and dirty" that I can wire up in a few days for CECC. I'm sure I'll think of something. I've never made a Youtube video though, so it would be my first time for this. ***I can't wait until all the students post their videos for all to view after the contest! Good luck to all!*** Jersey Mark FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-12-24T15:58:57Z SecondChildTAG: Sounds good :) But will it be correct involve into project microcontrollers?For my look project should show something very close related to the course matherial. Anyway, good luck! SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-24T16:34:41Z SecondChildTAG: Cool!:) I am so happy that all the Students from 6.002x Fall 2012 can Participate in this Contest. The Jury are Ex-Students of 6.002x Spring 2012. Good luck to all of you. P.D. Hi Serge, yes you can use microcontrollers :). We encourage to use your creativity, you are free to make your video as your way, we haven´t write tight rules :). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-24T17:05:22Z SecondChildTAG: Hi, Myriam!! I thought about small project, but it seems to be unreal even with new data because my preference is to show real device with explanation. I will think a little about this and may be I will need to ask you later. Serge PS I promise do not use BGA technologies in this simple project :) SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-24T18:52:22Z SecondChildTAG: Cool!:) I am happy too see your enthusiasm!:) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-24T18:58:34Z SecondChildTAG: I agree with both Myriam and Sergtronix. You can use a microcontroller, but your circuit should demonstrate something you learnt in 6002x. So, if you for instance just do the typical microcontroller blinky, you won't get my vote. If you can get an LED to blink without a microcontroller, you have a chance. (I'm not saying an LED blinking is going to win the contest, it's just an example.) SecondChildUserIdTAG: 90796 SecondChildUserNameTAG: ChaunceyGardiner SecondChildCreateTimeTAG: 2012-12-24T21:28:25Z FirstChildTAG: Now this is an extension I could use! I have been procrastinating! FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-12-24T15:46:47Z SecondChildTAG: Myrimit, por que no extienden ese premio también a quienes obtuvieron un 100% en el exámen final y que no tienen posibilidades económicas de adquirir componentes y herramientas para producir videos ? Las cámaras de video son muy caras, así como los componentes, especialmente en países de América Latína donde los impuestos son muy altos. Yo sugiero, pero está en ustedes. Creo que un libro de texto impreso sería un regalo muy valioso para un estudiante de Latinoamérica. SecondChildUserIdTAG: 183712 SecondChildUserNameTAG: colegiocientifico SecondChildCreateTimeTAG: 2012-12-24T16:24:32Z SecondChildTAG: Lol, I am not Myrimit, but perhaps you could borrow an inexpensive webcam or phone. Due your best, I'm sure they will judge content and not production quality. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-12-24T16:34:30Z SecondChildTAG: Do, not Due lol SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-12-24T16:34:43Z SecondChildTAG: Hi colegiocientifico, That is a nice idea colegiocientifico, but as there are many students worldwide with different conditions, and I am sure that there were many of them that made a lot of efford and scored 100% and it will be really difficult for us as Students, in our position, to decide who could receive the Signed Textbook...as in my point all deserves it. The Jury will not focus in the production quality,you can use any webcam, it is not necesary to be a cost camera... they will evaluate another things like if they are educative, if they are creative, etc... Also, some students asked us in this Fall if they could submit videos by group, and yes you can - the only problem will be who will stay with the signed Textbook haha- that would be a difficult decission internal of the Group :p . So, might if someone of the group don´t have a camera, but can colaborate in edition or simulating in sandbox the circuit that someone did in another part of the world or anything,yes, you can, you can use your creativity. And in that case, if that student will not have a camera or a web cam, or components, it would not be a restriction... and remember that the most important is that you can participate and have a lot of fun. Remember this Contest is organized by students to students, Prof. Agarwal kindly donated 3 Signed Textbook for our Contest and the edX will send them once we have the Winners like the previous Term-CECC 1. So, we have decided to open the doors to every student of this Fall 6.002x 2012 by equal to participate in this Contest - all the members of the CECC Team are ex-students of 6.002x Spring 2012, Jury and Organizers-, The main objective is the motivation present-future. If you see, in the Wiki you will find the last Term Winners of 6.002x, I hope that this motivates to all the Students of the Fall that is possible to put the Theory into Practise :). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-24T17:40:52Z SecondChildTAG: Myrimit I'm having problems to include subtitles in my video (the Window Movie Maker only has VERY big subtitles :( ), there's some problem if I add them using Youtube? SecondChildUserIdTAG: 298547 SecondChildUserNameTAG: AnthonyRF SecondChildCreateTimeTAG: 2013-01-05T19:10:37Z SecondChildTAG: Youtube subtitles are fine too. SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2013-01-07T09:05:27Z SecondChildTAG: Hi AnthonyRF, as ashwith said, Youtube subtitles are ok :). Sorry for the delay of the answer. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2013-01-07T13:38:02Z IndexTAG: 246 TitleTAG: From India Dear staff, I felt very happy after completing the course . I got 68% but full review of my Under graduation has happened during the course. I am working and during free time i do the course and i am feeling wonderful .The way the course was conducted is ***awesome.*** And i think due to this initiative we are exposed to global standard in education. I don't worry about my result, instead i am happy i am in .. My hearty congrads to **Prof. Agarwal and his whole team** and including the members who gave tutorial video. It was **Fun and learn** simultaneously. With regards T V Deepan UserIdTAG: 296303 UserNameTAG: DEEPatXUniv CreateTimeTAG: 2012-12-24T14:42:14Z VoteTAG: 8 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 0 IndexTAG: 247 TitleTAG: Great thanks from Ukrainian students Thanks to EdX Team for this beatiful course. This is great idea to provide science for all people, who like to investigate such amazing phenomenon of nature, like electricity and electromagnetic waves. This is great opportunity to test our level of knowledge in this branch of physics, being at a distance of thousands kilometers from MIT. With thanks to EdX Team from students of Faculty of Physics and Technology, Donetsk National University. UserIdTAG: 251960 UserNameTAG: PopkovV CreateTimeTAG: 2012-12-24T08:02:17Z VoteTAG: 8 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I want to add to this post special thanks to Mr. Agarwal, Myrian, Mr.Sussman, Mr. Mitros. Merry Xmas and Happy New Year everybody! P.S. I am from Faculty of Physics and Technology, Donetsk National University too :) FirstChildUserIdTAG: 394607 FirstChildUserNameTAG: Sergei_Zolotar FirstChildCreateTimeTAG: 2012-12-24T08:29:01Z SecondChildTAG: Kharkiv national university of radioelectroniks too. SecondChildUserIdTAG: 464744 SecondChildUserNameTAG: attache SecondChildCreateTimeTAG: 2012-12-24T14:14:55Z IndexTAG: 248 TitleTAG: To Myrian, Agarwall and the MIT staff. I'm adding my voice to the the number of students to extend my appreciation for the brilliant initiative of offering free education online and for your efforts. Thank You very much. I also would like to ask whether we will continue to have access to the site and for how long. I am trying to avoid hours of downloading the material for offline peruse as there is no one downloading package available. I learned a lot yet not enough. I am interested in learning the subject and I believe I could profit from another iteration on 6002x next time it is offered. Would that be possible even if I take it for no certificate? Thanks a Lot and my Best Regards -walter UserIdTAG: 11075 UserNameTAG: roncada CreateTimeTAG: 2012-12-23T20:35:26Z VoteTAG: 8 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hi Walter :), you are welcome. Based on my experience - I took 6.002x 2012 Spring-, we still have access to the Course of last Term (Forum Discussion, Videos, Textbook, etc...) :p, we still debating things there haha. Yes, you can take next offer of 6.002x, even if you take it not for the Certificate. I know people from the last term, that are registered here, just for taking a look ;) My best wish to you, Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-23T20:48:12Z FirstChildTAG: I too thank Dr. Agarwall!!! This is a great concept and I hope the digital class will be coming soon! Also thanks to the Community TA's, I would not have made it without them. Keep up the good work! Thurston FirstChildUserIdTAG: 176173 FirstChildUserNameTAG: SDdad FirstChildCreateTimeTAG: 2012-12-23T21:55:28Z IndexTAG: 249 TitleTAG: Exam deadline UPDATE The exam will now close at December 24th at 5:00am EST, December 24th 10:00am GMT The deadline as displayed on the final itself will not change. UserIdTAG: 292543 UserNameTAG: Pennypacker CreateTimeTAG: 2012-12-23T18:09:01Z VoteTAG: 8 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Thanks, Pennypacker. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-12-23T18:19:42Z SecondChildTAG: December 23, 1:30PM EST RE: Exam Deadline UPDATE Thanks, **Lyla**, for looking into this matter for me. I brought it up because, *at first*, the blue alarm clock icon that showed the deadline said December 23rd, 11:59PM. Then, last night I noticed that it **changed**, and it said December 24th, 5:00AM (note: not 4:59AM). *Neither* Greenwich Mean Time (GMT or UTC) nor Eastern Standard Time (EST) is specified in the sidebar. *It would be helpful to specify this in the future, especially with a global student body.* However, there was no official announcement from Staff, or indication on the Final Exam itself, or anywhere else on the MITx site, that the time had changed. Initially, I thought this a glitch specific to my PC, but I confirmed with other students. In a Discussion post, I advised students to take the "new" end time with a note of caution, and as another poster said, EST = GMT - 5, which could have accounted for the discrepancy. As for myself, I decided to risk it, instead of starting the Final at 11:59PM yesterday, I took some much needed sleep after studying for the exam, and opened the exam fresh in the morning. I am still working on it, and also on the Discussion boards to fulfill my CTA duties. I most likely will not need the full 24 hours, but it's nice to have that option, as I can work at a slower pace, take lunch and coffee/tea breaks, attend to CTA duties and monitor the Discussion boards while working on the exam, etc. If the end time / deadline listed next to the blue alarm clock did not change in the middle of a final exam, this whole issue could have been avoided. Though I surely appreciate that the exam is **confirmed extended to 5:00AM EST, 10:00AM GMT/UTC**. Thanks, Jersey Mark SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-12-23T18:38:11Z IndexTAG: 250 TitleTAG: MUCHAS GRACIAS AL EQUIPO DEL M.I.T. Y A TODOS LOS PARTICIPANTES Estimados Amigos de curso, Les deseo una feliz navidad y un muy venturoso año 2013, en estos espacios se demuestra que la unidad por una causa supera las fronteras politicas e idiomáticas, soy ingeniero electronico y trabajo con equipos médicos, muchos saludos desde Colombia... Dear Friends of course, I wish you a Merry Christmas and a very happy 2013, in these spaces is shown that the unit for a cause beyond the political and language frontiers , I am an electrical engineer and work with medical equipment, many greetings from Colombia ... 물론 친애하는 친구, 내가 콜롬비아 메리 크리스마스와 아주 좋은 2013이 공간에 정치적 경계와 언어 이외의 원인에 대한 장치가, 나 의료 장비를 갖춘 전기 기술자 및 작업임을 표시되며, 많은 인사를합니다 ... पाठ्यक्रम के प्रिय मित्र, मैं तुम्हें एक मेरी क्रिसमस और एक बहुत खुश 2013, इन स्थानों में दिखाया गया है कि राजनीतिक सीमाओं और भाषा से परे एक कारण के लिए इकाई, मैं एक बिजली इंजीनियर और चिकित्सा उपकरणों के साथ काम कर रहा हूँ, कोलम्बिया से कई बधाई चाहते हैं ... Дорогие друзья, конечно, Я желаю вам счастливого Рождества и счастливого 2013 года, в этих пространствах показано, что блок причине вне политических границ и языка, я инженер-электрик и работы с медицинским оборудованием, многие привет из Колумбии ... UserIdTAG: 459662 UserNameTAG: FRANCISCOG CreateTimeTAG: 2012-12-23T01:25:15Z VoteTAG: 8 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Muy de acuerdo contigo Francisco (Fully agree with you) Igualmente una Feliz Navidad y un próspero 2013!! Sandra FirstChildUserIdTAG: 342966 FirstChildUserNameTAG: SandraNavarro FirstChildCreateTimeTAG: 2012-12-23T11:46:19Z FirstChildTAG: Igualmente compañero! Thanks to all the course staff and Merry Christmas! FirstChildUserIdTAG: 368526 FirstChildUserNameTAG: JavierVara FirstChildCreateTimeTAG: 2012-12-23T16:33:42Z FirstChildTAG: Muchas gracias compañeros por sus comentarios... FirstChildUserIdTAG: 459662 FirstChildUserNameTAG: FRANCISCOG FirstChildCreateTimeTAG: 2012-12-24T03:56:56Z IndexTAG: 251 TitleTAG: The last thing !! This course was the best, that I had take. I am very glad, thanks to professor Annant and all the staff, I learnt a lot. Thanks again. The last thing that I have to say is: Merry christmas and a happy new year, my best wishes to everyone from Ecuador. UserIdTAG: 289949 UserNameTAG: manolito CreateTimeTAG: 2012-12-22T03:34:34Z VoteTAG: 8 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Merry Christmas and Happy New Year to you too manolito!:) My best wish to you too! Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-22T19:52:48Z IndexTAG: 252 TitleTAG: TO STAFF-H12P2 EXPLANATION NOT SATISFACTORY! hi edx staff, we are almost there on the verge of completing this course! thanks a lot for your commendable efforts! I hope to score well in the finals! I would just like to draw your attention to H12P2 explanation in HOMEWORK 12. The explanation is not at all satisfactory. One can not guess how the equations mentioned came into existense? Moreover a lot has been going on in discussion for this thing. Can you please update the explanation and express how to solve such questions in detail?? That would be really helpful for other students as well! I know there's little time left for final but still I would appreciate if we get something more on this problem! Thanks a ton in advance! Hope you listen to my plea! p.s- I request everyone to vote for the post so that we may get a good explanation to this horrible qs.! UserIdTAG: 368712 UserNameTAG: jmen CreateTimeTAG: 2012-12-19T11:19:43Z VoteTAG: 8 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I think the question also needs to be clarified: The iD vs vD diagram clearly shows that v_z is negative. Then part a asks what happens when v_in < v_z. It took me a while to see that the problem assumes v_z is positive, and the mark on the diagram should be labeled -v_z as in Figure 15.70 on page 882 of the textbook. So part a is asking about what happens when the diode is off, and part b is asking about what happens when the diode is reversed biased. I see there is a whole thread on this: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50c66ea655e05c1f00000061 FirstChildUserIdTAG: 339870 FirstChildUserNameTAG: rjlasota FirstChildCreateTimeTAG: 2012-12-21T06:36:16Z IndexTAG: 253 TitleTAG: Final Exam: What tricks and suprises do you expect? I know that lot of students put in a lot of effort, time and energy into this course. A lot of "aha moments", and of course a lot of tricky homework assignments, and a lot of surprises where we actually learned about upcoming elements that were not studied yet, just by doing the homework assignment! So we are down to the final exam: I am a CTA but also a current student so I will be taking it along with you all; I'm looking forward to it, and seeing what **tricks** and **surprises** Prof. Agarwal and the Staff put in the exam :-) Can you predict what tricks and surprises the Professor will throw at us? Or maybe list some of the homework problems that you found the most difficult, surprising, and/or rewarding? Help me complete my list, and by doing so, it will serve as sort-of an "unofficial" review, and perhaps help fellow students prepare for the Final Exam! To get started: - Will we have a complex filter with multiple capacitors, inductors, and resistors both in series and in parallel that we have analyze? One where computing the transfer function and getting it into $ \frac{\omega^2}{s^2 + 2 \alpha s + \omega^2}$ canonical form is a nightmare? Sort of like *H11P1* but more difficult? - Will we have op-amp circuits with crazy-difficult feedback loops, op-amp circuits with dependent current sources and nonlinear elements such as diodes wrecking havoc with your node analysis? Sort of like *H12P2*? Or maybe analysis of "real-world" op-amps where instead of $i^+=0$, $i^-=0$, and $v^+=v^-$ (the "virtual short") there is actual current leakage, and input impedance is actually finite? - Will we have to analyze circuits with "made-up" components like the *expoDweeb*, or small signal analysis of "new" MOSFETS where $ i_{DS} $ is **not** = $\frac{K}{2}(V_{GS}-V_T)^2$, but instead = $\frac{K}{2}(V_{GS})^3-V_T$ (for example) or some other non-traditional (Like *H6P1* and the *newFET*) value just to give our brains exercise? Of course, I don't know what the exam will actually contain, but if we, as students, list some of the more "trickier" problems, we can get a heads-up on this Final and hopefully we all get 100% :-) Jersey Mark UserIdTAG: 292546 UserNameTAG: JerseyMark CreateTimeTAG: 2012-12-13T20:47:04Z VoteTAG: 8 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 10 FirstChildTAG: H12P2 isnt hard for my opinion, but I had troubles with H8P1.. Tricky NewFETs ..yeah, it is require some time to problem understanding.. I think it will be hard to get statistical data.Each student will have own better and worst understanding parts of course.I do really worrying about final exam. It was pretty hard to read text book while the English isnt my native language. So Im afraid that I lost something very important... Good luck to all! FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-12-13T21:37:50Z SecondChildTAG: I believe you do quite well with English in your posts. I suspect (though of course I can't know) that your reading comprehension is better than your conversational English (jargon, idioms, etc. often throw people). In any case, try not to worry too much. :-) SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-12-13T22:22:50Z FirstChildTAG: just keep it simple ! - if you understand the principles teached , it will be very simple - if they give us a complex circuit and you messe up with, you just need to returne to the source (principles) - prof. Agarwal gave us a lot of trick that can help ,for example you can analyse the circuit by intuition and then do the calculation - ...... FirstChildUserIdTAG: 236136 FirstChildUserNameTAG: Hamid-ch FirstChildCreateTimeTAG: 2012-12-13T22:04:41Z FirstChildTAG: Take my word for it, it **will not** be very simple. It will be very, very hard. Expect a real MIT level exam. FirstChildUserIdTAG: 9673 FirstChildUserNameTAG: xp42 FirstChildCreateTimeTAG: 2012-12-14T04:39:45Z SecondChildTAG: I do expect this.And it is scary to me.. SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-14T06:38:48Z SecondChildTAG: Indeed. SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-12-14T07:45:38Z SecondChildTAG: I know that **analog-to-digital converters** can be implemented with op-amps. We never studied A/D converters in our university's equivalent to EECS 6.002x (Univ. of Connecticut's ECE 201). But we did study op-amps. So the Professor there threw in design of an A/D converter in on the final; his idea of a final was one where we learned something even there! I suspect Prof. Agarwal might try something like this. Right now I am scouring the web to look at all types of circuits using op-amps, inductors, capacitors, etc. to achieve a useful result. One practical circuit that often finds its way into an EECS curriculum is a first-order circuit that charges slowly and discharges suddenly; this has the effect to boost voltage or current momentarily. Examples of this are simplified **automotive ignition systems** (using a coil, basically a transformer, which can be modeled with inductors) to raise voltage enough to break the dielectric constant of air and allow a spark to bridge a gap and allow electric current to flow for a few milliseconds, or it's capacitive equivalent, the **electronic flash unit in cameras**, where capacitive discharge is necessary to activate the high-voltage flash tube. These are simple circuits, but often find their way into homeworks and exams. It's a good idea to review the analysis and design of such circuits as they may appear on the exam. If I think of any others, I'll list them. Jersey Mark SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-12-14T21:52:45Z SecondChildTAG: for DAC, go for R-2R ladder and binary weighted resistors, and for ADC go for SAR, dual slope integrating type,counter type and flash type. no need to search on net for other things !! SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-12-19T16:55:12Z SecondChildTAG: **jmen**: Thanks! R-2R ladder networks show up a lot on op-amp exam problems. SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-12-19T18:01:48Z FirstChildTAG: I studied this course at spring, and I had not enough time on it. Final exam was really difficult, because I didn't study weeks 10-14. So, I think, all of us **must** prepare attentively! Good luck, students from whole world. FirstChildUserIdTAG: 86420 FirstChildUserNameTAG: Aleksei_Katkov FirstChildCreateTimeTAG: 2012-12-14T06:12:20Z SecondChildTAG: Alex, but the final exam will cover weeks 1-12, isnt it? SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-14T06:40:18Z SecondChildTAG: It covered weeks 1-13 in the previous 6002x. SecondChildUserIdTAG: 90796 SecondChildUserNameTAG: ChaunceyGardiner SecondChildCreateTimeTAG: 2012-12-15T02:28:47Z FirstChildTAG: Are you aware of the review packet (and solutions) that they announced on the [course info tab][1] 10 days ago ? It gives you an idea of what to expect. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/info FirstChildUserIdTAG: 90796 FirstChildUserNameTAG: ChaunceyGardiner FirstChildCreateTimeTAG: 2012-12-14T06:22:57Z FirstChildTAG: If I understood correctly , last final exam have had 10 tasks, 47 questions total. FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-12-15T11:56:44Z SecondChildTAG: That is correct. SecondChildUserIdTAG: 90796 SecondChildUserNameTAG: ChaunceyGardiner SecondChildCreateTimeTAG: 2012-12-16T06:28:04Z FirstChildTAG: Why do you all think it will be more difficult than the mid-term? FirstChildUserIdTAG: 398594 FirstChildUserNameTAG: DaveyJC FirstChildCreateTimeTAG: 2012-12-14T12:56:52Z SecondChildTAG: That is a reasonable expectation, although it may depend on your skills, talent and background. I finished 6002x this spring with a total score of 97.4% What kept me from getting 100% was two missing ticks on the final. However, I know that many people got a better score than I did, so it is certainly possible. SecondChildUserIdTAG: 90796 SecondChildUserNameTAG: ChaunceyGardiner SecondChildCreateTimeTAG: 2012-12-15T02:35:38Z SecondChildTAG: It seemed like everyone and their dog managed 100% on the midterm. Approximately 350 people achieved the same on the final. SecondChildUserIdTAG: 9673 SecondChildUserNameTAG: xp42 SecondChildCreateTimeTAG: 2012-12-15T03:34:55Z SecondChildTAG: The finals was way harder than the midterms last time. I finished the midterms in about 3 hours. I needed the whole of 24 hours to get the finals right. A couple of them were right on the third attempt so I would say there was some luck involved in my case. I wouldn't have done well with just one attempt. SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-12-15T14:10:31Z FirstChildTAG: the labs were more simple and more helpful FirstChildUserIdTAG: 323963 FirstChildUserNameTAG: gtrf FirstChildCreateTimeTAG: 2012-12-14T17:07:46Z FirstChildTAG: Hi Guys! Did you see through again our homeworks and labs after publishing final review? And did you noticed that some tasks were taken from this exams? There is one of the tricks ;-) And about me, I really don`t want to see in final exam tasks with step impulse source. FirstChildUserIdTAG: 96873 FirstChildUserNameTAG: Santyaga FirstChildCreateTimeTAG: 2012-12-14T17:56:57Z SecondChildTAG: Ough, camrad..Im entirely agree with you about impulses :) SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-14T19:43:01Z SecondChildTAG: You should count on there being at least one such problem. SecondChildUserIdTAG: 9673 SecondChildUserNameTAG: xp42 SecondChildCreateTimeTAG: 2012-12-15T03:47:11Z SecondChildTAG: I am understanding this xp42. But there is nothing left except to wait and train on those examples which are in stock.Also can you tell sources where are examples of problems with step impulse source? SecondChildUserIdTAG: 96873 SecondChildUserNameTAG: Santyaga SecondChildCreateTimeTAG: 2012-12-15T14:32:26Z SecondChildTAG: Just make sure you understand the concept thoroughly and should be fine. Take the basic circuits (RL, RC) and re-derive the formulas, which professor Agarwal taught, yourself. The review packet and the textbook has a good selection of problems which I feel is more than enough. SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-12-15T14:56:40Z SecondChildTAG: Make up your own. That's what I do. Or, look in the textbook. SecondChildUserIdTAG: 9673 SecondChildUserNameTAG: xp42 SecondChildCreateTimeTAG: 2012-12-16T06:01:53Z FirstChildTAG: The [Vacuum Triode][1] scares me. If someone could check my reasoning [here][2], I would greatly appreciate it! Thanks in advance! [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_6/Week_6_Tutorials/ [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50cf74812b25f1230000001e FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-12-19T07:11:31Z SecondChildTAG: I remember coming across the vacuum triode. Using it in a circuit would be a challenging exam problem, though the final will likely use a device with slightly different parameters. You must be able to do DC and small-signal analysis for **any** $i_{DS}$ equation (not just the standard MOSFET one), and perhaps even be able to come up (derive) an $i_{DS}$ equation for non-standard devices similar to the vacuum triode tutorial that you reference. If the exam **does** include a plain, *standard* MOSFET, then there will be a trick thrown in somewhere else in the circuit, such as a circuit using multiple standard MOSFETs, or in non-standard amplification biasing. Remember to study up on the MOSFET connected in "diode configuration", as shown in [this MIT .pdf.][1] I have often seen these on exams in advanced courses, though not at the MIT-level. [1]: http://web.mit.edu/6.012/www/SP07-L25.pdf SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-12-19T18:16:04Z SecondChildTAG: Thanks, [JerseyMark][1]! [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/users/292546 SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-12-20T06:24:10Z IndexTAG: 254 TitleTAG: Final Exam Hi, I have a question regarding the final. Is the final exam cumulative? Thanks a lot. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-12-05T17:16:22Z VoteTAG: 8 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Why not weeks 13-14 too? FirstChildUserIdTAG: 398594 FirstChildUserNameTAG: DaveyJC FirstChildCreateTimeTAG: 2012-12-07T08:31:21Z SecondChildTAG: that's because the concepts taught there are more useful in researches and doesn't suit the syllabus for evaluation.. SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-12-07T13:24:35Z FirstChildTAG: The Final exam will cover topics from Weeks 1 - 12. FirstChildUserIdTAG: 391929 FirstChildUserNameTAG: RohanNagarkar FirstChildCreateTimeTAG: 2012-12-05T22:03:11Z SecondChildTAG: That's right SecondChildUserIdTAG: 60453 SecondChildUserNameTAG: TommyNittin SecondChildCreateTimeTAG: 2012-12-10T09:38:10Z SecondChildTAG: It was 1-13 last time. Has the staff decided to exclude Week 13 this time? SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-12-11T19:58:20Z IndexTAG: 255 TitleTAG: Lab10 Hints Requested by brainyash1990 [Read at the end for the Spanish Hints :)] **Lab10 Hints** In this lab we'll look at the response of first-order circuits as a function of frequency, the topic covered in Chapter 13 in the text. ![im][1] In this part, they give us a Resistive - Capacitive Circuit (RC) and other Resistive-Inductive Circuit (RL). The requested statement says that we have, as a data, the value of the resistor R; however, we do not know the values of L and C: - R=1kOhm - L=? - C=? Next, the problem says that we have two given expressions of H, once for the Resistive-Capacitive Circuit (RC) and the other for the Resistive-Inductive Circuit (RL): *Formula 1:* ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13372947795443318.png) To review this concept, you should read the example in the page 737 of the Textbook [read here][2] But, for those who want to know quicly what means H, it is quite simple: The H, is named as the transfer function. The transfer function, studies the behavior of a quadrupole (a quadrupole is a circuit that is analyzed as if it were inside a box and you anly have 4 terminals that you can see outside the box, that is to say, 2 terminals of the input and 2 terminals of the output). If you measure in the input terminals a voltage Vi and in the output terminals a voltage Vo, the transfer fuction is the relationship Vo/Vi. ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13372949874522077.png) Ok, they also give as the phase. A phase, is the arctan of the division of the imaginary part with the real part. ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13372950785645344.png) ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13372954102949819.png) If you are not dizzy yet haha (I hope that not ), let's go to the part 1 and 2: In the part 1, they ask us to calculate the break frecuency. First of all, we have to ask ourselves what is the break frecuency? In the page 738 of the textbook explains about this concept [read here][3] But, for those who want to know quicly what is the break frecuency, here it is: The break frecuency is, by definition, the frecuency, call it for eg. wc, where the transfer function decreases -3 decibels. H(wc)db=-3 decibels So, the question is, when does it down to -3 decibels? what does it means? Knowing that (remember that you need the value of H in other units than decibels, so you have to convert it - next is how you get the value in times (not in secons in quantity)) ![image description](https://mitx_askbot_stage.s3.amazonaws.com/133729556634612.png) So, with this simple concept, you can have an idea to solve the part 1 and 2. I will not continue with this because this will be a step by step solving and it is not allowed, so I let you this to you. **Part 1:** Having H(w) for the circuit RC, wich is the break frecuency? Hint: To wich value you have to compare the HRC(w) formula that they give us in problem to obtain the break frecuency? Isn't it 1/(√2 ) ? **Part 2:** Having H(w) for the circuit RL, wich is the break frecuency? Hint: To wich value you have to compare the HRC(w) formula that they give us in problem to obtain the break frecuency? Isn't it 1/(√2 ) ? Next, we will have 2 plots. One plot of magnitude and the other of phase. Here we have to pay attention (it took me a while to reach to that): *BE CAREFUL with the frecuencies given in the X- axis, they are in log!!! WARNING!!!* ![image description](https://mitx_askbot_stage.s3.amazonaws.com/1337295698625937.png) So, lets see the violet rect . For the value 8, means: ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13372957693275432.png) Log(f)=8 f=10^8 Hz The problem says: Your job is to figure out the unknown C and L values based on the information in the plot. Remember that the R value for both circuits is 1kΩ. You can perform your own experiments using the on-line lab tool or you can use the analysis above to determine the unknown component types and values. Note that the formulas for the phase use the frequency ω in radians/sec but that the plots show frequencies in Hz. Recall that 1 Hz=2π radians/sec. Part 3: See the red curve, what is that component that produces that type of curve? L or C ? Hint: Read the page 737 of the Textbook and compare it with the curves you have [read here][4]. **Part 4:** If you know what type of circuit produces that red curve (RL or RC), and knowing the transfer fuction given (HRL(w), HRC(w)) and having the data of the amplitude that is related to a frecuency that is given - see the dashed line and the related value of amplitude in the top of the curve - f = value (remember that in the plot that frecuency is given by log (f), that means if you see in the x-axis something like 7, that means that the frecuency it is not 7 hZ, it is 10 ^7 Hz ). If we know R, H (wvalue), wvalue=2pi*f, wich is the value of L or C? **Part 5:** See the blue curve, what is that component that produces that type of curve? L or C ? Hint: Read the page 737 of the Textbook and compare it with the curves you have.[read here][4] **Part 6:** If you know what type of circuit produces that blue curve (RL or RC), and knowing the transfer fuction given (HRL(w), HRC(w)) and having the data of the amplitude that is related to a frecuency that is given - see the dashed line and the related value of amplitude in the top of the curve - f = value (remember that in the plot that frecuency is given by log (f), that means if you see in the x-axis something like 7, that means that the frecuency it is not 7 hZ, it is 10 ^7 Hz ). If we know R, H (wvalue), wvalue=2pi*f, wich is the value of L or C? Be careful with the log frecuency... Finally, they ask us to design a Band-pass filter. The statement give us a clue that this band-pass filter is the result of two filters in cascade (one next to the other filter). Also, they say that we have previously analyze them both! Hint: This is a Test and error method. Such us we have seen before (red and blue curves of the part 3,4,5,6), we can find the values of L and C for a first aproximation (do not mistake while you are replacing in the formula, once belong to the rise and the other to the drop), choosing a value of R1=R2=value for a first aproximation. Once we have that values, click on TRAN and then, correct the value of one of the resistences to get the correct the final value. I can not tell you more. Now it is your turn! Design your own filter! I hope this can help you. Myriam. ---------- Now in Spanish! **Lab10 Hints** En esta parte, se nos da un circuito Resistivo Capacitivo (RC) y otro circuito Resistivo Inductivo (RL). El enunciado nos dice que tenemos como dato el valor de la resistencia R; sin embargo, no se conocen ni L ni C: ![im][1] - R=1kOhm - L=? - C=? A continuación, el mismo enunciado del problema nos arroja dos expresiones de H, tanto para el circuito Resistivo Capacitivo (RC) como así también para el circuito Resistivo Inductivo (RL). Fórmula 1: ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13372947795443318.png) Para repasar este concepto, es recomendable leer el ejemplo de la página 737 del Textbook[leer aquí][2] Pero para aquellos que quieren saber rápidamente qué significa H, es bastante sencillo: **H**, se denomina **función de transferencia**. La función transferencia, tiene por objeto el estudio del comportamiento de un **cuadripolo** (se entiende como cuadripolo a un circuito como si fuera analizado como si estuviera dentro de una caja y sólo se tuviera 4 terminales para analizar su comportamiento, es decir, dos terminales de entrada y dos terminales de salida). Si en los terminales de **entrada** se mide una tensión Vi y en los terminales de **salida** se mide una tensión Vo, la función transferencia es la relación **Vo/Vi**. También, se nos da la fase. Por fase, se entiende al arco tangente del cociente de la parte imaginaria respecto de la real. Fórmula 2: ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13372950785645344.png) ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13372954102949819.png) Bien, si no los he mareado hasta ahora, vayamos a la parte 1 y 2 : En la parte 1, nos solicitan calcular la **frecuencia de corte o break frecuency**. Primero que todo, debemos preguntarnos qué es la frecuencia de corte? En la página 738 se explica el concepto de break frecuency [leer aquí][3] Pero para aquellos que quieran saber rápidamente qué es la frecuencia de corte, se los explicaré a continuación: La frecuencia de corte, por definición, es aquella frecuencia, llamémosla wc, para la cual, la función de transferencia cae a -3 decibeles. H(wc)db=-3 decibeles Y cuándo cae a -3 decibeles? Sabiendo que: ![image description](https://mitx_askbot_stage.s3.amazonaws.com/133729556634612.png) Entonces, con este simple concepto, se puede tener una idea para resolver la parte 1 y 2. **Esto se los dejo a ustedes, sino estaría resolviendo paso a paso el ejercicio lo cual, no se puede.** **Parte 1:** Teniendo H(w) para el circuito RC, cuál es la frecuencia de corte? Pista: a qué valor tengo que igualar la HRC(w) dada en el enunciado para obtener la frecuencia de corte? No era igualarla a 1/(√2 ) ? **Parte 2:** Teniendo H(w) para el circuito RL, cuál es la frecuencia de corte? Pista: a qué valor tengo que igualar la HRL(w) dada en el enunciado para obtener la frecuencia de corte? No era igualarla a 1/(√2 ) ? A continuación de estas dos preguntas, nos dan dos gráficas. Una de magnitud y otra de fase. Aquí hay que prestar mucha atención a la siguiente aclaratoria (que me llevo mucho tiempo en darme cuenta): ![image description](https://mitx_askbot_stage.s3.amazonaws.com/1337295698625937.png) Veamos la **recta color violeta**. Para el valor **8**, quiere decir que : ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13372957693275432.png) Log(f)=8 f=10^8 Hz El enunciado dice: Your job is to figure out the unknown C and L values based on the information in the plot. Remember that the R value for both circuits is 1kΩ. You can perform your own experiments using the on-line lab tool or you can use the analysis above to determine the unknown component types and values. Note that the formulas for the phase use the frequency ω in radians/sec but that the plots show frequencies in Hz. Recall that 1 Hz=2π radians/sec **Parte 3:** Qué componente produce la curva roja? L o C ? pista: leer la página 737 del Textbook y comparar las gráficas para un RC y un RL y compararla con la figura que se nos da en el enunciado. **Parte 4:** Si se sabe qué tipo de circuito es (RL ó RC), teniendo el dato de la función transferencia (HRL(w), HRC(w)) y teniendo el dato del valor de la amplitud para dicha frecuencia f=valor (recordar que el valor del grafico esta en forma logarítmica- convertir. Ver línea de puntos y valor correspondiente en lado superior izquierdo). Si sabemos R, H(wvalor), wvalor=2.pi.f, Cuál es el valor de L o C? **Parte5:** Qué componente produce la curva azul? L o C ? pista: leer la página 737 del Textbook y comparar las gráficas para un RC y un RL y compararla con la figura que se nos da en el enunciado. **Parte 6:** Si se sabe qué tipo de circuito es (RL ó RC), teniendo el dato de la función transferencia (HRL(w), HRC(w)) y teniendo el dato del valor de la amplitud para dicha frecuencia f=valor (recordar que el valor del grafico esta en forma logarítmica- convertir. Ver línea de puntos y valor correspondiente en lado superior izquierdo). Si sabemos R, H(wvalor), wvalor=2.pi.f, Cuál es el valor de L o C? **Parte 7:** Finalmente, nos piden un diseño de un Filtro pasa banda. El mismo enunciado, nos da la pista que un filtro pasabanda es en sí, el resultado de dos filtros en cascada (uno al lado del otro). Nos dicen además que hemos analizado los dos! La idea general, es tratar de obtener la función de transferencia de la figura que nos dan. De inmediato, se pueden observar dos quiebres de la curva. Pista: Método de Prueba y error. Tal y como vimos anteriormente (curvas roja y azul de la parte 3,4,5,6), podemos hallar los valores de L y C como primera aproximación (no equivocarse al reemplazar en la fórmula, una corresponde a la subida y otra a la bajada de la curva), eligiendo un R1=R2=valor. Una vez que obtenemos dichos valores hacemos TRAN y luego, corregimos el valor de una de las resistencias para obtener el valor correcto. Más no puedo decirles. Ahora es su turno! A diseñar! Espero que les haya servido de ayuda. Myriam. [1]: https://www.edx.org/static/content-mit-6002x/images/circuits/Lab10_1.34655440d510.png [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/761 [3]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/762 [4]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/761 UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2012-11-24T18:35:03Z VoteTAG: 8 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: Having slight problem with reading this, some of the English section is also in Spanish...... FirstChildUserIdTAG: 375500 FirstChildUserNameTAG: Brej7665 FirstChildCreateTimeTAG: 2012-11-25T22:28:30Z FirstChildTAG: Thank you very much. I think that we must look in last video to solve the last question. Recall that in the band pass filters, we have two relations for magnitude w*R*C and R/(w*L). Build your circuit and done! I expect to have helped. FirstChildUserIdTAG: 366669 FirstChildUserNameTAG: pedroramus FirstChildCreateTimeTAG: 2012-11-25T02:26:56Z FirstChildTAG: Hi, Myriam! Thank you for all your hints! With all 6 parts it`s all clear, but with last part I have some confusion. We have series connection RL and RC filters as I understand from your hints. Now main theme choose values of elements. At first choose equal resistance. What next step in this part? Do we must make some equations or what? FirstChildUserIdTAG: 96873 FirstChildUserNameTAG: Santyaga FirstChildCreateTimeTAG: 2012-11-24T20:08:42Z SecondChildTAG: Hi Santyaga, see if my comment from this thread helps: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/undefined/threads/50b04b65e6f920270000003f SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-24T21:02:27Z SecondChildTAG: Let's say you have the RL first. That will be the high pass part of the filter, so you calculate L from R and the lower of the two break frequencies. Now go the the RC section. This will be the low pass part of the filter. The trick here is to realize that there are two resistors between the source and the capacitor (the inductor is effectively open circuit at high frequencies), so if the resistors are the same, you use 2*R when calculating the required value of C for the higher of the two break frequencies. There's no need for trial and error here, you can calculate the L and C values exactly once you've picked a resistor value. You'll probably have to enter the L and C values to 3 or 4 significant digits to get the green check. I know 3 worked for me. SecondChildUserIdTAG: 141000 SecondChildUserNameTAG: OrinE SecondChildCreateTimeTAG: 2012-11-24T21:17:24Z SecondChildTAG: Hi matiasgrodriguez! Unfortunately can`t open this link. It tells me page not found. SecondChildUserIdTAG: 96873 SecondChildUserNameTAG: Santyaga SecondChildCreateTimeTAG: 2012-11-25T09:46:19Z SecondChildTAG: Hmm... Maybe better try step by step. At first convertion frequency as 10^(break freq)*2*pi. next step for value of L element - *w*= L/R. So we have R, we have *w* and now after convertation we have L. For capasitor according previos post I took double R and used it to find C from equation for break frequency for RC circuit. Now substituting calculated values something wrong and red cross. Again qestion - where is mistake? SecondChildUserIdTAG: 96873 SecondChildUserNameTAG: Santyaga SecondChildCreateTimeTAG: 2012-11-25T10:50:20Z SecondChildTAG: Hi Santyaga, it looks like the page was removed :( Give me some time to post the hint again.. SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-25T13:02:02Z SecondChildTAG: I calculated the L and R values and put them right in to the circuit without tweaking. I did not use the angle formula. I used the magnitude formula. They are easy to calculate. If you did H10P1, you should have derived the magnitude formulas for circuits A and C. $|\frac{Vo}{Vi}|=FormulaInTermOf:w,L,R,$ $|\frac{Vo}{Vi}|=FormulaInTermOf:w,C,R,$ Look the given magnitude plot and substitute for the inductor and capacitor $-3dB=w:1KHz..L...R...$ and $-3dB=w:1MHz..C...R...$ Please remember to convert from decibels to Volts and from Hz to rad/sec Circuit layout. It is basically the hint of the problem but you only need one resistor (because we use the same resistor value for the two formulas) R+(L||C)... I'm using the same value as the Figure 1. I hope this helps. SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-25T13:42:28Z SecondChildTAG: Ok give me some time to try this. I hope it will be work. SecondChildUserIdTAG: 96873 SecondChildUserNameTAG: Santyaga SecondChildCreateTimeTAG: 2012-11-25T15:15:45Z SecondChildTAG: Also do we need to use value of magnitude from figure2? SecondChildUserIdTAG: 96873 SecondChildUserNameTAG: Santyaga SecondChildCreateTimeTAG: 2012-11-25T15:25:44Z SecondChildTAG: For the last part you only need to look figure 3 and the text above that (where they give you the problem). If something is not clear tell me so I can try to explain better. SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-25T16:03:32Z SecondChildTAG: Santyaga, sorry man. My hint was to the last part... but you can solve part 5 and 6 in the same way as I explained... then you only need to look the magnitude plot for fthe Image 2. Remember that the magnitude is in decibels and the freq is in log Hz. SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-25T16:26:59Z SecondChildTAG: Maybe this thread could also help https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50b240ca3930d82b00000025 SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-25T20:16:23Z SecondChildTAG: i did it !!! SecondChildUserIdTAG: 483548 SecondChildUserNameTAG: SAM_1993 SecondChildCreateTimeTAG: 2012-11-26T00:48:09Z SecondChildTAG: I'm glad to hear that SAM_1993 SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-26T01:53:03Z FirstChildTAG: MUCHAS GRACIAS MYRIMIT!!! FirstChildUserIdTAG: 113441 FirstChildUserNameTAG: De_Bonis FirstChildCreateTimeTAG: 2012-11-26T05:10:08Z SecondChildTAG: Aunque para la parte final, yo hubiera sacado la magnitud de la función de transferencia |H(jw)|, así que: |H(jw)|=10^-3. Ocupando superposición, se podía tomar el circuito RC y ocuparlo como pasabajos y hallar el valor de C (con un R arbitrario, supongamos R=1k) despejando la ecuación |H(j*10^6*2*pi)|=10^-3 (10^6 es la frecuencia de corte del pasa bajos). Análogamente, se hace con L, esta vez con la frecuencia de corte del pasa altos. Luego se ponen en cascada :P SecondChildUserIdTAG: 113441 SecondChildUserNameTAG: De_Bonis SecondChildCreateTimeTAG: 2012-11-26T05:14:28Z IndexTAG: 256 TitleTAG: H11P3 Last two parts (Checker accuracy?) Something curious concerning the the last two parts of H11P3. I used the information from the impedance magnitude plot, together with the impedance equation for the circuit, after understanding which element has been removed (I am not saying which one or giving any specific information as this homework is still due). I got an answer of the form, say, for example 10.000005 (pF or mH) and it was checked repeatedly as wrong. The correct answer was 10 pF or mH. Note that my first answer was the "exact" solution from the equation without any rounding of the parameters. Now, I am having a problem getting the check mark for the last part. I have double checked my algebra and all looks correct. I am wondering if it is again a similar issue as with the previous part. If there is something wrong with the method described above please do not give any info concerning a correct method. Anyone else has a similar problem? UserIdTAG: 97581 UserNameTAG: Asimakis CreateTimeTAG: 2012-11-21T14:44:55Z VoteTAG: 8 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: My answers are nice integers. Never thought to try anything else. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-21T14:52:25Z SecondChildTAG: Fro the final answer I was also getting the red X, but then, based on your comment, I rounded to the nearest integer (approximately 0.53% rounding) and I got the green check. Also, for the next to last answer I was also getting the red X until I rounded (~0.51%) to an integer and got the green check. SecondChildUserIdTAG: 428560 SecondChildUserNameTAG: MikeDayton SecondChildCreateTimeTAG: 2012-11-24T02:05:43Z FirstChildTAG: I had the same problem - I calculated the answer to be (say) 1.01 and it was marked wrong, but when I rounded to 1 I got a check mark. (Obviously 1.0 worked as well!). FirstChildUserIdTAG: 145194 FirstChildUserNameTAG: DerekH FirstChildCreateTimeTAG: 2012-11-21T22:25:06Z SecondChildTAG: same for me. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-11-22T10:57:35Z FirstChildTAG: I have also the same problem. My answer is 1.00503781526, but only 1 works. FirstChildUserIdTAG: 230701 FirstChildUserNameTAG: MEng_IIT FirstChildCreateTimeTAG: 2012-11-24T22:03:10Z FirstChildTAG: I have the same question,there must be something wrong here or something was neglected. FirstChildUserIdTAG: 361137 FirstChildUserNameTAG: Ericson FirstChildCreateTimeTAG: 2012-11-30T06:11:21Z FirstChildTAG: i went crazy over this and plugged math to it and was crazy getting it right and no green check, so based on this comment i rounded up and cutted off from the 3rd decimal, hence the accurate number of 1.0050378152592121*10^-9 was incorrect but 1.0 was correct. I spent like 2+ hours on this alone /sadface. I hardly could afford spending that time on this. STAFF, please fix this, ty. FirstChildUserIdTAG: 331483 FirstChildUserNameTAG: Doru FirstChildCreateTimeTAG: 2012-12-02T15:34:36Z IndexTAG: 257 TitleTAG: Funny I think we need to use superposition... ![enter image description here][1] I'd like to see the differential equation of a lumped electric eel. [1]: https://edxuploads.s3.amazonaws.com/1353493896134362.png UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-11-21T10:31:49Z VoteTAG: 8 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Classic. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-11-21T12:47:01Z SecondChildTAG: whats with the bat? SecondChildUserIdTAG: 357747 SecondChildUserNameTAG: kishores SecondChildCreateTimeTAG: 2012-11-21T13:58:46Z SecondChildTAG: Ballast SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-21T21:00:36Z FirstChildTAG: lol FirstChildUserIdTAG: 219780 FirstChildUserNameTAG: ReconIII FirstChildCreateTimeTAG: 2012-11-21T15:18:46Z FirstChildTAG: Holy water does that works. FirstChildUserIdTAG: 39980 FirstChildUserNameTAG: MarinoJr FirstChildCreateTimeTAG: 2012-11-21T16:31:55Z SecondChildTAG: Dunno, but you gotta pull that wire tight for it to work. SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-21T16:50:14Z SecondChildTAG: If it works, it would be a really big AHA! moment! ;) SecondChildUserIdTAG: 232157 SecondChildUserNameTAG: GayatriTR SecondChildCreateTimeTAG: 2012-11-21T17:05:42Z SecondChildTAG: hikz hikz.. SecondChildUserIdTAG: 509517 SecondChildUserNameTAG: randima SecondChildCreateTimeTAG: 2012-11-22T05:17:41Z IndexTAG: 258 TitleTAG: MOSFET or Op Amp? The speaker talks all the time about the MOSFET, but... should it not be rather the Op Amp? UserIdTAG: 341020 UserNameTAG: franjescribano CreateTimeTAG: 2012-11-19T19:26:09Z VoteTAG: 8 CoursewareTAG: Week 12 / S23V5 Op Amp characteristics CommentableIdTAG: 6002x_S23V5_Op_Amp_characteristics NumberOfReplyTAG: 1 FirstChildTAG: Thank you for catching this, there are indeed a few times where the word "MOSFET" was used instead of "Op Amp." We're working on fixing the transcript. - Rohan FirstChildUserIdTAG: 391929 FirstChildUserNameTAG: RohanNagarkar FirstChildCreateTimeTAG: 2012-11-19T20:11:17Z IndexTAG: 259 TitleTAG: edX is a FANTASTIC Learning Platform I couldn't help saying it. It was so helpful keeping my mind at the material. Thanks Dave and Lyla and all of edX. Thanks to all of MITx/6.002x for bringing us this course through this wonderful learning platform. Thanks to slashdot and the anonymous poster who helped me to discover EDx. Thank you so very much. UserIdTAG: 623210 UserNameTAG: preveen CreateTimeTAG: 2012-10-31T07:55:10Z VoteTAG: 8 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 260 TitleTAG: Hurricane Sandy Do you guys on the East Coast think the course will be disrupted by Hurricane Sandy? I hope you guys are all safe and sound during the storm. UserIdTAG: 21541 UserNameTAG: JSChambers CreateTimeTAG: 2012-10-29T03:47:04Z VoteTAG: 8 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: **Some thoughts about the storm: 4am EDT** I am in New Jersey, which, as of the 2:00AM Eastern Daylight Time *New York Times* hurricane forecast model, is right in the path of the storm's landfall. The model shows landfall at it's southern extreme in Delaware, and at it's northern extreme New York City's Lower Harbor. Winds have been blowing pretty steady all night here, but no rain yet (I expect it by 6:00AM local time). I'm in the 2nd floor of a brick building near New York's Lower Harbor, so I'll be OK. Flooding and power loss will be the major problems. The intensity of this storm is only Category 1 (75 mph or 120 km/hr winds maximum) so it is unlikely that we'll have any major structural damage (or even minor damage beyond downed power lines and maybe some old trees). *You guys in East and South-east Asia see typhoons way stronger than we get; plus the more agricultural-oriented use of the land on the coasts there promotes mudslides and other complications that we do not get here.* However, all rail and air transport in and out of NYC has been suspended for the duration of the storm, and looking outside the streets are deserted. I have to go out to the store at 7:30AM; I have gas heating so cooking and heat will not be a problem, and I bought food to last a few days; my biggest concern is anyone getting hurt (**my prayers go out for a mild, safe storm**), after that it's the lack of communication / news / edX if the internet / power goes down locally (...if I only had battery backup for my cable modem, and the old battery on my laptop is only good for an hour.) Jersey Mark FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-10-29T08:23:52Z SecondChildTAG: Whatever I see on the news looks really scary. I hope no one gets harmed. It's quite bad in the Caribbean. SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-10-29T16:15:19Z FirstChildTAG: We certainly hope not, and have no plans to interrupt anything. However, the entire eastern seaboard is supposed to be hit pretty hard, and we don't know how prepared many of the "cloud" data centers are for the storm. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-10-29T04:49:25Z FirstChildTAG: Here in midtown Manhattan the winds are starting to increase (it's almost 9AM local time). A problem in the city is the "canyon effect," the winds channeling through the spaces between buildings amplifies wind speeds and can cause windows to be sucked out of buildings (the higher the building, the more intense the effect). We're hoping for the best, preparing for the worst. FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-10-29T12:58:21Z FirstChildTAG: Be careful with all the hurricanes......... FirstChildUserIdTAG: 150120 FirstChildUserNameTAG: krishna1993 FirstChildCreateTimeTAG: 2012-10-30T07:11:25Z IndexTAG: 261 TitleTAG: where is the mid-term paper?? when will be the question paper for mid term will available?? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-25T04:41:40Z VoteTAG: 8 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 262 TitleTAG: What is the Value of the Certificate? I am wondering, assuming we finish the course, what is the **real world value of a MITx 6002x Circuits and Electronics Certificate**? UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-10-21T19:02:00Z VoteTAG: 8 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 6 FirstChildTAG: Right now its an honor code certificate so its not going to count for credits anywhere but I think it would look good on your resume/application. `If students were taking classes that were “MIT or Harvard hard,” Agarwal says he’d pay attention, later admitting, “Would I blindly hire a person? No. But I would conduct the rest of the interview.”` Thats what Prof. Anant Agarwal says about the importance of an edx certificate in an excerpt from an online interview. In the future they're planning on having proctored exams so then it will probably be worth a lot more. [https://www.edx.org/press/edX-announces-proctored-exam-testing][1] [1]: https://www.edx.org/press/edX-announces-proctored-exam-testing FirstChildUserIdTAG: 154440 FirstChildUserNameTAG: Aahlad FirstChildCreateTimeTAG: 2012-10-21T19:12:20Z FirstChildTAG: With a proctored exam it is worth University Credit. I suspect it is too early to tell just what the Certificate of Mastery will be worth. A Certificate of Mastery awarded from the ***world's best*** technical school ought be worth something. ;) It should help with entry level and mid-tiered type jobs. I would still recommend learning the basics though, like resistor color codes, tolerances, how to read capacitor number codes, understanding capacitor voltage ratings etc. Some of this may be in the second half of the course, I don't know. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-21T19:14:49Z SecondChildTAG: It's funny, but since I was 13 or so, I loved the practical side of electronic i.e resistors/capacitor codes and big MOS-FETs soldering etc... but now, learning the 'theory' behind it is very refreshing, even though I don't have the prerequisites. SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-21T19:24:13Z SecondChildTAG: Just MHO so take it for what it's worth. I'm an electronics hobbyist without a formal EE education. However, I did enter college 50 years ago this past September. I attended schools with less prestige than MIT and the work load was heavy and challenging. I've done the assignments through week 8 and have not seen anything challenging so far other that those made so by a poorly worded question or a poor explanation in lecture. For example, in the week 8 homework we have problems with voltage impulses after most of the lecture dealt with current impulses. The professor admits voltage impulses are much less intuitive than current pulses but goes ahead and gives a much briefer discussion. I found I needed to do my own derivation in the spirit of that done for current impulses! With so much help available on the forum, it's hard to place much value on homework or lab grades. Maybe the midterm and final will do some sorting out. However, with a perfect homework and lab score one can pass the course with less than a 50% average on the midterm and final. It will be interesting to see how it all shakes out. One thing worth noting is that there are probably thousands taking the course, so what is seen on the forum may not be representative of the typical student. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-21T20:16:29Z SecondChildTAG: Regarding your statement "With so much help available on the forum, it's hard to place much value on homework or lab grades.". This all depends. When going to college one has direct access to the professor and is able to ask questions of and work with other students, which is similar to us having the discussion forum. So, we each have our advantages and disadvantages. Sure, one can just slide by and get a certificate, but lets see how that works for them when sitting in a 4 to 5 hour interview with a company that asks hard technical questions. They will walk out of the interview with their head spinning wondering what happened. I'm taking the class for two reasons. One to be able to get a better job at work. I graduated from college in 1986 and this is a good refresher. The second is, I may seek to get a Masters degree, and this is a great way to revisit those classes that a student adviser from a university would recommend taking at the undergraduate level before entering the masters program. That is, if ones previous grades don't hurt them. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-22T04:37:33Z FirstChildTAG: For me as a dad of an MIT student who currently takes similar courses, the main value is to get insight into the quality of these classes. In addition it is a nice refresher. Will I frame my certificate in the end? Absolutely! (if I get it). FirstChildUserIdTAG: 12905 FirstChildUserNameTAG: Baer FirstChildCreateTimeTAG: 2012-10-21T20:22:48Z SecondChildTAG: "Will I frame my certificate in the end? Absolutely!" Me too! If I pass, we can go frame shopping together. lol SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-11-08T00:02:08Z FirstChildTAG: The value of the certificate is something that will take time to develop. If it turns out that people with the certificate can get the job done, it will be valuable, eventually. Until then, its value will very person to person. There is more than the certificate though. As You learn data is being collected. In time this data will be able to predict if you will be a good employee for a particular job. Employers will want to know who these future good employees are. FirstChildUserIdTAG: 202032 FirstChildUserNameTAG: KeithD FirstChildCreateTimeTAG: 2012-10-21T22:52:23Z SecondChildTAG: That is a very good point. SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-22T15:03:09Z FirstChildTAG: I see a value for people who work in industry where they are either encouraged or required to engage in continuing education. This course is certainly more demanding than the typical continuing education course. If I get the certificate, I definitely plan to take it to my next performance review. If I don't get the certificate, I still plan to tell the reviewer that I participated. I also see this as part of a plan to compete against second (and lower) tier schools. Why would a serious student go to a lesser school when they can get certification from MIT or similarly rated schools for the same or even less money than the local school. But I'm sure our local university will continue to thrive, since it is the number 1 party school and proud of it! FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-21T23:14:01Z FirstChildTAG: I already got the **credit** for Fundamentals of Circuit Analysis at the University of Connecticut - there it was called ECE 201 with ECE 209 for the "hands on, breadboards and oscilloscopes" lab - every university has a similar class for "intro circuits"; our topics were structured differently (i.e. we covered op-amps and AC analysis way before even talking about MOSFET circuits, which Agarwal seems fixated on - though UConn did stress amplifiers in later courses). Our math was just as rigorous, if not more (we had to learn linear algebra to solve matrices of simultaneous equations, and multivariable calculus was a prerequisite). My question, should I take a proctored exam? I'll wait and see for now, first see how the mid-term goes. If I'm looking at an A, I would definitely do it. For a C, not so much. I already have a decent grade from the 1st time around, though I started this class *to keep my mind busy while looking for work*. Since I never graduated with a full EE/CS degree (money ran out, 1 year shy out of 5), the certificate will definitely be part of my resume; but will employers distinguish between "proctored" and "regular" certificates? As I am in the job market now after being laid-off, this is important. Employers always ask: What have you been doing besides "collecting your unemployment check?", I say "I'm currently working part-time, and I go to school online." So this certificate is definitely worth something to me personally, but I think it varies from person-to-person. An electrician with a good business would not need the "cert". A student majoring in EECS probably doesn't, unless they are waiting to transfer to a better program, or they want to go beyond an associates' degree. It's definitely good for non-traditional students: Those upgrading skills, those high-schoolers stuck and bored of "too-easy" classes, those looking for a better job, and those who never got a B.S. in Engineering, or will one day hope to obtain one. I enjoyed reading what others thought of this credential, so, thanks all! Mark, New Jersey, USA FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-10-22T09:03:43Z IndexTAG: 263 TitleTAG: H6P1 Hints **H6P1: THE NEWFET DEVICE** This problem examines the behavior and application of a new field effect transistor (NewFET) with large-signal electrical characteristics as described in the Figure 1, where vDS≥0. Note that the coefficient K and the threshold voltage VT are both positive and constant. ![image1][1] An amplifier is constructed with the NewFET as shown in Figure 2. ![im2][2] **Part 1:** Derive an expression of vOUT as a function of vIN in terms of the power supply voltage VS, the resistance R, and the NewFET parameters, K and VT. Do so for the NewFET biased into the active region 00. Se debe notar que el coeficiente K y la tensión umbral VT son ambas positivas y de valor constante. ![image1][1] Se construye un amplificador NewFET como se muestra en la Figura 2. ![im2][2] **Parte 1:** Deducir una expresión de vOUT en función de vIN, en términos de la fuente de alimentación VS, la resistencia R, y los parámetros del NewFET, Ky VT. Realizarlo para que el NewFET tenga su punto de operación en la zona activa 0 UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2012-10-17T03:10:02Z VoteTAG: 8 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: I find small signal analysis really hard!..I dnt know if I am the only one!! FirstChildUserIdTAG: 122513 FirstChildUserNameTAG: Mona77 FirstChildCreateTimeTAG: 2012-10-17T06:30:12Z SecondChildTAG: I'm with you SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-17T16:50:04Z SecondChildTAG: even im finding it difficult SecondChildUserIdTAG: 337430 SecondChildUserNameTAG: aparna_b SecondChildCreateTimeTAG: 2012-10-20T12:44:50Z SecondChildTAG: Can I help you? SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-20T13:26:00Z SecondChildTAG: @Myrimit : i need hint for finding 'r0' in H6P1 SecondChildUserIdTAG: 337430 SecondChildUserNameTAG: aparna_b SecondChildCreateTimeTAG: 2012-10-21T02:01:31Z FirstChildTAG: Hello Myrimit Thanks for the info provided but I am still struggling to find answer for H6p1 part 1 (Vs-Vout)/R= K*(vIN-VT)*Vout^2 then re arranging and writing root [-1+sqrt(1+4*K*R*Vs*(vIN-VT))]/(2*K*R*(vIN-VT)) if i substitute vIN=VT i will get 0/0 so I am using L hospital Rule differentiating w.r.t vIN seperalty numerator and denominator I am getting [-1+4*K*R*Vs]/[2*K*R*sqrt(1+4*K*R*(vIN-VT)*Vs)] if I type above as answer I am not getting this as right can you please give me hint where am I going wrong thanks a lot FirstChildUserIdTAG: 352647 FirstChildUserNameTAG: praveenjugge FirstChildCreateTimeTAG: 2012-10-17T08:36:52Z SecondChildTAG: Thanks I got answer with hazel1919 hint SecondChildUserIdTAG: 352647 SecondChildUserNameTAG: praveenjugge SecondChildCreateTimeTAG: 2012-10-18T06:18:16Z SecondChildTAG: Well done praveenjugge! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-19T03:34:32Z SecondChildTAG: I am having the same problem..how did u solve? SecondChildUserIdTAG: 173147 SecondChildUserNameTAG: cruiser_rahit SecondChildCreateTimeTAG: 2012-10-20T20:34:36Z FirstChildTAG: Hello. I have solved the quadratic in part - 1,but when I click on the check button,it says "vOUT not permitted in answer". Wasnt THAT what we were working for..? What is happening??? FirstChildUserIdTAG: 137686 FirstChildUserNameTAG: Jaychandran FirstChildCreateTimeTAG: 2012-10-17T16:39:05Z SecondChildTAG: Jaychandran, vOUT is not permitted because you have to type the right part of your solution. I mean you don't have to type "vOUT=", it must be omitted. SecondChildUserIdTAG: 90765 SecondChildUserNameTAG: jdavidg SecondChildCreateTimeTAG: 2012-10-18T04:29:11Z SecondChildTAG: Thanks for the help.. But still if i type the solution in the quadratic form, -b +- sqrt(b^2 - 4*a*c)/2*a I am getting the error.."could not parse "...." as a formula.." Whats happening here??? SecondChildUserIdTAG: 137686 SecondChildUserNameTAG: Jaychandran SecondChildCreateTimeTAG: 2012-10-18T10:40:25Z SecondChildTAG: If I understand what you are saying, you have to choose which sign of the radical is the correct one. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-18T16:17:17Z SecondChildTAG: Hi Jaychandran, you have to choose the correct root, you can not write + - .... SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-19T03:37:04Z IndexTAG: 264 TitleTAG: Just for curious, Which is the difference between Staff TA's and Community TA's ? I was wondering that question haha. Thank you very much! :) ---- **EDIT:** read my answer response. UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2012-10-15T21:00:20Z VoteTAG: 8 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: The staff hopefully gets paid. FirstChildUserIdTAG: 90796 FirstChildUserNameTAG: ChaunceyGardiner FirstChildCreateTimeTAG: 2012-10-15T23:31:43Z SecondChildTAG: Thank you ChaunceyGardiner :). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-15T23:43:14Z SecondChildTAG: We are like prison trustees! SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-16T03:03:49Z SecondChildTAG: Hahaha ! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-16T03:34:14Z FirstChildTAG: Cool! haha, I have noticed that I have now the green poster too ;)! So happy! Thank you edX and kimt for the offering of being Community TA :), I will try to do my best here! I really enjoy this. Thank you for this opportunity! ---- **New things** that I am discovering of being a **Community TA**: 1) It is almost the same as being student but has some curious things: 1.1) All my posts has the green poster haha! So strange! (Past comments too!) 1.2) I can endorse answers (the light blue tick). Cool! 1.3) I can edit and delete and close any post, I can see that buttons in all the Post different from mine ones too(This will be a new responsability for me, because in the Prototype course I coudn't do this although I had the enough Karma Points to do that (Karma Points allowed you to do something like editing others, etc, if you had more Karma more thing that you could do)... Hmmm....even I couldn't downvote to any of my classmates, I have zero downvotes used till the present. This will be new for me! ). 2) Being Community TA is for free as ChaunceyGardiner wrote ;). I enjoy doing this, in fact, it is the same that I have been doing but with some others responsabilities that I have named before. 3) This Post I have asked it before becoming Community TA, I saw JSChambers green poster and as I am really curious I asked which were the difference of the light blue and green :). Cool! Thank you very much edX for offering me of being Community TA too. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-16T05:48:42Z SecondChildTAG: Also I can endorse! Cool! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-16T05:53:26Z SecondChildTAG: Congrats! You deserve the honor! SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-19T01:29:43Z SecondChildTAG: Thank you planetscape :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-19T03:39:31Z FirstChildTAG: Congrats Myrimit and JSChambers! :-) FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-10-16T08:40:23Z SecondChildTAG: Thank you ashwith :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-16T18:08:46Z FirstChildTAG: Yeap, Congratulations Myrimit and JSChambers :)) Myrimit, pls. correct me here if my english might not be the best: "Hard work pays off" ---> El Trabajo duro da sus frutos!! -- (correct??) This applies to both of you, guys!! Goooood!!!!!!!! Regards!! Sandra FirstChildUserIdTAG: 342966 FirstChildUserNameTAG: SandraNavarro FirstChildCreateTimeTAG: 2012-10-26T08:55:44Z SecondChildTAG: Thank you Sandra! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-27T02:34:18Z IndexTAG: 265 TitleTAG: Hint to problem 2 Use the slope derivative method here and find ids(small signal)= (dIDS/dvds)at vds=Vds*vds Here IDS=DC Bias, vds=small signal So we get vds = ids/K(Vds − VT) UserIdTAG: 582521 UserNameTAG: varunchabba CreateTimeTAG: 2012-10-11T17:05:41Z VoteTAG: 8 CoursewareTAG: Week 5 / Two Terminal Connection Exercise CommentableIdTAG: 6002x_2_term_con_e NumberOfReplyTAG: 1 FirstChildTAG: So, here we get vds=6mA/(90mA/V^2*(2.7V-0.5V))=147mV, it's not correct answer. Where am I wrong in calculations? FirstChildUserIdTAG: 324219 FirstChildUserNameTAG: Dmitry79 FirstChildCreateTimeTAG: 2012-10-18T17:29:37Z IndexTAG: 266 TitleTAG: Hint to problem 1 Use the relation Ids=k/2*(Vi-Vt)^2 Here Vi=Vds=Vr UserIdTAG: 582521 UserNameTAG: varunchabba CreateTimeTAG: 2012-10-11T16:51:02Z VoteTAG: 8 CoursewareTAG: Week 5 / Two Terminal Connection Exercise CommentableIdTAG: 6002x_2_term_con_e NumberOfReplyTAG: 1 FirstChildTAG: I used the same formula but the answer is wrong my answer is 328.05 mA FirstChildUserIdTAG: 226085 FirstChildUserNameTAG: Teto FirstChildCreateTimeTAG: 2012-10-13T16:17:18Z SecondChildTAG: i too get the same..any help? SecondChildUserIdTAG: 111540 SecondChildUserNameTAG: tksanthosh SecondChildCreateTimeTAG: 2012-10-25T06:56:53Z SecondChildTAG: see 417 of text book SecondChildUserIdTAG: 111540 SecondChildUserNameTAG: tksanthosh SecondChildCreateTimeTAG: 2012-11-22T06:12:16Z IndexTAG: 267 TitleTAG: v of input v is small cap In the picture, due to the choice of font it is not clear that VIN is actually vIN. I could have deduced this from the fact that in the question vOUT is written with a more appropriate font, but instead after a frustrating minute I had to push 'show answer'. Not a big deal, but as I said a different font in the picture would work much better. Thank you. UserIdTAG: 154396 UserNameTAG: mpg CreateTimeTAG: 2012-10-11T03:54:54Z VoteTAG: 8 CoursewareTAG: Week 5 / MOSFET Amplifier Exercise 1 CommentableIdTAG: 6002x_mosfet_amp_e1 NumberOfReplyTAG: 1 FirstChildTAG: for 2 days there is no "show answer" bottom ,,,, somebody help me. FirstChildUserIdTAG: 395257 FirstChildUserNameTAG: blackguitar FirstChildCreateTimeTAG: 2012-10-12T20:39:47Z IndexTAG: 268 TitleTAG: Submit pictures of your lab! We have all been using the virtual lab to submit our homework, but I'm curious about what your real life lab looks like, the place where the sparks fly and the soldering happens! I'll try and upload a photo or two after I clean my lab up a bit! UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-10-08T19:44:30Z VoteTAG: 8 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: Ha! My lab is pretty much anywhere I can set it up. Usually on the counter-top in the kitchen or my work desk. I have everything in a tool box so it is always ready to go! I guess I can post a pic of the tool box. I'll try that later on tonight. FirstChildUserIdTAG: 417864 FirstChildUserNameTAG: beauclark FirstChildCreateTimeTAG: 2012-10-08T21:51:22Z FirstChildTAG: Post it uncleaned! It's much more fun :D FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-08T23:43:30Z FirstChildTAG: Picture a space in desperate need of an intervention by *Hoarders: Buried Alive*. ;-) FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-10-08T23:42:39Z SecondChildTAG: lol! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-22T14:37:21Z FirstChildTAG: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50854ee96f3be627000000eb FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-10-22T14:37:12Z IndexTAG: 269 TitleTAG: LOST IN WEEK4 :/ Everything seems so difficult this week :(...its frustrated and depressing at the same time...feels like quieting!! UserIdTAG: 122513 UserNameTAG: Mona77 CreateTimeTAG: 2012-10-05T21:26:00Z VoteTAG: 8 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 6 FirstChildTAG: Hi Mona77! You are not alone! Be up! You can do it! I will help you. In which part are you lost? I will back in some hours (I am going now to a Class of the University haha - Classes starts at 19:00 pm , at night, and finishes late late at 23:00 pm commonly). I will back, please write what are your difficulties and when I back I will try to help you as far as I can :) and also I will not let you to quit , no, no. Again, you are not alone. We can try to find together how to reach to the answers. I am sure that you need, as I said in other post, a push. So here is mine to you: I will help you! Be up! See you! Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-05T21:39:27Z SecondChildTAG: I´m back Mona77! Anything-doubt please ask. I will be following this Post. I would like to help you. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-06T01:56:37Z SecondChildTAG: Myriam You are an angle :)...God Bless you!...I really appreciate the effort you are putting in!..thanks Alot! SecondChildUserIdTAG: 122513 SecondChildUserNameTAG: Mona77 SecondChildCreateTimeTAG: 2012-10-06T09:41:16Z SecondChildTAG: i can understand how to do lab and norton equivlent circuit.. i tried a lot of time but Norton equvilent circuit is not correct . ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13495844001343662.png SecondChildUserIdTAG: 438395 SecondChildUserNameTAG: ali_PU1 SecondChildCreateTimeTAG: 2012-10-07T04:33:31Z FirstChildTAG: Hi Mona77, Myrimit is right. Hang in there! I've started this course nearly 3 weeks late and I've got no background in physics at all and honestly, trying to catch up and understand new material made me want to quit 100 times, but I'm still here and it feels good to be finally getting the hang of things. Community here is great and people really help a lot, so if you stick with it, I'm sure you'll do just fine and just imagine how awesome it would feel when you fight to the end and get your certificate :) And also don't forget that when studying it's normal to reach a plateau every once in a while. Once you traverse that plateau, you'll be climbing up the knowledge mountain again in no time. FirstChildUserIdTAG: 502508 FirstChildUserNameTAG: amaher FirstChildCreateTimeTAG: 2012-10-05T22:07:23Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-06T01:56:59Z SecondChildTAG: thank you very much...i guess you are right:) SecondChildUserIdTAG: 122513 SecondChildUserNameTAG: Mona77 SecondChildCreateTimeTAG: 2012-10-06T09:39:37Z FirstChildTAG: Mona77, When I took the course last spring, every week, from week 4 onwards, there was a point that I wanted to bail out. Usually when I looked at the homework for the first time. It's a hard class. Don't give up. There is a process that they use, and your knowledge will gel together when you least expect it to. There are a lot of EE resources on the web. Calculators, circuit simulators, Wolfram, etc. Take full advantage of them. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-06T00:06:48Z SecondChildTAG: thanks Alot :) SecondChildUserIdTAG: 122513 SecondChildUserNameTAG: Mona77 SecondChildCreateTimeTAG: 2012-10-06T09:37:39Z FirstChildTAG: Well done Myrimit, Your encouragement will definitely boost moral. FirstChildUserIdTAG: 455950 FirstChildUserNameTAG: Wahabbaluch FirstChildCreateTimeTAG: 2012-10-06T04:11:20Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T02:11:07Z FirstChildTAG: awwww...thank you everyone for your encouraging comments...Means Alot!..I started studying after 11 years and now I am thinking how I gt the degree...Its seems everything just washed away from my head!! Well I am having trouble finding vo when Zener is connected to the circuit..RL and then Norton current! I havent touch the Lab yet :(....to much to do and there is no time left :/ FirstChildUserIdTAG: 122513 FirstChildUserNameTAG: Mona77 FirstChildCreateTimeTAG: 2012-10-06T06:03:09Z SecondChildTAG: okeyy..I have completed the circuit in lab as shown in the textbook..but I am not getting any plots!! SecondChildUserIdTAG: 122513 SecondChildUserNameTAG: Mona77 SecondChildCreateTimeTAG: 2012-10-06T11:55:19Z SecondChildTAG: I understand, Mona. I've been out of school for a quarter century. I do NOT have the requisite math background, and my dyslexia makes all these VIviVivIayeyiyi's almost impossible to read. Despite excellent explanations from fellow students (I know they are good even if I don't understand them, because others are being helped), I STILL do not understand how to get correct answers for H4P1 questions 2 and 4. I've put in 12+ hours A DAY for over a month; I'm exhausted, frustrated, depressed, and don't think I'll ever get it. I think I'm too old for this. Yet, here I am, on a Saturday morning... I dunno. ;-/ SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-06T15:12:05Z SecondChildTAG: Mona77 khassak chi7aja SecondChildUserIdTAG: 331441 SecondChildUserNameTAG: abdessamade SecondChildCreateTimeTAG: 2012-10-06T15:30:30Z SecondChildTAG: @ planetscape just look here http://forum.allaboutcircuits.com/showthread.php?t=75112 , me and you don't have the required mathematical prerequisites and Q2 and Q4 require differential calculus (**don't worry though**), but at the end of the day, the basic principles of differential calculus are easy enough to learn over the course. To get the required knowledge just check out the Khan academy http://www.youtube.com/watch?v=ANyVpMS3HL4&feature=BFa&list=EC19E79A0638C8D449 These videos will do you well. Remember, incrimental resistance is the "derivative" of the slope of the curve at the point 6V. Just punch the equation of the wierd element into wolfram alpha and your derivative is your incrimental resistance! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-06T17:05:20Z SecondChildTAG: Thank you, Hazel. I found your other thread earlier, and finally FINALLY managed to get the right incantation into Wolfram Alpha to get a derivative that gets me the coveted green checks for H4P1 parts 2 and 4. Yay! Now to begin beating head against the next question! SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-06T18:25:18Z SecondChildTAG: @planetscape...you are a FIGHTER :)...I am still trying to complete my HW and Lab though :).. SecondChildUserIdTAG: 122513 SecondChildUserNameTAG: Mona77 SecondChildCreateTimeTAG: 2012-10-06T18:49:50Z SecondChildTAG: lab circuit and plot done!...got a different graph though but got a GREEN TICK for that..dnt know how...First question done...but all the values i am putting for VT(after calculations) is not right :( SecondChildUserIdTAG: 122513 SecondChildUserNameTAG: Mona77 SecondChildCreateTimeTAG: 2012-10-06T18:53:04Z SecondChildTAG: can i post my graph here just to show how is it different from the graph shown in the lab? SecondChildUserIdTAG: 122513 SecondChildUserNameTAG: Mona77 SecondChildCreateTimeTAG: 2012-10-06T19:31:50Z SecondChildTAG: Hi Mona77 ! :) For Lab4 you can see here[Lab4 Hints][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5070e1b72210d02400000055 SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T02:13:26Z SecondChildTAG: Hi Mona77! Take a look to this Post for the Zener [H4P2][1] ;). [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5070be20fabaf62b0000004d SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T04:25:34Z SecondChildTAG: :)...thank you very much...I have already done my labs and almost all the home work except IN..which is giving me trouble...I dnt know how I did It but yaa I am through WEEK 4..:))...thank you everyone for your supportive comments and encouragement.. SecondChildUserIdTAG: 122513 SecondChildUserNameTAG: Mona77 SecondChildCreateTimeTAG: 2012-10-07T05:50:01Z SecondChildTAG: I am trying every method for In but couldnt get the answer :/ ... SecondChildUserIdTAG: 122513 SecondChildUserNameTAG: Mona77 SecondChildCreateTimeTAG: 2012-10-08T04:41:16Z FirstChildTAG: **Mona77 khassak chi7aja** FirstChildUserIdTAG: 331441 FirstChildUserNameTAG: abdessamade FirstChildCreateTimeTAG: 2012-10-06T13:27:35Z SecondChildTAG: I can only understand English! SecondChildUserIdTAG: 122513 SecondChildUserNameTAG: Mona77 SecondChildCreateTimeTAG: 2012-10-06T18:50:34Z IndexTAG: 270 TitleTAG: Late Penalty I came to know about this course a day before and today is my first day. Is there any penalty for late submission of homework and lab? UserIdTAG: 555841 UserNameTAG: AlokSheth CreateTimeTAG: 2012-10-05T09:24:20Z VoteTAG: 8 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: Please review the Syllabus in the "Course Info" section. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-05T14:06:46Z IndexTAG: 271 TitleTAG: Taylor Series Taylor series doesn't have to be complicated :) Taylor Series of f(x) : $T(x) =$ $f(a) \over 0!$$\cdot(x-a)^0$ + $f^1(a) \over 1!$$\cdot(x-a)^1$ + $f^2(a) \over 2!$$\cdot(x-a)^2$ + ... + $f^n(a) \over n!$$\cdot(x-a)^n$ With a = 0 this becomes $T(x) =$ $f(a)$ + $f^1(a) \over 1!$$\cdot(x)^1$ + $f^2(a) \over 2!$$\cdot(x)^2$ + ... + $f^n(a) \over n!$$\cdot(x)^n$ The more terms you have the more accurate the approximation. We can use only the first two terms for example ,so this becomes, $T(x) =$ $f(0)$ + $f^1(0) \over 1!$$\cdot(x)^1$ As an example, let's work out the taylor series of $e^x$ around the point a = 0 Note that the derivative of $e^x$ = $e^x$ So, $T(x) =$ $f(0)$ + $f^1(0) \over 1!$$\cdot(x)^1$ + $f^2(0) \over 2!$$\cdot(x)^2$ + ... + $f^n(0) \over n!$$\cdot(x)^n$ That is , $T(x) =$ $e^0$ + $e^0\over 1!$$\cdot(x)^1$ + $e^0\over 2!$$\cdot(x)^2$ + ... + $e^0 \over n!$$\cdot(x)^n$ which is $e^x = T(x) = $ $1$ + $x^1 \over1!$ + $x^2 \over2!$ + ... + $x^n \over n!$ Now say you want to know what $e^{0.5}$ is. Replace x with 0.5 in T(x). I’m only going to use a second order Taylor series, in other words, a Taylor polynomial with 3 terms. $e^{0.5} = $ $1$ + $0.5^1 \over1!$ + $0.5^2 \over2!$ $ = 1.625 $ The real answer of $e^{0.5}$ is 1.64872..., so even with only 3 terms, our approximation is pretty accurate. Side note: Choose the value of a to be close to the x value you are going to use. For example, I used a = 0 for x = 0.5. UserIdTAG: 236188 UserNameTAG: Rince CreateTimeTAG: 2012-10-01T16:20:51Z VoteTAG: 8 CoursewareTAG: Week 4 / Incremental Method Math CommentableIdTAG: 6002x_inc_method_math NumberOfReplyTAG: 2 FirstChildTAG: The Khan academy gives nice explanations: http://www.khanacademy.org/math/calculus/integral-calculus/v/taylor-polynomials FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-10-01T17:21:04Z SecondChildTAG: Thanks for the link that helped me so much as I have never done the Taylor series before SecondChildUserIdTAG: 404470 SecondChildUserNameTAG: LukeSki SecondChildCreateTimeTAG: 2012-10-02T14:16:14Z FirstChildTAG: Good, but exponentials are the easiest functions to differentiate and trivial to evaluate at X = 0. Of course, series expansions for common functions can be looked up in many sources if you don't know Taylor series. Expressions we are asked to evaluate are a bit more complex ... not hard but requiring more effort with the opportunity to make a mistake with the resulting algebra. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-01T21:45:43Z IndexTAG: 272 TitleTAG: Amused The fact that Prof. Agarwal chose to play Sublime in this demonstration made him 10x cooler. UserIdTAG: 381923 UserNameTAG: matteaton CreateTimeTAG: 2012-10-01T16:04:14Z VoteTAG: 8 CoursewareTAG: Week 4 / Incremental Analysis Demo CommentableIdTAG: 6002x_inc_analysis_demo NumberOfReplyTAG: 0 IndexTAG: 273 TitleTAG: Free USB sticks with videos You can download videos from [video downloadlink][1] .Videos are uploaded now to lesson 14 ,slides and wiki are included subtitles are in srt format; if you want to see subtitles use vlc mediaplayer i have 3.5TB/month but don't know how the brinksterserver behaves under heavy load. Also for people with bad internetconnections I have 50 8GB sticks available with all those videos for free. Just mail your postal address to 6002x@ruudoleo.com together with your alias and I will send the USB stick to the 50 first responders. No other use of address will be made. I make this offer because in the mitx 6002x course there were people who couldn't watch the videos by reason of bandwidth limitations and it seems the problem is not improving. [1]: http://www.ruudoleo.com/mitx/small/6002xvideosmall.htm UserIdTAG: 79885 UserNameTAG: ruudoleo CreateTimeTAG: 2012-09-27T00:14:07Z VoteTAG: 8 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Great! I'm doing some playlists for online viewing using JWPlayer, for people who cannot access YouTube. Maybe I can use your server. I've already downloaded the subtitles and I might add those to the playlists. FirstChildUserIdTAG: 276409 FirstChildUserNameTAG: IgnacioUY FirstChildCreateTimeTAG: 2012-09-27T02:39:26Z SecondChildTAG: IgnacioUY, I'm also using JWPlayer for a few videos for my website. But streaming video is not allowed by my provider because this causes heavy load on the servers. Also JWPlayer only works when run from inside the domain of the server where videos are stored. However if you open the video in your browser and start playing it the download normally keeps up with the playing so you should also be able to view the video online. Subtitles are then not supported. Also around midnight the server becomes slow (they probably make backups around that time). SecondChildUserIdTAG: 79885 SecondChildUserNameTAG: ruudoleo SecondChildCreateTimeTAG: 2012-09-27T06:17:48Z SecondChildTAG: Thanks a lot So We all are having one language one world "eDX " thanks for your good mind SecondChildUserIdTAG: 156694 SecondChildUserNameTAG: mkprasanth SecondChildCreateTimeTAG: 2012-09-27T11:25:31Z SecondChildTAG: Subscribers received for the USB stick 1. SHELLYB21 2. NIBU GTHOMAS 3. MAHFOUD 4. SSEMBAJJWE 5. SHUHUAN84 Sticks will be posted saturday SecondChildUserIdTAG: 79885 SecondChildUserNameTAG: ruudoleo SecondChildCreateTimeTAG: 2012-09-27T18:30:17Z SecondChildTAG: Subscriber received for the USB stick 6.Shivank Sharma SecondChildUserIdTAG: 79885 SecondChildUserNameTAG: ruudoleo SecondChildCreateTimeTAG: 2012-09-28T20:10:26Z SecondChildTAG: Subscriber received for USB stick 7.Memoona SecondChildUserIdTAG: 79885 SecondChildUserNameTAG: ruudoleo SecondChildCreateTimeTAG: 2012-10-02T07:46:15Z SecondChildTAG: What an amazing offer, ruudoleo! Thank you very much indeed for doing this. I have sufficient connectivity that I can download my own videos for when I'm off line so I'm not asking for one. I just want to thank you. <3 SecondChildUserIdTAG: 86632 SecondChildUserNameTAG: LCL SecondChildCreateTimeTAG: 2012-10-05T18:16:06Z SecondChildTAG: received it today by post...thanks ALOT :) SecondChildUserIdTAG: 122513 SecondChildUserNameTAG: Mona77 SecondChildCreateTimeTAG: 2012-10-16T15:29:16Z SecondChildTAG: Subscriber received for USB stick 8. Jain SecondChildUserIdTAG: 79885 SecondChildUserNameTAG: ruudoleo SecondChildCreateTimeTAG: 2012-11-02T05:18:02Z IndexTAG: 274 TitleTAG: H2P2 looks like... bug? or i didnt understand something... I have found **RL** and **Rth**. Test says that values is correct. But it didn't accept power value. I decided to use Norton equivalent scheme because its better in this case.![Norton equivalent scheme][1] [1]: https://edxuploads.s3.amazonaws.com/13481149316553046.png Since **Rth** = **RL**, current flowing through **RL** equals 1/2 **I**. So power delivered to **RL** equals **P** = **I^2** * **R** = 1/4 **I^2** * **RL** . Remind that I have **I** and **RL** (which is correct), but **P** value is wrong!!! Where is my mistake? or bug really exists? UserIdTAG: 189808 UserNameTAG: AlexanderZ CreateTimeTAG: 2012-09-20T04:37:27Z VoteTAG: 8 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Troubleshooting NumberOfReplyTAG: 10 FirstChildTAG: if you are replacing all the resistors with Rth then you should replace current source with I Norton then divide current with 2. FirstChildUserIdTAG: 908 FirstChildUserNameTAG: m_umair72001 FirstChildCreateTimeTAG: 2012-09-20T04:48:41Z SecondChildTAG: As I understand beauty of norton scheme is in fact that current source remains the same. SecondChildUserIdTAG: 189808 SecondChildUserNameTAG: AlexanderZ SecondChildCreateTimeTAG: 2012-09-20T07:13:13Z SecondChildTAG: no, it will be different in this case, take note of the current flowing only on the load side (or the current flowing only through the transmission lines). you might notice that won't be the same as your current source. SecondChildUserIdTAG: 312336 SecondChildUserNameTAG: drel SecondChildCreateTimeTAG: 2012-09-20T12:10:09Z SecondChildTAG: a mi se me dio 0.023 SecondChildUserIdTAG: 163395 SecondChildUserNameTAG: jasonlll88 SecondChildCreateTimeTAG: 2012-09-23T03:20:23Z FirstChildTAG: Too find out the power accurately you can use a formula which is quite easy to find the power value in load resistance ***IL = Vth/(RL + Rth)***. Here IL is the current passing through load resistance then ***P = IL^2 * RL*** and there you go the power value is found FirstChildUserIdTAG: 336001 FirstChildUserNameTAG: syd_buet12 FirstChildCreateTimeTAG: 2012-09-20T05:43:25Z SecondChildTAG: Thanks a lot!!!!!!!!!!!!!!!!!!! I was sure I get crazy after all the day of looking for this (f) power !!!! You safe my life!!! SecondChildUserIdTAG: 380287 SecondChildUserNameTAG: gayetan SecondChildCreateTimeTAG: 2012-09-23T20:22:42Z SecondChildTAG: jajaja funny :)) you're almost about to save mine as well :)) ANd I hope you guys to be more active from next onwards, since I'll be unemployee and I'll have more time to help back :)) SecondChildUserIdTAG: 342966 SecondChildUserNameTAG: SandraNavarro SecondChildCreateTimeTAG: 2012-09-23T21:37:07Z FirstChildTAG: The Norton method is fine for finding a suitable load resistance, but remember that the Norton circuit you devised does not incorporate the load resistance. Thus, it does not reflect the circuit with load resistance, so any power analysis on the load resistance is not valid using the devised Norton equivalent. Use the Rload you got from part 1, and plug it into the original circuit that they provided, not your simplified norton circuit. You'll notice that the current entering the load isn't 1/2 the source current, but something around 10% (I had Is=0.8A, Rp=0.75R, Rs=1.5R). FirstChildUserIdTAG: 141053 FirstChildUserNameTAG: Albert0ng FirstChildCreateTimeTAG: 2012-09-20T05:49:57Z SecondChildTAG: hm... U was right. I have inserted **RL** in initial scheme and calculated currents and voltage. All done. Another sad thing that I am not fully understand is borders of applications of equivalent circuits... SecondChildUserIdTAG: 189808 SecondChildUserNameTAG: AlexanderZ SecondChildCreateTimeTAG: 2012-09-20T07:10:15Z FirstChildTAG: Guys i am just too lost on finding RL.Can anyone help on how to find it. FirstChildUserIdTAG: 429851 FirstChildUserNameTAG: ssembajjwe FirstChildCreateTimeTAG: 2012-09-20T10:02:35Z SecondChildTAG: Power on a resistive load is `P = I^2*R`. So the higher the current, higher will be the power. So for which resistance will the current be higher in that circuit? Norton or Thevenin helps. SecondChildUserIdTAG: 308853 SecondChildUserNameTAG: fmorato SecondChildCreateTimeTAG: 2012-09-20T10:14:46Z SecondChildTAG: Thanks finally i got.Thanks 100 times fmorato. SecondChildUserIdTAG: 429851 SecondChildUserNameTAG: ssembajjwe SecondChildCreateTimeTAG: 2012-09-20T12:06:23Z SecondChildTAG: for max power Rth=RL SecondChildUserIdTAG: 266739 SecondChildUserNameTAG: satvikchugh SecondChildCreateTimeTAG: 2012-09-20T16:22:01Z FirstChildTAG: I deduced the crude network just as shown above. I calculated the Norton current as 0.075A and Norton resistance as 8 ohms. Please let me know if these are correct. Are we equating the load resistance to the norton resistance for maximum power to be delivered to the load? FirstChildUserIdTAG: 337926 FirstChildUserNameTAG: NeetaKumar FirstChildCreateTimeTAG: 2012-09-20T12:17:46Z FirstChildTAG: First you have to replace the norton existing of only I and Rp by thevenin. Now you know Vth and Rth as the replacement for the norton, but seen from the load, Rs is in series with Rth, so you get a new Rth ... FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-09-20T14:39:37Z SecondChildTAG: Can you please elaborate on that. I'm really confused. SecondChildUserIdTAG: 337926 SecondChildUserNameTAG: NeetaKumar SecondChildCreateTimeTAG: 2012-09-20T17:52:21Z FirstChildTAG: How did you calculate the Thévenin resistance? Those Rs resistors are making it pretty confusing for me. FirstChildUserIdTAG: 378726 FirstChildUserNameTAG: Shrapsmel FirstChildCreateTimeTAG: 2012-09-21T13:43:08Z FirstChildTAG: there are several ways for doing it. a number these have been discussed in the above discussions. Though i am doing the HW at the 11 hr so just a quick guide! first calculate the thevenin eq resistance or use the maximum power transfer theorem. then when you have Rth and R load. Convert the current source to a voltage source (V=IRp) Rp is 3ohm resistance. Now the total resistance in series is (Rp + Rs +Rs + Rl). Now the current throuth the whole circuit can be calculated by V\(Rp + Rs +Rs + Rl). Now use P=I^2*R (R= R load)to calculate power by Rl. FirstChildUserIdTAG: 232499 FirstChildUserNameTAG: Asim09 FirstChildCreateTimeTAG: 2012-09-22T19:38:24Z FirstChildTAG: This is crazy, I spent almost 2 h and the same result 0.0235, What is wrong? FirstChildUserIdTAG: 443442 FirstChildUserNameTAG: Nelutu FirstChildCreateTimeTAG: 2012-09-23T12:30:25Z SecondChildTAG: First find the Norton or Thevenin equivalent of the circuit and now Rth = RL. Construct the new circuit based on the equivalent circuit and calculate the current passing through RL. Power = SQ(V)/R or SQ(I)*R SecondChildUserIdTAG: 364851 SecondChildUserNameTAG: Varindra SecondChildCreateTimeTAG: 2012-09-25T21:23:58Z FirstChildTAG: You can do like this first find voltage across RL and write equation P =V^2 / R R can be represent has R = Rth+RL FirstChildUserIdTAG: 425325 FirstChildUserNameTAG: IranagoudaN FirstChildCreateTimeTAG: 2012-09-20T04:59:22Z IndexTAG: 275 TitleTAG: Wrong answer Node equation: (e-V1)/R1 + (e-V2)/R2 = 0 From the task it is clear that V2<0, so (e-5)/6800 + (e+7.2)/5600 = 0 Thus e = -1.6903 V. It's not 6.206 V UserIdTAG: 337855 UserNameTAG: flexo CreateTimeTAG: 2012-09-18T17:49:02Z VoteTAG: 8 CoursewareTAG: Week 2 / Node equation review CommentableIdTAG: 6002x_S3E1_NodeEquationReview NumberOfReplyTAG: 2 FirstChildTAG: You can try to find result using simulator.[CIRCUIT SANDBOX][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Overview/Circuit_Sandbox/ FirstChildUserIdTAG: 277168 FirstChildUserNameTAG: AndrewKiselev FirstChildCreateTimeTAG: 2012-09-18T17:57:31Z SecondChildTAG: notice that -7.2 source's polarity is reversed , so you can solve it as +7.2 with reversing the polarity on the Diagram "+ve source with + sign on the top node & - sign on the ground". SecondChildUserIdTAG: 185715 SecondChildUserNameTAG: amirengineer SecondChildCreateTimeTAG: 2012-09-20T04:49:43Z SecondChildTAG: If you consider the flow of current you notice that -7.2 will not change its sign as current flow is from the + to -. Thus when solving for e using the node method the equation will be (e-V1)/R1 + (e+V2)/R2 = 0 of which gives the correct answer. SecondChildUserIdTAG: 392760 SecondChildUserNameTAG: sonnie SecondChildCreateTimeTAG: 2012-09-23T16:00:42Z SecondChildTAG: so the correct answer must be -1.69... am i correct SecondChildUserIdTAG: 483548 SecondChildUserNameTAG: SAM_1993 SecondChildCreateTimeTAG: 2012-09-30T09:12:35Z FirstChildTAG: Through KCL,we have the equation *(e-V1)/R1 + (e-V2)/R2 = 0*, where the potential **V2** is **-7.2V** in the "+" to "-" of the source, but in the "-" to "+" of the source **V2** is **7.2V**. So, the result will be: *(e-5)/6800 + (e-7.2)/5600 = 0*. The solution for this equation is *e = 6.21V* FirstChildUserIdTAG: 338518 FirstChildUserNameTAG: PHLMenezes FirstChildCreateTimeTAG: 2012-09-18T18:38:39Z SecondChildTAG: How is this possible? In node analysis one should check node voltage with respect to ground, and if so then the node above V2 is -7.2, it can't be 7.2. SecondChildUserIdTAG: 120983 SecondChildUserNameTAG: mykisaacs SecondChildCreateTimeTAG: 2012-09-19T00:59:45Z SecondChildTAG: Ok..I see it now! If the V2 element was aligned/rotated like the V1 then node above V2 would be -7.2. But since it's reversed the '-' is cancelled. Thanks. SecondChildUserIdTAG: 120983 SecondChildUserNameTAG: mykisaacs SecondChildCreateTimeTAG: 2012-09-19T01:04:12Z SecondChildTAG: Is it like ***negative terminal of a negative voltage source = positive potential*** ?1 SecondChildUserIdTAG: 408765 SecondChildUserNameTAG: NeevGhodasara SecondChildCreateTimeTAG: 2012-09-19T08:22:19Z SecondChildTAG: we only have the potential difference so it could also be possible the potential of both terminals can be negative its just for reference SecondChildUserIdTAG: 451572 SecondChildUserNameTAG: rahulgarg11 SecondChildCreateTimeTAG: 2012-09-21T17:32:05Z SecondChildTAG: I follow combention and figure out it... SecondChildUserIdTAG: 804955 SecondChildUserNameTAG: alchy SecondChildCreateTimeTAG: 2012-12-03T19:38:18Z IndexTAG: 276 TitleTAG: Not particularly helpful I didn't find this exercise to be that beneficial. It might be improved by moving the text and so that the sliders and all 3 circuits can be viewed/controlled at the same time. UserIdTAG: 237421 UserNameTAG: Lmotloch CreateTimeTAG: 2012-09-18T11:17:38Z VoteTAG: 8 CoursewareTAG: Week 2 / Superposition experiment CommentableIdTAG: 6002x_Superposition_experiment NumberOfReplyTAG: 0 IndexTAG: 277 TitleTAG: Mixing of voltage signals ... LAb 2 ??? tutorial on mixing of signals ? UserIdTAG: 415376 UserNameTAG: imab90 CreateTimeTAG: 2012-09-17T12:51:48Z VoteTAG: 8 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 278 TitleTAG: I CAN'T UNDERSTAND AT ALL I can't understand this question at all. what does it mean? UserIdTAG: 128276 UserNameTAG: ChengBin CreateTimeTAG: 2012-09-17T03:49:35Z VoteTAG: 8 CoursewareTAG: Week 2 / Linear combinations of source strenghts CommentableIdTAG: 6002x_S3E2_Linear_Combinations_of_Source_Strengths NumberOfReplyTAG: 2 FirstChildTAG: Try to solve the task finding V3, i1 and i2 in terms of R1,R2,R3, V1 and V2. The coefficients must contain only R1, R2 and R3. FirstChildUserIdTAG: 308508 FirstChildUserNameTAG: Alex-Fion FirstChildCreateTimeTAG: 2012-09-17T08:59:06Z SecondChildTAG: Treat the node where R1, 2 and 3 meet as e. Use KCL to solve for the currents (I1 + I2 + I3 = 0) that go into the three resistors. Now, use Ohm's law (I = V/R) around the resistors to get these three currents in the KCL equation to be in terms of V and R (hint: V will be in terms of the element voltage and e). Solve for e, find the linear sum of sources and replace it with S, and get the equation into the three equation forms given for v3, i2 and i3. Then whatever junk fits the coefficient of the equation form is your answer. SecondChildUserIdTAG: 364798 SecondChildUserNameTAG: atari1994 SecondChildCreateTimeTAG: 2012-09-18T04:47:22Z FirstChildTAG: This question is really hard to understand, but when you discover the right way, you'll be able to do all the exercises about this circuit. So, initially, I recommend to you start revealing the nodes: Using the node method, you can find it (If have any difficult in find the nodes, you can ask later, but I think is not necessary show all the steps now, because the node method was explicated in another video): ![Circuit][1] [1]: https://edxuploads.s3.amazonaws.com/13478923361343655.png It was what I did, although probably other ways can be too efficient. To let you do the exercise and train your skills, I will not write all the exercise, but remember that you have to find the variables **according to the resistances**, so use the KVL, the node method and the v-i relationship to discover the answers based in the voltages (V1 and V2) and in the resistances (R1, R2 and R3). Do you want a tip for where you can start? Using the node method in the bottom points you can find "i1=(V3-V1)/R1", "i2=(V3-V2)/R2" and "i3=i1+i2", so thenceforth you can use the algebra to find the results. FirstChildUserIdTAG: 291362 FirstChildUserNameTAG: Gudson FirstChildCreateTimeTAG: 2012-09-17T14:45:52Z IndexTAG: 279 TitleTAG: Solve by using G instead of R If you solve the problem by "G"s instead of "R"s, The equations would be contain easier algebras and the answers would be: a1 = G1/(G1+G2+G3), a2 = G2/(G1+G2+G3) b1 = -(G1*G2+G1*G3)/(G1+G2+G3) b2 = (G1*G2)/(G1+G2+G3) c1 = (G1*G2)/(G1+G2+G3) c2 = -(G1*G2+G2*G3)/(G1+G2+G3) Now by considering following equations you can find the result in "R"s G1+G2+G3 = (R1*R2+R1*R3+R2*R3)/(R1*R2*R3) G1=1/R1, G2=1/R2 & G3=1/R3 and also G1*G2+G2*G3=(R3+R1)/(R1*R2*R3) UserIdTAG: 420339 UserNameTAG: AliJenabi CreateTimeTAG: 2012-09-16T13:53:01Z VoteTAG: 8 CoursewareTAG: Week 2 / Linear combinations of source strenghts CommentableIdTAG: 6002x_S3E2_Linear_Combinations_of_Source_Strengths NumberOfReplyTAG: 3 FirstChildTAG: very good thanks for this FirstChildUserIdTAG: 399292 FirstChildUserNameTAG: Qabali FirstChildCreateTimeTAG: 2012-09-16T14:23:56Z SecondChildTAG: I followed a similar path (though I hadn't looked at your post) and got the answer. Like you, I would recommend using conductance G and replacing it in the last step with resistance R. SecondChildUserIdTAG: 254346 SecondChildUserNameTAG: moijes12 SecondChildCreateTimeTAG: 2012-09-19T08:15:50Z FirstChildTAG: thanks it make me easy to get the concept.... FirstChildUserIdTAG: 372101 FirstChildUserNameTAG: Akkigupta FirstChildCreateTimeTAG: 2012-09-17T23:40:43Z FirstChildTAG: I set up a matrix of coefficients of resistance, a column vector of unknowns and equated them to a vector of V1 V2 0 and solved using cramers rule. FirstChildUserIdTAG: 373585 FirstChildUserNameTAG: radami1 FirstChildCreateTimeTAG: 2012-09-18T03:55:48Z IndexTAG: 280 TitleTAG: KCL and KVL is very easy and mind blowing for the start Thank you For MITx FOR CONDUCTING THESE PROGRAMS ONLINE. UserIdTAG: 396372 UserNameTAG: vsnarendran CreateTimeTAG: 2012-09-16T03:32:19Z VoteTAG: 8 CoursewareTAG: Week 1 / KVL and KCL exercises CommentableIdTAG: 6002x_kvl_kcl_exercises NumberOfReplyTAG: 1 FirstChildTAG: Yes. I agree :-) FirstChildUserIdTAG: 199839 FirstChildUserNameTAG: vijaykrishnankk FirstChildCreateTimeTAG: 2012-09-16T04:08:23Z IndexTAG: 281 TitleTAG: problems on paper On video part 2 i think the truth table of AND gate is inverted with NAND gate. What do you think? UserIdTAG: 159031 UserNameTAG: GabrielNeves CreateTimeTAG: 2012-09-14T18:02:14Z VoteTAG: 8 CoursewareTAG: Week 2 / Logic Gates CommentableIdTAG: 6002x_logic_gates_tutorial NumberOfReplyTAG: 1 FirstChildTAG: I agree. I think they should be written other way around. FirstChildUserIdTAG: 153760 FirstChildUserNameTAG: Kavka FirstChildCreateTimeTAG: 2012-09-17T03:42:27Z SecondChildTAG: It should say ![enter image description here][1] [1]: https://dl.dropbox.com/u/24096724/corrected.png SecondChildUserIdTAG: 276409 SecondChildUserNameTAG: IgnacioUY SecondChildCreateTimeTAG: 2012-09-18T18:20:04Z IndexTAG: 282 TitleTAG: Viewing written words with the videos Youtube recently changed their video hosting format, and we are working hard to adjust our video player customizations. Until then, we are using Youtube's embedded player without any modifications, which means that we cannot present our interactive transcript at the moment. However, the 'cc' button at the bottom of the youtube player will still give you access to some of the written words with the lectures, if you need them. ![location of closed captioning button][1] [1]: http://web.mit.edu/fischerl/Public/Screen%20Shot%202012-09-13%20at%208.37.01%20AM.png UserIdTAG: 96452 UserNameTAG: Lyla CreateTimeTAG: 2012-09-13T12:42:59Z VoteTAG: 8 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 5 FirstChildTAG: Thanks a lot! I hope you fix your system. I do not speak very well English. And for YouTube words "relax" and "002x" are identical (It's horrible for me). FirstChildUserIdTAG: 343092 FirstChildUserNameTAG: Rustam84 FirstChildCreateTimeTAG: 2012-09-13T13:20:47Z FirstChildTAG: Thanks lyly FirstChildUserIdTAG: 282351 FirstChildUserNameTAG: HalfDead FirstChildCreateTimeTAG: 2012-09-13T13:15:14Z FirstChildTAG: Thanks a lot! I was really wanting it =D FirstChildUserIdTAG: 291362 FirstChildUserNameTAG: Gudson FirstChildCreateTimeTAG: 2012-09-13T12:47:00Z FirstChildTAG: We hope that fix it!! Interactive transcript and video speed tools are very important to many people. Million Thanks Lyla!!!! FirstChildUserIdTAG: 149058 FirstChildUserNameTAG: sotoroman FirstChildCreateTimeTAG: 2012-09-13T16:11:37Z FirstChildTAG: Hi, please be sure to select "English" not "English (transcribed)" to get the accurate subtitles.![enter image description here][1] [1]: https://dl.dropbox.com/u/45575056/subtitleselection.png FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-13T16:50:08Z SecondChildTAG: Thank you very much! SecondChildUserIdTAG: 343092 SecondChildUserNameTAG: Rustam84 SecondChildCreateTimeTAG: 2012-09-13T16:59:23Z IndexTAG: 283 TitleTAG: The explicit solution Question 1 We've already known that: - v(t) = `120*sqrt(2)*cos(2*pi*60*t` V and - R = 110 Ohms. From math, we also know that minimum and maximum value of cos is -1 and 1. So the peak power (in Watts) is the absolute, maximum value of voltage v(t) divided by constant resistance, which means the absolute, maximum value of cos, which is 1. So the result of Question 1, using formula - `P=V^2/R` is -> P = 120^2 / 110 = 261.8 ---------- Question 2 To compute the average power dissipated in resistor, we have to solve the integral formula of the instantaneous power over one cycle of the waveform (as hint said). The simple way is: https://www.dropbox.com/s/d3j3lul9i3xyzir/Question%202.jpg ---------- Question 3 That's really simple exercises. We know that the value of resistor is constant, so the value of voltage as well. We just have to use the formula - `P = V^2/R` to compute the power (in Watts). Using this formula -> P = 120^2/110 = 130.9 UserIdTAG: 88649 UserNameTAG: synus_wroc CreateTimeTAG: 2012-09-13T05:19:03Z VoteTAG: 8 CoursewareTAG: Week 1 / AC power CommentableIdTAG: 6002x_S1E3_AC_power NumberOfReplyTAG: 6 FirstChildTAG: Please can the integration be broken down one step at a time FirstChildUserIdTAG: 382532 FirstChildUserNameTAG: emmanuelpeace FirstChildCreateTimeTAG: 2012-09-15T10:19:46Z FirstChildTAG: Thanks for the written out explanation on question 2. My calculus has gotten really rusty and your explanation was a big help to set things right. FirstChildUserIdTAG: 300453 FirstChildUserNameTAG: emcmahil FirstChildCreateTimeTAG: 2012-09-13T20:30:53Z FirstChildTAG: Gracias (: FirstChildUserIdTAG: 127825 FirstChildUserNameTAG: barbie93 FirstChildCreateTimeTAG: 2012-09-14T23:42:39Z FirstChildTAG: ..I think you left sqrt(2) out of your equation for power in question 1. P=(120*sqrt(2))^2/110=261.82 FirstChildUserIdTAG: 434357 FirstChildUserNameTAG: Malmsteen FirstChildCreateTimeTAG: 2012-09-14T14:17:01Z SecondChildTAG: Thanks for the correction, i was saying to myself that cant be right, because P = 120^2 / 110 = is definitely not 261.8. Here's the thing, how come we are ignoring the rest of the voltage 120⋅√2⋅cos(2π⋅60⋅t) Volts. as in the cos bit onwards? thanks :) SecondChildUserIdTAG: 208105 SecondChildUserNameTAG: jmunya SecondChildCreateTimeTAG: 2012-09-14T23:06:43Z FirstChildTAG: @ Hi synus wroc: ---------- S1E3: AC power: How the cancellation has taken place in the last step.. 1/60 canceled how how?????...in the solution of Question no.2 you posted. ---------- https://www.dropbox.com/s/d3j3lul9i3xyzir/Question%202.jpg ---------- kindly help me out.... its been mystery for me...:) thanks. Hassan FirstChildUserIdTAG: 107038 FirstChildUserNameTAG: Hassankhan FirstChildCreateTimeTAG: 2012-09-16T06:40:57Z FirstChildTAG: missing something for Q1 120^2 / 110 = 130.909 FirstChildUserIdTAG: 435704 FirstChildUserNameTAG: Gaby_64 FirstChildCreateTimeTAG: 2012-09-16T06:45:01Z IndexTAG: 284 TitleTAG: Forum Software & Interaction feedback/feature request First off I'd like to thank the techs for all their nice work with the system. Second, I'd like to ask a few questions and maybe suggest a feature or two. 1. It seems that the forum software in use isn't really designed for back and forth discussions, since it only supports 3 or 4 levels of replies. Is there a specific reason for this? I find it's much more helpful to the learning process to give a little hint at a time so that the learner can figure out as much as they can on their own. 2. It would be nice to have the ability to see all of the threads I'm following, rather than just five. 3. Would something like a chat functionality be possible? I feel like it would be very useful to quickly discuss resources or lectures with classmates who are online, or to set up a study group for example. 4. Is the discussion board the place for feedback on the course in general? I feel like it would be useful for staff and tech to have a separate place for that kind of information (just based on the UI that I'm seeing - there's not really a way to see a list of used tags as far as I can see). *EDIT: It would also be nice to tell if someone has replied to a thread I've followed since I last checked it. I just noticed that the numbers beside the threads is the number of up/down-votes the original item received, and not the number of replies.* *EDIT 2: It would seem that tags don't stay if you edit a post, meaning you have to re-add all the tags if you edit a post. It would be nice to have some persistence there.* UserIdTAG: 248902 UserNameTAG: Chanute CreateTimeTAG: 2012-09-13T05:16:12Z VoteTAG: 8 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: I agree with you. It is a really nice work but I think it could be even better ! ; ) It would be nice if the threads of "forum/discussion" were placed in different sections (at least, the weeks). I think that now it is a bit "messy" (it is not easy to know which exercise / howework / lab the discussion is about if the author does not write it). FirstChildUserIdTAG: 396446 FirstChildUserNameTAG: RousseauxS FirstChildCreateTimeTAG: 2012-09-13T08:59:19Z FirstChildTAG: Chanute, I'm not a staff member or moderator, but I can tell you that, yes, this is the place to put your feedback about the forums and the course in general. You might want to tag your post with the "forum" tag. Also, I posted your comments in the old forum where we have been discussing the new forum. I credited you, I hope you don't mind. The staff is actively looking for feedback about the discussion forums, so everyone who has a thought about it, chime in. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-09-13T18:24:03Z SecondChildTAG: Hi JSChambers! Where is the old forum? Is it possible to access? SecondChildUserIdTAG: 248902 SecondChildUserNameTAG: Chanute SecondChildCreateTimeTAG: 2012-09-14T00:14:13Z IndexTAG: 285 TitleTAG: How to find the current in S6E1 When there is a non-linear device whose property depends on a hybrid function of current and voltage at the terminal pair how do calculate the current without assuming that the voltage current is a certain value? More specifically how do you know what value of resistance to use in S6E1 without guessing? UserIdTAG: 85490 UserNameTAG: yashsavani CreateTimeTAG: 2012-09-11T09:28:21Z VoteTAG: 8 CoursewareTAG: Week 3 / Analyzing Nonlinear Circuits Exercise CommentableIdTAG: 6002x_analyzing_nonlinear_circuits_exercise NumberOfReplyTAG: 3 FirstChildTAG: I didn't use any method that avoided trial and error, but since there are only two values of resistance to check for the strange element, it wasn't really a big deal. Once you have Rth and Vth, you get a very simple expression to check if the resistance value you assumed gives an appropriate current. FirstChildUserIdTAG: 248902 FirstChildUserNameTAG: Chanute FirstChildCreateTimeTAG: 2012-09-13T04:24:29Z SecondChildTAG: how know which R without using trial and error ? SecondChildUserIdTAG: 39685 SecondChildUserNameTAG: galeano SecondChildCreateTimeTAG: 2012-09-15T02:28:20Z FirstChildTAG: You can assume that Vx and Ix are in some range, then you solve linear equation system, and then you check if that assumption was correct. FirstChildUserIdTAG: 345655 FirstChildUserNameTAG: roginvs FirstChildCreateTimeTAG: 2012-09-15T18:35:55Z FirstChildTAG: It seems that there are many misleading ideas out here on how to solve this problem (S6E1 - Q3 & Q4). Here is the approach. 1. Find the Vth 2. Find the Rth 3. Find the short-circuit current (ith). That is the maximum current that can ever flow in the given network. 4. Now, if you hook up the non-linear device, what current range will the nonlinear device operate in? Since the non-linear device operates like a piecewise-linear resistor, will you expect the current to increase, decrease, or remain the same when you connect the device? Select the resistance of the non-linear device for that range. 5. Now, use the selected resistance to determine your current and voltage. So, in essence, you don't need trial and error method to solve this problem. FirstChildUserIdTAG: 88460 FirstChildUserNameTAG: ioo FirstChildCreateTimeTAG: 2012-09-27T05:18:33Z IndexTAG: 286 TitleTAG: How many loops in the circuit? Four or seven? UserIdTAG: 337791 UserNameTAG: Roman_T CreateTimeTAG: 2012-09-10T18:11:51Z VoteTAG: 8 CoursewareTAG: Week 1 / KVL and KCL CommentableIdTAG: 6002x_kvl_and_kcl NumberOfReplyTAG: 2 FirstChildTAG: There are seven loops. It is a mistake in the lecture. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-09-11T19:58:08Z SecondChildTAG: Somehow that makes me feel better. SecondChildUserIdTAG: undefined SecondChildUserNameTAG: undefined SecondChildCreateTimeTAG: 2012-09-12T10:10:31Z SecondChildTAG: i need another explanation please ..................................................................................................................... wht i know about loop analysis is that : ............................................................................................................... this circuit consists of 7 loops , but we dont use them all because we choos a specific number of loops equals ( b-n+1 ) not all the 7 loops . and we have ( b = 6 ) and ( n = 4 ) . so i think that we should have just 3 equations not 7 or 4 because ( 6 - 4 +1 = 3 ) . SecondChildUserIdTAG: 293070 SecondChildUserNameTAG: Makary SecondChildCreateTimeTAG: 2012-09-12T15:33:52Z FirstChildTAG: Seven loops. a - c - d - a // a - b - c -a // a - b -c -d -a // b - d - c - b // a - b - d -a // a - b - d - c - a // a - d - b - c - a And we have 4 nodes, so... How many KVL equations are independent? B-N+1, where B = branches, N = nodes... 6-4+1 = 3 FirstChildUserIdTAG: 300169 FirstChildUserNameTAG: castrogfx FirstChildCreateTimeTAG: 2012-09-10T18:44:41Z SecondChildTAG: And why in the last exercice its says 7 loops and now he says 4 loops? Is he mistaking LOOP by Indepentend KVL? SecondChildUserIdTAG: 250790 SecondChildUserNameTAG: angelobuoro SecondChildCreateTimeTAG: 2012-09-11T02:27:05Z IndexTAG: 287 TitleTAG: Bug in videos Just a little bug in Chrome (in Win7)... when you watch the videos in fullscreen, it keeps showing the menu of videos. They don't dissapear. UserIdTAG: 298653 UserNameTAG: WilyP CreateTimeTAG: 2012-09-08T05:10:31Z VoteTAG: 8 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 5 FirstChildTAG: I have the same problem in Chrome on a Mac. Just tested it in Firefox (the latest version?) and that also seems not to work. It also puts the captions window in a really annoying spot that blocks a big part of the video. FirstChildUserIdTAG: 270160 FirstChildUserNameTAG: Sethhhh FirstChildCreateTimeTAG: 2012-09-08T07:40:26Z SecondChildTAG: It seems to be the same on other platforms, the site needs a few adjustments there. Maybe they could move the "screen" to the left, then the total left over "black bar" area can be used for closed captions.(Full-screen mode) The video menu also needs to be moved higher up, out of the way. I think it would be handy to keep the video menu, then you don't have to exit full-screen to start the next movie. I am impressed with the overall integration of components on this website, very slick. Thanks for the opportunity! SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-08T13:42:44Z FirstChildTAG: Just watch it on YouTube, with HD on. You should have no problems. FirstChildUserIdTAG: 193033 FirstChildUserNameTAG: NIslam FirstChildCreateTimeTAG: 2012-09-08T15:45:04Z FirstChildTAG: Please add an option "video quality" like on the you tube. FirstChildUserIdTAG: 821 FirstChildUserNameTAG: mpaluch FirstChildCreateTimeTAG: 2012-09-08T15:51:03Z FirstChildTAG: set it on full screen, right click on it, inspect element look on list for .sequence-list-wrapper and just uncheck z-index or change value to 1 FirstChildUserIdTAG: 319524 FirstChildUserNameTAG: RadeEric FirstChildCreateTimeTAG: 2012-09-10T14:34:48Z FirstChildTAG: I also have another problem with the video, is that when I put it in full screen the botton of the video is cut. FirstChildUserIdTAG: 250241 FirstChildUserNameTAG: DaniHerrBere FirstChildCreateTimeTAG: 2012-09-10T23:16:01Z IndexTAG: 288 TitleTAG: video mistake Not (e1-e3)G3!! (e1-e2)G3 , right? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-07T13:38:42Z VoteTAG: 8 CoursewareTAG: Week 1 / Node analysis CommentableIdTAG: 6002x_node_analysis NumberOfReplyTAG: 1 FirstChildTAG: ye, also noticed, its a small error, to check if we are paying attention :) FirstChildUserIdTAG: 329254 FirstChildUserNameTAG: KGabor FirstChildCreateTimeTAG: 2012-09-07T14:26:49Z SecondChildTAG: Yes2x e3 is not present in the current circuit. SecondChildUserIdTAG: 367072 SecondChildUserNameTAG: jansenlopez SecondChildCreateTimeTAG: 2012-09-10T09:35:33Z SecondChildTAG: lol SecondChildUserIdTAG: 230787 SecondChildUserNameTAG: Gabriel007 SecondChildCreateTimeTAG: 2012-09-10T17:01:04Z IndexTAG: 289 TitleTAG: There are 7 loops Loop does not need to contain voltage source, there are loops: V R1 R2; V R1 R3 R5; V R4 R5; V R4 R3 R2; R4 R3 R1; R5 R2 R3; R4 R5 R2 R1; UserIdTAG: 376468 UserNameTAG: jovan_serbia CreateTimeTAG: 2012-09-07T12:35:04Z VoteTAG: 8 CoursewareTAG: Week 1 / Circuit topology CommentableIdTAG: 6002x_S2E1_Circuit_Topology NumberOfReplyTAG: 2 FirstChildTAG: Thanks exactly the last had not seen.!! FirstChildUserIdTAG: 256189 FirstChildUserNameTAG: MarquezMario FirstChildCreateTimeTAG: 2012-09-08T04:54:45Z FirstChildTAG: That is right FirstChildUserIdTAG: 290270 FirstChildUserNameTAG: Diebo FirstChildCreateTimeTAG: 2012-09-08T16:02:53Z IndexTAG: 290 TitleTAG: Mathematica formula for Q2. Integrate[((120*\[Sqrt]2*Cos[2*\[Pi]*60* t])^2)/110, {t, 0, 1}] UserIdTAG: 149441 UserNameTAG: vgain CreateTimeTAG: 2012-09-06T22:01:48Z VoteTAG: 8 CoursewareTAG: Week 1 / AC power CommentableIdTAG: 6002x_S1E3_AC_power NumberOfReplyTAG: 4 FirstChildTAG: Also consider Root Mean Square. 1/SQRT(2) FirstChildUserIdTAG: 88550 FirstChildUserNameTAG: Neil_S_Berry FirstChildCreateTimeTAG: 2012-09-06T23:34:53Z FirstChildTAG: Integrate[((120*[Sqrt(2)]*Cos[2*[Pi]60 t])^2)/110, {t, 0, 1}] seems more like it (note the parentheses around 2). Input this on Wolfram Alpha give 130.909. Thanks for the tip. FirstChildUserIdTAG: 184153 FirstChildUserNameTAG: tthngn FirstChildCreateTimeTAG: 2012-09-07T17:21:51Z SecondChildTAG: yes, when I copied and pasted from mathematica it did that for some reason. SecondChildUserIdTAG: 149441 SecondChildUserNameTAG: vgain SecondChildCreateTimeTAG: 2012-09-07T20:56:26Z FirstChildTAG: How is the (t) entered on here vs a engineering calculator? FirstChildUserIdTAG: 359302 FirstChildUserNameTAG: GaryG FirstChildCreateTimeTAG: 2012-09-11T23:16:37Z FirstChildTAG: scilab expression `p=((60*(120^2)*2)/110)*integrate('(cos(2*%pi*60*t))^2','t',0,1/60)` integrate=integration of an expression by quadrature FirstChildUserIdTAG: 322050 FirstChildUserNameTAG: ryio25 FirstChildCreateTimeTAG: 2012-09-17T08:34:00Z IndexTAG: 291 TitleTAG: Current negative For problem 3, the current on the Simulator showed negative, I checked it, and it seemed like it should be positive, but I wrote the negative value, and it was wrong. I don't know why it said that UserIdTAG: 155972 UserNameTAG: sreicks CreateTimeTAG: 2012-09-06T02:47:24Z VoteTAG: 8 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 2 FirstChildTAG: I got the same thing. I think whoever programmed this either made a mistake or is using electron current instead of the standard positive current. The Kirchoff problem follows positive voltage drops though. Update: I tried the circuit sandbox in the next section and got the same thing so I think it's programmed to show electron current but only with the dc source. If you put a current probe (in the sandbox, next section) in line with the positive terminal of the dc source and the first resistor, it gives the correct current and polarity. FirstChildUserIdTAG: 366581 FirstChildUserNameTAG: ishango FirstChildCreateTimeTAG: 2012-09-06T04:58:11Z SecondChildTAG: Good to know that it's only the DC source that does this, that might have tripped me up later. I'll have to be extra careful. Thanks. SecondChildUserIdTAG: 180365 SecondChildUserNameTAG: Opus_723 SecondChildCreateTimeTAG: 2012-09-06T07:13:15Z FirstChildTAG: The current is negative when enter on positive terminal. See the arrow?(->-) FirstChildUserIdTAG: 294330 FirstChildUserNameTAG: JHRMatos FirstChildCreateTimeTAG: 2012-09-06T02:53:14Z SecondChildTAG: So I guess the simulator shows actual current flow but the question is asking for current relative to the source which would from the + to the -. Hopefully this will be elucidated @ some point. SecondChildUserIdTAG: 30508 SecondChildUserNameTAG: tmac SecondChildCreateTimeTAG: 2012-09-06T04:36:50Z IndexTAG: 292 TitleTAG: Missing button My Lab0 diagram is missing the AC, DC buttons and I can't use it without them. What do I do? UserIdTAG: 308260 UserNameTAG: andreasvitikan CreateTimeTAG: 2012-09-05T14:03:45Z VoteTAG: 8 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 3 FirstChildTAG: I'm having exactly the same problem. FirstChildUserIdTAG: 120654 FirstChildUserNameTAG: John_O_Shea FirstChildCreateTimeTAG: 2012-09-05T14:11:25Z SecondChildTAG: hi guys,use it in googlechrome browser, this problem comes with me 2 but in mozilla browser. SecondChildUserIdTAG: 369448 SecondChildUserNameTAG: rohitkrsingh SecondChildCreateTimeTAG: 2012-09-05T14:16:20Z FirstChildTAG: I have the same bug too, maybe is Firefox issue. FirstChildUserIdTAG: 34026 FirstChildUserNameTAG: Albovolt FirstChildCreateTimeTAG: 2012-09-05T14:20:03Z SecondChildTAG: in firefox it is not showing all three buttons "DC, AC, TRAN", and in google chrome it is showing "DC and TRAN" but not AC button... SecondChildUserIdTAG: 259693 SecondChildUserNameTAG: MehrozKhan SecondChildCreateTimeTAG: 2012-09-05T22:08:09Z FirstChildTAG: This is how I fixed it. First click on "Circuit Sandbox Lab " from the left sidebar. Then draw a simple schematic and click the check button at the bottom. The DC and TRAN buttons appeared for me, where they belong. FirstChildUserIdTAG: 207456 FirstChildUserNameTAG: Octomos FirstChildCreateTimeTAG: 2012-09-05T14:25:04Z SecondChildTAG: Using Firefox, this also works very well for me, Octomos. Thanks a lot SecondChildUserIdTAG: 125727 SecondChildUserNameTAG: derMatze SecondChildCreateTimeTAG: 2012-09-05T20:50:42Z IndexTAG: 293 TitleTAG: Signs of v and i I don't understand how gets the voltage a negative sign, and how is the current entering the network from the resistor, positive. UserIdTAG: 156661 UserNameTAG: balazsb CreateTimeTAG: 2012-09-05T13:57:39Z VoteTAG: 8 CoursewareTAG: Week 1 / Power CommentableIdTAG: 6002x_S1E2_Power NumberOfReplyTAG: 4 FirstChildTAG: If you look from the side of the network **N** (*left to right in the circuit*) the direction of the current must be opposite of what it is shown as there is a positive polarity of the voltage at the node from which current is entering the network **N**. FirstChildUserIdTAG: 137748 FirstChildUserNameTAG: Prabhash_j FirstChildCreateTimeTAG: 2012-09-05T14:19:24Z SecondChildTAG: OMG,I finally got what you meant, because in actuality the current is flowing in the opposite direction, the + and - in the diag. got me a little confused. AND the arrow got me even more confused :) Thanks SecondChildUserIdTAG: undefined SecondChildUserNameTAG: undefined SecondChildCreateTimeTAG: 2012-09-05T20:19:19Z SecondChildTAG: Thanks for your answer, I think I got it. SecondChildUserIdTAG: 156661 SecondChildUserNameTAG: balazsb SecondChildCreateTimeTAG: 2012-09-06T04:15:08Z SecondChildTAG: yeah you are correct ....... SecondChildUserIdTAG: 99583 SecondChildUserNameTAG: sanjeevkm0912 SecondChildCreateTimeTAG: 2012-09-06T10:22:38Z FirstChildTAG: the resistor is actually dissipating the power in the form of heat, which means that the resistor is drawing the current from the network so the current sign is negative of what is given in the diagram FirstChildUserIdTAG: 124534 FirstChildUserNameTAG: srihari46 FirstChildCreateTimeTAG: 2012-09-05T14:32:08Z SecondChildTAG: Don't think this is correct since the resistor would dissipate heat in whatever direction the current runs through it. SecondChildUserIdTAG: 141235 SecondChildUserNameTAG: ilitzroth SecondChildCreateTimeTAG: 2012-09-05T17:57:33Z FirstChildTAG: Actually, neither the voltage or the current "has" to be negative, they just have to be opposite of one another based on how they're labelled in the diagram. If the current is travelling in the direction indicated by the arrow, the voltage would have to be negative because conventional current flows from positive voltage to negative voltage. If the current is travelling opposite the direction indicated by the arrow (negative current), the voltage will be positive by the same logic. Try it, both work in the answer box as long as the two are opposite sign. I feel like the question is a bit misleading asking what the current 'entering' the network N is, because the current entering is the same as the current leaving since there are no branches. FirstChildUserIdTAG: 248902 FirstChildUserNameTAG: Chanute FirstChildCreateTimeTAG: 2012-09-06T01:48:12Z FirstChildTAG: Re: Sign (direction) of current flow. When I work out these problems, I look at the small **arrow** that shows the **direction** of the current (i). I also look at the orientation of the power source or sink. A sink dissipates power, while a source creates it. Hence, resistors are sinks. Next, I look at the **polarity** of the element. Here the resistor's positive (+) terminal is on **top**, and current is flowing **out**. Because the element is a sink, **i = negative**. *(Note if we reversed the situation, and the element used was a **source**, i.e. a battery, and the direction of i in the schematic was leaving the battery, the sign of i = positive. This is opposite from what a resistor, a sink does. Also note that **the directions of the arrows and the polarities are arbitrary;** you are expected to find the proper sign regardless. The schematic could have just as easily had the arrow reversed, pointing into the (+) terminal of the resistor, instead of out of it, and in this reversed situation, the sign of i would also be positive. )* FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-09-14T18:05:08Z IndexTAG: 294 TitleTAG: Exclusive Facebook group for the students attending Edx 6.002x. Hey Hi Guys, Firstly please do accept my well wishes. I have created this group exclusively for the students attending the Edx 6.002x Circuits and Electronics. This group is to share ur doubts, queries, suggestions and whatever u feel like. Lets get together and start exploring. Cheers.:) Below is the link- http://www.facebook.com/groups/483354858342165/ UserIdTAG: 253902 UserNameTAG: sajalok CreateTimeTAG: 2012-09-05T12:46:01Z VoteTAG: 8 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 295 TitleTAG: Unresponsive Link Thank you Dave. Just to let you know the COURSEWARE button/link on the home/introduction page (https://www.edx.org/courses/MITx/6.002x/2012_Fall/about) doesn't seem to lead anywhere. UserIdTAG: 79337 UserNameTAG: AppliedImagination CreateTimeTAG: 2012-09-05T12:11:45Z VoteTAG: 8 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 5 FirstChildTAG: On it. :-) Good to see you again. FirstChildUserIdTAG: 11 FirstChildUserNameTAG: dormsbee FirstChildCreateTimeTAG: 2012-09-05T12:13:34Z SecondChildTAG: Likewise - it's been too long! :-) SecondChildUserIdTAG: 79337 SecondChildUserNameTAG: AppliedImagination SecondChildCreateTimeTAG: 2012-09-05T12:25:00Z FirstChildTAG: Glad to see you here AppliedImagination! :-) FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-09-05T12:43:20Z SecondChildTAG: You, too, ashwith. SecondChildUserIdTAG: 79337 SecondChildUserNameTAG: AppliedImagination SecondChildCreateTimeTAG: 2012-09-05T13:16:24Z FirstChildTAG: We're rolling out the fix now. You should see it in a few minutes if you don't already. Thank you. :-) FirstChildUserIdTAG: 11 FirstChildUserNameTAG: dormsbee FirstChildCreateTimeTAG: 2012-09-05T13:03:11Z FirstChildTAG: Yea, May be some browsers doesn't support the site. Try using Chrome. FirstChildUserIdTAG: 158817 FirstChildUserNameTAG: Shubhamtomar FirstChildCreateTimeTAG: 2012-09-05T12:14:51Z SecondChildTAG: Good point, Shubhamtomar! I should have mentioned I was using Chrome, shouldn't I? SecondChildUserIdTAG: 79337 SecondChildUserNameTAG: AppliedImagination SecondChildCreateTimeTAG: 2012-09-05T12:18:35Z FirstChildTAG: Gone through it just now. It is nice addition. FirstChildUserIdTAG: 119861 FirstChildUserNameTAG: narejo FirstChildCreateTimeTAG: 2012-09-08T08:36:40Z IndexTAG: 296 TitleTAG: What is 6.003z? Read here 6.002x is over (or would be over soon for those taking the proctored exam). Many of you maybe wondering "What next?". Well there are two logical follow up courses - Signals & Systems (6.003) and Digital Design (6.004). There are definitely others as well but these two courses were the most popular requests during the pilot Spring 2012 session of 6.002x. Some of us couldn't wait so we thought of using the forums as a study group. We planned to use the OCW course by Prof. Alan V. Oppenheim for our main lectures and used his textbook for reference. Prof. Agarwal was very kind to allow us to use the forums and Piotr offered a lot of advice. Amol then took it further by creating a full website for it. What started as a "study group" ended up as an unofficial course. We learned a lot in the process and learned things beyond basic signals analysis (image processing, music). So what exactly is 6.003z? It's a student run initiative where we help each other and study Signals and Systems. We use the OCW lectures as our main resource but some of us create the assignments/labs and tutorials. If you think you're up for it, you can contribute too! So if you're interested head over to http://6003z.amolbhave.in to sign up. You can also see that there is a new category in the forums called "SignalsGroup". Use this category while posting questions so that it will be easier to find relevant posts. If you'd like to help out with making tutorials/labs/assignments, you can post ideas here. UserIdTAG: 14386 UserNameTAG: ashwith CreateTimeTAG: 2013-02-03T17:45:05Z VoteTAG: 7 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_6003z NumberOfReplyTAG: 4 FirstChildTAG: GREAT JOB! Ashwith will you give me your e-mail id? FirstChildUserIdTAG: 857413 FirstChildUserNameTAG: Naveenkrishna FirstChildCreateTimeTAG: 2013-02-04T14:04:02Z FirstChildTAG: I am definitely interested, but what will the material of the class cover? FirstChildUserIdTAG: 219780 FirstChildUserNameTAG: ReconIII FirstChildCreateTimeTAG: 2013-02-04T14:42:35Z SecondChildTAG: this course covers all the topics from the Alan V. Oppenheim "signals and systems" text book. SecondChildUserIdTAG: 857413 SecondChildUserNameTAG: Naveenkrishna SecondChildCreateTimeTAG: 2013-02-04T15:08:58Z SecondChildTAG: what are the main topics? SecondChildUserIdTAG: 219780 SecondChildUserNameTAG: ReconIII SecondChildCreateTimeTAG: 2013-02-04T18:01:21Z SecondChildTAG: @ReconIII Have a look at this: http://ocw.mit.edu/resources/res-6-007-signals-and-systems-spring-2011/index.htm We are following that OCW course. SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2013-02-04T19:31:26Z FirstChildTAG: **Syllabus from 6003z 2012** ***Topics Covered*** Signals and systems Discrete-time (DT) systems Feedback, poles, and fundamental modes Continuous-time (CT) systems Laplace transform Z transform Relations between CT and DT Convolution Frequency response CT frequency response and Bode plots Feedback and control CT feedback and control CT Fourier representations CT Fourier series CT Fourier transform DT frequency representations DT Fourier representations Relations among Fourier representations Sampling Sampling and quantization Quantization and modulation Modulation Applications of 6.003 ***Description*** Signal processing plays an extremely important and continually growing role in a wide variety of engineering systems. Furthermore, technology and algorithms for signal processing continue to develop rapidly. While only a short time ago signal processing systems were predominantly analog, integrated circuit technology has made digital signal processing often preferable and more cost-effective. This course is an introduction to the basic concepts and theory of analog and digital signal processing. The background assumed is calculus, experience in manipulating complex numbers, and some exposure to differential equations. Prior exposure to the fundamentals of circuits for electrical engineers or fundamentals of dynamics for mechanical engineers is helpful but not essential. Both for pedagogical reasons and as a reflection of the nature of modern signal processing systems, the concepts associated with continuous-time and with discrete-time signals and systems are treated together in a closely coordinated way. Among other things, this approach emphasizes both the similarities and the differences in the two classes of systems. Developing this video course has been an extremely enjoyable and rewarding experience. I hope that you also find it enjoyable, stimulating, and rewarding. 6.003z covers the fundamentals of signal and system analysis, focusing on representations of discrete-time and continuous-time signals (singularity functions, complex exponentials and geometrics, Fourier representations, Laplace and Z transforms, sampling) and representations of linear, time-invariant systems (difference and differential equations, block diagrams, system functions, poles and zeros, convolution, impulse and step responses, frequency responses). Applications are drawn broadly from engineering and physics, including feedback and control, communications, and signal processing. ***Prerequisites*** 6.02 Introduction to EECS II or 6.002 Circuits and Electronics ***Textbook*** Amazon logo Oppenheim, Alan, and Alan Willsky. Signals and Systems. 2nd ed. Upper Saddle River, NJ: Prentice Hall, 1996. ISBN: 9780138147570. FirstChildUserIdTAG: 79885 FirstChildUserNameTAG: ruudoleo FirstChildCreateTimeTAG: 2013-02-06T08:17:52Z FirstChildTAG: this link also have the video lectures on "signals and systems" by IITD Prof. S.C. Dutta Roy.these lectures may also be useful for 6.003z students http://www.youtube.com/playlist?list=PLC6210462711083C4 FirstChildUserIdTAG: 857413 FirstChildUserNameTAG: Naveenkrishna FirstChildCreateTimeTAG: 2013-02-07T15:16:01Z IndexTAG: 297 TitleTAG: 6.003z: Signals and Systems Hi everyone, during the last run of MITx 6.002x course, a follow-up course of 6.003z: Signals and Systems was organised by the students. It was a self-study course where students helped students. The Spring '13 run of the course is going to be run soon. You can see the website at: http://6003z.amolbhave.in We use video lectures which are available from MIT OCW. There are weekly homework assignments and progress tracking. The tentative start date is going to be announced soon. Leave a message if you are interested and your thoughts. UserIdTAG: 118131 UserNameTAG: ammubhave CreateTimeTAG: 2013-01-22T16:10:20Z VoteTAG: 7 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 7 FirstChildTAG: Excellent work! :) I like it! Congratulations for the Course! It is awesome! I am sure that the students of 6.002x Fall will enjoy this as the 6.002x Spring 2012 did last term :) Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2013-01-22T20:50:53Z SecondChildTAG: Hi Myriam, you always attentive to the news. I'm so glad to communicate with you again!!! I have to organize myself, I want to do Electromagnetism. Lewin is awesome professor!!! I'm taking now "Image Analysis" in Coursera and I'm very interesting in Robotics too. It's brilliant the idea that between students, we can help together to learn other courses, organized by the students themselves! I suscribed to this course but I'm not sure if I will do it. I wish you rest something in previous weeks, although you're very active! My greetings to you from Montevideo, and I'd like to comunicate with you again. Hola Myriam, tu siempre atenta a las noticias!!! Que bueno comunicarme contigo nuevamente!!! Tengo que organizarme, quiero hacer Electromagnetismo. Lewin es un profesor impresionante!!! Estoy cursando ahora "Image Analysis" en Coursera y me interesa robótica también. Es brillante la idea de que entre estudiantes nos ayudemos para aprender otros cursos, organizados por los propios estudiantes!!! Me anoté en este curso pero aún no estoy seguro si podré hacerlo. Deseo que hayas descansado las semanas anteriores, aunque se que eres muy activa!!! Mis saludos para tí desde Montevideo y deseo que podamos seguir en contacto. Luis SecondChildUserIdTAG: 227223 SecondChildUserNameTAG: luisbonelli SecondChildCreateTimeTAG: 2013-01-22T22:00:46Z SecondChildTAG: Reposting this up here as well: Another request by us (the people who organized it the last time). Each week we plan to have some tutorial videos and other kind of community content which should augment the OCW videos. If you are already an electrical/electronics engineer who has seen this subject before, a mathematics major or someone who ends up understanding the content really well, you're encouraged to try making videos and write ups which others would find useful. This doesn't *have to* be directly related to signals and systems. We touched upon communications, quite a bit of image processing and tried to connect 6.002x with this subject. Your content can cover areas like these as well (Control systems is another connected field). You could do something purely in mathematics - something with would provide more insight into what's going on in all the equations. SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2013-01-28T15:26:26Z FirstChildTAG: This is an excellent idea. Congratulations! ammubhave I know that to organize an activity like this, requires a lot of time, your time, and time of others students too. I really appreciate your work, and thanks for sharing it with us. I'm organizing my time for this year, doing others courses too, so I'm not sure if I will be in this course but I'd like to be here, however I subscribed to it. Greetings for you Es una excelente idea, felicitaciones!!! ammubhave se que organizar una actividad como esta requiere mucho tiempo, tiempo que es tuyo y tiempo de otros estudiantes también. Realmente aprecio tu trabajo y muchas gracias por compartirlo con nosotros. Estoy organizando mi tiempo para este año, estoy en otros cursos también, entonces no es seguro si estaré en este curso, aunque me gustaría estar, no obstante, me inscribí en el mismo. Un saludo para tí. Luis FirstChildUserIdTAG: 227223 FirstChildUserNameTAG: luisbonelli FirstChildCreateTimeTAG: 2013-01-22T22:16:46Z FirstChildTAG: Hi ammubhave, I registered but don't know if i will be able to keep up as I have a busy period. Need some refreshing of Laplace and Z-transforms (besides the name I don't recall much about it). Hope some other 6.002x join in. The lecturer is good, perhaps I'll miss the green ticks and AHA's of AA. FirstChildUserIdTAG: 79885 FirstChildUserNameTAG: ruudoleo FirstChildCreateTimeTAG: 2013-01-22T22:30:18Z FirstChildTAG: I I have registered and hopefully I'll manage to do till the end. Can you please tell us how many weeks is this course going to last? FirstChildUserIdTAG: 372321 FirstChildUserNameTAG: EnricoDona FirstChildCreateTimeTAG: 2013-01-23T10:37:50Z SecondChildTAG: it is 13 week course SecondChildUserIdTAG: 118131 SecondChildUserNameTAG: ammubhave SecondChildCreateTimeTAG: 2013-01-23T13:19:41Z FirstChildTAG: Hi all, I registered for Fundamentals of Electrical Engineering https://www.coursera.org/course/eefun It cover some topics of 6.003z: Signals and Systems, if anyone wants to sign up, welcome. FirstChildUserIdTAG: 296419 FirstChildUserNameTAG: oscman FirstChildCreateTimeTAG: 2013-01-24T19:29:05Z FirstChildTAG: Hi All, In response to a request by Myriam, and in order to help you study 6.003, we have added the tag SignalsGroup to the discussion board. We hope that you find it useful. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2013-01-25T18:46:27Z SecondChildTAG: Thanks Lyla. I'll have to review those complex math calculations. Looks like there is not much interest yet(a lot of people already subscribed to other courses). SecondChildUserIdTAG: 79885 SecondChildUserNameTAG: ruudoleo SecondChildCreateTimeTAG: 2013-01-26T11:27:04Z FirstChildTAG: Another request by us (the people who organized it the last time). Each week we plan to have some tutorial videos and other kind of community content which should augment the OCW videos. If you are already an electrical/electronics engineer who has seen this subject before, a mathematics major or someone who ends up understanding the content really well, you're encouraged to try making videos and write ups which others would find useful. This doesn't *have to* be directly related to signals and systems. We touched upon communications, quite a bit of image processing and tried to connect 6.002x with this subject. Your content can cover areas like these as well (Control systems is another connected field). You could do something purely in mathematics - something with would provide more insight into what's going on in all the equations. FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2013-01-28T15:25:27Z IndexTAG: 298 TitleTAG: Congratulations! Congratulations to everyone who got their certificates! It's really an honor to be the proud holder of an official MIT(x) certificate. Thanks to everyone who made this course be a reality, edX and MITx staff, Community TAs, and the awesome Prof. Agarwal. This was really a spectacular course, and I really hope to take other similar courses soon. BTW although I'm a medicine student, I really had fun and found it really worth it to take this course. Best wishes for everyone! P.S.: I got 94% on my final grade and it appears so in the progress tab, yet on my dashboard it says 97%. Is it like someone is giving off some free marks at the end of the course?! :) UserIdTAG: 129876 UserNameTAG: msarabi95 CreateTimeTAG: 2013-01-03T17:15:04Z VoteTAG: 7 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 0 IndexTAG: 299 TitleTAG: QUESTION What certificates were these great inventer holding. Eg Bill Gates,Henry Ford and Thomas Edison. For those crying over the grades of the certificates should find out from these great men. It is not about the certificate , it is about application of knowledge that makes the difference. I know and I definately know that these are the people Prof Agarwal is trying to create. There are brilliant people who do not have access to education. Prof is making it so accessible that he will develop the hidden potentials of brilliant but needy students. I think we should start new inventions with this great knowledge. UserIdTAG: 446722 UserNameTAG: Richmond CreateTimeTAG: 2012-12-31T15:59:24Z VoteTAG: 7 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: I think that a course like 6002x has not any kind of practical application. FirstChildUserIdTAG: 27399 FirstChildUserNameTAG: juancho FirstChildCreateTimeTAG: 2012-12-31T16:32:03Z SecondChildTAG: "I think that a course like 6002x has not any kind of practical application." => You are telling this after completing the course? That's a big "L.O.L." moment. SecondChildUserIdTAG: 405473 SecondChildUserNameTAG: Avisec SecondChildCreateTimeTAG: 2012-12-31T16:37:51Z SecondChildTAG: actually he has completed the course twice. (Happy New Year juancho!) SecondChildUserIdTAG: 126825 SecondChildUserNameTAG: aphoenix SecondChildCreateTimeTAG: 2012-12-31T19:47:15Z SecondChildTAG: Hi brilliant aphoenix, many thanks, you are very kind. Happy new year and all the best for 2013. SecondChildUserIdTAG: 27399 SecondChildUserNameTAG: juancho SecondChildCreateTimeTAG: 2012-12-31T23:47:43Z SecondChildTAG: juanchoooo, Feliz Año amigo!:) Te aprecio un montón! Que la pases muy pero muy lindo! Un abrazo, Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-31T23:57:24Z SecondChildTAG: Hola adorada y bellísima Ingeniera Jefe, me hacen muy feliz tus palabras. Siento un gran cariño, admiraciòn y respeto por tì. Tu eres alguien sensacional y me siento muy afortunado al tener la oportunidad de disfrutar de tu existencia. `Deseo que seas muy feliz y que tengas muchos éxitos en el 2013 y en todos los años venideros. Un super-abrazo. SecondChildUserIdTAG: 27399 SecondChildUserNameTAG: juancho SecondChildCreateTimeTAG: 2013-01-01T01:04:21Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2013-01-01T01:30:55Z FirstChildTAG: Actually, a very rich person needs no certificate. He can just go into his own lab, and experiment, using his own funds to buy all the equipment for the work. This is how it was long long ago. People actually kept secret what they knew, and would only share their discoveries and ideas with a few others. Part of the reason was that the Church frowned on scientific inquiry. A certificate of knowledge might get you burned at the stake. But, the world changed. People are no longer rich enough to do today's experiments, for the most part, anyway. It takes lots of funding, and convincing other people to invest their cash into your ideas. This is where things like certificates become important. The investors want to see some verified proof that you know what you're talking about. Having other people who are believed to know their stuff, recommend you, by issuing you with a certificate of achievement, is one way to help convince those picky investors. Well, that why the world needs certificates of some sort today. But, tomorrow things will change. There will be no such thing as certificates, diplomas, BA, MBA, PhD, etc..it will all be about what you currently know, not what you knew yesterday. Those investors will give you their own test. It's all possible now, with this online technology, and computer ed. A company seeking to hire a qualified candidate doesn't have to look at your certificate anymore. There should be no discrimination, anyway, where you got your degree. Right now, a doctor from China, immigrating to Canada, can't practice in Canada. It doesn't matter if he went to the top Chinese University, and graduated, Suma Suma Com Lada, etc..he still has to submit himself to being tested and examined by the Canadian standard medical system. Of course, given his knowledge, he should be able to fly through the requirements and requalify without issue. But, the employer still requires that "retesting" to occur. The same will be for all professions. The job applicant will simply sit the Corporate Exams for the particular Company he wishes to work at, and what he knows at the time of his application will be the critical factor. It's easy to do this today. The job seeker sits at a computer terminal, in the corporate headquarters, and punches some keys, and gets graded automatically. His grade is then sent on and collected by a software program that collates all the applicant scores, and ranks them, picking the best candidates for the position, for further interviewing. Whenever you apply for a computer programming job today, there's always that little test, they give you upfront, to verify you still have it, and haven't' forgotten it all by then. That's because they need to put programmers into production right away, and don't usually want to be training a person, just because he is supposed to be smart. Everything will be just like this, but all automated. It may be that you could even take the test from you own home, and if obtaining a high enough score, then you get to visit the co. to take the verifying test under supervision. The academic certificate is going the way old stock certificates and bond certificates went long ago, they are to become bookkeeping entries in some electronic data-center. Eventually, there will be "fractions" of a certificate, as the idea of diplomas will be broken down into smaller pieces, and people will collect "certificate fractions" the same way they collect money today. An academically rich person will be one that has collected many certificate fractions by taking courses all around the world, at academic institutions, and at various large corporations, and his "resume" will read like his "bank statement", showing his current academic wealth. Just as you show your bank statement, when applying for a loan, you'll deliver your academic statement when you apply for a job, or for investors to put money into your venture. People will continue to collect academic certificate fractions all through their life, just like they do with money today, learning will be lifelong, until the day you die. A person's "academic balance" will measure his potential to achieve things, while his "bank balance" will determine his power to put that into practice. One will be like "potential energy" and the other "kinetic energy". By academics we gain potential, and by money we gain kinetics. That's the way it will be, man, that's the wave of the future. FirstChildUserIdTAG: 4463 FirstChildUserNameTAG: pmj FirstChildCreateTimeTAG: 2012-12-31T17:34:53Z SecondChildTAG: **Very good post, pmj!** SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2013-01-02T18:26:30Z FirstChildTAG: To give one more little poem from the old viking poem Eddan *Wise shall he seem who well can question, And also answer well; Nothing is concealed that men may say, Among the sons of men.* A certificate can give you the possibility, but at the end it's what you can show that counts. A lot of development is still made by small groups and in many cases all your knowledge counts. In a lot of service another jobs knowledge of electronics is important. If a unit breaks down and you can make it work until new parts arrive the savings can be great. You are to blame if anything go wrong, so the risk is there. Those jobs can only be done by people with knowledge. FirstChildUserIdTAG: 151472 FirstChildUserNameTAG: Stensmed FirstChildCreateTimeTAG: 2013-01-01T13:02:15Z SecondChildTAG: Nice use of Viking poetry in your post (and earlier one), **Stensmed**. It broadened my horizons as I've never read such before. One is wise when he can tie in the old, and connect it with the modern. Ancient poetry has as much relevance today as it did then; brave men sailing through the unknown and fearing for their lives in the vicious storms of the North Atlantic had the same strong feelings as we do in today's uncertain economic world; where natural forces still cause widespread devastation and take life and treasure, and where human exploration of the unknown (i.e. in the depths of the ocean, and at the edges of the atmosphere and in space) still inspires us. Jersey Mark SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2013-01-02T18:41:46Z IndexTAG: 300 TitleTAG: Thank you Hoo!! We have successfully completed the MIT course. On behalf of everyone I thank EDX to give us this wonderful opportunity to strengthen our basics of electronics. It is great time for us in this course. We enjoyed a lot and learnt the electronics practically. We thank every staffs of 6.002x and Edx for giving us such a incredible course. We are looking forward for the next course and we like to end up our MITx electronic courses by making a device like a iPod or a Mobile Phone. UserIdTAG: 362299 UserNameTAG: anandbaskaran CreateTimeTAG: 2012-12-26T02:52:07Z VoteTAG: 7 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 301 TitleTAG: Proctored Exam: A few points. **All students** wondering about the **Proctored Exam**, such as **deepkar**: EdX was planning for a proctored exam and was partnering with [Pearson VUE][1] as outlined in this article in the [*Harvard Crimson*][2]; I have not heard anything official, though. The rumor I heard was that it would be for Fall 2012 6.002x, but I guess that there was not enough time to implement the logistics. I can think of many obstacles in order to break through the logistics barrier; **For example:** - How much would the exam cost? - What would be offered: A certificate, generic college credit for a course in circuit analysis, or specific credit for MIT's 6.002 course? - If credit would be issued, what institution would offer this credit? MIT? Or would edX apply for status to grant university credit? Note that this is a long difficult process in the U.S., and involves both state (Massachusetts) and Federal regulatory bodies and licensing. - Would U.S. university [accrediting bodies][3] such as the New England Association of Schools and Colleges and [ABET][4] (the Accreditation Board for Engineering & Technology) [recognize such credit][5] (or even credit from online education in general)? - Would edX be required (or would they seek) [recognition from the U.S. Department of Education][6]? If so, could Dept. of Education financial aid (i.e. grants and student loans) be used to cover expenses for the exam? So many unanswered questions; and edX was trying to resolve this in one semester. Of course more time is needed, in my opinion. It depends on the goal. Offering a **proctored certificate** is easy and faces less of a hurdle and can be implemented in months. However, offering **college credit** (in my opinion the only way for Massive open-enrollment online courses, MOOCs, to succeed in the long term and revolutionize global education) faces a greater bureaucratic hurdle and will take years to implement. Jersey Mark (***Note:** The above post does not reflect the views or official policies of edX, MITx, their affiliates, sponsors, etc. It is offered as my personal opinion concerning proctored exams that **may or may not** be offered in this course or any others, and should not be construed as a guarantee of this or any other online course.*) [1]: http://www.pearsonvue.com/programs/ [2]: http://www.thecrimson.com/article/2012/9/7/edx-offer-proctored-exams/ [3]: http://en.wikipedia.org/wiki/Educational_accreditation [4]: http://www.abet.org/ [5]: http://cihe.neasc.org/standards_policies/standards/standards_html_version [6]: http://www2.ed.gov/about/offices/list/ope/library.html UserIdTAG: 292546 UserNameTAG: JerseyMark CreateTimeTAG: 2012-12-25T18:10:03Z VoteTAG: 7 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: I love your posts JerseyMark! Good job! Happy holidays! :) FirstChildUserIdTAG: 304587 FirstChildUserNameTAG: Vasso FirstChildCreateTimeTAG: 2012-12-26T16:39:10Z FirstChildTAG: thanks for the elaboration. A good deal of time is definitely needed. Most important of all this is the pioneering year making the current Edx students/learners Edx pioneers. Lets give it time to "simmer". FirstChildUserIdTAG: 183507 FirstChildUserNameTAG: obiradaniel FirstChildCreateTimeTAG: 2012-12-25T21:53:17Z SecondChildTAG: This is no pioneering at all.There is coursera also. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-12-25T23:47:15Z SecondChildTAG: **AlexAlexandrescu**, I did not mean **pioneering in MOOCS**, I meant being the **second Edx students for 6.002x** and **pioneers in 6.00x,3.091x, CS169.1x and all other courses offered for the first time on Edx.** SecondChildUserIdTAG: 183507 SecondChildUserNameTAG: obiradaniel SecondChildCreateTimeTAG: 2012-12-26T07:49:11Z FirstChildTAG: Well, i wonder what decision the stuff will take, when there will be a "ambiguity" with the final exam and all of the students will start to beg for more chances, even if one of the subject is an allready done homework (H8P1) that ANYBODY could copy/paste the answers from. How can anybody from a education department or an employer can trust the certificates ? After what just happened + the fact that NOBODY will answer to a tutorial poorly explained , i really believe that a true certificate based on merit at an international level is an UTOPIA, at least at this level . I am really starting to believe that this is an INVESTMENT from their part, because no doctor will leave his department just to lead a free non profit organization ,and that the initiative will be on all news channels, if you catch my meaning. My advice : learn what you can and be thankful, as long as it is still free. There were already open courseware for the ones who want to learn . FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-12-26T00:00:28Z FirstChildTAG: The whole point of the MOOCs is to liberate education from the academic mafia. Why do you insist on going back? FirstChildUserIdTAG: 623210 FirstChildUserNameTAG: preveen FirstChildCreateTimeTAG: 2012-12-26T05:10:15Z SecondChildTAG: Academic mafia, isn't that a little harsh? I mean, if people from academia don't teach us, who will? They are the ones that have spent years and thousands of dollars studying and grasping the material. You also have to remember, teachers and professors have lives just like we do and have to make money too. The reason MOOC's are and will be successful is because of the volume of people it involves. Eventually Edx is going to charge for these courses, however the amount per person will be low, since there are so many of us. One other thing to remember, MOOC's aren't free either, someone is paying for them, somewhere, most likely from donations or government assistance. And you know where government gets their money! - Us. Happy New Year. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-12-26T07:24:24Z FirstChildTAG: I think an exam purpose is to make a distinction between students based on their exam degree. To produce a distribution of students based their degree, e.g. a form of histogram. For this purpose, no examination "rule" or "method" can have any inherent value or meaning by itself, or the particular problems given. The only thing that matters is to ensure all rules were applied to all equally without any possible distinction. **As long a certain exam produce appropriate acceptable "histogram", then it doesnt really matter in detail what was the exam procedure, as long as all rules were applied without distinction to each individual.** Nobody should ask for different exam type. When you ask for that, you actually ask for different set of rules for yourself or a grp of ppls. For one, the first purpose of an exam is not to accommodate grps of ppls in particular, but quite the opposite, to show the differences. Second, if you get population to split in two exams, then both parts results nullify each other and neither exam can be considered valid. I dont mean to be harsh, but I would say that those who cant understand this are more creatures of habit than of intelect. Btw, perhaps with the descriptive statistics of this course we can get that final degree distribution per se, e.g. graphically, coz its the image who is worth 1000 words and not vice versa. One image says all about the quality of the exam. FirstChildUserIdTAG: 331483 FirstChildUserNameTAG: Doru FirstChildCreateTimeTAG: 2012-12-26T08:37:08Z IndexTAG: 302 TitleTAG: The revolution has begun!! I found the edx initiative as being the real, unexpected, worldwide revolution of our century. It will certainly affect our thinking on education and learning for the next decades. OUR professor, Anant Agarwal will stand as one of the most important persons in history. I truly believe so! And i am honestly proud for being his student! UserIdTAG: 206669 UserNameTAG: dimitrios66 CreateTimeTAG: 2012-12-24T15:31:47Z VoteTAG: 7 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Yes I do believe the same. FirstChildUserIdTAG: 237462 FirstChildUserNameTAG: Amritkar FirstChildCreateTimeTAG: 2012-12-24T18:12:57Z FirstChildTAG: Yes, I think Prof. Agarwal, Prof. Thrun (Udacity) and Salman Kahn (Kahn Academy) are leaders in the Massive Open Online Courses. It's not only the format however, but especially the teachers that make this a big success. If I had teachers like these during my study, I'm sure I would have done MUCH better. These gents teach in way that makes the student enthusiastic and they explain everything so incredibly well. The only thing that I'm afraid of is that, as there are more and more MOOCs popping up, not all the teachers will live up to standards set by these gentlemen. FirstChildUserIdTAG: 114881 FirstChildUserNameTAG: xvink FirstChildCreateTimeTAG: 2012-12-24T19:31:22Z IndexTAG: 303 TitleTAG: Request to the course staff: Hi, Firstly, i would like to thank the edX team and the course instructors for the wonderful opportunity provided for us, and the impeccable way in which the course was organized. I would like to *request the course staff to provide solutions to the midterm and the final exam*, of course after everyone's final exam is over - because there are so many different ways of looking at the same problem, that although we might get the same answer in the end, our solutions might actually differ in many a ways. So, i feel it would really help us to know different ways of solving a problem. It could help us to think of different ways of approaching a problem(may be a real-world problem) in the future. UserIdTAG: 155793 UserNameTAG: Anil15 CreateTimeTAG: 2012-12-24T11:22:45Z VoteTAG: 7 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 304 TitleTAG: Thank you! - from Spain 6.002x team, I would like to say thank you to all of you guys! You created an incredible and magic environment, and I meant it. After several years studying at university, I have taken many courses but no one like this. The way you prepare the exercises, the labs, the tools you gave us, Dr.Agarwal's lectures, Myrimit's hints, etc. I think that you are doing a great job trying to put all this online. For many years, online education was expecting for academic investments, and by investments I am referring to human investments, teachers (their time and patience), programmers,.. Before this, there were only a couple of guys, making videos for youtube, explaining exercises, doing a great job, but it wasn't enough to give online education a strong impulse and get in touch with society. However, after this course, the entire academic world is focusing on online education. Most of the top universities are preparing courses, and offering their teachers very interesting tools to give their students the best resources. And I think that all of this is happening because of you. It is time to revolution, time to break the out of phase system, time to wake up the accommodated teachers and tell them that this is the time for people who love their job. They had defects but they weren't interested, now they have a mirror to look at. With this course, you made university classes accessible to everybody interested, and definitely, you encouraged us to keep learning new stuff with all the references to electronical devices. It has been a pleasure to share this time with you, good luck and thank you! Cannot wait for 6.003x or other electromechanical courses!! Un saludo, Rai, Spain UserIdTAG: 7057 UserNameTAG: raiabril CreateTimeTAG: 2012-12-24T11:02:19Z VoteTAG: 7 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Good luck to you too Rai! Un abrazo y te deseo lo mejor de lo mejor de aquí en más :) Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-24T14:56:26Z FirstChildTAG: I subscribe to everything said. Thanks a lot for this great opportunity Kind regards! Jaime (Spain) FirstChildUserIdTAG: 74074 FirstChildUserNameTAG: JCanton FirstChildCreateTimeTAG: 2012-12-24T15:06:23Z FirstChildTAG: I cannot wait until **6.003x (Signals and Systems)** either! Also, I believe that **6.002x** after polishing out the "rough spots" has the possibility to become even better! If all the suggestions we made throughtout this Fall semester are implemented, future students would have an even better experience! This course has a shot of becoming the **gold standard** for massive open-registration online courseware! FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-12-25T18:38:37Z IndexTAG: 305 TitleTAG: Внезапно кириллица ! (Cyrillic Suddenly!) Привет русскоязычным студентам! Привет всем, кто дошел до финального экзамена, кто решает его в данный момент и кто уже закончил! Как на счет устроить оффлайн-встречу по городам и отметить успешное окончание, посмотреть друг на друга и поделиться впечатлениями? Я из Москвы. --------------------------- Translation made by Sergtronix --- Hi to all Russian speaking students! Hi to all those who are before the final exam, trying to pass and who are passed it already! What is about "off-line" meeting in cities, we could see each other and to share our experiences about finished course? I'm from Moscow And by the way, who from what university? Who does what? Who is EE-student or is gray-haired programmer, a young doctor or who is elderly Circuit-engineer? I graduated a year ago MAI (Moscow Aviation Institute = State University of Aerospace Technologies), Faculty of electronics, now I'm doing projects on the FPGA (digital communication, DSP, interfaces). Went to this course to refresh knowledges that I got before; as a result it became clear that I do need to learn the new again instead to refresh ones. I am very glad that came through and got involved in all this. Me and my colleague are discussing involved in the following courses, we are looking at the "Statistics" and "AI". My email bwmuscat@gmail.com ----- UserIdTAG: 345464 UserNameTAG: Constantine_ru CreateTimeTAG: 2012-12-22T12:44:56Z VoteTAG: 7 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 20 FirstChildTAG: Политех, СПб, 1-й курс. Прошел курс ради интереса и за компанию с соседом по общежитию :) FirstChildUserIdTAG: 277739 FirstChildUserNameTAG: eclipsevl FirstChildCreateTimeTAG: 2012-12-30T00:15:14Z FirstChildTAG: И кстати, кто из какого ВУЗа? Кто чем занимается? Кто студент-электроншик, а кто седой программист, кто молодой врач, а кто пожилой схемотехник? Я год назад закончил МАИ, фак-т радиоэлектроники, сейчас делаю проекты на ПЛИСах. На курсы пошел освежить знания из института, в результате выяснилось, что многое надо не освежать, а учить по новой. Очень рад, что дошел до конца и вообще ввязался в это. Вот обсуждаем с коллегой участие в следующих курсах, смотрим на статистику и АИ. FirstChildUserIdTAG: 345464 FirstChildUserNameTAG: Constantine_ru FirstChildCreateTimeTAG: 2012-12-22T12:47:35Z SecondChildTAG: Constantine, it seems to be to incorrect, related to another students, write post by Russian only. With best regards, Сергей SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-22T13:01:31Z SecondChildTAG: My writing and talking English is rather poor, so if can translate correctly all of this text, I'll edit my topic :-) SecondChildUserIdTAG: 345464 SecondChildUserNameTAG: Constantine_ru SecondChildCreateTimeTAG: 2012-12-22T13:58:54Z SecondChildTAG: It will be pretty hard :) SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-22T14:03:23Z SecondChildTAG: What city are you from? SecondChildUserIdTAG: 345464 SecondChildUserNameTAG: Constantine_ru SecondChildCreateTimeTAG: 2012-12-22T14:37:03Z SecondChildTAG: If you want I will delete my post with translation SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-22T15:24:21Z SecondChildTAG: Updated But it seems to be useless post SecondChildUserIdTAG: 345464 SecondChildUserNameTAG: Constantine_ru SecondChildCreateTimeTAG: 2012-12-22T16:02:35Z SecondChildTAG: It is possible to meet more students from exUSSR here, we should wait IMHO. SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-22T18:08:28Z SecondChildTAG: Im from Kharkov, Ukraine. SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-22T20:18:17Z SecondChildTAG: HeHe - Kharkov is a very good city - educational center of Ukraine! I am from Kharkov also. SecondChildUserIdTAG: 261860 SecondChildUserNameTAG: Oleksander SecondChildCreateTimeTAG: 2012-12-22T20:42:39Z SecondChildTAG: And you still didn't write your email :-) SecondChildUserIdTAG: 345464 SecondChildUserNameTAG: Constantine_ru SecondChildCreateTimeTAG: 2012-12-22T20:50:13Z SecondChildTAG: I do use CPLDs in my projects somethimes :) Ok, I will write you email soon.. Serge SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-22T21:46:00Z SecondChildTAG: Best regards from Sevastopol. :) SecondChildUserIdTAG: 295103 SecondChildUserNameTAG: Syavick SecondChildCreateTimeTAG: 2012-12-24T22:08:38Z SecondChildTAG: :) And my regards to this glorious city! SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2013-01-06T17:01:18Z FirstChildTAG: Hello, I am from Vladivostok, Russia. This far from Moscow. Привет я из Владивосток-а, поэтому тоже весело провести время без меня. Так что с Новым Годом. Берклеевские курсы по крайней мере по SAAS гораздо хуже организованные. Много усилий, но мало отдачи, материалы для обучения вообще никакие, нужно постоянно носиться по Интернет-у без гарантии, что успеешь. На вторую часть я даже плюнул, не стал заворачиваться, времени убил массу практически без отдачи. Потом отдельно пройду и возьму сертификат. По химии неплохой курс, но там тоже проблема с материалами, на самом деле лекции и материалы отличаются и нужно смотреть по лекциям, иначе не решишь, это отнимает много времени. По Питону вариант видимо неплохой. Посмотрю, в качестве обновления знаний. Владимир Если что пишите на e-mail FirstChildUserIdTAG: 402617 FirstChildUserNameTAG: Vl FirstChildCreateTimeTAG: 2012-12-22T17:58:47Z SecondChildTAG: а где взять e-mail ? :) SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-22T18:06:40Z SecondChildTAG: sorokinv@hotmail.com sorokinboba@gmail.com This my e-mail-s, можно по этим е-мэйлить напропалую. Владимир SecondChildUserIdTAG: 402617 SecondChildUserNameTAG: Vl SecondChildCreateTimeTAG: 2012-12-22T18:25:40Z SecondChildTAG: I'm an engeener in digital communication, so I learned these topics in my institure courses. Some of materials were new for me, but most of all I learned earler, and almost forget :-) So it's interesting for to compare two schools. The next text in russia is about it, because I think I can't rigth translate it in englih Мне вообще интересно услышать впечатления от тех, кто обучался в России по этим курсам. У нас этот материал читался бОльшей частью в курсе "цепи и сигналы" (анализ цепей, переходные процессы, фильтры), частично в схемотехнике (операционники, простые усилители), частью в электронике (КМОП транзисторы, диоды). Поскольку данный курс вводный, то нельзя прямо оценить, что у нас было лучше или хуже. Например по фильтрам у них явно есть отдельный курс (мы считали фильтры в Матлабе, курсовая большая была). Явно ничего не сказано про спектры (у нас про спектры говорили еще в самом начале). Так же например задачи на node-method мы решали через матричные уравнения (курсовая), а не через алгебраические преобразования, тут это показано как "дополнительный материал" и вряд ли они вернуться к этому в дальнейшем. С другой стороны после наших курсов схемотехники практически ничего не оставалось в голове. Тут надеюсь останется больше. Так же радует практическая направленность, не сферические шары с вакууме, а реальные схемы, реальные задачи. Ну и качество подачи материала конечно несравнимо. У нас в институте преподавателей такого уровня за 5 лет я могу пересчитать по 1 руке. Работаю по специальности уже несколько лет, могу сказать, что тут реальные рабочие задачи рассматривались куда чаще, чем в наших курсах. Если меня большей частью учили делать электронику 80ых, то тут учат делать современную электронику. К сожалению даже на таких базовых вещах разница уже заметна. Возможно имеет смысл как то структурировать все выше сказанное и услышать мнение студентов других вузов. В любом случае надеюсь привлечь нынешних студентов и преподавателей к этому блестящему проекту, для получения великолепного опыта. --------------------- Best regards to all edx-team. You have made a great job, you help people to become clever, you give people one of the most expensive thing - knowledge, you are connecting countries and certain engineers whole the world. I have no words to explain all my feelings, you are wizzards of our technology days. SecondChildUserIdTAG: 345464 SecondChildUserNameTAG: Constantine_ru SecondChildCreateTimeTAG: 2012-12-22T19:54:35Z SecondChildTAG: Yes, you rigth. Good thing to refresh memories. Да, большую часть из этого мы проходили на ТОЭ (теоретические основы электротехники), но практическая направленность, а так же приминение диффур к физическим процессам - порадовало. Ну и естессно демонстрации (бензопила и цифра, и многие другие) вне всяких похвал ). SecondChildUserIdTAG: 394836 SecondChildUserNameTAG: v2g6ch4 SecondChildCreateTimeTAG: 2012-12-22T20:09:58Z SecondChildTAG: По поводу опытов смешанные чувства. Мне кажется, в танцах в очках и в разбиении компьютерной мышки было больше эпатажа, чем практической ценности. Но с другой стороны это конечно же фан, пресловутый разрыв шаблона, якоря и другие вещи, помогающие оставить хоть что то в голове. Мне особенно понравился стенд, где можно крутить параметры усилителя, менять рабочую точку и слушать искажения звуков. Вот то, что все время так и глядели в зеленые картинки на С1-61, вместо того, чтобы хоть раз пропустить через усилитель музыку из плеера, это большое упущение. На 5 курсе собирал такие штуки в матлабе just for fun, жаль что этого не было на младших курсах. ------ I think that some part of demonstation, like breaking of computer mouse and dancing was more targeting to shock people. But they still were usefull to fix some monents in out memmory, "A big WOW moment" :-) And I liked model of amplifer, were we could set parameters and listen music distortions. It amazing, one of the most interesting part for me, In my institute I did it manual in the older courses myself, and here it's part of beginning course. SecondChildUserIdTAG: 345464 SecondChildUserNameTAG: Constantine_ru SecondChildCreateTimeTAG: 2012-12-22T20:35:11Z FirstChildTAG: I dont know why, but my previous post was deleted. Чтобы выйти на связь - маякните на V2g6cH4@mail.ru ) FirstChildUserIdTAG: 394836 FirstChildUserNameTAG: v2g6ch4 FirstChildCreateTimeTAG: 2012-12-22T19:36:28Z SecondChildTAG: Because the English translation contained profanity. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-12-22T19:38:23Z SecondChildTAG: Where is your tolerance? Author wrote, that he watnt to meet with other persons in Russia and discuss cource. Also we want to compare quality and themes of this course with russian educational standarts. Now please tell me, why i need to translate all my messages into english? There are rule or something? If no - leave us alone ) SecondChildUserIdTAG: 394836 SecondChildUserNameTAG: v2g6ch4 SecondChildCreateTimeTAG: 2012-12-22T20:06:46Z SecondChildTAG: Pennypacker, I can tell you that the current thread does not contains any profanity in Russian yet SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-22T20:13:48Z SecondChildTAG: As I said the ***English translation*** of the word чертов is quite shocking, perhaps this was not the intention, nonetheless, it is easier to remove the post and let v2g6ch4 rephrase it. Thank you for your understanding. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-12-22T20:25:02Z SecondChildTAG: v2g6ch4, I think that Sergtronix was rigth and it would be polite for us to translate although themes of our talking to English. And I think it's roughly to tell someone of edx-team "leave us alone". SecondChildUserIdTAG: 345464 SecondChildUserNameTAG: Constantine_ru SecondChildCreateTimeTAG: 2012-12-22T20:27:15Z SecondChildTAG: May be Pennypacker and Constantine_ru is right. English as English SecondChildUserIdTAG: 402617 SecondChildUserNameTAG: Vl SecondChildCreateTimeTAG: 2012-12-22T20:33:26Z SecondChildTAG: In russian language word чертов (I hope you won't delete me comment for it) isn't shocking anyone. You can see it in one of the most popular bussines newpaper http://www.vedomosti.ru/search/?order=date&s=%D7%E5%F0%F2%EE%E2 SecondChildUserIdTAG: 345464 SecondChildUserNameTAG: Constantine_ru SecondChildCreateTimeTAG: 2012-12-22T20:45:14Z SecondChildTAG: Господа, давайте воздержимся от резких высказываний.То, что для нас обыденно и просто, может быть совершенно непонятно для людей других культур.И к слову, если есть большое желание, можно создать ветку или в скайпе, или на фейсбуке и тп, и там обсуждать что пожелаем. С уважением, Сергей. SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-22T21:40:51Z SecondChildTAG: Hi v2g6ch4, As CTAs we have some moderations responsabilities, although we still students like eveyone from here...we don´t have any privileges, also we are contributing free in this amazing iniciative. edX don´t give us a speech to say as CTA, we are just students with the extra responsability to keep a good student environment but anything else that makes us different from the rest... My apologies if the word that you wrote wasn´t what we thought that it was...I hope that you can understand that as CTA we make mistakes too and that we have the responsability of removing, editing things that we believe that are innapropiate - I don´t speak Russian, I wish I could, I always found interesting your language haha. But if I take that word to google translator it returns me an innapropiate word in English language...sorry if that wasn´t what you have mean by context...it happened also to me with one word in Spanish a few days in the old forum of 6.002x haha, luckily xp42 warned me about that and corrected me- but I hope that you understand that it is anything personal with you...and that you are free to write in your native language... I understand you, it happened to me frecuently, sometimes I write in Spanish out of the blue and then I say: Oops, I should write in english too, might the data that I am writting can help to others and so on. So I back and edit and write in english most of the times or viceversa...So, I understand you totally that translating all the things that you say in your native language might can be tedious...but believe me that I did it most of the time during this Fall...I could only have choosen to write only in Spanish but then I realized that if I wrote in English too I could help more students, not only to the Hispanic Community... That took me a double work this Fall but believe me that I were really happy to do that, I also speak and write in Japanese and speak a little bit of portuguese, believe me if I were confident writting Posts in that languages too I would have done that this Fall haha. Of Course that anyone of edX requested me to translate my Spanish Posts or Comments, they never told me that. So feel free to Post in your native languague if you want. Take care and my best wish to you, Greetings to my Russians Friends haha! Merry Christams and Happy New Year! Myriam. P.D. Just for curious v2g6ch4. How do you say Merry Christmas and Happy New Year in Russian?:P SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-22T21:59:10Z SecondChildTAG: Happy New Year -С Новым Годом SecondChildUserIdTAG: 402617 SecondChildUserNameTAG: Vl SecondChildCreateTimeTAG: 2012-12-22T22:11:38Z SecondChildTAG: Thank you VI :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-22T22:16:40Z SecondChildTAG: Sergtronix, выяснится, что у кого нет фейсбука, кто то скайп удалил, ЖЖ не было, вкотакте не регистируюсь и так далее. Почта и рега тут это то, что точно есть у каждого. Что касается корректности, испанцы вон в течении семестра между собой активно обсуждали домашки, выкладывали пояснения на испанском и вообще ничего не переводили. Мы же не обсуждаем здесь других людей, не нарушаем honor code. Не вижу ничего дурного устроить обсуждение в разделе discussion, кратко комментируя ее содержание на английском для интересующихся. Myrimit http://translate.google.ru/#en/ru/Happy%20new%20year http://translate.google.ru/#en/ru/Merry%20xmas In Russia New Year is more popular, because of USSR was atheistic country, so the was no celebration of Christmas at all. Nowadays everything change, but most of people give the gifts and make great celebration in New Year, and Christmas is more religeos date. SecondChildUserIdTAG: 345464 SecondChildUserNameTAG: Constantine_ru SecondChildCreateTimeTAG: 2012-12-22T22:18:26Z SecondChildTAG: Thank you for the data Constantine_ru :P, sorry for my ignorance, it happened me frecuently haha, I didn´t knew that - that is why I like this online worlwide Course because I can learn with the friendly people from here a lot of their Cultures and Countries- I really appreciate your kind explanation, thank you Constantine_ru. С Новым Годом!:) and greetings to your family and your friends in Russia! Take Care, Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-22T22:25:31Z SecondChildTAG: Let me say my opinion. None from EDX-team or CTA should explain why this or other post was moderated. But while here students from the entire world with different cultures ,we all here should remember that something may be understood in the wrong way, so we should use our sentences very carefully. Do you want to know about our greetings?Ok, the XMass is not common for non catolic Russia..but we DO LOVE New Year! I can tell you more - we have second New Year, it is OLD NEW YEAR (Jan, 14) :D ...(This is sample of difference with understanding). Our greetings are : "С Новым Годом!" or "Поздравляем с Новым Годом!" Marry Christmass Myriam! SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-22T22:29:19Z SecondChildTAG: Oops... Of course **MERRY CHRISTMASS**!!! :) SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-22T22:44:03Z SecondChildTAG: Serge, Thank you for explaining me more about the Russian Traditions too, I really enjoy knowing about other Cultures, sometimes I feel so ignorant haha. Cool Serge, I didn´t knew that of the second New Year, wow!:). Here in my Country, Argentina, we celebrate a lot december 24th night, also the 31th nigh haha. I always watch in TV news in the 31th - january 1st : how other Countries Celebrate it, and I always enjoy seeng that , all that artificial fire works, they are amazing, I will focus on Russia this time while watching the TV news in january 1st haha, now, I will remember of my Russian friends of 6.002x haha. Thank you Serge С Новым Годом for you! I wish you the best! Myriam. P.D. Don´t drink too much in the toast of december 31th celebration haha XD. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-22T22:53:16Z SecondChildTAG: I think you may search web cams from the Russia.Im sure it will be possible to find live cameras and to see New Year pictures.. [Kharkov's Liberty Square][1] sixth from the largest Europe's squares.. Im sorry, you cant see me there in this time :) [1]: http://velton.ua/ru/webcams/Wide1.shtml SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-22T23:04:52Z SecondChildTAG: Wow! That is cool, I will watch that camera haha! Thank you for sharing that! Will you be there in december 31th? SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-22T23:12:42Z SecondChildTAG: Im going to celebrate New Year with my family at home..Hmm..may be we will go to this square, but I'm not sure :) because AFTER toasts I will not drive car. So do walk 30 minutes in one side - I dont like this idea yet :) SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-22T23:25:59Z SecondChildTAG: Haha, yes it must be cold there. I could see that on that web cam link. A lot of snow. Better that you stay at home with calefaction haha, I would do that if here would be not summer at the end of the Year haha. Here we used to dinner outside home, in the garden, at open air and we make the taost at midnight 24th and 31th december, haha, so odd. For me this is normal since childhood haha. I can not imagine if it were cold :p SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-23T17:49:08Z SecondChildTAG: Hi Serge, now I am watching that web page, I can see many people in that place! Is it already 2013 in Russia? Here it is not yet haha. Happy 2013 for all of you! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-31T23:55:19Z SecondChildTAG: On all Russian language land is a new 2013 year SecondChildUserIdTAG: 402617 SecondChildUserNameTAG: Vl SecondChildCreateTimeTAG: 2013-01-01T01:45:48Z SecondChildTAG: Hi, Myriam! Yeah, I met New Year at Home, with part of my family :) As I told before, I was lazy to go to the square, but we saw fireworks from this place.It was pretty cool :) Serge SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2013-01-03T12:27:14Z FirstChildTAG: Hi everybody! Nice course! For me it was a rather good new experience, in spite of the fact i'm familiar with physics, but not with electronics. Привет, народ. Я из Москвы. Вообще я аспирант-физик, специализируюсь на нелинейной оптике. В электронике никогда не разбирался, поскольку в университете у нас была фундаментальная направленность, и грузили в основном суровой математикой и физикой. Инженерной направленности не было совсем. Потому было интересно попробовать. Всегда хотелось понимать принципы работы всяких электронных устройств, но почитать руки как-то не доходили. Хочется отметить, что в универе подобной подачи материала не было. Никогда реально не объяснялось, как, то, что мы изучаем в данный момент, будет применяться на практике и когда оно вообще понадобиться. Типа нате вам курс диффуров, интуров, линала (который реально только в квантах на полную используется) разгребайте сами. Конечно, когда они потом повылазили в последующих курсах, уже не так страшно было. Но вот это ощущение, что вся система преподавания материала построена как-то неверно, меня не покидает до сих пор. В общем умение понятно объяснять - это большая редкость. В этом курсе мне понравилось ясность изложения. Практически все понятно было сразу. Просто и красиво. Ну и вообще было бы круто, если в дальнейшем все курсы можно было бы осваивать в дистанционном режиме. Извините за сумбур. FirstChildUserIdTAG: 345452 FirstChildUserNameTAG: ale-yoman FirstChildCreateTimeTAG: 2012-12-22T20:59:47Z SecondChildTAG: Забыл добавить про вуз. Закончил физфак МГУ. Собиираюсь еще курс по химии послушать, поскольку остро ощущаю недостаток знаний в этой области. В годы моего обучение на физфаке химии не было вообще. Позже вроде ввели небольшой семистровый курс. Кто знает, сохранится ли доступ к этому курсу в январе и далее? Заходить иногда посмотреть кое-что. SecondChildUserIdTAG: 345452 SecondChildUserNameTAG: ale-yoman SecondChildCreateTimeTAG: 2012-12-22T21:05:45Z SecondChildTAG: Aren't you from MSU? I've some friends there. А не из МГУ? А то есть много знакомых с физфака и сам часто в районе МГУ бываю, в горный турклуб хожу. Даже нас инженеров, грубых ремесленников и низменных практикиков, все равно учат как теоретиков. При чем как плохих теоретиков. В итоге выходят специалисты в сдавании экзаменов и в знании отдельных учебников. SecondChildUserIdTAG: 345464 SecondChildUserNameTAG: Constantine_ru SecondChildCreateTimeTAG: 2012-12-22T21:08:53Z SecondChildTAG: Yes, i finished MSU, but now working in other institute. Физфак закончил в 2008. Сейчас в институте общей физики диссер пишу. В районе МГУ я бываю в последнее время только на день физика :) SecondChildUserIdTAG: 345452 SecondChildUserNameTAG: ale-yoman SecondChildCreateTimeTAG: 2012-12-22T21:14:57Z SecondChildTAG: Егор Юшков, Саша Новоселов? SecondChildUserIdTAG: 345464 SecondChildUserNameTAG: Constantine_ru SecondChildCreateTimeTAG: 2012-12-22T21:16:45Z SecondChildTAG: Сам курс выложен в ютубе, кто-то выкладывал даже плейлист составленый из роликов ютуба. Закроют ли доступ к этому мне тоже интересно, но они уже весенний объявили курс, ну а так впринципе тут еще доступен он пока.. https://6002x.mitx.mit.edu/courseware/ SecondChildUserIdTAG: 113561 SecondChildUserNameTAG: Illogical SecondChildCreateTimeTAG: 2012-12-22T21:17:04Z SecondChildTAG: Как закроют? Я планировал последние 2 недели досматривать в январе и доступ к вики, к базе задачек разве все это пропадет? SecondChildUserIdTAG: 345464 SecondChildUserNameTAG: Constantine_ru SecondChildCreateTimeTAG: 2012-12-22T21:22:52Z SecondChildTAG: нет, не знаком с ними. Надеюсь, что не закроют. Я специально на химию зарегистрировался заранее, чтобы можно было не дожидаться начала повтора в феврале. Надеюсь, моя комбинация сработает. И в этом курсе я тоже еще не досмотрел последние две недели. SecondChildUserIdTAG: 345452 SecondChildUserNameTAG: ale-yoman SecondChildCreateTimeTAG: 2012-12-22T21:43:31Z FirstChildTAG: Привет всем! Я студентка 5 курса физфака МГУ кафедры общей физики и волновых процессов. Решила пройти этот курс, потому что, как уже было сказано выше, фундаментальные курсы на факультете мало связаны с инженерным делом. Хотя я и прошла курс программирования микроконтроллеров(это было что-то вроде факультатива, но не совсем), но все равно была каша в голове. Хотелось систематизировать все знания, а вообще, многое уже конечно было известно, поэтому курс дался мне относительно легко. Очень понравилось, как организован материал, как он преподносится. Как бы мне хотелось, чтобы и у нас преподаватели так старались! FirstChildUserIdTAG: 129687 FirstChildUserNameTAG: Galiya91 FirstChildCreateTimeTAG: 2012-12-23T10:41:22Z SecondChildTAG: Галина, Вы проходили курс программирования микроконтроллеров в универе ?Или это было онлайн обучение ? SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-23T11:27:39Z SecondChildTAG: Галина,Ооо. Волна! Я кафедру радиофизики заканчивал. Сейчас ее переименовали в кафедру фотоники и свч. Или как-то так. А что за курс по микроконтроллерам? SecondChildUserIdTAG: 345452 SecondChildUserNameTAG: ale-yoman SecondChildCreateTimeTAG: 2012-12-23T14:28:39Z SecondChildTAG: Курс по программированию микроконтроллеров у меня был вместо стандартного программирования на 2м курсе и как дополнительная нагрузка на 3м, где мы уже делали электрические схемы и разводили платы. Я была семестр в группе 32-разрядных микроконтроллеров, а потом изучила 8-разрядные и на них сделала устройство для лаборатории в качестве курсовой работы по микроконтроллерам. Так же еще была возможность изучать вместо этого ПЛИС, но как-то не нашлось у нас практического применения этой технологии. Самая свежая новость - люди, которые организовали этот курс, открыли инженерное направление на физфаке, туда будут брать всего несколько человек, а в дипломе у них будет написано "инженер". SecondChildUserIdTAG: 129687 SecondChildUserNameTAG: Galiya91 SecondChildCreateTimeTAG: 2012-12-23T15:34:05Z SecondChildTAG: инженер- это плохо ? :) SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-23T16:11:20Z SecondChildTAG: ПЛИСы очень специфичная штука. Есть отрасли, где ПЛИСы сейчас вытеснили сигнальные процессоры и ASIC, но эти отрасли очень и очень узкие. Сам недавно менял работу, рынок труда маленький, но и специалистов очень мало. Короче много вопросов, о том, что будет с ПЛИС дальше. Сам временами думаю мигрировать либо в сторону 32битных процессоров либо совсем на верхние уровни программирования. SecondChildUserIdTAG: 345464 SecondChildUserNameTAG: Constantine_ru SecondChildCreateTimeTAG: 2012-12-23T17:07:28Z SecondChildTAG: Разве из моего поста следовало, что инженер это плохо? Как раз наоборот. Я вот интересуюсь этой тематикой, поэтому и прошла данный курс, поэтому радуюсь тому, что на физфаке хоть что-то предлагают для интересующихся. А ПЛИСы просто очень мощная штука, их сложно программировать и далеко не все задачи могут быть распараллелены. В большинстве случаев хватает обычных контроллеров. SecondChildUserIdTAG: 129687 SecondChildUserNameTAG: Galiya91 SecondChildCreateTimeTAG: 2012-12-23T17:21:48Z SecondChildTAG: я просто уточнил насчет "инженера" :) тк у самого диплом "инженер-конструктор-технолог" ,5 лет ВУЗа , правда еще some doctorate courses имеют место быть, те учился в аспирантуре.Но это было оччень давно :) SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-23T17:31:32Z SecondChildTAG: Привет, Галя! Давно уже где-то жалею, что не было у меня в мыслях к вам на физфак идти в свое время. Хорошо у вас там и интересно. Если бы мог сейчас, пнул бы себя тогдашнего, чтобы кончал фигней заниматься и поступал туда) SecondChildUserIdTAG: 447663 SecondChildUserNameTAG: brigadakuznetsov SecondChildCreateTimeTAG: 2012-12-26T15:46:56Z SecondChildTAG: >>Если бы мог сейчас, пнул бы себя тогдашнего, чтобы кончал фигней заниматься и поступал туда Плюс много :-( SecondChildUserIdTAG: 345464 SecondChildUserNameTAG: Constantine_ru SecondChildCreateTimeTAG: 2012-12-27T08:26:25Z SecondChildTAG: brigadakuznetsov, привет! А ты в итоге куда пошел? SecondChildUserIdTAG: 129687 SecondChildUserNameTAG: Galiya91 SecondChildCreateTimeTAG: 2012-12-27T19:02:08Z SecondChildTAG: а все, вижу пост. Но разве МАИ не ближе к инженерной специальности, чем физфак? Мы учим столько всего, что мне кажется потом вообще не пригодится... SecondChildUserIdTAG: 129687 SecondChildUserNameTAG: Galiya91 SecondChildCreateTimeTAG: 2012-12-27T19:04:30Z SecondChildTAG: Да ближе, но мне сейчас наука в целом и физика в частности больше интересна, чем более узкие инженерные специализации. Кроме того, физфаковское образование же, насколько я понимаю, может быть базой для всего вообще. То есть, потом я смог бы податься в любую заинтересовавшую меня область. Конечно, сложно это всё. Но и дико интересно вместе с тем. SecondChildUserIdTAG: 447663 SecondChildUserNameTAG: brigadakuznetsov SecondChildCreateTimeTAG: 2012-12-30T19:40:58Z FirstChildTAG: Приветствую! Я, как и @Constantine_ru, заканчиваю факультет радиоэлектроники МАИ, дописываю диплом. Поэтому, кстати, в последние недели уделял курсам чуть меньше внимания, чем нужно. Решил подтянуть знания, сравнить методы преподавания, узнать новое (а цепи я в свое время по некоторым причинам учил не очень). К слову, хоть английский знаю и не очень, в 6.002х всё для меня едва ли не понятнее преподносится, чем у нас. Ну и, как Константин уже писал, материал преподносится с более современной точки зрения и более наглядно. P.S. Привет, BWMuscat. Может, помнишь, kia я. Мы, порой, пересекались во всех этих интернетах. Если что, crowd.of.smiths@gmail.com :) FirstChildUserIdTAG: 447663 FirstChildUserNameTAG: brigadakuznetsov FirstChildCreateTimeTAG: 2012-12-23T11:21:20Z SecondChildTAG: Ох, интернет большая деревня :-) Напомни, ты с какой кафедры? Я писал диплом у Важенина, он очень хороший преподаватель, сейчас на 408 кафедре хорошее оборудование, я сам писал лабу для студентов по моделированию декодера Витерби в Matlab. Так же я точно помню, что Латышев (РТЦиС) делал отсылки на MIT-курс. Думаю надо предложить им как то внедрять идею прохождения этого курса в умы мотивированных студентов. SecondChildUserIdTAG: 345464 SecondChildUserNameTAG: Constantine_ru SecondChildCreateTimeTAG: 2012-12-23T16:56:36Z SecondChildTAG: Я с 403, пишу там с Лавровым. Вообще говоря, не самый плохой вариант, учитывая, кого нам предлагали на кафедре. В консультанты взял на предприятии грамотного человека. Хотел попробовать га тему "комплекс бортовой аппаратуры предотвращения столкновений спутников", но меня ещё в сентябре все дружно отговорили. В итоге всё свелось к довольно тривиальному комплексу исследования и оптимизации алгоритмов БПЛА (на Веге работаю, там это любят, вот и навязали). Впрочем, в итоге все равно есть, что делать. Сейчас вот думаю, что делать со всеми этими 170 листами пояснительной записки, надо сокращать. Цепи у нас вел Хачикян. Вел хорошо, но ничего такого не предлагал, при мне, по крайней мере. Я тоже считаю, что надо активнее заинтересовывать народ. Наши кафедры 403 и 404 в совсем пассивны в этом плане, хотя сейчас то, при желании, даже без особых средств можно набрать шикарную современную базу для обучения. SecondChildUserIdTAG: 447663 SecondChildUserNameTAG: brigadakuznetsov SecondChildCreateTimeTAG: 2012-12-23T17:11:25Z SecondChildTAG: Привет от 101-й (2000 г.) ! Не очень профильно, но это не мешает... :) А очень даже помогает. SecondChildUserIdTAG: 345502 SecondChildUserNameTAG: ElectroVova SecondChildCreateTimeTAG: 2012-12-24T13:13:00Z SecondChildTAG: Приветствую! Сложно тут судить, профильно или нет. Схемотехника для очень многих технических специальностей необходима, мне кажется) SecondChildUserIdTAG: 447663 SecondChildUserNameTAG: brigadakuznetsov SecondChildCreateTimeTAG: 2012-12-26T15:48:56Z SecondChildTAG: Это на 100% верно. А мне сейчас- уж и подавно. Сейчас пытаюсь скрестить 2 увлечения и профильное образование. Пробую свои силы в области авионики для СЛА. SecondChildUserIdTAG: 345502 SecondChildUserNameTAG: ElectroVova SecondChildCreateTimeTAG: 2012-12-27T10:32:51Z FirstChildTAG: Дорогие друзья, конечно, Я желаю вам счастливого Рождества и счастливого 2013 года, в этих пространствах показано, что блок причине вне политических границ и языка, я инженер-электрик и работы с медицинским оборудованием, многие привет из Колумбии ... FirstChildUserIdTAG: 459662 FirstChildUserNameTAG: FRANCISCOG FirstChildCreateTimeTAG: 2012-12-23T01:26:43Z SecondChildTAG: Hello, Franciscog! Thank you :) but in Russia Christmas is not widely celebrated. It is common to celebrate New Year here. In my turn i would like to wish you a Merry Christmas. Best regards. SecondChildUserIdTAG: 345452 SecondChildUserNameTAG: ale-yoman SecondChildCreateTimeTAG: 2012-12-23T08:52:29Z SecondChildTAG: Merry Christmas, have a nice year! SecondChildUserIdTAG: 345464 SecondChildUserNameTAG: Constantine_ru SecondChildCreateTimeTAG: 2012-12-23T17:01:38Z FirstChildTAG: http://www.youtube.com/watch?v=rp8hvyjZWHs FirstChildUserIdTAG: 394836 FirstChildUserNameTAG: v2g6ch4 FirstChildCreateTimeTAG: 2012-12-23T15:13:18Z SecondChildTAG: http://www.youtube.com/watch?v=j0ASj8R7wp4 - Can he live a normal life ? ... - No, he'll be an engineer SecondChildUserIdTAG: 345464 SecondChildUserNameTAG: Constantine_ru SecondChildCreateTimeTAG: 2012-12-23T17:04:16Z FirstChildTAG: Кстати забыл посоветовать, если хотите оставить лекции на будущее, купите книгу. Книга очень хорошая. Практически все собрано в одном месте, в том числе и необходимая математика. Страниц 700, толстая, с разобранными примерами. В принципе лекции, при наличии книги, не нужны. Только покупайте на amazone, она стоит около 70 USD плюс доставка. Доставляют за 3-5 дней. У нас на ozon-е, такая же книга, стоит 6000-7000 рублей и обещают доставить за 30 дней. By the way forgot to advise, if you want to leave a lecture on the future, buy a book. The book is very good. Virtually all in one place, including the necessary mathematics. 700 pages, thick, with a stripped down example. Basically the lecture, if the book is not needed. Just buy at amazone, it costs about 70 USD plus shipping. Delivered in 3-5 days. We in the ozon-e, the same book is 6000-7000 rubles and promise to deliver within 30 days. Владимир FirstChildUserIdTAG: 402617 FirstChildUserNameTAG: Vl FirstChildCreateTimeTAG: 2012-12-23T14:05:58Z SecondChildTAG: У меня книга в пдф формате.Кстати, если покупать на амазоне, то не забудьте про скидку! SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-23T15:16:46Z SecondChildTAG: My Amazon is trying to buy the book for the third or fourth time I have. All the while increasing the price. I do not know what it was. But I wrote this book in honor Demidomivichi. У меня амазон пытается купить книгу уже третий-четвертый раз у меня. Все время повышает цену. Даже не знаю в чем дело. Но я эту книгу записал в почетные Демидомивичи. VL SecondChildUserIdTAG: 402617 SecondChildUserNameTAG: Vl SecondChildCreateTimeTAG: 2013-01-06T18:31:32Z FirstChildTAG: По поводу книги - я готов ее купить. С учетом скидки выходит 35$, для книги такого уровня это вообще не деньги. Но доставка в РФ будет стоить порядка 1000р. Если наберется несколько человек из Москвы, плюс например те, кто бывает в Москве наездами, то можно собраться и заказать книгу всем вместе. В любом случае буду делать заказ в конце января (когда наша почта выпадет из коматоза) FirstChildUserIdTAG: 345464 FirstChildUserNameTAG: Constantine_ru FirstChildCreateTimeTAG: 2012-12-23T16:59:36Z SecondChildTAG: Я бы тоже прикупил. Соберешься заказывать - скинь на мыло, пожалуйста. karpo.ab@yandex.ru. У меня есть в электронном виде, но в печатном не помешает. SecondChildUserIdTAG: 345452 SecondChildUserNameTAG: ale-yoman SecondChildCreateTimeTAG: 2012-12-23T20:17:06Z SecondChildTAG: Как вы получаете со скидкой цену $35? Вижу цену 88,39 http://www.amazon.com/s/ref=nb_sb_ss_i_2_14?url=search-alias%3Daps&field-keywords=foundation%20of%20analog%20and%20digital%20electronic%20circuits SecondChildUserIdTAG: 325197 SecondChildUserNameTAG: Vitali_Jerin SecondChildCreateTimeTAG: 2012-12-28T09:14:23Z FirstChildTAG: General course in my estimation difficult for the average level of training. Somewhere on the second or third year of good technical institute or University. At least for the time when I was studying. But I learned a long time. For sure it requires a stable understanding of mathematical knowledge and skills in the following disciplines: ability to solve linear equations (linear algebra), familiarity with complex variables, solution of differential equations, knowledge of integral and differential calculus. In any case, we must be able to count. One desire and enthusiasm is not enough. I do not know how the rest of the people get out, but after regular school - is difficult. I think that all of Latin America with this course should Mirimitu samba to dance for his help and support. This is only my opinion. Вообще курс по моим оценкам сложный для среднего уровня подготовки. Где то на уровне второго-третьего курса хорошего технического института или государственного университета. По крайней мере для того времени, когда я обучался. А учился я давно. Для его уверенного понимания требуются устойчивые математические знания и навыки по следующим дисциплинам: умения решать линейные уравнения (линейная алгебра), навыки работы с комплексными переменными, решение дифференциальных уравнений, знания интегрального и дифференциального исчисления. На всякий случай нужно уметь считать. Одного желания и энтузиазма недостаточно. Я не знаю как остальной народ выкручивался, но после обычной школы - это трудновато. Я думаю, что вся Латинская Америка с этого курса должна Миримиту самбу сплясать за его помощь и поддержку. Но это лично мое мнение. Владимир FirstChildUserIdTAG: 402617 FirstChildUserNameTAG: Vl FirstChildCreateTimeTAG: 2012-12-24T03:53:39Z SecondChildTAG: Мне казалось, что в самом начале было прямо сказано, что требуются базовые знания по линейке, дифурам и комплексным числам и что без знания оных будет туговато. Что меня кстати удивило, так это то, что линейные уравнения предлагается решать как в 7 классе методом выражения переменных. Уже когда система 3 порядка (а такие задачки были и в midterm и в final), куда проще представить уравнение в матричной форме и решить в каком нибудь... Ну у меня маткад, но я думаю есть и фрифарный софт. Ну и в принципе для метода узловых потенциалов можно прямо составлять матричное уравнение. http://www-inst.eecs.berkeley.edu/~ee100/fa04/lecture_notes/EE100supplementarynoteweek03-2.pdf SecondChildUserIdTAG: 345464 SecondChildUserNameTAG: Constantine_ru SecondChildCreateTimeTAG: 2012-12-24T04:39:15Z SecondChildTAG: Понятно, что так быстрее. Но у людей здесь разный уровень подготовки. И для наглядности выводили именно так. Да что уж, если даже в тьюториалз ошибки математические были. SecondChildUserIdTAG: 345452 SecondChildUserNameTAG: ale-yoman SecondChildCreateTimeTAG: 2012-12-24T07:31:27Z SecondChildTAG: Myrimit - молодая девушка, и ее вовлеченность в процесс и понимание меня просто потрясли.Побольше бы таких студентов. SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-24T21:14:33Z SecondChildTAG: Да, удивительно. Я не думал, что это девушка. Обычно девушки не носятся наперевес с паяльником. Yes, amazing. I did not think it was a girl. Usually girls do not rush at the ready with a soldering iron. SecondChildUserIdTAG: 402617 SecondChildUserNameTAG: Vl SecondChildCreateTimeTAG: 2012-12-24T23:43:22Z SecondChildTAG: Most interesting is that Mirimit tracks and courses in chemistry 3.091x. At least I met her and comments on this course. Самое интересное, что МИРИМИТ также отслеживает и курс по химии 3.091х. По крайней мере я встретил ее комментарии и на этом курсе. VL SecondChildUserIdTAG: 402617 SecondChildUserNameTAG: Vl SecondChildCreateTimeTAG: 2013-01-06T18:49:23Z FirstChildTAG: Я понимаю, что результаты сдачи курса станут известны после праздников? Вообще существует практика публикации статистики? Я на других курсах этой практики не увидел, только косвенные оценки. Что слышно о монетизации курсов в новом году? I understand that the results of delivery of the course will be announced after the holidays? In general, there is a practice the publication of statistics? I'm on another course of this practice have not seen, only indirect estimates. What heard on the monetization of course in the new year? Vladimir FirstChildUserIdTAG: 402617 FirstChildUserNameTAG: Vl FirstChildCreateTimeTAG: 2012-12-24T18:08:30Z SecondChildTAG: так результаты же уже видны в Progress, разве нет? А дипломы обещали "до нового года". А вот статистику и правда хорошо бы посмотреть. Особенно интересно увидеть ее в виде анонимной таблицы "участник N - результаты за домашки-лабы-экзамены". Можно строить всякие зависимости интересные, например самая трудная домашка, или корреляцию оценок за домашки и лабы и результаты экзаменов. Про монетизацию я так понимаю в ближайшем году все останется бесплатным. Думаю, что для монетизации надо делать авторизованные оффлайновые центры сдачи экзаменов, что сразу повысит ценность диплома с уровня "бумажки" до официального документа. SecondChildUserIdTAG: 345464 SecondChildUserNameTAG: Constantine_ru SecondChildCreateTimeTAG: 2012-12-24T21:49:15Z SecondChildTAG: Yes, the results, I can see, I'm interested how many people were. How many people left after the exam. What is the result for the exams for other people. I saw the stats on Habre on results for spring 2012. Maybe the results will be available for Fall 2012. Да, свои результаты, я вижу, меня интересует сколько людей было. Сколько людей осталось после сдачи экзаменов. Каков результат по сдаче экзаменов для остальных людей. Я видел статистику на хабре по результатам за весну 2012. Может быть будут доступны результаты за осень 2012. Vladimir SecondChildUserIdTAG: 402617 SecondChildUserNameTAG: Vl SecondChildCreateTimeTAG: 2012-12-24T23:37:31Z SecondChildTAG: На 6.00 Computer science etc... публиковали статистику по результатам промежуточных экзаменов. Насчет общей статистики пока не знаю -- курс еще не закончился. SecondChildUserIdTAG: 97796 SecondChildUserNameTAG: meps SecondChildCreateTimeTAG: 2012-12-25T15:44:08Z FirstChildTAG: Всем ку! В ветке по ссылке ниже народ оставляет свои метки на карте. https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50d914bbef2ecd2b0000002a Касательно курса. Здорово. Наглядно. У Профессора Anant Agarwal явный талант к преподаванию. Уровень тоже серьезный для начального курса. Интересно было бы увидеть продолжение. Всех с Наступающим! P.S. Моя alma mater Московский Институт Электронной Техники Выпуск 1992. It was amazing and extremely fun course as prof. Anant Agarwal promised. Hope see new courses by prof. Anant Agarwal next year. Happy New Year! FirstChildUserIdTAG: 325197 FirstChildUserNameTAG: Vitali_Jerin FirstChildCreateTimeTAG: 2012-12-25T21:30:47Z FirstChildTAG: Народ, айда на 6.00х. Мне очень понравилось, дошел до первого Мидтерма, сдал на 91%, но потом пришлось выбирать между работой и учебой и оставить только один курс- 6.002х, а программинг принести в жертву работе. Записался по-новой, с 4-го февраля. Лектор зачетный первую часть ведет.. Рекомендую. In this post I invite our classmates to join 6.00x from 4th of Feb, 'cause I've already tried it and found it very interesting. FirstChildUserIdTAG: 345502 FirstChildUserNameTAG: ElectroVova FirstChildCreateTimeTAG: 2012-12-27T10:45:40Z SecondChildTAG: Спасибо за совет. Тоже думал взять еще один курс от MITX Thanks for the advice. I think to join to the next MITX cource next year. SecondChildUserIdTAG: 325197 SecondChildUserNameTAG: Vitali_Jerin SecondChildCreateTimeTAG: 2012-12-27T21:31:13Z FirstChildTAG: Мы с коллегой идем на статистику. FirstChildUserIdTAG: 345464 FirstChildUserNameTAG: Constantine_ru FirstChildCreateTimeTAG: 2012-12-28T06:36:33Z SecondChildTAG: I dont know yet, I want to take three courses. I would like 6.00, quantum computing, statistics, and to finish the second part of the course Saas. But it is quite difficult, time-consuming. Incidentally courses Mits I liked more than in Berkeley. They are more organized and are saturated. But this is my personal opinion. Я пока не знаю, хочу взять три курса. Хотелось бы 6.00, квантовые вычисления, статистику и закончить Сас курс вторую часть. Но это достаточно трудно, занимает много времени. Кстати курсы в Митс мне понравились больше, чем в беркли. Они более организованы и насыщены. Но это мое личное мнение. Vladimir SecondChildUserIdTAG: 402617 SecondChildUserNameTAG: Vl SecondChildCreateTimeTAG: 2012-12-28T20:31:59Z SecondChildTAG: ИИ не? я начал осенью, но на второй неделе понял, что не в этот раз. Времени не хватало катастрофически. SecondChildUserIdTAG: 447663 SecondChildUserNameTAG: brigadakuznetsov SecondChildCreateTimeTAG: 2012-12-30T19:35:18Z FirstChildTAG: Привет всем русскоязычным студентам! Я из Челябинска, Южно-Уральский Государственный университет (ЮУрГУ), Информационно-измерительная техника. Кто-нибудь знает, есть ли продолжение этого курса на более узкие направления в электронике? (например, силовая электроника, или оптоэлектроника) FirstChildUserIdTAG: 209041 FirstChildUserNameTAG: SmartEngine FirstChildCreateTimeTAG: 2012-12-28T11:04:08Z SecondChildTAG: На edx пока видимо нет и на весну не анонсировано. Мне кажется его скорее всего и не будет, по крайней мере бесплатно. Я так понимаю,что суть этого проекта (который кстати стоил больших денег) отработать методику для дальнейшего коммерческого применения. Статистика прошлого года такова, что на курс зарегестрировалось 150 000 человек, а диплом получило 7000. Посмотрим, что будет в этом. Но даже этот курс осилило несколько процентов от заявившихся. Соответственно на более сложный курс будет еще меньший спрос и еще меньше участников. Именно MITкурсы собраны тут. http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/ Силовая 6.334 , оптоэлектроника 6.637 Но это совершенно не то, что edx, это просто видеоматериалы для самостоятельного изучения. Еще можно поискать на курсере https://www.coursera.org/ SecondChildUserIdTAG: 345464 SecondChildUserNameTAG: Constantine_ru SecondChildCreateTimeTAG: 2012-12-28T12:23:11Z SecondChildTAG: You are probably right, if they do not improve the status or courses will not be paid a minimum, developing specialized training does not make sense. But of course it depends on the goal. Apparently did after the final exam a few of the course starts reached its end, and the specialized courses of people will be even less. Вы скорее всего правы, если они не повысят статус, либо курсы не будут минимально платными, то развивать специализированные курсы нет смысла. Но конечно это зависит от поставленной цели. Видимо действительно, после окончательного экзамена немногие из начавших курс дошли до его конца, а на специализированных курсах людей будет еще меньше. Vladimir SecondChildUserIdTAG: 402617 SecondChildUserNameTAG: Vl SecondChildCreateTimeTAG: 2012-12-28T20:25:24Z SecondChildTAG: Мне кажется, что следующим шагом (в США они его уже делают понемногу) это авторизованные экзаменационные центры. Так же, как с TOEFL и IELTS не имеет значения, где ты учил английский, важно лишь как ты сдал. Думаю тут будет тот же принцип - бесплатные, для вовлечения людей, курсы и платные экзамены в авторизованных центрах. Вместе с тем это сразу же многократно повышает стоимость сертификата с просто бумажки (защищенной только твоей честностью) во вполне себе солидный сертификат. Я бы заплатил. SecondChildUserIdTAG: 345464 SecondChildUserNameTAG: Constantine_ru SecondChildCreateTimeTAG: 2012-12-28T20:48:08Z SecondChildTAG: На courcera клевый курc - https://www.coursera.org/course/eefun SecondChildUserIdTAG: 394836 SecondChildUserNameTAG: v2g6ch4 SecondChildCreateTimeTAG: 2013-01-03T20:12:18Z SecondChildTAG: Judging from the description of the course, there are no certificates and examinations. This significantly lowers the bar course. But maybe I'm wrong. Судя по описанию курса, там нет никаких сертификатов и экзаменов. Это существенно снижает планку курсов. Но может быть я и не прав. Кстати что Вы делаете в Нерюнгри? VL SecondChildUserIdTAG: 402617 SecondChildUserNameTAG: Vl SecondChildCreateTimeTAG: 2013-01-06T19:10:10Z SecondChildTAG: В Нерюнгри я ставлю секретные эксперименты над слонами ![enter image description here][1] [1]: http://webdiscover.ru/uploads/comments/11352-1340990504.jpg А вообще заканчиваю местный вуз ). SecondChildUserIdTAG: 394836 SecondChildUserNameTAG: v2g6ch4 SecondChildCreateTimeTAG: 2013-01-08T16:14:34Z SecondChildTAG: +++ SecondChildUserIdTAG: 402617 SecondChildUserNameTAG: Vl SecondChildCreateTimeTAG: 2013-01-09T10:38:57Z FirstChildTAG: @Pennypacker, perhaps because of your prudish sensitivities you should stay out of the censorship business... чертов does not necessarily translate to anything rude in google translate - **if** you actually look beyond the sensational output box, and look at the actual probabilities, you will see that the sensational is probably less than 50% when translated as a single word. Many other choices are quite probable also. Don't you think it is wrong to assume that google translate is the final arbiter on the meaning of foreign languages? Unless you think you are beyond reproach, please explain why have you not issued an apology yet? FirstChildUserIdTAG: 9673 FirstChildUserNameTAG: xp42 FirstChildCreateTimeTAG: 2012-12-30T04:10:49Z FirstChildTAG: Is here anybody from Moscow who is going to take proctored? FirstChildUserIdTAG: 169228 FirstChildUserNameTAG: bashun FirstChildCreateTimeTAG: 2013-01-05T19:28:24Z SecondChildTAG: I'm not from Moscow, but the delivery of the exam is a problem. We have a point in Khabarovsk and Vladivostok. But these centers can not conduct these examinations. At the same time, I, for example, his taking the exam eight to nine hours, instead of two or three. In principle, I would have passed and two, but would be the worst results. Typically 70 percent of the time is one or two issues that do not go until you are bored. I do not need this test, I needed conceptual knowledge and assessment of himself. I have a university degree and working practices, which I received in the USSR. I need the knowledge, direction of travel in the future and systematization of knowledge and skills. Я не из Москвы, но сдача такого экзамена представляет проблему. У нас есть точки в Хабаровске и Владивостоке. Но эти центры могут и не проводить данные экзамены. В то же время я, например, его сдавал экзамен восемь-девять часов, вместо двух-трех. В принципе я бы сдал и за два, но были бы худшие результаты. Обычно 70 процентов времени занимает один два вопроса, которые не идут, пока не надоедают. Мне этот экзамен не нужен, мне нужны были понятийные знания и оценка себя. У меня есть высшее образование и практика работы, которое я получил еще в СССР. Мне нужны знания, направление движения в будущем и систематизация знаний и умений. VL SecondChildUserIdTAG: 402617 SecondChildUserNameTAG: Vl SecondChildCreateTimeTAG: 2013-01-06T18:44:17Z SecondChildTAG: Hi VL, as opposite of you i have completed nothing except a russian middle school. So i have one reason at least to have the proctored. Additionally i seriously hope staff intend to make the exam passable in order to advance the initiative. SecondChildUserIdTAG: 169228 SecondChildUserNameTAG: bashun SecondChildCreateTimeTAG: 2013-01-07T08:05:34Z SecondChildTAG: Hi bashun Then you good luck. In order to feel confident in these courses requires good knowledge of mathematics. For example the next course - 6,003 - is almost pure mathematics, in the form of the Fourier transform. The basis of these courses - Electromagnetism for engineers, is also an integral and differential calculus in the space for a surface and receiving partial solutions for specific surfaces and lines. So good luck to you. Тогда Вам удачи. Для того чтобы уверенно чувствовать на этих курсах необходимы устойчивые знания математики. Например следующий курс - 6.003 - это практически чистая математика, в виде преобразования Фурье. Основа этих курсов - Электромагнетизм для инженеров, тоже интегральное и дифференциальное исчисление в пространстве, на произвольной поверхности и получение частных решений для конкретных поверхностей и линий. Так что Вам удачи. VL SecondChildUserIdTAG: 402617 SecondChildUserNameTAG: Vl SecondChildCreateTimeTAG: 2013-01-07T11:08:43Z FirstChildTAG: Hi everybody! Merry Christmas and Happy New Year! Привет всем! С Новым годом и Рождеством! Мои поздравления всем, кто успешно закончил этот курс! Рад видеть здесь кириллицу ;) Я аспирант радиотехнического факультета Киевского политехнического. Решил посмотреть методику онлайн обучения, уровень материала, который дает МТИ, а также проверить свои остаточные знания (результат 96%). У нас на факультете ничуть не хуже уровень базового материала, только разбит он по разным предметам (напр., Основы Теории Цепей, Цифровые Устройства, Схемотехника и т.д.), но преподносят его как-то вяло (зависит от преподавателя). Как заметил студент выше, здесь преподносится с более современной точки зрения, а так же с уклоном, чтоб как-то заинтересовать студента. (Но это моя точка зрения). Видел здесь комментарии людей из Харькова… Интересно есть ли здесь киевские? Интересно Ваше мнение, отзывы. Нам есть к чему стремиться! FirstChildUserIdTAG: 168575 FirstChildUserNameTAG: AlexTron FirstChildCreateTimeTAG: 2013-01-07T16:27:49Z IndexTAG: 306 TitleTAG: THANKS TO MITX TEAM Just completed the Final Term. Quite happy and relaxed now. Was a bit tough paper but this is what you expect in Final term. Want to say thanks to professor Agrawal and other team members who have done a very great job. It was such a fun and a new experience of learning for me. And also special thanks to Myrimit for the help provided by her always. EDX and MITX Team it was an Great experience. Looking forward for new courses. Thanks!! And a question also, "Can we do two courses simultaneously?" Thank You! UserIdTAG: 443988 UserNameTAG: Dheeraj_Garg CreateTimeTAG: 2012-12-21T20:59:11Z VoteTAG: 7 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 3 FirstChildTAG: I agree about Myrimit, *she* was very helpful! FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-12-21T21:06:16Z FirstChildTAG: **RE:** *"Can we do two courses simultaneously?"* **Dheeraj_Garg**: If you mean that, for example for Spring 2013, you want to take both "**6.00x**: Introduction to Computer Science and Programming" and also "**3.091x**: Introduction to Solid State Chemistry" from ***MITx***, the answer is **yes**, you can. I'm pretty sure you can take three or even more! Then you can drop the classes you find uninteresting. That's the beauty of edX! This semester (Fall 2012) I was taking both 6.002x *and* 3.091x; I was doing well in both until we had that Hurricane Sandy in the United States, then I had a lot of work and I had to deal with the damage from the storm, so I decided to drop 3.091x right before the second exam, and I am currently taking the final for 6.002x. If you mean, can you take 6.002x now, and then re-take it in the Spring if you fail, **yes**, you can also do that. Even if you pass this semester, you can re-take it to attempt a higher score. Although I don't know yet if your certificate will be updated; you'll likely get a certificate with your old grade and your new (hopefully higher) grade. FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-12-21T22:28:05Z SecondChildTAG: well... the certificates don't contain the grades anyways ... right ? SecondChildUserIdTAG: 454609 SecondChildUserNameTAG: Alwahsh SecondChildCreateTimeTAG: 2012-12-21T23:12:16Z SecondChildTAG: Thank You Jersey Mark. I have got my query solved. SecondChildUserIdTAG: 443988 SecondChildUserNameTAG: Dheeraj_Garg SecondChildCreateTimeTAG: 2012-12-22T17:04:09Z FirstChildTAG: You are welcome Dheeraj_Garg ;) I am really happy that you could enjoy all this amazing experience, I wish you all the best for you ;) Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-24T01:30:33Z IndexTAG: 307 TitleTAG: Special Thanks to Myrimit Hi, Hola, I quit the course a month ago for personal reasons, which is a pity for me but I'm happy that I learned a lot (until week 8) and would like to finish it (do it again) by Spring :)) Thanks Myrimit for your very kind support!! Graciasssssss!!!! Regards, Sandra UserIdTAG: 342966 UserNameTAG: SandraNavarro CreateTimeTAG: 2012-12-21T16:47:48Z VoteTAG: 7 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: I hope to see you then! :-) FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-12-21T23:16:20Z SecondChildTAG: planetscape, will you take this course next Spring again? SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-22T12:21:21Z SecondChildTAG: Great, see u then :)) SecondChildUserIdTAG: 342966 SecondChildUserNameTAG: SandraNavarro SecondChildCreateTimeTAG: 2012-12-22T17:04:35Z FirstChildTAG: And of course, also special thanks to the edX Team for offering the course.... was a great experience!!!!!!! THANK YOU ALL!!!!!!!! Sandra FirstChildUserIdTAG: 342966 FirstChildUserNameTAG: SandraNavarro FirstChildCreateTimeTAG: 2012-12-22T17:05:27Z FirstChildTAG: That's correct, MyriMIT is the best FirstChildUserIdTAG: 418182 FirstChildUserNameTAG: euldji2005 FirstChildCreateTimeTAG: 2012-12-23T12:10:33Z SecondChildTAG: I'm blushing :p SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-24T01:26:18Z FirstChildTAG: Hola Sandra, La verdad es que disfruto poder poner mi granito de arena y ayudarles en todo lo que pueda. Me alegra mucho que estés feliz, eso es lo importante y te tomo la palabra, nos veremos entonces el próximo semestre, allí estaré ayudando nuevamente :p Que pases unas Felices Fiestas y que tengas un Próspero Año Nuevo, Un abrazo ;), Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-24T01:24:55Z SecondChildTAG: Si gracias, igualmente Felices Fiestas y Feliz año nuevo a todos :)) Un abrazo!! Sandra SecondChildUserIdTAG: 342966 SecondChildUserNameTAG: SandraNavarro SecondChildCreateTimeTAG: 2012-12-24T14:47:16Z IndexTAG: 308 TitleTAG: To EDX and MITX Team You guys have done a very great job and professor Agrawal and other team members have exactly delivered the best within the regime of this course. It was such a fun and very new experience of learning. I am, thankful to the whole team but these words are worthless for this "6.002x Circuits and electronics course". EDX and MITX Team it was an AWSOME experience. Looking forward for new courses. Thanks.. .. . . . UserIdTAG: 146930 UserNameTAG: fayzan_007 CreateTimeTAG: 2012-12-21T13:48:57Z VoteTAG: 7 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hope to see you around somewhere Fayzan! FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-12-21T14:06:05Z FirstChildTAG: [Referenced question] is so so so hard lol ***fares27**: I removed the specific question you mentioned as being difficult. This can discourage students, and we are prohibited from talking about the difficulty of particular questions, our scores, etc. until after the exam has concluded. - Jersey Mark (Community TA)* FirstChildUserIdTAG: 174229 FirstChildUserNameTAG: fares27 FirstChildCreateTimeTAG: 2012-12-21T18:20:12Z SecondChildTAG: Please do not discuss exam questions until after the exams. SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-12-21T22:09:49Z IndexTAG: 309 TitleTAG: it was a wondeful experience It was an excellent, very well organized course. I had an amazing experience. Thanks to Myriam for the help and all the hints. I am looking forward for other courses in spring 2013. I learned a lot from this course. I feel very very happy to have got this opportunity to get associated with MIT in any way. It is a real honor for me for this association with such a prestigious institute. UserIdTAG: 381797 UserNameTAG: Saira180 CreateTimeTAG: 2012-12-20T05:13:10Z VoteTAG: 7 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: completing 6.002x is the best thing a true engineer could do before the day of judgment, thanks FirstChildUserIdTAG: 323963 FirstChildUserNameTAG: gtrf FirstChildCreateTimeTAG: 2012-12-20T09:08:37Z IndexTAG: 310 TitleTAG: FINAL EXAM IS DELAYED Dear Students, There will be a delay in the release of the final exam, and we will let you know as soon as it has been released. Thank you for your patience! -- Rohan UserIdTAG: 391929 UserNameTAG: RohanNagarkar CreateTimeTAG: 2012-12-19T19:08:50Z VoteTAG: 7 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 311 TitleTAG: Big Thanks and God bless you Thank you very much for this wonderful course. I've really learned more than for 2 years in the university. Hope that trend for free education continues... Mary Christmas! UserIdTAG: 415375 UserNameTAG: ZWX CreateTimeTAG: 2012-12-17T18:32:08Z VoteTAG: 7 CoursewareTAG: Week 14 / S28V13_Voltage_Drop_across_the_Parasitic_Inductor CommentableIdTAG: 6002x_S28V13_Voltage_Drop_across_the_Parasitic_Inductor NumberOfReplyTAG: 0 IndexTAG: 312 TitleTAG: Popping up some Posts! Hi all, I am popping up some Posts that might got lost in the list. If you haven´t read them, you should haha ;) Here they are: ---------- **Evidences that proves that Prof. Agarwal is awesome, cool and a big hearted person. Thank you for being like this! :)** [Read Here][1] ---------- **Welcome to the CECC 2 ! Do you want to win a Textbook signed personally by Prof. Agarwal? Read here :)** [Read Here][2] ---- I wish you the best in the Final Exam, also we hope that you can participate in the Contest, that would be nice :) See you!;) [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50c08a8f2196ec2300000022 [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50b2892629b7b1270000003b UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2012-12-10T03:08:43Z VoteTAG: 7 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 313 TitleTAG: Thank you EDX and Anant Agarwal sir and other related people Thank you EDX and Anant Agarwal sir and other related people for providing this much for free of cost. Expecting more courses like thissssssssssssssssssssssssssssssssssssssssssssssssssssssssssssss UserIdTAG: 132685 UserNameTAG: deepakmurali CreateTimeTAG: 2012-12-05T06:01:30Z VoteTAG: 7 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 314 TitleTAG: H12P2 - how to compute efficiency and issues summary I spent most of the H12P2 time trying to figure out how to compute efficiency. Now I got all correct and here is the formula I used: efficiency = Pout/Pin, in which, Pin = Iin*Vin, Pout=Vout*Iout; That was straight forward and it is also easy to find Iout, which is Vout/RL. Here is the trick of the hard part: Iin = Iz+K*I (Iz is the current which goes through R0 and Rz and I is the BJT diode current). You can find I by knowing I = IL + Ir (IL is the current which goes through load resistor and Ir is the current which goes through R1 and R2). You also know that Vin = 20V and Vout is the value corresponding to that input. Be careful to compute Iz, remember you have to replace the zener diode by a voltage source + Rz. IMPORTANT: you don't need to design a super ultra good regulator for efficiency. Efficiency is not a design constraint, thus, it can be either good or bad. I hope it helps. ;-) UserIdTAG: 99238 UserNameTAG: arthurltc CreateTimeTAG: 2012-12-04T04:07:02Z VoteTAG: 7 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: As a summary of other issues: 1) don't use numerical abbreviations such as 1k or 1M. It only worked for me with 1000 and 1000000. 2) In order to check part a) and b) independently of the rest, you must supply some values to c) and d). Just fill with fake values and check a) and b). However, this does not work to check c) and d). To check c) and d) read issue 3) below. 3)You have to compute efficiency correctly according to the values you chose for R0, R1, R2. This means that even if your values for R0, R1 and R2 are correct, if the corresponding efficiency value is wrong, you will get all 4 wrong by the checker. FirstChildUserIdTAG: 99238 FirstChildUserNameTAG: arthurltc FirstChildCreateTimeTAG: 2012-12-04T04:14:16Z FirstChildTAG: If I, is the diode current, I is the output current too, since there's no current entering V-, right? FirstChildUserIdTAG: 244225 FirstChildUserNameTAG: lerimock FirstChildCreateTimeTAG: 2012-12-07T16:30:19Z SecondChildTAG: That's right. If I is the diode current, then I will go through the output resistors R1 and R2. Notice that if any other load was connected to Vout, there would be a part of this current I flowing through R1 and R2 and another part flowing through the new load connected. SecondChildUserIdTAG: 99238 SecondChildUserNameTAG: arthurltc SecondChildCreateTimeTAG: 2012-12-19T16:13:59Z FirstChildTAG: It shurely helped, arthurltc! thanks a lot! FirstChildUserIdTAG: 47372 FirstChildUserNameTAG: Digius FirstChildCreateTimeTAG: 2012-12-08T13:28:19Z SecondChildTAG: you're welcome ;) SecondChildUserIdTAG: 99238 SecondChildUserNameTAG: arthurltc SecondChildCreateTimeTAG: 2012-12-19T16:15:53Z FirstChildTAG: My efficiency is pretty low. I know it's required for this problem but how do you design for a higher efficiency? FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-12-09T20:12:08Z FirstChildTAG: There's a limit to what you can do with a linear regulator after you've made your resistors as large as practical. Better efficiency requires a switching regulator. Here is a nice hobbyist design: http://www.romanblack.com/smps/smps.htm Or you can buy a switching regulator ic: http://www.analog.com/static/imported-files/product_highlights/177228276DCtoDC_Regulator_E.pdf FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-12-10T01:19:42Z SecondChildTAG: Thanks skyhawk! :-) SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-12-10T04:56:16Z IndexTAG: 315 TitleTAG: the mouse pointer is indeed better sometimes in previous lectures i had to stop and go back the presentation to make sure i got the explanation right. Thanks for the change. UserIdTAG: 11075 UserNameTAG: roncada CreateTimeTAG: 2012-12-02T19:47:17Z VoteTAG: 7 CoursewareTAG: Week 12 / S23V5 Op Amp characteristics CommentableIdTAG: 6002x_S23V5_Op_Amp_characteristics NumberOfReplyTAG: 1 FirstChildTAG: The mouse is good. The profit is the most in situations when jumps to different parts of the screen are happening (mostly while explanation takes place). FirstChildUserIdTAG: 413065 FirstChildUserNameTAG: anikey FirstChildCreateTimeTAG: 2012-12-08T17:08:24Z IndexTAG: 316 TitleTAG: Question Concerning the Final Hello! First of all, I just wanted to thank MIT and our TAs for making 6002x such a valuable learning experience. As the course begins to wind down and we are getting closer to the holidays and closer to the end of the semester, I'm sure everyone is carefully balancing a variety of tasks and assignments. And, unfortunately, the Final takes place during the heat of this battle to keep up in our work. Like many others here, my primary motivation for taking this course was simply personal edification. Recently, however, it came to my attention that actually having a certificate would turn out to be quite nifty for some internships I am applying for. My plan is to take the final exam at some point, preferably after I am less busy (and, it seems, after the actual deadline for the course). If, however, the certificate was to state my grade, I probably would be significantly more motivated to take the exam during the specified period. And that brings me to my question. I was unable to find a definitive answer in the forum so I just wanted to ask: Will our certificates display our grades? I know it seems very trivial, but it seems it would really make a difference (especially now that I've already put so much effort into the course). Thank you very much in advance for your feedback. It's been a pleasure learning with you all. UserIdTAG: 153604 UserNameTAG: autohost CreateTimeTAG: 2012-11-29T05:44:43Z VoteTAG: 7 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: There will not be grades on the certificates. I think it's a pity, but that's the way they want it. FirstChildUserIdTAG: 90796 FirstChildUserNameTAG: ChaunceyGardiner FirstChildCreateTimeTAG: 2012-11-29T21:46:05Z SecondChildTAG: I see. Thank you kindly for your reply! SecondChildUserIdTAG: 153604 SecondChildUserNameTAG: autohost SecondChildCreateTimeTAG: 2012-11-29T22:18:09Z FirstChildTAG: There seem to be options, from the (costly) Proctored Exam to the (free) print a screenshot of your progress bars. I would think if you could prove that was your account (say by logging in), that might be enough for some prospective employers. FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-11-29T22:35:43Z IndexTAG: 317 TitleTAG: Course material WILL be available after this course finishes I see every so often someone asks whether the course materials will be accessible after the course finishes. This is an important question, because we (myself included) all want to get a head start in the spring, so... ...The answer is yes, all material will remain available, even this forum will remain accessible! So you can revise to your hearts content until the spring, when, if you do decide to retake the course, you will be given a fresh forum and set of assignments, labs etc... Hazel. UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-11-27T19:49:48Z VoteTAG: 7 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 318 TitleTAG: Thanks to edx Thank you EDX and edx staff for creating such a unique learning platform for us... Thank you very Much !!!! :) UserIdTAG: 178840 UserNameTAG: mehtanayanv CreateTimeTAG: 2012-11-22T13:29:00Z VoteTAG: 7 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 319 TitleTAG: [S18V21] - Error in the equation Hi, In the differential equation when differentiating the LHS and the RHS shouldn't the i also be differentiated and divided by C? which would give the final equation as: $\frac{d^2v}{dt^2}+\frac{1}{RC}\frac{dv}{dt}+\frac{v}{LC} = \frac{1}{C}\frac{di}{dt}$ UserIdTAG: 85490 UserNameTAG: yashsavani CreateTimeTAG: 2012-11-15T11:47:57Z VoteTAG: 7 CoursewareTAG: Week 9 / Driven_Parallel_RLC_Circuit CommentableIdTAG: 6002x_driven_parallel_RLC_circuit NumberOfReplyTAG: 3 FirstChildTAG: I found this entry from the old "Course Errors" wiki for S18V21: in characteristic equation w0 should be squared After divide by C and differentiate throughtout by d by dt, what we get should no current source i. 9:21: caption reads "developing the characteristic equation by substituting a e raised to sd in place of v for the homogeneous equation." It should be "a e raised to st" FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-11-15T16:28:29Z FirstChildTAG: See [my post from a couple of weeks ago][1] or search the discussions for S18V21. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/509602a35954722b00000014 FirstChildUserIdTAG: 141000 FirstChildUserNameTAG: OrinE FirstChildCreateTimeTAG: 2012-11-16T06:25:08Z FirstChildTAG: Check out page [778][1] of the textbook: it covers the parallel RLC example. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/802 FirstChildUserIdTAG: 391929 FirstChildUserNameTAG: RohanNagarkar FirstChildCreateTimeTAG: 2012-11-16T16:26:30Z IndexTAG: 320 TitleTAG: I think the anwers are wrong It seems that for the Q calculations, intead of taking omega/(R/L), they were calculated with the reciprocal, (R/L)/omega UserIdTAG: 450547 UserNameTAG: erikita CreateTimeTAG: 2012-11-14T02:59:33Z VoteTAG: 7 CoursewareTAG: Week 11 / S21E4: AM_Radio_Tuning CommentableIdTAG: 6002x_S21V1_AM_Radio_Tuning NumberOfReplyTAG: 4 FirstChildTAG: I also find the answers weird. FirstChildUserIdTAG: 138981 FirstChildUserNameTAG: Pr0bability FirstChildCreateTimeTAG: 2012-11-14T14:48:58Z SecondChildTAG: me too SecondChildUserIdTAG: 19003 SecondChildUserNameTAG: aamontoya SecondChildCreateTimeTAG: 2012-11-14T15:04:49Z SecondChildTAG: w/(R/L) and reciprocal of (R/L)/w.. Both are same... Just a simple maths SecondChildUserIdTAG: 477198 SecondChildUserNameTAG: Rajesh1993 SecondChildCreateTimeTAG: 2012-11-18T07:24:28Z SecondChildTAG: check your maths than.. SecondChildUserIdTAG: 352757 SecondChildUserNameTAG: Kerbyco SecondChildCreateTimeTAG: 2012-11-19T18:08:23Z SecondChildTAG: It's really weird, does anybody have an explanation for such thing?! SecondChildUserIdTAG: 197793 SecondChildUserNameTAG: HazelLarack SecondChildCreateTimeTAG: 2012-11-29T13:15:25Z SecondChildTAG: I think I've found out a plausible answer hahaha. The thing is that the Quality factor depends on the configuration of the circuit, so, if we're analyizing a series RLC Circuit, than the formula will be (omega0)/(R/L). Although, if the circuit is a Parallel LC in series with a resistor R (it could be a Parallel RLC as well), the formula changes to R/(omega0*L) which is equal to (omega0*R*C) SecondChildUserIdTAG: 197793 SecondChildUserNameTAG: HazelLarack SecondChildCreateTimeTAG: 2012-11-29T13:21:59Z FirstChildTAG: Hi, look at page 794, equation 14.47, Q = wo*R*C, that gives the expected result. FirstChildUserIdTAG: 220837 FirstChildUserNameTAG: Calsomus FirstChildCreateTimeTAG: 2012-11-15T00:19:57Z SecondChildTAG: Well... now delta w values are wrong... SecondChildUserIdTAG: 220837 SecondChildUserNameTAG: Calsomus SecondChildCreateTimeTAG: 2012-11-15T00:36:29Z SecondChildTAG: Ups silly mistake, they're fine SecondChildUserIdTAG: 220837 SecondChildUserNameTAG: Calsomus SecondChildCreateTimeTAG: 2012-11-15T08:47:37Z SecondChildTAG: Answer in Hz, not rad/s SecondChildUserIdTAG: 137421 SecondChildUserNameTAG: i026e SecondChildCreateTimeTAG: 2012-11-20T08:42:36Z FirstChildTAG: Hmm. lectures and textbook got different equations for Q? Quite strange FirstChildUserIdTAG: 352757 FirstChildUserNameTAG: Kerbyco FirstChildCreateTimeTAG: 2012-11-19T18:09:49Z FirstChildTAG: I think the circuit here is not the same as in the lectures, therefore derived formulas don't apply exactly. FirstChildUserIdTAG: 189680 FirstChildUserNameTAG: vnd FirstChildCreateTimeTAG: 2012-11-20T14:50:43Z SecondChildTAG: Yep, attenuation α is different here, plus you need to convert your answer to hertz, as somebody mentioned earlier. SecondChildUserIdTAG: 326409 SecondChildUserNameTAG: EugeneZ SecondChildCreateTimeTAG: 2012-11-21T18:52:07Z IndexTAG: 321 TitleTAG: Thanks For Extending Week 7 Just wanted to say thanks for extending the due dates for week 7. I am located in northern New Jersey and because of hurricane sandy I was without power from Monday until late Sunday and I was upset that I wouldn't be able to complete it. Now that It was extended until Tuesday, today, I was able to complete it all :D UserIdTAG: 219780 UserNameTAG: ReconIII CreateTimeTAG: 2012-11-07T03:19:10Z VoteTAG: 7 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Has eveverything settled now? FirstChildUserIdTAG: 99628 FirstChildUserNameTAG: Pranjal16 FirstChildCreateTimeTAG: 2012-11-07T11:25:14Z SecondChildTAG: Yeah, thanks for asking. Power is back for me and people are working to clean up all the trees down by me. Gas was in short supply and you would need to wait on lines for at least an hour but it is getting better by the day. SecondChildUserIdTAG: 219780 SecondChildUserNameTAG: ReconIII SecondChildCreateTimeTAG: 2012-11-07T15:29:21Z IndexTAG: 322 TitleTAG: Thank you Lyla and edX for the extension of Hw and Labs of Week 7 due the Hurricane Sandy ! :) Thank you very much Lyla and edX for giving the extension :)! It is really apreciated, Thank you from the heart! Myriam. ---------- Take a look at Course Info [here][1]: **NOVEMBER 4** *Due to the devastation caused by Hurricane Sandy, the deadline for week 7 assignments will extended by two days. Good luck to everyone who is still recovering from the storm.* And also the Post response of Lyla [here][2]. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/info [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50947eb7cde578270000000d UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2012-11-05T00:17:52Z VoteTAG: 7 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: it that for everyone in the class or just for the people in the area that were affected by Sandy FirstChildUserIdTAG: 160242 FirstChildUserNameTAG: Mlevins35 FirstChildCreateTimeTAG: 2012-11-05T02:54:24Z SecondChildTAG: everyone. SecondChildUserIdTAG: 96452 SecondChildUserNameTAG: Lyla SecondChildCreateTimeTAG: 2012-11-05T03:08:26Z SecondChildTAG: thank you having problems with the math glad for the extra time to figure it` ` out SecondChildUserIdTAG: 160242 SecondChildUserNameTAG: Mlevins35 SecondChildCreateTimeTAG: 2012-11-05T03:11:21Z SecondChildTAG: sweet! thanks! SecondChildUserIdTAG: 748857 SecondChildUserNameTAG: marcoatedx SecondChildCreateTimeTAG: 2012-11-05T21:16:31Z FirstChildTAG: Really appreciate edX for looking out for those affected (I'm not surprised though. You all were very responsive last time as well) :-) Thanks a ton! FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-11-05T15:29:34Z IndexTAG: 323 TitleTAG: howdy everyone Heyaaa Im new! UserIdTAG: 716537 UserNameTAG: andrewzzz CreateTimeTAG: 2012-10-24T16:21:48Z VoteTAG: 7 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Big welcome! Give us some background! FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-10-24T16:56:53Z IndexTAG: 324 TitleTAG: MidTerm Prep So is anyone else making models of often-seen circuits using tools like [CircuitLab][1] and plugging frequently used equations into Excel? I'm also reviewing revealed answers for homeworks and labs whose deadlines have passed - IMHO one of the most useful tools I've found yet. How are YOU preparing? [1]: https://www.circuitlab.com/ UserIdTAG: 148542 UserNameTAG: planetscape CreateTimeTAG: 2012-10-23T11:30:00Z VoteTAG: 7 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: planetscape, tell please, where is answers for homeworks and labs whose deadlines have passed here? I don't know why, but I can not find them. FirstChildUserIdTAG: 86420 FirstChildUserNameTAG: Aleksei_Katkov FirstChildCreateTimeTAG: 2012-10-23T13:24:30Z SecondChildTAG: Go to Week 1, for example. Click on Homework 1. Scroll down to where the "Check" (or "Submit", I've already forgotten) button WAS... now you will see "Show Answer." (Similar to Exercises.) Hope that helps! SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-23T14:01:05Z SecondChildTAG: OK, found it. Thanks! This course was spring 2012 first time, from march till june (6002x.mitx.mit.edu). When some week exersizes, home and lab deadline passed, course staff opened solutions (links to pdf close to links to lecture slides). Such solutions are very useful now, at second try to take this course. But - everyone who have solutions MUST be honest and use them only for finding correct way to solution, but not as ready answer. When I read your post about answers, I rememeber about such pdf's, - and didn't find them. P.S. Now pdf-files with solutions are removed! SecondChildUserIdTAG: 86420 SecondChildUserNameTAG: Aleksei_Katkov SecondChildCreateTimeTAG: 2012-10-23T19:47:28Z FirstChildTAG: Those are truely great tips for preparation. Thanks planetscape! FirstChildUserIdTAG: 443358 FirstChildUserNameTAG: torkelh FirstChildCreateTimeTAG: 2012-10-23T18:41:12Z SecondChildTAG: You are most welcome! I had the inspiration to use Excel (also [Wolfram Alpha][1] and others) because dumb math mistakes sink me every time! Good luck! [1]: http://www.wolframalpha.com/ SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-24T04:02:06Z SecondChildTAG: I've also had some valuable insights comparing [CircuitLab][1] models and their equations in Excel that really solidified some concepts. I think it's been time well spent, for me. [1]: https://www.circuitlab.com SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-24T04:43:49Z IndexTAG: 325 TitleTAG: To TheWiredBear and GladIDidThis, and others like them To those who don't know me, I took 6002x during the April-June 2012 cycle and busted my head doing the awesome-est (yeah, yeah grammar nazis, shoot me :P) circuits-and-electronics course ever. Oh, and I'm 15 and I live in New Delhi. I just received an email from TheWiredBear (nice forum name, btw ;), which prompted me to come back to 6002x and check out the discussion forum. I saw a few posts like [Probably not for high school kids][1] and [this one][2], and thought that "Gosh, these guys are going through the hell I was in barely 5 months back". To everybody like them, who are struggling with the deluge of partial derivatives and having dreams of killer algebraic equations threatening to strangle them (or maybe that was just me :P), please don't give up. This course will test you. Only a person who has just about zero background in this sort of thing, will understand how it feels like not having understood even a single word of an entire lecture sequence and a looming feeling of utter doom after seeing the week's homework problems. Please, please, don't give up. For the calculus, I watched the entire differential and integral calculus playlist from [Khan Academy][3] till multivariate calculus and the differential equations playlist, till the complex roots of characteristic equations. I also did a ton of chain rule problems from there. After that, I could follow along Prof. Agarwal's explanations, but only just. For the weekly problems, I mostly spent my time banging my head against the wall and tearing out page after page from my scrap notebook, scribbling crap and realizing that my answer is completely wrong. Other times, I didn't even know where to start from. In times like those, I took help from those blessed, Gods-of-the-forums, Aumatar and Myrimit. I still think I passed only because of their guidance and encouraging words to the idiots and me, the Grand Lord of the Idiots, haha :P Another thing I did was, use [Wolfram Alpha][4] for the extremely frustrating problems, which were simply too hard to do by hand. I don't know whether my opinion is right or wrong, but try not too use it too much, else you'll lose the habit of working by hand. And even in the rare cases you do use it in, be sure to check the steps of solution to understand how to do the problem the next time you encounter it. You cannot begin to imagine the satisfaction of completing this course, with its back-sloping wall of a learning curve. Weeks 6 through 9 will be hell. But weeks 10 through 14 will be bliss compared to them, with the elegant and oh-so-sexy (yes, I know I have problems :P) impedance method and holy $h!t, IMAGINARY numbers. You'll reap the benefits of working your butt off for the next 2 or 3 weeks and have a blast when you see that it all paid off. Sorry for writing so much and rambling on (yeah, Led Zeppelin, ftw !). Even if you do decide to drop this course, you would have learnt so so so much and realized how beautiful maths and physics can be. To Myrimit and Ashwith and the rest of the old-timers, hello :D Myrimit, you're doing such an amazing job here, just as before :D Oh, and I saw that you referenced me as a "he". I must point out that I am, in fact, a "she" and my name's Radhika ;) [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5071ccd71ed8e21f0000001c [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50847b728669712b00000022# [3]: http://www.khanacademy.org [4]: http://www.wolframalpha.com UserIdTAG: 84355 UserNameTAG: popoya CreateTimeTAG: 2012-10-22T13:49:49Z VoteTAG: 7 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: Thank you for this popoya!^_^ I am happy that Thewiredbear have put in contact with you! :) My best wish to you and all the kids! Myriam. P.D: I always thought that you were a boy! , apologies for my mistake Radhika :). FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-22T13:58:27Z SecondChildTAG: Hehe. No problem :) I think everybody from the old 6002x still imagines me as a guy :P SecondChildUserIdTAG: 84355 SecondChildUserNameTAG: popoya SecondChildCreateTimeTAG: 2012-10-22T14:14:14Z SecondChildTAG: Haha :). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-22T14:41:04Z FirstChildTAG: Thanks FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-22T14:08:45Z FirstChildTAG: Read your post from the old forum that Myriam (Myrimit) put up. You pretty much echoed the feelings I had (and still have) about the whole IIT-JEE prep program. Btw I'm a 12th grader in FIITJEE Hyderabad. Its pretty much the same as in Delhi but worse the whole school is changed into a coaching institute instead of after school/weekend classes. FirstChildUserIdTAG: 154440 FirstChildUserNameTAG: Aahlad FirstChildCreateTimeTAG: 2012-10-22T16:25:34Z FirstChildTAG: This course **is** possible for high-school kids, especially if you're looking for a challenge, and you've already taken high-school calculus. I actually recommend the first half of 6.002x for **all high-school seniors** going into an EE or CS major! When I first took this class, well, a similar one called "ECE 201: Fundamentals of Circuit Analysis" at the University of Connecticut that I attended, I did not take the "required" pre-requisites of electricity-and-magnetism in Physics (e.g. the second semester of most Physics programs) and linear algebra. I was in **shock**! Although I knew the calculus, there we had to solve simultaneous equations using matrices (unlike in this course). I was also working full-time evenings at the time in an engineering shop. So at midnights I would go to any open restaurant to eat dinner, and study on my linear algebra and do my homeworks. *Since circuit analysis was so new to me, the learning curve so steep, but the work so rewarding and intellectually stimulating, I put all my energy into this one university course* (at the expense of my programming and my elective courses). Our university's course was a bit more difficult than MIT's in some respects (we covered capacitors, first-order equations, AC analysis and phasors before midterms) but we were also graded on a curve. This was about 9 years ago, and we had no online learning back then. I wish we did. **If this class was available in high school, it would have helped me immensely!** In the United States, your typical Senior (last) Year in high-school is *boring and unproductive* because your AP courses are over, your SAT is over, and your acceptance to university is pretty much already set. There's nothing left to study, hence your mind slows down. Most U.S. seniors resort to over-socializing, partying, drinking, bad behavior, etc. If I had 6.002x in high school, I would have a **much easier transition** to rigorous university-level engineering courses! In conclusion, anyone saying "*probably not for high-schoolers*" is plain wrong to make such a blanket statement. Mark in New Jersey, USA FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-10-23T03:16:59Z IndexTAG: 326 TitleTAG: SUCCESS!! Started an edX marathon day-before-yesterday working through my backlog The goal: To get through week 3 to week 6 before week 6 deadlines. Submitted Week 6 Homework and Labs 40 minutes before the deadline!!! :D Didn't think I'd make it. Lesson for the future: Don't postpone things for later. Looking forward to the midterm :) UserIdTAG: 154440 UserNameTAG: Aahlad CreateTimeTAG: 2012-10-21T18:22:57Z VoteTAG: 7 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Good stuff! FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-21T18:38:27Z SecondChildTAG: Congratulations! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-21T19:12:37Z SecondChildTAG: Thanks! :) SecondChildUserIdTAG: 154440 SecondChildUserNameTAG: Aahlad SecondChildCreateTimeTAG: 2012-10-21T19:23:55Z SecondChildTAG: Wow! well done Aahlad! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-21T21:42:53Z SecondChildTAG: Wow, good work on the marathon session! I used to do that too, when I was a bit younger...now I can only do one HW/Lab per Sunday, but I need a few days in between to recuperate! With HW due in 3.091x as well (the Physical Chem course), and the Mid-Term for 6.002x (this class) all due this week, I will have to plan a **marathon study session** as well! SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-10-22T09:21:57Z SecondChildTAG: Thanx! @JerseyMark Good Luck :) SecondChildUserIdTAG: 154440 SecondChildUserNameTAG: Aahlad SecondChildCreateTimeTAG: 2012-10-22T13:14:53Z IndexTAG: 327 TitleTAG: To Get the Small Signal Gain 1st you have vgs = vi - vout Eq. 1 & ids = gm * vgs Eq. 2 substitute 1 in 2 you'll get ids = gm * (vi - vout) Hence ids = gm*vi - gm*vout Eq.3 Remember that vout = ids * RS ids = vout/RS Eq.4 Solve 3 & 4 simultaneously & you'll get the asnwer UserIdTAG: 185715 UserNameTAG: amirengineer CreateTimeTAG: 2012-10-20T09:37:00Z VoteTAG: 7 CoursewareTAG: Week 6 / Source-follower small-signal exercise CommentableIdTAG: 6002x_small_signal_source_follower NumberOfReplyTAG: 0 IndexTAG: 328 TitleTAG: WOW!!! MIT..i bow down to you..you are really THE best... :) UserIdTAG: 222911 UserNameTAG: bhaswardg CreateTimeTAG: 2012-10-15T11:01:24Z VoteTAG: 7 CoursewareTAG: Week 5 / Music Amplifier Demonstration CommentableIdTAG: 6002x_snd_d NumberOfReplyTAG: 0 IndexTAG: 329 TitleTAG: How to go to a specific page in the text book (textbook) Enter this URL in your browser (it won't work as is) https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/page# where page# = page number + 24. The +24 is because there is a lot of stuff at the front of the book before page 1. So, for example, to go to page 555 enter this URL (it will work) https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/579 Hopefully this is useful. UserIdTAG: 9673 UserNameTAG: xp42 CreateTimeTAG: 2012-10-11T22:08:32Z VoteTAG: 7 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: xp42: good to see you! FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-11T22:24:45Z FirstChildTAG: xp...heya. A useful pointer. FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-10-13T04:38:37Z IndexTAG: 330 TitleTAG: Possible bug There appears to be no captions for me for this video. UserIdTAG: 248807 UserNameTAG: 1kingsman CreateTimeTAG: 2012-10-10T19:14:11Z VoteTAG: 7 CoursewareTAG: Week 5 / Small Signal Mathematically Described CommentableIdTAG: 6002x_small_sig_math_des NumberOfReplyTAG: 1 FirstChildTAG: I see them. ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13499074201343649.png FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-10T22:17:14Z SecondChildTAG: Not S10V6, but S10V7 ![enter image description here][1] [1]: http://www.webpagescreenshot.info/i/295311-10112012104506am.png SecondChildUserIdTAG: 326135 SecondChildUserNameTAG: Uhimovich SecondChildCreateTimeTAG: 2012-10-11T08:48:11Z SecondChildTAG: I think it is s10v7 for me. It would appear that the comments get grouped under "mathematical view of small signal", rather than s10v6 or s10v7 SecondChildUserIdTAG: 248807 SecondChildUserNameTAG: 1kingsman SecondChildCreateTimeTAG: 2012-10-11T22:23:22Z IndexTAG: 331 TitleTAG: H5P1 I am totally lost, I mean I made a quadratic equation in Vi and solved it from wolfram alpha, and no luck!.......please some one help me here! the equation I got though was > 4x^2-3x-0.5=0 please anyone help me!.... FYI i got this quadratic from here: ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13498148824042299.png UserIdTAG: 169784 UserNameTAG: thewiredbear CreateTimeTAG: 2012-10-09T20:34:05Z VoteTAG: 7 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: Please just use the first equation you have written above alone and you will get the answer. Vo = vs-k/2(v1-VT)^2*RL So,by plugging in the parameters given,you will be fine. I hope this helps. FirstChildUserIdTAG: 11607 FirstChildUserNameTAG: deyoung36 FirstChildCreateTimeTAG: 2012-10-10T03:06:30Z SecondChildTAG: but then how to remove Vs SecondChildUserIdTAG: 169784 SecondChildUserNameTAG: thewiredbear SecondChildCreateTimeTAG: 2012-10-10T07:04:30Z SecondChildTAG: Vs=2V, isn't it? SecondChildUserIdTAG: 66669 SecondChildUserNameTAG: agarus SecondChildCreateTimeTAG: 2012-10-10T12:12:35Z SecondChildTAG: VS= 1 SecondChildUserIdTAG: 11607 SecondChildUserNameTAG: deyoung36 SecondChildCreateTimeTAG: 2012-10-10T18:58:27Z SecondChildTAG: i am using Vo as 0 then getting vs=k/2(v1-.5)*rl where rl was from previous but it is not accepting it. SecondChildUserIdTAG: 99628 SecondChildUserNameTAG: Pranjal16 SecondChildCreateTimeTAG: 2012-10-10T19:33:08Z SecondChildTAG: what should be Vo.? SecondChildUserIdTAG: 438395 SecondChildUserNameTAG: ali_PU1 SecondChildCreateTimeTAG: 2012-10-13T19:27:06Z FirstChildTAG: i am getting answer 1 volt, but its not right...help please FirstChildUserIdTAG: 535777 FirstChildUserNameTAG: amnaliaquat10 FirstChildCreateTimeTAG: 2012-10-11T15:11:03Z FirstChildTAG: Hi thewiredbear! You can take a look at this [Post][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_mosfet_amp/threads/507a1d6dd894f12300000039 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-14T04:18:34Z FirstChildTAG: Hey Wired bear this message is for you I am an Indian eleventh grader giving SAT next year, i want to talk to you about certain things. As for me i completely agree with your " I HATE COAching institutes" attitude. IF you want to meet me contact me at skype My id is Harvey.Specter13 FirstChildUserIdTAG: 378080 FirstChildUserNameTAG: GladIDidThis FirstChildCreateTimeTAG: 2012-10-21T22:50:37Z IndexTAG: 332 TitleTAG: Decided to be punctual I could complete only 60% of H W 4 , and could not even Crack the lab.It was my fault to start late on Homeworks.I listened to only half the lectures and was not able to attend the tutorials.I understood one thing ,do not wait foe the deadline to do the work, start doing when the week starts. Wishing all Good Luck UserIdTAG: 162670 UserNameTAG: charlesbabyt CreateTimeTAG: 2012-10-08T10:54:35Z VoteTAG: 7 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: That's true. But anyway, it is not a disaster - Homeworks do a little impact on the final grade. FirstChildUserIdTAG: 326787 FirstChildUserNameTAG: Angstrem FirstChildCreateTimeTAG: 2012-10-08T11:03:49Z SecondChildTAG: Or, preferably, a week or so ahead. I was a week or so ahead when I started; now I just meet the deadlines by 45 minutes or so... :-( SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-08T18:35:09Z SecondChildTAG: So am I =) but anyway, its amazing to attend such a course for me! SecondChildUserIdTAG: 326787 SecondChildUserNameTAG: Angstrem SecondChildCreateTimeTAG: 2012-10-08T21:36:01Z SecondChildTAG: yaa..just awesome course, we have to think a little beyond . It will help us all to gain practical learning experiance SecondChildUserIdTAG: 162670 SecondChildUserNameTAG: charlesbabyt SecondChildCreateTimeTAG: 2012-10-09T05:39:14Z IndexTAG: 333 TitleTAG: @Myrimit's help Please I need your help as soon as posible to solve h4p2. Just some hint UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-06T23:26:24Z VoteTAG: 7 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hi anonymous! Yes, sure :). I will write some hints for you of h4p2. Please wait a little. I will back here again and edit my post response with the hints. ---- EDIT: Here what I have promised, some hints of h4p2, I hope that this can be helpful for you :). Ok, in the statement they give you this circuit: ![image1][1] and it says: In the figure above, calculate the DC output voltage VO, and the output noise voltage vo for two values of the load resistance. Hint Part1: You should watch this video lecture [S7V5: Incremental method insight][2], it is important. You will find that vIN=VIN+vin and vOUT=VOUT+vout. So, you can divide your circuit analyzing it like two circuits separately: - The first one, with the voltage input VIN and the voltage output VOUT. - The other one, with the voltage input vin and the voltage output vout. Ok, recalling the concept of voltage dividers, if you have VI(VIN) and RL and RIN value, can you find VO? Yes! Hint part 2: idem part 1) but analyzing the second circuit (vin and vout). Hint part 3: idem part 1) but with a different value of RL. Hint part 4: idem part 2) but with a different value of RL. The statement says: **Now, we can insert a Zener diode into the circuit, as shown below. The Zener diode is a nonlinear device and a piecewise-linear approximation to its i-v characteristic is shown graphically below.** ![imagen2][3] Ok, here my first visual hint for you. Lets try to make the analysis but not in the small signal. Lets suppouse this conventions of signs :) ![imagen][4] Hint Part 5: lets recall the KCL (the sum of the current of a node it is equal zero) and KVL (the sum of voltages in a loop it is equal zero). If you have i1,i2 and iD can you obtain in someway VI expressed in function of VO, RL an RIN? Yes! Can you find your VO of a specific value of RL? Yes! Another hint for this part try to find i1, i2, iD in function of VO and vD. Another last hint VO it is opposite of your vD ;). Hint Part 6: Ok, now what happens if we derivate the expression of part 5? Isn't it our small signal? ;). Remember that in the case that you have x+constant the derivate will be dx+0 , that is to say dx. Hint Part 7: idem part 5) but evaluated in other point. Hint Part 8: idem part 6) but evaluated in other point. Hint Part 9: remember that the zener diode has to work in the reverse breakdown region . So, what it is your maximun VD? An how it is that VD related with your VIN? can you find your RL? Yes! I hope that this can help you. Be careful with signs ! :). Myriam. ---- Now in Spanish! En el enunciado se te da el siguiente circuito: ![image1][1] y nos dicen: "En la figura de arriba, calcular la tensión de salida VO en continua DC, y el voltaje de salida de ruido para ambos valores de resistencia de carga " Hint Parte 1: Deberías ver este video de las lecturas [S7V5: Incremental method insight][2], es importante que lo hagas porque ahí podrás ver que el análisis se divide en dos. Encontrarás que vIN=VIN+vin y vOUT=VOUT+vout. Es por ello, que debes analizar el circuito por separado: - El primero, con la tensión de entrada VIN y la tensión de salida VOUT. - El otro, con la tensión de entrada vin y la tensión de salida vout. Bien, recordando el concepto de divisores de tensión, si tienes el valor de VI (VIN), RL y RIN, puedes hallar VO? Sí! Hint parte 2: idem parte 1) pero analizando el segundo circuito, es decir, solo considerando vi (vin) y vo (vout). Hint parte 3: idem parte 1) pero con un valor distinto de RL. Hint part 4: idem parte 2) pero con un valor distinto de RL. El enunciado dice: **Ahora, podemos insertar un diodo zener en el circuito, tal y como se muestra a continuación. El diodo Zener es un dispositivo alineal y una aproximación por tramos de su característica i-v es la que se muestra abajo** ![imagen2][3] Aquí va mi primera ayuda visual para ti. Hagamos el análisis pero no en señal pequeña. Supongamos la siguiente convención de signos :). ![imagen][4] Hint Parte 5: Recordemos el KCL ( la suma de las corrientes en un nodo es cero) y el KVL (la suma de las tensiones en un loop es cero). Si tienes i1,i2 e iD, puedes de alguna forma obtener VI expresada en función de VO,RL y RIN? Sí! Puedes hallar el VO referido a un valor específico de RL? Sí! Otra hint para esta parte, intenta hallar i1,i2,iD en función de VO y VD. La última hint para esta parte, VO es opuesta a tu tensión vD ;). Hint Parte 6: Bien, ahora que pasaría si derivamos la expresión de la parte 5? No es nuestra señal pequeña? ;). Recuerda que en el caso que tengas una x+constante la derivada de dicha expresión será dx+0, es decir dx. Hint Parte 7: idem parte 5) pero evaluada en otro punto de carga. Hint Parte 8: idem part 6) pero evaluada en otro punto de carga. Hint Parte 9: Recuerda que el diodo zener tiene que trabajar en la zona de quiebre reverso. Entonces cuál es el máximo VD admisible? y cómo este VD se relaciona con tu VIN? Puedes entonces hallar tu RL? Sí! Espero que esto te haya sido de ayuda. Ten cuidado con los signos! :). Myriam. [1]: https://www.edx.org/static/content-mit-6002x/images/circuits/signal-source-with-load.d0abb6b53b4c.gif [2]: https://www.youtube.com/watch?v=MVDH7sF8zdQ&feature=player_embedded [3]: https://www.edx.org/static/content-mit-6002x/images/circuits/Zener-regulator.ce0e579b8571.gif [4]: https://edxuploads.s3.amazonaws.com/13495814791343616.png FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-07T02:10:00Z SecondChildTAG: how do i get id SecondChildUserIdTAG: 224318 SecondChildUserNameTAG: ashwin12312 SecondChildCreateTimeTAG: 2012-10-07T07:39:25Z SecondChildTAG: hi ashwin12312! You can find iD graphically. If you see the given image: ![here][1] you can use the equation of the line: y = m*x+b m=slop x= dependant variable y= independant variable. So, if you know the value of m (look at the triangle, which value it has?). Lets try to think the value of b by giving values to x and y. If you choose the value when x=-5 you will know the value of y. Isn't it? ;). Ok, I will let to you that you think this. Once you have your b value, you will see that your y=iD and your v=vD ;). See you! Myriam. [1]: https://www.edx.org/static/content-mit-6002x/images/circuits/Zener-diode.a7b4f06d4768.gif SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T08:10:24Z SecondChildTAG: don't forgot it's a forward voltage IV - in our schematic zener is reversed SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-07T08:37:08Z SecondChildTAG: Ok, if I take x=1 I know that m=1a/v and I get y=1a/v*(-5v)=-1a it will be Id? SecondChildUserIdTAG: 192114 SecondChildUserNameTAG: chuba SecondChildCreateTimeTAG: 2012-10-07T08:47:26Z SecondChildTAG: Myrimit Another hint for this part try to find i1, i2, iD in function of VO and vD. Another last hint VO it is opposite of your vD ;). It will be something like this -(Vin-Vo)/Rin-ID+Vo/RL? SecondChildUserIdTAG: 192114 SecondChildUserNameTAG: chuba SecondChildCreateTimeTAG: 2012-10-07T08:53:18Z SecondChildTAG: I assume x=-5 y ill be 0 till the cut in voltage is reached in this case .6v so the value of b ill be same as that of x but with a negative sign so b is also a variable its not constant SecondChildUserIdTAG: 224318 SecondChildUserNameTAG: ashwin12312 SecondChildCreateTimeTAG: 2012-10-07T11:42:42Z SecondChildTAG: I don't understand how can i used this, can you explain it? SecondChildUserIdTAG: 192114 SecondChildUserNameTAG: chuba SecondChildCreateTimeTAG: 2012-10-07T11:50:54Z SecondChildTAG: :0 SecondChildUserIdTAG: 127800 SecondChildUserNameTAG: MNB SecondChildCreateTimeTAG: 2012-10-07T14:45:29Z SecondChildTAG: The above solution is incorrect :) SecondChildUserIdTAG: 214085 SecondChildUserNameTAG: shohin SecondChildCreateTimeTAG: 2012-10-07T17:28:58Z SecondChildTAG: how do I get Id from that graph SecondChildUserIdTAG: 45307 SecondChildUserNameTAG: josejimenez2 SecondChildCreateTimeTAG: 2012-10-07T18:32:53Z SecondChildTAG: i have done all the homework for week 4 but still stuck with h4p2 part b where we have to find the noise voltage after inserting the zener diode :( SecondChildUserIdTAG: 278792 SecondChildUserNameTAG: sali SecondChildCreateTimeTAG: 2012-10-07T21:54:54Z SecondChildTAG: In this task I don't know how to find min Rl SecondChildUserIdTAG: 192114 SecondChildUserNameTAG: chuba SecondChildCreateTimeTAG: 2012-10-07T22:04:07Z SecondChildTAG: I used this formula to find y so m=(y1-y2)/(x1-x2), you know m= 1, then you pick a point, let's say x=-5, that line is a straight line which touches (0,0) plug in the values you now know solve for your unknown y and voila. then to solve v use the formula (y-y1)=m(x-x1)+b you know all the values except for b SecondChildUserIdTAG: 238005 SecondChildUserNameTAG: isisbocardo SecondChildCreateTimeTAG: 2012-10-08T04:10:25Z FirstChildTAG: Dear Myriam, I am not able to find the h4p2 only the circuit with zen er answer Please give the answer how to find any way the Home work submit time over I was score only 82% in the week 4 H/W Thanks MK.Prasanth FirstChildUserIdTAG: 156694 FirstChildUserNameTAG: mkprasanth FirstChildCreateTimeTAG: 2012-10-08T07:32:02Z SecondChildTAG: Hi mkprasanth! Can I help you? Where are you lost? SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-08T12:06:58Z SecondChildTAG: RL=2 kΩ the value of And the value of vo (in Volts) is: RL=4 kΩ the value of And the value of vo (in Volts) is: What is the minimum value of RL, in Ohms, that guarantees that the circuit will operate this way? SecondChildUserIdTAG: 156694 SecondChildUserNameTAG: mkprasanth SecondChildCreateTimeTAG: 2012-10-08T13:47:31Z SecondChildTAG: Hint: you know that the zener has to work in the reverse region. You will have a value for that reverse break (vD). Ok, also you know that vD is opposite to vO, so, if you know how to relate vI with vO (recall votage divider concept), can you find RL for that condition? ;). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-08T15:01:12Z SecondChildTAG: let me try SecondChildUserIdTAG: 156694 SecondChildUserNameTAG: mkprasanth SecondChildCreateTimeTAG: 2012-10-08T15:38:12Z SecondChildTAG: thanks a lot for your support SecondChildUserIdTAG: 156694 SecondChildUserNameTAG: mkprasanth SecondChildCreateTimeTAG: 2012-10-08T15:38:27Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-08T18:50:01Z IndexTAG: 334 TitleTAG: H5P1? Help me with 3 q. and if you explain the idea of zero offset it would be great, because im stuck UserIdTAG: 95280 UserNameTAG: Fortyq CreateTimeTAG: 2012-10-06T08:53:30Z VoteTAG: 7 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Zero offset refers to the DC offset at the output. E.g. if you have sinusoid with Vmin=-0.5V and Vmax=0.5V, it would have a zero offset. If that sinusoid would have the same amplitude, but between Vmin=1.5V and Vmax=2.5V, you could consider that as a DC offset of 2V. FirstChildUserIdTAG: 148015 FirstChildUserNameTAG: ralfh FirstChildCreateTimeTAG: 2012-10-06T14:52:37Z FirstChildTAG: finally got it. its like aha moment. the hint comes from EugenyL in his answer to the post https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50631b1199874e230000000e absolute values are nothing. You can add same voltage to every node, and it will be the same circuit! For example, if you add 10V FirstChildUserIdTAG: 95280 FirstChildUserNameTAG: Fortyq FirstChildCreateTimeTAG: 2012-10-08T15:57:16Z SecondChildTAG: Could you be so kind and tell something more about your solution. SecondChildUserIdTAG: 484236 SecondChildUserNameTAG: joteks SecondChildCreateTimeTAG: 2012-10-08T20:22:25Z IndexTAG: 335 TitleTAG: S7E1: LINEARIZ. beyond plug and chug I can see that node analysis would give me $$ i_A=\frac{5-v_A}{2}=10(1-e^{\frac{-v_A}{5}}) $$ and can [plot](https://www.desmos.com/calculator/kx8wa89dza) that and plug in my guesses into $$ v_A=20e^{\frac{-v_A}{5}}-15 $$ until I find the $ v_A $ that makes it true but I haven't got clue how to do it analytically. Any tips? (I didn't have trouble finding the resistance by taking the derivative.) UserIdTAG: 440714 UserNameTAG: mcktim CreateTimeTAG: 2012-10-02T19:48:48Z VoteTAG: 7 CoursewareTAG: Week 4 / Linearization Exercise 1 CommentableIdTAG: 6002x_linearization_ex_1 NumberOfReplyTAG: 2 FirstChildTAG: hi mcktim - you did what I did - you used kcl - the hint is kvl --- sum of voltages = zero --->>> vI - iA*R - vA = 0 --- plug in vI=5, R=2 and the iA expression and it falls out FirstChildUserIdTAG: 10512 FirstChildUserNameTAG: asicok FirstChildCreateTimeTAG: 2012-10-02T22:59:06Z SecondChildTAG: (asicok) Both equation gives same results. So we don't have to calculate again by KVL. ---- (Hint: Use the exponential term to solve for vA as a function of the assumed value of vA, and then iterate. Taking logs on both sides may facilitate convergence.) This is important clue to analyze your equation. We can't analyze equation iteratively with exponential term. So we can take logarithms on both sides to eliminate exponential term. Of course you should arrange constant term to do that. SecondChildUserIdTAG: 134925 SecondChildUserNameTAG: Hanbyul SecondChildCreateTimeTAG: 2012-10-03T07:54:21Z SecondChildTAG: Hi Hanbyul, Could you please explain this a little bit further? Thanks :) SecondChildUserIdTAG: 324642 SecondChildUserNameTAG: StNas SecondChildCreateTimeTAG: 2012-10-04T13:48:17Z SecondChildTAG: Thanks Hanbyul ! @StNas: look for the section called "iterative solution" in the following wikipedia article on [diode modelling][1] - hope this helps [1]: http://en.wikipedia.org/wiki/Diode_modelling SecondChildUserIdTAG: 87106 SecondChildUserNameTAG: Fabius SecondChildCreateTimeTAG: 2012-10-04T23:04:43Z SecondChildTAG: Hi, Isn't Delta(Va)/Delta(Ia) the dynamic resistance of the element? I suppose the questions 2 and 3 are asking for the same thing. Any suggestions, please? SecondChildUserIdTAG: 163264 SecondChildUserNameTAG: Aamir_edx SecondChildCreateTimeTAG: 2012-10-06T06:09:14Z FirstChildTAG: How are you getting 5-VA? I am getting (VA-VI)/R - IA=0 FirstChildUserIdTAG: 230787 FirstChildUserNameTAG: Gabriel007 FirstChildCreateTimeTAG: 2012-10-07T20:14:43Z IndexTAG: 336 TitleTAG: Terms There should be specified in what terms it should be expressed - it's not obvious! UserIdTAG: 342603 UserNameTAG: YakovO CreateTimeTAG: 2012-10-02T00:05:31Z VoteTAG: 7 CoursewareTAG: Week 5 / MOSFET Amplifier Exercise 1 CommentableIdTAG: 6002x_mosfet_amp_e1 NumberOfReplyTAG: 1 FirstChildTAG: Right, but they expect you to read discussion after all :) Hope there will be no such ambiguity at midterm exams. FirstChildUserIdTAG: 386636 FirstChildUserNameTAG: andy7panda FirstChildCreateTimeTAG: 2012-10-10T06:36:47Z IndexTAG: 337 TitleTAG: Can we please have the solutions of the homework and lab after the Due time is over?? Hello... I think it would be great if the staff can give us the solutions of the lab and homework after the due dates are passed. We really need to understand what we did wrong in order to progress. What do you guys think about this? UserIdTAG: 385948 UserNameTAG: sedamen03 CreateTimeTAG: 2012-10-01T01:50:45Z VoteTAG: 7 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Yes, I really think this would be a good idea... I had a bad time doing HW3, and I don´t know how to do some exercises... Also I don´t have anyone that can help me understand, so if staff could give us the solutions it would be great! FirstChildUserIdTAG: 371220 FirstChildUserNameTAG: dan10 FirstChildCreateTimeTAG: 2012-10-01T02:03:26Z SecondChildTAG: but HW3 is veeeery easy man :S come'n try harder.. SecondChildUserIdTAG: 310474 SecondChildUserNameTAG: aldaris565 SecondChildCreateTimeTAG: 2012-10-01T03:30:25Z FirstChildTAG: They do give the answers. Just go back to the homework and hit the "Show Answer" button. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-01T23:35:56Z IndexTAG: 338 TitleTAG: Necessary improvements to the discussion forum Some people can express their ideas cogently. Others cannot. Some people can speak English well. Others cannot. With the sheer number of people enrolled in 6.002x, I've found that the discussion forum has mostly devolved into a cacophony of confused, misunderstood, frustrated voices. Whenever I've found myself stumped on a problem and I've turned to the discussion forum for help, I've only ever encountered a lot of hot air and noise. I've only ever found a few posts helpful: those of the staff, and those of a handful of students. For the discussion forum to be a useful tool, there really ought to be some options available for filtering out the less useful discussions which proliferate the discussion forum and which only serve to obfuscate the learning process. I realize that the forums should be inclusive, but I would personally prefer one smaller 'study group' over a room full of squabblers. I realize that making posts 'private' (for only a select few students to read and participate in) would deprive others of the benefit of reading the discussions therein, so this approach probably goes too far. But, in general, I think it would be more useful for everyone using the discussion forums, if the dialogue were carried on primarily by a smaller number of English speakers while others contented themselves to reading and understanding. (I obviously also appreciate that someone who may or may not be a native english speaker might encounter some difficulty which hasn't been dealt with by anyone else in the forums, in which case he might choose to post a new topic and receive direct attention; but the truth is that the vast majority of difficulties that people might encounter will overlap. It would be best for everyone to be able to refer to one well-written, cogent answer, rather than for everyone to have to sift through 40 poorly-written, semi-cogent answers to questions that were not particularly well-posed.) All that being said, in the interest of improving clarity and usefulness, I have the following suggestions for encouraging that discussions are led by those students who are most well-suited to the task: -A personal option to favorite other students (or staff accounts) so that their comments and posts can be easily found. That way, if I find someone's comment, in one week, particularly well-written and helpful, I can look for what they might have to say on another issue some other week. -An option to better manage the responses to your own posts. Such as 1)being able to delete overly obfuscating responses; 2) being able to 'tag' or 'invite' 'friends' to respond to the question being posed (ie, an email notification would be sent to users who have accepted a 'friend request' or some such) I think if you make it easier for english speakers to 'find' other english speakers, they will naturally take over leadership of the discussion which should really be to the benefit of everyone using the forums. UserIdTAG: 287805 UserNameTAG: Beneficial CreateTimeTAG: 2012-09-30T18:29:02Z VoteTAG: 7 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 8 FirstChildTAG: Very good post. I am sad to see so much begging for answers rather than quests for understanding. It also seems reasonable to me that there is a point at which it is too late to start the course for credit. Begging for answers for week's two and three assignments at the last minute doesn't seem proper IMHO. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-09-30T19:01:56Z FirstChildTAG: This forum is better than what I've seen on other sites. FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-09-30T19:36:34Z FirstChildTAG: I, too, would like the ability to hide posts from some users (even if they're only hidden to me). I enjoy helping people with an interest in understanding the material, but there's just too many posts that I don't understand or asking questions in a way that discourage me from answering clogging up my board that I don't want to sift through it all. Usually from the same users Also, there should be sections to discuss the homework and labs specifically. I understand that they don't want us posting answers, but they're getting discussed anyways and it just makes it difficult to filter if there's no specific place for them. FirstChildUserIdTAG: 248902 FirstChildUserNameTAG: Chanute FirstChildCreateTimeTAG: 2012-10-04T01:46:43Z FirstChildTAG: Hi Beneficial! Thank you for your suggestions! :) I am sure that the edX Staff will improve day by day with the feedback of the students. So, it is important that you can give your point of view. But remember that this is an online worldwide nonprofit course and that there are a lot of students all over the world that makes a lot of effort by writting in other language from her/his mother tongue - I will include myself in the list - and I think that tolerance and colaboration it is needed with each others as classmates of the same Course, and nobody must not be excluded of this Forum Discussion for this reason or any. I guess that this is the point of this Discussion Forum, that all could participate and learn, isn't it? ;). See you! Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-05T03:37:00Z SecondChildTAG: If everyone put in as much effort as you, Myrimit, to express yourself clearly, in spite of the language barrier, there would be no issue whatsoever! :D I don't want to prevent anyone from benefiting from discussion. Rather, I think everyone -- native english speakers or not -- would benefit from clear, carefully communicated messages. I made a few suggestions above which I think would encourage the preponderance of good posts and filter out the noise, to everyone's benefit. SecondChildUserIdTAG: 287805 SecondChildUserNameTAG: Beneficial SecondChildCreateTimeTAG: 2012-10-05T19:57:18Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-05T20:18:43Z FirstChildTAG: Beneficial - While I understand your frustration at the trying to find useful information on this forum, I respectfully disagree with the idea of making posts "private" or otherwise hiding them. I feel it would make the forum seem exclusionary rather than inclusive, and the class cohort might degenerate into cliques or factions. It would make some students feel second-class, and that's not right. It might also facilitate breaches of the Honor Code. To quote a famous American jurist on holding public hearings by governmental bodies, "Sunshine is the best disinfectant." Although my native language is English, I often read posts in other languages with which I have some familiarity (Spanish, French). I haven't answered any posts in Spanish in this iteration of 6.002x, but I answered a few simple ones in Spanish in the pilot 6.002x. Of course, with Myrimit's technical expertise and her eloquence in at least two languages (and perhaps more, I don't know if she also speaks Japanese), I don't even need to think about answering the Spanish posts! There are vast differences in written and spoken English in the world (e.g. Commonwealth English vs. US English), and some of what sounds strange to one English-speaker may sound perfectly natural to another. I believe it was George Bernard Shaw who described the Americans and the British as "two peoples separated by a common language." Some of what you perceive as non-native English might actually be attributable to geographic differences in language usage (there are similar geographic differences in other languages, I'm aware of some of the Spanish and French differences). Even among native English speakers, their language skills vary widely. Myrimit is not a native English speaker, but she communicates in English far better than some of the so-called native English speakers I encounter in my daily life. People who are working on improving their English language skills, along with trying to master the difficult material of 6.002x, need exposure and access to everything here, not barriers and restrictions. The real issue here (in my opinion), as opposed to language barriers and issues, is the suboptimal feature set of the forum software and organization of the forum itself. The pilot 6.002x used Askbot, which the edX staff says does not scale easily. Indeed, we had some technical issues using Askbot, and we only had about 25,000 active students using the forum (my estimate, this is not an official MITx stat). The edX platform requires a robust forum that can support larger numbers of students. Askbot allowed us to create our own subject tags and easily view information about each post without having to tediously click on each subject to see if it was of interest. Users also earned "karma points" which, at specific levels, allowed them to perform some moderation of the forum, e.g. closing posts, marking offensive material, deleting or editing others' posts. All students could delete their own posts. The big drawback of the pilot 6.002x course forum was the lack of top-level subject areas, all the posts were dumped in one big pot. This new 6.002x forum has the same drawback of lack of structure and top-level organization, and none of the good features from Askbot. I particularly dislike that we cannot create our own subject tags and cannot tag a post with more than one. Having only a set of canned tags for use severely limits the usefulness of filtering by tag. No one is omniscient, and limiting the available subject tags to a canned list presumes that the list creator(s) have perfect knowledge of what subjects we might have questions about, and that's simply not possible. Perhaps what we should be asking for here is not ability to exclude some students from viewing or responding to others' posts, but improvements in the organization/structure of the posts, and feature improvements in the forum software. FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-10-06T17:21:45Z SecondChildTAG: I don't disagree with anything you've said. I think the whole language aspect of my post has come across as a bit over stated. I don't want to exclude anyone. As I said originally "I realize that making posts 'private' ... would deprive others of the benefit of reading the discussions therein, so this approach probably goes too far." It's not just language, but a simple lack of effort some people are putting into their posts to communicate clearly. It has the effect of muddying the waters and making the discussions forums, as a whole, not particularly useful, because it takes far too much time and effort to sift through the vast amounts of useless posts, in order to find the good stuff. You're right that "the real issue ... is the suboptimal feature set of the forum software and organization of the forum itself." Which is why I made suggestions about a) how to improve filtering and b) how to encourage the overall quality of posts to improve by instituting a few minor changes to the functionality of the forums. I think some of the concepts in this course can be quite complex, and one should be very clear and careful when discussing them, or else it's very easy to get muddled. I think a confused, muddled discussion actually does a disservice to the person trying to learn. I'm merely making suggestions on how to improve the discussion forums so that it's a tool to which someone can quickly refer in order to clarify a concept, rather than an all-inclusive, disorganized, collection of disjoint thoughts, the reading of which only serves to render the reader more confused than before he started. SecondChildUserIdTAG: 287805 SecondChildUserNameTAG: Beneficial SecondChildCreateTimeTAG: 2012-10-06T18:22:35Z SecondChildTAG: Hi g_hopper! Thanks for the compliment ;). I have to confess you that I understand Japanese language too haha! But what it is most difficult for me it is writting-reading and speaking in Japanese (although I can do it with a lot of effort, but a lot haha),so I have decided not to answer in japanese haha, but I undestand if someone talks with me in Japanese like my grandmother does haha :). I really like languages, I would like to learn Esperanto someday (although many people told me that it will be a waste of time because nobody talks it haha, but I will haha). See you! Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T02:38:33Z FirstChildTAG: Thanks Beneficial for the suggestion. I personally feel the same too. I would also like to appeal to everyone to read what topics have been already discussed and then start a new post. Although this has been mentioned in the discipline too but i feel that most of us have skipped it. It would a lot easier if everyone posted their problems related to one topic in the same thread. Cheers. FirstChildUserIdTAG: 310108 FirstChildUserNameTAG: Saakar FirstChildCreateTimeTAG: 2012-09-30T19:07:37Z FirstChildTAG: I give up... I have posted both a comment to the thread above (should be before Myriam's comment) and a separate response, and while both seemed to have been successfully submitted, each has disappeared from the thread into the bit bucket. Yet another suboptimal feature (or shall I call it what it is, a BUG), in the forum software. :( FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-10-07T02:56:49Z SecondChildTAG: yes, I happened to me once when I did a long post. Now I try to, before submitting , copy and paste it to a .docx for any case that I lost it ;). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T04:32:54Z SecondChildTAG: *It happened SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T04:35:14Z FirstChildTAG: Ok, I'm going to try again (this time I had the good sense to save my reply in a text file, I am tired of typing!) Beneficial - Ok, then we're on the same page. Communication skills do not always correlate well with one's ability to perform the work, and I think we have to have very wide latitude and tolerance for language and communication skill variance. Lack of "effort" may not be the problem, and simply exhorting folks to "try harder" isn't likely to resolve the problem. It's a larger issue beyond the scope of this course to resolve. In my work (I'm an engineering support manager), I have a polyglot and international staff whose command of English and accents (even the native speakers) can be very challenging and difficult to understand. Some of my best electronics technicians have trouble communicating in any language, but they are top-notch troubleshooters. One of my staff members who was born in the PRC has trouble making himself understood both in English and Chinese; his fellow ETs who speak his dialect of Chinese have trouble understanding him in both languages. But he's my go-to guy for difficult and intractable problems, so I know he understands English well enough to have mastered the material (all our technical training is in English), and he definitely knows how to perform the work, and that's all that really matters. Those of us with good communications skills need to ensure that we write as clearly and simply as possible to assist all students, especially those who are challenged by English. So let's focus on making suggestions to the edX staff for forum organization and feature improvements. If it were up to me, I'd make the forum software feature set more like Askbot or any of the myriad other platforms that are in widespread use, and have top-level topics, organization, user interface, and "sticky posts" that enhance ease of use. FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-10-07T02:59:34Z IndexTAG: 339 TitleTAG: Please check this solution If we will go node analysis we will get the following equation (e - V1)/R1 + (e - (-V2))/2 = 0 At present do not worry about the V1 and V2 values as 5 and -7.2. We will consider only V1 and V2. If you will see in above equation this part (e - (-V2)), Here we see two "-" 1. Voltage source V2 is connected in reverse way hence -V2. 2. second one is for the potential difference and e and -V2 hence (e - (-V2) Now if we will solve the equation, you will the following equation eR2 - V1R2 + eR1 + V2R1 = 0 therefore e = (R2V1 - R1V2)/(R1 + R2) Now start putting V1 = 5 and V2 = -7.2 values in the above equation, you will get the answer 6.206. Point need to understand is See do not confused with V2 = -7.2 -ve value assigned to V2 and this voltage source is connected in reverse way. Hence there are two point which need to understand is 1. Voltage source is -ve means it generates -ve voltage. 2. Voltage source is connected in reverse manner. Hence result is absolutely right and that is 6.206. UserIdTAG: 181935 UserNameTAG: vishsahare CreateTimeTAG: 2012-09-29T08:52:10Z VoteTAG: 7 CoursewareTAG: Week 2 / Node equation review CommentableIdTAG: 6002x_S3E1_NodeEquationReview NumberOfReplyTAG: 0 IndexTAG: 340 TitleTAG: Midterm preparation and pattern I am little worried about the midterm exam.as many students have previously taken this course and experienced it before,please tell me about the average number of questions,their difficulty level and the preparation.Do we also get a question where circuit is to be made like in labs. UserIdTAG: 99628 UserNameTAG: Pranjal16 CreateTimeTAG: 2012-09-28T17:52:53Z VoteTAG: 7 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: There were no labs for mid-terms and finals the last time. The exam is definitely challenging and very interesting. It pushes you apply what you learn and even discover things your self - which is a lot of fun. Don't bother too much about difficulty. Do the assignments and labs and you'll be fine :-) They'll hopefully release a few (actually many) practice exam papers a week or two before the exam so that you can get a feel of it. You can also discuss these practice questions on the forums. FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-09-28T18:42:22Z SecondChildTAG: :), Totally agree ashwith! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-28T20:13:36Z IndexTAG: 341 TitleTAG: fainally lab 2 is finished by "try and error":) UserIdTAG: 395257 UserNameTAG: blackguitar CreateTimeTAG: 2012-09-24T03:24:42Z VoteTAG: 7 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: Nice! FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-24T03:36:59Z FirstChildTAG: Lab 2 does'nt needed trial and error .............. It could have been approached logically to get to the answer.........Those figures (1/2V1+1/6V2) mean something ............ It was not brute force problem FirstChildUserIdTAG: 237236 FirstChildUserNameTAG: Roxxy FirstChildCreateTimeTAG: 2012-09-24T04:04:48Z FirstChildTAG: don't want hire blackguitar to wire my house - one of the "try and error" can set it on fire :) FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-09-24T07:50:04Z SecondChildTAG: Re: Setting Things on Fire - My Lab 2 Test graph looks 'right' but get's the red 'X'. The DC test indicates there's 17.5 Amps going through the circuit. Could be something burning something burning somewhere! SecondChildUserIdTAG: 150057 SecondChildUserNameTAG: Felinae SecondChildCreateTimeTAG: 2012-09-24T10:34:10Z FirstChildTAG: lab 2 is finished by "try and error":) But have you learned something?? FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-09-24T11:09:51Z SecondChildTAG: hhhhhh,you are right, but I've followed a "revers technique to get it" :) SecondChildUserIdTAG: 395257 SecondChildUserNameTAG: blackguitar SecondChildCreateTimeTAG: 2012-10-01T01:15:14Z FirstChildTAG: Yes, I've been working hard over lab 2 too... But it helped me to gain deeper understanding of Thevenin and Norton methods =) FirstChildUserIdTAG: 326787 FirstChildUserNameTAG: Angstrem FirstChildCreateTimeTAG: 2012-09-24T08:05:42Z IndexTAG: 342 TitleTAG: Staff At what point does a post become a violation of the Honor Code During the past week there have been those who have worked hard to give help solving Lab 2 without just giving the answer outright. What was the point? If posts like this are permitted, it would seem that the Honor Code is pointless. https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/505ec30cc162402b0000003f UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-23T17:54:30Z VoteTAG: 7 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: It is not permitted. Problem is that the staff is not here any time : /. Moreoever, couldn't we get more admins from other places in the world than the US ? (different time zones ?) FirstChildUserIdTAG: 396446 FirstChildUserNameTAG: RousseauxS FirstChildCreateTimeTAG: 2012-09-23T18:14:26Z FirstChildTAG: Thank you for highlighting the offending thread. It's been deleted and participants noted. FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-23T19:07:55Z IndexTAG: 343 TitleTAG: Can't get into Circuit Sandbox in Firefox browser This URL goes to https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Overview/Circuit_Sandbox/ and there is just a blank box - no circuit sandbox. It should work with Firefox? UserIdTAG: 228637 UserNameTAG: DCounsell CreateTimeTAG: 2012-09-21T19:28:13Z VoteTAG: 7 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Same issue.. FirstChildUserIdTAG: 156974 FirstChildUserNameTAG: ManojKumar FirstChildCreateTimeTAG: 2012-09-21T19:32:33Z FirstChildTAG: We're pushing a fix for this right now. Should be working in a few minutes. Thank you for your patience. FirstChildUserIdTAG: 11 FirstChildUserNameTAG: dormsbee FirstChildCreateTimeTAG: 2012-09-21T22:51:23Z IndexTAG: 344 TitleTAG: Trial and error sequence I think that trial and error sequence should be the following: Choose a value for vd and plug in the right hand side Get a new value for vd from the left hand side Plug in the value got from point 2 in the right hand side Repeat from point 2 If you plug 0.32V in the left hand side, as stated in the video, you get 1V again from the right hand side and not 0.65V. On the contrary, you must plug 0.32V in the right hand side to get 0.65V from the left hand side and repeat until convergence. UserIdTAG: 376877 UserNameTAG: AndBre CreateTimeTAG: 2012-09-21T11:02:52Z VoteTAG: 7 CoursewareTAG: Week 3 / Analyzing Nonlinear Circuits CommentableIdTAG: 6002x_analyzing_nonlinear_circuits NumberOfReplyTAG: 1 FirstChildTAG: Obviously there is a mistake with the right-left and vice versa substitution of the number found. What is needed is a convergence to a unique number as stated. It can be done if the new number found is put always at the same side. FirstChildUserIdTAG: 206669 FirstChildUserNameTAG: dimitrios66 FirstChildCreateTimeTAG: 2012-09-21T14:33:19Z IndexTAG: 345 TitleTAG: Tutorials very helpful If you're like me, and are having trouble with some of the homework questions, I have found that the tutorials are very helpful in putting you on the right path. UserIdTAG: 237167 UserNameTAG: Dug CreateTimeTAG: 2012-09-21T01:00:59Z VoteTAG: 7 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 346 TitleTAG: Homework Detailed Solution edX Staff - The 'Detailed Solution' visible in the homeworks after clicking on 'Show Answer' is terrific! Much better than the generic PDF handouts in the pilot course. I appreciate that the values used in the solutions are the actual randomized values assigned to the specific student's problem set, and that the solutions are presented on the same page as the homework (lots less paper-shuffling when it's time to study). Good job. UserIdTAG: 94557 UserNameTAG: g_hopper CreateTimeTAG: 2012-09-18T03:52:19Z VoteTAG: 7 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: The detailed solutions rock like socks in a box! Kudos to all involved. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-09-18T04:38:54Z FirstChildTAG: Hey i haven't got grade on the home work FirstChildUserIdTAG: 178115 FirstChildUserNameTAG: nehamakhija FirstChildCreateTimeTAG: 2012-09-18T04:34:12Z SecondChildTAG: If you submitted your Week 1 homework on time (prior to midnight your local time on 16-Sept), you should see green ticks/checks next to correct answers and/or red Xs next to incorrect ones. You can see your grades on the 'Progress' tab. Go to the homework and click on 'Show Answers' to see the correct answers and the detailed solutions for each. SecondChildUserIdTAG: 94557 SecondChildUserNameTAG: g_hopper SecondChildCreateTimeTAG: 2012-09-18T05:24:08Z IndexTAG: 347 TitleTAG: Two solutions According to VI-plot the unknown element may have two different resistance values. Why there is only one right solution for current and voltage which corresponds to the lowest resistance? There is also a solution for the bigger one. UserIdTAG: 345452 UserNameTAG: ale-yoman CreateTimeTAG: 2012-09-16T20:13:14Z VoteTAG: 7 CoursewareTAG: Week 3 / Analyzing Nonlinear Circuits Exercise CommentableIdTAG: 6002x_analyzing_nonlinear_circuits_exercise NumberOfReplyTAG: 1 FirstChildTAG: I have solved the problem for the larger resistance RD=2k and get i around 0.6mA, this solution is wrong because RD = 2k only if i > 1mA (according to the graph) FirstChildUserIdTAG: 146814 FirstChildUserNameTAG: humink FirstChildCreateTimeTAG: 2012-09-18T15:03:30Z SecondChildTAG: Agree. even i have same view SecondChildUserIdTAG: 111320 SecondChildUserNameTAG: jhalife SecondChildCreateTimeTAG: 2012-09-21T06:44:51Z SecondChildTAG: First try to find the Thevenin equivalent circuit and then analyze the circuit when you hook up a resistor. Would the current increase? decrease? SecondChildUserIdTAG: 138769 SecondChildUserNameTAG: ArturoPrado SecondChildCreateTimeTAG: 2012-09-24T16:32:18Z SecondChildTAG: humink, the solution you get for RD=2k is wrong not only because you end up out of the range where RD=2k is true, but because you cannot substitute the nonlinear element with a resistor of that value. If you try to use your method on the same circuit with VS=10V, you get a solution which falls in the range where the larger value of resistance holds, but is wrong nontheless. With VS=10V, you get VTH=6.36V and RTH=2.99k. Using your method, you get vD=VTH*2k/(RTH+2k)=1.27V (voltage divider). Assuming this value for vD, you get a current of 1.70mA on RTH but a smaller current on the nonlinear element (use the graph to determine the iD corrisponding to 1.27V). The two currents should be equal, so the solution is not correct. SecondChildUserIdTAG: 84801 SecondChildUserNameTAG: marcuspag SecondChildCreateTimeTAG: 2012-09-25T09:15:16Z IndexTAG: 348 TitleTAG: Will I be shocked? ![enter image description here][1] At node C if we apply KCL Sigma i towards or away at any instant is 0. So, will I be shocked? [1]: https://edxuploads.s3.amazonaws.com/13477324556275579.png UserIdTAG: 374590 UserNameTAG: ahope CreateTimeTAG: 2012-09-15T18:09:00Z VoteTAG: 7 CoursewareTAG: Week 1 / Circuit topology CommentableIdTAG: 6002x_S2E1_Circuit_Topology NumberOfReplyTAG: 1 FirstChildTAG: It depends. Node c is not explicitly tied to ground, but your feet will be. Between node c and ground is a capacitance, holding a voltage potential and a charge difference. When you touch node c, this voltage difference will discharge through you, and would give you an electric shock enough energy is stored in this capacitance. Schematically, this entire circuit is considered "floating", because none of the nodes are grounded. KCL is inadequate for this schematic because the ground potential is not modeled. FirstChildUserIdTAG: 151427 FirstChildUserNameTAG: RichardZ FirstChildCreateTimeTAG: 2012-09-15T18:14:09Z SecondChildTAG: If node C is grounded or at voltage equal to that of ground I hope I will not get a shock SecondChildUserIdTAG: 374590 SecondChildUserNameTAG: ahope SecondChildCreateTimeTAG: 2012-09-15T18:21:23Z IndexTAG: 349 TitleTAG: steps to solve I think the easy way to solve this problem is: 1. Find out v3 in terms of a1*V1+a2*V2. Use Nodal Analysis for the same. 2. express i1 = (V3 - V2)/R1 and here use earlier result and reduce it in terms of a1 and a2. which comes as : i1 = (a1-1)*(V1/R1) + (a2/R1)*V2 Simply put the values and you'll get the answer. 3. Similar to step 2 express i2 = (V3 - V2)/R2 and solve it. i2 = (a1/V2)*V1 + (a2 - 1)*(V2/R2) I hope this helps... UserIdTAG: 178840 UserNameTAG: mehtanayanv CreateTimeTAG: 2012-09-15T16:55:15Z VoteTAG: 7 CoursewareTAG: Week 2 / Linear combinations of source strenghts CommentableIdTAG: 6002x_S3E2_Linear_Combinations_of_Source_Strengths NumberOfReplyTAG: 1 FirstChildTAG: yeah.. but i1 = (V3-V2)/R1 (as above...) seems to be mistake.. what about i1 = (V3-V1)/R1...? FirstChildUserIdTAG: 380287 FirstChildUserNameTAG: gayetan FirstChildCreateTimeTAG: 2012-09-16T22:07:51Z IndexTAG: 350 TitleTAG: Q2 Hint confused I now know what to do for question 2. Thanks to: http://puu.sh/12InC But why i do it, and how it was figured out from the hint still remains a mystery. UserIdTAG: 208105 UserNameTAG: jmunya CreateTimeTAG: 2012-09-14T23:37:28Z VoteTAG: 7 CoursewareTAG: Week 1 / AC power CommentableIdTAG: 6002x_S1E3_AC_power NumberOfReplyTAG: 2 FirstChildTAG: thanx for both of ya!! helped a lot!! FirstChildUserIdTAG: 343615 FirstChildUserNameTAG: Ayoosha FirstChildCreateTimeTAG: 2012-09-15T20:43:14Z SecondChildTAG: Thanks, auch(for the calculus). SecondChildUserIdTAG: 141004 SecondChildUserNameTAG: Rafael_J SecondChildCreateTimeTAG: 2012-09-16T07:48:25Z FirstChildTAG: To answer your question, there are a couple of things happening with that equation. To calculate power, we have the equation: $P= \cfrac {V^2}{R}$ Since our $V$ is a sinusoid, in order to take the average, we have to integrate over one cycle (from $ 0 $ to $\frac {1}{60}$) and divide by the length of that cycle (that is, the period, which is $T= \frac {1}{60}$). The rest is just integral calculus using some clever trig identities to make it easier to compute. FirstChildUserIdTAG: 248902 FirstChildUserNameTAG: Chanute FirstChildCreateTimeTAG: 2012-09-15T00:54:01Z SecondChildTAG: Also it might be useful to note that there is an easy way to calculate the period. Generally the sine wave is written as: $sin(wt)$ And the period would be: $T=2 \cdot pi / w$ In this case: $w = 2 \cdot pi \cdot 60$ So: $T = 2 \cdot pi / (2 \cdot pi \cdot 60) = 1 / 60$ SecondChildUserIdTAG: 136253 SecondChildUserNameTAG: chemeng SecondChildCreateTimeTAG: 2012-09-16T07:15:27Z SecondChildTAG: Or of course if they give you the frequency already, like in this case **60Hz**, the period is just the inverse of the frequency SecondChildUserIdTAG: 136253 SecondChildUserNameTAG: chemeng SecondChildCreateTimeTAG: 2012-09-16T07:18:35Z IndexTAG: 351 TitleTAG: Joe shop some body please help in finding the value of heater 1 heater 2 and heater 3 and total power of Job shop ? UserIdTAG: 415376 UserNameTAG: imab90 CreateTimeTAG: 2012-09-14T17:37:12Z VoteTAG: 7 CoursewareTAG: Week 1 / AC power CommentableIdTAG: 6002x_S1E3_AC_power NumberOfReplyTAG: 5 FirstChildTAG: Hi imab90! How are you? Can I help you? ---------- *As this is a graded part I can colaborate with you in order that you can find by your own the answer. But I can not provide an explicit answer...only hints ;)* ---------- So let's try to figure it out what the Statement says: Joe wants to heat his 12'X20' workshop with electric heat. He has hired the HACME electrician company to build the system. **They propose to use three 1020.0W 240V** baseboard heaters to provide a **total heating capacity of 3060.0W**. (A **heater is basically a resistor**. This is not quite true, because there is a thermostatic switch incorporated into the heater and because the resistance of a heater varies a bit with its temperature. But we will use a linear resistor as a model of a heater.) ![enter image description here][1] (Remember that Students can have differents values in their assigments): **HINTS:** **Part 1:** How much current is expected to be drawn from the power line by this heating system when all three heaters are on? Ok, first of all they tell us that H behaves like a Resistor. So, basically they are three resistors in paralell, isn't it ? So, one way to find a power is Ptotal = (V)^2 / Rtotal Do we have Ptotal? yes, Do we have V? yes, Can we find Rtotal? yes, you can! ;) having Rtotal, And can we use the Ohm Law ? yes !!! So... How much current is expected to be drawn from the power line by this heating system when all three heaters are on?... I think that know you can answer this by your own. **PART 2:** If instead, HACME chose to implement the system with 120V heaters, how much current would have been needed? Hint: remember that the P it is V*I ... **PART 3:** As a consequence, Joe found his workshop too cold. H1 was weak; H2 and H3 barely warmed up. What power was being dissipated in H1? Lets try to think... Can we find H1?... If we have our Rtotal calculated in part 1) and if we now that H1=H2=H3=H... the Rtotal will be expressed only in terms of H...So, can we calculate H1? yes! hint: remember that P=(V)^2/R, what is the voltage in H1? what is the resistence in H1? Can we find the Power?, yes! **PART 4:** What power was being dissipated in H2 (or in H3)? really similar to PART 3. **PART 5:** So the total heating power in Joe's shop was? Do you have the power in part 3) and part 5) ? so, what is the total power? ;) ---- I hope this can help you a little bit.Try it! I am sure that you can solve it by your own. Myriam. [1]: https://www.edx.org/static/content-mit-6002x/images/circuits/heaters-parallel.2bfb1fc52110.gif FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-14T21:37:15Z SecondChildTAG: I calculated Rtotal=3H , from which i calculated H1=5.24 and took the voltage through H1 as 240V then the power =240*240/5.24=10992.3 ...i think this is wrong.Can any one help me out where i went wrong. SecondChildUserIdTAG: 429168 SecondChildUserNameTAG: arunprakashavm SecondChildCreateTimeTAG: 2012-09-15T09:30:20Z SecondChildTAG: Thanks myriam SecondChildUserIdTAG: 382532 SecondChildUserNameTAG: emmanuelpeace SecondChildCreateTimeTAG: 2012-09-15T10:23:26Z SecondChildTAG: ok so total resistance is 18.11 , there fore if H1=H2=H3=H then H =18.11/3= 6.03 ohms. P=I2R , my answer is still wrong . Please can some one help. SecondChildUserIdTAG: 178115 SecondChildUserNameTAG: nehamakhija SecondChildCreateTimeTAG: 2012-09-16T15:51:19Z SecondChildTAG: please help me SecondChildUserIdTAG: 178115 SecondChildUserNameTAG: nehamakhija SecondChildCreateTimeTAG: 2012-09-16T15:51:35Z SecondChildTAG: its urgent please help me SecondChildUserIdTAG: 178115 SecondChildUserNameTAG: nehamakhija SecondChildCreateTimeTAG: 2012-09-16T18:05:09Z FirstChildTAG: I calculated Rtotal=3H , from which i calculated H1=5.24 and took the voltage through H1 as 240V then the power =240*240/5.24=10992.3 ...i think this is wrong.Can any one help me out where i went wrong. FirstChildUserIdTAG: 429168 FirstChildUserNameTAG: arunprakashavm FirstChildCreateTimeTAG: 2012-09-15T09:30:52Z SecondChildTAG: Rtotal (from the first circuit) should be Rtotal = H / 3 Why: **1/Rtotal = 1/H + 1/H + 1/H** ==> **1/Rtotal = 3/H** ==> **Rtotal = H/3** SecondChildUserIdTAG: 305491 SecondChildUserNameTAG: lefam SecondChildCreateTimeTAG: 2012-09-15T11:48:48Z SecondChildTAG: ok so total resistance is 18.11 , there fore if H1=H2=H3=H then H =18.11/3= 6.03 ohms. P=I2R , my answer is still wrong . Please can some one help. SecondChildUserIdTAG: 178115 SecondChildUserNameTAG: nehamakhija SecondChildCreateTimeTAG: 2012-09-16T18:12:19Z FirstChildTAG: If Rtotal = H then how can u use the formula Rtotal = H/3. I calculated Rtotal = V^2/P (240 rms^2/3630) so if H1=H2=H3= Rtotal then H1 =Rtotal so then P=v^2/rtotal but this wrong and I don't know why. FirstChildUserIdTAG: 224302 FirstChildUserNameTAG: tcc626 FirstChildCreateTimeTAG: 2012-09-15T15:24:21Z SecondChildTAG: ok so total resistance is 18.11 , there fore if H1=H2=H3=H then H =18.11/3= 6.03 ohms. P=I2R , my answer is still wrong . Please can some one help. SecondChildUserIdTAG: 178115 SecondChildUserNameTAG: nehamakhija SecondChildCreateTimeTAG: 2012-09-16T18:12:26Z FirstChildTAG: i am not geeting ur last thrree problems answers pls he;p FirstChildUserIdTAG: 227777 FirstChildUserNameTAG: umairhassankhanniazi FirstChildCreateTimeTAG: 2012-09-16T06:39:52Z SecondChildTAG: ok so total resistance is 18.11 , there fore if H1=H2=H3=H then H =18.11/3= 6.03 ohms. P=I2R , my answer is still wrong . Please can some one help. SecondChildUserIdTAG: 178115 SecondChildUserNameTAG: nehamakhija SecondChildCreateTimeTAG: 2012-09-16T15:08:57Z SecondChildTAG: some one please help me SecondChildUserIdTAG: 178115 SecondChildUserNameTAG: nehamakhija SecondChildCreateTimeTAG: 2012-09-16T15:50:30Z SecondChildTAG: i just have the last three left SecondChildUserIdTAG: 178115 SecondChildUserNameTAG: nehamakhija SecondChildCreateTimeTAG: 2012-09-16T15:50:53Z FirstChildTAG: Solve it if you can, Joe wants to heat his 12'X20' workshop with electric heat. He has hired the HACME electrician company to build the system. They propose to use three 1300.0W 240V baseboard heaters to provide a total heating capacity of 3900.0W. (A heater is basically a resistor. This is not quite true, because there is a thermostatic switch incorporated into the heater and because the resista nce of a heater varies a bit with its temperature. But we will use a linear resistor as a model of a heater.) In the proposed system the heaters are connected in parallel with the 240V 60Hz AC power line (modeled by a voltage source) as shown in the diagram: Remember (from Exercise S1E3: AC power) that AC power-line voltages and currents are specified as RMS values. So 120V AC heats a given resistance exactly as much as 120V DC would heat that same resistance. How much current is expected to be drawn from the power line by this heating system when all three heaters are on? ..... If instead, HACME chose to implement the system with 120V heaters, how much current would have been needed? ..... Notice that this would require much heavier and more expensive wire to distribute the power. Back to the original plan with 240V power. Unfortunately, Sparky, who works for HACME, was a little sleepy that day. He accidentally connected the heaters as shown below: As a consequence, Joe found his workshop too cold. H1 was weak; H2 and H3 barely warmed up. What power was being dissipated in H1? ..... What power was being dissipated in H2 (or in H3)? .... So the total heating power in Joe's shop was: ..... No wonder Joe was cold. FirstChildUserIdTAG: 130118 FirstChildUserNameTAG: manis FirstChildCreateTimeTAG: 2012-09-16T21:03:19Z IndexTAG: 352 TitleTAG: Video Player Updates Hi All, Thank you all for your reports about the video player. While we are responding to some of the updates that Youtube, our video hosting service, has put out with their API, we have temporarily removed some of our customizations of the video player. Specifically, the speeds and transcripts are not currently available. We are aware of the issue, and we thank you for your patience. -Lyla UserIdTAG: 96452 UserNameTAG: Lyla CreateTimeTAG: 2012-09-13T17:04:18Z VoteTAG: 7 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: The videos don't work at all for me right now.. how long do you guys think it's going to take before it's back? FirstChildUserIdTAG: 270160 FirstChildUserNameTAG: Sethhhh FirstChildCreateTimeTAG: 2012-09-13T18:22:19Z SecondChildTAG: It's back! SecondChildUserIdTAG: 79361 SecondChildUserNameTAG: kimt SecondChildCreateTimeTAG: 2012-09-13T23:47:38Z SecondChildTAG: please give me the link of the online line player urgent SecondChildUserIdTAG: 136490 SecondChildUserNameTAG: Ali_PU SecondChildCreateTimeTAG: 2012-11-24T05:55:24Z IndexTAG: 353 TitleTAG: Midterm Dear MITx Team, The midterm exam is scheduled for the 25th of Oct. How will that work ? Is there a paticular time ? Cause there is a huge time differnce between MIT and my place. Regards UserIdTAG: 149844 UserNameTAG: amitaratnayake CreateTimeTAG: 2012-09-12T18:49:59Z VoteTAG: 7 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: I am not from, the edX team, but I suppose to solve this problem, they will have it at GMT time, maybe 12:00 noon, i.e., the same time when the course began, for us( I am from India too), it should be in the evening sometime...hope so FirstChildUserIdTAG: 169784 FirstChildUserNameTAG: thewiredbear FirstChildCreateTimeTAG: 2012-09-12T20:16:42Z IndexTAG: 354 TitleTAG: Did anyone else... Did anyone else hear the dog barking in the background or am I going insane? UserIdTAG: 175876 UserNameTAG: Sunden CreateTimeTAG: 2012-09-10T17:08:59Z VoteTAG: 7 CoursewareTAG: Week 1 / Lumped Element Abstraction CommentableIdTAG: 6002x_lumped_element_abstraction NumberOfReplyTAG: 4 FirstChildTAG: Yes and yes. ;) FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-10T17:15:12Z FirstChildTAG: ya i heard FirstChildUserIdTAG: 358302 FirstChildUserNameTAG: girishgowtham FirstChildCreateTimeTAG: 2012-09-10T17:50:51Z FirstChildTAG: Yes, it is the Dog of Prof. Anant Agarwal. The Dog´s name is "Rana" ;). FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-10T18:21:41Z FirstChildTAG: yes :) FirstChildUserIdTAG: 283286 FirstChildUserNameTAG: AliFarid717 FirstChildCreateTimeTAG: 2012-09-11T20:19:14Z IndexTAG: 355 TitleTAG: SOLUTION link Please keep this as reference and solve on your own http://dnovelz.webnode.com/album/photogallery/#img-2364-jpg UserIdTAG: 148523 UserNameTAG: span993 CreateTimeTAG: 2012-09-10T15:43:46Z VoteTAG: 7 CoursewareTAG: Week 1 / KVL/KCL example CommentableIdTAG: 6002x_S2E3_Using_KVL_KCL_and_VI_constraints NumberOfReplyTAG: 3 FirstChildTAG: thanks man for this, I couldn't solve the problem without your draw. it showed me that i2 in my schematic in the wrong direction. because I didn't notes the plus sign for R2. thanks again FirstChildUserIdTAG: 292360 FirstChildUserNameTAG: Loai FirstChildCreateTimeTAG: 2012-09-11T00:33:58Z FirstChildTAG: Thank you so much!!! thats was light to my darkness..... FirstChildUserIdTAG: 50947 FirstChildUserNameTAG: Picho FirstChildCreateTimeTAG: 2012-09-12T04:36:35Z FirstChildTAG: Thank you. I don't think circuits with more than one power supply were explained well in the material. Did I miss something? Hey guys, don't forgot the vote this thread up if you like the answer. FirstChildUserIdTAG: 332090 FirstChildUserNameTAG: WPurin FirstChildCreateTimeTAG: 2012-09-12T20:34:19Z IndexTAG: 356 TitleTAG: Vth Hi, I need help to resolve Vth! can someone shows me how can I do that? please! UserIdTAG: 166949 UserNameTAG: SaulOviedo CreateTimeTAG: 2012-09-10T03:24:52Z VoteTAG: 7 CoursewareTAG: Week 2 / Simple Thevenin CommentableIdTAG: 6002x_S3E4_Simple_Thevenin NumberOfReplyTAG: 3 FirstChildTAG: The point is that you should simple ignore the incoming current i (or assume i=0). FirstChildUserIdTAG: 326787 FirstChildUserNameTAG: Angstrem FirstChildCreateTimeTAG: 2012-09-13T16:14:41Z FirstChildTAG: You should calculate voltage v from the initial circuit, simply by node method. Let the node between Vs and R2 be grounded, and the top node voltage will be unknown e. Equation will be (e-5)/R1+e/R2=0. And e equals VTH. FirstChildUserIdTAG: 294712 FirstChildUserNameTAG: Blinchick FirstChildCreateTimeTAG: 2012-09-10T07:28:05Z SecondChildTAG: This is voltage divider. v=Vs*R2/(R1+R2) SecondChildUserIdTAG: 258500 SecondChildUserNameTAG: karas SecondChildCreateTimeTAG: 2012-09-10T16:03:58Z SecondChildTAG: You simply imagine that you have nothing on terminals and you see a cirquit with two series resistors, and the output voltage will be equal to the voltage drop on the R2 (you have to imagine that you take a voltmeter and measure a voltage on the terminals of R2, it really helps a lot and simplifies your life). Then all you have to do is just to use the Ohm's law and some algebra: Vr2=Vs(R1+R2)*R2 where Vs(R1+R2) is a current through this cirquit. SecondChildUserIdTAG: 446597 SecondChildUserNameTAG: garry_crannon SecondChildCreateTimeTAG: 2012-09-17T10:11:18Z SecondChildTAG: I did the same thing as karas and gary. I got the right answer. So maybe it was the right method. SecondChildUserIdTAG: 254346 SecondChildUserNameTAG: moijes12 SecondChildCreateTimeTAG: 2012-09-20T18:31:58Z FirstChildTAG: you can use voltage divider to find vr2 it very simple to do Vr2=R2*vs/(R2+R1) then vth is equal to Vr2. FirstChildUserIdTAG: 403660 FirstChildUserNameTAG: abuodeh FirstChildCreateTimeTAG: 2012-09-20T20:21:15Z IndexTAG: 357 TitleTAG: H1P3 HELP I decided to make this post so people can complete this part of the homework. For the first two questions (a and b) you need to consider that the heaters are conected in parallel, so what do they have in common? Also, remembering the basic equation of power should be useful in this part. For question b you need to ask yourself, what changes? To answer questions c and d, you first need to know the resistance of the heaters and after that it should be really easy to solve this circuit using the node method. How do you get the resistance of the heaters? Let's remember what they said at the beginning of this problem: "They propose to use three 1530.0W 240V baseboard heaters", what do you think that means? And for question e it's just a sum of the power of each heater. UserIdTAG: 118020 UserNameTAG: maikpaniagua CreateTimeTAG: 2012-09-10T01:44:52Z VoteTAG: 7 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: thanks! FirstChildUserIdTAG: 301736 FirstChildUserNameTAG: MOjangole FirstChildCreateTimeTAG: 2012-09-10T12:24:02Z FirstChildTAG: > For question b you need to ask yourself, what changes? This is the problem. 240V heater -> 1530.0W 240V 120V heater -> y W 120V 120V heater -> z W 240V y,z = ? Also in the proposed system's power line is still 240V 60Hz AC, isn't it? FirstChildUserIdTAG: 387070 FirstChildUserNameTAG: gfdtk FirstChildCreateTimeTAG: 2012-09-11T07:43:51Z IndexTAG: 358 TitleTAG: node analysis the tutorial was really helpful...:) UserIdTAG: 393582 UserNameTAG: faisalrahman CreateTimeTAG: 2012-09-09T06:25:04Z VoteTAG: 7 CoursewareTAG: Week 1 / Nodal analysis with floating voltage CommentableIdTAG: 6002x_FloatingVoltage NumberOfReplyTAG: 1 FirstChildTAG: Yes vey Help full we feel that we are in Lab FirstChildUserIdTAG: 156694 FirstChildUserNameTAG: mkprasanth FirstChildCreateTimeTAG: 2012-09-15T02:07:05Z IndexTAG: 359 TitleTAG: Negative I am also confused why the "entering the source" is negative. Shouldn't using it be negative and supplying it be positive? UserIdTAG: 230787 UserNameTAG: Gabriel007 CreateTimeTAG: 2012-09-07T20:55:02Z VoteTAG: 7 CoursewareTAG: undefined CommentableIdTAG: 5_Simple_Power NumberOfReplyTAG: 4 FirstChildTAG: Basically, what a voltage source is doing is "forcing" the voltage potential at one end to be V volts higher than at the other end. If this relative potential happens to be less than the potential at the other end of the resistor, then current will be flowing towards the voltage source, as in ee, as with most other subjects, energy moves from high to low potentials rather than vice versa. It IS, however, "supplying" power in that it is decreasing the "negative" flow of power through that route by 4/3 A, but this is just not enough to compensate for the larger potential for flow towards the voltage source. Hope this helps! FirstChildUserIdTAG: 26931 FirstChildUserNameTAG: Csscade FirstChildCreateTimeTAG: 2012-09-07T22:05:07Z FirstChildTAG: Because resister is doing consumption of power while source is giving out the power FirstChildUserIdTAG: 309803 FirstChildUserNameTAG: PurnenduK FirstChildCreateTimeTAG: 2012-09-08T05:52:16Z FirstChildTAG: hi, i am understanding this theme more easy, first a few elements generate energy for this reason your power is negative, second, a few elements use this energy an change the energy for example in hot, this elements use the energy and your power is positive; is the same concept that use KVL theorem energy conservation thank FirstChildUserIdTAG: 343291 FirstChildUserNameTAG: LuisEduardoH FirstChildCreateTimeTAG: 2012-09-11T03:04:19Z FirstChildTAG: Please refer to the textbook chapter 1.5.3, it is clearly explained using figures 1.19 and 1.20. FirstChildUserIdTAG: 214990 FirstChildUserNameTAG: Shuaibu FirstChildCreateTimeTAG: 2012-09-08T13:56:00Z IndexTAG: 360 TitleTAG: my recipe the simplest way is: 1. imagine there is no voltage source, then *two R in parallels* and current source, find voltage, since its in parallel each R would have this voltage. 2. then imagine there is no current source, so its simple voltage divider and you know voltage on R2 3. summ up voltage provided by current source (step 1) and voltage source (step 2) and youll get voltage on R2 the rest is simple KVL UserIdTAG: 95280 UserNameTAG: Fortyq CreateTimeTAG: 2012-09-07T10:56:34Z VoteTAG: 7 CoursewareTAG: Week 1 / KVL/KCL example CommentableIdTAG: 6002x_S2E3_Using_KVL_KCL_and_VI_constraints NumberOfReplyTAG: 1 FirstChildTAG: This is called superposition method, which we will learn later;p FirstChildUserIdTAG: 157610 FirstChildUserNameTAG: mradziwo FirstChildCreateTimeTAG: 2012-09-07T21:39:20Z IndexTAG: 361 TitleTAG: Signs I still don't completely understand the convention we are using and in essence the signs. Can someone explain. UserIdTAG: 311818 UserNameTAG: oyiemeke CreateTimeTAG: 2012-09-06T18:00:36Z VoteTAG: 7 CoursewareTAG: Week 1 / Associated reference directions CommentableIdTAG: 6002x_S2E2_Associated_Reference_Directions NumberOfReplyTAG: 4 FirstChildTAG: Remember when current acrosses the element from + to -, voltage is positive. Both loops have 2 elements so, both elements must have the same voltage but opposite cause KVL says, the sum of v's is zero. FirstChildUserIdTAG: 249375 FirstChildUserNameTAG: MarIgones FirstChildCreateTimeTAG: 2012-09-06T21:44:59Z FirstChildTAG: The +- outside the loop are the terminal labels, and the positive direction of current is always towards the positive labeled terminal for every element. I don't think the problem assumes a convention for labels on the terminals. I got them confused with the (electrical potential?) +- signs on the inside of the second loop. The first loop has them aligned so it wasn't an issue. FirstChildUserIdTAG: 252415 FirstChildUserNameTAG: ggerold FirstChildCreateTimeTAG: 2012-09-07T06:44:54Z FirstChildTAG: Basically the algorithm is as follows: 1) for each branch, pick + and - *arbitrarily*, so you get terminal variable v. 2) i must be directed in a way so that it enters "+", so that it's direction is uniquely determined by the step 1 The reason for that is simply a convenience and reduction of confusion (hopefully). From what I understand, you can actually assign everything at random and all equations will hold. The idea is just to use unified convention that everyone follows. FirstChildUserIdTAG: 189680 FirstChildUserNameTAG: vnd FirstChildCreateTimeTAG: 2012-09-08T16:58:50Z FirstChildTAG: We are using a (virtual) voltmeter here, we have to check the +- signs around v1, v2, v3 and v4 for they hint how the measurements are made. FirstChildUserIdTAG: 296965 FirstChildUserNameTAG: LGMailhos FirstChildCreateTimeTAG: 2012-09-08T23:58:03Z IndexTAG: 362 TitleTAG: Confusion about reporting the signed current in question 3 The DC simulation shows the signed current of -500uA; whereas the answer box to question 3 accepts 500uA as the correct answer. Any thoughts on this issue? UserIdTAG: 252722 UserNameTAG: vaboro CreateTimeTAG: 2012-09-06T17:05:25Z VoteTAG: 7 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 4 FirstChildTAG: The circuit simulator shows -500 uA going **into** the voltage source, which is equivalent to 500 uA going **out** of the voltage source. In turn, the specific question asks about the current going into the rest of the circuit from the voltage source. FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-06T17:12:12Z SecondChildTAG: Thanks for the clarification; I had also wondered about this. SecondChildUserIdTAG: 207014 SecondChildUserNameTAG: StMark SecondChildCreateTimeTAG: 2012-09-06T17:29:53Z SecondChildTAG: Excellent! Thank you! However, I'm still confused. Let's look at what the problem says: "The DC analysis annotates each voltage source with the current it is providing to the rest of the circuit. The simulator denotes **the sign of the source current as positive if the current is flowing out of the positive terminal of the source**, through the rest of the circuit, and then back in to the negative terminal. And the source current **sign is negative if the current is flowing out of the negative terminal of the source**, through the circuit, and into the positive terminal of the source. Please report the total resistance of the path connecting the positive and negative terminals of the source, the signed current flowing from the source into the circuit, and (using the voltage value for the source) verify that Ohm's law is obeyed." The simulator shows that the current flows out of the negative terminal into the positive terminal; thus, the sign is negative. The problem asks to report "the signed current flowing from the source into the circuit". The current flows from the negative terminal of the source into the circuit and then into the positive terminal of the source. The sign according to definition denotes the direction of the current. If one assumes one part of the current is positive and other is negative then by definition the currents should collide. Even thought your explanation seems brilliant I don't see how it complies with the definitions/assumptions... SecondChildUserIdTAG: 252722 SecondChildUserNameTAG: vaboro SecondChildCreateTimeTAG: 2012-09-06T18:49:02Z SecondChildTAG: No, no! Thank you!:) After a month of study I understand a lo-o-ot more about circuits:) If one considers this transition philosophically... even mathematically... it would result in infinity:) SecondChildUserIdTAG: 252722 SecondChildUserNameTAG: vaboro SecondChildCreateTimeTAG: 2012-10-06T01:21:59Z FirstChildTAG: I also do not understand this issue, as it would seem "-500u" would be the correct answer, whereas only "500u" is accepted as correct. FirstChildUserIdTAG: 266602 FirstChildUserNameTAG: XKaliber FirstChildCreateTimeTAG: 2012-09-06T23:48:26Z FirstChildTAG: The issue of sign is conventional but, it would help if that convention were a bit better described in early course material and if clarification was given to readings of the simulator. FirstChildUserIdTAG: 194450 FirstChildUserNameTAG: wrbuckley FirstChildCreateTimeTAG: 2012-09-07T01:22:32Z FirstChildTAG: DC Simulation shows the actual flow of current frm -ve to +ve but when we enter the data we should follow conventional flow i.e from +ve to -ve FirstChildUserIdTAG: 140076 FirstChildUserNameTAG: snarayana FirstChildCreateTimeTAG: 2012-09-06T21:02:51Z SecondChildTAG: It is nice to have this detail made explicit, however finding the explicit statement in the discussion is less efficient than were the explicit statement to be found in the lesson material. SecondChildUserIdTAG: 194450 SecondChildUserNameTAG: wrbuckley SecondChildCreateTimeTAG: 2012-09-07T15:35:37Z IndexTAG: 363 TitleTAG: S1E5: KVL-0 Please check S1E5: KVL-0 task. I suppose that point between resistors 2 and 3 (let call it point A) can not have the lowest potential (than potentials 1-2 and 2-3 ). In this case current will flow from point A in both sides, so dq/dt !=0 UserIdTAG: 191310 UserNameTAG: Kalashnikov CreateTimeTAG: 2012-09-06T10:17:22Z VoteTAG: 7 CoursewareTAG: Week 1 / KVL and KCL exercises CommentableIdTAG: 6002x_kvl_kcl_exercises NumberOfReplyTAG: 3 FirstChildTAG: So true. Situation described in S1E5 is not possible in "EESC playgound", imo. FirstChildUserIdTAG: 128561 FirstChildUserNameTAG: gusevoy FirstChildCreateTimeTAG: 2012-09-06T10:24:35Z SecondChildTAG: Maybe not, but KVL still applies. SecondChildUserIdTAG: 256880 SecondChildUserNameTAG: Jetbeard SecondChildCreateTimeTAG: 2012-09-06T14:33:51Z FirstChildTAG: Double-check your signs on the voltage sources FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-06T11:09:39Z SecondChildTAG: there is no voltage sources SecondChildUserIdTAG: 191310 SecondChildUserNameTAG: Kalashnikov SecondChildCreateTimeTAG: 2012-09-06T14:29:25Z FirstChildTAG: I've deleted wrong comment. FirstChildUserIdTAG: 191549 FirstChildUserNameTAG: Fodin FirstChildCreateTimeTAG: 2012-09-06T13:51:25Z IndexTAG: 364 TitleTAG: Explanation : why "-2" ! that source is a power source so it doesn't dissipate power. Otherwise, it generate power. so the power consumption must be negative. UserIdTAG: 213615 UserNameTAG: ahmeddaif1 CreateTimeTAG: 2012-09-06T08:49:13Z VoteTAG: 7 CoursewareTAG: undefined CommentableIdTAG: 5_Simple_Power NumberOfReplyTAG: 1 FirstChildTAG: The question of signs is always very confusing in physics. Especially when one uses vague terminology. One can summarize all the possible signs like that: - Power dissipated in the resistor = 2 Watts - Power supplied by the resistor = -2 Watts - Power dissipated in the source = -2 Watts - Power supplied by the source = 2 Watts But when it comes to *power, entering* something. I'm not really sure what does one actually mean by that... FirstChildUserIdTAG: 325358 FirstChildUserNameTAG: KKostya FirstChildCreateTimeTAG: 2012-09-06T12:43:10Z SecondChildTAG: Resistor is a load and it does not supply any power, but it always consumes. I am going to post a comment. Plz read it. SecondChildUserIdTAG: 378096 SecondChildUserNameTAG: Jamshaid271 SecondChildCreateTimeTAG: 2012-09-06T16:31:17Z IndexTAG: 365 TitleTAG: Using integration I used integration using the statement "(Hint: you compute the average power by integrating the instantaneous power over one cycle of the waveform.)", but I got the wrong answer. Then I discovered the hint should have said "(Hint: you compute the average power by integrating the instantaneous power over one cycle of the waveform and divide it by the period.) When I did this, the answer (133.2) was correct. UserIdTAG: 132828 UserNameTAG: rharris CreateTimeTAG: 2012-09-06T05:35:35Z VoteTAG: 7 CoursewareTAG: Week 1 / AC power CommentableIdTAG: 6002x_S1E3_AC_power NumberOfReplyTAG: 1 FirstChildTAG: Thanks for the hint rharris. This worked for me as well. Integrating V^2/R from t=0 to t=1/60 gives an answer of 2.1818. Dividing this by the period (1/60) provides the correct answer of 130.91. FirstChildUserIdTAG: 340194 FirstChildUserNameTAG: Shortbutlucky FirstChildCreateTimeTAG: 2012-09-06T17:13:02Z IndexTAG: 366 TitleTAG: signs the N network is a source - so the current direction is from source's plus to source's minus - so the current direction is opposite to showed (entering) direction - so it's negative for our case UserIdTAG: 342603 UserNameTAG: YakovO CreateTimeTAG: 2012-09-06T00:56:19Z VoteTAG: 7 CoursewareTAG: Week 1 / Power CommentableIdTAG: 6002x_S1E2_Power NumberOfReplyTAG: 0 IndexTAG: 367 TitleTAG: Help on H1P1 In network A all the resistor is in series. so adding up all the resistor will be the answer In network B all the resistor is in parallel connection .So here we have to apply law of parallel connected resistor 1/Rp=(1/R1+1/R2+.....+1/Rn) In network C last two resistor are connected in series then their combined resistor Rs=R+R is parallel connected with third resistor R and finally their combination (1/2*R+1/R)^-1 is series connected with forth one. It is obvious that is series connection resistance increases and in parallel connection resistance decreases.so largest-valued resistance will be obtained by adding the three resistors and smallest-valued resistor will be the resistance of parallel connection of the three resistor. if R1=4ohms R2=4 ohms and R3= 6 Ohms then largest-valued resistor= R1+R2+R3 and smallest-valued resistor=1/(1/R1+1/R2+1/R3). Now in the last question it is obvious that three resistor are connected in parallel. so pick the least one say R1=4. We know power W=V^2/R=I^2/R from the equation calculate V,from V^2/R1=1W. In parallel connection the voltage across the resistors and their combined resistor is same. So, by using the calculated V , determine the power by the equation V^2/smallest-valued resistor(which is previously calculated). Forgive my poor English UserIdTAG: 14730 UserNameTAG: Tanvir CreateTimeTAG: 2012-09-05T17:47:31Z VoteTAG: 7 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 4 FirstChildTAG: Thanks, last question became easy. FirstChildUserIdTAG: 18177 FirstChildUserNameTAG: MianSaad FirstChildCreateTimeTAG: 2012-09-06T08:21:21Z FirstChildTAG: Nice FirstChildUserIdTAG: 17862 FirstChildUserNameTAG: aemiliotis FirstChildCreateTimeTAG: 2012-09-06T11:18:54Z FirstChildTAG: for parallel resistors (R*R*R)/(R+R+R) is right or not? FirstChildUserIdTAG: 122513 FirstChildUserNameTAG: Mona77 FirstChildCreateTimeTAG: 2012-09-08T18:53:12Z SecondChildTAG: not SecondChildUserIdTAG: 14730 SecondChildUserNameTAG: Tanvir SecondChildCreateTimeTAG: 2012-09-09T16:52:20Z FirstChildTAG: The answer boxes will often also accept special circuit math symbols. R+R can be accepted. For parallel values the math symbol is ||, so R||R is accepted. Try it out! FirstChildUserIdTAG: 88550 FirstChildUserNameTAG: Neil_S_Berry FirstChildCreateTimeTAG: 2012-09-06T01:38:18Z IndexTAG: 368 TitleTAG: H1P1 question "Given that each individual resistor can dissipate up to 1 watt of power before burning up, how much total power in watts ( W ) can the smallest-valued composite resistor dissipate before burning up?" I really don't understand what we are after here.Somebody please tell me what to do, thanks. UserIdTAG: 64973 UserNameTAG: GabrielC CreateTimeTAG: 2012-09-05T16:54:00Z VoteTAG: 7 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 5 FirstChildTAG: [] FirstChildUserIdTAG: 300827 FirstChildUserNameTAG: PFonseca FirstChildCreateTimeTAG: 2012-09-05T16:56:59Z FirstChildTAG: Well it's pretty simple, just get the value of supply from the 1 W and 4 ohm resistor. Now using this value of supply get the power dissipation in the lowest value combination of resistances. FirstChildUserIdTAG: 367886 FirstChildUserNameTAG: KashwinKohli FirstChildCreateTimeTAG: 2012-09-05T17:18:35Z SecondChildTAG: That is more like it - thanks a million :). I got the right value in no time. I wish it would have been explained better. SecondChildUserIdTAG: 64973 SecondChildUserNameTAG: GabrielC SecondChildCreateTimeTAG: 2012-09-06T05:30:33Z SecondChildTAG: this question is not correctly understand yet any body help to understand this question? SecondChildUserIdTAG: 197422 SecondChildUserNameTAG: sheryar SecondChildCreateTimeTAG: 2012-09-06T14:11:10Z SecondChildTAG: Thx so much ! ^^ I didn't understandig it so well, but thx to you i calculated it in a moment ^^ SecondChildUserIdTAG: 200496 SecondChildUserNameTAG: Ranieri SecondChildCreateTimeTAG: 2012-09-06T16:58:49Z FirstChildTAG: ok FirstChildUserIdTAG: 369440 FirstChildUserNameTAG: masifbilal FirstChildCreateTimeTAG: 2012-09-09T12:37:20Z FirstChildTAG: explain please FirstChildUserIdTAG: 369440 FirstChildUserNameTAG: masifbilal FirstChildCreateTimeTAG: 2012-09-09T12:38:53Z FirstChildTAG: Start by calculating the current in each leg of the parallel circuit (obeying KCL). From those current values, you can calculate the power dissipated by each resistor. The maximum value for the power dissipation by each resistor (VI) is known from the statement of the problem. Once you have found the power dissipated by each resistor, you can find the total power dissipated in the circuit by summing the power dissipated by each resistor. FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-09T13:38:23Z IndexTAG: 369 TitleTAG: Help with program functionality My interactive tool does not show DC, tran and ac buttons. What do I have to do to see this tool properly? I'm using firefox, and explorer does not work as well. Thanks in advance UserIdTAG: 309359 UserNameTAG: abarea10 CreateTimeTAG: 2012-09-05T16:52:17Z VoteTAG: 7 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 4 FirstChildTAG: This issue was already adressed in some other pages of this discussion: The button is located on the top grey bar (as depicted in the introductory video), however sometimes the DC and transient analysis button do not appear. This seems to be your case. To correct this situation you should try to refresh the page and if that do not work get out of the page and try once again. I did not know that with Chrome works properly, bu it is good to know. Thanks :) FirstChildUserIdTAG: 300827 FirstChildUserNameTAG: PFonseca FirstChildCreateTimeTAG: 2012-09-05T16:58:09Z SecondChildTAG: I've already tried what you say, and it still doesn't work. Thanks a lot for your help but chrome wins, :) SecondChildUserIdTAG: 309359 SecondChildUserNameTAG: abarea10 SecondChildCreateTimeTAG: 2012-09-05T17:16:21Z FirstChildTAG: For all those having problems with the tool, I solved it using the one placed in the main screen, under "courseware", called "circuit sandbox": https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Overview/Circuit_Sandbox/ FirstChildUserIdTAG: 245908 FirstChildUserNameTAG: PabloLanga FirstChildCreateTimeTAG: 2012-09-05T17:26:22Z FirstChildTAG: Plz try using chrome, i also faced d similar prblm last time, but with chrome it works very nicely. FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-05T16:55:53Z SecondChildTAG: Thank you, I'm going to try chrome. SecondChildUserIdTAG: 309359 SecondChildUserNameTAG: abarea10 SecondChildCreateTimeTAG: 2012-09-05T17:12:00Z SecondChildTAG: Thank you for the solution. Even I am having the same problem and after your suggestion I am trying chrome. SecondChildUserIdTAG: 138310 SecondChildUserNameTAG: ganeshbhatis SecondChildCreateTimeTAG: 2012-09-05T18:16:34Z FirstChildTAG: Chrome does not work for me either. Its nopt totally down to the browser you are using. As everythign works fine on the MITx platform and also the sandbox is working. I only run macs I don't want to dig out an old PC just to do this. I hope they hurry up and fix it, or at least let us know what is happening. FirstChildUserIdTAG: 15344 FirstChildUserNameTAG: kob FirstChildCreateTimeTAG: 2012-09-05T21:51:53Z IndexTAG: 370 TitleTAG: thank u first of all i would like thank all the MITx team that they provide us with such a great opertunity UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-05T13:02:36Z VoteTAG: 7 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 371 TitleTAG: Using the Tools (lab tutorial) - tran analysis not working The transient analysis works fine in the sandbox, but when I'm supposed to do it in the last part of the "Using the Tools" tutorial, I set my probes, click on "TRAN", set the stop to 5m, click "OK", and then nothing happens, nothing pops out. Am I the only one experiencing this issue? I'm using Firefox 15.0 on Ubuntu. UserIdTAG: 16350 UserNameTAG: skedar CreateTimeTAG: 2012-09-05T12:44:37Z VoteTAG: 7 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 5 FirstChildTAG: I have the same setup but I see them. Can you check on Google Chrome? FirstChildUserIdTAG: 6977 FirstChildUserNameTAG: rocha FirstChildCreateTimeTAG: 2012-09-05T12:47:30Z SecondChildTAG: I am actually too lazy to install Chrome right now. However, it's not such a big deal, as it works fine in the sandbox and in the 2nd week lab. SecondChildUserIdTAG: 16350 SecondChildUserNameTAG: skedar SecondChildCreateTimeTAG: 2012-09-05T12:54:41Z SecondChildTAG: It is not a browser issue as if it was then the sandbox would also show same missing components. This is a bug in the simulator code. I expect they will have seen all these posts about it by now and are probably working on it. But it would be nice if we had confirmation of that as we all stuck now on this. Looks like its now working. A reset button has now appeared and that is able to show the correct compoents on the page. Hooray! SecondChildUserIdTAG: 15344 SecondChildUserNameTAG: kob SecondChildCreateTimeTAG: 2012-09-05T20:17:13Z FirstChildTAG: Same here. FirstChildUserIdTAG: 127195 FirstChildUserNameTAG: princeofsudan FirstChildCreateTimeTAG: 2012-09-05T12:56:15Z FirstChildTAG: I've got this same problem. I tried IE 9,FF 14.0.1 and Google Chrome and i don't see window chart. FirstChildUserIdTAG: 159547 FirstChildUserNameTAG: zyto FirstChildCreateTimeTAG: 2012-09-05T12:50:42Z FirstChildTAG: I have same problem. tried ie9, chrome, firefox. nothing is working. FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-05T16:41:24Z SecondChildTAG: It is not a browser issue as if it was then the sandbox would also show same missing components. This is a bug in the simulator code. I expect they will have seen all these posts about it by now and are probably working on it. But it would be nice if we had confirmation of that as we all stuck now on this. Looks like its now working. A reset button has now appeared and that is able to show the correct compoents on the page. Hooray! SecondChildUserIdTAG: 15344 SecondChildUserNameTAG: kob SecondChildCreateTimeTAG: 2012-09-05T20:16:23Z FirstChildTAG: Looks like its now working. A reset button has now appeared and that is able to show the correct compoents on the page. Hooray! FirstChildUserIdTAG: 15344 FirstChildUserNameTAG: kob FirstChildCreateTimeTAG: 2012-09-06T12:48:53Z SecondChildTAG: Tried again and still not working for me in circuit sandbox. Also no reset button displayed. SecondChildUserIdTAG: undefined SecondChildUserNameTAG: undefined SecondChildCreateTimeTAG: 2012-09-06T20:58:31Z IndexTAG: 372 TitleTAG: Discussion forum schedules Hi, Excited to begin the course, thanks for the opportunity. I realized that you published the discussion forum schedules without any indication of the time zone (or at least I couldn't find it) [Forum Schedule][1] As far as I am based in Europe (as many others), can you please ad some GMT reference for non-US students? Thanks in advance [1]: https://www.edx.org/static/content-mit-6002x/handouts/forum_schedules.47f0cd536066.pdf UserIdTAG: 144671 UserNameTAG: andresdans CreateTimeTAG: 2012-09-05T12:29:37Z VoteTAG: 7 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 4 FirstChildTAG: I agree, i'm from NL so GMT would be nice. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-09-05T12:34:42Z FirstChildTAG: Right - it is in EST. We will update it soon. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-09-05T12:36:44Z FirstChildTAG: I agree as well. FirstChildUserIdTAG: 253902 FirstChildUserNameTAG: sajalok FirstChildCreateTimeTAG: 2012-09-05T12:38:00Z FirstChildTAG: The forum hours handout should be updated now. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-09-05T20:06:01Z SecondChildTAG: Thanks for the Update!! SecondChildUserIdTAG: 144671 SecondChildUserNameTAG: andresdans SecondChildCreateTimeTAG: 2012-09-05T20:50:43Z IndexTAG: 373 TitleTAG: Proctored exam date. When the date of proctored exam will be announced? UserIdTAG: 360053 UserNameTAG: Ohedul CreateTimeTAG: 2013-01-19T07:47:20Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: It is planned ot be announced towards the end of the month. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2013-01-19T11:49:28Z SecondChildTAG: pls keep it after feb-10 2013 . pls SecondChildUserIdTAG: 91261 SecondChildUserNameTAG: mr_sophisticated SecondChildCreateTimeTAG: 2013-01-23T09:36:14Z SecondChildTAG: Are you from India? If yes, I'm guessing you're answering the Gate exam like me :-). If not, there is an entrance exam for us on the 10th of Feb. SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2013-01-28T15:35:11Z IndexTAG: 374 TitleTAG: Now I have another excuse to visit Brazil :) I went to check to Pearson VUE [here][1] if my Country , Argentina, had a Test Centre for edX(I was almost sure that had one)... ![im][2] The bad news is that Argentina don´t have a Test Centre at Pearson VUE :( , the most near Centre is in a neighboring Country (Brazil) that it is 1,433.0 Miles from where I live... ![im][3] Anyway, as I always try to find the possitive of the unfortunately things haha, and as I always wanted to go to Brazil since I have memory - as I always say, it is one of the six places that I have to visit yes or yes before I pass to another unknown state haha -, I will take it as a good news: Now I will have another excuse to visit Brazil! :). ![im5][4] But I guess that I will have to plan it with more time for the next January -next year 2014-, as I will have to go to a foreign Country to take the proctored exam, as now for me it is impossible for the Exams at University and I we will start another stage of the University Investigation Group of Education, yep, I work on Holidays too haha- I am volunteer since almost 3 years ago - haha, and it is possible that I will travel outside my Province-State to a National Congress of Photophysics this year - midle/end , to present a work of my Fellowship as undergraduate student of engineer - Research(it will be my first Congress of Photophysics , I hope to not be nervous. I were last year in the WEEF 2012 too-(World Engineering Education Forum)-but, it was different as it was for education, Sarma Sanjay´s Team of MIT was there too but I coudn´t say hello as I were a day before them presenting a Poster, the only day that I could go...), so I guess that I will have a busy schedule as engineering student this year, also I would like to continue helping to edX. But besides all that activities, I always wanted to take the Proctored Exam of 6.002x - since the Prototype Course-, for me is a challenge, so I will take it yes or yes, no mattering if I have to travel to another Country :) Myriam. P.D. That trip will hurt my pocket as student haha. Anyway, I most of the time don´t take holidays (I say summer holidays, but in fact, my summer holidays are staying studying at home or doing some activity related with academic things, so as I always say, once a time, if you can, it is not bad idea to go outside haha :P). I will try to practise my portuguese. [1]: http://www1.pearsonvue.com/Dispatcher?application=VTCLocator&action=actStartApp&v=W2L&cid=792 [2]: https://edxuploads.s3.amazonaws.com/13582023171343654.bmp [3]: https://edxuploads.s3.amazonaws.com/13582024201343622.bmp [4]: https://edxuploads.s3.amazonaws.com/13582025361343687.bmp UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2013-01-14T22:52:32Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hello, Mirian: I'm from Brazil, in São Paulo. When you come to São Paulo and need any help I can try to seek help. FirstChildUserIdTAG: 361823 FirstChildUserNameTAG: EliasOak FirstChildCreateTimeTAG: 2013-01-15T00:12:01Z SecondChildTAG: Cool! Thank you! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2013-01-15T00:34:02Z SecondChildTAG: Are there any news about the proctored examination dates, prices, etc? Thank you! :') SecondChildUserIdTAG: 87808 SecondChildUserNameTAG: pumoneon SecondChildCreateTimeTAG: 2013-01-15T10:22:16Z SecondChildTAG: Hi pumoneon, [take a look at here, it is a Post with doubts regarding the Proctored Exam ][1]. As far as I know, they will release soon the details. I know that it will cost 95USD but they haven´t provided a precise date of the Proctored. Also you can check if your Country have a Pearson VUE Centre - go to the page of it and clic on edX - and you can check if you have one in your Country. I hope this can help you. Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50e75b95a205372700000012 SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2013-01-15T12:20:15Z SecondChildTAG: Thank you Myrimit :') SecondChildUserIdTAG: 87808 SecondChildUserNameTAG: pumoneon SecondChildCreateTimeTAG: 2013-01-16T22:57:52Z SecondChildTAG: seja bem-vinda ao Brasil, Miriam ;) SecondChildUserIdTAG: 355503 SecondChildUserNameTAG: fabriciogs SecondChildCreateTimeTAG: 2013-01-21T00:57:33Z FirstChildTAG: I've always wanted to visit Brazil too. Should I use this as my excuse to travel to Brazil from Hong Kong? Sure, they have several centers in this city but Brazil.... Thanks for the idea! ;) FirstChildUserIdTAG: 467169 FirstChildUserNameTAG: Eyowzitgoin FirstChildCreateTimeTAG: 2013-01-17T05:00:18Z SecondChildTAG: Haha! 2x1! That will be cool = Holidays in Brazil+Proctored Exam! ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2013-01-17T11:35:26Z IndexTAG: 375 TitleTAG: I love you MIT I love you MIT UserIdTAG: 341358 UserNameTAG: akbar13 CreateTimeTAG: 2013-01-09T13:36:30Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: HAPPY NEW YEAR FirstChildUserIdTAG: 888444 FirstChildUserNameTAG: nikos2803 FirstChildCreateTimeTAG: 2013-01-09T16:22:04Z FirstChildTAG: Thanks for such great materials. FirstChildUserIdTAG: 991596 FirstChildUserNameTAG: Vicki818 FirstChildCreateTimeTAG: 2013-01-10T06:55:28Z SecondChildTAG: +1 SecondChildUserIdTAG: 204019 SecondChildUserNameTAG: Shmatko_Nazar SecondChildCreateTimeTAG: 2013-01-14T09:17:17Z IndexTAG: 376 TitleTAG: Statistics Do anyone know about the statistics of the course? UserIdTAG: 101902 UserNameTAG: DHEERAJK_VITS CreateTimeTAG: 2013-01-04T19:00:11Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 377 TitleTAG: Certificate-- Thanksgiving Today received certificate !! Thank you 6.002 and Anant Agarwal for the awsome course. UserIdTAG: 101902 UserNameTAG: DHEERAJK_VITS CreateTimeTAG: 2013-01-03T12:46:04Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 378 TitleTAG: Another way to solve Q6, if anybody is still interested Here are 2 alternative ways to solve Q6. This doesn't mean that i think it's better than the official explanation, it is just simply another way.Please forgive my writing.![enter image description here][1] ![enter image description here][2] ![enter image description here][3] And Sandbox is always an option to verify your answer. :) ![enter image description here][4] [1]: https://edxuploads.s3.amazonaws.com/13572080104333201.jpg [2]: https://edxuploads.s3.amazonaws.com/13572082771343647.jpg [3]: https://edxuploads.s3.amazonaws.com/13572084504954497.jpg [4]: https://edxuploads.s3.amazonaws.com/13572085074346896.jpg UserIdTAG: 240487 UserNameTAG: AlexAlexandrescu CreateTimeTAG: 2013-01-03T10:22:18Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Great Job! FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2013-01-03T17:32:27Z FirstChildTAG: I like it, Thanks. FirstChildUserIdTAG: 403987 FirstChildUserNameTAG: electricon FirstChildCreateTimeTAG: 2013-01-03T11:58:56Z SecondChildTAG: Thank you ! You are always welcome ! SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2013-01-03T20:55:27Z IndexTAG: 379 TitleTAG: HNY frm Belarus! My congratulations for everybody! This was realy great year, and I hope that 2013 would be 2013 times better, because we can use our new knowledge) Happy New Year! UserIdTAG: 200319 UserNameTAG: Virviil CreateTimeTAG: 2012-12-31T23:04:51Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: happy new year from algeria FirstChildUserIdTAG: 174229 FirstChildUserNameTAG: fares27 FirstChildCreateTimeTAG: 2012-12-31T23:14:24Z SecondChildTAG: Happy New Year fares27 and Virviil!:) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-31T23:46:22Z FirstChildTAG: С новым годом ) FirstChildUserIdTAG: 394836 FirstChildUserNameTAG: v2g6ch4 FirstChildCreateTimeTAG: 2013-01-01T19:01:57Z SecondChildTAG: С новым годом !!! SecondChildUserIdTAG: 297655 SecondChildUserNameTAG: Aljoska SecondChildCreateTimeTAG: 2013-01-02T00:18:35Z IndexTAG: 380 TitleTAG: Great Prof...! Get a look at this!!! http://www.washingtonpost.com/postlive/anant-agarwal-we-can-really-change-the-face-of-education/2012/12/11/3e8d839e-43a4-11e2-8061-253bccfc7532_video.html UserIdTAG: 277808 UserNameTAG: Hemanthmps CreateTimeTAG: 2012-12-27T17:57:38Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I think the prof missed an opportunity there. He began his answer pointing to all the great revolutionary advancements made in all theses other fields, which led me to believe that he was going to point out that with more access to education we'd expect to get even more dramatic revolutionary ideas, as the population of educated people expanded through free online ed. Which, of course, leads to quality of life improvements for all people. But, instead he focused on the narrow point of education itself having remained in the stone age, even while all other circles of activity around it have seen dramatic changes. We educate people today, the same way we did 400 years ago. A guy standing in front of an audience, in a lecture hall, delivering a speech. In fact, this paradigm could probably be traced back 2000 years, for even Jesus stood on the hillside, and delivered his "sermon on the mount" lecture to teach the peoples of his time. But, when the lady interviewer asked why is it good thing, to have free online ed, when she had to pay for her education, she was really looking for the kind of answer that justified giving this expensive service away for free. And the real answer to that, is that we expect dramatic improvement in all of our lifestyles with more educated people around. More problems will be solved, there should be less wars, as people are then capable in reasoning and debate, better health, with expected medical breakthroughs, travel to the stars, with lots of engineers to work on the future spacecraft, etc..The return on investment is huge. If only one guy has knowledge, only he can make advances. And we all have to wait on him to find solutions to problems. In the past, there were such men, when knowledge was young, and so much was yet to be discovered. So, history produced great men like Euler, and Hamilton, and Newton, and Faraday, etc.. Imagine if education was widespread back then, and not held tightly in a few aristocrats. We'd be on mars by now, farming and terraforming the planet. But we had to creep forward, at a snails pace, until recently. Look at China. The nation of China began a massive free education program decades ago, managed by a communist government, and now we buy almost all our products from the Chinese. And they lend us the money to buy all that stuff too. How could they have both the money and the technological know-how? What is their magic? Simple, massive open free education for the masses. And they did it without all this modern online technology. So, imagine what is going to happen, when the whole world comes online, and gets educated for free. A funny thing is happening in the world, if you haven't yet noticed. People are paying to produce and consuming for free. This turns capitalism on its head. Here is a characteristic example: Adobe Reader is free. If I want to read an existing PDF file, I can download the Adobe Reader and consume that product for free. But, if I want "to create" a PDF for other's to read, then I have "to pay" for the Adobe Acrobat software which will allow me "to produce" that PDF document. Funny, huh? Production is now more desirable than consumption. People are willing to pay good money to produce things, but are unwilling, in many cases, to pay anything at all to consume. We've all begun to expect free goods, and free services, and we're only willing to put hand in pocket to pull out the wallet, when we want to create or produce something ourselves, for others to consume. It's magic. Defies all sense of logic. This is not taught in the Economics 101. It's a paradigm that doesn't exist in the theoretical world of the economists. By lending the US the money to keep America afloat, so that the US can continue to buy things from China, even the Chinese are paying, in effect, to produce all the goods we in the west buy from them. Why are they paying us to buy their products? Because of a great secret: **It's more fun to produce.** ;) The hardest part, in all en-devours, is finding the consumers for our product. FirstChildUserIdTAG: 4463 FirstChildUserNameTAG: pmj FirstChildCreateTimeTAG: 2012-12-28T03:28:13Z SecondChildTAG: Hmmm.. very interesting and thought provoking. SecondChildUserIdTAG: 190670 SecondChildUserNameTAG: chauhan1955 SecondChildCreateTimeTAG: 2012-12-28T03:51:24Z SecondChildTAG: Great Comments SecondChildUserIdTAG: 455950 SecondChildUserNameTAG: Wahabbaluch SecondChildCreateTimeTAG: 2012-12-28T04:38:51Z IndexTAG: 381 TitleTAG: Special thanks from Kingdom of Saudi Arabia I would like to thanks DR. agarwal and all stuff in this course . I am happy to pass this course also in same time very sad because the challenging days are over . but is there any courses related to ELECTRICAL POWER because I want improve myself to go on in master degree . thank you very much ........ UserIdTAG: 156535 UserNameTAG: badawiIiIi CreateTimeTAG: 2012-12-27T03:18:36Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: Congratulation on passing the course I'm from Yemen and I'm glad to see Arabic person on course like this FirstChildUserIdTAG: 109941 FirstChildUserNameTAG: Alhashemy FirstChildCreateTimeTAG: 2012-12-27T04:34:37Z SecondChildTAG: me too brother i hope you life in peace and happy SecondChildUserIdTAG: 156535 SecondChildUserNameTAG: badawiIiIi SecondChildCreateTimeTAG: 2012-12-27T06:50:07Z SecondChildTAG: Salute from Egypt SecondChildUserIdTAG: 282561 SecondChildUserNameTAG: ShadyElbassiouny SecondChildCreateTimeTAG: 2012-12-27T21:35:42Z SecondChildTAG: Salamun Alaik, from Pakistan SecondChildUserIdTAG: 455950 SecondChildUserNameTAG: Wahabbaluch SecondChildCreateTimeTAG: 2012-12-28T12:44:40Z SecondChildTAG: Hello from Morocco SecondChildUserIdTAG: 324041 SecondChildUserNameTAG: sadrab SecondChildCreateTimeTAG: 2012-12-28T14:01:04Z SecondChildTAG: Salam from Sudan SecondChildUserIdTAG: 127195 SecondChildUserNameTAG: princeofsudan SecondChildCreateTimeTAG: 2012-12-28T15:40:55Z SecondChildTAG: Salamun Alaik, from Pakistan to. SecondChildUserIdTAG: 18177 SecondChildUserNameTAG: MianSaad SecondChildCreateTimeTAG: 2012-12-28T16:33:32Z SecondChildTAG: Walaikum aslam from UET Lahore SecondChildUserIdTAG: 37464 SecondChildUserNameTAG: DoubleA SecondChildCreateTimeTAG: 2012-12-28T23:28:03Z SecondChildTAG: WAALIKOM ASALAM WELCOMR MY BROTHERS IT IS VERY NICE TO MEET YOU HERE SecondChildUserIdTAG: 156535 SecondChildUserNameTAG: badawiIiIi SecondChildCreateTimeTAG: 2012-12-29T03:53:13Z FirstChildTAG: check out ocw.mit.edu. FirstChildUserIdTAG: 623210 FirstChildUserNameTAG: preveen FirstChildCreateTimeTAG: 2012-12-27T05:36:45Z FirstChildTAG: thank you very much preveen you really helped me FirstChildUserIdTAG: 156535 FirstChildUserNameTAG: badawiIiIi FirstChildCreateTimeTAG: 2012-12-27T07:11:19Z SecondChildTAG: Not at all. MIT deserves the credit. shukran SecondChildUserIdTAG: 623210 SecondChildUserNameTAG: preveen SecondChildCreateTimeTAG: 2012-12-27T09:29:25Z SecondChildTAG: of course preveen SecondChildUserIdTAG: 156535 SecondChildUserNameTAG: badawiIiIi SecondChildCreateTimeTAG: 2012-12-27T19:44:55Z FirstChildTAG: Salaam to my Arab Dudes from Pakistan,Could'nt help posting on seeing word "Saudi Arabia" ,my birth place. FirstChildUserIdTAG: 37464 FirstChildUserNameTAG: DoubleA FirstChildCreateTimeTAG: 2012-12-28T02:58:14Z SecondChildTAG: OOOOH welcome brother DoubleA the great salam to pakistan people from me SecondChildUserIdTAG: 156535 SecondChildUserNameTAG: badawiIiIi SecondChildCreateTimeTAG: 2012-12-28T06:18:52Z SecondChildTAG: Thanks, I am also studying electrical Power in Pakistan, I wonder in which university you studied? I may like to know about some Good Engineering Unversities in Saudi Arab ,will you be kind enough to mail at: aa.uetian@gmail.com SecondChildUserIdTAG: 37464 SecondChildUserNameTAG: DoubleA SecondChildCreateTimeTAG: 2012-12-28T11:30:25Z SecondChildTAG: I studied in YANBU industrial college in kingdom of saudi arabia http://www.yic.edu.sa/ SecondChildUserIdTAG: 156535 SecondChildUserNameTAG: badawiIiIi SecondChildCreateTimeTAG: 2012-12-29T04:02:23Z IndexTAG: 382 TitleTAG: Final Exam Answers and Explanations Posted! I hope you all had a great semester and hopes for the best in the coming year. I just logged in and found that the answers and explanations have been posted for the final. I feel rather silly about the questions I got wrong since they weren't the hard ones. I don't know about the rest of you, but I was not prepared for log-log plots. Now I am learning them well. Also, the question that got me was the shift in Q4 (t-T). I wonder if more EE courses will be offered on edX. Thank you. This was indeed a great experience. UserIdTAG: 349862 UserNameTAG: edsaf8 CreateTimeTAG: 2012-12-26T19:52:24Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hey! Thanks for that info. FirstChildUserIdTAG: 330357 FirstChildUserNameTAG: steryd FirstChildCreateTimeTAG: 2012-12-26T20:44:41Z FirstChildTAG: yes. the "show answer" button is now available in the final exam page. FirstChildUserIdTAG: 623210 FirstChildUserNameTAG: preveen FirstChildCreateTimeTAG: 2012-12-27T05:35:59Z IndexTAG: 383 TitleTAG: Thank you!! I would like to thank everyone who made this course possible and FREE!!! I am very glad I had to opportunity to take the course and I would highly recommend it!!!!!!!! Everyone who participated in the forum was must helpful and always had a great way of shining light when you felt stuck!! I would like to wish you a happy new year and that this new year is filled with many wonderful things for you and your loved ones!!! Best of all to everyone!!! Isis B :D UserIdTAG: 238005 UserNameTAG: isisbocardo CreateTimeTAG: 2012-12-26T04:11:05Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 384 TitleTAG: Thank You Thank You very much Dr.Anant Agarwal and whole team of EDX this course is really great. UserIdTAG: 40209 UserNameTAG: sadaf78 CreateTimeTAG: 2012-12-24T11:37:05Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 385 TitleTAG: What EdX Classes are you taking in 2013? Now that we bid adieu to an awesome semester of 6.002x; I've really enjoyed being a student as well as a Community TA, and with the Final Exam ending in less than five (5) hours, and personally just having finished it, I would like to end on a positive note by asking my fellow students "**What courses will you pursue in the Spring Semester for 2013?**" If not , why not? (i.e. too time consuming, not enough background in math, cannot find an interesting course...) *As for me:* There are several interesting offers from MITx. I am a hardware guy, and my programming has gotten rusty; and I've found the programming language Python interesting, so I have just signed up for **Programming 6.00x** in the Spring. I also had to drop **3.091x Physical Chemistry** this Fall about half-way through because of obligations due to Hurricane Sandy and the damage it caused; but Prof. Cima there teaches in such an interesting manner, I want to definitely finish that course. Be warned that his exams are tough as 6.002x's ( though the Homeworks are not). Just a tip. **Digital Signal Processing** looks interesting, but it's a tough course, at Coursera, as does **VLSI CAD**. I will take a look at these as well. Good luck students in all your future endeavors, both in the Electronics and Computer Engineering field, your careers *(I know we have an eclectic mix here: from high school students looking to get into university, university students looking for that first job, those laid off in this tough economy looking to pick up new skills, as well as retirees looking to keep busy)* and in enhancing your knowledge in general, whatever stage of life you may be in! May we meet again on the 'net, better yet, in an EdX course this Spring :-) Sincerely, Jersey Mark UserIdTAG: 292546 UserNameTAG: JerseyMark CreateTimeTAG: 2012-12-24T05:37:33Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 7 FirstChildTAG: Hi Mark, - I will take 6.00x and CS184.1x - I have registered this Term in order to take a look at them so that I can take an advance and see the material before the classes starts :p - I will re-take 3.091x :p. This Term I am CTA there too, I wish to help more there, also I would like to improve as I am a disaster in Chemistry - this is a personal challenge haha-. The Staff of 3.091x is really kind, I hope to not dissapoint them, so that is why I will re-take and do it better this time ;) - I will register again, for the third time, wow! haha, in 6.002x too in order to help the new students, as I really enjoy doing that. I hope to improve as CTA too. There are so interesting Courses comming in edX! I wish to do all of them. My friends of 6.002x Spring 2012 told me that they will register on Quantum Mechanics and Quantum Computation - CS191x. I am tempted to register in order to take the Course again with them haha. But I can not handle so many Courses at the same time... :/ Good luck Mark! I wish you the best to you! See you, Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-24T06:03:14Z SecondChildTAG: Myrimit! You are there always to help. Thanks! SecondChildUserIdTAG: 443988 SecondChildUserNameTAG: Dheeraj_Garg SecondChildCreateTimeTAG: 2012-12-24T09:18:19Z SecondChildTAG: I also registered 6.00x and cs814.1x ,hopw to see you there:) SecondChildUserIdTAG: 136959 SecondChildUserNameTAG: christerpher SecondChildCreateTimeTAG: 2012-12-24T12:40:10Z SecondChildTAG: You are welcome Dheeraj_Garg. Cool christerpher!:) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-24T15:11:19Z FirstChildTAG: Reasons for taking or not taking MOOC courses will be based mainly on my time constraints, so I opened up http://www.class-central.com/ and created a wish list that will keep me busy for the next couple of years, or longer ... :-) Having already taken Introduction to Artificial Intelligence (the very first course that caused the "avalanche" we see these days) from the ancestor of Udacity and Machine Learning from Coursera, I'm trying to be focused on the Computer Science-Artificial Intelligence-Robotics field but occasionally I took/will take other kind of courses like Circuits and Electronics (boom!) or Contraception (so curious about this course!). Here is the list: 1. Contraception: Choices, Culture and Consequences - Coursera 2. CS212 - The Design of Computer Programs - Udacity 3. Introduction to Databases - Class2Go 4. CS 373: Programming a Robotic Car - Udacity 5. CS188.1x: Artificial Intelligence - EdX 6. CS344 - Introduction to Parallel Programming - Using CUDA to Harness the Power of GPUs - Udacity 7. EP245 - Entrepreneurship: The Lean LaunchPad - How to Build a Startup - Udacity 8. Probabilistic Graphical Models - Coursera 9. Information Theory - Coursera All the best, Andrea from Italy FirstChildUserIdTAG: 376877 FirstChildUserNameTAG: AndBre FirstChildCreateTimeTAG: 2012-12-24T11:53:21Z FirstChildTAG: Hi there, Quantum Mechanics and Quantum Computation sounds very interesting. I had some quantum mechanics during my studies of physics and i always wanted to know more about those weird quantum computing stuff. So i guess i will take that one (come on Myri maybe we could need your help over there as well :D ;-), hm or maybe this time it can be me to be more of a help :D). Elsewise Introduction to Statistics could be very interesting (though maybe less engineering and more mathematics).Scientific Computing in coursera seems to be interesting as well. Man, so much to learn :-) The world seems to be interesting ;-) FirstChildUserIdTAG: 388514 FirstChildUserNameTAG: hdbam FirstChildCreateTimeTAG: 2012-12-24T09:55:35Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-27T18:02:43Z FirstChildTAG: This current offering I took four Edx courses, 6.00x, 6.002x, 3.091x and CS50x. I have good programming background so 6.00x and CS50x require less effort compared to 6.002x, 3.091x is relatively easier since I studied a good deal of univ chemistry for me. I literally spend most of my time on my computer since I have no job, just finished university this year. Here is the thing, if u can please offer both CS50x and 6.00x concurrently, CS50x is still open for registration. They introduce computer science from different perspectives and having both perspectives is very good. 6.00x- **(python and computation for sciences)** while CS50x- **(C and general purpose programming together with a deeper insight into how computers work and web technology(HTML, PHP and MYSQL))**. As for the next offering, I will retake **6.002x** (I missed this current midterm but I just passed the passmark) and **6.00x** to fully cement their principles. I have also registered for the following, the beauty is they have diferent timings; the most courses I will have concurrently are 4-5. **Stat 2.1x,** (to refresh and cement stastics); **CS 188.1x,** I had begun it but dropped out since I had little CS background but that has changed now after offering 6.00x. It has intriguing programming assignments. **CS 184.1x**, this is highly fundamental for me. **CS 169.1x** and eventually **CS 169.2x**. see you guys anywhere on Edx. **Merry Christmas.** FirstChildUserIdTAG: 183507 FirstChildUserNameTAG: obiradaniel FirstChildCreateTimeTAG: 2012-12-24T11:15:52Z FirstChildTAG: I have registered for CS191x and 6.00x. At coursera I am interested in Control of Mobile Robots, Fundamentals of Electrical Engineering, and Exploring Quantum Mechanics. If time becomes a constraint Control of Mobile Robots and Exploring Quantum Mechanics will receive priority if they turn out to be good courses. I am already watching the videos for 6.00x and enjoying them. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-12-24T23:44:35Z SecondChildTAG: Hello, Skyhawk, I registered the CS191 too. Let's meet there. SecondChildUserIdTAG: 361823 SecondChildUserNameTAG: EliasOak SecondChildCreateTimeTAG: 2012-12-26T21:20:03Z SecondChildTAG: Sounds like fun. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-12-27T00:44:35Z FirstChildTAG: I plan on re-taking 6.002x, hoping that with the math and time pressures eased somewhat, I'll have more time to think about the electronics. ;-) Also registered for 6.00x Introduction to Computer Science and Programming - we'll see if RL permits this. ;-) Looking forward to seeing familiar names in the Forums! FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-12-25T15:51:04Z FirstChildTAG: what is the code name for digital signal processing cause i don't see there as among of the offered course FirstChildUserIdTAG: 475448 FirstChildUserNameTAG: msamwelmollel FirstChildCreateTimeTAG: 2012-12-24T11:10:09Z IndexTAG: 386 TitleTAG: Greetings from Peru I studied informatics engineering in Peru a few years ago, and I got a good education; albeit, never delve into many of the topics covered in the course, in that respect, for me was a really challenging to accomplish with all exercises, homework and labs, mainly due to lack of time. In that regard, the great motivators were: 1. The incredible talent of Dr. Agarwal and the entire staff. 2. The feeling of taking a course at MIT, a university in which I have always wanted to study (I visited MIT 3 years ago, and simply I love it). 3. The interest and enthusiasm of many who took the course. Thank you very much to all, Dr. Agarwal for his great ability to teach and make students enjoy and are interested in the class, the entire course staff who provided support and acquitted doubts, and finally, many of my colleagues with whom I started a beautiful friendship. Greetings from Peru "Gracias Totales" UserIdTAG: 219727 UserNameTAG: eveliz CreateTimeTAG: 2012-12-23T01:45:30Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hola, soy de Perú también. **edX y MIT son super geniales.** Es un problema la educación en nuestro país. La mayoría de profesores solo enseña para poder ganar dinero y los estudiantes van a la universidad o institutos solo porque quieren tener un titulo para conseguir trabajo. FirstChildUserIdTAG: 138769 FirstChildUserNameTAG: ArturoPrado FirstChildCreateTimeTAG: 2012-12-23T12:47:46Z IndexTAG: 387 TitleTAG: thanks to EDX I wanna say my thanks to all the edx staff and particularly prof.anant agarwal....i felt very happy to complete the course...thanks a lot UserIdTAG: 118611 UserNameTAG: mitianhari CreateTimeTAG: 2012-12-22T09:42:12Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Indeed it was a great experience. FirstChildUserIdTAG: 272417 FirstChildUserNameTAG: MuKhan FirstChildCreateTimeTAG: 2012-12-22T12:36:22Z IndexTAG: 388 TitleTAG: thank you I am so happy that i took this course for the second time, i first took 64% but now I hit 100% just now, it was a great experience. I am truly thankful for everyone who created this course. And made it available for kids like me. UserIdTAG: 91934 UserNameTAG: Chingun CreateTimeTAG: 2012-12-22T07:21:51Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: congratulation FirstChildUserIdTAG: 226085 FirstChildUserNameTAG: Teto FirstChildCreateTimeTAG: 2012-12-22T09:24:20Z SecondChildTAG: Congratulations Chingun!:) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-22T19:54:42Z IndexTAG: 389 TitleTAG: extra attempts for Q1 & Q2 P Kindly give extra attempts for Q1 & Q2 UserIdTAG: 228440 UserNameTAG: hassanmuhuyuddin CreateTimeTAG: 2012-12-21T16:54:24Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 8 FirstChildTAG: agree FirstChildUserIdTAG: 400128 FirstChildUserNameTAG: RamshaAyub FirstChildCreateTimeTAG: 2012-12-21T17:06:44Z FirstChildTAG: extra attempts for Q1 & Q2 are must be given FirstChildUserIdTAG: 400242 FirstChildUserNameTAG: hassankhan_umt FirstChildCreateTimeTAG: 2012-12-21T17:03:33Z FirstChildTAG: i am also agree with u plz FirstChildUserIdTAG: 400125 FirstChildUserNameTAG: MoazzamKhan FirstChildCreateTimeTAG: 2012-12-21T17:05:06Z FirstChildTAG: agree FirstChildUserIdTAG: 71540 FirstChildUserNameTAG: Sergiohuma9 FirstChildCreateTimeTAG: 2012-12-21T18:42:37Z FirstChildTAG: Are there mistakes on our part in Q1 and Q2 that lead you to use all your attempts? If so, kindly bring them to our attention. --Rohan FirstChildUserIdTAG: 391929 FirstChildUserNameTAG: RohanNagarkar FirstChildCreateTimeTAG: 2012-12-21T18:52:28Z SecondChildTAG: no try i have solved SecondChildUserIdTAG: 342135 SecondChildUserNameTAG: vikash902 SecondChildCreateTimeTAG: 2012-12-21T21:42:07Z FirstChildTAG: I agree! FirstChildUserIdTAG: 90765 FirstChildUserNameTAG: jdavidg FirstChildCreateTimeTAG: 2012-12-21T20:12:22Z FirstChildTAG: Dear wise staff , what value will have a certificate to the people who put long hours into this course , when by begging you can practically redo the exam ? All the exercises were doable with all the tricks we have learned. An exam is an exam and the rules are rules. FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-12-21T22:26:53Z FirstChildTAG: I think Rohan is considering your request for extra attempts if a mistake is found, however he needs to have a reason. Can anyone explain why extra attempts are needed without violating the honor code? FirstChildUserIdTAG: 132828 FirstChildUserNameTAG: rharris FirstChildCreateTimeTAG: 2012-12-21T20:28:47Z SecondChildTAG: They can email their concerns to mit-6002x@edx.org without violating the honor code. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-12-21T20:34:50Z IndexTAG: 390 TitleTAG: FINAL EXAM IS UP! Good luck to all, and sorry once again for the delay. -- Rohan UserIdTAG: 391929 UserNameTAG: RohanNagarkar CreateTimeTAG: 2012-12-19T19:43:35Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 391 TitleTAG: Thanks for the amazing experience! In advance of the final, I just wanted to thank the teachers and staff for the opportunity to take this course. It's been tough, but I've gotten a lot out of it (being currently unemployed, I've been taking this class, 6.00x, and 3.091x simultaneously- this is definitely the hardest of the three, as well as the best organized), and I'm extremely grateful to edX for allowing me and others to do this. I look forward to taking additional courses, and can't wait to see what courses will be offered in the spring. I also wanted to say that Professor Agarwal is an amazing teacher. UserIdTAG: 340370 UserNameTAG: Alaric7 CreateTimeTAG: 2012-12-19T17:17:08Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: so true FirstChildUserIdTAG: 371220 FirstChildUserNameTAG: dan10 FirstChildCreateTimeTAG: 2012-12-19T19:12:04Z FirstChildTAG: Learn a lot of things from this class! It's a perfect class, because of the wanderful teacher--Professor Agarwal and these nice staffes work behind. Thanks!!! FirstChildUserIdTAG: 296511 FirstChildUserNameTAG: sky0917 FirstChildCreateTimeTAG: 2012-12-20T16:13:58Z IndexTAG: 392 TitleTAG: Thinking about H12P3 last part As background, look at circuit A at the top of the problem. We are told that is a low pass filter; that is, its transfer function is one for low frequencies. We can write H for circuit A using the voltage divider method. Hlp = 1/(j*w*C)/(R+1/(j*w*C)) = 1/(1+j*w*R*C) We are told circuit B is a high pass filter. Its transfer function is one for high frequencies. We can write its transfer function as: Hhp = R/(R + 1/(j*w*R)) = 1/(1+1/(j*w*R*C)) When solving the second half of H12P3 we want to write the result as H = A*Hhp*Hlp where A is the mid frequency gain. To solve the second part of the problem we call the circuit composed of R3 and C3 the input circuit with impedance Zin and the circuit composed of R4 and C4 the feedback circuit with impedance Zfb. We the apply the node method at there connection and make use of the fact that V- is a virtual ground. Therefore: (VI-0)/Zin + (VO-0)/Zfb = 0, which can be rearranged as: VO/Zfb = -Vi/Zin or H = VO/VI = - Zfb/Zin Zin = R3 + 1/(j*w*C3) = R3*(1+1/(j*w*R3*C3)) Zfb = R4/(j*w*C4)/(R4+(1/(j*w*C4)) = R4*(1/(1+(j*w*R4*C4))) Combining Zin and Zfb to form H and identifying the terms corresponding to Hlp, Hhp, and A will result in the answer. Note: Don't forget about the minus sign! UserIdTAG: 406202 UserNameTAG: skyhawk CreateTimeTAG: 2012-12-09T13:36:36Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Nice job, skyhawk! FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-12-09T17:23:56Z SecondChildTAG: definitelly SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-09T18:00:10Z FirstChildTAG: excellent! FirstChildUserIdTAG: 74193 FirstChildUserNameTAG: SpaceAgeRobot FirstChildCreateTimeTAG: 2012-12-09T22:14:09Z FirstChildTAG: @ skyhawk:: how wil u get vals of R4,R3,C4,C3 for second half of h12 p3? pls help FirstChildUserIdTAG: 356056 FirstChildUserNameTAG: Ganesh2810 FirstChildCreateTimeTAG: 2012-12-10T07:27:27Z SecondChildTAG: Use the same values as you did for the first part. The high pass filter has the low corner frequency (use big capacitor), and the low pass filter has the high corner frequency (use small capacitor). Choose R so that w*R*C = 1. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-12-10T10:43:34Z IndexTAG: 393 TitleTAG: To Staff: final exam and discussion! please, please, do turn off the post ability during the exam (read-only state). it was frustrating and annoying and very-very distracting during the mid-term!!! thank you!!!! UserIdTAG: 301962 UserNameTAG: labrinim CreateTimeTAG: 2012-12-06T16:46:51Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 394 TitleTAG: pencil vs mouse The mouse is certainly more visible than the dot of the pencil. It could be of another shape just to give a more "sneaky" impression but even as an arrow it serves quite well our needs. UserIdTAG: 206669 UserNameTAG: dimitrios66 CreateTimeTAG: 2012-12-01T12:06:48Z VoteTAG: 6 CoursewareTAG: Week 12 / S23V22 Inverting amplifier input resistance using virtual short method CommentableIdTAG: 6002x_S23V22_Inverting_amplifier_input_resistance_using_virtual_short_method NumberOfReplyTAG: 2 FirstChildTAG: I do absolutely definately agree. The arrow is more likely to be seen on the screen than the pencil. FirstChildUserIdTAG: 414462 FirstChildUserNameTAG: Pablo_C FirstChildCreateTimeTAG: 2012-12-03T12:17:58Z FirstChildTAG: The mouse is surely better than the pencil but I think that the yellow circle I saw in some previous videos (I don't remember the sequence number) should be far more visible. FirstChildUserIdTAG: 376877 FirstChildUserNameTAG: AndBre FirstChildCreateTimeTAG: 2012-12-03T14:25:39Z IndexTAG: 395 TitleTAG: Interesting expression Interesting expression: "very simple complex algebra" :) UserIdTAG: 266912 UserNameTAG: pietvo CreateTimeTAG: 2012-12-01T01:19:04Z VoteTAG: 6 CoursewareTAG: Week 12 / S23V1: Review Previous Amplifier Abstraction CommentableIdTAG: 6002x_S23V1_Review_previous_amplifier_abstraction NumberOfReplyTAG: 1 FirstChildTAG: Ha ha! But I prefer "very simple complex algebra" over "very complex simple algebra". FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-12-01T13:37:28Z SecondChildTAG: - So true! XD SecondChildUserIdTAG: 414462 SecondChildUserNameTAG: Pablo_C SecondChildCreateTimeTAG: 2012-12-03T12:22:12Z IndexTAG: 396 TitleTAG: visible pointer I am participating at edX-course 3.091x, too. There, in the sequences where the Prof explains materials on "a sheet of paper" (and not at the blackboard in class), for example @ S14V4, he uses a pointer surrounded permanentely by a yellow-filled circle regardless if he is writing or only pointing - so he doesn´t need to switch between writing- and pointing mode. This "cursor" allows to be followed very efficiently.... UserIdTAG: 388524 UserNameTAG: dasbinich CreateTimeTAG: 2012-11-30T20:17:46Z VoteTAG: 6 CoursewareTAG: Week 12 / S23V5 Op Amp characteristics CommentableIdTAG: 6002x_S23V5_Op_Amp_characteristics NumberOfReplyTAG: 0 IndexTAG: 397 TitleTAG: How to do a screen capture? Hallo, After watching videos from the first edx course, 6.002x and currently taking 6.00x and 3.091x I have really interested in how the lecturers do the screen video capture. What software are they using? I've tried looking for a software that can do that and I got Microsoft Expression trial and I've tried recording but it can't record powerpoint or word. Is there a way I can work around the non-recording of Word or PPT? I can't record whatever I'm typing in Word and I can't record the same in PPT nor can I record a powerpoint show. Or is there another software that I can use? I've heard of Camtesia but it's really expensive and I just want to do a few dozen tutorial videos for my classmates and university. Can anyone help? if the staff could reply with what software they've been using and even provide me with links for that and how to do videos like Eric Grimson or like the videos from www.powersearchingwithgoogle.com i would really appreciate that. Thank you Chintan UserIdTAG: 84248 UserNameTAG: Chintangohel CreateTimeTAG: 2012-11-29T08:46:15Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: Good question and I'm interested too. How are you doing with 6.00x? I'm following it too, but find it really tough, but very interesting and useful; until now my progress is only 12% Help!!! But I can advise everyone who's following 6.002x to try it at least. If I had done 6.00x before 6.002x, then I would have saved a lot of time on doing the calculations (only use pen, paper and simple calculator) Never new that Python and Pylab provides many of the facilities of MATLAB. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-11-29T12:17:29Z SecondChildTAG: I'm struggling as I've never tried programming and I guess you need to have a head or talent for it. I had done 6.002x in March when they first released it and I found it amazing considering I'm taking Electrical and Electronic Engineering at a local university in Kenya SecondChildUserIdTAG: 84248 SecondChildUserNameTAG: Chintangohel SecondChildCreateTimeTAG: 2012-12-03T07:48:53Z FirstChildTAG: Here is Prof. Agarwal's reply from the Spring discussion forums: > I use a lenovo tablet pc with a swivel screen, with powerpoint slides > as the background. Then write on them using camtasia. I use a > sennheiser headmounted microphone so that the sound quality is good > even if I move my head around, which I tend to do a lot when I get > animated! I also find it useful to have an external keyboard for > hotkeys. I've used a program called ActivePresenter to record screencasts. There is a free version available and it does a good job. FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-11-29T13:57:59Z SecondChildTAG: Camtasia is being sold which is a problem for me; let me look for Active Presenter. Thank you SecondChildUserIdTAG: 84248 SecondChildUserNameTAG: Chintangohel SecondChildCreateTimeTAG: 2012-12-03T07:49:45Z SecondChildTAG: I'm using activepresenter, seems to work well. Here is a video I did https://www.youtube.com/watch?v=cMEwRy2ggJg SecondChildUserIdTAG: 84248 SecondChildUserNameTAG: Chintangohel SecondChildCreateTimeTAG: 2012-12-07T05:57:50Z FirstChildTAG: Might also be worth asking at [Khan Academy][1] what they use over there. [1]: http://www.khanacademy.org/ FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-11-29T22:29:28Z FirstChildTAG: https://www.youtube.com/watch?v=cMEwRy2ggJg FirstChildUserIdTAG: 84248 FirstChildUserNameTAG: Chintangohel FirstChildCreateTimeTAG: 2012-12-07T05:57:56Z IndexTAG: 398 TitleTAG: H10P3 part1 and 2 Hints Requested by brainyash1990 **Hints - H10P3: AN L NETWORK** The inductor and capacitor in the diagram below are part of the output-coupling network of a radio transmitter. The rest of the transmitter (the source of radio-frequency energy) is represented as a Thevenin source, and the antenna load is represented by a resistor. ![image][1] In this problem we will examine some of the characteristics of this circuit. In the spaces provided below you will write algebraic expressions in terms of the part parameters L, C, R1, R2, and the angular frequency ω. (As usual, use w for ω in your expressions.) **Part 1. Visual Hints:** ![im1][2] ![IM2][3] **Part1. Math Hints.** 1- Voltage dividers :) .Recall previous weeks. 2- Capacitor impedance Zc= (1/(j*w*C)) 3- Inductor impedance ZL = j*w*L 4- Simplify the expression ;). 5- (j*w)*(j*w) = -(w^2) ---- **Part 2. Visual Hints.** What does it means VF divided by IF? Is it the impedance seen from that terminals? Take a look at the Box of the next image ;). ![IM3][4] - Simplify the expression ;). - Remember that (j*w)*(j*w) = -(w^2) [1]: https://www.edx.org/static/content-mit-6002x/images/circuits/LC-Lnetwork.deb8d32795bd.gif [2]: https://edxuploads.s3.amazonaws.com/1353784190134365.bmp [3]: https://edxuploads.s3.amazonaws.com/13537847101343688.bmp [4]: https://edxuploads.s3.amazonaws.com/13537857751343638.gif UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2012-11-24T19:46:13Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Hi there. Thank you very much for the hints!! In part 2, i get C²= something. Is that correct? In case yes, how do I represent then Cmatch in the equation field? with (equation)^(1/2)? I'm not getting a correct mark. Thank you very much! FirstChildUserIdTAG: 290028 FirstChildUserNameTAG: Fjuca FirstChildCreateTimeTAG: 2012-11-25T01:44:00Z SecondChildTAG: Nevermind. Got part 3 right. Working on part 4 now. (since I gave up part 1, haha). SecondChildUserIdTAG: 290028 SecondChildUserNameTAG: Fjuca SecondChildCreateTimeTAG: 2012-11-25T01:52:47Z SecondChildTAG: Hi, do you use the driving-point impedance to solve part 3, if yes, please give me a hint. SecondChildUserIdTAG: 296419 SecondChildUserNameTAG: oscman SecondChildCreateTimeTAG: 2012-11-25T02:39:52Z SecondChildTAG: Hi oscman, did you search the forum for H10P3, there are a lot of comments regarding part 3. For example: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/undefined/threads/50a688f55b77f62300000035 SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-25T03:28:45Z FirstChildTAG: Can someone tell me if this is a wrong approach to Part 1: - Create the impedance model, by replacing everything with boxes and complex sources. - Do a node analysis 1) I chose to collapse ZL and ZR2 in series as (ZL + ZR2) and 2) selected ground as below ZC, and 3) calculated e1, the unknown voltage above ZC. - Then I found Vo using Vo =(e1*ZR2)/(ZL + ZR2) - Then I just divided by Vi to find Vo/Vi and I simplified my equation Anyway, I'm getting a wrong answer, apparently. So if someone could tell me whether my analysis was flawed or whether I've taken the right approach and therefore I must have just made some algebraic mistake somewhere, that would be much appreciated. FirstChildUserIdTAG: 287805 FirstChildUserNameTAG: Beneficial FirstChildCreateTimeTAG: 2012-11-25T16:50:07Z SecondChildTAG: Sorry, I did not understand exactly what you did... But take a look at page 724 of the text book... there you have a problem with the same two voltage divider pattern as this exercise. I hope this helps. SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-25T17:33:44Z SecondChildTAG: siigh. This is turning into one of those really annoying problems where I spend hours repeating the solution and it's just some minor algebraic error that hangs me up. I've tried to solve this problem using the node method I sketched above, and now also using the method described on page 724 of the book. I get almost identical answers (answers which differ by only signs) and yet, both are marked wrong by the checker. I'm just getting plain bored of this. SecondChildUserIdTAG: 287805 SecondChildUserNameTAG: Beneficial SecondChildCreateTimeTAG: 2012-11-25T18:40:46Z SecondChildTAG: fiiiinally got it. Sigh. SecondChildUserIdTAG: 287805 SecondChildUserNameTAG: Beneficial SecondChildCreateTimeTAG: 2012-11-25T18:58:22Z SecondChildTAG: well done! SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-25T19:05:32Z FirstChildTAG: Alright. Part 2 is giving me just as many headaches. I would appreciate if someone could tell me if my approach here is correct. I'm looking for Vf/If. From my calculations for part A, I should already have an expression for Vf. If, by inspection, is = (Vf - Vi)/R1 . SO, Vf/If = Vf/( (Vf-Vi)/R1 ) = Vf*R1/Vf-Vi So now I just find the expression for that, and simplify. I won't spell out the whole thing here, but, a pretty messy expression ends up simplifying quite a bit, and I lose the Vi term altogether. Anyway... am I on the right track? FirstChildUserIdTAG: 287805 FirstChildUserNameTAG: Beneficial FirstChildCreateTimeTAG: 2012-11-25T19:54:57Z SecondChildTAG: I hope this helps :) https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50a825c23dae702300000056 SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-25T20:13:22Z SecondChildTAG: Yes. That was very helpful, thank you. Once again, the answer I had was off only by a negative sign. The method you provided was much easier and more intuitive though. Still, I can't find any error in my original calculations and see no reason why it shouldn't work. Don't know why I end up with this negative sign where it shouldn't be. Anyway... I'm getting sooo tired of this week's homework :D SecondChildUserIdTAG: 287805 SecondChildUserNameTAG: Beneficial SecondChildCreateTimeTAG: 2012-11-25T20:22:04Z SecondChildTAG: I guess I should have done Vi-Vf rather than Vf-Vi...maybe that's where my erroneous negative sign came from. Anyway, I don't care to check that fact, at this point. Thanks again for the help matiasgrodriguez SecondChildUserIdTAG: 287805 SecondChildUserNameTAG: Beneficial SecondChildCreateTimeTAG: 2012-11-25T20:23:51Z SecondChildTAG: Hey Beneficial, I'm very glad to help. If you see my question on the other post... It is a very complicated solution... but I did not understand why it didn't worked... Thanks to OrinE for shedding light on the right direction SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-25T20:37:40Z IndexTAG: 399 TitleTAG: Staff help needed: Grader isn't updating changes since edx changed H12P2 I have serious problems with the grader of H12P2. My problem is, that it doesn't matter what anwsers I give to this problem anymore: it never seems to update a good/wrong answer AFTER an INTENTIONALLY given wrong input, say for box1 where vout must be given as a function of vin and resistances. I have spend at least 8 hours to try out how I can CHANGE my inputs, but to no avail ... I've no problems with solving this HW, but only with not updating the input.... I give INTENTIONALLY the answer in box a: inf*vin*(R0/R0) (I use R0/R0 here, to not give away the answer) and why would I do that? Well, a few days ago, before this message was added to the problemset: **NOTE: Due to how these answer fields depend on each other, "H12P2 Linear Regulator" requires all 6 fields to contain answers for the "check" button to work. This means that writing the correct answer in the first two fields can yield a red X if the last part of the problem has not yet been answered. Also, as the last part of this problem is a design problem, the final four responses are related to each other; hence all four responses entered must agree with each other in order to receive a checkmark.** I, and several other people, had problems because we used vIN or VIN and vZ or Vz and so on. After our complaints, the before mentioned message was added, and also this this was changed: (a) Write an expression for the output voltage (vout) when the input voltage is below the Zener voltage (vin UserIdTAG: 83276 UserNameTAG: salsero CreateTimeTAG: 2012-11-22T14:03:17Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Troubleshooting NumberOfReplyTAG: 7 FirstChildTAG: All I can suggest is to clear your browser cache, reload the page, fix part (a) and (b), put something silly in for the part (c) answers - say 1000000 for all of them and a not-unreasonable value for the efficiency, say 0.1 - something like that eventually worked for me. FirstChildUserIdTAG: 141000 FirstChildUserNameTAG: OrinE FirstChildCreateTimeTAG: 2012-11-22T18:43:05Z SecondChildTAG: The problem is that I can't change 1 box and never get any kind of checkmark. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-11-23T09:28:30Z SecondChildTAG: I'm have the same problem I try change computer, browser and the problem continue. that's a bug... SecondChildUserIdTAG: 357453 SecondChildUserNameTAG: BrunoCanoso SecondChildCreateTimeTAG: 2012-11-23T10:53:42Z SecondChildTAG: Hi Bruno, I normally use Firefox and I also emptied the cache, reloaded etc. And I also tried with chrome, but also to no avail. The HW problem itself is easy, you don't need a calculator for this. We still have 2 weeks to go before this problem becomes really urgent, but it bothers me. I've done all the HW and Labs from the spring course and this one, always had 100% score, so in fact I'm already sure of a C for this one. BTW: did you ever see a red cross appear for H12P2? I've never seen it!! SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-11-23T11:54:33Z SecondChildTAG: I have the same problem. I can't check in H12P2! SecondChildUserIdTAG: 345967 SecondChildUserNameTAG: Tsipis SecondChildCreateTimeTAG: 2012-11-24T13:04:53Z SecondChildTAG: Don't use k or M in the boxes, but give the resistance in ohms. Normally I don't have this problem, but today I discovered that it was neccessary with HW12P2. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-11-28T21:14:46Z FirstChildTAG: This is extremely strange, and we are looking into it. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-11-24T03:52:34Z SecondChildTAG: Thank you! SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-11-24T14:02:04Z SecondChildTAG: Today I tried again and I changed my input in the boxes for the resistances from xM, yM, zM (M means mega=million isn't it? It's quite normal to use this notation in electronics) to just ohms and now my answers were accepted. So my problem is solved. But I wonder if you changed something in the s/w or not. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-11-28T13:30:38Z FirstChildTAG: To possibly assist in problem debugging: Same problem as above. If an incorrect answer is entered in any of the available text boxes, e.g. x, and 'check' pressed, the expected error message is returned viz. Invalid input: x not permitted in answer. If x is deleted and a valid formula is entered and 'check' pressed, the valid entry is displayed in the text box but no input to the server appears to happen i.e. the page does not update, (the previous error message continues to be displayed, but with a valid formula in the appropriate text box.) Closing the Browser down and using a different or same browser to access the page, the newly returned page displays x in the text box and Invalid input: x not permitted in answer. It appears as if the server content is not being updated for some reason, except under an error condition. e.g. However if x is changed to y, and check pressed, Invalid input: y not permitted in answer is returned and apparently stores on the server(?) for subsequent download i.e. closing the browser down and reloading page the y input and error is displayed. I am assuming that error checking occurs server side rather than client side. Clearing the Browser cache makes no difference. FirstChildUserIdTAG: 273287 FirstChildUserNameTAG: johnkh77 FirstChildCreateTimeTAG: 2012-11-25T13:26:04Z SecondChildTAG: Today I tried again and I changed my input in the boxes for the resistances from xM, yM, zM (M means mega=million isn't it? It's quite normal to use this notation in electronics) to just ohms and now my answers were accepted. So my problem is solved. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-11-28T13:31:06Z FirstChildTAG: **Please,** **Fix that problem fast, I Want to do my homework and I can't, I have the same problem!** **Thanks!** FirstChildUserIdTAG: 299251 FirstChildUserNameTAG: TheRedBlackOne FirstChildCreateTimeTAG: 2012-11-30T10:32:28Z FirstChildTAG: I have the same problem! FirstChildUserIdTAG: 129687 FirstChildUserNameTAG: Galiya91 FirstChildCreateTimeTAG: 2012-12-08T07:42:58Z FirstChildTAG: Same problem here. I tried various browsers (firefox, chrome) with cookies and cache cleared and I have no checkmarks (red or green) whatsoever. FirstChildUserIdTAG: 351871 FirstChildUserNameTAG: Lamarque FirstChildCreateTimeTAG: 2012-12-09T09:11:41Z FirstChildTAG: Can somebody help me. When I do homework 12 part 2, I made a mistake and put in one of the formulas uin instead of vin. You receive a message that the entry is invalid. And now I can not change the formula and check other tasks in part 2. I removed it, updated browser, cleaned cookies, tried go to another computer, but it did not work. As soon as I load my homework, in the entry box appears not right formula and the second part is not checked at all.I still have about 7 hours to enter my answers, but I can not do that. Please, help me. FirstChildUserIdTAG: 31407 FirstChildUserNameTAG: hippopotamus FirstChildCreateTimeTAG: 2012-12-09T21:53:48Z SecondChildTAG: I am having the same problem with H12P2. I entered an expression with some invalid variable names, and now I can't change it. I've tried clearing my cache and trying a different browser (Safari and Chrome). SecondChildUserIdTAG: 69075 SecondChildUserNameTAG: ErikR SecondChildCreateTimeTAG: 2012-12-10T02:11:12Z IndexTAG: 400 TitleTAG: Suggestion for Edx Staff I've noticed in these lectures and the ones in several other courses that during the slide presentation, all the lecturers tend to struggle for a moment and take a couple seconds just to switch colors of their pens. While watching today, I just thought how nice it might be for them to have a stylus for these courses similar to those little pens we have with the multi-colored ink selection. So, I thought I might suggest to someone to make a stylus for use with the equipment that simply had some buttons on the end for selecting some commonly used colors (black, red, blue, green or perhaps make them customizable) for the professors. This might require a wire (to keep the stylus battery free) or a small bluetooth connection. I don't believe the size would have to be bigger than those pens (or if designed right, you may even be able to use those pen shells as the stylus itself). It might be a bit of work but from the looks of things, it may be worth it as every lecturer I've seen so far has the same problem at some point and it may make it easier. Then again, the work required to make the stylus may be more than it's worth. Either way, just a thought! Come to think of it, that might be a great product for anyone with a drawing pad used for any purpose, particularly those who write or draw a lot on a drawing pad or tablet PC. Now if only I could make one myself to sell... and now you see why I am taking this course. ;) UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-11-11T10:30:28Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Quick! Patent it! FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-11-11T13:24:26Z SecondChildTAG: Unfortunately, patents require some technological know-how and... while I know how to calculate the capacitance of a MOSFET (yay!), I'm still a bit behind making a device as sophisticated as that. A quick glance at the interweb also shows some recent patents similar in function. One in particular would be the Microsoft E-Pen (which seems to have a lot of fancy junk that I'm not sure is really needed nor - to me - all that desirable. I don't think I'll ever get the concept of shaking something to make it change, perhaps because I shake too much on my own). So... I'm thinking I'll have to let this one go and just do more once I learn how. ^_^ SecondChildUserIdTAG: 467169 SecondChildUserNameTAG: Eyowzitgoin SecondChildCreateTimeTAG: 2012-11-12T03:47:51Z IndexTAG: 401 TitleTAG: Pease staff guys I've problems technical problems in Lab 7 When i clicked in the check button, i seen errors in my answers, so when i try to see the transient simulation, it was gone... the sistem doesn´t show me the graphs. Please help me... :/ UserIdTAG: 70519 UserNameTAG: Fipe CreateTimeTAG: 2012-11-06T16:18:58Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 9 FirstChildTAG: i'm scare >.< FirstChildUserIdTAG: 70519 FirstChildUserNameTAG: Fipe FirstChildCreateTimeTAG: 2012-11-06T17:15:26Z FirstChildTAG: Something of Staff guys? FirstChildUserIdTAG: 70519 FirstChildUserNameTAG: Fipe FirstChildCreateTimeTAG: 2012-11-06T19:26:11Z FirstChildTAG: Post screenshots of your problem. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-11-06T16:20:40Z SecondChildTAG: ![previously i checked the answers for some corrections... so when the graphs window doesn't appeared, I did a reset.][1] [1]: https://edxuploads.s3.amazonaws.com/13522194661343625.jpg SecondChildUserIdTAG: 70519 SecondChildUserNameTAG: Fipe SecondChildCreateTimeTAG: 2012-11-06T16:31:55Z SecondChildTAG: Try a different browser...? SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-06T16:33:18Z SecondChildTAG: No, in firefox i've the same problem.... :/ SecondChildUserIdTAG: 70519 SecondChildUserNameTAG: Fipe SecondChildCreateTimeTAG: 2012-11-06T16:50:20Z SecondChildTAG: :'( SecondChildUserIdTAG: 70519 SecondChildUserNameTAG: Fipe SecondChildCreateTimeTAG: 2012-11-06T16:53:35Z SecondChildTAG: Did you find a solution. i find the same problem in all labs. SecondChildUserIdTAG: 70519 SecondChildUserNameTAG: Fipe SecondChildCreateTimeTAG: 2012-11-06T17:33:53Z SecondChildTAG: Try logging out and logging in again. SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-06T17:52:23Z SecondChildTAG: doesn't work ! :´( SecondChildUserIdTAG: 70519 SecondChildUserNameTAG: Fipe SecondChildCreateTimeTAG: 2012-11-06T18:11:16Z SecondChildTAG: Yep, I get the same problem. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-11-06T20:34:19Z FirstChildTAG: I have the same problem. I press TRANS and the graph is not shown :/ FirstChildUserIdTAG: 35815 FirstChildUserNameTAG: Elendorial FirstChildCreateTimeTAG: 2012-11-06T16:22:01Z SecondChildTAG: I am also having the same problem. When I am pressing TRAN the graph is not shown SecondChildUserIdTAG: 333877 SecondChildUserNameTAG: gaurab SecondChildCreateTimeTAG: 2012-11-06T16:40:37Z FirstChildTAG: Hope it gets fixed soon FirstChildUserIdTAG: 35815 FirstChildUserNameTAG: Elendorial FirstChildCreateTimeTAG: 2012-11-06T17:31:48Z FirstChildTAG: Same problem here. I tried Safari and Firefox. There is JS bug in line 4830 ( var plist = content.$('.property');) FirstChildUserIdTAG: 153454 FirstChildUserNameTAG: Inshiqaq FirstChildCreateTimeTAG: 2012-11-06T17:08:01Z SecondChildTAG: tried everything that everyone had said nothing works SecondChildUserIdTAG: 160242 SecondChildUserNameTAG: Mlevins35 SecondChildCreateTimeTAG: 2012-11-06T20:19:35Z FirstChildTAG: Same here, i Tried multiple browsers and computers but cold not get the transient analysis to launch. FirstChildUserIdTAG: 371623 FirstChildUserNameTAG: rusticsankalp FirstChildCreateTimeTAG: 2012-11-06T17:58:11Z SecondChildTAG: I think then it is a website glitch! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-06T18:42:42Z FirstChildTAG: Yep, I get the same problems. I emailed the staff to alert them to it. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-11-06T20:34:05Z SecondChildTAG: Thanks for that. SecondChildUserIdTAG: 96452 SecondChildUserNameTAG: Lyla SecondChildCreateTimeTAG: 2012-11-06T21:57:08Z FirstChildTAG: Thanks for the reports. This should be fixed now. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-11-07T01:50:38Z IndexTAG: 402 TitleTAG: A visual aid for the norton circuit... ![enter image description here][1] Hazel. [1]: https://edxuploads.s3.amazonaws.com/13522037801343606.png UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-11-06T12:00:44Z VoteTAG: 6 CoursewareTAG: Week 8 / RC Response To A Falling Step Continued CommentableIdTAG: 6002x_RC_Response_To_Falling_Step_Continued NumberOfReplyTAG: 0 IndexTAG: 403 TitleTAG: How to make a living from electronics? For most of us electronics is a dear hobby, but also potentialy a way to make a living. I would love to hear some suggestions on how to do that. Many of you are already employed in the field of EE, so I am interested to hear your views and opinions! Manufacture, industry, reparation, design, troubleshooting etc.. Hazel. UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-11-05T22:09:46Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: Hi hazel1919! You are missing Research too haha :)! I won a fellowship as a engineering student and now I am in Research in Solid State Laser in a Institute of Scientific and Technical Research, I really love that! I am so happy!I think that this alternative is not that common in engineering, in fact, I have to be one of the few at my Career that decided that, the most part of my Classmates at the University decided to work in the Industry because of the money only and not because they like that as here scientists don't receive that much remuneration than industry, most of them ends working outside the Country, so in a few words, it is not that much atractive to be Scientist here, but I like that idea to become a Scientist someday haha, I guess that I will die poor anyway so I don't care haha! I will do what I like and research is what I love :) I am also in another investigation Group of Education at my University (colaborating freely since two years ago). I guess that this is one way, coming from heart, of returning to my University all that is giving to me as it is totally free in Argentina, so I owe them a lot because of my education studies. So I am really greateful to help and contribute with them and also to help the students, finding their difficulties that make them to abandone their studies, I really care about that... EE has a lot of places that you can work, I have been in internships before the fellowship to put in practise my EE (I decided to work at the same time of studying at the University, this becomes more difficult to finish your career so that is why some students decide not to do this and wait to work after the degree, but it is up to you, I guess) . In my point of view, I started working, because I always considered that it is important to have practise, to mistake and learn of your mistakes, to practise and practise a lot, to face the engineering barrier, that I guess that is the most difficult part because you get noticed about so much things of life that you didn't know, to avoid collisions with the wall once you receive your degree and not receiving your degree without even know how to handle situations, to be prepared for any challenge, you will get notice that you will face things that are not written, things that you have to solve by yourself and that were not teached in a Subject of the University, and all that practise becomes a part that forms you as a good engineer! I have learnt a lot in the internships (electomedicine, agro industry, ISO, etc..)! You should try with that in the furure ;). A good engineer has to know to do everything, from cutting a wire to design a project and of course, to know how to work in a Team, this is crucial. I am anxious to finally become engineer, so much effort I have put on this, a lot of sacrifice and perseverance! I guess that someday it will be rewarded :P So it is important that you can know all the alternatives and pick the ones that you like or that are convenient for you (money/because it is interesting/you like/etc...), but knowing all the possibilities it is really important for your decision,if you don't know all the alernatives your decision will be limited. Nice question by the way! Take care! :) Myriam. Advice: Keep studying always, be curious, learn and teach to others what you have learnt so the other one can teach it too to other one and so on :). Be happy and if you get mistake, just take it positive, like a way of learn a new thing. You have a future waiting you! My best wish to you! FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-06T00:38:57Z SecondChildTAG: Thanks Myrimit! That post just made my day! Really nice to hear you are doing something you love, not because it has a fat paycheck. SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-06T08:33:13Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-06T11:21:22Z FirstChildTAG: You forgot government. EE keep essential government services working: aircraft safely flying, mail sorted and delivered, radio spectrum free of interference, safe drinking water, reliable electrical power grid, established national technical standards, disaster response, weather and atmospheric monitoring, space and interplanetary exploration, communications and national security. You name it, there's a bunch of engineering folks (many of whom are EE) keeping all that stuff working for citizens to take for granted. In the US, government is also poorly paid compared to industry, but there is the intangible reward of service to fellow citizens and one's country. FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-11-07T13:43:51Z SecondChildTAG: I think there are many opportunities now, it helps if you are in a tech-minded city/place though. SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-07T21:47:26Z FirstChildTAG: Check out [silicon valley][1] [1]: http://www.glassdoor.com/Salary/Cisco-Systems-Hardware-Engineer-IV-San-Jose-Salaries-EJI_IE1425.0,13_KO14,34_IL.35,43_IM761.htm FirstChildUserIdTAG: 142857 FirstChildUserNameTAG: viking2 FirstChildCreateTimeTAG: 2012-11-09T06:48:43Z FirstChildTAG: I work in the renewable energy industry. It is the first job I got when I finished my studies and I have been doing it for about 10 years now. I get to design Power inverters to run off batteries, PV grid-connected inverters, MPPT solar chargers and all sorts of other cool power-electronics. This includes designing PCB's and writing the firmware for these devices. Renewable energy has become a very popular field in recent years. FirstChildUserIdTAG: 270011 FirstChildUserNameTAG: SnowmanZA FirstChildCreateTimeTAG: 2012-11-09T22:09:17Z IndexTAG: 404 TitleTAG: Will this course MITx: 6.002x run on Edx again? i am currently having a poor grade in this course, well it is due to my tight schedule so i missed a lot of homework and lab work including the mid-term exam, i would like to register for this course again if it would be rerun soon on Edx platform. Thanks UserIdTAG: 171289 UserNameTAG: harbey CreateTimeTAG: 2012-11-05T15:39:42Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Yes it will be offered again next year (mostly in March). FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-11-05T16:10:37Z IndexTAG: 405 TitleTAG: [STAFF] Midterm feedback Hi All, Mid term was great.. while preparing for the mid term and preparing notes for the same was able to get more broader and clearer picture of the topics.. in short it was really good.. How about you all? [STAFF] One suggestion, the submit button used during the exam probably needs some delay between two clicks actually seeing no response on my first click I clicked it twice only later I realized I lost 2 attempts :). Thank you edx team for all your efforts its a wonderful course! UserIdTAG: 111917 UserNameTAG: ashish_mit CreateTimeTAG: 2012-11-01T06:05:52Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 2 FirstChildTAG: How about a submit button for each answer form field? FirstChildUserIdTAG: 287805 FirstChildUserNameTAG: Beneficial FirstChildCreateTimeTAG: 2012-11-01T17:54:55Z FirstChildTAG: I totally agree with the delay thing. I too had experienced it ut thank god, I got it right on the third chance. FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-11-01T09:25:19Z SecondChildTAG: Also, I wish when you click the button it would immediately gray out the existing red x's (and I suppose green checkmarks), because too often I click it and then look back to see I'm still wrong. But then a second later it finally changes to a checkmark. It would also be nice if the error messages about "you can't use VIN" were next to the appropriate input box instead of near the check button. SecondChildUserIdTAG: 397804 SecondChildUserNameTAG: mdempsky SecondChildCreateTimeTAG: 2012-11-01T17:03:43Z SecondChildTAG: And marks could become green and red again when the page refreshes. SecondChildUserIdTAG: 208296 SecondChildUserNameTAG: OZ1 SecondChildCreateTimeTAG: 2012-11-07T20:03:46Z IndexTAG: 406 TitleTAG: to staff will edx offer any cources in microelectronic i'm interested in ic design so if yes which cources will be offered ? UserIdTAG: 221617 UserNameTAG: konan CreateTimeTAG: 2012-10-29T10:09:22Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Hope Staff Members answer to this !!! FirstChildUserIdTAG: 82597 FirstChildUserNameTAG: bondrajat FirstChildCreateTimeTAG: 2012-10-29T10:55:48Z FirstChildTAG: there are courses in digital system and vlsi design on ocw.mit.edu but not live. HTH FirstChildUserIdTAG: 623210 FirstChildUserNameTAG: preveen FirstChildCreateTimeTAG: 2012-10-29T11:44:27Z FirstChildTAG: **Courses in IC Design: Beyond an "Introduction to Digital Design"-type Class** **konan:** For the time being, edX does not post course offerings that far into the future. Everything's still relatively new; in fact, all of the courses besides MITx's 6.002x is new for this Fall. Since you say you are interested in IC design, I'm *assuming* that you've already taken a basic / introductory "Digital Design"-type class where topics like logic gates, Karnaugh maps, state machine design are covered. If not, then this is where you need to start. But beyond this: Coursera (I don't know if they are our competitor or not, but just taking a quick glance) has a **Digital Signal Processing** course (February 2013) through EPFL (a major Swiss university, though this particular class will be in English). The professors seem of high-caliber, and the textbook they've written for the course, [Signal Processing for Communications][1], is free online for download. Flip through the book and see if it's something you're interested in and can handle. Signal processing, communications systems, and digital design all cross over to a degree; though the EPFL course is more theoretical than practical. Also through Coursera, there is a class (date to be announced) called **VLSI CAD: Logic to Layout**, through the University of Illinois, Urbana-Champaign, which is known for it's innovations in computing. The tagline for the class is: *How it’s possible to synthesize logic and layout using cool algorithms and data structures. Let’s face it: you can build a Karnaugh map with about 6-7 variables, and beyond this your head explodes. Modern designs have a million gates. Clearly, something magical is happening somewhere, to pull this off. Come learn what that magic is.* This seems exactly like the type of course you are looking for, in designing digital ICs. These two courses are pretty **advanced** and math-and-programming intensive. The DSP class requires knowledge of linear algebra, calculus, probability theory (which is **not** the same as Statistics 101) and requires programming in Matlab or equivalent. The VLSI class prerequisites are not as intense: linear algebra, calculus, and introductory / basic digital design. Note that I am planning on taking one or both of these courses, myself. Both courses offer a certificate of completion signed by the instructor. Unlike edX, I am not familiar at all with the Coursera platform, so I cannot answer any questions about class format, etc. Jersey Mark [1]: http://www.sp4comm.org FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-10-29T15:02:00Z SecondChildTAG: Thanks JerseyMark for this valuable information... SecondChildUserIdTAG: 178840 SecondChildUserNameTAG: mehtanayanv SecondChildCreateTimeTAG: 2012-10-29T18:39:41Z SecondChildTAG: Thanks, you just opened a whole new world for me!! SecondChildUserIdTAG: 315681 SecondChildUserNameTAG: nahuja1 SecondChildCreateTimeTAG: 2012-10-30T16:56:10Z SecondChildTAG: thanks but i'm interested in analog ic design not digital dsign SecondChildUserIdTAG: 221617 SecondChildUserNameTAG: konan SecondChildCreateTimeTAG: 2012-10-31T08:53:47Z IndexTAG: 407 TitleTAG: [STAFF ] About the course If I can't continue ....is there repeat for this course? thanks advance UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-29T01:48:17Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Yes, it will be offered again in the spring. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-10-29T01:51:56Z SecondChildTAG: Thank you SecondChildUserIdTAG: 99053 SecondChildUserNameTAG: abdulnaser SecondChildCreateTimeTAG: 2012-10-29T01:53:29Z FirstChildTAG: Hi Lyla Do you know if there are there any plans to offer any follow on courses to this course? - Thanks. FirstChildUserIdTAG: 132828 FirstChildUserNameTAG: rharris FirstChildCreateTimeTAG: 2012-10-29T05:12:32Z IndexTAG: 408 TitleTAG: How can I get to be a teaching assistant on this forum?? I was just wondering about the criteria to be a teaching assistant in this course. Could someone shed some light on this? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-28T17:38:05Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: Simple;-) Assist others in Learning. Then staff will contact you if they are impressed with your efforts and will ask you if you are interested in becoming a community TA. FirstChildUserIdTAG: 623210 FirstChildUserNameTAG: preveen FirstChildCreateTimeTAG: 2012-10-28T18:03:40Z FirstChildTAG: Hi, I became Community TA because kimt(Staff) offered that to me in the old Forum of 6.002x (Spring) in a Post some days ago and I accepted - I registered mainly to this Course, for my own will, in order to help (Spanish/English) the New Students, in fact, I didn't knew about the Community TA's, I got noticed about that some days ago - I really don't know how they are based on choosing Community TA's , you should e-mail them if you are willing to be a Community TA. Might it is based on your Posts and Comments rates, but I am not sure... Remember that being Community TA is the same as helping here, so being Community TA or not is almost the same ;), the only difference is that you have some moderator responsability like: closing, deleting, editing, endorsing other students Posts, but anything else, eg. like having a magic button of showing answers hahaha. But if you feel that you would like to have that responsability, you should ask them by email [help e-mail][1] I hope this can help you, Myriam. [1]: https://www.edx.org/help FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-28T18:25:51Z SecondChildTAG: No magic button? Bummer! :) SecondChildUserIdTAG: 92346 SecondChildUserNameTAG: MobiusTruth SecondChildCreateTimeTAG: 2012-10-29T00:35:02Z SecondChildTAG: Hahaha! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-29T02:31:58Z SecondChildTAG: Thank you Myrimit for being our LA(Learning Assistant);0) SecondChildUserIdTAG: 623210 SecondChildUserNameTAG: preveen SecondChildCreateTimeTAG: 2012-10-29T05:45:50Z SecondChildTAG: **MoblusTruth:** Actually, we *do* have a "magic button." We have an "EDIT" button where we can edit / moderate other users' posts. For example, if a user gives out answers to Homework, or posts Exam questions, we have the responsibility to delete Exam questions / answers and to modify those Homework answers into a form where general concepts and/or hints still get across, without giving out the "exact" answer, so a student cannot just "copy and paste." I can also Endorse posts; but I don't think I can Delete and Close posts yet, as JSChambers says below (though I am new at this, maybe I have to figure it out). Note I will not edit a post under any circumstances, unless it's a violation of the Honor Code, or MITx / edX rules, or basic decorum. If I do decide to edit, I will post a notice that I did, and my reasons. Jersey Mark SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-10-29T06:31:18Z FirstChildTAG: Just to clear up any misconceptions, the Community TA's can only endorse, Edit, Delete and Close posts. That's it. On any other board, we would be called Moderators. Most, but not all, of the Community TA's took the class last spring, and all have proven themselves to be mature and responsible people, which I think is really the criteria that is most important. The skill in the subject varies. I, for example, got a C in the class last year, so I probably know most of the answers but you might want to check with someone else before following my advice. Most of the Community TA's know the subject better than I do. Which is why I'm taking the class again, and why Myrimit and others have to come to my rescue occasionally. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-28T20:36:49Z SecondChildTAG: I can only hope to do as well this semester. :-) SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-28T21:56:56Z SecondChildTAG: Seconded SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-28T21:58:26Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-28T23:37:03Z SecondChildTAG: Hey, JS...you are as modest as Myrimit is ebullient! :) SecondChildUserIdTAG: 92346 SecondChildUserNameTAG: MobiusTruth SecondChildCreateTimeTAG: 2012-10-29T00:38:28Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-29T02:31:50Z SecondChildTAG: Just like one cs50x guys have said, "It doesn't matter how you end up relative to your classmates, but how you end up with yourself in week0 of the course." Thank you so much to all LAs(Learning Assistant) for your valuable time and advice. SecondChildUserIdTAG: 623210 SecondChildUserNameTAG: preveen SecondChildCreateTimeTAG: 2012-10-29T05:49:01Z SecondChildTAG: **JSChambers:** Just for the record, I want to add that 6.002x Fall 2012 is my first edX course; I did not take it in the Spring, so I, personally, do not know the answers "ahead of time." As with you, **Myrmit** and others have come to my rescue in several Homeworks, especially when I do them last-minute due to work and family constraints. I did take a similar "Fundamentals of Circuits Analysis" course at my university about 8 years ago (the material being very similar to MIT's 6.002), for credit; so I'm definitely not a "beginner", and I also work as an Engineering Tech, so I can offer my insight from that experience. SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-10-29T06:38:45Z FirstChildTAG: **Re:** "How can I get to be a teaching assistant on this forum??" Just assist other users. If your answer (or question, even) is that popular, others will "up-vote" your answer. With enough "up-votes", you build points, and this is part of the criteria that brings you to the attention of staff as a potential moderator / Community TA. **My hint:** Post your answers as "Post a response:" instead of "Add a comment..." as the latter ends up in the "grey area" and other users cannot "up-vote" those answers. I was always posting in the "grey area", thus I became a Community TA rather late. **Also remember** that being a Community TA is an added responsibility; I am taking 6.002x this semester, and I have to study just as hard as everyone else (although I have prior work and university experience). I don't think Community TA is for everyone, especially the beginners with more questions than answers, but then again, beginners can still help if their math and physics background is strong. Anyways, when or if you get that email that asks you to become a Community TA, think about it; you can always say "no" if you are too busy, etc. If your grades are so-so, or below average, I would reconsider; although it's a personal choice for everyone. FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-10-29T06:49:30Z IndexTAG: 409 TitleTAG: Add a link to the Midterm to the main page I think you need to add a link to the midterm in the main page because being between lecture 6 and 7 some people may actually miss it. I for one looked at the Courseware and thought it had not been released yet, expecting to see it at the bottom of the lecture list, only realizing it was there after reading a post about it in the discussion. UserIdTAG: 370344 UserNameTAG: KostisL CreateTimeTAG: 2012-10-26T15:25:46Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: hmm. . perhaps you are right FirstChildUserIdTAG: 146930 FirstChildUserNameTAG: fayzan_007 FirstChildCreateTimeTAG: 2012-10-27T12:53:58Z FirstChildTAG: Can we have some statistics about the midterm from the course organizers? FirstChildUserIdTAG: 623210 FirstChildUserNameTAG: preveen FirstChildCreateTimeTAG: 2012-10-27T16:23:20Z IndexTAG: 410 TitleTAG: finished midterm with 97 percent i cant belive i forgot to put a decimal point and ruined my 100% streak :( UserIdTAG: 436749 UserNameTAG: waynebrown CreateTimeTAG: 2012-10-25T13:15:04Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: did it with 100%..:P FirstChildUserIdTAG: 266739 FirstChildUserNameTAG: satvikchugh FirstChildCreateTimeTAG: 2012-10-25T13:26:43Z FirstChildTAG: 93% FirstChildUserIdTAG: 297655 FirstChildUserNameTAG: Aljoska FirstChildCreateTimeTAG: 2012-10-25T13:31:06Z FirstChildTAG: party!!!!!!!!!!!!!! FirstChildUserIdTAG: 342135 FirstChildUserNameTAG: vikash902 FirstChildCreateTimeTAG: 2012-10-25T13:27:10Z FirstChildTAG: Lets leave this Post Discuss for monday , after the deadline, as this can might give some details of the Midterm Exam ... Congratulations!:) FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-25T13:51:00Z IndexTAG: 411 TitleTAG: Good luck everybody! The midterm will start soon and I wanted to wish good luck to each and every one of you... Give your best ;) UserIdTAG: 47372 UserNameTAG: Digius CreateTimeTAG: 2012-10-24T21:44:30Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Good luck Digius! Give your best in the Exam! ;) My best wish to all of you! Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-24T21:57:03Z IndexTAG: 412 TitleTAG: WIRED BEAR , ASHWITH , MYRIMIT PLEASE LOOK ITS FOR YOU Hey Wired bear this message is for you I am an Indian eleventh grader giving SAT next year, i want to talk to you about certain things. As for me i completely agree with your " I HATE COAching institutes" attitude. IF you want to meet me contact me at skype My id is Harvey.Specter13 Hey myrimit and ashwith i dont know if this guy is still hear if you know anything about him please tell me UserIdTAG: 378080 UserNameTAG: GladIDidThis CreateTimeTAG: 2012-10-21T22:47:14Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: Hi GladIDidThis! How are you? Yes, sure. I have not his e-mail. But if I see him in the Forum Discussion I will re-link him this Post to him ;). Might he is studying for the SAT . I remember as he wrote in this Post: [Probably not for high school kids][1] that he said that he had the SAT Exam near the Midterm of 6.002x . I hope that everything is ok with him. Good luck with the SAT @thewiredbear! :). See you! Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5071ccd71ed8e21f0000001c FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-21T23:05:46Z SecondChildTAG: hey myriam actually i am in an exactly same situation like him but i am a bit different, I have special permission to not go to school, and i do not go to coaching classes, to be honest i want to help him, because just an year ago i was exactly like him " OVERTENSED frustrated" and most importantly desparate. I am taking the course too and have got perfect score till sixth week. i just wanna talk to him once SecondChildUserIdTAG: 378080 SecondChildUserNameTAG: GladIDidThis SecondChildCreateTimeTAG: 2012-10-21T23:09:38Z SecondChildTAG: Hi @GladIDidThis, Yes that must be stressful for the kids... Now I am getting noticed that when I was in the School I lived inside a buble ! haha, in that time I didn't knew that existed the SAT, I just focused to be a kid and did all that I wanted: taking long nap haha, watching TV (specially Discovery Channel), I spent a lot of hours writting science fiction - I have finished some books haha,is my hobby till the present, I also when I was a kid I liked to relate/invent horror stories an scare others kids haha! so funny!, I liked magic, so I made magic shows as a kid, also I made a lot of experimets analyzing ants haha, so odd! but I don't remember in that days a pressure as I can read now, or things have changed or I always were living in a buble haha! By the way, in Which University you would like to study in the future? I hope that you can contact @thewiredbear :). My best wish to you @GladIDidThis! See you! Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-22T00:09:02Z SecondChildTAG: Thanks For the reply, I want to apply to MIT and Other than that, i wish to go to IVY league universities. I think these courses are for college undergrads but i want to take it because i like it. Although i have a number of science Olympiad credentials but i presume that there's nothing better than this, DO you think highly competitive colleges would find it noteworthy for high school students who complete this ?? SecondChildUserIdTAG: 378080 SecondChildUserNameTAG: GladIDidThis SecondChildCreateTimeTAG: 2012-10-22T02:19:44Z SecondChildTAG: Cool! Yes, I think it will sum up ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-22T02:35:25Z SecondChildTAG: Experiments analyzing ants? I think I did those same experiments as a kid... with a magnifying glass and the sun. Ouch! I am so impressed with the young kids taking this course and other online university-level classes. Even though these courses are not for credit, I believe most schools would look quite favorably upon any person who completed them. I wish this technology would have been available when I was in high school (early 1970s). I was so bored and unhappy in school, I would have loved to have had interesting and challenging classes without all the nasty social pressures that teenagers endure. SecondChildUserIdTAG: 94557 SecondChildUserNameTAG: g_hopper SecondChildCreateTimeTAG: 2012-10-22T05:24:22Z SecondChildTAG: Hi g_hopper! How are you? Haha! I liked to analyze the behaviour of the ants under different situations , odd! haha. Yes, totally agree with you :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-22T11:24:24Z FirstChildTAG: hey myriam actually i am in an exactly same situation like him but i am a bit different, I have special permission to not go to school, and i do not go to coaching classes, to be honest i want to help him, because just an year ago i was exactly like him " OVERTENSED frustrated" and most importantly desparate. I am taking the course too and have got perfect score till sixth week. i just wanna talk to him once FirstChildUserIdTAG: 378080 FirstChildUserNameTAG: GladIDidThis FirstChildCreateTimeTAG: 2012-10-21T23:20:49Z FirstChildTAG: Hey everyone. I'm yet another victim of the murderous coaching institutes. Luckily although I'm in FIITJEE Hyderabad which you guys probably heard of, I'm out of the whole IIT thing. In 12th grade now, did great on the SAT and hope to be in a university in America this time next year. Good luck and hope you get out of it too. :) FirstChildUserIdTAG: 154440 FirstChildUserNameTAG: Aahlad FirstChildCreateTimeTAG: 2012-10-22T16:20:12Z SecondChildTAG: Hey aahlad Can you give me the YOur Skype id or gmail I am an eleventh grader Please help me out,of this prison this asylum of monotonic,Banal, Endlessly TOrturous CRap I am form FIITJEE TOO. Thanks A Lot Buddy SecondChildUserIdTAG: 378080 SecondChildUserNameTAG: GladIDidThis SecondChildCreateTimeTAG: 2012-10-22T16:43:04Z FirstChildTAG: Hey aahlad Can you give me the YOur Skype id or gmail I am an eleventh grader Please help me out,of this prison this asylum of monotonic,Banal, Endlessly TOrturous CRap I am form FIITJEE TOO. Thanks A Lot Buddy FirstChildUserIdTAG: 378080 FirstChildUserNameTAG: GladIDidThis FirstChildCreateTimeTAG: 2012-10-22T16:43:00Z SecondChildTAG: Gmail is aahlad.asg@gmail.com. SecondChildUserIdTAG: 154440 SecondChildUserNameTAG: Aahlad SecondChildCreateTimeTAG: 2012-10-22T16:46:53Z IndexTAG: 413 TitleTAG: Trouble with the quadratics in H5P2 Q4, Q5 & Q6!! Using this equation here... **(vIN-VT)/RS+(1/(K*RS ^2) )*(1-sqrt(1+2*K*RS*(vIN-VT))) = IDS** I am substituting in the following values... **VIN = 6.3V RS = 16 ohms VT = 2v K = 0.5** I then get... **((6.3-2)/16)+(1/(0.5*16^2) )*(1-sqrt(1+2*0.5*16*(6.3-2)))= IDS = 0.211292A** Which is the current through RS, therefore: **VOUT=IDS*RS= 3.380672v** This is wrong! Where am I going wrong? UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-10-13T12:30:49Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: iDS=K/2*(vIN-iDS*RS-VT)^2 use this equation put the value given for V1N in this equation a quadrtc formula will come solve this for iDS value of iDS will come now put the value of iDS in this equation vOUT=RS*iDS. solve in this way u will find corect answer the way you are solving qustion also give me error.. i solved it using above way i hve told FirstChildUserIdTAG: 228871 FirstChildUserNameTAG: ELINAKHAN FirstChildCreateTimeTAG: 2012-10-13T15:05:29Z SecondChildTAG: I have done this, I have tried many results, but I can not find the correct answer... **x=0.5/2*(6.3-x*16-2)^2** Is **K = 2A/2V^2** or **2A/6.3V^2**? SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-13T16:36:00Z SecondChildTAG: Ok, I was putting the wrong value for k. SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-13T17:46:47Z SecondChildTAG: my iDS equation is "[{RS*(vIN-VT)+2/K}-sqrt{4*RS/K*(vIN-VT)+(2/K)^2}]/RS^2" but it says invalid input. why? SecondChildUserIdTAG: 259693 SecondChildUserNameTAG: MehrozKhan SecondChildCreateTimeTAG: 2012-10-13T18:04:55Z SecondChildTAG: Elinakhan, can you please explain to me where you came up with iDS=K/2*(vIN-iDS*RS-VT)^2? SecondChildUserIdTAG: 141376 SecondChildUserNameTAG: krtica SecondChildCreateTimeTAG: 2012-10-13T18:16:24Z SecondChildTAG: my iDS equation is "1/(K*RS^2)+(vIN-VT)/RS-1/RS^2*sqrt[1/K^2+2*RS*(vIN-VT)/K]" and it says could not parse as formula. I can't figure out what is wrong. SecondChildUserIdTAG: 319221 SecondChildUserNameTAG: stefan31i SecondChildCreateTimeTAG: 2012-10-13T18:20:36Z FirstChildTAG: Honor Code Mr hazel 1919 FirstChildUserIdTAG: 331441 FirstChildUserNameTAG: abdessamade FirstChildCreateTimeTAG: 2012-10-13T16:04:00Z FirstChildTAG: Try VIN = 6.2V and K=2....This values were given!!! FirstChildUserIdTAG: 164572 FirstChildUserNameTAG: Ranyeri_rocha FirstChildCreateTimeTAG: 2012-10-14T14:57:08Z IndexTAG: 414 TitleTAG: it is fantastic we really thank u for this funny learning for more progress and technology to our mankind and our world i repeat thank u UserIdTAG: 295983 UserNameTAG: qassam CreateTimeTAG: 2012-10-13T09:40:05Z VoteTAG: 6 CoursewareTAG: Week 5 / Music Amplifier Demonstration CommentableIdTAG: 6002x_snd_d NumberOfReplyTAG: 0 IndexTAG: 415 TitleTAG: h5p3 hi i found this formula for vout=(1-1/sqrt(1+2*K*RS(VIN-VT)))*vin i used it in solving the 2 other problems and it really really works so why he didn't agree that this formula is correct UserIdTAG: 84213 UserNameTAG: mohamed373 CreateTimeTAG: 2012-10-11T15:37:38Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: i am facing same problem from a week.i have tried to enter it in other formats like taking lcm or not taking out 2*K*RS from VIN VT term but it still showed could not parse(...............) as a formula. FirstChildUserIdTAG: 99628 FirstChildUserNameTAG: Pranjal16 FirstChildCreateTimeTAG: 2012-10-11T17:53:30Z SecondChildTAG: Remember that you have vin and VIN SecondChildUserIdTAG: 110486 SecondChildUserNameTAG: Mirthyn SecondChildCreateTimeTAG: 2012-10-15T04:49:57Z FirstChildTAG: I've the same error and to correct this is very easy. You have just do add * after RS. Every multiplication you've add * to avoid errors. FirstChildUserIdTAG: 300169 FirstChildUserNameTAG: castrogfx FirstChildCreateTimeTAG: 2012-10-11T21:06:22Z SecondChildTAG: aha thank you i forgot it SecondChildUserIdTAG: 84213 SecondChildUserNameTAG: mohamed373 SecondChildCreateTimeTAG: 2012-10-12T08:30:54Z FirstChildTAG: mohamed373 Can you tell have did you get this formula? FirstChildUserIdTAG: 500965 FirstChildUserNameTAG: barka0 FirstChildCreateTimeTAG: 2012-10-12T20:19:12Z SecondChildTAG: i cant understand u SecondChildUserIdTAG: 84213 SecondChildUserNameTAG: mohamed373 SecondChildCreateTimeTAG: 2012-10-12T20:33:32Z SecondChildTAG: sorry, mistake: how did you ger this formula? SecondChildUserIdTAG: 500965 SecondChildUserNameTAG: barka0 SecondChildCreateTimeTAG: 2012-10-12T20:40:00Z SecondChildTAG: here we go from the 2 previous equations u can sub vout by IR in the first equation u have quadratic function in ID solve it make sure when vt=vin ids=0 this is the key you will use in order to choose the right root SecondChildUserIdTAG: 84213 SecondChildUserNameTAG: mohamed373 SecondChildCreateTimeTAG: 2012-10-12T20:53:22Z IndexTAG: 416 TitleTAG: Proffesor Agarwal's WebSim I thought some of you would like to check out the Professor Agarwal's online laboratory. A Simulation-Based Online Electronic Circuits Laboratory. http://euryale.csail.mit.edu/ UserIdTAG: 292543 UserNameTAG: Pennypacker CreateTimeTAG: 2012-10-11T04:42:56Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 417 TitleTAG: Just about to give up! I spend 3 full days fooling around Week 4 exercises without success. I end up missing my window time frame to make week 4 homework. I'm completely frustrate. I just leave this comment because I think something went really wrong on week 4. Up to W4, with more or less effort things become understandable and somehow linked with theoretic/practice classes. On week 4 I simple could not find a natural progress path linking class to homework and I'm lost, hesitating on continue course. Does this only happen with me? UserIdTAG: 413869 UserNameTAG: nlfigueiredo CreateTimeTAG: 2012-10-08T12:29:07Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Do not give up if you do not perform well on LAB 4 HW 4 try hard learn what you do not know learn new thing and improve on what you have missed on last week. FirstChildUserIdTAG: 272523 FirstChildUserNameTAG: Jivraj FirstChildCreateTimeTAG: 2012-10-08T12:58:47Z SecondChildTAG: I'd agree. I also found it really hard to pull through this week's homework and spent upwards of 12 hours doing this. I do think that the homework was intentionally designed to make us work a lot harder than most would. After all, this is MIT! SecondChildUserIdTAG: 359310 SecondChildUserNameTAG: AaronYeoh SecondChildCreateTimeTAG: 2012-10-08T13:16:13Z SecondChildTAG: That's might be true. Guess what, I'm now starting W5 lectures, and I find a lot of information that would help me understand W4!!!! Frustration. 3 long days, 12 hours+ each, fooling around why use 1A additional source to solve the problem and ...because I miss the logic of the course!!!! That's bad... and I couldn't finish my homework in time. Frustration :( SecondChildUserIdTAG: 413869 SecondChildUserNameTAG: nlfigueiredo SecondChildCreateTimeTAG: 2012-10-08T13:53:09Z SecondChildTAG: You are not alone in this...I also find week 4 very difficult!..and I was up late for three nights!!..in the end I also wanted to give up!...but the encouraging people out there didn't let me...and I am through :).. SecondChildUserIdTAG: 122513 SecondChildUserNameTAG: Mona77 SecondChildCreateTimeTAG: 2012-10-08T14:55:50Z SecondChildTAG: Its too early to give up. I'm just curious about midterm. SecondChildUserIdTAG: 148523 SecondChildUserNameTAG: span993 SecondChildCreateTimeTAG: 2012-10-08T14:59:51Z SecondChildTAG: Totally agreed with all of the above. Three 12+ hour days for homework and lab this time, and only got several by guessing. SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-08T18:33:10Z SecondChildTAG: Me to, on night from Saturday to Sunday, I was awake until 4 clock in the morning :-( SecondChildUserIdTAG: 389333 SecondChildUserNameTAG: juergen SecondChildCreateTimeTAG: 2012-10-08T20:27:11Z FirstChildTAG: Based on my experience in the pilot 6.002x course, you will likely find that the information you need to solve the current week's homework and lab will be in the next week's material, or perhaps the week after that. Therefore, it really pays to read ahead in the textbook, even if you don't grasp all of it at first reading. This is MIT's method. They ask questions for which you have not already seen the material. They consider it brain-stretching (although students have other, unkinder, words to describe it!). Once you understand how the questions are presented, you just need to keep a week or so ahead of the deadlines in your reading, and then do the current homework and lab. Week 4 is the first slap in the face. By Weeks 8 and 9 you will be bruised and battered. Stick with it, even though the math will get really hard and ugly in coming weeks. After Weeks 9 and 10, you will absolutely know that you can make it to the end because you will look back at Week 4 and see how simple it seems in retrospect. You should also expect that the midterm and final will ask questions on topics which have not been explicitly covered in lectures. The idea is that if you have a good grasp of the methods and foundations, you can figure out how to solve the problems. That's actually applicable in solving real-world engineering problems, you don't always find yourself in simple easy-to-solve situations. After all, if they were easy to solve, they wouldn't need you on the payroll! FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-10-08T15:00:52Z SecondChildTAG: ... what a way to find out! Frankly, I review all previous lectures at least 3 times again, in slowwwwwly mode. I did not expected that. I really was in the edge to give up. And now I lost a precious time. I'm really irritate with this situation. Arrrghh! SecondChildUserIdTAG: 413869 SecondChildUserNameTAG: nlfigueiredo SecondChildCreateTimeTAG: 2012-10-08T15:51:22Z SecondChildTAG: What's the step after "stroke" ? SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-08T18:34:01Z SecondChildTAG: I have to second what g_hopper says. By week 9, you will really bruised. You'll probably hate the course and everything associated with it. But if you stick with it, even if you are doing poorly, my guess is it will snap into focus. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-08T18:36:38Z SecondChildTAG: JSChambers speaks the truth. Every week I swore I would drop this course, and after finishing the homework and lab, I would decide to hang on for another week. Before I knew it, I was beyond the "gnarly math" that Dr. Agarwal warned us about, and I knew I could get through whatever else was thrown at us. Of course, that final exam was a killer, but I didn't give up, I kept slogging away until my brain melted. And here I am back for another round of academic torture, hoping to learn even more (and perhaps help some of the first-timers through the course). SecondChildUserIdTAG: 94557 SecondChildUserNameTAG: g_hopper SecondChildCreateTimeTAG: 2012-10-08T20:27:05Z SecondChildTAG: Were any statistics provided about the grades on the midterms and finals. Was there information about percent of students who enrolled who successfully completed the course? SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-09T00:29:28Z SecondChildTAG: Here are the stats from the last class: Course statistics: 6.002x had 154,763 registrants. Of these, 69,221 people looked at the first problem set, and 26,349 earned at least one point on it. 13,569 people looked at the midterm while it was still open, 10,547 people got at least one point on the midterm, and 9,318 people got a passing score on the midterm. 10,262 people looked at the final exam while it was still open, 8,240 people got at least one point on the final exam, and 5,800 people got a passing score on the final exam. Finally, after completing 14 weeks of study, 7,157 people have earned the first certificate awarded by MITx, proving that they successfully completed 6.002x. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-09T14:47:36Z SecondChildTAG: Interesting! Less than 10% of those who registered looked at the midterm, but of those who did more than 50% got the certificate. On the other hand, even doing well on the midterm is no guarantee of success on the final. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-10T18:02:23Z IndexTAG: 418 TitleTAG: HW and lab due tonight Well, I got it done! :) The second time around is giving me some of the extra depth I was hoping for. This is nice. UserIdTAG: 92346 UserNameTAG: MobiusTruth CreateTimeTAG: 2012-10-08T02:34:46Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Nice! FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-08T02:41:06Z FirstChildTAG: Do you mean this is the second time through the course for you? How was the midterm the first time around? FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-10-08T03:24:50Z SecondChildTAG: I also would like to know. I could see myself taking this course again. It is starting to get tough now. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-08T04:07:30Z SecondChildTAG: If someone does the home works and exercises faithfully, they should be pretty ok for the midterm and final. SecondChildUserIdTAG: 92346 SecondChildUserNameTAG: MobiusTruth SecondChildCreateTimeTAG: 2012-10-08T12:09:12Z SecondChildTAG: But each of those will bring their own challenges and satisfactions. :) SecondChildUserIdTAG: 92346 SecondChildUserNameTAG: MobiusTruth SecondChildCreateTimeTAG: 2012-10-08T12:11:40Z FirstChildTAG: Well done MobiusTruth! ;) FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-08T12:00:13Z IndexTAG: 419 TitleTAG: Norton w/Dependent Sources Hello, When I was doing this week's homework, I experienced the unexpected difficulty of not being able to correctly find the Norton Equivalent of a circuit containing dependent sources. As I spent a few hours trying to figure out how to do this problem, I thought it might be potentially useful to present a methodology to easily solve these problems, which I stumbled upon while reading the Wikipedia article on Norton's Theorem. From Wikipedia... "... However, when there are dependent sources, the more general method must be used... Connect a constant current source at the output terminals of the circuit with a value of 1 Ampere and calculate the voltage at its terminals. This voltage divided by the 1 A current is the Norton impedance RNo. This method must be used if the circuit contains dependent sources, but it can be used in all cases even when there are no dependent sources." UserIdTAG: 153604 UserNameTAG: autohost CreateTimeTAG: 2012-10-08T02:20:48Z VoteTAG: 6 CoursewareTAG: Week 4 / Introduction to Dependent Sources CommentableIdTAG: 6002x_int_dep_src NumberOfReplyTAG: 1 FirstChildTAG: Good idea to post this. Most of our circuits have not required that approach, even though that is the way they are developed in the textbook. It caught me by surprise, too. FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-10-08T02:51:32Z SecondChildTAG: spend 3 full days fooling around this problem without success. I end up missing my window time frame to make week 4 homework. I'm completely frustrate. I just leave this comment because I think something went really wrong on week 4. Up to W4, with more or less effort things become understandable and somehow linked with theoretic/practice classes. On week 4 I simple could not find a natural progress path linking class to homework and I'm lost, hesitating on continue course. Does this only happen with me? SecondChildUserIdTAG: 413869 SecondChildUserNameTAG: nlfigueiredo SecondChildCreateTimeTAG: 2012-10-08T12:25:27Z SecondChildTAG: No. It does not happen only with you. For me it's about every other week. :) SecondChildUserIdTAG: 92346 SecondChildUserNameTAG: MobiusTruth SecondChildCreateTimeTAG: 2012-10-08T16:11:42Z IndexTAG: 420 TitleTAG: Probably not for high school kids Hey, everyone, I am making this post so as to say maybe my goodbye because I believe that now I won't be able to go further, because the scarcity of calculus is already killing me, I couldn't even find the incremental resistance, after one hour of efforts, going through many different results, and nothing turned out, and probably things are going to be more tough along the line, and also I have my PSAT just 4 days before the mid-year exam.....so I obviously won't be able to study for the exam, and having a bad score there is just as losing it all....... P.S.- I would be good if someone would tell me if I should or should not leave it for my PSAT and school studies..... UserIdTAG: 169784 UserNameTAG: thewiredbear CreateTimeTAG: 2012-10-07T18:41:27Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 8 FirstChildTAG: The amount of calculus required is minimal. You just need to know a few simple facts. If you start on the homework at the beginning of the week there is plenty of time to get help. Simple calculus formula: if y = x^p where p is any number then dy/dx = p*x^(p-1). Rinc = 1/(di/dv) FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-07T19:05:08Z FirstChildTAG: Hi thewiredbear, I think that is a personal decission (of course, I want you to stay here). But remember that school has a higher priority, remember that you can always retake 6.002x, it will be offered next spring ;). I did it last spring and there is a lot of people from the spring here that can tell you about their personal experience (some students couldn't follow in the spring for personal reasons and now they are here, I am sure that they are doing really good here), so, you can always re-take, in fact, you can follow this course, and then back in the spring :), you will have the advantage of having the videos for watch, all the material and study it with time. But be up thewiredbear! you are so young and smart! for me, I would like you to stay but as I have said, it is your personal decission and you should analyze your priorities that you have in the present. I think that school it is more important... All my support, my best wish to you, Good luck in your exams! Myriam. P.D: If anything I can do, please tell me. I would like to help :). FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-07T19:05:51Z SecondChildTAG: Hey, Myrimit, I would love to continue on....but seriously, it really was a major setback in week 4 as I just couldn't join the course with the homework, and I could again do it next year, but I am not sure, if the certificates would be free for next time, and plus I really don't have any info on how the course is going to open up further on, so you know what, I would for now do, week 5 videos, but then can you please give me a detailed outlook on what is going to be there further along the course, I mean how is tough is the Mid-Term, the Finals, and you know how much more is it going to ask further along, because if it does get too tough further along, then I just might leave it now because then going till that point and leaving would be senseless....and also to say, I have registered for 3 more edx courses, so if you could just tell me that would I be able to carry on with a 4 of these courses.... SecondChildUserIdTAG: 169784 SecondChildUserNameTAG: thewiredbear SecondChildCreateTimeTAG: 2012-10-08T17:36:29Z SecondChildTAG: As I have said it to you, it is up to you :). But If you decide to stay I will be here to help you. Here are some oficial statistics of the 6.002x prototype course: "*Course statistics: 6.002x **had 154,763 registrants**. Of these, 69,221 people looked at the first problem set, and 26,349 earned at least one point on it. 13,569 people looked at the midterm while it was still open, 10,547 people got at least one point on the midterm, and **9,318 people got a passing score on the midterm**. 10,262 people looked at the final exam while it was still open, 8,240 people got at least one point on the final exam, **and 5,800 people got a passing score on the final exam**. Finally, after completing 14 weeks of study, **7,157 people have earned the first certificate awarded by MITx, proving that they successfully completed 6.002x**.*" But, **you can be motivated with the "popoya" case**, **he was 15 years old kid** and he did it! I don't know if he is here now, I wish to be here so he can motivate to you. But here is a fragment of one of his post, I hope that popoya don't mind if I am writing here what he shared with us ;). He wrote at the end of the course: ["Before I start off this post, I’d just like to say that all these thoughts are mostly the ramblings of a 15 year old kid, but hear me out. I live in Delhi and like most kids my age, I enjoy doing anything but studying. Most teachers in my school don’t like me very much because I’m not exactly what you’d call “nice and disciplined”, I like to play pranks on people and I talk a lot. They expect someone like me to suck at academics and instead, I tend to score good marks without working hard, so they get irritated even more. Like most kids in India who are good at maths and physics, most people expect me to take the IIT-JEE (notoriously difficult examination to get into IIT). I waste a lot of time on Facebook, watching TV, listening to music and other random things, so my parents enrolled me in one of those weekend prep classes for IIT-JEE, so that I’d have something else to keep me occupied. There, I was (and still am) surrounded by ultra-competitive people who are absolutely dead-set on going to IIT. The teachers are quite unimaginative and uninspiring and take out much of the awesomeness of physics. I don’t really like going there much, because the people I’m surrounded by are too obsessed with getting more marks (probably the only way people here judge your intelligence) than the person sitting next to them (in every exam you take, they stick a "ranking sheet" on the door of the classroom, to show your rank compared to your classmates. This makes making friends there pretty difficult) Because of going there and the way they encouraged doing "grubby math" which wasn't even relevant to the real world, doing maths became tedious and boring to me. I did get the marks, but I didn't love maths anymore, which was painful to me, because I used to like maths and it was my favourite subject. I became (and still am) defined by marks and ranks. I realized, once I started to go to these prep classes, that these teachers, who didn't care whether I existed or not, would suck out all my joy of doing physics, like they did for maths. I didn't (and again, still don't) want that to happen. I developed an affinity for physics when I was around 12 and I thought that I might have a nice time making things. Since then, I've progressed to making more complicated circuitry like guitar amplifiers and pedals, but before taking this course, I never knew how they work and what really goes on inside them. I only knew Ohm's law and elementary physics. I had zero knowledge of calculus and only knew bits and pieces of trig picked up from here and there. When I signed up for this course, I thought "Well, if I manage to complete even half of the course, I have everything to gain and nothing to lose. What's the harm ?". From the period of March to end-April, I had to learn calculus from scratch from Khan Academy, along with the course material itself. Till week 4, I was floundering about and I felt absolutely hosed. When the Taylor Series was introduced, it was the first time I ever heard of such a thing. I barely knew how to take a derivative and suddenly here comes this ! I had to watch the entire small-signal analysis lecture twice (at 1x and 0.75x, respectively) to actually understand it and get the hang of the notation (VIN, vin, vIN was a nightmare !). From week 5 onwards, I began to get the hang of the pace and I began to really savour this course. Getting a 100% on the Midterms was very, very motivating for a person like me, who learnt advanced maths (for me at least) on the fly. I managed to pass this course with an A. This course gave me a lot more than merely having the knowledge to know what is going on inside the stuff I make. Prof. Agarwal is the first teacher I've ever respected. I felt that he really cares about his students and genuinely wants to make them fall in love with electrical engineering. I've never seen any teacher go the extra mile like he did, with his fun demos. I actually looked forward to every week, despite struggling (and banging my head against the wall) through every step (voltage source and otherwise). He made me fall in love with maths again (and like Physics even more !) and I really began to appreciate the beauty and elegance in maths and physics. I met people here on the discussion forum who appreciated what I said regardless of my age. Sorry for rambling on and writing so much. I just wanted to express how much this course meant to me. I had an absolute blast here. Thank you MIT."] P.D: So be up thewiredbear! :). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-08T18:18:55Z SecondChildTAG: Myrimit, I just cannot even express my thanks to you, I am just out of words!...... I mean for me its official now, you are the best forum poster ever!!! on the whole 6.002x network!...... I mean I loved the inspiration you gave, less of inspiration, this was something I could have just copy and pasted for my story!....the totally brain damaging IIT-JEE classes, and Im in delhi too!....and then the physics going from first love to the last book I ever pick up!.......I mean till half way, in the post, I could have said, that, I can relate a 100% and say it applies to me more than ever, and just hope I also manage to get the story same on my side, beginning from 100% on the mid-term, to getting an A in the course!....and rebuilding my love of physics........and please can you just shed some light on how to prepare for the Mid-Term, I mean tell me everything you can!....I hope I can anyhow contact, popoya! FYI i was same age when the course began!.. SecondChildUserIdTAG: 169784 SecondChildUserNameTAG: thewiredbear SecondChildCreateTimeTAG: 2012-10-08T18:57:16Z SecondChildTAG: There were a lot of high school students. Give you're school and PSAT higher preference however. They are hurdles you have to cross. I'm not sure what exactly you are finding hard. In case it is calculus, don't worry about how to do it (btw, if you're in class 12, the calculus you have there is more than enough for this course). If you don't know calculus yet (if you're in 11th) or if you're finding it hard, try to understand what the derivative means (rate, slope) and what the integral "means" (area, summation of very closely spaced values; that's how it's used in calculating averages). Once you know that, use a program like mathematica, maxima or wolfram alpha to do the calculus part for you. There were a lot of people who used these tools instead of hand calculations (one of the TA's suggested this as well during a tutorial). If you dislike physics you're either not reading the right book or you're reading only for an exam. Fundamentals of Physics by Halliday, Resnick & Walker is very interesting (I had a bad Physics teacher since class 8 so I won't buy the bad teacher excuse in this case :-)). Concepts of Physics by HC Verma is really good too. Especially if you want problems. I'm sure you would have heard of both these books however. Nevertheless, calculus is very important if you want to get through an electrical engineering course (so is probability & statistics and algebra). The only way to get it right is through tons of practice. I solved the integration problems from the NCERT textbook 3-4 times to be comfortable with it. I had forgotten most of calculus till a few months back but all that practice helped me get used to integration again in no time. SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-10-11T17:50:38Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-11T19:39:30Z FirstChildTAG: School's pretty important, especially if you want to get into a university later. So don't let this course compromise your results at school. At the same time, if calculus is your only problem, remember you're not required to solve equations manually for this course. Use Wolfram Alpha or other sites for solving equations. I'm sure electrical engineers in the field use all sorts of tools and software to simplify their tasks. Whatever you decide, best of luck. FirstChildUserIdTAG: 502508 FirstChildUserNameTAG: amaher FirstChildCreateTimeTAG: 2012-10-07T19:24:55Z FirstChildTAG: I believe that there are a number of people here who would be willing to help you. I am certainly willing, but you have to give us the time and opportunity. I don't know where you are located, but from other posts I suspect you are running out of time for week 4 assignments. That doesn't give us much to work with. p.s. I second the statements that school is most important for you at this point. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-07T19:16:48Z FirstChildTAG: Hello, WiredBear... I can't say if you should leave or not, but I can describe my opinion about this course from having completed it the first time it was offered. I recall thinking that the middle part was most math-intensive. That's what we're approaching now. There is a bit more with some differential equations for a short while. Then we return to mostly algebraic methods, although some of them may require some lengthy calculations. You should do what works best for you. There is no "one size fits all" path to success, in school or in life. Good luck! FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-10-07T19:00:28Z FirstChildTAG: @thewiredbear Hey, I use MathStudio on my iPad to save time with calculus (yet, one needs to understand the concepts of cause). So, this is not a big issue. The issue imho is that you may need a real company who could help you along the path with this course. Online videos and forums comprise another mode of learning I would say. One has to have some skills to use this mode of learning effectively and efficiently. So, you probably learn both along the way and this may make it harder for you. As already said, you can take this cause later. And moreover, its available on ocw.mit.edu FirstChildUserIdTAG: 252722 FirstChildUserNameTAG: vaboro FirstChildCreateTimeTAG: 2012-10-08T00:09:57Z FirstChildTAG: Hi thewiredbear! How are you? :) Take a look at this Post [popoya contact][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/507dabdc4a15a52b00000007 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-20T19:02:48Z FirstChildTAG: Hey Wired bear this message is for you I am an Indian eleventh grader giving SAT next year, i want to talk to you about certain things. As for me i completely agree with your " I HATE COAching institutes" attitude. IF you want to meet me contact me at skype My id is Harvey.Specter13 FirstChildUserIdTAG: 378080 FirstChildUserNameTAG: GladIDidThis FirstChildCreateTimeTAG: 2012-10-21T22:50:17Z IndexTAG: 421 TitleTAG: Frustrated with H4P3 - Part B? Here's the help you need This is the first problem that has really tied me up in knots. If you are struggling to the point of frustration, here's some help that will get you to the right answer: **Calculate Voc,** the open circuit voltage at B. Use your favorite method; I used the Node Method. It takes a bit of work because of the dependent source, but you will get there. Watch the signs; the current may end up running opposite to what your intuition might suggest. **Calculate Isc,** the Short Circuit Current. That's right, short the output port. Use the node method again to get the current in the short. Remember that a short is just that -- the voltage at port B is now at ground potential. **Calculate Rth = Rnorton = Voc/Isc.** Just a simple division at this point. Avoid the temptation to eliminate all the sources and use the parallel resistor rule to compute Rth. This will not work because of the dependent voltage source. Hope this helps. Don't give up! There is a great sense of satisfaction when you wrestle this one to the ground! UserIdTAG: 358873 UserNameTAG: PaulZ CreateTimeTAG: 2012-10-05T23:54:19Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: I followed this same procedure.The harder part to wrestle this problem to the ground was finding Isc. FirstChildUserIdTAG: 282828 FirstChildUserNameTAG: Chibolator FirstChildCreateTimeTAG: 2012-10-06T06:22:55Z SecondChildTAG: thanks man, that was some great advice right there! Really helped me solve it and you are right, it gives one a lot of satisfaction solving it! SecondChildUserIdTAG: 390558 SecondChildUserNameTAG: ilhan89bln SecondChildCreateTimeTAG: 2012-10-06T10:19:50Z SecondChildTAG: how you find the value of u? SecondChildUserIdTAG: 122513 SecondChildUserNameTAG: Mona77 SecondChildCreateTimeTAG: 2012-10-06T21:46:12Z SecondChildTAG: I followed Paul's advice above, along with everyone else's advice on the forum, and no matter which method I use I just can't get it to work. Here is what I did. **Find u and Voc:** Assign u to node between R1 and R2 and assign Voc to node at port B. **Eqn. 1:** (u-5)/1 + (u-Voc)/3 = 0, **Eqn. 2:** (Voc-u)/3 + (Voc-2*u)/5 = 0. From these 2 equations and 2 unknowns I get u = 5.74 giving A*u = 11.47 and Voc = 7.85. I then short port B to find Rn (Isc), which puts the negative terminal of the dependent source and the node Voc at ground. I then calculate the current from the 5V source to ground through R1 and R2 and from the positive terminal of the dependent source to ground through R3. From here I understand Voc = Vth, and Isc = In. Rn should then be Rn = Vth/In. I have also tried shorting the 5V source and applying a 1A current source or a 1V voltage source to port B, but it doesn't work either. Any help correcting my thinking would be appreciated. - Thanks SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-07T08:11:25Z SecondChildTAG: I finally got it. I read another post and they mentioned they used the current through R3 to find the resistance of the dependent source, and from there I was able to calculate the equivalent resistance, finding Rn. I still couldn't get In from finding Isc, as I couldn't seem to find the right current flowing through R1 and R2 from the 5V source. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-07T08:39:13Z SecondChildTAG: excellent rharris. You helped a lot. SecondChildUserIdTAG: 378096 SecondChildUserNameTAG: Jamshaid271 SecondChildCreateTimeTAG: 2012-10-07T16:21:56Z FirstChildTAG: thank you so much. FirstChildUserIdTAG: 171103 FirstChildUserNameTAG: hemanth463 FirstChildCreateTimeTAG: 2012-10-06T12:09:44Z FirstChildTAG: PaulZ Regarding your statement "**Avoid the temptation to eliminate all the sources and use the parallel resistor rule to compute Rth. This will not work because of the dependent voltage source.**", here is what I did and it worked. Please correct my thinking. I realized in the problem statement it said Z = 4 ohms, so I replaced the dependent source with a 4 ohm resistor, which was in series with the 2 ohm resistor, making the equivalent resistance 2 ohms, which worked. After this, I remembered the 2nd part of S8E1 was similar, of which I didn't include the 4 ohm resistance alpha, but got the correct answer. OK, so that was a dilemma, do I replace the dependent source with it's resistance, or not. Well what I found was, in S8E1 the addition of the 4 ohm resistance was so small compared to R1 and R2 (850 and 750 ohms), it didn't make enough of a difference to affect the answer, so either answers showed correct. In H4P3 this was not the case, since R1 and R4 are very close to Z=2, the dependent source had to be replaced by the 2 ohm resistance for the answer to be correct. Without it, the answer was wrong. So, is what I did a fluke and against the rules, or is this how to deal with a dependent source when calculating Rth? FirstChildUserIdTAG: 132828 FirstChildUserNameTAG: rharris FirstChildCreateTimeTAG: 2012-10-06T18:49:01Z SecondChildTAG: What you are saying seems to be completely logical and far from being a fluke. I must say it was a good observation. This seems to be quite good method and probably the easiest method to deal with dependent sources. SecondChildUserIdTAG: 315681 SecondChildUserNameTAG: nahuja1 SecondChildCreateTimeTAG: 2012-10-07T14:47:38Z IndexTAG: 422 TitleTAG: Ignorable dependent source I think with that small α, the dependent source is just ignorable. The independent current source gives 4mA, if - as a first guess - I presume that all this current goes to R1, then α*4mA=16 mV. The output voltage (ignored the dependent voltage source) is the voltage that the 4mA generates through the two resistors (R12 is about 400Ω), is 1.6V large, so that first guessed 16mV is only 1% of this: really ignorable "noise voltage." UserIdTAG: 156661 UserNameTAG: balazsb CreateTimeTAG: 2012-10-05T20:24:48Z VoteTAG: 6 CoursewareTAG: Week 4 / Dependent Sources Exercise 1 CommentableIdTAG: 6002x_dep_src_exsz_1 NumberOfReplyTAG: 0 IndexTAG: 423 TitleTAG: How should I enter it? How should I enter vB, iB and other variables? The system doesn't accept everything I try =( "Invalid input: vb not permitted in answer" UserIdTAG: 326787 UserNameTAG: Angstrem CreateTimeTAG: 2012-10-01T08:26:26Z VoteTAG: 6 CoursewareTAG: Week 4 / Dependent Sources Exercise 0 CommentableIdTAG: 6002x_dep_src_exsz_0 NumberOfReplyTAG: 2 FirstChildTAG: First of all in this exercise voltage must be presented with capital letter. Then you should check if there should indeed be a VB in your answer. FirstChildUserIdTAG: 304587 FirstChildUserNameTAG: Vasso FirstChildCreateTimeTAG: 2012-10-01T09:14:16Z SecondChildTAG: Use capital letter in the two answers then click "chick" SecondChildUserIdTAG: 464142 SecondChildUserNameTAG: wedadMA SecondChildCreateTimeTAG: 2012-10-01T12:32:46Z SecondChildTAG: dont use vB and iB , substitut them with VI and RI SecondChildUserIdTAG: 236136 SecondChildUserNameTAG: Hamid-ch SecondChildCreateTimeTAG: 2012-10-05T02:20:14Z FirstChildTAG: As a non native English speaker I thought Vo meant V[zero]. It appears it's the letter 'O' and not the digit '0'. And the opposite for V1 that is Vi. I should use my brain instead of my eyes on those diagrams...So you can try to combine VO and RI for example. FirstChildUserIdTAG: 10754 FirstChildUserNameTAG: clarkguenfu FirstChildCreateTimeTAG: 2012-10-03T13:51:35Z SecondChildTAG: I figured if they used RI (I=input) for the resistance of the input, they would use RO (cap letter O) for the resistance of the output. My question is, on the midterm, will these type of errors count against our 3 tries? I'm thinking not, as long as there are multiple ways to write the answer, as in this question. What they will probably do on the midterm is state in what terms they want the answer in, then there is no question on what to put in the answer. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-03T15:44:37Z IndexTAG: 424 TitleTAG: I am seeing a lot of discussion.. Well guys. I've seen lots of problems with this exercise. Then, I will tell the way I got it and expect it to be helpful. First of all, you need to find out the Vth. Secondly , find out the Rth. Right, now we've got a voltage source connected to 2 elements. One is linear (my Rth) and the another one is non-linear, my element x). Let's see what the greatest current is: So if we assume the resistance of the element x as 0, then the maximum current flowing will be 1.06 mA. So, it seems that , given a resistance for the element x in the order of kilos, the current flowing is going to be no greater than 1 mA. Since we 'know' that the current flowing is greater than 0 and less than 1mA, we do know what is the resistance of the element x: 1k ohms. Now we are able to find out the current flowing it and what is the voltage across it easily. Please notice: If we had chosen the resistance as 2K ohms , then the current flowing it would be less than 1 m A. But, for a less than 1mA current flowing the element x has a resistance as 1K ohms. Because of that, there is only one answer for this question. UserIdTAG: 125388 UserNameTAG: Glauber CreateTimeTAG: 2012-09-30T21:18:03Z VoteTAG: 6 CoursewareTAG: Week 3 / Analyzing Nonlinear Circuits Exercise CommentableIdTAG: 6002x_analyzing_nonlinear_circuits_exercise NumberOfReplyTAG: 1 FirstChildTAG: Thank you very much Glauber :) Good explanation FirstChildUserIdTAG: 329866 FirstChildUserNameTAG: Balaji_Gopal FirstChildCreateTimeTAG: 2012-10-06T05:47:45Z SecondChildTAG: Thanks so much SecondChildUserIdTAG: 359228 SecondChildUserNameTAG: EngSara SecondChildCreateTimeTAG: 2012-10-18T12:52:48Z IndexTAG: 425 TitleTAG: relationship between incremental resistance and element v-i properties I get the incremental resistance as the reciprocal of the slope of $i = v^3$ using the operating voltage, which gives $R= 0.14$ . However, at the operating point of the device, from $R_d=\frac{v}{i}=0.41$ and it is this value that I need if I want to check, for example, that the current flowing through the resistor and the non-linear device adds up to 4 A. I suppose that one is the effective operating resistance for the device (at the operating voltage) and the incremental resistance is the change in voltage that would arise from a small change in current, again at the specific operating point (ie $\Delta v = \Delta i \times R_{inc}$). I am a bit confused because they are both resistance values relating to the device and I expected them to be the same. Am I right about this? Is there a better way of thinking about it? UserIdTAG: 434869 UserNameTAG: patey CreateTimeTAG: 2012-09-30T20:23:24Z VoteTAG: 6 CoursewareTAG: Week 4 / Graphs Exercise CommentableIdTAG: 6002x_graphs_exercise NumberOfReplyTAG: 3 FirstChildTAG: I think this is a great question. We seem to have two kinds of resistances. I have vague memories of an "AC resistance" and a "DC resistance," but I just don't remember the details. I will speculate - but I do not know - that the original ohm's law, V = IR, comes from a differential equation. Something to do with Maxwell's equations. I'd love to see a really good, comprehensive discussion of this one. FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-09-30T22:07:57Z SecondChildTAG: Ohm's Law is called a constitutive relation. It is not a universal law but is true for a certain important class of materials. The microscopic version says that current density is equal to the conductivity times the electric field. Have a look here, equation 1.3.15 http://www.ece.rutgers.edu/~orfanidi/ewa/ch01.pdf SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-01T00:14:16Z SecondChildTAG: I am partial to Walter Lewin's physics course also from MIT. In this lecture he talks about Ohm's Law. http://ocw.mit.edu/courses/physics/8-02-electricity-and-magnetism-spring-2002/video-lectures/lecture-9-currents-resistivity-and-ohms-law/ SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-01T00:34:57Z SecondChildTAG: second W Lewin's lectures. They're incredible fun to watch! SecondChildUserIdTAG: 5325 SecondChildUserNameTAG: vkz SecondChildCreateTimeTAG: 2012-10-01T12:30:47Z FirstChildTAG: Your second paragraph is correct. Small change in current times incremental resistance gives change in voltage. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-09-30T23:55:22Z SecondChildTAG: second @skihawk. When you calculate the resistance at point you're doing so for this particular circuit. On the other hand incremental resistance is of interest when you begin to introduce small change (current or voltage) in your circuit which is the same as adding AC or periodic current source to your circuit. Remember why we're interested answering this question - we would like to get linear response from our non-linear device. Impossible in general but there're regions where linearity is good enough approximation of the behavior. SecondChildUserIdTAG: 5325 SecondChildUserNameTAG: vkz SecondChildCreateTimeTAG: 2012-10-01T12:44:09Z FirstChildTAG: I would say that the term "incremental resistance" means the effective resistance for an incremental voltage dífference at a specific point on the device's characteristic curve (in this case at the operating point). Using i=^3, I find di/dv = 3v^2. Being somewhat cavalier about the definition of the derivative at this point, I conclude that dv/di = 1/(3v^2). Substituting the voltage from the first part of the question (1.56v) this comes out to be 0.137 ohm, which is closer than I usually get to the required answer. I suppose one could get involved in calculating current through the device, but since it's in parallel with the 8.2 ohm "load" (?) resistor, this seems like a needlessly complicated way to get a result. FirstChildUserIdTAG: 149043 FirstChildUserNameTAG: ajhil FirstChildCreateTimeTAG: 2012-10-02T23:14:17Z SecondChildTAG: Yes, ajhil. You have correctly interpreted the question. The solution you have provide is absolutely correct. Keep it on. SecondChildUserIdTAG: 159427 SecondChildUserNameTAG: Suyog SecondChildCreateTimeTAG: 2012-10-05T11:04:00Z IndexTAG: 426 TitleTAG: Wiki Hints Hi my ClassmatesX haha! How are you doing in this amazing 6.002x (Fall)? I hope you are learning a lot and also having fun! If you are interested, I have added a section in [Myrimit's Guide to 6.002x][1]: "Myrimit's Hints" with Weekly Hints. I will try to update it weekly ;). So, that it is more easy that you can find them. See you! Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/MyrimitsGuideTo6002x/ UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2012-09-30T02:51:23Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: **Myrimit**, thank you for all the help you providing to new students like us. By the way would like to stay connected to you via some other network also, so that I can get constant feedback from you. Can i get your email id? FirstChildUserIdTAG: 159427 FirstChildUserNameTAG: Suyog FirstChildCreateTimeTAG: 2012-09-30T11:03:32Z SecondChildTAG: Thank you Suyog! ;). I really enjoy to help here! I have an account myrimit (at) gmail (dot) com but I prefer to answer in this Forum Discuss (remember that your doubt can be the doubt of others ;) ). And also, remember that the Staff are working hard with feedback by e-mail, so that you can get notice when someone post a comment, etc in your Post, etc ;). See you! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-30T16:19:38Z SecondChildTAG: thanks,amigo SecondChildUserIdTAG: 357747 SecondChildUserNameTAG: kishores SecondChildCreateTimeTAG: 2012-09-30T16:32:21Z SecondChildTAG: ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-30T16:51:44Z FirstChildTAG: **Myriam, you are awesome!** I often see you helping out more than the staff does! I try to do my part to help out, too; but sometimes work and life just leave me with too little time. I know when I do help others, with a homework hint for example, it helps me retain the material a lot better 'cause I go over it a second time. With all the help you give out, you stand a good shot at gettin' an A in the class; after a while you tend to analyze circuits in dreams and see schematics whenever you close your eyes! As for me, when I took Circuit Analysis the first time during university, there was so much information presented that I was overloaded! Taking it the second time for fun (and a cool certificate is good, too) lets me catch up on some of the finer details that I missed. Its nice to know that high-schoolers now have access to this material; if we had classes like this online back then...who knows...I'd choose online versus a large lecture hall with 100+ students anyday! The difference between MITs course and my university's EECS course is that here we cover a lot of material very quickly versus there we delved a lot deeper into the analog side. Here we're not required to use linear algebra or matrices to solve simultaneous equations for nodal analysis; there, it was a must. Here, we are covering logic gates and MOSFETs whereas at my university class Digital Analysis would be a separate class; we instead would talk about "supernodes", "phasors", "delta-wye", etc. It makes taking the class over seem like looking at it from a different perspective. Just my thoughts on a few things; keep up the great work! Mark in New Jersey, USA FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-09-30T05:12:02Z SecondChildTAG: Thank you! ;). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-30T16:08:19Z FirstChildTAG: In week 3, homework+lab had broken my neck and back. But your help revived my sustenance. Thank you Myriam. FirstChildUserIdTAG: 378096 FirstChildUserNameTAG: Jamshaid271 FirstChildCreateTimeTAG: 2012-09-30T22:56:56Z SecondChildTAG: You are welcome Jamshaid271 :)! How are you? Confession: Hahaha, when I first took a look to your comment I first read "broken my neck and back" and I said: Oh no! what happened to Jamshaid271??!! Poor, he have an accident!,nooo!(I got worried) and then I read it slow, haha. Again, you are welcome :). Nice to see that I am being helpul here. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-30T23:39:26Z SecondChildTAG: hahahahha...! SecondChildUserIdTAG: 378096 SecondChildUserNameTAG: Jamshaid271 SecondChildCreateTimeTAG: 2012-10-02T00:57:21Z SecondChildTAG: haha! XD SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-02T01:07:38Z IndexTAG: 427 TitleTAG: Feeling Like a boss Everything going to fine.its wonderful to learn. UserIdTAG: 518305 UserNameTAG: fathma_hafiz CreateTimeTAG: 2012-09-29T01:41:50Z VoteTAG: 6 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 0 IndexTAG: 428 TitleTAG: This is amazing! The way this course goes is just superb. Dr. Agrawal and the team involved in making this happen are amazing! I am glad to be a part here. A detailed expression follows soon! UserIdTAG: 372623 UserNameTAG: Amitraj CreateTimeTAG: 2012-09-28T12:20:29Z VoteTAG: 6 CoursewareTAG: Week 3 / Inside a Mouse CommentableIdTAG: 6002x_inside_a_mouse NumberOfReplyTAG: 0 IndexTAG: 429 TitleTAG: lab 3 Hi, I am getting the exact plot after transient analysis taking W/L=44 for each n-MOS, but after clicking check option it is showing my design wrong. Anybody solved that one? W/L = 44 or any other value? UserIdTAG: 113365 UserNameTAG: satyabrata CreateTimeTAG: 2012-09-27T16:00:52Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 7 FirstChildTAG: I have the same problem. Find the W / L and observing the graph I get perfectly as he asks ... but the check I get this wrong ... FirstChildUserIdTAG: 402650 FirstChildUserNameTAG: ManoloR21 FirstChildCreateTimeTAG: 2012-09-27T18:52:21Z SecondChildTAG: The checker will be verifying the voltage of the output node at several different times, so you'll earn a checkmark only after you've performed the transient simulation so that the checker will have a waveform to check SecondChildUserIdTAG: 232499 SecondChildUserNameTAG: Asim09 SecondChildCreateTimeTAG: 2012-09-29T19:06:31Z FirstChildTAG: Have you implemented (A+B)c properly FirstChildUserIdTAG: 272523 FirstChildUserNameTAG: Jivraj FirstChildCreateTimeTAG: 2012-09-27T16:45:29Z SecondChildTAG: Yes..The plot is coming the same as required too, still it is showing it's wrong. SecondChildUserIdTAG: 113365 SecondChildUserNameTAG: satyabrata SecondChildCreateTimeTAG: 2012-09-27T16:53:17Z SecondChildTAG: The checker will be verifying the voltage of the output node at several different times, so you'll earn a checkmark only after you've performed the transient simulation so that the checker will have a waveform to check. The waveform has to be exactly similar even the small spikes!! SecondChildUserIdTAG: 232499 SecondChildUserNameTAG: Asim09 SecondChildCreateTimeTAG: 2012-09-29T19:07:28Z FirstChildTAG: satyabrata I had the same problem a while ago, and i figured out what was wrong. in the transient analysis look at the value of the spike. its value must be lower than 0.25. If not, try adjusting the W/L values until you get a value lower than 0.25 for the spike. sorry if my explanation wasn;t clear. FirstChildUserIdTAG: 371000 FirstChildUserNameTAG: Joseph090892 FirstChildCreateTimeTAG: 2012-09-27T17:11:20Z SecondChildTAG: isn't the worst value equal to **2RON**? SecondChildUserIdTAG: 159427 SecondChildUserNameTAG: Suyog SecondChildCreateTimeTAG: 2012-09-28T12:40:40Z SecondChildTAG: Thanks..I got it. SecondChildUserIdTAG: 113365 SecondChildUserNameTAG: satyabrata SecondChildCreateTimeTAG: 2012-09-28T16:42:45Z FirstChildTAG: just put W/L more then you calculated - if you calculated 44 put it 50 FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-09-27T21:01:04Z SecondChildTAG: Thanx dear...:) SecondChildUserIdTAG: 319071 SecondChildUserNameTAG: Hashir34 SecondChildCreateTimeTAG: 2012-09-28T06:51:00Z SecondChildTAG: If I get L/W = 43.725, Would not be W/L the reciprocal?? I mean .02287 but when I enter 0.02287 the graph doen't match but using 43.725 match. Another questio is wouldn't be multiplies of 50s the factor L/W. ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13490326475864112.png SecondChildUserIdTAG: 99419 SecondChildUserNameTAG: Texada SecondChildCreateTimeTAG: 2012-09-30T19:17:59Z FirstChildTAG: how do i find W/L? FirstChildUserIdTAG: 357747 FirstChildUserNameTAG: kishores FirstChildCreateTimeTAG: 2012-09-28T17:35:02Z SecondChildTAG: Hint: they are multiples of 50 SecondChildUserIdTAG: 232499 SecondChildUserNameTAG: Asim09 SecondChildCreateTimeTAG: 2012-09-29T19:12:03Z SecondChildTAG: Rn is fixed to 25.5kohms and you have to calculate the equivalent Resistance of the network, once you done, it is too simple to calculate L/W using the equation 6.4 (see page 305 on Textbok) you should solve for (L/W) thinkink more like a division instead of multiplying (I guess)! [enter image description here][1] Notice the comment above, the value should be a multiple of 50 I guess this is for commercial/manufacture purposes. So any other value may not work for grade only for graph correctly. [1]: https://edxuploads.s3.amazonaws.com/13490316581343662.png SecondChildUserIdTAG: 99419 SecondChildUserNameTAG: Texada SecondChildCreateTimeTAG: 2012-09-30T19:03:57Z FirstChildTAG: Hi my edx friends please help me solving in LAB 3 1)How to implemented ~(A+B)c? 2)How set w/l? FirstChildUserIdTAG: 170636 FirstChildUserNameTAG: shanthasuresh29 FirstChildCreateTimeTAG: 2012-09-28T17:57:32Z SecondChildTAG: A+B can me like two parallel switches product acts as a series switch. This the whole system comes out to be like two parallels with a series or series with two parallels.!! SecondChildUserIdTAG: 232499 SecondChildUserNameTAG: Asim09 SecondChildCreateTimeTAG: 2012-09-29T19:10:35Z FirstChildTAG: A+B can be like two parallel switches product acts as a series switch. Thus the whole system comes out to be like two parallels with a series or series with two parallels.!! FirstChildUserIdTAG: 232499 FirstChildUserNameTAG: Asim09 FirstChildCreateTimeTAG: 2012-09-29T19:10:20Z IndexTAG: 430 TitleTAG: Improper units This is really a minor thing at it doesn't have impact on the actual solution, but i=v^3 would mean that the units of the current is (Volts)^3. Seems like there is a constant factor of 1 Ampere/(Volts^3) missing from the equation. UserIdTAG: 384467 UserNameTAG: dabdab CreateTimeTAG: 2012-09-26T19:47:14Z VoteTAG: 6 CoursewareTAG: Week 3 / Graphical Analysis of Nonlinear Circuits Exercise CommentableIdTAG: 6002x_graphical_analysis_of_nonlinear_circuits_exercise NumberOfReplyTAG: 2 FirstChildTAG: You're totally right. There is a sleight of hand here in the units. FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-26T20:13:24Z FirstChildTAG: I guess they meant : I = V^3 in matter of algebraic value, a possible correct expression may be : i = V^3 / R^3, where R = 1 ohm. You may find this often when you witness non-linear expressions. It may confuse you for a while, but you will learn how to adapt. FirstChildUserIdTAG: 418182 FirstChildUserNameTAG: euldji2005 FirstChildCreateTimeTAG: 2012-09-26T23:54:09Z SecondChildTAG: i think it means that the current is proportionally to v^3. if u want to be absolutely correct: i=k*v^3, where the constant k is Amper/Volt^3 SecondChildUserIdTAG: 65651 SecondChildUserNameTAG: Arnaldo_Martell SecondChildCreateTimeTAG: 2012-10-02T23:48:46Z IndexTAG: 431 TitleTAG: Two ways of getting Rth Prerequisite: solve KVL/KCL for i; i = R2*I0/(R1+alpha+R2). Calculate Vth from i. 1: Using Rth = Vth/I(short circuit): Substitute 0 for R2 in the formula for i and get i for the output short circuited. (It turns out rather obviously to be 0.) Short circuit current is therefore I0 and Rth = Vth/I0. 2: Calculate the resistance of the network with the independent sources turned off: Note that with I0 turned off, you have the voltage source in series with R1 with exactly i going through it. The voltage source's voltage being alpha*i is indistinguishable from a resistor with resistance alpha, so it can be combined with R1 by simple addition. So you have Rth = (R1 + alpha)||R2. In this particular case, it isn't necessary to force a current into the network to calculate it's resistance and if you do, you'll essentially end up deriving the formula for a resistor in parallel with two resistors in series :) UserIdTAG: 141000 UserNameTAG: OrinE CreateTimeTAG: 2012-09-25T06:25:14Z VoteTAG: 6 CoursewareTAG: Week 4 / Dependent Sources Exercise 1 CommentableIdTAG: 6002x_dep_src_exsz_1 NumberOfReplyTAG: 2 FirstChildTAG: 2 - ingenious :) FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-09-30T14:03:46Z SecondChildTAG: Thank You SecondChildUserIdTAG: 68114 SecondChildUserNameTAG: PrashanthGowda SecondChildCreateTimeTAG: 2012-10-05T17:39:09Z FirstChildTAG: brilliant idea ...........of solution no.1 FirstChildUserIdTAG: 477713 FirstChildUserNameTAG: ikm104 FirstChildCreateTimeTAG: 2012-10-06T08:16:17Z IndexTAG: 432 TitleTAG: Load Line Equation Apply KCL on the node i+v/8.2=4 so the equation of load line is **i=4-v/8.2** this gives, i=4 for v=0 and v=32.3 for i=0 UserIdTAG: 200355 UserNameTAG: bhavj CreateTimeTAG: 2012-09-24T08:00:13Z VoteTAG: 6 CoursewareTAG: Week 3 / Graphical Analysis of Nonlinear Circuits Exercise CommentableIdTAG: 6002x_graphical_analysis_of_nonlinear_circuits_exercise NumberOfReplyTAG: 1 FirstChildTAG: THANK YOU ! ! ! FirstChildUserIdTAG: 10512 FirstChildUserNameTAG: asicok FirstChildCreateTimeTAG: 2012-10-01T00:12:17Z IndexTAG: 433 TitleTAG: 7.2v or -7.2v? Well the correct answer is 6.20. But obtain this number is necessary to consider the -7.2v with a positive value. Because if we calculate with a negative voltage as indicated in the problem the voltage value will be in the node "e" = -1.69032V. Why we consider the value positive to calculate the voltage in the node? Thanks UserIdTAG: 166869 UserNameTAG: Vasco CreateTimeTAG: 2012-09-24T01:20:34Z VoteTAG: 6 CoursewareTAG: Week 2 / Node equation review CommentableIdTAG: 6002x_S3E1_NodeEquationReview NumberOfReplyTAG: 1 FirstChildTAG: Look back at the polarity of the voltage source: the negative just means that the potential difference is opposite of what is labeled as + and - on the diagram. FirstChildUserIdTAG: 248902 FirstChildUserNameTAG: Chanute FirstChildCreateTimeTAG: 2012-09-24T02:24:35Z IndexTAG: 434 TitleTAG: Link to a useful thread on Lab 2 Rather than repeat so much of what has already been written, have a look here: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5058f699cf5d7b2b00000005 UserIdTAG: 406202 UserNameTAG: skyhawk CreateTimeTAG: 2012-09-22T23:46:55Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Thanks for posting that to help out. +1 for that Bad news is I still have know idea as to what any of that means. I just got "a" correct answer, but I actually went to work, got 2 arb generators, 3 decade boxes, an o'scope, and just built the damn thing. Even with the correct answer I do not understand what the actual lesson was supposed to be. Everyone is passing around short-hand math formulas, that I just do not understand. Schematics I do understand. It might cost me some points in the class, but I will admit there are no simple math formulas that will explain my circuit. Is it against the rules for someone to post the answer after the homework is due? I would love to understand the point of the exercise. It was said before in another post. There is no real credit here, I just really want to learn it. FirstChildUserIdTAG: 349840 FirstChildUserNameTAG: Wilk FirstChildCreateTimeTAG: 2012-09-23T00:22:34Z SecondChildTAG: I believe that it is perfectly OK to post the solution after the due date. I found the problem very interesting and have formulated the problem a number of different ways. In my opinion it is a good problem to practice the basic methods of circuit analysis: Thevenin, superposition, the node method, etc. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-09-23T00:39:50Z SecondChildTAG: "I actually went to work, got 2 arb generators, 3 decade boxes, an o'scope, and just built the damn thing." Awesome. It is neat. In reality it is just a "fixed" potentiometer and resistor, kinda like a balance control of sorts. You could wire the circuit with a potentiometer and a single resistor two different ways. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-23T01:09:14Z SecondChildTAG: I believe your comment is most likely correct. My issue is I doubt I can pass the course just believing in you:) I want to know why you are arriving at the correct answer, and more importantly how. I was not able to solve using any of the methods being taught. I had to default to the "shotgun method". While this is very useful in real life, it probably won't get me past the final. All said lightly. You couldn't help me but did help many others, and is appreciated either way. We all learn and understand differently. SecondChildUserIdTAG: 349840 SecondChildUserNameTAG: Wilk SecondChildCreateTimeTAG: 2012-09-23T01:14:15Z SecondChildTAG: Putting it in "mathy terms" the two voltage sources and the two resistor mixer are equivalent to a Vth and Rth in series. Adding a load resistor (R3) then creates a voltage divider that attenuates the mixed signal to the degree that is desired. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-09-23T01:36:35Z IndexTAG: 435 TitleTAG: Lab 2 successful After working 4 hours, phewwwwww, I finally make the lab 2. All work done with only two resistors. No need of 3rd one. Just manipulate the volts in voltage source with square wave. Parameters are constant, for example type of wave, frequency etc. Voltage can be said as the parameter, but it is changing from 0 to 1 in square and -1 to +1 in sin. So we can choose any of the voltage in between them to solve over problem with only two resistors. I prefer to change square wave. UserIdTAG: 378096 UserNameTAG: Jamshaid271 CreateTimeTAG: 2012-09-22T23:11:00Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 6 FirstChildTAG: I thought the instructions say that we cannot change the parameters of the voltage sources? FirstChildUserIdTAG: 359310 FirstChildUserNameTAG: AaronYeoh FirstChildCreateTimeTAG: 2012-09-22T23:20:03Z SecondChildTAG: Don't change the parameters of voltage source, but we are told that voltage is in between o to 1. So we can hover over. SecondChildUserIdTAG: 378096 SecondChildUserNameTAG: Jamshaid271 SecondChildCreateTimeTAG: 2012-09-22T23:23:59Z SecondChildTAG: please give me the solution of H2P1 and n help me in doing lab2 ..actually i did same according to u bt invain..pleas heip me SecondChildUserIdTAG: 282892 SecondChildUserNameTAG: wiky SecondChildCreateTimeTAG: 2012-09-23T00:43:13Z SecondChildTAG: so wht changes u made in voltage source and what resistance u used? i am only stuck with the lab2 help SecondChildUserIdTAG: 403493 SecondChildUserNameTAG: aiai SecondChildCreateTimeTAG: 2012-09-23T17:18:15Z SecondChildTAG: huraaaaaaaaaaaaaaaiii!! SecondChildUserIdTAG: 403493 SecondChildUserNameTAG: aiai SecondChildCreateTimeTAG: 2012-09-23T18:17:12Z SecondChildTAG: dude it took 14 hours for me to solve the riddle and its aching my head now with homework its not saying correct even though its correct answer i.e, q2 in 2nd week rl =5.1 and rth eq =5.1 SecondChildUserIdTAG: 383699 SecondChildUserNameTAG: Gogineni SecondChildCreateTimeTAG: 2012-09-23T18:30:10Z SecondChildTAG: can u pls tell me how to submit homework.....no any link for submission SecondChildUserIdTAG: 278178 SecondChildUserNameTAG: sandy12992 SecondChildCreateTimeTAG: 2012-09-27T00:27:58Z SecondChildTAG: Just click check button and your homework will be checked as well as submitted. SecondChildUserIdTAG: 378096 SecondChildUserNameTAG: Jamshaid271 SecondChildCreateTimeTAG: 2012-09-27T05:01:29Z FirstChildTAG: You had me excited there for a second, until you mentioned you changed the voltage source. I'll let someone else rain on your parade. Good effort! FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-22T23:19:07Z SecondChildTAG: Parameters are called that what is constant in an equation, but voltages are not constant. We can choose from 0 to 1. So I did. Do it my way, you will oblige me. SecondChildUserIdTAG: 378096 SecondChildUserNameTAG: Jamshaid271 SecondChildCreateTimeTAG: 2012-09-22T23:27:26Z SecondChildTAG: Penny please help me solve it i solved the entire homework 2 but only lab 2 is left Thank you SecondChildUserIdTAG: 378150 SecondChildUserNameTAG: GladIDoThis SecondChildCreateTimeTAG: 2012-09-22T23:29:14Z SecondChildTAG: same here SecondChildUserIdTAG: 403493 SecondChildUserNameTAG: aiai SecondChildCreateTimeTAG: 2012-09-23T18:10:27Z FirstChildTAG: Hi Jamshaid271! **Be careful** with Lab 2... The statement says: "Enter your circuit below, using the appropriate configuration of resistors. Please **do not modify the wiring or parameters of the voltage sources** -- your goal is to take the signals they generate and combine them, not to change what is generated. Run a 5ms transient analysis to verify the correct operation of your circuit. We will be checking for the transient waveform at the "output" node." ![vsupply][1] So, you can **not change** anything of the given two sources... The statements says that: WARNING: V2 is a 5 kHz sine wave that varies between -1V and +1V. ![sin][2] WARNING: V1 is a 1 kHz square wave that varies between 0V and +1V. ![wave][3] So, the parameters of those given sources behaves live the images shown before... so, you don't have to change the voltages of the sources because they are already varying in the case of the sin from -1V to 1V and in the case of the square wave from 0V to 1V... So, be careful... Can I help you ? Myriam. [1]: https://edxuploads.s3.amazonaws.com/13483590712550784.png [2]: https://edxuploads.s3.amazonaws.com/13483596197979652.png [3]: https://edxuploads.s3.amazonaws.com/1348359994906241.png FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-23T00:31:48Z SecondChildTAG: so how can get the respective waveform..please tell me the whole solution..n also tell me the solution of home work 2 .H2P1..please SecondChildUserIdTAG: 282892 SecondChildUserNameTAG: wiky SecondChildCreateTimeTAG: 2012-09-23T00:41:59Z SecondChildTAG: So we have to add another resistor? SecondChildUserIdTAG: 229018 SecondChildUserNameTAG: Changming SecondChildCreateTimeTAG: 2012-09-23T00:54:41Z SecondChildTAG: Hi wiky. I only can give you hints so that you can solve it by yourself as this is a graded part...can I help you? in which part of Lab2 are you lost? for H2P1 you can read [here][1]. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/undefined/threads/505971f4dd2f4d1f00000004 SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-23T01:43:10Z SecondChildTAG: Hi changming. hint1: Try to find a resistive network, hint2: take a look at here [superposition method][1]. ![superposition][2] [1]: https://edxuploads.s3.amazonaws.com/13483651188187161.png [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/173 SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-23T01:52:46Z SecondChildTAG: ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13483652111343643.png SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-23T01:53:58Z SecondChildTAG: ya Myriam. u r right...)) SecondChildUserIdTAG: 93123 SecondChildUserNameTAG: Bilawal SecondChildCreateTimeTAG: 2012-09-23T07:51:01Z SecondChildTAG: Myrimit you are right and have wonderful posts. I really appreciate your effort. SecondChildUserIdTAG: 183507 SecondChildUserNameTAG: obiradaniel SecondChildCreateTimeTAG: 2012-09-23T12:49:23Z SecondChildTAG: Thank you obiradaniel :)! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-23T15:46:48Z SecondChildTAG: you are welcome SecondChildUserIdTAG: 183507 SecondChildUserNameTAG: obiradaniel SecondChildCreateTimeTAG: 2012-09-23T21:07:31Z FirstChildTAG: Hi jamshaid...i dit same accroding to your way..bt still invain. so please tell me the solution of lab2.and also H2P1...thnx FirstChildUserIdTAG: 282892 FirstChildUserNameTAG: wiky FirstChildCreateTimeTAG: 2012-09-23T00:54:39Z SecondChildTAG: Friend rules allow me to limit to hints. Plz don't mind. If allowed I must explain everyhing. SecondChildUserIdTAG: 378096 SecondChildUserNameTAG: Jamshaid271 SecondChildCreateTimeTAG: 2012-09-23T01:10:41Z SecondChildTAG: As far as H2P1 concerned, there is a list of resistors given by the E12. Choose to of them and do a voltage divider. That you get approximately 27V. Then the ranges or 10 % tolerance of those resistors you have chosen to figure out Vout will tell you the Vmax and Vmin. SecondChildUserIdTAG: 378096 SecondChildUserNameTAG: Jamshaid271 SecondChildCreateTimeTAG: 2012-09-23T01:17:43Z SecondChildTAG: so wht can i do SecondChildUserIdTAG: 282892 SecondChildUserNameTAG: wiky SecondChildCreateTimeTAG: 2012-09-23T01:18:49Z SecondChildTAG: jamshaid if i would change square wave voltage thn make it constant..and thn change resistance?? SecondChildUserIdTAG: 282892 SecondChildUserNameTAG: wiky SecondChildCreateTimeTAG: 2012-09-23T01:20:48Z SecondChildTAG: Yes after changing and making it fix, you have to change value of resistors. SecondChildUserIdTAG: 378096 SecondChildUserNameTAG: Jamshaid271 SecondChildCreateTimeTAG: 2012-09-23T01:24:18Z SecondChildTAG: n what about sin wave if we dnt change it to negative value..we wont get negative value is it rite...or jst remain it 0 volts SecondChildUserIdTAG: 282892 SecondChildUserNameTAG: wiky SecondChildCreateTimeTAG: 2012-09-23T01:27:00Z SecondChildTAG: why 27v in h2p1..we need vout 31.5??? jamshaid SecondChildUserIdTAG: 282892 SecondChildUserNameTAG: wiky SecondChildCreateTimeTAG: 2012-09-23T01:27:57Z SecondChildTAG: Hey don't touch sin wave. You will adhere to it. And 27V is because we need to provide. SecondChildUserIdTAG: 378096 SecondChildUserNameTAG: Jamshaid271 SecondChildCreateTimeTAG: 2012-09-23T01:31:06Z FirstChildTAG: Myriam you wrote true but you tell me that values of square wave is lying between 0 to 1 or not. After all I have tried every thing. I mean use 2,3,4,5,6... resistors. Attach them in series and parallel, but always I got -167mV and on the other side one thing remains constant and that was 1000 or 999mV. That was weird for me. So I think of doing this and I got "green symbol" of life. FirstChildUserIdTAG: 378096 FirstChildUserNameTAG: Jamshaid271 FirstChildCreateTimeTAG: 2012-09-23T01:07:04Z SecondChildTAG: I just got -167mv but V max always the same as 999mv as Jamshaid271. As I said, I feel we do have to have 3rd resistor. Don't know where. SecondChildUserIdTAG: 229018 SecondChildUserNameTAG: Changming SecondChildCreateTimeTAG: 2012-09-23T01:10:47Z SecondChildTAG: Hi Jamshaid271! Are you sure that you are choosing the correct resistive network? Hint 1: Try to take a look to the superposition method. Read this page of the Textbook [here][1]. Hint 2: [Take a look at this part of Textbook][2] . Hint 3: mean use more than 2 ;). [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/173 [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/99 SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-23T02:09:57Z SecondChildTAG: Do you solve with that are saying? SecondChildUserIdTAG: 378096 SecondChildUserNameTAG: Jamshaid271 SecondChildCreateTimeTAG: 2012-09-23T02:17:19Z SecondChildTAG: Hey Myrimit, something is wrong. I think you change your name spellings. Leave it whatever it was. If you are talking about superposition, then know that for superposition or any theoram we have to know the value of at least one power source. If you stick with superpostion, then after applying it and solving all problems we don't have our required solution. For that we have to walk between given voltages to find our necessity. SecondChildUserIdTAG: 378096 SecondChildUserNameTAG: Jamshaid271 SecondChildCreateTimeTAG: 2012-09-23T02:24:08Z SecondChildTAG: That are hints that can help you to find the answer... with that hints you can have some direction that can help you in order to find which is the resistive network that you need. You can use the superposition method mixing the concept of voltage dividers and choosing the correct values of the resistive network (finding how is it that network it is your task). Remember that when you use the concept of superposition you have to analyze the circuits by parts and sum all the effect in order to see how it is the output reffered two the both sources (they give you the condition of Vout≈...V1+...V2) and with the resistive network (that it is your task to find), you can have the equations and obtain the values that you need. Remember that you also can adopt values if the equations are not enough to find all the values ;). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-23T02:32:09Z SecondChildTAG: **Another last hint:** Suppoused that you have 3 voltage sources and they give you the output voltage Vout = (1/2)*V1 + 3*V2 + (3/2)*V3 this means that the output will be the sum of the different voltage sources (V1,V2 and V3) acting in the output... So, How do we find each contribution? we need to use the superposition method. What does it says that superposition method? it says that we can obtain, in this case, the output voltage by analyzing each contribution, that is to say sum: V1 contribution when V2=0 and V3=0 (Vo1); v2 contribution when V1=0 and V3=0 (Vo2); V3 contribution when V1=0 and V3=0 (Vo3). So, Vout = Vo1 + Vo2+ Vo3 Where Vout = function1(R1,R2,R3,R4)*V1 + function2(R1,R2,R3,R4)*V2 + function3(R1,R2,R3,R4)*V3 ![enter image description here][1] If you have that function1(R1,R2,R3,R4) = value1, function2(R1,R2,R3,R4) = value2, function3(R1,R2,R3,R4) = value3, and adopting some value etc... Can you find The components? yes! ;) I think that I can tell you more. I hope this can help you. Myriam. [1]: https://edxuploads.s3.amazonaws.com/13483695634954417.png SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-23T03:11:01Z SecondChildTAG: *can not tell you more. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-23T03:12:00Z SecondChildTAG: Thanx for your interest and I read all this. Actually I submitted my lab 2 with two ways. Best way I have told and other is similar to yours. Anyway I am pleased that you are keen to share knowledge, and today there are few who want to transfer information. Thanx again to show everyone here an excellent work. SecondChildUserIdTAG: 378096 SecondChildUserNameTAG: Jamshaid271 SecondChildCreateTimeTAG: 2012-09-23T03:36:41Z SecondChildTAG: Well. After filling all the text boxes in H2P1 when I click on "check", it showed that Vmax and Vmin are true and resistances are false. I didn't understand what was wrong. Anybody could explain it? SecondChildUserIdTAG: 378096 SecondChildUserNameTAG: Jamshaid271 SecondChildCreateTimeTAG: 2012-09-23T03:39:56Z SecondChildTAG: you are welcome Jamshaid271 ;). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-23T03:40:39Z SecondChildTAG: Jamshaid271 and Myriam. well done u r doing good job here..keep it up.. Bilawal SecondChildUserIdTAG: 93123 SecondChildUserNameTAG: Bilawal SecondChildCreateTimeTAG: 2012-09-23T07:54:16Z SecondChildTAG: Thank tou Bilawal! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-23T15:46:08Z SecondChildTAG: Appreciate you Bilawal! SecondChildUserIdTAG: 378096 SecondChildUserNameTAG: Jamshaid271 SecondChildCreateTimeTAG: 2012-09-23T19:24:36Z FirstChildTAG: To solve the second laboratory is necessary to apply the superposition theorem. Assume that a circuit is constiruido by a source of 1kHz and two resistors in series. Then reslover a circuit for the power 5kHz and two resistors in series. Then overlaps the results obtained from the two circuits. FirstChildUserIdTAG: 361823 FirstChildUserNameTAG: EliasOak FirstChildCreateTimeTAG: 2012-09-23T01:45:59Z IndexTAG: 436 TitleTAG: hw 2 and lab2 plz help me solve homework 2 and lab 2 m really confused plz help me UserIdTAG: 203092 UserNameTAG: shaan007 CreateTimeTAG: 2012-09-22T16:01:23Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: i am quite confused too in understanding the items needed to satisfied like the Vin/Vout ratio in the first part of the homework..is it allowed to share hints here? thanks FirstChildUserIdTAG: 458166 FirstChildUserNameTAG: Donai FirstChildCreateTimeTAG: 2012-09-22T16:48:59Z SecondChildTAG: *to be satisfied SecondChildUserIdTAG: 458166 SecondChildUserNameTAG: Donai SecondChildCreateTimeTAG: 2012-09-22T16:53:05Z SecondChildTAG: yeah u can share hints here SecondChildUserIdTAG: 376345 SecondChildUserNameTAG: Hassan-Ahmed SecondChildCreateTimeTAG: 2012-09-22T17:53:48Z SecondChildTAG: it is Vout/Vin SecondChildUserIdTAG: 452559 SecondChildUserNameTAG: kuz1toro SecondChildCreateTimeTAG: 2012-09-22T18:49:26Z SecondChildTAG: select resistors form E12 table e.g(56,68,82)k and (15,18,22)k than using voltage divider formula for min input voltage i.e 70 and than for max. voltage i.e 90 voltage calculate vout SecondChildUserIdTAG: 414201 SecondChildUserNameTAG: uzaifakram SecondChildCreateTimeTAG: 2012-09-22T19:09:30Z SecondChildTAG: the divider ratio Vout/Vin is within 10% of the requirement, it is mean that Vout/Vin must between 10% of the specified value. Remember that Vin is constant or can guarantee to be constant because there is no statement in the problem that say it has certain value range or tolerance. on the other hand Vout is definitely can change because its a function of current and resistor value. current also function of the total resistant and voltage (Vin). The problem state that the resistor have tolerance so it have value anywhere in its specific range. So it can say that the change in Vout is totally because the value of resistor that can change. SecondChildUserIdTAG: 452559 SecondChildUserNameTAG: kuz1toro SecondChildCreateTimeTAG: 2012-09-22T19:14:54Z FirstChildTAG: can you be more specific ? FirstChildUserIdTAG: 452559 FirstChildUserNameTAG: kuz1toro FirstChildCreateTimeTAG: 2012-09-22T16:25:22Z SecondChildTAG: can u plz tell me how can i find the values of vmax and vmin in h2p1 i have find the values of r1 and r2 SecondChildUserIdTAG: 376345 SecondChildUserNameTAG: Hassan-Ahmed SecondChildCreateTimeTAG: 2012-09-22T16:47:17Z SecondChildTAG: hassan-ahmed, i think values of vmax and vmin varies with assumed resistors values because i got vmax and vmin correct at first but as i changed the values of my resistors since i got them wrong at first, all my answers became wrong already.. SecondChildUserIdTAG: 458166 SecondChildUserNameTAG: Donai SecondChildCreateTimeTAG: 2012-09-22T16:52:05Z SecondChildTAG: to find Vmax - compute the minimum value of R1, remember this resistor have tolerance so its have a specific range value - compute the maximum value of R2 - Now you can compute the Vmax by replacing nominal value of R1 with it is minimum and nominal value of R2 with it's maximum and vice versa to find Vmin SecondChildUserIdTAG: 452559 SecondChildUserNameTAG: kuz1toro SecondChildCreateTimeTAG: 2012-09-22T18:47:48Z SecondChildTAG: select resistors form E12 table e.g(56,68,82)k and (15,18,22)k than using voltage divider formula for min input voltage i.e 70 and than for max. voltage i.e 90 voltage calculate vout SecondChildUserIdTAG: 414201 SecondChildUserNameTAG: uzaifakram SecondChildCreateTimeTAG: 2012-09-22T19:09:17Z SecondChildTAG: how to find Vmax and Vmin SecondChildUserIdTAG: 35623 SecondChildUserNameTAG: ARSHIYA25 SecondChildCreateTimeTAG: 2012-09-22T19:36:36Z FirstChildTAG: Hi shaan007! Now that the deadline has passed you can take a look at [here][1] :) [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_S3E2_Linear_Combinations_of_Source_Strengths/threads/506a280d036af91f000000c9 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-02T00:09:19Z IndexTAG: 437 TitleTAG: hahahah TOO GOOD... IF Only our lecturer acts the same.. it would be great doing lectures everyday :D UserIdTAG: 151205 UserNameTAG: jesher777 CreateTimeTAG: 2012-09-22T12:33:46Z VoteTAG: 6 CoursewareTAG: Week 2 / Chain saw CommentableIdTAG: 6002x_chain_saw NumberOfReplyTAG: 0 IndexTAG: 438 TitleTAG: Unable to access video lectures....I think its due to YouTube blockage I am unable to access video lectures pls advice.... UserIdTAG: 68760 UserNameTAG: akberbana CreateTimeTAG: 2012-09-21T07:32:28Z VoteTAG: 6 CoursewareTAG: Week 2 / Linearity CommentableIdTAG: 6002x_linearity NumberOfReplyTAG: 2 FirstChildTAG: What country are you in? If you are in a country that restricts youtube, you can use a program called 'web freer'. Contact me and I'll let you know how to get it. FirstChildUserIdTAG: 198563 FirstChildUserNameTAG: PaulGillett FirstChildCreateTimeTAG: 2012-09-21T07:38:16Z SecondChildTAG: i am also unable to access the lecture videos can you find me the best way how to get these videos??? SecondChildUserIdTAG: 472147 SecondChildUserNameTAG: ricco619 SecondChildCreateTimeTAG: 2012-09-21T08:09:51Z SecondChildTAG: YouTube is blocked in Pakistan as well, we are not able to access the videos !!!! SecondChildUserIdTAG: 121258 SecondChildUserNameTAG: RahimNoorani SecondChildCreateTimeTAG: 2012-09-21T10:15:59Z SecondChildTAG: How can i use 'web freer'? SecondChildUserIdTAG: 107465 SecondChildUserNameTAG: Abasit SecondChildCreateTimeTAG: 2012-09-22T09:04:30Z SecondChildTAG: I am using 'web freer' , but the video is not running, kindly help me. SecondChildUserIdTAG: 107465 SecondChildUserNameTAG: Abasit SecondChildCreateTimeTAG: 2012-09-22T09:10:19Z SecondChildTAG: It worked :) Thank u SecondChildUserIdTAG: 332941 SecondChildUserNameTAG: SabaSiddiqi SecondChildCreateTimeTAG: 2012-09-22T22:51:40Z FirstChildTAG: You can see in the Wiki [Lecture Videos Download][1]:). Also see more info [here][2]: **Due to the YouTube outage in certain parts of the world, we have provided direct links to download the lecture videos, and extended the Week 2 homework and lab deadlines to the coming Sunday (September 30). Please note that Week 3 deadlines have not changed.** [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/ [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/info FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-26T11:48:26Z IndexTAG: 439 TitleTAG: IN S2E1.. REGARDING LOOPS HOW 7 LOOPS ARE POSIBLE IN THE GIVEN CIRCUIT DIAGRAM????? UserIdTAG: 80970 UserNameTAG: cruse CreateTimeTAG: 2012-09-19T20:32:52Z VoteTAG: 6 CoursewareTAG: Week 1 / Circuit topology CommentableIdTAG: 6002x_S2E1_Circuit_Topology NumberOfReplyTAG: 1 FirstChildTAG: Yo can see this [Post][1] too (definition of Loop, Node, etc...) S2E1: CIRCUIT TOPOLOGY ![enter image description here][2] **Loop 1:** R1 -> R4 -> R3 -> R1 **Loop 2:** R2 -> R3 -> R5 -> R2 **Loop 3:** V-> R1 -> R2 -> V **Loop 4:** V -> R4 -> R5 -> V **Loop 5:** R1 -> R4 -> R5 -> R2 -> R1 **Loop 6:** V -> R5 -> R3 -> R1 -> V **Loop 7:** V-> R2 -> R3 -> R4 -> V [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/MITx_6_002x_2012_Fall/threads/504e785a08a190230000000a [2]: https://edxuploads.s3.amazonaws.com/13481089182550778.png FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-20T02:54:38Z IndexTAG: 440 TitleTAG: I am stucked with LAB #2 LAB #2 says that I need to design a MIXER circuit that combines two or more signals (V1 and V2) and that has this Vout function: Vout ≈ (1/2)V1+(1/6)V2 Where: V1 is a 1 kHz square wave that varies between 0V and +1V, and V2 is a 5 kHz sine wave that varies between -1V and +1V. And: The resulting output should be similar to that shown in Figure 1. The maximum value of the output is approximately 667mV and the minimum value is approximately −167mV. If you apply superposition to the circuit, you are going to see that the value "1/2" from the Vout equation, is equal to R2/(R1+R2). At the same way, the value "1/6" is equal to R1/(R1+R2). When I had those two equations ( 1/2=R2/(R1+R2) and 1/6=R1/(R1+R2) ), I just built the 2x2 matrix of equations in order to get the values of R1 and R2, but the answer to that system is R1=0 and R2=0 WHICH IS NOT TRUE! I don´t know what to do. I have no idea what else should I try. Help me please. Thanks to all. UserIdTAG: 181432 UserNameTAG: enriqueferreralcala CreateTimeTAG: 2012-09-18T22:32:57Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 8 FirstChildTAG: I figured out, as you say, that the coefficients of V1 and V2 are always going to add up to 1 in the simple mixer circuit pictured. Something else is required. I assume, because the coefficient of V1 is 3 times the coefficient of V2, that R1 will be 3 times the size of V2 (Edit:that's wrong). Third resistor to ground to get the proper reduction? I really have no idea what that means, however. FirstChildUserIdTAG: 208488 FirstChildUserNameTAG: moelarrycheese FirstChildCreateTimeTAG: 2012-09-19T03:55:38Z SecondChildTAG: Since the simple mixer gives w1 + w2 = 1, but 1/2 + 1/6 is a number less than one adding a third resistor from the intersection of the first two to ground creates a voltage divider that produces the proper reduction. Hint: what is Rth equivalent to the first two resistors? Can you calculate the reduction due to a voltage divider composed of Rth and the third resistor? SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-09-19T09:19:31Z SecondChildTAG: Thanks for your efforts to help, I'm afraid this may be a lost cause for myself. Rth would just seem to be R1||R2(||R3) but I don't see how this helps, I'm lost on this question. SecondChildUserIdTAG: 208488 SecondChildUserNameTAG: moelarrycheese SecondChildCreateTimeTAG: 2012-09-19T21:17:10Z SecondChildTAG: Please note that I said Rth for the **first two resistors** only. Then use that value along with the third resistor to form the divider to get the proper reduction. The numbers come out so simple I can do (and have done) the arithmetic in my head. Let me start walking you through the process: Let R1 = R then what is R2? Next what are w1 and w2 for that choice? What do you want w1 and w2 to be? What is the reduction necessary? What is Rth corresponding to your R1 and R2? Now choose R3 to give the necessary reduction. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-09-19T21:51:52Z SecondChildTAG: The simple divider circuit ends up with: V=[R2/(R1+R2)] V1 + [R1/(R1+R2)] V2 or {w1V1 + w2V2} I want w1=1/2 and w2=1/6, which is not possible with this circuit no matter how R1 and R2 are adjusted because w1 and w2 always add to 1. I want them to add up to 2/3 so I need to make an adjustment. Throw another resistor in somewhere. You say to let R1=R, I don't get it, just throwing in one unknown for another? I don't know how far away my thinking is from what you're trying to tell me but this question makes we wonder if I skipped some important lectures in the sequence. To go from simple superposition problems 101 to this. SecondChildUserIdTAG: 208488 SecondChildUserNameTAG: moelarrycheese SecondChildCreateTimeTAG: 2012-09-20T00:31:41Z SecondChildTAG: Go back to my **first** comment here!!! **Choose R1 and R2 to get the proper ratio.** The desired result is w1 = 1/2 and w2 = 1/6, so w1 = 3*w2. So the first step is to get that ratio using R1 and R2 only. After this step w1 and w2 still sum to one, but they are in the proper ratio. Now what reduction factor applied to both w1 and w2 will give the desired result? That's where you will use Rth computed from R1 and R2. The math to do this problem probably takes less than 5 minutes, and you don't need a calculator. The fact is that knowing what configuration of resistors works is more than half the battle. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-09-20T01:05:54Z SecondChildTAG: Ok, so I have the proper R1/R2 ratio and the proper "reduction factor" that will act on both w1 and w2 to bring it down to the proper values. So I can adjust R1 and R2 to maintain the proper ratio and come up with different values for Rth, even make it the inverse of the required reduction factor. I'm still not connecting the dots from there to finish the problem however (but I'm still mulling it over). Many thanks for the help. SecondChildUserIdTAG: 208488 SecondChildUserNameTAG: moelarrycheese SecondChildCreateTimeTAG: 2012-09-20T02:03:55Z SecondChildTAG: Pick R1 to be any convenient value (It doesn't matter in this problem because there is no other constraint.) Then use your ratio to calculate R2. Use R1 and R2 to calculate Rth. Rth is now fixed. R3 (the resistor to ground) will be some multiple of Rth that gives you the reduction you need, i.e, Rth and R3 constitute a voltage divider. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-09-20T02:49:25Z SecondChildTAG: Wonderful, got it done, thanks again for all your help (and patience). SecondChildUserIdTAG: 208488 SecondChildUserNameTAG: moelarrycheese SecondChildCreateTimeTAG: 2012-09-20T03:15:41Z SecondChildTAG: I'v also got it down. But how do we rotate the resistors? SecondChildUserIdTAG: 213452 SecondChildUserNameTAG: JoJosida SecondChildCreateTimeTAG: 2012-09-20T05:03:32Z SecondChildTAG: Select the resistor and press r SecondChildUserIdTAG: 208488 SecondChildUserNameTAG: moelarrycheese SecondChildCreateTimeTAG: 2012-09-20T11:34:25Z SecondChildTAG: Need help, whats with the voltage divider circuit that you are talking about? I can't figure how you are suggesting the use of the third resistor SecondChildUserIdTAG: 315681 SecondChildUserNameTAG: nahuja1 SecondChildCreateTimeTAG: 2012-09-20T14:37:52Z SecondChildTAG: The use of the third resistor is explained in the discussion above. What work have you already done and what did you find? SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-09-20T21:56:30Z SecondChildTAG: Thanks skyhawk you saved the day :D SecondChildUserIdTAG: 336001 SecondChildUserNameTAG: syd_buet12 SecondChildCreateTimeTAG: 2012-09-21T11:11:09Z SecondChildTAG: help.. i dont get how you are connecting this third resistor.. what do u mean, "to ground"? SecondChildUserIdTAG: 457713 SecondChildUserNameTAG: ASHWIN_DT SecondChildCreateTimeTAG: 2012-09-22T05:43:13Z SecondChildTAG: yes SecondChildUserIdTAG: 403493 SecondChildUserNameTAG: aiai SecondChildCreateTimeTAG: 2012-09-22T11:14:55Z SecondChildTAG: Hey everybody--posts like these NEED TO BE UPVOTED! With so many empty posts for Lab 2, it's nice to let others know, "Hey, look here!" Thanks again to everybody that contributed here and elsewhere on the forums! SecondChildUserIdTAG: 267702 SecondChildUserNameTAG: curio_city SecondChildCreateTimeTAG: 2012-09-22T18:42:54Z SecondChildTAG: yippy .finally my lab2 correctly completed.thanx aal for sharing concept.but truly speaking although i got result but still i need to understand the core concept. frnds jst try tis (r2=2r1, and hint for r3=133%). SecondChildUserIdTAG: 269641 SecondChildUserNameTAG: BAUWA SecondChildCreateTimeTAG: 2012-09-22T19:52:13Z SecondChildTAG: sorry r2=3r1 SecondChildUserIdTAG: 269641 SecondChildUserNameTAG: BAUWA SecondChildCreateTimeTAG: 2012-09-22T19:55:37Z SecondChildTAG: haha, me too, i got it done but still need to get the core of it SecondChildUserIdTAG: 254607 SecondChildUserNameTAG: werehenry SecondChildCreateTimeTAG: 2012-09-23T01:38:19Z SecondChildTAG: Thanks a lot for the info skyhawk..i think i got the values correct and vmax and vmin is also in the range it is still showing ![analysis][1]the incorrect answer [1]: https://edxuploads.s3.amazonaws.com/13483702551343661.png SecondChildUserIdTAG: 357440 SecondChildUserNameTAG: agadaria SecondChildCreateTimeTAG: 2012-09-23T03:18:23Z SecondChildTAG: please help me...thanks SecondChildUserIdTAG: 357440 SecondChildUserNameTAG: agadaria SecondChildCreateTimeTAG: 2012-09-23T03:18:47Z SecondChildTAG: hey skyhawk...don't worry now .i refreshed the page and put the values like 4K,12k and 6K and it showing me green light now...thanks a lot once again. by the way is it mentioned somewhere in the lecture about the stuff u told in the beginning that w1+w2 should be equal to 1? i just want to find out if i m missing something here? help will be appreciated... SecondChildUserIdTAG: 357440 SecondChildUserNameTAG: agadaria SecondChildCreateTimeTAG: 2012-09-23T03:38:03Z SecondChildTAG: agadaria , do not need the other probe u had put , take out 1 off they . and the r3 value is wrong . SecondChildUserIdTAG: 402850 SecondChildUserNameTAG: y1111 SecondChildCreateTimeTAG: 2012-09-23T12:21:30Z SecondChildTAG: I can't say that I heard it in lecture. I just looked at the form of the coefficients: R2/(R1+R2) and R1/(R1+R2). Clearly the sum is one. I probably have a lot more mathematical experience than most here, so it was obvious to me. I personally think that realizing that the problem can't be solved with two resistors is part of the learning experience. Telling someone that the problem can definitely be solved by connecting three resistors in a certain way "spoils" the problem a little, but what else can you do in the way of a hint? SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-09-23T12:33:04Z SecondChildTAG: hmm..i had also thought of connecting the 3rd resistor but was confused as where to connect it.i was trying to figure it out and your comment helped a lot. Thanks a lot once again... SecondChildUserIdTAG: 357440 SecondChildUserNameTAG: agadaria SecondChildCreateTimeTAG: 2012-09-24T03:59:23Z SecondChildTAG: thanks to all for your help specially skyhawk for your help all i was doing wrong was in wrong connection of ground. SecondChildUserIdTAG: 403493 SecondChildUserNameTAG: aiai SecondChildCreateTimeTAG: 2012-09-24T09:36:01Z FirstChildTAG: I also solved the problem using superposition to solve the circuit, but the solution I arrived at had a third resistor. FirstChildUserIdTAG: 373309 FirstChildUserNameTAG: njirving FirstChildCreateTimeTAG: 2012-09-19T08:05:11Z FirstChildTAG: Guys i am still stark on the reduction factor.Any help. FirstChildUserIdTAG: 429851 FirstChildUserNameTAG: ssembajjwe FirstChildCreateTimeTAG: 2012-09-20T15:16:55Z FirstChildTAG: hi,can any one show the method to solve the question? if can attach the picture and description. FirstChildUserIdTAG: 189766 FirstChildUserNameTAG: shuhuan84 FirstChildCreateTimeTAG: 2012-09-20T16:13:30Z FirstChildTAG: If you don't understand the two step approach using the a reduction factor, then solve the problem directly.You have three resistors: R1 to V1 R2 to V2 and R3 to ground. Analyze any way you want. You will get an expression for Vout that consists of an expression involving R1, R2, and R3 multiplying V1 and another expression involving R1, R2, and R3 multiplying V2. Set the expression multiplying V1 to 1/2 and the expression multiplying V2 to 1/6. Choose one resistor to have any value you like and solve for the other two. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-09-20T23:14:36Z SecondChildTAG: thanks skyhawk, the three resistor network was much easier to grasp and use. Thanks alot for your efforts. SecondChildUserIdTAG: 183507 SecondChildUserNameTAG: obiradaniel SecondChildCreateTimeTAG: 2012-09-21T13:22:49Z FirstChildTAG: Base on the two equations , you do not need to get the values but a equations that R2=3R1 instead . Then try to find the equivalent voltage source and resistor . The third step is to find another resistor that will share the voltage your needed Vout=(1/2)V1+(1/6)V2 FirstChildUserIdTAG: 232667 FirstChildUserNameTAG: KyleLiux FirstChildCreateTimeTAG: 2012-09-21T01:16:21Z SecondChildTAG: thanks skyhawk, the three resistor network was much easier to grasp and use. Thanks alot for your efforts. SecondChildUserIdTAG: 183507 SecondChildUserNameTAG: obiradaniel SecondChildCreateTimeTAG: 2012-09-21T13:22:16Z FirstChildTAG: Problem is the lab is misleading saying you can't change wiring. You can change wiring and add a third resistor. I tried different values of resistors to get right output waveform but can't figure out how I can use superposition to solve the problem FirstChildUserIdTAG: 175706 FirstChildUserNameTAG: rwskim FirstChildCreateTimeTAG: 2012-09-22T07:36:11Z SecondChildTAG: HAH! I don't know why I just read the instructions and didn't even think to try! Thanks rwskim! SecondChildUserIdTAG: 267702 SecondChildUserNameTAG: curio_city SecondChildCreateTimeTAG: 2012-09-22T18:41:15Z FirstChildTAG: Thanks skyhawk. Great explanation. finally got it right. FirstChildUserIdTAG: 332360 FirstChildUserNameTAG: srinivasav FirstChildCreateTimeTAG: 2012-09-23T05:53:40Z IndexTAG: 441 TitleTAG: Mathematics and Engineering It is wonderful to see the exchange between mathematics and engineering. The mathematical tool considered as an abstract, purely practical application located here, and offers another point of view to understand the concept. UserIdTAG: 129288 UserNameTAG: Tinchito CreateTimeTAG: 2012-09-18T18:15:59Z VoteTAG: 6 CoursewareTAG: Week 4 / Incremental Method Math CommentableIdTAG: 6002x_inc_method_math NumberOfReplyTAG: 0 IndexTAG: 442 TitleTAG: Textbook Companion Weblinks In the [Preface of the textbook][1] it is mentioned that there is a companion weblink that you can view additional problems/materials at. However, that link no longer works. The working link is [here][2]. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/23 [2]: http://www.elsevierdirect.com/v2/companion.jsp?ISBN=9781558607354 UserIdTAG: 248807 UserNameTAG: 1kingsman CreateTimeTAG: 2012-09-18T00:16:33Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Thanks! That's a good thing to have FirstChildUserIdTAG: 248902 FirstChildUserNameTAG: Chanute FirstChildCreateTimeTAG: 2012-09-22T00:35:56Z IndexTAG: 443 TitleTAG: The answer must be -1.69V The answer must be -1.69V. The answer is 6.2V if the voltage source is not rotated. UserIdTAG: 346381 UserNameTAG: palma81 CreateTimeTAG: 2012-09-16T19:13:36Z VoteTAG: 6 CoursewareTAG: Week 2 / Node equation review CommentableIdTAG: 6002x_S3E1_NodeEquationReview NumberOfReplyTAG: 2 FirstChildTAG: Voltage source is rotated but value of voltage source is also negative. So, answer = 6.2V. FirstChildUserIdTAG: 398903 FirstChildUserNameTAG: PriteshAmrelia FirstChildCreateTimeTAG: 2012-09-17T04:25:24Z SecondChildTAG: Through KCL,we have the equation (e-V1)/R1 + (e-V2)/R2 = 0, where the potential V2 is -7.2V in the "+" to "-" of the source, but in the "-" to "+" of the source V2 is 7.2V. So, the result will be: (e-5)/6800 + (e-7.2)/5600 = 0. The solution for this equation is e = 6.21V SecondChildUserIdTAG: 338518 SecondChildUserNameTAG: PHLMenezes SecondChildCreateTimeTAG: 2012-09-18T18:40:45Z FirstChildTAG: yes FirstChildUserIdTAG: 163395 FirstChildUserNameTAG: jasonlll88 FirstChildCreateTimeTAG: 2012-09-16T20:20:30Z SecondChildTAG: Use the Circuit Sandbox in Overview for verify! SecondChildUserIdTAG: 289132 SecondChildUserNameTAG: Nestor_Escala SecondChildCreateTimeTAG: 2012-09-16T23:38:38Z IndexTAG: 444 TitleTAG: second part was a little bit of tricky question... according to principle of conservation of energy: Energy consumed= Energy produced. since power can be given as Power=(energy/time). as power and energy are directly proportional we can modify the law of conservation of energy as 'Power produced= Power dissipated' i.e; net power in the circuit is zero. Hence the answer is -2. Also the negative symbol indicates that the power is entering through negative terminal of an element. UserIdTAG: 346016 UserNameTAG: muthukrishna CreateTimeTAG: 2012-09-16T05:19:02Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 5_Simple_Power NumberOfReplyTAG: 1 FirstChildTAG: heres a simple way to look at this stuff.............. remember the lecturer said V-I relation relates to when the current enters through the POSITIVE terminal. The current leaves the battery's positive terminal and enters the battery from the NEGATIVE terminal, thats where the negative sign comes from FirstChildUserIdTAG: 452690 FirstChildUserNameTAG: shreerajshrestha FirstChildCreateTimeTAG: 2012-09-17T15:59:42Z IndexTAG: 445 TitleTAG: HW1P1 Hello. There is a problem with the last question: > Given that each individual resistor can dissipate up to 1 watt of > power before burning up, how much total power in watts ( W ) can the > smallest-valued composite resistor dissipate before burning up? I think the answer should be 3 W because resistors are composed in parallel, so if 1 buned up, others will work until all 3 resisters burn. But this seems to be wrong answer... Anyone can help ? UserIdTAG: 138620 UserNameTAG: ZatoIchi CreateTimeTAG: 2012-09-15T13:45:45Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I think the aim is to not let anyone burn... try to keep everyone as alive as possible.. so if you let them dissipate 3W... the 4 Ohm would be emitting smoke... :P FirstChildUserIdTAG: 257806 FirstChildUserNameTAG: skoda FirstChildCreateTimeTAG: 2012-09-15T13:56:22Z SecondChildTAG: can u explain this in more detail SecondChildUserIdTAG: 126911 SecondChildUserNameTAG: sidhant7 SecondChildCreateTimeTAG: 2012-09-15T15:07:27Z SecondChildTAG: okay as you say i do connect the two 4 ohm resistor in series and this combination in parallel to the 6 ohm then the value of power dissipated should be 1.5 .but its wrong !! SecondChildUserIdTAG: 102529 SecondChildUserNameTAG: laksh SecondChildCreateTimeTAG: 2012-09-15T15:38:01Z SecondChildTAG: so the maximum current that the combination can allow without burning is that the current tat can pass thru the 4Ohm. So tat will help you in fixing the current thru 4Ohm.. based upon that you can find the current thru 6Ohm and finally the total current hence the power.. :) SecondChildUserIdTAG: 257806 SecondChildUserNameTAG: skoda SecondChildCreateTimeTAG: 2012-09-15T15:44:29Z SecondChildTAG: okay i got it ....i did not understand the question properly first ...its clear to me now :) SecondChildUserIdTAG: 102529 SecondChildUserNameTAG: laksh SecondChildCreateTimeTAG: 2012-09-15T15:46:49Z SecondChildTAG: 2.98ansr after calc....t its wrong :( SecondChildUserIdTAG: 59307 SecondChildUserNameTAG: irfan_s SecondChildCreateTimeTAG: 2012-09-15T16:22:07Z SecondChildTAG: wats answer as 2.98 calculated is also wron SecondChildUserIdTAG: 320074 SecondChildUserNameTAG: salmanumer76 SecondChildCreateTimeTAG: 2012-09-15T16:23:31Z SecondChildTAG: ans is 8/3 SecondChildUserIdTAG: 276808 SecondChildUserNameTAG: DEBASMITAMAJUMDER SecondChildCreateTimeTAG: 2012-09-16T05:33:24Z IndexTAG: 446 TitleTAG: How to find VL? Can someone explain why VL = Vth*(Rth + RL)/Rth is incorrect? I got that from doing node analysis after simplifying the network. ---- I figured it out, I just messed up my algebra. It should of been VL = Vth*RL/(Rth+RL) UserIdTAG: 208854 UserNameTAG: BrynnleeEaton CreateTimeTAG: 2012-09-15T06:54:12Z VoteTAG: 6 CoursewareTAG: Week 2 / Thevenin model CommentableIdTAG: 6002x_S3E5_Thevenin_Model NumberOfReplyTAG: 1 FirstChildTAG: Yeah, if you set the node at the positive terminal of the load, the circuit looks like a voltage divider with RTH and RL. FirstChildUserIdTAG: 339668 FirstChildUserNameTAG: chickwebb FirstChildCreateTimeTAG: 2012-09-16T22:20:56Z SecondChildTAG: Other way is by finding the current: i=Vth/(Rth+RL) Then you got VL=i*RL SecondChildUserIdTAG: 184827 SecondChildUserNameTAG: DiegoT SecondChildCreateTimeTAG: 2012-09-18T20:49:05Z IndexTAG: 447 TitleTAG: power for resistance in parallel.....net power is the sum of individual power then why am i getting my answer wrong in the question given in homework....?? the question in which power dissipated has to be calculate for the smallest composite resistance.... UserIdTAG: 340411 UserNameTAG: BIDYUT759 CreateTimeTAG: 2012-09-13T16:01:06Z VoteTAG: 6 CoursewareTAG: Week 1 / Circuit topology CommentableIdTAG: 6002x_S2E1_Circuit_Topology NumberOfReplyTAG: 1 FirstChildTAG: Not each resistance consumes 1 Watt, the question stated that consumption is (up to) 1 watt, meaning that the maximum consumed power(at the lowest resistance) is 1 watt and at larger resistances the power consumed is lower. what you have to do is 1'st get the unknown voltage given that power at the lowest resistance is 1 watt and power = (V^2 / R) then get the value of (I) then the power at the largest resistance(6 ohm) = (V * I) .. and you already have the power at the lowest resistance (4 ohm) total power = sum of dissipated power in each resistor FirstChildUserIdTAG: 398394 FirstChildUserNameTAG: Sakr85 FirstChildCreateTimeTAG: 2012-09-13T16:36:04Z SecondChildTAG: Okay, no more hints guys. Anything more than this and it's basically just telling you the answer. SecondChildUserIdTAG: 151427 SecondChildUserNameTAG: RichardZ SecondChildCreateTimeTAG: 2012-09-13T23:40:18Z IndexTAG: 448 TitleTAG: Week 3 Lecture Bug S6V7 appears to be the same video as S6V6, "Graphical Method". Was there suppose to be a different video for S6V7? UserIdTAG: 397595 UserNameTAG: mholin CreateTimeTAG: 2012-09-13T14:14:38Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: I am having the same issue. Would be nice to get a comment on this. FirstChildUserIdTAG: 248807 FirstChildUserNameTAG: 1kingsman FirstChildCreateTimeTAG: 2012-09-27T21:08:16Z IndexTAG: 449 TitleTAG: Solution Link Please use this solution as reference and solve on your own- vote me if this helps :P http://dnovelz.webnode.com/album/photogallery/#img-2365-jpg UserIdTAG: 148523 UserNameTAG: span993 CreateTimeTAG: 2012-09-13T09:21:23Z VoteTAG: 6 CoursewareTAG: Week 2 / Linear combinations of source strenghts CommentableIdTAG: 6002x_S3E2_Linear_Combinations_of_Source_Strengths NumberOfReplyTAG: 0 IndexTAG: 450 TitleTAG: BUG in S3E2 while writing algebraic expressions for a1,a2,b1,b2,c1 & c2 i wrote the denominator of each fraction like (R3*R1+R2*R3+R1*R2). i think what i wrote is correct.but the system showed it is wrong,and when i rearranged the terms like (R1*R2+R2*R3+R3*R1) this then it was accepted.i don't get it...why the sequence of the terms matter? UserIdTAG: 163407 UserNameTAG: shuvajit CreateTimeTAG: 2012-09-12T18:45:28Z VoteTAG: 6 CoursewareTAG: Week 2 / Linear combinations of source strenghts CommentableIdTAG: 6002x_S3E2_Linear_Combinations_of_Source_Strengths NumberOfReplyTAG: 1 FirstChildTAG: The sequence of the terms should not matter. Can you let me know the two exact inputs that you think to be equivalent? FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-12T18:48:47Z SecondChildTAG: (R3*R1+R2*R3+R1*R2) & (R1*R2+R2*R3+R3*R1) are not they equivalent? SecondChildUserIdTAG: 163407 SecondChildUserNameTAG: shuvajit SecondChildCreateTimeTAG: 2012-09-14T18:06:48Z SecondChildTAG: following basic math principles I'd think they were. SecondChildUserIdTAG: 426266 SecondChildUserNameTAG: TheNetImp SecondChildCreateTimeTAG: 2012-09-15T10:40:49Z IndexTAG: 451 TitleTAG: Video Player It would be much better to watch videos in a full screen mode with sequence navigator hidden and only displayed when pointer on it. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-10T13:45:18Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: If you use chrome right click on the navigation bar select "Inspect element". Choose " from the left panel and in the right css panel edit "z-index: 9999;" to "z-index: 1;" FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-10T14:30:02Z SecondChildTAG: It worked!Thank you a lot! SecondChildUserIdTAG: undefined SecondChildUserNameTAG: undefined SecondChildCreateTimeTAG: 2012-09-11T12:56:56Z FirstChildTAG: you are getting to learn all of this for free...how can you expect more?!! FirstChildUserIdTAG: 333107 FirstChildUserNameTAG: Mrugen FirstChildCreateTimeTAG: 2012-09-10T19:55:31Z SecondChildTAG: I am not complaining on the contrary I am grateful to edX for providing educational materials for free.I posted this just to make a suggestion for some small improvement in v-player. SecondChildUserIdTAG: undefined SecondChildUserNameTAG: undefined SecondChildCreateTimeTAG: 2012-09-11T13:01:57Z IndexTAG: 452 TitleTAG: Covered in the text If anyone is still stuck on this example, it is analysed in Sec. 2.5 (p.95) in the textbook. UserIdTAG: 142402 UserNameTAG: aaronrod CreateTimeTAG: 2012-09-10T05:06:56Z VoteTAG: 6 CoursewareTAG: Week 1 / KVL/KCL example CommentableIdTAG: 6002x_S2E3_Using_KVL_KCL_and_VI_constraints NumberOfReplyTAG: 1 FirstChildTAG: thanks :) FirstChildUserIdTAG: 155645 FirstChildUserNameTAG: aspirant_engineer FirstChildCreateTimeTAG: 2012-09-12T14:40:29Z IndexTAG: 453 TitleTAG: H1P1 How to solve H1P1 problem 7..regarding power dissipated by smallest composite resistor? UserIdTAG: 106395 UserNameTAG: Arunmozhi CreateTimeTAG: 2012-09-09T09:21:02Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 3 FirstChildTAG: Firs work out how to create the smallest composite resistor from the resistors available. Then you must find out at what voltage the smallest of the individual resistors will be dissipating 1W, that will be the maximum voltage applied. Then you should calculate the power dissipated in each resistor at that voltage. The sum of the 3 powers will be the correct answer. FirstChildUserIdTAG: 270011 FirstChildUserNameTAG: SnowmanZA FirstChildCreateTimeTAG: 2012-09-09T10:45:35Z SecondChildTAG: i tried to solve by that method but it still says wrong...what to do? SecondChildUserIdTAG: 333107 SecondChildUserNameTAG: Mrugen SecondChildCreateTimeTAG: 2012-09-09T14:45:03Z SecondChildTAG: i did that..and the answer concurs.Thank you. But can you tell me the idea behind the whole thing please? SecondChildUserIdTAG: 162671 SecondChildUserNameTAG: tuhin1991paul SecondChildCreateTimeTAG: 2012-09-10T16:20:35Z FirstChildTAG: still i could,nt solve this prolem :( what should i substitute for r? FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-11T15:13:50Z SecondChildTAG: the smallest resistance out of 4, 4 and 6..which is 4.. SecondChildUserIdTAG: 173147 SecondChildUserNameTAG: cruiser_rahit SecondChildCreateTimeTAG: 2012-09-11T16:17:47Z SecondChildTAG: Thankyou alot....was messing wid my mind & had really got exahausted...!!:P SecondChildUserIdTAG: 127800 SecondChildUserNameTAG: MNB SecondChildCreateTimeTAG: 2012-09-16T03:55:14Z FirstChildTAG: What is the value in Ohms ( Ω ) of the smallest-valued resistor that can be fabricated by combining these three resistors? FirstChildUserIdTAG: 376345 FirstChildUserNameTAG: Hassan-Ahmed FirstChildCreateTimeTAG: 2012-09-16T08:26:46Z IndexTAG: 454 TitleTAG: Don't understand the last two parts of S2E5 With e2 set to 0 I started from scratch: Writing the node equation around e3 I get: (e3-e2)/R2 + I + ? = 0, I don't know what to do with the voltage source, which I'm assuming whatever it is will be in place of the ? in the equation. With the remaining node, e1, I start with the equation: (e1-e0)/R1 + ? = 0. Here again, how do I deal with the voltage source? Do I add 5, which is the voltage from the voltage source? The only experience I have with dealing with a voltage source is When e3 is the reference node and the node equation at e2 is: (e2-e1)/R1 + e2/R2 - I = 0. With e1 as the voltage from the voltage source, it is easy. Can anyone explain how to write the equations for the whole circuit when e2 is the reference (ground) node? - Thanks. UserIdTAG: 132828 UserNameTAG: rharris CreateTimeTAG: 2012-09-09T02:15:59Z VoteTAG: 6 CoursewareTAG: Week 1 / Node method example CommentableIdTAG: 6002x_S2E5_Node_Method NumberOfReplyTAG: 1 FirstChildTAG: write equation for node 3.with voltage source:it is given e1=e3+V.therefore,current thru that part is(e3+V)/R1 FirstChildUserIdTAG: 55345 FirstChildUserNameTAG: sarmaji FirstChildCreateTimeTAG: 2012-09-09T05:09:15Z IndexTAG: 455 TitleTAG: Invalid input: v3 not permitted in answer Invalid input: v3 v1 not permitted in answer anybody know what am i doing wrong?how are we supposed to enter the variables? UserIdTAG: 112814 UserNameTAG: Joanne92 CreateTimeTAG: 2012-09-08T16:02:42Z VoteTAG: 6 CoursewareTAG: Week 1 / KVL and KCL exercises CommentableIdTAG: 6002x_kvl_kcl_exercises NumberOfReplyTAG: 1 FirstChildTAG: Write only one side of the equation... don't include v3 and the '=' sign. FirstChildUserIdTAG: 365201 FirstChildUserNameTAG: sirajmuhammad FirstChildCreateTimeTAG: 2012-09-08T16:23:51Z SecondChildTAG: dont write **v3 =** write only the write side of the equation and you must use the known voltage only SecondChildUserIdTAG: 236136 SecondChildUserNameTAG: Hamid-ch SecondChildCreateTimeTAG: 2012-09-09T21:28:11Z IndexTAG: 456 TitleTAG: Why change to AC voltage? What is the reason to change to an AC voltage in order to measure currents? (this happens slightly after the middle of the lecture). UserIdTAG: 189680 UserNameTAG: vnd CreateTimeTAG: 2012-09-08T09:03:51Z VoteTAG: 6 CoursewareTAG: Week 1 / KVL and KCL CommentableIdTAG: 6002x_kvl_and_kcl NumberOfReplyTAG: 5 FirstChildTAG: Ac has lower current FirstChildUserIdTAG: 336466 FirstChildUserNameTAG: kenstan FirstChildCreateTimeTAG: 2012-09-09T14:44:44Z FirstChildTAG: I think it's because the probe he is using can only measure AC currents. FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-09T20:27:17Z FirstChildTAG: Usually you have to put a current probe in series to make a current measurement. This involves taking a branch, breaking it, and inserting the current probe. I hadn't seen this trick before, and want to know how the oscilloscope is triggering to differentiate the two? Sorry if that's a stupid question. FirstChildUserIdTAG: 13614 FirstChildUserNameTAG: analognoise FirstChildCreateTimeTAG: 2012-09-10T04:50:55Z FirstChildTAG: My question too, why does he change it into AC? I have worked with DC currents my entire school life, and have no idea what an AC current does to the circuit. FirstChildUserIdTAG: 193033 FirstChildUserNameTAG: NIslam FirstChildCreateTimeTAG: 2012-09-10T16:04:28Z FirstChildTAG: The current probe he's using is basically a transformer, it only works with AC currents, he's using that probe so he doesn't need to break the connections and insert, let's say, a multimeter in series to measure the current, in order to use a similar probe with DC he needs one that uses a Hall effect sensor. FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-10T21:33:26Z IndexTAG: 457 TitleTAG: Suggestion: go back 30 seconds button for lecture viewer. Whether it's because I space off for a few seconds or just want to re-listen to something I didn't quite get, I find myself clicking back a few seconds quite often when watching lectures. The lecture viewer is really nice, and I think it could be made even nicer with a little "go back 30 seconds" button. UserIdTAG: 270160 UserNameTAG: Sethhhh CreateTimeTAG: 2012-09-08T07:33:02Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: I agree, I have the same manner. But I'd rather go for 15-20 seconds (you could always hit it twice). I also like the the video speed factor button and navigation by transcript. FirstChildUserIdTAG: 364822 FirstChildUserNameTAG: Strus FirstChildCreateTimeTAG: 2012-09-08T08:20:47Z FirstChildTAG: You could find exactly where you drifted off in the scrolling text and click on that. FirstChildUserIdTAG: 267993 FirstChildUserNameTAG: mavlijas FirstChildCreateTimeTAG: 2012-09-08T08:19:19Z IndexTAG: 458 TitleTAG: ¿numbers at left side? What do the numbers that are to the left of the comments? UserIdTAG: 268104 UserNameTAG: SaulCardenasG CreateTimeTAG: 2012-09-08T05:30:47Z VoteTAG: 6 CoursewareTAG: Week 1 / Lumped Element Abstraction CommentableIdTAG: 6002x_lumped_element_abstraction NumberOfReplyTAG: 1 FirstChildTAG: It shows the rating of the comment. If a user finds a comment or a response useful, can hit the up (^) button, and if the comment is not appropriate, down arrow... All these sums up and eventually, most useful comments/responses end up at the top of the list for other users convenience. FirstChildUserIdTAG: 43909 FirstChildUserNameTAG: hhomayouni FirstChildCreateTimeTAG: 2012-09-08T06:00:16Z IndexTAG: 459 TitleTAG: Thanks edX Hats off to edX.thanks to this venture,for bringing top class education to the whole world..let the learning revolution begin..today(8)is world literacy day and let's hope and try to make everyone literate..let education reign.. UserIdTAG: 55345 UserNameTAG: sarmaji CreateTimeTAG: 2012-09-08T01:41:50Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 460 TitleTAG: First two weeks complete thanks MIT for correcting your bugs so quickly After my struggles trying to get my lab session to work yesterday I am very happy to announce that within 24 hours MIT have corrected their mistakes allowing me to complete my first two weeks homeworks and labs quickly and thankfully...all correct. Nice one MITx. UserIdTAG: 80920 UserNameTAG: MichaelSturgess CreateTimeTAG: 2012-09-06T18:40:52Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: It was very nice to see tick marks at all places :) very good system to evaluate students :) nice work MIT :) now, i see, why MIT is world class institute :) FirstChildUserIdTAG: 823 FirstChildUserNameTAG: naveenatmit FirstChildCreateTimeTAG: 2012-09-06T18:52:23Z FirstChildTAG: Exact same for me to. Glad they finally got it sorted. FirstChildUserIdTAG: 15344 FirstChildUserNameTAG: kob FirstChildCreateTimeTAG: 2012-09-07T01:44:30Z IndexTAG: 461 TitleTAG: S1E9 #4 To answer 4, subtract the lower voltage from the higher voltage, and divide this by the sum of the internal resistances. UserIdTAG: 139780 UserNameTAG: Noloqy CreateTimeTAG: 2012-09-06T11:57:33Z VoteTAG: 6 CoursewareTAG: Week 1 / Battery model CommentableIdTAG: 6002x_S1E9_Battery_Model NumberOfReplyTAG: 2 FirstChildTAG: answer nahi aa raha hai FirstChildUserIdTAG: 375307 FirstChildUserNameTAG: imajeet FirstChildCreateTimeTAG: 2012-09-06T13:03:29Z SecondChildTAG: janab kis cheez ka answer nahin aa raha. SecondChildUserIdTAG: 908 SecondChildUserNameTAG: m_umair72001 SecondChildCreateTimeTAG: 2012-09-06T13:22:59Z SecondChildTAG: you have to sum up the resistances. (1.6-1.5)/(.25+.32) .1/(.57) SecondChildUserIdTAG: 338685 SecondChildUserNameTAG: varshaD SecondChildCreateTimeTAG: 2012-09-06T13:59:10Z FirstChildTAG: Thank you for the formula. Can you explain why is that? FirstChildUserIdTAG: 369969 FirstChildUserNameTAG: cakici FirstChildCreateTimeTAG: 2012-09-07T12:20:37Z SecondChildTAG: Ah ok, its the Ohms Law. I = V / R || I = 0.1 / (0.25+0.32) || I = .1754385964 SecondChildUserIdTAG: 369969 SecondChildUserNameTAG: cakici SecondChildCreateTimeTAG: 2012-09-07T12:23:51Z IndexTAG: 462 TitleTAG: Tool Wonderful interactive tool UserIdTAG: 235542 UserNameTAG: teto1988 CreateTimeTAG: 2012-09-06T00:14:14Z VoteTAG: 6 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 0 IndexTAG: 463 TitleTAG: Lab 1 circuit tool FIXED Looks like its now working. A reset button has now appeared and that is able to show the correct compoents on the page. Hooray! So all who are still having the problem, you'll now see a new reset button at bottim of page once you click check you'll see it. Just rest will then show you the compoents that are supposed to be on screen (So my post below is now obsolete) I am using a mac OSX with firefox browser. In lab 1 the voltage component is not visible in fact the only thing that is, is a resistor component. Can't do lab until its fixed. Tried chrome and same problem. But, all working in sandbox (Accept for tran code). So don't think its a browser issue as such. Else that would not show the components iether. UserIdTAG: 15344 UserNameTAG: kob CreateTimeTAG: 2012-09-05T19:51:16Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 4 FirstChildTAG: Same problem, lab1 tool does not work properly on OSX in Firefox and Safari, as well. FirstChildUserIdTAG: 137306 FirstChildUserNameTAG: Metallic FirstChildCreateTimeTAG: 2012-09-05T21:31:01Z SecondChildTAG: Ha ha! yes I checked those to. I wish they would post an update as this is a real show stopper. ;-( SecondChildUserIdTAG: 15344 SecondChildUserNameTAG: kob SecondChildCreateTimeTAG: 2012-09-05T21:38:48Z FirstChildTAG: You should only need to insert resistors to complete Lab 1 as long as you have the original 6V voltage source, ground, the bulb resistor for the first circuit, and the node A. The staff knows that if you delete something it can break the lab because you can only insert resistors. So DON'T DELETE ANYTHING. Or is nothing showing up? I found that by clearing my browser's cache/cookies (OSX+Firefox) and coming back to the page that I could see the original circuit. FirstChildUserIdTAG: 376895 FirstChildUserNameTAG: alecgg FirstChildCreateTimeTAG: 2012-09-05T22:37:55Z SecondChildTAG: I'll try that out. But no way should you have to jump through hoops like that on a graded part of the course. Daft. Don't know why they have changed it as it was working fine under their old MITx system. I'll try clearing cache but I don';t think it will make a difference. What is the system you are using the OS and machine? Looks like its now working. A reset button has now appeared and that is able to show the correct compoents on the page. Hooray! SecondChildUserIdTAG: 15344 SecondChildUserNameTAG: kob SecondChildCreateTimeTAG: 2012-09-05T23:23:21Z SecondChildTAG: No, did not make any difference. I amptied cach history the lot and restarted. I think it is partly due to combination of Mac OSX and the browser. But also it must be part of system code as the sandbox is working for me. It don't make much sense. Might have to give up and do this course without grades fun. Since if it is not fixed then a lot of us can't get graded properly. Very annoying really. Still its free so can't complain. SecondChildUserIdTAG: 15344 SecondChildUserNameTAG: kob SecondChildCreateTimeTAG: 2012-09-05T23:34:21Z FirstChildTAG: Do you have simillar problem in lab2? FirstChildUserIdTAG: 146390 FirstChildUserNameTAG: JGradzki FirstChildCreateTimeTAG: 2012-09-06T20:45:12Z FirstChildTAG: I am having a similar issue - I can see and place the components but the AC/DC/TRANS buttons do not appear. Also, even though I clicked 'check' to save my work, the circuit disappears when I move off the page. I am on a Mac - MacBookPro, OS 10.6.8, Firefox 15.0 FirstChildUserIdTAG: 248492 FirstChildUserNameTAG: Anne6238 FirstChildCreateTimeTAG: 2012-09-06T02:41:46Z SecondChildTAG: Interestingly, the sandbox appears to work well. SecondChildUserIdTAG: 248492 SecondChildUserNameTAG: Anne6238 SecondChildCreateTimeTAG: 2012-09-06T02:46:12Z SecondChildTAG: In my case I didn't had any analysis buttons (AC/DC/TRAN) and only resistors. So I build the resistors divider clicked check and then I got notification that the circuit is incorrect (X) but DC analysis button appeared :) So there is some progress :) SecondChildUserIdTAG: 146390 SecondChildUserNameTAG: JGradzki SecondChildCreateTimeTAG: 2012-09-06T10:18:36Z IndexTAG: 464 TitleTAG: Hello, from another MITx Alum Hey, folks... I did the inaugural 6.002x and just wanted to say I hope you guys have as much fun and learning and camaraderie as we did. It's a great course and a great experience. Best of luck, work hard and enjoy the ride! UserIdTAG: 92346 UserNameTAG: MobiusTruth CreateTimeTAG: 2012-09-05T19:39:49Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: Hi MobiusTruth I remember your posts from MITx. Nice to see you back for another go at it. FirstChildUserIdTAG: 15344 FirstChildUserNameTAG: kob FirstChildCreateTimeTAG: 2012-09-05T20:56:17Z FirstChildTAG: Welcome back! I'm glad to see so many MITx alumni here. We have to make sure we make the newcomers feel welcomed and not that we are a clique. For the new folks, this is will be an amazing 14 weeks of learning and brain-stretching. Also an amazing 14 weeks of uttering nasty words at the computer screen until the beloved green checkmarks appear. Warn anyone within earshot that you will be swearing at your computer, and not to take it personally. FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-09-06T01:10:21Z IndexTAG: 465 TitleTAG: Video - download option Is there an option to download the lecture videos other than streaming through you tube? UserIdTAG: 206294 UserNameTAG: Krishnaraj1987 CreateTimeTAG: 2012-09-05T12:41:17Z VoteTAG: 6 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: You can use internet download manager , i use it now , it's very well FirstChildUserIdTAG: 305879 FirstChildUserNameTAG: muhamedadel FirstChildCreateTimeTAG: 2012-09-05T12:45:33Z SecondChildTAG: I cannot download from youtube, as we have data restriction policies in our firm. however i can download free videos (anywhere other than youtube, metacafe etc) SecondChildUserIdTAG: 206294 SecondChildUserNameTAG: Krishnaraj1987 SecondChildCreateTimeTAG: 2012-09-05T12:58:53Z SecondChildTAG: what is the exact name of the "internet download manager"? Is this a windows or linux tool? (I have XP nad linux) SecondChildUserIdTAG: 10512 SecondChildUserNameTAG: asicok SecondChildCreateTimeTAG: 2012-09-06T16:38:36Z FirstChildTAG: Try using "YTD video downloader". You can google the name above, download it, run and use. How to use: 1. After opening the application, go to YouTube page where the video is located. 2. Copy the address on the address bar. 3. Go to the YTD video downloader application dialogue box and paste the address in "enter the URL of the video page" located under Download tab... and you are good to go. 4. Follow the instructions and ensure you save it as MPEG,MP3,MP4 or V7.WMV format. I am sure this will help. FirstChildUserIdTAG: 329879 FirstChildUserNameTAG: Ugochukwu FirstChildCreateTimeTAG: 2012-09-08T15:19:26Z IndexTAG: 466 TitleTAG: DC and other analysis How do I run this kind of analysis? During the tutorial video there were the buttons, but here there aren't! UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-05T12:37:25Z VoteTAG: 6 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 7 FirstChildTAG: Are you using Firefox? They didn't appear for me either. I switched to Chromium and it's working. FirstChildUserIdTAG: 312035 FirstChildUserNameTAG: philhiggins FirstChildCreateTimeTAG: 2012-09-05T12:43:59Z FirstChildTAG: What operating system and browser are you using? FirstChildUserIdTAG: 6977 FirstChildUserNameTAG: rocha FirstChildCreateTimeTAG: 2012-09-05T12:44:19Z FirstChildTAG: Hi, click the check button before this analysis can be completed. It checks the circuit is complete enough for this kind of analysis. FirstChildUserIdTAG: 80920 FirstChildUserNameTAG: MichaelSturgess FirstChildCreateTimeTAG: 2012-09-05T12:44:35Z FirstChildTAG: deleted FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-09-05T12:53:36Z FirstChildTAG: I had the same problem. But if you press check at the end of the page, then the analysis tools appear and than the DC analysis works, AC is still not working. I use Firefow 14.0.1 and windows xp sp3 FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-09-05T12:57:54Z FirstChildTAG: The only button that does not appear is the AC and it's because you do not need for this lab, I suggest you use mozilla firefox FirstChildUserIdTAG: 51696 FirstChildUserNameTAG: blasco23 FirstChildCreateTimeTAG: 2012-09-05T14:40:48Z FirstChildTAG: for people with compatibility problem use the circuit sandbox on the left top corner all the component are visible and usable just use a separate tab and do the analysis and fill in the answers.. FirstChildUserIdTAG: 375082 FirstChildUserNameTAG: satya1889 FirstChildCreateTimeTAG: 2012-09-05T13:49:15Z IndexTAG: 467 TitleTAG: CECC2 Prize received - Thanks to all of you Yesterday I received the CECC 2 prize. Thanks to the organizers and all the edX staff. [Here][1] is my Video Blog entry opening the mail box with the prize inside. Thanks again :') [1]: http://youtu.be/bfPYeZldljw UserIdTAG: 87808 UserNameTAG: pumoneon CreateTimeTAG: 2013-03-28T20:31:33Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: I am really happy for you Enrique! Nice video! Wow! I liked the edX T-Shirt! xD FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2013-03-29T22:17:15Z SecondChildTAG: Thanks Myrimit! You're some kind of a hero around this course xD Yeah, the T-Shirt is great. I will do my next VLog wearing it. I love it haha. SecondChildUserIdTAG: 87808 SecondChildUserNameTAG: pumoneon SecondChildCreateTimeTAG: 2013-03-30T11:29:50Z FirstChildTAG: Wow! Congrats! :D FirstChildUserIdTAG: 297370 FirstChildUserNameTAG: ayush3504 FirstChildCreateTimeTAG: 2013-03-30T21:39:21Z FirstChildTAG: Enrique, thank you for sharing the joy. It seems to me that you have a natural talent for video, you should keep at it no matter what you study or want to present to the rest of us. Just my two cents, whatever. Thanks and good luck, Sid FirstChildUserIdTAG: 90796 FirstChildUserNameTAG: ChaunceyGardiner FirstChildCreateTimeTAG: 2013-03-30T11:00:33Z SecondChildTAG: Thanks,you cheered me up a lot :') SecondChildUserIdTAG: 87808 SecondChildUserNameTAG: pumoneon SecondChildCreateTimeTAG: 2013-03-30T11:27:04Z IndexTAG: 468 TitleTAG: Some statistics on the course on January 31, 2013 http://web.mit.edu/newsoffice/2013/mitx-spring-offerings-0131.html UserIdTAG: 402617 UserNameTAG: Vl CreateTimeTAG: 2013-02-09T12:31:08Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Thank you for sharing VI :) FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2013-02-09T22:57:58Z FirstChildTAG: Thank you for sharing VI FirstChildUserIdTAG: 227223 FirstChildUserNameTAG: luisbonelli FirstChildCreateTimeTAG: 2013-02-10T06:12:56Z IndexTAG: 469 TitleTAG: humble request to staff as many of us will not able to give proctored exam because of several reasons..is it possible for you to give us question paper of proctored exam after 13th Feb.?? UserIdTAG: 114864 UserNameTAG: Vikaash CreateTimeTAG: 2013-02-07T15:24:20Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 470 TitleTAG: Just for curious, How many of you will take the Proctored Exam on February 13th? Just for curious, How many of you will take the Proctored Exam of 6.002x on February 13th 2013? If yes, where is your location and how are you preparing for the Exam? :) If no, why? Myriam. UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2013-02-06T17:58:18Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 8 FirstChildTAG: I wish I was saying yes. But there's no center nearby which is sad considering the fact that Bangalore is called India's IT capital :( FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2013-02-06T18:06:40Z SecondChildTAG: I wish to have a Pearson VUE Centre in Buenos Aires (Argentina) too ;). Anyway, if Pearson VUE don´t come to me I will go to Pearson VUE haha .I will go to Brazil´s Pearson VUE Centre, because is the most near one, and set it for the Pearson Exam next year-2014(I hope that it would be available in the future). So, I am planning to focus on helping all this year meticulously to the new students(6.002x spring 2013 and 6.002x fall 2013), working the difficulties with them together, reviewing all the topics and also preparing myself for the proctored Exam haha, also I will test myself allowing me to use only 2 hours to complete the Midterm and Final Exam without notes and without distracting with the cat, home, cat haha. Also I will focus in my English, I want to improve it haha. So, I will take an advance of the Pearson VUE Centre not available in my Country :p SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2013-02-06T20:06:13Z SecondChildTAG: Hey Miriam, why don't you start a Pearson VUE Centre in Argentina yourself someday? You could be your first client yourself and maybe earn a few pesos with that business too! SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2013-02-06T20:27:05Z FirstChildTAG: Hi Myriam, I won't take the exam for several reasons. First is that I don't need it. I'm not studying to start a career. I showed an old friend of mine, my 4 versions of the 6.002x certificates, and he congratulated me with my "Hamburger" certificates! Ha Ha! Therefore we are friends! Second is that I'm pretty sure that I'm not able to get the grades I've now, in only 2 hours without notes, and I don't need to prove that I'm right. Also I don't want to be a kind of "guinea pig" for this first time. I'll at least wait until I hear from other people how they experienced the proctored exam and maybe some day, I'll take the exam just for fun. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2013-02-06T19:08:15Z SecondChildTAG: Hi salsero! :). I understand your point. But it will be really challenging to set for the Proctored ;). In fact, I challenge you right now to take it next year, what do you think? Do you accept the challenge? By the way, Have you invited your friend to join edX's Courses? haha xD. Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2013-02-06T20:21:03Z SecondChildTAG: Now I'm sober, I don't think it's a good idea to accept your challenge. I'm over the hill, and my brains are as flexible as a brick. But I'll think it over when I'm drunk! About my friend: I told him last year about mitx/edx and than he looked at 6.002x but was afraid of the math, so he tried the course Artificial Intelligence on Coursera. He stopped after a few weeks, because of the math ... So, as his good friend, I pointed him out why HE needed A.I.! Ha ha! SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2013-02-06T20:45:09Z FirstChildTAG: Perhaps someone should organize an event where people from alover the world would gather to take proctored exams, but also to visit some lectures by American people from the industry even combined with a nice holiday visiting MIT, Boston and New York. And ofcourse we would all contribute for a plane ticket to get to meet Miriam there also. This is not so irrealistic as it sounds. Now a lot of companys take chartersflights to Germany or China to visit electronics or industrial exhibits. If professionally organized perhaps even our boss would see some benefit to sponsor our tickets. See you at the next CECWG (Circuits and Electronics Classmates World Gathering). FirstChildUserIdTAG: 79885 FirstChildUserNameTAG: ruudoleo FirstChildCreateTimeTAG: 2013-02-06T19:51:03Z SecondChildTAG: Nice idea!:) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2013-02-06T20:39:07Z FirstChildTAG: i am not able to give proctored exam as my college cul-fest is starting from 11 feb to 15 feb and i am one of the participant. moreover there is no "edx's proctored exam" center in my city although there is pearson's center in my city... it is quiet amazing!! FirstChildUserIdTAG: 114864 FirstChildUserNameTAG: Vikaash FirstChildCreateTimeTAG: 2013-02-07T15:14:11Z FirstChildTAG: I wish I could take it. I had planned on taking it and was so excited to hear the news but... it's the middle of Chinese New Year and I will be on vacation (in the Maldives no less!) on the 13th. So unless I could take it while sitting on a beach in the middle of the Indian Ocean, it's going to have to be a no for me. If it is offered again, I most certainly will take it and I'm super bummed that I won't be able to take it this time but ... C'est la vie! FirstChildUserIdTAG: 467169 FirstChildUserNameTAG: Eyowzitgoin FirstChildCreateTimeTAG: 2013-02-07T15:30:42Z FirstChildTAG: hi myriam no because the date is prohibitive for me :(( and the distance from my location to Athens is quite big FirstChildUserIdTAG: 319453 FirstChildUserNameTAG: leonidasGr FirstChildCreateTimeTAG: 2013-02-09T08:35:53Z FirstChildTAG: I've registered for the proctored exam. I live in New Delhi, India and there's a Pearson VUE test center in my city :) For preparations I'm going through the lecture sequences and practicing the sample questions which were provided before mid term and final exams. FirstChildUserIdTAG: 297370 FirstChildUserNameTAG: ayush3504 FirstChildCreateTimeTAG: 2013-02-09T14:00:57Z SecondChildTAG: Cool! :). I wish you all the best ayush3504 :)! Can you , after the deadline of the Proctored Exam, tell us how it was your experience in the Centre while doing the Exam? xD. Thank you for sharing this. Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2013-02-09T22:54:20Z SecondChildTAG: Thanks. Sure I will! :) SecondChildUserIdTAG: 297370 SecondChildUserNameTAG: ayush3504 SecondChildCreateTimeTAG: 2013-02-10T00:18:29Z SecondChildTAG: Gave the proctored exam yesterday. The center was easy to find. The behavior of the staff was very professional and courteous. The center was clean and tidy. Lockers were provided to keep personal belongings. Sign-in procedures included taking photograph, verifying signatures and taking palm vein-scans. It took time so reaching there 30 minutes early was a good idea. The environment was very secure and it is virtually impossible to cheat. Every thing was good except there were some minor technical issues with the test itself. The workstation they provided had the edX site opened in a very different environment. The on-screen calculator that is usually provided was not there. Luckily I had brought my own calculator. (BTW, they inspected my calculator to see whether it has not stored any formulas). The video lectures and textbook were made available but it did NOT work. The videos were not running, neither I could download slides. The textbook was opening, but I could not turn pages so I was stuck at a single page and it was not usable at all. The system was very slow to navigate through which wasted my time. The pages were not appearing correctly formatted. I guess this is because of an unsuppoted browser they were using. The pattern of the exam was identical to the online final exam that we gave. It had 6 questions and I think 150 minutes were pretty sufficient. I managed to answer 19 boxes correctly out of 24 boxes. Overall it was a nice experience and I enjoyed it. I hope these issues get addressed and are prevented in future exams. The center was: Pearson Professional Center 18,Ramanath House, Community Centre, Yusuf Sarai-Green Park, Delhi - 110016 SecondChildUserIdTAG: 297370 SecondChildUserNameTAG: ayush3504 SecondChildCreateTimeTAG: 2013-02-14T15:01:03Z SecondChildTAG: I forgot to mention that they also provided a nice writing pad with their pens. We could also ask for more if needed :) SecondChildUserIdTAG: 297370 SecondChildUserNameTAG: ayush3504 SecondChildCreateTimeTAG: 2013-02-14T15:06:54Z SecondChildTAG: Hi ayush, congrats. Btw: the things you mentioned, are, amongst others, where I was afraid of. Bad things can happen. I'm a bit surprised to see that they gave more than 2 hrs. And only 24 boxes this time. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2013-02-14T15:30:40Z FirstChildTAG: In my country or my region, there is no center to take the exam, at least in this area. My next exam reception centers in China, Japan and South Korea. In China, all the holidays up to 16 Febrary. But I may be mistaken. Maybe it's for the best. I still can not pass it on February 13, since going on a working trip from Wednesday to Sunday. Also, I'm not sure that without the week preparing for the exam, I can take the exam in two hours. FirstChildUserIdTAG: 402617 FirstChildUserNameTAG: Vl FirstChildCreateTimeTAG: 2013-02-10T13:09:54Z IndexTAG: 471 TitleTAG: This is my video for the contest Hi all, this is the video that I maked for the contest, enjoy it :D http://www.youtube.com/watch?v=8BB7_gcFKTo Anthony UserIdTAG: 298547 UserNameTAG: AnthonyRF CreateTimeTAG: 2013-01-16T00:43:52Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: Nice Video AnthonyRF! :) FirstChildUserIdTAG: 288174 FirstChildUserNameTAG: matiasgrodriguez FirstChildCreateTimeTAG: 2013-01-16T21:07:42Z FirstChildTAG: Sweet FirstChildUserIdTAG: 9673 FirstChildUserNameTAG: xp42 FirstChildCreateTimeTAG: 2013-01-16T22:19:25Z SecondChildTAG: Thanks to all! Best regards! _____________________________________________ Gracias a todos: Oscar, saludos desde Cumaná. Cómo hago para marcar el mapa? SecondChildUserIdTAG: 298547 SecondChildUserNameTAG: AnthonyRF SecondChildCreateTimeTAG: 2013-01-16T22:41:42Z FirstChildTAG: Thanks to all! Best regards! Gracias a todos: Oscar, saludos desde Cumaná. Cómo hago para marcar el mapa? FirstChildUserIdTAG: 298547 FirstChildUserNameTAG: AnthonyRF FirstChildCreateTimeTAG: 2013-01-16T22:41:49Z SecondChildTAG: Hola Anthony, te anexo los pasos: - Si no tienes, crea una cuenta en gmail - Abre el mapa, amplia tu ubicación, inicia sesión - Del lado izquierdo selecciona cualquier usuario (no modifiques sus datos) de la lista, arriba dale editar, dentro del mapa te aparecerán tres iconos, selecciona el globo azul que dice: agregar un marcador de posición - Arrastra el marcador a tu posición en Cumana y listo Te anexo otras instrucciones de un post anterior por si acaso: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50d914bbef2ecd2b0000002a SecondChildUserIdTAG: 296419 SecondChildUserNameTAG: oscman SecondChildCreateTimeTAG: 2013-01-17T20:06:55Z FirstChildTAG: Great video Anthony, never heard about the Manhattan method before (never to old to learn. FirstChildUserIdTAG: 79885 FirstChildUserNameTAG: ruudoleo FirstChildCreateTimeTAG: 2013-01-17T21:23:51Z FirstChildTAG: Hola Anthony, muy bueno y educativo tu vídeo, espero que ganes el libro. Veo que no te has registrado en el mapa, aquí te envió el link, ya que de Venezuela estoy yo nada mas.: https://maps.google.com/maps/ms?msid=218116803755853699395.0004d1a3f6e81df18faf5&msa=0&ll=36.173357,3.515625&spn=97.435884,228.339844 Saludos desde Caracas. Oscar FirstChildUserIdTAG: 296419 FirstChildUserNameTAG: oscman FirstChildCreateTimeTAG: 2013-01-16T20:16:47Z IndexTAG: 472 TitleTAG: MIT to MITx? How come the certificates went from saying MIT next to edX at the top to MITx? I know it's the same thing, I was just wondering. Personally I prefer MIT since even though it was online the class was all MIT material and not "watered down." ![enter image description here][1] ![enter image description here][2] [1]: https://edxuploads.s3.amazonaws.com/1358094740467095.png [2]: https://edxuploads.s3.amazonaws.com/1358094868134363.png UserIdTAG: 219780 UserNameTAG: ReconIII CreateTimeTAG: 2013-01-13T16:31:30Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: That's weird. Maybe they have updated it to be consistent with other course certificates they have already issued. Instead of updating thousands of certificates just for this minor issue they could have included a [QR code][1] for verification link that I suggested in this forum 13 days ago with post title ["Inclusion of 2D barcode for verification link in the certificate"][2]. It would have been a much useful effort. My post did get 25 votes quickly but the post is now buried away and still left unanswered. [1]: http://en.wikipedia.org/wiki/QR_code [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_Feedback/threads/50e310965cb26b2700000018 FirstChildUserIdTAG: 297370 FirstChildUserNameTAG: ayush3504 FirstChildCreateTimeTAG: 2013-01-14T13:38:14Z SecondChildTAG: It is a good suggestion, perhaps they did not have enough time to implement it. We may see it in the future. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2013-01-14T14:33:00Z SecondChildTAG: Only thing is, how many people have a barcode scanner? Also any other ideas why they did this? I liked the MIT logo better, as it seemed more professional. SecondChildUserIdTAG: 219780 SecondChildUserNameTAG: ReconIII SecondChildCreateTimeTAG: 2013-01-14T15:25:49Z SecondChildTAG: I think those QR codes are readable with cell phone cameras. I like the MIT logo better as well, glad I saved a copy of the original. Also the verification links have changed, though they both are still active so far. It is likely do to branding, MITx needs to be identified as a separate initiative of the brick and mortar MIT. Students that spend over $50,000 a year for MIT deserve that distinction.(My opinion only) So I guess it makes sense to have it's own unique logo. As long as it says MIT(x) somewhere on the ticket, I am satisfied. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2013-01-14T15:50:28Z SecondChildTAG: Well I hope that people who would use a certificate to judge us (employers/universities) look at the edX exams and assignments. They really are hard and of the same standard as the on-campus courses. That would really give a lot of value. SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2013-01-14T15:54:50Z SecondChildTAG: I find most people are understanding when I say that MITx is the online version of the actual MIT course. On the bright side, it still says MITx and mentions 6.002x, which would imply that it is the online version of 6.002. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2013-01-14T16:08:51Z SecondChildTAG: @ashwish Well the only thing is that most people have never heard of edx, where everyone has heard of MIT. SecondChildUserIdTAG: 219780 SecondChildUserNameTAG: ReconIII SecondChildCreateTimeTAG: 2013-01-14T21:27:32Z FirstChildTAG: BTW can anyone verify or find out if the first version's verification links will stay up? FirstChildUserIdTAG: 219780 FirstChildUserNameTAG: ReconIII FirstChildCreateTimeTAG: 2013-01-14T21:35:49Z SecondChildTAG: I will look into this for you. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2013-01-14T22:00:52Z SecondChildTAG: Great thanks!!!! It is really important to me. SecondChildUserIdTAG: 219780 SecondChildUserNameTAG: ReconIII SecondChildCreateTimeTAG: 2013-01-14T22:05:22Z SecondChildTAG: Why don't you just try it yourself ? SecondChildUserIdTAG: 90796 SecondChildUserNameTAG: ChaunceyGardiner SecondChildCreateTimeTAG: 2013-01-15T00:57:08Z SecondChildTAG: @ChaunceyGardiner Well I know it works currently, yet as I asked, I was wondering if the link will stay up or will it be taken down and replaced with the one from this new MITx version of the certificate. SecondChildUserIdTAG: 219780 SecondChildUserNameTAG: ReconIII SecondChildCreateTimeTAG: 2013-01-15T01:21:26Z SecondChildTAG: Checked now - verification links still working for both. SecondChildUserIdTAG: 394836 SecondChildUserNameTAG: v2g6ch4 SecondChildCreateTimeTAG: 2013-01-15T02:48:49Z SecondChildTAG: Reconlll, I have not heard anything back as of yet, just letting you know I have not forgotten. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2013-01-16T00:18:32Z SecondChildTAG: Either link will work for you permanently. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2013-01-16T11:21:23Z SecondChildTAG: Great, thanks so much! SecondChildUserIdTAG: 219780 SecondChildUserNameTAG: ReconIII SecondChildCreateTimeTAG: 2013-01-16T11:34:46Z FirstChildTAG: ![enter image description here][1] [1]: http:// FirstChildUserIdTAG: 60453 FirstChildUserNameTAG: TommyNittin FirstChildCreateTimeTAG: 2013-01-16T10:47:45Z IndexTAG: 473 TitleTAG: CECC - Four Days to Go That's right! So if you want the textbook with extra AA awesomeness make the weekend count! :-) UserIdTAG: 14386 UserNameTAG: ashwith CreateTimeTAG: 2013-01-10T12:59:38Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 474 TitleTAG: CECC 2 Updates - deadline due January 14th and Thank you!:) Hi Classmates! We have received entries from the students of 6.002x! As Organizers - ashwith, juancho and Myrimit- we will wait till the deadline - January 14th - to send to the Jury all the video links and files that we have received :). Of course, that as Organizers, we have taken a look to them in order to check if they were correctly submited, but the Jury have not watched the video links - files yet :p. *So, if you haven´t submited your video you have a week to do it! Remember that the most important is that you can participate :). And, if you result one of the 3 Winners you will receive a Textbook signed personally by Prof. Agarwal!* [Check out this - Popping up some old Posts][1] Remember that the Jury will be composed by ex-students of 6.002x Spring 2012: - Barrabas - ChaunceyGardiner - dantyrant - danik - JSChambers - komisz All the Integrants of the CECC Team- Organizers and Jury- are volunteer in this Contest , so I have to say to all the Team, thank you very much for making this possible. All the CECC Team is amazing! Luckily we all get well each others and that is really important. I am really grateful and I wish that the Students from this Fall can feel inspired and make a lot of projects in the Future, also we will leave the open doors for those of this Fall who wish to carry out the CECC Contest in the Spring ;), as we want to give to others the opportunity to participate as Jury and Organizers too. Because this involvement is a wonderful experience for the students, not only because you learn to colaborate and to work in Team with your worldwide Classmates, also because you can contribute, as student, to this amazing learning iniciative and the most important is that you can motivate to the future students and encourage them to participate, to get involved and the most important, to have a lot of fun ;). So all this movement is really important for you and the future of the education. Also, we will be always grateful with all edX for supporting us in this Contest, thank you very much. Thank you Prof. Agarwal, Piotr Mitros, Lyla, Rohan, Dianna and all the Staff from edX from 6.002x and from the other courses too, they are so many people involved behind the scene, like Tony, Lauren, Dave, Peter, Jennifer, John, Jorge, Prithwis and soo many more! Also my CTAs mates too , they are the best haha :p Good luck to all of the Contestants! :) the CECC Team. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50db0e95ad5e5e1f00000004 UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2013-01-08T06:20:16Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I agree with you, that participating is important, but can we have access to all the entries from the previous contest and not just the winners ? Just to make an ideea . FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2013-01-08T07:09:44Z SecondChildTAG: I too second Alex's idea. Would be great if we have access to much more entries apart from those of the winners. SecondChildUserIdTAG: 232157 SecondChildUserNameTAG: GayatriTR SecondChildCreateTimeTAG: 2013-01-08T07:13:49Z SecondChildTAG: What's in the wiki *is* all the entries. SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2013-01-08T07:52:13Z SecondChildTAG: Thank you ! SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2013-01-08T08:06:13Z SecondChildTAG: Also you asked a question in my CECC announcement post. I replied to it only recently. I'm sorry I replied to it very late. In case you haven't seen it, here's my reply > Hi AlexAlexandrescu, I'm really sorry I didn't see your question > before. Yes the project is eligible. You can explain how the circuit > works and also you should be able to show the motor speed change and > that should be more than enough. One of the reasons why Myriam started > this context (she did this last time as well) was to encourage future > students to join 6.002x by showing what they can do after they are > through with this course. Hopefully this time we'll be able to get the > winner's videos on the 6.002x homepage for people to see when they > sign up (no promises though). So your viewers will appreciate the > motor control aspect of it more than the oscilloscope output, which > would probably be quite cryptic unless they've seen this before. SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2013-01-08T08:38:54Z SecondChildTAG: I will send you my entry this weekend.Thank you again ! SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2013-01-10T07:48:50Z SecondChildTAG: Great! Will be waiting for it :-) SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2013-01-10T08:27:25Z IndexTAG: 475 TitleTAG: Vote of thanx... SHUKRIYA to Prof.Anant Agarwal..... UserIdTAG: 526154 UserNameTAG: RAJAT09 CreateTimeTAG: 2013-01-03T18:12:21Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: شكراً لكم جميعا ً FirstChildUserIdTAG: 41666 FirstChildUserNameTAG: AbdoMondy FirstChildCreateTimeTAG: 2013-01-03T18:54:39Z SecondChildTAG: اهلا و سهلا بك SecondChildUserIdTAG: 156535 SecondChildUserNameTAG: badawiIiIi SecondChildCreateTimeTAG: 2013-01-05T21:57:02Z IndexTAG: 476 TitleTAG: thank you i'm so happy because i could finish this course and i'm proud of that. i appreciated your situation when you extended the time of mid term exam because of Aid Aladha many thank for everyone who work on this course ^_^ UserIdTAG: 103719 UserNameTAG: soha CreateTimeTAG: 2013-01-03T18:10:01Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 477 TitleTAG: for STAFF: it'd be wonderful to count with the continuation of 6.002x!!! Hi, does anyone know if the staff is considering the idea of a more advanced course, as a second part, of 6.002x? It'd be great cause the last slide "violeting the abstraction barrier" really leaves you the feeling there is something else to discover! Please bear in mind this course in future! UserIdTAG: 99441 UserNameTAG: coyarce CreateTimeTAG: 2013-01-03T17:12:09Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: It was my understanding from the video S25V1 that there will be soon available 6.004x .So here you go . FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2013-01-03T20:52:47Z SecondChildTAG: for me too but since they haven't announced anything yet.....I was wondering it will come true or not...! SecondChildUserIdTAG: 99441 SecondChildUserNameTAG: coyarce SecondChildCreateTimeTAG: 2013-01-07T15:45:44Z FirstChildTAG: Friends,don't wait for 6.004x course to come,if you have interest in doing that course please follow this link.In this link you will have videos lectures(on the left side of page click on the handouts you can download that videos),homework etc, http://6004.mit.edu/Spring12/ FirstChildUserIdTAG: 857413 FirstChildUserNameTAG: Naveenkrishna FirstChildCreateTimeTAG: 2013-02-09T17:03:55Z IndexTAG: 478 TitleTAG: Greeting to Staff Hello, I just figure out that in the last 3 month many of my problems in understanding the electronic concepts have solved. I like to say "Thank you!" to all individuals who perform even a bit of the 6.002x course, and, I wish I could do that face to face. Nevertheless, I think maybe all of the students who pass the course could help staff in the next semester, therefore, if you are one of the staff and you think I can help you with the course sequence(No matter how). Please inform me! Best wishes, UserIdTAG: 420339 UserNameTAG: AliJenabi CreateTimeTAG: 2013-01-03T14:43:53Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: In my point of view, totally yes!You can help :) Also, if you are interested of being CTA take a look at course info: ---------- NOVEMBER 19 > **Community TA's:**[read here][1] > > In preparation for our spring offering, we are trying to identify > volunteer Community TA's from this semester's run. If you are > interested in the position, please send an email to mit-6002x@edx.org > with links to some of the posts that you are most proud of. ---------- [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/info FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2013-01-03T17:22:37Z IndexTAG: 479 TitleTAG: Final Exam Q1 - Video Explanation - Not Official Hi all, This is a Not Official Video Explanation - step by step - of the Q1 of the Final Exam 6.002 2012. [Watch Here - YouTube][1]. It is a little bit long haha. I hope that you don´t get bored. My hand writting it is not the best, in fact, it is horrible haha. I was planning to post this in some days when I have the others video explanations, but as PabloVic [here][2] was having some doubts with this Problem number 1 I have decided to post this before. I am also doing a video explanation of Q2 but with another format, with a video capture - I am trying to experiment with new didactic tools!:) - we are planning with ashwith and rharris to make some video explanations for the students of 6.002x Spring 2013 in order to help the new students based on their difficulties in this Fall. I hope this can help you if you didn´t do well the Q1 of the Final Exam :p Take care, Myriam. P.D1. Sorry for my English haha, I will try to improve it, I will take an English Course this year haha. P.D2. Don´t forget to participate in the Contest. The deadline was extended till January 14th 2013. The rules are in the Wiki. You can win a Textbook signed personally by Prof. Agarwal. [1]: http://www.youtube.com/watch?v=ZAYZJN-5tEg [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50e1e0e259d0cc2700000001 UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2013-01-01T03:27:24Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Great job Myriam, thanks..and Thanks for all the help and tips during the course too..:) FirstChildUserIdTAG: 173147 FirstChildUserNameTAG: cruiser_rahit FirstChildCreateTimeTAG: 2013-01-02T07:12:54Z FirstChildTAG: Great job Myriam. Thanks for that. Its clear my doubts. Thanks a lot. FirstChildUserIdTAG: 529515 FirstChildUserNameTAG: Low FirstChildCreateTimeTAG: 2013-01-03T01:16:52Z FirstChildTAG: Fantástico trabajo una vez más Myriam. De nuevo muchísimas gracias por todo. FirstChildUserIdTAG: 706676 FirstChildUserNameTAG: PabloVic FirstChildCreateTimeTAG: 2013-01-03T11:10:02Z IndexTAG: 480 TitleTAG: Job opportunity in Peru (2013). Happy New Year everyone! ***Peru needs professionals worldwide***, especially mining engineering, civil engineering and geological engineering. Peru will not cover the deal with Peruvian professionals. It is an opportunity for professionals from around the world especially for those in crisis. *Professionals around the world are highly valued in Peru*. **Peru is growing about 7%**. Peru is a mining country. Peru is 6th world’s biggest gold producer. **Peru** is **1st gold producer** in Latin America. ![Moche gold necklace depicting feline heads. Larco Museum Collection. Lima-Peru.][1] Moche gold necklace depicting feline heads. Larco Museum Collection. Lima-Peru. **Peru, the world’s biggest minerals-mining country.** ![Peru, the world’s biggest minerals-mining country][2] ![Peru flag][3] ![Peru map][4] **Happy New Year everyone!** [Job opportunity in Peru (2013)][5] [1]: https://edxuploads.s3.amazonaws.com/13570033581343604.jpg [2]: https://edxuploads.s3.amazonaws.com/13570223001343628.png [3]: https://edxuploads.s3.amazonaws.com/13570580021343664.png [4]: https://edxuploads.s3.amazonaws.com/13570598671343634.png [5]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/Job-opportunity-in-Peru/ UserIdTAG: 402193 UserNameTAG: Feramico CreateTimeTAG: 2013-01-01T00:31:44Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: spanish is a must, right? FirstChildUserIdTAG: 623210 FirstChildUserNameTAG: preveen FirstChildCreateTimeTAG: 2013-01-01T10:29:05Z SecondChildTAG: Any language, but you will have an advantage if you speak English, better if you have basic knowledge of Spanish. SecondChildUserIdTAG: 402193 SecondChildUserNameTAG: Feramico SecondChildCreateTimeTAG: 2013-01-01T15:55:17Z FirstChildTAG: Where is the site with job offers, or something like this? FirstChildUserIdTAG: 394836 FirstChildUserNameTAG: v2g6ch4 FirstChildCreateTimeTAG: 2013-01-01T19:04:18Z SecondChildTAG: http://www.infomine.com/careers/jobs/r6c106/peru.jobs.aspx SecondChildUserIdTAG: 402193 SecondChildUserNameTAG: Feramico SecondChildCreateTimeTAG: 2013-01-01T20:26:21Z SecondChildTAG: http://laborum.pe/postulante/buscar-ofertas?page=1 SecondChildUserIdTAG: 402193 SecondChildUserNameTAG: Feramico SecondChildCreateTimeTAG: 2013-01-01T20:32:06Z SecondChildTAG: http://empleo.universia.edu.pe/buscar-trabajo/?palabra=ingeniero+civil®ion=0 SecondChildUserIdTAG: 402193 SecondChildUserNameTAG: Feramico SecondChildCreateTimeTAG: 2013-01-01T20:39:39Z SecondChildTAG: http://aptitus.clasificados.pe/buscar/query/mina SecondChildUserIdTAG: 402193 SecondChildUserNameTAG: Feramico SecondChildCreateTimeTAG: 2013-01-01T20:44:26Z FirstChildTAG: Too many jobs and not enough engineers in Norway as well. A lot of open positions and salary for a design engineer is over EUR150k. This is better than Silicon Valley, CA where salary is around EUR125k Norway has cold weather but good and free universities for all. http://www.finn.no/finn/job/fulltime/advanced Use google translate FirstChildUserIdTAG: 142857 FirstChildUserNameTAG: viking2 FirstChildCreateTimeTAG: 2013-01-01T21:42:17Z SecondChildTAG: Could language be a bar? SecondChildUserIdTAG: 99628 SecondChildUserNameTAG: Pranjal16 SecondChildCreateTimeTAG: 2013-01-02T03:48:32Z SecondChildTAG: If you speak English, you are fine. Major Norwegian companies have English as a working language. Universities usually have the first 4 years available in English. Engineering and science use US English textbooks exclusively. SecondChildUserIdTAG: 142857 SecondChildUserNameTAG: viking2 SecondChildCreateTimeTAG: 2013-01-02T11:08:27Z IndexTAG: 481 TitleTAG: Any other courses related to electronics? As in the topic... Obviously I have noticed increased number of courses available at EDX, but unfortunately none of them covers more advanced topics than 6.002. Are there any other electronics-related courses going to be available in the nearest future? Kind regards UserIdTAG: 420332 UserNameTAG: gapa CreateTimeTAG: 2012-12-30T17:16:20Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: **gapa:** I believe there will be more courses announced; the new batch of courses is only the first-half of those to be announced and is more humanities-and-arts-based. I'm guessing that the second-half will be more science-and-engineering based. EdX simply wanted to broaden it's offerings beyond sci-and-eng to attract a more diverse, global student body! I think I will start a petition for ***MITx*** to offer **6.003x**, Signals and Systems, the next course after 6.002x, a logical progression, and one that I need, personally, to refresh my memory of, after taking it at University! Jersey Mark FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-12-30T18:13:20Z SecondChildTAG: I would also like to join for Signals and Systems. SecondChildUserIdTAG: 178840 SecondChildUserNameTAG: mehtanayanv SecondChildCreateTimeTAG: 2012-12-30T18:58:22Z SecondChildTAG: Basics of Analog Circuit Design(Amplifiers and Current sources etc.), Computer Organisation and Architecture, Digital Design(with Verilog/VHDL) SecondChildUserIdTAG: 178840 SecondChildUserNameTAG: mehtanayanv SecondChildCreateTimeTAG: 2012-12-30T19:01:37Z SecondChildTAG: There is a similar course on coursera (from Rice): [Fundamentals of Electrical Engineering][1] [1]: https://www.coursera.org/course/eefun SecondChildUserIdTAG: 45933 SecondChildUserNameTAG: Qubit SecondChildCreateTimeTAG: 2012-12-31T15:49:18Z SecondChildTAG: I would like it to be offered from MITx though as their format is a lot nicer. SecondChildUserIdTAG: 45933 SecondChildUserNameTAG: Qubit SecondChildCreateTimeTAG: 2012-12-31T15:50:10Z FirstChildTAG: Ashwith and a group of last semester students have put this amazing course together : http://6003z.amolbhave.in . It has the mit ocw lessons, but also a bunch of exercises to help you practicing . FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-12-30T18:59:45Z FirstChildTAG: I have the same feeling about 6.003x-that would be a natural next step. I understand that Undergraduate courses at MIT seem to be the most attractive for students as they introduce and familiarize with laws, tools or basic methods which facilitate solving some problems. What about EDX courses for Graduates at MIT? Is there a possibility to encourage Instructors of courses to participate in EDX? I am personally very much interested in Power Electronics thus 6.334 would be also highly interesting for me as well. I suppose that other EDX students have also different preferences. The best way would be to show the interest in particular courses which are available at http://ocw.mit.edu ...that could be also helpfull to understand which courses are the most desired by people. Kind regards, Tomasz Gapinski FirstChildUserIdTAG: 420332 FirstChildUserNameTAG: gapa FirstChildCreateTimeTAG: 2012-12-30T18:56:57Z SecondChildTAG: I would also love to see graduate-level courses on ***MITx*** as well! I have already taken most of the undergraduate electrical engineering courses offered at my university, however, I like taking them again as a review (I went to university 8+ years ago) and for the added challenge (the first time around I was a B- to B+ student, with the occasional A, now I strive for 100%'s!) However, I think that graduate-level courses will not be a priority, as edX is trying to reach as many students as possible. And I believe that graduate-level courses usually take place in a more intimate environment, with perhaps only 10 students to a Ph.D-level professor, and there is much more emphasis on research, partnering with industry, and for students to demonstrate that they can come up with their own breakthroughs and original thought in the field, than there is on grades and learning already-established theory. I would also love to see Power Electronics; as that is the backbone and workhorse of industry: I'm guessing that it covers 3-phase circuitry, power supply and quality, generation of electricity from devices such as alternators, stepping up/down voltages through transformers and solid-state circuitry, high current solid-state devices such as thyristors, etc. but I will take a look at the actual 6.334 syllabus. We had Power Electronics Lab at my university, so theory was only covered briefly; most of the time was spent building regulated power supplies, testing 3-phase motors and generators, and then building audio amplifiers and controlling high-voltage loads with control circuits using thyristors; hence why I brought all of this up above. I would love to go into the theory in more depth however! Thanks for the comments, Tomasz (edX: **gapa**)! Jersey Mark SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-12-30T21:30:34Z SecondChildTAG: I have graduated more than three years ago and I had exactly the same reasons for participating in edx: I wanted to recall some basis and verify what I have skipped during my studies :). Even if I am also affraid that graduate-level courses will not appear at EDX, I just wanted to give a signal that right now at least two people are highly interested (Mark and Tomasz). Maybe this could wake up other people who would also like to take part in graduate-level courses e.g. **6.334** . Summarizing, I am waiting for next **edx** courses oriented to electronics, but I would really love to see previously mentioned graduate-level course at **edx**. All the best to you Mark. Regards to staff and all edx students. SecondChildUserIdTAG: 420332 SecondChildUserNameTAG: gapa SecondChildCreateTimeTAG: 2012-12-31T10:36:44Z SecondChildTAG: count me in, I'm also interested in 6.334 Power Electronics Course. I really hope that edX MITx will to offer this course. SecondChildUserIdTAG: 185303 SecondChildUserNameTAG: NormanDP SecondChildCreateTimeTAG: 2013-01-01T12:37:11Z FirstChildTAG: i need it too (6.003), if you are interested in Signals and Systems here is an MIT video course : - http://ocw.mit.edu/resources/res-6-007-signals-and-systems-spring-2011/video-lectures/ FirstChildUserIdTAG: 236136 FirstChildUserNameTAG: Hamid-ch FirstChildCreateTimeTAG: 2012-12-30T19:06:38Z IndexTAG: 482 TitleTAG: Display of grade on Certificate ? Hi, I shall like to ask that whether the Grade will be displayed on the certificates ? Community TA members, Kindly let us know. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-12-28T17:08:35Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: I asked Lyla this question earlier, and the answer is (*unlike* 6.002x under MITx in the inaugural Spring 2012 semester where grades *could* be displayed as an option) is **no grades will appear on Fall 2012 certificates**. While I disagree with this policy, I am not in a position to change it. I will try to find out why this change was implemented, though I cannot guarantee a response. See the image below for Lyla's exact answer: ![enter image description here][1] [1]: http://i905.photobucket.com/albums/ac254/MarkPD/GradeonCert_zps39dc244e.png FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-12-28T18:47:29Z SecondChildTAG: Thanks Jersey, for your response. But this was the old story for spring 6.002x. Whereas, students who completed 6.002x in spring, they received certificates displaying their respective grades. SecondChildUserIdTAG: 211164 SecondChildUserNameTAG: Hamoodi SecondChildCreateTimeTAG: 2012-12-28T22:15:07Z SecondChildTAG: *Sorry for my previous comment, ignore that*. I got your point now. But the thing is that if this decision was to be made of not specifying the grades on the certificate, then why did they specified grade thing in the **Grading Section that a student has to total mark of 60% for a C, 70% for a B, and 87% for an A ?** And this criteria was written on 5th of September 2012 *(which is far ahead of spring 6.002x)* , which can be verified by going through the link specified below. If they had to do these amendments then why did they put forth this criteria ?They could had simply told the students that to certify the course just score upto 60%. I mean what about those students who studied hard and tried their best to achieve A Grade. But in the end what we are getting aware of, is that in fall 6.002x there will be no grade displayed. Is this just ? Kindly, consider this matter. [MITx 6.002x –Circuits and Electronics General Information][1] [1]: https://www.edx.org/static/content-mit-6002x/handouts/syllabus.a477535058a1.pdf SecondChildUserIdTAG: 211164 SecondChildUserNameTAG: Hamoodi SecondChildCreateTimeTAG: 2012-12-28T22:40:05Z SecondChildTAG: Hear Hear!!! there should be a difference between A and C SecondChildUserIdTAG: 37464 SecondChildUserNameTAG: DoubleA SecondChildCreateTimeTAG: 2012-12-28T23:14:35Z SecondChildTAG: **Hamoodi**: I am simply a student here; I just took the Final Exam just like you did. I also feel that there should be an option to display my grade. However, you (and anyone who feels a grade should be included on the certificate) should petition the **Staff**. I may be a Community TA, but that's nowhere near Staff. We do **not** have any special powers, we do not set course policy, and our opinions carry no official weight. The Staff Listing is included under the "Course Info" tab, so I would suggest you direct your petition there. If many people respond, maybe Prof. Agarwal will oblige. Also note that I am not privy to edX's operational model (except for that it is *not-for-profit*); I do not know why they took the "grade on your certificate" option away. All I can do is speculate: - Perhaps edX wants to keep the Honor Code certificates uniform among the edX universities. The other schools may not want to post grades on their certificates. Therefore MITx would have changed it's policy from Spring 2012 to keep in line with Harvard, Berkeleyx, etc. - Perhaps edX wants to reserve grades for only those who obtain certificates via a proctored exam (when or if it is offered) in the future. Who knows, certificates may not always be "free"; grades have a certain value, and perhaps this is a source of revenue being considered by edX. Anyways, as edX is still a work-in-progress, and massive open-enrollment online courses in their infancy, just expect to see various changes between the semesters as things are tried out by Staff, found to work or not work; think of edX and it's offerings as a "beta test" instead of "fully-accredited degree-granting university", even though it's members are! Jersey Mark SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-12-29T01:00:51Z SecondChildTAG: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50dd935856fba4290000003d SecondChildUserIdTAG: 183712 SecondChildUserNameTAG: colegiocientifico SecondChildCreateTimeTAG: 2012-12-29T15:19:39Z FirstChildTAG: They may be planning to experiment with grade less systems in the future. If you want to see what one looks like go here: https://www.khanacademy.org/exercisedashboard To give a brief description, a student does randomized questions until it is proven that the material is understood by getting a certain number of questions right in a row, then the student is marked proficient and moves on to the next topic. FirstChildUserIdTAG: 202032 FirstChildUserNameTAG: KeithD FirstChildCreateTimeTAG: 2012-12-29T03:03:29Z FirstChildTAG: Thanks for your guidance Jersey! Can you guide me how to do petition ? FirstChildUserIdTAG: 211164 FirstChildUserNameTAG: Hamoodi FirstChildCreateTimeTAG: 2012-12-29T15:00:15Z FirstChildTAG: Si, sin un nota, la verdad es que ninguno de tus amigos va a creer que te lo ganaste estudiando y quemándote las pestañas, van a pensar que es un simple certificado otorgado a todos los que se inscribieron, se hayan esforzado no, especialmente siendo de MIT, pues les va a ser imposible creer que alguien haya alcanzado los puntos para obtener una nota de aprobación. FirstChildUserIdTAG: 183712 FirstChildUserNameTAG: colegiocientifico FirstChildCreateTimeTAG: 2012-12-29T15:34:35Z SecondChildTAG: sí estoy de acuerdo SecondChildUserIdTAG: 342221 SecondChildUserNameTAG: deepkar SecondChildCreateTimeTAG: 2012-12-29T16:43:29Z SecondChildTAG: Porque no lo van a creer?? crees que hay personas que no capaces de hacerlo ??? Seguramente hacen ese comentario (colegiocientifico y deepkar) porque sus puntajes son bajos (diganlo sin miedo) !! Anda el mio es de 100 perfecto y creeme te duela o no, yo si me queme las pestañas estudiando. Es injusto que reciban solo certificados, sin poner la nota que cada uno se merece. SecondChildUserIdTAG: 628745 SecondChildUserNameTAG: albertclint SecondChildCreateTimeTAG: 2012-12-30T02:28:52Z SecondChildTAG: Todos somos capaces, lo importante es poner dedicacion en lo que uno quiere !! SecondChildUserIdTAG: 628745 SecondChildUserNameTAG: albertclint SecondChildCreateTimeTAG: 2012-12-30T02:29:59Z IndexTAG: 483 TitleTAG: @MYRIAM DEAR MYRIAM I'm again appear to bother you with some more query....as you are kind and helpful so i can expect it will be answered...:) 1) where are you come from??????? are you a student of MIT????? 2)how did you become the community member of 6.002x? 3)what is procedure to get admission in MIT after under graduation for international student ?? I've gone through the site for this but it was confusing enough so better would be if you could elaborate it...^_^ UserIdTAG: 108863 UserNameTAG: shiviz CreateTimeTAG: 2012-12-27T14:27:30Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Hi shiviz, 1) I am from South America (Argentina). I love MIT since my childhood! Unfortunately, I am not a student of MIT, oh my, that is the dream of all the students, is one of the most prestigious of the World with the best Scientists! The students that are there are so fortunate, I have envy (a well envy :p) to them haha. I am student of UTN (Universidad Tecnológica Nacional), It is a Public University in Argentina :), is one of the best of my Country :p 2) I took 6.002x Spring 2012. I registered again, for my own will, this Fall, in order to help the new students ;), I didn´t knew the possibility of being CTA haha. Then edX offered me to become one ;). 3) I wish to help you, but I really don´t know how is the procedure as I have never applied... Might you should send to MIT an e-mail requesting them information about the procedure of admission. I know a young talented student from India that wanted to get admission, Amol. I can ask him in the old forum how he did. But might it would be better if you send to MIT an e-mail. I am sure they will answer you, they are really kind people. And if in the future you see Prof. Agarwal there,give him my greetings haha! Take care, Myriam. P.D. If it helps, I will re-link this Post to the 6.002x Staff :). Might someone of MIT can answer you, I don´t promise anything :P. I hope this can help you. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-27T15:28:02Z SecondChildTAG: thanks a lot :)....that is really kind of you..... I'll mail to MIT,but if you got something related that;do inform me..... my id is "shivendraagrahari.7777@gmail.com" SecondChildUserIdTAG: 108863 SecondChildUserNameTAG: shiviz SecondChildCreateTimeTAG: 2012-12-27T16:02:58Z FirstChildTAG: Qué alegría jaja! Arriba el Río de la Plata! Saludos de Uruguay Myriam! Carlos FirstChildUserIdTAG: 220837 FirstChildUserNameTAG: Calsomus FirstChildCreateTimeTAG: 2012-12-27T18:06:22Z SecondChildTAG: better if you could write in english.........by the way which language is this????? SecondChildUserIdTAG: 108863 SecondChildUserNameTAG: shiviz SecondChildCreateTimeTAG: 2012-12-28T03:23:48Z SecondChildTAG: its spanish SecondChildUserIdTAG: 342221 SecondChildUserNameTAG: deepkar SecondChildCreateTimeTAG: 2012-12-28T03:31:03Z SecondChildTAG: Saludos Carlos! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-28T17:00:27Z SecondChildTAG: :) SecondChildUserIdTAG: 220837 SecondChildUserNameTAG: Calsomus SecondChildCreateTimeTAG: 2012-12-28T17:01:08Z FirstChildTAG: Hi shiviz, Admission to MIT Graduate programs (Master and PhD) is done by departments. Do you know which department you want to apply to? You can see the list here: http://web.mit.edu/education/ I applied to the EECS department (Electrical Engineering and Computer Science) as an international student and found the process easy but time consuming. I am not an expert in the admission process, but when I applied, among other things, I had to take the TOEFL exam, send 3 letters of recommendation and write a Statement of purpose. Other departments also require you to take the GRE. It roughly took me about 3-4 months to prepare everything. Also, keep in mind the deadlines. When I applied, the deadline for admission to start the next fall semester was mid-December, so probably you need to wait until next Dec to apply (so you can start in 2014). You can find more info about the process for the EECS department here: http://www.eecs.mit.edu/academics-admissions/graduate-program/admissions/dear-prospective-applicant I hope this helps and let me know if you have more specific questions! Jorge FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-12-28T16:40:51Z SecondChildTAG: Thank You for the response Jorge :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-28T16:59:48Z SecondChildTAG: thanks jelizon pardon for late reply because i see this today only.....so i do not have any further query till next year :D... so that basically means i need to focus for TOEFL first n all comes second.....thank you myrimit.:) SecondChildUserIdTAG: 108863 SecondChildUserNameTAG: shiviz SecondChildCreateTimeTAG: 2012-12-29T04:51:12Z SecondChildTAG: Jorge Simosa i got you....haha:) SecondChildUserIdTAG: 108863 SecondChildUserNameTAG: shiviz SecondChildCreateTimeTAG: 2012-12-29T04:55:26Z IndexTAG: 484 TitleTAG: Thanks edx and Professor Dr. Agarwal It has been a good and fun time with this mitx course and i have honestly learnt a lot .I would like to thank and recommend this online course to all college engineering students and my regards to Professor Dr Agarwal for his time and hard work.Is there any course about computer programming with C++? UserIdTAG: 391053 UserNameTAG: Ahsanuddin CreateTimeTAG: 2012-12-26T20:11:29Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 1 FirstChildTAG: I agree! FirstChildUserIdTAG: 148389 FirstChildUserNameTAG: chento FirstChildCreateTimeTAG: 2012-12-26T20:54:04Z SecondChildTAG: Mr. Agarwal thank you and your team for this very very useful, interesting and attractive course. Good luck you and Happy New Year! SecondChildUserIdTAG: 145655 SecondChildUserNameTAG: dimkaoz SecondChildCreateTimeTAG: 2012-12-29T17:22:12Z IndexTAG: 485 TitleTAG: Merry Christmas, Happy Christmas, Merry Xmas, Happy Holidays or Season's Greetings! ![Happy Holidays][1] From Peru. [1]: https://edxuploads.s3.amazonaws.com/13564895548474468.png UserIdTAG: 402193 UserNameTAG: Feramico CreateTimeTAG: 2012-12-26T02:40:00Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Pisco sour con lemon de pica ? FirstChildUserIdTAG: 373752 FirstChildUserNameTAG: Croak FirstChildCreateTimeTAG: 2012-12-26T02:55:38Z SecondChildTAG: Of course! SecondChildUserIdTAG: 402193 SecondChildUserNameTAG: Feramico SecondChildCreateTimeTAG: 2012-12-26T03:17:56Z SecondChildTAG: But better with peruvian lemon. SecondChildUserIdTAG: 402193 SecondChildUserNameTAG: Feramico SecondChildCreateTimeTAG: 2012-12-26T04:21:55Z FirstChildTAG: Thank you for the picture from the snowy Ukraine! Happy New Year! Serge FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-12-26T11:30:08Z IndexTAG: 486 TitleTAG: More Blessings to MITx staff from Ghana More blessings to the entire staff of MITx. I also acknowledge the effort of those behind the scene.Prof Agarwal may your blessings and wisdom continue to overflow so that we will continue to receive applicable knowledge from you. UserIdTAG: 446722 UserNameTAG: Richmond CreateTimeTAG: 2012-12-25T16:01:59Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 487 TitleTAG: Histogram of scores Hi, PLEASE, it's possible to publish the histogram of reached scores of all? Many thanks, Lukas UserIdTAG: 508733 UserNameTAG: burdal1 CreateTimeTAG: 2012-12-25T14:16:55Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Last semester the statistics were released shortly after the conclusion of the Final Exams; I'm pretty sure the statistics will again be released. Just keep an eye on the Discussion boards, and in the Course Info tab. Also remember that we are in the Holiday season here in the U.S., and many university Staff are off-of-work until after the New Year (for some January 2nd, or at least until December 26th.) Also depending on the university, U.S. schools have "Winter Break" and the Spring semester does not begin until late January or early February for most schools. In short, be patient! Jersey Mark FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-12-25T18:21:50Z SecondChildTAG: I'm patient. I didn't know that the statistics are usually published. SecondChildUserIdTAG: 508733 SecondChildUserNameTAG: burdal1 SecondChildCreateTimeTAG: 2012-12-25T20:19:44Z IndexTAG: 488 TitleTAG: chrismas Merry Christmas to all edx faculty members n all my friends...wishing you all a very happy year ahead..:) UserIdTAG: 108863 UserNameTAG: shiviz CreateTimeTAG: 2012-12-25T03:24:24Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: merry christmas!!!!! FirstChildUserIdTAG: 431237 FirstChildUserNameTAG: rohinraj FirstChildCreateTimeTAG: 2012-12-25T04:15:37Z IndexTAG: 489 TitleTAG: thanks to Agrawal Sir And other Staff thanks a lot UserIdTAG: 342135 UserNameTAG: vikash902 CreateTimeTAG: 2012-12-24T20:17:13Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 490 TitleTAG: Uruguay says: Present! Hi to all, well, we've come to the end of this great course. It was really amazing, to find such initiative, such quality, from nothing else than MIT, through EDx. To all staff, to Dr. Agarwal, to all classmates, MANY THANKS! Carlos UserIdTAG: 220837 UserNameTAG: Calsomus CreateTimeTAG: 2012-12-24T16:34:53Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 491 TitleTAG: Q3 in final... Now that the final is over, i can ask: WHY asking an already resolved problem, even with its solution given by staff and readable by all, in H8P1? At least, numbers should have been changed. It was already easy enough for anyone that knows how to read the solution to H8P1... Don't you think? UserIdTAG: 398594 UserNameTAG: DaveyJC CreateTimeTAG: 2012-12-24T10:45:48Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: it was the same question?..oh..i solved it all over again..:) FirstChildUserIdTAG: 173147 FirstChildUserNameTAG: cruiser_rahit FirstChildCreateTimeTAG: 2012-12-24T11:48:23Z SecondChildTAG: =) SecondChildUserIdTAG: 288637 SecondChildUserNameTAG: gotchi SecondChildCreateTimeTAG: 2012-12-24T12:22:55Z FirstChildTAG: I think if u genuinely worked that problem out, you would have remembered doing it, and would be able to retract the steps.. FirstChildUserIdTAG: 143823 FirstChildUserNameTAG: NathanNadeson FirstChildCreateTimeTAG: 2012-12-24T15:41:37Z SecondChildTAG: I guess it was an early Holiday gift from the Prof. for all that noticed :-) I'm glad I did! It did seem familiar! SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-12-24T16:02:45Z SecondChildTAG: That was my exact thinking! SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-12-24T16:14:26Z SecondChildTAG: True, but at least a change in numbers wouldn't have hurt. Anyway, nice Xmas present, easiest problem in the whole course. Took me one minute xD SecondChildUserIdTAG: 398594 SecondChildUserNameTAG: DaveyJC SecondChildCreateTimeTAG: 2012-12-24T16:31:28Z FirstChildTAG: It goes to show how important it is to complete as much of the homework as possible, as one never knows when it will show up again. It is probably a way for the professor to further reward those that completed all the homework, especially for people like me that struggle on exams and are able to make up the difference by completing as much as the homework as possible. FirstChildUserIdTAG: 132828 FirstChildUserNameTAG: rharris FirstChildCreateTimeTAG: 2012-12-24T16:41:02Z SecondChildTAG: I think the point is that you would have the answer of the question even if you would have not completed the homework. SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-12-24T17:25:52Z SecondChildTAG: I, too, struggle on exams, [rharris][1]. You are not alone. :-) [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/users/132828 SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-12-24T21:39:53Z IndexTAG: 492 TitleTAG: Thanks a lot ! It was Great! From Russia with love... :) It was amazing! It was the great pleasure an great honour to take a knowledge form you. I think, it will be very useful in my work. Thank you very much, Professor, many thanks to Gerald and Piotr, and to other stuff, that work hard and have done a really Cool job. UserIdTAG: 345502 UserNameTAG: ElectroVova CreateTimeTAG: 2012-12-24T10:35:57Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: It was GREAT.Thank you. FirstChildUserIdTAG: 354502 FirstChildUserNameTAG: Harmych FirstChildCreateTimeTAG: 2012-12-24T11:10:26Z FirstChildTAG: Preved medved! :D FirstChildUserIdTAG: 277739 FirstChildUserNameTAG: eclipsevl FirstChildCreateTimeTAG: 2012-12-24T12:35:34Z FirstChildTAG: :D +100500 ! FirstChildUserIdTAG: 345502 FirstChildUserNameTAG: ElectroVova FirstChildCreateTimeTAG: 2012-12-24T12:40:55Z IndexTAG: 493 TitleTAG: Thanks to MiT, Dr. Agrawal I am writing this to thank everybody for making this online quality education revolution successful. Special Thanks to Dr.Agrawal and entire EDX team for imparting knowledge through use of Internet. I actually work for one of the famour Tier 1 company and side by side with the help of EDX i could utilize my free time in excellent manner. Thanks a lot all :-) UserIdTAG: 111320 UserNameTAG: jhalife CreateTimeTAG: 2012-12-24T09:46:48Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 494 TitleTAG: Certificate When we will get our certificates? thank u UserIdTAG: 444204 UserNameTAG: hekan CreateTimeTAG: 2012-12-24T09:05:05Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50d343383c4ada2700000048 FirstChildUserIdTAG: 236136 FirstChildUserNameTAG: Hamid-ch FirstChildCreateTimeTAG: 2012-12-24T10:30:54Z IndexTAG: 495 TitleTAG: Thanks from Indian student Thank you very much to all the staff of edx UserIdTAG: 230108 UserNameTAG: u11ee044 CreateTimeTAG: 2012-12-24T09:04:20Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Merry Christmas! I could complete my exam before my Christmas eve. Thank you EDX. FirstChildUserIdTAG: 232076 FirstChildUserNameTAG: JOMANUSO FirstChildCreateTimeTAG: 2012-12-24T11:07:07Z SecondChildTAG: are you from SVNIT mr u11ee044 SecondChildUserIdTAG: 342135 SecondChildUserNameTAG: vikash902 SecondChildCreateTimeTAG: 2012-12-24T19:31:24Z IndexTAG: 496 TitleTAG: Thank you... A million thanks goes to the staff of edx most especially to Prof. Anant Agarwal for this wonderful course(Circuits and Electronics 6.002x)... and a Happy Holidays too! UserIdTAG: 127527 UserNameTAG: pepi2 CreateTimeTAG: 2012-12-24T00:33:45Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I'll just piggyback onto this post and echo the sentiments to also thank Prof. Agarwal and the staff, the amazing TA's, etc., fellow students and everyone associated with establishing and running 6.002x. An extraordinary experience. FirstChildUserIdTAG: 141108 FirstChildUserNameTAG: rl777 FirstChildCreateTimeTAG: 2012-12-24T01:16:34Z IndexTAG: 497 TitleTAG: Post final, Merry Christmas / Post final, Feliz Navidad Hello Many thanks to all staff of EDX and MIT that has given me the chance to learn, always wanted to learn online and in all institutions charge, and in them they do free work you entertain much, but as there are no dates, there is a strict regulation that follow a certificate to remember all those nights of study, or the fact staff this with attention to students, don't think that you end up discouraged it. I like much this course, your organization, the motivation they gave me, and that I learned new things and emphasized things I already knew. As suggestion, I added support in Spanish or the drafting of some things in English are more accurate (I had difficulties in some homework. videos and the final exam) in order to perform a translation and understand in a way more accurate, as well as also the tutorials that were only at the end put some in Spanish, and I would have been a great help having occupied. I hope that all other courses of EDX are like this, that left me with many expectations with each other. Finally give a Christmas greeting to all my classmates from Chile (Continental America) and Anant Agarwal and Gerald Sussman thanks for your excellent videos. I hope to not close the material to be able to go looking back if I need something. If someone I wish to contact dejo les my personal web site [www.zaybort.com][1] Zaybort PD: Sorry for my bad English [1]: http://www.zaybort.com **EN ESPAÑOL:** Hola Muchas gracias a todo el personal de EDX y el MIT que me ha dado la posibilidad de aprender, siempre quise aprender en linea y en todas las instituciones cobran, y en las que hacen una labor gratis te entretienes mucho, pero como no hay fechas, no hay un reglamento estricto que seguir ni un certificado para recordar todas esas noches de estudio, o el hecho que el personal no este con su atención a los estudiantes, creo que uno termina desmotivándose. Me agrado mucho este curso, su organización, la motivación que me dieron, y que aprendí cosas nuevas y reforcé cosas que ya sabia. Como sugerencia, me gustaría un mayor soporte en español o que la redacción de algunas cosas en ingles sean mas exactas (tuve dificultades en algunas homework. vídeos y en el examen final por esto) para poder efectuar una traducción y entender de manera mas precisa, así como también los tutoriales que ponían solo al final pusieron algunos en español, y me hubiese sido de gran ayuda haberlos ocupado. Espero que todos los demás cursos de EDX sean como este, por que me dejo con muchas expectativas con los otros. Finalmente dar un saludo de navidad a todos mis compañeros de clase desde Chile (América Continental) y las gracias Anant Agarwal y Gerald Sussman por sus excelentes vídeos. Espero que no cierren el material para poder ir repasando por si necesito algo. Si alguien me desea contactar les dejo mi sitio web personal [www.zaybort.com][1] Zaybort [1]: http://www.zaybort.com UserIdTAG: 378294 UserNameTAG: Zaybort CreateTimeTAG: 2012-12-23T19:22:31Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi Zaybort, I Speak Spanish as my native language. The first time that I have taken this Course - last Term-, I thought that it will be nice a support in Spanish. So, that is one of the main resons that I have re-taken this Course. I registered from my own will this Fall in order to help the New Students, focusing also in the Hispanic Community ;). Then, edX offered me to become Community TA, I am really happy that they could give me that opportunity. I hope to improve and to help more and more as far as I can in Spanish and English. I will re-take 6.002x for third time next term in order to help again. Also, if edX would like someday in the future to translate all the material in Spanish I would like to help, put me in the list haha. Take care and my best wish to you, Myriam. > Now in Spanish. Hola Zaybort, Yo hablo Español-Castellano como lenguaje nativo. La primera vez que tuve la gracia de haber cursado este maravilloso Curso - el semestre pasado-, pensé que sería genial si se tuviera un soporte en Español. Es por ello, que esta ha sido una de mis razones principales por la cual me he vuelto a suscribir en este Curso. Por propia voluntad e iniciativa, sin saber de la posibilidad de ser Community TA, me había suscrito en este Otoño con el objetivo de ayudar a los nuevos estudiantes, enfocándome más en la Comunidad Hispana ;). Luego, vino el ofrecimiento de edX para convertirme en Community TA, estoy muy contenta por esta oportunidad que me han dado. Espero mejorar día a día y poder ayudar cada vez más tanto en Español como en Inglés. Tal es así, que he decidido, por tercera vez, registrarme en el 6.002x para colaborar. También, si edX algún día quisiera, en el futuro, traducir todo el material en Español, estaría encantada en ayudarlos, así que ya me pueden poner en la lista de colaboradores jaja. Cuidate mucho y mis mejores deseos para ti, Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-23T21:13:12Z IndexTAG: 498 TitleTAG: Now that you have mastered circuits, ![enter image description here][1]Go build your own portable computer[pi-to-go][2] [1]: http://blog.parts-people.com/wp-content/uploads/2012/12/mobile-rpi-togo-08.jpg [2]: http://blog.parts-people.com/2012/12/20/mobile-raspberry-pi-computer-build-your-own-portable-rpi-to-go/ UserIdTAG: 623210 UserNameTAG: preveen CreateTimeTAG: 2012-12-23T17:17:09Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Or the cheaper, original option (£25 to £70) assuming you have a TV) Great for driving digital and analogue hardware and learning software. http://www.raspberrypi.org/faqs http://www.raspberrypi.org/ FirstChildUserIdTAG: 273287 FirstChildUserNameTAG: johnkh77 FirstChildCreateTimeTAG: 2012-12-23T18:34:58Z IndexTAG: 499 TitleTAG: Special thanks from Pakistani Students please accept my best wishes on behalf of all Pakistani students UserIdTAG: 430030 UserNameTAG: imali CreateTimeTAG: 2012-12-23T16:43:10Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: and what about indian students ? :p FirstChildUserIdTAG: 214085 FirstChildUserNameTAG: shohin FirstChildCreateTimeTAG: 2012-12-23T17:57:10Z FirstChildTAG: Thanks imali for your wishes on behalf of Pakistani students to edx staff. I would like to take this oppotunity to express my best wishes and special thanks to Sir Anand Agarwal and all community TAs, especially Miss Myriam. Merry Xmas and Happy New Year to All Staff and student community. Hope all other Pakistani Students will express their wishes to support imali's statement. Best Regards FirstChildUserIdTAG: 455950 FirstChildUserNameTAG: Wahabbaluch FirstChildCreateTimeTAG: 2012-12-24T06:02:28Z IndexTAG: 500 TitleTAG: Big Happiness & Great gratitude! After finishing the final exam,I could finally take a breath and sigh.Well done to myself! Though the exam is finished, the trip is not over as I have not learned week13 & 14,which I won't miss during the winter holiday.I really want to express my sincere gratefulness to Prof. Anant Agarwal and other course instructors,TAs and all staff inluding RohanNagarkar and Lyla Fischer and our closest friends----community TAs- Myrimit,Pennypacker,JerseyMark ... and my dear classmates who contribute their ideas and efforts to help others! Today when I log on the homepage of edx,what a big surprise!There are so many new courses provided in Spring 2013,the new year!I am so excited and can't wait.We will meet again! Happy Christmas and best wishes to all of you! UserIdTAG: 361137 UserNameTAG: Ericson CreateTimeTAG: 2012-12-23T15:05:45Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Thanks, **Ericson**! Merry Christmas and Happy Holidays to you as well! Perhaps we'll meet again in Spring 2013: I will be retaking 3.091x (I had to drop it mid-way due to being overloaded from Hurricane Sandy damage I had to fix) and taking 6.00x (the programming course) as well, since my C programming skills are rusty, and I want to learn Python. I will also try to push MITx to offer Signals and Systems, **6.003**, (the next logical step course after 6.002x at many universities; it focuses more on signals such as sinusoids, steps, ramps, etc., combining them as well as Laplace transforms and Fourier analysis / the frequency domain as opposed to actual circuit analysis). I may also take one of the ECE courses offered on Coursera. It all depends on my free time (I'll go easy if I start full-time work vs. part-time). Jersey Mark FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-12-23T17:22:31Z SecondChildTAG: Wow,a very big plan! I'm so excited to learn new ECE courses you mentioned, but I may not have enough time to take many courses at the same time.But I will try it :) Wish you good luck! SecondChildUserIdTAG: 361137 SecondChildUserNameTAG: Ericson SecondChildCreateTimeTAG: 2012-12-24T04:52:28Z IndexTAG: 501 TitleTAG: Thank you for the course Thanks a lot for providing such a great course. I am very much indebted to professor Agarwal sir. Thank you very much Myrimit for your suggestions to complete this course successfully. Thank you classmates. Awaiting for new courses, P S Maitrey UserIdTAG: 446691 UserNameTAG: maitrey CreateTimeTAG: 2012-12-23T12:33:46Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 502 TitleTAG: Thank you for the course!! Befor i'll start the final exam i want to thank you all MIT-team for a beautiful course. I've had something similar in the university, 5 years ago or something, but 6.002 is much much more interesting! UserIdTAG: 373498 UserNameTAG: Cheblan CreateTimeTAG: 2012-12-22T22:39:39Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 503 TitleTAG: WOO-HOOOO thank you, thank you, thank youuuuuu.. awesome program, amazing professor,great team(special thanks to Myrimit) wonderful experience n cool course-mates... thank you for making me feel supper smarttttttt MITx :) UserIdTAG: 285945 UserNameTAG: lindalapiso CreateTimeTAG: 2012-12-22T10:13:42Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You are welcome lindalapiso :)! I am so happy that you had a wonderful experience here ! My best wish to you! Take care, Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-22T19:58:07Z IndexTAG: 504 TitleTAG: To EDX and MITX Team Kindly give extra attempts for Q1 & Q2 UserIdTAG: 228440 UserNameTAG: hassanmuhuyuddin CreateTimeTAG: 2012-12-21T16:56:13Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: exactly FirstChildUserIdTAG: 400242 FirstChildUserNameTAG: hassankhan_umt FirstChildCreateTimeTAG: 2012-12-21T17:02:54Z FirstChildTAG: agree FirstChildUserIdTAG: 400128 FirstChildUserNameTAG: RamshaAyub FirstChildCreateTimeTAG: 2012-12-21T17:06:36Z FirstChildTAG: i agree .. please FirstChildUserIdTAG: 342221 FirstChildUserNameTAG: deepkar FirstChildCreateTimeTAG: 2012-12-21T17:03:36Z FirstChildTAG: i am also agree with u plz FirstChildUserIdTAG: 400125 FirstChildUserNameTAG: MoazzamKhan FirstChildCreateTimeTAG: 2012-12-21T17:04:54Z FirstChildTAG: Combined into duplicate thread here: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50d49440bae088230000000b FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-12-21T17:32:00Z IndexTAG: 505 TitleTAG: What after this course ? firstly i want to thank every one participated in making this course in this amazing view then, what do you think what can we do after finishing this course ? what shall be the next step to work in electronics field ? UserIdTAG: 41666 UserNameTAG: AbdoMondy CreateTimeTAG: 2012-12-20T11:24:20Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: signals and systems is required! FirstChildUserIdTAG: 368712 FirstChildUserNameTAG: jmen FirstChildCreateTimeTAG: 2012-12-20T12:02:38Z SecondChildTAG: Yup..I agree SecondChildUserIdTAG: 143823 SecondChildUserNameTAG: NathanNadeson SecondChildCreateTimeTAG: 2012-12-20T13:08:08Z SecondChildTAG: I agree for this. it will help us a lot. SecondChildUserIdTAG: 509592 SecondChildUserNameTAG: chinmaya24aug SecondChildCreateTimeTAG: 2012-12-20T16:04:53Z SecondChildTAG: Yes that would be wonderful.. ... also Analog circuits is an other option SecondChildUserIdTAG: 209930 SecondChildUserNameTAG: saikiraniitr SecondChildCreateTimeTAG: 2012-12-20T18:16:00Z IndexTAG: 506 TitleTAG: Final exam Hi! Before I take the exam I want to be sure which kind of legal helps are there? I am sure about white papers, pen, calculator. What about notes from lectures, lectures as well? Thank you for fast reply. Marek UserIdTAG: 405237 UserNameTAG: MarasK CreateTimeTAG: 2012-12-19T20:28:00Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: yes i wana to do this exame, with the same rigor as in the campus FirstChildUserIdTAG: 236136 FirstChildUserNameTAG: Hamid-ch FirstChildCreateTimeTAG: 2012-12-19T20:50:08Z FirstChildTAG: Hi MarasK, You can use all the material that you want. If your read the syllabus it says : Midterm and Final Exam [read here][1] > A significant portion of your final grade in 6.002x will be determined > by the midterm and the final exam. We will announce with the exams the > deadline by which you must complete them. Exercises, homework and labs > are critical to learning the material and for doing well on the exams. > It is very likely that one or more of the exercise, homework or lab > problems will appear in each of the exams. 2Once you view an exam, you > must work on your own till you have submitted all your work, and do > not discuss the exam until the deadline for exam submissions is past. > **While the exams will be open book, we encourage you to create a couple > of sheets of notes for each exam. These notes will not only help you > prepare, but they will also serve as a convenient reference during the > exam. You may use a calculator if needed.** You are not allowed to post > answers to exam problems. Collaboration of any form is strictly > forbidden in the midterm and the final exams. Take care, and Good luck in the Final Exam! :) Myriam [1]: https://www.edx.org/static/content-mit-6002x/handouts/syllabus.a477535058a1.pdf FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-19T21:35:44Z SecondChildTAG: Thank you for the reply - I checked the examplary exam which you have shown on the Courseware and there was a phrase "closed book". I just wanted to be sure before I start the exam. I don't want to break the MITx rules after all those great opportunities to learn! :) [PL]Wesołych Świąt/[EN]Merry Christmas Marek SecondChildUserIdTAG: 405237 SecondChildUserNameTAG: MarasK SecondChildCreateTimeTAG: 2012-12-22T14:34:27Z IndexTAG: 507 TitleTAG: Edx Mechanical Engineering courses Hi edX, I am hoping to see Mechanical engineering courses in the future,kindly respond on this prospect. UserIdTAG: 536922 UserNameTAG: arjshar CreateTimeTAG: 2012-12-13T06:54:17Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 0 IndexTAG: 508 TitleTAG: How to analyze (or develope) circuits of real devices? I have a problem with analyzing circuits of real-live devices. For example, I want analyze, how works my motion-detector (the circuit is below). I understand some pieces of circuit, but don't fully understand how it works complete! 1. Too many capicators in this circuit is big problem for me... 2.D1, D2, D3, D5, D10 - why we need them...? 3. I don't understand , how works VT1 and VT2 in this circuit... 4. How create (develope) circuits like this? It's really awful task for me now :))) ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13553389621343688.gif UserIdTAG: 104522 UserNameTAG: IgorNovice CreateTimeTAG: 2012-12-12T19:23:55Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: This device doing control for light ( night light).Only when luminosity is low, PIR will control load(light bulb) by relay. This device is pretty simple.You may analize it without any problem.If you have some troubles just try to divide entire schematic by few parts. VT1 is power switch for K1 relay.VT2 is doing more nice work - it block control sygnal from OA pin1 when luminosity is enough. Do you want an interesting question?What is D5 and C5 purpose here :) I see some "feature" here too :) Ok There is PIR sensor, two stage PIR-sygnal amplifier ( first stage has special frequency responce), sticky comparator :), "timer circuit", sticky bridge :D with light sensor and power relay control net (VT1 and VT2). Hmm..thats all PS If someone interested to learn more, [here][1] is link to the D203 datasheet. Good luck! [1]: http://blog.digit-parts.com/pdf/d203s.pdf FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-12-12T22:54:45Z FirstChildTAG: Hi Igor I think that the diode D10 is in the circuit to protect VT1. K1 is a relay coil and hence an inductive load. When VT1 switches off, a large change in current is implied and which in turn causes a huge reverse voltage across the coil. The diode is then forward biased and helps the current decay through the inductor. Sometimes its called a flywheel diode. You can bet your life that if you ever see a relay in a circuit, you will probably find a 1N400* across the coil. Some power MOSFETS have a diode across the drain and source. I think (maybe wrong) that its for this purpose.Switching inductive loads.. Hope that helps a little Smallblindchris FirstChildUserIdTAG: 259719 FirstChildUserNameTAG: Smallblindchris FirstChildCreateTimeTAG: 2012-12-12T22:44:40Z SecondChildTAG: Youre right about D10, this diode for protect VT1 SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-12T23:08:38Z SecondChildTAG: Thanks for tips, but I want know about big picture of circuit, how to understand? SecondChildUserIdTAG: 67648 SecondChildUserNameTAG: mgm SecondChildCreateTimeTAG: 2012-12-13T09:11:30Z SecondChildTAG: simply break entire schematic by few functional modules SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-13T10:31:44Z FirstChildTAG: Essentially this unit works as a passive Infrared (IR) detector. Generally they are set to detect IR energy between 9-10 micrometers. The detector uses a "CDS" cell(photo-resistor), which is set up to measure a *change* in IR energy, using the "sens" potentiometer,(Variable resistor) for calibration. You will see another potentiometer in the lower left labeled "Daylight", you use this to fine tune the device so that the device becomes active from dusk till dawn. The LM324N is a quad opamp. In section "1" of the opamp, you can see part of the circuit that controls the time that the light remains on once activated. If you look at the upper left of the diagram you will see the 500w Halogen light bulb labeled "EL1", which gets switched using the LS-T73 relay, connecting the light bulb directly to the 220V line. (Represented by the two little arrows "~220B") Diodes D6,D7,D8,D9 form a bridge rectifier which supplies DC power to the relay and other parts of the circuit, @+25v and +8v, using voltage regulators like the 78L08 labeled "DA1". Have fun FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-12-13T14:23:48Z SecondChildTAG: > Essentially this unit works as a passive Infrared (IR) detector. Generally they are set to detect IR energy between 9-10 micrometers. The detector uses a "CDS" cell(photo-resistor), which is set up to measure a change in IR energy This device uses TWO detectors: PIR and CDS with different purposes. While PIR do receive IR radiation from objects :), CDS just allow to make functionality of entire device more flexible. BTW link to the PIR datasheet in my post above. Good luck! SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-13T15:33:01Z FirstChildTAG: Thanks for help !!! I understand, how works parts of this circuit. "Divide and conquer" method win!)) What function of this individual elements: C6, C7, C8, C9, C10, C11, C14, C5, D1, D2, D3, D5??? FirstChildUserIdTAG: 104522 FirstChildUserNameTAG: IgorNovice FirstChildCreateTimeTAG: 2012-12-13T16:36:35Z IndexTAG: 509 TitleTAG: Really appreciate for your lecture Thank you very muck for your efforts! UserIdTAG: 345399 UserNameTAG: mybrand CreateTimeTAG: 2012-12-12T01:21:54Z VoteTAG: 5 CoursewareTAG: Week 14 / S28V13_Voltage_Drop_across_the_Parasitic_Inductor CommentableIdTAG: 6002x_S28V13_Voltage_Drop_across_the_Parasitic_Inductor NumberOfReplyTAG: 0 IndexTAG: 510 TitleTAG: A guide to understanding H12P2 At the opamp inputs, $V^+$ is given by a voltage divider, where one end is at $V_{in}$ and the other is at $V_z$, and $V^-$ is given by a voltage divider, where one end is at $V_{out}$ and the other end is at $Gnd = 0V$. The resistors in the first divider are $R_0$ and $R_z$; they are $R_1$ and $R_2$ in the other divider. My problem stated the max and min values for $V_{in}$ are $20V$ and $10V$. Your voltages may be different. Set $V_{in}$ to $20V$, and decide what maximum $V_{out}$ you wish to see with this input voltage. The problem states the maximum this can be is $5.1V$ in my problem statement. I chose $5.01V$ here instead. Your value may be different. Set $V_{in}$ to $10V$, and decide what minimum $V_{out}$ you wish to see with this input voltage. The problem states the minimum this can be is $4.9V$ in my problem statement. I chose $4.99V$ instead. Again, your value may be different. I figured these tighter tolerances might make the solution easier to find without falling foul (outside) of the problem's boundary conditions. Use the fact that the opamp is ideal to equate $V^+ = V^-$ at all times. This will give you one equation for $V_{in}=20V$ and one equation for $V_{in}= 10V$. Again, your voltages may be different. So now you have two equations involving $R_0$. These can be solved for a unique $R_0$ that does not depend upon $R_1$ or $R_2$. Solve by dividing one equation by the other, then rearrange to get $R_0$. This division removes the $R_1$ and $R_2$ terms. --- Now that you know $R_0$, go back to your two equations and substitute for $R_0$ and solve for $R_1$ as a function (multiple) of $R_2$. Now you know $R_0$, $R_1$, and $R_2$ (after picking a value for $R_1$ or $R_2$ and getting the other from the ratio you just calculated). --- Now to solve the efficiency, or likely prove the lack of it. The output current is the current $i$ flowing through the diode that is part of our model for the BJT. This is equal to the load current through $R_L$ plus the current through the divider network comprised of $R_1$ and $R_2$. The actual output power is $V_{out}$ times the current through $R_L$. But you still have to find the value for $i$ also. This is because the current through the BJT is $K * i$. The input current $i_{in}$ provided by the source is this $K*i$ current, plus any current through $R_0$. The efficiency = $ \frac{V_{out} * i_L}{V_{in} * i_{in}}$. Calculate this efficiency at the boundaries of our range of input voltages, and see what the minimum efficiency is. Fortunately the lesser of these two is accepted by the grader and we don't have to look for a minimum somewhere in between. Don't forget to use the appropriate $V_{out}$ for each $V_{in}$. --- When you are done, think about what you could do to improve the efficiency of your design. i.e. pick the optimal $R_1$ and $R_2$ values. Are they unique? Does the efficiency depend upon the value of $R_0$ you choose? --- P.S. the first two parts (which I have skipped over) should be easy for you to solve now. You don't need to do them first, and I recommend against it, so that you don't get confused by the special circumstances of those problems. UserIdTAG: 9673 UserNameTAG: xp42 CreateTimeTAG: 2012-12-10T08:38:24Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: I really liked this problem. Prof. Sussman should write a question bank (If he already has I'd love to buy it) either as a book or online with green ticks like over here. FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-12-10T09:20:22Z FirstChildTAG: Nice job, XP42. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-12-11T15:17:15Z FirstChildTAG: I did the efficiency calculation just like the way XP42 had explained and that got a green tick. But now, when I look at the explanation given, Pin consists of two parts: Pin,1 and Pin,2. It looks like XP42's Pin is same as Pin,2. Can somebody explain Pin,1? And how did that (1-K) sneak in? Looks like we cannot ignore it. It so happens that in this problem, Pin,1 seems to be very small because of huge K. But if K is small, then Pin,1 may also be significant. FirstChildUserIdTAG: 232157 FirstChildUserNameTAG: GayatriTR FirstChildCreateTimeTAG: 2012-12-12T09:09:43Z IndexTAG: 511 TitleTAG: H12P3 - Midband Frequency I would want to know more about midband frequency and How can I calculate it to the last question of H12P3? UserIdTAG: 366669 UserNameTAG: pedroramus CreateTimeTAG: 2012-12-08T20:44:31Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 7 FirstChildTAG: Thank you guys. I got it. It's the sign. Since it's an inverting amplifier, we need to take care of the gain. We can assume the capacitor being non existent.. FirstChildUserIdTAG: 8897 FirstChildUserNameTAG: sam1202 FirstChildCreateTimeTAG: 2012-12-08T23:51:52Z FirstChildTAG: I'm also stuck in the same question but I guess midband frequency is between 13 Hz and 18 KHz. I tried to do that though no luck so far. I believe vout/vin should be 1 or near to 1 at midband frequency . I still can't figure it out . FirstChildUserIdTAG: 8897 FirstChildUserNameTAG: sam1202 FirstChildCreateTimeTAG: 2012-12-08T22:13:15Z FirstChildTAG: Also can't find the answer here. I'm puzzled. I tried the obvious approach (using the same formula as in circuit 1, since the configuration is the same), first with only resistances, then the exact approach, using impedances and the mid band frequency 1khz (and some painful algebra). Both answers were marked as wrong despite being very similar. I checked the other discussion threads but the help out there is cryptic at best. FirstChildUserIdTAG: 351871 FirstChildUserNameTAG: Lamarque FirstChildCreateTimeTAG: 2012-12-08T22:17:55Z SecondChildTAG: Well I guess I was really tired. The opamp configuration in the second circuit is NOT the same as in the first. The answer follows easily with that in mind. SecondChildUserIdTAG: 351871 SecondChildUserNameTAG: Lamarque SecondChildCreateTimeTAG: 2012-12-09T07:36:57Z FirstChildTAG: Write out the transfer function. You should be able to identify a low pass part and a high pass part. At mid band frequencies both of those terms should be approximately one. The remaining term, which depends on the resistances only is the gain. Be careful about the sign! FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-12-08T22:34:48Z FirstChildTAG: Remember that you have one of the capacitors limiting the low frequencies (below 16Hz) and the other one limiting the very high frequencies (above 19kHz). When they say midband frequency, they mean that both capacitors have the same kind of behavior - it will be either short or open. I'll won't explicitly say which behavior it is but it can be easily (and even intuitively) figured out. As soon as you find out how both capacitors are behaving in the so called midband frequencies, it should be pretty straight forward to figure out the gain. Remember that is is an INVERTING amplifier, meaning that there is a phase inversion of 180 degrees - and this need to be accounted for in your answer. The analysis is extremely simplistic and very intuitive. Just think about how the capacitors would be behaving in frequencies far away from both the low and high boundaries. FirstChildUserIdTAG: 342148 FirstChildUserNameTAG: gustavopereira FirstChildCreateTimeTAG: 2012-12-08T22:43:39Z FirstChildTAG: Thanks, friends FirstChildUserIdTAG: 366669 FirstChildUserNameTAG: pedroramus FirstChildCreateTimeTAG: 2012-12-09T12:58:28Z FirstChildTAG: Hi pedroramus, Can I help you? You have two alternatives, one is mathematical. The other one is with sandbox. Take a look at [here][1] Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50c3b82f1f09202b0000002a FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-09T04:20:24Z SecondChildTAG: Thanks, I got green. SecondChildUserIdTAG: 366669 SecondChildUserNameTAG: pedroramus SecondChildCreateTimeTAG: 2012-12-09T12:57:39Z IndexTAG: 512 TitleTAG: AhA! It is clear for me - definitely the virtual short method for op amps IS AN AHA! Moment. UserIdTAG: 385692 UserNameTAG: tpfslima CreateTimeTAG: 2012-12-04T21:38:26Z VoteTAG: 5 CoursewareTAG: Week 12 / S23V22 Inverting amplifier input resistance using virtual short method CommentableIdTAG: 6002x_S23V22_Inverting_amplifier_input_resistance_using_virtual_short_method NumberOfReplyTAG: 2 FirstChildTAG: Count another Yea vote for me. Aha! FirstChildUserIdTAG: 428560 FirstChildUserNameTAG: MikeDayton FirstChildCreateTimeTAG: 2012-12-08T04:57:09Z SecondChildTAG: yes...surely virtual short metho is an Aha!!!!!moment...no doubt about it...otherwise analyzing the opamp circuits will be complex SecondChildUserIdTAG: 249575 SecondChildUserNameTAG: badimalamadhu SecondChildCreateTimeTAG: 2012-12-11T05:15:56Z FirstChildTAG: Definitely Agree!! Working in electronics for 30 years with a much less rigorous understanding than I now have (Thanks so much Professor), the virtual short was THE way to understand OpAmp circuits. FirstChildUserIdTAG: 369519 FirstChildUserNameTAG: perpstud FirstChildCreateTimeTAG: 2012-12-13T04:36:15Z IndexTAG: 513 TitleTAG: Negative Feedback Path Through Capacitor Hi, I was just wondering how there was a negative feedback path through the capacitor. For negative feedback a portion of the output has to be redirected into the negative terminal of the Op-Amp. Doesn't having a capacitor block that feedback path? I would be much obliged if someone could help clear my doubt. Thank you in advance. UserIdTAG: 85490 UserNameTAG: yashsavani CreateTimeTAG: 2012-12-04T17:35:54Z VoteTAG: 5 CoursewareTAG: Week 12 / S24V6 Op amp integrator CommentableIdTAG: 6002x_S24V6_Op_amp_integrator NumberOfReplyTAG: 4 FirstChildTAG: In reality, most of the time an integrator is implemented with a resistor, with a very large resistance (say 1M to 10M ohm) parallel to the C, so you'll get a large RC-time, that discharges the C very slowly, but also creates a dc-feedback/connection. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-12-04T19:37:05Z SecondChildTAG: So where did the prof. put that resistor? It seems to me that he has no 'proper negative feedback' as he forgot to put the resistor in. Or am I wrong here? SecondChildUserIdTAG: 114881 SecondChildUserNameTAG: xvink SecondChildCreateTimeTAG: 2012-12-07T19:29:45Z SecondChildTAG: He didn't put it there, because he gives a theoretical approach. The idea is, that the current through the capacitor is creating the voltage over the capacitor, because it's being charged. That causes the feedback. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-12-08T18:53:02Z FirstChildTAG: This is something I'd like to have a clarification to also. I guess one possible explanation might be that since the Op-amp has a very large gain, it's action can be thought as something of a step response for which the cap looks like short. But I'm definitely not sure if this idea applies here. FirstChildUserIdTAG: 105523 FirstChildUserNameTAG: eero FirstChildCreateTimeTAG: 2012-12-07T14:30:49Z SecondChildTAG: The minus input of the OA wants to stay at the same potential as the plus input. Because the input current must flow somehow through the C, the OA is continuously compensating the output voltage, trying to create a current through the C and so causing the voltage over the C. In fact the AC-feedback resistance, is the impedance of the C. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-12-08T20:29:59Z FirstChildTAG: same question here.. FirstChildUserIdTAG: 136959 FirstChildUserNameTAG: christerpher FirstChildCreateTimeTAG: 2012-12-09T06:57:23Z FirstChildTAG: Another question: what is the reason to use an inverting op-amp config instead of non-inverting? if non-inv there'd be no minus and the need for a reverting sub-circuit... (unless i remember incorrectly all VSM assumptions for op-am with feedback apply in both configs). FirstChildUserIdTAG: 184153 FirstChildUserNameTAG: tthngn FirstChildCreateTimeTAG: 2012-12-11T00:49:03Z IndexTAG: 514 TitleTAG: diff op amp 5v both, output =0? differential op amp, as input is v+=5v and v-=5v, isnt the output =0? steady state, 0v instead of 10v that is? so vo is being pulled up 0->+2 instead, then vi becomes +1 ... etc? somebody pls correct me. UserIdTAG: 331483 UserNameTAG: Doru CreateTimeTAG: 2012-12-04T13:32:51Z VoteTAG: 5 CoursewareTAG: Week 12 / S23V11 Negative feedback CommentableIdTAG: 6002x_S23V11_Negative_feedback NumberOfReplyTAG: 1 FirstChildTAG: if you will tie both inputs toghether $V_O$ will be equal zero. This is one of the most important OA features : to reject sinphase voltage. FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-12-04T13:47:11Z SecondChildTAG: i dont tie them together, i just look at the picture and both have +5v labels on them, so the function is A * zero_difference. What am i missing pls? SecondChildUserIdTAG: 331483 SecondChildUserNameTAG: Doru SecondChildCreateTimeTAG: 2012-12-04T13:58:42Z SecondChildTAG: ok, your question is about feedback. You may imagine this problem by the different way. Let $v^+$ has some potential.Next, you should brake connection between Output pin and Feedback resistor.Re-connect this free resistors pin to the second (new) voltage source V2.Now, at which V2 value $V_O$ will be zero? This value will be equal $V_O$ for normal connection. May be Im correct here :)) SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-04T14:12:52Z SecondChildTAG: The way i see it its a VCCS with a feedback that makes the input (difference) go to zero. I cant get the A*zero=10v there. I think its coz R=R1=R2 so maybe that shouldnt happen? I cant figure this yet, so will follow the rest sequences maybe things fall into place maybe ill get that later. SecondChildUserIdTAG: 331483 SecondChildUserNameTAG: Doru SecondChildCreateTimeTAG: 2012-12-04T14:35:44Z SecondChildTAG: I think i am starting to get it in s23v13, "with resistor feedback v+ tends to equal v-" coz A being large then the difference is equal number of magnitudes lower. SecondChildUserIdTAG: 331483 SecondChildUserNameTAG: Doru SecondChildCreateTimeTAG: 2012-12-04T14:49:13Z SecondChildTAG: I'm stuck with this too. I'll watch it a little bit further. SecondChildUserIdTAG: 290028 SecondChildUserNameTAG: Fjuca SecondChildCreateTimeTAG: 2012-12-09T22:59:18Z IndexTAG: 515 TitleTAG: What is this small creature? I keep seeing this tiny creature in the lecture slides. What is it? Does it have a name? Is it the 6.002x mascot? ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13544480191343612.png UserIdTAG: 148542 UserNameTAG: planetscape CreateTimeTAG: 2012-12-02T11:34:01Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: the name is roadkill! FirstChildUserIdTAG: 11075 FirstChildUserNameTAG: roncada FirstChildCreateTimeTAG: 2012-12-02T20:27:49Z FirstChildTAG: i'd rather go with "brainworm", these guys tend to get on surface when the surface is watered :) FirstChildUserIdTAG: 331483 FirstChildUserNameTAG: Doru FirstChildCreateTimeTAG: 2012-12-02T20:40:11Z SecondChildTAG: Water on the Brain? LOL SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-12-03T09:15:26Z SecondChildTAG: doing the course might look something like exercising it, to make it grow, so watering was a metaphore :) or something - how its called in english :) SecondChildUserIdTAG: 331483 SecondChildUserNameTAG: Doru SecondChildCreateTimeTAG: 2012-12-04T14:03:13Z SecondChildTAG: :-) It's a GOOD metaphor! SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-12-09T23:16:19Z IndexTAG: 516 TitleTAG: Is Week-13 also included for Final Exam? Staff or one of the TAs, please answer: Are week-13 contents also part of final exam syllabus? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-11-28T14:55:47Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: The final will cover weeks 1 - 12. FirstChildUserIdTAG: 391929 FirstChildUserNameTAG: RohanNagarkar FirstChildCreateTimeTAG: 2012-11-28T15:40:40Z SecondChildTAG: Then, why week-12 Home work is not released? SecondChildUserIdTAG: 114473 SecondChildUserNameTAG: Veerabasanagouda SecondChildCreateTimeTAG: 2012-12-16T09:45:19Z IndexTAG: 517 TitleTAG: S21E3: there must be some error, I think I agree with part 1, but: When w -> 0, if we don't neglect j*w*L in the numerator, why do we neglect it in the denominator? Shouldn't it be (j*w*L)/(j*w*L+R) ? For w -> inf, shouldn't it be 1/(-w^2*R*L*C) ? It's a mistery! Thank you! UserIdTAG: 398594 UserNameTAG: DaveyJC CreateTimeTAG: 2012-11-26T22:43:27Z VoteTAG: 5 CoursewareTAG: Week 11 / S21E3: Thevenin Tank CommentableIdTAG: 6002x_Thevenin_Tank NumberOfReplyTAG: 4 FirstChildTAG: in math, they proof that: if lim(x->a) f(x) = F and lim(x->a) g(x) = G <> 0 then: lim(x->a) f(x)/g(x) = F/G Here: a = 0, f(w) = j*w*L and g(w) = R+j*w*L-w^2*L*R*C lim(w->0) g = R ==> lim(w->0) f/g = (lim(w->0) j*w*L) / R = lim(w->0) j*w*L/R (=0) FirstChildUserIdTAG: 388524 FirstChildUserNameTAG: dasbinich FirstChildCreateTimeTAG: 2012-11-27T02:08:03Z SecondChildTAG: Thanks dasbinich, but I think we are not using plain limits here. Because then, the answer would be 0 (in the limit, extreme). We are not asked for the limit, but for the behaviour near w=0, which I think is not the same thing. We are asked for an approximation form of the function when it approaches w=0. In the previous exercises, we had done it the following way: - For w nearing 0, by neglecting higher order terms (squared ones) which would be one order of magnitude smaller than lower order terms. - For w nearing infinity, by neglecting lower order terms, which would be one order of magnitude smaller than higher order ones (squared ones). I can't explain why in this case the solution is different. Besides, I can't see the logic applied to get to the solutions. Can anyone help me? SecondChildUserIdTAG: 398594 SecondChildUserNameTAG: DaveyJC SecondChildCreateTimeTAG: 2012-11-27T08:16:17Z SecondChildTAG: There is none.For w->0, (j*L*w)/R would be even smaller, because is w already small divided by R, which is even smaller , not even speaking of j. And they are comparing imaginary parts of complex numbers, a silly way to make a point. Calculus and limits was invented to study this kind of situations. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-11-28T14:00:11Z FirstChildTAG: Thanks dasbinich, but I think we are not using plain limits here. Because then, the answer would be 0 (in the limit, extreme). We are not asked for the limit, but for the behavior near w=0, which I think is not the same thing. We are asked for an approximation form of the function when it approaches w=0. In the previous exercises, we had done it the following way: * For w nearing 0, by neglecting higher order terms (squared ones) which would be one order of magnitude smaller than lower order terms. * For w nearing infinity, by neglecting lower order terms, which would be one order of magnitude smaller than higher order ones (squared ones). I can't explain why in this case the solution is different. Besides, I can't see the logic applied to get to the solutions. Can anyone help me? FirstChildUserIdTAG: 398594 FirstChildUserNameTAG: DaveyJC FirstChildCreateTimeTAG: 2012-11-27T08:17:14Z FirstChildTAG: You neglect a term when there is another term to compare with it: so for the denominator when w->0 you neglect R*L*C*w^2 and j*w*L with respect to R but at the numerator there is nothing to compare with j*w*L and so you leave it to get j*w*L/R; that explains why for low frequencies the magnitude plot increases linearly with w; the same for w->inf: j*w*L at the numerator and -w^2*R*L*C at the denominator to get j/(-w*R*C); that explains why for high frequencies the magnitude plot decreases like an hyperbola (1/w). FirstChildUserIdTAG: 376877 FirstChildUserNameTAG: AndBre FirstChildCreateTimeTAG: 2012-11-27T16:36:53Z SecondChildTAG: You made my day, man! Now I understand the logic. Much thanks! :) SecondChildUserIdTAG: 398594 SecondChildUserNameTAG: DaveyJC SecondChildCreateTimeTAG: 2012-11-27T17:53:55Z SecondChildTAG: for w->inf it's become much obvious if numerator and denominator divide by j*w*L - you will get: 1/(1+R/(j*w*L)+j*w*R*C) so we can neglect 1 because it's too small in compare to j*w*R*C and neglect R/(j*w*L) because it ->0 and we come with: 1/(j*w*R*C) SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-11-27T19:35:24Z SecondChildTAG: thanks YakovO,I got it. I first got: for w->0,the ratio is 1/(1-j*R/(j*w)), for w->inf,the ratio is 1/(1+j*w*R*C), For both of them,I did not consider the 1. Now I find where I was wrong. SecondChildUserIdTAG: 361137 SecondChildUserNameTAG: Ericson SecondChildCreateTimeTAG: 2012-11-28T12:16:03Z FirstChildTAG: For w->inf or w-> 0, you calculate the limit and that does not include w . This is just a silly way to make a point, because they don't want you to use the analytical way . That would involve limits calculus and not everybody is familiar with that. In my opinion AndBre 's algorithm seems to work , but he is forgetting that he explains a modulus behaviour looking at an expression which contains j. FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-11-28T13:49:10Z SecondChildTAG: I'm not forgetting anything: Vo/Vi at low frequencies approximates to jwL/R and at high frequencies to 1/jwRC; now, normally you plot imaginary transfer functions using magnitude and phase: magnitudes of the above functions are wL/R and 1/wRC, while their phases are pi/2 and -pi/2. Simply I consider the magnitude plot more interesting. SecondChildUserIdTAG: 376877 SecondChildUserNameTAG: AndBre SecondChildCreateTimeTAG: 2012-11-28T15:02:58Z SecondChildTAG: Ok. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-12-06T09:01:22Z IndexTAG: 518 TitleTAG: Spring Classes Hello All, Does anyone know where to find the courses they will be offering in the spring? Or at the very least does anyone know what time they will be available to view? UserIdTAG: 366427 UserNameTAG: agaidis1 CreateTimeTAG: 2012-11-26T13:00:22Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: If you log out and visit the edX homepage, you will see the nine current offerings. I would imagine most of these and more will be available in the spring. I would also expect to see courses from the University of Texas this spring or next fall, as they have recently joined the edX consortium. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-11-27T14:24:45Z FirstChildTAG: Berkley's CS70 will apparently be included at some point, and given the fact that it was announced back in august, I would assume that it could easily be present this spring Source: [here][1] [1]: http://forums.udacity.com/st101/questions/11834/edx-to-offer-discrete-mathematics-and-probability-theory FirstChildUserIdTAG: 26931 FirstChildUserNameTAG: Csscade FirstChildCreateTimeTAG: 2012-11-28T03:45:05Z SecondChildTAG: Interesting, thanks. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-11-29T00:03:57Z IndexTAG: 519 TitleTAG: STAFF: Pruning the Wiki - guidance requested Dear people in charge of the 6.002x Wiki, I'd like to ask for some advice regarding the netiquette of editing the Wiki. Aside from blank pages with no content, one also runs across pages that don't belong, pages that are almost-but-not-quite-totally useless, and pages that are in need of a redirect or a "what links here" button. For instance: [This page][1] is useless except for the fact that it shows how to embed a circuit in the Wiki. Is there a dedicated page that explains how to embed circuits in the wiki? **Should this be moved there?** The page [Bioelectronics][2] is a straight-up copy-paste of [the Wikipedia page][3] and [this Columbia University page][4]. **Should pages like this one be deleted?** The page [Digital Logic Design][5] is completely blank, and has been since its inception 6 weeks ago, but seems like it belongs in the 6.002x Wiki. **Should it be deleted?** The page [Globally Controlled Advance Ambulance System][6] is no doubt fascinating but has nothing whatsoever to do with 6.002x. Either a student of some other edX course misplaced it, or it's another copy-paste off the web. **Should it be deleted?** There are currently 3 pages titled "Myrimit's Guide to 6.002x". Two of them ([this][7] and [this][8]) are blank and do not have so much as a link to [the actual guide][9]. **Can you set up a redirect? Or should they be deleted?** Finally there are pages like the one titled [wilson][10], which contain one useful tidbit (in this case a link to a discussion of KVL) but nothing else. **For pages such as these, is it appropriate to move their content into the appropriate article (e.g. in the "See also:" section) and then remove them?** Thanks for reading. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/hope-2-nurture/ [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/bioelectronics/ [3]: http://en.wikipedia.org/wiki/Bioelectronics [4]: http://www.bioee.ee.columbia.edu/research/ [5]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/CS302/ [6]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/globally-controlled-advance-ambulance-system/ [7]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/myrimits-guide-6002x/ [8]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/myrimits-guide/ [9]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/MyrimitsGuideTo6002x/ [10]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/wilson/ UserIdTAG: 96595 UserNameTAG: Eris CreateTimeTAG: 2012-11-15T23:22:38Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 0 IndexTAG: 520 TitleTAG: Interactive Graph Interactive graph here: https://www.desmos.com/calculator/nziwc2olh9 UserIdTAG: 366083 UserNameTAG: smath CreateTimeTAG: 2012-11-11T22:11:41Z VoteTAG: 5 CoursewareTAG: Week 10 / Series RLC frequency response plot CommentableIdTAG: 6002x_Series_RLC_frequency_response_plot NumberOfReplyTAG: 1 FirstChildTAG: smath, very nice. Thanks for that. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-11-12T01:36:32Z IndexTAG: 521 TitleTAG: "t" missing (At 8:00): "t" is missing in both exponents of VH. UserIdTAG: 282828 UserNameTAG: Chibolator CreateTimeTAG: 2012-11-11T17:36:18Z VoteTAG: 5 CoursewareTAG: Week 9 / S18V10 Homogeneous Solution Pt 2 CommentableIdTAG: 6002x_S18V10_Homogeneous_Solution_pt2 NumberOfReplyTAG: 0 IndexTAG: 522 TitleTAG: Mistake in currents for t=infty. The current through the inductance at t=infty is NOT 0, it will be i_1=10V/R_2... UserIdTAG: 341020 UserNameTAG: franjescribano CreateTimeTAG: 2012-11-10T07:46:27Z VoteTAG: 5 CoursewareTAG: Week 9 / A&L Problem 12.3 CommentableIdTAG: 6002x_AL_P_12_3 NumberOfReplyTAG: 0 IndexTAG: 523 TitleTAG: HW 8 help required by Myrimit!!! It's look quite difficult for me to solve Home work 8. I have also not been able to find the topic and formulae from text book . Please i need some hints from Myrimit . Regards UserIdTAG: 145239 UserNameTAG: Maheenjd CreateTimeTAG: 2012-11-09T10:35:51Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hi Maheenjd, Can I help you? In which part are you lost? Ok, I will Posts Hints of HW8 at night when I back to home :). Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-09T12:55:18Z SecondChildTAG: We are awaiting you, Myriam! SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-11-09T18:43:24Z SecondChildTAG: Facing similar problem, awaiting your support, Thanks in advance. SecondChildUserIdTAG: 455950 SecondChildUserNameTAG: Wahabbaluch SecondChildCreateTimeTAG: 2012-11-10T05:56:25Z SecondChildTAG: As I have promised to you Maheenjd, here you can read the [h8p1, h8p2, h8p3 Hints][1] I hope this can help you, Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/509e68568e407d1f0000001b SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-10T17:00:02Z SecondChildTAG: Thanks Myriam :) God Bless you !!! SecondChildUserIdTAG: 145239 SecondChildUserNameTAG: Maheenjd SecondChildCreateTimeTAG: 2012-11-11T09:33:17Z SecondChildTAG: You are welcome Maheenjd ! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-11T19:17:03Z FirstChildTAG: You can review the video sequence S15V18 about [initial conditions and superposition][1], basically the impulse source sets an initial condition for the inductor AND the capacitor and then you can apply superposition treating the initial conditions as independent sources; notice that after t=0 vin is zero so you have a short circuit in parallel with the two branches (R-L and R-C) and this lets you solve the two branches independently (the hint says that they are decoupled). HTH. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_8/Ramps_Steps_and_Impulses/ FirstChildUserIdTAG: 376877 FirstChildUserNameTAG: AndBre FirstChildCreateTimeTAG: 2012-11-09T11:54:34Z IndexTAG: 524 TitleTAG: wrong formula for angle angle(a + jb) = arctan(b/a) [page 948 of textbook][1] [1]: https://s3.amazonaws.com/mitxstatic/agarwal_lang/p972.png UserIdTAG: 137421 UserNameTAG: i026e CreateTimeTAG: 2012-11-08T12:14:09Z VoteTAG: 5 CoursewareTAG: Week 10 / Complex Numbers CommentableIdTAG: 6002x_Complex_numbers NumberOfReplyTAG: 2 FirstChildTAG: Exactly. tan^-1 is is the inverse tangent operation, which is the cotangent. FirstChildUserIdTAG: 361823 FirstChildUserNameTAG: EliasOak FirstChildCreateTimeTAG: 2012-11-08T16:18:01Z FirstChildTAG: You may not like the notation, but it is nevertheless standard. http://en.wikipedia.org/wiki/Inverse_tangent FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-08T16:44:57Z SecondChildTAG: No, it's not. If you read carefully at the link you posted, you will notice that arctan is a ONE argument function (so you must use it as arctan(b/a)) whilst atan2 is a TWO arguments function (so you must use it as atan2(b,a)). SecondChildUserIdTAG: 376877 SecondChildUserNameTAG: AndBre SecondChildCreateTimeTAG: 2012-11-18T11:58:05Z IndexTAG: 525 TitleTAG: A toast to everyone who just passed the class! Congratulations! If you have done everything so far, you should have 60% and just barely a passing "C" now, before even doing the final or any more homework. And the final will just be gravy. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-11-08T04:06:18Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I also Have Yet 43% with incomplete H10. :( got 45% in MID....:( > **Regards:** asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-11-08T16:21:48Z SecondChildTAG: Don't feel bad; not everyone aced the MidTerm. I didn't. Just hang in there! You are still learning, and you always have the option to re-take the course and try for a higher grade. :-) SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-11-09T00:22:32Z FirstChildTAG: Congratulations! I only have 49 percent right now, unfortunately I haven't completed H8. FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-11-08T10:09:44Z IndexTAG: 526 TitleTAG: Inductor vs Capacitor Hi, This question has no relation with any of the homework or the midterm whatsoever. I'm just curious: I always noticed that in the electronics world capacitors are used more often than inductors. Why is that? For example, why are there no memories built with inductors? If they are duals to each other, couldn't you design the same circuit that uses a capacitor to perform the same task using an inductor? Thanks in advance UserIdTAG: 318577 UserNameTAG: takeuchi CreateTimeTAG: 2012-11-05T13:30:57Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: I think perhaps because inductors have an inherent resistance associated with them they are harder to implement, I remember in one of the lecture sequences super inductors were mentioned, but they have to be frozen to almost absolute zero for the associated resistance to 'au revoir'. FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-11-05T13:35:56Z SecondChildTAG: I think, inductors are harder to manufacture in ICs. For capacitor you just draw one polygon next to another. And it is planar. But for inductor the picture would be more complex and use more IC space. Maybe inductor should not be planar either to have a significant inductance. SecondChildUserIdTAG: 208296 SecondChildUserNameTAG: OZ1 SecondChildCreateTimeTAG: 2012-11-05T14:15:16Z SecondChildTAG: I think you are right! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-06T10:04:54Z FirstChildTAG: also i'd think incudctors would be way more vulterable to electro magnetic fields and feromagnetic materials in close proxility FirstChildUserIdTAG: 263693 FirstChildUserNameTAG: Coldberg FirstChildCreateTimeTAG: 2012-11-05T15:15:45Z FirstChildTAG: Historically, another reason is that capacitors could be manufactured on silicon directly (kind of like 2-D) while inductors needed 3-D space so difficult to build it in silicon directly initially. Even though things have changed in recent times, its difficult to unstick history. FirstChildUserIdTAG: 213493 FirstChildUserNameTAG: StudentZ FirstChildCreateTimeTAG: 2012-11-05T16:21:45Z SecondChildTAG: Why upset the status quo! If it aint broke, don't fix it. SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-06T10:06:19Z FirstChildTAG: A capacitor can store data statically. An inductor requires a current with its inherent dissipation in order to store data. In reality a capacitor has leakage, so it must be refreshed. But that intermittent current has less dissipation less than a constant current. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-05T15:40:55Z FirstChildTAG: MAGNETIC ENERGY VS ELECTROSTATIC ENERGY .....11 WHICH ONE YOU WILL FAVOUR ONE IS DYNAMIC OTHER STATIC ... ACC. TO ME IT'S ELECTROSTATIC FirstChildUserIdTAG: 285222 FirstChildUserNameTAG: dhaval24 FirstChildCreateTimeTAG: 2012-11-08T08:28:50Z IndexTAG: 527 TitleTAG: Solutions to problems in textbook I found a full set of solutions for the textbook problems here: *edited* It won't let me download it without signing up for a "premium account", but you can view it online in the browser window. It's a bit strange, because I haven't seen this book for sale anywhere and I can't find it for download anywhere else. Perhaps if anyone out there has a scribd account then they could download it and share it with others on the course? Not sure whether that would be illegal since no-one seems to be trying to sell it ?! UserIdTAG: 434869 UserNameTAG: patey CreateTimeTAG: 2012-11-01T18:57:14Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Post of the month. Thanks! FirstChildUserIdTAG: 329254 FirstChildUserNameTAG: KGabor FirstChildCreateTimeTAG: 2012-11-02T08:10:51Z FirstChildTAG: Solutions for the textbook are available to instructors only. The link you provided does not appear to be a legal source of the of the documents. If you would like a copy of the book *without* solutions, it is available at many places, for example: http://store.elsevier.com/specialOffer.jsp?offerId=EST_MIT&utm_source=EMAP&utm_medium=text_link&utm_term=Circuits_Electronics&utm_campaign=edx#oid=1003_4 Thank you for understanding. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-11-02T13:05:30Z SecondChildTAG: OK, fair enough. I didn't mean to break any rules. The document I previously linked to seems to be incomplete anyway so maybe it is an unofficial set of answers. Out of interest, what is the reason for keeping the answers to half the problems in the book a secret? From my point of view it is frustrating to have a go at some of the problems in the book only to find that I can't check my answers. SecondChildUserIdTAG: 434869 SecondChildUserNameTAG: patey SecondChildCreateTimeTAG: 2012-11-02T14:11:12Z SecondChildTAG: In my opinion, it is because the textbook is used specifically for teaching, as opposed to a general reference book. Students in a course may be asked to solve some or all of the questions for a grade. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-11-02T14:22:05Z IndexTAG: 528 TitleTAG: H9P1 Why UC(t-) not equal UC(t+)??? UserIdTAG: 464744 UserNameTAG: attache CreateTimeTAG: 2012-10-31T09:32:09Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Because of the current impulse applied at t=1.0s FirstChildUserIdTAG: 127195 FirstChildUserNameTAG: princeofsudan FirstChildCreateTimeTAG: 2012-10-31T16:47:19Z SecondChildTAG: dUC(t)=∞ then IC(t)=∞ LOL... SecondChildUserIdTAG: 464744 SecondChildUserNameTAG: attache SecondChildCreateTimeTAG: 2012-10-31T17:48:23Z SecondChildTAG: When computing the capacitor voltage after the impulse, I thought the charge created from the impulse would go to the capacitor only. But it's not the correct answer. Any hint ? Thanks. SecondChildUserIdTAG: 169461 SecondChildUserNameTAG: jnd77 SecondChildCreateTimeTAG: 2012-11-01T05:36:26Z SecondChildTAG: Find the voltage on the capacitor just before the impulse. Work out how much energy is stored in the capacitor. Then find the change in energy created by the pulse. (The polarity of the charge on the capacitor before the pulse is important). Once you have the energy after the pulse, you can calculate the voltage on the capacitor after the pulse. SecondChildUserIdTAG: 270011 SecondChildUserNameTAG: SnowmanZA SecondChildCreateTimeTAG: 2012-11-01T18:11:13Z SecondChildTAG: I think that other way is... Since all the charge Q goes to the capacitor, you can calculate the voltage increase Q/C and add it to the capacitor voltage just before the impulse (you already have this vc(t-) from previous questions) SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-11T23:23:49Z FirstChildTAG: Apparently the data in the problem varies by individual. In my case, it was important to take note of the polarity of the charge on the capacitor before the arrival of the impulse. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-01T12:26:37Z SecondChildTAG: skyhawk, Your anwer UC(t-) = UC(t+)? SecondChildUserIdTAG: 464744 SecondChildUserNameTAG: attache SecondChildCreateTimeTAG: 2012-11-01T12:58:14Z SecondChildTAG: The impulse changes the charge on the capacitor. The voltage changes. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-11-01T13:35:49Z IndexTAG: 529 TitleTAG: Accepted answers to Homework 9 (f and g of H9P2) are incorrect I think the answers accepted as correct for items f and g of H9P2 are incorrect. Those seems to be the values for a series RLC circuit, whereas the circuit in the problem is a parallel RLC when it is not driven by a source. In fact, Eq. 12.108 (in the supplemental book) should be different than Eq. 12.66 in the book. This fact is even stated prior to Eq. 12.108. But they are both the same, $Q=\frac{1}{R}\sqrt{\frac{L}{C}}$. So it seems that 12.108 is wrong, which can be easily checked by substituting 12.86 and 12.87 into 12.107. And the answer to H9P2, Item f should be the correct version of Eq. 12.108. UserIdTAG: 153760 UserNameTAG: Kavka CreateTimeTAG: 2012-10-30T01:30:47Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 1 FirstChildTAG: Found the same issue, eq.12.108 seems to be wrong in the book. And as the result - H9P2 f/g are wrong too, they accept answers based on eq. 12.108 only. FirstChildUserIdTAG: 372582 FirstChildUserNameTAG: atarlykov FirstChildCreateTimeTAG: 2012-10-30T10:10:13Z SecondChildTAG: i will helps to you the response is t*w0 SecondChildUserIdTAG: 174229 SecondChildUserNameTAG: fares27 SecondChildCreateTimeTAG: 2012-11-15T17:57:10Z IndexTAG: 530 TitleTAG: help TA about midterm Is it necessary to click on check 3 times, if u got correct result in first attempt. Because if you got all correct in 1st click, it saying your answers are not graded, click on check to grade them. UserIdTAG: 726377 UserNameTAG: g_goel CreateTimeTAG: 2012-10-27T12:42:38Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: you have to save your result. . . FirstChildUserIdTAG: 146930 FirstChildUserNameTAG: fayzan_007 FirstChildCreateTimeTAG: 2012-10-27T12:51:04Z FirstChildTAG: Hi g_goel, ---------- **Save button:** Only saves your answer, not check them... **Check button:** It does 3 things at the same time: Checks your answer (if it is incorrect or correct), saves your answer and submit it, all at the same time. ---------- If you have a correct answer in your first attempt (you had click on Check and you got the green tick), you don't need to click on the check and use the others two attempts if you already have your correct answer in your 1st attempt... ----- *your answers are not graded, click on check to grade them.* Might you are saving your answers and not checking them... Remember that saves button only saves your answer and not grade/submit/check them. Remember also, that you have only 3 opportunities to click on check button... ---- Also, you can see your score in the Progress Tab... ---- I hope this can help you, See you, Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-27T13:24:54Z SecondChildTAG: Thank u very much... SecondChildUserIdTAG: 316761 SecondChildUserNameTAG: gk_goel SecondChildCreateTimeTAG: 2012-10-27T16:01:56Z IndexTAG: 531 TitleTAG: 24 hours done My 24 hours just finished, pleased with my effort but know there is more to learn. Good luck to those still working. Fellow U.S. east coasters, stay safe! UserIdTAG: 11730 UserNameTAG: RonFlip CreateTimeTAG: 2012-10-27T10:15:50Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 532 TitleTAG: Midterm exam is on the courseware section Hey! Just log out your account. Once you have logged in again, it will appear in the Courseware Section, just after Week 6. UserIdTAG: 320933 UserNameTAG: serdigom CreateTimeTAG: 2012-10-25T05:33:54Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: got it.. FirstChildUserIdTAG: 111540 FirstChildUserNameTAG: tksanthosh FirstChildCreateTimeTAG: 2012-10-25T05:35:04Z SecondChildTAG: yes SecondChildUserIdTAG: 428658 SecondChildUserNameTAG: Moza SecondChildCreateTimeTAG: 2012-10-25T05:43:27Z IndexTAG: 533 TitleTAG: mid terms just wanted to confirm.. that can i log in and begin test then continue it afterwards but finish it within 24 hours.. UserIdTAG: 266739 UserNameTAG: satvikchugh CreateTimeTAG: 2012-10-24T15:13:32Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: yep. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-10-24T16:51:48Z FirstChildTAG: Does that really mean that I can do half of the test in the morning and the rest in the evening? FirstChildUserIdTAG: 131726 FirstChildUserNameTAG: chemiboy11 FirstChildCreateTimeTAG: 2012-10-24T17:22:55Z SecondChildTAG: I count on that.... SecondChildUserIdTAG: 342966 SecondChildUserNameTAG: SandraNavarro SecondChildCreateTimeTAG: 2012-10-24T21:18:00Z SecondChildTAG: Yes. As long as you finish within 24 hours or before 23:59 (11:59 pm) on Sunday, October 28th, Boston time, whichever comes first. SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-25T05:42:05Z IndexTAG: 534 TitleTAG: mid term i am very thankful to MITx admin...i have little bit confusion regarding the **duration of submission**. for eg. **if i login on 25 october then what is the deadline to submit it? is it 24 hours after the login or sunday 11:59pm?** UserIdTAG: 114864 UserNameTAG: Vikaash CreateTimeTAG: 2012-10-24T08:16:17Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Its after 24 hours from the time you have started taking the exam or 11:59 Pm on sunday,Boston Time (Which ever comes first) :D FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-10-24T09:36:53Z SecondChildTAG: really? it's a good news SecondChildUserIdTAG: 395257 SecondChildUserNameTAG: blackguitar SecondChildCreateTimeTAG: 2012-10-24T19:49:15Z IndexTAG: 535 TitleTAG: FOR STAFF The show discussion button is not working for week 7! UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-10-23T12:42:10Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Thank you for pointing that out. A fix is forthcoming. It might take a little bit more time than usual due to the Amazon outages. http://www.techspot.com/news/50573-amazon-web-services-outage-takes-down-reddit-netflix-pinterest.html Thank you for your patience. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-10-23T14:49:40Z SecondChildTAG: Thanks a ton! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-23T16:09:21Z SecondChildTAG: Sweet! It's working now, I appreciate your guys hard work! Thanks again! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-23T20:34:51Z FirstChildTAG: It appears to have disappeared again! FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-10-24T09:33:46Z IndexTAG: 536 TitleTAG: LAB6 How to find RON ,Hint plz How to find RON ,Hint plz UserIdTAG: 97359 UserNameTAG: azeem179 CreateTimeTAG: 2012-10-18T16:54:33Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: It's just a voltage divider problem. Measure the voltage and then calculate the required resistance. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-18T17:10:56Z SecondChildTAG: i am doing by voltage divider but do not getting the green tick mark SecondChildUserIdTAG: 97359 SecondChildUserNameTAG: azeem179 SecondChildCreateTimeTAG: 2012-10-18T17:23:50Z SecondChildTAG: What is your measured voltage? SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-18T17:26:51Z SecondChildTAG: ''Measure final output voltage of the inverter when the input is high and use that to estimate RON for the mosfet switch.'' im confused about this statement plz help me SecondChildUserIdTAG: 97359 SecondChildUserNameTAG: azeem179 SecondChildCreateTimeTAG: 2012-10-18T17:33:41Z SecondChildTAG: i have measured voltage 2.533 SecondChildUserIdTAG: 97359 SecondChildUserNameTAG: azeem179 SecondChildCreateTimeTAG: 2012-10-18T17:38:53Z SecondChildTAG: Is that after a long time? Mine is much lower. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-18T17:58:42Z SecondChildTAG: got it the correct ans. . SecondChildUserIdTAG: 97359 SecondChildUserNameTAG: azeem179 SecondChildCreateTimeTAG: 2012-10-18T18:46:24Z SecondChildTAG: thanks skyhawk SecondChildUserIdTAG: 97359 SecondChildUserNameTAG: azeem179 SecondChildCreateTimeTAG: 2012-10-18T18:46:51Z SecondChildTAG: My pleasure! SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-18T18:50:15Z SecondChildTAG: Should we use the formula.....Vout=(Vs*Ron)/(Rl+Ron) but I am not able to understand when do we get the Vout....it is the blue line but when the output is high? SecondChildUserIdTAG: 182470 SecondChildUserNameTAG: nitesh2703 SecondChildCreateTimeTAG: 2012-10-19T11:34:00Z SecondChildTAG: The same thing i don't understand that how to see the output voltage on the graph :( SecondChildUserIdTAG: 145239 SecondChildUserNameTAG: Maheenjd SecondChildCreateTimeTAG: 2012-10-20T20:12:37Z SecondChildTAG: I am exhausted finding the Ron, i have tried each and every Vout value but still no green tick, atleast tell whether Ron is less than or greater than Rl. And also would like to ask where are we suppose to get Vout between Voh and Vol or above Voh??? Please help because of this question i m not able to proceed further in this lab, i have completed the last question of this lab but stucked in the first one. SecondChildUserIdTAG: 182470 SecondChildUserNameTAG: nitesh2703 SecondChildCreateTimeTAG: 2012-10-21T06:09:59Z IndexTAG: 537 TitleTAG: H7P3 part 3 stuck with time I`m trying to find the time that Vo reach 0.9V I`m doing this: i(t) = Io(1-e^(-t*R/L)) I opperate and: t = -ln((i(t)-Io)/(-Io)) with i(t) = 0.9/77A; Io = 1.8/77A; tau = 15.44nH/77omh What am I doing wrong? UserIdTAG: 299251 UserNameTAG: TheRedBlackOne CreateTimeTAG: 2012-10-18T04:49:45Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: t = - tau * ln( ( Ii - i(t) ) / ( Ii ) ) Tau = L/(Rs+Ro) Ii = V/(Rs+Ro) i(t) = Vo/Ro. I dnt know what im making wrong but the solution is not working. FirstChildUserIdTAG: 78992 FirstChildUserNameTAG: chuso06pdm FirstChildCreateTimeTAG: 2012-10-18T13:42:58Z FirstChildTAG: Read pages 518 to 520 in the textbook. Here is a hint: the current going through all the items will be the same and we know what voltage we need across the resistor. FirstChildUserIdTAG: 375824 FirstChildUserNameTAG: KMatariyeh FirstChildCreateTimeTAG: 2012-10-18T15:35:48Z FirstChildTAG: I cant understand where am i going wrong!!! in the very last part of this section we are asked to calculate time for light to travel 2cm. It is dam easy II grade HW. but i cant get the answer i am doing the following calculation : time= distance/speed speed =299792458m/s distance =2cm =0.02m thus t=6.67*10^-11 sec Which is marked wrong!!! Please help me finish 100% HW FirstChildUserIdTAG: 232499 FirstChildUserNameTAG: Asim09 FirstChildCreateTimeTAG: 2012-11-01T04:46:42Z SecondChildTAG: You need to answer the question in nano seconds. Think about that detail. SecondChildUserIdTAG: 433574 SecondChildUserNameTAG: endika86 SecondChildCreateTimeTAG: 2012-11-01T14:09:07Z SecondChildTAG: i am getting -ln(0.23)/5 i=2.5/50 il(0)=5/77 il(t)=il(0)*(1-e^(77*t/15.44)) SecondChildUserIdTAG: 185921 SecondChildUserNameTAG: Ravi_Teja SecondChildCreateTimeTAG: 2012-11-01T18:22:10Z SecondChildTAG: il(t)=il(0)*(1-e^-1\*(77*t/15.44)) SecondChildUserIdTAG: 185921 SecondChildUserNameTAG: Ravi_Teja SecondChildCreateTimeTAG: 2012-11-01T18:23:05Z SecondChildTAG: time is in **nanoseconds** not seconds SecondChildUserIdTAG: 266739 SecondChildUserNameTAG: satvikchugh SecondChildCreateTimeTAG: 2012-11-03T06:24:13Z SecondChildTAG: on solving my answer is 0.29XXX it says it is wrong any other hints please SecondChildUserIdTAG: 185921 SecondChildUserNameTAG: Ravi_Teja SecondChildCreateTimeTAG: 2012-11-03T11:58:43Z IndexTAG: 538 TitleTAG: [Staff] Course Info page missing Week 8 lecture slide links [Staff] The 'Course Info' page is missing links to the Week 8 lecture slides. The links are available within the Week 8 courseware itself, but that's inconvenient. UserIdTAG: 94557 UserNameTAG: g_hopper CreateTimeTAG: 2012-10-16T03:22:27Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 0 IndexTAG: 539 TitleTAG: Thoughts on Week 5 **Hello, all:** Just want to see your thoughts on Homework 5 and Lab 5. Simple or Difficult? I had work IT this week Thursday through Saturday, doing installs, so a lot of standing and walking around. These are usually the days I do homework but I was very tired and could not do homework **until Sunday** because after work I went straight to bed! I did Lab 5 early in the week and it was the *easiest so far*, easier than 1-4! I thought H5 would be easy as well... So I was in a shock Sunday (tonight) with the homework! Very difficult with the algebraic equations. Stupid mistakes on H5P3 with the differential equation really got me. First I forgot to multiply out $R_S$ (H5P2 gives $i_{DS}$, but H5P3 wants $V_O$.) Then I realized that small signal analysis requires two types of voltage in the equation, small-signal and bias (I repeatedly used $V_{IN}$, finally I tried $v_{IN}$, error, not allowed...frustration, then a desperate me finally tried $($*Long derivative equation with a* $V_{IN}$ *term* $) \cdot$ $ v_{in}$ = Success! I was so happy that the wasted hours resulted in a 100% to show for it. Now, despite taking 6 hours on this assignment (longest yet), and it's the first time I really had to look in-depth in the book for help, and to check the Discussion board, too. **To those who posted help on H5, thanks!**, I really appreciated it, especially this week doing homework at the last minute! Jersey Mark UserIdTAG: 292546 UserNameTAG: JerseyMark CreateTimeTAG: 2012-10-15T06:45:06Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: I would say, for me, the hardest homework so far was week 4, and the hardest lab was **lab 2**. Lab 5 was easy in that it was just visual analysis, everything was nicely given on a silver platter, guess work kind of. Homework 5 however, was definitely among the harder home-works, and like you I really struggled with *H5P2 Q3* and *H5P3 Q1*! But with the help of **Wolfram Alpha**, I pulled through to **100 percent**!!! What scores did the rest of you guys get? Has anyone managed 100 percent on all of your problems so far? FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-10-15T10:23:38Z SecondChildTAG: No I am stuck at H5P2 ids equation....can u please help me.... SecondChildUserIdTAG: 223286 SecondChildUserNameTAG: Swathi4941 SecondChildCreateTimeTAG: 2012-10-15T10:27:40Z SecondChildTAG: Swathi... Just look here https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50795ef945649d270000005f I put in the wrong value for K (it should be 2 not 0.5) because I had a misunderstanding... but otherwise the info is good! I hope this helps you! Hazel1919. SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-15T10:42:32Z FirstChildTAG: I did the homework a couple weeks ago when it first came out. I had forgotten what a pain it was to learn how to use the equation parser. I tried to help others navigate the difficulty. I saw a common problem of people not understanding the difference between VGS and VIN. I attribute this to people's lack of practical experience. People complain about the math, as far as I can remember for week 5 one only needed the power rule and chain rule to do the calculus. Next week requires being able to differentiate exponential functions. For week 7 you need to be able to integrate a sine function. All really basic stuff. IMHO the lab was entirely too subjective. There was no criterion was determining how much distortion was too much distortion. Completed it in only a few minutes and learned nothing. Lab 2 was certainly the most fun ... so many ways to analyze the problem. And to answer Hazel's question, I haves 100's for the first seven weeks. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-15T11:19:29Z SecondChildTAG: same here skyhawk and hazel! SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-24T15:47:34Z FirstChildTAG: Well done JerseyMark!:) FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-15T11:48:00Z IndexTAG: 540 TitleTAG: Week 5 and Honor Code The deadline for HW5 hasn't passed for everyone yet. The honor code is still in effect. Do not post actual answers to the problems. UserIdTAG: 21541 UserNameTAG: JSChambers CreateTimeTAG: 2012-10-15T03:21:12Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: No, the deadlines are at 12 midnight local time. So the deadline in CA is, as of now, 2 hours and 15 minutes away. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-15T04:45:39Z SecondChildTAG: Thank you. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-15T05:30:11Z FirstChildTAG: Isn't the deadline 12 AM US Eastern Standard Time? In other words, in California it is 9 PM? FirstChildUserIdTAG: 132828 FirstChildUserNameTAG: rharris FirstChildCreateTimeTAG: 2012-10-15T04:27:30Z FirstChildTAG: In the last course the real deadline was 12 midnight at the international dateline. So anyone could keep submitting as long as it was Sunday somewhere. FirstChildUserIdTAG: 9673 FirstChildUserNameTAG: xp42 FirstChildCreateTimeTAG: 2012-10-15T05:08:21Z FirstChildTAG: Xp42 is right, it's the international date line. For some reason I thought they changed it. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-15T15:21:24Z IndexTAG: 541 TitleTAG: Mature age students Who is the oldest student doing 6.002x? I am 54, hopefully not me!! UserIdTAG: 345671 UserNameTAG: cbjerregaard CreateTimeTAG: 2012-10-14T23:33:54Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: No, sadly it is not you. I'm quite sure. :) FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-10-14T23:35:35Z SecondChildTAG: Hi cbjerregaard! I remember in the Prototype Course a similar Post of yours, there was students more than 70 years old , so, I am totally sure that you are not the oldest of 6.002x ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T23:53:28Z SecondChildTAG: Wow. Juggling this course with home, family...kids....work is rough but my son at 8 watches the lectures with me :) SecondChildUserIdTAG: 244115 SecondChildUserNameTAG: Pedro1969 SecondChildCreateTimeTAG: 2012-10-15T01:21:50Z SecondChildTAG: This course plus that pesky thing called "Real Life" can be a tough juggle... Hang in there! SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-15T03:31:54Z FirstChildTAG: I'm actually quite curious who the youngest is. Any child geniuses among us? FirstChildUserIdTAG: 230226 FirstChildUserNameTAG: zoliking FirstChildCreateTimeTAG: 2012-10-15T00:56:40Z FirstChildTAG: Here is really a big community , also a big family!Fantastic! FirstChildUserIdTAG: 361137 FirstChildUserNameTAG: Ericson FirstChildCreateTimeTAG: 2012-10-15T02:47:03Z FirstChildTAG: 74 FirstChildUserIdTAG: 231764 FirstChildUserNameTAG: PaulP4881 FirstChildCreateTimeTAG: 2012-10-15T03:57:55Z SecondChildTAG: Wow! To continue studying for so long is amazing. I never want to stop learning but I don't know how long I can go on. You just showed me it's possible :-) SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-10-15T05:42:29Z FirstChildTAG: Cbjerregaard - You're definitely not the oldest. You and I are the same age, and this is my second time through 6.002x. It's good to stretch the brain at every age! I think I saw some 15 year-olds in the class this time, and someone in his 70s. Last time, I think the youngest was about 14, but he had help from his engineer mom with some of the complex math. I think the oldest was about 83 or so. FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-10-16T03:04:10Z IndexTAG: 542 TitleTAG: The More You Know: iCircuit Did you know that iCircuit will show you when a MOSFET goes from Off, to Linear, and to Saturation Mode? I didn't know this until this morning. Just highlight the MOSFET. ![enter image description here][1] ![enter image description here][2] [1]: https://edxuploads.s3.amazonaws.com/13499737641343611.png [2]: https://edxuploads.s3.amazonaws.com/1349973807134361.png UserIdTAG: 21541 UserNameTAG: JSChambers CreateTimeTAG: 2012-10-11T16:44:22Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 543 TitleTAG: H5P1 Could anyone help me to find out the value of maximum input voltage Vin (Third sub question)? I used Vo = Vs-Ids*Rl equation by using Ids=k/2(Vin-VT)^2. I plugged in values for Vo as 0, Vs as 1 and VT 0.5. I got Vin =1, but its wrong. I also saw the other hints, but not able to figure out how to get green tick... UserIdTAG: 21687 UserNameTAG: Benadicta CreateTimeTAG: 2012-10-10T10:11:54Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: me too. I have tried every variation I can think of FirstChildUserIdTAG: 18073 FirstChildUserNameTAG: gburkhart FirstChildCreateTimeTAG: 2012-10-11T00:37:43Z SecondChildTAG: And me too. Have enyone got the green tick on 3rd question? SecondChildUserIdTAG: 190057 SecondChildUserNameTAG: Anastasia00435 SecondChildCreateTimeTAG: 2012-10-11T14:55:03Z SecondChildTAG: I am sailing in the same boat. SecondChildUserIdTAG: 99628 SecondChildUserNameTAG: Pranjal16 SecondChildCreateTimeTAG: 2012-10-11T18:28:08Z SecondChildTAG: me too plz help.. SecondChildUserIdTAG: 438395 SecondChildUserNameTAG: ali_PU1 SecondChildCreateTimeTAG: 2012-10-13T19:21:48Z FirstChildTAG: The correct expression is: iDS=K/2(vGS-VT)^2 Although Vin is often the same as vGS, it is not in this case. Notice the VS- connected to the source. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-11T19:40:54Z FirstChildTAG: Its okay, I will wait for the answer after the deadline FirstChildUserIdTAG: 21687 FirstChildUserNameTAG: Benadicta FirstChildCreateTimeTAG: 2012-10-12T01:43:10Z SecondChildTAG: Hi Benadicta! How are you? You can take a look at this Hints [Post][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_mosfet_amp/threads/507a1d6dd894f12300000039 SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T04:17:04Z IndexTAG: 544 TitleTAG: for leonidasGr can't post it as a response for some reason for discharge VI will not be zero, it will be VS*RB/(R+RB)=92*10k/(1.5M+10k)=0.613 also R will not be 10k, but 1.5M||10k=9933.77 it's not a big deal for numbers, but for understanding :) UserIdTAG: 342603 UserNameTAG: YakovO CreateTimeTAG: 2012-10-10T08:50:01Z VoteTAG: 5 CoursewareTAG: Week 6 / Neon relaxation oscillator exercise CommentableIdTAG: 6002x_neon_relaxation_oscillator_exercise NumberOfReplyTAG: 1 FirstChildTAG: "it's not a big deal for numbers, but for understanding :)" yes thanks for crunching the numbers for me. the error is in the order of 10000/1500000 that's why I mentioned the voltage dividers in my post I am also posting your answer there for completeness FirstChildUserIdTAG: 319453 FirstChildUserNameTAG: leonidasGr FirstChildCreateTimeTAG: 2012-10-10T12:44:37Z IndexTAG: 545 TitleTAG: [Staff] Questions about mid-term test Hello everyone, As usual, sorry for my English :D I have some questions about the mid-term test. 1 - The test will be released October 25th until October 27th, but over this 2 days we will have only 24 hours to finish the test. Am I right? 2 - The homeworks, labs, videos, textbook and exercises will be available during this two days? 3 - The test will have some questions where we will use the lab tools? 4 - The test will have check buttons? To the others students who want to help, please only reply my questions if you are ABSOLUTELY RIGHT about your answers. I don't want start a discussion topic where everybody has a opinion but no one knows the right answer. (Hope this don't sound too rude =/) Little off-topic but how will we get our certificates? It will be available online or will we receive in paper? Don't be ashamed to correct my English, I'm here to learn it too and I will be very thankful. UserIdTAG: 310147 UserNameTAG: ildomarcarvalho CreateTimeTAG: 2012-10-10T06:15:47Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Don't apoligize for your English. It isn't as bad as you think. A few minor mistakes, so don't worry about it :) FirstChildUserIdTAG: 114881 FirstChildUserNameTAG: xvink FirstChildCreateTimeTAG: 2012-10-10T06:32:13Z SecondChildTAG: An Honor Of Code Certificate will be given online. SecondChildUserIdTAG: 309238 SecondChildUserNameTAG: ArunRawat SecondChildCreateTimeTAG: 2012-10-10T09:07:02Z FirstChildTAG: An Honor Of Code Certificate will be given online. FirstChildUserIdTAG: 309238 FirstChildUserNameTAG: ArunRawat FirstChildCreateTimeTAG: 2012-10-10T09:07:39Z FirstChildTAG: Ildomar: de que lugar do Brasil você é? FirstChildUserIdTAG: 361823 FirstChildUserNameTAG: EliasOak FirstChildCreateTimeTAG: 2012-10-10T12:20:48Z SecondChildTAG: Sou de Joinville, e você? SecondChildUserIdTAG: 310147 SecondChildUserNameTAG: ildomarcarvalho SecondChildCreateTimeTAG: 2012-10-10T17:07:29Z SecondChildTAG: Sou de São Paulo. Também tenho o sobrenome Carvalho. Meu nome é Elias Felipe de Carvalho. SecondChildUserIdTAG: 361823 SecondChildUserNameTAG: EliasOak SecondChildCreateTimeTAG: 2012-10-10T17:36:48Z SecondChildTAG: Bom saber :) Já é da área de eletrônica ou está vindo meio de curioso como eu? SecondChildUserIdTAG: 310147 SecondChildUserNameTAG: ildomarcarvalho SecondChildCreateTimeTAG: 2012-10-10T17:48:40Z SecondChildTAG: Sou professor no SENAI. Conheci esse curso pelo jornal. Fiquei curioso e me inscrivi. Achei que ia só começar e não terminar por falta de tempo, mas acho que vou até o fim. Gostei desse curso. Você tem alguma formação em eletrônica? Eu me graduei em Física, mas fiz o colégio técnico em São Paulo. SecondChildUserIdTAG: 361823 SecondChildUserNameTAG: EliasOak SecondChildCreateTimeTAG: 2012-10-11T10:40:49Z SecondChildTAG: Faço Ciência da Computação na UDESC mas não tenho formação em eletrônica.. o que tenho é uma pequena experiência trabalhando com automação e com hardware livre. Também achei que não ia ter tempo pra terminar o curso, ainda mais por estar no fim da graduação e estar na correria do TCC, mas gostei tanto que estou dando um jeito para arrumar tempo hehe Afinal, quem precisa dormir mais do que 5h por dia? xD SecondChildUserIdTAG: 310147 SecondChildUserNameTAG: ildomarcarvalho SecondChildCreateTimeTAG: 2012-10-12T01:54:00Z IndexTAG: 546 TitleTAG: Wolfram Tips I thought I would post some tips about how I use Wolfram Alpha. I would love to see other people's tips on how to use it more effectively. If you sign up with Wolfram, you can store "Favorites". I love the favorites, as I can save and reuse equations. When you save a favorite, it goes into a list like this: ![enter image description here][1] I like to label with the homework or lab or example number, so I can find them if I need them. I also like to put the original formula in the notes, so I can see exactly what it is, like this: ![enter image description here][2] Wolfram also has a bunch of built-in electric circuit functions, that you can see here: http://www.wolframalpha.com/examples/ElectricCircuits.html That's all for now. Please, add your own tips on how you use Wolfram Alpha to the fullest. [1]: https://edxuploads.s3.amazonaws.com/13498000262224599.png [2]: https://edxuploads.s3.amazonaws.com/13498001871343682.png UserIdTAG: 21541 UserNameTAG: JSChambers CreateTimeTAG: 2012-10-09T16:34:39Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Cool. Great tips. I used Wolfram Alpha extensively during the pilot 6.002x, and I didn't know about the built-in circuits functions. FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-10-09T16:43:53Z SecondChildTAG: Yeah they don't go out of their way to advertise them. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-09T16:48:35Z FirstChildTAG: Can Wolfram Alpha evaluate derivatives? In other words, if I entered the function $f(x)=x^2$ into Wolfram Alpha, would I get $f'(x)=2\cdot x$? If the answer is "yes", I would appreciate your telling me how to do it. Until now I have used MathStudio for iPad to evaluate derivatives. FirstChildUserIdTAG: 252722 FirstChildUserNameTAG: vaboro FirstChildCreateTimeTAG: 2012-10-09T18:48:20Z SecondChildTAG: Hi, vaboro. Just write (x^2)' - and you got it. SecondChildUserIdTAG: 324219 SecondChildUserNameTAG: Dmitry79 SecondChildCreateTimeTAG: 2012-10-09T19:22:31Z IndexTAG: 547 TitleTAG: Name of the Classical Music piece? Anybody know the name of the classical music piece. I liked it a lot. UserIdTAG: 153760 UserNameTAG: Kavka CreateTimeTAG: 2012-10-09T00:49:00Z VoteTAG: 5 CoursewareTAG: Week 5 / Music Amplifier Demonstration CommentableIdTAG: 6002x_snd_d NumberOfReplyTAG: 1 FirstChildTAG: Hi Kavka! ![imagen][1] **Johann Sebastian Bach** (31 March [O.S. 21 March] 1685 – 28 July 1750) was a German composer, organist, harpsichordist, violist, and violinist of the Baroque Period. He enriched many established German styles through his skill in counterpoint, harmonic and motivic organisation, and the adaptation of rhythms, forms, and textures from abroad, particularly from Italy and France. Many of Bach's works are still known today, such as the Brandenburg Concertos, the Mass in B minor, the The Well-Tempered Clavier, and his cantatas, chorales, partitas, passions, and organ works – and his music is revered for its intellectual depth, technical command, and artistic beauty.[read more - Wikipedia][2] **MUSIC of Lab:** [J.S. Bach - Partita nr.2 - Gigue][3] See you! Myriam. P.D: Don't think that I have a good ear or that I am an erudite haha **jmc** username got the answer in the Prototype course ( I should have paid attention in the primary school when the Teacher passed CD's of Classical Music and told us to listen and then he asked us Who was the Interpreter by listening only and I always wondered why we did that,why? haha and years years later, now, I have the answer haha: for the 6.002x haha). [1]: http://upload.wikimedia.org/wikipedia/commons/thumb/6/6a/Johann_Sebastian_Bach.jpg/220px-Johann_Sebastian_Bach.jpg [2]: http://en.wikipedia.org/wiki/Johann_Sebastian_Bach [3]: http://www.youtube.com/watch?v=NdzGKR0KUC0 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-09T04:34:01Z SecondChildTAG: Great! Thanks for the info. SecondChildUserIdTAG: 153760 SecondChildUserNameTAG: Kavka SecondChildCreateTimeTAG: 2012-10-11T13:12:43Z SecondChildTAG: You are welcome Kavka! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T04:27:26Z IndexTAG: 548 TitleTAG: Pratice Questions Will the staff be releasing any practice questions to do in advance of the midterm exam in order to better prepare? I'm just thinking ahead a few weeks and trying to decided how best to study. I think, obviously, reviewing some problems would be a good strategy, but I wouldn't know which to tackle. UserIdTAG: 287805 UserNameTAG: Beneficial CreateTimeTAG: 2012-10-08T17:46:44Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 549 TitleTAG: About transient analysis this the first day when i join it and i am very happy and i am very thankful to whole team. I do not know about Transient analysis, please help me. UserIdTAG: 580208 UserNameTAG: shivamyadav10 CreateTimeTAG: 2012-10-07T10:29:32Z VoteTAG: 5 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: Hi shivamayadav10, Welcome to 6.002x! :) How are you? [This is a Test - as JSChambers says, I couldn't write you in your other post, there is a problem with your post] JSChambers wanted to reply your post but he couldn´t reply you here ... [read this Post][1]. Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5071b40bc0a2fa1f00000015 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-07T18:33:15Z IndexTAG: 550 TitleTAG: H4P2 zener regulator smal vo problem hello all, i have done almost all the parts of 2nd question but i am confused in vo (small single at output when zener is connected.. how to solve this part i am completely lost in it ... kindly guide me thanks. UserIdTAG: 438395 UserNameTAG: ali_PU1 CreateTimeTAG: 2012-10-06T12:00:45Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: This is my problem too! I thought it was zero... but.. :| FirstChildUserIdTAG: 385692 FirstChildUserNameTAG: tpfslima FirstChildCreateTimeTAG: 2012-10-06T12:54:23Z FirstChildTAG: I got it! Think the zener diode as an 1 ohm resistor... 1ohm in parallel with 2k is aprox. 1 ohm. 1ohm in parallel with 4k is aprox. 1 ohm. So, solve the voltage divisor. FirstChildUserIdTAG: 385692 FirstChildUserNameTAG: tpfslima FirstChildCreateTimeTAG: 2012-10-06T13:02:38Z SecondChildTAG: Do you use the source as only the small signal one or the complete addition of two volate sources in this ? SecondChildUserIdTAG: 97340 SecondChildUserNameTAG: AnkitRana SecondChildCreateTimeTAG: 2012-10-06T15:08:04Z SecondChildTAG: Nevermind ! Got it :D thanks ! SecondChildUserIdTAG: 97340 SecondChildUserNameTAG: AnkitRana SecondChildCreateTimeTAG: 2012-10-06T15:12:26Z SecondChildTAG: But why it is acting like just a resistor? Can u explain the reason please? Although it is giving accurate answer now. SecondChildUserIdTAG: 329015 SecondChildUserNameTAG: farah_sarwar SecondChildCreateTimeTAG: 2012-10-06T17:58:15Z SecondChildTAG: ^ Because the slope of Zener diode is 1 Ampere per Volt in the region of operation. This means that it is working as 1ohm resistor and a DC voltage source. SecondChildUserIdTAG: 382253 SecondChildUserNameTAG: sameerlatif SecondChildCreateTimeTAG: 2012-10-06T18:05:52Z SecondChildTAG: But what's the operating region in this one? How is the Zener replaced by a resistor ONLY and NOT a DC source? SecondChildUserIdTAG: 278301 SecondChildUserNameTAG: prateektaneja SecondChildCreateTimeTAG: 2012-10-06T18:52:55Z SecondChildTAG: Why zener diode is 1 ohm? SecondChildUserIdTAG: 196404 SecondChildUserNameTAG: Yel1owstone SecondChildCreateTimeTAG: 2012-10-06T20:54:44Z SecondChildTAG: I can't get the answer yet... what do you replace and in which way do you do it? how will the voltage divider look like after replacing the zener with the 1 ohm resistor? SecondChildUserIdTAG: 45307 SecondChildUserNameTAG: josejimenez2 SecondChildCreateTimeTAG: 2012-10-06T22:24:03Z SecondChildTAG: ok, so I replace the zener with a 1 ohm resistor and a 5V DC source. But then, being only 1 ohm, there is no longer a sufficient voltage drop across the zener diode to make it conduct current, so it will stop and act like an open circuit! When it's an open circuit, there is sufficient voltage drop across the diode, so that it now starts conducting current again. ???? What am I doing wrong here? SecondChildUserIdTAG: 339870 SecondChildUserNameTAG: rjlasota SecondChildCreateTimeTAG: 2012-10-06T23:41:48Z SecondChildTAG: oh I see, there will always be at least 5V across the diode b/c of the 5V DC source. SecondChildUserIdTAG: 339870 SecondChildUserNameTAG: rjlasota SecondChildCreateTimeTAG: 2012-10-06T23:55:50Z SecondChildTAG: please clearly state the reason for why you have considered the zener diode as only a 1 ohm resistor. it doesnt make any sense of your consideration. though it's giving a right answer, the assumption is wrong, i think. please explain SecondChildUserIdTAG: 467186 SecondChildUserNameTAG: santoshbitwaala14 SecondChildCreateTimeTAG: 2012-10-07T17:26:57Z SecondChildTAG: it's because R = V/I and V/I = 1/1 =1 from the graph. Cheers tpfsima! thanks for the info! SecondChildUserIdTAG: 247968 SecondChildUserNameTAG: tdd5024 SecondChildCreateTimeTAG: 2012-10-08T02:07:16Z IndexTAG: 551 TitleTAG: staff : video release dates for wiki section Due to the outage of youtube in certain parts of world direct video links are provided in the wiki section. these videos are available up to week 4. can video release dates for this section be provided please by the staff ? we have video lectures up to week 6 in courseware section but up to week 4 only are present there UserIdTAG: 358539 UserNameTAG: syed_abdullah CreateTimeTAG: 2012-10-05T11:08:23Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Please also add tutorial videos in each weekly sections as well FirstChildUserIdTAG: 455950 FirstChildUserNameTAG: Wahabbaluch FirstChildCreateTimeTAG: 2012-10-06T03:34:13Z FirstChildTAG: Now I have access to you tube with the **Hotspot Shield** VPN or **Spotflux** VPN. everyone can download them by free. I know that it works at least it works in Iran and Pakistan. FirstChildUserIdTAG: 374393 FirstChildUserNameTAG: rmaleki FirstChildCreateTimeTAG: 2012-10-06T19:51:34Z IndexTAG: 552 TitleTAG: Online videos for weekly tutorials To staff, Kindly note that the ban on Youtube is still imposed. Weekly Lectures are seen from wiki but they do not contain videos of tutorials. Kindly arrange to post all tuturial video from week 1. Thanks and best regards Abdul Wahab UserIdTAG: 455950 UserNameTAG: Wahabbaluch CreateTimeTAG: 2012-10-04T04:22:28Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 553 TitleTAG: It's truly a big "aha!" moment I'm amazed ! It's so awesome watching the theory work so perfectly and beautifully in real life. I couldn't believe my ears when I heard such a huge difference in the audio quality before and after applying the small signal trick. UserIdTAG: 56851 UserNameTAG: Alonso_sh CreateTimeTAG: 2012-10-04T03:56:21Z VoteTAG: 5 CoursewareTAG: Week 4 / Incremental Analysis Demo CommentableIdTAG: 6002x_inc_analysis_demo NumberOfReplyTAG: 0 IndexTAG: 554 TitleTAG: About exams i know it's too early to ask about exam as m in week 5 right now..but still wanna know how much time we'll get to complete the midterm and final exam, i mean what is deadline for submitting the ans..(1 or 2 days)???? UserIdTAG: 219442 UserNameTAG: aki31 CreateTimeTAG: 2012-10-03T04:30:53Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi aki31! Based on my experience in the Prototype Course 6.002x (spring), you will have a range of days, like 4 days to set for the Exam. The Midterm option will appear in the same column where it is the Week1, week2, week3,etc.., week6 option. Once you click on it, it will appear in the web page a warning telling you that once you accept to start the Exam you will have 24hs to complete it. You accept it by clicking in a sort of arrow that is pointing to the right. If you click on it you will start the Midterm Exam, you only have 24hs to complete it or to submit the answers. You will have limited check times (3 times by each check buttom). Also, you can not discuss the Exam in the Forum Discussion, they are really severe with this. See you! Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-03T04:50:24Z SecondChildTAG: thanx Myrimit..it was really a helpful advice.. SecondChildUserIdTAG: 219442 SecondChildUserNameTAG: aki31 SecondChildCreateTimeTAG: 2012-10-03T06:17:28Z SecondChildTAG: you are welcome ! ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-03T11:18:10Z IndexTAG: 555 TitleTAG: Honor Code I believe there should be a report button so that there aren't any direct answers of Homework or Lab's questions in the discussion forum. Not trying to moan, this is just a thought to keep things fair. UserIdTAG: 304587 UserNameTAG: Vasso CreateTimeTAG: 2012-09-30T20:57:21Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: I agree. We are working on that button so that we can delete the posts with answers. Also, the accounts of users posting answers in the discussion forum will be deleted. FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-30T23:34:33Z FirstChildTAG: about the certification , is it will be delivered by hand to my address FirstChildUserIdTAG: 395257 FirstChildUserNameTAG: blackguitar FirstChildCreateTimeTAG: 2012-10-01T00:59:52Z SecondChildTAG: By hand? Bwhahahaha! Like MIT will send a personal representative to give you a certificate by hand? No way man :D The certificate should be via email in .pdf format, and you can print it out on a nice color printer! It has an URL on it so you can type that in to any internet browser to confirm it's authenticity (many governments are starting to do this as well, the apostillate for my birth certificate has an URL on it for confirmation)... SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-10-01T02:09:46Z SecondChildTAG: They will send a plane ticket to you and invite you to come to collect the cirtificate yourself. They'll collect you at the airport and drive you to the campus. All the professors will bow for you and your name will be carved in stone and added to mit's walk of fame .... SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-10-01T10:36:27Z SecondChildTAG: Cool salsero! funny! haha! you make my day! - you have to post something in the ruudoleo humor section [see in the down Post-comment of Ruudoleo][1] haha! Seriously talking, blackguitar, based on my experience in the Prototype Course, once you have passed the Course, it will appear in your Progress a buttom where you can download your Certificate. This certificate will be in a .pdf with an authentification link that re-link you to edx where your complete name it is shown and says that you have the certificate with your grade :). See you! [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/MITx_6_002x_2012_Fall/threads/50474faf685941270000000d SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-02T00:48:57Z SecondChildTAG: just stop it...... i didn't mean it by that way...i thought they will send the certificate as a package to my mail box in my country ( hard copy not soft) .. did you get it? ..... So,every one owe me an apology for that way he thinks about :P ,,,,, i'm not Stupid to think by that way guys SecondChildUserIdTAG: 395257 SecondChildUserNameTAG: blackguitar SecondChildCreateTimeTAG: 2012-10-08T06:34:32Z SecondChildTAG: Sorry blackguitar, I never meant to be not polite... My apologies if you offended. My apologies, I didn't want it that... I wondered the same in the Prototype course too.. My best wish to you. Just for curious, How are you doing with the course? Are you in other edX courses too? ;) See you! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-09T03:40:20Z FirstChildTAG: I see your point, but, as there is no actual academic credit for doing this course, then my feeling on the matter, is that anybody actively breaching the honor code, is ultimately fooling nobody but themselves. I do object though to the blatant posting of answers to the forum. If I was stuck on something, I might look for hints to point me in the right dirction but wouldn't like to stumble across the answer by accident. That would ruin my chance of finding an answer myself. Often it the journey to discovering the correct answer where the real learning takes place. FirstChildUserIdTAG: 385677 FirstChildUserNameTAG: Michaelc1 FirstChildCreateTimeTAG: 2012-09-30T22:27:27Z SecondChildTAG: Exactly! SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-10-01T02:10:16Z IndexTAG: 556 TitleTAG: Recommended Exercises From The Text Before edx was born, I often studied on my own by simply reading textbooks on various subjects. For Maths and Sciences, the only means of testing my knowledge of the material was by doing the questions at the end of each chapter and checking my answers against the answer keys. I actually found this very valuable. The questions were always ordered in such a way as to facilitate learning: the early questions were quite rudimentary and took very little time to complete, but working through them really reinforced the fundamentals of the material and developed an intuition; the later questions would be more complex, but they wouldn't take too long to complete, so long as you had a grasp of the fundamentals and could apply them in combination; the final questions in a chapter would be a bit more challenging, but still not absolutely confounding, and they would point towards areas of interest, thought, study, or practical use slightly beyond the scope of the course but where the techniques being learned had some application. Through the first few weeks of 6.002x, between the homework assignments, labs, and exercises included in the lecture sequences, I haven't found time to do hardly any of the exercises contained in the textbook (especially with deadlines! :D). As such, I've found it difficult at times to grasp some of the homework assignments and labs, and I have found myself wasting many hours toiling over them at the end of each week. I think this frustration comes from the fact that, in any given week, I lack a good intuitive grasp of the fundamentals of whatever subject was taught because I wasn't able to work through several very rudimentary problems, as I described above. Truly challenging problems shouldn't be challenging simply because the student hasn't yet had an opportunity to develop the tools and techniques he will be implementing in order to solve the problem. My typical study patterns from week-to-week is to 1) read all the required readings near the start of the week, as I find time; 2) watch all the lecture sequences and do the problems provided therein; 3) watch the additional tutorial videos ; 4) tackle the homework and lab. After all that, I've still found some of the labs and homework assignments frustratingly time-consuming. This is, I believe, because there's a disproportionate leap in difficulty between the examples provided in the materials, and the questions posed in the assignments. The surest, most efficient, way to *learn* the material and *develop* an intuition, without wanting to bash my head in out of confusion and frustration, is if the materials were presented in a step-wise manner of ascending difficulty, with no problem being *tremendously* more difficult than the previous. The questions in the textbook, I imagine, would fill the gap quite nicely, but, as I've said, with the deadlines, I have not found time to do any of them, let alone all of them. (And if I set aside some time to do *some* of them, I wouldn't know which to try first, or which would be most instructive.) As such, **I think it would be very useful if the course staff could provide, each week, a list of recommended problems from the text (or elsewhere**), which are primarily straight-forward (ie. not very time-consuming) and serve to develop an intuition and confidence for the subject matter. I think there is a lot to be learned by thinking through a problem on one's own, but I think there is very little to be gained by pulling one's hair out trying to solve something altogether foreign. As educators, I believe it is the course staff's responsibility to facilitate the learning process by helping the student along. You don't have to hold my hand, but please pave the road. I would really appreciate it if the course staff could take the time and direct their minds to ordering problems incrementally from easiest to most difficult or in such a way that the student finds each question no more time-consuming or confusing than the next because he approaches each one armed with the knowledge he gained from the previous. Oh, and a set of solutions against which we could check our answers would be invaluble...If not worked-solutions, at the very least a list of final answers. This would be a really helpful tool! And really very much appreciated, and I think would really fill a rather glaring hole in the course materials. Thanks. UserIdTAG: 287805 UserNameTAG: Beneficial CreateTimeTAG: 2012-09-30T19:42:32Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 1 FirstChildTAG: In respect to what you explained very well, i know that being given some direction from the tutor would be really convinient, but it is an engineer's job to find answers to difficult matters. So as an engineer you must learn how to research, use a few basic tools and solve the problem. Offcourse some guidance is always welcome, but i believe that making us research is the best way to learn how to solve a real life problem Because we will not get any kind of "pointing to the right direction" from anyone there. Best wishes FirstChildUserIdTAG: 304587 FirstChildUserNameTAG: Vasso FirstChildCreateTimeTAG: 2012-09-30T20:07:34Z SecondChildTAG: I see your argument, and you're right. Obviously, the guidance available to a practicing engineer in the real world would be far more limited than what could potentially be available to students in a course. That being said, no engineer is ever solving any problem truly 'from scratch'. The engineer has studied, and, through years of practice, has developed an intuition, confidence, and fluency for the subject matter. He doesn't have to derive his toolset from first principles every single time he is presented with a new problem. On a day-to-day basis, he merely has to turn his mind to applying them in a useful ways. I'm not really asking for every single problem to have another worked analogue with which we can follow along step-by-step. I'm just asking for a more coherent set of exercises and problems which betters serves to reinforce the fundamentals, train the techniques, develop an intuition, and foster confidence in the student's fluency in the topic. I don't think doing that causes the student to lose out on the practice of developing general problem-solving skills. SecondChildUserIdTAG: 287805 SecondChildUserNameTAG: Beneficial SecondChildCreateTimeTAG: 2012-09-30T20:34:35Z SecondChildTAG: > I'm just asking for a more coherent set of exercises and problems > which betters serves to reinforce the fundamentals, train the > techniques, develop an intuition, and foster confidence in the > student's fluency in the topic... I hasten to add "...in the most efficient, least time-consuming, and least frustrating manner!" These problems should be like eating popcorn. You jump from one to the next quickly and with ease. In fact, the course staff should feel free to go ahead and call them "Popcorn Questions." I make no claim to copyright :D . SecondChildUserIdTAG: 287805 SecondChildUserNameTAG: Beneficial SecondChildCreateTimeTAG: 2012-09-30T20:39:12Z SecondChildTAG: I agree with Beneficial. I have tried some of the exercises at the end of the chapters but frustratingly the answers to many of these exercises are not provided. SecondChildUserIdTAG: 345671 SecondChildUserNameTAG: cbjerregaard SecondChildCreateTimeTAG: 2012-09-30T23:35:04Z SecondChildTAG: I agree with @Beneficial. I am having great difficulty with this course (yes, I didn't take calculus, but that's another story) mostly because I just don't understand all the problems. The leaps between the givens in the videos and the homeworks and labs seem to be very great, and not always fully explained. (For example, on of the problems says as a hint to connect another current source and use superposition to solve it, but doing so only complicates the circuit, which can be solved without any modifications.) SecondChildUserIdTAG: 248807 SecondChildUserNameTAG: 1kingsman SecondChildCreateTimeTAG: 2012-10-05T04:30:07Z IndexTAG: 557 TitleTAG: Another way of visualising... Here are the two signals superimposed on the same graph...![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13490121042081366.jpg UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-09-30T13:35:11Z VoteTAG: 5 CoursewareTAG: Week 4 / Incremental Analysis Demo CommentableIdTAG: 6002x_inc_analysis_demo NumberOfReplyTAG: 1 FirstChildTAG: Nice graph. Which tool did you use to generate it? FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-09-30T15:24:45Z SecondChildTAG: Just Paintbox SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-09-30T15:29:21Z IndexTAG: 558 TitleTAG: Happy to be part of this course. Oh Dr Agarwal, IIT madras has filled loads of talent in you sir! ! I'm happy to be part of this course! UserIdTAG: 209164 UserNameTAG: KarthikHegde CreateTimeTAG: 2012-09-29T17:37:47Z VoteTAG: 5 CoursewareTAG: Week 3 / Inside a Mouse CommentableIdTAG: 6002x_inside_a_mouse NumberOfReplyTAG: 0 IndexTAG: 559 TitleTAG: Amazing course! The course is amazing! Every homework or lab forces me to pass to a completely new level of understanding the material, and every lecture video is very enthralling and extremely easy to understand. Respected organizers, thank you for it! UserIdTAG: 326787 UserNameTAG: Angstrem CreateTimeTAG: 2012-09-29T16:01:05Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I totally agree with you.i am still in high school so after second week everything is new and something I have never heardd before.There are highs and lows but overall I am enjoying it. FirstChildUserIdTAG: 99628 FirstChildUserNameTAG: Pranjal16 FirstChildCreateTimeTAG: 2012-09-29T16:08:10Z IndexTAG: 560 TitleTAG: Did Well Had problem only with the signs but everything was ok i hope I can find time playing with this...:-) UserIdTAG: 263040 UserNameTAG: rtabish CreateTimeTAG: 2012-09-28T05:56:06Z VoteTAG: 5 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 2 FirstChildTAG: Hi rtabish! Nice to see that you are having fun! :). See you! Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-28T20:09:39Z SecondChildTAG: Hi, i have the same problems, haha is fun! SecondChildUserIdTAG: 110730 SecondChildUserNameTAG: powerrush SecondChildCreateTimeTAG: 2012-09-30T15:35:25Z FirstChildTAG: I have a problem with the sign too. The simulator give me a minus sign in the current circuit from the source (voltage supply)to the circuit, this was my mistake. FirstChildUserIdTAG: 146473 FirstChildUserNameTAG: pedro_rs FirstChildCreateTimeTAG: 2012-10-08T20:46:41Z IndexTAG: 561 TitleTAG: 6.004x ? Will be 6.004x available on next spring? So we can continue with that once we finish this course?? UserIdTAG: 154373 UserNameTAG: JReyes87 CreateTimeTAG: 2012-09-27T23:54:07Z VoteTAG: 5 CoursewareTAG: Week 2 / Digital logic circuits CommentableIdTAG: 6002x_digital_logic NumberOfReplyTAG: 1 FirstChildTAG: It is very interesting question. FirstChildUserIdTAG: 158740 FirstChildUserNameTAG: Ili FirstChildCreateTimeTAG: 2012-09-29T12:35:32Z IndexTAG: 562 TitleTAG: Recomendation I recommend read chapter 8.2 of the text book to understand small signal method of Mosfet transistor. UserIdTAG: 362201 UserNameTAG: Albertbano CreateTimeTAG: 2012-09-27T20:55:46Z VoteTAG: 5 CoursewareTAG: Week 5 / Two Terminal Connection Exercise CommentableIdTAG: 6002x_2_term_con_e NumberOfReplyTAG: 0 IndexTAG: 563 TitleTAG: Now this was fun! Logic gates and math best combination for fun learn. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-27T20:05:20Z VoteTAG: 5 CoursewareTAG: Week 3 / Switch Resister Model Exercise CommentableIdTAG: 6002x_switch_resistor_model_exercise NumberOfReplyTAG: 0 IndexTAG: 564 TitleTAG: H5P1 Zero Offset Amplifier Hi all, I am stuck with the 3rd Question, in the Lectures we always had the Source Terminal of the MOSFET connected to Ground. Can somebody give me a hint , how to calculate this? Any help is appreciated. Thanks UserIdTAG: 149844 UserNameTAG: amitaratnayake CreateTimeTAG: 2012-09-26T15:11:13Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Hello! Only voltage difference matters, not the absolute value. The most interesting thing of MOSFET is voltage between gate and source. You simply subtract values: $$V_{GS} = V_{in} - V_{S-}$$ It's OK, if one value is negative, for example, Vs- = -10: $$V_{GS} = V_{in} - (-10) = V_{in} +10$$ ---------- And the other way: absolute values are nothing. You can add same voltage to every node, and it will be the same circuit! For example, if you add 10V: | circuit node | default value | value with offset | | -VS | -10 | 0 | | node with ground symbol | 0 | 10 | | Vin | Vin | 10 + Vin | | +VS | 10 | 20 | Now you have MOSFET connected to the zero node =) FirstChildUserIdTAG: 261503 FirstChildUserNameTAG: EugenyL FirstChildCreateTimeTAG: 2012-09-26T17:32:02Z SecondChildTAG: Oh great , Thank you very much ! Thanks to the offsets , it does not confuse me anymore. SecondChildUserIdTAG: 149844 SecondChildUserNameTAG: amitaratnayake SecondChildCreateTimeTAG: 2012-09-26T19:00:20Z SecondChildTAG: You're a life saver, EugenyL. It seems intuitive now but I just couldn't see it before. Thank you. SecondChildUserIdTAG: 331664 SecondChildUserNameTAG: dwmnctrh3 SecondChildCreateTimeTAG: 2012-09-26T21:33:20Z SecondChildTAG: Thanks in advance. It works. SecondChildUserIdTAG: 366841 SecondChildUserNameTAG: imono SecondChildCreateTimeTAG: 2012-09-28T00:56:58Z SecondChildTAG: I agree with what you are doing in respect to VGS. Can you please explain what you are doing for VDS? SecondChildUserIdTAG: 249325 SecondChildUserNameTAG: lcjenkins SecondChildCreateTimeTAG: 2012-10-04T08:56:41Z SecondChildTAG: My understanding that vDS = +Vs SecondChildUserIdTAG: 149923 SecondChildUserNameTAG: Odessa SecondChildCreateTimeTAG: 2012-10-04T15:20:54Z SecondChildTAG: thank you! SecondChildUserIdTAG: 95280 SecondChildUserNameTAG: Fortyq SecondChildCreateTimeTAG: 2012-10-08T15:57:30Z SecondChildTAG: Cant thank you enough for the superb clarification! SecondChildUserIdTAG: 232157 SecondChildUserNameTAG: GayatriTR SecondChildCreateTimeTAG: 2012-10-09T14:30:00Z SecondChildTAG: Can someone explain better? i know that VIN max = VT+ (-1+sqrt(1+2kRlVs))/KRl But i know i have wrong the answer becouse the offset. I have tried for all the ways and i couldnt take the right answer. Plz if someone can explain me this... SecondChildUserIdTAG: 78992 SecondChildUserNameTAG: chuso06pdm SecondChildCreateTimeTAG: 2012-10-09T14:35:28Z SecondChildTAG: many thanks i knew VGS = VIN-Vs- but i forgot VS =Vs+ - Vs- .That's why im always calculate wrong result. SecondChildUserIdTAG: 114913 SecondChildUserNameTAG: ngoctuan SecondChildCreateTimeTAG: 2012-10-10T12:13:24Z SecondChildTAG: Thank you very much ngoctuan, that was my mistake too!! SecondChildUserIdTAG: 299251 SecondChildUserNameTAG: TheRedBlackOne SecondChildCreateTimeTAG: 2012-10-10T14:34:27Z SecondChildTAG: Can you clarify it to me ? You are using this equation VIN max = VT+ (-1+sqrt(1+2kRlVs))/KRl ? VS = Vs+ - Vs- in this case equals 2 ? where are you using VGS ? SecondChildUserIdTAG: 447792 SecondChildUserNameTAG: Icex_ SecondChildCreateTimeTAG: 2012-10-11T19:50:32Z SecondChildTAG: Excellent post. Using this idea of offset, I was able to get the correct answer for all 3 parts. SecondChildUserIdTAG: 248807 SecondChildUserNameTAG: 1kingsman SecondChildCreateTimeTAG: 2012-10-11T22:58:26Z FirstChildTAG: I want to know the minimum value of Vout in Saturation region! ids=K/2{Vout)^2? FirstChildUserIdTAG: 89440 FirstChildUserNameTAG: yuk FirstChildCreateTimeTAG: 2012-09-30T04:14:16Z FirstChildTAG: I am stuck with part 2 here. What am i supposed to do? substitute all V's zeros ? plz guide. FirstChildUserIdTAG: 97340 FirstChildUserNameTAG: AnkitRana FirstChildCreateTimeTAG: 2012-10-07T09:54:03Z SecondChildTAG: +1 SecondChildUserIdTAG: 297960 SecondChildUserNameTAG: sebseb95 SecondChildCreateTimeTAG: 2012-10-07T17:42:03Z SecondChildTAG: Vo=Vs-k/2*Rl*(Vi-Vt)^2 use this equation where vi & vo both are 0 SecondChildUserIdTAG: 92895 SecondChildUserNameTAG: shihab2555 SecondChildCreateTimeTAG: 2012-10-10T08:10:37Z IndexTAG: 565 TitleTAG: For Mac users! We can plot the functions using the Grapher App. It's very simple when dealing with transcendental equations, like here. Take a look! ![Grapher plot][1] [1]: https://edxuploads.s3.amazonaws.com/13486032911343631.jpg UserIdTAG: 278353 UserNameTAG: vgmariucci CreateTimeTAG: 2012-09-25T20:04:46Z VoteTAG: 5 CoursewareTAG: Week 3 / Graphical Analysis of Nonlinear Circuits Exercise CommentableIdTAG: 6002x_graphical_analysis_of_nonlinear_circuits_exercise NumberOfReplyTAG: 0 IndexTAG: 566 TitleTAG: Explicit Solution (1) Download GraphCalc http://www.graphcalc.com/ . (2) Find the two coordinates for the Loadline, they are (0,4) on the Y-axis (ordinate) and (32.8,0) on the X- Axis (Abscissa) (3) Follow the steps explicitly detailed out here using our coordiantes (0,4) and (32.8,0) http://www.mathsisfun.com/algebra/line-equation-2points.html to obtain the Equation for the load line. (4) Type in the two equations (I=v^3 and y=-(0.121*x)+4) into the two separate fields and tick them both so that they appear superimposed over one another. (5) Hover your mouse over the point at which the two lines intersect. y is your current and x is your voltage at that point. Voila! UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-09-25T11:08:35Z VoteTAG: 5 CoursewareTAG: Week 3 / Graphical Analysis of Nonlinear Circuits Exercise CommentableIdTAG: 6002x_graphical_analysis_of_nonlinear_circuits_exercise NumberOfReplyTAG: 1 FirstChildTAG: Still dont get it FirstChildUserIdTAG: 331441 FirstChildUserNameTAG: abdessamade FirstChildCreateTimeTAG: 2012-09-25T19:08:36Z SecondChildTAG: What don't you get? SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-09-25T20:10:48Z SecondChildTAG: strange I got 1.61 volts...got a tick but 4.2 Amperes is wrong..why?....Thank you today I learn how to use Google graphing calculator by following your instructions :) SecondChildUserIdTAG: 122513 SecondChildUserNameTAG: Mona77 SecondChildCreateTimeTAG: 2012-10-24T10:34:07Z IndexTAG: 567 TitleTAG: SOLUTION FOR ACCESSING YOUTUBE IN THE COUNTRIES WHERE IT IS BLOCKED. http://www.unblockyoutube.co.uk/permalink.php?url=vDNG0dxURune3%2BS1NGBmjuY2QOXqgb7%2BfW43lAW22zU%3D use above given link to access youtube if its blocked in your country. now go on to watch any video on your account of edx, it will show "an error occurred please try again later" don't be annoyed. click on the bottom on the right side of video, it shows the option of "watch on youtube" click that. it will show error again, now click on the share button which is right under the video on youtube, it will show a link, copy that link & paste it to the quick browse links option on the website for accessing youtube which stated above & click go. now you are ready to watch any video of edx classes. thanks. hope i will solve your solve your problem. for example: going through above given process i got the access to the videos of 5th week. final generated link of a video of 5th week is given below. http://www.unblockyoutube.co.uk/permalink.php?url=xIQ0pQ5KYfPxWSu2MueIUEqG6h4V1AQY88f71wGygrWKVG85N8c%2ByrGEUv6xmPb93JgFkx3Jt7XbC53qwhLsPyeWxG2gnyeijwy%2FnyX7NjQ%2BMVFu0TpC35xCo7Plm513 UserIdTAG: 72709 UserNameTAG: sarwansagar CreateTimeTAG: 2012-09-24T14:31:08Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: it works :) thanks for contributing a valuable link. FirstChildUserIdTAG: 460906 FirstChildUserNameTAG: comsat002 FirstChildCreateTimeTAG: 2012-09-24T18:08:09Z SecondChildTAG: Take care. SecondChildUserIdTAG: 372623 SecondChildUserNameTAG: Amitraj SecondChildCreateTimeTAG: 2012-09-25T11:55:04Z FirstChildTAG: You can also, see in the Wiki [Lecture Videos Download][1]:). Also see more info [here][2]: **Due to the YouTube outage in certain parts of the world, we have provided direct links to download the lecture videos, and extended the Week 2 homework and lab deadlines to the coming Sunday (September 30). Please note that Week 3 deadlines have not changed.** [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/ [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/info FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-26T11:43:37Z IndexTAG: 568 TitleTAG: Goggle Graph Search in google 4-​x/​8.2, x^​3 The graph of above two lines will give the exact operating point i.e. intersection of the two curves. UserIdTAG: 200355 UserNameTAG: bhavj CreateTimeTAG: 2012-09-24T07:44:00Z VoteTAG: 5 CoursewareTAG: Week 3 / Graphical Analysis of Nonlinear Circuits Exercise CommentableIdTAG: 6002x_graphical_analysis_of_nonlinear_circuits_exercise NumberOfReplyTAG: 1 FirstChildTAG: This worked, the one time I'd seen it posted before, but it's not working now. Would anyone know a reason? FirstChildUserIdTAG: 280220 FirstChildUserNameTAG: pilgrimCycle FirstChildCreateTimeTAG: 2012-09-25T19:00:43Z IndexTAG: 569 TitleTAG: Lab2 Seems broken I've produced virtually the same graph as is given as an example and yet it is still giving a wrong answer. The image below shows the graph I produced and the 'x' after checking it. If I'm doing something wrong I have know idea what. ![The graph of the circuit I designed][1] [1]: https://edxuploads.s3.amazonaws.com/1348452270174491.jpg UserIdTAG: 306110 UserNameTAG: AEJ CreateTimeTAG: 2012-09-24T02:03:48Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: The voltages are close but it does look slightly distorted. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-24T02:14:08Z SecondChildTAG: Example on the lower part of your waveform, it peaks out a little too high, yours looks like around 190mV. It should be closer to 166mV. Same with the top part of your waveform. Your sinusoid signal looks "stretched". SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-24T02:18:47Z SecondChildTAG: Mine kept giving me a red X over and over. I tweaked some values and finally was getting correct (or really close) voltages and still got a red X. I clicked on reset, re-drew the circuit from scratch and the next time I ran the transient analysis and then clicked "check", I got the coveted green checkmark.... SecondChildUserIdTAG: 375500 SecondChildUserNameTAG: Brej7665 SecondChildCreateTimeTAG: 2012-09-24T02:31:46Z SecondChildTAG: Right on. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-24T02:37:22Z FirstChildTAG: Never mind, Apparently appearances can be deceiving. Got it to work FirstChildUserIdTAG: 306110 FirstChildUserNameTAG: AEJ FirstChildCreateTimeTAG: 2012-09-24T02:17:21Z FirstChildTAG: Every time you guys say that the circuit simulator is broken, I freak out :( Please be gentle FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-24T04:27:45Z FirstChildTAG: it's not broken just leave the transit open and click on check while it open if you closed it and clicked check it will give you X even if your numbers right FirstChildUserIdTAG: 340227 FirstChildUserNameTAG: yaser_x FirstChildCreateTimeTAG: 2012-09-24T07:59:42Z IndexTAG: 570 TitleTAG: Improvement suggestion It would be extremely beneficial for students, if implemented calculator would have a feature of remembering few last calculations. It could be done as a drop down list. UserIdTAG: 157610 UserNameTAG: mradziwo CreateTimeTAG: 2012-09-23T20:31:33Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 571 TitleTAG: Consult Greetings. I'm having trouble finding "e", my result is -1.69 respect to the ground given. I want to know if someone else share the same result that me. If different, please explain how did you got the result? Thanks in advance. Heaklig UserIdTAG: 471327 UserNameTAG: Heaklig CreateTimeTAG: 2012-09-23T19:57:17Z VoteTAG: 5 CoursewareTAG: Week 2 / Node equation review CommentableIdTAG: 6002x_S3E1_NodeEquationReview NumberOfReplyTAG: 1 FirstChildTAG: you have maybe overseen the +/-direction of on of the sources? but can you maybe first post your equations for 'e'? FirstChildUserIdTAG: 131747 FirstChildUserNameTAG: hasi FirstChildCreateTimeTAG: 2012-09-23T20:32:51Z IndexTAG: 572 TitleTAG: Search Specifics. I notice if I do a search for homework 4, or "homework 4", that I get results with either homework or 4, just not homework 4. Same as when searching for Lab 4, anything with the word Lab shows up. Maybe the search engine could be improved slightly as time goes on. Thank you for your consideration. UserIdTAG: 292543 UserNameTAG: Pennypacker CreateTimeTAG: 2012-09-23T18:08:56Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Just search using lab4 or lab2, then it works fine. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-09-23T18:31:06Z SecondChildTAG: You are right, if I don't put a space in there, I get a result. I can't help but wonder if I am still missing results with a space in them though. This should do for now, thanks. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-23T18:37:45Z IndexTAG: 573 TitleTAG: Hi I don`t understand, I write a algebraic expresion, but say "Invalid input: v3 a1 v1 v2 not permitted in answer" UserIdTAG: 129492 UserNameTAG: alejomorenogomez CreateTimeTAG: 2012-09-21T08:24:41Z VoteTAG: 5 CoursewareTAG: Week 2 / Linear combinations of source strenghts CommentableIdTAG: 6002x_S3E2_Linear_Combinations_of_Source_Strengths NumberOfReplyTAG: 2 FirstChildTAG: write the expression in terms of R1,R2 and R3. In addition, don't write a1= .... but write the expression directly without a1= FirstChildUserIdTAG: 463816 FirstChildUserNameTAG: theKnight FirstChildCreateTimeTAG: 2012-09-21T09:18:43Z SecondChildTAG: U NEED TO MENTION THE MULTIPLICATION SIGN FOR EXAMPLE..IF IT SAYS R1R2 INVALID...U NEED TO WRITE IT THIS WAY R1*R2 SecondChildUserIdTAG: 433368 SecondChildUserNameTAG: SurbhiMahajan SecondChildCreateTimeTAG: 2012-09-21T10:47:18Z FirstChildTAG: Hola alejomorenogomez! Eso significa que no se permiten ingresar dichas variables en las respuestas, es decir, que en los casilleros blancos debes ingresar, de tener una ecuación v1=i1*R1+i2*R2, y de requerirte v1 expresada en función de ciertas variables (i1,R1,i2,R2), lo que se encuentra del lado derecho del = , o sea, sin escribir v1 en la respuesta: i1*R1+i2*R2 (esto es un ejemplo arbitrario). También el mensaje de error se puede deber porque te encuentras expresando las ecuaciones en función de variables no solicitadas. Saludos, Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-21T11:31:07Z IndexTAG: 574 TitleTAG: homogeneity and superposition what is difference between homogeneity and superposition? they are similar in the sense that in the linear network as per input output changes. UserIdTAG: 374547 UserNameTAG: VenkateshKurminla CreateTimeTAG: 2012-09-20T05:30:26Z VoteTAG: 5 CoursewareTAG: Week 2 / Superposition CommentableIdTAG: 6002x_superposition NumberOfReplyTAG: 0 IndexTAG: 575 TitleTAG: Where is the Admin???? Why not RESPONDING...?? **The YouTube has been BLOCKED by the Pakistan's government in our country PAKISTAN due to a wide range of protest against hated Anti-Islam movie........ Kindly tell us the alternate way to view the video.... or do something for the students here ... It will be great kindness for us by you... Regards Hassan** UserIdTAG: 107038 UserNameTAG: Hassankhan CreateTimeTAG: 2012-09-19T17:38:30Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Kindly upload it on some alternative website as videolectures.net or any alternative plan. We shall be tankful for that. FirstChildUserIdTAG: 408848 FirstChildUserNameTAG: Sher11 FirstChildCreateTimeTAG: 2012-09-19T18:23:53Z SecondChildTAG: Hassan you are going bit harsh, and it should not be here. SecondChildUserIdTAG: 408848 SecondChildUserNameTAG: Sher11 SecondChildCreateTimeTAG: 2012-09-19T18:25:28Z SecondChildTAG: Hassan is not getting harsh. In my opinion he wants quick action by the admin or staff or whatever is responsible to resolve this issue as it is serious issue otherwise all students in this area will not able to catch the course in sequence. SecondChildUserIdTAG: 358539 SecondChildUserNameTAG: syed_abdullah SecondChildCreateTimeTAG: 2012-09-20T15:16:44Z SecondChildTAG: sher1: I'm not of that mind...As 23rd Sep is the last date of submission of homework 2 and lab 2... and only 2 days to go ..... yet no action by admin.... if quick action not taken then all will be astray....:P Thanks...:) SecondChildUserIdTAG: 107038 SecondChildUserNameTAG: Hassankhan SecondChildCreateTimeTAG: 2012-09-21T09:28:21Z FirstChildTAG: please take quick action regarding this issue otherwise the students of this area will not be able to follow the course completely due to this restriction FirstChildUserIdTAG: 358539 FirstChildUserNameTAG: syed_abdullah FirstChildCreateTimeTAG: 2012-09-20T15:17:42Z FirstChildTAG: You can see in the Wiki [Lecture Videos Download][1]:). Also see more info [here][2]: **Due to the YouTube outage in certain parts of the world, we have provided direct links to download the lecture videos, and extended the Week 2 homework and lab deadlines to the coming Sunday (September 30). Please note that Week 3 deadlines have not changed.** [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/ [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/info FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-26T11:48:48Z IndexTAG: 576 TitleTAG: answer is correct my equation is (5-e)/6800 + (7.2-e) / 5600 =0 solve this equation and you get e=6.2064516129 just mark the node before R1 with +5 which is the positive value of V1 and mark the node after R2 with (+7.2) which is the negative value of V2 and make node analysis at ( e ) and no matter your current direction is , it will solve it UserIdTAG: 285675 UserNameTAG: Ascot CreateTimeTAG: 2012-09-18T19:05:18Z VoteTAG: 5 CoursewareTAG: Week 2 / Node equation review CommentableIdTAG: 6002x_S3E1_NodeEquationReview NumberOfReplyTAG: 1 FirstChildTAG: thanks FirstChildUserIdTAG: 67238 FirstChildUserNameTAG: guisaojulian FirstChildCreateTimeTAG: 2012-09-20T00:51:55Z IndexTAG: 577 TitleTAG: Problems in H2P1 Hello, first, I apologize if this question was made before, but I have been looking for information and I wasn't answered... The problem I found with this exercise is the same than other people found. I proved the differents combinations of resistors using an Excel table and I managed to get the voltage of 12V with exactly the 10% error I was asked by using R1 of 82K (82000 O) and R2 of 18K (18000 O). However the web sais that it's wrong. I would think I am wrong, but when I fill the gaps of Vmin and Vmax I get it is a correct answer. I don't know what I didn't well... The only thing I'm not completely sure to have done well is calculating RThevening, so I ask: Is it simply the parallel between R1&R2? Thank you PD: If I dind't make any misstake calculating Rth, it is of 14.7K, that is in the range of 10K-30K UserIdTAG: 347507 UserNameTAG: pedquiig CreateTimeTAG: 2012-09-17T20:02:46Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hello, If R1 were 82k and R2 18k, Rth = R2 // R1 = 82k // 18k = 14.76V. Vth = R2/(R2+R1) * Vin = 7.2! <= wrong... ========= Rth = R2 // R1. Rth = 39k // 22k = 14.06V Vth = R2/(R2+R1) * Vin = 22k/(39k+22k) * 40 = 14.426V. Va = 1.1 R2/(1.1 R2 + 0.9 R1) * Vin Va = 1.1*22k / (1.1*22k+0.9*39k)*40 = 16.32 Vb = 0.9 R2/(0.9 R2 + 1.1 R1) * Vin Vb = 0.9*22k/(0.9*22k + 1.1*39k) * 40 = 12.63 ========== FirstChildUserIdTAG: 385692 FirstChildUserNameTAG: tpfslima FirstChildCreateTimeTAG: 2012-09-18T01:08:06Z SecondChildTAG: Thank you, tpfslima! I appreciate your explanation. It works. SecondChildUserIdTAG: 149923 SecondChildUserNameTAG: Odessa SecondChildCreateTimeTAG: 2012-09-18T03:49:07Z SecondChildTAG: hellow mate sorry to distrube u ....my Q is that i have to set my vout = 7.5 V .An additional requirement is that the Thevenin resistance as seen from the output terminals is between 10kΩ and 30 K ohm,,,which seem to me it describe me that my R2 must be in the range of 10kΩ and 30 K ohm ?? pleae some one help me SecondChildUserIdTAG: 415376 SecondChildUserNameTAG: imab90 SecondChildCreateTimeTAG: 2012-09-18T05:10:59Z SecondChildTAG: how u add 1.1 in and 0.9 in this explaination and how 14.06 and 14.26 is correct SecondChildUserIdTAG: 164689 SecondChildUserNameTAG: muhammadfaizan SecondChildCreateTimeTAG: 2012-09-18T16:13:07Z SecondChildTAG: 1.1 represents the 110% and 0.9 the 90% because there is 10% of tolerance SecondChildUserIdTAG: 253807 SecondChildUserNameTAG: DCan10 SecondChildCreateTimeTAG: 2012-09-19T00:38:56Z SecondChildTAG: Why Vin is = 40V? SecondChildUserIdTAG: 306534 SecondChildUserNameTAG: sonchu SecondChildCreateTimeTAG: 2012-09-19T03:23:46Z SecondChildTAG: the thing which i dnt understand is how R1||R2? it seems in series to me SecondChildUserIdTAG: 122513 SecondChildUserNameTAG: Mona77 SecondChildCreateTimeTAG: 2012-09-19T06:36:26Z SecondChildTAG: Thanks, but my value for Vin is of 60V... SecondChildUserIdTAG: 347507 SecondChildUserNameTAG: pedquiig SecondChildCreateTimeTAG: 2012-09-19T16:57:07Z SecondChildTAG: when I calculate the Vout, I obtain 10.8V 10.8/12=0.9---->It's exatly a 10% error... SecondChildUserIdTAG: 347507 SecondChildUserNameTAG: pedquiig SecondChildCreateTimeTAG: 2012-09-19T16:58:41Z SecondChildTAG: thanks tpfslima... :) SecondChildUserIdTAG: 222911 SecondChildUserNameTAG: bhaswardg SecondChildCreateTimeTAG: 2012-09-20T06:46:14Z IndexTAG: 578 TitleTAG: Where are the solutions to the week 1 homework? Unlike last term, we are releasing the detailed solutions via the "show solution" button instead of PDF. You will find that the "show answer" button below the week 1 problems is visible and will explain how to find the answer to a problem. UserIdTAG: 96452 UserNameTAG: Lyla CreateTimeTAG: 2012-09-17T17:18:19Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: That's really neat! The formatting of the solutions is so much better because of the Latex code. Thanks! :-) FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-09-17T18:26:14Z SecondChildTAG: Also it looks like the solutions use the randomized values for each person's assignment. Is this correct (or was it by chance that my values matched)? SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-09-17T18:29:38Z SecondChildTAG: Will the labs' solutions be provided like this as well? (I don't see Lab 1's solutions posted yet) SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-09-17T19:41:33Z IndexTAG: 579 TitleTAG: MANAGMENT HELP ME.. BECAUSE OF ELECTRICITY CRISES IN MY CITY (COUNTRY: PAKISTAN) + YOUTUBE PROBLEM I WAS NOT ABLE TO COMPLETE MY HOME WORK WITHIN DUE DATE. NOW SUGGEST ME WHAT SHOULD I DO? UserIdTAG: 146171 UserNameTAG: fayyazalijandan CreateTimeTAG: 2012-09-17T16:32:21Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: lol.....nthing now concentrate on 2nd assignment :) FirstChildUserIdTAG: 430030 FirstChildUserNameTAG: imali FirstChildCreateTimeTAG: 2012-09-17T17:40:41Z SecondChildTAG: mamu.....you tube is also not working SecondChildUserIdTAG: 146171 SecondChildUserNameTAG: fayyazalijandan SecondChildCreateTimeTAG: 2012-09-17T19:19:03Z FirstChildTAG: The course grading drops the two lowest grades for homeworks and labs. So you are not in any trouble until you either miss more than two of each, or get less than 100% on any of the remaining 10 labs and homeworks that will constitute part of your grade (the remaining part of your grade will consist of midterm and final exam grades). Youtube is a huge problem, it's blocked in many countries and workplaces. You can try using a proxy server or VPN to bypass country restrictions, or simply download the MIT OCW version of the videos, which show Dr. Agarwal lecturing in front of a live class. I happen to prefer the OCW live class video. It's hosted on archive.org, which is not normally blocked in most places. You can find the MIT OCW 6.002 materials here: http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-electronics-spring-2007/ FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-09-18T02:15:22Z FirstChildTAG: You can see in the Wiki [Lecture Videos Download][1]:). Also see more info [here][2]: **Due to the YouTube outage in certain parts of the world, we have provided direct links to download the lecture videos, and extended the Week 2 homework and lab deadlines to the coming Sunday (September 30). Please note that Week 3 deadlines have not changed.** [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/ [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/info FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-26T11:50:27Z IndexTAG: 580 TitleTAG: Operations Symbols Staff: Is it correct to use * for multiplication? My keyboard does not have symbol used in correct homework answers for H1P1. UserIdTAG: 211150 UserNameTAG: Bernie1961 CreateTimeTAG: 2012-09-17T15:55:58Z VoteTAG: 5 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 2 FirstChildTAG: You can find symbols on internet and use copy & paste !!!! Another option use a virtual keyboard on your screen FirstChildUserIdTAG: 149058 FirstChildUserNameTAG: sotoroman FirstChildCreateTimeTAG: 2012-09-17T16:18:11Z FirstChildTAG: For writing things down we will often omit the multiplication sign, but you should always include it when typing in answers to homework. If your keyboard does not have the * sign by itself, you can reconfigure your keyboard to work as a US keyboard. In windows this is in Control Panel->Region and Language->Keyboards and Languages. FirstChildUserIdTAG: 151427 FirstChildUserNameTAG: RichardZ FirstChildCreateTimeTAG: 2012-09-17T16:17:56Z IndexTAG: 581 TitleTAG: Direction of flow of current The problem i always encountered here is the direction of flow of current.Can anyone explain how the current will flow? UserIdTAG: 171289 UserNameTAG: harbey CreateTimeTAG: 2012-09-17T14:54:47Z VoteTAG: 5 CoursewareTAG: Week 2 / Node equation review CommentableIdTAG: 6002x_S3E1_NodeEquationReview NumberOfReplyTAG: 0 IndexTAG: 582 TitleTAG: awesome hello there ,i want to tell you that the interface is very user friendly and the system is ultimate....super like for this UserIdTAG: 230108 UserNameTAG: u11ee044 CreateTimeTAG: 2012-09-17T08:04:16Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 0 IndexTAG: 583 TitleTAG: Infinite resistor ladder, Fibonacci & Golden Ratio - a longer but pretty method You can use the recursive definition of the resistance of an $ N $-runged ladder derived from the continued fraction form in this [stackexchange discussion][1] to prove by induction that the equivalent resistance, $ R_{eq}(N) $, is equal to the ratio of the $ (2N+1)^{th} $ to the $ (2N)^{th} $ Fibonacci terms multiplied by the input resistance, $ R $. Then you can use the [Euler-Binet formula][2] for the $ N^{th} $ Fibonacci term to [reduce][3] $ R_{eq}(N) $ to $ \left(\phi + \displaystyle\frac{\sqrt{5}}{(-1-\phi)^{2N}-1} \right) \cdot R $, which tends to $ \phi \cdot R = \displaystyle\frac{1+\sqrt{5}}{2} \cdot R $ as $ N $ tends to infinite. Quite a beautiful problem! In fact, one way of *defining* the golden ratio is the ratio $ \displaystyle\frac{A}{B} $ such that $ \displaystyle\frac{A}{B} = \displaystyle\frac{A+B}{A} $, i.e. $ \phi = 1 + \displaystyle\frac{1}{\phi} $, which leads by infinite self-substitution to $ \phi = 1 + \displaystyle\frac{1}{1+\displaystyle\frac{1}{1 + \displaystyle\frac{1}{1+\displaystyle\frac{1}{\ddots}}}} $, i.e. the continued fraction expression for the infinite-runged ladder. So this circuit is an electronic definition of the golden ratio! [1]: http://physics.stackexchange.com/questions/10615/what-would-be-the-effective-resistance-of-the-ladder-of-resistors-having-n-steps/ [2]: https://www.proofwiki.org/wiki/Euler-Binet_Formula [3]: https://www.proofwiki.org/wiki/Ratio_of_Consecutive_Fibonacci_Numbers UserIdTAG: 419029 UserNameTAG: resonate CreateTimeTAG: 2012-09-16T22:02:21Z VoteTAG: 5 CoursewareTAG: Week 1 / Long Resistor Chains CommentableIdTAG: 6002x_long_resistor_chains NumberOfReplyTAG: 1 FirstChildTAG: Goosebumps. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-16T22:39:14Z SecondChildTAG: I remember years ago with the "infinite ladder". It's what EECS professors all over the globe use to **scare** their students on midterms. I also remember starting to work it out the long way, using the parallel resistor rule, and remember by the third set "I think I see a pattern here...I've seen it before in Calculus class when studying sequences and series." I took a quick look in my old, tattered high school calculus textbook and saw the same pattern under the Golden Ratio and said "These EECS professors are **twisted** if they expect students to think this deep." SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-09-16T23:03:27Z SecondChildTAG: What a bunch of sadists. lol I also noticed that the midterm is not mid-term, it's pretty much phi. 12 weeks /1.618 = ~7 1/2 weeks, which coincidentally is the date of the midterm. Spooky. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-16T23:33:46Z IndexTAG: 584 TitleTAG: Power at best load ???? H2P2: Solar Power in fact my Thevenin equivalent resistance and optimum value is right but i currently mess up with second part mean Power at best load ...some 1 please explain this from instructor .... UserIdTAG: 415376 UserNameTAG: imab90 CreateTimeTAG: 2012-09-16T16:41:09Z VoteTAG: 5 CoursewareTAG: Week 2 / Thevenin Example 2 CommentableIdTAG: 6002x_thevenin_example_2 NumberOfReplyTAG: 1 FirstChildTAG: Best load resistance means the the value of the resistance (**RL**) for which it dissipates maximum power. Its value is same as the **Thevenin equivalent resistance** you are talking about. Now for the question H2P2 you just have calculate the power dissipated in that load resistor. FirstChildUserIdTAG: 95058 FirstChildUserNameTAG: KeshawP FirstChildCreateTimeTAG: 2012-09-16T17:18:35Z SecondChildTAG: ok by which formula i^2 R or V^2 / R .....but i 1000 time calculated this but at the end wrong answer ... kindly if u explain me through formula as well .. Indian ? SecondChildUserIdTAG: 437230 SecondChildUserNameTAG: somibahrian SecondChildCreateTimeTAG: 2012-09-16T17:28:15Z SecondChildTAG: You have **Rth=2*Rs+Rp**. So load resistance, **Rl=Rth** (maximum power transfer condition). Now you'll be having a circuit like **Vth----Rth----Rl**. Find the current in the circuit using the formula **i=Vth/(Rl+Rth)** (All in series). Now having the current in hand you can find the power dissipated in the resistor by using the formula **P=i^2*Rth**. SecondChildUserIdTAG: 95058 SecondChildUserNameTAG: KeshawP SecondChildCreateTimeTAG: 2012-09-16T17:46:52Z SecondChildTAG: **Vth----Rth----Rl** (All in series). SecondChildUserIdTAG: 95058 SecondChildUserNameTAG: KeshawP SecondChildCreateTimeTAG: 2012-09-16T17:48:09Z SecondChildTAG: that is plus sign ???? or divide ?? SecondChildUserIdTAG: 430030 SecondChildUserNameTAG: imali SecondChildCreateTimeTAG: 2012-09-16T17:49:59Z SecondChildTAG: Yes. SecondChildUserIdTAG: 95745 SecondChildUserNameTAG: RohanMathur SecondChildCreateTimeTAG: 2012-09-16T18:20:54Z SecondChildTAG: Plus + Divide / Multiplication * Equality = SecondChildUserIdTAG: 95745 SecondChildUserNameTAG: RohanMathur SecondChildCreateTimeTAG: 2012-09-16T18:21:50Z SecondChildTAG: I think KeshawP has posted the right solution. There should not be any doubt now. SecondChildUserIdTAG: 95745 SecondChildUserNameTAG: RohanMathur SecondChildCreateTimeTAG: 2012-09-16T18:24:21Z SecondChildTAG: basically precision in round off digit is the only problem in this Q :) any how thanks somi bahrian , imali , rohan , and specially KeshawP thnaks alot SecondChildUserIdTAG: 415376 SecondChildUserNameTAG: imab90 SecondChildCreateTimeTAG: 2012-09-16T18:27:27Z SecondChildTAG: why and how RL is equal to RTH in this question? SecondChildUserIdTAG: 122513 SecondChildUserNameTAG: Mona77 SecondChildCreateTimeTAG: 2012-09-17T12:42:01Z SecondChildTAG: That is because of the maximum power transfer theorem. The object here is to figure that out for yourself by differentiating $I^2 \cdot R_L$ with respect to $R_L$ to find the maximum power. SecondChildUserIdTAG: 392100 SecondChildUserNameTAG: glenng SecondChildCreateTimeTAG: 2012-09-18T00:27:37Z SecondChildTAG: how to calculate vth SecondChildUserIdTAG: 444708 SecondChildUserNameTAG: jagadeeshv16 SecondChildCreateTimeTAG: 2012-09-21T13:38:39Z SecondChildTAG: what is the value of vth? SecondChildUserIdTAG: 403493 SecondChildUserNameTAG: aiai SecondChildCreateTimeTAG: 2012-09-21T14:26:21Z SecondChildTAG: what's the method i can use to calculate vth??? SecondChildUserIdTAG: 206794 SecondChildUserNameTAG: Abdofarrag SecondChildCreateTimeTAG: 2012-09-21T14:41:21Z IndexTAG: 585 TitleTAG: Notification Is there a way to receive notification when another person replied to thread/comment? Any tutorial for posting mathematical formula in the Discussion Forum? UserIdTAG: 89084 UserNameTAG: anakluhur CreateTimeTAG: 2012-09-16T15:42:34Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: If you place a dollar sign, (Shift 4) at each end of your equation, it will show up mathy like. :) $Pennypacker + coffee /time = weeee$ FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-16T15:55:49Z SecondChildTAG: 1) I don't think so but it should be possible... 2) The mathematical formula follow the "LaTeX language" via MathJax. You must enclose your commands between two dollars `$$` symbols. The most useful commands are : - **Division / Fraction**: `$\frac{A}{B}$` gives $\frac{A}{B}$ - **Subscript**: `$A_B$` gives $A_B$ - **Superscript / Exponent**: `$A^B$` gives $A^B$ - **Multiplication**: `$A \cdot B$` gives $A \cdot B$ Note that if you want to use more letters in a subscript you can do it using {}, e.g. "A_{BC}" gives $A_{BC}$. SecondChildUserIdTAG: 396446 SecondChildUserNameTAG: RousseauxS SecondChildCreateTimeTAG: 2012-09-16T16:20:23Z SecondChildTAG: Let me give it a try: $P =$ $I^2 \cdot R$ and $I_1 =$ $\frac{{v}_{1}}{R}$ Cool, man! Thanks, **RousseauxS!** Is there a website you can recommend that gives more LaTeX commands like this in the tabular format like you pasted? SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-09-16T16:34:57Z SecondChildTAG: http://en.wikibooks.org/wiki/LaTeX/Command_Glossary You can try this, though I have not actually tried it myself. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-16T18:05:30Z SecondChildTAG: I've posted the most useful commands here ; ) : [Link][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/undefined/threads/505611d8a717032b00000016 SecondChildUserIdTAG: 396446 SecondChildUserNameTAG: RousseauxS SecondChildCreateTimeTAG: 2012-09-16T18:26:55Z SecondChildTAG: On the top corner of each thread there are star sign. And when you point to it, it says "follow." Still, I didn't receive any notification when someone replied to my thread :( SecondChildUserIdTAG: 89084 SecondChildUserNameTAG: anakluhur SecondChildCreateTimeTAG: 2012-09-17T07:33:41Z FirstChildTAG: On the top corner of each thread there are star sign. And when you point to it, it says "follow." Still, I didn't receive any notification when someone replied to my thread :( FirstChildUserIdTAG: 89084 FirstChildUserNameTAG: anakluhur FirstChildCreateTimeTAG: 2012-09-17T07:34:12Z IndexTAG: 586 TitleTAG: week 1 homework can somebody please give me a hint on the question no. 6 about the resistors having each 1 watt power (week 1 homework) UserIdTAG: 413613 UserNameTAG: shaliesh CreateTimeTAG: 2012-09-16T15:42:24Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: i too have the same problem can any one help FirstChildUserIdTAG: 383699 FirstChildUserNameTAG: Gogineni FirstChildCreateTimeTAG: 2012-09-16T16:31:07Z SecondChildTAG: I am not even getting a green-check for my calculated value for question 5 - the smallest resistor-value - also tried some other numbers, but nothing seems to be it SecondChildUserIdTAG: 403571 SecondChildUserNameTAG: HeikoS SecondChildCreateTimeTAG: 2012-09-16T16:50:42Z SecondChildTAG: for q.no5 make parallel combinations of all the resistor. SecondChildUserIdTAG: 353405 SecondChildUserNameTAG: karthickks SecondChildCreateTimeTAG: 2012-09-16T17:42:43Z FirstChildTAG: I don't know either FirstChildUserIdTAG: 150641 FirstChildUserNameTAG: DungTienTran FirstChildCreateTimeTAG: 2012-09-16T16:11:15Z FirstChildTAG: the smallest resistor would dissipate 1W.consider that and find out voltage and how much power would be dissipated by the other resistors FirstChildUserIdTAG: 219204 FirstChildUserNameTAG: vsriram FirstChildCreateTimeTAG: 2012-09-16T15:47:24Z SecondChildTAG: what does it mean by "the smallest valued composite resistor" SecondChildUserIdTAG: 413613 SecondChildUserNameTAG: shaliesh SecondChildCreateTimeTAG: 2012-09-16T15:53:20Z SecondChildTAG: im wondering the same SecondChildUserIdTAG: 210641 SecondChildUserNameTAG: niharikab SecondChildCreateTimeTAG: 2012-09-16T16:08:51Z SecondChildTAG: But the thing is ..how many resistors are there in circuit?? SecondChildUserIdTAG: 395816 SecondChildUserNameTAG: SushantRoy SecondChildCreateTimeTAG: 2012-09-16T16:26:14Z SecondChildTAG: I'm taking smallest-valued composite resistor to mean the lowest possible Ohm resistor you can make using the 3 resistors given. So basically the configuration from the question just above this one. Find the watts of the circuit of lowest resistance using all three resistors. SecondChildUserIdTAG: 358764 SecondChildUserNameTAG: BrianKolb SecondChildCreateTimeTAG: 2012-09-16T16:59:35Z SecondChildTAG: in Q.3 m typing 5R/3 but itz saying 'counldn't parse' i've typed the algebraic expression in terms of R also but then also it's showing the same error...y its so?? SecondChildUserIdTAG: 383375 SecondChildUserNameTAG: ashuskywalker SecondChildCreateTimeTAG: 2012-09-16T17:59:11Z FirstChildTAG: What I did since I was quite lost I looked at what values I had and which formulas I could use, so we know that P=(V^2)/R and you got two of the values the R= which is 4 and the P = 1 so then plug it in the formula and find the V, that voltage will be of the two 4 ohms resistors and as for the 6 ohms well i got quite lost so i use percentage, (4/x=6/100) found x and subtracted from 100% which is 33.33% more than the value you got from 4 ohm resistor then from there I worked my answer out, hope that helps, there was a post that explain it in a more detail manner, you can search for it with h1p1 and all the posts about it will come up :D FirstChildUserIdTAG: 238005 FirstChildUserNameTAG: isisbocardo FirstChildCreateTimeTAG: 2012-09-16T18:03:00Z FirstChildTAG: For the previous question you found "the smallest valued composite resistor" right? Use that value to continue your calculation. Relate that composite R value to the formalas for Power. P= V^2/R So right now the only variable you're missing is V. Calculate your V for both 4R and 6R, and there'll only be a certain max, because one of those will burn out before the other. You'll find the V you need without burning even one of the R's out. And apply that back into the Power formula for the TOTAL power. P.S. don't get confused between the individual values and total values Hope it helped ^^ FirstChildUserIdTAG: 151030 FirstChildUserNameTAG: Eskamo FirstChildCreateTimeTAG: 2012-09-16T18:12:17Z IndexTAG: 587 TitleTAG: S6E2 graphic problem The shown graphic is not in the proper scale. You could use this one to apply the Graphical Method (click 'View discussion' to see the graphic): ![VIC-Load Line graphic][1] [1]: https://edxuploads.s3.amazonaws.com/13478061546481068.png UserIdTAG: 294348 UserNameTAG: Pere CreateTimeTAG: 2012-09-16T14:36:15Z VoteTAG: 5 CoursewareTAG: Week 3 / Graphical Analysis of Nonlinear Circuits Exercise CommentableIdTAG: 6002x_graphical_analysis_of_nonlinear_circuits_exercise NumberOfReplyTAG: 1 FirstChildTAG: Finally I solved out the problem using analytic method. FirstChildUserIdTAG: 143268 FirstChildUserNameTAG: jborrego FirstChildCreateTimeTAG: 2012-09-16T14:51:06Z IndexTAG: 588 TitleTAG: This is not correct! The chainsaw is the node method... :) I haven't expected using the chainsaw this way. It was really really funny! Thanks! UserIdTAG: 375724 UserNameTAG: sergestus CreateTimeTAG: 2012-09-16T11:40:19Z VoteTAG: 5 CoursewareTAG: Week 2 / Chain saw CommentableIdTAG: 6002x_chain_saw NumberOfReplyTAG: 0 IndexTAG: 589 TitleTAG: information Why the independent voltage graph is vertical?As this voltage is supply to resistor and also use for the flow of current. UserIdTAG: 442829 UserNameTAG: saadzulfiqar CreateTimeTAG: 2012-09-16T05:57:17Z VoteTAG: 5 CoursewareTAG: Week 1 / Various V-I characteristics CommentableIdTAG: 6002x_S1E1_Various_V-I_characteristics NumberOfReplyTAG: 2 FirstChildTAG: the graph indicates that irrespective of the flow of current in the circuit of yours you have a constant voltage generated across the terminals of the voltage source FirstChildUserIdTAG: 102529 FirstChildUserNameTAG: laksh FirstChildCreateTimeTAG: 2012-09-16T10:46:40Z SecondChildTAG: take it this way......... no matter what the current the independent voltage source (like the battery) will have the same volt no matter the current. So for every practical value of I the V of the independent source will always be the same (= V). For example: A battery of 1.5 volt is always 1.5 volt although it may get weak due time (i.e. the current decreases), a 1.5 volt battery is always a 1.5 volt battery If you are confused with the graph itself, just invert the x-y axes and you will get the point SecondChildUserIdTAG: 452690 SecondChildUserNameTAG: shreerajshrestha SecondChildCreateTimeTAG: 2012-09-17T15:24:19Z FirstChildTAG: the X axis represents voltage; the voltage from the voltage source should remain the same at all currents represented on the Y axis FirstChildUserIdTAG: 232178 FirstChildUserNameTAG: Jtangowski FirstChildCreateTimeTAG: 2012-09-16T22:19:01Z IndexTAG: 590 TitleTAG: video lectures, hw and lab as reviewers for midterm exam hi! I just want to ask, will the video lectures, lab and homework be removed after the deadline has passed? Will I still be able to view them days before midterm exam and use them as reviewer? thanks! UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-16T01:40:54Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi anonymous! Based on my experience in the Prototype Course 6.002x I can tell you that the video lectures, lab and homework will **not be** removed after deadline has passed ;). Also, during the exam, you can see all (Hw, labs, video lectures and Textbook). The only thing that is **strictly not allowed** is to discuss the Exam in the Forum. ;) See you! Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-16T02:30:43Z SecondChildTAG: thanks for the clear response! :) SecondChildUserIdTAG: 371000 SecondChildUserNameTAG: Joseph090892 SecondChildCreateTimeTAG: 2012-09-16T02:52:24Z SecondChildTAG: You are welcome joseph090892! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-16T03:24:00Z SecondChildTAG: like :-) SecondChildUserIdTAG: 199839 SecondChildUserNameTAG: vijaykrishnankk SecondChildCreateTimeTAG: 2012-09-16T04:03:48Z SecondChildTAG: thanks :) SecondChildUserIdTAG: 361991 SecondChildUserNameTAG: manvitha SecondChildCreateTimeTAG: 2012-09-16T05:10:33Z SecondChildTAG: Thanks ^_^ SecondChildUserIdTAG: 358712 SecondChildUserNameTAG: emansabah SecondChildCreateTimeTAG: 2012-09-16T05:49:10Z IndexTAG: 591 TitleTAG: SOLVING FINISH!!!YEHUUUUUUUUUUU UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-15T23:48:29Z VoteTAG: 5 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 2 FirstChildTAG: Might as well go for a soda! FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-16T00:02:47Z SecondChildTAG: Just a side note, you don't get graded for completing the overview, but Homework 1 is due in a matter of hours. Sorry for being Captain Buzzkill. lol Have fun. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-16T00:04:47Z SecondChildTAG: Is it? I thought all homework was due by midnight Sunday? SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-09-16T00:17:01Z SecondChildTAG: yes, correct. Midnight Sunday. I just finished too. ;) I want to get a drift as to where u guys are from... I'm from Brazil. SecondChildUserIdTAG: 349595 SecondChildUserNameTAG: danvisk SecondChildCreateTimeTAG: 2012-09-16T01:21:15Z FirstChildTAG: Hello Good evening, I'm from Brazil, I am having difficulty in using the interactive tool FirstChildUserIdTAG: 327787 FirstChildUserNameTAG: REINALDOPARANHOS FirstChildCreateTimeTAG: 2012-09-16T03:33:37Z IndexTAG: 592 TitleTAG: many thanks!!!! thank you everyone! finally,i figured it out UserIdTAG: 285945 UserNameTAG: lindalapiso CreateTimeTAG: 2012-09-15T23:38:09Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Nice to know that lindalapiso ;). Congratulations! FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-16T02:32:18Z IndexTAG: 593 TitleTAG: HOMEWORK1 HI GOT A PROBLEM IN MY HOMEWORKS1 I REALLY SURE THE ANSWER ABOUT THE EQUIVALENT RESISTANCE AS A ALGEBRAIC EXPRESSION BUT MY ANSWERED ARE INVALED COZ I CANT USE R1 + R2 AND SO ON I DONT KNOW WHAT IS THE VALID ANSWER BUT WHEN I SET ASIDE COZ I PROCEED ANOTHER QUESTION THE QUESTION ABOUT NETWORK A TO C ARE GONE I TRY TO FIND THE VALED KEYWORK BUT HOW TO ANSWER IF THE QUESTION GONE UserIdTAG: 235727 UserNameTAG: amor CreateTimeTAG: 2012-09-15T14:06:57Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 7 FirstChildTAG: i have the same problem cannot proceed due to invalid syntax please help FirstChildUserIdTAG: 324463 FirstChildUserNameTAG: Junbacs FirstChildCreateTimeTAG: 2012-09-15T17:04:57Z SecondChildTAG: Please make sure that you add a mathematical function when writing your answer. For example: 5R is going to give you an error while 5*R is considered to be a correct answer. SecondChildUserIdTAG: 167413 SecondChildUserNameTAG: TeTAn SecondChildCreateTimeTAG: 2012-09-16T02:41:47Z FirstChildTAG: I WANT TO GET START BUT I AM LATE.CAN I GET ANY HELP HOW TO START? FirstChildUserIdTAG: 442115 FirstChildUserNameTAG: safiantaseer FirstChildCreateTimeTAG: 2012-09-15T14:22:59Z SecondChildTAG: what you mean that your ;late the lecture? even me i dont know the exact time coz its different the time here SecondChildUserIdTAG: 235727 SecondChildUserNameTAG: amor SecondChildCreateTimeTAG: 2012-09-15T14:32:53Z SecondChildTAG: In the top left corner pick Courseware. See all the videos from Week 1, answer all the exercises and then solve the homework and the lab. Those are due tomorrow, so hurry up. SecondChildUserIdTAG: 73251 SecondChildUserNameTAG: vhvidall SecondChildCreateTimeTAG: 2012-09-15T14:54:57Z FirstChildTAG: I AM ALSO EXPERIENCING SYNTAX INVALID HOW TO WRITE INALGEBRaic FirstChildUserIdTAG: 324463 FirstChildUserNameTAG: Junbacs FirstChildCreateTimeTAG: 2012-09-15T14:54:36Z SecondChildTAG: nothing dude ,instead writing the answer as 0.666667 write it as 2/3 and then check your answer SecondChildUserIdTAG: 102529 SecondChildUserNameTAG: laksh SecondChildCreateTimeTAG: 2012-09-15T15:33:00Z FirstChildTAG: ha what is the value of v1 ,i trying head but ,its give me wrong ans ,i use KCL ,method ,but cant find ANS , -10v ,i guss, bu give me some help , and clue , so then i do it FirstChildUserIdTAG: 133891 FirstChildUserNameTAG: junaidkarim FirstChildCreateTimeTAG: 2012-09-15T14:28:44Z SecondChildTAG: what you mean the homework ? that ask the value of v1? ah thats makes me crazy also try to calculate my answer is 1 but my answer is wrong :( SecondChildUserIdTAG: 235727 SecondChildUserNameTAG: amor SecondChildCreateTimeTAG: 2012-09-15T14:35:37Z FirstChildTAG: Dont use any numeric values just use R FirstChildUserIdTAG: 225562 FirstChildUserNameTAG: lasyapinky FirstChildCreateTimeTAG: 2012-09-15T14:13:56Z SecondChildTAG: yah tried to use R but to right my answer the network A that is series right? so R= R1 + R2 + R3 so i only right R = R + R + R? and also the question are gone never found it and about the lab i dont know how to rotate the ohms SecondChildUserIdTAG: 235727 SecondChildUserNameTAG: amor SecondChildCreateTimeTAG: 2012-09-15T14:20:22Z SecondChildTAG: just write 3*R for series and for parallel write R/3. jst connect the wires from resistors end to the point wherever you want it.no need to rotate it. SecondChildUserIdTAG: 359813 SecondChildUserNameTAG: Uppili SecondChildCreateTimeTAG: 2012-09-15T14:26:36Z SecondChildTAG: ah by the way i dont know what the exact time there im here in the philippines the dealine of my lab and homeworks sept. 16 for i know sept 16 there is night here so i have a chance to answer my lab? if have how hours left? or i mean what exact time my dealine coz until now never answer my lab :( SecondChildUserIdTAG: 235727 SecondChildUserNameTAG: amor SecondChildCreateTimeTAG: 2012-09-15T14:28:55Z SecondChildTAG: really just only write 3*R? but the all questions the 1st part is gone how to answer :(( SecondChildUserIdTAG: 235727 SecondChildUserNameTAG: amor SecondChildCreateTimeTAG: 2012-09-15T14:40:20Z SecondChildTAG: and the parallel only write R/3? wow thanx for that but thats what i said my question are gone so how to correct my answer :( ah about the network C that is series and parallel so how to write a valed keyword Rp = R1 + Rp? or Rp = R2(R3+R4)/R2+(R3+R4) this is the series and parallel right? how to write the correct keyword? and how to back the question in my homework coz are gone means deadline? so no chance to pass my homework? SecondChildUserIdTAG: 235727 SecondChildUserNameTAG: amor SecondChildCreateTimeTAG: 2012-09-15T14:50:58Z FirstChildTAG: its makes me crazy im so discouraged :( the question of my homeworks about the algebraic are gone and i know the answer :( how to answer those ? becasue of those i cant concentrate my lab so until now never answer im looking my progress below 60% so really discourged only lab that never answer and my homeworks the problem is some question are gone hoooo FirstChildUserIdTAG: 235727 FirstChildUserNameTAG: amor FirstChildCreateTimeTAG: 2012-09-15T15:14:25Z FirstChildTAG: Can I change the answer? FirstChildUserIdTAG: 156835 FirstChildUserNameTAG: kphariprakash1968 FirstChildCreateTimeTAG: 2012-09-16T16:12:11Z IndexTAG: 594 TitleTAG: Cool! What happened to the Forum Discussion? Hi, I have noticed that something changed with the Design of the Plataform ;) Creating New Posts: ![enter image description here][1] Organization by topics: ![enter image description here][2] I have to explore this more haha. It is really cool! The thing is that I can not discover where is the recent Posts...? [1]: http://s13.postimage.org/c5ei4ctav/forum1.png [2]: http://s8.postimage.org/4bjx0c5id/forum2.png UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2012-09-15T02:56:48Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Image uploads should work now as well. FirstChildUserIdTAG: 11 FirstChildUserNameTAG: dormsbee FirstChildCreateTimeTAG: 2012-09-15T03:02:53Z SecondChildTAG: Thank you Dave! Yes, it works now. Thank you!!! Good work! ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13476786567720766.png SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-15T03:12:36Z FirstChildTAG: Oh, here!! ;) I figured out! haha! Nice! Thanks edX! FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-15T02:58:02Z SecondChildTAG: you're awesome, myrimit :) SecondChildUserIdTAG: 96452 SecondChildUserNameTAG: Lyla SecondChildCreateTimeTAG: 2012-09-15T03:01:15Z SecondChildTAG: Hi Lyla! you are super-awesome too ;)! Thank you for all this! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-15T03:08:30Z SecondChildTAG: welcome change ! but I cant find where to locate my followings Followings? SecondChildUserIdTAG: 314624 SecondChildUserNameTAG: Owais001 SecondChildCreateTimeTAG: 2012-09-15T05:29:13Z SecondChildTAG: We still need to build that page but it's coming very soon. We wanted to get this redesign out first to get some feedback and all that good stuff. SecondChildUserIdTAG: 323387 SecondChildUserNameTAG: ibrahimawwal SecondChildCreateTimeTAG: 2012-09-15T06:09:10Z SecondChildTAG: The new forum is way better than the old one ; ). Thank you. Another suggestion, could it be possible to get the subdivision written in the left (with a header) instead of clicking on each subject in the dropdown menu ? For example, by clicking on "Week 2" we could see headers with "Boolean functions", "Cardiac experiments", etc. Under each header we would get the related threads. The header could be clickable to show or mask its threads. SecondChildUserIdTAG: 396446 SecondChildUserNameTAG: RousseauxS SecondChildCreateTimeTAG: 2012-09-15T09:43:14Z SecondChildTAG: I agree with RousseauxS. Dormsbee: please implement. SecondChildUserIdTAG: 79361 SecondChildUserNameTAG: kimt SecondChildCreateTimeTAG: 2012-09-15T16:16:33Z SecondChildTAG: Hi, I have two other suggestions: 1. **Same questions:** shouldn't we find a technique to force people to search if their question has not been answered yet ? I really like helping people but it is boring to see the exact same questions over and over. It could be done by forcing "tags" on each thread like the reference of the exercise / HW / work. The name of the thread could look like "[H1P1] Syntax error". Or we could get "reference threads" if one thread reaches a dedicated number of votes or validation by the staff, it becomes different (appears in the "Sort By" ?) ensuring its visibility. 2. **New answers:** as it is done by many forums (phpBB, vBulletin, and so on), would it be possible to get a different highlight if there is a new answer in the threads since the last visit ? SecondChildUserIdTAG: 396446 SecondChildUserNameTAG: RousseauxS SecondChildCreateTimeTAG: 2012-09-16T14:00:25Z SecondChildTAG: We have things in the works for both of those issues: one will automatically search for you as you're creating a post, suggesting posts that may be asking the same thing as you. We'll also think about a method for allowing people to easily mark threads as duplicates in a better way than having to open the thread and link to the thread that it's a duplicate of. Regarding #2, it's coming very soon :) Thanks for the feedback! SecondChildUserIdTAG: 12663 SecondChildUserNameTAG: arjun SecondChildCreateTimeTAG: 2012-09-16T23:53:48Z SecondChildTAG: Thank you ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-21T03:12:49Z FirstChildTAG: I really like the new layout so far! A big thank you to the techs! A couple suggestions: 1. I'd like to also say that I agree with RousseauxS's 3 feature requests, and I am looking forward to seeing a page that lists the threads I'm following. 2. It would also be nice to have a persistent button somewhere that links back to my user-page, rather than what I currently have to do, which is find a post I made and click on my hyper-linked name. 3. And I mentioned this one before, it would be nice to have more than 2 layers of replies (although I understand that can turn into a UI disaster if each new level is indented) C'est tout pour maintenant! This is definitely a vast improvement over what it was just a couple short days ago =) I'm impressed with how responsive you guys are! FirstChildUserIdTAG: 248902 FirstChildUserNameTAG: Chanute FirstChildCreateTimeTAG: 2012-09-16T23:36:15Z SecondChildTAG: Thanks for the feedback! #1 and #2 are being addressed (see my reply above to RousseauxS's post, and #3 is a subjective thing. We'll keep evaluating, but we're not currently thinking of changing it. SecondChildUserIdTAG: 12663 SecondChildUserNameTAG: arjun SecondChildCreateTimeTAG: 2012-09-16T23:55:21Z SecondChildTAG: Nice, thank you arjun! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-21T03:13:52Z SecondChildTAG: Thanks Arjun! Now I've noticed that there seems to be an issue with displaying matrices in mathjax. When I type \$ \left [ \matrix {a&b \cr c&d} \right ] \$ it renders correctly in the preview, but not once I hit submit! I'm guessing this has something to do with the input pre-processing to the comment database (I know MySQL likes to escape some special characters, maybe they need to be de-escaped on the way back out to solve this problem)? I can't imagine why else the preview would render properly but not the submitted comment. I escaped the \$'s because the same issue arises in using the 'code' escape - the &'s get replaced with &'s. See result below: $ \left [ \matrix {a&b \cr c&d} \right ] $ SecondChildUserIdTAG: 248902 SecondChildUserNameTAG: Chanute SecondChildCreateTimeTAG: 2012-09-21T23:57:53Z IndexTAG: 595 TitleTAG: New forum coming soon. Hold onto your hats! just wanted to give a warning... UserIdTAG: 96452 UserNameTAG: Lyla CreateTimeTAG: 2012-09-15T02:04:18Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 3 FirstChildTAG: Looks great! Now if only I could still follow posts, and easily look at my own profile. FirstChildUserIdTAG: 248807 FirstChildUserNameTAG: 1kingsman FirstChildCreateTimeTAG: 2012-09-15T02:36:40Z SecondChildTAG: Yeah, we're working to get those improved and out to you folks sometime next week. :-) Please stay tuned. SecondChildUserIdTAG: 11 SecondChildUserNameTAG: dormsbee SecondChildCreateTimeTAG: 2012-09-15T02:45:51Z FirstChildTAG: I like how this looks better. Having categories divided up into "weeks" might help minimize duplicate questions and posts. Anything else could fit into a "General" category. Good job! FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-15T02:22:31Z FirstChildTAG: WooooW ! amazing one :) thnX EDX all staff :) FirstChildUserIdTAG: 392663 FirstChildUserNameTAG: matto0o FirstChildCreateTimeTAG: 2012-09-15T02:17:44Z IndexTAG: 596 TitleTAG: Why Vs=10? Is there a reason the lecturer sets Vs=10 at 5:35? Or is it just a trick to make the calculation easier? UserIdTAG: 373220 UserNameTAG: oohall CreateTimeTAG: 2012-09-14T09:09:04Z VoteTAG: 5 CoursewareTAG: Week 3 / Switch Resistor Model of a MOSFET CommentableIdTAG: 6002x_switch_resistor_model NumberOfReplyTAG: 2 FirstChildTAG: I think it was a mistake, in the minute 9.05 Vs = 5. FirstChildUserIdTAG: 39685 FirstChildUserNameTAG: galeano FirstChildCreateTimeTAG: 2012-09-14T16:50:24Z SecondChildTAG: It's a miststake; Vsshould be 5V for Vout=0.45V (10/11 is NOT 0.45) SecondChildUserIdTAG: 277787 SecondChildUserNameTAG: kirilaska SecondChildCreateTimeTAG: 2012-09-16T14:48:50Z FirstChildTAG: the Vs value is 5 here...by mistake,he said as 10...Vs is taken as a value for example purpose only...3V,5V,12V,24V are some standard values of the supply FirstChildUserIdTAG: 118611 FirstChildUserNameTAG: mitianhari FirstChildCreateTimeTAG: 2012-09-21T16:52:44Z IndexTAG: 597 TitleTAG: Anyone out there for help Hey, guys, is anyone out there, trying to map out this lovely course without REAL knowledge of calculus and AP Level Physics( i am Still in 11th standard), I mean I have just skimmed over these things, in like just 3-4 days, and that is all I know, I just want to know if anyone is out there, please reach out, we both or more can study as a team, and maybe even pass this course, and yes even people who have knowledge of all this stuff, are also welcome, in simple this is an INVITATION FOR A STUDY GROUP, don't worry, if it is required i would even set up a small website, chat and everything required to communicate, just need study buddies..... Waiting for your word.... P.S. - Provide your e-mails so I can contact you( only if you are interested) UserIdTAG: 169784 UserNameTAG: thewiredbear CreateTimeTAG: 2012-09-12T20:14:04Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: I am getting ups, but I don't need ups, I need e-mails or something of the sort, please guys, reach out! I really need help here, and yes I can't tell who you are from your ups so please reach out through the comments section FirstChildUserIdTAG: 169784 FirstChildUserNameTAG: thewiredbear FirstChildCreateTimeTAG: 2012-09-12T21:07:37Z FirstChildTAG: Given that each individual resistor can dissipate up to 1 watt of power before burning up, how much total power in watts ( W ) can the smallest-valued composite resistor dissipate before burning up? explain it what is asking in it FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-12T21:45:43Z SecondChildTAG: this is not what i meant.... SecondChildUserIdTAG: 169784 SecondChildUserNameTAG: thewiredbear SecondChildCreateTimeTAG: 2012-09-13T13:30:32Z IndexTAG: 598 TitleTAG: Performing integration and calculating peak power. The integration of calculating average power is performed [here][1]: Here frequency f=60Hz converted into period T=1/f=1/60 s. The peak power can be calculated first by converting V_rms (root mean square) voltage into amplitudal voltage V, that is multiplying it by sqrt(2)=0.707... And then using the appropriate equation for power, that is P=V^2/R. [1]: http://www.wolframalpha.com/input/?i=60/110*%20Int%5Bfrom%200%20to%201/60%5D%20%28120*sqrt2*cos%282pi*60x%29%29%5E2dx UserIdTAG: 211206 UserNameTAG: Augustinas CreateTimeTAG: 2012-09-12T08:11:19Z VoteTAG: 5 CoursewareTAG: Week 1 / AC power CommentableIdTAG: 6002x_S1E3_AC_power NumberOfReplyTAG: 1 FirstChildTAG: Just a think that may confuse: The "60" of "(60/110)" is because when you calculate the average power you have to take a part divided by a time. Sorry for the english. => FirstChildUserIdTAG: 294330 FirstChildUserNameTAG: JHRMatos FirstChildCreateTimeTAG: 2012-09-12T12:23:32Z IndexTAG: 599 TitleTAG: Question 2 simple answer ... Assuming that everything is ideal ... the electric current will take the lowest resistance path ( the ideal wire ) and it won't take the path with resistance ... in that case the two batteries will act as they are in series and the total current will be the sum ... I1 + I2 = 1.5/0.25 + 1.5/0.32 = 10.6875A UserIdTAG: 340031 UserNameTAG: Content CreateTimeTAG: 2012-09-12T00:53:19Z VoteTAG: 5 CoursewareTAG: Week 1 / Battery model CommentableIdTAG: 6002x_S1E9_Battery_Model NumberOfReplyTAG: 1 FirstChildTAG: Maybe you are correct, I don't mean to interfere, just mentioning a method I tried, and maybe is correct :) Since both batteries are in parallel, the voltage would be same (voltage is same in parallel branches), and resistances can be combined in parallel, which gives 0.1403 Ohms. So current I can be given as: I= 1.5/0.1403 = 10.691, which is close to 10.6875 and is marked correct ! FirstChildUserIdTAG: 358606 FirstChildUserNameTAG: st1046 FirstChildCreateTimeTAG: 2012-09-12T09:29:23Z SecondChildTAG: you missed some digits, so 1.5/0.14035087719298245614035087719298=10.6875 SecondChildUserIdTAG: 97529 SecondChildUserNameTAG: ivanesses SecondChildCreateTimeTAG: 2012-09-14T20:18:25Z SecondChildTAG: Can you guys explain how you got this 0.14035087719298245614035087719298 number?!? SecondChildUserIdTAG: 277281 SecondChildUserNameTAG: bmovlanov SecondChildCreateTimeTAG: 2012-09-21T03:22:43Z IndexTAG: 600 TitleTAG: placing of lumped elements Can anyone tell me how to place a resistor or an inductor or a capacitor horizontally?? UserIdTAG: 357885 UserNameTAG: rahulshetty CreateTimeTAG: 2012-09-10T13:56:45Z VoteTAG: 5 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 6 FirstChildTAG: press R on your keyboard FirstChildUserIdTAG: 34341 FirstChildUserNameTAG: carlos1208 FirstChildCreateTimeTAG: 2012-09-10T14:26:16Z FirstChildTAG: "r" on keyboar will rotate FirstChildUserIdTAG: 319524 FirstChildUserNameTAG: RadeEric FirstChildCreateTimeTAG: 2012-09-10T14:30:34Z FirstChildTAG: r FirstChildUserIdTAG: 132396 FirstChildUserNameTAG: pandu1993 FirstChildCreateTimeTAG: 2012-09-10T15:30:51Z FirstChildTAG: Hi rahulshetty, When you have the component (resister, capacitor etc...) selected (the component should be green when selected), just press 'r' on your keyboard to rotate it. FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-09-10T14:00:45Z FirstChildTAG: Sometimes the "R" didnt work to me ... so, i refresh the page. Try it. FirstChildUserIdTAG: 370989 FirstChildUserNameTAG: Menendez91 FirstChildCreateTimeTAG: 2012-09-10T15:04:04Z FirstChildTAG: select the element. Then press "r" from the keyboard. FirstChildUserIdTAG: 234636 FirstChildUserNameTAG: KAYBEE FirstChildCreateTimeTAG: 2012-09-10T17:23:21Z IndexTAG: 601 TitleTAG: S1E6 : KVL v5 please explain the loop that you are using for solving expression for v5 UserIdTAG: 272417 UserNameTAG: MuKhan CreateTimeTAG: 2012-09-10T04:59:01Z VoteTAG: 5 CoursewareTAG: Week 1 / KVL and KCL exercises CommentableIdTAG: 6002x_kvl_kcl_exercises NumberOfReplyTAG: 4 FirstChildTAG: I am also confused with v5 ,did u got a conclusion ? FirstChildUserIdTAG: 162670 FirstChildUserNameTAG: charlesbabyt FirstChildCreateTimeTAG: 2012-09-10T05:49:32Z FirstChildTAG: you can use this loop v5-v1-v4 = 0 then v5 = v1 + v4 Then substitute v4 = v2-V the answer will be v1+v2-V FirstChildUserIdTAG: 247612 FirstChildUserNameTAG: Radwan FirstChildCreateTimeTAG: 2012-09-10T22:35:03Z SecondChildTAG: I'm confused, why is v4's polarity different from the other elements? SecondChildUserIdTAG: 272059 SecondChildUserNameTAG: Kalamgish SecondChildCreateTimeTAG: 2012-09-11T20:34:55Z FirstChildTAG: They are asking for a potential difference (potential one minus potential two) so if you put the positive probe of a voltmeter between v4 and v2 you can see that in both elements that it is a positive point, if you put the positive negative probe between v1 and v3 in that point v1 is negative but v3 is positive so the voltage metered is: v5=v4+v1 because is the diference of the positive potential of v4 minus the negative potencial of v1 so you need to know the value of v4. Also v5=v2-v3 (the positive potential of v2 minus the positive potential of v3) so you need to knov v3. I know that my english is bad but I hope you understand my answer. FirstChildUserIdTAG: 340568 FirstChildUserNameTAG: witedonkey FirstChildCreateTimeTAG: 2012-09-10T22:36:08Z FirstChildTAG: ![enter image description here][1] Hope this helps. [1]: http://82.imagebam.com/download/Awth0txqhfbiGnMN-ron_w/21007/210062165/Capture.PNG FirstChildUserIdTAG: 193033 FirstChildUserNameTAG: NIslam FirstChildCreateTimeTAG: 2012-09-10T05:53:06Z SecondChildTAG: For sure, it doesnt matter what loop to use. For ex, you can use v5+v3-v2=0. you already know v3. Or v5-v1-v4=0. you already know v4 SecondChildUserIdTAG: 200319 SecondChildUserNameTAG: Virviil SecondChildCreateTimeTAG: 2012-09-10T06:46:02Z SecondChildTAG: Yes, i was confused about what branch to take but your diagram so turn on the bulb... Thks SecondChildUserIdTAG: 472876 SecondChildUserNameTAG: otakudany SecondChildCreateTimeTAG: 2012-09-23T13:02:45Z IndexTAG: 602 TitleTAG: kcl what is independent equation in kcl??? UserIdTAG: 210333 UserNameTAG: akshay10uec012 CreateTimeTAG: 2012-09-09T05:20:16Z VoteTAG: 5 CoursewareTAG: Week 1 / Circuit topology CommentableIdTAG: 6002x_S2E1_Circuit_Topology NumberOfReplyTAG: 2 FirstChildTAG: An independent equation is one that cannot be derived from another two or more equations. So, if in a circuit you can write X number of equations usually a few of them are simply a combination of the others. Good luck! FirstChildUserIdTAG: 303762 FirstChildUserNameTAG: leocarrasco FirstChildCreateTimeTAG: 2012-09-09T05:24:39Z SecondChildTAG: In general cases - the number of independent equation is equal to number of unknown variables. SecondChildUserIdTAG: 394836 SecondChildUserNameTAG: v2g6ch4 SecondChildCreateTimeTAG: 2012-09-09T05:29:32Z SecondChildTAG: Independent equation in KCL-Quite confusing. **@v2g6ch4**- If it is equal to unknown variable means I have five in the above example (ica,iba,ida,icd,ibd).Plz correct me,if I'm wrong.Thanks!!! SecondChildUserIdTAG: 147577 SecondChildUserNameTAG: rithi3 SecondChildCreateTimeTAG: 2012-09-16T05:50:04Z FirstChildTAG: Independent - has not lineary dependence with other equations. Read Linear Algebra to know more about this. Example: 3x+3y=3 (1) -x-y=-1 (2) -x+y=1 (3) Equations 1 and 2 are dependent. We have only 2 independent： 2 and 3 or 1 and 3. FirstChildUserIdTAG: 394836 FirstChildUserNameTAG: v2g6ch4 FirstChildCreateTimeTAG: 2012-09-09T05:25:21Z IndexTAG: 603 TitleTAG: Advice from the last class Hello all, This post is a good place to post advice from the last class members. Not answers to anything, just advice about the class to people who are new to the people who attended the inaugural course. Here's my advice to start off: Use Wolfram Alpha. It's easy to use, but has some of it's own quirks, so look through the examples that are given. It helped me so much during the last class. So people who are returning, give some advice to everyone. UserIdTAG: 21541 UserNameTAG: JSChambers CreateTimeTAG: 2012-09-08T17:10:22Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Thank you JSChambers! How are you? :). Yes Wolfram Alpha was really useful ! Thank you for the advice . Can I ask you a Question? Can you Post in the Old Forum? I have Problems and I can not Post anything... I have 505 Bad Gateway ...Do you have the same Problem? Can you Check? Thank youuu. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-08T18:10:34Z SecondChildTAG: Myrimit, No I am not having trouble posting in the old forum. I just posted there this morning. Maybe you should email the admins from the "help" page. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-09-08T18:13:32Z SecondChildTAG: Hola, Miriam... I was able to post a comment a couple days ago. SecondChildUserIdTAG: 92346 SecondChildUserNameTAG: MobiusTruth SecondChildCreateTimeTAG: 2012-09-08T22:20:18Z IndexTAG: 604 TitleTAG: Pickle as a lumped element Next monday we are going to v-i characterize a pickle. I understand you just connected 110 ac voltage, is it correct? The experiment will take place in a University laboratory so I think we can control safety. UserIdTAG: 92800 UserNameTAG: carlos_sv CreateTimeTAG: 2012-09-08T02:23:22Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 3 FirstChildTAG: Be careful Carlos! FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-08T03:32:06Z FirstChildTAG: Wish you best luck. Be careful ^_^ FirstChildUserIdTAG: 211715 FirstChildUserNameTAG: pitankar FirstChildCreateTimeTAG: 2012-09-08T05:30:13Z FirstChildTAG: Next Monday, I will blog our result on using pickles as lumped elements. In two figures, I show the [pickle circuits][1] we are going to test. [1]: http://6002x-sv.blogspot.com/2012/09/the-burning-of-pickle-ii.html FirstChildUserIdTAG: 92800 FirstChildUserNameTAG: carlos_sv FirstChildCreateTimeTAG: 2012-09-08T22:47:53Z IndexTAG: 605 TitleTAG: transient analysis I get the graph and all the values at 1.25 ms.But they are not same as the answer.I am not getting where am I wrong? I get 999.513m ,832.928m and 333.171m. Any suggestions please! UserIdTAG: 281159 UserNameTAG: ManasiS CreateTimeTAG: 2012-09-08T00:24:13Z VoteTAG: 5 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 2 FirstChildTAG: You get that values if you don't put offset value=1 FirstChildUserIdTAG: 250179 FirstChildUserNameTAG: diegoCmC FirstChildCreateTimeTAG: 2012-09-08T03:30:48Z SecondChildTAG: It is still same. I don't get that.lemme clear this.Do we need to measure by our self or just we put the cursor on 1.25 ms and it shows the value?? SecondChildUserIdTAG: 281159 SecondChildUserNameTAG: ManasiS SecondChildCreateTimeTAG: 2012-09-08T20:48:45Z FirstChildTAG: Check ur conditions of voltage source FirstChildUserIdTAG: 55345 FirstChildUserNameTAG: sarmaji FirstChildCreateTimeTAG: 2012-09-08T01:27:41Z SecondChildTAG: I got the same problem with manasis, My voltage source is sin 1 amp 1 offset 1000Hz any idea with this? SecondChildUserIdTAG: 317060 SecondChildUserNameTAG: paulwong87 SecondChildCreateTimeTAG: 2012-09-08T03:09:19Z IndexTAG: 606 TitleTAG: error at 1:34 at this moment the professor starts to analyse the Z output, based on X and Y. but in some moments he mistakes the X and Y values, calling them 0(low) when they are 1(high), here is the phrase "Then in this region here, notice that y is 0 and so is x, so z is going to be a 1." the drawn output is correct, but i spotted this mistake UserIdTAG: 264841 UserNameTAG: Jcert CreateTimeTAG: 2012-09-07T22:43:25Z VoteTAG: 5 CoursewareTAG: Week 2 / Digital logic circuits CommentableIdTAG: 6002x_digital_logic NumberOfReplyTAG: 1 FirstChildTAG: make corrections, that was a mistake FirstChildUserIdTAG: 440453 FirstChildUserNameTAG: Asandyz FirstChildCreateTimeTAG: 2012-09-22T18:58:30Z IndexTAG: 607 TitleTAG: "bug" hello thank you very much for extra edX tutorial it is really very helpful. But problem is still there I am using windows 7 I have download latest versions of Firefox and Google chrome. when I started working in lab 0 I found that there were no wiring even i tried a lot to connect components as I learned by tutorial but wires are still not appear, after that I tried Fire fox , in Fire Fox wiring problem has solved and I have done the task till Dc analysis but in Fire fox TRANSIENT analysis is not working,some times button are not appeared and if they appear transient analysis doesn't work ,another thing when I closed the session or exit my schematic disappeared and I have to make another one for analysis,this problem is in lab 1 and lab 2 also,hope you have understand the problem and it will be fixed soon before the deadline of homework 1 and 2. UserIdTAG: 40209 UserNameTAG: sadaf78 CreateTimeTAG: 2012-09-07T18:36:44Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 7 FirstChildTAG: now my transient analysis has started working and I have completed lab 0. FirstChildUserIdTAG: 40209 FirstChildUserNameTAG: sadaf78 FirstChildCreateTimeTAG: 2012-09-07T19:05:06Z SecondChildTAG: Excellent :) SecondChildUserIdTAG: 79361 SecondChildUserNameTAG: kimt SecondChildCreateTimeTAG: 2012-09-07T19:05:28Z SecondChildTAG: How did you solved ur problem? SecondChildUserIdTAG: undefined SecondChildUserNameTAG: undefined SecondChildCreateTimeTAG: 2012-09-07T19:40:50Z FirstChildTAG: I switched from Firefox to Chrome yesterday because of problems like the ones you describe. I am also on W7 currently. Things work fine for me so far on Chrome 21.0.1180.89, so as another commenter says try to clear the cache and re-login. Transient analysis is working fine. If there is no window it could be unable to analyse your circuit. Remove any superfluous probes and remove any unused circuit elements, make sure that your circuit diagram is the only circuit in the sandbox, then try a short time such as 0.1 seconds. Let us know if that works. In the old circuit editor (prev 6.002x course) unused or too many extra circuit elements could sometimes be a problem. If you want to keep multiple circuits try disconnecting any current or voltage source in the circuit by dragging it one grid space away from the circuit. Sometimes analysis seems to get confused. Also if doing long sessions of circuit editing regularly 'check' the circuit to save it because that seems to stop prob;ems building up (I've suspected a code bug that only occurs after many edits). In the previous version of this course I also had to use Chrome. It looks like there are still problems with Firefox. FirstChildUserIdTAG: 9147 FirstChildUserNameTAG: SueP FirstChildCreateTimeTAG: 2012-09-07T19:15:11Z SecondChildTAG: Hello SueP, I recognize you from the old 6.002x forums :) SecondChildUserIdTAG: 79361 SecondChildUserNameTAG: kimt SecondChildCreateTimeTAG: 2012-09-07T19:19:49Z SecondChildTAG: I tried logging out, clearing browser's cache and re-logging but still not working. there are no unused elements. I also tried in sandbox building simple circuit. but transient analysis is not working. SecondChildUserIdTAG: undefined SecondChildUserNameTAG: undefined SecondChildCreateTimeTAG: 2012-09-07T19:34:58Z FirstChildTAG: Dear Sir same problem with me al so but I am able to do work with other simulators i can make circuits I was used yenka circuit stimulator Thanks MK.Prasanth FirstChildUserIdTAG: 156694 FirstChildUserNameTAG: mkprasanth FirstChildCreateTimeTAG: 2012-09-08T04:19:36Z SecondChildTAG: Dear Sir I for got tell my OS is win XP Version 5.1.2600 thanks Please help.... SecondChildUserIdTAG: 156694 SecondChildUserNameTAG: mkprasanth SecondChildCreateTimeTAG: 2012-09-08T04:26:43Z FirstChildTAG: hi, i m using windows 7 and fire fox browser. i m trying to do lab 0.but problem is there i cant be able to join the ends of components as wire is not working.what can i do? thanks in advance FirstChildUserIdTAG: 1453 FirstChildUserNameTAG: muhammadusmanbashir FirstChildCreateTimeTAG: 2012-09-09T07:51:19Z FirstChildTAG: Hello sir, I am also facing same problem. I am using windows 7 and the latest version of google chrome. When i started working on lab 0 , wires are not appearing. I tried to clear cache and re-logged still wiring is not appearing. Please help.. FirstChildUserIdTAG: 330918 FirstChildUserNameTAG: pratapchintu333 FirstChildCreateTimeTAG: 2012-09-09T18:37:10Z FirstChildTAG: I am also having the same problem.... in chrome i can't draw the wires, and in firefox it is not showing all the components and buttons..... another thing is, in chrome it shows already drawn diagram of lab 1 but i can't wire that and in firefox i can draw wires but it is not showing already drawn diagrams and also not showing other components except resistor.... please sir make it correct before the due date of Lab 1.... FirstChildUserIdTAG: 259693 FirstChildUserNameTAG: MehrozKhan FirstChildCreateTimeTAG: 2012-09-11T11:17:13Z FirstChildTAG: I am having same problem. I also have windows 7 and firefox 15.0.1 and chrome 21.0.1180.89. In firefox dc tran buttons do not appear. In chrome buttons appear. But in both browsers transient analysis is not showing any window after putting stop time.It doesn't work in sandbox also. FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-07T18:55:00Z SecondChildTAG: With Google Chrome, can you try logging out, clearing the browser's cache, then re-logging in to the website? SecondChildUserIdTAG: 79361 SecondChildUserNameTAG: kimt SecondChildCreateTimeTAG: 2012-09-07T19:01:03Z IndexTAG: 608 TitleTAG: YouTube Access Denied Hi.Thanks for offering this course Unfortunately, We can't have access to YouTube. Can you please upload video lectures on another server. Thank you again [UPDATE]: **For students in Iran:** In order to download videos: 1.Right-click on the video (which is showing the error message) 2.click "Copy video URL" 3.Paste into http://cyber-tube.ir 4.Transfer and download Best wishes ;) UserIdTAG: 91706 UserNameTAG: PooyaM CreateTimeTAG: 2012-09-07T15:10:13Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: Hi Mate Thanks a lot and I'm glad to find a friend and actually a mate from iran [ my contry :) ] FirstChildUserIdTAG: 133033 FirstChildUserNameTAG: ay3 FirstChildCreateTimeTAG: 2012-09-07T15:52:27Z SecondChildTAG: i am from Pakistan can you suggest me how to watch these videos as youtube is also blocked in our country. SecondChildUserIdTAG: 403493 SecondChildUserNameTAG: aiai SecondChildCreateTimeTAG: 2012-09-26T09:36:07Z FirstChildTAG: You can see in the Wiki [Lecture Videos Download][1]:). Also see more info [here][2]: **Due to the YouTube outage in certain parts of the world, we have provided direct links to download the lecture videos, and extended the Week 2 homework and lab deadlines to the coming Sunday (September 30). Please note that Week 3 deadlines have not changed.** [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/ [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/info FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-26T11:51:54Z IndexTAG: 609 TitleTAG: New schematic tutorial is coming We saw that a lot of people were confused about how to use the schematic editor, and we have a new tutorial in the pipeline. We hope to have it published and viewable in the next several hours. UserIdTAG: 96452 UserNameTAG: Lyla CreateTimeTAG: 2012-09-07T12:52:19Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: Thank you Lyla! FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-07T14:54:59Z FirstChildTAG: Need help Lyla Even if its updated and corrected! i still have the same problem!! no options except resistance icon!! screenshot attached below https://rapidshare.com/#!download|26p9|57535273|mitx%202.jpg|227|1366|768 Thanks Asim FirstChildUserIdTAG: 232499 FirstChildUserNameTAG: Asim09 FirstChildCreateTimeTAG: 2012-09-08T19:00:14Z IndexTAG: 610 TitleTAG: The easy way This one totally got me confused first, but here is some explanation on my side, w/o any integrals. If you look at the function itself, its clear, that its peak value of the voltage is 120*sqrt(2) [V]. Therefore it can be used to get the maximum dissipation, which is calculated from P=(V*V) / R. On the second one, without integrating anything, hit up wiki for RMS, and you will see, that the RMS value for a sinusoidal (there for cosinus as well) is peakvalue*sqrt(2). So you get ( 120*sqrt(2) / sqrt(2) ) = 120V as rms value, and using the same formula as before you calculate the dissipation. This method is maybe not so mathematical as others beautifully stated under me, but it works. UserIdTAG: 329254 UserNameTAG: KGabor CreateTimeTAG: 2012-09-07T08:27:53Z VoteTAG: 5 CoursewareTAG: Week 1 / AC power CommentableIdTAG: 6002x_S1E3_AC_power NumberOfReplyTAG: 2 FirstChildTAG: KGabor your answer is very elegant and simple so it is parsimonious. I can do some differential and integral calculus but this is the preferred way for me when calculus is not required. To add to your post you then use the equation I = V/R where the I is the current in amps, V is the voltage (120) and R is the resistance given, which is 110. Thus 120/110 = 1.09 rounded to 3 significant digits from the rational number, and then we multiply the 120 volts by 1.09 to get 130.909. FirstChildUserIdTAG: 8026 FirstChildUserNameTAG: jcbmack FirstChildCreateTimeTAG: 2012-09-09T03:00:43Z SecondChildTAG: RMS = CV SecondChildUserIdTAG: 8026 SecondChildUserNameTAG: jcbmack SecondChildCreateTimeTAG: 2012-09-09T03:02:36Z FirstChildTAG: that looks nice, Good work FirstChildUserIdTAG: 388747 FirstChildUserNameTAG: Azeem1721 FirstChildCreateTimeTAG: 2012-09-09T17:11:50Z IndexTAG: 611 TitleTAG: Don't equal to zero Don't equal to zero this expression otherwise it won't work. UserIdTAG: 359677 UserNameTAG: guillermb CreateTimeTAG: 2012-09-07T07:07:04Z VoteTAG: 5 CoursewareTAG: Week 1 / Node analysis practice, part 3 CommentableIdTAG: 6002x_L2Node2 NumberOfReplyTAG: 1 FirstChildTAG: thanks FirstChildUserIdTAG: 171378 FirstChildUserNameTAG: ABHISHEKFROMINDIA FirstChildCreateTimeTAG: 2012-09-07T18:41:33Z IndexTAG: 612 TitleTAG: Remember Remember use the calc in radians... UserIdTAG: 118022 UserNameTAG: E_Dorf CreateTimeTAG: 2012-09-07T04:15:42Z VoteTAG: 5 CoursewareTAG: Week 1 / AC power CommentableIdTAG: 6002x_S1E3_AC_power NumberOfReplyTAG: 0 IndexTAG: 613 TitleTAG: Independent Loop What dose it mean by independent, as in what is the loop equation 'not dependant' on ? UserIdTAG: 204745 UserNameTAG: allwynmendes CreateTimeTAG: 2012-09-06T20:22:11Z VoteTAG: 5 CoursewareTAG: Week 1 / Circuit topology CommentableIdTAG: 6002x_S2E1_Circuit_Topology NumberOfReplyTAG: 4 FirstChildTAG: Look I think I got it by myself See the Video S2 v4 in the video in some parts professor will say that by circuit topology there are 3 independent equation .. REal meaning - notice in the video that if you only take 3 out of four equations that are given the you can get the fourth equation which inturn means the fourth equation is useless and only the three equations are usefull since they cannot be derived by any other equation combination. i have got my own answer too FirstChildUserIdTAG: 378150 FirstChildUserNameTAG: GladIDoThis FirstChildCreateTimeTAG: 2012-09-06T21:30:18Z SecondChildTAG: GladIDoThis explained the physical meaning. There is a precise, mathematical, meaning, but you have to know the first rudiments of linear algebra to appreciate it. It involves the concept of the rank of a matrix SecondChildUserIdTAG: 346056 SecondChildUserNameTAG: fiatlux SecondChildCreateTimeTAG: 2012-09-15T21:19:49Z FirstChildTAG: You can ensure that the loops are independent by having at least one component in each loop that is not contained in any other loop. FirstChildUserIdTAG: 15730 FirstChildUserNameTAG: DonC FirstChildCreateTimeTAG: 2012-09-07T00:20:50Z FirstChildTAG: Equations for the 3 loops inside and for the outer loop: Loop 1: -V + i1R1 + i2R2 = 0 Loop 2: -i1R1 + i4R4 -i3R3 = 0 Loop 3: -i2R2 +i3R3 + i5R5 = 0 Loop 4: -V + i4R4 + i5R5 = 0 You can substitute the values of V, i4R4 and i5R5 in Loop 4 from Loop 1,2 and 3. So, from Loop 4, we can write, -i1R1 - i2R2 + i1R1 + i3R3 + i2R2 - i3R3 = 0 or, 0 = 0 that is, you can derive the equation for Loop 4 from the other three equations. So Loop 4 has a dependent equation. FirstChildUserIdTAG: 56434 FirstChildUserNameTAG: Fahim FirstChildCreateTimeTAG: 2012-09-08T08:00:30Z FirstChildTAG: I think cause one is the ground node: voltage = 0 FirstChildUserIdTAG: 249375 FirstChildUserNameTAG: MarIgones FirstChildCreateTimeTAG: 2012-09-06T20:59:03Z IndexTAG: 614 TitleTAG: Spinning my wheels I've gotten to the point where I'm looking at v1=i1*4, v2=i2*5, v1+v2=2, i2-i1=3, but I feel like I've just gotten myself stuck and I don't know what to do next. I'm sure I'm just not thinking well, but anything that might spur me along would be appreciated. UserIdTAG: 342518 UserNameTAG: RobO2112 CreateTimeTAG: 2012-09-06T13:41:41Z VoteTAG: 5 CoursewareTAG: Week 1 / KVL/KCL example CommentableIdTAG: 6002x_S2E3_Using_KVL_KCL_and_VI_constraints NumberOfReplyTAG: 4 FirstChildTAG: c [here][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/MITx_6_002x_2012_Fall/threads/5047a8af6859412700000021 FirstChildUserIdTAG: 82571 FirstChildUserNameTAG: Nidhi3 FirstChildCreateTimeTAG: 2012-09-06T13:48:20Z FirstChildTAG: put parameters instead of each other FirstChildUserIdTAG: 244954 FirstChildUserNameTAG: ghotbi FirstChildCreateTimeTAG: 2012-09-06T13:48:48Z FirstChildTAG: push yourself FirstChildUserIdTAG: 359504 FirstChildUserNameTAG: Sisbboy FirstChildCreateTimeTAG: 2012-09-07T07:14:27Z FirstChildTAG: It's a correct start. Combine the equations and solve for you unknowns. FirstChildUserIdTAG: 310007 FirstChildUserNameTAG: ErinF FirstChildCreateTimeTAG: 2012-09-06T23:07:04Z IndexTAG: 615 TitleTAG: Question 5 - transient analysis didn't work in any browser. Completely assembled the whole setup that question 5 asks for. I used Google Chrome, Safari, Firefox and Opera to try to get the transient analysis to work. Re-installing and clearing memory everywhere I could. Hasn't worked. Had to resort to clicking show answer for it. Didn't want to reveal the answer without solving it but had to so I could not be held back by the software itself. UserIdTAG: 179353 UserNameTAG: virtualpsycho CreateTimeTAG: 2012-09-06T13:05:06Z VoteTAG: 5 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 7 FirstChildTAG: Try updating Flash. FirstChildUserIdTAG: 141987 FirstChildUserNameTAG: lawrence26 FirstChildCreateTimeTAG: 2012-09-06T19:19:44Z FirstChildTAG: When you got such sort of any problem, don't blame anyone or anything, just download latest version of your browser. I experienced it from my first certificate. FirstChildUserIdTAG: 908 FirstChildUserNameTAG: m_umair72001 FirstChildCreateTimeTAG: 2012-09-06T13:21:11Z FirstChildTAG: Try in Google Chrome. At first it didnt work on lastest Mozilla Firefox, but seems to work fine on Chrome. FirstChildUserIdTAG: 250790 FirstChildUserNameTAG: angelobuoro FirstChildCreateTimeTAG: 2012-09-06T15:38:26Z FirstChildTAG: Would check all values esp. Vsupply values and re-run analysis. I input ohms for Kohms, for example, before catching my error. Transient analysis worked for me in Chrome. FirstChildUserIdTAG: 253337 FirstChildUserNameTAG: Remguy FirstChildCreateTimeTAG: 2012-09-06T14:20:53Z FirstChildTAG: I had the exact same problem. The "Circuit Sandbox" works, but the "Using the Tools" Lab does not work. The DC and TRAN does not show up until after hitting [Check] for the first time. TRAN does not bring up any window with the graph. I tried Firefox, Safari, and Chrome on my Mac. "Circuit Sandbox" always works, "Using the Tools" NEVER works (for TRAN). I just tried on Windows and I got the same problem in Firefox. Internet Explorer was even worse -- neither of the two work windows (where the circuit diagram to be reproduced is displayed and where the circuit design tool is displayed) had anything in them -- they were blank windows with a border around them and no contents. FirstChildUserIdTAG: 332090 FirstChildUserNameTAG: WPurin FirstChildCreateTimeTAG: 2012-09-06T16:34:26Z FirstChildTAG: I'm wondering if the problem isn't 32-bit machines vs. 64-bit machines. So much software made for 32-bit machines doesn't work on 64-bit. Maybe everyone could post whether or not the transient analysis worked, and whether their processor is 32 or 64-bit. It did NOT work for me, and I'm using Firefox on a 64-bit computer running Windows7. FirstChildUserIdTAG: 283726 FirstChildUserNameTAG: ByronInLawrence FirstChildCreateTimeTAG: 2012-09-07T03:11:51Z FirstChildTAG: https://6002x.mitx.mit.edu/courseware/6.002_Spring_2012/Overview/Circuit_Sandbox/ it helped to me. FirstChildUserIdTAG: 295103 FirstChildUserNameTAG: Syavick FirstChildCreateTimeTAG: 2012-09-11T08:16:36Z IndexTAG: 616 TitleTAG: I can't do chapter number 5. TRAN isn't work, window is not open. I check in FF and Chrome and Opera. Some idea ? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-06T12:41:40Z VoteTAG: 5 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 5 FirstChildTAG: The same here! Can anyone suggest the solution? FirstChildUserIdTAG: 301141 FirstChildUserNameTAG: Smaragda FirstChildCreateTimeTAG: 2012-09-06T13:58:28Z SecondChildTAG: I used Google Chrome. In firefox, under windows7 it didnt work. Any way, you could always give a try to VMWARE Operating system emulator. Try a prebuild virtual machine if you have any problem with that specific machine. SecondChildUserIdTAG: 264320 SecondChildUserNameTAG: israel05 SecondChildCreateTimeTAG: 2012-09-06T14:24:44Z FirstChildTAG: What operating system are you using? FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-06T13:02:48Z FirstChildTAG: When you got such sort of any problem, don't blame anyone or anything, just download latest version of your browser. I experienced it from my first certificate. FirstChildUserIdTAG: 908 FirstChildUserNameTAG: m_umair72001 FirstChildCreateTimeTAG: 2012-09-06T13:21:54Z SecondChildTAG: I have new version of everything so it isn't problem in by my side. I see that others have the same problem. SecondChildUserIdTAG: 136519 SecondChildUserNameTAG: damiS SecondChildCreateTimeTAG: 2012-09-06T13:28:40Z FirstChildTAG: same here .. FirstChildUserIdTAG: 18287 FirstChildUserNameTAG: saurav FirstChildCreateTimeTAG: 2012-09-06T16:10:37Z FirstChildTAG: Me too. This seems to be a common problem. Every time I open the forum I see a new complaint about the same thing. I have tried Windows and Mac and Safari, Firefox, Chrome, and Internet Explorer. TRAN does not work on ANY of them. But TRAN (and everything else) works fine in the next lab "Circuit Sandbox". FirstChildUserIdTAG: 332090 FirstChildUserNameTAG: WPurin FirstChildCreateTimeTAG: 2012-09-06T16:36:57Z SecondChildTAG: Yes... try Circuit Sandbox. :-) SecondChildUserIdTAG: 236810 SecondChildUserNameTAG: jmendego SecondChildCreateTimeTAG: 2012-09-07T05:38:14Z IndexTAG: 617 TitleTAG: The layman's review In my opinion it would be nice if you could check your answers as you go along, without having to see a series of obtrusive red X's after only attempting to check only the answers you have filled out so far. I was so proud to get past step 2 but was quickly brought back to reality regarding my inexperience with math and electronics, haha. Joking aside, system seems great and user friendly or as user friendly as a course can be that is so analytical in nature. I'm sure my improvement will be slow and steady, but "exponential", relatively speaking of course. UserIdTAG: 380622 UserNameTAG: JamesTucker CreateTimeTAG: 2012-09-06T08:32:09Z VoteTAG: 5 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 0 IndexTAG: 618 TitleTAG: Remember the abstract. Consider the circuit loop to be a closed system. i.e.: in order for something to gain , something must lose , net gain of 0 so if the resistor has 2 Watts of power to dissipate, then the battery has to have 2 Watts to start with, and then will lose when the current flows out. UserIdTAG: 372489 UserNameTAG: Estymagic CreateTimeTAG: 2012-09-05T20:51:09Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: 5_Simple_Power NumberOfReplyTAG: 2 FirstChildTAG: I almost went wrong in the power entering the source. Consumed power = positive sign. ==> Generated power = negative sign. FirstChildUserIdTAG: 250780 FirstChildUserNameTAG: miguelpom FirstChildCreateTimeTAG: 2012-09-05T21:42:34Z FirstChildTAG: In this case the source should also "dissipate". For word "entering" it definitely should be positive 2 - as "entering" already means opposite action to "dissipate". This part is understandable in math terms but not in linguistic. FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-09-06T00:40:24Z IndexTAG: 619 TitleTAG: Lab1 hi everybody. I'm try to submit de lab1 homework but i'm not able to insert a voltage source into de circuit. Can anybody help me? UserIdTAG: 242529 UserNameTAG: antpc CreateTimeTAG: 2012-09-05T19:47:15Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: It is broken. As sandbox showing it but not on lab. So its not due to browser as such. I would imagine they will be working on it. Hopefully they will post an update soon. FirstChildUserIdTAG: 15344 FirstChildUserNameTAG: kob FirstChildCreateTimeTAG: 2012-09-05T20:22:09Z FirstChildTAG: You should not need to insert a voltage source into the Lab 1 circuits in order to complete them. Please DO NOT DELETE ANYTHING that is in the original circuit (ground, voltage source, etc.) because you can only insert resistors. The staff does know that this is an issue. FirstChildUserIdTAG: 376895 FirstChildUserNameTAG: alecgg FirstChildCreateTimeTAG: 2012-09-05T22:01:06Z SecondChildTAG: There is no voltage or ground component showing?? This is weird as obviously it works for some and not others. I think it must be the way they distribute the servers between students. Probably so many hundred per server or virtual server. With the one I happen to be on is one of the ones with problems. You must have been assigned to a good server! Mind you that is a wild guess. Has to be something like that though. SecondChildUserIdTAG: 15344 SecondChildUserNameTAG: kob SecondChildCreateTimeTAG: 2012-09-05T23:39:22Z IndexTAG: 620 TitleTAG: lab1 Hint For lab plz see the hint written in the last paragraph i.e. use a two-resistor voltage divider to create the voltage for node A. You'll have two unknowns (R1 and R2) which can be determined by solving the two equations for vs derived from the constraints above: one involving R1, R2 and Rbulb where vs=1.5, and one involving R1 and R2 where vs=2. UserIdTAG: 82571 UserNameTAG: Nidhi3 CreateTimeTAG: 2012-09-05T19:45:14Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 621 TitleTAG: Why does the study of V-I characteristics of an electrical circuit element so important ? We apply the voltage across the element and study the current flowing through it. There can be many other electrical properties we could be concerned with. What so special with the V-I characteristics ? I suppose its because it is directly related to the resistance of the material the element is made up of, which is the most fundamental property. UserIdTAG: 158817 UserNameTAG: Shubhamtomar CreateTimeTAG: 2012-09-05T13:20:11Z VoteTAG: 5 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: The V-I characteristics is important because, it tells us the behavior of the element to an applied input. for ex a semiconductor diode V-I characteristics gives us a firm knowledge about forward resistance, the the minimum forward voltage to be applied so that the diode starts conducting. Based on these parameters, the diode can be used for a miriad of applications FirstChildUserIdTAG: 357885 FirstChildUserNameTAG: rahulshetty FirstChildCreateTimeTAG: 2012-09-05T13:26:45Z FirstChildTAG: I've always thought it's because current and voltages are the two responses we can measure at terminals. We can also add up voltages and currents easily. Good question! FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-09-05T13:28:15Z IndexTAG: 622 TitleTAG: inquiry regarding taking midterm exam or final exam the direction for attempting the mid-term or the final exam is that we have attempt the exams at one go, but say for example if there is a power failure or internet dis connectivity then do we have to begin all over again? UserIdTAG: 171378 UserNameTAG: ABHISHEKFROMINDIA CreateTimeTAG: 2012-09-05T12:55:28Z VoteTAG: 5 CoursewareTAG: Week 1 / Course overview CommentableIdTAG: 6002x_S1V1toS1V4 NumberOfReplyTAG: 1 FirstChildTAG: sir videos are not running plzz help me and topics in text form r not appear plzz guide and help me FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-06T21:42:16Z IndexTAG: 623 TitleTAG: Vannevar Bush Differential machine Before the digital computers MIT build mechanical ones ![Differential analyzer][1] As they haven't the mechanical anymore they rebuild a virtual one solving differential equations. [mit museum][4] Marshall University build a physical one. [Lizzie][2] Hmm, I'll have to build also one of these if I want to succeed in 6003z. The military still seem to use them [vannevar youtube][3] [1]: https://edxuploads.s3.amazonaws.com/13605210821343667.jpg [2]: http://www.youtube.com/watch?v=NmX151Jd3_o [3]: http://www.youtube.com/watch?v=TQj3PsSDoUo [4]: http://web.mit.edu/klund/www/analyzer/ UserIdTAG: 79885 UserNameTAG: ruudoleo CreateTimeTAG: 2013-02-10T19:01:37Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_6003z NumberOfReplyTAG: 0 IndexTAG: 624 TitleTAG: Start date for 6.003z announced. Hi, 6.003z is going to start from Feb 18th. Wish you all the best. http://6003z.amolbhave.in UserIdTAG: 118131 UserNameTAG: ammubhave CreateTimeTAG: 2013-02-03T02:15:13Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_6003z NumberOfReplyTAG: 3 FirstChildTAG: Put some more info Amol. Your previous post will be buried deep down the forums. Put all the info every time you post (We had to do this with CECC as well). FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2013-02-03T07:11:58Z FirstChildTAG: hello what is this? I am interested too. FirstChildUserIdTAG: 319453 FirstChildUserNameTAG: leonidasGr FirstChildCreateTimeTAG: 2013-02-03T15:07:01Z SecondChildTAG: This is a student run initiative. After the last session of 6.002x (Spring 2012), many of us wanted MITx to offer Signals and Systems. That is apparently going to take a while be cause they are more focussed on quality than just offering a large number of courses (which is definitely a good thing!). But we couldn't wait so we formed a study group and used the forums to coordinate. Amol took it further and built a website and that became the place to view the course material. The course material is basically MIT OCW videos by Prof. Alan V. Oppenheim (you would recognize his book on the subject as 'Oppenheim & Willisky'). We also created our own assignments and tutorials to supplement the main videos. Amol also was able to arrange an exam because someone at MIT was kind to offer a previous exam question paper. SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2013-02-03T17:30:21Z FirstChildTAG: Thank you Amol FirstChildUserIdTAG: 857413 FirstChildUserNameTAG: Naveenkrishna FirstChildCreateTimeTAG: 2013-02-03T15:48:51Z IndexTAG: 625 TitleTAG: Pearson Proctored Exam Logistics Great news. We have set the date for the proctored final exam we are offering for the current version of 6.002x: Wednesday, February 13th. Registration for the exam is now open and there are two steps to follow if you are interested: Step 1: Register on edX. Registration for the proctored examination is open now and will remain open until Monday, February 11th at 9:00am ET (GMT -5). Please go to your student dashboard if you are interested in signing up for the exam with edX. Step 2: Schedule your seat with Pearson VUE. In addition to registering for the exam with edX, the proctored exam requires that you schedule a seat at a Pearson VUE proctoring center. You will be able to register on the Pearson VUE site after we have communicated your registration request with Pearson, which may take up to a day. Please check your edX dashboard for up-to-date status and a link to the edX Pearson VUE site. As a reminder, it will cost $95 USD to take this proctored exam (You make payment on the Pearson site). We wanted to remind you of the details about how this prototype proctored exam process is going to work, and what kind of certificates will be available for you. As a student taking this course you had the option of taking the final exam online in a non-proctored setting on December 20th, 2012 in a format that was identical to that of the midterm exam. If you passed the course, you’re able to receive an honor code certificate free of charge, and will not be required to take a final under proctored conditions. BTW, all of the certificates from our university partners issued through edX in the past have been honor code certificates. In addition, you now have an opportunity to take a different exam that is proctored on Wednesday, February 13th and become eligible to receive a proctored certificate based entirely on your performance on this exam. As you might expect, taking the two-hour exam at a Pearson VUE location will cost some money – the pilot price is $95 USD for this proctored exam. You can take the proctored exam regardless of whether or not you took the honor code exam and regardless of how you did on it. It is possible to get both an honor code certificate and a proctored certificate. Neither certificate will have a grade on it. Have questions? Ask them here. UserIdTAG: 96452 UserNameTAG: Lyla CreateTimeTAG: 2013-01-29T14:51:43Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: There wouldn't be a chance of another date to take the exam would there? The 13th of February is the middle of the Chinese New Year holiday here in China (the actual New Year this year is the 10th). I, and I expect many of the students from China, will be on vacation on that day. FirstChildUserIdTAG: 467169 FirstChildUserNameTAG: Eyowzitgoin FirstChildCreateTimeTAG: 2013-01-29T16:17:19Z SecondChildTAG: Unfortunately, this is the only day that we can offer the exam. We don't currently have many Chinese users because our video is primarily hosted from YouTube, which is blocked in China. We have provided downloadable versions of all of our videos, but we are unable to accomodate the Chinese New Year in scheduling the exam. SecondChildUserIdTAG: 96452 SecondChildUserNameTAG: Lyla SecondChildCreateTimeTAG: 2013-01-29T17:00:02Z SecondChildTAG: Bummer. Well, I'll see if I am able to do it while on vacation then. Oh, and it's not just China, it's anyone Chinese so a large part of Singapore, Taiwan, Macau, Hong Kong, etc. But, as you said, if that's the date, then that's the date! I and those like me will just have to work around it or wait for the next offering. Thanks! SecondChildUserIdTAG: 467169 SecondChildUserNameTAG: Eyowzitgoin SecondChildCreateTimeTAG: 2013-01-30T06:06:48Z SecondChildTAG: Hi,How will you take this exam?I'm in China too.And how much is it?Is such a certificate valuable?Thanks.. SecondChildUserIdTAG: 136959 SecondChildUserNameTAG: christerpher SecondChildCreateTimeTAG: 2013-02-01T01:08:45Z SecondChildTAG: Sorry for getting back to this so late Christerpher. I'm not going to be taking this exam as I'll be sitting on a beach in the Maldives on that day. :) If you wish to take the exam, I know there are testing centers in Hong Kong and Shanghai (possibly also Beijing but I'm not sure, I really stopped looking after I saw HK since that's where I live). As for the cost... read the original post. ;) The value of it, to me, would be worth a lot as I would take it as another major accomplishment. To other people, it would be valuable to some, it just depends on who I guess. In other words... I don't think I could answer that. :D SecondChildUserIdTAG: 467169 SecondChildUserNameTAG: Eyowzitgoin SecondChildCreateTimeTAG: 2013-02-07T15:09:19Z FirstChildTAG: Thank you for the announcement Lyla! There were quite a few questions over here: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50e75b95a205372700000012 1. I wanted to know if MIT will keep our scores and if we can ask for transcripts later on if we need them (I think I remember reading that these won't have grades either). 2. Can we take notes to the exam? EDIT: Are there any chances that the exam will be available in more centres before the registration deadline? I can't believe it's not there in Bangalore! :( ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13594768051343683.png FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2013-01-29T15:03:18Z SecondChildTAG: 1) Yes, we will keep a record of your scores. We do not currently have the infrastructure to provide transcripts upon request, but if we implement that feature later, we will have the data to be able to provide transcripts from the fall 2012 classes and proctored exams. 2) Unfortunately, no. The exam will include access to the entire course website (except the wiki and the discussion forum), but the tool that you can take with you to the exam will be a calculator. SecondChildUserIdTAG: 96452 SecondChildUserNameTAG: Lyla SecondChildCreateTimeTAG: 2013-01-29T17:07:12Z SecondChildTAG: Also - sorry about the lack of test center availability. We partnered with one of the most expansively available testing centers in the world, but even that infrastructure has room to grow and new educational models develop. SecondChildUserIdTAG: 96452 SecondChildUserNameTAG: Lyla SecondChildCreateTimeTAG: 2013-01-29T17:08:16Z SecondChildTAG: Well the thing is Pearson VUE has test centres in my city. The problem is that they aren't offering the test in any of the centres here. If it's not going to happen this time I hope that Pearson does offer this in more centres for the next run. SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2013-01-29T17:42:13Z SecondChildTAG: I also have a problem with the exam being taken in the middle of the week. It will be hard for me going to work then drive 360km on narrow roads, have a clear mind to answer the questions and then drive another 360km back to my pregnant wife :) SecondChildUserIdTAG: 319453 SecondChildUserNameTAG: leonidasGr SecondChildCreateTimeTAG: 2013-01-30T17:40:58Z SecondChildTAG: Lyla, In Bangladesh there is no exam centre. Our neighbouring country has exam centre but VISA processing and other staffs take time... now what should we do? If it was announced a bit earlier then we could have manage to attend the exam even after moving to another country. Please consider our circumstances. SecondChildUserIdTAG: 154172 SecondChildUserNameTAG: mofassair SecondChildCreateTimeTAG: 2013-01-31T17:26:24Z SecondChildTAG: We do appreciate your situation, and those are the kinds of things that we definitely want to know about in our pilot offering. However, at this point, we can't expand availability more than we already have for this particular exam. SecondChildUserIdTAG: 96452 SecondChildUserNameTAG: Lyla SecondChildCreateTimeTAG: 2013-02-04T01:21:27Z IndexTAG: 626 TitleTAG: CECC - Contest Closed It's 15th Jan in every time zone so we're closing the contest. I hope everyone who wanted to enter has submitted their entries. We'll let you know about the result (i.e. when we'll announce it) in another post. All the best to all the participants :-) UserIdTAG: 14386 UserNameTAG: ashwith CreateTimeTAG: 2013-01-15T10:41:01Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: All the best to all the participants! Thank you very much for participating, that is really important and very valuable. Thank you, thank you. Now, the Jury (ex-students of 6.002x Spring 2012) > (dantyrant, ChaunceyGardiner, JSChambers,Barrabas, komisz and Danik) will evaluate your videos - this may take 2 weeks-. So, once they have the result we will post it in the Forum Discussion and also we will comunicate with you by email. If you result one of the 3 winners, you will have to provide your personal data -complete name, adress, country, etc..- by email to an edX Staff who we will put in contact with you, so that you can receive your Signed Texbook at your home :). Good Luck! **Updated. January 16th** I forgot to include Barrabas too, my apologies. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2013-01-15T12:11:49Z IndexTAG: 627 TitleTAG: Only one edX account per individual. Just a reminder. The honor code states "Maintain only one user account and not let anyone else use my username and/or password." By having more then one account, an individual would have more attempts to answer the exams correctly, which could be interpreted as cheating. If anyone has been using two accounts, I suggest you take measures to close the duplicate(s) account(s) promptly, if there is such a provision. If not, unregister from the classes and abandon the duplicate accounts. Thank you for understanding. UserIdTAG: 292543 UserNameTAG: Pennypacker CreateTimeTAG: 2013-01-14T14:50:10Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 628 TitleTAG: Aaron Swartz commits suicide http://tech.mit.edu/V132/N61/swartz.html UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2013-01-13T16:45:53Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Sir Tim Berners-Lee, inventor of the web, wrote: "Aaron dead. World wanderers, we have lost a wise elder. Hackers for right, we are one down. Parents all, we have lost a child. Let us weep." I think all of us who enjoy this free knowledge delivery are in some way indebted to people like Aaron who fought and died for freeing knowledge from being locked up by a few greedy people. Dear Aaron, please do not rest! May your soul guide us forever. FirstChildUserIdTAG: 232157 FirstChildUserNameTAG: GayatriTR FirstChildCreateTimeTAG: 2013-01-18T09:22:43Z IndexTAG: 629 TitleTAG: locate pearsonvue centers in your country here you can locate som pearsonvue centers in your country : http://www.pearsonvue.com/servlet/vue.web2.core.Dispatcher?webContext=CandidateSite&webApp=TestCenterLocator&requestedAction=register&cid=346 UserIdTAG: 236136 UserNameTAG: Hamid-ch CreateTimeTAG: 2013-01-04T21:11:27Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Like most others: I am out of the word! :) There is no Iran in the website. FirstChildUserIdTAG: 420339 FirstChildUserNameTAG: AliJenabi FirstChildCreateTimeTAG: 2013-01-05T06:10:33Z IndexTAG: 630 TitleTAG: Course in sequency When will be the course in sequency to the 6.002x ? UserIdTAG: 260994 UserNameTAG: capvl CreateTimeTAG: 2013-01-04T11:12:40Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 631 TitleTAG: how many? anyone here knows how many have earned d certificates(passed d course)..?..jst for info.. UserIdTAG: 169416 UserNameTAG: subramanya26shin CreateTimeTAG: 2013-01-03T17:00:22Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: yeah, I am curiuos too FirstChildUserIdTAG: 207728 FirstChildUserNameTAG: RossyFromBulgaria FirstChildCreateTimeTAG: 2013-01-03T17:10:52Z IndexTAG: 632 TitleTAG: distinction and many thanks for the amazing course Another edx course (artificial intelligence) issued a separate signed letter to students that achieved distinction. Personally, the course has been extremely rewarding regardless of certificates and grades. I really hope that MIT considers offering more of their brilliant ocw courses on edx. UserIdTAG: 180393 UserNameTAG: anikitas CreateTimeTAG: 2013-01-03T11:01:27Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I'm hoping that MITx 6.002x team consider issuing a separate letter for those who achieved distinction in the course. Similiar with Berkeleyx CS188.1x Artificial Inteligence course signed by professors. Here's the sample Letter ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13572798631343699.png FirstChildUserIdTAG: 185303 FirstChildUserNameTAG: NormanDP FirstChildCreateTimeTAG: 2013-01-04T06:11:39Z IndexTAG: 633 TitleTAG: AMAZING CERTIFICATE. THE CERTIFICATE IS AS GOOD AS THE AMAZING COURSE.THANKS EDX , MITx ,AND EVERYONE ! UserIdTAG: 230076 UserNameTAG: ANANDAM CreateTimeTAG: 2013-01-03T06:05:38Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: I am still waiting for it.... FirstChildUserIdTAG: 529515 FirstChildUserNameTAG: Low FirstChildCreateTimeTAG: 2013-01-03T07:38:53Z SecondChildTAG: mee too SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2013-01-03T08:15:01Z SecondChildTAG: Check https://www.edx.org/dashboard and u can see " Download your PDF certificate" button. SecondChildUserIdTAG: 230076 SecondChildUserNameTAG: ANANDAM SecondChildCreateTimeTAG: 2013-01-03T08:18:07Z SecondChildTAG: nothing there SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2013-01-03T08:18:42Z SecondChildTAG: The same to me. SecondChildUserIdTAG: 364126 SecondChildUserNameTAG: shunyi SecondChildCreateTimeTAG: 2013-01-03T10:22:31Z SecondChildTAG: same here :( SecondChildUserIdTAG: 120850 SecondChildUserNameTAG: Nitin1A SecondChildCreateTimeTAG: 2013-01-03T12:29:41Z FirstChildTAG: when did u get that FirstChildUserIdTAG: 611666 FirstChildUserNameTAG: Shaminder420 FirstChildCreateTimeTAG: 2013-01-03T06:30:40Z SecondChildTAG: 35 minutes back in the dashboard ! Maybe i saw it the moment it came there...:) luv you all ! SecondChildUserIdTAG: 230076 SecondChildUserNameTAG: ANANDAM SecondChildCreateTimeTAG: 2013-01-03T06:34:24Z FirstChildTAG: congrats anandam :) i too got.happy and satisfied,thanx edx and mitx. but there's no grade mentioned on it. FirstChildUserIdTAG: 269641 FirstChildUserNameTAG: BAUWA FirstChildCreateTimeTAG: 2013-01-03T08:32:02Z SecondChildTAG: The way i thought was like its a dream to get a certificate with " completed course of study offered by MITx, an online learning initiative of The Massachusetts Institute of Technology" printed on it.I feel its more than a grade.The real grade come when we apply what we have learned so far and this will stand as a beginning for the best things thats yet to come. SecondChildUserIdTAG: 230076 SecondChildUserNameTAG: ANANDAM SecondChildCreateTimeTAG: 2013-01-03T08:53:49Z IndexTAG: 634 TitleTAG: Happy new year to all **Happy new year** to all edxian and associated people.. Let this New Year be the one, where all your dreams come true, so with a joyful heart, put a start to this year anew. Wishing you a happy and prosperous New Year 2013. UserIdTAG: 114864 UserNameTAG: Vikaash CreateTimeTAG: 2013-01-01T08:11:02Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: All the best to you and yours. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2013-01-01T12:57:44Z IndexTAG: 635 TitleTAG: Happy New Year from Estonia Happy New Year to all the Edx members. Wish you all stroooong stroooong health and patience in a new 2013 year. I wish that all your dreams come true !!! Happy new 2013 year !!! С новым 2013 годом !!! Head uut 2013 aastat !!! UserIdTAG: 269412 UserNameTAG: andrei836 CreateTimeTAG: 2012-12-31T20:38:31Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Happy New Year andrei836 :) FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-31T23:49:20Z IndexTAG: 636 TitleTAG: Happy New Year from Far East of Russia Happy New Year from Far East of Russia Всех с Новым 2013 Годом с дальнего Востока!!! UserIdTAG: 402617 UserNameTAG: Vl CreateTimeTAG: 2012-12-31T13:53:27Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 6 FirstChildTAG: Примите мои поздравления с Новым Годом! Всех благ и успехов, любви и радости! Сергей, Харьков. [HAPPY NEW YEAR!][1] [1]: http://www.youtube.com/watch?v=LRa8g-m4UNk Please accept my sincere greetings with the New Year! I wish you all the best and success, love and joy! FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-12-31T14:08:28Z SecondChildTAG: Спасибо, thank you. Вас с Наступающим!!! SecondChildUserIdTAG: 402617 SecondChildUserNameTAG: Vl SecondChildCreateTimeTAG: 2012-12-31T14:17:39Z FirstChildTAG: Happy new year Vladivostok ! :) FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-12-31T14:26:57Z FirstChildTAG: Примите поздравления из старого года:) В Донецке еще 2012. FirstChildUserIdTAG: 464744 FirstChildUserNameTAG: attache FirstChildCreateTimeTAG: 2012-12-31T14:33:01Z FirstChildTAG: С Новым Годом, Дальний Восток!!! Всего только самого лучшего! А мы его пока только ждем, еще почти 5 часов. Чебоксары, Чувашия, Россия:) FirstChildUserIdTAG: 328920 FirstChildUserNameTAG: TisO FirstChildCreateTimeTAG: 2012-12-31T15:17:26Z FirstChildTAG: same 2 u.:) FirstChildUserIdTAG: 266739 FirstChildUserNameTAG: satvikchugh FirstChildCreateTimeTAG: 2012-12-31T14:07:16Z SecondChildTAG: Happy New Year to all of you! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-31T15:21:41Z FirstChildTAG: Happy New Year from Moscow! FirstChildUserIdTAG: 345502 FirstChildUserNameTAG: ElectroVova FirstChildCreateTimeTAG: 2012-12-31T17:00:53Z IndexTAG: 637 TitleTAG: Happy New Year NewZeland and Pacific Islands !!!!! Happy New Year for all edx's in the pacific islands, Australia and New Zeland.!!!!!!!!! http://www.guardian.co.uk/lifeandstyle/video/2012/dec/31/new-zealand-new-year-auckland-fireworks-video UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-12-31T12:44:36Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Happy New Year from Spain too! Although we still have 11 hours of 2012 left yet ;) FirstChildUserIdTAG: 414462 FirstChildUserNameTAG: Pablo_C FirstChildCreateTimeTAG: 2012-12-31T12:56:53Z SecondChildTAG: Hmm.. I have less then 9 hours here.. Happy New Year! SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-31T13:20:06Z SecondChildTAG: just 4.20 to go..here in india..:) happy new yr world.:) SecondChildUserIdTAG: 266739 SecondChildUserNameTAG: satvikchugh SecondChildCreateTimeTAG: 2012-12-31T14:08:35Z FirstChildTAG: I haven't even had my morning coffee yet! T-16 hours here. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-12-31T13:13:28Z FirstChildTAG: Happy new year to you from Morocco ! and I'm glad to have classmates from all around the world ! FirstChildUserIdTAG: 324041 FirstChildUserNameTAG: sadrab FirstChildCreateTimeTAG: 2013-01-01T00:13:47Z IndexTAG: 638 TitleTAG: Jersey mark I know that this course is Priceless in its true sense ............. Nuff said But....... buddy in amrican university admission if i As apply as an undergraduate will it be considered noteworthy . To be honest My financial conditions do not allow me to take aP courses so i took 4 edx Courses SAAS = 55% Circuits and electronis 97 % Solid state chemistry perfect score in all exams and homeworks till now SAme case in health and numbers DOes taking this course instead of AP course hurt me .... What do you think ......Thanks a lot to you .......... And thanks a lot to the god of Circuits and electronics .....Mr Anant Agarwal (May your dreams come true /// If they havent Already !! LOL) UserIdTAG: 378150 UserNameTAG: GladIDoThis CreateTimeTAG: 2012-12-29T11:47:35Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: **GladIDoThis** (and those interested in "course credit by examination" in U.S. universities, may be **TL;DR** for others): If you apply to a U.S. university, you should definitely present the edX certificates along with a printout of your "Progress" to the Admissions Office along with your application. It will certainly help in that respect. Now as for credit, you would want to get credit that counts to your degree. So you would go to your prospective university's website and look at their course catalog. Here, you would see if there would be equivalent courses in which you have sufficient knowledge in (e.g. enough to retake and pass an university-level exam in) to qualify to take "course credit by examination." As for your particular example, and as most U.S. universities require you to take at least one science course no matter what the major (even if you are taking Art History or German as a major) and most U.S. universities offer a course in **general chemistry**; it would be 3.091x Solid State Chemistry that would be the most relevant in almost all cases. Once you find out your university offers "course credit by examination", which 99% should, you would go to the Chemistry department of your university, and 99% of U.S. universities have a chemistry department. You would talk to the department head professor (you would already be enrolled in a major and taking other classes at this point) and arrange with him/her a time and place to take this exam, the credit you would receive, the fee, and sometimes, the remedial or equivalency portion that the professor would require of you. With a "perfect score" in edX's Physical Chemistry, you should get an 'A' on almost any general chem exam, and maybe even pass a physical chemistry (a more advanced course) exam. Here, though, you would be deficient in the laboratory, or hands-on, portion, of a general and/or physical chemistry course, if that particular university structured its courses to have a laboratory component included with the lecture portion. The professor could remedy this in several ways, the most common being to take just the lab portion the following semester. If the particular university offers the lab portion *separately*, that is even better, as you would just get credit for the theory portion, and then have the option to take the lab separately whenever you wish. Note such a scenario is also applicable when talking about an introductory circuits analysis course with a lab, too; as I discuss next: As for 6.002x, you can only gain credit for this course if your university has a department of Electrical Engineering (or Electrical and Computer Engineering) available to give you a "course credit by examination." If you apply to a small, liberal-arts college, they will have no such department, and thus you will have no way of translating your knowledge into course credit. Not that it would matter for your degree anyways as these small colleges do not offer engineering degrees. But I'm sure every student that does take 6.002x considers taking Electrical Engineering, Computer Engineering, Engineering Physics, or one of the other majors at a major university where a **beginning course in circuit analysis** is one of the degree requirements. At -my- university, even the Electrical and Computer engineering students had to take at least one Civil Engineering course (Analysis of static forces; statics here as in load-bearing; analysis of trusses, etc. not static electricity!) and Civil and Mechanical vise-versa had to take a circuits course. In short, you should get "course credit by examination" for Fundamentals of Circuit Analysis, Introduction to Circuits and Electronics, etc. (as every university has a different name for it's first course on circuit analysis). And it should count towards your engineering degree. However, see the caveat about the lab portion that I pointed out. My particular university, University of Connecticut, where I took the course many years ago, had a separate circuits theory and simulation portion (EE 201) and hands-on lab portion (EE 209). The latter involved using oscilloscopes, bread-boarding and prototyping circuits, actually building op-amp amplifiers that amplify music, etc. Recent trends at major universities are to combine the two courses, so you will have to do a bit of talking with the professor. Also note that many universities have arcane requirements regarding courses that count for degree credit; one popular one being that once you start taking junior- and senior- level engineering courses in any discipline, you cannot get credit for the freshman and sophomore level courses after this point. Thus, depending on the university, if you take a junior-level course in Thermodynamics or Structural Analysis (if you are a ChemE or CE major), or in Electronic Control/Feedback Systems (if you are a EE major), you would not get credit for a sophomore-level Circuits course to count towards your degree. You would still get the degree, yes, as your prerequisite would be filled by the more-advanced course, and your grade would appear on the transcript, but you would have to take an additional, higher-level, course to get that degree. What may be applicable in your case, is that if you **do** get credit for a 6.002x-style course, and if your university has prerequisites for this course that are similar, e.g. differential equations and physics; then you should be able to skip all your math courses from calculus through differential equations, skip physics (not that you would get credit for them anyways in most cases), and *start off at a higher level than most of your other students!* You may have to make up for the credit for these lower-level courses by taking additional higher-level courses to replace these, but this looks excellent to employers and on your transcript. *Think how good graduate-level courses taken by a senior undergrad seem!* You talk about **AP (Advanced Placement)**, but AP only gives you credit in the low-level calculus and physics courses (any monkey can pass these!) and 6.002x is at a **much higher** level than this! However, it is *very important* to talk to your admissions counselor whatever you choose, since university policy varies so widely that you don't want to make costly mistakes; in the U.S. credit costs $$$. The 'bible', or the university guide that has the final say, is the Course Catalog (and every university should have one accessible online); so make sure you read it thoroughly before and after enrolling and when doing the ever-important task of choosing classes that count towards your major / degree. Finally, as for the **SAAS courses**, they seem more like "professional development" courses aimed at those already in the industry, and most universities do not give transfer credit, much less "credit by examination", for such courses, especially towards an undergraduate degree in any type of major. You could get lucky if you apply to the rare university that has a course similar to these, but I wouldn't worry too much; just take the SAAS courses as knowledge you gained, and as a boost for your C.V. (resume) and not as a means for college credit. Good luck in your educational endeavors, Jersey Mark FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-12-29T22:31:35Z SecondChildTAG: Thanks Jersey Mark! SecondChildUserIdTAG: 402193 SecondChildUserNameTAG: Feramico SecondChildCreateTimeTAG: 2012-12-29T23:21:05Z IndexTAG: 639 TitleTAG: Why wont grade be mentioned in Certificate? Most of the threads state that grade wont be mentioned in the certificate. I would like to know why? Wouldn't it dilute the efforts put in? I know that learning is a reward in itself, but, Grading would have added flavor to the efforts put in. I would really like to know the thought process behind not including the grade... UserIdTAG: 148007 UserNameTAG: adityashekhr CreateTimeTAG: 2012-12-28T08:36:09Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: are you sure???????????????????????? if it is a right then what was the logic to put grade in the course everywhere?????? is that for self assessment? if it is,then it doesn't create any sense at all.... I'm not sure if you are right but if it is, then i totally agree with the point you made... at least it deserve an answer....... FirstChildUserIdTAG: 108863 FirstChildUserNameTAG: shiviz FirstChildCreateTimeTAG: 2012-12-28T08:57:14Z SecondChildTAG: It contains grade of A, B or C. There are some example of Certificate at websites. SecondChildUserIdTAG: 374393 SecondChildUserNameTAG: rmaleki SecondChildCreateTimeTAG: 2012-12-28T12:43:41Z SecondChildTAG: no it doesn't they have mentioned earlier. what you have seen must be of prototype course SecondChildUserIdTAG: 342221 SecondChildUserNameTAG: deepkar SecondChildCreateTimeTAG: 2012-12-28T13:09:15Z FirstChildTAG: See https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50dd935856fba4290000003d FirstChildUserIdTAG: 183712 FirstChildUserNameTAG: colegiocientifico FirstChildCreateTimeTAG: 2012-12-28T13:47:25Z FirstChildTAG: Here is a possible thought process to explain the MIT "no grade" policy. I don't know if they actually think this way, but it's something that occurred to me, when I came to know about the grading policy. When a student attends MIT, he generally pays a fee, receives instruction, gets tested, passes the minimum requirements, and is granted a diploma certificate establishing that fact. Generally, all students "pay" the "same fee". So, they receive the "same diploma." If a student works harder than all the other students, and obtains 100% in a course, he gains more knowledge than others. However, that's his own effort. He didn't "pay" for that. MIT didn't give him anything there. He gave himself that extra. He didn't need to get 100% to meet the MIT requirements, he only needed 60%. What he paid for, is the certification that he met the minimum requirements for the MIT diploma. Since all students pay the same money for that certification, they should all get the same certificate. To make additional distinctions between students, and grant some of them more prestigious certificates than others, when they all pay exactly the same money for the service, could be viewed as unfair. So, they set a policy of not putting the grade, or actual performance report, in the certificate itself. For the real classroom students, the time for distinction would come when the student graduates and seeks employment, and has to offer up 3 references. The same certificate gets the student into the door, and granted the first job interview. But, in completing the hiring process, the employer checks with the Professors, whome the student mentions as references. At that point, the Professor can recall the actual grade, and make a comment, like "He was an excellent student," or "He was a pretty good student," or "He applied himself, and made it through all our rigorous course requirements, showing dedication and ambition," etc..by not putting the grade, the employer has to call the professor to check on the quality of the student he is thinking of hiring. So, the reward comes when you get a good reference report from your professor, in the job application process. The professor may not even remember the student, but most likely has that actual grade stored somewhere, which he will look up, and use as the basis of his appraisal. Other universities take a different approach, and even include such things as "valedictorian" etc..various Latin "Honors" which MIT shuns. ;) It's just a different philosophy. FirstChildUserIdTAG: 4463 FirstChildUserNameTAG: pmj FirstChildCreateTimeTAG: 2012-12-28T17:23:41Z SecondChildTAG: I like your thought process, but I can hardly imagine that the universities are able to process all the calls from potential employers. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-12-29T19:54:12Z IndexTAG: 640 TitleTAG: Suggestion to improve the course Thanks to all of you for providing this opportunity to us. I mention some point for improvement (I apologize if they are not elaborated enough): 1- The upper and small alphabet letter should be completely different in shape. for example (V and v) were same & confusing for inserting during some drills. 2- Subtitle should be available in videos that can be downloaded by the provided link (like movies) 3- The volume of downloaded videos from provided link is too higher than videos downloaded directly from YouTube or download manager software (200 Mb vs. 1 GB for same video). For countries that have filtration problem (such as India & Iran), it is again a bigger problem. 4- The behind of exercises can be covered with pictures from campus of the university presenting this course followed with some folk music of the university. Of course it should be an optional choice. 5- There should be a PDF file containing the summery of all sections from each week with the page number as a reference to each section (Week12-v5). During this course I took lots of print screen from desktop and saved all of them as a note for final term. Each image was saved with the name of each video, so I was able to watch related video in case of misunderstanding from the saved pictures (for example when the handwriting was not readable for me). ![enter image description here][1] 6- The idea of having a question similar to HW in final exam was great. I was both sick & busy. So when I saw it I was so happy. I just copied the answer and after getting B I left the exam happily. During the first minuets of final exam I made a mistake and I lost A. So when I passed the B grade, I left the exam. There should be some motivation for students to complete the entire exam. For example there could be some questions for extra point. 7- Statistics should be available, for example the number of students attending the drills for each week. 8- We should be able to add our picture to our accounts, discussion part and our certificate. It makes the class livelier. 9- Students who help others should take some extra point and their name should be announced weekly. (Like the first weeks of this semester) 10- Finally, please give us a memorial picture from all people who prepared this wonderful course for us. [1]: https://edxuploads.s3.amazonaws.com/13564616288474456.jpg UserIdTAG: 374393 UserNameTAG: rmaleki CreateTimeTAG: 2012-12-25T19:01:18Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 1 FirstChildTAG: All the lecture slides are available (with and without notations) on the Course Info page. Just look on the RHS panel for the link "Lecture Slides". You did not need to take all those screen shots. FirstChildUserIdTAG: 9673 FirstChildUserNameTAG: xp42 FirstChildCreateTimeTAG: 2012-12-26T00:27:55Z SecondChildTAG: That's exact ! SecondChildUserIdTAG: 373752 SecondChildUserNameTAG: Croak SecondChildCreateTimeTAG: 2012-12-26T01:39:33Z SecondChildTAG: Look at Lecture Slides Handout [Clean] [Annotated] under each videos ! SecondChildUserIdTAG: 373752 SecondChildUserNameTAG: Croak SecondChildCreateTimeTAG: 2012-12-26T01:41:24Z IndexTAG: 641 TitleTAG: Thank you Thank you Sir Agarwal and the entire EDX team for providing such a beautiful blend of theory and practical understanding of the subject .Hope to see the advanced versions of this course in the near future. Wishing all of you, Merry Christmas and a Happy New Year. UserIdTAG: 343388 UserNameTAG: bijojoseph CreateTimeTAG: 2012-12-25T04:53:35Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 0 IndexTAG: 642 TitleTAG: Feliz Navidad MIT 6.002x, from Costa Rica [Feliz Navidad y Próspero Año Nuevo (Click Aquí)][1] [1]: http://www.sotojohnson.net/mit.html UserIdTAG: 183712 UserNameTAG: colegiocientifico CreateTimeTAG: 2012-12-24T17:15:40Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Muchas Gracias, Igualmente para ti Carlos ! Me gustó mucho el arbolito de navidad animado, que lindo! Gracias! Te deseo lo mejor, que tengas una Feliz Navidad y un Próspero Año Nuevo!:) ¸¸.•*¨*•*´¨) ¸.•´¸.•*´¨) ¸.•*¨) (¸.•´ (¸.•` ¤ Merry Christmas and Happy New Year Carlos! ¸.•*¨*•*´¨) ¸.•´¸.•*´¨) ¸.•*¨) (¸.•´ (¸.•` ¤ Un abrazo! Y un cálido saludo a toda Costa Rica! Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-24T18:43:40Z IndexTAG: 643 TitleTAG: would the final exam answers be posted? I was a bit lax in my course work towards the end..there were a few difficult portions i had. .well, but t has given me an insight into what all m yet to be thorough about. .came out with an 81%. . t wud b really helpful if the answers of the final exam be posted. . Thanx to the Edx team , esp, Anant Sir for such a great initiative. .hats off! :-) i m eagerly waiting to do the other courses offered by the Edx team :-) Merry Christmas & Happy New year to all! P.S- is t possible 2 do d same course agn , as an attempt to better our grades? lol. . UserIdTAG: 167618 UserNameTAG: annvilla CreateTimeTAG: 2012-12-24T17:02:10Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hi annvilla, Your score is good! Congratulations!:) I were of the happy B Club too in the last Spring haha, I have shown my Certificate to all my family, I remember that haha. Yes, you can re-take the Course as many times you wish. Merry Christmas and Happy New year to you, Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-24T18:29:39Z SecondChildTAG: Hi Myriam, did you do the exam again? SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-12-24T21:39:01Z SecondChildTAG: Yes, I did Midterm Exam, Final Exam and all the Homeworks again :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-24T21:56:33Z SecondChildTAG: Do we get the certificate after the 25th? SecondChildUserIdTAG: 298775 SecondChildUserNameTAG: surja SecondChildCreateTimeTAG: 2012-12-25T00:28:01Z SecondChildTAG: Thanx a lot Myriam! :) & yea..mostly i wl b bk wd a b8r score in ds course nex yr! lol .. SecondChildUserIdTAG: 167618 SecondChildUserNameTAG: annvilla SecondChildCreateTimeTAG: 2012-12-25T04:32:22Z FirstChildTAG: for me this is my first time taking this course and I get 89% that is really cool does that mean I am smart...... :P (no but mr. Anant Agarwal is good lecturer and thanks to our friend Myriam ) FirstChildUserIdTAG: 226085 FirstChildUserNameTAG: Teto FirstChildCreateTimeTAG: 2012-12-25T04:03:41Z SecondChildTAG: an A agn ! congozz ! :) SecondChildUserIdTAG: 167618 SecondChildUserNameTAG: annvilla SecondChildCreateTimeTAG: 2012-12-25T04:32:59Z IndexTAG: 644 TitleTAG: Final Exam Solutions? Will there be final exam solutions posted at some point? Just curious to see what I did wrong. Buckets of thanks to Prof and staff for this course!!! UserIdTAG: 397595 UserNameTAG: mholin CreateTimeTAG: 2012-12-24T16:59:50Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 645 TitleTAG: Thanks a Lot to edx....,Waiting for More " Aha!!! Moments" I would Like to thank entire edx staff related to 6.002x course. Special thanks to Prof. Agarwal for the fun based learning and a bunch of Aha!!! moments. I really like those "Aha!!! moments" in this course. UserIdTAG: 178840 UserNameTAG: mehtanayanv CreateTimeTAG: 2012-12-24T15:35:12Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 646 TitleTAG: Thank you I just wanted to say thank you for this wonderful educational opportunity you have provided for all of us here! This is probably the closest thing to MIT level of education I will ever experience so I can't thank you enough! I learned a lot during this course and this knowledge will be extremely valuable in my future academic/professional career. Mr Agarwal you are the best professor ever, THANK YOU! Also, MANY THANKS to all the staff here and especially Myrimit who's hints saved me a couple of times from hours of frustration! Finishing this course is definitely THE BIGGEST AHA MOMENT!!! :) Merry Christmas to all! UserIdTAG: 145658 UserNameTAG: Wh1t3w0lf CreateTimeTAG: 2012-12-24T08:16:11Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You are welcome Wh1t3w0lf, I am really happy for you Aha Moment!:) I wish you all the best, Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-24T14:51:18Z IndexTAG: 647 TitleTAG: Special thanks from Pakistani Students Thanks imali for your wishes on behalf of Pakistani students to edx staff. I would like to take this oppotunity to express my best wishes and special thanks to Sir Anand Agarwal and all community TAs, especially Miss Myriam. Merry Xmas and Happy New Year to All Staff and student community. Hope all other Pakistani Students will express their wishes to support imali's statement. Best Regards UserIdTAG: 455950 UserNameTAG: Wahabbaluch CreateTimeTAG: 2012-12-24T06:04:03Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You are welcome Wahabbaluch ;) I wish you the best to you and a wonderful start for the 2013! Keep studying always Wahabbaluch XD, Greetings to Pakistan, Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-24T06:19:35Z SecondChildTAG: Great Thanks to EDX staff for this wonderful course.... SecondChildUserIdTAG: 368981 SecondChildUserNameTAG: asimakhtar SecondChildCreateTimeTAG: 2012-12-24T19:48:48Z IndexTAG: 648 TitleTAG: It's time to say goodbye! Thank you Prof. Anant Agarwal and all the professionals/staff that made this course a wonderful experience! Thiago Lima. UserIdTAG: 385692 UserNameTAG: tpfslima CreateTimeTAG: 2012-12-24T00:53:30Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 649 TitleTAG: Thank you for the knowlegde The course is excellent. I recommend it in all social network because EDX is good a source of knowledge. Thanks a lot. Now this course is famous until in the Amazon. I am of there. Knowledge is here!! UserIdTAG: 366669 UserNameTAG: pedroramus CreateTimeTAG: 2012-12-23T19:17:06Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hay, Pedro. I´m from Brasil. São Paulo. FirstChildUserIdTAG: 361823 FirstChildUserNameTAG: EliasOak FirstChildCreateTimeTAG: 2012-12-23T21:17:36Z FirstChildTAG: - Thank you for EDX courser to the provide a excellent initiative of offer the Circuits and Electronics course, very important to dissemination and liberty of knowledge. - I also thank Professor Anant Agrawal for excellent teaching and good humor in the most complicated of course, thanks to the aha-moments, I could understand more satisfactory. - Myriam, thank you for your patience and graciousness for helping us often in to resolve the most difficult issues. - All other community members and students who have helped us. - I will mention in my university here in Brazil, for my friends and teachers by the quality of courses EDX, and most importantly without paying anything! Thanks a lot and Merry Christmas. **PS:** Myriam, if you go to Brazil, talk me jejej.. :) best regards, Fabrício. ---------- **Em português - Brasil:** Agradeço ao EDX por proporcionar a excelente iniciativa de promover o curso de Circuitos e Eletrônica, de grande importância para a divulgação e libertação do conhecimento. Agradeço também ao professor Anant Agrawal pela excelente didática e bom humor nos momentos mais complicados do curso, graças ao aha-moments, pude compreender de maneira mais satisfatória. Agradeço a Myriam pela sua paciência e graciosidade por ter nos ajudado muitas vezes a resolver as questões mais difíceis. A todos os outros membros da comunidade e aos alunos que nos ajudaram. Indicarei na minha universidade aqui no Brasil, para meus amigos e professores da qualidade dos cursos EDX, e o mais importante sem pagar nada! Muito obrigado e Feliz Natal. PS: Myriam, se for ao Brasil, avise-me kkkk (jejeje...) :) Atenciosamente, Fabrício. FirstChildUserIdTAG: 355503 FirstChildUserNameTAG: fabriciogs FirstChildCreateTimeTAG: 2012-12-23T21:34:34Z SecondChildTAG: Tá paquerando a Myriam, Fabríciao!!! SecondChildUserIdTAG: 361823 SecondChildUserNameTAG: EliasOak SecondChildCreateTimeTAG: 2012-12-23T21:39:53Z SecondChildTAG: kkkkkkk q isso... apenas sendo gentil ;) SecondChildUserIdTAG: 355503 SecondChildUserNameTAG: fabriciogs SecondChildCreateTimeTAG: 2012-12-23T22:05:05Z SecondChildTAG: Muito obrigada Fabrício ;). I have learnt a little bit of Portuguese in my High School haha. Thank you for your nice words. This is so awesome, here I have met a lot of friendly people, thank you. It is funny and really cool, Classmates of 6.002x have invited me to visit their Countries if I were someday near there haha: I am welcome in India, Spain and now Brazil, what else can I ask haha ;) , I am so happy to have meet you to all of you. I wish you the best! Feliz Natal, Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-23T23:55:40Z SecondChildTAG: Hi,Myrimit,welcome to China!I'm very glad to show you around my country.haha~ SecondChildUserIdTAG: 361137 SecondChildUserNameTAG: Ericson SecondChildCreateTimeTAG: 2012-12-24T03:55:43Z SecondChildTAG: Cool Ericson!;) Thank you. China must be a beautiful place. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-24T05:00:13Z IndexTAG: 650 TitleTAG: Wow.....Final exam finished !!!!! Thank you so much professor Anant Agrawal for this beautiful journey of electronics...!!!! UserIdTAG: 410204 UserNameTAG: hkaushalya CreateTimeTAG: 2012-12-23T18:57:03Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Dear Prof. Agrawal Anant, You are a pearl among pearls. The way you took the course made electronics and circuit analysis so so simple to understand and your flow as well as apt in this course was key. As you migrated from one topic to the other, all were linking. It was amazing! You made it look like I crammed at my undergraduate days to get my exams solved. But now I see circuits and electronics in another light. Most especially using nodal analysis to get things done the "aha" way. Sir, I am very grateful. More grease to your elbows. Long live circuit & electronics, long live Prof Agrawal. Best wishes this yuletide season. Regards, Ugo From Nigeria. FirstChildUserIdTAG: 329879 FirstChildUserNameTAG: Ugochukwu FirstChildCreateTimeTAG: 2012-12-24T10:59:11Z IndexTAG: 651 TitleTAG: Thanks a lot Well done MIT!!!! Excelet work!!! I would like to sincerely thank all the people who made ​​this possible. You are doing a great job. It was not easy, but very interesting and realy exciting. Hope to see more courses about electronics from MIT in future. UserIdTAG: 379741 UserNameTAG: Tolcheev_Art CreateTimeTAG: 2012-12-23T18:34:04Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 0 IndexTAG: 652 TitleTAG: STAFF: Is the Final extended to Dec. 24th 5:00AM EST? I was just looking at the exam deadline and it now says December 24th at 5:00AM EST, instead of the previous December 23rd at 11:59PM EST. Is the exam extended? Can we get an official confirmation? I will email Staff, as nothing so far is posted in the Discussion, the Course Info, or has been emailed to us as Community TAs. I am just about to open the exam, and was wondering if I should delay so in order to study more / get some sleep until 5:00AM December 23rd, so I get the benefit of the full 24 hours just in case I need it! Sincerely, Jersey Mark ***EDIT: I have received official confirmation from Staff that the exam will be extended to December 24th, 5:00AM Eastern Standard Time (EST) / 10:00AM Greenwich Mean Time (GMT or UTC).*** UserIdTAG: 292546 UserNameTAG: JerseyMark CreateTimeTAG: 2012-12-23T05:07:27Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: I also see the same INFO you mentioned above. FirstChildUserIdTAG: 99441 FirstChildUserNameTAG: coyarce FirstChildCreateTimeTAG: 2012-12-23T05:23:07Z FirstChildTAG: I believe the difference is explained by Boston vs. Greenwich Mean Time. I would not gamble on the deadline being beyond December 23rd at 11:59PM Boston. (Note: I do NOT have official word; an alternate explanation COULD be a few additional hours were added because the release was delayed... but personally, I wouldn't bet on it, especially since the Dec. 24th date/time does NOT give a time zone AFAICT.) FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-12-23T11:19:08Z SecondChildTAG: **planetscape:** I live in the Boston time zone (in New Jersey, it's the same for the U.S. East Coast), and previously my PC showed December 23rd 11:59 (PM Eastern Standard Time) right next to the little "alarm clock" icon by where it says "Final Exam." Last night that changed to December 24th, 5:00 (AM Eastern Standard Time, but this is not explicitly noted). Another user confirmed that they saw the same time (5:00 AM) posted, so I know this is not specific with my PC. Anyways, I just woke up (7:00AM Eastern Time) and Staff has **not** confirmed anything nor responded to my emails to them as of this time. I did, however, decide to go to bed last night instead of opening the exam at 11:59 on the 22nd. If the exam indeed ends at 11:59 tonight, I will have lost the 5 hours, but at least I gained sleep! If the clock is correct, however, I should still be able to submit answers until 5:00AM after midnight on the 24th. Hopefully my brain is in good operating mode and I finish in 12 hours, and not have to need the whole 24 anyways :-) Jersey Mark SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-12-23T12:54:43Z SecondChildTAG: Good luck Mark! SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-12-23T13:14:50Z SecondChildTAG: Thanks **Pennypacker**! SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-12-23T14:09:46Z SecondChildTAG: I'm glad you got some decent sleep, Mark - that's essential, IMHO. Hey, if *I* finished in under 24 (13.5 to be specific ;-/) I'm sure you will not need the entire time. Good luck! SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-12-23T15:02:35Z SecondChildTAG: The displayed deadline was due to the difference between GMT and EST. However, because of the potential confusing that this may have caused, we have now extended the true deadline to Dec 24th 5:00 am EST, Dec 24th, 10:00am GMT. The display should remain the same. I hope that you can get some good sleep, JerseyMark. SecondChildUserIdTAG: 96452 SecondChildUserNameTAG: Lyla SecondChildCreateTimeTAG: 2012-12-23T18:22:46Z FirstChildTAG: I think they have extended the deadline, so as to compensate the delay occured during the time of release of exam :) FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-12-23T16:36:31Z IndexTAG: 653 TitleTAG: Thanks for the course I wanna to thanks to all 6.002x team that are really great. Was a really nice experience!! My big Merry Christmas and Happy New Year!!! to the 6.002x team and for all the people that took the course all over the world. UserIdTAG: 298612 UserNameTAG: fabianh CreateTimeTAG: 2012-12-22T23:38:50Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 654 TitleTAG: is there any continuation of the course may i ask if there is any continuation of the course. UserIdTAG: 475448 UserNameTAG: msamwelmollel CreateTimeTAG: 2012-12-22T20:26:32Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: While I have no news about it's offering at edX, you could have a look at courses like 6.003 http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-003-signals-and-systems-spring-2010/index.htm , although it depends on what you want to do. They have lots to choose from, if/when then become available here, you will be set. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-12-22T20:51:10Z SecondChildTAG: Is there a course about automatic control on mitOcw ? I talk processes, regulators , pd, pi, pid ? Thanks SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-12-23T00:00:09Z FirstChildTAG: There will be other couses . Watch S25S1 . FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-12-28T23:20:58Z IndexTAG: 655 TitleTAG: Thank you for extra tries , got a perfect 100% ! Thank you for extra tries , got a perfect 100% Now my soul is at peace : ) UserIdTAG: 263693 UserNameTAG: Coldberg CreateTimeTAG: 2012-12-22T14:24:27Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 656 TitleTAG: doubt about certificate hi..i want to know about the certificate..will it show the exact grade or just that you have passed the course?..can someone from the TAs please answer? UserIdTAG: 173147 UserNameTAG: cruiser_rahit CreateTimeTAG: 2012-12-22T14:22:37Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: It will not have the grade. This was stated by the staff in a couple of threads. FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-12-22T14:24:52Z SecondChildTAG: thanks SecondChildUserIdTAG: 173147 SecondChildUserNameTAG: cruiser_rahit SecondChildCreateTimeTAG: 2012-12-22T14:27:39Z SecondChildTAG: I may have said at one point that it will not have a grade, since then however I have come across previous examples that did show a *letter* grade. I guess we will see. If you feel strongly either way about this, you could start a new feedback thread or send your feedback to mit-6002x@edx.org . Have fun SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-12-22T14:42:52Z FirstChildTAG: I think it would be fine so show grades, but equally fine and fair would be to close forum (read only) during exams time, and giving just one try in exams. In real exams, how many tries do you have? I think having Internet and doing from the comfort of your home is already a nice advantage. I suppose what MIT doesn't want is people to desert real MIT. Anyway, this has been the best learning experience I've ever had. If all other subjects and professors are half as good as this one, studying at MIT must be a pleasure. Imagine how many good students that have done this course will want to study in real MIT! THANK YOU all for this magnificent course! FirstChildUserIdTAG: 398594 FirstChildUserNameTAG: DaveyJC FirstChildCreateTimeTAG: 2012-12-22T16:31:14Z SecondChildTAG: thanks to our prof agrawal thanks to mush and to mit team for there tutorials and this courses. SecondChildUserIdTAG: 174229 SecondChildUserNameTAG: fares27 SecondChildCreateTimeTAG: 2012-12-22T16:40:11Z SecondChildTAG: There's an important reason for giving more than one try. In a real MIT exam you would be able to show your work and get partial credit even if you got the wrong answer. Also if you got part 1 of a problem one and then propagated that answer into part b, you could possibly get 100% on part b as long as you did the work correctly for part b (this would depend on the course policy). The extra attempts do a good job of compensating for not being able to show your work. SecondChildUserIdTAG: 174100 SecondChildUserNameTAG: markpolak SecondChildCreateTimeTAG: 2012-12-23T04:42:27Z SecondChildTAG: That's true, I hadn't thought it that way... SecondChildUserIdTAG: 398594 SecondChildUserNameTAG: DaveyJC SecondChildCreateTimeTAG: 2012-12-23T13:24:58Z SecondChildTAG: Markpolak's reasoning is a very valid and an important one for giving multiple attempts (at least 2 to 3). Adding to that, in places like where I live, there are frequent and uninformed power cuts and power fluctuations. Even when I was taking the final exam this time, the connection suddenly got cut due to a sudden voltage glitch, which almost gave me a heart-attack. To be on a safer side, just having the assurance that there are 2 to 3 extra attempts prevents heart-attacks! :-) Also, more than anything else, the extra attempts thing acts like charm on the general psyche of nerve-wrecks like me when it comes to exams. In fact, if real exams (offline exams) have this option, I'm sure it would work wonders on the students. SecondChildUserIdTAG: 232157 SecondChildUserNameTAG: GayatriTR SecondChildCreateTimeTAG: 2012-12-24T08:36:47Z FirstChildTAG: I think it would be good to have the grade letter in the certificate. Looks like last time (pilot course in spring) the certificates contained the grade. I saw that in one of the ppt slides of Dr.Agarwal on the net. Hope the same format is followed this time as well. Any comments, Myriam?? FirstChildUserIdTAG: 232157 FirstChildUserNameTAG: GayatriTR FirstChildCreateTimeTAG: 2012-12-22T17:36:00Z SecondChildTAG: If grades are added to the certificate, will it also include the plus or minus? ie. B+ or A- SecondChildUserIdTAG: 284781 SecondChildUserNameTAG: DavidShen SecondChildCreateTimeTAG: 2012-12-22T17:44:42Z SecondChildTAG: are there seperate + and minus grades?.. SecondChildUserIdTAG: 173147 SecondChildUserNameTAG: cruiser_rahit SecondChildCreateTimeTAG: 2012-12-22T19:18:39Z FirstChildTAG: "If grades are added to the certificate, will it also include the plus or minus? ie. B+ or A- " - David Shen. Last time it was a letter grade A,B,C. "are there seperate + and minus grades?.." -cruiser_rahit. It would be the same as viewed under your progress tab. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-12-22T19:43:37Z IndexTAG: 657 TitleTAG: Thanks to all the MITx team I really enjoyed the classes and I did learn so much from all of you, so I'd like to thank you all... all the team that made this possible. Prof. Anant Agarwal,Prof. Gerald Sussman,Dr. Piotr Mitros,Prof. Chris Terman,Prof. Khurram Afridi, Richard Zhang,Jorge Elizondo Martinez, Kahlil Dozier, Alec Garza-Galindo, Lyla Fischer, Rohan Nagarkar, and anyone envolved in the development of this course. Thank you all for your efforts! It has been an amazing journey across the electronic basics with all of you. Greetings from Spain! UserIdTAG: 414462 UserNameTAG: Pablo_C CreateTimeTAG: 2012-12-22T12:55:24Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Yes, Thank You for all of them, they are doing a great work in edX :) FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-22T20:01:28Z SecondChildTAG: indeed. SecondChildUserIdTAG: 418182 SecondChildUserNameTAG: euldji2005 SecondChildCreateTimeTAG: 2012-12-22T23:40:35Z SecondChildTAG: without to forget your assistance, Myriam ! SecondChildUserIdTAG: 418182 SecondChildUserNameTAG: euldji2005 SecondChildCreateTimeTAG: 2012-12-22T23:41:10Z IndexTAG: 658 TitleTAG: Another Thank You A very big thank you! to the whole team of 6.002x, especially Professor Agarwal. It's been great fun- I'm reasonably happy with my result, but incredibly happy at being allow to take part in something as amazing as edX. I hope that free education is always around for those who want to learn something new. Thanks for all your energy, help, patience, and thank you so much for your all your time! Hans UserIdTAG: 403503 UserNameTAG: jmohrmann CreateTimeTAG: 2012-12-20T23:41:52Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 1 FirstChildTAG: I concur with these sentiments. Happy Holidays. FirstChildUserIdTAG: 260299 FirstChildUserNameTAG: Jack1947 FirstChildCreateTimeTAG: 2012-12-21T02:44:22Z IndexTAG: 659 TitleTAG: Thanks a lot Dear staff, thanks for excellent course and new experience. Hope to have time for other MIT courses. UserIdTAG: 195218 UserNameTAG: Al_Incognito CreateTimeTAG: 2012-12-20T14:43:22Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 660 TitleTAG: Please refrain from asking about the exam (Q6) A few students have reported issues about question 6. The details have been passed on to Staff. Try to refrain from discussing the question in this forum. Thanks! UserIdTAG: 292543 UserNameTAG: Pennypacker CreateTimeTAG: 2012-12-20T14:16:31Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: My question intended to be seen by the staff because I was reporting an conflict in the way the question was enunciated. It seems that there is also an issue in Q4. Please check. My time limit is near. Thanks FirstChildUserIdTAG: 401503 FirstChildUserNameTAG: locounpoco FirstChildCreateTimeTAG: 2012-12-20T15:06:50Z FirstChildTAG: Also in -=Q4=-!!Must somethings wrong,the expression makes me confused. I am sure with my answer both in Q4 and last 2 question of Q6. Please Check!!! FirstChildUserIdTAG: 296511 FirstChildUserNameTAG: sky0917 FirstChildCreateTimeTAG: 2012-12-20T16:11:33Z FirstChildTAG: there is the problem me also i think in the question 6 .so staff must check it and i don't know when they find the mistake they are going to compensate on all of us or FirstChildUserIdTAG: 475448 FirstChildUserNameTAG: msamwelmollel FirstChildCreateTimeTAG: 2012-12-20T16:18:16Z FirstChildTAG: OK the issue with Question 6 should be resolved. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-12-20T17:37:15Z SecondChildTAG: Not that I can see... Is there a direct email for such problems? SecondChildUserIdTAG: 141000 SecondChildUserNameTAG: OrinE SecondChildCreateTimeTAG: 2012-12-20T18:17:35Z SecondChildTAG: Let me see, I will get back to you. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-12-20T18:43:52Z SecondChildTAG: OK your best bet is to email mit-6002x@edx.org to voice any concerns. Thanks for your understanding. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-12-20T19:01:06Z SecondChildTAG: Thanks! I sent a note and the problem I noticed is fixed. SecondChildUserIdTAG: 141000 SecondChildUserNameTAG: OrinE SecondChildCreateTimeTAG: 2012-12-20T23:12:19Z IndexTAG: 661 TitleTAG: solder smoke what about the smoke which is generated when melting the solder? It is harmful? What is the composition of this smoke? UserIdTAG: 388524 UserNameTAG: dasbinich CreateTimeTAG: 2012-12-17T20:49:13Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: Soldering NumberOfReplyTAG: 5 FirstChildTAG: Yes, the smoke from solder may contain lead which if inhaled in very large quantities can lead to lead poisoning. At many electronics workbenches, there are [Solder Fume Extractors][1] which reduce the risk of inhaling lead. [1]: http://www.markertek.com/Tools-Test-Equipment/Soldering-Equipment-Solder/Tenma/21-7960.xhtml?utm_medium=shoppingengine&utm_source=googlebase&cvsfa=3786&cvsfe=2&cvsfhu=54454e2d534645&gclid=CMyPu5O3orQCFemiPAodHD4A3A%22 FirstChildUserIdTAG: 391929 FirstChildUserNameTAG: RohanNagarkar FirstChildCreateTimeTAG: 2012-12-17T22:01:51Z FirstChildTAG: People, and also don't forget that in Week 13 tutorials we have a solder lesson from a person involved on the Apollo project https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_13/wk13_solder They even talk about these risks. FirstChildUserIdTAG: 288174 FirstChildUserNameTAG: matiasgrodriguez FirstChildCreateTimeTAG: 2012-12-18T00:18:09Z FirstChildTAG: I have read that the risk of getting lead poisoning from solder fumes is minuscule compared to the risk of handling the solder with your bare hands - especially if you don't have a strict habit of always washing your hands thoroughly after soldering. I also read that the major risk with solder fumes is the evaporated flux, not the lead. Whether that is correct or not - it makes sense to get rid of the smoke *and* wash your hands after soldering. FirstChildUserIdTAG: 90796 FirstChildUserNameTAG: ChaunceyGardiner FirstChildCreateTimeTAG: 2012-12-17T23:05:51Z SecondChildTAG: this is mentioned in the tutorial video SecondChildUserIdTAG: 623210 SecondChildUserNameTAG: preveen SecondChildCreateTimeTAG: 2012-12-23T18:30:58Z FirstChildTAG: In Europe it's not possible to buy lead solder any more. I haven't got my own soldering iron setup yet, but I would be interested to hear if anyone has some experience with lead-free solder. I have heard it's not as good as leaded solder, but I guess it's what we're stuck with. Is there a type which is better than others? Does it require a different soldering technique? e.g. different soldering iron temperature? FirstChildUserIdTAG: 434869 FirstChildUserNameTAG: patey FirstChildCreateTimeTAG: 2012-12-18T12:47:43Z SecondChildTAG: Get a weller TCP, with out a doubt the best soldering Iron (socond only to the METCAL with is out of most peoples pricerange, works with Radio Frequency Induction). The weller TCP uses the Curi effect, so each soldering tip has a magnetic slug at the bottom that switches a magnetic relay, regulating the temperature. It really works. SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-12-19T14:11:44Z SecondChildTAG: OK, thanks. I will look into this. SecondChildUserIdTAG: 434869 SecondChildUserNameTAG: patey SecondChildCreateTimeTAG: 2012-12-24T14:43:46Z FirstChildTAG: In my experience, working with lead-free solder really means a hot air rework station. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-12-20T15:52:54Z SecondChildTAG: sorry, but what's a hot air rework station? Something to do with desoldering? (wikipedia didn't help me!) SecondChildUserIdTAG: 434869 SecondChildUserNameTAG: patey SecondChildCreateTimeTAG: 2012-12-24T14:46:44Z IndexTAG: 662 TitleTAG: Just a recommendation We all take this course online.And we don't have a real book. But we all need to take notes.If we take notes of the lectures on our own notebooks,it won't be easy for us to review..If there is a section for notes beside every video,the notes will relate to lectures closely.Besides,we also can share notes with others,isn't it cool? Sorry for my English... UserIdTAG: 136959 UserNameTAG: christerpher CreateTimeTAG: 2012-12-16T11:43:39Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: you can download notes here![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13556794611343695.jpg FirstChildUserIdTAG: 236136 FirstChildUserNameTAG: Hamid-ch FirstChildCreateTimeTAG: 2012-12-16T17:38:05Z SecondChildTAG: I knew that,but it's not personal enough.. SecondChildUserIdTAG: 136959 SecondChildUserNameTAG: christerpher SecondChildCreateTimeTAG: 2012-12-17T02:10:10Z SecondChildTAG: I've been using [TiddlyWiki][1] with the [MathJax][2] [plugin][3]. [1]: http://www.tiddlywiki.com/ [2]: http://www.mathjax.org/ [3]: http://tiddlyspace.com/bags/math-template_public/tiddlers/PluginMathJax%20v1.3/revisions.wiki SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-12-17T07:38:32Z SecondChildTAG: Write them by hand . This way you will memorize them better, and you can take your notebook everywhere. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-12-19T00:42:20Z IndexTAG: 663 TitleTAG: Current IC Process Technology We just past 22nm technology heading down to 14nm. UserIdTAG: 132828 UserNameTAG: rharris CreateTimeTAG: 2012-12-15T00:12:31Z VoteTAG: 4 CoursewareTAG: Week 14 / S27V8_How_to_further_Reduce_Power_Consumption_ CommentableIdTAG: 6002x_S27V8_How_to_further_Reduce_Power_Consumption_ NumberOfReplyTAG: 2 FirstChildTAG: Oh yes, you are well aware of this because you work for Intel! If I recall correctly... ;-) FirstChildUserIdTAG: 376877 FirstChildUserNameTAG: AndBre FirstChildCreateTimeTAG: 2012-12-15T17:33:39Z SecondChildTAG: Really? Are we already at the level of Intel employees? xD SecondChildUserIdTAG: 398594 SecondChildUserNameTAG: DaveyJC SecondChildCreateTimeTAG: 2012-12-16T21:19:47Z SecondChildTAG: Yes, that is correct. Oh, and by the way, I confirmed those numbers on an independent internet site before disclosing them to insure I wasn't disclosing confidential information to the public before Intel did. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-12-16T21:28:48Z SecondChildTAG: Sure, you can be at that level, it just depends on one's level of education. We have admins with no degree, technicians with AA degrees and engineers with BS, MS and PhD's degrees. Heck, I heard we even have employees with PhD's in Anthropology, who study the complexity of human cultures around the world. If you are going to sell to multiple cultures you have to understand them. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-12-17T00:12:50Z FirstChildTAG: Geez, and I just installed my 28nm video card. It's true what they say it's obsolete when you buy it! :) FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-12-17T16:27:44Z IndexTAG: 664 TitleTAG: ZENER in H12P2! I can't understand the behavior of the zener diode in the two basic situations: * vin < vz * vin > vz I thought that if (vin < vz) then, according to the zener v-i graphic, vin is negative, so on the plus side of the diode we got 0 potential and on the minus side we got something negative lower than vz. So vD = 0 - (something negative lower than vz) = (something positive greater than vz in magnitude). So I thought the zener diode would be conducting in this first (vin UserIdTAG: 398594 UserNameTAG: DaveyJC CreateTimeTAG: 2012-12-10T23:22:14Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: This had me stumped for while, too. Until I realized that the diode is in the circuit "backwards" (also called reverse-biased). Because of that, the voltage on the anode (positive terminal) is at ground and is *always* going to be less than or equal to the value on the cathode (which is at $v_+$ if $R_Z=0$). So, you need to look at the portion of the I-V graph for the diode that is to the left of the origin. In that case, when $v_{in} \lt v_z$, the voltage across the diode will be zero, and when $v_{in}$ increases to the point that it’s greater than $v_z$, the voltage will be pegged at (positive) 2.5 volts (the breakdown voltage of the zener diode). It’s a key point and a fundamental element of a simple voltage clamp like this, so it’s worth taking a few minutes to fully understand it, as you have. FirstChildUserIdTAG: 339668 FirstChildUserNameTAG: chickwebb FirstChildCreateTimeTAG: 2012-12-10T23:51:05Z SecondChildTAG: At least, I think that's correct! :-) SecondChildUserIdTAG: 339668 SecondChildUserNameTAG: chickwebb SecondChildCreateTimeTAG: 2012-12-10T23:51:26Z SecondChildTAG: yes, it is. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-12-11T16:56:12Z FirstChildTAG: Isn't voltage just a relative measure? 0 volts is not always going to be less than anything else, you can have -5V that's even less voltage. And vin < vz means for me that the positive terminal of the zener is at 0V and its negative terminal is at -(something)V. And 0-(-something) = +something, isn't it? I think the key must be the relationship between vin & vD. Could someone elaborate, please? Thank you! FirstChildUserIdTAG: 398594 FirstChildUserNameTAG: DaveyJC FirstChildCreateTimeTAG: 2012-12-11T13:28:32Z FirstChildTAG: Found this: "A zener diode is designed to allow a current to flow through it in a direction that is reverse to the normal flow of current that would occur if it were used as a rectifier. Current can flow through a zener diode in both directions. In the forward direction, current will flow at a low voltage, usually about 1 volt. In the reverse direction, no current will flow until the voltage impressed across it is equal to the zener voltage. At this point, a current will flow and an extremely small increase in voltage will cause a large increase in current. Most importantly, it should be noted that the current flow through the zener diode is in the reverse direction to that of a normal rectifier." This maybe explains things here: in zener voltage, only its magnitude is usually considered, taking into account there's only a zener voltage if the diode is reverse biased. So if vin < vz (supposing vz is positive), means that 0-(positive vz) = -ve but higher than vz in the graphic (negative), so it doesn't conduct. I keep on thinking that the problem is somewhat poorly specified. What do you think? FirstChildUserIdTAG: 398594 FirstChildUserNameTAG: DaveyJC FirstChildCreateTimeTAG: 2012-12-11T13:52:00Z SecondChildTAG: In the graph you see that Id>0 is the forward current (normally flowing through anode(vd+) to kathode(vd-) in forward direction of the diode, you can see that because the triangle points to the vd-minus), causes vf=vforward. But the reverse current Id, flowing from the kathode to anode, causes vz, because in fact you reversed the diode. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-12-11T17:05:35Z FirstChildTAG: I'm also having trouble understanding parts a and b and how to deal with Vz with respect to Vin, so I simulated it with LT SPICE IV from Linear Technologies, which use models of actual devices with real parameters. Looking at the results from the waveform, I still believe I would have a hard time figuring out which region the zener diode is operating in based on Vin, without it actually being stated in the homework problem, or simulating it to find out. Comments are welcome. Currently the circuit is a little different from our homework problem, because I don't know that much about BJT's, so I just picked one, manufactured by Philips. It seems to affect Vout as Vout basically stays constant at 0. Also, the breakdown voltage for the Zener diode is Vz = 4.7, manufactured by OnSemi, which is the closest to 2.5V I could find. Same goes for the Op Amp, I just picked one manufactured by Linear Technologies that had an offset current in the uA range. Regarding the waveform, I swept the source Vin from -10V to 20V DC, shown on the bottom horizontal axis. The left vertical axis is for voltages V(vd), V(vin) (dark blue) and V(out) and the right axis is for the current through the diode and R0. The odd thing I found was the direction of current flow through the diode and R0 were opposite to each other, but equal in magnitude. The current flow through R0 was from Vin down through the resistor towards the -eg terminal of the Op Amp and the current through the diode is from ground towards the -eg terminal. I guess that is how the Op Amp maintains zero current into it's terminals, their magnitudes cancel it out. Can anybody suggest the proper BJT and Op Amp to use, so I can get a Vout of 5V like the homework? At least until I have time to research how to select the proper components. ![enter image description here][1] ![enter image description here][2] [1]: https://edxuploads.s3.amazonaws.com/13552730671343673.jpg [2]: https://edxuploads.s3.amazonaws.com/13552733551343644.jpg FirstChildUserIdTAG: 132828 FirstChildUserNameTAG: rharris FirstChildCreateTimeTAG: 2012-12-12T01:40:20Z SecondChildTAG: How about half of a LM358 and a 2N2222/PN2222A. http://www.ti.com/lit/ds/symlink/lm158-n.pdf http://www.fairchildsemi.com/ds/PN/PN2222A.pdf SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-12-12T02:09:58Z SecondChildTAG: The op-amp inputs would appear to be backwards... the zener/R0 should go to the non-inverting input and the output voltage divider to the inverting input. SecondChildUserIdTAG: 141000 SecondChildUserNameTAG: OrinE SecondChildCreateTimeTAG: 2012-12-12T02:19:45Z SecondChildTAG: Good catch, thank you. That is my Dyslexia kicking in - so frustrating. Hmmmmm, I get to see a positive feed back scenario and can compare it to the the negative feedback of the homework problem. Good for comparisons. Thanks Skyhawk, I'll look and see if models exist for the devices you suggest. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-12-12T04:56:30Z SecondChildTAG: Let us know how your LTSpice simulation turns out. A PN2222A will probably work for a 100 ohm load, but calculate the power it will dissipate with Vin of 20V; you might be letting the magic smoke out! I'd be thinking of a transistor in a TO220 package and a heatsink in this case. Orin. SecondChildUserIdTAG: 141000 SecondChildUserNameTAG: OrinE SecondChildCreateTimeTAG: 2012-12-12T08:27:05Z SecondChildTAG: > The current flow through R0 was from Vin down through the resistor towards the -eg terminal of the Op Amp and the current through the diode is from ground towards the -eg terminal. ???? You sure about that? Sounds like it would violate KCL to me. No current flows through the Op Amp inputs because they have very high input impedance. The Op Amp is sensing the voltage (relative to ground) at the input terminals using internal circuitry. Good diagram of the internals of one here on Wikipedia - http://en.wikipedia.org/wiki/File:OpAmpTransistorLevel_Colored_Labeled.svg. Note that the inputs are to the Gates of BJTs, which have low base input current. Besides all that, if current is flowing from ground through the diode, then it is forward biased (and the portion of the diode I-V curve that is relevant is the part to the right of the origin). But we’re told that’s not the case (and can confirm by inspection), since the voltage on the negative terminal (cathode), is by the design of the circuit always going to be $\ge 0$ (i.e., the diode is reverse biased). If it’s always reverse biased, then only the portion of the diode I-V curve that’s to the left of the origin applies, and the voltage across the diode is (ideally, anyway) either zero or $v_Z$. SecondChildUserIdTAG: 339668 SecondChildUserNameTAG: chickwebb SecondChildCreateTimeTAG: 2012-12-12T17:23:13Z SecondChildTAG: You just have to be careful to note which way the arrow points on LTSpice's current probes. In this case, all seems fine as the graph shows the current through the zener to be negative and LTSpice is most likely considering positive current through the zener to be when it's forward biased. SecondChildUserIdTAG: 141000 SecondChildUserNameTAG: OrinE SecondChildCreateTimeTAG: 2012-12-12T17:51:25Z SecondChildTAG: Ok, I believe I corrected my mistake and added some of the components suggested. For the LM358 from TI I used a LT1013 from Linear Technologies. I looked at the Absolute Maximum Ratings and at first glance they seem similar. Now Vout is 9V, a whole lot closer to the 5V required by the homework, with negative feedback this time. I reduced VS for he Op Amp to +/- 22V as the datasheet specifies. Also, I'll look for a component similar to TO220 that has a model for LT SPICE One thing I noticed there is never an instance where Vin < Vz [V(vd) in the case of this simulation] with the diode current at 0A, which was the scenario when calculating the equation in part a of the HW. By the way, this is an excellent way to study and learn electronics!! Thanks, Dr. Agarwal, students, TA's and Staff, great job. ![enter image description here][1] ![enter image description here][2] [1]: https://edxuploads.s3.amazonaws.com/13553383391343664.jpg [2]: https://edxuploads.s3.amazonaws.com/13553383631343622.jpg SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-12-12T18:59:00Z SecondChildTAG: A 1N750 has a Zener voltage of 4V7 so an output of 9V4 is expected. http://www.futurlec.com/Diodes/1N750.shtml A BZX79-C2V4 should work better. A tweak of R1 will put you right on. http://www.futurlec.com/Diodes/BZX79.shtml OrinE is correct that the power dissipation limits for the 2N2222 are exceeded for a vin between 16V and 17V. The TIP31 and TIP41 are two inexpensive transistors with high power ratings that should work. http://www.futurlec.com/Transistors/TIP31C.shtml http://www.futurlec.com/Transistors/TIP41C.shtml SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-12-12T19:39:27Z SecondChildTAG: Regarding the current flow through R0 and VD, actually, KCL seems to hold up, as LT SPICE shows current flow into the Op Amp to be 0V, which makes sense, since the current through R0 and VD is equal in magnitude, but opposite in direction, both flowing towards the V+ node connected to the Op Amp. I see the direction of the current flow in LT SPICE by placing the cursor over the component in question. And, come to think of it, looking at the datasheet for the LT1013, the Input Offset Current is 0.8nA, which is probably why Dr. Agarwal calls the connection between V+ and V- to be a virtual short. There is actually current flow, but very very small if I'm reading the datasheet right. On an interesting note, according to the documentation for LT SPICE, it can be used to draw ones own component, assign SPICE commands to it and simulate anything electronically as long as industry standard SPICE is used. I thought LT SPICE was limited, however it seems it is a lot more powerful than I expected. Try it out, it's free. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-12-12T20:51:30Z IndexTAG: 665 TitleTAG: Is 100% possible without studying weeks 13-14? That's my doubt. I'd like to know if topics in weeks 13-14 are somewhat important for the final exam? Or can I safely leave them to review 1-12 for the final and read 13-14 after the final more thoroughly? Thank you! UserIdTAG: 398594 UserNameTAG: DaveyJC CreateTimeTAG: 2012-12-09T21:10:00Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50bf8166c67c571f00000031 FirstChildUserIdTAG: 361137 FirstChildUserNameTAG: Ericson FirstChildCreateTimeTAG: 2012-12-10T02:45:57Z FirstChildTAG: In the previous 6002x, the final covered material from weeks 1-13. It seems very unlikely that they have changed that. EDIT: They **did** change that. Read the topic linked to by Ericson above. FirstChildUserIdTAG: 90796 FirstChildUserNameTAG: ChaunceyGardiner FirstChildCreateTimeTAG: 2012-12-09T22:34:31Z FirstChildTAG: I'm going to suggest you study it. From the previous course, here are some statistics ![enter image description here][1] One missing statistic is that of all these people, I recall seeing that only around 350 achieved 100% on the final (or was that on the course as a whole). Anyways, expect it to be hard. Don't make it harder by skipping material. [1]: https://edxuploads.s3.amazonaws.com/13551152481343679.png FirstChildUserIdTAG: 9673 FirstChildUserNameTAG: xp42 FirstChildCreateTimeTAG: 2012-12-10T05:00:04Z FirstChildTAG: Well, it was possible in the end! :D FirstChildUserIdTAG: 398594 FirstChildUserNameTAG: DaveyJC FirstChildCreateTimeTAG: 2012-12-23T13:26:37Z IndexTAG: 666 TitleTAG: edX Translator Community (ETC) Can You help to translate EDX content? If answer is positive and edX allows us. We would form edX Translator Community (ETC). We would translate to all language. Please, let's translate in wiki or here! UserIdTAG: 402193 UserNameTAG: Feramico CreateTimeTAG: 2012-12-08T19:02:28Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: oh, great idea. I can translate into Russian FirstChildUserIdTAG: 328920 FirstChildUserNameTAG: TisO FirstChildCreateTimeTAG: 2012-12-08T19:26:35Z SecondChildTAG: We learn more. Sign up here, in Google groups: https://groups.google.com/forum/?hl=es-419&fromgroups=#!topic/edxtranslatorcommunity/32vQOF25VL0 SecondChildUserIdTAG: 402193 SecondChildUserNameTAG: Feramico SecondChildCreateTimeTAG: 2012-12-08T19:32:00Z SecondChildTAG: Search in Google groups so: edX Translator Community (ETC) SecondChildUserIdTAG: 402193 SecondChildUserNameTAG: Feramico SecondChildCreateTimeTAG: 2012-12-08T19:35:09Z SecondChildTAG: I joined the group SecondChildUserIdTAG: 328920 SecondChildUserNameTAG: TisO SecondChildCreateTimeTAG: 2012-12-08T19:40:45Z SecondChildTAG: Meet the Russians in the edX Translator Community (ETC) and let's translate! SecondChildUserIdTAG: 402193 SecondChildUserNameTAG: Feramico SecondChildCreateTimeTAG: 2012-12-08T19:43:38Z SecondChildTAG: Thanks! SecondChildUserIdTAG: 402193 SecondChildUserNameTAG: Feramico SecondChildCreateTimeTAG: 2012-12-08T19:44:48Z SecondChildTAG: It wil be only in google group or anywhere else? SecondChildUserIdTAG: 96873 SecondChildUserNameTAG: Santyaga SecondChildCreateTimeTAG: 2012-12-08T19:45:42Z SecondChildTAG: Please, give +1 to the community. SecondChildUserIdTAG: 402193 SecondChildUserNameTAG: Feramico SecondChildCreateTimeTAG: 2012-12-08T19:59:06Z SecondChildTAG: I'm sorry. This is on Google +1 SecondChildUserIdTAG: 402193 SecondChildUserNameTAG: Feramico SecondChildCreateTimeTAG: 2012-12-08T20:29:15Z SecondChildTAG: Anywhere! SecondChildUserIdTAG: 402193 SecondChildUserNameTAG: Feramico SecondChildCreateTimeTAG: 2012-12-08T20:39:04Z SecondChildTAG: For rules of edX here I don't request you +1, but if You like, You can have it. SecondChildUserIdTAG: 402193 SecondChildUserNameTAG: Feramico SecondChildCreateTimeTAG: 2012-12-08T20:47:17Z FirstChildTAG: I can translate into Spanish. FirstChildUserIdTAG: 402193 FirstChildUserNameTAG: Feramico FirstChildCreateTimeTAG: 2012-12-08T20:52:38Z FirstChildTAG: Please, translate one word and world will be more happy! FirstChildUserIdTAG: 402193 FirstChildUserNameTAG: Feramico FirstChildCreateTimeTAG: 2012-12-08T20:30:17Z FirstChildTAG: H12P1: FUENTE DE CORRIENTE Nosotros a menudo necesitamos una fuente de corriente de precisión dentro de un circuito. Un amplificador operacional nos da una manera de hacer un flujo de corriente controlada con precisión a través de una carga cuyas propiedades no se conoce con precisión. El circuito práctico siguiente es a menudo la respuesta. ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13550061781343607.gif Por ejemplo, supón que nosotros necesitamos poner precisamente i = 10mA a través de una carga que presenta una resistencia pero el valor de la resistencia no está bien controlada. Supón también que nosotros tenemos una fuente de voltaje de precisión de v = 5V, pero esta fuente de tensión es bastante débil: ella no puede abastecer más de 0,1 mA sin humo que sale de ella. Con el circuito que se muestra arriba nosotros podemos elegir una resistor de resistencia R que pueden lograr este objetivo. En primer lugar, suponiendo que el amplificador operacional es ideal (A = ∞) da el valor de la resistencia R, en ohmios, que nosotros necesitamos. FirstChildUserIdTAG: 402193 FirstChildUserNameTAG: Feramico FirstChildCreateTimeTAG: 2012-12-08T22:33:05Z SecondChildTAG: Por supuesto, nuestro amplificador operacional real es imperfecto, de muchas maneras. Por ejemplo, él se suspende entre dos rieles de fuentes de alimentación con voltaje ± VS= 15V, por lo que el circuito real se parece más a esto: ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13550105811343661.gif SecondChildUserIdTAG: 402193 SecondChildUserNameTAG: Feramico SecondChildCreateTimeTAG: 2012-12-08T23:50:11Z SecondChildTAG: Un amplificador operacional no puede poner un voltaje que excede las guías de suministro de energía. Pero asumamos nuestro amplificador operacional puede llegar a las guías. Dado el valor de R que usted dedujo anteriormente, ¿cuál es la resistencia de carga máxima, en Ohmios, que esta fuente de corriente puede manejar? SecondChildUserIdTAG: 402193 SecondChildUserNameTAG: Feramico SecondChildCreateTimeTAG: 2012-12-09T00:00:21Z SecondChildTAG: Otro problema es que la ganancia del amplificador operacional es realmente finito. Supón que A = 100. Usando el valor de R que usted calculó anteriormente, y suponiendo una resistencia de carga de 500 Ω, ¿Cuál es la corriente real, en amperios, que pasará a través de la carga? Por favor dé su respuesta hasta 0,1% de la respuesta exacta. SecondChildUserIdTAG: 402193 SecondChildUserNameTAG: Feramico SecondChildCreateTimeTAG: 2012-12-09T00:29:45Z FirstChildTAG: Let's translate here. FirstChildUserIdTAG: 402193 FirstChildUserNameTAG: Feramico FirstChildCreateTimeTAG: 2012-12-09T00:47:06Z IndexTAG: 667 TitleTAG: It's time for the finals I have got an overall of 58% which will get past 60 when i get past final exam.All the best for everyone in the world for the final exam UserIdTAG: 150120 UserNameTAG: krishna1993 CreateTimeTAG: 2012-12-08T13:56:38Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: congratulations. I have little if idont make it ine the final i hope ill be allowed to try again. Im slow in this subject but very interested. FirstChildUserIdTAG: 498458 FirstChildUserNameTAG: jefwa FirstChildCreateTimeTAG: 2012-12-08T14:39:57Z SecondChildTAG: It's ok...do ur best next time. SecondChildUserIdTAG: 329725 SecondChildUserNameTAG: ammuniran SecondChildCreateTimeTAG: 2012-12-09T06:46:32Z SecondChildTAG: i know exactly how you feel, I am going to try again the next go around. I got lost on the math but everything else was fun SecondChildUserIdTAG: 160242 SecondChildUserNameTAG: Mlevins35 SecondChildCreateTimeTAG: 2012-12-09T14:04:10Z SecondChildTAG: I'm with [Mlevins35][1] - I wasn't prepared for the math, and it was brutal. Hopefully, next semester the time pressures will be eased and I'll be able to actually think about the electronics part. ;-) Plus, just think how smart we'll appear to the newbies next time around! ;-) [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/users/160242 SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-12-09T22:54:16Z IndexTAG: 668 TitleTAG: EE courses on Coursera https://www.coursera.org/category/ee On the other online education platform there is some courses on electrical engineering. Such as: Digital Signal Processing; Fundamentals of Electrical Engineering (hmmm.:); MOS Transistors; etc. It will be very interesting, to compare this courses with 6.002x... UserIdTAG: 104522 UserNameTAG: IgorNovice CreateTimeTAG: 2012-12-05T18:09:09Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: But dates? FirstChildUserIdTAG: 132685 FirstChildUserNameTAG: deepakmurali FirstChildCreateTimeTAG: 2012-12-07T09:09:43Z SecondChildTAG: I think, they will be later... SecondChildUserIdTAG: 104522 SecondChildUserNameTAG: IgorNovice SecondChildCreateTimeTAG: 2012-12-10T16:52:31Z SecondChildTAG: Fundamentals of Electrical Engineering starts jan 21 2013 SecondChildUserIdTAG: 104522 SecondChildUserNameTAG: IgorNovice SecondChildCreateTimeTAG: 2012-12-29T10:03:08Z IndexTAG: 669 TitleTAG: Strange symbol on circuit I see two strange symbols on KA7905 IC. What they mean? ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13546504291343607.jpg UserIdTAG: 104522 UserNameTAG: IgorNovice CreateTimeTAG: 2012-12-04T19:47:21Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Current source FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-12-04T20:18:53Z FirstChildTAG: Yes, it's a current source. In all the data books I have, I've always almost only seen that symbol, and sometimes with a small line with a small inclination, to indicate a variable current source. But there are also other symbols in use, like a circle, with a horizontal line in the middle from left to right. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-12-04T20:29:18Z FirstChildTAG: Thanks for help! FirstChildUserIdTAG: 104522 FirstChildUserNameTAG: IgorNovice FirstChildCreateTimeTAG: 2012-12-05T17:29:09Z IndexTAG: 670 TitleTAG: Noise on IC pins I'v got simple circuit, based on ATmega8 microcontroller. My oscillograph, connected to VCC and GND pins of ATmega8, show noise about 100 mV (fig 1) ![enter image description here][1]. In the next step I connect 0.1 uF capicator on VCC and GND pins and connect oscillograph again. Noise decrease to 25 mV (fig 2) ![enter image description here][2]. Why? (When I connect 20- Ohm resistor instead microcontroller to measure self-noise of power supply, I got 15 mV) [1]: https://edxuploads.s3.amazonaws.com/13546500811343663.jpg [2]: https://edxuploads.s3.amazonaws.com/13546500921343617.jpg UserIdTAG: 104522 UserNameTAG: IgorNovice CreateTimeTAG: 2012-12-04T19:42:35Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: You should block Hi freq ripples by capasitors.Decencies are required do use X7R ceramic (generally 10-100nF) connected as close as possible to the VSS and VDD pins.Depended from design you may need to add 100pF too and it is very recomended to add tantalum capasitor 22-100uF too. FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-12-04T20:15:45Z FirstChildTAG: The microcontroller is constantly switching inside, when it runs a program. Those short impulses are caused by the switches inside and the inductance of the leads to the processor. If you replace the processor by a resistor, nothing switches, so the ripple is smaller (ripple can be caused because the powersupply itself is a switched pws, or because the scope probes or wires to the resistor pickup noise from something switching in the vicinity). Even voltage regulators can oscillate if you don't use decoupling capacitors, so it's always advised to use decoupling caps. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-12-04T20:38:26Z FirstChildTAG: You have just discovered why decoupling caps are necessary. The decoupling caps should be as close to the chip pins as possible, and you should have them on every VCC/GND pair. So one on VCC, one on AVCC and one on AREF. Also make the ground connection as short as possible, i.e. put your three caps right next to the chip between Pin 7 and 8, Pin 20 and 22, Pin 21 and 22 (assuming you use the 28-pin PDIP package). The 100 nF capacitors you are using is a good choice for the mcu pins. FirstChildUserIdTAG: 90796 FirstChildUserNameTAG: ChaunceyGardiner FirstChildCreateTimeTAG: 2012-12-04T22:16:35Z FirstChildTAG: 1. What equivalent circuit of my power suply and it connection to microcontroller (include all parasite inductors and capicators)? 2. But why I must choose 100 nF capicators? Maybe I want choose 10000 uF capicator :)... By what equations I can calculate value of decoupling capicator? Or this task very hard / impossible? FirstChildUserIdTAG: 104522 FirstChildUserNameTAG: IgorNovice FirstChildCreateTimeTAG: 2012-12-05T17:28:30Z SecondChildTAG: You can read alot if you want it [here][1] or [here][2] or [here][3] and etc. You can use different values, for example 82nF :D . Value 100nF is common as well as 10nF or 150nF. Here is alot of different reasons which should be considered. Yeah, you may use 10000uF if you want..But will it help you? [1]: http://download.micron.com/pdf/technotes/TN0006.pdf [2]: http://www-micrel.deis.unibo.it/~augusto/bypass1.pdf [3]: http://www.synopsys.com/tools/verification/lowpowerverification/capsulemodule/decoupling_cap2.pdf SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-05T17:54:35Z SecondChildTAG: A 10000 uF capacitor would be an electrolytic capacitor, which doesn't have sufficiently high frequency response. Use a ceramic capacitor. Experience has shown that 100nF (.1uF) is usually adequate. .1uF ceramic capacitors are readily available and cheap. Why fight experience? SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-12-05T18:05:18Z SecondChildTAG: Very useful links, thanks for help!!! SecondChildUserIdTAG: 104522 SecondChildUserNameTAG: IgorNovice SecondChildCreateTimeTAG: 2012-12-05T18:20:02Z SecondChildTAG: Thats why I have asked "Will it help you?" :)) SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-05T18:51:54Z IndexTAG: 671 TitleTAG: Is it over ? :( ? UserIdTAG: 373752 UserNameTAG: Croak CreateTimeTAG: 2012-12-03T17:16:04Z VoteTAG: 4 CoursewareTAG: Week 14 / S27V7_CMOS_Logic_DEMO CommentableIdTAG: 6002x_S27V7_CMOS_Logic_DEMO NumberOfReplyTAG: 2 FirstChildTAG: No way mate, your life has just begun! FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-12-03T19:14:55Z SecondChildTAG: I am totally agree with hazel1919 ;). I have to confess that felt the same feeling last semester. But the good is that we still in touch with some students in the old forum and we debate interesting things there. So be up, this will not end here. My best wish to you, Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-03T23:06:30Z FirstChildTAG: I was wondering when this would start :-) I felt the same last time (many of us did). That's part of the reason I signed up again (and will sign up again). The green ticks are magnetic. FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-12-05T16:03:51Z IndexTAG: 672 TitleTAG: Is w_0 always 1/sqrt(LC), or just for our simple circuits? I've started in H11P1 and I didn't get the usual $1/LC$ in the $w_0^2$ part of the transfer function denominator. There are some resistance values ($R_C$, $R_L$) in the expression as well. Rob UserIdTAG: 468623 UserNameTAG: RobNik CreateTimeTAG: 2012-12-02T18:42:34Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: I think if you didnt get that might be because the expression isnt in the standard form, e.g. s^2+x*s+y, that is the coefficient of s^2 is not =1. I think if you rewrite / shuffle terms to make that coefficient 1, it should work out, as the remaining x and y will only then turn out to be 2*alpha and w0^2 respectively. Also i think its just the denominator of the fraction that counts because the nominator gives zeroes who nobody really cares about, while denominator gives the poles, something like that. Somebody pls correct me if i am wrong on this. FirstChildUserIdTAG: 331483 FirstChildUserNameTAG: Doru FirstChildCreateTimeTAG: 2012-12-02T20:44:52Z SecondChildTAG: I did try to put it in standard form like you said. I got the $2\alpha$ part right, so my coefficient of $s$ was good. Weird. I got the required answers, so I guess I'll wait until the solutions are posted, and hopefully they show the algebra. SecondChildUserIdTAG: 468623 SecondChildUserNameTAG: RobNik SecondChildCreateTimeTAG: 2012-12-02T22:43:19Z FirstChildTAG: The constant term in the denominator may, in fact, not be equal to 1/LC when the coefficient of S^2 is one. The constant term is, nevertheless, equal to w0^2, where w0 is the resonant (angular) frequency. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-12-02T21:58:13Z FirstChildTAG: A nice question! I have noticed this before,in my view,w0 is not always 1/sqrt(L*C),it depends on the circuits form. But it applies for simple RLC in series and simple RLC in parallel. H11P1 is an example that shows w0 is not 1/sqrt(L*C). FirstChildUserIdTAG: 361137 FirstChildUserNameTAG: Ericson FirstChildCreateTimeTAG: 2012-12-03T02:34:09Z FirstChildTAG: In most practical situations, you'll see that only, as a rule of thumb, 1/(sqrt(L*C)) is being used, because so many other things are unknown, like wiring capacitance and inductance of the wires. But to answer the questions of this course correctly, it's always advisible to use the standard form s^2+x*s+y to obtain the proper formula for the natural resonance frequency. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-12-03T16:47:00Z IndexTAG: 673 TitleTAG: Wrong checker I think there is a error in the checker. The right result should be 2*alpa = 1/CR, so according to my calculations that's 273.97 KHZ bandwidth for the upper band edge and 2782.5854 KHZ for the lower band edge. Please prove me wrong! UserIdTAG: 415375 UserNameTAG: ZWX CreateTimeTAG: 2012-11-28T18:40:29Z VoteTAG: 4 CoursewareTAG: Week 11 / S21E4: AM_Radio_Tuning CommentableIdTAG: 6002x_S21V1_AM_Radio_Tuning NumberOfReplyTAG: 1 FirstChildTAG: I just proved myself wrong :D. The thing is that 2*alpha is in rad/s so we need to divide by 2*pi :P FirstChildUserIdTAG: 415375 FirstChildUserNameTAG: ZWX FirstChildCreateTimeTAG: 2012-11-28T18:45:47Z IndexTAG: 674 TitleTAG: After the "End" of the cource... Hello. What is going to be with this course after it reach the final date? All material will be closed? I had lost too many time, and now I will not get in time... I am not interesting in certificate, but in knowledge... So, what is going to be after final exam? UserIdTAG: 756526 UserNameTAG: QUS CreateTimeTAG: 2012-11-27T16:33:41Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hmm... It is really interesting question. I do really want to save matherials..but how? FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-11-27T17:02:03Z FirstChildTAG: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50b5195c5b28952b00000027 I hope this helps you... FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-11-27T19:50:26Z IndexTAG: 675 TitleTAG: H12P2, we meet again. I have completed every single question up to H12 and LAB12, except H12P2 c and d. My head is going to explode from overthinking it, and i don't seem to find what my error is. Let's hope there isn't anything like that in the final exam because i feel really stupid right now. UserIdTAG: 304587 UserNameTAG: Vasso CreateTimeTAG: 2012-11-25T13:11:19Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: It is pretty straight forward, set up two equations with the two unknowns R0 and opamp gain and the endpoint values of input voltage and output voltage. Model the zener as a voltage source and series resistance RZ. FirstChildUserIdTAG: 397595 FirstChildUserNameTAG: mholin FirstChildCreateTimeTAG: 2012-11-25T13:47:14Z SecondChildTAG: Thank you mholin...but still... :/ SecondChildUserIdTAG: 304587 SecondChildUserNameTAG: Vasso SecondChildCreateTimeTAG: 2012-11-25T19:22:54Z SecondChildTAG: Don't forget that there is a minimum current you need to flow through R0 for the BJT to turn on. SecondChildUserIdTAG: 520029 SecondChildUserNameTAG: subterfuge SecondChildCreateTimeTAG: 2012-11-26T09:55:44Z SecondChildTAG: I almost spend 3 days for (mathematicaly and practicaly) calculs R0, R1 and R2 and... no result yet :-(! SecondChildUserIdTAG: 261860 SecondChildUserNameTAG: Oleksander SecondChildCreateTimeTAG: 2012-11-26T12:08:38Z SecondChildTAG: Due to the wide tolerances in performance that are allowed, you can get by with a rather sloppy design. With a little thought you can write down values for the three resistors with practically no calculation. Note that the feedback circuit is the same as the non-inverting amplifier explained in the lectures. The key to getting the check marks is doing the proper calculation of the efficiency for your design. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-11-26T22:51:01Z SecondChildTAG: For questions a and b, I'm having trouble with the signs when modeling the zener diode as a voltage source and series resistance. In an earlier homework it was said to set the polarity of the voltage source in the same orientation as the polarity for the diode. My question: Is Vz then negative, since it is on the –VD axis, like shown below, or does setting the polarities on the voltage source like I’ve shown take care of it? ![enter image description here][1] The next questions are related to: **Write an expression for the output voltage (vout) when the input voltage is below the Zener voltage (vin < vz).** This says Vin is less than Vz, however in order to model the zener diode as a voltage source and resistor, VD ( voltage across the zener diode) must be less than Vz. How is it determined if the diode is in the breakdown region or not, since they say vin < vz? For question b, it says vin>vz. Is Vin greater than or less than VF? Thanks. [1]: https://edxuploads.s3.amazonaws.com/13549892208474459.jpg SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-12-08T18:04:18Z FirstChildTAG: Since a 2.5V Zener diode is providing the reference voltage, you know that you will need a gain of approximately two. Also, because of Rz the reference voltage will be a maximum for vIN = 20V and a minimum when vIN = 10V. Choose R0 so that the reference voltage times the gain stays within the design specifications for the output. Note that you will not get any check marks until your computed efficiency is consistent with your component values. Edited to correct misstatement. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-25T20:23:37Z SecondChildTAG: Yes, its importante! I spend two days :-) Note that you will not get any check marks until your computed efficiency is consistent with your component values. Thanks! SecondChildUserIdTAG: 261860 SecondChildUserNameTAG: Oleksander SecondChildCreateTimeTAG: 2012-11-26T12:59:38Z FirstChildTAG: [Vasso][1], Please don't beat your head too much. Your brain is a very sensitive instrument. ;-) Seriously, I am feeling pretty stupid too right about now, and I'm still in Week 10. ;-/ [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/users/304587 FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-11-26T09:44:53Z SecondChildTAG: Yes, perhaps i need a break :) SecondChildUserIdTAG: 304587 SecondChildUserNameTAG: Vasso SecondChildCreateTimeTAG: 2012-11-26T20:50:00Z SecondChildTAG: Vasso, i too faced the same problems.i did as mholin said.i got an equation in k i.e R1/R2 ,RO,vin and vout.then got two equations .even after solving the two equations i had no luck.then i read a previous post on H12P2 in which it was suggested to use 4.99 instead of 4.9 and 5.01 instead of 5.1.i thought it wouldn't make much difference.but there was a significant change in R0(about 10 times the previous value) and a slight change in R1/R2.i guessed the efficiency though .luckily i got all green ticks.although i do want to know how to calculate efficiency and need a help on Lab 12. SecondChildUserIdTAG: 357747 SecondChildUserNameTAG: kishores SecondChildCreateTimeTAG: 2012-11-27T10:40:18Z SecondChildTAG: LAB12 is pretty simple.You do have all that you need in the textbook. With small algebra you will get some required values very fast. Note that is Q>5, not equal and more.. Unfortunatelly, Im stuck with H12P2 too.. And I need some help.. SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-11-30T15:16:25Z IndexTAG: 676 TitleTAG: H10P3 part 3 and 4 not accepting answers Hi guys, I have calculated Lmatch and Cmatch and the corresponding values in part 5 and 6 are even correct. But when I input the equation for both, the parser is not accepting it.It says."Invalid input: Could not parse 'xxxxxxx' as a formula".Kindly Help UserIdTAG: 18877 UserNameTAG: johnkhiangte CreateTimeTAG: 2012-11-25T09:23:29Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: I have a more strange problem: the expressions are wrong but the results are correct when I calculate Cmatch and Lmatch I will next try to derive alternative expressions .... I tried. All good now :)) FirstChildUserIdTAG: 319453 FirstChildUserNameTAG: leonidasGr FirstChildCreateTimeTAG: 2012-11-25T09:30:44Z SecondChildTAG: check whether you have included multiplication sign between necessary parameters. SecondChildUserIdTAG: 111540 SecondChildUserNameTAG: tksanthosh SecondChildCreateTimeTAG: 2012-11-25T10:28:38Z FirstChildTAG: not sure, but try to check the brackets in your answer equation and dot/semicolon as well FirstChildUserIdTAG: 325197 FirstChildUserNameTAG: Vitali_Jerin FirstChildCreateTimeTAG: 2012-11-25T09:39:32Z FirstChildTAG: OK. Thanks for the reply guys. I got it now. It was a mistake with the brackets. :-) FirstChildUserIdTAG: 18877 FirstChildUserNameTAG: johnkhiangte FirstChildCreateTimeTAG: 2012-11-25T10:39:43Z IndexTAG: 677 TitleTAG: Error in LAB11/Task1 only the undamped resonance frequency f0 = 10kHz (NOT w0!!!!) gives the correct results. UserIdTAG: 388524 UserNameTAG: dasbinich CreateTimeTAG: 2012-11-25T00:28:53Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: grrrrrrr........ FirstChildUserIdTAG: 301962 FirstChildUserNameTAG: labrinim FirstChildCreateTimeTAG: 2012-11-27T16:00:15Z SecondChildTAG: I've been there too!... SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-12-03T23:42:54Z IndexTAG: 678 TitleTAG: H10P2 B Hi all! I cant get green mark for part B.. In this part we have three impendances Zr=R1+R2, ZL and ZC in the parallel. So, ZB=ZR||ZL||ZC. After substituting equates I had ZB=((R1+R2)*L/C)/(R1+R2+j*w*L+1/(j*w*C)) But--I have red mark :( Where am I wrong? Thanks! UserIdTAG: 205311 UserNameTAG: Sergtronix CreateTimeTAG: 2012-11-23T12:23:06Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hi Sergtronix, A Hint that might help you: > The admitance Y is the inverse of the impedance Z. > > So Y=(1/Z) Also, if the elements are in paralell the total admitance will be the sum of the admitances of each branch, suppouse that you have 2 branches: Z1 and Z2, each branch admitances will be : > Y1= (1/Z1) > > Y2=(1/Z2) > > So, Y = Y1 + Y2 And what is your total Z ? ;). Isn´t it the inverse? Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-23T13:28:46Z SecondChildTAG: I do have 3 branches Zr=R1+R2;Zc=1/Cs and ZL=Ls.I can find Yr, Yc and YL. Impendance of circuit B will be ZB=1/(Yr+Yc+YL)... Am I correct here? If so- it is strange, but I dont have green mark yet.. Any hints where I may be wrong? SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-11-23T17:56:28Z SecondChildTAG: Hi, I tried with the admittance method. It simplified the algebra a lot, but still having the wrong answer. The value calculated by my for Zb = (R1+R2) - j (w*L) / (w^2*L*C+1) Can some one help us a little bit? SecondChildUserIdTAG: 87808 SecondChildUserNameTAG: pumoneon SecondChildCreateTimeTAG: 2012-11-24T23:58:19Z FirstChildTAG: Hi Sergtronix! Do you remember how looks equation for 3 parallel conected resistors? Use it for your problem! FirstChildUserIdTAG: 96873 FirstChildUserNameTAG: Santyaga FirstChildCreateTimeTAG: 2012-11-23T18:03:19Z SecondChildTAG: sure :) Zb=Zr*Zc*Zl/(Zr+Zc+Zl) or 1/Zb=1/Zr+1/Zc+1/ZL SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-11-23T18:14:27Z SecondChildTAG: As you like Z or 1/Z ;-) SecondChildUserIdTAG: 96873 SecondChildUserNameTAG: Santyaga SecondChildCreateTimeTAG: 2012-11-23T18:18:02Z SecondChildTAG: wow I forget to add (R1+R2) into denominator... a;; is fine.. just stupid error Thank you to all! SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-11-23T18:22:58Z SecondChildTAG: I'm stuck where you were Sergtronix... what got you past it? I have tried about 10 different ways of entering the equation and I can't get it marked correct. Following the above method, I get: L/C*((R1+R2)/(R1+R2+1/jwC +jwL)) What did you do to get this corrected? SecondChildUserIdTAG: 166031 SecondChildUserNameTAG: krebryna SecondChildCreateTimeTAG: 2012-11-24T03:34:39Z SecondChildTAG: Actually, I got it. The general expression for three parallel "resistances" should be: $Z_{b} = \cfrac {Z_{r} \cdot Z_{c} \cdot Z_{l}}{Z_{r} \cdot Z_{c} + Z_{r} \cdot Z_{l} + Z_{c} \cdot Z_{l}}$ I see now. Just a bit more complicated than I thought, but straightforward when I sit down and work it out from first principles. SecondChildUserIdTAG: 166031 SecondChildUserNameTAG: krebryna SecondChildCreateTimeTAG: 2012-11-24T04:13:25Z SecondChildTAG: krebryna , Youre correct. My solution was made through admittance. Good luck! SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-11-24T09:14:56Z SecondChildTAG: After resolve it I got the Expression ZB=((R1+R2)*j*w*L)/((j*w)²*L*C*(R1+R2)+j*w*L+R1+R2), however I get the message: onvalid input: Could not parse '((R1+R2)*j*w*L)/((j*w)²*L*C*(R1+R2)+j*w*L+R1+R2)' as a formula. What do I have to do? SecondChildUserIdTAG: 90765 SecondChildUserNameTAG: jdavidg SecondChildCreateTimeTAG: 2012-11-24T19:01:30Z IndexTAG: 679 TitleTAG: HELP I am going crazy Myriam Guys i already have a 60% and will obviously get a degree, but i am going to quit coming to edx because my exams are going on. i cant miss a second. I still want to take the proctored exams, Obviously i wont be able to score even 30% on them . DO you think That getting a C Grade is pointless or should i take proctored exams without studying I cant take them next fall because of certain circumstances. What do you think ?? UserIdTAG: 817034 UserNameTAG: Runneed CreateTimeTAG: 2012-11-22T22:07:31Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I think you have a point but i am not going to take the protored ones cant afford FirstChildUserIdTAG: 817136 FirstChildUserNameTAG: ramanlal1 FirstChildCreateTimeTAG: 2012-11-22T23:29:39Z SecondChildTAG: Hi ramanlal1, I don´t know how much it will cost the proctored Exam... I haven´t inquire that to Pearson VUE as I would like to take it at the end of the next year... I guess it would be accessible (I took in my Country International Exams and they were accessible, they weren´t cheap, but at least I could afford them, I correct myself, my parents could afford that in that time haha). I don´t know where you live... But I understand you totally... I will start to save for the Exam from now haha. In this cases I would like to have money and help you...but I am a South American engineering student and I don´t even have a credit card, oh my... I have to do something with this. Anyway, as I have said in a previous Post I am more than sure that I will die poor if I continue doing what I like - Research- haha, but I don´t care, at least I will die happy ;). Take care. Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-23T02:38:46Z SecondChildTAG: Dear Myrim , Sorry my teacher I am working and getting good salary but I am al so same like you all buying books doing research no money for saving so money is sad happiness is more important SecondChildUserIdTAG: 156694 SecondChildUserNameTAG: mkprasanth SecondChildCreateTimeTAG: 2012-11-23T12:18:38Z SecondChildTAG: :). Hi mkprasanth! How are you? I haven´t seen you recently in this Forum Discussion. How are you doing with the others Courses? Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-23T13:06:19Z FirstChildTAG: Hi Runneed, First of all, congratulations! You have passed this Course with that score (60% percent of total score = Grade C). I have decided that I will take the Proctored Exam but at the end of the next year because I want to be 100 % sure and confident haha. I can not tell you what you should or you shouldn´t do...That is a personal decission that you have to take... I can give you some ideas or alternatives, but I can not tell you what you should or you shouldn´t do: 1 - You can still following this Course and stop working and prioritizing your other exams, as you already have the minimun requisite to Pass this Course, you will have your Certificate. 2 - Alternative 1 + and Set proctored Exam at the end of the Course without studying (for me, this is not recommended, unless you are a kamikaze student haha). And based on my experience the Final Exam wasn´t a piece of cake... remember that will have all the content of the Course, from Week 1 to Week 12... unless, of course, that you already know the material from that remaining Weeks... 3 - Alternative 1 + following the material more relaxed in the Holidays (You will have the Certificate + you will learn what you have missed). 4 - Alternative 3 + Re-taking again this course in other opportunity as this Course will be offered again + study completely all the material of this Course + set, if is it your will, the Proctored Exam . As I have said before, is your personal decission... Whatever you choose, my best wish to you :). If there is anything that I can do in order to help you, I will be here. Take care, Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-23T01:31:22Z IndexTAG: 680 TitleTAG: Proctored exam for 6.002X? Just read in the Edx News that they are planning to offer a proctored exam for one class this fall. Not sure if they mean this year or next nor which class, but I'm hoping it's this class and this year. Don't get me wrong, I'm super glad and happy with just taking the course period and getting the knowledge I've gained but I'd still love a chance to take the proctored exam and get some credit for it! :D Keeping my fingers crossed! UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-11-22T05:18:52Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 681 TitleTAG: Staff :is it required to pass the final exam? Congratulations for the amazing course! I have 60% end of week 10. Is it required to take part during the final in order to get the certificate passing grade? UserIdTAG: 206669 UserNameTAG: dimitrios66 CreateTimeTAG: 2012-11-18T17:53:23Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: You are required to get a 60%. I don't know of any requirement to take the final. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-11-18T19:27:32Z SecondChildTAG: Correct. SecondChildUserIdTAG: 96452 SecondChildUserNameTAG: Lyla SecondChildCreateTimeTAG: 2012-11-19T19:47:57Z SecondChildTAG: What do you mean, "correct" ? He didn't answer the question. SecondChildUserIdTAG: 90796 SecondChildUserNameTAG: ChaunceyGardiner SecondChildCreateTimeTAG: 2012-11-20T00:17:34Z SecondChildTAG: I'm not on staff, and so my answer reflects only what I understand to be true. Answers from anyone who are not staff members, including you, should be taken as unoffical. I said I don't know of any requirement to take the final, Lyla said correct, so I think everyone understands what that means: there is no requirement to take the final, only to get a 60%. Personally, I don't think it should be the case. I would to see the grading system changed so that the highest score is like a 58%, so you at least have to answer a question on the final. But that's just my personal opinion, and has really no other validity. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-11-20T17:57:08Z FirstChildTAG: From what I understand, it is not required to pass the Final exam, all that matters is your final score, if you have completed all the assignments including the midterm with 100 percent scores, you will have 59.9999 percent, but you need 60 percent to pass, so you will have to answer at least one Q on the final to get a certificate. FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-11-19T09:13:33Z SecondChildTAG: well, my intention is to continue with week 11 and 12 and also pass the Final provided i have the time for that. However your answer has confused me. I thought that 30+15+15 should count as 60, not as 59+. If you were right, noone would ever get 100% even though had scored 40+15+15+30. In any case if someone is certain that knows the answer please let me know. SecondChildUserIdTAG: 206669 SecondChildUserNameTAG: dimitrios66 SecondChildCreateTimeTAG: 2012-11-20T00:57:54Z FirstChildTAG: You don't have to take the final. Your total score is what matters. FirstChildUserIdTAG: 90796 FirstChildUserNameTAG: ChaunceyGardiner FirstChildCreateTimeTAG: 2012-11-19T03:25:49Z IndexTAG: 682 TitleTAG: excellent course 6002x is an excellent course it is sharpening the concepts of electrical engineering students i request to include more courses relevant to electrical engineering UserIdTAG: 153584 UserNameTAG: SalmanShahzad CreateTimeTAG: 2012-11-18T07:33:19Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: here here FirstChildUserIdTAG: 18073 FirstChildUserNameTAG: gburkhart FirstChildCreateTimeTAG: 2012-11-18T23:03:59Z IndexTAG: 683 TitleTAG: Anyone else having trouble with HW and Lab using Chrome browser? Weird problem only experienced today, when I check answers in HW and Lab I occasionally (~25%) of the time have trouble with the page reloading. Values are saved but the math functions and text are fully or partially hidden. Hopefully this is just a temporary error with either my browser or edX, but I would love to know if other people are experiencing the same issues. UserIdTAG: 238627 UserNameTAG: Dwilds CreateTimeTAG: 2012-11-18T04:12:56Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Troubleshooting NumberOfReplyTAG: 5 FirstChildTAG: Even i too encountered the same problem in my chrome browser... FirstChildUserIdTAG: 477198 FirstChildUserNameTAG: Rajesh1993 FirstChildCreateTimeTAG: 2012-11-18T06:11:47Z FirstChildTAG: I encountered this problem yesterday and it persisted for about an hour. It has something to do with the web service response at edx. It rectified itself after about an hour. FirstChildUserIdTAG: 150267 FirstChildUserNameTAG: vwsingh FirstChildCreateTimeTAG: 2012-11-18T07:29:03Z SecondChildTAG: Yes, we were working hard to fix the errors. If these kinds of these come up again, please post just as you did. SecondChildUserIdTAG: 96452 SecondChildUserNameTAG: Lyla SecondChildCreateTimeTAG: 2012-11-19T15:20:22Z FirstChildTAG: I have frequent problems with chrome. Especially when math text is involved. Broswer hangs and becomes unresponsive. I have to close the browser and log back into edx to continue, so I make sure I save all work in a text file along the way, so I can refill all the answer fields etc..when I restart the browser and log back in. FirstChildUserIdTAG: 4463 FirstChildUserNameTAG: pmj FirstChildCreateTimeTAG: 2012-11-18T16:33:12Z SecondChildTAG: yo tambien SecondChildUserIdTAG: 18073 SecondChildUserNameTAG: gburkhart SecondChildCreateTimeTAG: 2012-11-18T23:08:40Z FirstChildTAG: I had very similar problems with Firefox when the course started, often solved by clearing the browser cache and reloading. Though sometimes it took several iterations of clear+reload. Try that; it may help. Good luck! FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-11-18T17:10:10Z FirstChildTAG: I also have problems with chrome. Tryed to reload, but it seems to me wasn`t very helpful FirstChildUserIdTAG: 96873 FirstChildUserNameTAG: Santyaga FirstChildCreateTimeTAG: 2012-11-21T12:12:02Z IndexTAG: 684 TitleTAG: H9P1 Q1 w0 - natural frequency/resonance frequency/angular frequency? Througout the lectures w0 (omega naught) is called natural frequency. So I spent about an hour trying to figure out why equation: w0 = 1/sqrt(L*C) doesn't give me the correct answer for Q1. And it looked like noone else had problem with that according to existing posts. So in the end I figured out that I needed to give the answer for f0 (f naught) which is called resonance frequency (by Wikipedia at least). The equation is: w0 = 2*pi*f0 So, if someone from staff reads that post, make something to remove confusion and don't make people waste their time. Hope that also helps to some1 like me. P.S. How do I make equations to look like equations in my post? Not like sequence of chars entered from keyboard... UserIdTAG: 413065 UserNameTAG: anikey CreateTimeTAG: 2012-11-17T10:19:00Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: visit https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/mathjax-tutorial-write-formula-forum/ FirstChildUserIdTAG: 357747 FirstChildUserNameTAG: kishores FirstChildCreateTimeTAG: 2012-11-17T10:34:44Z FirstChildTAG: Thanks for the tip, I was trying all sorts of crazy answers after my initial answer didn't work! I understood the lectures and I read the entire chapter 12 in the textbook and throughout they refer to "natural frequency" in many places and never once say that it is given by $\omega_0/(2*\pi)$. I agree that the wording in the question is confusing... FirstChildUserIdTAG: 367878 FirstChildUserNameTAG: spoida FirstChildCreateTimeTAG: 2012-11-17T13:09:44Z SecondChildTAG: The "natural frequency of oscillation" is given in [S17E5: An ILC Circuit][1] as $\displaystyle f=\frac{\omega_0}{2\pi}$. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_9/Undamped_Second-Order_Systems/ SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-11-17T16:31:41Z IndexTAG: 685 TitleTAG: lab9 boost convertor problem Can anyone tell how to calculate L.I had calculated C but now stucked with L.any hints UserIdTAG: 611666 UserNameTAG: Shaminder420 CreateTimeTAG: 2012-11-17T07:41:47Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: i am also in the same situation.. FirstChildUserIdTAG: 114864 FirstChildUserNameTAG: Vikaash FirstChildCreateTimeTAG: 2012-11-17T09:11:42Z SecondChildTAG: [Lab9 Hints][1] ;). [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50a78f0dce1ad22700000024 SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-17T13:47:04Z IndexTAG: 686 TitleTAG: grader issues with S19E4 - staff? I get the middle two answers (of six) correct. All I should have to do to get the other answers right is substitue in the new frequency. For the other four questions I get different answers than the grader; however, these answers show up as the answers for different questions! For prob. 1 I get 0.99996, that's the value the grader want for prob. 5. For prob. 2 I get 0.00875; the grader wants that for prob. 1. For prob. 5 I get 0.00111; the grader wants that for prob. 6 For prob. 6 I get 1.56; the grader wants that for prob. 2 Thanks for checking this out. UserIdTAG: 174100 UserNameTAG: markpolak CreateTimeTAG: 2012-11-16T04:50:49Z VoteTAG: 4 CoursewareTAG: Week 10 / Magnitudes And Angles CommentableIdTAG: 6002x_Magnitudes_and_angles NumberOfReplyTAG: 1 FirstChildTAG: Remember, you can not simply use the formulas for magntude & phase from the previous lecture slide, where: ${V_p\over V_i}={1\over1+j\omega RC}$ Here, ${V_p\over V_i}={j\omega RC\over1+j\omega RC}$ ... so you have to recalculate both formulas using complex algebra. FirstChildUserIdTAG: 714237 FirstChildUserNameTAG: PaxPolaris FirstChildCreateTimeTAG: 2012-11-23T05:28:04Z SecondChildTAG: Thanks! I was confused because I seemed so close to having it right. I came back to this problem today and did it without reference to any past formulas and did much better. Makes total sense! SecondChildUserIdTAG: 174100 SecondChildUserNameTAG: markpolak SecondChildCreateTimeTAG: 2012-11-23T23:55:11Z IndexTAG: 687 TitleTAG: H8P3 - mistake in answer explanation? I have doubt about correct answer in H8P3. Please check the red lines in this picture: ![https://edxuploads.s3.amazonaws.com/13527318131343653.png][1] [1]: https://edxuploads.s3.amazonaws.com/13527318901773561.png UserIdTAG: 508733 UserNameTAG: burdal1 CreateTimeTAG: 2012-11-12T14:52:24Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: i even didn't get that i used the same formula or method rather and i got the green tick in all upper 4 but last one showing incorrect.. FirstChildUserIdTAG: 108863 FirstChildUserNameTAG: shiviz FirstChildCreateTimeTAG: 2012-11-12T16:01:57Z FirstChildTAG: in 5: why for vc(t) used ZIR? is the vc(t) final value zero volt, but not (Vs*Ron)/(Rpu+Ron)? FirstChildUserIdTAG: 188778 FirstChildUserNameTAG: xsaq FirstChildCreateTimeTAG: 2012-11-12T16:23:08Z SecondChildTAG: Yes, this is the right remark. I am afraid the model of situation for 5. is not so simple: ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13527486661343666.png SecondChildUserIdTAG: 508733 SecondChildUserNameTAG: burdal1 SecondChildCreateTimeTAG: 2012-11-12T19:31:32Z FirstChildTAG: I think you are right, something is messed up here. When I do the calculations with your numbers in Wolfram, it doesn't return the correct answer. I emailed the staff to let them know about it. UPDATE: The staff see the problems and are working on it. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-11-12T17:23:32Z FirstChildTAG: You are right ... the explanations for 4 and 5 are inaccurate. I'll try to correct it: - 4) The gate capacitance will be charged by the supply voltage. However, the resistors $R_{PU}$ and $R_{ON}$ (due to the MOSFET Q3) are in series between the supply and the capacitor. This is a RC circuit again, with the capacitor voltage given by the expression: $ v_C(t) = V_{s} + (V_{OL} - V_{s}) \cdot e^{-\frac{t}{R \cdot C}} $ We are asked to find the time when $v_C(t) = V_{OH}$, so that we need to solve for t from the above equation. This gives: $\large t = -R \cdot C \cdot ln \left( \frac{V_{OH} - V_s}{V_{OL} - V_s} \right) $ Using the known values for $R=\color{grey}{R_{ON}+(R_{PU} \parallel R_{OFF})}$ *(Note: you can still use $R=R_{ON}+R_{PU}$ to get a close enough answer)* , $C$, $V{OH}$ and $V{OL}$, we get: $R\approx R_{ON}+R_{PU} = (2,150+10,000)\Omega=\color{blue}{12,150\Omega}\\ $ $\large t = -(\color{blue}{12,150} \Omega)(3.5 \cdot 10^{-15}) \cdot ln \left( \frac{-1.5V}{-4.5V} \right) = \color{blue}{46.72} ps \\ $ ----- - 5) In this case, the gate capacitance of Q2 will be dicharged through the on resistances of Q3 and Q1 ... ***until it finally reaches the voltage at the drain of*** $Q_1$: $\color{blue}{V_{Q1}} $. $ \large \color{blue}{V_{Q1}= V_S\times{R_{ON}\over R_{PU}+R_{ON}}=0.884773662551V} \\ $ This is an RC circuit with the capacitor voltage given by the expression: $ \large v_C(t) = V_{Q1}+ (V_{OH}-V_{Q1}) \cdot e^{-\frac{t}{R \cdot C}} \\ $ Using the known values for $R = \color{grey}{R_{ON} + (R_{ON}\parallel R_{PU})}$ , $C$, $V_{IL}$ and $V_{OH}$ we can solve for t to find: $\Large t = -R \cdot C \cdot \ln \left( \color{blue}{V_{IL}-V_{Q1}\over V_{OH}-V_{Q1}} \right)\\ = -(\color{blue}{3,919.55} \Omega) \cdot (3.5 \cdot 10^{-15}) \cdot ln \left( \color{blue}{ 0.9-0.88477\over 3.5-0.88477} \right) \\ = 70.59 ps \\ $ FirstChildUserIdTAG: 714237 FirstChildUserNameTAG: PaxPolaris FirstChildCreateTimeTAG: 2012-11-12T17:39:46Z SecondChildTAG: Everyone got different value for resistances ... so, the equivalent resistance $R$, anV SecondChildUserIdTAG: 714237 SecondChildUserNameTAG: PaxPolaris SecondChildCreateTimeTAG: 2012-11-12T19:48:13Z SecondChildTAG: so, the equivalent resistance $R$, and $V_{Q1}$ will differ. SecondChildUserIdTAG: 714237 SecondChildUserNameTAG: PaxPolaris SecondChildCreateTimeTAG: 2012-11-12T19:49:54Z SecondChildTAG: and about different values!... i have seen two different values for Ron, 1900 and 2150 at two different moments of time, its looks likes a crazy, but i swear SecondChildUserIdTAG: 188778 SecondChildUserNameTAG: xsaq SecondChildCreateTimeTAG: 2012-11-12T20:49:52Z SecondChildTAG: The values of the resistors are different from person to person. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-11-12T22:03:02Z SecondChildTAG: mee to, first I had 1800 later I get 2000 for RPU. I noticed it very late. SecondChildUserIdTAG: 389333 SecondChildUserNameTAG: juergen SecondChildCreateTimeTAG: 2012-11-12T22:20:01Z SecondChildTAG: yes..had the same problem..I had one particular set of values..and I calculated the answers and it was wrong..later I noticed that the resistance values had changed..luckily I had enough time to calculate using new values and get it right.. SecondChildUserIdTAG: 173147 SecondChildUserNameTAG: cruiser_rahit SecondChildCreateTimeTAG: 2012-11-13T05:30:33Z IndexTAG: 688 TitleTAG: Staff question Because of some other projects, I had to quit this course. I plan, however, to re-enroll the next time it is offered. My question is, can I take up where I left off or do I have to start again at square 1? UserIdTAG: 237167 UserNameTAG: Dug CreateTimeTAG: 2012-11-11T17:26:53Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: Hi Dug, As far as I have read in the Forum Discussion, some students of 6.002x spring asked the same question of you, that is to say, they left 6.002x spring and they wanted to know if they could start from where they have left the course in the previous term but in this 6.002x fall... and the answer was that they couldn't.... So, unfortunately, if you re-take this course again you will have to start again all from zero... Myriam. P.D. If there is anything that I can do for you in order that you can continue, please tell me, I would like to help you...of course, that if you are willing to re-take due your time it is up to you... FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-11T19:13:07Z FirstChildTAG: Myriam, Thank you for answering my question. Having to start over will be a good review. I had originally planned to take this course and then finish the book I'm currently working on (I'm a writer and a retired college prof.). But unfortunately, it didn't work out that way. Because of deadlines, I'm going to have to finish the book first, and then take this course. (BTW, I don't remember when it was, but you were really helpful in my having one of those "ah-ha" moments.) Doug (I had to use "Dug" because "Doug" was already taken.) FirstChildUserIdTAG: 237167 FirstChildUserNameTAG: Dug FirstChildCreateTimeTAG: 2012-11-11T20:38:17Z SecondChildTAG: Ho Doug, You are welcome :) Oh, so nice! I like to write too, but Science Fiction, I have finished a draft Book of more than 400 pages haha, that is master piece haha, It took me a lot of years, might I someday decide to publish it under an unknown Author's name and in Spanish because I am a dissaster in English haha, I write since my childhood haha, that is one of the things that I have to do, I also Ilustrated the draft book as I like drawing too haha. Just for curious, What kind of Books do you write ? Haha, thank you, I like helping here, I guess that the thing that I enjoy most is when a Classmate tells me: I got it ! I am really happy for them. Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-11T21:14:38Z SecondChildTAG: *Hi SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-11T21:15:01Z SecondChildTAG: Don't be afraid to put your name on it! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-11T21:24:53Z SecondChildTAG: Hahaha ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-11T22:21:34Z SecondChildTAG: Myrimit, I think you should publish under your real name, then come back here and tell us all where to get a copy - you'll be an instant bestseller! :-) SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-11-12T00:59:06Z SecondChildTAG: Hahaha planetscape, might I will consider that, good idea haha! Knowing myself I guess that I will publish the Book and donate all the founds to Public Hospitals for Kids or Education non-profit like edX haha and also, make it a downloable .pdf for free for those that can not afraid the costs. As I always have said I guess that I will die poor but happy, so I don't care ;) haha. See you, Take care. Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-12T01:40:07Z SecondChildTAG: afford haha SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-12T01:41:15Z FirstChildTAG: Care to share the subject matter or title of the book? FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-11T23:42:06Z FirstChildTAG: Myrian, I like your attitude. Half of My books are novels, most of them set in the Appalachian Mountains of North Carolina, although the one I'm working on now is set in Libya, where I lived as a child (I'm an American; my father, a former career Air Force sergeant, was stationed in Libya). The other half of my books are local-history books, that are sold at the local museum,with all profits going to the museum. Doug FirstChildUserIdTAG: 237167 FirstChildUserNameTAG: Dug FirstChildCreateTimeTAG: 2012-11-12T17:04:34Z SecondChildTAG: Cool, so nice Doug! Thank you for sharing this ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-12T21:47:31Z IndexTAG: 689 TitleTAG: Quick comment about Lecture 8 videos Just wanted to say that, in general, I normally don't enjoy lectures as a format for learning. I find them to be a slow, less coherent rehashing of what is contained in the text. And I always hated going to class when I was in school, for that reason. But the lecture 8 videos were REALLY very helpful! So ...well done team... on putting together a lecture sequence that wasn't completely dull and useless. The videos really helped me have a better understanding of what the heck was going on, and built up a good intuition. UserIdTAG: 287805 UserNameTAG: Beneficial CreateTimeTAG: 2012-11-10T14:21:50Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Totally agreed FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-11-10T18:06:08Z IndexTAG: 690 TitleTAG: Frequency Calculator Here is a free frequency/angler frequency/wavelength/period calculator that I found useful around this time in the course. http://ekalk.eu/flt_en.html For example: you can enter the frequency f (don't forget to choose the proper unit), select the units for the results and click [Compute others] button and the calculator will compute the period, wavelength and angular frequency. UserIdTAG: 21541 UserNameTAG: JSChambers CreateTimeTAG: 2012-11-06T16:33:57Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Thanks! FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-11-06T23:20:01Z SecondChildTAG: Yes, these are very good resources!! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-07T21:43:04Z IndexTAG: 691 TitleTAG: Hello Everything is cool. I started with this course from yesterday. I know I am late but still happy that I started anyway :) UserIdTAG: 504574 UserNameTAG: Abhyudha CreateTimeTAG: 2012-11-05T09:28:49Z VoteTAG: 4 CoursewareTAG: Week 1 / KVL and KCL exercises CommentableIdTAG: 6002x_kvl_kcl_exercises NumberOfReplyTAG: 2 FirstChildTAG: ~Welcome!~ FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-11-05T11:17:57Z SecondChildTAG: awesome! SecondChildUserIdTAG: 748857 SecondChildUserNameTAG: marcoatedx SecondChildCreateTimeTAG: 2013-01-16T16:23:17Z FirstChildTAG: Good that you are starting from the beginning, don't feel rushed, take in the information at your own pace. Then you will be double prepared for the spring. FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-11-05T12:53:18Z IndexTAG: 692 TitleTAG: Too Fast She goes way too fast through these problems. I don't understand her logic. UserIdTAG: 278463 UserNameTAG: zniazi CreateTimeTAG: 2012-11-05T09:18:26Z VoteTAG: 4 CoursewareTAG: Week 7 / Constitutive Laws for Capacitors and Inductors CommentableIdTAG: 6002x_Constitutive_Laws_Caps_Inductors NumberOfReplyTAG: 1 FirstChildTAG: Use the speed control to slow down the playback; that will help. Remember you can pause the video as well. Perhaps you want to go through the videos twice; once as a preview so you know what to expect and watch for, and a second time to take notes. FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-11-05T20:55:26Z IndexTAG: 693 TitleTAG: @Myrimit I saw in your recent response to a post that we could take this same course with proctored exam. What exactly the exam here refers to?(Midterm or Final). I'm surprised that the Staff people never mentioned about the proctored run of this course. UserIdTAG: 148523 UserNameTAG: span993 CreateTimeTAG: 2012-11-03T19:50:37Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Oh! Thanks for the info. Well, Is it only for US citizens? FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-11-04T09:24:52Z SecondChildTAG: Pearson VUE has testing centers in 175 countries. I do not know if they mention which edX course will be the *first* one available for proctoring. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-11-04T11:45:04Z SecondChildTAG: But I cant see any of the Edx courses on their site (testing program scroll list). http://www.pearsonvue.com/vtclocator/ SecondChildUserIdTAG: 148523 SecondChildUserNameTAG: span993 SecondChildCreateTimeTAG: 2012-11-04T12:55:46Z SecondChildTAG: It has not been determined yet which edX class will be the *first* to be offered examination. If I was a betting man, I would say that this course would be the first one offered. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-11-05T11:30:59Z FirstChildTAG: Hi span993, You can set for the Final Exam but proctored , in a center. This was announced [here][1]. I really don't know much about that, but I guess I will take the proctored Exam next year in my Country ;). I will prepare more :p . Of course, that if you pass the Course this fall, you will obtain a free of charge Honor-Code Certificate. But in the case that you would like to obtain a proctored Certificate, you should send an e-mail to Amanda Keane, akeane@webershandwick.com , in order to clarify your doubts about it. I hope this can help you. Myriam. ---- EDX ANNOUNCES OPTION OF PROCTORED EXAM TESTING THROUGH COLLABORATION WITH PEARSON VUE Online learning venture edX continues to transform higher education by announcing today its agreement with Pearson VUE to offer learners the option of taking a proctored final exam. "Our online learners who want the flexibility to provide potential employers with an independently validated certificate may now choose to take the course exam at a proctored test site," said Anant Agarwal, president of edX. "This option enhances the value of our courses in the real world, helps us maintain our goal of making high-quality education both accessible and practical and thus is a natural evolution of ed's core philosophy of transforming lives through education." Pearson VUE, a Pearson business, is the global leader in computer-based testing. Due to this new agreement, edX learners now have the option of taking a course final exam at one of over 450 Pearson VUE test centers in more than 110 countries. Proctors at the centers will verify the identity of the examinee and administer the tests. Examinees using the Pearson VUE centers will take the same rigorous exam as online learners and will be charged a modest fee for the proctoring service. EdX will offer the option to test takers for one of its online courses this Fall. Robert Whelan, president and chief executive officer of Pearson VUE said, "We are thrilled to collaborate with edX and help its learners improve their employability and career prospects. Pearson VUE is also pleased to help edX protect the integrity of their courses by providing the option of proctored exams." ABOUT EDX EdX is a not-for-profit enterprise of its founding partners Harvard University and the Massachusetts Institute of Technology that features learning designed specifically for interactive study via the web. Based on a long history of collaboration and their shared educational missions, the founders are creating a new online-learning experience with online courses that reflect their disciplinary breadth. Along with offering online courses, the institutions will use edX to research how students learn and how technology can transform learning-both on-campus and worldwide. Anant Agarwal, former Director of MIT's Computer Science and Artificial Intelligence Laboratory, serves as the first president of edX. EdX's goals combine the desire to reach out to students of all ages, means, and nations, and to deliver these teachings from a faculty who reflect the diversity of its audience. EdX is based in Cambridge, Massachusetts and is governed by MIT and Harvard. ABOUT PEARSON VUE Pearson VUE (www.pearsonvue.com) is the global leader in computer-based testing for information technology, academic, government and professional testing programs around the world. Pearson VUE provides a full suite of services from test development to data management, and delivers exams through the world's most comprehensive and secure network of test centers in more than 175 countries. Pearson VUE is a business of Pearson (NYSE: PSO; LSE: PSON), the international education and information company, whose businesses include the Financial Times Group, Pearson Education and the Penguin Group. Contact: Amanda Keane akeane@webershandwick.com ---- [1]: https://www.edx.org/press/edX-announces-proctored-exam-testing FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-03T20:32:46Z IndexTAG: 694 TitleTAG: Questions on impulse response Given for a capacitor that $i_C = C\frac{dV_C}{dt}$ shouldn't changes in voltage be continuous and not change abruptly? Otherwise the current through a capacitor would be infinite which is not physically possible. Similarly for an inductor since $v_L = L\frac{di_L}{dt}$ changes in current should be continuous. This seems to contradict the impulse responses discussed in lecture where capacitors have instantaneous changes in voltage and inductors have instantaneous changes in current. UserIdTAG: 370247 UserNameTAG: Dijkstra CreateTimeTAG: 2012-11-02T19:59:24Z VoteTAG: 4 CoursewareTAG: Week 8 / Response To Impulse Limit Case CommentableIdTAG: 6002x_Response_To_Impulse_Limit_Case NumberOfReplyTAG: 1 FirstChildTAG: I think impulses are easier to think about using the integral forms. For the capacitor, $V_C = \frac{1}{C} \int_{-\infty}^ti_C dt$. So if $i_C = Q\delta(t)$, then $V_C(0_{-}) = 0$ and $V_C(0_{+}) = Q/C$. So, just as you say, the voltage jumps discontinuously. And as you point out, this requires an infinite current. But the delta function *does* have infinite amplitude. So there is no contradiction between the behavior of the capacitor voltage and the amplitude of the current. The possible contradiction arises because an infinite current "is not physically possible." That is technically true, but the point of the impulse is that it idealizes a signal with a finite area that has a very short duration with respect to a characteristic time of interest, like the time constant of the circuit. You could carefully model the details, including the finite rise time of $V_C$, but it would be a ton of work, and in the end you wouldn't learn anything interesting, at least on the longer time scale. So it is easier to pretend that you have an infinitesimally short, infinitely large, current pulse with a specified area. If the fast time scale transients are what you care about, then the impulse approximation is not for you! If infinite amplitudes bother you, only think of the impulse using the integral form, and focus on the fact that the integral has a finite non-zero value. Don't even try to give the delta function a separate existence outside the integral sign. FirstChildUserIdTAG: 174848 FirstChildUserNameTAG: jberg FirstChildCreateTimeTAG: 2012-11-03T04:59:47Z IndexTAG: 695 TitleTAG: Q6 MidTerm Thevenin Resistance by Dependent Resistor Method This problem always seems to trip me up: finding the Thevenin Resistance when shorting the output terminals would effectively short out a dependant source placed across those terminals inside the circuit. I know there are various ways to solve this problem. But, I have my own method, which always works for me. Maybe someone else will find it useful also. I call it the "Dependent Resistor" method. The trick is to notice that all dependent sources can be replaced by resistors. So, we just turn off the independent sources, replace all dependent sources by their effective resistors, and solve for the resistance using the usual intuitive methods of combining series and parallel resistances. The thing to note, is that the effective resistance of a dependent source is "dependent" on the values of other fixed resistors in the circuit. So, we have to first find this resistance in terms of those other resistors. Then observe that the resistance can take on positive or negative resistance values, since they are not real resistors, only replacements that act like resistors when the independent sources are all turned off. ![enter image description here][1] So in the circuit above, we replace VCVS "A.va" with a resistor rA, and replace the VCCS "B.vb" with a resistor rB, set Vin = 0, let v(R) be the voltage across any given resistor "R", and i(R) be the current through "R", then obviously R = v(R)/i(R), so we can find rA and rB, and solve as follows: ![enter image description here][2] This method leads to less errors for me, when computing the Thevenin Resistance. [1]: https://edxuploads.s3.amazonaws.com/13517080071343688.jpg [2]: https://edxuploads.s3.amazonaws.com/13517081661343673.jpg UserIdTAG: 4463 UserNameTAG: pmj CreateTimeTAG: 2012-10-31T18:30:23Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: **Note** For me, R1 = R2, so Va = Vb, so I just used Va in the description below. The dependent current source is actually = B*Vb in the drawing above, and was for me also, but it was equal to B*Va for my values (and when I wrote this I wasn't looking at the diagram). **Step 1. Find VTh** VTh = Voc, the open circuit output voltage. Oops, they already had you calculate Vout, so just copy the answer from part 2 of the problem (from the answer box above). $$V_{Th} = V_{out}$$. **Step 2. Find RTh** There are two ways to do this. One is to find IN and use RTh = VTh / IN to get the desired result. Leave independent sources in for this approach. The other is to inject a test voltage Vtest at the output and measure (calculate) the current, Itest say, flowing through the test source, then use RTh = Vtest/Itest to get the desired result. But don't forget to remove the independent sources first. **step 2a. Find I_Norton** The Norton current, IN, is the current that flows through a short that has been applied so as to short-circuit the output. So redraw the circuit with a shorting wire placed parallel to R3, across the output. Now calculate the currents flowing through R1 and through the voltage controlled current source. There will be no current through R3, because it is parallel to a zero ohm shorting wire. I had R1 = 4 ohms, so using i1 for the current through R1, I can see that Va = i1 * 4. Now the total current through R1 splits, and some goes through the dependent current source (=BVa = B*i1*4), and the remainder goes through the short circuit across the output (=INorton = i1* (1-4*B) ). For me, B = 1/8, so the current splits evenly, and BVa = IN = 1/2 * i1. This description is just to convince you that there is current flowing through the short circuit we applied. You need to do KVL/KCL to find Va and thereby, i1. But this can probably be done by inspection since you have 2 resistors plus one dependent voltage source between Vs and the ground created by the short on the output. $$R_{Th} = \frac {V_{Th}}{I_N} = \frac {V_{out}}{i1*(1-R_1 \cdot B)} = \frac {V_{out}}{i1/2}$$ where the last equivalence only work for my R1 and B values. **step 2b. Find Vtest and Itest - remove independent sources first** I did this too to check my answers. After removing VS, the only independent source, I looked at the circuit and decided to inject a test voltage Vtest across Vout of 4V, but any voltage will do. By choosing 4V I didn't have to do a lot of involved KVL/KCL calculations because the ratios of components made it fairly straight forward to see what the currents in the different branches were. I first thought of inject 1A for Itest, and solving for Vtest, but with VS gone from the circuit, the polarity of Va will be reversed, so the dependent current source will actually be sending current upwards, against the direction of the arrow drawn in the problem diagram. What this is saying, it the value of Va is negative, and so will B*Va will be negative also. If not for the reversal, if I had 1A for Itest, (you see I had R3 = 7ohms) then i3 = 4V/7ohms, leaving 1A - 4/7A to go through the dependent current source and through the R1 circuit branch. And I know from above in part 2a that the current ratios would be 2:1. So 1A was going to be pretty close, but not exact since the ratio now (with VS removed) would be (-2):(-1), meaning the currents were flowing in the reverse direction and the sum of the currents at the top center node would be different. So I used Vtest = 4V for a nice round number, and the currents were easy to solve for, so I got $$R_{Th} = \frac {V_{test}}{I_{test}} = \frac {4V}{I_{test}}$$ and the value for RTh here matched that from step 2a, so all was ok. There is no need to make good/lucky choices for Vtest or Itest. Commonly you will choose Vtest = 1V, or Itest = 1A, and use KVL/KCL to solve for the other. If Itest = 1A, then RTh = Vtest/1, so the math at the end is easy. FirstChildUserIdTAG: 9673 FirstChildUserNameTAG: xp42 FirstChildCreateTimeTAG: 2012-10-31T21:19:20Z FirstChildTAG: Thanks @pmj - that's a useful method. I was wondering if it was possible to convert the dependent sources to effective resistances. FirstChildUserIdTAG: 434869 FirstChildUserNameTAG: patey FirstChildCreateTimeTAG: 2012-11-02T20:23:48Z IndexTAG: 696 TitleTAG: YouTube Playlists for Videos To anybody who is interested and want to watch the whole lectures without interruption, here i have made 2 playlists on youtube: http://www.youtube.com/playlist?list=PLTgEiAIXLilKxLeUqkqPzfBUP0eScqqnF my first playlist until week 8 https://www.youtube.com/playlist?list=PLTgEiAIXLilJ73IwCn-S9U9_7CFxp_345 the second playlist, week 9 included. UserIdTAG: 240487 UserNameTAG: AlexAlexandrescu CreateTimeTAG: 2012-10-31T18:21:17Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: great. Just what i am looking for! thanks FirstChildUserIdTAG: 623210 FirstChildUserNameTAG: preveen FirstChildCreateTimeTAG: 2012-10-31T18:29:56Z IndexTAG: 697 TitleTAG: H8P3 part 4 I have completed rest parts of P3 but badly stuck on this.Please point out mistakes.using VOH = VG2max + (VOL - VG2max) * exp(-t/k) where k is time constant I got t=1.94k .now for K using product of C=3.5fF and R as Ron+(Rpu parallel Roff) the final answer does not give me green tick.secondly I noticed that the system accepts value of VG2max in a range of about +-0.2 of the formula.using different values of VG2max from previous problem I tried to redo whole process.But those answers were also incorrect.Am I not calculating R correct ? thank you in advance. UserIdTAG: 99628 UserNameTAG: Pranjal16 CreateTimeTAG: 2012-10-30T18:49:49Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Pranjal6, I think you expression for the capacitor voltage is incorrect. Your supply voltage is Vs not VG2max, and that should be the steady state value (i.e. the value when t tends to infinity) FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-10-31T13:10:36Z SecondChildTAG: So, am I correct that the $R_{ON}$ of Q3 is the same as the other MOSFET? Is it equal to $2000 \Omega$? SecondChildUserIdTAG: 331664 SecondChildUserNameTAG: dwmnctrh3 SecondChildCreateTimeTAG: 2012-11-05T00:36:59Z SecondChildTAG: Why $2000\Omega$? It is given that $R_{ON}=1900\Omega.$ SecondChildUserIdTAG: 213452 SecondChildUserNameTAG: JoJosida SecondChildCreateTimeTAG: 2012-11-08T02:46:12Z SecondChildTAG: yeah u are correct Its Vs not VG2max SecondChildUserIdTAG: 108929 SecondChildUserNameTAG: namit SecondChildCreateTimeTAG: 2012-11-08T12:59:55Z SecondChildTAG: @jelizon what will be the equation for H8P3 part 5? I am n't getting what is the steady state voltage in that problem. Please help me as soon as possible. SecondChildUserIdTAG: 156859 SecondChildUserNameTAG: PrithviMonangi SecondChildCreateTimeTAG: 2012-11-10T11:15:07Z FirstChildTAG: Can anyone give tips on how to solve this one exactly ? Nomatter what i try i'm not getting the correct time .. FirstChildUserIdTAG: 263693 FirstChildUserNameTAG: Coldberg FirstChildCreateTimeTAG: 2012-11-06T02:02:28Z SecondChildTAG: Do a Thevenin equvalent of Vs (5V), Rpu, Roff (for part4), last two are in parallel) and Ron (which is the switch resistance) and Cgs. You will find that Vth=maximum value of voltage on the drain of Q1 (as in part1). Rth is easy, Rpu in parallel with Roff and series with Ron. That will give you a simple Vth, Rth and Cgs series circuit. Use Vgs=Vol+(Vth-Vol)*(1-exp(-t/(Rth*Cgs))). From that equation you can find t when Vgs=3.5V. SecondChildUserIdTAG: 226643 SecondChildUserNameTAG: AJ1 SecondChildCreateTimeTAG: 2012-11-09T15:47:55Z SecondChildTAG: that really helped..:) SecondChildUserIdTAG: 266739 SecondChildUserNameTAG: satvikchugh SecondChildCreateTimeTAG: 2012-11-11T04:54:20Z SecondChildTAG: thank you SecondChildUserIdTAG: 452559 SecondChildUserNameTAG: kuz1toro SecondChildCreateTimeTAG: 2012-11-11T15:51:57Z SecondChildTAG: Thanks SecondChildUserIdTAG: 413784 SecondChildUserNameTAG: Dorotka SecondChildCreateTimeTAG: 2012-11-11T23:48:15Z IndexTAG: 698 TitleTAG: H9P2-d The question should specify the frequency in Hz. Also there is a typo error in the given differential equation (+o). Many Thanks to all the staff. UserIdTAG: 446822 UserNameTAG: HAwad CreateTimeTAG: 2012-10-30T17:19:42Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: Thanks for posting about the answer in Hz instead of rad/s (question d). I was staring at this question for 5 minutes. Certainly, the question should specify the unity of the answer. FirstChildUserIdTAG: 39980 FirstChildUserNameTAG: MarinoJr FirstChildCreateTimeTAG: 2012-11-01T13:56:05Z FirstChildTAG: Also the answers accepted as correct for H9P2 f & g are wrong. They just copied Eq. 12.108 from the supplemental book, and that equation is incorrect as I posted before in the forum. FirstChildUserIdTAG: 153760 FirstChildUserNameTAG: Kavka FirstChildCreateTimeTAG: 2012-10-30T17:54:47Z SecondChildTAG: Thanks Kavka. I was getting ready to start throwing things ;-) And thanks HAwad... saved me some time better spent elsewhere. SecondChildUserIdTAG: 397247 SecondChildUserNameTAG: pcbolt SecondChildCreateTimeTAG: 2012-10-31T04:27:00Z SecondChildTAG: What page is "12.108" on that you are referring to? I can see the schematic in the book,but mine goes from 12.80 and jumps to 12.116. Thanks SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-31T11:27:47Z SecondChildTAG: There is a supplemental booklet for the main textbook available on the internet that has this equation in it. Please see http://www.elsevierdirect.com/companions/9781558607354/casestudies/001~All_supplemental_sections_and_examples_in_one_file.pdf SecondChildUserIdTAG: 153760 SecondChildUserNameTAG: Kavka SecondChildCreateTimeTAG: 2012-10-31T17:43:32Z SecondChildTAG: Thank you. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-11-01T13:09:11Z FirstChildTAG: Thank you for catching that. We're looking into it now. Answers which have been incorrectly awarded credit will retain their credit, and when the question has been edited, those who have yet to answer it will only be given credit for the correct answer. FirstChildUserIdTAG: 391929 FirstChildUserNameTAG: RohanNagarkar FirstChildCreateTimeTAG: 2012-10-31T14:00:24Z FirstChildTAG: Hint: ω=2*π*f FirstChildUserIdTAG: 198948 FirstChildUserNameTAG: leoayrton FirstChildCreateTimeTAG: 2012-11-13T22:42:08Z SecondChildTAG: the answzeer is sqrt(1/l*c)/(2*pi) SecondChildUserIdTAG: 174229 SecondChildUserNameTAG: fares27 SecondChildCreateTimeTAG: 2012-11-15T17:25:55Z IndexTAG: 699 TitleTAG: Help for those struggling with H7P1 P2 and P3!! ![enter image description here][1] ![enter image description here][2] ![enter image description here][3] ![enter image description here][4] ![enter image description here][5] [1]: https://edxuploads.s3.amazonaws.com/13515025021343693.png [2]: https://edxuploads.s3.amazonaws.com/13515025121343631.png [3]: https://edxuploads.s3.amazonaws.com/13515025423779162.png [4]: https://edxuploads.s3.amazonaws.com/13515025581343695.png [5]: https://edxuploads.s3.amazonaws.com/13515049791343683.png UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-10-29T09:23:21Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: thanks hazel what is the app you use to make these notes FirstChildUserIdTAG: 623210 FirstChildUserNameTAG: preveen FirstChildCreateTimeTAG: 2012-10-29T10:03:45Z SecondChildTAG: Hi preveen! I use paintbox mainly. SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-04T21:30:56Z FirstChildTAG: i am having problem with part 3 question 3. could you please explain? thanks! FirstChildUserIdTAG: 222911 FirstChildUserNameTAG: bhaswardg FirstChildCreateTimeTAG: 2012-10-29T11:29:04Z SecondChildTAG: I ma going to be away for a week! NO 6.002x! Sorry, perhaps someone else can assist? SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-29T12:20:59Z SecondChildTAG: Even I'm in trouble for the same question! Can anyone help us!? SecondChildUserIdTAG: 362299 SecondChildUserNameTAG: anandbaskaran SecondChildCreateTimeTAG: 2012-10-29T13:53:46Z SecondChildTAG: The answers are like theory in the textbook in chapter 9.2.2 and the time constant and transiens in chapter 10. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-10-29T17:16:24Z SecondChildTAG: what is the R of time constant being used here??? SecondChildUserIdTAG: 127800 SecondChildUserNameTAG: MNB SecondChildCreateTimeTAG: 2012-10-30T04:57:48Z SecondChildTAG: sorry, but just can't get through this part! a little more help please. SecondChildUserIdTAG: 222911 SecondChildUserNameTAG: bhaswardg SecondChildCreateTimeTAG: 2012-10-30T06:02:54Z SecondChildTAG: nevermind..i got it! :) SecondChildUserIdTAG: 222911 SecondChildUserNameTAG: bhaswardg SecondChildCreateTimeTAG: 2012-10-31T10:51:29Z SecondChildTAG: I can't find solution for part3 Q3 and Q4!! please, can anybody help me? SecondChildUserIdTAG: 314294 SecondChildUserNameTAG: victormp SecondChildCreateTimeTAG: 2012-11-02T10:39:44Z SecondChildTAG: I got it!! Thanks anyway!! :) SecondChildUserIdTAG: 314294 SecondChildUserNameTAG: victormp SecondChildCreateTimeTAG: 2012-11-02T11:27:02Z SecondChildTAG: I have problems with part 3 Q3, i tied almos everything, can anybody help me? SecondChildUserIdTAG: 250473 SecondChildUserNameTAG: pajaropica SecondChildCreateTimeTAG: 2012-11-02T17:30:17Z SecondChildTAG: Hi pajaropica, Notice that current is the same throughout the circuit. So you need to find out the current and then it becomes simple. I guess you know how to calculate the voltage drop across a resistor! Use that formula to find t. Look through the pages mentioned by hazel1919 and you will know how to find the current. :) SecondChildUserIdTAG: 222911 SecondChildUserNameTAG: bhaswardg SecondChildCreateTimeTAG: 2012-11-03T07:53:21Z IndexTAG: 700 TitleTAG: Midterm Comparison / Easier / Harder / About The Same? How do you think this midterm compared with the last one? UserIdTAG: 9673 UserNameTAG: xp42 CreateTimeTAG: 2012-10-29T07:49:21Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 7 FirstChildTAG: About The Same. FirstChildUserIdTAG: 378096 FirstChildUserNameTAG: Jamshaid271 FirstChildCreateTimeTAG: 2012-10-29T08:02:52Z SecondChildTAG: I agree. I had the same result too. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-10-29T11:05:27Z SecondChildTAG: easier! SecondChildUserIdTAG: 219204 SecondChildUserNameTAG: vsriram SecondChildCreateTimeTAG: 2012-10-29T13:55:54Z FirstChildTAG: **Midterm Exam Thoughts:** I did not take the course last semester, but I found that many of the Questions in the Midterm were either exactly the same, or nearly the same, as Exercise Problems /Homework Problems / Practice Midterm / Tutorials. Thus if you followed the Courseware, you should have gotten an A or B without even cracking the textbook. *I did take a similar course when I was in the School of Engineering at my university, but that was over 8 years ago. There, we had three "monthly" exams and one "final", so I cannot compare the two (there we studied op-amps long before transistors, AC analysis/phasors before first-order, and BJT instead of MOSFET, we covered MOSFETs in later courses). Different professors = different teaching style.* Certainly for such things as Shared Current Sources between two Loops / Mesh Analysis, and for Source Transformation I had to refer to my old university textbook "Fundamentals of Electric Circuits" by Alexander and Sadiku, and before I took the exam I scoured the internet, my old textbooks, our current textbook, and lecture notes to compile a "useful equations" sheet, with hints on how to quickly analyze ladder networks, different MOSFET amplifier combinations such as source-follower and common-base, etc. But all this was not absolutely necessary: It made the exam less stressful and simplified it a bit, but it created more work in preparation ahead of the exam. I would say this prepared me well, and my grade speaks of this. I made stupid mistakes (such as in Mid-Term Question 1 I confused VA with VB, so I got all red Xs, I panicked, then I switched the values and corrected the rest, and got all green checks!) of course. So I decided to slow down and not "race" to complete the Exam in 2 hours. I took a 15-minute break between questions, and an hour for lunch, so the exam took longer, but it was more relaxed this way! No where did I have to use all 3 of my submissions. I used 2 at maximum. Jersey Mark FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-10-29T08:59:26Z FirstChildTAG: Didn't check the other midterm, but I believe this one was rather easy. FirstChildUserIdTAG: 392546 FirstChildUserNameTAG: JohnWayne FirstChildCreateTimeTAG: 2012-10-29T10:20:22Z FirstChildTAG: I joined the course 13th oct only. I took some time for the midterm. first 5 questions I could complete at the first attempt itself. 6th question gave me really a hard time. (actually the last of last) But in the end, after referring to answers and advice from previous home work and all I could get 100%. I had a great time scribbling a lot for the last question. Almost 25 pages of scribbling and drawing circuits. I was a bit hesitant joining the course which is already 6 weeks old. But glad I joined. I could only finish some of the hw6 after many hints from colleagues on the forum. I am sure I am learning something. Grades are great but learning is more important. Those who took the trouble to get through the home work and labs, the midterm should be completed with in max 90 mintues. Thanks everybody. FirstChildUserIdTAG: 623210 FirstChildUserNameTAG: preveen FirstChildCreateTimeTAG: 2012-10-29T10:13:53Z SecondChildTAG: That's the spirit!! SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-10-29T12:54:21Z FirstChildTAG: I thought this midterm was easier than the Spring 2012 version for several reasons. First, unlike last semester, there were no homework and lab assignments due the same weekend, thus giving us an opportunity to study. Second, the timing of the midterm last semester was in Week 8, and by then, my brain was totally fried from the "grubby math" of impulses, steps, and ramps in first-order circuits. Third, the homework problem which made its way to the midterm last semester (ideal diodes) was one that many people had trouble with. I think the homework problem selected for this semester's midterm (SR model of logic circuit) was much more straightforward. Fourth, the logic gate problem last semester included working out truth tables for the circuit and selecting the appropriate logic diagram, none of which appeared on this semester's midterm. Many people had trouble with the Boolean math and the logic. Last, although this semester's midterm had several imaginary devices, the circuits themselves were a configuration (source follower) that we had studied. Last semester's midterm sprang a common gate configuration on us (which we had not studied), and neither the book nor the lectures covered it. Having gone through this material once before, it does seem a bit easier the second time, although I am still learning (and re-learning!) the material. I've also since amassed a collection of other textbooks that I bought used on eBay, which helped me (including the Alexander/Sadiku textbook that Jersey Mark referenced above). On the other hand, I did have to interrupt my exam to prepare for the hurricane, so I had stress of a different and unexpected kind! Just my two cents... FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-10-29T12:37:36Z SecondChildTAG: Good thoughts, g_hopper. In some ways, it was easier, like you said. Mostly because of the timing. Last course we had the homework and the mid-term due at the same time so you didn't get a break. That made it very difficult. Brain-fry was a real factor. On the other hand, I knew the Boolean stuff very well. I nailed that part. Dependent sources, though, I still don't have that down. So I blew that part. So it was kind of a mixed bag. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-29T13:40:26Z SecondChildTAG: I thought the hardest part of my week was going to be the midterm. I arrived at work on Monday just hoping I could get home and sleep after working on the midterm over the weekend. Little did I know how my life would be turned upside down from Hurricane Sandy. I posted a description here: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50947eb7cde578270000000d I'm going to miss week 7 labs and homework, I am back at work and simply don't have time and energy (or the brain cells) to attempt 6.002x and/or 3.091x homework and labs before the deadline. I'm really grateful the midterm was ***BEFORE*** the hurricane, or I would have missed the midterm exam. Be well, I'll check in later in the week. SecondChildUserIdTAG: 94557 SecondChildUserNameTAG: g_hopper SecondChildCreateTimeTAG: 2012-11-04T15:49:37Z FirstChildTAG: I found it easier. I'm not sure if it is because this is the second time I'm taking the course. I'm sure the Spring batch would agree with me when I say that when I saw the Final Exam, I thought nothing could be tougher than that (does not apply to those with a lot of prior experience :-))! I got all but one answer right on first attempt and finished it in one and a half hours. However, there were things that were simpler. For example, I struggled with circuits containing capacitors and was able to get a good hang of it only after the finals because I took time to practice more problems. There were no questions related to capacitors this time. The last part of question four was exactly the same as one of the homework problems. The only difference being that the MOSFET was replaced by another device. Question 5 was also a direct application of the MOSFET amplifier taught in the lecture. Only the device was changed. Like last time, there was one homework problem which was presented directly. The one that was picked this time carried a lot more points. Personally, I feel it should carry the least points as one know how to do it before hand. Nevertheless I did have fun solving the problems :-). I do wish it was a bit more difficult. FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-10-29T16:05:18Z FirstChildTAG: Easier. > **Regards:** asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-11-04T07:30:55Z IndexTAG: 701 TitleTAG: Aaaha moment.. Finishing mid term is an "Aaha" moment...!! UserIdTAG: 356391 UserNameTAG: anoopkrishnan CreateTimeTAG: 2012-10-29T04:45:31Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: more like ahh for me!! FirstChildUserIdTAG: 357747 FirstChildUserNameTAG: kishores FirstChildCreateTimeTAG: 2012-10-29T12:27:59Z FirstChildTAG: AAAAHHHH AAAHHH moment....!!! :D FirstChildUserIdTAG: 127800 FirstChildUserNameTAG: MNB FirstChildCreateTimeTAG: 2012-10-30T04:35:45Z FirstChildTAG: I was sad there weren't more questions. It was fun to solve them. :( FirstChildUserIdTAG: 397804 FirstChildUserNameTAG: mdempsky FirstChildCreateTimeTAG: 2012-10-29T15:37:03Z FirstChildTAG: :) yes it was abig aha moment FirstChildUserIdTAG: 221617 FirstChildUserNameTAG: konan FirstChildCreateTimeTAG: 2012-10-29T10:02:13Z IndexTAG: 702 TitleTAG: Additional Problem sets please I guess it would be great to have additional problems sets so we can further savor the experience of the "Green Tick" and of course deepen our understanding of these materials. UserIdTAG: 623210 UserNameTAG: preveen CreateTimeTAG: 2012-10-27T17:15:48Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hi preveen, If you take a Look at the Textbook Preamble, you will find a Web link to some additional Examples - Web supplemts ([here][1]). I hope this can help you. See you, Myriam. [1]: http://www.elsevierdirect.com/v2/companion.jsp?ISBN=9781558607354 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-27T19:16:00Z SecondChildTAG: thanks but those are sections that is not included in the text proper. What I mean is other courses like "solid state chemistry" on mitx/edx has additional problems sets in courseware (not graded). It would be great if we can have some in 6.002x. SecondChildUserIdTAG: 623210 SecondChildUserNameTAG: preveen SecondChildCreateTimeTAG: 2012-10-28T16:16:18Z FirstChildTAG: There are a lot of additional problems at the end of each chapter of the book. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-27T17:17:23Z SecondChildTAG: but not the greek tick;-) SecondChildUserIdTAG: 623210 SecondChildUserNameTAG: preveen SecondChildCreateTimeTAG: 2012-10-27T17:37:06Z SecondChildTAG: yes, there is lack of solutions in the textbook- only half - without knowing is your solution correct it's almost useless SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-27T19:08:22Z SecondChildTAG: True, although you could do the examples and then build the examples in some simulator to check your work. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-27T19:58:42Z IndexTAG: 703 TitleTAG: COMMUNICATION COURSE I wish there was an communications related course also! -- Electronics and Communications Engineering student. UserIdTAG: 277808 UserNameTAG: Hemanthmps CreateTimeTAG: 2012-10-27T15:12:52Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: check opencourseware. they have everything. HTH. FirstChildUserIdTAG: 623210 FirstChildUserNameTAG: preveen FirstChildCreateTimeTAG: 2012-10-27T15:49:19Z SecondChildTAG: link please ? SecondChildUserIdTAG: 41666 SecondChildUserNameTAG: AbdoMondy SecondChildCreateTimeTAG: 2012-10-27T16:58:57Z SecondChildTAG: Here are just a few examples. Have a look around, you should find something that interests you. http://ocw.mit.edu/courses/aeronautics-and-astronautics/16-36-communication-systems-engineering-spring-2009/ http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-450-principles-of-digital-communications-i-fall-2006/ http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-452-principles-of-wireless-communications-spring-2006/ SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-27T17:10:30Z IndexTAG: 704 TitleTAG: How much would it cost to take a proctored exam? Roughly? Will there be any printed certificate with a stamp/signature at the end for those who would succeed in this exam? UserIdTAG: 94899 UserNameTAG: cls33 CreateTimeTAG: 2012-10-25T14:01:38Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Maybe around $70 - 80 At the end of the course if you have 60% or greater, you can print your Certificate of Mastery for this course. There is also a link you can use, should you need to print it again, verification etc. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-25T14:25:45Z SecondChildTAG: Well, yeah, thanks, i know that. I've already completed this course last semester. But they might also make some special certificates for the ones who would take proctored exam. That would be nice. SecondChildUserIdTAG: 94899 SecondChildUserNameTAG: cls33 SecondChildCreateTimeTAG: 2012-10-25T17:05:23Z SecondChildTAG: Not just an online certificate but maybe real paper certificate with real MIT stamp :) I would even pay the delivery to Russia :) SecondChildUserIdTAG: 94899 SecondChildUserNameTAG: cls33 SecondChildCreateTimeTAG: 2012-10-25T17:07:04Z SecondChildTAG: **Pennypacker**: Do you have a source for that price? Did you take a proctored exam or other test at a [Pearson VUE][1] or a competitor testing-center for a non-edX class? I am thinking about taking a proctored exam in the future and I'm exploring the options at this time. Thanks! [1]: http://www.pearsonvue.com/faqs/ SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-10-25T18:30:44Z SecondChildTAG: I have not taken a proctored test with Pearson VUE. Here is a link referring to edX. http://www.pearsonvue.com/about/release/12_09_06_edx.asp This second link has prices for the other tests that Pearson VUE offers. These are "vouchers" that you can purchase for specific tests that can be used at your convenience. http://www.pearsonvue.com/vouchers/pricelist// SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-25T19:55:10Z FirstChildTAG: In this [recent edX press release][1], Amanda Keane (akeane@webershandwick.com) announced that "edX learners now have the option of taking a course final exam at one of over 450 [Pearson VUE test centers][2] in more than 110 countries. Proctors at the centers will verify the identity of the examinee and administer the tests ... EdX will offer the option to test takers for *one of its online courses this Fall*. " I am not affiliated with MIT nor am I affiliated with edX, so I do not know the exact cost for a proctored exam, nor do I know the course that will have it's final exam proctored as a choice (out of 6.002x, 6.00x, 3.091x, CS50x, PH207x, CS169.1x, CS184.1x, and CS188.1x) but as Ms. Keane's email appears in the press release, if you are interested in the cost and any updates, you should definitely **contact her** and share with us what you hear. Because edX and massive online education is just so new, nothing is "set in stone" yet; and we are on the frontier of this revolution in education. EdX is not-for-profit, which is a good sign as this generally keeps student expenses at a minimum compared to for-profit endeavors, such as Coursera. As for the benefits of taking a proctored final exam versus a "Honor Code" final exam, that is still uncertain. It also depends on what you're taking the course for. If you just want a certificate to add to your resume as a job-seeker, the "Honor Code Certificate" is probably the way to go. However, if you plan to further your education, taking the proctored exam **may** allow the educational institution you plan on attending or transferring to, to give you credit for the course you take a proctored exam in. In the United States, the school you plan to attend in the future, or transfer to, sets the standards for transfer credits and acceptance; so **first check with that institution** to see if they will accept an edX/MITx online course with a Pearson proctored final for credit. Some will, some won't, and others may want you to take another, separate exam at their institution (instead of Pearson, thus taking a proctored exam duplicative) instead. [1]: https://www.edx.org/press/edX-announces-proctored-exam-testing [2]: http://www.pearsonvue.com/faqs/ FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-10-25T18:26:31Z SecondChildTAG: Thank you! SecondChildUserIdTAG: 362299 SecondChildUserNameTAG: anandbaskaran SecondChildCreateTimeTAG: 2012-10-25T18:47:34Z FirstChildTAG: Do I need to take a proctored exam in order to complete the course? Or can I just pass the final exam and all the HWs and Labs and that's all? FirstChildUserIdTAG: 326787 FirstChildUserNameTAG: Angstrem FirstChildCreateTimeTAG: 2012-10-25T15:42:01Z SecondChildTAG: you don't need to take a proctored exam for this semster. SecondChildUserIdTAG: 454609 SecondChildUserNameTAG: Alwahsh SecondChildCreateTimeTAG: 2012-10-25T17:38:41Z IndexTAG: 705 TitleTAG: STAFF - MIDTERM EXAM - IMPORTANT (OR MYRIMIT) Dear Staff, It's 8.15 am in Barcelona Spain right now when I just woke up a few minutes ago. I log in, go to Courseware and I see: "You were most recently in Midterm exam. If you're done with that, choose another section on the left" But I didn't click on the MIDTERM EXAM YET.... I went to sleep at about 01.00 am of today, the 25th in Barcelona, so it was about 07.00 pm Boston time of the 24th.... The last thing I did as far as I remember was try to solve the Q2 Blackbox of LAST YEAR MID TERM.... SO I hope it doesn't count yet, I need time to get prepared until the 27th or 28th!! I didn't see anything of the actual MIDTERM EXAM.... Please could you tell me if the 24 hours for some reason accidentally already started?? I hope no!!!!!! My e-mail: sandranavarro00@gmail.com Pls. help, Thanks in advance!! Gracias Myriam también!! Regards, Sandra UserIdTAG: 342966 UserNameTAG: SandraNavarro CreateTimeTAG: 2012-10-25T06:27:49Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: same thing it was showing for me also FirstChildUserIdTAG: 365309 FirstChildUserNameTAG: chetnasinghaldas FirstChildCreateTimeTAG: 2012-10-25T06:56:17Z SecondChildTAG: yea,me too..what's happening? SecondChildUserIdTAG: 386004 SecondChildUserNameTAG: ravinarv SecondChildCreateTimeTAG: 2012-10-25T07:00:26Z SecondChildTAG: Thanks, that relieves me.... waiting from an answer, then. SecondChildUserIdTAG: 342966 SecondChildUserNameTAG: SandraNavarro SecondChildCreateTimeTAG: 2012-10-25T07:00:48Z FirstChildTAG: I guess that the way they used to make it clear that the midterm was posted. I found the same this morning and was initially doubtful about whether the time had started accidentally for me or not. Since I had done nothing else but just logging in... But I think there was an information somewhere about this explaining that after clicking on the midterm link an intermediate information page will appear explaining the rules and that the time will only start counting after explicitely agreeing on that and accessing to the following page, where the actual exam will be displayed. I would like to have that confirmed by the staff, anyway... FirstChildUserIdTAG: 366320 FirstChildUserNameTAG: Galli FirstChildCreateTimeTAG: 2012-10-25T07:14:42Z FirstChildTAG: Hi Sandra, If you click on the Midterm you should see a warning message like this: Have you seen this? ![image][1] If you click on accepting start, the counter starts for you :) So, don´t worry,it was the last semester Midterm Exam were you were more recently not the actual one :) See you! Myriam. [1]: https://edxuploads.s3.amazonaws.com/13511645821343618.png FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-25T11:32:54Z SecondChildTAG: Hi Myriam, No I haven't seen this.... cause I really didn't intend to start the exam. Hope it will be fine. Thank you Myriam and folks!! Sandra SecondChildUserIdTAG: 342966 SecondChildUserNameTAG: SandraNavarro SecondChildCreateTimeTAG: 2012-10-25T14:48:17Z FirstChildTAG: No panic, it is ok ! By clicking on the “Midterm Exam” button between “week 6” and “week 7” you are led to the exam presentation page (see below) and from there you can finally start the exam (and the timer). **Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science** 6.002x - Circuits & Electronics Midterm Exam, Fall 2012 Out 10/25/12 00:01 (12:01 am) - Due 10/28/12 23:59 (11:59 pm) Boston time This exam is designed to take two hours; however, to compensate for any Internet or power outages in your area you will have 24 hours to finish this exam. You can start the exam when it is convenient for you, but you must complete this examination by 11:59 pm (almost midnight) Boston time on October 28th. Please look up what time this is in your local time zone. When you open the next page, you will have started the examination. You do not need to start now: you will not be timed until you open the next page. Once you have opened the next page you must complete the exam and make your final submission within twenty four hours of starting the exam. You may use any notes, computational, or auxiliary materials that you think can help you. However, you may not communicate with any person about this examination while working on it. Furthermore, you may not communicate about the exam until the exam has been closed for everyone. For each problem you will be allowed exactly three submissions. Your answers to that problem will be checked after each submission, so you can fix mistakes you may have made, within the three-submission limit. If you want to go back and study some more before starting this exam you can do so. Good Luck! Heading FirstChildUserIdTAG: 372321 FirstChildUserNameTAG: EnricoDona FirstChildCreateTimeTAG: 2012-10-25T09:01:19Z SecondChildTAG: Thanks Enrico :)) SecondChildUserIdTAG: 342966 SecondChildUserNameTAG: SandraNavarro SecondChildCreateTimeTAG: 2012-10-25T14:48:37Z IndexTAG: 706 TitleTAG: MIDTERM TEST............ i am sleeping Where is the exam.... in my country is very dificult pay for the internet... is like 12 am in my country... please release the midterm.... Saludos a todos los espanioles y exitos en sus pruebas !! UserIdTAG: 628745 UserNameTAG: albertclint CreateTimeTAG: 2012-10-25T05:22:24Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: wow FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-25T09:45:57Z IndexTAG: 707 TitleTAG: Midterm is the midterm open book and open internet? UserIdTAG: 365225 UserNameTAG: doublegj526 CreateTimeTAG: 2012-10-23T20:53:52Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Yes. The only restriction is that you cannot communicate with any other human being about the exam while the exam is open. That includes people on this forum. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-10-23T21:12:53Z SecondChildTAG: Hi,and congratulations for the good work you are doing. I'd like to know if are you planning to publish a spanish version of this course in the near future. Thanks SecondChildUserIdTAG: 369339 SecondChildUserNameTAG: vargaslen SecondChildCreateTimeTAG: 2012-10-23T22:03:58Z SecondChildTAG: Not in the near future, but it is part of our general aspirations. SecondChildUserIdTAG: 96452 SecondChildUserNameTAG: Lyla SecondChildCreateTimeTAG: 2012-10-23T22:06:49Z SecondChildTAG: That would be great! :). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-24T00:30:41Z SecondChildTAG: hey but if we start working on the exam, it will be omy 24 jurs open after the starting point? SecondChildUserIdTAG: 308861 SecondChildUserNameTAG: Ignaas SecondChildCreateTimeTAG: 2012-10-24T01:45:15Z SecondChildTAG: I think that translating to Spanish could beneficiate a lot of intelligent people who haven't had the chance to learn English. May be if MIT calls for collaboration to this end,many people would respond. I would be glad to collaborate. SecondChildUserIdTAG: 369339 SecondChildUserNameTAG: vargaslen SecondChildCreateTimeTAG: 2012-10-24T18:02:21Z FirstChildTAG: Well, I think that translate the entire course into Spanish would be an error and also a huge effort, I think that the time and money necessary to translate it could be used it to make other better courses. I am from Spain FirstChildUserIdTAG: 418303 FirstChildUserNameTAG: AlexNadal FirstChildCreateTimeTAG: 2012-10-24T08:18:20Z SecondChildTAG: I think that translating to Spanish could beneficiate a lot of intelligent people who haven't had the chance to learn English. May be if MIT calls for collaboration to this end,many people would respond. I would be glad to collaborate. SecondChildUserIdTAG: 369339 SecondChildUserNameTAG: vargaslen SecondChildCreateTimeTAG: 2012-10-24T18:01:46Z IndexTAG: 708 TitleTAG: CONFUSING!!!! watching this video has confused me all the more..I couldn't get through after the first part... its purely mathematical but not easily comprehensible. UserIdTAG: 368712 UserNameTAG: jmen CreateTimeTAG: 2012-10-23T14:42:43Z VoteTAG: 4 CoursewareTAG: Week 6 / Two-stage MOSFET amplifier CommentableIdTAG: 6002x_Two_Stage_MOSFET_amplifier NumberOfReplyTAG: 2 FirstChildTAG: I find a lot of Dr. Argawal's lectures very confusion. Went back to the text I used in electronics school (copyright 1971) and found some very simple explanations of the same concepts. FirstChildUserIdTAG: 237167 FirstChildUserNameTAG: Dug FirstChildCreateTimeTAG: 2012-10-23T16:47:14Z SecondChildTAG: Right there with you guys! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-23T16:53:57Z SecondChildTAG: hazel1919, can you elaborate? SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-23T17:02:14Z SecondChildTAG: well my post was concerning one of the tutorials videos solved for week 6! SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-23T17:18:21Z SecondChildTAG: @ JSChambers: For me, personally, the differential equations encountered in week 6 - capacitors and first order circuits are too much! The one video did show how to solve the differential equation, but I really was lost. The equations require, in the words of the Prof, "guess work", but clearly you need to know what you are guessing about. That knowledge, I assume is held in The Calculus, which I do not have a grasp of. Just the sheer mathematical gymnastics made me feel like I was 100 years old and had to do the hurdles (maybe you have a better illustration ;) ). I suppose the best I can do is... ..carry on learning from the Khan Academy, using Wolfram and hoping they don't ask me to solve a first order diff calc in the midterm! Hazel. SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-23T17:20:16Z SecondChildTAG: I find Dr. Argawal's lecturing style to be very annoying. He will say something and then immediately say it again in almost exactly the same words. Sometimes he will repeat even a third time. To meaningfully clarify the subject it needs to be repeated using different terms that add information. Otherwise he is just wasting time. In the trial run of the course the subject of differential equations must have created confusion and discussion on the forum. Therefore, this presentation of the course should have tried a new approach to explain. If I were teaching this subject and I knew it was causing problems, I would try a discretized time approach and see if that would get the concept across. It just might work! SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-23T18:56:49Z SecondChildTAG: If Agarwal repeats something he is doing so because of some wisdom (repeating not parroting)in it, I find it very easy to remember if some important facts and concepts are repeated it become our part of memory and handy when solving problems instead consulting books abd reading it again and again (I am repeating)....learning.... SecondChildUserIdTAG: 272523 SecondChildUserNameTAG: Jivraj SecondChildCreateTimeTAG: 2012-10-23T21:44:37Z SecondChildTAG: I strongly disagree. The type of repeating of phases I am talking about has nothing to do with repetition to enforce learning. I have had some excellent teachers in my life, and I know what I am talking about. I hope you will agree that each of us has a right to his own opinion, and I wish you well in your learning endeavors. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-23T22:07:42Z SecondChildTAG: I can tell you right now, I could do with more **explicit** explanations when dealing with the math, not repetitions! Now, I am not accusing Agarwal of this, but I refer back to my book Calculus made easy (a notable book on calculus by a notable author). It contains this quote: "Considering how many fools can calculate, it is surprising that it should be thought either a difficult of tedious task for any other fool to learn how to master the same tricks. Some calculus-tricks are quite easy. Some are enormously difficult. The fools who write the textbooks of advance mathematics - and they are mostly clever fools - seldom take the trouble to show you how easy the calculations are. On the contrary, they seem to desire to impress you with their tremendous cleverness by going about it in the most difficult way. Being myself a remarkably stupid fellow, I have had to unteach myself the difficulties and now beg to present to my fellow fools the parts that are not hard. Master these thoroughly, and he rest will follow. What one fool can do, another can." SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-24T08:57:53Z SecondChildTAG: Thanks for the book recommendation! SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-24T12:11:03Z SecondChildTAG: Yep, a really good book! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-24T20:25:41Z FirstChildTAG: I had never heard of the book, but I did find a PDF on line: http://djm.cc/library/Calculus_Made_Easy_Thompson.pdf I also understand that there is an updated version of the book available. I took a look at the PDF, and all I can say is that I am glad I didn't have to learn calculus from that book. So many words to explain such a simple concept. I would say that the author is guilty of exactly what he accuses others of in the preface. He finally gets around to the geometrical concept of a derivative in chapter 10. The geometrical concept should have been the starting point. At least the basic results are obtained in the earlier chapters. Having said the above, I understand different people learn in different ways and if this satisfies a student of the subject then who am I to criticize. As the saying goes, the proof of the pudding is in the eating." So, eat up! FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-24T22:16:12Z SecondChildTAG: Haha, so true... " The geometrical concept should have been the starting point" That is why I also value the Khan academy **highly**. SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-25T18:16:52Z IndexTAG: 709 TitleTAG: Midterm preperations Hi all, Any suggestions for a good preperation for our midterm exam....Myrimit do you have any tips like always?:)) UserIdTAG: 308861 UserNameTAG: Ignaas CreateTimeTAG: 2012-10-23T11:59:23Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hi Ignaas! :) Some Tips: 1) Review Homework problems, might one point of them can appear in the Exam ;). 2) I suggest to print the Exam and solve it in your desk :). Be relaxed and take your time. 3) Think your answers before submit them (submit button), because you will have only 3 attemps per problem (not by each question). You also will have a save button (this will not submit your answer, only save it so that later you can submit it). 4) Be careful with the reset button! This will erase your answers (correct or incorrect ones)! 5) They are really severe: Don´t ask about Midterm Exam problems during the period of the Exam. Remember that you have signed an Honor Code agreement, you musn´t post your results in any place (inside/outside the Forum Discussion)during the Exam. Your account can be deleted... That is all :). My best wish to all of you! Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-23T12:59:30Z SecondChildTAG: I would add that for the Mid-term and Final that the honor code is a little different than for the homework. Here is what is says: "You are not allowed to post answers to exam problems. **Collaboration of any form is strictly forbidden** in the midterm and the final exams." So unlike the homework, you are not allowed any form of collaboration on the mid-term. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-23T15:04:04Z SecondChildTAG: Oh and Myrimit, could you take a look at this question? https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/508573b08669711f0000010c I'm having a heck of a time with it. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-23T15:06:49Z SecondChildTAG: Hi JSChambers ! Yes, sure. I will take a look and answer you at night when I back home :). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-23T15:34:49Z SecondChildTAG: thanks for the tips :) SecondChildUserIdTAG: 226085 SecondChildUserNameTAG: Teto SecondChildCreateTimeTAG: 2012-10-23T16:16:57Z SecondChildTAG: Good news Guys..... The deadline to submit the midterm exam is extended to 23:59 (11:59 pm) on Sunday, October 28th, Boston time. The exam will still be released on Thursday, October 25th at 00:01 (12:01 am) Boston time. SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-23T16:31:50Z SecondChildTAG: Hi JSChambers ;). I have answered you in your Post as I have promesed you. I hope that can help you. See you! Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-23T23:06:37Z FirstChildTAG: Remember your circuit patterns, such as the Voltage Divider Rule, too! FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-10-24T06:37:45Z IndexTAG: 710 TitleTAG: How can I access 6.002x once the course has ended? I would like to save the resources available in 6.002x to my hard drive, so that I may use them even after the course has ended. I hope that edX doesn't decide to do this, but I am assuming that they will remove all of the course material once the course has been completed. The reason I will need these resources for a later date is because I simply don't have the mathematical aptitude quite yet to begin studying circuits and electronics. I gave the course a try a few weeks ago when it began, but I simply didn't have any clue of what was going on and decided to learn mathematics up to differential equations as well as learn the fundamentals of physics before beginning a course on circuits and electronics. If anyone can help me to save the content to my hard drive, or somehow access the content even if edX removes it, then thank you. UserIdTAG: 213706 UserNameTAG: TheLibrarian- CreateTimeTAG: 2012-10-23T02:55:30Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I believe, as with past years, that the policy will be to leave the materials online and accessible. I do not speak for staff, just reiterating what was done before. FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-10-23T03:27:48Z SecondChildTAG: how can download the text book........?????????? SecondChildUserIdTAG: 346003 SecondChildUserNameTAG: anupamshakya SecondChildCreateTimeTAG: 2012-10-23T08:55:33Z FirstChildTAG: how can download the text book........?????????? FirstChildUserIdTAG: 346003 FirstChildUserNameTAG: anupamshakya FirstChildCreateTimeTAG: 2012-10-23T08:55:45Z SecondChildTAG: I believe that you can't, except if you buy it... SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-10-23T09:53:19Z SecondChildTAG: you can view html code and use get to download it SecondChildUserIdTAG: 231676 SecondChildUserNameTAG: Roosemberth SecondChildCreateTimeTAG: 2012-10-23T11:50:55Z SecondChildTAG: How do you do that? SecondChildUserIdTAG: 154440 SecondChildUserNameTAG: Aahlad SecondChildCreateTimeTAG: 2012-10-23T12:23:24Z IndexTAG: 711 TitleTAG: H6P2 Small Signal Diagram Can someone show me the correct small-signal diagram for this circuit? Mine is obviously wrong since I got the wrong answer from it. UserIdTAG: 21541 UserNameTAG: JSChambers CreateTimeTAG: 2012-10-22T16:26:24Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: Deriving the small signal circuit is basically about replacing all the elements in the circuit with their equivalent linearized ("small-signal") models. We've already discussed what the small signal model of a MOSFET looks like extensively-- it is just a dependent current source, with a value of current given by gm*vgs. And what do we do with the sources? Well, if it's a purely DC source, like VDD, it gets shorted out and acts like a connection to ground in the small signal model. But if it's a DC biased source with a tiny AC variation on top of it, like vIN, the small signal model completely shorts out the DC bias and all that's left is the AC source, vin. So, simply redraw the circuit, but replace the MOSFETs with dependent current sources and the voltage sources with their small signal equivalents. FirstChildUserIdTAG: 390034 FirstChildUserNameTAG: kahlil FirstChildCreateTimeTAG: 2012-10-22T23:44:39Z SecondChildTAG: So it's just two VCCSs in series, each with a current of gm*vgs. One of them had the gate tied to the drain, and the other has the AC source $vin$ between the gate and the source? SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-23T00:56:11Z SecondChildTAG: Is that right? SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-23T01:51:01Z SecondChildTAG: Q2 should be replaced with resistor 1/gm SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-23T16:17:44Z SecondChildTAG: my records in comment: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5085744f6f3be62300000126 SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-23T17:06:03Z SecondChildTAG: Yeah, I saw the example on page 417. I guess my question is why? On page 416, the small-signal model equivalent of the MOSFET is a VCCS. Why does that change when the gate is tied to the drain? SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-23T17:36:24Z SecondChildTAG: because there is no voltage control anymore - it becomes depends on itself drain-source voltage: ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13510141134028244.png SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-23T17:42:10Z SecondChildTAG: hope it will help: ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13510164851343686.png SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-23T18:21:45Z FirstChildTAG: Hi JSChambers! Ok, here what I have promised to you :) (as this is after the deadline of this Homework, I guess I can explain you How to obtain the Small Signal Circuit for the H6P2). They are giving you this Circuit: ![IMAGE][1] If you go to page 417 of the Textbook [Read here][2], you will noticed that the Q2 Is like the Figure 8.11, the Gate and Drain terminals are conected together. If you keep reading, you will observe that you can represent that Q2 as a resistance in the small signal model (as G and D are conected together). Also, remember that a constant value voltage source in large signal is a groud in a small signal. ![im1][3] So, parcially your circuit will be like this: ![im2][4] Now, lets see what happens with Q1 . Go to page 417 [read here][5] and take a look the small signal for the mosfet and replace that model in Q1 (pay attention to the green lines: G,D and S correspondent terminals): ![IM3][6] Now, as we are studying this in small signal, we will have in our input our vin only. If you see, our input is in the terminals Gate Source of Q1. If you see in the orange box in the right, you will see the small signal representation of a MOSFET, in order to understand this better, I re-draw that representation as you can see (see right to the blue arrow circuit). ![IM4][7] Now, lets see this more friendly JSChambers, lets rotate our resistor (Q2 behaves like a resistor in small signal), ![im5][8] We will obtain something like this: ![IM6][9] Lets clean this, and get the following. Finally, we have our small circuit model! ;) ![final][10] I hope this can help you, See you, Myriam. P.D 1: Sorry for my english haha. I have to back to my classes of english :P. But more or less that is the idea of how can you obtain the small circuit model :). [1]: https://www.edx.org/static/content-mit-6002x/images/circuits/phase-inverter.3bb223a10913.gif [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/441 [3]: https://edxuploads.s3.amazonaws.com/13510325701343632.png [4]: https://edxuploads.s3.amazonaws.com/13510326591343676.png [5]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/440 [6]: https://edxuploads.s3.amazonaws.com/13510327441343659.png [7]: https://edxuploads.s3.amazonaws.com/13510329435648297.png [8]: https://edxuploads.s3.amazonaws.com/13510331401343659.png [9]: https://edxuploads.s3.amazonaws.com/13510332401343613.png [10]: https://edxuploads.s3.amazonaws.com/13510333051343617.png FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-23T23:04:42Z SecondChildTAG: Thank you. Myriam, don't worry about your english. Your skill in the class more than makes up for it. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-24T00:13:55Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-24T00:24:32Z SecondChildTAG: Agarwal better watch his back. Soon Myrimit will be teaching this class. ;-) SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-24T03:29:34Z SecondChildTAG: Do you chosen a dog yet Miriam? SecondChildUserIdTAG: 9673 SecondChildUserNameTAG: xp42 SecondChildCreateTimeTAG: 2012-10-24T07:43:17Z SecondChildTAG: Haha @ planetscape and @xp42! :). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-24T20:31:24Z FirstChildTAG: Just wanted to say thank you to kahlil, xp42, YakovO, and of course Myrimit. I don't know why I had such a mental stumbling block with this issue, but I could not clear it up without your help. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-24T00:16:18Z SecondChildTAG: You are welcome JSChambers! The important is that you have clarified your doubts, that is very valuable :). My best wish to you! Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-24T00:28:58Z FirstChildTAG: **HINT:** The MOSFETs in this problem **are not** NewFET's. FirstChildUserIdTAG: 9673 FirstChildUserNameTAG: xp42 FirstChildCreateTimeTAG: 2012-10-23T16:00:44Z SecondChildTAG: I know, the question has nothing to do with the question above it. What I am having trouble with is visualizing the small-circuit model for the phase inverter circuit. I guessed the answers, easily, so I got all the questions right, but I can't seem to go back and draw the small signal model correctly. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-23T17:11:12Z SecondChildTAG: The upper mosfet is a pull-up resistor! Notice that $v_{ds} = v_{gs}$ and $i_{ds} = g_mv_{gs}$, and since the gate and drain are connected, we can replace the small-signal 3-terminal mosfet model with a 2-terminal model (drop either the drain or gate, your choice: just keep the notation straight for the 2 remaining terminals you keep). So the upper mosfet appears/behaves as a 2-terminal device with a current flowing through it equal to $g_m$ times the voltage across the two (retained) terminals. Therefore, $$r_{ds} = \frac{ v_{ds}}{i_{ds}} = \frac{ v_{gs}}{i_{ds}} = \frac{ v_{gs}}{g_mv_{gs}} = \frac{1}{g_m}$$ SecondChildUserIdTAG: 9673 SecondChildUserNameTAG: xp42 SecondChildCreateTimeTAG: 2012-10-23T20:01:43Z SecondChildTAG: Ohhhhhh. Because the Gate and Drain are connected you no longer have a 3 terminal device, so now you just drop the VCCS model and go to a resistor. So now the small signal model of this circuit would be a pullup resistor in series with a VCCS with that has a small AC signal. Is that right? SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-23T20:15:30Z SecondChildTAG: You still have a 3-terminal device, but when you short two of the terminals some of it's behavior changes, and you can consider it a two-terminal device. Analyze what is left, and that must be true. What I didn't discuss is whether it is a variable resistor. For that you need to see the equation for $g_{m}$. Since $G_{m}$ has only DC voltage terms, it is fixed for a given operating point, but will change if the DC operating point is shifted. If you used the NewFET in this configuration, then the small-signal model would have the $r_o$ resistor in parallel with the $g_mv_{gs}$ current source and the analysis would be something different. SecondChildUserIdTAG: 9673 SecondChildUserNameTAG: xp42 SecondChildCreateTimeTAG: 2012-10-23T20:36:26Z SecondChildTAG: Lyla, we really need to be able to edit comments. That should be $g_m$ above, not $G_m$. SecondChildUserIdTAG: 9673 SecondChildUserNameTAG: xp42 SecondChildCreateTimeTAG: 2012-10-23T20:37:38Z SecondChildTAG: look VI in my comment above SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-23T22:06:50Z FirstChildTAG: Although the topic of the thread is small signal analysis, It should be pointed out that the circuit functions as a phase inverter for either small or large signals as long as Q1 stays in saturation. When Q1 is in saturation VDS of Q2 follows VIN if the MOSFETS are matched. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-24T11:44:02Z IndexTAG: 712 TitleTAG: By what age did you start learning the Calculus? By what age did you start learning the Calculus? UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-10-22T15:53:50Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 8 FirstChildTAG: Ummm what time is it? FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-22T15:58:55Z SecondChildTAG: Haha, wise answer! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-22T16:16:38Z SecondChildTAG: Were you just replying with an arbitrary question because the original question was random or was there something I didn't get? SecondChildUserIdTAG: 154440 SecondChildUserNameTAG: Aahlad SecondChildCreateTimeTAG: 2012-10-22T16:39:17Z SecondChildTAG: It means I'm just learning calculus now. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-22T17:12:14Z SecondChildTAG: Subtle British wit SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-22T17:17:07Z SecondChildTAG: I'm not British, I'm American. I hope to achieve Britishness one day, though. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-22T17:21:21Z SecondChildTAG: Me, too, learning as I go through this course. Hope it doesn't end with this course, though. SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-23T03:32:35Z SecondChildTAG: Ohh, lol. SecondChildUserIdTAG: 154440 SecondChildUserNameTAG: Aahlad SecondChildCreateTimeTAG: 2012-10-23T08:15:03Z FirstChildTAG: 14-ish maybe 15. Isn't that kind of random though? FirstChildUserIdTAG: 154440 FirstChildUserNameTAG: Aahlad FirstChildCreateTimeTAG: 2012-10-22T16:06:57Z FirstChildTAG: 17, senior year of high school. It was a very good applied course, then freshman year of college had the rigorous theory with the epsilons and deltas. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-22T16:11:18Z FirstChildTAG: I took calculus in the 9th grade in the U.S., so that would be 13-14 years old. I skipped two grades in math, so most of my peers didn't start calculus until 11th grade in the U.S. Math was my strong point, but I skipped out on Physics until university (wish I took it earlier!) Weird thing is that I did not take the Advanced Placement (AP) exam in High School for Calculus (I did take the class though). Passing the exam would mean skipping a year of calculus in University. I had to re-take "calc" again at the university. Since I had to re-take the material I already knew in my Freshman College year it gave me **two unexpected benefits**: (1) instead of just getting the credit via the AP exam, I also got 2 A's (2 x 4.0) to boost my grade point average (GPA); and (2) I got a lot of free time since I could skip my "calc" class whenever they discussed material I already knew. If it was a nice day, I'd go on a picnic with my girlfriend instead of staying in class! Since I was going full-time, tuition was the same whether I would take 12, 15, or 18 credits. So for you U.S. high-schoolers thinking about taking the AP exam, there are *hidden drawbacks* (you don't get a grade that counts to your university GPA) as well as the *obvious benefits* (you get the 3-6 university credits depending on your exam results, and if you are paying on a per-credit basis, e.g. part-time students, you save money). Just something to think about...Mark in N.J., USA FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-10-23T02:47:29Z SecondChildTAG: I never recommend to 'test out' of Calculus. For Science & Engineering majors, it is the bedrock that almost all of your courses will rely upon. As JerseyMark pointed out, if you know it, use it to pad your GPA, and practice. I know for EE students, the Jr year can shred a GPA. SecondChildUserIdTAG: 13125 SecondChildUserNameTAG: JMigdal SecondChildCreateTimeTAG: 2012-10-23T04:20:59Z FirstChildTAG: So I guess my next question is: How hard was it? FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-10-23T08:53:45Z SecondChildTAG: not hard at all..:p SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-23T09:56:01Z FirstChildTAG: Round about **15-years**. FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-10-23T10:03:20Z FirstChildTAG: Here in India we learn it in the beginning of 11th grade. (for students o fage -around 15 years) FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-10-22T16:35:24Z SecondChildTAG: sometimes even standard 10th...span993 SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-23T09:55:36Z FirstChildTAG: I think answers to the question about how hard calculus is should be taken with caution because calculus courses vary greatly with regard to both content and rigor. My high school course was strong on the mechanics of differentiation and integration and their basic applications like maxima and minima, areas, volumes, arc lengths, etc. My college course was more rigorous with theorems and their proofs and a lot of epsilon-delta analysis. Both were good, but they were different. The amount of abstract thinking required for my college course was a challenge at age 18, and I could not have handled it when I was in high school. I don't think there are many 15 and 16 year olds that could have handled that type of course. I would further note that both the high school and college course were both honors type courses with only the top 5% or so of the class enrolled The mechanics of taking derivatives is easy, as is the integration of those basic functions (anti-derivatives). Beyond that integral calculus becomes a bag of tricks with various substitutions. Some schools include differential equations as part of their calculus sequence calling it calculus IV or similar; others list it as a separate course. I found found my differential equations course to be very difficult because so much theory was mixed with the basic techniques. Nevertheless, it gave me a good foundation that I hold on to nearly fifty years later. I have looked at the MIT video courses on calculus and differential equations and find them to have much less content that when I started college. I also note that the European students that I met in graduate school were generally better prepared in math than the American students. At least that was the case in chemical engineering. I can't speak for American electrical/electronic engineers. Finally, as far the amount of calculus that you need for this course ... very little. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-23T15:39:29Z SecondChildTAG: Yeah, and they vary widely in what I think is the most important category: an ability to communicate the information. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-24T17:10:47Z SecondChildTAG: Of course, this is not limited to basic calculus courses. At the university I had my share of truly horrible courses. There were occasions when I was able to audit the course later for free when the course was taught by a better teacher. There were others that I was glad to be finished with and never wanted to see them again. In the last category was a course in algebraic topology that I was talked into by a professor who had previously taught me. I cannot begin to tell anyone what that course was about. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-24T18:17:06Z IndexTAG: 713 TitleTAG: Midterm Info In another post someone remarked that we get three tries in the midterm for our answers, does anyone know where they mentioned that? Note: In 3.091x they mention that in the edx tutorial lecture sequence. UserIdTAG: 154440 UserNameTAG: Aahlad CreateTimeTAG: 2012-10-22T12:22:28Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: hi, **Aahlad** it's not mention anywhere but this news is very authentic that we have only 3 attempts to solve any query in **MIDTERM EXAM**, [MIDTERM Info of MITx Spring FALL-2012][1] [1]: https://6002x.mitx.mit.edu/wiki/view/MidtermInfo after checking above link it is very clear that v have only 3 attempts... > **Regards:** asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-10-22T12:38:20Z FirstChildTAG: Hmm...maybe. By the way in 3.091x you have 7 attempts. FirstChildUserIdTAG: 154440 FirstChildUserNameTAG: Aahlad FirstChildCreateTimeTAG: 2012-10-22T13:12:28Z SecondChildTAG: they like prime numbers :) SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-22T15:43:56Z SecondChildTAG: Lol. SecondChildUserIdTAG: 154440 SecondChildUserNameTAG: Aahlad SecondChildCreateTimeTAG: 2012-10-22T16:26:14Z SecondChildTAG: It will be completely unfair if the answers were expression type, coz I never get it in 3 chances SecondChildUserIdTAG: 148523 SecondChildUserNameTAG: span993 SecondChildCreateTimeTAG: 2012-10-22T16:37:46Z FirstChildTAG: We're giving three chances again in this one. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-10-23T01:08:36Z SecondChildTAG: Thanks for the confirm! SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-23T03:36:37Z SecondChildTAG: Thanks Lyla! SecondChildUserIdTAG: 154440 SecondChildUserNameTAG: Aahlad SecondChildCreateTimeTAG: 2012-10-23T08:15:45Z SecondChildTAG: Thanks Lyla SecondChildUserIdTAG: 179140 SecondChildUserNameTAG: JoydeepSil SecondChildCreateTimeTAG: 2012-10-23T08:27:34Z SecondChildTAG: Good news Guys..... The deadline to submit the midterm exam is extended to 23:59 (11:59 pm) on Sunday, October 28th, Boston time. The exam will still be released on Thursday, October 25th at 00:01 (12:01 am) Boston time. SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-23T16:35:28Z IndexTAG: 714 TitleTAG: @ STAFF / MYRIMIT Hi teachers, Myrimit, I have a couple of doubts: 1. GRADING: for the collection of the 10 out of 12 best marks, what does exactly count: A) pack of homework + lab (10 best out of 12) OR B) Homework (10 best out of twelve)and LABS (10 best out of twelve) independently ? 2. "Students must know basic calculus and linear algebra and have some background in differential equations" - Is it going to me MUCH HARDER on the second half of the course? 3. I've heard Midtearm exam will be ok (not very easy, not very hard) but that the final exam will become very very very very HARD, that even PhD's couldn't sometimes solve. Is it true? it's just to know what will expect me and if I really have a chance keeping with the course. It also depend on my effort, of course, it's my responsibility!!!! Thanks in advance, gracias de nuevo Myriam!! Regards, saludos, Sandra UserIdTAG: 342966 UserNameTAG: SandraNavarro CreateTimeTAG: 2012-10-21T15:17:23Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: The labs and the homework are worth 15% each for a total of 30%, so that (for example) if you didn't do any of the homeworks, but did all the labs, you'd get 15%. If you do all the labs and homework you get 30%. There is more info in the syllabus under the "Course Info" tab. Much harder is, of course, a matter of opinion. Week 6 is hard, especially with mid-terms so close. Week 9 is also hard, math-wise. The later weeks seemed harder to me, simply because my brain was begging for death by that point. I know lots of people who got 100s on both the mid-term and final, so I think the rumors of the final being out of reach for PhD's are exaggerated. Unless their PhD is in Literature. That being said, I found both exams to be quite challenging. The final exam was harder for me because it covered a wider range of topics. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-21T15:52:16Z SecondChildTAG: Hi JS Chambers, thanks for helping. What do you mean "it covered a wider range of topics"? may I ask what are you studying or what did you study? For me it's difficult 'cause I studied business science :-P but still i?m following the course. "quite challenging" --- more than homework of every week? Thanks a lot, Regards, Sandra SecondChildUserIdTAG: 342966 SecondChildUserNameTAG: SandraNavarro SecondChildCreateTimeTAG: 2012-10-21T16:06:55Z SecondChildTAG: "Covered a wider range of topics" means that the Midterm only covers topics up to and including the first half of week 6. The final, of course, can test for subjects from any week. It's about the same as a hard homework week, but you only have 24 hours to do it in instead of a week. Of course, that's just my personal opinion. I would say, don't fret about it so much. Just do your homework and labs and review for it, and try not to worry about how hard it will be. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-21T16:23:04Z SecondChildTAG: Thank u!!!!!!!! You helped a lot :)) Regards! SecondChildUserIdTAG: 342966 SecondChildUserNameTAG: SandraNavarro SecondChildCreateTimeTAG: 2012-10-21T17:23:20Z FirstChildTAG: Hola Sandra! Cómo estás? Respecto a tus dudas, básicamente es lo que te ha respondido JSChambers :). **1-** En cuanto a la ponderación de los Labs y Homeworks, es la siguiente: - Si bien tienes 12 Homeworks, ellos te perdonan que no hagas 2 Homeworks. La palabra perdonar es media confusa: perdonar significa que el Alumno que haya entregado 10 Homeworks al 100% podrá recibir la misma nota que aquél que ha entregado los 12 Homeworks al 100%. En resumidas palabras, puedes saltearte 2 Homeworks y no influirá en nada en tu calificación. Aquí también se debe aclarar, que puede suceder que tengas 10 Homeworks al 100% pero 2 Homeworks con 70%, como te perdonan 2 Homeworks, los de nota baja no se te consideran en tu ponderación final. - Lo mismo ocurre con los Labs, aquí, independientemente de los Homeworks, también puedes salterte 2 Labs y no influirá en nada en tu calificación final, es decir, a lo largo del curso, puedes tener 2 Labs sin entregar más otros 2 Homeworks sin entregar. **2-** Si bien este curso menciona que debes tener ciertos pre-requisitos, debo decirte que hubieron muchos alumnos en el curso prototipo que no lo han tenido, lo que sí es que les ha costado más y tuvieron que hacer más esfuerzo, pero no les ha sido un impedimento para terminar el Curso y obtener el Certificado. **3-** En cuanto a los Exámenes, no es tan difícil/complejo como para decir que un PhD no lo pueda resolver, tampoco es fácil: tiene su pizca de sabor para pensar, además algunas cosas se basan en lo que vimos de los Homeworks. Sólo , por curiosidad, de dónde has escuchado eso del PhD? jaja ;) Mi mejor deseo para tí, trata de practicar los ejercicios que hemos visto, repasar conceptos y hacer tranquila el Examen. Te irá bien :). Cualquier duda que tengas avísame, estaré aquí para ayudarte. Un abrazo! Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-21T22:21:26Z SecondChildTAG: Hola Myriam! Muchas gracias por tu apoyo y consejos! no lo había visto hasta ahora!! Lo de que los "PhD tenían serias dificultades para realizar el final exam" según recuerdo lo leí, creo con bastante certeza en el Discussion Forum o si no en alguno relacionado. Después del exámen intentaré buscarlo, jeje Me apunté al curso consciente de los pre-requisitos :)) pero ví que aun y así era factible y me puse a revisar antes del curso conceptos básicos (amplios) de electricidad y electrotécnica. Pero sí, aun y así me cuesta más de lo normal porque no es mi "especialidad". Así que si apruebo el curso(bueno, mejor el 70%, la B) estaría más que contenta!!!! y si no bueno, no pasa nada So, thanks again a lot for your support!! You're making a double effort in helping students (spanish - english).... BTW do you know somewhere (link, for instance), where I can practice some derivatives? I found someones but are not really good. Un abrazo, gracias! Sandra SecondChildUserIdTAG: 342966 SecondChildUserNameTAG: SandraNavarro SecondChildCreateTimeTAG: 2012-10-26T08:35:05Z SecondChildTAG: :). Seguro que te irá bien! Mi mejor deseo para ti! Suerte! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-27T02:33:16Z IndexTAG: 715 TitleTAG: H6P2 I am stuck on parts 5 and 6... any hints? UserIdTAG: 266739 UserNameTAG: satvikchugh CreateTimeTAG: 2012-10-20T10:53:57Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Formulas I enput using the given symbols are delared invalid. Please help. FirstChildUserIdTAG: 498458 FirstChildUserNameTAG: jefwa FirstChildCreateTimeTAG: 2012-10-20T13:06:54Z SecondChildTAG: maybe you arent using * where u want to multiply or check your parenthesis and remember double check the symbols you are using SecondChildUserIdTAG: 266739 SecondChildUserNameTAG: satvikchugh SecondChildCreateTimeTAG: 2012-10-20T13:10:21Z FirstChildTAG: Do you know how to draw the small signal model of the circuit? That is really the key to this question. Doing so is simple-- replace the MOSFETs with their small-signal models (current sources with the value gm*vgs), replace the total signal source vIN with the small signal source vin, and short out VDD to ground. Now, using the fact that 1: gm1 = gm2 and 2: the voltage at the vout node is also the voltage at the source of Q2, we should be able to write an equation for this circuit involving vout, which will give us the answer. For the resistance, we have to do is find the thevenin resistance looking in through the output port. So that involves injecting a test current i_test into vout and solving for the ratio vout/i_test. FirstChildUserIdTAG: 390034 FirstChildUserNameTAG: kahlil FirstChildCreateTimeTAG: 2012-10-20T12:07:27Z SecondChildTAG: cant digest gm1=gm2.. help.:( SecondChildUserIdTAG: 266739 SecondChildUserNameTAG: satvikchugh SecondChildCreateTimeTAG: 2012-10-20T13:20:04Z SecondChildTAG: We solved for this already in the previous parts of the problem, don't you remember? Unless you were doing the parts out of order in which case, you can just safely assume that. As for why it makes sense... well, remember that both devices are constrained to operate in their saturation region, and because they're identical devices, connected in series (which means they have to have the same current), it should make sense that they have the same transconductance SecondChildUserIdTAG: 390034 SecondChildUserNameTAG: kahlil SecondChildCreateTimeTAG: 2012-10-20T14:09:32Z SecondChildTAG: hell yeah..ans was right there..i was just messing up the signs Thanks a ton kahlil..:) SecondChildUserIdTAG: 266739 SecondChildUserNameTAG: satvikchugh SecondChildCreateTimeTAG: 2012-10-20T14:22:07Z SecondChildTAG: "phase inverter" u dont even need to solve it..:P SecondChildUserIdTAG: 266739 SecondChildUserNameTAG: satvikchugh SecondChildCreateTimeTAG: 2012-10-20T14:24:19Z SecondChildTAG: WOW, do you mean to tell me it was that simple??? Now I'm able tie the answer to to why the circuit is called a phase inverter. I guess I should have put the circuit in the Sandbox. And now for the reason why everyone is having so much difficulty!! I want to just SCREAM - AHAHAH. Did anybody hear that? LOL How did we come to the point thinking the answer must include all the variables from the question? Probably because up to this point that is exactly what we did, at least I did. Until now. So, I guess what this means is the variables given in the questions are just hints, but it doesn't necessary mean they must be in the answer. This makes it more difficult to find the answers, however it makes one understand the material better, instead of just trying to manipulate the variables by plugging and chugging. Now, a question for you kahlil. For this problem, can the answer include all the variables included from the question and still be correct, or must it be exactly what I got from following your directions above? Can the staff put out an explanation on this topic so everybody doesn't spend hours and hours trying to get the variables in the answer to match that of the variables in the question, or has it already been addressed somewhere? Thanks for your help kahlil, you're outstanding. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-21T18:47:44Z FirstChildTAG: Kahlil - THANKS ALOT! I got it Vth..Thats crazy.. FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-10-20T15:26:07Z SecondChildTAG: @Sergtronix- can u give me a hint to find vth? im unable to solve. SecondChildUserIdTAG: 337430 SecondChildUserNameTAG: aparna_b SecondChildCreateTimeTAG: 2012-10-21T13:11:58Z IndexTAG: 716 TitleTAG: H8P3 Hello, I've got a difficulty with the exercice 3 of H8. Can someone explain me how to abstract the MOSFET circuit to a Series RC Circuit? Or how should I resolve this exercice. Thanks. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-19T17:21:19Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Model FET's that are on as resistors with given Ron values and FET's that are off as open circuit. (Don't forget the gate-source capacitance of Q2) FirstChildUserIdTAG: 270011 FirstChildUserNameTAG: SnowmanZA FirstChildCreateTimeTAG: 2012-10-19T19:44:43Z FirstChildTAG: The correct answer for VGS2Max doesn't have any connection with the capacitor Cgs2 and it's an extremely simple problem. We know that the gate of Q3 is connected to the drain of Q1.If you draw it you will notice that you will have a diode connected mosfet, with VGS3 = VDS3. We know that VGS3 > VT, so VDS3 >=1 .If you write KVL you will get that VGS2 = Vs -VDS3 => VGS2max = Vs - VDS3min .So, VGS2max = Vs - 1. Vs is not really 5V due to Roff.It is 4.99947, but insignificant smaller . ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13513596791343676.gif FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-10-27T17:49:23Z IndexTAG: 717 TitleTAG: GOOD NEWS:both tutorials and lecture sequences in courseware can be downloaded ! For those who have video problems due to kinds of reasons,I have just noticed that both tutorials and lecture sequences can be downloaded.It is really a good news for me!There is a link "Download video **here**" below the video sequence. UserIdTAG: 361137 UserNameTAG: Ericson CreateTimeTAG: 2012-10-18T15:59:38Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: how?? FirstChildUserIdTAG: 128409 FirstChildUserNameTAG: arijitbme FirstChildCreateTimeTAG: 2012-10-18T16:02:25Z SecondChildTAG: click" courseware",for example,"week 6",then"Week 6 Tutorials",you will find "download video **here**" SecondChildUserIdTAG: 361137 SecondChildUserNameTAG: Ericson SecondChildCreateTimeTAG: 2012-10-18T16:11:21Z SecondChildTAG: Hi arijitbme, Take a look to the red circle in the following image. If you click on it, it will download that video. ![enter image description here][1] I hope this can help you! :) [1]: https://edxuploads.s3.amazonaws.com/13506155391343676.png SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-19T03:00:48Z SecondChildTAG: thanks for the help... SecondChildUserIdTAG: 428658 SecondChildUserNameTAG: Moza SecondChildCreateTimeTAG: 2012-10-24T06:11:55Z IndexTAG: 718 TitleTAG: Hello Does someone lives here? UserIdTAG: 155265 UserNameTAG: OBazan CreateTimeTAG: 2012-10-18T04:23:04Z VoteTAG: 4 CoursewareTAG: Week 6 / Norton circuit capacitor exercise CommentableIdTAG: 6002x_norton_capacitor_circuit_exercise NumberOfReplyTAG: 1 FirstChildTAG: Sure.. People do live in cyberspace. FirstChildUserIdTAG: 678074 FirstChildUserNameTAG: pentest FirstChildCreateTimeTAG: 2012-10-18T07:22:58Z IndexTAG: 719 TitleTAG: H6P2 - Minimum VDD Hi everybody!!! I´m lost. I don´t know what are the conditions I have to consider to get the minimum value of VDD to ensure that both mosfet are working in the saturation region. I thought that VDD>(VGS2-VT2)+(VGS1-VT1) but I get a not very logical solution. Any hint or advice? Thanks in advance. Daniel from Spain UserIdTAG: 261081 UserNameTAG: DanielSP CreateTimeTAG: 2012-10-17T14:30:38Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: First thing to note is that the answer to the second part has nothing to do with the answer to the first part. I wasted much time thinking that it did. IMHO the question is not well worded. Things to consider: 1. The sum of the voltage drops across the two MOSFETs is equal to VDD. 2. iDS is the same for both MOSFET's. 3. vIN determines iDS. 4. For Q2 VDS = VGS, which is determined by iDS. 5. For Q1 VDS >= VGS-VT in order to remain in saturation. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-17T15:34:00Z SecondChildTAG: I got you!!! Thanks for the advice. So easy!!! Thank you very much, you really showed me the way!!!! SecondChildUserIdTAG: 261081 SecondChildUserNameTAG: DanielSP SecondChildCreateTimeTAG: 2012-10-17T16:06:13Z SecondChildTAG: My pleasure! SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-17T16:12:45Z SecondChildTAG: Fellas, please can you help me. I still got an indetermined expression, something like Vdd=Vdd+ Vt, and there's no possible answer. I feel I'm trapped in this point SecondChildUserIdTAG: 376173 SecondChildUserNameTAG: nacho110987 SecondChildCreateTimeTAG: 2012-10-17T19:14:45Z SecondChildTAG: Think about the list that I gave and ask any questions that you have about them. I will be away from the computer about two hours, but I will check when I return. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-17T19:59:22Z SecondChildTAG: > **skyhawk** why are u taking VDS=VGS in step 4 SecondChildUserIdTAG: 97359 SecondChildUserNameTAG: azeem179 SecondChildCreateTimeTAG: 2012-10-18T14:00:26Z SecondChildTAG: Because the gate is connected to the drain. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-18T16:15:03Z SecondChildTAG: Que buena recomendación. SecondChildUserIdTAG: 197569 SecondChildUserNameTAG: darksamaro SecondChildCreateTimeTAG: 2012-10-18T22:07:05Z SecondChildTAG: Hey Skyhawk, I can only get as far as (vIN - VT)^2 = (vDS2 - VT)^2 From there it's just a mess of quadratic equations that isn't helpful. What's the equality that I'm missing. Your hints, while useful, just don't quite help my brain make the leap. SecondChildUserIdTAG: 166031 SecondChildUserNameTAG: krebryna SecondChildCreateTimeTAG: 2012-10-19T20:55:54Z SecondChildTAG: Nevermind Skyhawk, I was being a retard, as usual. Figured out the equality I was missing. SecondChildUserIdTAG: 166031 SecondChildUserNameTAG: krebryna SecondChildCreateTimeTAG: 2012-10-19T22:37:09Z SecondChildTAG: krebryna: try to solve equation for Vds2, you will get one correct root. SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-10-20T10:00:14Z SecondChildTAG: I went for that steps 1) VDD=VDS1+VDS2 ---------------- 2) ID=ID1=ID2 ------------- 3) ID1=K/2*(Vin-VT)^2 --------------------- 4)ID2=K/2*(VDD)^2 ----------------- K/2*(Vin-VT)^2=K/2*(VDD)^2 --------------------------------- Vin-VT=VDD ---------- But it isn't correct (((( HELP PLZ SecondChildUserIdTAG: 192114 SecondChildUserNameTAG: chuba SecondChildCreateTimeTAG: 2012-10-20T17:29:18Z SecondChildTAG: I understood) just remember that Vgs2=VDD-Vout; Vout = Vin - Vt SecondChildUserIdTAG: 192114 SecondChildUserNameTAG: chuba SecondChildCreateTimeTAG: 2012-10-20T17:36:45Z SecondChildTAG: vdd=vgs1+vgs2 now put values as said by skyhawk in point 4 and 5 and solve but to remember vgs= vout+vt and vout=vin-vt then solve inshallah u will get the answer SecondChildUserIdTAG: 164689 SecondChildUserNameTAG: muhammadfaizan SecondChildCreateTimeTAG: 2012-10-21T15:20:31Z FirstChildTAG: Thanks Buddy,,, FirstChildUserIdTAG: 408534 FirstChildUserNameTAG: kkashyap FirstChildCreateTimeTAG: 2012-10-19T04:38:43Z SecondChildTAG: Hi I don't understand. Please help me. So, I try used list. 1) VDD=VDS1+VDS2 2) iDS=iDS1=iDS2 3) iDS1=K/2*(VIN-VT)^2 4) iDS2=K/2*VGS2^2 It's right? 5) VDS2=VGS2 if iDS1=iDS2 then we have VDS2=VIN-VT Then need to find VDS1. I know iDS1 but how find resistance? I can find gm1 whis analising small signal. Then I fall into a stupor, because i don't understand what to do. Sorry for my english SecondChildUserIdTAG: 413517 SecondChildUserNameTAG: mari_safonova SecondChildCreateTimeTAG: 2012-10-19T12:21:27Z SecondChildTAG: 4) iDS2=k/2*(VGS2-VT)^2 it might be helpful. thanks SecondChildUserIdTAG: 21193 SecondChildUserNameTAG: Oboneel SecondChildCreateTimeTAG: 2012-10-19T18:00:10Z SecondChildTAG: Thanks!!! SecondChildUserIdTAG: 362299 SecondChildUserNameTAG: anandbaskaran SecondChildCreateTimeTAG: 2012-10-20T18:12:50Z SecondChildTAG: Hi. Could anyone explain me how Q2 may work in the saturation region where VDS >= VGS-VT. I thihk it will always be in the triode region. SecondChildUserIdTAG: 213452 SecondChildUserNameTAG: JoJosida SecondChildCreateTimeTAG: 2012-10-22T04:58:16Z IndexTAG: 720 TitleTAG: That hellish expression in H6P1 part 1 We are asked to derive an expression, result from solving a 2nd grade equation, and then write it down on that field, in an algebraic manner. There's a lot of ways of expressing such a huge formula, does the grade consider that? I've been trying in all ways, even deriving with the computer, but it won't say it's right. Am I supposed to put everything in one fraction, eliminate square roots, or something? UserIdTAG: 482646 UserNameTAG: elgambitero CreateTimeTAG: 2012-10-16T17:45:09Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 4 FirstChildTAG: I'm still getting nothing. I tried to do the whole calculations with a calculus program. Nothing. Grader will not say it's correct. I don't know if I'm misinterpreting the problem or what, I don't see any other way of solving it, it's pretty simple... FirstChildUserIdTAG: 482646 FirstChildUserNameTAG: elgambitero FirstChildCreateTimeTAG: 2012-10-16T19:26:26Z SecondChildTAG: Why don't you give us your starting equation, and we can step you through the derivation. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-16T19:31:50Z SecondChildTAG: Yes, it is simple. It takes me four lines. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-16T19:32:46Z FirstChildTAG: Hey i have the same problem....can you give some hints, what you need to do...I don't understand the hint they gave us.. FirstChildUserIdTAG: 308861 FirstChildUserNameTAG: Ignaas FirstChildCreateTimeTAG: 2012-10-16T17:50:19Z FirstChildTAG: I think grader has equation invariant calculator, so any equivalent equation will be considered. FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-16T17:52:36Z FirstChildTAG: Hi elgambitero, Derive it means in this context that you should find/deduct in terms of ... and not derive $\frac{ \delta y }{\delta x }$ literally a expression ;). I hope this can help you. Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-17T01:52:26Z SecondChildTAG: I think it will help. I mean, a lot. English is not my native language and I was convinced that I needed to do the derivative. I'll try when I arrive home, I can still remember the expression I got on the quadratic expression from derivating it too much times. SecondChildUserIdTAG: 482646 SecondChildUserNameTAG: elgambitero SecondChildCreateTimeTAG: 2012-10-18T07:05:55Z SecondChildTAG: It worked! Thank you!!! SecondChildUserIdTAG: 482646 SecondChildUserNameTAG: elgambitero SecondChildCreateTimeTAG: 2012-10-18T07:22:55Z SecondChildTAG: Well done elgambitero! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-19T03:38:24Z IndexTAG: 721 TitleTAG: Look here for help with the calculus https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/507d2b565ae43d1f00000140 UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-10-16T09:42:34Z VoteTAG: 4 CoursewareTAG: Week 6 / Small-signal mosfet exercise CommentableIdTAG: 6002x_small_signal_mosfet_exercise NumberOfReplyTAG: 1 FirstChildTAG: Thank you so much! FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-10-19T00:53:02Z IndexTAG: 722 TitleTAG: Who is the speaker Who is her? she's not in course staffs i think. Am I right? UserIdTAG: 285241 UserNameTAG: Kazemi CreateTimeTAG: 2012-10-15T12:05:44Z VoteTAG: 4 CoursewareTAG: Week 5 / Large Signal Analysis of MOSFET Amplifier CommentableIdTAG: 6002x_large_sig_MOSFET_amp_t NumberOfReplyTAG: 0 IndexTAG: 723 TitleTAG: STAFF: H5P2...correct results for vOUT but won't accept my formula for iDS. Ok, folks...could be worse, I suppose. The input form will NOT accept my formula for iDS but all three of my answers for vOUT ARE being accepted. It seems to be a problem with formula length. Once it gets too long it resets my screen and some of what I've typed in is just gone. My other formulas were accepted. I do suspect a problem of some kind with the software. EDIT: one thing I could try is to look for some terms to cancel, but that will only make a difference of...say...a half dozen characters? I wouldn't expect the buffer space to be so precious as to make a difference of a cancelled term or two. UserIdTAG: 92346 UserNameTAG: MobiusTruth CreateTimeTAG: 2012-10-13T23:29:38Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 2 FirstChildTAG: Hi MobiusTruth! How many terms do you have in your equation? Are getting the minimus expression of the equation? FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-14T02:54:04Z SecondChildTAG: Hi, Myriam. I don't think I cancelled everything I could cancel. That's the only thing can think of for now. I'll look at it again tomorrow. Always appreciate a good prompt! It's interesting: this is where I first ran into some bumps first time around, too. I'm going to have to decide what it's trying to tell me. My expression-manipulation skills aren't what they used to be, I guess. Ack! ;) SecondChildUserIdTAG: 92346 SecondChildUserNameTAG: MobiusTruth SecondChildCreateTimeTAG: 2012-10-14T03:07:42Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T03:42:31Z SecondChildTAG: I have a similar problem too , its showing : Could not parse '(vIN-VT) / RS + (1 / (K * RS^2)) * (1-sqrt(1+2K * RS * (vIN-VT) ))' as a formula SecondChildUserIdTAG: 91261 SecondChildUserNameTAG: mr_sophisticated SecondChildCreateTimeTAG: 2012-10-14T08:30:54Z SecondChildTAG: Put an operator between 2 and K (2*K) and it will work! parsing error is because of forgotten operators! good luck SecondChildUserIdTAG: 394240 SecondChildUserNameTAG: ManU81 SecondChildCreateTimeTAG: 2012-10-14T10:17:41Z FirstChildTAG: Tonight I did cancel my RS terms. That shortened the expression by about a half-dozen characters. I did get the same error message as last night, but now I'm getting the message mentioned above, "Cannot parse...." The 'fancy font' display below the line I enter is reading EXACTLY as it should AND the numbers I ran yesterday were all accepted by the grader and I DID perform the obligatory Happy Emerald Checkmark Dance, so the Fates should not be against me...unless one of YOU GUYS made'em mad! :) My next attempt - unless I just sacrifice the points (I'm feeling impatient) would be to see if I can simplify the expression under the radical sign...anyone care to offer a thought if that's worth pursuing? Thanks. FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-10-14T23:47:39Z SecondChildTAG: Hi MobiusTruth, Haha, sorry for asking this... I didn't understand it, so, could you get the green check finally? I got confused with your second paragraph :p SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-15T00:00:24Z SecondChildTAG: Hi, Myrimit...no. I did not finally get the formulas to be accepted. I did run one of them through Wolfram for the derivative and it came back even more complicated looking than my own result for that part. (No way I'd try to put that one into the text box!) I'm thinking I'll just have to take a hit on the grade for the last part of Lesson 5. I know what I want to do but it's just not working out. I am quite confident of my starting point expression for iDS. I am sure that's not the problem. Oh, well. "This too shall pass." :) SecondChildUserIdTAG: 92346 SecondChildUserNameTAG: MobiusTruth SecondChildCreateTimeTAG: 2012-10-15T01:04:35Z SecondChildTAG: Oh... Can I help you? SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-15T01:09:29Z SecondChildTAG: I'm not sure. I've done the algebra a few times. I think I'm ok, conceptually. Maybe I'll ask for some info after the due date, just so I can rehearse things properly. :) SecondChildUserIdTAG: 92346 SecondChildUserNameTAG: MobiusTruth SecondChildCreateTimeTAG: 2012-10-15T01:40:45Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-15T03:03:26Z IndexTAG: 724 TitleTAG: When will we/will we have a mock exam? Are MITx planning on releasing a mock for the midterm exam? UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-10-13T19:25:50Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: You can find prior years' sample tests at [Spring 2012][1] and [Spring 2007][2]. [1]: https://6002x.mitx.mit.edu/ [2]: http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-electronics-spring-2007/ FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-10-15T04:23:39Z SecondChildTAG: Actually I don't know that you can see the Spring 2012 midterm at the old site. I logged on there as a guest and the mid-term and finals are not visible to guests. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-19T16:04:04Z SecondChildTAG: Could someone please kindly post the past sample tests for those who does not have the access? :) SecondChildUserIdTAG: 264887 SecondChildUserNameTAG: joyce428 SecondChildCreateTimeTAG: 2012-10-21T19:46:40Z SecondChildTAG: I must have been thinking of [this midterm][1] and [this final][2], which I seem to be able to access even while logged out. [1]: https://6002x.mitx.mit.edu/static/handouts/6002x-MidTermReview-S2012.pdf [2]: https://6002x.mitx.mit.edu/static/handouts/6002x-FinalReview-S2012.pdf SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-22T13:39:37Z SecondChildTAG: Ah, I do NOT see those links under Course Info while logged in as a guest, but am able to access as a guest if I use the direct links... SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-22T13:42:12Z FirstChildTAG: If they do like last term, they will release a practice mid-term soon. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-19T16:04:45Z SecondChildTAG: Wow! I hope so! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-19T16:17:17Z SecondChildTAG: Well, as I say, I only have the last class to judge by. I don't have any official word from the staff. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-19T16:19:13Z SecondChildTAG: Thanks for keeping us informed!! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-19T16:24:58Z SecondChildTAG: I don't know why but I hear the theme music from the TV show "Community" every time I see the "Community TA" label next to my name. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-19T16:40:15Z SecondChildTAG: Better to hear music than to not! :) SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-19T16:59:06Z SecondChildTAG: I think you are losing it JS ;-D SecondChildUserIdTAG: 9673 SecondChildUserNameTAG: xp42 SecondChildCreateTimeTAG: 2012-10-19T21:00:14Z SecondChildTAG: And why can't we edit comments? SecondChildUserIdTAG: 9673 SecondChildUserNameTAG: xp42 SecondChildCreateTimeTAG: 2012-10-19T21:02:03Z FirstChildTAG: There are 7 past ones here: http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-electronics-spring-2007/exams/ In the last class practice questions were posted before the mid-term so everyone could become acquainted with having limited attempts per question. The following link should show for everyone. https://6002x.mitx.mit.edu/section/MidtermFormatExamples/ I notice that if you hit grade repeatedly, the question values change. This is probably an artifact of not being logged in as a registered user. FirstChildUserIdTAG: 9673 FirstChildUserNameTAG: xp42 FirstChildCreateTimeTAG: 2012-10-19T20:40:36Z SecondChildTAG: Thanks a lot! SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-10-26T09:21:11Z FirstChildTAG: Yes. it should be coming out shortly. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-10-23T15:43:10Z IndexTAG: 725 TitleTAG: H5P3 Q1 I got VOUT in H5P2. In order to find vout I have differentiated VOUT and multiplied with vin. Its showing wrong. What might be gone wrong? UserIdTAG: 362299 UserNameTAG: anandbaskaran CreateTimeTAG: 2012-10-13T15:54:56Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: vin*(1-1/sqrt(1+2*K*RS*(VIN-VT))) use this.. FirstChildUserIdTAG: 228871 FirstChildUserNameTAG: ELINAKHAN FirstChildCreateTimeTAG: 2012-10-13T15:58:53Z SecondChildTAG: Thank you.. It worked!!! SecondChildUserIdTAG: 362299 SecondChildUserNameTAG: anandbaskaran SecondChildCreateTimeTAG: 2012-10-13T16:01:14Z SecondChildTAG: ur welcome SecondChildUserIdTAG: 228871 SecondChildUserNameTAG: ELINAKHAN SecondChildCreateTimeTAG: 2012-10-13T16:05:20Z SecondChildTAG: help me in H5P1 pls SecondChildUserIdTAG: 228871 SecondChildUserNameTAG: ELINAKHAN SecondChildCreateTimeTAG: 2012-10-13T16:05:36Z SecondChildTAG: See the comment in HWP1. SecondChildUserIdTAG: 362299 SecondChildUserNameTAG: anandbaskaran SecondChildCreateTimeTAG: 2012-10-13T16:14:10Z SecondChildTAG: how is above equation correct? i differentiated vOUT and it gives "(vin/RS)[RS-(2RS/K){1/sqrt{4(RS/K)(VIN-VT+(2/K)^2)}]" and it didn't worked but when i puted numbers in my equation it gave correct answer. i am confused :/ SecondChildUserIdTAG: 259693 SecondChildUserNameTAG: MehrozKhan SecondChildCreateTimeTAG: 2012-10-13T17:14:17Z SecondChildTAG: Isn't it against the honor code to post a direct answer? SecondChildUserIdTAG: 359310 SecondChildUserNameTAG: AaronYeoh SecondChildCreateTimeTAG: 2012-10-14T04:58:21Z SecondChildTAG: ELINAKHAN, remember that you can't help by giving direct answers. SecondChildUserIdTAG: 302713 SecondChildUserNameTAG: EVega SecondChildCreateTimeTAG: 2012-10-14T09:02:50Z SecondChildTAG: ELINA gr88 job..... could u plz brief....how u got dis....!! anyways thanks!!! :D :D :D SecondChildUserIdTAG: 127800 SecondChildUserNameTAG: MNB SecondChildCreateTimeTAG: 2012-10-14T17:40:21Z IndexTAG: 726 TitleTAG: the biggest problem how to write the answer in the specified place with out see the word "Invalid input"? I don't want to stuck in the midterm. UserIdTAG: 226085 UserNameTAG: Teto CreateTimeTAG: 2012-10-13T15:07:54Z VoteTAG: 4 CoursewareTAG: Week 5 / Incremental Voltage Exercise CommentableIdTAG: 6002x_inc_volt_e NumberOfReplyTAG: 2 FirstChildTAG: Same is the case here ! Even if our concepts are right, even then there are chances that our answer might not be shown correct due to the **Invalid Input** error. Even I was also considering the same, what if we would stuck in such weird state through the midterms. Also, I have heard that there would be only three chances in exam to post answers. Quite confused ! FirstChildUserIdTAG: 211164 FirstChildUserNameTAG: Hamoodi FirstChildCreateTimeTAG: 2012-10-13T16:15:33Z FirstChildTAG: Hi Teto and Hamoodi! Based on my experience in the Prototype Course 6.002x, we hadn't to enter expressions in the Midterm and in the Final Exam :). Invalid input, might be because of the case sensitive: try to check if you are writting correcly the variables K and not k, VT and not vt, etc... See you! Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-13T20:33:53Z SecondChildTAG: I did what you said ...... it didn't work thank you very much for your response and concern SecondChildUserIdTAG: 226085 SecondChildUserNameTAG: Teto SecondChildCreateTimeTAG: 2012-10-13T21:11:46Z SecondChildTAG: Hi Teto, can I help you? SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T00:56:31Z SecondChildTAG: yes, please. to solve the above exercise I wrote -(RL K) (VI-VT) vi and this message show up Invalid input: Could not parse '-(RL K) (VI-VT) vi' as a formula what should I do to write it correctly? SecondChildUserIdTAG: 226085 SecondChildUserNameTAG: Teto SecondChildCreateTimeTAG: 2012-10-14T09:52:56Z SecondChildTAG: Hi Teto, try writing in this form: -(RL*K)*(VI-VT)*vi Always use the asterisk mark(*) when you are writing multiplied terms. Hope this helps you. :) SecondChildUserIdTAG: 222911 SecondChildUserNameTAG: bhaswardg SecondChildCreateTimeTAG: 2012-10-14T14:32:50Z SecondChildTAG: Hi Teto, take a look at what bhaswardg, wrote you, that is true... SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T14:50:25Z IndexTAG: 727 TitleTAG: LATE Extremely sorry for finishing it late.I joined late,so I will try to cover all the stuff as soon as possible. UserIdTAG: 504997 UserNameTAG: IKJOT CreateTimeTAG: 2012-10-12T03:36:37Z VoteTAG: 4 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: [For Late Joining][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_Troubleshooting/threads/5077401011e2811f00000034 hi IKJOT you should join this course in next session..... it will be very helpful..... FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-10-12T13:29:49Z IndexTAG: 728 TitleTAG: Masterschaft That was exactly what I wanted while watching the demos and there it is! I know it's probably the biggest cliché to say it to MIT, but you are really the best! UserIdTAG: 415375 UserNameTAG: ZWX CreateTimeTAG: 2012-10-11T20:17:20Z VoteTAG: 4 CoursewareTAG: Week 5 / Music Amplifier Demonstration CommentableIdTAG: 6002x_snd_d NumberOfReplyTAG: 0 IndexTAG: 729 TitleTAG: H5P2 getting a very long result for iDS..feeling kind of stuck..any clues? UserIdTAG: 266739 UserNameTAG: satvikchugh CreateTimeTAG: 2012-10-11T18:36:29Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: The right answer is indeed a long result, so you must persist :) FirstChildUserIdTAG: 304587 FirstChildUserNameTAG: Vasso FirstChildCreateTimeTAG: 2012-10-11T19:15:59Z SecondChildTAG: did it..:))) SecondChildUserIdTAG: 266739 SecondChildUserNameTAG: satvikchugh SecondChildCreateTimeTAG: 2012-10-12T14:28:29Z SecondChildTAG: the answer to this question should be in this format (xx-xx)/xx + (1/(xx^x*x)*(1-sqrt(1+x*x*xx*(xx-xx))) finally after hours of head banging and lots if note book pages i was sure that my answer is correct & i finally got the green holy tick!! Just make sure that you put a * for every multiplication!!! This was the mistake that I was doing SecondChildUserIdTAG: 232499 SecondChildUserNameTAG: Asim09 SecondChildCreateTimeTAG: 2012-10-13T15:51:24Z IndexTAG: 730 TitleTAG: You might be a good 6.002x student if.. -you have ever tried to repair a $15.00 radio. -you think of the gadgets in your office as "friends." -you think your computer looks better without the cover. -you have ever purchased an electronic appliance "as is." -you have ever saved the power cord from a broken appliance. -you think jokes about being unable to program a VCR are stupid. -the salespeople at Circuit City can't answer any of your questions. -the microphone at a meeting doesn't work and you rush up to fix it. -you own a set of small screwdrivers and you actually know where they are. -you just don't have the heart to throw away the 100-in-1 electronics kit you got for your ninth birthday. UserIdTAG: 292543 UserNameTAG: Pennypacker CreateTimeTAG: 2012-10-11T00:46:38Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: I once purchased a radio receiver at a thrift store for a $1. The guy said, "It doesn't work". Turns out the lights behind the LCD display were burned out. 5 minutes after I got it home it worked fine. I have to say, you can fix most electronics without knowing any of this stuff. If you can recognize bad caps and clean things well, you can fix most electronics. Certainly most stereo equipment. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-11T01:03:40Z SecondChildTAG: Nice! SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-11T01:07:33Z FirstChildTAG: Thomas Edison's Mother once said... " I'm proud that you invented the electric light bulb. Now turn it off and get to bed!" FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-11T00:48:25Z SecondChildTAG: thud thud..."Is this thing on?" SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-11T00:56:55Z FirstChildTAG: hi! I'm currently in my second year as an electronics student. I must say that when it comes to the practical side of electronics engineering, I still lack the knowledge and experience. I want to be as enthusiastic as you guys when it comes to practical application. Can you suggest books or kits than can be found online which do not cost that much? Please. I don't want to become an engineer who only knows the theories and concepts behind. I want to be someone who can apply these concepts in actual situation. thanks! :) FirstChildUserIdTAG: 371000 FirstChildUserNameTAG: Joseph090892 FirstChildCreateTimeTAG: 2012-10-11T03:28:36Z SecondChildTAG: A great useful project is the cMoy headphone amplifier. http://tangentsoft.net/audio/cmoy-tutorial/ Why do you need an amplifier for an iPod? It is about quality, not just volume. You will hear things in songs you never heard before. I still remember getting goosebumps the first time I heard mine. Very cool little project. Use it with your stereo or computer too. Now, I was able to build mine from salvaged parts, so it does not have to cost anything. On the other hand the parts-count is low if you want to go new. There are even lot's of people selling circuit boards and/or kits. I will let you google "cMoy kit" if you want. Don't get caught up in buying fancy exotic parts for now, unless you want to. I even built one without a circuit board! For best results use with decent over-the-ear type headphones, those little earbuds cannot compete or reproduce bass properly. These do not have to be expensive, something like the Sony MDR-XD200 will do fine. Comfortable too. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-11T04:20:19Z SecondChildTAG: breadboard, components, power supply and Internet - everything you need fast and expensive - Radioshack slow and cheap - DigiKey, Mouser and other internet distributers SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-11T05:35:49Z SecondChildTAG: thanks pennypacker and yokovo for the suggestions! :) now, this may sound silly, but what equipments hat electronics engineering students like us must have (primary necessities)? aside from multimeter breadboard and powersupply? thanks! btw, where can i buy a cheap power supply? last question, what electrnic books for hobbyists do you recommend? sorry if i have a lot of questions. i just want to have these info from people with experience. thanks again! SecondChildUserIdTAG: 371000 SecondChildUserNameTAG: Joseph090892 SecondChildCreateTimeTAG: 2012-10-11T06:17:55Z SecondChildTAG: >but what equipments hat electronics engineering students like us must have http://www.youtube.com/watch?v=R_PbjbRaO2E SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-11T06:44:12Z FirstChildTAG: cool So I am a superstar. I did that all plus I put computer liquid cooling on my car engine, yup don't need a radiator anymore :))) FirstChildUserIdTAG: 286954 FirstChildUserNameTAG: ododo FirstChildCreateTimeTAG: 2012-10-11T01:17:39Z SecondChildTAG: Are you cooling your car with PC fans?, or do you have a 3'x2' radiator full of bugs bolted to your computer tower? Either way I think it's great!!! SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-11T01:23:34Z FirstChildTAG: I would highly recommend buying an Arduino. Then only your mind sets the limits for what you can do:) FirstChildUserIdTAG: 329453 FirstChildUserNameTAG: Even83 FirstChildCreateTimeTAG: 2012-10-11T07:53:25Z SecondChildTAG: I agree, I recently purchased as Raspberry pi to play with as well. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-17T13:08:28Z IndexTAG: 731 TitleTAG: Does anyone feel the same as me? This course is fantastic from a technical/mathematical point of view, but There are a few aspects which I don't feel are as coherent as they could be, especially in light of the fact that this is an *introductory* course on electronics. I somehow feel that the practical side of electronics is not stressed as much as it could be... ...for example, in some electronics books that I have, the progression of understanding is very nicely laid out, we come across Ohm's law first. Then the resistor, and the concepts of voltage and current are introduced. After that we learn abut pn/np junctions and are given at least some overview about the concept of conductors, semiconductors and insulators, from which we can understand the diode and Zener diode. In 6.002x homework 4 we received an assignment to use the piecewise linear method to model a zener diode. Before this I don't believe we learned about even the basic concept of a zener diode in the course, let alone pn junctions and their practical applications in electronics. Of course, the objective of the homework assignment was to test our understanding of using the mathematics on any non linear element, but it would be good to build up an arsenal of real world components (ones you would buy from RS), components circuit designers would use, and understand what they are. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-10T14:36:56Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 6 FirstChildTAG: This is what I thought when I took the class in the spring. I thought this was an ass-backwards way of teaching electronics. Throughout the class, I thought it and said it. I was wrong. They know what they are doing. By the time you finish the course, you will be with us. It's hard or impossible to see right now. In particular, the "brain-stretching" thing they do by asking you about things they haven't even mentioned is painful and frustrating. It's also very effective in the end. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-10T15:14:36Z SecondChildTAG: 100% agreed SecondChildUserIdTAG: 127195 SecondChildUserNameTAG: princeofsudan SecondChildCreateTimeTAG: 2012-10-10T17:06:02Z FirstChildTAG: I agree, it is a little overwhelming. I am drowning in equations! Sometimes it feels more like a math course then anything resembling electronics. I hope JSChambers is right! (I'm sure he is.) FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-10T16:37:53Z FirstChildTAG: I'm not sure about effectiveness of "brain-stretching" technique - electronics is my hobby and for many HWs and labs I have feeling how it works on physical level due to my experience (however without exact math base), and watching people trying to solve problems with naked math makes me think that "brain-stretching" technique is very contradictory. It would be nice to add as many as possible physics information (texts and, much better, videos) to the tutorials. FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-10T17:05:06Z SecondChildTAG: Wouldn't more physics just mean more equations? SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-10T22:12:26Z SecondChildTAG: I mean something in [HowStuffWirks][1] style. [1]: http://www.howstuffworks.com/ SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-10T23:47:11Z FirstChildTAG: "I am just here to learn as much as I can." Me too, I finally unplugged the soldering iron halfway through week four. lol I am fine with it being harder then I expected, I expected it to be harder then expected, what do you expect from MIT. For the more basic things, I still have my trusty Forrest M. Mims booklet. Shout-out to Forrest M. Mims. No affiliation. http://www.forrestmims.com/ FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-11T01:18:38Z SecondChildTAG: I have that book too! SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-11T02:47:14Z FirstChildTAG: This thread reminds me of this little diddy. Do you feel like we do? http://www.youtube.com/watch?v=BzBQ4OKcNNU FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-11T01:54:56Z FirstChildTAG: I am a high school physics and (very basic) electronics teacher. I am taking the course to improve my understanding of electronics. I am already familiar with the components - the math is kicking my butt - last calculus class was in 1993. It is nice to be taking a challenging, interesting class without concern for GPA - I am just here to learn as much as I can. FirstChildUserIdTAG: 18073 FirstChildUserNameTAG: gburkhart FirstChildCreateTimeTAG: 2012-10-11T00:34:56Z SecondChildTAG: Welcome gburkhart. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-13T04:24:58Z IndexTAG: 732 TitleTAG: Midterm - 24hrs = 48 hrs ? Just noticed something odd in the midterm announcement. Students can take the exam from 25th 00:01am till 27th 11:59pm. That's 48 hrs and not 24. So what is the correct deadline here ?? Or does one have to finish it within 24 hrs after starting it (or 27th 11:59pm, whatever comes first) and is there a 48hr time frame to start? [edit] It's not a 48hr but a 72hr time frame to start the exam. [/edit] UserIdTAG: 114881 UserNameTAG: xvink CreateTimeTAG: 2012-10-09T19:57:18Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: It's a 24-hour timeframe. You have 24-hours to finish it from the moment you go past the front page. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-09T20:33:18Z IndexTAG: 733 TitleTAG: H5P1 part C I set the two equations for Vout = to each other for the bottom end of the saturation curve - got a quadratic and solved it got vIN but no green check mark. Any ideas where I am going wrong? UserIdTAG: 18073 UserNameTAG: gburkhart CreateTimeTAG: 2012-10-09T02:26:07Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5076e36fccb9e81f00000082 check this out..:) FirstChildUserIdTAG: 266739 FirstChildUserNameTAG: satvikchugh FirstChildCreateTimeTAG: 2012-10-13T04:23:05Z IndexTAG: 734 TitleTAG: Mid-term exam do you happen to know if the lecture sequences, homeworks and labs are going to be open while taking the mid-term exam? UserIdTAG: 45307 UserNameTAG: josejimenez2 CreateTimeTAG: 2012-10-09T01:22:30Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: No FirstChildUserIdTAG: 404188 FirstChildUserNameTAG: jishudasorissa FirstChildCreateTimeTAG: 2012-10-09T04:30:24Z SecondChildTAG: Are they not going to open?? SecondChildUserIdTAG: 122513 SecondChildUserNameTAG: Mona77 SecondChildCreateTimeTAG: 2012-10-09T10:21:56Z SecondChildTAG: I've joined the course yesterday,if i complete my homework and labs now,then i will be allowed for mid exam and my homeworks will be accepted............? plz post a message if any one know........... SecondChildUserIdTAG: 593858 SecondChildUserNameTAG: luckynomi SecondChildCreateTimeTAG: 2012-10-09T13:25:50Z SecondChildTAG: no i think your homework till the first four week wouldn't be accepted as they have crossed their deadlines..but yes you can surely give your mid term exam..... SecondChildUserIdTAG: 413613 SecondChildUserNameTAG: shaliesh SecondChildCreateTimeTAG: 2012-10-09T16:17:31Z SecondChildTAG: luckynomi your first four homeworks and labs are not going to be accepted, because they've crossed the deadline, also you won't be able to submit them since you can't click on the check buttom. But if you do all the homeworks and labs from this week, and you take the mid-term and the final exam and get good scores on them maybe you'll be able to earn a certificate. Ask staff for better information SecondChildUserIdTAG: 45307 SecondChildUserNameTAG: josejimenez2 SecondChildCreateTimeTAG: 2012-10-09T19:07:12Z FirstChildTAG: Yes, they are all going to be open. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-11T04:56:44Z SecondChildTAG: oh, so is jishudasorissa confused about this? SecondChildUserIdTAG: 45307 SecondChildUserNameTAG: josejimenez2 SecondChildCreateTimeTAG: 2012-10-13T00:21:07Z IndexTAG: 735 TitleTAG: Midterm Pattern?? Any one having any idea about the midterm examination pattern(total number of questions?). UserIdTAG: 148523 UserNameTAG: span993 CreateTimeTAG: 2012-10-08T18:48:59Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 736 TitleTAG: What is the limit for submitting the Midterm answers ? Hi all, I need to ask that what would be the limit to submit answers for the questions asked in midterm exam ? As in homework, we can submit the answer as many times as we want. But is this the case with midterms as well or there are amendments ? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-08T14:54:29Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: In the pilot 6.002x, there was a limit of three submission attempts per problem set. Note that the limit is per ***set***! If a problem requires four answers, you cannot click 'Submit' after each one, you will run out of attempts for that problem. You get the red 'X' or green tick after each submission attempt, then you can go back and try to redo the ones that were marked wrong. Nothing is more nerve-wracking than to submit a second time and still get the red 'X' on answers that are needed for subsequent parts of the problem, because you know your next submission is your last chance, and a wrong intermediate answer kills the rest of the answers. FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-10-08T15:37:36Z SecondChildTAG: So what's your recommendation? Is there a chance of "assuming this previous result, this answer will be..." SecondChildUserIdTAG: 413869 SecondChildUserNameTAG: nlfigueiredo SecondChildCreateTimeTAG: 2012-10-08T15:54:54Z SecondChildTAG: I'm having a stroke. And that's official now :) SecondChildUserIdTAG: 304587 SecondChildUserNameTAG: Vasso SecondChildCreateTimeTAG: 2012-10-08T17:33:34Z SecondChildTAG: @g_hopper Do you know if the questions in the Midterm exam are going to be similar to those in the homework assignments? SecondChildUserIdTAG: 304587 SecondChildUserNameTAG: Vasso SecondChildCreateTimeTAG: 2012-10-08T17:42:47Z SecondChildTAG: I decided to have my stroke yesterday... or the day before. SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-08T18:31:20Z SecondChildTAG: I didn't have my stroke in time, now it's past it's due date! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-08T19:38:34Z SecondChildTAG: Thanks for your reply Hopper. But the thing u quoted was for the previous one. Is this the same case as well this time ? SecondChildUserIdTAG: 211164 SecondChildUserNameTAG: Hamoodi SecondChildCreateTimeTAG: 2012-10-09T00:57:27Z FirstChildTAG: I'm not a staff member, just a student from the pilot 6.002x who is taking this class for the second time (there are many of us from the pilot 6.002x doing this). The edX staff are the authoritative source for info, but here are my observations and recommendations from the pilot 6.002x regarding midterm and final exams. 1) Do the all homeworks, labs, and sequence exercises, and review the detailed answers given for the homeworks and labs (click on the 'Show Answer' button after the weekly deadline has passed). The worked problems in the tutorials look like a good resource, they are a new feature to this iteration of 6.002x. 2) In general, one problem on each exam was very similar to a homework or sequence exercise. I found it helpful to review my notes made while working on the homeworks and labs, so I could quickly locate needed formulas and concepts during the exam. If you don't take or retain your notes while working on the weekly assignments, you will waste time and energy searching through the textbook during the exams. 3) Some students made study sheets for themselves before the exam which contained important formulas, concepts, etc. I didn't do that last time, but I will this time. It saves time during the exam, and reinforces the material in your head as you prepare for the exam. 4) Each exam had one problem that contained material not explicitly covered in the lectures. Don't go screaming "Not fair!" The point of these problems was to apply the general methods and techniques to a novel situation. For example, if the covered lecture and homework problems have the circuit in a specific configuration, expect to see an exam problem with a different circuit configuration, or using a different type of component. 5) The exams are open book, and open Internet. That means you can use online calculators and equation solvers such as Wolfram Alpha http://www.wolframalpha.com, Number Empire http://www.numberempire.com/equationsolver.php, etc. You can also use the discussion forum to review posts. What you ***CAN'T*** do is cheat or collaborate with others. You can't ask questions that pertain to the exam material on this forum or elsewhere, or reveal exam material until the exam deadline has closed for everyone. The staff monitor other online venues and use statistical techniques to identify plagiarism and dishonest students. Don't embarrass yourself and get banned from edX. You only cheat yourself if you aren't making honest original effort in these classes. 6) Set aside as much time as possible to take the exam. It's open for 24 hours, and many of us used all or nearly all of that for the final. I can't imagine taking the final as an on-campus MIT student with a two or three-hour limit, and no access to online resources, but they do it. Of course, the typical on-campus MIT student is much younger than me, and probably a lot smarter, so perhaps some of us won't need all 24 hours to complete the exams. 7) Consider reviewing material from similar classes online or from different textbooks as you go through this course, along with more general electronics learning sites. Google is your friend; you may find alternative explanations for the concepts that make everything clear to you. All About Circuits http://www.allaboutcircuits.com/ and Electronics Tutorials http://www.electronics-tutorials.ws/ are two good places to start. Also look at the MIT OpenCourseware version of 6.002 http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-electronics-spring-2007/. Good luck! FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-10-08T20:10:55Z SecondChildTAG: And also important: don't wait till the last day to open your exam, because your internet connection can fail, your pc can stop working and even an earthquake can move you to another timezone .... Try to be sure about at what time the exam is closing in your timezone. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-10-09T16:29:56Z IndexTAG: 737 TitleTAG: Page 404 Error on week 6's tutorial problem 9.7. I got the page link missing for problem 9.7. UserIdTAG: 172304 UserNameTAG: Demohunter CreateTimeTAG: 2012-10-08T08:28:45Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Troubleshooting NumberOfReplyTAG: 1 FirstChildTAG: I emailed technical@edx.org on October 8th, but I got no reply as of now (Oct. 14th) and the link is still 404. Not sure how we can contact them. FirstChildUserIdTAG: 160020 FirstChildUserNameTAG: antoineleclair FirstChildCreateTimeTAG: 2012-10-14T20:39:17Z IndexTAG: 738 TitleTAG: Thank you IgnacioUY for uploading the videos in the Wiki - Week 5 and Week 6! :) Thank you **IgnacioUY** for uploading the videos! You are doing a awesome work! ![imagen][1] Thank you! :) I am sure many worldwide classmates are really thankful with you! How are you doing with this course? I hope that you are doing good :). Again, thank you for your effort! My best wish to you! Myriam. P.D: Can you also add the video tutorials of the week ;). That will be nice too. Thank you, thank you! [1]: https://edxuploads.s3.amazonaws.com/13496392961343631.png UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2012-10-07T19:50:54Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 739 TitleTAG: A problem with voltage notations. Hello 6.002x staff Your course is just fantastic. But I am facing a little problem. As I have English as 2nd language I have to follow through the captions and consequently I get confused sometimes between the notations of voltages and current such as vd, VD, id, ID etc. as in the captions all are written as 'VD' or 'ID'. So, it will be better if the captions are updated according to the notations used by the professor like 'vd', 'Vd', 'vD' (similarly for currents also). Thank You. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-07T19:17:58Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 2 FirstChildTAG: The captions are not always correct, especially when it comes to $V_D$ (which of course you know is bias / DC voltage) and $v_d$ (which is the small signal component). 1. I am not staff, but I do not think that the captioning system supports subscripts and superscripts and mathematical language at all. So this entire component needs upgrading. 2. Even if MITx had a captioning system that supported subscripts and math notation (i.e. LaTeX formatting, like these discussion boards support), then one person would have to go through EACH lecture one-by-one, and verify that everything in the captioning agrees with everything the Professor is writing on the board / screen. So, such a system would be very difficult to implement, but I think it would be worth it. I speak English as a first language, but I even read the captions; it helps memorize things the Professor says, and it helps me understand him better as even his English is not 100% - his grammar is excellent but he has a South-Asian accent. FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-10-07T20:51:23Z SecondChildTAG: And, like everyone else, Prof. Agarwal occasionally misspeaks. I find a COMBINATION of reading captions and careful listening (you can also adjust your playback speed) to the videos is the best. SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-08T04:29:29Z FirstChildTAG: I'm also having a lot of difficulty with the notation. I would like to continue this discussion and ask about some specific examples, so I/we can get on with solving problems regarding the large and small signal models for MOSFET Amplifiers. Take for example S10E2, in the schematic the voltage from drain to source is VR, but in the question it is vR=2.7V. I would normally think these were different, causing me to use different equations (small vs. large signal model), but in calculating them I find they are the same. I say this, because I calculated iD (or in this case iR) with the equation iD = (K/2)(VIN - VT)^2, where VIN=VR (or vR) and got the right answer. The second part of this question is another matter, I haven't figured it out yet as the notation is confusing me. Since I was concerned about VR vs. vR, I then looked up the equation for iD, and low an behold, on page 368 of the book the current is ID, but it refers back to Equation 7.8 which uses iDS (referring to the subscript in caps not D vs DS). Is it the fact that iDS is being used when discussing the SCS MOSFET model in general and ID is used when using the SCS MOSFET model for amplifier applications? Or, is this just an oversight? I believe what the staff needs to do is on a case by case basis, clarify any ambiguities that may exist or that may arise when they are discussing topics. I realize the difficulty in preventing misstatements, dealing with multiple fonts from multiple applications (the book vs. the answer boxes in homework/labs etc. vs. the discussion forum etc.) and dealing with all parts of the world and clarifying on the spot will help. FirstChildUserIdTAG: 132828 FirstChildUserNameTAG: rharris FirstChildCreateTimeTAG: 2012-10-10T18:17:14Z IndexTAG: 740 TitleTAG: S12E5 - bit off for th time on answer 1? Am i the only one who gets a little off value like 2.72sec for the first value? Like the given answer gives me 67v threshold instead of 77v? Any clue what is wrong with my math? UserIdTAG: 331483 UserNameTAG: Doru CreateTimeTAG: 2012-10-06T23:29:24Z VoteTAG: 4 CoursewareTAG: Week 6 / Neon relaxation oscillator exercise CommentableIdTAG: 6002x_neon_relaxation_oscillator_exercise NumberOfReplyTAG: 1 FirstChildTAG: I got the same as you at first. But then i realised that the question asks for the time it takes to the capacitor from one thresold to the other one. So, your initial voltage (V0) can't be 0 ;) FirstChildUserIdTAG: 193982 FirstChildUserNameTAG: Wfk FirstChildCreateTimeTAG: 2012-10-07T02:13:09Z SecondChildTAG: Oh yes you are totally right, its the initial voltage v0 ... darn i must be blind to the wording of most texts. SecondChildUserIdTAG: 331483 SecondChildUserNameTAG: Doru SecondChildCreateTimeTAG: 2012-10-07T11:43:49Z IndexTAG: 741 TitleTAG: What interesting electronic devices have you ever built? Hello everyone! It's quite boring for me to simple follow the lectures and do homeworks and labs, I'd like to apply my knowledges in practice. I want to build something interesting and electronic, but nothing comes up to mind. Have you ever built something like that? I would be very grateful for your stories and advices! UserIdTAG: 326787 UserNameTAG: Angstrem CreateTimeTAG: 2012-10-06T19:38:37Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: I've built many things at work and through kits. But experimenting with op-amps and inductors one day, I built this high gain amplifier where $Vout$ would vary depending on the distance from the inductor to coins nearby; a metal-detector. I then tried various metals - brass, aluminum, iron, etc. I used another op-amp as a threshold detector, and coupled with logic, I had the device identify metals by illuminating one of 3 LEDs, green for nickel, yellow for zinc, and red for iron. I was going to fine-tune the device to work with aluminum, copper, brass (copper + zinc alloy), etc. but I ran out of time. Maybe one day I can build a device that can tell me % metal content (i.e. 5% copper 95% zinc); I saw a commercial device that had a numerical readout of metal composition, how it did it, I don't know...but wow! FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-10-07T18:27:53Z SecondChildTAG: Nice JerseyMark! :). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T18:51:23Z SecondChildTAG: I made a very sensitive metal detector once by driving a parallel LC circuit at it's resonant frequency so that the Voltage and current would be in phase. Then I turned the voltage and current waveforms into square waves and fed them into a nand gate. This meant that the output of the nand gate would be high for some time if the voltage and current were not exactly in phase. By filtering the output of the nand gate you can get a nice DC value that varies as the resonant frequency of the LC circuit changes (which happens when the inductance of the coil changes as metals are brought near it). You can set the sensitivity of the detector by feeding the filtered DC value into a comparator and have an adjustable reference to compare it to. SecondChildUserIdTAG: 270011 SecondChildUserNameTAG: SnowmanZA SecondChildCreateTimeTAG: 2012-10-07T19:06:37Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T19:30:28Z SecondChildTAG: Edit: NAND should have read XOR, I had NAND gates on the brain when I typed that. SecondChildUserIdTAG: 270011 SecondChildUserNameTAG: SnowmanZA SecondChildCreateTimeTAG: 2012-10-08T16:05:25Z FirstChildTAG: A bulbterry. It is made of a battery and a bulb. After hundreds of tries it finally works, thanks to 6.002x FirstChildUserIdTAG: 190618 FirstChildUserNameTAG: Kirbabaev FirstChildCreateTimeTAG: 2012-10-06T22:55:00Z SecondChildTAG: A phase-locked loop solid state Tesla coil, hydrogen generator, 500VA MOS-FET Full-bridge (using IRFP450s), delta radio controlled aircraft, solder fume extractor, toroidal step up transformer (1000 turns of fine gauge wire through a 1@ hole!), power supplies of various types, signal generators etc. Electronics is even more fun when you get into the practical side! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-07T12:49:34Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T18:50:14Z FirstChildTAG: I used to be interested in radio and so I used to make simple AM radios that just used a variable capacitor for tuning in parallel with a small ferrite core inductor and a germanium diode as a detector. They work very well and can drive a small piezo earphone without any power supply. You could also feed the output into an audio amplifier. I also used to make my own carbon microphones using the carbon rods that I took out of old alkaline AA cells. I made many small single transistor FM transmitters (you can find many examples on the web) they are great fun to play with. I later started playing with microcontrollers such as Microchip's PIC's they are quite easy to use and you can do quite a lot. I have also played with CPLD's and FPGA's quite a bit, they are very interesting devices and programming in VHDL takes a bit of getting used to. FirstChildUserIdTAG: 270011 FirstChildUserNameTAG: SnowmanZA FirstChildCreateTimeTAG: 2012-10-07T19:13:34Z SecondChildTAG: Nice SnowmanZA! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T19:29:58Z FirstChildTAG: Thanks everybody for your answers! It seems that I want to construct a hydrogen generator =) I haven't googled for it's concept yet, but it seems to me that it should work by simply transmitting current through the water... FirstChildUserIdTAG: 326787 FirstChildUserNameTAG: Angstrem FirstChildCreateTimeTAG: 2012-10-07T22:22:34Z SecondChildTAG: Good luck Angstrem! Just for curious: What is the particular application for which you want it? ;). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T22:35:05Z SecondChildTAG: Hmm, I haven't thought about it yet... Well, maybe I would be able to create some balloons filled with hydrogen! It would be pretty nice to attach a device like Raspberry Pie to it and, for example, to make photos from the sky =) SecondChildUserIdTAG: 326787 SecondChildUserNameTAG: Angstrem SecondChildCreateTimeTAG: 2012-10-07T23:04:36Z SecondChildTAG: That will be really creative ;)!! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T23:08:51Z SecondChildTAG: Here is the one I built, but you might want to use stainless steel plates for yours. ![enter image description here][1] But remember, this is not hydrogen gas, this is a perfect mixture of 2:1 hydrogen and oxygen (not to mention all the toxic gasses if you use powerful electrolytes) highly combustible and dangerous. Be safe! [1]: http:// SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-08T08:22:20Z SecondChildTAG: ![enter image description here][1] [1]: http:// SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-08T08:50:55Z SecondChildTAG: But actually, as I've read over the internet, hydrogen will appear only on one electrode, while oxygen on another one. So, in theory, it's possible to gather them in 2 different containers. By the way, hazel1919, you've written "Here is the one I built", but I see nothing... There might were some troubles while you were posting your staff. SecondChildUserIdTAG: 326787 SecondChildUserNameTAG: Angstrem SecondChildCreateTimeTAG: 2012-10-08T10:29:40Z SecondChildTAG: I don't know why my picture is not coming up. If you want a useful amount of hydrogen you don't want to use a setup like this: http://en.wikipedia.org/wiki/Electrolysis You want, at minimum, a 2mm gap betwixt the electrodes, and a hefty power source (car battery charger). And a setup like this: http://www.youtube.com/watch?v=RLXDgFUz3W4 SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-08T17:03:21Z SecondChildTAG: Thank you hazel1919! Can you post it? I am curious about your hydrogen generator :). How much does it cost you to develop it (materials, etc..)? SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-08T18:54:18Z SecondChildTAG: Here is a link to the picture...http://forum.allaboutcircuits.com/showthread.php?p=531259#post531259 SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-08T19:50:49Z SecondChildTAG: Thank you hazel1919! I have suscribed to that forum in order to see the pic of your generator, it is really prolix, nice work! :). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-09T02:01:19Z FirstChildTAG: I'm currently in the middle of a project I have slowly been putting together. The big picture here is I have an entertainment center that has no cooling built in so my devices get hot. My project consists of 4 120mm fans (2 intake/2 exhaust) to cool the system. Where it gets fun is I took a ~15v ac->dc adapter and ran it to a circuit I have put together. This particular circuit uses a TIP120 transistor (I think, I'll get the actual part numbers tonight when I get home) as a switch to power the fans on/off when given a 5v input. The circuit also uses voltage regulators to power the 5v temp/humidity sensors and the 9v Arduino that reads the sensors and kicks the fans on/off depending on the actual temp and the desired temp. They are set up using Pulse Width Modulation (PWM) so that I can control the fan speeds. I don't want them on at 100% all the time because they do cause some noise. Finally the temps and fan speeds are displayed on an lcd I wired into that circuit as well. It has been a long project because I don't have much time to work on it but I have learned a lot and thanks to this course I am understanding even more! I'll try and get some pics to post up when I get a chance. FirstChildUserIdTAG: 417864 FirstChildUserNameTAG: beauclark FirstChildCreateTimeTAG: 2012-10-08T13:12:14Z SecondChildTAG: I will like to see that pics of your project! :). Can you post it? SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-08T15:02:56Z SecondChildTAG: I have a few pics from my phone I can post ![Main Power Circuit][1] This is the main power circuit. The top power and ground rails are where the power adapter connects to. I have LEDs connected to different parts to show me when they are getting power. The top green LED is lit when the power adapter is plugged in (showing I have the starting 15v or so). The top redLED is lit when the fans are connected to power (this light will pulse and blink depending on how fast the fans are being switched on and off). The bottom red LED (near that rectangular blue box) lights up when I have 5v. The bottom positive rail will be all 5v due to the power regulator I have set up. Finally the yellow LED on the right lights up when I get 9v. That one column on the breadboard will put out 9v due to another voltage regulator I have set up. I don't need as many connections to the 9v since it is solely for an Arduino. The other elements on the board are as follows (I will not mention the jumpers, diodes, resistors or caps since those are somewhat obvious. I can give the values of them later tonight). The first black box on the bottom left is the switching transistor that switches the fans. Its output is connected to the red LED at the moment. The black box directly to the right of that is the first voltage regulator that drops the 12-15v down to 5v. The output of this powers the bottom positive rail all with that 5v. Connected to the 5v rail is a potentiometer that allows for adjustment of the desired temp (or the temp that the fans will try and get to). The POT value is read by the Arduino and it makes the adjustments via software and ultimately PWM. The blue box next to that is the humidity/temp sensor. It runs off of 5v and has a digital connection to the Arduino. The top black box is another voltage regulator that drops the 12-15v to 9v to power the Arduino. The various resistors and caps and diodes are there to try and protect the connecting devices. ![LCD circuit][2] This is hard to see I know but it is the only close shot I have for now. It is a really simple circuit that powers the LCD and displays the text. The POT there adjusts contrast. It is the basic schematic for an LCD with an Arduino. I have another pic of it all connected together to the Arduino but it is having issues uploading right now. This should do for now. I'll try and edit this post later with cap values and better descriptions. [1]: https://edxuploads.s3.amazonaws.com/13497293231343655.jpg [2]: https://edxuploads.s3.amazonaws.com/1349731441380842.jpg SecondChildUserIdTAG: 417864 SecondChildUserNameTAG: beauclark SecondChildCreateTimeTAG: 2012-10-08T21:27:31Z SecondChildTAG: It looks like I can't upload pics higher than 1MB. I'll resize and upload later when I can. Hope you guys like! Please point out any possible improvements etc. I'm open to criticism. SecondChildUserIdTAG: 417864 SecondChildUserNameTAG: beauclark SecondChildCreateTimeTAG: 2012-10-08T21:32:00Z SecondChildTAG: Thank you beauclark! Great work! I am tempted to reveal something that happened in the Prototype Course at the end of the course haha! :p , But I have to talk with the Team and with the edX Staff again ;). I suggest to all of you to be alert ! ^_^ . SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-09T02:16:21Z IndexTAG: 742 TitleTAG: Lab 4 I have constructed a circuit that is similar to the one given in the textbook. I got a green tick for it as well. But am unable to obtain the graph for it on running the transient analysis. Kindly help. Am stuck with it for a long time now. UserIdTAG: 226266 UserNameTAG: anjanasgf CreateTimeTAG: 2012-10-06T07:24:55Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: ya. i am also having same problem. FirstChildUserIdTAG: 171103 FirstChildUserNameTAG: hemanth463 FirstChildCreateTimeTAG: 2012-10-06T12:08:42Z FirstChildTAG: Is there a "RESET" button at the bottom of the page? FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-06T17:52:51Z FirstChildTAG: Hi anjanasgf! Take a look at this hints of Lab4 [here][1], might they can help you... [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5070e1b72210d02400000055 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-07T06:00:51Z SecondChildTAG: I managed to solve it! Thanks anyway!:) SecondChildUserIdTAG: 226266 SecondChildUserNameTAG: anjanasgf SecondChildCreateTimeTAG: 2012-10-07T08:51:16Z IndexTAG: 743 TitleTAG: My weird way that made sense to me current through resistor = current through N vi-va/r = 10*(1-e^(-va/5)) Put (5.0-x)/2-(10*(1-e^(-x/5))) into google which plots a nice graph, find where it intersects 0 on the x axis and you have Va UserIdTAG: 329964 UserNameTAG: SmartMike CreateTimeTAG: 2012-10-06T05:42:04Z VoteTAG: 4 CoursewareTAG: Week 4 / Linearization Exercise 1 CommentableIdTAG: 6002x_linearization_ex_1 NumberOfReplyTAG: 1 FirstChildTAG: And it works good for number 2 as well: va @ Vi = 5V*1.02 : ((5*1.02-x) - (10*(1-e^(-x/5)))) = 0; FirstChildUserIdTAG: 351044 FirstChildUserNameTAG: mvniekerk FirstChildCreateTimeTAG: 2012-10-06T14:30:05Z IndexTAG: 744 TitleTAG: Facing problems yet i am new here facing many problem in lab tutorials....don't know what to do??? UserIdTAG: 549863 UserNameTAG: Darshan129 CreateTimeTAG: 2012-10-04T09:54:21Z VoteTAG: 4 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: Hi Darshan129! Can I help you? Where are you lost? FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-04T11:15:30Z IndexTAG: 745 TitleTAG: H6P1 Partial Derivatives For the NewFET device given in H6P1 we are required to find the transconductance. Unfortunately, unlike that of the MOSFET, the NewFET's equation to calculate $i_{DS}$ is dependent on both $V_{GS}$ and $V_{DS}$. Does this mean, to solve the small circuit model for the NewFET we will have to use partial derivatives? If this is so, I would be much obliged if someone could explain how to use partial derivatives or give a link to a site which could teach this. Thank you in advance. UserIdTAG: 85490 UserNameTAG: yashsavani CreateTimeTAG: 2012-10-03T12:52:58Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Just solved both for $gm$ and $ro$. I used the method shown in the first video of the tutorial. Apparently you do have to use partial derivatives. It seems to be a multivariable calculus question. FirstChildUserIdTAG: 85490 FirstChildUserNameTAG: yashsavani FirstChildCreateTimeTAG: 2012-10-03T13:08:21Z SecondChildTAG: I got that. Thanks. SecondChildUserIdTAG: 39980 SecondChildUserNameTAG: MarinoJr SecondChildCreateTimeTAG: 2012-10-03T13:21:24Z FirstChildTAG: It is not partial derivatives, as the unique variable is vI. The other are constants. You must solve the derivative of K*(vI-VT)*vO^2 in respect of vI and apply vI=VI. Expand the terms and it will be easy. FirstChildUserIdTAG: 39980 FirstChildUserNameTAG: MarinoJr FirstChildCreateTimeTAG: 2012-10-03T12:59:34Z SecondChildTAG: Hey i got gm but how to go about r0? SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-06T07:33:22Z SecondChildTAG: I got r0 easly (just look at tutorial with vacuum triode), but cant find gm. SecondChildUserIdTAG: 352757 SecondChildUserNameTAG: Kerbyco SecondChildCreateTimeTAG: 2012-10-10T17:02:09Z SecondChildTAG: the derivative of K*(vgs-VT)*vO^2 |vgs=VGS equal K*(VGS-VT)*vO^2*vgs where is a mistake? SecondChildUserIdTAG: 352757 SecondChildUserNameTAG: Kerbyco SecondChildCreateTimeTAG: 2012-10-10T17:10:25Z SecondChildTAG: Oh. got it. :) SecondChildUserIdTAG: 352757 SecondChildUserNameTAG: Kerbyco SecondChildCreateTimeTAG: 2012-10-11T08:43:23Z SecondChildTAG: (gm= K*VDS^2) THIS IS THE RIGHT ANSWER SecondChildUserIdTAG: 106816 SecondChildUserNameTAG: Laith SecondChildCreateTimeTAG: 2012-10-14T11:54:54Z SecondChildTAG: Honor code! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-18T20:46:10Z FirstChildTAG: If you would like to learn about partial derivatives, or generally anything, I recommend this site: http://www.khanacademy.org/ FirstChildUserIdTAG: 329254 FirstChildUserNameTAG: KGabor FirstChildCreateTimeTAG: 2012-10-18T15:31:08Z SecondChildTAG: just use its search function at the top of the page: http://www.khanacademy.org/math/calculus/multivariable-calculus/v/partial-derivatives SecondChildUserIdTAG: 329254 SecondChildUserNameTAG: KGabor SecondChildCreateTimeTAG: 2012-10-18T15:31:51Z IndexTAG: 746 TitleTAG: Calculation question 1 As this isn't a question that is used to assess the students, I think it is in line with the honor code to post a solution to the question (but please correct me if I'm wrong) Right, I think I did this the hard way but here's how I solved the first question: Rt = R1 + Ra Ra = Va / Ia Ia = 10*(1-e^(-Va/5)) Rt = 2 + (Va / (10*(1-e^(-Va/5)))) Va = Vi * Ra / Rt Va = Vi * Ra / (2 + Ra) Va = Vi * (Va / Ia) / (2 + (Va / Ia)) Va = Vi * Va / (Ia * (2 + (Va / Ia))) Va / Va = Vi / (Ia * (2 + (Va / Ia))) 1 / Vi = 1 / (Ia * (2 + (Va / Ia))) Vi = Ia * (2 + (Va / Ia)) Vi = 2*Ia + (Va*Ia)/Ia Vi = 2*Ia + Va 2*Ia + Va - Vi = 0 2*(10*(1-e^(-Va/5))) + Va - Vi = 0 20 - 20*e^(-Va/5) + Va - Vi = 0 20 - 20*e^(-Va/5) + Va - 5 = 0 15 - 20*e^(-Va/5) + Va = 0 Makes Va = 1.088V for Vi = 5V Now the second question... I totally read incorrect (several times :() as being dVa/dVi instead of dVa/dVi... UserIdTAG: 114881 UserNameTAG: xvink CreateTimeTAG: 2012-10-02T19:11:04Z VoteTAG: 4 CoursewareTAG: Week 4 / Linearization Exercise 1 CommentableIdTAG: 6002x_linearization_ex_1 NumberOfReplyTAG: 1 FirstChildTAG: so long solution kvl : -Vi + Vr + Va =0 Vr= R*Ia= 20(1-e^(-Va/5)) Va= Vi-Vr = 5 -20*(1-e^(-Va/5)) Va = -15 +20*e^(-Va/5) FirstChildUserIdTAG: 236136 FirstChildUserNameTAG: Hamid-ch FirstChildCreateTimeTAG: 2012-10-03T23:09:47Z SecondChildTAG: yeah i too wasted time reading the above post SecondChildUserIdTAG: 129030 SecondChildUserNameTAG: vjbhatt786 SecondChildCreateTimeTAG: 2012-10-07T06:56:13Z IndexTAG: 747 TitleTAG: How was the signal shrunk ? I don't understand how the signal was meant to be shrunk, unless vi(t) it just meant to be an already shrunk signal of vI ? UserIdTAG: 404470 UserNameTAG: LukeSki CreateTimeTAG: 2012-10-01T12:53:41Z VoteTAG: 4 CoursewareTAG: Week 4 / Incremental Method Insight CommentableIdTAG: 6002x_inc_method_insight NumberOfReplyTAG: 1 FirstChildTAG: I think I know what you are asking, I think it is. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-01T15:00:51Z IndexTAG: 748 TitleTAG: Late but never too late Hey, I made this post to tell a little about my feelings to EDx, well lets see, I hated school because of my boring teachers, so I studied physics through OCW and BTW Prof. Walter Lewin is legendary, then I started off with 6.002x and truthfully it was highly more challenging than my 11th grade syllabus as I had no calculus or maxwell background, but then suddenly I was getting the lectures, was able to just make through week 1 and week 2 but then this week I got flooded with school homework, still every night I woke till 4am so I could complete my week 2 course ( I got late at week 2 with on going exams!) then today evening I did my week 2 complete, as yesterday my internet stopped working, right now I plan to stay up all night and atleast get a C grade on week 3, I mean the point here is I hate school, that does not mean I hate studying, as I am loving College Level computers, and that too just because of MIT faculty..... MIT has been my dream college ever since the true image of Indian College system opened in front of me and that is about 3 years, now I am doing MIT courses to pursue my passion, putting aside school, sleep, everything..... I was so in love with MIT before just with OCW and now I can't wait to ace my PSAT on 20th of October, ace the SATs next year and finally apply for my dream, my way out of misery, my passion, my love MIT.... I don't expect most of you to like this post, as I know I creeped out a bit, but think of it as that moment when your father let you go cycling on your own when you were 3 years old, you had dreamt of cycling since so many weeks and now you were doing it, though when you looked back and saw your father back there, you got scared and stopped, but then realized you can do it and continued on!... for me MIT is the cycle.......... I always wanted MIT and now that I am just kinda getting through its homeworks and still loving the lectures..... I just can't wait to apply! UserIdTAG: 169784 UserNameTAG: thewiredbear CreateTimeTAG: 2012-09-30T17:48:20Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: I hate to "rain on your parade", but the material in this course is typical of material taught in circuits courses in hundreds of universities in the US as well as around the world. It's even taught in some associate degree (2 year) programs in the US. What does seem to be different is the order in which material is presented (and I don't think it's good!!), non-linear and active elements presented before covering the other linear, passive elements: capacitors and resistors. I was very surprised how simple the homeworks and labs have been. I expected more from the "MIT brand". I am also surprised by the lack of depth. While I plan to continue with the course, I am quite disappointed. It is very disturbing seeing students begging for answers without a concern for a true understanding of the material. Just my 2 cents. Take it for what it's worth! FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-09-30T18:45:04Z SecondChildTAG: I suspect that by the time this course is over you will have found some value that you are not finding at this point. SecondChildUserIdTAG: 92346 SecondChildUserNameTAG: MobiusTruth SecondChildCreateTimeTAG: 2012-09-30T19:41:29Z SecondChildTAG: you yourself said.... these are typical university courses while you overlooked the fact I am still in school!, in 11th grade, the course has per-requisites of AP level Physics and Calculus, while I know none officially, as my school teaches them next year and I did them, rather just skimmed over them in less than a week.... SecondChildUserIdTAG: 169784 SecondChildUserNameTAG: thewiredbear SecondChildCreateTimeTAG: 2012-09-30T20:01:26Z SecondChildTAG: I have already watched all the videos from when the course was recorded live in 2007, so I think I have a good handle for what is in store. I am more than half way through the videos a second time. I watch them at work while eating my lunch. I was drawn to the videos in the first place because I was working on a hobby project, and I wanted to here about linearization about an operating point. When I found out the course was being offered online for a certificate I jumped at the chance in order to get "the MIT experience". Because this was from MIT, I expected the homework and labs to be really challenging and to require a deep understanding, but they aren't. I am really distressed by what I see with people begging for the answer at the last minute. Based on what I've seen so far, getting the certificate is no guarantee of proficiency in the subject. I remind you that I got involved with the course because I am building actual hardware as part of a hobby and wanted to do a better job of analyzing the design. My real goal is to build a circuit that performs well not to brag about getting a certificate from MIT. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-09-30T20:18:32Z SecondChildTAG: I think the homework *is* pretty challenging. They do test understanding of a concept. If that's not hard enough, I'm sure the exams will be. There wasn't one single person who said the exam was easy during the Spring session of the course. We all found it hard. Also, MIT structures it's course a little differently. If you look through OCW, 6.002 is more like a tour of the entire electronic devices and circuits field - it gives you a feel of everything. Next they have follow up courses which get into details. Personally, I feel this makes sense. The course here seems to follow a simplified version of the structure in the book Integrated Electronics by Millman and Halkias. That book introduces everything in the subject - analog, digital and network analysis (the book does go quite deep however). I like that. Finally, what I liked about 6.002x in the last session was that, knowing how small signal analysis of a MOSFET is done, I could figure out how to do the same for a vacuum tube. That's there in Week 4 homework. This course seems to give a base from where it is easy to self explore. SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-09-30T20:42:32Z SecondChildTAG: **skyhawk**: I agree with you that, at my university as well (a large U.S. state university) we got more in-depth with AC analysis using complex numbers and phasors to analyze networks with capacitors and inductors first. Digital design (i.e. logic gates, Karanaugh maps, MOSFETs) was a seperate course, and it made sense that way. Here, there is only a superficial touch (so far) on DC analysis. No mention of "supernodes". No requirement to actually solve systems of linear circuit equations using matrices or other problem-solving methods. The schematics we had to analyze for DC were mind-boggling at my university. Here, everything is simple and homeworks sometimes take less than an hour! Many answers can be gotten by inspection, or via trial and error! A mess. I love how in Week 3, they present a very important circuit, **the clamping circuit**, for the first time, in Homework! With no theory or anything behind it's operation, just solve using node method. That only has limited value. This circuit appears time and time again in EECS, and we only get a cursory view in a homework (maybe next week's lecture will cover it in more depth - I'm hoping!) Diodes are very important nonlinear elements - but why not cover capacitors, linear elements (in AC analysis at least) which are just as important, first? It's more logical that way! You need to understand both to understand power electronics and to design power supplies, preferably studied together, not apart! I know MIT is investing a big deal in the "abstraction method" because it is "cool", but the classical method of teaching circuits is the way generations of electrical engineers, techs, and electricians were trained in. SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-10-01T03:17:35Z SecondChildTAG: **thewiredbear**: Is the *Indian educational system* at fault? You point toward that suggestion, but do not actually mention it. Something has to be wrong there if you love this course, but you hate school in real-life. I like the online system because it's convenient (no driving to school or finding parking for the car, no worries about rain or snow, no problem if I'm working during the day since I can watch lectures any time). In the traditional university setting, I worried more about a social life, about partying/drinking, about appearance, atc. Here I don't have to. Or maybe because I'm older now and things like that don't matter as much. You should realize that you may do great at **any** Western-style university; because the electrical engineering curriculum is standardized (pretty much). Maybe it's the online aspect you like. Your post just had me wondering on how your day-to-day life is like in India, and how it is the same as when I was your age, and how it must be different, too. SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-10-01T03:27:43Z SecondChildTAG: > @Jersey Mark well the thing I meant about Indian Education System was, that I can go through the snow or rain, or reaching the classes on time, and FYI in India you get driver's license at 18, so no driving to school here, its all school buses! But the main thing is that from ground up or rather day one, the mindset of all the teachers is that they are teaching the students so that they get a job, the ideal mindset of teaching to teach, is now vanished from all the maths and science and even computer teachers I have heard of and who have taught me, and I am not targeting just my school, ask any Indian student that "is your physics teacher ready to jump up on a rope at an elderly age and show you the natural phenomena that a pendulum has the same frequency no matter how much weight you add on it?" ( as Prof. Walter Lewin,an MIT professor does it every year!).... and not even this, I mean in the US, if you go to your teacher and ask her why voltage does divide in a series circuit and not in a parallel circuit, well he/she would obviously try to help you giving the example of maybe a hosepipe ( saw it on a US high school website and even Khan Academy) or I mean any other way so that **YOUR CONCEPTS BECOME CLEAR** while in India the teacher is only concerned about you mugging up the formulas and jotting them down on the test again, from day one our teachers lacked the fire to teach, and because of that now most of my classmates lack the fire to learn, I mean its like a brainwash, the teachers from day 1 teach you so you get a good job rather than learning the subject, because of which most students now care more about their future jobs than Science, the main motive should be that you learn science and build stuff and not go sot on a desk, signing papers!.... I mean see this, more than 96% of Indian engineers think of completing a business course after engineering, and most of them do it, not because they may want to start their own company, but rather to build a better profile for a better job, I mean I see this and I go like " Why did you even do engineering when you always wanted to do business? " This Indian Mindset kills innovation, go in any college in India, and the first thing they tell you is that if you study from their college they would get jobs in good companies, google, intel, yahoo, blah , blah , blah..... everyone here needs money at the end, the fire to invent and progress is OFF in general in the country, in the US, you start making a website on your own to pursue your passion in softwares and programming, your parents would praise you for your efforts, your school teachers would come forward to help you, everyone would encourage you to continue, you may even get credits for that, and you will have many competitions where you could show the world your talent, here in India, this is what happened to me when I made my own website, " My parents told to stop wasting time and finish my syllabus and homework, I ask my teachers for help, and they tell me to complete the boring syllabus ( they teach MS word in computers in india, so in the end, the computer class was boring as hell and I didnt like the syllabus, I mean who enjoys MS Word in exams ???? ) and yes credits for making that website are a faaaaar faaaar off dream and we dont have many competitions for anything to show our talent....... The story was long, but now I suppose you know why I want to be in MIT, because I can't let my fire of knowledge and passion die off! SecondChildUserIdTAG: 169784 SecondChildUserNameTAG: thewiredbear SecondChildCreateTimeTAG: 2012-10-01T10:22:17Z SecondChildTAG: Which board did you study in? Also, aren't you preparing for IIT-JEE/The new entrance exam (where mugging cannot help you)? I studied in CBSE and the focus was never on mugging things up. We had a ton of problems we needed to practice. We practiced so much that we could figure out how to do half the integration problems in our head (even ones we've not seen before). I think that was a really good thing. I was good at the basic stuff and could focus on bigger and better things. I also feel it is incorrect to criticize something if we do not even try to do something about it. In college we didn't have good teachers so a few of us decided to learn what we were good at and teach the rest of the class. We learned better ourselves in the process. In class 11 and 12, we had a subject in computer programming and RDBMS (Visual Basic and Oracle). Our teacher knew very little. She was a foxpro teacher and the syllabus just changed. Not her fault. One of my classmates however was a pro. He taught me and we taught the whole class. Half the reason why the education system is going down here is because all the good people are old and slowly retiring. The new one engineers who are very good at what they know are not interested in education at all so the quality will suffer. My point is that if you see a problem, do something about it or don't complain. Complaining is very very easy. Trying to fix things isn't. One of my classmates in engineering was very dissatisfied with the fact that no one in the faculty cared about projects in engineering. He knew professors from IISc and he got them to college. We had a one week course on basic nanotechnology and soon some of my friends were working on smart materials and MEMs sensors. I think that is a pretty big achievement considering the fact that we did nothing before this. This particular friend of mine is now pursuing his MS at Stanford only because of the work he did. No one in the college cared about projects so some friends and I started an IEEE student branch in college and conducted workshops. I don't mean to show off. My point again is that we did something about the problem. Now people are participating (and winning) in contests conducted by IEEE, Texas Instruments and Freescale. In engineering mugging could have easily helped us score full marks. There were books written by some authors which were tailor-made for the exam. We decided not to use them. We used standard textbooks prescribed by the university. We didn't get a 90%. But we understood the subject better and got better jobs.Which board did you study in? I understand why people aren't interested in education here. Research opportunities don't seem great. But that's incorrect too. I've interacted with professors from IISc (Indian Institute of Science). They are doing a lot of good work and need young people to take things to the next level. You don't have to wait to get into MIT (it would be awesome if you get into MIT. I'm not saying you shouldn't. I really hope you do.) to keep your passion going. Bad teachers didn't kill my love for DSP - the reason why I picked the stream. I find signals and systems extremely hard to understand but I still love the subject. If you really love what you are doing, it doesn't matter where you study. What matters is what you know - this something that was said at a keynote of a set of talks. The person studied in a government engineering college which has buildings which could fall off any moment. But he worked hard during his B.E. He ended up at, you guessed it, MIT for his masters and is now working as a nano-electronics engineer here in Bangalore. SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-10-01T12:31:43Z SecondChildTAG: Wish I could edit or delete that comment. I see at least one grammar mistake and I forgot to delete a line which I pasted twice. :-| SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-10-01T13:31:46Z SecondChildTAG: @ashwith yes, I am preparing for IIT-JEE, but just because my parents want me to, I have no intentions of going there, because I have many close relatives who have done IIT, and what they say is that even in their third and fourth year, they felt they had not done any practical work, my brother is in USA currently, and has done Electrical Engineering from IIT Delhi, I talk to him and he says that give him an equation now, and he would solve it, tell him to build the circuit whose equation he was solving, and he won't have " THAT " confidence, so if the best engineers of our country are such then my outlook towards the system changes, and you may have heard of the coaching institutes for IIT-JEE prep, probably you would have joined one too, I currently go to vidyamandir, the most reputed coaching for iit in Delhi, I happen to be in their top 300 students so now I study in their, as they say it, " highly sought after school integrated program", where the Vidyamandir teachers teach us the school premises itself, now isnt that time saving? I would say, at Vidyamandir, its a big NO! because they hardly care about CBSE, they teach us the IIT way!, solving big questions in just a few lines, with a bag full of tricks, now whats the problem in that? Well try solving a 6 mark question in a CBSE exam in just 4-5 lines, and failure is bound! , I hope you know what I mean, because CBSE builds you up from ground up, I mean I just had my Mid-Year Exams, and 3 days before the exam our teachers told us to start studying CBSE, and we thought we could do it all very easily, but we opened our books and it was all new, everything was new for me and majority of my class, it was like we were told to do 6 months of syllabus in 3 days, I mean it had the same topics, same everything but we had just forgot how to do CBSE, we were used to our formulas, and CBSE needs detailed answers which we always skipped for IIT prep, now me any nearly 90% of the class just barely passed in a few exams where I got a 700 something rank out of 20,000 students in a IIT level exam which is held nation wide, so that should tell you I am pretty clear on my concepts, but CBSE just doesnt jingle with IIT, that is now the biggest reason I hate school, I mean if I don't score good in school, how am I to go to any good college, and these Vidyamandir guys, give so much homework you don't have any time to study CBSE at home, for example 150 IIT level math problems in 2 and a half days, where in those days you have 30-50 questions of physics and chemistry each!......and every question takes more than 5-10 minutes to solve! You my friend were lucky enough to be in an environment, where you could do stuff to solve your problems, here if I go ahead to solve one problem I would have a 100 more, and about teaching other students of my class!?? Well I would say that would never happen as it would hurt their "I am the smartest guy" ego. Try putting 10 CGPA from 10th class guys and girls in one class and seriously speaking they don't jingle well because in our previous schools ( we all had to change schools for the integrated program) we were the smartest ones and here, that image washes off, and when you barely pass in CBSE chemistry, your confidence goes down the drain!........you were seriously lucky to have an environment, where you could solve your problems and that's why you don't hate the system, but here all the PRESTIGIOUS TEACHERS OF VIDYAMANDIR want from us is IIT so we get a good job, all of the class besides me would go in IIT for any field, they just need IIT no matter if they get Mechanical from Computers all they want is the IIT brand on their head so they get a good job, and I find that really depressing! Thats why I want MIT!!! SecondChildUserIdTAG: 169784 SecondChildUserNameTAG: thewiredbear SecondChildCreateTimeTAG: 2012-10-02T07:33:21Z SecondChildTAG: Well my point is that complaining won't help. I'm surprised about the kind of feedback you've got about the IITs. I have friends who've studied there and their feedback is different. These were from IITB and IITM however. Both say that there is a lot of importance given to practical application of concepts. For example, the one who studied at IITM had to fabricate a diode in one of his terms. They had to do everything from scratch. That's pretty awesome compared to only learning what the equations are (which mind you, is also very important. You can't make a diode if you don't know the equations). I was lucky enough to see a clean room through a window. My classmates have been to IITM tech fests. I've visited an IITB tech fest when I was in school and trust me there were some (all of them were) really cool projects. Also, in case you didn't know IITB has a nano design center - they make real sensors. IITM is building one and IISc has one as well. I wasn't lucky at all. The computer teacher I told you about, failed the friend of mine who was good at computers just because he was good - ego clash. In engineering I've had a ton of problems thanks to ego clashes with teachers. My science teachers thought I asked questions to test them (which was absolutely false). My chemistry teacher would insult people who prepared for entrance exams and said we would get nowhere. I never had a good Physics teacher from class 7 but it's my favorite science subject. I've heard of Vidyamandir. My friend's brother went there a few years back. I agree the fact that school way and IIT-JEE way differ too much. But the IIT's had to do it because of the amount of competition. The coaching institutes took advantage of the situation and found a way to make money. Students were able to crack the exam easily so the IITs had to raise their standards. So there are two causes here - the coaching classes and us who join these places in a rush to get into the IITs (I did that too). BTW, you can do an integration in 4-5 lines at your board exam. Don't always listen to what people tell you. CBSE releases a booklet (they did when I was in school) which contains sample papers and the kind of answers they expect. You'll be surprised to see how less Anyways, my 2 cents. Don't complain. Things will be nasty everywhere. There are people who have achieved a lot in very hard circumstances - not because they were lucky. It's because they worked hard and didn't complain. You'll meet a lot of them here or already have (nope, I'm not one of them.). The Indian education system does have issues. But none of them are perfect. The reason for my two long posts is that I had a similar outlook towards Indian education as you do right now. However, in my case I thought the IITs were the only good institutes followed by a few NITs. I couldn't handle IIT coaching and gave up. Didn't do well enough to get electronics through AIEEE. I wrote Karnataka CET and got a good rank and managed an electronics seat. But I hated where I was. Things changed when I did something about it. I also realized that I was in the same city as DRDO, HAL, ISRO and most importantly IISc. Please do give it all you have at PSAT and do your best to get into MIT. However, when you see a problem do something about it no matter what others do or say. The fact that you, being a high school student, can do well in this course means that you have what it takes. There were people like you in the Spring session of this course and they were all really talented. Do something about what's not right - set up a blog/website with tutorials (if you haven't yet) for example. Your classmates may think you are showing off. But there will be a 1000 times more people around the world who will be grateful that you shared what you know. Go to an electronics shop, buy some parts (not kits) and build something (I'm guessing you're already doing this however). Then tell the world how you did it. SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-10-02T14:30:34Z SecondChildTAG: @ashwith you sir, are one of those few people who have inspired me to do stuff, even my father hasn't got in that list, you are right, I should stop cribbing about my problems and rather solve them, because its when I solve them, the world will recognize me, yup, you are right, and mind you, you are rare if I am saying you are right, thanks a lot man, and yes, I did build a website, made codes in javascript, CSS, HTML, took to making a full fledged video player out of sheer javascript and CSS so the world won't have any branding on their videos, and to be true the video player is already done, but I need to polish it, you know, people won't like just buttons and solid colors hanging everywhere, seriously, it took me months to learn javascript for that and build it, 206 lines of code and 5706 length to be precise for now, and still needs polishing as I said.....but man, you know how the coaching institutes are, they don't leave you time to do anything!..... you were lucky you could drop out from the coaching, my dad has already done a payment of 100k Rs, to pay the coaching fees and change my school, so I cant bluntly leave it, again man, I would just say that you had different circumstances which built you, my circumstances were not that good, never had been, as far as I remember,...... I won't say more about my personal problems, but they have built me as such, that I just feel, I will be better off in MIT or Harvard or someplace like that, as their people really live what they do, which due to some reasons I have never been able to do here, and I don't want MIT coz its famous, there is princeton, Caltech, NYC university, there are many, but MIT and Harvard fall into my list as they have need-blind aid, so there atleast my father won't have to worry about his money.......I wish I could share my problems with you, because you seem like a guy who could help me, but the discussion ain't no place good for that, and yes IITs may go into details of as such building rather fabricating diodes, I still don't feel I would happy there....again I cant tell the reasons here......and yup, on a whole our system may have worked fine for you but, for me, the system + my family problems + my aspirations = failure, these things don't go hand in hand......I have to let go of the first two for now to atleast solve one of them in the long run, you say I should do something about it, well I have it all planned for the long run ,and believe me I have tried stuff, I have moved many people to solve my problems, but they all just added to the pain in certain parts of my anatomy, so I choose the long run path....hope you could help me.... and yes, thanks, please hope I get to MIT...and yup my PSAT is on this 20th and I am already grinding for it! and again, thanks a bunch!...I owe you one! SecondChildUserIdTAG: 169784 SecondChildUserNameTAG: thewiredbear SecondChildCreateTimeTAG: 2012-10-02T19:09:05Z SecondChildTAG: You own me nothing. Just do your best wherever you are and don't worry about where you are once you get somewhere. All the best! :-) P.S. I didn't "drop" out of coaching in that sense. I just stopped preparing seriously. We didn't have a refund policy either. But that ended up being a good thing. There were a few good teachers there and one of them got me to like inorganic chemistry and find it easy (I found it hard and hated it). It wasn't this costly though! 100k is a lot! SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-10-02T19:22:39Z FirstChildTAG: @thewiredbear...our passions are often misunderstood by those who don't share them. Just keep learning. And your line, "I know I creeped out a bit...", great stuff! :) FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-09-30T19:46:38Z SecondChildTAG: with a such a great community, how can I stop even though I just pass the HWs with a C or B grade......... and BTW.... I have done the week 3 homework with 70% and the lab with a 100%..... could do only 70% just because i started week 3, 3 hours ago!...and completed it 1 hour ago....and couldnt do the 2nd lecture completely....though I took help for the LAB... :( will try to do it on my own tomorrow, coz now its about 2 am here...... and now I have to study for my PSAT! SecondChildUserIdTAG: 169784 SecondChildUserNameTAG: thewiredbear SecondChildCreateTimeTAG: 2012-09-30T19:58:35Z FirstChildTAG: You are not late! There are people who haven't realised their dreams. For you to realise your dream now is actually very commendable. FirstChildUserIdTAG: 254325 FirstChildUserNameTAG: bondablack FirstChildCreateTimeTAG: 2012-09-30T18:10:53Z SecondChildTAG: with a such a great community, how can I stop even though I just pass the HWs with a C or B grade......... and BTW.... I have done the week 3 homework with 70% and the lab with a 100%..... could do only 70% just because i started week 3, 3 hours ago!...and completed it 1 hour ago....and couldnt do the 2nd lecture completely....though I took help for the LAB... :( will try to do it on my own tomorrow, coz now its about 2 am here...... and now I have to study for my PSAT! SecondChildUserIdTAG: 169784 SecondChildUserNameTAG: thewiredbear SecondChildCreateTimeTAG: 2012-09-30T19:58:33Z IndexTAG: 749 TitleTAG: WRITE THE ANSWER WITH CAPITAL LETTER Use the capital letter to get a correct answer in this exercise. UserIdTAG: 257736 UserNameTAG: LuisPalacios CreateTimeTAG: 2012-09-30T13:31:47Z VoteTAG: 4 CoursewareTAG: Week 3 / Isolation Via Thevenin CommentableIdTAG: 6002x_isolation_via_thevenin NumberOfReplyTAG: 1 FirstChildTAG: Thanks... :) FirstChildUserIdTAG: 95362 FirstChildUserNameTAG: asvictor FirstChildCreateTimeTAG: 2012-10-01T15:35:09Z IndexTAG: 750 TitleTAG: Tip: parallel resistor calculator Wow, I've just noticed that the calculator in this course supports a || expression to calculate the resistance for any number of resistors in parallel. Just enter the resistance values like `1k||2k` or `1000||2000||3000`. A really handy shortcut! Especially when you have a bunch of those to count in a Thevenin network :) UserIdTAG: 378267 UserNameTAG: artfwo CreateTimeTAG: 2012-09-30T09:05:00Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 751 TitleTAG: Study Group in Central Jersey (NJ, USA) Anyone interest to study for the Midterm and willing to meet living in Central Jersey please contact me ...Skype nieveskaku or just add your comment.... UserIdTAG: 164471 UserNameTAG: BillNieves CreateTimeTAG: 2012-09-29T20:22:42Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 6 FirstChildTAG: **Hello Bill:** I would love to meet to study and practice before the Midterm. If we can get a nice-sized group together that would be great. I find myself more disciplined if there are others around pursuing a common goal. I work Part-time so I am free a lot. I do not have Skype set up on my PC, however please email me and I will let you know where I live and what areas are convenient for me to drive to. Let me know what town you live in as well. **My email is MarkGT (at) Gmail (dot) com** Hope to hear from you soon, Mark in N.J. FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-09-29T20:48:01Z FirstChildTAG: Nice iniciative BillNieves! ;). I am sure that you will find a lot of Classmates from NJ! I am in Argentina, a little far from there haha. But if I can help you, I will be here in the Forum! See you! Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-29T21:49:20Z FirstChildTAG: hello I am also interested but far away.I study alone and study alone is really difficult. my Skype ID is snomi79. FirstChildUserIdTAG: 40209 FirstChildUserNameTAG: sadaf78 FirstChildCreateTimeTAG: 2012-09-29T22:09:34Z SecondChildTAG: ok mi to msamwelmollel SecondChildUserIdTAG: 475448 SecondChildUserNameTAG: msamwelmollel SecondChildCreateTimeTAG: 2012-09-29T22:21:57Z SecondChildTAG: I might be interested. I am from northern NJ though, Bergen County, so It might or might not be too far. Love the idea, just let me know where you where thinking of meeting. SecondChildUserIdTAG: 219780 SecondChildUserNameTAG: ReconIII SecondChildCreateTimeTAG: 2012-09-30T00:02:35Z SecondChildTAG: i have added you on Skype . . fayzan_ahmed172 SecondChildUserIdTAG: 146930 SecondChildUserNameTAG: fayzan_007 SecondChildCreateTimeTAG: 2012-10-21T08:38:06Z FirstChildTAG: I might be interested. I am from northern NJ though, Bergen County, so It might or might not be too far. Love the idea, just let me know where you were thinking of meeting. FirstChildUserIdTAG: 219780 FirstChildUserNameTAG: ReconIII FirstChildCreateTimeTAG: 2012-09-30T00:02:52Z FirstChildTAG: Thanks all for your comments...took just few hours to get some input. I am thinking to meet in a public library, or some coffee place like Panera ...We need to bring homework , labs and practice material. Unfortunately, our schedule may not overlap. I am wondering if Saturday Oct 6th is a good date for a first meeting, no more than 2 hours during the afternoon...Meeting options: 1. Panera at 1551 US Highway 1 Edison, NJ 08837 2. Anyone related with a community center or church willing to ask if we can use their facility ? If Saturday is not a good option, what about one night during weekdays? I will like to know your potential "sharing-time", in my case I'm a full time worker so weekends is my best option or no more than one night Monday thu. Friday after 6 PM.... We may need to meet 2 times before Midterm. This is a good opportunity to expand your professional network, and to bring ideas for further classes... I will contacting you by email with further details so please add your email address. you can reach me at wn5@njit.edu or in SKYPE nieveskaku. Thanks again for your attention, William FirstChildUserIdTAG: 164471 FirstChildUserNameTAG: BillNieves FirstChildCreateTimeTAG: 2012-09-30T02:00:35Z SecondChildTAG: **Bill:** I got your email. Thanks! I know that old exams are available somewhere on the MIT site, and they will be good for practice exams. **I wonder if MITx will give out an "official" practice exam that we can work on.** Maybe we can request one via this board? About the "format" of the sessions (I am just throwing an idea out there, it may or may not work): I think we should designate a certain set of easier problems that we will all work on prior to meeting, and then we will discuss our strategies for solving the problems to those in the "group" that did not understand them. We should also designate one or two problems, especially challenging ones, to work on "during" the study session. We won't have the time to work on them through to completion, but we should discuss strategy on how to begin the 'tough' problems, finish them at home, and discuss the answers at the final study session. Mark SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-09-30T04:40:10Z FirstChildTAG: Hi guys . .. This is Fayzan Ahmed and i am from Pakistan . . i would also like to join your study group. my skype id is "fayzan_ahmed172@ FirstChildUserIdTAG: 146930 FirstChildUserNameTAG: fayzan_007 FirstChildCreateTimeTAG: 2012-10-21T08:17:39Z SecondChildTAG: "fayzan_ahmed172" SecondChildUserIdTAG: 146930 SecondChildUserNameTAG: fayzan_007 SecondChildCreateTimeTAG: 2012-10-21T08:18:04Z IndexTAG: 752 TitleTAG: That song... For anyone interested in the song that is played. It's Santeria by Sublime. Without distortion, of course. [http://www.youtube.com/watch?v=AEYN5w4T_aM][1] [1]: http://www.youtube.com/watch?v=AEYN5w4T_aM UserIdTAG: 359310 UserNameTAG: AaronYeoh CreateTimeTAG: 2012-09-28T11:34:23Z VoteTAG: 4 CoursewareTAG: Week 4 / Incremental Analysis Demo CommentableIdTAG: 6002x_inc_analysis_demo NumberOfReplyTAG: 1 FirstChildTAG: Loving the Mexican gangsta music to liven up a lecture on the oscilloscope. I was just disappointed they cut off Bad to the Bone. FirstChildUserIdTAG: 348408 FirstChildUserNameTAG: pilot_project FirstChildCreateTimeTAG: 2012-09-28T20:56:01Z IndexTAG: 753 TitleTAG: Why polymer engineers have to learn EECS Polymers have viscoelastic behaviour. Sometimes they behave like springs, sometimes like a plasticine. Much of those problems are approached by simplifications very similar to EECS. Spring (Youngs Modulus) has reply similar to a resistor. St Venant model for viscoelasticity could be approached as an imperfect diode. UserIdTAG: 320612 UserNameTAG: Ezechiel CreateTimeTAG: 2012-09-27T22:29:18Z VoteTAG: 4 CoursewareTAG: Week 2 / Speakers CommentableIdTAG: 6002x_speakers_tutorial NumberOfReplyTAG: 1 FirstChildTAG: I never knew circuits could be used for modelling polymers like this. Thanks for sharing! :-) Could you please provide some more information on this? FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-09-28T02:50:29Z SecondChildTAG: Look for some basic rheology courses in polymers. St Venant / Newton / Rouse model act for engineers as a simplification of Navier-Stokes eqts, similiar as EECS acts as a application of Maxwell's. SecondChildUserIdTAG: 320612 SecondChildUserNameTAG: Ezechiel SecondChildCreateTimeTAG: 2012-10-02T08:26:19Z IndexTAG: 754 TitleTAG: Taylor's series Hello, I've got a question. Is there any way to understand incremental method without knowing Taylor's series? It's the first time I hear of Taylor's series, I never learned about it. Is it necessary to know and understand that? Or it's a must and I can't proceed without it? UserIdTAG: 405039 UserNameTAG: Zaknafein CreateTimeTAG: 2012-09-27T18:27:11Z VoteTAG: 4 CoursewareTAG: Week 4 / Incremental Method Math CommentableIdTAG: 6002x_inc_method_math NumberOfReplyTAG: 2 FirstChildTAG: Yes, there is. Consider the straight line that is tangent to a function f(), at a given point, x. Now consider x + delta_x, where delta_x is very small. The point on that straight line at x+delta_x will be close to the value of the function f(x+delta_x). That is, they are "approximately equal." Some books give this approximation to f(x+delta_x) the name of "differential." I'll scout around for a picture of what I'm talking about (I've never posted an image to the forum) but it boils down to the same approximation we get from our use of the Taylor Series. EDIT +16 days: See text, Pg. 411, figure at bottom of page. FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-09-27T19:12:58Z FirstChildTAG: >It's the first time I hear of Taylor's series, I never learned about it. probably it's time to learn ;) FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-09-27T21:27:57Z IndexTAG: 755 TitleTAG: How to get VL I dont understand how to get VL, can someone help me please? UserIdTAG: 154373 UserNameTAG: JReyes87 CreateTimeTAG: 2012-09-26T22:45:58Z VoteTAG: 4 CoursewareTAG: Week 2 / Thevenin model CommentableIdTAG: 6002x_S3E5_Thevenin_Model NumberOfReplyTAG: 1 FirstChildTAG: Ok I got it now... using the voltage divider approach FirstChildUserIdTAG: 154373 FirstChildUserNameTAG: JReyes87 FirstChildCreateTimeTAG: 2012-09-26T23:08:32Z IndexTAG: 756 TitleTAG: Cardano's method for solving cubic equations [Wiki][1] [1]: http://en.wikipedia.org/wiki/Cubic_function#Cardano.27s_method UserIdTAG: 137421 UserNameTAG: i026e CreateTimeTAG: 2012-09-25T11:56:56Z VoteTAG: 4 CoursewareTAG: Week 3 / Graphical Analysis of Nonlinear Circuits Exercise CommentableIdTAG: 6002x_graphical_analysis_of_nonlinear_circuits_exercise NumberOfReplyTAG: 0 IndexTAG: 757 TitleTAG: Offline use of the course material(Including Video) Hi - I do not have internet access all the time. Is it possible to post compressed form of this course (Including videos) every week, so that I may download and work on it even without internet access (offline) - Thanks UserIdTAG: 350697 UserNameTAG: Khalil_Awan CreateTimeTAG: 2012-09-24T13:54:19Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 4 FirstChildTAG: use firefox video download helper,it will help you to download the videos and convert it FirstChildUserIdTAG: 112406 FirstChildUserNameTAG: MKG FirstChildCreateTimeTAG: 2012-09-24T17:57:44Z FirstChildTAG: But videos are for week 2 only, Can it will be done for all the coming weeks. It will enable us to work off line especially where we have slower internet speed. FirstChildUserIdTAG: 350697 FirstChildUserNameTAG: Khalil_Awan FirstChildCreateTimeTAG: 2012-09-25T12:19:28Z FirstChildTAG: You can download videos from [download videos][1] videos are uploaded now to lesson 14 ,slides and wiki are included subtitles are in srt format; if you want to see subtitles use vlc mediaplayer i have 3.5TB/month but don't know how the brinksterserver behaves under heavy load. Also for people with **bad internetconnections** I have 50 8GB sticks available with all those videos for free. Just post your postal address to 6002x@ruudoleo.com together with your alias and I will send the USB stick to the 50 first responders. No other use of address will be made. I make this offer because in the mitx 6002x course there were people who couldn't watch the videos by reason of bandwidth limitations and it seems the problem is not improving. [1]: http://www.ruudoleo.com/mitx/small/6002xvideosmall.htm "download videos" FirstChildUserIdTAG: 79885 FirstChildUserNameTAG: ruudoleo FirstChildCreateTimeTAG: 2012-09-26T21:53:31Z FirstChildTAG: The most recent solution posted for students in countries where YouTube is blocked should also make it slightly simpler to watch videos off-line. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-09-25T03:32:11Z IndexTAG: 758 TitleTAG: i did it i just could solve lab 2 before the due time by 1 min i couldn't believe it UserIdTAG: 453422 UserNameTAG: MennatAllah CreateTimeTAG: 2012-09-23T22:03:26Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Ha. Congratulations! :) FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-09-23T22:27:37Z SecondChildTAG: give me solution please SecondChildUserIdTAG: 331441 SecondChildUserNameTAG: abdessamade SecondChildCreateTimeTAG: 2012-09-23T23:55:52Z FirstChildTAG: I couldn't solve it and the time is up. How I can get the solution? FirstChildUserIdTAG: 437236 FirstChildUserNameTAG: Hossameldinradwan FirstChildCreateTimeTAG: 2012-09-23T23:02:41Z SecondChildTAG: Are you sure you can't continue trying? Here it is midnight (UK) and I know I still have another 10h at least. I think someone said; as long it is 23rd somewhere in the world you can still hand in your work. SecondChildUserIdTAG: 207204 SecondChildUserNameTAG: hgustavii SecondChildCreateTimeTAG: 2012-09-23T23:08:20Z SecondChildTAG: Here in Brazil, are 23:00 pm. SecondChildUserIdTAG: 375517 SecondChildUserNameTAG: CristianeTeburcio SecondChildCreateTimeTAG: 2012-09-24T02:05:44Z IndexTAG: 759 TitleTAG: I don't Understand the last item It says that V2= 1.6 but i don't know why the result of the current that will flow is 0.17513859.... i need someone tell me the procedure.! UserIdTAG: 472876 UserNameTAG: otakudany CreateTimeTAG: 2012-09-23T14:02:52Z VoteTAG: 4 CoursewareTAG: Week 1 / Battery model CommentableIdTAG: 6002x_S1E9_Battery_Model NumberOfReplyTAG: 1 FirstChildTAG: You have to think in terms of the Kirchhoff's Voltage Law (KVL), since there are no nodes. By doing so, you will be able to find out the value of the current in the loop. FirstChildUserIdTAG: 9100 FirstChildUserNameTAG: dmgongora FirstChildCreateTimeTAG: 2012-09-24T04:37:44Z SecondChildTAG: explain... SecondChildUserIdTAG: 486510 SecondChildUserNameTAG: RamanRJN SecondChildCreateTimeTAG: 2012-09-27T20:42:28Z SecondChildTAG: Remember that the circuit only consists of (battery+resistance) in parallel with (battery+resistance), no connections are made for the load. You know that V1 = 1.5, and V2 = 1.6, and you know from the question that because of this voltage difference between the batteries, the current will flow from V2 to V1, which is what we consider charging the battery. You also know that the voltage across V1 + RV1 has to equal V2 + RV2, but V1 and V2 are constant values. Hopefully that is enough to help you solve it. SecondChildUserIdTAG: 3571 SecondChildUserNameTAG: ifoughtsharks SecondChildCreateTimeTAG: 2012-09-28T17:27:01Z SecondChildTAG: I get the 0.0175438596491 as the answer V2-V1 = 0.1 R1+r2 = 5.7 ohms 0.1/5.7 = 0.017543A SecondChildUserIdTAG: 672329 SecondChildUserNameTAG: PeterHunt SecondChildCreateTimeTAG: 2012-10-18T13:39:29Z SecondChildTAG: R1+R2=0.57ohm correct this and u will get correct answer SecondChildUserIdTAG: 482042 SecondChildUserNameTAG: utkarshpar SecondChildCreateTimeTAG: 2012-12-24T17:11:04Z SecondChildTAG: how can you add both resister, even they are parallel to each other?? SecondChildUserIdTAG: 1069402 SecondChildUserNameTAG: Kamleshpri SecondChildCreateTimeTAG: 2013-01-27T17:10:21Z SecondChildTAG: They are parralell reletively to V. But for current that flows from V2 to V1 they are connected in series. And it is sayed that circuit is opened, so current don't flows anywhere else except the direction i have mentioned above. SecondChildUserIdTAG: 1254417 SecondChildUserNameTAG: egor_g SecondChildCreateTimeTAG: 2013-02-23T09:03:43Z IndexTAG: 760 TitleTAG: Lab4 [Solved] Suddenly got stuck on Lab4. I'm have no problem with understanding the task, system accepted my schematic, but numerical answer on first question is not accepted. It's hard to describe things without violating Honor Code, but I'll try: - in conditions, described in Lab I've got the current through DS port is around 10.5u Is this value close to right value, or I have missed something? UserIdTAG: 329361 UserNameTAG: ikrukov CreateTimeTAG: 2012-09-23T09:22:09Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: I believe the system provides different values for each student. However, what you have to do is to measure ids twice for the circuit you made for uds=max value for you. You have to change the value for ugs twice (to the values you are given), and measure ids in each case. If you get it right it is a piece of cake. :) FirstChildUserIdTAG: 304587 FirstChildUserNameTAG: Vasso FirstChildCreateTimeTAG: 2012-09-23T09:36:10Z SecondChildTAG: Vasso, thank you for comment. Not sure that I'm uderstand you correctly. I'm asking about the first numerical question, on computing effective RON. My understanding - I have to apply given voltage to GS gate, measure the current through DS port with specified voltage on DS port and calculate the appropriate resistance. Am I wrong? SecondChildUserIdTAG: 329361 SecondChildUserNameTAG: ikrukov SecondChildCreateTimeTAG: 2012-09-23T09:46:16Z SecondChildTAG: i have a common doubt i have started course right now i don't know how to answer the questions can u please help me o solve problems SecondChildUserIdTAG: 473143 SecondChildUserNameTAG: joyson SecondChildCreateTimeTAG: 2012-09-23T10:30:31Z SecondChildTAG: I was asked to find VT in first numerical question of Lab4. Sorry for my missunderstanding! Glad you found it! SecondChildUserIdTAG: 304587 SecondChildUserNameTAG: Vasso SecondChildCreateTimeTAG: 2012-09-23T12:08:20Z FirstChildTAG: Found a bug in my schematic. Everything is OK now. FirstChildUserIdTAG: 329361 FirstChildUserNameTAG: ikrukov FirstChildCreateTimeTAG: 2012-09-23T10:32:13Z SecondChildTAG: please tell the bug, . iam stuck on the same problem SecondChildUserIdTAG: 33266 SecondChildUserNameTAG: MBialaschik SecondChildCreateTimeTAG: 2012-09-26T18:05:51Z SecondChildTAG: forgett it. i see it now. some floating messed it up^^ SecondChildUserIdTAG: 33266 SecondChildUserNameTAG: MBialaschik SecondChildCreateTimeTAG: 2012-09-26T18:08:31Z SecondChildTAG: I am having similar problem . When I apply the given vgs i get ids same as what you mentioned in your post . How did you find what was wrong? SecondChildUserIdTAG: 127488 SecondChildUserNameTAG: iwannalearn SecondChildCreateTimeTAG: 2012-09-30T08:22:45Z SecondChildTAG: problem solved . irukov , MBialaschik I think I also did the same mistake you did . I did not ground my dc voltage of 3 volt. MBialaschik , your comment really helped me to find my mistake. Thank you guys. SecondChildUserIdTAG: 127488 SecondChildUserNameTAG: iwannalearn SecondChildCreateTimeTAG: 2012-09-30T08:33:41Z FirstChildTAG: Probably it's make sense to point out my mistake - it was the same as mentioned in iwannalearn post. So for all who have troubles with numeric answers in this lab - check that ALL voltage sources in your schematic connected to the same ground level! P.S. To stuff: I hope this post not violates HC. If it is not so, please delete it. FirstChildUserIdTAG: 329361 FirstChildUserNameTAG: ikrukov FirstChildCreateTimeTAG: 2012-09-30T11:30:39Z SecondChildTAG: Hi everyone! I have problem with Ron. I find Ids=0.042mA in point Vds=1V. Using Ron=Vds/Ids but answer is incorrect.. can't find where is error. SecondChildUserIdTAG: 387154 SecondChildUserNameTAG: Junglist SecondChildCreateTimeTAG: 2012-10-01T11:28:09Z SecondChildTAG: thnx, solved SecondChildUserIdTAG: 387154 SecondChildUserNameTAG: Junglist SecondChildCreateTimeTAG: 2012-10-01T13:05:43Z SecondChildTAG: i find the same value for ids as junglist,whats wrong. SecondChildUserIdTAG: 400832 SecondChildUserNameTAG: panosbr SecondChildCreateTimeTAG: 2012-10-06T16:07:19Z IndexTAG: 761 TitleTAG: Buttons Showing solutions and answers to the practice questions I first of all I thank the wonderful work done by the developers of this site and the courses that is mounted. Thanks alot.. please I will like to suggest that a button should such as show answer will not only give the answer but also the solution to the answer . or a button showing solutions to the practice questions so that we will discuss only area of difficulty and or other related issues thanks UserIdTAG: 382532 UserNameTAG: emmanuelpeace CreateTimeTAG: 2012-09-23T05:13:18Z VoteTAG: 4 CoursewareTAG: Week 2 / Node equation review CommentableIdTAG: 6002x_S3E1_NodeEquationReview NumberOfReplyTAG: 0 IndexTAG: 762 TitleTAG: Lab 2 Help I am still stuck on lab 2, I am really confused on where to start with the problem. Can anyone point me in the right direction of what I should do? UserIdTAG: 219780 UserNameTAG: ReconIII CreateTimeTAG: 2012-09-23T00:59:43Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/505e4def893f19270000002e FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-23T01:10:01Z SecondChildTAG: FINALLY!!!!!!! got it thanks soo much for the help! SecondChildUserIdTAG: 219780 SecondChildUserNameTAG: ReconIII SecondChildCreateTimeTAG: 2012-09-23T01:52:10Z FirstChildTAG: I used superposition with node analysis. I used 3 resistors. One of the resistors kind of alternates to parallelism when I short the voltage for superposition with the corresponding voltage supply that I short. FirstChildUserIdTAG: 247286 FirstChildUserNameTAG: ronald_borunda FirstChildCreateTimeTAG: 2012-09-23T04:55:47Z IndexTAG: 763 TitleTAG: I solved this in this way: the operating point is $I_d = f(v_d)$ and this is the operating point so we know that $I_d = 4$, so $4=v^3$ that means that $v=1.587401052 volt$ is that right? UserIdTAG: 292360 UserNameTAG: Loai CreateTimeTAG: 2012-09-22T16:31:06Z VoteTAG: 4 CoursewareTAG: Week 4 / Graphs Exercise CommentableIdTAG: 6002x_graphs_exercise NumberOfReplyTAG: 1 FirstChildTAG: Loai, your answer is close, and actually the system will consider it correct. However, it is not the exact answer. The problem is that not all the 4 Amps are going through the non-linear device, and part of it will go through the resistor. Can you find a way to make your answer more precise? FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-22T19:22:32Z SecondChildTAG: You just need to write KCL, use the given condition i=v^3 and remember that voltage drop on R and the circuit is the same. Based on these equations you will find the operating voltage. SecondChildUserIdTAG: 271458 SecondChildUserNameTAG: haykp SecondChildCreateTimeTAG: 2012-10-06T14:22:57Z IndexTAG: 764 TitleTAG: lab 2 Thanks for pointing me in the right direction. I was determined to figure this out one way or another. It only took me I dont know how many tries, but if anyone is having trouble with it do not give up, you will get it through determination. UserIdTAG: 160242 UserNameTAG: Mlevins35 CreateTimeTAG: 2012-09-22T14:59:26Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 765 TitleTAG: Values of homework change I hammered the check button like 10e10 times, then suddenly my values of the homework changed (Resistance was lower and current was doubled). Is there a limit on checking before values change? Edit: values are back to normal (must have been some noise signal ;O) UserIdTAG: 388555 UserNameTAG: 4lk4tr43 CreateTimeTAG: 2012-09-21T13:11:17Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: There's an answer to this question in the tutorial section. Some questions can be checked unlimited number of times, others have a limited number of trials after which the values can be changed no more. FirstChildUserIdTAG: 252722 FirstChildUserNameTAG: vaboro FirstChildCreateTimeTAG: 2012-09-21T14:50:09Z SecondChildTAG: Yea I knew that, but that's not the problem. I can still enter any values, what happened was, that the question parameters changed. The progress still shows the answers as correct. BTW I thought the limited answer count will only happen at the exams. SecondChildUserIdTAG: 388555 SecondChildUserNameTAG: 4lk4tr43 SecondChildCreateTimeTAG: 2012-09-21T15:01:09Z IndexTAG: 766 TitleTAG: e= 6.20 answer is wrong 6.20v is correct without the earth-sign below the circuit. With earth-sign I think must be e= 1.21v UserIdTAG: 410943 UserNameTAG: JPalaciosRoman CreateTimeTAG: 2012-09-21T12:59:31Z VoteTAG: 4 CoursewareTAG: Week 2 / Node equation review CommentableIdTAG: 6002x_S3E1_NodeEquationReview NumberOfReplyTAG: 2 FirstChildTAG: how is dat go wrong pls mail me dat to aathirajaramp@gmail.com FirstChildUserIdTAG: 325182 FirstChildUserNameTAG: Aathi FirstChildCreateTimeTAG: 2012-09-21T14:01:09Z SecondChildTAG: i got 1.69 and i think this might sound SecondChildUserIdTAG: 475448 SecondChildUserNameTAG: msamwelmollel SecondChildCreateTimeTAG: 2012-09-22T11:17:17Z SecondChildTAG: i mean -1.69V SecondChildUserIdTAG: 475448 SecondChildUserNameTAG: msamwelmollel SecondChildCreateTimeTAG: 2012-09-22T11:18:39Z SecondChildTAG: Even i dont understand why the given answer has to be wrong.. Can someone pls explain?? SecondChildUserIdTAG: 413948 SecondChildUserNameTAG: raswini93 SecondChildCreateTimeTAG: 2012-09-22T13:46:54Z SecondChildTAG: actually + of V2 is grounded. so it is -(-7.2)v = +7.2v SecondChildUserIdTAG: 310108 SecondChildUserNameTAG: Saakar SecondChildCreateTimeTAG: 2012-09-22T19:25:05Z SecondChildTAG: i too got the same -1.69 ..by Nodal analysis SecondChildUserIdTAG: 342886 SecondChildUserNameTAG: hthimz SecondChildCreateTimeTAG: 2012-09-23T07:09:57Z SecondChildTAG: i got the same answer of -1.69v by nodal analysis and superposition SecondChildUserIdTAG: 251230 SecondChildUserNameTAG: esle SecondChildCreateTimeTAG: 2012-09-23T19:37:39Z FirstChildTAG: 6.2 is a correct answer. Looks like you consider V1 "+" side potential is zero, but its not. V1 "-" is ground, zero potential. V1 "+" is not zero, and defined by V1 power. The difference of potentials you found should be added to V1 "+" potential. FirstChildUserIdTAG: 66669 FirstChildUserNameTAG: agarus FirstChildCreateTimeTAG: 2012-09-22T15:05:48Z IndexTAG: 767 TitleTAG: S4E2 explanation (with graph) ![enter image description here][1] ![enter image description here][2] ![enter image description here][3] #truth tables= 2^(2^2)=16 see more details: http://mathworld.wolfram.com/BooleanFunction.html [1]: https://edxuploads.s3.amazonaws.com/13482167683819458.bmp [2]: https://edxuploads.s3.amazonaws.com/13482173052691641.bmp [3]: https://edxuploads.s3.amazonaws.com/1348217343269160.bmp UserIdTAG: 138445 UserNameTAG: jih43711 CreateTimeTAG: 2012-09-21T08:53:04Z VoteTAG: 4 CoursewareTAG: Week 2 / Boolean functions CommentableIdTAG: 6002x_S4E2_Boolean_Functions NumberOfReplyTAG: 1 FirstChildTAG: f1 is the and gate f7 is the or gate f14 is the nand gate f8 is the nor gate I know of exclusive and , exclusive or, exclusive nand + exclusive nor but do the other combinations have names? That I don't know FirstChildUserIdTAG: 145945 FirstChildUserNameTAG: FrankBishop FirstChildCreateTimeTAG: 2012-09-21T18:26:56Z SecondChildTAG: all 16 combos have names. refer a book on digital circuits and logic design by Morris Mano and u'll get all 16 names SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-12-19T13:42:20Z IndexTAG: 768 TitleTAG: Python code to solve H2P1 Since H2P1 involves choosing from a combination of several resistors, I decided to make a python program that iterates through all those possible combinations and test them against the constraints and specifications of the circuit. The program then presents a set of resistor pairs to choose from and their respective max and min Vouts. Although that was very useful to me, and I had to think the problem through to come up with the right formulas, the program presents a set of correct answers to the user, so I don't know if it's proper to post the code here. Will I break the honor code by doing so? Should I wait for the homework's due date to post it here? What do you think? Cheers. UserIdTAG: 185338 UserNameTAG: mgaldieri CreateTimeTAG: 2012-09-21T07:17:13Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Great mgaldieri! But I think that you should Post it **after** the due date of the Homework 2 :). Posting it now it would be like giving precise answers. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-21T11:15:46Z SecondChildTAG: It is fun trying different pair resistance and come up with right solution after discarding first and try another strategy and when you go to text book and answer and solution is already solved in form of example just what you need to have to do is to find the pattern in your problem and apply the solution.... great have fun.. SecondChildUserIdTAG: 272523 SecondChildUserNameTAG: Jivraj SecondChildCreateTimeTAG: 2012-09-21T12:44:47Z SecondChildTAG: You're right! I'll wait for the due date to post the code. After all, why take the fun out of solving these problems? :) SecondChildUserIdTAG: 185338 SecondChildUserNameTAG: mgaldieri SecondChildCreateTimeTAG: 2012-09-21T15:40:53Z FirstChildTAG: can i have the answer of h2p1 for vin=20v and vout=6v. i have try many logics but still invain. any one who can help me before due date FirstChildUserIdTAG: 328820 FirstChildUserNameTAG: saifee FirstChildCreateTimeTAG: 2012-09-21T12:46:40Z FirstChildTAG: Actually there is s logical provable answer within the given constraints.Nice idea though writing the program to cut off the manual work when using the trial and error method. FirstChildUserIdTAG: 396673 FirstChildUserNameTAG: anthonypraveen FirstChildCreateTimeTAG: 2012-09-26T07:40:05Z IndexTAG: 769 TitleTAG: H4P3/S8E2 Thevenin resistance Hi, I struggled with S8E2, but got the green ticks eventually, and was fairly confident about my method. Now I'm on H4P3 part A, and while I can calculate the thevenin voltage and get marked correct, I can't find the correct answer for the thevenin resistance, using the same method as S8E2. My method: To find $V_{TH}$ I use the node method at the node connecting $I_0$, $R_1$ and $Z$$i$. Then I solve for $i$ to get its value. Then I substitute the value of $i$ into the KVL equation for the loop including the two resistors, solving to find the voltage across $R_2$, which is the thevenin voltage. The site says my answer is correct. Now to find $R_{TH}$... I turn off the independent source $I_0$, replacing it with an open circuit. Now as there is no current flowing, since I just turned off the only independent source, the dependent source must be off, so I replace that with a short. I'm left with 2 resistors connected in parallel, so I use the $R_1$ || $R_2$ calculation to find the equivalent resistance. Right answer on S8E2, wrong answer on H4P3. What am I missing? I know that S8E2 hints at adding a current source at the port and using superposition. I spent a long time on that with S8E2, wasn't getting anywhere, then I arrived at the above method. Maybe what I'm missing is something on that front, any hints on what I should be measuring after adding the extra current source? UserIdTAG: 2620 UserNameTAG: dsd CreateTimeTAG: 2012-09-21T02:24:23Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Important point that I haven't seen covered is that the current controlled voltage source looks like a resistor when calculating Rth. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-09-21T09:06:52Z SecondChildTAG: Thanks. When I treated the dependent source like that, the calculation worked. I'd be interested in the background/explanation here, if anyone can offer it. SecondChildUserIdTAG: 2620 SecondChildUserNameTAG: dsd SecondChildCreateTimeTAG: 2012-09-22T22:29:25Z SecondChildTAG: Thank you. I had tried this earlier out of desperation and apparently entered the wrong numbers. I made sure I entered the correct numbers this time and I got the green checkmark (whew!). HOWEVER...like you, I do not understand WHY I should treat the CCVS this way (for getting Rth, Guess I'll have to review how dependent sources are treated, again.). Thanks again. SecondChildUserIdTAG: 92346 SecondChildUserNameTAG: MobiusTruth SecondChildCreateTimeTAG: 2012-10-01T02:30:50Z SecondChildTAG: I may have had a 'moment' for this one. The problems I'm having stem from the CCVS. If I DO use a test source to find the Thevenin equivalent then I think I should have a current with which to evaluate the dependent CCVS. I don't usually use that test source but that IS the way the equivalents are defined, so it's an honest approach. :) I'll have to try it tonight after work. SecondChildUserIdTAG: 92346 SecondChildUserNameTAG: MobiusTruth SecondChildCreateTimeTAG: 2012-10-01T14:36:51Z FirstChildTAG: HI, can you help me in finding Vth and Vn.U said u already solved it. I suck at thevenen and norton so i am finding trouble solving them.. Can u tell me step by step with a little detail. Thanks FirstChildUserIdTAG: 18177 FirstChildUserNameTAG: MianSaad FirstChildCreateTimeTAG: 2012-09-22T19:00:37Z FirstChildTAG: Wait wait, if you turn off the independent source, it does not mean that the dependent source is off too ;) Try to think on that ^^ FirstChildUserIdTAG: 200496 FirstChildUserNameTAG: Ranieri FirstChildCreateTimeTAG: 2012-09-21T07:41:34Z SecondChildTAG: its me Harvey I really need your help please come back on skype.......................................................... Remember me HArvey Specter SecondChildUserIdTAG: 378080 SecondChildUserNameTAG: GladIDidThis SecondChildCreateTimeTAG: 2012-11-18T22:27:42Z IndexTAG: 770 TitleTAG: Incremental Resistance How do you get that? UserIdTAG: 245291 UserNameTAG: rlicas CreateTimeTAG: 2012-09-20T17:17:03Z VoteTAG: 4 CoursewareTAG: Week 4 / Linearization Exercise 1 CommentableIdTAG: 6002x_linearization_ex_1 NumberOfReplyTAG: 2 FirstChildTAG: Hi. please see 221 page on the book. FirstChildUserIdTAG: 378778 FirstChildUserNameTAG: happylife FirstChildCreateTimeTAG: 2012-09-20T19:54:26Z FirstChildTAG: It's the slope of the curve at the operational point (1.088 V). FirstChildUserIdTAG: 220837 FirstChildUserNameTAG: Calsomus FirstChildCreateTimeTAG: 2012-09-21T22:27:35Z SecondChildTAG: actually,it's the reciprocal of the slope SecondChildUserIdTAG: 222175 SecondChildUserNameTAG: harjot SecondChildCreateTimeTAG: 2012-09-24T08:28:53Z SecondChildTAG: How do you find that the operational point is 1.088V ? SecondChildUserIdTAG: 37280 SecondChildUserNameTAG: zour SecondChildCreateTimeTAG: 2012-09-24T08:39:43Z IndexTAG: 771 TitleTAG: Effected by the youtube block I am just stuck can't play any thing please tell me what should I do ? UserIdTAG: 40209 UserNameTAG: sadaf78 CreateTimeTAG: 2012-09-19T11:15:14Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Download the videos from OCW. http://ocw.mit.edu/index.htm FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-19T12:37:17Z SecondChildTAG: Sorry, it looks like those are youtube videos as well. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-19T13:21:39Z SecondChildTAG: i read it somewhere For those who can't watch the lecture videos and also want to archive them. 1. Right click on the video and choose "Copy link address." 2. Go to "keepvid" website. 3. Paste the video link in the box and press the download button in front of it. 4. Choose the resolution of the video and 5. SAVE IT! SecondChildUserIdTAG: 145503 SecondChildUserNameTAG: chirag3553 SecondChildCreateTimeTAG: 2012-09-19T13:50:02Z FirstChildTAG: You can see in the Wiki [Lecture Videos Download][1]:). Also see more info [here][2]: **Due to the YouTube outage in certain parts of the world, we have provided direct links to download the lecture videos, and extended the Week 2 homework and lab deadlines to the coming Sunday (September 30). Please note that Week 3 deadlines have not changed.** [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/ [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/info FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-26T11:49:10Z IndexTAG: 772 TitleTAG: Wrong captions There are wrong captions next to the video. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-19T10:52:25Z VoteTAG: 4 CoursewareTAG: Week 3 / Graphical Analysis of Nonlinear Circuits CommentableIdTAG: 6002x_graphical_analysis_of_nonlinear_circuits NumberOfReplyTAG: 0 IndexTAG: 773 TitleTAG: video lecture problem you tube is blocked in my region,so i can't see the lectures ,please five me somr alternate soloution to see the videos. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-18T19:02:29Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: You can download the MIT OCW version of 6.002, it was taped live in front of actual MIT students. The link is here: http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-electronics-spring-2007/ You can also try to use a proxy server or VPN to bypass country restrictions on Youtube. FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-09-18T23:08:58Z SecondChildTAG: thanks for the link SecondChildUserIdTAG: 272417 SecondChildUserNameTAG: MuKhan SecondChildCreateTimeTAG: 2012-09-19T07:44:54Z FirstChildTAG: can you tell me how can I use VPN to bypass country restrictions on YouTube. FirstChildUserIdTAG: 40209 FirstChildUserNameTAG: sadaf78 FirstChildCreateTimeTAG: 2012-09-19T11:29:01Z SecondChildTAG: Basically, vpn is a a way of connecting securely to a company server. eg. I user vpn when working from home to connect to my office network. Once I connect via vpn, it's as if I am on the lan internally. What you would need to do is obtain a VPN service from another country than your own, where YouTube is not restricted. Then, when you start the vpn client program on your computer, you connect to the vpn server in that country. From that point on, as far as the rest of the internet is concerned, you are browsing from that country. Search for commercial vpn services on the internet, you should be able to find something suitable - but please be aware, that depending on your own countries laws, it may be illegal to circumvent the ban that is in place. Regards. M SecondChildUserIdTAG: 385677 SecondChildUserNameTAG: Michaelc1 SecondChildCreateTimeTAG: 2012-09-22T00:02:07Z IndexTAG: 774 TitleTAG: S7E3: LINEARIZATION I don't know how to solve the last part of this exercise. I have calculated RD, but I am having problems to calculate VI. Is it supposed that Vi=VI or are they different? I suppose that RD=100, then I obtain VD with the equation of the first part of the exercise. Using KLC I obtain this ecuation: VI-R*ID-RD*VD=0 but when I obtain VI the solution is wrong. Could someone give me a hint? UserIdTAG: 370769 UserNameTAG: PabloFCid CreateTimeTAG: 2012-09-18T13:32:06Z VoteTAG: 4 CoursewareTAG: Week 4 / Linearization Exercise 2 CommentableIdTAG: 6002x_linearization_ex_2 NumberOfReplyTAG: 5 FirstChildTAG: I also have problem with the second part. I tried to calculate what is DC voltage on diode to give 100 Ohm resistance. So I used equation from first part (1/(I0*(1/VT)*e^(VD/VT))). Then I calculated that voltage (VD) as VD = ln( VT/(R*I0) ) * VT, where R is 100 Ohm resistance, or 1000 Ohm resistance in last exercise. Then I calculated that VD is 1/40 of VI (because of resistor divider). So VI = 40 * VD. Where I made a mistake? FirstChildUserIdTAG: 146390 FirstChildUserNameTAG: JGradzki FirstChildCreateTimeTAG: 2012-09-18T16:15:36Z SecondChildTAG: JGradzki, I initially did the exact same calculation as you. I was going in circles for a long time! I was finally able to get the correct answers by remembering that I can't directly back-substitute the (small-signal) equation for the incremental resistance into a (large-signal) KVL equation for the bias voltage. Try taking your correct expression for the incremental resistance and substituting it into the original equation for the diode current. Many terms cancel out, but you should be left with something nontrivial. Then, you can use that expression with nodal analysis to solve for the remaining unknown, VI--I did it with a KCL equation at the output port. HTH, tmcnulty SecondChildUserIdTAG: 139646 SecondChildUserNameTAG: tmcnulty SecondChildCreateTimeTAG: 2012-09-18T21:19:48Z SecondChildTAG: Thank you! That really helped! SecondChildUserIdTAG: 146390 SecondChildUserNameTAG: JGradzki SecondChildCreateTimeTAG: 2012-09-18T21:42:24Z SecondChildTAG: Very good. I was doing the same mistake. It is important not to mix small signal circuit with large signal circuit. If I'm not wrong, the incremental resistor is only valid like a resistor model of a diode in the linearized small signal circuit, because in the large signal circuit, a diode is not like a resistor. SecondChildUserIdTAG: 362201 SecondChildUserNameTAG: Albertbano SecondChildCreateTimeTAG: 2012-09-20T19:52:05Z FirstChildTAG: Thanks for your replies. I was having a problem with the rounding off. I have solved it. FirstChildUserIdTAG: 370769 FirstChildUserNameTAG: PabloFCid FirstChildCreateTimeTAG: 2012-09-19T08:32:53Z FirstChildTAG: tmcnulty please explain me what substituting incremental value in original current equation means? FirstChildUserIdTAG: 211715 FirstChildUserNameTAG: pitankar FirstChildCreateTimeTAG: 2012-09-26T12:29:25Z FirstChildTAG: I spent quite a bit of time not only on solving this problem, but on understanding how exactly the "correct" answers appeared in the check boxes. I decided to share my approach in order to initiate the discussion on whether it was actually correct. First, I computed and drew the load line for resistor R: ![The load line for resistor R][1] The extreme points (1) and (2) on horizontal and vertical axes correspond to circuits (1) and (2) on the following Figure where $V_I$ corresponds to the DC bias voltage.: ![Circuits corresponding to extreme points of the load line][2] Now we introduce the diode into the circuit and the v-i relationship changes to that depicted in the following figure: ![enter image description here][3] Due to the variable component $v_i$ the operating point $v_I$ wobbles around some value on the horizontal axis. This wobbling of operating point means that the whole load line wobbles and the tangent to the v-i curve of the diode oscillates. Wobbling of the load line means that the bias voltage $V_I$ also changes. To find the $v_D$ the equation $r_d=\cfrac {V_T}{I_0\cdot e^\cfrac{v_D}{V_T}}$ needs to be solved for $v_D$. Solving this equation for $v_D$ gives $v_D=V_T \cdot \ln \frac {V_T}{r_d\cdot {I_0}}$ For $r_d=100$ Ohms we get with wolframalfa.com $v_D=$![Operating voltage][4]=0.56945 For $r_d=1000$ Ohms we get with wolframalfa.com $v_D=$![Operating voltage][5]=0.509583 Since $I_0\cdot(e^\frac{v_D}{V_T} - 1)=\frac{V_I}{R} - \frac{v_D}{R}$ we can write $I_0\cdot {R}\cdot {(\frac{V_T}{r_d \cdot {I_0}}-1)}=V_I-V_T \cdot \ln \frac {V_T}{r_d\cdot {I_0}}$ and, finally, $V_I=I_0\cdot {R}\cdot {(\frac{V_T}{r_d \cdot {I_0}}-1)}+V_T \cdot \ln \frac {V_T}{r_d\cdot {I_0}}$ Now using wolframalfa.com to compute the value of $V_I$ for $R=100$ Ohms we get: ![The value][6]=1.58345 For $R=1000$ Ohms we get: ![The bias voltage][7]=0.610983 Even though these numbers are not quite what are shown under the "Show answer" option, they are accepted as correct. This is why I would like to actually discuss the correctness of my approach to solving this exercise. [1]: https://edxuploads.s3.amazonaws.com/13490406754260579.jpg [2]: https://edxuploads.s3.amazonaws.com/13490408442610838.jpg [3]: https://edxuploads.s3.amazonaws.com/13490434703160001.jpg [4]: https://edxuploads.s3.amazonaws.com/13490473572038805.gif [5]: https://edxuploads.s3.amazonaws.com/13490675763107371.gif [6]: https://edxuploads.s3.amazonaws.com/13490467627180945.gif [7]: https://edxuploads.s3.amazonaws.com/13490483037180957.gif FirstChildUserIdTAG: 252722 FirstChildUserNameTAG: vaboro FirstChildCreateTimeTAG: 2012-09-30T23:39:29Z SecondChildTAG: your approach is absolutely correct. SecondChildUserIdTAG: 133068 SecondChildUserNameTAG: swarn99 SecondChildCreateTimeTAG: 2012-10-04T07:10:40Z SecondChildTAG: Can you explain why $i_D = \frac{V_I-v_D}{R}$ instead of $i_D = \frac{V_D-v_I}{R}$? SecondChildUserIdTAG: 4266 SecondChildUserNameTAG: soariez SecondChildCreateTimeTAG: 2012-10-04T08:27:48Z SecondChildTAG: I always mess up the signs. SecondChildUserIdTAG: 4266 SecondChildUserNameTAG: soariez SecondChildCreateTimeTAG: 2012-10-04T08:30:47Z SecondChildTAG: Nevermind, I got it! SecondChildUserIdTAG: 4266 SecondChildUserNameTAG: soariez SecondChildCreateTimeTAG: 2012-10-04T08:41:45Z SecondChildTAG: shoot me someone please))) SecondChildUserIdTAG: 373498 SecondChildUserNameTAG: Cheblan SecondChildCreateTimeTAG: 2012-10-04T13:12:03Z SecondChildTAG: I've also solved it using a closed formula [wolframalpha.com][ T*\ln(T/(U * D)) + (R*T)/D - RU, where T=0.026, U=8e-14, D=1000, R=3900 ] and got the same numbers as @vaboro. I think the model solution uses $\Delta v_d / \Delta i_d$ for a small change in $V_I$ to estimate small signal $R_d$. SecondChildUserIdTAG: 4076 SecondChildUserNameTAG: damians SecondChildCreateTimeTAG: 2012-10-06T22:31:28Z FirstChildTAG: $V_I$ is the large signal DC voltage, and $V_i$ is the small signal incremental voltage. Incremental resistance $v_d / i_d $ refers to the slope of the diode iv curve, and you need calculus to obtain it. I suggest you watch the videos again more carefully to understand these concepts. "Incremental" is a keyword here with a very special meaning. FirstChildUserIdTAG: 151427 FirstChildUserNameTAG: RichardZ FirstChildCreateTimeTAG: 2012-09-18T13:55:52Z IndexTAG: 775 TitleTAG: easy step to find find Rth.. 1st open the terminals and find open circuit voltage that is Vth. 2nd close the terminal and find short circuit current Isc. Now Rth will be Vth/Isc. UserIdTAG: 279136 UserNameTAG: rudra_ece CreateTimeTAG: 2012-09-18T11:28:31Z VoteTAG: 4 CoursewareTAG: Week 2 / Thevenin model CommentableIdTAG: 6002x_S3E5_Thevenin_Model NumberOfReplyTAG: 1 FirstChildTAG: how do i find the Vth? FirstChildUserIdTAG: 185317 FirstChildUserNameTAG: sunny30 FirstChildCreateTimeTAG: 2012-09-18T14:33:06Z SecondChildTAG: there must be 1 or more active independent source in the network. for current source use source transformation then find the Open Circuit voltage. KCL can also be used. As we know to apply KCL to find Voltage and apply KVL to find current. SecondChildUserIdTAG: 279136 SecondChildUserNameTAG: rudra_ece SecondChildCreateTimeTAG: 2012-09-18T17:06:10Z IndexTAG: 776 TitleTAG: The reason why we have multiple solutions is because they have placed V2 source upside down and then assigned a negative value to it. So, it neutralized the fact that it was turned upside down. The reason why we have multiple solutions is because they have placed V2 source upside down and then assigned a negative value to it. So, it neutralized the fact that it was turned upside down. UserIdTAG: 160683 UserNameTAG: Jinish CreateTimeTAG: 2012-09-18T10:27:00Z VoteTAG: 4 CoursewareTAG: Week 2 / Node equation review CommentableIdTAG: 6002x_S3E1_NodeEquationReview NumberOfReplyTAG: 1 FirstChildTAG: Although your explanation is correct, there are no multiple solutions to this problem, only one. I prefer thinking on a negative voltage source as one in which the "-" has a higher voltage than the "+" terminal, regardless of the source orientation. FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-18T14:10:49Z SecondChildTAG: Jelizon, according to your explanation, is it since the **"+"** is taken reference node, hence **"-"** is taken as having higher potential ??! SecondChildUserIdTAG: 408765 SecondChildUserNameTAG: NeevGhodasara SecondChildCreateTimeTAG: 2012-09-19T08:25:24Z SecondChildTAG: You actually don't need a reference node. What you need is to keep in mind that a voltage source will provide a voltage difference "V" between two nodes. The symbols "+" and "-" are used to indicate how is that difference provided. So if you start at the "-" node and move to the "+" node, you will add "V" to your voltage. However if "V" is negative, then you are actually reducing your absolute voltage (regardless of the reference position) SecondChildUserIdTAG: 381746 SecondChildUserNameTAG: jelizon SecondChildCreateTimeTAG: 2012-09-19T21:10:42Z SecondChildTAG: I think the question was fair but tricky. The top two nodes are 5V (5V at the plus side is 5V), and 7,2V (-7.2 at the NEGATIVE side is -(-7.2), thus 7.2). No error on their side... SecondChildUserIdTAG: 324332 SecondChildUserNameTAG: petsol SecondChildCreateTimeTAG: 2012-09-23T12:31:59Z SecondChildTAG: The mistake I made is when I wrote up the equation I assumed that the R2 part is negative. When writing up the equation one should assume every part as positive I guess... SecondChildUserIdTAG: 324332 SecondChildUserNameTAG: petsol SecondChildCreateTimeTAG: 2012-09-23T12:37:29Z IndexTAG: 777 TitleTAG: Youtube Banned Kindly upload videos in some other links also i cannot see videos in my country Pakistan due to protest UserIdTAG: 294370 UserNameTAG: Naif1125 CreateTimeTAG: 2012-09-18T03:59:08Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: You can download the MIT OCW version of 6.002, it was taped live in front of actual MIT students. The link is here: http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-electronics-spring-2007/ You can also try to use a proxy server or VPN to bypass country restrictions on Youtube. FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-09-18T05:20:15Z SecondChildTAG: try use google dns setup, http://chungliwen.com/bypass-malaysias-piratebay-censorship-using-google-dns/ SecondChildUserIdTAG: 189766 SecondChildUserNameTAG: shuhuan84 SecondChildCreateTimeTAG: 2012-09-18T12:08:12Z FirstChildTAG: not working??? :( FirstChildUserIdTAG: 127800 FirstChildUserNameTAG: MNB FirstChildCreateTimeTAG: 2012-09-23T18:26:40Z FirstChildTAG: You can see in the Wiki [Lecture Videos Download][1]:). Also see more info [here][2]: **Due to the YouTube outage in certain parts of the world, we have provided direct links to download the lecture videos, and extended the Week 2 homework and lab deadlines to the coming Sunday (September 30). Please note that Week 3 deadlines have not changed.** [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/ [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/info FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-26T11:50:07Z IndexTAG: 778 TitleTAG: Double Trouble Just like many of you, I got -1.69v as well, but eventually I just wrote down exactly what was in front of me, instead of what I *thought* was in front of me, and I saw the double minus for the (-)7.2 V for V2. Very sneaky. To prove it to myself, I used the sandbox, and got 6.2V - try it yourself, just remember to rotate V2 180 Degrees, and type in -7.2 v as the voltage, and you should get the right answer. UserIdTAG: 337858 UserNameTAG: garyk CreateTimeTAG: 2012-09-18T01:04:44Z VoteTAG: 4 CoursewareTAG: Week 2 / Node equation review CommentableIdTAG: 6002x_S3E1_NodeEquationReview NumberOfReplyTAG: 2 FirstChildTAG: I got 6.20 using algebra and the sandbox. I'm considering the current going hat direction, once I get the answer the sign will tell me if that's the real direction. So I did it this way: ![enter image description here][1] - -V1 +VR1 +VR2 -V2 = 0 - V1 + V2 = VR1 + VR2 - V1 + V2 = I(R1 +R2) - I = (V1+V2)/(R1+R2) Now if you try to replace values and solve for I, you'll get a negative current which tells us that the current direction is opposite to what I thought it was. Of course we can use that current **(I)** equation X and get the final answer (**e**) right away. Then I worked with e: - e = -VR1 +V1 - e = -(R1*I) +V1 .....................(X) - e = -(R1*((V1+V2)/(R1+R2))) +V1 If you do some funny algebra with the previous equation, you'll end up with this for e: - **e = (V1*R2 - V2*R1)/(R1+R2)** Now you can replace all values and get the answer, at this point it doesn't matter what direction voltage source V2 is set up, we just use given values. ---------- **Now the big dilemma V2=-7.2V works this way:** Imagine you have made your own voltage source and also made a mistake soldering terminals to a battery you put inside it. ![enter image description here][2] The next diagram will help a little bit. By KVL that means that V2 = - Vbaterry; V2 = -(7.2V) So that's why V2 = -7.2V ![enter image description here][3] If somebody agrees or disagrees with this idea, please feel free to leave your comments. I'm not so sure if this is the right way to explain it, I need some help from MITx staff. [1]: https://edxuploads.s3.amazonaws.com/13479361713770793.png [2]: https://edxuploads.s3.amazonaws.com/13479418593949176.png [3]: https://edxuploads.s3.amazonaws.com/13479419171343605.png FirstChildUserIdTAG: 138769 FirstChildUserNameTAG: ArturoPrado FirstChildCreateTimeTAG: 2012-09-18T03:28:14Z SecondChildTAG: in the equation e = (V1*R2 - V2*R1)/(R1+R2), if you will replace V1 = 5 and V2 = -7.2 then you will get value e = 1.69 not 6.206. Please check your equation once again. In the equations -V1 +VR1 +VR2 -V2 = 0 V1 + V2 = VR1 + VR2 V1 + V2 = I(R1 +R2) I = (V1+V2)/(R1+R2) e = -VR1 +V1 e = -(R1*I) +V1 .....................(X) e = -(R1*((V1+V2)/(R1+R2))) +V1 If you wil plcae V1 = 5 and V2 = -7.2 its gives you the same result. SecondChildUserIdTAG: 181935 SecondChildUserNameTAG: vishsahare SecondChildCreateTimeTAG: 2012-09-29T08:25:40Z FirstChildTAG: What you guys are doing here works, but you might be over-thinking it. All you have to do is redraw the circuit without one voltage source(just draw it as a wire so the circuit is still connected). Then redraw it with just the other voltage source(so you only have one at a time). So it's just a simple voltage divider equation twice, once with each voltage source. Then add together. You can almost do it just in your head. Of course the double-negative means that it's 7.2 and not -7.2 for V2. Like so: ![enter image description here][1] And then like so: ![enter image description here][2] Then it's just 2.26 + 3.95 = 6.21(the simulator is rounding a bit, but it's within the margin to be acceptable). [1]: https://edxuploads.s3.amazonaws.com/1347945477877087.png [2]: https://edxuploads.s3.amazonaws.com/134794552887706.png FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-09-18T04:52:21Z IndexTAG: 779 TitleTAG: HELLO GUYS! PLEASE PETITION WITH ME FOR MATHS COURSES. Big Grateful Congratulations for the whole EDx Team for their unique and tremendous work. However, I fell like an important piece of the puzzle is still missing: MATHEMATICS COURSES. SO, I would like to invite everybody to petition with me and ask the EDx team to include some Maths Courses too. A Maths Certificate from EDx would be really, really COOL, don't you think so? Respectfully yours! UserIdTAG: 84260 UserNameTAG: tebiezou12 CreateTimeTAG: 2012-09-18T00:03:13Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: that's a good idea FirstChildUserIdTAG: 395257 FirstChildUserNameTAG: blackguitar FirstChildCreateTimeTAG: 2012-09-18T00:09:22Z SecondChildTAG: yes it is SecondChildUserIdTAG: 428157 SecondChildUserNameTAG: MRAHIL SecondChildCreateTimeTAG: 2012-09-18T00:43:38Z SecondChildTAG: yah really good idea SecondChildUserIdTAG: 235727 SecondChildUserNameTAG: amor SecondChildCreateTimeTAG: 2012-09-18T01:29:05Z FirstChildTAG: http://ocw.mit.edu/courses/#mathematics FirstChildUserIdTAG: 404284 FirstChildUserNameTAG: tomdrifmach FirstChildCreateTimeTAG: 2012-09-19T04:25:18Z FirstChildTAG: Give Khan Academy at http://www.khanacademy.org/ a try. FirstChildUserIdTAG: 66042 FirstChildUserNameTAG: rarciniegas FirstChildCreateTimeTAG: 2012-09-18T00:26:39Z FirstChildTAG: And you can watch lots of maths classes from MIT on iTunes University. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-09-18T05:02:07Z IndexTAG: 780 TitleTAG: Clubs Dear **edx** management team, This project to educate a thunderous multitude with your resources and visionary eyes, is a great and welcome development as I am really benefiting maximally. Sequel to this, I wish the team my best wishes and warm regards. But, for the classes to have a read-and-fun session, I would like to suggest mind-developing clubs that introduce games that help build thinking faculties. For example **"Chess Club" and "Circuits/Robotic Design Club"**. I feel these two clubs out of many other suggestions would go a long way in creating fun while studying. Thank you. Regards, Ugochukwu. UserIdTAG: 329879 UserNameTAG: Ugochukwu CreateTimeTAG: 2012-09-17T19:44:22Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Good idea! FirstChildUserIdTAG: 6977 FirstChildUserNameTAG: rocha FirstChildCreateTimeTAG: 2012-09-17T20:59:31Z FirstChildTAG: I'd like to find people from Perú. FirstChildUserIdTAG: 138769 FirstChildUserNameTAG: ArturoPrado FirstChildCreateTimeTAG: 2012-09-18T04:42:36Z IndexTAG: 781 TitleTAG: time bar for mid term and final exams??? what is criteria about mid term and final paper ? UserIdTAG: 415376 UserNameTAG: imab90 CreateTimeTAG: 2012-09-17T04:49:51Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: There is no midterm and final papers to research and write (like in History class). There is only a midterm and final *exam*, and the dates of these tests is given in the calendar under the "course info" tab at top. I am not sure myself either, but I think that the exams will be similar to the homeworks that we do online, except we only get 1 day (or maybe a few hours...IDK) to complete it. Also I think we only get a limited number of times to "check" your answer before it's considered a submittal (e.g. right or wrong). Maybe someone who took the course last spring (when it was offered for the first time) can explain the format of the exams more in detail; I would like to know as well. FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-09-17T12:30:42Z FirstChildTAG: Imab90, JerseyMark - The definitive answer as to the timelimits for the exams will have to come from staff, but here's how we did it in the pilot version of 6.002x in Spring 2012: The midterm and final exams had limits of three submission attempts per problem (a group of answers counted as one problem). There was a 'Save' button, which allowed the student to save answers without submission, and a 'Submit' button which used up one attempt each time it was clicked. Feedback was immediate, green ticks (checks) or red Xs, and a note indicating how many submission attempts remained for that problem. As for timing, the exam opened at time/date X (in GMT/UTC), and closed a few days later. Once the student clicked past the intro page of the exam, s/he had 24 hours to complete the exam. Absolutely no collaboration during the exams, either on this forum or outside, the staff monitored Facebook, Google study groups, IRC, etc., along with other forensic metrics which might indicate plagiarism or dishonest answers. The final was BRUTAL. Even the best students took 8-10 hours on it. One student said it was the hardest exam he had ever taken aside from his Physics PhD exams. On the other hand, one of the perfect scores on the final was earned by a 15-year old in Mongolia, in Tony Kim's class. So your mileage may vary. My suggestion is to take really good notes along the way, and keep your worksheets for the homeworks, labs, and sequence exercises. Some of the exam questions were very similar to previous assignments, and it was helpful to see how I solved similar problems. Also note that on each exam, they gave at least one question that pertained to material not explicitly covered in class. If you understood the methodology, you could solve the problem, even though the material was new. If you relied on "plug-and-chug" blindly plugging numbers into formulas, you were doomed. FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-09-18T02:54:33Z IndexTAG: 782 TitleTAG: Lab 1 and HW 1 due today Hi all, Remember that Lab 1 and Homework 1 are due today (Sunday). The 6.002x teaching staff will be online between 6pm and 8pm (Boston time) helping with those assignments! Feel free to ask questions, but also use the search tool of the discussion forum to see if your question has been answer (that will allow us to be more efficient) Good luck with the assignments! UserIdTAG: 381746 UserNameTAG: jelizon CreateTimeTAG: 2012-09-16T21:14:47Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: hello can you help me with this assignment for c and e answers. H1P3: POOR WORKMANSHIP Joe wants to heat his 12'X20' workshop with electric heat. He has hired the HACME electrician company to build the system. They propose to use three 1300.0W 240V baseboard heaters to provide a total heating capacity of 3900.0W. (A heater is basically a resistor. This is not quite true, because there is a thermostatic switch incorporated into the heater and because the resistance of a heater varies a bit with its temperature. But we will use a linear resistor as a model of a heater.) FirstChildUserIdTAG: 130118 FirstChildUserNameTAG: manis FirstChildCreateTimeTAG: 2012-09-16T21:43:49Z SecondChildTAG: Sure, what problem are you having? Notice that the equivalent resistance in the new configuration of resistances is larger than initially. What will this do to the current leaving the source? SecondChildUserIdTAG: 381746 SecondChildUserNameTAG: jelizon SecondChildCreateTimeTAG: 2012-09-16T22:01:40Z SecondChildTAG: yes, i have a problem with Lab 1, i am making it, but there are problems, i am sure of my answers, and there are correct, but exist a problem with this. SecondChildUserIdTAG: 257168 SecondChildUserNameTAG: Valenciamc SecondChildCreateTimeTAG: 2012-09-16T22:50:42Z SecondChildTAG: Valenciamc, have you run the DC simulation to verify the answers? Remember you have to click "Check" after running the simulation SecondChildUserIdTAG: 381746 SecondChildUserNameTAG: jelizon SecondChildCreateTimeTAG: 2012-09-16T22:58:34Z FirstChildTAG: hello master of this course I would like to say that my case I have problems with the simulator but most of all it gives me the practica0 as did the circuit raised but I trasient analysis showed graphs can tell me what is wrong please FirstChildUserIdTAG: 339154 FirstChildUserNameTAG: Gerson2012 FirstChildCreateTimeTAG: 2012-09-16T21:45:13Z SecondChildTAG: Gerson, I didn't understand your question. Are you having problem with the simulator? Make sure that all the elements are wired and that there is ground symbol in the circuit. You should be able to sun the simulator in that case. Notice that in Lab 1 you don't need a transient analysis but a DC analysis. SecondChildUserIdTAG: 381746 SecondChildUserNameTAG: jelizon SecondChildCreateTimeTAG: 2012-09-16T22:04:21Z FirstChildTAG: Logistical question: I assume that we "turn in" our homework by entering answers into each blank and clicking each "check" button? Unlike the Lab 1 page, there is no "save" option on the HW 1 page. Thanks! FirstChildUserIdTAG: 211352 FirstChildUserNameTAG: rgbiv FirstChildCreateTimeTAG: 2012-09-16T21:52:16Z SecondChildTAG: Yes, just click the "Check" button and your homework will be saved. You can click as many times as you want to update your solutions SecondChildUserIdTAG: 381746 SecondChildUserNameTAG: jelizon SecondChildCreateTimeTAG: 2012-09-16T22:05:01Z IndexTAG: 783 TitleTAG: HINT! (If you keep getting a negative value for e.) Remember that V1 and V2 give the difference in potential from the (+) side of the source to the (-) side of the voltage source. If V1 is positive it means that the voltage at the (+) side is greater than the voltage at the (-) side. V2 is a negative value, meaning the (-) side is greater than the (+) side. If the (+) side is defined as 0V (because it's grounded), and (+) - (-) = -7.2V, substituting for (+) we get (-) = 0V + 7.2V. UserIdTAG: 164898 UserNameTAG: jbparkes CreateTimeTAG: 2012-09-16T16:38:55Z VoteTAG: 4 CoursewareTAG: Week 2 / Node equation review CommentableIdTAG: 6002x_S3E1_NodeEquationReview NumberOfReplyTAG: 1 FirstChildTAG: Nice. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-16T17:08:49Z IndexTAG: 784 TitleTAG: finding Voltage across R2 sorry im stuck here... how do u find the Voltage across R2...? UserIdTAG: 133084 UserNameTAG: Warrensiggs CreateTimeTAG: 2012-09-16T02:30:54Z VoteTAG: 4 CoursewareTAG: Week 1 / Series and parallel example CommentableIdTAG: 6002x_S2E4_Series_and_Parallel NumberOfReplyTAG: 3 FirstChildTAG: voltage across R2 = V - (the potential drop across R1) i.e voltage across R2 = V - (i1*R1) FirstChildUserIdTAG: 119355 FirstChildUserNameTAG: A_KUMAR FirstChildCreateTimeTAG: 2012-09-16T02:46:46Z SecondChildTAG: thank you... i think i might need to read the textbook now... SecondChildUserIdTAG: 133084 SecondChildUserNameTAG: Warrensiggs SecondChildCreateTimeTAG: 2012-09-16T04:19:57Z FirstChildTAG: You can picture R3 and R4 together as one half of a voltage divider, R1 being the other half. The tap in the between those resistances is where your R2 gets it's voltage. Have fun. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-16T02:45:51Z FirstChildTAG: Hi First you must find the total resistance to find the currnt will flow in this circuit (i). After that you must determine the current will flow in resistace R2, then you can find the volt across it. If you don't understand replay to me FirstChildUserIdTAG: 318037 FirstChildUserNameTAG: Maher-84 FirstChildCreateTimeTAG: 2012-09-16T03:37:51Z IndexTAG: 785 TitleTAG: What is the proper method of typing the expression? What is the proper method of typing the expression? I keep getting: "`Invalid input: R1 R4 not permitted in answer`" UserIdTAG: 306110 UserNameTAG: AEJ CreateTimeTAG: 2012-09-15T17:46:00Z VoteTAG: 4 CoursewareTAG: Week 1 / Node analysis practice, part 1 CommentableIdTAG: 6002x_L2Node0 NumberOfReplyTAG: 1 FirstChildTAG: You should take that to mean that you can give the answer without using the variables R1 and R4. FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-15T17:47:16Z SecondChildTAG: Alright, I thought we needed to solve for all input sources, aka a=V0-R1-R4. SecondChildUserIdTAG: 306110 SecondChildUserNameTAG: AEJ SecondChildCreateTimeTAG: 2012-09-15T17:51:50Z SecondChildTAG: I don't get it when you say "you can give the answer without using the variables R1 and R4" Kimt. SecondChildUserIdTAG: 287520 SecondChildUserNameTAG: Equbay SecondChildCreateTimeTAG: 2012-09-20T09:08:48Z SecondChildTAG: @Equbay...well that means R1 and R4 are not requiered to find the answer.The answer requires a different approach. SecondChildUserIdTAG: 156225 SecondChildUserNameTAG: sanjana_m SecondChildCreateTimeTAG: 2012-09-23T18:21:21Z IndexTAG: 786 TitleTAG: Mistake There is no e3 node ... But in the video at 0:36 it is said like (e1-e3)/R3.... I think this is not a big mistake but it might confuse others UserIdTAG: 296303 UserNameTAG: DEEPatXUniv CreateTimeTAG: 2012-09-15T13:54:51Z VoteTAG: 4 CoursewareTAG: Week 1 / Node analysis CommentableIdTAG: 6002x_node_analysis NumberOfReplyTAG: 2 FirstChildTAG: Thank you, we'll add a note underneath that lecture video. FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-15T14:24:22Z FirstChildTAG: me too im so comfuse about the node but try to understand FirstChildUserIdTAG: 235727 FirstChildUserNameTAG: amor FirstChildCreateTimeTAG: 2012-09-15T14:30:03Z IndexTAG: 787 TitleTAG: H1P1 anyone help me with H1P1 last question please UserIdTAG: 254607 UserNameTAG: werehenry CreateTimeTAG: 2012-09-15T11:44:43Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Could you please try to be more specific with your conceptual question? This looks like a request for an answer, which is against the honor code. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-09-15T13:34:57Z IndexTAG: 788 TitleTAG: Independent eqautions Can anyone explain me what is the meaning of using INDEPENDENT EQUATION while analyzing circuits??? This term is new to me!!! UserIdTAG: 127800 UserNameTAG: MNB CreateTimeTAG: 2012-09-15T07:52:51Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Independent Equations are the number of algebriac equations formed for finding the Values of Voltage and current. It depends upon the type of analysis you choose to solve, it can b 1)node analysis 2) Mesh analysis FirstChildUserIdTAG: 335979 FirstChildUserNameTAG: premkumar403 FirstChildCreateTimeTAG: 2012-09-15T08:30:12Z SecondChildTAG: then what is the difference between independent eqn and dependent eqn in nodal and mesh analysis?? can anyone explain clearly ,i can't get the exact thing!! SecondChildUserIdTAG: 435803 SecondChildUserNameTAG: nalim135 SecondChildCreateTimeTAG: 2012-09-15T09:50:47Z IndexTAG: 789 TitleTAG: homework submission hi, how are we supposed to submit the homeworks and labs ??? please help UserIdTAG: 364991 UserNameTAG: coolvishal CreateTimeTAG: 2012-09-15T07:17:08Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: just complete all your's homeworks and labs and hit check button. FirstChildUserIdTAG: 114864 FirstChildUserNameTAG: Vikaash FirstChildCreateTimeTAG: 2012-09-15T10:44:10Z SecondChildTAG: Can we submit multiple times? SecondChildUserIdTAG: 143786 SecondChildUserNameTAG: Matjaz SecondChildCreateTimeTAG: 2012-09-15T11:57:35Z IndexTAG: 790 TitleTAG: How do I find my own posts which i posted earlier? How do I find my own posts which i posted earlier? i am having trouble with this UserIdTAG: 135072 UserNameTAG: Shriniwas CreateTimeTAG: 2012-09-15T05:37:35Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: If you click on your user name, you will find your earlier Posts ;) https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/users/135072 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-15T05:39:32Z IndexTAG: 791 TitleTAG: Week 2 Lab 2 - Is there correct input data? Hello all who tried to design the signal mixer in Lab 2. Don't you think that it's impossible to combine the resistors to get the required mixed voltage (1/2)*V1+(1/6)*V2 ? It's impossible to get the output voltage 0.667V at both inputs equal to 1V. Look, let's write the equation for the output when both inputs arise their max values (1V): Vout = R2/(R1+R2)*V1 + R1/(R1+R2)*V2 = R2/(R1+R2)*1V + R1/(R1+R2)*1V = (R1 + R2)/(R1 + R2)*1V. Reasonable question: how can this equation be equal to 0.667V? It's strange... I think here we have wrong input data. Any third resistor can't help out. UserIdTAG: 209041 UserNameTAG: SmartEngine CreateTimeTAG: 2012-09-14T17:22:23Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 3 FirstChildTAG: Thank you all, I have got the correct answer! FirstChildUserIdTAG: 209041 FirstChildUserNameTAG: SmartEngine FirstChildCreateTimeTAG: 2012-09-15T16:29:13Z SecondChildTAG: if you have found the answer can you give me some little hints please smartengine SecondChildUserIdTAG: 266386 SecondChildUserNameTAG: zakzak200 SecondChildCreateTimeTAG: 2012-09-15T21:36:10Z FirstChildTAG: Nobody said you must use only 2 resistors ; ). FirstChildUserIdTAG: 396446 FirstChildUserNameTAG: RousseauxS FirstChildCreateTimeTAG: 2012-09-14T18:49:06Z FirstChildTAG: This is a slightly tricky problem and you need to do a little bit more than the obvious to solve it. But I assure you it is 100% correct and solvable with what you already know from the course. FirstChildUserIdTAG: 370344 FirstChildUserNameTAG: KostisL FirstChildCreateTimeTAG: 2012-09-14T19:00:05Z SecondChildTAG: could you suggest to get an answer SecondChildUserIdTAG: 149154 SecondChildUserNameTAG: santhosh1993 SecondChildCreateTimeTAG: 2012-09-15T17:56:16Z IndexTAG: 792 TitleTAG: 2nd question Hi, Why Paverage = 1/T * Integral(v²/R) and not just the integral ? I don't understand 1/T. Someone can explain me ? Thank you ! UserIdTAG: 206962 UserNameTAG: JulienL CreateTimeTAG: 2012-09-14T12:38:12Z VoteTAG: 4 CoursewareTAG: Week 1 / AC power CommentableIdTAG: 6002x_S1E3_AC_power NumberOfReplyTAG: 2 FirstChildTAG: its because 1/T refers to the average of the complete cycle so 1/T is necessary. FirstChildUserIdTAG: 170475 FirstChildUserNameTAG: sandeshacharya FirstChildCreateTimeTAG: 2012-09-14T16:10:15Z FirstChildTAG: integral mean continuous summation. for taking average, you must divide by total no of observations, which in this case is every instant of time. therefore divide by T. FirstChildUserIdTAG: 364086 FirstChildUserNameTAG: jordan3110 FirstChildCreateTimeTAG: 2012-09-15T15:12:53Z SecondChildTAG: I had the same doubt, thank you!! SecondChildUserIdTAG: 445949 SecondChildUserNameTAG: EMillan SecondChildCreateTimeTAG: 2012-09-18T03:57:41Z IndexTAG: 793 TitleTAG: S6E2 - Grapic Graphics ploted with different scales, so it's impossible to graphicaly solve this, using only figure in this task. Plot your own graphics to solve it. UserIdTAG: 128561 UserNameTAG: gusevoy CreateTimeTAG: 2012-09-14T11:33:06Z VoteTAG: 4 CoursewareTAG: Week 3 / Graphical Analysis of Nonlinear Circuits Exercise CommentableIdTAG: 6002x_graphical_analysis_of_nonlinear_circuits_exercise NumberOfReplyTAG: 0 IndexTAG: 794 TitleTAG: Power supplied Hello! I was having a lot of difficulty to find the power supplied by the voltage source and the current source. Searched in the discussion, but I didn't find anything that could help me. So, I did from a different way, and I would like to know if its correct... Voltage source: The voltage of the voltage source was 2.0V. I just applied it in he formula "P=V.I", multiplying to the current i1, and the result was approximately -2.88W. Current source: I knew that the current source had a value of 3.0A, and using the KVL with V2, I found approximately 7.77V to the current source, and, again, multiplied using the formula "P=V.I" and got 23.33W. Somebody did it? I just wanna know if my method is correct =D UserIdTAG: 291362 UserNameTAG: Gudson CreateTimeTAG: 2012-09-14T11:13:29Z VoteTAG: 4 CoursewareTAG: Week 1 / KVL/KCL example CommentableIdTAG: 6002x_S2E3_Using_KVL_KCL_and_VI_constraints NumberOfReplyTAG: 2 FirstChildTAG: Hi Gudson! Can I help you? Can you tell me the values of the components of all your Circuit. Remember that the Students have different values. Isource, Vsource, R_v2 and R_v3? FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-14T11:47:51Z SecondChildTAG: Thanks a lot! I'm just really confused about the correct method to get the power supplied by the Current and by the Voltage... I searched in the discussion for more details, but it seems don't have a fix formula. My biggest doubt was in S2E3, when I had the voltage source's value and the current source's one, but I didn't find the way to apply these values at the formula of the power supplied/dissipated, "P=V.I", so I tried to improvise, using the nearer current or voltage from each source. Fortunately, the result was correct, but I still don't know how to do this correctly. There is a standard way to follow? SecondChildUserIdTAG: 291362 SecondChildUserNameTAG: Gudson SecondChildCreateTimeTAG: 2012-09-14T13:07:21Z FirstChildTAG: Hi ishpanec! You asked me in this Post where did I get the step 1) that I explained to Gudson: *"And how you received formula V-i1*R1-(I + i1) * R2 = 0? In what lesson it is? Thanks."* I got it from KVL. Remember that KVL says that the sum of the voltages in a loop it is zero. If you go to the blue loop: ![enter image description here][1] You will see that: V-v1-v2=0 But v1=i1*R1 and v2=i2*R2 and also with KCL i2=(I+i1) So, that is why V-i1*R1-(I + i1) * R2 = 0 ;) I hope that this can help you! ;) . Any doubt please ask I will pleasured to help you. Myriam. [1]: http://s12.postimage.org/z7u3bqx7x/loop.png FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-14T21:02:57Z SecondChildTAG: Thanks, I understood all the exercise now =D SecondChildUserIdTAG: 291362 SecondChildUserNameTAG: Gudson SecondChildCreateTimeTAG: 2012-09-16T19:01:57Z SecondChildTAG: thnx a lot myriam for the detailed solution.but y did u assume the polarity of v1 the way u did and not its reverse? SecondChildUserIdTAG: 156225 SecondChildUserNameTAG: sanjana_m SecondChildCreateTimeTAG: 2012-09-23T11:01:03Z IndexTAG: 795 TitleTAG: v2 i could not understand how v2=-8.75 UserIdTAG: 429168 UserNameTAG: arunprakashavm CreateTimeTAG: 2012-09-14T05:22:24Z VoteTAG: 4 CoursewareTAG: Week 1 / Node method example CommentableIdTAG: 6002x_S2E5_Node_Method NumberOfReplyTAG: 2 FirstChildTAG: e3 as a ground node, we would have the KCL at e2 : i2 - i1 - I = 0; (e2 - e3)G2 - (e1 - e2)G1 - I = 0; e2(G1 + G2) = e1G1 + I It should be 8.75 volts. FirstChildUserIdTAG: 201508 FirstChildUserNameTAG: datle FirstChildCreateTimeTAG: 2012-09-15T15:56:53Z FirstChildTAG: me too!!I got answer for v2=(e2-0)/R2 i.e v2=(8.75-0)/5.0=1.75Volts this is the answer i got!! help plz FirstChildUserIdTAG: 151205 FirstChildUserNameTAG: jesher777 FirstChildCreateTimeTAG: 2012-09-15T06:01:49Z SecondChildTAG: me too....got t answer to be 1.75..can someone pls help us to know why it is same as the nodal voltage?? SecondChildUserIdTAG: 414118 SecondChildUserNameTAG: raje93 SecondChildCreateTimeTAG: 2012-09-15T15:11:14Z SecondChildTAG: V2 is the same as the nodal voltage, because voltage (by definition) is a difference between potentials, so V2 = e2-e3 = 8.75 - 0 = 8.75. (e2 - 0)/R2 - is a formula for current, not for voltage SecondChildUserIdTAG: 322444 SecondChildUserNameTAG: koluch SecondChildCreateTimeTAG: 2012-09-15T22:46:53Z IndexTAG: 796 TitleTAG: H3P2 Anybody figure out H3P2? How do you even approach the problem? Thanks. UserIdTAG: 397595 UserNameTAG: mholin CreateTimeTAG: 2012-09-13T22:19:09Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 5 FirstChildTAG: I just figured it out graphically. i.e. I used the graph provided and plotted a couple of points on it that seemed to be logical to me. Then I entered the voltages at those points into the answers and they were correct. The graph has a couple of clues in it, for example it is made of three straight lines and the points where the gradient changes are important. FirstChildUserIdTAG: 270011 FirstChildUserNameTAG: SnowmanZA FirstChildCreateTimeTAG: 2012-09-14T05:23:52Z SecondChildTAG: what is the values?? i mess up with this Q for last 3 days :( SecondChildUserIdTAG: 415376 SecondChildUserNameTAG: imab90 SecondChildCreateTimeTAG: 2012-09-14T17:22:51Z SecondChildTAG: That's an entirely different problem. Just as easy though. They tell you there are three heaters rated for a certain power each. Since you know the power rating, and that they work at 240Vac, then you can calculate the resistance using P=V^2/R. Substitute the rated power for P and 240V for V, then you can calculate R. Once you have the resistance of the heaters, then you can calculate all the other answers. SecondChildUserIdTAG: 270011 SecondChildUserNameTAG: SnowmanZA SecondChildCreateTimeTAG: 2012-09-15T09:16:19Z SecondChildTAG: Yes, I see those gradient changes in graph. But I don`t see any reasons why we shall restrict input voltages at their values. SecondChildUserIdTAG: 330638 SecondChildUserNameTAG: systemfault SecondChildCreateTimeTAG: 2012-09-25T19:16:52Z FirstChildTAG: I think, it's all about noise immunity. If you connect a few inverters in a row, and send '0' signal to the first one, then the outputs should be 1-0-1-0-1-0..., no matter, how many devices are connected. Without the noise it will be fine: | logical values | 0 | 1 | 0 | 1 | ... | voltage values | 0V | 5V | 0V | 5V | ... With the 0.1 V noise situation is also good: | logical values | 0 | 1 | 0 | 1 | ... | 0 | 1 | ... | voltage values | 0V | 4.95 | 0.075 | 4.91 | ... | 0.1 | 4.9 | ... The voltage levels will be stabilized at 4.9 V for ones and 0.1 V for zeros. Now the 1.5 V noise: | logical values | 0 | 1 | 1 | | voltage values | 0V | 3.88 | 2.72 | (error, 2.72 > 2.5!) The last device failed to invert logical 1 with 3.88 V, the output value is also logical 1. ---------- As you can see, there is a noise level, which will break our inverter. This will be the noise margin of the inverter, which we need to find out. Some tips to do that: 1. The worst-case noise will increase logical 0 and reduce logical 1 voltages. 2. You can model this inverter with noise using any programming language . 3. Or you can solve this task only using the math. There are few ideas: 3.1. The flat Input-Output curve rejects noise better than the steep one, so we need to keep voltage in some limits 3.2. If the noise can't break the inverter, then the voltage will will never exceed the critical value: $V_{zero} = double\_invert\_with\_noise(V_{zero})$. FirstChildUserIdTAG: 261503 FirstChildUserNameTAG: EugenyL FirstChildCreateTimeTAG: 2012-09-15T16:01:59Z FirstChildTAG: I figured out the correct voltages for H3P2; however, I don't understand why and how exactly gradients are important for figuring out the voltages that the problem accepts as correct? Without mentioning the correct voltages for the problem why the following voltages are unacceptable: $V_{IL}=2.0$, $V_{OL}=1.0$, $V_{IH}=3.0$ and $V_{OH}=4.0$ The forbidden region would this way be $V_{IH}-V_{IL}=3.0-2.0=1.0V$ and the noise margins would be: $NM_H=NM_L=V_{IL}-V_{OL}=2.0-1.0=V_{OH}-V_{IH}=4.0-3.0=1.0V$ FirstChildUserIdTAG: 252722 FirstChildUserNameTAG: vaboro FirstChildCreateTimeTAG: 2012-09-22T22:44:27Z SecondChildTAG: I don`t understand this either. I think those gradients in graph can`t influence the noise immunity. Also, what must the forbidden region value be? Why not 0.1V or zero? Noise margins would be even greater. SecondChildUserIdTAG: 330638 SecondChildUserNameTAG: systemfault SecondChildCreateTimeTAG: 2012-09-25T19:12:40Z SecondChildTAG: A forbidden region of zero, leaves zero margin for noise. Review the very first lecture on the static discipline, which explains why a non zero forbidden region is necessary. SecondChildUserIdTAG: 385677 SecondChildUserNameTAG: Michaelc1 SecondChildCreateTimeTAG: 2012-09-29T19:18:21Z SecondChildTAG: Zero forbidden region does not mean zero noise margin since we have a more strict standard for Vout than for Vin (VOH > VIH and VOL < VIL). Am not I right? SecondChildUserIdTAG: 330638 SecondChildUserNameTAG: systemfault SecondChildCreateTimeTAG: 2012-10-02T16:11:29Z SecondChildTAG: Actually, the forbidden region can't be zero because of the ambiguity of the signal this way (one can find a detailed discussion of this issue in the textbook). However, it is unclear to me why not to make forbidden region as narrow as possible to maximize noise margin (this probably has to do with non-linearity of MOSFET v-i characteristic). SecondChildUserIdTAG: 252722 SecondChildUserNameTAG: vaboro SecondChildCreateTimeTAG: 2012-10-03T05:11:47Z FirstChildTAG: Make in excel an matrix for graph B (see picture)![enter image description here][1]. [1]: https://edxuploads.s3.amazonaws.com/13489364531343674.png FirstChildUserIdTAG: 410943 FirstChildUserNameTAG: JPalaciosRoman FirstChildCreateTimeTAG: 2012-09-29T16:35:52Z FirstChildTAG: extensive analysis is only just look at the graph, for VIH and VIL values​​, you look where it cuts the values ​​plotted in Vin (v); For values ​​VOL and VOH look graph values ​​for Vout (V) ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/1348961883958847.gif FirstChildUserIdTAG: 343291 FirstChildUserNameTAG: LuisEduardoH FirstChildCreateTimeTAG: 2012-09-29T23:38:14Z SecondChildTAG: Thank you friend for this help. But there is one mistake that you have to take notice of. Where you wrote V(IH) is basically V(IL). Because as you draw, Vin is x-axis and Vout is y-axis. And x-axis is ascending from left to right. Overall your post was very helpful. SecondChildUserIdTAG: 378096 SecondChildUserNameTAG: Jamshaid271 SecondChildCreateTimeTAG: 2012-09-30T05:43:17Z SecondChildTAG: Why did you place your points at the place where the slope of the line changes? Why not a bit to the left o right from there? SecondChildUserIdTAG: 330638 SecondChildUserNameTAG: systemfault SecondChildCreateTimeTAG: 2012-10-02T16:13:02Z IndexTAG: 797 TitleTAG: Question: Independent KVL Equations According to the textbook, there is no problem answering the question. We have 6 branches (B) and 4 nodes (N), so we get 6-4+1=3 independent KVL equations. But I found in several other books another method to find out how many independent KVL equations there are. Basically it says: L-(N-1) where L is the number of loops of the circuit. So using this for answering the question, we would get 4 independent KVL equations: 7-(4-1)=4. So why is there a difference? Am I missing a point? Thank you for your help :-) UserIdTAG: 394599 UserNameTAG: Albie CreateTimeTAG: 2012-09-13T12:52:16Z VoteTAG: 4 CoursewareTAG: Week 1 / Circuit topology CommentableIdTAG: 6002x_S2E1_Circuit_Topology NumberOfReplyTAG: 1 FirstChildTAG: there are 6 loops instead of 7 FirstChildUserIdTAG: 572704 FirstChildUserNameTAG: mitali1994 FirstChildCreateTimeTAG: 2012-10-05T16:28:27Z IndexTAG: 798 TitleTAG: The explicit solution Ex. S1E5 KVL notion, that sum of the voltages in the loop must equal 0. Thus sum of the depict circuit: - `v1-v2-v3 = 0` have to be 0. Then we must convert the equation: - `v1-v2 = v3` so the result is `v3 = 0.5 V` ---------- Ex. S1E6 Case 1 We have to create the loop, containing 3 branches with the source V, resistors with voltage v1 and v3. Thus, as in ex. S1E5, we can write: - `v1+v3-V = 0`(clockwise direction), after converting the formula, the result is `v3 = V-v1`. Case 2 There we also create the loop, but in this case it contains 4 branches and 4 resistors with voltages v1, v2, v3 and v4. The equation for this loop is: - `-v4+v2-v3-v1=0` (clockwise direction), where v3 is equal to V-v1 (from Case 1). After converting the equation, the right answer is: - `v4=v2-V` Case 3 At the end, we create the loop containing 3 branches and resistors with voltage v1, v4 and v5 or either with voltage v3, v2 and v5. There's the same solution for both of them. So the equation is: - `v5+v3-v2=0` (counterclockwise direction), after converting (as earlier v3=V-v1), the right answer is: - `v5=v2-V+v1`. UserIdTAG: 88649 UserNameTAG: synus_wroc CreateTimeTAG: 2012-09-13T09:28:17Z VoteTAG: 4 CoursewareTAG: Week 1 / KVL and KCL exercises CommentableIdTAG: 6002x_kvl_kcl_exercises NumberOfReplyTAG: 0 IndexTAG: 799 TitleTAG: Can the video settings be saved? Also Auto-advance? It would be great if the video settings could be saved for each video - the speed seems to but the volume, captions on/off, and the full screen reset each video. Also - an auto-advance option for continuing through the videos without having to hit next would be very nice. UserIdTAG: 323230 UserNameTAG: Xango5346 CreateTimeTAG: 2012-09-13T00:24:00Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 800 TitleTAG: What if... What about if for logical 1 you get more than 5V at the reciver? UserIdTAG: 303787 UserNameTAG: Gukasian CreateTimeTAG: 2012-09-12T19:38:44Z VoteTAG: 4 CoursewareTAG: Week 2 / Static Discipline CommentableIdTAG: 6002x_static_discipline NumberOfReplyTAG: 1 FirstChildTAG: In practice (i.e. with real chips like [7400][1]) you can destroy the chip by going over the nominal $V_S$ at the inputs by a few volts. [1]: http://en.wikipedia.org/wiki/7400_series FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-12T19:53:15Z SecondChildTAG: Well, in practice we need some additional filters on input, but we're now talking about concept, yes? So Gukasian's question have sense for me, I dont understand it too. As I think, we should have output in the middle of VH and 5V (with some margins up and down), and input range stays from VIH to 5V. The same for 0. SecondChildUserIdTAG: 324219 SecondChildUserNameTAG: Dmitry79 SecondChildCreateTimeTAG: 2012-09-14T10:00:27Z SecondChildTAG: The text talks about this on p 248 - in practice, "circuits are generally able to correctly interpret values outside the extremum points (below 0 volts for logical 0 and above 5 V for a logical 1)"...assuming you don't blow up or melt down the circuit due to too much voltage or current. SecondChildUserIdTAG: 142402 SecondChildUserNameTAG: aaronrod SecondChildCreateTimeTAG: 2012-09-17T05:31:05Z IndexTAG: 801 TitleTAG: power supplied by the voltage and current source By making some algebra, a arrived to the correct answers for almost every question, but, in the power supplied by the voltage and the current source, the only mistake was the sign. I read some of the discussions about this same problem, but what i can't understand is why by making the algebra, i arrived to the opposite signs? and why are those positive in current source and negative in voltage source correct? UserIdTAG: 394345 UserNameTAG: Yarai CreateTimeTAG: 2012-09-12T19:08:09Z VoteTAG: 4 CoursewareTAG: Week 1 / KVL/KCL example CommentableIdTAG: 6002x_S2E3_Using_KVL_KCL_and_VI_constraints NumberOfReplyTAG: 4 FirstChildTAG: I had the same confusion. However, following the associated variables convention we get that positive power means the element is consuming, and negative power means it is supplying. Consuming power can also be seen as supplying negative power. FirstChildUserIdTAG: 106229 FirstChildUserNameTAG: jc_lounge FirstChildCreateTimeTAG: 2012-09-14T04:07:42Z FirstChildTAG: If the power is supplied by the source (voltage or current source) the sing is negative, and if the power is consumed by the source the sing is positive. As you, I can't get the result in positive for the current source, but I'm pretty sure that it is because our sing rules has a mistake somewhere. Sorry for my English . FirstChildUserIdTAG: 305497 FirstChildUserNameTAG: Baxter_87 FirstChildCreateTimeTAG: 2012-09-13T10:35:20Z SecondChildTAG: It's funny, I have exactly the same problem. I assume that the current source, like the voltage source, is giving power!! Therefore if power is going out, the Wattage should be negative. My intuitions are all over the place right now! SecondChildUserIdTAG: undefined SecondChildUserNameTAG: undefined SecondChildCreateTimeTAG: 2012-09-13T11:21:48Z FirstChildTAG: Power is given by P=V*I. By doing the circuit analysis, you find that the current i1 is really going in the opposite direction, thus it's negative. So, by applying the formula for the voltage source you get a positive voltage times a negative current, and as a result the power supply has a negative value. For the current source it's the opposite, you get a positive voltage times a positive current which gives you a positive power supply. I might be making this up, but the current source should have a positive power supply 'cause you need an input of energy to to make electrons flow (ie make current), you're need energy to make the device work. In the case of the voltage source, your taking energy out of the device so the power supply is negative. Think of it this way, you don't need energy to discharge a battery, but you do need it to recharge it. Hope this helped a bit. FirstChildUserIdTAG: 256249 FirstChildUserNameTAG: And90 FirstChildCreateTimeTAG: 2012-09-14T02:03:12Z FirstChildTAG: Using the Associated Variables Convention, the currents (and thus the powers) will all be expressed as the current flowing in to the element. In terms of power, this will give you the power consumed by the device. However, the question asks for the power supplied by each of the sources, which require you to change the sign. FirstChildUserIdTAG: 21491 FirstChildUserNameTAG: AndyDovin FirstChildCreateTimeTAG: 2012-09-14T09:18:33Z IndexTAG: 802 TitleTAG: Wiki Textbook Errata I copied the wiki about the textbook errata from the old MiT wiki to edx. https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/Book/ UserIdTAG: 83276 UserNameTAG: salsero CreateTimeTAG: 2012-09-12T11:40:40Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: You are a good man (or woman). FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-12T14:12:54Z IndexTAG: 803 TitleTAG: Regarding question #4 I think it's important to emphasize that in this exercise we are no longer working with 2 loops! It's just one loop formed by the 2 sets of batteries and resistors. I tried applying KVL to the 2 loops on #2 and I was getting 6 A for loop 1 and 11 A for loop 2. I tried doing substitution and then using determinants, and these two values kept popping up - does anyone know why? Applying KVL to just the first loop gave me the right value straight away. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-12T00:47:35Z VoteTAG: 4 CoursewareTAG: Week 1 / Battery model CommentableIdTAG: 6002x_S1E9_Battery_Model NumberOfReplyTAG: 0 IndexTAG: 804 TitleTAG: Q:2 ; Average power Why do we have to perform a long solution using integration , if we could have the answer of "average power" by a simple approach. P avg = P peak / 2. UserIdTAG: 146930 UserNameTAG: fayzan_007 CreateTimeTAG: 2012-09-11T19:42:53Z VoteTAG: 4 CoursewareTAG: Week 1 / AC power CommentableIdTAG: 6002x_S1E3_AC_power NumberOfReplyTAG: 3 FirstChildTAG: That only works for sinusoidal signals, if the signal is triangular for example the factor is not 2. FirstChildUserIdTAG: 244841 FirstChildUserNameTAG: eyubero FirstChildCreateTimeTAG: 2012-09-11T19:46:36Z SecondChildTAG: The question is about sinusoidal. What is the point of looking for difficult complex solutions in simple situations? Simplicity is the ultimate sophistication (Leonardo Da Vinci). SecondChildUserIdTAG: 413065 SecondChildUserNameTAG: anikey SecondChildCreateTimeTAG: 2012-09-12T08:39:14Z FirstChildTAG: I did the same. That's just natural. FirstChildUserIdTAG: 413065 FirstChildUserNameTAG: anikey FirstChildCreateTimeTAG: 2012-09-12T08:35:18Z FirstChildTAG: what is the value of P peak ??? i think peak value is 130.90 W FirstChildUserIdTAG: 415376 FirstChildUserNameTAG: imab90 FirstChildCreateTimeTAG: 2012-09-12T17:57:09Z SecondChildTAG: yes it is . . P peak = 130.9 W SecondChildUserIdTAG: 146930 SecondChildUserNameTAG: fayzan_007 SecondChildCreateTimeTAG: 2012-09-13T14:01:04Z SecondChildTAG: As we have already calculated the P_peak in Q.1.......i-e P_peak = 262 W So by the formula of sinusoidal wave: P_avg= (P_peak)/2 P_avg=131 W SecondChildUserIdTAG: 390271 SecondChildUserNameTAG: AzharMehmood SecondChildCreateTimeTAG: 2012-09-16T10:31:30Z IndexTAG: 805 TitleTAG: value entry how to enter the values,wit suffix 'R' for resistor?'v' for voltage? i m gettin this dialog Invalid input: could not interpret '1.76v' as a number UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-11T14:27:23Z VoteTAG: 4 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 3 FirstChildTAG: You only have to input the number, leaving out the unit. FirstChildUserIdTAG: 9806 FirstChildUserNameTAG: AneHCraM FirstChildCreateTimeTAG: 2012-09-11T14:43:35Z FirstChildTAG: If you're entering simply a number value, leave off units. If it specifies to answer in terms of something, it will usually say so. (For example, write an equation for V using I and R) FirstChildUserIdTAG: 414313 FirstChildUserNameTAG: staticvoid FirstChildCreateTimeTAG: 2012-09-11T14:44:06Z FirstChildTAG: You don´t need to write the units (again) once they wrote it in the question. FirstChildUserIdTAG: 304082 FirstChildUserNameTAG: frjeremias FirstChildCreateTimeTAG: 2012-09-12T05:03:11Z IndexTAG: 806 TitleTAG: Independent Equations and S2E4 - Requested by Mona77 Hi Mona77!Sorry for the delay, I hope this can help you. You ask me [in this Post][1] an explanation of Independent Equations and also if I Could help you in S2E4. **Independent Equations KVL and KCL:** ---------- *It is important that you can watch the S1V12 in order to understand the Concepts of Node N , Branch B and Loop.[s1v12][2]* **Nodes [N]:** The junction points at witch the terminals of two or more elements are connected. **Branch [B]:** The connections between the nodes are referred to as branches of a Circuit. **Loop:** Are defined to be closed Paths through a Circuit along its branches. ---------- In general, in a Circuit you will have 2*B independent equations where: B equations will come from elements Law and the other B from KCL and KVL. So, you can obtain the independent equations of a Circuits with the two following expressions: N = Nodes in the Circuit B = Branches of the Circuit. **Independent Equations KCL =** N - 1 **Independent Equations KVL =** B - N + 1 ---------- The Dependant Equations can be get as a lineal combination of the Independant equations. That is the different between Independent and Dependent equations. ----- [Read more in the Textbook][3] I suggest you to do excercise S2E1 ---- Exercise S2E4: ![enter image description here][4] Here Mona77, try to take a look to the Textbook: Resistors in Series [Here][5] . You will see that Rseries = Rx + Ry Resistors in Paralell [Here][6] . You will see that Rparalell = Rx*Ry/(Rx+Ry) So, Mona77 Lets try with the First Part together: 1) What is the equivalent resistance (in Ohms) of the series combination of R3 and R4? If you see R3 and R4 are in series combination. As you can see in the textbook series combinations sums the resistances in series. So what is the result here? ;) 2) What is the equivalent resistance (in Ohms) of the parallel combination of R2 and the series combination of R3 and R4? Now that you have Rseries from part 1 (this is the combination of R3 and R2, we called Rseries or Rs). You can see that Rs is in paralell with R2... So, what is the result? ;) Try it. 3) What is the equivalent resistance (in Ohms) of the series combination of R1 and the parallel combination of R2 and the series combination of R3 and R4? It is really similar to part 1 and 2. I let this to you. 4) What is the current (in Amperes) i1? Do you have the resistence obtained in 3) ? Isn't that the total resistence? and i1 isn't it V devided the total resistence? So, what is the result? ;) Question 5) and 6) I let it to you :) See you! I hope this can help you a little bit. Anything you need you can ask. I will try Edit this Post later so it is in Spanish too. Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/MITx_6_002x_2012_Fall/threads/50474faf685941270000000d [2]: https://www.youtube.com/watch?v=SaieZYN_WR0&feature=player_embedded#! [3]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/93 [4]: https://www.edx.org/static/content-mit-6002x/images/circuits/ladder.43051f019474.gif [5]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/100 [6]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/107 UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2012-09-10T23:31:38Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: Myrimit: you deserve a medal :) FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-10T23:41:26Z SecondChildTAG: Thank you kimt! How are You? ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-11T11:27:25Z FirstChildTAG: Thanks a lot!Really was confused about nodes but definition u wrote made it clear. FirstChildUserIdTAG: 373262 FirstChildUserNameTAG: nwtn_pro FirstChildCreateTimeTAG: 2012-09-13T10:01:21Z SecondChildTAG: You are welcome nwtn_pro! ;) . Nice that you have cleaned your doubt ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-16T06:47:57Z IndexTAG: 807 TitleTAG: Textbook Error Pg 72 The discussion of figure 2.28 in the text on page 72 suggests a voltage of 0.5V as opposed to the correct value of 2V. UserIdTAG: 384766 UserNameTAG: CodeSF CreateTimeTAG: 2012-09-10T23:09:50Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Looks like all the working is correct, just the evaluation of the final expression is wrong. FirstChildUserIdTAG: 312035 FirstChildUserNameTAG: philhiggins FirstChildCreateTimeTAG: 2012-09-11T00:37:40Z IndexTAG: 808 TitleTAG: a different direction I do not understand how the i1 and i2 can have a different direction in a closed loop. After all, it's one and the same current. Please. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-10T20:24:42Z VoteTAG: 4 CoursewareTAG: Week 1 / Associated reference directions CommentableIdTAG: 6002x_S2E2_Associated_Reference_Directions NumberOfReplyTAG: 5 FirstChildTAG: they both are the same but with opposite signs. FirstChildUserIdTAG: 287792 FirstChildUserNameTAG: Robinhood FirstChildCreateTimeTAG: 2012-09-10T23:14:33Z FirstChildTAG: It's all a question of the sign. FirstChildUserIdTAG: 144694 FirstChildUserNameTAG: johndoe31415 FirstChildCreateTimeTAG: 2012-09-10T20:53:01Z FirstChildTAG: You can also think of it as adding an arbitrary node between the battery and the resistor, and from this node you could measure i1 and i2 and apply KCL to get i1 + i2 = 0. FirstChildUserIdTAG: 390175 FirstChildUserNameTAG: mdlima FirstChildCreateTimeTAG: 2012-09-10T22:33:01Z FirstChildTAG: The direction of a current is a totally arbitrary decision of what direction to consider "positive" when talking about a circuit. If the current happens to be going in a particular direction on a branch, you will talk about it in either positive or negative terms depending on which direction you decided "positive" would be when you were setting up the circuit. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-09-11T20:02:40Z FirstChildTAG: Even I thought like you the first time I saw the figure but then The way I solved this problem was~ i1 and i2 ,although representing the same current can go to opposite directions because here they can be of the opposite sign.When you solve the answer you will get i1 = - i2 which means i1 + i2 = 0 which clearly shows that they are the same current going in and out of any point. And i1 and i2 representing the same current, the direction of i1 and i2 being different does not matters as long as they keep satisfying the relation i1= -i2. ~ Sanjana FirstChildUserIdTAG: 156225 FirstChildUserNameTAG: sanjana_m FirstChildCreateTimeTAG: 2012-09-23T10:23:34Z IndexTAG: 809 TitleTAG: Algebric expression Well i have seen many post here and seems like there is a confusion with the correct answer but actually their answer is right if you just write the algebric expression correctly you then you will get the answer. Notice here they have given only one node but there are more two nodes in between R1 and V1 and R2 and V2, those nodes are know so the equation will be simple (e-5)/R1 + (e-7.2)/R2 = 0 you will get 6.20 UserIdTAG: 336001 UserNameTAG: syd_buet12 CreateTimeTAG: 2012-09-10T18:27:56Z VoteTAG: 4 CoursewareTAG: Week 2 / Node equation review CommentableIdTAG: 6002x_S3E1_NodeEquationReview NumberOfReplyTAG: 1 FirstChildTAG: for this equation also we are getting only -1.690....could u pls explain how u got 6.20..... FirstChildUserIdTAG: 361329 FirstChildUserNameTAG: manjuhasinivenkatachalal FirstChildCreateTimeTAG: 2012-09-18T03:55:10Z IndexTAG: 810 TitleTAG: Q2 answer simplified! average power is the product of rms values of voltage and currrent. therefore this answer will be same as the answer to the 3rd question UserIdTAG: 372535 UserNameTAG: _Infinity CreateTimeTAG: 2012-09-10T11:53:50Z VoteTAG: 4 CoursewareTAG: Week 1 / AC power CommentableIdTAG: 6002x_S1E3_AC_power NumberOfReplyTAG: 0 IndexTAG: 811 TitleTAG: Confused about reference versus actual direction of current I don't understand always how we know which way current REALLY flows. For example, on page 6 of the textbook, we see that current is flowing away from the positive end of the voltage source. Even on the NEXT exercise we have current positive leaving the positive end of the voltage source. Yet here we have it always flowing into the positive end. Which way is correct, how do we know which way it's really going? I thought that voltage always increases as we leave the positive terminal of a voltage source. If so, isn't current also flowing in that direction? Are we intentionally putting the direction of current as negative with voltage sources? If I decide to put current as leaving the positive end of a battery then the power value is positive. Is that ok if I just say positive energy = energy moving into the circuit? The one thing that can't change is that voltage jumps when we leave the positive terminal of a voltage source, right? UserIdTAG: 357225 UserNameTAG: MJBoa CreateTimeTAG: 2012-09-10T09:04:53Z VoteTAG: 4 CoursewareTAG: Week 1 / Associated reference directions CommentableIdTAG: 6002x_S2E2_Associated_Reference_Directions NumberOfReplyTAG: 3 FirstChildTAG: according to old concept thei direction of current is from +ve to -ve (this is called conventional flow) but according to modern concept current can flow from -ve to +ve.... in solving ckt it doesnt matter that which rule you are following.... but the rule should be same througout the solution.... FirstChildUserIdTAG: 285548 FirstChildUserNameTAG: Faizan89 FirstChildCreateTimeTAG: 2012-09-10T11:16:22Z FirstChildTAG: Direction of flow of electronics is electronics current....and the arrows we make on circuit as direction of current is the conventional current...i.e., opposite of electron flow....thats the convention. FirstChildUserIdTAG: 381619 FirstChildUserNameTAG: siddhantmishra007 FirstChildCreateTimeTAG: 2012-09-10T11:25:01Z FirstChildTAG: If memory serves, Prof Agarwal mentions in Lect #2 that this is the convention he uses. ie He always labels/shows current flowing into the positive terminal of a lumped component. By doing this, the calculation of the power consumed by that lumped component will always be positive. FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-12T00:00:50Z IndexTAG: 812 TitleTAG: What happens when 1.6V comes into action? What exactly happens? Why does it start charging the one with smaller voltage, I mean, isn't a capacitor a thing that is charged? And why does it flow to the 1.5V resistor? Because the bit V thingy still has 0V as its potential.... UserIdTAG: 193033 UserNameTAG: NIslam CreateTimeTAG: 2012-09-09T20:51:07Z VoteTAG: 4 CoursewareTAG: Week 1 / Battery model CommentableIdTAG: 6002x_S1E9_Battery_Model NumberOfReplyTAG: 1 FirstChildTAG: use the KVL : V1 vR1 vR2 V2 form a loop -V2 +vR2 -vR1 +V1 = 0 ==> vR2 - vR1 = V2 - V1 = 1.6 - 1.5 = 0.1 vR2 - vR1 > 0 ==> the curent flow from R2 to R1 i = 0.1/(R2 +R1) = 0.1/0.57 = 0.17amps FirstChildUserIdTAG: 236136 FirstChildUserNameTAG: Hamid-ch FirstChildCreateTimeTAG: 2012-09-09T22:28:24Z SecondChildTAG: That was not what I was asking. What I wanted to know is why, the circuit in the first loop starts going in the opposite direction?? SecondChildUserIdTAG: 193033 SecondChildUserNameTAG: NIslam SecondChildCreateTimeTAG: 2012-09-10T05:35:49Z SecondChildTAG: hey but why u do this i=0.1/(R2+R1)?? why it is not like this i=0.1/(equivalent resistance)? since the 2 resistors are in parallel!!! therefore they are not in series!!! help plz SecondChildUserIdTAG: 151205 SecondChildUserNameTAG: jesher777 SecondChildCreateTimeTAG: 2012-09-11T18:01:55Z IndexTAG: 813 TitleTAG: Homework1 How to get the answer for "Given that each individual resistor can dissipate up to 1 watt of power before burning up, how much total power in watts ( W ) can the smallest-valued composite resistor dissipate before burning up?" Where three resistors are 4 , 4 and 6 ohms. UserIdTAG: 269109 UserNameTAG: Sanjana16 CreateTimeTAG: 2012-09-09T18:54:54Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: The power will be limited by the smallest *single* resistor in your circuit. We must then determine how this single resistor limits the voltage throughout the entire circuit. What can you say about the voltage in our circuit of parallel resistors? Will the voltage across each resistor be different or the same? Once we have determined this we may proceed to determine the power dissipated in our composite resistor. P_composite = V_max^2 / R_composite FirstChildUserIdTAG: 183711 FirstChildUserNameTAG: MMatheson FirstChildCreateTimeTAG: 2012-09-09T19:40:28Z IndexTAG: 814 TitleTAG: For question 2 ANOTHER WAY ( WITHOUT INTEGRATION) p avg = v^2 rms / R we know that v rms = 120 volts therefore p avg = 120^2/110 UserIdTAG: 140065 UserNameTAG: Madhav92 CreateTimeTAG: 2012-09-09T18:25:55Z VoteTAG: 4 CoursewareTAG: Week 1 / AC power CommentableIdTAG: 6002x_S1E3_AC_power NumberOfReplyTAG: 0 IndexTAG: 815 TitleTAG: SIMPLE TRICK! firstly ignore the current and voltage polarity/directions given. take current from +ve terminal of battery to -ve terminal so if the given current direction is opposite, take it as -ve and +ve if it is of same direction similarly, if the +ve of battery coincides with +ve of v1/v3 then it is +ve else -ve. also remenber, there is always potential drop across resistor. UserIdTAG: 148523 UserNameTAG: span993 CreateTimeTAG: 2012-09-09T17:01:25Z VoteTAG: 4 CoursewareTAG: Week 1 / Associated reference directions CommentableIdTAG: 6002x_S2E2_Associated_Reference_Directions NumberOfReplyTAG: 1 FirstChildTAG: is it true in all cases FirstChildUserIdTAG: 168651 FirstChildUserNameTAG: naval59 FirstChildCreateTimeTAG: 2012-09-13T08:44:24Z SecondChildTAG: Yes it is. You will get the same final answer eventhough the intermediate parameters varies(mostly in terms of sign). SecondChildUserIdTAG: 148523 SecondChildUserNameTAG: span993 SecondChildCreateTimeTAG: 2012-09-13T09:35:07Z IndexTAG: 816 TitleTAG: Change video quality button Please add change resolution button to the video playback controlls. It will be very usefull. UserIdTAG: 207456 UserNameTAG: Octomos CreateTimeTAG: 2012-09-08T15:01:11Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 817 TitleTAG: Using your body to short circuit a battery S1E9 got me curious. Why don't we electrocute ourselves when our body connects two poles of the same battery? Our bodies should be decent conductors, right? What does it mean to electrocute oneself anyway? Or using the model from S1E9 I have two simple off the shelf batteries 1.2V each. Can I use my body to put them in parallel? I have no intention of doing it but curious if my calculations are correct. So how would I go about doing it then? Connect + and + with one palm and - and - with my other palm? Then my body should act like a wire conducting roughly 9.6A since effectively I have a short circuit. Where am I wrong? 10A is huge right? UserIdTAG: 5325 UserNameTAG: vkz CreateTimeTAG: 2012-09-07T18:15:37Z VoteTAG: 4 CoursewareTAG: Week 1 / Battery model CommentableIdTAG: 6002x_S1E9_Battery_Model NumberOfReplyTAG: 1 FirstChildTAG: Your body is not a perfect conductor, and you would not be short-circuiting the battery with your palms. The resistance of your body depends on variety of conditions, such as whether you are soaked in water or not, but I've heard numbers like 100 kOhm and higher. A current of 10 A is huge and would very easily kill a person. FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-07T18:38:01Z SecondChildTAG: In my automotive class in high school, our teacher took an old lawnmower and connected two wires to the coil that created a spark for the spark plug. He made us stand in a circle holding hands with the lawn mower wires as part of the circle. In other words, person 1 held the positive wire in one hand and the person's hand next to him in the other. The last person in the circle had their free hand holding the negative wire. The teacher then pulled the lawn mower's starter cable and it sent a good shock through all of us in the circle. He went on the explain that the coil produced a very high voltage (40,000 maybe? I don't remember). He explained that it's not volts that kill you but instead the amperage; Which the coil had very low amps. You can recreate this by putting your tongue across the terminals of a 9 volt battery. You'll get a shock but not enough amps to hurt you. SecondChildUserIdTAG: 265346 SecondChildUserNameTAG: k_candiotti SecondChildCreateTimeTAG: 2012-09-07T19:58:17Z IndexTAG: 818 TitleTAG: Edit personal profile Anyone know how to edit username? UserIdTAG: 364364 UserNameTAG: agungsantoso CreateTimeTAG: 2012-09-07T13:21:55Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: You should send an e-mail to the edX Staff. I am not sure if you can change your username... Based my experience in the Prototype 6.002x, once we have finished the Course some Students have Problems because they wanted to modify their names because they missed something , eg: " albert einstein " for " Albert Einstein ". Luckily they could get their certificates with the requested modidications,they e-mailed to the Staff in order to solve that. Fortunately, the Staff worked really hard and kindly responded to the Students request each by each...so I suggest that if the New Students want to change something that is misspelled or anything else, you should try to do this now, at the beginning ;) ---------- Help email System-related questions: technical@edx.org Content-related questions: content@edx.org Bug reports: bugs@edx.org Suggestions: suggestions@edx.org ---------- Deberías enviarle un e-mail al Staff de edX. No estoy muy segura si puedes cambiar tu nombre de usuario...Basada en mi experiencia del Curso Prototipo 6.002x, una vez que finalizó el Curso algunos Estudiantes tuvieron dificultades a la hora de la visualización de sus nombres en los Certificados, algunos necesitaban modificar sus nombres por algún problema de tipeo o por razones como por ejemplo, que sus nombres no aparecían con letra mayúscula en sus certificados porque al inicio se habían inscripto por ejemplo como "albert einstein" y no como "Albert Einstein"... Afortunadamente, con mucho trabajo, esfuerzo y amabilidad el Staff pudo hacer frente a cada pedido individual... Por eso, sugiero a los Nuevos Estudiantes que de tener algún error de deletreo en sus nombres o cualquier inconveniente de esta magnitud, que deberían tratar de resolverlo ahora, desde el comienzo del Curso, para evitar tener dificultades al momemto de la Certificación ;) Aquí les dejo los e-mail de ayuda: System-related questions: technical@edx.org Content-related questions: content@edx.org Bug reports: bugs@edx.org Suggestions: suggestions@edx.org FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-07T14:50:28Z IndexTAG: 819 TitleTAG: Internet Access Problem Internet access in Nigeria is expensive and at times network is not stable. How can I download the lecture videos and textbook on my pc. This will enable me watch video and read the textbook directly on my computer.Kindly help. UserIdTAG: 315669 UserNameTAG: dejiroje CreateTimeTAG: 2012-09-07T12:13:38Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: they are available on youtube or use a plugin in your browser called download helper. download flv files if since they are smaller. As for the book you just have to use it online, using the book shouldn't be a problem. FirstChildUserIdTAG: 183507 FirstChildUserNameTAG: obiradaniel FirstChildCreateTimeTAG: 2012-09-07T12:21:51Z SecondChildTAG: Thank you, please How can I download the lecture videos from you tube SecondChildUserIdTAG: 315669 SecondChildUserNameTAG: dejiroje SecondChildCreateTimeTAG: 2012-09-07T13:54:29Z IndexTAG: 820 TitleTAG: Kirchhoff Gustav Kirchhoff http://en.wikipedia.org/wiki/Gustav_Kirchhoff UserIdTAG: 389424 UserNameTAG: jmlietaer CreateTimeTAG: 2012-09-07T08:46:57Z VoteTAG: 4 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 0 IndexTAG: 821 TitleTAG: Speed of light constraint There are three assumptions mentioned in the video, while only first two of them are explained. Why is that speed of signals can not be close to the speed of light? Why it has to be WAY lower? UserIdTAG: 203445 UserNameTAG: x13n CreateTimeTAG: 2012-09-07T06:31:07Z VoteTAG: 4 CoursewareTAG: Week 1 / Lumped Element Abstraction CommentableIdTAG: 6002x_lumped_element_abstraction NumberOfReplyTAG: 6 FirstChildTAG: because the laws like Newtons laws, Maxwell's equations are'nt applicable if speed of signals exceed speed of light.A different analysation is needed for such signals. FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-07T07:57:53Z FirstChildTAG: Read the textbook...... FirstChildUserIdTAG: 150267 FirstChildUserNameTAG: vwsingh FirstChildCreateTimeTAG: 2012-09-07T10:01:07Z FirstChildTAG: Is it is to allow a wide enough interval to ensure that the system is in sync? Like when you have slow music, it is easier for dancers to get their act together. If the music a lot faster, not everyone would have time to be in sync. FirstChildUserIdTAG: 254233 FirstChildUserNameTAG: jalba313 FirstChildCreateTimeTAG: 2012-09-07T15:35:28Z FirstChildTAG: Like with the other two abstractions, I believe it is to make calculations easier. When the signal is not close to c you don't have to worry about using wave equations. (my opinion. I don't have the text book) FirstChildUserIdTAG: 167370 FirstChildUserNameTAG: alphathreethree FirstChildCreateTimeTAG: 2012-09-08T00:52:36Z FirstChildTAG: Nothing can be faster than light. If you are very close to the speed of light you would have to use quantum calculations. FirstChildUserIdTAG: 268104 FirstChildUserNameTAG: SaulCardenasG FirstChildCreateTimeTAG: 2012-09-08T05:25:19Z SecondChildTAG: Remember that the speed of light was exceeded in CERN Geneva, I think in September 2011. So, is not anymore an universal constant. SecondChildUserIdTAG: 370201 SecondChildUserNameTAG: Andrey26 SecondChildCreateTimeTAG: 2012-09-09T22:36:54Z SecondChildTAG: yeah write said - experimental order SecondChildUserIdTAG: 539003 SecondChildUserNameTAG: amitgupta25121993 SecondChildCreateTimeTAG: 2012-10-04T23:38:14Z FirstChildTAG: If the speeds of the signals are close to the speed of light the newton's laws breakdown.. Einstein's theory of relativity comes into play there.So this constraint is necessary if we are to avoid quantum physics which makes everything more complex FirstChildUserIdTAG: 363825 FirstChildUserNameTAG: FARAN1195 FirstChildCreateTimeTAG: 2012-09-09T19:00:10Z IndexTAG: 822 TitleTAG: awesome this course is awesome. some of the problems are challenging and it always feels great when you answered a problem and a check beside it appears! Thanks edX for this opportunity! UserIdTAG: 371000 UserNameTAG: Joseph090892 CreateTimeTAG: 2012-09-07T01:48:31Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 823 TitleTAG: For the problem #4 Don't try to invent fancy tricks. Just go with Kirchgoffs: - KVL (clockwise): i1 r1 - i2 r2 = -V2 + V1 - KCL (upper wire): i1 - i2 = 0 Currents i1 and i2 go up. Solve the system and you're done. UserIdTAG: 325358 UserNameTAG: KKostya CreateTimeTAG: 2012-09-06T14:07:19Z VoteTAG: 4 CoursewareTAG: Week 1 / Battery model CommentableIdTAG: 6002x_S1E9_Battery_Model NumberOfReplyTAG: 3 FirstChildTAG: but it's still closed circuit and a current flows, so Ohm law is easyer=) FirstChildUserIdTAG: 210380 FirstChildUserNameTAG: MurdocRus FirstChildCreateTimeTAG: 2012-09-06T15:52:11Z FirstChildTAG: I forgot that I need to sum up the voltage when using KVL. I just used i1r1-i2r2=0 and the best part is I just had my A2 trials earlier this week. Lol Anyway, thanks for the tips. FirstChildUserIdTAG: 157273 FirstChildUserNameTAG: ongchihang FirstChildCreateTimeTAG: 2012-09-07T06:18:39Z FirstChildTAG: best answer so far. i see a lot of ppl commenting about quick tricks to solve the problems FirstChildUserIdTAG: 163686 FirstChildUserNameTAG: caioaao FirstChildCreateTimeTAG: 2012-09-11T19:13:54Z IndexTAG: 824 TitleTAG: Circuit Simulator still broken on one of the platforms/servers. Can't do any of the labs at moment. As the server or platform I and others are running on is still broken. When we did this course with MITx they posted regular updates when problems like this were found. It is working for many others but others are seeing the same bugs. Lots are assuming it is a browser issue. It is NOT the browser. I have logged into my old MITx course and my old course-ware still shows the simulator working correctly. So, the problem is probably due to how they allocate resources as obviously not all students will be sharing same platform/server. My guess is there are Lots of platforms/servers running identical code but all linked to a central Database. Those of us seeing these bugs are on a server which is flaky or with simulator software corrupted on it in some way (probably needs latest patch or something). Everyone else who are not seeing these problems are almost certainly on platforms which are stable. It would be nice, like the staff did at MITx, if they could possibly let us know what is happening. Anyone figured it out yet???? UserIdTAG: 15344 UserNameTAG: kob CreateTimeTAG: 2012-09-06T10:25:20Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 825 TitleTAG: lab 1 showing nothing My lab 1 is not showing any circuit. Just two graphs are there, and on the right side a symbol of resistance. I have tried different PCs and browsers, but nothing worked. PLZ help? How to overcome this problem. UserIdTAG: 378096 UserNameTAG: Jamshaid271 CreateTimeTAG: 2012-09-06T09:02:20Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: Its is still broken for many of us. When we did this with MITx they posted regular updates when problems like this were found. It is working for many others but I, you and others are seeing the same bugs. Lots assume it is the browser. It is not the browser. I have logged into my old MITx course and my old coursware still shows the simulator working correctly. The problem is probably due to how they allocate resources as not all students will be sharing same platform. Lots of platforms running identical code but all linked to a central Database. Those of us seeing this problem are on some server which is flaky. They should, like the staff did at MITx, let us know what has happening. Anyone figured it out yet???? THEY FIXED THIS BY INCLUDING THE RESET BUTTON A COUPLE DAYS AGO. JUST CLICK THE RESET BUTTON AT BOTTOM WHEN IT APPEARS. THEN YOU'LL SEE CORRECT COMPONENTS SHOWING. FirstChildUserIdTAG: 15344 FirstChildUserNameTAG: kob FirstChildCreateTimeTAG: 2012-09-06T10:16:11Z SecondChildTAG: nopes. in mine only resister is showing. SecondChildUserIdTAG: 72647 SecondChildUserNameTAG: NEEL11 SecondChildCreateTimeTAG: 2012-09-06T13:01:58Z SecondChildTAG: even i have only resistor no reset button is available dude SecondChildUserIdTAG: 78647 SecondChildUserNameTAG: shiva24 SecondChildCreateTimeTAG: 2012-09-11T15:55:05Z FirstChildTAG: even i have the same problem dude FirstChildUserIdTAG: 78647 FirstChildUserNameTAG: shiva24 FirstChildCreateTimeTAG: 2012-09-09T14:30:20Z IndexTAG: 826 TitleTAG: question 2 can any on explain me 2nd question UserIdTAG: 309525 UserNameTAG: balamuralimanoghar CreateTimeTAG: 2012-09-06T05:21:36Z VoteTAG: 4 CoursewareTAG: Week 1 / Battery model CommentableIdTAG: 6002x_S1E9_Battery_Model NumberOfReplyTAG: 3 FirstChildTAG: make short-circuit, then apply KCL,, current flow from batteries to the ground or negative probe of battery over nodes. FirstChildUserIdTAG: 60496 FirstChildUserNameTAG: tomipiriyev FirstChildCreateTimeTAG: 2012-09-06T12:22:36Z FirstChildTAG: I used KCL at the upper node. 1. Calculate the current I1 = V1/R1 going up from the first battery 2. Calculate the current I2 = V2/R2 going up from the second battery 3. As the circuit is shorted, the current flowing through the shorted wire must be the sum of I1 and I2 leaving the upper node. Hope that helps FirstChildUserIdTAG: 35642 FirstChildUserNameTAG: caled FirstChildCreateTimeTAG: 2012-09-06T20:21:07Z FirstChildTAG: me too .As for diagram , just join the + and - with a zero resistance wire . that's short circuiting .but if it's zero resistance , won't the current in it be infinite , since it has a non zero potential difference across it ? FirstChildUserIdTAG: 211840 FirstChildUserNameTAG: edxian FirstChildCreateTimeTAG: 2012-09-06T06:41:18Z IndexTAG: 827 TitleTAG: S1E6: KVL question Why in S1E6: KVL is v5 = v1+v2-V ? Thanks. UserIdTAG: 373629 UserNameTAG: Gupu25 CreateTimeTAG: 2012-09-06T03:13:42Z VoteTAG: 4 CoursewareTAG: Week 1 / KVL and KCL exercises CommentableIdTAG: 6002x_kvl_kcl_exercises NumberOfReplyTAG: 1 FirstChildTAG: The loop you use to calculate it is the loop from V through v1, through v5, then through v2 and back to V. Also make sure you are checking which way the + and - are pointing along the the circuit because this will determine the sign of the voltage(v#) in the equation. For S1E6: 1 - Lets assume the + - direction for V are positive in the equations, so for V we get: V + ... = 0 2 - Since v1's + - are in the opposite direction from V's + - the sign in front v1 is negative. Now the equation is V - v1 + ... = 0 3 - Since v5's + - are in the same direction as V's the sign is positive. Now the equation is: V - v1 + v5 + ... = 0 4 - Lastly v2's + - are in the opposite directions as V the sign is negative. The final equation for the loop is: V - v1 + v5 - v2 = 0 Rearranging the equation, the equation becomes v5 = v1 + v2 - V Hope this help you out. FirstChildUserIdTAG: 5242 FirstChildUserNameTAG: Pmurph FirstChildCreateTimeTAG: 2012-09-06T04:24:56Z SecondChildTAG: Thank you for this. I've been struggling a lot with the signs, and you just made it so clear! SecondChildUserIdTAG: 208854 SecondChildUserNameTAG: BrynnleeEaton SecondChildCreateTimeTAG: 2012-09-06T13:58:47Z SecondChildTAG: Very helpful! SecondChildUserIdTAG: 265346 SecondChildUserNameTAG: k_candiotti SecondChildCreateTimeTAG: 2012-09-07T03:04:40Z SecondChildTAG: help me!!! hoe write the rta i cant!!!! need help please SecondChildUserIdTAG: 81972 SecondChildUserNameTAG: andres857 SecondChildCreateTimeTAG: 2012-09-16T02:23:04Z IndexTAG: 828 TitleTAG: "strength"? "the strength of the source is VS=10 V"... Really? Strength, not voltage? UserIdTAG: 188586 UserNameTAG: FLara CreateTimeTAG: 2012-09-06T00:09:42Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 5_Simple_Power NumberOfReplyTAG: 0 IndexTAG: 829 TitleTAG: Circuit simulation broken In lab1 there is no voltage or ground component showing?? This is weird as obviously it shows for some and not others. I think it must be the way they distribute the servers between students. Probably so many hundred per server or virtual server. With the one I happen to be on is one of the ones with problems. Where as others must have been assigned to good servers! Mind you that is a wild guess. Has to be something like that though. THEY FIXED IT YESTERDAY BY ADDING A RESET BUTTON. JUST CLICK IT AND YOU"LL SEE ALL COMPONENTS NOW APPEAR UserIdTAG: 15344 UserNameTAG: kob CreateTimeTAG: 2012-09-05T23:41:40Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: Maybe everybody should add the problem to some sort of dedicated comment, so that staff can find out on what server(s) the problem exist and which are working ok. The sandbox on my machines (PC with WinXP64 and IE8 and MacBook with OSX Lion Safari/Firefox/Chrome - all updated) does not work for Transient analysis functionality. Rest of functionality seems to be OK. Weird ... FirstChildUserIdTAG: 389424 FirstChildUserNameTAG: jmlietaer FirstChildCreateTimeTAG: 2012-09-07T12:57:01Z FirstChildTAG: The idea is that for various labs, you do not need the full set of components. For example, in Lab 1, you only need resistors to be able to solve the problem. FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-06T00:29:51Z SecondChildTAG: I have zero components showing, that is the problem. I have done this course at MITx so I know how it all works. The problem here is that nothing is showing in the lab at all. Accept for single resistor. I have just completed the two homeworks but looks like I can't do labs because of this bug. Very frustrating. I tried chrome, firefox, Safari and opera. None are showing the components so it has to be due to the segment of the server I (and others) have been allocated to. So although you may well be able to use the lab those of us who are unlucky enough to be on this particular server (VMware Guessing???) will all be seeing this same problem. I just hope they can fix it. THEY FIXED IT YESTERDAY BY ADDING A RESET BUTTON. JUST CLICK IT AND YOU"LL SEE ALL COMPONENTS NOW APPEAR SecondChildUserIdTAG: 15344 SecondChildUserNameTAG: kob SecondChildCreateTimeTAG: 2012-09-06T01:45:22Z IndexTAG: 830 TitleTAG: 2 similar answers, both accepted as correct. Using the 120⋅2√⋅cos(2π⋅60⋅t) formula I got the following answers: 268.55, 134.28 and 134.28. Using (v^2)/R I got the following: 261.82, 130.91 and 130.91. I didn't quite understand the difference in working out the values between the second and third answers, which are identical in my answers. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-05T21:42:25Z VoteTAG: 4 CoursewareTAG: Week 1 / AC power CommentableIdTAG: 6002x_S1E3_AC_power NumberOfReplyTAG: 2 FirstChildTAG: Look - page 42 (textbook) ... the rms value of a periodic signal is the value of DC signal that would have resulted in the same average power dissipation. FirstChildUserIdTAG: 307779 FirstChildUserNameTAG: Izabella FirstChildCreateTimeTAG: 2012-09-06T02:20:40Z FirstChildTAG: Hi anonymous good work! The reason they equate is due to the circumstances of the resistor. FirstChildUserIdTAG: 8026 FirstChildUserNameTAG: jcbmack FirstChildCreateTimeTAG: 2012-09-09T03:04:00Z IndexTAG: 831 TitleTAG: Greetings.. Hello friends and teachers!!! My name is Jesus Eduardo Ortiz Sandoval, I´m from Colombia, I'm electronic engineer graduate of the University of Pamplona in April this year. For me is a placer be here!! I hope to make many friends, enhance my skills and maybe if I can pull information request for a scholarship in the United States and doing my graduate studies in that country. I hope that we all help and enjoy these four months together. -------------------------------------------------------------------------------- Hola amigos y profesores, Mi nombre es Jesus Eduardo Ortiz Sandoval, soy Colombiano, yo soy Ingeniero Electronico egresado de la Universidad de Pamplona y obtuve mi titulo en Abril de este año. Para mi es un gran placer estar aca con ustedes. Espero hacer muchos amigos, reforzar mis conocimientos y tal vez si puedo lograrlo solicitar informacion para obtener una beca en los Estados Unidos y hacer mi estudios de postgrado en ese pais. Espero que todos nos ayudemos y poder disfrutar estos 4 meses juntos. UserIdTAG: 331801 UserNameTAG: ingjesusortiz CreateTimeTAG: 2012-09-05T21:39:59Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 4 FirstChildTAG: Hola Jesus, es un gusto de mi parte saber que hay latinoamericanos en este curso, saludos. FirstChildUserIdTAG: 363033 FirstChildUserNameTAG: JesusAlvarez FirstChildCreateTimeTAG: 2012-09-05T22:00:09Z SecondChildTAG: Hola Jesus, ojala nos unamos para ayudarnos, un abrazo. SecondChildUserIdTAG: 331801 SecondChildUserNameTAG: ingjesusortiz SecondChildCreateTimeTAG: 2012-09-08T02:51:36Z FirstChildTAG: Hola Jesus, estaba buscando un aliado venezolano pero no consegui ninguno. Ya realizastes todos los ejercicios introductorios (del overview)? Te ayudaran a entender y familiarizarte con el sistema. Espero que podamos ayudarnos todos los latinoamericanos, tambien quisiera terminar mis estudios en EEUU. FirstChildUserIdTAG: 298547 FirstChildUserNameTAG: AnthonyRF FirstChildCreateTimeTAG: 2012-09-05T21:42:37Z SecondChildTAG: Que tal Anthony, si ya los hice todos, me faltan el HW, y el Lab. Cualquier duda con gusto te ayudo. SecondChildUserIdTAG: 331801 SecondChildUserNameTAG: ingjesusortiz SecondChildCreateTimeTAG: 2012-09-08T02:52:17Z FirstChildTAG: Hola Jesus, saludos desde Honduras. Veo que andamos por caminos parecidos, porque yo me gradué como ingeniero en telecomunicaciones en junio de este año, y también estoy en el proceso de aplicar a postgrado en universidades de EEUU, Canadá, Inglaterra, Corea, etc. Quise entrar a este curso para desempolvar mis conocimientos de electrónica y circuitos, y espero poder encontrar aún más hermanos latinoamericanos por aquí. FirstChildUserIdTAG: 379669 FirstChildUserNameTAG: jcperez90 FirstChildCreateTimeTAG: 2012-09-05T22:24:29Z SecondChildTAG: Hola jcperez, un gusto saludarte y empezar a sellar alianzas latinoamericanas ojala, si tienes alguna informacion la puedas compartir, lo mismo que yo la compartire contigo, un abrazo. SecondChildUserIdTAG: 331801 SecondChildUserNameTAG: ingjesusortiz SecondChildCreateTimeTAG: 2012-09-08T02:54:18Z FirstChildTAG: Nice ingjesusortiz! I wish you the best and I hope that you can go to United States! I got my Passaport almost 3 years ago but it don't have any stamp haha, I have never traveled outside my Country in my life, I am Student but happy haha! But I am sure that you will! You are an engineer and you will have a lot of opportunities! Give your best and you will be rewarded! My best wish to you! ---- Hola ingjesusortiz! Te deseo lo mejor y espero que puedas ir a Estados Unidos! Yo me he sacado el Pasaporte hace tres años y todavía ni siquiera tengo un sello, está en blanco. Todavía nunca salí de mi País, es que como verás soy Estudiante jaja, pero feliz! Pero estoy segura que lo lograrás ya que estás graduado y eso te abre muchas puertas! Da lo mejor y serás recompensado! Mi mejor deseo para tí. Fuerzas desde Argentina! FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-06T00:26:53Z SecondChildTAG: Hola gracias por tus saludos, feliz de encontrar amigos Latinoamericanos!!! Un abrazo. SecondChildUserIdTAG: 331801 SecondChildUserNameTAG: ingjesusortiz SecondChildCreateTimeTAG: 2012-09-08T02:54:38Z IndexTAG: 832 TitleTAG: HW Question If you get a question wrong the first time on the HW, does it subtract from the overall score? For example say I got question 1 wrong, but then answered it correctly; is my grade impacted? I only ask this because I "checked" my answers and didn't realize it would mark blank ones wrong... UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-05T21:11:45Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: Hi anonymous! :). Based my experience in the Circuits and Electronics Course, no. You have unlimited Checks in your Homeworks and Labs. This don't impact in your grade.So, don't be upset... But, remember that in the Midterm and Final, yes. Because you only have 3 check opportunities, only 3... I am here to help New Students, if you have any doubt please ask me, I will be pleasured to help you. See you! Myriam. ---- Hola anonymous!:). En base a mi experiencia, ya que he cursado este mismo Curso, pero el Prototipo, el que fue lanzado previo a este, te puedo responder que no modifica tu Nota de Calificación. Es decir, tu tienes ilimitadas oportunidades para Chequear (Check) tus Homeworks y Labs. Así es que no te pongas mal... Lo que sí, es que en el Examen Medio y Final, sí que impacta el contestar mal tres veces, ya que solo tienes 3 oportunidades para presionar el botón Check... Estoy aquí, cursando nuevamente este Curso principalmente para ayudar a los Nuevos Estudiantes (también para brindar soporte a la Comunidad Hispanohablante). Es por ello, que si tienes una duda, por favor pregúntame, estaré encantada de ayudarte. Nos vemos! Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-05T23:26:29Z SecondChildTAG: HI myriam,my ans r not being accepted in ans box.for eg h1p1- 1st que 3R OR R+R+R,AND HOW TO ANS CIRCUIT C QUE, MY ANS 5R/3is not being expected.pls help. SecondChildUserIdTAG: 269641 SecondChildUserNameTAG: BAUWA SecondChildCreateTimeTAG: 2012-09-10T20:52:17Z FirstChildTAG: No it's not. Your grades on the homeworks are what you can do before the due date, basically. At the end of the day, though, during the original MITx Prof. Agarwal said that > "We encourage collaboration on homeworks and labs since they are primarily learning exercises. However the midterm should be completed on your own without any help from any other person." My point is that the real part of the grade comes from the exams anyway, and, if they do the same thing this time around as last time, you'll only get 3 entries to get it right on the exams. FirstChildUserIdTAG: 70103 FirstChildUserNameTAG: KMayne FirstChildCreateTimeTAG: 2012-09-05T21:59:20Z IndexTAG: 833 TitleTAG: signs ¿? i can't understand the problem with the signs, + or -, i can't get it. I'm sure isn't difficult UserIdTAG: 149549 UserNameTAG: Java_Dido CreateTimeTAG: 2012-09-05T21:00:01Z VoteTAG: 4 CoursewareTAG: Week 1 / Power CommentableIdTAG: 6002x_S1E2_Power NumberOfReplyTAG: 2 FirstChildTAG: Current is opposite to voltage, so it is negative. Curent goes from "+" to "-" by convention FirstChildUserIdTAG: 279867 FirstChildUserNameTAG: Lisenik FirstChildCreateTimeTAG: 2012-09-05T21:14:53Z FirstChildTAG: In an element that dissipates power, as current passes through it there's a decrease of potential. For a source, as current passes through it there's an increase of petential. As the drawing shows the current i in the opposite way you would expect for a resistor, its sign is negative. FirstChildUserIdTAG: 188586 FirstChildUserNameTAG: FLara FirstChildCreateTimeTAG: 2012-09-06T00:28:59Z SecondChildTAG: I got it!!!! thank you both. It's really simple SecondChildUserIdTAG: 149549 SecondChildUserNameTAG: Java_Dido SecondChildCreateTimeTAG: 2012-09-06T16:18:28Z IndexTAG: 834 TitleTAG: Hints for H1P2 1) Jst count the no. of nodes (Remember a node is a junction point between two elements). 2) No of indepenedent KCL euations are N-1; N=no. of nodes (from previous question). 3) jst count no. of loops i. e. hw many closed paths are there. 4) Jst count the no. of branches (Remember a branch is a connection between the node) and now we know the no. of nodes from first part of the question so the no. of independent KVL eqnts are: B-N+1 ; B= no. of branches 5)As given in question:The total number of independent equations needed to solve a circuit using the KCL/KVL method is twice the number of elements in the circuit (eight in this example) and out of these 8 questions jst subtract the no. of independent KVL & KCL equations. 6)& 7) jst apply KVL & KCL for the given figure. remember i1=-8A,v4=4V; Also u have to use v2=i2*1, v3=i3*3.using this procedure calculate all d unknown variables. 8) power coming out of the source =v1*i1 9) power coming out of voltage source = v4*i4 10)power dissipated in 1ohm resistor =1*i1*i1 11) power dissipated in 3ohm resistor=3*i3*i3. UserIdTAG: 82571 UserNameTAG: Nidhi3 CreateTimeTAG: 2012-09-05T19:31:59Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: for the 9)--> v4 * i4 = -ive number, but when i4 is clearly a +ive number, why is the answer -ive FirstChildUserIdTAG: 204745 FirstChildUserNameTAG: allwynmendes FirstChildCreateTimeTAG: 2012-09-07T04:37:41Z IndexTAG: 835 TitleTAG: Help on H1P2 In the Image there are 3 nodes. One of them are ground node. So, number of node is 3 . ![enter image description here][1] [1]: https://mail-attachment.googleusercontent.com/attachment/?ui=2&ik=327d8f4ebb&view=att&th=13997a3e08800825&attid=0.1&disp=inline&realattid=f_h6qqx51a0&safe=1&zw&saduie=AG9B_P9dUEua0Ht-fBe8wMzIlM6B&sadet=1346868895697&sads=gZG_tyoSo6XdpXE3tzceQ6GctFk If we consider an incoming current is positive and outgoing current in negative . Then from the KCL outgoing current=incoming current.At node 1 from KCL i1+i2+i3=0.......(1) at node 2 from KCL: i3=i4.............(2) replacing (1) by (2): i1+i2+i4=0.........(3) but (3) is dependent on (1) and (2) so number of independent KCL equations is 2 Notice there are 3 loops in the circuit 2 little and 1 big ;) so it is clear that 2 KVL equation can be derived from two loops and one equation by combining them so number of independent KVL equations is 2. so 2 KVL + 2 KCL ;) So we assume that the current flow through left loop is I1 and current flow through the right loop is I2. Then from left loop by KVL: I1=8A, since we have a current source in the loop. and from the right loop: 3*I1+4+1*(I2-I1)=0 solving the two equations we get value of I2 now V1=V2 as they are parallal connected. now V2=i2*1ohms=(I1-I2)*1 i4=I2 as same current flow through the loop power coming out of the current source,W=V1*i1=V2*I1 power coming out of the voltage source,W=V4*i4=V4*I2 (Here comes a negative sign guess why ;)) dissipated in the 1Ω resistor=V2*i1=V2*(I1-I2)=V2^2/1 dissipated in the 3Ω resistor=i3^2*3=I2^2*3 "so here comes y conservation of power, the net power coming out of the two sources must equal the total power dissipated in the two resistors. If this is not true, then you made a mistake." UserIdTAG: 14730 UserNameTAG: Tanvir CreateTimeTAG: 2012-09-05T19:09:37Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: This is just not done!..I mean one is not supposed to post answers directly before the stipulated time limit is over!!!.... kindly respect the honor code!!!... FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-05T19:42:32Z IndexTAG: 836 TitleTAG: LAB0 HEY,I finding little problem at naming A,B,C,D. some one help me out here? UserIdTAG: 48151 UserNameTAG: santoshhalagatti CreateTimeTAG: 2012-09-05T19:04:41Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Double click on the ??? of the node, to edit the properties of the node. FirstChildUserIdTAG: 85490 FirstChildUserNameTAG: yashsavani FirstChildCreateTimeTAG: 2012-09-05T19:05:42Z SecondChildTAG: where will be that ??? SecondChildUserIdTAG: 48151 SecondChildUserNameTAG: santoshhalagatti SecondChildCreateTimeTAG: 2012-09-05T19:16:24Z IndexTAG: 837 TitleTAG: Need help finding a book. Hey everyone, I'm currently a student in grade 12, and my physics isn't very sound...i'm pretty much just learning the basics these days...I joined this course because i'm extremely interested in electrical engineering stuff, and I was hoping I'd get some idea about the course or learn some really cool stuff...but I don't understand much, if not anything in this course...can someone recommend a good physics book or something which I can first read and practice, which will help me to survive this course? Thanks a lot...I know I sound like a dumb person, but I really want to study in this field, and it would be great if you guys could help me out :) UserIdTAG: 362302 UserNameTAG: Bhaiyo95 CreateTimeTAG: 2012-09-05T18:51:01Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 5 FirstChildTAG: Do NOT underestimate your chance of doing well in this course. "The more you learn, the more you *can* learn." FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-09-06T02:29:32Z SecondChildTAG: And check out khanacademy.org. They have a nice set of video lectures on physics. SecondChildUserIdTAG: 92346 SecondChildUserNameTAG: MobiusTruth SecondChildCreateTimeTAG: 2012-09-23T04:59:24Z FirstChildTAG: I too am in grade 12, and at first had a hard time understanding the byzantine physics concepts underlying this course. I found the material on MIT OCW, course 8.002 by prof. Walter Lewin extremely helpful. In addition I used the physics textbook, University Physics with modern physics Young And Freedman, to supplement what I learn't with MIT OCW's 8.002 course. FirstChildUserIdTAG: 85490 FirstChildUserNameTAG: yashsavani FirstChildCreateTimeTAG: 2012-09-05T19:03:41Z FirstChildTAG: Hi I am Iranian student and in my country every one want to study engineering must read Fundamentals of Physics By David Halliday, Robert Resnick and Jearl Walker this book are very good FirstChildUserIdTAG: 133121 FirstChildUserNameTAG: loe FirstChildCreateTimeTAG: 2012-09-05T19:08:15Z FirstChildTAG: This is the second couse, not the first. You probably want to take the 6.001 course at open courseware first... FirstChildUserIdTAG: 25252 FirstChildUserNameTAG: cwm9 FirstChildCreateTimeTAG: 2012-09-05T19:21:05Z SecondChildTAG: Although 6.001x is technically the first course in the curriculum, based on the description, it is not a prerequisite. 6.002x is all about understanding basic circuits, not so much about the low-level code that can run on top of computer architectures. The most important prerequisite for this course is a basic understanding of electricity and magnetism, most of which is covered by Unit 6 of Udacity's PH100 [here][1]. Alternatively, 8.002 provides a much more in depth and complete understanding, at the cost of a longer learning curve. Whatever you choose, I hope you learn a lot and wish you the utmost success in electrical engineering. [1]: http://www.udacity.com/view#Course/ph100/CourseRev/1/Unit/418001/Nugget/420001 SecondChildUserIdTAG: 26931 SecondChildUserNameTAG: Csscade SecondChildCreateTimeTAG: 2012-09-07T23:16:34Z FirstChildTAG: I think the point of this course is that you can just throw away all the complicated Maxwells equations with div grad and curl math, and just use the simplified formulas, so you do not need any of that physics. So far, only one simple derivatives for finding maximums. Just follow the lectures, and you will see it is quite intuitive! Good luck. FirstChildUserIdTAG: 142857 FirstChildUserNameTAG: viking2 FirstChildCreateTimeTAG: 2012-09-12T02:27:03Z IndexTAG: 838 TitleTAG: I can't watch the lecture sequence videos I can't watch the lecture sequence videos. Can somebody help me with this please? UserIdTAG: 45307 UserNameTAG: josejimenez2 CreateTimeTAG: 2012-09-05T16:32:44Z VoteTAG: 4 CoursewareTAG: Week 1 / Course overview CommentableIdTAG: 6002x_S1V1toS1V4 NumberOfReplyTAG: 3 FirstChildTAG: Do you have flash installed? You can download and install chrome it's a fast web browser that has flash built in. FirstChildUserIdTAG: 306208 FirstChildUserNameTAG: Creamsaw FirstChildCreateTimeTAG: 2012-09-05T18:37:38Z FirstChildTAG: try to reload the page,if you still have the problem,try to use another browsers,dont forget about adobe flash,you must download and install it before watching video FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-06T09:50:12Z FirstChildTAG: Even i am unable to watch the lecture videos..i tried with both google chrome and firefox and also installed the latest adobe flash player but theres still an error. I could watch the previous video but not this and some of them. FirstChildUserIdTAG: 294497 FirstChildUserNameTAG: Saurabhkotian FirstChildCreateTimeTAG: 2012-09-06T17:10:50Z SecondChildTAG: well the same happened to me and now I'm using windows internet explorer. I know that it's far slower than chrome and many others browsers, but to me it's the only one that worked to watch the videos SecondChildUserIdTAG: 45307 SecondChildUserNameTAG: josejimenez2 SecondChildCreateTimeTAG: 2012-09-09T22:39:35Z IndexTAG: 839 TitleTAG: Transient Analysis I can't activate the transient analysis. I don't understand what is wrong. UserIdTAG: 140842 UserNameTAG: srvz39 CreateTimeTAG: 2012-09-05T15:30:54Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 840 TitleTAG: -2 power Can anyone explain in more detail why the incoming power for the voltage source is -2? It makes sense intuitively, but we were hoping to figure out a more concrete reason. Thanks! - Den Haag UserIdTAG: 197521 UserNameTAG: suenteus CreateTimeTAG: 2012-09-05T14:48:24Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: 5_Simple_Power NumberOfReplyTAG: 7 FirstChildTAG: Simply because its a power "SOURCE" therefore incoming power is negative, outgoing power is positive. If incoming power is positive it is a SINK not a SOURCE. FirstChildUserIdTAG: 80920 FirstChildUserNameTAG: MichaelSturgess FirstChildCreateTimeTAG: 2012-09-05T14:54:07Z SecondChildTAG: Don't really see this. If I connect the negative terminals of 2 batteries with voltages V1 and V2 and the positive terminals with a resistor R in between do they both have outgoing power positive because they are "SOURCES"? SecondChildUserIdTAG: 141235 SecondChildUserNameTAG: ilitzroth SecondChildCreateTimeTAG: 2012-09-05T17:42:45Z FirstChildTAG: I believe it has to do with the word "entering", to make it algebraically correct since the power is not created, the total sum must be zero. If the word used was "leaving" then it would be +2, but intuitively we would know the value wold not be added to the equation but subtracted. my 2 cents... FirstChildUserIdTAG: 371118 FirstChildUserNameTAG: Cameus FirstChildCreateTimeTAG: 2012-09-05T16:39:17Z SecondChildTAG: I also dont quite get the term of watts *entering* the psu. My pov is 2 watts get out and and get dissipated over the resistor, hence its lost, so 2-2=0. Dissipation means its lost in thermal energy around resistor. It doesnt mean it flows out of resistor and enters psu. Power doesnt *enter* a psu but it exits it. If it enters, then its a charging battery not psu. Saying: power *entering* is minus 2w is weird at least. Poor choice of words. Some teachers will try double negations to try catch a student with its pants off but i never get it why. Is this about grammar logic and double negations or is it about electronics? Coz not all languages allow double negations in the same way it appears in english. I suspect native language syntax is why some ppls are ok with this wording and why others arent quite so. SecondChildUserIdTAG: 331483 SecondChildUserNameTAG: Doru SecondChildCreateTimeTAG: 2012-09-06T02:51:36Z FirstChildTAG: First, calculate the current, I= V/R, I= 10/50 = 0.2. Now, P= V*I = 10*0.2=2 FirstChildUserIdTAG: 185168 FirstChildUserNameTAG: pionoor FirstChildCreateTimeTAG: 2012-09-05T17:17:44Z SecondChildTAG: tks SecondChildUserIdTAG: 184556 SecondChildUserNameTAG: rhod SecondChildCreateTimeTAG: 2012-09-06T22:44:26Z FirstChildTAG: I believe it has to do with the direction of the current and the voltage difference. Through the resistor the current flows from + to - and the resistor dissipates heat (does work) as a result of that. Inside the battery current flows from - to + and something has to do work to make that happen. FirstChildUserIdTAG: 141235 FirstChildUserNameTAG: ilitzroth FirstChildCreateTimeTAG: 2012-09-05T17:48:53Z SecondChildTAG: well it is clearly told in the lectures that whenever the current is entering the circuit element (via +ve terminal) the power will be consumed by the element and when it is going out of the +ve terminal, it is delivered by the element SecondChildUserIdTAG: 129030 SecondChildUserNameTAG: vjbhatt786 SecondChildCreateTimeTAG: 2012-09-05T19:13:12Z FirstChildTAG: The power incoming must be equal in magnitude but opposite in sign to the power dissipated/used up. -2 is a reflection of + 2, representing power coming in and not yet used/dissipated. FirstChildUserIdTAG: 8026 FirstChildUserNameTAG: jcbmack FirstChildCreateTimeTAG: 2012-09-05T20:45:44Z FirstChildTAG: the input power calculation done by = voltage drop across * current entering the element = 10V * -0.2amps(bcoz current is entering through negative terminal) =-2W. This is my analogy. correct me if i am wrong. Regards Hari FirstChildUserIdTAG: 123484 FirstChildUserNameTAG: hudkmr FirstChildCreateTimeTAG: 2012-09-06T05:40:07Z FirstChildTAG: This is what makes most sense to me, not sure how correct it is: 1) current i in clock-wise direction is always positive (0.02) since it flows from plus to minus, in particular when it enters the source. 2) voltage v between - and + is negative (-10) since it's the reverse of normal situation. So at the point of entering the source p = v*i = -0.2 FirstChildUserIdTAG: 189680 FirstChildUserNameTAG: vnd FirstChildCreateTimeTAG: 2012-09-06T19:41:10Z IndexTAG: 841 TitleTAG: Video Lectures For those who want the video lectures of this course, taken in 2000. You can find those at http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-electronics-spring-2007/video-lectures/ I've read that a video will be added in the next few days or so. However, most of you will probably be impatient like I am. UserIdTAG: 272660 UserNameTAG: RonnieDouma CreateTimeTAG: 2012-09-05T13:34:15Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 842 TitleTAG: Welcome Hello! Welcome to 6.002x. I just want to say that this course is amazing, and I didn't miss a lecture in Spring 2012. Explore all of the cool features online education has to offer and be part of this revolution! UserIdTAG: 85819 UserNameTAG: Stelaras0 CreateTimeTAG: 2012-09-05T12:32:59Z VoteTAG: 4 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: Welcome Stelaras0! FirstChildUserIdTAG: 26931 FirstChildUserNameTAG: Csscade FirstChildCreateTimeTAG: 2012-09-06T00:36:42Z FirstChildTAG: Good day, Welcome on board! FirstChildUserIdTAG: 329879 FirstChildUserNameTAG: Ugochukwu FirstChildCreateTimeTAG: 2012-09-10T10:37:11Z IndexTAG: 843 TitleTAG: Has anyone received the proctored exam certificate? Has anyone received their proctored exam certificate? Just wanted to confirm whether everyone who participated in Pearson exam is waiting for the certificate or is it just me. Sorry for posting this again and again but I'm not getting any answer from edX. I'm worried because I had spent ~Rs.5200 on it, it was supposed to be available by 13 March and there's still no update. UserIdTAG: 297370 UserNameTAG: ayush3504 CreateTimeTAG: 2013-03-30T13:26:50Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Ayush, this must be very stressing to be left in the dark about the certificate and no one from staff caring. I googled a bit about Pearson certificates. So probably you will get a electronic certificate that will come from EDx not Pearson. Pearson can normally give testresults shortly after the exam is taken (So I assume you know already your results?). I do hope they do their evaluation more intelligible than with the online MOOC's where you are right or wrong just looking at the math results without looking at the road you followed. As we have had the same problem of certificates late delivered in some online courses I think it will come soon, probably they are still discussing the format of their certificate. I found complaints from people waiting two year for their certificate from SAP after doing a Pearson exam, I am sure EDx is more professional. I think you could try to post your question in the forum of the present running 6002x course (you would have to signup first) were there is still Staff monitoring the forum. Or you could post your question to: mit-6002x@edx.org or agarwal@edx.org I'm sure you are not the only one impatiently waiting for your certificate but EDx should care about their students certainly if you paid a fee. If you get an answer let us know, I'm interested to see how the certificate looks but first make it unusable for forgery before posting it here. FirstChildUserIdTAG: 79885 FirstChildUserNameTAG: ruudoleo FirstChildCreateTimeTAG: 2013-03-30T17:35:33Z FirstChildTAG: Hi, The test results haven’t been declared, at least for me. The Proctored exam bar in Progress tab still appears vacant. I just remember that I had 19/24 boxes correct, so it would be reasonable to assume that the percentage is nearly 79% if each box has equal weight-age. The course staff has not updated us after the Pearson exam on Feb 13 and this is not a fair way to treat those few participants who had spent on this pilot project. **Their deadline of Mar 13 (mentioned in the exam at Pearson test centre) has passed long ago and they ought to give an update on this.** Also, I still have doubts and need to confirm whether this is the case with other examinees because I don't see others participants asking about the Pearson exam results or certificate. I wonder if they had received their test report and certificates through email. I will be graduating in few months and this certificate is meant to be mentioned in my CV. It will be absolutely useless for me if it comes out too late as in SAP’s case you just mentioned. I will definitely share the certificate (low-res.) when I get it. And thank you so much for the reply. I really appreciate all the info you found for me. Will follow your advice. FirstChildUserIdTAG: 297370 FirstChildUserNameTAG: ayush3504 FirstChildCreateTimeTAG: 2013-03-30T21:35:00Z SecondChildTAG: Hi ayush3504, I have been reported this to the Staff of 6.002x Spring 2013 some days ago. I am sure that they are working on it. I hope that you can receive your Certificate soon :). Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2013-03-31T15:35:57Z SecondChildTAG: Oh! Thank you so much!! :) :) SecondChildUserIdTAG: 297370 SecondChildUserNameTAG: ayush3504 SecondChildCreateTimeTAG: 2013-03-31T16:43:08Z IndexTAG: 844 TitleTAG: H1P4 some hints In the homework they make big jumps if you have to start what you learned from the lecture. For H1P4 I made some graphs in excel (just adding the signals and putting them in a graph) to visualize it, still have to solve it in math. ![H1P4a][1] ![H1P4b][2] [1]: https://edxuploads.s3.amazonaws.com/13611282661343657.jpg [2]: https://edxuploads.s3.amazonaws.com/13611282971343609.jpg UserIdTAG: 79885 UserNameTAG: ruudoleo CreateTimeTAG: 2013-02-18T03:28:49Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_6003z NumberOfReplyTAG: 1 FirstChildTAG: two sinusoidal waves superimpose each other to give a resultant sinusoidal wave of single amplitude and frequency.this is called superposition theorem.equation of a sinusoidal wave travelling in +ve x direction with phase difference 'phi' is given as y=a sin(w*t-phi) FirstChildUserIdTAG: 357747 FirstChildUserNameTAG: kishores FirstChildCreateTimeTAG: 2013-02-18T14:16:41Z IndexTAG: 845 TitleTAG: Discussion Forum and Wiki planned outages In order to provide a secure testing environment and the maximum amount of credibility for the Pearson Proctored Exam, we will be taking down the discussion forum and wiki on Tuesday Feb 12th and Wednesday Feb 13th. We apologize for any inconvenience this may cause. UserIdTAG: 96452 UserNameTAG: Lyla CreateTimeTAG: 2013-02-11T16:41:01Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I have a request. Could you please release the exam questions, after the proctored exam, here so that the some of us who couldn't attempt it can have a go at it on our own? FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2013-02-11T18:54:02Z SecondChildTAG: Sure. SecondChildUserIdTAG: 96452 SecondChildUserNameTAG: Lyla SecondChildCreateTimeTAG: 2013-02-11T19:46:24Z SecondChildTAG: Thanks a ton Lyla! :-) SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2013-02-11T21:03:03Z FirstChildTAG: Looks like the Pearson exam will be done from the edx website. There appeared a pearson button in the course material. FirstChildUserIdTAG: 79885 FirstChildUserNameTAG: ruudoleo FirstChildCreateTimeTAG: 2013-02-12T02:14:11Z IndexTAG: 846 TitleTAG: أينكم يا عرب أنا سمحمد من المغرب..أدرس هندسة الميكاتروس أرجو أن أجد هنا بعض الاخوان العرب. تعيش لغة الضاد. الله ولى التوفيق UserIdTAG: 853213 UserNameTAG: EINSTEIN-00 CreateTimeTAG: 2013-01-31T13:15:10Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: اهلاً فيك اخوك ناظم من اليمن ادرسة هندسة ميكا ترونكس وانا سعيد بروئيتك هنا FirstChildUserIdTAG: 109941 FirstChildUserNameTAG: Alhashemy FirstChildCreateTimeTAG: 2013-01-31T20:50:18Z SecondChildTAG: يا اهلا وسهلا بكم اخواني انا اخوكم من الاردن وادرس هندسة كهربائية وانا جديد هنا SecondChildUserIdTAG: 1133708 SecondChildUserNameTAG: stereo SecondChildCreateTimeTAG: 2013-02-02T07:50:15Z SecondChildTAG: الجزائر حاضرة محمد الله يوفق الجميع SecondChildUserIdTAG: 1156437 SecondChildUserNameTAG: bhamid20 SecondChildCreateTimeTAG: 2013-02-05T13:16:08Z SecondChildTAG: السلام عليكم انا رشيد من المغرب ادرس هندسة كهرباء SecondChildUserIdTAG: 1175861 SecondChildUserNameTAG: Rachidfriend SecondChildCreateTimeTAG: 2013-02-08T12:20:01Z FirstChildTAG: السلام عليكم انا اخوكم احمد من مصر ادرس هندسة الالكترونيات FirstChildUserIdTAG: 221617 FirstChildUserNameTAG: konan FirstChildCreateTimeTAG: 2013-02-02T09:03:06Z FirstChildTAG: يا اهلا وسهلا بكم اخواني انا اخوكم عبدالقادر من الاردن وادرس هندسة كهربائية وانا جديد هنا FirstChildUserIdTAG: 1133708 FirstChildUserNameTAG: stereo FirstChildCreateTimeTAG: 2013-02-02T18:52:48Z FirstChildTAG: و انا اخوكم محمود بدوي من السعودية و اتشرف بيكم FirstChildUserIdTAG: 156535 FirstChildUserNameTAG: badawiIiIi FirstChildCreateTimeTAG: 2013-02-03T01:15:17Z FirstChildTAG: السلام عليكم انا نورالدين من الجزائر هذا حسابي في فايس بوك للتواصل mecher_1989@hotmail.fr FirstChildUserIdTAG: 1144701 FirstChildUserNameTAG: nounoutorino FirstChildCreateTimeTAG: 2013-02-03T22:54:43Z SecondChildTAG: مرحبا انا كرم من سوريا لهلق مالي عرفان شي انا عم ادرس هندسة نظم حاسوب ياريت كم نصيحة منك مشان هذا الموقع SecondChildUserIdTAG: 1164910 SecondChildUserNameTAG: kemo21 SecondChildCreateTimeTAG: 2013-02-06T16:58:24Z IndexTAG: 847 TitleTAG: edX ID is not your username :p (Pearson VUE). Hi all, As I am curious I clicked on the magenta button ![image][1] Then I completed my personal data and appeared something like this in my dashboard ![im2][2] After some time a large code appeared there ![ima][3] Then I click on schedule pearson Exam ![ima3][4] The first mistake here is that I thought that the username was Myrimit and the Password my password . But, it wasn't. Then I got noticed that I had to click on create a web account. Then it appeared something like this ![im][5] I though that my edX ID was Myrimit haha. So, I entered that and a warning post said that it was an invalid ID. So, I got really confused haha. I click again on what is this ![im][6] Then I realized that I had to put the long number instead of my username - I found this confused... edX ID for me was Myrimit...., might if the edX ID could be called as edX ID code would be less confusing :p.... ![cod][7] Then , after some steps more - I had to complete my username in Pearson VUE and check availability and enter a appropiate password / yes or yes it has to be with uppercase, downcases , numbers and symbols, if not it is considered invalid... After that I reached to this ![ima][8] I will take it next year haha[reasons][9] I hope this can help you. Myriam. [1]: https://mitx_askbot_stage.s3.amazonaws.com/13593838691343668.png [2]: https://mitx_askbot_stage.s3.amazonaws.com/13595805452669143.png [3]: https://edxuploads.s3.amazonaws.com/13595971524134919.png [4]: https://edxuploads.s3.amazonaws.com/13595972668474412.png [5]: https://edxuploads.s3.amazonaws.com/13595974961343615.png [6]: https://edxuploads.s3.amazonaws.com/13595976531343676.png [7]: https://edxuploads.s3.amazonaws.com/13595979018474477.png [8]: https://mitx_askbot_stage.s3.amazonaws.com/13595966554640874.png [9]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50f48c302d34f61f00000012 UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2013-01-31T02:09:39Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Dear Myrim , I need a support is any idea about Taiwan university my brother sister want to study there if you have any friend please send the contact details i want to ask a couple of doubt FirstChildUserIdTAG: 156694 FirstChildUserNameTAG: mkprasanth FirstChildCreateTimeTAG: 2013-01-31T13:45:07Z SecondChildTAG: Yes, sure I will let you know. But currently I don´t have any friend from the Taiwan University... Have you asked in the Forum Discussion if some student is attending in that University? SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2013-01-31T14:20:08Z SecondChildTAG: Hey you can be friends with this guy Allen Shang https://www.facebook.com/KuangMing.S you can also find Harry Hsu They are both from NTU Taiwan university SecondChildUserIdTAG: 378150 SecondChildUserNameTAG: GladIDoThis SecondChildCreateTimeTAG: 2013-01-31T15:18:11Z SecondChildTAG: Hi Arafat, Thank you for the link for mkprasanth :). But do you know who is Allen Shang? Is student from the edX Community? Is a reliable person to contact ? Is he really from the NTU Taiwan University? Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2013-01-31T18:29:26Z SecondChildTAG: I do not know him personally He is not from edx community He is MAYBE a reliable person to contact, In reality he is young person around 20 , who went to international science fairs and even MIT , he also lost half of his hair due to radioactivity in research experiments , you can see that in his cover page I cannot assure if he could really help, but he does have dozens of people in his own friend list who went to NTU Taiwan University SecondChildUserIdTAG: 378150 SecondChildUserNameTAG: GladIDoThis SecondChildCreateTimeTAG: 2013-01-31T19:47:15Z SecondChildTAG: Thank you for the data Arafat :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2013-02-01T13:37:06Z FirstChildTAG: Hello, Mirian, my name is Elias. I'm from Sao Paulo and work near the site that will make you the exam. If you need any help you can ask me. My address is: elias.felipe @ hotmail.com My daughter also studies near the location you will take the exam. We know very well the city of Sao Paulo. Good luck. FirstChildUserIdTAG: 361823 FirstChildUserNameTAG: EliasOak FirstChildCreateTimeTAG: 2013-02-01T03:16:15Z SecondChildTAG: Done :). e-mailed you xD. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2013-02-01T13:36:31Z IndexTAG: 848 TitleTAG: Pearson Exam FAQ We are collecting some of the questions asked on this forum on this wiki page. https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/pearson-exam-faq/ UserIdTAG: 96452 UserNameTAG: Lyla CreateTimeTAG: 2013-01-29T17:33:44Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I'm looking for answer to "The Proctored Certificate will be sent by Post to our home?" which is unanswered in the FAQ above. FirstChildUserIdTAG: 297370 FirstChildUserNameTAG: ayush3504 FirstChildCreateTimeTAG: 2013-02-01T10:28:14Z IndexTAG: 849 TitleTAG: Register for Pearson Exam button appeared!... And then disappeared. Back again :p ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13593840701343605.png UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2013-01-28T14:41:19Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi Myrimit, I didn't do the Fall exams and no button appears so probably only people who passed this course get that button. FirstChildUserIdTAG: 79885 FirstChildUserNameTAG: ruudoleo FirstChildCreateTimeTAG: 2013-01-28T15:19:48Z SecondChildTAG: Hmm... It dissapeared in my dashboard too ... As I am curious I clicked on the button... The button re-linked you to a page where you have to complete things.... Might I shouldn´t have to see that.... Anyway, I will not disclose what I have seen xD SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2013-01-28T15:27:34Z SecondChildTAG: Oh, I forgot a bondgirl never discloses a secret :<) (a joke only you and I understand). SecondChildUserIdTAG: 79885 SecondChildUserNameTAG: ruudoleo SecondChildCreateTimeTAG: 2013-01-28T17:57:55Z SecondChildTAG: I pass the course, but in my case there wasn't any button at any time XD. Perhaps they are still establishing the date for the exam. SecondChildUserIdTAG: 250241 SecondChildUserNameTAG: DaniHerrBere SecondChildCreateTimeTAG: 2013-01-28T21:23:49Z SecondChildTAG: There was a little blip and the registration button was released slightly prematurely. It is fully available now, and you can register for the exam on Feb 11th. SecondChildUserIdTAG: 96452 SecondChildUserNameTAG: Lyla SecondChildCreateTimeTAG: 2013-01-29T14:56:44Z IndexTAG: 850 TitleTAG: "Fundamentals of Electrical Engineering " , Rice University "Fundamentals of Electrical Engineering " course from the Rice University have started on Coursera . UserIdTAG: 240487 UserNameTAG: AlexAlexandrescu CreateTimeTAG: 2013-01-27T22:01:44Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: It seems to be shaping up as a good course. It is primarily a signals and systems course rather than a circuits course. I especially like the participation of both the professor and staff on the forum. Questions get answered, and problems get fixed in a timely manner. The course doesn't seem "fixed in stone", but seems as though it will evolve according to student needs. The professor has already added practice exercises at student request. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2013-01-28T02:09:42Z SecondChildTAG: I agree. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2013-01-28T07:21:05Z IndexTAG: 851 TitleTAG: Thank you Thank you very much for everything. It was a wonderful experience. I hope to hear soon from the following courses 6.004x, 6.846x and 6.033x Best regards UserIdTAG: 132591 UserNameTAG: juantronic CreateTimeTAG: 2013-01-09T09:52:18Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 852 TitleTAG: I thought i passed the course? To the Staff, I got 59% on the progress report but when i viewed my grade last week Jan 3rd it was 65% (passed) on the webpage and so there appears a button for my certificate and i was able to print it. Now i just viewed it again today and my grade went back to 59% and i don't have that certificate button anymore and saying i didn't passed.. What happened here..which is which, please explained why is this inconsistent.. Thank you. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2013-01-09T01:58:03Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: I think there was some sort of software glitch that morning- my grade temporarily went up 3 points (giving me an A), then was back down later that day. FirstChildUserIdTAG: 340370 FirstChildUserNameTAG: Alaric7 FirstChildCreateTimeTAG: 2013-01-09T04:34:01Z FirstChildTAG: my grade was 77% but now 80% loooooool but still in B grade FirstChildUserIdTAG: 395257 FirstChildUserNameTAG: blackguitar FirstChildCreateTimeTAG: 2013-01-12T00:09:23Z FirstChildTAG: apparently the guys who program this site learned software engineering elsewhere;-) FirstChildUserIdTAG: 623210 FirstChildUserNameTAG: preveen FirstChildCreateTimeTAG: 2013-01-13T16:51:16Z IndexTAG: 853 TitleTAG: certificate tremendous thank you for the certificate UserIdTAG: 462861 UserNameTAG: edxforme CreateTimeTAG: 2013-01-07T16:14:53Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 854 TitleTAG: Thanks for making this possible Thank you to all edX staff and Professor Agarwal for making possible this awesome course. I really learned a lot of new stuff and remembered some things from my career that I had forgotten. This intiative of MIT - edX is great and I hope that many courses more will be open to continue my learning path. Thank you very much and special greetings from Bogota - Colombia UserIdTAG: 376173 UserNameTAG: nacho110987 CreateTimeTAG: 2013-01-04T04:23:27Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 855 TitleTAG: What Will you take in 2013 What courses are you guys will you guys take in year 2013 on edx Please write down UserIdTAG: 378150 UserNameTAG: GladIDoThis CreateTimeTAG: 2013-01-03T17:08:58Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 7 FirstChildTAG: Statistics from Berkeley, Stat2.1x. FirstChildUserIdTAG: 132828 FirstChildUserNameTAG: rharris FirstChildCreateTimeTAG: 2013-01-03T18:29:51Z FirstChildTAG: AI :) FirstChildUserIdTAG: 263693 FirstChildUserNameTAG: Coldberg FirstChildCreateTimeTAG: 2013-01-03T22:32:30Z FirstChildTAG: Hi, I have already anrolled in the following 3 courses: CS191x Quantum Mechanics and Quantum Computation ER22x Justice 14.73x The Challenges of Global Poverty FirstChildUserIdTAG: 99441 FirstChildUserNameTAG: coyarce FirstChildCreateTimeTAG: 2013-01-03T17:14:28Z FirstChildTAG: I've enrolled in 3 courses on [coursera][1]: - Calculus: Single Variable - Clinical Problem Solving - Introductory Human Physiology Yet I don't know if I'm going to get along with all of them together. :) [1]: http://www.coursera.org/ FirstChildUserIdTAG: 129876 FirstChildUserNameTAG: msarabi95 FirstChildCreateTimeTAG: 2013-01-04T05:55:39Z FirstChildTAG: CS191x Quantum Mechanics and Quantum Computation FirstChildUserIdTAG: 207728 FirstChildUserNameTAG: RossyFromBulgaria FirstChildCreateTimeTAG: 2013-01-03T17:16:05Z SecondChildTAG: Me too. SecondChildUserIdTAG: 364126 SecondChildUserNameTAG: shunyi SecondChildCreateTimeTAG: 2013-01-03T21:16:37Z FirstChildTAG: AI (CS188.1x) Introduction to computer science (6.00x) FirstChildUserIdTAG: 509592 FirstChildUserNameTAG: chinmaya24aug FirstChildCreateTimeTAG: 2013-01-04T07:35:08Z FirstChildTAG: *3.091x Introduction to Solid State Chemistry* FirstChildUserIdTAG: 327710 FirstChildUserNameTAG: Teo2502 FirstChildCreateTimeTAG: 2013-01-04T21:57:24Z IndexTAG: 856 TitleTAG: Raise Up the Thread: Add your marker on the Google map! An initial thread is [here][1] This is the link to the new map for the fall course 6002x: http://goo.gl/maps/kR6AK [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50d914bbef2ecd2b0000002a UserIdTAG: 205311 UserNameTAG: Sergtronix CreateTimeTAG: 2013-01-03T12:38:03Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 857 TitleTAG: Statistics Can we get some statistics about this cource? For example, how many people were registered on it and how many people had finished it successfully? Just for fun... UserIdTAG: 323416 UserNameTAG: Alexxkr CreateTimeTAG: 2013-01-01T02:34:04Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi Alexxkr, This was the Last Term Official Statistics. It was published after the end of the Course, might this Fall they will do the same... > Course statistics: 6.002x had 154,763 registrants. Of these, 69,221 > people looked at the first problem set, and 26,349 earned at least one > point on it. 13,569 people looked at the midterm while it was still > open, 10,547 people got at least one point on the midterm, and 9,318 > people got a passing score on the midterm. 10,262 people looked at the > final exam while it was still open, 8,240 people got at least one > point on the final exam, and 5,800 people got a passing score on the > final exam. Finally, after completing 14 weeks of study, 7,157 people > have earned the first certificate awarded by MITx, proving that they > successfully completed 6.002x. Happy New Year!:) Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2013-01-01T02:37:22Z SecondChildTAG: Thank you! It funny to be in 4.62% оf lucky guys (or obstinate guys), who could made it. Happy new Year! Nice to learn your course. SecondChildUserIdTAG: 323416 SecondChildUserNameTAG: Alexxkr SecondChildCreateTimeTAG: 2013-01-01T13:07:28Z IndexTAG: 858 TitleTAG: TO ALL THE STAFF HAPPY NEW YEAR 2013 UserIdTAG: 150120 UserNameTAG: krishna1993 CreateTimeTAG: 2012-12-31T16:05:50Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Happy new year! 21minutes has past the new year in my time zone. FirstChildUserIdTAG: 361137 FirstChildUserNameTAG: Ericson FirstChildCreateTimeTAG: 2012-12-31T16:22:17Z SecondChildTAG: not yet in my contry 6 houre and 20 minut for new year 2013 SecondChildUserIdTAG: 331441 SecondChildUserNameTAG: abdessamade SecondChildCreateTimeTAG: 2012-12-31T18:42:06Z IndexTAG: 859 TitleTAG: CECC 2 competition deadline The deadline for CECC 2 competition have really been extended until 14 January ? Because i saw a thread announcing that the deadline have been extended, but there is no activity for a week. Thank you ! UserIdTAG: 240487 UserNameTAG: AlexAlexandrescu CreateTimeTAG: 2012-12-31T13:08:47Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: yes, it is extended.. I think we will meet New Year in the same time :) Happy New Year! FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-12-31T13:18:47Z SecondChildTAG: You're right Serghei ! Happy New Year ! SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-12-31T14:25:44Z FirstChildTAG: Hi AlexAlexandrescu, Yes, it has been extended due New Year and Christmas. You have till January 14th for submitting your video :) I hope that you can participate. Take care and Happy New Year! Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-31T15:17:11Z SecondChildTAG: Thank you and Happy New Year ! SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-12-31T15:19:33Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-31T15:25:51Z IndexTAG: 860 TitleTAG: Thank you all very much!!!! I wanna express my deep gratitude for the excellent course which I had the honor to receive. To all EDX personnel, and especially to Professor Anant Agarwal who has displayed an admirable teaching methodology and a contagious sense of humor, thank you very much. I am a graduate in Physics from Havana University and this course was very useful me to deepen and systematize my knowledge of electronics. I say goodbye with sadness as it is always a pleasure to learn with such good teachers. Karel Negrin UserIdTAG: 138934 UserNameTAG: KarelNN CreateTimeTAG: 2012-12-30T20:51:02Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 861 TitleTAG: TO STAFF We have been waiting so long but u have still not given the certificate.When it will be available? UserIdTAG: 150120 UserNameTAG: krishna1993 CreateTimeTAG: 2012-12-30T05:36:42Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: i hope they will be providing it after new year not before new year FirstChildUserIdTAG: 342221 FirstChildUserNameTAG: deepkar FirstChildCreateTimeTAG: 2012-12-30T05:39:02Z FirstChildTAG: edX: I will give you free cookies, just be patient. you: I want it now !!!!! (do you also want it with little pieces of chocolate or without? - because its not for you to decide, its free cookie) FirstChildUserIdTAG: 329254 FirstChildUserNameTAG: KGabor FirstChildCreateTimeTAG: 2012-12-30T12:49:24Z SecondChildTAG: :) SecondChildUserIdTAG: 126825 SecondChildUserNameTAG: aphoenix SecondChildCreateTimeTAG: 2012-12-30T14:37:49Z IndexTAG: 862 TitleTAG: Exam Q1 d In Q1 d of the exam we are asked for the Thevenin resistence of the circuit from the terminals of in1. The answer is 12 ohms. But shoulndt it be 12 kohms, since th resistors are 6k resistors? UserIdTAG: 400567 UserNameTAG: JoseMartins CreateTimeTAG: 2012-12-29T17:26:09Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi JoseMartins, Yes, you are all right, I haven´t noticed this, I will report this to the Staff. It is missing in the solution explanation the k in the last answer explanation... Thank you very much for pointing this, Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-29T17:40:09Z SecondChildTAG: Wasn't this mistake present in the correction itself? I'm pretty shure I checked 12k in the second submission and got it wrong SecondChildUserIdTAG: 400567 SecondChildUserNameTAG: JoseMartins SecondChildCreateTimeTAG: 2012-12-29T17:45:22Z SecondChildTAG: Hi JoseMartins, The answer it must be entered in kOhms units... The statement says. **Assuming that Input 2 is left as an open circuit, what is the Thevenin equivalent resistance (in kOhms**) seen from Input 1? So, if Rth is 12kOhms you have to enter only the number **12**... I hope this can help you... P.D. Anyway,in the Final Explanation is missing the kilo... ![im][1] [1]: https://edxuploads.s3.amazonaws.com/13568037681343618.png SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-29T17:56:47Z SecondChildTAG: Thank you! I am my only enemy, for doing things in such a rush. SecondChildUserIdTAG: 400567 SecondChildUserNameTAG: JoseMartins SecondChildCreateTimeTAG: 2012-12-29T18:05:42Z SecondChildTAG: ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-29T18:12:49Z IndexTAG: 863 TitleTAG: 6.002x and edX: congratulations and constructive feedback **TL;DR**: Wonderful course, so good I took it twice! However, I found the pacing was a problem. I would love to see an alternative version, which is the same course but just spread over double the amount of time. Due to pacing issues of this course, I'd like a head-start on another edX course, but edX seems to be witholding both the autumn and spring learning materials from new enrollers. Would be lovely if at least past materials were available, to improve people's chances of making it through the course when it's next open. --- **Congratulations!** Firstly, I'd like to thank and congratulate the 6.002x team for their marvellous work on this course. You have made the world a better place. And well done to everyone out there who put in the effort of filtering out unimportant distractions, and learning the things this course had to offer - hopefully this experience will have improved our lives, and the lives of those around us. **My experience** This is the second time I have enrolled for this course, and it's the first time I've completed it successfully. The first time I took this course, I spent a lot of time learning how to learn. I didn't make notes, I watched all the lecture videos, and I read all of the relevant textbook material. After about 2 weeks I realised that it just wasn't going to work out. It was taking far too much time (WAY more than the estimated effort of 10 hours/week) and, when it came to homework, I didn't have my notes to fall back on when I got stuck. I would have to watch the lectures again (which took too much time) or read the textbook again. The textbook had its own problems. I couldn't make good use of my lunch hours, because I don't have internet access at this time, so the online textbook is inaccessible. It's also a textbook where backward navigation over many pages is quite common, in order to find a diagram that is mentioned in a subsequent paragraph. It's a problem with the textbook itself, but it's hugely exacerbated by the clunky web interface. So, I ditched the textbook, and decided to just watch lectures and write notes based on them. This still took a lot of time. I made it to the midterm exam... and made the decision to quit. Not quit the course, mind you, but to quit the scheduling and pace of the course. I felt a bit of a failure for not being able to hack the pace of the course, but I was comforted by similar stories I was reading on the discussion forum. I remember feeling burnt out, but relieved, and my partner giving me a huge hug, and telling me how pleased she was that she'd be seeing me in the evenings again. The course started again this autumn, and I eased myself back in slowly, going over my notes from the previous course, wanting to make sure my foundations were solid, and submitting homeworks as early as possible. I began the second half of the course in a much better position than I had been in during the previous course. Even so, I quickly hit the same pacing problem. As weeks went by, I was becoming less certain about reaching my deadline targets. It was almost a relief to have the option of dropping two homeworks and two labs, due to the way the course is marked. However, I felt I needed the experience of all the homeworks to prepare me for the final, and therefore felt obliged to complete them. Anyway, I got to the end (yay! although I've still yet to look at the week 13 & 14 material, and many, many tutorials). I'm pleased with myself, I can answer questions about a subject I was clueless about last year. And I'd like to take this further, join a hackspace, start a project, learn more stuff, help myself and others to improve the world, and be able to help my daughter out if she decides that circuits and electronics are worth investigating. **However!** Right. So. Would I do this, or another edX course, again? I probably wouldn't. Here's why: whilst I was doing this course, it easily filled all my evenings. I have a full time job, I have a toddler, evenings are all I can spare. I don't *think* I'm unusual, in this regard. Despite having a strong background in mathematics, and giving every evening I had, I was able to maintain the pace of the course, but no more. This means that every time I was ill, or had flaky internet, or was visiting (or being visited by) relatives, I would slip a bit further behind. There was no catching up. This is a *wonderful* course! But it's not an accomodating course. It took a physical toll on me (sedentary studying on top of a sedentary job is not a good thing), the house became a bit of a mess, and my partner and I felt estranged, almost. I don't think it's a family-friendly course. The pacing of this course appears to suit many people. That's brilliant, and for that reason I don't think the pacing should be changed (I'm also sure that there are many advantages to having the course synchronised with the real world MIT course). However, if an alternative version of the course, lasting double the time (28 weeks rather than 14 weeks) were available, I would dive at the chance to attend that. And, before asserting that 28 weeks is too long, remember that I had to quit during the first attempt, so it ended up taking me 36 weeks to successfully complete the course. Sometimes, taking your time will save you time. :) As for other edX courses - I am interested in taking one, but I'm wary given the pacing of this one. I could mitigate the pacing issue by starting on a new edX course right now, to give me a decent head-start. However, enrolling for a spring course does not give me access to the learning materials (why not??), and I am prevented from enrolling for the autumn courses, and so I cannot see past materials either (why not??). Stranded between autumn and spring semesters, edX feels very much like it's closed at the moment, which does not feel very... internet-y. It would be fantastic if the full course-worth of material were always available for each course. Anyway, thanks again everyone. I really appreciate it. Have a good 2013! Archie UserIdTAG: 7334 UserNameTAG: Hiacre CreateTimeTAG: 2012-12-27T15:14:11Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: ocw.mit.edu has courses that you can take at your own pace. FirstChildUserIdTAG: 623210 FirstChildUserNameTAG: preveen FirstChildCreateTimeTAG: 2012-12-27T15:49:13Z SecondChildTAG: Hi Preveen, Yes, Open Courseware might have to be where I go next. My reservations with that are the apparent absence of interactivity and community. What I like about an edX course is that it's almost like you're travelling with a group of people, and you're all battling the same deadlines. And, whilst I had trouble with the spacing of the deadlines, in general I like them because they keep me on the straight and narrow! Oh, and green ticks. OCW lacks green ticks. :) SecondChildUserIdTAG: 7334 SecondChildUserNameTAG: Hiacre SecondChildCreateTimeTAG: 2012-12-27T16:20:37Z SecondChildTAG: coursera has some self paced courses. almost all of udacity's courses are selfpaced. I agree that there is nothing more peaceful than seeing those green ticks;-) SecondChildUserIdTAG: 623210 SecondChildUserNameTAG: preveen SecondChildCreateTimeTAG: 2012-12-27T16:25:57Z IndexTAG: 864 TitleTAG: too close!!!! just gone through the solution of Final exam and realized the silly mistakes which i committed without keeping patience and lost check button..hope i will learn lesson from this incident!! UserIdTAG: 114864 UserNameTAG: Vikaash CreateTimeTAG: 2012-12-27T13:15:27Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 865 TitleTAG: A Big Thank you and God Bless When I was first introduced to the course through IEEE Spectrum articles, I knew I had to check it out. Going through the course with Prof Agarwal's insightful and generous teaching style, great TA's, collaborators and fellow students, I was awed by a great sense of having learnt at the feet of masters. The overall EDX delivery was truly exceptional and is clearly a demonstration of a radicle change that is doubtlessly going to improve the world for the better, with most of its population unable to access this kind of best class education for countless reasons. Your generosity and goodwill will be constantly rewarded by the Master of all Masters and along with it our prayers for all of you for your true selflessness in enabling mankind to do better. Shabbir Moosajee, Melbourne, Australia UserIdTAG: 168038 UserNameTAG: smoosajee CreateTimeTAG: 2012-12-27T05:21:00Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 866 TitleTAG: Contacting Sir Can i mail Dr agarwal personally......i would like to thank sir and the whole edx team for bringing out such an awesome course.Sir taught the subject just the way i ever dreamt of.If anytime i get an oppurtunity to meet u i would like to meet sir. Happy holidays and a happy new year ahead to all of u. UserIdTAG: 358198 UserNameTAG: Jay1492 CreateTimeTAG: 2012-12-26T17:43:27Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You can try emailing mit-6002x@edx.org perhaps he will see it or they can forward it for you. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-12-26T17:51:29Z SecondChildTAG: Thanks.... SecondChildUserIdTAG: 358198 SecondChildUserNameTAG: Jay1492 SecondChildCreateTimeTAG: 2012-12-27T16:52:22Z SecondChildTAG: You can contact him directly from his web page: http://www.csail.mit.edu/user/723 SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-12-27T17:28:56Z IndexTAG: 867 TitleTAG: Gratitude & Respect Thank you very much sir from core of heart. It was great learning experience from best professor i have ever listen. UserIdTAG: 431908 UserNameTAG: navneet_ipu CreateTimeTAG: 2012-12-25T16:59:29Z VoteTAG: 3 CoursewareTAG: Week 14 / S28V13_Voltage_Drop_across_the_Parasitic_Inductor CommentableIdTAG: 6002x_S28V13_Voltage_Drop_across_the_Parasitic_Inductor NumberOfReplyTAG: 0 IndexTAG: 868 TitleTAG: To staff and Myrim Final exam was skip due to family member not well , is any chance to attend the same again i am scored with out final exam 58% is any possibility to get the certificate UserIdTAG: 156694 UserNameTAG: mkprasanth CreateTimeTAG: 2012-12-25T12:39:26Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi mkprasanth, Unfortunately, you can not submit the Final Exam after deadline... I am sure that you are an excellent person as you were with your family member that really needed you in his or her dificult time instead of settig for the Exam. And I have to tell you, by heart, that your gesture was a wonderful one. Also remember that this Course will be offered again, you can re-take it as many times you wish. I can tell you this because this is the second time that I am taking this Course. The first time I were from the happy B Club and this time I could improve :p. So, be up mkprasanth! You can get the Certificate in other opportunity, now your family is your priority and I am sure that that family member is really proud of having a family member as you are. My best wish to you and for she or he to recover soon, Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-25T14:00:25Z SecondChildTAG: Dear Myriam , Thanks a lot keep in touch if possible i registered for 2013 spring bye mk Prasanth SecondChildUserIdTAG: 156694 SecondChildUserNameTAG: mkprasanth SecondChildCreateTimeTAG: 2012-12-27T17:55:36Z IndexTAG: 869 TitleTAG: Thank you and Compliments of the Season Thank you edx and Dr. Agarwal for starting this which is nothing short of a revolution in education. The course content, delivery and design were simply top class. We are humbled and ever so grateful for your hard work towards the betterment of humanity. Personally, this has been a dream come true for me. Thank you so much! :) UserIdTAG: 298775 UserNameTAG: surja CreateTimeTAG: 2012-12-25T00:37:29Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 870 TitleTAG: Next Classes at MIT If I were to go through Open Courseware at MIT, what would be the next appropriate classes for me to work on? Thanks, UserIdTAG: 214701 UserNameTAG: alanmatson CreateTimeTAG: 2012-12-24T21:07:38Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I think you could take a look at MIT EECS curriculum --> http://www.eecs.mit.edu/academics-admissions/undergraduate-programs/course-6-2-electrical-eng-computer-science On the field of electronics, probably 6.012 (devices & circuits) or 6.003 (signals & systems). FirstChildUserIdTAG: 398594 FirstChildUserNameTAG: DaveyJC FirstChildCreateTimeTAG: 2012-12-24T23:15:19Z SecondChildTAG: Where do I go next or read next if my goal is to understand and repair PC motherboards or cell phones? :) Would be great to apply this wonderful course knowledge in practice. SecondChildUserIdTAG: 298775 SecondChildUserNameTAG: surja SecondChildCreateTimeTAG: 2012-12-25T00:25:44Z IndexTAG: 871 TitleTAG: When will the Spring course start? I enjoyed this semester so much. I wish to take the course again, when will the spring course start? And also when will the advance courses that include topics like digital circuits and microprocessors be released? UserIdTAG: 211715 UserNameTAG: pitankar CreateTimeTAG: 2012-12-24T17:21:36Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: https://www.edx.org/courses/MITx/6.002x/2013_Spring/about FirstChildUserIdTAG: 183712 FirstChildUserNameTAG: colegiocientifico FirstChildCreateTimeTAG: 2012-12-24T17:25:29Z SecondChildTAG: Keep checking the above link. As of today (Dec-25-2012) it says "Spring 2013", but that should change once a firm date is established. Note that I am registered for 3.091x and 6.00x for Spring 2013, and they start at the beginning of February. So a **guess** would be late January or early February. I do not have official info, so I may be wrong on this. Keep checking the link for any updates, and enjoy 6.002x Spring 2013! Best Wishes, Jersey Mark SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-12-25T16:35:11Z IndexTAG: 872 TitleTAG: A special thanks to MIT and Professor Agarwal! A special thanks to Professor Agarwal and his staff for a wonderful learning experience! They have managed to make advanced learning fun! I might have missed the smell of the frying pickle, but loved the videos and demonstrations. I also appreciated the tutorials, and I also offer a special thanks to Myrimit for helping me gain intuition. Overall, this course has sparked an electronics love, and has prepared me to tackle practical electronic engineering problems. I look forward taking other courses offered by EDX, and especially Professor Agarwal! UserIdTAG: 393045 UserNameTAG: SrChasJC CreateTimeTAG: 2012-12-24T16:21:28Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You are welcome SrChasJC!:) I am so happy that all this sparked an electronics love. The 6.002x is so amazing, all the edX Staff involved are awesome and also my CTAs mates, they are really excellent people, all of them. I also have to say that JSChambers was helping a lot behind the scene, he suggested a lot of improvements, reported bugs and all that might you didn´t notice, he did a lot of hard work :) My best wish to you and I hope that you suscribe in many future edX Courses :) Take care, Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-24T17:50:52Z IndexTAG: 873 TitleTAG: Thank you from Colombia I have finished the exam, and I have lost one question (the last one). It doesn't matter at all :) (my final grade was 99%). Thank you very much to MIT, edx staff, Mr Agarwal and all the people that made this possible. Thank you from Colombia, I'm waiting for more courses :) UserIdTAG: 221050 UserNameTAG: jucapini CreateTimeTAG: 2012-12-24T16:15:31Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Congratulations! FirstChildUserIdTAG: 361137 FirstChildUserNameTAG: Ericson FirstChildCreateTimeTAG: 2012-12-24T17:05:47Z SecondChildTAG: hats off jucapini SecondChildUserIdTAG: 342135 SecondChildUserNameTAG: vikash902 SecondChildCreateTimeTAG: 2012-12-24T19:48:22Z IndexTAG: 874 TitleTAG: Thank you so much from Chile :D I'm student of Electronic engeneer in Universidad Tecnológica Metropolitana del Estado de Chile (UTEM). I'm so grateful and i'll be waiting for new courses of electronics. Thank you so much from Chile for the M.I.T staff and proffesor Anat Agarwal. Was a very good experience. UserIdTAG: 70519 UserNameTAG: Fipe CreateTimeTAG: 2012-12-24T16:04:12Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 875 TitleTAG: FINAL STATS Now that the final is over, I'd like to ask: - Your grade in the midterm: - Your grade in the final: - Your grade in the whole course: - How long did it take you to finish midterm: - How long did it take you to finish final: I can begin: - My grade in the midterm: 100% - My grade in the final: 100% - My grade in the whole course: 100% - How long did it take you to finish midterm: ~7 hours - How long did it take you to finish final: ~8 hours I think I couldn't have done it in a 2 hour exam. Maybe I could have got 50-70% there. Or maybe it's my compulsive obsession of checking, double checking and triple checking before answering xD UserIdTAG: 398594 UserNameTAG: DaveyJC CreateTimeTAG: 2012-12-24T10:56:15Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 14 FirstChildTAG: Some students worked hard and got, for example, 75%.. FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-12-24T11:41:07Z SecondChildTAG: Hi DaveyJC. My stats are these: - My grade in the midterm: 97% - My grade in the final: 75% - My grade in the whole course: 88% - How long did it take you to finish midterm: ~6 hours - How long did it take you to finish final: ~8 hours I couldn't study all topics for this final exam because I had too much work this month, but I do think the final exam was a really fair one. Nonetheless, I would need to study much more to have done it in just 2 hours. gotchi SecondChildUserIdTAG: 288637 SecondChildUserNameTAG: gotchi SecondChildCreateTimeTAG: 2012-12-24T12:22:11Z FirstChildTAG: My grade in the midterm: 100% My grade in the final: 100% My grade in the whole course: 100% How long did it take you to finish midterm: ~2.5 hours How long did it take you to finish final: ~3.5 hours FirstChildUserIdTAG: 209930 FirstChildUserNameTAG: saikiraniitr FirstChildCreateTimeTAG: 2012-12-24T12:57:04Z SecondChildTAG: Wow! That's **fast**! I need my coffee breaks during an exam, and I like to make neat schematics with **colored pencils** to keep track of all the different currents and voltages ($v_R$ in red, $v_C$ in blue, $V_S$ in black, etc), and to sketch out the waveforms at different points; this adds to my time, but makes it MUCH easier to spot mistakes, and more difficult to make them! SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-12-25T18:45:13Z FirstChildTAG: Surely EDX could provide official stats on all students who registered for this course. FirstChildUserIdTAG: 270011 FirstChildUserNameTAG: SnowmanZA FirstChildCreateTimeTAG: 2012-12-24T13:25:07Z FirstChildTAG: Labs and Assignment --- 25/30 Midterm --- 0.0/30 (missed midterm) Final --(25/28)~=89.28% (took me about 6 hours) thus final~=35.7/40 Total mark=61% (60.6% to be exact) FirstChildUserIdTAG: 183507 FirstChildUserNameTAG: obiradaniel FirstChildCreateTimeTAG: 2012-12-24T13:22:45Z SecondChildTAG: Lucky! Just *barely passing* is still passing, and you still get a certificate! Since the Final is cumulative, getting almost 90% on it means you **really learned a lot**! Congrats, Jersey Mark SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-12-25T18:47:44Z SecondChildTAG: Yeah true **JerseyMark,** I have to **retake the next course**. I missed many assignments and labs due to family issues but will be more settled next time with a much deeper foundation. **And won't miss a single exam, lab or assignment again.** SecondChildUserIdTAG: 183507 SecondChildUserNameTAG: obiradaniel SecondChildCreateTimeTAG: 2012-12-25T22:26:02Z FirstChildTAG: Homework: 88; Labs: 94; Midterm: 97 (~5H); Final: 93 (~5H); Total: 94 FirstChildUserIdTAG: 414306 FirstChildUserNameTAG: YN300 FirstChildCreateTimeTAG: 2012-12-24T13:41:39Z FirstChildTAG: My grade in the midterm: 100% My grade in the final: 100% My grade in the whole course: 100% How long did it take you to finish midterm: ~3 hours How long did it take you to finish final: ~3 hours FirstChildUserIdTAG: 289949 FirstChildUserNameTAG: manolito FirstChildCreateTimeTAG: 2012-12-24T13:45:52Z FirstChildTAG: - My grade in the midterm: 100% - My grade in the final: 96% - My grade in the whole course: 99% - How long did it take you to finish midterm: ~3 hours - How long did it take you to finish final: ~2 hours FirstChildUserIdTAG: 377452 FirstChildUserNameTAG: Ded FirstChildCreateTimeTAG: 2012-12-24T13:57:43Z FirstChildTAG: My grade in midterm :83 My grade in final :96 My grade in the whole course :93 How long did it take you to finish midterm :~ 6 hours how long did it take you to finish final:~ 6 hours FirstChildUserIdTAG: 18877 FirstChildUserNameTAG: johnkhiangte FirstChildCreateTimeTAG: 2012-12-24T14:53:20Z FirstChildTAG: Homework: 100; Labs: 100; Midterm: 100 (~4H); Final: 93 (~5H); Total: 97 FirstChildUserIdTAG: 97359 FirstChildUserNameTAG: azeem179 FirstChildCreateTimeTAG: 2012-12-24T15:36:18Z FirstChildTAG: UNTIL midterm everything was 100 final 64 total 86 How long did it take you to finish midterm: ~3 hours How long did it take you to finish final: ~3 hours however until the final it took me almost 20 hours per week for the course, not 12 I think 12 hours is just enough to pass with a C FirstChildUserIdTAG: 206669 FirstChildUserNameTAG: dimitrios66 FirstChildCreateTimeTAG: 2012-12-24T15:41:04Z FirstChildTAG: Homework:100; Labs:100; Midterm: 100 (several hours (more than two anyway :D) including a break in between); Final: 100 (Again maybe 6 hours excluding breaks); Total:100 (Took me many tens of hours of fun and pain :D) FirstChildUserIdTAG: 388514 FirstChildUserNameTAG: hdbam FirstChildCreateTimeTAG: 2012-12-24T15:42:15Z SecondChildTAG: Me too!! SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-12-24T16:40:08Z FirstChildTAG: Ded, can you publish your progresspage you have never posted before on 6002x and the only one who claims to finish the final exam in 2 hours and that would be a first in the last two 6002x courses. If you really did it you are a professor or a genius. FirstChildUserIdTAG: 79885 FirstChildUserNameTAG: ruudoleo FirstChildCreateTimeTAG: 2012-12-24T16:20:37Z SecondChildTAG: preview page screenshot - https://edxuploads.s3.amazonaws.com/13563744691343646.png Im neither professor nor a genius, the exam is actually designed to be taken within 2 hours, so it is doable. I believe there are a lot of people our there who managed to pass it in time. I never posted on forums here, because there was no need for it, I found all the answers I need in lecture slides, but now I was curious to see the results of other people, so I end up in this topic :) SecondChildUserIdTAG: 377452 SecondChildUserNameTAG: Ded SecondChildCreateTimeTAG: 2012-12-24T18:45:00Z SecondChildTAG: Congratulations, my doubt was because in the Spring exam there was a problem tubeamplifier wich took most people at least two hours. But I understand from a posting that it was dropped in the Fall exam together with some other questions. I remember I did about 10 hours over that exam and even didn't found the right solution for that tubeamplifier (I gave up after the first calculation as I was exhausted). SecondChildUserIdTAG: 79885 SecondChildUserNameTAG: ruudoleo SecondChildCreateTimeTAG: 2012-12-25T01:19:52Z SecondChildTAG: And don't forget the problem about Increasing the Q that resulted in adding/creating a negative resistance or the Trapping Noise. That were really interesting problems! In my opinion this final was much easier compared to that of the spring. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-12-25T02:39:30Z SecondChildTAG: ruudoleo, thanks. This is the first time I took 6.002x, so I don't know about previous final exam problems. I think the staff has adjusted the final according to community feedback. Current final was not a super-hard one, and was totally doable in 2 hours. SecondChildUserIdTAG: 377452 SecondChildUserNameTAG: Ded SecondChildCreateTimeTAG: 2012-12-25T05:22:24Z SecondChildTAG: I also believe that the Fall 2012 Final is do-able in 2 hours, because of the following: - If you realize that Q3 is exactly the same as Homework 8 P1, then you need **5 minutes** or less. - If you use intuition and/or the circuit simulator to plot the outputs and solve Q5, then you also save a massive amount of time. This problem should take about **10 minutes**. - Q6 can be solved using the circuit simulator (as long as you know how to properly get it to create a square wave using the PULSE function) and then use the graphical method to measure the $v_{min}$ and $v_{max}$ (bypassing any calculations) and takes about **15 minutes**. - Q1 is a simple node analysis / superposition (which the question makes plainly obvious!) and should take **15 minutes** also. That's 2/3 of the exam in about 45 minutes! Now for the two more difficult questions: - Q4 may be difficult if you do not understand what's happening, but if you've encountered this circuit before, it's actually very easy to analyze! At $0 After $T>t$, you get a simple LC circuit, but if you forget that it is offset $(t-T)$, you will get it wrong! This question should take **30-45 minutes** depending on your understanding. - That leaves you **30-45 minutes** for Q2, which if you've practiced your small-signal analysis, gives you more than enough time. Half of this question asks about using the standard MOSFET common-source configuration, so it's a simple plug-in with the equations given in the text. The ExpFET part simply involves finding $i_d = \frac{\partial}{\partial v_{GS}}$ of $i_D = K \cdot (e^\frac{v_{GS}-V_T}{V_N}-1) \cdot v_{gs}$ evaluated at $v_{GS}=V_{IN}$ and using $v_{out} = i_d \cdot R$ and $v_{in} = v_{gs}$. Simple as long as you don't mix up $v_{gs}$ with $v_{GS}$ and because $\frac{d}{dx} e^x = e^x$! 2 hours! - SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-12-25T20:01:10Z SecondChildTAG: I thought that by 2 hours they meant that is the same time that regular MIT students get and I imagine that they don't have the same online tools that we have... SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-12-26T00:34:07Z FirstChildTAG: Homework:100 Labs:100 Midterm: 100 (4 hours, including breaks) Final: 93 (4 hours. if only i were a bit patient.. :( ) Total:97 anyway, thanks a lot to MIT staff (especially Prof. Agarwal), TAs and everybody here! It has been a long time since I felt that I really learned a lot. FirstChildUserIdTAG: 201818 FirstChildUserNameTAG: ThreeHundred FirstChildCreateTimeTAG: 2012-12-24T16:22:00Z FirstChildTAG: HW: 100 Labs: 100 Midterm: 93 (3 hours) Final: 93 (3 hours) Total: 95 FirstChildUserIdTAG: 411504 FirstChildUserNameTAG: taubrafi FirstChildCreateTimeTAG: 2012-12-25T03:08:46Z IndexTAG: 876 TitleTAG: (One more) Thank you very much post (from germany) Hey there, i will try to keep it short ;-) Thank you all (staff and my fellow students as well as TAs) **very** much for this cool ride through electronics! I had a lot of fun (and some painful moments as well :D) going through the topics of this course. Having finished it was very satisfying and worth the effort. Btw I didnt see too many people (revealing themselves) being from germany. So maybe this can be the thread to build up a big Thank you from germany :-) Hope to see some of you in upcoming courses! UserIdTAG: 388514 UserNameTAG: hdbam CreateTimeTAG: 2012-12-24T10:01:33Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Here comes another huge "Thank you" from Germany :). It was an awesome experience, and compared to other online courses it was by far the best! Merry Christmas to all of you! FirstChildUserIdTAG: 219823 FirstChildUserNameTAG: Karsroh FirstChildCreateTimeTAG: 2012-12-24T11:34:48Z FirstChildTAG: Let me join this thread as well an send a big thank you from Germany to all who made this course possible :-) Last year I took the AI online course by Thrun and Norvig and the Machine Learning course by Andrew Ng. Both had their unique motivating factors that made me complete both courses. I liked the (maybe illusion of) direct contact to the teachers in the AI course because of their Q&A sessions on youtube. And they had very challenging exercises and homework. The Machine Learning course had excellent programming exercises combined with the joy of learning a new programming language (new for me at least). But 6.002x was definitely the best among these. The lectures were well structured and presented in a motivating way. Having unlimited submissions for the homework was definitely a good thing, because that made me continue until I had my personal "big aha moment" and got the correct answer. Thanks a lot all :-) FirstChildUserIdTAG: 148015 FirstChildUserNameTAG: ralfh FirstChildCreateTimeTAG: 2012-12-24T14:33:03Z IndexTAG: 877 TitleTAG: error in q4 of final exam [***Editor's Note:** Content deleted due to discussion of an exam question while the exam is still ongoing. To directly answer your question,* **gpb1984**, *the variable you stated as missing is included in the list in Q4. Take a closer look. **I have personally verified this.** We cannot make exceptions and give students extra chances on an individual basis. If you feel necessary to appeal this deletion, please contact a Staff member as listed in the Staff directory* - Jersey Mark, Community TA] UserIdTAG: 315378 UserNameTAG: gbp1984 CreateTimeTAG: 2012-12-24T06:18:24Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 1 FirstChildTAG: Good day, edX team and any concerned: I had internet issue today while inputting values for Q6. For my 6a. I was to input 0.2ms as time constant for VR to rise as well as 6b. 0.2ms for the fall, then Vmax as 33.8V and Vmin as 12.5V.. before the exam cut me off. It was supposed to be my final check to put me above 70% which is a B. Kindly look into it if it is possible for me to cross from 68% to >= 70% with my solution as stated above. Thank you. Ugo. FirstChildUserIdTAG: 329879 FirstChildUserNameTAG: Ugochukwu FirstChildCreateTimeTAG: 2012-12-24T10:45:13Z IndexTAG: 878 TitleTAG: Thank you for the exceptional course Hi, I am a grade 12 student studying from Mumbai, India who just finished the course. These 15 weeks have been some of the most illuminating of my life. Much of the mathematics that was connected with this course was not taught to me during my high school career, but with help from the discussion forum and with a little of my own initiative I was able to overcome the obstacles and complete the course. So, once again I just want to say thank you everybody for this fantastic experience. UserIdTAG: 85490 UserNameTAG: yashsavani CreateTimeTAG: 2012-12-24T06:05:14Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You are so young yashsavani. You have a lot of future! I am really happy to see so many young students willing to study, that is awesome! You remember me to popoya username, she is a young student like you, and she did excellent in 6.002x Spring 2012, I wish hadn´t spend so many time watching TV while I was a teenanger haha, you are really inspiring for all the young and brilliant students that will be the future of the World :) Keep always studying that is a really valuable tool for the life ;). My best wish to you. Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-24T06:28:05Z IndexTAG: 879 TitleTAG: Great thanks to Mr Agarwal ! I thought i would not be able to do that , but i did it ... (To be honest I think I spent more time than a young student !) (I'm very found on physics and if a course of this quality is available on EdX, please, staff, warn me... I have big lacks in maths too . ) But now, I just dont want to stop ! Merry Christmas to everybody . And future full of happiness. UserIdTAG: 373752 UserNameTAG: Croak CreateTimeTAG: 2012-12-24T03:21:46Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: http://ocw.mit.edu/courses/ocw-scholar/#phys FirstChildUserIdTAG: 252722 FirstChildUserNameTAG: vaboro FirstChildCreateTimeTAG: 2012-12-24T06:20:52Z IndexTAG: 880 TitleTAG: Yes, I passed! Yesssssss! This class was not easy, however it was worth it. As and electronics technician it has deepened my understanding of electronics. Thanks to everyone at MITx and edx for the opportunity. Blessings. UserIdTAG: 77598 UserNameTAG: Spacecadet1974 CreateTimeTAG: 2012-12-24T03:06:31Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 881 TitleTAG: Final Exam end Time At what time exactly (BOSTON Time) does the final exam ends. Thanks.. UserIdTAG: 111917 UserNameTAG: ashish_mit CreateTimeTAG: 2012-12-23T13:59:38Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: I'm just really glad I was wrong. Not often I say that. ;-) FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-12-23T18:46:07Z FirstChildTAG: What time do you see in the Courseware tab, on the right-hand side where it says under Final Exam next to the blue **alarm clock** icon. Mine is as follows, hence my earlier post to STAFF: ![enter image description here][1] ***EDIT: I have received official confirmation from Staff that the exam will be extended to December 24th, 5:00AM Eastern Standard Time (EST) / 10:00AM Greenwich Mean Time (GMT or UTC).*** [1]: http://i905.photobucket.com/albums/ac254/MarkPD/Clock_zps474a9a7b.png FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-12-23T14:17:57Z SecondChildTAG: However, remember that the Final Exam itself still lists December 23rd 11:59PM Eastern Standard/Boston Time (EST) as the finish time on the front cover. I have **not** gotten an *official confirmation* of December 24th 5:00AM EST as of right now, despite an email to Staff as soon as I saw the time change last night. Until we get a confirmation from Staff, we should take the new end time of 5:00AM with a "grain of salt", meaning that we should rely on it only with a great deal of caution, and suspect that it *may* **not** be accurate. SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-12-23T14:42:23Z SecondChildTAG: I agree. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-12-23T14:46:49Z SecondChildTAG: yeah mine courseware tab shows the same.. better to be safe.. Thanks.. SecondChildUserIdTAG: 111917 SecondChildUserNameTAG: ashish_mit SecondChildCreateTimeTAG: 2012-12-23T17:13:59Z FirstChildTAG: I think the time listed on the Courseware Final refers to GMT. In other words, I believe the Course Info time of "23:59 (11:59 pm) on Sunday, October 28th, Boston time" holds. FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-12-23T14:54:53Z SecondChildTAG: That might just be correct, considering Eastern Standard Time (EST) = Greenwich Mean Time (GMT, or UTC) - 5. But why the change, and shouldn't it be listed as 4:59AM instead of 5:00AM then? SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-12-23T17:06:11Z SecondChildTAG: It would have been good if they would have mentioned GMT or EST against the mentioned time in Courseware tab.. it would have been clear than.. SecondChildUserIdTAG: 111917 SecondChildUserNameTAG: ashish_mit SecondChildCreateTimeTAG: 2012-12-23T17:16:48Z IndexTAG: 882 TitleTAG: If I were born again I'd take this course. **Would you do the same? I know you would.** UserIdTAG: 138769 UserNameTAG: ArturoPrado CreateTimeTAG: 2012-12-23T12:50:27Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: May you be reincarnated in time for the spring course of 6.002x! FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-12-23T13:13:52Z FirstChildTAG: I would have done it about 15 years earlier if I could, but better late than never :) FirstChildUserIdTAG: 114881 FirstChildUserNameTAG: xvink FirstChildCreateTimeTAG: 2012-12-23T14:51:39Z SecondChildTAG: Ditto! (But 25 in my case ;-)) SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-12-23T14:55:37Z SecondChildTAG: wow..planetscape, was it 25 years ago? SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-23T15:23:10Z FirstChildTAG: I'd take a calculus course first ;) FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-12-23T18:55:47Z SecondChildTAG: Yes, definitely. Even better would be to integrate the calculus course into the electronics course so the practical use of it would be clear. SecondChildUserIdTAG: 114881 SecondChildUserNameTAG: xvink SecondChildCreateTimeTAG: 2012-12-26T10:07:12Z IndexTAG: 883 TitleTAG: Caltech machine learning class starts Jan 8th Professor Yaser Abu-Mostafa of Caltech is offering his machine learning class starting Jan 8th. "This time it will be the exact homeworks for both Caltech and non-Caltech students, and both will be graded online. The discussions on the forum, which used to be conducted by non-Caltech students when the homeworks were different, will now be conducted by both about the same homeworks, with TA's and myself contributing as well." Prof. Yaser Abu-Mostafa sign up at http://work.caltech.edu/telecourse UserIdTAG: 9673 UserNameTAG: xp42 CreateTimeTAG: 2012-12-21T09:31:51Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Thanks for this info! FirstChildUserIdTAG: 330357 FirstChildUserNameTAG: steryd FirstChildCreateTimeTAG: 2012-12-21T16:57:07Z IndexTAG: 884 TitleTAG: Final exam There is several hours till that moment when final exam will be published. I think there isn`t need to compare our scores before final exam. As about me I fill a little bit worry. And the mane question... Guys are you ready? UserIdTAG: 96873 UserNameTAG: Santyaga CreateTimeTAG: 2012-12-19T11:54:01Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: Who knows, ready we are or not.. My head is overloaded already.. I will try to start my exam after 11.00PM East time, because it will be late morning at my time zone, so.. I'll see... Good luck to all! FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-12-19T12:30:54Z SecondChildTAG: Good luck to you Sergtronix! SecondChildUserIdTAG: 96873 SecondChildUserNameTAG: Santyaga SecondChildCreateTimeTAG: 2012-12-19T18:16:25Z SecondChildTAG: And you too! Thanks! SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-19T18:33:03Z FirstChildTAG: I don't know if I am ready... I hope so. Main thing I need is a really good night's sleep before I start... FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-12-19T12:43:59Z SecondChildTAG: Then I can wish you good night! SecondChildUserIdTAG: 96873 SecondChildUserNameTAG: Santyaga SecondChildCreateTimeTAG: 2012-12-19T18:15:24Z FirstChildTAG: The final comes at a bad time for me. I will be travelling for Christmas tomorrow. Until then I will be preparing for the trip. The earliest I will be able to work on the final is Friday. I need to make sure I will have the full time available before I open the final and start the clock ticking. I hope to do well, but I realize I might "hit a wall" and become dispirited and quit. Wishing for everyone to do their best and show what they can do. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-12-19T17:36:28Z SecondChildTAG: merry christmas and a happy 'final' to you! SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-12-19T17:59:56Z SecondChildTAG: I wish you good traveling skyhawk! ;-) SecondChildUserIdTAG: 96873 SecondChildUserNameTAG: Santyaga SecondChildCreateTimeTAG: 2012-12-19T18:14:38Z SecondChildTAG: merry Christmas SecondChildUserIdTAG: 40209 SecondChildUserNameTAG: sadaf78 SecondChildCreateTimeTAG: 2012-12-19T21:34:36Z FirstChildTAG: The Mayans predicted that I would do poorly on this test. FirstChildUserIdTAG: 202032 FirstChildUserNameTAG: KeithD FirstChildCreateTimeTAG: 2012-12-19T18:23:55Z SecondChildTAG: the mayans predicted the world would end! forget them! :p good luck! SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-12-19T18:31:46Z SecondChildTAG: Maybe they were wrong? SecondChildUserIdTAG: 96873 SecondChildUserNameTAG: Santyaga SecondChildCreateTimeTAG: 2012-12-19T18:32:32Z SecondChildTAG: Wow..Have begun shaman's dancing with a tambourine :) KeithD, good luck with the final exam! SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-19T18:38:13Z IndexTAG: 885 TitleTAG: When are we supposed to use pull-up or pull-down resistors ? I have tried to make a small experiment with digital cmos gates from 4000 dip family . My first lesson was that , we should always have some potential at the inputs .I am stuck on three questions : 1) should i use a 10k pull-up resistor, or a 470 pulldown resistor on the input ? 2) Do i have to use a pullup resistor at the output also ? 3) If so, should i use a pullup resistor even if i connect for example the output of a NAND gate from one dip to the A input of another NAND gate of a different dip ? The last question is trivial, because if so, a lot of pcbs were full of resistors, but anyway. Thank you ! UserIdTAG: 240487 UserNameTAG: AlexAlexandrescu CreateTimeTAG: 2012-12-12T14:41:05Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Simplified.Pull-up and pull-down resistors are used when it is necessary to avoid floating pins(inputs).For example, when inputs are connected to the MCU pins, to switches, when PCB may be disconnected from other pcb and etc.. 4000th Family that you want to use is CMOS.It means inputs leakage current is very very small so you dont need to use relatively low-ohm resistors for pull-down or pull-up in the most cases. In your design simply do connect inputs to the output.And do not connect outputs toghether (except open drain and tri-state). FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-12-12T15:44:34Z SecondChildTAG: Thank you Sergtronix ! I've found in the manual that there must not be opened pins, and i also have found this : https://www.youtube.com/watch?v=KllMUO_pHxk . Thanks again ! SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-12-12T19:03:21Z SecondChildTAG: This video isnt correct for values. This man uses 74HCT series- CMOS high speed with TTL levels.As I told you above same gate's inputs has very very small input current.Datasheet [here][1]. You may ask me "but why in that video talking about small resistors values?" Because these values are correct for TTL or LS-TTL series, where for example : $I_0$ is :$I_{IL}$ LOW Level Input Current -0.36 mA (at VCC = Max, VI = 0.4V) and $I_1$ is :$I_{IH}$ HIGH Level Input Current 20 uA (at VCC = Max, VI = 2.7V). Try to compare 74HCT08 and 74LS08 datasheets. Good luck! [1]: http://www.nxp.com/documents/data_sheet/74HC_HCT08.pdf SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-12T23:35:12Z SecondChildTAG: Thank you ! SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-12-13T10:45:47Z FirstChildTAG: In my experience, there are only a few times when it matters whether to use a pull-down instead of a pull-up, or vice versa. One such instance for example is if you're driving coaxial switch relays from an MCU, and you don't want those switches to be confused by floating inputs when you switch your MCU on and it hasn't initialized yet. So, you put a pull-down on those pins, to make sure that that any floating inputs get pulled to ground, and the relay doesn't go clatter-clatter. Here's a relay: http://d1k4x70rbnswqm.cloudfront.net/media/blfa_files/tds/ramses/r574882425.pdf FirstChildUserIdTAG: 291046 FirstChildUserNameTAG: lunarplasma FirstChildCreateTimeTAG: 2012-12-16T20:46:23Z FirstChildTAG: Ok. If you want to test a mosfet transistor, you will realize that the gate remains polarized even after you remove the GS voltage, due to the GS capacitor. You have to connect the gate to the ground with a 1k resistor so that when you remove the GS voltage, the capacitor will discharge through the resistor to the ground. i wanted to make a video on this thing, but i found one here : http://www.youtube.com/watch?v=Q8R4FVRgB4Q .Same thing with cmos logic gates.I have actually tested this on a IRF 630, IRF 510, IRF 9510 and CD 4013 . FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-12-23T16:04:56Z IndexTAG: 886 TitleTAG: thanks great UserIdTAG: 805245 UserNameTAG: polson CreateTimeTAG: 2012-12-12T01:19:27Z VoteTAG: 3 CoursewareTAG: Week 2 / Digital Abstraction CommentableIdTAG: 6002x_digital NumberOfReplyTAG: 0 IndexTAG: 887 TitleTAG: H12P2 LINEAR REGULATOR c.) ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13552200248474479.jpg how did he get this equation? Anybody? UserIdTAG: 237941 UserNameTAG: per2x CreateTimeTAG: 2012-12-11T10:01:38Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Suppose that vin > vz: Than R0 is in series with the rz from the zener, so the the current iz will be iz=(vin-vz)/(R0+rz). The voltage drop over rz is: vrz=iz*rz, therefore the v+ input of the OA will be v+=vz+vrz and the gain of the OA is A=(1+R1/R2), so vout=A*v+. You can also write: v+=A*(vz+iz*rz)=A*(vz+rz*((vin-vz)/(R0+rz))). If you work that out, you get the formula. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-12-11T16:55:02Z FirstChildTAG: thanks salsero, i got it FirstChildUserIdTAG: 237941 FirstChildUserNameTAG: per2x FirstChildCreateTimeTAG: 2012-12-12T09:03:06Z SecondChildTAG: you're welcome SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-12-12T15:07:36Z IndexTAG: 888 TitleTAG: Fall 2003 Final Exam Problem 1C In this problem it is requested to draw the output voltage for an op amp having a diode in its negative feedback path.This is a logarithmic amplifier. Can please someone explain how this circuit is working at the rising part of the sinusoidal input and before the input reaches 0.6 Volts where the diode starts to conduct ? In the provided solutions Vout is drawn as -0.6 Volts for the the whole interval between 0 and pi,but to my understanding diode does not conduct when Vi <0.6 Volts Thanks for your help. UserIdTAG: 38331 UserNameTAG: janadel CreateTimeTAG: 2012-12-08T13:00:29Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You almost gave the answer yourself. The ideal diode is not conducting below 0.6V,so it's resistance is infinite and therefore the gain of the circuit will be infinite for an inputvoltage greater than 0V. So as soon as vin>0, the output goes directly to -0.6V. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-12-08T18:26:25Z SecondChildTAG: Thank you very much SecondChildUserIdTAG: 38331 SecondChildUserNameTAG: janadel SecondChildCreateTimeTAG: 2012-12-08T18:47:42Z IndexTAG: 889 TitleTAG: Aha? I'm not sure if it's quite an aha moment simply because it's only an approximation for large A. I think most aha moments provide a deeper understanding of what's actually going on. UserIdTAG: 317744 UserNameTAG: Trey92 CreateTimeTAG: 2012-12-08T04:04:10Z VoteTAG: 3 CoursewareTAG: Week 12 / S23V22 Inverting amplifier input resistance using virtual short method CommentableIdTAG: 6002x_S23V22_Inverting_amplifier_input_resistance_using_virtual_short_method NumberOfReplyTAG: 0 IndexTAG: 890 TitleTAG: regarding final exam does the lecture sequences be available during the final xam ..??? UserIdTAG: 183166 UserNameTAG: yogeshk CreateTimeTAG: 2012-12-07T12:50:04Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: They were available last time so it should probably be the same again. FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-12-07T17:40:35Z IndexTAG: 891 TitleTAG: LAB12 Tool shows wrong plot My LAB12 Circuit Tool shows the wrong plot for the right circuit. edX circuit sandbox also shows wrong plot. The old MITx Circuit Sandbox shows correct plot. So I was able to confirm solution with the old site's circuit tool. The current checker accepts the input and marks it correct. However, it is not possible to verify the plot using the current circuit tools. UserIdTAG: 4463 UserNameTAG: pmj CreateTimeTAG: 2012-12-03T05:51:13Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: i think i have the same problem as you. The plot i obtained was not similar to the sample, but my numerical answers were correct :) FirstChildUserIdTAG: 259653 FirstChildUserNameTAG: CKW FirstChildCreateTimeTAG: 2012-12-03T12:19:49Z FirstChildTAG: I have the same problem as you, I found the right values according to the old circuit sandbox, as you suggested, but the numbers were not correct when I checked... ![AC Analysis in the old sandbox][1] From the textbook, it is easy to get the values of $\omega_0$, $\alpha$ and Q from $C_1$, $C_2$, $g_1$ and $g_2$. [1]: https://edxuploads.s3.amazonaws.com/13545725331343674.png FirstChildUserIdTAG: 7057 FirstChildUserNameTAG: raiabril FirstChildCreateTimeTAG: 2012-12-03T22:04:03Z SecondChildTAG: Solved, I just set a value for $R_1$, play with $C_1$ and $C_2$ until f$\approx10kHz$ and then adjust with $R_2$. It is useful if you derive the expressions for $Q$ and $f$ from the equation $s^2+2\alpha s+\omega_0^2$. Important tips: $Q=\frac{\omega_0}{2\alpha}$ and $2\pi f=\omega_0$ SecondChildUserIdTAG: 7057 SecondChildUserNameTAG: raiabril SecondChildCreateTimeTAG: 2012-12-03T22:37:41Z SecondChildTAG: Don't forget to use only standard values of the resistors and capacitors. SecondChildUserIdTAG: 7057 SecondChildUserNameTAG: raiabril SecondChildCreateTimeTAG: 2012-12-03T22:39:38Z IndexTAG: 892 TitleTAG: lab11 i calculated the values of l and c correctly n got my ckt green tick...bt when i insert them below i m getting red tick...ne idea why m i facing this problem!!!!! UserIdTAG: 214277 UserNameTAG: yadsam CreateTimeTAG: 2012-12-01T18:30:41Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: Aside from the obvious that L is in henry and that C is in farad, what did you enter? I got mine to work so I could compare with your answer and perhaps make a suggestion. FirstChildUserIdTAG: 569199 FirstChildUserNameTAG: neogerald FirstChildCreateTimeTAG: 2012-12-01T18:40:54Z FirstChildTAG: Hi yadsam, Can I help you? That is weir... If you have already your correct values, you should obtain the green tick in the simulator after you replace that values... Are you sure that you have replaced correctly the values ? Have you click on the AC button before click on the check button? Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-01T19:09:24Z SecondChildTAG: Hi Myriam, I have the same problem and i don't know how to get green tick for the value of the L and C I have a Green tick for the figure one circuit but when i enter my calculated value in the value of L and C section (L in Henries and C in Farad) i get the red tick. Please Give me some suggestion SecondChildUserIdTAG: 229979 SecondChildUserNameTAG: Hafezi SecondChildCreateTimeTAG: 2012-12-02T16:43:45Z SecondChildTAG: There was something wrong with the math and i manage to get the correct answer. Now i have the green tick for Value of L and C and figure one :-) SecondChildUserIdTAG: 229979 SecondChildUserNameTAG: Hafezi SecondChildCreateTimeTAG: 2012-12-02T17:02:43Z SecondChildTAG: Hi, Hafezi. I believe that if you don't get the green mark on the L and C values then they are incorrect independent of the green mark of the image. To find the C and L values put your transfer function in the canonical form: $\frac{V2}{V1}=\frac{numerator}{S^2+2\alpha S+\omega o^2}$ then with the given values you will be able to get two equations to find the two unknowns. As always remember to convert from Hz to rad/s for $\omega$. SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-12-02T17:07:51Z FirstChildTAG: Was your green check directly below the circuit simulator? I have long suspected that the checker in the simulator is not quite as rigorous as is, say, the one for algebraic expressions. I've found a green check sometimes appears as long as there's something, anything, in the simulator that's even vaguely correct. And think about it - I doubt there is the sort of routing capability in our little lab simulator as there would be in, for example, a professional design tool like Eagle. Anyway - sounds like your math is at fault somewhere. Double- and triple-check your numbers. Write back if you need more help. FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-12-02T11:39:05Z SecondChildTAG: Thank you for your suggestion There was something wrong with the math and i manage to get the correct answer. Now i have the green tick for Value of L and C and figure one :-) SecondChildUserIdTAG: 229979 SecondChildUserNameTAG: Hafezi SecondChildCreateTimeTAG: 2012-12-02T17:02:12Z FirstChildTAG: You will get Check-Mark for circuit(figure1) in Lab 11 as long as it is a working circuit... just like in circuit-sandbox This is because you have to change the configuration of the circuit to answer later question. When you press the check button you simply save the current configuration of circuit. You only have the right answer if AC-Analysis of figure 1 matches figure 2. FirstChildUserIdTAG: 714237 FirstChildUserNameTAG: PaxPolaris FirstChildCreateTimeTAG: 2012-12-03T05:19:14Z IndexTAG: 893 TitleTAG: H12P3 My trouble is with the midband gain for the second part of this question. I calculated the correct resistors and capacitors, but when I check the gain at 1kHz, the grader does not like my answer. I confirmed the answer by building the circuit in the sandbox, and I've also tried expressing is as gain and in dB, but grader still doesn't like it. Thoughts? Thanks! UserIdTAG: 198899 UserNameTAG: trids64 CreateTimeTAG: 2012-11-30T18:12:40Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Does your gain have the correct sign? FirstChildUserIdTAG: 174100 FirstChildUserNameTAG: markpolak FirstChildCreateTimeTAG: 2012-11-30T18:21:41Z SecondChildTAG: Yes, my gain is positive. SecondChildUserIdTAG: 198899 SecondChildUserNameTAG: trids64 SecondChildCreateTimeTAG: 2012-11-30T19:29:21Z SecondChildTAG: Yeah, I see my error now. Thanks! SecondChildUserIdTAG: 198899 SecondChildUserNameTAG: trids64 SecondChildCreateTimeTAG: 2012-11-30T19:30:57Z SecondChildTAG: I suppose we take magnitude of transfer function H(jw) to find the gain. And magnitude can't be positive. SecondChildUserIdTAG: 154016 SecondChildUserNameTAG: Albatross SecondChildCreateTimeTAG: 2012-12-07T08:06:31Z SecondChildTAG: I am sorry for my mistake in the comment above. Rectification-'The magnitude should always be positive'. SecondChildUserIdTAG: 154016 SecondChildUserNameTAG: Albatross SecondChildCreateTimeTAG: 2012-12-07T08:21:04Z SecondChildTAG: What is that question supposed to mean ? The gain is positive. Is there something else to see ? SecondChildUserIdTAG: 351871 SecondChildUserNameTAG: Lamarque SecondChildCreateTimeTAG: 2012-12-08T22:12:57Z SecondChildTAG: Is it? SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-12-09T03:09:56Z FirstChildTAG: Please give me hint how to find gain! FirstChildUserIdTAG: 96873 FirstChildUserNameTAG: Santyaga FirstChildCreateTimeTAG: 2012-12-01T10:38:27Z SecondChildTAG: The simplest way to solve the last part that I can think of, is to use most of the values that you found in the first part. In the last figure, you see that the MINUS input of the OA is a virtual ground, so the values for the HPF can remain the same. But the LPF section becomes important now. Try to calculate the break freq for that part with one of the given C's. Since the midband freq is far away from the break freqs, now imagine what happens if the input freq goes from very low to inf. What happens with the impedances of the C's? Use this info, since we earlier supposed that the midband freq was far away from the break freqs, to calculate the midband gain. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-12-01T15:39:36Z SecondChildTAG: Compare the transfer functions (TF) for parts 1 and 2. In the TF for part 1 there is an expression for the gain in terms of RS and RF. Look at the TF for part 2, and you will see the term for gain in terms of R3 and R4. Part 2 will use the component values from part 1 except for RS and RF. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-12-01T16:13:48Z FirstChildTAG: Thank you guys! Solve it! FirstChildUserIdTAG: 96873 FirstChildUserNameTAG: Santyaga FirstChildCreateTimeTAG: 2012-12-02T13:46:21Z IndexTAG: 894 TitleTAG: S21E3: Thevenin Tank Could someone explain to me how they've analyzed the circuit here? I can't get my expression to match theirs. We have Zr in series with ZL || ZC . So the equivalent impedance of the circuit is (ZL*ZC)/(ZL+ZC) + ZR. To get the complex current, take Vi and divide by the equivalent impedance. Then ohm's law relation to determine the voltage across any given element. So for Vo: I note that ZC and ZL are || and so have the same voltage across them, so multiply ZC || ZL with the complex current through them: Vo = [Vi*(ZL+ZC)/( (ZL*ZC) + ZR*(ZL + ZC) )]*[(ZL*ZC)/(ZL + ZC)] You lose the Vi when take the transfer function, but I have worked and reworked this expression and I CANNOT simplify it to the form given by the "show answer" button. What am I doing wrong? Sidenote: I at first multiplied my complex current by ZC rather than ZC || ZL, which I now think was erroneous, but, doing it that way, I was able to arrive at a final answer which differed from the given answer by only a single term in the numerator, while our denominator terms matched exactly. So where am I going astray. Much appreciated. Thank you. UserIdTAG: 287805 UserNameTAG: Beneficial CreateTimeTAG: 2012-11-30T01:24:20Z VoteTAG: 3 CoursewareTAG: Week 11 / S21E3: Thevenin Tank CommentableIdTAG: 6002x_Thevenin_Tank NumberOfReplyTAG: 4 FirstChildTAG: bump FirstChildUserIdTAG: 287805 FirstChildUserNameTAG: Beneficial FirstChildCreateTimeTAG: 2012-11-30T13:59:12Z SecondChildTAG: I think your specified Vo equation is wrong, numeric error. Start again, work out your current as stated V/Ztotal, and then multiply by your appropriate impedances, to give Vo. SecondChildUserIdTAG: 273287 SecondChildUserNameTAG: johnkh77 SecondChildCreateTimeTAG: 2012-11-30T15:37:16Z SecondChildTAG: What is the appropriate impedance to multiply by? SecondChildUserIdTAG: 287805 SecondChildUserNameTAG: Beneficial SecondChildCreateTimeTAG: 2012-11-30T23:20:31Z FirstChildTAG: You can use limits to solve this. It's not stated that w=0, but that w is GOING towards zero and GOING towards inf. It's important how you write you formulas. I use: s=jw and call C parallel to L: Z=sL/(1+LCs^2) and vo/vi=t=Z/(Z+R) and divide all the terms of t by Z so we will get: t1=1/(1+R/Z). I'll use the w->x as w GOING to x (not the same as w=x !!!) For (w->0)=>(s->0): You can see that in Z, that s^2 is going faster to ZERO then s, therefore write s^2=0 than the limit goes to Z=sL/(1+0)=sL and t1=1(1+R/sL) becomes t1=1/(R/sL)=sL/R=jwL/R because R/sL is increasing fast, so neglect the 1 in (1+R/sL). Now for (w->inf)=>(s->inf) we see that in this case the s^2 is INCREASING faster than s, so we neglect the 1 in (1+LCs^2) and we get: Z=sL/(LCs^2)=1/sC as a limit for w to inf. Now use the original t1 again and write Z=1/sC so we'll get: So for w->inf, t=1/(1+R/(1/sC))=1/sRC=1/jwRC (neglected the 1 again in (1+R/(1/sC)) Hope this will do. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-11-30T22:53:11Z SecondChildTAG: Not asking about that part of the question. Just asking about the first part where the transfer function is derived. SecondChildUserIdTAG: 287805 SecondChildUserNameTAG: Beneficial SecondChildCreateTimeTAG: 2012-11-30T23:22:28Z SecondChildTAG: the equation you wrote for V_o is right. \large V_o= V_i \times {Z_l+Z_c \over Z_l\cdot Z_c+R(Z_l+Z_c)} \times {Z_l\cdot Z_c\over Z_l+ Z_c} You can cross out the Z_l+Z_c. After substituting the values of the impedances ... your answer should get the checkmark even if it doesn't match exactly the given answer's format. Z_c = {1\over j\omega C} \\ Z_l = {j\omega L} \\ Z_r = R ... To get the given answer it might be easier to use admittance instead of impedance Z_c || Z_l= {1\over 1/Z_c+1/Z_l} \\ = {1\over j\omega C+{1\over j\omega L}} \\ = {j\omega L \over 1- omega^2 LC} SecondChildUserIdTAG: 714237 SecondChildUserNameTAG: PaxPolaris SecondChildCreateTimeTAG: 2012-12-01T08:21:20Z FirstChildTAG: You're trying to do it the hard way. Look carefully to the circuit. Doesn't it look like a voltage divider? You konw the expression for the voltage in a resistive voltage divider. Here is the same thing, just with complex numbers. FirstChildUserIdTAG: 188586 FirstChildUserNameTAG: FLara FirstChildCreateTimeTAG: 2012-11-30T18:53:41Z SecondChildTAG: I can see the voltage divider relationship. I have tried to apply the voltage divider relationship to the circuit. The result of such an analysis repeatedly ends up differing from the given answer. I have outlined my attempt above. SecondChildUserIdTAG: 287805 SecondChildUserNameTAG: Beneficial SecondChildCreateTimeTAG: 2012-11-30T23:21:47Z FirstChildTAG: the equation you wrote for $V_o$ is right. $\large V_o= V_i \times {{Z_l+Z_c} \over Z_l\cdot Z_c+R(Z_l+Z_c)} \times {Z_l\cdot Z_c\over {Z_l+Z_c} }$ You can cross out the $Z_l+Z_c$. After substituting the values of the impedances ... your answer should get the checkmark even if it doesn't match exactly the given answer's format. $Z_c = {1\over j\omega C} \\ Z_l = {j\omega L} \\ Z_r = R$ ... To get the given answer it might be easier to use admittance instead of impedance $ \large{Z_c || Z_l= {1\over 1/Z_c+1/Z_l} \\ = {1\over j\omega C+{1\over j\omega L}} \\ = {j\omega L \over 1- \omega^2 LC}}$ FirstChildUserIdTAG: 714237 FirstChildUserNameTAG: PaxPolaris FirstChildCreateTimeTAG: 2012-12-01T08:30:00Z SecondChildTAG: Thanks for your response PaxPolaris, somehow in my calculations I managed to miss being able to cancel the (Zl+ZC) terms every single time I tried to work the equation. It's always the most trivial oversights that will hang me up for DAYS on a question. SecondChildUserIdTAG: 287805 SecondChildUserNameTAG: Beneficial SecondChildCreateTimeTAG: 2012-12-01T15:41:31Z IndexTAG: 895 TitleTAG: Find Q? I don't understand how to calculate Q. I tried with the formula Q = (1/sqrt(L*C)) * (L/R) and also with Q = wo * L / R and the result is not even close. Help please!! UserIdTAG: 433574 UserNameTAG: endika86 CreateTimeTAG: 2012-11-29T17:00:18Z VoteTAG: 3 CoursewareTAG: Week 11 / S21E4: AM_Radio_Tuning CommentableIdTAG: 6002x_S21V1_AM_Radio_Tuning NumberOfReplyTAG: 2 FirstChildTAG: This is not the standard RLC circuit for which 2*alpha = R/L. You need to set up the complex equation for the circuit gain and pull the appropriate value for 2*alpha out of it. FirstChildUserIdTAG: 174100 FirstChildUserNameTAG: markpolak FirstChildCreateTimeTAG: 2012-11-29T17:21:31Z FirstChildTAG: mmm... the $\frac{L}{R}$ part does not look correct... what is your bandwidth $ \Delta\omega$? (recall that $2\alpha=\Delta\omega$) FirstChildUserIdTAG: 288174 FirstChildUserNameTAG: matiasgrodriguez FirstChildCreateTimeTAG: 2012-11-29T17:26:15Z SecondChildTAG: 1/(R*C) SecondChildUserIdTAG: 194717 SecondChildUserNameTAG: kaa SecondChildCreateTimeTAG: 2012-11-30T19:29:23Z SecondChildTAG: mmm that looks correct :) SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-30T20:23:57Z SecondChildTAG: And dont forget to divide by 2*pi SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-12-01T18:34:22Z IndexTAG: 896 TitleTAG: STAFF about final exam Will be published review of final exam at previous years as it was with midterm exam? UserIdTAG: 96873 UserNameTAG: Santyaga CreateTimeTAG: 2012-11-28T17:41:09Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 897 TitleTAG: WolframAlpha vs Maxima Has anyone tried to solve Homework 10 Part 3 using WolframAlpha? I found Lmatch and Cmatch using Maxima, but in WA I don't know how to solve a system of equations with parameters (basically I need to solve the equations: real(z)=R1 and imaginary(z)=0 where z is the impedance seen by the final amplifier). In the following image there are the commands (marked in yellow) and the results using Maxima. ![System of equations with parameters solved using Maxima][1] [1]: https://edxuploads.s3.amazonaws.com/13539512975604769.png UserIdTAG: 376877 UserNameTAG: AndBre CreateTimeTAG: 2012-11-26T17:37:15Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Inspired by my own success at typing "[how to plot log scale with wolfram alpha][1]" into [teh Googlez][2], I tried "[how to solve for real and imaginary with wolfram alpha][3]", and the [very first hit][4] looks promising. Scroll down to "Do you want to solve an equation over the reals? Just tell Wolfram|Alpha to restrict the domain. ..." [1]: https://www.google.com/search?q=how%20to%20plot%20log%20scale%20with%20wolfram%20alpha [2]: http://www.google.com/ [3]: https://www.google.com/search?q=how%20to%20solve%20for%20real%20and%20imaginary%20with%20wolfram%20alpha [4]: http://blog.wolframalpha.com/2010/01/14/solving-equations-with-wolframalpha/ FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-11-27T04:27:17Z SecondChildTAG: Thanks planetscape, I think that the following command should be fine: re(1/(i*w*C+1/(i*w*L+R2)))=R1,im(1/(i*w*C+1/(i*w*L+R2)))=0 for C and L but I get "Standard computation time exceeded". I will try to search more. SecondChildUserIdTAG: 376877 SecondChildUserNameTAG: AndBre SecondChildCreateTimeTAG: 2012-11-27T10:55:55Z IndexTAG: 898 TitleTAG: The "Virtual Short" method DOES NOT RAISE to an aha moment (Vote if you believe) If you think the short method is not very important and does not cause an aha moment,vote here. UserIdTAG: 259238 UserNameTAG: omidsadeghi CreateTimeTAG: 2012-11-26T14:09:01Z VoteTAG: 3 CoursewareTAG: Week 12 / S23V21_Inverting_amplifier_input_resistance_-_2 CommentableIdTAG: 6002x_S23V21_Inverting_amplifier_input_resistance_-_2 NumberOfReplyTAG: 1 FirstChildTAG: It is a useful and *important* simplifying trick when dealing with op amps, but it doesn't provide an avenue to understand the circuit element behaviour at a more fundamental level. The small signal method qualifies (linearisation of a fundamentally non-linear circuit element); LMD qualifies (elimination of differential forms of Maxwell's Equations, allowing all of modern EECS); static discipline for noise immunity qualifies (billions of interacting circuit elements can handle and eliminate noise to correctly evaluate logic problems in digital design - that gets and extra wow(!)); impedance method qualifies (essentially allows AC analysis using DC mathematics); feedback qualifies (instantaneous noise cancellation in analog amplification circuits). There are all OMG Aha! moments. Do you see what I mean? It's big picture stuff that almost transcends EECS. Allowing simpler mathematics, while also losing recourse to describe some of the circuit behaviour, i.e. not providing the relationships with respect to "A", does not fall in the same category as those incredible moments of insight highlighted above. FirstChildUserIdTAG: 166031 FirstChildUserNameTAG: krebryna FirstChildCreateTimeTAG: 2012-12-06T20:23:45Z IndexTAG: 899 TitleTAG: H10P3 - An L Network question, Request to Skyhawk Dear Skyhawk, Help needed, as you explained for H10P3 Q C, Cmatch, The denominator of the complex Impedance also carries 1, that is somewhat like (1-XXw^2+XXjw). Now which part will be Re and Img among these. Whether removal of 1 will affact our calculation? UserIdTAG: 455950 UserNameTAG: Wahabbaluch CreateTimeTAG: 2012-11-23T18:50:48Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: The denominator has the form D = 1 - w^2*E + j*w*F; therefore Re(D) = 1 - w^2*E and Im(D) = w*F. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-23T19:53:28Z SecondChildTAG: Thanks! SecondChildUserIdTAG: 455950 SecondChildUserNameTAG: Wahabbaluch SecondChildCreateTimeTAG: 2012-11-24T02:05:28Z IndexTAG: 900 TitleTAG: H10P2 After the end of the work week back to edx and solving homeworks on weekend!!! Impedance for circuit C. I Have gren mark for impedance of circuit C, ZC=... But can`t get it for ω→0 & ω→∞, ZC. As I understand from figure 13.9 in textbook in branch witw capasitor when ω→∞ we take into account only R, And for ω→0 we have the same situation with inductor. Where is the mistake? Some hints please! UserIdTAG: 96873 UserNameTAG: Santyaga CreateTimeTAG: 2012-11-23T11:57:02Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: For ω→0 you should ignore capasitor and for ω→∞ inductance. Next, for ω→0 inductance impendance will be zero and for ω→∞ capasitor impendance will be zero too. FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-11-23T12:18:03Z FirstChildTAG: Hi Santyaga, Can I help you? The key of this exercises is to find the total impedance of the Circuit with laplace. Then, substitute the s for j*w. Hints: - For a Capacitor, the impedance will be Zc = (1/s*C)=(1/j*w*C) - For a Inductor, the impedance will be ZL= s*L = j*w*L Then, try to in some way to obtain the total impedance (recall series impedances combinations and parallel impedances combinations from previous Weeks). And then, try to think, what happens for a high value of w and for a low value of w. Replace it in your formula :). Remember that: - 1 divided by a huge huge number tends to zero . - 1 divided by a small small number tends to infinite. I hope this can help you, Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-23T13:16:37Z FirstChildTAG: Hi Myriam! As I remember for R1||R2 it`ll be like (R1*R2) /(R1+R) series for series R and L will be R+j*w*L; for series R and C R+1/(j*w*C). With it all clear. In my homework I also have green mark for impedance of whole curcuit, but when try to find Z for w-> 0 or infinite and writing some like (x*(x+"Z of element"))/(x+(x+"Z of element")) I have red dead mark And such pucture with another part. Where is the mistake? FirstChildUserIdTAG: 96873 FirstChildUserNameTAG: Santyaga FirstChildCreateTimeTAG: 2012-11-23T17:49:57Z SecondChildTAG: there are only two possible answers : 0 or inf SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-11-23T18:01:11Z SecondChildTAG: for curcuit C 0 or inf are not grading as a right answer SecondChildUserIdTAG: 96873 SecondChildUserNameTAG: Santyaga SecondChildCreateTimeTAG: 2012-11-23T18:13:49Z SecondChildTAG: And it is not right answer! Get it! SecondChildUserIdTAG: 96873 SecondChildUserNameTAG: Santyaga SecondChildCreateTimeTAG: 2012-11-23T18:14:52Z SecondChildTAG: Ah, sorry.My hints above are correct. More price. At w->0 you should ignore entire RC branch.In this case Zl=0.So, your answer will be Rx. At w->inf you should ignore entire RL branch.In this case ZC=0.So, your answer will be Ry. SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-11-23T18:29:41Z SecondChildTAG: Ok, more hints. What is DC capasitor impendance? (∞) What is inductance impendance at very very high freq? ( ∞) SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-11-23T18:35:40Z IndexTAG: 901 TitleTAG: H11P1, I can calculate parts 1,2,3,4 but I can't calculate part 5?, why? Hi, We calculate the C (in part 2) and with that C we calculate bandwith (part 3) with equation from part 1. So I calculated C from part 4 and It's a green check but when I use that C to calculate the bandwidth for the part 5 It give me a red check? I desided attack the problem by brute force and I discover the bandwidth! but is given for a different C from part 4, why? Thanks! UserIdTAG: 299251 UserNameTAG: TheRedBlackOne CreateTimeTAG: 2012-11-23T11:00:35Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I also had that problem, but I found out that I made a mistake. The formulas that I used were correct, and I had the correct C for part 4 too, but I accidentally took the value from part 2, and then I forgot to add the pF, because in the box it's without ... (I was tired). But in the end it all worked out OK. So maybe you should check for a typo somewhere. Another point is important: It is possible that if your calculations are wrong somewhere, that the answer you get, will be accepted by the grader, if it is not to far off from the correct value. And is even possible, that a few answers get the green check, BUT!! that the last answer in that case is too far off, so you'll get the red mark! It happened to me in the previous course! I made, with the same problem as this one, somewhere a mistake and all the answers were correct, but not the last one. I discovered later that my calculation for the Q was wrong, but because I made a type in the box, (forgot 1 digit!!) and the answer was by accident close enough, so I got the green mark. So I assumed that my calculations were OK! NOT SO! So my next answers depended on that Q, and since I used the Q from the box (not from my paper!!) for the next calculation, that answer was also by accident close enough, and I got the green mark. But the answers on the following questions were marked wrong! After 6 hours, I discovered that my formula was wrong! (lack of concentration again, I copied my own formula wrong to another piece of paper!!) And that I made that typo in the box! So what I mean to say is: A green mark, doesn't always mean that you have the exact answer!! And it doesn't guarantee that your formulas or calculations were correct! Hope this will help. Good luck! FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-11-23T13:20:18Z SecondChildTAG: So if I get a green check in the part 1 my equation for bandwidth is right, the bandwidth is the squareroot of the spoiled equation here: [spoiled equation][1] and to calculate C I'm using the next C = (RC*RL)/(L*RC*w^2) which I suppose it's wrong! Any comment? [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50acc00d7cc2352700000013 SecondChildUserIdTAG: 299251 SecondChildUserNameTAG: TheRedBlackOne SecondChildCreateTimeTAG: 2012-11-23T14:03:30Z SecondChildTAG: Yes, it's wrong. it's the first term: RC*RL should be RC+RL Let me know if it works. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-11-23T14:36:24Z SecondChildTAG: This is exactly the reason why you have to watch out for the green marks! Don't ever trust your own answers to the first questions ... SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-11-23T14:40:04Z SecondChildTAG: Sorry I typed bad, C =(RC+RL)/(L*RC*w^2) doesn't work for me! SecondChildUserIdTAG: 299251 SecondChildUserNameTAG: TheRedBlackOne SecondChildCreateTimeTAG: 2012-11-23T16:09:44Z SecondChildTAG: Didn't you forget that (omega)^2=(2*pi*f)^2 and f in Hertz? (I'm very sloppy myself, so sometimes forget the 2*pi) Or forgotten to use the new frequency/capacitance and instead used the old one to solve the last BW (happened also to me)? SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-11-23T17:07:54Z SecondChildTAG: C = ((490+4)/(0.65m*490*(2*pi*1150k)^2))*10^12 in pico farads C = 29.7072256349 pF w (bandwidth) = sqrt((0.65m+4*490*29.7p)/(0.65m*490*29.7p))/1000 in KHz w (bandwidth) = 8.28978336718 in KHz and the solution of the bandwidth is 12!!! any comment? Thanks! SecondChildUserIdTAG: 299251 SecondChildUserNameTAG: TheRedBlackOne SecondChildCreateTimeTAG: 2012-11-23T21:08:26Z SecondChildTAG: I calculated in Python: a2 is 2*alpha: a2=(L+C*rc*rl)/(L*C*rc) # BW in rad/s ; print a2 ; a2=75411.71901436518 # rad/s ; a0=a2/(2*pi) # is BW freq in Hz ; print a0 ; a0= 12002.147848193292 # Hertz so: 12kHz ; so I think you changed Q and BW ... BTW: You left in your formulas the k (1000) for RC. In the first it worked out but not in the second and in the second you divided by 1000 ... Hmmm! Asking for problems. But ok, Hope this helps. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-11-24T13:54:14Z SecondChildTAG: Ok, Thanks a lot! I forgot the "k", that's right! and the square root left over! SecondChildUserIdTAG: 299251 SecondChildUserNameTAG: TheRedBlackOne SecondChildCreateTimeTAG: 2012-11-24T15:49:48Z FirstChildTAG: Are you paying attention to the units for RL and RC? FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-23T23:45:55Z IndexTAG: 902 TitleTAG: New "Aha moment" ? With reference to S23V22, should the "Virtual Short trick" rise to the level of an "Aha moment" ? PLEASE VOTE! Prof.Agarwal is interested in knowing your opinion UserIdTAG: 372321 UserNameTAG: EnricoDona CreateTimeTAG: 2012-11-21T22:03:34Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Its the concept of the negative feedback. That concept made the virtual short trick applicable. FirstChildUserIdTAG: 127195 FirstChildUserNameTAG: princeofsudan FirstChildCreateTimeTAG: 2012-11-21T22:54:30Z FirstChildTAG: I'm starting to feel sick when i hear "Aha moment" .It's like a stupid marketing campaign. FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-11-22T08:59:27Z SecondChildTAG: You are quite right, but remember how difficult it must be for the teaching staff, and Prof Agarwal in particular, to gauge feedback to the material without a live lecture hall in front of them. Getting more of us involved in the discussion forum, even if it means using tricks like "aha moments" is a very valid technique, IMHO. Maybe we should go a step further and vote for the "Top Ten Aha Moments" in 6.002x? I will put the OpAmp virtual ground up there. SecondChildUserIdTAG: 145194 SecondChildUserNameTAG: DerekH SecondChildCreateTimeTAG: 2012-11-25T17:38:42Z SecondChildTAG: I agree. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-12-06T09:02:33Z FirstChildTAG: Yes, it is! FirstChildUserIdTAG: 84801 FirstChildUserNameTAG: marcuspag FirstChildCreateTimeTAG: 2012-12-03T22:05:26Z IndexTAG: 903 TitleTAG: !!! Impossible!!!! can someone explain this problem!!!! plz!!!! >o< UserIdTAG: 70519 UserNameTAG: Fipe CreateTimeTAG: 2012-11-21T21:52:07Z VoteTAG: 3 CoursewareTAG: Week 10 / Trigonometry Isint so bad CommentableIdTAG: 6002x_Trigonometry_isnt_so_bad NumberOfReplyTAG: 2 FirstChildTAG: I need help too. FirstChildUserIdTAG: 337799 FirstChildUserNameTAG: Evandilson FirstChildCreateTimeTAG: 2012-11-21T22:37:30Z FirstChildTAG: v'=-A*w*cos(wt)+B*w*cos(wt) -R*C*A*w*cos(wt)+R*C*B*w*cos(wt)+A*cos(wt)+B*sin(wt)=Vi*cos(wt) sin(wt)=0 then cos(wt)=1 : B*R*C*w+A=Vi; sin(wt)=1 then cos(wt)=0 : -A*R*C*w+B=0; B=A*R*C*w (A*R*C*w)*R*C*w+A=Vi A*(R*C*w)^2+A=Vi A=Vi/((R*C*w)^2+1) B=(Vi*R*C*w)/((R*C*w)^2+1) FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-11-21T23:09:36Z SecondChildTAG: thx a lot SecondChildUserIdTAG: 70519 SecondChildUserNameTAG: Fipe SecondChildCreateTimeTAG: 2012-11-22T04:13:03Z SecondChildTAG: Hello guys. YakovO, -x'sin(x) is not the derivate of cos(x)? SecondChildUserIdTAG: 138585 SecondChildUserNameTAG: dfourier SecondChildCreateTimeTAG: 2012-11-23T15:55:54Z SecondChildTAG: d*sin(x)/dx=cos(x) d*cos(x)/dx=-sin(x) SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-11-24T16:43:43Z SecondChildTAG: Dude, v' is wrong, it should be: v'= -A*w*sin(wt)+B*w*cos(wt) SecondChildUserIdTAG: 61423 SecondChildUserNameTAG: libo654 SecondChildCreateTimeTAG: 2012-11-24T18:31:08Z IndexTAG: 904 TitleTAG: I don't understand expressions with complex numbers (jw) I understand complex algebra, but I don't understand what means complex numbers in our world? What difference between, for example e^wt and e^jwt? I confused and can't jump into world of complex numbers:) Please, give me pictures, video or other stuff to understand it...(PS: I understand all operations under complex numbers, but don't understand why?..., sorry for my english...) UserIdTAG: 104522 UserNameTAG: IgorNovice CreateTimeTAG: 2012-11-21T19:44:34Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Hi IgorNovice, Can I help you? You should take a look at page 941 of the Textbook [read here][1] - Complex Numbers. Why exists Complex numbers? I asked to myself that same question when I saw that strange numbers for the first time and then years later I can´t live without complex numbers as engineer student hahaha. In fact, sometimes I debate that with a Historian University Teacher, my Co-Director of the Investigation Group of Education at the University, haha. She says that it is important to know all, without knowing if that is practical or not and I say the opposite, I have some kind of inercia to change my point of view - I guess that that is normal for an engineer student that is more practical than theorical- but I have meditated that a lot because of her context and arguments, so I guess that all that new Filosophy is opening my mind haha. I have concluded that is important to know all, that opens you your mind, might you will not know precisely now when you will apply it :)but you know that they exists haha, odd. ---------- Ok, complex numbers born to solve originally this question: - What is the square root of a negative number? $\sqrt(-1)$ -And what is the square root of a number? Lets suppouse $\sqrt(4)$, the answer is 2, because 2*2 is 4. Lets suppouse $\sqrt(25)$, the answer is 5, because 5*5 is 25. . . ---------- Hmmm , but **what is the product of a number that multiplied gives a negative number???** $\sqrt(-1)$ , -1*-1=1 and not -1 . So, -1 can not be the solution…. Neither 1. So, there was a need to invent that strange complex numbers to solve certain problems . ---------- This have to be your : ((Aha Moment!)) haha! ---------- Ok, so, they invented the complex numbers. They say that $\sqrt(-1)$ = j And then it comes the others operations like, So $(\sqrt(-1))^2= j^2 $ = -1 And so on. ---------- So, that was the start of the complex numbers. It was in order to solve that conflict. Then It came a lot of useful things with the complex numbers! like that you can synthesize Circuits based on data that are in complex numbers ! so odd! But it is true hahaha. Now I can’t live without complex numbers haha. The exponential with the j, is another way to express the complex number, it comes from the Euler identity. You can read that more described in the Textbook. But the concept of the complex number is the same. The only thing, is that you will have different ways to express a complex number. I hope that this can help you. See you, Myriam. P.D. If you have any doubt I will be here :). [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/971 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-21T20:28:34Z SecondChildTAG: I will practice with complex numbers next few days, and ask again if not understand something! Thanks for help! SecondChildUserIdTAG: 104522 SecondChildUserNameTAG: IgorNovice SecondChildCreateTimeTAG: 2012-11-22T07:53:15Z SecondChildTAG: Complex numbers were invented by mathematicians to solve equations. Physics uses mathematics as a tool. Newton did that to understand nature. One of the pioneers in the use of complex numbers in electrical circuits was Charles Proteus Steinmetz. Find out about this scientist and discover about complex numbers in electrical circuits. SecondChildUserIdTAG: 361823 SecondChildUserNameTAG: EliasOak SecondChildCreateTimeTAG: 2012-11-22T22:10:22Z SecondChildTAG: Thanks! I found "Lectures on electrical engineering Volume III". Chapter IV "Symbolic method" was very useful for me! SecondChildUserIdTAG: 104522 SecondChildUserNameTAG: IgorNovice SecondChildCreateTimeTAG: 2012-11-23T18:01:37Z FirstChildTAG: http://betterexplained.com/articles/a-visual-intuitive-guide-to-imaginary-numbers/ FirstChildUserIdTAG: 127195 FirstChildUserNameTAG: princeofsudan FirstChildCreateTimeTAG: 2012-11-21T23:13:48Z SecondChildTAG: Thanks, very useful link! SecondChildUserIdTAG: 104522 SecondChildUserNameTAG: IgorNovice SecondChildCreateTimeTAG: 2012-11-22T08:45:08Z FirstChildTAG: I think, I understand it. My problem was that I imagine complex exponent (e^jw) similar real exponent(e^w). But now I understand that e^jw is periodical function vs e^w is grow up infinitly fucntion. (Picture 1 was big Aha for me) ![e^jw][1] ![enter image description here][2] If I see e^jw like a function of time (picture 1) - it seems like a "spring":) When I see 3d "spring" model of complex exponent - I understand it very well!!! Now, according to Euler's formula sin(x) = (e^jw - e^-jw)/2j, I can imagine sin(wt) function like 2 "springs". First rotate CCW and second rotate CW. If I get projection of them on Imaginary axis-time plane - i see picture 3. I make substraction between them and divide on 2j (rotate CW Pi/2) and get source sin(wt)! ![enter image description here][3] [1]: https://edxuploads.s3.amazonaws.com/13538376631343631.png [2]: https://edxuploads.s3.amazonaws.com/13538376861343663.png [3]: https://edxuploads.s3.amazonaws.com/13538387593047467.bmp FirstChildUserIdTAG: 104522 FirstChildUserNameTAG: IgorNovice FirstChildCreateTimeTAG: 2012-11-25T10:15:13Z SecondChildTAG: Wow man, indeed picture A is very spectacular! :) SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-25T13:57:50Z SecondChildTAG: I'm agree with you:) SecondChildUserIdTAG: 104522 SecondChildUserNameTAG: IgorNovice SecondChildCreateTimeTAG: 2012-12-10T16:53:43Z IndexTAG: 905 TitleTAG: Circuit Program? Is there any program out there that is similar to what we have used in the labs? Where you can place basic components and design circuits? The only program I have found was the CEDAR Logic Simulator http://sourceforge.net/projects/cedarlogic/ but that only is good for using the abstract logic gates to build something. UserIdTAG: 219780 UserNameTAG: ReconIII CreateTimeTAG: 2012-11-21T01:38:52Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 7 FirstChildTAG: I have sometimes used QUCS to verify results from homework assignments. [http://qucs.sourceforge.net/][1] It's free software, available for all major operating systems including Windows and MacOS. The schematics capture part works similar to the circuit simulator used in the labs. To run a simulation, you need to place a "simulation component" into the schematics first (dc, ac, transient ...). After running the simulation you get a sheet where you can place your diagrams. Fairly intuitive. [1]: http://qucs.sourceforge.net/ FirstChildUserIdTAG: 148015 FirstChildUserNameTAG: ralfh FirstChildCreateTimeTAG: 2012-11-21T07:08:42Z FirstChildTAG: [CircuitLab][1] was designed by MIT grads. I find it indispensable. [1]: https://www.circuitlab.com/ FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-11-21T07:33:28Z FirstChildTAG: I would highly recommend the Linear Technology software LTSpice. http://www.linear.com/designtools/software/ The software is very easy to use, and does everything our circuit sim does. There are a lot of people out there using it, so you can always find the help that you need with google. FirstChildUserIdTAG: 151427 FirstChildUserNameTAG: RichardZ FirstChildCreateTimeTAG: 2012-11-21T03:23:59Z SecondChildTAG: LTSpice is free! It also has the advantage that if you have an exotic component and can find a spice model for it, LTSpice can in most cases, use that model. Finally, there is a yahoo group named (no surprise here) [LTspice ][1] dedicated to it. The group files contain tutorials, device models, FAQs and much more. [1]: http://tech.groups.yahoo.com/group/LTspice/ SecondChildUserIdTAG: 141000 SecondChildUserNameTAG: OrinE SecondChildCreateTimeTAG: 2012-11-21T19:27:50Z FirstChildTAG: "electronics workbench" is a powerful soft for modelling of circuits with real elements. I used it to check my calculations for homeworks and midterm exam. FirstChildUserIdTAG: 323416 FirstChildUserNameTAG: Alexxkr FirstChildCreateTimeTAG: 2012-11-22T05:50:48Z FirstChildTAG: There is a free program called TINA-TI which can be downloaded from Texas Instruments' website: http://www.ti.com/tool/tina-ti. It features schematic capture and simulation. It also comes with a large library of SPICE models. FirstChildUserIdTAG: 118046 FirstChildUserNameTAG: Bill24 FirstChildCreateTimeTAG: 2012-12-04T17:50:04Z FirstChildTAG: There is also the [NGSPICE Online simulator][1] that allows you to simulate circuits using only your browser - no install required. :-) [1]: http://www.ngspice.com/ FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-12-14T01:42:39Z FirstChildTAG: Lab 0 became my favourite :) FirstChildUserIdTAG: 208296 FirstChildUserNameTAG: OZ1 FirstChildCreateTimeTAG: 2012-12-22T23:26:45Z IndexTAG: 906 TitleTAG: H12P2: Anybody successfully completed the problem? I have spent almost 4 hours on this problem. No matter what I put in the last four answer areas, I cannot get green check marks. My solution approach was to solve for R0 and R1/R2, then pick an R2 to compute R1. I wrote two equations for 10V input to 4.9V output and 20V input to 5.1V outputs. So I had two equations and two unknowns. I think this should have worked, but I cannot pass the grader. And I computed the efficiency as power output (vO^2/Rl) divided by power input (from both vIN and Op Amp). UserIdTAG: 153760 UserNameTAG: Kavka CreateTimeTAG: 2012-11-21T00:08:07Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: Same problem! Try all day, but the grader won't accept the changes ... Did you notice that staff added this to H12P2? **NOTE: Due to how these answer fields depend on each other, "H12P2 Linear Regulator" requires all 6 fields to contain answers for the "check" button to work. This means that writing the correct answer in the first two fields can yield a red X if the last part of the problem has not yet been answered. Also, as the last part of this problem is a design problem, the final four responses are related to each other; hence all four responses entered must agree with each other in order to receive a checkmark.** Update: read my complete findings in my new post: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50ae30a5a2185c2700000014 FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-11-21T00:26:37Z SecondChildTAG: Yes, I put all four numbers at once, but still no check marks. I tried to double check my answers every way I know, but no success with the grader. SecondChildUserIdTAG: 153760 SecondChildUserNameTAG: Kavka SecondChildCreateTimeTAG: 2012-11-21T00:48:09Z SecondChildTAG: Don't over think the problem! In computing the efficiency there are losses in the two voltage dividers as well as the Op Amp and BJT. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-11-21T00:58:18Z SecondChildTAG: G@R!R*! The grader doesn't accept my changes.... If I put 1/-0 in the first box and all the other boxes filled with the answers, it says it can't evaluate 1/-0, that's ok. But when I change it to the correct answer, press the check button, it says nothing! and when I refresh the page, I get my 1/-0 in the first box back ... SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-11-21T11:42:48Z SecondChildTAG: BTW: I'm sure it has something to do with the grader. I think it is because the s/w to grade this, is a bit different from the normal s/w because the answers depend on each other now. This problem is so simple, that you don't need a calculator and certainly not for the first 2 formulas. Somehow it's not updating the changes. Maybe when one puts in the correct answers at the first time, one won't encounter this problem. But because I had the problem with the wrong capitals, it might some how be related. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-11-21T11:52:23Z FirstChildTAG: Can anyone give hints regarding this question! I can't pass through it :( FirstChildUserIdTAG: 277808 FirstChildUserNameTAG: Hemanthmps FirstChildCreateTimeTAG: 2012-11-21T17:26:42Z FirstChildTAG: There are hints earlier in the thread. What is your thinking? What have you tried? FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-21T18:09:48Z FirstChildTAG: Are we supposed to get exactly 4.9V and 5.1 or these are min/max ? I have 3 resistors that give an output voltage of 5.001V (Vin=10V) or 5.003V (Vin=50V) but the answer is not correct. I don't understand why except if we want a worse regulator... FirstChildUserIdTAG: 396446 FirstChildUserNameTAG: RousseauxS FirstChildCreateTimeTAG: 2012-11-23T13:09:20Z SecondChildTAG: I believe that it's all or nothing. Either the regulator works within tolerance and the efficiency is calculated correctly or all answers for the design are marked wrong. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-11-24T00:31:03Z SecondChildTAG: Thanks! It works with my values and the correct efficiency ; ). SecondChildUserIdTAG: 396446 SecondChildUserNameTAG: RousseauxS SecondChildCreateTimeTAG: 2012-11-24T10:40:47Z FirstChildTAG: ok It works now. Some hints : The opamp's inputs consume absolutely zéro current, so you can choose R0 greater than the value that gives vout = 5.0 V and 5.1V ( if you see what i mean , it begins by 19. :) The precision will be greater, and power loss smaller. For same reason, you can choose the divider R2/(R1+R2) with greater values for less power loss. output power is in the load total input power = v(in)*i(in) + ( power coming out of the opamp ) efficiency= Power out/Power in FirstChildUserIdTAG: 373752 FirstChildUserNameTAG: Croak FirstChildCreateTimeTAG: 2012-11-28T02:02:58Z SecondChildTAG: I get Ro = 94. What did I do wrong? SecondChildUserIdTAG: 245291 SecondChildUserNameTAG: rlicas SecondChildCreateTimeTAG: 2012-12-06T19:28:24Z IndexTAG: 907 TitleTAG: Differential equation solution Because of the simple form of the equation it is easy to use differential equations. First we have to write i or v equation form by using node analysis. In this case: ![enter image description here][1] Now change the derivatives in to equivalent power of them: ![enter image description here][2] Then solve the equation for i: ![enter image description here][3] If i had one or two real answer(s), (no j appears in the answers) the homogenous part would have the following figure: ![enter image description here][4] But here we have i1=0+w0*j, i2=0-w0*j so the real part is ‘0’ (R) and the imaginary part is w0 (J): So the form of homogeneous part should be as fallows: ![enter image description here][5] Which in this case we have: ![enter image description here][6] And the partial part is easy to write because the left side of the equation is constant. ![enter image description here][7] Thus: ![enter image description here][8] So the delta would give us the total shape the answer and you don’t need to know the rest of it! Now just use initial condition to solve for As and you will have the total answer. [1]: https://edxuploads.s3.amazonaws.com/13534517511343687.bmp [2]: https://edxuploads.s3.amazonaws.com/13534518221343686.bmp [3]: https://edxuploads.s3.amazonaws.com/13534518441343626.bmp [4]: https://edxuploads.s3.amazonaws.com/13534518976120091.bmp [5]: https://edxuploads.s3.amazonaws.com/13534524681343639.bmp [6]: https://edxuploads.s3.amazonaws.com/13534526231343627.bmp [7]: https://edxuploads.s3.amazonaws.com/13534526641343678.bmp [8]: https://edxuploads.s3.amazonaws.com/13534526831343609.bmp UserIdTAG: 420339 UserNameTAG: AliJenabi CreateTimeTAG: 2012-11-20T23:05:32Z VoteTAG: 3 CoursewareTAG: Week 9 / An ILC circuit CommentableIdTAG: 6002x_an_ILC_circuit NumberOfReplyTAG: 0 IndexTAG: 908 TitleTAG: Laplace Transform Hi, I was trying to analyze the LC circuit below using the Laplace transform. I am relatively new to Laplace Transforms, so I would be grateful if someone could let me know whether the method I have used is correct and whether there is an easier way. Circuit ![Series LC Circuit][1] Variables v = Voltage across capacitor i = current through the circuit Initial Conditions $v(0)=0,\ i(0)=0$ Differential Equation $LC\frac{d^2v}{dt^2}+v=V_I$ Laplace Transform $\mathscr{L}\left\{LC\frac{d^2v}{dt^2}+v\right\}=\mathscr{L}\left\{V_I\right\}$ $\therefore LC\left(s^2\mathscr{L}\left\{v\right\} - sv(0) - v'(0)\right) + \mathscr{L}\left\{v\right\} = \frac{V_I}{s}$ Because of initial conditions $\therefore LCs^2\mathscr{L}\left\{v\right\} + \mathscr{L}\left\{v\right\} = \frac{V_I}{s}$ $\therefore \left(LCs^2+1\right)\mathscr{L}\left\{v\right\} = \frac{V_I}{s}$ $\therefore \mathscr{L}\left\{v\right\} = \frac{V_I}{s\left(LCs^2+1\right)}$ $\therefore v = \mathscr{L}^{-1}\left\{ \frac{V_I}{s\left(LCs^2+1\right)} \right\}$ $\therefore v = \mathscr{L}^{-1}\left\{ \frac{V_I}{s}\right\}*\mathscr{L}^{-1}\left\{\frac{1}{\left(LCs^2+1\right)}\right\}$ $\therefore v = V_I*\mathscr{L}^{-1}\left\{\frac{1}{LCs^2+1}\right\}$ $\therefore v = V_I*\mathscr{L}^{-1}\left\{\frac{\frac{1}{LC}}{s^2+\frac{1}{LC}}\right\}$ $\therefore v = V_I*\mathscr{L}^{-1}\left\{\frac{\sqrt{LC}}{LC}\frac{\sqrt{\frac{1}{LC}}}{s^2+\frac{1}{LC}}\right\}$ $\therefore v = V_I*\frac{\sqrt{LC}}{LC}\mathscr{L}^{-1}\left\{\frac{\sqrt{\frac{1}{LC}}}{s^2+\frac{1}{LC}}\right\}$ $\therefore v = V_I*\frac{\sqrt{LC}}{LC}sin(\omega _0t),\ \omega _0=\sqrt{\frac{1}{LC}}$ Solving Convolution $v = V_I*\frac{\sqrt{LC}}{LC}sin(\omega _0t)$ $\therefore v = \frac{\sqrt{LC}}{LC}sin(\omega _0t)*V_I$ $\therefore v = \frac{\sqrt{LC}}{LC}\int_0^tsin(\omega _0(t-\tau))\cdot V_I(\tau) d\tau$ $\therefore v = \frac{\sqrt{LC}}{LC}\int_0^tsin(\omega _0(t-\tau))\cdot V_I(\tau) d\tau$ Integration By Parts $\therefore v = \frac{\sqrt{LC}}{LC}\left[\frac{V_I(\tau)}{\omega_0}cos(\omega_0(t-\tau)) - \frac{1}{\omega_0}\int 0\times cos(\omega_0(t-\tau))d\tau\right]_0^t$ $\therefore v = \frac{\sqrt{LC}}{LC}\left[\frac{V_I}{\omega_0}cos(\omega_0(t-t)) - \frac{V_I}{\omega_0}cos(\omega_0(t-0))\right]$ $\therefore v = \frac{\sqrt{LC}}{LC}\frac{V_I}{\omega_0} - \frac{\sqrt{LC}}{LC}\frac{V_I}{\omega_0}cos(\omega_0t)$ $\therefore v = \frac{\sqrt{LC}}{LC}V_I\sqrt{LC} - \frac{\sqrt{LC}}{LC}V_I\sqrt{LC}cos(\omega_0t)$ $\therefore v = V_I - V_Icos(\omega_0t)$ Final Answer $v = V_I - V_Icos(\omega_0t)$ [1]: https://edxuploads.s3.amazonaws.com/1353420682134360.png UserIdTAG: 85490 UserNameTAG: yashsavani CreateTimeTAG: 2012-11-20T14:56:04Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Its usually a lot easier to break the formula for u(s) into simple expressions in the s domain than to calculate the convolution in the time domain. $U(s)=\frac{\frac{V_I}{LC}}{s(s^2+\frac{1}{LC})} = \frac{A}{s}+\frac{Bs+C}{s^2+\frac{1}{LC}}$ and by calculating A B and C (refresh your calculus ;) ) you get A = Vi, B = -Vi, C = 0 so $U(s)=\frac{\frac{V_I}{LC}}{s(s^2+\frac{1}{LC})} = \frac{V_I}{s}-\frac{V_Is}{s^2+\frac{1}{LC}}$ and from there you just take the inverse transform and get $u(t) = V_I-V_Icos(\frac{1}{\sqrt{LC}}t)$ Also note that generally when you find $U(s) = Y(s)V_I(s)$ the denominator of Y(s) is the characteristic equation of the model and its roots end up determining the form of the solution. FirstChildUserIdTAG: 370344 FirstChildUserNameTAG: KostisL FirstChildCreateTimeTAG: 2012-11-20T16:28:52Z SecondChildTAG: Thank you very much. This has helped me a lot. SecondChildUserIdTAG: 85490 SecondChildUserNameTAG: yashsavani SecondChildCreateTimeTAG: 2012-11-20T17:47:22Z IndexTAG: 909 TitleTAG: H10P3Q3&4 How to calculate Cmatch and Lmatch and could anyone please rephrase or explain the question better Thnx.. UserIdTAG: 425976 UserNameTAG: P_V_Rohith CreateTimeTAG: 2012-11-19T22:45:47Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Perhaps my comments in this thread will help: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50a688f55b77f62300000035 FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-19T23:03:13Z SecondChildTAG: thnx :) SecondChildUserIdTAG: 425976 SecondChildUserNameTAG: P_V_Rohith SecondChildCreateTimeTAG: 2012-11-20T09:44:56Z FirstChildTAG: Another way to look at this problem: We know that if you have a non-ideal voltage source, then max power is delivered to the load, when Rload=Rthevenin. The same is (almost) true when Rthevenin and Rload are complex, call them Zt and Zl. However, this is only true when you use the conjugate of jb. If Zt=a+jb, then Zl should be Zl=a-jb, not Zl=a+jb. The point is, that always abs(Zt)=abs(Zl), for abs(a+jb)=abs(a+jb) but also for abs(a+jb)=abs(a-jb)! The reason for using the conjugate is, that the imaginary parts are disappearing at the desired frequency, and that is why the max power is transferred, but ONLY at that specific frequency. So what you want, is that you transfer max REAL power from Zt to Zl, and you have to take care that the REAL parts of the POWER in Zt and Zl become the same, but also that the complex parts of the power are the same. So then you get Zt=a+jb and Zl=c-jd and the power Pa=Pc and abs(Pb)=abs(Pd). What you can do, is converting a circuit Zt with resistor=a parallel with an imaginary component=jb to an equivalent circuit Zl consisting of a resistor=c with in series a imaginary component=jd.(and also the other way) Hope this helps. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-11-20T20:16:04Z FirstChildTAG: All very nice, but the problem simply states: **It is possible to find values of L and C that make the driving-point impedance you just computed exactly R1, if R1>R2. This will "match the antenna to the amplifier".** That's all one needs to know to answer the question. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-20T20:23:57Z IndexTAG: 910 TitleTAG: H12P2 : VZ not permitted in answer Hi, I've just tried to complete the last Homework but I'm puzzled with the H12P2 (b). It is asked to give a relation for the output voltage depending on $V_Z$ but when I check my answer I get : "VZ not permitted in answer" : so what ? : ) Thanks! Note: same problem with the (a) part, we must use Vin but it is not allowed... UserIdTAG: 396446 UserNameTAG: RousseauxS CreateTimeTAG: 2012-11-19T21:32:34Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: I have the same issue. I have tried with different letter sizes, but I get still the same error. FirstChildUserIdTAG: 146390 FirstChildUserNameTAG: JGradzki FirstChildCreateTimeTAG: 2012-11-19T21:59:36Z SecondChildTAG: Me too, i got the same problems with Vin and Vz. Where to fill in the values for R0, R1, R2 are not clear either. I guess it has to be in the same order? SecondChildUserIdTAG: 107802 SecondChildUserNameTAG: protonman71 SecondChildCreateTimeTAG: 2012-11-20T09:20:38Z SecondChildTAG: Your value for R0 should go in the first blank, R1 in the second, and R2 in the third. SecondChildUserIdTAG: 391929 SecondChildUserNameTAG: RohanNagarkar SecondChildCreateTimeTAG: 2012-11-20T15:56:45Z FirstChildTAG: Problem 2 of Homework 12 seems to require that all fields contain an answer before the grader checks it correctly. FirstChildUserIdTAG: 347268 FirstChildUserNameTAG: atratrs FirstChildCreateTimeTAG: 2012-11-20T22:00:59Z SecondChildTAG: Yes, that is exactly what I had in mind. So I solved and put the answers in all the boxes, but still nada! No green/red marks ... And no error notification! SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-11-20T23:52:09Z FirstChildTAG: With the proper symbols for voltages and answers in all the boxes I am able to get the check marks. Thanks to all for the information. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-21T00:09:52Z SecondChildTAG: hello, skyhawk: Can you post your answers in H12P2 with the process. I have some puzzle about this problem. Thank you for your help. Rainbow. SecondChildUserIdTAG: 371617 SecondChildUserNameTAG: rainbow12 SecondChildCreateTimeTAG: 2012-12-10T22:31:10Z FirstChildTAG: Dear all, "vz" and "vin" should work. Sorry for the confusion with variable capitalization, in the problem prompt both "vin" and "vz" were incorrectly shown to be "Vin" and "Vz." FirstChildUserIdTAG: 391929 FirstChildUserNameTAG: RohanNagarkar FirstChildCreateTimeTAG: 2012-11-20T15:55:51Z SecondChildTAG: I still can't get a checkmark. The "Check" button does nothing (it works in the other problems but not this one). When using $V_Z$ I get my error but not with $v_z$. SecondChildUserIdTAG: 396446 SecondChildUserNameTAG: RousseauxS SecondChildCreateTimeTAG: 2012-11-20T17:59:20Z SecondChildTAG: Same with me, the check button doesn't do anything ... I use vin and vz and don't get an error message, but no red/green tics either .... SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-11-20T18:50:02Z SecondChildTAG: I put bogus values in the part (c)/part (d) boxes and then the Check button worked. I hadn't filled them out before as I was checking out the problems with the "Check" button. A random guess is that the check code is dividing by zero somewhere... so try reloading the page, then non-zero values in the part (c) and (d) boxes. SecondChildUserIdTAG: 141000 SecondChildUserNameTAG: OrinE SecondChildCreateTimeTAG: 2012-11-20T19:49:42Z SecondChildTAG: I filled in all the boxes of H12P2, but still no marks. Even worse: when I later close the browser and start with a new browser, than I don't see the last entered values in the boxes, but the values from this afternoon. So somehow the grader doesn't do what it is supposed to do .... SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-11-20T21:35:59Z SecondChildTAG: Well, the grader is doing something for me, I think... but it doesn't like my values for R0,R1,R2. I assumed we are supposed to model the zener by a 2.5V voltage source in series with a 1ohm resistor and so on. Doing that, I get an R0 that will give the right outputs for the given values of vin and appropriate R1,R2, but the grader doesn't like it. SecondChildUserIdTAG: 141000 SecondChildUserNameTAG: OrinE SecondChildCreateTimeTAG: 2012-11-20T22:15:49Z SecondChildTAG: That should have been a $v_z$ voltage source in series with an $R_z$ resistor - I guess everyone doesn't get the same $v_z$ and $R_z$. SecondChildUserIdTAG: 141000 SecondChildUserNameTAG: OrinE SecondChildCreateTimeTAG: 2012-11-20T22:23:38Z IndexTAG: 911 TitleTAG: Hint for H9P1 Q1 Hi all, sorry I haven't posted this sooner, but wolfram has a great LC circuit calculator. http://www.wolframalpha.com/input/?i=resonance+frequency+20uF%2C+5H&lk=3 Just plug in L and C. Hazel. UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-11-18T15:42:02Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Thank you for sharing this hazel1919!:) Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-18T21:22:26Z SecondChildTAG: :) SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-19T07:57:05Z IndexTAG: 912 TitleTAG: Are we fewer than before the midterm?? Is it just me, or there is indeed a smaller number of us that are still here after the midterm? Can anyone provide us with any statistics about successful completion of the course and the progress of the student's rates? UserIdTAG: 304587 UserNameTAG: Vasso CreateTimeTAG: 2012-11-17T23:33:50Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: I am also curious, I would speculate that there is less then 25% of those who enrolled. I would imagine Professor Agarwal will mention it after the course is over. On a side note, did that one student ever finish his survey of the midterm? Sorry his name eludes me at the moment. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-11-17T23:51:27Z SecondChildTAG: I can't recall his name either.. I guess we will know after the course is over! :) Some weeks ago i found an article on the net that stated that, many students do not make it till the end of the course, nevertheless the final exam's scores of the remaining students are higher than their midterm was! This is encouraging! :) SecondChildUserIdTAG: 304587 SecondChildUserNameTAG: Vasso SecondChildCreateTimeTAG: 2012-11-18T00:28:57Z SecondChildTAG: so yeah I scored a 100% in midterm... Does that means I'm going to get like 101% in the finals? :D I like the sound of it! SecondChildUserIdTAG: 499671 SecondChildUserNameTAG: maliha266 SecondChildCreateTimeTAG: 2012-11-23T16:51:03Z FirstChildTAG: *This was the official Statistics of **6.002x Spring 2012** at the end of the Course:* Course statistics: 6.002x had 154,763 registrants. Of these, 69,221 people looked at the first problem set, and 26,349 earned at least one point on it. 13,569 people looked at the midterm while it was still open, 10,547 people got at least one point on the midterm, and 9,318 people got a passing score on the midterm. 10,262 people looked at the final exam while it was still open, 8,240 people got at least one point on the final exam, and 5,800 people got a passing score on the final exam. Finally, after completing 14 weeks of study, 7,157 people have earned the first certificate awarded by MITx, proving that they successfully completed 6.002x. ---- I were one of the 7,157 fortunately students of 6.002x 2012 that got that Certificate last Spring, I am really proud of that, 6.002x was and is really amazing. I am really happy :). Yes, we have noticed the same in 6.002x Spring after the Midterm Exam. But, believe it or not, we are still in contact with some students via the old Forum. Although the Course have already ended in June 2012, we still debating things and giving our point of view, is like a big family! That is really strange! I don´t know if that will happen here after this Course ends... I hope that yes :), that you can keep in touch all together. Also It would be nice that some of you can get involved with 6.002x in someway, like helping in the future 6.002x 2013 as I am doing here this Fall ;), I hope that you can feel inspired and give that big step in the next year, I am sure that many new students will need your help, I encourage to feel confident to do that ;). See you Vasso, Myriam. P.D. I will like to take the Proctored Exam of 6.002x at the end of the next year ;). FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-18T03:02:11Z SecondChildTAG: Thanks Myriam. Gracias (I hope that is correct for "thank you") SecondChildUserIdTAG: 623210 SecondChildUserNameTAG: preveen SecondChildCreateTimeTAG: 2012-11-18T11:04:17Z SecondChildTAG: Thank you very much Myriam!! :) That was really inspiring!! I was thinking about taking the course again next time it is available so that i can also help others! Furthermore i would like to better my grade as i did not have the time that was needed and i already have lost 8 per cent of the final grade. I don't know if i will get a passing score after the final exam, but i will try to do my best. I am just so surprised about the fact that from all those people that were willing to follow the course, only a small percentage of them finally made it! I know that circuits are not an easy subject to study, but i can't understand why someone would not "take advantage" of such a great teacher, that offers his knowledge for free! On the other hand, i am an Electrical Engineering Educator myself (a new one, but still)and i use this opportunity so that i can be a better teacher. So i have a special interest in Dr. Agarwal's methods :) I hope i will be able to help as much as i can :) SecondChildUserIdTAG: 304587 SecondChildUserNameTAG: Vasso SecondChildCreateTimeTAG: 2012-11-18T11:25:56Z SecondChildTAG: Hi @preveen, Yes, that is correct in spanish haha. Por nada :)(you are welcome in spanish)! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-18T14:52:44Z SecondChildTAG: Hi @Vasso, I am sure you will succeed in this Course :). I am happy that you are feeling inspiring, you will help a lot in 6.002x Spring 2013 :)! My best wish to you, Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-18T14:55:04Z SecondChildTAG: Thank you Myrimit, you have helped me along the way even when I almost gave up. I am not as smart as most of the people that take this course I took it because it is something that I am truly interested in. I have made a few friends from this class and hope to keep in touch. Everyone that has taken this course has been helpful and not judging people on some of the questions that have been asked, and explaining problems that come up in a way that makes sense. This has been a great class with a great professor and I am glad that I took this class. SecondChildUserIdTAG: 160242 SecondChildUserNameTAG: Mlevins35 SecondChildCreateTimeTAG: 2012-11-18T18:57:30Z SecondChildTAG: You are welcome Mlevins35! Yes, this is an amazing Class, I am glad to be here too :). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-18T21:02:43Z FirstChildTAG: For some of the reasons Vasso enumerated, I will probably be here next semester as well. I don't think I can provide the level of help Myrimit does, but I will help as I can. :-) FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-11-18T17:15:29Z SecondChildTAG: Cool planetscape! That would be awesome! I am happy for that! I am sure that you will help a lot there and many students will be grateful with you :). My best wish to you! Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-18T21:05:17Z SecondChildTAG: Thank you! My best wishes to you, as well. :-) I do hope you continue to hang around the forums, though. :-) SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-11-18T23:57:43Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-19T13:20:27Z FirstChildTAG: as we are moving towards the end of the course , hope that i will complete the course successfully... ------ and a big thanks to @Myrimit,@skyhawk and many more..i will remember this experience throughout my life..hope our MITx family goes bigger and bigger..and always remain in touch with each other through this platform.. FirstChildUserIdTAG: 114864 FirstChildUserNameTAG: Vikaash FirstChildCreateTimeTAG: 2012-11-19T08:11:29Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-19T13:20:57Z IndexTAG: 913 TitleTAG: I am not quite sure if the teacher was right or wrong, but I have an alternative proposition Hello, If we think about this circuit as an anloguous case to the previously mentioned circuit, then we should find the solution such that we seek to find the current passing by the inductance. Let me give my proposition (I hope I didn't made any stupid mistake): ![my proposition of the initial 2nd Order DE][1] [1]: https://edxuploads.s3.amazonaws.com/13531144004409834.png We can notice easily that every term is analoguous to the previously mentioned circuit: 1. vC in the first circuit is analoguous to iL. 2. R in the first circuit is analoguous to 1/R. 3. L is analoguous to C. 4. C is analoguous to L. 5. V is analoguous to I. UserIdTAG: 418182 UserNameTAG: euldji2005 CreateTimeTAG: 2012-11-17T01:07:44Z VoteTAG: 3 CoursewareTAG: Week 9 / Driven_Parallel_RLC_Circuit CommentableIdTAG: 6002x_driven_parallel_RLC_circuit NumberOfReplyTAG: 1 FirstChildTAG: Thank you for this analysis. The teacher's solution for voltage here is also correct (though there was a smal mistake in the analysis). So, I guess the formula for voltage here should be analogous to the formula for current for the previous (Series-RLC)circuit. :) FirstChildUserIdTAG: 714237 FirstChildUserNameTAG: PaxPolaris FirstChildCreateTimeTAG: 2012-11-19T01:45:52Z IndexTAG: 914 TitleTAG: H10P3 - An L Network question So I figured out the first two parts of this, and got the driving point impedance. When I set this equal to R1 (as per the question), I get an equation for C-match that includes R1, R2, w and L. Similarly I get an equation for L-match that includes R1, R2, w and C. But how do I get to a value for C-match and L-match that does not depend on the other one?? UserIdTAG: 145194 UserNameTAG: DerekH CreateTimeTAG: 2012-11-16T18:41:57Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 7 FirstChildTAG: Use any technique for solving simultaneous linear equations for L and C. For example, solve the equation for C in terms of R1, R2, w, and L. Then substitute into the equation for L and solve in terms of R1, R2, and w. Finally substitute back into the equation for C to obtain C in terms of R1, R2, and w. Or in the opposite order, if that's more convenient. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-16T19:32:33Z SecondChildTAG: derekh and skyhawk, thank u very much for ur hints of obtaining C and L SecondChildUserIdTAG: 414616 SecondChildUserNameTAG: local_hero SecondChildCreateTimeTAG: 2012-11-26T07:02:55Z FirstChildTAG: OK - I get that, but I have one equation ("R1 = driving point impedance", which I got a correct answer on) and two unknowns (L and C). Solving for one (C, for example) and then substituting that answer for C and solving for L will just take me round in a big circle. Or is there another equation linking L and C that I am missing? Thanks for the quick response by the way, Skyhawk. Much appreciated. FirstChildUserIdTAG: 145194 FirstChildUserNameTAG: DerekH FirstChildCreateTimeTAG: 2012-11-16T20:50:19Z FirstChildTAG: Set your **complex** impedance equal to R1. That gives you two equations, one for the real part and one for the imaginary part. Two equations, two unknowns ... solve simultaneously. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-16T22:02:06Z SecondChildTAG: Hi, I'm in the loop! I'm stuck! I don't understand how Zth = R1 can give 2 ec. These are my loop results: C = -(I*L*w-R1+R2)/((L*w-I*R2)*R1*w), L = L Please help! SecondChildUserIdTAG: 299251 SecondChildUserNameTAG: TheRedBlackOne SecondChildCreateTimeTAG: 2012-11-18T14:21:32Z SecondChildTAG: wait I dont get it how do u know which part is the real part and which part is the imaginary part???? SecondChildUserIdTAG: 435197 SecondChildUserNameTAG: RichmondRichter SecondChildCreateTimeTAG: 2012-11-26T10:05:44Z FirstChildTAG: OK RedBlack Have you gotten a check mark for the second part of the problem? My expression for the complex impedance is a fraction that I will write as N/D, where N and D are both complex. Let N/D = R1. Clear fractions to get N = R1*D Therefore: Re(N) = Re (R1*D) and Im(N) = Im(R1*D) The second equation immediately gives L as a function of C, R1, and R2. Substitute this expression for L into the first equation and solve for C in terms of R1 and R2. Use that in the second equation to solve for L in terms of R1 and R2. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-18T23:59:11Z SecondChildTAG: Skyhawk, Thank you very much for this help. I could not have finished this without it. SecondChildUserIdTAG: 428560 SecondChildUserNameTAG: MikeDayton SecondChildCreateTimeTAG: 2012-11-19T05:15:49Z SecondChildTAG: Ok, skyhawk, Thanks a lot, I have gotten a green check, I have done exercices 2 and 3! but I'm lost in exercice 1, I suppose that's easy because people don't ask about it, can you help me with any hint? SecondChildUserIdTAG: 299251 SecondChildUserNameTAG: TheRedBlackOne SecondChildCreateTimeTAG: 2012-11-19T09:31:57Z SecondChildTAG: TheRedBlackOne, are you lost on part one of H10P3? If yes, It is just one voltage divider inside another one... lot of fun with algebra :) the pattern is the same as the exercise on page 724 of the text book. If your doubt is about H10P1, you have to match magnitude plot with the phase plot. For each circuit figure out if the magnitude goes up or down (low $\omega$ and high $\omega$ analysis). Also figure out what is the angle for low/high $\omega$. Calculate $\omega$ for the break frequency. This will tell you if it is the left or right plot... Also note that if you think that the **p** magnitude graph matches the **u** phase graph then your answer should be **p∗u** literally. SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-20T11:54:54Z SecondChildTAG: I think we got that all worked out yesterday here: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50a9f563608f322300000017 SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-11-20T15:53:37Z SecondChildTAG: Skyhawk, Where can I find information on the difference between the real and imaginary parts of the complex impedance? I cannot find this is the text. I got H10P3 b correct, but I don't understand what you mean by the real part of the numerator and denominator and the imaginary parts. Any help would be appreciated. SecondChildUserIdTAG: 145915 SecondChildUserNameTAG: WhoLetTheDougsOut SecondChildCreateTimeTAG: 2012-11-23T15:26:21Z SecondChildTAG: Can't I just set the solution for H10P3(b) = R1 and then solve for C and then for L? SecondChildUserIdTAG: 145915 SecondChildUserNameTAG: WhoLetTheDougsOut SecondChildCreateTimeTAG: 2012-11-23T15:27:34Z SecondChildTAG: Dear Skyhawk, Help needed, sa you explained for H10P3 Q C, Cmatch, The denominator of the complex Impedance also carries 1, that is somewhat like (1-XXw^2+XXjw). Now which part will be Re and Img among these. Whether removal of 1 will affact our calculation? SecondChildUserIdTAG: 455950 SecondChildUserNameTAG: Wahabbaluch SecondChildCreateTimeTAG: 2012-11-23T18:48:36Z SecondChildTAG: Thanks a lot for your hint, Skyhawk! I was completely lost until I saw it. SecondChildUserIdTAG: 324931 SecondChildUserNameTAG: pgromovikov SecondChildCreateTimeTAG: 2012-11-23T18:50:44Z SecondChildTAG: Thanks alot! SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-11-23T20:50:50Z SecondChildTAG: Thank you skyhawk for help on this question! Wahabbaluch, make the answer of part 2 equals to R1. So you will have something like ... $\frac{RealPart1+j(ImaginaryPart1)}{RealPart2+j(ImaginaryPart2)}=R1$, pass the denominator to the right side. Then you will be able to get two equations, RealPart1=RealPart2 and ImaginaryPart1=ImaginaryPart2 (you have two unknowns) SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-23T22:08:52Z SecondChildTAG: Thanks a lot skyhawk SecondChildUserIdTAG: 452559 SecondChildUserNameTAG: kuz1toro SecondChildCreateTimeTAG: 2012-11-25T19:27:28Z SecondChildTAG: skyhawk, you are truly awesome! Thanks a lot for the hints. SecondChildUserIdTAG: 194509 SecondChildUserNameTAG: blackcompe SecondChildCreateTimeTAG: 2012-11-25T23:45:06Z FirstChildTAG: thanks for your help guys (especially skyhawk) FirstChildUserIdTAG: 435193 FirstChildUserNameTAG: ManosP FirstChildCreateTimeTAG: 2012-11-25T23:08:36Z FirstChildTAG: The value for Cmatch is asked in pF, but the system is returning wright in nF, instead. Is it me, or the staff should correct this detail? Congrats to everybody and thank you all for the invaluable help! FirstChildUserIdTAG: 381024 FirstChildUserNameTAG: QZ FirstChildCreateTimeTAG: 2012-11-24T10:36:28Z SecondChildTAG: I agree with Qz. It is wright in nF!! SecondChildUserIdTAG: 90765 SecondChildUserNameTAG: jdavidg SecondChildCreateTimeTAG: 2012-11-25T00:27:23Z SecondChildTAG: Seems that now is OK. System accept answer in pF. Thanks! SecondChildUserIdTAG: 325197 SecondChildUserNameTAG: Vitali_Jerin SecondChildCreateTimeTAG: 2012-11-25T09:26:50Z FirstChildTAG: Thanks Skyhawk :) FirstChildUserIdTAG: 64656 FirstChildUserNameTAG: AbuZar FirstChildCreateTimeTAG: 2012-11-25T11:26:33Z IndexTAG: 915 TitleTAG: I was able to register.But is closing enrollment date is over?? Closing Enrollment: Although we usually like to give students the chance to do impressive and unlikely feats of learning, the due date for **week 7 assignments marks the time** when it is impossible to earn enough points for a certificate, and we will therefore not be accepting any new registrations for the course. Good luck to everyone who is still in the running! Now what I do..... **Can I enroll later.. or apply for a certificate later**??.. I enrolled today... but home work 7 assingmet due was on november 4th... what shall I do.. will I be able to submit homeworks from beginning (Hw1 onwards).. what should I follow next.. If my enrollment is canceled now, will I still be able to access edx? UserIdTAG: 715014 UserNameTAG: rosh007 CreateTimeTAG: 2012-11-16T13:54:15Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: you are late rosh007 .however you can come back in the spring session :) FirstChildUserIdTAG: 357747 FirstChildUserNameTAG: kishores FirstChildCreateTimeTAG: 2012-11-16T14:26:41Z SecondChildTAG: :0 So what I have to do for that SecondChildUserIdTAG: 715014 SecondChildUserNameTAG: rosh007 SecondChildCreateTimeTAG: 2012-11-16T14:59:32Z SecondChildTAG: and when (which month) spring session will begin? SecondChildUserIdTAG: 715014 SecondChildUserNameTAG: rosh007 SecondChildCreateTimeTAG: 2012-11-16T15:00:47Z SecondChildTAG: We don't have a spring schedule yet. If it's like the last class, it would be March, I'm not sure if that's a good guide or not. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-11-16T15:04:48Z FirstChildTAG: Due to the overwhelming number of people that are interested in looking at 6.002x material before the start of next semester, we have temporarily re-opened registration. However, it is impossible for new students to earn enough points for a certificate, and if you would like to participate in the course, you will need to wait until next semester. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-11-16T15:17:12Z SecondChildTAG: Do we have any idea of when the Spring course will be? SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-11-16T15:26:58Z FirstChildTAG: Hi, [Check This Thread][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50a3f629fc2c2c2b00000058 > **Regards:** asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-11-20T10:35:29Z IndexTAG: 916 TitleTAG: What just happened? Maths occurred. I have absolutely no idea what just happened! :S UserIdTAG: 56024 UserNameTAG: onidaito CreateTimeTAG: 2012-11-16T12:23:46Z VoteTAG: 3 CoursewareTAG: Week 10 / Particular Solution To Cosine Input Part 2 CommentableIdTAG: 6002x_Particular_solution_to_cosine_input_part_2 NumberOfReplyTAG: 1 FirstChildTAG: A) I am guessing you are English, because you said "Maths". B) You are English and yet you have not understood the central message of Hitchhikers Guide to the Galaxy, i.e. Don't Panic. C) There are lots of online calculators that will do all the stuff for you. Ones specifically for things like magnitude and phase and ones like Wolfram Alpha that will do all kinds of stuff. Read this: http://www.physicsforums.com/showthread.php?t=315513 D) If you are not English then please become so as soon as possible. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-11-16T16:07:48Z IndexTAG: 917 TitleTAG: Lab10 grader error? ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13530635683920288.png Made a calculation, builds a circuit, made AC analys. Always gets error for last question. UserIdTAG: 394836 UserNameTAG: v2g6ch4 CreateTimeTAG: 2012-11-16T11:00:42Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: You have to be quite accurate. FirstChildUserIdTAG: 142857 FirstChildUserNameTAG: viking2 FirstChildCreateTimeTAG: 2012-11-16T11:11:19Z FirstChildTAG: Same problema here man. FirstChildUserIdTAG: 98262 FirstChildUserNameTAG: rafaelmarques FirstChildCreateTimeTAG: 2012-11-22T01:57:06Z FirstChildTAG: I am struggling with the same problem. The AC analysis gives exactly the same characteristic like it is required, but it is not accepted as a correct answer. Have any of you succeed to get rid of this problem? FirstChildUserIdTAG: 319324 FirstChildUserNameTAG: SIZUSKA FirstChildCreateTimeTAG: 2012-11-25T21:13:09Z IndexTAG: 918 TitleTAG: Video is unavailable I'm able to download, but not to watch only this video from site. UserIdTAG: 378778 UserNameTAG: happylife CreateTimeTAG: 2012-11-15T13:01:15Z VoteTAG: 3 CoursewareTAG: Week 9 / Total Solution Part 1 CommentableIdTAG: 6002x_total_solution_part_1 NumberOfReplyTAG: 1 FirstChildTAG: I can view it. Have you tried re-loading? Have you tried another browser? What about the next video, does it also say "unavailable"? FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-11-15T16:14:10Z IndexTAG: 919 TitleTAG: Laplace Hi, I was just wondering that it would have been much easier to use the Laplace transform to solve the differential equations over the Homogenous and Particular method. In addition the Laplace transform also caters to several of the forms of input that we use, for example we could use it to solve for the step, impulse and ramp inputs. So why then do we not use it in this course? Thank You in advance. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-11-15T10:49:35Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Yup, Laplace transforms are way much easier in solving DE's. FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-11-15T11:03:51Z FirstChildTAG: Well, in your coursework you are free to use whatever method works for you. I don't know why they chose to use a particular solution in a particular setting. Maybe there is a reason and maybe it is what they are comfortable with. With these problems, there are like six different ways to skin this cat, as the saying goes. I was just watching a video where he says to use Laplace transforms, so he certainly says to use it. Also, here's a table of Laplace transforms for common wave types(pulse, square, step, etc): http://people.seas.harvard.edu/~jones/es154/Laplace/laplace_cwf/laplace_cwf.html FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-11-15T16:33:41Z IndexTAG: 920 TitleTAG: Absolutly fantastic!! I just would like to say thank you, this is fantastic, I now understand why, when I was experimenting with high power MOS-FETs, I would see these crazy waveforms http://www.richieburnett.co.uk/temp/gdt/gdt2.html ! I always had an inkling that is was capacitance/parasitic inductance casing the problems, so I would spend my time shortening my wires etc. Now, I may not understand the mathematical interpretations completely, but my intuition is **certainly more developed**! UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-11-15T10:05:30Z VoteTAG: 3 CoursewareTAG: Week 9 / S18V1 Review CommentableIdTAG: 6002x_S18V1_Review NumberOfReplyTAG: 1 FirstChildTAG: A real "Aha!" moment for me as well. Also, on another thread I suggested that you make a tutorial to go over doing Week 8 equations in Wolfram, because you are the Wolfram Master. I'm sure I wouldn't be the only one to benefit from your knowledge. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-11-15T16:19:42Z SecondChildTAG: Ah, yes! Last week was really busy for me. Trying to get back on top ;) . To be honest I am lost when it comes to the differential equations. oh well! How are things going for you? SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-15T16:55:57Z SecondChildTAG: Good, busy as well, but keeping on top of it. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-11-15T17:05:42Z SecondChildTAG: Yeah I am lost on DEs, too. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-11-15T17:06:18Z SecondChildTAG: Haha! Great. SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-15T17:10:12Z SecondChildTAG: Sucking at DEs certainly makes the course harder. Oh well. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-11-15T17:55:31Z SecondChildTAG: I don't know if this is much help... http://au.answers.yahoo.com/question/index?qid=20100117160310AA7nYBf SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-15T18:35:08Z SecondChildTAG: I also found this: http://www.math.uci.edu/~baskaran/HiMCM_2011/Compartmental_Analysis_Wolfram_Alpha.pdf SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-11-16T01:29:02Z IndexTAG: 921 TitleTAG: H9 and Lab 9 Hello all, I have posted about 3 posts but no one has replied. Can anyone tell me if my posts are reaching you all? Also -to Myrimit when would you be able to post Homework 9 hints and the Lab 9 hints? Please respond.-arjshar UserIdTAG: 536922 UserNameTAG: arjshar CreateTimeTAG: 2012-11-14T23:50:58Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: Which question in H9/lab 9 You require Assistance? Also There are still three days left for the deadline so I think you should first try solving it personally and ask for assistance if you are constrained by time. FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-11-15T06:22:30Z FirstChildTAG: Hi guys! Any hints to deal with the 3rd part of the lab? In particular, how do I obtain the value of del t to obtain C? FirstChildUserIdTAG: 143823 FirstChildUserNameTAG: NathanNadeson FirstChildCreateTimeTAG: 2012-11-15T13:51:59Z SecondChildTAG: del t of 0.5ms (10-9.5) somehow worked for me. SecondChildUserIdTAG: 184153 SecondChildUserNameTAG: tthngn SecondChildCreateTimeTAG: 2012-11-16T04:30:05Z SecondChildTAG: thanks! SecondChildUserIdTAG: 143823 SecondChildUserNameTAG: NathanNadeson SecondChildCreateTimeTAG: 2012-11-16T09:00:25Z SecondChildTAG: how can i find the value of L tho? SecondChildUserIdTAG: 285945 SecondChildUserNameTAG: lindalapiso SecondChildCreateTimeTAG: 2012-11-16T09:49:05Z SecondChildTAG: The instructions say start by fixing C; that's where you need *dt* (*i* and *dv* are given already). I guess L could be approximated using one of the formula of 2nd-order circuits knowing: - value of R, from diagram in the box. - approximate value of C, from last step. - this is basically a parallel config of RLC, specifically a low pass filter pattern (see HW9P2). - the L value needed would bring the circuit to a slightly over-damped setting, where there is no ringing.![final analysis][1] That's only a guess, as I completed the last part of the lab mostly by experiments, before HW9. Even if L is calculated there's still one more step before Aha!, which is to adjust the duty cycle, or the last parameter of the square wave.This too can probably be calculated, but not by a common mortal like me. [1]: https://edxuploads.s3.amazonaws.com/13530914481343676.png SecondChildUserIdTAG: 184153 SecondChildUserNameTAG: tthngn SecondChildCreateTimeTAG: 2012-11-16T19:11:57Z FirstChildTAG: Hi arjshar, Ok, I will Post HW9 Hints too :). I also have promised in other Post to make Lab9 Hints too. I hope to post them today at night or tomorrow. But I will, you have my word haha. Might I will be a little bit disconnected this weekend, because I have the 3.091x Exam :P . Ouch, I have to study , opens tomorrow... Good luck to all of you in the 3.091x Exam! See you! Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-15T15:34:13Z SecondChildTAG: Myriam, how have found 3091x? I dropped out because the person I signed up with dropped out and I didn't want to do it alone. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-11-15T16:23:09Z SecondChildTAG: Hi JSChambers, I like 3.091x. But you have to do the exercises, at least me that I don´t know much chemistry haha. Can you back? I am there too, so you would not be alone hahaha. The Exam is tomorrow, you still have time :). Also you can skip 1 homework without penalty. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-15T18:23:43Z SecondChildTAG: Myrimit, If you like the class, I urge you to watch the 3091 lectures by Sadoway, on OCW or iTunes U. They are best lectures I have ever seen, hands down. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-11-16T00:53:52Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-16T03:32:21Z SecondChildTAG: Good luck in 3.091 Myriam. SecondChildUserIdTAG: 366786 SecondChildUserNameTAG: LuisFernandoMurguia SecondChildCreateTimeTAG: 2012-11-17T12:11:07Z SecondChildTAG: Thank you Luis :)- By the way, here are the [Lab9 Hints][1]. Sorry for the delay :p [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50a78f0dce1ad22700000024 SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-17T13:51:05Z FirstChildTAG: Best of Luck Myrimit for your exam. thanks for responding.I really appreciate you helping us all with utmost dedication.Thanks FirstChildUserIdTAG: 536922 FirstChildUserNameTAG: arjshar FirstChildCreateTimeTAG: 2012-11-16T02:35:41Z SecondChildTAG: Thank you arjshar :). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-16T03:33:07Z IndexTAG: 922 TitleTAG: H11P3 Please give me hints for the last two questions of H11P3... Thanks! UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-11-13T17:52:58Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: For the penultimate question, the graph gives you a value of $|Z(j\omega)|$ for a given $\omega$. You have to draw the modified circuit, work out its transfer function, plug in the values from the graph and solve for the value of the remaining component. The last question requires that you know some of the formulae for a series LCR circuit. The ones that should be of use are (and you know the formula for $\omega_0$): $Q=\omega_0/(2\alpha)$ $\alpha=R/(2L)$ From the above, you can derive a formula for Q in terms of R, L and C. You know Q, R and one of L and C... (You might find the formula for Q in terms of R, L and C for a series LCR circuit in the textbook, but the above is enough to work it out). FirstChildUserIdTAG: 141000 FirstChildUserNameTAG: OrinE FirstChildCreateTimeTAG: 2012-11-13T20:45:57Z SecondChildTAG: done all of that still can't get the correct value for the last component. I have the correct values for everything else and think I did the algebra correctly. SecondChildUserIdTAG: 18073 SecondChildUserNameTAG: gburkhart SecondChildCreateTimeTAG: 2012-11-18T23:16:08Z SecondChildTAG: there is a missing element so Q as a function of R,L,C can\t work , it must be in terms of R,C or R,L only SecondChildUserIdTAG: 285675 SecondChildUserNameTAG: Ascot SecondChildCreateTimeTAG: 2012-12-01T17:33:47Z SecondChildTAG: ok i am sorry for the last comment , now i have done , i just couldn't understand the question right , as it said it wants the value of the ( REMOVED ELEMENT ) i just thought we still working on the circuit with the remove element SecondChildUserIdTAG: 285675 SecondChildUserNameTAG: Ascot SecondChildCreateTimeTAG: 2012-12-01T17:38:48Z IndexTAG: 923 TitleTAG: Video Download Bundle Request to STAFF Would it be possible to bundle all the individual downloadable videos that are available now under each youtube video into one big file, for each lecture sequence, either in compressed or uncompressed format and post that as an additional downloadable option? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-11-13T13:55:04Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 1 FirstChildTAG: Hi, I have made similar request with options for lower resolution and caption notes also to be made available. Please check and vote for below link. Staff please take a note. Thanks https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/509e6af8179cf31f0000003d FirstChildUserIdTAG: 111917 FirstChildUserNameTAG: ashish_mit FirstChildCreateTimeTAG: 2012-11-14T10:03:14Z IndexTAG: 924 TitleTAG: Happy Diwali Happy Diwali buddies..:) hope you all have fun.:) UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-11-13T10:19:06Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: same to u :) FirstChildUserIdTAG: 269641 FirstChildUserNameTAG: BAUWA FirstChildCreateTimeTAG: 2012-11-13T14:06:28Z FirstChildTAG: same to u too FirstChildUserIdTAG: 285222 FirstChildUserNameTAG: dhaval24 FirstChildCreateTimeTAG: 2012-11-13T10:29:50Z FirstChildTAG: same to you FirstChildUserIdTAG: 342221 FirstChildUserNameTAG: deepkar FirstChildCreateTimeTAG: 2012-11-13T10:59:35Z SecondChildTAG: Best wishes to all! SecondChildUserIdTAG: 99628 SecondChildUserNameTAG: Pranjal16 SecondChildCreateTimeTAG: 2012-11-13T11:09:00Z FirstChildTAG: same to you FirstChildUserIdTAG: 114864 FirstChildUserNameTAG: Vikaash FirstChildCreateTimeTAG: 2012-11-13T12:12:13Z IndexTAG: 925 TitleTAG: Material en español Para quienes tengan dificultades con el material del curso en inglés. Existen las notas del MIT OpenCourseWare para Circuitos y Electrónica 6.002 en español en: [Universia][1]. Lastimosamente, parece que los videos ya no están disponibles. Saludos. [1]: http://mit.ocw.universia.net/6-002/OcwWeb/Electrical-Engineering-and-Computer-Science/6-002Circuits-and-ElectronicsFall2000/LectureNotes/index.htm UserIdTAG: 366786 UserNameTAG: LuisFernandoMurguia CreateTimeTAG: 2012-11-12T07:19:13Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Gracias Luis, No sabía que existía este tipo de información en Español. Gracias por la data!;) Saludos, Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-12T21:50:31Z IndexTAG: 926 TitleTAG: H8P3 I'm still stucking in part 4 and 5, and it drives me crazy. At the Moment I see that the drain from Q1 has the voltage calculated in part 2. This must be the Thevenin voltage. With Q1Roff||RPU + Q3Ron I get the Rth. For charging, I'm using the formula from S16V2. But I don't get the check-mark. UserIdTAG: 389333 UserNameTAG: juergen CreateTimeTAG: 2012-11-11T22:47:24Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Me too, I'm following the same procedure :( FirstChildUserIdTAG: 298547 FirstChildUserNameTAG: AnthonyRF FirstChildCreateTimeTAG: 2012-11-12T01:34:43Z SecondChildTAG: Part 4 is the rising form (bottom graph) of S16V2 in terms of VOL, VOH, VQ1D (from part 2) and e^-t/Rth*C. Remember that you want t when voltage reaches VOH. SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-12T02:20:57Z SecondChildTAG: I'm having problems with part 5.. I had all day working in that particular question.. SecondChildUserIdTAG: 298547 SecondChildUserNameTAG: AnthonyRF SecondChildCreateTimeTAG: 2012-11-12T02:24:26Z SecondChildTAG: I'm using t=(Rth_Q1+RON_Q2)*Cgs2*Ln(Vil/Voh) but never appear the green mark SecondChildUserIdTAG: 298547 SecondChildUserNameTAG: AnthonyRF SecondChildCreateTimeTAG: 2012-11-12T02:27:25Z SecondChildTAG: Part 5 does not look correct, try to look the S14V8-V10 but note that you have to calculate a specific Rth ((rpu||ron)+ron) and Vth (voltage divider rpu, ron) for this exercise. SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-12T02:35:40Z SecondChildTAG: ... and then draw the VCgsQ2(t) x t graph to help you find your equation in terms of VIL, VTH and VOH and e^(-t/RthC). Good luck SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-12T02:38:30Z SecondChildTAG: But look that VCgsQ2 is discharging through RthQ1 and Ron2.. SecondChildUserIdTAG: 298547 SecondChildUserNameTAG: AnthonyRF SecondChildCreateTimeTAG: 2012-11-12T02:43:08Z SecondChildTAG: I don't know if the question has the same conditions: ... "Now, suppose DIN is high, the drain of Q1 is low, and the gate of Q2 is at VOH. A STORE pulse comes in and turns on Q3" ... DIN high means the mosfet is ON and its drain has a logical "0". Then the Q3 mosfet gets ON too (store pulse). I think that the resistance as seen from Q2CGS is ron(q3)+(rpu||ron(q1)) SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-12T02:53:05Z SecondChildTAG: Exactly, that's what I'm telling to you. But I think the problem is the fraction Vil/Voh.. I still have a red mark :( SecondChildUserIdTAG: 298547 SecondChildUserNameTAG: AnthonyRF SecondChildCreateTimeTAG: 2012-11-12T02:58:23Z SecondChildTAG: yes, ...(Vil/Voh) does not look correct... It looks that something is missing did you watch the S14V8-V10 videos? SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-12T03:02:34Z SecondChildTAG: Finally I FOUND IT! Thanks a lot matiasgrodriquez! It sounds like you live in some place in suramerica (Like me!) Good night! SecondChildUserIdTAG: 298547 SecondChildUserNameTAG: AnthonyRF SecondChildCreateTimeTAG: 2012-11-12T03:06:45Z SecondChildTAG: cool man! yep, I live in Brasil. Now I can go to sleep :) SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-12T03:08:20Z FirstChildTAG: I slept over it, values in task are changed, I updated my calculating for task, now 4 I'm right, but 5 is still waiting, and I must go to work:-( FirstChildUserIdTAG: 389333 FirstChildUserNameTAG: juergen FirstChildCreateTimeTAG: 2012-11-12T07:17:52Z IndexTAG: 927 TitleTAG: A lecture content suggestion to EDx staff In "S17V3: Motivation .. Fast Case," we see for the first time an under-damped circuit with the problem of ringing caused by trying to speed up a circuit with the insertion of a small resistor in parallel with the pull-up resistor. After some wonderful analytical explanations in subsequent lectures we learn a great deal about characteristic equations, damping coefficient, etc. and it would seem that we might return to that "ringing" example and show that there is an appropriate resistor to use in parallel with the pull-up resistor that would actually provide the best case critical damping. I don't know if you could analytically predict this R value given that the inductance is parasitic but a demo of the critically damped circuit would provide closure on the "motivation" scenario that was used to very effectively (in my opinion) start this entire discussion. UserIdTAG: 569199 UserNameTAG: neogerald CreateTimeTAG: 2012-11-11T11:56:28Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 928 TitleTAG: Initial conditions for part 2 Hi everybody. Everything is fine, but I don't understand why $I(0_+)=0.0 A$. It seems it should be $I_0$ UserIdTAG: 213452 UserNameTAG: JoJosida CreateTimeTAG: 2012-11-11T05:31:05Z VoteTAG: 3 CoursewareTAG: Week 9 / An ILC circuit CommentableIdTAG: 6002x_an_ILC_circuit NumberOfReplyTAG: 4 FirstChildTAG: Hi, [H8P1,H8P2, H8P3 Hints][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/509e68568e407d1f0000001b > **Regards:** asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-11-11T08:57:50Z FirstChildTAG: Hi, I didn't understand why as well. I thought $I(0^+)=3.00V$ since it was directly after the step. FirstChildUserIdTAG: 85490 FirstChildUserNameTAG: yashsavani FirstChildCreateTimeTAG: 2012-11-13T10:28:32Z FirstChildTAG: My guess for a possible explanation is as follow, Io(0)=3A gives the initial condition for iL(t)=3A, then question then states *suppose* Io(0+)= 0A. So the system with the 2nd order differential equation will have to satisfy initial conditions of iL(t)=3A and Io(0+)= 0A in its oscillations. FirstChildUserIdTAG: 544644 FirstChildUserNameTAG: Ethanaung FirstChildCreateTimeTAG: 2012-11-14T07:56:56Z FirstChildTAG: I think instead of $I(0_+)=0.0A$ they mean $i_L(0_+)=0.0A$. FirstChildUserIdTAG: 266912 FirstChildUserNameTAG: pietvo FirstChildCreateTimeTAG: 2012-11-14T21:12:47Z SecondChildTAG: Agree. SecondChildUserIdTAG: 398594 SecondChildUserNameTAG: DaveyJC SecondChildCreateTimeTAG: 2012-11-14T22:29:26Z SecondChildTAG: I was just thinking the same! SecondChildUserIdTAG: 373752 SecondChildUserNameTAG: Croak SecondChildCreateTimeTAG: 2012-11-15T00:45:39Z SecondChildTAG: TYPO SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-12-18T07:52:13Z IndexTAG: 929 TitleTAG: H8P1 - decoupled WRT Part c of H8P1, the hint states that the L-R1 and the C-R2 branches are decoupled. Does that mean that when analyzing the behavior of one of the branches, the other branch can be treated as if it didn't exist? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-11-10T23:35:20Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Yes, notice that both ends of both branches are held at ground except at the instant the impulse occurs. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-11T00:42:53Z SecondChildTAG: Thanks, Skyhawk. SecondChildUserIdTAG: 141108 SecondChildUserNameTAG: rl777 SecondChildCreateTimeTAG: 2012-11-11T04:41:22Z FirstChildTAG: Hi, [H8P1,H8P2, H8P3 Hints][1] [1]: http://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/509e68568e407d1f0000001b > **Regards:** asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-11-11T08:55:19Z IndexTAG: 930 TitleTAG: To Staff : Progress window Hi all! What is happened with all bars in the Progress window? Homework,Labs, Midterm, Overall - all these bars are absent. But if I do put cursor into some position I can see percentage. Thanks. UserIdTAG: 205311 UserNameTAG: Sergtronix CreateTimeTAG: 2012-11-10T16:04:29Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I have the same problem in chrome, but in firefox the bars are shown in progress. Also I have problems to visualize the circuits sandbox in the labs (The lines who connect the components are not shown) FirstChildUserIdTAG: 370769 FirstChildUserNameTAG: PabloFCid FirstChildCreateTimeTAG: 2012-11-10T19:24:56Z IndexTAG: 931 TitleTAG: [STAFF] Can the lecture video file size reduced by providing option for reduced resolution? Hi I have been finding it tough to view the lecture videos as speed of my net connection is slow most of the time, I think many taking the course in developing nations might be facing this problem. Is there any chance to reduce the size of online videos and downloads further by giving us options for reduced resolution as course for CS50 does? Even for download the file size is too big when compared to download at lower resolution from CS50 offering ? I had the same problem while taking the course by MITX, one the reasons I was not able to complete than, gave same as feedback. If anything can be done in this regard, I suppose it will help many. UserIdTAG: 111917 UserNameTAG: ashish_mit CreateTimeTAG: 2012-11-10T14:55:52Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Found a way to reduce size by going to youtube link for the lecture and reducing resolution but would like the feature be embedded in edx circuits and electronics course. Also it would be good if speaker captions can be made available for download. It would be helpful to go along with the lecture notes and the captions.. without the video. Should be great while revising. FirstChildUserIdTAG: 111917 FirstChildUserNameTAG: ashish_mit FirstChildCreateTimeTAG: 2012-11-10T15:41:25Z IndexTAG: 932 TitleTAG: Low speed of the internet I think it is better to make all tutorials in a virtual paper with virtual pen like this one. Because when you film the real paper and pen and hand of the teacher, It makes the size of the video much more and in my country, Iran, for most of the people it is hard to watch those videos because we have 128 kbps internet and with this speed we must wait a lot until the video is downloaded UserIdTAG: 259238 UserNameTAG: omidsadeghi CreateTimeTAG: 2012-11-08T22:35:30Z VoteTAG: 3 CoursewareTAG: Week 10 / Complex Numbers Math Review CommentableIdTAG: 6002x_Complex_Numbers_Math_Review NumberOfReplyTAG: 0 IndexTAG: 933 TitleTAG: H8P1 - c and e Hi everyone! I have a trouble with this part... I get the same result with maths and with sandbox... But the computer say that it is incorrect... What's happen?? Any hint?? UserIdTAG: 239608 UserNameTAG: guillegf84 CreateTimeTAG: 2012-11-08T21:14:45Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: I'll recommend seeking textbook.It helped me a lot. FirstChildUserIdTAG: 477713 FirstChildUserNameTAG: ikm104 FirstChildCreateTimeTAG: 2012-11-09T07:07:03Z SecondChildTAG: What part of the book? SecondChildUserIdTAG: 239608 SecondChildUserNameTAG: guillegf84 SecondChildCreateTimeTAG: 2012-11-09T11:07:06Z SecondChildTAG: 10.6.x SecondChildUserIdTAG: 477713 SecondChildUserNameTAG: ikm104 SecondChildCreateTimeTAG: 2012-11-09T17:35:47Z SecondChildTAG: Oh thanks! It helped me as well with the RC circuits! Now I try with the RL. I didn't know the meaning of the area. I thought that the amplitude pulse was 1, but 1 is the area... Thanks! SecondChildUserIdTAG: 239608 SecondChildUserNameTAG: guillegf84 SecondChildCreateTimeTAG: 2012-11-10T00:42:14Z FirstChildTAG: Try searching this forum for "H8P1" - there are hints that will help you, as they helped me. :-) FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-11-09T00:17:17Z FirstChildTAG: VOILA V1(1ms) = (R/L)*e^(-R*t/L) ET VC(1ms)= 1/(RC)*e^(-t/RC) FirstChildUserIdTAG: 316899 FirstChildUserNameTAG: elou FirstChildCreateTimeTAG: 2012-11-10T10:13:40Z IndexTAG: 934 TitleTAG: to staff i'm looking forward to study and work in analog integrated circuits i need help showing me the way and the track to this field may any one help me ?? UserIdTAG: 221617 UserNameTAG: konan CreateTimeTAG: 2012-11-07T19:27:14Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi Konan, I am not staff, but I did notice your thread the last time. Have you had a look at some of the courses offered by MIT's OpenCourseWare? I would imagine you will eventually see courses like these migrate over to edX just like 6.002(x) While I cannot guide you down an "official" path, these courses would be very valuable in one way or another. Here are a just a few examples of courses you may be interested in. Please check out the site as I just grabbed these quickly, there are other cool courses waiting for you. http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-101-introductory-analog-electronics-laboratory-spring-2007/ http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-331-advanced-circuit-techniques-spring-2002/ FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-11-07T23:55:28Z IndexTAG: 935 TitleTAG: H8P3 - Some one please give me a hint Hi I am able to solve part1,2,3 of H8p3 but unable to solve Part4 and 5 can any one give me hint to solve this problem UserIdTAG: 352647 UserNameTAG: praveenjugge CreateTimeTAG: 2012-11-07T08:15:42Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I got stuck with part 3 myself FirstChildUserIdTAG: 143823 FirstChildUserNameTAG: NathanNadeson FirstChildCreateTimeTAG: 2012-11-07T16:04:45Z SecondChildTAG: check discussions on h8p3 there is hint to solve part3 SecondChildUserIdTAG: 352647 SecondChildUserNameTAG: praveenjugge SecondChildCreateTimeTAG: 2012-11-08T06:29:37Z FirstChildTAG: There is an equation (the Zero State Response (ZSR)) in a memory cell, from which you can find the final Vout. In H8P3, 4th part you are told the initial and final voltage, so you can use this equation in order to solve over t. The same equation applies in the 5th part as well but you must be careful with the Resistance in this case, as it is not the same as in the 4th. Good Luck! FirstChildUserIdTAG: 304587 FirstChildUserNameTAG: Vasso FirstChildCreateTimeTAG: 2012-11-07T23:30:23Z SecondChildTAG: Mmm..Im fighting with part 5 , without any success..Any hints, Vasso ? PS I do use Thevenin equivalent for Q1 drain voltage, I do remember about Q3 too.. Unfortunatelly I cant get correct value SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-11-08T19:53:33Z SecondChildTAG: P5 is solved.. SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-11-08T20:41:33Z SecondChildTAG: i am still stuck in part 4 and 5. SecondChildUserIdTAG: 368981 SecondChildUserNameTAG: asimakhtar SecondChildCreateTimeTAG: 2012-11-08T20:49:36Z SecondChildTAG: For questions 4 and 5: - the example of [this][1] Wikipedia article has a configuration similar to the resistors involved (ignore R2 in the example). Calculate Vth and Rth following the method described, with R4=RPU, R3=ROFF in question 4 (since Q1 Vds is high) and R3=RON in question 5, and Rth involves Q3's RON in response to STOR signal. See diagrams. ![diagrams][2] - MOSFETs' characteristics including VT and parasitic resistances are not factors, "for simplicity". [1]: http://en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem [2]: https://edxuploads.s3.amazonaws.com/13524158737929062.png SecondChildUserIdTAG: 184153 SecondChildUserNameTAG: tthngn SecondChildCreateTimeTAG: 2012-11-08T23:42:13Z SecondChildTAG: solved part 4 but not getting part 5 any hint pl SecondChildUserIdTAG: 352647 SecondChildUserNameTAG: praveenjugge SecondChildCreateTimeTAG: 2012-11-09T07:35:30Z SecondChildTAG: Hi Praveen, Solve it exactly the way you solved part-4. Just redraw the circuit for the new condition: Q1 is turned on. And, thanks to tthngn for the hint to solve part4&5. SecondChildUserIdTAG: 232157 SecondChildUserNameTAG: GayatriTR SecondChildCreateTimeTAG: 2012-11-09T08:31:27Z SecondChildTAG: I am using the formula Vil=Voh*e^(t/(R*C)) but not able to solve SecondChildUserIdTAG: 352647 SecondChildUserNameTAG: praveenjugge SecondChildCreateTimeTAG: 2012-11-09T11:12:55Z SecondChildTAG: Hope you got the correct Vth and Rth. That is very important. After that, you need to get the correct equation for Vc(t). Hope you are using the logic explained by the professor for the voltage curve. You may want to see a quick review in video S16V1. If you have doubts regarding what you derived intuitively based on the lecture, you may cross-verify that using the canonical form: Vc(t) = Final Value + (Initial Value - Final Value)*e^(-t/RC) SecondChildUserIdTAG: 232157 SecondChildUserNameTAG: GayatriTR SecondChildCreateTimeTAG: 2012-11-09T14:42:19Z SecondChildTAG: hi gayatri did you get answer for the part 5 , if yes then please give hint SecondChildUserIdTAG: 352647 SecondChildUserNameTAG: praveenjugge SecondChildCreateTimeTAG: 2012-11-09T15:07:05Z SecondChildTAG: I am totally confused SecondChildUserIdTAG: 352647 SecondChildUserNameTAG: praveenjugge SecondChildCreateTimeTAG: 2012-11-09T15:27:09Z SecondChildTAG: me too is totally confused in part 4&5, although i have done all homework questions and lab as well. SecondChildUserIdTAG: 368981 SecondChildUserNameTAG: asimakhtar SecondChildCreateTimeTAG: 2012-11-09T18:13:05Z SecondChildTAG: **praveenjugge**: simplified schematics above by **GayatriTR** is clearly show what is happen when Q1 is turned ON. You do have: Thevenin resistance for Q1(Q1on||Rpu); voltage source: divider of Vs by Q1 Ron and Rpu(later you should add Q3 Ron too). Next, please use this : Vc(t) = Final Value + (Initial Value - Final Value)*e^(-t/RC) as described above,where Vc(t) is VIL, Final Value is Thevenin voltage :), Initial Value is VOH; R is sum of Rth and Q3 Ron.Cgs you do know. If you want, you may Plot this function within 0..100pS.You should see exponentially fallen Vgs. SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-11-09T19:10:20Z SecondChildTAG: Thank you Sergtronix , GayatriTR and tthngn Finally I could solve Part5 Thaks once again for your guidance and hints SecondChildUserIdTAG: 352647 SecondChildUserNameTAG: praveenjugge SecondChildCreateTimeTAG: 2012-11-10T07:52:35Z SecondChildTAG: thanks all,I finally solve the part3 Q4&5 after couples of hours SecondChildUserIdTAG: 361137 SecondChildUserNameTAG: Ericson SecondChildCreateTimeTAG: 2012-11-11T19:01:10Z SecondChildTAG: Thank you very much, Sergtronix. You gave me a very good hint. SecondChildUserIdTAG: 418182 SecondChildUserNameTAG: euldji2005 SecondChildCreateTimeTAG: 2012-11-12T00:01:13Z IndexTAG: 936 TitleTAG: 60% Today finished the homework 10 and lab 10, totalizand the final average by 60%. I read the note that is the minimum to get the certificate. The certificate will be available only at the end of the course? on a specific date? or when you finish the final exam? UserIdTAG: 357453 UserNameTAG: BrunoCanoso CreateTimeTAG: 2012-11-06T21:02:21Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: If they are based on 6.002x Spring, It will be available a few days after the Final Exam, It will appear a button of downloading the Certificate in the Progress section ;) Congrats, you have the minimun requisite to pass this Course 60%, C grade! Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-07T00:50:29Z SecondChildTAG: Thank you. SecondChildUserIdTAG: 357453 SecondChildUserNameTAG: BrunoCanoso SecondChildCreateTimeTAG: 2012-11-07T09:49:04Z SecondChildTAG: I just checked my progress nd it was 49%....:0 :( does it show the % of the work done or does it get the % from all the home works....!! IS IT 49 % of the course or 49% of the work done?? SecondChildUserIdTAG: 127800 SecondChildUserNameTAG: MNB SecondChildCreateTimeTAG: 2012-11-18T01:52:07Z IndexTAG: 937 TitleTAG: Question about listing this course in a college application Hello! As the application deadlines of many universities are approaching, I too have to fill up a few of them. I wanted to ask you (preferably a TA) if it's okay to list this course (and other EdX courses, and perhaps OCW courses) on my college application. Thanks for your help! UserIdTAG: 138981 UserNameTAG: Pr0bability CreateTimeTAG: 2012-11-06T20:53:34Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: of course. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-11-07T01:48:24Z IndexTAG: 938 TitleTAG: Circuit Sandbox Not Functioning in Any Lab Everyone, The circuit sandbox is not functioning(which many people already figured out). I alerted the staff. It's not just Lab 7. Hopefully it will be fixed soon. UserIdTAG: 21541 UserNameTAG: JSChambers CreateTimeTAG: 2012-11-06T20:38:45Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Right, I can't run any simulation. I'm using Firefox 16.0.2, IE ain't work either. FirstChildUserIdTAG: 34026 FirstChildUserNameTAG: Albovolt FirstChildCreateTimeTAG: 2012-11-06T21:26:37Z FirstChildTAG: There was a bug. It should be fixed now. Thanks for the report. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-11-07T01:48:00Z SecondChildTAG: what options do we have if we can not get to work I do not want the staff to think that I dropped out if not able to get homework on SecondChildUserIdTAG: 160242 SecondChildUserNameTAG: Mlevins35 SecondChildCreateTimeTAG: 2012-11-07T02:32:51Z IndexTAG: 939 TitleTAG: to staff i have aproblem with lab 7 it's not responding to me when i click on the tran butt or elements they don't respond UserIdTAG: 221617 UserNameTAG: konan CreateTimeTAG: 2012-11-06T20:37:24Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: The labs are temporarily broken. We are working on fixing it. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-11-06T21:05:32Z SecondChildTAG: i am not able to get on to the labs to get them to respond i tried to reboot my computer a little bit ago and still nothing What do i do if I can not get them by the time the homework is due SecondChildUserIdTAG: 160242 SecondChildUserNameTAG: Mlevins35 SecondChildCreateTimeTAG: 2012-11-07T02:17:10Z IndexTAG: 940 TitleTAG: totally baffled Could anyone explain this problem in more detail? I do not understand it at all to the point where I am not even sure how to frame a question. I would guess the answer is, take a Taylor series expansion (first two terms only) like in the video, but I don't understand why, nor did I follow the expansion in the video (it looked like maybe just the second term of the series and not the first? and not the (x-a) multiplier?) And I don't know what equation to take a Taylor series OF anyway, none of the ones I tried end up looking like Agarwal's in the video. Any help appreciated! UserIdTAG: 91356 UserNameTAG: mnrsiat CreateTimeTAG: 2012-11-06T19:23:09Z VoteTAG: 3 CoursewareTAG: Week 8 / Area CommentableIdTAG: 6002x_Area NumberOfReplyTAG: 2 FirstChildTAG: The problem asks for discharging equation.With Taylor series expansion you will find the current "IS";then multiply with inductor discharging equation. Hope it helps you! Sorry for my english FirstChildUserIdTAG: 127444 FirstChildUserNameTAG: Covidiu FirstChildCreateTimeTAG: 2012-11-06T20:52:36Z FirstChildTAG: I explained Taylor series expansion for the IRC circuit in [this link][1]. This IRL circuit is analogous. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Response_To_Impulse_Limit_Case/threads/50923ec3baf4ab2b0000000d FirstChildUserIdTAG: 398594 FirstChildUserNameTAG: DaveyJC FirstChildCreateTimeTAG: 2012-11-07T14:33:11Z IndexTAG: 941 TitleTAG: [News from World]Congratulation to Zac and all the work around him! I just want say contratulation to the MIT and also a lot of others Univesitys for the great work: [Zac Vawter Climb Chicago's Willis Tower][1] [1]: http://wxerfm.com/blogs/post/bolson/2012/nov/02/video-watch-bionic-stair-climber-zac-vawter-climb-/ UserIdTAG: 200496 UserNameTAG: Ranieri CreateTimeTAG: 2012-11-05T10:06:56Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: This is a great result of Robotic and people that a lot of time agò studyed 6.002 circuits :P FirstChildUserIdTAG: 200496 FirstChildUserNameTAG: Ranieri FirstChildCreateTimeTAG: 2012-11-05T10:09:23Z SecondChildTAG: This is Great By the Way its me Harvey I really need your help please come back on skype.......................................................... SecondChildUserIdTAG: 378080 SecondChildUserNameTAG: GladIDidThis SecondChildCreateTimeTAG: 2012-11-18T22:26:04Z SecondChildTAG: Remember me HArvey Specter SecondChildUserIdTAG: 378080 SecondChildUserNameTAG: GladIDidThis SecondChildCreateTimeTAG: 2012-11-18T22:26:25Z IndexTAG: 942 TitleTAG: algebra equations i guess that this is about the only part of this class that is throwing me. I never have been good at math, kinda of just getting by, I understand most of the course that the professor has said, but the math is what throws me. Does any one know what i can do for help with algebra equations. UserIdTAG: 160242 UserNameTAG: Mlevins35 CreateTimeTAG: 2012-11-04T17:45:22Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: The [Khan Academy][1] has a lot of great videos for improving your math skills. [1]: http://www.khanacademy.org/ FirstChildUserIdTAG: 90796 FirstChildUserNameTAG: ChaunceyGardiner FirstChildCreateTimeTAG: 2012-11-04T18:06:24Z SecondChildTAG: Try Wolfram Alpha! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-05T13:03:30Z FirstChildTAG: [WolframAlpha][1] is indispensable, IMHO. [1]: http://www.wolframalpha.com FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-11-05T05:06:37Z IndexTAG: 943 TitleTAG: about TAs can some one tell me how can i join to TA group UserIdTAG: 509517 UserNameTAG: randima CreateTimeTAG: 2012-11-04T16:41:32Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi randima, You can read this post [here][1] and also take a look at the Course Info ;)[here][2]: **Community TA's:** *Some of you may have noticed our distinguished Community TA's on the forum, and we hope that you will learn from their example in how they have shaped the social space. We are going to need similarly dedicated Community TA's for the spring offering of 6.002x, and we like to draw from the capable, helpful, and kind people that inhabit the increasingly mature cohort of previous terms.* I hope that this Info can be useful, Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/508d6d7da8f7a21f0000005f [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/info FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-04T16:50:46Z SecondChildTAG: Hi Myrim.. thank yu very much... i'll follow that.. SecondChildUserIdTAG: 509517 SecondChildUserNameTAG: randima SecondChildCreateTimeTAG: 2012-11-04T16:52:48Z SecondChildTAG: Cool @randima! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-04T17:14:30Z SecondChildTAG: Hi Myriam, One of my friend who took the course during March-June this year wants to join the Community TA. His final score was 89%. Is is it possible for him to join the community. Thanking you, Joydeep SecondChildUserIdTAG: 179140 SecondChildUserNameTAG: JoydeepSil SecondChildCreateTimeTAG: 2012-11-07T15:04:03Z SecondChildTAG: Hi @JoydeepSil, I think that is not important the score to be Community TA :). If he has the skills that they wrote in the Course Info, I guess he can be one. He should send an email to edX Staff. See you, Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-07T17:47:22Z SecondChildTAG: thats great news!! SecondChildUserIdTAG: 509517 SecondChildUserNameTAG: randima SecondChildCreateTimeTAG: 2012-11-11T12:16:42Z IndexTAG: 944 TitleTAG: To "Staff"-----------Still 'No Reply'? ![enter image description here][1] it was said that at the moment,capacitor decays ;we consider THEVENIN EQ. CKT which looks like C connected with parallel combination of R1 and R2.So,it must get discharged thro both of them simultaneously,but at the point 'A' after a moment potential will be greater than that of potential at C.Then how could the C get discharged thro that path for whole of the discharge duration? [1]: https://edxuploads.s3.amazonaws.com/13520094711343647.jpg UserIdTAG: 477713 UserNameTAG: ikm104 CreateTimeTAG: 2012-11-04T06:19:36Z VoteTAG: 3 CoursewareTAG: Week 7 / Fall Time CommentableIdTAG: 6002x_Fall_Time NumberOfReplyTAG: 4 FirstChildTAG: Through R2, if I understand you correctly. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-11-04T11:24:28Z SecondChildTAG: yes SecondChildUserIdTAG: 477713 SecondChildUserNameTAG: ikm104 SecondChildCreateTimeTAG: 2012-11-05T13:07:31Z FirstChildTAG: During the discharge time, the voltage of the source Vdc will be 0, so it will be the same as if the voltage source is a short circuit. If you re-draw the circuit with the voltage source as a short, then you will see that the capacitor is in parallel with R2 and R1, and so it will discharge through the parallel combination of the two resistors. FirstChildUserIdTAG: 270011 FirstChildUserNameTAG: SnowmanZA FirstChildCreateTimeTAG: 2012-11-04T13:20:51Z SecondChildTAG: Your idea seems wrong,please don't mind what i'm gonna write further.....is it possible that battery would come to know...oh,its the time for capacitor to discharge so i should short myself? You are pretty intelligent so can think about it that source voltage never changes. SecondChildUserIdTAG: 477713 SecondChildUserNameTAG: ikm104 SecondChildCreateTimeTAG: 2012-11-05T05:37:39Z SecondChildTAG: If there was a battery permanently connected, then the capacitor would never discharge. If there was a battery connected and then removed, then the capacitor would discharge through R2 only, since there would be an open circuit where the battery (Voltage source) was. If the voltage source was a power supply that was set to produce a pulse or a square wave output, then the capacitor would charge through R1 during the high period and discharge through the parallel combination of R1 and R2 during the low period. SecondChildUserIdTAG: 270011 SecondChildUserNameTAG: SnowmanZA SecondChildCreateTimeTAG: 2012-11-05T16:51:18Z SecondChildTAG: thats the dc source not a pulsed output and in digital circuit Vss is always a dc source. SecondChildUserIdTAG: 477713 SecondChildUserNameTAG: ikm104 SecondChildCreateTimeTAG: 2012-11-06T06:29:57Z FirstChildTAG: Hi ikm104, Sorry for the delay of this post response, I haven't seen this before... I were thinking on how will be practical to show you this, so I went to the Sandbox ;) in order that you could understand the behaviour. --- Ok, lets make your circuit in Sandbox: ![im1][1] I put a Step source (As this is the better way to see the charge and discharge of the capacitor. This will be like having a power source and pressing "ON" and you will see that will go from 0V to 10V DC, in this case). Configure the voltage source like this: ![IM2][2] Ok, know, click on TRAN (50n), you will see that the Capacitor will charge till reach to a constant value, and then it will remain like that for an infinite time. ![SAN3][3] Ok, this will remain like this unless you press "OFF"to your voltage source, so lets, simulate this. You can do it changing the settings like this: ![san3][4] Ok, know, click on TRAN (50n), you will see that the Capacitor will discharge till reach to zero, and then it will remain like that for an infinite time unless you press "ON" again to the voltage source. ![san4][5] So, that is all. In order to discharge the Capacitor you should put to zero your voltage source ("OFF").In order to charge your Capacitor, just "ON" your voltage source ;) See you! I hope this can help you, Myriam. P.D: You should take a look at page 512 of the Textbook [here][6] [1]: https://edxuploads.s3.amazonaws.com/13522557555746698.png [2]: https://edxuploads.s3.amazonaws.com/1352255929560204.png [3]: https://edxuploads.s3.amazonaws.com/13522560581343682.png [4]: https://edxuploads.s3.amazonaws.com/13522561931343686.png [5]: https://edxuploads.s3.amazonaws.com/13522563481343645.png [6]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/536 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-07T02:49:36Z FirstChildTAG: Does it mean that we don't use constant DC in digital circuits? As if we use constant DC ,then "Cgs" can't be discharged thro both "RL" and "Ron" but can be discharged with Ron only. I simply want to ask that at the discharge moment,we consider both the resistors in Thevenin equivalent circuit but we should consider only one resistor whose path goes to ground. FirstChildUserIdTAG: 477713 FirstChildUserNameTAG: ikm104 FirstChildCreateTimeTAG: 2012-11-07T18:36:27Z SecondChildTAG: > Does it mean that we don't use constant DC in digital circuits? Answer: **NO**, it doesn't mean that. You actually have to use DC for digital circuits. You consider both RL and Ron because if you look at the node you have two incoming currents: "current from voltage source" and "discharge current", and the only way out is through Ron, and that becomes a bottle neck for both currents. So that means that not all of the discharge current can pass through Ron, and the part of the current that is able to pass is the equivalent to the value that you would get if Ron and RL were in parallel to the ground, kind of "like if the voltage source became 0V". ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13524487251343697.png SecondChildUserIdTAG: 304535 SecondChildUserNameTAG: hugo_h SecondChildCreateTimeTAG: 2012-11-09T08:36:13Z SecondChildTAG: dear, there is an imperative condition for charge to flow that charge always flows from higher potential to lower potential and at point A potential will become higher after a moment,the discharge process starts,so this idea doesn't seem feasible. SecondChildUserIdTAG: 477713 SecondChildUserNameTAG: ikm104 SecondChildCreateTimeTAG: 2012-11-09T14:00:45Z SecondChildTAG: Potential at point A is always the same, it ***never*** changes, and it **will always be greater or equal to the potential on the capacitor**. The discharge process only happens when the mosfet is on and the mosfet is then *modeled* as Ron, but when the mosfet is off there is no R2 on the circuit therefore all the current from the voltage source (point A) goes into the capacitor charging it. SecondChildUserIdTAG: 304535 SecondChildUserNameTAG: hugo_h SecondChildCreateTimeTAG: 2012-11-10T02:18:54Z SecondChildTAG: Just to clarify something that might be confusing about my previous post. And Vdc=Vs which is the potential I said never changes. ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13525186611343631.png SecondChildUserIdTAG: 304535 SecondChildUserNameTAG: hugo_h SecondChildCreateTimeTAG: 2012-11-10T03:45:30Z SecondChildTAG: I didn't get it,would u please elaborate more on it! SecondChildUserIdTAG: 477713 SecondChildUserNameTAG: ikm104 SecondChildCreateTimeTAG: 2012-11-10T05:43:13Z IndexTAG: 945 TitleTAG: Blue pulse. I think that the decay equation for the blue pulse should be : (QR/T)e^(-(t-T1)/RC). UserIdTAG: 282828 UserNameTAG: Chibolator CreateTimeTAG: 2012-11-04T05:45:51Z VoteTAG: 3 CoursewareTAG: Week 8 / Response To A Pulse As Pulse Gets Narrower Continued CommentableIdTAG: 6002x_Response_To_A_Pulse_As_Pulse_Gets_Narrower_Continued NumberOfReplyTAG: 1 FirstChildTAG: Agree FirstChildUserIdTAG: 214584 FirstChildUserNameTAG: Askeroff FirstChildCreateTimeTAG: 2012-11-08T11:06:17Z IndexTAG: 946 TitleTAG: LAB7 TASK 1 Last question in TASK 1. The formula 9.18 (Cv(t)^2)/2) in the textbook is not giving right answer. Can anyone help here? UserIdTAG: 239736 UserNameTAG: khatal CreateTimeTAG: 2012-11-04T02:41:15Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hi khatal, Can I help you? Hint 1: Do you have the voltage of v for the time t? Take a look at the previous question ;). Hint 2: Be careful if you have $AB^2$ is $B^2$ and then multiplied by A and not $(AB)^2$ ... I hope this can help you. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-04T02:54:28Z SecondChildTAG: I got every other question right, except this one. If u look at the formula carefully, it should be: (q^2(t)/2C = Cv^2(t)/2), so the square should be on v not on t, right? SecondChildUserIdTAG: 239736 SecondChildUserNameTAG: khatal SecondChildCreateTimeTAG: 2012-11-04T03:27:20Z SecondChildTAG: Okey so we substitute the variable, our x shall be(1000*2*pi) so vc(t)=(1/0.001)*(1/1000*2*pi)*[integral from to t of sin(x) dx] vc(t)=(1000/(1000*2*pi))*[integral from sin(x) dx] derivative of sin is -cos so vc(t)=(1000/(1000*2*pi))*-cos(x) substituting again vc(t)=(1000/(1000*2*pi))*-cos(1000*2*pi*t), what are the next simplifications? it keeps telling me that it's wrong and according to your hints, and the rules of derivatives and and integration this should be correct.... SecondChildUserIdTAG: 317836 SecondChildUserNameTAG: VitorRodrigues SecondChildCreateTimeTAG: 2012-11-05T07:00:12Z FirstChildTAG: @Khatal: I think you interpreted the formula in a wrong way. v(t) is one single entity in the formula implying voltage as a function of time. Please see that there is no t term in the expression. FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-11-04T09:31:56Z SecondChildTAG: Thanks Span993, I was making that mistake of multiplying 0.5ms, because they mention the time in the question. Thanks again SecondChildUserIdTAG: 239736 SecondChildUserNameTAG: khatal SecondChildCreateTimeTAG: 2012-11-04T16:54:52Z IndexTAG: 947 TitleTAG: Self Study Mode after Course End Date I started the course too late due to family commitments. I really am not concerned at this point about a certificate but would very much like to learn the material and continue to use the courseware for this purpose. Will the class still be open after the due date for those of us who wish to continue to self-study? UserIdTAG: 534915 UserNameTAG: jturner421 CreateTimeTAG: 2012-11-03T17:51:54Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Yup, I believe. Because the courseware of the last run of this course is still in open. FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-11-03T19:44:22Z SecondChildTAG: Fantastic!! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-08T19:13:38Z FirstChildTAG: True, but some of the functions do not work. For example, will you still will be able to show answers on homework and labs? I'm hoping that this becomes the equivalent of MIT OCW scholar for those of us less interested in certification. FirstChildUserIdTAG: 534915 FirstChildUserNameTAG: jturner421 FirstChildCreateTimeTAG: 2012-11-04T12:32:54Z IndexTAG: 948 TitleTAG: @STAFF - Community TA - Myrimit :)) Hello!! One question: if i quit/drop the course now, should I notify this or can I just keep on learning and do the exams and homework on spring again (and enroll again for that)? Thanks!! UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-11-02T16:32:40Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi, I took this from the last semester Course 6.002x. If they are based on the Prototype Course anything will happen if you drop, so don´t worry about it. You can follow and re-take it next semester. But, can I help you? In which part are you lost? Can I do something in order that you don´t abandone? Myriam. ---------- **How do I drop the course?** You do not have to do anything. You can simply stop working on the course at any time you choose to do so. **What happens if I drop the course?** For the prototype course, learners achieving grades of "A," "B," or "C" will receive an electronic Certificate of completion with the learner's name and grade on it. If you receive a grade below a "C" or do not complete the course, you will not receive a Certificate and no grade record attaching your name to your participation in the class will be disclosed outside of MITx. You can also choose to opt for a no record at any time. However, the posts you make while enrolled in the class will remain visible. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-02T17:36:03Z SecondChildTAG: Hi Myriam, thanks for your answer!! so "If they are based on the prototype course..." so, are they based on the prototype course? so, if I drop this time, could I take it up next spring again?? Thank you Myriam!! don't worry, it's just for personal reasons. You are already doing a lot. I'm not deceided yet :)) so it's just to "kind" of make sure I would have an opportunity again with the advantage that I will have already learned something before (now, in Fall) Thank you!! I'm not anonymous anymore :)) Sandra Navarro SecondChildUserIdTAG: 342966 SecondChildUserNameTAG: SandraNavarro SecondChildCreateTimeTAG: 2012-11-02T18:06:35Z SecondChildTAG: Cuenta conmigo para lo que necesites, Sandra. Espero que puedas continuar con el Curso :). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-02T18:35:35Z SecondChildTAG: Gracias Myriam, lo haré. Pero si lo dejara, podría hacer el de spring sin problemas? me dejarían registrarme e intentarlo otra vez? Gracias!!!! SecondChildUserIdTAG: 342966 SecondChildUserNameTAG: SandraNavarro SecondChildCreateTimeTAG: 2012-11-02T23:28:42Z SecondChildTAG: Sí, por su puesto, puedes intentarlo las veces que quieras, es más, hay mucha gente que ha dejado el Curso pasado (por distintos motivos) y se ha vuelto a re-inscribir en este Curso sin ningún tipo de problemas. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-03T01:17:45Z SecondChildTAG: Ohhhh gracias por la aclaración!! Una pregunta más (a mi desventaja, jeje): hasta cuando se pueden entregar los deberes semanales? 23.59 de los domingos, BOSTON TIME o GTM time? Por degracia, creo que a vece trabajo bajo presión.... y cuando se va acercando el momento me pongo las pilas. Tus hints me ayudan muchíssimo!! Espero, gracias a tu ayuda a tantissimos, TU trabajo duro den sus frutos!!!! Un abrazo, Sandra SecondChildUserIdTAG: 342966 SecondChildUserNameTAG: SandraNavarro SecondChildCreateTimeTAG: 2012-11-03T12:54:12Z SecondChildTAG: Hasta el momento, en este Curso, la entrega de los Hw y Labs ha sido hasta las 23:59 del domingo de tu hora local, es decir de tu País. Sin, embargo, en los Exámenes ha sido respecto del Boston Time. Por nada, me alegro mucho ser de ayuda a la Comunidad Hispanohablante. Gracias! :) Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-03T13:21:55Z SecondChildTAG: ok... yo siempre entregaba a las 01.59 hora de España (GTM, antes de que atrasaran la hora en europa, ahora serían las 00.59) y me lo aceptaban en la barra del "progress"... hmmmm ok pues mejor así :)) Claro que eres de ayuda a la comunidad hispanoparlante :)) Tu vives en Argentina o EEUU? vi en un post que eres de Argentina :)) Un abrazo, Sandra SecondChildUserIdTAG: 342966 SecondChildUserNameTAG: SandraNavarro SecondChildCreateTimeTAG: 2012-11-03T17:12:25Z SecondChildTAG: Por cierto, si vienes por Barcelona, avisa, tengo sitio en mi humilde morada :)) Te dejo mi e-mail en señal de agradecimiento: sandranavarro00@gmail.com SecondChildUserIdTAG: 342966 SecondChildUserNameTAG: SandraNavarro SecondChildCreateTimeTAG: 2012-11-03T17:56:56Z SecondChildTAG: Hola Sandra! Sí, vivo en argentina :)! jaja, gracias por la bienvenida a Barcelona XD!(igual mi ayuda siempre fue desinteresada jaja, lo hago porque disfruto mucho ayudando aquí, sin querer recibir nada a cambio jaja)! Bueno, agendo tu e-mail por si me voy algún día de vacaciones a España (jajja, nah, no soy aprovechadora jaja, pero gracias gracias por ofrecerme quedarme en tu casa jaja):). Si quieres también puedes escribirme a myrimit(at)gmail(dot)com por cualquier cosa que necesites. Un abrazo! Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-04T01:01:00Z SecondChildTAG: Gracias :)) No, encantada!!!! solo me gusta ser agradecida.... me siento bien así :)) Hasta pronto!!!! Un abrazo, Sandra SecondChildUserIdTAG: 342966 SecondChildUserNameTAG: SandraNavarro SecondChildCreateTimeTAG: 2012-11-04T11:07:43Z IndexTAG: 949 TitleTAG: Suggestion for Lecture Videos Download Hi! First of all, since this is my first time posting in the discussion forums, I would like to say a huge Thank You to all the MITx staff for this course, as it has tons of valuable knowledge for free, and you guys are doing a great job at providing it to everybody! Now for the main point of my post: I would suggest you guys have a link where we could download all the lecture videos for each sequence at the same time, like one big "zip" file or something. I'm saying this because I'm going out of town for the weekend to a place where I will not have internet access, and I want to watch the lecture videos there. But I'm taking a really looooooong time going to each video's page and downloading them one by one, so I think it would be much more practical if we could get all the videos for a particular sequence in a single file. By the way, it would also be nice to have some kind of "offline form" of the homework too (like a PDF file of the homework questions), so I could solve it when I'm out, then just come back home on sunday night to fill out the answers in the courseware page. Renato Piovesana Sao Paulo, Brazil UserIdTAG: 225639 UserNameTAG: repiovesana CreateTimeTAG: 2012-11-02T13:59:12Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 2 FirstChildTAG: Hello Renato, try saving the pages for offline use in Firefox. Have fun. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-11-02T14:28:29Z FirstChildTAG: Olá Renato. Também sou de São Paulo. Estou tentando fazer uma biblioteca com o material desse curso. Tem muito filme para baixar no YouTube, mas tem que ter muita paciência. Um abraço. Elias. FirstChildUserIdTAG: 361823 FirstChildUserNameTAG: EliasOak FirstChildCreateTimeTAG: 2012-11-02T15:30:11Z IndexTAG: 950 TitleTAG: Neon Relaxation Oscillator with Thevenin The problem can also be solved by doing a Thevenin equivalent of the curcit when the neon ligt is on and using an equaion we did on the class. ![Circuit diagram][1] R_th = R || R_s = (R*R_s)/(R+R_s) = 9933,77 V_th = V_Rs = V_s*(R_s)/(R_s + R) = 0,609V After we have this, we can calculate the time via this equation: v_C = V_I + (V_0 - V_I)*e^(-t/R*C) V_0 = 77V V_I = 0,609V v_C = 35V R = R_th = 9933,77 OHM When you crunch the numbers, you get: t = 0,007932s Doesn't match the solution 100%, but I think it's OK. [1]: https://edxuploads.s3.amazonaws.com/13516886766735177.bmp UserIdTAG: 294766 UserNameTAG: dejanst CreateTimeTAG: 2012-10-31T13:03:02Z VoteTAG: 3 CoursewareTAG: Week 6 / Neon relaxation oscillator exercise CommentableIdTAG: 6002x_neon_relaxation_oscillator_exercise NumberOfReplyTAG: 1 FirstChildTAG: This is very true- you should always look to take advantage of the concepts you learned previously, you never know when they can help! FirstChildUserIdTAG: 390034 FirstChildUserNameTAG: kahlil FirstChildCreateTimeTAG: 2012-10-31T14:54:38Z IndexTAG: 951 TitleTAG: H9P2 THERE ANSWER IS NOT CORRECT the circuit is clearly RLC parallel. why the answer accepted by the grader is Q for RLC series? thanks for all your effort 6.002x team!!!!! UserIdTAG: 9460 UserNameTAG: aloreta CreateTimeTAG: 2012-10-31T12:55:00Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: aloreta, you are right. There was a mistake in the homework grader, it should be corrected very soon. Thanks for pointing that out! FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-10-31T14:52:03Z SecondChildTAG: Please see H9P1. Why UC(t-) not equal UC(t+)??? SecondChildUserIdTAG: 464744 SecondChildUserNameTAG: attache SecondChildCreateTimeTAG: 2012-10-31T18:18:44Z SecondChildTAG: Because of additional charged pumped into the capacitor at time t=5. SecondChildUserIdTAG: 208296 SecondChildUserNameTAG: OZ1 SecondChildCreateTimeTAG: 2012-11-05T15:55:16Z IndexTAG: 952 TitleTAG: Final exam release date Dear staff Could you please confirm if the 20th December is the date the final exam needs to be submitted or the date the exam is due? Please release both the release and due date. With the holiday season being so close to the exam date, a few days will make a big difference. Thank you cwmb UserIdTAG: 107899 UserNameTAG: cwmb CreateTimeTAG: 2012-10-31T12:06:35Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: It seems obvious to me that the plan is to release the final on December 20th. The due date will probably be December 23rd. FirstChildUserIdTAG: 90796 FirstChildUserNameTAG: ChaunceyGardiner FirstChildCreateTimeTAG: 2012-11-01T00:06:44Z IndexTAG: 953 TitleTAG: S18E3 Graph An interactive graphical solution can be found at https://www.desmos.com/calculator/qhdnbat5ef. The values are accurate but time has been scaled by 1000 to cope with the resolution limit. Move the R slider to change the damping. UserIdTAG: 366083 UserNameTAG: smath CreateTimeTAG: 2012-10-31T09:59:25Z VoteTAG: 3 CoursewareTAG: Week 9 / S18E3 Total Solution CommentableIdTAG: S18E3_Total_Solution NumberOfReplyTAG: 3 FirstChildTAG: Big Thanks for this FirstChildUserIdTAG: 623210 FirstChildUserNameTAG: preveen FirstChildCreateTimeTAG: 2012-10-31T10:58:17Z FirstChildTAG: Beautiful. How did you actually find this desmos? FirstChildUserIdTAG: 252722 FirstChildUserNameTAG: vaboro FirstChildCreateTimeTAG: 2012-11-16T00:44:20Z FirstChildTAG: Excellent post! Thanks for sharing... FirstChildUserIdTAG: 342148 FirstChildUserNameTAG: gustavopereira FirstChildCreateTimeTAG: 2012-11-18T22:47:53Z IndexTAG: 954 TitleTAG: Midterm Q4 Expanation video - download link 52.6 MB - requested by Wahabbaluch Hi Wahabbaluch, As I have promised you I have uploaded the video for those who are facing problems with YouTube... you can take a look [here][1] See you! Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/undefined/threads/508f43a96d0c542500000007 UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2012-10-30T22:38:09Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Nice explaination. Gracias Wahab. FirstChildUserIdTAG: 455950 FirstChildUserNameTAG: Wahabbaluch FirstChildCreateTimeTAG: 2012-10-31T04:28:43Z SecondChildTAG: You are welcome :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-31T10:59:25Z IndexTAG: 955 TitleTAG: H7P3 Q1 I solve. But i don't understand. Why i=V/(RS+RO)? UserIdTAG: 196404 UserNameTAG: Yel1owstone CreateTimeTAG: 2012-10-30T21:09:23Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Remove the inductor and you will see why;-) FirstChildUserIdTAG: 623210 FirstChildUserNameTAG: preveen FirstChildCreateTimeTAG: 2012-10-31T07:48:20Z FirstChildTAG: Hi, in that question you're asked final voltage across resistor RO. Remember that the inductance, after some time, behaves as a short circuit, so you end up with a voltage divider. That gives you the final voltage. FirstChildUserIdTAG: 220837 FirstChildUserNameTAG: Calsomus FirstChildCreateTimeTAG: 2012-10-30T22:46:28Z SecondChildTAG: got it.u r genius calsomus:) SecondChildUserIdTAG: 97359 SecondChildUserNameTAG: azeem179 SecondChildCreateTimeTAG: 2012-11-01T15:35:23Z SecondChildTAG: clear. SecondChildUserIdTAG: 196404 SecondChildUserNameTAG: Yel1owstone SecondChildCreateTimeTAG: 2012-11-01T21:44:50Z SecondChildTAG: mmm stop. Inductance, after some time, behaves as a short circuit with DC or AC? there is a difference? SecondChildUserIdTAG: 196404 SecondChildUserNameTAG: Yel1owstone SecondChildCreateTimeTAG: 2012-11-01T21:46:25Z IndexTAG: 956 TitleTAG: I need a hint for H8P1 ... because I'm obvious on the wrong track here; every time I do the math, I have, for example, vc(1ms)=0.952V Even the sandbox simulation shows the same value but I'm getting an wrong answer... Thanks UserIdTAG: 277787 UserNameTAG: kirilaska CreateTimeTAG: 2012-10-30T17:59:06Z VoteTAG: 3 CoursewareTAG: Week 8 / Response To Impulse Limit Case CommentableIdTAG: 6002x_Response_To_Impulse_Limit_Case NumberOfReplyTAG: 3 FirstChildTAG: Is this the only part of the problem that's giving you difficulty? There are various ways to attack the problem. I think the easiest way is to solve the problem for a unit step, and then differentiate to get the response to a unit impulse. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-30T18:09:46Z SecondChildTAG: try solving it using laplace it will be soooooooooo easy SecondChildUserIdTAG: 84213 SecondChildUserNameTAG: mohamed373 SecondChildCreateTimeTAG: 2012-10-31T00:11:45Z SecondChildTAG: Thanks skyhawk; I had a blonde moment :) that's what I did but using the wrong formula... I don't understand why sandbox simulation didn't show the right response... Oh, well... SecondChildUserIdTAG: 277787 SecondChildUserNameTAG: kirilaska SecondChildCreateTimeTAG: 2012-10-31T01:33:25Z FirstChildTAG: try solving it using laplace it will be soooooooooo easy FirstChildUserIdTAG: 84213 FirstChildUserNameTAG: mohamed373 FirstChildCreateTimeTAG: 2012-10-31T00:11:55Z FirstChildTAG: So what formula should we use? I can't get the right answer for Vr1(1ms). Thanks for your help! FirstChildUserIdTAG: 245291 FirstChildUserNameTAG: rlicas FirstChildCreateTimeTAG: 2012-11-07T01:42:30Z SecondChildTAG: First, you should think about the hint given in (c), that the branches with L respectively C are decoupled. Now try to solve this as you were to calculate the voltage across the inductor and neglecting the C-Branch with $V_L=\frac{R}{L}*(\text{Formula for decay})$. I hope that this helps you without anticipating to much of the answer. SecondChildUserIdTAG: 219823 SecondChildUserNameTAG: Karsroh SecondChildCreateTimeTAG: 2012-11-07T19:45:38Z IndexTAG: 957 TitleTAG: playlist for online viewing... i wish we have the "playlist for online viewing" options for all the weeks' classes.... it's present only for the 1st six weeks.... in fact it greatly help me avoid buffering problems.... i could just open all videos... Allow them to buffer fully and then sit at a stretch.... UserIdTAG: 316353 UserNameTAG: Vivekanand123 CreateTimeTAG: 2012-10-29T16:45:45Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: You can make yourself a playlist.Copy video youtube's url to another tab, create a youtube playlist, then copy url and add.It's a tedious process, but if it's ok, i will share my playlists.I have 2 by now, since max nr of videos from a playlist is 200, at least for me. FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-10-29T17:22:52Z FirstChildTAG: http://www.youtube.com/playlist?list=PLTgEiAIXLilKxLeUqkqPzfBUP0eScqqnF my first playlist until week 8 https://www.youtube.com/playlist?list=PLTgEiAIXLilJ73IwCn-S9U9_7CFxp_345 the second playlist, week 9 included. FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-10-29T18:06:36Z SecondChildTAG: Hi, *Alexalexander*, but wht is the solution for those who can't access **"Youtube"** SecondChildUserIdTAG: 329051 SecondChildUserNameTAG: asadbhatti42 SecondChildCreateTimeTAG: 2012-11-05T12:39:49Z SecondChildTAG: Sorry, but i cannot help with that. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-12-06T09:05:44Z IndexTAG: 958 TitleTAG: Symbol ǁ Please?? Hi, can anybody pls. explain what does the symbol ǁ mean?? Thanks.... Regards from Barcelona, Sandra UserIdTAG: 342966 UserNameTAG: SandraNavarro CreateTimeTAG: 2012-10-27T18:37:11Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: A shorthand for parallel connection, i.e. R1 || R2 might mean R1*R2/(R1+R2) in calculations. FirstChildUserIdTAG: 189680 FirstChildUserNameTAG: vnd FirstChildCreateTimeTAG: 2012-10-27T18:38:15Z SecondChildTAG: Great, thanks!! SecondChildUserIdTAG: 342966 SecondChildUserNameTAG: SandraNavarro SecondChildCreateTimeTAG: 2012-10-27T18:56:01Z SecondChildTAG: for parallel SecondChildUserIdTAG: 410823 SecondChildUserNameTAG: ALIMURTAZA007 SecondChildCreateTimeTAG: 2012-10-27T20:23:23Z IndexTAG: 959 TitleTAG: Misplaced worked problems? Step and impulse inputs with those delta functions are not covered in the lectures and don't appear in the following HW (Week 7). So it is actually strange to even see problem statement -- I had no idea what u(t) and delta(t) means and had to look it up in the textbook. It would be completely weird if they are not needed in the following sequences as well. UserIdTAG: 189680 UserNameTAG: vnd CreateTimeTAG: 2012-10-27T18:33:42Z VoteTAG: 3 CoursewareTAG: Week 7 / Unknown Capacitance CommentableIdTAG: 6002x_Unknown_Capacitance NumberOfReplyTAG: 1 FirstChildTAG: The **impulse** and **step** functions, among others, usually do not appear until a course on Signal Analysis. For the other students wondering, these functions appear starting on **Page 482** of our Textbook. They are useful as a teaching aid, so Prof. Agarwal introduces them early. These functions are useful when studying first-order circuits with capacitors (and second-order with inductors as well) , as follows: What happens when you apply DC to a circuit with a **first-order circuit** (a circuit with a capacitor and resistor in parallel or in series, or both)? It will charge to it's steady state, and nothing interesting happens. If you apply AC to a first-order circuit, it acts as a frequency-dependent resistor in the time-domain (e.g. a simple filter). Somewhat interesting, but this does not show the full response of such a circuit. Finally, with an impulse function applied, the capacitor in a first-order circuit charges instantaneously, and then begins it's discharge until the first-order circuit reaches steady-state. Thus, the full dynamic of these circuits can be studied in this manner. Later courses will teach you how to manipulate a signal (and multiple signals) in all sorts of interesting ways. FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-10-28T01:44:12Z IndexTAG: 960 TitleTAG: Staff and All: How Does Midterm 24-Hour Countdown Start? Hello Staff and All, While in the Courseware section printing some of the past exercises, homeworks, and labs for review last night, Friday night, at about 9 PM Boston time, I clicked on the Midterm Exam link by mistake. I thought that some of the review materials and the practice test might be in there, as well as the actual midterm exam. I saw a page with some circuit diagrams and figured that I might have been looking at the actual exam before I intended to do so. In a panic, I got right out of there. I don't know if I was on an introductory information or cover page or in the actual exam. Now I am wondering if I have to get back in there and finish by 9 PM Boston time tonight on Saturday or if I can start and finish tomorrow on Sunday after continuing some review. Much thanks for any urgent guidance, please! Laureen UserIdTAG: 188609 UserNameTAG: Laureen CreateTimeTAG: 2012-10-27T15:23:45Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Hi Laureen, Have you click on here? ![menu][1] Have you seen this warning window message once you clicked on it? ![image][2] If you have seen this and you accepted by clicking on the down arrow of that web message , after you accepted to start the Exam, you will see the Exam and that means that you have started the Midterm Exam (and the clock starts counting for you, you have to control your time by yourself)... If you haven't click on the arrow (down this message), you haven't started your Exam. I hope this can help you, Myriam. P.D: Remember that once you have started the Midterm Exam you can not Discuss about the Exam content... [1]: https://edxuploads.s3.amazonaws.com/13513519741343694.png [2]: https://edxuploads.s3.amazonaws.com/13511645821343618.png FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-27T15:40:34Z FirstChildTAG: Hi Myriam, I only clicked the first Courseware - Midterm Exam button, but not the one that says Midterm Exam - Midterm due October 28, 23:59 with the little clock - because I did not see the introductory page that you pasted above. Also, I did not click any arrow. I think that the circuits that were showing on my screen were the ones from the previous section were I had been printing. So, now I am pretty sure that I did not start the exam countdown. What a relief! Another question, once one starts the countdown, can one log out and log back in before the countdown ends? Someone else in the discussion forum was asking that. Thank you very, very much! Laureen FirstChildUserIdTAG: 188609 FirstChildUserNameTAG: Laureen FirstChildCreateTimeTAG: 2012-10-27T17:01:32Z SecondChildTAG: Hi Laureen! You are welcome! :) I was worried about your issue haha, nice to know that you didn't click the button by accident :p Unfortunately, the clock don't stops if you log out... So, if you start your Exam, independently of what you have done in the 24hs (log out, log in, etc...), the clocks still counting your time (You have to count your own time, as there will not be any counter in your Exam web page). Good luck! I hope this can help you, Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-27T17:21:19Z FirstChildTAG: Hi Myriam, Thanks for your help with the logistics of the midterm exam! Laureen FirstChildUserIdTAG: 188609 FirstChildUserNameTAG: Laureen FirstChildCreateTimeTAG: 2012-11-02T03:05:22Z SecondChildTAG: You are welcome! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-02T11:49:31Z IndexTAG: 961 TitleTAG: Second Order Differential Equations Hi, I am aware that week 9 of 6.002x will include second order circuits. Hence I guess that we will be asked to solve second order differential equations. I would be grateful if someone could recommend a site where I could learn how to solve second order differential equations so I may be better prepared for the coming weeks. Thank you in advance. UserIdTAG: 85490 UserNameTAG: yashsavani CreateTimeTAG: 2012-10-27T14:37:55Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Try this https://6002x.mitx.mit.edu/wiki/view/MathReview FirstChildUserIdTAG: 9673 FirstChildUserNameTAG: xp42 FirstChildCreateTimeTAG: 2012-10-27T21:12:56Z FirstChildTAG: If you would like to use a platform similar to this, you could try. http://www.khanacademy.org/math/differential-equations If not a google search will turn up many abbreviated alternatives. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-27T14:52:53Z SecondChildTAG: Ohh thanks, I'll have a look as well :)) SecondChildUserIdTAG: 342966 SecondChildUserNameTAG: SandraNavarro SecondChildCreateTimeTAG: 2012-10-27T16:25:31Z SecondChildTAG: Thank you ! SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-10-27T18:33:40Z IndexTAG: 962 TitleTAG: Schematic editor Hi, Since most of the people in this group are engineers, I was wondering what type of schematic editors do you prefer. I know many people use Cadsoft's Eagle however it seems weird that there would only be one such software. I would be much obliged if you could tell me what some of the other schematic editors that you use. Thanks in advance UserIdTAG: 85490 UserNameTAG: yashsavani CreateTimeTAG: 2012-10-27T09:58:15Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: proteus FirstChildUserIdTAG: 146930 FirstChildUserNameTAG: fayzan_007 FirstChildCreateTimeTAG: 2012-10-27T12:51:37Z FirstChildTAG: EAGLE is for PCB manufacture. There are many others, but they come at a high price $1000<, EAGLE is the most affordable/free, but also has the same core features. FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-10-27T10:47:56Z FirstChildTAG: Electronic Workbench. FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-10-27T11:27:49Z FirstChildTAG: Proteus, NI Multisim, TinyCAD (open source), ExpressPCB (free), KiCad (open source), Circuit Maker, Dip Trace, Circuit Lab (onLine), EDWinXP, PCB123, CirCAD, OrCAD, FreePCB (free). FirstChildUserIdTAG: 182664 FirstChildUserNameTAG: Glaucus FirstChildCreateTimeTAG: 2012-10-27T13:03:44Z FirstChildTAG: Eagle and LTSpice (for simulation) FirstChildUserIdTAG: 411504 FirstChildUserNameTAG: taubrafi FirstChildCreateTimeTAG: 2012-10-28T00:45:42Z IndexTAG: 963 TitleTAG: LAB 7 question 2 I have got the correct answer for the equation of Vc and my calculated value is extremely close to the simulated value. my calculator is in radians. What is the problem? It is showing me that I'm wrong. (my answers are same till 3 decimal digits after the decimal point) UserIdTAG: 135072 UserNameTAG: Shriniwas CreateTimeTAG: 2012-10-27T07:37:26Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: you should type 6 digits precision (rounded) FirstChildUserIdTAG: 482256 FirstChildUserNameTAG: Nus FirstChildCreateTimeTAG: 2012-10-27T10:30:46Z FirstChildTAG: Try to compute in built-in calculator using built-in "pi" constant (just copy your formula for vC(t) there and replace "t" with 0.0005), and then copy the result exactly as it appears. Worked for me. FirstChildUserIdTAG: 192703 FirstChildUserNameTAG: voffch FirstChildCreateTimeTAG: 2012-10-27T10:27:15Z IndexTAG: 964 TitleTAG: Good luck! Good luck to all those sitting for the Midterms! UserIdTAG: 143823 UserNameTAG: NathanNadeson CreateTimeTAG: 2012-10-26T17:59:28Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: thankss! FirstChildUserIdTAG: 127800 FirstChildUserNameTAG: MNB FirstChildCreateTimeTAG: 2012-10-26T19:25:05Z FirstChildTAG: Despite joining the course 5 weeks late I could finish midterm with 100%. Though I take a long time the experience was well worth it. Thank you so much Edx and MITx Thank you all for your all your messages. It has been a very nice experience. But I do feel the gaps knowledge because of late joining and not being able to watch/download the videos because of bandwidth issues of my EDGE connection. For those who took the trouble to go through the homeworks and labs It should only take one hour to score 100%. Thank you all once again. FirstChildUserIdTAG: 623210 FirstChildUserNameTAG: preveen FirstChildCreateTimeTAG: 2012-10-27T16:18:55Z IndexTAG: 965 TitleTAG: Integration with Wolfram Alpha I (basically) understand how to do integration with WA, but I have to split equations with integration into separate pieces to get them done. Is there a way to enter whole equations with integration into WA? UserIdTAG: 21541 UserNameTAG: JSChambers CreateTimeTAG: 2012-10-26T16:23:58Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Please show us! I am currently trying to understand how to put the equation in Lab seven into Wolfram Alpha, and am in the process of creating a tutorial/Hints thing on that (lab 7), but the wolfram integral part is giving me a hard time. I am going to see how far I can get with the Khan academy tonight. I'll post some screen shots to show how far I got. Thanks JSChambers!! FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-10-26T17:16:16Z SecondChildTAG: ![enter image description here][1] Here are my efforts thus far, I don't know what I'm doing ;) [1]: https://edxuploads.s3.amazonaws.com/13512724891343671.png SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-26T17:28:47Z SecondChildTAG: The lower limit on the integral needs to be 0. i = 0 for t<= 0. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-26T19:22:44Z SecondChildTAG: Awesom! Thanks. SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-26T19:51:13Z SecondChildTAG: Oh my goodness! It worked!!! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-26T19:55:25Z SecondChildTAG: Try putting the following into Wolfram Alpha: **1/c integrate sin(1000*2*pi*t) from t=0 to t** ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13512818219273546.png SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-26T20:03:45Z SecondChildTAG: good job hazel1919! SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-26T22:59:39Z FirstChildTAG: You can very simply use your calculus and the variable change method, since sin(1000*2*PI*t) is a composed function : g(x)= sin(f(x)), f(x)=2000*PI*x. The integration interval starts from 0 to t. You can think of an integral from -inf to 0 (which is 0 since i(t)=0 for t<=0) , plus our integral from 0 to t. FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-10-27T11:43:43Z FirstChildTAG: A mathematical note for future reference: In order for an integral with lower limit minus infinity to exist, the integrand must go to zero sufficiently rapidly as its argument goes to infinity. Since sin(x) does not have a limit as x goes to minus infinity (It oscillates between -1 and +1.), the integral is not defined with a lower limit of minus infinity. The point to take away from this comment is that it is possible to ask Wolfram Alpha to solve a problem that is not properly posed. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-28T21:16:34Z IndexTAG: 966 TitleTAG: I succeeded on the midterm I finished the exam, I worked hard and the results are noticeable. I learned a lot. Thank you all! UserIdTAG: 110802 UserNameTAG: leoblack CreateTimeTAG: 2012-10-26T02:04:30Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: I'm agree with you! :-) thanx a lot to Adnan! I can prove that my education is as well as MIT education!! FirstChildUserIdTAG: 299251 FirstChildUserNameTAG: TheRedBlackOne FirstChildCreateTimeTAG: 2012-10-26T05:10:01Z FirstChildTAG: Well, it was awesome. Finished my midterm too, need some time for relaxation now :) FirstChildUserIdTAG: 81712 FirstChildUserNameTAG: dzhon FirstChildCreateTimeTAG: 2012-10-26T05:10:12Z FirstChildTAG: Finished the midterm too. It was a good exam! Time for some relaxation! FirstChildUserIdTAG: 443988 FirstChildUserNameTAG: Dheeraj_Garg FirstChildCreateTimeTAG: 2012-10-26T05:57:54Z IndexTAG: 967 TitleTAG: The Humble 555 timer chip Are we going to see a tutorial on that humble of components, the NE555 timer chip? UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-10-25T20:41:01Z VoteTAG: 3 CoursewareTAG: Week 6 / Ring Oscillator CommentableIdTAG: 6002x_Ring_Oscillator NumberOfReplyTAG: 1 FirstChildTAG: I think that 555 timers find more application in home-brew and "quick and dirty" circuitry that's constructed in the lab. That's not to say that there's no theory behind the 555, as Week 6 especially relates to capacitors and the time constant, and the 555 timer uses the same time constant for it's timing mechanism. The problem with using a capacitor-resistor pair for timing is that it's not that accurate when used as a clock for digital circuitry; temperature, low-precision components, and leakage play a big role in reducing the accuracy of the 555. Crystal oscillators are much more stable. With micro-controllers becoming so prevalent, cheap, and easier and easier to program and use, they will take the place of 555 timers in about a generation or so, if not yet already. I remember using a 555 timer to create a cheap pulse-width-modulated (PWM) motor controller, along with a relay. This taught me the importance of diodes to clamp inductive feedback. I also found out that there is a CMOS flavor of the 555 (555CN), and that it's less sensitive to operating noise/conditions than the BJT-based 555. FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-10-25T22:47:54Z SecondChildTAG: the interesting thing that it's almost impossible to find 50% duty cycle schematics - the most common schematic with diode parallel one of charge/discharge resistor - it gives almost 50% duty cycle. I found exactly 50% only here: http://www.aldinc.com/pdf/time_15002.0.pdf SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-25T23:10:51Z SecondChildTAG: What I really really like is the fact that the 555 time has a relatively simple circuit - 3 resistors, 2 flip flops, a comparator and a transistor. Yet the number of applications of that one chip is huge! SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-10-26T15:34:20Z SecondChildTAG: That is why it is so accessible to hobbyists. SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-26T17:18:47Z SecondChildTAG: Where do you shop for components like the 555 besides Radio Shack? thanks for tips. SecondChildUserIdTAG: 184153 SecondChildUserNameTAG: tthngn SecondChildCreateTimeTAG: 2012-12-22T15:03:10Z IndexTAG: 968 TitleTAG: complete just completed with 97% still trying for 100% UserIdTAG: 342135 UserNameTAG: vikash902 CreateTimeTAG: 2012-10-25T18:24:23Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: finally completed 100% FirstChildUserIdTAG: 342135 FirstChildUserNameTAG: vikash902 FirstChildCreateTimeTAG: 2012-10-25T18:42:11Z SecondChildTAG: Total 54% 100% in Midterm 100% in all homeworks (1-8) SecondChildUserIdTAG: 357453 SecondChildUserNameTAG: BrunoCanoso SecondChildCreateTimeTAG: 2012-10-25T19:13:11Z FirstChildTAG: Now you can relax for the weekend! FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-25T19:43:29Z FirstChildTAG: I also got 100% FirstChildUserIdTAG: 216684 FirstChildUserNameTAG: Taimoor1017 FirstChildCreateTimeTAG: 2012-10-28T18:41:32Z IndexTAG: 969 TitleTAG: Putting it all together... ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13511820191343627.png UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-10-25T16:20:24Z VoteTAG: 3 CoursewareTAG: Week 7 / Falling Delay Part 3 CommentableIdTAG: 6002x_Falling_Delay_Part_3 NumberOfReplyTAG: 1 FirstChildTAG: You got it! FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-10-25T20:10:29Z IndexTAG: 970 TitleTAG: mid term exam i still have doubt that the moment i click on the MID-TERM exam link... from that moment my time will start or there is any further confirmation i have to make??? please reply as soon as possible ... UserIdTAG: 219442 UserNameTAG: aki31 CreateTimeTAG: 2012-10-25T06:57:14Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I think the moment you see the question paper, your time starts. FirstChildUserIdTAG: 446691 FirstChildUserNameTAG: maitrey FirstChildCreateTimeTAG: 2012-10-25T07:01:36Z SecondChildTAG: thanx maitrey..but it would be really helpful if anyone of staff give the confirm answer..what say SecondChildUserIdTAG: 219442 SecondChildUserNameTAG: aki31 SecondChildCreateTimeTAG: 2012-10-25T07:06:03Z SecondChildTAG: You can click the link and not have your timer start. You see a cover page first, once you advance beyond the cover page, your 24 hour timer begins. SecondChildUserIdTAG: 213386 SecondChildUserNameTAG: dmascenik SecondChildCreateTimeTAG: 2012-10-25T07:23:27Z FirstChildTAG: Hi. [Just Check It][1] > **Regards:** asadbhatti42 [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5088dbe5959a592b00000024 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-10-25T14:55:36Z IndexTAG: 971 TitleTAG: I hope its already 00:00:01 in boston? Is my time calculations right? Waiting for the exam to start..... UserIdTAG: 594747 UserNameTAG: vikashkumar140 CreateTimeTAG: 2012-10-25T04:48:45Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: yes u r right and we are all eagerly waiting too** FirstChildUserIdTAG: 124534 FirstChildUserNameTAG: srihari46 FirstChildCreateTimeTAG: 2012-10-25T04:52:00Z FirstChildTAG: I'm also waiting for the exam! But it's not available yet, I think. FirstChildUserIdTAG: 277808 FirstChildUserNameTAG: Hemanthmps FirstChildCreateTimeTAG: 2012-10-25T04:52:18Z FirstChildTAG: its more than 45 mins than 00:01.............still exam not released...... FirstChildUserIdTAG: 326759 FirstChildUserNameTAG: karthickviswan FirstChildCreateTimeTAG: 2012-10-25T04:53:21Z SecondChildTAG: yes it is late ...?????? SecondChildUserIdTAG: 428658 SecondChildUserNameTAG: Moza SecondChildCreateTimeTAG: 2012-10-25T05:10:59Z IndexTAG: 972 TitleTAG: midterm exam if we go in and print the exam we have to submit the exam 24 hours after that or will we have till Sunday at midnight to turn it in. am just a little scared about the exam I do not do good on tests. I want to wish everyone good luck UserIdTAG: 160242 UserNameTAG: Mlevins35 CreateTimeTAG: 2012-10-24T22:27:43Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hi Mlevins35, You can set for the Midterm Exam from a period of days. If you take a look at course info [here][1] it says : *In order to accommodate the Muslim holiday of Eid al-Adha, we have extended the deadline to submit the midterm exam to 23:59 (11:59 pm) on Sunday, October 28th, Boston time. The exam will still be released on Thursday, October 25th at 00:01 (12:01 am) Boston time.* I have made for you this Diagram, that might can help you: ![enter image description here][2] My best wish to you! Good luck! Myriam. P.D: Click with the right button of the mouse and open the image in other window, you will see it bigger than this one ;). [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/info [2]: https://edxuploads.s3.amazonaws.com/1351119783134361.png FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-24T23:03:41Z FirstChildTAG: I do not do well on tests, either. I blame Calculus. ;-) FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-10-25T05:31:11Z IndexTAG: 973 TitleTAG: @Staff I see that there would be a question or so (with numericals changed of course)From homework/ lab's repeating in the Mid term exam. So will we be able to access our past Homework / Lab's submissions, solutions. If so are we allowed to insert our numeric's in the very same formulas given in solution and deduce the answer? UserIdTAG: 148523 UserNameTAG: span993 CreateTimeTAG: 2012-10-24T09:42:45Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 974 TitleTAG: OLD MIDTERM EXAM Dear staff, Is it possible not to limit the number of time to check the answer? I do check after each question and I realized after that it there is a limit to the number of time you can check your answers. Sincerely! UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-24T00:35:47Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Hi anonymous, Unfortunately, no... You will have only 3 attempts per each problem as we had in the 6.002x (spring). You can read [this Post][1]. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/508618f0ffe2512b00000058 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-24T00:49:23Z SecondChildTAG: And Do you haveany idea about sample/old mid term papers? SecondChildUserIdTAG: 281159 SecondChildUserNameTAG: ManasiS SecondChildCreateTimeTAG: 2012-10-24T01:05:13Z SecondChildTAG: Myrimit, there is a sample exam: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Midterm_Spring_2012/Midterm_Exam_1121/ Does it reflect real size and complexity of midterm? SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-24T01:08:13Z SecondChildTAG: Hi Manasis, Take a Look at Course info :). We had this Midterm Exam in 6.002x Spring [read here][2]. It would be good if you can practise with it :) See you! Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/info [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Midterm_Spring_2012/Midterm_Exam_1121/ SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-24T01:11:58Z SecondChildTAG: Yes @Yakovo, In fact that was our Midterm Exam in 6.002x Spring ;). The only difference is that when we did the Midterm Exam we didn't knew the Format of the Midterm Exam of the previous Course because we were the Prototype of 6.002x haha, so we were a little bit unclueless haha, so now that you have the chance of knowing the previous Midterm Exam, you should try to practise with it. My best wih to you in the Exam! Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-24T01:23:54Z SecondChildTAG: thanks looks pretty simple :) SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-24T01:27:46Z FirstChildTAG: Hi! The solutions for the last Midterm are there, but there is no working. I need help on question 3, parts (c) , (d) and (e) FirstChildUserIdTAG: 143823 FirstChildUserNameTAG: NathanNadeson FirstChildCreateTimeTAG: 2012-10-24T12:56:44Z SecondChildTAG: I have some difficulty getting the small signal model SecondChildUserIdTAG: 143823 SecondChildUserNameTAG: NathanNadeson SecondChildCreateTimeTAG: 2012-10-24T13:02:59Z FirstChildTAG: Hi, It's not mention anywhere but this news is very authentic that we have only 3 attempts to solve any query in **MIDTERM EXAM**, [MIDTERM Info of MITx Spring FALL-2012][1] [1]: https://6002x.mitx.mit.edu/wiki/view/MidtermInfo after checking above link it is very clear that v have only 3 attempts... > **Regards:** asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-10-24T15:07:22Z IndexTAG: 975 TitleTAG: STAFF::BJT IN COURSE? while going through the book I found a considerable discussion on the bipolar junction transistors and its analysis. I wished to ask whether it is part of this course 6.002x or not? Shouldn't we focus on Mosfets only? UserIdTAG: 368712 UserNameTAG: jmen CreateTimeTAG: 2012-10-23T17:22:03Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: theory could be applied to any device - MOSFET is (partially) VCCS and BJT is CCCS FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-23T17:28:34Z IndexTAG: 976 TitleTAG: will edx be free? so for how many more semesters edx will be free UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-23T14:33:18Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: The basic offerings of edX will always be free. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-10-23T14:46:59Z SecondChildTAG: what do "basic offerings" correspond to? SecondChildUserIdTAG: 278792 SecondChildUserNameTAG: sali SecondChildCreateTimeTAG: 2012-10-23T20:36:52Z SecondChildTAG: **sali**: You must understand that everything at edX is very new at this stage. One thing that we do know at this stage is that edX is following a **nonprofit** business model, compared to the for-profit model envisioned by its competitors, one of which is Coursera. I am not affiliated with MIT nor am I affiliated with edX, so I don't know what Lyla means by "basic offerings." My guess is that students will be able to *choose* between a "**free**" Honor Code certificate, or a "**paid**" proctored final exam. In a [recent edX press release][1], Amanda Keane (akeane@webershandwick.com) announced that "edX learners now have the option of taking a course final exam at one of over 450 [Pearson VUE test centers][2] in more than 110 countries. Proctors at the centers will verify the identity of the examinee and administer the tests." One edX course this Fall (not announced yet) will have this option. With a proctored exam, there comes the possibility of getting **college transfer credits** (*maybe* even MIT credit for another fee) in addition to just a "Certificate of Mastery." But this is not definite as far as I know, **just my own speculation**. Ms. Keane has provided an email address, so feel free to contact her for more details, such as **cost**... [1]: https://www.edx.org/press/edX-announces-proctored-exam-testing [2]: http://www.pearsonvue.com/faqs/ SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-10-25T17:12:59Z IndexTAG: 977 TitleTAG: LAB 7 Why is this formula incorrect?? (-1/(X*X))*X(X*X*X*t) (**Note:** I have removed the incorrect formula that **Ignaas** has provided, because although incorrect, it is very close to the actual answer, and that's against the Honor Code. I have replaced terms like PI, sin, cos, and numbers with an **X**. I have preserved the *form* of **Ignaas**'s incorrect formula so that other readers may assist him in finding the problem. -**JerseyMark**/Community TA) UserIdTAG: 308861 UserNameTAG: Ignaas CreateTimeTAG: 2012-10-23T12:09:49Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: me too i have same problem FirstChildUserIdTAG: 331441 FirstChildUserNameTAG: abdessamade FirstChildCreateTimeTAG: 2012-10-23T12:24:13Z SecondChildTAG: i got it you forget -1 SecondChildUserIdTAG: 331441 SecondChildUserNameTAG: abdessamade SecondChildCreateTimeTAG: 2012-10-23T12:27:34Z SecondChildTAG: hey i too cant get an answer for this one.. same problem. any hints? SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-23T12:33:17Z SecondChildTAG: Why, -1 is present here. SecondChildUserIdTAG: 213452 SecondChildUserNameTAG: JoJosida SecondChildCreateTimeTAG: 2012-10-23T12:35:20Z SecondChildTAG: It's a definite integral. Don't forget the contribution from the lower limit. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-23T12:40:06Z SecondChildTAG: Yes skyhawk you put integral to sommation of 2 integrals don't gorhet the inital condition SecondChildUserIdTAG: 331441 SecondChildUserNameTAG: abdessamade SecondChildCreateTimeTAG: 2012-10-23T12:49:47Z SecondChildTAG: not getting it skyhawk..i mean value of sine at -infinity? SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-23T12:49:57Z SecondChildTAG: skyhawk i got this (1/2*PI)*((-cos(1000*2*PI*t))+1) evaluating the integral on the upper and the lower values SecondChildUserIdTAG: 320871 SecondChildUserNameTAG: seb1256 SecondChildCreateTimeTAG: 2012-10-23T12:53:39Z SecondChildTAG: could anyone advise me why i am still getting the wrong answer,although I got green tick on everything else. SecondChildUserIdTAG: 320871 SecondChildUserNameTAG: seb1256 SecondChildCreateTimeTAG: 2012-10-23T12:58:45Z SecondChildTAG: me i got (1/2*PI)*((-cos(1000*2*PI*t))-1) SecondChildUserIdTAG: 331441 SecondChildUserNameTAG: abdessamade SecondChildCreateTimeTAG: 2012-10-23T12:58:53Z SecondChildTAG: Signal is zero before t = 0. Lower limit is 0. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-23T13:06:40Z SecondChildTAG: Be careful! The factor out front is 1/(2*PI). SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-23T13:11:17Z SecondChildTAG: If you are having trouble getting the correct answer check all your signs. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-23T13:18:40Z SecondChildTAG: hi iwas able to find the correct formula but when i sub in it by t= 0.5ms i got a incorrect answer why? SecondChildUserIdTAG: 84213 SecondChildUserNameTAG: mohamed373 SecondChildCreateTimeTAG: 2012-10-23T13:38:05Z SecondChildTAG: Maybe your value for cos(PI) is incorrect. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-23T13:44:02Z SecondChildTAG: i should sub insie the cos in radien rigt? thats what i did SecondChildUserIdTAG: 84213 SecondChildUserNameTAG: mohamed373 SecondChildCreateTimeTAG: 2012-10-23T14:56:56Z SecondChildTAG: The value should be one you memorized when you learned trig. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-23T15:15:07Z SecondChildTAG: i hv the same problem as mohamed373. i hv the formula right but wrong when evaluating for t=.5ms? i dont get it??? SecondChildUserIdTAG: 285945 SecondChildUserNameTAG: lindalapiso SecondChildCreateTimeTAG: 2012-10-30T08:07:16Z FirstChildTAG: skyhawk, Why is my answer incorrect...the lower limit is 0.... FirstChildUserIdTAG: 308861 FirstChildUserNameTAG: Ignaas FirstChildCreateTimeTAG: 2012-10-23T19:33:35Z SecondChildTAG: Last time I checked cos(0) /= 0. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-23T22:09:44Z SecondChildTAG: hmm ok...but there it is stated infinity SecondChildUserIdTAG: 308861 SecondChildUserNameTAG: Ignaas SecondChildCreateTimeTAG: 2012-10-24T01:48:25Z SecondChildTAG: In the problem statement it says: Assume that I(t)=0 for t≤0 and simplify your equation accordingly. Therefore the lower limit for the integral is t = 0. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-24T11:00:35Z FirstChildTAG: SKYHAWK.....the lower limit is zero....what is wrong with my expression...i don't get the green tick FirstChildUserIdTAG: 308861 FirstChildUserNameTAG: Ignaas FirstChildCreateTimeTAG: 2012-10-23T19:28:24Z SecondChildTAG: See above. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-23T22:10:37Z SecondChildTAG: Be sure that you considered the chain rule properly. Regrouping terms and integrating again may help you. SecondChildUserIdTAG: 482646 SecondChildUserNameTAG: elgambitero SecondChildCreateTimeTAG: 2012-10-24T17:39:24Z FirstChildTAG: Check out [this Wikipedia article][1], [1]: http://en.wikipedia.org/wiki/Lists_of_integrals#Tables_of_integrals and pay particular attention to the initial condition as assumed in the instructions. FirstChildUserIdTAG: 184153 FirstChildUserNameTAG: tthngn FirstChildCreateTimeTAG: 2012-10-24T17:36:28Z FirstChildTAG: 1/(X*X)*((-X(X*X*X*X))+X) 100% CORRECT FirstChildUserIdTAG: 228440 FirstChildUserNameTAG: hassanmuhuyuddin FirstChildCreateTimeTAG: 2012-10-25T16:09:56Z SecondChildTAG: Dear **hassanmuhuyuddin**: Please **do not** post Homework and/or Lab answers in Discussion per the Honor Code that governs your grade. I have edited your response to *remove the exact answer*, while still preserving the *form* of your equation as that may be of some assistance to students needing a proper hint. SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-10-25T16:33:37Z IndexTAG: 978 TitleTAG: Ouch That was the longest 5 minutes of my life :( UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-10-23T10:30:31Z VoteTAG: 3 CoursewareTAG: Week 6 / First-order differential equation CommentableIdTAG: 6002x_first_order_differential_equation NumberOfReplyTAG: 1 FirstChildTAG: My head exploded. Brains are difficult to clean from ceilings. FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-10-23T11:23:09Z SecondChildTAG: Haha, glad to hear i'm not alone. All I can do is hope they don't ask us to find Vc any time soon! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-23T11:38:02Z SecondChildTAG: hhhhhhhhhh SecondChildUserIdTAG: 428658 SecondChildUserNameTAG: Moza SecondChildCreateTimeTAG: 2012-10-23T22:04:32Z IndexTAG: 979 TitleTAG: Help with mid-term examples! I can't figure out the mid-term example 1 in https://6002x.mitx.mit.edu/section/MidtermFormatExamples/ The first question asks for VTH, right? I think you should first calculate voltage drop through speaker (P=V^2/R), then current in closed circuit (P=I^2*R) and at last VTH=V_DROP_SPEAKER+(RTH*I) to calculate "open-circuit voltage". Maybe I'm getting wrong something related to RMS values? Please, tell me where I am wrong! HELP! Thanks a lot! UserIdTAG: 398594 UserNameTAG: DaveyJC CreateTimeTAG: 2012-10-23T00:26:20Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: every time I click Grade button values are changing!!! probably it's a bug of example, but it will be very frustrating on exam! for first problem: V=sqrt(R*P)*(1+Rth/R) FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-23T06:40:14Z SecondChildTAG: Why are you using Rth? My solution would be `sqrt(P / R)`, as the question states that the amplifier pumps P Watt through his R ohm speaker (at output terminal). So that makes `I = V / R` & `P = V * I` `V^2 = P / R` `V = sqrt( P / R)` Did I misinterpret the question? SecondChildUserIdTAG: 114881 SecondChildUserNameTAG: xvink SecondChildCreateTimeTAG: 2012-10-23T08:36:29Z SecondChildTAG: Hmm.. well, it looked a LOT nicer in the preview pane. Where is my markup? SecondChildUserIdTAG: 114881 SecondChildUserNameTAG: xvink SecondChildCreateTimeTAG: 2012-10-23T08:37:19Z SecondChildTAG: First, it will be sqrt(P*R) - check your math Second, sqrt(P*R) is a voltage on the amplifier output when speaker is connected, so it's the voltage on voltage divider (Rth and R in series, and voltage across R). So: v=sqrt(P*R) v=V*R/(R+Rth) sqrt(P*R)=V*R/(R+Rth) V=sqrt(R*P)*(1+Rth/R) SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-23T14:43:52Z SecondChildTAG: Hi fellows! well, I've done it in a graphical way, and my answer is equal to YakovO. But... when I entered the result didn't get the green tick. After trying many times (by the way, more than 3), got the green tick, but the problem was with other values and the result was of the previous ones! Seems buggy! SecondChildUserIdTAG: 220837 SecondChildUserNameTAG: Calsomus SecondChildCreateTimeTAG: 2012-10-23T17:48:51Z SecondChildTAG: Could someone of the staff helps us with this problem? SecondChildUserIdTAG: 220837 SecondChildUserNameTAG: Calsomus SecondChildCreateTimeTAG: 2012-10-23T17:50:47Z SecondChildTAG: Calsomus, on last try you were lucky to get the same problem's parameters after click "Grade" button - notice that after hit "Grade" button parameters changes - it's why most cases your solution will be for different parameters and you get red SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-23T18:33:42Z SecondChildTAG: That's right YakovO, parameters repited when clicking again, so my alredy written solution fit. Hope mid term is not like that!!!!! Best luck for all! two days left. SecondChildUserIdTAG: 220837 SecondChildUserNameTAG: Calsomus SecondChildCreateTimeTAG: 2012-10-23T23:45:35Z SecondChildTAG: I've just noticed there's another example, just posted in Course Info. Let's try with that one. SecondChildUserIdTAG: 220837 SecondChildUserNameTAG: Calsomus SecondChildCreateTimeTAG: 2012-10-23T23:48:22Z SecondChildTAG: Looks pretty simple, but doesn't solve it yet though. Is it real size and complexity of exam? SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-24T01:05:21Z SecondChildTAG: It's last year's > Midterm Study Materials: > > If you would like some extra problems to help you study for the > midterm, as well as an example of the midterm format, you can use last > year's midterm. Just to be clear, the real midterm will not have the > 'show answer' button available during the examination period, but the > rest of the format is the same as the practice midterm. SecondChildUserIdTAG: 220837 SecondChildUserNameTAG: Calsomus SecondChildCreateTimeTAG: 2012-10-24T01:24:44Z SecondChildTAG: @YakovO: thanks for the explanation.I checked my math and you are absolutely right. At one point I multiplied by P instead of dividing. No wonder I find this course so hard :) My math is quite rusty after so many years not being used SecondChildUserIdTAG: 114881 SecondChildUserNameTAG: xvink SecondChildCreateTimeTAG: 2012-10-24T18:16:54Z FirstChildTAG: I don't know, but that practice exam has a bug in it if you ask me. Tried to grade it and the question parameters change (and my answer is wrong). If that is how the mid term will be, then everyone will get a 0% (= D- ?) FirstChildUserIdTAG: 114881 FirstChildUserNameTAG: xvink FirstChildCreateTimeTAG: 2012-10-23T07:12:33Z SecondChildTAG: Unless we're fortune tellers xD SecondChildUserIdTAG: 231676 SecondChildUserNameTAG: Roosemberth SecondChildCreateTimeTAG: 2012-10-23T11:49:08Z FirstChildTAG: Has anyone managed to actually get a green tick on these practice questions?? FirstChildUserIdTAG: 145420 FirstChildUserNameTAG: pmac12345 FirstChildCreateTimeTAG: 2012-10-23T13:57:52Z SecondChildTAG: I was able to get green on first question of first problem, after that that it screwed up and started changing values on every click "Grade" button SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-23T14:46:08Z IndexTAG: 980 TitleTAG: Forum is Back And no one had to get staked through the heart. Good job team. UserIdTAG: 21541 UserNameTAG: JSChambers CreateTimeTAG: 2012-10-22T22:53:20Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Must have been this one: http://venturebeat.com/2012/10/22/amazon-cloud-outage-takes-down-reddit-airbnb-flipboard-more/ That crazy Cloud. I can't imagine why I don't want to put all my data there. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-22T23:06:25Z SecondChildTAG: Thanks for understanding SecondChildUserIdTAG: 96452 SecondChildUserNameTAG: Lyla SecondChildCreateTimeTAG: 2012-10-23T01:07:22Z SecondChildTAG: LOL! SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-23T03:29:30Z FirstChildTAG: Sorry for the delay. Amazon broke the internet today, and while most of the site weathered the storm, the discussion forums sadly did not. FirstChildUserIdTAG: 11 FirstChildUserNameTAG: dormsbee FirstChildCreateTimeTAG: 2012-10-22T22:57:21Z SecondChildTAG: Stuff happens. That was a Buffy quote, if you were wondering what I was on about. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-22T23:03:58Z SecondChildTAG: Damn, and I missed it. SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-23T03:30:09Z IndexTAG: 981 TitleTAG: Chrome goes Aw Snap! Whenever I'm making a post in the forum, if it involves math formulas then chrome usually goes Aw Snap! Anyone having the same problem? Any suggestion on how to fix this would be appreciated. UserIdTAG: 154440 UserNameTAG: Aahlad CreateTimeTAG: 2012-10-22T16:37:53Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: It has occurred for me. It is annoying. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-22T16:48:25Z SecondChildTAG: Yeah. Sometimes I retype the same post nearly a dozen times. I've taken to copying what I've wrote to notepad at regular intervals so that I don't have to write it all again. SecondChildUserIdTAG: 154440 SecondChildUserNameTAG: Aahlad SecondChildCreateTimeTAG: 2012-10-22T16:50:27Z SecondChildTAG: If you are using / decide to use FireFox, I highly recommend the add-on [It's All Text!][1] [1]: https://addons.mozilla.org/en-US/firefox/addon/its-all-text/?src=userprofile SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-24T13:20:15Z FirstChildTAG: So I am not the only one! FirstChildUserIdTAG: 86976 FirstChildUserNameTAG: kavita FirstChildCreateTimeTAG: 2012-10-23T03:22:34Z FirstChildTAG: Mine does that when I'm typing a really long answer in the homework section. VERY annoying. That's why I use Firefox just for 6.002x. No problems there so far. FirstChildUserIdTAG: 359310 FirstChildUserNameTAG: AaronYeoh FirstChildCreateTimeTAG: 2012-10-22T22:53:07Z IndexTAG: 982 TitleTAG: Downloading subtitles I have found that the subtitles serve a very good purpose - having ready made notes for annotations, finding doubts, discussions, keeping comments and so on. My main aim in downloading subtitles is not so much of viewing them alongside downloaded videos but instead maintaining them in a Word file or a Google Document and use the comments feature over there. How can I download these subtitles? Preferably in bulk! All help and proactive problem solving very much appreciated :-) UserIdTAG: 78396 UserNameTAG: dharav CreateTimeTAG: 2012-10-22T15:55:00Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: select, ctrl C and paste it in WRITER libreoffice or word with ctrl V FirstChildUserIdTAG: 327787 FirstChildUserNameTAG: REINALDOPARANHOS FirstChildCreateTimeTAG: 2012-10-23T01:47:30Z SecondChildTAG: Haha! Yes, I am gonna stick with that for now. It's convenient too. Just wondering if it is possible to download in bulk. Till then, what you said works. SecondChildUserIdTAG: 78396 SecondChildUserNameTAG: dharav SecondChildCreateTimeTAG: 2012-10-23T03:01:05Z IndexTAG: 983 TitleTAG: Why I think is is why I fell in love with electronics actually... UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-10-22T11:23:11Z VoteTAG: 3 CoursewareTAG: Week 6 / Capacitor energy storage demo CommentableIdTAG: 6002x_capacitor_energy_storage_demo NumberOfReplyTAG: 0 IndexTAG: 984 TitleTAG: What does your progress look like? Here's mine!![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13509030161343697.png UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-10-22T10:50:29Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: This is mine. Its pretty bad but will get better. ![][1] [1]: https://edxuploads.s3.amazonaws.com/13509058111343622.png FirstChildUserIdTAG: 154440 FirstChildUserNameTAG: Aahlad FirstChildCreateTimeTAG: 2012-10-22T11:38:24Z SecondChildTAG: Did you arrive late? SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-22T11:39:00Z SecondChildTAG: Na..got caught up in some stuff and told myself I was too busy...luckily I've done a pretty advanced course in school so it wasn't too hard to catch up. SecondChildUserIdTAG: 154440 SecondChildUserNameTAG: Aahlad SecondChildCreateTimeTAG: 2012-10-22T11:44:11Z SecondChildTAG: *Aahlad* don't give up keep it up....:) > Regards: **asadbhatti42** SecondChildUserIdTAG: 329051 SecondChildUserNameTAG: asadbhatti42 SecondChildCreateTimeTAG: 2012-10-22T12:32:00Z SecondChildTAG: Yeah, i will, thanks! SecondChildUserIdTAG: 154440 SecondChildUserNameTAG: Aahlad SecondChildCreateTimeTAG: 2012-10-22T16:28:13Z FirstChildTAG: Mine...![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13509235731343611.png FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-10-22T16:33:02Z SecondChildTAG: Nice. Perfect scores. :) SecondChildUserIdTAG: 154440 SecondChildUserNameTAG: Aahlad SecondChildCreateTimeTAG: 2012-10-22T16:46:10Z SecondChildTAG: similar to mine :-) SecondChildUserIdTAG: 111320 SecondChildUserNameTAG: jhalife SecondChildCreateTimeTAG: 2012-10-22T16:53:52Z SecondChildTAG: l00kin pretty! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-22T17:15:45Z SecondChildTAG: Thanks!I guess This wont look same next week after midterm :P SecondChildUserIdTAG: 148523 SecondChildUserNameTAG: span993 SecondChildCreateTimeTAG: 2012-10-23T04:04:42Z FirstChildTAG: Mine is a flatline. Of course, I just registered today. I'm curious to read through the content, but I have barely enough time to do the other classes I'm registered for. FirstChildUserIdTAG: 13125 FirstChildUserNameTAG: JMigdal FirstChildCreateTimeTAG: 2012-10-23T04:17:02Z FirstChildTAG: ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13509689811343685.png Well this is my progress chart but I think its difficult to maintain this progress as my school final exams are going take place this November :( FirstChildUserIdTAG: 346016 FirstChildUserNameTAG: muthukrishna FirstChildCreateTimeTAG: 2012-10-23T05:10:10Z IndexTAG: 985 TitleTAG: STAFF Sorry... one more question: As an example, I just reached for the Lab6 43%. Let's say, in the end, it is ONE OF THE TEN BEST out of twelve (because I missed two others, let's say) Will it count for the average, or will it not because it doen't reach the 60% minimum? Thank you.... Regards, Sandra UserIdTAG: 342966 UserNameTAG: SandraNavarro CreateTimeTAG: 2012-10-21T21:47:32Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: In that scenario, it will count. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-21T21:50:50Z SecondChildTAG: Thanks a lot :)) SecondChildUserIdTAG: 342966 SecondChildUserNameTAG: SandraNavarro SecondChildCreateTimeTAG: 2012-10-22T03:01:20Z IndexTAG: 986 TitleTAG: H6P2 ... minimum VDD could some1 help me please.... I am taking Vgs>=VT, Vds>=Vgs-VT.... for Q2 the mosfet would always be in saturation for Vgs=(Vdd-Vout)>=VT and for Q1... i have Vin>=Vt and Vout>=Vin-Vt what to do next UserIdTAG: 117319 UserNameTAG: iqramalik CreateTimeTAG: 2012-10-21T21:07:47Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Concentrate on Vin>=VT to make sure the transistor doesn't go into cutoff. All you have to do is make sure whatever your input value is it doesn't violate the conditions. FirstChildUserIdTAG: 132828 FirstChildUserNameTAG: rharris FirstChildCreateTimeTAG: 2012-10-21T21:17:50Z FirstChildTAG: You can't solve that without writing the node equation for Vout node. You got the condition for Q1 right, now figure out what are vGS and vDS for Q2. You will find Vout in these too hence the node method. FirstChildUserIdTAG: 405936 FirstChildUserNameTAG: Whible FirstChildCreateTimeTAG: 2012-10-21T23:58:33Z FirstChildTAG: What is the minimum value that the power supply voltage VDD must be to ensure that both transistors are operating in the saturated region? Write an algebraic expression involving VIN and VT for this minimum value of VDD in the space provided below i real don't know to solve this question help me please to find the minimum vddd FirstChildUserIdTAG: 475448 FirstChildUserNameTAG: msamwelmollel FirstChildCreateTimeTAG: 2012-10-22T08:43:19Z IndexTAG: 987 TitleTAG: What are your thought on week six? Wonder what you guys thought of the course ware week six? Is there anything you found particularly difficult in this weeks lectures? I know I found the homework pretty hard. I could have done with a tutorial involving a circuit similar to H6P1, other than the Triode example. But all in all, the lectures were fantastic! UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-10-21T19:05:10Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I think serious things are beginning ...differential equations, etc ... !!! FirstChildUserIdTAG: 373752 FirstChildUserNameTAG: Croak FirstChildCreateTimeTAG: 2012-10-21T22:34:29Z IndexTAG: 988 TitleTAG: h6p2 thevenins whatevr i solve i get vin as the thevenins answer ...bt cannot get the green tick ...can ne1 help... this is the last 1 left for me..... i hv got evrything rite...pls help UserIdTAG: 214277 UserNameTAG: yadsam CreateTimeTAG: 2012-10-21T09:45:28Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Mabe current flows in opposite direction? FirstChildUserIdTAG: 95280 FirstChildUserNameTAG: Fortyq FirstChildCreateTimeTAG: 2012-10-21T13:39:25Z FirstChildTAG: [Check this][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5082ba3a7021a8230000009d FirstChildUserIdTAG: 268444 FirstChildUserNameTAG: Marlonabeykoon FirstChildCreateTimeTAG: 2012-10-21T14:37:33Z IndexTAG: 989 TitleTAG: Inquiry I'm a new member & I've signed up a few minutes ago what about labs , H.Ws and exams ?? UserIdTAG: 696107 UserNameTAG: MegaMind_78 CreateTimeTAG: 2012-10-21T07:27:32Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: I believe it would be in your best interest to "audit" the course for now, and take it when it is offered again in the Spring. FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-10-21T08:06:10Z FirstChildTAG: It's my first edx course ^_^ but i see that midterm is 25/10 O.o I need somebody's help to catch up with you :D FirstChildUserIdTAG: 696107 FirstChildUserNameTAG: MegaMind_78 FirstChildCreateTimeTAG: 2012-10-21T07:57:18Z SecondChildTAG: its of no use now,you're late. If u r not hoping for a certificate just go through the videos of lecture and solve it on paper... GOOD LUCK! SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-21T08:07:40Z SecondChildTAG: Thanks for your advice ^_^ SecondChildUserIdTAG: 696107 SecondChildUserNameTAG: MegaMind_78 SecondChildCreateTimeTAG: 2012-10-21T08:27:41Z SecondChildTAG: If you have the time you could do a 6.002X Marathon and complete Weeks 1-6 by 25th but that isn't going to be easy. I would know, considering I'm in the middle of a marathon Week 3 - Week 6, currently on Week 5 :). You'll miss the grades for about 4 homeworks and 4 labs tho. But youmight be able to do the midterm. Good Luck! SecondChildUserIdTAG: 154440 SecondChildUserNameTAG: Aahlad SecondChildCreateTimeTAG: 2012-10-21T10:25:03Z SecondChildTAG: Hopefully I'll finish week 6 in time to submit homework today. SecondChildUserIdTAG: 154440 SecondChildUserNameTAG: Aahlad SecondChildCreateTimeTAG: 2012-10-21T10:26:56Z FirstChildTAG: To get the full benefit of the course, you should start at the beginning, and forget about the grade this semester. You should take it again for the grade another time. In the spring, this course will be offered again hopefully, and in the spring, there's a good chance **MITx** will offer a proctored exam certificate in addition to the free honor code certificate of completion that is currently offered. if you have a little bit interest in Computer then join **CS50x** to get good knowledge of Computer. Regards: asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-10-21T10:58:25Z FirstChildTAG: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/507f58641323942b0000003c FirstChildUserIdTAG: 154440 FirstChildUserNameTAG: Aahlad FirstChildCreateTimeTAG: 2012-10-21T11:54:01Z IndexTAG: 990 TitleTAG: Wrong Link The link on the top of the page which says "[WEEK 7 TUTORIALS][1]" points to a wrong place. And Also this is not Week 7. We are in Week 6 Tutorials. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_6/Week_7_Tutorials/ UserIdTAG: 259238 UserNameTAG: omidsadeghi CreateTimeTAG: 2012-10-20T15:29:12Z VoteTAG: 3 CoursewareTAG: Week 6 / Parallel plate capacitor CommentableIdTAG: 6002x_Parallel_Plate_Capacitor NumberOfReplyTAG: 1 FirstChildTAG: Thank you omidsadeghi :). Now it seems to be fixed [WEEK 7 TUTORIALS][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_7/Week_7_Tutorials/ FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-20T17:13:16Z SecondChildTAG: Still not fixed. It should be pointing to [WEEK 6 TUTORIALS][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_6/Homework_6/ SecondChildUserIdTAG: 304535 SecondChildUserNameTAG: hugo_h SecondChildCreateTimeTAG: 2012-10-21T06:20:46Z IndexTAG: 991 TitleTAG: Enunciado del Lab 6 Traducido al Español (This is an Spanish translation of Lab 6 statement due some difficulties of some students in understanding the statement in English) - This is one part of the request of dandradet [here][1], now I am going to a Class of my University , I will back later and try to Post what I have promised to you, sorry for the delay - Lab 6 Hints :). I hope that this translation can help you a little bit more in order to understand the statement but in spanish. ---- **LAB 6** El objetivo de este laboratorio es medir el delay de propagación de un inversor. Quizás consideres útil repasar la [Sección 10.4][2] del Textbook. El circuito de abajo contiene un inversor que se diseñó para que pudiera ser utilizado en un sistema en donde Vs=3v, VOL=0.25V y vOH=2.5V. La entrada del inversor se encuentra enganchada a una fuente de tensión que hace una transición de cero a uno en t=0. El rendimiento de un inversor se puede medir ya que posee una capacidad de carga de 200fF, que representa las capacidades parásitas de los cables y las entradas de otras compuertas lógicas enganchadas a la salida del inversor. ![fig1][3] Ejecutar el análisis transitorio sobre este circuito en 2ns y utilizar las mediciones del plot para responder las siguientes preguntas: 1- Medir el voltaje final de salida del inversor cuando la entrada se encuentra en estado alto y utilice dicho dato para estimar la RON para el mosfet. Estime el valor de RON en ohms: ![cuadro][4] 2- Cuando la entrada del inversor se encuentra en estado alto, utilice el equivalente Thevenin para este circuito tal como se muestra en la Figura 10.22 en el libro. Utilizando las mediciones, debe estimar VTH y RTH para el circuito equivalente Thevenin. Estime el valor de VTH en volts: ![cuadro][5] Estime el valor de RTH en Ohms: ![cuadro][6] 3- Si se utiliza la Ecuación 10.66 del libro y los parámetros de abajo, se debe calcular un valor estimado para tpd,0->1, es decir, el tiempo estimado que tarda la salida del inversor para caer de VS a VOL cuando posee una capacidad de carga de 200fF. Estimar el valor de tpd en nanosegundos: ![cuadro][7] 4- Al utilizar el plot de la tension de salida del inversor en su análisis transitorio, ingresar el valor medido de tpd,0->1. Medición del tpd en nanosegundos para una entrada creciente: ![cuadro][8] Notarás que el valor que se ha medido es más grande que el valor estimado por cálculo. En orden de entender por qué, hay que recordar que has utilizado un valor de RON en los calculus que se han basado en el estado estable de la tensión de salida del inversor luego que la transición de salida se ha completado, en otras palabras, cuando el mosfet switch se encuentra en la zona de operación de triodo. No obstante, la transición de mosfet switch en su region de saturación, en donde su Resistencia efectiva (VDS/IDS) era alta. Por lo tanto, el verdadero delay de propagación será de alguna forma más grande que la calculada por fórmula. 5- Finalmente, se debe cambiar la fuente de alimentación Vin para que la entrada realice una transición 0->1 (solo debes intercambiar las tensiones de plateu e initial) y medir el valor de tpd, 1->0, es decir, el tiempo que tarda la salida del inversor en llegar a VOH. Medición de tpd en nanosegundos para una entrada decreciente : ![cuadro][9] Si computas un valor estimado para tpd, 1->0 utilizando la ecuación 10.71 junto con los apropiados parámetros, encontrarás que encajará bien con las mediciones. En este caso la Resistencia utilizada en el cálculo (RL) fue exactamente la Resistencia en el circuito a lo largo de toda la transición. Estas mediciones y cálculos del delay de propagación de un inversor son los valores en el peor caso. Cuando hay muchas compuertas lógicas en serie, el delay real de una señal se considera la más pequeña que la suma de los peores casos de todos los delays de cada compuerta. Esto es porque las compuertas comenzarán su transición un largo tiempo después que han arribado a su valor final, por ejemplo, el mosfet switch en un inversor se encuentra en estado lógico 1 cuando su entrada excede VT (tensión umbral), mucho antes que la entrada llegue a VOH. Para ver este efecto, construyamos un oscilador anillo de 9 etapas, es decir una cadena de nueve inversores enganchados en un loop tal y como se muestra en el circuito de abajo. Funciona de la siguiente manera: 1- Se debe pensar como 9 inversores que se enganchan en serie: si se setea la primera entrada del inversor a un valor lógico "0", produce un "1" en su salida luego de un corto delay. Esto causa que el siguiente inversor produzca un "0" luego de otro corto delay,y así sucesivamente hasta que el noveno inversor produzca un "1" en su salida luego de un cierto delay que refleja cuán largo tiempo tardó desde la entrada que produjo el origen hasta haber llegado a la salida luego de haber atravesado los nueve inversores. Ya que existe un número impar de números de inversores, la señal de salida será la negación de la señal de entrada. 2- Ahora si se supone que la salida del noveno inversor se engancha al primero de todos. Ocurre lo explicado anteriormente, hasta que llegar al noveno inversor en donde se produce un "1" en su salida. Este ciclo se repite una y otra vez: el valor de cada entrada/salida se alterna entre "0" y "1" ya que cada nuevo valor se propaga a través de la cadena y luego, vuelve a comenzar. 3 - Qué sucede cuando el circuito alimentado es un poco más complicado, y si todos los nodos comienzan en cero volts, cada inversor tendrá un "0" en su entrada si se desea llevar su salida a "1". Pero cualquier pequeña alteración de su simetría causará que algunos inversores realicen esa inversión más lenta que otras . Esta falta de simetría ocurre naturalmente en los circuidos manufacturados, por ejemplo, el cable largo que completa el loop tiene más capacidad que el resto de los cables. Se requerirá de un ciclo para que ellos puedan acomodarse. Se podrá ver que sucederá como un pequeño sobresalto al principio del análisis transitorio del circuito de abajo. Si quieres ver mayor información adicional, agrega puntas de prueba de tensión si quieres ver en forma más detallada qué es lo que está sucediendo. Se tienen que utilizar dos trucos para asegurarnos que el oscilador oscilará en el mundo ideal de la simulación: introduzca una pequeña asimetría en las capacidades parásitas y "enciende" el circuito luego del tiempo cero utilizando una fuente de tensión escalón como fuente de alimentación. ![fig2][10] Debido a que hay un número impar de inversores en el loop, el voltaje de nodo oscilará entre el nivel alto y bajo, el período de oscilación se determinará por el tiempo que tardará la señal en propagarse luego de dos veces alrededor del loop(piensa porqué). Si se considera una tensión para un nodo particular, el periodo de oscilación representa una transición entre 0->1 y otra de 1->0, separada del tiempo de propagación de los 9 inversores. Compila la simulación transitoria a 50 ns y mide el período de oscilación. Sé paciente! Puede ser que tarde un momento hasta que se complete la simulación. Divide el resultado por nueve y obtén un tiempo estimativo para el cual el inversor tarde en realizar una transición de 0->1 seguida de una transición de 1->0. Estimar el tiempo de transición para ambas transiciones, en nanosegundos: ![cuadro][11] Comparar este tiempo con la suma de tus respuestas para las preguntas (4) y (5) de arriba. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/507ef87ef4d89e1f00000137 [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/549 [3]: https://edxuploads.s3.amazonaws.com/13506774791343639.png [4]: https://edxuploads.s3.amazonaws.com/13506775901613435.png [5]: https://edxuploads.s3.amazonaws.com/13506775901613435.png [6]: https://edxuploads.s3.amazonaws.com/13506775901613435.png [7]: https://edxuploads.s3.amazonaws.com/13506775901613435.png [8]: https://edxuploads.s3.amazonaws.com/13506775901613435.png [9]: https://edxuploads.s3.amazonaws.com/13506775901613435.png [10]: https://edxuploads.s3.amazonaws.com/13506804581343643.png [11]: https://edxuploads.s3.amazonaws.com/13506775901613435.png UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2012-10-19T21:21:21Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 992 TitleTAG: Practice midterm ? Hi All, Many thanks to all and specially to the course staff. Is it possible to have access to a practice midterm exam ? Thanks ! UserIdTAG: 140136 UserNameTAG: bsall CreateTimeTAG: 2012-10-19T11:04:27Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Please see [this thread][1]. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/undefined/threads/5079c03eca6daa2300000016 FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-10-19T15:35:31Z IndexTAG: 993 TitleTAG: about certificat this program is vary good for us. technology that you are using is OSM.. i would like to thank for all team who started this thing.. BTW i want to know is it possible that we can have some certificate after this cores. or is it only email?? UserIdTAG: 509517 UserNameTAG: randima CreateTimeTAG: 2012-10-19T08:25:35Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: can we get a hard copy of it?? FirstChildUserIdTAG: 509517 FirstChildUserNameTAG: randima FirstChildCreateTimeTAG: 2012-11-11T12:18:10Z FirstChildTAG: I think there will be a digital certificate, similar to this one: ![edX certificate][1] [1]: http://1.bp.blogspot.com/-PYTkDqIDsQU/T9l2IfdQUnI/AAAAAAAABvU/z2OMvC9BN8M/s1600/mitx_certificate.png FirstChildUserIdTAG: 466665 FirstChildUserNameTAG: raresmihaipopa FirstChildCreateTimeTAG: 2012-10-19T10:01:30Z SecondChildTAG: How can you upload a snapshot? I tried but failed. SecondChildUserIdTAG: 128276 SecondChildUserNameTAG: ChengBin SecondChildCreateTimeTAG: 2012-10-19T16:03:49Z SecondChildTAG: yep, that's what it looks like. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-19T17:20:49Z SecondChildTAG: There is a little IMG button above the post while you are writing it. Next to the "Code Sample" button. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-19T17:22:03Z SecondChildTAG: Thats great.. Do we get a hard copy of it? SecondChildUserIdTAG: 362299 SecondChildUserNameTAG: anandbaskaran SecondChildCreateTimeTAG: 2012-10-20T17:53:14Z SecondChildTAG: yeah!! thats great. can we get hard copy of it... SecondChildUserIdTAG: 509517 SecondChildUserNameTAG: randima SecondChildCreateTimeTAG: 2012-10-22T05:17:41Z IndexTAG: 994 TitleTAG: PI Dear all, Please help me how to put the pi symbol in the answer Thanks UserIdTAG: 156694 UserNameTAG: mkprasanth CreateTimeTAG: 2012-10-19T08:06:40Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: just copy and paste.. FirstChildUserIdTAG: 509517 FirstChildUserNameTAG: randima FirstChildCreateTimeTAG: 2012-10-19T08:20:02Z SecondChildTAG: Enter PI SecondChildUserIdTAG: 466665 SecondChildUserNameTAG: raresmihaipopa SecondChildCreateTimeTAG: 2012-10-19T09:59:53Z FirstChildTAG: Hi mkprasanth! If you see the calculator of the course, click on "i" symbol and you will see that: ![enter image description here][1] if yo need pi, just write pi, eg.: 4*$\pi$ is 4*pi See you! :) Myriam [1]: https://edxuploads.s3.amazonaws.com/13506455341343683.png FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-19T11:20:20Z SecondChildTAG: That is so cool! I never knew that! Thank you! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-19T17:59:06Z SecondChildTAG: You are welcome hazel1919! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-19T20:00:07Z SecondChildTAG: Dear Myriam , When ever I need SOS you are coming and saving me Thanks a lot SecondChildUserIdTAG: 156694 SecondChildUserNameTAG: mkprasanth SecondChildCreateTimeTAG: 2012-10-20T17:08:54Z SecondChildTAG: Hahaha! you are welcome mkprasanth :). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-20T17:14:56Z IndexTAG: 995 TitleTAG: Perfect timing for the Midterm - October 25th! What a coincidence! My Analog Electronics class at college also schedule the midterm to be October 25th. So i'll have to take the brick-and-motar midterm first and then rush out for the online one here. So the time spent preparing for our midterm is time doubly well spent. Wish me luck! UserIdTAG: 99971 UserNameTAG: tungai CreateTimeTAG: 2012-10-18T11:57:02Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: i think 20 oct 'll be best time for mid term FirstChildUserIdTAG: 681348 FirstChildUserNameTAG: alizarrar FirstChildCreateTimeTAG: 2012-10-18T16:46:45Z FirstChildTAG: Good luck on both! FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-10-18T23:45:24Z FirstChildTAG: You can take the mid term from 25-10-2012 00:01 (Boston time) till 27-10-2012 23:59, so there is no need to rush out :) FirstChildUserIdTAG: 114881 FirstChildUserNameTAG: xvink FirstChildCreateTimeTAG: 2012-10-18T17:40:10Z SecondChildTAG: can anyone convert it to Indian Time and write the date and time.. some confusion this side.. Thanks! SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-18T17:45:51Z SecondChildTAG: For Indians exams will be activated from 25-10-2012 9:31 am till 28-10-2012 9:31 am.. Wishing you all success!!! SecondChildUserIdTAG: 362299 SecondChildUserNameTAG: anandbaskaran SecondChildCreateTimeTAG: 2012-10-18T17:54:55Z SecondChildTAG: thanx anandbaskaran..wishing u good luck too!! SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-19T03:04:33Z SecondChildTAG: deadline extended to monday 29th 9.30 am...yipee! SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-23T15:54:00Z IndexTAG: 996 TitleTAG: MYRIMIT AYUDA BUENAS TARDES ESPERO TE ENCUENTRES BIEN OYE ME ATORE EN EL LAB 6 NO LOGRO COMPRENDER QUE ES LO QUE PIDEN ESTOY TENIENDO PROBLEMAS CON LA TRADUCCION ESPERO PUEDAS ORIENTARME . GRACIAS UserIdTAG: 321559 UserNameTAG: dandradet CreateTimeTAG: 2012-10-17T18:27:10Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hola dandradet! Cómo estás? Sí, estoy bien. Es que estuve media colgada y también estuve ocupada con otro proyecto :). Algo interesante, ayer estuve en la WEEF 2012 (World Engineering Education Forum)presentando un Póster sobre un trabajo sobre el abandono de estudios en el que estoy colaborando Ad Honorem hace ya casi dos años en mi Universidad porque siempre quise devolver/ayudar de alguna manera a mi Universidad, hoy justo iban a estar los de la MIT :), saludos al equipo de Sarma Sanjay!, hubiera querido ir a saludarlos/conocerlos pero como me encuentro trabajando como Becaria no graduada en Ingeniería en otro Proyecto de Investigación independiente del Grupo de la Facultad, no pude ir hoy también :'/ jaja. Pero me hubiera encantado conocerlos! Sí, estaré encantada en ayudarte. Bueno, ahora voy a ver si lo puedo traducir al Español y realizar unas Hints para ti :) Saludos, Myriam. ------ Hi dandradet! How are you? Yes, I am fine. I was a little bit dissapeared and also I busy with other project :). Something interesting, yesterday I was in the WEEF 2012 (World Engineering Education Forum) we were showing a Poster of our work about the abandoned studies of the students in which I am colaborating Ad Honorem in almost two years ago in my University because I always wanted to return something to my University, just today some of MIT's reasearchers was there :), greetings to Sarma Sanjay and Team!, I wish to were there today and meet/greeting them personally but as I am working in a Ungraduate Engineering Fellownship in other research project apart from my University Group I couldn't go today too :'/ haha. But I wished to meet them! Yes, I will be pleasured to help you . Ok, I will see if I can translate the statement to spanish and make some hints for you :) See you! Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-17T20:56:51Z SecondChildTAG: Hi dandradet! :) Here what I have promised to you :) [Lab 6 Hints ][1] and the Spanish translation of Lab 6 [here][2] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Ring_Oscillator/threads/5082c6039e78031f000000c0 [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5081c45132aa871f00000002 SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-20T15:43:05Z IndexTAG: 997 TitleTAG: H6P1 Transconductance gm = Transconductance gm =? really stuck. any hints UserIdTAG: 18073 UserNameTAG: gburkhart CreateTimeTAG: 2012-10-16T22:59:26Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Take the expression for iD, express iD, vIN, and vOUT as a DC bias plus small signal. Linearize (You don't need to use Taylor series, but you can). Compare your result with the answer to preceding part and identify gm. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-16T23:44:31Z FirstChildTAG: Hi gburkhart! How are you? hint 1 : You should take a look at this video tutorial of week 6 [watch here][1]. It is explained for the vacuum triode model ... but it can help you to understand a little bit how to obtain a gm . hint 2: if you have iD, so can you find gm ? ;) I hope that this can help you. See you! Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_6/Week_6_Tutorials/ FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-17T02:10:00Z FirstChildTAG: Thanks for the help guys - I got it! FirstChildUserIdTAG: 18073 FirstChildUserNameTAG: gburkhart FirstChildCreateTimeTAG: 2012-10-17T03:02:14Z SecondChildTAG: Well done gbukhart! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-17T03:13:07Z IndexTAG: 998 TitleTAG: Provide download links for week4 onwards ! Hi Staff, We really have difficulty watching the videos. Download links for week 4 made our lives easy. Please provide the links for week5 and all other weeks! Otherwise it is really impossible for us to carry on ! UserIdTAG: 314624 UserNameTAG: Owais001 CreateTimeTAG: 2012-10-16T12:11:07Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: We want videos' links............ FirstChildUserIdTAG: 199839 FirstChildUserNameTAG: vijaykrishnankk FirstChildCreateTimeTAG: 2012-10-16T13:05:16Z FirstChildTAG: [JWplayer Video Link Week(1-6)][1] [1]: https://dl.dropbox.com/u/24096724/6002xMP4/menu.html# Regards: asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-10-18T15:24:56Z IndexTAG: 999 TitleTAG: Week 6 mp4 lectures Dear, kindly upload in mp4 the lectures of Week 6, I really had a bad time doing week 5. I had to do it at the last time after i watched all the movies which was difficult due to bad connection. Thanks a lot! Charbel UserIdTAG: 141709 UserNameTAG: charbelantonios CreateTimeTAG: 2012-10-16T08:41:21Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: [JWplayer Video Link Week(1-6)][1] [1]: https://dl.dropbox.com/u/24096724/6002xMP4/menu.html# Regards: asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-10-18T15:22:23Z IndexTAG: 1000 TitleTAG: STAFF: Getting "Page Not Found" 404 Error on week 6 tutorial problem 9.7 Page not found The page that you were looking for was not found. Go back to the homepage or let us know about any pages that may have been moved at technical@edx.org. [https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_6/wk7_9_7/][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_6/wk7_9_7/ UserIdTAG: 148542 UserNameTAG: planetscape CreateTimeTAG: 2012-10-16T05:24:04Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Thanks for pointing this out, it should be fixed now. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-10-18T16:15:32Z SecondChildTAG: Thanks! SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-19T01:27:58Z IndexTAG: 1001 TitleTAG: Request to Staff Dear Staff, Can it be possible to provide sequence notes in Wiki in pdf format.The sequence notes would be good for revision but if i try to copy in MS word the formula changes. Kindly provide a link to download in pdf format if possible. Regards, Gopal UserIdTAG: 111320 UserNameTAG: jhalife CreateTimeTAG: 2012-10-15T14:07:54Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Agreed! The Wiki does not print well either to my laser printer nor when I use [PDFCreator][1]. Thanks in advance! [1]: http://sourceforge.net/projects/pdfcreator FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-10-15T21:41:53Z IndexTAG: 1002 TitleTAG: Сoncrete math Hello I'd like to thank a lot stuff of the course! I learned these themes in high school some years ago, but I've forgotten recently almost today :-) It's really interesting and breathtaking course. In Russia, one of the most famous MIT course is Concrete Math of Donald Knuth. There is selfstudy books, and tryed to learn it myself 3 times, but I didn't get after 4th lesson :) Don't you planning to begin "Concrete mathematic" course, may be on commercial? And what another courses do you planning to begin in future for ee-engeneers? UserIdTAG: 345464 UserNameTAG: Constantine_ru CreateTimeTAG: 2012-10-15T06:42:45Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1003 TitleTAG: Exam Date Change I want to request the staff to post the exam on 26th October and the due date should be October 28th so that people who have missed lectures & HWs because of circumstances not under their control can finish their preparation by going through missed lectures , tutorials and HWs .. UserIdTAG: 30155 UserNameTAG: electricalbest CreateTimeTAG: 2012-10-14T20:52:11Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I second that..... FirstChildUserIdTAG: 229430 FirstChildUserNameTAG: MuhammadAli201 FirstChildCreateTimeTAG: 2012-10-15T10:21:53Z FirstChildTAG: I also agree with electricalbest... I am a student from West Bengal, India and the mid term exam has been scheduled just the day after the bigggest religious festival in this part of the country... Im pretty sure others will also agree that its a tremendous inconvenience for all students from West Bengal. FirstChildUserIdTAG: 314654 FirstChildUserNameTAG: Mandar001 FirstChildCreateTimeTAG: 2012-10-15T11:51:59Z SecondChildTAG: I agree with u mandar001...Btw I am frm Delhi. SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-16T02:44:49Z SecondChildTAG: Thanks for your support jmen... I read in another post that they have scheduled the exam so that it will end on 28th Oct 9:30 AM IST. Can somebody please confirm this or point me to a thread on this? SecondChildUserIdTAG: 314654 SecondChildUserNameTAG: Mandar001 SecondChildCreateTimeTAG: 2012-10-20T14:05:00Z IndexTAG: 1004 TitleTAG: Awesome Tutorials These tutorials are what makes this course way better than the other courses. UserIdTAG: 19863 UserNameTAG: Samir CreateTimeTAG: 2012-10-14T06:30:46Z VoteTAG: 3 CoursewareTAG: Week 3 / Load Line Tutorial CommentableIdTAG: 6002x_load_line_tutorial NumberOfReplyTAG: 2 FirstChildTAG: I am totally agree with you Samir! :) FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-14T19:53:39Z FirstChildTAG: IMHO, there could be more wrt the Math. ;-) FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-10-15T03:55:27Z IndexTAG: 1005 TitleTAG: HW5P2 Question 3 in which order should I type the equation...I have found the correct root, but i have some problems typing the solution...it says that it cant process the equation or sotmeting like that UserIdTAG: 308861 UserNameTAG: Ignaas CreateTimeTAG: 2012-10-14T02:18:57Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hi Ignaas! Are you reducing the equation to the minimun expression? FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-14T02:45:24Z SecondChildTAG: hmm i don't really understand what you mean by your comment SecondChildUserIdTAG: 308861 SecondChildUserNameTAG: Ignaas SecondChildCreateTimeTAG: 2012-10-14T02:50:33Z SecondChildTAG: Hi Ignaas reducing an equation to the minimun expression is for example: if you have this equation: X+ 3X - X*Y =0 4*X-X*Y=0 X*(4-Y)=0 minimun expression ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T03:25:17Z SecondChildTAG: same problem with me. SecondChildUserIdTAG: 259693 SecondChildUserNameTAG: MehrozKhan SecondChildCreateTimeTAG: 2012-10-14T06:55:21Z SecondChildTAG: My expression was not the most reduced form, but was accepted as correct. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-14T13:02:59Z SecondChildTAG: but must i use the fact that Vin=VT for reducing it???I get the comment that it cannot parse the equation SecondChildUserIdTAG: 308861 SecondChildUserNameTAG: Ignaas SecondChildCreateTimeTAG: 2012-10-14T18:03:11Z FirstChildTAG: There are many possible reasons for this error: Unbalanced parens, missing operators, wrong signs, wrong case... Keep trying! FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-10-15T03:57:22Z IndexTAG: 1006 TitleTAG: HW5 & Lab5: Finally done! I've finally done that Week 5 HW and Week 5 Lab! Did everyone got that enormous big equations at Part 2 and Part 3? I feel like I have to speed up a little, because that's not a good idea to do all the homeworks at last night before deadline. And, also, it seems, that I better understand the covered topics in mathematical sense better, than in a circuit, so I'll have to read the textbook a little... Oh, yes, thanks to the guy, who's wifi-network name is teo (although he might not attend the course :) ) for free wifi access to the Internet: my "official" internet channel was broken while I was doing the HW. Completing this HW was really amazing! Dear staff, thanks for that experience! UserIdTAG: 326787 UserNameTAG: Angstrem CreateTimeTAG: 2012-10-13T22:40:37Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Congrats! FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-10-13T23:30:50Z SecondChildTAG: Well done!:) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T00:54:52Z SecondChildTAG: I also finished it last night. SecondChildUserIdTAG: 99628 SecondChildUserNameTAG: Pranjal16 SecondChildCreateTimeTAG: 2012-10-14T06:49:54Z SecondChildTAG: Well done Pranjal16! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T19:54:38Z IndexTAG: 1007 TitleTAG: H5P2 Source Follower Large Signal I have developed an equation for IDS. I have checked it on paper that for Vin=VT the current is equal to 0 i.e. IDS=0. But the same equation is shown as wrong in the answer...what is the problem? UserIdTAG: 64618 UserNameTAG: ashfaq2419 CreateTimeTAG: 2012-10-13T22:34:36Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi ashfaq2419! Can I help you? Try to express the equation to the minimun expression... Also remember that the answer is case sensitive VIN is not the same of vIN... FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-14T02:57:26Z IndexTAG: 1008 TitleTAG: Help for Math Processing Error hello I am unable to do home work due to Math Processing error. Please guide me what to do. thanks UserIdTAG: 49048 UserNameTAG: Mumtaz CreateTimeTAG: 2012-10-13T22:12:49Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: That error is due to how your browser interacts with MathJax. MathJax is used by the server to present mathematical symbols on your screen. Here's how to fix it: 1) Place your cursor on the [Math Processing Error] message and right-click. 2) You will see a pull-down MathJax menu. 3) Select Math Settings, Math Renderer, and choose either MathML or SVG. I prefer the way MathML looks, but it drops some symbols when I print, so I have to use SVG. 4) You can also play with the zoom settings on the MathJax menu to adjust the zoom functionality. FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-10-14T19:25:20Z FirstChildTAG: I had that error when I was trying to write RS^3. I simply wrote (RS)^3 and then removed brackets. FirstChildUserIdTAG: 326787 FirstChildUserNameTAG: Angstrem FirstChildCreateTimeTAG: 2012-10-13T22:32:18Z IndexTAG: 1009 TitleTAG: H5P2 Part 3 IDS Equation Ok, so for the life of me, I cant get the equation for part 3 sorted. I fail at basic algebra! :( I've been trying for literally 4 hours now and still no joy. I get that: (K*(vIN-vOUT-VT)^2)/2 and that: iDS*RS We need to solve in terms of K,vIN, RS & VT for iDS So I end up with the first equation as: K/2 * ( (vIN - VT) - iDS*RS)^2 = iDS This expands to: K/2 *( iDS^2*RS^2 - 2(vIN-VT)*iDS*RS + (vIN-VT)^2) - iDS = 0 The problem I'm having is where do I go from here in order to solve what looks like a quadratic equation? I cant seem to spot the A,B and C terms. Actually, more accurately, I cant seem to spot the B term. A is K/2*RS^2 C is K/2*(vIN-VT)^2 but what is B? And is this part even correct? I've gone through reams of paper and still I cant simplify it down. I know what Im aiming for as someone has posted the final solution but thats no good without knowing how to get there. UserIdTAG: 56024 UserNameTAG: onidaito CreateTimeTAG: 2012-10-13T22:02:33Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: I'm having the very same problem...can't relate to a b and c FirstChildUserIdTAG: 256543 FirstChildUserNameTAG: sidney23 FirstChildCreateTimeTAG: 2012-10-13T22:09:35Z SecondChildTAG: Can I help you sidney23? SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T03:05:51Z SecondChildTAG: ITS gonna be a looooong equation... iDS is the x of the equation. Collect parts that are multiplied with iDS and iDS^2 and other. Try to combine them into one a, b and c value. Don't expect it to be short... SecondChildUserIdTAG: 324332 SecondChildUserNameTAG: petsol SecondChildCreateTimeTAG: 2012-10-14T12:44:03Z FirstChildTAG: i got B and it is [ -1 - K*RS*(Vin-Vt) ] the +1 is for the ids on the right and the rest is for the terms beside ids in the pracket i continued the soultion and i got that solution for ids (1+(K*RS*(vIN-VT))-sqrt(1+2*K*RS*(vIN-VT)^2))/(K*RS^2) continue on my way , if you got the same result tell me please FirstChildUserIdTAG: 285675 FirstChildUserNameTAG: Ascot FirstChildCreateTimeTAG: 2012-10-13T22:11:03Z FirstChildTAG: Hi onidaito! Can I help you? Try to replace (vIN-VT) = X - and then in someway to agrupate the iDS^2: value_1*iDS^2 + value_2*iDS^2 + so on with all the terms that you have with iDS^2. - Do the same with the terms that have iDS: value_3*iDS + value_4*iDS + so on with all the terms that you have with iDS^2. - Agrupate the constant value all together (the ones that don't have iDS): value_5+value_6+so on..... FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-14T03:05:29Z FirstChildTAG: K/2 *( iDS^2*RS^2 - 2(vIN-VT)*iDS*RS + (vIN-VT)^2) - iDS = 0 ====> K/2 *( iDS^2*RS^2 - 2(vIN-VT)*iDS*RS + (vIN-VT)^2 - 2/K *iDS) = 0 ====> K/2 *( iDS^2*RS^2 - 2(vIN-VT)*iDS*RS - 2/K *iDS + (vIN-VT)^2) = 0 ====> K/2 *( iDS^2*RS^2 - (2(vIN-VT)*RS + 2/K) *iDS + (vIN-VT)^2) = 0 so ====> B = k/2 *(- (2(vIN-VT)*RS + 2/K)) FirstChildUserIdTAG: 296511 FirstChildUserNameTAG: sky0917 FirstChildCreateTimeTAG: 2012-10-14T15:25:46Z IndexTAG: 1010 TitleTAG: H5P2 Q7 Hi, I have a trouble with calculation VDD. I know that vDS ≥ vGS-VT and VDD-vOUT ≥ vIN-vOUT-VT simplify VDD ≥ vIN-VT But, how I should find correct vIN for this equation? I know that it must be vIN maximum. I can't find equation for it. UserIdTAG: 269368 UserNameTAG: StAlex CreateTimeTAG: 2012-10-13T12:13:05Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: I love you svenkat :-) FirstChildUserIdTAG: 184153 FirstChildUserNameTAG: tthngn FirstChildCreateTimeTAG: 2012-10-13T20:00:06Z FirstChildTAG: iDS=K/2*(vIN-iDS*RS-VT)^2 use this equation put the value given for V1N in this equation a quadrtc formula will come solve this for iDS value of iDS will come now put the value of iDS in this equation vOUT=RS*iDS.. you will find vOut value then the Vout value that come from this equation vOUT=RS*iDS use this value do Vout-VT you wil find correct answer FirstChildUserIdTAG: 228871 FirstChildUserNameTAG: ELINAKHAN FirstChildCreateTimeTAG: 2012-10-13T14:48:44Z SecondChildTAG: vIN(Max)-VT will yeild correct ans SecondChildUserIdTAG: 171007 SecondChildUserNameTAG: svenkat91 SecondChildCreateTimeTAG: 2012-10-13T19:32:06Z SecondChildTAG: I try to solve equation for iDS and vOUT but answer is still wrong, I'am still can't undestand how find vIN(Max). SecondChildUserIdTAG: 269368 SecondChildUserNameTAG: StAlex SecondChildCreateTimeTAG: 2012-10-14T10:19:12Z SecondChildTAG: I find it! Thanks, but yesterday I few times try substitute the right value into checkbox and it was wrong. It's very strange. My brain almost exploded, when I try find the right answer, although I already have it :) SecondChildUserIdTAG: 269368 SecondChildUserNameTAG: StAlex SecondChildCreateTimeTAG: 2012-10-14T10:45:54Z FirstChildTAG: Before you get to this question you are asked to calculate $V_{OUT}$ for three different values of $V_{IN}$. One of these requires a larger value of $VDD$ than the others. Use that one for you calculation of $VDD$ FirstChildUserIdTAG: 12419 FirstChildUserNameTAG: rpm FirstChildCreateTimeTAG: 2012-10-13T14:56:38Z SecondChildTAG: Why? SecondChildUserIdTAG: 298867 SecondChildUserNameTAG: JNPH SecondChildCreateTimeTAG: 2012-10-14T04:17:11Z SecondChildTAG: that doesn't work here... and I tried the 3 values SecondChildUserIdTAG: 310147 SecondChildUserNameTAG: ildomarcarvalho SecondChildCreateTimeTAG: 2012-10-14T04:33:55Z SecondChildTAG: forget it, I was using the vOUT values instead of the vIN values, now it works! Thank you a lot! But like JNPH I don't understand why use this value? SecondChildUserIdTAG: 310147 SecondChildUserNameTAG: ildomarcarvalho SecondChildCreateTimeTAG: 2012-10-14T04:36:42Z IndexTAG: 1011 TitleTAG: Backspace = Back page? Arrgg If I highlight some text in a post I am writing, I want to press the backspace button to delete the highlighted text. The problem is that it turns the page to the previous page and I lose **everything**. Hard as I try, I have done this a half a dozen times and it is always on responses that I have spent a bit of time on. Sometimes I want to delete a sentence to rephrase then POOF it's gone. I just did it again a few moments ago, I almost folded my keyboard across my head. This makes me sad. Another thing that is very awkward.> Sometimes I would like to press "enter" so I can start a sentence on a new line. (Bullet style) I can't though, I can only press it twice, then it starts a new paragraph. This also makes me sad. While I am here, there is some funky scroll thing that goes on with these pages. You know what I mean. It is not very intuitive. I think I am getting used to it though. Say I am looking or hover over the left hand column and I want to scroll down, well the bottom of the page comes up, doing the opposite of what I want, actually showing me less. Other then that, good job everyone! UserIdTAG: 292543 UserNameTAG: Pennypacker CreateTimeTAG: 2012-10-13T04:02:14Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I would suggest typing your responses in notepad or something and then copy pasting them to avoid those headaches. FirstChildUserIdTAG: 213386 FirstChildUserNameTAG: dmascenik FirstChildCreateTimeTAG: 2012-10-13T12:44:20Z IndexTAG: 1012 TitleTAG: Fault tolerance For staff members: you should work on fault tolerance, specially in formulae or expressions, for example, VIN must be equal to Vin or vIN, or K"O" must be equal to K"zero (0)" . Just a recomendation :) UserIdTAG: 268104 UserNameTAG: SaulCardenasG CreateTimeTAG: 2012-10-13T04:01:36Z VoteTAG: 3 CoursewareTAG: Week 5 / MOSFET Amplifier Exercise 1 CommentableIdTAG: 6002x_mosfet_amp_e1 NumberOfReplyTAG: 0 IndexTAG: 1013 TitleTAG: H5P2 Hey all, Here's a good equation solver that I used for H5P2. For some reason Wolfram wouldn't do it. http://www.numberempire.com/equationsolver.php UserIdTAG: 21541 UserNameTAG: JSChambers CreateTimeTAG: 2012-10-12T23:47:39Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: This solver works for any value, besides vIN = 6.1. What is wrong? FirstChildUserIdTAG: 96873 FirstChildUserNameTAG: Santyaga FirstChildCreateTimeTAG: 2012-10-13T18:24:44Z SecondChildTAG: Thank you for the data JSChambers! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T03:19:02Z SecondChildTAG: Santyaga, I didn't actually use it to "solve" the equation. I used it to get the algebraic formula. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-15T02:56:53Z FirstChildTAG: Thank YOU, JSChambers, you saved my tush with this one! FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-10-15T00:18:52Z IndexTAG: 1014 TitleTAG: Looks Good! I logged for the first time today and it seems to be fine.I had a few problems at calculations but i'm ok now! UserIdTAG: 620830 UserNameTAG: Lazaros1 CreateTimeTAG: 2012-10-12T19:56:58Z VoteTAG: 3 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: Welcome to 6.002x Lazaros1! :) FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-14T03:20:46Z IndexTAG: 1015 TitleTAG: Midterm examp attempts Hello! Could please anyone explain me, how mane attempts would I have during passing midtermexam. I mean would it be like a big homework and I'll be able to check number of question? Or it would be like real exam, I will write answers and then press "check" once a time and would see my result? UserIdTAG: 345464 UserNameTAG: Constantine_ru CreateTimeTAG: 2012-10-12T16:03:33Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: As I understand, it will be *somewhat* similar to homework and exercises in that there will be a "check" button. From what I gather, you will have three "attempts" *PER PROBLEM SET* to get the right answer. FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-10-12T17:47:13Z SecondChildTAG: Thank you for answer! And are you shure about three attempts? I tried to find information in course info, but I found nothing about form of exam :-( SecondChildUserIdTAG: 345464 SecondChildUserNameTAG: Constantine_ru SecondChildCreateTimeTAG: 2012-10-12T18:40:57Z SecondChildTAG: Please see [this thread][1]. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5072e92588c9422200000097 SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-22T12:06:38Z IndexTAG: 1016 TitleTAG: Homeworks and Labs Submission Hi, I would like to know how to submit the homework and lab assignments. I only see a "Check" button which reveals the correct answer but I cannot find a button to submit my works. Previously, in the quizzes, the system will provide feedback to my answers with ticks or a crosses when I click check. But it is not the case in the homework and lab sessions. I have encountered the following issues. 1. The system does not save my answers in the homework and lab pages 2. I cannot know if I have submitted (despite clicking "check") because in the progress section, homework and labs are without scores whereas the system is reflecting my scores for quizzes. UserIdTAG: 544644 UserNameTAG: Ethanaung CreateTimeTAG: 2012-10-12T14:25:23Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: The "Check" button is the submit button. If you are looking at Weeks 1-4, the deadline has passed for the Homework and Labs. So you cannot submit work for those. Look at the homework for Week 5, that should still have a check button that records your answers. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-12T15:17:52Z SecondChildTAG: Thank you. I submitted for week1 assignments. I see what you mean. SecondChildUserIdTAG: 544644 SecondChildUserNameTAG: Ethanaung SecondChildCreateTimeTAG: 2012-10-12T19:39:57Z SecondChildTAG: I have just joined the course, so I don't have a chance to compensate for the marks I've lost so far? Do I have a chance to pass after losing all those homework and lab marks? SecondChildUserIdTAG: 622465 SecondChildUserNameTAG: Layali SecondChildCreateTimeTAG: 2012-10-13T02:17:08Z FirstChildTAG: Very occasionally, I have had problems with checking, etc. Sometimes the math equations will not display in typeset form. Clearing my browser cache and reloading has solved these issues, though it has sometimes taken more than one clear/reload. If you are working on a homework or lab whose deadline has not yet passed, try that. FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-10-12T17:49:38Z SecondChildTAG: Thank you. I will try this too. SecondChildUserIdTAG: 544644 SecondChildUserNameTAG: Ethanaung SecondChildCreateTimeTAG: 2012-10-12T19:40:38Z IndexTAG: 1017 TitleTAG: H5P1 Part C I am using equation 7.32 for vIN maximum and consider VS=(VS+)-(VS-)...but not getting the answer...Must be missing something!..any help! UserIdTAG: 122513 UserNameTAG: Mona77 CreateTimeTAG: 2012-10-12T06:16:32Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: got it :)...I was missing something but not before calculating vIN...after that there is a FINAL STEP! FirstChildUserIdTAG: 122513 FirstChildUserNameTAG: Mona77 FirstChildCreateTimeTAG: 2012-10-12T06:30:46Z SecondChildTAG: i still could not get the right answer.though i am using the same formula.plz help me this part.. SecondChildUserIdTAG: 368981 SecondChildUserNameTAG: asimakhtar SecondChildCreateTimeTAG: 2012-10-12T18:07:28Z SecondChildTAG: Well done Mona77! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-13T16:45:15Z IndexTAG: 1018 TitleTAG: What happens to vi^2? How do you get rid of the vi^2 term? I know the Vs- RLK/2*(Vi-Vt)^2 is substituted for VO and removed from the equation, leaving me with v0 -(RL*K)/2*(2*(VI-VT)*vi + vi^2).... But the answer is with out the vi^2.... where does it go? UserIdTAG: 381923 UserNameTAG: matteaton CreateTimeTAG: 2012-10-11T15:39:01Z VoteTAG: 3 CoursewareTAG: Week 5 / Incremental Voltage Exercise CommentableIdTAG: 6002x_inc_volt_e NumberOfReplyTAG: 1 FirstChildTAG: (Vi-Vt)^2 is where it goes. (Vi-Vt)^2 will have the unit V^2, which will cancel with the A/(V^2) in K. FirstChildUserIdTAG: 248807 FirstChildUserNameTAG: 1kingsman FirstChildCreateTimeTAG: 2012-10-11T22:21:53Z SecondChildTAG: I've used the Tailor's series here. But I'm getting 2VI instead of VI inside the brackets... Now I'll go to bed and find out the bug tomorrow. I really want to sleep. SecondChildUserIdTAG: 326787 SecondChildUserNameTAG: Angstrem SecondChildCreateTimeTAG: 2012-10-11T22:34:30Z SecondChildTAG: vi is very small, therefore viˆ2 is even smaller and can be despised. That's why the answer is called Linear Approximation. SecondChildUserIdTAG: 177566 SecondChildUserNameTAG: AndreH SecondChildCreateTimeTAG: 2012-10-11T23:49:45Z SecondChildTAG: thank you AndreH SecondChildUserIdTAG: 201508 SecondChildUserNameTAG: datle SecondChildCreateTimeTAG: 2012-10-12T18:24:39Z SecondChildTAG: thank you AndreH SecondChildUserIdTAG: 226085 SecondChildUserNameTAG: Teto SecondChildCreateTimeTAG: 2012-10-13T15:03:05Z IndexTAG: 1019 TitleTAG: Boom! For anyone who hasn't seen the week 7 capacitor tutorial, it's explosive and awesome. How many Doctors of Electrical Engineering does it take to blow up a capacitor? UserIdTAG: 21541 UserNameTAG: JSChambers CreateTimeTAG: 2012-10-11T00:15:55Z VoteTAG: 3 CoursewareTAG: Week 7 / Types of Capacitors CommentableIdTAG: 6002x_Types_of_Capacitors NumberOfReplyTAG: 2 FirstChildTAG: He must hook one up backwards does he? Can't wait. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-11T00:20:07Z FirstChildTAG: I did the same thing when I was in College with a 470uF/10V capacitor. It was so cool FirstChildUserIdTAG: 184827 FirstChildUserNameTAG: DiegoT FirstChildCreateTimeTAG: 2012-11-01T03:36:57Z IndexTAG: 1020 TitleTAG: H5P2 Part A Here is how I tried the question, still my answer is wrong!....please help me! ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13498171241343667.png and same help in H5P3 Part A would be great, I think I am missing one really important point, here, if you guys could just tell me that!...then the rest is pure algebra....I mean I got to the point shown above, but obviously it is wrong.....because in the original equations in the "Known To Us" part, probably have capital V's and not small v's....but then how should I convert...... UserIdTAG: 169784 UserNameTAG: thewiredbear CreateTimeTAG: 2012-10-09T21:06:49Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: iDS is a function of VGS - VT. VGS is not the same as vIN in this circuit. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-09T23:42:09Z SecondChildTAG: Thanks, now i realize what i was doing wrong SecondChildUserIdTAG: 40865 SecondChildUserNameTAG: guilima10 SecondChildCreateTimeTAG: 2012-10-10T03:06:26Z SecondChildTAG: hint maybe? SecondChildUserIdTAG: 182653 SecondChildUserNameTAG: monkeyfoahead SecondChildCreateTimeTAG: 2012-10-13T18:15:54Z IndexTAG: 1021 TitleTAG: H7P3 I solved the equation given in the textbook and I've got my answer for Vo but It keeps saying me that 't'is not permitted in my answer. To solve the other parts I need 't' to be in my answer. Can someone help ? UserIdTAG: 206648 UserNameTAG: TsvetanGeorgiev CreateTimeTAG: 2012-10-09T15:05:46Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: 't' is not required in any of the answers to H7P3. The first question asks for the final voltage (or steady state voltage), which is not time dependant. The second question asks for the time constant which also doesn't require 't' in the answer. FirstChildUserIdTAG: 270011 FirstChildUserNameTAG: SnowmanZA FirstChildCreateTimeTAG: 2012-10-09T16:33:14Z SecondChildTAG: Equation given in the textbook? Can you tell me a page? I am having problem with this one. SecondChildUserIdTAG: 286954 SecondChildUserNameTAG: ododo SecondChildCreateTimeTAG: 2012-10-10T13:01:45Z SecondChildTAG: You don't need any fancy equations to solve the first question. The final voltage means the voltage after a long time, after which any time constants will no longer have any effect. SecondChildUserIdTAG: 270011 SecondChildUserNameTAG: SnowmanZA SecondChildCreateTimeTAG: 2012-10-10T16:22:31Z SecondChildTAG: In other words, Final voltage at t>>0+ is the voltage when L becomes short circuit (in case of C when in becomes open circuit). To find the time constant try to find RTH as seen from the L terminals. Hope that would help simplify things. SecondChildUserIdTAG: 127195 SecondChildUserNameTAG: princeofsudan SecondChildCreateTimeTAG: 2012-10-10T21:09:24Z SecondChildTAG: как е цецо върви ли ученето яко :) SecondChildUserIdTAG: 277027 SecondChildUserNameTAG: nikidp SecondChildCreateTimeTAG: 2012-10-19T10:13:12Z SecondChildTAG: here u have consider t=infinity (t>>>o) so that your exp. term reduced 0. SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-10-28T08:09:24Z IndexTAG: 1022 TitleTAG: Symbolic algebraic expressions It's so funny to bruteforce all possible combinations of letters (vIN, Vin, VIN, even VIn) as an answer and to understand finally, that you have to give your answer in form of "X+Y", and not "Z>X+Y"... UserIdTAG: 326787 UserNameTAG: Angstrem CreateTimeTAG: 2012-10-08T22:26:38Z VoteTAG: 3 CoursewareTAG: Week 5 / MOSFET Amplifier Exercise 1 CommentableIdTAG: 6002x_mosfet_amp_e1 NumberOfReplyTAG: 2 FirstChildTAG: I spent the majority of the time on this problem figuring out I needed VT not Vt, etc. Oh well. FirstChildUserIdTAG: 264596 FirstChildUserNameTAG: Nuru FirstChildCreateTimeTAG: 2012-10-11T20:48:30Z FirstChildTAG: No brute forcing at all, if you read the question right. If the question says "write algebraic expression for minimum value of Z", it is obvious that there is no need in "Z>" part. And the right case of the parameters you can see in the there exactly as they should appear in the answer. I am a student also, and never faced such a problems. FirstChildUserIdTAG: 194098 FirstChildUserNameTAG: ZhekaS FirstChildCreateTimeTAG: 2012-10-09T00:01:16Z SecondChildTAG: It's obvious now to me, but at 01:30 AM, when I was doing all this, it wasn't... SecondChildUserIdTAG: 326787 SecondChildUserNameTAG: Angstrem SecondChildCreateTimeTAG: 2012-10-09T13:07:40Z SecondChildTAG: Agree'd. It'd be helpful if an example of the format wanted for the correct parsing of the answer was supplied. I'm still bruteforcing the possibilities now... SecondChildUserIdTAG: 281858 SecondChildUserNameTAG: aspringv SecondChildCreateTimeTAG: 2012-10-10T03:51:51Z IndexTAG: 1023 TitleTAG: H7P2 It seems that decay time constant associated with network B looks weird. Have anybody here already tried it? The answer marked as correct has wrong unit. Mike UserIdTAG: 157610 UserNameTAG: mradziwo CreateTimeTAG: 2012-10-08T18:30:46Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: You're not first. It was already mentioned here: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5072c742b524d627000000a3 FirstChildUserIdTAG: 203445 FirstChildUserNameTAG: x13n FirstChildCreateTimeTAG: 2012-10-08T18:32:13Z FirstChildTAG: I agree. I managed to get a tick, but I had to use an incorrect answer. Instead of L/R it took L*R as correct. FirstChildUserIdTAG: 270011 FirstChildUserNameTAG: SnowmanZA FirstChildCreateTimeTAG: 2012-10-08T19:31:21Z SecondChildTAG: I agree too. [Henry*Ohm] is not equal [Second] SecondChildUserIdTAG: 391745 SecondChildUserNameTAG: we_told SecondChildCreateTimeTAG: 2012-10-08T21:42:02Z FirstChildTAG: Sorry, there was a typo that has now been fixed. FirstChildUserIdTAG: 42833 FirstChildUserNameTAG: KKA FirstChildCreateTimeTAG: 2012-10-08T23:39:19Z IndexTAG: 1024 TitleTAG: LINK TO DOWNLOAD VIDEOS OF WEEK 5 AND WEEK 6 Plz post the link to download the videos of week 5 and 6 UserIdTAG: 178713 UserNameTAG: AlbertExist CreateTimeTAG: 2012-10-08T14:20:23Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: You mean the ones that are posted here: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/ ? FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-08T18:43:30Z SecondChildTAG: They've only posted the playlists, not the links to the videos for those weeks. Also, the Week 5 and 6 playlists are hosted on dropbox, which is problematic, that's another site that is often restricted. SecondChildUserIdTAG: 94557 SecondChildUserNameTAG: g_hopper SecondChildCreateTimeTAG: 2012-10-08T20:17:18Z FirstChildTAG: Plz post the link to download the videos of week 5 and 6 FirstChildUserIdTAG: 568795 FirstChildUserNameTAG: Ramkrishna1623 FirstChildCreateTimeTAG: 2012-10-08T15:12:52Z IndexTAG: 1025 TitleTAG: Lab in Midterm?? Will there be a Lab in the Midterm examination, or will it be like a Homework assignment? I would really appreciate it if someone could give me a clue about the Midterm exam structure. Thanks UserIdTAG: 304587 UserNameTAG: Vasso CreateTimeTAG: 2012-10-08T09:10:35Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi Vasso In the earlier edition of 6.00x both the mid-term and final exams were like the homeworks - only more questions; no Labs Melbur FirstChildUserIdTAG: 95673 FirstChildUserNameTAG: melbur FirstChildCreateTimeTAG: 2012-10-08T13:50:54Z SecondChildTAG: Hi Melbur! Thank you for your answer! I just hope there aren't so many that we won't make it in time! :) SecondChildUserIdTAG: 304587 SecondChildUserNameTAG: Vasso SecondChildCreateTimeTAG: 2012-10-08T17:30:07Z IndexTAG: 1026 TitleTAG: Solution to HW Will we have the solution to the home works and Labs which their dead lines have passed. This will be very helpful whan it come to revising for mid-term and final exams. UserIdTAG: 47733 UserNameTAG: Solmaz CreateTimeTAG: 2012-10-08T07:45:05Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Yeah you can have the solutions. Just go to those HW & Labs whose deadlines have passed and you will find a 'Show Answer' button in place of the 'Check' buttons. FirstChildUserIdTAG: 279303 FirstChildUserNameTAG: Raja91 FirstChildCreateTimeTAG: 2012-10-08T13:11:12Z IndexTAG: 1027 TitleTAG: I hate HW4P2B I finished with all problem excluding v calculation. I don't understand what's wrong: we have equation when V< -5 id = (V + b)*slope. slope = 1, b = -5. df(V) = slope = 1 Rd = 1/1 = 1 Create Th for small signal. Vth = 0.025, Rth = 1/2. Vth*RL / (RL + RTH) = 0.22 But it'e incorrect. Could someone help. Where have I been wrong UserIdTAG: 240170 UserNameTAG: seca CreateTimeTAG: 2012-10-08T07:42:10Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: i think y should be: y = mx +b. with y = Id, and m = slope, b is a contanst. both m and b can take from graph. 1. Id = Slope * Vd + b <=> Id = 1* Vd - 5 (this take from graph) Apply this to node Expression you will find the Vd. That is all have to do. Tuan FirstChildUserIdTAG: 114913 FirstChildUserNameTAG: ngoctuan FirstChildCreateTimeTAG: 2012-10-08T08:02:53Z SecondChildTAG: I don't understand how it can help to find vD(RL). SecondChildUserIdTAG: 240170 SecondChildUserNameTAG: seca SecondChildCreateTimeTAG: 2012-10-08T12:01:26Z SecondChildTAG: The deadline is over. Could someone provide explanation about ZDiod problem? I spent a day but still have no answer how it must be calculated SecondChildUserIdTAG: 240170 SecondChildUserNameTAG: seca SecondChildCreateTimeTAG: 2012-10-08T13:03:27Z SecondChildTAG: 1. from graph : we have 3 part: 2. Vd > 0.6 volt 3. Vd < -5 volt 4. -5< Vd < 0.6 ---> this region Id always be 0, that mean dont worry about it. 5. the behavior of zenner when Vd >0.6 and Vd < -5Volt is look like a RESISTOR. that means the relationship Id and Vd in the GRAPH must be something like this : **id= slope*Vd + b** 6. luckily on the graph we have slope = 1A/V then we have **Id=Vd+b** 7. now thinking about it: we write the expression for a RESISTOR, because the behavior of zenner like a resistor. mean we can not choose B between (-5, 0.6), why in this region in the graph Id always 0 and it is not a line o resistor. --> B >= 0.6 or B < -5. 8. ok Zenner is resistor and wat value of Vd that make the Id = 0. that means we find the starting point of Resistor line on the graph. and we only have 2 : Vd = -5 => B = 5 ( 0 = -5 + B) // and Vd = 0.6 => B = -0.6 9. now we have: Id = Vd + B with B = 5 or B = -0.6 10. with KVL we have : **Vo + Vd =0** (be careful with the sign) ==> **V0 = -Vd** 11. thinking about the starting point of curve, means Id = 0 when?. **Vd = -B** or **-Vd = B** ==> **B = VO** 12. but **V0** can not be **negative** because it plug right sign (-Source+)------>(+Vd-)----- ==> V0=B = 5. 13. Ok now: **Id = Vd + 5** this only work correct when **Vd< -5** 14. Node analysis: **(V0-Vi)/RIN + (V0)/RL - Id =0** (be care full with sign of Id) 15. replace Id = Vd + 5 to NOde expression we have: 16. **(V0-Vi)/RIN + (V0)/RL - (Vd+5) =0** but we know that Vd = -V0 then 17. **(V0-Vi)/RIN + (V0)/RL - (-V0+5) =0** NOW you can find V0 SecondChildUserIdTAG: 114913 SecondChildUserNameTAG: ngoctuan SecondChildCreateTimeTAG: 2012-10-08T13:49:56Z IndexTAG: 1028 TitleTAG: Why my answers to (b) and (c) accepted as correct don't match the answers given? Solution for (a) is obtained by directly substituting the given values into the first-order differential equation for RC circuit: $77=92+(35-92)\cdot e^\frac{-t}{1.5\cdot 10^6 \cdot 10^{-6}}$ $t=-\cfrac{3\cdot ln\frac{15}{57}}{2}=2.0025016001$ And this solution exactly matches the correct answer. However, my solution to (b) - 0.00792789066321, and (c) - 0.00394338160062, even though accepted as correct, do not match the answers given. In order to find solution to (b) and (c) I used the following circuit model: ![enter image description here][1] By solving differential equation $\cfrac{v_C(t)-V_S}{R}+\cfrac{v_C(t)}{R_B}+i_C(t)=0$ I obtained solution for $v_C$: $v_C=V_S\cdot \cfrac{R_B}{R+R_B} + \left(V_0-V_S\cdot\cfrac{R_B}{R+R_B}\right)\cdot e^\frac{-t}{RC\frac{R_B}{R+R_B}}$ Substituting the values into the solution equation I obtained: $35=92\cdot\cfrac{10^4}{1.5\cdot10^6+10^4}+\left(77-92\cdot\cfrac{10^4}{1.5\cdot10^6+10^4}\right)\cdot e^\frac{-t}{1.5\cdot10^6\cdot 10^{-6}\cdot\frac{10^4}{1.5\cdot10^6+10^4}}$ This resulted in: $t=-\cfrac{3\cdot ln\frac{5193}{11535}}{302}=0.00792789066321$, which is accepted as correct even though 0.00783235788441 is provided in the 'Answer:' field. My answer to (c): $\cfrac{0.00792789066321}{0.00792789066321+2.0025016001}=0.00394338160062$ is also accepted as correct even though 0.0038960481433 is provided in the 'Answer:' field. Note that if the initial differential equation is written as: $\cfrac{v_C(t)-V_S}{R}+\cfrac{v_C(t)}{R_B}-i_C(t)=0$ (the current flowing into the node), the answer to (b) will be $-0.00792789066321$ (there will be no 'minus' in the power index of $e$). Hmmm... [1]: https://edxuploads.s3.amazonaws.com/1349647219106381.jpg UserIdTAG: 252722 UserNameTAG: vaboro CreateTimeTAG: 2012-10-07T21:57:31Z VoteTAG: 3 CoursewareTAG: Week 6 / Neon relaxation oscillator exercise CommentableIdTAG: 6002x_neon_relaxation_oscillator_exercise NumberOfReplyTAG: 3 FirstChildTAG: Wow...we have oscillators coming up...cool! And it looks like we have to play with differential equations...scary! I'm still in Week 5; I'll save your post for next week and have a look :) FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-10-07T22:07:52Z SecondChildTAG: Right, I don't even want to think about solving an R(t)C circuit:))) or whatever is the correct way to say about nonlinear resistance like a diode in combination with a capacitor:))) SecondChildUserIdTAG: 252722 SecondChildUserNameTAG: vaboro SecondChildCreateTimeTAG: 2012-10-07T23:09:43Z SecondChildTAG: Vaboro, (that's quite an extensive post you've written up) Eventually, getting to it independently, I've got the same result (using Excel) as you do. That's 0.01% difference with the published result. Perhaps if the reference calculations were done with less precision than Excel and whatever calculator you've used provide, then 0.01% difference is quite OK and not a reason to doubt our solution formulas. Thanks! SecondChildUserIdTAG: 321686 SecondChildUserNameTAG: klisitsyn SecondChildCreateTimeTAG: 2012-10-09T07:23:02Z SecondChildTAG: Right! Nonetheless, when the result is not quite what it should be I think it's always better to check if the path taken was indeed correct. SecondChildUserIdTAG: 252722 SecondChildUserNameTAG: vaboro SecondChildCreateTimeTAG: 2012-10-09T18:53:56Z SecondChildTAG: ...Prof. Grimson of MIT mentioned on one of his lectures: "A small bias can end up making a huge difference over time" SecondChildUserIdTAG: 252722 SecondChildUserNameTAG: vaboro SecondChildCreateTimeTAG: 2012-10-09T18:57:00Z SecondChildTAG: I get vC=VS*RB/(R+RB)+((V0−VS)*RB*e^(−t/RC))/(R+RB), no idea why SecondChildUserIdTAG: 380703 SecondChildUserNameTAG: vargy SecondChildCreateTimeTAG: 2012-10-19T17:38:37Z SecondChildTAG: vaboro, would you be so kind and post some details about solving the differential equation. I just can't get my head arround it and need some tips on how to get the correct result ... SecondChildUserIdTAG: 294766 SecondChildUserNameTAG: dejanst SecondChildCreateTimeTAG: 2012-11-01T10:14:36Z FirstChildTAG: The $vc$ equation you found is exactly the same as me, and you solved $t$ for $vc=35V$ the same way I did. It must be rounding errors. FirstChildUserIdTAG: 160020 FirstChildUserNameTAG: antoineleclair FirstChildCreateTimeTAG: 2012-10-08T22:10:01Z FirstChildTAG: The beginning part of you question seems legit. Maybe some rounding errors somewhere? I came up with the same numbers as you. At the end - when you say (vC(t)−VS)/R+vC(t)/RB−iC(t)=0 (the current flowing into the node),` If you set iC(t) to d(-vC)/dt or rewrite the equation with ALL currents flowing into the nodes as (VS-vC(t))/R-vC(t)/RB−iC(t)=0 where iC(t) = dvC(t)/dt then you should still get a negative exponential and (thankfully), a stable system. ` FirstChildUserIdTAG: 274263 FirstChildUserNameTAG: CoreyO FirstChildCreateTimeTAG: 2012-10-08T03:07:10Z SecondChildTAG: Actually, what I wanted to say was when the lamp is an open circuit the model is the following: ![enter image description here][1] You know the direction of the current $i_C$ and voltages on the capacitor's 'plates.' Yet, when the current flows through the lamp the circuit model is the following: ![enter image description here][2] The voltages on the capacitor 'plates' didn't change, but the current now flows in another direction. Thus, positive time may represent capacitor's charging and negative - capacitor's discharging. This is just a hypothesis. Correct me if I'm wrong. [1]: https://edxuploads.s3.amazonaws.com/13497866061343684.jpg [2]: https://edxuploads.s3.amazonaws.com/1349786800315573.jpg SecondChildUserIdTAG: 252722 SecondChildUserNameTAG: vaboro SecondChildCreateTimeTAG: 2012-10-09T12:47:36Z IndexTAG: 1029 TitleTAG: LAB4 can't plot iv trace for mosfet ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13496308781343666.jpg When I try plot vi characteristic for mosfet, and give Vgs 3V and Vds 5V. Transient analys show me empty screen. What I do wrong? UserIdTAG: 104522 UserNameTAG: IgorNovice CreateTimeTAG: 2012-10-07T17:30:41Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: having the same problem.....the current probe cannot detect the response from the circuit.... ![enter image description here][1] i know it is not giving the response, what the question is demanding, but when i checked this. the right mark is shown.. [\[1\]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_4/Curve_Tracer/][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_4/Curve_Tracer/ FirstChildUserIdTAG: 308571 FirstChildUserNameTAG: Sayantani FirstChildCreateTimeTAG: 2012-10-07T17:34:24Z FirstChildTAG: You currently ploted GS, But must plot DS FirstChildUserIdTAG: 240170 FirstChildUserNameTAG: seca FirstChildCreateTimeTAG: 2012-10-07T17:57:00Z SecondChildTAG: i have done it what u r telling....but it is also not giving the desired curve.... SecondChildUserIdTAG: 308571 SecondChildUserNameTAG: Sayantani SecondChildCreateTimeTAG: 2012-10-07T18:31:48Z FirstChildTAG: Your VDS needs to supply a triangle wave as in the example pic. VGS is just a constant DC voltage. FirstChildUserIdTAG: 502508 FirstChildUserNameTAG: amaher FirstChildCreateTimeTAG: 2012-10-07T18:34:44Z SecondChildTAG: Oh, and as seca said you need to plot _VDS_ SecondChildUserIdTAG: 502508 SecondChildUserNameTAG: amaher SecondChildCreateTimeTAG: 2012-10-07T18:43:23Z SecondChildTAG: Thanks!!! SecondChildUserIdTAG: 104522 SecondChildUserNameTAG: IgorNovice SecondChildCreateTimeTAG: 2012-10-07T18:45:46Z SecondChildTAG: thanks. SecondChildUserIdTAG: 308571 SecondChildUserNameTAG: Sayantani SecondChildCreateTimeTAG: 2012-10-07T20:24:47Z FirstChildTAG: **seca** and **amaher** are correct. Do **not** put your probe on $V_{gs}$ because that is a constant; hence all it will plot is a straight line. Also note that in your drawing $V_{ds}$ needs to be changed from a DC source to "triangle." Simple mistakes like this should not be made. It means you are either rushing too fast and not reading the directions, or you don't understand the language that well. Not to be a wise-azz, but this lab (Week 4) was the easiest we've had yet. FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-10-07T20:58:50Z IndexTAG: 1030 TitleTAG: H4P2 ZENER REGULATOR how to solve the H4P2 when zener diode is inserted? UserIdTAG: 114864 UserNameTAG: Vikaash CreateTimeTAG: 2012-10-07T11:29:25Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: I am still on it but the first thing is to follow the HINT...get your LOAD LINE writen in this way: iz=-slope*V0+constant. Once u have it u can decide wether u are in the 1st of 3rd quadrant working...then you can apply all what lecture sequence 7 says about nonlinear devices FirstChildUserIdTAG: 99441 FirstChildUserNameTAG: coyarce FirstChildCreateTimeTAG: 2012-10-07T11:52:08Z SecondChildTAG: How to build this LOAD LINE ? SecondChildUserIdTAG: 192114 SecondChildUserNameTAG: chuba SecondChildCreateTimeTAG: 2012-10-07T12:09:06Z FirstChildTAG: I've given some hints in this thread https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5070bcde0584032300000047 And Myrimit's given a brilliant explanation in https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5070be20fabaf62b0000004d HTH FirstChildUserIdTAG: 502508 FirstChildUserNameTAG: amaher FirstChildCreateTimeTAG: 2012-10-07T12:09:00Z SecondChildTAG: after spending 3hours i got the ans......thanx @amaher SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-10-07T12:25:58Z FirstChildTAG: How to build this LOAD LINE I maid Thevenin equivalent, but I don't understand how to build LOAD LINE... HELP.. FirstChildUserIdTAG: 192114 FirstChildUserNameTAG: chuba FirstChildCreateTimeTAG: 2012-10-07T12:44:02Z SecondChildTAG: give me a few secs and I help u with an example!!! SecondChildUserIdTAG: 99441 SecondChildUserNameTAG: coyarce SecondChildCreateTimeTAG: 2012-10-07T12:48:49Z SecondChildTAG: in general, the equation of a line is represented by: y=mx+b where "y" is the dependent variable, "x" is the independent variable, "m" is the slope of the line and "b" is the point where the line cut the y-axis. In the specific case of our exercise, "y" is the current of the diode "iD", "x" is the voltage drop of the diode "vD" and the the slope "m" and the "b" should be calculated...HOW? just use the NODE analysis in the circuit and you will get one equation.....right? so....just simple order in a convenient way such equation in order that looks like y=mx+b...that's how u construct a load line. A good example can be found in lecture S6V4 where it is explained the load line use! SecondChildUserIdTAG: 99441 SecondChildUserNameTAG: coyarce SecondChildCreateTimeTAG: 2012-10-07T12:57:07Z SecondChildTAG: I find Thevenin equivalent and Get Vth=8 Rth = 666.666 ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13496174232453941.jpg SecondChildUserIdTAG: 192114 SecondChildUserNameTAG: chuba SecondChildCreateTimeTAG: 2012-10-07T13:44:34Z SecondChildTAG: Am I riht? SecondChildUserIdTAG: 192114 SecondChildUserNameTAG: chuba SecondChildCreateTimeTAG: 2012-10-07T13:44:56Z SecondChildTAG: ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13496198246758752.jpg Id=V1-V/Rin+V1/RL SecondChildUserIdTAG: 192114 SecondChildUserNameTAG: chuba SecondChildCreateTimeTAG: 2012-10-07T14:23:55Z SecondChildTAG: check it out the signs of your expression. Check how voltage and current are defined for the diode....sorry for the late answer! SecondChildUserIdTAG: 99441 SecondChildUserNameTAG: coyarce SecondChildCreateTimeTAG: 2012-10-07T21:30:52Z SecondChildTAG: Hi coyarce . I done all but I don't know how to find min RL SecondChildUserIdTAG: 192114 SecondChildUserNameTAG: chuba SecondChildCreateTimeTAG: 2012-10-07T22:24:38Z IndexTAG: 1031 TitleTAG: First problem help Using the node method you get (5-Vo)/(850)=0.088/V^2 or Vo= 5-74.8/V^2 Then also i know that 5-Vo =V Replaicing in the Other equation i can fino the values for Vo UserIdTAG: 284184 UserNameTAG: Rodrigo10 CreateTimeTAG: 2012-10-07T07:57:19Z VoteTAG: 3 CoursewareTAG: Week 4 / Dependent Sources Exercise 1 CommentableIdTAG: 6002x_dep_src_exsz_1 NumberOfReplyTAG: 1 FirstChildTAG: Nice. I needed your help. Thank you. FirstChildUserIdTAG: 330170 FirstChildUserNameTAG: H_Litzroth FirstChildCreateTimeTAG: 2012-10-07T12:41:47Z IndexTAG: 1032 TitleTAG: lab4 : value of k is not comin even though i got vt correctly . can someone help me out where i went wrong help me asap UserIdTAG: 230347 UserNameTAG: andrewwinney CreateTimeTAG: 2012-10-07T05:33:45Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Make sure you use the correct units for Ids. You need to convert the value obtained from the graph to Ampere. FirstChildUserIdTAG: 226266 FirstChildUserNameTAG: anjanasgf FirstChildCreateTimeTAG: 2012-10-07T08:38:57Z FirstChildTAG: Hi andrewwinney! Can I help you? Are you sure that you are choosing the values of the saturated region? Be careful with that. Remember that the first formula that they give you is for that condition. P.D: In the case that you are needing more hints of Lab4 you can take a look at [here][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5070e1b72210d02400000055 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-07T06:10:54Z FirstChildTAG: in order to get the value of K ; just put the value of Vgs to 2.5(approx.) and see the current Ids into the VI graph keeping Vds fixed at~ 3V(approx.).After getting the value of Ids & Vt ,put Vgs=2.5 in the given formula ,you will get the desired result. we have taken the value of Vgs =2.5 to get in the saturation region. FirstChildUserIdTAG: 181793 FirstChildUserNameTAG: mitrahul FirstChildCreateTimeTAG: 2012-10-07T06:10:51Z SecondChildTAG: i am getting VT in comlex number. Why..?? SecondChildUserIdTAG: 259693 SecondChildUserNameTAG: MehrozKhan SecondChildCreateTimeTAG: 2012-10-07T08:54:39Z IndexTAG: 1033 TitleTAG: Minimum value of Vin Hi everyone, I'm doing the first part of the exercise and I don't get why the correct answer is VT... I mean, VIN (VGS) has to be at least VT to be the MOSFET in its ON state, thats ok, but just that value doesn't mean that we are working in the saturated region, it has also to be true that: VOUT >= VIN - VT. So the minimum value for VIN would be VOUT + VT, that is higher than just VT... am I wrong? Thank you for helping me with this issue :) UserIdTAG: 414462 UserNameTAG: Pablo_C CreateTimeTAG: 2012-10-06T19:23:14Z VoteTAG: 3 CoursewareTAG: Week 5 / MOSFET Amplifier Exercise 1 CommentableIdTAG: 6002x_mosfet_amp_e1 NumberOfReplyTAG: 1 FirstChildTAG: as you said: VOUT >= VIN - VT. so if I arrange it this way VIN<=VT+VOUT. Notice that VT is a fixed characteristic of MOSFET and can not be zero, but Vout can be zero. Therefore the VIN<=VT+0=VT and this is the least value for VIN. FirstChildUserIdTAG: 374393 FirstChildUserNameTAG: rmaleki FirstChildCreateTimeTAG: 2012-10-06T19:40:39Z SecondChildTAG: Now I see! :D Thank you! SecondChildUserIdTAG: 414462 SecondChildUserNameTAG: Pablo_C SecondChildCreateTimeTAG: 2012-10-07T10:37:07Z SecondChildTAG: Then it is .5,,,but its wrong SecondChildUserIdTAG: 379867 SecondChildUserNameTAG: rbgowda SecondChildCreateTimeTAG: 2012-10-07T19:47:22Z SecondChildTAG: A simple way to look at it is, if you set Vin = VT, the requirement of Vin >= VT is met. So then, plug in VT for Vin in the equation Vout=Vin-VT, what do you get? SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-09T00:10:35Z IndexTAG: 1034 TitleTAG: h4p2-noise voltage hello friends although i got ans for almost all part in h4p2 but i did not get correct ans for any cases for small vo ie noise voltage across load.please help. whats the use of this sinusoid 30mv source in this circuit.please help. UserIdTAG: 269641 UserNameTAG: BAUWA CreateTimeTAG: 2012-10-06T18:07:50Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: step 1.you have to take into account the sinusoid of 40mv in your VI step 2.repeat the steps you did to calculate V0 step 3. after that just take the difference of V0 and the voltage you will get after step2 which will be you vo(noise voltage ) im not sure how to calculate noise voltage after the zener has been plugged in ...if anyone knows please help FirstChildUserIdTAG: 278792 FirstChildUserNameTAG: sali FirstChildCreateTimeTAG: 2012-10-06T18:54:02Z SecondChildTAG: @sali. i did the same.so u should get same noise voltage in 2k and 4k ohm load voltage.but its not.only my first part is correct. SecondChildUserIdTAG: 269641 SecondChildUserNameTAG: BAUWA SecondChildCreateTimeTAG: 2012-10-06T19:02:08Z SecondChildTAG: no RL is differnt in both cases so you will get different v0 SecondChildUserIdTAG: 278792 SecondChildUserNameTAG: sali SecondChildCreateTimeTAG: 2012-10-06T22:24:39Z IndexTAG: 1035 TitleTAG: tutorial link require due to you tube blocking plz give us tutorial link UserIdTAG: 134103 UserNameTAG: saad5114 CreateTimeTAG: 2012-10-06T07:26:30Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You should find what you are looking for [here][1] [1]:https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/ FirstChildUserIdTAG: 248807 FirstChildUserNameTAG: 1kingsman FirstChildCreateTimeTAG: 2012-10-06T08:15:09Z SecondChildTAG: these are not there SecondChildUserIdTAG: 134103 SecondChildUserNameTAG: saad5114 SecondChildCreateTimeTAG: 2012-10-12T14:01:58Z IndexTAG: 1036 TitleTAG: H5P1 part 1 A)how VIN can be calculated without given RL. B)in part 3 i put vo as 0 and used 0=Vs-K/2(vin-vt)^2*rl, using Rl from answer of part 2.but not getting the correct answer.also I tried to find it using calculus by simply differentiating above equation but getting 0=K(vin-vt)Rl this gives vo=vt but this is also wrong.Please guide. UserIdTAG: 99628 UserNameTAG: Pranjal16 CreateTimeTAG: 2012-10-05T12:33:16Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: When I am stuck on problems like this that seem to need a value I don't possess, I just guess what that number might be, put it into the equations and see what happens. Or I try a progression, like 1, then 10, then 100, etc, etc. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-05T17:56:41Z FirstChildTAG: My understanding is: Vs - ids RL >= Vi - VT ids = K/2 * (Vi - VT)^2 But seems to be wrong. Anyone knows why? FirstChildUserIdTAG: 210986 FirstChildUserNameTAG: JorgeRmz FirstChildCreateTimeTAG: 2012-10-06T22:33:19Z FirstChildTAG: Hi Pranjal! You can take a look at this Post [Hints][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_mosfet_amp/threads/507a1d6dd894f12300000039 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-14T04:20:26Z SecondChildTAG: I understood my mistake and rectified it when I reread the textbook.still I found hints benificial. SecondChildUserIdTAG: 99628 SecondChildUserNameTAG: Pranjal16 SecondChildCreateTimeTAG: 2012-10-14T12:23:44Z IndexTAG: 1037 TitleTAG: H4P1 q1 Dear all, I am having confusion with H4P1. The current **Ip** is given by the expression **P*(Vpk^(3/2))**. **P = 0.002A/V^(3/2)** Therefore if **Vpk=6V** then **Ip = (0.002A/6^(3/2))*(6V^(3/2))= 0.002A** ***which is wrong.*** This is assuming, of course, that V is Vpk is *vpk*. I think the notation in the question is badly explained, unless I am missing something, which is probably the case. **If so, please could someone kindly point out what I am missing?** Many thanks in advance. UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-10-05T10:10:32Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: You must multiply, and not divide because Ip is function from the set parameters FirstChildUserIdTAG: 96873 FirstChildUserNameTAG: Santyaga FirstChildCreateTimeTAG: 2012-10-05T11:01:47Z SecondChildTAG: So the equation should be **Ip = (0.002A*6^(3/2))*(6V^(3/2))= 0.432**? This too is wrong. SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-05T11:14:53Z SecondChildTAG: Why do you make this second multiply? You need only one part(half) of this... SecondChildUserIdTAG: 96873 SecondChildUserNameTAG: Santyaga SecondChildCreateTimeTAG: 2012-10-05T11:45:28Z SecondChildTAG: "A/V" is unit is this equation, don't replace V with voltage value. SecondChildUserIdTAG: 405984 SecondChildUserNameTAG: dpedro SecondChildCreateTimeTAG: 2012-10-06T21:00:49Z SecondChildTAG: more precisely, "A/V^3/2" is unit SecondChildUserIdTAG: 405984 SecondChildUserNameTAG: dpedro SecondChildCreateTimeTAG: 2012-10-06T21:03:38Z FirstChildTAG: in the definition of P.... A and V refer to some unit of measure (Ampere and Volt) respectively. for values of Vpk larger than zero; current ip is equal to P*(Vpk)^1.5. say Vpk = 6V; then ip = 0.002*6^1.5. ip = .0294 Ampere for vpk = 6Volts FirstChildUserIdTAG: 236810 FirstChildUserNameTAG: jmendego FirstChildCreateTimeTAG: 2012-10-05T12:37:18Z FirstChildTAG: You are understandably confusing units with terms here. Ask yourself, what is the unit that perveance is expressed in? That should simplify matters for you. I'm sure this has confused more than a few folks who are not used to this kind of material. FirstChildUserIdTAG: 141108 FirstChildUserNameTAG: rl777 FirstChildCreateTimeTAG: 2012-10-05T12:47:26Z SecondChildTAG: Indeed. SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-06T17:43:20Z FirstChildTAG: Hi hazel1919! I think that you are confusing units with variables... P = 0.002A/V^(3/2) means P=0.002 **[Ampere/Volt^(3/2)]** so what you are seeing in the **[ ]** it is the unit of the Perveance **P** and **you should not confuse the V [Volt] with the voltage variable vPK** while replacing in the formula ... So, the expression that they give you it is: iPK=P*(vPK)^(3/2) **value_of_iP**[Ampere]= **value_of_P** [Ampere/Volt^(3/2)]* **value_of_vPK ^(3/2)** [Volt^(3/2)] Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-05T15:08:42Z SecondChildTAG: Thank you, this had me confused too, this explanation really helped. SecondChildUserIdTAG: 176173 SecondChildUserNameTAG: SDdad SecondChildCreateTimeTAG: 2012-10-08T01:20:09Z FirstChildTAG: Thank, thank you, thank you, Myriam. Thanks to you, I finally got the right answers! FirstChildUserIdTAG: 237167 FirstChildUserNameTAG: Dug FirstChildCreateTimeTAG: 2012-10-06T16:31:41Z SecondChildTAG: Nice! :). Well done! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-06T22:28:22Z IndexTAG: 1038 TitleTAG: H5P3 problem statment mistake? Hi everyone. Solving H5P3 task, I met some interesting thing. We have to write an algebraic expression for SOURCE FOLLOWER's vout in terms of parameters K, VT, and RS, the input bias voltage VIN, and the incremental input voltage vin. But I don't understand, how I can derive this expression without VOUT term. Is that statment mistake? Or, maybe, I'm missing something important. Thanks in advance UserIdTAG: 81344 UserNameTAG: DmitriyL CreateTimeTAG: 2012-10-04T15:31:11Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Do you mean H6P3? You don't need the VOUT term. If you read it carefully, you will see that they are asking for the vout which is the small signal output, so you can ignore the VOUT and just see how vout changes with vin. FirstChildUserIdTAG: 270011 FirstChildUserNameTAG: SnowmanZA FirstChildCreateTimeTAG: 2012-10-04T16:47:31Z SecondChildTAG: But equation for vout is K(VIN-VT-VOUT)*(vin-vout)*RS, isn't it? So i do need VOUT anyway to express incremental vout. Or what? Tell me another way please. SecondChildUserIdTAG: 81344 SecondChildUserNameTAG: DmitriyL SecondChildCreateTimeTAG: 2012-10-04T17:14:42Z SecondChildTAG: Remember incremental vout is del Vout / del vin. SecondChildUserIdTAG: 270011 SecondChildUserNameTAG: SnowmanZA SecondChildCreateTimeTAG: 2012-10-04T19:14:53Z FirstChildTAG: In H5P2 you actually should have solved the quadratic equation for V_out and now you have to find the derivative of V_out=f(V_in). They gave us a hint that is perfectly valid. H5P3 nicely resolves if you solved H5P2 successfully. FirstChildUserIdTAG: 252722 FirstChildUserNameTAG: vaboro FirstChildCreateTimeTAG: 2012-10-04T20:21:45Z SecondChildTAG: I am getting some expression(which i feel correct) but it always displays could not parse (..) as a solution. SecondChildUserIdTAG: 99628 SecondChildUserNameTAG: Pranjal16 SecondChildCreateTimeTAG: 2012-10-05T13:36:42Z SecondChildTAG: VABORO can u help us with this problem H5P3 part 1? SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-06T03:24:40Z SecondChildTAG: In H5P2 you wrote an expression for ids. Vout is ids * Rs. vout is vin * del Vout / del vIN. So you need to multiply solution of H5P2 by Rs. Then take the derivative of that function with respect to vIN, then multiply that with vin. SecondChildUserIdTAG: 270011 SecondChildUserNameTAG: SnowmanZA SecondChildCreateTimeTAG: 2012-10-06T07:44:38Z SecondChildTAG: yeah that's it, but it is more troublesome than my expression written above, which is correct too accotding to the chapter 8.2.4, page 436, expretion 8.55 SecondChildUserIdTAG: 81344 SecondChildUserNameTAG: DmitriyL SecondChildCreateTimeTAG: 2012-10-06T13:00:07Z FirstChildTAG: OK guys am stuck again. So my equation for vOUT looks like this: vOUT=(a+b(vIN-VT)-sqrt((a+b(vIN-VT))^2-(b(vIN-VT))^2))/b or vOUT=(a+b(vIN-VT)-sqrt(a+2b(vIN-VT)))/b it is sticked green mark so i derive it for vout: vout=dvOUT/dvIN*vin|vIN=VIN so from that i have: vout=(1-1/(2*b*sqrt(1+2b(VIN-VT)))*vin and this is still incorrect answer. what can i not see? I'm sure that is stupid think as usual. thanks in advance FirstChildUserIdTAG: 81344 FirstChildUserNameTAG: DmitriyL FirstChildCreateTimeTAG: 2012-10-06T15:47:30Z SecondChildTAG: It does look slightly incorrect. I think you made a mistake when taking the derivative dvOUT/dvIN. Your denominator is wrong, my answer had nothing outside the sqrt() in the denominator i.e. it was vout = vin * 1-1/sqrt(1+....). SecondChildUserIdTAG: 270011 SecondChildUserNameTAG: SnowmanZA SecondChildCreateTimeTAG: 2012-10-06T18:26:52Z SecondChildTAG: Oh man. Thanks alot. this happens when you think about trouble 'that's small school's task'. thanks again.)) SecondChildUserIdTAG: 81344 SecondChildUserNameTAG: DmitriyL SecondChildCreateTimeTAG: 2012-10-06T18:56:52Z IndexTAG: 1039 TitleTAG: wow nice tool :) I like it UserIdTAG: 342603 UserNameTAG: YakovO CreateTimeTAG: 2012-10-04T06:21:39Z VoteTAG: 3 CoursewareTAG: Week 5 / Music Amplifier Demonstration CommentableIdTAG: 6002x_snd_d NumberOfReplyTAG: 1 FirstChildTAG: Great tool!!! Love this course!!! FirstChildUserIdTAG: 324332 FirstChildUserNameTAG: petsol FirstChildCreateTimeTAG: 2012-10-14T08:27:50Z IndexTAG: 1040 TitleTAG: thank you I feel really thanks for this course, such feeling as i've been there by myself) UserIdTAG: 523129 UserNameTAG: AlexMuravl CreateTimeTAG: 2012-10-03T22:14:00Z VoteTAG: 3 CoursewareTAG: Week 1 / Lumped Element Abstraction CommentableIdTAG: 6002x_lumped_element_abstraction NumberOfReplyTAG: 0 IndexTAG: 1041 TitleTAG: started today Guys, I am so late, just started today. I am sure, I can complete the backlogs very soon. But, now what to do regarding penalty? How can I increase my grades, as I have not yet started with the homework and lab work? Any wise suggestions plz.. UserIdTAG: 546022 UserNameTAG: supersakshi CreateTimeTAG: 2012-10-03T14:55:17Z VoteTAG: 3 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: doesn't matters much ,,,just take out all lecture starting from week1 ,surely ll help u for the midterm xm .. FirstChildUserIdTAG: 183166 FirstChildUserNameTAG: yogeshk FirstChildCreateTimeTAG: 2012-10-03T17:37:06Z SecondChildTAG: k.. thnx !! SecondChildUserIdTAG: 546022 SecondChildUserNameTAG: supersakshi SecondChildCreateTimeTAG: 2012-10-04T11:34:06Z IndexTAG: 1042 TitleTAG: Incremental Resistance Solving for the third answer, I took the derivative of id and got -2*e^(-va/5). From my understanding, I should evaluate this at the operating point, namely the answer to question 1 (1.08825V). This would lead me to dRa = va/dia = 1.08825/(-2*e^(-va/5)) = 0. 676. However, to get the correct answer, it is just 1/(-2*e^(-va/5). Do you not evaluate it at the operating point? Thanks for any help. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-03T00:53:01Z VoteTAG: 3 CoursewareTAG: Week 4 / Linearization Exercise 1 CommentableIdTAG: 6002x_linearization_ex_1 NumberOfReplyTAG: 1 FirstChildTAG: The incremental resistance is the inverse of diA/dvA at the operating point, i.e. 1/(2*e^(-vA/5)) at vA=1.088 = 0.62. You are doing dRa, which is not correct; it is not the increment of a resistance that you are calculating, but the incremental resistance, i.e. the ratio between the increment of the voltage and the increment of the current. Besides you had a - sign that shouldn't be there. FirstChildUserIdTAG: 266912 FirstChildUserNameTAG: pietvo FirstChildCreateTimeTAG: 2012-10-03T03:35:45Z SecondChildTAG: The reason your leading -2 should be +2 is because i_a = 10* (1 - e^(-v_a/5)) so when you take the derivative of that and evaluate at v_a, you must do the chain rule, so -e^stuff brings out a leading coefficient of -1, and then when you take the derivative of -v_a/5, the -.2 cancels out the - sign of the -1. SecondChildUserIdTAG: 164898 SecondChildUserNameTAG: jbparkes SecondChildCreateTimeTAG: 2012-10-03T17:40:43Z SecondChildTAG: incremental resistance is jst the inverse of the coefficient vA in the linearized expression(becoz d inverse of the slope in i-v curve is resistance)...so take the derivative of iA that is f(va) with respect to va at the point VA(operating point) the inverse of which wil be the answer SecondChildUserIdTAG: 169416 SecondChildUserNameTAG: subramanya26shin SecondChildCreateTimeTAG: 2012-10-07T06:45:22Z IndexTAG: 1043 TitleTAG: Can't understand the equation Hi there. I follow the tutorial for week 4 and ended up in this equations: Va = -5 Ln ( (va/2+ Iin+10)/10) But I can't find what is the "Iin" value to replace on this equation. Someone can help me to find out where I went wrong? Thanks. UserIdTAG: 316902 UserNameTAG: JesseTeixeira CreateTimeTAG: 2012-10-02T23:03:49Z VoteTAG: 3 CoursewareTAG: Week 4 / Linearization Exercise 1 CommentableIdTAG: 6002x_linearization_ex_1 NumberOfReplyTAG: 1 FirstChildTAG: Hi JesseTeixeira! If you see in the video at time 3:35 [here][1], you will see that the statement says: "(b) solve for the voltage VA when **iIN = 5A**...." So, the iIN that you need it is 5A :). See you! Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_4/wk4t3/ FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-03T00:09:46Z SecondChildTAG: yes, in the video there is a given iIN ... but in the homework I'm finding no Iin to replace it on the equation. I'm really lost about that! :( Thanks for your help. SecondChildUserIdTAG: 316902 SecondChildUserNameTAG: JesseTeixeira SecondChildCreateTimeTAG: 2012-10-03T14:56:45Z IndexTAG: 1044 TitleTAG: is the book really important? Hey, I know I may look like a fool asking this, but is the book really important, because I haven't touched anything of that book, but now I feel I should do it, but just to be sure, can you please tell me if it is trivial or not! UserIdTAG: 169784 UserNameTAG: thewiredbear CreateTimeTAG: 2012-10-02T19:34:12Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: I would suggest solving the problems from the textbook if you feel you aren't confident with the concepts. I had problems in understanding a few things properly and solving problems helped. You don't need to attempt every problem though. However, it is possible to do well in this course without the book as well. The lectures and tutorials seem to cover everything you need. The book does have a lot of other topics which are not part of this course and you could explore if you're interested. FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-10-02T19:45:54Z SecondChildTAG: thanks man!......thanks a lot, I suppose I should see the book now, and waiting for your reply on my previous post...........okay gtg, have to wake up 5:30 am today, its 1:30 am already! SecondChildUserIdTAG: 169784 SecondChildUserNameTAG: thewiredbear SecondChildCreateTimeTAG: 2012-10-02T19:49:45Z SecondChildTAG: Hi thewiredbear! You should go to sleep haha! In which country do you live? here now it is afternoon haha. See you! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-02T20:14:32Z FirstChildTAG: I've found that the problems in the book develop the ideas more gradually than in the lectures. In general, I like that. Every now and then it's very helpful. I did ok the first time around. This is my 2nd go at 6.002x and this time I'm paying more attention to the book. I'm getting some of the added depth that I was hoping to get. I'm beginning to think that books just work better for me, but the amount of information that can come through the videos is really hard to beat. FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-10-02T21:47:51Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-02T22:07:06Z FirstChildTAG: I watch the old videos that were recorded live and then attempt the homework and labs. If I have any questions I may go to the book, but I am more likely to do an internet search for the information. I suppose that I am just too impatient to sit and read the book. My approach may rise up and bite me when it comes to the exams. We shall see!!! FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-03T02:00:31Z SecondChildTAG: Great skyhawk! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-03T04:28:14Z FirstChildTAG: The book is absolutely awesome! FirstChildUserIdTAG: 252722 FirstChildUserNameTAG: vaboro FirstChildCreateTimeTAG: 2012-10-03T04:51:40Z FirstChildTAG: Hey Wired bear this message is for you I am an Indian eleventh grader giving SAT next year, i want to talk to you about certain things. As for me i completely agree with your " I HATE COAching institutes" attitude. IF you want to meet me contact me at skype My id is Harvey.Specter13 FirstChildUserIdTAG: 378080 FirstChildUserNameTAG: GladIDidThis FirstChildCreateTimeTAG: 2012-10-21T22:50:31Z IndexTAG: 1045 TitleTAG: Lab6 - can't modify circuit components Working through lab 6, one of the questions asks me to modify the Vin voltage source. However, the circuit view is totally unresponsive to my clicks - I can't click on anything except for the TRAN button. Is anyone else seeing this or is there a problem at my end? UserIdTAG: 2620 UserNameTAG: dsd CreateTimeTAG: 2012-10-02T02:42:24Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: Yes, it seems a bug... neither do I... FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-02T02:46:29Z FirstChildTAG: dsd, you are right! Let me report this bug right now. Sorry about that. FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-10-02T14:20:05Z SecondChildTAG: Thank you jelizon! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-02T14:58:32Z FirstChildTAG: maybe, we can use the sandbox and make the modifications. Go to Overview and open the sandbox. I hope this work!!! FirstChildUserIdTAG: 9460 FirstChildUserNameTAG: aloreta FirstChildCreateTimeTAG: 2012-10-02T15:36:10Z SecondChildTAG: Thank you aloreta for the suggestion :). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-02T22:23:25Z FirstChildTAG: Ihave the same problem. Finished lab, modelling the circuit in sandbox. FirstChildUserIdTAG: 195218 FirstChildUserNameTAG: Al_Incognito FirstChildCreateTimeTAG: 2012-10-02T16:01:19Z FirstChildTAG: It's ready now. You should be able to modify the first circuit. FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-10-02T21:19:37Z SecondChildTAG: Thank you jelizon! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-02T22:22:52Z IndexTAG: 1046 TitleTAG: Lab 4 - wrong value for Vt Hi all! I calculated Ron - it's value is right(according to system's sign). Then I calculated Vt - it's value is wrong. Then I calculated constant K on base of calculated early Vt(wrong???) - and K's value is right(as system reported). How can it be? UserIdTAG: 192489 UserNameTAG: IgorKostyuk CreateTimeTAG: 2012-10-01T19:27:33Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Troubleshooting NumberOfReplyTAG: 4 FirstChildTAG: hey can u please help me with the connections of lab 4?where do u connect the current probe?between d and s terminals? FirstChildUserIdTAG: 156585 FirstChildUserNameTAG: AdiRanga FirstChildCreateTimeTAG: 2012-10-02T10:56:56Z SecondChildTAG: the probe will be between the output of VDS and D terminal SecondChildUserIdTAG: 143823 SecondChildUserNameTAG: NathanNadeson SecondChildCreateTimeTAG: 2012-10-02T12:21:05Z FirstChildTAG: Hi, I seem to have a problem with RON. However, I could get a value for VT. All you have to do is to get 2 iDS values for the same VGS value (the value on the x - axis) and plug into the equation provided FirstChildUserIdTAG: 143823 FirstChildUserNameTAG: NathanNadeson FirstChildCreateTimeTAG: 2012-10-02T12:20:17Z SecondChildTAG: From the Lab Description: At vDS≈3V, well into the saturation region, measure iDS twice, once when vGS=3V and once when vGS=2.5 So, at the same Vds=3V we should get Ids for Vgs=2.5 and 3.0 respectively. Not same Vgs value. Unfortunately, I cant get correct Vt. I do try to understand where is mistake, but have no positive result yet. SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-10-02T19:43:24Z SecondChildTAG: Ok, I have Green sign. Excel is very helpful. SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-10-02T20:05:46Z FirstChildTAG: Same problem. VT shows up as wrong number but calculated with it K is alright. FirstChildUserIdTAG: 376572 FirstChildUserNameTAG: sergy_kan FirstChildCreateTimeTAG: 2012-10-02T15:40:54Z SecondChildTAG: I've recalculated Vt but defined iDS1 and iDS2 more precise. The answer has been approved as right. So the decision is to find out from plot values of iDS1 and iDS2 more precise. SecondChildUserIdTAG: 192489 SecondChildUserNameTAG: IgorKostyuk SecondChildCreateTimeTAG: 2012-10-02T18:32:31Z FirstChildTAG: I have opposite story, My VT is true but when I plug in in the equation trying to calculate K, value is wrong. FirstChildUserIdTAG: 140867 FirstChildUserNameTAG: DarkWishMaster FirstChildCreateTimeTAG: 2012-10-06T13:56:40Z SecondChildTAG: same here SecondChildUserIdTAG: 134124 SecondChildUserNameTAG: anadelta SecondChildCreateTimeTAG: 2012-10-06T16:21:41Z IndexTAG: 1047 TitleTAG: S7E2 graphs online tool some user in the past gave an online tool which I don’t remember so you can also use this: http://www.wolframalpha.com/ and give "plot x/8.2 and 4-x^3 from 1.5 to 1.7" to plot the load line and the element characteristic simultaneously it would really help if the axis had numbers... UserIdTAG: 319453 UserNameTAG: leonidasGr CreateTimeTAG: 2012-10-01T14:34:02Z VoteTAG: 3 CoursewareTAG: Week 4 / Graphs Exercise CommentableIdTAG: 6002x_graphs_exercise NumberOfReplyTAG: 2 FirstChildTAG: Was the tool you were looking for Desmos? FirstChildUserIdTAG: 270160 FirstChildUserNameTAG: Sethhhh FirstChildCreateTimeTAG: 2012-10-05T03:06:46Z FirstChildTAG: yes sethhh FirstChildUserIdTAG: 319453 FirstChildUserNameTAG: leonidasGr FirstChildCreateTimeTAG: 2012-10-09T12:55:09Z IndexTAG: 1048 TitleTAG: WOW!!!!! Superb demonstration!!!Loved it!!! Truly an **ahaa** moment!!! UserIdTAG: 11538 UserNameTAG: trishul CreateTimeTAG: 2012-10-01T09:41:59Z VoteTAG: 3 CoursewareTAG: Week 4 / Incremental Analysis Demo CommentableIdTAG: 6002x_inc_analysis_demo NumberOfReplyTAG: 0 IndexTAG: 1049 TitleTAG: I PASSED MY FIRST LITTLE TEST AND I M STEEL IN MISSLE SCHOOLE YES UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-30T21:52:07Z VoteTAG: 3 CoursewareTAG: Week 1 / Various V-I characteristics CommentableIdTAG: 6002x_S1E1_Various_V-I_characteristics NumberOfReplyTAG: 1 FirstChildTAG: That is great, I am sure this course will help you in missile school. :-) FirstChildUserIdTAG: 219780 FirstChildUserNameTAG: ReconIII FirstChildCreateTimeTAG: 2012-09-30T22:56:16Z SecondChildTAG: Apparently the missile was on target! SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-09-30T23:33:09Z SecondChildTAG: I would use carbon fiber or aluminum for my "missle", not **steel**, as it is too heavy. You may have passed your missile test, but you should repeat your aerospace engineering course :D SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-10-01T02:01:03Z SecondChildTAG: Ha! SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-01T15:46:26Z SecondChildTAG: +1 SecondChildUserIdTAG: 740649 SecondChildUserNameTAG: hqjb SecondChildCreateTimeTAG: 2012-10-30T12:09:51Z IndexTAG: 1050 TitleTAG: Superposition? Where???? Don't see where we can use superposition... UserIdTAG: 342603 UserNameTAG: YakovO CreateTimeTAG: 2012-09-30T14:06:38Z VoteTAG: 3 CoursewareTAG: Week 4 / Dependent Sources Exercise 1 CommentableIdTAG: 6002x_dep_src_exsz_1 NumberOfReplyTAG: 5 FirstChildTAG: Amplifier design is based on superposition - you isolate the input signal into two components - the one which varies and the one which is constant (which is required for biasing). You design each part separately and then "add" the two together. Thevenin and Norton methods are derived from superposition. Superposition also makes circuit analysis easy. Suppose you have a circuit where you need to find the current through a particular resistor. In many cases, you can use superposition to calculate the current due to one source at a time and then add the result. At first this looks complicated because you have to redraw the circuit each time you shut off a source. However with practice, you'll not need to do any of that - you'll see you can just write down the answer in one line. When you use circuits to process signals (let's say you want to build a circuit which amplifies the bass guitar in any audio you pass through it), the theory behind the signals analysis is heavily based on superposition (and linearity). Hope this helps :-) FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-09-30T14:54:39Z SecondChildTAG: The question was about S8E2. Hint says: "Add an independent current source at the port and use superposition of independent sources. Do not suppress the dependent source!". But I don't see a place for applying superposition. SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-30T18:06:19Z SecondChildTAG: Oh ok. My bad. I thought you wanted to know where superposition is applied. Sorry about that :-) You don't *have* to use superposition there. Maybe it simplifies calculations if you do. I didn't use it either. SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-09-30T19:25:49Z SecondChildTAG: I calculated it with no problem initially (node method for Vth and test current source for Rth), but after that I realized that did't use superposition anywhere as hint mentioned. And I still don't see where it can be applied. If I don't see it - I don't understand something - that's not good. Could you clarify please how to apply superposition method anywhere in this problem? SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-30T19:52:04Z SecondChildTAG: Well, I think this is how you can use superposition. 1. You first connect a test source as they suggest in the hint. 2. Shut off the independent source that was already part of the circuit. Calculate the current/voltage you need. 3. Now shut off the test source. Calculate the voltage/current you need. Add the results from step 2 and 3. You should get the same answer. SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-09-30T20:46:07Z SecondChildTAG: That's exactly what I did except "Add the results from step 2 and 3" - add for what??? On step (2) you will get Rth, on step (3) you will get Vth. Adding will not work - what we want to get with it? Current (where?) and voltage (where?) when additional source is connected? These values are irrelevant to any purpose. These questions are driving me totally crazy :) SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-30T22:18:42Z SecondChildTAG: Have a look at my solution before. SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-10-01T13:11:29Z FirstChildTAG: To YakoVO: the superposition is in my solution introduced -see my message to Ashwith- because I introduced my VS. And if you introduce VS+deltaVS, then the deltaVS is superimposed on VS. And after differentiating the terms with VS in it, the terms with VS*something become zero, because VS is constant. I'm sure that if you introduce an external CS and a deltaCS superimposed on it, you'll get the same result, because converting CS+deltaCS to thevenin with R2 will also give VS+deltaVS. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-10-01T18:39:05Z SecondChildTAG: Your solution is too complicated but looks interesting to me. Though I don't see any point to apply differentiating, I would like to see your solution. Could you please post it. Formulas will be enough - I understand your circuit. SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-01T21:26:20Z FirstChildTAG: To Ashwith: I think you made a mistake somewhere, because you have the term (IO+1) in your result and I don't have the 1 in it, only IO. The rest of your last formula is the same as mine, and my answer is the same on all digits as the check button gives. I also added the external CS, but I changed the combination of my CS||R2 into a VS+deltaVS and from there I did the calculations, with some differentiation of a term with my VS in it, to find the Rth seen from VS in series with R2 and the rest of the circuit. In the end I simply said VS=0 to get rid of it. I don't see any differentiation in your solution and I'm sure that is why you get the 1 in it. You'll have to superimpose the external CS with a deltaCS. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-10-01T18:15:50Z SecondChildTAG: I see my mistake. I've been solving a lot of problems and have made the same mistake :-| Taking up 6.002x again really was a good idea after all. Thanks!! SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-10-01T19:36:55Z FirstChildTAG: Here is how I think you can use superposition. Here is the circuit: ![Problem Circuit][1] Here $I_O=0.004A, R_1=850.0\Omega, R_2=750.0\Omega$ and $\alpha=4.0\Omega$. **Step 1** The hint tells us to add an independent current source to the terminals. So lets do that next: ![Add the current source][2] **Step 2** Shut of the left independent source and find the voltage across $R_2$. Here is the resulting circuit: ![Left Source Shut Off][3] Lets calculate $V_1'$. To do this, we'll figure out what $i$ is and then move on from there. From the circuit it's clear that $\alpha i = V_1' - V_2'$ $\ \ \ \ \ = R_2(I_o - i) - R_1i$ With a little bit of algebra we get the following. $i = \frac{R_2I_O}{\alpha + R_2 + R_1}$ So, $V_1' = R_2(I_O - i) = \frac{R_2(R_1 + \alpha)I_O}{\alpha + R_2 + R_1}$ **Step 3** Shut off the right source (keep the left source on this time) and calculate the voltage across $R_2$. Here's the circuit: ![Right Source Off][4] We'll follow the same strategy. $\alpha i = V_1'' - V_2''$ $\ \ \ \ \ = R_2(1 - i) - R_1i$ We get the following expression for $i$: $i = \frac{R_2}{\alpha + R_2 + R_1}$ Therefore, $V2'' = R_2(1 - i)\frac{R_2(R_1 + \alpha)}{\alpha + R_2 + R_1}$ Finally using superposition, $V_1 = V_1' + V_1''$ $\ \ \ \ = \frac{(I_O + 1)(\alpha + R_1)R_2}{\alpha + R_2 + R_1}$ Substitute the values of the components according to what's given in the problem. This is the value of $V_1$ I got: $400.9114713V$. Therefore, $R_{TH} = \frac{V_1}{1} = 400.9114713\Omega$ which is supposed to be the correct answer. However the Show answer button is giving a different value and I know why. Bug report time :-) I think this is what they mean by superposition but I hope I'm wrong because finding $V_1$ is straightforward nodal analysis even without going through all this. Calculating $V_{TH}$ is simple. It doesn't even need the $1A$ source so I won't detail that here. [1]: https://edxuploads.s3.amazonaws.com/13490907061343688.png [2]: https://edxuploads.s3.amazonaws.com/13490908606818485.png [3]: https://edxuploads.s3.amazonaws.com/13490910076818439.png [4]: https://edxuploads.s3.amazonaws.com/13490916153948095.png FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-10-01T12:46:58Z SecondChildTAG: >However the Show answer button is giving a different value and I know why. Bug report time :-) You don't know why. I know - your answer is not correct - error somwhere in formulas - I'll take a look later and say. And I still don't see any application of superposition - where is adding??? :) SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-01T15:29:03Z SecondChildTAG: Oh yes, I see. As I said before superposition is totally irrelevant, so >Finally using superposition, V1=V1′+V1′′ is totally wrong. Please notice, that on step (2) you already have V1' with no unknowns: >V1′=R2(Io−i)=R2(R1+α)Io/(α+R2+R1) V1' - is the Vth For Rth - you should be able to find it directly from step (3) by applying i - which can be found from (2). But didn't check your formulas here, though. SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-01T16:38:51Z SecondChildTAG: oh, sorry, made a mistake - for Rth i can be find the same way you find it in (2) - NOT from (2) of course SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-01T16:46:21Z SecondChildTAG: About step (3) - I meant before applying superposition - V1"/1 is Rth SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-01T16:52:46Z FirstChildTAG: Thank you ashwith, you saved me big time. FirstChildUserIdTAG: 418182 FirstChildUserNameTAG: euldji2005 FirstChildCreateTimeTAG: 2012-10-05T03:27:13Z IndexTAG: 1051 TitleTAG: edX certificates Are grades (A,B,C) written on the certificates or only that the student has passed the course? UserIdTAG: 263602 UserNameTAG: elekov CreateTimeTAG: 2012-09-30T10:42:40Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: ![Sample Certificate of "edX"][1] [1]: https://edxuploads.s3.amazonaws.com/13490038651386343.jpg FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-09-30T11:18:29Z SecondChildTAG: They gave 2 certificates: 1 without the grade and 1 with the grade, Because some people don't like to show off with their grade. And to be honest, sometimes showing grades on a certificate can be a disadvantage ... Once, about 35 yrs ago, I applied for a job and I had, amongst others, 3 diploma's from a certain institute,with the grades printed on it, in total 8 times an A, nothing less, and the guy who interviewed me for the job, really had a problem with it, and he said something like: does that institute give grade A to every student? Wasn't a smart remark of him! I refused the job and they went broke .... Ha ha!! SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-09-30T13:23:14Z FirstChildTAG: yea,they write the grade FirstChildUserIdTAG: 395257 FirstChildUserNameTAG: blackguitar FirstChildCreateTimeTAG: 2012-09-30T10:49:13Z FirstChildTAG: They gave 2 certificates: 1 without the grade and 1 with the grade, Because some people don't like to show off with their grade. And to be honest, sometimes showing grades on a certificate can be a disadvantage ... Once, about 35 yrs ago, I applied for a job and I had, amongst others, 3 diploma's from a certain institute,with the grades printed on it, in total 8 times an A, nothing less, and the guy who interviewed me for the job, really had a problem with it, and he said something like: does that institute give grade A to every student? Wasn't a smart remark of him! I refused the job and they went broke .... Ha ha!! FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-09-30T13:30:41Z IndexTAG: 1052 TitleTAG: Tool Nice work with the schematic tool. .. enjoyed this lab !! quite impressed with its working and curious about its technology & programming :) UserIdTAG: 483548 UserNameTAG: SAM_1993 CreateTimeTAG: 2012-09-30T10:13:36Z VoteTAG: 3 CoursewareTAG: Week 2 / Superposition experiment CommentableIdTAG: 6002x_Superposition_experiment NumberOfReplyTAG: 0 IndexTAG: 1053 TitleTAG: H5P2 SOURCE FOLLOWER LARGE SIGNAL See page 349 of the book! UserIdTAG: 299251 UserNameTAG: TheRedBlackOne CreateTimeTAG: 2012-09-29T19:49:07Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1054 TitleTAG: Links to Textbook Sections are Incorrect I'm not sure if this has been reported before, but I found that the links to the textbook in the (as given in each of the lecture sequences) are incorrect. For example, S6V1 is supposed to link to page 193. The link to this page doesn't account for the initial pages such as the table of contents and the preface. You need to add 24 to the page number in the URL to get to the correct page. UserIdTAG: 14386 UserNameTAG: ashwith CreateTimeTAG: 2012-09-29T16:06:19Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 1 FirstChildTAG: correct, ive found this out myself. FirstChildUserIdTAG: 248807 FirstChildUserNameTAG: 1kingsman FirstChildCreateTimeTAG: 2012-10-05T04:25:32Z IndexTAG: 1055 TitleTAG: question 2 i sloved it this way...first i calculated incremental resistance and then the required ratio= 1/(1+R(=2)/incremental resistance )=0.237100357081, bt actual answer differs a bit...why it is so??? m i doing something wrong ...if yes then what?? UserIdTAG: 432151 UserNameTAG: rgdixit CreateTimeTAG: 2012-09-29T08:35:36Z VoteTAG: 3 CoursewareTAG: Week 4 / Linearization Exercise 1 CommentableIdTAG: 6002x_linearization_ex_1 NumberOfReplyTAG: 1 FirstChildTAG: I solved this way you only need to derivate ia = f(va) ia' = K*va' R = 1/K FirstChildUserIdTAG: 110214 FirstChildUserNameTAG: jeffsjunior FirstChildCreateTimeTAG: 2012-09-30T02:45:59Z IndexTAG: 1056 TitleTAG: simulator model for this tutorial http://www.falstad.com/circuit/#%24+1+5.0E-6+15.472767971186109+66+5.0+50%0Ar+272+144+400+144+0+600.0%0Ad+400+144+400+224+1+0.8%0Aw+400+144+496+144+0%0Ar+272+304+400+304+0+600.0%0Ai+400+224+272+224+0+7.800000000000001E-4%0Ad+400+304+400+224+1+0.8%0Aw+400+304+496+304+0%0Ar+496+144+496+304+0+100000.0%0Ar+272+144+272+224+0+10000.0%0Ar+272+224+272+304+0+10000.0%0Av+192+144+192+304+0+1+400.0+0.2+0.0+0.0+0.5%0Aw+272+144+192+144+0%0Aw+272+304+192+304+0%0Ao+10+16+0+290+0.29230032746618057+9.134385233318143E-5+0+-1%0Ao+1+32+0+98+1.1692013098647223+7.307508186654515E-4+1+-1%0Ao+5+64+0+98+1.1692013098647223+7.307508186654515E-4+2+-1%0Ao+7+16+0+291+0.03653754093327257+3.5681192317648997E-7+3+-1%0A UserIdTAG: 342603 UserNameTAG: YakovO CreateTimeTAG: 2012-09-29T01:16:12Z VoteTAG: 3 CoursewareTAG: Week 4 / Attenuator Tutorial CommentableIdTAG: 6002x_attenuator_t NumberOfReplyTAG: 1 FirstChildTAG: I try importing this url but does not work FirstChildUserIdTAG: 410943 FirstChildUserNameTAG: JPalaciosRoman FirstChildCreateTimeTAG: 2012-09-29T12:56:48Z SecondChildTAG: it's a Java applet - your browser should support it and you should allow java applets to run SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-29T17:38:48Z IndexTAG: 1057 TitleTAG: Textbook Link to Supplementary Sections and Examples Are you wondering what the www with a number behind it means? E.g.: www 12.6 It means that it is a Supplementary Section and www example 2.20 means it is a Supplementary Example. Look here for all Supplementary Sections and Examples - not the book itself. You can download all in one pdf or all chapters seperatly. http://www.elsevierdirect.com/v2/companion.jsp?ISBN=9781558607354 I also added the link at the Wiki: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/book-link-supplementary-sections-and-examples/ UserIdTAG: 83276 UserNameTAG: salsero CreateTimeTAG: 2012-09-28T17:20:26Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: @salsero Thanks a lot.I was knowing the materials were available with the publisher but was little lazy to search them myself. FirstChildUserIdTAG: 99628 FirstChildUserNameTAG: Pranjal16 FirstChildCreateTimeTAG: 2012-09-28T17:54:37Z SecondChildTAG: Thanks salsero! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-28T20:12:24Z SecondChildTAG: Thank you. I was needing this information. SecondChildUserIdTAG: 361823 SecondChildUserNameTAG: EliasOak SecondChildCreateTimeTAG: 2012-09-29T00:11:57Z IndexTAG: 1058 TitleTAG: Basic doubt to Staff First of all I thank to MIT for providing such a good learning program, i have a doubt that if suppose we want to learn the basics of Electronics and Communication engineering then how to do that. When suppose someone asks about how a transistor works we are not able to tell so,but we are getting good marks instead, so please tell me from where can I make my basics strong. UserIdTAG: 182470 UserNameTAG: nitesh2703 CreateTimeTAG: 2012-09-28T08:57:24Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: 1. Best way for you now is, Learn 6.002x, try to understand everything around 6.002x and get 100%. When you finish this, Edx might be starting Electronic Devices course. In that, you can learn from best professors directly. 2. Good books are the second option if you can understand books easily. 3. Ask questions. In particular, Do ask your friend or your colleague or your lecturer. Note, discuss with them. or Ask yourself!! find the answers through Google-which gives you access to all materials. Caution: TIME is very important, It will take considerable amount of time when you learn new concepts for the first time. That's the reason why khanacademy/udacity/coursera/edX is shaping up to simplify the way we learn. FirstChildUserIdTAG: 823 FirstChildUserNameTAG: naveenatmit FirstChildCreateTimeTAG: 2012-09-28T09:13:56Z IndexTAG: 1059 TitleTAG: Mesh analysis The dual to this method and powerful as this is the Mesh Method. In which we will find a impedance matrix [Z], a currents vector [i], and the voltage vector [v]--> [Z]*[i]=[v] UserIdTAG: 155605 UserNameTAG: JaimeLopezCerezo CreateTimeTAG: 2012-09-28T06:10:29Z VoteTAG: 3 CoursewareTAG: Week 1 / Node analysis CommentableIdTAG: 6002x_node_analysis NumberOfReplyTAG: 1 FirstChildTAG: Mesh analysis is definitely a lot easier because we don't have fractions. I use it most of the time. However, mesh analysis fails when the circuit is non-planar i.e. when you have wires overlapping each other no matter how you re-arrange the circuit. Nodal works even on non-planar circuits. FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-09-28T16:09:04Z IndexTAG: 1060 TitleTAG: Equivalent MOSFET ![][1] [1]: https://edxuploads.s3.amazonaws.com/13488115747245433.bmp UserIdTAG: 268104 UserNameTAG: SaulCardenasG CreateTimeTAG: 2012-09-28T05:54:19Z VoteTAG: 3 CoursewareTAG: Week 3 / Switch Resister Model Exercise CommentableIdTAG: 6002x_switch_resistor_model_exercise NumberOfReplyTAG: 1 FirstChildTAG: how is R(off) 10^7? FirstChildUserIdTAG: 133084 FirstChildUserNameTAG: Warrensiggs FirstChildCreateTimeTAG: 2012-12-09T13:51:13Z IndexTAG: 1061 TitleTAG: Typo list for S6V11-S6V12 (Sorry for such a long and, in places, nitpicky post, but I really feel that the transcripts should be improved; in quite a few places they include mistakes that could confuse people relying on them. So from now on I'll start posting typo lists, in the hope that someone can incorporate the changes.) S6V11: At 2:14 in the transcript, "light emitting dial" should be "light emitting diode". At 5:25, "between the leading V and I" should be "relating V and I". At 7:40, "And I asked you" should be "And I ask you". S6V12: At 1:29, "into sound and an actual linear amplifier" should be "into sound through a linear amplifier". At 1:37, "we learn how to" should be "we'll learn how to". At 2:28, "If the node voltage here is VI, it is equal to the node" should be "The node voltage here is VI, which is equal to the node" At 2:31, "They're two of the same." should be "The two are the same." At 2:50, "pictorially, and look at that" should be "pictorially, look at that". At 3:00, "Here is IV relationship" should be "Here is the IV relationship". At 3:08, "So this my" should be "So this is my". At 3:19, "And VD equals to VI" should be "And VD equals VI". (Also note that the transcripts in general are very cavalier about capitalisation: these should be vD and vI. Ideally, the transcript system should also be upgraded to allow for subscripts.) At 5:03, "an input right here of iD" should be "an input right here of vD". (Or should that be VD? See previous note.) At 6:33, "And they can line up." should be "And take a line up." At 7:38, "that a lot of" should be "there's a lot of" At 8:55, "I definitely want to" should be "I really want to". UserIdTAG: 280731 UserNameTAG: QuantumCaffeine CreateTimeTAG: 2012-09-27T16:22:18Z VoteTAG: 3 CoursewareTAG: Week 3 / Incremental Method Motivation CommentableIdTAG: 6002x_incremental_method_motivation NumberOfReplyTAG: 1 FirstChildTAG: Thank you very much! I have createad a [Post in the Wiki][1] : QuantumCaffeine's Typo lists I hope you don't mind :). See you! Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-27T21:44:09Z IndexTAG: 1062 TitleTAG: H5P3 SOURCE FOLLOWER SMALL SIGNAL BUG ? Hi iClassmates & iTeachers, Regarding the algebraic expression for vout (incremental output): Isnt it right if i compute the total output voltage with sum of the input bias voltage(VIN) and the incremental input voltage(vin) And than substract it from the total output voltage for the input bias voltage only. I should get the incremental output voltage. It looks right for the second question where i should compute it for a certain incremental input, but the algebraic expression is marked as wrong. Any ideas? Regards UserIdTAG: 149844 UserNameTAG: amitaratnayake CreateTimeTAG: 2012-09-26T19:50:23Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: You are being asked to differentiate the expression you got for vOUT In H5P2 and evaluate it at vIN = VIN. You get a rather simple expression! FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-09-26T21:45:31Z FirstChildTAG: Write the differentiate expression multiplied by vin, with small letters for vin. I spent a couple of hours until i observed that was vin not vIN. Good luck ! FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-09-27T15:03:32Z SecondChildTAG: can u help me with this problem?? My H5P2 is correct and I have differentiated correctly..But still coudn't get an answer..However my answer for lower parts are correct! Please help! SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-06T03:29:20Z IndexTAG: 1063 TitleTAG: Operating point I understand that for the third question we must find the slope of the i-v curve at the operating point of N. But how do you find the operating point ? It is not the answer to the first question ? UserIdTAG: 37280 UserNameTAG: zour CreateTimeTAG: 2012-09-26T17:22:45Z VoteTAG: 3 CoursewareTAG: Week 4 / Linearization Exercise 1 CommentableIdTAG: 6002x_linearization_ex_1 NumberOfReplyTAG: 2 FirstChildTAG: Hi. Yes indeed, but remember R=V/I therefore you should power your answer to -1. Regards FirstChildUserIdTAG: 378522 FirstChildUserNameTAG: Alejo_Velasquez FirstChildCreateTimeTAG: 2012-09-27T03:26:51Z FirstChildTAG: The operating point is indeed the same as the answer of the first question, but according to the textbook, page 221, (4.75), the incremental resistante is 1/slope, where va=VI FirstChildUserIdTAG: 318577 FirstChildUserNameTAG: takeuchi FirstChildCreateTimeTAG: 2012-09-27T17:14:05Z SecondChildTAG: va=VI - 2iA SecondChildUserIdTAG: 373752 SecondChildUserNameTAG: Croak SecondChildCreateTimeTAG: 2012-09-27T21:41:23Z IndexTAG: 1064 TitleTAG: H4P1 VACUUM DIODE hello everyone.. I am facing a problem in this question in homework of week 4 regarding vacuum diode. my first and third part of the question have got a check but unfortunately m not able to get through the 2nd and 4th parts of the question.Can someone explain me the approach to find out the incremental resistance in both cases. Although i have gone through the video again,still m not able to find the mistake. i calculated derivative of vd w.r.t id by first finding vd in terms of id and then substituted value of id from my 1st and 3rd parts respectively... Please help.Thanks in advance!! UserIdTAG: 368712 UserNameTAG: jmen CreateTimeTAG: 2012-09-26T05:25:20Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hi. From Wiki rd=1/(df(vd)/dvd)|vd=VD So, to solve 2nd question You had to calculate derrivative from diode id function in point of VD. Notice that P is in **milli** ampers/volts^3/2 FirstChildUserIdTAG: 195218 FirstChildUserNameTAG: Al_Incognito FirstChildCreateTimeTAG: 2012-09-26T06:18:07Z SecondChildTAG: hey! i got it... Thank you very much... one more thing..if instead i calculate derivative of vd w.r.t id and substitue values of id from part 1 and 3 respectively, i think we'll get incremental resistance .am i right? i mean is df(vd)/did = (df(id)/dvd)^-1 ?? SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-09-26T06:33:48Z SecondChildTAG: I think Yes, it seem so. SecondChildUserIdTAG: 195218 SecondChildUserNameTAG: Al_Incognito SecondChildCreateTimeTAG: 2012-09-26T07:45:11Z SecondChildTAG: Graffically r is slope of function in point (dV/DI) or seems to be tan of corner between tangent and v-axis. If we transform graph to v(i), this corner must be ctg (di/di). SecondChildUserIdTAG: 195218 SecondChildUserNameTAG: Al_Incognito SecondChildCreateTimeTAG: 2012-09-26T07:51:53Z SecondChildTAG: Can't say anything for certain, but seems so. SecondChildUserIdTAG: 195218 SecondChildUserNameTAG: Al_Incognito SecondChildCreateTimeTAG: 2012-09-26T07:52:24Z SecondChildTAG: hmm. i'll go online and find something related to this.Anyways thanks man! Where are you from? India? SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-09-26T16:13:25Z SecondChildTAG: Ukraine SecondChildUserIdTAG: 195218 SecondChildUserNameTAG: Al_Incognito SecondChildCreateTimeTAG: 2012-09-26T18:11:33Z SecondChildTAG: i cant understand such above eq plzzz simplify it SecondChildUserIdTAG: 164689 SecondChildUserNameTAG: muhammadfaizan SecondChildCreateTimeTAG: 2012-09-28T12:39:19Z SecondChildTAG: Good morning can someone help Figuring NOT IN ID H4P1. THANKS SecondChildUserIdTAG: 320715 SecondChildUserNameTAG: maraivette SecondChildCreateTimeTAG: 2012-10-01T15:06:03Z SecondChildTAG: Hi Guys, i have derived the equation and came up with the formula for 2nd and 4rd question as dVpk/dIp=2/3(Ip/P)^-1/3 but on substituting the values that i have found to be correct for question 1 and 3, the answer is wrong? kindly help me out with this? Can the answer checkbox not accept forms like e^-05? SecondChildUserIdTAG: 18877 SecondChildUserNameTAG: johnkhiangte SecondChildCreateTimeTAG: 2012-10-05T07:21:08Z SecondChildTAG: thank u SecondChildUserIdTAG: 231318 SecondChildUserNameTAG: moutasem SecondChildCreateTimeTAG: 2012-10-05T20:47:43Z FirstChildTAG: Good morning can someone help Figuring NOT IN ID H4P1. THANKS FirstChildUserIdTAG: 320715 FirstChildUserNameTAG: maraivette FirstChildCreateTimeTAG: 2012-10-01T18:17:26Z IndexTAG: 1065 TitleTAG: H2P1 Im stuck with V_min and V_max help please I have v_in=80v and v_out=28v,I picked up R1=82k and R2=47k but still have no answer help please UserIdTAG: 331441 UserNameTAG: abdessamade CreateTimeTAG: 2012-09-25T17:48:54Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hi, You know that Rth must be between 10kOhm and 30 kOhm, so let's suppose it's 20kOhm, the half of the margin. And asumme R2 something like, 56kOhm. Rth = 56k * R2 /(56k + R2) Resolve for R2, and it's done. There's another thing, and that is you're using resistances with 10% tolerance, so that means you won't have exactly the values you need, they can be, for the 56k resistor, from 50.4 kOhm to 61.6 kOhm. From that practical fact comes Vmax and Vmin. You've got Vmax when the series resistor of the voltage divider is in its lower value, and the parallel resistor is in its higher value. And the inverse for Vmin. Hope it helps FirstChildUserIdTAG: 220837 FirstChildUserNameTAG: Calsomus FirstChildCreateTimeTAG: 2012-09-25T20:49:39Z SecondChildTAG: can you be more clear please SecondChildUserIdTAG: 344037 SecondChildUserNameTAG: HAMISS SecondChildCreateTimeTAG: 2012-09-25T21:10:28Z FirstChildTAG: I GOT THE SAME PROBLEM MAN BUT I MANAGED TO WORK OUT Vmax some how FirstChildUserIdTAG: 344037 FirstChildUserNameTAG: HAMISS FirstChildCreateTimeTAG: 2012-09-25T18:03:31Z IndexTAG: 1066 TitleTAG: A resource for help I have posted a question about part two of this exercise at the following link. http://forum.allaboutcircuits.com/showthread.php?t=74849 I hope this helps! UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-09-24T17:49:33Z VoteTAG: 3 CoursewareTAG: Week 3 / Analyzing Nonlinear Circuits Exercise CommentableIdTAG: 6002x_analyzing_nonlinear_circuits_exercise NumberOfReplyTAG: 1 FirstChildTAG: Why not posting your questions here ? ; ) FirstChildUserIdTAG: 396446 FirstChildUserNameTAG: RousseauxS FirstChildCreateTimeTAG: 2012-09-24T19:02:36Z SecondChildTAG: Good idea! How did you Graph the solution to the question? SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-09-24T19:44:29Z SecondChildTAG: [enter link description here][1] [1]: http://forum.allaboutcircuits.com/attachment.php?attachmentid=46593&d=1348515292 SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-09-24T19:45:13Z IndexTAG: 1067 TitleTAG: H2P1 Solution oddity. Error in question? Am I the only one that worked this question assuming Vout=20V because the question indicates "we need to provide an open-circuit output voltage of Vout≈20.0V. The answer provided however assumed Vout=7.5V UserIdTAG: 30705 UserNameTAG: Step CreateTimeTAG: 2012-09-24T16:28:11Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: "For this solution, we assume that V in =30V,V out =7.5V...." However in the question Vin = 20V and Vout = 5V. Why the values given in the question were different than those used in solution? As I was using the same method as mentioned in the solution but obviously ending up with different R1 and R2. Was there really error in the question? FirstChildUserIdTAG: 329015 FirstChildUserNameTAG: farah_sarwar FirstChildCreateTimeTAG: 2012-09-24T16:42:15Z FirstChildTAG: I am just realizing that both Vin and Vout used in the solution are assumed values. Maybe just to explain the concept and not really to provide the actual required answers. In my question it specified Vin=50V and Vout=20V. Maybe we all had different voltages in the question hence the reason the voltages in the solution may not match those in the question. Sorry if jumped to conclusion that question had an error FirstChildUserIdTAG: 30705 FirstChildUserNameTAG: Step FirstChildCreateTimeTAG: 2012-09-24T18:05:04Z FirstChildTAG: Because the students get different values in their HW en Lab, the values in the solutions given by the staff can differ from the values in your personal solution. However, the way to solve the problems remains the same. The reason is simple: if the values given in the HW/Lab for each student are the same, then some people copy the values from other students. This system was also used for the midterm and final. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-09-24T18:58:06Z IndexTAG: 1068 TitleTAG: Scrolling Textbook Viewer Jeff Kent created a scrolling version of the textbook viewer during the Spring session of this course. I've found it a lot easier to use while reading through the book. This book is available for online viewing through this link: https://6002x.mitx.mit.edu/static/contrib/xbook.html I've added a link at the wiki as well. Hit 'f' on the keyboard to widen the pages. UserIdTAG: 14386 UserNameTAG: ashwith CreateTimeTAG: 2012-09-24T06:48:06Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Thank you ashwith ;) FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-25T02:39:47Z IndexTAG: 1069 TitleTAG: Linear Circuit We call every circuit whose v-i Graph is straight line passing through origin , but what is the physical significance of it ?Please help me out.. UserIdTAG: 284172 UserNameTAG: mayankg175 CreateTimeTAG: 2012-09-24T05:45:18Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: f(ax+by)=af(x)+bf(y) - linear all others - not FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-09-24T07:59:49Z FirstChildTAG: Linear circuits do **not** have to have the v-i graphical characteristics of a straight line passing through the origin (x=0, y=0). Think of a battery. The v-i relationship plots as a vertical straight line at voltage V, and as such it does not cross the origin. Linear circuits are defined as being composed of only linear elements. Linear elements are defined to be independent voltage sources, independent current sources and resistors for DC and AC analysis, and also including inductors and capacitors for AC analysis. Dependent sources may be linear *or not* based on their defining equation. The voltage source $v_x=(i_2-i_1)\cdot R_x$ is linear, but the voltage source $v_y=v_2 \cdot v_1$ is **not**. Review algebra to see examples of linear and nonlinear equations. For a resistor to be linear, it must have a linear algebraic equation: For example, resistors with the v-i characteristic: $v=i \cdot R$ or $v=(i-i_L)\cdot R$, where $i_L$ is a variable leakage current, are **linear**, but resistors with the v-i characteristics: $v=i^2 \cdot R$ or $v = i /(1-i^3) \cdot R$ are **not** linear. If I missed something, let me know, it's 3:00AM here and I am going to bed late... FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-09-24T06:58:33Z SecondChildTAG: Linear circuits don't necessarily have those characteristics, yes. According to what we've been taught, linear circuits are made up of linear elements *and independent sources*. Linear **elements** on the other hand, *do* have to have v-i characteristics that pass through the origin. That's why Prof A makes the distinction that dependent sources are not strictly speaking linear elements. SecondChildUserIdTAG: 248902 SecondChildUserNameTAG: Chanute SecondChildCreateTimeTAG: 2012-09-24T11:05:39Z SecondChildTAG: Agh, I mis-typed. Prof A makes the distinction that *independent* sources are not strictly speaking linear elements. See the tutorial videos from week 2 for confirmation. Specifically https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_2/Week_2_Tutorials/8 should contain the written explanation! SecondChildUserIdTAG: 248902 SecondChildUserNameTAG: Chanute SecondChildCreateTimeTAG: 2012-09-25T00:23:32Z IndexTAG: 1070 TitleTAG: Do something about this youtube banned issue Dear professor, please do something as my 3rd week homework,labs and tutorials are still pending. Thanks and god bless you all for this amazing course i love it the way you make us learn things. UserIdTAG: 294370 UserNameTAG: Naif1125 CreateTimeTAG: 2012-09-24T03:16:02Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: i agree mine r pending too as i can't watch the video lectures...pls post the videos again using some other video player. FirstChildUserIdTAG: 48053 FirstChildUserNameTAG: nadra123 FirstChildCreateTimeTAG: 2012-09-24T06:18:06Z SecondChildTAG: where u from nadra123? SecondChildUserIdTAG: 294370 SecondChildUserNameTAG: Naif1125 SecondChildCreateTimeTAG: 2012-10-07T06:34:32Z FirstChildTAG: You can see in the Wiki [Lecture Videos Download][1]:). Also see more info [here][2]: **Due to the YouTube outage in certain parts of the world, we have provided direct links to download the lecture videos, and extended the Week 2 homework and lab deadlines to the coming Sunday (September 30). Please note that Week 3 deadlines have not changed.** [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/ [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/info FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-26T11:45:03Z FirstChildTAG: thanks to all dear professors FirstChildUserIdTAG: 294370 FirstChildUserNameTAG: Naif1125 FirstChildCreateTimeTAG: 2012-09-29T08:35:56Z IndexTAG: 1071 TitleTAG: Lab 2 - hmm Hello, I have been trying to figure out how to set up this lab but I am having a hard time figuring out how to handle the two types of waves. I don't really know what to do with square and sine waves... Also, the figure above uses 2 resistors, everyone in the discussion forum have posted that you need three... Can someone please give me a clue of what to do with those types of waves? Thanks! UserIdTAG: 230787 UserNameTAG: Gabriel007 CreateTimeTAG: 2012-09-24T02:27:48Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 7 FirstChildTAG: you should not change anything with the "waves" the sources are set. basically the circuit adds the two waveforms, how much of each is dependent on the resistor combination you choose. what you need to find is a resistor combination that gives Vout = 1/2 of V1 + 1/6 of V2. FirstChildUserIdTAG: 231206 FirstChildUserNameTAG: Bradley_B FirstChildCreateTimeTAG: 2012-09-24T03:09:26Z SecondChildTAG: refer the text book under network theorems..you will find that useful to better utilise voltage divider principles to this SecondChildUserIdTAG: 443809 SecondChildUserNameTAG: Duvindu SecondChildCreateTimeTAG: 2012-09-24T04:11:14Z FirstChildTAG: Thanks Chanute and Bradley_B. I have just been staring at it for like 1.5-2hrs and I have no idea how to even start it. So, when I am making the calculations, I should use 1 volt for both sources when at maximum? I don't know what V to use. The two types of sources are throwing me off. FirstChildUserIdTAG: 230787 FirstChildUserNameTAG: Gabriel007 FirstChildCreateTimeTAG: 2012-09-24T03:43:55Z SecondChildTAG: Don't change the voltage sources from default! Leave them as is. All you have to do is build a network of resistors between them and your output to give you $ \frac {1}{2} \cdot \left ( V_1 + \frac {1}{3} \cdot V_2 \right ) $ Maybe writing it this way will be enough of a hit to help you: $ \frac {1}{2} \cdot V_1 + \frac {1}{2} \cdot \left ( \frac {1}{3} \cdot V_2 \right )$ SecondChildUserIdTAG: 248902 SecondChildUserNameTAG: Chanute SecondChildCreateTimeTAG: 2012-09-24T03:54:13Z FirstChildTAG: The figure is showing you a single voltage divider. The output they want can't be achieved using a single voltage divider; you need two of them! FirstChildUserIdTAG: 248902 FirstChildUserNameTAG: Chanute FirstChildCreateTimeTAG: 2012-09-24T02:46:04Z FirstChildTAG: refer the text book under network theorems..you will find that useful to better utilise voltage divider principles to this FirstChildUserIdTAG: 443809 FirstChildUserNameTAG: Duvindu FirstChildCreateTimeTAG: 2012-09-24T04:11:38Z FirstChildTAG: ;( I don't have a clue.hahah FirstChildUserIdTAG: 230787 FirstChildUserNameTAG: Gabriel007 FirstChildCreateTimeTAG: 2012-09-24T04:07:00Z FirstChildTAG: Ok! I got it. But I don't fully understand it. Can someone please post the answer verbatim tomorrow after the due date? Should not violate honor code since we can't submit it any more. I really would like to learn this. I used three R because of hints and ratios because of hints. Thanks! :D FirstChildUserIdTAG: 230787 FirstChildUserNameTAG: Gabriel007 FirstChildCreateTimeTAG: 2012-09-24T05:24:08Z SecondChildTAG: I'll be happy to explain it in detail once the due date is passed =) SecondChildUserIdTAG: 248902 SecondChildUserNameTAG: Chanute SecondChildCreateTimeTAG: 2012-09-24T06:15:48Z SecondChildTAG: Thank you Chanute! SecondChildUserIdTAG: 230787 SecondChildUserNameTAG: Gabriel007 SecondChildCreateTimeTAG: 2012-09-27T22:53:41Z FirstChildTAG: I've got the correct graph, but still can't get V1 lower. It is stunned at max 1v, and any attempts to rearrange the circuit down to 600mV is useless( FirstChildUserIdTAG: 205077 FirstChildUserNameTAG: Altleonidas FirstChildCreateTimeTAG: 2012-09-24T08:25:03Z IndexTAG: 1072 TitleTAG: Lab2 and Analysis I am impressed by how easily the solution "fell out" once I settled on a good way of representing the parallel resistors in an equation. All along I had the right understanding, and my math was ok but it became too cluttered with symbols for me to use it in the best way. How did I find this "just right" formulation? I just kept trying. The formulation I found so helpful was to use "R1 || R2" instead of (R1*R2)/(R1+R2). The latter was just too busy-looking for me to parse things out in the most convenient way. UserIdTAG: 92346 UserNameTAG: MobiusTruth CreateTimeTAG: 2012-09-24T01:44:55Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Notation can help or hinder. Glad you found a notation that helped! FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-09-24T03:23:04Z IndexTAG: 1073 TitleTAG: Lab 2 - explanation after deadline? I really don't get Lab 2. Getting a 1/6 of V2 is no problem, and so is 1/2 of V1, but adding these two together seems impossible. As I don't see how I can solve this problem in the next 1.5hrs, I'll leave it blank and HOPE that someone can explain this lab when after the deadline. UserIdTAG: 114881 UserNameTAG: xvink CreateTimeTAG: 2012-09-23T20:22:00Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: You can use a third one! And connect it to ground!! FirstChildUserIdTAG: 244841 FirstChildUserNameTAG: eyubero FirstChildCreateTimeTAG: 2012-09-23T20:28:26Z FirstChildTAG: If you have 1/2 of V1 and 1/6th of V2, that part is done. Now you will notice that your overall voltage is not suitable for the voltage requested. In my Lab, it asked for 0.5 of (V1+V2). Experiment in the circuit sandbox, it may be easier. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-23T20:29:07Z FirstChildTAG: You should look into the equations to see the truth. Using the simple mixer you will get output like $\frac{R_2}{R_1+R_2}V_1 + \frac{R_1}{R_1+R_2}V_2$. You need $\begin{cases}{\frac{R_2}{R_1+R_2}=\frac{1}{2} \\ \frac{R_1}{R_1+R_2}=\frac{1}{6}}\end{cases}$ but it's impossible. First one decrypts into $R_1=R_2$ so the second one can't be satisfied simultaneously. So pure math says: simple mixer is not what you're asked for. What you're asked for is your brain work. Connect a third resistor directly to the output node (another resistors' wire goes into ground). This way you get the next mixing $\frac{R_2R_3}{R_1R_2+R_1R_3+R_2R_3}V_1 + \frac{R_1R_3}{R_1R_2+R_1R_3+R_2R_3}V_2$. Look what has changed. If you divide V1's coefficient by V2's one you get $\frac{R_2}{R_1}$ and it's the same as before. BUT. Now you've got a correcting multiplyer $\frac{R_3}{R_1R_2+R_1R_3+R_2R_3}$ that you can use to put your coefficients into the desired values. So they say $\frac{1}{2}V_1 + \frac{1}{6}V_2$? No problem AT ALL. $\frac{1}{2} : \frac{1}{6}=3 $ so you take ANY $R_1$ and $R_2=3R_1$. Next you insert that picked values into $\frac{R_2R_3}{R_1R_2+R_1R_3+R_2R_3}$ and find such $R_3$ that $\frac{R_2R_3}{R_1R_2+R_1R_3+R_2R_3} = \frac{1}{2}$. It's just a linear equation with one variable, as $R_1$ and $R_2$ are picked values. As only you find $R_3$ from the equation, another condition $\frac{R_1R_3}{R_1R_2+R_1R_3+R_2R_3} = \frac{1}{6}$ will be satisfied automaticly cause of the way you picked $R_1$ and $R_2$. So that's it, yes FirstChildUserIdTAG: 190618 FirstChildUserNameTAG: Kirbabaev FirstChildCreateTimeTAG: 2012-09-23T22:33:06Z SecondChildTAG: Thanks a lot. I have been experimenting with more than two resistors (as it is impossible to do with two), but I just couldn't get it right. SecondChildUserIdTAG: 114881 SecondChildUserNameTAG: xvink SecondChildCreateTimeTAG: 2012-09-24T08:01:13Z SecondChildTAG: Thank you SecondChildUserIdTAG: 283808 SecondChildUserNameTAG: Hanboushy SecondChildCreateTimeTAG: 2012-09-28T18:58:45Z FirstChildTAG: S3E2 FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-09-25T19:04:56Z IndexTAG: 1074 TitleTAG: [TIP] H2P2 - DO NOT ROUND THE VALUE Sorry for the poor english :) Just a tip for the people who are having trouble to solve the power of H2P2. DO NOT ROUND THE VALUE. I had to use 4 numbers after the comma to make it works! When I use 2 numbers after the comma, the system doesn't accepted my answer. UserIdTAG: 310147 UserNameTAG: ildomarcarvalho CreateTimeTAG: 2012-09-23T20:12:44Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: you mean: 'point' ;) everytime i copy the result from calc over to the side i forget to convert it, it's a lil' bit bagging, but at least h2p2 is solved now. FirstChildUserIdTAG: 131747 FirstChildUserNameTAG: hasi FirstChildCreateTimeTAG: 2012-09-23T20:21:13Z SecondChildTAG: Would you [lease give me some hint about how you solve H2P2? Thanks. SecondChildUserIdTAG: 229018 SecondChildUserNameTAG: Changming SecondChildCreateTimeTAG: 2012-09-23T23:09:16Z SecondChildTAG: Use the Discussion Forum's Search function. I bet you'll find some help. SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-09-23T23:59:26Z FirstChildTAG: as I had noticed for most answers there is 5% error tolerance FirstChildUserIdTAG: 157610 FirstChildUserNameTAG: mradziwo FirstChildCreateTimeTAG: 2012-09-23T20:33:18Z IndexTAG: 1075 TitleTAG: missing the midterm exam (October 25th) When looking at the Calendar, I realised I won't be able to do the midterm exam in the date of 25th October - I am on vacation from 22nd October to 29th October. I doubt I will be able to access the Internet that week. Does that mean I am left with 70% rating and with the B grade at best? That would be very unlucky and I would prefer doing this course in the next semester. Or will the midterm exam be open for longer time period than just one day? Thank you in advance UserIdTAG: 371860 UserNameTAG: zsojka CreateTimeTAG: 2012-09-23T18:55:49Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: From reading an earlier post about a similar matter, I believe that you can chose any day during that week to have the exam. However it is very unlucky that your vacation covers the whole week. Maybe you can get a extension from the course providers? Otherwise try and bring a laptop and buy/bring a local mobile broadband (pay as you go) dongle/sim to cover you where you are? You have a whole day in which to complete the exam in. Hope you find a solution. FirstChildUserIdTAG: 207204 FirstChildUserNameTAG: hgustavii FirstChildCreateTimeTAG: 2012-09-23T19:31:16Z FirstChildTAG: Dear Myriam and hgustavii, thank you for replies. I am leaving on Monday evening, are arriving back on Monday morning. So if the exam will be open for the whole week, I might be able to finish it before leaving. It will be my first whole-week vacation this year, so it's very unlucky the date collides... oh well :-) Anyway, your replies gave me a chance I will be able to do the exam. Have a nice evening, Zdenek FirstChildUserIdTAG: 371860 FirstChildUserNameTAG: zsojka FirstChildCreateTimeTAG: 2012-09-23T19:40:10Z SecondChildTAG: As Myrimit said: we had 4 days, not a week. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-09-23T22:48:48Z FirstChildTAG: Hi zsojka, Based on my experience in the Circuits and Electronics 6.002x Prototype Course The Midterm Exam were released on Wednesday, April 25th and closed at the end of Sunday, April 29th... We had 4 days range to choose, eg., if you chosed to set it on friday you had 24hs from the moment that you have opened the Exam... So, I guess that if it is on October 25th, it will end on October 29th... In the period of days that it is the Exam you can not discuss it in Forum, they are really severe with this. See you! Myriam. P.D: Have a nice Holidays! I hope that you can back on time on October 29th so that you can set for the Exam ;). FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-23T19:29:28Z FirstChildTAG: Hello there..i have a question for all of you.. What is the pattern for the mid term exams? i mean whether it allows us to check and recheck as in the homeworks or we can check it only once or may be a limited no of times? i am feeling really anxious on this thing..please relieve me from this anxiety... Help! FirstChildUserIdTAG: 368712 FirstChildUserNameTAG: jmen FirstChildCreateTimeTAG: 2012-09-26T07:03:16Z FirstChildTAG: Good news zsojka.... The deadline to submit the midterm exam is extended to 23:59 (11:59 pm) on Sunday, October 28th, Boston time. The exam will still be released on Thursday, October 25th at 00:01 (12:01 am) Boston time. FirstChildUserIdTAG: 368712 FirstChildUserNameTAG: jmen FirstChildCreateTimeTAG: 2012-10-23T16:33:36Z IndexTAG: 1076 TitleTAG: Think hardware I know it is supposed to be worked out in terms of boolean gates(=functions), but still. Look at the scheme and notice: there are three big branches IN PARALLEL. No matter what is happening to the left most one or the centre one, your cookie is resided in the right most branch. So if you try X=1 and see the left most branch becomes grounded that is not the end of the story as the branches are IN PARALLEL. Moving on, use the node method. For example, when X=1 then the node above the left most switch becomes grounded so there is no potential there to short the switch next to it. Next you try Y=1 and see that the center resistor becomes grounded so the right most switch stays opened. That means 1 from Vs will "flow" into Z so we say Z(1,1) = 1. Next let Y=0 so now both center switches are opened and the center resistor is not grounded. Great! So we've got some tasty potential to short the right most switch so 1 from Vs will "flow" into ground so we say Z(1,0)=0. And so on for X=0. You could also start from looking at Y saying "When Y=1 then the center resistor is grounded so the right most switch stays opened no matter what X is. So Z(X,1)=1" UserIdTAG: 190618 UserNameTAG: Kirbabaev CreateTimeTAG: 2012-09-23T13:15:04Z VoteTAG: 3 CoursewareTAG: Week 3 / Switch Model Exercise CommentableIdTAG: 6002x_switch_model_exercise NumberOfReplyTAG: 0 IndexTAG: 1077 TitleTAG: Youtube not working Staff Youtube is not working in my country.... Please help..... UserIdTAG: 190545 UserNameTAG: saadi_sheikh CreateTimeTAG: 2012-09-23T05:37:51Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Troubleshooting NumberOfReplyTAG: 3 FirstChildTAG: Me also from Pakistan. None of the edx course's video is playing. Is there any other way to see those videos. FirstChildUserIdTAG: 908 FirstChildUserNameTAG: m_umair72001 FirstChildCreateTimeTAG: 2012-09-23T05:40:58Z FirstChildTAG: try chrome + goagent Or VPN FirstChildUserIdTAG: 136959 FirstChildUserNameTAG: christerpher FirstChildCreateTimeTAG: 2012-09-23T06:09:10Z FirstChildTAG: You can see in the Wiki [Lecture Videos Download][1]:). Also see more info [here][2]: **Due to the YouTube outage in certain parts of the world, we have provided direct links to download the lecture videos, and extended the Week 2 homework and lab deadlines to the coming Sunday (September 30). Please note that Week 3 deadlines have not changed.** [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/ [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/info FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-26T11:46:02Z IndexTAG: 1078 TitleTAG: YOUTUBE ALTRNATIVE?? I want to see the video lectures....Youtube blocked.....does anybody have any other alternative like vimeo or metacafe links or something else???? :( UserIdTAG: 127800 UserNameTAG: MNB CreateTimeTAG: 2012-09-22T12:57:26Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: you can use the torproject. If you search in the discussion there are another users that put alternatives. FirstChildUserIdTAG: 308853 FirstChildUserNameTAG: fmorato FirstChildCreateTimeTAG: 2012-09-22T14:08:31Z SecondChildTAG: In case of blocking by state authorities using TOR is unlawful. Some countries have brainless government. Sad but true. SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-22T16:14:03Z SecondChildTAG: you are 100% right I agree with you YakovO. SecondChildUserIdTAG: 40209 SecondChildUserNameTAG: sadaf78 SecondChildCreateTimeTAG: 2012-09-22T17:38:33Z SecondChildTAG: you can even use another advanced proxy service called psiphon. SecondChildUserIdTAG: 183507 SecondChildUserNameTAG: obiradaniel SecondChildCreateTimeTAG: 2012-09-23T13:01:56Z FirstChildTAG: You can see in the Wiki [Lecture Videos Download][1]:). Also see more info [here][2]: **Due to the YouTube outage in certain parts of the world, we have provided direct links to download the lecture videos, and extended the Week 2 homework and lab deadlines to the coming Sunday (September 30). Please note that Week 3 deadlines have not changed.** [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/ [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/info FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-26T11:46:52Z IndexTAG: 1079 TitleTAG: Rotating an element in a Lab tool How to rotate an element such as resistor in a lab tool ? UserIdTAG: 87309 UserNameTAG: Vijayenthiran CreateTimeTAG: 2012-09-22T05:14:43Z VoteTAG: 3 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: Hi, press R key FirstChildUserIdTAG: 220837 FirstChildUserNameTAG: Calsomus FirstChildCreateTimeTAG: 2012-09-22T05:18:43Z SecondChildTAG: thanks a lot :) SecondChildUserIdTAG: 87309 SecondChildUserNameTAG: Vijayenthiran SecondChildCreateTimeTAG: 2012-09-22T06:54:17Z SecondChildTAG: You're welcome! I discovered it by chance SecondChildUserIdTAG: 220837 SecondChildUserNameTAG: Calsomus SecondChildCreateTimeTAG: 2012-09-22T23:37:49Z IndexTAG: 1080 TitleTAG: Iteration method Hi to all, by applying KVL, I got 5 - 20*(1 - e^(-vA/5)) = vA, and after many tries I got the answer, 1.088 V. I didn't understand how to solve it by iteration, someone can explain it? Thanks!!! UserIdTAG: 220837 UserNameTAG: Calsomus CreateTimeTAG: 2012-09-21T22:26:01Z VoteTAG: 3 CoursewareTAG: Week 4 / Linearization Exercise 1 CommentableIdTAG: 6002x_linearization_ex_1 NumberOfReplyTAG: 3 FirstChildTAG: They give you one equation: ia=10(1-e^(-va/5), which explains ia over the non-linear device. There is other device so there is another equation. That would be ia=? Do them simultaneously and there is your answer. You can do calculus if you like. Maybe it would help to think of it as a Node method problem. That would mean KCL. Take the first equation, add it to the second one and make them sum to zero(so firsteq+secondeq=0.) That's KCL, all currents must sum to zero. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-09-22T00:22:15Z FirstChildTAG: This equation can't be solved in radicals, so you have to use numerical methods, such as http://en.wikipedia.org/wiki/Newton's_method There are many other methods, some of them have better convergence, than Newton's method. What they call "iteratively apply KVL" is really an application of Newton's method without naming it and check for it applicability. That's why in lection it was said that sometimes it works and sometimes it doesn't. If it was a math cource you would be given a strict proof and all needed checks would be performed. FirstChildUserIdTAG: 400884 FirstChildUserNameTAG: kibergus FirstChildCreateTimeTAG: 2012-09-22T08:30:33Z SecondChildTAG: Thanks! SecondChildUserIdTAG: 220837 SecondChildUserNameTAG: Calsomus SecondChildCreateTimeTAG: 2012-09-22T23:39:10Z FirstChildTAG: I used a lazy man's solution. I set up a spreadsheet that calculated the iA for the two equations with incrementing values of VA. Then I manually looked for the point where the two calculated results are equal. It's a numerical version of the graphical solution. It took about 4 minutes. FirstChildUserIdTAG: 202518 FirstChildUserNameTAG: MCN FirstChildCreateTimeTAG: 2012-09-24T21:23:31Z SecondChildTAG: I did almost the same thing except for I used Solver module of Excel, it converged equation for me. :) SecondChildUserIdTAG: 326409 SecondChildUserNameTAG: EugeneZ SecondChildCreateTimeTAG: 2012-09-25T17:43:10Z IndexTAG: 1081 TitleTAG: I can't use Lab 2 I cannot use Lab2. There is only a rectangular box, without any tools in it. I am using Firefox 15.0.1 on Mac OS X 10.6.8. I have cleared the cache, restarted Firefox, reloaded the page, but nothing works. I think I have the solution of the exercise but there is no way to try it, let alone to submit it. Help, what can I do to submit before the deadline expires? UserIdTAG: 266912 UserNameTAG: pietvo CreateTimeTAG: 2012-09-21T22:23:30Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Troubleshooting NumberOfReplyTAG: 2 FirstChildTAG: Everything was working yesterday. Today none of the labs show the simulator. Nothing was changed and I checked in another PC. I also had trouble to check H2P2 but no problems with H2P1 or H2P3 in the same page. Please help the deadline is near. Tried again and it is working now. Thanks FirstChildUserIdTAG: 392699 FirstChildUserNameTAG: lu2adw FirstChildCreateTimeTAG: 2012-09-21T22:57:57Z FirstChildTAG: Today it is working and I could submit my lab exercise. Thanks to whoever changed it. FirstChildUserIdTAG: 266912 FirstChildUserNameTAG: pietvo FirstChildCreateTimeTAG: 2012-09-22T07:38:39Z IndexTAG: 1082 TitleTAG: Herb I assume that what is required here is the incremental change in vA - not the dvA/dVi calculation. I arrived at an incorrect response by simply repeating part one of this problem at vI = 5.1volts and computing the difference between the the two vI results. Where did I miss the boat? UserIdTAG: 286880 UserNameTAG: herbsteiner CreateTimeTAG: 2012-09-21T21:32:00Z VoteTAG: 3 CoursewareTAG: Week 4 / Linearization Exercise 1 CommentableIdTAG: 6002x_linearization_ex_1 NumberOfReplyTAG: 1 FirstChildTAG: You started correctly,you already have $V_a$ for $V_I = 5$ now try to get the new value of $V_a$ for the new value of $V_I$ which is 5.1 Volt. now you have to get the incremental change from the old and new values. that means : $d V_a = V_a1 - V_a2$ and for the source voltage it's: $ d V_i = 5.1-5 =0.1$ now go and get the final answer by dividing both values ... and have fun. FirstChildUserIdTAG: 292360 FirstChildUserNameTAG: Loai FirstChildCreateTimeTAG: 2012-09-22T14:01:00Z SecondChildTAG: I used this method and my answer was marked as correct but my value was high by 0.005263. Not a big deal but I wonder where the difference came from. SecondChildUserIdTAG: 202518 SecondChildUserNameTAG: MCN SecondChildCreateTimeTAG: 2012-09-24T21:32:38Z IndexTAG: 1083 TitleTAG: Log problem I was trying to solve ( $Va$ ) by starting like this: KVL >> $V_a= V_I - V_R$ and when $V_R = I_a *R^2$ I ended up with: $15 + V_a = 20e^-v/5$ now when I apply log on both sides I get this: $Log(15 + V_a) = -V_a/5 * Log(20e)$ how to solve this from here on, and did I made mistakes from the previous steps UserIdTAG: 292360 UserNameTAG: Loai CreateTimeTAG: 2012-09-21T20:50:32Z VoteTAG: 3 CoursewareTAG: Week 4 / Linearization Exercise 1 CommentableIdTAG: 6002x_linearization_ex_1 NumberOfReplyTAG: 1 FirstChildTAG: This equation can't be solved in radicals. Thy to feed it to wolfram alpha and you would see a special function in analytical answer. You have to use numeric methods, such as Newton's method. Or you can feed to some of available solvers, for example to wolfram alpha I've mentioned above and it would use these methods to find a solution for you. FirstChildUserIdTAG: 400884 FirstChildUserNameTAG: kibergus FirstChildCreateTimeTAG: 2012-09-22T08:40:32Z SecondChildTAG: thanks a lot man for pointing at wolframalpha.com, I got the result... but I actually didn't know how to solve it http://www.wolframalpha.com/input/?i=+5+-+20*%281+-+e%5E%28-v%2F5%29%29+%3D+v SecondChildUserIdTAG: 292360 SecondChildUserNameTAG: Loai SecondChildCreateTimeTAG: 2012-09-22T13:17:03Z SecondChildTAG: Use this: http://en.wikipedia.org/wiki/Newton's_method When they say to iteratively use KVL you actually find tangent line to your equation, find place where it intersects oX and use it as new point where to apply KVL. That's what Newton method is. Of-course you can use other numerical methods which converge faster, but this is the simplest one. SecondChildUserIdTAG: 400884 SecondChildUserNameTAG: kibergus SecondChildCreateTimeTAG: 2012-09-22T21:14:55Z SecondChildTAG: Correct me if im wrong but : Vr should be Ia*R and not R^2 Also ln(a*b) = ln(a) + ln(b) also instead of multiplying with ln(20), you should have written +ln(20) SecondChildUserIdTAG: 329254 SecondChildUserNameTAG: KGabor SecondChildCreateTimeTAG: 2012-10-05T11:43:02Z IndexTAG: 1084 TitleTAG: Why do we apply superposition priniciple if we say the voltage source is a non-linear element? Hi there, I was wondering why we used the superposition principle, which is used only for linear circuits.In the above problem, the instructor was seen to prove the voltage source to be a non-linear circuit while later he used the superposition principle to solve the network consisting of voltage sources and resistors?I will be waiting you feedback, please. UserIdTAG: 294329 UserNameTAG: EZRAMELES CreateTimeTAG: 2012-09-21T20:15:59Z VoteTAG: 3 CoursewareTAG: Week 2 / Superposition CommentableIdTAG: 6002x_superposition_tutorial NumberOfReplyTAG: 3 FirstChildTAG: the system itself is linear because it s composite of resistors ( linear elements ) FirstChildUserIdTAG: 382505 FirstChildUserNameTAG: AhmedGalal2 FirstChildCreateTimeTAG: 2012-09-22T01:20:33Z FirstChildTAG: I am also wondering how it could be linear if it contains a voltage source and a current source. It is not going to have a 0,0 intersect or 2F(i) will not work. FirstChildUserIdTAG: 230787 FirstChildUserNameTAG: Gabriel007 FirstChildCreateTimeTAG: 2012-09-23T19:58:32Z FirstChildTAG: ## Why do we apply superposition priniciple if we say the voltage source is a non-linear element? ## I still have the same doubt of Ezrameles, this look like a little bit confusing. FirstChildUserIdTAG: 149549 FirstChildUserNameTAG: Java_Dido FirstChildCreateTimeTAG: 2012-09-23T12:46:52Z IndexTAG: 1085 TitleTAG: I laughed to this. > So if our VCCS is a device that can supply power. In other words, if our VCCS is not a passive device. In other words, if it is a device that can supply power, or it can amplify power, if it's a device that can amplify power. In other words, if it's not a passive device, it is a device that can amplify power or can supply power. Then the mathematically predictive behavior will be observed. Man I loled at this. Well, in fact I just giggled. It is at 2:35 until 3:00. Just to point it out. UserIdTAG: 194232 UserNameTAG: PedroTeles CreateTimeTAG: 2012-09-21T14:17:49Z VoteTAG: 3 CoursewareTAG: Week 4 / Passive and Active Devices CommentableIdTAG: 6002x_psv_n_active_dev NumberOfReplyTAG: 0 IndexTAG: 1086 TitleTAG: Can't Access Videos: Youtube Blockage If someone has managed to access the videos even after having youtube blockage in his country , do tell me how ... UserIdTAG: 332941 UserNameTAG: SabaSiddiqi CreateTimeTAG: 2012-09-21T11:50:37Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: same problem here FirstChildUserIdTAG: 403493 FirstChildUserNameTAG: aiai FirstChildCreateTimeTAG: 2012-09-21T14:16:15Z FirstChildTAG: You can see in the Wiki [Lecture Videos Download][1]:). Also see more info [here][2]: **Due to the YouTube outage in certain parts of the world, we have provided direct links to download the lecture videos, and extended the Week 2 homework and lab deadlines to the coming Sunday (September 30). Please note that Week 3 deadlines have not changed.** [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/ [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/info FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-26T11:47:32Z IndexTAG: 1087 TitleTAG: Explanation of boolean-valued functions Follow next link http://mathworld.wolfram.com/BooleanFunction.html UserIdTAG: 164471 UserNameTAG: BillNieves CreateTimeTAG: 2012-09-21T03:11:17Z VoteTAG: 3 CoursewareTAG: Week 2 / Boolean functions CommentableIdTAG: 6002x_S4E2_Boolean_Functions NumberOfReplyTAG: 0 IndexTAG: 1088 TitleTAG: lab 2 i have tried to figure out the lab every which way that i know how and i can not figure out what i am doing wrong it does not show me my mistake. If anyone can help point me in the right direction that would be great i am not looking for the answer just wht i am doing wrong UserIdTAG: 160242 UserNameTAG: Mlevins35 CreateTimeTAG: 2012-09-21T01:00:35Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: 电池 FirstChildUserIdTAG: 471183 FirstChildUserNameTAG: snow_yxx FirstChildCreateTimeTAG: 2012-09-21T06:06:07Z FirstChildTAG: Search in others post. Many people have helped to solve this lab. Look the "hint" Vo equation and the Vo equation of the problem and find the relationship, Thats helped me a lot. FirstChildUserIdTAG: 184827 FirstChildUserNameTAG: DiegoT FirstChildCreateTimeTAG: 2012-09-21T01:55:47Z IndexTAG: 1089 TitleTAG: bug section S6V7 is the same as S6V6 UserIdTAG: 372321 UserNameTAG: EnricoDona CreateTimeTAG: 2012-09-20T19:29:36Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1090 TitleTAG: S3V2 Linear circuits and not linear elements ... S3V2 says: "circuits composed of linear elements were, themselves linear", but in week 2 tutorials I see that voltage sources and current sources a not linear. That means that circuits with nonlinear elements may be linear or may be not linear? I'm confused. UserIdTAG: 104522 UserNameTAG: IgorNovice CreateTimeTAG: 2012-09-20T16:52:17Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1091 TitleTAG: Video is not playing I got a problem i am trying to open the 2nd week videos but it just stuck everytime i open it didn't show a thing.. help UserIdTAG: 198382 UserNameTAG: Rizi2k5 CreateTimeTAG: 2012-09-20T16:52:03Z VoteTAG: 3 CoursewareTAG: Week 2 / Linearity CommentableIdTAG: 6002x_linearity NumberOfReplyTAG: 1 FirstChildTAG: use google chrome FirstChildUserIdTAG: 410943 FirstChildUserNameTAG: JPalaciosRoman FirstChildCreateTimeTAG: 2012-09-20T18:46:07Z IndexTAG: 1092 TitleTAG: Definition of "linear circuit" I think I understood the implications of linearity and what it does in practice, i.e. superposition and so on... But what would be a precise definition of "linear circuit"? UserIdTAG: 346056 UserNameTAG: fiatlux CreateTimeTAG: 2012-09-20T16:47:50Z VoteTAG: 3 CoursewareTAG: Week 2 / Linearity CommentableIdTAG: 6002x_linearity NumberOfReplyTAG: 1 FirstChildTAG: I think as the name implies, its linear, in this case the relation between voltage and current are linear, or V = a.I + b where a and b are any constant, and the graph are always form a straight line FirstChildUserIdTAG: 452559 FirstChildUserNameTAG: kuz1toro FirstChildCreateTimeTAG: 2012-09-20T19:34:00Z SecondChildTAG: like the equation of a straight line: y=mx+b SecondChildUserIdTAG: 184827 SecondChildUserNameTAG: DiegoT SecondChildCreateTimeTAG: 2012-09-20T21:53:00Z SecondChildTAG: Adding: That straight line should pass through the origin (0,0). SecondChildUserIdTAG: 127195 SecondChildUserNameTAG: princeofsudan SecondChildCreateTimeTAG: 2012-09-21T00:54:26Z SecondChildTAG: Yes, but you are describing a *circuit element*, not a whole circuit. A circuit does not have two terminals necessarily. It is just a bunch of circuit elements connected together. There must be a more existential property that makes a circuit "linear" SecondChildUserIdTAG: 346056 SecondChildUserNameTAG: fiatlux SecondChildCreateTimeTAG: 2012-09-23T23:50:39Z IndexTAG: 1093 TitleTAG: Times can one submit an answer? How many times can one submit an answer to a question. UserIdTAG: 125652 UserNameTAG: Adedayo CreateTimeTAG: 2012-09-20T16:33:03Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Thanks. FirstChildUserIdTAG: 125652 FirstChildUserNameTAG: Adedayo FirstChildCreateTimeTAG: 2012-09-21T11:59:35Z FirstChildTAG: Good question, for one of the questions I have submitted wrong answers for about 6 7 times and finally got the right answer by fixing my calculations. but not sure if there is any limit on this or any negative consequense. FirstChildUserIdTAG: 210954 FirstChildUserNameTAG: Shahrouz FirstChildCreateTimeTAG: 2012-09-20T21:01:40Z FirstChildTAG: Depends.. **Midterm and Final Exam:** Limited (3 chances). **Homeworks/Labs:** Unlimited (infinite chances till the deadline date). [More info - Checking and submitting an answer exercise][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Overview/edx_introduction/ FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-20T16:46:21Z IndexTAG: 1094 TitleTAG: H3P4. How to attempt this question? I have read from another post that when Vs is positive, V2 and D2 need to be removed and Vs is negative V1 and D1 need to be removed. Taking that, Vs has maximum positive voltage of 10V. Then, removing V2 and D2, I am left with V1 and D1. How to find the Vth using Thevenin? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-20T15:34:38Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: It's fairly simple. When a diode is forward biased, then it conducts, when it is reverse biased then it does not. So when Vs > (V1 + Vd) where Vd is the diode voltage drop, D1 will conduct and the Output voltage will be V1 + Vd. When Vs < (-V2 - Vd) then D2 will conduct and the output voltage will be -V2 - Vd. FirstChildUserIdTAG: 270011 FirstChildUserNameTAG: SnowmanZA FirstChildCreateTimeTAG: 2012-09-20T18:24:57Z FirstChildTAG: The fact is when Vs is reaching V1 in positive voltage D1 will be open and therefore there will not be any more increment on v. same applies when VS turning negative but this time when it reaches to V2 D2 will be open. Back to currents on D1 and D2, any additional voltage after V1 and V2 will be directed through D1 and D2. FirstChildUserIdTAG: 210954 FirstChildUserNameTAG: Shahrouz FirstChildCreateTimeTAG: 2012-09-20T20:57:10Z IndexTAG: 1095 TitleTAG: PC as free Oscilloscope, Function generator, and logic analyzer Hi, I am using the sound-card on the PC as as simple Oscilloscope and function generator, as well as using the printer port as a logic analyzer. With a simple breadboard, a lot of simple circuits can be tested and analyzed. The soundcard will operate at 32 bit and 96kHz in two channels. Some card as AC coupled, so it can not be used for DC signals. These apps use very little resources, so often an old laptop or PC can be a good dedicated electronics lab. You can find it here: http://www.zeitnitz.de/Christian/scope_en?mid=4.01 ![oscilloscope][1] As a function generator, you can use http://heliso.tripod.com/download/generator/dsg.htm ![function generator][2] The logic analyzer uses the printer port as a 17 channel analyzer, and can operate up to 500 k samples/s http://www.codeproject.com/Articles/75279/17-Channel-Logic-Analyzer ![logic analyzer][3] PS: http://audacity.sourceforge.net/ is good for audio with generator, spectrum analyzer etc. [1]: http://puu.sh/16TRv [2]: http://puu.sh/16TRR [3]: http://puu.sh/16TSH UserIdTAG: 142857 UserNameTAG: viking2 CreateTimeTAG: 2012-09-20T13:17:25Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: What kind of interface do you need to construct to safely use it with your computer? What are it's practical limitations? It's worth mentioning before people download this and start plugging things into their computers. It sounds great, but are there any caveats? FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-20T14:52:56Z SecondChildTAG: You need to read the description. Main issue is that the hardware is accessed directly on many versions of Windows. This will disable the use of this hardware for other apps. i.e. No sound. Secondly an app that hogs hardware directly may not release it cleanly if it crashes. You may also want to make sure you dont fry your computer by plugging the "Line In" into a 110V wall outlet, but as a future EE, you can handle this just fine. I would not recommend this to philosophy majors. And again, find an old PC for your lab, not the one with your baby pictures. Here is the [missing ling to logic analyzer][1] [1]: http://www.codeproject.com/Articles/75279/17-Channel-Logic-Analyzer SecondChildUserIdTAG: 142857 SecondChildUserNameTAG: viking2 SecondChildCreateTimeTAG: 2012-09-20T15:21:49Z SecondChildTAG: Have used the oscilloscope for a long time. No problems. On this win7 x64 PC, I have HDMI with sound built in, so the sound card is unused. It also have an auto-detect when I plug in devices. If nothing is plugged in, the oscilloscope will not run. Plug a mic, or any other signal in to the line-in port, and the scope will run just fine on sound-cards like this. SecondChildUserIdTAG: 142857 SecondChildUserNameTAG: viking2 SecondChildCreateTimeTAG: 2012-09-20T15:36:42Z SecondChildTAG: I mainly asked for the benefit of others that may try it without understanding it's limitations. I will look further into this particular example and might give it a go. I had researched them in the past, at the time they fell short of my needs. Thanks for the links! SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-20T15:43:53Z IndexTAG: 1096 TitleTAG: Awesome Learning That's was so funny and awesome! UserIdTAG: 97545 UserNameTAG: HeshamZaghloul CreateTimeTAG: 2012-09-20T12:39:23Z VoteTAG: 3 CoursewareTAG: Week 2 / Chain saw CommentableIdTAG: 6002x_chain_saw NumberOfReplyTAG: 0 IndexTAG: 1097 TitleTAG: problem H2P1 Iam unable to determine the values of r1 and r2 correctly. When i type r1=33k and r2=27k the values of vout(max)=48v is marked wrong and v1(min)=32v is marked correct . when i type the values of r1=22k and r2=27k both are marked wrong but vout(max) =48v is marked correct.The value of vmin=32v is marked wrong . please help me to proceed. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-20T06:09:04Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: same problem here. i tried r1=22k, r2=5600.. r1+r2=27.6k.. i got the resistor values wrong but a "correct" Vmax and Vmin.. Vmax=280/59 Vmin=2520/7.. what am i doing wrong? btw, i didn't quite get this ". Come up with resistors R1 and R2 such that the divider ratio Vout/Vin is within 10% of the requirement." FirstChildUserIdTAG: 312336 FirstChildUserNameTAG: drel FirstChildCreateTimeTAG: 2012-09-20T12:20:46Z FirstChildTAG: I don't get those values for Vmax and Vmin. From my point of view, with R1=33K and R2=27K the second condition (the ratio Vout/Vin) is not achieved. The best thing to get the values is to do an excel with R1, R2 and the two conditions. In the second part of the exercise you have to take into account the minimum and maximum values of the resistances. FirstChildUserIdTAG: 370769 FirstChildUserNameTAG: PabloFCid FirstChildCreateTimeTAG: 2012-09-20T14:24:45Z FirstChildTAG: The best way is to start with choosing Rth seen from Vout in the middle of the given range for Rth. Write the expression for Vout, this is also Vth. Write the expression for Ith. Then you can calculate Ith and from that, R1 and R2. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-09-20T14:29:40Z IndexTAG: 1098 TitleTAG: transient analysis? what is the significance of offset value in transient analysis? UserIdTAG: 464459 UserNameTAG: prashant725664 CreateTimeTAG: 2012-09-20T04:35:27Z VoteTAG: 3 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 0 IndexTAG: 1099 TitleTAG: Literally confusing question! Hi, I cannot understand what "How many distinct boolean-valued functions are there of two boolean-valued signals?" literally means. Can anybody describe it in words and then with an example??? UserIdTAG: 153707 UserNameTAG: masoud_np CreateTimeTAG: 2012-09-19T20:25:26Z VoteTAG: 3 CoursewareTAG: Week 2 / Boolean functions CommentableIdTAG: 6002x_S4E2_Boolean_Functions NumberOfReplyTAG: 3 FirstChildTAG: AndrewKiselev had posted a link on the discussion on that page. Can your browser get the "show discussion" on the bottom of the "S4E2: Boolean Functions" page? He said: 'Hi, look at this post:[Difference between "Distinct value"and "Distinct boolean valued function"][1].' [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_S4E2_Boolean_Functions/threads/505626a378dbb71f0000002c FirstChildUserIdTAG: 280220 FirstChildUserNameTAG: pilgrimCycle FirstChildCreateTimeTAG: 2012-09-19T23:02:28Z SecondChildTAG: I agree its confusing it confused me but I think one simple way to look at it is that if a and gate is one type of boolean value function and an or gate is another what is the total number of functions we could have with 2 inputs an and gate is one way to describe the 4 outputs since there are 4 different ways to describe all the combination of 2 inputs (2^2 = 4) and since the 4 different ways can be combined in 16 unique patterns that is were the answer comes from SecondChildUserIdTAG: 145945 SecondChildUserNameTAG: FrankBishop SecondChildCreateTimeTAG: 2012-09-21T18:22:35Z FirstChildTAG: Imagine a box that accepts two boolean inputs and has one boolean output. We don't know what's in the box, but we know it has two inputs and one output. There are 4 possible combinations of the input values. The box will have a boolean output of 1 or 0 for each of these input values. How many unique boxes are possible? Example: A box containing a "nand" gate is distinguishable from a box containing an "or" gate, so those are two of the possible boxes. FirstChildUserIdTAG: 10523 FirstChildUserNameTAG: Barrabas FirstChildCreateTimeTAG: 2012-09-20T03:57:26Z FirstChildTAG: In Colombia we call it "Cascarita" that means that something is asked with the intention to make you confus, but the question have the answer inside, so you have to read very carefuly. FirstChildUserIdTAG: 184827 FirstChildUserNameTAG: DiegoT FirstChildCreateTimeTAG: 2012-09-20T00:57:18Z IndexTAG: 1100 TitleTAG: Unable to see followed posts and comments I've mentioned it before too that I'm not able to see even my own previous posts and comments on other posts. It just make discussion forum useless. Kindly either let me know how to see comments and our followed posts or switch to the previous layout. "staff" "bug" UserIdTAG: 329015 UserNameTAG: farah_sarwar CreateTimeTAG: 2012-09-19T18:54:40Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: lolllzz and I think I still do not know how to tag staff or bug. :) FirstChildUserIdTAG: 329015 FirstChildUserNameTAG: farah_sarwar FirstChildCreateTimeTAG: 2012-09-19T18:56:13Z SecondChildTAG: The forum is still under construction (see https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_Feedback/threads/5053ee70125ee41f00000001). The staff is aware of the problems and tries to take the suggestions into account. It's not handy now but should be better in the future. Anyway, to find your posts, you have to click on your own name in an existing thread (to find one, you must remember the title of one of your threads and use the search function...). SecondChildUserIdTAG: 396446 SecondChildUserNameTAG: RousseauxS SecondChildCreateTimeTAG: 2012-09-19T19:18:27Z FirstChildTAG: there seems to be another comment but I still cannot see it.. can someone email me the reason? Here is my email id: farah_sarwar_28@yahoo.com FirstChildUserIdTAG: 329015 FirstChildUserNameTAG: farah_sarwar FirstChildCreateTimeTAG: 2012-09-19T19:44:49Z FirstChildTAG: m 2 facing d same problem....jus noticed :-( "staff" "bug" FirstChildUserIdTAG: 375305 FirstChildUserNameTAG: sandy07 FirstChildCreateTimeTAG: 2012-09-19T20:06:40Z IndexTAG: 1101 TitleTAG: Sandbox wish list 1. To save circuit as a a file (so I can look back at earlier solutions) 2. Copy and paste between sandboxes. (i.e. assembling macro blocks) 3. Undo (and maybe redo) 4. Easier to grab a part to move or delete. Grab and move small components leave the component, and add a wire instead. With no undo this is a constant hassle. If you agree mark it up, and change it if you like. UserIdTAG: 142857 UserNameTAG: viking2 CreateTimeTAG: 2012-09-19T14:28:19Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1102 TitleTAG: Same diagram OK.If you go to the textbook page 134 you can find a same diagram UserIdTAG: 422394 UserNameTAG: KreshMech CreateTimeTAG: 2012-09-19T12:56:33Z VoteTAG: 3 CoursewareTAG: Week 2 / Node equation review CommentableIdTAG: 6002x_S3E1_NodeEquationReview NumberOfReplyTAG: 2 FirstChildTAG: Thanks! FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-19T13:59:30Z FirstChildTAG: Thanks... FirstChildUserIdTAG: 394374 FirstChildUserNameTAG: Eubyse FirstChildCreateTimeTAG: 2012-09-21T16:48:08Z IndexTAG: 1103 TitleTAG: Resistance RTH i RNT in H4P3 and S8E2 I had some problems with dependent sources in homework for week 4. I did not now how to get resistance in Thevenin and Norton methods (RTH i RNT). I had wrong answers. Then I took dependent source voltage and current flowing across it and I calculated resistance for dependent source. I placed dependent source resistance during calculations of RTH and RNT in place of dependent source. It looks like it worked but I am not sure if it's good method. It may be correct because of error margin in answer box... They should show correctly calculated answer when you put correct answer in homework... I looked up in text book but there are examples with current source and it's open circuit when counting RTH... It may have been somewhere on lectures but my English isn't perfect so I may have missed that. UserIdTAG: 302188 UserNameTAG: Pepek CreateTimeTAG: 2012-09-19T11:31:03Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: If you read section 3.6 of your textbook you can find the answer to your problem. For finding Rth or Rnt you need to set all **independent** sources to zero (zero volts for a voltage source and zero current for a current source). For the dependent sources take a look at example 3.23. What did they do with the dependent source? (In that case it is a dependent current source, but the same will be done with a dependent voltage source) FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-19T13:42:57Z SecondChildTAG: I know that. "RTH can be found by calculating or measuring the resistance of the open-circuit network seen from the designated terminal pair with all independent sources internal to the network set to zero. That is, with independent voltage sources replaced with short circuits, and independent current sources replaced with open circuits. (Dependent sources must be left intact, however.)" at the beginning of 3.6 chapter. I have problems with part B of H4P3. When independent source is gone u=0 and Au=0. So resistance = 0... However correct answer is when you take resistance from calculating Norton current and use it... SecondChildUserIdTAG: 302188 SecondChildUserNameTAG: Pepek SecondChildCreateTimeTAG: 2012-09-19T14:30:52Z SecondChildTAG: Thank you. I've figured it out now. I had to look at the circuit with port B short circuited and open. After node analysis in each situation, I was able to determine both the current and resistance. Thanks again. SecondChildUserIdTAG: 331664 SecondChildUserNameTAG: dwmnctrh3 SecondChildCreateTimeTAG: 2012-09-19T16:05:45Z IndexTAG: 1104 TitleTAG: Signal speeds I couldn't understand the bit about signal speeds of interest should be lower than that of electromagnetic waves. It seemed that by signal speed they meant frequency rather than actual speed of the wave. As in frequency of signal should be less than that of electromagnetic waves. Can anyone explain this to me? UserIdTAG: 154440 UserNameTAG: Aahlad CreateTimeTAG: 2012-09-19T11:24:49Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: This is a response I wrote to a similar question last week. This is just my interpretation, and may not be completely correct. "The signals need to be slower then the speed at which the system operates. If signals are too fast, the the system cannot react fast enough to changes, or keep up with the frequency of changes. Ex, a fast (high frequency) signal would switch polarities too fast. By the time the first positive swing reached the other side of the device, the source may have switched polarities many times, thus rendering the device useless in "real time". Newer processors are coming close to operating at light's frequencies, which will eventually be a problem. Wavelength size is also a consideration, if the device is to run "instantaneous". The device needs to be physically smaller then the wavelength used. "Slow" wavelengths, like 60Hz used in our power grids are huge, that's why it is able to stay in sync over a large area. "Fast" wavelengths like the ones used in our computers are much, much smaller and cannot travel more then a few centimeters before changing phase. Frequency implies that it is an "Alternating Current" (AC), so the signal changes from positive to negative, over and over again. If the distance is too far compared to the wavelength, A positive voltage at the source, may be viewed as a negative voltage by the time the signal reaches the destination." FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-19T12:35:48Z SecondChildTAG: So its true that by speed they are actually referring to frequency? SecondChildUserIdTAG: 154440 SecondChildUserNameTAG: Aahlad SecondChildCreateTimeTAG: 2012-09-19T13:44:42Z SecondChildTAG: So basically the voltage source or other elements should not have a frequency so high that the voltage changes significantly in the time taken for propagation of the electromagnetic field in the circuit. Because then our abstraction would fail. Is that right? Or the propagation delay should be a very small time interval in comparison to the time period of oscillation. So small that there is no significant change in the phase of the signal in that time. SecondChildUserIdTAG: 154440 SecondChildUserNameTAG: Aahlad SecondChildCreateTimeTAG: 2012-09-19T13:55:54Z SecondChildTAG: So basically the frequency of the signals in the circuit should be low enough that there is no significant change in phase during the time it takes for the propagation of changes in the electromagnetic field to all parts of the circuit. Because otherwise our abstraction would fail. Is that it? SecondChildUserIdTAG: 154440 SecondChildUserNameTAG: Aahlad SecondChildCreateTimeTAG: 2012-09-19T13:59:14Z SecondChildTAG: "So basically the frequency of the signals in the circuit should be low enough that there is no significant change in phase during the time it takes for the propagation of changes in the electromagnetic field to all parts of the circuit. Because otherwise our abstraction would fail. Is that it?" That's the way I see it, sorry for the delay in response. The frequency is dependent on the propagation speed to work, so in effect it really comes down to the speed of the wavefront through a particular material. This could be a pulse, steady state or sinusoid etc. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-19T14:17:00Z SecondChildTAG: Example: A radio wave can travel at the speed of light in a vacuum. In a cable however, it can travel around say around 80% the speed of light, more or less depending on the quality of wire used. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-19T14:19:55Z IndexTAG: 1105 TitleTAG: S1E5 STAFF The voltages in the circuit in the question don't seem to make sense. Since v1 = 1.4 Volts and v2 = 0.9 volts but in opposite direction that would mean that the same current caused a potential drop in one resistor and a gain in the other implying that one of the resistances is negative. Which isn't possible. Can anyone explain whether there's something i'm doing wrong or if the question is wrong? UserIdTAG: 154440 UserNameTAG: Aahlad CreateTimeTAG: 2012-09-19T10:31:44Z VoteTAG: 3 CoursewareTAG: Week 1 / KVL and KCL exercises CommentableIdTAG: 6002x_kvl_kcl_exercises NumberOfReplyTAG: 2 FirstChildTAG: That's a good question. It might seem like a wrong set-up but notice that the problem mention that the diagram is only part of the entire circuit. There may be other elements connected in parallel with the ones shown or even an extra connection at the node between the two resistors (meaning that current is different in each resistor). However, that doesn't change the fact that the sum of all voltages around the loop should be zero. FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-19T13:31:33Z SecondChildTAG: Thanks, I didn't think of that. :-) SecondChildUserIdTAG: 154440 SecondChildUserNameTAG: Aahlad SecondChildCreateTimeTAG: 2012-09-19T13:41:51Z FirstChildTAG: i Think It is Just Assumption FirstChildUserIdTAG: 341818 FirstChildUserNameTAG: Elshazly FirstChildCreateTimeTAG: 2012-09-19T11:48:19Z IndexTAG: 1106 TitleTAG: question Dear classmates, In the last question_ at situation O_ where both A and B are zero,the voltage is (3*10^7/2)/(2000+10^7/2)=2.99880047981 But the answer at the website is : 2.99970003- Could you tell me where I am wrong, Please? This happens for P and q, too. UserIdTAG: 374393 UserNameTAG: rmaleki CreateTimeTAG: 2012-09-19T10:16:48Z VoteTAG: 3 CoursewareTAG: Week 3 / Switch Resister Model Exercise CommentableIdTAG: 6002x_switch_resistor_model_exercise NumberOfReplyTAG: 1 FirstChildTAG: I get different answers depending on which calculator I use. (edX vs Windows calculator.) I can't tell you what's wrong, both answers work, both answers are 3. You can either round up a little or trim a few numbers two or three spots past the decimal place, if you run into a situation were you think you have the right answer that is not being accepted. There seems to be a generous margin of error. Have fun. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-19T13:14:30Z SecondChildTAG: rmaleki i think like you. The correct answer is 2.9988 V. SecondChildUserIdTAG: 270769 SecondChildUserNameTAG: 2214sanchez SecondChildCreateTimeTAG: 2012-09-20T11:16:21Z SecondChildTAG: rmaleki: I agree with your equation and your arithmetic. I think the posted answer is wrong. SecondChildUserIdTAG: 202518 SecondChildUserNameTAG: MCN SecondChildCreateTimeTAG: 2012-09-20T22:33:32Z SecondChildTAG: I was wrong. The circuit is parallel. So the current go through the easiest way that is r=50. we should just consider one of the resistance. 3*50/(2000+50)= 0.0731707317073 SecondChildUserIdTAG: 374393 SecondChildUserNameTAG: rmaleki SecondChildCreateTimeTAG: 2012-09-22T15:19:10Z SecondChildTAG: 3*(10M^2/20M)/((10M^2/20M)+2k) = 2.99880047981 is my solution too. As you can see I took 10M in parallel and then used simple v-divider. Ok, where am I being an idiot again ;O. SecondChildUserIdTAG: 388555 SecondChildUserNameTAG: 4lk4tr43 SecondChildCreateTimeTAG: 2012-09-23T20:36:40Z SecondChildTAG: Getting 2.99880, too. should be correct. SecondChildUserIdTAG: 394240 SecondChildUserNameTAG: ManU81 SecondChildCreateTimeTAG: 2012-09-27T19:16:01Z SecondChildTAG: I think they just didn't bother with calculating a new solution, since it's pretty much all the same. 2.9988 is more accurate though. SecondChildUserIdTAG: 320040 SecondChildUserNameTAG: Gigityghost SecondChildCreateTimeTAG: 2012-09-29T12:01:59Z SecondChildTAG: I also think that you are correct and the exact answer is 2.99880047981. SecondChildUserIdTAG: 145658 SecondChildUserNameTAG: Wh1t3w0lf SecondChildCreateTimeTAG: 2012-09-30T22:42:18Z SecondChildTAG: The equation that I used for question O was 3*(20M/10M^2)/((20M/10M^2)2k)=2.9970003 (Using the reciprocal of the parallel resistor expressions)..better late then never SecondChildUserIdTAG: 144672 SecondChildUserNameTAG: leeisagenius SecondChildCreateTimeTAG: 2012-10-03T02:42:52Z IndexTAG: 1107 TitleTAG: YouTube ban in my country Hi I am from Pakistan. Since yesterday you Tube has been banned by the government. Is any one experiencing this same problem. I request the Edx team to please use html5 for the video streaming UserIdTAG: 272417 UserNameTAG: MuKhan CreateTimeTAG: 2012-09-19T07:41:34Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I tried using proxy .. But the streaming take too much time FirstChildUserIdTAG: 272417 FirstChildUserNameTAG: MuKhan FirstChildCreateTimeTAG: 2012-09-19T07:44:30Z SecondChildTAG: if you join the face book group https://www.facebook.com/groups/483354858342165/ sarwan kumar has the videos SecondChildUserIdTAG: 436749 SecondChildUserNameTAG: waynebrown SecondChildCreateTimeTAG: 2012-09-19T07:52:07Z FirstChildTAG: You can see in the Wiki [Lecture Videos Download][1]:). Also see more info [here][2]: **Due to the YouTube outage in certain parts of the world, we have provided direct links to download the lecture videos, and extended the Week 2 homework and lab deadlines to the coming Sunday (September 30). Please note that Week 3 deadlines have not changed.** [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/ [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/info FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-26T11:49:48Z IndexTAG: 1108 TitleTAG: Possible mistake in the lecture I think that at 5:35 when you state that Vs is 10, you mean that Vs is 5. Then you will get a low output voltage of 0.45. UserIdTAG: 373309 UserNameTAG: njirving CreateTimeTAG: 2012-09-18T17:53:45Z VoteTAG: 3 CoursewareTAG: Week 3 / Switch Resistor Model of a MOSFET CommentableIdTAG: 6002x_switch_resistor_model NumberOfReplyTAG: 1 FirstChildTAG: Just marking this with "staff". I agree - Vs should be 5V not the 10V noted and mentioned by the professor. Perhaps it is an inserted mistake to make sure we are paying attention:-) FirstChildUserIdTAG: 179027 FirstChildUserNameTAG: Gretchen FirstChildCreateTimeTAG: 2012-09-24T16:40:05Z IndexTAG: 1109 TitleTAG: Easier way to do it? What I did was: 1) Short circuit and draw it, I get $Rp$ in parallel with 2 $Rs$'s 2) Find equivalent resistence $Re=(Rp*2Rs)/(Rp+2Rs)$ 3) Then I found the equivalent Voltage $Ve=I*Re$ 4) Went back to circuit drawn in 1) and found the $In$ as $In=I*Re/2Rs$ ---------- So in this case it looks like I did 1 less step aplying the **Thevenin** method than the **Norton** method. And when finding $Vl$ I have to do one more step in **Thevenin** than in **Norton**. So it looks like the easiest way to do this exercise is to first apply **Norton** method, convert it into **Thevenin** by $In=Vth/Rth$ and then find $Vl$. But if there is an easier way I would like to know how! Thanks UserIdTAG: 136253 UserNameTAG: chemeng CreateTimeTAG: 2012-09-18T15:25:34Z VoteTAG: 3 CoursewareTAG: Week 2 / Norton method example CommentableIdTAG: 6002x_S3E6_Norton_Model NumberOfReplyTAG: 2 FirstChildTAG: You can easily find *In* using 'Current divider'. In = I*Rp/(Rp+2Rs) FirstChildUserIdTAG: 360883 FirstChildUserNameTAG: raiden00 FirstChildCreateTimeTAG: 2012-09-19T10:56:45Z SecondChildTAG: Cool, thanks! That would be easier. I think that was not explained in lectures. 'Voltage Divider' was explained but not 'Current Divider' SecondChildUserIdTAG: 136253 SecondChildUserNameTAG: chemeng SecondChildCreateTimeTAG: 2012-09-19T21:38:09Z SecondChildTAG: 2Rs ? SecondChildUserIdTAG: 146638 SecondChildUserNameTAG: aphtab SecondChildCreateTimeTAG: 2012-09-22T14:22:21Z SecondChildTAG: chemeng: the current divider was also explained, it is on the following slide to the voltage divider (number 15) SecondChildUserIdTAG: 131747 SecondChildUserNameTAG: hasi SecondChildCreateTimeTAG: 2012-09-23T10:34:56Z SecondChildTAG: aphtab: yes. as you see in the schematic you have twice Rs, once in the upper terminal, and once in the lower one. For the Norton equivalent you short circuit the wire at the right hand side between them, bringing them in series which just lead the sum of the both. SecondChildUserIdTAG: 131747 SecondChildUserNameTAG: hasi SecondChildCreateTimeTAG: 2012-09-23T10:35:30Z SecondChildTAG: I slove it in the same manner, but there is a thing which confuse me. When I drew short circuit to find In, the direction of In is opposite to the Norton equivalent scheme. SecondChildUserIdTAG: 269639 SecondChildUserNameTAG: ishalyapin SecondChildCreateTimeTAG: 2012-10-27T14:13:14Z FirstChildTAG: short the port then you'll see I(N), and make I = 0 you will see RN as 3 resistors in series. As the load attached to, the Norton equivalent is adding another parralel resitor (RL). Very easy to find VL. FirstChildUserIdTAG: 201508 FirstChildUserNameTAG: datle FirstChildCreateTimeTAG: 2012-09-19T10:57:36Z SecondChildTAG: Well I guess that's pretty much what I said unless you are suggesting that $I=In$ which was a mistake I did at first! The computer thought my answer was valid though, so that's why I got confused. But then with the comments here I realized that $I$ is different of $In$. SecondChildUserIdTAG: 136253 SecondChildUserNameTAG: chemeng SecondChildCreateTimeTAG: 2012-09-19T21:43:46Z IndexTAG: 1110 TitleTAG: Missed the HW1 and Lab1- what can i do? hey, please is there anything i can do about the homework and Lab 1, i missed it because i resumed late and was trying to understand the environment. UserIdTAG: 450481 UserNameTAG: blueink CreateTimeTAG: 2012-09-18T01:47:18Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: out of the 12 homeworks and labs only 10 will be graded.. the other 2 labs and homeworks with the lowest scores will be dropped. FirstChildUserIdTAG: 371000 FirstChildUserNameTAG: Joseph090892 FirstChildCreateTimeTAG: 2012-09-18T03:01:49Z FirstChildTAG: I posted a reply to a similiar question here: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5058581a5dc8a52300000002 The course grading drops the two lowest grades for homeworks and labs. So you are not in any trouble until you either miss more than two of each, or get less than 100% on any of the remaining 10 labs and homeworks that will constitute part of your grade (the remaining part of your grade will consist of midterm and final exam grades). I suggest you review the course syllabus https://www.edx.org/static/content-mit-6002x/handouts/syllabus.a477535058a1.pdf and other materials on the 'Course Info' page. The Course Syllabus explains how the course is graded, among other things. Course grades are computed as follows: Homeworks - 15% Labs - 15% Midterm - 30% Final - 40% Grades are distributed as follows: A - 87% B - 70% C - 60% You need a minimum of a C to receive a certificate in this course. Each of the homeworks and labs has equal weight. Therefore, since the two lowest grades of homeworks and labs are dropped, only ten of each count toward your overall grade. That means that each homework and lab can contribute a maximum of 1.5 points toward your overall grade (assuming you get all the questions correct on each lab or homework). FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-09-18T13:10:02Z IndexTAG: 1111 TitleTAG: Lab 2 Tips and Tricks Hey there! If you are still wondering how are you supposed to do the Lab 2, I decided (because I spend some time on it myself) to write a few tips for you. 1. Vout = 1/2 * V1 + 1/6 * V2. Think about the methods covered in week 2 and figure out which one is applied here. 2. You are supposed to get a fraction of the input voltage(s), remember Lab 1. (Keyword: divider) 3. If you are not sure how to do the problem on paper, remember the circuit sandbox. Trick: If you are not a fan of waves, simplify the circuit by using DC voltages (and analysis) BUT only in your own testing circuit, not the one in Lab (you are not supposed to modify the sources). 4. If you cannot solve the Lab with X elements, you can always try X+1 (or X-1 for that matter), but stay reasonable. 5. Once again - circuit sandbox is your friend. If everything fails, try to build up from the basic blocks, ie. basic Voltage-Resistor circuit, then fitting it into the figure given in the Lab, etc. Hope this helps someone. If you have any questions or more tips and tricks to add, feel free to comment. UserIdTAG: 138981 UserNameTAG: Pr0bability CreateTimeTAG: 2012-09-17T19:00:46Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Don't forget, the voltage sources are a sine wave and a square wave, so the voltages change with time. We're used to analyzing circuits with constant voltages, and those methods have to be modified to account for varying voltages. FirstChildUserIdTAG: 132828 FirstChildUserNameTAG: rharris FirstChildCreateTimeTAG: 2012-09-19T04:01:30Z IndexTAG: 1112 TitleTAG: Lab1 My mistake Dear collegues I could not make lab1 due to simple mistake- pointing "," not "." System could not recognise this difference. i am not living in USA - we write 3,1415..... Be careful ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13478795432550797.jpg UserIdTAG: 202249 UserNameTAG: ABZ CreateTimeTAG: 2012-09-17T11:04:03Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: [Edited by moderator] FirstChildUserIdTAG: 164611 FirstChildUserNameTAG: IgorStrelnikov FirstChildCreateTimeTAG: 2012-09-17T12:55:37Z SecondChildTAG: Do not post answers in the forum. SecondChildUserIdTAG: 79361 SecondChildUserNameTAG: kimt SecondChildCreateTimeTAG: 2012-09-17T13:05:36Z SecondChildTAG: To my mind subject is mistake (or bug) in Lab1. I just wrote about it and have provide proofs.Excuse me for incorrect form of a statement, but ABZ has shown right answer. SecondChildUserIdTAG: 164611 SecondChildUserNameTAG: IgorStrelnikov SecondChildCreateTimeTAG: 2012-09-17T20:29:31Z SecondChildTAG: but could you please upload the lab ans too moderator...i didn know how to do the first lab... :( SecondChildUserIdTAG: 400239 SecondChildUserNameTAG: LincyG SecondChildCreateTimeTAG: 2012-09-19T17:23:54Z SecondChildTAG: LincyG, line 6V-A must be straight.I guess it's bug of Lab pattern. SecondChildUserIdTAG: 164611 SecondChildUserNameTAG: IgorStrelnikov SecondChildCreateTimeTAG: 2012-09-19T20:45:59Z IndexTAG: 1113 TitleTAG: Tutorial - How good is the lightbulb model? If they need to draw straight line from (0,0) to the end point, why bother measuring the other points? What I am trying to say, when the instructors made the curve to be a line, they should consider all the points and choose a middle gradient that represent all points. Not just the end point. UserIdTAG: 89084 UserNameTAG: anakluhur CreateTimeTAG: 2012-09-17T07:30:58Z VoteTAG: 3 CoursewareTAG: Week 2 / Lightbulb model CommentableIdTAG: 6002x_lightbulb_model NumberOfReplyTAG: 0 IndexTAG: 1114 TitleTAG: Difference between dependent/independent KVL and KCL equations? What's an example of an independent and dependent KCL equation for a node? An independent and dependent KVL equation for a loop? Thanks! UserIdTAG: 364798 UserNameTAG: atari1994 CreateTimeTAG: 2012-09-17T01:58:47Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi atari1994! How are you? If it helps you here is a similar [Post][1] ;) [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/MITx_6_002x_2012_Fall/threads/504e785a08a190230000000a FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-17T02:46:33Z SecondChildTAG: Thank you so much. Sorry for asking a question that was already answered. SecondChildUserIdTAG: 364798 SecondChildUserNameTAG: atari1994 SecondChildCreateTimeTAG: 2012-09-17T02:59:28Z SecondChildTAG: You are welcome atari1994 :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-17T02:59:59Z IndexTAG: 1115 TitleTAG: What the heck happened to the forum format? What a difference a week makes! I took a week off from logging in for business reasons, and tonight I see all sorts of strange things have happened to the forum. Instead of a list of the posts I was following on the left, I now see a sidebar box which ostensibly can be sorted by date, vote, or comment. HOWEVER (and you *knew* there was one), all that is displayed are the first few words of each title, which is not enough info to figure out what each post is about. The column widths cannot be adjusted so I can see the full title, nor can the entire sidebar width be adjusted. And there's no date showing at all. Boo on this "improvement." Can we please have something like a normal forum, with top-level topics, and individual posts under each top-level topic where the full title (and perhaps the first few lines of each post) show up? Even the pilot 6.002x forum, in all its glorious disorganization, was better than this. FYI, I'm using Portable Firefox v.10.0.2 under various versions of WinXP (depending on which computer I am using), currently SP3 at the moment. Feh. UserIdTAG: 94557 UserNameTAG: g_hopper CreateTimeTAG: 2012-09-16T22:44:45Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 5 FirstChildTAG: I think both forums still would be nice. This one you don't have to rifle back through many many posts you have already seen. But the other is more directly located toward the problem you have. Or Q& A you might be seeking Just a thought. FirstChildUserIdTAG: 359302 FirstChildUserNameTAG: GaryG FirstChildCreateTimeTAG: 2012-09-16T23:22:54Z SecondChildTAG: yah ur right gary and also this forum to encourage us to create excitement to our heart thats why even its so hard to learn this course that can felt discouraged or want to drop this course but because of this forum we encourage :) SecondChildUserIdTAG: 235727 SecondChildUserNameTAG: amor SecondChildCreateTimeTAG: 2012-09-17T00:29:18Z FirstChildTAG: P.S. How do we add tags to our posts under this new format? FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-09-16T22:45:55Z SecondChildTAG: Hi g_hopper! Hahah! Same happened to me !If you see at the left you will see an arrow: All, General, Feedback, Trouble Shoting, etc... You Post by Topics ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-16T22:54:07Z SecondChildTAG: Also If you clic on your name, you can see your active threads ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-16T22:54:42Z SecondChildTAG: You also can see this Post https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_Feedback/threads/5053ee70125ee41f00000001 and this too https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_Feedback/threads/5053e222c6f4462700000009 SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-16T22:57:11Z SecondChildTAG: Those aren't really tags as such, and that whole list is going to change over time. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-09-16T23:33:54Z FirstChildTAG: g_hopper, The forum is still in flux, as it were. I personally think it's an improvement over the old system(as in the forum that was on edx when we got here) but it's still got a ways to go. This system has an organizing principle that I believe was a mistake, but I think it's getting better. Having talked to PM I can tell you that for a variety of reasons, they can't go back. It's actively being worked on and you should definitely make suggestions. All the suggestions you made in this thread are good as far as I am concerned. Since you have access to the old forum you can post in there too, and I know that PM reads those threads. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-09-16T23:30:43Z FirstChildTAG: For me it was also strange when I saw the changes, however after getting use to it I feel it's better organized now. I do have some suggestions though: put back the "follow" button, make it easier to find our own comments/responses/initiated discussions without the need to follow them (it could be as simple as a little symbol next to the discussion title) and, as g_hopper says, make the whole view of the discussion title available. FirstChildUserIdTAG: 256249 FirstChildUserNameTAG: And90 FirstChildCreateTimeTAG: 2012-09-17T01:42:46Z FirstChildTAG: yea rathered the old version aswell. is there a way to find the threads i am following? FirstChildUserIdTAG: 388273 FirstChildUserNameTAG: kilmarta FirstChildCreateTimeTAG: 2012-09-19T12:06:35Z IndexTAG: 1116 TitleTAG: help me out in H 2 P 1 ??? what is the problem basically ?? UserIdTAG: 437230 UserNameTAG: somibahrian CreateTimeTAG: 2012-09-16T19:27:39Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1117 TitleTAG: Lucid KCL dreams? I had a vivid lucid dream this morning in which I could visualize KCL, it was quite remarkable. Perhaps I need some fresh air. If only I could take the exams while in that state. ;) UserIdTAG: 292543 UserNameTAG: Pennypacker CreateTimeTAG: 2012-09-16T18:47:51Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Oh that is totally normal :-). It's happened to a lot of us who took this course in Spring. Just hang on and you'll reach a point where you'll be able to solve some circuits in your head. This course is really that good! Hope I'm not spoiling the suspense. Oh and the green ticks will soon get addictive :-) FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-09-16T18:55:52Z SecondChildTAG: You're not spoiling anything. Yes the green check marks are addicting, some of these problems are taking on Warcraft quest proportions. I don't know what I will do come exam time when I can't check answers 32 times. :s SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-16T19:08:08Z SecondChildTAG: Its really addicting!!! I have left all my other studies & are always running after a few red crosses in the quest to turn them into green ticks. SecondChildUserIdTAG: 317445 SecondChildUserNameTAG: arghya33 SecondChildCreateTimeTAG: 2012-09-16T19:59:58Z SecondChildTAG: I have a feeling I will be sitting here at 11:59pm tonight waiting for the fourth week expansion pack! SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-16T20:03:35Z SecondChildTAG: Its already 2 am here & I am still in here solving the problems... SecondChildUserIdTAG: 317445 SecondChildUserNameTAG: arghya33 SecondChildCreateTimeTAG: 2012-09-16T20:31:50Z SecondChildTAG: guys help me out.... Solve it if you can, Joe wants to heat his 12'X20' workshop with electric heat. He has hired the HACME electrician company to build the system. They propose to use three 1300.0W 240V baseboard heaters to provide a total heating capacity of 3900.0W. (A heater is basically a resistor. This is not quite true, because there is a thermostatic switch incorporated into the heater and because the resista nce of a heater varies a bit with its temperature. But we will use a linear resistor as a model of a heater.) In the proposed system the heaters are connected in parallel with the 240V 60Hz AC power line (modeled by a voltage source) as shown in the diagram: Remember (from Exercise S1E3: AC power) that AC power-line voltages and currents are specified as RMS values. So 120V AC heats a given resistance exactly as much as 120V DC would heat that same resistance. How much current is expected to be drawn from the power line by this heating system when all three heaters are on? ..... If instead, HACME chose to implement the system with 120V heaters, how much current would have been needed? ..... Notice that this would require much heavier and more expensive wire to distribute the power. Back to the original plan with 240V power. Unfortunately, Sparky, who works for HACME, was a little sleepy that day. He accidentally connected the heaters as shown below: As a consequence, Joe found his workshop too cold. H1 was weak; H2 and H3 barely warmed up. What power was being dissipated in H1? ..... What power was being dissipated in H2 (or in H3)? .... So the total heating power in Joe's shop was: ..... No wonder Joe was cold. SecondChildUserIdTAG: 130118 SecondChildUserNameTAG: manis SecondChildCreateTimeTAG: 2012-09-16T21:01:41Z SecondChildTAG: @manis: we all have already solved it! If u require any help we can surely help u nut we can't solve it for u because that's against the Honor Code. SecondChildUserIdTAG: 317445 SecondChildUserNameTAG: arghya33 SecondChildCreateTimeTAG: 2012-09-17T18:06:16Z IndexTAG: 1118 TitleTAG: Finally took me some time to get used to the "SAND-BOX".....lets move ahead....:P UserIdTAG: 448671 UserNameTAG: anirudhsom3000 CreateTimeTAG: 2012-09-16T18:45:41Z VoteTAG: 3 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 0 IndexTAG: 1119 TitleTAG: Open violation of Honor Code I am afraid to notice that some people are openly distributing direct answers to the problems while others are putting links to some social sites where they upload answers to the whole course. Can any staff member justify "What they are doing regarding this?". Its not a good practice to use such answers and what will be the purpose then, in appearing this course? just to get green tick and a certificate which will be valueless without concepts of the course? UserIdTAG: 62232 UserNameTAG: mohsinaliarif CreateTimeTAG: 2012-09-16T17:57:57Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/505604ef369cc52300000025 FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-16T18:06:40Z SecondChildTAG: yha u r right :( Liya where r u ?? we need ur help please :) SecondChildUserIdTAG: 430030 SecondChildUserNameTAG: imali SecondChildCreateTimeTAG: 2012-09-16T18:10:25Z FirstChildTAG: Dear i have not observed any thing like that . Any how you are right. It should not be done. But please guide me in H1P2. FirstChildUserIdTAG: 347789 FirstChildUserNameTAG: engr_zohaib FirstChildCreateTimeTAG: 2012-09-16T18:07:01Z SecondChildTAG: yaha zohaib what happend ?? SecondChildUserIdTAG: 430030 SecondChildUserNameTAG: imali SecondChildCreateTimeTAG: 2012-09-16T18:10:59Z SecondChildTAG: Sory by mistaken i write H1P1. plz guide me how i solve H2P1? SecondChildUserIdTAG: 347789 SecondChildUserNameTAG: engr_zohaib SecondChildCreateTimeTAG: 2012-09-16T18:27:47Z SecondChildTAG: Basically you need to use a pair of resistors that have a Thevenin equivalent(Voltage source included) between 10K and 30K, while at the same time producing the required 21v. Your numbers or values may be different. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-16T18:51:23Z FirstChildTAG: Only homework problems (complete) solutions that add up to our overall progress are banned. FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-09-16T18:48:27Z SecondChildTAG: Good point. Exercises do not count. Also, after the deadline for the homework passes, feel free to discuss it openly with the answers. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-09-16T19:22:29Z IndexTAG: 1120 TitleTAG: Anyone else studying in a school? Am i the only one here, currently studying in a school? please tell me, atleast i would have someone of my age UserIdTAG: 169784 UserNameTAG: thewiredbear CreateTimeTAG: 2012-09-16T14:41:50Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Next year i'm a graduate from the institute. What I can say? You are on the right way, continue studying, don't look to another people ) Good Luck! ;) FirstChildUserIdTAG: 394836 FirstChildUserNameTAG: v2g6ch4 FirstChildCreateTimeTAG: 2012-09-16T15:35:48Z SecondChildTAG: thanks a lot, join the group, https://www.facebook.com/groups/mitxfallsemester/ SecondChildUserIdTAG: 169784 SecondChildUserNameTAG: thewiredbear SecondChildCreateTimeTAG: 2012-09-16T15:38:38Z IndexTAG: 1121 TitleTAG: Resistance of battery Hei! Internal resistance it's matter only when current flow, if not dont think about it, because is not there then. When you measure voltage on open circuit, current dosn't flow. Hope it help :-) UserIdTAG: 97792 UserNameTAG: SzymonJozef CreateTimeTAG: 2012-09-16T11:28:20Z VoteTAG: 3 CoursewareTAG: Week 1 / Battery model CommentableIdTAG: 6002x_S1E9_Battery_Model NumberOfReplyTAG: 0 IndexTAG: 1122 TitleTAG: How many attempts do we have for Homework? For example, in H1P1 i wrote 0.33*R and they said is wrong. Then I wrote (2/6)*R and t said it's right, but actually it's almost the same. So, Could anybody please say, how many attempts do we have to writte the right answer? Thanks.... UserIdTAG: 342966 UserNameTAG: SandraNavarro CreateTimeTAG: 2012-09-16T11:02:11Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I believe that homework allows for unlimited attempts while the number of attempts will be limited in the exams. FirstChildUserIdTAG: 12419 FirstChildUserNameTAG: rpm FirstChildCreateTimeTAG: 2012-09-16T11:19:43Z FirstChildTAG: Hi SandraNavarro! Based on my experience in the Prototype Course 6.002x, **you have unlimited Check attempts in the Homework and Lab** , that means that no matter if you have pressed 1 time, 2 times, 10 times the buttom check in order to get the correct answer. **But,remember that in the Midterm and Final Exam you will not have unlimited check, only three attemps...** For more information you can go to Overview->edX Tutorial->CHECKING AND SUBMITTING AN ANSWER (This is a nice way to understand this). Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-16T15:38:50Z IndexTAG: 1123 TitleTAG: -ve sign Is that necessary to always check the polarity in such type of questions ?? Its really confusing at tymes . . . UserIdTAG: 379483 UserNameTAG: annjoseph9159 CreateTimeTAG: 2012-09-16T11:01:08Z VoteTAG: 3 CoursewareTAG: Week 1 / Power CommentableIdTAG: 6002x_S1E2_Power NumberOfReplyTAG: 1 FirstChildTAG: (-ve sign means that direction of current in the diagram is wrong and so direction of the current is opposite to that given in diagram) this problems teaches a concept --- 1)+ sign and - sign are for voltage difference across resistor And as we know that current flows from current(positive charge) flow from positive voltage area to negative voltage area(taking voltage at infinity as reference point) 2) do not consider resistor as a battery . i thought the current direction given in diagram was right .i thought that in a diagram of current flowing out of battery - the current always flow from positive terminal so here also the current is coming out from that + sign. I mistook resistor as battery not really but because always seeing current flowing from + sign.3) the direction of current in diagram is wrong.if we use +1.17 this means the direction given in the diagram is write but that is non the case so the answer is -1.17 (-ve sign means that direction of current in the diagram is wrong and so direction of the current is opposite to that given in diagram) FirstChildUserIdTAG: 539003 FirstChildUserNameTAG: amitgupta25121993 FirstChildCreateTimeTAG: 2012-10-04T00:54:07Z IndexTAG: 1124 TitleTAG: Voltage Solution To find v1,I used KVL So,v1-v2=0 v1=v2=6V,same for v3=v4=6V. I don't know where I can differentiate these.How it comes -6V. Also near v3 there are different polarities. which one to consider? i1=-i2.i1 crossing + to - . So my answer was i1=-.33 & i2=.33 & i3=-i4. i3=.33 (- to +)& i4=-.33. Multiplication of corresponding values give p*. I need to be clear on the voltages.Plz anyone explain me? Thanks.. UserIdTAG: 147577 UserNameTAG: rithi3 CreateTimeTAG: 2012-09-16T09:21:49Z VoteTAG: 3 CoursewareTAG: Week 1 / Associated reference directions CommentableIdTAG: 6002x_S2E2_Associated_Reference_Directions NumberOfReplyTAG: 1 FirstChildTAG: both are 6 volts in magnitude,but when u measuring V1 the relative voltmeter terminals are in phase with V1 where for V3 the terminals are connected opposite. thats why it measures -6 volts. FirstChildUserIdTAG: 279136 FirstChildUserNameTAG: rudra_ece FirstChildCreateTimeTAG: 2012-09-16T16:11:48Z SecondChildTAG: Perspective from other angle: + (pluses) and - (minuses) shows where you stick your multimeter + and - probes. As you can see v1=v2 because you stick + to the top and - to the bottom in both v1 and v2 mesurements. The same with v3=v4 - its the same circuit but you change probes positions to opposite. In first circuit you stick + probe where battery gives you +, and minus probe go to batery -. In second circuit probe - go to batery +, and + go to -. So signes are opposit. SecondChildUserIdTAG: 66669 SecondChildUserNameTAG: agarus SecondChildCreateTimeTAG: 2012-09-17T10:10:20Z IndexTAG: 1125 TitleTAG: Wauuu,,, Finish Home Work Week1 is less the 10 hour :) Made lot of Stupid mistakes. Took me almost 8 hour's UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-16T07:56:55Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: please part 3 last 3 question's answers please........ FirstChildUserIdTAG: 169784 FirstChildUserNameTAG: thewiredbear FirstChildCreateTimeTAG: 2012-09-16T08:04:51Z SecondChildTAG: just add all the powers from the heaters in the new arrangement you'll get the answer.. regards Neeraj SecondChildUserIdTAG: 210992 SecondChildUserNameTAG: neerajnatu SecondChildCreateTimeTAG: 2012-09-16T08:06:26Z SecondChildTAG: I am obviously not able to do that, because the arrangement has H1 in series to the combination of other 2 in parallel, now we don't even have any resistance or current given in this scenario, if i find current or resistance from previous values.....it does not go right....please help...I beg you SecondChildUserIdTAG: 169784 SecondChildUserNameTAG: thewiredbear SecondChildCreateTimeTAG: 2012-09-16T08:08:48Z SecondChildTAG: Total P = Total V^2/R ( you can calculate Total R and I) Calculate R of 1 element P(1element)=V^2/R(1 element) SecondChildUserIdTAG: 410943 SecondChildUserNameTAG: JPalaciosRoman SecondChildCreateTimeTAG: 2012-09-16T08:31:26Z IndexTAG: 1126 TitleTAG: H1P1: RESISTOR COMBINATIONS help needed Can someone give me a hint. Given that each individual resistor can dissipate up to 1 watt of power before burning up, how much total power in watts ( W ) can the smallest-valued composite resistor dissipate before burning up? I know the replacement R voor R4 R4 R6 parallelle is 1.5 Ohm UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-15T20:27:49Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Given that each individual resistor can dissipate up to 1 watt of power before burning up, how much total power in watts ( W ) can the smallest-valued composite resistor dissipate before burning up? i don't understand this ques- :( FirstChildUserIdTAG: 284515 FirstChildUserNameTAG: SMSMA FirstChildCreateTimeTAG: 2012-09-15T23:46:54Z FirstChildTAG: home come you calculate 1.5 Ohm ?? FirstChildUserIdTAG: 284144 FirstChildUserNameTAG: reem28 FirstChildCreateTimeTAG: 2012-09-15T20:36:02Z SecondChildTAG: 1/4+1/4+1/6 = 3/12+3/12+2/12 ==>> 12/8=1.5 ohm SecondChildUserIdTAG: 410943 SecondChildUserNameTAG: JPalaciosRoman SecondChildCreateTimeTAG: 2012-09-15T20:40:16Z FirstChildTAG: If you use the search function, you will find these questions have been asked many, many times. It would be faster for you to search, then to wait for someone to write out a response, yet again. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-15T20:50:34Z SecondChildTAG: How to search SecondChildUserIdTAG: 444708 SecondChildUserNameTAG: jagadeeshv16 SecondChildCreateTimeTAG: 2012-09-16T05:05:25Z IndexTAG: 1127 TitleTAG: Why there are 7 loops? Hi, everyone I found that the answer (7 loops) of question "how many loops" in the above circuit contradicts with that of lecture's explanation later, which is 4. Can anyone help me clarify this? Thanks a lot! Ling UserIdTAG: 298870 UserNameTAG: rogerouyang CreateTimeTAG: 2012-09-15T18:08:13Z VoteTAG: 3 CoursewareTAG: Week 1 / Circuit topology CommentableIdTAG: 6002x_S2E1_Circuit_Topology NumberOfReplyTAG: 3 FirstChildTAG: The other 3 loops are abdca, acbda and abcda. Hope this helps! FirstChildUserIdTAG: 395953 FirstChildUserNameTAG: dustinge FirstChildCreateTimeTAG: 2012-09-15T18:38:45Z FirstChildTAG: The 7 loops are: abca, adcba, adba, bdcb, adca, adbca, abdca FirstChildUserIdTAG: 57012 FirstChildUserNameTAG: sam747 FirstChildCreateTimeTAG: 2012-09-15T21:24:17Z FirstChildTAG: I also can count the loops either way - i.e., loops which include the voltage source or loops which do not necessarily include the voltage source. However, since the circuit in S2E1 is identical to the circuit in S2V4, having two conflicting official answers is at best confusing. In the case of electronic circuits, how is a loop defined? (It would be helpful if staff could correct or at least notate the errant explanation.) Thanks... FirstChildUserIdTAG: 86632 FirstChildUserNameTAG: LCL FirstChildCreateTimeTAG: 2012-09-16T19:29:26Z IndexTAG: 1128 TitleTAG: H2P1... how to solve it? This is what I did. 1)I found out that R1=(13/7)R2. 2)I found all the pairs of R1 and R2 that has Thevenin resistor between 10k and 30k 3)There were six of them {18,33}....{47,82}. Is this correct? what should I do after this if it is correct or where is it wrong? ---------- UserIdTAG: 295240 UserNameTAG: Jack_my CreateTimeTAG: 2012-09-15T15:10:18Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Take one (R1, R2) combination from that 6.. and calculate the maximum and minimum voltage for that combination by considering the tolerance values of the resistors.. :) FirstChildUserIdTAG: 257806 FirstChildUserNameTAG: skoda FirstChildCreateTimeTAG: 2012-09-15T15:54:06Z SecondChildTAG: hii..how will i calculate maximum voltage? SecondChildUserIdTAG: 171170 SecondChildUserNameTAG: rinutituschakkattil SecondChildCreateTimeTAG: 2012-09-17T19:03:12Z FirstChildTAG: Thank you guys. Its a lot of trial and error. Got the answer in the end. :) FirstChildUserIdTAG: 295240 FirstChildUserNameTAG: Jack_my FirstChildCreateTimeTAG: 2012-09-15T16:16:14Z SecondChildTAG: then pls post the answer that you got...pls SecondChildUserIdTAG: 182470 SecondChildUserNameTAG: nitesh2703 SecondChildCreateTimeTAG: 2012-09-18T07:16:19Z FirstChildTAG: The first two things are perfect, now you just have to look which of these pairs fits the first condition APROXIMATIVELY(Vout≈3.5V). you'll find out that one of these pairs makes Vout/Vin 0.3529411765 which is really close to 0.35. Good luck! FirstChildUserIdTAG: 413236 FirstChildUserNameTAG: ignacioRosario FirstChildCreateTimeTAG: 2012-09-15T15:33:11Z SecondChildTAG: I think Vout is appx. 2.5 not 3.5 I guess.. :) SecondChildUserIdTAG: 257806 SecondChildUserNameTAG: skoda SecondChildCreateTimeTAG: 2012-09-15T15:52:23Z SecondChildTAG: Thank you guys. Its a lot of trial and error. Got the answer in the end. :) SecondChildUserIdTAG: 295240 SecondChildUserNameTAG: Jack_my SecondChildCreateTimeTAG: 2012-09-15T16:15:49Z SecondChildTAG: no skoda, each one of us has different Vout and i have 3.5 SecondChildUserIdTAG: 413236 SecondChildUserNameTAG: ignacioRosario SecondChildCreateTimeTAG: 2012-09-16T08:38:40Z SecondChildTAG: Well, did anybody with Vin=40 and Vout=8 ever get green marks? Is this even possible? SecondChildUserIdTAG: 446597 SecondChildUserNameTAG: garry_crannon SecondChildCreateTimeTAG: 2012-09-17T10:59:12Z SecondChildTAG: hi..i am a bit confused..in the question its given Vout is appx. 8.0V..so how are you getting Vout appx. 3.5V ? SecondChildUserIdTAG: 222911 SecondChildUserNameTAG: bhaswardg SecondChildCreateTimeTAG: 2012-09-17T12:46:41Z SecondChildTAG: I get Vout = 8.45V when using nominal values. SecondChildUserIdTAG: 392100 SecondChildUserNameTAG: glenng SecondChildCreateTimeTAG: 2012-09-17T21:35:13Z SecondChildTAG: i have 2 set the value 7.5 kindly some 1 explain !!!! my vin = 30 v ...... some one please SecondChildUserIdTAG: 430030 SecondChildUserNameTAG: imali SecondChildCreateTimeTAG: 2012-09-18T20:34:30Z IndexTAG: 1129 TitleTAG: I did solve I'm done with all correct answers UserIdTAG: 442098 UserNameTAG: Alizai CreateTimeTAG: 2012-09-15T12:38:47Z VoteTAG: 3 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 6 FirstChildTAG: can u help for lab 1 FirstChildUserIdTAG: 431253 FirstChildUserNameTAG: Nandhyalaramaraju FirstChildCreateTimeTAG: 2012-09-15T12:58:37Z SecondChildTAG: IN A UPPER 1 IS 155 OHM AND IN BOTTOM 50 OHM AND IN B KEPT 1 RESISTOR 8 OHM AND OTHER 2 R SAME VALUE SecondChildUserIdTAG: 123153 SecondChildUserNameTAG: ghulammd SecondChildCreateTimeTAG: 2012-09-16T20:16:37Z FirstChildTAG: For lab 1, add 1 resistor each in top and bottom, and one in parallel to the bulb(resistor). It will help you. FirstChildUserIdTAG: 199839 FirstChildUserNameTAG: vijaykrishnankk FirstChildCreateTimeTAG: 2012-09-15T13:02:11Z SecondChildTAG: wt shd be the value of those 3 resistances? SecondChildUserIdTAG: 356056 SecondChildUserNameTAG: Ganesh2810 SecondChildCreateTimeTAG: 2012-09-15T13:33:17Z FirstChildTAG: WHAT WILL BE THE VALUE OF THE RESISTANCES? FirstChildUserIdTAG: 276808 FirstChildUserNameTAG: DEBASMITAMAJUMDER FirstChildCreateTimeTAG: 2012-09-15T13:43:42Z FirstChildTAG: alizai can you please let me know your email address?? FirstChildUserIdTAG: 383229 FirstChildUserNameTAG: nafeesahmed FirstChildCreateTimeTAG: 2012-09-15T13:00:25Z FirstChildTAG: alizai help me for lab1 FirstChildUserIdTAG: 356056 FirstChildUserNameTAG: Ganesh2810 FirstChildCreateTimeTAG: 2012-09-15T13:02:55Z FirstChildTAG: both homework and lab please FirstChildUserIdTAG: 169784 FirstChildUserNameTAG: thewiredbear FirstChildCreateTimeTAG: 2012-09-15T17:37:28Z IndexTAG: 1130 TitleTAG: Request to Edx team. Hello, I thank you for such a great opportunity of learning. I am really loving the edx Lab tool. Extremely easy to use. Is there any offline version of the same tool? can you please give us the link. Is it a open source. Thanks Manoj Kumar S UserIdTAG: 156974 UserNameTAG: ManojKumar CreateTimeTAG: 2012-09-15T10:09:25Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: For offline use, you can use LTSpice. I haven't used it myself but I've heard it's quite easy to use. It's free and you can get it from http://www.linear.com/designtools/software/ Personally, I use the gEDA (http://www.gpleda.org/) set of tools. These are free and open source. They aren't as *user friendly* as the commercial ones but it fits very well in a Linux (read command line intensive) setup. FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-09-15T11:56:16Z SecondChildTAG: LT Spice is a good tool, but for small circuits only. SecondChildUserIdTAG: 199839 SecondChildUserNameTAG: vijaykrishnankk SecondChildCreateTimeTAG: 2012-09-15T13:17:09Z SecondChildTAG: Thanks ashwith. I follow your blogs.:) SecondChildUserIdTAG: 156974 SecondChildUserNameTAG: ManojKumar SecondChildCreateTimeTAG: 2012-09-15T17:57:01Z IndexTAG: 1131 TitleTAG: I can solve the problem in English! I find this small test a very good experience for me while we studied all these concept and kind of problems in Arabic and French. So I find that the tool is very good and use of canvas element of HTML5 in optimized way so my old computer can handle calculations and tool's events without whining or crying. thank for the developer(s). UserIdTAG: 285398 UserNameTAG: mohessaid CreateTimeTAG: 2012-09-15T09:43:25Z VoteTAG: 3 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 0 IndexTAG: 1132 TitleTAG: Consider the sign that's what I learned in this activity UserIdTAG: 260189 UserNameTAG: gelo_14 CreateTimeTAG: 2012-09-15T05:38:42Z VoteTAG: 3 CoursewareTAG: Week 1 / Associated reference directions CommentableIdTAG: 6002x_S2E2_Associated_Reference_Directions NumberOfReplyTAG: 0 IndexTAG: 1133 TitleTAG: VTH WHAT U CAN DO IS APPLY NODE METHOD :(VTH-VS)/R1+VTH/R2=0 U WILL GET THE ANSWER. UserIdTAG: 133068 UserNameTAG: swarn99 CreateTimeTAG: 2012-09-15T04:36:22Z VoteTAG: 3 CoursewareTAG: Week 2 / Simple Thevenin CommentableIdTAG: 6002x_S3E4_Simple_Thevenin NumberOfReplyTAG: 0 IndexTAG: 1134 TitleTAG: staff Sir/ Madam , If after completing homework, we click "check" and our answer is wrong, can we retry with the new answers ( will it accept the new answer)?? UserIdTAG: 365309 UserNameTAG: chetnasinghaldas CreateTimeTAG: 2012-09-14T16:42:31Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 3 FirstChildTAG: Yep, you can try as many times as you'd like. FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-14T16:58:55Z SecondChildTAG: But for me, it is not taking the changed answer... What should I do? SecondChildUserIdTAG: 365309 SecondChildUserNameTAG: chetnasinghaldas SecondChildCreateTimeTAG: 2012-09-15T13:10:04Z FirstChildTAG: i'm not part of the staff team but the answer to your question is: yes you can re-try as many times as you want until you get the correct answer. :) FirstChildUserIdTAG: 371000 FirstChildUserNameTAG: Joseph090892 FirstChildCreateTimeTAG: 2012-09-14T16:59:06Z FirstChildTAG: follow the instruction given in the introduction of course FirstChildUserIdTAG: 154418 FirstChildUserNameTAG: TECHJNKI FirstChildCreateTimeTAG: 2012-09-14T19:52:22Z IndexTAG: 1135 TitleTAG: V0 I do not understand why V0 UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-14T09:14:14Z VoteTAG: 3 CoursewareTAG: Week 1 / Node analysis practice, part 1 CommentableIdTAG: 6002x_L2Node0 NumberOfReplyTAG: 1 FirstChildTAG: Check out the next lecture, you will understand everything. FirstChildUserIdTAG: 317445 FirstChildUserNameTAG: arghya33 FirstChildCreateTimeTAG: 2012-09-14T20:31:13Z IndexTAG: 1136 TitleTAG: Is there any way you could post links to the YouTube videos underneath the player? I am using a device which does not support Flash and it looks like the HTML5 version doesn't work for me. I would love a link to the YouTube videos underneath the player. That way, I can pop open the YouTube player app directly from the page. UserIdTAG: 107219 UserNameTAG: AlexandreZ CreateTimeTAG: 2012-09-13T14:54:50Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Hi, will follow up on this. :) FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-13T15:21:08Z SecondChildTAG: I would also like to request this change if possible. My internet speeds are very unreliable, and the ability to download elsewhere would be very helpful- and would also allow me to view them/study while offline. I'm sure this is a common problem others. Perhaps some similar download facility as done with the OCW videos. Many thanks. M. SecondChildUserIdTAG: undefined SecondChildUserNameTAG: undefined SecondChildCreateTimeTAG: 2012-09-13T22:15:13Z SecondChildTAG: @kimt: Thanks. SecondChildUserIdTAG: 107219 SecondChildUserNameTAG: AlexandreZ SecondChildCreateTimeTAG: 2012-09-15T14:57:53Z IndexTAG: 1137 TitleTAG: Confused.. I was confused on the 2nd question because of the hint about doing an integrating blah blah blah. I already know well how to compute Average Power as here in our college, we term it as RMS Power. The hint made it confusing. Lol. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-13T08:06:25Z VoteTAG: 3 CoursewareTAG: Week 1 / AC power CommentableIdTAG: 6002x_S1E3_AC_power NumberOfReplyTAG: 0 IndexTAG: 1138 TitleTAG: Sequence 6 - Example 2 (S6E2) - What am I missing? It looks like the graph is plotted wrong if you accept the correct answers for questions 1 and 2. Maybe the x-axis is current not voltage. UserIdTAG: 397247 UserNameTAG: pcbolt CreateTimeTAG: 2012-09-13T08:02:27Z VoteTAG: 3 CoursewareTAG: Week 3 / Graphical Analysis of Nonlinear Circuits Exercise CommentableIdTAG: 6002x_graphical_analysis_of_nonlinear_circuits_exercise NumberOfReplyTAG: 2 FirstChildTAG: The axis don't have the same scale and the x axis is definitely voltage FirstChildUserIdTAG: 22258 FirstChildUserNameTAG: ClareLouise86 FirstChildCreateTimeTAG: 2012-09-13T09:08:51Z FirstChildTAG: I don't understand the answers like you... Both operating point coordinates are strange (too low for the voltage and too high for the current). FirstChildUserIdTAG: 396446 FirstChildUserNameTAG: RousseauxS FirstChildCreateTimeTAG: 2012-09-13T17:28:07Z SecondChildTAG: The scale of the graph is misleading. If you graph the two functions on a proper graphing calculator, you can see that the results make sense. Try Google's graphing calculator - just enter the two functions separated by a comma into the Google search box and it wil plot them. SecondChildUserIdTAG: 339668 SecondChildUserNameTAG: chickwebb SecondChildCreateTimeTAG: 2012-09-23T20:13:38Z IndexTAG: 1139 TitleTAG: Future Electronics Hello, dear Professor Anant Agarwal, Thank you so much for your course I have a question, we are learning about Electronic in this era, many electronic devices are made from anorganic material, but how we can make an electronic devices from organic material? thank you best wishes for EdX team Arip Nurahman From Indonesia UserIdTAG: 51896 UserNameTAG: Arip CreateTimeTAG: 2012-09-13T02:17:24Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 1140 TitleTAG: Feedback- Program needs better examples I think the 6.002x program would be improved if in the lecture series there would be more examples of the introduced methods. In the videos they show how to do different the methods with only variables in fairly simple examples and then they throw you into these complex problems in the worksheets. That is just my feeling, anyone else feel the same way? UserIdTAG: 219780 UserNameTAG: ReconIII CreateTimeTAG: 2012-09-13T01:34:13Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 1141 TitleTAG: videos freezing all videos in week 3 freezing randomly - Windows 7 with Firefox 15.0.1 have to refresh each time it happens - typically only lasts about 2 min before it freezes and refresh is required. Please help UserIdTAG: 105732 UserNameTAG: nunchucksuka CreateTimeTAG: 2012-09-12T23:56:56Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 4 FirstChildTAG: same with Mac OS X Lion running Safari/Chrome. Latest available flash installed. You can watch the videos on youtube (click on the bottom left corner of video) where this problem is non existant. FirstChildUserIdTAG: 27851 FirstChildUserNameTAG: Arman FirstChildCreateTimeTAG: 2012-09-13T00:04:52Z FirstChildTAG: Yep...I think its happening to all. I have Win7 and Chrome. I am watching the videos form Week 1 (slacker!)...and they are crapping out. Nothing seems to help... FirstChildUserIdTAG: 25084 FirstChildUserNameTAG: waf102 FirstChildCreateTimeTAG: 2012-09-13T00:08:42Z FirstChildTAG: We're working on it. Thanks for the report. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-09-13T01:15:43Z FirstChildTAG: YouTube made a recent change to their Flash player that broke our video embeds. As a temporary workaround while we figure things out, you can use YouTube's HTML5 player. To do so, please go to YouTube's [Join the HTML5 Trial][1] page and click "Join the HTML5 Trial" at the bottom. Then reload our site in your browser. Thank you for your patience as we work on this. FirstChildUserIdTAG: 11 FirstChildUserNameTAG: dormsbee FirstChildCreateTimeTAG: 2012-09-13T02:27:18Z IndexTAG: 1142 TitleTAG: Real problem in my head Okay, the idea of making an EE playground was amazing and limiting the conditions of that playground was also cliche, but think about it for a second, that there will be many real life situations that would stay outside that playground, then how will we deal with those situations? UserIdTAG: 169784 UserNameTAG: thewiredbear CreateTimeTAG: 2012-09-12T20:05:27Z VoteTAG: 3 CoursewareTAG: Week 1 / KVL and KCL CommentableIdTAG: 6002x_kvl_and_kcl NumberOfReplyTAG: 3 FirstChildTAG: You are correct in that there are real-life situations which fall outside the "playground" so to speak, but the fantastic thing about the particular approximations we have made is, they are very good approximations for a large class of real-world systems. Typically if we're building a circuit in real-life lab, a term like the time rate of change of magnetic flux is an extremely small quantity, so we can carry out KVL as normal and get the expected results. But what happens if we do encounter a real-world system that doesn't behave as we expect? In general, what engineers do is figure out what assumptions in their model of reality might no longer hold, and then attempt to alleviate their model to more accurately describe the phenomena. Basically, they have to make a new "playground," one that works again. But as said before, the particular playground we've decided to limit ourselves to is a just fine approximation of reality in most circumstances. FirstChildUserIdTAG: 390034 FirstChildUserNameTAG: kahlil FirstChildCreateTimeTAG: 2012-09-12T20:53:16Z SecondChildTAG: thanks for the amazing reply, I would have been more confident reading your comment, had I been an engineer, I am still a high school student, still appreciate the effort, and I did get hold of your base point. SecondChildUserIdTAG: 169784 SecondChildUserNameTAG: thewiredbear SecondChildCreateTimeTAG: 2012-09-12T21:03:18Z FirstChildTAG: I guess we have to deal with the basics first. And then they will get to that point. FirstChildUserIdTAG: 308853 FirstChildUserNameTAG: fmorato FirstChildCreateTimeTAG: 2012-09-12T20:54:02Z SecondChildTAG: thanks for replying, though you also hold a strong point, still right now this all may seem easy to you, but its not THAT easy for a high school boy like me, even though I get the base of the course videos,.... I just am not able to solve some questions at all SecondChildUserIdTAG: 169784 SecondChildUserNameTAG: thewiredbear SecondChildCreateTimeTAG: 2012-09-12T21:05:07Z FirstChildTAG: @thewiredbear...Since u are a high school student, and say you have encountered a chapter in mathematics called FUNCTIONS, you will find that a particular function is say defined for only an interval.I mean we have got the REAL AXIS,say the x-axis containing all possible real numbers.right? Now thers a function ,for example, which only exists for ,say, the numbers from zero to infinity, or say (-2 to +2)...so then in that case (o to infinity) or (-2,2) are called intervals.Say the function exists only for that DOMAIN. Similarly, in LMD and in this EECS playground, we are restricting ourselves to a particular domain .Maybe later we may require and we might make extensions.And with time we will aim for that. When we go to that point, we will BUILD another set of equations or theories still TRYING TO make an EXTENSION of what we knew previously,so that all that we knew before, still exist...but for a certain boundary. FirstChildUserIdTAG: 156225 FirstChildUserNameTAG: sanjana_m FirstChildCreateTimeTAG: 2012-09-23T07:27:03Z IndexTAG: 1143 TitleTAG: Where does P=VI come from? Experimental or derived from V=IR? Quite possibly I missed it in the lectures or textbook, but I did not see a discussion of where the equation P = VI comes from. That is, can it be derived from V = IR or Maxwell's equestions? My understanding is that V = IR is basically an experimental observation (and not always true!) and can't be derived from Maxwell's equations. Please correct me if I'm wrong! Does P = VI fall into this category? Many Thanks, Dave UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-11T23:04:38Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: actually they took asum of values of the power that is dissipitad in arisistor and avalues of the current they found arelationship that P=i^2*R and since R=v/i then p=vi FirstChildUserIdTAG: 326507 FirstChildUserNameTAG: mohamed200 FirstChildCreateTimeTAG: 2012-09-11T23:31:48Z FirstChildTAG: we can derive it from basic laws.generally power is defined as work done/time and voltage is the work done per unit charge implies p=w/t,v=w/q,p=v*q/t,p=v*i as current i=q/t FirstChildUserIdTAG: 397298 FirstChildUserNameTAG: ZA90 FirstChildCreateTimeTAG: 2012-09-11T23:36:00Z SecondChildTAG: Thank you ZA90. That makes sense. ZA90 (or anyone else): Was this discussed in the lectures or book? I did not see it. -Dave SecondChildUserIdTAG: undefined SecondChildUserNameTAG: undefined SecondChildCreateTimeTAG: 2012-09-12T15:20:38Z IndexTAG: 1144 TitleTAG: How to enter my expressions in the Box? Requested by BAUWA and Owais001 Hi BAUWA and Owais001! I hope you are enjoying this Course as much as I did in the Prototype Course! Owais001, you told me in this [Post][1] that you were having difficulty in answering some Questions. You said that you got a message **"it can parse 3R"** when you entered 3*R in somewhere. Ok, this is because, if you have a product, you have to use an asterisc "*" between the two factors, that is to say, **if you have to do a product of a and b, you have to write in the box a*b and not ab**. So, if you have 3R, you should write: 3*R I hope that this can help you a little bit Owais001. You can also see more about this in this [answer][2]. ---- Ok, Now the turn of BAUWA! BAUWA, you told requested me helping you in H1P1. You wrote me: *h1p1- 1st que 3R OR R+R+R,AND HOW TO ANS CIRCUIT C QUE, MY ANS 5R/3is not being expected.pls help.* This is really similar of what I have answered to Owais001. Try to use asterisc * in the Product. Also, try to express with the minimun expression as possible. For example, an arbitrary example, if you have R+R , you should write 2*R and not 2R or R+R in the box.If you have a fraction and your variable, fraction*variable, you should enter in the box (fraction)*variable, eg., (2/4)*R .Try it ;). You can do it! ---- [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/MITx_6_002x_2012_Fall/threads/50474faf685941270000000d [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_S2E2_Associated_Reference_Directions/threads/5047d8853c82c1230000002c#5047e841715ee6270000001b UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2012-09-11T22:13:50Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 1145 TitleTAG: Attention Please dont try to type A+B=C; You need do type A+B UserIdTAG: 205311 UserNameTAG: Sergtronix CreateTimeTAG: 2012-09-11T19:54:38Z VoteTAG: 3 CoursewareTAG: Week 1 / Node analysis practice, part 3 CommentableIdTAG: 6002x_L2Node2 NumberOfReplyTAG: 1 FirstChildTAG: Thnx. FirstChildUserIdTAG: 401175 FirstChildUserNameTAG: Raghav12 FirstChildCreateTimeTAG: 2012-09-14T15:04:07Z IndexTAG: 1146 TitleTAG: [FIX] Transient analysis If anybody having problems with it - use this sandbox https://6002x.mitx.mit.edu/courseware/6.002_Spring_2012/Overview/Circuit_Sandbox/ Good luck! ;) P.S. For time input use "m" for milliseconds. Ex: 10m, 20m (not ms) UserIdTAG: 295103 UserNameTAG: Syavick CreateTimeTAG: 2012-09-11T08:10:51Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: Thanks so much for the link! It was the only way to use the Transient Analysis. Now I can continue with the class! Cheers! FirstChildUserIdTAG: 354910 FirstChildUserNameTAG: Lee_Sosuley FirstChildCreateTimeTAG: 2012-09-11T18:48:47Z FirstChildTAG: Did not work for me. The transient analysis window opens, but no graph is displayed. Dragging the mouse across shows a dotted line with 'NaN' displayed for the value. Ugh. it helps to enter the time period without the 's' for seconds. Now it works. FirstChildUserIdTAG: 29468 FirstChildUserNameTAG: randallroman FirstChildCreateTimeTAG: 2012-09-14T05:59:13Z SecondChildTAG: I'm also having trouble displaying the graph. No waveform is plotted, and it almost seems like the curve is the same color as the background, therefore I can't see the curve. I checked and I didn't use an 'S' when entering the time for a transit analysis. I ran a transit analysis on the sandbox homework, and it worked fine. Any ideas why this is happening? SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-09-19T06:34:05Z SecondChildTAG: I found my problem. I was evaluating over 1 second, so analyzing a sine wave with a period in the milliseconds for 1 second creates so many oscillations that the screen is filled with one color, the color of the probe. It is like having the wrong record length set on a scope, and in order to see the waveform the time scale needs to be changed. I set the time duration to 5ms (5m for the simulator) and I saw the waveform. Thank you Syavick. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-09-20T19:36:39Z SecondChildTAG: You are welcome! ;) SecondChildUserIdTAG: 295103 SecondChildUserNameTAG: Syavick SecondChildCreateTimeTAG: 2012-09-22T23:11:58Z IndexTAG: 1147 TitleTAG: Voltage increase at 3:06 It is a bit hard to see what happens during this increase. Perhaps by having the camera pointing at the scope prior to the increase, would help. UserIdTAG: 378522 UserNameTAG: Alejo_Velasquez CreateTimeTAG: 2012-09-11T03:07:11Z VoteTAG: 3 CoursewareTAG: Week 2 / Superposition CommentableIdTAG: 6002x_superposition NumberOfReplyTAG: 0 IndexTAG: 1148 TitleTAG: How to: set the bottom of the diagram as the ground Set the top as the node potential - this happens to be equal to V3. UserIdTAG: 253630 UserNameTAG: PhillipAdkins CreateTimeTAG: 2012-09-11T02:54:41Z VoteTAG: 3 CoursewareTAG: Week 2 / Linear combinations of source strenghts CommentableIdTAG: 6002x_S3E2_Linear_Combinations_of_Source_Strengths NumberOfReplyTAG: 1 FirstChildTAG: @Philip Adkins Your post just happened to be a real help for me... FirstChildUserIdTAG: 118462 FirstChildUserNameTAG: Naino FirstChildCreateTimeTAG: 2012-09-17T14:19:28Z IndexTAG: 1149 TitleTAG: A matter of conventions? So, the convention is that currents flow from positve to negative terminals and therefore, since this is the opposite case, the current is negative? UserIdTAG: 256249 UserNameTAG: And90 CreateTimeTAG: 2012-09-10T23:24:41Z VoteTAG: 3 CoursewareTAG: Week 1 / Power CommentableIdTAG: 6002x_S1E2_Power NumberOfReplyTAG: 3 FirstChildTAG: > So, the convention is that currents flow from positve to negative > terminals Yes, this is known as the *Associated Variables Convention*. > and therefore, since this is the opposite case, the current is > negative? To the best of my knowledge, it depends on the actual direction of the current flow. See Sect. 1.5.3 of the text. FirstChildUserIdTAG: 194509 FirstChildUserNameTAG: blackcompe FirstChildCreateTimeTAG: 2012-09-11T01:14:18Z FirstChildTAG: Only a matter of energy flow. The question refers to the *incoming current toward the network N* from the resistor, but because **the power dissipated in the resistor is positive**, it can not deliver power to the grid, ie, the current (and power) flows the resistor, or what is the same, the current "enter" to the network N *is negative* ... FirstChildUserIdTAG: 129288 FirstChildUserNameTAG: Tinchito FirstChildCreateTimeTAG: 2012-09-11T16:47:49Z FirstChildTAG: In Germany we've learned to handle this kind of confusing conventions like this: There is an universal convention (like "blackcompe" said it may be called *Associated Variables Convention*) saying that current always flows from plus to minus - we are calling this *Technical Current Flow Direction*. I think someone said, that this was determined right at the beginning of exploring electricity before they discovered that the current actually flows in the other direction (so they lost their 50/50-chance), meaning the electrons are going to the spot with less negative potential and leaving the spot with high negative potential (current flows from minus to plus). This is called *Physical Current Flow Direction* (now you can forget about this, it's always the first one). Anyway, **all** the schematics and circuits are using the first convention, where current is supposed to be **always** flowing from plus to minus. In this particular question, the direction of the current is inversed - it goes obviously from minus to plus. But we have to abide the convention and therefore we say the current is negative (or has been multiplied with -1). Don't try to put "negative current" in a logical scale or chart in your head as an opposit of the "normal current". "Positive" or "negative" just indicates the **direction** (maybe it's better not to use these two words in this context at all). Is the current flowing according to the convention? Fine, then multiply with +1 (or just leave it as it is). Is the direction inverse to the convention? Then multiply with -1. The reason we have to do this is because it's important later on, when we (I assume) come to more complex network analysis. Then you have to point out which elements are dissipating power and which are applying power to the circuit (defining loads and generators). With lumped elements you have to pay attention to that and with the right current prefix, there will be no problems in doing so. FirstChildUserIdTAG: 394599 FirstChildUserNameTAG: Albie FirstChildCreateTimeTAG: 2012-09-11T13:48:57Z SecondChildTAG: I just want to add, that this also counts for the voltage. If a voltage becomes negative, then it just switches terminals. The absolute value of the voltage can be interpreted as usual. SecondChildUserIdTAG: 394599 SecondChildUserNameTAG: Albie SecondChildCreateTimeTAG: 2012-09-11T14:03:19Z SecondChildTAG: Thanks for the explanation, specially the Technical and Physical Current Flow part. Danke schön :) SecondChildUserIdTAG: 256249 SecondChildUserNameTAG: And90 SecondChildCreateTimeTAG: 2012-09-12T02:24:25Z SecondChildTAG: I think this clear now SecondChildUserIdTAG: 427173 SecondChildUserNameTAG: macupanda SecondChildCreateTimeTAG: 2012-09-13T09:04:04Z SecondChildTAG: Direction of current is the direction of flow of '**hole**'s as they would say. One thing still is confusing. Since, the power supply is *not* to be seen outside, it must be somewhere in network N. Now, the voltage across R could be like that only if a battery (say) was placed with its (+)ve terminal up. In that case, current is seen to be heading towards the +ve direction. It's illusory. SecondChildUserIdTAG: 285616 SecondChildUserNameTAG: AmiyaX SecondChildCreateTimeTAG: 2012-09-15T13:14:39Z SecondChildTAG: Thanks for the explanation SecondChildUserIdTAG: 483990 SecondChildUserNameTAG: RobertoGea SecondChildCreateTimeTAG: 2012-09-23T23:44:51Z IndexTAG: 1150 TitleTAG: Q 3 I first assumed current through branch which is i1=e1/R1=(e3+V)/R1 apply KCL for node e3 i.e.e3/R2+I+(e3+V)/R1=0 then i got e3=-8.75 UserIdTAG: 247933 UserNameTAG: shalinidug CreateTimeTAG: 2012-09-10T21:45:05Z VoteTAG: 3 CoursewareTAG: Week 1 / Node method example CommentableIdTAG: 6002x_S2E5_Node_Method NumberOfReplyTAG: 1 FirstChildTAG: Thank you! FirstChildUserIdTAG: 312043 FirstChildUserNameTAG: Osmar17 FirstChildCreateTimeTAG: 2012-09-16T16:46:13Z IndexTAG: 1151 TitleTAG: the correct answer you have to choose the currents out the node e for the current of the voltage source one (e-V1)/R1 for the current of the voltage source two (e-v2)/R2 AND THE SIGN OF V2 SHOULD BE positive because the current is on the opposite direction with the voltage it is very easy good luck. UserIdTAG: 266386 UserNameTAG: zakzak200 CreateTimeTAG: 2012-09-10T21:10:14Z VoteTAG: 3 CoursewareTAG: Week 2 / Node equation review CommentableIdTAG: 6002x_S3E1_NodeEquationReview NumberOfReplyTAG: 0 IndexTAG: 1152 TitleTAG: **Current in DC Analysis** Current flowing into the circuit after analysis shows 500 microamps but answers show 500e-6, **why is that?** UserIdTAG: 176701 UserNameTAG: raashid_khan CreateTimeTAG: 2012-09-10T16:21:08Z VoteTAG: 3 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 3 FirstChildTAG: Change the resistor values to k ohms, i.e. 1 kilo ohm instead of just 1 ohm. FirstChildUserIdTAG: 263488 FirstChildUserNameTAG: tdejongh FirstChildCreateTimeTAG: 2012-09-14T02:08:19Z FirstChildTAG: This is scientific notation. The "E" always stands for "times ten to the power of". So 500e-6 is 500 times 10 to the power of -6, i.e. 500 * 10^(-6) or 500 micro${unit}. FirstChildUserIdTAG: 144694 FirstChildUserNameTAG: johndoe31415 FirstChildCreateTimeTAG: 2012-09-10T16:29:47Z SecondChildTAG: I got every other question correct, but I found this very confusing. The diagram shows -500uA. If I enter -0.0005 shouldn't that be correct? SecondChildUserIdTAG: 42528 SecondChildUserNameTAG: Pauly SecondChildCreateTimeTAG: 2012-09-11T17:00:27Z FirstChildTAG: e-6 means 10 power to -6 that is micro unit FirstChildUserIdTAG: 387473 FirstChildUserNameTAG: sridharjaya FirstChildCreateTimeTAG: 2012-09-13T06:51:21Z IndexTAG: 1153 TitleTAG: ADMINSTRIVIA Please I Wanna Know the meaning of "**ADMINSTRIVIA**"?? UserIdTAG: 36823 UserNameTAG: Hesham CreateTimeTAG: 2012-09-10T15:09:52Z VoteTAG: 3 CoursewareTAG: Week 1 / Course overview CommentableIdTAG: 6002x_S1V1toS1V4 NumberOfReplyTAG: 1 FirstChildTAG: Matters of administration, such as registering and understanding how the course works. FirstChildUserIdTAG: 193033 FirstChildUserNameTAG: NIslam FirstChildCreateTimeTAG: 2012-09-10T15:35:37Z IndexTAG: 1154 TitleTAG: Lab - 1 PROBLEM My LAB - 1 Circuit simulator requires me to run a DC analysis, but there is only a resistor in the component bin!! No capacitor, gnd, voltage source, nothing. Is this a problem anyone else is experiencing? UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-09-10T15:09:31Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 9 FirstChildTAG: If you have saved your work, try clicking in reset at the bottom. It worked for me, they added the probe and the GND. FirstChildUserIdTAG: 302490 FirstChildUserNameTAG: carlosramirez FirstChildCreateTimeTAG: 2012-09-14T03:26:49Z SecondChildTAG: P.S: I'm using Firefox 15.0.1 in Windows 7 SecondChildUserIdTAG: 302490 SecondChildUserNameTAG: carlosramirez SecondChildCreateTimeTAG: 2012-09-14T03:28:18Z SecondChildTAG: I only saw the resistor too, also how do you place the resistor so that it is horizontal? SecondChildUserIdTAG: 264614 SecondChildUserNameTAG: NDziva SecondChildCreateTimeTAG: 2012-09-17T03:03:42Z FirstChildTAG: Hi hazel1919! Hint Lab1: You have to try to solve this Lab with that component ;). It is not a Bug. I hope this little hint can help you. ---- Hola hazel1919! Aqué te doy una pequeña Pista del Lab1: Tienes que tratar de resolver este Lab con aquél componente ;). No es un Bug. Espero que esto te haya sido de ayuda. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-10T15:56:25Z FirstChildTAG: I'm having the same problem. If I find any answer I'll tell you FirstChildUserIdTAG: 302490 FirstChildUserNameTAG: carlosramirez FirstChildCreateTimeTAG: 2012-09-10T16:09:52Z FirstChildTAG: I have that same problem too! I´m using mozilla firefox 15.0.1 , should I use google chrome instead? FirstChildUserIdTAG: 79854 FirstChildUserNameTAG: AlbertoGT FirstChildCreateTimeTAG: 2012-09-10T18:58:12Z SecondChildTAG: Ya solucionaron el problema, dale click al boton reset abajo de la pagina y te carga de nuevo un circuito con el tester y la puesta a tierra SecondChildUserIdTAG: 302490 SecondChildUserNameTAG: carlosramirez SecondChildCreateTimeTAG: 2012-09-14T03:30:50Z FirstChildTAG: It's ok. You have to make this just with resistors. Look what the problem is requesting. FirstChildUserIdTAG: 300169 FirstChildUserNameTAG: castrogfx FirstChildCreateTimeTAG: 2012-09-10T19:01:28Z SecondChildTAG: Yes, but i can't do a DC analysis because i don't have a Voltage probe either a gnd SecondChildUserIdTAG: 302490 SecondChildUserNameTAG: carlosramirez SecondChildCreateTimeTAG: 2012-09-10T20:18:50Z FirstChildTAG: Try using a different browser. It was the same for me in FF, but when I switched to Chrome I could see the already-made parts of the lab. FirstChildUserIdTAG: 267993 FirstChildUserNameTAG: mavlijas FirstChildCreateTimeTAG: 2012-09-10T20:49:40Z FirstChildTAG: I can confirm as well... lab did not work on FF (15), had to switch to Chrome to complete. Ugh. FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-11T07:10:47Z SecondChildTAG: i'am using ff 15 and with that i completed my lab. SecondChildUserIdTAG: 378967 SecondChildUserNameTAG: krish2012 SecondChildCreateTimeTAG: 2012-09-11T08:08:31Z SecondChildTAG: lab 01![enter image description here][1] [1]: http:// SecondChildUserIdTAG: 378967 SecondChildUserNameTAG: krish2012 SecondChildCreateTimeTAG: 2012-09-11T08:22:44Z FirstChildTAG: hey don't worry about component missing.in the lab they have given only the component that required to solve problem. you are asked to solve the problem with resistor network, not with capacitor or inductor. for that DC analysis is helpful.place the resistor,connect the circuit and run the DC analysis.you will get node voltage.try to make it 2v at node A by changing the resistor value or by adding some more resistor.i think Lab 1 is the easiest one to solve. FirstChildUserIdTAG: 378967 FirstChildUserNameTAG: krish2012 FirstChildCreateTimeTAG: 2012-09-11T04:06:35Z FirstChildTAG: i got the answer by only using the given components.no need for any extra component.in the first chapter we have only learned about resistor network.so they have given only resistor. FirstChildUserIdTAG: 378967 FirstChildUserNameTAG: krish2012 FirstChildCreateTimeTAG: 2012-09-11T04:10:09Z IndexTAG: 1155 TitleTAG: H2P1 - wrong answers counted in Progress? For H2P1, I submitted my answers (well, several, but I'm pretty sure the current ones are right) and it never marked them as correct. However, the progress bar is counting them as completed. Help? UserIdTAG: 267993 UserNameTAG: mavlijas CreateTimeTAG: 2012-09-10T07:49:17Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 4 FirstChildTAG: i am using chrome and the same thing is happening to me.. FirstChildUserIdTAG: 371000 FirstChildUserNameTAG: Joseph090892 FirstChildCreateTimeTAG: 2012-09-12T20:24:20Z FirstChildTAG: try to correct it as soon as possible(before the due date). only your correct answers is counted for progress indication FirstChildUserIdTAG: 378967 FirstChildUserNameTAG: krish2012 FirstChildCreateTimeTAG: 2012-09-10T10:50:41Z SecondChildTAG: You're not getting me - the progress counter is showing 100%. But on the homework page, the answers are marked as wrong. Which is right? SecondChildUserIdTAG: 267993 SecondChildUserNameTAG: mavlijas SecondChildCreateTimeTAG: 2012-09-10T16:20:45Z FirstChildTAG: I have the same problem!!! FirstChildUserIdTAG: 244841 FirstChildUserNameTAG: eyubero FirstChildCreateTimeTAG: 2012-09-11T16:51:05Z FirstChildTAG: Tried Firefox and Chrome, same thing. FirstChildUserIdTAG: 267993 FirstChildUserNameTAG: mavlijas FirstChildCreateTimeTAG: 2012-09-11T18:47:43Z IndexTAG: 1156 TitleTAG: This convention Seems silly to me. I've seen circuits taught many taught many times with a different (and all the same) convention - in many different classes from high school to college. I teach this myself. This convention seems odd. I think it's easier to just think that the voltage drops across a resistor and bumps up across a voltage source. I'd put the current in the direction of the increase in voltage around the voltage source and in arbitrary or obviously "good" directions around resistors. I understand why the instructor chose this - kinda - probably because it seems more platonically correct to him in some way, but I don't know if it's very intuitive. UserIdTAG: 253630 UserNameTAG: PhillipAdkins CreateTimeTAG: 2012-09-09T22:42:29Z VoteTAG: 3 CoursewareTAG: Week 1 / Associated reference directions CommentableIdTAG: 6002x_S2E2_Associated_Reference_Directions NumberOfReplyTAG: 2 FirstChildTAG: The problem is that when the circuit has numerous elements and sources it might be hard to determine the polarities of all voltages and the direction of all currents before actually doing the calculations. Although in this simple exercise it might be obvious, it won't be in future exercises. It is a good idea to understand what a negative voltage or current actually means because they are very common in circuit analysis (when using KVL and KCL for example) FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-09T22:53:16Z FirstChildTAG: That's also how I was taught. In this case the voltage bump is positive leaving the positive end of the voltage source, right? And the power is then positive as well? FirstChildUserIdTAG: 357225 FirstChildUserNameTAG: MJBoa FirstChildCreateTimeTAG: 2012-09-10T09:04:26Z IndexTAG: 1157 TitleTAG: Device's assembly Thumbs UP for placing capacitors on the beginning of the device assembly - not after hours of debugging :) UserIdTAG: 405237 UserNameTAG: MarasK CreateTimeTAG: 2012-09-09T21:18:09Z VoteTAG: 3 CoursewareTAG: Week 2 / Cardiac Experiment CommentableIdTAG: 6002x_cardiac NumberOfReplyTAG: 1 FirstChildTAG: Also, it's better to check hardware before 3 days of debugging firmware ) FirstChildUserIdTAG: 394836 FirstChildUserNameTAG: v2g6ch4 FirstChildCreateTimeTAG: 2012-09-10T00:20:34Z IndexTAG: 1158 TitleTAG: lab 2 ( signal mixer) Okay, so I'm stuck here because i have got the right ratios of the resistance values to be used by applying some simple logic but got the same result no matter even if i used 1ohm and 3 ohm or 1kohm and 3kohm UserIdTAG: 135072 UserNameTAG: Shriniwas CreateTimeTAG: 2012-09-09T19:56:54Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: The result depends only on the ratio of the resistances so that 1 ohm and 3 ohm will give the same result than 1 kohm and 3 kohm. However, using a 1 to 3 ratio resistors will give an answer Vo = (1/4)*V1 + (3/4)*V2... can you see why? Notice that the coefficients in the question doesn't add to 1. Can that be achieved using only two resistors? Can you think of another configuration to achieve this? FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-09T21:03:10Z SecondChildTAG: Thanks for the tip! That one took a while, but getting the correct answer was oh so sweet. ^_^ SecondChildUserIdTAG: 267993 SecondChildUserNameTAG: mavlijas SecondChildCreateTimeTAG: 2012-09-10T05:54:03Z SecondChildTAG: I cant seem to get this right, I used parallel and series resistors, for V1 I had the parallel of R2 and R3 to be equivalent to that of R1 so that it comes V1/2, but I cant get V2/6. Please help. SecondChildUserIdTAG: undefined SecondChildUserNameTAG: undefined SecondChildCreateTimeTAG: 2012-09-10T16:21:50Z IndexTAG: 1159 TitleTAG: Circuit Simulators Hello All, In case you haven't found them yet, there are a few other circuit simulators out there for you to use if you like. Some are free, some are not. iCircuit for Mac OSX and Ipad, not free. Works great on iPad, not so great on iTouch. The screen is too small. http://icircuitapp.com/ CircuitLab, online, crossplatform, free. Developed by guys who also took this course(at the MIT campus). https://www.circuitlab.com/ LTSPICE, Windows, Free. Most professional simulators are based on the SPICE simulator, so this is a free way to engage with a pro system. http://www.linear.com/designtools/software/ Logic.ly, Digital Logic Circuits, has free online demo which works great and also paid app for Windows and OS X. http://logic.ly/ There are lots more out there, but these seem to be on everyone's list. I've used all of these and they all work well. In case you are wondering, no I don't have any financial interest in any of these. And yes, they are all legal to use on your homework/midterms/finals. I don't know if they will be ok to use with the proctored exams but they were allowed in the last class. UserIdTAG: 21541 UserNameTAG: JSChambers CreateTimeTAG: 2012-09-09T15:30:37Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: thanx,it ws indeed useful ..i used circuit lab site to do my lab1 simulation. .but plz tell me if the work i did & saved in my circuit lab profile be copied or linked to the lab1 worksheet? because, i am not being able to complete my diagram in the lab1 schematic sheet provided in the courseware,as just a resistor component alone is available in the toolbox to the right of the schematic sheet. .what is to be done? FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-09T16:45:40Z IndexTAG: 1160 TitleTAG: loops are in clock wise and anticlockwise why you take a loop1 and loop3 are in clock wise direction and loop2 is in anticlockwise direction? UserIdTAG: 276803 UserNameTAG: srinivasaram CreateTimeTAG: 2012-09-09T13:34:59Z VoteTAG: 3 CoursewareTAG: Week 1 / KVL and KCL CommentableIdTAG: 6002x_kvl_and_kcl NumberOfReplyTAG: 4 FirstChildTAG: to make them all in the same direction you change the sign for example L1 and L3 have the same direction but L3 not then to find L4 (depend on the three KVL) : L4 = L1 - L2 + L3 FirstChildUserIdTAG: 236136 FirstChildUserNameTAG: Hamid-ch FirstChildCreateTimeTAG: 2012-09-10T00:18:47Z FirstChildTAG: hi friend,"It's not necessary that you should take loop 1 and loop 3 in clock wise or loop 2 in anticlockwise,you can take all in clockwise or all in anti-clockwise but what matters is solving the equations correctly and get correct outputs that's it". FirstChildUserIdTAG: 396562 FirstChildUserNameTAG: nikky94 FirstChildCreateTimeTAG: 2012-09-09T13:49:53Z SecondChildTAG: what really matters is observing the sign you assign to each loop as you move SecondChildUserIdTAG: 336466 SecondChildUserNameTAG: kenstan SecondChildCreateTimeTAG: 2012-09-09T14:42:51Z SecondChildTAG: The sign of results at the end will tell you is the sense of the voltage inverts SecondChildUserIdTAG: 231676 SecondChildUserNameTAG: Roosemberth SecondChildCreateTimeTAG: 2012-09-10T10:57:56Z FirstChildTAG: you can assume the direction of a loop to solve for the current. if you assigned loop 1 and 3 clock wise, but your answer gives you a (-) current, then you know that in fact, the current is flowing in the opposite direction than what you assigned to the loop FirstChildUserIdTAG: 154366 FirstChildUserNameTAG: frankyalfaro FirstChildCreateTimeTAG: 2012-09-10T02:29:17Z FirstChildTAG: still i cant understand plz some one help me how can take the direction of loop either positive side or negative side ? FirstChildUserIdTAG: 431408 FirstChildUserNameTAG: sathishkumar FirstChildCreateTimeTAG: 2012-09-15T15:26:38Z SecondChildTAG: The direction of the loop an either be in clockwise or in anticlockwise direction. It does not matter. What matters is the "sign convention" that we use while writing down the equations for KVL and KCL using the directions. Here you can use a simple trick, Starting from the 1st loop, that has a power source, consider the direction of the current as the power source yields it. i.e. positive to negative. Now taking this direction into account, put an arbitrary sign on + or - on the nodes it has. Then for the next loop, consider those signs as base and put the current directions in the loop. Ok? SecondChildUserIdTAG: 453281 SecondChildUserNameTAG: chaiku SecondChildCreateTimeTAG: 2012-09-23T15:37:33Z IndexTAG: 1161 TitleTAG: Lumped Element ![Lumpy!][1] Why on earth are electrical components called 'lumped elements'? Why not electrical components or even better, 'thingies' ?? I'm just asking because I'm curious. [1]: http://84.imagebam.com/download/VuQ_cIzm9X4nREAOTZKfcw/20982/209819592/lumpy.png UserIdTAG: 193033 UserNameTAG: NIslam CreateTimeTAG: 2012-09-08T16:34:23Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 3 FirstChildTAG: Yes, yes. I like your drawings very much. As you delve further into electrical engineering and practice, you'll find that there are a lot of terms that are roughly analogous to "lumped element." When folks talk about "electrical components," it's often implied that the component is a lumped element. The contrast to lumped elements would be ["distributed elements"][1], which is just slightly beyond the scope of 6.002x. I think "thingy" is too informal :) [1]: http://en.wikipedia.org/wiki/Distributed_element_model FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-08T16:40:45Z SecondChildTAG: I guess so...i mean resistors look pretty lumpy! :) SecondChildUserIdTAG: 193033 SecondChildUserNameTAG: NIslam SecondChildCreateTimeTAG: 2012-09-08T17:36:12Z FirstChildTAG: What method did you use to copy your picture into this discussion forum? Regarding "lumped elements", page 5 and 6 of the text book has a good explanation. A good example is with a light bulb, if you were to use Maxwell's equations to analyze all the physical properties of the internal workings of the light bulb, it would be very complicated. It would take into consideration what metal the filament was made of (titanium etc), the resistivity, the length of wire and what medium the filament was in inside the light bulb etc. Instead of doing this, lump all those properties into one element, the resistor, as this is a good model to represent all the properties inside the light bulb. In other words, lump all the properties together and represent them with a resistor. I hope this helps. Thanks. FirstChildUserIdTAG: 132828 FirstChildUserNameTAG: rharris FirstChildCreateTimeTAG: 2012-09-08T17:25:23Z SecondChildTAG: I simply uploaded it to an image host, and then just pasted the URL. And thank you for the explanation. :) SecondChildUserIdTAG: 193033 SecondChildUserNameTAG: NIslam SecondChildCreateTimeTAG: 2012-09-09T06:35:53Z FirstChildTAG: Hello, Nlslam... Cute drawing. Nice change of pace. I think it's lumpy because it's not distributed. FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-09-08T22:23:38Z SecondChildTAG: I was just thinking about this, and I realized that in a circuit, all of a sudden, you have these lumpy things like a bulb and a resistor popping up, connected to it...so they are just lumpy things in the middle of the circuit. That is why, our great forefathers, decided that it would be a good idea to call these things lumped elements...well thats the explanation that I came up with anyways! :P SecondChildUserIdTAG: 193033 SecondChildUserNameTAG: NIslam SecondChildCreateTimeTAG: 2012-09-09T05:51:39Z IndexTAG: 1162 TitleTAG: Clearing the basics of how current flows ![A picture of the problem][1] I have always thought of electrons as little men who carry sacs of energy. But I know that, this analogy is overly simplified. What exactly happens after current has flown through a resistor? And this idea of flowing, is it really accurate, how does current really behave in a circuit? Thanx in advance. :) EDIT : I was hoping to know the other theories behind how current flows, because sacs of little men carrying energy just does not work under difficult constraints. I understand what you are saying, but I wanted to know, if since there is a potential difference between two ends, would there potentially be an electromagnetic force within the wire, and if so, there should not no current at all, the thing that we call current should be 'always present'. So, is it more like little men passing sacs of energy among them? And if so, why does the resistor (lets say a bulb), light up instantly, if we used say a longer wire, would it take (a little) more time for the bulb to light up? You've explained how current flows, I want to understand how current really behaves. [1]: http://64.imagebam.com/download/uCxZk8MO9W4ZK3CZtQLbEQ/20982/209815768/What%20exactly%20happens1.png UserIdTAG: 193033 UserNameTAG: NIslam CreateTimeTAG: 2012-09-08T16:00:30Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: Hi, your drawing style is very cool. You should do more of them :) To answer your question, if your voltage source has voltage $V$ and the resistor has resistance $R$, then there will be a current of $I=V/R$ *throughout* the circuit. This means that the current $I$ is flowing clockwise in the top part of the wire ("before" the resistor), and that the same current $I$ is flowing clockwise in the bottom part of the wire ("after" the resistor). Overall, in your simple diagram, the current that leaves the positive terminal of the voltage source is returned to the negative terminal of the voltage source. Another approach is to note that the resistor has two terminals: "before" and "after". The current that enters through one terminal must exit through the other. This is part of the "lumped matter abstraction." FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-08T16:10:03Z SecondChildTAG: Thank you for complementing my drawing, I'll use more colors next time. *grin* But, I was hoping to know the other theories behind how current flows, because sacs of little men carrying energy just does not work under difficult constraints. I understand what you are saying, but I wanted to know, if since there is a potential difference between two ends, would there potentially be an electromagnetic force within the wire, and if so, there should not no current at all, the thing that we call current should be 'always present'. So, is it more like little men passing sacs of energy among them? And if so, why does the resistor (lets say a bulb), light up instantly, if we used say a longer wire, would it take (a little) more time for the bulb to light up? You've explained how current flows, I want to understand how current really behaves. -- Kind Regards as always. SecondChildUserIdTAG: 193033 SecondChildUserNameTAG: NIslam SecondChildCreateTimeTAG: 2012-09-08T16:25:13Z SecondChildTAG: And btw...how did you use latex?? Those Vs and Is look cool! :P SecondChildUserIdTAG: 193033 SecondChildUserNameTAG: NIslam SecondChildCreateTimeTAG: 2012-09-08T16:27:12Z FirstChildTAG: The electrons "flow" in the opposite direction of the direction of the current flow. It is accepted that the direction of the electric current has the direction of the positive charges. FirstChildUserIdTAG: 207456 FirstChildUserNameTAG: Octomos FirstChildCreateTimeTAG: 2012-09-08T16:13:39Z IndexTAG: 1163 TitleTAG: S2E6: Ans given: distance from the barn 113 feeet 12 AWG copper wire have resistance of 1.588 ohm per 1000 feet therefore, 12 Awg copper wire have resis of 1.588 /1000 per feet total distance = 113, thus tot rest = 1.588/1000 * 113 =0.17944 ohm since, circuit requires 2 lengths of the wire, thus total resis=0.17944*2=0.358 2) p=vi=v^2/R, 1000*R=v^2, v=240, R=57.6 3) i=v/Rt= 670.391, p=vi, v=1000/670.391=1.491 UserIdTAG: 132474 UserNameTAG: m_abid CreateTimeTAG: 2012-09-08T15:43:55Z VoteTAG: 3 CoursewareTAG: Week 1 / Power delivery modeling CommentableIdTAG: 6002x_S2E6_Modeling NumberOfReplyTAG: 1 FirstChildTAG: It gives correct answer for voltage drop, but I don't get why. Because by calculating current this way there is no resistance in the barn. I don't get it. FirstChildUserIdTAG: 406479 FirstChildUserNameTAG: alonzo07 FirstChildCreateTimeTAG: 2012-09-11T21:00:34Z IndexTAG: 1164 TitleTAG: regarding Practice Questions and "check" Hi.. I have done my homework. Do i need to do practice questions as well? and can we use "check" any number of times or is there any limit? does it affect my progress if i use it more than that limit? UserIdTAG: 242676 UserNameTAG: LinjharaSahil CreateTimeTAG: 2012-09-08T14:20:52Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Hi Linjhara, Only the homework and lab sections are for credit, as you can see in the progress bars under the "Progress" tab. The practice questions are **not** for credit, and are there for your benefit only. You can "Check" homework and lab problems as many times as you want without limit. :) FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-08T15:11:13Z IndexTAG: 1165 TitleTAG: node analysis practice part-3 it should be written as....find the kcl eqn. at node e2, since we can choose which branch current should enter or leave,except if current directions are already given..like I1. UserIdTAG: 153869 UserNameTAG: rabindra CreateTimeTAG: 2012-09-08T12:36:07Z VoteTAG: 3 CoursewareTAG: Week 1 / Node analysis practice, part 3 CommentableIdTAG: 6002x_L2Node2 NumberOfReplyTAG: 1 FirstChildTAG: Yes very true. The question has been poorly written....not very clear. Either the directions should be given or the question should be differently written. FirstChildUserIdTAG: 314624 FirstChildUserNameTAG: Owais001 FirstChildCreateTimeTAG: 2012-09-16T19:07:09Z IndexTAG: 1166 TitleTAG: Question about the certificate Hey everyone! Really excited about this class and thank you to all those who made this possible!!! On the syllabus it says we can get an honor-code certificate. I was wondering if it is from edx or the schools providing the classes. Thanks UserIdTAG: 247224 UserNameTAG: Engr_Girl CreateTimeTAG: 2012-09-08T12:30:09Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: You will simply get a certificate that you can print out from your home. It will have the course title, your grade and the University offering it. For example, MIT will give you the certificate under MITx FirstChildUserIdTAG: 193033 FirstChildUserNameTAG: NIslam FirstChildCreateTimeTAG: 2012-09-08T15:00:07Z FirstChildTAG: From what I understand of the FAQs, it's going to be a little of both: "Online learners who demonstrate mastery of subjects can earn a certificate of completion. Certificates will be issued by edX under the name of the underlying "X University" from where the course originated, i.e. HarvardX, MITx or BerkeleyX." FirstChildUserIdTAG: 333348 FirstChildUserNameTAG: easherly FirstChildCreateTimeTAG: 2012-09-08T14:11:59Z IndexTAG: 1167 TitleTAG: Homework and Labs What is the procedure to 'submit' the homework and labs? I have completed all of them and have got green tick marks next to the answers... now what do I do? UserIdTAG: 135072 UserNameTAG: Shriniwas CreateTimeTAG: 2012-09-08T07:12:14Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: I thiiiink it auto-submits. Check your 'progress' tab and see if it's given you credit to be sure, though. FirstChildUserIdTAG: 270160 FirstChildUserNameTAG: Sethhhh FirstChildCreateTimeTAG: 2012-09-08T07:34:49Z FirstChildTAG: Hi Shriniwas! The Check Buttom has 3 functions at the same time: Save+ Check +Submit ;). So, you don't need to do anything if you already clicked on Check and everything was correct. You can also see in the tab Progress. There you will notice your own progress. So, if you already have done all the Hw1, 2 and Lab 1 and 2. You should see there 100% in the Graph. Congratulations! ;) ---- Hola Shriniwas! El botón Check posee 3 funciones, es decir, cuando haces clic sobre él hace tres cosas a la vez: Guarda+Corrige+Envía ;). Es decir, que si ya has hecho clic en Check y todo ha salido correcto, no debes hacer nada extra para enviar/submit tu tarea. También si te sirve de ayuda, puedes observar que arriba de todo hay un tab de Progreso. Allí podrás, sin ser redundantes, ver tu progreso. Entonces, si ya has hecho todo el Hw1,2 y Lab1 y 2. Te debería aparecer en la gráfica un 100% en cada caso. Felicitaciones! ;) FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-08T19:34:23Z SecondChildTAG: Thank you so much for the reply :) Got it :) SecondChildUserIdTAG: 135072 SecondChildUserNameTAG: Shriniwas SecondChildCreateTimeTAG: 2012-09-09T19:53:44Z IndexTAG: 1168 TitleTAG: Theory vs reality I loved the Pickle Test, of course it has resistor characteristics, capacitance properties will work too :). EE will focus on functional stuff and practical solutions. But we will allways need a return to Theory just until our Paradigms change. UserIdTAG: 124360 UserNameTAG: nlama CreateTimeTAG: 2012-09-08T00:46:08Z VoteTAG: 3 CoursewareTAG: Week 1 / Lumped Element Abstraction CommentableIdTAG: 6002x_lumped_element_abstraction NumberOfReplyTAG: 0 IndexTAG: 1169 TitleTAG: Whats wrong? Dissipated power is more then the supplied! There is some contradiction in this problem. According to the correct answers shown, it appears that dissipated power is more then the supplied. The correct answers display positive values for power dissipated from R1 and R2. Also power supplied by the current source is positive value while the power supplied by the voltage source is negative. As stated in the lectures supplied power should always be negative. And the dissipated power should always be positive. This should mean that the current source dissipates power as resistor! UserIdTAG: 229896 UserNameTAG: Svilen CreateTimeTAG: 2012-09-07T15:06:02Z VoteTAG: 3 CoursewareTAG: Week 1 / KVL/KCL example CommentableIdTAG: 6002x_S2E3_Using_KVL_KCL_and_VI_constraints NumberOfReplyTAG: 1 FirstChildTAG: Power is positive when it is dissipated by a resistor or provided by a source. What's happening here is that the voltage source is being charged by the current source - power is being dissipated by the voltage source so it is negative. FirstChildUserIdTAG: 267993 FirstChildUserNameTAG: mavlijas FirstChildCreateTimeTAG: 2012-09-07T15:47:07Z SecondChildTAG: So, That's why i1 is -ve....i was very confused about i1 because as mentioned previously a current should be delivered by the source "not consumed" but i think you give a reasonable answer ....but now , the question is how to distinguish between a source that delivers power & a source that is charged by another one ? SecondChildUserIdTAG: 185715 SecondChildUserNameTAG: amirengineer SecondChildCreateTimeTAG: 2012-09-08T09:15:25Z SecondChildTAG: aha :) , so if we separate the current source the voltage source will deliver higher power but the reason it conduct just "2.8W" is because the power delivered from the current source is countering the power of the voltage source..."if I'm wrong ,kindly correct for me". SecondChildUserIdTAG: 185715 SecondChildUserNameTAG: amirengineer SecondChildCreateTimeTAG: 2012-09-08T09:26:00Z IndexTAG: 1170 TitleTAG: S1E 1.5 i am really tied up in these the three terminologies... **power delivered by the source ** **power dissipated by load** **power entering the source** and why is the "-ve" sign in answer of S1E1.5 UserIdTAG: 279621 UserNameTAG: MuhammadArshadshamsher CreateTimeTAG: 2012-09-07T11:32:45Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 5_Simple_Power NumberOfReplyTAG: 4 FirstChildTAG: source deliver power and load consumes it.easy terminology:power delivered by source=power consumed by load.just add and check whether they are equal.if not,just think where will be-ve sign. FirstChildUserIdTAG: 55345 FirstChildUserNameTAG: sarmaji FirstChildCreateTimeTAG: 2012-09-07T11:53:08Z FirstChildTAG: Agreed, "power entering the source" is poor terminology. I would stick to Power generated/delivered BY and power dissipated/consumed BY. If you find that a generated power is negative, then in reality, it is being consumed, not generated and similarly a negative dissipated power is actually a positive delivered power. It gets much more interesting, and possibly confusing, when you get into real, reactive and complex power in AC circuits where there are phase shifts between the V and I. But, this course probably will not get into that. FirstChildUserIdTAG: 208891 FirstChildUserNameTAG: tjwuth FirstChildCreateTimeTAG: 2012-09-07T14:53:52Z SecondChildTAG: I think that "power entering the source" means "power from source point of view" thing is it is always negative? or considering charging cycle it could be positive? SecondChildUserIdTAG: 298612 SecondChildUserNameTAG: fabianh SecondChildCreateTimeTAG: 2012-09-07T16:22:46Z FirstChildTAG: What determines the charge is either the voltage polarity or the direction of the current flow. Kindly refer to chapter 1.5.3 figure 1.19 and 1.20 FirstChildUserIdTAG: 214990 FirstChildUserNameTAG: Shuaibu FirstChildCreateTimeTAG: 2012-09-08T13:58:47Z FirstChildTAG: P=i*v For resistor, i flows from +ve to -ve potential hence according to our convention it is assumed to be +ve and hence power is positive. For source, voltage is positive, but due to emf of source current is carried out from -ve terminal to +ve terminal inside source, so power is negative. FirstChildUserIdTAG: 309803 FirstChildUserNameTAG: PurnenduK FirstChildCreateTimeTAG: 2012-09-08T05:56:49Z IndexTAG: 1171 TitleTAG: I love Gerry's bolo tie Thanks for being so awesome :-) UserIdTAG: 222931 UserNameTAG: tnyander CreateTimeTAG: 2012-09-07T09:30:54Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Thank you Gerry! :) FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-21T03:10:25Z IndexTAG: 1172 TitleTAG: Why is it.... .... (e1-V0) and not (V0-e1)? Thanks in advance... UserIdTAG: 214275 UserNameTAG: TheodoreGr CreateTimeTAG: 2012-09-07T06:42:03Z VoteTAG: 3 CoursewareTAG: Week 1 / Node analysis practice, part 2 CommentableIdTAG: 6002x_L2Node1 NumberOfReplyTAG: 4 FirstChildTAG: He said in the former video to take the outgoing currents. But you can take all the incoming currents too. Notice the difference is only a multiplier of -1, which is the convention itself. Either way you get good results. ( Sum(I_outgoing) = 0, or Sum(I_incoming) = 0 - does not matter) cheers FirstChildUserIdTAG: 329254 FirstChildUserNameTAG: KGabor FirstChildCreateTimeTAG: 2012-09-07T14:24:31Z FirstChildTAG: Current is taken in upward direction FirstChildUserIdTAG: 55345 FirstChildUserNameTAG: sarmaji FirstChildCreateTimeTAG: 2012-09-07T07:00:30Z FirstChildTAG: Normally current flows from (+)ve to (-)ve. They are asking for the current going up. So Vo is (+)ve. Actually the curent is flowing from V0 to e1. Thatz why it has become minus FirstChildUserIdTAG: 268444 FirstChildUserNameTAG: Marlonabeykoon FirstChildCreateTimeTAG: 2012-09-08T05:16:45Z SecondChildTAG: NO, current flows from e1 to V0. SecondChildUserIdTAG: 159427 SecondChildUserNameTAG: Suyog SecondChildCreateTimeTAG: 2012-09-16T08:28:24Z FirstChildTAG: When you're writing the KCL equation for a node you assume that each branch current is leaving the node. If a current is flowing out of a node then the voltage at the node (in our case, e1) must be higher than the voltage at the destination node (V0). If it was the other way around we'd be assuming that the current was flowing into the node. FirstChildUserIdTAG: 373220 FirstChildUserNameTAG: oohall FirstChildCreateTimeTAG: 2012-09-07T07:47:35Z SecondChildTAG: So we can just assume the voltage is higher at e1 because current is flowing out of e1? Always in cases like this? SecondChildUserIdTAG: 357225 SecondChildUserNameTAG: MJBoa SecondChildCreateTimeTAG: 2012-09-10T09:51:04Z IndexTAG: 1173 TitleTAG: Second part Can someone explain how we are suppose to find e3 in the second part? Are we suppose to use the information from the first part? Or are we suppose to do a node analysis at e3. If we do node analysis, how do we account for the voltage source and current? UserIdTAG: 208854 UserNameTAG: BrynnleeEaton CreateTimeTAG: 2012-09-07T05:23:41Z VoteTAG: 3 CoursewareTAG: Week 1 / Node method example CommentableIdTAG: 6002x_S2E5_Node_Method NumberOfReplyTAG: 3 FirstChildTAG: e1=e3+V.therefore,current through e3>source > ground becomes (e3+V)/R1.current source:look:I is leaving from e3.it should be added to equation.otherwise,do KVL and KCL,find current thru R2,findv2 and e3 from there FirstChildUserIdTAG: 55345 FirstChildUserNameTAG: sarmaji FirstChildCreateTimeTAG: 2012-09-07T06:47:29Z FirstChildTAG: I tried to write the node expression for e3, and failed. Then did it for e2, and got the answer by substituting e1=e3+V. Hope you had the same problem, and I could help. FirstChildUserIdTAG: 329254 FirstChildUserNameTAG: KGabor FirstChildCreateTimeTAG: 2012-09-07T14:53:15Z FirstChildTAG: I was also confused at first because I thought that we were supposed to determine e3 just by looking at the diagram and intuition, but then I realized that you do have to do the node analysis at e3 to solve for e3. You should end up with an equation that looks the same as the one for question 2. FirstChildUserIdTAG: 310007 FirstChildUserNameTAG: ErinF FirstChildCreateTimeTAG: 2012-09-08T02:41:57Z SecondChildTAG: I can't get the answer. I got this equation at e3 (e3+V)/R1+I+e3/R2=0. Is this correct? SecondChildUserIdTAG: undefined SecondChildUserNameTAG: undefined SecondChildCreateTimeTAG: 2012-09-08T14:59:31Z IndexTAG: 1174 TitleTAG: Any body have done with H2P1.... Ha dears any one of you have done with Homework 2 question 1. Hint needed. Thanks in Advance. UserIdTAG: 113109 UserNameTAG: Akif CreateTimeTAG: 2012-09-07T02:54:38Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 3 FirstChildTAG: applied voltage divider formula Vout = Vin * R2 / (R1 + R2), one can assume the values ​​of R1 and R2 in the vector that E12, when Vout is apx 3.5 v, are taken parallel resistors and alleged should be between 10k and 30k I am from Colombia, sorry for my English so poor FirstChildUserIdTAG: 58618 FirstChildUserNameTAG: ingeniero13 FirstChildCreateTimeTAG: 2012-09-07T22:13:59Z SecondChildTAG: How did you get Vout apx 3.5 V? The ratio Vout/Vin is 10%. How. Please help me. SecondChildUserIdTAG: 171674 SecondChildUserNameTAG: wajahat1 SecondChildCreateTimeTAG: 2012-09-16T15:24:37Z SecondChildTAG: may i know,how u got Vout=3.5v? SecondChildUserIdTAG: 171170 SecondChildUserNameTAG: rinutituschakkattil SecondChildCreateTimeTAG: 2012-09-17T18:59:53Z FirstChildTAG: I've been working on that problem, but when I evaluate my voltage using the entire range of resistances for a given pair, it causes the Vout to vary by more than 10%. I'm not sure where I'm going wrong. I know how to calculate the Thevenin resistance but I keep getting voltages that exceed 11 volts across R2. FirstChildUserIdTAG: 331664 FirstChildUserNameTAG: dwmnctrh3 FirstChildCreateTimeTAG: 2012-09-10T19:55:03Z SecondChildTAG: Same here. I've tried every pair that satisfies the Vin/Vout and Thevenin resistance requirements, but even the best possible answer is outside the 10% range. Then a funny thing happened - at some point the progress check page gave me credit for the answers, but I never got green ticks for them. Huh. SecondChildUserIdTAG: 267993 SecondChildUserNameTAG: mavlijas SecondChildCreateTimeTAG: 2012-09-10T20:31:45Z FirstChildTAG: If Rth is range calculate R1 and R2 also in ranges and then select the best possible combinations from E12. There are multiple correct answers to this question. FirstChildUserIdTAG: 170349 FirstChildUserNameTAG: amitGuliya FirstChildCreateTimeTAG: 2012-09-15T11:22:36Z IndexTAG: 1175 TitleTAG: software used? What software/hardware are they using for displaying the slopes of the various lumped elements on the projector screen? UserIdTAG: 348624 UserNameTAG: raaay CreateTimeTAG: 2012-09-07T02:35:33Z VoteTAG: 3 CoursewareTAG: Week 1 / Lumped Element Abstraction CommentableIdTAG: 6002x_lumped_element_abstraction NumberOfReplyTAG: 1 FirstChildTAG: I'm really interested as well. FirstChildUserIdTAG: 374 FirstChildUserNameTAG: mturilin FirstChildCreateTimeTAG: 2012-09-10T05:38:31Z IndexTAG: 1176 TitleTAG: Cheat Sheet apparently some of us could use this: http://2.bp.blogspot.com/_IGPyEGm6pT0/TEr7980y-9I/AAAAAAAAAh4/Via21LTOjgM/s1600/OhmsWattsLaws.png UserIdTAG: 369511 UserNameTAG: Year CreateTimeTAG: 2012-09-07T00:06:34Z VoteTAG: 3 CoursewareTAG: Week 1 / Power CommentableIdTAG: 6002x_S1E2_Power NumberOfReplyTAG: 2 FirstChildTAG: Looks like a great summary of equations. However if you continue your education in EE you will find it necessary to simply "remember" these equations off by heart. Don't worry, it will come to you naturally if you keep doing practice problems. FirstChildUserIdTAG: 151427 FirstChildUserNameTAG: RichardZ FirstChildCreateTimeTAG: 2012-09-07T00:51:15Z FirstChildTAG: Year, Another good source of info pertaining to this course is the original wiki from the MITx 6.002x spring course: https://6002x.mitx.mit.edu/wiki/view/ Tons of information including Ohms and Joules laws about halfway down the page: https://6002x.mitx.mit.edu/wiki/view/JoulesLaw FirstChildUserIdTAG: 11730 FirstChildUserNameTAG: RonFlip FirstChildCreateTimeTAG: 2012-09-07T01:30:47Z IndexTAG: 1177 TitleTAG: Homework power Given that each individual resistor can dissipate up to 1 watt of power before burning up, how much total power in watts ( W ) can the smallest-valued composite resistor dissipate before burning up? i think that 3 w but it is wrong help me please? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-06T19:59:21Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 4 FirstChildTAG: Notice that not all the resistors have the same value. Do you think that makes a difference? Would the smallest or the largest resistor dissipate more power in the configuration for smallest combined resistance? Remember that resistors in series have all the same current and in parallel they have all the same voltage. Jorge FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-06T20:13:12Z FirstChildTAG: If you think about it carefully you will realise that it is not 3W. The resistors do not all have the same resistance, so by the time the voltage is high enough to cause one of them to burn up, another may have burned up already. FirstChildUserIdTAG: 270011 FirstChildUserNameTAG: SnowmanZA FirstChildCreateTimeTAG: 2012-09-06T20:18:49Z FirstChildTAG: First of all, you have to calculate the voltage for the 4-ohm resistor. Then, using that voltage, plug in the value of the three resistors in parallel (I'm assuming your answer to that question was correct). The power value you get is the correct answer. FirstChildUserIdTAG: 237167 FirstChildUserNameTAG: Dug FirstChildCreateTimeTAG: 2012-09-06T22:11:14Z FirstChildTAG: thank people, the help that they gave me, this was very important FirstChildUserIdTAG: 58618 FirstChildUserNameTAG: ingeniero13 FirstChildCreateTimeTAG: 2012-09-06T23:22:54Z IndexTAG: 1178 TitleTAG: In this circuit, no power is entering the source A series current source could also burn up this voltage source. A parallel voltage source with a higher voltage output could burn up this voltage source. BUT no power is coming from the resistor, and to say "power is entering the source from this resistor" sounds like an amateur. Real engineers would never say this. UserIdTAG: 145676 UserNameTAG: CathyPK CreateTimeTAG: 2012-09-06T19:47:51Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 5_Simple_Power NumberOfReplyTAG: 3 FirstChildTAG: About What is the power entering the source I agree with u that there is no power is interning the voltage source, if they should ask for the -2.0 Watt it must be what the power generated from the voltage source referred to the load FirstChildUserIdTAG: 218038 FirstChildUserNameTAG: sonic_sie FirstChildCreateTimeTAG: 2012-09-06T20:24:30Z SecondChildTAG: @cathypk-according to me, power can be negative since it is a dot product..and current does pass through a voltage source..DO NOT look it as a current source in series with a voltage source...by saying that u are implying that there is an extra current(when actually it is not the case)..the current is goin back to the voltage source(u can check it by applying KCL)..so it does make sense when say power is entering a source.. hope this helps u ;) SecondChildUserIdTAG: 625118 SecondChildUserNameTAG: alex_003 SecondChildCreateTimeTAG: 2012-10-17T17:58:08Z FirstChildTAG: **i think it is well said because power is entering the source in the negative entry of the battery after passing through the resistor FirstChildUserIdTAG: 184556 FirstChildUserNameTAG: rhod FirstChildCreateTimeTAG: 2012-09-06T22:34:16Z FirstChildTAG: I agree. Saying that some power is entering the source on that circuit sounds at least confusing. FirstChildUserIdTAG: 413065 FirstChildUserNameTAG: anikey FirstChildCreateTimeTAG: 2012-09-11T20:38:27Z IndexTAG: 1179 TitleTAG: why 7.7.??? can someone tell me why 7.7v in r2? and the relationship with the 3a current source please.! UserIdTAG: 156060 UserNameTAG: radeon9550 CreateTimeTAG: 2012-09-06T19:39:20Z VoteTAG: 3 CoursewareTAG: Week 1 / KVL/KCL example CommentableIdTAG: 6002x_S2E3_Using_KVL_KCL_and_VI_constraints NumberOfReplyTAG: 3 FirstChildTAG: look do a loop analysis ! equation will become -2+I1*4+(I1+3)5=O ............ EQU 1 SIMPLIFY IT ! -2 +4I1+5I1+15=0 9I1=-13 I1=-13/9 NOW SEE EQU I (I1+3)5=0 put value of I1 in it we will get 7.77 FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-07T07:40:02Z SecondChildTAG: Where did you get (I1+3) on the first EQN? Why do you take into account the I from the other loop? I thought sum of one loop is zero. SecondChildUserIdTAG: 89084 SecondChildUserNameTAG: anakluhur SecondChildCreateTimeTAG: 2012-09-15T13:22:03Z FirstChildTAG: let's say v0 is voltage across voltage source with value V=2 take a first loop including the voltage source will give: -v0+i1R1+v2=0 so v2=2-4i1 (substituting values)----(1) ------- next: current into v2 is i2=I+i1 still v2=i2R2 therefore v2=(i1+I)x5 ----(2) ----- merging the two equations (i1+I)x5=2-4i1 5i1+15=2-4i1 9i1= =-15+2=-13 i1=-13/2 = 1.44 ------- substituting into v2=2-4i1 gives v2=2-4(-1.44) = 2+4x1.44 =7.76 FirstChildUserIdTAG: 314386 FirstChildUserNameTAG: jid FirstChildCreateTimeTAG: 2012-09-08T15:09:28Z SecondChildTAG: Nice! thanks I got every answer but power in R2 ! SecondChildUserIdTAG: undefined SecondChildUserNameTAG: undefined SecondChildCreateTimeTAG: 2012-09-08T21:44:52Z SecondChildTAG: That's great thanks! SecondChildUserIdTAG: 154726 SecondChildUserNameTAG: tizianococcio SecondChildCreateTimeTAG: 2012-09-15T17:05:40Z SecondChildTAG: Good thinking.You are a Guru SecondChildUserIdTAG: 291422 SecondChildUserNameTAG: LIN44 SecondChildCreateTimeTAG: 2012-09-17T19:08:57Z FirstChildTAG: I cant tell you why, but i could help you get there. get the equation for all the loops possible using KVL, in fact just one is enough. Then get the current inflow using the KCL...the 3A current source eventually splits up into i1 and i2(which is the current across r2).... Thats all I needed. Jiggled with those few equations and BAM !!!, didnt get the last one though, but the wrest was awesome....i answered the wrong question by mistake initially, but this is for you radeon9550 FirstChildUserIdTAG: 204745 FirstChildUserNameTAG: allwynmendes FirstChildCreateTimeTAG: 2012-09-06T21:55:35Z IndexTAG: 1180 TitleTAG: Answer S3E1 Hello I think the answer posted is incorrect, it is correct if V2 is reversed UserIdTAG: 369098 UserNameTAG: Marcelog CreateTimeTAG: 2012-09-06T13:27:50Z VoteTAG: 3 CoursewareTAG: Week 2 / Node equation review CommentableIdTAG: 6002x_S3E1_NodeEquationReview NumberOfReplyTAG: 2 FirstChildTAG: I'm sorry, it can be convention problem, if you put the psu reversed but voltage also negative, the negative voltage indicates that the psu in the schematic is reversed or are they independent? Best regards FirstChildUserIdTAG: 369098 FirstChildUserNameTAG: Marcelog FirstChildCreateTimeTAG: 2012-09-06T13:30:18Z SecondChildTAG: Marcelog, I think the answer is correct. Notice that the source is reversed and also that it's voltage is negative. SecondChildUserIdTAG: 381746 SecondChildUserNameTAG: jelizon SecondChildCreateTimeTAG: 2012-09-06T13:36:19Z FirstChildTAG: I too feel Answer is incorrect .... Circuit which can be represented by (5-e)/6800 - (e-(-7.2))/5600 = 0 But it does not yield answer mentioned where as equestion (5-e)/6800 - (e-7.2)/5600 = 0 gives desired answer but did not got logic ..... FirstChildUserIdTAG: 306294 FirstChildUserNameTAG: swpnljoshi FirstChildCreateTimeTAG: 2012-09-15T03:41:14Z IndexTAG: 1181 TitleTAG: Course Material It is possible to download easily all the Material of the course??? I dont have internet access in my house, and I need the material to learn Regards UserIdTAG: 219742 UserNameTAG: gustavoM CreateTimeTAG: 2012-09-06T08:05:03Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: I agree, it would also be nice if I could download the text book to a pad or some other handheld device for reference. FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-06T08:20:23Z IndexTAG: 1182 TitleTAG: Input power calculation HI, Is the input power calculation done by = voltage drop across * current entering the element = 10V * -0.2amps(bcoz current is entering through negative terminal) =-2W UserIdTAG: 123484 UserNameTAG: hudkmr CreateTimeTAG: 2012-09-06T05:38:45Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: 5_Simple_Power NumberOfReplyTAG: 1 FirstChildTAG: Hi,You used P=V*I try using P=sqrt(I)*R, it will not depends on I's direction right ? FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-09T09:51:55Z SecondChildTAG: it does depend on current's direction since POWER IS DOT product.. SecondChildUserIdTAG: 625118 SecondChildUserNameTAG: alex_003 SecondChildCreateTimeTAG: 2012-10-17T18:01:57Z IndexTAG: 1183 TitleTAG: Auto-Next. Perhaps there should be Auto-Next like feature of Udacity in EdX. UserIdTAG: 2956 UserNameTAG: akshayb CreateTimeTAG: 2012-09-06T04:05:05Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 1184 TitleTAG: facing a problem My tool board is not showing the DC, Ac and Trans analysis that is why i made mistakes in AC. My all calculations are manually calculated. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-06T03:29:22Z VoteTAG: 3 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 4 FirstChildTAG: Having the same problem. Has the solution actually been found so far? FirstChildUserIdTAG: 252722 FirstChildUserNameTAG: vaboro FirstChildCreateTimeTAG: 2012-09-06T18:45:01Z FirstChildTAG: I had the same problem. I am using Firefox on a Mac. The toolbar got stuck in editor mode, showing only the grid,cut,copy,and paste tools. Only after I refreshed the page (and I lost my circuit diagram because of the refresh), did I get the DC and other tools to show. Now I have everything done, but the TRAN button only brings up the dialog. When I enter the values and click [OK], no analysis graph appears. FirstChildUserIdTAG: 332090 FirstChildUserNameTAG: WPurin FirstChildCreateTimeTAG: 2012-09-06T05:57:21Z SecondChildTAG: I have problems in the Using the Tools Lab, but everything looks like it works in the Circuit Sandbox Lab. I tried both Firefox and Safari and neither worked well or correctly with the Using the Tools Lab. The Circuit Sandbox worked for me fine, including TRAN, but it has the same problem as the Using the Tools Lab when I use Firefox. SecondChildUserIdTAG: 332090 SecondChildUserNameTAG: WPurin SecondChildCreateTimeTAG: 2012-09-06T06:11:56Z FirstChildTAG: I had the same issue, should we post our browsers/OS to a bug report? Iceweasel 11.0/Crunchbang (based on Debian Squeeze) Worked on Chrome/same OS FirstChildUserIdTAG: 369511 FirstChildUserNameTAG: Year FirstChildCreateTimeTAG: 2012-09-06T05:56:02Z FirstChildTAG: The circuit simulator tutorial shows 'the minimum number of time points' to be entered in transient analysis tool box; however, the actual tool board doesn't have an option of entering 'the minimum number of time points'; I guess there's a bug somewhere in the code that doesn't allow to perform the transient analysis properly... **Transient analysis works in the Sandbox though**. Even without 'the minimum number of time points' value. FirstChildUserIdTAG: 252722 FirstChildUserNameTAG: vaboro FirstChildCreateTimeTAG: 2012-09-06T19:57:51Z IndexTAG: 1185 TitleTAG: Forum tag weirdness, et alia Forum tags are not behaving as expected. The error message says that words, numbers, spaces, and dashes are acceptable ("Tag can consist of words, numbers, dashes and spaces only and cannot start with dash"), but it rejects tags consisting of letter and number combinations, such as S1 and S4 (for Sequence 1 and 4, respectively). How else to tag posts referring to H1P2 or S1E3, if not alphanumeric? Despite claiming to allow spaces, it appears to reject tags with embedded spaces. It also does not accept underscores such_as_this. Lastly, although the preview shows italic when the HTML codes are used, the actual post shows the HTML tags literally, and does not display the text in *italic* (note that using a single asterisk on either side of the text displays it in italic). Postscript: Editing a post does not retain prior tagging, and all tags are lost unless reentered during edit. Foo. I prefer the old forum. UserIdTAG: 94557 UserNameTAG: g_hopper CreateTimeTAG: 2012-09-06T03:02:54Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: I found I could not tag with a capital letter, but lowercase works. Example, try h1p1 instead of H1P1. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-13T14:19:50Z IndexTAG: 1186 TitleTAG: Reply Count Poll: How many took/started this class in the spring? I would like this to be a reply poll. Do not reply if you did not take the course last spring. You may add comments (does affect the count). Did you take this course when it was first offered as Massive Open Online Course in the spring MIT/x? UserIdTAG: 88550 UserNameTAG: Neil_S_Berry CreateTimeTAG: 2012-09-05T22:25:07Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: Hi Neil_S_Berry ;) ! Here I am again! You can also read this post There are some people from the Prototype Corse re-taking it ;) [read here][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/MITx_6_002x_2012_Fall/threads/50474faf685941270000000d FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-05T22:37:39Z FirstChildTAG: I'm back for more fun, and hopefully to learn some more new stuff. Great to see so many of the old familiar users back here. A big shout out to Myrimit... I'm sorry I didn't have time in my work schedule to enter the Classmates Contest, I look forward to seeing the winners. FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-09-06T00:47:48Z SecondChildTAG: ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-29T21:55:11Z IndexTAG: 1187 TitleTAG: Video settings and fullscreen problems The volume setting isn't saving between each video but the speed setting saves. When clicking to the next video the fullscreen drops back to small screen. Also, in fullscreen the captions covers part of the video and the 'tool bar' doesn't move. Using Firefox 15, Flash 11.2, Win XP, Resolution = 1024x768. UserIdTAG: 147226 UserNameTAG: jaybiz3 CreateTimeTAG: 2012-09-05T22:04:42Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 1188 TitleTAG: Technical problems clicking on icons When I double click on the circuit components, the edit box is not coming up. UserIdTAG: 358929 UserNameTAG: NehaBh CreateTimeTAG: 2012-09-05T20:13:58Z VoteTAG: 3 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 2 FirstChildTAG: Hello. Please to try an other browser. I have encouter the same problem today. FirstChildUserIdTAG: 165141 FirstChildUserNameTAG: fariik FirstChildCreateTimeTAG: 2012-09-05T20:39:34Z FirstChildTAG: I am using chrome browser and I had no problem. You may use Chrome. FirstChildUserIdTAG: 259238 FirstChildUserNameTAG: omidsadeghi FirstChildCreateTimeTAG: 2012-09-06T05:42:56Z IndexTAG: 1189 TitleTAG: Measure current Hi, I was checking my errors, and I note that I wrote incorrect the current flow, in specific I wrote like a notation of amperes and with negative sign, and in the check answer it's wrote in exponetial notation, it could be a little problem to have a good point, I will be more careful the next time. UserIdTAG: 291037 UserNameTAG: ZackC CreateTimeTAG: 2012-09-05T20:13:30Z VoteTAG: 3 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: I am still do not have any idea about the reason that the total current should be positive 500u instead of -500u like what it is on my schematic! Plz give me your idea.. FirstChildUserIdTAG: 76415 FirstChildUserNameTAG: ELECCE FirstChildCreateTimeTAG: 2012-09-06T10:17:21Z IndexTAG: 1190 TitleTAG: question to the linear resistance because not is the graph c? is because the line does not pass through zero? UserIdTAG: 187265 UserNameTAG: ferrerfad CreateTimeTAG: 2012-09-05T20:11:01Z VoteTAG: 3 CoursewareTAG: Week 1 / Various V-I characteristics CommentableIdTAG: 6002x_S1E1_Various_V-I_characteristics NumberOfReplyTAG: 8 FirstChildTAG: As you say, a simple resistor must be a straight line that passes through the origin ("zero"), since $v = i\cdot R$ implies that when $v=0$ then $i=0$ as well. FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-05T22:02:26Z SecondChildTAG: I think it has to do more with the fact that V and I are directly proportional to one-another. SecondChildUserIdTAG: 397407 SecondChildUserNameTAG: BurimD SecondChildCreateTimeTAG: 2012-09-15T20:51:03Z SecondChildTAG: As we can see that V=I.R it implies V and I are directly proportional and as directly proportional means that as V increases gradually I also increases gradually. in graph "C" we can clearly see that as V is increasing I is decreasing. This contradicts the V-I relationship SecondChildUserIdTAG: 346016 SecondChildUserNameTAG: muthukrishna SecondChildCreateTimeTAG: 2012-09-16T04:34:44Z SecondChildTAG: resistance is always positive and in C the resistance a.k.a the slope of the line is negative. SecondChildUserIdTAG: 452690 SecondChildUserNameTAG: shreerajshrestha SecondChildCreateTimeTAG: 2012-09-17T15:26:45Z FirstChildTAG: I am just making an assumption here, but I think that c is not a correct answer because it would mean that the resistance of the element would have a negative value and its very difficult to think of an element that responds that way. FirstChildUserIdTAG: 210561 FirstChildUserNameTAG: Alejovillapar FirstChildCreateTimeTAG: 2012-09-05T23:40:40Z FirstChildTAG: The answer isn't C because in graph C the voltage decreases as current increases, where as in a resistor if current is increased across it then voltage is also increased. The origin, in this case, isn't relevant. FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-05T23:30:42Z FirstChildTAG: As you and kimt said, a resistor's curve must pass through the origin. There's no problem of having a negative resistor. It doesn't exist as a physical device, but can be easily done with an association of devices. Remember the word "abstraction"! FirstChildUserIdTAG: 188586 FirstChildUserNameTAG: FLara FirstChildCreateTimeTAG: 2012-09-06T00:04:40Z FirstChildTAG: As you see on [c] plot, there is still some current [i] through this device even at zero voltage [v] — normal resistor doesn't behave like this AFAIK. FirstChildUserIdTAG: 386636 FirstChildUserNameTAG: andy7panda FirstChildCreateTimeTAG: 2012-09-06T13:46:51Z FirstChildTAG: Yes because this is the "characteristic of a resistor (linear graph that passes through zero coordinates (0,0) and the line should be slanting up from left to the right; the Graph C doesn't have this characteristic only graph a & f. Passing zero means on the simple equation V=IR, when voltage V is zero applied to the linear resistor, the current will be zero. I = V/R; I = 0 volt/ R ( in ohms); I = 0 Ampere. If we continue to plot the iv graph with different values applied to a certain resistor, we will come up with such graph as a & f only. My idea is: There is no such negative value of the resistor here. This means only v and i changes signs either both negative or positive depending on the what the cycle is being applied for the time being...for example if we applied an ac (alternating signal) on linear resistor element, the straight line on the 1st quadrant or on the right of the vertical y-axis is graph due to the the positive cycle of the ac signal, while the straight line on the 3rd quadrant or left side of the the y-axis is due to the negative cycle of the ac signal. FirstChildUserIdTAG: 309255 FirstChildUserNameTAG: mamba747 FirstChildCreateTimeTAG: 2012-09-06T06:37:16Z FirstChildTAG: Due to Ohms low we assume that the resistance is constant so,if the voltage applied on the element (R) increased,the current which passes throw the resistance must increased. FirstChildUserIdTAG: 218038 FirstChildUserNameTAG: sonic_sie FirstChildCreateTimeTAG: 2012-09-06T20:14:39Z FirstChildTAG: Yes FirstChildUserIdTAG: 189808 FirstChildUserNameTAG: AlexanderZ FirstChildCreateTimeTAG: 2012-09-06T10:21:02Z IndexTAG: 1191 TitleTAG: Flow of electrons verse flow of "holes". This might be located in the wrong area but in terms of positive/negative current flow, are we using the idea of the current being "holes" flowing from positive to negative or the idea of electrons flowing from negative to positive? Since I wrote the wrong sign value I suspect we will be viewing it as a flow of "holes" but I want to make sure. UserIdTAG: 371819 UserNameTAG: Lokiie CreateTimeTAG: 2012-09-05T19:57:03Z VoteTAG: 3 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 2 FirstChildTAG: Yes, it is apparent that we must use the positive current convention, "holes" if you like. NOT the direction of electron flow. FirstChildUserIdTAG: 347289 FirstChildUserNameTAG: kellrobinson FirstChildCreateTimeTAG: 2012-09-05T20:08:32Z SecondChildTAG: Ok, then why does the DC analysis show -500uA through the DC PS? I put in -500u for the current and it was marked wrong. SecondChildUserIdTAG: 154526 SecondChildUserNameTAG: silentquasar SecondChildCreateTimeTAG: 2012-09-05T20:43:59Z FirstChildTAG: Holes is a very good way to think about it. It always confused me until I got that. When you think of it in terms of hole movement then it does not really matter what convention they choose to use. But since this course is taught using conventional current flow, holes is the way to imagine it. FirstChildUserIdTAG: 15344 FirstChildUserNameTAG: kob FirstChildCreateTimeTAG: 2012-09-05T21:06:49Z IndexTAG: 1192 TitleTAG: S1E3 Integration Insights Did anybody have any insight that allowed them to avoid integrating this to the bitter end? UserIdTAG: 182168 UserNameTAG: arthurdent CreateTimeTAG: 2012-09-05T18:07:29Z VoteTAG: 3 CoursewareTAG: Week 1 / AC power CommentableIdTAG: 6002x_S1E3_AC_power NumberOfReplyTAG: 2 FirstChildTAG: Look at this image on the left you have the voltage and on the right you have the peak power which is only positive (because of the V^2 in the peak power equation) so if you want have to have the average power you just have to dividade the peak power by 2!! ![][1] [1]: http://img269.imageshack.us/img269/469/sanstitreeea.jpg FirstChildUserIdTAG: 151444 FirstChildUserNameTAG: mael FirstChildCreateTimeTAG: 2012-09-05T22:22:41Z SecondChildTAG: If you are really going to integrate as suggested in the exercise you have to do the whole thing $$\int cos^2(2\,\pi\,60\,t)\:dt = \dfrac{t}{2} + \dfrac{sin(240\,\pi\,t)}{480\,\pi} + c$$ The confusion in this series of posts is that some people say integrate, others are using the 120 as the average voltage (that is given), and others are using RMS/peak formula. They all give the same result, but one does not really help explain the other. SecondChildUserIdTAG: 206049 SecondChildUserNameTAG: Ed66 SecondChildCreateTimeTAG: 2012-09-06T03:23:44Z FirstChildTAG: All i can see is that the integration knocked out the sqrt(2) constant (which is 2 when squared). That accounts for it being half the peak power. The sine wave calculation also gets knocked out in that process. Ah me, my math is ancient got to go remedial. FirstChildUserIdTAG: 190331 FirstChildUserNameTAG: Sangye FirstChildCreateTimeTAG: 2012-09-06T22:04:20Z IndexTAG: 1193 TitleTAG: where is the error ? my last four answers aren't right , why ? ![][1] [1]: http://www.mediafire.com/conv/b9c61c56c449130fdac01e2ca685e08f0874f035415c77d713a81a175581fd704g.jpg UserIdTAG: 283383 UserNameTAG: Yasmin_Mohamed CreateTimeTAG: 2012-09-05T14:34:02Z VoteTAG: 3 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 3 FirstChildTAG: Read carefully the text, you are doing a DC analysis instead TRANS analysis... ¨Add scope probes to nodes A, B and C and edit their properties so that the plots will be different colors. Now run a transient analysis for 5ms.¨ FirstChildUserIdTAG: 144671 FirstChildUserNameTAG: andresdans FirstChildCreateTimeTAG: 2012-09-05T14:39:48Z SecondChildTAG: Do it as andresdans said. In this case you did a DC analysis when a Transient analysis was required. SecondChildUserIdTAG: 30011 SecondChildUserNameTAG: euler SecondChildCreateTimeTAG: 2012-09-05T15:54:58Z SecondChildTAG: why to run transient analysis for 5ms? SecondChildUserIdTAG: 347775 SecondChildUserNameTAG: praveen_yvs SecondChildCreateTimeTAG: 2012-09-05T17:50:01Z FirstChildTAG: The solution is done with 2V value this is the mistake! FirstChildUserIdTAG: 299251 FirstChildUserNameTAG: TheRedBlackOne FirstChildCreateTimeTAG: 2012-09-06T00:10:45Z SecondChildTAG: I mean the transient analisis. SecondChildUserIdTAG: 299251 SecondChildUserNameTAG: TheRedBlackOne SecondChildCreateTimeTAG: 2012-09-06T04:01:26Z FirstChildTAG: You are supposed to do the Transient Analysis (TRAN button) and get the answers off the plot/graph that will pop up. But its not working for me... FirstChildUserIdTAG: 66241 FirstChildUserNameTAG: Ephexis FirstChildCreateTimeTAG: 2012-09-06T00:16:11Z IndexTAG: 1194 TitleTAG: Labeling points? How do I label points A, B, C on the schematic? UserIdTAG: 236064 UserNameTAG: AllenWoolfrey CreateTimeTAG: 2012-09-05T14:05:05Z VoteTAG: 3 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 5 FirstChildTAG: Second symbol in the interactive tool. 'R' to rotate it, double click to change '???' with label you want. FirstChildUserIdTAG: 128561 FirstChildUserNameTAG: gusevoy FirstChildCreateTimeTAG: 2012-09-05T14:10:51Z FirstChildTAG: to the left of the box is a small line item as a drag to the start screen and press R until it is as in Figure 1.1 then give double click on this and put a capital A, sorry for the English but I do not know much of this is a google translation I hope you understand something FirstChildUserIdTAG: 51696 FirstChildUserNameTAG: blasco23 FirstChildCreateTimeTAG: 2012-09-05T14:32:59Z SecondChildTAG: thanks, I did not know how to rotate the label SecondChildUserIdTAG: 430872 SecondChildUserNameTAG: rrvrkch SecondChildCreateTimeTAG: 2012-09-29T08:28:00Z FirstChildTAG: It's not needed. Just to you know how point them are talking about. FirstChildUserIdTAG: 300169 FirstChildUserNameTAG: castrogfx FirstChildCreateTimeTAG: 2012-09-05T14:35:58Z FirstChildTAG: Hi, In right panel,use node label for this FirstChildUserIdTAG: 168602 FirstChildUserNameTAG: chittiprolu FirstChildCreateTimeTAG: 2012-09-05T14:45:39Z FirstChildTAG: You realize how yours functions is somewhat dependent on what program or server you are using to access this. Try all of them. FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-06T15:56:53Z IndexTAG: 1195 TitleTAG: Sandbox Help Does anyone know how to add the inverted T for the ground connection in the lab. It doesn't appear to be there? Regards Mike. UserIdTAG: 80920 UserNameTAG: MichaelSturgess CreateTimeTAG: 2012-09-05T12:31:31Z VoteTAG: 3 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: It appears on the grey strip at the right of the "circuit board" on mine - it's the topmost symbol. I'm using Chrome, by the way. :-) FirstChildUserIdTAG: 79337 FirstChildUserNameTAG: AppliedImagination FirstChildCreateTimeTAG: 2012-09-05T12:34:22Z SecondChildTAG: It appears on the sandbox practice section but not on lab1. Is it there on your lab1 question? Thanks SecondChildUserIdTAG: 80920 SecondChildUserNameTAG: MichaelSturgess SecondChildCreateTimeTAG: 2012-09-05T12:36:44Z IndexTAG: 1196 TitleTAG: Tutorials week1 Some tutorials for week 1 as posted in the Fall version of 6003z: Open youtubepages in 6003z on a new tabpage (rightclick mouse) [Course Introduction youtube][1] by cclaunch [Geometric Series youtube][2] by dantyrant [Continuous vs Discrete vs Digital youtube][3] by ashwith [Continuous vs Discrete vs Digital mp4 denoised for download ][4] by ashwith [Signal Transformations youtube][5] by ashwith [Signal Transformations mp4 denoised for download][6] by ashwith [Introduction to GNU Octave youtube][7] by ashwith [Introduction to GNU Octave mp4 denoised for download][8] by ashwith [How to load mp3 files to Octave youtube][9] by Dmitry_Kh [1]: http://www.youtube.com/watch?v=zuDNlMBc3P0 [2]: http://www.youtube.com/watch?v=k5bBCrskoOU [3]: http://www.youtube.com/watch?v=CAPJMA__xQA [4]: http://www.ruudoleo.com/edx/6003z/videos/Continuous_vs_Discrete_vs_Digital.mp4 [5]: http://www.youtube.com/watch?v=ACL0qexzDDA [6]: http://www.ruudoleo.com/edx/6003z/videos/Signal_Transformations.mp4 [7]: http://www.youtube.com/watch?v=SMkkBfSdm1E [8]: http://www.ruudoleo.com/edx/6003z/videos/Introduction_to_GNU_Octave.mp4 [9]: http://www.youtube.com/watch?v=LQy0qgYuQeE UserIdTAG: 79885 UserNameTAG: ruudoleo CreateTimeTAG: 2013-02-19T11:45:25Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_6003z NumberOfReplyTAG: 0 IndexTAG: 1197 TitleTAG: H1BP2 After being flabbergasted for about a week I found a interesting post by Agcopa in the old forum of 6002x. [physnet wave motion][1] ![enter image description here][2] ![enter image description here][3] [1]: http://www.physnet.org/modules/pdf_modules/m201.pdf [2]: https://edxuploads.s3.amazonaws.com/1361271349909944.jpg [3]: https://edxuploads.s3.amazonaws.com/13612713771343636.jpg UserIdTAG: 79885 UserNameTAG: ruudoleo CreateTimeTAG: 2013-02-19T10:58:04Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_6003z NumberOfReplyTAG: 0 IndexTAG: 1198 TitleTAG: Bookchapters for 6003z As the chapters have changed in the second edition of the book Signals and Systems a pdf file with the contentsoverview and indication of the corresponding videolectures. [contents book][1] [1]: http://www.ruudoleo.com/Uploads/6003z/signals_book_readings.pdf UserIdTAG: 79885 UserNameTAG: ruudoleo CreateTimeTAG: 2013-02-16T15:41:25Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1199 TitleTAG: Feedback on Proctored exam Gave the proctored exam yesterday. The center was easy to find. The behavior of the staff was very professional and courteous. The center was clean and tidy. Lockers were provided to keep personal belongings. Sign-in procedures included taking photograph, verifying signatures and taking palm vein-scans. It took time so reaching there 30 minutes early was a good idea. Instructions were made clearly and nothing was confusing. The environment was very secure and it is virtually impossible to cheat. We were also monitored continuously through audio and video. They also provided nice writing pads with pens. We could ask for more if needed. Every thing was good except there were some minor technical issues with the test itself. The workstation they provided had the edX site opened in a very different environment. The on-screen calculator that is usually provided was not there. Luckily I had brought my own calculator. (BTW, they inspected my calculator to check whether it has any stored formulas). The video lectures and textbook were made available but it did NOT work. The videos were not running, neither I could download slides. The textbook was opening, but I could not turn pages so I was stuck at a single page and it was not usable at all. The system was very slow to navigate through which wasted some time. The pages were not appearing correctly formatted. I guess all this happened because of an unsuppoted browser and a 3:4 resolution (1024x768 I guess) they were using. The pattern of the exam was identical to the online final exam that we gave. It had 6 questions and I think 150 minutes were pretty sufficient. I expect my score to be nearly 80%, which is not very different from what I got in the online final exam. After the exam it said the certificate will be available on 13th March. Overall it was a nice experience for me and I would love to give more exams like these in future. I hope these issues get addressed by the staff and are prevented in future exams. FYI, The center was: Pearson Professional Center 18,Ramanath House, Community Centre, Yusuf Sarai-Green Park, Delhi - 110016 UserIdTAG: 297370 UserNameTAG: ayush3504 CreateTimeTAG: 2013-02-14T15:39:13Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 1 FirstChildTAG: Thank you for sharing ayush3504 :) Will your certificate be downloadle or will be sent to your home? FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2013-02-14T16:38:20Z SecondChildTAG: That is something I eagerly want to know. It just said that that the certificates will be available on 13 March. SecondChildUserIdTAG: 297370 SecondChildUserNameTAG: ayush3504 SecondChildCreateTimeTAG: 2013-02-14T17:05:53Z SecondChildTAG: And since they gave me a specific date I think it will be downloadable. SecondChildUserIdTAG: 297370 SecondChildUserNameTAG: ayush3504 SecondChildCreateTimeTAG: 2013-02-14T19:54:40Z IndexTAG: 1200 TitleTAG: Subscribed at 6.003z Add your location at the google map Add your location on the google map We will be a small community so let's get to know each other better. Attention:Open this link in a new tabpage ( via rightclick mouse) [google map 6.003z][1] Look at following post if you don't know how to add your marker [post about how to add a marker][2] [1]: http://goo.gl/maps/ym83D [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50d914bbef2ecd2b0000002a UserIdTAG: 79885 UserNameTAG: ruudoleo CreateTimeTAG: 2013-02-10T19:03:57Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_6003z NumberOfReplyTAG: 0 IndexTAG: 1201 TitleTAG: Help Myrimit! Can´t create a web account at pearson vue. I Myrimit, I need you help. I get the edx ID, then I press the button for schedule pearson exam and I go to create web account...First name....last name....edx ID.... and when I get to this page: ![enter image description here][1] All ok for now, but whe I select my country of residence (Spain): ![enter image description here][2] And I clik it, this error message appears: ![enter image description here][3] I tried many time in different days. I alsow tried calling the center in my country...but they don't answer. Please help me :). Thanks in advance. [1]: https://edxuploads.s3.amazonaws.com/1360255691134367.png [2]: https://edxuploads.s3.amazonaws.com/13602557187637857.png [3]: https://edxuploads.s3.amazonaws.com/13602557381343653.png UserIdTAG: 250241 UserNameTAG: DaniHerrBere CreateTimeTAG: 2013-02-07T16:51:30Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Thanks for reporting this. We will work with Pearson in order to fix this as soon as possible. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2013-02-07T17:14:23Z SecondChildTAG: Thank you Lyla! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2013-02-07T17:18:29Z SecondChildTAG: Thanks!!! :) SecondChildUserIdTAG: 250241 SecondChildUserNameTAG: DaniHerrBere SecondChildCreateTimeTAG: 2013-02-07T18:49:30Z SecondChildTAG: Any news on this issue? SecondChildUserIdTAG: 22419 SecondChildUserNameTAG: dbscoach SecondChildCreateTimeTAG: 2013-02-11T02:58:19Z SecondChildTAG: It didn`t change. It seems I can not take the exam after all. SecondChildUserIdTAG: 250241 SecondChildUserNameTAG: DaniHerrBere SecondChildCreateTimeTAG: 2013-02-11T23:27:31Z SecondChildTAG: Too bad. I'd have taken the exam if it had worked. Maybe next time... SecondChildUserIdTAG: 22419 SecondChildUserNameTAG: dbscoach SecondChildCreateTimeTAG: 2013-02-13T15:00:27Z FirstChildTAG: Hi DaniHerrBere, I will report inmediately to the Staff. Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2013-02-07T17:02:31Z SecondChildTAG: Thanks a lot ^^ SecondChildUserIdTAG: 250241 SecondChildUserNameTAG: DaniHerrBere SecondChildCreateTimeTAG: 2013-02-07T18:49:24Z IndexTAG: 1202 TitleTAG: 6.004 course video lectures Friends, 6.004 "computational structures" course video lectures are available in this link http://6004.mit.edu/Spring12/ UserIdTAG: 857413 UserNameTAG: Naveenkrishna CreateTimeTAG: 2013-02-06T18:18:12Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Where are the video lectures? The link is to a page saying 6.004 is over and to have a good summer and I can't seem to find them on the rest of the site? :L FirstChildUserIdTAG: 219780 FirstChildUserNameTAG: ReconIII FirstChildCreateTimeTAG: 2013-02-07T00:37:34Z SecondChildTAG: on left side of page there is a option "handouts" click on that. SecondChildUserIdTAG: 857413 SecondChildUserNameTAG: Naveenkrishna SecondChildCreateTimeTAG: 2013-02-07T13:05:44Z SecondChildTAG: Any idea if there is a download for the videos? Since they keep on playing for like 6 seconds in the start then crashing? I have great internet and a great pc so I think its the site? SecondChildUserIdTAG: 219780 SecondChildUserNameTAG: ReconIII SecondChildCreateTimeTAG: 2013-02-07T20:00:43Z SecondChildTAG: nvm... lol google chrome tries to play the file which doesn't work out well, with internet explorer it save it and works fine. SecondChildUserIdTAG: 219780 SecondChildUserNameTAG: ReconIII SecondChildCreateTimeTAG: 2013-02-07T20:22:22Z FirstChildTAG: These video lectures are very easy to follow. Would be a nice follow up for 6.002x. FirstChildUserIdTAG: 79885 FirstChildUserNameTAG: ruudoleo FirstChildCreateTimeTAG: 2013-02-06T19:27:18Z SecondChildTAG: Cool, thank you Naveenkrishna! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2013-02-06T20:22:47Z IndexTAG: 1203 TitleTAG: 6.00x on edx?? Is anyone taking 6.00x on edx? i am taking it as no other advance course in electronics is available on edx, moreover i am interested in programming and i have also registered on coursera for MOSFET course which is going to start soon. ---------- need reviews from you all regarding the Python programming language. As i am new to electronics so i wanna know that will Python programming help me in future in the electronics field? UserIdTAG: 114864 UserNameTAG: Vikaash CreateTimeTAG: 2013-02-03T11:23:03Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 6 FirstChildTAG: Hi Vikaash, I followed the first few weeks of 6.00x, but stopped after a few weeks because lack of time (I did 6.002x also and had to prepare for the exam). It's really a very interesting course, but my main goal was to learn some Python. Python was totally new to me, but it is a very interesting language. What I like about it, is that it's a lot more user friendly than Java (my personal opinion). It's much easier to create graphs (plots), and the math package of python also supports standard imaginary numbers, so it's ideal for using it with electronics. Python has a lot of supporting packages: e.g. for math and plotting graphs. But I advise you to buy the book from the teacher if you want to do 6.00x, because it will save you a lot of time. Or, if you want to get acqainted with Python, try beforehand the course on Udacity: Introduction to Computer Science, that introduces Python and has no (homework) pressure, because you can do it at your own pace. The HW problems of 6.00x were very interesting, but also very time consuming if you have to learn the language from zero and have no programming experience at all. But I'm glad that I learned at least some Python, because I noticed that a lot of on-line courses use it also, and instead of a calculator, I now use Python more often. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2013-02-05T15:21:48Z SecondChildTAG: i thought it may be little bit easy for me to learn python as i know c++ very well. I want to learn python because it is a prerequisite for **artificial intelligence** course. SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2013-02-05T16:14:34Z SecondChildTAG: Precisely that course was the reason I decided first to do 6.00x. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2013-02-05T19:22:38Z FirstChildTAG: I am there too. I have seen you there Vikaash :) I don't know python but I am familiar with other programming languages :). I registered to 6.00x Fall but I didn't follow. Now I am following it. Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2013-02-05T21:19:53Z FirstChildTAG: I am a student there also. FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2013-02-07T08:18:28Z FirstChildTAG: I am there too! FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2013-02-07T14:31:17Z FirstChildTAG: I took the course last Fall and it was quite good. There were a few glitches (like all things new) but the course material on the whole was quite informative and useful I'd say. Personally, I'd never had any formal training on programming. I had taught myself VBA by designing and building a database system for a company. Python is a bit different than VB so it was worth learning the little things that were different and how the language varied. But if you're really good at programming, I would assume you could easily learn that by just reading some Python FAQs. What was useful to me was learning how to program recursively and learning about certain algorithms that I could use and ways to think of new algorithms. Also, you learn Big O notation in the course which was something I had never seen before. In the end, it has brought me to a level of proficiency in Python where I can write my own efficient programs for my personal use. It helped me take another course on Computational Financing where I'm now tweaking a program that will help me simulate trades and then really trade stocks (if the results are good) all in Python. And now I'm looking forward to the AI class which will start here in a couple weeks. I'd recommend it personally. If you've got a lot of programming knowledge, the latter part of the course may not be as useful, but the first part of the course will at least teach you the differences in the coding language. FirstChildUserIdTAG: 467169 FirstChildUserNameTAG: Eyowzitgoin FirstChildCreateTimeTAG: 2013-02-07T15:21:43Z FirstChildTAG: it's great to see my old classmate on 6.00x.. FirstChildUserIdTAG: 114864 FirstChildUserNameTAG: Vikaash FirstChildCreateTimeTAG: 2013-02-07T15:15:48Z IndexTAG: 1204 TitleTAG: Proctored Exam for Bangladeshi Bangladeshi students...hi to all. It is to decide and survey that how many of the Bangladeshi students are going to seat for Proctored exam. As there is no PEARSON exam centre in Bangladesh so we have to goto to india or thailand to seat for 6002x . I am going to india(delhi) to attend the exam. if we can go as a group that would be better. I hope each and every student will comment on this topic so that we can know each other. atleast know the name. while giving note please state educational status, and where you are from. thank you very much MOFASSAIR HOSSAIN BHUIYAN BSc in EEE(3rd year)AIUB , DHAKA UserIdTAG: 154172 UserNameTAG: mofassair CreateTimeTAG: 2013-01-31T06:37:49Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Hi, i am also a Bangladeshi.But now i feel very disappointed to know here is no Pearson exam centre.I am unable to go India or Thailand to attend the proctored exam.But still i want to attend this exam, is there any alternative way? FirstChildUserIdTAG: 360053 FirstChildUserNameTAG: Ohedul FirstChildCreateTimeTAG: 2013-01-31T13:10:39Z FirstChildTAG: I am, MOFASSAIR HOSSAIN BSc in EEE @ AIUB lives inDHAKA FirstChildUserIdTAG: 154172 FirstChildUserNameTAG: mofassair FirstChildCreateTimeTAG: 2013-01-31T06:39:03Z FirstChildTAG: Nice iniciative Mofassair, I will go to another Country for the proctored Exam as here we don´t have a Proctored Centre, but I will go next year XD. That idea of the grupal trip is really nice Mofassair. How many hours are from Bangladesh to New Delhi or Thailand? I have from Argentina- Buenos Aires- to Brazil - Sao Pablo - 36 hours by bus more or less. The thing is that the route is dangerous, as it is possible that you have a bus accident... So, it is possible that I will go by airplane...also I will have to book the Hotel to stay some days xD , because I will go to the Beach , Turism, eating fruits, going to the Discoteques and of course taking the Proctored Exam haha. I can not go this February 13th as I have Exams at University and also need time, I like my Country but here is not allowed to buy any foreign currency unless you request permission to the Government, minimun I would need a month, more time, and they don´t give you all that you want, they restrict you the amount per day :/ and they also corroborate if you went outside the Country ... Also I guess the Hotel and tickets I had to have more time to make the booking - now is summer here and a lot of people worldwide goes to Brazil for Holidays and the Hotels are collapsed and I bet airplane tickets too ... Also I will have to practice my Portuguese too, I will not feel confident if I go to a foregin Country without knowing previously the basic of their language... Anyway, I will take the Proctored but next year, so I am happy haha. Take care, Then let us know how it was your experience in the Proctored Exam, I am anxious to know that XD. Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2013-01-31T14:57:41Z SecondChildTAG: Hi Myriam, It is nice to see you and your plan. Our country is more or less similar to yours. Bangladesh is developing country so PEARSON is not available here. But our neighbouring country INDIA has 5 Exam centre in different states. The one very near from me is Delhi. and if you wanna save some money you have to go through train. and which take almost 3day. 1day to cross our border and then again change train from colcata. But its possible to manage Air trip but that would be costly. We have to face the same types of problem as yours. Its takes time to process visa, then endorsing foreign currency will be issued depending upon your visa approval. I am not actually worried about hotels and all that, as I know their language and they also understand english a little. TRIP and TOUR!!! definitely!!!:D There are lots of place you can go to in INDIA. But the thing is, Its my passion to roam around the tourist spots since my childhood.From my very childhood i used to go for a trip whenever i got chance. Though still 50% of my county is undiscovered my myself:D. And i am very proud to say our country may not be a rich country but we are very rich in natural beauty. You can check this out [Beautiful Bangladesh][1]. It is my humble invitation to all of you to visit Bangladesh. I would be honered and very thankful if you come to my country. I promise that, I will personally receive you and show your the beauty of my country. Lastly, Myriam,Can you give the information about taking the exam next year. As you have said you will attend the same exam next year. Is it verified that you will have the opportunity to signup with PEARSON next your with the certification of this year. or you have to do the same course again. I need this information not only for me but also for the whole country. I would be pleased if you provide the reguarding information in detail. thank you very much. [1]: http://www.youtube.com/watch?v=Ppf-UA36ljE SecondChildUserIdTAG: 154172 SecondChildUserNameTAG: mofassair SecondChildCreateTimeTAG: 2013-01-31T17:18:46Z SecondChildTAG: Hi mofassair, Wow 3 days is really far! Bangladesh it sounds a beautiful place, I definitely have to go there in the Future :). Of course, you are welcome to Argentina too. I am really not sure if this Proctored Exam will be available next year... I hope so...but I am not sure...I guess that this will be the first time and in my point of view, they will analyze this impact and depending on that, they will decide, but that is my opinion, I really not sure...I guess that if I had a Pearson VUE Centre in Buenos Aires I would have taken the Exam on February 13th... SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2013-01-31T18:24:40Z SecondChildTAG: I Could not manage to go to india for the pearson exam. So i hope, noxt time:( SecondChildUserIdTAG: 154172 SecondChildUserNameTAG: mofassair SecondChildCreateTimeTAG: 2013-02-06T11:30:44Z SecondChildTAG: Thanks SecondChildUserIdTAG: 154172 SecondChildUserNameTAG: mofassair SecondChildCreateTimeTAG: 2013-02-06T11:30:48Z IndexTAG: 1205 TitleTAG: what is it? so,hi all ,how r u? i miss u so much,all of u so,what is the pearson exam? i'm sorry i didn't understand what they need from me to do,am why? please provide me with explanation thanks UserIdTAG: 395257 UserNameTAG: blackguitar CreateTimeTAG: 2013-01-28T23:20:55Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1206 TitleTAG: A certificate with grade!!?? As I read on the posts of the forum there isn`t available any certificate with the corresponding grade. But today searching on the net I found this link: http://www.anu.edu.au/physics/Savage/images/6002x_Certificate-8149-1.pdf And the corresponding link in the certificate goes to: https://verify.edxonline.org/cert/2ed292083d2740e8ae56472261a94508 In that page we can see clearly the corresponding grade from this person. I also get an A in the final results so.... there is any method to get this certificate? Is this certificate fake? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2013-01-28T22:25:54Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: No, it is not fake... I got one with grade of last term too. I took the Prototype Course of 6.002x Spring 2012. My Certificate it is similar to that one that you are showing in that Link. Due a confusion in the Prototype Course: the FAQ of the 6.002x Prototype 2012 Spring said that they will provide a Certificate with grade when it was not suppoused... The Staff took the fair decission of providing one Certificate with grade and another one without grade due this confusion by the request of the students. But I guess that it was an exception as it was not intended to display a grade. So, there is no method to get that Certificate unless you were passed the Prototype Course of 6.002x. That students only have that type of Certificates, with grades, as it was a exception... I hope this can help you. Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2013-01-28T22:51:23Z SecondChildTAG: aah! ok. Thanks for the clarification Myriam :) SecondChildUserIdTAG: 250241 SecondChildUserNameTAG: DaniHerrBere SecondChildCreateTimeTAG: 2013-01-29T18:26:29Z IndexTAG: 1207 TitleTAG: Someone knows about a inline denoiser program for videos There is a lot of humming in the videos of 6003z. I already tried Magix video but it takes forever to render the video. What I was wondering can't we use a inline denoiser or dehummer while playing the video without rerendering it. Someone already used such a program? UserIdTAG: 79885 UserNameTAG: ruudoleo CreateTimeTAG: 2013-01-28T14:50:32Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_6003z NumberOfReplyTAG: 4 FirstChildTAG: I discovered you can easily suppress the noise by playing in VLC media player and tweaking the settings in menu extra-effects and filters-audio effects-equalizer- select normal and put the sliders of 3,6 and 12 kHz to -12dB and activate the filter. Now that mosquito is gone buzzing in my ears. FirstChildUserIdTAG: 79885 FirstChildUserNameTAG: ruudoleo FirstChildCreateTimeTAG: 2013-01-28T15:06:46Z SecondChildTAG: If you know of any free software (as in "free beer") for Windows or Linux which can edit the audio, do let me know. I can run my videos through an equalizer and suppress the issue. I could go one step further. Tell me how to load the video into gnu octave or scilab so that I can edit the audio. I'll run it through a filter and suppress it even more B-) SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2013-01-28T15:32:30Z FirstChildTAG: Those are my videos. I was using the open source driver on Fedora so my CPU fan would, for some reason, make a lot of noise. In later videos it's less (because I switched to Windows for recording) but I can't do anything to remove it entirely. :-( I will re-record the first few *really bad* ones when I'm free after the 10th of Feb. FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2013-01-28T15:29:20Z FirstChildTAG: Aswith, with the videos from 6003z I mean the OCW videos from 6-007S not the videos who were made afterwards. I'm still trying with the trial version of http://www.magix.com/us/video-sound-cleaning-lab/ which seems to do a very good job but takes a long time. When one is finished I'll post the link so you can compare sound quality. FirstChildUserIdTAG: 79885 FirstChildUserNameTAG: ruudoleo FirstChildCreateTimeTAG: 2013-01-28T16:53:21Z SecondChildTAG: Oh ok. Yeah I used an equalizer too. SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2013-01-28T17:53:07Z FirstChildTAG: Aswith, I made some tests to denoise your files you can download or play them at http://www.ruudoleo.com/BB_6003z/viewtopic.php?f=4&t=12 FirstChildUserIdTAG: 79885 FirstChildUserNameTAG: ruudoleo FirstChildCreateTimeTAG: 2013-02-03T22:34:11Z SecondChildTAG: Thanks ruudoleo! How did you do it? I'll denoise them before uploading next time. This isn't a big deal but it's As**h**with :-) SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2013-02-05T07:24:13Z SecondChildTAG: I did it with the Magix video sound cleaning lab. I rerendered them also to PAL 720 and changed streamrate to 300bits/sec to keep filesize small. But I have now found you can do better and faster with free tools. First demux the video and audio with [mp4 demux/muxer][1] Then denoise the soundfile with [Audacity][2] and export as ac3 format (in Audacity you can record a noise profile and use this to clean the track (in effects noise removal). Ofcourse you can probably do this in Octave also. Then you can mux the video and soundfile again with the previous program (This all only 10 minutes and the video stays untouched unlike the Magix program wich rerenders the file and take hours to complete--> Looks like I better do the job again). After this you can also enhance the playing in browsers, so that the file does not download first but can directly start playing by placing the timing metadata at the beginning of the file. This can be done with a free tool [qtindexswapper][3] but for this to run you need Adobe Air. Hope you can follow my gibberish explanation. [1]: http://www.videohelp.com/tools/MP4Muxer [2]: http://portableapps.com/apps/music_video/audacity_portable [3]: http://renaun.com/blog/code/qtindexswapper/ SecondChildUserIdTAG: 79885 SecondChildUserNameTAG: ruudoleo SecondChildCreateTimeTAG: 2013-02-07T00:11:42Z SecondChildTAG: Don't forget if you mix to put in the frames per second the same as in your video otherwise the sound will be out of sync. SecondChildUserIdTAG: 79885 SecondChildUserNameTAG: ruudoleo SecondChildCreateTimeTAG: 2013-02-08T08:15:31Z SecondChildTAG: Alright so it's audacity. Actually I was wondering how you took the audio out of the video and put it back. I was under the impression that the software you have can denoise the audio in place. Speaking of audacity, did you sign up for the digital sound design course in Coursera? SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2013-02-08T11:09:49Z SecondChildTAG: No, I will have my hands full on 6003z as I have a busy period now. I redid the cleaning of sound for all the videos without rerendering, that keeps the video sound in sync and video quality is not changed. SecondChildUserIdTAG: 79885 SecondChildUserNameTAG: ruudoleo SecondChildCreateTimeTAG: 2013-02-08T23:53:45Z IndexTAG: 1208 TitleTAG: Jury duty - CECC 2 Just in case anybody wondered what it's like being in the jury for the Circuits and Electronics Classmates Contest this time around - [this video](http://youtu.be/7WmMcqp670s) pretty much sums it up for me. The results will hopefully be in no later than Monday evening EST. UserIdTAG: 90796 UserNameTAG: ChaunceyGardiner CreateTimeTAG: 2013-01-28T04:34:14Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I am anxious to see the result of the Jury xD. All entries were so good!I imagine that the decission it is being difficult :) Thank you for all this. I know that all the Jury is really responsible and it is taking this the most fair as possible. Thank you :) FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2013-01-28T11:31:44Z SecondChildTAG: I am anxious too lol well still it was great to participate in this and thanks to everyone involved! SecondChildUserIdTAG: 219780 SecondChildUserNameTAG: ReconIII SecondChildCreateTimeTAG: 2013-01-28T15:21:16Z IndexTAG: 1209 TitleTAG: CECC - Last Day. Time To Hurry Up! If you haven't submitted your entry yet, do so soon. You have one more day. All the best! UserIdTAG: 14386 UserNameTAG: ashwith CreateTimeTAG: 2013-01-14T04:23:29Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1210 TitleTAG: I would like to help in learning Mechatronics Hi guysI'm Ahmed from EgyptI would like to learn MechatronicsI do not know anything about them, but I am looking forward to اتعلمها and learn all Mavera of expertiseOr to learn sports learn physics ..... etc.And also ZGeek to learn all the material Taihlna to start aOr that Get Sakeshn extended Physics and Sports ..... etc. As I saidIn the end,Thank you very much for much Hmae site uniforms introduced me to youYour friend Ahmed UserIdTAG: 999815 UserNameTAG: Mr70o0 CreateTimeTAG: 2013-01-11T17:36:25Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: مانصحكش يا أحمد بصراحه :) FirstChildUserIdTAG: 395257 FirstChildUserNameTAG: blackguitar FirstChildCreateTimeTAG: 2013-01-12T00:04:19Z SecondChildTAG: hi ahmed its Deep, i am from India, i have read your message ,even i want to learn about mechatronics but mostly i know about it is combination of electronics with programmimg the whole circuits and also allowing the michanical components or parts to move in the desired way . SecondChildUserIdTAG: 1020269 SecondChildUserNameTAG: dcm03 SecondChildCreateTimeTAG: 2013-01-15T20:21:25Z IndexTAG: 1211 TitleTAG: Question How do you know that i1, i2, i3 are leaving the node, not entering? UserIdTAG: 702959 UserNameTAG: Sashkow CreateTimeTAG: 2013-01-06T16:12:52Z VoteTAG: 2 CoursewareTAG: Week 1 / Node analysis practice, part 3 CommentableIdTAG: 6002x_L2Node2 NumberOfReplyTAG: 2 FirstChildTAG: Would *you* enter the node if everybody else was leaving ? Seriously, though - if you want to, you can have each current enter the node if you want to. It doesn't matter - just pay attention to the signs of each current as you define them. FirstChildUserIdTAG: 90796 FirstChildUserNameTAG: ChaunceyGardiner FirstChildCreateTimeTAG: 2013-01-06T16:14:55Z FirstChildTAG: when you asume that the current is leaving the node, and after solving for i you get negative value "-" that mean ===> the current is entering the node not leeaving; very simple ;) - what this mean : we dont know that the current leave or enter, we just assume FirstChildUserIdTAG: 236136 FirstChildUserNameTAG: Hamid-ch FirstChildCreateTimeTAG: 2013-01-06T21:12:38Z SecondChildTAG: But whatever your assumption - keep things consistent! If you "decide" current always enters an element at the positive terminal, don't decide tomorrow to have it entering the negative. Pick a method and stick with it. SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2013-01-07T05:24:27Z IndexTAG: 1212 TitleTAG: To staff I am very grateful for the certificate. I would like to payback by accepting to be a community TA. Please give me the guidelines. UserIdTAG: 446722 UserNameTAG: Richmond CreateTimeTAG: 2013-01-06T06:06:48Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Yes, it is an interesting question to me too. FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2013-01-06T09:35:39Z FirstChildTAG: For guidelines please take a look link below (Myrimit post) https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/undefined/threads/50e5992934c7bc2700000007 the same info in Course Updates & News chapter posted November 19 FirstChildUserIdTAG: 325197 FirstChildUserNameTAG: Vitali_Jerin FirstChildCreateTimeTAG: 2013-01-06T12:50:22Z SecondChildTAG: Vitaly, do you want join to the TA community? SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2013-01-06T16:44:43Z FirstChildTAG: Richmond, you should try to participate more in the forum. I only see a few posts from you all within the last 12 days. I think Sergtronix would do a good job as a TA, as long as it isn't *my* job. :) Sergtronix would make the top 3 in my list for potential TAs. You both can apply on the course info page under November 19th, or use Vitali's link above. They put the request out over a month ago, so I cannot guarantee that the positions have been not been filled already. Good luck! FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2013-01-06T14:18:19Z SecondChildTAG: I have never thought... To be TA community member is a big responsibility. SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2013-01-06T16:51:10Z SecondChildTAG: What caught cunning observation and retentive people. SecondChildUserIdTAG: 402617 SecondChildUserNameTAG: Vl SecondChildCreateTimeTAG: 2013-01-06T18:14:11Z IndexTAG: 1213 TitleTAG: money in exam i m from india ...95$ is too huge amount ...may be it may be moderate in usa ....can amount not vary from country to country... UserIdTAG: 285222 UserNameTAG: dhaval24 CreateTimeTAG: 2013-01-06T05:39:35Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: Well it's all up to you then. Even I'm from India and for me 95 dollars is a little bit on the costly side, considering that this isn't going to be a major certificate for me. This certificate will indeed have some value, maybe a lot, but since 'circuits and electronics' is a basic course I don't know if it's really worth the price. One the plus side, we'll be getting a certificate from the GREAT MIT! and on the negative side, its quite costly. Also this certificate won't carry a grade. So getting around 60 or so in the end would be same as getting 95+. So finally it comes down to how much you would want to pay for the certificate fro MIT. FirstChildUserIdTAG: 131526 FirstChildUserNameTAG: nithin_243 FirstChildCreateTimeTAG: 2013-01-06T06:50:47Z SecondChildTAG: I do agree with some your words.General question is "What will give you proctored exam certificate?" SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2013-01-06T09:39:08Z FirstChildTAG: I agree it's costly for us but it's cheaper than taking the course at MIT. Also you have to pay only if you want the proctored certificate. The learning is still free. I wouldn't be able to take the proctored exam as a student. I can now only because I can fund myself. FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2013-01-06T07:23:52Z FirstChildTAG: i have a doubt in here.. what would the certificate say now.(for the proctored exam. again ONLINE through EDX.. coz that wouldn't make any difference as long as we have this certificate and we can prove it ourselves. FirstChildUserIdTAG: 36121 FirstChildUserNameTAG: nikhilkumar FirstChildCreateTimeTAG: 2013-01-06T07:48:41Z FirstChildTAG: In Norway you can take the whole course with auditorium, instructors, lab, and proctored exam and points towards a degree for $17. Not sure what a proctored exam would add to this course. FirstChildUserIdTAG: 142857 FirstChildUserNameTAG: viking2 FirstChildCreateTimeTAG: 2013-01-06T09:16:22Z SecondChildTAG: Sounds good :) Dear Viking2, may you tell us , as example, about which Norway Universities are you talking? SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2013-01-06T09:42:04Z SecondChildTAG: Yeah, but then you would have to pay 50% income tax when you start using your knowledge. And on what's left they'll charge you 25% sales tax. And gas taxes making gasoline cost $10 a gallon in a country that is one of the biggest oil exporters in the world. If you have anything left after that, they'll slap several hundred percent in taxes on the price of a car. That car will wear down in a hurry because they can't afford to take care of their roads. Face it, Norway is a paradise for poor people and others that want to take advantage of the system without doing anything. For anybody with half a brain, it's a living tax Hell. SecondChildUserIdTAG: 90796 SecondChildUserNameTAG: ChaunceyGardiner SecondChildCreateTimeTAG: 2013-01-06T11:03:11Z SecondChildTAG: ChaunceyGardiner, You are exactly right. It is cold also. SecondChildUserIdTAG: 142857 SecondChildUserNameTAG: viking2 SecondChildCreateTimeTAG: 2013-01-06T14:32:59Z FirstChildTAG: They have said in the presentation of eDX in the summer of 2012, that this FIRST courses will be charge free. FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2013-01-06T15:55:24Z SecondChildTAG: so that we are among the lucky ones SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2013-01-06T16:12:23Z IndexTAG: 1214 TitleTAG: Proctored Examination As i have also received the email regarding the **Proctored Examination** like others. So i have lot of doubts regarding it, as some are similar that others posted in the discussion forum but still have few doubts as follows: 1.**EXAMINATION FEE** - what is the mode of paying the fee? in which currency we have to paid i.e. in USD or equivalent amount in my country's currency? as i am from INDIA,$95 USD is a huge amount for any examination fee..is this fee will differ from country to country? 2 . **Syllabus for exam**- what is the syllabus for Proctored exam? is it same as our 6.002x syllabus? UserIdTAG: 114864 UserNameTAG: Vikaash CreateTimeTAG: 2013-01-05T14:25:39Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: 1- 95 USD will be the pilot price. Pearson Vue accepts VISA, MasterCard, American Express and JCB. (Due to international verification issues, debit cards are not accepted.) 2- The Syllabus should be the same, but we will wait to see if there are any changes. The estimated price for 2012-13 for undergraduates at MIT is $57,010 USD. While I understand that this only one class, it is still a bargain no matter how you look at it. Do your best, if you need the proctored test it will be a wise investment to do so. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2013-01-06T14:56:34Z SecondChildTAG: Yes, the syllabus is exactly the same. SecondChildUserIdTAG: 96452 SecondChildUserNameTAG: Lyla SecondChildCreateTimeTAG: 2013-01-29T14:59:42Z FirstChildTAG: At this point it is not clear who the target audience is. Companies routinely pay "big bucks" to get their employees certified in specialties. $95 would be inconsequential to a company if this sort of training served their needs. I guess time will tell. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2013-01-06T15:18:31Z IndexTAG: 1215 TitleTAG: Proctored Examination Hi all, I just received an email asking me whether I want to participate in the proctored examination for 6.002x at a Pearson Vue test center. I have a couple of doubts regarding this issue. 1. How to I register for the proctored examination (I checked http://www.pearsonvue.com/programs/) but there was no mention of any proctored edX course. 2. After sitting for the proctored examination will we receive an official transcript? With college credits? 3. Will the portion of the exam be the same as the portion of the unproctored final exam? 4. How can I find the closest Pearson Vue center? Thank you in advance. UserIdTAG: 85490 UserNameTAG: yashsavani CreateTimeTAG: 2013-01-04T20:51:00Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hi! I believe that more details will be given in the following days about how to register. I find it really impossible to get college credit, but i think that it will worth a little bit more in general, as it will prove that we have indeed passed the exam ourselves. Cheers FirstChildUserIdTAG: 304587 FirstChildUserNameTAG: Vasso FirstChildCreateTimeTAG: 2013-01-04T20:59:41Z SecondChildTAG: I find it reasonable that a educational facility set up by MIT faculty, associates and proctored by Pearson Vue to be effective enough for a generic college credit. Whether or not this happens remains to be seen. In my opinion an organization that does not recognize a proctored edX certificate is doing so for an alternative reason. ie: Political or business*emphasized text* reasons. A few months ago MIT took first place in the top 200 Universities in the *World*, based on QS World University Rankings, followed by Cambridge and Harvard. That ranking should satisfy any reasonable institution as to the quality and content of the course taken. If not then that is their prerogative. Remember this particular course is taught by the fellow who literally co-authored the textbook that MIT uses with it's electrical studies. The rest of the teaching staff, also world-class. "They" didn't put a man on the moon, M.I.T. did. http://en.wikipedia.org/wiki/Apollo_Guidance_Computer In earlier attempts Professor Draper also invented the inertial guidance system while researching the Apollo mission. He was able to successfully navigate an airplane from the east to west coast (USA) using nothing but mechanical gyroscopes that he created with consulting from the Waltham watch company. He is also credited for inventing the "Clean room" which is still used today in the manufacturing of computers. Very exciting times indeed. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2013-01-05T00:25:49Z SecondChildTAG: **Ahhh...** back when the U.S. actually manufactured stuff! Brings back memories of when my dad worked in tool-rooms and machine shops before computerization (e.g. CNC programmed machining and tool-die making), and actually made a lot of money doing so since it was a "craft." Back when the Northeast U.S. was a manufacturing powerhouse instead of the "Rust Belt." I wonder about Waltham Watch...looked it up and they went out of business. My dad worked for Remington Repeating Rifles and for Pratt & Whitney; they're still open but the tech used now is so different from how it used to be done. I was going to go into CNC and mechanical engineering, but my dad pushed me in a different direction. I apprenticed in a few small shops during university, and worked the night shift as the QC-inspector to pay for school; these shops machined the subminiature connector ends for RF cables (the same type as your wireless antenna uses to connect to your WLAN router; but these were used in the F-22 Fighter plane for the U.S. military) out of metal stock; I remember a roomful of ladies being paid minimum wage just to solder the wire into the small groove (that was actually hand-milled) on the other end of the center pin on the inside of those connectors. Then a brass plug was placed on the back as a "cover", and brazed on; then plated to look like one piece. I guess no "crimping" because you don't want an F-22 jet to fail from a loose connection (vibration), so an all-soldered and 100% tested connector cable. Each day at work I'd see waste, and room for improvement where one computerized machine could take the job of 40-or-so employees (for example the room full of ladies I mentioned). Between that and seeing what happened to my dad (working his whole life just to become obsolete) was the "aha"-moment I needed to find out that the manufacturing industry wasn't for me; though I am still very fascinated with old machinery! Engineering new and faster methods of production is the way to go, and the financially lucrative way; hence the study of electrical engineering for me. There is much romance, though, in creating your own tooling, fixtures, jigs, setting up a machine correctly, etc. - where a worker can just sit down and **manually** create mass-produced components using a setup you created - the same way Henry Ford's workers did it a 100+ years ago! SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2013-01-05T19:13:44Z FirstChildTAG: 1: Soon we will have more information. 2:"College Credits" would be determined by the institution which you are applying to. Pearson Vue is a very respected leader in the electronic testing industry and does testing for school, business, government and military establishments. 3: I would expect that we should be prepared for different questions and perhaps a significantly longer exam. (Since it is only one test, it would be heavily weighted when you consider it covers midterm exam topics as well) 4: You can locate a Test Center here: http://www.pearsonvue.com/vtclocator/ FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2013-01-05T01:14:54Z IndexTAG: 1216 TitleTAG: To Staff: What is that verification and signature? Hi, Thank you very much for your great effort :) I have downloaded my certificate, and clicked on that link in the bottom of it, as I expected it was a way of verification... but I noticed that there is something like a digital signature, I followed the steps provided in the link "What's this" but nothing special happened to my certificate! Let me tell you what I thought... The link provided in the cert. is to ensure that the student has a valid certificate number and he did successfully completed a course through edX. The PDF of the cert is digitally signed and those steps are just to ensure that the PDF file is really signed by edX. Am I right about those thoughts? Thank you very much in advance... I'm just curious to understand the details :D UserIdTAG: 365201 UserNameTAG: sirajmuhammad CreateTimeTAG: 2013-01-04T17:33:42Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Yes Sir, you are correct. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2013-01-04T18:25:33Z FirstChildTAG: Correct...though I'm surprised you haven't encountered this before! Our state government (New Jersey U.S.A.) is using such links as digital signatures / online verification on more and more documents (example: Business Registrations) it issues. I'm not sure about other states, much less the rest of the world, but it seems like the best and easiest way to implement one level of fraud-prevention. With currency (and also goods) counterfeiting so prevalent, one day there will be a URL link (likely embedded via RFID or barcode) so you can verify if your Gucci handbag is real, if your Nike sneakers are not from some Chinese counterfeiter, if Intel really made that processor in your PC; and most importantly, if your paper money is not made by some printing press in a thief's basement or garage! FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2013-01-05T18:33:57Z SecondChildTAG: COOOOOOL ! thank you for these tremendous info SecondChildUserIdTAG: 418182 SecondChildUserNameTAG: euldji2005 SecondChildCreateTimeTAG: 2013-01-09T00:28:17Z IndexTAG: 1217 TitleTAG: Where to continue from here? This course was an exhilarating experience for me :) If I had to continue to more advanced topics in Electronics, which courses could I take up. There are different courses listed on the OCW website under the EECS department (which are of course are in a different format). Is 6.003 the continuation of the 6.002 course? UserIdTAG: 298775 UserNameTAG: surja CreateTimeTAG: 2013-01-04T02:20:36Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Yep,signals and systems. FirstChildUserIdTAG: 136959 FirstChildUserNameTAG: christerpher FirstChildCreateTimeTAG: 2013-01-04T02:52:35Z FirstChildTAG: From what I've been able to tell, there's a lot of overlap between 6.003 and [this course][1] on coursera, if you're happy trying a different platform and want some distinct recognition for your learning. Without the need or want for recognition, you could just as easily follow along the OCW course, or alternatively see if the the [site created for it][2] is still functional. [1]: https://www.coursera.org/course/dsp [2]: http://6003z.amolbhave.in/ FirstChildUserIdTAG: 26931 FirstChildUserNameTAG: Csscade FirstChildCreateTimeTAG: 2013-01-05T04:00:34Z IndexTAG: 1218 TitleTAG: תודה לכם בס“ד תודה לפרופסור ולכל האירגון UserIdTAG: 346056 UserNameTAG: fiatlux CreateTimeTAG: 2013-01-03T19:00:38Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1219 TitleTAG: Certificate verification How can i verify my certificates on OSX system? I read about the process for "GPG keychain Access" program but i don't get it! I tried it too, but i'm not sure because i used only the one file. Any help? UserIdTAG: 345967 UserNameTAG: Tsipis CreateTimeTAG: 2013-01-03T17:09:47Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: The GPG verification is the public distributed verification protocol that you can use to verify certificates completely independently of any service. However, because it is kind of complicated, edX also provides its own verification service using the link at the bottom of the certificate. If you just want an easy way to verify your certificate use that link. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2013-01-03T19:41:01Z FirstChildTAG: so we don't have to go to GPG complicated process,,whenever we have to show validation of our certificate ,we just have to go to that link? FirstChildUserIdTAG: 267220 FirstChildUserNameTAG: ABHINAVSAXENA318619 FirstChildCreateTimeTAG: 2013-01-04T05:57:59Z IndexTAG: 1220 TitleTAG: hard copy of certificate will the hard copy of certificate be sent through d post or only electronic certificate(PDF) UserIdTAG: 169416 UserNameTAG: subramanya26shin CreateTimeTAG: 2013-01-03T16:04:45Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Im not STAFF, but Im sure that it will not be send by the Post. Try to print it on the preferred paper by Color Laser Printer :) FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2013-01-03T16:12:15Z SecondChildTAG: no hard copy!! SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2013-01-03T16:14:24Z SecondChildTAG: okie :) SecondChildUserIdTAG: 169416 SecondChildUserNameTAG: subramanya26shin SecondChildCreateTimeTAG: 2013-01-03T16:15:07Z SecondChildTAG: proctored exam soon..... SecondChildUserIdTAG: 353709 SecondChildUserNameTAG: anshu10750 SecondChildCreateTimeTAG: 2013-01-05T06:26:51Z FirstChildTAG: We only distribute the electronic versions. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2013-01-03T19:39:16Z SecondChildTAG: okie sir.. SecondChildUserIdTAG: 169416 SecondChildUserNameTAG: subramanya26shin SecondChildCreateTimeTAG: 2013-01-04T04:34:08Z IndexTAG: 1221 TitleTAG: To update name in certificate to my full name. I changed the name to my full name now but it is showing only the previous one in the certificate. How to solve it.Please Help. UserIdTAG: 369051 UserNameTAG: praveencbe0 CreateTimeTAG: 2013-01-03T15:04:56Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Unfortunately the course has been completed. Name changes should have been done before the generations of Certificates. You could try contacting mit-6002x@edx.org, although I am not making any promises. For obvious reasons it is not desirable to have students change their names after a course has completed. Generally speaking, things of this nature should have your full legal name, just like any drivers license, library card, ownership of property etc. etc. edX also enabled the use of a screen name, which should have led to less confusion about what to fill in under the heading "Full Name". See what you can do, worst case scenario you take it again in the spring or take a proctored exam if it becomes available. Have a good New Year. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2013-01-03T15:23:40Z SecondChildTAG: Thank you for your response. Happy New year. SecondChildUserIdTAG: 369051 SecondChildUserNameTAG: praveencbe0 SecondChildCreateTimeTAG: 2013-01-03T16:17:09Z SecondChildTAG: That's what you get for putting "Mickey M. Mouse" or "Mr. Homer Simpson" under your Full Name...you think it's funny and will look good hanging on your wall...until you realize that you can't really just print out multiple certificates with multiple name changes! SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2013-01-04T15:52:47Z IndexTAG: 1222 TitleTAG: Certificates are on the way! Certificates are being generated at this moment. Some have already received their certificates while some are still waiting. Please be patient, the wheels are in motion. (I do not know if they are distributing alphabetically, random, by country or times zone.) Thanks for your understanding. UserIdTAG: 292543 UserNameTAG: Pennypacker CreateTimeTAG: 2013-01-03T14:00:11Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: yes,they have generated for me..This is the first time i'm taking this course and it all ended in success.Thanks for AnanthAgarwal and team.Hope they conduct many more wonderful courses in future FirstChildUserIdTAG: 150120 FirstChildUserNameTAG: krishna1993 FirstChildCreateTimeTAG: 2013-01-03T14:03:51Z SecondChildTAG: do they contain grades? SecondChildUserIdTAG: 446998 SecondChildUserNameTAG: kumar_vikash SecondChildCreateTimeTAG: 2013-01-03T17:06:16Z FirstChildTAG: Yes. I got mine. I noticed that there is a change in my final exam mark which they accept my wrong answer in Q6. Anyway thanks for providing such a great course. I learnt a lot. Thanks and see you all soon. FirstChildUserIdTAG: 529515 FirstChildUserNameTAG: Low FirstChildCreateTimeTAG: 2013-01-03T14:17:33Z SecondChildTAG: Congrats and nice that they changed the grade in your advantage.. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2013-01-03T14:21:19Z FirstChildTAG: Just got mine. FirstChildUserIdTAG: 99628 FirstChildUserNameTAG: Pranjal16 FirstChildCreateTimeTAG: 2013-01-03T14:52:17Z FirstChildTAG: I'm glad I've some spare batteries for my pacemaker ... At fist there was this text: Final course details are being wrapped up at this time. Your final standing will be available shortly. Later it changed to: Your Certificate is Generating And finally after 2 hours or so: Download Your PDF Certificate Where is the champagne? FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2013-01-03T14:15:40Z SecondChildTAG: How much time does it take "You certificate is Generating."??? SecondChildUserIdTAG: 77046 SecondChildUserNameTAG: Sarwar20 SecondChildCreateTimeTAG: 2013-01-03T14:28:51Z SecondChildTAG: I have no idea, but it can be long. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2013-01-03T15:02:57Z SecondChildTAG: I don't have a certificate yet. SecondChildUserIdTAG: 43957 SecondChildUserNameTAG: Rafael_Nunes SecondChildCreateTimeTAG: 2013-01-03T19:12:14Z IndexTAG: 1223 TitleTAG: I didn't get the grade on the certificate. Hello. I just downloaded the certificate but the grade(99%) is not on it. I saw several previous certificates and they had the grade on them. Am I the only one or it's a new procedure? Regards. UserIdTAG: 359677 UserNameTAG: guillermb CreateTimeTAG: 2013-01-03T11:29:25Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: No... FirstChildUserIdTAG: 207728 FirstChildUserNameTAG: RossyFromBulgaria FirstChildCreateTimeTAG: 2013-01-03T11:38:59Z SecondChildTAG: No, what? SecondChildUserIdTAG: 359677 SecondChildUserNameTAG: guillermb SecondChildCreateTimeTAG: 2013-01-03T11:40:13Z SecondChildTAG: No, you won't be the only one. At the las moment from edx.org told us that they won't include the grades in the certificates like previous year. So... congratulations. Aniway, you can show us your certificate, part of us are still waiting for ours certificates :) SecondChildUserIdTAG: 207728 SecondChildUserNameTAG: RossyFromBulgaria SecondChildCreateTimeTAG: 2013-01-03T12:05:42Z SecondChildTAG: P.S.: Sorry for the mistakes, my keyboard at work sucks. SecondChildUserIdTAG: 207728 SecondChildUserNameTAG: RossyFromBulgaria SecondChildCreateTimeTAG: 2013-01-03T12:06:53Z SecondChildTAG: Oh Don't worry fot the mistakes. I don't know why the grades won't be included so...what's the difference between getting 60% and 100%? SecondChildUserIdTAG: 359677 SecondChildUserNameTAG: guillermb SecondChildCreateTimeTAG: 2013-01-03T12:11:43Z SecondChildTAG: ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13572151891343666.png SecondChildUserIdTAG: 359677 SecondChildUserNameTAG: guillermb SecondChildCreateTimeTAG: 2013-01-03T12:13:22Z SecondChildTAG: nice one! SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2013-01-03T12:14:23Z SecondChildTAG: Thank you but I expected the grade on it... :( SecondChildUserIdTAG: 359677 SecondChildUserNameTAG: guillermb SecondChildCreateTimeTAG: 2013-01-03T12:16:30Z SecondChildTAG: Wow, it looks great! Congratulations again. It's not the same without the grades, but anyway, it's better than nothing. I got an A and really wanted to see it on the certificate. SecondChildUserIdTAG: 207728 SecondChildUserNameTAG: RossyFromBulgaria SecondChildCreateTimeTAG: 2013-01-03T12:18:01Z FirstChildTAG: I wish they reconsider showing the grades. The course has been tough. FirstChildUserIdTAG: 398594 FirstChildUserNameTAG: DaveyJC FirstChildCreateTimeTAG: 2013-01-03T14:32:35Z FirstChildTAG: Guiilermb, congratulations for your grade. No you don´t get specific grades. The following is stated in the course description presented in September 2012: Will certificates be awarded? Yes. Online learners who achieve a passing grade in a course can earn a certificate of mastery. These certificates will indicate you have successfully completed the course, but will not include a specific grade. Certificates will be issued by edX under the name of either HarvardX, MITx or BerkeleyX, designating the institution from which the course originated. For the courses in Fall 2012, honor code certificates will be free. Happy 2013. FirstChildUserIdTAG: 403987 FirstChildUserNameTAG: electricon FirstChildCreateTimeTAG: 2013-01-03T12:15:47Z SecondChildTAG: Thank you for the info, electricon. I didn't read that. What a pity!! I really wanted the grade otherwise what's the point to get more than 60%? SecondChildUserIdTAG: 359677 SecondChildUserNameTAG: guillermb SecondChildCreateTimeTAG: 2013-01-03T12:18:56Z SecondChildTAG: The point is that you learned a bit more than a person who got 60% :p SecondChildUserIdTAG: 370343 SecondChildUserNameTAG: JJarquin SecondChildCreateTimeTAG: 2013-01-03T18:02:36Z SecondChildTAG: nicely thought SecondChildUserIdTAG: 433368 SecondChildUserNameTAG: SurbhiMahajan SecondChildCreateTimeTAG: 2013-01-04T14:34:50Z IndexTAG: 1224 TitleTAG: certificate certificates out UserIdTAG: 285222 UserNameTAG: dhaval24 CreateTimeTAG: 2013-01-03T07:42:46Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1225 TitleTAG: how does this circuit work? Hi everyone, Happy New Year. For Christmas I got a little soldering kit [kit information][2]. I am trying to analyse how it works before I build it. Can anyone help me?? I have copied the circuit diagram below. It is a "crawling microbug". It has 2 little motors that are controlled by two photoresistors so that it walks towards light. There is a switch (SW2) to change between 2 different types of "walk". The two parts on the right seem to be for switching the two motors on when light shines on the photoresistors. I don't really understand what the left part of the circuit does / how it might work. Also, what are the capacitors for? ![circuit diagram][1] [1]: https://edxuploads.s3.amazonaws.com/13571345151343653.png [2]: http://www.vellemanprojects.eu/products/view/?id=346576 UserIdTAG: 434869 UserNameTAG: patey CreateTimeTAG: 2013-01-02T13:57:44Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Left part of this schematic is "unstable" multivibrator.It has two outputs, first is Collector (C) of T1 and the second is C of T2.These outputs are in antiphase to each other.This multivibrator do generate pulses (square wave) with time and period depended from RV1, R3, R4 and С1 and C2 values(when C1=C2 and R3=R4 duty cycle is 2). When switch SW2 is in upper position, Bases (B) of T3 and T5 are tied together(Note here, that T3 and T5 are simply switch which allow to control light sensor signal amplifier:when B of T3/T5 is "High" amplifier is disabled).So, it allow to move your toy periodically. In the upper position motors may be turned on simultaneously , in the lower position - alternately ( step left, step right movement). FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2013-01-02T14:38:39Z SecondChildTAG: Thanks! Why are the motors driven from T7 and T8, and not by T4 and T6? Is that so that there is more amplification of the light sensor control? SecondChildUserIdTAG: 434869 SecondChildUserNameTAG: patey SecondChildCreateTimeTAG: 2013-01-02T16:04:25Z SecondChildTAG: Yep..npn+pnp there give us complementary Darlington transistor which has VERY large $h_{21}$ coefficient and low ON voltage (~0.7V) SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2013-01-02T16:35:49Z SecondChildTAG: Motors are drived not from collectors of T1\T2 because of R1\R2 - useless loss of power. So, multivibrator used as "control element", 2 transistors - power output cascade, with low internal resistance. It's make scheme more efficient. SecondChildUserIdTAG: 394836 SecondChildUserNameTAG: v2g6ch4 SecondChildCreateTimeTAG: 2013-01-03T09:57:34Z SecondChildTAG: Where is mentioned that T1/T2 are used for direct driving motors? About "power output cascade, with low internal resistance"-not resistance, it is wrong, but **with LOW SATURATION VOLTAGE, and high current gain**. SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2013-01-03T10:33:34Z SecondChildTAG: Ok, ok, it's my fault ) SecondChildUserIdTAG: 394836 SecondChildUserNameTAG: v2g6ch4 SecondChildCreateTimeTAG: 2013-01-03T10:34:41Z FirstChildTAG: Dear Patey, let me try... The left combination of transistors, capacitors and resistors is a typical example of an **astable multivibrator**. It means that it is intended to oscilate. It works at a frequency specified by the R-C elements. Signals at each collector of each transistor (T1 and T2) are complemented -> if one is hight, then the other one is low and vice versa. I suppose that applying the SW1 switch in the upper (**bold**) position gives you a simultanous operation of motors (like a butterfly), while the lower position of SW1 (dashed line) gives you alternating operation of motors (like crawl). Regards FirstChildUserIdTAG: 420332 FirstChildUserNameTAG: gapa FirstChildCreateTimeTAG: 2013-01-02T14:50:18Z SecondChildTAG: OK, thanks, now I know what it's called I can look up how it works. [Wikipedia][1] has a page about multivibrators. [1]: https://en.wikipedia.org/wiki/Multivibrator SecondChildUserIdTAG: 434869 SecondChildUserNameTAG: patey SecondChildCreateTimeTAG: 2013-01-02T16:13:52Z FirstChildTAG: Hi patey ! Just search on the web for a 555 based version. You will use less components and replace T7 and T8 with BD 140 if you intend to use rechargeable batteries, because BC 557 will overheat. FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2013-01-02T15:50:21Z SecondChildTAG: Thanks for the tip. I will leave the 555 timer for now since it comes with a printed circuit board and all the components. Is a BD140 just a higher power rated PNP transistor than the BC 557? SecondChildUserIdTAG: 434869 SecondChildUserNameTAG: patey SecondChildCreateTimeTAG: 2013-01-02T16:17:33Z SecondChildTAG: BC557 will be enough for the **micropower DC motor** SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2013-01-02T16:37:31Z SecondChildTAG: Not if you use rechargeable batteries, which provide more current and your motor will get stuck. :) SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2013-02-11T08:32:46Z IndexTAG: 1226 TitleTAG: Please explain Final course details are being wrapped up at this time. Your final standing will be available shortly. **What it means** UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2013-01-01T08:54:08Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: This means that Staff working on the certificates at this time. They ask to be patience a bit. FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2013-01-01T09:58:29Z SecondChildTAG: I'm thankful for all their hard work. I'm curious to know what is meant by "we still need a couple of days to get the certificates **signed and distributed**". Do they mean digital signatures? SecondChildUserIdTAG: 297370 SecondChildUserNameTAG: ayush3504 SecondChildCreateTimeTAG: 2013-01-01T16:28:06Z SecondChildTAG: No, it is the signatures that the relevant institute and course staff put on the displayable version of the certificate. SecondChildUserIdTAG: 96452 SecondChildUserNameTAG: Lyla SecondChildCreateTimeTAG: 2013-01-02T16:10:40Z SecondChildTAG: Okay SecondChildUserIdTAG: 297370 SecondChildUserNameTAG: ayush3504 SecondChildCreateTimeTAG: 2013-01-03T00:27:20Z IndexTAG: 1227 TitleTAG: HNY! Have a happy and safe new year everyone:D UserIdTAG: 148523 UserNameTAG: span993 CreateTimeTAG: 2013-01-01T08:22:48Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You too, fine Sir! FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2013-01-01T12:57:11Z IndexTAG: 1228 TitleTAG: About the final exam I was under the impression that the final was rather easy. Does anybody know if it was a watered down version of the real exam at MIT? UserIdTAG: 346056 UserNameTAG: fiatlux CreateTimeTAG: 2012-12-31T19:37:26Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1229 TitleTAG: Enjoy Learning in 2013 Wish you all a $$H^AP_PY\ N_EW \ YEAR$$ UserIdTAG: 623210 UserNameTAG: preveen CreateTimeTAG: 2012-12-31T17:53:51Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1230 TitleTAG: Happy New Year from India Happy New Year to all the edx participants. Wish you all a wonderful and prosperous new year. UserIdTAG: 199839 UserNameTAG: vijaykrishnankk CreateTimeTAG: 2012-12-31T17:03:13Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Happy New Year from Indonesia (technically I am in Japan tho :)) And also, although this might be conflicting with new year spirit, please let me express my deepest condolences to Delhi rape case victims and their families. FirstChildUserIdTAG: 201818 FirstChildUserNameTAG: ThreeHundred FirstChildCreateTimeTAG: 2013-01-01T14:37:07Z FirstChildTAG: Happy New Year from Budapest, Hungary for all awaken pupils of this course :) Cheers, -G FirstChildUserIdTAG: 151942 FirstChildUserNameTAG: GyuriK FirstChildCreateTimeTAG: 2012-12-31T18:37:07Z IndexTAG: 1231 TitleTAG: happy online learning and successful year. This year has been online life for me,means i spent most of my days learning one or the other new things and enrolling in many courses(after aug) ,but completed sucessfully only two of them wit **6.002x** and **Power searching with google**(some other still in progress)!!. with so many useful courses online, i can't give away learning and wish aal frnds to be motivated and keep learning. but then friends to quote respected Anant sir'S WAY, as to enjoy **eecs playground**,the ckts need to abide rules n disciplene,though i enrolled many courses, but what i practically felt that every sphere of life is playground and all subjects are like playground and to enjoy the most out of playground one needs to abide disciplene. this was best demonstrated by **Myrimit** in our playground and many others too. myrimit and other community TA u really showed how to help others and enjoy the most of learning.hopefully i wish to do the same in coming year in other courses and not just getting enrolled. thanx **edx, mitx** for this awesome 6.002x demoful course!! UserIdTAG: 269641 UserNameTAG: BAUWA CreateTimeTAG: 2012-12-31T14:08:39Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: That is really nice BAUWA! Cool iniciative! I like your enthusiasm . I am sure that you will help a lot of students, I am really happy for that :) I wish you the best! Happy New Year 2013! Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-31T15:21:06Z IndexTAG: 1232 TitleTAG: to staff It is possible that we come over to MIT for our certificates as regular students do? UserIdTAG: 446722 UserNameTAG: Richmond CreateTimeTAG: 2012-12-31T11:53:54Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1233 TitleTAG: Mesh Analysis I dont remember mesh analysis from any of the lecture videos, can anyone direct me to them? also, I still understand why the thevenin resistance for the first question was 2R. I thought you had to use that whole test voltage source at the input terminals and calculate it that way..... UserIdTAG: 381923 UserNameTAG: matteaton CreateTimeTAG: 2012-12-28T15:12:40Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: "the thevenin resistance for the first question was 2R" what problem number (or week) you are talking about? FirstChildUserIdTAG: 325197 FirstChildUserNameTAG: Vitali_Jerin FirstChildCreateTimeTAG: 2012-12-28T16:45:37Z SecondChildTAG: The Q1 of the final, sorry SecondChildUserIdTAG: 381923 SecondChildUserNameTAG: matteaton SecondChildCreateTimeTAG: 2012-12-28T20:35:30Z SecondChildTAG: Hi Matteaton,about the mesh menthod. Here may help you http://en.wikipedia.org/wiki/Mesh_analysis SecondChildUserIdTAG: 361137 SecondChildUserNameTAG: Ericson SecondChildCreateTimeTAG: 2012-12-29T02:36:23Z SecondChildTAG: ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13567512127848768.jpg After you understand the concepts in wikipedia,the following is the picture of part 1. We have iA=i1-i2 , iA=i1, 2*R*i2 - Vout=0, and Vout= rm* iA, then we get we want. SecondChildUserIdTAG: 361137 SecondChildUserNameTAG: Ericson SecondChildCreateTimeTAG: 2012-12-29T03:25:35Z FirstChildTAG: Hi Matteaton. I think you mean the thevenin resistance from the last question of problem 1 of the final. That resistance is 2*R, because there is no voltage drop caused by a resistance on the place where iA was measured, so that is effectively a short. And seen from input 1, with input 2 not connected, you see effectively 2 R's in series, because input 1 is floating if you try to measure a voltage OVER it (not measured with respect to ground!), which will be 0 volt in this case, whatever the voltage at the place of iA is. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-12-28T17:26:03Z SecondChildTAG: What do you mean floating? Why dont you have to put a dummy voltage across it like it explains in the sequence on dependent voltage sources. When would you have to put a, say 1V test voltage across the terminals, calculate the current at the input, and us V/I to find R? SecondChildUserIdTAG: 381923 SecondChildUserNameTAG: matteaton SecondChildCreateTimeTAG: 2012-12-28T20:39:04Z SecondChildTAG: That is, because the thevenin voltage is measured without a load. And seen from the open input1, you look into the network. This was a tricky question, that's all. All you need to see, is that it doesn't matter whatever voltage is at point iA. If you use ground as reference then at the upper input1 the voltage is the same as on lower input1, (with input2 not connected) so the difference will be zero between the terminals. But in this case you have to determine thevenin R, not referenced to ground, but between the two terminals itself, and therefore it's floating. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-12-28T21:24:07Z IndexTAG: 1234 TitleTAG: To edx please include more courses on electronics or programming in c++(or java) for 2013.. UserIdTAG: 114864 UserNameTAG: Vikaash CreateTimeTAG: 2012-12-27T14:40:41Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: If I'm not wrong then they have already launch that as CS50x-Introduction to Computer Science and 6.00x for python.......have a look that..... FirstChildUserIdTAG: 108863 FirstChildUserNameTAG: shiviz FirstChildCreateTimeTAG: 2012-12-27T14:49:41Z SecondChildTAG: i have already gone through the above mentioned courses but looking for c++ (or java) programming or more on electronics SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-12-27T14:52:52Z SecondChildTAG: i agree with vikaash and if they give some topic on electric machines it will be nice SecondChildUserIdTAG: 342221 SecondChildUserNameTAG: deepkar SecondChildCreateTimeTAG: 2012-12-27T14:56:29Z SecondChildTAG: kk...i hope they will.:) SecondChildUserIdTAG: 108863 SecondChildUserNameTAG: shiviz SecondChildCreateTimeTAG: 2012-12-27T14:58:30Z IndexTAG: 1235 TitleTAG: Va a venir la letra del grado tal y como fue establecido en un principio ? Lo anterior lo pregunto porque viendo la información para el spring del 2013, se observa que a diferencia de la que en principio se dió para el fall 2012 hay un gran contraste con respecto a lo que se ofrece para el próximo año pues las personas que quieran superar su nota no van a poder hacerlo porque según se dice textualmente en https://www.edx.org/courses/MITx/6.002x/2013_Spring/about "certificates will indicate you have successfully completed the course, but will not include a specific grade"- Es una lástima pues yo quería ver si podía superar mi 80% para obtener una A con un 87% o más el próximo año. UserIdTAG: 183712 UserNameTAG: colegiocientifico CreateTimeTAG: 2012-12-27T00:11:20Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hola colegiocientifico, Espero que esto no sea un impedimento y que puedas desafiarte nuevamente, en forma personal, y poder lograr superar tu puntuación anterior, por más que el Certificado no posea una A,B o C. Porque estoy más que segura que lo harás :) Jamás pierdas tu entusiasmo y tu ánimo para mejorar en lo que te propongas día a día, Saludos, Myriam. > Now in English Hi Colegiocientifico, I hope this will not be a deterrent and that you could be able to challenge again yourself, in a personal way, in order to break your last score, no mattering if the Certificate will not have written an A,B or C. I am sure that you will do it :) Never lost your enthusiasm and your encouragment to improve on what you set every day, Greetings, Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-27T00:39:50Z IndexTAG: 1236 TitleTAG: Thanks a lot to EDX and MITx Thaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaank you very much UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-12-25T22:01:29Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1237 TitleTAG: Power electronics converter interesting Hello Interested in power electronics. Is there such a course? And you can read? thanks UserIdTAG: 156233 UserNameTAG: Gleb1 CreateTimeTAG: 2012-12-25T07:09:12Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Probably, I can post you some materials in russian if you'll manage to understand that. Studied power electronics in university, cause i'm power engineer. Also, can post some materials in power system protection(in russian also). FirstChildUserIdTAG: 200319 FirstChildUserNameTAG: Virviil FirstChildCreateTimeTAG: 2012-12-25T08:30:21Z SecondChildTAG: Oh? it's interesting SecondChildUserIdTAG: 156233 SecondChildUserNameTAG: Gleb1 SecondChildCreateTimeTAG: 2012-12-25T14:48:02Z IndexTAG: 1238 TitleTAG: Merry Christmas everyone! Merry Christmas and Happy New Year everyone! UserIdTAG: 143823 UserNameTAG: NathanNadeson CreateTimeTAG: 2012-12-25T00:50:26Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Have a Merry Christmas and Happy New Year for you and your family and friends as well! Jersey Mark FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-12-25T16:36:14Z IndexTAG: 1239 TitleTAG: Thanks from Venezuela I want to thanks to teacher Anant Agarwal and all the edX team by the excellent course, a very good opportunity for learning electronics in a different way. I am studing engineering of telecommunications at the UBTJR Cantv Los Cortijos, so for me was a very challenging task because I had difficulties in math so I pushed harder and finally got the grade. I encourage to all fellows to continue taking courses even if they are tough, all in life has sacrifice, never give up. Merry Xmas UserIdTAG: 296419 UserNameTAG: oscman CreateTimeTAG: 2012-12-24T22:12:30Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I'm agree with you oscman: "I encourage to all fellows to continue taking courses even if they are tough, all in life has sacrifice, never give up." That is the way!!! Luis FirstChildUserIdTAG: 227223 FirstChildUserNameTAG: luisbonelli FirstChildCreateTimeTAG: 2012-12-25T04:50:29Z SecondChildTAG: También estoy muy agradecido con la plataforma edX y la gran cantidad de información gratuita que está ofreciendo el MIT! Saludos oscman (hombre oscilador?) desde Cumaná, Estado Sucre! SecondChildUserIdTAG: 298547 SecondChildUserNameTAG: AnthonyRF SecondChildCreateTimeTAG: 2013-01-05T23:43:33Z IndexTAG: 1240 TitleTAG: thx & congratulattion from Morocco thnx for everyone who used to make this happen .. and congratulattion to every one who finished the course with atleast a C grade :pp !! i got 99% hope that i enroll in more other lessons here !! i liked the way that this course work !! homeworks n LABs. Good luck to every one in next 6.002x spring course !!! thx alot c u later :)) UserIdTAG: 392654 UserNameTAG: Hamzah_bakhti CreateTimeTAG: 2012-12-24T21:35:01Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1241 TitleTAG: Happy New Year! Merry Christmas.....!!!!!!!! Happy New Year! Merry Christmas.....!!!!!!!! Wish u all bright future.. UserIdTAG: 154189 UserNameTAG: nebi_nane CreateTimeTAG: 2012-12-24T19:31:09Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Same to you ! FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-12-24T22:50:01Z IndexTAG: 1242 TitleTAG: edX Alumni on Linkedin Join the "Friends and Alumni of edX" group on linkedin to share experiences and opinions. best regards UserIdTAG: 372321 UserNameTAG: EnricoDona CreateTimeTAG: 2012-12-24T18:53:44Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1243 TitleTAG: Merry X'mas & Happy New Year Wish you all $$Merry\ X'mas\ and\ Happy\ New\ Year!$$ UserIdTAG: 623210 UserNameTAG: preveen CreateTimeTAG: 2012-12-24T16:40:35Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Merry Christmas and Happy New Year preveen!:) My best wish to you ! Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-24T17:52:42Z SecondChildTAG: Same to you! All the best wishes to you too.. SecondChildUserIdTAG: 443988 SecondChildUserNameTAG: Dheeraj_Garg SecondChildCreateTimeTAG: 2012-12-24T19:09:31Z FirstChildTAG: Any hints for broken Christmas tree lights? ;-) I keep thinking- What would The Professor do? FirstChildUserIdTAG: 259719 FirstChildUserNameTAG: Smallblindchris FirstChildCreateTimeTAG: 2012-12-24T18:14:31Z SecondChildTAG: +1 SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-12-24T18:19:37Z SecondChildTAG: S'okay, we have light! SecondChildUserIdTAG: 259719 SecondChildUserNameTAG: Smallblindchris SecondChildCreateTimeTAG: 2012-12-24T18:24:48Z SecondChildTAG: a-ha!!! lights, my treasure SecondChildUserIdTAG: 244706 SecondChildUserNameTAG: Miguel_Angel SecondChildCreateTimeTAG: 2012-12-24T18:36:39Z SecondChildTAG: I think the Prof. would just buy new lights as he is too busy to fix it himself :) SecondChildUserIdTAG: 114881 SecondChildUserNameTAG: xvink SecondChildCreateTimeTAG: 2012-12-24T19:18:00Z SecondChildTAG: Its easy, and I learn't it all here. I tweaked all the bulbs, nothing happened... so I said the magic word "CHAINSAW" and boom! Lights come on! ;-) Merry Christmas Peoples SBC SecondChildUserIdTAG: 259719 SecondChildUserNameTAG: Smallblindchris SecondChildCreateTimeTAG: 2012-12-24T21:10:16Z IndexTAG: 1244 TitleTAG: CECC Query Hello there, Can we use Microcontrollers in our circuits. UserIdTAG: 273912 UserNameTAG: raj2691 CreateTimeTAG: 2012-12-24T16:12:41Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Yes you can :) FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-24T17:44:26Z SecondChildTAG: I have one question when is the delay is it exactly today at 12 pm SecondChildUserIdTAG: 91934 SecondChildUserNameTAG: Chingun SecondChildCreateTimeTAG: 2013-01-05T16:15:03Z SecondChildTAG: Hi Chingun, sorry for the delay of the answer. You have time to submit your video till January 14th -the deadline it was extended- SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2013-01-07T13:38:59Z FirstChildTAG: You can use a microcontroller, but your circuit should demonstrate something you learned in 6002x. So, if you for instance just do the typical microcontroller blinky, you won't get my vote. If you can get an LED to blink without a microcontroller, you have a chance. (I'm not saying a blinking LED is going to win the contest, it's just an example.) FirstChildUserIdTAG: 90796 FirstChildUserNameTAG: ChaunceyGardiner FirstChildCreateTimeTAG: 2012-12-24T22:38:58Z SecondChildTAG: Understood. SecondChildUserIdTAG: 273912 SecondChildUserNameTAG: raj2691 SecondChildCreateTimeTAG: 2012-12-25T04:00:25Z IndexTAG: 1245 TitleTAG: edx staff: pass book it would be nice to have a pass book online with names of all studend who get a certificate each semester. once again thanks for the discipline you gave me to stick to a goal and finish UserIdTAG: 462861 UserNameTAG: edxforme CreateTimeTAG: 2012-12-24T15:54:55Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1246 TitleTAG: thanks thanks for such nice course and i would like to take up many more in future. UserIdTAG: 35623 UserNameTAG: ARSHIYA25 CreateTimeTAG: 2012-12-24T14:21:14Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1247 TitleTAG: To the staff - I cannot submit my answers to the final exam Hello, When I first checked the final date to submit my answers to the final exam on my computer. It said that I was able to submit them until 5:00 of the 25th of December. This morning I've started to do the exam and the submit buttons were available, however when I pulse them no action happened. So I reloaded the page and suddenly the buttons disappeared. Moreover, when I've re-checked the final date to submit my answers, now it says that I had to submit them until the 24th of December at 10:00. Why this happened?? Thank you anyway for letting me participate in this edx program. Kind regards, Juan Antonio UserIdTAG: 55097 UserNameTAG: constarint CreateTimeTAG: 2012-12-24T11:44:30Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: i am unable to submit my exam as per indicated , due date is 24th 10 o'clock and indian time is 10hrs ahead of boston time . pls help i am a regular student and i have always sticked to my deadlines pls help immediately FirstChildUserIdTAG: 91261 FirstChildUserNameTAG: mr_sophisticated FirstChildCreateTimeTAG: 2012-12-24T11:55:07Z SecondChildTAG: The exam will now close at December 24th at 5:00am EST, December 24th 10:00am GMT.. So final exam is closed 2 hours ago SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-24T12:02:44Z SecondChildTAG: 10 am gst was not mentioned it was always wrt boston time hence it was misunderstood is thr anyway to submit it plzzzzzzzzzzzz SecondChildUserIdTAG: 91261 SecondChildUserNameTAG: mr_sophisticated SecondChildCreateTimeTAG: 2012-12-24T12:21:31Z SecondChildTAG: many r thr like mw who misunderstood the time as gst was not mentioned in the tab pls SecondChildUserIdTAG: 91261 SecondChildUserNameTAG: mr_sophisticated SecondChildCreateTimeTAG: 2012-12-24T12:23:15Z SecondChildTAG: correction by gst i ment gmt SecondChildUserIdTAG: 91261 SecondChildUserNameTAG: mr_sophisticated SecondChildCreateTimeTAG: 2012-12-24T12:27:28Z SecondChildTAG: I thought the deadline is 10am eastern time of Dec 24 too. It confused me. SecondChildUserIdTAG: 309255 SecondChildUserNameTAG: mamba747 SecondChildCreateTimeTAG: 2012-12-24T12:44:39Z SecondChildTAG: me also>>>>> SecondChildUserIdTAG: 156535 SecondChildUserNameTAG: badawiIiIi SecondChildCreateTimeTAG: 2012-12-24T12:59:19Z SecondChildTAG: Same here! I had exams at my university so I kept this final to for today morning but it seems like i've missed it :( SecondChildUserIdTAG: 444155 SecondChildUserNameTAG: samer1 SecondChildCreateTimeTAG: 2012-12-24T14:01:53Z SecondChildTAG: me tooo SecondChildUserIdTAG: 509517 SecondChildUserNameTAG: randima SecondChildCreateTimeTAG: 2012-12-24T15:46:53Z FirstChildTAG: Unfortunately, the deadline of Final Exam have already finished.... End of Final Exam due December 24th at 5:00am EST, December 24th 10:00am GMT [Take a look at here][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50d748bd1f858b2700000015 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-24T15:05:31Z SecondChildTAG: Thank you for your early answer Myrimit. I'm a bit sad because I wanted to do my best in the final exam due to all the effort applied to this course. Anyway, Merry Christmas and a Happy New Year to everyone! :-) SecondChildUserIdTAG: 55097 SecondChildUserNameTAG: constarint SecondChildCreateTimeTAG: 2012-12-24T15:25:21Z SecondChildTAG: I am so sorry for what happened to you constarint...I undestand you totally...the time zone is really confusing...the deadline it should be based on your local zone in order to no get confused or they should add a world clock in the page...I hope they implement in the future... Merry Christmas and Happy New Year constarint. I wish you the best too :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-24T15:32:30Z SecondChildTAG: better, users should register their time zones so the course deadlines are reflected as properly time shifted to their tz It is very sad that people missed their exams due to this confusion. edx do lose some "grade credit" for confusing the already panic . SecondChildUserIdTAG: 623210 SecondChildUserNameTAG: preveen SecondChildCreateTimeTAG: 2012-12-24T15:48:26Z SecondChildTAG: To give edX credit, we had a *4-day* window to take the exam, spanning two weekdays and a whole weekend plus at least one extension. This is very accommodating. If I was to take the exam on the last day, I may have asked ahead of time for clarification if I did not understand what my local time was with respect to MIT's clock. (Actually I would have sorted that out before the first assignment was due, let alone an exam.) :p Nonetheless we may see some new and exciting features added in the spring. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-12-24T16:07:40Z IndexTAG: 1248 TitleTAG: Thank you from Indonesia student to All the staff in 6.002x specially professor Agarwal, thank you very much for the great course... ---------- kuswantoro jakarta, Indonesia UserIdTAG: 452559 UserNameTAG: kuz1toro CreateTimeTAG: 2012-12-24T08:50:44Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1249 TitleTAG: compliant against Q4 in final exam [***Editor's Note:** Content deleted due to discussion of an exam question. To directly answer your question, the variable you stated as missing is included in the list in Q4. Take a closer look. **I have personally verified this.** We cannot make exceptions and give students extra chances on an individual basis. If you feel necessary to appeal this deletion, please contact a Staff member as listed in the Staff directory* - Jersey Mark, Community TA] UserIdTAG: 315378 UserNameTAG: gbp1984 CreateTimeTAG: 2012-12-24T06:17:42Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 0 IndexTAG: 1250 TitleTAG: University Applications Hi, I am a student studying in grade 12 from Mumbai, India. I am currently in the process of applying to my universities and was wondering whether I could add edX as a college level course I have completed. I know that the certificate does not show us our grades but perhaps I can take a screenshot of my progress page and clip it with the certificate before sending it to the universities. Does anybody know whether universities would accept edX as a certified online course? Thank you in advance. UserIdTAG: 85490 UserNameTAG: yashsavani CreateTimeTAG: 2012-12-24T06:12:11Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: strongly advise against it FirstChildUserIdTAG: 623210 FirstChildUserNameTAG: preveen FirstChildCreateTimeTAG: 2012-12-24T06:25:01Z SecondChildTAG: Why? SecondChildUserIdTAG: 85490 SecondChildUserNameTAG: yashsavani SecondChildCreateTimeTAG: 2012-12-24T06:57:10Z SecondChildTAG: I too am surprised about preveen's comment. I would personally be thoroughly impressed if a 12th grader took this course and passed with bright colors, assuming that you passed the exams sincerely thru your own efforts. I think this course is not too easy for an average 12th grader. So, you'll definitely stand out. SecondChildUserIdTAG: 232157 SecondChildUserNameTAG: GayatriTR SecondChildCreateTimeTAG: 2012-12-24T08:52:34Z FirstChildTAG: There's nothing wrong with mentioning what you have done. It illustrates your interests and activities. All this is relevant to your application to university. If you did it, and are proud of it, then definitely mention it. This demonstrates the kinds of things you are interested in. You won't get credit for it, but it will get their attention. Just as you would mention your hobbies, clubs you may have participated in, leadership roles you may have played in those clubs, etc..Universities want to know what kinds of things you do, when you're not "required" to do so. i.e. you were not required to take this course. It was not part of any school curriculum. It is thus an outside activity. You took in on your own initiative. Prof. Agarwal himself mentioned that he would certainly take a look at anyone mentioning having taken an edX course. So, if you apply to MIT, or Harvard, you're guaranteed to get their attention by mentioning this course. At the very least, your application will probably go into their "second review" pile, rather than just being tossed out with the vast majority of also well qualified candidates on first glance. Semd in a print out of your edX certificate, clipped onto your application, and you'll get noticed, for sure. The whole trick in this life, is getting the eyeballs to look your way. Lot's of very smart and capable people walk around this planet totally unnoticed. You need a flag in this competitive world, a flag to wave about and get attention. Well, your edX certificate is an excellent flag to wave about. Use it. FirstChildUserIdTAG: 4463 FirstChildUserNameTAG: pmj FirstChildCreateTimeTAG: 2012-12-24T08:03:21Z IndexTAG: 1251 TitleTAG: Amazing experience First of all, i thank Prof. Anant Agarwal and his team for giving me the opportunity to gain tremendous knowledge about the subject that i love to the core. Its been an amazing experience to learn so much about the circuits that i did study, but never learned what they actually did and how they behaved! Secondly, i find myself lucky to have been a part of this course and i am simply gained a lot out of this course. Thank you once again to the whole team. Rishikant Mishra UserIdTAG: 201462 UserNameTAG: rishi19923 CreateTimeTAG: 2012-12-23T21:38:19Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1252 TitleTAG: Thank you all and Merry Christmas! but questions remain. I finally finished and got A in this course, which is my first total English course, since I was born and study in China. I would like to thank you all for this excellent course, and appreciate the idea of offering free access to courses by top universities.Compared to similar lectures in China, I found this course more up-to-date and systematic. As a senior student, I also found some relations in this course with other courses (e.g. control theory). A minor pity is I found the course not difficult. I had thought the contents, homework and exams to be hard, but eventually they are easy. I wonder if this course is of same level in difficulty, compared to what MIT undergraduates learn. And another question. In one sequence, Prof.Agarwal said that from characteristic equation Vc * (s^2 + 2*alpha*s+ omega0^2) = omega0^2 * Vi, we can get its ODE d^2(Vc)/dt^2 + 2*L*d(Vc)/dt + Vc*omega0^2 = omega0^2 Viewing from complex analysis, this is Laplace transform, but the Laplace transform of f''(t) is s^2*F(s)-s*f(0)-f'(0), and that of f'(t) is s*F(s)-f(0). See Wikipedia http://en.wikipedia.org/wiki/Laplace_transform . That means the initial condition( namely, function value of f and f' at t=0) has to be zero for the method to be valid. However I did not see any clue for this to happen. This problem has baffled me quite a while, would you please help me? Thanks a lot. Last but not least, Merry Christmas and best wishes to all of you! UserIdTAG: 161054 UserNameTAG: zhiyuan90 CreateTimeTAG: 2012-12-23T17:27:02Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: from page 2 of the text book ‘‘Finally, an introductory circuit analysis book has been written that truly unifies the treatment of traditional circuit analysis and electronics. Agarwal and Lang skillfully combine the fundamentals of circuit analysis with the fundamentals of modern analog and digital integrated circuits. **I applaud their decision to eliminate from their book the usual mandatory chapter on Laplace transforms,** ***a tool no longer in use by modern circuit designers. I*** expect this book to establish a new trend in the way introductory circuit analysis is taught to electrical and computer engineers.’’ - T I M T R I C K , University of Illinois at Urbana-Champaign FirstChildUserIdTAG: 623210 FirstChildUserNameTAG: preveen FirstChildCreateTimeTAG: 2012-12-23T17:34:13Z SecondChildTAG: I know Laplace is beyond course requirement. But I just want an answer to this. It seems not rigid mathematically. SecondChildUserIdTAG: 161054 SecondChildUserNameTAG: zhiyuan90 SecondChildCreateTimeTAG: 2012-12-23T17:50:40Z SecondChildTAG: No. He was not taking inverse Laplace transforms. SecondChildUserIdTAG: 623210 SecondChildUserNameTAG: preveen SecondChildCreateTimeTAG: 2012-12-23T18:17:16Z FirstChildTAG: Well, why not do something like this; Since, Vc(0) and Vc'(0) are unknown, define a new variable, Vc1(t) = Vc(t) - Vc(0) - t.Vc'(0). Then, we'd have a function, Vc1(t), with, Vc1(0) = 0, and, Vc1'(0) = 0. Rearranging, Vc(t) = Vc1(t) + Vc(0) + t.Vc'(0) Substitute the r.h.s. expression for Vc(t) in the differential eqn. We get a new differential equation in Vc1(t), with one extra constant "2a.Vc'(0)", which we move over to the r.h.s. Now we apply Laplace transform, with the new function variable Vc1(t), instead of the old Vc(t). Then, solve the s-algebra equation, invert the Laplace transform, for the Vc1(t), then add back the "initial terms", Vc(0) and t.Vc'(0), to get Vc(t). Then, use "initial conditions" to set the final values of Vc(0) and Vc'(0). Wouldn't that solve the problem? FirstChildUserIdTAG: 4463 FirstChildUserNameTAG: pmj FirstChildCreateTimeTAG: 2012-12-23T19:09:11Z SecondChildTAG: Don't know! Looks like a formula from Harry Potter to me .... SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-12-23T21:25:10Z FirstChildTAG: Hi,this course is a basic introduction to EECS world in MIT's courses,so it won't be so difficult, and actually the course is registerd by many people who may not have taken similar courses like high school students. Professor's excellent organization shows us a much easier perspective to understand the concepts, and if you read the textbooks in detail, you may find the book very good!Best wishes to you ! FirstChildUserIdTAG: 361137 FirstChildUserNameTAG: Ericson FirstChildCreateTimeTAG: 2012-12-24T04:45:11Z IndexTAG: 1253 TitleTAG: Appreciate and Wish you all Happy X'mas What a hard exam it is! Any way I got the XX% score. It's inconceivable for me to reach it. I proud of myself. I am so exiting, hardly to tear. Thank you Sir Agarwal and edX team. Thanks for your hard work. I wish you all Happy Christmas! UserIdTAG: 128276 UserNameTAG: ChengBin CreateTimeTAG: 2012-12-23T16:24:58Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: My congratulations! But please, dont post score while exam is open. FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-12-23T16:29:36Z IndexTAG: 1254 TitleTAG: Final Review Packet, Problem 3, page 25 In part (A) of this problem, can someone explain why the voltage across the capacitor, vc(t), is zero when the voltage across the resistor is (R*lambda)/L? Is the current discharging from the inductor in a counterclockwise fashion? Thank you in advance for your consideration. UserIdTAG: 345398 UserNameTAG: xost33 CreateTimeTAG: 2012-12-22T18:25:04Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I believe it's because at t=0+ the capacitor doesn't charge up immediately, but rather acts as a short-circuit. FirstChildUserIdTAG: 443358 FirstChildUserNameTAG: torkelh FirstChildCreateTimeTAG: 2012-12-22T18:50:22Z FirstChildTAG: Thank you for the response. It makes that the vc(t) voltage would be zero now but I'm still not sure of the direction of the current. According to the explanation for Homework 8 - Problem 1 - part c/3, after an impulse through an RLC circuit the source voltage also acts like a short circuit. Therefore the current could travel counterclockwise and dissipate through the resistor? Regardless, good luck on the final if you haven't taken it yet! FirstChildUserIdTAG: 345398 FirstChildUserNameTAG: xost33 FirstChildCreateTimeTAG: 2012-12-23T01:12:52Z IndexTAG: 1255 TitleTAG: Typo in final Q6? (Inconsistent T specification) EDIT: Unfortunately we cannot discuss the exam at this here at this time. Please report any issues with the exam to mit-6002x@edx.org, make sure you are using a current version of Firefox or Chrome, and include browser and version in your e-mail, as well as screenshots or other pertinent details. Thank you for understanding. UserIdTAG: 369355 UserNameTAG: CaptChes CreateTimeTAG: 2012-12-20T23:51:45Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Alarm! Alarm! Specs are even worse now (after correction), they're misleading! Grader grades your answers ***EDITED.*** I heard the above, but there're people's lives out there you know. FirstChildUserIdTAG: 332194 FirstChildUserNameTAG: IronAlex FirstChildCreateTimeTAG: 2012-12-21T05:08:53Z SecondChildTAG: I understand your concerns, therefor please report any issues with the exam to mit-6002x@edx.org and do not discuss specifics in the forum at this time. Thanks! SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-12-21T05:32:07Z SecondChildTAG: Ok, but I thought I didn't give anything out, so no EDITions were needed. I sent a mail. Let's hope no one will spoil their 3 attempts until the problem is corrected. SecondChildUserIdTAG: 332194 SecondChildUserNameTAG: IronAlex SecondChildCreateTimeTAG: 2012-12-21T06:03:04Z SecondChildTAG: iv used all 3 of my attepmts :S SecondChildUserIdTAG: 448869 SecondChildUserNameTAG: SlR_ar SecondChildCreateTimeTAG: 2012-12-21T06:08:15Z SecondChildTAG: Clearly you did, plus the post above yours had implicit instructions of where to direct your exam concerns in private. Alex, others have no idea if you are correct or not in your claims as they have not taken the exam. I have faith that the edX staff will resolve any issues with questions that cannot be answered due to clerical errors *if* that is the case. They have done a fine job so far! SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-12-21T06:21:53Z IndexTAG: 1256 TitleTAG: Suggestion for course filters With the recent addition of all the new courses in the edX main page the site became even more interesting but at the same time already a little bit overwhelming! I think that as the number of courses available will probably be increasing there should be an option to filter out the courses you have registered / completed in the Courses page as well as the courses you have completed in the Dashboard page. Thus we will be able to more easily overview the courses we may consider taking as well as our obligations towards our running courses. This will make life easier for people like me that are looking to acquire as much knowledge as possible out of this amazing initiative! Thank you! UserIdTAG: 370344 UserNameTAG: KostisL CreateTimeTAG: 2012-12-20T20:13:00Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1257 TitleTAG: Thank You everybody..... Thank You so much MIT, EDX and specially Sir ANANT AGARWAL and whole MIT staff for arranging this kind of Course.... Best of Luck to everyone for future..... UserIdTAG: 51259 UserNameTAG: Wardah CreateTimeTAG: 2012-12-20T18:39:52Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1258 TitleTAG: How to write sin and cos as a right formula Hi , Could anyone tell me that how to write in the answer sheet that Sin(...) or cos(...)as a right formula . It seems the grade system do not recognize what i wrote . UserIdTAG: 232667 UserNameTAG: KyleLiux CreateTimeTAG: 2012-12-20T06:48:54Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Remember to "*t" in these quotes FirstChildUserIdTAG: 136959 FirstChildUserNameTAG: christerpher FirstChildCreateTimeTAG: 2012-12-20T07:29:31Z SecondChildTAG: sin and cosine functions are written as sin(x) and cos(x), just as they would be written on paper. SecondChildUserIdTAG: 391929 SecondChildUserNameTAG: RohanNagarkar SecondChildCreateTimeTAG: 2012-12-20T14:03:19Z SecondChildTAG: They should have wrote : "Answer the following questions in terms of the parameters: V, R1 R2, C, L, t and **T** as applicable" . It is confusing to some persons, and was not my intention to give up anything or to brake the rules. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-12-20T14:04:43Z IndexTAG: 1259 TitleTAG: Thanks a lot this final sequence was my personal big **Aha-moment** and I hope to learn more about it in the future. Thank you very much for all this lessons... UserIdTAG: 388524 UserNameTAG: dasbinich CreateTimeTAG: 2012-12-19T21:18:20Z VoteTAG: 2 CoursewareTAG: Week 14 / S28V13_Voltage_Drop_across_the_Parasitic_Inductor CommentableIdTAG: 6002x_S28V13_Voltage_Drop_across_the_Parasitic_Inductor NumberOfReplyTAG: 0 IndexTAG: 1260 TitleTAG: CheckReset & ShowAnswer buttons I would of liked to still have a Check and a Reset button on homeworks now, and those could of been useful. Like the lectures exercises. For reviews, i would of re-do most of the homeworks but .... i cant coz Check button is gone. Having only a ShowAnswer button kinda takes the fun out and puts the fire off re-doing homeworks. Ofc as it is now, i can just read both problem and ShowAnswer, but its not quite the same thing. Also the same thing with the review packet: that annotated stuff would of been of so much better value in the format that we already got used to ... a review like a big homework ... again with little check buttons and show answer. On a second thought maybe i am getting too attached to a little button ... any1 else? :) UserIdTAG: 331483 UserNameTAG: Doru CreateTimeTAG: 2012-12-18T04:17:25Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: There's always the low-tech solution of printing something out and using another piece of paper to "hide" certain sections... ;-) FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-12-18T05:09:40Z IndexTAG: 1261 TitleTAG: Final Exam Review Tiny Vacuum Tube Problem I don't understand where we learned to put rout in parallel with RLoad for the small signal analysis. For all the mosfet amplifier problems rout was equal to RLoad and thus interchangeable. Can anyone point out some more info on this? Thanks. UserIdTAG: 132086 UserNameTAG: parbro CreateTimeTAG: 2012-12-17T19:37:37Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: I agree with you... I would also like to know the concept behind this... FirstChildUserIdTAG: 435193 FirstChildUserNameTAG: ManosP FirstChildCreateTimeTAG: 2012-12-18T00:25:59Z FirstChildTAG: Not sure i understand correctly but if you are talking about that 400kohms resistor which i think its r0 in small model, then notice in normal model its a pullup. It doesnt show but that means its implicitly tied to a voltage source, which in the small model disappears coz it doesnt have implicit small signal component, and when disappears its replaced with a short to ground. Sort of pullups in the large model become pull-downs in small model because the source its tied to becomes a short, i guess coz that particular source has a small signal component=0. FirstChildUserIdTAG: 331483 FirstChildUserNameTAG: Doru FirstChildCreateTimeTAG: 2012-12-18T06:21:18Z FirstChildTAG: I think the small signal model of the vacuum tube by itself contains an output resistance rout because if you set vin which is vgc to zero there can still be current flow depending on the vpc voltage. This is in contrast to the mosfet where if vgs = 0 there can be no current flow. So for the vacuum tube amplifier with a rload resistor that goes in parallel with rout in the small signal model. So rout of amplifier becomes vacuum tube rout in parallel with rload. FirstChildUserIdTAG: 132086 FirstChildUserNameTAG: parbro FirstChildCreateTimeTAG: 2012-12-18T16:35:50Z FirstChildTAG: I, too, have been concerned with the Vacuum Triode as I did not fully understand the [Tutorial][1]. I found that there is a vacuum tube-related problem beginning on Page 4 in the [review packet][2], which shows the small-signal model. Now we know from the tutorial that: $$\displaystyle i_p=v_{gk}\cdot \frac{\delta i_p}{\delta v_{GK}}|_{v_{GK}=V_{GK}} + v_{pk}\cdot \frac{\delta i_p}{\delta v_{PK}}|_{v_{PK}=V_{PK}}$$ We also know that the first term is $v_{gk}$ multiplied by a transconductance, $g_m$ , and the second term is $v_{pk}$ multiplied by 1 over the incremental plate resistance: $\displaystyle\frac{1}{r_p}$. $\displaystyle\frac{1}{r_p} = g_p$ where $g_p$ is the incremental plate *conductance*. (Observe $g_m$ and $r_o=r_p$ in the small signal model on [page 6][3]) Remember that conductances in parallel *add*; in the equation, the two terms are being added, therefore, I believe, the circuit diagram based on the equation should show the corresponding elements in *parallel*, which it does. Now I have enough trouble going from diagram to equation, much less the reverse ;-) , so if I have misunderstood, I do hope someone corrects me. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_6/Week_6_Tutorials/ [2]: https://www.edx.org/static/content-mit-6002x/handouts/6002x-FinalReview-S2012-clean.19c690685471.pdf [3]: https://www.edx.org/static/content-mit-6002x/handouts/6002x-FinalReview-S2012-clean.19c690685471.pdf FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-12-19T06:57:27Z SecondChildTAG: Hi, you can find info here: http://books.google.co.ve/books?id=n0rf9_2ckeYC&pg=PA26&lpg=PA26&dq=vacuum+triode+characteristics&source=bl&ots=--_wGwE7AA&sig=zGMOTcZh2owcOEle0sbM725riRU&hl=es&sa=X&ei=d5ppUPiVMIvF0AGu3YCYDw&ved=0CEgQ6AEwBQ#v=onepage&q=vacuum%20triode%20characteristics&f=false SecondChildUserIdTAG: 296419 SecondChildUserNameTAG: oscman SecondChildCreateTimeTAG: 2012-12-20T04:42:01Z FirstChildTAG: The vacuum tube by itself has the small signal circuit shown in figure 3(v). Problem c is asking for the small signal of the amplifier circuit in 3(iv). The second circuit is the vacuum tube plus some extra stuff. When you do the small signal model for the circuit, put a short in place of the DC voltage source (turn it off.) You will do this for a MOSFET circuit as well. The first lecture of week six covers this if you need review. S11V8 is a good video to watch. FirstChildUserIdTAG: 202032 FirstChildUserNameTAG: KeithD FirstChildCreateTimeTAG: 2012-12-19T18:43:23Z SecondChildTAG: Thanks! SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-12-20T06:20:57Z IndexTAG: 1262 TitleTAG: After the course ends Does anyone know if after the course has ended we will still have access to course material? I think I would like to consult some of this material in the future, and if we loose access, I will need to store it. Thanks in advance UserIdTAG: 400567 UserNameTAG: JoseMartins CreateTimeTAG: 2012-12-15T20:05:03Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Those of us that took 6002x in the spring still have access to all the material. Except they recently removed stuff from the previous final, for obvious reasons. FirstChildUserIdTAG: 90796 FirstChildUserNameTAG: ChaunceyGardiner FirstChildCreateTimeTAG: 2012-12-15T22:55:39Z SecondChildTAG: Thank you for your help! SecondChildUserIdTAG: 400567 SecondChildUserNameTAG: JoseMartins SecondChildCreateTimeTAG: 2012-12-17T16:58:13Z IndexTAG: 1263 TitleTAG: the energy dissipated by capacitor and resistor are the same ,coincidence? Is it coincidence? Or any insight? UserIdTAG: 136959 UserNameTAG: christerpher CreateTimeTAG: 2012-12-15T12:13:21Z VoteTAG: 2 CoursewareTAG: Week 13 / S26V5_Energy_from_source_during_T1_continued CommentableIdTAG: 6002x_S26V5_Energy_from_source_during_T1_continued NumberOfReplyTAG: 1 FirstChildTAG: the capacitor store energy ( not dissipate) FirstChildUserIdTAG: 236136 FirstChildUserNameTAG: Hamid-ch FirstChildCreateTimeTAG: 2012-12-15T15:13:09Z IndexTAG: 1264 TitleTAG: H2P1 Hi Friends please tell me in H2P1 how did we derive the following formula for Vmax and Vmin. Vmax=Vin*(1.1)(R2)/(0.9(R1)+(1.1)(R2)) UserIdTAG: 40209 UserNameTAG: sadaf78 CreateTimeTAG: 2012-12-14T18:16:46Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hi sadaf78! As I remember in task was said that resistance can defferiniate for +-10% so considering this moment in formula for voltage devider we geting final formula. Maybe it is not clear explaining, but don`t know how to explain it more clearly FirstChildUserIdTAG: 96873 FirstChildUserNameTAG: Santyaga FirstChildCreateTimeTAG: 2012-12-14T19:14:35Z FirstChildTAG: to have a Vmax we should maximise the nominator and minimise the denominator - we have a resistance within 10% of nominal value - Vout = Vin * R2/(R1+R2) - max R2 --> R2 + 10% R2 = 1.1*R2 - min R1 --> R1 - 10% R1 = 0.9 *R1 - Vmax = Vin *R2max/(R1min + R2max) = Vin * 1.1 *R2/(0.9 * R1+ 1.1 * R2) FirstChildUserIdTAG: 236136 FirstChildUserNameTAG: Hamid-ch FirstChildCreateTimeTAG: 2012-12-14T20:11:05Z SecondChildTAG: i hope this will help SecondChildUserIdTAG: 236136 SecondChildUserNameTAG: Hamid-ch SecondChildCreateTimeTAG: 2012-12-14T20:12:39Z IndexTAG: 1265 TitleTAG: H12P2 C and D part help please I have tried over and ove and I don't understand why i'm wrong. My value for R0 = 94 , R1 = 90 and R2 = 100. Efficiency = 0.0457. Help Please!! UserIdTAG: 433574 UserNameTAG: endika86 CreateTimeTAG: 2012-12-09T22:37:33Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I need voltages, K, and load resistance for calculation FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-12-09T22:51:30Z SecondChildTAG: Load resistance 100ohms. k = 0.95. Vin 20v and vout 5.1V SecondChildUserIdTAG: 433574 SecondChildUserNameTAG: endika86 SecondChildCreateTimeTAG: 2012-12-09T23:03:57Z SecondChildTAG: Im sorry, I cant get correct output with your resistor values SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-10T00:14:40Z SecondChildTAG: with these values Vo=4.9/5.1V but I think it is wrong Please try to re-calculate for 4.92/5.08 ( as example) SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-10T00:18:30Z SecondChildTAG: check [here][1] last posts about tolerances. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50b8d3275516b62b00000010 SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-10T00:34:59Z SecondChildTAG: i am leaving H12P2 c and d.....its enough SecondChildUserIdTAG: 368981 SecondChildUserNameTAG: asimakhtar SecondChildCreateTimeTAG: 2012-12-10T01:51:19Z IndexTAG: 1266 TitleTAG: query regarding homeworks and labs..... helo guys... hope you doing fine. i wanted to ask about week 13 n 14 tasks. the problem is i missed 2 homeworks n 2 labs. so could week 13 n 14 tasks make up for the missed ones..??? also are they inluded in finals....??? UserIdTAG: 117319 UserNameTAG: iqramalik CreateTimeTAG: 2012-12-09T20:04:32Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: If you missed two homeworks and two labs, you still have 100% if you scored 100% on them. Only the 10 best out of 12 counts. For week 13 and 14 there is no homework or labs. FirstChildUserIdTAG: 90796 FirstChildUserNameTAG: ChaunceyGardiner FirstChildCreateTimeTAG: 2012-12-09T22:36:33Z IndexTAG: 1267 TitleTAG: QUERY RELATED TO CERTIFICATE I hadn't submitted homework of week 3 and week 6 due to some unavoidable reasons...but my progress bar is 52% out of 60%(of-course excluding final exam)...whether I will get certificate or not...(supposing I score good in final exam)?? UserIdTAG: 410204 UserNameTAG: hkaushalya CreateTimeTAG: 2012-12-09T16:12:59Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: **In syllabus** ---------- *9 Grading* ---------- Letter grades will be based on the following weighting: homework 15%, labs 15%, midterm 30%, and final exam 40%. Each of the homework and labs carries equal weight. You will need to get a total mark of 60% for a C, 70% for a B, and 87% for an A. Homework and labs will be graded based on the best ten out of twelve individual grades. Therefore, two homework assignments and two labs may be missed in total without a grade penalty FirstChildUserIdTAG: 402193 FirstChildUserNameTAG: Feramico FirstChildCreateTimeTAG: 2012-12-09T16:28:53Z FirstChildTAG: You need 20% on the final to get a passing grade. FirstChildUserIdTAG: 90796 FirstChildUserNameTAG: ChaunceyGardiner FirstChildCreateTimeTAG: 2012-12-09T18:11:39Z SecondChildTAG: Where is that written? Thanks. SecondChildUserIdTAG: 398594 SecondChildUserNameTAG: DaveyJC SecondChildCreateTimeTAG: 2012-12-09T21:11:08Z SecondChildTAG: Oh, 20% for hkaushalya in the final, cause he's left an 8%, and 20% of 40 is 8. I see. SecondChildUserIdTAG: 398594 SecondChildUserNameTAG: DaveyJC SecondChildCreateTimeTAG: 2012-12-09T22:12:04Z SecondChildTAG: Phew...I was worrying already :) SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-09T22:18:28Z FirstChildTAG: Thank you so much !!!!!! FirstChildUserIdTAG: 410204 FirstChildUserNameTAG: hkaushalya FirstChildCreateTimeTAG: 2012-12-10T05:11:44Z FirstChildTAG: Oh and don't forget, this course is not like other corses, if you only get 60 percent, that's niot only 60 percent, that is a lot of interesting and difficult concepts you understand and can be proud of. Forget about D, C, B's and As :) FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-12-10T08:39:55Z FirstChildTAG: The worst two homework scores and the worst two lab scores don't count, so by the time you finish all the homeworks (and labs) those scores for HW3 and HW6 will be discarded. The final is worth 40%, so you can still get 98% for your total grade for the class. FirstChildUserIdTAG: 9673 FirstChildUserNameTAG: xp42 FirstChildCreateTimeTAG: 2012-12-10T09:02:18Z IndexTAG: 1268 TitleTAG: yet another current source cofiguration with an op-amp we can also make a current source with an op-amp alone using the op-amp in the inverting configuration At http://en.wikipedia.org/wiki/File:Op-Amp_Inverting_Amplifier.svg simply take Rf as the load, Rin=1k and Vin=1V. Op-amp's - terminal is effectively at 0Volts so Rin develops a current of 1mA. As op-amp input impedance is infinite to a first aproximation , the current is forced to flow through our load. UserIdTAG: 319453 UserNameTAG: leonidasGr CreateTimeTAG: 2012-12-09T15:50:21Z VoteTAG: 2 CoursewareTAG: Week 12 / How to build a current source CommentableIdTAG: 6002x_How_to_build_a_current_source NumberOfReplyTAG: 0 IndexTAG: 1269 TitleTAG: H12P3 How Can I find Rf? I found R1, R2, C1, C2, but I didn't manager to find Rf. I need help, please. UserIdTAG: 366669 UserNameTAG: pedroramus CreateTimeTAG: 2012-12-08T18:37:13Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: RF and RS determine the gain. This is just a non-inverting amplifier configuration. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-12-08T18:51:42Z SecondChildTAG: I was doing this, but there was a mistake in my calculations. Thank for your help. SecondChildUserIdTAG: 366669 SecondChildUserNameTAG: pedroramus SecondChildCreateTimeTAG: 2012-12-08T19:11:07Z IndexTAG: 1270 TitleTAG: gates because we are using two gates for same logic we must calculate for 10million gates for fare comparison. UserIdTAG: 307137 UserNameTAG: kishen CreateTimeTAG: 2012-12-07T09:48:11Z VoteTAG: 2 CoursewareTAG: Week 14 / S27V6_Power_Dissipation_in_CMOS_Logic CommentableIdTAG: 6002x_S27V6_Power_Dissipation_in_CMOS_Logic NumberOfReplyTAG: 2 FirstChildTAG: Or use 2 fF for C capacity. FirstChildUserIdTAG: 258674 FirstChildUserNameTAG: i-rinat FirstChildCreateTimeTAG: 2012-12-18T11:03:27Z FirstChildTAG: no, kishen, we are using 2 MOSFETs for each gate. And the formula for dynamic power is for one gate. So, it is ok to calculate with 5 millions units (= gates)... FirstChildUserIdTAG: 388524 FirstChildUserNameTAG: dasbinich FirstChildCreateTimeTAG: 2012-12-19T00:53:42Z IndexTAG: 1271 TitleTAG: Final exam: Are you ready? Hi all! Are you ready for the Final exam? Do you have some troubles with matherials? UserIdTAG: 205311 UserNameTAG: Sergtronix CreateTimeTAG: 2012-12-05T22:14:00Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: hi Scared to see problems like H12p2 if such problems appear in exam it would be difficult good luck to all FirstChildUserIdTAG: 352647 FirstChildUserNameTAG: praveenjugge FirstChildCreateTimeTAG: 2012-12-06T10:08:08Z SecondChildTAG: For my opinion H12P2 isnt so hard. Our course has more awful parts like impulse and depend sources, imho. SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-06T10:13:34Z IndexTAG: 1272 TitleTAG: H12P1 Fix To all who may have experienced issues with problem 1 of week 12 (giving correct answers a green check, and later giving the same response a red X), the issue should now be resolved. Please verify that your "old" correct answers, for which you initially received credit, are currently valid. If not, please reply! Thanks, --Rohan UserIdTAG: 391929 UserNameTAG: RohanNagarkar CreateTimeTAG: 2012-12-05T14:01:33Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I have problems with H12P1 part 1. It gives me a red X. I am using the exact equation that appears in the book at pagie 856,which is the exact same circuit topology. I dont know what is happening. FirstChildUserIdTAG: 181432 FirstChildUserNameTAG: enriqueferreralcala FirstChildCreateTimeTAG: 2012-12-06T16:58:17Z IndexTAG: 1273 TitleTAG: Digital ground Is there any difference between normal groung and digital ground? If so what is that difference? UserIdTAG: 132685 UserNameTAG: deepakmurali CreateTimeTAG: 2012-12-05T08:11:50Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: What is "Normal" ground ? Do you mean an Analog ground? In the term "Normal" all is relative :) Ok. Simplified: Separate ground (power lines) is technique to decrease unwanted noise and hum on the important signal lines. This is allow currents flow by the defined paths without impact to specified lines. We may use few "digital grounds" as well as VCC lines to supply for example MCU core/MCU peripheral drivers/Ethernet drivers and etc. In cases when separate grounds must be galvanically connected, this connection is made in one point, usually close to the voltage regulators.Note here : plain not always may be considered as equipotential point. FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-12-05T19:26:04Z SecondChildTAG: Digital ground is usually an upside down triangle or little horizontal line, all that means is that you connect all the grounds together! Power grounds are usually symbolized by three horizontal lines, each getting smaller. SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-12-06T12:15:24Z SecondChildTAG: thank u SecondChildUserIdTAG: 132685 SecondChildUserNameTAG: deepakmurali SecondChildCreateTimeTAG: 2012-12-07T09:04:13Z IndexTAG: 1274 TitleTAG: To the staff..... About Final Exam Respected sir Kindly tell when the final exam is going to be held and what will be the duration of submission of the Exam.Also it will be good for all the students if the final will be on Weekend. Regards UserIdTAG: 51259 UserNameTAG: Wardah CreateTimeTAG: 2012-12-04T20:53:21Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: The calendar indicates that the final will be on the 20th, which unfortunately is a Thursday. [https://www.edx.org/static/content-mit-6002x/handouts/calendar.c7b2799155f8.pdf][1] [1]: https://www.edx.org/static/content-mit-6002x/handouts/calendar.c7b2799155f8.pdf FirstChildUserIdTAG: 334855 FirstChildUserNameTAG: JimMonty FirstChildCreateTimeTAG: 2012-12-05T05:25:30Z FirstChildTAG: Is it just for one day? I mean we will have only one day when it will be published and have to solve it the same day? or will it be like midterm? FirstChildUserIdTAG: 51259 FirstChildUserNameTAG: Wardah FirstChildCreateTimeTAG: 2012-12-05T11:47:05Z FirstChildTAG: https://www.edx.org/courses/MITx/6.002x/2012_Fall/info FirstChildUserIdTAG: 236136 FirstChildUserNameTAG: Hamid-ch FirstChildCreateTimeTAG: 2012-12-05T23:28:46Z FirstChildTAG: Looks like it's the same as the midterm. From the link above: > The final exam will be released on December 19th at 14:01 (2:01 pm) Boston time and will close on December 23rd at 23:59 (11:59 pm) Boston time. It is designed to take two hours, but you will have 24 hours to finish the exam. You can start the exam when it is convenient for you, but you must complete this examination by 23:59 Boston time on December 23rd. I think that the "you will have 24 hours to finish the exam bit" is incorrect and a holdover from the originally announced schedule. FirstChildUserIdTAG: 339668 FirstChildUserNameTAG: chickwebb FirstChildCreateTimeTAG: 2012-12-10T23:27:13Z SecondChildTAG: I believe you will have 24 hours from the point at which you first view the exam until the point at which it is closed. Once you open it, the clock is ticking. You will, however, have a roughly 3.5 day "window" in which to start. But if you start at 23:58 on Dec 23rd, you'd better be very very fast... SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-12-13T23:58:04Z IndexTAG: 1275 TitleTAG: H11P1 Q1 please explain I've spent quite some hours on the first question of H11P1, and finally gave up. Now that the answer is available, I still don't get this: $(jwR_cL+R_cR_l) \cdot$ $\displaystyle\frac{1}{LCR_c}\) \(= jw \cfrac{R_cL+R_cR_l}{LCR_C}$ ???? Shouldn't this be: $(jwR_cL+R_cR_l) \cdot$ $\displaystyle\frac{1}{LCR_c}\) \(= \cfrac{jwR_cL+R_cR_l}{LCR_C}$ Can someone please explain this to me? UserIdTAG: 114881 UserNameTAG: xvink CreateTimeTAG: 2012-12-04T19:09:31Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You are right,the expression in the given explaination must be wrong. FirstChildUserIdTAG: 361137 FirstChildUserNameTAG: Ericson FirstChildCreateTimeTAG: 2012-12-05T04:39:04Z SecondChildTAG: But doesn't this change the outcome of the whole answer? And if it does, what is the right answer? SecondChildUserIdTAG: 114881 SecondChildUserNameTAG: xvink SecondChildCreateTimeTAG: 2012-12-05T10:52:08Z SecondChildTAG: Yes, it doesn't affect the answer of the given questions.Because the denominator expression is right, what we want is delta w,which is also known as 2a=(Rl*Rc*C+L)/(L*C*Rc) and ω0^2=(Rl+Rc)/(L*C*Rc). SecondChildUserIdTAG: 361137 SecondChildUserNameTAG: Ericson SecondChildCreateTimeTAG: 2012-12-05T11:18:35Z SecondChildTAG: AHA! :) We don't care about the nominator here. That's probably the part I was missing here. Thanks. SecondChildUserIdTAG: 114881 SecondChildUserNameTAG: xvink SecondChildCreateTimeTAG: 2012-12-05T11:28:06Z SecondChildTAG: You are welcome! By the way, I didn't notice that mistake in the very beginning, you are eagle-eyed! You can report that to our staff :) SecondChildUserIdTAG: 361137 SecondChildUserNameTAG: Ericson SecondChildCreateTimeTAG: 2012-12-05T13:10:23Z IndexTAG: 1276 TitleTAG: To STAFF : H12P1 bug ? Hi! Looking [this thread][1] I tried to "check" and got problem.Successfully solved few days ago H12P1 has red cross now instead green mark for the last question. It seems that something was changed in this task. Please explain what is going. Thanks! [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50bd9cb645ea1f2a00000070 UserIdTAG: 205311 UserNameTAG: Sergtronix CreateTimeTAG: 2012-12-04T13:54:01Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: WOW! It works again! Thanks ! :) FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-12-04T19:17:15Z IndexTAG: 1277 TitleTAG: To Staff: Tuning Inductor? (Staff): Why can't we use inductor as a variable inductor in AM Radio Receiver? UserIdTAG: 526154 UserNameTAG: RAJAT09 CreateTimeTAG: 2012-12-04T04:02:29Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 6 FirstChildTAG: Can't make one!. To vary the inductance,you'd either have to variably increase/ decrease the number of turns on the core of the inductor, or vary the core length/ area. Physically impossible. Alternatively switch in inductors in series/parallel combinations. Don't even go there. Variable capacitors are simple to make and do a great job. Simples. FirstChildUserIdTAG: 273287 FirstChildUserNameTAG: johnkh77 FirstChildCreateTimeTAG: 2012-12-04T12:50:46Z SecondChildTAG: I absolutely disagree. See my post. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-12-04T19:01:58Z FirstChildTAG: I should have mentioned that switching inductors with associated tuning capacitors has been implemented in the past as a "Turret Tuner" in both radios and TVs, the latter because of a limited number of national definite fixed transmission frequencies. In the 1970s I bought a multi-band Russian radio with a Turret Tuner. What a nightmare, over time corrosion built up ( producing very high resistive joints) on the Turret's connecting contact points. Had to eventually bin it. Link to the culprit below. http://www.richardsradios.co.uk/vef.html FirstChildUserIdTAG: 273287 FirstChildUserNameTAG: johnkh77 FirstChildCreateTimeTAG: 2012-12-04T13:19:08Z FirstChildTAG: It may be a difference in size and precision. Inductors can be tuned by moving their entire core by the action of a screw, or by having another linearly moving part coupling into first inductor field, stuff like that. I think that implies big size, as if it were to be small, it wouldnt be too reliable, not to mention it may couple its field into neighboring pcb parts (inductor pcb layout or traces at 45 degrees). Variable capacitors are used nowadays, i think because the size of a variable cap is probably 10-100+ times smaller than the size of a variable inductor. Also probably because the inductors arent so precise & standardized in construction and maybe so even in variation, at least when compared to caps. FirstChildUserIdTAG: 331483 FirstChildUserNameTAG: Doru FirstChildCreateTimeTAG: 2012-12-04T13:48:43Z FirstChildTAG: Thanks Doru, you are right. I had forgotten about tunable inductors used extensively in communication equipment. They do tend however to be tunable "presets" within the body of the equipment rather than accessible to external users. FirstChildUserIdTAG: 273287 FirstChildUserNameTAG: johnkh77 FirstChildCreateTimeTAG: 2012-12-04T16:36:05Z FirstChildTAG: We can make variabele inductors in AM radio. In fact, if I remember well, it was done especially in AM car-radios. I discovered that to my surprise about 40 years ago, when I repaired my first car radio. How do you make an inductor variable? Simple, they used a ferrite-core that could be moved into or out of the coils centre, like you can turn a screw, so it's induction changed. I suppose they didn't use a variable capacitor, because the mechanical vibrations of the car, could change the capacitance of a variable capacitor (it consists of small plates that can vibrate and other mechanical problems). Another reason to use the variable inductors, was because you could preset the radio easily to different channels/frequencies with a few cores in one block. It's much harder to do that with variable capacitors. At that time, varicaps for use at AM frequencies, were not yet in existence or easily available. I even remember that when I was about 14 years old (around 1969), I heard on a technical program on the radio, that Philips had invented or started to produce the BA102 varicap, for use in FM radio and TV-tuners. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-12-04T19:00:45Z FirstChildTAG: Thanks a lot for all the clarifications. :-) FirstChildUserIdTAG: 526154 FirstChildUserNameTAG: RAJAT09 FirstChildCreateTimeTAG: 2012-12-11T20:45:56Z IndexTAG: 1278 TitleTAG: Increase number of execises in each homework How other students and administration 6.002x think about increase number of exercises in each homework (in spring session)...:) Maybe make addition little quize after every video lecture (quizes can influence on total progress like homeworks and labs)...:) Sorry for my english:( UserIdTAG: 104522 UserNameTAG: IgorNovice CreateTimeTAG: 2012-12-02T19:31:45Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Try looking at the text book for more exercises. For me, I am just barely surviving, more exercises would be tough for me. FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-12-03T09:14:07Z SecondChildTAG: I'm with you on this one, Hazel. SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-12-03T17:46:53Z SecondChildTAG: Haha, thanks planetscape. SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-12-03T19:15:59Z FirstChildTAG: I would like to see homework for weeks 13 and 14. Homeworks help to learn the material. FirstChildUserIdTAG: 397595 FirstChildUserNameTAG: mholin FirstChildCreateTimeTAG: 2012-12-03T10:24:21Z FirstChildTAG: they should post a bonus homework that will increase our score by say 1 or 2%!! FirstChildUserIdTAG: 368712 FirstChildUserNameTAG: jmen FirstChildCreateTimeTAG: 2012-12-04T15:41:16Z SecondChildTAG: Good idea! SecondChildUserIdTAG: 104522 SecondChildUserNameTAG: IgorNovice SecondChildCreateTimeTAG: 2012-12-04T18:44:19Z SecondChildTAG: Maybe bonus homework for every week. Total influence of all bonus homeworks very low (1-5%) but students, who want get 100% score should make this homeworks... SecondChildUserIdTAG: 104522 SecondChildUserNameTAG: IgorNovice SecondChildCreateTimeTAG: 2012-12-04T18:50:46Z IndexTAG: 1279 TitleTAG: CHECKER IS TRUE! there is a point: first of all Q=W0/2alpha but Alpha is different in series and parallel RLC circuts. in parallel alpha=1/2RC and series alpha=R/2L secondly wo=frequency*2pi it means when you want to use frequency instead of angular frequency which is Omega(w) you must first multiply it by 2*pi. all together: Q=(fr*2pi)/(2*RC) will satisfy the checker. UserIdTAG: 733429 UserNameTAG: EhsanMokhtari CreateTimeTAG: 2012-12-02T14:55:18Z VoteTAG: 2 CoursewareTAG: Week 11 / S21E4: AM_Radio_Tuning CommentableIdTAG: 6002x_S21V1_AM_Radio_Tuning NumberOfReplyTAG: 0 IndexTAG: 1280 TitleTAG: Anti-Lock Brake System I just wanted to add: Due to ice or water on the road, which are moving on a certain speed, they tend to have a phenomenon which is called Aqua Planning or Hydro Planning due to which the tyre surface looses contact with the road and tends to move. This is also as roads suffer from rutting and depressions which cause the water to stand there and cause a trough of water to be developed due to which this happens. If brakes are applied, which would happen if there is friction, but in this case there is no friction, so the tyres tend to skid. There is also another test called the Skid Resistance Test which tests the coarsity of the road, if the road is coarse the water drains qucikly and provides for the required friction. UserIdTAG: 171674 UserNameTAG: wajahat1 CreateTimeTAG: 2012-12-02T08:56:18Z VoteTAG: 2 CoursewareTAG: Week 12 / S23V12 Antilock brakes and negative feedback CommentableIdTAG: 6002x_S23V12_Antilock_brakes_and_negative_feedback NumberOfReplyTAG: 0 IndexTAG: 1281 TitleTAG: PROBLEM WITH LAB TOOL Can anyone please help me, my lab tool is not working, i am not able to edit circuit values, please help me out......... UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-12-01T17:58:33Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Hi, Can I help you? In which lab are you lost? Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-01T19:11:55Z FirstChildTAG: I've had problems in chrome. Firefox seems to work better. FirstChildUserIdTAG: 174100 FirstChildUserNameTAG: markpolak FirstChildCreateTimeTAG: 2012-12-01T19:23:28Z SecondChildTAG: Mmm, that is weir, I am using Chrome an I don´t have problems with the Lab tools .... Have you cleared your caché with chrome? - try to clean your Historial. Some students of 3.091x had some similar problems, might is because of this... Have you tried to logg out and in? Have you tried to refresh your page with F5? Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-01T19:48:36Z FirstChildTAG: I have had similar issues in the past, with Firefox. Solved by one or more iterations of clearing cache, restarting browser, and reloading page. FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-12-02T01:33:12Z IndexTAG: 1282 TitleTAG: S21E4: AM RADIO TUNING How do you calculate the first part? I cannot get the answer. UserIdTAG: 374393 UserNameTAG: rmaleki CreateTimeTAG: 2012-11-30T21:32:14Z VoteTAG: 2 CoursewareTAG: Week 11 / S21E4: AM_Radio_Tuning CommentableIdTAG: 6002x_S21V1_AM_Radio_Tuning NumberOfReplyTAG: 1 FirstChildTAG: Here is the formula and a useful calculator. Go to the section "Resonant Frequency of an LC Circuit", then "Enter any two of three variables. Select and Click on a button to solve for the third." No affiliation. http://www.midnightscience.com/formulas-calculators.html FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-12-01T00:47:58Z SecondChildTAG: that is easy to get numbers from the website. But how the formula f=1/(2pi*sqrt(L*c)) is derived from: (j*L*w)/((R-C*L*R*w^2)+j*L*w)? Why R is omitted from the formula? SecondChildUserIdTAG: 374393 SecondChildUserNameTAG: rmaleki SecondChildCreateTimeTAG: 2012-12-01T11:50:22Z SecondChildTAG: For the first two questions I just isolated the resonant part of the circuit. In this application it should resonate at the same frequency regardless of the voltage across the antenna, there for I ignored R. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-12-01T14:45:28Z IndexTAG: 1283 TitleTAG: non-sinusoidal signals These questions are not directly related to 6.002x, but I do not know a better place to ask this question. 1) Does anyone use non-sinusoidal carrier to transfer information? (e.g. triangle, sawtooth waveform carriers, etc.) 2) Is it possible to get an "ideal" sinusoidal carrier signal or the waveform will always differ from sine? (I am particularly interested in VCO waveforms) 3) Where can I read about it? (Books / articles) UserIdTAG: 386688 UserNameTAG: renatyv CreateTimeTAG: 2012-11-30T17:35:33Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I think a sinusoidal wave is used for two reasons: 1. It is always changing (a square wave, for example, isn't). This is important because an antenna can only transmit when the signal is always changing. 2. A pure sine wave has only a single frequency component. The waves you mentioned have an infinite Fourier series (and therefore use a big chunk of spectrum) and that is not desirable when you think about bandwidth and multiple carriers in a channel. I doubt an ideal sinusoid is possible. We are limited by the devices we use and they will have deviations in behavior. However I think what we get is close enough for practical purposes. I should run an oscillator through a spectrum analyzer when I get the opportunity :-) FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-12-01T04:10:06Z FirstChildTAG: Pulses are also used to carry information, in the early days of digital signaling, in the so called binary baseband coding. For example what is called pulse-position modulation (PPM) and pulse-duration modulation (PDM), delay modulation (DM), polar return-to-zero (PRZ) and many others. However, you must differentiate between carrying info via a carrier or not. Modems convert the baseband signals by modulating a carrier, to for example FM, AM, SSB. You want a sinoid, because you don't want harmonics, and in many cases prefer a small bandwidth. You can create an almost ideal sinoid, but as soon as you modulate it, it's not pure anymore. A VCO can generate an almost perfect sine, triangle or whatever, no problem, but it requires also stability. So small fluctuations in phase, frequency, amplitude, are in fact a kind of unwanted modulation. I think an ideal sinoid carrier/signal has an infinite small bandwith, i.e. no modulation. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-12-01T18:20:15Z SecondChildTAG: Hi Salsero I wish I could do the things you said... I mean how to encode, send and receive information using a carrier wave using electronic components... For me is still black magic... :) Regards. SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-12-01T21:16:21Z SecondChildTAG: It's actually quite doable. If you don't have equipment you can at least use a simulator like this one: https://www.circuitlab.com/ While you can't send intelligible voice data, you can use a stream of bits, encode it and modulate it on a carrier on the simulator. SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-12-02T03:08:50Z SecondChildTAG: Hi ashwith. Do you have reference material on how to do the actual modulation/demodulation? SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-12-02T04:14:27Z SecondChildTAG: I have books (the books by Simon Haykin a pretty good). Don't know what's available online. Modulation systems are taught in communications. I'll build a circuit on circuit lab if I get time today so that you can look at a circuit. Would you like analog modulation or digital? SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-12-02T08:04:57Z SecondChildTAG: Oh I recorded a little tutorial on digital modulation. See if it helps: http://www.youtube.com/watch?v=pZbczyghP8Y We had a signals and systems group study on the 6.002x forums after the Spring session. This is one of the tutorials we made during that time :) I'm not very good at using graphics tablets :-/ SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-12-02T08:07:30Z SecondChildTAG: The "lectures" I talk about in the video are these: http://ocw.mit.edu/resources/res-6-007-signals-and-systems-spring-2011/video-lectures/lecture-13-continuous-time-modulation/ and http://ocw.mit.edu/resources/res-6-007-signals-and-systems-spring-2011/video-lectures/lecture-14-demonstration-of-amplitude-modulation/ I don't show "actual implementations" but I show blocks that should be easy to translate into circuits. SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-12-02T08:11:22Z SecondChildTAG: Hi ashwith, thank you very much for your video and links... My goal is to send digital information over a mechanic wave. Ex: using a smartphone speaker ultra sound and being able to decode that with a circuit with the help of an arduino. Regards. SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-12-02T14:08:32Z SecondChildTAG: Oh I can give you some ideas there. The ultrasound transmitter should have some sort of an enable pin (if not it should be easy to improvise with some circuitry). Use your digital message to drive the enable pin (you can do something similar if you are using your smartphone's speaker). On the receiver side, you need to make sure that the receiver's resonant frequency is the same as the carrier you are using. Just curious. Is this for fun? A mechanical wave won't travel very far. SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-12-02T14:35:21Z SecondChildTAG: Yes this is for fun... My goal is to unlock my front door key with my smartphone :) SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-12-02T14:49:31Z SecondChildTAG: Looks like a really neat CECC2 project :) All the best! SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-12-02T17:08:42Z SecondChildTAG: Hey nice idea ;) I hope to have some time for this. Thanks! SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-12-02T18:09:54Z IndexTAG: 1284 TitleTAG: H12P2 Hi to all! Im trying to solve H12P2 and cant get correct answer for c and d questions. Ok, next is how to define R0.I may use more strict requirements for $V_{out}$ within specified range, ie I can use +-0.05V tolerance or better. Or, I should use equation for $V_{out}$=4.9V@$V_{in}$=10V and $V_{out}$=5.1V@$V_{in}$=20V. What is correct way? Edited : Solved, we need all 4 values Thanks UserIdTAG: 205311 UserNameTAG: Sergtronix CreateTimeTAG: 2012-11-30T15:39:19Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 7 FirstChildTAG: Your equation for v+ looks OK. Choose R0 so that the variation in v+ is less than .05V as vin goes from 10V to 20V. I just chose a relatively large R0, so that my variation was approximately .001V. That makes the rest of the design easy. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-30T16:39:05Z SecondChildTAG: Thank you, skyhawk! To choose R0 manually was my first attemption(In the real design we will look into Zener's datasheet for understanding current ranges through the diode). I wrote equation that simplified R0 calculation: $R0={R_z}*(\frac{{V_{in}}-v_z}{\frac{V_o}{G}-{v_z}}-1)$ This equation clearly show that the narrow tollerance for $V_O$ will result larger R0 value. I should note here that $v^+$ will be greater then 2.5V that may be compensated by G less then 2. Ok, let I have R0, R1 and R2..I cant get green marks :( Should I put correct value for the efficiency too for getting all greenn marks? SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-11-30T19:40:51Z SecondChildTAG: It's all or nothing. You need the correct efficiency for your design to get check marks! SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-11-30T20:00:58Z SecondChildTAG: Wow! I made stupid error : I placed R1 value into R2 box and vice versa.. thats crazy..life isnt simple..ok. I have all green marks. Skyhawk, thank you very much for help! PS I do worry about final exam. SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-11-30T20:07:28Z SecondChildTAG: Very good, and it was my pleasure! SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-11-30T21:46:04Z SecondChildTAG: It's impossible. I can't get the values. With R0=94, R1=90 and R2=100 equations are fulfilled, but efficiency i suppose is not ok. Choosing a large R0 i obtain two equations with the relationship (R1+R2)/R2 with two different values. I can not understand it. SecondChildUserIdTAG: 244225 SecondChildUserNameTAG: lerimock SecondChildCreateTimeTAG: 2012-12-06T16:18:26Z SecondChildTAG: What's that G in the equation? SecondChildUserIdTAG: 244225 SecondChildUserNameTAG: lerimock SecondChildCreateTimeTAG: 2012-12-06T16:21:01Z SecondChildTAG: $G$= 2 for first approximation.Please understand THAT equation for R0 is only for the first attempt in the design path.When you do use NARROW tollerance for $v_O$ it is not so important will be G=2 or 1.95 or something else.Your TASK is understand WILL $v_O$ be within initial requirements or not.It is simple. SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-07T19:41:12Z SecondChildTAG: I cant make sense of that formula, how did you get it, and what is G meaning? Coz if i think of G as circuit gain, ratio of vout/vin, then vout/G = vin, then fraction numerator=denominator and formula simplifies to R0 = Rz*(1-1). Which makes no sense to me. Cheers. SecondChildUserIdTAG: 331483 SecondChildUserNameTAG: Doru SecondChildCreateTimeTAG: 2012-12-08T19:32:57Z SecondChildTAG: Really? :) Try to catch idea : when $v_{out}$=5.001V and $v_z$=2.5V, $V_{in}$=10/20V..what will you get? Now change $v_{out}$ to 5.1V, compare both R0. Cheers SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-09T22:35:13Z FirstChildTAG: V+ is maximum when Vin = 20V and V+ is minimum when Vin = 10V. Then using the the equation in the first post, Delta_V+ = 10*Rz/(R0+Rz). Delta_V+ must be less that .05V, but can be smaller, for example .01V or even .001V. Choose a Delta_V+ and calculate R0. Using that R0 calculate minimum and maximum V+ and use those values to calculate the average V+, then calculate the gain required for average V+ to produce an output of 5V. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-12-06T18:35:52Z SecondChildTAG: @skyhawk, what is the equation for V+? Thanks! SecondChildUserIdTAG: 245291 SecondChildUserNameTAG: rlicas SecondChildCreateTimeTAG: 2012-12-06T19:45:26Z SecondChildTAG: See the first post in this thread. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-12-06T20:52:34Z SecondChildTAG: And for R1 and R2, how can I get these values? SecondChildUserIdTAG: 310147 SecondChildUserNameTAG: ildomarcarvalho SecondChildCreateTimeTAG: 2012-12-07T02:55:37Z SecondChildTAG: desired gain = (R1+R2)/R2 SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-12-07T12:56:34Z SecondChildTAG: SKYHAWK, I did exactly what you said: 1) I calculated R0 supposing delta V+ was 0.01, then I calculated V+ MAX and V+ MIN with the value of R0 that I found Before. 2) After, I calculated the V+ Average as ((V+MAX)+(V+MIN))/2. I replaced that V+ average in the V- formula that comes from the voltage divider of Vout in R1 and R2. Assuming R2 equal to something, you could solve for R1. 3) Now I had the 3 resistor values. I replaced those values in a formula of Vout that invoves Vin (that formula can be found just equating V+ and V-) and I got correctly 5.1 Volts at Vout for 20 Volts of Vin, and 4.9 Volts for Vout for 10 Volts of Vin. 4) Then I wanted to find the efficiency, efficiency is basically Pout/Pin. Pout = Vout*Iout Pin = Vin*Iin Vout = 5.1 volts Iout = Vout/RL Vin = 20 Volts Iin = Iz + K*i Where Iz is the current that flows trought the Zener. Iz = (Vin-2.5)/(R0+Rz) i is the current that goes through the diode: i = Ir + iOUT Ir is the current that goues trought the resistors R1 and R2 Ir = Vout/(R1+R2) Iout is the current that goes trought RL Iout = Vout/RL Now I can replace Ir and Iout in i, and replace i in the formula of Iin, and finally replace Iin in Pin. When you calculate Pout and Pin, you just have to divide Puot/Pin to get the efficiency. What I am not sure is if the efficiency, the ratio of Pout/Pin has to be multiplied by 100. Are my formulas correct? Do I have to take care of the zener voltage source as another part of Pin so that Pin would be equal to: Pin = Vin*Iin + Pzener Where Pzener would be equal to Vz*Iz Or what about the voltage source that the OpAmp has as an equivalent circuit? Is that another power that I have to consider for Pin ? Im not getting the green checks at all. I need help! SecondChildUserIdTAG: 181432 SecondChildUserNameTAG: enriqueferreralcala SecondChildCreateTimeTAG: 2012-12-07T14:42:25Z SecondChildTAG: hey thanks for the help. I got green checks by your hints :) SecondChildUserIdTAG: 120850 SecondChildUserNameTAG: Nitin1A SecondChildCreateTimeTAG: 2012-12-09T08:30:32Z FirstChildTAG: Enrique Your formulas look correct. For this problem, efficiency is expressed as a fraction not as a percent. But I don't understand this this statement: **3) Now I had the 3 resistor values. I replaced those values in a formula of Vout that invoves Vin (that formula can be found just equating V+ and V-) and I got correctly 5.1 Volts at Vout for 20 Volts of Vin, and 4.9 Volts for Vout for 10 Volts of Vin.** If you choose R0 so that delta V+ is .01V, I would expect about 5.01V at Vout for Vin = 20V, and about 4.99V for Vout when Vin = 10V. I would also expect R0 to be about 1000 ohm. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-12-07T16:17:03Z SecondChildTAG: Here is a thread that deals with efficiency: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50bd76e645ea1f1f0000007b SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-12-07T16:38:39Z SecondChildTAG: What I wanted to say with that statement, is that once I had the 3 resistor values, which are: R0 : calculated from the delta V+ formula R2 : supposed to be 100 ohms (thats what I decideed to suppose) R1 : calculated from the Vout formula that involves Vin , R1 and R2, using the V+ average. What I wanted to say was that, after having those 3 resisores values, I replaces them in the Vout(Vin) formula and got the correct values as expected: for 20 Volts at Vin, I got 5.1 Volts at Vout. And for 10 volts at Vin I got 4.9 Volts at Vout. The question is, Why I suposed a value for R2? Basically because we only have 1 equation, with 2 unknowns, so I we assume that R2 has an X value, in this case I said that R2 was 100 ohms, so at the end I could just solve for R1. Im not sure if that assumption is valid in order to solve an equation with 2 unknowns. Is it valid? After doing all of the Pin and Pout calculations, I took the ratio Pot/Pin but the result is not giving me green check. SecondChildUserIdTAG: 181432 SecondChildUserNameTAG: enriqueferreralcala SecondChildCreateTimeTAG: 2012-12-07T18:28:04Z SecondChildTAG: you dont have to assume , just make for Vin 20 --> Vout 5.1 and for Vin 10 --> Vout 4.9 1. Vz+(20-Vz/(R0+rz))= 5.1*R2/(R1+R2) 2. Vz+(10-Vz/(R0+rz))= 4.9*R2/(R1+R2) 3. then equation1/equation2 will result with 1 unknown R0,you solve for R0 SecondChildUserIdTAG: 236136 SecondChildUserNameTAG: Hamid-ch SecondChildCreateTimeTAG: 2012-12-07T20:00:59Z SecondChildTAG: Since the amplifier gain is almost exactly 2 then delta Vout = 2*delta V+. You can choose one or the other. You discovered for yourself that a delta Vout of .2V may not get a check mark, so why not just start with a small delta V+ and save yourself the algebra. A check mark is a check mark! SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-12-07T22:19:49Z SecondChildTAG: the gain is vout/vin, and is NOT 2: it actually varies between roughly 2 and 4. SecondChildUserIdTAG: 331483 SecondChildUserNameTAG: Doru SecondChildCreateTimeTAG: 2012-12-08T20:12:07Z SecondChildTAG: actually to correct myself on the above, its inverse like 1/2 and 1/4 :) as the output stays roughly at 5v, the input varies between 10v and 20v, so its roughly 5/10 and 5/20 :) SecondChildUserIdTAG: 331483 SecondChildUserNameTAG: Doru SecondChildCreateTimeTAG: 2012-12-08T20:14:03Z SecondChildTAG: makes sense sort of it stepping down the voltage its gain is less than 1 SecondChildUserIdTAG: 331483 SecondChildUserNameTAG: Doru SecondChildCreateTimeTAG: 2012-12-08T20:16:19Z SecondChildTAG: The gain of which I speak is with reference to V+ that is set by the Zener diode and R0. It is fixed by the values of R1 and R2. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-12-09T14:45:46Z FirstChildTAG: Of course the gain is going to be approximately 2. The exact value is going to depend on the values of V+ you end up with. So R1 is approximately equal to R2. Since the purpose of R1 and R2 is just to provide a voltage divider for feedback, all the current through them is wasted. I see no reason to make that current more than a fraction of a milliamp. Values of R1 and R2 around 10000 ohms seemed reasonable to me. I still do not understand how you can be getting the Vout values you are getting. If you chose delta V+ to be .01V, the values you are getting are impossible. Is your R0 approximately 1000 ohms? In that case, your worst case value of Iz should be about 17.5 milliamps. That should give a V+ max of about 2.518V. Are these the kind of numbers you are seeing? FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-12-07T19:16:23Z SecondChildTAG: For : Vout=5.01V and Vin=10V I have :R0 between 1400 and 1600 Ohm, Iz=5mA, vz=2.505V.In the worst case zeners current is twice more.its ok for both cases; it is possible to improve this situation if someone need it, but I dont see any sense for this-we just do solve home task. Equation for R0 works well, problem is only HOW to use it :) And again, this is first aproximation in the design. Thank you Skyhawk for attention, I have rechecked my equation and spreadsheet. SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-09T23:22:54Z FirstChildTAG: **important**: when i make the output 5V with +-0.1V i get red x check but when i make it with a value < 0.1V ( i make it 0.01V) i get green check - if this help you then let others know it FirstChildUserIdTAG: 236136 FirstChildUserNameTAG: Hamid-ch FirstChildCreateTimeTAG: 2012-12-07T20:27:07Z SecondChildTAG: Same here with 4.9-5.1 (0.20 variation). I investigated and i think i discovered why. Checking my math backwards, and it seems to be coz of the upper limit which my values got to 5.10002 instead of 5.1, therefore a little over that it should be, but again as the procedure was correct the checker shouldnt have been *that* sensitive. Doing it again for 4.91-5.09 (0.18 variation instead of 0.20), fits right in with extra room for the legs. Imo, the checker needs a check and increase in the top interval a little, like fifth decimal to be 3. SecondChildUserIdTAG: 331483 SecondChildUserNameTAG: Doru SecondChildCreateTimeTAG: 2012-12-09T00:38:44Z SecondChildTAG: on a second thought, this looks more important than at first, it may be why many ppls dont get their green marks, and i think it should be fixed asap. SecondChildUserIdTAG: 331483 SecondChildUserNameTAG: Doru SecondChildCreateTimeTAG: 2012-12-09T00:49:23Z FirstChildTAG: Re-check your equation for $V^+$. FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-12-07T22:57:37Z SecondChildTAG: Do you know where is mistake ? Show please SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-10T00:16:04Z SecondChildTAG: My equation is correct, check explanation :D SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-10T16:10:13Z FirstChildTAG: Can you Sergtronix help me? I found this equation: 4.9=(R1+R2)/R2*2.5 + [(R1+R2)/(R2*(R0+1))]7.5 5.1=(R1+R2)/R2*2.5 + [(R1+R2)/(R2*(R0+1))]17.5 Then I found that (R1+R2)/R2 have to be 1.9 and (R1+R2)/(R2*(R0+1)) 0.02. Solving I found R0=94 R1=90 ans R2=100. Can you point me what I am doing wrong? Thank you. FirstChildUserIdTAG: 98262 FirstChildUserNameTAG: rafaelmarques FirstChildCreateTimeTAG: 2012-12-10T02:06:29Z IndexTAG: 1285 TitleTAG: just question :) i`d like to know when the course will be available again :) UserIdTAG: 834357 UserNameTAG: MahmoudEG91 CreateTimeTAG: 2012-11-30T00:26:08Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: As far as I have read, it will be offered again next Spring, approximately on March :) Take care, Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-30T00:36:57Z SecondChildTAG: Thnx alot :) is there any courses about Mechanical Engineering Here?! SecondChildUserIdTAG: 834377 SecondChildUserNameTAG: mahmoudeg1991 SecondChildCreateTimeTAG: 2012-11-30T00:44:41Z SecondChildTAG: You are welcome :), You can take a look at Courses offered in edX [here][1]. Take care, Myriam. P.D. I remember a student from 6.002x Spring that at the end of the Course offered a Course of Mechanical Vibration ME3521, without certificates as he did that for free for us haha, - username ID, siliconbronze- I don´t have his email... But if you are interested I can try to get his email, my friend juancho from 6.002x Spring took that Course... [read syllabus ME3541][2] [1]: https://www.edx.org/courses [2]: https://dl.dropbox.com/u/24426498/Course%20Description.pdf SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-30T01:12:42Z SecondChildTAG: thnx alot :) in fact iam not used to online education .. its my first time with 6.002x and iam waitin to try it in spring :) So, if things went good .. and if i get used to ED iwill ask again for mechanical engineering .. but any way thnx alot for helping :) just another question : is thw online education is Good and Fixable process or difficult ?! SecondChildUserIdTAG: 834377 SecondChildUserNameTAG: mahmoudeg1991 SecondChildCreateTimeTAG: 2012-11-30T01:48:19Z SecondChildTAG: Hi mahmoudeg1991, Based on my experience. I really liked this online learning iniciative from edX. It is amazing! You can learn a lot and be in Contact with a lot of friendly person worldwide. I did a lot of friends from 6.002x Spring 2012, we still in touch ;). Then I re taked this same Course in this Fall in order to help the New Students and edX offered me to became Community TA, so I am really happy to be here helping freely to my Classmates haha. In my point of view, the online education is great. For me, edX has quality in their Courses - they don´t paid me to saying this haha, I am really telling you this by heart. Not only for the Course material, also for the good people behind the Scene, that you might not know but they are there, they are really involved with the Students concerns, believe me that this is true. Have you seen to Prof. Agarwal in the videos of this Course, well, he is the coolest ever Prof. . Many of the ex-students of the 6.002x Spring miss him so much, so that says a lot !;) As all learning Process, it is not easy, you have to do the exercises and watch the videos, read the Textbook, etc ;). You will learn a lot, and the most important you will have fun! Take care, Myriam. P.D. Sorry for my english haha. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-30T02:06:27Z SecondChildTAG: ocw.mit.edu has many courses pls check SecondChildUserIdTAG: 623210 SecondChildUserNameTAG: preveen SecondChildCreateTimeTAG: 2012-12-01T18:04:24Z IndexTAG: 1286 TitleTAG: certificate? I am confused about how they determine the final score about the certificate? Is it the total score or does it go by the total score for homework, the total score for labs, and the total score for the finals, or is it just the total overall score? Can someone help me with this because I am confused. UserIdTAG: 160242 UserNameTAG: Mlevins35 CreateTimeTAG: 2012-11-30T00:06:19Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hi Mlevins35, Might this can help you, Visual help --- - PASS and NOT PASS Table: ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13374846943979601.png) - Percentages of total score will be based on: ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13374850316133229.png) So, if you have : Example 1. - 10 Hw =100% you will get a 15% points for that. - 10 Labs=100% you will get a 15% points for that. - If you solved 100% the Midterm you will get 30% points for that. - If you solve 100% the Final Exam you will get 40% points for that. - So, the total score will be 15%+15%+30%+40% = 100% total score . - You will pass with grade A. Example 2. - 10 Hw = 70% you will get 10.5% points for that. - 10 Labs= 80% you will get 12% points for that. - If you solved 60% of the Midterm exam you will get 18% points for that. - If you solve 80% of the Final exam you will get 32% points for that. - So the total score will be 10.5%+12%+18%+32%=72.5% - You will pass with the grade B. Example X: - 10 Hw = H % you will get H*0.15= h [%] - 10 Labs=L% you will get L*0.15= l [%] - Midterm Exam= ME% you will get ME*0.30= me [%] - Final Exam = FE% you will get FE*0.40= fe[%] - Total Score =ts= h + l + me + fe Pass ts high or equal 60% Not pass ts less than 60% FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-30T00:34:36Z SecondChildTAG: I have not done that great so far I got a 9% on homework, a 39% on labs and a 24% on me. so with these scores I will not get a certificate. My total score has not moved up for a while. I am not the smartest tool in the shed but I do understand most of the class. I plan on trying again in the spring, I am determined to do better and try to get that certificate SecondChildUserIdTAG: 160242 SecondChildUserNameTAG: Mlevins35 SecondChildCreateTimeTAG: 2012-11-30T00:44:03Z SecondChildTAG: Hi Mlevins35. It is really valuable what you are saying, the important thing is that you are understanding and learning. So don`t worry if you can not get the Certificate this time, try to make all the best that you can this Fall, try to not be upset, have fun because that is important in the process of learning and don´t feel alone, because I will be here to help you. I am sure that next time you will do it better and I am really happy that you are planning to try it again in the spring, that says a lot of good things about you, and of course, I am happy that you want to challenge yourself and get next time the certificate, that is a role model to follow, really insipiring :). My best wish to you Mlevins35. Be up always! Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-30T15:14:25Z SecondChildTAG: *inspiring :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-30T15:15:42Z FirstChildTAG: Thank you all so much I am going to take the class in the next session and see if I can improve . I have learned so much and the professor makes it sound so easy. I would tell anyone about this class. FirstChildUserIdTAG: 160242 FirstChildUserNameTAG: Mlevins35 FirstChildCreateTimeTAG: 2012-12-03T18:36:10Z IndexTAG: 1287 TitleTAG: Oh! man, what a storm of ideas. This definitively puts everything we have learned so far into a whole new perspective. UserIdTAG: 369339 UserNameTAG: vargaslen CreateTimeTAG: 2012-11-29T09:57:02Z VoteTAG: 2 CoursewareTAG: Week 12 / How to build a current source CommentableIdTAG: 6002x_How_to_build_a_current_source NumberOfReplyTAG: 1 FirstChildTAG: Yep!! And 6.002x is just only a basic introduction! But a very good one! FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-11-29T12:22:29Z IndexTAG: 1288 TitleTAG: Final Exam To the Staff: Could you please give us more information about the final exam dates? It is planned to be released on 20th December but what about the due date? Will it be like in the midterm exam when we had 3 or 4 days to choose when to start? It is important to know the due date because of the Christmas holidays being so close. Thanks in advanced. UserIdTAG: 255161 UserNameTAG: mlovelle CreateTimeTAG: 2012-11-28T07:41:35Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I am not staff, but I would imagine the exam will be 1, maybe 2 days. (The midterm was extended to 4 days to accommodate a Muslim holiday.) Have fun. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-11-28T13:07:06Z SecondChildTAG: Makes sense to offer ppl possibilty to take the exam on saturday and sunday since lots of us actually only got these days to devote so much time and efforts into an exam SecondChildUserIdTAG: 326409 SecondChildUserNameTAG: EugeneZ SecondChildCreateTimeTAG: 2012-11-28T18:29:57Z SecondChildTAG: That was a common request during the midterm as well. I suppose it is due to the fact that behind the scenes there is a brick and mortar school, tradition and support staff that need to be available during the exam. I am confident that they would accommodate this if or when it becomes practical to do so. Essentially an exam period that is during the week and on the weekend would be best, similar to the midterm. There are lot's of people that work on the weekend as well and may prefer to take it during the week. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-11-28T23:59:27Z IndexTAG: 1289 TitleTAG: S23E3: Not so painful (possible spoilers if you want to figure it out yourself) The professor claimed that this would be painful, but I found it fairly easy. I solved it algebraically as a function of gain and the two resistances (v0 = G*vI, iI = (G*vI - vI)/(R1 + R2) ) and plugged in the values of gain determined on the last problem. All the pain was handled by doing the previous problem (which I found to be much harder than this one). UserIdTAG: 174100 UserNameTAG: markpolak CreateTimeTAG: 2012-11-28T05:08:54Z VoteTAG: 2 CoursewareTAG: Week 12 / S23E3_L23AmplifierInputResistance CommentableIdTAG: 6002x_S23E3_L23AmplifierInputResistance NumberOfReplyTAG: 0 IndexTAG: 1290 TitleTAG: H12P2 One little detail For me the normal way to give answers is to use k as 1000, but in this answer you must give the figures in ohm. 1000 not 1k. UserIdTAG: 151472 UserNameTAG: Stensmed CreateTimeTAG: 2012-11-28T00:08:30Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Yes, I found that out today too! I'm also used to give answers in k (ohm, frequency) or M (ohm, frequency), but it was not accepted in HP12P2!!! FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-11-28T14:05:02Z IndexTAG: 1291 TitleTAG: H11P3 Dear all, Can anybody please give a hint to find Q from the voltage graph of RLC Circuit? UserIdTAG: 156835 UserNameTAG: kphariprakash1968 CreateTimeTAG: 2012-11-27T16:16:13Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: just count the glitches...watch S22V6 and you will understand... FirstChildUserIdTAG: 301962 FirstChildUserNameTAG: labrinim FirstChildCreateTimeTAG: 2012-11-27T19:08:30Z FirstChildTAG: Can I help you? Hint 1. Take a look at page 748 of the Textbook :) Hint 2. Take a look at Lab 11 ;) Take a look at [H11P3 Hints :)][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50b54fce5fb7411f00000006 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-27T23:15:42Z SecondChildTAG: Thank you very much SecondChildUserIdTAG: 156835 SecondChildUserNameTAG: kphariprakash1968 SecondChildCreateTimeTAG: 2012-11-28T11:15:59Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-28T12:00:22Z IndexTAG: 1292 TitleTAG: please help i cant take lectures online hello dears i cant take online lectures.. where is the link to online player for video lectures... youtube is still blocked and i cant download the videos due to there huge size.. please help i want to get certificate.. please help thanks UserIdTAG: 136490 UserNameTAG: Ali_PU CreateTimeTAG: 2012-11-27T12:26:20Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: I agree that the downloadable files are big. I myself have been using the textbook the last while to save on bandwidth. Fortunately the whole textbook has been offered to us online as well. It would be nice if the videos were offered for download in a smaller format. I do not think they should replace the HD, but have both available. I get my internet via wireless (WISP), there simply is not any bigger package available, so essentially I have 20 Gb a month to share with three people. If I go over they charge around $10 a Gb :0. It seems like enough but we have found we can use that up easily, without any significant downloading. It is fast though, we got a bi-quad type antenna and attached it to a parabolic satellite dish, this really helped because we are almost in a fringe area, without the home made dish the new service was almost unusable. Up until a few months ago we used to have unlimited internet with a different provider, but the speeds became so slow I could not really watch videos properly anymore. (Too many neighbors on our local loop.) It's too bad because I wanted to take more classes at the same time, but I just don't have the bandwidth to do it. So now I am just enrolled in this course and CS50x which has a flexible timeline. I will focus on that course over the winter. I will likely be moving this winter, hopefully the new place has cable internet. :) If anyone out there is having difficulty watching videos due to poor internet *reception* I recommend you build your self a parabolic antenna such as I have. The first little one uses a USB Wifi dongle inside of a 2.5Ghz waveguide attached to the dish. I have been successful in gaining access to a WEP protected router that is around 8km line of sight away. This was for educational purposes only. You can attach a balloon around the USB dongle to make it weather resistant. The second bigger antenna is a little different. It uses a 2.4Ghz Biquad style antenna but also has a router running DD-WRT in repeater mode. Basically it pics up a Wifi signal, and boosts/repeats it wirelessly to multiple devices, this is nice because the smaller dish has to be relatively close to the computer. (USB cables cannot be very long) It can run on a car battery or a 12v solar panel array. If it's close enough to the house, I just use a 12v adapter. I get the best distance out of the smaller dish, I can use multiple devices with the repeater type dish. I may be able to increase the range of the repeater dish through experimentation. Right now it only works within a kilometer or two. These are "offset" type dishes, they look like they are pointed up, they are actually looking straight ahead. ![enter image description here][1] ![enter image description here][2] ![enter image description here][3] [1]: https://edxuploads.s3.amazonaws.com/13540234081343659.jpg [2]: https://edxuploads.s3.amazonaws.com/1354023477134362.jpg [3]: https://edxuploads.s3.amazonaws.com/13540234971343603.jpg FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-11-27T13:35:23Z SecondChildTAG: You can download the complete lecture at the OCW site. http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-electronics-spring-2007/video-lectures/ The videos here are in pieces. But its the same. You can get lecture notes at: http://www.mediafire.com/?lxzhh9wox4hq3 SecondChildUserIdTAG: 171674 SecondChildUserNameTAG: wajahat1 SecondChildCreateTimeTAG: 2012-11-27T14:17:01Z SecondChildTAG: there are parts of lectures which are new and you still have to download their pieces, just found out. SecondChildUserIdTAG: 171674 SecondChildUserNameTAG: wajahat1 SecondChildCreateTimeTAG: 2012-11-27T14:38:39Z FirstChildTAG: Please also see these notes from Spring 2012: [FSanchez's Lecture Summaries][1] [Thomas Backman's (exscape) Notes][2] They will help! (They are also linked in the course [Wiki][3]) [1]: http://www.mediafire.com/?lxzhh9wox4hq3 [2]: http://exscape.org/perm/6002x-notes-exscape.pdf [3]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/course_wiki FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-11-28T04:30:35Z SecondChildTAG: That is a nice set of notes. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-11-29T00:10:25Z FirstChildTAG: You can download medium format 2012 spring lectures at http://www.ruudoleo.com/mitx/small/ The subtitles are in separate srt files. They are the same lectures as used here. FirstChildUserIdTAG: 79885 FirstChildUserNameTAG: ruudoleo FirstChildCreateTimeTAG: 2012-11-27T21:28:00Z SecondChildTAG: Thanks. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-11-27T23:45:10Z SecondChildTAG: Thanks a lot,it is very helpful! SecondChildUserIdTAG: 361137 SecondChildUserNameTAG: Ericson SecondChildCreateTimeTAG: 2012-11-28T01:16:20Z IndexTAG: 1293 TitleTAG: LAB 11 red cross :( Why do I have a green mark in Figure 1 after placing the calculated values ​​of L and C but to putting these values ​​in their boxes I have a red mark :( when I'm wrong? UserIdTAG: 244706 UserNameTAG: Miguel_Angel CreateTimeTAG: 2012-11-27T01:26:57Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: I suspect that whatever value you put in the simulator gives you a a green mark :( I just tested that... FirstChildUserIdTAG: 288174 FirstChildUserNameTAG: matiasgrodriguez FirstChildCreateTimeTAG: 2012-11-27T02:57:28Z FirstChildTAG: You need find anywhere right values of L and C! It will need to get right answers in next part of this lab. All equations what you need you will find in text of lab and chapter 14 if i`m not wrong! ;-) FirstChildUserIdTAG: 96873 FirstChildUserNameTAG: Santyaga FirstChildCreateTimeTAG: 2012-11-27T11:37:10Z FirstChildTAG: Hi Miguel_Angel, Can I help you? Have you click on the AC analysis button? Was it similar to the Figure given in the statement? Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-27T23:20:31Z SecondChildTAG: Always the same silly mistake :P w0 = 2 * pi * f I forget pi :( . I'm back to do the calculations and everything is green as Shrek (moment haha!) Gracias Miriam :) Eres tu la del video de youtube?, (momento guauuu) SecondChildUserIdTAG: 244706 SecondChildUserNameTAG: Miguel_Angel SecondChildCreateTimeTAG: 2012-11-28T00:15:06Z SecondChildTAG: Ojalá que sea eso del 2pi :). Sip, soy yo jaja . Espero que puedas participar del Concurso :). Saludos, Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-28T00:58:18Z SecondChildTAG: Thank you Miguel_Angel, you gave me the answer I desired. SecondChildUserIdTAG: 418182 SecondChildUserNameTAG: euldji2005 SecondChildCreateTimeTAG: 2012-11-30T22:38:20Z SecondChildTAG: To you euldji2005 :) SecondChildUserIdTAG: 244706 SecondChildUserNameTAG: Miguel_Angel SecondChildCreateTimeTAG: 2012-12-03T17:56:44Z IndexTAG: 1294 TitleTAG: a new plus! Thanks Dr. Agarwall, for at last I came to understand the way that antilock break system in my car works! This is a great course in many many different angles! UserIdTAG: 369339 UserNameTAG: vargaslen CreateTimeTAG: 2012-11-27T00:17:19Z VoteTAG: 2 CoursewareTAG: Week 12 / S23V12 Antilock brakes and negative feedback CommentableIdTAG: 6002x_S23V12_Antilock_brakes_and_negative_feedback NumberOfReplyTAG: 0 IndexTAG: 1295 TitleTAG: H12P2 evil details Hello, I am currently finished H12P2 after long hours of trying. The evil is in the details. This message can help students to focus to solution of problem, not to autograder syntactic details. a. and b.) the function is written with (vin,vz - with all small letters). The first evil is that the current source is producing current in painted direction in all circumstances (vout can be greater than vin). This can be seen only in this simplified model. In real situation, the BJT is more complex and have another diode between base and collector, that shorts output from operational amplifier to vin in this operation point allowing output voltage of maximum of vin (+vf). c-e) Don't forget that word "satisfy" means "be better than the requirement". This sentence very simplifies solution. The result is accepted in Ohms. f) The evil is in in carefully describing all of currents flowing through the circuit. Do not forget that operational amplifier has its own evil :-) The result is coefficient between 0 and 1 (not percents) Cheers, UserIdTAG: 505903 UserNameTAG: Burian CreateTimeTAG: 2012-11-26T18:31:50Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 6 FirstChildTAG: Not sure how you conclude that vout can be greater than vin. vin must also power the op amp (although the connection is not shown), so it is impossible for vout to be greater than vin. In the problem statement, 10V <= vin <= 20V, but 4.9V <= vout <= 5.1V. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-26T18:55:59Z SecondChildTAG: The accepted answer for part (a) gives $v_{out} > v_{in}$. The obvious answer to part (a) (to me) was $v_{in}$, but they never specified what powers the op-amp, just that it was ideal, as was the diode, so no $v_f$ across the diode in the BJT model. For a more realistic BJT and rail-rail op-amp powered by $v_{in}$, I'd go for $v_{out} = v_{in}-v_f$. SecondChildUserIdTAG: 141000 SecondChildUserNameTAG: OrinE SecondChildCreateTimeTAG: 2012-11-26T19:55:19Z SecondChildTAG: In the real situation, with opamp powered with different power source, the truth is vin + vf (collector-base diode opened and flowing current). In real situation, with opamp powered with vin, vout will be maximum voltage on opamp output rail (on some opamps it is vin - 2V) minus vf. To raise this voltage, the opamp must be rail-to-rail on output. Both situations is not the problem described in H12P2 SecondChildUserIdTAG: 505903 SecondChildUserNameTAG: Burian SecondChildCreateTimeTAG: 2012-11-27T07:04:37Z SecondChildTAG: I agree that the real situation is the one you described. However, this is a homework problem and if you stick to the circuit shown, then you should get the correct answer SecondChildUserIdTAG: 381746 SecondChildUserNameTAG: jelizon SecondChildCreateTimeTAG: 2012-11-27T16:09:19Z SecondChildTAG: Ok i tried with values of R0,R1,R2 just to have vout= 5v with vin=10V vout= 5.1V with vin = 20V doesn`t works I think ill try with 5 and 5.05 V instead and with greater values for R1/R2 in order to consume less useless power SecondChildUserIdTAG: 373752 SecondChildUserNameTAG: Croak SecondChildCreateTimeTAG: 2012-11-27T19:22:16Z SecondChildTAG: Ah and I considered that the (1-K)*i*5V power coming out of the opamp is an input power ... Do you think its correct ? SecondChildUserIdTAG: 373752 SecondChildUserNameTAG: Croak SecondChildCreateTimeTAG: 2012-11-27T19:26:28Z SecondChildTAG: Input power to the op amp is (1-K)*i*Vs where Vs is the op amp power supply voltage, which is assumed to be the same as vin. The input power to the BJT is K*i*vin. Finally there is the small amount of power consumed by R0 and the zener diode. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-11-27T23:41:52Z SecondChildTAG: ok it works now ! I confirm I did'nt use the value of power supply of the opamp. So there is a little approximation : The power consumed by the opamp itself is neglected. Another view : The power coming from the external alim. of the amp is considered to entirely come out of his output . I took the current value (1-K)i and the tension=Vout - 0.6V ( 0.6V= the tension drop across the diode ) That's enough to compute the power due to vcc of the opamp. SecondChildUserIdTAG: 373752 SecondChildUserNameTAG: Croak SecondChildCreateTimeTAG: 2012-11-28T01:47:35Z SecondChildTAG: According to the problem description, the diode is ideal, therefore there's no such drop. SecondChildUserIdTAG: 318577 SecondChildUserNameTAG: takeuchi SecondChildCreateTimeTAG: 2012-11-30T17:08:12Z FirstChildTAG: Since (for my data set) .05 of the BJT emitter current is supplied by the op amp by way of the base-emitter current, it is necessary to know the power supply voltage for the op amp in order to calculate the losses in the op amp, (vs-vout)*ib. The only available choice for vs is vin. This seems reasonable since the usual three pin regulators just have vin, gnd, and vout pins. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-26T20:29:41Z SecondChildTAG: No, this is not true. In this practice, the opamp has infinity voltage (ie no saturation on output). The losses in the opamp is not necessary to compute effeciency. SecondChildUserIdTAG: 505903 SecondChildUserNameTAG: Burian SecondChildCreateTimeTAG: 2012-11-27T07:01:23Z SecondChildTAG: I think this is a very unhealthy way to calculate efficiency! If the OA has infinite voltage, then it consumes infinite power, even for small currents, like in this problem. Would you buy a pws with an efficiency of say, 0.5 for P(vout)/P(VIN), but need an external pws to deliver the voltage/current for the OA, that needs lots more power than will ever be delivered by VIN??? SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-11-27T15:13:23Z SecondChildTAG: I think we don't have to know the supply voltage of opamp. Only the U*I at his output, and count this as an input power... the problem is we know I : = ( 1-K ) ( because K<1 ) But i supposed U was 5V ( 5.1 ) and that didn't worked.. SecondChildUserIdTAG: 373752 SecondChildUserNameTAG: Croak SecondChildCreateTimeTAG: 2012-11-27T19:42:53Z FirstChildTAG: Hello ... Help in H12P2 I wonder what is " efficiency" of the regulator !!?? FirstChildUserIdTAG: 373752 FirstChildUserNameTAG: Croak FirstChildCreateTimeTAG: 2012-11-27T04:11:03Z SecondChildTAG: This is simple, power delivered to resistor divide by power that enters circuit from vin :-) SecondChildUserIdTAG: 505903 SecondChildUserNameTAG: Burian SecondChildCreateTimeTAG: 2012-11-27T06:58:34Z SecondChildTAG: The best way to calculate is to compute input power (input terminal and Op Amp) and output power. You don't need to know what is powering the Op Amp to calculate the power it is delivering to the circuit. SecondChildUserIdTAG: 381746 SecondChildUserNameTAG: jelizon SecondChildCreateTimeTAG: 2012-11-27T16:25:17Z SecondChildTAG: OK thank's I I'll say if it works :) SecondChildUserIdTAG: 373752 SecondChildUserNameTAG: Croak SecondChildCreateTimeTAG: 2012-11-27T17:33:17Z FirstChildTAG: why there will be three combinations? and how? plz post some hints. I ve already spent more than two days in this problem. FirstChildUserIdTAG: 414616 FirstChildUserNameTAG: local_hero FirstChildCreateTimeTAG: 2012-11-27T12:51:11Z SecondChildTAG: Me too, someone can help us please? SecondChildUserIdTAG: 353589 SecondChildUserNameTAG: klaramun SecondChildCreateTimeTAG: 2012-12-04T20:48:32Z FirstChildTAG: The problem is in fact unclear and therefore misleading, because of a lack of info. There are several possibilities, depending on where the supply voltage Vs, for the OA AND for the CS is coming from. Take question a for example. One possibility is : If VIN is also the Vs for the OA AND the CS, then vout can never exceed VIN, when you convert the first figure (with the BJT) to the second figure (with Ki and diode). If you replace the BJT in that case by an ideal dependent CS, then still vout cannot exceed VIN, because ideal in this case would mean that the collector/emitter voltage of the BJT could drop to 0V, the diode is ideal, so also 0V drop. For simplicity: replace the zener with a variable resistor Ra and say K=0 (not i). In that case it's easy to see, that, if v+=+input of OA between R0 and Ra, the total voltage gain Gv caused by the OA in combination with R1 and R2, is ALWAYS >=1. Suppose Gv=4, than, for VIN=30V, for (0 <= v+ <= 7.5V), it results in: ((vout=4*v+) <= (VIN=30V)) and (vout/+v)=4 is constant, thus linear. But for (+v > 7.5V) the vout wants to be (vout > 30V) and that is not possible in this case! So in this last case, vout limits to VIN for (+v > 7.5V). Now we take Ra away and leave the zener out and the Gv=4. Than +v=VIN, so vout=VIN always! So in fact, in this case Gv=1, would also give the same result. (I'm not discussing what happens when you load the output or K>0) So the term for causing the gain in question a, doesn't neccessarily need to be, and in fact shouldn't be, the same as in question b. Take question d: You can only solve this question properly, if you assume that the Vs of the OA and CS is the same as VIN. In that case it's simple, and, near realistic. If Vs is not the same as VIN, but delivered by another source, then there is also power delivered by an external source connected to Vs!! And in my opinion, it's NOT reasonable NOT to include that power into the efficiency. If I would do that, not including it, then I won't last long as an engineer ... So in fact problem a and d are fishy!! FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-11-27T13:47:41Z SecondChildTAG: Agree for par (a). However, this is true for all the circuits we have studied (adder, multipliers, etc), and usually we assumed that power to the op amp was given by external sources. If you assume that the op amp is powered by an external source and use the BJT model given (very simplified), then gain in (a) is the same as gain in (b). We will be more specific on that. For part (d), I believe the answer won't change if the supply is from external sources or from Vin. The power that the op amp is delivering to the circuit is the same. SecondChildUserIdTAG: 381746 SecondChildUserNameTAG: jelizon SecondChildCreateTimeTAG: 2012-11-27T16:31:43Z SecondChildTAG: I am going to disagree with your comment about part (d). The Op amp supplies the base-emitter current to the BJT (5% of the total current for the data given). If Vs for the op amp is different from vin then separate calculations need to be made for the power inputs to the op amp and the BJT using the respective voltages and currents. If Vs = vin then a single computation for input power can be made for those devices. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-11-27T17:55:03Z SecondChildTAG: Agree with you that power should be calculated at the input terminals. Notice, however, that if the op amp is ideal, the input power should be the same as the output power, regardless of where the power is coming from... SecondChildUserIdTAG: 381746 SecondChildUserNameTAG: jelizon SecondChildCreateTimeTAG: 2012-11-27T19:32:21Z SecondChildTAG: Shouldn't one simply calculate the power dissipated by the resistors, since all other components are ideal and do not dissipate power? i.e. efficency = P(RL)/ P(R0, R1, R2, RL, Rz) SecondChildUserIdTAG: 372321 SecondChildUserNameTAG: EnricoDona SecondChildCreateTimeTAG: 2012-11-27T21:46:52Z SecondChildTAG: Ideal doesn't mean power isn't dissipated... the resistors are ideal too or we'd be concerned about selecting the right wattage rating for the resistors so as not to let the magic smoke out! Power is dissipated by the transistor (current source in the model). SecondChildUserIdTAG: 141000 SecondChildUserNameTAG: OrinE SecondChildCreateTimeTAG: 2012-11-27T22:41:29Z SecondChildTAG: Ideal can mean many things, that's right. In the case of the Op Amp, however, it does mean that power is not dissipated because the input impedance is infinity and output impedance is zero. The BJT, on the other hand, does dissipate power. SecondChildUserIdTAG: 381746 SecondChildUserNameTAG: jelizon SecondChildCreateTimeTAG: 2012-11-28T05:21:28Z SecondChildTAG: I disagree with jelizon. See why in my second response. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-11-28T12:30:52Z FirstChildTAG: Second response, I disagree with jelizon because of how the ideal OA is described in the text book, see chapter 15.2 page 841, figure 15.6 says: The ouput voltage vo=A*([v+]-[v-}) .. The gain A goes to infinity, the output resistance = 0. . The input currents [i+]=0 and [i-]=0. Accordingly, the input resistance is infinite. And chapter 15.3.2 page 844 first part states: If we use the OA-model in figure 15.6, HEREAFTER referred to as the IDEAL OA, then ... not important. On page 843 at the bottom, it's mentioned that the power supplies are OBVIOUSLY NECCESSARY for OA operation (THEY POWER THE VOLTAGE-CONTROLLED VOLTAGE SOURCE), their inclusion in the CIRCUIT MODEL we use for ANALYSIS is not very helpful... To me all this means that modelling the gain, input impedance and output impedance are defined to calculate the behaviour of the OA, but NOT it's power consumption and/or dissipation! You may see the INTERNAL idealized voltage source in the OA as ideal one, but IT IS powered by the supply voltage, and I've never read, or seen being stated, that the OA itself doesn't consume power or dissipate energy!! So if the powersupply of an OA is VS and the vout of the OA is vout and the current of the output is iout, then the OA delivers to the load Pout=vout*iout, but the powersupply of the OA delivers at least P(VS)=VS*iout, so the dissipation in the OA is at least P(VS)-P(vout) And therefore I'm still convinced that H12P2 questions a, b and f are fishy!! It should be stated that Vsupply for the OA is vin, otherwise my assumption that one cannot calculate the efficiency still holds. For Vs is VIN, also vout cannot exceed vin, since also the BJT replaced by a ideal CS cannot cause a higher vout, because the OA would compensate it. If I'm wrong, please correct me. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-11-28T13:07:38Z SecondChildTAG: I am in agreement with Salsero on this, specifically: **You may see the INTERNAL idealized voltage source in the OA as ideal one, but IT IS powered by the supply voltage, and I've never read, or seen being stated, that the OA itself doesn't consume power!! So if the powersupply of an OA is VS and the vout of the OA is vout and the current of the output is iout, then the OA consumes at least VS*iout.** I would hope for clarification on this point. Thanks SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-11-28T14:34:23Z SecondChildTAG: I think the little power that the OA consumes for it's own functionment is neglected ( It's true that we would need to know +-Vcc to calculate it. ) But all the remaining power coming in the OA goes out at his output under the form P=U*I ( I is known, and U is not Vcc) SecondChildUserIdTAG: 373752 SecondChildUserNameTAG: Croak SecondChildCreateTimeTAG: 2012-11-28T16:00:11Z SecondChildTAG: In general, I don't believe that the power consumed by the OA is small. It is (VS-vout)*iout. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-11-28T16:16:10Z SecondChildTAG: Salsero, Why do you think (b) is fishy? Also, the power consumed by the op-amp is not Vs*Iout. (Vs - Vout)*Iout is a better assumption, but still requires the input current to be the same as the output current, which is not necessarily true. If Zin is infinity and Zout is zero, then the op-amp will perhaps consume power in the circuit powering the voltage source, agree. Let's assume this source is ideal (it is not in real life of course) so that its consumed power is zero. I will push to clarify this if you think that will make the problem clearer. SecondChildUserIdTAG: 381746 SecondChildUserNameTAG: jelizon SecondChildCreateTimeTAG: 2012-11-28T20:58:26Z SecondChildTAG: Yes, I agree with skyhawk. For the people who don't see what is happening, a little clarification: The so called internal idealized voltage source INSIDE the OA, is artificially created, because the idealized gain=A is supposed to be infinite, and therefore it looks like its small signal ri=0, SEEN FROM THE OUTPUT and therefore it SEEMS seen from the output that the delta dvout/diout=ri=0. But that doesn't mean, that the powersupply of the OA isn't delivering the current AND energy to the internal voltage source, e.g.: If the powersupply VS=20V and the vout is set to 8V and we use a simple feedback from the output to the negative input and output is loaded with 8k then iout=1mA and the load consumes 8mW, but the powersupply delivers the iout, therefore P(VS)=20V*1mA=20mW, not, as some people seem to think (20V-8V)*1mA=12mW, or, even worse, only 8mW. The formula (VS-vout)*iout tells what the OA is at least dissipating, and vout*iout tells us the power delivered to the load. But the power dilivered by the powersupply is the sum of these powers. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-11-28T21:08:30Z SecondChildTAG: To jelizon. Yes you are right, I was wrong about the consumed power by the OA, I meant delivered power by the powersupply of the OA. I suppose I was correcting this and writing the clarification, while you were updating this. I have no problem with solving H12P2, but I just disagree with the accepted answers, because it implies weird things. In my opinion the expected answer a, is not correct when the powersupply for the OA is the same as VIN,(but then answer f can be correct) Answer a is correct if the OA has an external powersupply at least equal to the expected vout (but then answer f could be different). And the reasons are explained in my interpretation, based on the textbook, above. But I'm in for a better one! Answer f is ok as long SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-11-28T21:54:23Z SecondChildTAG: The expected answer b is also fishy in my opinion, because it also implies that vout could be greater than VIN. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-11-28T22:15:28Z SecondChildTAG: Salsero and skyhawk, thanks for your comments. Proper modifications have been made to make the problem less ambiguous. I don't think the solutions are wrong, but definitively they can be interpreted in different ways. Part (a) and (b) are now assumed to have an external power supply. For part (f), ambiguity comes from the word "ideal", which may or may not imply that power is dissipated. Bear in mind that, in the ideal case, the input and output currents to the op-amp might not be the same and that is why (Vs-Vout)*Iout might not be the dissipated power. Consider, for example, a case in which an ideal voltage control voltage source is used to convert voltage from Vs to Vout, so that no power is dissipated. To avoid ambiguity, now power dissipated in the op-amp is ignored. Changes are visible in the problem now. SecondChildUserIdTAG: 381746 SecondChildUserNameTAG: jelizon SecondChildCreateTimeTAG: 2012-11-28T23:58:15Z SecondChildTAG: Ok, jelizon, thanks for the clarification and taking the effort. And I agree that an ideal VCVS that converts VS to Vout dissipates no power, but that's a different subject. Here some other reasons why I called it fishy: On page 841 at the top, says: guote "eqn 15.2 states that the output current comes from the power supplies. Important, but not very useful EXCEPT POSSIBLY for the CALCULATION OF POWER DISSIPATION." and chapter 15.7 page 867 says: "How do we model the OA when it is in saturation?" The answer: a near short circuit between the output and the powersupply. If I try to solve problems like HW12P2, especially when converting the first figure to the second figure, I try to look at the figures as a black box and then keep in mind that the ideal represention is bounded by, in this case, VIN and calculating efficiency incorporates also the OA in this case. Thanks again! SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-11-29T13:09:44Z IndexTAG: 1296 TitleTAG: Tacoma Narrows Minutephysics did a rather interesting video on the disaster. Have a look at the [video on Youtube][1]. [1]: http://www.youtube.com/watch?v=6ai2QFxStxo UserIdTAG: 359310 UserNameTAG: AaronYeoh CreateTimeTAG: 2012-11-26T11:44:04Z VoteTAG: 2 CoursewareTAG: Week 11 / S21V13: RLC capacitor response peakiness and Q CommentableIdTAG: 6002x_S21V13_RLC_capacitor_response_peakiness_and_Q NumberOfReplyTAG: 2 FirstChildTAG: If you ask an engineer what resonance is, Tacoma is always sited as THE example of natural resonance. Pretty good example I guess then FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-11-26T15:34:56Z FirstChildTAG: Ah HA! I finally understand what the Tacoma Narrows Bridge has to do with resonance. I had heard the canonical example over and over, but it never made any sense when compared to other forms of resonance because "wind" is not a waveform, and therefore things can't resonate with it. However, if you start listening at 1:02 of the video, it finally makes sense where the source waveform is coming from. My life is now complete. FirstChildUserIdTAG: 776701 FirstChildUserNameTAG: Aviendha FirstChildCreateTimeTAG: 2012-11-26T20:35:21Z IndexTAG: 1297 TitleTAG: I need to know how to convert the "log" to frequency even if the homework time is over I need it for the future can any one help please Myriam :) UserIdTAG: 226085 UserNameTAG: Teto CreateTimeTAG: 2012-11-26T05:17:58Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: You need to make use of the following information: $\log_b(y)=x$ is the same as $y=b^x$. Here, $b$ is the base of the logarithm and unless you are told otherwise you can assume that $b=10$. That is, $\log_{10}(y)$ is normally just written as $\log(y)$. If you are given that $\log(f)=a$, you can quicky find that $f=10^a$. Finally, for most of the problems in electronics we use the angular frequency $\omega$ which is related to the frequency $f$ by $\omega = 2\pi f$. So if you want to find $\omega$ from the value of $\log(f)$ you can use $\omega = 2\pi 10^{\log(f)}$. Note: The other common base is $e$ but normally this log function is written as $\ln(y)$ to distinguish it from the log function $\log(y)$ with base of 10. Hope that helps you out :) FirstChildUserIdTAG: 367878 FirstChildUserNameTAG: spoida FirstChildCreateTimeTAG: 2012-11-26T05:54:25Z SecondChildTAG: Thank you! SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-11-26T11:08:39Z SecondChildTAG: All my life I thought the same as you, until I reach university and had to change my view of the world :(. The logarithm is Log, and always is base e. If it were base 10, would be Log10. The difference between log and ln are devised for the primary school, or so I'm told. When you see in a book, log is base e :P SecondChildUserIdTAG: 244706 SecondChildUserNameTAG: Miguel_Angel SecondChildCreateTimeTAG: 2012-11-27T01:47:46Z SecondChildTAG: thank you very much spoida that was helpful I am sorry that I am a little bit late on the response I have midterm test in my university SecondChildUserIdTAG: 226085 SecondChildUserNameTAG: Teto SecondChildCreateTimeTAG: 2012-12-01T02:34:50Z FirstChildTAG: Note that the frequency axis is in log. Each number you see in this axis corresponds to the exponent of the power of 10. For example, if you take the number 3, is the number 10^3, or 10 with exponent 3, ie 1000Hz. If catch the number 4, corresponds with the number 10 exponent 4, or 10000Hz. FirstChildUserIdTAG: 361823 FirstChildUserNameTAG: EliasOak FirstChildCreateTimeTAG: 2012-11-26T10:18:36Z SecondChildTAG: thank you SecondChildUserIdTAG: 226085 SecondChildUserNameTAG: Teto SecondChildCreateTimeTAG: 2012-12-01T02:35:08Z FirstChildTAG: Hi Teto, Ok, I will make a cool video explanation for you soon :). Take care, Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-26T17:58:19Z SecondChildTAG: I wager if you put up a website with a Paypal "donate" button on it and advertised its existence here, you would become modestly wealthy in short order. ;-) SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-11-26T23:14:37Z SecondChildTAG: Hahaha, I haven´t thought on that ;). That explains why I am not good in bussiness haha, I will die poor haha. Anyway, I guess , knowing myself, that I would put a paypay donate button to edX instead of me if I would have a website haha ;). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-27T14:38:24Z SecondChildTAG: that will be a big generous from you. thank you and I am sorry that I am a little bit late on the response I have midterm test in my university. all the best to you :) SecondChildUserIdTAG: 226085 SecondChildUserNameTAG: Teto SecondChildCreateTimeTAG: 2012-12-01T02:38:20Z IndexTAG: 1298 TitleTAG: I cant access my courseware even after 24 Hrs of site update. Ever since the beginning of at 11:00PM, Saturday November 24 Boston time (4AM November 25 th GMT) for performing system maintenance i am unable to access my course-ware and hence cant finish my home work!! I can access all other features of the site except course-ware, the site says "We’re busy building a brand new edX. We’ll be right back! " Please help! is this problem only with me as the maintenance was to last 90 Min!! UserIdTAG: 232499 UserNameTAG: Asim09 CreateTimeTAG: 2012-11-26T03:50:05Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi Asim09, I have no problems... Can you Post a Screen shot? Have you tried to log out and then log in? Try to do that... If after doing that still saying "We’re busy building a brand new edX. We’ll be right back! " , I will report that to the Staff... Please let me know if you are still having issues... Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-26T04:13:33Z IndexTAG: 1299 TitleTAG: Chapter 13 of textbook questions 1. I don't understand, how we get equation 3.19 from equation 3.17... 2. Why we use equation 3.20 to find arg? How we get eq. 3.20? 3. If I have input signal as A1*cos(wt) I got output |Vc|*cos(wt+Ф). But if I have input signal as A1*sin(wt) I also got output |Vc|*cos(wt+Ф)(in other words cos input -> cos output; sin input -> cos output)? UserIdTAG: 104522 UserNameTAG: IgorNovice CreateTimeTAG: 2012-11-25T16:42:44Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: **1)** Every complex number can be represented as a point in the complex plane, and can therefore be expressed by specifying either the point's Cartesian coordinates (called rectangular or Cartesian form) or the point's polar coordinates (called polar form). The complex number z can be represented in rectangular form as z = x + jy or can alternatively be written as z = r*e^(j*θ) Where: e is Euler's number and r= sqrt(x^2+y^2) so, 1 of the numerator is written as 1*e^0 , and the denumerator as sqrt(1+(wRC)^2)*e^jθ where θ is the equation 13.20. When we divide numbers in such forms we get: z= (r1/r2)*e^(θ1-θ2) in general form, and thus: z= (1/(sqrt(x^2+y^2))*e^jθ FirstChildUserIdTAG: 304587 FirstChildUserNameTAG: Vasso FirstChildCreateTimeTAG: 2012-11-25T23:25:36Z FirstChildTAG: **2)** Because θ = arctan(y/x) in general FirstChildUserIdTAG: 304587 FirstChildUserNameTAG: Vasso FirstChildCreateTimeTAG: 2012-11-25T23:29:54Z FirstChildTAG: **3)** if you mean in page 710..then in general, from maths, if z = r*e^jθ , then going backwards from euler's equation that i explained above, we get x = r*cosθ and y = r*sinθ so we can write it in the form z=x+jy FirstChildUserIdTAG: 304587 FirstChildUserNameTAG: Vasso FirstChildCreateTimeTAG: 2012-11-25T23:34:08Z IndexTAG: 1300 TitleTAG: I probably need a refresher on the linearity principle The further in the course we go the more often we hear about the linearity principle. Since this becomes the basic assumption of the impedance model I decided to refresh the idea of linearity and recovered the following on pp.129-130 of the text book: "A network is linear if the response to an input ax1 + bx2 is the same as a times the response to x1 alone plus b times the response to x2 alone. That is, if f (x) is the response to some excitation x, then the system is linear if and only if f(ax1 + bx2) = af(x1) + bf(x2)." I also remember the tutorials about linearity of circuit elements emphasizing the idea that an element can be considered linear if it's v-i relationship curve is coming through the origin. All of the above is fine until I started dealing with circuits containing more of the complicated elements: MOSFETs, diodes, capacitors and inductors. Now I see circuits that contain the whole bunch of these elements and there's the basic assumption of linearity in the foundation of the analysis methods covered. So, my question is how can I be sure the networks are indeed linear? I understand that this is kind of the basic principle and this is why it was covered early in the course, but it was introduced on simple circuits, and now I feel I need a refresher on this principle for more complicated circuits because every time I read in the textbook or hear in the lecture the phrase containing the work 'liner' or 'linearity' when they are talking about transients and capacitors and inductors and the bunch of other seemingly nonlinear staff, the question of 'why is this network is linear' bumps up in my mind. UserIdTAG: 252722 UserNameTAG: vaboro CreateTimeTAG: 2012-11-25T03:55:08Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: If a circuit is linear, it will conform to **Homogeneity** and **Superposition**. http://en.wikipedia.org/wiki/Linearity FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-11-25T09:24:16Z SecondChildTAG: that's right, but can one be sure that a circuit is, indeed, linear? SecondChildUserIdTAG: 252722 SecondChildUserNameTAG: vaboro SecondChildCreateTimeTAG: 2012-11-25T15:32:25Z FirstChildTAG: A circuit is linear if the individual component models in the circuit are linear. Ideal resistors, capacitors, and inductors are linear. Diodes and active devices such as bipolar transistors, MOSFETs, vacuum tubes, etc. are non-linear. If the circuit has non-linear components, the typical engineering analysis consists of linearizing the the non-linear components around the operating point (small signal analysis), i.e. replacing the non-linear component models with approximate linear models. There are no general analysis techniques for non-linear systems, so the usual methods employ either linear approximations or simulation (numerical solutions). FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-25T19:33:39Z SecondChildTAG: @skyhawk: so, if I understand correctly using SCS or SU model of a MOSFET in transients analysis would be too complicated or even impossible without numerical methods? SecondChildUserIdTAG: 252722 SecondChildUserNameTAG: vaboro SecondChildCreateTimeTAG: 2012-11-25T20:38:25Z SecondChildTAG: Yes, current is a quadratic (non-linear) function of voltages. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-11-25T20:47:51Z SecondChildTAG: You can do a small signal analysis if that is appropriate. Also, adding negative feedback may make the system approximately linear. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-11-25T20:53:24Z IndexTAG: 1301 TitleTAG: H10P3, Part 4 Hi all, I'm at the Week 10 and got stuck in this part of the homework: > Now let's look at some real numbers. For a big transmitting amplifier > the output resistance may be R1=1000.0Ω. A typical antenna has a > radiation resistance of R2=50.0Ω. Consider an AM broadcast transmitter > at f=1500.0kHz. In the spaces provided below, write the numerical > values of the capacitance (in picoFarads) and inductance (in > microHenrys) for match. I've already obtained the Cmatch and Lmatch equations and got it right (submitted them and they were correct), but, when I replace the variables with the given values (R1 = 1000, R2 = 50 and w = 1500*10^3), I got the answer wrong. Then, assuming I was calculating it wrong, I tried Wolfram Alpha: > (Eq. of Cmatch)/10^-12 where R1 = 1000 and R2 = 50 and w = 1500*10^3 and > (Eq. of Lmatch)/10^-6 where R1 = 1000 and R2 = 50 and w = 1500*10^3 And the answers are still wrong. Have you faced this same problem, or is it just an(other) error of mine? UserIdTAG: 147694 UserNameTAG: FilhoJoel CreateTimeTAG: 2012-11-25T02:56:43Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: w is in rad/sec FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-25T03:13:00Z SecondChildTAG: Thank you. That was the mistake. SecondChildUserIdTAG: 147694 SecondChildUserNameTAG: FilhoJoel SecondChildCreateTimeTAG: 2012-11-25T03:29:58Z IndexTAG: 1302 TitleTAG: staff Help Hi I can't get green tick with the expression H10P2. Also give some hints for H10P3 and Lab. UserIdTAG: 227508 UserNameTAG: bhavyab CreateTimeTAG: 2012-11-25T01:57:29Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi bhavyab, Can I help you? In wich part of H10P2 are you lost? Be careful with the expressions, w is not the same as W. Also jw is j*w .... Lab10 Hints [Post][1] Myriam [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50b11357f6c1e02700000045 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-25T15:11:06Z IndexTAG: 1303 TitleTAG: More about H12P2 In order to arrive at the "correct" response for both parts a) and b) it must be assumed that the operational bias voltage is sufficiently high in magnitude to deliver the required output voltage (and power) regardless of how low the input voltage, vin, might be. This should have been clearly stated as one might reasonably assume that the op amp bias level is tied to the vin terminal which would result in a very different response (i.e., it would be impossible to get 5 volts out of the system until vin>5.) UserIdTAG: 286880 UserNameTAG: herbsteiner CreateTimeTAG: 2012-11-24T18:47:00Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: This is true. All problems so far have assume an external supply that allows the circuit to reach any desired voltage, but it is better to state it clearly FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-11-27T16:18:14Z IndexTAG: 1304 TitleTAG: H10P1: Confused about break frequencies The break frequency for R-L circuit is approximately 16000Hz(R/2*pi*L) and that for R-C circuit is approximately 32000Hz(1/2*pi*R*C). Now the plots are showing the break frequency of R-L circuit about 10 times higher than that of R-C circuit...What is the reason for this? Please clear my confusion. Am I misinterpreting the plots???? UserIdTAG: 381797 UserNameTAG: Saira180 CreateTimeTAG: 2012-11-24T08:01:12Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I think your calculation for the frequency of the R/L circuit is wrong. I don't know your values for R and L but mine give the result that the frequency is about 160kHz, and your freq for RC is the same as mine. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-11-24T14:22:34Z IndexTAG: 1305 TitleTAG: H11P2: SCOPE PROBE, Totaly lost! Hi, I don't have idea how to calculate Cp? Please any hint! Thanks! UserIdTAG: 299251 UserNameTAG: TheRedBlackOne CreateTimeTAG: 2012-11-23T13:16:20Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: If your first answer is correct for omega=0, then you have the value for the DC attennuation. Than try to calculate vo/vi with the Zprobe from the probe and the Zosc from the oscilloscope, and make it the same as in question one. Look at Figure 13.46 of the book. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-11-23T13:32:45Z SecondChildTAG: Dímelo en Español mejor! SecondChildUserIdTAG: 299251 SecondChildUserNameTAG: TheRedBlackOne SecondChildCreateTimeTAG: 2012-11-23T14:13:16Z SecondChildTAG: **Should I calculate again the polynom s^2+2*a*s+w^2?** SecondChildUserIdTAG: 299251 SecondChildUserNameTAG: TheRedBlackOne SecondChildCreateTimeTAG: 2012-11-23T14:26:31Z SecondChildTAG: Sorry, my Spanish is very bad! Vista Fig 13.46 in el libro y el texto. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-11-23T14:31:23Z SecondChildTAG: No, you don't need the polynom! SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-11-23T14:42:16Z SecondChildTAG: Ok, Thanks! SecondChildUserIdTAG: 299251 SecondChildUserNameTAG: TheRedBlackOne SecondChildCreateTimeTAG: 2012-11-24T21:00:43Z SecondChildTAG: Equation 13.124 in the book makes this section REALLLY easy. SecondChildUserIdTAG: 428560 SecondChildUserNameTAG: MikeDayton SecondChildCreateTimeTAG: 2012-11-25T01:15:46Z FirstChildTAG: Hi TheRedBlackOne, Try to think it as a voltage divider :). Take a look at here [Hints H11P2][1] :) I also invite you to participate in the CECC 2 Contest :). You can be one of the 3 that will win the Textbook signed personally by Prof. Agarwal :). Will you participate? Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50b54fce5fb7411f00000006 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-28T01:09:48Z IndexTAG: 1306 TitleTAG: R/L? Doesn't work for me! Hi, In this case shouldn't bandwith=R/L be applied? R/L 375/(242*10^-6)=1549.58677686 KHz , and as for Q, I've applied 1/sqrt(LC)/R/L (1/sqrt(242*10^-6*98.6621210846*10^-12))/(375/242*10^-6), which gives me 4.17639138578e+12 rings. please tell me what is wrong here? UserIdTAG: 369339 UserNameTAG: vargaslen CreateTimeTAG: 2012-11-23T11:04:48Z VoteTAG: 2 CoursewareTAG: Week 11 / S22E2: The filter is ringing CommentableIdTAG: 6002x_S22E2_The_filter_is_ringing NumberOfReplyTAG: 1 FirstChildTAG: R/L doesn't give the frequency in Hz, but in rad/sec. You must divide by 2*pi to get the correct answer. As for Q, the denominator is wrong. you are calculating (375/242*10^-6), but should be doing (375/(242*10^-6)) FirstChildUserIdTAG: 318577 FirstChildUserNameTAG: takeuchi FirstChildCreateTimeTAG: 2012-11-24T01:02:50Z IndexTAG: 1307 TitleTAG: Lab 10 - bandpass I am stuck with the last exercise. I don't know how to make a bandpass circuit by "cascading" those subcircuits. I am always getting very attenuated (-60 dB) results like these: ![enter image description here][1] Any hints? [1]: https://edxuploads.s3.amazonaws.com/13536200231343602.png UserIdTAG: 346056 UserNameTAG: fiatlux CreateTimeTAG: 2012-11-22T21:34:06Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: You must deduce the equation for |Vo/Vi| and then you can use the fact that |Vo/Vi| = (sqrt2)/2 at the two break frequencies. After that, i chose the resistance that suited the circuit best and solved the equation that i found earlier. I am not sure if that's the right way to find R though. Good luck :) FirstChildUserIdTAG: 304587 FirstChildUserNameTAG: Vasso FirstChildCreateTimeTAG: 2012-11-22T22:59:32Z SecondChildTAG: Is it an horrendous equation like [this][1]? [1]: http://www.wolframalpha.com/input/?i=%7C%28%28%28b%2bi*e*d%29*%281/%28i*e*c%29%29%29/%28b%2bi*e*d%2b1/%28i*e*c%29%29%29/%28%28%28b%2bi*e*d%29*%28%281/%28i*e*c%29%29%29/%28b%2bi*e*d%2b1/%28i*e*c%29%29%29%2bb%29%7C SecondChildUserIdTAG: 346056 SecondChildUserNameTAG: fiatlux SecondChildCreateTimeTAG: 2012-11-23T08:51:39Z SecondChildTAG: Not really..it looks quite clean. SecondChildUserIdTAG: 304587 SecondChildUserNameTAG: Vasso SecondChildCreateTimeTAG: 2012-11-23T10:23:43Z FirstChildTAG: If I look at your plot, then I notice that your attenuation of -60dB in the midband must be caused somehow, by heavvy loading of the first network by the second network, or, that you use a resistive attenuator somewhere, or load the output with a resistor. What to do is: calculate the break frequency for the first circuit, but use a low value for the resistor , so that you get a relatively low output impedance(seen from the input of the second circuit). Since the second circuit is loading the first circuit, you have to take into account that load impedance of the second circuit. But you can almost neglect this, by making the input impedance from the last circuit (seen by the output of the first), very high. So pick a resistor with a high value for the second circuit and calculate the break frequency. Hope it helps. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-11-23T14:10:52Z SecondChildTAG: I solved my problem, thanks. I will post my guess (which worked) after the homework deadline, so I will avoid ambiguities SecondChildUserIdTAG: 346056 SecondChildUserNameTAG: fiatlux SecondChildCreateTimeTAG: 2012-11-23T15:07:06Z IndexTAG: 1308 TitleTAG: CHECK BUTTON:-( whenever i hit check button in homework, it responds too late and sometime the whole web page become unresponsive and i have too reload it, then i encounters "AW snap"..i am experiencing it for the last two weeks..i am using google chrome..is anyone encountering the same?? UserIdTAG: 114864 UserNameTAG: Vikaash CreateTimeTAG: 2012-11-22T16:33:54Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Happens to me as well but only on Windows (Windows 7 Professional 64 bit). I don't know why. I usually see webpage crashes on my Fedora setup on other websites. It looks like it's got something to do with Mathjax because the crash occurs when I'm entering something in the answer box. FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-11-22T16:48:08Z SecondChildTAG: i am using windows 7 ultimate..still facing the same problem..hope it will not occur on final exam where "check" is restricted to 3 times only. SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-11-22T17:06:17Z SecondChildTAG: Try Firefox. See if the issue doesn't occur there. SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-11-22T17:17:10Z SecondChildTAG: I have the same problem and although I hate to say it, everything works fine in Internet Explorer. SecondChildUserIdTAG: 411504 SecondChildUserNameTAG: taubrafi SecondChildCreateTimeTAG: 2012-11-23T05:47:41Z IndexTAG: 1309 TitleTAG: unfair to give certificate without grade..... If the grade will not be mentioned in the certificate, then it would be very discouraging for the ones scoring A grade. I feel very disappointed with this.If someone is not taking the course seriously, just got 60%, and another person is working hard scoring more than 90% , then the certificate should reflect this. If my score is 59%, then why should I prepare for finals. In that situation I would answer one or two questions and quit. Its too unfair. UserIdTAG: 381797 UserNameTAG: Saira180 CreateTimeTAG: 2012-11-22T05:01:00Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 6 FirstChildTAG: Most people take this course for the knowledge that they can gain from this..Being one of them, I dont mind if grades are there or not there..but I feel anyone who manages a 60% would not be taking the course lightly..it needs a lot of hard work to get that itself..with atleast 10 HWs and labs and the midsem exam..so I dont feel it is unfair.. FirstChildUserIdTAG: 173147 FirstChildUserNameTAG: cruiser_rahit FirstChildCreateTimeTAG: 2012-11-22T05:10:02Z SecondChildTAG: It's not unfair. I don't believe most of "us" are doing this for a grade. Desire for learning, and a desire for developing, and using your mind; these are worth more than any grade. Cheers to all my classmates. Thanks to 6.002X crew! You are providing a great opportunity to all. SecondChildUserIdTAG: 61923 SecondChildUserNameTAG: MikeJones SecondChildCreateTimeTAG: 2012-11-22T09:54:50Z SecondChildTAG: I completely agree with [MikeJones][1]. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/users/61923 SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-11-22T19:54:07Z SecondChildTAG: I agree with MikeJones too! I'm happy to see my progress bars growing. Print the progress bars: a picture says more than 1000 words ... SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-11-23T18:06:45Z FirstChildTAG: Is this a new thing this year ... no grade on the certificates? I have found previous 6.002x certificates that mention A, B, or C on certificateswall.com ... FirstChildUserIdTAG: 714237 FirstChildUserNameTAG: PaxPolaris FirstChildCreateTimeTAG: 2012-11-22T16:43:13Z FirstChildTAG: I think the biggest gain I got this course was the knowledge. Could achieve the average score of 59%, ie, only 1% failure to achieve the level C. But that means nothing if I did not feel satisfied with the learning that I got. FirstChildUserIdTAG: 361823 FirstChildUserNameTAG: EliasOak FirstChildCreateTimeTAG: 2012-11-22T11:29:43Z FirstChildTAG: I agree with everyone who says they are here for the knowledge. You should not work towards the grade, but towards the knowledge you will acquire. If you realize that the real reward is the knowledge, then you'll remain motivated despite the "grade game"! I already got a score high enough that will make me pass, and despite that I still feel very motivated, because of the high quality of the course and the knowledge they are providing. FirstChildUserIdTAG: 318577 FirstChildUserNameTAG: takeuchi FirstChildCreateTimeTAG: 2012-11-23T11:10:13Z SecondChildTAG: So am I. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-11-23T18:07:54Z FirstChildTAG: I am so grateful to have this opportunity that even if I had a chance at acing this (I don't because my Midterm score isn't high enough), I wouldn't care one wit whether my certificate had a grade on it, or not. Just think - a FREE course from one of the most prestigious Universities in the *world*... How many have that chance? A cert is just a piece of paper - but the knowledge I am gaining is *priceless*. FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-11-22T09:30:25Z FirstChildTAG: We are not at school and it is not a competition! We are just taking the course to learn and doing the exam to confirm ourselves we have really understood the matter. FirstChildUserIdTAG: 372321 FirstChildUserNameTAG: EnricoDona FirstChildCreateTimeTAG: 2012-11-22T07:32:25Z IndexTAG: 1310 TitleTAG: congratulations. I thank the organizers of this course. This course is excellent and helped me in updating my knowledge. I hope I can do other courses like this. Thank you all. UserIdTAG: 361823 UserNameTAG: EliasOak CreateTimeTAG: 2012-11-21T17:52:10Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi ! Can you help me with H12P2 part c. Hope you would.! FirstChildUserIdTAG: 277808 FirstChildUserNameTAG: Hemanthmps FirstChildCreateTimeTAG: 2012-11-21T17:58:01Z SecondChildTAG: Make the item d to get the correct answer in item c. SecondChildUserIdTAG: 361823 SecondChildUserNameTAG: EliasOak SecondChildCreateTimeTAG: 2012-11-21T18:51:49Z SecondChildTAG: I'm stuck at part c only..! Don't know how to move further! SecondChildUserIdTAG: 277808 SecondChildUserNameTAG: Hemanthmps SecondChildCreateTimeTAG: 2012-11-22T06:17:59Z IndexTAG: 1311 TitleTAG: H11P1 Any hint for question 1? "bandwidth Δω of the impedance Z looking into this tank circuit" UserIdTAG: 39980 UserNameTAG: MarinoJr CreateTimeTAG: 2012-11-21T11:50:37Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: Put the denominator of Z in canonical form and read off the answer. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-21T12:16:59Z SecondChildTAG: Can you be more specific???? @skyhawk SecondChildUserIdTAG: 277808 SecondChildUserNameTAG: Hemanthmps SecondChildCreateTimeTAG: 2012-11-21T12:45:19Z SecondChildTAG: I did the same procedure that lecture S21V15 does. I found the Z as (1/RC + j*(w*C-1/(w*L+RL)))^(-1). Then it should be equalized with (1 +/- j1)^(-1). Should RC be considered as 1? SecondChildUserIdTAG: 39980 SecondChildUserNameTAG: MarinoJr SecondChildCreateTimeTAG: 2012-11-21T12:57:04Z FirstChildTAG: I found that Z=$\frac{\frac{s \cdot L \cdot R_C+R_L \cdot R_C}{R_C \cdot C \cdot L}}{s^2+ \frac{L+R_C \cdot C \cdot R_L}{R_C \cdot C \cdot L} \cdot s+\frac{R_C+R_L}{R_C \cdot C \cdot L}}$ but result A = $\frac{L+R_C \cdot C \cdot R_L}{R_C \cdot C \cdot L}$ does not match also. I think I'm getting dizzy. FirstChildUserIdTAG: 39980 FirstChildUserNameTAG: MarinoJr FirstChildCreateTimeTAG: 2012-11-21T14:05:17Z SecondChildTAG: Don't put over a common denominator. Leave as two fractions. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-11-21T14:10:38Z SecondChildTAG: The A=... you have written above is the correct answer. Are you sure you do not get a green check mark? SecondChildUserIdTAG: 97581 SecondChildUserNameTAG: Asimakis SecondChildCreateTimeTAG: 2012-11-21T16:06:58Z SecondChildTAG: I told you I was getting dizzy. Typo. GREEN CHECK ACHIEVED! Thanks. SecondChildUserIdTAG: 39980 SecondChildUserNameTAG: MarinoJr SecondChildCreateTimeTAG: 2012-11-21T16:21:22Z SecondChildTAG: Congratulations! SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-11-21T16:39:38Z SecondChildTAG: Yes, as skyhawk says, leave "A" as two fractions. The two fractions should look rather familiar... In the lectures, A would have been $2\alpha$ and B would have been $\omega _0$, i.e. the denominator would have been of the form: $s^2+2\alpha s+\omega_0$ The bandwidth, $\Delta\omega$ is simply $2\alpha$. SecondChildUserIdTAG: 141000 SecondChildUserNameTAG: OrinE SecondChildCreateTimeTAG: 2012-11-21T17:59:24Z SecondChildTAG: A question,please: As long as I get the canonical form (characteristic equation for the denominator), doesn't matter what I have in the numerator? the rule always apply? thanks! SecondChildUserIdTAG: 369339 SecondChildUserNameTAG: vargaslen SecondChildCreateTimeTAG: 2012-11-23T23:10:23Z SecondChildTAG: I get the same ecuation for A. But if I make 2α= (RL*RC*C + L) / RC*L*C . My result is 54.03 KHz. What is my mistake? SecondChildUserIdTAG: 433574 SecondChildUserNameTAG: endika86 SecondChildCreateTimeTAG: 2012-11-30T13:41:51Z SecondChildTAG: rad/sec SecondChildUserIdTAG: 623210 SecondChildUserNameTAG: preveen SecondChildCreateTimeTAG: 2012-12-01T10:14:50Z FirstChildTAG: Form the complex impedance for the circuit. Simplify the denominator so that it has the form: s^2 + A*S + B where s = jw. The bandwidth is given by A. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-21T13:10:16Z SecondChildTAG: Z = s ⋅ L + R C + R L ⋅ R C R C ⋅ C ⋅ L s 2 + ( L + R C ⋅ C ⋅ R L R C ⋅ C ⋅ L ) ⋅ s + R C + R L R C ⋅ C ⋅ L SecondChildUserIdTAG: 39980 SecondChildUserNameTAG: MarinoJr SecondChildCreateTimeTAG: 2012-11-21T13:48:35Z SecondChildTAG: Well, this does not work. SecondChildUserIdTAG: 39980 SecondChildUserNameTAG: MarinoJr SecondChildCreateTimeTAG: 2012-11-21T13:49:49Z SecondChildTAG: In my opinion, I don't think the videos and book prepared me for this problem, but in the end it turned out very easy ... about 10 minutes of work ... half a page of notes. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-11-21T13:57:18Z SecondChildTAG: If the correct procedure is that you told, certainly I was not prepared to solve that question. SecondChildUserIdTAG: 39980 SecondChildUserNameTAG: MarinoJr SecondChildCreateTimeTAG: 2012-11-21T13:58:23Z FirstChildTAG: Hi guys! Do we simply substitute the values of C, RL, RC into the first answer the obtain the bandwidth values? I have tried and I did not obtain the answer. Thanks! FirstChildUserIdTAG: 143823 FirstChildUserNameTAG: NathanNadeson FirstChildCreateTimeTAG: 2012-11-26T13:18:17Z FirstChildTAG: Take a look at this Hints [here][1] :) P.D: Will you participate in the CECC 2. Isn´t it? ;) [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50b54fce5fb7411f00000006 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-28T01:11:52Z IndexTAG: 1312 TitleTAG: More accurate shape of diagrams. For better view I drawn Vc=Sin(w0) and ic=Cos(w0) and also vc^2 and ic^2 and summation of those tow vs w0 which I think could help to have better view of the equations. have fun! ![enter image description here][1] ![enter image description here][2] [1]: https://edxuploads.s3.amazonaws.com/13534435731343681.png [2]: https://edxuploads.s3.amazonaws.com/13534436077637891.png UserIdTAG: 420339 UserNameTAG: AliJenabi CreateTimeTAG: 2012-11-20T20:37:26Z VoteTAG: 2 CoursewareTAG: Week 9 / Undriven LC Network Response CommentableIdTAG: 6002x_undriven_LC_network_response NumberOfReplyTAG: 0 IndexTAG: 1313 TitleTAG: helpout a 6.002x bro! =) Hey guys! Amazing program this is, it has been awesome! well I dont want to spam but in one of my classes of college (Im from chile) I had to make a video to explain the zero law of thermodinamycs to high school students, it is in youtube, and here is the thing: the one who get more comments on youtube will get a whole point in his grade (like from B to A), so I was wondering if you guys can helpme out with some comments (in any language), I will be very grateful! =P thanks! here is the link:https://www.youtube.com/watch?v=dNrWE9pjA4Y&feature=BFa&list=ULdNrWE9pjA4Y UserIdTAG: 199008 UserNameTAG: matuko CreateTimeTAG: 2012-11-20T16:41:39Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1314 TitleTAG: What do you all think of this? http://www.youtube.com/watch?v=YhuSn6sc7sc UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-11-19T15:01:32Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: I think the rabbit hole of electrical engineering is almost limitless. Here's another example: http://www.youtube.com/watch?v=HqZtptHnC2I FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-11-19T15:36:49Z SecondChildTAG: I think the work of Nikola Tesla is as interesting as it is important. SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-19T16:07:17Z SecondChildTAG: oh yeah it is. Tesla is one of the things that drew me to electronics. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-11-19T17:00:11Z SecondChildTAG: Which aspects of his work? Tesla has been a long-term project for me, the central theme for me seems to be the Tesla Coil resonant transformer. SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-19T17:08:23Z SecondChildTAG: Well if I understand correctly, he basically invented resonant inductive coupling, which is the basis for all kinds of technologies. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-11-19T21:04:14Z FirstChildTAG: we think it's a bull*** - when ppl say about something greater then c these ppl just don't understand basic physics concepts of our Universe - causality and time dimension anisotropy - it's not a laws - it what our Universe IS!!! - the limit of speed of light just a consequence of these two fundamental phenomenon. FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-11-19T17:34:49Z SecondChildTAG: http://www.scientificamerican.com/article.cfm?id=particles-found-to-travel SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-19T18:42:46Z SecondChildTAG: CERN has retracted their claim. http://www.asiantribune.com/news/2012/06/10/einstein-triumphs-cern-retracts-%E2%80%93-nothing-travels-faster-light-after-all SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-11-19T18:52:51Z SecondChildTAG: But the point is that they are thinking about it, no one should fear a paradigm shift. BTW Are any of you guys on other forums? SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-19T19:37:52Z SecondChildTAG: I found this really interesting! http://www.youtube.com/watch?v=9ckpQW9sdUg Skip to the second half ;) SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-19T19:42:37Z SecondChildTAG: the point is that speed of light is not a sacred cow - there is much much more behind it - as I said it's just a one small appearance of huge fundamental thing - it's why most of scientists were more then skeptical about this news - there should be much more evidences including violation of theory of relativity principles' SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-11-19T23:34:59Z FirstChildTAG: I don't fear a paradigm shift, but the shift has to be compatible with what we know that works. Maxwell's equations describe electromagnetic phenomena including electrostatics and magneto-statics. Likewise special relativity places constraints on the form any new theory can take. See for example http://en.wikipedia.org/wiki/Lorentz_invariance FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-19T19:51:57Z SecondChildTAG: But we know that's not the whole story, electricity, I'm sure, is not done and dusted. SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-19T19:56:47Z IndexTAG: 1315 TitleTAG: If there was one thing I learned from week 9... If there was one thing I learned from week 9...which thankfully there was, it would be that I would intuitively know how to deal with bad waveforms when driving MOS-FETs, or in circuits in general. The ringing means you have parasitic inductance/capacitance and parasitic inductance can be remedied by shorter wires. Fantastic! Another thing that gets me, largely because I would play around with resonant circuits when I was a little younger, is that that ripple you see is the natural resonant frequency of the circuit! So, I'm going to ask, what nuggets of wisdom did you mine from Week 9?? UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-11-19T09:06:53Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: There is so much in this course that I've yet to digest... I'm just hanging on for the ride, trying to get assignments done... That is one reason (of many) that I believe I will be taking this course again in the Spring. Hopefully then I will know enough that the time pressures will be eased, and I will actually have the luxury to **think** about things... FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-11-19T18:14:48Z SecondChildTAG: Oh yeah, I feel the same way. SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-19T19:43:37Z SecondChildTAG: I feel the same way too! SecondChildUserIdTAG: 122513 SecondChildUserNameTAG: Mona77 SecondChildCreateTimeTAG: 2012-11-20T18:59:53Z IndexTAG: 1316 TitleTAG: LAB10 - BPF W0=1/sqrt(LC) B.W=(R/L) where w0=sqrt(w1*w2) B.W=w2-w1 Even though i am using these formula to calculate L&C values by assuming R value.... I am not getting correct answer.... Can anyone tell me what i am doing wrong...? UserIdTAG: 477198 UserNameTAG: Rajesh1993 CreateTimeTAG: 2012-11-19T07:44:37Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Finally got the answer..... (w) must be in radians/second FirstChildUserIdTAG: 477198 FirstChildUserNameTAG: Rajesh1993 FirstChildCreateTimeTAG: 2012-11-19T08:07:00Z SecondChildTAG: Can you give me some help? Didn't understand how you solve it... SecondChildUserIdTAG: 98262 SecondChildUserNameTAG: rafaelmarques SecondChildCreateTimeTAG: 2012-11-22T01:58:34Z IndexTAG: 1317 TitleTAG: Not getting booster circuit Please refer to Pg 672 of text book 5th paragraph. The lines goes as: "Let us now consider the case whereS1 is opened and S2 is kept closed only for a short amount of time. This case starts out just like the previous case. In other words,at the instant thatS1 is opened and S2 is closed, the inductor has a current equal to V/R flowing through it. Since the current flowing through the inductor cannot change instantaneously, the same current will flow into the capacitor through the closed switch S2. As the capacitor charges, its voltage rises past V(again, assuming a small enough capacitor, or a large enough initial voltage on the capacitor). Now, an interesting scenario arises if we closeS1and openS2after the capacitor voltage has risen past V, but before the current reverses direction. Since S2is now open, the capacitor has no path to discharge, and so it holds the final value of its voltage. **Then when S1 is opened and S2 is closed again, the inductor current charges up the capacitor further, thereby further increasing its voltage. If this process is repeated, notice that the capacitor voltage will keep rising indefinitely over time.**" I am not getting why the capacitor will keep on charging indefinitely?, does that mean that it can have infinite charge (if considered an ideal capacitor with no dielectric breakdown). UserIdTAG: 213503 UserNameTAG: jyotech CreateTimeTAG: 2012-11-19T07:24:26Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I know that this is throwing me to a point that i could not get it right . I tried and did what it said, i dont know if i was just adjusting to high or too little, i could not get a check mark on this no matter how hard i tried. i just do not understand boost circuits FirstChildUserIdTAG: 160242 FirstChildUserNameTAG: Mlevins35 FirstChildCreateTimeTAG: 2012-11-19T08:28:13Z IndexTAG: 1318 TitleTAG: To "Staff"-----------Still 'No Reply'? ![enter image description here][1] it was said that at the moment,capacitor decays ;we consider THEVENIN EQ. CKT which looks like C connected with parallel combination of R1 and R2.So,it must get discharged through both of them simultaneously,but at the point 'A' after a moment potential will be greater than that of potential at C.Then how could the C get discharged thro that path for whole of the discharge duration? Last time,i got the response as if i were to discharge the capacitor fully then DC source must be kept OFF ,thereafter i posed another question that 'Is there any procedure to put the source off in digital circuits at the moment Cgs discharges?' [1]: https://edxuploads.s3.amazonaws.com/13533085211343627.jpg UserIdTAG: 477713 UserNameTAG: ikm104 CreateTimeTAG: 2012-11-19T07:03:07Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: probably there is a way to put the source off in digital circuits at the moment Cgs discharges. Use a MOSFET switch in parallel with the battery and apply at its gate a square wave with duty cycle suited to your need.MOSFET will keep the battery connected till there is sufficient threshold voltage at its gate, and for rest it will short the battery. Now you just need to adjust the duty cycle. FirstChildUserIdTAG: 213503 FirstChildUserNameTAG: jyotech FirstChildCreateTimeTAG: 2012-11-19T07:34:48Z FirstChildTAG: Your expectations are off - the staff isn't there to answer every single question a student may have. FirstChildUserIdTAG: 90796 FirstChildUserNameTAG: ChaunceyGardiner FirstChildCreateTimeTAG: 2012-11-19T08:40:18Z IndexTAG: 1319 TitleTAG: Lab 9 well 100% Indeed the Lab 9 UserIdTAG: 471282 UserNameTAG: juancarlosxd CreateTimeTAG: 2012-11-19T02:19:53Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Well done! FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-11-19T13:41:27Z IndexTAG: 1320 TitleTAG: H11P3 part g I determined the value for Q correctly. I determined the correct value for R and (L or C) from the graph and have beloved green check marks for all that. I understand the relationship between Q and omega not and 2 alpha and how omega not and 2 alpha are related to R C and L. I cannot seem to calculate an acceptable value for the component that was removed. Anyone have any luck with this and be willing to share a hint? UserIdTAG: 18073 UserNameTAG: gburkhart CreateTimeTAG: 2012-11-18T23:02:14Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Write the relation for Q in terms of the component vales R, L, and C and equate that to the previously determined integer value for Q. Solve the algebraic expression for L or C and plug in the known values for R and C or L. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-18T23:34:24Z SecondChildTAG: i did all that - for some reason I got the inverse of the correct answer. SecondChildUserIdTAG: 18073 SecondChildUserNameTAG: gburkhart SecondChildCreateTimeTAG: 2012-11-19T04:39:26Z SecondChildTAG: **yes, me too! any opinion is welcome!** SecondChildUserIdTAG: 299251 SecondChildUserNameTAG: TheRedBlackOne SecondChildCreateTimeTAG: 2012-11-24T22:24:19Z SecondChildTAG: @gburkhart so am I did it twice and still have wrong answer SecondChildUserIdTAG: 447663 SecondChildUserNameTAG: brigadakuznetsov SecondChildCreateTimeTAG: 2012-11-27T16:23:05Z SecondChildTAG: But we don't know value of C nor L. How are we supposed to find one of them without knowing the other? SecondChildUserIdTAG: 380703 SecondChildUserNameTAG: vargy SecondChildCreateTimeTAG: 2012-12-02T09:59:40Z SecondChildTAG: Vargy, here are some hints that I put on the myrimit post about this question... Note that to resolve the last part you need to know which element was removed and which one stayed in the circuit * If the inductor was removed you get an RC circuit. If the capacitor was removed you get an LC circuit. * Write the transfer function for the LR circuit (S, R and L) and other for the CR circuit (S, R and C). * Look at both the graph and the two formulas... what happens when $\omega$ tends to infinity? (you will be able to match the graph with one of the equations. * You just found which element should have been removed and which one still is in the circuit. * Looking at the graph you should be able find R. Using the graph and the magnitude formula you should be able to find C or L. For the last part, please take a look at video S22V4. There we worked on the transfer function for an LCR circuit like in this problem. There you will understand the formula $Q=\frac{\omega o}{2\alpha}$ in terms of L, C and R. Based on the previuous answers you already have the value to Q, R, and (C or L). Using the formula above you can calculate the remaining unknown. SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-12-02T15:42:55Z FirstChildTAG: Regarding "the inverse of the correct answer": I believe you mean the reciprocal, i.e., $\frac{1}{x}$. I believe that is because we are not looking for $\mid\frac{V_P}{V_I}\mid$, we are looking for **$\mid\frac{V_I}{V_P}\mid$**, because of where we are taking the measurement: to wit, "looking from the terminals of the voltage source". If I am wrong, I hope someone corrects my misapprehension. FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-12-01T07:47:12Z IndexTAG: 1321 TitleTAG: Lab 0 I am very surprised. This a great course. UserIdTAG: 804492 UserNameTAG: lborges CreateTimeTAG: 2012-11-18T13:24:54Z VoteTAG: 2 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: welcome to edX FirstChildUserIdTAG: 623210 FirstChildUserNameTAG: preveen FirstChildCreateTimeTAG: 2012-11-18T13:53:43Z IndexTAG: 1322 TitleTAG: Staff: The grader is not working for HW9 I have entered the answers for parts d and f that: * I know are correct and * have been said to be correct by many people in the discussion forum I get the red X. The problem seems to be that the grader is not grading the values I've entered. If I refresh the page, a different formula appears in the answer box. That formula is one that I tried previously because the correct answer didn't work. It is really frustrating that nothing has been posted to warn us that there is a problem. I have already restarted my computer to see if the problem was on my end. It did not help. I don't have time to waste. The CS188.1x final and 3.091x midterm are due at the same time as HW and Lab 9. UserIdTAG: 126825 UserNameTAG: aphoenix CreateTimeTAG: 2012-11-18T08:20:01Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: I also tried switching to Chrome from Firefox. Same problem. FirstChildUserIdTAG: 126825 FirstChildUserNameTAG: aphoenix FirstChildCreateTimeTAG: 2012-11-18T08:23:18Z FirstChildTAG: You are correct about the fact that there is something on edx.org side about those check requests. I just encountered same problem with LAB 9. Upon sending check request to server I was getting 500 response - internal server error. I was able to fix that though by clearing browser's cache. It can be done by hitting Ctrl+Shift+Del. You'll see new window opened different for different browsers. In chrome u will have to choose clear cache and pick for all the time in the dropbox, then press clear button to finish the process (not sure about all options I named because at home I have russian version of chrome). Hope that will help. FirstChildUserIdTAG: 413065 FirstChildUserNameTAG: anikey FirstChildCreateTimeTAG: 2012-11-18T10:38:20Z FirstChildTAG: I'm having the same problem with HW9 :( FirstChildUserIdTAG: 298547 FirstChildUserNameTAG: AnthonyRF FirstChildCreateTimeTAG: 2012-11-18T21:21:16Z FirstChildTAG: Try a different browser if you can; if not, try clearing your browser cache and restarting your browser. I've had similar problems that were solved this way, though sometimes it took more than one clear/restart. FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-11-18T23:53:58Z FirstChildTAG: There were some mysterious transient errors on the platform that got cleaned up. Please inform us if these errors return again. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-11-19T15:19:36Z IndexTAG: 1323 TitleTAG: H9Q3 last part Can someone help with the value of b in critical damping. It is probably a dumb question because I didn't find it anywhere in the discussion :( UserIdTAG: 499671 UserNameTAG: maliha266 CreateTimeTAG: 2012-11-17T16:37:44Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi, the coefficient b (viscous-friction) represents our R (resistance). Remember that the value of the quality factor Q=$\omega$o/2*$\alpha$ tells us whether the system is under damped, over dumped or critically dumped. Look videos S18V18, S18V21 (7:40) So you have $\alpha$, $\omega$o and the desired value to Q (to be critically dumped). Just find b. FirstChildUserIdTAG: 288174 FirstChildUserNameTAG: matiasgrodriguez FirstChildCreateTimeTAG: 2012-11-17T19:29:53Z SecondChildTAG: Gosh I can be so stupid sometimes... I was taking b as R*C the whole time... Thanks a lot :) SecondChildUserIdTAG: 499671 SecondChildUserNameTAG: maliha266 SecondChildCreateTimeTAG: 2012-11-17T19:42:21Z SecondChildTAG: you're welcome! SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-17T20:27:48Z IndexTAG: 1324 TitleTAG: LAB9 Please give an idea of boost converter problem....... UserIdTAG: 156835 UserNameTAG: kphariprakash1968 CreateTimeTAG: 2012-11-17T14:20:57Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi kphariprakash1968, Take a look at here, [Hints Lab9][1] :) [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50a78f0dce1ad22700000024 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-17T14:37:37Z IndexTAG: 1325 TitleTAG: H9P2 d again So, As I (and many others) derived : $w_0= ...$(....edited...). If the answer is required on Hz, should it not be $f=\frac{w_0}{2\pi}$ ? Why is it not accepted? I need serious help here, my frustration is on extrem levels. Thanks UserIdTAG: 329254 UserNameTAG: KGabor CreateTimeTAG: 2012-11-17T13:38:00Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hi KGabor, Yes you are right. Your 2nd approach is correct ... so as $2*\pi*f = \omega_0$ $f=\frac{w_0}{2\pi}$ [Hz] The question should be formulated like this: ---------- d) The circuit would ring when excited with a step response vi=u(t). What is the natural frequency **in Hertz** of this circuit, in terms of the component values? ---------- Remember that - $ \omega_0$ is in [rad/s] - **F is in [Hz]** - And [Hz] is the same as [$\frac{1}{s}$] - 2*$\pi$ is in [rad] So the dimentional analysis, 2*$\pi$*f = $\omega_0$ [rad]*[Hz]=[rad/s] [rad]*[1/s]=[rad/s] [rad/s]=[rad/s] FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-17T14:03:07Z SecondChildTAG: what is the value of 1/(2*PI)??..write that value.. SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-11-17T14:09:13Z SecondChildTAG: OK In homework is not correct question. In book natural frequence is w0. SecondChildUserIdTAG: 402617 SecondChildUserNameTAG: Vl SecondChildCreateTimeTAG: 2012-11-17T14:11:05Z SecondChildTAG: I almost started to rant again, as it didnt accept 0.159 * w0. Then I entered 0.1591549 * w0, and it was accepted. The question is poorly formulated, and the accapteance of the answer is also very bad. However it is my first negative experience on this course, so, well... Thanks Vikaash SecondChildUserIdTAG: 329254 SecondChildUserNameTAG: KGabor SecondChildCreateTimeTAG: 2012-11-17T14:15:07Z SecondChildTAG: @Myrimit - thanks for your answer as well, I think editing out stuff was really not necessary, as this answer can be found in like 2-3 other forum posts as well. The real challenge in this question is, not to figure out the formula of w0, but to enter it correctly. SecondChildUserIdTAG: 329254 SecondChildUserNameTAG: KGabor SecondChildCreateTimeTAG: 2012-11-17T14:16:58Z SecondChildTAG: You are welcome :). yep, pi = $\pi$ in the text box. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-17T14:26:14Z SecondChildTAG: The answer is in the text but the grader will not accept it. :( SecondChildUserIdTAG: 126825 SecondChildUserNameTAG: aphoenix SecondChildCreateTimeTAG: 2012-11-18T00:54:26Z FirstChildTAG: Thanks for the comments above. I would go one step further and say the problem should be stated: d) The circuit would ring when excited with a step response vi=u(t). What is the **undamped** natural frequency in Hertz of this circuit, in terms of the component values? Several places in the text refer to the roots of the characteristic equation as the natural frequency (p. 507, 630, 642). On p. 646 the distinction is made between the undamped natural frequency w0 and the damped natural frequency wd. So that gives 4 different possible values to the term 'natural frequency': 1) one root of characteristic equation, s1 2) second root of characteristic equation, s2 3) undamped natural frequency, w0 and 4) damped natural frequency, wd. Not to mention whether the answer should be Hz or radians. FirstChildUserIdTAG: 339870 FirstChildUserNameTAG: rjlasota FirstChildCreateTimeTAG: 2012-11-17T19:59:09Z IndexTAG: 1326 TitleTAG: lab9 boost convertor can anybody help me regarding this problem?? how to choose the value of L and C? UserIdTAG: 114864 UserNameTAG: Vikaash CreateTimeTAG: 2012-11-17T09:09:42Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi Vikaash, Take a look at here [Hints of Lab9][1]. You have to use the plot analysis with the test and error. Try to vary each component of the Circuit per time in order tat they verify the requirements :). I hope that the Hints of Lab9 can help you. See you, Myriam. [1]: http://%20https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50a78f0dce1ad22700000024 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-17T13:46:20Z IndexTAG: 1327 TitleTAG: H9P1 Question 7 I have used the following formula to calculate the voltage on the capacitor after the impulse (N.B. My impulse was given at time 5 seconds): $v_C(5_+) = \frac{2}{\pi\cdot C}+v_C(5_-)$ Why am I not getting the green tick? Any help would be much appreciated. Thank You in advance UserIdTAG: 85490 UserNameTAG: yashsavani CreateTimeTAG: 2012-11-16T22:25:30Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Assuming that you got the current at 5- correct, I would check the sign of vC(5-). FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-16T22:37:36Z SecondChildTAG: It should be 2/pi, not 2/(pi*C) SecondChildUserIdTAG: 143823 SecondChildUserNameTAG: NathanNadeson SecondChildCreateTimeTAG: 2012-11-17T00:24:38Z SecondChildTAG: I have got the sign of vC(5-) right, at least according to the green tick. I think it is supposed to be negative. SecondChildUserIdTAG: 85490 SecondChildUserNameTAG: yashsavani SecondChildCreateTimeTAG: 2012-11-17T06:01:06Z SecondChildTAG: I thought that it was q/C which gave the voltage. 2/PI only gives the number of coulombs. I had to multiply C in the denominator to give the voltage. SecondChildUserIdTAG: 85490 SecondChildUserNameTAG: yashsavani SecondChildCreateTimeTAG: 2012-11-17T06:02:31Z SecondChildTAG: If you are using the integer period for the oscillation, exactly 2/pi for the strength of the impulse, and have the correct sign for the charge on the capacitor before the impulse then you should get the correct answer. It's a "nice" number! SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-11-17T12:41:59Z SecondChildTAG: add the charges before and after impulse with correct sign SecondChildUserIdTAG: 473933 SecondChildUserNameTAG: rahul_pradhan SecondChildCreateTimeTAG: 2012-11-17T13:20:27Z SecondChildTAG: I cant get green tick on question 7, ive done correctly question 5 that ask about Vc(9-) and solve using energy conservation but i dont understand how must do question 7. Could give us some hints?? Million thanks sorry for my bad english SecondChildUserIdTAG: 149058 SecondChildUserNameTAG: sotoroman SecondChildCreateTimeTAG: 2012-11-17T14:04:19Z SecondChildTAG: sotoroman. OK, you have the voltage on the capacitor just before the impulse. At the impulse all that energy Q (area) will go to the capacitor in the form of voltage. You can calculate this "delta" voltage as Q/C. Take a look to this post and see if it helps https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/undefined/threads/50a7d5a6a50e3e260000002a SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-18T17:55:20Z FirstChildTAG: I had to round the two voltages to whole number, and then add them ... to get the check mark FirstChildUserIdTAG: 714237 FirstChildUserNameTAG: PaxPolaris FirstChildCreateTimeTAG: 2012-11-18T17:42:25Z IndexTAG: 1328 TitleTAG: H9P1: RESPONSE TO A DELAYED IMPULSE , hints Hi, People ples think! Instantaniusly, básicaly at time 0, you dont have to make any calculus in the half problems, only think carefuly. Some times the curent cant change instataneusly or the voltage, so at moment 0 probable some things can be nothing! And remember the energy total in the circuit is C(V^2)/2!!! And check you sings I cant say more just think please! It's easy! Bye! UserIdTAG: 299251 UserNameTAG: TheRedBlackOne CreateTimeTAG: 2012-11-16T20:38:38Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1329 TitleTAG: Sound quality seems low Speech sounds somewhat strange in this video -- as if it is both slowed down a bit and there's some distortion. UserIdTAG: 189680 UserNameTAG: vnd CreateTimeTAG: 2012-11-16T19:46:14Z VoteTAG: 2 CoursewareTAG: Week 9 / LCOT CommentableIdTAG: 6002x_LCOT NumberOfReplyTAG: 1 FirstChildTAG: The 0.75x speed and the 1x speed got inverted. It seems like a bug. I think the staff should look into it. I'm using Firefox 16.0.2 and winXP. FirstChildUserIdTAG: 304535 FirstChildUserNameTAG: hugo_h FirstChildCreateTimeTAG: 2012-11-18T08:45:22Z SecondChildTAG: ...or they both just drunk ;))) SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-11-19T18:48:34Z IndexTAG: 1330 TitleTAG: H9P3 For question B we have that wo=1/sqrt(L*C) substituting parametrs for shown model i get equation for w0= 1/sqrt(x*y) But still have red cross. The same for qestion C f=2*pi*w0 and still red cross. Where is the mistake or bug? UserIdTAG: 96873 UserNameTAG: Santyaga CreateTimeTAG: 2012-11-16T18:55:32Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: HI! You should get the characteristic equation from the homogenous equation md2y/dt+b dy/dt+ky = 0. Your characteristic polynomial would be in the form ms2+bs+k = 0. You then should compare the characteristic polynomial with the canonical form of s2+2as+w02 = 0.From here, you can solve for alpha, and omega. Hope this helps! FirstChildUserIdTAG: 143823 FirstChildUserNameTAG: NathanNadeson FirstChildCreateTimeTAG: 2012-11-16T19:05:27Z SecondChildTAG: I will try that! SecondChildUserIdTAG: 96873 SecondChildUserNameTAG: Santyaga SecondChildCreateTimeTAG: 2012-11-16T19:28:12Z SecondChildTAG: 0.1591549*(1/sqrt(m*(1/k))) This is the right answer for the last part of this example SecondChildUserIdTAG: 106816 SecondChildUserNameTAG: Laith SecondChildCreateTimeTAG: 2012-11-18T07:28:26Z SecondChildTAG: @Laith, be carefull about posting solutions. That is against the Honor Code. SecondChildUserIdTAG: 248807 SecondChildUserNameTAG: 1kingsman SecondChildCreateTimeTAG: 2012-11-19T04:21:46Z FirstChildTAG: Skip the R, L and C stuff. How would you get the spring's equation into the characteristic equation of s^2 + 2as + w^2 = 0 ? (hint: stare at the coefficients). Add the 2pi only to convert w to Hertz, i.e. if not asked, just forget it. FirstChildUserIdTAG: 184153 FirstChildUserNameTAG: tthngn FirstChildCreateTimeTAG: 2012-11-16T20:04:37Z IndexTAG: 1331 TitleTAG: H9P1 Easy numbers, but if I calculate by octave or maxima answers seems to be wrong for example L= 160.0H and C=2.53mF. f0=0.2515 not 0.25 and so in next steps zero no is zero there are small values when time 1s. and in next steps answers wrong UserIdTAG: 406023 UserNameTAG: neitrino CreateTimeTAG: 2012-11-16T18:10:05Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Troubleshooting NumberOfReplyTAG: 2 FirstChildTAG: I think the intent is to round off to a number for the frequency that gives the period as an integer number of seconds. Try answering the rest of the questions with that assumption. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-16T19:23:46Z FirstChildTAG: I lost quite some time for the same reason ... 0.0013842749128 A would have been the correct answer (L = 145H, C = 2.8mF), but it is not recognized. Just putting 0 works, and I was looking for my fault... FirstChildUserIdTAG: 406420 FirstChildUserNameTAG: Picolo FirstChildCreateTimeTAG: 2012-11-16T21:15:57Z IndexTAG: 1332 TitleTAG: lab9 boost convertor problem i m unable to understand the problem .can any one help UserIdTAG: 611666 UserNameTAG: Shaminder420 CreateTimeTAG: 2012-11-16T15:13:07Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: From Lab 9: **This cycle repeats itself indefinitely. Let the duration of the first state be DT, and the duration of the second state be (1−D)T , where D is the fraction of a cycle for which the square wave controlling the MOSFET is high (sometimes called the duty cycle) and T is the switching period (set to 1/30kHz in the circuit above).** **Try a 2ms TRAN simulation to see how the circuit works, paying attention to the current through the inductor and the output voltage. Start by adjusting the capacitance C so that the ripple on the output voltage is 0.1V or less using the equation iΔt=CΔv to help choose the appropriate value. Here i is the desired load current 6V/1kΩ, Δt is the discharge time in one cycle DT and Δv is the desired ripple 0.1V.** **Now choose values for the inductance L and duty cycle D to produce the desired output voltage. As explained in the text, the average output voltage is roughly VIN/(1−D), so increasing the duty cycle will raise the output voltage. One wants to choose D and the cycle time T so that the inductor current reaches zero just as the next cycle begins.** **Experiment with the various parameters until the output voltage is 6V with a maximum 0.1V ripple. As a final check, perform a 10ms TRAN simulation and then click CHECK. The on-line system will be veryifying that the output voltage meets the specification for 9.5ms So they want you to change the values of the capacitor, inductor and the duty cycle of the source until you hit the specified parameters(6V output, .1V ripple). You can either do that with an equation or by just changing the values in the provided sandbox. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-11-16T15:53:52Z SecondChildTAG: [Lab9 Hints][1]:) - Spanish and English. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50a78f0dce1ad22700000024 SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-17T13:48:10Z IndexTAG: 1333 TitleTAG: h9p1 part 7 i m using q/c + v9- .but not getting the right ans. am i missing something UserIdTAG: 611666 UserNameTAG: Shaminder420 CreateTimeTAG: 2012-11-16T09:11:28Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: be careful with the sign of vC(t = 9-)..by the way, how do u solve for the iL value after the impulse is applied? FirstChildUserIdTAG: 143823 FirstChildUserNameTAG: NathanNadeson FirstChildCreateTimeTAG: 2012-11-16T09:59:27Z SecondChildTAG: For a short time the inductor acts like an open circuit. SecondChildUserIdTAG: 85490 SecondChildUserNameTAG: yashsavani SecondChildCreateTimeTAG: 2012-11-16T21:58:08Z SecondChildTAG: In h9p1 to find the last three answers,u need 2 consider capacitor as short circuit and inductor as open circuit...! SecondChildUserIdTAG: 436350 SecondChildUserNameTAG: seetharam1992 SecondChildCreateTimeTAG: 2012-11-17T15:46:15Z SecondChildTAG: thnks it was of g8 help... bt i hv a doubt ...wont initial vondition of t=9- will affect the situation SecondChildUserIdTAG: 214277 SecondChildUserNameTAG: yadsam SecondChildCreateTimeTAG: 2012-11-18T09:43:36Z IndexTAG: 1334 TitleTAG: H9 P1 Please any hints how to solve C and E questions! UserIdTAG: 96873 UserNameTAG: Santyaga CreateTimeTAG: 2012-11-15T22:50:14Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1335 TitleTAG: IDEAL CAPACITOR(To Myrimit) can anyone give detailed explanation for this question?![enter image description here][1] [1]: http:// UserIdTAG: 378967 UserNameTAG: krish2012 CreateTimeTAG: 2012-11-15T04:53:54Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: ![IDEAL CAPACITOR][1] [1]: https://edxuploads.s3.amazonaws.com/13529555201343633.png FirstChildUserIdTAG: 378967 FirstChildUserNameTAG: krish2012 FirstChildCreateTimeTAG: 2012-11-15T04:59:05Z SecondChildTAG: Its gate2012 question.i need much detailed explanation for this problem SecondChildUserIdTAG: 378967 SecondChildUserNameTAG: krish2012 SecondChildCreateTimeTAG: 2012-11-15T05:01:02Z SecondChildTAG: it should be impulse. due to conservation of charge ,charge on the upper plates will be distributed according to their capacitance and then no charge flows, as the system is ideal this will happen within no time so theoretically infinite current will flow for a very small time.hence the answer is impluse SecondChildUserIdTAG: 251805 SecondChildUserNameTAG: JasmeenPatel SecondChildCreateTimeTAG: 2012-11-15T06:30:08Z FirstChildTAG: Hi krish2012, I am agree with JasmeenPatel too, the answer should be an impulse. Lets see this together, - The statement tells you that the C1 is previoulsly charged to 12V. So, you know that the Capacitor will have a charge Q1i. ![ci][1] - Ok, now, we know that the other capacitor it is not charged previously. Also, you know that your Q2i = 0. So, what happends when you try to conect both?. In t=0, the charges of the Capacitor 1 will flow and both capacitors charges will be re-acomodated in order to balance the charges . This is because of the conservation of energy. Why? ![im2][2] Initially you have a total charge Q. at the end, you should have that same total Q. But, you know that once you conect the capacitor 2, that capacitor will be to a V2=V1 and so, it should have as C2*V2=Q2f, so now you will have a charge in C2. So, now your total Q will be distributed to Q1f and Q2f to be according the conservation Law. After that acomodation, it will not allow the flow of the charges, as they will be balanced. This charge flow will happen in a small portion of time, in t=0. Then the charges will be balanced, constant, and because the current is the vary of the charges respect the time, it will be zero. So, your current will be behaving like this for all t. ![cur][3] So, the answer will be D), and impulse function. See you, Myriam. [1]: https://edxuploads.s3.amazonaws.com/13529815558868319.png [2]: https://edxuploads.s3.amazonaws.com/13529834601343661.png [3]: https://edxuploads.s3.amazonaws.com/1352983862134369.png FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-15T12:52:51Z SecondChildTAG: Miriam, super buena la explicación. Has hecho excelentes aportes al curso. Henry SecondChildUserIdTAG: 175734 SecondChildUserNameTAG: hestrada SecondChildCreateTimeTAG: 2012-11-16T02:11:56Z SecondChildTAG: Por nada hestrada, me alegro mucho :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-16T03:31:41Z FirstChildTAG: Thank you JasmeenPatel and Myrimit for your intuitive explanation.I am sure that myrimit will give us a detailed explanation with figures and graphs.I'm having more questions and i hope you will answer it all :) .. FirstChildUserIdTAG: 378967 FirstChildUserNameTAG: krish2012 FirstChildCreateTimeTAG: 2012-11-16T08:16:52Z IndexTAG: 1336 TitleTAG: QUERRY... hello sir, I'm B.E fresher ECE engineer.I have registered for Electronic Circuits course but can u plz tel me whether i can cover up the syllabus that i have missed...?? I'm very excited to learn new things but unfortunately i have come to know about this course now.Please tel me if i can continue with it now.. or when is the next batch course?? Thank YOU. Sincerely, UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-11-14T19:51:05Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: You can continue with the course for gaining knowledge as all the previous lectures and homeworks and all are available..but you wont be able to get a pass grade since the time for first 8 HWs and labs and the midterm exam are already over..and you wont be able to get the required 60% with the remaining..so if your aim is to get pass grade and the certificate I suggest you try next time..if you do learn now, it will make it a lot easier the next time you take it.. :) FirstChildUserIdTAG: 173147 FirstChildUserNameTAG: cruiser_rahit FirstChildCreateTimeTAG: 2012-11-14T19:57:03Z FirstChildTAG: Unfortunately it is too late to complete the course as most of the homework deadlines and the midterm exam have passed; however, this course will be offered again in the Spring semester. Rohan FirstChildUserIdTAG: 391929 FirstChildUserNameTAG: RohanNagarkar FirstChildCreateTimeTAG: 2012-11-14T19:56:18Z IndexTAG: 1337 TitleTAG: Can anyone tell me please, why this doesn't work H9P1? I've calculated iL(t)=cos(wo*t) for t<9 (double checked) , from where I can know iL(9-).This doesn't give me t(9-) as for the grader. Please If your answer is analysys by inspection, which I don't know how to do for t(9-),please elaborate a little. Thanks! UserIdTAG: 369339 UserNameTAG: vargaslen CreateTimeTAG: 2012-11-14T18:04:53Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: If you correctly calculated the frequency in part 1 then you can calculate the period for the oscillation. Compare that with 9. This will give you a number of complete periods plus a fraction of a period. It should be a simple fraction of a period for which you know the value of cos. It is a "nice" number. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-14T18:23:45Z SecondChildTAG: could u elaborate meaning of comparing with 9 .i m nt gettng it SecondChildUserIdTAG: 611666 SecondChildUserNameTAG: Shaminder420 SecondChildCreateTimeTAG: 2012-11-16T08:30:20Z FirstChildTAG: Just a thought ... If you aren't thinking but just plugging numbers into your calculator, you could be having problems if your calculator is not in radians mode. You shouldn't need a calculator to do this part of the problem! FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-14T20:23:41Z SecondChildTAG: Thanks skyhawk!, belive me, I'm doing a lot of thinking! I'm 56 years old and long ago since I left College, so I'm I little rusty, but doing my best effort, Thanks for your support!. SecondChildUserIdTAG: 369339 SecondChildUserNameTAG: vargaslen SecondChildCreateTimeTAG: 2012-11-14T22:41:37Z SecondChildTAG: I know that wo is the oscillation frequency for the circuit and is in radians per second. So my error is in the 9 seconds? I thought that it should give me the radians at that time and obtain iL by applying cos to this.Thanks again for your kind response. SecondChildUserIdTAG: 369339 SecondChildUserNameTAG: vargaslen SecondChildCreateTimeTAG: 2012-11-14T22:49:05Z FirstChildTAG: The answer to the first part is in Hertz. Just invert and that's the period in seconds. For my data, the period was a nice integer number of seconds. Then all I had to do was compare with 9 and write down the answer at that point, no calculator needed! BTW I am 68. I started college 50 years ago this fall. This is my golden anniversary, so to speak. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-14T23:39:42Z SecondChildTAG: Good to know I'm not the only veteran here! SecondChildUserIdTAG: 369339 SecondChildUserNameTAG: vargaslen SecondChildCreateTimeTAG: 2012-11-15T00:34:53Z SecondChildTAG: Thanks again skyhawk!, I've got my correct answer now(0), even tough my previous answer wasn't so far from the good one (0.012),so I'm still in doubt, excuse me if I insist in the same question: why cos(wo*9s) where wo in in radians per second is not right?. Do you belong in this course's staff? I am from Venezuela, and have a degree in Computer Science. SecondChildUserIdTAG: 369339 SecondChildUserNameTAG: vargaslen SecondChildCreateTimeTAG: 2012-11-15T00:48:31Z SecondChildTAG: @skyhawk and @vargaslen..i am impressed by seeing your's passion for studying at this age..hope i can also keep the same passion like you both. SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-11-16T07:14:43Z FirstChildTAG: No, I'm not staff, probably more a pain to the staff! I suppose I was just lucky when I first computed the frequency. I decided that the intent was for it to be a "nice" number to work with. When I saw the follow on questions that confirmed my view. Technically speaking your answer is the correct one but not the one the grader was looking for! I have wasted my share of time working on problems that in my opinion were not well stated ... part of H9P2 comes to mind. P.S. My original education is in theoretical physics, but I recycled myself into petroleum reservoir engineering. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-15T01:14:26Z SecondChildTAG: I'm trying to recycle myself as well, that's my motivation for doing this course, what is yours? By the way, I just calculated my vc(9-), by applying the balance equation for the energy CV^2/2=Cvc^2/2+LiL^2/2 and got 274.647109001, but the grader doesn't agree!. SecondChildUserIdTAG: 369339 SecondChildUserNameTAG: vargaslen SecondChildCreateTimeTAG: 2012-11-15T01:44:12Z FirstChildTAG: If you got the answers to parts 3 and 4 correct, you shouldn't have a problem with part 5, but do watch the sign. I'm just taking the course for fun. I am an electronics hobbyist. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-15T02:31:55Z SecondChildTAG: Thanks again! SecondChildUserIdTAG: 369339 SecondChildUserNameTAG: vargaslen SecondChildCreateTimeTAG: 2012-11-15T12:26:53Z IndexTAG: 1338 TitleTAG: S17E5, first question How in the heck did they get the answer to the first question? Using the node method, I can't seem to get rid of C*dVc(t)/dt from the capacitor out of the equation. In the answer the only remnants from the capacitor is the "C". How did they do that? UserIdTAG: 132828 UserNameTAG: rharris CreateTimeTAG: 2012-11-14T07:01:20Z VoteTAG: 2 CoursewareTAG: Week 9 / An ILC circuit CommentableIdTAG: 6002x_an_ILC_circuit NumberOfReplyTAG: 1 FirstChildTAG: If you use the node method, you will end up with voltages in v in the equation. Use KVL, to get equation in current i. FirstChildUserIdTAG: 544644 FirstChildUserNameTAG: Ethanaung FirstChildCreateTimeTAG: 2012-11-14T07:46:51Z SecondChildTAG: Thanks Ethanaung. I actually tried KVL starting with the loop through the inductor and capacitor, but I didn't know how to calculate the KVL for the loop through the current source and the capacitor. Since there was no resistance in the circuit, how was I going to find the voltage across the current source to use in the KVL equation. As it turns out after getting your reply, I didn't have to write a KVL equation for the loop containing the current source and capacitor. Once I wrote the KVL equation for the loop containing the Capacitor and inductor, I discovered I could use a KCL equation to eliminate the iC variable in the d^2*iL/dt^2 = 1/c*1/L*iC equation. The other confusing part was knowing when to use iL vs. iL(t) = 1/L*Integral(VL(t)). I guess it is a game in semantics. Thanks again. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-11-14T16:16:14Z IndexTAG: 1339 TitleTAG: Discussion board help (staff) Is there a way for me to see just the posts related to week 10 or the HW for week 10 (for example)without plowing through all of them? UserIdTAG: 18073 UserNameTAG: gburkhart CreateTimeTAG: 2012-11-14T02:25:31Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: The convention is for people to tag their posts in helpful fashion like so: "H9P1" for Homework (week) 9, Problem 1. Similarly with Labs: try searching for "Lab9" for example. For posts more generally pertaining to a given week, I would try "week9", "S18E1" for a particular exercise, or even "S17V1" for a particular video. Hope this helps! FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-11-14T04:05:54Z IndexTAG: 1340 TitleTAG: [S17E2] Plus or minus In fact $\omega_0 = \pm$ the value accepted by the exercise UserIdTAG: 346056 UserNameTAG: fiatlux CreateTimeTAG: 2012-11-13T23:06:01Z VoteTAG: 2 CoursewareTAG: Week 9 / Characteristic Equation CommentableIdTAG: 6002x_characteristic_equation NumberOfReplyTAG: 0 IndexTAG: 1341 TitleTAG: TROUBLE USING SANDBOX AND LABS!!!!! this is something i observed when i open the sandbox or any of the labs in general 1. the navigation bar on the right disappears/does not appear. 2. the components do not appear properly. 3. i can see the components /circuit if i minimize the circuit window . [1]: https://mail-attachment.googleusercontent.com/attachment/?ui=2&ik=7900ccb76a&view=att&th=13afacefbab3114e&attid=0.1&disp=safe&realattid=f_h9hama1j0&zw&saduie=AG9B_P_tTbXa46lPInxLiASQLHwy&sadet=1352827672495&sads=P4_V4JLLzs6q5RaAbL7Rubiwh0c&sadssc=1 UserIdTAG: 357747 UserNameTAG: kishores CreateTimeTAG: 2012-11-13T17:19:48Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Can you post a screenshot? What OS are you on? What browser are you using, and what version number is that? Did you try another browser? We have gotten a few scattered reports of this, but the staff cannot reproduce it. So we need more information. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-11-13T18:43:19Z FirstChildTAG: ![on minimising][1] on minimising circuit window [1]: https://mail-attachment.googleusercontent.com/attachment/?ui=2&ik=7900ccb76a&view=att&th=13afad047c402951&attid=0.2&disp=safe&realattid=f_h9haoeen1&zw&saduie=AG9B_P_tTbXa46lPInxLiASQLHwy&sadet=1352873748409&sads=TlJ0l9FactRYsSAPHUCAdt-Y9hI FirstChildUserIdTAG: 357747 FirstChildUserNameTAG: kishores FirstChildCreateTimeTAG: 2012-11-14T06:18:51Z SecondChildTAG: without minimising circuit window [1]: https://mail-attachment.googleusercontent.com/attachment/?ui=2&ik=7900ccb76a&view=att&th=13afad047c402951&attid=0.1&disp=safe&realattid=f_h9hama1j0&zw&saduie=AG9B_P_tTbXa46lPInxLiASQLHwy&sadet=1352873741221&sads=p8UxPzc_LmcLjxwXaIKCqNaRrww&sadssc=1 SecondChildUserIdTAG: 357747 SecondChildUserNameTAG: kishores SecondChildCreateTimeTAG: 2012-11-14T06:24:24Z SecondChildTAG: ![enter image description here][1] [1]: https://mail-attachment.googleusercontent.com/attachment/?ui=2&ik=7900ccb76a&view=att&th=13afad047c402951&attid=0.1&disp=safe&realattid=f_h9hama1j0&zw&saduie=AG9B_P_tTbXa46lPInxLiASQLHwy&sadet=1352873741221&sads=p8UxPzc_LmcLjxwXaIKCqNaRrww&sadssc=1 SecondChildUserIdTAG: 357747 SecondChildUserNameTAG: kishores SecondChildCreateTimeTAG: 2012-11-14T06:27:07Z SecondChildTAG: my os is WINDOWS XP PROFESSIONAL VERSION 2002. my browseris GOOGLE CHROME. SecondChildUserIdTAG: 357747 SecondChildUserNameTAG: kishores SecondChildCreateTimeTAG: 2012-11-14T06:31:24Z SecondChildTAG: no issues with firefox. SecondChildUserIdTAG: 357747 SecondChildUserNameTAG: kishores SecondChildCreateTimeTAG: 2012-11-14T07:05:06Z FirstChildTAG: I can't see your screenshots, can you resend? FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-11-14T14:28:47Z IndexTAG: 1342 TitleTAG: Could anybody explain the answer to question 5 I don't see how the answer to question 5 could be $-V_I*(w_0/w_d)$. I think there should be a cosine term there. I think the answer should be $-V_I/\cos(w_d/w_0)$. UserIdTAG: 153760 UserNameTAG: Kavka CreateTimeTAG: 2012-11-13T05:19:10Z VoteTAG: 2 CoursewareTAG: Week 9 / S18E3 Total Solution CommentableIdTAG: S18E3_Total_Solution NumberOfReplyTAG: 4 FirstChildTAG: From i(0)=0 you get -acos(phi)=wd*sin(phi), then sin(phi)=sqrt(1-cos(phi)^2) and after playing around with math you get the result. FirstChildUserIdTAG: 376877 FirstChildUserNameTAG: AndBre FirstChildCreateTimeTAG: 2012-11-13T11:01:14Z FirstChildTAG: Use this trigonometric relation: cos(x) = sqrt( 1/(1+tan^2(x)) ). If you substitute x for arctan(-a/wd) it is straight forward to get cos(x) = wd/w0, which inverts to w0/wd when passing to the other side of the equation. FirstChildUserIdTAG: 99238 FirstChildUserNameTAG: arthurltc FirstChildCreateTimeTAG: 2012-11-16T01:33:24Z FirstChildTAG: One can also use the trig identity of cos(arctan(x)) = 1/sqrt(x^2+1) FirstChildUserIdTAG: 132828 FirstChildUserNameTAG: rharris FirstChildCreateTimeTAG: 2012-11-16T06:09:52Z FirstChildTAG: From v(0)=0 we are getting VI+A*cos(phi) Remembering previous solution phi = -arctan(a/wd) we are also getting VI+A*cos(-arctan(a/wd))=0 And then you may find A as: A=-VI*sqrt(a^2/wd^2 + 1) = **-VI*w0/wd** FirstChildUserIdTAG: 220304 FirstChildUserNameTAG: sergei_m FirstChildCreateTimeTAG: 2012-11-13T17:12:40Z SecondChildTAG: Use this trigonometric relation: cos(x) = sqrt( 1/(1+tan^2(x)) ). If you substitute x for arctan(-a/wd) it is straight forward to get cos(x) = wd/w0, which inverts to w0/wd when passing to the other side of the equation. SecondChildUserIdTAG: 99238 SecondChildUserNameTAG: arthurltc SecondChildCreateTimeTAG: 2012-11-16T01:33:08Z IndexTAG: 1343 TitleTAG: H8P3 Answers Being Revised The "Show Answer" in the homework for H8P3 has some errors in it(specifically for Q4 and Q5) and is being revised. Check back soon and the answers should be updated. UserIdTAG: 21541 UserNameTAG: JSChambers CreateTimeTAG: 2012-11-12T18:16:28Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Yep, it is fixed now. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-11-12T20:07:44Z FirstChildTAG: Looks like it has been fixed now... Thank you. FirstChildUserIdTAG: 714237 FirstChildUserNameTAG: PaxPolaris FirstChildCreateTimeTAG: 2012-11-12T20:06:11Z IndexTAG: 1344 TitleTAG: Suggestion for Discussion forum Hi all! I am sure you agree this platform is one of the smoothest out there! The design is more or less impeccable. I just have a suggestion, why don't we have 'profiles', then we could have descriptions of our background etc, just like a "real" forum. Hazel UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-11-11T16:39:47Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: hay Hazel!! that's a good idea.. if there is our profile then it could be more easy to communicate and do other things.. FirstChildUserIdTAG: 509517 FirstChildUserNameTAG: randima FirstChildCreateTimeTAG: 2012-11-11T18:13:53Z SecondChildTAG: Yep, like PMs etc... SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-11T20:32:20Z SecondChildTAG: While I agree with that we could use a few minor adjustments, in a time of extreme social media, I like the clutter free and *quiet* interface here. I also rather enjoy the absence of a social hierarchies and distractions that things like profiles may create. In my opinion, a Personal Message (PM) system may be counter productive to the forum for obvious reasons. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-11-12T01:14:14Z SecondChildTAG: That is very true SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-12T08:14:14Z IndexTAG: 1345 TitleTAG: Certificate Requirements On how much marks(in %age)i will be able to get certificate?? UserIdTAG: 217686 UserNameTAG: saifullahshafiq99 CreateTimeTAG: 2012-11-11T16:31:01Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: 60 percent, you can verify this by clicking on the progress tab above your post. FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-11-11T16:42:52Z FirstChildTAG: hi saifullahshafiq99!! how are you,, end of the program result should be greater than 60%.. FirstChildUserIdTAG: 509517 FirstChildUserNameTAG: randima FirstChildCreateTimeTAG: 2012-11-11T18:15:41Z FirstChildTAG: Hi, [GRADING SYSTEM & CERTIFICATE REQUIREMENT][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/509f6a7bb6f29f1f00000020 > **Regards:** asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-11-16T10:37:37Z SecondChildTAG: I just checked mine nd it was 49%....:0 :( does it show the % of the work done or does it get the % from all the home works....!! like I have done till H9 so does the total go for 9 HW or 12 HW???? :( IS IT 49 % of the course or 49% of the work done?? SecondChildUserIdTAG: 127800 SecondChildUserNameTAG: MNB SecondChildCreateTimeTAG: 2012-11-18T01:51:24Z IndexTAG: 1346 TitleTAG: Why do we suppose that voltage is falling down from 3V? It could be falling from 10V or 20V or more a few sec ago and turned 3V at t=0.001. In this cases Voltage reduction speed would be very different and at t=0 would be slightly less then 3V. UserIdTAG: 66669 UserNameTAG: agarus CreateTimeTAG: 2012-11-11T15:01:56Z VoteTAG: 2 CoursewareTAG: Week 8 / Initial Conditions CommentableIdTAG: 6002x_Initial_Conditions NumberOfReplyTAG: 0 IndexTAG: 1347 TitleTAG: GRADING SYSTEM ON DEMAND FOR 'maliha266' **GRADING SYSTEM** ----------------- - **A** for 87% - **B** for 70% - **C** for 60% > *Note:* **C** is mandatory to get certificate. For Further query check your progress bar... Thank you. > **Regards:** asadbhatti42 UserIdTAG: 329051 UserNameTAG: asadbhatti42 CreateTimeTAG: 2012-11-11T09:06:03Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Thankyou very much asad! :) FirstChildUserIdTAG: 499671 FirstChildUserNameTAG: maliha266 FirstChildCreateTimeTAG: 2012-11-12T04:46:29Z IndexTAG: 1348 TitleTAG: LAB 9 Please help me, I am unable to solve LAB9, BOOST CONVERTER problem. UserIdTAG: 456732 UserNameTAG: anilkumar6745 CreateTimeTAG: 2012-11-11T07:02:34Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: Yeah right! =) Me too =) FirstChildUserIdTAG: 340272 FirstChildUserNameTAG: ezekielbrizuela FirstChildCreateTimeTAG: 2012-11-11T15:25:17Z FirstChildTAG: What are we supposed to observe in Booster converter lab? output of 6V with +-0.1V ripple and zero inductance current? I like to know if there is another goal other than the 6.0V output. FirstChildUserIdTAG: 265795 FirstChildUserNameTAG: krishnakm FirstChildCreateTimeTAG: 2012-11-11T21:34:38Z FirstChildTAG: No, there are no other requirements, only output and ripple voltages. Just follow the instructions from the text; you'll have to change values for C and L to obtain the required output. FirstChildUserIdTAG: 277787 FirstChildUserNameTAG: kirilaska FirstChildCreateTimeTAG: 2012-11-12T03:04:01Z SecondChildTAG: Hi! How do I calculate the value of L? SecondChildUserIdTAG: 143823 SecondChildUserNameTAG: NathanNadeson SecondChildCreateTimeTAG: 2012-11-15T14:12:57Z FirstChildTAG: You can get the required output just choosing the right value of the duty cicle D (: FirstChildUserIdTAG: 198948 FirstChildUserNameTAG: leoayrton FirstChildCreateTimeTAG: 2012-11-13T20:48:21Z FirstChildTAG: Take a look at here [Lab9 Hints][1] :)- Spanish and English. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50a78f0dce1ad22700000024 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-17T13:49:32Z IndexTAG: 1349 TitleTAG: S15E3: formula of unit pulse? Shouldn't the formula of a unit pulse from $t_0$ to $t_1$ be: $$\large u(t-t_0)-u(t-t_1)$$ UserIdTAG: 714237 UserNameTAG: PaxPolaris CreateTimeTAG: 2012-11-11T05:59:51Z VoteTAG: 2 CoursewareTAG: Week 8 / Pulse Is A Step Up Then Step Down CommentableIdTAG: 6002x_Pulse_Is_A_Step_Up_Then_Step_Down NumberOfReplyTAG: 0 IndexTAG: 1350 TitleTAG: Tutorial .... not quite I found the delivery of this tutorial rushed and lacking depth of explanation in places. Wikipedia: "A tutorial is a method of transferring knowledge and may be used as a part of a learning process. More interactive and specific than a book or a lecture...." UserIdTAG: 345671 UserNameTAG: cbjerregaard CreateTimeTAG: 2012-11-11T01:54:40Z VoteTAG: 2 CoursewareTAG: Week 8 / Exercise 10.19 CommentableIdTAG: 6002x_Ex10_19 NumberOfReplyTAG: 2 FirstChildTAG: Hi cbjerregaard, Based on my experience of making Tutorials this Fall 6.002x, it took me a lot of time, the last video had a 28 min duration [This Post][1] and [This one][2] had a 19 min duration haha, a long boring time haha. I wish to make more complementary Tutorials explanations videos but unfortunately, I don't have too much time hahaha, but I will be loved to make more ones, might I would prepare materials in my summer Holidays - by the way here is now spring haha and not fall, I always get confused-, I will try to do that because I want to set the proctored Exam at the end of the next year and I want to be 100 percent sure and confident haha. - If I can and - if I still welcome here again and - if any asteroid falling on earth before march of 2013, I guess will upload more videos tutorials in the next 6.002x Spring - 2013 to help the New Students haha. I guess that I will need suggestions of the topics/excercises to be done. Do you have suggestions? Then I guess I will finally retire from 6.002x, I don't want to do that because I like to help and motivate students, but after a long meditation, I guess that I should leave, giving to others the opportunity to help and to participate more here, staying here a lot of time might will not leave to others students of previous terms to participate as Community TA's and from heart, I don't want that. See you, Take care :), Myriam. P.D. Might the explanations of the videos Tutorials are fast because imagine if the Tutorials, each ones, had like more than 30 min each - like mines ones haha- each Course will demand - watching lectures videos, tutorials, etc..- a lot of time I guess... [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/508f43a96d0c542500000007 [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_S3E2_Linear_Combinations_of_Source_Strengths/threads/506a280d036af91f000000c9 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-11T03:31:07Z FirstChildTAG: Well, the lecturer is quite executive - that's her style - and sometimes, when she says, for instance, "So it's option B" you perhaps think "How was that?". (It was my case a couple of times with other tutorials.) But there is a trick... consisting on hitting the pause button and spending a couple of minutes on it, or reviewing the last part... and usually is then easy to see the logic behind. I like the way she explains and keeps the videos short. Best, Jordi FirstChildUserIdTAG: 366320 FirstChildUserNameTAG: Galli FirstChildCreateTimeTAG: 2012-11-11T13:01:47Z IndexTAG: 1351 TitleTAG: sandbox I am not finding Sandbox? How I may get that? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-11-11T01:01:01Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You can go to - Overview - Courseware - Sandbox ![im][1] Or Click on [here][2] [1]: https://edxuploads.s3.amazonaws.com/13525971681404509.png [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Overview/Circuit_Sandbox/ FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-11T01:26:59Z IndexTAG: 1352 TitleTAG: Sandbox HI every one How I may use sandbox? Help me.. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-11-11T00:30:17Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi, Can I help you with the Sandbox? Take a look at here [Wiki of the Course - click on here][1]. You also can read [this - click on here][2] from the Course Info. Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/MyrimitsGuideTo6002x/ [2]: https://www.edx.org/static/content-mit-6002x/handouts/schematic_tutorial.ba422f80d72b.pdf FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-11T00:43:04Z IndexTAG: 1353 TitleTAG: s15e5-graph/staff can somebody draw the I-t and V-t graph of this question from minus infinity to positive infinity? I don't understand the question. UserIdTAG: 374393 UserNameTAG: rmaleki CreateTimeTAG: 2012-11-10T19:00:04Z VoteTAG: 2 CoursewareTAG: Week 8 / Initial Conditions CommentableIdTAG: 6002x_Initial_Conditions NumberOfReplyTAG: 1 FirstChildTAG: Just a straight line (I0=10mA) for any t from minus infinity to positive infinity, with an impulse (Q(t)) in t=0. FirstChildUserIdTAG: 66669 FirstChildUserNameTAG: agarus FirstChildCreateTimeTAG: 2012-11-11T15:15:24Z IndexTAG: 1354 TitleTAG: H8P2 for 1st part, i am using the relation q=cv..still getting wrong answer..plz help. UserIdTAG: 114864 UserNameTAG: Vikaash CreateTimeTAG: 2012-11-10T15:47:51Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Do you mean for "initial concentration"? If so, look carefully at the problem definition: you are given everything you need to solve the 1st question very easily. FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-11-10T16:05:34Z SecondChildTAG: thanx @planetscape for giving response..but i am trying but getting wrong answer..am putting the value in equation v=q/c. SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-11-10T16:12:40Z SecondChildTAG: finally got it.. thinking out of the box!!!!!!! SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-11-10T16:51:08Z SecondChildTAG: Good! SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-11-10T17:23:18Z FirstChildTAG: Hi, [H8P2 Hint][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/509e68568e407d1f0000001b > **Regards:** asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-11-11T08:50:37Z IndexTAG: 1355 TitleTAG: Where is download link? Where is download link? UserIdTAG: 341358 UserNameTAG: akbar13 CreateTimeTAG: 2012-11-10T15:47:31Z VoteTAG: 2 CoursewareTAG: Week 4 / Schematic Neatness Tutorial CommentableIdTAG: 6002x_schematic_neatness_t NumberOfReplyTAG: 1 FirstChildTAG: Are you restricted from YouTube? It does not look like there is a direct download link for this video as there are, for example, for recent lectures. You may want to edit your subject line to include "STAFF" to bring it to their attention. FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-11-10T16:10:26Z IndexTAG: 1356 TitleTAG: h8p2 part 3 any hints for h8p2 part 3.. UserIdTAG: 266739 UserNameTAG: satvikchugh CreateTimeTAG: 2012-11-10T11:27:51Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: it is asked If the doctor wants to keep the concentration between 1/4 and 2/3 of the initial concentration, what is longest time, in hours, that may be allowed to elapse before a new injection is required? it takes more time to reach the concentration to 1/4 of initial concentration, so from that we can calculate the time elapsed from natural decay methods PAVAN FirstChildUserIdTAG: 131558 FirstChildUserNameTAG: PAVAN6 FirstChildCreateTimeTAG: 2012-11-10T16:10:52Z SecondChildTAG: thnks.:) SecondChildUserIdTAG: 266739 SecondChildUserNameTAG: satvikchugh SecondChildCreateTimeTAG: 2012-11-11T04:22:45Z IndexTAG: 1357 TitleTAG: Assumptions for the first part of H8P3 ![enter image description here][1] Hi herbsteiner Can you please tell me where you got 115M ohms as the Roff resistance for Q3? Also, given that the problem gives both Ron and Roff, shouldn't they be used instead of infinite resistance? Question for anyone who can help: I am having trouble with the first part of the homework problem. I believe I'm almost there, but not quite. Can someone please suggest where I'm going wrong? To find the parasitic resistance (Rp), my assumptions are Dout was read as a 1, and rewritten to Din as a 1 on the gate of Q1. With this configuration implemented in the above circuit, Q1 has resistance Ron, Q2 is Roff and Q3 is Roff after Store is de-asserted when the capacitor starts to discharge. I also made the assumption reading and rewriting a 0 has the same affect, but with Q2 set to Ron and Q1 set to Roff. The approach I used is to remove the capacitor and take the Thevenin equivalent of the circuit, calculating Rth and Vth as functions of Rp (p for parasitic), then substituting in the equation Vc = VIN + (V0-VIN)e^-(t/(R*C)). Of course this didn't work due to Rp being in the exponent and the body of the Vc equation. I believe the idea of using the Thevenin equivalent is correct, but I don't quite know how to handle Rp once the Thevenin equivalent is achieved. Below is the modified circuit I used, so picking the nodes to calculate the Thevenin equivalent was easier. I calculated Rth and Vth with resistor Rp connected to the output terminals, which is not shown. Thanks. ![enter image description here][2] [1]: https://edxuploads.s3.amazonaws.com/13524984625676028.jpg [2]: https://edxuploads.s3.amazonaws.com/13525001558615252.jpg UserIdTAG: 132828 UserNameTAG: rharris CreateTimeTAG: 2012-11-09T22:35:15Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Search the discussion for "magic Q3". Just assume Q3 is open circuit for this part of the problem. FirstChildUserIdTAG: 141000 FirstChildUserNameTAG: OrinE FirstChildCreateTimeTAG: 2012-11-09T23:36:23Z SecondChildTAG: Where did you get that Q3 has a resistance of Roff = 105M ohms? Doesn't the question give Roff = 90M ohms? I've seen different values in other discussions and I can't figure out why. - Thanks. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-11-09T23:45:39Z SecondChildTAG: I am still troubled with how to deal with finding Rp. When finding Vth and Rth, having Rp connected to the output makes it seem unsolvable? SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-11-09T23:54:49Z SecondChildTAG: Roff was 105M ohms for me. The system gives different values to different people. There is no Vth or Rth to calculate here. The question is assuming the gate has capacitance to the source and that there is parasitic resistance to the source. They give you the capacitance. You are told the capacitance is charged to $V_{OH}$ and discharges through the parasitic resistance to the source to $V_{IH}$ in the time they give you. You have to assume that this is the only parasitic resistance that the capacitance discharges through to get the right answer. SecondChildUserIdTAG: 141000 SecondChildUserNameTAG: OrinE SecondChildCreateTimeTAG: 2012-11-10T07:18:25Z SecondChildTAG: Ok, the 105M ohms makes sense, I didn't know they did that. I understand the voltage discharging through the parasitic capacitance, however I'm having trouble finding the right equation to use. With using Vc=VIN+(Vo-VIN)*e^(-t/(R*C)), I'm used to using VIN, Vo, t, R and C and then finding Vc, but this time I'm thinking Vo is VOH, VIN is VIH, however I'm not sure what to set Vc to in order to solve for R. I thought maybe if the capacitor was let to completely discharge it would discharge to 0, so that way I set Vo to VOH, VIN to 0 and Vc to VIH, then solve for R. Obviously it didn't work. Thanks SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-11-10T09:27:28Z FirstChildTAG: Follow the post (H8P3 - first question) FirstChildUserIdTAG: 477713 FirstChildUserNameTAG: ikm104 FirstChildCreateTimeTAG: 2012-11-10T05:40:57Z FirstChildTAG: Hi rharris, Take a look at here [Hints HW8][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/509e68568e407d1f0000001b FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-10T17:19:41Z SecondChildTAG: I had the correct answer all along. No wonder I was so confused, the question is so simple. I have converted so many milliseconds and microseconds to seconds that in this case I was moving the decimal point to the left, where I should have been moving it to the right. My answer from my calculator was similar to .yyy X 10^14, so my answer should be in the form yy.y X 10^12. I guess it doesn't pay to work until 1:30 AM in the morning. Thanks for putting up with my convoluted brain. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-11-10T20:20:10Z IndexTAG: 1358 TitleTAG: H9P2 (d) For question (d) please indicate that you are requiring the natural frequency in Hz. UserIdTAG: 269325 UserNameTAG: Fed_Ang CreateTimeTAG: 2012-11-09T08:33:17Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1359 TitleTAG: Impulse not yet applied I don't get this at all. We're asked to find the voltage across the capacitor at a time BEFORE t = 0. ...When the impulse hasn't yet been applied. The question tells us that we measured the voltage across the capacitor at t < 0 to be 3.0V. So... why is the answer to part A not 3.0V? I feel like I've completely misunderstood what is being asked here. I'm imagining the application of a normal impulse as in all the lecture videos prior to this, except that we have a constant I0 current which keeps us boosted up from the x-axis by an amount equal to its value. UserIdTAG: 287805 UserNameTAG: Beneficial CreateTimeTAG: 2012-11-09T04:45:10Z VoteTAG: 2 CoursewareTAG: Week 8 / Initial Conditions CommentableIdTAG: 6002x_Initial_Conditions NumberOfReplyTAG: 2 FirstChildTAG: the capacitor is discharging to the resistor connected to it... FirstChildUserIdTAG: 269325 FirstChildUserNameTAG: Fed_Ang FirstChildCreateTimeTAG: 2012-11-09T08:50:19Z SecondChildTAG: Yes, I agree. The steady state is 1v for the capacitor since the capacitor is at 3 V it will discharge for 1ms and reach a value between 1v and 3v. Then the impulse which delivers a negative charge (causes fall in current) further decrease the voltage across the capacitor and then the capacitor has to rise from that value to steady state of 1v. SecondChildUserIdTAG: 22953 SecondChildUserNameTAG: Linus SecondChildCreateTimeTAG: 2012-11-10T19:28:41Z FirstChildTAG: (I had a longer answer composed, but the system lost it...) In brief, replace the current source in parallel with the resistor with its Thevenin equivalent. At t=-1, you have 3V on the capacitor, the Thevenin equivalent resistance and the Thevenin equivalent voltage source, all in series. You have to find how much the capacitor discharges before t=0. FirstChildUserIdTAG: 141000 FirstChildUserNameTAG: OrinE FirstChildCreateTimeTAG: 2012-11-09T05:15:41Z IndexTAG: 1360 TitleTAG: NOT "Directly" related to 6.002x... 1. Why is that we don't mention Amps rating in power supply when we give/draw a schematic of [some circuit][1]? We just assign a voltage, say 6VDC. I mean, may I connect a power supply from a Lead-acid 12 V battery terminals capable of 'putting' in some 30 amps into the supply rail/port, for example, of my laptop? Won't it burn the resistors (if power ratings exceed) or burn the IC/transistors downstream? [Obviously, if more power is required for the same voltage, then we have to increase the current.] 2. We calculate & say that a parallel equivalent capacitor is 'theoretically' the same as a combination. Yet, in real life circuitry, a 100 nf (or 1 mF) is often connected in parallel with 47 pF (or 10 nF); the logic being the larger value one will bypass lower frequency signals while the lower value caps will 'short' high frequency ones (due to reactance being related with frequency). Why this divergence between theory & practice? Why don't we use a single capacitor to do the job? 3. I am interested in building small signal circuits which are very prone to noise (signal to noise ratio being very low) e.g. ECG/EKG, EMG etc. I have even built some; with some circuits available on the net, my own experience & some bit of knowledge in this field. But they are very prone to noise (hum). Will this course give us this 'holistic' view of circuits, so that I may eventually come-up with the one Piotr demonstrated (in his tutorial--bag of salt water) or we only get stuck in Calculus and don't venture beyond? Any relevant suggestion from anyone is welcome. Thanks & best wishes in advance. [1]: http://www.google.co.in/imgres?imgurl=http://www.circuitstoday.com/wp-content/uploads/2008/04/fm-transmitter-circuit.JPG&imgrefurl=http://www.circuitstoday.com/tag/radio-circuits&h=315&w=624&sz=29&tbnid=O1pcVtW4Tiz9mM:&tbnh=68&tbnw=135&zoom=1&usg=__u19tGViajkTs0O6jIvozKLl-Rzs=&docid=RYKmLpGfTos9uM&sa=X&ei=MHybUM32NMr7rAeYg4GgBw&ved=0CDgQ9QEwBw&dur=19611 UserIdTAG: 285616 UserNameTAG: AmiyaX CreateTimeTAG: 2012-11-08T09:31:12Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: In answer to you first question, many of these applications will not draw more current than the power supply can handle (i.e. the secondary windings of the power transformer), usually in the mA range. Also, if the terminal resistance of whatever your connecting the power supply to is known, then current draw is also known. Further, in the case of a short circuit over the power-supply output terminals, many devices have inbuilt current-limiters http://en.wikipedia.org/wiki/Current_limiting Hazel, FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-11-08T10:25:23Z SecondChildTAG: Thanks [Hazel][1], I 'almost' grasp it! [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/users/365551 SecondChildUserIdTAG: 285616 SecondChildUserNameTAG: AmiyaX SecondChildCreateTimeTAG: 2012-11-08T12:28:47Z FirstChildTAG: In answer to your second question, an ideal capacitor has no internal resistance or inductance, so it is not necessary to show a large capacitor in parallel with a small one. In practice, however, large capacitors are usually electrolytic and have a substantial impedance at high frequencies, so it is common practice to put a smaller ceramic or polyester cap in parallel with the large electrolytic one. FirstChildUserIdTAG: 270011 FirstChildUserNameTAG: SnowmanZA FirstChildCreateTimeTAG: 2012-11-08T18:29:26Z SecondChildTAG: Thanks Snowman, you have explained it well. Regarding *"however, large capacitors are usually electrolytic and have a substantial impedance at high frequencies"*: **higher** the capacitance, **lower** the reactance, [given by 1/(2*Pi*f*C).] At high frequencies, reactance (impedance) *decreases* (for all values, as the equation shows). But still somewhere things don't really match, e.g. why should we calculate parallel 'equivalent' values, if in reality, we **have** to do it with two (or more) separate ones,& not one single parallel capacitor, as predicted theoretically. SecondChildUserIdTAG: 285616 SecondChildUserNameTAG: AmiyaX SecondChildCreateTimeTAG: 2012-11-10T08:49:28Z IndexTAG: 1361 TitleTAG: Circuit Sandbox Not Functioning in Any Lab can't get the labs to work. It may be this laptop I am using. UserIdTAG: 18073 UserNameTAG: gburkhart CreateTimeTAG: 2012-11-08T00:32:28Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Exactly where are you getting stuck. Can you enter a trivial curcuit? FirstChildUserIdTAG: 142857 FirstChildUserNameTAG: viking2 FirstChildCreateTimeTAG: 2012-11-08T04:08:43Z SecondChildTAG: Are you having trouble with the transient button? Could you post a screenshot? SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-08T09:57:44Z SecondChildTAG: it was the lap top - chrome is buggy at work. internet exploder worked fine. SecondChildUserIdTAG: 18073 SecondChildUserNameTAG: gburkhart SecondChildCreateTimeTAG: 2012-11-08T21:46:20Z SecondChildTAG: how did you solve this. I have same issue. I tried using internet explorer, does not work for me. can not log in. SecondChildUserIdTAG: 171674 SecondChildUserNameTAG: wajahat1 SecondChildCreateTimeTAG: 2012-11-09T16:41:13Z SecondChildTAG: HI every one How I may use sandbox? Help me.. SecondChildUserIdTAG: 227508 SecondChildUserNameTAG: bhavyab SecondChildCreateTimeTAG: 2012-11-11T00:29:46Z IndexTAG: 1362 TitleTAG: Please leave the syringe alone! Personally, I found drug delivery example very exciting! Drug delivery relates to the medicine. What the heck it is doing in 6.002x? How a human body could be related to electronics? was my thoughts at the beginning. And the last task in the homework is about a drug dissipation. I enjoyed that task so much! Regret it ended so fast. Everyone of us at least once I suppose was ill. It gave an insight of what the doctors think when they cure us. At least they should to :). Applying 6.002x knowledge to other areas (medicine, mechanics) is definitely the strong side of the course! Keep it that way! As for me, such thing prevent me from being bored of this course. Especially when it comes to differential equations and trigonometry. ---------- I wanted to reply to anonymous post, but it was closed. Wierd anonymous' fears or whatever should not push this course to be an ordinary math blah-blah-blah-zzzzz... You're doing it right! Keep going that way! UserIdTAG: 208296 UserNameTAG: OZ1 CreateTimeTAG: 2012-11-07T19:40:18Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I want to know if the professor is the artist too...;) FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-11-07T21:49:24Z FirstChildTAG: in russia english word "drug" usually means, usually translated as narcotic FirstChildUserIdTAG: 188778 FirstChildUserNameTAG: xsaq FirstChildCreateTimeTAG: 2012-11-15T16:25:20Z IndexTAG: 1363 TitleTAG: if you listen very carefully... you can hear a dog barking at 0:45 :) UserIdTAG: 345452 UserNameTAG: ale-yoman CreateTimeTAG: 2012-11-07T19:34:03Z VoteTAG: 2 CoursewareTAG: Week 8 / Storage Time CommentableIdTAG: 6002x_Storage_Time NumberOfReplyTAG: 1 FirstChildTAG: Haha! A 6.002x dog. Probably forgot to lock the **gate**, I hope he hasn't **bit** anyone on **impulse**! I'm sure you can think up some bad jokes too... FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-11-07T21:32:46Z IndexTAG: 1364 TitleTAG: Proctored Exam and value of this course certificate (To staff) I have read many news articles about one course this year offering proctored exam in collaboration with Pearson Inc. Is there any update on whether this course MITx 6.002x will have the final exam offered on a proctored centre? Also, I wish to know the value of the certificate obtained from edX and whether or not it will serve as a big catch to my resumé since MIT is a renowned institution? Yet there are some concerns because this operates on an Honour Code. UserIdTAG: 390093 UserNameTAG: 5riram CreateTimeTAG: 2012-11-07T16:54:49Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Take a look here for your second question... https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/508446a8d210431f0000012c FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-11-07T17:00:29Z FirstChildTAG: Thanks for that. I didn't see that one. Hope someone answers the first question too. FirstChildUserIdTAG: 390093 FirstChildUserNameTAG: 5riram FirstChildCreateTimeTAG: 2012-11-11T14:14:01Z IndexTAG: 1365 TitleTAG: plot I think the plot drawn by the prof is not quite correct: at small t, it is a sinusoid curve with constant amplitude but shifted (not scaled!) by the term cos(phi)*exp(-t/RC). UserIdTAG: 388524 UserNameTAG: dasbinich CreateTimeTAG: 2012-11-07T16:15:01Z VoteTAG: 2 CoursewareTAG: Week 10 / Sinusoidal Steady State SSS Solution CommentableIdTAG: 6002x_Sinusoidal_Steady_State_SSS_Solution NumberOfReplyTAG: 0 IndexTAG: 1366 TitleTAG: staff:textbook link not accessible textbook link not accessible. pls help UserIdTAG: 288210 UserNameTAG: Rajubalaji CreateTimeTAG: 2012-11-07T02:36:53Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: After reading your other thread, I tried reading the textbook in Firefox and Chrome and it worked in both. I am on OSX 10.6.8 Snow Leopard. I also tried it on Windows 7 in Firefox and Chrome. It worked fine in both of those. If I was in your position, I would find another computer and try it on that, and see if you get the same result. If that one works ok, then the problem is specific to your machine. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-11-07T13:46:09Z SecondChildTAG: I'll check in other machine. Meanwhile i've attached screenshot of textbook weblink ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13522994221343671.jpg SecondChildUserIdTAG: 288210 SecondChildUserNameTAG: Rajubalaji SecondChildCreateTimeTAG: 2012-11-07T14:44:25Z FirstChildTAG: Which textbook link? Do you mean this one: ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13522870301343606.png FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-11-07T11:17:21Z IndexTAG: 1367 TitleTAG: circuit sandbox doesnt work why tools in circuit sandbox dont work? UserIdTAG: 60496 UserNameTAG: tomipiriyev CreateTimeTAG: 2012-11-06T20:21:37Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: We don't know, and we are working on it. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-11-06T21:05:13Z SecondChildTAG: should be fixed. Thanks for the report. SecondChildUserIdTAG: 96452 SecondChildUserNameTAG: Lyla SecondChildCreateTimeTAG: 2012-11-07T01:48:47Z SecondChildTAG: Thank you Lyla. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-07T02:06:53Z IndexTAG: 1368 TitleTAG: S19E3: Formula checker bugs For magnitude wrong Vi/sqrt(1+(w*R*C)^2) is accepted while the same expression Vi/(sqrt(1+(w*R*C)^2)) is not. UserIdTAG: 208296 UserNameTAG: OZ1 CreateTimeTAG: 2012-11-06T18:00:16Z VoteTAG: 2 CoursewareTAG: Week 10 / Complex Numbers CommentableIdTAG: 6002x_Complex_numbers NumberOfReplyTAG: 2 FirstChildTAG: Vi/sqrt(1+w^2*R^2*C^2) also fails. FirstChildUserIdTAG: 366083 FirstChildUserNameTAG: smath FirstChildCreateTimeTAG: 2012-11-09T22:06:45Z SecondChildTAG: Vi/sqrt(1+(w*R*C)^2) is not accepted,while Vi/sqrt(1+w^2*R^2*C^2) is accepted.There must be something wrong here.Both should not be accepted. SecondChildUserIdTAG: 361137 SecondChildUserNameTAG: Ericson SecondChildCreateTimeTAG: 2012-11-23T13:24:51Z FirstChildTAG: > For magnitude wrong Vi/sqrt(1+(w*R*C)^2) is accepted No, it is not accepted also.The correct answer have abs(Vi), since Vi can take negative values. FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-11-22T09:23:59Z IndexTAG: 1369 TitleTAG: Experiment with BA56-12GWA I buy 7-seg 3-digit led display BA56-12GWA and experiment with one segment of it. I'm download datasheet for it and plot "load line" for Vs 2.5 V and R = 110 Ohm's. Intersection between load line and led characteristic give me theoretical current about 7.5 mA. Now I connect my Power supply 2.5 V, resistor 110 Ohm, ampermeter and one segment of BA56-12GWA in loop. And my ampermeter show current about 4.4 mA. Why I got this big difference between theoretical and prattical values (over 2 times)? In the next step I use trashhold voltage Vt of the led (1.95 V) and resistence of the led (Rled ~13 Ohm)(picewise-lonear approximation). Using formula (Vs-Vt)/(R+Rled) = 0.55/123 = 4.4 mA. Why in the "load line method" I have big difference between theoretical and practical results? And in the "picewise linear approximation method" I have no difference between theoretical and practical results? Sorry for my english:) ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13521907357654894.jpg UserIdTAG: 104522 UserNameTAG: IgorNovice CreateTimeTAG: 2012-11-06T08:30:05Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: From what I can see, you have two possible reasons... * Your LED display is subject to manufacturing tolerances, however this is unlikely as the piecewise linear method yielded a correct result exactly. * The graph is not accurate If the device works (which I am assuming it does) then I would not worry too much about it. By the way, well done with your English, I can only imagine how difficult it must be to understand 6.002x in a Foreign language!! FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-11-06T10:02:59Z SecondChildTAG: Thanks for help! I think, that my ampermeter show wrong value on low-current measurment mode! I measure another diode (1N4148) and got right result! SecondChildUserIdTAG: 104522 SecondChildUserNameTAG: IgorNovice SecondChildCreateTimeTAG: 2012-11-06T14:06:54Z SecondChildTAG: Awesome! Well done! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-06T14:20:59Z IndexTAG: 1370 TitleTAG: percentage score in progress page is wrong Just finished Week 9 homework and lab. In the bar for the total mark in the progress page the total percentage is given as 56%, but it should be 57%. This is obvious by even adding the individual percentages as they are given in the progress page: [2*(13.5/100)+30/100]*100 =57%. Anyone else has the same issue? UserIdTAG: 97581 UserNameTAG: Asimakis CreateTimeTAG: 2012-11-05T19:22:32Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Troubleshooting NumberOfReplyTAG: 5 FirstChildTAG: Me too. FirstChildUserIdTAG: 288323 FirstChildUserNameTAG: fatslow FirstChildCreateTimeTAG: 2012-11-06T04:40:49Z FirstChildTAG: Yes FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-05T19:34:26Z FirstChildTAG: Same here FirstChildUserIdTAG: 99628 FirstChildUserNameTAG: Pranjal16 FirstChildCreateTimeTAG: 2012-11-05T20:10:45Z FirstChildTAG: I guess its meant to be like this so you have get at least 1% at the final exam to pass. No pass without final. FirstChildUserIdTAG: 127195 FirstChildUserNameTAG: princeofsudan FirstChildCreateTimeTAG: 2012-11-05T22:16:35Z FirstChildTAG: It seems to ignore fractions in the individual values that it adds up. After HW and LAB 10, mine say 60%. FirstChildUserIdTAG: 270011 FirstChildUserNameTAG: SnowmanZA FirstChildCreateTimeTAG: 2012-11-06T15:45:54Z IndexTAG: 1371 TitleTAG: Completed Completed week 7 lab and Hw.Having a lot of fun in this course,and i especially want to thank all the staffs for extending the deadline of week 7 for those who are affected by hurricane sandy . UserIdTAG: 150120 UserNameTAG: krishna1993 CreateTimeTAG: 2012-11-05T13:39:04Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Congrats...:) > **Regards:** asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-11-06T06:33:21Z IndexTAG: 1372 TitleTAG: Solar Cells, Fuel Cells and Batteries Course I just discovered this new online course from Stanford called, Solar Cells, Fuel Cells and Batteries http://class.stanford.edu/ Class started Oct 8th, so I missed the beginning, and the first assignment. However, better late than never. The course seems like a lot of fun. Check it out if if catches your fancy. Yeah, Stanford starts up yet another online site...this one is internal though. More things to do, more places to check out. UserIdTAG: 4463 UserNameTAG: pmj CreateTimeTAG: 2012-11-05T04:02:22Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Gud Job...:) > **Regards:** asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-11-05T13:15:43Z IndexTAG: 1373 TitleTAG: STAFF Based on the latest tasks for homework and laboratory as seen mathematical part of job is becoming more difficult. Could you give us more information and exercises to train than those that already is in the textbook. Maybe some links. Maybe in tutorials. UserIdTAG: 96873 UserNameTAG: Santyaga CreateTimeTAG: 2012-11-04T20:13:51Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: [WolframAlpha][1] is your friend. [1]: http://www.wolframalpha.com FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-11-05T05:03:01Z SecondChildTAG: Definitely! Also, check out the Khan academy to become familiar with the concepts http://www.khanacademy.org/ Take a look here for some help with LAB 7 Q-1 https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/508da86cd8a1b11f0000006c SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-05T13:02:36Z SecondChildTAG: check out Paul's Online Math Notes too SecondChildUserIdTAG: 882923 SecondChildUserNameTAG: jfv33 SecondChildCreateTimeTAG: 2012-12-18T07:06:33Z FirstChildTAG: Thank you! It`s realy helpfull! FirstChildUserIdTAG: 96873 FirstChildUserNameTAG: Santyaga FirstChildCreateTimeTAG: 2012-11-11T17:53:46Z IndexTAG: 1374 TitleTAG: Late! Hi,fellows.... i just started in edx and see that it's so much good related to the methodology of this course, I hope to complete all the assignments on time... UserIdTAG: 737053 UserNameTAG: jhonrodriguez92 CreateTimeTAG: 2012-11-04T19:06:26Z VoteTAG: 2 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: Hi jhonrodriguez92! Welcome to 6.002x! Unfortunately, is really late ... you have missed the deadline of Homewors and Labs from Week 1 to Week 7 and also the Midterm Exam too ... If you take a look at Course info, they wrote this: **Closing Enrollment:** *Although we usually like to give students the chance to do impressive and unlikely feats of learning, the due date for week 7 assignments marks the time when it is impossible to earn enough points for a certificate, and we will therefore not be accepting any new registrations for the course. Good luck to everyone who is still in the running!* [read here][1] But you can follow this Course, remember that this Course will be offered again in Spring. My best wish to you, Myriam. P.D: by the way an if you are interested, there is another Course on edX that Starts today :). CS184.1x: Foundations of Computer : ![enter image description here][2] You can check that [here][3] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/info [2]: https://www.edx.org/static/content-berkeley-cs184x/images/course_image.71b14e30694a.jpg [3]: https://www.edx.org/courses/BerkeleyX/CS184.1x/2012_Fall/about FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-04T20:31:56Z IndexTAG: 1375 TitleTAG: Completed Just completed 100% both the Lab and homework UserIdTAG: 342135 UserNameTAG: vikash902 CreateTimeTAG: 2012-11-04T18:21:19Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Well done vikash902! :) FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-04T18:53:11Z IndexTAG: 1376 TitleTAG: Music... Your probably too old for this course if you recognized the rock music.... Released March 8, 1994. Mr. Self Destruct, The Downward Spiral, Nine Inch Nails. The former song is Britney Spears, You Drive me Crazy (I believe this is the song - I've never heard it before... I swear) UserIdTAG: 299258 UserNameTAG: bannerts CreateTimeTAG: 2012-11-04T09:44:45Z VoteTAG: 2 CoursewareTAG: Week 6 / Amplifier operating point CommentableIdTAG: 6002x_amplifier_operating_point NumberOfReplyTAG: 0 IndexTAG: 1377 TitleTAG: Lab 7-probably just doing something stupid as usual... I'm evaluating the integral, and I got something like -cos(1000*2*PI*t)/(1000*2*PI*C) I don't see what's wrong with that... probably something with the negative infinity. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-11-04T00:21:14Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hi, Can I help you? A little help, hints: In the statement they give you this integral: $v_C(t) = \frac{1}{C} \int^{t}_{-\infty} i(t)dt$ Ok, if you keep reading they tell you : "Assume that **I(t)=0 for t≤0** and **simplify your equation** accordingly" So, what does it mean for t≤0? isn't it the same to say from $t= -\infty$ to t=0 ? ;). So, now, can you split the integral? like something this? $v_C(t) = \color{red}{\int^{0}_{-\infty} somethingdt} + \int^{t}_{0} somethingdt$ And... so if I(t)=0 for t≤0 can you anulate one term and simplify your integral? (take a look at the red color)yes! ;) Also, you can take a look at Lab7 Hints [here][1]. I hope this can help you :). Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5091dd5d3fc09f2b0000018a FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-04T00:45:12Z SecondChildTAG: I have the same result for the integral other than also subtracting the value at 0. This equation gives me the right answer when substituting 0.0005, but the checker won't accept it. Any idea what I'm doing wrong? SecondChildUserIdTAG: 308617 SecondChildUserNameTAG: jrosenberger SecondChildCreateTimeTAG: 2012-11-04T01:12:33Z SecondChildTAG: Hi @jrosenberger, Can I help you? Have you obtained the correct expression of vC(t)? Ok, Now, the following part is requesting you the value of vC in the time t. But here you have to be careful as they are requesting you to write the answer with 6 digits of precision (try to not round the valu of $\pi$) and also be careful of radians, they are not in sexagesimal degrees , so if you are using a calculator at your home you should change DEG mode to RAD mode... or you can convert radians to degrees and calculate the resul... I hope this can help you... SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-04T01:21:52Z SecondChildTAG: *value and *result SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-04T01:22:54Z SecondChildTAG: OK, I see what you are saying. So, just using the second integral, I get the same result as above for the t part, and then just 1 for the 0 part. It still says it's wrong. SecondChildUserIdTAG: 344331 SecondChildUserNameTAG: intellectualwanderer SecondChildCreateTimeTAG: 2012-11-04T01:25:32Z SecondChildTAG: Hi @ intellectualwanderer, I can tell you if that is correct or incorrect as this is a graded part... but, have you replaced the value of C in your result? SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-04T01:30:39Z SecondChildTAG: *can not SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-04T01:32:59Z SecondChildTAG: D'oh! SecondChildUserIdTAG: 344331 SecondChildUserNameTAG: intellectualwanderer SecondChildCreateTimeTAG: 2012-11-04T01:38:53Z SecondChildTAG: Hahaha! ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-04T01:46:46Z SecondChildTAG: Ok, before i saw this posts, I did the same with the two integrals and I get this answer: ((-cos(2000*pi*t))/(2*pi))+1 and still can't see what I'm doing wrong... Any hints? Thanks.... SecondChildUserIdTAG: 342966 SecondChildUserNameTAG: SandraNavarro SecondChildCreateTimeTAG: 2012-11-04T17:13:41Z SecondChildTAG: Sandra.. you must also keep in mind the C term and its value. Please re check your equation..there is one term being multiplied throughout the eqn itself. SecondChildUserIdTAG: 338685 SecondChildUserNameTAG: varshaD SecondChildCreateTimeTAG: 2012-11-04T17:40:33Z SecondChildTAG: I did.... anyway, i finally got it, thanks!!!! SecondChildUserIdTAG: 342966 SecondChildUserNameTAG: SandraNavarro SecondChildCreateTimeTAG: 2012-11-04T17:56:23Z FirstChildTAG: I had the same issue, The answer I got by substituting pi value as 3.14 was not accepted. But then I tried calculating with pi as 22/7 without rounding . Hurray! I got the answer. FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-11-04T09:29:02Z SecondChildTAG: "But then I tried calculating with pi as 22/7 without rounding" I find this very useful, thanks. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-11-04T11:39:57Z SecondChildTAG: :-) SecondChildUserIdTAG: 148523 SecondChildUserNameTAG: span993 SecondChildCreateTimeTAG: 2012-11-04T12:53:55Z IndexTAG: 1378 TitleTAG: another lab 7 formula checker problem the formula is green , but substituting with 0.0005 gives wrong answer ,what shall i do!!? UserIdTAG: 185715 UserNameTAG: amirengineer CreateTimeTAG: 2012-11-03T20:06:12Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi amirengineer, Are you calculating correctly the expression? Be careful, your angle is expressed in radians... and also the answer has to have 6 digits of precision (try to not round the value of $\pi$)... I hope this can help you. Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-03T20:39:35Z SecondChildTAG: i though the angel in degree...:), thanks alot SecondChildUserIdTAG: 185715 SecondChildUserNameTAG: amirengineer SecondChildCreateTimeTAG: 2012-11-03T20:46:50Z SecondChildTAG: You are welcome @amirengineer :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-03T21:12:49Z IndexTAG: 1379 TitleTAG: H8P1 According to the equation №10.137 in textbook we have Vc(t)= Vin(t)*(1-e^(-t/"tau"))? where tau is a time constant = R*C. But when I substitute numbers for this equation I have red dead mark unfortunatly. So where is a mistake? Also please give me some hints about Vr! UserIdTAG: 96873 UserNameTAG: Santyaga CreateTimeTAG: 2012-11-03T19:51:27Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: il faut metre V= 10^(5) ET T= 10^(-5)POUR LA QUESTION B ET D FirstChildUserIdTAG: 316899 FirstChildUserNameTAG: elou FirstChildCreateTimeTAG: 2012-11-03T23:31:43Z SecondChildTAG: what do you mean? SecondChildUserIdTAG: 96873 SecondChildUserNameTAG: Santyaga SecondChildCreateTimeTAG: 2012-11-04T13:12:30Z SecondChildTAG: POUR CALCULER VC(0+) A T = 0+ CHOIX VIN(t)= 100000V ET t= 0.00001S SecondChildUserIdTAG: 316899 SecondChildUserNameTAG: elou SecondChildCreateTimeTAG: 2012-11-04T18:37:22Z SecondChildTAG: merci SecondChildUserIdTAG: 96873 SecondChildUserNameTAG: Santyaga SecondChildCreateTimeTAG: 2012-11-04T19:46:13Z SecondChildTAG: but, steel get wrong answer :-( SecondChildUserIdTAG: 96873 SecondChildUserNameTAG: Santyaga SecondChildCreateTimeTAG: 2012-11-04T20:04:50Z SecondChildTAG: elou, sorry! Last post is incorrect it was my big mistake :-( Get it! SecondChildUserIdTAG: 96873 SecondChildUserNameTAG: Santyaga SecondChildCreateTimeTAG: 2012-11-04T20:22:35Z SecondChildTAG: vc(t)= vin(t)*(1-e^(-t/RC)) SecondChildUserIdTAG: 316899 SecondChildUserNameTAG: elou SecondChildCreateTimeTAG: 2012-11-05T01:17:29Z IndexTAG: 1380 TitleTAG: no graded certificate ? is this true ? Could someone please confirm this, because I already know this stuff, and if there is no graded certificate, i dont think there is anything left for me in this course once i cross the passing score ? UserIdTAG: 214085 UserNameTAG: shohin CreateTimeTAG: 2012-11-03T15:12:39Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hi shohin, This Course will give you a honor-code certificate free of charge if you pass the Course. You can [read here][1], part 10. Also, you can have another alternative, edX offer learners the option of taking a proctored final exam (this will not be free of charge) in Pearson VUE centers. You can read more [here][2] I hope this can help you. Myriam. [1]: https://www.edx.org/static/content-mit-6002x/handouts/syllabus.a477535058a1.pdf [2]: https://www.edx.org/press/edX-announces-proctored-exam-testing FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-03T15:21:36Z SecondChildTAG: Hey Myrimit Will the certificate have the grade?..or just that we took the course? SecondChildUserIdTAG: 173147 SecondChildUserNameTAG: cruiser_rahit SecondChildCreateTimeTAG: 2012-11-03T17:53:04Z SecondChildTAG: Hi @cruiser_rahit, Last semester they gave us a certificate with grade and another one without showing the grade ;). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-03T21:10:29Z SecondChildTAG: thats convenient :) ..thanks for replying.. SecondChildUserIdTAG: 173147 SecondChildUserNameTAG: cruiser_rahit SecondChildCreateTimeTAG: 2012-11-10T08:42:25Z FirstChildTAG: Well, one way to "prove" to yourself that you already know this stuff is to try to obtain 100% on everything. As Ronald Reagan would say, "Trust, but verify!" You trust you know, so verify it ! Otherwise, it's just a belief. Many people believe they can climb mount Everest, but few have verified it for themselves.;) Besides, it's good practice. Refresh your memory. The certificate is just wallpaper, but the neurons in your brain will have been "fixed" to the harmonic tune of the electric circuit resonance, having being struck once again by the challenges of computing the circuit parameters, making recall easier, perhaps even Total Recall this time around the bend. FirstChildUserIdTAG: 4463 FirstChildUserNameTAG: pmj FirstChildCreateTimeTAG: 2012-11-05T03:45:05Z IndexTAG: 1381 TitleTAG: Doubt Regarding Capacitor Acting Short Ckt For Sudden Changes Hi All, In the lecture sequence S15V14 it is said that capacitor acts as a short ckt for the sudden changes and it is shown that voltage across the capacitor is vC = Q/C after taylor series simplification. But my doubt is(intuitively looking at the circuit), if capacitor is acting as a short than the voltage across it should be zero. Kindly clarify whether is it true or is it something else ? so when we say a short ckt are we really saying it at t=0- and (as mentioned in the lecture) the voltage vC=Q/C is at time t=0+ or t=0(when the impulse is applied)? UserIdTAG: 178840 UserNameTAG: mehtanayanv CreateTimeTAG: 2012-11-02T15:33:35Z VoteTAG: 2 CoursewareTAG: Week 8 / Response To Impulse Limit Case CommentableIdTAG: 6002x_Response_To_Impulse_Limit_Case NumberOfReplyTAG: 4 FirstChildTAG: They mean that the capacitor acts like a short circuit only to the sudden change. We know that Ic = C.dv/dt so if dv/dt is very high then the current into the cap will be very large, making it appear like a short. FirstChildUserIdTAG: 270011 FirstChildUserNameTAG: SnowmanZA FirstChildCreateTimeTAG: 2012-11-02T16:04:09Z SecondChildTAG: what about the voltage across the capacitor ? will it rise at t=0 to Q/C value within no time ? SecondChildUserIdTAG: 178840 SecondChildUserNameTAG: mehtanayanv SecondChildCreateTimeTAG: 2012-11-02T16:14:07Z SecondChildTAG: If the source can supply infinite current and the capacitor is ideal (i.e. it has no internal resistance or inductance), then it would be able to rise in no time. In reality, the rise time of the voltage across the capacitor would be limited by the source impedance and the capacitors internal resistance and lead inductances etc. SecondChildUserIdTAG: 270011 SecondChildUserNameTAG: SnowmanZA SecondChildCreateTimeTAG: 2012-11-02T20:15:48Z FirstChildTAG: It only acts like a short during the rapid change, but the charge doesn't flow "through" the capacitor, it collects there. Therefore, it "builds up" a voltage quickly. If it were a true wire short, then the charge would flow through and the voltage would remain 0, like a real short. So, it's a short for changes, but not for stable currents and voltages. FirstChildUserIdTAG: 4463 FirstChildUserNameTAG: pmj FirstChildCreateTimeTAG: 2012-11-02T16:10:10Z SecondChildTAG: so the charge varies twice here in this case, from 0 to Q and then from Q to 0(short impulse).Then it should react to zero change also rapidly.. SecondChildUserIdTAG: 178840 SecondChildUserNameTAG: mehtanayanv SecondChildCreateTimeTAG: 2012-11-02T16:21:44Z SecondChildTAG: It does. It rapidly stays there. ;) SecondChildUserIdTAG: 4463 SecondChildUserNameTAG: pmj SecondChildCreateTimeTAG: 2012-11-02T22:41:28Z FirstChildTAG: Here's a somewhat crude explanation for the operation of a capacitor but it may address the conceptual difficulty (it does for me, at least) of that instantaneous 'short': 1) Consider an actual short circuit. We could say the electrons pass all the way through it, so the same amount of charge 'leaves' the short circuit as enters it. 2) A capacitor accomplishes this by a somewhat different means. The dielectric material allows the electric field of an electron 'entering' the capacitor to cause the displacement an electron from the other side of the capacitor. This is by way of an electric field across the dielectric, rather than by conduction, as with the short circuit. 3) What a capacitor has that a short circuit does not have is the capacity to accumulate charges on one side of the dielectric which displace those like-charges on the other side. This accumulated abundance of negative charge potential on one side and the accumulated lack of negative (i.e., positive) potential on the other side give rise to the voltage across the capacitor. 4) The 'short circuit' condition of a capacitor exists for only an instant. In that same instant the accumulation of charge also begins and the voltage across the capacitor begins to build up and the short circuit disappears in that same instant. 5) All this happens at something near the speed of light, so don't blink or you'll miss it. :) FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-11-04T19:09:28Z SecondChildTAG: Thanks MobiusTruth... SecondChildUserIdTAG: 178840 SecondChildUserNameTAG: mehtanayanv SecondChildCreateTimeTAG: 2012-11-08T17:42:38Z FirstChildTAG: You Can use *L’Hôpital’s Rule* to show that $V_C(T)= \frac QC $ when T is really really small. $$ \large \lim_{T \rightarrow 0} V_C(T) = \lim_{T \to 0}{RQ\left(1-e^{-T/RC}\right)\over T}$$ Since, at T=0 this evaluates to 0/0 we can apply l'hopital's rule: $$= \lim_{T \to 0}{RQ\left(-{1\over RC} \times-e^{-T/RC}\right)\over 1}= \frac QC \cdot e^0 = \color{green}{\frac QC}$$ FirstChildUserIdTAG: 714237 FirstChildUserNameTAG: PaxPolaris FirstChildCreateTimeTAG: 2012-11-11T04:13:40Z IndexTAG: 1382 TitleTAG: Can I still cope up with the lessons? I just enrolled this day. Can I still make it to finish the course? UserIdTAG: 751933 UserNameTAG: ecbeljeda CreateTimeTAG: 2012-11-02T09:42:30Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Even if you complete remaining hw and labs (from 7-12) you would get 18% and since final exam contributes 40% you have maximum 58% which falls short of minimum grade by 2%. But this course will start again next year so this experience will no go waste FirstChildUserIdTAG: 99628 FirstChildUserNameTAG: Pranjal16 FirstChildCreateTimeTAG: 2012-11-02T09:48:54Z SecondChildTAG: Is it probable to make up for the previous lab reports and homework (from 1-6) by submitting them all by next week? SecondChildUserIdTAG: 751933 SecondChildUserNameTAG: ecbeljeda SecondChildCreateTimeTAG: 2012-11-02T10:12:24Z SecondChildTAG: Unfortunately the due date for weeks 1-6 has expired. Hope to see you in the spring. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-11-02T11:58:09Z SecondChildTAG: I understand. Perhaps, I'll enroll in other courses. Thank you for your answers. I really appreciate it. SecondChildUserIdTAG: 751933 SecondChildUserNameTAG: ecbeljeda SecondChildCreateTimeTAG: 2012-11-02T14:21:09Z SecondChildTAG: Check out the CS50x - Introduction to Computer Science offered by Harvard here on edX. If it interests you, it does not have a fixed timeline as long as it is completed before mid-April. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-11-02T14:35:55Z IndexTAG: 1383 TitleTAG: Can i collect all my homeworks and labs in one pdf file Hi, I'm interested in collecting all homeworks and labs done by me in a single pdf file. Sudhakar UserIdTAG: 318832 UserNameTAG: sudhakarV CreateTimeTAG: 2012-11-02T09:13:46Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: I am not sure about PDF, but I have been saving pages in Firefox as "Complete Web Page". It seems to work OK. It will save it as an HTML document and have a folder accompanying to store the information in. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-11-02T12:01:54Z FirstChildTAG: Get the tool "PDF Creator" and install in on your machine, if you're in Windows. Then you can "print" anything to a PDF. So, just print the web page, and choose "PDFCreator" as your printer, and it will print to a pdf file instead of to an actual printer. Then get "Foxit PDF Editor", which will allow you to combine several pdf documents into a single document, and you can make a book of your work, all stored in a single pdf. You can also get an "Image Capture" tool, to save parts of the screen as a .jpg or .gif, and then import these images into your PDF in Foxit PDF Editor. That way you can make any kind of book of your work you'd like. For Mac OS X and/or Linux, there should be similar tools out there. FirstChildUserIdTAG: 4463 FirstChildUserNameTAG: pmj FirstChildCreateTimeTAG: 2012-11-02T16:17:39Z SecondChildTAG: [PDFCreator][1] allows you to "wait" and "collect" pages into one large PDF, so an additional tool should not be necessary. [1]: http://sourceforge.net/projects/pdfcreator/ SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-11-02T17:43:39Z FirstChildTAG: You can also simply print it to a pdf using Firefox. I normally use Chrome, but the formulas and special symbols are not correctly saved to a pdf. Firefox works fine for this if you just want to save you progress (and the explanations provided by the teachers after the due time) in pdf format. FirstChildUserIdTAG: 366320 FirstChildUserNameTAG: Galli FirstChildCreateTimeTAG: 2012-11-02T23:09:15Z FirstChildTAG: A good, free print-to-PDF app I use for Windoze is doPDF. It's free for personal and commercial use. It can be downloaded from: http://www.dopdf.com/download.php A student in the pilot version of 6.002x said he used CutePDF. It can be downloaded from: http://www.cutepdf.com/ Another good print-to-PDF application is PDF Creator: http://sourceforge.net/projects/pdfcreator/ Any other "print-to-PDF" application will likely work, there are many to choose from for nearly every modern operating system. Use Google to find them. The application behaves as if it is a printer driver, it will show up as a printer when you click on 'Print'. After selecting the print-to-PDF application, it will ask you for a filename to save the page. You can then view the PDF online or print it out. I start the week by saving the sequence exercises and homeworks to PDF, and print them out so I can work on them at my leisure. After I complete them and the deadline has passed, I click on 'Show Answer' for each problem and save them to PDF again. For practice problems, I do a 'Check & Show Answer' prior to printing to PDF, so the solutions are on the same page. You might want to save two versions, one with and one without the solutions. You can merge the individual PDFs into one file. There are several good free open-source tools to do this. PDF Binder: http://code.google.com/p/pdfbinder/ PDF Split and Merge: http://www.pdfsam.org/ FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-11-05T00:29:07Z IndexTAG: 1384 TitleTAG: A trick: if you consider VI = Initial value VF = Final value then the equation is often the same ( no matter if it climbs or decays ) : VF + (VI - VF )* e^(-t/RC) VI is given ( most of time ) VF easy to guess ( capacitor is open after long time ) UserIdTAG: 373752 UserNameTAG: Croak CreateTimeTAG: 2012-11-02T04:25:14Z VoteTAG: 2 CoursewareTAG: Week 8 / S16V1 Review CommentableIdTAG: 6002x_S16V1_Review NumberOfReplyTAG: 0 IndexTAG: 1385 TitleTAG: H8 p1 Even after Using all my resources and watching the video lectures twice ,I could not proceed after the first part(a).can anyone give some hints or the textbook pages which could be useful .Very confused after trying to solve it in almost all ways I could think of.thank you in advance. UserIdTAG: 99628 UserNameTAG: Pranjal16 CreateTimeTAG: 2012-11-01T19:28:59Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 6 FirstChildTAG: same here.... u can read 10.1.3 Series RC circuit step input n 10.2.1 series RL circuit step input.... but i am not getting how to form this kind of equation for impulse response.... FirstChildUserIdTAG: 51259 FirstChildUserNameTAG: Wardah FirstChildCreateTimeTAG: 2012-11-01T19:55:14Z FirstChildTAG: The voltage impulse will deliver flux linkage to the inductor establishing an initial condition current on it with the value (Flux Linkage/Inductance). FirstChildUserIdTAG: 127195 FirstChildUserNameTAG: princeofsudan FirstChildCreateTimeTAG: 2012-11-01T20:36:31Z FirstChildTAG: There are various ways to solve the problem. The easiest, in my opinion, is to solve for a unit voltage step input and then differentiate with respect to time. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-01T21:34:02Z FirstChildTAG: Test it out in the sandbox. That will give you a good qualitative understanding of the behavior. The result was a little unintuitive, but covered in the text. Compare the behavior to the current impulse. FirstChildUserIdTAG: 142857 FirstChildUserNameTAG: viking2 FirstChildCreateTimeTAG: 2012-11-02T09:52:04Z FirstChildTAG: Like some of you i have been very confused as to what concepts apply here, but i don't recommend the sandbox for accepted answers (because i didn't get them with the sandbox). I did it out of the same kind of desperation as can be felt in some of the posts. I'll get to a suggestion in 0+, but here's an example of my sandbox attempts: ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13521778671343699.png Here's my suggestion: to me the critical issue has been how to find out the A area for the C charge and/or the L flux linkage. In retrospect the reasoning is understandable (merci particulièrement à *elou*), but i found nowhere in the textbook or lecture videos a formula to calculate it. To be plain: the voltage unit of 1V-sec gives us, for example, Vs=1e5 (or 10ˆ5) **and** T(Vs)=1e-5 (Vs*T(Vs)=1). Use those values in the formula we know well by now, e.g vc=Vs.(1-eˆ(-t/RC) as Vo=0 and with t0+=T, and you should find acceptable answers for everything (for L use the correct tau formula instead of RC) and every time (i.e. 0+ or 1ms). Note that acceptable answers require both rising and falling time formula. If you get rejects or have any doubt, try the other (truth be told, i'm still struggling with understanding this part clearly myself). FirstChildUserIdTAG: 184153 FirstChildUserNameTAG: tthngn FirstChildCreateTimeTAG: 2012-11-06T05:17:13Z SecondChildTAG: L value is 5mH,it is not 5nH.Right? SecondChildUserIdTAG: 282828 SecondChildUserNameTAG: Chibolator SecondChildCreateTimeTAG: 2012-11-06T08:21:18Z SecondChildTAG: Take the equation of iL(t) equation(10.59) and then differentiate it w.r.t (t)... while in 2nd part the t=0 and in 3rd part it is given to be some ms..... then multiply it with R1 = VR1..... In the 4th n 5th part.... take the equation of Vc(t) equation (10.37) and differentiate it.... Hint is that the Vo is zero for impulse as it is 0 for all the value before 0.... V is the instantaneous value..... i.e. given value for delta..... SecondChildUserIdTAG: 51259 SecondChildUserNameTAG: Wardah SecondChildCreateTimeTAG: 2012-11-09T11:46:22Z FirstChildTAG: Take the equation of iL(t) equation(10.59) and then differentiate it w.r.t (t)... while in 2nd part the t=0 and in 3rd part it is given to be some ms..... then multiply it with R1 = VR1..... In the 4th n 5th part.... take the equation of Vc(t) equation (10.37) and differentiate it.... Hint is that the Vo is zero for impulse as it is 0 for all the value before 0.... V is the instantaneous value..... i.e. given value for delta..... FirstChildUserIdTAG: 51259 FirstChildUserNameTAG: Wardah FirstChildCreateTimeTAG: 2012-11-09T11:49:50Z IndexTAG: 1386 TitleTAG: Voltage Divider I don't find that way of explanation with voltage divider really clear and intuitive. First of all if we want to consider a voltage divider viewed from that circuit topology it should be something like 1000/1001 -> drop of 4.995 Volts. If we have a switch on 100 MHz that means we have now the power voltage swinging between 5 V and 4.995 V, which is the 5mV tutors are talking about. I'm fine with the rest. UserIdTAG: 415375 UserNameTAG: ZWX CreateTimeTAG: 2012-10-31T18:50:29Z VoteTAG: 2 CoursewareTAG: Week 7 / Bypass Capacitor CommentableIdTAG: 6002x_Bypass_Capacitor NumberOfReplyTAG: 1 FirstChildTAG: Very much agreed. Not very intuitive problem outline. FirstChildUserIdTAG: 398594 FirstChildUserNameTAG: DaveyJC FirstChildCreateTimeTAG: 2012-11-02T09:14:56Z IndexTAG: 1387 TitleTAG: "' Unofficial Survey Deadline '4 Nov 2012' "' Follow post [here][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/508e0829ba8f10260000000c UserIdTAG: 374590 UserNameTAG: ahope CreateTimeTAG: 2012-10-30T14:07:34Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1388 TitleTAG: Mistake I think in the second problem the answer is incorrect. That time is from 5V to 2V or from 4V to 1V. From 4V to 2V it should be 6.93ps. UserIdTAG: 359677 UserNameTAG: guillermb CreateTimeTAG: 2012-10-30T13:16:23Z VoteTAG: 2 CoursewareTAG: Week 8 / Charging And Discharging CommentableIdTAG: 6002x_Charging_And_Discharging NumberOfReplyTAG: 3 FirstChildTAG: Rethink! FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-10-30T16:04:00Z FirstChildTAG: In the second case, VOH corresponds to the initial state of the capacitor. Remember that vc(0)=VI+(VO-VI)*e^(-t/R*C) then t=-C*R*ln((vc(0)-VI)/(VO-VI)) where VI=**dIN** vc(0)=**VOH** VO=**VIL** After that you can find **t** very simple FirstChildUserIdTAG: 220304 FirstChildUserNameTAG: sergei_m FirstChildCreateTimeTAG: 2012-11-02T20:45:20Z SecondChildTAG: Please note that I think sergei_m's formula is wrong, it should be: Vc(0)=VO+(VI-VO)*e^(-t/R*C) and assign: Vc(O) = VIL, VO = VOH, VI = dIN If using the formula: VIL = VOH + (dIN - VOH)*e^-t/RC we have: 2 = 4 + (1 - 4) * e^t/RC and t will be: RC*ln(2/3) or 10*ln(2/3), yielding a wrong result. If however we use (1 - e^-t/RC) instead of just e^-t/RC, we got RC*ln(1/3) and it's OK. Is it? Lectures suggest that (1 - e^-t/RC) should be used for raising voltage while e^-t/RC correspond to the case of a discharging circuit. SecondChildUserIdTAG: 296965 SecondChildUserNameTAG: LGMailhos SecondChildCreateTimeTAG: 2012-11-10T17:26:21Z FirstChildTAG: the question has asked you to calculate the time between 1 and 4 volt. so you should calculate T once for 1 volt and once for 4 volt. then T4-T1 is the answer. it is a ZSI situation 4=5(1-e^(-T4/RC)), 1=5(1-e^(-T1/RC)) => T4-T1=13.86*e^-12 I hope it does not violate the honor code. FirstChildUserIdTAG: 374393 FirstChildUserNameTAG: rmaleki FirstChildCreateTimeTAG: 2012-11-13T16:32:57Z IndexTAG: 1389 TitleTAG: Continuity Um, did I just miss some videos? UserIdTAG: 740649 UserNameTAG: hqjb CreateTimeTAG: 2012-10-30T12:52:30Z VoteTAG: 2 CoursewareTAG: Week 1 / AC power CommentableIdTAG: 6002x_S1E3_AC_power NumberOfReplyTAG: 0 IndexTAG: 1390 TitleTAG: Where do I find solutions to Mid term exam? When will be the progress published? UserIdTAG: 318832 UserNameTAG: sudhakarV CreateTimeTAG: 2012-10-30T09:23:54Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/508e8e5225f1ef2700000063 See if this helps.. FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-10-30T11:58:53Z SecondChildTAG: I should have linked to your thread Span, thanks. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-30T12:07:26Z SecondChildTAG: :) SecondChildUserIdTAG: 148523 SecondChildUserNameTAG: span993 SecondChildCreateTimeTAG: 2012-10-30T12:14:21Z FirstChildTAG: And regarding the progress, Don't worry it gets updated instantaneously if you submit your answers/ work. FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-10-30T12:03:09Z FirstChildTAG: They should be available some time today. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-30T11:22:30Z IndexTAG: 1391 TitleTAG: Large download video size I can play the videos over my EDGE connection okay, but when I try to download them they are HUGE. Can anybody help me to find a way to download the HTML5 video rather than the .mov files provided in the download links. Thanks UserIdTAG: 623210 UserNameTAG: preveen CreateTimeTAG: 2012-10-30T09:10:49Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I was using IDM,(Internet Download Manager) and downloading them as a .flv file. The average size ends up being around 10KB. No affiliation. While it is nice to have it available in detail, I agree that it would be good to have an alternate smaller file size for direct download. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-30T11:09:35Z SecondChildTAG: 10kB? SecondChildUserIdTAG: 623210 SecondChildUserNameTAG: preveen SecondChildCreateTimeTAG: 2012-10-30T12:11:27Z SecondChildTAG: I think He intended it to be 10 MB or something, Defenitely not in Kb SecondChildUserIdTAG: 148523 SecondChildUserNameTAG: span993 SecondChildCreateTimeTAG: 2012-10-30T12:13:17Z SecondChildTAG: Just about the right size. Lyla Please, can you this for us, $bandwidth$ $challenged$ students? Thanks SecondChildUserIdTAG: 623210 SecondChildUserNameTAG: preveen SecondChildCreateTimeTAG: 2012-10-30T12:18:07Z SecondChildTAG: Sorry about that, yes 10,000KB or around 10MB. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-30T12:24:57Z IndexTAG: 1392 TitleTAG: To STAFF When will the answers for the midterm exam will be available sir? UserIdTAG: 150120 UserNameTAG: krishna1993 CreateTimeTAG: 2012-10-30T07:24:18Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: They should be available sometime today. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-30T11:23:26Z FirstChildTAG: Should be out now, thanks for your patience FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-10-30T15:43:50Z FirstChildTAG: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/508e8e5225f1ef2700000063 This is not official, but this might give you some idea.. FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-10-30T11:59:47Z IndexTAG: 1393 TitleTAG: How to write the expression Can anyone tell me how to write the expression for diL/dt? UserIdTAG: 150120 UserNameTAG: krishna1993 CreateTimeTAG: 2012-10-30T06:38:43Z VoteTAG: 2 CoursewareTAG: Week 7 / Thevenin Inductor Circuit CommentableIdTAG: 6002x_Thevenin_Inductor_Circuit NumberOfReplyTAG: 1 FirstChildTAG: If you meant how to calculate it: Write KVL : $V = v_R + v_L = Ri_L + L\frac{di_L}{dt}$ Now solve this and you get $\frac{di_L}{dt} = \frac{V-Ri_L}{L}$ FirstChildUserIdTAG: 329254 FirstChildUserNameTAG: KGabor FirstChildCreateTimeTAG: 2012-10-30T10:51:15Z SecondChildTAG: If we pick the equation: R/L*($di_L$/dt) + $i_L$ = $I_I$ and replaces $I_I$ for an expression related to voltage, we can easily write it. SecondChildUserIdTAG: 296965 SecondChildUserNameTAG: LGMailhos SecondChildCreateTimeTAG: 2012-10-31T14:39:25Z SecondChildTAG: Don't we also need the analog to I0 (the current through the inductor at t = 0)? SecondChildUserIdTAG: 278463 SecondChildUserNameTAG: zniazi SecondChildCreateTimeTAG: 2012-10-31T21:10:21Z IndexTAG: 1394 TitleTAG: Midterm answers When will the answers of midterm exams will be realesed? UserIdTAG: 150120 UserNameTAG: krishna1993 CreateTimeTAG: 2012-10-30T04:38:41Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/508e8e5225f1ef2700000063 Hope this might be of some help. FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-10-30T12:02:14Z FirstChildTAG: They should be available sometime today. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-30T11:24:11Z SecondChildTAG: You are saying this for the past few days...but still its's not available.please do it soon SecondChildUserIdTAG: 150120 SecondChildUserNameTAG: krishna1993 SecondChildCreateTimeTAG: 2012-10-30T15:13:06Z FirstChildTAG: should be out now. Thanks for your patience. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-10-30T15:43:35Z SecondChildTAG: Thankyou Sir SecondChildUserIdTAG: 150120 SecondChildUserNameTAG: krishna1993 SecondChildCreateTimeTAG: 2012-11-01T04:35:03Z IndexTAG: 1395 TitleTAG: Real midterm grade In order to find out if I was able to get through the real midterm, I simulated it. 2 hours and first check grade (+another answers that I did not check but I had already completed before the end). I scored 75% Has anybody else simulated the midterm? UserIdTAG: 309359 UserNameTAG: abarea10 CreateTimeTAG: 2012-10-29T15:35:39Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: No, I took advantage of the allotted time frame. I suppose I was around 55% after the first 2 hours, if that helps. If I knew I only had 2 hours, I may have approached the questions in a different order, which may have changed my score at the 2-hour point. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-29T15:47:14Z SecondChildTAG: I would have approached it differently as well. I would have made much more stupid algebraic and sign (+/-) errors if I had to rush. I enjoyed the mid-term particularly because I got to take a 15-minute break after each question, and an hour break for lunch. Approximately. Part of the reason why I do not do so well on timed exams is focus. If I have time out to clear my head and think of a problem from a fresh perspective, I can solve many difficult problems. If I am forced to "plow through" an timed exam at an unnatural pace, I lose focus and I am more prone to error and giving up. Having a full stomach and snacks during the exam definitely helps, too. But there is a point where taking too long would tire you out. For me, the perfect amount of time is four hours. I can approach six questions leisurely, and not tire or bore myself out, yet not rush and finish with less than a perfect score. SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-10-29T16:33:48Z SecondChildTAG: About 40% in the first two hours and finally I just scored 62%. SecondChildUserIdTAG: 342966 SecondChildUserNameTAG: SandraNavarro SecondChildCreateTimeTAG: 2012-10-30T15:29:08Z FirstChildTAG: I certainly didn't. I made sure that when I opened the exam I would have the full 24 hours if I needed it. I opened the exam about 7:30 Saturday morning and looked over the exam to see how hard it was. I decided that I could probably do all the problems and started working on the last one, which looked like it was the most difficult. After about 30 minutes I put the whole thing aside and went about my normal Saturday morning errands. About 11:30 I again took a look at the exam while eating my lunch. I finally got around to devoting my full attention to the exam around 12:30 in the afternoon. During that time I did a consistency check on problem 1, then did problems 2, 3, 4, and 5 rather quickly and completed the calculation of Rth for problem 6. I held off entering all my final answers until I attended to more personal business. Finally, a little past 6:30 in the evening I entered my final answers. All my answers were correct on the first try with the exception of the amplifier gain where I neglected to initially enter the minus sign. That was corrected in less that 30 seconds. So, my elapsed time for the exam was about 11 hours. Why rush? there were no extra points given for finishing quickly. Knowing that I had the time available, I did as many checks as I could image on my answers before entering them for scoring. If I had been limited to exactly two hours my stategy would have been different. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-29T16:17:02Z FirstChildTAG: how to calculate the real score if there is one ?? FirstChildUserIdTAG: 309933 FirstChildUserNameTAG: miramar FirstChildCreateTimeTAG: 2012-10-29T17:11:11Z SecondChildTAG: He meant the score you got after 2hours making the exam SecondChildUserIdTAG: 308853 SecondChildUserNameTAG: fmorato SecondChildCreateTimeTAG: 2012-10-29T23:49:44Z IndexTAG: 1396 TitleTAG: missed midterm Did anybody else miss midterm like me, for unexplainable reasons. Working hard to get at least 60% out of the remaining 70% UserIdTAG: 183507 UserNameTAG: obiradaniel CreateTimeTAG: 2012-10-29T13:50:39Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: Good Luck with the remain hw, lab and the final exam. FirstChildUserIdTAG: 623210 FirstChildUserNameTAG: preveen FirstChildCreateTimeTAG: 2012-10-29T14:01:24Z FirstChildTAG: sorry to hear it bro FirstChildUserIdTAG: 92895 FirstChildUserNameTAG: shihab2555 FirstChildCreateTimeTAG: 2012-10-29T13:54:04Z FirstChildTAG: don't give up.. FirstChildUserIdTAG: 114864 FirstChildUserNameTAG: Vikaash FirstChildCreateTimeTAG: 2012-10-29T14:04:58Z FirstChildTAG: **obiradaniel:** Sorry to hear that you missed the mid-term. The course will be offered again in the next semester, Spring. If you do not pass this semester's class, there's always next semester, where you can be more careful about the midterm deadline. If you do pass this year's class with a 60-70%, then you can improve your grade in the Spring, also. You will be at an advantage because you already know some or most of the material, depending on your mastery of it. FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-10-29T15:11:39Z SecondChildTAG: thanks JerseyMark and all the rest. will try but will definitely have to offer the course again. SecondChildUserIdTAG: 183507 SecondChildUserNameTAG: obiradaniel SecondChildCreateTimeTAG: 2012-11-01T18:24:13Z FirstChildTAG: Try sticking by to this course. Im sure it could help you gain some experience, if you are planning to take this course ne xt time FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-10-29T15:31:53Z IndexTAG: 1397 TitleTAG: [STAFF] Validating the certificate Hi staff! I'd like to suggest to you to implement some kind of validating process to the 6.002x future certificates. I study at Unicamp, Brazil, and here we can do this using an alphanumeric code. It's really useful sometimes! Thanks! UserIdTAG: 288637 UserNameTAG: gotchi CreateTimeTAG: 2012-10-29T11:13:09Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: **Document Security / Validation** **gotchi:** I believe that each 6.002x "certificate" has a unique url on it (e.g. web address) where if you type in that url into a browser, you will see that the certificate is valid. I know that here, where I live, the State of New Jersey has recently been using the same technology to validate *apostilles*. This is the attesting documentation that we receive to attach to our birth certificates and other State documents to validate that they are genuine (when we travel abroad) and that present the information in a common standard (according to the Apostille convention) so that foreign nations can interpret them. Note that the birth certificates and other State documents still have the "old-fashioned" but "better-looking" embossed seal and are printed on special paper, while the *apostille* for these certificates is printed on ordinary paper and looks like anyone can print it out on a good color laser printer. But once you type in the url, it gives the serial number and an online validation through the internet. We can argue whether mechanical means (i.e. seals, imprints, special paper, ultraviolet ink, holograms) or electronic means (i.e. encryption, online validation, alphanumeric code) are better; forgers can easily duplicate the former, while hackers can get around the latter. In the U.S. (at least in N.J. as far as I know), we use both methods in combination as I just explained. Jersey Mark FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-10-29T11:50:30Z SecondChildTAG: There's indeed a lot to think about it. But at least one kind of validation is better than none. =) Thank you for your answer. SecondChildUserIdTAG: 288637 SecondChildUserNameTAG: gotchi SecondChildCreateTimeTAG: 2012-10-29T11:55:54Z SecondChildTAG: cisco provides a number and you can check the number on their website to check to whom the certificate is awarded to. Also from their website you can send a reference to any email about your certification (after you login) SecondChildUserIdTAG: 623210 SecondChildUserNameTAG: preveen SecondChildCreateTimeTAG: 2012-10-29T11:56:42Z IndexTAG: 1398 TitleTAG: Tutor needed to explain maths part of this course. Hi EECS, Midterm done and dusted, congrats to everyone who did well and still here past the half way mark. I did fine for questions 1, 2, 3. but failed miserably for Fet amplifier and dependant source problems, looking back over homework, I see the values I scored in these sections (weeks 5 and 6) were heavily math related, even the ones I got right I didn't get the exact answer and for the wrong reason (despite spending days revising this.) My total now after 6 weeks and midterm stands at 34%, at the end I'd be happy with a pass, and delighted to snatch a B grade. Analogue electronics was never my strong point so to pull a pass out of this would be a great achievement for me. That said I'm sure the course is not going to get any easier, and while the content videos, reading materials, and tutorials are great, I find this course differs from a brick and mortar course in that I can't directly interact to overcome ones shortcommings, even on the forums which are great for hints etc. I want to be clear that I adhere to the honour code and am not looking for answers, but for a friendly study group or tutor via skype/google docs to guide me through the sticking points. I can pay for tuition, or repay tuition for nearly any subject IT or computer related. e-mail me on Spacedog [at] gubuwire.com if interested. UserIdTAG: 61469 UserNameTAG: Spacedog CreateTimeTAG: 2012-10-29T10:23:29Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: Hi Spacedog, more or less the same for me. Mid Term 62% and Overall 30%. BUT I admit, even if I had just a just a little knowdlege of EECS, I could have done more EFFORT. I guess for personal reasons I didn't. BUT I have the same problem with Maths. I'm very autodidact (self-taught) and THERE I put effort indeed. But was still having some difficulties.... So I would join you. Regards, Sandra FirstChildUserIdTAG: 342966 FirstChildUserNameTAG: SandraNavarro FirstChildCreateTimeTAG: 2012-10-29T11:12:57Z SecondChildTAG: The solutions involve minimal math, refer to it and please feel free to ask if you get stuck anywhere. https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/508e8e5225f1ef2700000063 SecondChildUserIdTAG: 148523 SecondChildUserNameTAG: span993 SecondChildCreateTimeTAG: 2012-10-29T14:19:03Z FirstChildTAG: Please specify where you are getting stuck. There are a lot of very kind senior people here. I am sure they are willing to help. Personally, I don't think there are any math so far. (No offense) Please do not try to do your steps mentally. Do it on paper/notebook and show the forum where you are getting stuck FirstChildUserIdTAG: 623210 FirstChildUserNameTAG: preveen FirstChildCreateTimeTAG: 2012-10-29T11:43:18Z SecondChildTAG: Thanks, I'll do it like this. But some prerequisites are supposed, so, I don't really "dare" to ask many things in the discussion forum. Thank you!! SecondChildUserIdTAG: 342966 SecondChildUserNameTAG: SandraNavarro SecondChildCreateTimeTAG: 2012-10-29T13:19:55Z SecondChildTAG: Please don't feel there exists that sort of restriction on what you can ask in the forum. Obviously, you can't ask for solutions, but I see NO REASON why you can't ask anything math-related you want. Prerequisites be damned, no one here has exactly the same educational background. Ask away! SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-29T15:25:04Z SecondChildTAG: Wowwww I'm really having fun in this course. Thanks a lot :)) I'll do!! SecondChildUserIdTAG: 342966 SecondChildUserNameTAG: SandraNavarro SecondChildCreateTimeTAG: 2012-10-30T00:10:11Z FirstChildTAG: If I can be of help to anyone, I would be glad to help/tutor for free. Just let me know. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-29T18:01:04Z SecondChildTAG: Thay is nice skyhawk! :). Me too Spacedog, I will try to help too. Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-29T23:44:16Z FirstChildTAG: Try this https://6002x.mitx.mit.edu/wiki/view/MathReview FirstChildUserIdTAG: 9673 FirstChildUserNameTAG: xp42 FirstChildCreateTimeTAG: 2012-10-29T21:13:04Z IndexTAG: 1399 TitleTAG: To EDX Dear EDX , Link(https://www.edx.org/book-shifted/465) to Equation 9.18 in LAB 7 Task 1 doesn't work . Regards . UserIdTAG: 194717 UserNameTAG: kaa CreateTimeTAG: 2012-10-28T23:33:24Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Thanks. We will fix it. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-10-29T01:52:56Z SecondChildTAG: Lyla, it worked fine for me 17 days ago, and it is still working OK now. SecondChildUserIdTAG: 9673 SecondChildUserNameTAG: xp42 SecondChildCreateTimeTAG: 2012-10-29T04:48:03Z IndexTAG: 1400 TitleTAG: Notepad with Math capability Can you guys recommend any notepad app with the math capability like that of our forum. Thanks UserIdTAG: 623210 UserNameTAG: preveen CreateTimeTAG: 2012-10-28T17:51:15Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: use wolfram alpha .com or app for iOS or Android FirstChildUserIdTAG: 95280 FirstChildUserNameTAG: Fortyq FirstChildCreateTimeTAG: 2012-10-28T20:14:44Z FirstChildTAG: For actual calculations, I highly recommend [Wolfram Alpha][1]. For my own notes, I have been looking into [TiddlyWiki][2] with the [MathJax Plugin][3]. ([MathJax][4] is what both this Forum and the 6.002x Wiki uses to display equations.) [1]: http://www.wolframalpha.com [2]: http://tiddlywiki.com/ [3]: http://math-template.tiddlyspace.com/#%5B%5BPluginMathJax%20v1.3%5D%5D [4]: http://www.mathjax.org/ FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-10-28T21:03:22Z SecondChildTAG: thanks planetscape. please give me some pointers to learn $LaTeX$ and $TeX$ SecondChildUserIdTAG: 623210 SecondChildUserNameTAG: preveen SecondChildCreateTimeTAG: 2012-10-29T07:05:20Z SecondChildTAG: I started with 6.002x' own [Wiki Editing Guide][1]; that links to [Wiki MathJax Syntax][2]; and of course the [documentation at the MathJax website][3]. More than you could ever want can also be found by Goggling "MathJax cheat sheet". Hope this helps! [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/edX/wiki-editing-guide/ [2]: http://www.suluclac.com/Wiki+MathJax+Syntax [3]: http://www.mathjax.org/resources/docsindex/ SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-29T15:49:07Z SecondChildTAG: Heh. "Googling", even. ;-) SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-29T15:51:52Z FirstChildTAG: **planetscape:** Thanks! I also have been looking into a way to store mathematical notes and formulas on my PC. So far I've been using the "Math Input Panel" that comes with Windows 7 or Office (I forgot which, you have to enable it as an option though upon install). But that requires a tablet and converts your handwriting, so it's not 100% accurate, and you have to go back and fix any errors. I enjoy inputting in MathJax and using the keyboard; it is faster for me on the laptop. I will try that plug-in. As for Wolfram-Alpha, it is just plain awesome. It's not just for solving equations; but for almost any kind of serious research. I used to use MATLAB a lot, but I lost my copy and a new license, especially with the optional signal processing package, is expensive, even for the student edition; at least to me since I usually get my software free and open-source when I can! FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-10-29T09:53:52Z SecondChildTAG: You are most welcome! So glad when I can help others! SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-29T15:41:55Z SecondChildTAG: JerseyMark: Have you tried Octave? http://www.gnu.org/software/octave/ It's a free MATLAB clone, with very similar syntax. SecondChildUserIdTAG: 270011 SecondChildUserNameTAG: SnowmanZA SecondChildCreateTimeTAG: 2012-11-01T20:23:10Z IndexTAG: 1401 TitleTAG: Problem with the midterm exam Today I have started the exam. This exam had got 5 questions. I did it. I could see under each question "Check", "Save" and "Show answer". When I finished all the question I went to see my progress, but it saw that my midterm is in 0%. Surprised, I returned to the midterm exam and now it appears another exam. Where is the problem? What can I do? Thanks. UserIdTAG: 61050 UserNameTAG: Ferdyeh CreateTimeTAG: 2012-10-28T14:08:05Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Ok, thanks for your answers! I have finished the real exam. FirstChildUserIdTAG: 61050 FirstChildUserNameTAG: Ferdyeh FirstChildCreateTimeTAG: 2012-10-28T18:10:29Z SecondChildTAG: Great to know you made it in time! I forgot to mention that we received and email about the Midterm in the address used for registering to the course. That email summarized the main aspectes about the exam, thus making it quite difficult to get confused. I'd suggest you to check that email address (you probably used one that do not regularly check), unless some days before the final exam, to avoid further problems. ;-) Best, Jordi SecondChildUserIdTAG: 366320 SecondChildUserNameTAG: Galli SecondChildCreateTimeTAG: 2012-10-28T19:20:39Z FirstChildTAG: Yes, Galli is right, the mid-term is in the Courseware section, between Week 6 and Week 7. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-28T15:24:42Z FirstChildTAG: The bad news are that it seems that you did the sample Midterm (just for practice, and not the real one). The real midterm has six questions and of course does not have the "Show answer" button. You can find it in the Coursware section, between weeks 6 & 7. The good news are that you still have almost 15 hours to complete the real exam. Time enough for it. Good luck! FirstChildUserIdTAG: 366320 FirstChildUserNameTAG: Galli FirstChildCreateTimeTAG: 2012-10-28T14:17:22Z SecondChildTAG: can u help me with answers? SecondChildUserIdTAG: 414461 SecondChildUserNameTAG: akashrandev SecondChildCreateTimeTAG: 2012-10-28T16:22:18Z SecondChildTAG: The midterm is still currently active. We cannot discuss it at this time. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-28T16:28:55Z IndexTAG: 1402 TitleTAG: think without limits now after we riched to this level what do you think what have we to do to increase our knowledge in elecronics ? i want to work in this field specially may be in digital electronics design have you any recommends for me ? and what about you ? what do you want to do after finishing this course ? specially if you're interested in electronics UserIdTAG: 41666 UserNameTAG: AbdoMondy CreateTimeTAG: 2012-10-27T16:57:49Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Keep learning. The knowledge we possess is increasing more and more quickly. There are more and more opportunities. The more we do, the more we know what we would like to do next. At the stage we're at, here with this course, it's hard to go wrong following any of the major paths in an EE curriculum. FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-10-29T04:22:50Z IndexTAG: 1403 TitleTAG: S15E5: Week 8 Can anyone help me with this? I'm not able to understand how to do it! :( https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_8/Ramps_Steps_and_Impulses/ UserIdTAG: 277808 UserNameTAG: Hemanthmps CreateTimeTAG: 2012-10-27T03:30:36Z VoteTAG: 2 CoursewareTAG: Week 8 / Initial Conditions CommentableIdTAG: 6002x_Initial_Conditions NumberOfReplyTAG: 8 FirstChildTAG: Here's some snippets from earlier posts in no particular order. Hope you find what you want here. --- IIRC the formula was posted for a falling signal in the beginning parts, where $V_{final} \lt V_{initial}$. When you have a rising signal, as in the last part, the formula is slightly different. I used it, but didn't point that out. For a rising signal we have $$V_C(t) = V_{initial} + ( V_{final} - V_{initial} ) \cdot ( 1 - e^{{-t} \over \tau} )$$ I think I skipped a few steps also, but hopefully this is what you need. --- I fixed it and put the extra -ve in the time exponential calculation, it now reads $$V_C(t) = V_{final} + (V_{initial}-V_{final})e^{{-(t_{final}-t_{initial}) }\over {RC}}$$ therefore, if $t_{initial} = -0.001$ and $t_{final} = 0$ $$V_C(0-) = 1 + (3-1)e^{{-(0-(-0.001)) }\over {RC}}$$ --- Something to get you started $$i_L(t) = i_{final} + ( i_{initial} - i_{final} )e^{{-t} \over {L/R}}$$ so for the first part $$i_L(0-) = I_0 + ( i_L(t=-0.001) - I_0 )e^{{-(t=0.001)} \over {L/R}}$$ --- Actually this may be confusing because I'm mixing up time references here, so I'm adding an explanation here. The $i_L(t=-0.001)$ time reference refers to the global time scale for the problem, and is an initial condition we have been given. Using that point in time as a new time base, the time on the exponential is correct, but what I probably should have written is $$i_L(0-) = I_0 + ( i_L(t=-0.001) - I_0 )e^{{-((t=0-) -(t=-0.001))} \over {L/R}}$$ FirstChildUserIdTAG: 9673 FirstChildUserNameTAG: xp42 FirstChildCreateTimeTAG: 2012-10-27T04:39:42Z FirstChildTAG: **SPOILER - solutions below** Part 1. $$V_C(t) = V_{final} + (V_{initial}-V_{final})e^{{-(t_{final}-t_{initial}) }\over {RC}}$$ therefore, if $t_{initial} = -0.001$ and $t_{final} = 0$ $$V_C(0-) = 1 + (3-1)e^{{-(0-(-0.001)) }\over {RC}}$$ And I have $R = 100\Omega$, $C = 10 \mu F$ So we get $$V_C(0-) = 1 + 2e^{{-0.001 }\over {100\cdot10\mu}}$$ which reduces to $$V_C(0-) = 1 + {2\over e} = 1.7357 V$$ ----- Part 2. Impulse at $t = 0$ is $Q = -20\mu C$, thus add this $Q \over C$ voltage to $V_C(0-)$ and we get $$V_C(0+) = 1 + {2 \over e} + {{-20\mu C}\over {10\mu F}}$$ which reduces to $$V_C(0+) = 1 + {2 \over e} + - 2 = -1 + {2 \over e} = -0.2643 V$$ --- Now, $V_C(t), t \gt 0 $ becomes $$V_C(t) = V_C(0+) \cdot e^{{-t} \over {RC}} + V_{final} \cdot (1 - e^{{-t} \over {RC}})$$ which is $$V_C(t) = ( - 1 +{2 \over e}) \cdot e^{{-t} \over {RC}} + 1 \cdot (1 - e^{{-t} \over {RC}})$$ So, at time $t=0.001$ and with $R=100\Omega$ and $C=10\mu F$ we get $$V_C(0.001) = {{{2 \over e}}\over e} - {1 \over e} + 1 - {1 \over e} = 0.5349$$ --- **Answer** to @mikecngan's question about part 1: We know $V_{final} = 1.0V$ because the final steady state as $t \rightarrow \infty $ has the capacitor reaching some saturation voltage, and all the current passing through R. When all the current passes through R, we have $$V_C = I \cdot R = I_0 \cdot R = 10mA \cdot 100\Omega = 1.0V$$ --- **Answer** to @ragarwal's question about part 2: When the current impulse = $Q \cdot \delta(t)$ is applied at t=0, *all* the current will flow in to the (ideal) capacitor. When the charge Q is added to the capacitor, it will increase the capacitor's voltage by ${Q \over C} Volts = {{-20 \mu C} \over {10 \mu F}} = -2 V$. So we see the impulse actually decreases the voltage across the capacitor. FirstChildUserIdTAG: 9673 FirstChildUserNameTAG: xp42 FirstChildCreateTimeTAG: 2012-10-27T04:40:30Z SecondChildTAG: Thanks! @xp42 SecondChildUserIdTAG: 277808 SecondChildUserNameTAG: Hemanthmps SecondChildCreateTimeTAG: 2012-10-27T04:48:17Z SecondChildTAG: This forum is so fcukd up. Start reading this post at **SPOILER - solutions below**. Then it seems to wrap back on to the top once you get to the bottom. But not quite. The last line is really the last line. It follows the last line *before* **SPOILER - solutions below**. So it follows the line ending with **= -2V** SecondChildUserIdTAG: 9673 SecondChildUserNameTAG: xp42 SecondChildCreateTimeTAG: 2012-10-27T04:49:18Z SecondChildTAG: Now i have understood everything. :) SecondChildUserIdTAG: 277808 SecondChildUserNameTAG: Hemanthmps SecondChildCreateTimeTAG: 2012-10-27T04:54:56Z SecondChildTAG: Actually, it's all messed up. It's not so simple as I first thought - things are all over the place. It's not a simple wrap around. SecondChildUserIdTAG: 9673 SecondChildUserNameTAG: xp42 SecondChildCreateTimeTAG: 2012-10-27T05:07:04Z SecondChildTAG: **SKIP THIS POST AND CONTINUE BELOW** SecondChildUserIdTAG: 9673 SecondChildUserNameTAG: xp42 SecondChildCreateTimeTAG: 2012-10-27T05:13:54Z FirstChildTAG: **Answer** to @mikecngan's question about part 1: We know $V_{final} = 1.0V$ because the final steady state as $t \rightarrow \infty $ has the capacitor reaching some saturation voltage, and all the current passing through R. When all the current passes through R, we have $$V_C = I \cdot R = I_0 \cdot R = 10mA \cdot 100\Omega = 1.0V$$ FirstChildUserIdTAG: 9673 FirstChildUserNameTAG: xp42 FirstChildCreateTimeTAG: 2012-10-27T05:12:33Z FirstChildTAG: **SPOILER - solutions below** Part 1. $$V_C(t) = V_{final} + (V_{initial}-V_{final})e^{{-(t_{final}-t_{initial}) }\over {RC}}$$ therefore, if $t_{initial} = -0.001$ and $t_{final} = 0$ $$V_C(0-) = 1 + (3-1)e^{{-(0-(-0.001)) }\over {RC}}$$ And I have $R = 100\Omega$, $C = 10 \mu F$ So we get $$V_C(0-) = 1 + 2e^{{-0.001 }\over {100\cdot10\mu}}$$ which reduces to $$V_C(0-) = 1 + {2\over e} = 1.7357 V$$ ----- FirstChildUserIdTAG: 9673 FirstChildUserNameTAG: xp42 FirstChildCreateTimeTAG: 2012-10-27T05:08:32Z SecondChildTAG: Again the order is messed up, but this is part 1 only. See if you can figure it out. The first 2 lines follow the first Vc eqn, the third line follows the Vc(0-) eqn. SecondChildUserIdTAG: 9673 SecondChildUserNameTAG: xp42 SecondChildCreateTimeTAG: 2012-10-27T05:10:13Z SecondChildTAG: lol SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-10-27T11:33:24Z FirstChildTAG: Part 2. Impulse at $t = 0$ is $Q = -20\mu C$, thus add this $Q \over C$ voltage to $V_C(0-)$ and we get $$V_C(0+) = 1 + {2 \over e} + {{-20\mu C}\over {10\mu F}}$$ which reduces to $$V_C(0+) = 1 + {2 \over e} + - 2 = -1 + {2 \over e} = -0.2643 V$$ --- FirstChildUserIdTAG: 9673 FirstChildUserNameTAG: xp42 FirstChildCreateTimeTAG: 2012-10-27T05:11:53Z FirstChildTAG: Now, $V_C(t), t \gt 0 $ becomes $$V_C(t) = V_C(0+) \cdot e^{{-t} \over {RC}} + V_{final} \cdot (1 - e^{{-t} \over {RC}})$$ which is $$V_C(t) = ( - 1 +{2 \over e}) \cdot e^{{-t} \over {RC}} + 1 \cdot (1 - e^{{-t} \over {RC}})$$ So, at time $t=0.001$ and with $R=100\Omega$ and $C=10\mu F$ we get $$V_C(0.001) = {{{2 \over e}}\over e} - {1 \over e} + 1 - {1 \over e} = 0.5349$$ FirstChildUserIdTAG: 9673 FirstChildUserNameTAG: xp42 FirstChildCreateTimeTAG: 2012-10-27T05:12:01Z SecondChildTAG: Thanks for you explanations xp42! They were really helpful. For part three though, could you elaborate on why your equation for VC(t) contains the term Vfinal⋅(1−e^−tRC) ?? I understand that after the impulse the capacitor will begin discharging, but why do we have a term for the capacitor to be charging up again? SecondChildUserIdTAG: 287805 SecondChildUserNameTAG: Beneficial SecondChildCreateTimeTAG: 2012-11-10T02:17:13Z FirstChildTAG: See if you can figure it out using these answers 1 + (3-1)\*e^(-0.001/(100\*.00001)) 1 + (3-1)\*e^(-0.001/(100\*.00001)) - 0.00002/0.00001 (-1) \* ( 1 + (3-1)\*e^(-0.001/(100\*.00001)) - 0.00002/0.00001 ) + ( 1 + ( 1 + (3-1)\*e^(-0.001/(100\*.00001)) - 0.00002/0.00001 ) ) \*e^(-0.001/(100\*0.00001)) FirstChildUserIdTAG: 9673 FirstChildUserNameTAG: xp42 FirstChildCreateTimeTAG: 2012-10-27T03:46:51Z SecondChildTAG: How did you got all those??? I got only the first one! Can you elaborate it more? SecondChildUserIdTAG: 277808 SecondChildUserNameTAG: Hemanthmps SecondChildCreateTimeTAG: 2012-10-27T04:03:29Z SecondChildTAG: The mathjax feature is not working properly. The second post below, read the comment before reading the answer. The text order is not displayed correctly. SecondChildUserIdTAG: 9673 SecondChildUserNameTAG: xp42 SecondChildCreateTimeTAG: 2012-10-27T04:56:46Z FirstChildTAG: **Answer** to @ragarwal's question about part 2: When the current impulse = $Q \cdot \delta(t)$ is applied at t=0, *all* the current will flow in to the (ideal) capacitor. When the charge Q is added to the capacitor, it will increase the capacitor's voltage by ${Q \over C} Volts = {{-20 \mu C} \over {10 \mu F}} = -2 V$. So we see the impulse actually decreases the voltage across the capacitor. FirstChildUserIdTAG: 9673 FirstChildUserNameTAG: xp42 FirstChildCreateTimeTAG: 2012-10-27T05:12:48Z SecondChildTAG: so the circuit has 10mA output all the time but impulses at zero thats why you get -2v drop then increases back up because you still have have the 10mA supply. Is that the just of it, if not im completly lost. SecondChildUserIdTAG: 262875 SecondChildUserNameTAG: GrantDennison SecondChildCreateTimeTAG: 2012-11-08T21:00:01Z IndexTAG: 1404 TitleTAG: save button Can anyone tell that why this "SAVE" button is in midterm exam? How do I use it? UserIdTAG: 378096 UserNameTAG: Jamshaid271 CreateTimeTAG: 2012-10-26T19:41:52Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: The Save button records your answer, but does not grade them. Useful if you want to record your answers and go do something else. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-26T19:43:18Z SecondChildTAG: I did not get the word record. I think it is useful when doing labs, but how record works in midterm. SecondChildUserIdTAG: 378096 SecondChildUserNameTAG: Jamshaid271 SecondChildCreateTimeTAG: 2012-10-27T10:03:07Z SecondChildTAG: For example, if you want to use a different PC, such as if your laptop runs out of batteries, then you can "SAVE", turn off the computer, sign-in on the new computer, and you can continue your exam without having to re-type your answers. It's especially useful for Questions with many parts to them. SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-10-28T02:11:44Z SecondChildTAG: Thanks. Good feature. SecondChildUserIdTAG: 378096 SecondChildUserNameTAG: Jamshaid271 SecondChildCreateTimeTAG: 2012-10-28T05:00:49Z IndexTAG: 1405 TitleTAG: Grateful Muslim ![enter image description here][1] I'm a Muslim and I want to say thanks a lot to all the staff of Edx. [Non-course relevant material removed] [1]: https://edxuploads.s3.amazonaws.com/1351259828117229.jpg UserIdTAG: 109941 UserNameTAG: Alhashemy CreateTimeTAG: 2012-10-26T13:57:52Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: i m also muslim thank u FirstChildUserIdTAG: 688321 FirstChildUserNameTAG: zeekhan FirstChildCreateTimeTAG: 2012-10-26T14:50:55Z FirstChildTAG: While edX gladly welcomes anyone and everyone to learn on our website, we ask that the forums be used for discussions that are at least semi-related to the course or course materials. - Rohan FirstChildUserIdTAG: 391929 FirstChildUserNameTAG: RohanNagarkar FirstChildCreateTimeTAG: 2012-10-26T15:31:00Z FirstChildTAG: As this Post is going in a wrong direction and might some comments can hurt to someone as this is a worldwide Course and there is a lot of people with different belief and Religions, I will close this Post... My apologies.... I deleted comments responses from this Post... ---------- Also, remember that: [read here Terms of Service][1] THE FOLLOWING LIST OF ITEMS IS STRICTLY PROHIBITED ON THE SITE: Content that defames, harasses or threatens others; Content that discusses illegal activities with the intent to commit them; Content that infringes another's intellectual property, including, but not limited to, copyrights or trademarks; Profane, pornographic, obscene, indecent or unlawful content; Advertising or any form of commercial solicitation; Content related to partisan political activities; Viruses, trojan horses, worms, time bombs, corrupted files, malware, spyware or any other similar software that may damage the operation of another's computer or property; and Content that contains intentionally inaccurate information or that is posted with the intent of misleading others. [1]: https://www.edx.org/tos FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-26T15:34:10Z IndexTAG: 1406 TitleTAG: last year midterm can anyone please help me in Q3.2,3.4& Q.4.6 of last year midterm sample questions.thanks in advance.. UserIdTAG: 443660 UserNameTAG: itsme71 CreateTimeTAG: 2012-10-26T13:07:21Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: What do you not understand here? Where are you stuck (i.e. at which point in 3.2/3.4/4.6 are you stuck)? I could help you better from that point. FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-10-26T13:42:09Z FirstChildTAG: I have no idea how to do 3.2/4.6 i've been stuck on it for a while. For 3.4, Vin/Vout = -gm*Vgs*Rl/-vgs = gmRL, so assuming you got the right gm in 3.3 you would just plug in the gm and the given RL. FirstChildUserIdTAG: 23555 FirstChildUserNameTAG: NickFrei FirstChildCreateTimeTAG: 2012-10-26T16:15:44Z FirstChildTAG: Alrighty, here's how you do it. I'll start from 3.1. My values for this question are: $V_T=1.0V$, $K=0.017A/V^2.$ $R = 120.0 \Omega$ **3.1** To keep the MOSFET out of cut-off, we must have $v_{GS} \geq V_T$ Now $V_S = v_{IN}$ $\therefore V_G - v_{IN} \geq V_T$ $\implies V_G \geq V_T + v_{IN}$ In my case, it's given that $v_{IN}$ swings between $1.0V$ to $2.0V$. You can try plugging in both numbers but obviously, the limit on $V_G$ is set by the $1.0V$. Substituting the values given, we get the minimum limit on $V_G$ as $1.0 + 1.0 = 2.0V$ ---> 3.1 --- **3.2** To keep the MOSFET in saturation, it is necessary that $v_{DS} \geq (v_{GS} - V_T)$. In this case we must take $v_{IN} = -2$ as $v_{GS} = V_G - v_{IN}$ is the maximum value of $v_{GS}$ and therefore imposes a "higher minimum" on $V_{DD}$ We know both terms on the right hand side, namely $v_{GS}$ and $V_T$. Lets concentrate on $V_{DS}$. We know that, Apply KVL to the loop: $V_{DD}$ -> $R_L$ -> Drain -> Source -> $v_{IN}$. You should get the following: $-V_{DD} + i_D R_L + v_{DS} + v_{IN} = 0$ or, $v_{DS} = V_{DD} - i_D R_L - v_{IN}$ Put this in the condition I stated at the beginning of this section. This will give us: $V_{DD} - i_D R_L - v_{IN} \geq v_{GS} - V_T$ $\implies V_{DD} \geq i_D R_L + v_{IN} + v_{GS} - V_T$ However, $i_D = \frac{K}{2}(v_{GS} - V_T)^2$ Hence: $V_{DD} \geq \frac{K}{2}(v_{GS} - V_T)^2 R_L + v_{IN} + v_{GS} - V_T$ Substitute the values as given. You should get $V_{DD} \geq 18.32$ ---> 3.2 --- **3.3** This part should be simple enough. From lecture we know that $g_m = K(V_{GS} - V_T)$ You can derive this by taking the partial derivative of $i_D$ with respect to $v_{GS}$. You will have to put the operating point to the result. Since $v_{IN} = 0, V_{GS} = V_G = 3$ Hence, $g_m = 0.034$ ---> 3.3 --- **3.4** For this part, replace the MOSFET with it's small signal model. This is a voltage controlled current source. This is what the circuit looks like (sorry I wasn't neat enough) ![enter image description here][1] So, $v_{out} = - R_L g_m v_{gs}$ However, $v_{gs} = v_g - v_{in} = 0 - v_{in} = -v_{in}$. Hence, $v_{out} = - R_L g_m (-v_{in})$ or $\frac{v_{out}}{v_{in}} = R_L g_m = 4.08$ ---> 3.4 --- **3.5** This one's really easy. The incremental resistance is the reciprocal of the transconductance ($g_m$)) i.e. $\frac{1}{g_m}$. We know $g_m$. The reciprocal gives $29.412\Omega$ This post is pretty long so I'll do 4.6 in another one (After a break :-)). Hope this helps :-) [1]: https://edxuploads.s3.amazonaws.com/13512833022886131.png FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-10-26T20:36:14Z SecondChildTAG: Ugh. I hope the Mathjax is rendering correctly for you. It was OK in the preview for the post (still is). But it doesn't look right in the actual post. SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-10-26T20:40:43Z SecondChildTAG: Don't use > and <, use \gt and \lt, etc SecondChildUserIdTAG: 9673 SecondChildUserNameTAG: xp42 SecondChildCreateTimeTAG: 2012-10-27T01:11:12Z SecondChildTAG: Ah I didn't know that. Thanks xp! I wish this showed up in the preview too. I would have looked for the right way to do it in that case. SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-10-27T06:16:43Z FirstChildTAG: Here's question 4. I guess sections 1 to 5 are quite straightforward. However, if something is unclear over here, I will be glad to help. So lets jump right to section 6 of the problem. Here's what the question says for me: "We now have $R_{ON}=5.0k\Omega$. Let's choose $R_{PU}=17.0k\Omega$. What is the maximum amount of power, in milliWatts, that is consumed by this circuit for legitimate logic levels applied at A and B?" To get this right, we need to figure out the values for A and B which would cause the maximum number of MOSFETs to turn on. There is a catch however. We'll see what that is in a moment. Lets number the transistors from left to right as $T_1$, $T_2$, $T_3$ and $T_4$. Construct a table which checks the transistors that are ON for various values of A and B. You'll that two transistors are ON for the cases AB = {01}, {10} & {11}. Of these, the second one is special. When A = 1 and B = 0, the two center transistors $T_2$ and $T_3$ turn ON. These are in parallel and hence we will have to combine their individual ON resistances using the parallel combination. Using this idea, we get, $P = \frac{V^2}{R_{PU} + R_{ON}||R_{ON}} = 1.282\ mW$ Lets look at the other two conditions. In either case, we get two ON resistances which are not in parallel like before. In both cases the power, $P = \frac{V^2}{R_{PU} + R_{ON}} +\frac{V^2}{R_{PU} + R_{ON}} = 2.273\ mW$ Which is the maximum of the two cases. --- Here are the last two sections. In my case, $C_{GS} = 50.0\ pF$: **4.7** The rising edge time constant refers to the point when the inverter is outputting a 1 i.e. 5V. At this point, the MOSFET is off. So only the pull up resistance $R_{PU}$ exists in the circuit. Therefore the time constant, $\tau = R_{PU}C_{GS} = (17\times10^3) \times (50 \times 10^{-15}) = 0.85 ns$. --- **4.8** The falling edge time constant refers to the point when the inverter is giving out a 0 i.e $V_{OL}$. In this case, both $R_{PU}$ and $R_{ON}$ exist. Replace the MOSFET with it's model. Then calculate the Thevenin resistance. Use this to calculate the time constant. $R_{th} = R_{PU}||R_{ON}$. Therefore, $\tau = R_{th}\times C_{GS} = R_{PU}||R_{ON} \times C_{GS}$ $\tau = (17\times10^3)||(5\times10^3) \times (50 \times 10^{-15}) = 0.193 ns$ Hope this helps as well :-) FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-10-26T21:38:20Z SecondChildTAG: They are. Could you help, please? SecondChildUserIdTAG: 359214 SecondChildUserNameTAG: Aduial SecondChildCreateTimeTAG: 2012-10-27T09:45:37Z SecondChildTAG: Done :-) I added it to this post. SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-10-27T18:36:32Z IndexTAG: 1407 TitleTAG: Query regarding question Shoudn't the value of Vs be given in order to find out the correct waveform in the problem S14E1? The value till which the capacitor would charge up would depend on the supply voltage and not on the input voltage(at the gate terminal) UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-26T06:33:21Z VoteTAG: 2 CoursewareTAG: Week 7 / Response To A Step Down CommentableIdTAG: 6002x_Response_To_A_Step_Down NumberOfReplyTAG: 1 FirstChildTAG: I suppose in the context of digital circuits, it would be an inversion of the input. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-26T14:40:00Z SecondChildTAG: true but how do we know its choice C and not choice D? SecondChildUserIdTAG: 285945 SecondChildUserNameTAG: lindalapiso SecondChildCreateTimeTAG: 2012-10-29T08:20:51Z SecondChildTAG: maybe it has to do with the 5V mark? like the signal has to rise up to 5volts? SecondChildUserIdTAG: 285945 SecondChildUserNameTAG: lindalapiso SecondChildCreateTimeTAG: 2012-10-29T08:22:43Z SecondChildTAG: Easy, babe! :) SecondChildUserIdTAG: 385692 SecondChildUserNameTAG: tpfslima SecondChildCreateTimeTAG: 2012-10-29T20:38:30Z IndexTAG: 1408 TitleTAG: If you have a problem with algebra in Q6 [Edited for content - Do not discuss questions on the mid-term] UserIdTAG: 345464 UserNameTAG: Constantine_ru CreateTimeTAG: 2012-10-25T21:03:52Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Do not discuss the midterm. Don't give advice, don't offer suggestions. Just leave it until the deadline passes. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-25T21:27:30Z IndexTAG: 1409 TitleTAG: Algebra The algebra tasks are very complicated eg: it's good:-RO*K1*VI but these are bad: RO*-K1*VI or RO*K1*-VI or RO*-(K1-V1) This problem is annoying because I know the correct answer but I can't type it in. UserIdTAG: 58468 UserNameTAG: Hollady CreateTimeTAG: 2012-10-25T19:42:37Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 1 FirstChildTAG: If you have a negative sign, it needs to be put into parentheses in order for the grader to understand that you are not trying to subtract something. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-10-25T20:32:41Z SecondChildTAG: you need parentheses not for grader only - human beings also expect parentheses here :) SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-25T23:16:47Z SecondChildTAG: I use a lot of parenthesis when I enter the equations. That way, I don't make a stupid mistake based on mathematical order of operations. Just make sure you balance all your left and right parentheses. Too many sets of parentheses are never a problem, too few are. SecondChildUserIdTAG: 94557 SecondChildUserNameTAG: g_hopper SecondChildCreateTimeTAG: 2012-10-26T23:38:18Z IndexTAG: 1410 TitleTAG: About the second question... Could somebody help me out about this question. Why 4.4? What mathematical formula should we use? Thanks in advance! UserIdTAG: 214275 UserNameTAG: TheodoreGr CreateTimeTAG: 2012-10-25T19:06:51Z VoteTAG: 2 CoursewareTAG: Week 7 / Inductors Store Energy CommentableIdTAG: 6002x_Inductors_Store_Energy NumberOfReplyTAG: 4 FirstChildTAG: Hi, remember that E = W*t, so if you divide the energy you've got by the transaction time (0.25 useg), that gives you 4.4 Watts. FirstChildUserIdTAG: 220837 FirstChildUserNameTAG: Calsomus FirstChildCreateTimeTAG: 2012-10-25T19:35:40Z SecondChildTAG: But this is not quite the *peak* power, it is more like the *average* power SecondChildUserIdTAG: 346056 SecondChildUserNameTAG: fiatlux SecondChildCreateTimeTAG: 2012-11-04T21:53:54Z FirstChildTAG: Divide energy you got by time given! FirstChildUserIdTAG: 277808 FirstChildUserNameTAG: Hemanthmps FirstChildCreateTimeTAG: 2012-10-26T04:38:16Z FirstChildTAG: Thanks a lot! FirstChildUserIdTAG: 214275 FirstChildUserNameTAG: TheodoreGr FirstChildCreateTimeTAG: 2012-10-27T09:07:55Z FirstChildTAG: On the other hand, we can get (di/dt) from the current change over time, which is 1A over .25us, which gives di/dt = 4*1e6. Using the formula P=Li(di/dt) would give 8.8 watts. I guess the factor of 1/2 could come from the integral of P over the .25us interval. FirstChildUserIdTAG: 402144 FirstChildUserNameTAG: jsage FirstChildCreateTimeTAG: 2012-11-05T06:07:17Z IndexTAG: 1411 TitleTAG: certificate reg What is the minimum marks to get the mit certificate. How will we get the certificat e? UserIdTAG: 149154 UserNameTAG: santhosh1993 CreateTimeTAG: 2012-10-25T14:08:29Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: my current total is around 48%. can i get the certificate! FirstChildUserIdTAG: 149154 FirstChildUserNameTAG: santhosh1993 FirstChildCreateTimeTAG: 2012-10-25T14:09:22Z SecondChildTAG: Myrimit please reply if u could SecondChildUserIdTAG: 149154 SecondChildUserNameTAG: santhosh1993 SecondChildCreateTimeTAG: 2012-10-25T14:11:00Z SecondChildTAG: Click on the **Progress** tab at the top of this page. Look on the LHS of your histogram (bar chart) and you will see the scores required for each letter grade. The bar on the right is your total score. SecondChildUserIdTAG: 9673 SecondChildUserNameTAG: xp42 SecondChildCreateTimeTAG: 2012-10-25T14:27:45Z SecondChildTAG: when ur progress will 60% u can recve certificate SecondChildUserIdTAG: 688321 SecondChildUserNameTAG: zeekhan SecondChildCreateTimeTAG: 2012-10-25T14:52:05Z FirstChildTAG: 60% FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-25T14:22:29Z FirstChildTAG: You'll get certificate if you have minimum of 60% **Grade C- 60%** Grade B- 70% **Grade A- 87%** FirstChildUserIdTAG: 277808 FirstChildUserNameTAG: Hemanthmps FirstChildCreateTimeTAG: 2012-10-25T15:09:41Z FirstChildTAG: Hi santhosh1993! **You will need to sum** a Final Score of **60%** in order to Pass this Course and get the Certificate . This sum, **is based on**: - Homeworks (15%). - Labs (15%). - Midterm Exam (30%). - Final Exam (40 %). Lets see this visual help (I wrote this on the old Forum - 6.002x Spring, might this can help you): - PASS and NOT PASS Table: ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13374846943979601.png) - Percentages of total score will be based on: ![image description](https://mitx_askbot_stage.s3.amazonaws.com/13374850316133229.png) So, if you have : Example 1. - 10 Hw =100% you will get a 15% points for that. - 10 Labs=100% you will get a 15% points for that. - If you solved 100% the Midterm you will get 30% points for that. - If you solve 100% the Final Exam you will get 40% points for that. - So, the total score will be 15%+15%+30%+40% = 100% total score . - You will pass with grade A. Example 2. - 10 Hw = 70% you will get 10.5% points for that. - 10 Labs= 80% you will get 12% points for that. - If you solved 60% of the Midterm exam you will get 18% points for that. - If you solve 80% of the Final exam you will get 32% points for that. - So the total score will be 10.5%+12%+18%+32%=72.5% - You will pass with the grade B. Example X: - 10 Hw = H % you will get H*0.15= h [%] - 10 Labs=L% you will get L*0.15= l [%] - Midterm Exam= ME% you will get ME*0.30= me [%] - Final Exam = FE% you will get FE*0.40= fe[%] - Total Score =ts= h + l + me + fe Pass ts high or equal 60% Not pass ts less than 60% I hope this can help you. See you! Myriam. ---- P.D: Yes! you can get the certificate :). It matters Final the percentage resulting of the sum of: Hws, Labs and Both Exams score. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-25T19:41:23Z FirstChildTAG: I don't need any 60% or 70%, I need only 100%! FirstChildUserIdTAG: 326787 FirstChildUserNameTAG: Angstrem FirstChildCreateTimeTAG: 2012-10-25T15:40:24Z SecondChildTAG: **Yes**, that is what I am shooting for! If I get 60% I will be **disappointed**! SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-10-25T15:46:59Z SecondChildTAG: i am also thinking the same way. SecondChildUserIdTAG: 99628 SecondChildUserNameTAG: Pranjal16 SecondChildCreateTimeTAG: 2012-10-25T16:12:50Z IndexTAG: 1412 TitleTAG: Another way of visualising... This is also true, correct me if I am wrong, because of the laws of conservation energy.![enter image description here][1] The area of the curved waveforms due to the capacitor has to be the same as that of the original square wave. So the rise slope will mirror the fall slope. [1]: https://edxuploads.s3.amazonaws.com/13511663713306408.png UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-10-25T11:59:56Z VoteTAG: 2 CoursewareTAG: Week 7 / Faling Delay Continued CommentableIdTAG: 6002x_Falling_Delay_Continued NumberOfReplyTAG: 3 FirstChildTAG: Cool! FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-25T14:28:25Z FirstChildTAG: nice catch :) FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-25T23:49:56Z FirstChildTAG: thanks for the visual FirstChildUserIdTAG: 623210 FirstChildUserNameTAG: preveen FirstChildCreateTimeTAG: 2012-10-27T16:37:08Z SecondChildTAG: Absolute pleasure! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-27T19:55:32Z IndexTAG: 1413 TitleTAG: I finish, i like the problems... XD Cool i like the test !! UserIdTAG: 628745 UserNameTAG: albertclint CreateTimeTAG: 2012-10-25T09:02:34Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi, Did you study from the textbook for the exam? FirstChildUserIdTAG: 85490 FirstChildUserNameTAG: yashsavani FirstChildCreateTimeTAG: 2012-10-25T09:03:56Z IndexTAG: 1414 TitleTAG: midterm results atlast finished the exam with 83% UserIdTAG: 149154 UserNameTAG: santhosh1993 CreateTimeTAG: 2012-10-25T08:16:45Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: hi can you tell what to do for the final submission. I am also done with all the questions. FirstChildUserIdTAG: 365309 FirstChildUserNameTAG: chetnasinghaldas FirstChildCreateTimeTAG: 2012-10-25T08:31:37Z SecondChildTAG: For me it's 100%...... :) SecondChildUserIdTAG: 277808 SecondChildUserNameTAG: Hemanthmps SecondChildCreateTimeTAG: 2012-10-25T08:32:17Z SecondChildTAG: For me also 100% SecondChildUserIdTAG: 365309 SecondChildUserNameTAG: chetnasinghaldas SecondChildCreateTimeTAG: 2012-10-25T08:43:46Z SecondChildTAG: wow amazing guys. but i think there's some bug in Q3.The first question i answered it correctly but it went wrong other question is correct. help please. SecondChildUserIdTAG: 340180 SecondChildUserNameTAG: JohnErich SecondChildCreateTimeTAG: 2012-10-25T09:06:17Z SecondChildTAG: There is no bug and it's not possible to help. (Honor code: do you remember?). SecondChildUserIdTAG: 376877 SecondChildUserNameTAG: AndBre SecondChildCreateTimeTAG: 2012-10-25T10:12:24Z SecondChildTAG: sorry.. my bad SecondChildUserIdTAG: 340180 SecondChildUserNameTAG: JohnErich SecondChildCreateTimeTAG: 2012-10-25T10:56:41Z FirstChildTAG: After the exam I thought to start with week 7. but week 7 onwards videos are also not opening. it means to start with week 7 and 8 we have to wait till sunday?? FirstChildUserIdTAG: 365309 FirstChildUserNameTAG: chetnasinghaldas FirstChildCreateTimeTAG: 2012-10-25T08:54:00Z SecondChildTAG: No, the videos are working fine. At least for me.. SecondChildUserIdTAG: 97581 SecondChildUserNameTAG: Asimakis SecondChildCreateTimeTAG: 2012-10-25T08:56:40Z SecondChildTAG: I just checked now and first lecture video for week 8 run as usual. It was probably a temporary problem or something with your connection. Best, Jordi SecondChildUserIdTAG: 366320 SecondChildUserNameTAG: Galli SecondChildCreateTimeTAG: 2012-10-25T09:03:33Z IndexTAG: 1415 TitleTAG: [STAFF] - MYRIMIT - PLEASE CHECK THE SCORING SCRIPT ON THE MIDTERM I entered the right answer en the box, it was my final check, everyting was fine, just the last (Q2,9) i entered ..xyz and the system got put it wrong, and when I checked what was the right answer it was ...xyza just by "a" and the a was hugely lower than cero, please take a look at it, my email is roosemberth@hotmail.com.ar UserIdTAG: 231676 UserNameTAG: Roosemberth CreateTimeTAG: 2012-10-25T08:05:26Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Now that would've been irritating! FirstChildUserIdTAG: 108454 FirstChildUserNameTAG: Raven7281 FirstChildCreateTimeTAG: 2012-10-25T10:22:21Z FirstChildTAG: Hi Roosemberth, I didn't understand what happened to you... Did you mean that your answer was correct and the system did not check it correctly? If that is true, you should e-mail to the edX Staff: bugs@edx.org I would like to help you ... but, remember that Collaboration of any form is strictly forbidden in the midterm and the final exams...[syllabus][1] [1]: https://www.edx.org/static/content-mit-6002x/handouts/syllabus.a477535058a1.pdf FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-25T19:53:41Z SecondChildTAG: Yes, my answer was right SecondChildUserIdTAG: 231676 SecondChildUserNameTAG: Roosemberth SecondChildCreateTimeTAG: 2012-10-25T23:55:07Z SecondChildTAG: I've sent the e-mail to bugs@edx.org, I hope you to check it out Best Regards Roosemberth Palacios SecondChildUserIdTAG: 231676 SecondChildUserNameTAG: Roosemberth SecondChildCreateTimeTAG: 2012-10-26T00:01:47Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-26T11:11:08Z IndexTAG: 1416 TitleTAG: mid term login shall I login any number of times after I start my exam within 24 hours? UserIdTAG: 446691 UserNameTAG: maitrey CreateTimeTAG: 2012-10-25T06:58:01Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi. [Just Check It][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5088dbe5959a592b00000024 > **Regards:** asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-10-25T14:56:08Z IndexTAG: 1417 TitleTAG: Midterm I saw the link of Midterm exam in the courseware.If I open that link,will my time start from that moment? UserIdTAG: 156835 UserNameTAG: kphariprakash1968 CreateTimeTAG: 2012-10-25T06:20:07Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: No. When you open the link information on midterm will come. If you go to the next only your time start. You will get 24 hours after you start. FirstChildUserIdTAG: 260154 FirstChildUserNameTAG: jjmattam FirstChildCreateTimeTAG: 2012-10-25T06:27:58Z IndexTAG: 1418 TitleTAG: Please send an email notification mentioning the test's link in it as soon as the test is released. Please send an email notification mentioning the test's link in it as soon as the test is released. am waiting UserIdTAG: 694092 UserNameTAG: emanurag CreateTimeTAG: 2012-10-25T05:31:10Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: It's posted. But we have 24h from the time we open the page after the cover. Last time to start is thus Oct 27 11:59 pm EDT (deadline - 24h). FirstChildUserIdTAG: 184153 FirstChildUserNameTAG: tthngn FirstChildCreateTimeTAG: 2012-10-25T05:32:56Z SecondChildTAG: you can start any time before the dead line. Either you get 24 hours or maximum up oct 28 11:59pm EDT SecondChildUserIdTAG: 623210 SecondChildUserNameTAG: preveen SecondChildCreateTimeTAG: 2012-10-27T16:44:31Z FirstChildTAG: its released now... FirstChildUserIdTAG: 169181 FirstChildUserNameTAG: NINI FirstChildCreateTimeTAG: 2012-10-25T05:33:04Z FirstChildTAG: Hi. Mid term available at **course ware** just under the **Week6** and **edX** also e.mail each n every candidate. > **Regards:** asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-10-25T14:25:01Z IndexTAG: 1419 TitleTAG: MId term sir dont upload the mid term exam???????????????????????? UserIdTAG: 116764 UserNameTAG: faisal82 CreateTimeTAG: 2012-10-25T04:23:10Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I'm thinking the Amazon Cloud outages are still causing problems for the 6.002x team. I could be wrong. FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-10-25T05:24:35Z IndexTAG: 1420 TitleTAG: Will the staff be available to discuss at the time of mid-term exam? HI As we are not allowed to discuss our doubts regarding the mid-term exam questions. Will the staff be available if we want to clear any doubt during mid-term? thanks. UserIdTAG: 365309 UserNameTAG: chetnasinghaldas CreateTimeTAG: 2012-10-25T03:39:34Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: We will be able to clarify questions, but we will not be able to help you understand any of the subject material that a question needs in order to solve it. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-10-25T14:27:05Z IndexTAG: 1421 TitleTAG: Inductor Diagram ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13511137164796128.jpg UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-10-24T21:22:00Z VoteTAG: 2 CoursewareTAG: Week 7 / Ideal Linear Inductors CommentableIdTAG: 6002x_Ideal_Linear_Inductors NumberOfReplyTAG: 2 FirstChildTAG: That is extremely pretty, and it makes me wish that our discussion forum had a like button. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-10-25T14:34:33Z SecondChildTAG: Haha! Thanks! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-25T14:37:19Z FirstChildTAG: why you wrap donut with spaghetti? dont think it will be tasty... FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-26T01:36:07Z IndexTAG: 1422 TitleTAG: To STAFF! Midterm_Exam !!!!!!!!!!!!!!!!! Plz tell me, when will start the deadline of Midterm_Exam ? And second. How mach time we have to done this exam? UserIdTAG: 192114 UserNameTAG: chuba CreateTimeTAG: 2012-10-24T19:46:40Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You may find ALL infos on the Course Info Page. "The mid-term exam will be released on October 25th at 00:01 am Boston time. The exam is designed to take 2 hours; however, in order to compensate for any Internet or power outages in your area you will have 24 hours to finish this exam. You can start the exam when it is convenient for you, but you must complete this exam by 11:59 pm (almost midnight) Boston time on October 27th*, even if that is less than 24 hours after you started the exam. The exam will cover material through Small-Signal Analysis of Transistor Circuits (end of first lecture video sequence of Week 6). The exam must be completed online. Please understand that we cannot accept submissions in any format past the deadline, or delay the deadline for individual students for to any reason. *The exam deadline has been extended to 23:59 (11:59 pm) Boston time on Sunday, October 28th." FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-10-24T20:17:09Z SecondChildTAG: Thanks SecondChildUserIdTAG: 192114 SecondChildUserNameTAG: chuba SecondChildCreateTimeTAG: 2012-10-24T20:44:15Z IndexTAG: 1423 TitleTAG: YAMTQ (Yet Another Mid Term Question) Maybe someone has asked this question before, but I couldn't find it on the forum. Can I still login and do all the homework and other practices between the 25th and 28th without starting the 24hr mid term time frame? Or in other words: when does the 24hr counter start? When logging in or on actually starting the mid term (is there a confirmation dialog or something?). UserIdTAG: 114881 UserNameTAG: xvink CreateTimeTAG: 2012-10-24T18:21:36Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: The 24 hour counter starts when you access the problems in the midterm. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-10-24T19:52:06Z IndexTAG: 1424 TitleTAG: qs 2 part 2.....MINIMUM VDD in mock mid term i guess this minimum Vdd qs will haunt me in the actual mid term as well.. couldn't solve it for the mock mid term... can somebody give a quick reply for the solution to this.. please help! UserIdTAG: 368712 UserNameTAG: jmen CreateTimeTAG: 2012-10-24T15:31:24Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: do you mean Q3? struggling too, how did you do first part? i am getting 2.2 as ans while it giving 2.4 part 2: i get 23.936 what are your Vgs and Vds values? mine: Vgs =Vg - Vin and Vds=Vout - Vin FirstChildUserIdTAG: 388273 FirstChildUserNameTAG: kilmarta FirstChildCreateTimeTAG: 2012-10-24T17:21:44Z SecondChildTAG: i tried it this way for out of cutoff region Vgs>=VT so Vg-Vin>=VT Vg>=Vin+VT so min value of Vg is Vin+VT and given Vin oscillates between 1 and -2 so min Vg=1+1=2 (i am confused whether to take Vin value 1 or -2) SecondChildUserIdTAG: 124534 SecondChildUserNameTAG: srihari46 SecondChildCreateTimeTAG: 2012-10-24T17:31:31Z SecondChildTAG: way i see it Vg>=Vin+VT at all points between Vin =1 and vin = -2 therefore take vin = 1 as if vg bigger at that point it will be bigger others therefore smallest vg = 1+vt =2.2 but that coming up wrong SecondChildUserIdTAG: 388273 SecondChildUserNameTAG: kilmarta SecondChildCreateTimeTAG: 2012-10-24T17:38:12Z SecondChildTAG: but given VT(THRESHOLD)=1V so your smallest vg=1+VT=1+1=2 SecondChildUserIdTAG: 124534 SecondChildUserNameTAG: srihari46 SecondChildCreateTimeTAG: 2012-10-24T17:40:35Z SecondChildTAG: hmm seems numbers changed and now i am right =) SecondChildUserIdTAG: 388273 SecondChildUserNameTAG: kilmarta SecondChildCreateTimeTAG: 2012-10-24T17:40:38Z SecondChildTAG: but my doubt is why can't i take Vin=-2 so that min Vg becomes -2+1=-1 which will become min value than that of Vin=1 min Vg=2 SecondChildUserIdTAG: 124534 SecondChildUserNameTAG: srihari46 SecondChildCreateTimeTAG: 2012-10-24T17:43:06Z SecondChildTAG: and min VDD in Q3 part 2, for mosfet to be in saturation VDS>=VGS-VT VDD-IDS*RL-Vin>=VGS-VT VDD>=IDS*RL+Vin+Vg-Vs-VT VDD>=IDS*RL+Vin+3-Vin-VT VDD>=IDS*RL+3-1 VDD>=IDS*RL+2 VDD>=6*10^-3*(VGS-VT)^2*120+2 VDD>=.720*(Vg-Vin-VT)^2+2 VDD>=.720*(3-Vin-1)^2+2 VDD>=.720*(2-Vin)^2+2 now Vin can take either 1 or -2 on taking Vin=1 we get VDD>=2.72 and on taking Vin=-2 we get VDD>=13.72 but in the question they have asked for min VDD but taking Vin=1 gives min VDD=2.72 which i am getting wrong so please correct me SecondChildUserIdTAG: 124534 SecondChildUserNameTAG: srihari46 SecondChildCreateTimeTAG: 2012-10-24T17:52:43Z SecondChildTAG: sorry for the formatting SecondChildUserIdTAG: 124534 SecondChildUserNameTAG: srihari46 SecondChildCreateTimeTAG: 2012-10-24T17:56:05Z SecondChildTAG: something weird happening numbers on this question, keep changing. refresh page i think they correct now. we will take -2 < Vin < 1 and VT =1V now Vg >= Vin + VT therefore Vg>= 1+1 and Vg >= -2+1 Vg >= 2 and Vg >= -1 smallest number bigger or equal than both -1 and 2 is 2 SecondChildUserIdTAG: 388273 SecondChildUserNameTAG: kilmarta SecondChildCreateTimeTAG: 2012-10-24T17:57:37Z SecondChildTAG: now i get it thank u SecondChildUserIdTAG: 124534 SecondChildUserNameTAG: srihari46 SecondChildCreateTimeTAG: 2012-10-24T17:59:15Z SecondChildTAG: can u help me with the min VDD part which is in Q3 part 2 with my analysis above SecondChildUserIdTAG: 124534 SecondChildUserNameTAG: srihari46 SecondChildCreateTimeTAG: 2012-10-24T18:01:06Z SecondChildTAG: and min VDD in Q3 part 2, for mosfet to be in saturation VDS>=VGS-VT VDD-IDS*RL-Vin>=VGS-VT VDD>=IDS*RL+Vin+Vg-Vs-VT VDD>=IDS*RL+Vin+3-Vin-VT VDD>=IDS*RL+3-1 VDD>=IDS*RL+2 VDD>=6*10^-3*(VGS-VT)^2*120+2 VDD>=.720*(Vg-Vin-VT)^2+2 VDD>=.720*(3-Vin-1)^2+2 VDD>=.720*(2-Vin)^2+2 now Vin can take either 1 or -2 on taking Vin=1 we get VDD>=2.72 and on taking Vin=-2 we get VDD>=13.72 but in the question they have asked for min VDD but taking Vin=1 gives min VDD=2.72 which i am getting wrong so please correct me SecondChildUserIdTAG: 124534 SecondChildUserNameTAG: srihari46 SecondChildCreateTimeTAG: 2012-10-24T18:02:25Z SecondChildTAG: in the preview i am seeing the new line but on posting it it is showing weird format SecondChildUserIdTAG: 124534 SecondChildUserNameTAG: srihari46 SecondChildCreateTimeTAG: 2012-10-24T18:03:07Z SecondChildTAG: same, with your numbers vdd has to be bigger than both 2.72 and 13.72 therefore vdd > 13.72 hit enter twice to start new para. SecondChildUserIdTAG: 388273 SecondChildUserNameTAG: kilmarta SecondChildCreateTimeTAG: 2012-10-24T18:05:33Z SecondChildTAG: When you consider Vin = 1 and Vin = -2, remember that the two inequalities you get need to be satisfied simultaneously (imagine putting an AND between them). Hence you need to use the larger value for VDD SecondChildUserIdTAG: 502508 SecondChildUserNameTAG: amaher SecondChildCreateTimeTAG: 2012-10-24T18:07:38Z SecondChildTAG: the same logic right thanks again SecondChildUserIdTAG: 124534 SecondChildUserNameTAG: srihari46 SecondChildCreateTimeTAG: 2012-10-24T18:08:14Z SecondChildTAG: help with Q3 part 5 of midterm practice questions finding incremental input resistance?? SecondChildUserIdTAG: 124534 SecondChildUserNameTAG: srihari46 SecondChildCreateTimeTAG: 2012-10-24T18:10:59Z SecondChildTAG: need help on that bit my self SecondChildUserIdTAG: 388273 SecondChildUserNameTAG: kilmarta SecondChildCreateTimeTAG: 2012-10-24T18:21:55Z SecondChildTAG: See answer below. Really hope it helps. SecondChildUserIdTAG: 502508 SecondChildUserNameTAG: amaher SecondChildCreateTimeTAG: 2012-10-24T18:28:03Z FirstChildTAG: Given: $V_{GS} \ge V_T$ $V_{DS} \ge V_{GS} - V_T$ From circuit (KCL + KVL): $V_{GS} = V_G - V_{IN}$ $V_{DS} = V_{OUT} - V_{IN}$, and then using KCL $V_{DS} = V_{DD} - i \cdot R_L - V_{IN}$ From these: $V_{DD} \ge V_G - V_T + i\cdot R_L$ $V_{DD} \ge V_G - V_T + \frac{(K\cdot R_L)}{2}\cdot(V_G - V_{IN} - V_T)^2$ FirstChildUserIdTAG: 502508 FirstChildUserNameTAG: amaher FirstChildCreateTimeTAG: 2012-10-24T18:18:09Z FirstChildTAG: Incremental Resistance: $r_{in} = \cfrac{v_{in}}{i_{in}} = \cfrac{v_{in}}{g_m \cdot v_{gs}}$ And since $V_{IN} = 0$, then $v_{gs}$ is simply $v_{in}$, so we get $r_{in} = \cfrac{1}{g_m}$ FirstChildUserIdTAG: 502508 FirstChildUserNameTAG: amaher FirstChildCreateTimeTAG: 2012-10-24T18:25:46Z SecondChildTAG: can you explain it with diagram SecondChildUserIdTAG: 124534 SecondChildUserNameTAG: srihari46 SecondChildCreateTimeTAG: 2012-10-24T18:31:34Z SecondChildTAG: can you explain me the meaning of this for Q3 part 5 where iin is the current taken from the input voltage source for the voltage vin where is iin and where is vin i am unable to understand that part of question and not able to visualize SecondChildUserIdTAG: 124534 SecondChildUserNameTAG: srihari46 SecondChildCreateTimeTAG: 2012-10-24T18:43:55Z SecondChildTAG: Uhm, I'm not sure how to display a diagram here, but basically you analyze the Small Signal model of the circuit. For SS model, you need to replace all elements with their SS equivalents. MOSFET becomes a dependent current source with i = gm*vgs (as discussed in lectures). Independent VG and VDD are shorted and vIN becomes vin. Then the question asks you to give $r_{in}$ as $\cfrac{v_{in}}{i_{in}}$ and $i_{in}$ is $g_m\cdot v_{gs}$ SecondChildUserIdTAG: 502508 SecondChildUserNameTAG: amaher SecondChildCreateTimeTAG: 2012-10-24T18:47:49Z SecondChildTAG: my doubt is how does input current become ids i.e gm*vgs SecondChildUserIdTAG: 124534 SecondChildUserNameTAG: srihari46 SecondChildCreateTimeTAG: 2012-10-24T18:50:05Z SecondChildTAG: any way thanks buddy for trying to explain me in many ways but if possible try to post a diagram of the solved problem SecondChildUserIdTAG: 124534 SecondChildUserNameTAG: srihari46 SecondChildCreateTimeTAG: 2012-10-24T18:59:04Z SecondChildTAG: ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13511062571343669.png SecondChildUserIdTAG: 502508 SecondChildUserNameTAG: amaher SecondChildCreateTimeTAG: 2012-10-24T19:18:06Z SecondChildTAG: now i got it clear really thanks buddy for your help and patience thanks a lot again SecondChildUserIdTAG: 124534 SecondChildUserNameTAG: srihari46 SecondChildCreateTimeTAG: 2012-10-24T19:35:45Z SecondChildTAG: No problem. Good luck on the exam :) SecondChildUserIdTAG: 502508 SecondChildUserNameTAG: amaher SecondChildCreateTimeTAG: 2012-10-24T19:38:09Z SecondChildTAG: thanks and same to u SecondChildUserIdTAG: 124534 SecondChildUserNameTAG: srihari46 SecondChildCreateTimeTAG: 2012-10-24T19:39:21Z SecondChildTAG: thanks for the tips. but i still get vOUT/vIN = -12.122 lol. SecondChildUserIdTAG: 157273 SecondChildUserNameTAG: ongchihang SecondChildCreateTimeTAG: 2012-10-25T09:06:40Z SecondChildTAG: what if `VIN` wasnt zero?? what would be incremental resistamce in that case??? SecondChildUserIdTAG: 126705 SecondChildUserNameTAG: arpitchugh SecondChildCreateTimeTAG: 2012-10-25T09:11:21Z FirstChildTAG: please explain how *`vgs=vin`* ---------- in last part??? FirstChildUserIdTAG: 126705 FirstChildUserNameTAG: arpitchugh FirstChildCreateTimeTAG: 2012-10-25T09:45:15Z FirstChildTAG: whoaa!! such a healthy discussion!! good luck all of you!! i completed my mid term successfully!! thanks to you guys too!! FirstChildUserIdTAG: 368712 FirstChildUserNameTAG: jmen FirstChildCreateTimeTAG: 2012-10-25T16:23:06Z IndexTAG: 1425 TitleTAG: current direction during nodal analysis when going upwards i.e; opposite to the direction of the current we need to take negative sign. but why didn't sir take a negative sign? UserIdTAG: 156713 UserNameTAG: HARITEJAREDDYP CreateTimeTAG: 2012-10-24T15:29:23Z VoteTAG: 2 CoursewareTAG: Week 4 / Dependent Sources Example 2 CommentableIdTAG: 6002x_dep_src_ex_2 NumberOfReplyTAG: 1 FirstChildTAG: during nodal analysis assume that current in each branch goes out from the node and take it the positive current irrespective of whether you know the direction or not... if you get a negative answer for the current it just means that the direction of flow of current is opposite... don't get confused at the last moment! Good luck! FirstChildUserIdTAG: 368712 FirstChildUserNameTAG: jmen FirstChildCreateTimeTAG: 2012-10-24T15:37:40Z IndexTAG: 1426 TitleTAG: staff:solution for mock mid term thanks for uploading the previous course's mid term...i have tried to solve it.. from where can we get the solutions for the same... thanks! UserIdTAG: 368712 UserNameTAG: jmen CreateTimeTAG: 2012-10-24T14:28:12Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: thnx got the solutions.... FirstChildUserIdTAG: 368712 FirstChildUserNameTAG: jmen FirstChildCreateTimeTAG: 2012-10-24T16:06:38Z SecondChildTAG: did you get solutions?? if so where SecondChildUserIdTAG: 388273 SecondChildUserNameTAG: kilmarta SecondChildCreateTimeTAG: 2012-10-24T17:22:23Z SecondChildTAG: please find the below link for prevous midterm solved paper https://6002x.mitx.mit.edu/static/handouts/6002x-MidTermReview-S2012.pdf SecondChildUserIdTAG: 124534 SecondChildUserNameTAG: srihari46 SecondChildCreateTimeTAG: 2012-10-24T18:18:56Z IndexTAG: 1427 TitleTAG: test i just signed in one week ago... can i do the test? i am a litle bit confused... UserIdTAG: 594667 UserNameTAG: Mateso CreateTimeTAG: 2012-10-23T19:51:58Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Of course you can. FirstChildUserIdTAG: 9673 FirstChildUserNameTAG: xp42 FirstChildCreateTimeTAG: 2012-10-23T20:23:46Z SecondChildTAG: What xp42 said it is true Mateso, However, you will not have chances to receive a 100% in the Final Score as you have missed some asignments of the previous Weeks ...(Remember that Final Score will be based on: 15% Homeworks, 15% Labs, 30% Midterm Exam and 40% Final Exam). Also, remember that the Midterm Exam is this Week... and it will be based from Week 1 to Week 6 content. So, you will have to study almost 6 Weeks in few days... Remember that you can always re-take this Course, I am sure that it will be another one in spring, but it is up to you, you decide... My best wish to you. Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-24T01:01:47Z FirstChildTAG: > **As Myrimit Said:** There are chances to get the certificate but you will not get the full grade, that is to say, you will not have chances to get a final score of 100% but yes to get an A with 91% (in the case that you do Midterm and Final Exam with 100% and also submit 100% of WEEK 6,7,8,9,10,11 and 12 assignments)... GENERAL Remember that this Course has: - 12 Homeworks (but if you read syllabus here, they tell you that you can skip two without penality, that means that you will still have 100% of Homework score if you make only 10). If you get 100% in your Homework score, that homework score it will contribute your final score only a 15%. So, if you have done so far 50% of your homework, it will contribute with your final score with (50%*15%)/100% = 7.5%. - 12 Labs (but if you read syllabus, they tell you that you can skip two without penality, that means that you will still have 100% of Labs score if you make only 10). - 1 Midterm Exam. If you get 100% in the Midterm, it will contribute with your final score with 30%. - 1 Final Exam.If you get 100% in the Final, it will contribute with your final score with 40%. So, it is not that late to get the certificate :), but you will have to do a lot of effort and work from now to the end of the Course... Also, as some students have said it to you, you can follow this Course and re-take it again, but that is always up to you :). See you! I hope this can help you. > **Regards:** asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-10-24T15:05:19Z SecondChildTAG: Thanks for helping me, i will re-take this course. Best regards God bless you SecondChildUserIdTAG: 594667 SecondChildUserNameTAG: Mateso SecondChildCreateTimeTAG: 2012-10-27T16:22:20Z IndexTAG: 1428 TitleTAG: Course Info not being updated anymore? I'm wondering what's happening to this great Circuits 6.002x course. The course info isn't being updated. And week 9 isn't even posted yet. Did we lose some staff team members? UserIdTAG: 4463 UserNameTAG: pmj CreateTimeTAG: 2012-10-23T01:58:15Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: If you check the calendar you will see that week nine is scheduled to be released Oct, 29. We just have the midterm this week. FirstChildUserIdTAG: 202032 FirstChildUserNameTAG: KeithD FirstChildCreateTimeTAG: 2012-10-23T02:22:23Z SecondChildTAG: Thanks. I missed that. SecondChildUserIdTAG: 4463 SecondChildUserNameTAG: pmj SecondChildCreateTimeTAG: 2012-10-23T03:17:02Z FirstChildTAG: Sorry about the delays in the course info. Week 9 is not scheduled to be released until next week in order to give you time to prepare for the midterm, and the [amazon outages][1] are making it difficult for us to update the announcements. [1]: http://www.techspot.com/news/50573-amazon-web-services-outage-takes-down-reddit-netflix-pinterest.html FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-10-23T14:52:57Z IndexTAG: 1429 TitleTAG: Hi everyone nice to see all of you UserIdTAG: 667997 UserNameTAG: tientran957 CreateTimeTAG: 2012-10-22T09:10:36Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Welcome! FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-10-22T10:54:50Z IndexTAG: 1430 TitleTAG: What do you think was the hardest question of week six? What do you all think was the hardest question of Homework 6 and Lab 6? UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-10-22T07:28:11Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: For me: * Easiest was definitely H6P3. * Most math-intensive, but straightforward was H6P1 * Problem with the most tricks, and with an "easy way out" was definitely H6P2. But if I had to do the computation instead of realizing "it's a phase shifter, so output is negative of input", it would be the most difficult. * Lab was pretty easy, just plugging into equations and measuring graphs. I made stupid math mistakes, but that's almost for granted when rushing at the last minute. (Although if you don't know your quadratic formula and partial derivatives, this week would be extremely difficult, and I believe English comprehension often presents more difficulty in the Labs than the Homeworks. If English is not your native language, I think Lab 6 would be difficult to understand.) Jersey Mark FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-10-22T08:12:08Z FirstChildTAG: Problem two, part 2 because IMO it was poorly worded. I initially believed that the problem was requesting the minimum VDD so that the circuit would work with the voltage swing given in part 1. Only after wasting much time and "sleeping on it" did I come back and compute the problem for the DC bias alone. I got the check but the problem doesn't make sense! Any negative signal swing and Q1 goes out of saturation. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-22T09:17:12Z FirstChildTAG: I spent a lot of time on H6P2 minimum VDD ! A stupid math error , but I had the correct answear right uder my nose ! The most amazing is that I have a proof that the answear is 2VIN ! : -The 2 mosfets have same K, same VT and the same IDS => They have the same VGS => VGS1=VGS2=VIN => They have the same VDS => VDS1=VDS2=VDD/2 VDS2=VGS2( By circuit) => VDS2=VIN .... =VDD/2 !!! Where's the bug ? FirstChildUserIdTAG: 373752 FirstChildUserNameTAG: Croak FirstChildCreateTimeTAG: 2012-10-22T14:17:57Z SecondChildTAG: VDS2 is fixed by the current because the gate is tied to the drain. VDS1 is not fixed by the current, it can be any value greater than or equal to VGS1 - VT. VDS2 is always VIN but VDS1 = VDD - VDS2. The minimum value for VDS1 is VIN - VT. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-22T15:57:09Z SecondChildTAG: Ok I understand . One find again 2VIN-VT . But is the characteristic ID=f(VDS) of a mosfet perfectly horizontal ?! SecondChildUserIdTAG: 373752 SecondChildUserNameTAG: Croak SecondChildCreateTimeTAG: 2012-10-24T01:14:27Z FirstChildTAG: definitely h6p2... h6p3 was so simple that it took me just a few seconds to do it.!! nd i finished homework 7 in exactly 4.30 minutes.....so happy fr it!! FirstChildUserIdTAG: 368712 FirstChildUserNameTAG: jmen FirstChildCreateTimeTAG: 2012-10-22T14:32:07Z SecondChildTAG: Yes, week 7 was quick. But I wasn't that quick. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-22T15:47:50Z SecondChildTAG: That's amazing! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-22T15:49:55Z IndexTAG: 1431 TitleTAG: Question to everyone.. Which week do you Think was the Toughest till this point???? Also, For those who had taken this course earlier, Will there be any more of such weeks? UserIdTAG: 148523 UserNameTAG: span993 CreateTimeTAG: 2012-10-22T05:31:21Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: I don't think I have the toughest week. But every week was very interesting! FirstChildUserIdTAG: 326787 FirstChildUserNameTAG: Angstrem FirstChildCreateTimeTAG: 2012-10-22T06:11:09Z SecondChildTAG: Sure It was interesting.. Thats the whole point, why Im still on this course :D SecondChildUserIdTAG: 148523 SecondChildUserNameTAG: span993 SecondChildCreateTimeTAG: 2012-10-22T16:30:17Z FirstChildTAG: Week 8 homework. Not impossibly hard, I needed to slow down and think. I don't think the lectures quite prepared the student for problems 1 and 3. Problem 2 has IMO a poorly worded part but is otherwise easy. Week 8 lab was fun. One of the better assignments in the course. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-22T12:58:17Z SecondChildTAG: Hmm, I just completed week6 and nearly fainted upon completion(especially h6p2 Phase inverter, though I sorta figured it out from the forum). Thanks everyone :-) SecondChildUserIdTAG: 148523 SecondChildUserNameTAG: span993 SecondChildCreateTimeTAG: 2012-10-22T16:28:00Z FirstChildTAG: till no week 4 was harrowing... fingers crossed.do not want anything like that again... FirstChildUserIdTAG: 368712 FirstChildUserNameTAG: jmen FirstChildCreateTimeTAG: 2012-10-22T14:37:18Z IndexTAG: 1432 TitleTAG: "I’ve shared this question with various engineers through the years" A 1V AC source is connected to a 1Ω resistor in series with a 1Ω reactance capacitor. What is the AC voltage across the capacitor? http://e2e.ti.com/blogs_/b/thesignal/archive/2012/10/02/oh-that-interview-question-a-reprise.aspx UserIdTAG: 323963 UserNameTAG: gtrf CreateTimeTAG: 2012-10-22T04:35:07Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: That's a good question especially when introducing someone to Power Analysis. With phasors it's easy to why there's the same drop across the resistor and the capacitor. In my university EECS course, AC and phasors were introduced early on (imaginary number, cool) but in this EECS course, people will ask, what's reactance? What's a phasor? How is a capacitor measured in ohms, not farads? I don't know when we get to that part in this class. We haven't had op-amps yet, either. But we already discussed MOSFETs without even touching BJTs! I guess all EECS classes are structured differently; every professor has his "formula" for teaching "intro circuits." FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-10-22T08:31:11Z SecondChildTAG: Just checked the link...He calculates the answer with a Bode plot! That's an engineer for you! SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-10-22T08:34:48Z SecondChildTAG: I believe his course covers the topic under the title sinusoidal steady-state. Then come filters and finally op amps. I have no idea why this course introduces MOSFETS so early and why BJT get a mention in the book (small signal analysis) but not in lecture. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-22T11:55:08Z IndexTAG: 1433 TitleTAG: circuit sandbox i know this a really stupid question at this point of the course ...but can anyone please tell me how we rotate an element(resistor, voltage source etc) while making a circuit ? UserIdTAG: 278792 UserNameTAG: sali CreateTimeTAG: 2012-10-22T03:33:40Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: "R" key works for me. FirstChildUserIdTAG: 405936 FirstChildUserNameTAG: Whible FirstChildCreateTimeTAG: 2012-10-22T03:35:50Z SecondChildTAG: Hi @sali! What Whible said it is true :). In order to rotate an element, click on that element and then press the "R" from your keyboard. For more information about you can take a look at here [Interactive Laboratory Usage Summary][1] P.D: All questions are welcome, remember that if you ask you learn and your doubt can be the doubt of others ;). [1]: https://www.edx.org/static/content-mit-6002x/handouts/schematic_tutorial.ba422f80d72b.pdf SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-22T03:50:36Z SecondChildTAG: thanx alot Whible and Myrimit :) SecondChildUserIdTAG: 278792 SecondChildUserNameTAG: sali SecondChildCreateTimeTAG: 2012-10-22T03:57:36Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-22T04:01:47Z SecondChildTAG: it took me 3 weeks to find that...:p SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-22T14:32:47Z IndexTAG: 1434 TitleTAG: Out of luck with Lab6... Hi everyone. Reading the textbook at page 526 I think I should set $v_{OUT}$'s value to $V_{OL}$ in Equation 6.5 at page 306 or use some approximation to it read from the transient analysis, like 250.330m or 248.148m, and then solve for $R_{ON}$. But it didn't work at all. After the deadlines, could someone please clarify this calculation? Thanks. ![][1] [1]: https://edxuploads.s3.amazonaws.com/13508696731343629.png UserIdTAG: 288637 UserNameTAG: gotchi CreateTimeTAG: 2012-10-22T01:35:28Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi gotchi! Can I help you? **Hint 1:** In your plot, you will observe your output voltage. What value does the $\color{blue}{blue}$ curve **tends to** ... after some time... ? **Hint 2:** Now, if you go to page 306 and review the RON of the inverter, can you get the RON? ;) FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-22T01:43:53Z SecondChildTAG: Thank you Myrimit. "Tends to" was the expression I was looking for! SecondChildUserIdTAG: 288637 SecondChildUserNameTAG: gotchi SecondChildCreateTimeTAG: 2012-10-22T02:23:28Z SecondChildTAG: You are welcome gotchi :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-22T02:46:28Z IndexTAG: 1435 TitleTAG: H8p3 - last question [solved] Can't get right answer for last question. My understanding of this problem is: - Q2 gate capacitance discharged through resistance Ron ( of Q3 MOSFET ) - Initial voltage of Q2 gate is VOH - Source voltage for RC circuit defined by voltage divider RPU and RON ( of Q1 MOSFET ) - It's necessary to determine the discharging time to voltage VIL Could someone point me out where my understanding is wrong? UserIdTAG: 329361 UserNameTAG: ikrukov CreateTimeTAG: 2012-10-21T19:38:35Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 6 FirstChildTAG: What is the minimum value that the power supply voltage VDD must be to ensure that both transistors are operating in the saturated region? Write an algebraic expression involving VIN and VT for this minimum value of VDD in the space provided below: HELP FirstChildUserIdTAG: 475448 FirstChildUserNameTAG: msamwelmollel FirstChildCreateTimeTAG: 2012-10-21T20:20:23Z FirstChildTAG: it seems that you understand the question correctly. Initial value is VOH, Final value is Vdrain of Q1 (voltage divider relation). Maybe you have to check for what you consider R in the time constant RC? FirstChildUserIdTAG: 127195 FirstChildUserNameTAG: princeofsudan FirstChildCreateTimeTAG: 2012-10-21T20:43:50Z SecondChildTAG: Thank you for answer. I think R == RON (for Q3 "store" MOSFET ) Am I wrong? SecondChildUserIdTAG: 329361 SecondChildUserNameTAG: ikrukov SecondChildCreateTimeTAG: 2012-10-21T21:10:59Z SecondChildTAG: Yes It Is 190 SecondChildUserIdTAG: 106816 SecondChildUserNameTAG: Laith SecondChildCreateTimeTAG: 2012-11-11T06:50:14Z FirstChildTAG: Q2 gate capacitance discharged through resistance Ron ( of Q3 MOSFET ) and Thevenin Rth ( of Q1 MOSFET, RPU) PS: You from Russia? FirstChildUserIdTAG: 464744 FirstChildUserNameTAG: attache FirstChildCreateTimeTAG: 2012-10-21T21:09:13Z SecondChildTAG: Attache, thank you for hint! I missed this Rth in my calculations. Everything is Ok now. P.S. Yes, I'm from Russia. SecondChildUserIdTAG: 329361 SecondChildUserNameTAG: ikrukov SecondChildCreateTimeTAG: 2012-10-21T21:19:13Z SecondChildTAG: I'm from Ukraine :) SecondChildUserIdTAG: 464744 SecondChildUserNameTAG: attache SecondChildCreateTimeTAG: 2012-10-21T21:26:07Z SecondChildTAG: Спасибо из Санкт-Петербурга! SecondChildUserIdTAG: 208296 SecondChildUserNameTAG: OZ1 SecondChildCreateTimeTAG: 2012-10-21T22:54:29Z SecondChildTAG: But i still wrong here. It seems I took everything into account. I have VOH=Vol+(VGSQ2-VOL)*(1-exp(-t/R*CGS)), where R=RON+RPU||ROFF, VGSQ2=3.995 V (this value is accepted) P.S: Привет из Томска. SecondChildUserIdTAG: 213452 SecondChildUserNameTAG: JoJosida SecondChildCreateTimeTAG: 2012-11-04T07:45:17Z SecondChildTAG: Sorry, I have written about part 4 here. SecondChildUserIdTAG: 213452 SecondChildUserNameTAG: JoJosida SecondChildCreateTimeTAG: 2012-11-04T07:47:06Z SecondChildTAG: Hi friends, Could someone tell me why shouldn't I consider just Ron in Q3 for my RC given that my final voltage (from the voltage divider Rpu , Ron in Q3)is known(.5)?. Shouldn't I take this last one as my voltage source for the RC? This is driving me crazy. Thanks for your help. SecondChildUserIdTAG: 369339 SecondChildUserNameTAG: vargaslen SecondChildCreateTimeTAG: 2012-11-05T09:23:33Z FirstChildTAG: Thanks pronceofsudan and attache - their answers help me to find right solution. FirstChildUserIdTAG: 329361 FirstChildUserNameTAG: ikrukov FirstChildCreateTimeTAG: 2012-10-21T21:20:34Z FirstChildTAG: Hi friends, Could someone tell me why shouldn't I consider just Ron in Q3 for my RC given that my final voltage (from the voltage divider Rpu , Ron in Q3)is known(.5)?. Shouldn't I take this last one as my voltage source for the RC? This is driving me crazy. Thanks for your help. FirstChildUserIdTAG: 369339 FirstChildUserNameTAG: vargaslen FirstChildCreateTimeTAG: 2012-11-05T09:24:28Z FirstChildTAG: Hi I am able to solve part1,2,3 of H8p3 but unable to solve Part4 and 5 can any one give me hint to solve this problem FirstChildUserIdTAG: 352647 FirstChildUserNameTAG: praveenjugge FirstChildCreateTimeTAG: 2012-11-07T08:17:23Z SecondChildTAG: Part 4: you should use simple net : Vs-Rpu-Cgs; in this case you will get simple equation VOH=VS*(1-e(-T/(Rpu*Cgs)))-->T=-Cgs*Rpu*ln((VOH/VS)-1) Part 5: I cant solve it yet SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-11-08T18:43:09Z SecondChildTAG: More pricesely: net is Vs-Rpu-Ron-Cgs; so Voh=Vs+(Vol-Vs)e(-T/((Rpu+Ron)*C)) SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-11-08T19:20:49Z SecondChildTAG: Part 5 solved.. SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-11-08T20:40:49Z SecondChildTAG: Dear Sergtronix Thank you very much I could finally solve part 4 with your guidance and hints Part5 not solved yet SecondChildUserIdTAG: 352647 SecondChildUserNameTAG: praveenjugge SecondChildCreateTimeTAG: 2012-11-09T05:45:47Z SecondChildTAG: I am using the formula Vil=Voh*e^(t/(R*C)) but not able to solve SecondChildUserIdTAG: 352647 SecondChildUserNameTAG: praveenjugge SecondChildCreateTimeTAG: 2012-11-09T07:32:39Z SecondChildTAG: To solve Part 5: Just imagine entire net. You do have previously charged capasitor from the right side, and discharge net from the left side.This left side contain Ron for Q3, Thevenin resistance for Q2 ( it is open now) and voltage divider based on opened Q1 and Rpu. Final equation is pretty simply and well known. I may tell that My mistake was crazy - Excel required more "brackets" :( SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-11-09T18:52:38Z SecondChildTAG: "Thevenin resistance for Q2"- here is typo Q1 of course SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-11-09T18:54:56Z SecondChildTAG: sergotronix please explain in easy way SecondChildUserIdTAG: 115703 SecondChildUserNameTAG: munimunna SecondChildCreateTimeTAG: 2012-11-09T19:07:32Z SecondChildTAG: Im ready to explain, but what? :) OK; What do you have? Voltage source, Rtotal and C=Cgs.Please find Voltage Source value and Rtotal.Hints are above. Initial conditions are: Vcgs, Vs, Q1 state is ON, Q3 state is ON Also, you have values for all components. Now your Cgs is discharged throw Rsummary by the Voltage source. Note here - Voltage source value not equal to VIL. While you have all required values you may write equation. This equation is pretty simple- it is for discharge previously charged capasitor with initial value V0 by some voltage source to the required value. You may find more [**here**][1] Sorry for my English :) [1]: http:// "https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/509a18aedb6fed1f0000000d" SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-11-10T16:43:13Z SecondChildTAG: [here][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/509a18aedb6fed1f0000000d SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-11-10T16:44:06Z IndexTAG: 1436 TitleTAG: Help with Formula Parsing I realize the grading equation parser may not be intuitive, but it is correct. If the grader will not parse your answer, then there is no doubt that you have missed a * sign, or a ( sign, or added a ) sign somewhere, or something else that the grader doesn't like. Go back and look, carefully! If you still need help, and want someone to look at it, you need to remove ALL the variables in your equation when you post them and replace them with X's, so you don't violate the honor code by giving out the answer. If you don't, then your equation/post could be edited or removed, and then you could be back to square one. UserIdTAG: 21541 UserNameTAG: JSChambers CreateTimeTAG: 2012-10-21T18:04:09Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: actually it would be nice if for exams grader will not count such tries FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-21T18:40:26Z IndexTAG: 1437 TitleTAG: @Staff Regarding Midterm examination schedule I've got college from 8am till 5pm and moreover It takes an hour for me to get back home from college. So could you consider my request and change the Midterm examination date to sunday. I live in Chennai,India so according to my local time the test concludes at 9:30 am on sunday which only leaves me an option to take it on saturday evening. It could be much better if I could take it on sunday afternoon or so. Please pardon my poor english and do consider my request. Thanks in Advance :-) UserIdTAG: 148523 UserNameTAG: span993 CreateTimeTAG: 2012-10-21T17:43:39Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Sorry that I have started and ended abruptly.. FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-10-21T17:44:07Z SecondChildTAG: Hey fellow student from chennai. No worries you can do your exam in evening time with no worries. SecondChildUserIdTAG: 433352 SecondChildUserNameTAG: venky_10 SecondChildCreateTimeTAG: 2012-10-21T17:47:10Z FirstChildTAG: The exam's actual duration is only 2 hours so you should be fine. FirstChildUserIdTAG: 154440 FirstChildUserNameTAG: Aahlad FirstChildCreateTimeTAG: 2012-10-21T18:16:41Z FirstChildTAG: Good news span993.... The deadline to submit the midterm exam is extended to 23:59 (11:59 pm) on Sunday, October 28th, Boston time. The exam will still be released on Thursday, October 25th at 00:01 (12:01 am) Boston time. FirstChildUserIdTAG: 368712 FirstChildUserNameTAG: jmen FirstChildCreateTimeTAG: 2012-10-23T16:39:56Z IndexTAG: 1438 TitleTAG: H6P2 Full Discussion with numerical details after deadline PLEASE!!! I'm so frustrated with this particular HW question. I managed to get the Min Vdd with a lot of "guided guesswork". Maybe Im thinking way too much. I feel like my mind has reached saturation at the moment, though thanks to MITx that threshold has increased over the past 5 weeks. So please, show me your ways, thanks :) -Allwyn UserIdTAG: 204745 UserNameTAG: allwynmendes CreateTimeTAG: 2012-10-21T17:01:56Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: After the deadline passes, there will be a complete explanation of the problem in the homework. Just go back to the homework and hit the "Show Answer" button where the "Check" button used to be. It is there for all homework where the deadline has passed. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-21T17:54:07Z FirstChildTAG: Wow thats amazing !!! Thanks a lot FirstChildUserIdTAG: 204745 FirstChildUserNameTAG: allwynmendes FirstChildCreateTimeTAG: 2012-10-21T18:34:54Z FirstChildTAG: I got Vth, but now I'm stuck on getting Rth. I short the output port and find Itest, but it equates to zero due to gm1 being equal to gm2. I used VIN = vgs for Q1 and -Vout = vgs for for Q2, but when I use them to calculate Itest, Itest becomes 0. It seems so simple, what am I doing wrong? FirstChildUserIdTAG: 132828 FirstChildUserNameTAG: rharris FirstChildCreateTimeTAG: 2012-10-21T20:23:46Z IndexTAG: 1439 TitleTAG: Lab 6 Part3 Hello, I have problems to calculate the falling time. As an example I tried to solve the task on page 533 in the textbook. Maybe I have something on my eyes, I cant find the failure. I got the right numbers, but one decimal power below. The correct answer is 1.928*10^-5 Anyone who can give me a hint? ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13508251841343685.jpg UserIdTAG: 389333 UserNameTAG: juergen CreateTimeTAG: 2012-10-21T13:14:32Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Hi juerguen, Be careful with the units, **C is in femto** units **f = 10^-15** and not in nano 10^-9 as you wrote :) See you, Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-21T13:24:46Z SecondChildTAG: Thanks, it was hard work to get to this point. therefore, for times I lost the survey. SecondChildUserIdTAG: 389333 SecondChildUserNameTAG: juergen SecondChildCreateTimeTAG: 2012-10-21T17:32:38Z SecondChildTAG: I got the same answer but it is not giving me the check mark what am I doing wrong SecondChildUserIdTAG: 160242 SecondChildUserNameTAG: Mlevins35 SecondChildCreateTimeTAG: 2012-10-21T19:43:04Z FirstChildTAG: juerguen, What is the program you are using do to the equations on? It looks nice. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-21T16:24:41Z SecondChildTAG: MathCAD Prime 2.0 but you can have an eye on smath-studio SecondChildUserIdTAG: 389333 SecondChildUserNameTAG: juergen SecondChildCreateTimeTAG: 2012-10-21T17:35:56Z FirstChildTAG: I am also not able to get the answer on page 533 in the textbook. What is the significance of ln? What is it, and how is it calculated? FirstChildUserIdTAG: 393045 FirstChildUserNameTAG: SrChasJC FirstChildCreateTimeTAG: 2012-10-21T19:13:40Z IndexTAG: 1440 TitleTAG: New to the class Am just joining the class and hope i can still make, please advise. UserIdTAG: 696379 UserNameTAG: Salawe CreateTimeTAG: 2012-10-21T11:43:47Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5083a3e4ce23fe2b0000006d FirstChildUserIdTAG: 154440 FirstChildUserNameTAG: Aahlad FirstChildCreateTimeTAG: 2012-10-21T11:53:23Z FirstChildTAG: check ut this post https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/507f58641323942b0000003c FirstChildUserIdTAG: 82651 FirstChildUserNameTAG: Ayazkazi FirstChildCreateTimeTAG: 2012-10-21T11:49:11Z FirstChildTAG: Hi Salawe! Welcome to 6.002x! Take a look at my answer [here][1] Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/507f58641323942b0000003c FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-21T13:28:57Z SecondChildTAG: I think Ayazkazi posted your answer as well, lol. SecondChildUserIdTAG: 154440 SecondChildUserNameTAG: Aahlad SecondChildCreateTimeTAG: 2012-10-21T18:01:14Z IndexTAG: 1441 TitleTAG: for instructors - course info I am sorry for posting with a hell of delay this happened b/c i registered late.i want to know that if this course would be repeated again or given at other time. UserIdTAG: 671224 UserNameTAG: mekdimafework CreateTimeTAG: 2012-10-21T11:16:56Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: ya...if you are late, don't be disappoint..it will be repeat in spring i.e. in march but its better to keep yourself up todate. FirstChildUserIdTAG: 114864 FirstChildUserNameTAG: Vikaash FirstChildCreateTimeTAG: 2012-10-21T11:35:12Z FirstChildTAG: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5083a3e4ce23fe2b0000006d https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/507f58641323942b0000003c FirstChildUserIdTAG: 154440 FirstChildUserNameTAG: Aahlad FirstChildCreateTimeTAG: 2012-10-21T11:55:42Z IndexTAG: 1442 TitleTAG: H6P1 <....deleted...> What is wrong with this expression?? i cant see a green tick UserIdTAG: 332360 UserNameTAG: srinivasav CreateTimeTAG: 2012-10-21T08:30:47Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Case Sensitive. Make 'K' Capital FirstChildUserIdTAG: 367886 FirstChildUserNameTAG: KashwinKohli FirstChildCreateTimeTAG: 2012-10-21T08:43:42Z SecondChildTAG: K should be in capital. SecondChildUserIdTAG: 107612 SecondChildUserNameTAG: gunpowder SecondChildCreateTimeTAG: 2012-10-21T13:58:34Z FirstChildTAG: Hi, **srinivasav** I requested to you that please avoid from uploading the exact answers of HW Questions.... Thnx... FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-10-22T11:49:33Z IndexTAG: 1443 TitleTAG: please tell me this question i cannot understand Now, suppose we short-circuit the compound battery. (This is very dangerous. NEVER do this to a large battery, such as a lead-acid battery in a car, or to a lithium-ion battery from your laptop. You MAY live to regret it, but you may not.) What is the current (in Amperes) you should expect to go through the short circuit? incorrect UserIdTAG: 572704 UserNameTAG: mitali1994 CreateTimeTAG: 2012-10-21T04:20:48Z VoteTAG: 2 CoursewareTAG: Week 1 / Battery model CommentableIdTAG: 6002x_S1E9_Battery_Model NumberOfReplyTAG: 2 FirstChildTAG: Could have specified the problem that you have. Incorrect answer is fine, but what did you do? What approach did you take? What answer did you get. When you do that you will eventually get a lot of helpful answers. :-) FirstChildUserIdTAG: 78396 FirstChildUserNameTAG: dharav FirstChildCreateTimeTAG: 2012-10-26T03:26:14Z FirstChildTAG: I think because a very high current will flow throw a wire which can't stand this ! Maybe FirstChildUserIdTAG: 1515242 FirstChildUserNameTAG: Muradalfgeeh FirstChildCreateTimeTAG: 2013-04-05T11:02:57Z IndexTAG: 1444 TitleTAG: @ Staff I don't understand how the assumption that the time varying part is small signal has been made. Here if we take the partial deivative (del V/del I) its zero right so shouldn't small signal circuit be a short circuit at any operating point at t=T and v=VI VI(t)=VI + Vi(T)?? Until now I was under the impression that small signal meant th e linearity of the v Vs. i curve over a small interval. I'm really confused. Please Help. Thanks in advance. UserIdTAG: 154440 UserNameTAG: Aahlad CreateTimeTAG: 2012-10-20T17:07:18Z VoteTAG: 2 CoursewareTAG: Week 4 / Small Signal Model CommentableIdTAG: 6002x_small_signal NumberOfReplyTAG: 1 FirstChildTAG: My bad, sorry for the trouble, didn't realize it was small signal with respect to time. :| FirstChildUserIdTAG: 154440 FirstChildUserNameTAG: Aahlad FirstChildCreateTimeTAG: 2012-10-20T17:15:24Z IndexTAG: 1445 TitleTAG: nice presentation This was very well explained. Thanks, well done. UserIdTAG: 12905 UserNameTAG: Baer CreateTimeTAG: 2012-10-19T16:35:01Z VoteTAG: 2 CoursewareTAG: Week 6 / Parallel plate capacitor CommentableIdTAG: 6002x_Parallel_Plate_Capacitor NumberOfReplyTAG: 0 IndexTAG: 1446 TitleTAG: STAFF: Trouble seeing the cursor during lectures. **Dear Dr. Agarwal** I'm watching your S12V13 lecture and as you point to equations and variables your cursor is so tiny it can't be found. So, at one point you say "and I join them by this exponential here", however there are 4 instances of exponential's on the page and you can't tell which one you are pointing to. If your lectures are already generated and you can't change them, I understand, however please consider changing them for the next class. Also, if you can't change them please reply and let me know so I don't re-post this. Sincerely, rharris Is anybody else having trouble with this? UserIdTAG: 132828 UserNameTAG: rharris CreateTimeTAG: 2012-10-19T15:07:50Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: You can email the staff using the email address "bugs@edx.org" and file a bug report with them, or you can put it in the Wiki. Dr. Agarwal will probably not personally read this post. The staff will be much more likely to fix the problems if you do either of the above. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-19T16:52:56Z SecondChildTAG: Thank you, I'll do that. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-19T21:07:47Z FirstChildTAG: please can any one help with question 1 i have tried to no success please i have no time thanks FirstChildUserIdTAG: 660899 FirstChildUserNameTAG: faithjumbo FirstChildCreateTimeTAG: 2012-10-20T07:42:03Z SecondChildTAG: Hi faithjumbo!, Can I help you? Of which problem are you referring about? Lab or Homework? SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-20T16:19:09Z IndexTAG: 1447 TitleTAG: Previous material guys i have just joined the course. Can anyone tell me where can I get the previous material..? UserIdTAG: 648723 UserNameTAG: saxenaanki CreateTimeTAG: 2012-10-18T16:12:09Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Welcome! Just look under [Courseware][1]! [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-10-18T23:43:48Z SecondChildTAG: Hi saxenaanki! Welcome to 6.002x! Nice to meet you! :). As planetscape said, go to the **Courseware Tab** [click on here][1], you will find the weekly materials. Also, you can read also *The Course Handouts* in the **Course Info** tab [click on here][2] . [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_6/ [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/info SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-19T03:06:56Z FirstChildTAG: To get the full benefit of the course, you should start at the beginning, and forget about the grade this semester. You should take it again for the grade another time. In the spring, this course will be offered again hopefully, and in the spring, there's a good chance **MITx** will offer a proctored exam certificate in addition to the free honor code certificate of completion that is currently offered. Regards: asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-10-19T17:59:42Z IndexTAG: 1448 TitleTAG: LAB 6. The 9 Inverter part.(Last queston) I can't seem to get the concept of 'period of oscillation' through 9 inverters to come up to an answer. Can anyone please help me thanks :) UserIdTAG: 367072 UserNameTAG: jansenlopez CreateTimeTAG: 2012-10-18T07:00:12Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi! Measure the time pick to pick and then divide by 9 FirstChildUserIdTAG: 299251 FirstChildUserNameTAG: TheRedBlackOne FirstChildCreateTimeTAG: 2012-10-18T07:10:52Z SecondChildTAG: I got it thanks :) SecondChildUserIdTAG: 367072 SecondChildUserNameTAG: jansenlopez SecondChildCreateTimeTAG: 2012-10-18T10:01:07Z IndexTAG: 1449 TitleTAG: Lab 6, Part 3, Propagation Time Delay Using Equation 10.67 derived in text from Eq: 10.66, with estimated Vth and Rth the time is coming to be 0.16 ns, but this indicating a wrong answer.Can anyone explain, Why? UserIdTAG: 455950 UserNameTAG: Wahabbaluch CreateTimeTAG: 2012-10-18T04:07:14Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi! Your Vc(t) in equation 10.66 should be equal to the VOL value from the schematic, i.e. 250 mV. Using equation 10.66, you can obtain the decay time. Hope that helps! FirstChildUserIdTAG: 143823 FirstChildUserNameTAG: NathanNadeson FirstChildCreateTimeTAG: 2012-10-18T14:26:30Z SecondChildTAG: Thanks alot! That really helped. :) SecondChildUserIdTAG: 288881 SecondChildUserNameTAG: Maxisokol SecondChildCreateTimeTAG: 2012-10-18T18:01:30Z SecondChildTAG: i have solved the question in the same way as you suggested,but not getting the green tick.plz help SecondChildUserIdTAG: 97359 SecondChildUserNameTAG: azeem179 SecondChildCreateTimeTAG: 2012-10-19T07:18:22Z SecondChildTAG: anybody here to help me SecondChildUserIdTAG: 97359 SecondChildUserNameTAG: azeem179 SecondChildCreateTimeTAG: 2012-10-19T15:35:43Z SecondChildTAG: same problm ed me :( SecondChildUserIdTAG: 228871 SecondChildUserNameTAG: ELINAKHAN SecondChildCreateTimeTAG: 2012-10-19T16:54:44Z SecondChildTAG: same here :/ SecondChildUserIdTAG: 108454 SecondChildUserNameTAG: Raven7281 SecondChildCreateTimeTAG: 2012-10-20T11:41:23Z SecondChildTAG: Just answer it upto one decimal point. Worked for me :/ SecondChildUserIdTAG: 108454 SecondChildUserNameTAG: Raven7281 SecondChildCreateTimeTAG: 2012-10-20T12:17:47Z IndexTAG: 1450 TitleTAG: H8P3: MEMORY help me please i have problem, in H8P3 these are the equations that I am using Q3 VG2(t) = VG2(inf) + (VG2(0) - VG2(inf)) * exp(-t/τ) VOH = VG2max + (VOL - VG2max) * exp(-t/τ) R = RON3 + RPU || ROFF1 but is wrong Q4 VG2min = RON / (RON+RPU) * VS VG2(t) = VG2(inf) + (VG2(0) - VG2(inf)) * exp(-t/τ) VG2(t) = VG2min + (VOH - VG2min) * exp(-t/τ) R = RON3 + RPU || RON1 but is wrong UserIdTAG: 58618 UserNameTAG: ingeniero13 CreateTimeTAG: 2012-10-18T03:37:37Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 7 FirstChildTAG: I have exactly the same problem:VOH = VG2max + (VOL - VG2max) * exp(-t/τ) R = RON3 + RPU || ROFF1 but is wrong And I don't know why. FirstChildUserIdTAG: 244225 FirstChildUserNameTAG: lerimock FirstChildCreateTimeTAG: 2012-10-18T10:34:02Z FirstChildTAG: I have been working in this problem for a whole day. it seems that there is a problem. If you calculate based on VG2max=5V Which is the VS voltage instead of VG2max=4V as you get from previous section of the problem the grader accept your answer as right. So. is it a grader problem or we are doing it wrong?? FirstChildUserIdTAG: 127195 FirstChildUserNameTAG: princeofsudan FirstChildCreateTimeTAG: 2012-10-18T11:08:15Z FirstChildTAG: As long as Q3 is conducting then VS is the driving voltage for the circuit. The maximum for VGS2 below VS occurs because Q3 cuts off so that charge is no longer supplied to the capacitor (Q2 gate). FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-18T11:31:35Z SecondChildTAG: So. The grader is wrong. SecondChildUserIdTAG: 127195 SecondChildUserNameTAG: princeofsudan SecondChildCreateTimeTAG: 2012-10-18T11:45:52Z SecondChildTAG: No SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-18T12:48:37Z SecondChildTAG: Don't get it. Skyhawk, could you describe more? SecondChildUserIdTAG: 208296 SecondChildUserNameTAG: OZ1 SecondChildCreateTimeTAG: 2012-10-21T22:32:45Z SecondChildTAG: I think that they should have mentioned that you should not consider the constraint from the previous part of the question which is (the store line is held at the same voltage as the drain of Q1). SecondChildUserIdTAG: 127195 SecondChildUserNameTAG: princeofsudan SecondChildCreateTimeTAG: 2012-10-22T11:38:32Z FirstChildTAG: Same problem with the last part with Vf, etc. What are the values? FirstChildUserIdTAG: 244225 FirstChildUserNameTAG: lerimock FirstChildCreateTimeTAG: 2012-10-18T12:39:13Z SecondChildTAG: Vf=? SecondChildUserIdTAG: 39980 SecondChildUserNameTAG: MarinoJr SecondChildCreateTimeTAG: 2012-10-18T19:07:22Z SecondChildTAG: Vth with discharge through Rth. Note the target level is VIL. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-18T19:33:27Z SecondChildTAG: Vf= Vfinal SecondChildUserIdTAG: 244225 SecondChildUserNameTAG: lerimock SecondChildCreateTimeTAG: 2012-10-18T20:26:51Z FirstChildTAG: The correct answer for VGS2Max doesn't have any connection with the capacitor Cgs2 and it's an extremely simple problem. We know that the gate of Q3 is connected to the drain of Q1.If you draw it you will notice that you will have a diode connected mosfet, with VGS3 = VDS3. We know that VGS3 > VT, so VDS3 >=1 .If you write KVL you will get that VGS2 = Vs -VDS3 => VGS2max = Vs - VDS3min .So, VGS2max = Vs - 1. Vs is not really 5V due to Roff.It is 4.99947, but insignificant smaller . ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13513596791343676.gif FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-10-27T17:38:36Z FirstChildTAG: Hi I am able to solve part1,2,3 of H8p3 but unable to solve Part4 and 5 can any one give me hint to solve this problem FirstChildUserIdTAG: 352647 FirstChildUserNameTAG: praveenjugge FirstChildCreateTimeTAG: 2012-11-07T08:18:18Z SecondChildTAG: part 4 also solved but not finding way how to solve part 5 any hint? pl SecondChildUserIdTAG: 352647 SecondChildUserNameTAG: praveenjugge SecondChildCreateTimeTAG: 2012-11-09T07:34:34Z SecondChildTAG: plz tell me the exact ans of H8P3 part 4,5 SecondChildUserIdTAG: 466478 SecondChildUserNameTAG: Akshay1990 SecondChildCreateTimeTAG: 2012-11-10T16:46:48Z FirstChildTAG: what is meant by parasitic resistance which is used in h8p3? FirstChildUserIdTAG: 115703 FirstChildUserNameTAG: munimunna FirstChildCreateTimeTAG: 2012-11-09T18:26:01Z IndexTAG: 1451 TitleTAG: Ring oscillator question “introduce a small asymmetry in the parasitic capacitances and turn "on" the circuit after time 0 by using a step voltage source for the power supply.” why? And why should we increase the gate voltage and resistor voltage to make it faster? Thanks UserIdTAG: 136959 UserNameTAG: christerpher CreateTimeTAG: 2012-10-18T01:31:36Z VoteTAG: 2 CoursewareTAG: Week 6 / Ring Oscillator CommentableIdTAG: 6002x_Ring_Oscillator NumberOfReplyTAG: 0 IndexTAG: 1452 TitleTAG: help please Hi every body, I have just registered in that course , i just want to ask about the penalities of homework and labs ,because i already missed the homeworks and labs of the seven fisrt weeks . thanks. UserIdTAG: 651832 UserNameTAG: adnane235 CreateTimeTAG: 2012-10-17T22:27:53Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: To get the full benefit of the course, you should start at the beginning, and forget about the grade this semester. You should take it again for the grade another time. In the spring, this course will be offered again hopefully, and in the spring, there's a good chance **MITx** will offer a proctored exam certificate in addition to the free honor code certificate of completion that is currently offered. Regards: asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-10-18T15:37:51Z IndexTAG: 1453 TitleTAG: Textbook formula 10.73?? Can someone please explain to me where the 4V VOH disappeared to between formula 10.72 and 10.73 in the textbook? The example should calculate how long it takes to go from 5/11V to 4.0V, but I don't see the 4V in formula 10.73. Also, at the end of page 534, it says > tpd, 1->0 But shouldn't that be tpd, 0 -> 1, as we were calculating the delay of the capacitor to go from 5/11V (low aka 0) to 4V (high aka 1)? UserIdTAG: 114881 UserNameTAG: xvink CreateTimeTAG: 2012-10-17T20:41:12Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Is there no one who can explain this to me? FirstChildUserIdTAG: 114881 FirstChildUserNameTAG: xvink FirstChildCreateTimeTAG: 2012-10-18T17:42:20Z FirstChildTAG: (10.72) gives a general relationship between vC, RL and CGS2. 3 lines below vC is set equal to 4V. In the calculation into (10.73) some arithmetic must be performed before taking the log. A part of this is to subtract 4 from 5. When log is taken, this term disappears (ln(1)=0) and (10.73) entails. Thus the 4 V is still there, but not plainly visible. ---------- I had the same notion and had to think about it to accept the books version as correct. Notice that on page 532 the determination of the propagation delay for a low to high transition at the **input** of the inverter is termed *tp,0-->1*. Thus subscript refers to the input. The increase in the input corresponds to a decrease in the condensator. On page 534 *tp,1-->0* again refers to the input which decreases as vc increases. FirstChildUserIdTAG: 144056 FirstChildUserNameTAG: Pesnfy FirstChildCreateTimeTAG: 2012-10-19T13:09:51Z SecondChildTAG: Thanks SecondChildUserIdTAG: 114881 SecondChildUserNameTAG: xvink SecondChildCreateTimeTAG: 2012-10-23T09:18:33Z IndexTAG: 1454 TitleTAG: can you help me, please? Hi every body, I registered in that course from the start but because of some complexities i couldn't join you effectively until now. here is my question ,can i join now? or am i too late? if i can, would you tell me what to do so i can start Immediately? otherwise, tell me when can i join again? by the way i can study all past lectures in a week or less. answer me as fast as you can? thanks a lot.... UserIdTAG: 192228 UserNameTAG: alsakhry CreateTimeTAG: 2012-10-17T19:25:23Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: in the "course info" section you can find all the information you need. IF you start now you'll just lose the points for the first assignments and Labs FirstChildUserIdTAG: 372321 FirstChildUserNameTAG: EnricoDona FirstChildCreateTimeTAG: 2012-10-18T13:09:44Z SecondChildTAG: thank you for responding to my question .but could you tell me how much points will i lose. thank you again SecondChildUserIdTAG: 192228 SecondChildUserNameTAG: alsakhry SecondChildCreateTimeTAG: 2012-10-18T16:27:56Z FirstChildTAG: To get the full benefit of the course, you should start at the beginning, and forget about the grade this semester. You should take it again for the grade another time. In the spring, this course will be offered again hopefully, and in the spring, there's a good chance **MITx** will offer a proctored exam certificate in addition to the free honor code certificate of completion that is currently offered. Regards: asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-10-18T15:37:24Z SecondChildTAG: thanks a lot for your concern. but do you know when will the next course start over again? regards:alsakhry SecondChildUserIdTAG: 192228 SecondChildUserNameTAG: alsakhry SecondChildCreateTimeTAG: 2012-10-18T16:33:26Z IndexTAG: 1455 TitleTAG: h6p1- guidance required Hello Myrimit Thanks for the info provided but I am still struggling to find answer for H6p1 part 1 (Vs-Vout)/R= K*(vIN-VT)*Vout^2 then re arranging and writing root <...deleted...> if i substitute vIN=VT i will get 0/0 so I am using L hospital Rule differentiating w.r.t vIN seperalty numerator and denominator I am getting [4*K*R*Vs]/[2*K*R*sqrt(1+4*K*R*(vIN-VT)*Vs)] if I type above as answer I am not getting this as right can you please give me hint where am I going wrong thanks a lot UserIdTAG: 352647 UserNameTAG: praveenjugge CreateTimeTAG: 2012-10-17T15:25:36Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: [2*Vs]/[sqrt(1+4*K*R*(vIN-VT)*Vs)] FirstChildUserIdTAG: 352647 FirstChildUserNameTAG: praveenjugge FirstChildCreateTimeTAG: 2012-10-17T16:06:11Z FirstChildTAG: It looks like you used Vs for VS. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-17T17:45:09Z SecondChildTAG: Hi praveenjugge! You are welcome ! What shyhawk says it is true :). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-17T21:17:27Z SecondChildTAG: Take a look again to your 1st Formula Post ;). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-17T21:20:29Z SecondChildTAG: Thanks I got answer with hazel1919 hint SecondChildUserIdTAG: 352647 SecondChildUserNameTAG: praveenjugge SecondChildCreateTimeTAG: 2012-10-18T06:17:34Z SecondChildTAG: A pleasure! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-18T19:10:30Z IndexTAG: 1456 TitleTAG: The offset voltage The offset voltage was like 4.9975 volts. When Peter said it's about 5 volts and drew 2.5mv up, it sounds like the output voltage is bigger than 5 volts. 7:33 - And here it will discharge, tending to *4.9925* volts. I think the number should be 4.995. 1000/(1+1000)*5 = 4.995004995 UserIdTAG: 564606 UserNameTAG: ArequipaRadio CreateTimeTAG: 2012-10-17T13:38:04Z VoteTAG: 2 CoursewareTAG: Week 7 / Bypass Capacitor CommentableIdTAG: 6002x_Bypass_Capacitor NumberOfReplyTAG: 1 FirstChildTAG: Exactly what I thought when they drew this graph with mid voltage 5V. Surely it should be 4.995-5 peak-to-peak FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-18T11:41:45Z IndexTAG: 1457 TitleTAG: links for textbook and videos not accessible Hi, When i follow the links for course videos or textbook in mozilla firefox version 16.0 i see only empty space in the video or textbook section of webpage. Pls help. UserIdTAG: 288210 UserNameTAG: Rajubalaji CreateTimeTAG: 2012-10-17T10:33:57Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Try using Chrome for now, or Firefox 15. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-17T13:05:17Z SecondChildTAG: I see a similar problem with chrome also.. pls let me know if there is some other way. Thanks in advance SecondChildUserIdTAG: 288210 SecondChildUserNameTAG: Rajubalaji SecondChildCreateTimeTAG: 2012-10-17T14:48:02Z SecondChildTAG: I could download videos for lecture and tutorials from Download video here link. Textbook is also accesible from today in firefox version 16. Thanks SecondChildUserIdTAG: 288210 SecondChildUserNameTAG: Rajubalaji SecondChildCreateTimeTAG: 2012-10-18T06:25:21Z FirstChildTAG: [Online Textbook Link][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/ Regards: asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-10-18T15:45:12Z SecondChildTAG: [JWplayer Video Link Week(1-6)][1] [1]: https://dl.dropbox.com/u/24096724/6002xMP4/menu.html# Regards:asadbhatti42 SecondChildUserIdTAG: 329051 SecondChildUserNameTAG: asadbhatti42 SecondChildCreateTimeTAG: 2012-10-19T18:06:58Z SecondChildTAG: again textbook is not accessible in either of the the links. pls help. videos are accessible. SecondChildUserIdTAG: 288210 SecondChildUserNameTAG: Rajubalaji SecondChildCreateTimeTAG: 2012-10-20T16:29:19Z IndexTAG: 1458 TitleTAG: H6P2 What is the Thevenin equivalent voltage What is the Thevenin equivalent open-circuit voltage of the small-signal circuit? Don't understand how K can be in the expression when Vth needs to have units of volts. worked on this endlessly - got all the rest of the parts done but stumped on this one. UserIdTAG: 18073 UserNameTAG: gburkhart CreateTimeTAG: 2012-10-16T23:08:54Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: The problem statement tells you what parameters **can** appear in the answer, but doesn't say that they must. Think about the name of the circuit! FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-16T23:48:36Z SecondChildTAG: OK thanks SecondChildUserIdTAG: 18073 SecondChildUserNameTAG: gburkhart SecondChildCreateTimeTAG: 2012-10-17T01:05:34Z SecondChildTAG: so much time spent on something so simple I have been getting the signs wrong for 17 years. SecondChildUserIdTAG: 18073 SecondChildUserNameTAG: gburkhart SecondChildCreateTimeTAG: 2012-10-17T01:39:38Z SecondChildTAG: I still could not get this part. only this part is left.plz help me. SecondChildUserIdTAG: 368981 SecondChildUserNameTAG: asimakhtar SecondChildCreateTimeTAG: 2012-10-19T01:10:52Z SecondChildTAG: it is a small circuit model - take out the large circuit components ... SecondChildUserIdTAG: 18073 SecondChildUserNameTAG: gburkhart SecondChildCreateTimeTAG: 2012-10-19T03:51:02Z SecondChildTAG: Thanks alot skyhawk, didn't know the circuit name could actually give you the answer. I get it now entirely. SecondChildUserIdTAG: 183507 SecondChildUserNameTAG: obiradaniel SecondChildCreateTimeTAG: 2012-10-19T11:04:10Z SecondChildTAG: Thanks a lot ***SKYHAWK***!!! Ssly appreciate ur efforts to help all of us,by posting in almost all the discussions :-) n so very true @obiradaniel :-P SecondChildUserIdTAG: 375305 SecondChildUserNameTAG: sandy07 SecondChildCreateTimeTAG: 2012-10-19T14:59:55Z SecondChildTAG: when we replace the small signal equivalent of MOSFET it will be a current source and a small voltage input .. now there is no resistor in the circuit then how i may find vo? SecondChildUserIdTAG: 227508 SecondChildUserNameTAG: bhavyab SecondChildCreateTimeTAG: 2012-10-19T22:01:01Z SecondChildTAG: Should I get the resistance of Q1 and Q2? And the value of VDD here is equal to minimum value? any hints.... SecondChildUserIdTAG: 182624 SecondChildUserNameTAG: Daoling SecondChildCreateTimeTAG: 2012-10-20T03:15:10Z SecondChildTAG: @bhavyab When u short the gate and drain of a mosfet, it behaves as a resistor.. says so in the question itself.. refer to page 417 of text for the resistance value.. SecondChildUserIdTAG: 457713 SecondChildUserNameTAG: ASHWIN_DT SecondChildCreateTimeTAG: 2012-10-20T17:21:40Z IndexTAG: 1459 TitleTAG: part n°4 Hey, I'm also new. I don't understand the part n°4 UserIdTAG: 632770 UserNameTAG: hunno_ayan CreateTimeTAG: 2012-10-16T19:23:03Z VoteTAG: 2 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 0 IndexTAG: 1460 TitleTAG: MOFSET amplifier experiment Hands on.This is great :) UserIdTAG: 443809 UserNameTAG: Duvindu CreateTimeTAG: 2012-10-16T19:18:40Z VoteTAG: 2 CoursewareTAG: Week 5 / Music Amplifier Demonstration CommentableIdTAG: 6002x_snd_d NumberOfReplyTAG: 0 IndexTAG: 1461 TitleTAG: A message from popoya to thewiredbear! :) Hi thewiredbear, Do you remember this Post? [Probably not for high school kids][1] Ok, the good thing is that I wrote about you to popoya in the old Forum (6.002x Spring) and he wrote me this after that: " *Hello Myrimit :D Long time no see :) Wow, it feels wonderful that someone felt nice after hearing my story :) I'd love to share my experiences with this student. My email id is radhikaghosal (at) gmail (dot) com* " I hope that you can put in contact each others as both have almost the same age and share similar things! Hahaha! This is so funny! How small is the World haha!Thank you to Internet this is possible haha! I hope that you can become friends! Popoya is an excelent student and a good kid! My best wish to both of you! See you! Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5071ccd71ed8e21f0000001c UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2012-10-16T18:47:56Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1462 TitleTAG: I just joined Hi there I have just joined the course, is there any point in me participating since I have missed like 6 weeks plus? UserIdTAG: 663176 UserNameTAG: jadjayen CreateTimeTAG: 2012-10-16T17:33:28Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You cannot get full marks as you can only drop 2 of the assignments. The mid terms start next week so you'll need to cover 6 weeks of material (unless you've already studied this course before) Alternatively, you could just audit the course. I'm pretty sure it will be offered again in Spring and you can join from the beginning that time. FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-10-16T18:29:29Z SecondChildTAG: Hi jadjayen! Nice to meet you! Welcome to 6.002x :) Unfortunately, what ashwith said it is true... and the Midterm Exam is the next week.... But remember that you can stay here, viewing the videos, reading the materials and participating in the Forum Discuss while you are waiting for the next 6.002x ;) . SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-16T19:04:00Z SecondChildTAG: To get the full benefit of the course, you should start at the beginning, and forget about the grade this semester. You should take it again for the grade another time. In the spring, this course will be offered again hopefully, and in the spring, there's a good chance **MITx** will offer a proctored exam certificate in addition to the free honor code certificate of completion that is currently offered. Regards: asadbhatti42 SecondChildUserIdTAG: 329051 SecondChildUserNameTAG: asadbhatti42 SecondChildCreateTimeTAG: 2012-10-18T15:30:43Z IndexTAG: 1463 TitleTAG: A MOSFET datasheet - how to get K, and understand? I'd like to create a constant current source for 3W LEDs. (Following [this][1]) Here is a MOSFET [datasheet][2] I'm looking at. I think I see $V_T$ ("gate threshold voltage") and is not exactly known (between 1.0 and 2.5V). Is there a way to get $K$ from the numbers in the tables? I looked at the first graphic (further down in the spec) and am I right that it's showing $I_{DS}$ over 10A for a $V_{GS}$ of 3.5V, so I guess this is a MOSFET designed for high currents? Rob [1]: http://electronics-diy.com/power-led-driver-circuit.php [2]: http://www.mouser.com/ds/1/149/FQP50N06L-56273.pdf UserIdTAG: 468623 UserNameTAG: RobNik CreateTimeTAG: 2012-10-16T15:09:07Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Sometimes you can calculate K from given transconductance for specific VGS by gm=K*(VGS-VT), but as I can see there is no VGS given, but VDS. In practice you will rarely use formulas for transistors - usually graphs is using. So lets take look on VI graph. As you can see the minimum VGS given is 3V, so min VDS=VGS-VT=3-1.75=1.25 (using VT=1.75). At this VDS current is 12A - your target current for 3W LED at 3V I=3/3=1A. As soon there is no smaller VGS on graph we can come to a conclusion that this MOSFET is too powerful for this application. But if you really want to figure out what K is, you can use this graph to calculate K from IDS=K/2*VGS^2 on triod-saturation transfer points for a few VGS and average result - I got about 3A/V^2 :). After calculation I got approximately VGS=2.327V - so you can try. But be aware that switch-current source model we use is an abstraction, and actual MOSFET characteristics is little different - it would be negligible for working ranges of specific transistor, but for this case you are using MOSFET on its range's boundary - so probably it will not work, most likely because VGS I calculated actually is little lower and become less then threshold voltage. FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-16T16:58:40Z SecondChildTAG: Thanks for the help! > we can come to a conclusion that this MOSFET is too powerful for this application Odd that he's got it in a circuit with 1W LEDs (~.3A). (The other link in my post.) I want to start this project but I guess I should wait to learn more and build a better power source. Apparently there is something called a *buck converter* which is more efficient, and maybe I can figure that out after studying capacitors and inductors. Is it correct that a fair amount of power is dissipated/wasted in the transistor? Ie, $P = I_{DS}V_{DS} $ Rob SecondChildUserIdTAG: 468623 SecondChildUserNameTAG: RobNik SecondChildCreateTimeTAG: 2012-10-16T23:01:21Z SecondChildTAG: There is some kind of mistake in that circuit - he said "The Q2 transistor specified will work up to about 18V power supply", but from specs this transistor is working up to 60V. I suspect he gave us wrong transistor. Also I suspect as soon he is using negative feedback to control current, power transistor can be used, because he relies on feedback, which switchs VGS, instead of constant current of saturation which set by specific constant VGS, so transistor works in its triode and saturation regions uniformly. Now about buck converter - it's just a switching regulator - it's more effective because regulates power consumed from source by switching on and off, instead, as first variant, just dissipating excess energy through transistor. It will be more effective, and more expensive. Last question - yes - just said about it :) SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-17T01:23:34Z SecondChildTAG: and a little note why use this schematic, instead of simpler MOSFET saturation mode - this schematic will work with not stable voltage source - it's tricky to achieve stable VGS with such source. But it's possible with zener diode :) SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-17T01:41:11Z SecondChildTAG: MOSFET current driver with zener stabilized gate voltage :) : http://www.falstad.com/circuit/#%24+1+5.0E-6+1.1685319768402522+50+5.0+50%0Ar+224+208+224+288+0+100.0%0Az+224+352+224+288+1+0.805904783+5.0%0Ag+224+352+224+400+0%0AO+320+288+384+288+1%0AO+224+288+176+288+1%0Ar+320+288+320+352+0+100000.0%0Aw+320+352+224+352+0%0A172+224+208+112+208+0+6+14.120000000000001+20.0+8.0+0.0+0.5+Source+voltage+deviation%0Af+400+240+464+240+0+1.0%0Ag+464+256+464+320+0%0Aw+464+80+224+80+1%0Aw+320+288+320+240+0%0Aw+320+240+400+240+0%0A174+224+288+320+288+0+100000.0+0.5198+Current+select%0Aw+272+272+272+240+0%0Aw+272+240+320+240+0%0Aw+224+80+224+208+0%0A174+464+80+464+144+0+10.0+0.5099+LED+resistance+deviation%0A162+464+144+464+224+1+2.1024259+1.0+0.0+0.0%0Aw+464+80+512+80+0%0Aw+512+80+512+112+0%0Aw+512+112+480+112+0%0A SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-17T02:06:41Z FirstChildTAG: And before using LED (which is not cheap :) put powerful resistor and make sure that current through it is stable. When choosing resistor consider, that due approximation in calculations and fact that we are working on the MOSFET range's boundary - current can become much bigger, let's say 15A, so your power supply should give at least 15A at 3V - P=15*3=45W, and resistor should stand this power. The value of the resistor should be approximately the same as a resistance of LED in it's working state, so 3V/1A=3ohm. Well, don't know where you can find 3ohm resistor 45W power :). But you can try with hope it will work and current will be 1A, so you need resistor 3W :) - do you love the smell of burnt electronics in the morning? ;) FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-16T17:19:07Z SecondChildTAG: Love the smell of solder in the morning, yes! Fried components... not so much... ;-) SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-19T00:49:35Z IndexTAG: 1464 TitleTAG: can I know when will this course commence again in on spring date ? I have joined this course but one month over of course started,can I know the course will commence again in spring and when details. UserIdTAG: 592021 UserNameTAG: fahimahmed12 CreateTimeTAG: 2012-10-16T14:37:16Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: it may be in mid-march...better you should visit the site regularly.. FirstChildUserIdTAG: 114864 FirstChildUserNameTAG: Vikaash FirstChildCreateTimeTAG: 2012-10-16T14:44:40Z FirstChildTAG: Hi fahimahmed12, Based on my experience in the Prototype Course of Circuits and Electronics 6.002x (Spring), we started on March 5th ... So, I guess that they will start the next course more less in that date :) FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-16T19:56:20Z IndexTAG: 1465 TitleTAG: H6P1: value of r0???? I got the value of r0 in terms of id which is not permitted.. plz help.. UserIdTAG: 346003 UserNameTAG: anupamshakya CreateTimeTAG: 2012-10-16T11:47:17Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: refer tutorial of vacuum triode in week 6 tutorials..u'll definitely get the answer.. FirstChildUserIdTAG: 368712 FirstChildUserNameTAG: jmen FirstChildCreateTimeTAG: 2012-10-16T12:44:44Z FirstChildTAG: Hi anupamshakya! Can I help you? Hint 1: Remember that you have to find the ro in terms of the large-signal model parameters K and VT and the bias voltages VGS and VDS. Hint 2: You can take a look also at the video that jmen suggested you, it is really good. Watch it [here][1]. Hint 3: Do you have your iD? ;). See you! Myriam. [1]: https://www.youtube.com/watch?feature=player_embedded&v=qaOljUCNMFs FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-17T02:22:48Z SecondChildTAG: @Myrimit i have solved this question for ro and got the correct answer but i have little bit doubt in it. in tutorial the resistance we got is the plate resistance but here we are finding the output resistance. SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-10-20T08:00:00Z SecondChildTAG: i got it.. SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-10-20T10:13:59Z SecondChildTAG: Well done vikaash :)! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-20T13:39:38Z SecondChildTAG: (2*k*VDS*(VGS-VT))^-1 SecondChildUserIdTAG: 224318 SecondChildUserNameTAG: ashwin12312 SecondChildCreateTimeTAG: 2012-10-21T04:03:10Z SecondChildTAG: what is wrong with that SecondChildUserIdTAG: 224318 SecondChildUserNameTAG: ashwin12312 SecondChildCreateTimeTAG: 2012-10-21T04:03:24Z SecondChildTAG: plz help SecondChildUserIdTAG: 224318 SecondChildUserNameTAG: ashwin12312 SecondChildCreateTimeTAG: 2012-10-21T04:13:24Z SecondChildTAG: you are writing k instead of K.. o SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-10-21T07:47:49Z IndexTAG: 1466 TitleTAG: How do we use Wolfram alpha for calculating Derivatives (S11V7)? How do we take an expression like **IDS=K/2*(VGS-VT)^2** and use Wolfram Alpha to take a 'partial derivative with respect to vgs of k over 2 vgs minus VT, all squared?' Like in the example discussed here: S11V7 https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_6/Small-Signal_Circuit_Models/ UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-10-16T08:07:47Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Okay, I have found the answer, I will post a tutorial. FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-10-16T08:36:28Z SecondChildTAG: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_small_signal_mosfet_exercise/threads/507d2c0a4e51ca2b00000167 SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-16T09:43:08Z IndexTAG: 1467 TitleTAG: Answers to homework and Labs Can Edx post the Solution for homeworks and Lab works after there deadline? It would be helpful to us UserIdTAG: 272417 UserNameTAG: MuKhan CreateTimeTAG: 2012-10-15T05:35:58Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: they do, tomorrow after everyone's deadline passes they post them FirstChildUserIdTAG: 238005 FirstChildUserNameTAG: isisbocardo FirstChildCreateTimeTAG: 2012-10-15T06:00:43Z FirstChildTAG: Go to the homework and lab after the deadline has passed. The 'Check' box will be replaced with a 'Show Answer' box, which gives a full explanation of the answers, including the specific parameters for each student's set of questions. FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-10-16T02:58:02Z IndexTAG: 1468 TitleTAG: Honor code question Am I allowed to use an online tool for derivating and other miscellaneous high school math operations that I have not done in years, or do I have to figure them out on my own? Either way is fine, I'm just checking to make sure I follow the rules. UserIdTAG: 230226 UserNameTAG: zoliking CreateTimeTAG: 2012-10-14T22:01:38Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: I posted a link to an online derivative calculator which I discovered and used in solving a problem, much as I use the calculator tool. I felt this was well within the bounds of acceptable behavior and not in any way a violation of the honor code. If this was a calculus course, that would be another matter. If this is a violation or questionable in any way, I will surely remove the posting, if it hasn't been done so already. However, I highly doubt that informing folks of the existence/availability of this type of tool is at all inappropriate. I stand to be corrected. FirstChildUserIdTAG: 141108 FirstChildUserNameTAG: rl777 FirstChildCreateTimeTAG: 2012-10-15T01:15:53Z FirstChildTAG: Calculating tools are fine. Wolfram, Microsoft Math, Mathematica, Maple, Matlab, etc. FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-10-15T01:29:16Z SecondChildTAG: Thanks. SecondChildUserIdTAG: 141108 SecondChildUserNameTAG: rl777 SecondChildCreateTimeTAG: 2012-10-15T01:37:11Z FirstChildTAG: You can use any tool you want to. Complete the midterm and final on your own and you're within the honor code. FirstChildUserIdTAG: 90796 FirstChildUserNameTAG: ChaunceyGardiner FirstChildCreateTimeTAG: 2012-10-15T02:24:44Z IndexTAG: 1469 TitleTAG: Useful Tool - Derivative Calculator In working H5P3, I found the following tool to be most useful - http://www.derivative-calculator.net/ I suspect that I will find it useful again. I also find the 6.002x calculator tool to be indispensable. I was scratching my head for a long time on H5P3, trying to figure out why I wasn't getting the correct numerical answer from a a simple manual calculation I was performing. I evaluated the expression I was working with using the calculator and discovered I'd made a simple, stupid arithmetic error. But I just couldn't see it, though it was staring me in the face. The calculator revealed the problem. UserIdTAG: 141108 UserNameTAG: rl777 CreateTimeTAG: 2012-10-14T21:51:28Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: My McAfee anti-virus blocked part of that site. I have no specifics on why...all I got was an advisory. FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-10-14T23:53:41Z SecondChildTAG: I had no problem with the site. It appears legit to me. It certainly worked mathematically. SecondChildUserIdTAG: 141108 SecondChildUserNameTAG: rl777 SecondChildCreateTimeTAG: 2012-10-15T01:20:47Z SecondChildTAG: My anti-virus didn't kill the site. It just said some parts were 'questionable' and gave no further information. I dunno. SecondChildUserIdTAG: 92346 SecondChildUserNameTAG: MobiusTruth SecondChildCreateTimeTAG: 2012-10-15T01:31:01Z FirstChildTAG: I swear by Wolfram Alpha http://www.wolframalpha.com. FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-10-16T03:05:39Z IndexTAG: 1470 TitleTAG: Mid term exam Guys, Something that I don't understand is that the mide term exam is from 25th to 27th of october....how do we get only 24 hours (1 day) instead of 2 days???... and only till lecture sequence 11 will be part of the test right? UserIdTAG: 308861 UserNameTAG: Ignaas CreateTimeTAG: 2012-10-14T19:14:33Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You are expected to complete the test within 24 hours of opening it. FirstChildUserIdTAG: 90796 FirstChildUserNameTAG: ChaunceyGardiner FirstChildCreateTimeTAG: 2012-10-15T02:27:41Z SecondChildTAG: thnx.. SecondChildUserIdTAG: 308861 SecondChildUserNameTAG: Ignaas SecondChildCreateTimeTAG: 2012-10-15T14:22:39Z IndexTAG: 1471 TitleTAG: Week 5 done Hey everyone, Thank you all for your support..I just finished my homework of week 5 and lab 5....100%.. also special thanks to myrimit for your brilliant explanations.... UserIdTAG: 308861 UserNameTAG: Ignaas CreateTimeTAG: 2012-10-14T19:12:56Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: hey can you help me with H5P3 1st question.?? thanks :) FirstChildUserIdTAG: 120850 FirstChildUserNameTAG: Nitin1A FirstChildCreateTimeTAG: 2012-10-14T19:17:04Z SecondChildTAG: In which part are you lost Nitin1A? Can I help you? SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T19:35:05Z SecondChildTAG: i got the expression for vout as (1+(2/(K*RS*sqrt(1+2*K*RS*(VIN-VT)))))*vin). i don't know where i have gone wrong can you help me ?? SecondChildUserIdTAG: 120850 SecondChildUserNameTAG: Nitin1A SecondChildCreateTimeTAG: 2012-10-14T19:58:57Z SecondChildTAG: Hi Nintin1A, Hage you got your iDS correctly from H5P2? Ok. The H5P3 ask you to find the small signal of vOUT, that is to say vout: Hint: From H5P2, you know from part 2 how is your vOUT. From H5P2 part 3 you have your iDS... NOTE: I have noticed that you have 5 parenthesis in the left and 6 parenthesis in the right... SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T21:04:37Z SecondChildTAG: *Have SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T21:04:53Z SecondChildTAG: like Myrimit said, first of all it is important to have the third question of the second problem correct then you can make the first question. SecondChildUserIdTAG: 308861 SecondChildUserNameTAG: Ignaas SecondChildCreateTimeTAG: 2012-10-15T00:50:10Z SecondChildTAG: Missing parens were a huge problem for me; double-check those. Also: Signs (+/-), case (VIN vs vIN etc), and missing "*" operators. Good luck! SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-15T03:37:16Z SecondChildTAG: thank u all :) SecondChildUserIdTAG: 120850 SecondChildUserNameTAG: Nitin1A SecondChildCreateTimeTAG: 2012-10-18T12:05:08Z FirstChildTAG: So happy for you Ignaas! Well done! :) FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-14T19:33:20Z SecondChildTAG: thnx Myrimit SecondChildUserIdTAG: 308861 SecondChildUserNameTAG: Ignaas SecondChildCreateTimeTAG: 2012-10-15T00:49:14Z IndexTAG: 1472 TitleTAG: solution of homework exercises Anyone please tell where I can find the solutions of the homework problems of earlier weeks. In the course info page I only found lecture slides from s1 to s11 UserIdTAG: 127488 UserNameTAG: iwannalearn CreateTimeTAG: 2012-10-14T18:05:11Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: No problem... Just go in the homework section you are interested with and press "Show Answer". FirstChildUserIdTAG: 297279 FirstChildUserNameTAG: ZlatkoF FirstChildCreateTimeTAG: 2012-10-14T18:19:33Z IndexTAG: 1473 TitleTAG: How to set a probe to x-axis as shown here? Please help me with this: How to set the probe to x-axis in transient analysis?![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13502337081343658.png UserIdTAG: 358549 UserNameTAG: Harshainertia CreateTimeTAG: 2012-10-14T16:55:20Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Double click on probe......... Ull get option for colors......in that...the last option will be this.... " X-axis" FirstChildUserIdTAG: 209930 FirstChildUserNameTAG: saikiraniitr FirstChildCreateTimeTAG: 2012-10-14T16:57:36Z SecondChildTAG: Hi Harshainertia! What saikiraniitr says it is true :). In the case that you need it, here are some Hints of lab 5 [Wiki Hints][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/myrimits-hints-links/ SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T19:50:32Z SecondChildTAG: I thank you. SecondChildUserIdTAG: 358549 SecondChildUserNameTAG: Harshainertia SecondChildCreateTimeTAG: 2012-10-16T04:45:34Z IndexTAG: 1474 TitleTAG: homework answers where can i find the answers for past homeworks? UserIdTAG: 285945 UserNameTAG: lindalapiso CreateTimeTAG: 2012-10-14T15:04:25Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: If you just go back to them in "Courseware", at the top, and click through to the problem set, you can click "show answers" at the bottom. This "show answers" button appears after the homework is due. When you click it, the full solutions are shown. FirstChildUserIdTAG: 393689 FirstChildUserNameTAG: etindell FirstChildCreateTimeTAG: 2012-10-14T15:16:14Z SecondChildTAG: Hi lindalapiso! What etindell said it is true :). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T19:51:56Z IndexTAG: 1475 TitleTAG: H5P2 3rd one after applying so many efforts not getting the holy green tick. please somebody help. i have calculated ids root and i know that i have to take the negative root even then not getting the green tick. its saying that Invalid input: Could not parse 'xxxxxx' as a formula.what to do ?? UserIdTAG: 90309 UserNameTAG: ashish06121991 CreateTimeTAG: 2012-10-14T10:20:17Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Be patient. Did you get right answers by applying your equation for other questions? If you do, you are on the right track and pay attention to letters. You should only use vIN, VT, RS, and K. FirstChildUserIdTAG: 172304 FirstChildUserNameTAG: Demohunter FirstChildCreateTimeTAG: 2012-10-14T10:36:09Z SecondChildTAG: Put * sign in proper places. I was ignoring vin(....) in my answer and was getting same warning. SecondChildUserIdTAG: 99628 SecondChildUserNameTAG: Pranjal16 SecondChildCreateTimeTAG: 2012-10-14T12:14:48Z SecondChildTAG: Unbalanced parens, missing operators ("*"), signs (+/-), case (vIN vs. VIN), all can give you grief, and sites like Wolfram Alpha may auto-correct and give you good numeric answers... KEEP TRYING! SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-15T03:51:21Z IndexTAG: 1476 TitleTAG: HW P1 last question!! I still cannot get da answer....plz help wid some hints...!! I had used Vs= Vs+(−Vs−).... but still wrong??? :( VT+ [sqrt(1+2VsKRl)]/KRl used here UserIdTAG: 127800 UserNameTAG: MNB CreateTimeTAG: 2012-10-14T06:53:29Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: use formula vt+(-1+sqrt(1+2vs.R.K))/(Rl.k) and get the answer FirstChildUserIdTAG: 611666 FirstChildUserNameTAG: Shaminder420 FirstChildCreateTimeTAG: 2012-10-14T07:00:43Z SecondChildTAG: after that do vin-vt SecondChildUserIdTAG: 611666 SecondChildUserNameTAG: Shaminder420 SecondChildCreateTimeTAG: 2012-10-14T07:01:18Z SecondChildTAG: from where this formula came?? SecondChildUserIdTAG: 526734 SecondChildUserNameTAG: darkknight90 SecondChildCreateTimeTAG: 2012-10-14T08:52:46Z SecondChildTAG: check the textbook chapter 7 SecondChildUserIdTAG: 611666 SecondChildUserNameTAG: Shaminder420 SecondChildCreateTimeTAG: 2012-10-14T09:05:28Z SecondChildTAG: 0.093 is the answer SecondChildUserIdTAG: 429149 SecondChildUserNameTAG: divyakavi11 SecondChildCreateTimeTAG: 2012-10-14T13:52:39Z IndexTAG: 1477 TitleTAG: What is V^2? What is V in K=A/V^2? Thanks, Hazel. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-13T12:20:18Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: It's a unit. Amps over volts squared. FirstChildUserIdTAG: 213386 FirstChildUserNameTAG: dmascenik FirstChildCreateTimeTAG: 2012-10-13T13:25:17Z SecondChildTAG: How do I know which voltage to use? SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-13T13:29:21Z SecondChildTAG: Its a unit of a number. Do not substitute any values for it. Take K as 1 in HWP1 SecondChildUserIdTAG: 362299 SecondChildUserNameTAG: anandbaskaran SecondChildCreateTimeTAG: 2012-10-13T16:12:19Z SecondChildTAG: I really do not understand that. Why 1? SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-13T16:32:26Z SecondChildTAG: I understand now, thank you! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-13T17:46:14Z SecondChildTAG: Thank you dmascenik ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T03:16:47Z IndexTAG: 1478 TitleTAG: Something I don't quite understand... When the equation... ***VO=VS-(RL*K/2)(VI-VT)^2*** ...is subtracted from **VO+vo=VS-(RL*K/2)(VI-VT)+2(VI-VT)vi+vi^2** Why is the term " **-(RL*K/2)** " left? This leaves... **vo=-(RL*K/2)(2(VI-VT))vi+vi** ...instead of... **vo=2(VI-VT))vi+vi** UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-10-13T08:46:06Z VoteTAG: 2 CoursewareTAG: Week 5 / Small Signal Mathematically Described CommentableIdTAG: 6002x_small_sig_math_des NumberOfReplyTAG: 1 FirstChildTAG: you forgot the big bracket: VO + vo = VS - (RL*K/2) * [(VI-VT) + 2(VI-VT)vi + vi^2] VO + vo = VS - (RL*K/2)*(VI-VT) - (RL*K/2)*2(VI-VT)vi - (RL*K/2)*vi^2 FirstChildUserIdTAG: 389333 FirstChildUserNameTAG: juergen FirstChildCreateTimeTAG: 2012-10-13T20:53:38Z IndexTAG: 1479 TitleTAG: Overdrive guitar pedal Does anyone know if the triode region distortion is the one generated by an overdrive? Maybe a fuzz? UserIdTAG: 241012 UserNameTAG: InakiL CreateTimeTAG: 2012-10-12T21:13:28Z VoteTAG: 2 CoursewareTAG: Week 6 / Amplifier operating point CommentableIdTAG: 6002x_amplifier_operating_point NumberOfReplyTAG: 2 FirstChildTAG: A distortion pedal for a guitar usually uses two diodes to "clip" the tops of the waveform. One diode is opposite orientation to the other, they shunt the peaks of the signal to ground, creating an effect that sounds similar to a tube guitar amp clipping. (They flatten the tops of the waveform). If I recall, these can be on the output of an opamp or in the feedback loop. A "real" tube amplifier has two types of distortion. One is a more "modern" sound were distortion is created by a big signal that does not fit into the next stage, this "clips" the waveform peaks. The second type of distortion is in the power stage, when the power tubes start to run out of power, it cannot recreate the peaks, in effect clipping the waveform. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-12T22:06:08Z SecondChildTAG: You can also use LEDs in a distortion pedal, as they are really just diodes. You can also use different types of diodes for different sounds, due to their different forward voltage thresholds.(VT) Silicon diodes create harder clipping sound (distortion) then say germanium diodes which clip a little softer and sooner, sounding a little more "tube" like then the silicon diodes. If I recall correctly. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-12T22:11:07Z FirstChildTAG: Marshall amps bias the push-pull (tube) pre-amp so that the two half-amps overlap at what should be the zero crossing instead of a clean transition between the two halves of the output signal. This creates some distortion and a sound that is (or was) unique to their brand. I've been away from the music scene for quite awhile so this may be old information. The same technique can be used with transistors but never in the final stage of any type of amplifier. The main issue with clipping in the output stage (finals) of power amplifiers is the resulting square wave generates a significant DC component that will turn the speaker voice coil into a very poor and short-lived heater...... The other stages are usually capacitor-coupled to block the DC but the final power output rarely has them, often being transformer-coupled. FirstChildUserIdTAG: 375500 FirstChildUserNameTAG: Brej7665 FirstChildCreateTimeTAG: 2012-10-14T02:05:46Z SecondChildTAG: Interesting. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-15T00:13:37Z IndexTAG: 1480 TitleTAG: Help me with H5P1 part3 please [Solved] After simulating and counting a lot, still cant get the answer for this one. Here are my formulas I begin with: $V_0=V_s-\frac{KR_l(V_{GS}-V_t)^2}{2}$ from textbook 7.14 $V_0=V_{in}-V_T$ and $V_{GS}=V_{in}-V_s$ as saturation condition and Vgs. Now counting $\frac{KR_l}{2}=A$ i conclude to this quadratic equ: $V_{in}-V_T = V_s-A(V{in}^2 + V_s^2 - 2V_{in}V_s)$ Solving this gives me the wrong solution. Pls aid me. I use R that was accepted, K as 0.001, and $V_s=1$ everywhere Thanks p.s. edited A=KR/2 for better understanding [SOLVED] Thank you both skyhawk(especially) and Khanth for your help. I have made my calculations once more, with both of your method, and I got the solution. Note if someone follows my equations and skyhawks help considering $V_{GS} = V_{in}+V_S$, then Vs = 1 is ok, however with Khanths method, as he nicely explains, Vs=2. (https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5076e36fccb9e81f00000082) Thanks once more. (tbh, as i graphically evaluated a solution x, and with algebra i got x-0.01), this part of the homework was extremely annoying for me. :( ) UserIdTAG: 329254 UserNameTAG: KGabor CreateTimeTAG: 2012-10-12T12:13:57Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 6 FirstChildTAG: I believe that VGS = Vin + VS. See if that works. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-12T12:29:56Z SecondChildTAG: Thanks for your comment, but it did not work out. Got -1.9 and -0.3, however my simulation says it has to be positiv. :( Any other tips? SecondChildUserIdTAG: 329254 SecondChildUserNameTAG: KGabor SecondChildCreateTimeTAG: 2012-10-12T12:57:36Z SecondChildTAG: VGS=VIN-(VS-), and to find max VGS you need to use formula from here https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/MOSFETAmplifiersSmallsignalmodels/ (in this ling VI=VGS) SecondChildUserIdTAG: 500965 SecondChildUserNameTAG: barka0 SecondChildCreateTimeTAG: 2012-10-12T13:06:21Z SecondChildTAG: No other ideas, but I can confirm that the answer is a small positive number. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-12T13:14:34Z SecondChildTAG: Do you suggest that I use: $V_T + 1/KR_l * \sqrt{1+2V_sKR_l} -1 $ ? Because there I get $0.5 + \frac{\sqrt{17}-1}{8}$ which is not accepted... this one is driving me crazy... SecondChildUserIdTAG: 329254 SecondChildUserNameTAG: KGabor SecondChildCreateTimeTAG: 2012-10-12T13:16:09Z SecondChildTAG: also Vs=(vs+)-(vs-) SecondChildUserIdTAG: 500965 SecondChildUserNameTAG: barka0 SecondChildCreateTimeTAG: 2012-10-12T13:33:52Z SecondChildTAG: I have recalculated the upper and my original equations with $V_S = 2V$, still no green tick. I have reached the point where I calculate $\sqrt{9}$, cause I am not sure its 3 anymore. :-( SecondChildUserIdTAG: 329254 SecondChildUserNameTAG: KGabor SecondChildCreateTimeTAG: 2012-10-12T13:47:14Z SecondChildTAG: Am I at least close to the answer? SecondChildUserIdTAG: 329254 SecondChildUserNameTAG: KGabor SecondChildCreateTimeTAG: 2012-10-12T14:08:42Z SecondChildTAG: I just get the answer, look harder in your digits and signs SecondChildUserIdTAG: 286954 SecondChildUserNameTAG: ododo SecondChildCreateTimeTAG: 2012-10-12T14:10:04Z SecondChildTAG: very close got formula, put your NUMBERS in it SecondChildUserIdTAG: 286954 SecondChildUserNameTAG: ododo SecondChildCreateTimeTAG: 2012-10-12T14:50:27Z SecondChildTAG: I am not going anywhere with this. Do I get this right, that the formula $V_T + \frac{\sqrt{1+2V_SKR_l}-1}{KR_l}$ is the right one, and I need to use 2*Vs = 2Volts there? Because if thats so, i gottat get a new calculator or a new brain SecondChildUserIdTAG: 329254 SecondChildUserNameTAG: KGabor SecondChildCreateTimeTAG: 2012-10-12T15:14:40Z SecondChildTAG: The VS supply voltage is the total voltage across the mosfet from source to drain. This one gets complicated by the different uses of VI and VS subscripts so be careful with the labeling. SecondChildUserIdTAG: 256543 SecondChildUserNameTAG: sidney23 SecondChildCreateTimeTAG: 2012-10-12T15:41:17Z SecondChildTAG: Please see this post https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5076e36fccb9e81f00000082 SecondChildUserIdTAG: 82571 SecondChildUserNameTAG: Nidhi3 SecondChildCreateTimeTAG: 2012-10-12T16:58:11Z SecondChildTAG: THANK YOU Nidhi3! SecondChildUserIdTAG: 384339 SecondChildUserNameTAG: nstrento SecondChildCreateTimeTAG: 2012-10-12T17:36:31Z SecondChildTAG: numbers and sign - SecondChildUserIdTAG: 286954 SecondChildUserNameTAG: ododo SecondChildCreateTimeTAG: 2012-10-12T23:36:15Z SecondChildTAG: I still cannot get da answer....plz help wid some hints...!! I had used Vs= Vs+(−Vs−).... but still wrong??? :( SecondChildUserIdTAG: 127800 SecondChildUserNameTAG: MNB SecondChildCreateTimeTAG: 2012-10-14T06:48:59Z SecondChildTAG: 0.5+(√33−1)/8 correct?? SecondChildUserIdTAG: 127800 SecondChildUserNameTAG: MNB SecondChildCreateTimeTAG: 2012-10-14T06:56:57Z SecondChildTAG: 1. List item SecondChildUserIdTAG: 127800 SecondChildUserNameTAG: MNB SecondChildCreateTimeTAG: 2012-10-14T06:57:14Z SecondChildTAG: I have just got the right answer, but I don't know why Vs = 2 Volts? SecondChildUserIdTAG: 653 SecondChildUserNameTAG: ziyou SecondChildCreateTimeTAG: 2012-10-14T09:43:15Z SecondChildTAG: how to calculate Vs SecondChildUserIdTAG: 276808 SecondChildUserNameTAG: DEBASMITAMAJUMDER SecondChildCreateTimeTAG: 2012-10-14T09:47:59Z SecondChildTAG: well dat is bcoz it is the Vs+ - (-Vs-) SecondChildUserIdTAG: 127800 SecondChildUserNameTAG: MNB SecondChildCreateTimeTAG: 2012-10-14T18:12:11Z FirstChildTAG: When you expanded (VGS-VT)^2 did you drop terms in VT? FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-12T14:04:31Z SecondChildTAG: yes + just noticed I forgot the ^2 from my 7.14 equation, so updated my original post. I think I am giving up on this one. At least for today SecondChildUserIdTAG: 329254 SecondChildUserNameTAG: KGabor SecondChildCreateTimeTAG: 2012-10-12T14:49:44Z SecondChildTAG: I just ran a check. If you make the sign correction that I first pointed out (VGS = Vin+1) and do the expansion of (VGS-VT)^ correctly, you will get the correct answer. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-12T16:01:47Z FirstChildTAG: One more comment: I found the combination VIN-VT occurring several times so I let x = VIN-VT and got a simpler quadratic expression IMHO. After solving for x, Vin = x + VT. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-12T21:04:27Z FirstChildTAG: Here's a factor to consider: The actual value of Vt in reference to -Vs. FirstChildUserIdTAG: 77598 FirstChildUserNameTAG: Spacecadet1974 FirstChildCreateTimeTAG: 2012-10-13T03:14:32Z FirstChildTAG: I don't want to try this anymore...no aha moment for me as well :) FirstChildUserIdTAG: 21687 FirstChildUserNameTAG: Benadicta FirstChildCreateTimeTAG: 2012-10-12T14:23:03Z FirstChildTAG: Glad to see it all worked out! FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-14T19:18:44Z IndexTAG: 1481 TitleTAG: Test for now, preview is faulty - Attention Dormsbee **General Comments** Be wary about using $v_{IN}$, $V_{IN}$, or $v_{in}$ too freely. Remember at all times that the equations for the mosfet model refer to $v_{GS}$, $V_{GS}$ and $v_{gs}$. Stick with these variables until you have to substitute for them. The later you replace them the less likely you are to get confused. Recall that $$v_{GS} = V_{GS} + v_{gs}$$ where $v_{GS}$ is the total signal across the gate and source of the mosfet. $V_{GS}$ is the DC component of $v_{GS}$ $v_{gs}$ is the AC component of $v_{GS}$ **H5P1** The previous comment about $v_{GS}$ applies to $v_{DS}$ and its AC and DC components also. Thus, $$v_{DS} = v_{OUT} - V_{S-}$$ UserIdTAG: 9673 UserNameTAG: xp42 CreateTimeTAG: 2012-10-12T06:56:51Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1482 TitleTAG: RL Here the value of Rl is not given then how it is possible to find the current through the circuit as it always make the value of Ids = 0 amp. UserIdTAG: 227508 UserNameTAG: bhavyab CreateTimeTAG: 2012-10-11T21:31:29Z VoteTAG: 2 CoursewareTAG: Week 5 / Two Terminal Connection Exercise CommentableIdTAG: 6002x_2_term_con_e NumberOfReplyTAG: 2 FirstChildTAG: See [this tutorial][1]. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_5/wk4t8/ FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-10-12T05:14:18Z SecondChildTAG: I saw tuturial, but without it I had the same result: vr=ir/(K*(VI-VT)), ir=6mV, 90.0 mA/V^2, VI=2.7V, VT=0.5V So I get 147V, but this is wrong. SecondChildUserIdTAG: 324219 SecondChildUserNameTAG: Dmitry79 SecondChildCreateTimeTAG: 2012-10-18T18:36:22Z FirstChildTAG: Remember that for a certain input voltage at the gate vGS the current iDS = (K/2).(VI - VT)^2 FirstChildUserIdTAG: 133606 FirstChildUserNameTAG: smadi FirstChildCreateTimeTAG: 2012-10-12T13:16:55Z IndexTAG: 1483 TitleTAG: Small Signal Analysis For those struggling with small-signal analysis(me!), here's a link to an interesting PDF: http://www.ittc.ku.edu/~jstiles/312/handouts/Steps%20for%20MOSFET%20Small%20Signal%20Analysis.pdf Sometimes it helps to have the same information presented in a slightly different way. Hope it helps. UserIdTAG: 21541 UserNameTAG: JSChambers CreateTimeTAG: 2012-10-11T15:50:30Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You will be introduce to this way on Week 6. FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-11T17:50:15Z IndexTAG: 1484 TitleTAG: What does Dv mean in H6P2? Dv? UserIdTAG: 299251 UserNameTAG: TheRedBlackOne CreateTimeTAG: 2012-10-10T23:03:02Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: derivative? FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-10T23:43:52Z FirstChildTAG: If this is with reference to H6P2, it is the amplitude of the input voltage swing. It probably means Delta voltage. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-11T00:16:15Z SecondChildTAG: any idea how to solve it? sky? SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-16T13:54:14Z SecondChildTAG: GOT IT! SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-16T16:43:16Z SecondChildTAG: And... it is? It's delta voltage? SecondChildUserIdTAG: 70519 SecondChildUserNameTAG: Fipe SecondChildCreateTimeTAG: 2012-10-22T05:58:45Z IndexTAG: 1485 TitleTAG: Error! I calculated everything and confirmed that all my calculations are correct. However, I still get my Vout volume lower than Vin volume although no distortion occurs. Since this is not an exercise, can anyone just give me the correct value of Vbias and I can backtrack my own workings to see where I wrong. Thanks in advance. UserIdTAG: 157273 UserNameTAG: ongchihang CreateTimeTAG: 2012-10-10T14:10:45Z VoteTAG: 2 CoursewareTAG: Week 5 / Music Amplifier Demonstration CommentableIdTAG: 6002x_snd_d NumberOfReplyTAG: 2 FirstChildTAG: My Vout in the graph are all higher values than Vin. FirstChildUserIdTAG: 157273 FirstChildUserNameTAG: ongchihang FirstChildCreateTimeTAG: 2012-10-10T14:12:25Z FirstChildTAG: How did you do the calculation? If you take VO = VS - RL*K/2*(VI-VT)^2; with VI=1.5, VS=1.6, RL=1000, K=10^-3, VT=1 VO becomes 1.475 This is < VI but that's OK. The amplification is only relevant for the small signals. FirstChildUserIdTAG: 266912 FirstChildUserNameTAG: pietvo FirstChildCreateTimeTAG: 2012-10-12T20:50:45Z IndexTAG: 1486 TitleTAG: Invalid Input? What does the message "Invalid Input" mean? Is the answer wrong? Below is the exact message I am getting: Invalid input: Could not parse 'VT+ (-1+sqrt(1+2K*RL*VS))/(K*RL)' as a formula Thanks. UserIdTAG: 135674 UserNameTAG: sowmyas CreateTimeTAG: 2012-10-10T08:34:36Z VoteTAG: 2 CoursewareTAG: Week 5 / MOSFET Amplifier Exercise 12 CommentableIdTAG: 6002x_mosfet_amp_e2 NumberOfReplyTAG: 1 FirstChildTAG: It might be the "2K", try "2*K". FirstChildUserIdTAG: 397595 FirstChildUserNameTAG: mholin FirstChildCreateTimeTAG: 2012-10-10T09:32:13Z SecondChildTAG: the correct is 2*K SecondChildUserIdTAG: 361823 SecondChildUserNameTAG: EliasOak SecondChildCreateTimeTAG: 2012-10-10T12:16:27Z SecondChildTAG: Thanks :) SecondChildUserIdTAG: 135674 SecondChildUserNameTAG: sowmyas SecondChildCreateTimeTAG: 2012-10-12T09:18:50Z SecondChildTAG: But how can you get that equatiion? SecondChildUserIdTAG: 398903 SecondChildUserNameTAG: PriteshAmrelia SecondChildCreateTimeTAG: 2012-10-14T05:33:05Z SecondChildTAG: Look at the previous lecture in the box in the bottom lefthand corner. SecondChildUserIdTAG: 270160 SecondChildUserNameTAG: Sethhhh SecondChildCreateTimeTAG: 2012-10-14T06:21:50Z IndexTAG: 1487 TitleTAG: h5p2 NEGATIVE ROOT Hello. I have soluted (green tick) the iDS for H5P2, but when i introduce the values for Vi, Rs, etc i have negative root. What happen with this? Thanks. UserIdTAG: 244225 UserNameTAG: lerimock CreateTimeTAG: 2012-10-10T08:34:27Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: how u find rs plzzz help us FirstChildUserIdTAG: 164689 FirstChildUserNameTAG: muhammadfaizan FirstChildCreateTimeTAG: 2012-10-12T13:31:42Z FirstChildTAG: can you give me a hint as to how you solved this - can't get the green tick FirstChildUserIdTAG: 18073 FirstChildUserNameTAG: gburkhart FirstChildCreateTimeTAG: 2012-10-11T00:39:24Z SecondChildTAG: how u find rs plzzz help us SecondChildUserIdTAG: 164689 SecondChildUserNameTAG: muhammadfaizan SecondChildCreateTimeTAG: 2012-10-12T13:31:22Z IndexTAG: 1488 TitleTAG: Page 28 lecture slide hanout Ids = Vs/RL + Vo/RL should be Ids = Vs/RL - Vo/RL Page 28 lecture slide hanout Ids = Vs/RL + Vo/RL should be Ids = Vs/RL - Vo/RL UserIdTAG: 352647 UserNameTAG: praveenjugge CreateTimeTAG: 2012-10-09T11:41:26Z VoteTAG: 2 CoursewareTAG: Week 5 / MOSFET Amplifier Graphically Described CommentableIdTAG: 6002x_mosfet_amp_graph NumberOfReplyTAG: 0 IndexTAG: 1489 TitleTAG: Simulation error The simulation is not working for me. UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-10-09T08:13:10Z VoteTAG: 2 CoursewareTAG: Week 5 / Music Amplifier Demonstration CommentableIdTAG: 6002x_snd_d NumberOfReplyTAG: 0 IndexTAG: 1490 TitleTAG: thanks in advance Dear, Please can you upload all the movies for week 5. I am having problem with the network connection. Each start of the week, i have to wait for middle of the week in order to start watching the movies and doing the corresponding HW. the last days for solving the HW are being hectic. I really appreciate if all the movies are uploaded in mp4 before the middle of the week. thanks a lot! You are doing an excellent Job, i really feel lucky to be able to take a course that is being offered in MIT. UserIdTAG: 141709 UserNameTAG: charbelantonios CreateTimeTAG: 2012-10-09T07:55:00Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: hey!! me too.. I'm still waiting for the videos. links aren't available. FirstChildUserIdTAG: 509517 FirstChildUserNameTAG: randima FirstChildCreateTimeTAG: 2012-10-09T08:01:37Z FirstChildTAG: Totally agree. Could you please upload the videos? Thank you. FirstChildUserIdTAG: 255161 FirstChildUserNameTAG: mlovelle FirstChildCreateTimeTAG: 2012-10-09T09:57:26Z FirstChildTAG: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/ FirstChildUserIdTAG: 122513 FirstChildUserNameTAG: Mona77 FirstChildCreateTimeTAG: 2012-10-09T12:05:59Z SecondChildTAG: The links for Weeks 5 and 6 lack the videos themselves, the only thing posted is the playlist for each week. That is not helpful for downloading. SecondChildUserIdTAG: 94557 SecondChildUserNameTAG: g_hopper SecondChildCreateTimeTAG: 2012-10-09T16:12:07Z SecondChildTAG: You are right, it's not helpful for downloading. Can you play them when you follow the links? SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-09T17:04:29Z SecondChildTAG: that's correct... pleas make it easier... SecondChildUserIdTAG: 509517 SecondChildUserNameTAG: randima SecondChildCreateTimeTAG: 2012-10-10T08:42:01Z FirstChildTAG: dear, we are still waiting for the movies to be uploaded in mp4, here, where i live, the connection is really bad... i have to wait an hour to watch a lecture of 5 minutes!!! thanks in Advance, FirstChildUserIdTAG: 141709 FirstChildUserNameTAG: charbelantonios FirstChildCreateTimeTAG: 2012-10-12T06:08:26Z IndexTAG: 1491 TitleTAG: Solutions to previous homeworks Can someone point me to where I can find solutions for the previous homeworks especially Week 4 HW ? UserIdTAG: 31767 UserNameTAG: GVNK CreateTimeTAG: 2012-10-09T03:58:37Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Go back to the homework, and hit the "Show Answer" button, where the check button used to be. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-09T04:40:24Z SecondChildTAG: What JSChambers says it is true :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-09T04:46:15Z SecondChildTAG: I think JSChambers thinks that HW is alike Practice Question.... SecondChildUserIdTAG: 329051 SecondChildUserNameTAG: asadbhatti42 SecondChildCreateTimeTAG: 2012-10-09T12:07:16Z SecondChildTAG: JSChambers is correct. After the submission deadline has passed for each homework and lab, the 'Check' button disappears and is replaced by a 'Show Answer' button. Click on it and you will get the full explanation of the answers, customized for your set of homework values. (Values are randomized for each student. For example, your question might have the voltage as 12V and the resistance as 120 ohms, my question might have 7.6V and 90 ohms.) SecondChildUserIdTAG: 94557 SecondChildUserNameTAG: g_hopper SecondChildCreateTimeTAG: 2012-10-09T16:09:31Z SecondChildTAG: Also, the explanations that are shown are much better than the ones they had for the last class. For the last class they just gave you the numerical answer without any explanation. Kudos to whoever did them this time! SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-09T16:16:27Z IndexTAG: 1492 TitleTAG: FAIL: H4P2 Could not find RL I could not find RL!! This is the only question in the entire homework and lab of week 4 I could not get! Now that it's past the due date, could someone explicitly give me the solution for my understanding? Many thanks! UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-08T09:30:39Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I had a hard time with this problem as well. I ended up getting it though, thankfully. For the circuit to work correctly you need to be in a region that the diode will work. The particular region that my diode worked (in the manner the problem was talking about) was if it had at least 5v going across it. Because the diode was opposing this voltage it was shown as a -5v on the diode graph. I then used VI, and the RIN + RL in series to find out what value of RL would cause 5v across the RL resistor. 5=RL*(VI/(RIN+RL)) We know VI and RIN. Solve for RL. I used them in series because until right at 5v the diode will resist all current and act like an open circuit. (Please correct me if I am wrong anyone) FirstChildUserIdTAG: 417864 FirstChildUserNameTAG: beauclark FirstChildCreateTimeTAG: 2012-10-08T13:22:18Z IndexTAG: 1493 TitleTAG: Simple question about H4P2 In H4P2 we are supposed to model the Zener diode as a 1 ohm resistor. How do I get that 1 ohm from the slope of the graph 1A/V? How do the two relate to one another? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-08T09:28:30Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: The slope (both of them) is 1 A/V and Om's law is I=V/R, so it kinda looks like the slope is 1/R. Thus R=1/(1/R). I think... FirstChildUserIdTAG: 416659 FirstChildUserNameTAG: Jonas3000 FirstChildCreateTimeTAG: 2012-10-08T11:29:57Z IndexTAG: 1494 TitleTAG: algebraic ninja I'm not following the cancelling here, most of this course skips the details of the maths as if it's nothing, yet it's impossible to get anything right without it, would be of great help to explain these parts more carefully as opposed to cancelling multiple elements in one line. spare a thought for us old fogies haven't been in maths class in decades. UserIdTAG: 61469 UserNameTAG: Spacedog CreateTimeTAG: 2012-10-07T23:59:52Z VoteTAG: 2 CoursewareTAG: Week 4 / Dependent Sources Example 2 CommentableIdTAG: 6002x_dep_src_ex_2 NumberOfReplyTAG: 1 FirstChildTAG: Hang tight. It's a careful balancing act...finding that 'just right' Goldilocks point in a lecture for so many. Don't hesitate to use a calculation tool...Wolframalpha dot com, for instance, or Microsoft Math (free download, I believe), or even Google (with Chrome browser...just type in the formula and it will show you a graph and solve for the variable, etc.. FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-10-08T02:56:25Z SecondChildTAG: I can't hype Wolfram Alpha enough - it's an amazing tool. I've also used [SAGE][1], which is similar. Don't forget that you can use the Circuit Sandbox to simulate circuits! Good luck! (fellow old fogie) [1]: http://www.sagemath.org/ SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-08T04:24:56Z SecondChildTAG: I have been using Microsoft Math, it is a free download (Thanks Microsoft), and it has been very useful because I have also not had a maths lecture for more than 10 years and most of the stuff I don't use very often has fled my mind. SecondChildUserIdTAG: 270011 SecondChildUserNameTAG: SnowmanZA SecondChildCreateTimeTAG: 2012-10-08T16:18:28Z SecondChildTAG: EDIT: Wolframalpha just became my new favourite. Microsoft Maths couldn't solve the integration in Lab7, but Wolframalpha could. SecondChildUserIdTAG: 270011 SecondChildUserNameTAG: SnowmanZA SecondChildCreateTimeTAG: 2012-10-08T19:37:04Z IndexTAG: 1495 TitleTAG: Week 4 **I can't understand How do we get voltage source from music ?** UserIdTAG: 283808 UserNameTAG: Hanboushy CreateTimeTAG: 2012-10-07T22:20:09Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I remember from our High-School physics course, that the microphone, the device for converting sound waves to voltage fluctuations, works on the principles of electromagnetic induction. That is, if you have a magnet and an coil, you can move the magnet through the coil and the current will appear in the latter. In the microphones they use a very tender membrane with a little magnet attached to it. If there's some sound fluctuations near the membrane, it starts vibrating too. If the membrane is vibrating, the attached magnet is vibrating too. If the magnet is vibrating, than you can put a coil near it, so that it would react on that magnet fluctuations. And this is it, you have a current, that fluctuates in the same frequency as the sound does. If you're interested in the concept, I recommend you to read about it on Wikipedia, because I don't remember it in details and could make some mistakes in the upper text. FirstChildUserIdTAG: 326787 FirstChildUserNameTAG: Angstrem FirstChildCreateTimeTAG: 2012-10-07T22:32:10Z SecondChildTAG: actually most of the modern microphones are electret: http://en.wikipedia.org/wiki/Electret_microphone SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-08T01:18:49Z IndexTAG: 1496 TitleTAG: Forum Reply Problem I'm having trouble replying to this post: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/507169160584032700000080 When I hit the submit button, nothing happens. Anyone else having this problem? Can someone else try replying to the post above and see if it works? Thanks. UserIdTAG: 21541 UserNameTAG: JSChambers CreateTimeTAG: 2012-10-07T16:55:39Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hi JSChambers, I couldn't too... So, I decided to look at his/her recent threads and I found that he/she has another Post, so I tried to Post there thinking that might his/her account had an issue, but not, it worked perfectly [Other Post][1]... So, it seems that it is a problem of that particular Post... See you! Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/undefined/threads/5071598c460b1d2700000090 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-07T18:36:46Z SecondChildTAG: Thanks for checking Myriam. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-09T15:20:30Z FirstChildTAG: ya same here. not able to reply on the post link you provided FirstChildUserIdTAG: 111320 FirstChildUserNameTAG: jhalife FirstChildCreateTimeTAG: 2012-10-07T17:11:09Z SecondChildTAG: Thanks jhalife. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-07T17:20:54Z IndexTAG: 1497 TitleTAG: H4P3, Part B: Something to think about with dependent sources I spent 5 hours on this problem yesterday and I finally got the answers. In the last 3 circuits involving dependent sources in which I have analyzed, the resistance of the dependent source has played a role in calculating Rth and Rn. It is either given in the problem statement, as with the last 2 problems we have solved, or it has to be calculated with the voltage across and the current through the dependent device. In solving this problem, everything I have calculated for u and Voc at port B were correct, so when I considered the resistance of the dependent source I got the answer. I believe this should hold true for any situation with a dependent source as long as the dependent source operates in the linear region, i.e., the resistance is held constant and doesn't change. I would sure like DR. Agarwal to address this topic and verify this. UserIdTAG: 132828 UserNameTAG: rharris CreateTimeTAG: 2012-10-07T15:47:00Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: But this resistance depends on the chosen operating point on the linear part of the slope, and it's the whole idea for small signal model.Watch please the week 6 lessons. FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-10-07T18:08:02Z SecondChildTAG: I agree. If you notice, I said "as long as the dependent source operates in the linear region, i.e., the resistance is held constant and doesn't change.", which is the same thing as choosing an operating point on the liner part of the slope. Do you agree? I'll definitely watch the week 6 lessons to gather further insight on the issue. - Thanks SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-07T18:23:06Z SecondChildTAG: Yes Harris, i agree it is constant as long as the bias point values remain constant.If you change the bias point , the resistor value changes with the new bias.Was this your observation ? Sorry if i wasn't much help. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-10-08T07:48:27Z SecondChildTAG: Yes, this was my observation. It helps just to talk about it, so yes you did help. - Thanks SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-10T18:19:54Z IndexTAG: 1498 TitleTAG: LAB 4 HELP NEEDED hello evryone, i got my circuit marked correct but on running TRANS ,it doen't show any plot.so plz help me in completing lab 4 by telling where i'm comitting mistake. plzzz UserIdTAG: 137918 UserNameTAG: jaisneha CreateTimeTAG: 2012-10-07T15:17:33Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: you have to take VDS value as triangular wave or any continous wave, not dc wave form. FirstChildUserIdTAG: 171103 FirstChildUserNameTAG: hemanth463 FirstChildCreateTimeTAG: 2012-10-07T15:22:15Z SecondChildTAG: to add 0 to 3v SecondChildUserIdTAG: 111320 SecondChildUserNameTAG: jhalife SecondChildCreateTimeTAG: 2012-10-07T16:51:46Z IndexTAG: 1499 TitleTAG: H4P3 PART B: Rth I am able to calculate Vth. I tried calculating Isc (In) but couldn't get the right answer. Then tried Rth since In = Vth/Rth i got the right answer for Rth by (R2||R3)+R1 but i feel that it is not right. i cant ignore the dependent source. can someone tell me why am i getting the right answer when i do (R2||R3)+R1. UserIdTAG: 244670 UserNameTAG: ravikiran1201 CreateTimeTAG: 2012-10-07T10:01:39Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You should not eliminate the dependent source because Norton and Thevinin are special cases of superposition theorem. You might be getting the correct answer due to numerals. Try solving the problem by giving a known voltage(Vt) and known current(It) and try to eliminate them finally.Hope this works well. FirstChildUserIdTAG: 273912 FirstChildUserNameTAG: raj2691 FirstChildCreateTimeTAG: 2012-10-07T11:06:19Z IndexTAG: 1500 TitleTAG: S8E0 allowed symbols If in S8E0 system doesn't accept your symbols, use this: **K1**, **K2**, **VI**, **RI**, **RO** don't use something like: R0, V1, r1, k1... UserIdTAG: 95521 UserNameTAG: Gatling CreateTimeTAG: 2012-10-07T10:00:55Z VoteTAG: 2 CoursewareTAG: Week 4 / Dependent Sources Exercise 0 CommentableIdTAG: 6002x_dep_src_exsz_0 NumberOfReplyTAG: 1 FirstChildTAG: are you sure? it's not working, VI,RI, even RO FirstChildUserIdTAG: 317836 FirstChildUserNameTAG: VitorRodrigues FirstChildCreateTimeTAG: 2012-10-07T23:37:07Z IndexTAG: 1501 TitleTAG: Norton/Thevenin Resistance for circuit w/ dependant sources H4P3 and S8E2 are almost identical circuits, and yet, the method I successfully used for calculating Rth in S8E2 didn't work for H4P3. What am I missing? In S8E2, I just looked from the perspective of the target port, and said, I've got two resistors in parallel and calculated the equivalent resistance. 850(750)/1600 If I try the same for H4P3, I get a wrong answer. What's going on? UserIdTAG: 287805 UserNameTAG: Beneficial CreateTimeTAG: 2012-10-07T08:08:26Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Because you're not allowed to turn off the dependent source. You can turn off only independent sources. FirstChildUserIdTAG: 326787 FirstChildUserNameTAG: Angstrem FirstChildCreateTimeTAG: 2012-10-07T08:19:42Z SecondChildTAG: Both examples feature dependent sources, so you haven't addressed the discrepancy. SecondChildUserIdTAG: 287805 SecondChildUserNameTAG: Beneficial SecondChildCreateTimeTAG: 2012-10-07T08:40:26Z SecondChildTAG: Yes, you're right, I'm sorry for useless answer. Actually, I've solved S8E2 by the same technique. It's really interesting to know, why for S8E2 it is correct... SecondChildUserIdTAG: 326787 SecondChildUserNameTAG: Angstrem SecondChildCreateTimeTAG: 2012-10-07T08:48:54Z SecondChildTAG: Probably, the difference between a correct and an incorrent answer in S8E2 is really small, and the autograder fails to see the difference? :) SecondChildUserIdTAG: 378267 SecondChildUserNameTAG: artfwo SecondChildCreateTimeTAG: 2012-10-07T10:00:28Z SecondChildTAG: PLZ say how to find In?(( I found Rn but In is always wrong SecondChildUserIdTAG: 192114 SecondChildUserNameTAG: chuba SecondChildCreateTimeTAG: 2012-10-07T11:17:13Z SecondChildTAG: How to find Rn PLZ SecondChildUserIdTAG: 106816 SecondChildUserNameTAG: Laith SecondChildCreateTimeTAG: 2012-10-07T11:24:28Z SecondChildTAG: Even i'm facing the same problem. I found the relation between Vth and U but unable to find U.. SecondChildUserIdTAG: 309722 SecondChildUserNameTAG: kaushikraghavan1992 SecondChildCreateTimeTAG: 2012-10-07T12:05:30Z SecondChildTAG: What relation between Vth and U you find? Can you write it, may be I can help you. SecondChildUserIdTAG: 192114 SecondChildUserNameTAG: chuba SecondChildCreateTimeTAG: 2012-10-07T12:12:34Z SecondChildTAG: @artfwo , I thought that might be the case. But if you compare my result with the answer which appears when you click the "show answer" button, you see that my answer agrees with theirs to within 1/1000. That strongly indicates that we arrived at answers by implementing the same methods, rather than the autograder being overly liberal in its application. So, I still wonder...why, apparently, are S8E2 and H4P3 treated differently. SecondChildUserIdTAG: 287805 SecondChildUserNameTAG: Beneficial SecondChildCreateTimeTAG: 2012-10-07T13:59:51Z SecondChildTAG: Actually, artfwo, I took a closer look at the questions and now I think maybe you were right and that it is just a case of an overly generous autochecker. Based on a comparison of these two questions, I gather that I'm supposed to treat the CCVS as a resistor, with a value which corresponds to the given constants (alpha and Z in each question respectively). Then, viewing from the target port, I have to resistors in series, in parallel with the other resistor. Solving this way in S8E2, I arrived at an answer which corresponds with the checker's to within 1/1000000000and-more-zeroes-than-I-care-to-count. It's just that the 4ohms of alpha is so much smaller than the 750 and 850 of the other resistors, it can almost be disregarded. You can't disregard Z in H4P3, though. Solving in a similar manner results in a correct answer that makes sense to me. Hope this helped anyone else who was confused by this. Take away: for calculating the equivalent resistance, treat the dependent source as having a resistive value. SecondChildUserIdTAG: 287805 SecondChildUserNameTAG: Beneficial SecondChildCreateTimeTAG: 2012-10-07T14:18:39Z SecondChildTAG: "I have two* resistors..." not "to resistors" SecondChildUserIdTAG: 287805 SecondChildUserNameTAG: Beneficial SecondChildCreateTimeTAG: 2012-10-07T14:20:36Z SecondChildTAG: Actually though, this approach is not giving me proper results for part B of H4P3, with the VCVS. Anyone care to help me out with that one? SecondChildUserIdTAG: 287805 SecondChildUserNameTAG: Beneficial SecondChildCreateTimeTAG: 2012-10-07T14:34:46Z SecondChildTAG: In H4P3 part A and S8E2 the resistance of the dependent source is given in the problem statement (it doesn't specifically say this, but the units of Z and Alpha are in Ohms), but in H4P3 part B you have to calculate the resistance once you calculate the voltage across the dependent source and the current through the dependent source. I then got In by using Vth/Rth, with Rth consisting of R1, R2 and the resistance of the dependent source in a series parallel resistive network. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-07T16:18:19Z IndexTAG: 1502 TitleTAG: dead line How can I calculate the dead line time in my country ?? UserIdTAG: 290966 UserNameTAG: AhmedImam CreateTimeTAG: 2012-10-06T22:29:41Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I believe deadlines are midnight in _your_ own timezone. This certainly seemed to be the case last week. I've submitted HW few minutes before midnight European time :) FirstChildUserIdTAG: 502508 FirstChildUserNameTAG: amaher FirstChildCreateTimeTAG: 2012-10-06T23:03:47Z SecondChildTAG: So you are saying that computers on east coast time get most time to finish the assignment? (Assuming all timezones get new material at the same time) SecondChildUserIdTAG: 142857 SecondChildUserNameTAG: viking2 SecondChildCreateTimeTAG: 2012-10-06T23:52:02Z FirstChildTAG: I BELIEVE ITS THE MIDNIGHT TIME IN BOSTON i.e.~(UTC-05:00) FirstChildUserIdTAG: 443660 FirstChildUserNameTAG: itsme71 FirstChildCreateTimeTAG: 2012-10-07T04:35:42Z IndexTAG: 1503 TitleTAG: Lab4 Help Hi, I need some help with the lab. I plotted the Mosfets and I get a green check for how I plotted it. However, my graph doesn't look the same as the picture. The slopes of my curves start small and then increase (Unlike the picture where the slopes of the curves start high and then decrease). Anybody have any ideas? And why does it give a green check if the graphs are different? UserIdTAG: 181745 UserNameTAG: Yoavrott CreateTimeTAG: 2012-10-06T18:05:29Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I think that the green check is for saving the draw. No matter if it`s wrong or not. The important thing is that the other exercises are corrects. About your problem, maybe you did put the Vtriangle how to Vgs, and Vdc how to Vds. Try in the other way, Vtriangle = Vds, and Vdc = Vgs. Excuse my bad english. I hope you understand me. FirstChildUserIdTAG: 87950 FirstChildUserNameTAG: neoacademic FirstChildCreateTimeTAG: 2012-10-06T18:33:38Z SecondChildTAG: Thanks neoacademic. That was in fact the problem. Silly mistake on my part. SecondChildUserIdTAG: 181745 SecondChildUserNameTAG: Yoavrott SecondChildCreateTimeTAG: 2012-10-06T19:20:50Z SecondChildTAG: the green check is that the graph is right (not only saved).-------------------- neoacademic, about your english, when you say "x how to y" the right is "x like y" . it's a normal error when your begginnig and your first language is portuguese(that is my case) or spanish because the meaning traduced apears to be the same, its only a syntax error. SecondChildUserIdTAG: 402850 SecondChildUserNameTAG: y1111 SecondChildCreateTimeTAG: 2012-10-07T20:48:55Z FirstChildTAG: I don't know for sure, but here's what I think: The grader in this particular lab doesn't care about what your curves look like. The grader only cares about your answers in the boxes below the circuit box. It does seem a little confusing, since usually the grader has cared about what happens in the circuit box. There should be a "RESET" button at the bottom of the page, which you can use to reset the whole thing if you need to. EDIT: When I say there should be a "RESET" button, I mean that it is already a feature. If you don't have it, that's a problem. It should be at the bottom of the page(the whole page, not just under the circuit box). FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-06T18:15:03Z SecondChildTAG: Yes, I agre with that, they should put a "reset" button at the end of every HW and lab! SecondChildUserIdTAG: 366267 SecondChildUserNameTAG: sergioglz_n SecondChildCreateTimeTAG: 2012-10-06T18:30:37Z SecondChildTAG: I gt the same graph and tick for my lab too!..my first question is right but the second one is wrong!..dnt know WHY? SecondChildUserIdTAG: 122513 SecondChildUserNameTAG: Mona77 SecondChildCreateTimeTAG: 2012-10-06T18:41:30Z SecondChildTAG: Mona, does your graph look exactly like the example? SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-06T19:25:18Z IndexTAG: 1504 TitleTAG: H4P3 RN what is wrong with the last problem ? I keep calculating RN and it keep telling me that I'm doing wrong isn't it calculated the same way as RTH ? (R1+R2) in parallel with R3 ? UserIdTAG: 528897 UserNameTAG: vepoo CreateTimeTAG: 2012-10-06T17:49:10Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You're right, it's calculated like Rth. But Rth is not equal to (R1+R2) in parallel with R3: you forget about a dependent voltage source. You can't turn it off while computing Rth (or Rn, that is the same)! You can turn off only independent sources. So, in order to compute Rth, you turn off only the dependent source, than you connect some current source I to the target port, than you compute the voltage across the target port. Rth would be equal to the target voltage over the connected current source I. FirstChildUserIdTAG: 326787 FirstChildUserNameTAG: Angstrem FirstChildCreateTimeTAG: 2012-10-06T19:41:47Z SecondChildTAG: what exactly do you mean by target voltage? SecondChildUserIdTAG: 173147 SecondChildUserNameTAG: cruiser_rahit SecondChildCreateTimeTAG: 2012-10-06T19:43:55Z SecondChildTAG: sorry..what exactly do u mean by the target port?..I am really unable to solve this problem SecondChildUserIdTAG: 173147 SecondChildUserNameTAG: cruiser_rahit SecondChildCreateTimeTAG: 2012-10-06T19:44:43Z SecondChildTAG: How about replacing the dependent voltage source with its resistance, Z = 2 ohms. That is what the problem statement says Z is. You then have 3 resistors to calculate Rth with. Does anybody dispute this as illegal? SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-06T20:20:37Z SecondChildTAG: I used this method in S8E1 also. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-06T20:22:48Z SecondChildTAG: Z=2 is given in the first part..where we need to find thevinin eqvt..what about the second part? SecondChildUserIdTAG: 173147 SecondChildUserNameTAG: cruiser_rahit SecondChildCreateTimeTAG: 2012-10-06T20:47:20Z SecondChildTAG: The "target port" is the port, corresponding to which you should find your Norton's I. For example:![enter image description here][1] here e1 and e0 terminals form the entity which I call "the target port". [1]: https://edxuploads.s3.amazonaws.com/1349558511648486.png SecondChildUserIdTAG: 326787 SecondChildUserNameTAG: Angstrem SecondChildCreateTimeTAG: 2012-10-06T21:22:25Z SecondChildTAG: thanks a lot..that really helped..is there a similar procedure to find norton current? SecondChildUserIdTAG: 173147 SecondChildUserNameTAG: cruiser_rahit SecondChildCreateTimeTAG: 2012-10-06T22:08:32Z SecondChildTAG: No problem) In = Vth/Rth - a trivial Ohm's low. For me, it was more comfortable to find Thevenin voltage first, and than Norton's current by Ohm's low. SecondChildUserIdTAG: 326787 SecondChildUserNameTAG: Angstrem SecondChildCreateTimeTAG: 2012-10-06T22:18:19Z SecondChildTAG: thanks a lot..I was trying to directly get In and screwing it up..finding Vth and then dividing by R is so much easier..:) SecondChildUserIdTAG: 173147 SecondChildUserNameTAG: cruiser_rahit SecondChildCreateTimeTAG: 2012-10-06T22:30:16Z SecondChildTAG: I got the current but can't figure out the resistance SecondChildUserIdTAG: 18073 SecondChildUserNameTAG: gburkhart SecondChildCreateTimeTAG: 2012-10-06T23:22:12Z SecondChildTAG: When you say, "Rth would be equal to the target voltage over the connected current source I". How can this be? The current I from the connected current source would divide between the leg with R1 and R2 and the leg with the dependent source. Therefore, by this logic, you would divide the voltage found through the node method by the current through the leg with the dependent source. I actually tried both, but neither worked. Pointers? SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-07T00:02:35Z SecondChildTAG: Now, 4 hours ago when I started this problem, I found "u" using the nodal method, with a node "u" between R1 and R2 and node V1 at port B. From this I found 2 equations with 2 unknowns ("u" and V1), then solved for both u and V1. From u, I found the voltage from the dependent source (2u), then used the resulting circuit to try and find Vth and In, then calculate Rth. It is not working, no matter what I do. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-07T00:11:23Z SecondChildTAG: I finally figured it out. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-07T08:28:12Z SecondChildTAG: what is R4 in that circuit SecondChildUserIdTAG: 346960 SecondChildUserNameTAG: priya91 SecondChildCreateTimeTAG: 2012-10-07T13:08:54Z IndexTAG: 1505 TitleTAG: Some Help for those struggling with H4P1 http://forum.allaboutcircuits.com/showthread.php?t=75112 UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-10-06T16:58:38Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: For what Prof. Agarwal would probably call those "funky" exponents like "3/2" ;-), see [Fractional Exponents][1] [1]: http://www.mathsisfun.com/algebra/exponent-fractional.html "Fractional Exponents" FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-10-06T17:54:31Z SecondChildTAG: I use Maths is fun all the time! A surprisingly good resource! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-06T18:38:20Z FirstChildTAG: hazel1919, is that your post on AAC? FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-06T19:16:23Z IndexTAG: 1506 TitleTAG: LAB 4 HINTS REGARDING LAB 4: we have to plot the VI characteristic of MOSFET (non ideal model) using circuit mixer in order to built the vi characteristic,we can do this way, step 1:applying any variable dc voltage sources(eg. rectangular ) to DS(drain-source) & constant dc voltage source GS(gate-source). step 2: the dc voltage for GS should be fixed for any particular value(0-3V). step 3: using voltage probe & current probe, by applying along the mosfet,we can measure the required parameters. step 4: set the voltage probe to x-axis ,which can be done this way ( double click on voltage probe & selecting x-axis from the choose option) step 5: when we run transient analysis ,we will get the graph b/w the Vds v/s Ids **whole procedure is for getting a single VI characteristic for a fixed value of GS,say GS1** in order to get multiple VI characteristic graph step 1: just copy the whole circuit we have made for GS1, for 3-4 times on your schematic graph by simply selecting the whole model and pasting it by taking cursor at different position. step 2:simply change the voltage along gate-source ,say Vgs2, Vgs3 , vgs4,etc step 3: run transient analysis and you will get multiple VI characteristic on your graph. UserIdTAG: 181793 UserNameTAG: mitrahul CreateTimeTAG: 2012-10-06T16:42:38Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1507 TitleTAG: Missing download links to lecture slides on 'Course Info' page edX Folks - The download links to the lecture slides on the 'Course Info' page are missing the Week 5 and Week 6 slides. I did find the links buried in the Courseware for Weeks 5 and 6, but that's very inconvenient. The Wiki page with direct video download links (no Youtube) are also missing Weeks 5 and 6. Our students who cannot access Youtube are missing out on this critical material. Additionally, the Wiki page with direct video download links lacks links to the tutorial videos. Lastly, some spell-check is in order for the Week 4 link page, the proper spelling for the headers is "dependent." Note the final 'E'. UserIdTAG: 94557 UserNameTAG: g_hopper CreateTimeTAG: 2012-10-06T16:25:51Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 2 FirstChildTAG: I see the lecture slides for Week 5, which are the MOSFET ones. But Week 6 slides are missing. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-06T17:56:20Z FirstChildTAG: Looks like they put up the slides for week 6. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-07T21:32:34Z IndexTAG: 1508 TitleTAG: lab 4 i am getting Voltage time characteristic instead of VI? what i do? please help.... UserIdTAG: 114864 UserNameTAG: Vikaash CreateTimeTAG: 2012-10-06T13:46:00Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: read how they plotted the first graph : Please observe that The Vtest voltage source generates a triangle waveform that ramps from 0V to 3V then back again over a period of 1ms. The stop time of the transient analysis is chosen to be 0.5ms so that the voltage on node vtest will be just the rising ramp from 0V to 3V. The voltage probe on the vtest node has its "Plot color" set to "x-axis", which asks the tool to use the values sampled by the probe as the x-coordinate when plotting. So the horizontal axis of the plot showing the transient results will be the voltage vtest instead of time. A current probe -- a short wire segment with a colored chevron showing the reference direction -- is used to plot the current flowing into the resistor. As with a voltage probe, one can double-click a current probe to select the plot color that will be used when plotting the current. You can use the label component (looks like a short straight line in the parts bin) to add names to circuit nodes. Nodes with the same name are considered to be electrically connected. FirstChildUserIdTAG: 454609 FirstChildUserNameTAG: Alwahsh FirstChildCreateTimeTAG: 2012-10-06T13:50:06Z SecondChildTAG: Hi vikaash, what Alwahsh said it is true. In order to see a voltage and not time in the x-axis, try to change the voltage probe with the option x-axis ;). If you are needing more hints of Lab4 you can read [here][1]. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5070e1b72210d02400000055 SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T05:47:15Z SecondChildTAG: thanks to both @Myrimitand @Alwahsh SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-10-07T09:09:13Z SecondChildTAG: :) You are welcome vikaash! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T09:47:44Z IndexTAG: 1509 TitleTAG: H4P2 RL!! Dear all, the following is my attempt to find the minimum value of RL. So, I formulated the following equation, putting it into wolfram alpha to solve for x, where x is RL: ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13495212946921393.gif The hope is that we can find the minimum value of RL where the Zener will be acting in the same piecewise linear region. But alas the answer of RL = -0.99999 is obviously wrong! How do I formulate the minimum value of RL correctly? UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-10-06T11:03:42Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: you have 13 V in source and 5 volts in RL and 8 in Ri. Voltage divider 13v*1k/(1k+RL)=8 so you find Rl FirstChildUserIdTAG: 152106 FirstChildUserNameTAG: JoaoBR FirstChildCreateTimeTAG: 2012-10-06T13:31:59Z SecondChildTAG: I still don't get it! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-06T16:52:22Z SecondChildTAG: its a bit tricky I guess. :) SecondChildUserIdTAG: 329015 SecondChildUserNameTAG: farah_sarwar SecondChildCreateTimeTAG: 2012-10-06T18:06:21Z SecondChildTAG: It's all so cryptic! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-06T18:32:55Z IndexTAG: 1510 TitleTAG: HOW to get Va? I am trying to solve this from last 2 hours but still have no idea how to?? I don't know the iterating Method and maximum i was able to solve was : Va/g+logVa=log20-log15 (**by substituting the the eua of Ia to KVL equation..**) How to get Va??? Please **Help** by telling me the steps i am missing UserIdTAG: 303485 UserNameTAG: iparitosh CreateTimeTAG: 2012-10-06T09:58:56Z VoteTAG: 2 CoursewareTAG: Week 4 / Linearization Exercise 1 CommentableIdTAG: 6002x_linearization_ex_1 NumberOfReplyTAG: 1 FirstChildTAG: 1. Express the eqn in the form Va=f(Va), i.e., Va only in LHS, don worry if another Va is present in RHS. 2. Substitute Va=1, in RHS, and see whether the expn on the RHS turns out to be 1. If so, Va=1! ese, plug in different values of Va, around 1 and see that the resulting value of the expn closes in to ur assumption. (Eg: Try Va=1. ull get some value on RHS. if its not equal, try Va=1.1 If the value on RHS further deviates from 1.1, then try 0.9 and so on.) 3. Va is correct when the expression on the right converges to the assumption on the left. FirstChildUserIdTAG: 357292 FirstChildUserNameTAG: RaghavAbboy FirstChildCreateTimeTAG: 2012-10-06T11:39:47Z SecondChildTAG: thanq raghav SecondChildUserIdTAG: 303485 SecondChildUserNameTAG: iparitosh SecondChildCreateTimeTAG: 2012-10-07T04:16:23Z SecondChildTAG: okey your answer is good, i did like this va=f(va)=5-(2*(10(1-e^(-va/5)))) the i made the table: va | ans 1 | 1.37 diverged a lot 1.1 | 1.05 converged near 1.05 | 1.21 diverged again must be under 1.1 1.09 | 1.08 converged it's really near so it must be beetween 1.09 and 1.085 | 1.095 diverged mas be beetween 1.085 and 1.09 1.087 | 1.0921 diverged again must be beetween 1.087 and 1.09 1.088 | 1.088943 converged must be this value... SecondChildUserIdTAG: 317836 SecondChildUserNameTAG: VitorRodrigues SecondChildCreateTimeTAG: 2012-10-07T05:49:01Z IndexTAG: 1511 TitleTAG: What difference between mesh and loop? What means mesh and what means loop? And what is difference between them? (I don't understand english very well). UserIdTAG: 104522 UserNameTAG: IgorNovice CreateTimeTAG: 2012-10-06T04:16:32Z VoteTAG: 2 CoursewareTAG: Week 4 / Schematic Neatness Tutorial CommentableIdTAG: 6002x_schematic_neatness_t NumberOfReplyTAG: 2 FirstChildTAG: Thanks!!! FirstChildUserIdTAG: 104522 FirstChildUserNameTAG: IgorNovice FirstChildCreateTimeTAG: 2012-10-06T12:07:11Z FirstChildTAG: Both Mesh and Loop are closed circuits however a loop may contain another loop inside it whereas a mesh can not have a loop inside it.It implies that the Mesh is itself the shortest possible loop and any other path for current flow is not possible in a mesh apart from the path of mesh. FirstChildUserIdTAG: 368712 FirstChildUserNameTAG: jmen FirstChildCreateTimeTAG: 2012-10-06T11:15:51Z IndexTAG: 1512 TitleTAG: possible way of looking at the problem Lets look at it this way. In the first case, the battery is the voltage source and the resistor is the dissipation. So, the power supplied by the battery to the resistor/power dissipated in the resistor is +2 W. In the second case, to find the power entering the resistor,we have to look at the resistor as the power source and the battery as the dissipation. Since the resistor does not provide any power, but only consumes it, the power entering the source/power provided by the resistor to the battery is -2 V UserIdTAG: 332470 UserNameTAG: Nithinsp CreateTimeTAG: 2012-10-06T04:09:47Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 5_Simple_Power NumberOfReplyTAG: 0 IndexTAG: 1513 TitleTAG: Why is the V_GS = V_I - V_O assumption made? The assumption that $V_{GS} = V_I-V_O$ means that $V_I$ and $V_O$ are measured to a common ground or node. Why and how is this assumption made? Is it implied? Is it always implied? UserIdTAG: 252722 UserNameTAG: vaboro CreateTimeTAG: 2012-10-06T02:00:12Z VoteTAG: 2 CoursewareTAG: Week 5 / MOSFET Amplifier with Source Degeneration CommentableIdTAG: 6002x_MOSFET_amp_w_source_degeneration_t NumberOfReplyTAG: 2 FirstChildTAG: Voltage always measures between nodes. If the second node is not specified then second node is the common ground. FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-06T03:03:33Z SecondChildTAG: All right, clear. Thank you! SecondChildUserIdTAG: 252722 SecondChildUserNameTAG: vaboro SecondChildCreateTimeTAG: 2012-10-06T11:18:08Z FirstChildTAG: I disagree that ids = (Vo-Vs)/Rs. This is the current source, no? FirstChildUserIdTAG: 457034 FirstChildUserNameTAG: Ichihara FirstChildCreateTimeTAG: 2012-10-10T23:21:13Z IndexTAG: 1514 TitleTAG: youtube link Sir please provide a separate link outside flash player to watch videos on youtube. It will help S60 users .I use nokia c5 and opera mini. UserIdTAG: 14726 UserNameTAG: Sajilck CreateTimeTAG: 2012-10-05T19:16:49Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Sir I am also finding it difficult to watch the video in flash player, please advise some alternative for watching tutorials. Thanks FirstChildUserIdTAG: 139589 FirstChildUserNameTAG: Adilyousaf FirstChildCreateTimeTAG: 2012-10-06T11:42:03Z IndexTAG: 1515 TitleTAG: How DC voltage source becomes short circuit and DC current source becomes open circuit? I got the point that for small signal model, v_s for voltages source is zero and i_s for current source is zero. But I didn't get why voltage source changes to short circuit and current source to open circuit. Can somebody help me? Myrimit,are you there for this problem? UserIdTAG: 159427 UserNameTAG: Suyog CreateTimeTAG: 2012-10-05T12:05:42Z VoteTAG: 2 CoursewareTAG: Week 4 / Small Signal Model CommentableIdTAG: 6002x_small_signal NumberOfReplyTAG: 5 FirstChildTAG: As of my knowledge... In order to make 0 volt and amp respectively. FirstChildUserIdTAG: 375373 FirstChildUserNameTAG: Siddappa FirstChildCreateTimeTAG: 2012-10-05T15:14:28Z FirstChildTAG: Voltage source do not prevent current from flowing (current on both of its terminals the same and comes from rest of the circuit) - so for superposition we replace it with short circuit. Current source prevents current from rest of the circuit from flowing through it - that is, current through current source is only its own current - so for superposition we replace it with open circuit. FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-05T16:05:09Z FirstChildTAG: Another explanation: If you take a look at IV of voltage source - you will see it's a line parallel to I axis. Now lets find its incremental resistance at any point - it will be deltaV/deltaI. As soon as it's parallel to I axis - deltaV=0, so incremental resistance will be 0/deltaI=0 - it's a short circuit. If you take a look at IV of current source - you will see it's a line parallel to V axis. Now lets find its incremental resistance at any point - it will be deltaV/deltaI. As soon as it's parallel to V axis - deltaI=0, so incremental resistance will be deltaV/0=infinite - it's an open circuit. FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-05T17:59:24Z FirstChildTAG: On suppression of a DC voltage source we may think it as having no effect on the circuit that is as a 0 voltage DC source so we can think it as a short circuit in case of current source when it is suppressed no current is suppose to flow through it so we must open circuit it . FirstChildUserIdTAG: 14726 FirstChildUserNameTAG: Sajilck FirstChildCreateTimeTAG: 2012-10-05T20:05:13Z FirstChildTAG: Hi Suyog! I haven't seen your Post before. Sorry for the delay :P. Lets see if I have understanded correctly your question. Did you mean when we considered a DC Voltage source as a short circuit and when a DC current source as an open circuit? **Current source:** Open circuit **when i=0 A**. Lets see this with an example: Suppose that you have this circuit, a simple resistive circuit. You can go to Textbook page 36 [here][1] ![imagen][2] If you see, a simple resistive circuit, it behaves like v=R*i Ok, if you see in the Graph, the current vs the voltage have the shape that you see in the green curve. But why? v=R*i for the Ohm's Law, if we do i/v=1/R (the slope of our curve);). Ok, lets see. If we open a resistive circuit, like the figure 1.43 (c) you will notice that as there is nothing to the right and the circuit doesn't close in a loop, you will have that the current it is zero i=0 A. So, this is the condition of open circuit, when i=0 A. Also if you see, at this point in the output you will have a voltage of vl = V because in the R you will not have voltage (0*R=0). If you back to the graph you will see that at the vl=V it is when it produces the open circuit, that is to say, if you take a look in the Graph when i=0 A (open circuit condition). ---- **Voltage Source:** In the voltage source the condition of short cut it is **when v=0 V**. Take a look at Figure 3.78 and Figure 3.79 [169][3]. See you! Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/ [2]: https://edxuploads.s3.amazonaws.com/13495928342651307.png [3]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/ FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-07T07:43:07Z IndexTAG: 1516 TitleTAG: Some help... 1)va=... I'm using google search to plot. For i=v^3(for example) I'm just typing y=x^3 - it will be our 1st graph. For 2nd graph I find 2 points, as in video - when x=0 and y=0. When y=0 - x=Vi When x=0 - y=Vi\R (or =I) After I make load line - y=I-(I/V)*x And the end I zoom graph and find a cross point of graphics. - x will be Va, y will be Ia. 2)Va/Vi - it's simple, just add 2% for Vi and find Va, using above method. Then divide Va(new)-Va/Vi(new)-Vi. 3) After lots of hours I found a solution - as dI=dV/Rinc. So Rinc=dV/dI, and 1/Rinc=dI/dV. So 1/Rinc=(10⋅(1−e^(−vA/5))' For deriving you can use online derivative calculator, and make sure it works ok. UserIdTAG: 295103 UserNameTAG: Syavick CreateTimeTAG: 2012-10-05T09:14:25Z VoteTAG: 2 CoursewareTAG: Week 4 / Linearization Exercise 1 CommentableIdTAG: 6002x_linearization_ex_1 NumberOfReplyTAG: 2 FirstChildTAG: Sayvick, wich calculator did you use? Can you give link? FirstChildUserIdTAG: 96873 FirstChildUserNameTAG: Santyaga FirstChildCreateTimeTAG: 2012-10-05T11:10:28Z SecondChildTAG: [Here][1] is an example. [1]: https://www.google.com/search?q=y=x%5E3&aq=f&oq=y=x%5E3&sugexp=chrome,mod=19&sourceid=chrome&ie=UTF-8#hl=pt-PT&sclient=psy-ab&q=y=-5*ln%281-x/10%29%20,%20y=5-2*x&oq=y=-5*ln%281-x/10%29%20,%20y=5-2*x&gs_l=serp.3...7646.13398.5.13996.8.8.0.0.0.2.460.2859.2-2j4j2.8.0...0.0...1c.1.shzRwI20ojo&pbx=1&bav=on.2,or.r_gc.r_pw.r_cp.r_qf.&fp=d326ebc8bcaa07f8&biw=1366&bih=667 SecondChildUserIdTAG: 305491 SecondChildUserNameTAG: lefam SecondChildCreateTimeTAG: 2012-10-05T15:57:03Z FirstChildTAG: For problem 3, if you use that equation for I wouldnt that imply that you have not linearized I? FirstChildUserIdTAG: 311818 FirstChildUserNameTAG: oyiemeke FirstChildCreateTimeTAG: 2012-10-05T18:51:24Z IndexTAG: 1517 TitleTAG: answer of 3rd qes if we have a constant value of voltage then we need not to evaluate rms value.. so power dissipated by resistor= v^2/r =120*120/110 =130.9090 UserIdTAG: 551747 UserNameTAG: arunamangal CreateTimeTAG: 2012-10-05T07:13:06Z VoteTAG: 2 CoursewareTAG: Week 1 / AC power CommentableIdTAG: 6002x_S1E3_AC_power NumberOfReplyTAG: 0 IndexTAG: 1518 TitleTAG: answer of 2nd qes.. average power= peak power/2 =261.8181/2 =130.9090 watt UserIdTAG: 551747 UserNameTAG: arunamangal CreateTimeTAG: 2012-10-05T07:08:52Z VoteTAG: 2 CoursewareTAG: Week 1 / AC power CommentableIdTAG: 6002x_S1E3_AC_power NumberOfReplyTAG: 0 IndexTAG: 1519 TitleTAG: answer of 1st qes.. for solving peak power we should use this formula.. peak power=Vrms*Vrms/R so peak power= 120sqrt2*120sqrt2/110 =261.8181 watt which is answer... i hope it will help u. UserIdTAG: 551747 UserNameTAG: arunamangal CreateTimeTAG: 2012-10-05T06:59:52Z VoteTAG: 2 CoursewareTAG: Week 1 / AC power CommentableIdTAG: 6002x_S1E3_AC_power NumberOfReplyTAG: 1 FirstChildTAG: Or you can simply look at the wave equation. Peak Power will be when the cosine element is at 1 (as cos elements cannot exceed), this is at intervals, one of which is at t=0. Therefore you can simply say, the peak of this equation occurs when cos(f(a)) is equal to 1. Therefore peak V =120 * (2)^(1/2). This is what I did. Hope I'm not taking shortcuts I shouldn't be! FirstChildUserIdTAG: 734170 FirstChildUserNameTAG: nahoskins FirstChildCreateTimeTAG: 2012-10-30T22:32:26Z IndexTAG: 1520 TitleTAG: Amplifier produces drop, but how does that amplify? If the amplifier reduces $V_o$, then how does that amplify, as in make the output greater? I do not understand this. UserIdTAG: 248807 UserNameTAG: 1kingsman CreateTimeTAG: 2012-10-05T04:57:15Z VoteTAG: 2 CoursewareTAG: Week 4 / Amplifier with Dependent Source CommentableIdTAG: 6002x_amplifier_w_dep_src NumberOfReplyTAG: 1 FirstChildTAG: It's an inverting amplifier - it amplify the signal and subtract it from voltage source (by pulling it down to ground) FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-05T05:06:06Z SecondChildTAG: So in other words, it only produces a voltage drop, rather than a increase in voltage SecondChildUserIdTAG: 248807 SecondChildUserNameTAG: 1kingsman SecondChildCreateTimeTAG: 2012-10-05T05:10:05Z SecondChildTAG: yes, exactly SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-05T06:28:00Z SecondChildTAG: You put a bias UI=2V and your little signal uI=0.1V*sin(t) and get v0=-sin(t) (minus, because it's inverted relatively to uI) biased by V0=5V. It realy is an amplifier. SecondChildUserIdTAG: 66669 SecondChildUserNameTAG: agarus SecondChildCreateTimeTAG: 2012-10-08T08:30:56Z SecondChildTAG: So, it does amplify, but it inverts at the same time. Ok, that makes sense. SecondChildUserIdTAG: 248807 SecondChildUserNameTAG: 1kingsman SecondChildCreateTimeTAG: 2012-10-11T22:24:11Z IndexTAG: 1521 TitleTAG: H6P1 Issue So, I'm using the node method and setting $(v_{OUT} - V_S)/R = K*(v_{IN} - V_T)*v_{OUT}^2$ and it keeps telling me that I've got an incorrect equation for $v_{OUT}$. Am I approaching this the wrong way? UserIdTAG: 331664 UserNameTAG: dwmnctrh3 CreateTimeTAG: 2012-10-05T00:54:51Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: It looks almost correct. The voltage across the resistor will be (Vs-vOUT). So id = (Vs-vOUT)/R. You also have vOUT^2 in the equation for id so you should get a quadratic equation that you can solve with the normal method. FirstChildUserIdTAG: 270011 FirstChildUserNameTAG: SnowmanZA FirstChildCreateTimeTAG: 2012-10-05T05:00:42Z SecondChildTAG: no matter how I right vi it is not permit it. SecondChildUserIdTAG: 286954 SecondChildUserNameTAG: ododo SecondChildCreateTimeTAG: 2012-10-05T13:52:23Z SecondChildTAG: I've used the quadratic formula and come up with $(1-\sqrt{1-4*K*R*V_S*(v_{IN} - V_T)})/(2*K*R*(v_{IN} - V_T))$. I'm told that's incorrect. I wish I knew where I went wrong. SecondChildUserIdTAG: 331664 SecondChildUserNameTAG: dwmnctrh3 SecondChildCreateTimeTAG: 2012-10-05T14:10:15Z SecondChildTAG: I got same answer still wrong SecondChildUserIdTAG: 286954 SecondChildUserNameTAG: ododo SecondChildCreateTimeTAG: 2012-10-05T14:25:37Z SecondChildTAG: dwmnctrh3, try re-reading the problem and the hint and see what you come up with. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-05T14:33:15Z SecondChildTAG: You are very close. I think you just have the wrong root. SecondChildUserIdTAG: 270011 SecondChildUserNameTAG: SnowmanZA SecondChildCreateTimeTAG: 2012-10-05T14:52:33Z SecondChildTAG: Thanks, I figured it out. I summed up my currents incorrectly. SecondChildUserIdTAG: 331664 SecondChildUserNameTAG: dwmnctrh3 SecondChildCreateTimeTAG: 2012-10-05T17:47:40Z SecondChildTAG: To begin I thin that the current is inverted so: -id=(VS-vOUT)/2 but I have`t got good results. If somone can tell us how big or looks like the solution, please! SecondChildUserIdTAG: 299251 SecondChildUserNameTAG: TheRedBlackOne SecondChildCreateTimeTAG: 2012-10-05T18:01:28Z SecondChildTAG: the solution that dwmnctrh3 posted is almost correct except for a few signs. So that is a good example of how the solution should look without giving you the actual solution. SecondChildUserIdTAG: 270011 SecondChildUserNameTAG: SnowmanZA SecondChildCreateTimeTAG: 2012-10-05T18:11:44Z SecondChildTAG: can anyone help me with the H5P3 part 1 where we need to differentiate vout w.r.t vin? My all other questions are correct but I am not able to get a tick for this one? BTW HOW DO WE WRITE THE EQUATIONS AS ONE OF THE STUDENT HAS WRITTEN ABOVE IN COMMENTS? I MEAN HOW TO USE MATHJAX IN COMMENTS TO MAKE THE EQUATIONS READABLE? SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-06T03:02:29Z SecondChildTAG: Hi jmen, please don't use all capitals next time, it makes me think you're shouting :o) That said, there's a nifty mathjax tutorial in this comment thread: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/505611d8a717032b00000016 And also on the wiki: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/mathjax-tutorial-write-formula-forum/ SecondChildUserIdTAG: 248902 SecondChildUserNameTAG: Chanute SecondChildCreateTimeTAG: 2012-10-10T09:57:30Z SecondChildTAG: Chanute first of all m really sorry if you thought that ways.. Second,its not that writing in caps always means shouting as its stated in the net etiquettes they mentioned here.. While stressing on a fact, one can use it. Anyways,I'll pay heed to your advice. And finally Thanx a ton for your links, I kept on trying earlier but couln't get it..But now i thnk i'll be able to do it. Thankx again!! SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-10T18:26:25Z FirstChildTAG: the expression is correct you must change sign before 4 INTO + FirstChildUserIdTAG: 174229 FirstChildUserNameTAG: fares27 FirstChildCreateTimeTAG: 2012-10-15T18:55:55Z SecondChildTAG: before sqrt? SecondChildUserIdTAG: 127800 SecondChildUserNameTAG: MNB SecondChildCreateTimeTAG: 2012-10-20T19:51:56Z IndexTAG: 1522 TitleTAG: cannot solve for vA i got vA = 20*e^-vA/5 -15 BUT DO NOT KNOW TO SOLVE FOR vA PLS HELP mE UserIdTAG: 162670 UserNameTAG: charlesbabyt CreateTimeTAG: 2012-10-04T18:12:42Z VoteTAG: 2 CoursewareTAG: Week 4 / Linearization Exercise 1 CommentableIdTAG: 6002x_linearization_ex_1 NumberOfReplyTAG: 1 FirstChildTAG: I've given this same answer to a question just below yours. But before you can try that, you need to re-arrange your equation so that you're solving for the VA in the exponent (e^VA/5), i.e re-arrange your equation so that it looks like VA = -5*ln((VA+15)/20). Now, take a look at S6V5 lecture video again, where a nonlinear equation is solved iteratively using trial and error method. Basically, start with VA = 1 for VA on RHS of equation VA = -5*ln((VA+15)/20) and solve for VA on LHS. You'll get some value that isn't correct, but you can then plug it again into RHD of eq and keep solving until you get a value that's the same for RHS and LHS. Note: You really need to use this with VA = -5*ln((VA+15)/20) as for the first equation form the values are more likely to diverge while you're relying on convergence for this solution to work. HTH FirstChildUserIdTAG: 502508 FirstChildUserNameTAG: amaher FirstChildCreateTimeTAG: 2012-10-04T22:21:33Z SecondChildTAG: thanks amaher, but how can i reach VA=-5*ln((va+15)/20) SecondChildUserIdTAG: 162670 SecondChildUserNameTAG: charlesbabyt SecondChildCreateTimeTAG: 2012-10-05T07:08:53Z SecondChildTAG: Okay, currently you have ${v_A} = 20 \cdot e^{-v_A/5} - 15$. You need to rearrange the equation so that you're solving for $v_A$ on the right hand side (RHS), i.e. the one in the exponent to $e$. Start with getting $e^{-v_A/5}$ on LHS. You get $20 \cdot e^{-v_A/5} = {v_A} + 15$ and then $e^{-v_A/5} = \cfrac{{v_A} + 15}{20}$. Now to get exponent to e, you need to apply $log_e$, which is the same thing as $\ln$. Thus, you get: ${-v_A/5} = log_e(\cfrac{{v_A} + 15}{20})$ or ${-v_A/5} = \ln(\cfrac{{v_A} + 15}{20})$. From here getting $v_A$ is simple: ${vA} = -5 \cdot\ln(\cfrac{{v_A} + 15}{20})$ HTH SecondChildUserIdTAG: 502508 SecondChildUserNameTAG: amaher SecondChildCreateTimeTAG: 2012-10-05T13:52:35Z IndexTAG: 1523 TitleTAG: "H4P3-PART B" hello everyone, please help me in solving In as soon as possible.. UserIdTAG: 137918 UserNameTAG: jaisneha CreateTimeTAG: 2012-10-04T13:56:30Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: The Norton current, $I_N$, is the current flowing through a short placed directly across Port B. This will be the sum of the current flowing through $R_2$ and $R_3$. As the node at $R_2$ / $R_3$ is now grounded (by the short) working out the $R_2$ current should be easy, from which you can then get the control voltage u, then the VCVS device voltage and then the current through $R_3$. To get the Norton resistance, $R_N$, replace the independent voltage source with a short, apply an arbitrary test voltage $V_t$ to Port B and work out the test current $I_t$ (current through $R_2$ -> u -> VCVS -> current through $R_3$ -> and sum). The ratio of $V_t$ to $I_t$ is the Norton resistance, $R_N$. I don't know whether this is the way it should be done, but it gives the right answers! FirstChildUserIdTAG: 366083 FirstChildUserNameTAG: smath FirstChildCreateTimeTAG: 2012-10-05T11:03:58Z SecondChildTAG: Looks right to me. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-05T17:30:35Z SecondChildTAG: This - *To get the Norton resistance, RN, replace the independent voltage source with a short, apply an arbitrary test voltage Vt to Port B and work out the test current It (current through R2 -> u -> VCVS -> current through R3 -> and sum).* THANK YOU! If you take that approach you should be able to express the equation for the test current in terms of known quantities, re-group terms, and solve directly for Vt/It. SecondChildUserIdTAG: 339668 SecondChildUserNameTAG: chickwebb SecondChildCreateTimeTAG: 2012-10-05T23:10:30Z SecondChildTAG: Thank you so much for your help! Now I found the answers :D SecondChildUserIdTAG: 310147 SecondChildUserNameTAG: ildomarcarvalho SecondChildCreateTimeTAG: 2012-10-06T01:48:40Z SecondChildTAG: SMATH..*-*As the node at R2 / R3 is now grounded (by the short) working out the R2 current should be easy, from which you can then get the control voltage u, then the VCVS device voltage and then the current through R3.** Lot of thanks!!! SecondChildUserIdTAG: 298612 SecondChildUserNameTAG: fabianh SecondChildCreateTimeTAG: 2012-10-06T12:54:01Z SecondChildTAG: SMATH: you rock.:) SecondChildUserIdTAG: 266739 SecondChildUserNameTAG: satvikchugh SecondChildCreateTimeTAG: 2012-10-07T04:27:02Z SecondChildTAG: Wonderful, thank you so much smath, very helpful. Now to go over what I did and see if I can make sense of it. SecondChildUserIdTAG: 403503 SecondChildUserNameTAG: jmohrmann SecondChildCreateTimeTAG: 2012-10-07T11:37:53Z SecondChildTAG: thanks a lot smath SecondChildUserIdTAG: 97359 SecondChildUserNameTAG: azeem179 SecondChildCreateTimeTAG: 2012-10-07T13:21:42Z FirstChildTAG: BTW, there were more replies in this thread, but they were removed because someone posted an actual answer to the question in their comments. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-05T17:32:19Z IndexTAG: 1524 TitleTAG: I don't know how to get the value, I'm stuck in the middle of calculation. I apply KVL with VA around the loop, and comes down to the equation, I can write it in 2 ways: -15 = VA - 20*e^(-VA/5) VA = -5*ln(3/4+VA/20) But I can't get the value for VA from either one of them, their technically are the same one just organized in different ways. EDIT: I could throw the equation into wolframalpha and get a solution out, but how did it do it? UserIdTAG: 98259 UserNameTAG: rogerloh0 CreateTimeTAG: 2012-10-04T01:49:13Z VoteTAG: 2 CoursewareTAG: Week 4 / Linearization Exercise 1 CommentableIdTAG: 6002x_linearization_ex_1 NumberOfReplyTAG: 2 FirstChildTAG: excel, like someone said in previous lectures.. A1=va, A2=right part of equation, A3=A1-A2 FirstChildUserIdTAG: 373498 FirstChildUserNameTAG: Cheblan FirstChildCreateTimeTAG: 2012-10-04T06:15:10Z FirstChildTAG: Take a look at S6V5 lecture video again, where a nonlinear equation is solved iteratively using trial and error method. Basically, start with VA = 1 for VA on RHS of equation VA = -5*ln(3/4+VA/20) and solve for VA on LHS. You'll get some value that isn't correct, but you can then plug it again into RHD of eq and keep solving until you get a value that's the same for RHS and LHS. _Note:_ You really need to use this with you 2nd equation i.e. VA = -5*ln(3/4+VA/20) as for the first the values are more likely to diverge while you're relying on convergence for this solution to work. HTH FirstChildUserIdTAG: 502508 FirstChildUserNameTAG: amaher FirstChildCreateTimeTAG: 2012-10-04T22:14:08Z IndexTAG: 1525 TitleTAG: How to Calculate the Average power AS per my understanding, the power dissipated = V^2/R. The complete cycle takes 1/60 sec because we have a 60 Hz voltage. so the Power P would be = Intig ((120*sqrt(2)cos(2*pi*60*t))^2 dt)/R from t=0 till t=1/60 After my calculations, i get a wrong result = 2.1818 Watts. kindly tell me what did i do wrong? The solution steps are in the attached image. ![Solution][1] [1]: https://edxuploads.s3.amazonaws.com/13492980171343631.jpg UserIdTAG: 557705 UserNameTAG: Ahmed2316 CreateTimeTAG: 2012-10-03T21:10:15Z VoteTAG: 2 CoursewareTAG: Week 1 / AC power CommentableIdTAG: 6002x_S1E3_AC_power NumberOfReplyTAG: 2 FirstChildTAG: Your calculation corresponds to energy transferred: $\Delta E = \int_0^t P \cdot dt = \int_0^t \frac{V^2}{R} dt$. You want average power: $\langle P \rangle = \frac{\Delta E}{\Delta t}$. FirstChildUserIdTAG: 9253 FirstChildUserNameTAG: strakus FirstChildCreateTimeTAG: 2012-10-04T00:28:48Z FirstChildTAG: But what if I integrate the voltage curve from 0 to 1/60? I get zero! And after substitution to the formula W=V^2/R I get average W=0! Where am I mistaken? FirstChildUserIdTAG: 702959 FirstChildUserNameTAG: Sashkow FirstChildCreateTimeTAG: 2012-10-25T20:51:34Z IndexTAG: 1526 TitleTAG: LAB 6 Hints Hy to all guys !^^ I ask for a little help in LAB 6, and i explain how i worked aroud it ! I'm stuck on the first question, about the RON. I associate the curve of the plot, to an exponential that decrease from the start voltage value(3V) and after 2ns become 192.823mV. It decrease with the law -(2*10^-9)/(RON*200*10^(-15)) i combined all this informations to extract the value of RON but i don't understand where is the error :S Can you give me an hint please ?:P UserIdTAG: 200496 UserNameTAG: Ranieri CreateTimeTAG: 2012-10-03T19:30:53Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: To find RON, you must use the concept of resistor divider. This helps? FirstChildUserIdTAG: 39980 FirstChildUserNameTAG: MarinoJr FirstChildCreateTimeTAG: 2012-10-03T20:00:38Z SecondChildTAG: Oh thx you so much MarinoJr !!! I didn't think about it, and i never thought that... i was too much focused on the exponential way ^^ SecondChildUserIdTAG: 200496 SecondChildUserNameTAG: Ranieri SecondChildCreateTimeTAG: 2012-10-03T22:05:10Z SecondChildTAG: Is a resistor divider the same as a voltage divider? SecondChildUserIdTAG: 370769 SecondChildUserNameTAG: PabloFCid SecondChildCreateTimeTAG: 2012-10-04T10:33:05Z SecondChildTAG: Yes, they are the same. SecondChildUserIdTAG: 270011 SecondChildUserNameTAG: SnowmanZA SecondChildCreateTimeTAG: 2012-10-04T16:58:33Z SecondChildTAG: This is Great By the Way its me Harvey I really need your help please come back on skype.......................................................... Remember me HArvey Specter SecondChildUserIdTAG: 378080 SecondChildUserNameTAG: GladIDidThis SecondChildCreateTimeTAG: 2012-11-18T22:26:39Z IndexTAG: 1527 TitleTAG: S7E3: checking solution by voltage divider doesn't work I got the right answers but they don't make sense, particularly for the $100 \Omega$ model. By volatage divider, only $ \frac{100}{4000} $ woud show up on $v_D$ , some .04 volts. The $v_D$ that got me $R_D =100 \Omega$ and then the answer was nowhere near .04. How is my thinking wrong? UserIdTAG: 440714 UserNameTAG: mcktim CreateTimeTAG: 2012-10-03T17:20:15Z VoteTAG: 2 CoursewareTAG: Week 4 / Linearization Exercise 2 CommentableIdTAG: 6002x_linearization_ex_2 NumberOfReplyTAG: 1 FirstChildTAG: No, the 100 Ohm is the incremental resistance, that is the resistance for the small signal, not the resistance for the large signal. For the large signal you have to use the formula with the e^ in it. FirstChildUserIdTAG: 266912 FirstChildUserNameTAG: pietvo FirstChildCreateTimeTAG: 2012-10-03T23:29:20Z IndexTAG: 1528 TitleTAG: Importance of Studying from Textbook Hi all, I was just wondering what the importance or benefits of studying from the textbook are. Up till now I have been able to follow everything the Prof. Anant Agarwal has taught, and have been able to complete all the homework problems and labs without much issue. Whenever I have had a doubt I would clarify with the textbook or in the discussion forum, however I have not read more than a few pages. I would be much obliged if someone could explain to me the benefits of using the textbook in tandem with this course. Thank you in advance. UserIdTAG: 85490 UserNameTAG: yashsavani CreateTimeTAG: 2012-10-03T16:24:44Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I asked the same question once,.......... some said to read it, some said it was awesome, some said google helped them with all the queries do they didnt use they book, some said they didnt need the book at all........ hope I find a definite answer here....still if nothing comes up, I am going to read the book.... FirstChildUserIdTAG: 169784 FirstChildUserNameTAG: thewiredbear FirstChildCreateTimeTAG: 2012-10-03T18:22:44Z FirstChildTAG: If you have the time and patience to read the book, then reading the book would be a good thing. But the book is 1000 pages long and reading will take much time. Also, I personally do not like the electronic format. I prefer a hard copy, but I am unwilling to spend the money to buy one. i realize that I need to take things more seriously as we approach the midterm. I have started printing out the lecture notes to study when I am away from the computer. I have also found that when I go to the book to find an answer to a specific question I often can't find it in the book and need to go elsewhere. I can usually find a number of good answers on the internet. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-03T22:54:09Z IndexTAG: 1529 TitleTAG: H6P1 Hint for finding ro Anybody? UserIdTAG: 397595 UserNameTAG: mholin CreateTimeTAG: 2012-10-03T12:46:12Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Fellow skyhawk gave a hint in: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_small_signal_ckt_model/threads/506b4893d677191f0000006f If you can manage this.. FirstChildUserIdTAG: 39980 FirstChildUserNameTAG: MarinoJr FirstChildCreateTimeTAG: 2012-10-03T13:00:55Z FirstChildTAG: The first video of week 6's tutorial of the week gives you the exact method of solving for both $gm$ and $ro$. FirstChildUserIdTAG: 85490 FirstChildUserNameTAG: yashsavani FirstChildCreateTimeTAG: 2012-10-03T13:19:52Z SecondChildTAG: Thank you! SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-10-05T08:45:10Z SecondChildTAG: You mean the vacuum triode video and the partial derivatives method? SecondChildUserIdTAG: 252722 SecondChildUserNameTAG: vaboro SecondChildCreateTimeTAG: 2012-10-06T21:24:44Z SecondChildTAG: I got the correct answer for $r_o$! Partial derivatives should definitely be used! Excitation should be applied to the output port. When applying excitation, the value of the input voltage should be noted, and some circuit elements should be substituted with others based upon the value of input voltage using the superposition principle since the small-signal circuit is linear. SecondChildUserIdTAG: 252722 SecondChildUserNameTAG: vaboro SecondChildCreateTimeTAG: 2012-10-06T22:42:23Z IndexTAG: 1530 TitleTAG: Lab 3 I don't agree with the solution. Can someone please take a look at my equation? https://www.dropbox.com/s/0djk1l8h0avu7s5/Lab%203.tiff UserIdTAG: 230787 UserNameTAG: Gabriel007 CreateTimeTAG: 2012-10-03T03:39:18Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hi Gabriel007! I have seen your excel... As I could interpret, the fx it belongs to the C10 , that it is Vol , isn't it? Ok, You are saying that C10= C5*(C5*C8/C8+C4), where - C5= Vdd =... Volts - C8= Req= ...Ohm - C4= Rpull= ... Ohm - C10= Vol= ... Volts If we do the dimentional analysis of your formula: [Volt]=[Volt] * [Volt] * [Ohm]/[Ohm] [Volt]=[Volt]^2 So there is something wrong in your excel formula... Volt can not be equal Volt squared , it should be equal to a Volt unit... ---- The Correct formula in your excel for C10 would be: C10=C5*(C8/C8+C4) [Volt]= [Volt]*([Ohm]/[Ohm]) [Volt]= [Volt] So, here verifies the dimentional analysis... Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-03T04:19:10Z SecondChildTAG: Hey Myriam! Thanks for the help...again. hahah I uploaded a different pic. I am not sure why I entered c5 twice. But, the problem is still there even with your fix. To be honest that is the answer that I got all day Sunday... Do you mind taking a look again? - Same link. SecondChildUserIdTAG: 230787 SecondChildUserNameTAG: Gabriel007 SecondChildCreateTimeTAG: 2012-10-03T04:46:55Z SecondChildTAG: Haha! I have found it! The solution of the Lab that was posted it is wrong, it should be Rpullup in the voltage divider and not RPU=Rn... I will report it, I haven't noticed that... So, change the value of RPU = Rpullup and will be ok ;). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-03T05:07:01Z SecondChildTAG: [mistake reported][1]. Thanks Gabriel007! :). [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/506bcafb079bd21f000000b8 SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-03T05:21:00Z FirstChildTAG: Yeah I was using Rn instead of Rpullup...haha Ok I posted the pic with everything working.Same link. Thank You! FirstChildUserIdTAG: 230787 FirstChildUserNameTAG: Gabriel007 FirstChildCreateTimeTAG: 2012-10-03T12:28:34Z SecondChildTAG: :). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-03T12:37:51Z IndexTAG: 1531 TitleTAG: a hard math problem I know that this has nothing to do with this course, but my daughter came home from school with her math homework and asked me to help her with it. The problem is this: (3 + 2)squared - 3 squared + 2 * 3. The answer that she comes up with is 8 but I dont know if that is right. I know that this is not part of the course but I need some help trying to figure it out. Please help me I am lost with what she calls new math. UserIdTAG: 160242 UserNameTAG: Mlevins35 CreateTimeTAG: 2012-10-02T20:48:22Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 9 FirstChildTAG: Hey Pennypacker I got another formula. It's "PEMDAS". Parenthesis, Exponential, Multiplication, Division, Addition, Subtraction. FirstChildUserIdTAG: 378096 FirstChildUserNameTAG: Jamshaid271 FirstChildCreateTimeTAG: 2012-10-03T09:58:24Z FirstChildTAG: k FirstChildUserIdTAG: 543990 FirstChildUserNameTAG: sunny89 FirstChildCreateTimeTAG: 2012-10-02T21:05:40Z FirstChildTAG: Hi, I am stuck with HP4P3 Part B. The Norton resistance supposed to be (R1+R2)||R3 assuming short circuit to replace two voltage sources? where am I going wrong? Would appreciate a pointer to this problem. I have computed "u" to be 5.714V and open voltage to be 7.857v using node method. However, I am not sure of the accuracy. FirstChildUserIdTAG: 246991 FirstChildUserNameTAG: spatra FirstChildCreateTimeTAG: 2012-10-02T21:12:44Z FirstChildTAG: I assume the equation is as follows: $(3+2)^2-3^2+2*3.$ Which is $5^2-3^2+2*3$. Finally $25-9+6=22.$ FirstChildUserIdTAG: 213386 FirstChildUserNameTAG: dmascenik FirstChildCreateTimeTAG: 2012-10-02T21:28:35Z SecondChildTAG: Totally agree with you dmascenik! :) Thank you for helping Mlevins and her daughter. See you! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-02T21:32:46Z FirstChildTAG: For melvins35 There is a priority calculation, you must follow a rule. The priorities are simple: the First () Then the deepest X (multiplication) and / (divide) And finally the + (addition) and - (subtraction) Ex: A = 8 +3 (2 +3) A = 8 +3 x 5 A = 8 +15 A = 23 Voila! Your case is the following (3 + 2) squared - 3 squared + 2 * 3 = (3 +2) ² -3 ² +2 * 3 = (5) ² -9 +5 = (25) -9 +5 = 16 +5 = 22 FirstChildUserIdTAG: 544325 FirstChildUserNameTAG: MEDI FirstChildCreateTimeTAG: 2012-10-02T21:32:23Z SecondChildTAG: Thank you MEDI! :). Thank you for the explanation! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-02T21:34:24Z SecondChildTAG: 16 +5 = 22 ... really? SecondChildUserIdTAG: 64512 SecondChildUserNameTAG: OZSorescu SecondChildCreateTimeTAG: 2012-10-03T09:31:24Z FirstChildTAG: thank everyone so much for the help with this it is greatly appriecated. FirstChildUserIdTAG: 160242 FirstChildUserNameTAG: Mlevins35 FirstChildCreateTimeTAG: 2012-10-02T21:40:52Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-02T22:05:43Z FirstChildTAG: I find this useful: "BEDMAS" Brackets, exponents, division, multiplication, addition, subtraction. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-02T22:39:42Z SecondChildTAG: Haha! I vote for BEDMAS :) ! Thank you! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-03T00:31:25Z SecondChildTAG: You are welcome! SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-03T02:13:12Z FirstChildTAG: I want to thank all of you for all the help that you all gave me I saw what solution you came up with and it made perfect sense. I see where the answer came from. I did it the old school way, added what was in () had the rest right but did the first part wrong. Thank you all so much for the help FirstChildUserIdTAG: 160242 FirstChildUserNameTAG: Mlevins35 FirstChildCreateTimeTAG: 2012-10-03T00:04:35Z SecondChildTAG: You need to go to the Khan Academy, it is free. He must solve thousands of equations from very beginning Algebra to advanced Calculus. He also does physics, statistics, finance, chemistry etc. I believe Mr. Khan got his masters at MIT and MBA at Harvard. Type it into Google and check it out. Good luck. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-03T01:26:59Z FirstChildTAG: old school???? when math ops priorities had changed? I lost something??? OMG, say it was just a joke FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-03T01:37:34Z IndexTAG: 1532 TitleTAG: S7V6: DEMO - MUSIC OVER A LIGHT BEAM, DISTORTION AND NO DISTORTION this demo is epic! :) truly an AHA moment! :) UserIdTAG: 371000 UserNameTAG: Joseph090892 CreateTimeTAG: 2012-10-02T20:06:27Z VoteTAG: 2 CoursewareTAG: Week 4 / Incremental Analysis Demo CommentableIdTAG: 6002x_inc_analysis_demo NumberOfReplyTAG: 0 IndexTAG: 1533 TitleTAG: H6P1 Output resistance ro In the item that asks the "Output resistance ro", any hint in how to find that? The small signal model (given by the exercise) also contains ro and, following what was done in Exercise S11E3, I would connect a current independent source to inject current in the output port and then obtain the voltage across the output port. That ratio would be the output resistance. Though, that calculus become confuse and does not achieve the correct result. Thank, in advance, for any hint. UserIdTAG: 39980 UserNameTAG: MarinoJr CreateTimeTAG: 2012-10-02T20:03:31Z VoteTAG: 2 CoursewareTAG: Week 6 / Small-signal circuit model CommentableIdTAG: 6002x_small_signal_ckt_model NumberOfReplyTAG: 3 FirstChildTAG: Assuming that you have the correct answers for parts 1 and 2: Take the expression that you got in part 1 and express vIN and VS in terms of a DC operating point plus a small signal. Do a Taylor series expansion in **both** variables (first order only). Compare the result of the Taylor series with what you got for part 2 and identify ro and gm accordingly. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-02T21:53:42Z SecondChildTAG: If we take the first part of the exercise to solve the second part, Is not the solution of ro related with R and VS, which don't appear in the second part? SecondChildUserIdTAG: 370769 SecondChildUserNameTAG: PabloFCid SecondChildCreateTimeTAG: 2012-10-03T09:18:10Z SecondChildTAG: Expression in part 1 gives vOUT and in part 2 gives id. It would be necessary to find iD in part 1 before. OK. But VS just has DC component. In what point would I apply the Taylor series? SecondChildUserIdTAG: 39980 SecondChildUserNameTAG: MarinoJr SecondChildCreateTimeTAG: 2012-10-03T12:51:35Z FirstChildTAG: Video 1 of the "Week 6 Tutorials" shows how to solve that. FirstChildUserIdTAG: 39980 FirstChildUserNameTAG: MarinoJr FirstChildCreateTimeTAG: 2012-10-03T13:22:14Z SecondChildTAG: Can you please tell which are d videos that explain this. I couldn't find it.. If possible please send the link.. Thanx.. SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-06T07:16:52Z SecondChildTAG: It's here: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_6/Week_6_Tutorials/ SecondChildUserIdTAG: 309359 SecondChildUserNameTAG: abarea10 SecondChildCreateTimeTAG: 2012-10-10T15:25:46Z FirstChildTAG: Notice that the intensity of the 'Newfet', not only depends on vgs, it also depends on vds. So you have to find the small signal intensity wich will depend on both small signal voltages. Like MarinoJr has said before me, video 1 of tutorials gives you how to achieve the answer. FirstChildUserIdTAG: 309359 FirstChildUserNameTAG: abarea10 FirstChildCreateTimeTAG: 2012-10-10T15:24:43Z IndexTAG: 1534 TitleTAG: H4P1 I need help on understanding questions of H4P1. I am stuck right the way on the first question, how to calculate current. What is the V referring to in the formula of P? Thanks! Julie UserIdTAG: 240389 UserNameTAG: julieqpt CreateTimeTAG: 2012-10-02T18:06:07Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hm. Don't understand Your problem. You have formula, describing i from v. You have v, You have coefficient P. Just put v and p in formula and get result. FirstChildUserIdTAG: 195218 FirstChildUserNameTAG: Al_Incognito FirstChildCreateTimeTAG: 2012-10-02T18:27:11Z SecondChildTAG: mmmmmm the problem is put in the formula vyp not understand how you can help me thanks SecondChildUserIdTAG: 320715 SecondChildUserNameTAG: maraivette SecondChildCreateTimeTAG: 2012-10-02T18:37:48Z SecondChildTAG: Ok, here is IP = 2.0 * 10 ^ (-3) / V ^ (3/2) * (VPK) ^ (3/2). What is V here? Is V same as VPK? Looks like V is not VPK, if it is, then IP = 0.002, but it shows as wrong answer. So, what is V? -Julie SecondChildUserIdTAG: 240389 SecondChildUserNameTAG: julieqpt SecondChildCreateTimeTAG: 2012-10-02T18:42:09Z SecondChildTAG: OK, I got it now. I put p = 2.0 * 10 ^ (-3) / V ^ (3/2) mistakenly. Thanks! Julie SecondChildUserIdTAG: 240389 SecondChildUserNameTAG: julieqpt SecondChildCreateTimeTAG: 2012-10-02T19:49:23Z SecondChildTAG: i cant figure out the answer,i find as an answer 0.002 thats wrong of course SecondChildUserIdTAG: 400832 SecondChildUserNameTAG: panosbr SecondChildCreateTimeTAG: 2012-10-02T20:26:47Z SecondChildTAG: V is not the same as Vpk,how i find V? SecondChildUserIdTAG: 400832 SecondChildUserNameTAG: panosbr SecondChildCreateTimeTAG: 2012-10-02T20:44:52Z SecondChildTAG: Oh. Now I understand what You are meaning. P=2ma/v(2/3) is not a formula! A/V^(2/3) is dimentions of constant P. So formula is ip=2*10^-3*VPK SecondChildUserIdTAG: 195218 SecondChildUserNameTAG: Al_Incognito SecondChildCreateTimeTAG: 2012-10-03T14:27:07Z SecondChildTAG: I already did so and ERROR NOT MAKE ME I'm doing wrong IP = 2e-3 * VPK someone can help me THANKS SecondChildUserIdTAG: 320715 SecondChildUserNameTAG: maraivette SecondChildCreateTimeTAG: 2012-10-03T16:20:28Z SecondChildTAG: Al_incognito is right except for one thing... P= 0.002 A/V^(2/3) where A/V^(2/3) IS THE UNIT OF THE PARAMETER P then the final equation is gonna be just: ip= 0.002*Vpk^(2/3) SecondChildUserIdTAG: 45307 SecondChildUserNameTAG: josejimenez2 SecondChildCreateTimeTAG: 2012-10-06T17:49:15Z FirstChildTAG: Hi Julie! You can take a look at [here][1]! :) [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_graph_interp_inc_method/threads/506b7c614f5ec3270000007d FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-02T23:51:18Z SecondChildTAG: thanks goodness you are always helping us Myrimit! SecondChildUserIdTAG: 45307 SecondChildUserNameTAG: josejimenez2 SecondChildCreateTimeTAG: 2012-10-06T16:10:53Z SecondChildTAG: :) Thank you josejimenez2! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-06T22:29:04Z IndexTAG: 1535 TitleTAG: What went wrong ? I've spent considerable time for solution of last 2 problems,but i'm unable to figure out the mystery...as my method seems to be right but answers doesn't match. UserIdTAG: 477713 UserNameTAG: ikm104 CreateTimeTAG: 2012-10-02T12:48:44Z VoteTAG: 2 CoursewareTAG: Week 4 / Linearization Exercise 2 CommentableIdTAG: 6002x_linearization_ex_2 NumberOfReplyTAG: 1 FirstChildTAG: I posted my solution to linearization excercise which gives answers that are not exactly the same as those shown under the "Check answer" option but are nonetheless acceptable FirstChildUserIdTAG: 252722 FirstChildUserNameTAG: vaboro FirstChildCreateTimeTAG: 2012-10-02T13:16:56Z IndexTAG: 1536 TitleTAG: please help me in lab 4!!! I am getting my Ids as zero always..where do you connect the current probe?and what should Vds be? UserIdTAG: 156585 UserNameTAG: AdiRanga CreateTimeTAG: 2012-10-02T12:05:33Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: The current probe should be connected to the source terminal FirstChildUserIdTAG: 143823 FirstChildUserNameTAG: NathanNadeson FirstChildCreateTimeTAG: 2012-10-02T12:15:37Z FirstChildTAG: ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13491806135446738.png FirstChildUserIdTAG: 156585 FirstChildUserNameTAG: AdiRanga FirstChildCreateTimeTAG: 2012-10-02T12:23:45Z SecondChildTAG: what is the mistake in my circuit? SecondChildUserIdTAG: 156585 SecondChildUserNameTAG: AdiRanga SecondChildCreateTimeTAG: 2012-10-02T12:24:13Z SecondChildTAG: you have grounded both the exits of the mosfet by putting the ground probe there... SecondChildUserIdTAG: 301962 SecondChildUserNameTAG: labrinim SecondChildCreateTimeTAG: 2012-10-02T12:32:28Z SecondChildTAG: try putting the current probe just between the upper exit and the voltage probe SecondChildUserIdTAG: 301962 SecondChildUserNameTAG: labrinim SecondChildCreateTimeTAG: 2012-10-02T12:33:23Z SecondChildTAG: and connect the dc source with the "left" exit o the mosfet! SecondChildUserIdTAG: 301962 SecondChildUserNameTAG: labrinim SecondChildCreateTimeTAG: 2012-10-02T12:35:48Z FirstChildTAG: ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13491819511343607.png FirstChildUserIdTAG: 156585 FirstChildUserNameTAG: AdiRanga FirstChildCreateTimeTAG: 2012-10-02T12:46:00Z SecondChildTAG: the transient response for the above circuit is shown below SecondChildUserIdTAG: 156585 SecondChildUserNameTAG: AdiRanga SecondChildCreateTimeTAG: 2012-10-02T12:47:54Z SecondChildTAG: when i checked my circuit, i got a TICK for it..which means my circuit is proper but still i dont get a proper transient analysis!! SecondChildUserIdTAG: 156585 SecondChildUserNameTAG: AdiRanga SecondChildCreateTimeTAG: 2012-10-02T12:50:33Z SecondChildTAG: hey, your circuit seems OK! check the wiring! try connecting the node 'a' directly to the ground and check the other wiring as well (you current probe doesn't seem connected properly) SecondChildUserIdTAG: 252722 SecondChildUserNameTAG: vaboro SecondChildCreateTimeTAG: 2012-10-02T13:12:43Z SecondChildTAG: ...directly to the GATE of the MOSFET, sorry! SecondChildUserIdTAG: 252722 SecondChildUserNameTAG: vaboro SecondChildCreateTimeTAG: 2012-10-02T13:13:36Z FirstChildTAG: ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13491819707701703.png FirstChildUserIdTAG: 156585 FirstChildUserNameTAG: AdiRanga FirstChildCreateTimeTAG: 2012-10-02T12:46:18Z SecondChildTAG: i made the corrections to the circuit but still the transient response is as shown... SecondChildUserIdTAG: 156585 SecondChildUserNameTAG: AdiRanga SecondChildCreateTimeTAG: 2012-10-02T12:47:24Z SecondChildTAG: am not getting it!!! SecondChildUserIdTAG: 156585 SecondChildUserNameTAG: AdiRanga SecondChildCreateTimeTAG: 2012-10-02T12:48:09Z SecondChildTAG: connect the two a's... SecondChildUserIdTAG: 301962 SecondChildUserNameTAG: labrinim SecondChildCreateTimeTAG: 2012-10-02T13:03:56Z SecondChildTAG: You can use the label component (looks like a short straight line in the parts bin) to add names to circuit nodes. Nodes with the same name are considered to be electrically connected. SecondChildUserIdTAG: 156585 SecondChildUserNameTAG: AdiRanga SecondChildCreateTimeTAG: 2012-10-02T13:08:52Z SecondChildTAG: then, use the same ground for everything! SecondChildUserIdTAG: 301962 SecondChildUserNameTAG: labrinim SecondChildCreateTimeTAG: 2012-10-02T13:43:16Z SecondChildTAG: i think your 'a' source is not right for TransAn. Re-read the instructions about curve tracer for a resistor, or look at the specifications of the vGS source there. SecondChildUserIdTAG: 184153 SecondChildUserNameTAG: tthngn SecondChildCreateTimeTAG: 2012-10-02T20:28:10Z FirstChildTAG: A tip from my experience... always hit Check before running Tran. It seems that after changes in the circuit the x-axis show proper vGS voltages only if Check is ok. In hindsight that is sort of obvious, but i wasted time trying to reconcile a new config with one that worked previously (showing all vGS increments). FirstChildUserIdTAG: 184153 FirstChildUserNameTAG: tthngn FirstChildCreateTimeTAG: 2012-10-02T20:45:09Z IndexTAG: 1537 TitleTAG: Staff: Learning disabilities Does MIT give reasonable accommodations for learning disabilities on exams? Through testing my Doctor recommends I be given time and a half for taking exams. For instance, if the exam is an hour, it will take me 1-1/2 hours to take the exam. I don't have a short term memory and I have dyslexia, which makes it difficult to go fast. Somebody suggested we will be given a few days to complete the exam, such as starting it on a Wednesday or Thursday and having it due on Sunday. If this is the case, it shouldn't be a problem. Please let me know if I have to contact the staff privately. - Thanks **It sounds like 24 hours should be OK, as long as the exam is not intended to take 24 hours to complete. I can't imagine anyone designing an exam to take 24 hours to complete, then only giving 24 hours to complete it. One has to sleep etc. sometime in there (chuckle). So, 24 hours to complete the exam will work. Where was this type on online learning when I went to college 26 years ago? It is great. - Thanks. ** UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-02T03:19:50Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Actually you get 24 hours to do the exam. And you have a choice to start it over a few days. But as soon as you click to open the second page of the online exam paper, then you have just 24 hours to do it, which is plenty :) FirstChildUserIdTAG: 99941 FirstChildUserNameTAG: Nexus1974 FirstChildCreateTimeTAG: 2012-10-02T08:04:40Z SecondChildTAG: What Nexus1974 says it is true :). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-02T11:21:19Z IndexTAG: 1538 TitleTAG: Solutions for closed assignments It looks like some people are looking for solutions to closed assignments, and I wanted to make sure that everybody knew that the show answer button should be revealed after the due date for all assignments (for both homeworks and labs) under the problem statement, instead of the "check" button. UserIdTAG: 96452 UserNameTAG: Lyla CreateTimeTAG: 2012-10-01T19:34:14Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Yes we know I guess :) thanks anyway! FirstChildUserIdTAG: 365201 FirstChildUserNameTAG: sirajmuhammad FirstChildCreateTimeTAG: 2012-10-01T20:00:15Z IndexTAG: 1539 TitleTAG: Analytical solution for lab2 Can someone post an analytical solution for lab 2? I completed it but it is more or less by guessing, I would like to see it written out mathematically. UserIdTAG: 371819 UserNameTAG: Lokiie CreateTimeTAG: 2012-10-01T19:13:29Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: The show answer button for lab 2 should be visible now that it is closed. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-10-01T19:31:15Z FirstChildTAG: Hi Lokiie! You can take a look at [here][1] :) See you! Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_S3E2_Linear_Combinations_of_Source_Strengths/threads/506a280d036af91f000000c9 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-01T23:43:47Z FirstChildTAG: I solved it by relying on S3E2, as I've already [posted][1] here. Given that $v_3 = a_1*V_1 + a_2*V_2$ from the excercise solution we know: $a1=(1/R1)/(1/R1+1/R2+1/R3).$ $a2=(1/R2)/(1/R1+1/R2+1/R3).$ We also have the following prerequisite $V_\mathrm{out} \approx \frac{1}{2}V_1 + \frac{1}{6}V_2.$ That gives us $a1=1/2$ and $a2=1/6$. So we have 2 equations and 3 unknowns here. Bummer. But anyway, we're designing a voltage divider here and we're not looking for specific resistor values so let's just assume $R3=100\Omega$. Now the system can be solved and we'll get: $R3=100\Omega.$ $R2=200\Omega.$ $R1=66.6\Omega.$ If you missed S3E2, remember that is tedious to solve through resistances. In order to find the algebraic expressions for the coefficients just use conductances, as a friend of mine pointed out to my huge relief. UPD: oops, it seems the official solution is pretty close to the above calculations anyway. Sorry, didn't read it first. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5067e012e745e91f00000032 FirstChildUserIdTAG: 378267 FirstChildUserNameTAG: artfwo FirstChildCreateTimeTAG: 2012-10-02T01:49:30Z IndexTAG: 1540 TitleTAG: how many courses of EDX could i take at once??? how many courses of EDX could i take at once??? UserIdTAG: 146930 UserNameTAG: fayzan_007 CreateTimeTAG: 2012-10-01T17:42:00Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: The question should be: How many courses at edX ***can*** I take at once? As many as you can handle! I have signed up for all 7. I do have concerns with some as I don't have the prerequisites, however now that some have started, they might not be so bad. This course, 6.002x is a big course, some of the others ***might*** be a little easier to digest. For example I was going to drop SAAS (CS169.1x), however I did manage to make it through the first quiz, so I am going to stick around for another week to see how it unfolds. So I recommend you try them all, the price is right! FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-01T19:05:40Z SecondChildTAG: Wow, 7! I admire you! My best wish to you Pennypacker!:) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-01T23:45:22Z SecondChildTAG: Thankz SecondChildUserIdTAG: 146930 SecondChildUserNameTAG: fayzan_007 SecondChildCreateTimeTAG: 2012-10-03T17:14:00Z FirstChildTAG: Depends on how much time per week you can spent. I'm following 6.002x and can just manage it having a fulltime job, one wife and two kids  FirstChildUserIdTAG: 114881 FirstChildUserNameTAG: xvink FirstChildCreateTimeTAG: 2012-10-01T20:26:58Z SecondChildTAG: :). I imagine that having kids, family and job it is really hard. My best wish to you xvink! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-01T23:47:08Z SecondChildTAG: my best wihses to you XVINK SecondChildUserIdTAG: 146930 SecondChildUserNameTAG: fayzan_007 SecondChildCreateTimeTAG: 2012-10-03T17:13:47Z IndexTAG: 1541 TitleTAG: S7E2 incremental resistance someone can say how to find incremental resistance. I got 0.41 by it isn't correct( I see the 4.75 for rd but I don't understand how to used it((( UserIdTAG: 192114 UserNameTAG: chuba CreateTimeTAG: 2012-10-01T13:20:48Z VoteTAG: 2 CoursewareTAG: Week 4 / Graphs Exercise CommentableIdTAG: 6002x_graphs_exercise NumberOfReplyTAG: 1 FirstChildTAG: Did you find the operation voltage? Remember that the incremental resistance is $r_d = \frac{1}{\frac{di}{dv}}$, evaluated at the operation point (i.e. the result from part (a)). Since you have the current as a function of voltage, you can take the derivative, evaluate and find rd FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-10-01T14:31:36Z SecondChildTAG: Yes, I found operation voltage = 1.56 .derivative I=3*v^2 =3*1.56 = 0.2136 But it isn't correct ((( Help PLZ SecondChildUserIdTAG: 192114 SecondChildUserNameTAG: chuba SecondChildCreateTimeTAG: 2012-10-01T16:46:28Z SecondChildTAG: ooooo))))) Done) Thanks) SecondChildUserIdTAG: 192114 SecondChildUserNameTAG: chuba SecondChildCreateTimeTAG: 2012-10-01T16:50:49Z SecondChildTAG: same problem here,getting 0.2136,now how to get the correct answer? SecondChildUserIdTAG: 171378 SecondChildUserNameTAG: ABHISHEKFROMINDIA SecondChildCreateTimeTAG: 2012-10-07T04:52:05Z SecondChildTAG: got it!!!! SecondChildUserIdTAG: 171378 SecondChildUserNameTAG: ABHISHEKFROMINDIA SecondChildCreateTimeTAG: 2012-10-07T05:15:39Z IndexTAG: 1542 TitleTAG: thats a fine tool! (errors) Thank you for this simulation opportunity - its very valuable. (errors: 1. the K-slider doesn't work here 2. there is no difference in volume of music for different vIN and vOUT-amplitudes ) UserIdTAG: 388524 UserNameTAG: dasbinich CreateTimeTAG: 2012-10-01T13:05:54Z VoteTAG: 2 CoursewareTAG: Week 5 / Music Amplifier Demonstration CommentableIdTAG: 6002x_snd_d NumberOfReplyTAG: 0 IndexTAG: 1543 TitleTAG: Textbook Error Pg 195 The text on page 195 begins: "We are given that iDS=4mA." Then goes on to solve for 8mA. UserIdTAG: 384766 UserNameTAG: CodeSF CreateTimeTAG: 2012-10-01T09:16:41Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1544 TitleTAG: Mid term exam includes syllabus up to week 8 content ? Can someone kindly confirm if the Mid term exam (@ 25 Oct 2012) includes syllabus up to week 8 content (last release before mid term exam) OR up to week 6 (last homework due) OR up to week 7 (last week number as per calendar) ? Many thanks in advance and will help in my planning :) UserIdTAG: 170349 UserNameTAG: amitGuliya CreateTimeTAG: 2012-10-01T08:45:02Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: More details please. FirstChildUserIdTAG: 162670 FirstChildUserNameTAG: charlesbabyt FirstChildCreateTimeTAG: 2012-10-01T11:04:47Z SecondChildTAG: Hello, I wanted to know the syllabus for mid term exam on 25 Oct. i.e. do i need to study till week 7 contents ? At that time week 8 content will also be available so not sure if the exam will contain questions containing week 8 content also. Thanks in advance :) SecondChildUserIdTAG: 170349 SecondChildUserNameTAG: amitGuliya SecondChildCreateTimeTAG: 2012-10-01T11:59:56Z SecondChildTAG: In my opinion, it would be up to the last homework due. I base this on the fact that upcoming material seems to be released "early". Right now we are on the fourth week, but 5 and 6 are available. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-01T15:20:12Z SecondChildTAG: Thanks Penny !! SecondChildUserIdTAG: 170349 SecondChildUserNameTAG: amitGuliya SecondChildCreateTimeTAG: 2012-10-02T11:13:56Z FirstChildTAG: It was week 1 to 6 the last time. It should be the same this time as well. The final exam will contain material from week 1 to 13. FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-10-01T12:50:38Z SecondChildTAG: Thanks Ashwith !! SecondChildUserIdTAG: 170349 SecondChildUserNameTAG: amitGuliya SecondChildCreateTimeTAG: 2012-10-01T15:20:58Z SecondChildTAG: Hi I think the same.. can someone from [STAFF] confirm the syllabus for Mid term and if any practice sets can be made available? SecondChildUserIdTAG: 111917 SecondChildUserNameTAG: ashish_mit SecondChildCreateTimeTAG: 2012-10-23T13:23:31Z IndexTAG: 1545 TitleTAG: What you need to solve H3P4: DIODE LIMITER You only need to solve these circuits: D1: ![enter image description here][1] D2: ![enter image description here][2] [1]: https://edxuploads.s3.amazonaws.com/13490764694873105.png [2]: https://edxuploads.s3.amazonaws.com/13490765721872176.png Obviously with the correct parameters UserIdTAG: 299251 UserNameTAG: TheRedBlackOne CreateTimeTAG: 2012-10-01T07:30:40Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Indeed, it's that easy. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-10-01T10:04:55Z IndexTAG: 1546 TitleTAG: Printing the HW pages? Trying to print HW5 didn't work so well for me - there were a lot of missing characters and text (I'm using Chrome on Windows.) It would be nice to be able to print out the HW problems. It would save having to reproduce the schematics and also allow me to work on them away from the computer. UserIdTAG: 69075 UserNameTAG: ErikR CreateTimeTAG: 2012-10-01T03:11:58Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Do you know how to capture screen shots in Windows by pressing the "PrtSc" button, and then opening Paint, or another graphics editing program, and clicking Edit>Paste (or pressing Ctrl-V); and scaling and printing from there? That way you have an exact copy of what is on the screen. There are other ways, too; I use a simple program that prints everything to a .pdf file so I can view it later on without printing or wasting paper. Copying and pasting into Microsoft Word or another text editor is a pain and is very difficult unless you cut and paste chunks at a time and keep the images and text separate. (If certain characters are missing or show up as "?" then you may have to change your browser's encoding scheme; in Firefox it's under View > Character Encoding, and varies by country, language and OS, I think. Chrome may be similar.) FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-10-01T03:34:50Z SecondChildTAG: Good tip! SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-01T15:40:08Z IndexTAG: 1547 TitleTAG: defeated in lab 3 I have tried everything and every hint that has been given and i am defeated. I can not do this, I am at a point that if i can not do this, how am I going to make it through the rest of the course, I do not want to drop out dont plan to but this has got me beat. Maybe it is just this one that has me but I dont know what to do Just to let everyone know am not looking for answers just trying to find out where I am making mistakes UserIdTAG: 160242 UserNameTAG: Mlevins35 CreateTimeTAG: 2012-09-30T22:48:06Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: Just because you can't get one little lab doesn't mean you fail, labs only count for 15%. What helped me is just first concentrate on building something that follows the truth table rather then if it is a Nand or a Nor gate ect. Plus this discussion on lab 3 should really help you: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50673e4c2193ea2300000009 FirstChildUserIdTAG: 219780 FirstChildUserNameTAG: ReconIII FirstChildCreateTimeTAG: 2012-09-30T22:57:41Z FirstChildTAG: Hi Mlevins35! Sometimes we need a push. So here it is mine to you: "Persevere and you will succeed" and "you are not alone here, you have a lot of TA's and Classmates that can help you to become this more friendly ;)" Be up! You can do it! Sometimes this take a lot of effort and time. But remember that in the end of the Course you will get a reward :)! My best wish to you, Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-30T22:58:56Z SecondChildTAG: thank you for the vote of encouragement. I really need it. I will get it, it may just take me a little bit. I just hate not being able to get it and be aable to turn it in on time. SecondChildUserIdTAG: 160242 SecondChildUserNameTAG: Mlevins35 SecondChildCreateTimeTAG: 2012-09-30T23:13:54Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-30T23:24:30Z SecondChildTAG: That's a nice phrase: "Sometimes we need a push" :) SecondChildUserIdTAG: 79361 SecondChildUserNameTAG: kimt SecondChildCreateTimeTAG: 2012-10-01T00:17:34Z SecondChildTAG: Thank you kimt! ;). I hope that you are having fun as a staff now ^_^ ! My best wish to you! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-01T00:38:31Z FirstChildTAG: Mlevin35s, Labs and Homeworks for this course are not easy at all... I found myself struggling when solving them (especially with some of the labs)! Keep trying and practicing, and I am sure that you will get the concepts. Just as an additional comment, answers for the Labs are going to be posted after the due date, so you should be able to see the solutions for Lab 3 soon. FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-30T23:59:32Z SecondChildTAG: Thank you jelizon! ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-01T00:03:31Z SecondChildTAG: **I don't know.** I found this week's (Week 3) Homework and Lab extremely easy, but that's because I have a background in this material. I think that *high school students* should have a difficult time with this course. However, if you are an EECS major (or even a math/physics major) just starting in college/university, and you have a good grasp of algebra, have a basic understanding of calculus, and passed the electromagnetics portion of university-level physics (which are the pre-requisites, after all) you should have a moderate-to-easy time with this course (e.g. 30 minutes to 4 hours to complete Homework and Lab). If you are *struggling*, you are either not reading the text, not watching the videos (you are too busy) or you lack the fundamental knowledge of the pre-requisites and need help that is beyond the scope of this course. This course, after all, is used as a "weed out" course for EECS students (to identify the students that do not have the potential to succeed in electrical engineering, and to notify them that another line of study is better suited to their skill set or their abilities. That is not to say that everyone who fails this course should forget about engineering as a career; it's just to say that they should consider mastering the pre-requisite material first (i.e. for **non-English speakers**, perhaps taking a *similar* course in the native language first will help them in *this* course immensely). SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-10-01T02:27:09Z FirstChildTAG: Mlevins35... There is a huge difference between seeing this material for the first time and seeing it a second (or third) time. Also, this may be one of the differences between online and on-ground courses. Online may really require more personal initiative and careful, critical thinking, just because the other students and professors are not so easily available for talking things out. In which case, it may actually provide a superior learning experience if one is determined enough to 'tough it out.' FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-10-01T05:05:56Z FirstChildTAG: Don't worry about missed check marks - and careful with being critical of others. I completed courses like these Edx courses (including this one) many years ago. I am a software engineer/consultant with a sucessful career under my belt. I am spending too many hours on the mechanical details of using 6.002x course tools. And even more troubling is the busy work calculating. What has helped tremendously is to look at the HW and go read about that but not necessarily in the course text. For example: one home work was a "mystery" circuit that elsewhere is called a "double shunt clipper" that paragraph from outside materials made that HW easy. That I was helped by a "cookbook circuits course" is derisive but so is "ivory tower". Edx anything is a non credit course. enjoy it. the green check marks are not everything; and, not a good measure of ones abilities and knowledge. I was frustrated by the HW's and LAB's and was planning to stop turning them in. Now I have decided missing green check marks won't dissuade me. Also Prof. Agarwal's lectures and demos are a hoot; and, the circuits sandbox is total fun until you are pressed by a deadline. Sincerely flamed, hofacker27@netscape.com Victoria,TX. FirstChildUserIdTAG: 331542 FirstChildUserNameTAG: james77901 FirstChildCreateTimeTAG: 2012-10-01T14:25:01Z IndexTAG: 1548 TitleTAG: lab 3 problem After building the circuit,when i performed the transient analysis,the graph totally matched with the one given.Still when i tried to submit the result by clicking "check",it was showing wrong.Please help me. UserIdTAG: 366793 UserNameTAG: ipratiknandi CreateTimeTAG: 2012-09-30T18:42:30Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: the same is happening to me ... FirstChildUserIdTAG: 157259 FirstChildUserNameTAG: JulianTerzyk FirstChildCreateTimeTAG: 2012-09-30T18:44:50Z SecondChildTAG: how do you build the circuit? SecondChildUserIdTAG: 137918 SecondChildUserNameTAG: jaisneha SecondChildCreateTimeTAG: 2012-09-30T19:27:44Z SecondChildTAG: the same thing i am getting.... SecondChildUserIdTAG: 368981 SecondChildUserNameTAG: asimakhtar SecondChildCreateTimeTAG: 2012-09-30T19:44:29Z SecondChildTAG: same issue SecondChildUserIdTAG: 509366 SecondChildUserNameTAG: shikhagoel003 SecondChildCreateTimeTAG: 2012-09-30T19:46:17Z SecondChildTAG: what is the value of w/L ? i am getting 31.85 but answer is still wrong. can any one tell me the exact value. SecondChildUserIdTAG: 368981 SecondChildUserNameTAG: asimakhtar SecondChildCreateTimeTAG: 2012-09-30T20:26:11Z SecondChildTAG: I can't tell you the exact value, but I can tell you that that it is not the value I got. I'll also add, it's possible that not all of your MOSFETs will have the same W/L value. SecondChildUserIdTAG: 287805 SecondChildUserNameTAG: Beneficial SecondChildCreateTimeTAG: 2012-09-30T20:51:59Z FirstChildTAG: I had that issue too. The graph I had seemed to be almost identical to the example graph...as far as I could tell by examining it visually. After making some changes to my numbers, I was eventually able to get that check mark. The checker is actually checking values. In all likelihood, you're very close the solution but your numbers are slightly off. Two things that helped for me. 1) Note that the pullup resistor is 10 KILO ohms, not 10 ohms, which messed me up for a while. Also note that the switch variable that the schematic tool permits you to change is W/L , whereas the formula provided in the problem and in the text is L/W. So you might have to take the reciprocal of your answers once you've solved for it. Also, just be careful with your decimal places. I was stumped until I realized I had inadvertently used a value that was missing one more zero after the decimal! :D FirstChildUserIdTAG: 287805 FirstChildUserNameTAG: Beneficial FirstChildCreateTimeTAG: 2012-09-30T18:57:46Z SecondChildTAG: i have done all of these points as you have mentioned correctly..but still i was facing the same problem SecondChildUserIdTAG: 366793 SecondChildUserNameTAG: ipratiknandi SecondChildCreateTimeTAG: 2012-10-02T04:22:56Z FirstChildTAG: did u close the transient analysis window before clicking check? I think the window needs to remain open while the system checks the results... FirstChildUserIdTAG: 74193 FirstChildUserNameTAG: SpaceAgeRobot FirstChildCreateTimeTAG: 2012-09-30T18:59:37Z FirstChildTAG: In the problem there is a hint that says: "Hint: you'll only need 3 mosfet switches to implement the gate." You must think of a way to use exactly 3 switces so make sure you didn't use more! The equation provided in the problem helps a lot. FirstChildUserIdTAG: 304587 FirstChildUserNameTAG: Vasso FirstChildCreateTimeTAG: 2012-09-30T20:36:05Z FirstChildTAG: My stupid mistake was that that in "Vth=0.5V, so Vol<0.25V" I considered that V stands for Vs. When i got it was volts, just took hi enough W/L parameters so that Vol would be less then 0.25 volts. And that answer received "right", even with glitches exceeding 0.25 volts, where they shouldn't be. So I doubt that graphs have to be identical. FirstChildUserIdTAG: 66669 FirstChildUserNameTAG: agarus FirstChildCreateTimeTAG: 2012-09-30T21:58:50Z IndexTAG: 1549 TitleTAG: LAB 3 HELP!!! someone please tell me how can i start doing lab 3..i'm not getting anything.. PLEASE HELP ME AS SOON AS POSSIBLE!! UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-30T18:24:53Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: i would love to help you, but any help I suppose would be like telling the answer right away in this LAB so cant help here....... though If you contact me or anyone else outside this group...then you maybe saved FirstChildUserIdTAG: 169784 FirstChildUserNameTAG: thewiredbear FirstChildCreateTimeTAG: 2012-09-30T20:15:39Z SecondChildTAG: There are useful ways of providing guidance, within the scope of the honour code, to help someone solve a problem for themselves, without just telling to covertly break the honour code outside of the MITx system. If I could down vote your comment, I would. It was really of no help whatsoever to anyone. SecondChildUserIdTAG: 287805 SecondChildUserNameTAG: Beneficial SecondChildCreateTimeTAG: 2012-09-30T20:54:27Z FirstChildTAG: This should help you get started. You're asked to create a network of MOSFET switches which implements the logic outlined in the truth table provided. The boolean logic form is provided for you. You're told you will need 3 MOSFET switches. A lot of the possibilities are limited by that hint which simplifies things greatly. You're also asked to find the values for W/L which will allow your MOSFETs to abide by the static discipline of the network (ie, communicate with each other as desired). Divide the question into two parts: 1) What arrangement should my MOSFETs take in order to implement the logic. 2) What values of W/L should my MOSFETs have in order to satisfy the static discipline. (The key point here is that Vol < 0.25V ) If you're confused by the way the schematic tool is set up for you. This might help. The three voltage sources, A, B, and C, represent the input voltages for your MOSFET switches. You don't need to alter any values related to them. The Vpwr source is what is referred to elsewhere in the materials as Vs. The reason you're given square waves, with different frequencies (that is, they go on and off in regular intervals), is just for convenience, so that you can see what your output is for various conditions (think of your output as the combination of all the inputs). In other words, the square waves are a convenient way to allow you to see, at a glance, what your output is when, A is on, B is on, and C is off; or when A is on, B is on, and C is on; or when ...etc. In this way, you can use the transient analysis to verify that your logic is performing as it should. Use the formulas provided in the text and in the lab in order to solve for the values of W/L that you want. FirstChildUserIdTAG: 287805 FirstChildUserNameTAG: Beneficial FirstChildCreateTimeTAG: 2012-09-30T21:04:43Z IndexTAG: 1550 TitleTAG: stuck with the equation i solved the problem till we get the equation ia=(vi-va)/2 then i added this eqn with the given eqn and quated it to zero, how should i proceed then? please help thanks in advance. UserIdTAG: 160677 UserNameTAG: ARMACK CreateTimeTAG: 2012-09-30T16:23:13Z VoteTAG: 2 CoursewareTAG: Week 4 / Linearization Exercise 1 CommentableIdTAG: 6002x_linearization_ex_1 NumberOfReplyTAG: 1 FirstChildTAG: ARMACK, are you using the equation to find the voltage "va"? Remember to use an iterative method because you won't be able to solve the exercise by analytic methods. The idea is to estimate a value for "va" and then use the expression for the non-linear device to compute "ia". Then compute "va" using your equation. If they are equal, then you have the correct "va", but if not, then use the new value of "va" and repeat the process. FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-30T23:31:28Z SecondChildTAG: so basically you suggest that first i assume a value of va and calculate the value of ia from the given eqn and then use this ia to calculate again va and if it is equal to my assumed va then the answer is correct. SecondChildUserIdTAG: 160677 SecondChildUserNameTAG: ARMACK SecondChildCreateTimeTAG: 2012-10-01T04:23:34Z SecondChildTAG: thanks JELIZON finally got the answer. SecondChildUserIdTAG: 160677 SecondChildUserNameTAG: ARMACK SecondChildCreateTimeTAG: 2012-10-01T05:27:14Z SecondChildTAG: Hi, I solved it as follows by taking log of both sided: LN([(Va-Vi)/10*R]-1)=-Va/5. I am converging to a wrong solution i.e. I am getting Va = -1.385. This solution satisfies (Va-Vi)?R + 10*(1-exp(-Va/5))=0. Where am I going wrong? This does not match the given solution. SecondChildUserIdTAG: 246991 SecondChildUserNameTAG: spatra SecondChildCreateTimeTAG: 2012-10-01T17:13:07Z SecondChildTAG: I think I found the problem. For the solution given, Ia = -10*(1-exp(-va/5)) and NOT 10*(1-exp(-va/5)) SecondChildUserIdTAG: 246991 SecondChildUserNameTAG: spatra SecondChildCreateTimeTAG: 2012-10-01T17:19:46Z IndexTAG: 1551 TitleTAG: Prerequisites **I do not have the mathematical prerequisites**, *i.e. calculus and differential equations. Just simple algebra.* I have had relatively little trouble completing the homework and lab assignments for the past 3 weeks with the exception of the parts on maxwell's equations, but week four '***the small signal method***' requires differential equations. How will this impact my ability to finish the course? I think badly. ***(1) - Is anyone else in my situation*** ***(2) - Will MITx be offering courses like this one that covers the prerequisites.*** I have looked at the **Khan Academy** sections on calculus and have a book, Calculus made easy (good book I think). But I don't think I will be able to gain the understanding I need for week 4 homework. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-30T15:15:19Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: You may be surprised at how much you can do. Chances are you will still learn a lot. FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-09-30T15:20:56Z SecondChildTAG: Totally agree MobiusTruth ;)! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-30T15:51:48Z SecondChildTAG: Hey, Miriam...nice to see ya'! ;) SecondChildUserIdTAG: 92346 SecondChildUserNameTAG: MobiusTruth SecondChildCreateTimeTAG: 2012-09-30T16:17:09Z FirstChildTAG: To be completely frank, it's ours fault we took this course without the required prerequisites. I have never done calculus or differential equations in school, as I'm only at the beginning of Czech equivalent to 12 grade (no AP / A levels etc. here). This leaves us in a difficult position, either accept that our grade is going to be affected, or try to understand the Math. So far we haven't encountered differential equations, but that is coming right next week. The only thing I can suggest is to try and go with the flow, see what you understand, use Wolfram Alpha for problems that you cannot do yourself and learn a bit every day on the side. FirstChildUserIdTAG: 138981 FirstChildUserNameTAG: Pr0bability FirstChildCreateTimeTAG: 2012-09-30T16:02:05Z SecondChildTAG: Hang tight. Your approach - using Wolfram - sounds good. SecondChildUserIdTAG: 92346 SecondChildUserNameTAG: MobiusTruth SecondChildCreateTimeTAG: 2012-09-30T16:16:01Z FirstChildTAG: Just to emphasize what has been said, The material in week four did not require differential equations, only simple differential calculus. Differential equations will be required to analyze the time dependent behavior of circuits containing capacitors and/or inductors. To follow the math only requires familiarity with exponentials and their derivatives along with the ability to work with complex numbers. Knowledge of the Euler equation and trig identities also helps, but you can look up what you need to know on the internet. Google is your friend! FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-09-30T18:08:49Z FirstChildTAG: Use wiki. Differential equations which used in this course are essential and very easy. http://en.wikipedia.org/wiki/Derivative http://en.wikipedia.org/wiki/Differentiation_rules FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-01T08:37:08Z IndexTAG: 1552 TitleTAG: S7E1 - Linearization Hello guys! I have some trouble with this exercise. I was able to find the first two answers, so they are well understood. But I can't figure out how to do the third, which is: **Find the value for the incremental resistance of the nonlinear element $N$ by linearizing the expression for $i_A$ about the operating point when $v_I = 5.0V$.** I do understand that the operation point is the value that was found in the first question, right? So.. How do I linearize it? What does that mean? is it derivative and then equals to operating point? Thanks in advance! UserIdTAG: 308853 UserNameTAG: fmorato CreateTimeTAG: 2012-09-30T13:01:45Z VoteTAG: 2 CoursewareTAG: Week 4 / Linearization Exercise 1 CommentableIdTAG: 6002x_linearization_ex_1 NumberOfReplyTAG: 3 FirstChildTAG: ra= delta rA =1/(diA/dvA) evaluated at vA = VA d(iA)/dvA (eval @vA = VA) = d/dvA (10 * (1-exp^(-vA/5))) = 4* exp^(-1.0882/5) ra = 1/above eqn FirstChildUserIdTAG: 98462 FirstChildUserNameTAG: kishorekumar FirstChildCreateTimeTAG: 2012-09-30T19:08:08Z FirstChildTAG: The hint suggests that you use Tailor expansion like the one in the lecture sequence. This a somewhat longer way but it does the trick. Lets take it first: - Using Tailor: Va at point (5+dVi) is (const1 + const2 * dVi), where const1 = Va(5) and const2 = dVa/dVi(5) = derivative of Va by Vi at 5 volts. Note that you already know these constants from problems 1 and 2. - Using Tailor: get analogous expansion for the current Ia at point (5+dVi) of the form (const3 + const4 * dVi) - What you now have is expressions for voltage and current (Va and Ia) around across your device. One way to look at it is as if you set your Vi at constant value of 5Volts and added an AC source in series with it which produces sinusoidal wave small enough that total voltage remains in a small enough region of 5Volts (you want to stay in the linear region of the V-I curve) - Incremental resistance is the one of the device operating in the small neighborhood of 5Volts. Note that your expressions already have offsets (const1 and const3) that take you to that region. You're interested in the response of your device to increments of voltage dVa and current dIa. These are the parts of your Tailor expansions that aren't constant. So: dVa = const2*dVi and dIa = const4*dVi. - As usual resistance R = V/I. Incremental resistance = dVa/dIa = const2/const4 ---------- Going through these steps is straightforward and instructive, but with a little intuition you can cut their number to just one. - iA=10⋅(1−e^(−vA/5)) as stated in the task - Note that incremental current diA = derivative(of iA with respect to vA taken at Vi=5volts)*dvA. Ohm's law dictates that this derivative is the reciprocal of resistance: diA = 1/R * dvA. There you have it R = 1/derivative FirstChildUserIdTAG: 5325 FirstChildUserNameTAG: vkz FirstChildCreateTimeTAG: 2012-09-30T16:48:49Z SecondChildTAG: And derivative of iA with respect to VA in point VI is just 2/e, and it's reciprocal which is resistance is e/2 = 1.36, it's false answer SecondChildUserIdTAG: 140867 SecondChildUserNameTAG: DarkWishMaster SecondChildCreateTimeTAG: 2012-09-30T18:42:33Z SecondChildTAG: Hey, it's not in point VI, it's in point VA(1.088) when VI=5 V, and it's done :) (whew) SecondChildUserIdTAG: 140867 SecondChildUserNameTAG: DarkWishMaster SecondChildCreateTimeTAG: 2012-09-30T18:45:36Z FirstChildTAG: there's a full explanation of how to solve an almost identical problem in the week 4 tutorials: [small signal problem][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_4/wk4t3/ FirstChildUserIdTAG: 195933 FirstChildUserNameTAG: Amitbe FirstChildCreateTimeTAG: 2012-10-01T07:30:36Z IndexTAG: 1553 TitleTAG: Why is RTH the parallel resistance? Can someone tell me why the answer is parallel equation for 2 resistors? Surely Rs and Rp are in series with each other, and therefore the RTH = sum of Rs and Rp? UserIdTAG: 409867 UserNameTAG: shazmiah CreateTimeTAG: 2012-09-30T11:15:03Z VoteTAG: 2 CoursewareTAG: Week 3 / Isolation Via Thevenin CommentableIdTAG: 6002x_isolation_via_thevenin NumberOfReplyTAG: 1 FirstChildTAG: You need to find RTH from the point of view of the load. For that, remove your load, short circuit all voltage sources and open circuit all current sources and see how the circuit looks like when the terminals are seen from the load's viewpoint. FirstChildUserIdTAG: 128924 FirstChildUserNameTAG: arjun392 FirstChildCreateTimeTAG: 2012-09-30T11:36:56Z SecondChildTAG: Take also a look on figure 3.79 (p. 169) in the textbook SecondChildUserIdTAG: 131747 SecondChildUserNameTAG: hasi SecondChildCreateTimeTAG: 2012-09-30T19:24:37Z SecondChildTAG: Thanks. I forgot about short-circuiting the VS. Of course, after that the 2 register become parallel. SecondChildUserIdTAG: 409867 SecondChildUserNameTAG: shazmiah SecondChildCreateTimeTAG: 2012-10-06T11:23:45Z IndexTAG: 1554 TitleTAG: Vt How Vt for MOSFET is determined? UserIdTAG: 99628 UserNameTAG: Pranjal16 CreateTimeTAG: 2012-09-30T11:07:07Z VoteTAG: 2 CoursewareTAG: Week 5 / MOSFET Amplifier CommentableIdTAG: 6002x_mosfet_amp NumberOfReplyTAG: 2 FirstChildTAG: Vil < Vt < Vih, is a margin between Vil and Vih FirstChildUserIdTAG: 322923 FirstChildUserNameTAG: Serhii_S FirstChildCreateTimeTAG: 2012-09-30T11:20:12Z FirstChildTAG: It depends on how the device is fabricated. FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-09-30T15:33:26Z IndexTAG: 1555 TitleTAG: Bad mistake I've spent a half of the day because of thinking, that the non-linear element can be replaced with a resistor of resistance R=[answer on question 1] and it's current can be computed by i = Vd/R... But it can't. UserIdTAG: 326787 UserNameTAG: Angstrem CreateTimeTAG: 2012-09-30T10:39:59Z VoteTAG: 2 CoursewareTAG: Week 4 / Linearization Exercise 2 CommentableIdTAG: 6002x_linearization_ex_2 NumberOfReplyTAG: 1 FirstChildTAG: use the smal signal circuit technique to find VD (1/derevitive f(vD)= R) then u replace VD in the fomula 1 of the diod to find id then u can use the KVL method to the large signal circuit (-Vi +(R*id)+VD=0) to get Vi FirstChildUserIdTAG: 236136 FirstChildUserNameTAG: Hamid-ch FirstChildCreateTimeTAG: 2012-10-04T03:40:09Z SecondChildTAG: Thanks Hamid-ch. Your comment help me a lot. SecondChildUserIdTAG: 159031 SecondChildUserNameTAG: GabrielNeves SecondChildCreateTimeTAG: 2012-10-05T19:26:49Z IndexTAG: 1556 TitleTAG: first day first day, a little bit late but very excited... hopefully i'll catch up soon UserIdTAG: 523697 UserNameTAG: marc_alex CreateTimeTAG: 2012-09-30T04:05:07Z VoteTAG: 2 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 2 FirstChildTAG: Welcome to 6.002x marc_alex! ;) Something that you should read/know first: 1- [syllabus][1] 2- Deadline of assignments ([read here][2]): Week 1: september 16th (time limit has passed)... Week 2: september 30th (today it is the limit - this assignment was extended from september 23th to today because of some issues with some places of the world with YouTube...) Week 3: september 30th (today) 3- Use the Search, eg., if you want to see if there are Posts about some homework or lab, you can tag for the case of homework h2p1 (means Homework 1 problem 1) or lab2 (all together, not lab separated 2 ;), you will find nice stuffs ). 4- Also, if you clic on your name you can go to your active threads. 5- If you go to the left where it says all and if you press the down arrow, you can find the option following. The following can be set clicking on the triange with the start that you see up-right ;). 6- Remember that you can skip two labs and homework without penality. But, as the week 1 deadline already passed, you can skip only one Homework and Lab without penality (because you already lost the homework 1 and lab 1). You can do it! Try to use this Forum, watch the week tutorials, also you can see here some hints [of labs and homework][3]. See you! Myriam. P.D: If you have any doubt I will try to do my best and help you. Sorry for my English. [1]: https://www.edx.org/static/content-mit-6002x/handouts/syllabus.a477535058a1.pdf [2]: https://www.edx.org/static/content-mit-6002x/handouts/calendar.c7b2799155f8.pdf [3]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/myrimits-hints-links/ FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-30T04:24:54Z SecondChildTAG: Hi Myriam You don't have to apologize for your English, you are doing great. I have been speaking and writing English for 56 years now, and you do better than I do. Thanks for all your help and for the assistance you give to others. You rock. rharris SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-09-30T04:39:42Z SecondChildTAG: My first day also! Thank you for the information Myriam! SecondChildUserIdTAG: 241379 SecondChildUserNameTAG: CrisAzc SecondChildCreateTimeTAG: 2012-09-30T09:18:58Z SecondChildTAG: You are welcome rharris and CrisAzc! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-30T15:06:37Z FirstChildTAG: Hi...I hav joined lately to this forum...will it affect my grading and score? Which 1 is preferred? Continuing with the current status or join newly for spring session? FirstChildUserIdTAG: 704220 FirstChildUserNameTAG: Shruthi91 FirstChildCreateTimeTAG: 2012-10-22T15:07:19Z IndexTAG: 1557 TitleTAG: Mixer is messing up my mind Hey guys! Can someone help me out with lab 2. Just a clue on how to calculate the resistors. UserIdTAG: 499671 UserNameTAG: maliha266 CreateTimeTAG: 2012-09-29T19:23:23Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: We discussed extensively here including why certain things work and others don't. https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5058f699cf5d7b2b00000005 FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-09-29T23:35:42Z FirstChildTAG: Hi maliha266! Can I help you? You can read some hints [here][1] I also made a video tutorial of Lab2, but I can not upload it till the end of the deadline of Week 2 ... See you! Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/505f41be015492230000001c FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-29T19:36:52Z IndexTAG: 1558 TitleTAG: still nothing in lab3 some help please with lab3 UserIdTAG: 344037 UserNameTAG: HAMISS CreateTimeTAG: 2012-09-29T18:30:05Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: use 3 gates bit i don't now how to connect with other please help FirstChildUserIdTAG: 331441 FirstChildUserNameTAG: abdessamade FirstChildCreateTimeTAG: 2012-09-29T18:35:07Z SecondChildTAG: Look in the book, chapter 6, there is a circuit in there that is very close and only needs a simple modification. When I saw it, it was enlightening. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-09-29T22:17:22Z FirstChildTAG: not (C and (A or B)) = (not C) or (not(A or B)) = (not C) or ((not A) and (not B)). Z = (not C) or ((not A) and (not B)) The point here to notice, is that if C = 0 than Z will ALWAYS be 1. You know, you'll have to connect each outputs (A,B,C) with MOSFETs. So, what can you say about C's MOSFET? If you don't apply current to it's gate, it will be closed and Z will be 1. No matter what A and B are, if C = 0, than Z = 1. So the most reasonably would be to connect this C's MOSFET directly to the ground with one terminal. And the A's and B's MOSFETs will be connected to the ground ONLY throught C's MOSFET. So it would always have it's final word. Now assume, that C = 1. Now Z = (not A) and (not B). That's a NOR gate! You see, if either A or B is 1 than Z is 0. What does it mean? It means, that if we open one of this 2 MOSFETs, than the current from the source can reach the ground (of course, through the opened C MOSFET). Here the NOR pattern should be used, that is, we connect A's and B's MOSFETs in parallel. FirstChildUserIdTAG: 326787 FirstChildUserNameTAG: Angstrem FirstChildCreateTimeTAG: 2012-09-29T19:15:03Z SecondChildTAG: thakx i do it SecondChildUserIdTAG: 331441 SecondChildUserNameTAG: abdessamade SecondChildCreateTimeTAG: 2012-09-29T20:15:14Z FirstChildTAG: Hi HAMISS! Can I help you? In which part are you lost? Myriam. P.D: You can also see this post [here][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50673e4c2193ea2300000009 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-29T22:27:24Z IndexTAG: 1559 TitleTAG: Overlapping wires in circuit simulator - Lab 3 Hi, After looking at the intro videos I haven't been able to work out how to get wires to overlap in the simulator without them interacting. Is there a key you need to hold down when creating a new wire? Thanks, Robert UserIdTAG: 335803 UserNameTAG: bobbanovski CreateTimeTAG: 2012-09-29T15:08:02Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Troubleshooting NumberOfReplyTAG: 3 FirstChildTAG: Don't worry about their interactions. If there's no node symbol in the place of their intersection, than they wouldn't interact. FirstChildUserIdTAG: 326787 FirstChildUserNameTAG: Angstrem FirstChildCreateTimeTAG: 2012-09-29T15:18:06Z SecondChildTAG: please help me how to calculate w/l value SecondChildUserIdTAG: 284628 SecondChildUserNameTAG: jumana_mp SecondChildCreateTimeTAG: 2012-09-29T15:22:52Z SecondChildTAG: That's pretty easy. There's an equation, Ron = Rn * (L/W). So you just evaluate Ron (the resistance of the MOSFETs in their ON state) and solve this equation, considering W/L is unknown. It seems, that W/L = Rn/Ron. Rn is given in the lab task text, and Ron is computed by you, so you should deal with it easily. SecondChildUserIdTAG: 326787 SecondChildUserNameTAG: Angstrem SecondChildCreateTimeTAG: 2012-09-29T16:32:47Z FirstChildTAG: Hi Robert! Imagine that you have the following Circuit: And you want to conect/wiring: B to the up MOSTFET A to the down MOSFET ![circ1][1] CORRECT WAY: 1) You have to go to the B point and without releasing, reach to the up MOSFET: ![circ2][2] 2) Do the same as 1) but don't release the mouse from A to the down MOSFET. Don't stop in the interection of the two wires ;). ![circ3][3] ---- Ok, lets see what we shouldn't do... INCORRECT WAY: 1) Conect B to the up MOSFET. ![circ2][4] 2) Conect A to the down MOSFET but you stopped in the intersection and then continue to the up MOSTFET. Incorrect, the two wires will overlap! ![circ3][5] ---- I hope this can help you. Myriam. [1]: https://edxuploads.s3.amazonaws.com/13489487115276365.png [2]: https://edxuploads.s3.amazonaws.com/13489489223013087.png [3]: https://edxuploads.s3.amazonaws.com/13489490717143543.png [4]: https://edxuploads.s3.amazonaws.com/13489489223013087.png [5]: https://edxuploads.s3.amazonaws.com/13489494478474493.png FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-29T20:12:53Z FirstChildTAG: Thanks for the solution, Myriam. I do wonder, however, if there isn't another way. Add just a few more components to the problem and it will become practically impossible to both avoid unintentional interconnections and have a readable diagram using only this technique. Anybody know another way? FirstChildUserIdTAG: 182168 FirstChildUserNameTAG: arthurdent FirstChildCreateTimeTAG: 2012-09-29T23:08:23Z SecondChildTAG: ;). You are welcome! . Hmmm, I dont know... SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-30T00:22:51Z IndexTAG: 1560 TitleTAG: Linearity Current sources and voltage sources are non-linear yet we use thevenin and norton even when they are present in the circuit.WHY?? UserIdTAG: 334777 UserNameTAG: AAYUSH007 CreateTimeTAG: 2012-09-29T15:04:27Z VoteTAG: 2 CoursewareTAG: Week 2 / Linearity CommentableIdTAG: 6002x_linearity NumberOfReplyTAG: 1 FirstChildTAG: I think you have to rephrase this into: in the real world they behave in *a certain range* like they were linear. If Vth and Rth are known, then if Vth=Vsource without a load, with Vsource being an *ideal driving source*, we suppose that Vsource is constant, independent of the current, meaning it has zero resistance, because deltaVsource/deltaCurrent is zero. However, in the real world we notice an internal resistance, and in our Thevenin *model* we express that resistance as an ideal resistance Rth, which is also a *model* and the *modeled voltage source* as Vth. So for the Thevenin model, deltaVth/deltaCurrent=Rth is constant, thus linear. You need these models as a starting point to explain first linear behavior and to learn to apply KVL and KCL. And than, later, you introduce the nonlinearity, and that is also done with models too, like using the Taylorseries to *approach* the nonlinear behavior of a circuit, and than, with the first derivative of that nonlinear model, we try to approach again an almost ideal linear model for *a small* signal deviation, because we than can use KVL and KCL again, because we linearized for small signals. And for learning purposes, we don't exactly define how small, small is, as long as we suppose that it is in *an almost linear part* of the original nonlinear function of the circuit. For many purposes these simple linearized models work, but there are many cases where it won't and than you can't always ignore the terms from the first derivative of the function. See chapter8 pages 405-408 . FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-09-29T17:03:13Z SecondChildTAG: Thank you salsero! ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-29T20:16:17Z IndexTAG: 1561 TitleTAG: H4P3 part A - Frustration - I'm not getting anywhere Dear community, I usually post stuff when I'm hopelessly lost - and no other thread regarding H4P3 explains the problem I'm facing with the task. So, the trick is to add a current source, let's call it i_{i}. Then we should use a superposition of two solutions - so here are the questions I'm facing: - Which solution do we have in mind? Are we supposed to node-analyze the circuit and then add the appropriate node voltages together? (thats basically the definition of superposition, but I want to be sure) -How do I go about the nodes? Am I supposed to treat the dependent voltage source as a super node? If so, then it would mean that the current Io is the same as the current i, that drives the voltage source. -I believe that the main issue here is to discover the expressions at every node. Or am I supposed to analyze it using KCL/KVL? I guess it doesn't matter in this situation, but as before - I'm not certain. The main problem is that the dependent source is a voltage source - otherwise the currents would be specified explicitly. And also some recaps, just for someone to check whether my train of thought is correct: -The open loop voltage at the terminal will be Vth -Vth/Rth=In, thus if I calculate a short-circuit current at the output terminal and divide Vth by that current, I'll arrive at Rth I haven't touched part B yet - however I'm hoping that a proper answer will shed some light here. I'm counting on you guys, thanks UserIdTAG: 7114 UserNameTAG: piotroxp CreateTimeTAG: 2012-09-29T14:47:34Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi, I think it's the current controlled voltage source that is giving you trouble. I started by writing out the equations for the voltage at port A which is just the voltage across R2. So Vr2 = Ir2.R2. We also know that the voltage across R2 will be the Voltage across R1 + the voltage across the dependant source Zi. So Vr2 = Vr1 + Z.Ir2 Substituting eqn 2 into eqn 1, we get: Vr1 = R2.Ir2 - Z.Ir2 Enter the values for Z and R2 to simplify this to Vr1 = x.Ir2 Where x = R2-Z We also Know that Vr1 = R1.Ir1 So R1.Ir1 = (R2-Z)Ir2 Since we know that Ir1 + Ir2 must equal the current from our current source I0, we can now solve for both currents. Once you have the currents, the rest is easy. I hope this helps you. FirstChildUserIdTAG: 270011 FirstChildUserNameTAG: SnowmanZA FirstChildCreateTimeTAG: 2012-09-29T18:46:47Z SecondChildTAG: Thank you so much! SecondChildUserIdTAG: 344331 SecondChildUserNameTAG: intellectualwanderer SecondChildCreateTimeTAG: 2012-09-29T19:40:17Z SecondChildTAG: > We also know that the voltage across R2 will be the Voltage across R1 + the voltage across the dependant source Zi. So Vr2 = Vr1 + Z.Ir2 AAAAH MAN this is exactly it!!! Thanks a bunch. SecondChildUserIdTAG: 7114 SecondChildUserNameTAG: piotroxp SecondChildCreateTimeTAG: 2012-09-30T09:55:54Z SecondChildTAG: thank you !! SecondChildUserIdTAG: 157259 SecondChildUserNameTAG: JulianTerzyk SecondChildCreateTimeTAG: 2012-10-06T22:36:23Z IndexTAG: 1562 TitleTAG: slope why the slope = -1/R..... i know it's a ridiculous question ,but excuse me UserIdTAG: 395257 UserNameTAG: blackguitar CreateTimeTAG: 2012-09-29T14:40:12Z VoteTAG: 2 CoursewareTAG: Week 3 / Graphical Analysis of Nonlinear Circuits CommentableIdTAG: 6002x_graphical_analysis_of_nonlinear_circuits NumberOfReplyTAG: 2 FirstChildTAG: He says: from equation 1 prime, iD=vD/R. So the tangent of the drawn loadline has a negative value, because delta iD divided by delta vD is negative. If vD=V, then for increasing vD starting at 0, delta vD=V-0=V, and than, because the maximum of iD occurs at vD=0, than iD=V/R and it is going down to iD=0 at vD max=V, then delta iD=0-V/R=-V/R so delta [iD/vD]=-1/R and that is the tangent of the line and it is negative because the current in the circuit goes down, when the voltage vD is increasing. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-09-29T15:06:06Z SecondChildTAG: perfect ........... thnks SecondChildUserIdTAG: 395257 SecondChildUserNameTAG: blackguitar SecondChildCreateTimeTAG: 2012-09-29T20:37:34Z FirstChildTAG: Given an i-v coordinate system, the slope of a line is delta_i/delta_v. Resistance is v/i. So the slope is 1/R. Positive and negative is the usual. FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-09-30T16:05:31Z SecondChildTAG: thanks a lot ,MobiusTruth SecondChildUserIdTAG: 395257 SecondChildUserNameTAG: blackguitar SecondChildCreateTimeTAG: 2012-10-01T01:09:56Z IndexTAG: 1563 TitleTAG: I was so dumb I lost a huge amount of time just because I didn't think carefully before I start calculating. I saw 26 mV as 26 MV !!! :( And in addition, this is not my first occasion in which I make this stupid mistake. Once I wrote 90/2 = 30 !!! this costed me a huge price. UserIdTAG: 418182 UserNameTAG: euldji2005 CreateTimeTAG: 2012-09-29T13:03:20Z VoteTAG: 2 CoursewareTAG: Week 4 / Linearization Exercise 2 CommentableIdTAG: 6002x_linearization_ex_2 NumberOfReplyTAG: 0 IndexTAG: 1564 TitleTAG: lab 3 hurray done with lab 3 UserIdTAG: 335979 UserNameTAG: premkumar403 CreateTimeTAG: 2012-09-29T09:34:37Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: how u did it plz explain FirstChildUserIdTAG: 444708 FirstChildUserNameTAG: jagadeeshv16 FirstChildCreateTimeTAG: 2012-09-29T09:40:59Z SecondChildTAG: wait untill monday.. there should be solutions available then SecondChildUserIdTAG: 157610 SecondChildUserNameTAG: mradziwo SecondChildCreateTimeTAG: 2012-09-29T09:43:22Z FirstChildTAG: minimize the given equation and based on the equation design ur circuit FirstChildUserIdTAG: 335979 FirstChildUserNameTAG: premkumar403 FirstChildCreateTimeTAG: 2012-09-29T09:48:00Z SecondChildTAG: PLEASE HELP SecondChildUserIdTAG: 284628 SecondChildUserNameTAG: jumana_mp SecondChildCreateTimeTAG: 2012-09-29T13:58:54Z IndexTAG: 1565 TitleTAG: lab3: how to pick values of w/l i've read the text part too and i still cant understand hw to figure out the w/l ratio..help.. UserIdTAG: 364086 UserNameTAG: jordan3110 CreateTimeTAG: 2012-09-29T02:38:03Z VoteTAG: 2 CoursewareTAG: Week 3 / Switch Resistor Model of a MOSFET CommentableIdTAG: 6002x_switch_resistor_model NumberOfReplyTAG: 2 FirstChildTAG: Your $R_O$$_N$ is equal to $R_n\cdot(\frac{L}{W})$ which is the inverse of $\frac{W}{L}$. It's just a coefficient to your $R_n$. For example if your $R_n$ was 100$\Omega$ and you chose 10 (which is $\frac{10}{1}$) for $\frac{W}{L}$ then your $R_O$$_N$ would be $100\cdot(\frac{1}{10})$ which would of course be 10$\Omega$. FirstChildUserIdTAG: 213386 FirstChildUserNameTAG: dmascenik FirstChildCreateTimeTAG: 2012-09-29T04:54:39Z FirstChildTAG: to find the values ​​of (w/l)=(Rn/Ron) is necessary to consider the arrangement of the mosfet and the condition neg((1+0)*(1)) being the worst case is taken to calculate the value of the pullup resistor which shall 2.75 v and the equivalent resistance of the MOSFET 0.25. of the equation Ron clears the value and the value of Rn already have 26.5k ohm is a matter of analyzing it and make a division, Ron resistance is half the resistance value found. ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13489703543490216.jpg FirstChildUserIdTAG: 343291 FirstChildUserNameTAG: LuisEduardoH FirstChildCreateTimeTAG: 2012-09-30T01:59:38Z IndexTAG: 1566 TitleTAG: midterm proctored? I read that edX is going to allow the option of taking a proctored exam at various testing locations and I was wondering if that option will be offered for this class and whether or not it will be an option for the upcoming midterm. UserIdTAG: 184529 UserNameTAG: tfors CreateTimeTAG: 2012-09-29T02:06:12Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: There is no plan for a proctored midterm. FirstChildUserIdTAG: 42833 FirstChildUserNameTAG: KKA FirstChildCreateTimeTAG: 2012-09-29T02:19:40Z SecondChildTAG: About the proctored final examination. Do you know where can i get some information about the available locations and fees? I would deeply appreciate some information about the whole process. Thank you! SecondChildUserIdTAG: 304587 SecondChildUserNameTAG: Vasso SecondChildCreateTimeTAG: 2012-09-29T05:15:35Z IndexTAG: 1567 TitleTAG: S9E2 What i have to type for my answer will be permitted? UserIdTAG: 354502 UserNameTAG: Harmych CreateTimeTAG: 2012-09-28T19:24:13Z VoteTAG: 2 CoursewareTAG: Week 5 / MOSFET Amplifier Exercise 1 CommentableIdTAG: 6002x_mosfet_amp_e1 NumberOfReplyTAG: 1 FirstChildTAG: use VT and vIN FirstChildUserIdTAG: 354502 FirstChildUserNameTAG: Harmych FirstChildCreateTimeTAG: 2012-09-28T19:27:04Z IndexTAG: 1568 TitleTAG: Last two task 1. Get the Thevenin equivalent 2. With Vth and Rth, is easy to obtain i=Vth / R+Rth (R resistance of the strange element) 3. As R = Vx / i, then Vx + Rth * i = Vth 4. Create a table giving values to Vx and getting i, the relation in between Vx and i give us R. 5. With R, is easy to compute Vx and i. UserIdTAG: 470530 UserNameTAG: leandrorey CreateTimeTAG: 2012-09-28T17:36:20Z VoteTAG: 2 CoursewareTAG: Week 3 / Analyzing Nonlinear Circuits Exercise CommentableIdTAG: 6002x_analyzing_nonlinear_circuits_exercise NumberOfReplyTAG: 2 FirstChildTAG: is the slope of the line equal to the resistance of the diode or to 1/R? Do you use node analysis to compute the i? FirstChildUserIdTAG: 263602 FirstChildUserNameTAG: elekov FirstChildCreateTimeTAG: 2012-09-29T02:17:03Z SecondChildTAG: it is the slope of the line SecondChildUserIdTAG: 12144 SecondChildUserNameTAG: Sonam SecondChildCreateTimeTAG: 2012-09-29T05:03:05Z SecondChildTAG: i mean the slope of the line is equal to Resistance in this case as u can see the y axis is the voltage and x axis is the current axis. So according to the slope formula m = (y1-y2)/(x1-x2) #### here which gives us the Resistance If x axis was taken to be voltage axis and y axis was taken to be current axis then we would have Resistance = 1/R SecondChildUserIdTAG: 12144 SecondChildUserNameTAG: Sonam SecondChildCreateTimeTAG: 2012-09-29T05:07:17Z FirstChildTAG: Ok, to get the Thévenin equivalent: Vx = β + α.i = V * Rp/(Rp+Rs) + Rp*Rs/(Rp+Rs) * i Thus VTH = V * Rp/(Rp+Rs) = 3.17829457364 Thus RTH = Rp*Rs/(Rp+Rs) = 2987.59689922 FirstChildUserIdTAG: 351044 FirstChildUserNameTAG: mvniekerk FirstChildCreateTimeTAG: 2012-09-29T14:59:48Z IndexTAG: 1569 TitleTAG: Bug: S6V6 and S6V7 are the same video at 1.00x Only at speed 1.00x, S6V6 and S6V7 are the same video. It seems like it didn't get properly cut at the point where you're supposed to figure out the slope. The direct link videos in the wiki are also like this. At the speeds 0.75x, 1.25x, and 1.50x the video is correct, however the closed captioning is wrong for S6V7 at 0.75x, 1.25x, and 1.50x as it is based on the full length video. UserIdTAG: 213386 UserNameTAG: dmascenik CreateTimeTAG: 2012-09-28T15:17:57Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You can choose any speed you like, in .1 increments. See: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/505f2ee8786ce92b00000014 FirstChildUserIdTAG: 4076 FirstChildUserNameTAG: damians FirstChildCreateTimeTAG: 2012-10-02T17:09:38Z IndexTAG: 1570 TitleTAG: Enrollment No. Where can I find my Enrollment Number of EDX of this course? UserIdTAG: 234636 UserNameTAG: KAYBEE CreateTimeTAG: 2012-09-28T11:57:25Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I am following your instruction! Still I could not find it. I am using firefox 13.01 . Instead I am getting the follwing! ![https://edxuploads.s3.amazonaws.com/13488621754950774.jpg][2] [2]: https://edxuploads.s3.amazonaws.com/13488621754950774.jpg FirstChildUserIdTAG: 234636 FirstChildUserNameTAG: KAYBEE FirstChildCreateTimeTAG: 2012-09-28T19:48:34Z SecondChildTAG: I don't know Firefox. Sorry. SecondChildUserIdTAG: 92346 SecondChildUserNameTAG: MobiusTruth SecondChildCreateTimeTAG: 2012-09-29T01:48:48Z FirstChildTAG: Right click on your name (at the top of your question) and then click on 'inspect element.' Your number is '234636'. I *believe* that is the enrollment number you're asking about. FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-09-28T16:34:20Z SecondChildTAG: What browser are you using? I tried various ways to find out my enrollment no but in vain!! SecondChildUserIdTAG: 234636 SecondChildUserNameTAG: KAYBEE SecondChildCreateTimeTAG: 2012-09-29T19:46:39Z IndexTAG: 1571 TitleTAG: H4P2 Minimum Value of R_L The last part of H4P2 asks for the minimum value of $R_L$ that guarantees a certain way of circuit operation? What way of operation? I figured out the answers to all eight previous questions of H4P2 and now stuck on this minimum value of $R_L$. UserIdTAG: 252722 UserNameTAG: vaboro CreateTimeTAG: 2012-09-28T05:45:45Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: All right! I figured it out, great! The idea was that for $v_D$ between -5V and 0.6V the Zener diode kind of "doesn't operate" FirstChildUserIdTAG: 252722 FirstChildUserNameTAG: vaboro FirstChildCreateTimeTAG: 2012-09-28T06:09:35Z SecondChildTAG: It's not the voltage that is important with zener diodes. Zener diodes need a certain amount of current through them to reach their breakdown voltages. SecondChildUserIdTAG: 61923 SecondChildUserNameTAG: MikeJones SecondChildCreateTimeTAG: 2012-10-02T09:07:11Z SecondChildTAG: right, and if you look at the v-i curve for the Zener diode that is given there's no current through it in the range between -5V and 0.6V SecondChildUserIdTAG: 252722 SecondChildUserNameTAG: vaboro SecondChildCreateTimeTAG: 2012-10-02T13:20:27Z FirstChildTAG: I have formed a general expression for `v_D` in terms of `R_L` and tried to equate it to the margin values, it didn't work. Can someone please guide me? FirstChildUserIdTAG: 383316 FirstChildUserNameTAG: Srujana_K FirstChildCreateTimeTAG: 2012-10-02T11:31:44Z SecondChildTAG: If you're calculating the minimum value of R_L note that when v_D is -5V there's no current through the Zener diode. What type of element does this imply? What is the potential of the node between R_IN and R_L? Note what voltage is considered positive for the diode in order to assign the correct sign to the node voltage. SecondChildUserIdTAG: 252722 SecondChildUserNameTAG: vaboro SecondChildCreateTimeTAG: 2012-10-02T14:20:10Z SecondChildTAG: I got the answer, thank you :) SecondChildUserIdTAG: 383316 SecondChildUserNameTAG: Srujana_K SecondChildCreateTimeTAG: 2012-10-02T16:30:32Z SecondChildTAG: $I have formed a general expression for v_D in terms of R_L and tried to equate it to the margin values, it didn't work. Can someone please guide me?$ SecondChildUserIdTAG: 248807 SecondChildUserNameTAG: 1kingsman SecondChildCreateTimeTAG: 2012-10-06T04:40:40Z SecondChildTAG: ooops sorry lol SecondChildUserIdTAG: 248807 SecondChildUserNameTAG: 1kingsman SecondChildCreateTimeTAG: 2012-10-06T04:40:51Z IndexTAG: 1572 TitleTAG: Great explanation of static power in digital circuits! Hearing this man one realizes he has really worked for the industry, and not just studied the theory like most professors. So he knows how to get to the bottom real-life important thing and link them to theory. So now I understand why a little bump in CPU voltage makes so much impact on power consumption! Thanks! UserIdTAG: 398594 UserNameTAG: DaveyJC CreateTimeTAG: 2012-09-26T20:01:03Z VoteTAG: 2 CoursewareTAG: Week 3 / Static Power of a MOSFET CommentableIdTAG: 6002x_mosfet_static_power NumberOfReplyTAG: 1 FirstChildTAG: I agree, in fact He answered many questions that occured in my mind in a very brilliant way. For example : I used to think that logical "1" is discribed by exact 5 Volt, so How can I find a component that react exactly and stably like that if we take in concideration the additional noise ? The answer of this professor was brilliant. FirstChildUserIdTAG: 418182 FirstChildUserNameTAG: euldji2005 FirstChildCreateTimeTAG: 2012-09-26T23:48:41Z IndexTAG: 1573 TitleTAG: Slowpoke I find this course right now, but i understand that I am late. So I have no chans to get sertificate of mastery, because i already missed 2 homeworks and labs(as i read it in previous posts). Am I right? UserIdTAG: 500965 UserNameTAG: barka0 CreateTimeTAG: 2012-09-26T19:31:32Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: Second week was extended FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-09-26T19:45:30Z FirstChildTAG: No. You missed just the first week, and can still get up to 100%. Please lock forward to the posting with the title "NEW STUDENT", it's perfectly explained there. And maybe it should be posted somewhere on the course info-page too, the question dents to repeat quit often today. FirstChildUserIdTAG: 131747 FirstChildUserNameTAG: hasi FirstChildCreateTimeTAG: 2012-09-26T19:46:37Z FirstChildTAG: Even if we suppose that you have 0% in the first week, you have lost only 3% in the total average. So you still can slam dunk it. FirstChildUserIdTAG: 418182 FirstChildUserNameTAG: euldji2005 FirstChildCreateTimeTAG: 2012-09-27T00:00:37Z FirstChildTAG: Hi barka0! You are not late :). Remember that week2 assignaments has been extended to september 30th [read here][1]. You have time to do it ;). Also, two las and homeworks can be missed without penality [read here syllabus][2]. Also, if you see at your left, you can seach for hints that will help you with your Homeworks and Labs of the Weeks just write : h2p1, h2p2, etc..., lab2. See you! Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/info [2]: https://www.edx.org/static/content-mit-6002x/handouts/syllabus.a477535058a1.pdf FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-27T12:05:13Z IndexTAG: 1574 TitleTAG: staff.. Is the ans for 4th qstn correct..? according to me it should be "0"... plz clarify it, if 'm wrong anywhere... thanks..!!! UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-26T17:18:05Z VoteTAG: 2 CoursewareTAG: Week 3 / Switch Model Exercise CommentableIdTAG: 6002x_switch_model_exercise NumberOfReplyTAG: 2 FirstChildTAG: The question is correct. FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-26T18:44:41Z SecondChildTAG: the answer too. SecondChildUserIdTAG: 365792 SecondChildUserNameTAG: pabhijeet SecondChildCreateTimeTAG: 2012-09-29T05:57:38Z FirstChildTAG: The answer is 1. The best way imo to do this question is too simply break up the circuit into its different gates. X goes into an inverter, the X + Y go into the NOR gate, and that output goes through an inverter. FirstChildUserIdTAG: 355556 FirstChildUserNameTAG: seribus FirstChildCreateTimeTAG: 2012-09-29T19:26:27Z IndexTAG: 1575 TitleTAG: Value of the intesection I don't understand how he find the value of the intersection of the two graphic without any maths... UserIdTAG: 216115 UserNameTAG: Orsi CreateTimeTAG: 2012-09-26T17:07:25Z VoteTAG: 2 CoursewareTAG: Week 3 / Graphical Analysis of Nonlinear Circuits CommentableIdTAG: 6002x_graphical_analysis_of_nonlinear_circuits NumberOfReplyTAG: 2 FirstChildTAG: ANALYTICAL METHOD FirstChildUserIdTAG: 230943 FirstChildUserNameTAG: samsung FirstChildCreateTimeTAG: 2012-09-26T18:20:14Z SecondChildTAG: Just read off the reading in the axes of the graph that correspond to the point of intersection and you have it. if you plot the two curves on the same graph paper or use MSExcel graph to plot the equations on the same graph. SecondChildUserIdTAG: 314386 SecondChildUserNameTAG: jid SecondChildCreateTimeTAG: 2012-09-26T23:02:57Z FirstChildTAG: As samsung says, it's just a case of drawing the two graphs out carefully and measuring the position of the intersection e.g. using a ruler. However, in this case what he actually did (I suspect) is repeat the value he already calculated using the analytical method a couple of videos ago, since both methods have to give the same answer. FirstChildUserIdTAG: 280731 FirstChildUserNameTAG: QuantumCaffeine FirstChildCreateTimeTAG: 2012-09-27T15:38:58Z IndexTAG: 1576 TitleTAG: Assumption that i can't do Why can't i assume that if the resistance of the non-linear element is 1KOhms than the current through it is 1mA and the voltage across is 1V? I tell this looking at the graphic of its v-i relation, but the correct result of the exercise is 0.796mA and 0.796V. I solved the exercise and found the correct value of V and I but i still don't understand why i couldn't do those assumption. UserIdTAG: 216115 UserNameTAG: Orsi CreateTimeTAG: 2012-09-26T16:13:10Z VoteTAG: 2 CoursewareTAG: Week 3 / Analyzing Nonlinear Circuits Exercise CommentableIdTAG: 6002x_analyzing_nonlinear_circuits_exercise NumberOfReplyTAG: 1 FirstChildTAG: For example, take 1K resistor and take it out of circuit. It will have 0 mA and 0V on it, so you can't say that V=1V and I=1mA without a knowledge of remaining circuit. It is true for any element, not only about resistors. But you can replace this element with 1K resistor, that's right! Now you can calculate regular resistor network and get the result. Then you can check: $$ -1 \le V_x \le 1$$ and make sure that your assumption is correct. FirstChildUserIdTAG: 261503 FirstChildUserNameTAG: EugenyL FirstChildCreateTimeTAG: 2012-09-26T16:59:53Z IndexTAG: 1577 TitleTAG: Outside range resistance Notice for every increment in i - voltage is increasing by 1 greater then previous value, so for i=3 v will be one more then the previous segment UserIdTAG: 350697 UserNameTAG: Khalil_Awan CreateTimeTAG: 2012-09-26T08:25:43Z VoteTAG: 2 CoursewareTAG: Week 3 / Analyzing Nonlinear Circuits Exercise CommentableIdTAG: 6002x_analyzing_nonlinear_circuits_exercise NumberOfReplyTAG: 0 IndexTAG: 1578 TitleTAG: H5P2: SOURCE FOLLOWER LARGE SIGNAL I have troubles submitting iDS answer in the H5P2 task. I think that I got it right, as my answers to the several following questions are marked correctly and I got them using this formula. However I cannot get a check mark on the iDS part of the HW. Please, check it. UserIdTAG: 59903 UserNameTAG: Lumag CreateTimeTAG: 2012-09-26T07:31:31Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Troubleshooting NumberOfReplyTAG: 3 FirstChildTAG: Be sure that you have not started your equation with "iDS=". FirstChildUserIdTAG: 304587 FirstChildUserNameTAG: Vasso FirstChildCreateTimeTAG: 2012-09-26T09:34:01Z FirstChildTAG: Solved: Tried multiple times with "Math parsing error". Then it was accepted FirstChildUserIdTAG: 142857 FirstChildUserNameTAG: viking2 FirstChildCreateTimeTAG: 2012-09-26T11:14:41Z FirstChildTAG: It seems I have found something strange. I mistyped K like k (small letter). Usually in such case I would see an error message (like vin instead of VIN, etc.). However in this case just 'k' was silently accepted. FirstChildUserIdTAG: 59903 FirstChildUserNameTAG: Lumag FirstChildCreateTimeTAG: 2012-09-26T10:30:07Z SecondChildTAG: I had this exact same problem - thanks so much for posting this! I suspect this happens because 'k' is accepted as a multiplier; you can type 2k for 2000, etc. Seems like a bug in the equation parser to take it without a preceding numeric value, though. SecondChildUserIdTAG: 209001 SecondChildUserNameTAG: cjj SecondChildCreateTimeTAG: 2012-09-29T23:20:30Z IndexTAG: 1579 TitleTAG: Forum Bug - Re-sizing of the left Pane The left pane acts a little weird sometimes. I'm not sure how to describe the issues so I recorded a short screencast for it. 1. Suppose I click on New Post and then scroll down to read the something. Now suppose I click cancel after that, the left pane remains at the small size that it was re-sized to. This is the first issue shown in the video. 2. Scroll down but not to the bottom of the page. Next scroll to the bottom of the list of threads on the left. Next scroll to the bottom of the page. You see that the scroll bar on the pane jumps up. It has been a bit of hindrance for me. Here is the screencast: http://www.youtube.com/watch?v=z6SdPTHZcN8 Thanks! :-) BTW, why are some topics highlighted with that yellow background now? UserIdTAG: 14386 UserNameTAG: ashwith CreateTimeTAG: 2012-09-26T06:40:18Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 1 FirstChildTAG: The yellow background thing is an unfortunate bug; when you search for something it highlights the text you searched for in yellow, but the markup for highlighting is inserted into the body of the post, so if you edit the post and save it makes the highlight permanent. Thanks for the feedback! FirstChildUserIdTAG: 323387 FirstChildUserNameTAG: ibrahimawwal FirstChildCreateTimeTAG: 2012-09-26T19:38:54Z IndexTAG: 1580 TitleTAG: Rth would we give external excitation "i"? when we measure Rth and when we turn off all the internal voltages and currents of network N. UserIdTAG: 40209 UserNameTAG: sadaf78 CreateTimeTAG: 2012-09-25T23:43:57Z VoteTAG: 2 CoursewareTAG: Week 2 / Thevenin method CommentableIdTAG: 6002x_thevenin NumberOfReplyTAG: 0 IndexTAG: 1581 TitleTAG: Very nice Very nice lab I really enjoyed . UserIdTAG: 40209 UserNameTAG: sadaf78 CreateTimeTAG: 2012-09-25T22:42:02Z VoteTAG: 2 CoursewareTAG: Week 2 / Superposition experiment CommentableIdTAG: 6002x_Superposition_experiment NumberOfReplyTAG: 0 IndexTAG: 1582 TitleTAG: lab 3 How can i connect two mosfet in parallell, I mean the wires are getting crossed when i take from A and B and feed it to the gates of the above said mosfets?? Please help!! UserIdTAG: 148523 UserNameTAG: span993 CreateTimeTAG: 2012-09-25T20:37:15Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Try to position them like this: ![MOSFETS in Parallel][1] [1]: https://edxuploads.s3.amazonaws.com/13486062881343609.png FirstChildUserIdTAG: 40235 FirstChildUserNameTAG: mrabe FirstChildCreateTimeTAG: 2012-09-25T20:52:24Z SecondChildTAG: Thanks! Now I got the same o/p plot(including the spikes) but it shows wrong why? SecondChildUserIdTAG: 148523 SecondChildUserNameTAG: span993 SecondChildCreateTimeTAG: 2012-09-25T21:10:17Z SecondChildTAG: Check that the z voltage is low enough. (move cursor it shows in upper left of plot) SecondChildUserIdTAG: 440714 SecondChildUserNameTAG: mcktim SecondChildCreateTimeTAG: 2012-09-25T23:57:45Z SecondChildTAG: Hello , I an having the same problem. I have the same plot. and my voltage at z is 0.20 volts but the checker considers it wrong . why SecondChildUserIdTAG: 423976 SecondChildUserNameTAG: Sirbrevis SecondChildCreateTimeTAG: 2012-09-26T08:44:13Z SecondChildTAG: how many the pullup resistor used please ? SecondChildUserIdTAG: 331441 SecondChildUserNameTAG: abdessamade SecondChildCreateTimeTAG: 2012-09-26T18:22:52Z FirstChildTAG: @span993 can you please share with me your email address so that i can talk to you regarding the problem im facing in solving. FirstChildUserIdTAG: 223378 FirstChildUserNameTAG: hadi89 FirstChildCreateTimeTAG: 2012-09-30T20:50:01Z IndexTAG: 1583 TitleTAG: Show me how to use The Hint to solve, not the graphical technique Hint: use guess and check to solve for the operating point. Assuming this means the trial and error method from S6V5. vx = 32.8 - 8.2vx^3 Try 1 on the right, get 24.6 on the left. Then try 24.6 on the left, get -1 for vx on the right. etc What am I missing, how is this helpful? UserIdTAG: 437586 UserNameTAG: sl1ck CreateTimeTAG: 2012-09-25T16:12:53Z VoteTAG: 2 CoursewareTAG: Week 3 / Graphical Analysis of Nonlinear Circuits Exercise CommentableIdTAG: 6002x_graphical_analysis_of_nonlinear_circuits_exercise NumberOfReplyTAG: 2 FirstChildTAG: just google it (example only): [y=x^2 and y=12-2*x][1] [1]: http://www.google.com/search?hl=en&q=y=x%5E2%20and%20y=12-2*x&oq=y=x%5E2%20and%20y=12-2*x&gs_l=serp.3...14217.16688.0.18065.3.3.0.0.0.0.670.852.1j1j5-1.3.0.les;..0.0...1c.1.OmyWXwPGLP8 FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-09-25T19:51:14Z SecondChildTAG: What **YakovO** means is that Google has a built-in graphing calculator if you type the equations into the "search" box in the format he/she gave. I did not know this! I knew there was a basic calculator and a conversion calculator (i.e. imperial gallons to millileters) in Google; but now we have a free, online graphing solution :) SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-09-26T04:49:40Z SecondChildTAG: Very nice info. Thanks YakovO! :) SecondChildUserIdTAG: 352373 SecondChildUserNameTAG: keyholder SecondChildCreateTimeTAG: 2012-09-26T14:01:30Z SecondChildTAG: Hello , friends . Can someone point me in the right direction of how to do : What is the operating voltage (in Volts) of the device in this circuit? I don't know is correct or not : -I+v/R+i = 0 , i=v^3.. I=4.. R=8.2, ---------- v^3 + v/8.2 -4 =0 , v~ 1.55 .. , i=I-v/8.2 ...i~3.81 SecondChildUserIdTAG: 194717 SecondChildUserNameTAG: kaa SecondChildCreateTimeTAG: 2012-09-26T15:26:07Z SecondChildTAG: A simple way to do this is to use MS Excel. 1. Using the node method, we see that I=i+v/R so v=-R*i+I*R substituting values: v=-8.2*i+32.8 we have been given that i=v^3 so: v=-8.2*v^3+32.8 USING EXCEL in cell A1 put any arbitrary (guess number) an cell A2, type =-8.2*A1^3+32.8 in cell A3, type =A1-A2 keeping cell A3 selected, go to the "Data" menu, under "Data tools" toolbar, select "what-if analysis", "goal seek", fill in the dialog with: set cell A3 to 0.00001 (just a very small number~0); by changing cell A1 whatever you get in A1 or A2, which should be the same is the correct answer. I like easy stuff. ------- Graphical method: Still using MS Excel: plot a graph of i=v^3 (simply by populating a vertical range of , say 40 cells with numbers 0 to 39 in steps of 1, 1. in cell A5, type 1 2. In cell B5, type =A4^3 3, in cell C5, type =4-(A5/8.2) This is from v=-Ri+IR. this is equivalent to i=I-(v/R) select all cells A5..C5, Drag the cell handle down all the way down 40 cells for auto-fill. From the "Insert" menu, select line graph Add two series V3 and the load line by selecting the appropriate ranges (Columns B and C in this case) then the Horizontal axis should be column A. Experiment by changing the values on the horizontal axis. I got a good result by autofilling with 1 to 2 in steps of 0.025 ie 1.0, 1.025, 1.05... SecondChildUserIdTAG: 314386 SecondChildUserNameTAG: jid SecondChildCreateTimeTAG: 2012-09-27T10:01:07Z SecondChildTAG: Someone can help me to understand how to use the try error method? in the examples the professor always gave the value for B, but now I got no clue about which value to put in the equation. Thanks for helping. SecondChildUserIdTAG: 316902 SecondChildUserNameTAG: JesseTeixeira SecondChildCreateTimeTAG: 2012-09-30T16:26:48Z FirstChildTAG: Besides the methods you have shown here, you can do: **Id = IWillNotPutTheAnswerHere (Equation 1)** **Id = Vd³ (Equation 2)** **Vd³ = IWillNotPutTheAnswerHere Vd³ - IWillNotPutTheAnswerHere = 0** And solve a wonderfull cubic equation =D FirstChildUserIdTAG: 310147 FirstChildUserNameTAG: ildomarcarvalho FirstChildCreateTimeTAG: 2012-09-28T02:39:12Z SecondChildTAG: he he! SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-12-19T16:19:22Z IndexTAG: 1584 TitleTAG: What is the current (in Amperes) through the strange object when it is in operation in the circuit? here if we assume the resistance of that element to be 1 kohm, we r getting the correct answer.... why shud i not consider the resistance of that element as 2kohm.... if i take it as 1kohm, the voltage across it falls inside the range 0 to 1 volt.... if i take it as 2kohm, the voltage across it falls inside the range 1 to 2 volt..... so why r we restricting to 1 kohm.... ??? anyone please explain... UserIdTAG: 316353 UserNameTAG: Vivekanand123 CreateTimeTAG: 2012-09-25T13:47:20Z VoteTAG: 2 CoursewareTAG: Week 3 / Analyzing Nonlinear Circuits Exercise CommentableIdTAG: 6002x_analyzing_nonlinear_circuits_exercise NumberOfReplyTAG: 2 FirstChildTAG: You have the behaviour in the graph.This is piecewise like. FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-09-25T16:11:22Z SecondChildTAG: because if we consider in both cases that resistance is 2kohm or 1 kohm the voltage across the strange object will be less than 3 volts which means that resistance will be 1KOhm SecondChildUserIdTAG: 140867 SecondChildUserNameTAG: DarkWishMaster SecondChildCreateTimeTAG: 2012-09-27T19:45:10Z FirstChildTAG: Because otherwise the device's law, or in other words, the v-i characteristic for the device doesn't hold: if you assume $R_D=2k$ you will end up with voltage greater than 1V and current less than 1mA which is not true for the device. FirstChildUserIdTAG: 252722 FirstChildUserNameTAG: vaboro FirstChildCreateTimeTAG: 2012-09-25T23:02:21Z IndexTAG: 1585 TitleTAG: Something that might help I would suggest using wolfram alpha to help find the slope of the load line and the intersection of the two curves. UserIdTAG: 302713 UserNameTAG: EVega CreateTimeTAG: 2012-09-25T12:56:40Z VoteTAG: 2 CoursewareTAG: Week 3 / Graphical Analysis of Nonlinear Circuits Exercise CommentableIdTAG: 6002x_graphical_analysis_of_nonlinear_circuits_exercise NumberOfReplyTAG: 1 FirstChildTAG: I use it for the intersection too! But can it be solved without computers? FirstChildUserIdTAG: 98259 FirstChildUserNameTAG: rogerloh0 FirstChildCreateTimeTAG: 2012-09-26T00:49:17Z IndexTAG: 1586 TitleTAG: Difficulty seeing where you are writing on the screen. In the future, is it possible to use a larger pointer when writing on the screen? I believe in other videos I have seen your cursor highlighted with a transparent yellow circle around it. That would be helpful here. - Thanks. UserIdTAG: 132828 UserNameTAG: rharris CreateTimeTAG: 2012-09-25T06:59:01Z VoteTAG: 2 CoursewareTAG: Week 3 / Incremental Method Motivation CommentableIdTAG: 6002x_incremental_method_motivation NumberOfReplyTAG: 0 IndexTAG: 1587 TitleTAG: Discussion account/profile page link by menu or similar function If I wish to see a listing of all topics that I have posted to (contrasting with [following a post][1], due to various reasons [possibly] not following every topic that I post to, or accidentally un-following a topic that I've posted to), I must search for and find a topic that I've already posted to, and click on my username in that list. Once there, I can find a list of all topics that I've posted to. My suggestion is that edX provide some method of navigating to my Discussion Profile page by an actual menu while within any discussion page, regardless of whether or not I've posted to that topic previously. (A working workaround: Once you've found your username by searching as I've outlined above, you can right-click your username, and select "Bookmark this link" in the menu that appears. As I use Firefox as my browser, this may differ from the procedure for your browser.) [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_Feedback/threads/5060f7fc4f44872300000052 UserIdTAG: 248807 UserNameTAG: 1kingsman CreateTimeTAG: 2012-09-25T01:40:17Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 1 FirstChildTAG: Below the "All" dropdown is the "Following" option that lets you see posts that you're following. But yes, that button should be placed somewhere more visible. FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-25T03:42:59Z SecondChildTAG: yes, I found that Following option. However, I was looking for a way to access only the topic that I've posted in. SecondChildUserIdTAG: 248807 SecondChildUserNameTAG: 1kingsman SecondChildCreateTimeTAG: 2012-09-26T23:54:00Z IndexTAG: 1588 TitleTAG: LAB 4 (wrong result) Hi, I have got a wrong result from my circuits in sandbox as the attached snapshot. Any idea where have I gone wrong? ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13485197261343655.jpg UserIdTAG: 210954 UserNameTAG: Shahrouz CreateTimeTAG: 2012-09-24T20:48:50Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Hi, Found the cause but will leave the post for the possible mistake from others :) if you end up having such a funny result then you need to check your Vgs and Vds. one must be DC as mentioned in the lab explanation. FirstChildUserIdTAG: 210954 FirstChildUserNameTAG: Shahrouz FirstChildCreateTimeTAG: 2012-09-24T21:25:51Z SecondChildTAG: Hi! That is good thinking to help others :) SecondChildUserIdTAG: 304587 SecondChildUserNameTAG: Vasso SecondChildCreateTimeTAG: 2012-09-24T21:32:05Z FirstChildTAG: Are you sure you got the right polarities? What about the direction of the current? FirstChildUserIdTAG: 304587 FirstChildUserNameTAG: Vasso FirstChildCreateTimeTAG: 2012-09-24T21:26:22Z FirstChildTAG: hi, please may somebody tell me how to modify x axis to show voltage values? I got time on it and no idea how to change it, many thanks FirstChildUserIdTAG: 298612 FirstChildUserNameTAG: fabianh FirstChildCreateTimeTAG: 2012-09-25T01:48:22Z SecondChildTAG: actually it is explained in the description of the problem, you just have to change the color property of the probes for the x-axis. "The voltage probe on the vtest node has its "Plot color" set to "x-axis", which asks the tool to use the values sampled by the probe as the x-coordinate when plotting. So the horizontal axis of the plot showing the transient results will be the voltage vtest instead of time." SecondChildUserIdTAG: 370343 SecondChildUserNameTAG: JJarquin SecondChildCreateTimeTAG: 2012-09-25T05:36:06Z IndexTAG: 1589 TitleTAG: Solution to Question 2 Use this slope calculator http://www.basic-mathematics.com/slope-calculator.html and plot the coordinates of the beginning of the slope (1,1) and the end of the slope (2,3) and you will get your answer. UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-09-24T16:44:04Z VoteTAG: 2 CoursewareTAG: Week 3 / Analyzing Nonlinear Circuits Exercise CommentableIdTAG: 6002x_analyzing_nonlinear_circuits_exercise NumberOfReplyTAG: 0 IndexTAG: 1590 TitleTAG: Lab 2 Can we get a detailed explanation of how to work Lab 2? - I spent like 4-5 hours trying to figure it out and I got it only because of the hints that peers and TAs posted. - I don't fully understand the problem. ---Like why do we normalize and use 1ohm as default for one of the resistors. ---How did we know we need three resistors? Can we get a video on creating your own circuit? - None of the videos show how to design a circuit, they only show how to analyze a circuit. Also, on the tutorials we were explained what is linear and what is not linear. -- They showed that a resistor + a voltage source as a system is not linear. --- Why can we use superposition on systems like that? Thanks!! UserIdTAG: 230787 UserNameTAG: Gabriel007 CreateTimeTAG: 2012-09-24T15:23:06Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Yes, for me too. I spent more than 10 hours to figure it out. But, I wouldn't test it out in the sandbox. I want to understand the problem. Also, I like it to solve by equations, not by testing. Its strange, if I fix R1 with 1 and check R2 and R3 till I done. For me, that isn't science. FirstChildUserIdTAG: 394042 FirstChildUserNameTAG: Gurgussum FirstChildCreateTimeTAG: 2012-09-24T16:32:55Z SecondChildTAG: I agree! SecondChildUserIdTAG: 230787 SecondChildUserNameTAG: Gabriel007 SecondChildCreateTimeTAG: 2012-09-24T18:22:30Z SecondChildTAG: Take a look at this [thread][1] where I gave all the math [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/505f6f684baa441f0000003c SecondChildUserIdTAG: 190618 SecondChildUserNameTAG: Kirbabaev SecondChildCreateTimeTAG: 2012-09-24T19:34:42Z FirstChildTAG: About the second question about linear system, the composite voltage source is a non linear element, but it is a linear circuit. And as far a I know, we can apply superposition on linear circuit. For more explaination, you can visit: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_2/Week_2_Tutorials/ ,in the video current source answer section, scroll down to see more detail on linear element and linear circuit. I hope my answer helps you out. FirstChildUserIdTAG: 146814 FirstChildUserNameTAG: humink FirstChildCreateTimeTAG: 2012-09-24T16:25:34Z SecondChildTAG: Got it! Tank you humink! SecondChildUserIdTAG: 230787 SecondChildUserNameTAG: Gabriel007 SecondChildCreateTimeTAG: 2012-09-24T18:26:10Z IndexTAG: 1591 TitleTAG: H3P4 when do diodes flip? How to determine the exact voltage(s) Vs when diodes change their state? Mostly people erroneously (just check via simulator) assume that if Vs is positive first diode is on and second is off and vice versa if Vs is negative. It's just pure luck that answers derived from this assumption are correct. Nevertheless, what is the correct approach? UserIdTAG: 189680 UserNameTAG: vnd CreateTimeTAG: 2012-09-24T15:19:43Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: That was a bit confusing for me also.I actually made the circuit in sandbox to convince myself.The diodes are ideal, what means that when vD > 0, their resistance is 0, which is like a straight line.A diode is direct connected, when the voltage across them is positive.The diode symbol is an arrow that shows the direction of the current when is conducting.If the diode has a negative vD, which means that the diode is invert connected a, it's not conducting.A diode conducts current only in the arrow's direction.That is the way i remember it anyway. Hope i could be some help, since i explain in a terrible way. FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-09-24T18:21:28Z SecondChildTAG: Thank you, but unfortunately it doesn't explain why both diodes are blocked at +1V voltage (for example). Intuitively it's clear that you have to build up some voltage first to "break through" but what is the exact value is unclear for me. SecondChildUserIdTAG: 189680 SecondChildUserNameTAG: vnd SecondChildCreateTimeTAG: 2012-09-24T18:37:24Z SecondChildTAG: I don't know how much i am supposed to reveal, but think simple.An ideal diode is like a switch.If the voltage across it is positive the switch is closed and if it's negative, the switch is opened.It does not contain anything else.The triangle wave is positive and negative, so there are positive and negative cases.Just draw the switches in both cases. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-09-24T20:13:25Z SecondChildTAG: I understand what you mean (if not let's just wait till the HW deadline), but the problem is, there are other voltage sources that affect diode behavior too. It's not as simple as: Vs is positive voltage => diode opens, Vs is negative => diode closes. For example, 1V is not enough to open the first diode. SecondChildUserIdTAG: 189680 SecondChildUserNameTAG: vnd SecondChildCreateTimeTAG: 2012-09-25T06:41:28Z SecondChildTAG: Yes, it's just as simple.The diodes are IDEAL.Just search for ideal diodes, and if you don't believe me, just make the circuit in sandbox.There is an option to make the diode ideal in sandbox. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-09-25T10:47:02Z SecondChildTAG: See here: http://imgur.com/Hjxg2 If what you say was true, D1 would be open but it's closed. SecondChildUserIdTAG: 189680 SecondChildUserNameTAG: vnd SecondChildCreateTimeTAG: 2012-09-25T11:32:45Z SecondChildTAG: You must see that both diodes are blocked. The potential at the D1 anode is 1 v, but at it's cathode is 2V.In which way you suppose that the current will flow ? I made the source of having 2V, to see clearer, then of 2.01 and the diode is opened. http://postimage.org/image/q3xf5xzvz/ 10fA is a simulated residual current because in their application , you can't have a 0 resistor to avoid division by 0. You must keep it very simple, and keep in mind that sandbox is an application , and ideal conditions are hard to simulate sometimes due to numeric problems.Look at the curents ..KA . Use a sinusoidal current source for the problem, not a dc source. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-09-25T14:51:19Z SecondChildTAG: What you say doesn't make sense and at some point I even thought that you are trolling (sorry and thank you for your effort if not). This is summary of what you wrote: 1) trivial facts about ideal diodes, doesn't answer the original question 2) "The triangle wave is positive and negative, so there are positive and negative cases" which is just **wrong** 3) suggest to test it in a sandbox 4) suddenly you do admit that both diode can be blocked 5) some non-relevant comments about implementation of ideal diode in the sandbox. What I wanted to know can be rephrased as: we have three states for diodes (off/off, on/off, off/on), when do they transition into each other *exactly*? Or to put it in yet another way, I needed a method to find out V1 and V2 such that: when Vs crosses V1 state changes from on/off to off/off, when Vs crosses V2 state changes from off/off to off/on). **jelizon** answers this below perfectly. SecondChildUserIdTAG: 189680 SecondChildUserNameTAG: vnd SecondChildCreateTimeTAG: 2012-09-25T15:50:53Z SecondChildTAG: Hey man, i wanted to help. Use sandbox to see what's happening but do not expect very accurate currents when R ==0 . There is a case when Vs is negative, and one when is positive.Calculate for each case using superposition. Look how much it took to convince you they are switches. :) SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-09-25T16:28:01Z SecondChildTAG: You just turn around the tail, and i realize that you think in term of Vs , V1, and V2, when the graph is ID / VD .What you don't seem to get it is that when VD is positive the switch is opened, and in rest is closed. Calculate Vd in each case to convince you, use sandbox etc. And after i try to explain and i waste my time, you come to say i am trolling. I think i will just mind my business from now on. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-09-25T16:49:13Z SecondChildTAG: Oh these ungrateful people.. Thank you, it's uncommon to see such an effort and patience, sadly we misunderstood each other. SecondChildUserIdTAG: 189680 SecondChildUserNameTAG: vnd SecondChildCreateTimeTAG: 2012-09-25T17:13:03Z SecondChildTAG: Sadly, i'm not a very bright bulb at explanations ... SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-09-25T18:45:05Z SecondChildTAG: I think i know what is confusing for you.There is another thread where somebody told to use 2 cases for Vs : positive and negative.I agree, that doesn't seem to be right and that's why you posted that image. From my side ,i assumed that you thought that ideal diodes have a resistance.(Like i said,i am not the brightest bulb). I was speaking all the time about VD1 and VD2, as i treat them separately, positive and negative, and finally i added the results through superposition. A funny thing is that , if you treat the 2 cases for the supply voltage source, you get the same results, and here is where i don't get it. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-09-25T19:02:44Z FirstChildTAG: In fact, the diode D1 won't turn on immediately when Vs becomes positive. For the diode D1 to be on the voltage at the top node should be larger than V1. The most common way to solve these kind of problems is by the method of assumed states: 1. For a particular point in time, assume the diodes to be on or off. 2. Replace assumed off diodes with open circuits, and assumed on diodes with short circuits 3. Find the voltage in off diodes and the currents though on diodes. 4. If the conditions for an ideal diode are not met, then change your assumptions and repeat from step 2. 5. If the conditions are met, then calculate the required currents and voltages in your circuit, and move on to the moment in time you want to analyze next (typically, to the moment when the conditions of the ideal diode won't be satisfied using the same assumptions). This may seem like a tedious method but it is safe. Practice will allow you to quickly visualize which diodes are on and off at a particular time in simple circuits. In the particular case of the homework problem, let's assume that the diode D1 turns on at the time when Vs is positive but less than 2V, let say at the time when Vs is 1V. For completeness, let's assume D2 is off. If we replace D1 with a short circuit and D2 with an open circuit, we will see that current will flow out of positive terminal of the V1 source and hence $i_D$ is negative. However, that violates the ideal diode conditions and therefore our assumptions are incorrect!! As a final note, if you have more than 1 diode in your circuit, finding that the ideal diode conditions are not met for one diode doesn't necessarily mean that the assumed state of that particular diode is wrong. It means that the combined assumptions are wrong (i.e. the problem might be in the assumption of another diode). Hope this helps! FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-25T13:46:56Z SecondChildTAG: The only real problem is to determine if they are positive or negative polarized. Why are you trying to complicate things, i don't get it. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-09-25T15:01:37Z SecondChildTAG: PS: There are bigger fishes to fry.. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-09-25T15:01:59Z SecondChildTAG: Thank you, this is just what I needed. SecondChildUserIdTAG: 189680 SecondChildUserNameTAG: vnd SecondChildCreateTimeTAG: 2012-09-25T15:52:24Z SecondChildTAG: Glad to help. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-09-25T16:12:15Z SecondChildTAG: AlexAlexandrescu, we are basically saying the same thing. The method I showed above is useful if you don't have much experience solving circuit with diodes. If you can see easily when are the diodes open or close, then you may skip a few steps! However, when you analyze complicated circuits, you won't be able to see all the diode configurations without making some calculations. SecondChildUserIdTAG: 381746 SecondChildUserNameTAG: jelizon SecondChildCreateTimeTAG: 2012-09-26T13:44:34Z SecondChildTAG: Yes i realized it after, but i already embarrassed myself enough...I will keep this in mind as it will prove useful at some point. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-09-26T20:16:54Z IndexTAG: 1592 TitleTAG: Is there a way to follow my own posts? Is there a way to follow, track, or find my own posts? The posts I was following used to show up to the left, but I'm not seeing that anymore. Surely there is a way to keep track of answers to my post other than remember the exact subcategory I put them under and searching through that (sometimes lengthy) subcategory? Thanks, Dave UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-24T14:56:05Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: In the dropdown menu below "All" is a section called "Following" :) FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-24T15:23:09Z SecondChildTAG: Thank you kimt! Now I can see my following Posts ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-24T15:31:34Z IndexTAG: 1593 TitleTAG: H4P3/S8E2 method for calculating Thevenin Resistance For both H4P3 and S8E2 I calculated the Thevenin Resistance $R_{TH}$ of a linear circuit with dependent sources By first calculating the Thevenin Voltage $V_{TH}$, followed by calculating the Norton Current $I_N$ and then using the equation $R_{TH}=\frac{V_{TH}}{I_N}$ to find $R_{TH}$. This method is rather unwieldy and I would be much obliged if someone could explain to me a better way of calculating $R_{TH}$. Thanks in advance. UserIdTAG: 85490 UserNameTAG: yashsavani CreateTimeTAG: 2012-09-24T07:57:08Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: Hope I'm not giving off answers here. As far as I know, there isn't any other way to solve it. There are three kinds of circuits here based on the sources they contain. 1. **Independent Sources Only:** The is the easiest type. You can calculate either the open circuit voltage or the short circuit current. Next you need to shut off all sources and find the Thevenin (or Norton) resistance. 2. **Independent + Dependent Sources:** In this case, you cannot calculate the Thevenin resistance directly because you cannot shut off dependent sources. So you calculate the open circuit voltage, the short circuit current and then use Ohm's law. 3. **Dependent Sources Only:** Like the previous case, you can't shut off the dependent sources. So you add a test source at the terminals and find the current and voltage at the terminals. Ohm's law will give you the equivalent resistance. As S8E2 shows, the third method works for category 2 as well. FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-09-24T10:57:53Z SecondChildTAG: In S8E2 they have provided us with a hint to add another independent current source, perhaps that is another method of finding $R_{TH}$. If you know anything about that method please let me know. P.S. I don't think there is any risk of giving away the answer. You have just provided a method of doing the problem and not given the answer itself. SecondChildUserIdTAG: 85490 SecondChildUserNameTAG: yashsavani SecondChildCreateTimeTAG: 2012-09-24T11:10:33Z SecondChildTAG: You nail it prety well from my part.Nice ! SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-09-25T16:52:16Z FirstChildTAG: It's all got to do with the equation $ V_{terminal} = V_{TH}+I_{excitation} \cdot R_{TH} $ and what happens when you do different things to the circuit. If all you do is leave the terminal as an open circuit, then $I_{excitation}=0$ and $ V_{terminal}=V_{TH} $. On the other hand, if all you do is short-circuit the terminal, then $V_{terminal}=0$ and $V_{TH}=I_{excitation} \cdot R_{TH}$ where $I_{excitation} = I_{SC}$. I know it should be 'negative', but that's just a matter of which direction you take to be the positive reference direction. What effect does zero-ing all the independent sources have on this equation? As far as I have been able to determine that sets $V_{TH}=0$. Which is great for determining the resistance in a purely resistive system, but less elucidating when there are dependent sources in the mix. Adding an $I_{excitation}$ seems to solve this problem nicely, because then you have $V_{terminal}=I_{excitation} \cdot R_{TH}$ and if you use $I_{excitation}=1A$, $R_{TH}=V_{terminal}$ (I'm omitting the unit conversion for the sake of the simplicity of the equation). FirstChildUserIdTAG: 248902 FirstChildUserNameTAG: Chanute FirstChildCreateTimeTAG: 2012-09-24T11:19:52Z SecondChildTAG: Where did you get your Iexcitation=1A?, please clarify, Thanks! SecondChildUserIdTAG: 349139 SecondChildUserNameTAG: 1977ROYELMER SecondChildCreateTimeTAG: 2012-10-06T07:28:25Z SecondChildTAG: @1977ROYELMER, it's from the video on Thevenin Voltage. SecondChildUserIdTAG: 248807 SecondChildUserNameTAG: 1kingsman SecondChildCreateTimeTAG: 2012-10-06T21:22:20Z FirstChildTAG: I also had trouble with H4P3 and compared it to S8E2. What I found was, in the problem statement it says Z = 2 ohms for the function for the dependent source, i.e., Z*i. So, keep that in mind when calculating Rth, hint, hint. In S8E2, alpha was so much smaller than R1 and R2, considering it in calculating Rth didn't matter, with our without the answer showed correct. FirstChildUserIdTAG: 132828 FirstChildUserNameTAG: rharris FirstChildCreateTimeTAG: 2012-10-06T19:30:27Z FirstChildTAG: Take a look at example 3.23 of the textbook where they calculate a Thevenin resistance of a circuit containing a dependent source, using a simpler method (in my opinion) FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-24T13:31:55Z SecondChildTAG: Hi Jelizon. I feel example 3.23 is different. In that example, the control voltage is isolated from any external circuit that is connected to the network terminals. So if we shut off the independent source, the control voltage will remain $0V$ no matter what we connect at the terminals. However in S8E2 and H4P3, an external voltage or current will change the control current. Therefore we cannot shut off the independent source and then shut off the dependent source as an external circuit can activate that dependent source even though the independent source is off. SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-09-24T15:07:57Z SecondChildTAG: I had the same issue with trying to use that example as ashwith. It seems like the only example they give in the textbook of finding the Thevenin/Norton equivalent when a dependent source is involved is this exceptional case where the dependent source actually works out to having a zero value... Scouring the internet for an example didn't help, and I thought it'd be very useful to have one! So I've tried to work one out here: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5060f4a8a3f6d21f00000050 SecondChildUserIdTAG: 248902 SecondChildUserNameTAG: Chanute SecondChildCreateTimeTAG: 2012-09-25T00:04:17Z IndexTAG: 1594 TitleTAG: about studying online Is there more sites like EDX to studying online and get certifications for free, any sites? UserIdTAG: 395257 UserNameTAG: blackguitar CreateTimeTAG: 2012-09-24T04:22:50Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: yes. i know one> https://www.coursera.org/ FirstChildUserIdTAG: 369431 FirstChildUserNameTAG: pkaistha FirstChildCreateTimeTAG: 2012-09-24T04:50:48Z SecondChildTAG: thanks a lot is there more guys? SecondChildUserIdTAG: 395257 SecondChildUserNameTAG: blackguitar SecondChildCreateTimeTAG: 2012-09-24T04:52:56Z SecondChildTAG: **Thanks as well!** We did not have edX or Coursera when I went to university. I just looked at Coursera's offerings and they have a lot more universities and as a result, a lot more classes listed than edX. Note that some universities on Coursera have a policy of **not** offering "certificates of completion", while others **do**. I also read in the *New York Times* that Coursera is a **for-profit** entity, which means they may start charging money soon, after they "try" the courses out first, similar to beta software. I think edX is still non-profit. A lot of their courses start in October; I may take **one** extra class at Coursera; as I am already signed up for MIT's Physical Chemistry 3.091x, and we all know 6.002x takes a lot of time :) Once these online classes start offering proctored exams and credit for a fee, they will overtake regular university classes, some of which are boring, and for some people, like older students and those that work during the day, it will be a lot easier to get a degree. This is only the beginning. Certain things, though, like Chemistry Lab, and Electrical Circuits Lab, have to be done the traditional way in a university lab setting, though. Most cannot afford to buy a full set of chemicals and glassware to do Chemistry Lab at home, and likewise can neither buy an oscilloscope, frequency generator, power supplies, multimeters, etc. (even though resistors and a breadboard are cheap) just to do Electrical Circuits Lab at home. And do you trust the kids, friends, pets, etc. in your household messing with or touching your dangerous chemicals or your 120/240V AC mains project? SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-09-24T06:24:30Z FirstChildTAG: Thanks for that info, very valuable . FirstChildUserIdTAG: 428554 FirstChildUserNameTAG: swimming3744 FirstChildCreateTimeTAG: 2012-09-24T17:10:39Z IndexTAG: 1595 TitleTAG: related to lab2 post Staff action it was my first post ,honestly i had not read the honor code before my post for lab 2 , but after reading it i accept my mistake and i do apologize for mistakenly posting answer without hints . UserIdTAG: 13188 UserNameTAG: rakeshpagi CreateTimeTAG: 2012-09-24T03:34:19Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: do you have any hints FirstChildUserIdTAG: 233700 FirstChildUserNameTAG: ANDREW77 FirstChildCreateTimeTAG: 2012-09-24T04:49:04Z SecondChildTAG: i m worried about my progress being erased or not .. really sorry could'nt post, i fear of violating honor code now ... i learnt a lesson .. SecondChildUserIdTAG: 13188 SecondChildUserNameTAG: rakeshpagi SecondChildCreateTimeTAG: 2012-09-24T06:02:00Z FirstChildTAG: Hi rakeshpagi, I really apreciate your apologies with your classmates, that says a lot about you :). I hope that they forgive you and give you another chance considering that you haven't read the Honor Code [(read here)][1] before your Post... Remember that you can not post precise answers before the deadline of the Homeworks and Labs. Myriam. [1]: https://www.edx.org/honor FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-24T06:19:49Z SecondChildTAG: thankx Myriam for your support SecondChildUserIdTAG: 13188 SecondChildUserNameTAG: rakeshpagi SecondChildCreateTimeTAG: 2012-09-24T08:17:54Z IndexTAG: 1596 TitleTAG: Lab2 Can someone help me? I'm kinda stuck here in this exercise, I'm not getting what exactly I'm supposed to do. UserIdTAG: 145108 UserNameTAG: Wlozano CreateTimeTAG: 2012-09-24T01:57:19Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: The key is to add some resistor to the example that is given, and after you must calculate the 2 voltage divisors. (sorry for my English) FirstChildUserIdTAG: 221050 FirstChildUserNameTAG: jucapini FirstChildCreateTimeTAG: 2012-09-24T02:24:55Z SecondChildTAG: Please don't apologize for your language (unless you're swearing ;-)). I believe everyone should recognize this is a multilingual forum, and frankly, your English is at least as good as mine, given the above example. :-) SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-09-24T03:24:44Z FirstChildTAG: refer the text book under network theorems..you will find that useful to better utilise voltage divider principles to this FirstChildUserIdTAG: 443809 FirstChildUserNameTAG: Duvindu FirstChildCreateTimeTAG: 2012-09-24T04:12:00Z FirstChildTAG: Hi Wlozano! You can see an explanation in this Post [Here][1]. :). [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_S3E2_Linear_Combinations_of_Source_Strengths/threads/506a280d036af91f000000c9 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-02T00:03:15Z IndexTAG: 1597 TitleTAG: Is the lab 2 deadline affected by your time zone? After an all-nighter, I just completed Lab 2 (a lot simpler than I envisaged, just took a while to get my head around it). Will it be accepted or is it too late? It's around 2am GMT, so 2 hours into the 24th September, and the deadline was the 23rd September, but I'm not sure if the deadline changes depending on your timezone, or if there is one objective deadline for everyone, regardless of timezone. Any clarification? Thanks :) UserIdTAG: 393120 UserNameTAG: irnbro CreateTimeTAG: 2012-09-24T01:01:41Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: You might check your progress in: https://www.edx.org/courses/MITx/6.002x/2012_Fall/progress to see if you have the points. FirstChildUserIdTAG: 276409 FirstChildUserNameTAG: IgnacioUY FirstChildCreateTimeTAG: 2012-09-24T01:11:15Z FirstChildTAG: Submit your solution and see if it's accepted. FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-09-24T02:09:05Z IndexTAG: 1598 TitleTAG: How was chainsaw affecting the wire? How was electrical noise from the chainsaw getting to the gate circuit, which I assume was over on the table to the left? Are there EM waves going through the air? I didn't catch that part, if it was explained? UserIdTAG: 468623 UserNameTAG: RobNik CreateTimeTAG: 2012-09-23T23:57:47Z VoteTAG: 2 CoursewareTAG: Week 2 / Chain saw CommentableIdTAG: 6002x_chain_saw NumberOfReplyTAG: 2 FirstChildTAG: I don't think it was explained in the demo, but I believe it's the EM waves generated by the motor, which powers the chain, which is causing the noise on the circuit. I think in general when motors run, the sparks then cause internally when they spin, "*pollutes*" (i.e electromagnetically speaking), the surroundings, which is picked up by nearby circuits. FirstChildUserIdTAG: 78 FirstChildUserNameTAG: Viraj FirstChildCreateTimeTAG: 2012-09-24T00:13:48Z SecondChildTAG: Viraj is absolutely right. It has to do with the methods we use to convert electrical power into motion. Notice that the chainsaw he's using is powered by an electric motor and not a gas motor. There's two things that could be happening in the video: 1. The electromagnets that are actuating the motor are also inducing some random currents in the wire, or, 2. The chain saw and the circuit are connected to the same power source, and the power draw spikes from the motor as the brushes move over the electromagnets causes the circuit's power sources to fluctuate voltages. I think 2. is more likely based on where he's standing in relation to the circuit bench and the non-reaction of the other input and the output. That's the main reason for always having your motor power sources electrically isolated from your logic power sources. Electric motors induce so much electrical noise! SecondChildUserIdTAG: 248902 SecondChildUserNameTAG: Chanute SecondChildCreateTimeTAG: 2012-09-24T02:18:36Z SecondChildTAG: We "old people" remember when turning a blender or vacuum cleaner on in the next room would cause "static" on our television screens. ;-) SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-09-24T03:34:17Z SecondChildTAG: Also notice that they likely took the cutting chain off of the chainsaw for **safety**. I thought the Prof. was really going to cut the laptop on the desk in half! But then I realized that the liability of having a chainsaw with a functional cutting chain on it in class would be too great. Simply put, the Prof. was running a large AC motor, that by **induction**, created electromotive interference (EMI), or noise, in the circuit. I wonder if the saws had a two-terminal AC plug connection to the 60Hz mains, and thus it also induced that AC component, called 'hum', into the circuit as well, or if the saws had an extra **ground terminal**. I also wonder if the DC power supply to the demonstration logic circuit had a ground terminal, and if the circuit itself was tied into that ground. Most lab-bench DC power supplies have the black (-) and red (+) terminals, and also a green (ground) terminal. Usually the black (-) and the green (ground) terminals are tied together, so the circuit is not "floating", and has the same reference ground as the oscilloscope, and in this case, maybe even the chainsaws. In ideal circuits, this is not that important; but in real life, and even in EECS hands-on lab, it is. Many times my circuits would not work properly due to the lack of a good proper ground. I think this course covers **transformers** when we get to the AC analysis portion, and that is where we must be careful with grounds. One purpose of a 1:1 transformer is to provide isolation and AC coupling in a circuit; this does not allow a DC component to carry over. Hence, it cancels DC noise, but AC noise remains. On step-up transformers 1:2 and so forth, AC noise is amplified. Thus carefully valued and placed capacitors are used to **clamp** (removal via path to ground) this AC noise to ground if it is above a certain frequency. So while noise can be controlled in analog circuits, it cannot be eliminated, something the Prof. stressed in this week's logic lectures. Can't wait until **AC analysis** and transformers! As the Prof. loves to say, it's "lots of fun!" SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-09-24T04:08:37Z SecondChildTAG: Okay, but anyone explained why only one of the inputs to the AND gate was receiving the noise, and the other was not? Anyways, I liked the comic touch to the class. Entering the classroom with the Men in Black song... cool. SecondChildUserIdTAG: 349595 SecondChildUserNameTAG: danvisk SecondChildCreateTimeTAG: 2012-09-25T03:44:15Z FirstChildTAG: Electrical noise can also be conducted by ordinary power outlet wiring. FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-09-24T02:16:51Z SecondChildTAG: Correct. In North America, a major component is the 60Hz hum that is inherent from AC power transmission. In Europe, and the Middle East I believe, it is 50Hz. This AC noise can be filtered out (a simple design uses capacitors) if one designs a power supply correctly. Sensitive instruments also must be shielded to ground; this uses the Faraday cage principle. 1:1 coupling transformers take care of the DC noise, and nonlinear elements such as metal oxide varistors take care of excessive transients, like voltage spikes. Many universities have "power supply design" or "power circuit design" as an elective in an EECS curriculum. If the power supplies in our PCs and digital TVs did not correctly filter out this noise, the logic elements (i.e. microprocessors) in them would not get a nice clean supply voltage. Week 2 did not cover this; maybe next week they will cover how logic gates are actually constructed from discreet components (i.e. transistors, resistors, capacitors). SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-09-24T04:22:44Z IndexTAG: 1599 TitleTAG: Good summary of thevinin voltage/resistance You know, that thing with e. http://hyperphysics.phy-astr.gsu.edu/hbase/electric/thevenin.html Hope this helps someone. UserIdTAG: 364798 UserNameTAG: atari1994 CreateTimeTAG: 2012-09-23T22:58:42Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Yoink! FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-23T23:04:14Z SecondChildTAG: Yoink? SecondChildUserIdTAG: 92346 SecondChildUserNameTAG: MobiusTruth SecondChildCreateTimeTAG: 2012-09-24T17:50:41Z IndexTAG: 1600 TitleTAG: Urgent..!! What if didn't finish my first week homework and my second one too?? if i solved them what is the grade im gonna take? UserIdTAG: 36823 UserNameTAG: Hesham CreateTimeTAG: 2012-09-23T22:34:18Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: The lowest 2 homework scores will be dropped. :) FirstChildUserIdTAG: 197764 FirstChildUserNameTAG: d_artcuffs FirstChildCreateTimeTAG: 2012-09-23T22:39:21Z SecondChildTAG: thanks alot SecondChildUserIdTAG: 36823 SecondChildUserNameTAG: Hesham SecondChildCreateTimeTAG: 2012-09-25T22:41:55Z IndexTAG: 1601 TitleTAG: SUPER COOL Man I wish I had the brains and cash to go to MIT UserIdTAG: 230787 UserNameTAG: Gabriel007 CreateTimeTAG: 2012-09-23T19:44:30Z VoteTAG: 2 CoursewareTAG: Week 2 / Chain saw CommentableIdTAG: 6002x_chain_saw NumberOfReplyTAG: 2 FirstChildTAG: You can do anything you want to with a bit of application. FirstChildUserIdTAG: 195916 FirstChildUserNameTAG: pflynn FirstChildCreateTimeTAG: 2012-09-23T20:02:29Z FirstChildTAG: I'll never make it to MIT but I am free to read and think. FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-09-24T02:32:13Z IndexTAG: 1602 TitleTAG: About the truth table and simplification I would like to now if this step is possible; A*B*En+En*C == En(A*B+C) UserIdTAG: 264320 UserNameTAG: israel05 CreateTimeTAG: 2012-09-23T19:05:39Z VoteTAG: 2 CoursewareTAG: Week 3 / Circuit Truth Table Tutorial CommentableIdTAG: 6002x_circuit_truth_table_tutorial NumberOfReplyTAG: 1 FirstChildTAG: Yes, that is correct, or better: En*(A*B+C) FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-09-23T22:43:08Z IndexTAG: 1603 TitleTAG: staff kindly help as you know that youtube is banned in our region. we cannot access lectures and tutorials. so plz kindly extend the homework dates. please.. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-23T19:04:07Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1604 TitleTAG: A tip If confused with wrong answer to "What is the resistance (in kOhms) when the current is outside that range?", remember what is an IV curve for a true resistor and what is the meaning of the slope of that curve UserIdTAG: 190618 UserNameTAG: Kirbabaev CreateTimeTAG: 2012-09-23T18:44:54Z VoteTAG: 2 CoursewareTAG: Week 3 / Analyzing Nonlinear Circuits Exercise CommentableIdTAG: 6002x_analyzing_nonlinear_circuits_exercise NumberOfReplyTAG: 1 FirstChildTAG: "That range" is just a game about referent and reference (Using Demonstrative Pronouns to Create Referential Cohesion). The slope of a line segment is rise over run. But in this case we have [V] in the numerator and [I] in the denominator, we can remember Ohm's Law and find that [V]/[I] = [R], therefore the slope means resistance. As you may have noticed now current is in the x axis and voltage is in the y axis. That was a **good tip Kirbabaev**. Students hardly ever ask for meaning. That's why they have problems later on in Calculus. FirstChildUserIdTAG: 138769 FirstChildUserNameTAG: ArturoPrado FirstChildCreateTimeTAG: 2012-09-24T16:26:32Z SecondChildTAG: It wasn't a question. It was a TIP for people confused. And i'm not "Atip" )) SecondChildUserIdTAG: 190618 SecondChildUserNameTAG: Kirbabaev SecondChildCreateTimeTAG: 2012-09-24T19:25:21Z IndexTAG: 1605 TitleTAG: This class takes me way longer than the estimated 12 hours I must drop this class. I don't have the free time required to do the work. I found the first week took me about 30 hours of work. This second week I have stumbled around and have spend at least 12 hours already and I am still in the dark on most of the work. I cannot afford this kind of time. It is too bad, because I find the subject interesting, but I would only be able to complete this course if it were delivered at about 1/2 the pace of the current course. This is a very interesting idea and has opened up an opportunity for me to learn. I think it is very promising. I will check back at a later date and maybe give it another go. Thank you for presenting this opportunity. UserIdTAG: 332090 UserNameTAG: WPurin CreateTimeTAG: 2012-09-23T17:45:09Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: It's taking me a lot longer too, just do it. By christmas you will have a nice little certificate. FirstChildUserIdTAG: 195916 FirstChildUserNameTAG: pflynn FirstChildCreateTimeTAG: 2012-09-23T19:18:56Z FirstChildTAG: Hi WPurin! You are not alone! Don't give up! Sometimes it is normal to feel down when some exercises take a lot of effort and time ... But I am sure that you can do it! Be up! You need to give this another chance ;). Also, You have here a lot of TAs and Students that can help you and become this more friendly! Again, Be up that you are not alone and you can do it! Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-23T18:18:30Z FirstChildTAG: I hate to admit this, but it's been more like 12 hrs A DAY for me. I've been out of school for a quarter century, and never was that good at math. Not sure what to tell you that will help, but do realize the course is free, you are, in effect, "auditing" it because you will not get transferable credits in any case, and you can re-take it later at no cost (except your time and effort). Don't give up just yet (this is what I tell myself ;-)). Keep trying. Ask questions. Look for other tools/material online that will help. If nothing else, you're building a foundation for your next run at this course. ;-) Good luck! FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-09-24T00:13:59Z IndexTAG: 1606 TitleTAG: i dont get it so far the circuit analyses are just becoming complex for me; having problem analysing the circuit. UserIdTAG: 411318 UserNameTAG: chinelo CreateTimeTAG: 2012-09-23T17:38:40Z VoteTAG: 2 CoursewareTAG: Week 2 / Superposition experiment CommentableIdTAG: 6002x_Superposition_experiment NumberOfReplyTAG: 1 FirstChildTAG: You can always use the Circuit Sandbox to help. Also, there's no substitute for practice. Try working through the examples in the book, covering remaining steps. IOW, work one step at a time to the best of your ability, then move the paper down to check if you're on the right track. Good luck! FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-09-24T00:16:34Z IndexTAG: 1607 TitleTAG: problem:USING LAB TOOLS m not able to get*connecting wires* even after clicking on the connection points...pll help me to overcum thios prblm UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-23T17:15:11Z VoteTAG: 2 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: **STEP 1:** We will choose a Resistor. Go to that element. And select it with the mouse. Click on that element without releasing and dragg it to the work place. ![1][1] **STEP 2:** You will see something like this. The resistor will be appear note in blue, but yes in green. Ok, if you click on the work place you will find that the resistor will change to blue. When it is green is because we are selecting that element, if not not. ![2][2] **STEP 3:** Lets try to select another resistor and connect them with a wire. So, repeat the STEP 1 again. And dragg the other resistor in other different place from the first resistor. ![3][3] **STEP 4:** Now the difficult part. Lets use the wire ;) ![4][4] **STEP 5:** oK! Done. You should see something like this. But wait a minute... I made a mistake, if I want to delet that wire??? How can I do that? Ok, see STEP 6. ![5][5] **STEP 6:** Select the wire with the mouse and press delete in your keyboard ;). You will delete the wire that you have created before. ![6][6] This is a comment that I wrote it to mkprasanth [here][7] ;). See you! P.D: I think that I wrote mouse as mousse in step 4 image haha (a little mistake, sorry for my english) ![imagen de recetas varias][8] Myriam. [1]: http://s7.postimage.org/xiogau7e3/STEP_1.png [2]: http://s16.postimage.org/z4c1erqg5/STEP2.png [3]: http://s16.postimage.org/vfqxt15t1/step_3.png [4]: http://s13.postimage.org/mri1a7b7r/STEP4.png [5]: http://s9.postimage.org/t0ubwrlxr/STEP5.png [6]: http://s9.postimage.org/et98luymn/step6.png [7]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/MITx_6_002x_2012_Fall/threads/50474faf685941270000000d [8]: http://www.recetasdiarias.com/wp-content/uploads/2010/01/mousse-de-chocolate.jpg FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-23T17:33:10Z SecondChildTAG: Dude I know that already but its not working... Is there anyone else having the same? SecondChildUserIdTAG: 355420 SecondChildUserNameTAG: parkavi SecondChildCreateTimeTAG: 2012-09-23T17:43:35Z SecondChildTAG: A few others, are you using an iPad? If not you could try a different browser, just until they get it sorted. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-23T17:54:34Z SecondChildTAG: hello there! do you know how to change the orientation of the components? i mean using them in vertical makes the circuit a bit clumsy. can't we use them horizontally like a resistor like this (-------)? SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-09-23T17:55:40Z SecondChildTAG: "R" = rotate. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-23T17:59:06Z SecondChildTAG: click on the resistor then press `R`. SecondChildUserIdTAG: 357747 SecondChildUserNameTAG: kishores SecondChildCreateTimeTAG: 2012-09-23T17:59:14Z SecondChildTAG: Hi Jmen Click on the element and then press R :) You can rotate the element. Hi parkavi, oops!, I thought that might you didn't know..., do you have an updated version of chrome or firefox? SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-23T18:04:39Z IndexTAG: 1608 TitleTAG: How do we tell if a circuit is linear? How do we tell, a priori, if a circuit is linear. I understand that if we solve a circuit and see that each branch acts as if it would have if we just set the other sources to 0, then it is linear, but how do you tell if a circuit is linear without solving it? I'm guessing I missed this key point somewhere in the lectures or book. Please point me towards it. Thanks, Dave P.S. How the heck to we see the answers to our post now? I can no longer see the posts I am following (on MacBook Air). UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-23T15:52:34Z VoteTAG: 2 CoursewareTAG: Week 2 / Linearity CommentableIdTAG: 6002x_linearity NumberOfReplyTAG: 2 FirstChildTAG: http://en.wikipedia.org/wiki/Linear FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-09-23T17:42:36Z FirstChildTAG: I saw the week 2 tutorials... Now I'm more confused. A resistor is linear, so does that imply a circuit made of just resistors is linear? The tutorials state a current source is NOT linear. But I'm quite sure that examples were given of circuits claimed to be linear that had current sources. Thanks, FirstChildUserIdTAG: 94009 FirstChildUserNameTAG: David1956 FirstChildCreateTimeTAG: 2012-09-23T16:52:01Z SecondChildTAG: a resistor is linear over a some range simply because their V I characteristic that follow ohm law, $V=I\cdot R$ which is linear equation on the other hand voltage source and current source do not obey ohm law, their Voltage and current relation not linear, why ? well maybe it is natural phenomena or maybe have to ask chemist or physician if resistor and current source or voltage source are wire to make a circuit or system. By assuming that current source or voltage source are ideal and we are analyzing it in static way (not varying with time) by applying the KVL and KCL technique, one can prove it is a form of linear equation SecondChildUserIdTAG: 452559 SecondChildUserNameTAG: kuz1toro SecondChildCreateTimeTAG: 2012-09-23T18:20:00Z IndexTAG: 1609 TitleTAG: Fine controll over video speed -- YouTube Extension I find it rather unpleasant to be limited to only a few video speeds. So I've written simple YouTube player hack, that allows setting custom video speed in 0.1 increments using square bracket keys [ ]. It works on all embedded YouTube videos, not only in edX :) It should work in Chrome. How to use it: 1. Opt-in HTML5 YouTube player: www.youtube.com/html5 2. Install my userscript: http://userscripts.org/scripts/show/148132 It works fine under TamperMonkey add-on: https://chrome.google.com/webstore/detail/dhdgffkkebhmkfjojejmpbldmpobfkfo 3. That's it. Use [ ] keys to change video speed on the fly. No need to use edX video speed controls. (YouTube video must be in focus = active) I've also implemented some simple external API based on messages, so it is possible to integrate this hack in the edX player. Hope someone would find it useful. UserIdTAG: 4076 UserNameTAG: damians CreateTimeTAG: 2012-09-23T15:46:48Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1610 TitleTAG: Now that's creative :) Thanks for this lesson Dr. Agarwal. You're just as cool as the MIB :) UserIdTAG: 298775 UserNameTAG: surja CreateTimeTAG: 2012-09-23T11:19:23Z VoteTAG: 2 CoursewareTAG: Week 2 / Chain saw CommentableIdTAG: 6002x_chain_saw NumberOfReplyTAG: 1 FirstChildTAG: yes surja. I am also accept your comments on Dr.Agarwal FirstChildUserIdTAG: 329525 FirstChildUserNameTAG: udhayaraj12 FirstChildCreateTimeTAG: 2012-09-23T13:43:31Z IndexTAG: 1611 TitleTAG: Look at Y y = 0 => z = z(x) y = 1 => z = 1 UserIdTAG: 337855 UserNameTAG: flexo CreateTimeTAG: 2012-09-23T09:03:47Z VoteTAG: 2 CoursewareTAG: Week 3 / Switch Model Exercise CommentableIdTAG: 6002x_switch_model_exercise NumberOfReplyTAG: 0 IndexTAG: 1612 TitleTAG: H2P1 Answer Issue My R1 and R2 are perfectly alright but Vmin and Vmax are not coming correct. Assistance needed. UserIdTAG: 293566 UserNameTAG: Ayush2117 CreateTimeTAG: 2012-09-23T03:49:10Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 6 FirstChildTAG: My Vmin and Vmax are perfectly right but R1 and R2 are not coming correct. Assistance needed. FirstChildUserIdTAG: 378096 FirstChildUserNameTAG: Jamshaid271 FirstChildCreateTimeTAG: 2012-09-23T04:37:15Z SecondChildTAG: Hi Jamshaid271, How did you calculate Vmin and Vmax without determining R1 and R2? SecondChildUserIdTAG: 11538 SecondChildUserNameTAG: trishul SecondChildCreateTimeTAG: 2012-09-23T04:39:37Z SecondChildTAG: I just don't know what happened. I am also wondering. SecondChildUserIdTAG: 378096 SecondChildUserNameTAG: Jamshaid271 SecondChildCreateTimeTAG: 2012-09-23T04:44:40Z SecondChildTAG: How did you calculate R1 and R2? SecondChildUserIdTAG: 309722 SecondChildUserNameTAG: kaushikraghavan1992 SecondChildCreateTimeTAG: 2012-09-23T04:54:02Z SecondChildTAG: i got it my all answer correct SecondChildUserIdTAG: 373281 SecondChildUserNameTAG: Gaurav025 SecondChildCreateTimeTAG: 2012-09-23T05:02:01Z SecondChildTAG: We require the thevenin equivalent in between 10k to 30k. So I did for example 33->3300 and 56->560 etc. SecondChildUserIdTAG: 378096 SecondChildUserNameTAG: Jamshaid271 SecondChildCreateTimeTAG: 2012-09-23T05:02:54Z SecondChildTAG: I get the R1/R2 ratio as 1.5. So i could assume more that one value for R1 itself. How should i choose the resistance value?? I can assume R1=15 and R2=10 both of which are within the range. I can also choose R1=18 and R2=12 which is also inside the range specified?? How should i solve this issue?? SecondChildUserIdTAG: 309722 SecondChildUserNameTAG: kaushikraghavan1992 SecondChildCreateTimeTAG: 2012-09-23T05:10:26Z SecondChildTAG: The key to the answer is " An additional requirement is that the Thevenin resistance as seen from the output terminals is between 10kΩ and 30kΩ" SecondChildUserIdTAG: 27898 SecondChildUserNameTAG: adebon SecondChildCreateTimeTAG: 2012-09-23T05:18:08Z SecondChildTAG: i agree with adebon, so you have the Rth then, find a way to calculate R2 and R1 but dont forget that you have constraints Vmin The answers are easy and simple. SecondChildUserIdTAG: 370343 SecondChildUserNameTAG: JJarquin SecondChildCreateTimeTAG: 2012-09-23T06:00:50Z SecondChildTAG: i got my Vmax and Vmin correct and i'm not getting my R1 and R2 correctly?? SecondChildUserIdTAG: 309722 SecondChildUserNameTAG: kaushikraghavan1992 SecondChildCreateTimeTAG: 2012-09-23T06:34:45Z SecondChildTAG: I have a problem with the R1 and R2. So Vth is r2/r1+r2*vin and rth=r1+r2. I am confused how to select the resistors SecondChildUserIdTAG: 178115 SecondChildUserNameTAG: nehamakhija SecondChildCreateTimeTAG: 2012-09-23T08:20:59Z SecondChildTAG: Please help SecondChildUserIdTAG: 178115 SecondChildUserNameTAG: nehamakhija SecondChildCreateTimeTAG: 2012-09-23T08:22:56Z SecondChildTAG: Hi nehamakhija, $R_{th}$ is not the series combination of the two resistances that being said you should guess what combination they are in.Next equate $R_{th}$ to any of the standard values between 10k and 30k.Preferably ones closer to the mean of 10k and 30k.Call this equation 1. Next write the expression for voltage across $R_2$.Move $V_{in}$ to lhs.You will get $V_{out}/V_{in}$ equal to an expression in $R_1$ and $R_2$.You know $V_{out}/V_{in}$ from the question use it.Call this equation 2. Solve for $R_1$ and $R_2$ using equations 1 and 2. SecondChildUserIdTAG: 11538 SecondChildUserNameTAG: trishul SecondChildCreateTimeTAG: 2012-09-23T11:03:26Z SecondChildTAG: hey trishul, ur explanation is very clear.. thanks for that.. but at the same time, It is mentioned in the question that vout/vin is within the 10 % of the requirement... what does that mean? SecondChildUserIdTAG: 288715 SecondChildUserNameTAG: vnsudar SecondChildCreateTimeTAG: 2012-09-23T11:23:37Z SecondChildTAG: so for equation 1 rth= r1*r2/r1+r2 right all the values are unknown SecondChildUserIdTAG: 178115 SecondChildUserNameTAG: nehamakhija SecondChildCreateTimeTAG: 2012-09-23T11:23:44Z SecondChildTAG: any problem with H2P2, question part 2? ie "What is the power (in Watts) that is delivered to this best load resistance?" not getting the ans. pls help. SecondChildUserIdTAG: 226103 SecondChildUserNameTAG: piusroshan SecondChildCreateTimeTAG: 2012-09-23T11:28:15Z SecondChildTAG: vnsundar, vout/vin= 0.1 as it has to be 10%, I hope i an right SecondChildUserIdTAG: 178115 SecondChildUserNameTAG: nehamakhija SecondChildCreateTimeTAG: 2012-09-23T11:29:09Z SecondChildTAG: nope nehamakhija... its not giving right answer if i try the way u said SecondChildUserIdTAG: 288715 SecondChildUserNameTAG: vnsudar SecondChildCreateTimeTAG: 2012-09-23T11:46:42Z SecondChildTAG: Even i faced the same problem SecondChildUserIdTAG: 178115 SecondChildUserNameTAG: nehamakhija SecondChildCreateTimeTAG: 2012-09-23T11:49:18Z SecondChildTAG: Can someone help SecondChildUserIdTAG: 178115 SecondChildUserNameTAG: nehamakhija SecondChildCreateTimeTAG: 2012-09-23T11:49:31Z SecondChildTAG: m also having same problem..not able to get correct r1 and r2.. whereas i have found correct vmax and vmin.. SecondChildUserIdTAG: 219950 SecondChildUserNameTAG: sumit_010 SecondChildCreateTimeTAG: 2012-09-23T12:00:26Z SecondChildTAG: i have got the answer for r1 and r2. let me share what is the mistake i have committed in solving that... actually the method i followed is correct and hopefully most of us may do that correctly.. then where i have slipped? the answers should be in terms of "resistance values in that e12 set".. i mean if u get answer around 25 or 26, check for the nearest value in the e12 set, huh! its 27.. so take 27 as the resistance value and solve for the next resistance. hope this may help SecondChildUserIdTAG: 288715 SecondChildUserNameTAG: vnsudar SecondChildCreateTimeTAG: 2012-09-23T12:16:29Z SecondChildTAG: @nehamakhija and @vnsundar....vout/vin gives a value if you take the values given in the question.take 10% of that value and solve. SecondChildUserIdTAG: 11538 SecondChildUserNameTAG: trishul SecondChildCreateTimeTAG: 2012-09-23T12:37:48Z FirstChildTAG: Hi Ayush, The voltage across $R_2$,i.e,$V_{out}$ will be maximum when $R_2$ is maximum and $R_1$ is minimum.Similarly,$V_{out}$ is minimum when $R_2$ is minimum and $R_1$ is maximum. Here minimum and maximum in terms of resistance means 10% less than standard value and 10% higher than standard value respectively. I think this should help! FirstChildUserIdTAG: 11538 FirstChildUserNameTAG: trishul FirstChildCreateTimeTAG: 2012-09-23T04:43:47Z SecondChildTAG: Thanks trishul!! SecondChildUserIdTAG: 291362 SecondChildUserNameTAG: Gudson SecondChildCreateTimeTAG: 2012-09-23T15:48:17Z FirstChildTAG: Hi Ayush, How did you calculate R1 and R2?? Please help me out. FirstChildUserIdTAG: 309722 FirstChildUserNameTAG: kaushikraghavan1992 FirstChildCreateTimeTAG: 2012-09-23T04:55:50Z SecondChildTAG: You have Rth, and i assume that you know how to obtain it (matematically, according to circuit) then find R1 and R2 that satisfy the Vout constraint SecondChildUserIdTAG: 370343 SecondChildUserNameTAG: JJarquin SecondChildCreateTimeTAG: 2012-09-23T06:02:17Z SecondChildTAG: I forgot to say that: The answers are easy and simple. SecondChildUserIdTAG: 370343 SecondChildUserNameTAG: JJarquin SecondChildCreateTimeTAG: 2012-09-23T06:03:03Z FirstChildTAG: hi, you will have to take care while calculating thevenin's resistance. FirstChildUserIdTAG: 373281 FirstChildUserNameTAG: Gaurav025 FirstChildCreateTimeTAG: 2012-09-23T05:06:24Z FirstChildTAG: Vmax=when R2 is max and R1 is min.(according to given 10% tolerance ) Vmin=when R2 is min and R1 is max.(according to given 10% tolerance ) FirstChildUserIdTAG: 93123 FirstChildUserNameTAG: Bilawal FirstChildCreateTimeTAG: 2012-09-23T07:42:34Z SecondChildTAG: It can be said that Vmax= Vout max(according to given 10 % tolerance) and Vmin= Vout min (according to 10% tolerance). SecondChildUserIdTAG: 378096 SecondChildUserNameTAG: Jamshaid271 SecondChildCreateTimeTAG: 2012-09-23T10:15:55Z SecondChildTAG: any 1 need LAB2 answer? SecondChildUserIdTAG: 345688 SecondChildUserNameTAG: asif100 SecondChildCreateTimeTAG: 2012-09-23T17:35:41Z FirstChildTAG: First you have to follow the constraints: 1. Vout/Vin is within 10% of the requirement. This lead you to an **inequality** (a range) involving a voltage divider. 2. Thevenin resistance as seen from the output terminals is between 10kΩ and 30kΩ. This gives you another inequality. Now you solve this system of inequalities an will have several choices **At first I thought my R1 and R2 choices were perfect**, until I calculated Vout/Vin and It was over the maximum specified. So, I had to choose another set of values for R1 and R2. FirstChildUserIdTAG: 164171 FirstChildUserNameTAG: PacoJuarez FirstChildCreateTimeTAG: 2012-09-23T20:48:38Z IndexTAG: 1613 TitleTAG: Homeworks and labs issue, please read. Hello 6.002x community, this is the second time that I take this course since I was not able to finish it the first time for personal reasons. The thing is, I remember that in the first one, homeworks and labs had a limmited number of attemps (I think there were 3) to check your answers before submitting.. I see now you can check unlimitted times until you get the right answer... doesn´t that makes it a lil bit less "believable" if you pass the course with good grades? I mean, besides the honor code, I think students would have more "credit" if there were limited times to check the answer, don't you think? Thanks for reading! (sorry if my english is not completely right, I haven't used it for a long while xD ) UserIdTAG: 42787 UserNameTAG: rafuk CreateTimeTAG: 2012-09-23T00:25:21Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: I can see your point about the homework, but when I think it through I can see why it is better to have unlimited checking in the homework/lab stage. The homework stage is for learning, the way to maximize learning is to facilitate it. I myself have entered answers dozens of times. If I was sent home the first wrong answer, I would not have learned anything. Now the exams let you have 3 attempts, which is nice should you make one or two clerical mistakes in your answers, in lieu of the interpretive eye of a Professor or Assistant. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-23T00:53:46Z SecondChildTAG: hmmm I agree with your argument, but still I think a limited number of attemps (not send you home the first wrong answer, that would be bad xD, but say maybe five attemps) can help us students to try a little harder before submitting an answer... Its easy to get used to check and correct - check and correct - check and correct (maybe its just me heheh). Thanks for your answer. SecondChildUserIdTAG: 42787 SecondChildUserNameTAG: rafuk SecondChildCreateTimeTAG: 2012-09-23T02:51:10Z SecondChildTAG: Me too! SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-23T04:00:36Z FirstChildTAG: Hi rafuk! How are you? I took The Prototype Course in 6.002x and we had unlimited check buttoms in Homeworks and Labs... might you are cnfused with the Midterm and Final Exam where we only had 3 chances by each exercise ;). Nice to see you again on board! I hope that you are now fine and that everything it is ok with your personal issue ... I am happy that you are here! *"persevere and succeed"* Good luck! Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-23T01:09:56Z SecondChildTAG: Hi Myriam! Im fine thank you! =) Well now that I think of it, I might be confused as you pointed out! lol. Thanks for your response, and good luck to you too! SecondChildUserIdTAG: 42787 SecondChildUserNameTAG: rafuk SecondChildCreateTimeTAG: 2012-09-23T02:53:39Z SecondChildTAG: You are welcome rafuk! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-23T03:14:24Z FirstChildTAG: IMHO, If you do have like 3 attempts on hw/labs, then you fail at 3rd and go away. If you have unlimited times, you take your time after the 3rd and spend more time understanding the problem. So i think more time spent on the issue is better for learning. Could have a limited attempts, but it has to be quite generous, say like 20 to 50 of them or something like that. Just to discourage mindless punching of check button. Or on the other side, could have a counter for how many times check was submitted. This could be also a good way to self-evaluate, even later on to a recapitulation before an exam i could see way back that i needed 10 answers to a problem, so i need to look more at that subject later. Also could use that make some nice little statistics. And also i think homework should have more problems, like say 5 to 10. Plenty in book to go around, but less incentive to go there. I want to do them all, but i always end up postponing that. At least +2 +3 more in homework wouldnt hurt. FirstChildUserIdTAG: 331483 FirstChildUserNameTAG: Doru FirstChildCreateTimeTAG: 2012-09-23T05:24:22Z SecondChildTAG: Its a good idea to count the times one checks! even better than having limited attemps I think youre right! thanks. I agree that homeworks should have more excersises too, or maybe make some sort of "excercise bank" in the courseware would be nice too. SecondChildUserIdTAG: 42787 SecondChildUserNameTAG: rafuk SecondChildCreateTimeTAG: 2012-09-23T20:18:18Z IndexTAG: 1614 TitleTAG: Resistance 1/R should be R The slide 2 says "We can replace ... with a resistance of value $1/R_D$". I believe it should actually say "resistance of value $R_D$". UserIdTAG: 153760 UserNameTAG: Kavka CreateTimeTAG: 2012-09-22T23:22:13Z VoteTAG: 2 CoursewareTAG: Week 3 / Piecewise Linear Analysis CommentableIdTAG: 6002x_piecewise_linear_analysis NumberOfReplyTAG: 1 FirstChildTAG: Kavka, you are right. However, keep in mind that RD is the reciprocal of the slope, not the slope itself FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-23T18:40:19Z IndexTAG: 1615 TitleTAG: Just joined... Unfortunately, I was unaware that these courses were available until last night, so I am a bit behind. Is there any way I can get caught up rather quickly? mind you, I am still in school from 6:30-10:30 every weekday and I do work, so I have about 3 hours a day to be online, I suppose. UserIdTAG: 480310 UserNameTAG: noellaluciotti CreateTimeTAG: 2012-09-22T23:12:02Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hi there. Basically you can miss 2 out of 12 homeworks/labs without consequence. The second one is due tomorrow. If you have a little experience you could knock out the second one by tomorrow night. If not, start at week one, get yourself caught up. You have until the 30th, when week 3 is due. It's doable, but it will be tight for time. The lectures don't take all that long, actually the homework would be fast if someone picks it up quick, for the rest of us, it takes time. The best thing you can do is search for answers to any questions, many questions get asked over and over again. The earlier questions have replies. You could still get 100%, the only thing you lost were your two get out of homework cards. Have fun. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-22T23:30:54Z SecondChildTAG: Welcome noellaluciotti, I agree with Pennypacker. the only problem is that u missed the easier homework and lab. The homeworks and labs are getting harder at an exponential rate but with effort you can be flawless. Best regards. SecondChildUserIdTAG: 183507 SecondChildUserNameTAG: obiradaniel SecondChildCreateTimeTAG: 2012-09-23T21:13:58Z FirstChildTAG: welcome...... FirstChildUserIdTAG: 276450 FirstChildUserNameTAG: ritece FirstChildCreateTimeTAG: 2012-09-22T23:47:10Z IndexTAG: 1616 TitleTAG: Course Content Will the content of this course (lectures, problem sets, labs, and exams) be accessible after the course has ended? It would be nice to have it here for reference in the future if possible. Thanks. UserIdTAG: 213386 UserNameTAG: dmascenik CreateTimeTAG: 2012-09-22T19:39:10Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I agree, I hear that the last class still has access to the forums etc. You can download the Youtube videos with a program or keep the private links. You can save web pages with firefox.(labs etc.) I don't know about the exams. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-22T19:52:43Z SecondChildTAG: I know that I can download the videos and such manually. I was just wondering if going through all that trouble would be a waste or not. SecondChildUserIdTAG: 213386 SecondChildUserNameTAG: dmascenik SecondChildCreateTimeTAG: 2012-09-22T19:59:37Z IndexTAG: 1617 TitleTAG: How to watch TUTORIALS with out YOUTUBE? is there any way to watch tutorials without YouTube? UserIdTAG: 40209 UserNameTAG: sadaf78 CreateTimeTAG: 2012-09-22T17:33:11Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Do a search, there was a guy on here sharing the videos. This question gets asked all day long. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-22T18:08:50Z FirstChildTAG: You can see in the Wiki [Lecture Videos Download][1]:). Also see more info [here][2]: **Due to the YouTube outage in certain parts of the world, we have provided direct links to download the lecture videos, and extended the Week 2 homework and lab deadlines to the coming Sunday (September 30). Please note that Week 3 deadlines have not changed.** [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/ [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/info FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-26T11:46:27Z IndexTAG: 1618 TitleTAG: H2P1 "how to" discussion after the deadline Frustrated, I figured out the answer by designing a C++ program to compute the entire thing for me. Coz unlike me, my computer can process a lot without banging its head on the wall. But I'm sure there's a much easier way to do so since so many of ya'll got the answer simply using pen and paper. So please show me your ways, after the deadline of course :))) UserIdTAG: 204745 UserNameTAG: allwynmendes CreateTimeTAG: 2012-09-22T12:09:12Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: For me the easy way is to see Vout as a voltagesource and to apply Thevenin to find its Vth [that one is in fact given] and Rth. I think most people have problems here with understanding Thevenin. Rth depends on R1 and R2, but they are not known yet. So you need to find eqn for the short circuit current Is for the given circuit. If you take Rth in the given range, then you can find the eqn for the shortcircuit current for Vth and Rth. From there it's relatively easy if you express R1 in Vth,Vin and Rth. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-09-22T15:41:11Z SecondChildTAG: select resistors form E12 table e.g(56,68,82)k and (15,18,22)k than using voltage divider formula for min input voltage i.e 70 and than for max. voltage i.e 90 voltage calculate vout SecondChildUserIdTAG: 414201 SecondChildUserNameTAG: uzaifakram SecondChildCreateTimeTAG: 2012-09-22T18:52:45Z SecondChildTAG: resistances values are ok but how to calculate max voltage and min? SecondChildUserIdTAG: 182470 SecondChildUserNameTAG: nitesh2703 SecondChildCreateTimeTAG: 2012-09-23T04:09:05Z SecondChildTAG: Any formula to calculate v max or v min?? SecondChildUserIdTAG: 182470 SecondChildUserNameTAG: nitesh2703 SecondChildCreateTimeTAG: 2012-09-23T04:27:42Z SecondChildTAG: to Find the Max. & Min. , 1st calculate your output voltage range by finding 10% above & below "that's your output tolerance". 2nd after you choose your resistance values , calculate the tolerance of each one , so you'll have a Max. & a Min. Value for each resistance apply the values that gives you the max. output "that will be your max. output" to the voltage divider eq. , apply the values that will gives you the Min. to the same Eq. "your values depends upon the tolerance that you calculate for each resistance"...good luck SecondChildUserIdTAG: 185715 SecondChildUserNameTAG: amirengineer SecondChildCreateTimeTAG: 2012-09-23T19:15:17Z SecondChildTAG: THANK U BRO'S FINALLY I GOT ALL THE ANS. AFTER 10 BRAIN STRETCHING HOURS......JUST BCOZ OF U...THANK U A LOT...... SecondChildUserIdTAG: 307088 SecondChildUserNameTAG: PRAKHAR2012 SecondChildCreateTimeTAG: 2012-09-25T10:29:13Z IndexTAG: 1619 TitleTAG: homework 2 **Ihav got some values of R1 AND R2 and my output voltage is 18, so i want to know whether this output is correct or we have to get the 10% of the output and then the ans should be 18,m a little confused** UserIdTAG: 35623 UserNameTAG: ARSHIYA25 CreateTimeTAG: 2012-09-22T08:46:03Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: **R1/R2= should be 2.2 approx witch is the ratio of Vout/Vin** FirstChildUserIdTAG: 399292 FirstChildUserNameTAG: Qabali FirstChildCreateTimeTAG: 2012-09-22T09:43:41Z SecondChildTAG: could u please tell me the exact procedure of solving the whole question with proper formulas.....please SecondChildUserIdTAG: 35623 SecondChildUserNameTAG: ARSHIYA25 SecondChildCreateTimeTAG: 2012-09-22T19:44:35Z IndexTAG: 1620 TitleTAG: what does operating volt¤t mean? i dont know what is operating volt and current and how to find the values??can somebody help me UserIdTAG: 118611 UserNameTAG: mitianhari CreateTimeTAG: 2012-09-22T06:25:53Z VoteTAG: 2 CoursewareTAG: Week 3 / Graphical Analysis of Nonlinear Circuits Exercise CommentableIdTAG: 6002x_graphical_analysis_of_nonlinear_circuits_exercise NumberOfReplyTAG: 1 FirstChildTAG: It says: turn the current source ON and find I and V for that alien device. It **IS** operating when there is an active power source in the circuit so its **operating** voltage and current are those i told about above. FirstChildUserIdTAG: 190618 FirstChildUserNameTAG: Kirbabaev FirstChildCreateTimeTAG: 2012-09-23T20:23:36Z IndexTAG: 1621 TitleTAG: "Keeps momentum" > So if you look at this speaker, you'll see that it has this thing over here that's called a port. And what that is, is a tube that goes through the speaker, and it **keeps momentum**. Could you please explain what this phrase means? Is it the same as "stores momentum" (energy)? UserIdTAG: 410033 UserNameTAG: sagitta CreateTimeTAG: 2012-09-22T05:09:19Z VoteTAG: 2 CoursewareTAG: Week 2 / Speakers CommentableIdTAG: 6002x_speakers_tutorial NumberOfReplyTAG: 1 FirstChildTAG: It is not the same as "stores momentum". Good question though. Basically a certain speaker in a box can only play sound down to a certain frequency. Now think of this tube as a whistle. This whistle is designed to make sound just below the point of the speaker/box without the whistle. It is a way to extend the frequency response lower. More (Lower) Bass. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-23T01:24:18Z IndexTAG: 1622 TitleTAG: Make sure to put the ground node I made a very stupid mistake. I was hitting dead ends and scratching my head to solve Lab 2. It turned out that I forgot to put the ground node at the third resistor. UserIdTAG: 172304 UserNameTAG: Demohunter CreateTimeTAG: 2012-09-22T00:26:09Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1623 TitleTAG: Problem with check button "bug" "bug" The check button is not working, it doesn't submit the answers neither for this exercise nor in the S4E3. Today I've seen other posts in the forum about the same problem with the check button on other sections/exercises. I'm using winXP and Firefox 15.0.1, already tried cleaning cache, cookies and history but still doesn't work. Thanks in advance for any help. #bug UserIdTAG: 304535 UserNameTAG: hugo_h CreateTimeTAG: 2012-09-21T21:58:55Z VoteTAG: 2 CoursewareTAG: Week 2 / Boolean functions CommentableIdTAG: 6002x_S4E2_Boolean_Functions NumberOfReplyTAG: 1 FirstChildTAG: I am also using FireFox 15.0.1 and Windows XP, but have tried IE 8 as well with same results. There is some problem with **H2P1: Voltage-Divider Design** the check-button for that also does not work. All of the rest of the check-buttons on HW #2 does work. There also seem to be errors in the text as well, all of the OHMS designation and numbers are not formatted correctly. I just noticed this today (Sept 21, 2012). And finally, LAB#2 does not work for me either, where you are suppose to enter the circuit in does not show up. FirstChildUserIdTAG: 406848 FirstChildUserNameTAG: CyberTempest FirstChildCreateTimeTAG: 2012-09-21T22:13:59Z SecondChildTAG: I had problems with H2P2 but not with the others in the same page. I restarted the PC and worked. But I don see the simulator on the lab2. I already posted in the lab2 post. SecondChildUserIdTAG: 392699 SecondChildUserNameTAG: lu2adw SecondChildCreateTimeTAG: 2012-09-21T23:02:58Z IndexTAG: 1624 TitleTAG: For Anybody that doesn't want to watch the video for answers before trying part B Attached is Figure 4.50 ![Figure 4.50][1] [1]: https://edxuploads.s3.amazonaws.com/13482640572550703.png UserIdTAG: 248902 UserNameTAG: Chanute CreateTimeTAG: 2012-09-21T21:48:09Z VoteTAG: 2 CoursewareTAG: Week 4 / Textbook Load Line Tutorial CommentableIdTAG: 6002x_txtbk_load_line_t NumberOfReplyTAG: 0 IndexTAG: 1625 TitleTAG: Lab2- I'm stuck Hello guys! I did this: ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13482463539017856.jpg then i did a voltage divider for v1 and v2 and got two equations: [R1R3/(R1+R3)]*5=R2 and [R2R3/(R2+R3)]= R1 Basically it's the same as R1*5=R2 and R2=R1, if you consider R2 as if the parallel between R1 and R3 when the V1 is short circuited and R1 being R2//R3 when V2 is short circuited. My question is: How am i suppose to proceed? Because 3 unknowns and 2 equations is impossible to solve, should i try random values till i got it right? Or there is a way to find the exactly value of R1, R2 and R3? Thanks guys! UserIdTAG: 122637 UserNameTAG: Caetano CreateTimeTAG: 2012-09-21T19:33:58Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: You should choose the value for one of the resistors and then the others will appear from your equations. FirstChildUserIdTAG: 308853 FirstChildUserNameTAG: fmorato FirstChildCreateTimeTAG: 2012-09-21T19:41:38Z SecondChildTAG: these equations give you a ratio between 3 resistances - you can choose any values - obvious way, as fmorato said, - pick any value for one and calculate two others SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-21T19:59:37Z SecondChildTAG: I have the same equation for R1 but a different one for R2. I set R3 but when I do that the other values reset to 0! Still no green check :( SecondChildUserIdTAG: 35642 SecondChildUserNameTAG: caled SecondChildCreateTimeTAG: 2012-09-22T19:31:00Z FirstChildTAG: I am not confirming your equations but you are right three unknown and two equations are impossible to solv. Trick is just simply pick a value like 1 ohm or 10 ohm for R3 and then you will have two unknown only. FirstChildUserIdTAG: 210954 FirstChildUserNameTAG: Shahrouz FirstChildCreateTimeTAG: 2012-09-22T00:15:41Z SecondChildTAG: Thanks for the tip, it really worked. But it could be used with any value or just 1 to 10? If is just 1 to 10, why only this range? SecondChildUserIdTAG: 122637 SecondChildUserNameTAG: Caetano SecondChildCreateTimeTAG: 2012-09-23T14:37:27Z FirstChildTAG: Hey, I took the same approach as you but couldn't get the right waveform. Were you any luckier? FirstChildUserIdTAG: 107219 FirstChildUserNameTAG: AlexandreZ FirstChildCreateTimeTAG: 2012-09-22T01:22:49Z SecondChildTAG: OK, scratch that. A tip: the lab system does not like fractions. It must take decimals. I covered a dozen sheets of paper solving this problem every way I could conceive of, banging my head against the wall finding the same answer every single time and having the lab tell me it was wrong when it was right all along. Don't write 5/6, write 0.833333. (The numbers are not answers to the problem) On the up side, I got lots of practice. SecondChildUserIdTAG: 107219 SecondChildUserNameTAG: AlexandreZ SecondChildCreateTimeTAG: 2012-09-22T04:23:47Z FirstChildTAG: Thank you guys for this wonderful disussion...... i got the green check finally after banging my head for hours.... if this would not have been here i think it would have taken few more hours today..... thank you all again..... FirstChildUserIdTAG: 276450 FirstChildUserNameTAG: ritece FirstChildCreateTimeTAG: 2012-09-22T22:30:37Z SecondChildTAG: oh, you got it, get me some powerful hints please SecondChildUserIdTAG: 254607 SecondChildUserNameTAG: werehenry SecondChildCreateTimeTAG: 2012-09-22T23:03:55Z FirstChildTAG: Hi Caetano! Now that the deadline has passed you can take a look at [here][1] :) [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_S3E2_Linear_Combinations_of_Source_Strengths/threads/506a280d036af91f000000c9 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-02T00:29:37Z IndexTAG: 1626 TitleTAG: bug The 'Check' button doesn't submit the responses. There's no green check mark, nor red 'X' and 'show answers' button. Thank you kind sirs/madams. UserIdTAG: 280220 UserNameTAG: pilgrimCycle CreateTimeTAG: 2012-09-21T19:23:47Z VoteTAG: 2 CoursewareTAG: Week 3 / Switch Resister Model Exercise CommentableIdTAG: 6002x_switch_resistor_model_exercise NumberOfReplyTAG: 2 FirstChildTAG: I'm having de same problem on the [Static Discipline and Boolean Logic][1] sequence exercises. I'm using windows XP and Firefox 15.0.1 "bug" [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_2/Static_Discipline_and_Boolean_Logic/ FirstChildUserIdTAG: 304535 FirstChildUserNameTAG: hugo_h FirstChildCreateTimeTAG: 2012-09-21T20:13:17Z SecondChildTAG: use Google Chrome SecondChildUserIdTAG: 374393 SecondChildUserNameTAG: rmaleki SecondChildCreateTimeTAG: 2012-09-22T15:20:27Z SecondChildTAG: I'm using Chrome and the Java is still a little buggy, it usually works after a 'Refresh'. SecondChildUserIdTAG: 141469 SecondChildUserNameTAG: stavross29 SecondChildCreateTimeTAG: 2012-09-25T23:40:15Z FirstChildTAG: Yesterday this certainly seems to have cleared up. Great work, debuggers! FirstChildUserIdTAG: 280220 FirstChildUserNameTAG: pilgrimCycle FirstChildCreateTimeTAG: 2012-09-25T01:14:52Z IndexTAG: 1627 TitleTAG: YouTube banned in country, can't watch videos! There's an indefinite ban on youtube in my country, so im not being able to watch any of the videos. Any solutions?? UserIdTAG: 467203 UserNameTAG: jennifer515 CreateTimeTAG: 2012-09-21T18:13:06Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Go for Tor: https://www.torproject.org/ FirstChildUserIdTAG: 308853 FirstChildUserNameTAG: fmorato FirstChildCreateTimeTAG: 2012-09-21T19:00:55Z SecondChildTAG: +1 for tor project. Although you may experience a slowdown in your connection. Another solution could be the free (ad-ware) [HotspotShield][1]. [1]: http://www.anchorfree.com/ SecondChildUserIdTAG: 185338 SecondChildUserNameTAG: mgaldieri SecondChildCreateTimeTAG: 2012-09-21T19:44:08Z SecondChildTAG: while using tor project i get a message the video is currently unavailable what is solution to this problem SecondChildUserIdTAG: 358539 SecondChildUserNameTAG: syed_abdullah SecondChildCreateTimeTAG: 2012-09-23T18:42:42Z FirstChildTAG: There is a ban on youtube in my country, so im not able to watch any of the videos. Any solutions?? FirstChildUserIdTAG: 118756 FirstChildUserNameTAG: abdulrazzaq789 FirstChildCreateTimeTAG: 2012-09-21T18:30:12Z SecondChildTAG: https://www.torproject.org/ SecondChildUserIdTAG: 308853 SecondChildUserNameTAG: fmorato SecondChildCreateTimeTAG: 2012-09-21T19:00:34Z FirstChildTAG: You can see in the Wiki [Lecture Videos Download][1]:). Also see more info [here][2]: **Due to the YouTube outage in certain parts of the world, we have provided direct links to download the lecture videos, and extended the Week 2 homework and lab deadlines to the coming Sunday (September 30). Please note that Week 3 deadlines have not changed.** [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/ [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/info FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-26T11:47:11Z IndexTAG: 1628 TitleTAG: Week 3 Tutorials Week 3 Tutorials doesn't work ! PAGE NOT FOUND UserIdTAG: 194717 UserNameTAG: kaa CreateTimeTAG: 2012-09-21T16:41:01Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Thanks for this bug report. We are working on it. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-09-21T17:38:54Z FirstChildTAG: if you put a slash at the end of the url you can access to the contents, for example currently the url for the problem 1 is: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_3/week3_ex_5_5 with a slash at the end now it works: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_3/week3_ex_5_5/ and that will work the same with the rest of the week 3 tutorials (and week 2 too). FirstChildUserIdTAG: 393284 FirstChildUserNameTAG: tomasbolano FirstChildCreateTimeTAG: 2012-09-21T19:05:23Z SecondChildTAG: thanks! SecondChildUserIdTAG: 194717 SecondChildUserNameTAG: kaa SecondChildCreateTimeTAG: 2012-09-21T19:40:19Z IndexTAG: 1629 TitleTAG: Excellent Great tool! Completed exercise no probs. Tip - make sure you observe the units i.e. k - kilo 1000's or m - milli - thousandth's etc. and input/convert for use in the web app. UserIdTAG: 250150 UserNameTAG: V3RN CreateTimeTAG: 2012-09-21T15:33:33Z VoteTAG: 2 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 0 IndexTAG: 1630 TitleTAG: i can't use the lab environment please can somebody help me? up till now, I am yet to see how to use the lab environment (software) with respect to setting the resistance value to what I want. UserIdTAG: 433857 UserNameTAG: Chexzy CreateTimeTAG: 2012-09-21T14:55:51Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1631 TitleTAG: Problem with Mathematical Equation Display "Staff" "Bug" Hello, Staff - When I logged in this morning (approx 0940EDT), I found that the display of mathematical formulas was non-functional. For example from S6E3: ---------- "The characteristic of the nonlinear element shown in the circuit is made up of two linear segments. The resistance of one segment is \(R_1=1.0\)M\(\Omega\) and the resistance of the other segment is \(R_2=1.0\Omega\). (Of course, the graph is not to a uniform scale. If it were, we would not see the slope of the segment with slope \(1/R_1\).)" ---------- I checked in both Firefox and IE and found the same behavior. Is there an issue with the course software or is this problem on my end? Thanks. **EDIT**: I can neither check work on exercises nor view their solutions. Is there a problem at the EDX end or my end? Thanks. UserIdTAG: 348141 UserNameTAG: TomTrieb CreateTimeTAG: 2012-09-21T13:51:48Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: It's working for me, though we did do a recent code update this morning and that might have broken something for some users. Could you please try it in Chrome, as well as giving us your exact version of Firefox? Thank you. FirstChildUserIdTAG: 11 FirstChildUserNameTAG: dormsbee FirstChildCreateTimeTAG: 2012-09-21T16:35:52Z SecondChildTAG: FF 15.0.1 or IE 8.0.7601.17514 on Windows 7 Debugging techniques: tried logging out and back in from both FF and IE. IE showed a "Math Processing Error" Will try on Chrome and FF from my Mac over the weekend. FYI, the labs are broken as well (in that the actual lab working areas do not appear on the screen) SecondChildUserIdTAG: 348141 SecondChildUserNameTAG: TomTrieb SecondChildCreateTimeTAG: 2012-09-21T18:56:37Z SecondChildTAG: Did you just do an update around 1510 EDT on 9/21? HW4 and S7E1 now shows formulas; however, the Lab 4 still does not show the formulas or the circuit working area. Thanks. SecondChildUserIdTAG: 348141 SecondChildUserNameTAG: TomTrieb SecondChildCreateTimeTAG: 2012-09-21T19:18:08Z SecondChildTAG: Tried in both Chrome and FF on my Mac; everything seems fine. SecondChildUserIdTAG: 348141 SecondChildUserNameTAG: TomTrieb SecondChildCreateTimeTAG: 2012-09-23T20:17:34Z IndexTAG: 1632 TitleTAG: Build an Equation I was able to build a truth table for the circuit above, however, I was wondering if there were an easy way to look at the picture and build an equation. If so, has anybody come of with a fairly easy way to do this? UserIdTAG: 237421 UserNameTAG: Lmotloch CreateTimeTAG: 2012-09-21T11:54:20Z VoteTAG: 2 CoursewareTAG: Week 3 / Switch Model Exercise CommentableIdTAG: 6002x_switch_model_exercise NumberOfReplyTAG: 4 FirstChildTAG: That would be cool and give me added insight. Either that or being able to build a logic equation from some manipulation of a truth table FirstChildUserIdTAG: 440714 FirstChildUserNameTAG: mcktim FirstChildCreateTimeTAG: 2012-09-21T16:25:04Z SecondChildTAG: I found this boolean [sum-of-products thing][1]. ![enter image description here][2] [1]: http://www.allaboutcircuits.com/vol_4/chpt_7/9.html [2]: https://lh6.googleusercontent.com/-Owzq0uJenWQ/UFyi2Irtz5I/AAAAAAAABTY/eKpXkWyUcew/s320/2012-09-21.jpg SecondChildUserIdTAG: 440714 SecondChildUserNameTAG: mcktim SecondChildCreateTimeTAG: 2012-09-21T17:52:24Z SecondChildTAG: i got the same truth table with the relation Z= [inverse(X)]+Y SecondChildUserIdTAG: 316353 SecondChildUserNameTAG: Vivekanand123 SecondChildCreateTimeTAG: 2012-09-23T09:53:48Z SecondChildTAG: oh sorry... not the same... SecondChildUserIdTAG: 316353 SecondChildUserNameTAG: Vivekanand123 SecondChildCreateTimeTAG: 2012-09-23T09:55:43Z FirstChildTAG: INVERTER(INVERTER(X) NOR Y ) = INVERTER(X) OR Y , look at the right to left and compare with a ( inverter ) and b ( nor ) figures of examples. FirstChildUserIdTAG: 333731 FirstChildUserNameTAG: Timophei_NhaTrang FirstChildCreateTimeTAG: 2012-09-22T09:42:14Z FirstChildTAG: Just look at the four switches from left to right, name them as S1, S2, S3, S4, and the corresponding input (as the arrows pointing at) as I1, I2, I3, I4. Following the basic gate logic, we have: I1=X, I2=bar(X), I3=Y, I4=bar(I2+I3), Z=bar(I4). So finally we can get: Z=bar(bar(bar(X)+Y))=bar(X)+Y. Then is easy to find the answer using this expression. FirstChildUserIdTAG: 372709 FirstChildUserNameTAG: mulinsen FirstChildCreateTimeTAG: 2012-09-22T21:45:43Z FirstChildTAG: ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13483913057281093.jpg FirstChildUserIdTAG: 194646 FirstChildUserNameTAG: Rojer FirstChildCreateTimeTAG: 2012-09-23T09:08:46Z IndexTAG: 1633 TitleTAG: Lab2 question Are the parameters of V1 and V2 voltage sources in LAB2 correct. I tried to use superposition to figure out R1 and R2 values and I come up with something like R2=5*R1, yet this always results in Vout(max)=1V, not 0.667V. Vout(min)=-0.167V (as expected). I tried several times and not sure if I am doing anything wrong or there is a typo in the Voltage source parameters. Any feedback will be appreciated. Thanks a lot, GT UserIdTAG: 350183 UserNameTAG: gt94539 CreateTimeTAG: 2012-09-21T11:06:04Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 6 FirstChildTAG: Along with GT's question, I'd like to know if it is legal to introduce an additional ground point in the circuit to bring Vout below 1V. FirstChildUserIdTAG: 142896 FirstChildUserNameTAG: Halekulani FirstChildCreateTimeTAG: 2012-09-21T11:08:50Z FirstChildTAG: Hi GT You are right with your figures - so how do you get the results required? try putting a third resistor to ground into the equation Melbur FirstChildUserIdTAG: 95673 FirstChildUserNameTAG: melbur FirstChildCreateTimeTAG: 2012-09-21T11:32:18Z FirstChildTAG: I have the same Problem FirstChildUserIdTAG: 448321 FirstChildUserNameTAG: Keroo FirstChildCreateTimeTAG: 2012-09-21T13:34:04Z FirstChildTAG: put the third resistor along the output side with ground to get req waveform FirstChildUserIdTAG: 149154 FirstChildUserNameTAG: santhosh1993 FirstChildCreateTimeTAG: 2012-09-21T13:45:01Z FirstChildTAG: Yes, the voltages are correct. Two resistors are not enough pro mix the two signals. Also, you can use fractional resistance values of 0 < R < 1 Ohm FirstChildUserIdTAG: 252722 FirstChildUserNameTAG: vaboro FirstChildCreateTimeTAG: 2012-09-21T15:09:55Z FirstChildTAG: Hi gt94539! Now that the deadline has passed you can take a look at [here][1] :) [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_S3E2_Linear_Combinations_of_Source_Strengths/threads/506a280d036af91f000000c9 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-02T00:28:16Z IndexTAG: 1634 TitleTAG: electrons flow in a Neutral & Line. What use of Earth. 1. Normally transformers are few KM away from our houses. If I consider electron at start of transformer, will it travel all the way to house & go back in a loop. 2. There are 3 Lines & 1 common neutral. Line voltages have 120 degree phase with each other & so be there respective currents. At neutral all three current merges since there is only one neutral. So total current in neutral is I= I1 at 0 + I2 at 120 + I3 at 240. Is it ? Just my stupid theory. Don't know what exactly happens. 3. What role of earthing in this. Is it just for protecting devices/human. 4. Normally there is some voltage b/w earth & neutral. Why is it , is there any special reason. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-21T10:37:44Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Neutral and earth are the same thing once you get into your circuit panel. 1. The extra uninsulated "ground" wire is an added safety device. It is connected to the metal around appliances etc, in the event that there is a fault, the electricity can travel straight to the ground rather then a person. 2. I don't think there should be any potential between ground and neutral, unless you have a bad connection somewhere. What kind of voltages are you reading? (The earth and neutral are connected together at the panel.) They should be virtually identical. I just measured mine, there is less then 1/10th of a volt between the ground and neutral, measured at a distant outbuilding. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-21T14:35:14Z SecondChildTAG: In 3 three pin socket in houses for 220V/110V ac , I checked there is around 2V b/w earth & neutral. Why is that voltage & how come 3 V come across SecondChildUserIdTAG: 137709 SecondChildUserNameTAG: DeepakBansal SecondChildCreateTimeTAG: 2012-09-22T10:16:15Z FirstChildTAG: I suppose the electrons don't travel in a loop from KM as you write to the house and back. They wobble because the voltage oscillates. What travels is the signal. It's like the wave on water: matter stays on its position but the wave goes in all directions outside from the source. FirstChildUserIdTAG: 252722 FirstChildUserNameTAG: vaboro FirstChildCreateTimeTAG: 2012-09-21T15:40:44Z SecondChildTAG: I am unsure where the return path really goes, although my guess is that it would follow the path of least resistance, which would be through the grounding plate buried beneath your home. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-21T16:55:42Z IndexTAG: 1635 TitleTAG: Internet blocked in China :( Can't watch youtube videos Hi, I'm in mainland China, where utube isn't accessible.. Any chance these videos will be reposted elsewhere? I've been using 'uueb freer' (rhymes with 'neb' starts with 'w') I posted it to mediafire http://www.mediafire.com/?0r5jm0mr7yijvpq You can download it and try it out, but it's not 100% reliable. Does anyone have a totally reliable solution for people in countries with restricted internet? Thanks for your help! UserIdTAG: 198563 UserNameTAG: PaulGillett CreateTimeTAG: 2012-09-21T08:52:39Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Did you find any solution to that youtube problem ? I'm suffering from it too. FirstChildUserIdTAG: 332941 FirstChildUserNameTAG: SabaSiddiqi FirstChildCreateTimeTAG: 2012-09-21T11:40:09Z FirstChildTAG: You can see in the Wiki [Lecture Videos Download][1]:). Also see more info [here][2]: **Due to the YouTube outage in certain parts of the world, we have provided direct links to download the lecture videos, and extended the Week 2 homework and lab deadlines to the coming Sunday (September 30). Please note that Week 3 deadlines have not changed.** [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/ [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/info FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-26T11:47:55Z IndexTAG: 1636 TitleTAG: homework 1 detailed answers (step tp step) H1P1 (1)3×R=3R (2)R/3 (3)R+1/(1/R+1/2R)=R+2R/3=5R/3 (4)![4+4+6=14][1] 4+4+6=14 (5)![enter image description here][2] 3/2 (6)2^2/(3/2)=8/3 H1P2 ![enter image description here][3] (1)3 (2)3-1=2 (3)3 (4)3-1=2 (5)#Variables-#independent equs.=4+4-(2+2)=4 euqs: -8+v2/1+(v2-4)/3=0 -(v2-4)/3+i4=0 ams: v1=7; v2=7; v3=3; i1=-8; i2=7; i3=1; i4=1; (6)7 (7)1 (8)-i1×v1=56 (9)-i4×v4=-4 (10)7^2/1=49 (11)3^2/3=3 H1P3 prob. description: Heater: 1270W 240V AC: 240V 60Hz Q1:When H1, H2, H3 on, I from power line=? ![enter image description here][4] R(heater)=Vh^2/Ph=240^2/1270=45.3543307087(ohm) Rh1||Rh2||Rh3=Rh/3=15.1181102362(ohm) I=Vh/(Rh1||Rh2||Rh3)=240/15.1181102362=15.875(A) Q2: Heater→120V others unchanged, I=? Rh=Vh^2/Ph=120^2/1270=11.3385826772(ohm) Rh1||Rh2||Rh3=Rh/3=3.77952755906(ohm) I=Vh/(Rh1||Rh2||Rh3)=120/3.77952755906=31.75(ohm) ![enter image description here][5] Q3: power E1 dissipated in H1? E1=I1^2*Rh I1=Vh/(Rh1+Rh2||Rh3)=2Vh/3Rh=3.52777777778(A) E1=I1^2*Rh1=564.444444445(W) Q4: power E2 dissipated in H2? E2=V2^2/Rh V2=Vh*[(Rh2||Rh3)/(Rh1+Rh2||Rh3)]=Vh/3=80(V) E2=141.111111111(W) Q5: total power ΣE=？ ΣE=Vh^2/(Rh1+Rh2||Rh3)=(2/3)*(Vh^2/Rh)=846.666666667(W) [1]: https://edxuploads.s3.amazonaws.com/13482130741833289.bmp [2]: https://edxuploads.s3.amazonaws.com/13482131391833219.bmp [3]: https://edxuploads.s3.amazonaws.com/1348213272804789.bmp [4]: https://edxuploads.s3.amazonaws.com/13482138971343604.bmp [5]: https://edxuploads.s3.amazonaws.com/13482145842691684.bmp UserIdTAG: 138445 UserNameTAG: jih43711 CreateTimeTAG: 2012-09-21T08:15:12Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1637 TitleTAG: What's my mistake? hello everyone, I just can't the last two questions right and as far as I know my procedure is good. what I to solve them is to first replace Ro=100 in the equation of the first question: $$R=\frac{VT}{I0* e^{\frac{VD}{VT}}}$$ and then I get the value of VD=.5694499416. after that I replace the value of VD on the diode equation: $$I0⋅(e^{vD/VT}−1)$$ and I get that the current throught the diode is ID=0.2599999990e-3. finally to get the value of VI I use the following equation: $$ VI=ID*(R+Ro)$$ and I get that for Ro=100 VI=1.04. Please help. UserIdTAG: 385046 UserNameTAG: SergioRod CreateTimeTAG: 2012-09-21T00:09:11Z VoteTAG: 2 CoursewareTAG: Week 4 / Linearization Exercise 2 CommentableIdTAG: 6002x_linearization_ex_2 NumberOfReplyTAG: 2 FirstChildTAG: Can we apply voltage divider to find Vi aftr calculating Vd? I got Vd but Vi doesnt come that way FirstChildUserIdTAG: 230347 FirstChildUserNameTAG: andrewwinney FirstChildCreateTimeTAG: 2012-10-02T05:06:29Z SecondChildTAG: We cant cause the incremental resistance doesnt deal with the very bias voltage. It only deals with fluctuations around the bias. If you only power the scheme with a bias voltaged DC source, you know the output. If you add another small source, the incremental resintance comes into the stage because now $V_i=V_{bias} + V_{fluctuation}$ SecondChildUserIdTAG: 190618 SecondChildUserNameTAG: Kirbabaev SecondChildCreateTimeTAG: 2012-10-02T17:31:41Z FirstChildTAG: 1. rd = VT/(I0*e^(vd/VT)), rd = 100; rd = 1000; 2. Find vd 3. VI = R I0 (e^(vd/VT)-1)+vd. FirstChildUserIdTAG: 197569 FirstChildUserNameTAG: darksamaro FirstChildCreateTimeTAG: 2012-09-21T07:26:55Z SecondChildTAG: thanks, the problem is that I used R to find VI. nice. SecondChildUserIdTAG: 385046 SecondChildUserNameTAG: SergioRod SecondChildCreateTimeTAG: 2012-09-22T00:45:01Z SecondChildTAG: I'm still getting the wrong answer here: rd=100 ==> vd=0.8089 What value am I supposed to be plugging in for R in step 3? If I plug in 100, I get VI=260.8 which is several orders of magnitude too large. the original R=3.9k is even worse. SecondChildUserIdTAG: 366165 SecondChildUserNameTAG: silicon_ghost SecondChildCreateTimeTAG: 2012-10-04T00:02:49Z SecondChildTAG: Ok, I think I figured out my problem. I got the R=V/I equation upside down somewhere in my calculations. rd=100 ==> vd=0.56945. darksamaro's equation now makes perfect sense. The voltage drop of a diode looks like an independent source in the circuit (hence the "+ vd" in darksamaro's equation 3). darksamaro equation #3 is simply VI=R * ID + vd, where ID is evaluated at vd. Or I could be totally wrong :) SecondChildUserIdTAG: 366165 SecondChildUserNameTAG: silicon_ghost SecondChildCreateTimeTAG: 2012-10-04T01:41:04Z IndexTAG: 1638 TitleTAG: Vth question Could someone explain how did he find the relationship (1k / (1k+1k)) * 1v on the fly? To me it's not that obvious. UserIdTAG: 270284 UserNameTAG: nkukushkin CreateTimeTAG: 2012-09-20T13:53:15Z VoteTAG: 2 CoursewareTAG: Week 2 / Thevenin Example CommentableIdTAG: 6002x_thevenin_example_1 NumberOfReplyTAG: 2 FirstChildTAG: Well, 1k+1k is 2k, and 1k / 2k is a half. It mostly come from experience - after enough practice you'll be able to do it too! FirstChildUserIdTAG: 151427 FirstChildUserNameTAG: RichardZ FirstChildCreateTimeTAG: 2012-09-20T18:55:51Z FirstChildTAG: If you have a voltage source and two equal resistors in series, AND you know that the voltage source will equal the total voltage drop across both resistors, AND you know that current into each resistor has to be the same; THEN the equations of voltage splitting will allow you to model the physics for all circuits that match the conditions. You might try using some of the special circuit math functions in the online calculator. "+" for resitances is pretty obvious. "||" for parallel resistance values is very useful FirstChildUserIdTAG: 88550 FirstChildUserNameTAG: Neil_S_Berry FirstChildCreateTimeTAG: 2012-09-21T02:06:33Z IndexTAG: 1639 TitleTAG: discussion forum how does we come to know when someone will give response to our question? earlier interface has "follow " section but now where to find it? UserIdTAG: 114864 UserNameTAG: Vikaash CreateTimeTAG: 2012-09-20T11:14:30Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: It's coming... the feature has been programmed, just waiting on a few things to get sorted out first. Edit: It's live now! FirstChildUserIdTAG: 323387 FirstChildUserNameTAG: ibrahimawwal FirstChildCreateTimeTAG: 2012-09-20T12:31:44Z IndexTAG: 1640 TitleTAG: Boosting and Amplification Sir, What is the difference between boosting and amplification of a signal ? UserIdTAG: 14726 UserNameTAG: Sajilck CreateTimeTAG: 2012-09-20T02:18:15Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: If you took a 1 volt AC signal and made it 10 volt AC, you could consider that amplification. Now with boosting, the signal may be brought up to 10 volt(DC), but the actual signal would only fluctuate +-1 volt AC(9-11 volt). It is still considered a "small signal", it just rides along the DC at a higher voltage potential. This is done at times so a device can be used in it's linear range which is usually desirable. I believe you could apply the term "bias" also to this situation. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-20T02:37:17Z SecondChildTAG: Thank you very much. So boosting is only the addition of DC shift right? 1.Then Can the DC shift be negative also? 2.Is clamping is same as boosting?, If yes then why we have different names for same concept?. If not how they differ ? SecondChildUserIdTAG: 14726 SecondChildUserNameTAG: Sajilck SecondChildCreateTimeTAG: 2012-09-25T05:18:56Z IndexTAG: 1641 TitleTAG: Lecture Terms This lecture was quiet confusing as there were some terms left unexplained like alpha,beta,how is beta is form of resisters n not alpha.And the last last equivalent thevn. network.Please help me in this UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-19T18:32:09Z VoteTAG: 2 CoursewareTAG: Week 2 / Thevenin method CommentableIdTAG: 6002x_thevenin NumberOfReplyTAG: 1 FirstChildTAG: $\alpha_m$ and $\beta_n$ are simply constants / coefficients depending on the circuit. This is a generic mathematical formula with $m$ and $n$ not always the same, so with two different names. If you take the S3V5 video, the final equation is : $e=\frac{R_2}{R_1+R_2}\cdot V+\frac{R_2 R_1}{R_1+R_2}\cdot I$ Now compare this formula to $v=\displaystyle\sum_m \alpha_m V_m+\sum_n \beta_n I_n$ *(I did not mention $Ri$ because it is external to the circuit - or given by the second term here)* In the S3V5 example there is only one voltage source which can be renamed $V_1$ and only one current source $I_1$. So you find $m=1$ and $n=1$. $e=\frac{R_2}{R_1+R_2}\cdot V_1+\frac{R_2 R_1}{R_1+R_2}\cdot I_1= \alpha_1 V_1+ \beta_1 I_1$ You find $\alpha_1=\frac{R_2}{R_1+R_2}$ and $\beta_1=\frac{R_2 R_1}{R_1+R_2}$ Another example could be S3E5: $v_3=a_1*V_1+a_2*V_2$. In this case, $m=2$ (you have two voltage sources) and $n=0$ (you have no current source). You have now: $\alpha_1=a_1$ and $\alpha_2=a_2$ FirstChildUserIdTAG: 396446 FirstChildUserNameTAG: RousseauxS FirstChildCreateTimeTAG: 2012-09-19T18:53:12Z SecondChildTAG: I was also wondering if the courseware would point out the relevance of Thevenin's general equation in regards to an actual circuit. Thanks for the excellent writeup/example. SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-09-20T14:02:39Z IndexTAG: 1642 TitleTAG: Found X1 but not so sure Hi everyone. I found the value of X1 doing this: after isolate V1, I make some simplifications with the resistor R2 and R3 and found an equivalent Resistor having 1.87 Ohm After that I just used the concept of Voltage Divider and my values was 2 * (1.87/(7+1.87)) = 0.4228. Seems to me that its right, but I can't apply the same technique to find X2. Can anyone tell me if I'm in the right direction? Thanks a lot. UserIdTAG: 316902 UserNameTAG: JesseTeixeira CreateTimeTAG: 2012-09-19T18:20:53Z VoteTAG: 2 CoursewareTAG: Week 2 / Superposition example CommentableIdTAG: 6002x_S3E3_superposition NumberOfReplyTAG: 2 FirstChildTAG: You're going ok. I too used voltage divider to solve it. FirstChildUserIdTAG: 254346 FirstChildUserNameTAG: moijes12 FirstChildCreateTimeTAG: 2012-09-19T18:29:55Z SecondChildTAG: I Finally found out what was wrong. I forget that when I isolate the V2, the resistors in parallel mode is just the R1//R3. I was messing up with this information. Thanks for your help. SecondChildUserIdTAG: 316902 SecondChildUserNameTAG: JesseTeixeira SecondChildCreateTimeTAG: 2012-09-19T18:53:11Z SecondChildTAG: To solve Y1 I just made that way: Y1 = (0.4228 - 2) / 7 = -0.2253 I know its the right answer (the value), but I'm afraid that its not the wai I should do this. SecondChildUserIdTAG: 316902 SecondChildUserNameTAG: JesseTeixeira SecondChildCreateTimeTAG: 2012-09-19T18:58:08Z SecondChildTAG: 1- turn off V2----> replace with a short circuit. 2- Make resistances simplifications to the circuit. 3-applay ohm's law "you'll get the Value in +ve, Anyway as the current going into the source in the direction of voltage drop , you put the -ve sign". SecondChildUserIdTAG: 185715 SecondChildUserNameTAG: amirengineer SecondChildCreateTimeTAG: 2012-09-20T20:37:22Z SecondChildTAG: & i think your way is also right SecondChildUserIdTAG: 185715 SecondChildUserNameTAG: amirengineer SecondChildCreateTimeTAG: 2012-09-20T20:54:17Z SecondChildTAG: sorry how did you get 1.87? SecondChildUserIdTAG: 133084 SecondChildUserNameTAG: Warrensiggs SecondChildCreateTimeTAG: 2012-09-23T12:56:54Z FirstChildTAG: seems reasonable. much better than my approach: i'm trying to use nodal analysis, setting node under the short circuited V2 to ground and top node to e. Then trying to do KCL on e getting wrong answers. [...] as i was writing this realised what i'd done wrong. i have the correct answer through node analysis. would like to understand the voltage divider approach too. FirstChildUserIdTAG: 149469 FirstChildUserNameTAG: wileyx FirstChildCreateTimeTAG: 2012-09-23T18:10:55Z IndexTAG: 1643 TitleTAG: What does the tick (check mark) mean? What does the tick (check mark) mean? That staff have answered the question or....? UserIdTAG: 154440 UserNameTAG: Aahlad CreateTimeTAG: 2012-09-19T15:57:32Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: The blue check mark on forum posts, means that the Staff endorses, or generally agrees with a response to a question. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-19T16:06:23Z SecondChildTAG: endorsed! SecondChildUserIdTAG: 151427 SecondChildUserNameTAG: RichardZ SecondChildCreateTimeTAG: 2012-09-19T16:31:19Z IndexTAG: 1644 TitleTAG: sandbox -how to measure current can any one tell me how to measure current, with the given system we can measure only voltage UserIdTAG: 352647 UserNameTAG: praveenjugge CreateTimeTAG: 2012-09-19T15:22:10Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: There are many ways to measure current if you have known voltages and resistor values. :) FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-19T15:31:07Z FirstChildTAG: Right at the bottom of the sandbox list of items is a line with an arrow through it. Use this in a circuit just like the Probe for voltages - ie in the TRAN mode. You can change the colours by clicking on it. Hope this makes sense Melbur FirstChildUserIdTAG: 95673 FirstChildUserNameTAG: melbur FirstChildCreateTimeTAG: 2012-09-19T15:47:13Z FirstChildTAG: Hi praveenjugge! I have answered you [here][1] ;) [1]: http://praveenjugge FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-20T04:57:00Z IndexTAG: 1645 TitleTAG: Exercise H3P3 Is the value of R correct ? It seems that 1k would be more reasonable. UserIdTAG: 169511 UserNameTAG: voxbcn CreateTimeTAG: 2012-09-19T14:08:39Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Troubleshooting NumberOfReplyTAG: 1 FirstChildTAG: You can almost do this one visually. If you look at the lower graph, you can clearly see the point of maximum wattage. Now, look straight up from that point to the chart that shows the current vs voltage curve. At that point apply ohms law to get the resistor value. There is even a bit of an error margin, you can pick an intersection of graph lines that is close to the spot on the curve and it seems to accept it. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-19T15:25:54Z SecondChildTAG: Thanks! I actually got confused on the first part with the given 100 Ohm resistor due to a typing error - but solved both parts now. SecondChildUserIdTAG: 169511 SecondChildUserNameTAG: voxbcn SecondChildCreateTimeTAG: 2012-09-19T17:46:36Z SecondChildTAG: pleae help me with this part!!!! SecondChildUserIdTAG: 414516 SecondChildUserNameTAG: randevakash SecondChildCreateTimeTAG: 2012-09-28T05:51:10Z SecondChildTAG: From the second graph of the question it is very much understood that the maximum power is at .3 Volts and is about .00027 W. Thus from The relation P=V^2/R R can be easily calculated. SecondChildUserIdTAG: 232499 SecondChildUserNameTAG: Asim09 SecondChildCreateTimeTAG: 2012-09-29T19:24:00Z IndexTAG: 1646 TitleTAG: Is it possible to rotate resistors in the circuit simulator? Is it possible to rotate resistors in the circuit simulator? The circuits either get unnecessarily large or look ugly when you have to make them all vertical. UserIdTAG: 154440 UserNameTAG: Aahlad CreateTimeTAG: 2012-09-19T08:37:43Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: press r FirstChildUserIdTAG: 436749 FirstChildUserNameTAG: waynebrown FirstChildCreateTimeTAG: 2012-09-19T09:03:22Z SecondChildTAG: Thanks a lot! SecondChildUserIdTAG: 154440 SecondChildUserNameTAG: Aahlad SecondChildCreateTimeTAG: 2012-09-19T09:14:42Z SecondChildTAG: I was thinking about this yesterday. Thanks! SecondChildUserIdTAG: 276409 SecondChildUserNameTAG: IgnacioUY SecondChildCreateTimeTAG: 2012-09-19T12:51:54Z IndexTAG: 1647 TitleTAG: For all effected by the youtube block go to face book join this group https://www.facebook.com/groups/483354858342165/488894611121523/ and there is a man who will send you the videos his name is sarwan kumar UserIdTAG: 436749 UserNameTAG: waynebrown CreateTimeTAG: 2012-09-19T08:25:41Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: please tell me how he send the videos? FirstChildUserIdTAG: 40209 FirstChildUserNameTAG: sadaf78 FirstChildCreateTimeTAG: 2012-09-19T11:06:12Z FirstChildTAG: now many students will have received videos from him so if they please can share videos with each other so that those people who cant access face book also get videos FirstChildUserIdTAG: 358539 FirstChildUserNameTAG: syed_abdullah FirstChildCreateTimeTAG: 2012-09-19T11:24:01Z FirstChildTAG: thx FirstChildUserIdTAG: 138095 FirstChildUserNameTAG: zobzyer FirstChildCreateTimeTAG: 2012-09-19T09:27:31Z IndexTAG: 1648 TitleTAG: Proof for KVL/KCL independent equations? STAFF It's stated in the textbook that the number of independent equations that you can get through KCL is `N - 1` and through KVL is `B - (N - 1)`. The first one makes sense intuitively, but the second doesn't (to me) so I was wondering if anyone knew how to prove these? UserIdTAG: 154440 UserNameTAG: Aahlad CreateTimeTAG: 2012-09-19T08:20:56Z VoteTAG: 2 CoursewareTAG: Week 1 / KVL and KCL CommentableIdTAG: 6002x_kvl_and_kcl NumberOfReplyTAG: 1 FirstChildTAG: B-(N-1) is the number of independent loops. And each loop equation is independent of the other. COnsider the example of the circuit in the video : B=6 ; N=4 Hence, no of independent loops = 6-(4+1) = 3 which is equal to the no. of independent equations. FirstChildUserIdTAG: 151751 FirstChildUserNameTAG: Senade FirstChildCreateTimeTAG: 2012-09-20T10:49:55Z IndexTAG: 1649 TitleTAG: To Myriam for Help..... Hi > @Myrimit , My Name is ASAD UR REHMAN. I belong form PAKISTAN. I registered myself for edx online courses CS50x & 6.002x. Unfortunately I forget to enter my Mailing Address Properly. Can u guide me "How can I change my Mailing Address?" Or Mailing Address doesn't matter in account? waiting for your favorable reply.. UserIdTAG: 329051 UserNameTAG: asadbhatti42 CreateTimeTAG: 2012-09-19T07:57:47Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi asadbhatti42! I suggest to send an e-mail to the edX Staff in order to solve your inconvenient [Help e-mail][1]: **System-related questions:** technical@edx.org UPDATED 24/10/12: They have added an option, you can edit your name, e-mail and reset password [here -dashboard][2]. [1]: https://www.edx.org/help [2]: https://www.edx.org/dashboard FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-19T11:29:04Z SecondChildTAG: I send e.mail 2times about 2,3 weeks ago but no reply from there... SecondChildUserIdTAG: 329051 SecondChildUserNameTAG: asadbhatti42 SecondChildCreateTimeTAG: 2012-09-20T05:20:58Z SecondChildTAG: Hi asadbhatti42! Sometimes they have a lot of requests and they delay to answer you and also says that migh not answer all the emails...In the last version of this Course we had an edit option in our Profile ...I don´t see it here in this new version ... I guess that might they will implement that soon (if you see [here][1] they say that you can change your account information in Progress ). [1]: https://www.edx.org/static/content-mit-6002x/handouts/SystemTools.401f97bf87ce.pdf SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-20T18:40:55Z SecondChildTAG: Thnx Myrimit They repond me on my post that I post on Fb Page of edX.. :) SecondChildUserIdTAG: 329051 SecondChildUserNameTAG: asadbhatti42 SecondChildCreateTimeTAG: 2012-09-21T11:31:39Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-21T19:45:58Z IndexTAG: 1650 TitleTAG: battery voltage What is the practical significance of negative voltage V2? UserIdTAG: 268966 UserNameTAG: avinash1947 CreateTimeTAG: 2012-09-19T06:35:49Z VoteTAG: 2 CoursewareTAG: Week 2 / Node equation review CommentableIdTAG: 6002x_S3E1_NodeEquationReview NumberOfReplyTAG: 1 FirstChildTAG: To understand how we should consider different connected polarities ) FirstChildUserIdTAG: 66669 FirstChildUserNameTAG: agarus FirstChildCreateTimeTAG: 2012-09-22T15:08:10Z IndexTAG: 1651 TitleTAG: correct VI for 100 Ohms, incorrect for 1000 Ohms Something strange is going on with my calculation. I've created a spreadsheet with necessary formulas to calculate vD, iD, RD and, finally VI. The answer (value for VI) with 'vd/id' equaling 100 Ohms matches the provided answer. However, changing 'vd/id' to 1000 gives the answer differing by ~10% from the provided one. That'd completely weird. If my spreadsheet is wrong how could it give correct value for the problem with value A and the wrong one for IDENTICAL problem with just one of the inputs changed to B? It's a closed-form solution, so numerical stability considerations aren't an issue. I guess, that's a rhetorical question anyway, since I can't easily provide the spreadsheet. Hopefully the answer will eventually present itself. Perhaps, as another A-HA moment. UserIdTAG: 321686 UserNameTAG: klisitsyn CreateTimeTAG: 2012-09-19T06:13:51Z VoteTAG: 2 CoursewareTAG: Week 4 / Linearization Exercise 2 CommentableIdTAG: 6002x_linearization_ex_2 NumberOfReplyTAG: 2 FirstChildTAG: If you've got an answer differing by 10%, it might be a rounding issue. If you solve the equation you got in the first part for $V_D$ (since you have an expression that $= \frac {v_d}{i_d} = 100\Omega \text { or } 1000 \Omega $) and realize that $V_D=V_O$ in the circuit. Once you have that you should just be able to plug it in to a simple KCL equation to get the solution. To be fair, using exact solutions with no rounding (as in, actually plugging the equations with given numbers directly into the answer boxes) gives some variation from the answers they give (but only by about 0.6%) I'd be happy to take a look at your spreadsheet if you want. You could post it on google docs, make it visible to anyone with the link, and post the link maybe? It's not homework so it shouldn't violate the honour code. FirstChildUserIdTAG: 248902 FirstChildUserNameTAG: Chanute FirstChildCreateTimeTAG: 2012-09-19T06:57:08Z SecondChildTAG: Thanks Chanute - this helped me find the answer for both 100 and 1000 ohms. I had already used most of your method without having read your post first, but fallen down at the end. Just curious, what was wrong with my final step? I had determined $V_O$ and then set about to use the resistor divider equation to find $V_I$ as follows: $V_O$ = $\frac{100 \cdot V_I}{100+R}$ Solving for $V_I$ gave me the wrong answer (very high). Using the same value for $V_O$ I then used KCL according to your suggestion and then got a different value and the green tick. SecondChildUserIdTAG: 2620 SecondChildUserNameTAG: dsd SecondChildCreateTimeTAG: 2012-09-20T00:02:59Z SecondChildTAG: Hello Chanute, thank you for your offer to take a look at the spreadsheet. It's here: [link to the spreadsheet][1]. As I've entered the data, I've noticed that my problem answer A, which is accepted, actually is wrong by the 3%. Initially I was under the impression that the answer gets accepted with much stricter tolerance. So my main question, "how it could be possible that the formulas give correct answer for A but not for B" is now addressed - it turns out the tolerances are pretty relaxed, and A is actually wrong, too. Now I can get off the hook of the bedazzlement of that A-accepted/B-wrong, and start thinking again. Thanks again! [1]: https://docs.google.com/spreadsheet/ccc?key=0AtjuTAnJ4DNWdHhZajBDVHBBT0hyZFNRUm5Yb2dJVGc SecondChildUserIdTAG: 321686 SecondChildUserNameTAG: klisitsyn SecondChildCreateTimeTAG: 2012-09-20T06:13:29Z SecondChildTAG: @dsd: I think the problem with using the voltage divider equation in that way is that it is trying to use some variables from the small signal model in combination with some values from the large signal biasing of the circuit. It still confuses me from time to time. The $ \frac {v_d}{i_d} = 100 \Omega$ approximation is only valid for small signal variations around the bias point, so you can't use the voltage divider equation which assumes the incremental resistance's i/v characteristic passes through the origin (which it most certainly does not!). You might be able to use it if you figured out the *actual* resistance the diode exhibits, similar to what klisitsyn did. @klisitsyn: Fixed it for you in a column beside. You simply had $I_O$ a factor of 10 too big =) Also, check out doing some of the algebra before you plug it into excel or add numbers: if you plug in the symbolic answer you get for $V_D$ you get a really simple expression for $i_D= \frac{V_T}{R_i}-I_O$ SecondChildUserIdTAG: 248902 SecondChildUserNameTAG: Chanute SecondChildCreateTimeTAG: 2012-09-21T23:16:21Z SecondChildTAG: @dsd I also made this mistake, but after that I realized that we looking for a specific slope, NOT resistance - that is, vd/id=1/slope=Rd SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-27T08:41:28Z FirstChildTAG: Hello, For the last two questions: Based on 1st answer, I calculated VD=VT*Ln(VT/(RD*I0)) Then I calculated ID=I0*(e^(VD/VT) -1) Then I got VI=R*ID + VD Hope this helps. FirstChildUserIdTAG: 384914 FirstChildUserNameTAG: fbarnel FirstChildCreateTimeTAG: 2012-09-24T23:04:15Z IndexTAG: 1652 TitleTAG: The description of the H2P1 is too complex to understand I am a Chinese. The description of the H2P1 is too complex for me to understand. I wholly confused what I should carry out. I hope the author can simplify the interpretation of the question so that the foreigner could follow the question clearly. Such as " Come up with resistors R1 and R2 such that the divider ratio Vout/Vin is within 10% of the requirement.". It would cause extremely difficulty to comprehend. UserIdTAG: 128276 UserNameTAG: ChengBin CreateTimeTAG: 2012-09-19T02:32:48Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: The first part of the question says that only some resistor values are available to buy. These are given in the table (the E12 set). The question asks you to calculate R1 and R2 values so that the output, Vout, is approximately 2V. The resistance values must be chosen so that the Thevenin resistance is between 10k-30k ohms. The resistor values you use must be those available in the table. After that you must calculate the maximum and minimum Vout when you take into account that the R values may be 10% different. Hope that helps a bit. FirstChildUserIdTAG: 398786 FirstChildUserNameTAG: cww501 FirstChildCreateTimeTAG: 2012-09-19T03:23:59Z SecondChildTAG: @cww501 I think Vout=20V not v.... Am I right? SecondChildUserIdTAG: 329051 SecondChildUserNameTAG: asadbhatti42 SecondChildCreateTimeTAG: 2012-09-19T05:53:36Z IndexTAG: 1653 TitleTAG: dimensional analysis In i=v^3, how can Amper be equal to Volt^3 ? UserIdTAG: 376468 UserNameTAG: jovan_serbia CreateTimeTAG: 2012-09-18T16:05:37Z VoteTAG: 2 CoursewareTAG: Week 4 / Graphs Exercise CommentableIdTAG: 6002x_graphs_exercise NumberOfReplyTAG: 2 FirstChildTAG: Technically the formula should be $i = Av^3$ where A is a constant with appropriate units which will take care of the dimensions. For the problem, $A = 1$ FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-09-18T17:32:06Z FirstChildTAG: I think that exercise is not aimed at applying dimensional analysis, but rather the graphical analysis of a circuit with nonlinear elements ... Obviously, missing a proportionality constant with appropriate units, for $ i = k \cdot v ^ 3 $, but this is not the purpose of the approach. FirstChildUserIdTAG: 129288 FirstChildUserNameTAG: Tinchito FirstChildCreateTimeTAG: 2012-09-18T19:09:01Z IndexTAG: 1654 TitleTAG: A composite voltage source is non-linear but is it a linear circuit? "We often say that a circuit containing only linear elements and independent sources is a "linear circuit." So, in the informal sense, a linear circuit is one where we can apply the Thevenin or Norton theorems to summarize the behavior at a pair of exposed terminals." From the statement above, can I say that a composite voltage source is a linear circuit? And when the composite voltage source is said to be non-linear, is it supposed to mean a non-linear element, and non-linear mathematically ? UserIdTAG: 146814 UserNameTAG: humink CreateTimeTAG: 2012-09-18T14:13:32Z VoteTAG: 2 CoursewareTAG: Week 2 / Is it linear? CommentableIdTAG: 6002x_is_it_linear NumberOfReplyTAG: 0 IndexTAG: 1655 TitleTAG: How can I calculate ev? Hello! In the video, when the current source was disconnected the ev had the value of "R2*V/(R1+R2)", Dr Agarwal said its just a pattern, but there is a way to discover this value? UserIdTAG: 291362 UserNameTAG: Gudson CreateTimeTAG: 2012-09-18T13:23:32Z VoteTAG: 2 CoursewareTAG: Week 2 / Superposition CommentableIdTAG: 6002x_superposition NumberOfReplyTAG: 2 FirstChildTAG: Try to calculate $e_{v}$ using KCL: $\frac{(e_{v}-V)}{R_{1}}+\frac{e_{v}}{R_{2}}=0$ FirstChildUserIdTAG: 343633 FirstChildUserNameTAG: lexder FirstChildCreateTimeTAG: 2012-09-18T16:10:58Z SecondChildTAG: You can also think about it in this way: The total resistance of this circuit is given by R1+R2, because they are in series. Then you can obtain the current provided by the voltage source: I=V/(R1+R2). Then, you know that ev is the voltage drop across R2, so ev=I*R2. Replacing I in this last expression you get: ev=V*(R2/(R1+R2)). SecondChildUserIdTAG: 349903 SecondChildUserNameTAG: jaumefilba SecondChildCreateTimeTAG: 2012-09-18T18:23:52Z SecondChildTAG: Whoa, thanks! It helped me a lot! SecondChildUserIdTAG: 291362 SecondChildUserNameTAG: Gudson SecondChildCreateTimeTAG: 2012-09-19T01:50:41Z FirstChildTAG: Sorry, I typed my last comment in the wrong place, and I can't find a way to change it. I was responding to Gudson's question. FirstChildUserIdTAG: 349903 FirstChildUserNameTAG: jaumefilba FirstChildCreateTimeTAG: 2012-09-18T18:34:04Z IndexTAG: 1656 TitleTAG: H2P2: I've run out of ideas how to solve this. I'm not even very sure what's being asked. Could someone suggest how to begin? . . . (Maybe I'd better be more specific.) I believe I can calculate the values for the circuit elements, but to maximize the power using R(L) as the variable seems to me like a linear equation. It's maximum would occur at zero resistance. This is pretty clearly not what's being asked, is it? UserIdTAG: 280220 UserNameTAG: pilgrimCycle CreateTimeTAG: 2012-09-18T12:39:02Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: Hi, :) "It's maximum would occur at zero resistance." - wrong! :( Maximum current over load resistor is not corresponding to maximum power dissipated by the resistor. (in this task RL is supposed to be variable , so maximum available current (in load) is not corresponding to maximum power in RL) FirstChildUserIdTAG: 277168 FirstChildUserNameTAG: AndrewKiselev FirstChildCreateTimeTAG: 2012-09-18T13:27:29Z FirstChildTAG: I'm only slightly more confused now . . . although I'm appreciative of the attention. Isn't power expressed by $P=V\cdot I$? FirstChildUserIdTAG: 280220 FirstChildUserNameTAG: pilgrimCycle FirstChildCreateTimeTAG: 2012-09-18T14:26:25Z SecondChildTAG: Yes, it's right! But, V (оn RL) not a constant and I (over RL)not a constant and R is not constant!! If you will using RL with different resistance you will see different V and I over RL (and different power!!) For example If you'll take very little resistor R=0.000001 Om it will like shortage of solar cell . So I will near to possible maxium, but V will near to 0.000 , and power will near to 0.00 too. If you'll take very big resistor R=10000000 Om solar cell will not loaded , so V will near possible maximum (of V on load) but I (over the load) will near 0. And power will near 0 againe. But if you'll take RL=250 Om (for example) you'll get power significantly greater. SecondChildUserIdTAG: 277168 SecondChildUserNameTAG: AndrewKiselev SecondChildCreateTimeTAG: 2012-09-18T15:23:06Z SecondChildTAG: Power = $V\cdot I$ or $I^2 \cdot R$ or $V^2 / R$ May be a small typo in is reply, unless I am missing something too, the rest holds true though. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-18T15:35:24Z FirstChildTAG: Power is $P=V\cdot I$ or $P = I^2 \cdot R$ and so to maximize power would require you to find the balance between maximum voltage and maximum current. FirstChildUserIdTAG: 151427 FirstChildUserNameTAG: RichardZ FirstChildCreateTimeTAG: 2012-09-18T13:50:08Z SecondChildTAG: RicardZ , From my point of view you make a missprint. P=V*I (and becouse V=I*R ) we have P=I*I*R. Or, maybe you want to write P=V*V/R and P=I*I*R SecondChildUserIdTAG: 277168 SecondChildUserNameTAG: AndrewKiselev SecondChildCreateTimeTAG: 2012-09-18T15:37:27Z SecondChildTAG: RicardZ , From my point of view you have made a misprint. P=V*I (and because V=I*R ) we have P=I*I*R. Or, maybe you wanted to write P=V*V/R and P=I*I*R (Excuse for my English...) SecondChildUserIdTAG: 277168 SecondChildUserNameTAG: AndrewKiselev SecondChildCreateTimeTAG: 2012-09-18T15:44:25Z SecondChildTAG: Whoops, my sincere apologies. Thanks for catching that one Andrew! SecondChildUserIdTAG: 151427 SecondChildUserNameTAG: RichardZ SecondChildCreateTimeTAG: 2012-09-18T19:45:02Z FirstChildTAG: http://en.wikipedia.org/wiki/Maximum_power_transfer_theorem#Calculus-based_proof_for_purely_resistive_circuits FirstChildUserIdTAG: 252722 FirstChildUserNameTAG: vaboro FirstChildCreateTimeTAG: 2012-09-18T20:15:14Z SecondChildTAG: thanks, that helped me a lot SecondChildUserIdTAG: 292360 SecondChildUserNameTAG: Loai SecondChildCreateTimeTAG: 2012-09-19T19:44:35Z SecondChildTAG: Thanks a lot ! SecondChildUserIdTAG: 370538 SecondChildUserNameTAG: samcore2804 SecondChildCreateTimeTAG: 2012-09-23T00:49:03Z SecondChildTAG: thanks alot SecondChildUserIdTAG: 285024 SecondChildUserNameTAG: nigelgomes SecondChildCreateTimeTAG: 2012-09-28T19:50:28Z FirstChildTAG: Thank you, good fellows, for the advice and company on this ride . . . I resorted at last to calculating and plotting possible solutions by hand! It did indeed work. FirstChildUserIdTAG: 280220 FirstChildUserNameTAG: pilgrimCycle FirstChildCreateTimeTAG: 2012-09-18T23:05:25Z IndexTAG: 1657 TitleTAG: Basic HTML doesn't work properly in the forum edX Folks - Basic HTML does not work properly in the forum. For example, using the break tag to create line breaks previews properly, but the actual post displays the raw tags without processing the breaks. Same story for the subscript tag, the raw tag is displayed in the post. Markup tags using single asterisk for italics, and double asterisk for bold, display properly. Examples: Break 1 Break 2 Break 3 Numbersub1 Numbersub2 Numbersub3 *Italic* **Bold** ***Bold Italic*** UserIdTAG: 94557 UserNameTAG: g_hopper CreateTimeTAG: 2012-09-18T12:06:05Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 2 FirstChildTAG: Thanks, we'll look into this. In the mean time, you shouldn't be needing to use HTML for anything. Anything placed between the dollar sign is interpreted as latex and marked up accordingly. For example $Number_{sub1}$ turns into $Number_{sub1}$ FirstChildUserIdTAG: 151427 FirstChildUserNameTAG: RichardZ FirstChildCreateTimeTAG: 2012-09-18T13:47:23Z FirstChildTAG: Thanks Richard. I know basic HTML, now I'm going to have to learn basic $La_{Tex}$. Or is it $L_a{T}_{ex}$? Where's Donald Knuth when you need him? ;) FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-09-18T20:53:47Z SecondChildTAG: Actually, you can use [Markdown](http://daringfireball.net/projects/markdown/syntax) for your formatting needs. It's a pretty simple mapping to HTML. I would avoid using $\latex$ for basic formatting. SecondChildUserIdTAG: 323387 SecondChildUserNameTAG: ibrahimawwal SecondChildCreateTimeTAG: 2012-09-18T22:58:09Z SecondChildTAG: For LaTeX, you can take a look at my tutorial : [Forum][1] or [Wiki][2] Don't hesitate to ask more commands ; ). [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/505611d8a717032b00000016 [2]: https://www.edx.org/wiki/6.002x/mathjax-tutorial-write-formula-forum SecondChildUserIdTAG: 396446 SecondChildUserNameTAG: RousseauxS SecondChildCreateTimeTAG: 2012-09-19T18:35:28Z IndexTAG: 1658 TitleTAG: troubles with lab2 I'm having trouble solving lab 2. I figured the circuit should look some thing like that (Ignore resistors Ohms)- ![lab2 circuit][1] [1]: https://edxuploads.s3.amazonaws.com/13479566135364216.jpg but this comes to finding 5 unknown resistors. and using node analysis I'm getting equations that will take about a week to solve. so I was wondering if I'm looking at this lab the wrong way. UserIdTAG: 271081 UserNameTAG: YonJah CreateTimeTAG: 2012-09-18T08:26:39Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: OK never mind. I was trying to build the wrong circuit. if someone have the same problem with this lab the most important hint I can give you is that there is an option to use ground for a reason FirstChildUserIdTAG: 271081 FirstChildUserNameTAG: YonJah FirstChildCreateTimeTAG: 2012-09-18T09:24:39Z SecondChildTAG: the circuit can be implemented by using only 3 resistors... SecondChildUserIdTAG: 183169 SecondChildUserNameTAG: JerinS SecondChildCreateTimeTAG: 2012-09-18T10:00:26Z SecondChildTAG: hii,will u please explain the concept of 3 resisitors.. SecondChildUserIdTAG: 171170 SecondChildUserNameTAG: rinutituschakkattil SecondChildCreateTimeTAG: 2012-09-18T18:31:20Z SecondChildTAG: Just draw two separate voltage dividers, when connect them. And you will see the shem SecondChildUserIdTAG: 195503 SecondChildUserNameTAG: andreytan SecondChildCreateTimeTAG: 2012-09-19T11:21:58Z SecondChildTAG: Hi rinutituschakkattil! Now that the deadline has passed you can take a look at [here][1]:) [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_S3E2_Linear_Combinations_of_Source_Strengths/threads/506a280d036af91f000000c9 SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-02T00:26:57Z IndexTAG: 1659 TitleTAG: help please! Staff, you... Is there any reference online where one can see research papers (original) of scientists. for example: If I wish to see the paper of Albert Einstein on his General relativity (a paper in his own writing...) Google Scholar, SCIRUS are a real mess. They show .................................................... many things UserIdTAG: 374590 UserNameTAG: ahope CreateTimeTAG: 2012-09-18T08:01:48Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I don't know about Einstein's papers but you can check out http://arxiv.org/ . In their own words - "arXiv.org is a highly-automated electronic archive and distribution server for research articles. Covered areas include physics, mathematics, computer science, nonlinear sciences, quantitative biology and statistics" FirstChildUserIdTAG: 311022 FirstChildUserNameTAG: GordanS FirstChildCreateTimeTAG: 2012-09-18T11:40:06Z SecondChildTAG: Thnxs! but it didn't work for me SecondChildUserIdTAG: 374590 SecondChildUserNameTAG: ahope SecondChildCreateTimeTAG: 2012-09-18T11:58:33Z SecondChildTAG: Sorry. I used this site few months ago, links worked well. SecondChildUserIdTAG: 192653 SecondChildUserNameTAG: Quaz SecondChildCreateTimeTAG: 2012-09-18T18:48:35Z FirstChildTAG: If you wish to see the paper of Albert Einstein:) you may look here http://www.alberteinstein.info/ . But I don't know about sites, for other scientists. I think trere must be some. FirstChildUserIdTAG: 192653 FirstChildUserNameTAG: Quaz FirstChildCreateTimeTAG: 2012-09-18T15:23:23Z SecondChildTAG: Thanks but...! SecondChildUserIdTAG: 374590 SecondChildUserNameTAG: ahope SecondChildCreateTimeTAG: 2012-09-18T16:25:42Z IndexTAG: 1660 TitleTAG: my teachers also brought a chainsaws but only on final exams=) UserIdTAG: 210380 UserNameTAG: MurdocRus CreateTimeTAG: 2012-09-18T06:53:52Z VoteTAG: 2 CoursewareTAG: Week 2 / Chain saw CommentableIdTAG: 6002x_chain_saw NumberOfReplyTAG: 0 IndexTAG: 1661 TitleTAG: Remember S3E2 Do you remember S3E2 exercise? So, well, you can find that x1 = a1*V1 and x2 = a2*V2 Then y1 = X1/R1 and y2 = x2/R1 or viseversa. And y1 is negative because the current direction. UserIdTAG: 184827 UserNameTAG: DiegoT CreateTimeTAG: 2012-09-17T23:17:14Z VoteTAG: 2 CoursewareTAG: Week 2 / Superposition example CommentableIdTAG: 6002x_S3E3_superposition NumberOfReplyTAG: 1 FirstChildTAG: the result of y2 i cant obtain i reached to result=1.3 FirstChildUserIdTAG: 295983 FirstChildUserNameTAG: qassam FirstChildCreateTimeTAG: 2012-09-18T09:11:03Z SecondChildTAG: First obtain X2. V1: short circuit. Use node method. The node is in the top of R3 branch. Reference node in the bottom of the branch. Once you get X2, then Y2=X2/1 SecondChildUserIdTAG: 184827 SecondChildUserNameTAG: DiegoT SecondChildCreateTimeTAG: 2012-09-18T19:45:20Z SecondChildTAG: try this one y1= V1*(1/r1+1/r2+1/r3)/(1/r2+1/r3) SecondChildUserIdTAG: 178082 SecondChildUserNameTAG: muhammedvaseem007 SecondChildCreateTimeTAG: 2012-09-22T14:49:28Z IndexTAG: 1662 TitleTAG: layout of variables in examples and questions it may be just me but i have difficultly equating the variables to the values given, for example in S3E4 VS=5.0V, R1=56000.0Ω, and R2=18000.0Ω. would be much easier to read as: VS=5.0V R1=56000.0Ω R2=18000.0Ω. think this is due to being dyslexic and having trouble getting the info out of writing in this form UserIdTAG: 388273 UserNameTAG: kilmarta CreateTimeTAG: 2012-09-17T20:54:04Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 0 IndexTAG: 1663 TitleTAG: Why does the DC offset turn the volume up? I am wondering why applying a DC offset seems to turn up the volume... I had the idea that speakers cannot reproduce DC components... I thought the volume depended on the peak to peak signal which is applied to the speaker, but somehow this DC offset is also making the music louder. Any idea? UserIdTAG: 370084 UserNameTAG: jmiguelsalgado CreateTimeTAG: 2012-09-17T18:43:05Z VoteTAG: 2 CoursewareTAG: Week 4 / Incremental Analysis Demo CommentableIdTAG: 6002x_inc_analysis_demo NumberOfReplyTAG: 2 FirstChildTAG: In this particular case (professor Agarwal's expoDweep :)), notice that the slope of the function $i_D=f(v_D)=ae^{bv_D}$ increases as $v_D$ increases, thus the small signal response current $i_d$ is more "stretched" for higher values of operating volume. Hope this helps. FirstChildUserIdTAG: 338813 FirstChildUserNameTAG: djapens FirstChildCreateTimeTAG: 2012-09-17T19:26:45Z SecondChildTAG: It does really help :), thanks! SecondChildUserIdTAG: 370084 SecondChildUserNameTAG: jmiguelsalgado SecondChildCreateTimeTAG: 2012-09-18T20:04:35Z FirstChildTAG: How the music can be applied as an input voltage (V(t))? Music is sound and not a voltage source. How the prof. change it to a voltage from a CD player? FirstChildUserIdTAG: 374393 FirstChildUserNameTAG: rmaleki FirstChildCreateTimeTAG: 2012-09-24T07:42:20Z SecondChildTAG: Music is an accoustic wave. This wave can be generated eg. inside the speaker which converts variable voltage into oscillations of the membrane what affects the pressure change in an environment which can be named accoustic wave. Have a look eg. on this page: http://en.wikipedia.org/wiki/Loudspeaker SecondChildUserIdTAG: 176228 SecondChildUserNameTAG: PawelB1 SecondChildCreateTimeTAG: 2012-09-27T18:27:50Z SecondChildTAG: Sound waves can be converted to electrical signals using transducers: http://en.wikipedia.org/wiki/Transducer You'll note that microphone is listed under "Electroacoustic" on the Wikipedia page. The professor is replaying a previously recorded voltage signal that stores the information that was in the pressure wave when it recorded the song. I hope this helps :) SecondChildUserIdTAG: 13614 SecondChildUserNameTAG: analognoise SecondChildCreateTimeTAG: 2012-10-10T02:55:26Z IndexTAG: 1664 TitleTAG: H2P1 Isn't the Voltage Divider 50*R2/(R1+R2)=20? So R1= 1.5*R2, right? Or am I getting it all wrong? UserIdTAG: 19774 UserNameTAG: Gor131313 CreateTimeTAG: 2012-09-17T18:19:36Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You are right about the ratio, and should be able to get the right Vmin and Vmax, however I am too having problems getting the correct values for R1 and R2 themselves. Do they have to be from the E12 set? I've tried that too, and I suppose they should be in kilo-ohms, but can't get the right R1 and R2 values, even though I got the right Vmin and Vmax with those 'incorrect' values. FirstChildUserIdTAG: 395349 FirstChildUserNameTAG: bgr FirstChildCreateTimeTAG: 2012-09-17T20:46:57Z IndexTAG: 1665 TitleTAG: Why 4 loops and not 7 loops In the prev discussion it was mentioned that for this circuit we can consider 7 loops. But in this case only four is considered. Why so? UserIdTAG: 87309 UserNameTAG: Vijayenthiran CreateTimeTAG: 2012-09-17T17:56:44Z VoteTAG: 2 CoursewareTAG: Week 1 / KVL and KCL CommentableIdTAG: 6002x_kvl_and_kcl NumberOfReplyTAG: 1 FirstChildTAG: Hello Vijayenthiran, well...even I was wondering about this.But I guess, its true that there are 7 loops but out of which 4 loop has been used in this solution to solve the numerical completely.Since its suffiecient,in this example, with 4 loops to equate all the unknowns and get our answer, that's why.Out of the 7, only 3 of the loops are independent. FirstChildUserIdTAG: 156225 FirstChildUserNameTAG: sanjana_m FirstChildCreateTimeTAG: 2012-09-23T06:57:20Z IndexTAG: 1666 TitleTAG: Personal info!!! How to change,update or edit personal details after log in to edx ? Please post some kind information ...... UserIdTAG: 295278 UserNameTAG: souravfn7 CreateTimeTAG: 2012-09-17T17:44:42Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I don't think you can right now, edX platform is still fresh and they'll probably add the functionality soon. FirstChildUserIdTAG: 395349 FirstChildUserNameTAG: bgr FirstChildCreateTimeTAG: 2012-09-17T20:48:49Z IndexTAG: 1667 TitleTAG: lab 2 can anyone plzz help me with lab 2 UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-17T14:47:54Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Please give us some info about what areas where you need help. We can't do your work for you, but we can give you some hints and advice. FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-09-18T02:20:10Z IndexTAG: 1668 TitleTAG: Vs the value of Vs is 5v but why are you saying that Vs is 10 because if Vs is 10 Vout will be equal to 0.90909 UserIdTAG: 286165 UserNameTAG: nchibwe CreateTimeTAG: 2012-09-17T12:45:56Z VoteTAG: 2 CoursewareTAG: Week 3 / Switch Resistor Model of a MOSFET CommentableIdTAG: 6002x_switch_resistor_model NumberOfReplyTAG: 1 FirstChildTAG: Agree, Vs should be 5V and not 10V. It is a mistake FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-17T13:22:54Z IndexTAG: 1669 TitleTAG: Just a matter of signs... The current through R1 is equal to e minus 5V divided by R1. The current through R2 is ***equal to e minus the minus 7.2*** or, plus 7.2, divided by R2. So we have: (e-5)/(6800) + (e-(-7.2))/(5600) = 0, or (e-5)/(6800) + (e+7.2)/(5600) = 0 UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-09-17T11:22:30Z VoteTAG: 2 CoursewareTAG: Week 2 / Node equation review CommentableIdTAG: 6002x_S3E1_NodeEquationReview NumberOfReplyTAG: 1 FirstChildTAG: If You reslolve this, You will not get 6.2 V. You will get -1.69. I have the same equation, like your first - (e-5)/(6800) + (e-(-7.2))/(5600) = 0 FirstChildUserIdTAG: 413784 FirstChildUserNameTAG: Dorotka FirstChildCreateTimeTAG: 2012-09-17T12:21:24Z SecondChildTAG: Is there scheme changed? SecondChildUserIdTAG: 413784 SecondChildUserNameTAG: Dorotka SecondChildCreateTimeTAG: 2012-09-17T12:35:59Z SecondChildTAG: I think, I get it. Just source i rotated and value is -. It`s like source will be not rotated and value +. SecondChildUserIdTAG: 413784 SecondChildUserNameTAG: Dorotka SecondChildCreateTimeTAG: 2012-09-17T13:36:58Z SecondChildTAG: I agree I resolved that to -1.69 myself, but I modeled the circuit in the sandbox and it shows 6.2V at that node so I must be doing something wrong... Back to the textbook ;-) SecondChildUserIdTAG: 334127 SecondChildUserNameTAG: EHarrigan SecondChildCreateTimeTAG: 2012-09-17T13:57:22Z SecondChildTAG: thinking in e as a node the current that come from v2 is equal the current that goes to v1 so (e-5)/6800=(7.2-e)/5600 so e=6.2V SecondChildUserIdTAG: 152106 SecondChildUserNameTAG: JoaoBR SecondChildCreateTimeTAG: 2012-09-17T22:30:04Z IndexTAG: 1670 TitleTAG: HW1? Hello everyone! I can not tell of solutions, because I resolve the problem again and again but no thing goes well. I am talking about the second problem in homework 1. When he ask me to find the value of v1 I used the KCL/KVL method but nothing is correct! my solution: First: i1 + i2 + i3 =0 ---(1) and v1 = v2 and v4 + v3 - v2 = 0 Then: v1 = v2 <=> v1 = R2 * i2 <=> i2 = v1 / R2 ---(2) Also: v4 + v3 - v2 = 0 <=> v3 = v2 - v4 <=>(by replacing v2 by v1) v3 = v1 - v4 <=> R3 * i3 = v1 - v4 <=> i3 = v1 / R3 - v4 / R3 ---(3) So by using (2) and (3) in the (1) we will find: i1 + (v1 / R2) + (v1 / R3 - v4 / R3) = 0 <=> v1 / 1 + v1 / 3 = 4 / 3 - 8 <=> 4 * v1 / 3 = -20 / 3 <=> v1 = -20 / 3 * 3 / 4 <=> v1 = -5 but its not correct when I submit it! Why? +5 do not work either. UserIdTAG: 285398 UserNameTAG: mohessaid CreateTimeTAG: 2012-09-17T05:53:51Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: the "<=>" thing is the "<=>" the script that threat comments conert the <> to there code to avoid ambiguity in with the html of page FirstChildUserIdTAG: 285398 FirstChildUserNameTAG: mohessaid FirstChildCreateTimeTAG: 2012-09-17T05:57:10Z FirstChildTAG: I got the same results as you. I am curious to find out what the correct way to do the problem was. -Gwen FirstChildUserIdTAG: 311920 FirstChildUserNameTAG: ghowell FirstChildCreateTimeTAG: 2012-09-17T07:02:40Z FirstChildTAG: This is wrong: First: i1 + i2 + i3 =0 ---(1) And the reason is, that i1 is delivered by a current source, so i1 is always independent of i2 and/or i3. You have to sum the currents through the node, but according to the diagram, all the currents flow out the node and that is impossible, so one has at least to flow into the node So you should write: First: i1 = (i2 + i3) => i1 - (i2 + i3)= 0 ---(1) or -i1 = -(i2 + i3) => -i1+(i2 + i3)= 0 ---(1)If you work that out, than you will get the right answer. Your other calculations are ok. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-09-17T11:34:04Z IndexTAG: 1671 TitleTAG: The great lecture example that is! I wish I had lectures like that while studying in Bauman university. If they were as finely structured like these and had such remarkable, spectacular and funny examples of theory application, the discipline would have been way more fun and way more memorizable than it was (we had some examples, in the laboratories, that had AC and DC engines, but we were super restricted on what we could do there). So, thanks a lot for the course! A number of my friends, who did not understand anything in electronics at all are also into the course and finally started to understand, how things work! Many thanks from them also! UserIdTAG: 196898 UserNameTAG: Undead CreateTimeTAG: 2012-09-17T03:04:24Z VoteTAG: 2 CoursewareTAG: Week 2 / Chain saw CommentableIdTAG: 6002x_chain_saw NumberOfReplyTAG: 0 IndexTAG: 1672 TitleTAG: Solve this way Find out V3 in terms of V1,V2,R1,R2 and R3. Use Node Analysis, once you find it, do the same with i1 and i3((V3-V1)/R1 for i1 and (V3-V2)/R2 for i2).a1, a2, ... are algebraic expressions in terms of R1,R2 and R3. UserIdTAG: 225977 UserNameTAG: diego_ruiz CreateTimeTAG: 2012-09-17T00:48:43Z VoteTAG: 2 CoursewareTAG: Week 2 / Linear combinations of source strenghts CommentableIdTAG: 6002x_S3E2_Linear_Combinations_of_Source_Strengths NumberOfReplyTAG: 0 IndexTAG: 1673 TitleTAG: Negative resistance in Lab 1 When I solve for R1 and R2 in Lab 1, I get negative values for the resistance. The system marks it as a correct answer, but that doesn't seem right to me. Is it valid to have a negative resistance in this model or not? Thank you! UserIdTAG: 346777 UserNameTAG: fishmael CreateTimeTAG: 2012-09-16T21:11:31Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: No. Negative resistances are likely impossible within the lumped matter discipline. Your intuition serves you well. FirstChildUserIdTAG: 108926 FirstChildUserNameTAG: kayakMike FirstChildCreateTimeTAG: 2012-09-16T21:27:02Z FirstChildTAG: Negative resistances do exist, but not in these simple resistor networks. http://en.wikipedia.org/wiki/Negative_resistance You probably have a sign problem on one of your voltages, or you're looking at a current direction backwards. FirstChildUserIdTAG: 131970 FirstChildUserNameTAG: elucify FirstChildCreateTimeTAG: 2012-09-17T03:49:21Z IndexTAG: 1674 TitleTAG: Precesion in Answer ?? in H 2 P2 admin please clarify the precision of digits in this Question :( UserIdTAG: 415376 UserNameTAG: imab90 CreateTimeTAG: 2012-09-16T18:22:40Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: For the numbers I got (the values of resistance and the current source are randomized so they could be different for you), I didn't get recurring decimals in my answers. If you feel your answer is incorrect due to precision issues, try using fractions. The answer box accepts that. So for example, if the answer to the first box is 1/3, you can enter 1/3 instead of 0.333. The answer box also accepts the parallel operation. So if you find that the equivalent resistance is the parallel of a $1\Omega$ and a $2\Omega$ resistance, you can enter 1||2 in the answer box.. FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-09-16T19:02:24Z SecondChildTAG: thankss ...... SecondChildUserIdTAG: 415376 SecondChildUserNameTAG: imab90 SecondChildCreateTimeTAG: 2012-09-16T19:09:29Z FirstChildTAG: I am also frustrated on this one. Wasted 20mins on this, as it did not accept my 3 digit answer, only when I entered 4, after recalculating 4 times, and reading all comments here. Pls dear edX staff, clarify precisely what kind of answer do you want... FirstChildUserIdTAG: 329254 FirstChildUserNameTAG: KGabor FirstChildCreateTimeTAG: 2012-09-21T12:32:17Z IndexTAG: 1675 TitleTAG: Staff: So the check box at the end of each section in the homework : does hitting any one of these submit ALL the homework or just the section you have finished for submission? You can only submit ONCE so it is very important to submit the maximally correct homework. N'est ce pas? And I am thinking why isn't the button marked "submit" like the button on this page? But perhaps I haven't located the handout that explains all this. In which case accept my apologies in advance. UserIdTAG: 241170 UserNameTAG: Mauette CreateTimeTAG: 2012-09-16T17:48:43Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I am not staff, but this recent thread may answer your questions. https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50560f50578a8d2300000035 FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-16T18:17:02Z SecondChildTAG: i started that thread - I read your answer. My point is you can check your answer ad infinitum by hitting the button over and over, yes- but the first check is what submits therefore the check button needs to be SUBMIT as that is actually whats going on. And I would prefer not having to guess about what the button does. SecondChildUserIdTAG: 241170 SecondChildUserNameTAG: Mauette SecondChildCreateTimeTAG: 2012-09-16T18:21:20Z SecondChildTAG: [EdX Overview][1] The first exercise in that lecture sequence (The Overview sections) explains that the check button also submits the answers for grading. I think the name "Check" makes sense because you are actually checking your answers. You have unlimited attempts to get it right. I would suggest not getting used to it however as the exams will give you just three attempts and that can be a big pain if you are used to checking results too many times. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Overview/edx_introduction/ SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-09-16T18:46:47Z SecondChildTAG: are you saying you also have unlimited chances to have your homework graded until it is correct???? SO I check and it says all are wrong, I redo the HW and then I check and it says none are worng are you saying the grade applied will be the best grade?? SecondChildUserIdTAG: 241170 SecondChildUserNameTAG: Mauette SecondChildCreateTimeTAG: 2012-09-16T18:59:34Z SecondChildTAG: Yes that's right. They've done this because many of the homework problems guide you. Suppose a particular question has 4 answer boxes. In many cases, each answer box will depend on the answer of the previous box. So you can work through the problem in steps. But like I said, try not getting used to it if possible. SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-09-16T19:11:50Z SecondChildTAG: wow thanks - good to know. SecondChildUserIdTAG: 241170 SecondChildUserNameTAG: Mauette SecondChildCreateTimeTAG: 2012-09-16T20:59:43Z IndexTAG: 1676 TitleTAG: Dont understand Hello mi question is why the power (in Watts) supplied by the voltage source (-2.8864) is negative and the power (in Watts) supplied by the current (23.3) source is positive UserIdTAG: 163395 UserNameTAG: jasonlll88 CreateTimeTAG: 2012-09-16T17:46:34Z VoteTAG: 2 CoursewareTAG: Week 1 / KVL/KCL example CommentableIdTAG: 6002x_S2E3_Using_KVL_KCL_and_VI_constraints NumberOfReplyTAG: 3 FirstChildTAG: In that circuit power is being supplied by the current source (Current and voltage are positive here) and power is being absorbed by the voltage source so it's negative (Current is negative here). You better go through the **associated variables convention** carefully. FirstChildUserIdTAG: 95058 FirstChildUserNameTAG: KeshawP FirstChildCreateTimeTAG: 2012-09-16T18:01:38Z SecondChildTAG: Conceptually, figure that power is being spoken of with positive power being power released by the system (heat). Since the V and I sources are adding power to the system, they are negative. SecondChildUserIdTAG: 154526 SecondChildUserNameTAG: silentquasar SecondChildCreateTimeTAG: 2012-09-17T05:38:55Z SecondChildTAG: How is the voltage source **adding** power to the system? The powers don't add up. 12.1 W + 8.35 W + (-2.89 W) + 23.3 W =/= 0 W However, if the voltage source **consumes** power (is positive) and the current source is adding power (is negative) then it both makes sense conceptually and the powers add up. 12.1 W + 8.35 W + 2.89 W + (-23.3) W = 0 W SecondChildUserIdTAG: 483746 SecondChildUserNameTAG: RyanToll SecondChildCreateTimeTAG: 2012-09-24T01:11:58Z SecondChildTAG: Never mind, I just read it backward. The wording switches it around. I was keeping positive power as consumption and negative power as production. However, if it asks how much power was **added**, then it switches the sign around. :P SecondChildUserIdTAG: 483746 SecondChildUserNameTAG: RyanToll SecondChildCreateTimeTAG: 2012-09-24T03:28:30Z FirstChildTAG: This schema looks like battery (V) charging from the current source and in this case negative power suppliance means consumption. Guess when i1 appeared negative this could be a key to conclusion of consumption. FirstChildUserIdTAG: 66669 FirstChildUserNameTAG: agarus FirstChildCreateTimeTAG: 2012-09-17T16:45:05Z FirstChildTAG: In the circuit as current I1 is negative this shows current is flowing into voltage source.so power is absorbed by voltage source. FirstChildUserIdTAG: 66364 FirstChildUserNameTAG: maheshwar FirstChildCreateTimeTAG: 2012-09-23T04:20:49Z IndexTAG: 1677 TitleTAG: MID TERM EXAM WILL OUR MID TERM EXAM BE ON 25TH OCT OR CAN WE GIVE THE EXAM ON ANY DAY IN THAT WEEK AS PER OUR CONVINIENCE UserIdTAG: 126911 UserNameTAG: sidhant7 CreateTimeTAG: 2012-09-16T14:14:03Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: 25th FirstChildUserIdTAG: 169784 FirstChildUserNameTAG: thewiredbear FirstChildCreateTimeTAG: 2012-09-16T14:42:19Z FirstChildTAG: Can we have any kind of sample test for mid term and finals?? FirstChildUserIdTAG: 330759 FirstChildUserNameTAG: Osho FirstChildCreateTimeTAG: 2012-09-16T15:08:29Z SecondChildTAG: View wiki of MITx You can find link in edx wiki SecondChildUserIdTAG: 374590 SecondChildUserNameTAG: ahope SecondChildCreateTimeTAG: 2012-09-16T15:20:14Z IndexTAG: 1678 TitleTAG: wrong word at transcription At 2:13 minutes, the transcriptions says "decomposition rules" instead of "composition rules", it's a detail, but there is. UserIdTAG: 149549 UserNameTAG: Java_Dido CreateTimeTAG: 2012-09-16T12:13:44Z VoteTAG: 2 CoursewareTAG: Week 2 / Linearity CommentableIdTAG: 6002x_linearity NumberOfReplyTAG: 0 IndexTAG: 1679 TitleTAG: How Can I change my Mailing Address given in edX account? hi, if anyone from staff or technical member can help me How can I change my ''Mailing Address'' that demanded on 'edX' account? UserIdTAG: 329051 UserNameTAG: asadbhatti42 CreateTimeTAG: 2012-09-16T10:52:03Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I want to change my full name.. FirstChildUserIdTAG: 136959 FirstChildUserNameTAG: christerpher FirstChildCreateTimeTAG: 2012-09-19T08:03:13Z IndexTAG: 1680 TitleTAG: Request for Detailed Solutions to S2E3 and S2E5 Those of us who are having some difficulty might well benefit from an excruciatingly detailed solution to the problems S2E3 and S2E5. I know that the current sources are giving me the most difficulty; I could use some help in this area. UserIdTAG: 194450 UserNameTAG: wrbuckley CreateTimeTAG: 2012-09-16T08:45:35Z VoteTAG: 2 CoursewareTAG: Week 1 / KVL/KCL example CommentableIdTAG: 6002x_S2E3_Using_KVL_KCL_and_VI_constraints NumberOfReplyTAG: 1 FirstChildTAG: S2E3 is detailed in section 2.5 of the text, p. 95. FirstChildUserIdTAG: 154526 FirstChildUserNameTAG: silentquasar FirstChildCreateTimeTAG: 2012-09-17T04:34:14Z IndexTAG: 1681 TitleTAG: Resistance equation as algebra in 6.002X I am not possible to write equivalent circuit in algebraic equations. Pls help me today is the last date of h.w submission. UserIdTAG: 396372 UserNameTAG: vsnarendran CreateTimeTAG: 2012-09-16T04:44:14Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Hi All, First of I would like to say greetings to all of you. Your posts are most helpful and it is great to see so may people diving in to this programme. Secondly I'm having an issue with the homework assignment...specifically the entry of algebraical expressions for the all the Network types. Every time I enter one it says cannot parse express. I'm not looking for your answers just a solution for what this system will parse. Thanks, Scorliss FirstChildUserIdTAG: 267537 FirstChildUserNameTAG: scorliss FirstChildCreateTimeTAG: 2012-09-16T04:50:41Z SecondChildTAG: Common mistakes: - Write 5*R instead of 5R - Don't include the equals sign of an equation. Just "one side" of the equation is asked for. SecondChildUserIdTAG: 79361 SecondChildUserNameTAG: kimt SecondChildCreateTimeTAG: 2012-09-16T04:54:26Z SecondChildTAG: Thank you for your quick reply. I will try the "one sided" equation. SecondChildUserIdTAG: 267537 SecondChildUserNameTAG: scorliss SecondChildCreateTimeTAG: 2012-09-16T04:56:30Z SecondChildTAG: When like terms are used in the diagrams....such as 3 elements named R is there a standard that we can use where by the system will parse it correctly? like R1+R2 or R(1)+R(2)? SecondChildUserIdTAG: 267537 SecondChildUserNameTAG: scorliss SecondChildCreateTimeTAG: 2012-09-16T05:08:08Z SecondChildTAG: They're all R, without distinction of R1, R2, etc. (That will come later) SecondChildUserIdTAG: 79361 SecondChildUserNameTAG: kimt SecondChildCreateTimeTAG: 2012-09-16T05:52:31Z SecondChildTAG: Ok thank you Scorliss SecondChildUserIdTAG: 396372 SecondChildUserNameTAG: vsnarendran SecondChildCreateTimeTAG: 2012-09-16T06:49:02Z FirstChildTAG: In making an algebraic equation it is best that you write in paper first using the circuit analysis concepts in the textbook or in the week1 notes. Make sure that you are mindful of the the parenthesis quotations in the equation. For example: voltage equations we have v1 = v2+v3 with i2 passing through r2 and i3 passing through r3 we can find the equation for v1 we have: (i2*R2)+(i3*R3) Please see S1E6 for reference. :) FirstChildUserIdTAG: 209082 FirstChildUserNameTAG: Nikemurton FirstChildCreateTimeTAG: 2012-09-16T04:54:10Z SecondChildTAG: Thanks SecondChildUserIdTAG: 267537 SecondChildUserNameTAG: scorliss SecondChildCreateTimeTAG: 2012-09-16T04:56:46Z SecondChildTAG: ok ok thanks, nikemurton . SecondChildUserIdTAG: 396372 SecondChildUserNameTAG: vsnarendran SecondChildCreateTimeTAG: 2012-09-16T06:58:40Z FirstChildTAG: which question r u referring to FirstChildUserIdTAG: 233221 FirstChildUserNameTAG: JPShukla FirstChildCreateTimeTAG: 2012-09-16T06:22:01Z IndexTAG: 1682 TitleTAG: Unable to upload images - "TypeError: data is undefined" Hey guys, I am unable to upload images into my posts from my computer. I can use URLs from other websites, it's just the uploads that result in a dialog box saying "TypeError: data is undefined" P.S. You and your team are doing an amazing job handling all of this on the fly. Have a great weekend. UserIdTAG: 292543 UserNameTAG: Pennypacker CreateTimeTAG: 2012-09-16T00:23:18Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Troubleshooting NumberOfReplyTAG: 6 FirstChildTAG: Thank you for letting us know, and for putting it in the Troubleshooting section. :-) 1. What browser are you using? 2. What kind of image file are you uploading (JPEG, PNG, TIFF?) 3. How big is your image file? Thank you. FirstChildUserIdTAG: 11 FirstChildUserNameTAG: dormsbee FirstChildCreateTimeTAG: 2012-09-16T01:23:55Z SecondChildTAG: Opps, I am on Windows 7 with Firefox 15.0.1 SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-16T03:07:44Z SecondChildTAG: I tried bmp and Jpegs of unknown size, I also tried a GIF that was 7KB SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-16T03:10:27Z SecondChildTAG: ![enter image description here][1] [1]: http:// SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-16T03:11:06Z SecondChildTAG: ![test][1] [1]: http:// SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-16T03:11:38Z SecondChildTAG: I just tried to upload a couple of small PNGs, the error did not show up, but the posts above are blank. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-16T03:12:28Z SecondChildTAG: Yup, we can reproduce the problem using Firefox. We'll create a fix, but it's probably not going to come out this weekend. Thank you for reporting this to us. :-) SecondChildUserIdTAG: 11 SecondChildUserNameTAG: dormsbee SecondChildCreateTimeTAG: 2012-09-16T04:09:09Z SecondChildTAG: Good stuff, have a good weekend. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-16T15:21:15Z SecondChildTAG: This should be working with Firefox now. Thank you for your patience. SecondChildUserIdTAG: 11 SecondChildUserNameTAG: dormsbee SecondChildCreateTimeTAG: 2012-09-17T23:37:06Z SecondChildTAG: Can you check this, please? https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_S3E2_Linear_Combinations_of_Source_Strengths/threads/5057e7d588beb22800000013 Did i break the honor code? If i did, i will delete the post. Thanks SecondChildUserIdTAG: 268104 SecondChildUserNameTAG: SaulCardenasG SecondChildCreateTimeTAG: 2012-09-19T05:43:51Z FirstChildTAG: How about http://www.flickr.com/ ? I can't upload either. I think if we could upload the picture,it's better to type millions words. Thanks! FirstChildUserIdTAG: 282748 FirstChildUserNameTAG: larryzhu FirstChildCreateTimeTAG: 2012-09-16T01:54:01Z FirstChildTAG: I have no problems uploading images in edxuploads (image icon that is in the tool tab in this Post response)... Do you still have the that issue? Can I help you? ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13477607042631679.png FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-16T02:01:32Z SecondChildTAG: Yes, still have the issue using the edxupload icon in the post response section. Maybe I will try an older version of Internet Exploder. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-16T03:14:18Z SecondChildTAG: I just tried Exploder, I can't even view responses to the posts with that browser. I have not used it for a while, so it may not be up to date. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-16T03:19:52Z SecondChildTAG: Hi Pennypacker! I am using Chrome Browser You can download it here. I have no problems uploading images... https://www.google.com/intl/us/chrome/browser/?hl=es SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-16T05:03:18Z SecondChildTAG: Maybe I will take Chrome for another test drive, thanks! I do like Firefox with it's Ad-Block, but there is no reason why I could not use Chrome for the duration of this course. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-16T15:23:05Z FirstChildTAG: Hi Myrimit, Thanks!I tried to use this image icon in the Post a response bar.Then I use the "browse" to point a BMP file and click OK. I can see " [enter image description here][1] [1]: http:// " But I can't see any picture show up. How you can make it success? you use http://**** link ? Thanks!! FirstChildUserIdTAG: 282748 FirstChildUserNameTAG: larryzhu FirstChildCreateTimeTAG: 2012-09-16T02:34:14Z SecondChildTAG: 1) Clic on the Image Icon. ![enter image description here][1] 2) Select a file. ![enter image description here][2] 3) You will see that once you selected a file, it will be uploaded and automatically will appear in the box the http. It will look like this: ![enter image description here][3] 4) Now, the ![...] means what description you want to attach to your uploaded image, eg. you can write ![hi larryzhu] haha! remember that the other [] will have a number. Down your Post response will be something like [....] : https://.... , the web page of that respective description. ![enter image description here][4] Also , you can put a lot of times one image repeating ![...][...] . Try it, cut and pase it a lot of time! Have fun! ;) ![enter image description here][5] ![enter image description here][5] ![enter image description here][5] [1]: https://edxuploads.s3.amazonaws.com/13477709218085651.png [2]: https://edxuploads.s3.amazonaws.com/13477710302550742.png [3]: https://edxuploads.s3.amazonaws.com/13477708171237814.png [4]: https://edxuploads.s3.amazonaws.com/13477708318474415.png [5]: https://edxuploads.s3.amazonaws.com/13477705208474402.bmp SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-16T04:59:52Z FirstChildTAG: Thanks! I installed the Chrome Browser and finally it works now. But I still can't let it work on Firefox. ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13478073333359425.jpg FirstChildUserIdTAG: 282748 FirstChildUserNameTAG: larryzhu FirstChildCreateTimeTAG: 2012-09-16T14:57:17Z FirstChildTAG: ![test4][1] [1]: http:// FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-16T03:22:43Z SecondChildTAG: Just a test. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-16T03:23:04Z IndexTAG: 1683 TitleTAG: The second part of this question... How you interpret this second question is very important. At first I had entered zero as the answer because it makes alot of sense; power is leaving the source not entering the source... its logic right? But think of it. When you take out two watts you are left with two watts less. UserIdTAG: 445228 UserNameTAG: Gichumbi CreateTimeTAG: 2012-09-15T21:24:33Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 5_Simple_Power NumberOfReplyTAG: 2 FirstChildTAG: **I think its -4W.** FirstChildUserIdTAG: 328336 FirstChildUserNameTAG: prince1 FirstChildCreateTimeTAG: 2012-09-16T02:46:27Z SecondChildTAG: great SecondChildUserIdTAG: 548713 SecondChildUserNameTAG: veenu_ SecondChildCreateTimeTAG: 2012-10-10T10:33:06Z FirstChildTAG: Hi, It far past midnight in Moscow, & I could not recognize only one thing: The power (in watts) coming out of the voltage source - what it is All other values are OK but in this case must be 4W ( as my opinion) FirstChildUserIdTAG: 202249 FirstChildUserNameTAG: ABZ FirstChildCreateTimeTAG: 2012-09-15T22:31:30Z SecondChildTAG: Тоже не спиться? Так же получается 4 Вт. Сколько получилось рассеивание на 1 омном и 3 омном? SecondChildUserIdTAG: 202275 SecondChildUserNameTAG: Valya SecondChildCreateTimeTAG: 2012-09-15T23:34:12Z SecondChildTAG: Появилась родная раскладка клавиатуры SecondChildUserIdTAG: 402617 SecondChildUserNameTAG: Vl SecondChildCreateTimeTAG: 2012-09-16T00:30:43Z IndexTAG: 1684 TitleTAG: Node Method I found it so easy to solve this exercise using node analysis. You end up with only one node equation and it is a piece of key to find the parameters for each branch. UserIdTAG: 350458 UserNameTAG: Aaron_PSX CreateTimeTAG: 2012-09-15T17:53:15Z VoteTAG: 2 CoursewareTAG: Week 1 / KVL/KCL example CommentableIdTAG: 6002x_S2E3_Using_KVL_KCL_and_VI_constraints NumberOfReplyTAG: 2 FirstChildTAG: anyone please tell me how to solve lab 1 soon.... FirstChildUserIdTAG: 220425 FirstChildUserNameTAG: rishienggju FirstChildCreateTimeTAG: 2012-09-15T20:36:56Z FirstChildTAG: ok.. thank you... FirstChildUserIdTAG: 133084 FirstChildUserNameTAG: Warrensiggs FirstChildCreateTimeTAG: 2012-09-15T18:14:03Z IndexTAG: 1685 TitleTAG: Lab 2 I think I'm conceptually lost. How can I figure out what type of circuit I need to build? I see everyone saying 3 resistors but I'm having trouble building a circuit and understanding what's happening. I think I'm also a bit confused conceptually by the ground symbol. As well as by the open circuit voltage. Is the open circuit voltage always equal to the voltage from the shortest path back to ground? UserIdTAG: 357225 UserNameTAG: MJBoa CreateTimeTAG: 2012-09-15T17:47:28Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I seem to have figured out how the network looks but how do I figure out the correct values? I seem to have two equations with 3 variables... FirstChildUserIdTAG: 357225 FirstChildUserNameTAG: MJBoa FirstChildCreateTimeTAG: 2012-09-15T19:17:12Z IndexTAG: 1686 TitleTAG: what's my mistake? ![][1] [1]: https://edxuploads.s3.amazonaws.com/13477297168769748.png If we apply KVL in loop abcd (direction clockwise) then it turns out > iR2+V2-V1+iR1=0 Since V2=V1. It further turns out R2=-R1 But, it's wrong. Please tell me my mistake UserIdTAG: 374590 UserNameTAG: ahope CreateTimeTAG: 2012-09-15T17:27:56Z VoteTAG: 2 CoursewareTAG: Week 1 / Battery model CommentableIdTAG: 6002x_S1E9_Battery_Model NumberOfReplyTAG: 4 FirstChildTAG: check out the last paragraph on pg 17 of the text. it starts like "if one wishes to increase the current capacity...." make sense?? FirstChildUserIdTAG: 364086 FirstChildUserNameTAG: jordan3110 FirstChildCreateTimeTAG: 2012-09-16T02:10:25Z SecondChildTAG: the current in R2 is not the same in R1 SecondChildUserIdTAG: 474111 SecondChildUserNameTAG: achref01 SecondChildCreateTimeTAG: 2012-09-21T19:58:28Z SecondChildTAG: ![enter image description here][1] [1]: http:// Because the current can't flow to the positif boundary : it means that electrons can not flow to the same point from two opposit direction; it will be a catastroph ! SecondChildUserIdTAG: 526248 SecondChildUserNameTAG: Sair SecondChildCreateTimeTAG: 2012-10-03T16:44:53Z SecondChildTAG: ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13492827331343662.jpg SecondChildUserIdTAG: 526248 SecondChildUserNameTAG: Sair SecondChildCreateTimeTAG: 2012-10-03T16:45:47Z SecondChildTAG: There we have i>=0 SecondChildUserIdTAG: 526248 SecondChildUserNameTAG: Sair SecondChildCreateTimeTAG: 2012-10-03T16:46:29Z FirstChildTAG: You assume that $i\neq 0$. Do you think that's correct? FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-15T17:45:00Z SecondChildTAG: Thanks! SecondChildUserIdTAG: 374590 SecondChildUserNameTAG: ahope SecondChildCreateTimeTAG: 2012-09-15T17:53:28Z FirstChildTAG: I am totally messed it will be great if someone answers it fast.. FirstChildUserIdTAG: 374590 FirstChildUserNameTAG: ahope FirstChildCreateTimeTAG: 2012-09-15T17:39:10Z SecondChildTAG: But what do we have to find in it here. SecondChildUserIdTAG: 443695 SecondChildUserNameTAG: Physicistsandy1 SecondChildCreateTimeTAG: 2012-09-15T17:43:32Z SecondChildTAG: Voltage across terminals. SecondChildUserIdTAG: 374590 SecondChildUserNameTAG: ahope SecondChildCreateTimeTAG: 2012-09-15T17:54:11Z FirstChildTAG: well is it specified in the q that v2=v1? if not then you cant say whethr they are equal.. the voltge sources v1 n v2 r nt i a parallel combination , and thre r respective voltage drops across r1 and r2. however the brances ac n bd have equal voltages = v FirstChildUserIdTAG: 390803 FirstChildUserNameTAG: Spurty FirstChildCreateTimeTAG: 2012-09-16T19:43:36Z IndexTAG: 1687 TitleTAG: Lab 1 How can i submit the lab ? I check it so it is right but when i save it it becomes wrong and not checked . UserIdTAG: 309933 UserNameTAG: miramar CreateTimeTAG: 2012-09-15T15:43:31Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: then keep a screen print(just to remember what u have done) then log out from your account then log in again...i hope problem will be solved FirstChildUserIdTAG: 154172 FirstChildUserNameTAG: mofassair FirstChildCreateTimeTAG: 2012-09-15T16:33:04Z SecondChildTAG: It automatically saves after checking. Use google chrome [chrome link][1] [1]: http://www.google.com/chrome SecondChildUserIdTAG: 374590 SecondChildUserNameTAG: ahope SecondChildCreateTimeTAG: 2012-09-15T17:45:32Z IndexTAG: 1688 TitleTAG: CAN I ENROLL NOW? AS WE KNOW THAT MITx: 6.002x Circuits and Electronics HAS STARTED FROM 5TH SEPTEMBER. CAN I TAKE COURSE NOW PLEASE?I AM IN GRADE 12. I LIVE IN PAKISTAN. I HAVE STUDIED PHYSICS SINCE CLASS 6TH.I AM REALLY INTERESTED IN IT. I WILL WORK HARD. I WILL WORK FOR THE THINGS WHICH I LEFT BEHIND. KINDLY GUIDE ME PLEASE REGARDS SAFIAN TASEER UserIdTAG: 442115 UserNameTAG: safiantaseer CreateTimeTAG: 2012-09-15T15:23:03Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: The fact that you're in this discussion forum means that you are enrolled :) First assignment is due this Sunday. Good luck! P.S. Please don't use all capital letters FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-15T15:25:47Z SecondChildTAG: kimt, I didn't know you were a staff member. Cool. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-09-15T16:04:15Z SecondChildTAG: My secret's out! :o SecondChildUserIdTAG: 79361 SecondChildUserNameTAG: kimt SecondChildCreateTimeTAG: 2012-09-15T16:08:24Z SecondChildTAG: can i submit it by Monday? SecondChildUserIdTAG: 442115 SecondChildUserNameTAG: safiantaseer SecondChildCreateTimeTAG: 2012-09-15T17:25:03Z IndexTAG: 1689 TitleTAG: Question 3 (bug ?) Why for the voltage VL all these answers are shown as correct: 1287/9200, 1287/9300, 1287/9400, 1287/9500, 1287/9600, 1287/9700, 1287/9800, 1287/9900, 1287/10000 and I don't know how many more? UserIdTAG: 301141 UserNameTAG: Smaragda CreateTimeTAG: 2012-09-15T11:37:36Z VoteTAG: 2 CoursewareTAG: Week 2 / Norton method example CommentableIdTAG: 6002x_S3E6_Norton_Model NumberOfReplyTAG: 1 FirstChildTAG: *Possibly* because the values are within 10%. In real life average cost resistors are accurate to within around 10%. If you solved the problem in a physical lab, one or more of your answers may be correct. The fourth band to the right, if present represents the tolerance of resistors. Silver = 10% Gold = 5% No band = 20% FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-15T14:11:28Z IndexTAG: 1690 TitleTAG: Home work submission deadline query Hi For the Sunday deadlines for the lab and home work, what time is final submission Grenwich mean time? Thanks UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-15T08:51:52Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: midnight I think, I'm rushing as I only found this course yesterday haha. FirstChildUserIdTAG: 436749 FirstChildUserNameTAG: waynebrown FirstChildCreateTimeTAG: 2012-09-15T10:01:12Z SecondChildTAG: Right on! Anyhow the homework will be accepted as long as it is the 16th of September somewhere on Earth. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-15T15:08:57Z IndexTAG: 1691 TitleTAG: About Work Submission Respected, What is the method of submitting the homework and lab ? Whenever I check the homework and Lab, It uploaded the progress bar. And how many times I'll check my home work and My next midterm exams.Please help me about this... UserIdTAG: 271448 UserNameTAG: UsmanRashid CreateTimeTAG: 2012-09-15T07:50:24Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I think when you press "check" and the marks are recorded under the "progress" tab, the homework and lab are submitted. FirstChildUserIdTAG: 157273 FirstChildUserNameTAG: ongchihang FirstChildCreateTimeTAG: 2012-09-15T12:35:33Z FirstChildTAG: large value in ohms which fabricated n small value in ohms which fabricated ?? didn't understand this part, can anyone explain me FirstChildUserIdTAG: 131206 FirstChildUserNameTAG: ahmediu FirstChildCreateTimeTAG: 2012-09-15T08:02:20Z SecondChildTAG: Could not get you, be clear please. SecondChildUserIdTAG: 222234 SecondChildUserNameTAG: ivsprasad SecondChildCreateTimeTAG: 2012-09-15T08:19:55Z SecondChildTAG: You shouldn't hijack a thread, but they mean: when you combine the resistors, what is the largest resistance you can get? What is the smallest? SecondChildUserIdTAG: 25252 SecondChildUserNameTAG: cwm9 SecondChildCreateTimeTAG: 2012-09-15T09:27:24Z IndexTAG: 1692 TitleTAG: Currents When current was first known about, nobody knew which way it was actually flowing. It was thus decided to treat it as flowing from the positive to the negative terminals (positive and negative also being arbitrarily chosen). In fact, in a metal wire, it is the negative charge-carriers, the electrons, that move. Mathematically, negative charges moving from B to A is equivalent to positive charges moving from A to B. Consider two buckets of green and white tennis balls. Bucket A has more green tennis balls than B, which has exactly the same number more white balls. To equalise the numbers in the buckets, we can move green balls from A to B (positive charge). If the green balls are fixed (analogous to the protons in a metallic lattice), we can move the white balls from B to A, and equalise the numbers that way (electrons flowing in a wire). If we artificially move white balls back to B, they will keep flowing (analogous to a voltage source). Thus there is only one current, but we can consider it as a positive current going one way, or a negative current going the other. Thus i1 = -i2, but as i2 is negative to i1, we are really saying i1 = -(-i1). As for power conventions, think of it this way. Energy over time (power) is dissipated in the resistor (or whatever it is), which is positive. The energy had to come from somewhere. It came from the voltage source, so the source lost the energy, and the power is thus negative. I hope that this is helpful. UserIdTAG: 261378 UserNameTAG: es2377 CreateTimeTAG: 2012-09-14T22:19:34Z VoteTAG: 2 CoursewareTAG: Week 1 / Associated reference directions CommentableIdTAG: 6002x_S2E2_Associated_Reference_Directions NumberOfReplyTAG: 1 FirstChildTAG: I came up with a mathematical approach that may help to 'ground' one's interpretations (pun intended). Notice that KVL loop analysis is setup so that if the polarity of the v-i variables are set the same (+ then -), the sum to zero outcome forces the sign of voltages to be the same. In contrast KCL signs must be opposite since we are evaluating from a zero point node. I'm still struggling with the intuitive interpretation and have seen so many different variations including the one above with the white and green balls. Looking at it from a mathematical approach at least gives me an understanding of what the sign must be to meet the Lump Matter Discipline requirements. Comments? FirstChildUserIdTAG: 23573 FirstChildUserNameTAG: mktfax FirstChildCreateTimeTAG: 2012-09-15T19:48:34Z IndexTAG: 1693 TitleTAG: Lab values When i use exact value with fractions in labs, e.g. 2/3 Ohms for resistance, behavior is very strange and differs from rounded 0.6666 value. Is it possible to handle fractions instead nerest numeric values ? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-14T21:28:43Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: 2/3 is easy just connect set of two three series registers in parallels..... FirstChildUserIdTAG: 306294 FirstChildUserNameTAG: swpnljoshi FirstChildCreateTimeTAG: 2012-09-15T03:43:48Z SecondChildTAG: It's workaround! Values like "2/3" works perfectly in lecture sequences tasks and homeworks but not for nominals of elements in labs. SecondChildUserIdTAG: 81712 SecondChildUserNameTAG: dzhon SecondChildCreateTimeTAG: 2012-09-15T05:17:07Z IndexTAG: 1694 TitleTAG: I can't get this. (Help) I've tried, and I even watched the nodal video, and couldn't get it to work. the picture attached is where I'm stumped. I get an e1 of 0 & an e2 of 15, but it doesn't work out for the answers. Well.. I would upload my image, turns out the image uploader doesn't work for for the site yet -.- So here's my "attempt" in words. The "node" between R1 & R2 I labeled e1. The "node" between R2 and V0 I labeled e2. Did the equation, placed a grounder at the top (as 0V) and used these as my equations: (e1-0V)/4.0(ohms) + (e1-e2)/5.0(ohms) - 3.0A = 0 - AND - (e2 + 2V)/0(ohms) + (e2-e1)/5.0(ohms) + 3.0A = 0 I know, my second equation does that whole, never divide by 0 you dip!@#$. I plug in numbers, find e1 as 0 and e2 as 15, but how could potential go through a resistor (R1) without changing? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-14T04:25:11Z VoteTAG: 2 CoursewareTAG: Week 1 / KVL/KCL example CommentableIdTAG: 6002x_S2E3_Using_KVL_KCL_and_VI_constraints NumberOfReplyTAG: 4 FirstChildTAG: http://postimage.org/image/sph3p3nf7/6f678f7f/ Thanks FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-14T13:28:03Z SecondChildTAG: Hi jGeorge! I haven't seen this before. Sorry for the delay of the answer. The Cool think it is that now, with the new Plataform we can upload images (now it works!). Ok, I will show the image that you have uploaded in postimage: ![enter image description here][1] With the ground node at where you place it: KVL: **V - e1 - v2 =0** V - i1*R1 - i2*R2=0 5 - i1*4 - i2*5 =0 KCL: i1-i2+I=0 i1= (0-e1)/R1 i2= (e1-e2)/R2 I = 3 So, **[(0-e1)/R1] - [(e1-e2)/R2]+3=0** (If you see e2=-V=-5) If it helps you , you can see also this post [Power Supplied][2]. It is almost the same but I used here the Ground node in the negative of V supply... [1]: https://edxuploads.s3.amazonaws.com/13476793496762562.png [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/undefined/threads/50531159ee13321f00000003 SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-15T04:57:46Z SecondChildTAG: That's great, I don't know why I used V = 5.0V when the problem clearly states otherwise (probably because the tutorial problem was @ 5.0V). SecondChildUserIdTAG: 365894 SecondChildUserNameTAG: jGeorge SecondChildCreateTimeTAG: 2012-09-15T15:06:01Z FirstChildTAG: Hi! Can I help you? Could you Post your circuit image here? You can upload it at www.postimage.org and then put the link in the image icon?;) FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-14T12:58:52Z FirstChildTAG: Hello I have the same example, did you solve it? FirstChildUserIdTAG: 233394 FirstChildUserNameTAG: ishpanec FirstChildCreateTimeTAG: 2012-09-14T17:29:03Z SecondChildTAG: By, example do you mean problem? haha unfortunately not. SecondChildUserIdTAG: undefined SecondChildUserNameTAG: undefined SecondChildCreateTimeTAG: 2012-09-14T17:47:50Z FirstChildTAG: We initially have two variables i1 and i2. Using KVL for the first loop and KCL for Node e1, solve for i1 and i2. You'll get the rest. I tried solving it through node analysis but i had the same 0 dividing the numerator. Solution looks something like this: KVL: i1R1+i2R2-V (substituting the values of R1, R2 and V, you get an equation in terms of i1 and i2) KCL: i2-i1-I Solve for i1 and i2. Hope this helps! FirstChildUserIdTAG: 395953 FirstChildUserNameTAG: dustinge FirstChildCreateTimeTAG: 2012-09-15T09:40:31Z IndexTAG: 1695 TitleTAG: Better Rs=R1+R2+R3 1/Rp=1/R1+1/R2+1/R3 UserIdTAG: 337902 UserNameTAG: Dan_Pop CreateTimeTAG: 2012-09-14T03:44:54Z VoteTAG: 2 CoursewareTAG: Week 1 / Three series resistors CommentableIdTAG: 6002x_three_series_resistors NumberOfReplyTAG: 0 IndexTAG: 1696 TitleTAG: lab 2 hi guys! i have been figuring out how to answer this question and the farthest i have gotten was to arrive with equations 5R1=R2 and R1=R2 and it is impossible for me to get the values to make the equations true.Can someone give me a hint on what to do next? please! Thanks so much! UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-14T00:01:12Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: This is true at R1=R2=0. You should check to see if your work up to this point is correct since this is never true in nature (unless you include super conductors which have an R almost=0). FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-14T00:43:20Z SecondChildTAG: His work is correct so far - next step is to work out you need to add to the circuit a bit. SecondChildUserIdTAG: 312035 SecondChildUserNameTAG: philhiggins SecondChildCreateTimeTAG: 2012-09-14T01:02:01Z IndexTAG: 1697 TitleTAG: What is a node in fact? I know that it is a little late to ask, but I wanna know why the concept of this course for node is different from that one that I had learned? I always thought that a node was the junction of 3 or more branches. But here, a simple corner in the circuit is called a node... Can somebody explain it for me? Thanks a lot! UserIdTAG: 265133 UserNameTAG: Elienai CreateTimeTAG: 2012-09-13T21:24:44Z VoteTAG: 2 CoursewareTAG: Week 1 / Nodal analysis example CommentableIdTAG: 6002x_NodalAnalysisExample NumberOfReplyTAG: 1 FirstChildTAG: Both definitions are good enough. A '3-wire' node is useful, if you do the work with pen and paper, a '2-wire' node is better for computer-aided calculations. If you use only **3+ wire nodes**, then you have to optimize a circuit. For example, replace a serial-connected resistors with one equivalent. This will reduce the number of equations, but will add some work. If we allow a **nodes with 2 branches**, then we need only few formalized rules. This approach will give a little bit more equations, but it is completely unified, and relatively easy may be given to the computer. The time of Nodal Analysis lecture is also reduced :) FirstChildUserIdTAG: 261503 FirstChildUserNameTAG: EugenyL FirstChildCreateTimeTAG: 2012-09-13T22:19:59Z IndexTAG: 1698 TitleTAG: Slightly wider inputs It would be nice if text inputs were slightly bigger as some people, me included, may like to insert whole equation - not just calculated answer. It helps when you want to revise stuff and figure out where the value came from. Cheers! ![slightly wider inputs][1] [1]: http://i.imgur.com/S5Rtl.png UserIdTAG: 322552 UserNameTAG: riv CreateTimeTAG: 2012-09-13T20:15:10Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Okay thanks for the feedback :D FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-13T20:26:51Z IndexTAG: 1699 TitleTAG: edx team whats happen to the speech converts to words and the video become youtube where is the edx system and speed control of the video pls solve the problem UserIdTAG: 282351 UserNameTAG: HalfDead CreateTimeTAG: 2012-09-13T11:24:14Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Please see [Dave's (edX developer) post][1]. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/MITx_6_002x_2012_Fall/threads/5051442739934a2b0000001a FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-13T12:44:45Z SecondChildTAG: thanks Kimt SecondChildUserIdTAG: 282351 SecondChildUserNameTAG: HalfDead SecondChildCreateTimeTAG: 2012-09-13T13:20:00Z IndexTAG: 1700 TitleTAG: Video Stops The lecture sequences stop randomly and require a page refresh to restart. This is extremely frustraiting. Windows 7 with IE9. Please advise as to how to correct this. UserIdTAG: 397595 UserNameTAG: mholin CreateTimeTAG: 2012-09-12T23:19:16Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 4 FirstChildTAG: Exactly the same for me. Ubuntu 12.04 with Firefox 15 and Chrome 13. FirstChildUserIdTAG: 368419 FirstChildUserNameTAG: LuciaDR FirstChildCreateTimeTAG: 2012-09-13T00:04:50Z FirstChildTAG: same with Mac OS X Lion running Safari/Chrome. Latest available flash installed. You can watch the videos on youtube (click on the bottom right corner of video) where this problem is non existant. FirstChildUserIdTAG: 27851 FirstChildUserNameTAG: Arman FirstChildCreateTimeTAG: 2012-09-13T00:05:03Z FirstChildTAG: YouTube made a recent change to their Flash player that broke our video embeds. As a temporary workaround while we figure things out, you can use YouTube's HTML5 player. To do so, please go to YouTube's [Join the HTML5 Trial][1] page and click "Join the HTML5 Trial" at the bottom. Then reload our site in your browser. Thank you for your patience as we work on this. FirstChildUserIdTAG: 11 FirstChildUserNameTAG: dormsbee FirstChildCreateTimeTAG: 2012-09-13T02:27:27Z FirstChildTAG: This is a bug. We are working on it. Thanks for the report. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-09-13T01:16:03Z IndexTAG: 1701 TitleTAG: Bug? Ubuntu 11.10 Chromium. For dome reason tutorials just stop in the middle. It's not a buffering, it looks like the player thinks it's the end of the clip. A. UserIdTAG: 213170 UserNameTAG: AdamKuch CreateTimeTAG: 2012-09-12T22:29:36Z VoteTAG: 2 CoursewareTAG: Week 1 / Three parallel resistors CommentableIdTAG: 6002x_parallel_resistors2 NumberOfReplyTAG: 3 FirstChildTAG: Now I can see that EVERY clip stops at 1:08. Hmmm... FirstChildUserIdTAG: 213170 FirstChildUserNameTAG: AdamKuch FirstChildCreateTimeTAG: 2012-09-12T22:38:19Z SecondChildTAG: BUT when I have watched he clip on YT (and therefore loaded into memory/buffer) the clip, watched for the second time here on page of edx, run from start to end. SecondChildUserIdTAG: 213170 SecondChildUserNameTAG: AdamKuch SecondChildCreateTimeTAG: 2012-09-12T22:40:00Z FirstChildTAG: YouTube made a recent change to their Flash player that broke our video embeds. As a temporary workaround while we figure things out, you can use YouTube's HTML5 player. To do so, please go to YouTube's [Join the HTML5 Trial][1] page and click "Join the HTML5 Trial" at the bottom. Then reload our site in your browser. Thank you for your patience as we work on this. FirstChildUserIdTAG: 11 FirstChildUserNameTAG: dormsbee FirstChildCreateTimeTAG: 2012-09-13T02:27:46Z FirstChildTAG: We are working on it. Thanks for the bug report. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-09-13T01:18:15Z IndexTAG: 1702 TitleTAG: How does open circuiting work? Voltage without current? I don't understand something conceptually. When I is 0 or is open circuited. How is there any voltage flowing between A and B then, without a current? And why doesn't Rth affect it? UserIdTAG: 357225 UserNameTAG: MJBoa CreateTimeTAG: 2012-09-12T18:29:58Z VoteTAG: 2 CoursewareTAG: Week 2 / Thevenin method CommentableIdTAG: 6002x_thevenin NumberOfReplyTAG: 1 FirstChildTAG: You shouldn't think of voltage as something that *flows* in a circuit, but something that is statically *set up* (at least in the case of DC circuits). In the case of the open circuit, there is no current flowing. This means that the voltage difference across $R_\mathrm{th}$ will be zero (i.e. the voltage on its two ends are the same). Since the $R_\mathrm{th}$ node on the side of the voltage source is $v_\mathrm{th}$ with respect to the circuit's ground, it follows that the open end of $R_\mathrm{th}$ is also $v_\mathrm{th}$. FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-12T18:40:05Z SecondChildTAG: True, Voltage is a measure of potential difference. But how do we measure that difference without creating a circuit and current through Rth? SecondChildUserIdTAG: 357225 SecondChildUserNameTAG: MJBoa SecondChildCreateTimeTAG: 2012-09-12T19:06:14Z IndexTAG: 1703 TitleTAG: S2v8,S2V9,S2V10 PLEASE TELL ME IN STEP THREE OF NODE METHOD IS IT e2 OR e3???? ANOTHER LITTLE CONFUSION IS THAT,IN KCL AT NODE e1 WE HAVE CURRENT GOING UP SO WE TOOK **(e1-V0)/R1** **BUT** WHEN WE MOVE TO S2V10 AND IN STEP FIVE IN BRANCH VOLTAGE WE TOOK V1 AS **(V0-e1)/R1** I HAVE LITTLE CONFUSION CURRENT IS GOING UP OR COMING TO THE NODE????? UserIdTAG: 40209 UserNameTAG: sadaf78 CreateTimeTAG: 2012-09-12T17:21:44Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: My standard way to do KCL method is consider the branch voltage as positive and the adjacent others as negative, producing something like the first equation you typed (e1-V0)/R1. This equation is indepedent of the current direction, as e1 or V0 may be positive or negative. Hence, (v0-e1)/R1 is not wrong or different from the 1st. If both the voltages (in the 1st eq) are positive, then current is leaving the node. Nevertheless, if you define a good standard for you (liked that), you will have no troubles. FirstChildUserIdTAG: 39980 FirstChildUserNameTAG: MarinoJr FirstChildCreateTimeTAG: 2012-09-12T17:33:21Z SecondChildTAG: thank you marinojr. SecondChildUserIdTAG: 40209 SecondChildUserNameTAG: sadaf78 SecondChildCreateTimeTAG: 2012-09-12T17:46:38Z IndexTAG: 1704 TitleTAG: Lab 1 bug? In Firefox 15.0.1 power source and bulb don't appear in the sandbox. In IE9 works well. UserIdTAG: 337791 UserNameTAG: Roman_T CreateTimeTAG: 2012-09-12T12:24:44Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: As I see It's common bug. Press "Check" button and then "reset" and scheme will be reset. FirstChildUserIdTAG: 378778 FirstChildUserNameTAG: happylife FirstChildCreateTimeTAG: 2012-09-12T12:46:22Z FirstChildTAG: Hi, can you try the Lab 1 again in Firefox? A fix for the circuit simulator was released today. FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-12T14:10:09Z SecondChildTAG: It's ok now. SecondChildUserIdTAG: 337791 SecondChildUserNameTAG: Roman_T SecondChildCreateTimeTAG: 2012-09-14T14:02:10Z IndexTAG: 1705 TitleTAG: Assistance with parse error When a parse error appears, kindly solve by not putting = RHS which is in the *emphasized text* e.g. Invalid input: Could not parse 'e2*(1/R4+1/R3+1/R5)=*I1+V0/R4+e1/R3*' as a formula This I figured out myself now UserIdTAG: 329879 UserNameTAG: Ugochukwu CreateTimeTAG: 2012-09-12T09:49:40Z VoteTAG: 2 CoursewareTAG: Week 1 / Node analysis practice, part 3 CommentableIdTAG: 6002x_L2Node2 NumberOfReplyTAG: 1 FirstChildTAG: thanks FirstChildUserIdTAG: 273339 FirstChildUserNameTAG: FaranAB FirstChildCreateTimeTAG: 2012-09-12T17:09:56Z IndexTAG: 1706 TitleTAG: watching at 1.5 speed video caption and navigation doesn't work watching at 1.5 speed video caption and navigation doesn't work UserIdTAG: 271081 UserNameTAG: YonJah CreateTimeTAG: 2012-09-11T20:15:05Z VoteTAG: 2 CoursewareTAG: Week 2 / Static Discipline CommentableIdTAG: 6002x_static_discipline NumberOfReplyTAG: 1 FirstChildTAG: same here at 1.25 FirstChildUserIdTAG: 25655 FirstChildUserNameTAG: Alois FirstChildCreateTimeTAG: 2012-09-19T07:48:35Z SecondChildTAG: Yeah! Run at 1.5x no timer, change back to 1.25x timer OK. Back to 1.5x: Invalid parameters (Youtube) SecondChildUserIdTAG: 324332 SecondChildUserNameTAG: petsol SecondChildCreateTimeTAG: 2012-09-23T16:39:30Z SecondChildTAG: Using Chrome... SecondChildUserIdTAG: 324332 SecondChildUserNameTAG: petsol SecondChildCreateTimeTAG: 2012-09-23T16:40:02Z IndexTAG: 1707 TitleTAG: how how can 10k+10k = 10V ???? and how can 10k+20k = 20k ???? UserIdTAG: 371328 UserNameTAG: abdo2 CreateTimeTAG: 2012-09-11T19:59:24Z VoteTAG: 2 CoursewareTAG: Week 1 / Combination Rules CommentableIdTAG: 6002x_CombinationRules NumberOfReplyTAG: 1 FirstChildTAG: hey its explained at the end of the video. FirstChildUserIdTAG: 443805 FirstChildUserNameTAG: Rash5mi FirstChildCreateTimeTAG: 2012-09-17T17:25:10Z IndexTAG: 1708 TitleTAG: Technical problem with video There is a problem with this video : video controls do not work ( firefox browser ) and also text does not move forward ... UserIdTAG: 408116 UserNameTAG: todorp CreateTimeTAG: 2012-09-11T19:11:55Z VoteTAG: 2 CoursewareTAG: Week 1 / Lumped Element Abstraction CommentableIdTAG: 6002x_lumped_element_abstraction NumberOfReplyTAG: 1 FirstChildTAG: iam using google chorme and it working well ,try it . FirstChildUserIdTAG: 418898 FirstChildUserNameTAG: Mostafa_Elboss FirstChildCreateTimeTAG: 2012-09-12T03:20:10Z SecondChildTAG: Thanks it works with google chrome SecondChildUserIdTAG: 382532 SecondChildUserNameTAG: emmanuelpeace SecondChildCreateTimeTAG: 2012-09-13T06:10:54Z SecondChildTAG: also i use google chrome it work very well SecondChildUserIdTAG: 515333 SecondChildUserNameTAG: ramankakasalihaskary SecondChildCreateTimeTAG: 2012-09-30T08:59:31Z IndexTAG: 1709 TitleTAG: VIDEO COULD U PLS TELL HOW TO DOWNLOAD THE VIDEOS FROM THIS WEBSITE..PLS UserIdTAG: 351262 UserNameTAG: subbucmr CreateTimeTAG: 2012-09-11T14:34:47Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: I use the Flash Video Downloader plugin for Firefox. I don't have enough bandwidth where I stay so I have to download the videos elsewhere otherwise I will not be able to do this course. FirstChildUserIdTAG: 360872 FirstChildUserNameTAG: ScorpKing FirstChildCreateTimeTAG: 2012-09-11T18:02:38Z FirstChildTAG: I do not believe that video downloads are supported. However, you can access and download just the slides from the right-hand panel at the main screen. FirstChildUserIdTAG: 414313 FirstChildUserNameTAG: staticvoid FirstChildCreateTimeTAG: 2012-09-11T14:55:29Z SecondChildTAG: you can use YTD to do it SecondChildUserIdTAG: 282351 SecondChildUserNameTAG: HalfDead SecondChildCreateTimeTAG: 2012-09-11T16:01:14Z IndexTAG: 1710 TitleTAG: Second Part of S2EF Can anyone help me with clear explanation of this equation e1 = e3 + V and that too -8.75 for V2 UserIdTAG: 150120 UserNameTAG: krishna1993 CreateTimeTAG: 2012-09-11T13:43:08Z VoteTAG: 2 CoursewareTAG: Week 1 / Node method example CommentableIdTAG: 6002x_S2E5_Node_Method NumberOfReplyTAG: 1 FirstChildTAG: Actually, V is a potential difference between the terminals of the voltage source. So, if e1 is a positive terminal and e3 is a negative terminal, then e1-e3=V, hence, e1=e3+V. FirstChildUserIdTAG: 326787 FirstChildUserNameTAG: Angstrem FirstChildCreateTimeTAG: 2012-09-11T16:51:32Z SecondChildTAG: very nice SecondChildUserIdTAG: 273339 SecondChildUserNameTAG: FaranAB SecondChildCreateTimeTAG: 2012-09-12T17:37:52Z SecondChildTAG: good answer SecondChildUserIdTAG: 370538 SecondChildUserNameTAG: samcore2804 SecondChildCreateTimeTAG: 2012-09-21T22:43:48Z IndexTAG: 1711 TitleTAG: lab1 i know there are a few about this already, downvote/delete if rehashing. didnt really know what i was doing but through trial and error got it to give me 2 ticks, but with answers that were a bit off, no idea how i did this. was feel guilty so went to look at it again, but once checked i cant see old ones. so i restarted and figured it out properly this time (anyone struggling read hint and look up two-resistor voltage dividers). still a few problems with the labs though, when checking answer i often forgot 1 little thing and then had to rebuild the hold thing, annoying. UserIdTAG: 388273 UserNameTAG: kilmarta CreateTimeTAG: 2012-09-11T13:38:20Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Yeah, the lab should allow multiple checks without resetting :D FirstChildUserIdTAG: 321830 FirstChildUserNameTAG: kimth FirstChildCreateTimeTAG: 2012-09-11T21:09:46Z IndexTAG: 1712 TitleTAG: H3P1 Very confused about static discipline The more I think about H3P1, the more confused I am. I earned checkmarks for the NOT and the NAND, but can't get the NOR done at all. But the more I think about it, the more I believe that the (marked as correct) answers that I gave for the NOT and NAND are also false (although they're marked correctly). Let me try to explain with the simple NOT. When the output is HIGH, V_O = V_S = 5V (this alone, IMHO, should violate the static discipline, since V_S > V_OH). In the case of A = HIGH, there's a voltage divider, dividing 5V through x (Pullup) and then 13k. So we can say V_O = 13k / (13k + x) * 5V. This means the higher the pullup x, the lower the output voltage (and vice versa), since x is in the denominator. The question asks for the lowest value of x. This means logically the highest value for the output voltage V_O. So when I lookup the maximum allowed output voltage, this is V_IL and should be 1.5V. When I set V_O = 1.5V and solve for x, I get 30.33 k. This is the wrong answer, however. Obviously I cannot write the right answer here, but the right answer can be calculated when setting V_O = 1.0V and solving for x. This all seems wrong and confusing. I'm either misunderstanding something completely or somethings off about the question. As I said, I could successfully solve the NOR and NAND. The NOR I can't solve at all. Here's my attempt: For the NOR, we have (in contrast to the other two) two different cases: Either one or two FETs are on. This means we have two different pulldowns, 13k and 6.5k. So we need to ensure that in both cases 1.5V <= V_O <= 1.0V. Let's call the V_O for the 13k case (one FET active) v_1 and the V_O for the 6.5k case (two FETs active) v_2. Immediately obvious: v2 < v1. So let's solve for the higher one (v1) to the upper margin (1.5V). This gives x = 30.33k. For this value of x, however, v2 would be 0.88V, i.e. < 1.0V. It violates the SD. Then let's try solving for the lower one (v2) and the lower margin (1.0V). This gives x = 26k. However, for this choice of x, v1 would be 1.66V, i.e. > 1.5V. It also violates the SD. I've simulated it in Maple and whichever values of x I choose, either V_OL or V_IL is violated. Can somebody enlighten me what I'm doing wrong? Thank you very much. UserIdTAG: 144694 UserNameTAG: johndoe31415 CreateTimeTAG: 2012-09-11T10:11:44Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: I'm going to try to push you in the right direction here. Think about static discipline again. It sounds like you need to review S4V8, hopefully that clears up the issues you're talking about in your second paragraph. All of your issues stem from a misunderstanding of the static discipline. You aren't looking for a signal in between VOH and VIH. Neither are you looking for a signal in between VOL and VIL. *Sent* signals have to be either above VOH, or below VOL. *Received* signals have to be either above VIH or below VIL. Hopefully that clears a few things up? If not, I can try to give a hint in a different way. FirstChildUserIdTAG: 248902 FirstChildUserNameTAG: Chanute FirstChildCreateTimeTAG: 2012-09-12T06:18:29Z SecondChildTAG: You are correct. When watching S4, I was taking notes and put this down in my notes. Since the first two answers were correct, I assumed that I was doing it correctly. Now looking at S4V8 again, it's clear that he talks about voltages >= V_OH for a high output, I don't know how I got to the crap I wrote down. Thank you for your hint! Edit: Got it. All correct now. Thank you very much again! SecondChildUserIdTAG: 144694 SecondChildUserNameTAG: johndoe31415 SecondChildCreateTimeTAG: 2012-09-12T11:07:19Z SecondChildTAG: How did you figure out the NOR gate? I did parallel but that didn't work... Help? This is the last problem I have. SecondChildUserIdTAG: 344331 SecondChildUserNameTAG: intellectualwanderer SecondChildCreateTimeTAG: 2012-09-16T04:21:27Z SecondChildTAG: @Chanute I have been using the correct idea like what you explained since I started doing H3P1. I can calculate the correct value of the resistor for the inverter and NAND except for NOR gate. I first combine the two parallel R **ON** together and I get 5500 ohm. Then I use the voltage divider formula 5500/(5500+R)*5 < 1 since V **OL** is 1V. And I get 22k ohm for the pullup resistor which is wrong. Can you guide me from here? SecondChildUserIdTAG: 157273 SecondChildUserNameTAG: ongchihang SecondChildCreateTimeTAG: 2012-09-19T08:41:12Z IndexTAG: 1713 TitleTAG: textbook as pdf or similar is there any way to download full textbook as a pdf or other filetype? i wanted to read it when i have not acces to the Internet. if someone could help me please write grzegorz.pietrzak321 at google mail. thx for help UserIdTAG: 408796 UserNameTAG: GrzegorzPietrzak CreateTimeTAG: 2012-09-11T08:11:01Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 1714 TitleTAG: LaTeX Is latex form available for comments? What tags do i use? UserIdTAG: 404284 UserNameTAG: tomdrifmach CreateTimeTAG: 2012-09-11T07:32:18Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Yes Mathjax is used. You need to add code within two $ symbols FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-09-11T18:06:53Z SecondChildTAG: ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-30T01:45:12Z IndexTAG: 1715 TitleTAG: Calculating currents For getting the current you have to remember the ohm law. V = IR, you have the potencial that crosses the resistence (for b1 is v3-v1) and to get the current you have to divide by that resistence, do that and after lot of mathematical manouvers you get what you need. UserIdTAG: 401611 UserNameTAG: FedericojjjkQkkuevedo CreateTimeTAG: 2012-09-11T03:48:31Z VoteTAG: 2 CoursewareTAG: Week 2 / Linear combinations of source strenghts CommentableIdTAG: 6002x_S3E2_Linear_Combinations_of_Source_Strengths NumberOfReplyTAG: 1 FirstChildTAG: The math is rather simply. as you said i1 = (v3-V1)/R1 thus b1 = (a1-1)/R1, b2 = (a2-1)/R1. Convert 1 to (R1*R2+R1*R3+R2*R3)/(R1*R2+R1*R3+R2*R3) and the just substitute for a1 from the first answer. FirstChildUserIdTAG: 66101 FirstChildUserNameTAG: bluefin FirstChildCreateTimeTAG: 2012-09-11T04:22:43Z SecondChildTAG: No disagreement with your method - but I think b2 is a2/R1 not (a2-1)/R1, for anyone else who might be following this. SecondChildUserIdTAG: 48907 SecondChildUserNameTAG: jskelton1979 SecondChildCreateTimeTAG: 2012-09-12T12:15:58Z IndexTAG: 1716 TitleTAG: Texbook Reading vs. Courseware Progress It would fine it helpful to state the proper textbook reading chapters/pages as I progress through the Courseware, so that I could better keep textbook reading in sync with the course details. ( I am taking this course 100% online. ) UserIdTAG: 330688 UserNameTAG: chota300 CreateTimeTAG: 2012-09-11T00:10:19Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: The assigned readings can be found at **Course Info->6.002x At-A-Glance**. FirstChildUserIdTAG: 194509 FirstChildUserNameTAG: blackcompe FirstChildCreateTimeTAG: 2012-09-11T00:56:47Z SecondChildTAG: Thanks. It would still be nice to see this information in the courseware section. SecondChildUserIdTAG: 107219 SecondChildUserNameTAG: AlexandreZ SecondChildCreateTimeTAG: 2012-09-11T16:30:05Z IndexTAG: 1717 TitleTAG: The burning of a pickle For those who want a try on burning a pickle see our results [here][1]: http://6002x-sv.blogspot.com/2012/09/the-burning-of-pickle-iii.html The success of a course like 6.002x is based on capturing students attention. Anant Agarwal and his team have become masters of their craft. [1]: http://6002x-sv.blogspot.com/2012/09/the-burning-of-pickle-iii.html UserIdTAG: 92800 UserNameTAG: carlos_sv CreateTimeTAG: 2012-09-10T21:52:47Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 1718 TitleTAG: Week 3 videos, HW and labs schedule Hi, The class calendar shows that today (Sep. 10th) HW3 and Lab3 are out. Will the videos be out at the same time? Do I have to finish the previous HWs and videos before I can see week 3? or do they come out at midnight? Does anyone have any idea? Thanks! UserIdTAG: 303762 UserNameTAG: leocarrasco CreateTimeTAG: 2012-09-10T21:11:01Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Hey! Hola Leo, I was wondering the same... it's weird, There's nothing for week 3 posted yet... FirstChildUserIdTAG: 342966 FirstChildUserNameTAG: SandraNavarro FirstChildCreateTimeTAG: 2012-09-10T21:36:43Z SecondChildTAG: Given your enthusiasm to start looking at Week 3 material we have decided to release Week 3 videos along with Week 3 HW and Lab. We will do this in the future weeks as well. Week 3 material should get released soon. SecondChildUserIdTAG: 42833 SecondChildUserNameTAG: KKA SecondChildCreateTimeTAG: 2012-09-10T23:23:33Z SecondChildTAG: to @KKA: please suggest any alternative to see the videos as you tube is blocked in our country. SecondChildUserIdTAG: 403493 SecondChildUserNameTAG: aiai SecondChildCreateTimeTAG: 2012-09-26T09:26:06Z IndexTAG: 1719 TitleTAG: HomeWork Can someone helpe me with this question: "Given that each individual resistor can dissipate up to 1 watt of power before burning up, how much total power in watts ( W ) can the smallest-valued composite resistor dissipate before burning up?". My first thought was that, since all dissipate 1W each, the equivalent resistor would dissipate the same value.. UserIdTAG: 304905 UserNameTAG: SergioSilva CreateTimeTAG: 2012-09-10T19:09:57Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 3 FirstChildTAG: Hello, when resistors are connected in parallel and have different resistance they are dissipate a different power. FirstChildUserIdTAG: 277168 FirstChildUserNameTAG: AndrewKiselev FirstChildCreateTimeTAG: 2012-09-10T19:27:42Z SecondChildTAG: And pls. distinguish "nominal" power of a resistor and power which is really dissipated by the resistor (due to current ) . SecondChildUserIdTAG: 277168 SecondChildUserNameTAG: AndrewKiselev SecondChildCreateTimeTAG: 2012-09-10T19:47:27Z FirstChildTAG: That's a homework question. First you should make sure you've got the smallest-valued composite resistor, then figure out which resistor will dissipate 1W first. From there you should be able to figure it out! FirstChildUserIdTAG: 248902 FirstChildUserNameTAG: Chanute FirstChildCreateTimeTAG: 2012-09-10T19:36:58Z SecondChildTAG: Solved! Thanks!! SecondChildUserIdTAG: 304905 SecondChildUserNameTAG: SergioSilva SecondChildCreateTimeTAG: 2012-09-11T01:11:45Z FirstChildTAG: The lower value resistor (4 ohm) will hit its 1 Watt power limit sooner- the higher value resistors (6 ohm) are still dissipating less than 1 Watt when the 4 ohm resistor hits 1 Watt. FirstChildUserIdTAG: 228637 FirstChildUserNameTAG: DCounsell FirstChildCreateTimeTAG: 2012-09-10T20:01:11Z SecondChildTAG: Solved! Thanks!! SecondChildUserIdTAG: 304905 SecondChildUserNameTAG: SergioSilva SecondChildCreateTimeTAG: 2012-09-11T01:11:50Z IndexTAG: 1720 TitleTAG: can some one please explain what I'm doing wrong I'm trying to solve this question using the Node analysis method - I'm using this equation -> (e-ev1)/R1 + (e-ev2)/R2 = 0 (ev1 = 5, ev2 = 7.2 ) this should be working but I can't get the right result. UserIdTAG: 271081 UserNameTAG: YonJah CreateTimeTAG: 2012-09-10T18:35:05Z VoteTAG: 2 CoursewareTAG: Week 2 / Node equation review CommentableIdTAG: 6002x_S3E1_NodeEquationReview NumberOfReplyTAG: 3 FirstChildTAG: You have written the expression correctly may be you are making mistake in the calculation FirstChildUserIdTAG: 336001 FirstChildUserNameTAG: syd_buet12 FirstChildCreateTimeTAG: 2012-09-10T18:38:47Z SecondChildTAG: Expression is correct, but ev2= **-** 7.2V , not +7.2V SecondChildUserIdTAG: 258500 SecondChildUserNameTAG: karas SecondChildCreateTimeTAG: 2012-09-10T19:31:55Z FirstChildTAG: Yes, I agree with syd_buet12 , You have written the expression correctly. FirstChildUserIdTAG: 277168 FirstChildUserNameTAG: AndrewKiselev FirstChildCreateTimeTAG: 2012-09-10T19:01:42Z SecondChildTAG: Thanks I'll try to solve the expression again SecondChildUserIdTAG: 271081 SecondChildUserNameTAG: YonJah SecondChildCreateTimeTAG: 2012-09-11T05:22:11Z FirstChildTAG: Hi, it was discussed in the post: [V2 sign confusion][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_S3E1_NodeEquationReview/threads/504bdb642a555b1f0000000a FirstChildUserIdTAG: 277168 FirstChildUserNameTAG: AndrewKiselev FirstChildCreateTimeTAG: 2012-09-10T18:44:16Z IndexTAG: 1721 TitleTAG: signal and it's speed what do u mean by signals speed of interest ??? and why it's speed shud be way lower than speed of light UserIdTAG: 343262 UserNameTAG: rgdixitt CreateTimeTAG: 2012-09-10T13:11:13Z VoteTAG: 2 CoursewareTAG: Week 1 / Lumped Element Abstraction CommentableIdTAG: 6002x_lumped_element_abstraction NumberOfReplyTAG: 2 FirstChildTAG: Physics... Read it ) FirstChildUserIdTAG: 394836 FirstChildUserNameTAG: v2g6ch4 FirstChildCreateTimeTAG: 2012-09-10T13:26:30Z SecondChildTAG: thats not fair! He's already done reading! All he needs is for you too explain! SecondChildUserIdTAG: 117661 SecondChildUserNameTAG: BUNDAY SecondChildCreateTimeTAG: 2012-09-10T15:03:07Z FirstChildTAG: The signals need to be slower then the speed at which the system operates. If signals are too fast, the the system cannot react fast enough to changes, or keep up with the frequency of changes. Ex, a fast (high frequency) signal would switch polarities too fast. By the time the first positive swing reached the other side of the device, the source may have switched polarities many times, thus rendering the device useless in "real time". (Our devices run at a speed slower then light) Newer processors are coming close to operating at light's frequencies, which will eventually be a problem. Wavelength size is also a consideration, if the device is to run "instantaneous". The device needs to be physically smaller then the wavelength used. "Slow" wavelengths, like 60Hz used in our power grids are huge, that's why it is able to stay in sync over a large area. "Fast" wavelengths like the ones used in our computers are much, much smaller and cannot travel more then a few centimeters before changing phase. Frequency implies that it is an "Alternating Current" (AC), so the signal changes from positive to negative, over and over again. If the distance is too far compared to the wavelength, A positive voltage at the source, may be viewed as a negative voltage by the time the signal reaches the destination. Hope this helps a bit, it's kinda vague, and probably wrong, but it's how I kinda understand it. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-10T14:46:42Z IndexTAG: 1722 TitleTAG: HTML5 video not playing Ironically, the video from the edX tutorial about this very discussion forum isn't playing in HTML5 (YouTube's WebM). I am using Linux and do not want to have to install proprietary software and/or unsupported plugins in order to follow this course. I haven't yet found another video here with the same problem. UserIdTAG: 204217 UserNameTAG: smar CreateTimeTAG: 2012-09-10T12:38:28Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 3 FirstChildTAG: "using Linux (Fedora 17) and do not want to have to install proprietary software and/or unsupported plugins in order to follow this course. " If you so smart, why u can't get direct link to video? ) "I am using freeBSD + links portable and do not want to have to install any another software. So i can't watch video. And it's YOUR problem!" - it souns like this... FirstChildUserIdTAG: 394836 FirstChildUserNameTAG: v2g6ch4 FirstChildCreateTimeTAG: 2012-09-10T13:01:28Z SecondChildTAG: I did get the direct link to the video, but can't get it playing without the flash plugin or similar. Of course it's my problem, but I do think that the videos should be platform/plugin-independent. There is no point in supporting a dying proprietary technology even though there are some work-around solutions which are distributed under open source licences. **Note:** As I can't comment further I will extend this post below without changing its upper part. edX is about innovation and knowledge sharing. It is open to everyone and there is no mention that I need a particular plugin in order to play the content of this course. As the case should be! HTML5 is the future and before voting down comments and writing unnecessary ones, you could at least read better what the original poster has said. I don't mean to be rude, for me this really is a **serious problem**. It's not that I *can't* install a certain plugin. In fact, I may have done so, you can't know that for sure. It's just hard for many and even not always possible (e.g. when you are using a public workstation in a library). No one should get only a limited access to the tremendous content of edX just because they don't know how or aren't allowed to install the necessary software when most of the parts of the website work perfectly with only a modern web browser. P.S. Have fun exploring the beauty of electronics with this class! **Last note:** If you want to attract the masses, yes, you can consider even a text version. The percentage is, however, really low in comparison with the flash problem. And citation from http://fedoraproject.org/: > Fedora is a fast, stable, and powerful operating system for everyday > use built by a worldwide community of friends. It's completely free to > use, study, and share. Please stop trolling and allow for a purposeful discussion here. SecondChildUserIdTAG: 204217 SecondChildUserNameTAG: smar SecondChildCreateTimeTAG: 2012-09-10T13:28:23Z FirstChildTAG: I'm having similar problems. I had to opt-out of the HTML5 experiment on my laptops and I can't play the videos on my Nexus7. Yes, I know I could install the flash apk on my tablet, but I would really appreciate it if edX brought their HTML5 support out of beta. Flash has a nasty tendency to crash my browsers and not depending on it makes life happier. FirstChildUserIdTAG: 107219 FirstChildUserNameTAG: AlexandreZ FirstChildCreateTimeTAG: 2012-09-11T15:47:46Z FirstChildTAG: Do you have HTML5 youtube enabled here http://www.youtube.com/html5 ? FirstChildUserIdTAG: 207456 FirstChildUserNameTAG: Octomos FirstChildCreateTimeTAG: 2012-09-12T19:24:29Z IndexTAG: 1723 TitleTAG: KVL and KCL If i understood: Kvl=**-V+v1+v2**=0--> **-V+i1R1+i2R2**=0 --> **-2+4i1+5i2**=0 **4i1+5i2=2(1)** KCL=**-i1+i2-I**=0--> -i1+i2=I --> -i1+i2=3----> **-i1+i2=3(2)** equation(1) and equation(2) give: i2=1.55555555 A i1=-1.4444444 A V2=i2R2=7.77 What is the power (in Watts) dissipated by the resistor with resistance R2 **P=(V2^2)/R2** power (in Watts) dissipated by the resistor with resistance R1 **P=(V1^2)/R1** power (in Watts) supplied by the voltage source **P=V*i1=2*1.4444** power (in Watts) supplied by the current source? **P=v2*I=7.77*3** UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-10T01:28:24Z VoteTAG: 2 CoursewareTAG: Week 1 / KVL/KCL example CommentableIdTAG: 6002x_S2E3_Using_KVL_KCL_and_VI_constraints NumberOfReplyTAG: 2 FirstChildTAG: i did exactly as you did , same results FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-12T16:34:26Z FirstChildTAG: Don't forget that $i_1$ is negative for the Power equation of $V$. FirstChildUserIdTAG: 359452 FirstChildUserNameTAG: Panoplos FirstChildCreateTimeTAG: 2012-09-21T02:34:44Z IndexTAG: 1724 TitleTAG: Use superposition Since the circuit is linear, you can figure out all the linear variables first. First do the circuit like the voltage source is not there (but the path is unblocked - - - a voltage source ideally has no resistance). Then do the circuit like the current source isn't there (but the path is totally blocked - an ideal current source has infinite resistance). Whatever currents/voltages you get across the resistors in each case can be added together to get the total interaction. Then you can get the powers by using the normal formulas (power isn't linear so you can't get it by superposition in the same way). That's it :) UserIdTAG: 253630 UserNameTAG: PhillipAdkins CreateTimeTAG: 2012-09-09T23:14:40Z VoteTAG: 2 CoursewareTAG: Week 1 / KVL/KCL example CommentableIdTAG: 6002x_S2E3_Using_KVL_KCL_and_VI_constraints NumberOfReplyTAG: 0 IndexTAG: 1725 TitleTAG: Lab1 sheets completely empty - cleaned - cleared out. Lab 1 - sheets cleaned out: i submitted my hw & lab 1 data earlier, and now went back to check it, and now lab sandbox sheets are empty - and i mean totally empty, clear, cleaned out of everything. That just doesnt look normal so i am worried. At progress i do have 100% on both homework and lab 1, but the thing is that i am just anxious to see the lab sheets gone. It was easy and i dont want to score zero on this one. Is that normal or is a bug or something? Can anybody confirm it either way? Tyvm for any kind of reply. UserIdTAG: 331483 UserNameTAG: Doru CreateTimeTAG: 2012-09-09T23:12:37Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: i am facing same problem and have complained in this forum FirstChildUserIdTAG: 40209 FirstChildUserNameTAG: sadaf78 FirstChildCreateTimeTAG: 2012-09-09T23:42:28Z FirstChildTAG: Follow-up on myself. Seems to be a local bug something with firefox. If i open the same lab1 page with Chrome then both sheets display correctly with components and solution on them. Sigh :) FirstChildUserIdTAG: 331483 FirstChildUserNameTAG: Doru FirstChildCreateTimeTAG: 2012-09-09T23:22:14Z SecondChildTAG: This is a known bug with Firefox :( SecondChildUserIdTAG: 79361 SecondChildUserNameTAG: kimt SecondChildCreateTimeTAG: 2012-09-09T23:49:34Z IndexTAG: 1726 TitleTAG: S3E2: Bs and Cs I'm confused as to where we get the linear expressions for i1 and i2. I understand the voltage one just fine from node method, but from where do the current expressions come? UserIdTAG: 183711 UserNameTAG: MMatheson CreateTimeTAG: 2012-09-09T19:20:47Z VoteTAG: 2 CoursewareTAG: Week 2 / Linear combinations of source strenghts CommentableIdTAG: 6002x_S3E2_Linear_Combinations_of_Source_Strengths NumberOfReplyTAG: 4 FirstChildTAG: Hi, Really, all you need to do, is find the expressions for v3, i1 and i2, to do so, you could apply the node method and find that the currents i1 and i2 are related to v3. Apply the node method, and solve first v3 in function of V1 and V2, and you'll be able to see it from there. FirstChildUserIdTAG: 83726 FirstChildUserNameTAG: Enriquejgc FirstChildCreateTimeTAG: 2012-09-09T20:16:57Z SecondChildTAG: could show how you do it SecondChildUserIdTAG: 462861 SecondChildUserNameTAG: edxforme SecondChildCreateTimeTAG: 2012-09-26T14:39:07Z FirstChildTAG: I think you could see it better thinking in terms of superposition. Regarding the linearity of the circuit, you can solve the circuit in two steps, removing one of the voltage sources on each step. So for b1, you will have only V1 (V2 is shorted), and you can find the current i1 by dividing V1 by the equivalent resistor in the loop, which is R1 plus the parallel of R2 and R3 (and be careful with the sign of the current). Same procedure applies to the rest of the parameters. I hope it helps! FirstChildUserIdTAG: 370084 FirstChildUserNameTAG: jmiguelsalgado FirstChildCreateTimeTAG: 2012-09-09T22:12:22Z SecondChildTAG: Also note that the coefficient b1 does not have V1 in the algebraic expression ;) SecondChildUserIdTAG: 370084 SecondChildUserNameTAG: jmiguelsalgado SecondChildCreateTimeTAG: 2012-09-09T22:16:45Z FirstChildTAG: Once you calculate a1 & a2 by superposition you know the voltage of the top node i.e. a1*V1+a2*V2. Now use the node method to calculate the current leaving the node, For example, i1 = (v3-V1)/R1. You know v3 in terms of V1 & V2, so b1 = (a1-1)/R1. FirstChildUserIdTAG: 66101 FirstChildUserNameTAG: bluefin FirstChildCreateTimeTAG: 2012-09-10T03:25:54Z SecondChildTAG: I got correct answer for v3, then i done the above method for b1, bt i am not getting the correct answer. SecondChildUserIdTAG: undefined SecondChildUserNameTAG: undefined SecondChildCreateTimeTAG: 2012-09-12T16:30:00Z FirstChildTAG: I'm a little confused too. Does anyone else believe the model answers for b2 and c1 are wrong? I worked b2 out by calculating the v2 current component as voltage dividers current through one leg of parallel resistors. eg the part of i1 due to V2 as =v2*[(r3*r1)/(r3+r1)+r2]/r1 which reduces to v2*r3/(r1+r2+r3) ie b2 = r3/(r1+r2+r3) FirstChildUserIdTAG: 345958 FirstChildUserNameTAG: PWilson123 FirstChildCreateTimeTAG: 2012-09-11T21:23:50Z IndexTAG: 1727 TitleTAG: h2p2 Anyone got a hint concerning the way to answer this question: "What is the power (in Watts) that is delivered to this best load resistance?" Dont need the answer just a hint - maybe I'm not sure if I understanding the question... UserIdTAG: 222931 UserNameTAG: tnyander CreateTimeTAG: 2012-09-09T17:25:17Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 4 FirstChildTAG: the load resistance and the transmission line resistance (Rp+2Rs) are together consider as the best load resistance FirstChildUserIdTAG: 130187 FirstChildUserNameTAG: Yeshwanthjpr FirstChildCreateTimeTAG: 2012-09-09T18:04:01Z SecondChildTAG: Sorry but i understand if best load resistance is a sum of 2Rs and Rl, then i compute P system tell me wrong result, may be its bug in system? SecondChildUserIdTAG: 321414 SecondChildUserNameTAG: Brash SecondChildCreateTimeTAG: 2012-09-12T13:27:28Z SecondChildTAG: at max. power transfer Rl=Rth SecondChildUserIdTAG: 349139 SecondChildUserNameTAG: 1977ROYELMER SecondChildCreateTimeTAG: 2012-09-16T13:33:17Z SecondChildTAG: maximum power transfer theorem. Help with that. dont know it. SecondChildUserIdTAG: 171674 SecondChildUserNameTAG: wajahat1 SecondChildCreateTimeTAG: 2012-09-16T16:12:22Z FirstChildTAG: The voltage across the two loops is the same. Express this voltage in term of the current through each loop. For the load loop you then want to maximize i^2*RL FirstChildUserIdTAG: 66101 FirstChildUserNameTAG: bluefin FirstChildCreateTimeTAG: 2012-09-09T18:45:47Z FirstChildTAG: You should comverter the circuit without load in its Thevenin equivalent, with a voltage source and a resistor in series. Then you must apply the law of maximum energy transfer. FirstChildUserIdTAG: 361823 FirstChildUserNameTAG: EliasOak FirstChildCreateTimeTAG: 2012-09-09T18:46:22Z FirstChildTAG: I'm also having trouble here. By impedance mismatching, I understand how we obtain the best load resistance. For maximum power transfer, with respect to what would we take the derivative? dP/d? FirstChildUserIdTAG: 183711 FirstChildUserNameTAG: MMatheson FirstChildCreateTimeTAG: 2012-09-13T00:35:36Z SecondChildTAG: dPload/dRload SecondChildUserIdTAG: 396446 SecondChildUserNameTAG: RousseauxS SecondChildCreateTimeTAG: 2012-09-13T15:36:37Z SecondChildTAG: Wouldn't that just leave i^2 ? Or are we to take i=i(Rload) ? SecondChildUserIdTAG: 183711 SecondChildUserNameTAG: MMatheson SecondChildCreateTimeTAG: 2012-09-16T17:47:12Z IndexTAG: 1728 TitleTAG: Sintaxis suggestion It should allow to input the equation in terms of conductance as well. In this case, (e2-e1)*G3+(e2-V0)*G4+e2*G5-I1 should be possible. Regards! UserIdTAG: 378522 UserNameTAG: Alejo_Velasquez CreateTimeTAG: 2012-09-09T17:03:06Z VoteTAG: 2 CoursewareTAG: Week 1 / Node analysis practice, part 3 CommentableIdTAG: 6002x_L2Node2 NumberOfReplyTAG: 2 FirstChildTAG: Your answer is right in terms of conductance. But here it requires with resistance and not with conductance. They should have mentioned that though. Hope this helps :) FirstChildUserIdTAG: 269109 FirstChildUserNameTAG: Sanjana16 FirstChildCreateTimeTAG: 2012-09-09T18:24:19Z SecondChildTAG: Hi. (Now)They have mentioned it explicitly (to enter in Resistance and not in Conductance). While the end equation (whether in conductance or resistance, will make no difference, I think it is more about the complexity of implementation of Parsing our answers that MIT developers wish to avoid, hence making it simple to implement also! SecondChildUserIdTAG: 372623 SecondChildUserNameTAG: Amitraj SecondChildCreateTimeTAG: 2012-09-16T07:12:34Z FirstChildTAG: Yes you are right, But you should also flow the question given. FirstChildUserIdTAG: 159427 FirstChildUserNameTAG: Suyog FirstChildCreateTimeTAG: 2012-09-16T08:35:44Z IndexTAG: 1729 TitleTAG: The 120 Vac is already the Root Mean Square of the AC The 120 V AC is the RMS Value. The AVERAGE voltage can be found by multiplying the Peak Voltage by 0.636. The AVERAGE Voltage is NOT used to determine the same power as the DC equivalent voltage. This is the second mistake I have found in the FIRST WEEK!!!! UserIdTAG: 145676 UserNameTAG: CathyPK CreateTimeTAG: 2012-09-09T16:25:41Z VoteTAG: 2 CoursewareTAG: Week 1 / AC power CommentableIdTAG: 6002x_S1E3_AC_power NumberOfReplyTAG: 3 FirstChildTAG: Actually the average voltage would be 0 since it's AC. FirstChildUserIdTAG: 270011 FirstChildUserNameTAG: SnowmanZA FirstChildCreateTimeTAG: 2012-09-09T16:28:20Z SecondChildTAG: SnowmanZA, You might want to read this: https://6002x.mitx.mit.edu/discussion/question/16744/isnt-the-average-power-zero SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-09-09T16:43:11Z SecondChildTAG: right SecondChildUserIdTAG: undefined SecondChildUserNameTAG: undefined SecondChildCreateTimeTAG: 2012-09-09T16:57:54Z SecondChildTAG: Dear SnowmanZA, The average voltage appears that it would be zero, but because it does cause heat in the resistor, the average cannot be zero. SecondChildUserIdTAG: 145676 SecondChildUserNameTAG: CathyPK SecondChildCreateTimeTAG: 2012-09-09T19:21:47Z FirstChildTAG: CathyPK, I assume you are referring to S1E3, AC Power? Is that right? Or can you otherwise point to where you are referring to? FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-09-09T16:31:18Z SecondChildTAG: Dear JSChambers, 0.636 * Vmax = average voltage, and 1/(SQRT 2) * Vmax = Root Mean Square. The Voltage RMS is exactly equivalent to the DC Voltage as far as Power is concerned. How to derive them, I forgot. 0.636 does not equal cos(45 degrees), Maybe someone can help us there. I can't rememb SecondChildUserIdTAG: 145676 SecondChildUserNameTAG: CathyPK SecondChildCreateTimeTAG: 2012-09-09T19:37:21Z FirstChildTAG: Hello. Is it 0.636 or 1/sqrt(2) which would equal 0.707? FirstChildUserIdTAG: 119440 FirstChildUserNameTAG: WG FirstChildCreateTimeTAG: 2012-09-12T03:36:44Z SecondChildTAG: 1/SQRT(2) = .7071, .7071*Vmax = Vrms 0.636*Vmax = average V SecondChildUserIdTAG: 145676 SecondChildUserNameTAG: CathyPK SecondChildCreateTimeTAG: 2012-09-12T22:34:33Z IndexTAG: 1730 TitleTAG: Current Flow Why the current of the circuit has de negative sign in DC analysis? UserIdTAG: 344485 UserNameTAG: Thiagolb CreateTimeTAG: 2012-09-09T15:54:37Z VoteTAG: 2 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 2 FirstChildTAG: 1 FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-09-09T16:04:22Z SecondChildTAG: 1 SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-09-10T18:05:35Z FirstChildTAG: The probe is showing a direction of current flow to the source through the circuit not directionally from the source to the circuit FirstChildUserIdTAG: 404685 FirstChildUserNameTAG: SamGS FirstChildCreateTimeTAG: 2012-09-09T18:16:31Z IndexTAG: 1731 TitleTAG: H1P2, Help with questions 7,8,9,10. Can anyone lend a few hints, or point me in the right direction for solving questions 7,8,9,10 in H1P2? It is all I have left to do for homework 1, I have been stuck on these questions for over a day now. Thanks, I'm having a blast! UserIdTAG: 292543 UserNameTAG: Pennypacker CreateTimeTAG: 2012-09-09T15:37:58Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 3 FirstChildTAG: Pennypacker, The instructions say to use KVL/KCL or the Node Method to solve, and that if you choose the appropriate ground node, you have only one node equation to solve. You can use either method, but I know which one I would use(hint: only one equation). If you want to use that method, the first step is choosing a ground node. Have you done that? I urge you to review video lecture S2V6(and subsequent videos), Node Method Analysis, where it lays out the five steps of the Node Method. The Node Method is one of the most important things to learn in this class, as you will use it all the time. You must know how to use it to finish successfully. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-09-09T15:44:09Z SecondChildTAG: Thanks for the reply. I will go and review S2V6, onwards. That's what I was after. (I was able to get 7 and 8 using a simulator, lol, I will try and learn to do 9,10 the proper way.) SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-09T16:03:25Z SecondChildTAG: Start with the node method, it's quite easy. From KCL, we know I2 + I3 = 8A. V2 = I2 (1R Resistor) V2 = V3 + 4 (3R resistor + 4V source in parallel with R2) I3 = V3 / 3 (3R resistor) so 8 = V3 + 4 + V3/3 Solving for V3 is then very simple. Once you have V3, you also have V2, I2 and I3. Then you can answer all the questions. SecondChildUserIdTAG: 270011 SecondChildUserNameTAG: SnowmanZA SecondChildCreateTimeTAG: 2012-09-09T16:25:23Z SecondChildTAG: Eureka I have finally done it. I was not picturing the 4V source as consuming power, simply adding a negative sign to an answer I had two days ago seems to have done the trick. -.- As far as the other question it was more of a shotgun approach, I will just sum it up to a clerical error in my calculations. I will need to review this again, at least now I feel better starting week 2 having completed week 1. Thanks again for your help guys. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-12T03:51:45Z SecondChildTAG: Hi Pennypacker and Snowman, Thanks, I could also get some help here, even If I read the text book and saw the videos. I had the same equations, just I think I get nervous. Will try to make it more in advance next time. Thanks again and cheers from Spain :)) SecondChildUserIdTAG: 342966 SecondChildUserNameTAG: SandraNavarro SecondChildCreateTimeTAG: 2012-09-16T18:04:21Z FirstChildTAG: Thanks guys, I am still hammering away. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-10T00:24:24Z FirstChildTAG: Not the only one: it happened badly the same to me. Sometimes you miss some assumptions and you can go very far away from the beginning :-( We'll have to keep on trying: doing all the exercises you can is the best cure for this. FirstChildUserIdTAG: 314477 FirstChildUserNameTAG: ignos FirstChildCreateTimeTAG: 2012-09-16T01:54:48Z IndexTAG: 1732 TitleTAG: DOUBT IN LAB1 In lab1 question, 1a, we can get node voltage A as 1.5 by just connecting top by 9ohm resister and other by just wire, by voltage division rule we will get 1.5v at node A, but its showing circuit is wrong, can please tell me......... UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-09T13:08:41Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Click on DC after making the circuit. it will show current and voltage values. then check again. FirstChildUserIdTAG: 242676 FirstChildUserNameTAG: LinjharaSahil FirstChildCreateTimeTAG: 2012-09-09T13:28:03Z IndexTAG: 1733 TitleTAG: What's behind an iPhone In the introductory video the teacher says "you'll know what's behind an iPhone" but the picture shows an Android device :) UserIdTAG: 401592 UserNameTAG: netghost CreateTimeTAG: 2012-09-08T19:45:23Z VoteTAG: 2 CoursewareTAG: Week 1 / Course overview CommentableIdTAG: 6002x_S1V1toS1V4 NumberOfReplyTAG: 3 FirstChildTAG: OMG you recognised that!!! According to USPO & iShit you shouldn't have FirstChildUserIdTAG: 198567 FirstChildUserNameTAG: rafee1344 FirstChildCreateTimeTAG: 2012-09-08T21:36:23Z SecondChildTAG: LOLLLLLLLLLLLL SecondChildUserIdTAG: 317009 SecondChildUserNameTAG: LucioAlex SecondChildCreateTimeTAG: 2012-12-19T17:25:18Z FirstChildTAG: Maybe they wanted to convey a message. FirstChildUserIdTAG: 364822 FirstChildUserNameTAG: Strus FirstChildCreateTimeTAG: 2012-09-08T20:12:48Z FirstChildTAG: @rafee1344, we don't use foul language in the discussion forum. Please be considerate of others, this is a learning exchange. Thanks. FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-09-09T02:43:16Z IndexTAG: 1734 TitleTAG: Q3 Help There are two wires. So when finding the total resistance it's true that we must consider the resistance of both the lines (Q1). But in Q3 to find Voltage drop we must consider only 1 wire right? because wires are situated parrellely to each other. UserIdTAG: 268444 UserNameTAG: Marlonabeykoon CreateTimeTAG: 2012-09-08T06:06:50Z VoteTAG: 2 CoursewareTAG: Week 1 / Power delivery modeling CommentableIdTAG: 6002x_S2E6_Modeling NumberOfReplyTAG: 3 FirstChildTAG: do not confuse yourself. the resistance is measured for the whole wiring system (ie)2*113*1.588*10^3 you will get r=0.358, mult FirstChildUserIdTAG: 114721 FirstChildUserNameTAG: vinayv FirstChildCreateTimeTAG: 2012-09-08T12:51:24Z FirstChildTAG: Yes, because it asks "from the house to the barn," not for both directions. EDIT: I take that back. It would seem that way, but to get the right answer you must consider both wires. I guess they meant the drop due to the line "from the house to the barn AND from the barn to the house." FirstChildUserIdTAG: 325730 FirstChildUserNameTAG: gadzin1203 FirstChildCreateTimeTAG: 2012-09-09T21:02:17Z SecondChildTAG: Agreed, the question might need re-wording. SecondChildUserIdTAG: 230787 SecondChildUserNameTAG: Gabriel007 SecondChildCreateTimeTAG: 2012-09-10T18:22:42Z FirstChildTAG: Thank you everyone FirstChildUserIdTAG: 268444 FirstChildUserNameTAG: Marlonabeykoon FirstChildCreateTimeTAG: 2012-09-15T04:28:13Z IndexTAG: 1735 TitleTAG: Previous knowledge Following the videos, previous to this exercise we see nothing being said about POWER and POWER CONSUMPTION or DISSIPATION, so this must be understood as a previous knowledge. UserIdTAG: 175378 UserNameTAG: Ivanildo CreateTimeTAG: 2012-09-08T04:12:00Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 5_Simple_Power NumberOfReplyTAG: 2 FirstChildTAG: Remember the AP level physics requirement for this course? FirstChildUserIdTAG: 364822 FirstChildUserNameTAG: Strus FirstChildCreateTimeTAG: 2012-09-08T04:39:34Z FirstChildTAG: is @ textbook :) ![DISSIPATION][1] [1]: https://lh6.googleusercontent.com/-G4SH-Lev6YU/UExy9rWgFtI/AAAAAAAAAIU/2rg6tu_OPXc/s391/powerDissip.png FirstChildUserIdTAG: 263290 FirstChildUserNameTAG: JaoCarSan FirstChildCreateTimeTAG: 2012-09-09T10:47:58Z IndexTAG: 1736 TitleTAG: S1E3 Does anyone know how to calculate the answer of the second exercise in this section? UserIdTAG: 248905 UserNameTAG: AJGP CreateTimeTAG: 2012-09-07T20:26:50Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: We know the resistance is 110 ohms and the voltage is 120 * sqrt(2) * cos(2pi * 60 * t). Since power(P) can be calculated using P = vi and i = v/r. We can substitute v / r into the equation for power. P = vi = v^2/r Substituting the values of V and R: P = (120 * sqrt(2) * cos(2pi * 60 * t))^2 / 110 = 28800 * cos^2(2pi * 60 * t) / 110 = 261.818181.... * cos^2(2pi * 60 * t) To find the average power integrate the equation for power we determined above(this will give us an equation for the total power dissipated over a particular length of time): So we now have 261.8181... * integral(cos^2(2pi * 60 * t)dt): If we were to integrate using this we would get an equation that wouldn't be useful so we use the trigonometric identity cos^2(x) = (1 + cos(2x)) / 2 Now the equation is: 261.8181.... * integral((1 + cos(4pi*60*t))/2 * dt) Now evaluate the integral. The result is: 261.8181... * [t/2 + 240pi * sin(4pi * 60 * t) / 2] Now evaluate the integral over one wave cycle ie t=0 to t=1/60. This will cause the sin term to be zero at both t's, which leaves us the equation: 261.8181... * [(1/60)/2 - 0/2] = 261.8181.../120 Since this equation is the power dissipated over a t= 1/60s, we can divide it by 1/60 to get the average power, which give us: (261.8181.../120)/(1/60) = 261.8181.../2 = 130.9091 Hopefully this helps. FirstChildUserIdTAG: 5242 FirstChildUserNameTAG: Pmurph FirstChildCreateTimeTAG: 2012-09-07T20:59:31Z SecondChildTAG: Thank you for the detailed explanation but regarding the last step while calculating the average power, you have already integrated the power over one cycle (1/60 s), the integration output should be the average power without the need to add anymore steps, why did you divide again by 1/60? ---------- sorry for bothering you, I have looked it up and found that I am mistaken, Integration itself gives the energy which needs to be divided by time. SecondChildUserIdTAG: 398394 SecondChildUserNameTAG: Sakr85 SecondChildCreateTimeTAG: 2012-09-08T03:12:14Z SecondChildTAG: can you tell me why we are integrating from 0 to 1/60? SecondChildUserIdTAG: 134904 SecondChildUserNameTAG: ASHWIN07 SecondChildCreateTimeTAG: 2012-09-12T17:00:13Z SecondChildTAG: Thank you Pmurph for your help. You saved all my night :) Anakluhur thank for this explanetion. I gave you a like, boys SecondChildUserIdTAG: 821 SecondChildUserNameTAG: mpaluch SecondChildCreateTimeTAG: 2012-09-17T03:22:08Z IndexTAG: 1737 TitleTAG: editing on dashboard i am unable to correct my details on dashboard/ profile . please help UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-07T18:48:40Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 1738 TitleTAG: Negative Current The diagramm shows a current value of -500uA after the DC analyses because the current is flowing from the circuit into the positive terminal of the source. But why the current is not flowing out of the negative terminal into the circuit? UserIdTAG: 230701 UserNameTAG: MEng_IIT CreateTimeTAG: 2012-09-07T17:38:03Z VoteTAG: 2 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 3 FirstChildTAG: ye sawal konxe chpter ya lecture ka hay?? FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-07T17:41:23Z SecondChildTAG: Example lab hai. Aise toh koi bhi chapter ka lecture nahin. Just online lab se familiarise hone ke liye. SecondChildUserIdTAG: undefined SecondChildUserNameTAG: undefined SecondChildCreateTimeTAG: 2012-09-07T18:57:03Z FirstChildTAG: The diagram actually shows a negative value for the current because it actually flows in the other direction. When you do a circuit analysis you simply assume the direction that the current is flowing in, if your assumption is correct, then the value for the current will be positive, if you are wrong and the current's actually flowing in the opposite direction you initially assumed, then it will have a negative value. In the example lab, the current is assumed to be flowing from the negative to the positive side of the source, indicated by the arrow flowing into the positive terminal. Since the current displayed has a negative value, the assumption is wrong, and the current is actually +500uA flowing out of the positive terminal and into the negative one. Hope I could clear it up a bit for you. FirstChildUserIdTAG: 156608 FirstChildUserNameTAG: luiseramos FirstChildCreateTimeTAG: 2012-09-07T18:07:03Z SecondChildTAG: yups thats right. current is actually flow from positive to negative :) SecondChildUserIdTAG: 396953 SecondChildUserNameTAG: raisfikri18 SecondChildCreateTimeTAG: 2012-09-07T18:26:10Z SecondChildTAG: But doesn't current always flow from the positive terminal of the source, through the rest of the circuit and back into the negative terminal of the source? Because, electrons move from the negative terminal to the positive terminal so current flows in the exact opposite direction. So isn't it pointless to "**assume**" that the current flows from the negative terminal to the positive one, like in the example lab above? Because it always flows from the positive to negative terminal? SecondChildUserIdTAG: 281154 SecondChildUserNameTAG: NoelThomasNicholas SecondChildCreateTimeTAG: 2012-09-07T18:56:05Z SecondChildTAG: Thanks a lot, i´ve had that dude. I had forgotten the KLV, but now i remember. SecondChildUserIdTAG: 68468 SecondChildUserNameTAG: Ricap SecondChildCreateTimeTAG: 2012-09-08T03:52:32Z SecondChildTAG: Thank you for your excellent explanation as to why the current was negative, as it was (wrecking my head) annoying me very much. So the current is actually going from + to -, which you would expect, and the negative current implies that the arrow direction should be "turned around", so that it is is pointing away from the positive source, instead of towards it. So if the arrow had of been pointing away from the positive source, it would have been calculated as +500uA Thanks SecondChildUserIdTAG: 337858 SecondChildUserNameTAG: garyk SecondChildCreateTimeTAG: 2012-09-09T22:42:03Z FirstChildTAG: when the current was defined, it was defined as the movement of positive charges because they didn't know about the electron that's what it flows from the positive terminal I'm sorry about me English FirstChildUserIdTAG: 250179 FirstChildUserNameTAG: diegoCmC FirstChildCreateTimeTAG: 2012-09-08T03:36:40Z IndexTAG: 1739 TitleTAG: About proctored exam Hi, everyone. I wanna know more about proctored exam testing. Thanks. UserIdTAG: 21193 UserNameTAG: Oboneel CreateTimeTAG: 2012-09-07T17:37:27Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: So does everyone else :). Unfortunately, they haven't released more information than [this][1], though Pearson VUE (the company that will be organizing the testing) has also posted [this][2] article, and you can read through the site to see how they do their testing. That's all the information I could find, and you probably won't find more right now other than through direct contact with edX so be patient- We're all hoping with you. [1]: https://www.edx.org/press/edX-announces-proctored-exam-testing [2]: http://www.pearsonvue.com/about/release/12_09_06_edx.asp FirstChildUserIdTAG: 26931 FirstChildUserNameTAG: Csscade FirstChildCreateTimeTAG: 2012-09-08T01:43:56Z SecondChildTAG: Thanks :) SecondChildUserIdTAG: 21193 SecondChildUserNameTAG: Oboneel SecondChildCreateTimeTAG: 2012-09-10T05:24:49Z IndexTAG: 1740 TitleTAG: Adding probe to the diagram How can I add probe to the diagram ? UserIdTAG: 258932 UserNameTAG: ricardouy21662 CreateTimeTAG: 2012-09-07T15:24:36Z VoteTAG: 2 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 2 FirstChildTAG: Click on probe icon and drag it to the position in the diagram FirstChildUserIdTAG: 228637 FirstChildUserNameTAG: DCounsell FirstChildCreateTimeTAG: 2012-09-07T15:56:36Z FirstChildTAG: I can do it only using Internet explorer, if you are using another browser, could have problem like me. Try out again but using IE. FirstChildUserIdTAG: 174548 FirstChildUserNameTAG: andrepereira FirstChildCreateTimeTAG: 2012-09-07T15:53:25Z IndexTAG: 1741 TitleTAG: Q3 How did you calculate the drop? I assumed the source in his house is 240V, then calculated the current, and looked at the voltage on the barn-res, and took the difference. Its almost okay, but i think its wrong. Pls enlighten me. UserIdTAG: 329254 UserNameTAG: KGabor CreateTimeTAG: 2012-09-07T15:20:38Z VoteTAG: 2 CoursewareTAG: Week 1 / Power delivery modeling CommentableIdTAG: 6002x_S2E6_Modeling NumberOfReplyTAG: 3 FirstChildTAG: The source in his house isn't 240V - he needs to supply the 240V to the barn, and adjust the house source so it compensates for the drop in the wire. You can calculate the current via the power and voltage given for the barn load. FirstChildUserIdTAG: 267993 FirstChildUserNameTAG: mavlijas FirstChildCreateTimeTAG: 2012-09-07T15:41:03Z SecondChildTAG: Thanks for your reply - I calculated the current this way, and its ok. SecondChildUserIdTAG: 329254 SecondChildUserNameTAG: KGabor SecondChildCreateTimeTAG: 2012-09-07T17:28:06Z FirstChildTAG: you have to do is, find the current through the line trasmicion, and then multiply this current by the resistance of the line I=240/57.6 = 4.1666666 Vline= 4.1666*0.358 = 1.4916 FirstChildUserIdTAG: 64183 FirstChildUserNameTAG: jesuslayton FirstChildCreateTimeTAG: 2012-09-08T03:40:31Z FirstChildTAG: The source in the house is more than 240V so that the voltage supplied to the resistive load will be exactly 240V. Solution to Q3: I=P/V=25/6 A. Voltage drop=I*R(wire)=(25/6)*0.35888=1.495... FirstChildUserIdTAG: 227139 FirstChildUserNameTAG: qiuyi FirstChildCreateTimeTAG: 2012-09-08T06:02:28Z IndexTAG: 1742 TitleTAG: Negative Value Why the answer of the current is -1.17A.I don't understand because the value is negative. Can anyone tell me? UserIdTAG: 288842 UserNameTAG: Paulofilho CreateTimeTAG: 2012-09-07T15:06:22Z VoteTAG: 2 CoursewareTAG: Week 1 / Power CommentableIdTAG: 6002x_S1E2_Power NumberOfReplyTAG: 4 FirstChildTAG: When the circuit is solved, the circuit element currents may have positive or negative values. A negative value means that the actual direction of current through that circuit element is opposite that of the chosen reference direction. In electronic circuits the reference current directions are usually chosen so that all currents flow toward ground. This often matches conventional current direction, because in many circuits the power supply voltage is positive with respect to ground.(http://en.wikipedia.org/wiki/Electric_current) I think this might help. FirstChildUserIdTAG: 173585 FirstChildUserNameTAG: habib_akbar FirstChildCreateTimeTAG: 2012-09-07T16:04:11Z SecondChildTAG: As given in the S1E2 Problem, we must first find out the voltage across the resistor labeled 'R' having 8 ohms.However, we must first find out the current flowing towards the resistor 'R' using the derived formula I = sqrt(P/R), where I is current(amp), P is power(watt), and R is resistance(ohms). Therefore, this formula gives us a positive current of 1.17 Amperes flowing through the resistor 'R'. Thus, to find out the voltage we use the voltage formula V = R x I, where 8 ohms times 1.17 amperes gives us 9.38 voltage dissipated across resistor 'R'. Remember now, all of these values for voltage and current are positive because we are defining the direction of current to be moving towards our resistor 'R', which gives us a positive value of 1.17 ampere current entering the resistor 'R', not the Network N. However, if the same positive value of 1.17 ampere current turns to its opposite direction and entered Network N instead, than it will give us a negative value of (-1.17) ampere. Therefore, the value of current i entering Network N from resistor is a negative value: -1.17 ampere. SecondChildUserIdTAG: 181732 SecondChildUserNameTAG: Anthony341994 SecondChildCreateTimeTAG: 2012-09-11T02:51:38Z FirstChildTAG: When analyzing circuits we need to follow certain conventions. The convention we choose is to label current going *into* a node as positive, and current going *out of* a node as negative. There's nothing special about this choice, it just puts everyone on the same page. So when doing analysis, you total all the currents for a node and expect that they will total to zero. Since current is flow of electrons and electrons have mass which is conserved, the number going in must equal the number going out. In other words, the current must total to zero, which makes some currents be positive (into the node) and some negative (out of the node). FirstChildUserIdTAG: 10523 FirstChildUserNameTAG: Barrabas FirstChildCreateTimeTAG: 2012-09-07T16:15:01Z FirstChildTAG: Simply if you carefully examine the "direction of current" it is from negative to positive which is against the "CONVENTIONAL THEORY of FLOW of CURRENT". That's why there is a negative sign. FirstChildUserIdTAG: 164513 FirstChildUserNameTAG: MZeeshanSheikh FirstChildCreateTimeTAG: 2012-09-07T21:08:42Z FirstChildTAG: The arrow in the diagram shows the current flowing the opposite direction that what a positive current would flow, therefore it's negative current. The math won't tell you that, but the diagram does. FirstChildUserIdTAG: 426266 FirstChildUserNameTAG: TheNetImp FirstChildCreateTimeTAG: 2012-09-12T18:04:32Z IndexTAG: 1743 TitleTAG: TEXTBOOK I HAVE LOW SPEED BROWSING, KINDLY LET ME KNOW, HOW TO DOWNLOAD TEXTBOOK FOR READING. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-07T12:09:52Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 4 FirstChildTAG: Although I do not agree with the use of all caps to make the point, I would kindly request Mr. Agarwal and the concerned teams to allow us to download the book. It takes an immense amount of time to load for me as well. FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-07T12:42:22Z FirstChildTAG: Please do not post links to pirated material. This is illegal. The file in the 4shared link below has been deleted because it violated the 4shared user agreement terms. You can view the textbook free online via the edX courseware. If your bandwidth or Internet access is limited, you can purchase a discounted copy of the book (hard copy or softcopy) directly from the publisher using the discount code shown on the 'Course Info' tab. You can also purchase inexpensive international versions in much of the world. For example, in India, Flipkart sells the legal Indian reprint inexpensively. You can also purchase legal international versions on eBay, they are legal for importation for personal use in the USA, I do not know what the laws are in your country. FYI, I live in the USA, and I am using a legal Indian reprint that I purchased on eBay. The only difference between the US version and the Indian version is the text is all black/white and the paper is less expensive and non-glossy, the US version uses green color for some of the text and better quality glossy paper. The cheaper paper does have advantages besides being free of glare, it makes the book a little lighter to carry around. After a few weeks of lugging the book and your notes around, you will appreciate every little bit of weight that you don't have to carry! FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-09-10T05:12:55Z SecondChildTAG: The book that you are using is a good photocopy of the original Indian reprint. The Indian reprint has glossy pages and is also partially colored. There is virtually no difference between the Indian reprint and the original version of the book. Be very careful when buying the Indian versions, because a lot of the time, people buy a pirated version and not the real reprint. Secondly, people in India would pay a max of Rs. 500 behind a book, and that is something like 10 USD. Also there is no discount for the Indian reprint. Whereas with discount here, you can buy with $60. For most people in the India subcontinent, because people have lower incomes (and frankly their goods are a lot cheaper), they do not really have the kind of cash to buy these books. And another thing, (I am not using an illegal copy btw, but thats only because I have the kind of cash to afford the original), using an Indian reprint in the US is actually a breaking of copyright. You will see that the Indian version is only allowed in the Indian subcontinent (India, Pakistan, Sri-Lanka). I do not mean under any circumstances to belittle you, or what your stance is on piracy or copyright. However, please see that different people have different circumstances. Not so long ago, my financial situation was not favorable, so I needed to use pirated versions of books. Not because I wanted to, but because I had to. No one wants to pirate something, they only do it because the alternatives are not feasible. SecondChildUserIdTAG: undefined SecondChildUserNameTAG: undefined SecondChildCreateTimeTAG: 2012-09-10T05:32:34Z FirstChildTAG: the download link above is great - you have to make an account but it looks safe. the pdf download is pretty quick. I bought the text as I like to work in my books but the pages are shiny and the glare hurts my eyes - I don't have that problem with the pdf-laptop FirstChildUserIdTAG: 10512 FirstChildUserNameTAG: asicok FirstChildCreateTimeTAG: 2012-09-08T22:57:45Z FirstChildTAG: kindly check this link you can download the book from there http://www.4shared.com/office/s7km9HXd/Foundations_Of_Analog_And_Digi.html FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-07T13:20:11Z IndexTAG: 1744 TitleTAG: Random Query In the navigation bar where you click my this page icon is not fully black but i answered all questions, any reason why? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-07T10:27:26Z VoteTAG: 2 CoursewareTAG: Week 1 / KVL and KCL exercises CommentableIdTAG: 6002x_kvl_kcl_exercises NumberOfReplyTAG: 3 FirstChildTAG: I have the same problem, does anyone now what this is? FirstChildUserIdTAG: 300043 FirstChildUserNameTAG: klushcheva FirstChildCreateTimeTAG: 2012-09-07T12:57:02Z FirstChildTAG: same problem here... FirstChildUserIdTAG: 306244 FirstChildUserNameTAG: kevinsysum FirstChildCreateTimeTAG: 2012-09-09T18:40:24Z FirstChildTAG: here too. But probably just a bug. Unless there's some invisible question :P FirstChildUserIdTAG: 387218 FirstChildUserNameTAG: DiogoLopes FirstChildCreateTimeTAG: 2012-09-10T10:44:41Z IndexTAG: 1745 TitleTAG: hemu ques where is hw1 &2 UserIdTAG: 155590 UserNameTAG: hemulion CreateTimeTAG: 2012-09-07T09:23:10Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Courseware -> Week 1 -> Homework 1 [Here][1] Courseware -> Week 1 -> Lab 1 [Here][2] ---- Courseware -> Week 2 -> Homework 2 [Here][3] Courseware -> Week 2 -> Lab 2 [Here][4] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_1/Basic_Circuit_Analysis/ [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_1/Resistor_Divider/ [3]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_2/Linearity_Thevenin_and_Norton_Digital/ [4]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_2/Mixing_Two_Signals/ FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-07T13:17:11Z IndexTAG: 1746 TitleTAG: we Chinese can not visit youtube I notice maybe the video lectures are stored on the youtube website,but we can not watch them. Would someone else can help? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-07T05:12:25Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 3 FirstChildTAG: Unfortunately, edX hosts its videos on Youtube, which China blocks. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-09-07T21:55:56Z SecondChildTAG: I thought you could VPN to bypass the great firewall. Not sure what your resources are, but you should be able to get VPN for 10 USD or so per month. SecondChildUserIdTAG: 107219 SecondChildUserNameTAG: AlexandreZ SecondChildCreateTimeTAG: 2012-09-15T15:02:46Z SecondChildTAG: Not that I'm advocating you bypass the measures your government has put in place for your protection of course. SecondChildUserIdTAG: 107219 SecondChildUserNameTAG: AlexandreZ SecondChildCreateTimeTAG: 2012-09-15T15:04:03Z FirstChildTAG: same problems FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-08T13:06:32Z FirstChildTAG: You can see in the Wiki [Lecture Videos Download][1]:). Also see more info [here][2]: **Due to the YouTube outage in certain parts of the world, we have provided direct links to download the lecture videos, and extended the Week 2 homework and lab deadlines to the coming Sunday (September 30). Please note that Week 3 deadlines have not changed.** [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/ [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/info FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-26T11:52:38Z IndexTAG: 1747 TitleTAG: Is there's a difference between the current entrering the resistor and that leaves it ? I mean, resistors consume power. power is transferred from a source, as far as I understand the current can be looked at as something that "carry" that power to the load ( a resistor here) which consumes it. so this current is now somehow "unload" the power and moves on back to the source...Is that description right ? UserIdTAG: 34401 UserNameTAG: ROBOTIZER CreateTimeTAG: 2012-09-06T23:07:51Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 5_Simple_Power NumberOfReplyTAG: 3 FirstChildTAG: You can imagine it as a circular tube completely filled with marbles. Because it's completely filled, no one marble can move faster than the other - they all travel at the same speed. That speed would be the current. Now imagine a part of the tube is filled with mud which cannot escape that portion of the tube - that's our resistor. The marbles would still travel at the same speed - because there's no empty space in the tube the first marble has to push the second one out of the way before it can move, which pushes the next one and so on through the entire tube. What changes is the pressure on either end of the mud portion of the tube. That's the voltage. A voltage source provides a constant amount of "pressure" to the marbles. That pressure is then "spent" pushing the marbles through the mud portion, so you have a difference in pressure from one end of the mud portion to the other, like the difference in electric potential from one end of the resistor to the other. So the current (speed) wouldn't change, but electric potential (the pressure) would. The difference between potentials on either end of the resistor is the voltage drop on that resistor. FirstChildUserIdTAG: 267993 FirstChildUserNameTAG: mavlijas FirstChildCreateTimeTAG: 2012-09-06T23:30:40Z SecondChildTAG: Nice one, really cool analogy. There are few like this that I like. An amusing one I like to think of, is of people all stuffed together walking through different diameter sized tunnels. On entering the extra narrow tunnels the first person entering tries to push through but only manages to push person in front of him. So people entering tunnel from say a larger tunnel are going to find it much harder to get through but they can't stop as the people behind them are touching them and so pushing them through. So they smack up against the sides of tunnel which causes tunnel to warm up with all the bodies hitting it from different sides. They all move through tunnels at same speed (as no gaps between the people) they just get beaten up more than when they were in the wider tunnel sections. So you can think of narrow tunnels as high value resistors. And when you are at start of a narrow tunnel your going to need to use up a lot more energy banging into walls etc when moving through at same speed by time you get to other end. That is your potential energy difference (voltage drop), so you are expending much less energy when you first enter into a narrow tunnel but much more is used up by time you are beaten about against walls etc, at same speed to reach the other end. And the tunnel gets all that energy as heat. SecondChildUserIdTAG: 15344 SecondChildUserNameTAG: kob SecondChildCreateTimeTAG: 2012-09-07T00:43:50Z SecondChildTAG: It's like friction. When there is friction around, and you move at a certain speed, you lose kinetic energy (and thus speed) in the form of heat. If you want to go at a constant speed (like marbles in a tube), you need a force equal to that friction, facing opposite direction. That's at least how I imagine the power dissipation in a circuit. SecondChildUserIdTAG: 364822 SecondChildUserNameTAG: Strus SecondChildCreateTimeTAG: 2012-09-08T02:02:55Z FirstChildTAG: You're right. The current is carrying that power. And it's gets unloaded in the resistor. Remember that there is a potential difference between one end of the resistor and the other end of the resistor. That voltage drop means that the electric charges change their potential energy as they flow through the resistor. That loss in potential energy emerges as heat, as the charges battle against the "resistance" to proceed through the circuit element. The charges collide with the "vibrating atoms" and release heat that way, and this phenomenon is what we call "resistance". If we cool the resistor down sufficiently, so that the atoms are vibrating less, then the charges of the current collide less, and resistance decreases. So, resistance generally increases with rising temp and decreases with falling temp. FirstChildUserIdTAG: 4463 FirstChildUserNameTAG: pmj FirstChildCreateTimeTAG: 2012-09-07T01:23:12Z SecondChildTAG: Nice one pmj that is a perfect description. SecondChildUserIdTAG: 15344 SecondChildUserNameTAG: kob SecondChildCreateTimeTAG: 2012-09-07T01:50:49Z SecondChildTAG: Nice description.. Thanks for this.. SecondChildUserIdTAG: 138310 SecondChildUserNameTAG: ganeshbhatis SecondChildCreateTimeTAG: 2012-09-16T14:24:56Z FirstChildTAG: The difference is the charge being positive or negative and this is based on the voltage polarity or the current flow direction. FirstChildUserIdTAG: 214990 FirstChildUserNameTAG: Shuaibu FirstChildCreateTimeTAG: 2012-09-08T14:02:37Z IndexTAG: 1748 TitleTAG: how can there be 7 loops when the video S2V4: Method 1 - KVL, KCL method says 4 loops S2V4: Method 1 - KVL, KCL method after looking at this video i answered the questions and person in the video says that there are 4 loops and the circuit is identical . can anyone explain this? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-06T22:42:15Z VoteTAG: 2 CoursewareTAG: Week 1 / Circuit topology CommentableIdTAG: 6002x_S2E1_Circuit_Topology NumberOfReplyTAG: 4 FirstChildTAG: It's a little bit tricky because you don't move in square loops, you just start from any nod and come back to it. So there are the following loops : abd ,bdc, adc, abc, ac (those are loops in square figures), adbca, abdca. FirstChildUserIdTAG: 289472 FirstChildUserNameTAG: OsamaAdel FirstChildCreateTimeTAG: 2012-09-06T23:25:13Z SecondChildTAG: It seems that the original question is not what the 7 loops are, but why there is a discrepancy between S2V3 and S2V4 (which counts 4 loops total) and S2E1 (which counts 7 loops total). SecondChildUserIdTAG: 26279 SecondChildUserNameTAG: tavislhall SecondChildCreateTimeTAG: 2012-09-07T05:09:48Z FirstChildTAG: I think that Dr. Anant Agarwal only forget the others 3 loops. It was a simple mistaken. PS: Sorry if I wrote anything wrong. FirstChildUserIdTAG: 125390 FirstChildUserNameTAG: GabrielGentil FirstChildCreateTimeTAG: 2012-09-09T22:42:43Z SecondChildTAG: he didn't forget...all he was trying to say is that there can be more no. of loops than what we actually see....and he never said that the circuit had only 4 loops SecondChildUserIdTAG: 343262 SecondChildUserNameTAG: rgdixitt SecondChildCreateTimeTAG: 2012-09-11T10:26:18Z FirstChildTAG: U see there are 3 independent loops/meshes. 7 loops are the maximum combination of loops that can be formed from given circuit.if u solve for other 4 loops you will eventually get same answer! try it out ! FirstChildUserIdTAG: 372535 FirstChildUserNameTAG: _Infinity FirstChildCreateTimeTAG: 2012-09-10T13:32:23Z FirstChildTAG: i still dnt know how der r 7 loops....? pls sm1 xplain FirstChildUserIdTAG: 404902 FirstChildUserNameTAG: preeti123 FirstChildCreateTimeTAG: 2012-09-12T14:41:06Z SecondChildTAG: maximum possible loops are 7 ,,but 3 loops equations are enough to find out ckt parameters SecondChildUserIdTAG: 183166 SecondChildUserNameTAG: yogeshk SecondChildCreateTimeTAG: 2012-09-15T13:21:30Z SecondChildTAG: ![7 loops - illustration][1] [1]: https://edxuploads.s3.amazonaws.com/13511864642557189.gif SecondChildUserIdTAG: 714237 SecondChildUserNameTAG: PaxPolaris SecondChildCreateTimeTAG: 2012-10-25T17:35:10Z IndexTAG: 1749 TitleTAG: Definition of resistance In the old dear Halliday-Resnick, resistance is *defined* as $\frac{V}{I}$. Then things like diodes are simply devices that do not have a constant resistance. In this course, however, how is resistance defined? It seems to be more like an intrinsic property of some materials... UserIdTAG: 346056 UserNameTAG: fiatlux CreateTimeTAG: 2012-09-06T22:07:03Z VoteTAG: 2 CoursewareTAG: Week 1 / Lumped Element Abstraction CommentableIdTAG: 6002x_lumped_element_abstraction NumberOfReplyTAG: 4 FirstChildTAG: There are some weird concepts associated with the definition of "resistances" when you look at AC analysis as well as devices like diodes. Currently we're only looking at DC circuits (without capacitors and inductors), and under these conditions, resistance can also be *defined* as V/I. To answer your question about the intrinsic property of materials. This is a huge topic outside the scope of this course, but conductors and insulators tend to have relatively stable *conductivities*, which means that the resistance of a material is some magic number (derived or from complicated physics or measured) multiplied by the dimensions of the material. When you look at semiconductor devices like diodes, then the conductivity (hence resistance) can be drastically changed by the introduction of impurities and the presence of electric and magnetic fields - i.e. variable resistance. FirstChildUserIdTAG: 151427 FirstChildUserNameTAG: RichardZ FirstChildCreateTimeTAG: 2012-09-07T01:10:30Z FirstChildTAG: Resistance is defined in just the way you described it: V/I . In things like resistors, this is defined as a constant ratio, and thus they have linear V-I graphs, while in other nonlinear devices, I as a function of V is more complex and difficult to work with. Resistance is also pretty intrinsic to most materials, this depends mostly on chemical structure (how fast electrons are pushed along the atomic lattice) and physical properties of the resistor, mainly the length and width of the material the electrons can travel through. FirstChildUserIdTAG: 26931 FirstChildUserNameTAG: Csscade FirstChildCreateTimeTAG: 2012-09-06T22:55:47Z FirstChildTAG: The name "resistor" is the clue here. And if you think about it logically, it is really not too difficult to understand. Below is just an analogy which explains this logic. And all though never specified directly, its Its all about energy conservation really. For example, If you (acting as a human store of energy) keep pushing, to maintain a constant speed, a series of objects all close together like a long row of snooker balls all touching end to end on each other, through a series of joined tunnels just wide enough for them to get through. Then if a section of tunnel's walls become narrower and even touch the balls, then the balls will scrap the sides and hence will be losing energy to the walls. But because they all in a long line touching end to end they won't slow down much at all. You'll notice the extra energy release as you will need to push much harder to maintain the constant speed. You'll be behaving like a battery of constant voltage, ie, pushing at a constant speed. So although the walls are resisting this push they can not stop the progress of the balls. All that can happen is a movement of energy from the balls to the sides of the walls. (That energy comes from what ever it is that is pushing them all along against each other. "Battery" your tired arms) Larger resistors then, are like tighter tunnel sections for the balls to push through. This is just like what resistors do. And its why they get hot. So if you make a material that is hard for the balls (holes/electrons) to pass through then you have effectively, a large value resistor. But the tight section of tunnel (resistor) will never stop the progress of the balls (holes/electrons). A higher resistance only means more energy is given up by them as the ball pushes through. So you have a higher potential energy from entry end of the "tighter section of the tunnel" with respect to the exit end of the "tighter section of the tunnel". That is your voltage. And that is exactly what resistance means. It is all with respect to what direction the current moves in. So current always moves from a higher energy potential through to a lower energy potential. And if there is no resistance then the potential energy in that particular section of tunnel no longer exists (in other words, voltage zero = short circuit) For electrons think of lots of tiny tunnels all staked on top of each other parallel to each other. If some tunnels are hard to get through more balls will move through the less congested ones. So long sections of wide tunnels from start to finish will mean that you can push many more balls through at same speed. Hence a higher current. You can even push this analogy a bit further and explain linear versus non linear devices. Tight tunnel sections (Similar to Resistors) are linear if the resistance is equal through all parts of that tighter section of the tunnel. But lets assume one of these tight sections has sticky stuff on walls at one end and in different areas within this particular tighter section. Now when the balls are pushing through the energy absorbed will be different in different spots along same section. So the potential will be ever changing non linearly along this non standard tight section. Ha ha! FirstChildUserIdTAG: 15344 FirstChildUserNameTAG: kob FirstChildCreateTimeTAG: 2012-09-07T01:08:04Z FirstChildTAG: This course is dealing with the abstraction of the resistance, we have not to care about resistance itself or any kind of materials. We use a model based on the abstraction of resistance not the resistance itself. The resistance should be defined as V/I instead of being defined by V/I, that is, we have V/I and then call it "resistance". FirstChildUserIdTAG: 296965 FirstChildUserNameTAG: LGMailhos FirstChildCreateTimeTAG: 2012-09-07T16:54:57Z SecondChildTAG: You don't need to know the details. But they are not hard to learn and its nice when you know it as it helps to give a deeper understanding. SecondChildUserIdTAG: 15344 SecondChildUserNameTAG: kob SecondChildCreateTimeTAG: 2012-09-07T20:43:12Z IndexTAG: 1750 TitleTAG: edX en español para personas de habla hispana que no tengan un dominio total del ingles se creo el grupo en facebook edX en español el cual les perimitira hacer preguntas en su idioma nativo y nos ayudaremos entre todos en este gran reto por el aprendizaje UserIdTAG: 51696 UserNameTAG: blasco23 CreateTimeTAG: 2012-09-06T21:10:29Z VoteTAG: 2 CoursewareTAG: Week 1 / KVL and KCL CommentableIdTAG: 6002x_kvl_and_kcl NumberOfReplyTAG: 3 FirstChildTAG: No os encuentro poniendo "edx en español" en el buscador de fb, ¿alguna sugerencia para encontraros? FirstChildUserIdTAG: 249375 FirstChildUserNameTAG: MarIgones FirstChildCreateTimeTAG: 2012-09-06T21:16:34Z FirstChildTAG: Yo tampoco los encuentro . por favor me informarian exactamente donde los puedo localizar FirstChildUserIdTAG: 45303 FirstChildUserNameTAG: brenda FirstChildCreateTimeTAG: 2012-09-06T22:26:47Z FirstChildTAG: puedes facilitar el link FirstChildUserIdTAG: 155756 FirstChildUserNameTAG: vmendez FirstChildCreateTimeTAG: 2012-09-07T19:17:24Z IndexTAG: 1751 TitleTAG: Tutorial was helpful In figuring out these problems I found it most helpful to watch the Parallel resistors video on the Week 1 tutorials link. This helped me understand how those equations were created and why you would use them. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-06T18:37:20Z VoteTAG: 2 CoursewareTAG: Week 1 / Battery model CommentableIdTAG: 6002x_S1E9_Battery_Model NumberOfReplyTAG: 0 IndexTAG: 1752 TitleTAG: Difficulty in lab tools Dear Sir , I am not able to connect the wires in between the components kindly help me even I click on the components no highlight of green line visible but I am able to change the components values Is there any short cut key is available to get the wire thanks Prasanth UserIdTAG: 156694 UserNameTAG: mkprasanth CreateTimeTAG: 2012-09-06T17:28:51Z VoteTAG: 2 CoursewareTAG: Week 1 / Course overview CommentableIdTAG: 6002x_S1V1toS1V4 NumberOfReplyTAG: 2 FirstChildTAG: In the lab 1, I can put the letters A B C D yet...please help. Thanks. FirstChildUserIdTAG: 377637 FirstChildUserNameTAG: Maguana FirstChildCreateTimeTAG: 2012-09-06T18:30:40Z FirstChildTAG: Hi mkprasanth :)! Take a look to this .pdf of this Course https://www.edx.org/static/content-mit-6002x/handouts/schematic_tutorial.ba422f80d72b.pdf it might help you. Hola mkprasanth :)! Si quieres puedes ver este .pdf https://www.edx.org/static/content-mit-6002x/handouts/schematic_tutorial.ba422f80d72b.pdf quizás te sea de ayuda. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-06T17:37:39Z SecondChildTAG: hi Myrimit! i am also facing this problem! i hve visted the link tht u posted but could'nt find the solution SecondChildUserIdTAG: 400892 SecondChildUserNameTAG: ahmadyousaf SecondChildCreateTimeTAG: 2012-09-08T17:59:47Z SecondChildTAG: @ahmadyousaf: i think the lab tools are coded with java so u need to install java runtime environment on ur machine...though i may be wrong.... SecondChildUserIdTAG: 532062 SecondChildUserNameTAG: monty12345 SecondChildCreateTimeTAG: 2012-10-01T16:24:13Z IndexTAG: 1753 TitleTAG: Homework How and when to submit homework? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-06T16:57:35Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 5_Simple_Power NumberOfReplyTAG: 1 FirstChildTAG: You can submit homework anytime before the deadline. FirstChildUserIdTAG: 373262 FirstChildUserNameTAG: nwtn_pro FirstChildCreateTimeTAG: 2012-09-12T14:49:54Z IndexTAG: 1754 TitleTAG: Answering questions hi can anyone help me regarding this problem ? when i enter the answer for the homework questions it displays as ***Invalid input: Could not parse '3R' as a formula*** UserIdTAG: 131920 UserNameTAG: Viswajpr CreateTimeTAG: 2012-09-06T16:19:21Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 6 FirstChildTAG: you are missing the multiplication sign. FirstChildUserIdTAG: 370089 FirstChildUserNameTAG: Quarck FirstChildCreateTimeTAG: 2012-09-06T17:04:11Z FirstChildTAG: use 3*R FirstChildUserIdTAG: 315378 FirstChildUserNameTAG: gbp1984 FirstChildCreateTimeTAG: 2012-09-06T16:33:56Z FirstChildTAG: You must use an asterisk (*) as a multiply operator. E.g. 3*R FirstChildUserIdTAG: 9147 FirstChildUserNameTAG: SueP FirstChildCreateTimeTAG: 2012-09-06T16:21:54Z FirstChildTAG: try this 3*R FirstChildUserIdTAG: 131426 FirstChildUserNameTAG: M_Inam_Ul_Haq FirstChildCreateTimeTAG: 2012-09-06T16:24:16Z FirstChildTAG: In the question its asked for algebraic expression... therefore answer is R+R+R FirstChildUserIdTAG: 383167 FirstChildUserNameTAG: padmanemi FirstChildCreateTimeTAG: 2012-09-06T16:57:41Z FirstChildTAG: Try 3*R instead. FirstChildUserIdTAG: 66042 FirstChildUserNameTAG: rarciniegas FirstChildCreateTimeTAG: 2012-09-06T17:09:14Z IndexTAG: 1755 TitleTAG: Answer of th quest. y -2 in 2nd question.. -2 because of the negative terminal. UserIdTAG: 380912 UserNameTAG: Gaurav_Anand CreateTimeTAG: 2012-09-06T15:53:39Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: 5_Simple_Power NumberOfReplyTAG: 1 FirstChildTAG: i dont thnik so..u have to check the current direction and it is negative because power is a dot product.. FirstChildUserIdTAG: 625118 FirstChildUserNameTAG: alex_003 FirstChildCreateTimeTAG: 2012-10-17T17:59:34Z IndexTAG: 1756 TitleTAG: Video Lectures always available? Hi I have a question: in Week 2, will still be Video Lectures available from Week 1, and so on? or will it be removed? Or for example, in Week 3, will still be video lectures available from week 1 and 2? Thanks for your answer!! UserIdTAG: 342966 UserNameTAG: SandraNavarro CreateTimeTAG: 2012-09-06T15:24:29Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: On the MITx platform (where the same course was offered in the last semester) the course materials were available for the whole length of the course, and also later - for people who just had another pace of learning ;) I guess it's gonna be the same here. FirstChildUserIdTAG: 147593 FirstChildUserNameTAG: knack FirstChildCreateTimeTAG: 2012-09-06T15:38:59Z IndexTAG: 1757 TitleTAG: problem number 2 in SIE3 For the second answer I Integrated the the instantaneous power and got the answer as zero. how is it same as the power in the third case?? UserIdTAG: 357885 UserNameTAG: rahulshetty CreateTimeTAG: 2012-09-06T14:49:54Z VoteTAG: 2 CoursewareTAG: Week 1 / AC power CommentableIdTAG: 6002x_S1E3_AC_power NumberOfReplyTAG: 4 FirstChildTAG: if you are integrating V/R instead of V^2 / R you would get zero. FirstChildUserIdTAG: 388273 FirstChildUserNameTAG: kilmarta FirstChildCreateTimeTAG: 2012-09-06T15:49:32Z FirstChildTAG: It is not 0.how did u integrate?.it is (integral of(v_max*i_max*cos^2(2*pi*60*t)).it should be divided by one cycle.that is,divide integral by (2*pi).u will get answer close to 3rd one.bcoz this integral indeed defines RMS value.. FirstChildUserIdTAG: 55345 FirstChildUserNameTAG: sarmaji FirstChildCreateTimeTAG: 2012-09-06T16:39:35Z SecondChildTAG: could you plz further elaborate the solution for this problem. thanks SecondChildUserIdTAG: undefined SecondChildUserNameTAG: undefined SecondChildCreateTimeTAG: 2012-09-06T17:17:18Z FirstChildTAG: Power (p) = V^2/R.....so p=120^2/110=130.91 FirstChildUserIdTAG: 268104 FirstChildUserNameTAG: SaulCardenasG FirstChildCreateTimeTAG: 2012-09-08T06:07:11Z FirstChildTAG: thanks...i have open books after 11 years and I am kinda lost :/ FirstChildUserIdTAG: 122513 FirstChildUserNameTAG: Mona77 FirstChildCreateTimeTAG: 2012-09-08T07:35:18Z IndexTAG: 1758 TitleTAG: Help I want to clarify matters calculate voltage, current, and how and solution steps UserIdTAG: 318869 UserNameTAG: blackknight0 CreateTimeTAG: 2012-09-06T13:38:20Z VoteTAG: 2 CoursewareTAG: Week 1 / KVL/KCL example CommentableIdTAG: 6002x_S2E3_Using_KVL_KCL_and_VI_constraints NumberOfReplyTAG: 1 FirstChildTAG: c [here][1] hope it helps [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/MITx_6_002x_2012_Fall/threads/5047a8af6859412700000021 FirstChildUserIdTAG: 82571 FirstChildUserNameTAG: Nidhi3 FirstChildCreateTimeTAG: 2012-09-06T13:49:14Z IndexTAG: 1759 TitleTAG: Explanation of Answer of Question S1E2: POWER. Could anybody please explain the Answer of Question S1E2: POWER? UserIdTAG: 2956 UserNameTAG: akshayb CreateTimeTAG: 2012-09-06T11:33:47Z VoteTAG: 2 CoursewareTAG: Week 1 / Power CommentableIdTAG: 6002x_S1E2_Power NumberOfReplyTAG: 1 FirstChildTAG: Power is taken from the voltage source and dissipated in the resistor. We can think of power into the circuit (negative by convention) as given by the source, and power out of the circuit (positive) as taken by the resistor. Every time current is going out from the positive terminal of an element you will have a negative power, and when current is going into the positive terminal you will have a positive power. To calculate the value just remember the equation P = V*I and Ohms law V = R*I. Just be careful with the signs! FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-06T13:14:07Z SecondChildTAG: p=V*I and I=V/R means p=(V^2)/R. And V=sqrt(p*R). sqrt(8*ll)=89.3808 SecondChildUserIdTAG: 283726 SecondChildUserNameTAG: ByronInLawrence SecondChildCreateTimeTAG: 2012-09-08T23:09:12Z IndexTAG: 1760 TitleTAG: scroll wheel I don't like how the zoom function on the tool responds to the scroll wheel on my mouse. I much prefer that zoom is controlled only within the tool window by clicking on the + or - UserIdTAG: 166989 UserNameTAG: chrshammer CreateTimeTAG: 2012-09-06T06:15:50Z VoteTAG: 2 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: I has had exactly the same feeling when using the simulator. FirstChildUserIdTAG: 252722 FirstChildUserNameTAG: vaboro FirstChildCreateTimeTAG: 2012-09-06T18:43:29Z IndexTAG: 1761 TitleTAG: thanks mit. thanks mit. u awesome, i can learn electronics now. keep it up. cant wait for my AI class by BerkelyX UserIdTAG: 215287 UserNameTAG: yomi CreateTimeTAG: 2012-09-06T03:39:38Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 1762 TitleTAG: resistor power entering network current *entering* network from resistor ... srsly, is resistor a power source now? source is supposed to be in the cloud somewhere therefore the power enters resistor, doesnt enter the cloud; what is it with these logic markers that get reversed and double negated all the time, or is it just me that thinks so? UserIdTAG: 331483 UserNameTAG: Doru CreateTimeTAG: 2012-09-06T03:03:54Z VoteTAG: 2 CoursewareTAG: Week 1 / Power CommentableIdTAG: 6002x_S1E2_Power NumberOfReplyTAG: 3 FirstChildTAG: i think that's why the current is negative. FirstChildUserIdTAG: 341230 FirstChildUserNameTAG: anuragrattan FirstChildCreateTimeTAG: 2012-09-07T07:31:51Z FirstChildTAG: I find it all a bit confusing too, but it seems like frames of reference in physics. Is the train moving, or is the earth moving? Well, they are both moving relative to each other. Is the ball moving positively or negatively? Depends on whether up or down is defined as positive. If I understand correctly, it is **current**, not power, that is 'entering' the network at what we are calling its positive terminal. This current is passing through the resistor, not emanating from it. See Chanute's post about how there are two solutions, because we aren't told exactly which way the actual current/voltage is oriented. FirstChildUserIdTAG: 106229 FirstChildUserNameTAG: jc_lounge FirstChildCreateTimeTAG: 2012-09-06T08:28:57Z FirstChildTAG: And the cloud as you put it represents an unknown network and the things you do know are stated. Not everything you need too know . but certain other variables can be figured from either laws or given known and unknowns. The current i and direction is known variable and what it is going to inside does't matter to answer the Questions. FirstChildUserIdTAG: 359302 FirstChildUserNameTAG: GaryG FirstChildCreateTimeTAG: 2012-09-11T21:33:31Z IndexTAG: 1763 TitleTAG: Good synthesis Flight of a bird, from Nature observation experiments unto the iPad, is always good to know the purpose just before beginning. UserIdTAG: 124360 UserNameTAG: nlama CreateTimeTAG: 2012-09-06T00:45:02Z VoteTAG: 2 CoursewareTAG: Week 1 / Course overview CommentableIdTAG: 6002x_S1V1toS1V4 NumberOfReplyTAG: 0 IndexTAG: 1764 TitleTAG: Speed up OUR progress by self-study - The Spring course materials The 2012 Spring course can be accessed here https://6002x.mitx.mit.edu/courseware/6.002_Spring_2012/Week_14/Energy_and_CMOS_Design/ I believe they're just the same, and anybody has abundant time can increase their progress, without having to wait for 1 whole semester. This university and course thing is so time-consuming anyway. UserIdTAG: 342241 UserNameTAG: hieuxlu CreateTimeTAG: 2012-09-05T23:37:58Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 1765 TitleTAG: first question why is the open-circuit voltage the same with the one battery's voltage ? UserIdTAG: 37280 UserNameTAG: zour CreateTimeTAG: 2012-09-05T23:34:58Z VoteTAG: 2 CoursewareTAG: Week 1 / Battery model CommentableIdTAG: 6002x_S1E9_Battery_Model NumberOfReplyTAG: 2 FirstChildTAG: it's Kirchhoff; imagine as a 'load' a resistor with infinite value... meaning the current i that node is zero... meaning zero voltage drop on R1 or R1 FirstChildUserIdTAG: 277787 FirstChildUserNameTAG: kirilaska FirstChildCreateTimeTAG: 2012-09-06T00:40:16Z FirstChildTAG: They're in parallel, so the voltage drop across either branch is the same. FirstChildUserIdTAG: 267993 FirstChildUserNameTAG: mavlijas FirstChildCreateTimeTAG: 2012-09-06T00:37:36Z IndexTAG: 1766 TitleTAG: experimenting did anyone had fun changing values to form different waveforms? funny and nice start UserIdTAG: 134593 UserNameTAG: ilsalai CreateTimeTAG: 2012-09-05T23:19:40Z VoteTAG: 2 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 0 IndexTAG: 1767 TitleTAG: Formatting Question (S1E6) Can someone help me with the formatting? I keep getting "Invalid input: v v not permitted in answer" for my answer in S1E6. UserIdTAG: 366427 UserNameTAG: agaidis1 CreateTimeTAG: 2012-09-05T22:13:37Z VoteTAG: 2 CoursewareTAG: Week 1 / KVL and KCL exercises CommentableIdTAG: 6002x_kvl_kcl_exercises NumberOfReplyTAG: 1 FirstChildTAG: Nevermind, I deleted my answers and checked how they wanted me to write it. FirstChildUserIdTAG: 366427 FirstChildUserNameTAG: agaidis1 FirstChildCreateTimeTAG: 2012-09-05T22:16:37Z SecondChildTAG: For the sake of people who might be reading this question, I will point out that answers are case sensitive, and v is different from V. SecondChildUserIdTAG: 96452 SecondChildUserNameTAG: Lyla SecondChildCreateTimeTAG: 2012-09-05T23:15:49Z IndexTAG: 1768 TitleTAG: iOS Hello, all! A am amused that the tools are so great and useful. However, do you plan some iOS integration? Maybe some other handheld platform. Most of my (and I believe many others) spare time is in the subways (no WiFi there in Russia), or other travelling, so the ability to download materials and view them offline could be precious. For the schematics we have another 3rd parties offline tools, but they are not integrated with courses (and not that straightforward sometimes). And calendar downloading (in the "Calendar" readable format) would be very awesome too (maybe subscribing to Google calendars?). Any way you've done a really good job so far, so that's just the hint to make it even better. UserIdTAG: 208296 UserNameTAG: OZ1 CreateTimeTAG: 2012-09-05T21:56:31Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 1769 TitleTAG: beginning I had a few mistakes, but it was a great start! UserIdTAG: 279321 UserNameTAG: dahd416 CreateTimeTAG: 2012-09-05T21:06:50Z VoteTAG: 2 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: Nice dahd416! Good start! FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-05T23:29:04Z IndexTAG: 1770 TitleTAG: System Best Practices Please use the "system best practices" tag with all the posts that contain best practices for this platform. The staff will also do the same. So this tag will become the go-to site for frequently asked questions related to system best practices, e.g., which browser works best? (the answer to which in my opinion is Google Chrome). UserIdTAG: 42833 UserNameTAG: KKA CreateTimeTAG: 2012-09-05T19:37:33Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: All the browsers should work properly as the developers used Javascript. Maybe Ruby on Rails as it's more famous at MIT. But when I check the source code, I see more Javascripts. If things are not working correct or if you believe they should be otherwise, I suggest you ask EdX experts Lyla Fischer and/or Rohan Nagarkar at: fischerl@MIT.EDU &&/|| <"no email repository at MIT for Rohan Nagarkar"> before you do something stupid ... like running Windows on Mac :) FirstChildUserIdTAG: 152317 FirstChildUserNameTAG: easyMIT FirstChildCreateTimeTAG: 2012-09-05T20:23:18Z FirstChildTAG: I tell you that I am using firefox v 16.0 and up to the moment do not having any problem. FirstChildUserIdTAG: 337887 FirstChildUserNameTAG: kevil FirstChildCreateTimeTAG: 2012-09-06T06:00:10Z IndexTAG: 1771 TitleTAG: Conventions What convention are we using, the arrow leaving a node is that positive or negative current ? UserIdTAG: 371118 UserNameTAG: Cameus CreateTimeTAG: 2012-09-05T19:26:11Z VoteTAG: 2 CoursewareTAG: Week 1 / KVL and KCL exercises CommentableIdTAG: 6002x_kvl_kcl_exercises NumberOfReplyTAG: 3 FirstChildTAG: I like to think of node as the master and all othe rnodes as the salves. So I use current moving away from node as positive and currents moving into nodes as negative. But either way will do. But remember this only applies when you are doing KCL. Once you have found the unknowns you then need to make sure the correct sign is on the resultant currents in the circuit, this you can do but checking the voltage drops across the branches. (eg. for this course which uses conventional flow, current moving from high down to low voltage is assumed to be positive). FirstChildUserIdTAG: 15344 FirstChildUserNameTAG: kob FirstChildCreateTimeTAG: 2012-09-05T21:26:14Z FirstChildTAG: You can chose whichever convention you like!..But then you just have to follow the same convention throughout the question... I always choose current leavinga node as negative and currentt entering a node as positive!... :D FirstChildUserIdTAG: 73762 FirstChildUserNameTAG: sidd_myself FirstChildCreateTimeTAG: 2012-09-05T19:34:34Z FirstChildTAG: friend, to make it correct... all currents entering a node minus all currents leaving a node equals to zero. FirstChildUserIdTAG: 349139 FirstChildUserNameTAG: 1977ROYELMER FirstChildCreateTimeTAG: 2012-09-07T13:35:43Z SecondChildTAG: how does it work for S1E7 i2, if all currents are exiting the node? SecondChildUserIdTAG: 230787 SecondChildUserNameTAG: Gabriel007 SecondChildCreateTimeTAG: 2012-09-07T23:20:14Z IndexTAG: 1772 TitleTAG: Dependent vs Independent nodes and loops In S2E1: Circuit Topology we are asked to differentiate between the total number of nodes and the independent nodes; and the same with the number of loops. Unfortunately even after reading the write up below the questions I am still unclear on what the difference between the dependent and independent nodes and loops is. I would be much obliged if someone could clarify this. Thank you in advance. UserIdTAG: 85490 UserNameTAG: yashsavani CreateTimeTAG: 2012-09-05T19:16:26Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Imagine triangle. Let the sides would be the rezistances, and the tops would be the nodes. It has 3 nodes. Currents: from 1 to 2, from 2 to 3, from 3 to 1. So, KCL equations: 1: I12-I31=0 2: I23-I12=0 3: I31-I23=0 As you see, you have 3 nodes, and 3 equations. The problem is that, you have 3 variables, and 3 equations, but cant solve the problem, because one of this nodes is DEPENDENT. That means, that for example if you add 1st equation to 2nd - you'l get third. And at the end of solving you will get smthing like I31=I31. Thats why in every curcuit if there are m nodes - m-1 are independent. And you can use equations only for independant nodes. The same is with loops. FirstChildUserIdTAG: 200319 FirstChildUserNameTAG: Virviil FirstChildCreateTimeTAG: 2012-09-05T20:06:36Z SecondChildTAG: apply the concept of RANK in matrix could also help. suppose the KCL equations is A*x=0 let x = [i12 i23 i31]', which is a column let A = [1 0 -1;-1 1 0;0 -1 1] calculate the rank of A, rank(A) = 2, which is less than the column of A, which is 3. so there is one independent variable. SecondChildUserIdTAG: 386237 SecondChildUserNameTAG: forlsy SecondChildCreateTimeTAG: 2012-09-06T12:18:17Z IndexTAG: 1773 TitleTAG: how to change the answers? Hi everyone, i know the answer of questions in the homework but when i checked them most of them went wrong.I guess i have mis read the question and the way to answer them. I mean the syntax..Can i change the answers? please help! UserIdTAG: 368712 UserNameTAG: jmen CreateTimeTAG: 2012-09-05T17:29:28Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: May b u r putting d answers in a wrong fromat like if you have to write 6R, then u should enter 6*R, this is the correct format. Try with this. FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-05T17:47:29Z IndexTAG: 1774 TitleTAG: Has anyone successfully completed Lab1? I find this impossible to do as to run a DC analysis I need a ground point. On the right side of my circuit I only have an option to place resistors and not the ground point (inverted T). Anyone else got around this issue? UserIdTAG: 80920 UserNameTAG: MichaelSturgess CreateTimeTAG: 2012-09-05T17:23:47Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 8 FirstChildTAG: I've got this same problem, but i click reset button at the bottom of page and now, everything is ok FirstChildUserIdTAG: 159547 FirstChildUserNameTAG: zyto FirstChildCreateTimeTAG: 2012-09-06T07:56:02Z SecondChildTAG: yea.... SecondChildUserIdTAG: 399041 SecondChildUserNameTAG: Salohiddin SecondChildCreateTimeTAG: 2012-09-28T10:32:35Z FirstChildTAG: use chrome only,in other browsers it may not work FirstChildUserIdTAG: 343068 FirstChildUserNameTAG: Born_in_USSR FirstChildCreateTimeTAG: 2012-09-05T19:42:35Z SecondChildTAG: I've tried Chrome, Firefox and IE all to the same effect. SecondChildUserIdTAG: 80920 SecondChildUserNameTAG: MichaelSturgess SecondChildCreateTimeTAG: 2012-09-05T19:45:49Z FirstChildTAG: The inverted 'T' symbol is already placed in the figure at the bottom-left corner!!.... :D FirstChildUserIdTAG: 73762 FirstChildUserNameTAG: sidd_myself FirstChildCreateTimeTAG: 2012-09-05T19:46:14Z SecondChildTAG: Its not anymore haha :-o SecondChildUserIdTAG: 80920 SecondChildUserNameTAG: MichaelSturgess SecondChildCreateTimeTAG: 2012-09-05T19:49:02Z FirstChildTAG: Hi, i've got this same problem. FirstChildUserIdTAG: 159547 FirstChildUserNameTAG: zyto FirstChildCreateTimeTAG: 2012-09-05T17:39:24Z FirstChildTAG: Jst c the figure, T is aready there , u jst have to put the correct resistors there. dts all. FirstChildUserIdTAG: 82571 FirstChildUserNameTAG: Nidhi3 FirstChildCreateTimeTAG: 2012-09-06T08:04:02Z SecondChildTAG: Thats because they've fixed it now SecondChildUserIdTAG: 80920 SecondChildUserNameTAG: MichaelSturgess SecondChildCreateTimeTAG: 2012-09-06T17:38:49Z FirstChildTAG: YES SIR FirstChildUserIdTAG: 156694 FirstChildUserNameTAG: mkprasanth FirstChildCreateTimeTAG: 2012-09-13T18:53:17Z FirstChildTAG: Its busted. Not totally a browser issue, as components showing in sandbox but not on lab. If it was browser then we would have same problem on both. I expect they are working on it. My guess is the Lab version of the code is newer than sandbox version. The one in lab is probably broken due to browser issues. Easy fix is to use sandbox version of the code. THEY FIXED THIS BY INCLUDING THE RESET BUTTON. JUST CLICK THE RESET BUTTON AT BOTTOM WHEN IT APPEARS. THEN YOU'LL SEE CORRECT COMPONENTS SHOWING. FirstChildUserIdTAG: 15344 FirstChildUserNameTAG: kob FirstChildCreateTimeTAG: 2012-09-05T20:26:56Z FirstChildTAG: I have successfully completed Lab1 on Sept 6th only by using latest chrome. I didn't find any problem in chrome. But the lab is not working in Internet Explorer. FirstChildUserIdTAG: 188460 FirstChildUserNameTAG: Lokesh17 FirstChildCreateTimeTAG: 2012-09-10T22:25:21Z IndexTAG: 1775 TitleTAG: DC button Can anyone tell me where is the DC button? UserIdTAG: 308571 UserNameTAG: Sayantani CreateTimeTAG: 2012-09-05T16:21:08Z VoteTAG: 2 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 9 FirstChildTAG: its on top corner FirstChildUserIdTAG: 177128 FirstChildUserNameTAG: MARYAAM FirstChildCreateTimeTAG: 2012-09-05T16:29:55Z FirstChildTAG: im using firefox and dont see any dc buttons =( FirstChildUserIdTAG: 322518 FirstChildUserNameTAG: Ruslan_ua FirstChildCreateTimeTAG: 2012-09-05T16:50:36Z SecondChildTAG: ok now im in chrome and cant write wires, when i try, i can only move elevents, but dc button now is visible SecondChildUserIdTAG: 322518 SecondChildUserNameTAG: Ruslan_ua SecondChildCreateTimeTAG: 2012-09-05T17:16:15Z FirstChildTAG: it is located on uppermost left side beside train FirstChildUserIdTAG: 382821 FirstChildUserNameTAG: 91 FirstChildCreateTimeTAG: 2012-09-05T16:41:51Z FirstChildTAG: it's on the top of the gray region FirstChildUserIdTAG: 292093 FirstChildUserNameTAG: NourKittawi FirstChildCreateTimeTAG: 2012-09-05T16:42:12Z FirstChildTAG: The button is located on the top grey bar (as depicted in the introductory video), however sometimes the DC and transient analysis button do not appear. This seems to be your case. To correct this situation you should try to refresh the page and if that do not work get out of the page and try once again. This issue was alredy address by other students and the solution found only a fee pages away in this discussion... FirstChildUserIdTAG: 300827 FirstChildUserNameTAG: PFonseca FirstChildCreateTimeTAG: 2012-09-05T16:53:44Z SecondChildTAG: I read elsewhere that this problem do not occur in Chrome... SecondChildUserIdTAG: 300827 SecondChildUserNameTAG: PFonseca SecondChildCreateTimeTAG: 2012-09-05T17:02:38Z FirstChildTAG: I can't see it either...Is it because of the browser? I am currently using firefox. **New**: I've tried to use safari browser and the 'DC button' appears. FirstChildUserIdTAG: 364043 FirstChildUserNameTAG: AliciaSpain FirstChildCreateTimeTAG: 2012-09-05T17:22:40Z FirstChildTAG: I couldn't find the buttons at first too, then I scrolled to the bottom of the page and clicked "Check" and the buttons appeared for me. Hope this helps. FirstChildUserIdTAG: 141462 FirstChildUserNameTAG: xubuohao FirstChildCreateTimeTAG: 2012-09-06T03:07:12Z FirstChildTAG: DC button is on the top of ckt sandbox FirstChildUserIdTAG: 85617 FirstChildUserNameTAG: Anuragindia FirstChildCreateTimeTAG: 2012-09-05T16:31:17Z FirstChildTAG: which browser u using ..?? is it firefox FirstChildUserIdTAG: 177128 FirstChildUserNameTAG: MARYAAM FirstChildCreateTimeTAG: 2012-09-05T16:38:55Z IndexTAG: 1776 TitleTAG: 4-th task can anyone explain 4-th task UserIdTAG: 192114 UserNameTAG: chuba CreateTimeTAG: 2012-09-05T15:57:41Z VoteTAG: 2 CoursewareTAG: Week 1 / Battery model CommentableIdTAG: 6002x_S1E9_Battery_Model NumberOfReplyTAG: 1 FirstChildTAG: it is just Ohm's law. V = R*I. The Voltage difference is 0.1. Hence we get: V = (0.1)/(0.25+0.33) FirstChildUserIdTAG: 327015 FirstChildUserNameTAG: Fabi2607 FirstChildCreateTimeTAG: 2012-09-05T16:36:58Z SecondChildTAG: has to be I not V, I am sorry ;) SecondChildUserIdTAG: 327015 SecondChildUserNameTAG: Fabi2607 SecondChildCreateTimeTAG: 2012-09-05T16:37:42Z SecondChildTAG: thanks) SecondChildUserIdTAG: 192114 SecondChildUserNameTAG: chuba SecondChildCreateTimeTAG: 2012-09-05T17:16:53Z IndexTAG: 1777 TitleTAG: Problem in S1E3 Lecture sequence(Administrivia and circuit elements)..... i think the average power should be zero..... the rms power is 130.9 W..... UserIdTAG: 76418 UserNameTAG: ankitarora CreateTimeTAG: 2012-09-05T15:23:16Z VoteTAG: 2 CoursewareTAG: Week 1 / AC power CommentableIdTAG: 6002x_S1E3_AC_power NumberOfReplyTAG: 9 FirstChildTAG: I can't seem to get the Average Power either, could someone please elaborate on this one. FirstChildUserIdTAG: 135675 FirstChildUserNameTAG: JohnGreiner FirstChildCreateTimeTAG: 2012-09-05T17:05:45Z SecondChildTAG: If you have $v=120*sqrt(2)*cos(2*pi*60*t) $ where f=60, then 1 cycle is T=1/60 s $i=v/R$ $i=(120*sqrt(2)/110)*cos(2*pi*60*t)$ instantaneous power is $p=v*i=(120*sqrt(2)*cos(2*pi*60*t))*((120*sqrt(2)/110)*cos(2*pi*60*t))$ $p=((120^2)*2/110)*cos^2(2*pi*60*t)$ Average power is the integral of the last equation over 1 cicle (1/60). and you get $P=[(120^2)*2/110]*1/2$ $P=(120^2)/110=130.91$ SecondChildUserIdTAG: 133032 SecondChildUserNameTAG: alfredox72 SecondChildCreateTimeTAG: 2012-09-05T18:35:15Z FirstChildTAG: Edit: I think I've got it. One cycle is NOT a wave length (silly mistake), and it equals to one period. Period T = 1/f. The unit of Power is VA, and if we look on the integral it's unit is VAs, so we need to supply the frequency to get the desired unit of VA. FirstChildUserIdTAG: 138981 FirstChildUserNameTAG: Pr0bability FirstChildCreateTimeTAG: 2012-09-05T16:48:16Z SecondChildTAG: One full cycle is 2 * Pi So, the next one after 0 is when 2*Pi*60*t = 2*Pi So you need to integrate over 0 and 1/60 and you'll get the result of 130.9 SecondChildUserIdTAG: 250776 SecondChildUserNameTAG: RoamingBlue SecondChildCreateTimeTAG: 2012-09-05T18:59:19Z FirstChildTAG: The power cant be zero, cause current always flow through the resistor, one way or another. So resistor is always heating. If power will be zero, than resistor should be always cold, and we know that this is wrong) FirstChildUserIdTAG: 324823 FirstChildUserNameTAG: Lerk FirstChildCreateTimeTAG: 2012-09-05T16:51:42Z SecondChildTAG: To a resistor, the direction of the current is always associated with the polarity of the voltage. That is when the voltage across a resistor is positive, the current through it is forward, and when the voltage across a resistor is negative, the current through it is backward. So a resistor always dissipate energy in a circuit. That means the power cannot be zero. If the power was zero, that means the resistor sometimes (the positive half circle) absorb energy and sometimes (the negative half circle) emit energy. It is not the characteristic of a resistor. SecondChildUserIdTAG: 325624 SecondChildUserNameTAG: ZhiDong SecondChildCreateTimeTAG: 2012-09-06T12:12:34Z FirstChildTAG: Average power = 1/T * [integral v^2(t)/R from t1 to t1+T dt]. It would be zero if there was v(t) instead of v^2(t). FirstChildUserIdTAG: 201151 FirstChildUserNameTAG: irossales FirstChildCreateTimeTAG: 2012-09-05T17:09:04Z SecondChildTAG: By the way, you can use standard Latex by enclosing Latex in dollar signs: $P_\mathrm{avg} = \frac{1}{T}\int_0^T V(t)^2/R \cdot dt$ SecondChildUserIdTAG: 79361 SecondChildUserNameTAG: kimt SecondChildCreateTimeTAG: 2012-09-05T18:26:37Z FirstChildTAG: v(t) = 120*sqrt(2)*cos(2*pi*60*t) T= 1/60 average power = 1/T * int(v(t)^2/R dt) from O to T That should give you 130.9 FirstChildUserIdTAG: 208854 FirstChildUserNameTAG: BrynnleeEaton FirstChildCreateTimeTAG: 2012-09-05T17:59:25Z FirstChildTAG: The average power for resistive circuit is P=(1/2)*V*I = V^2/(2*R) = {[120*sqrt(2)]^2}/(2*110) = 130.9 W FirstChildUserIdTAG: 181973 FirstChildUserNameTAG: ciceroalisson FirstChildCreateTimeTAG: 2012-09-05T22:05:54Z FirstChildTAG: any one give the exact explanation for average power FirstChildUserIdTAG: 380107 FirstChildUserNameTAG: shravanipadma FirstChildCreateTimeTAG: 2012-09-06T11:19:25Z SecondChildTAG: If you apply a sinusoidal voltage ( $v(t)=A \sin(2 \pi \omega t)$ ) to a resistor for a period of time, e.g. T, the resistor will dissipate a certain thermal energy $E_1$. If you replace the sinusoidal voltage with a direct voltage ( $v(t)=V$ ), and make it across the same resistor for the same period time T, the resistor will also dissipate a certain thermal energy $E_2$. Let us assume $E_1 = E_2$, and we will finally find the only suitable V which makes our assumption correct. We call $v(t)=V$ is the RMS of $v(t)=A \sin(2 \pi \omega t)$. So is it to power. SecondChildUserIdTAG: 325624 SecondChildUserNameTAG: ZhiDong SecondChildCreateTimeTAG: 2012-09-06T12:37:24Z SecondChildTAG: When an AC source is supplied to a resistor, then the current through the resistor will be in phase with the voltage, i.e. When the voltage is positive, the current will also be positive, and when the voltage is negative, the current will also be negative. The instantaneous power is V*I, which will always give a positive result due to the fact of the previous statement. The average power is then just the average of the instantaneous power over a period of time. This could be any multiple of 1/2 cycles and would still give the same result. i.e. the Average power over 1/2 cycle will be the same as the average power over 1 or 1.5 or 2 cycles. SecondChildUserIdTAG: 270011 SecondChildUserNameTAG: SnowmanZA SecondChildCreateTimeTAG: 2012-09-09T11:02:05Z FirstChildTAG: The formal concepts in AC power relevant at this stage of the course are: "instantaneous power" and "average power". Please note that "rms power" is not a formalized concept and you will not find the term anywhere in an electrical engineering textbook. Please see the picture on the top of page 760 in Chapter 13. Instantaneous power is proportional to the squared voltage or current for a resistor, that is p = v^2/R, or p = i^2*R. Since the square of a negative number is a positive number, p can never be negative in a resistor. The average power is the average of p taken over its period. Since p is always positive, its average must also be positive and non-zero. Disclaimer: all of the above for "resistive" systems without components like capacitors and inductors. FirstChildUserIdTAG: 151427 FirstChildUserNameTAG: RichardZ FirstChildCreateTimeTAG: 2012-09-06T15:00:34Z FirstChildTAG: time period here is pi. therefore the average power is the same as the RMS or DC power which is 130.9090 FirstChildUserIdTAG: 242035 FirstChildUserNameTAG: Varundev FirstChildCreateTimeTAG: 2012-09-05T15:27:03Z SecondChildTAG: how time period is pi? SecondChildUserIdTAG: 353634 SecondChildUserNameTAG: imumair SecondChildCreateTimeTAG: 2012-09-05T16:43:45Z SecondChildTAG: Time period isn't pi. w=2*pi*f f=60 Hz T=1/f T=1/60 seconds. SecondChildUserIdTAG: 133032 SecondChildUserNameTAG: alfredox72 SecondChildCreateTimeTAG: 2012-09-05T18:38:03Z IndexTAG: 1778 TitleTAG: Why "-" not equal to "ground"? Sorry for my stupid question. Why do we have to place the "Ground" component on the scheme? I really thought, that "Minus" pin is the same thing. UserIdTAG: 327785 UserNameTAG: Kirill_Suvorov CreateTimeTAG: 2012-09-05T14:56:32Z VoteTAG: 2 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 3 FirstChildTAG: Because voltage source doesnt give you potential of, for example 5 volts. It gives you PD (potential difference) of 5 volts. Thats why if you connect 2 sources in series, "+" of the first would not anigilate with "-" of the second. Also it means, that "+" is "- add n volts". Thats why only if you mark "-" as zero, "+" will be something. To say more, not every time this is "ground" - this is global bus. In some schemas can be global bus AND ground. Ground is +0V to ground, and global bus - not. FirstChildUserIdTAG: 200319 FirstChildUserNameTAG: Virviil FirstChildCreateTimeTAG: 2012-09-05T15:04:50Z FirstChildTAG: This is because in actuality the voltage source is defined by its potential difference. It is not necessary that the - of the cell is at 0V it may be at 3V but the + terminal would then not be 9V but 12V for a 9V cell. By putting a ground we assume that the negative terminal is 0V even though it is probably not. FirstChildUserIdTAG: 85490 FirstChildUserNameTAG: yashsavani FirstChildCreateTimeTAG: 2012-09-05T15:06:23Z FirstChildTAG: They are the same only when connected together, so that ground is your reference. Say for example your battery supplies is 6V. That only means 6V difference between its terminals not one is 0V and the other is 6V. If you have your ground connected to the positive terminal the voltages, with reference to ground become 0V and -6V so the minus terminal does not indicate ground. FirstChildUserIdTAG: 80920 FirstChildUserNameTAG: MichaelSturgess FirstChildCreateTimeTAG: 2012-09-05T15:01:39Z IndexTAG: 1779 TitleTAG: Circuit diagram disappeared In lab0 my drawn circuit diagram has disappeared -I had checked it several times, after returning from looking at some other things on the site, but the green mark is still there under the toolbox in lab0 , and with with all the answers that I had given. Hmm ... Update: I used firefox 14.0.1 and windows xp sp3. I noticed that a lot of people seem to have problems with firefox. I downloaded and installed Chrome [I don't want anything from google on my computer, but this is an emergency} and my circuit in the toolbox appeared ok. But still not in firefox: I deleted the cache, I checked allowing cookies, disabled all addons. And btw: internet explorer 8 is not even showing the toolbox at all ... So for the moment it's at least working in chrome, but the real problem is not solved. And, on the the mitx site at 6.002x firefox is working ok. So I suppose it has something to do with edx. UserIdTAG: 83276 UserNameTAG: salsero CreateTimeTAG: 2012-09-05T13:55:38Z VoteTAG: 2 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 2 FirstChildTAG: See if clicking the tiny square on the left helps. It's there with the zoom controls. FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-09-05T14:30:37Z SecondChildTAG: no, doesn't work. lab0 and the sandbox don't store the diagram, but they store the green checks ... SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-09-05T16:55:57Z FirstChildTAG: I think you should press the "check" button at the end of the page. I suppose it keeps your work in progress. Anyway, it must be confirmed. FirstChildUserIdTAG: 245908 FirstChildUserNameTAG: PabloLanga FirstChildCreateTimeTAG: 2012-09-05T14:48:10Z SecondChildTAG: I did that several times, and my answers are still checked with the green mark. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-09-05T16:04:46Z IndexTAG: 1780 TitleTAG: Suggestion for developer If the Discussion forum could have its threads organized by categories, it would be fairly simple for all the students attending this course. I can see that the people have already started posting "links" and other "thank you greetings" and "how great they are for the course" inside this discussion ... as this course progresses, it would be hard to keep track of forum elements as more people begin to add questions, suggestions and course related topics. Please add Discussion sub-categories such as Suggestions, Questions ??? But No Answers Please, Documents, Problems accessing Course, Extras ... or any other organized categories so that students can utilize this Discussion efficiently and don't have to waste their time clicking 1,2,3,4,5,6 ... Next Page ... to see all the threads Thank You UserIdTAG: 152317 UserNameTAG: easyMIT CreateTimeTAG: 2012-09-05T13:55:27Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 1781 TitleTAG: Exclusive Facebook group for the students attending Edx 6.002x. Hey Hi Guys, Firstly please do accept my well wishes. I have created this group exclusively for the students attending the Edx 6.002x Circuits and Electronics. This group is to share ur doubts, queries, suggestions and whatever u feel like. Lets get together and start exploring. Cheers.:) Below is the link- http://www.facebook.com/groups/483354858342165/ UserIdTAG: 253902 UserNameTAG: sajalok CreateTimeTAG: 2012-09-05T13:30:58Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: hello class today is my first day in the class what do i do iam having a hard time getting started FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-10T21:35:28Z IndexTAG: 1782 TitleTAG: Best Wishes to all Best wishes to all for the course, I am checking out the Courseware's overview section . UserIdTAG: 254685 UserNameTAG: abhisheknag CreateTimeTAG: 2012-09-05T13:08:46Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 1783 TitleTAG: Having Problems with Video I am not able to watch the videos. They do not load. Does this platform use youtube by any chance. I am in China and youtube is blocked here. UserIdTAG: 181745 UserNameTAG: Yoavrott CreateTimeTAG: 2012-09-05T12:58:43Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 6 FirstChildTAG: I think it does use you tube! FirstChildUserIdTAG: 301736 FirstChildUserNameTAG: MOjangole FirstChildCreateTimeTAG: 2012-09-05T12:59:49Z SecondChildTAG: Hi Yavrott! Where in China are you? I'm also in China, in Shandong province. I'm using 'web freer' to access the videos. Or as Hejinjie suggested, you can also try freegate. If you search in any search engine from within china you will not be able to download it. I will upload a link to a file sharing website and post it here for you. The trouble is, neither of these programs are 100% reliable for overcoming the firewall. If any EDx moderators see this discussion, please post videos to youku, or something else that is available in China! - I'll post a link to download freegate and web freer soon. SecondChildUserIdTAG: 198563 SecondChildUserNameTAG: PaulGillett SecondChildCreateTimeTAG: 2012-09-21T08:20:30Z FirstChildTAG: With the exception of CS50x, all edX courses this term use YouTube. FirstChildUserIdTAG: 11 FirstChildUserNameTAG: dormsbee FirstChildCreateTimeTAG: 2012-09-05T13:05:43Z SecondChildTAG: Thank you SecondChildUserIdTAG: 181745 SecondChildUserNameTAG: Yoavrott SecondChildCreateTimeTAG: 2012-09-05T13:08:11Z FirstChildTAG: Hey,I'm from China too,have you solved the problem? FirstChildUserIdTAG: 361137 FirstChildUserNameTAG: Ericson FirstChildCreateTimeTAG: 2012-09-06T10:53:49Z SecondChildTAG: same here SecondChildUserIdTAG: 176234 SecondChildUserNameTAG: hejinjie SecondChildCreateTimeTAG: 2012-09-07T02:06:17Z SecondChildTAG: same issue here SecondChildUserIdTAG: 91706 SecondChildUserNameTAG: PooyaM SecondChildCreateTimeTAG: 2012-09-07T15:00:59Z SecondChildTAG: Hi, I'm in mainland China, where utube isn't accessible.. Any chance these videos will be reposted elsewhere? I've been using 'uueb freer' (rhymes with 'neb' starts with 'w') I posted it to mediafire http://www.mediafire.com/?0r5jm0mr7yijvpq You can download it and try it out, but it's not 100% reliable. Does anyone have a totally reliable solution for people in countries with restricted internet? Thanks for your help! SecondChildUserIdTAG: 198563 SecondChildUserNameTAG: PaulGillett SecondChildCreateTimeTAG: 2012-09-21T08:52:03Z FirstChildTAG: try freegate, hope that will be hlepful. FirstChildUserIdTAG: 176234 FirstChildUserNameTAG: hejinjie FirstChildCreateTimeTAG: 2012-09-07T02:05:31Z SecondChildTAG: Hi hejinjie. Has freegate worked for you? I'm using web freer, and it works, but not all the time. I'm afraid the only reliable solution will be if EDx posts the videos elsewhere. SecondChildUserIdTAG: 198563 SecondChildUserNameTAG: PaulGillett SecondChildCreateTimeTAG: 2012-09-21T08:21:49Z FirstChildTAG: These are some solutions: Download and install UUeb freer (rhymes with 'neb' starts with a 'w') - I uploaded a file to mediafire. http://www.mediafire.com/?0r5jm0mr7yijvpq This software will allow you to surf the web somewhat /normally/ and access yewtube in China. I find this to be the most reliable solution so far. There is other software. Please let me know what you are using as this is not 100% reliable. FirstChildUserIdTAG: 198563 FirstChildUserNameTAG: PaulGillett FirstChildCreateTimeTAG: 2012-09-21T08:39:06Z FirstChildTAG: You can see in the Wiki [Lecture Videos Download][1]:). Also see more info [here][2]: **Due to the YouTube outage in certain parts of the world, we have provided direct links to download the lecture videos, and extended the Week 2 homework and lab deadlines to the coming Sunday (September 30). Please note that Week 3 deadlines have not changed.** [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/ [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/info FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-26T11:53:33Z IndexTAG: 1784 TitleTAG: Changing Parameters of dc voltage source Hi Dave, There is a problem with the circuit sandbox.I am not able to change the parameters of the voltage source.Double clicking on it is not having any effect,I mean not a getting a window where I can change its parameters.Resistor values can be changed as shown in the video but the voltage source we are not able to.Please check! UserIdTAG: 11538 UserNameTAG: trishul CreateTimeTAG: 2012-09-05T12:48:42Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: What browser are you using? Seems to work in firefox 15.0 without any issues. I changed from 1 V dc to 3 V dc without any issues. FirstChildUserIdTAG: 271670 FirstChildUserNameTAG: moncapitane FirstChildCreateTimeTAG: 2012-09-05T12:55:07Z SecondChildTAG: even i am using mozilla firefox but an older will shift to the latest and check out!!thank you!! SecondChildUserIdTAG: 11538 SecondChildUserNameTAG: trishul SecondChildCreateTimeTAG: 2012-09-05T13:15:31Z IndexTAG: 1785 TitleTAG: Is anybody from Russia(or ex USSR)? :) Is anybody from Russia(or ex USSR)? Please, join group http://vk.com/edxrussia UserIdTAG: 295103 UserNameTAG: Syavick CreateTimeTAG: 2012-09-05T12:48:17Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 4 FirstChildTAG: Hi Yes there are some people from Estonia FirstChildUserIdTAG: 269412 FirstChildUserNameTAG: andrei836 FirstChildCreateTimeTAG: 2012-09-05T12:49:54Z FirstChildTAG: Hello, I am from Russia. Skype: avspostbox FirstChildUserIdTAG: 151867 FirstChildUserNameTAG: AndreiStoiakin FirstChildCreateTimeTAG: 2012-09-05T12:54:00Z FirstChildTAG: hello, world! Oh, man..before I even clicked on the "Discussion" link I already knew that the would be a comment like this here :-) Привет из сибири! FirstChildUserIdTAG: 188680 FirstChildUserNameTAG: corsairnv FirstChildCreateTimeTAG: 2012-09-05T13:01:01Z SecondChildTAG: *there would be SecondChildUserIdTAG: 188680 SecondChildUserNameTAG: corsairnv SecondChildCreateTimeTAG: 2012-09-05T13:03:05Z SecondChildTAG: Привет! Из Сибири это как... Привет, я из Америки (по территории). Новосибирск? SecondChildUserIdTAG: 208296 SecondChildUserNameTAG: OZ1 SecondChildCreateTimeTAG: 2012-09-05T22:03:28Z SecondChildTAG: А что нужно? SecondChildUserIdTAG: 402617 SecondChildUserNameTAG: Vl SecondChildCreateTimeTAG: 2012-09-16T06:08:39Z FirstChildTAG: Of course we are here! Greetings from Ukraine :) FirstChildUserIdTAG: 190466 FirstChildUserNameTAG: kop FirstChildCreateTimeTAG: 2012-09-05T16:13:21Z SecondChildTAG: I'm from Ukraine too! ;))) SecondChildUserIdTAG: 295103 SecondChildUserNameTAG: Syavick SecondChildCreateTimeTAG: 2012-09-05T20:00:24Z SecondChildTAG: Тогда и из Севастополя! =) SecondChildUserIdTAG: 295103 SecondChildUserNameTAG: Syavick SecondChildCreateTimeTAG: 2012-09-11T12:40:26Z SecondChildTAG: I'm from Ukraine too! ;))) +1 SecondChildUserIdTAG: 228846 SecondChildUserNameTAG: BoykoA SecondChildCreateTimeTAG: 2012-09-20T22:31:20Z SecondChildTAG: Донецк присоединяется :) SecondChildUserIdTAG: 464744 SecondChildUserNameTAG: attache SecondChildCreateTimeTAG: 2012-10-22T08:16:42Z IndexTAG: 1786 TitleTAG: My First post: Let's try again I'm just trying to experiment with how things work here. Hope to see a lot of activity, helpful people and to read interesting answers on questions. UserIdTAG: 83276 UserNameTAG: salsero CreateTimeTAG: 2012-09-05T12:17:58Z VoteTAG: 2 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Actually you raise a good point: is there a sandbox where people can experiment with using this forum "offline"? FirstChildUserIdTAG: 79337 FirstChildUserNameTAG: AppliedImagination FirstChildCreateTimeTAG: 2012-09-05T12:23:07Z SecondChildTAG: The first reaction on my post shows something new: the indentation. Looks nice this way, less messy. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-09-05T12:31:21Z IndexTAG: 1787 TitleTAG: Proctored exam status update My proctored exam score isn't being displayed at Progress tab and I'm not seeing the any link for downloading the Proctored exam certificate. It was said to be available by 13th March when I gave the exam. So is it just me facing the problem or is it because the results haven't been declared yet? UserIdTAG: 297370 UserNameTAG: ayush3504 CreateTimeTAG: 2013-03-25T21:11:36Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 0 IndexTAG: 1788 TitleTAG: Homework1 and Lab1 I cannot find any "Check" button so how can I put my answers? Instead of "Check" button, I can only find "Show Answer" button and Explanations. Or is "check" button appeared on the day of 24th March(submission date)? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2013-03-23T14:47:25Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Yes check buttons are not available once the deadlines have been passed. FirstChildUserIdTAG: 297370 FirstChildUserNameTAG: ayush3504 FirstChildCreateTimeTAG: 2013-03-24T19:57:56Z IndexTAG: 1789 TitleTAG: Proctored exam certificates? The proctored exam certificates were said to be made available by 13th March. Any news? UserIdTAG: 297370 UserNameTAG: ayush3504 CreateTimeTAG: 2013-03-14T18:45:20Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Anyone from staff? FirstChildUserIdTAG: 297370 FirstChildUserNameTAG: ayush3504 FirstChildCreateTimeTAG: 2013-03-18T21:36:24Z IndexTAG: 1790 TitleTAG: The Text Book why can't I download the textbook ?? please I need to download it UserIdTAG: 299546 UserNameTAG: Gedo CreateTimeTAG: 2013-03-08T01:51:07Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Only the Sun rises for free. FirstChildUserIdTAG: 79885 FirstChildUserNameTAG: ruudoleo FirstChildCreateTimeTAG: 2013-03-09T16:14:14Z IndexTAG: 1791 TitleTAG: Proctored Exam when will the results of procotored exam be declared????? my result has still not come in progress graph..... UserIdTAG: 126370 UserNameTAG: BrijeshModasara CreateTimeTAG: 2013-03-02T10:22:13Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Neither my result shows on the Progress tab. It's supposed to be declared on 13th March. FirstChildUserIdTAG: 297370 FirstChildUserNameTAG: ayush3504 FirstChildCreateTimeTAG: 2013-03-03T19:47:56Z IndexTAG: 1792 TitleTAG: 6.002x spring 2013 this year spring course is going to be start from march13th UserIdTAG: 857413 UserNameTAG: Naveenkrishna CreateTimeTAG: 2013-02-26T10:40:36Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: thanks for your inform FirstChildUserIdTAG: 1012181 FirstChildUserNameTAG: ChuKee FirstChildCreateTimeTAG: 2013-03-02T06:10:49Z SecondChildTAG: how can i take part as course SecondChildUserIdTAG: 1388249 SecondChildUserNameTAG: chandrabhansilawat SecondChildCreateTimeTAG: 2013-03-13T16:53:24Z FirstChildTAG: Yep and will end Dec 20, 2012, as I see :) FirstChildUserIdTAG: 325197 FirstChildUserNameTAG: Vitali_Jerin FirstChildCreateTimeTAG: 2013-03-02T07:24:33Z FirstChildTAG: how can I apply in this course..??? FirstChildUserIdTAG: 969964 FirstChildUserNameTAG: Waqar789 FirstChildCreateTimeTAG: 2013-03-08T10:15:38Z IndexTAG: 1793 TitleTAG: 6.003z has started! Hey everyone! 6.003z, the unofficial Signals & Systems (can't wait for an official one by MIT :-) ) course has begun! Please head over to http://6003z.amolbhave.in/ to begin. We will still use the 6.002x forums for discussion. Please also feel free to contribute. You can make videos, circuit demos (If you've never seen this subject before, you'll find a a lot of connections with 6.002x in terms of concepts) and write ups! Big thanks to Amol for redoing website! See you all there :-) UserIdTAG: 14386 UserNameTAG: ashwith CreateTimeTAG: 2013-02-18T15:17:55Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_6003z NumberOfReplyTAG: 0 IndexTAG: 1794 TitleTAG: Future of Proctored Exams? Would we be able to take proctored exam with future offering of this course? without having to enroll again? There seem to be no seats available in any Pearson Vue Centers in my vicinity. (Some may be closed due to last weeks blizzard :( ) UserIdTAG: 714237 UserNameTAG: PaxPolaris CreateTimeTAG: 2013-02-11T21:00:33Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I believe you will be able to take the proctored exam at a later date, without having to enroll again. (Probably after the spring 2013 offering) I myself was under the assumption that we could take the exam at any time once it became available, I was wrong and not prepared for this round, nor was I able to attend that week. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2013-02-16T22:52:55Z IndexTAG: 1795 TitleTAG: 6.002x Circuits & Electronics guys!! i want to know about 6.002x Circuits & electronics. As i've seen that the course 6.002x has been finished & final exam is also over. But, can i start the course now?? and can i get the certficate of course accomplishment??? please help me out!!!! UserIdTAG: 1188221 UserNameTAG: NEERAJCUSAT CreateTimeTAG: 2013-02-10T12:18:20Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hello Neerajcust, Sorry for the bad news, you cannot get the certificate but you can still register for the next season course which will start around 2-4 months from now, so you have alot of time to study for it and get the certificate then Yours truly, FirstChildUserIdTAG: 91934 FirstChildUserNameTAG: Chingun FirstChildCreateTimeTAG: 2013-02-10T15:19:08Z IndexTAG: 1796 TitleTAG: Cant open Proctored Exam link on Course Info page. It shows PAGE NOT FOUND on going to the "Click Here" link for Proctored Exam. UserIdTAG: 112682 UserNameTAG: Pratik94 CreateTimeTAG: 2013-02-03T16:13:37Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: It seems broken...Thank you for pointing this out, I will report it to the Staff. Have you tried to click on the button of your Dashboard? There is a magenta button where you can register for the proctored Exam. I hope this can help you. Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2013-02-03T19:08:34Z FirstChildTAG: Fixed. Thanks for reporting it. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2013-02-04T02:22:49Z SecondChildTAG: Thank you Lyla :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2013-02-04T02:59:22Z IndexTAG: 1797 TitleTAG: The Pearson exam I'm new here and I can't understand alot of things.....I'm taking the course on the 5th of september. First What is this Pearson exam and should I take it now or should I wait there will be another one before the begining of my course ? Second The Course info is not organized and I cann't understand if these are for the new students or the ones that have taken this course before. I hope to replay fast....Thank you UserIdTAG: 1084194 UserNameTAG: Jongaur CreateTimeTAG: 2013-02-03T13:23:58Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Troubleshooting NumberOfReplyTAG: 1 FirstChildTAG: Hi Jongaur, *A proctored certificate is given to students who take and pass an exam under proctored conditions.* You can take it on February 13th in a Pearson VUE Centre . It is open to the students that took this 6.002x, didn't took it, passed this course, didn't pass this course, took the prorotype course, didn't take it, etc. *So, it is open for all the people that wants to take it*. It costs 95USD. --- You can take a look at the Final Exam that it is in the Courseware, in my point of view, I guess that the Proctored Exam will have the same level of complexity. [Final Exam][1] You have to be aware that you will have only 2 hours to complete it under proctored conditions and you can only take to the exam a calculator. Notes and other things are not allowed to take it to the Exam. Also, as far as I have read you will have access to the course material while doing the exam but will not be available this Forum discussion and the wiki. If you have questions regarding the Exam, you can write them here **exam-help@edx.org** --- The Spring Course have not started yet, I estimate that it will start on march. You can register here: https://www.edx.org/courses/MITx/6.002x/2013_Spring/about I hope this can help you. Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Final_Exam/Final2012Seq/ FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2013-02-03T19:41:45Z SecondChildTAG: Yes it helped me in understanding many things......I'm very Greatfull but unfortunately I'm in 3rd Sec. and in march I will be finishing my studies and preparing for my final exams so I'll have to wait till September to take the following course. Karim. SecondChildUserIdTAG: 1084194 SecondChildUserNameTAG: Jongaur SecondChildCreateTimeTAG: 2013-02-03T22:57:26Z SecondChildTAG: But I have another question and I wish I'm not annoing.....what is the differnce between the person and the final exam ? is it just the money ? SecondChildUserIdTAG: 1084194 SecondChildUserNameTAG: Jongaur SecondChildCreateTimeTAG: 2013-02-03T23:20:38Z SecondChildTAG: No problem :). The Official answer is here https://www.edx.org/help But, in my point of view, is more than the money... A Proctored Certificate will have more value than the Honor Code Certificate. One reason is because: 1- Your Proctored Exam will be done yes or yes by you. The Honor Code Certificate is only based in your Honor. I know that many people cheated during the Exams in other Courses, so having a Honor Code Certificate does not mean necesarily, unfortunately, that you have done it by your own merit: eg. you can perfectly gave me your password and I could perfectly done your Exam for you with my results - by the way, I will never accept to do that haha-, or you could have copy and paste it all the answer by cheating in other web pages, which I know that exists and I will not name...so, this will definitely decrease the value of it in the case that you want to show it in your resume if you wish to apply in a job... 2- Evaluation conditions will be different. You will have to solve the Exam in 2 hours only and without notes. So, a person that takes the Proctored Exam and pass it, you will make sure that the student knows the content and can solve the problems of the Exam quickly. So, in my point of view, a Proctored Certificate will be more serious, professionally talking ... I hope this can help you. Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2013-02-04T02:38:21Z SecondChildTAG: Please i know that the time is almost up ,but i have one question about it I live and study in Mongolia so is it possible to give this exam from here?? please reply :) SecondChildUserIdTAG: 91934 SecondChildUserNameTAG: Chingun SecondChildCreateTimeTAG: 2013-02-07T16:29:40Z SecondChildTAG: Hi Chingun, Take a look at here http://www2.pearsonvue.com/Dispatcher?application=VTCLocator&action=actStartApp&v=W2L&cid=792 and enter your Country. It will return you the most near Centre where you can take it :). In my case, my Country don´t have one... :/. I hope this can help you. Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2013-02-07T17:14:02Z IndexTAG: 1798 TitleTAG: Digital signal processing course on coursera Guys how many of you taking this course which is going to be start from Feb 18th ? UserIdTAG: 857413 UserNameTAG: Naveenkrishna CreateTimeTAG: 2013-02-02T17:14:00Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: i have already enrolled FirstChildUserIdTAG: 154172 FirstChildUserNameTAG: mofassair FirstChildCreateTimeTAG: 2013-02-03T12:35:21Z FirstChildTAG: I will FirstChildUserIdTAG: 288174 FirstChildUserNameTAG: matiasgrodriguez FirstChildCreateTimeTAG: 2013-02-02T20:05:03Z FirstChildTAG: Don't forget there is a DSP course here also assumed to start around half februari: [6003z][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50feb9ec6c25a3230000002a FirstChildUserIdTAG: 79885 FirstChildUserNameTAG: ruudoleo FirstChildCreateTimeTAG: 2013-02-02T18:51:32Z SecondChildTAG: This will be signals and systems, not DSP :-) S&S is a prerequisite for DSP. However, we will do a few tutorials and have Dantyrant's amazing labs on DSP and image processing. SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2013-02-02T20:10:39Z FirstChildTAG: I'm up for it.And also in Fundamentals of Electrical Engineering, Robots Design and Image Processing.The courses are amazing. FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2013-02-07T08:08:38Z IndexTAG: 1799 TitleTAG: MOS Transistors course MOS Transistors course by prof.Yannis Tsividis is going to be start from Mar 11th 2013 UserIdTAG: 857413 UserNameTAG: Naveenkrishna CreateTimeTAG: 2013-02-01T11:43:43Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Signed up too, can't wait! FirstChildUserIdTAG: 219780 FirstChildUserNameTAG: ReconIII FirstChildCreateTimeTAG: 2013-02-01T19:27:43Z FirstChildTAG: I'm looking forward to it. See you there! FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2013-02-01T13:06:14Z SecondChildTAG: ok! Ashwith SecondChildUserIdTAG: 857413 SecondChildUserNameTAG: Naveenkrishna SecondChildCreateTimeTAG: 2013-02-01T13:45:28Z SecondChildTAG: ashwith what you doing? SecondChildUserIdTAG: 857413 SecondChildUserNameTAG: Naveenkrishna SecondChildCreateTimeTAG: 2013-02-01T13:46:47Z SecondChildTAG: *Hopefully*, I'll be doing my masters in electronics this year. SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2013-02-02T06:51:49Z SecondChildTAG: ok. i'm doing my under graduation in EE (3rd year) SecondChildUserIdTAG: 1078452 SecondChildUserNameTAG: Radhika7 SecondChildCreateTimeTAG: 2013-02-03T05:19:55Z FirstChildTAG: Isn't this the textbook for the class: http://www.oup.com/us/catalog/general/subject/EngineeringTechnology/ElectricalComputerEngineering/?view=usa&ci=9780195170153 I want to start reviewing the material early. FirstChildUserIdTAG: 219780 FirstChildUserNameTAG: ReconIII FirstChildCreateTimeTAG: 2013-02-04T04:14:16Z SecondChildTAG: Yes that's the one. Sadly, that edition isn't available here yet :-/ SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2013-02-04T05:17:54Z SecondChildTAG: yes,i'm lucky because that text was there in my college library SecondChildUserIdTAG: 857413 SecondChildUserNameTAG: Naveenkrishna SecondChildCreateTimeTAG: 2013-02-04T14:07:35Z IndexTAG: 1800 TitleTAG: HELP!!!!! ,Myrimit I have a robot with a horizontally rotating hammer.I want the hammer to rotate clockwise when i press switch 1 and vice-versa when i press switch 2.how do i make the connections?i have ON/OFF switches and one power supply. UserIdTAG: 357747 UserNameTAG: kishores CreateTimeTAG: 2013-01-30T17:40:07Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hi kishores. To reverse a DC motor, you need to be able to reverse the direction of the current in the motor. Might you should use H-bridge circuit ... Here is a interesting Web page with some data that might can be useful for you , it provides the circuit[here][1]. Also you can take a look at here too [here][2] I hope that this can help you. Myriam. [1]: http://itp.nyu.edu/physcomp/Labs/DCMotorControl [2]: http://www.instructables.com/id/Super-Easy-Reversible-Motor-Control-for-Arduino-/ FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2013-01-31T19:38:33Z SecondChildTAG: hey,thanks SecondChildUserIdTAG: 357747 SecondChildUserNameTAG: kishores SecondChildCreateTimeTAG: 2013-02-01T05:22:46Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2013-02-01T13:37:31Z SecondChildTAG: You can also turn around the magnets in the motor, than you don't need a switch ... SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2013-02-05T19:25:41Z FirstChildTAG: Have a look at [the winning video from the CECC2 contest by AlexAlexandrescu.][1] What you are specifically looking for starts at around the 16:20 mark. [1]: http://ttp://www.youtube.com/watch?v=Jk6OIjlnBUs&feature=youtu.be FirstChildUserIdTAG: 9673 FirstChildUserNameTAG: xp42 FirstChildCreateTimeTAG: 2013-02-12T00:16:56Z IndexTAG: 1801 TitleTAG: proctored exam some of my friends want to give proctored exam but they neither registered on edx nor taken 6.002x . is there any option available that they can also give proctored exam ? UserIdTAG: 114864 UserNameTAG: Vikaash CreateTimeTAG: 2013-01-30T08:01:04Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: It is actually still possible to register for edx and the fall offering of 6.002x. They just won't have gotten any of the online points. Feel free to register for the class very late, and then register for the exam. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2013-01-30T20:12:37Z IndexTAG: 1802 TitleTAG: proctored exam: Cannot find the Candidate id? for proctored exam schedualing there is a singup option and it require candidate id... which is candidate id?i Have provided that registration number but pearson dosent accept that... can any one please explain the whole process... UserIdTAG: 154172 UserNameTAG: mofassair CreateTimeTAG: 2013-01-29T15:45:13Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: anybody schedule for the exam? FirstChildUserIdTAG: 154172 FirstChildUserNameTAG: mofassair FirstChildCreateTimeTAG: 2013-01-29T15:46:28Z IndexTAG: 1803 TitleTAG: "Control of Mobile Robots" course started The "Control of Mobile Robots" course offered by Georgia Institute of Technology has started on coursera. Like Myriam would say : " Like , like !" . :)) UserIdTAG: 240487 UserNameTAG: AlexAlexandrescu CreateTimeTAG: 2013-01-28T18:06:48Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I'm in this one too Alex. So far so good! :) FirstChildUserIdTAG: 467169 FirstChildUserNameTAG: Eyowzitgoin FirstChildCreateTimeTAG: 2013-01-30T06:12:57Z IndexTAG: 1804 TitleTAG: Myriam You have been a great help, just one more request Can you give me the stats of this course ?? or previous course UserIdTAG: 378150 UserNameTAG: GladIDoThis CreateTimeTAG: 2013-01-28T13:27:44Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi Arafat :) **6.002x Spring 2012 Official statistics (previous Course):** > Course statistics: 6.002x had 154,763 registrants. Of these, 69,221 > people looked at the first problem set, and 26,349 earned at least one > point on it. 13,569 people looked at the midterm while it was still > open, 10,547 people got at least one point on the midterm, and 9,318 > people got a passing score on the midterm. 10,262 people looked at the > final exam while it was still open, 8,240 people got at least one > point on the final exam, and 5,800 people got a passing score on the > final exam. Finally, after completing 14 weeks of study, 7,157 people > have earned the first certificate awarded by MITx, proving that they > successfully completed 6.002x. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2013-01-28T15:19:59Z IndexTAG: 1805 TitleTAG: about last question.... How to find power, supplied by the current source?? UserIdTAG: 1069402 UserNameTAG: Kamleshpri CreateTimeTAG: 2013-01-27T19:33:28Z VoteTAG: 1 CoursewareTAG: Week 1 / KVL/KCL example CommentableIdTAG: 6002x_S2E3_Using_KVL_KCL_and_VI_constraints NumberOfReplyTAG: 1 FirstChildTAG: Hi kamleshpri, Can I help you? Lets see this together. You know that by definition, the power P is the current multiplied by a voltage. P = V*I So, what is the current of the current source? Do we have this data in the statement ;) Ok, now. What is the voltage in the terminals of thhat current source? Isn´t it the blue one? ![im][1] But wait a minute, isn´t it V the same as the v2? V = v2 So, what is the value of the power P? ;) I hope this can help you. Myriam. [1]: https://edxuploads.s3.amazonaws.com/13593236321343682.png FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2013-01-27T21:56:31Z SecondChildTAG: sorry , i am not able to get the answer...please give me full solution....!!! SecondChildUserIdTAG: 1069402 SecondChildUserNameTAG: Kamleshpri SecondChildCreateTimeTAG: 2013-01-28T16:57:45Z IndexTAG: 1806 TitleTAG: 8.02x: Electricity and Magnetism This group is for all students who enrolled in 8.02x: Electricity and Magnetism https://www.facebook.com/groups/147450595411337/ UserIdTAG: 42624 UserNameTAG: AndersonCaires CreateTimeTAG: 2013-01-27T00:05:09Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1807 TitleTAG: proctored exam when will the proctored exam announced will it be before febraury or on the begining plz tell us as we have exams UserIdTAG: 221617 UserNameTAG: konan CreateTimeTAG: 2013-01-26T08:29:24Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: We finally got the proctored exam logistics figured out! You can register from your dashboard for an exam on Feb 13th. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2013-01-29T14:58:13Z IndexTAG: 1808 TitleTAG: 8.02x: Electricity and Magnetism [New course from MITx ][1] [1]: http://8.02x:%20Electricity%20and%20Magnetism UserIdTAG: 60453 UserNameTAG: TommyNittin CreateTimeTAG: 2013-01-23T14:23:17Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1809 TitleTAG: Thanks for this great experience ! i'm a Tunisian electrical and automation engineering student, i really enjoyed this course and appreciate all the work the edx team did! thank you all for this experience!i hope to see you soon in other courses especially more developed ones for electrical engineering! UserIdTAG: 125131 UserNameTAG: TRUTHSEEKER CreateTimeTAG: 2013-01-13T22:21:23Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1810 TitleTAG: CECC Update [IMPORTANT!] Hey Everyone, We noticed that two CECC entries went to our spam folder. Myriam luckily noticed the first one recently (the entry was sent two weeks back) and I found one today. So if you have entered the contest and did not get a reply from any of us saying that we received your video, please let me know either here or by sending an email to ceccmitx@gmail.com And do keep sending your videos! :) You have one more day to go. Best Regards The CECC Team UserIdTAG: 14386 UserNameTAG: ashwith CreateTimeTAG: 2013-01-13T03:43:20Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1811 TitleTAG: Next course schedule Hi - I would like to when is the next classes starting? Plz let me know the date UserIdTAG: 992713 UserNameTAG: Chinna007 CreateTimeTAG: 2013-01-10T11:33:48Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1812 TitleTAG: Thank you so much for this wonderful experience and for the certificate Hello, Thank you everybody for this wonderful experience. It was extremely fun and full of extremely useful information. Thank you so much for the certificate. It will be one of my best souvenirs in my lifetime. UserIdTAG: 418182 UserNameTAG: euldji2005 CreateTimeTAG: 2013-01-09T00:33:38Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1813 TitleTAG: No grades in certificate Why doesn't the certificate have the grade I obtained in this course ?? Plz help. is it suppose to be like that or its kinda a mistake in mine ? UserIdTAG: 336308 UserNameTAG: karangadhia CreateTimeTAG: 2013-01-07T20:46:52Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Congratulations! It is supposed to be like that, the Honor Code Certificate does not contain a grade. You may want to use your progress page for reference if needed. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2013-01-07T21:45:03Z IndexTAG: 1814 TitleTAG: Proctored Examination what is Proctored Examination.....oh tremendous thank you for the 6.002x Circuits and Electronics certificate UserIdTAG: 462861 UserNameTAG: edxforme CreateTimeTAG: 2013-01-07T16:12:59Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: A proctored exam takes place in a controlled, supervised environment. Your identity will also be confirmed. This gives educational facilities or potential employers more confidence when considering the certificate you present them. Good luck if you decide to take it! FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2013-01-07T16:42:44Z SecondChildTAG: You really think tha you will get a job with a certificate from an introductory course to the circuits? SecondChildUserIdTAG: 27399 SecondChildUserNameTAG: juancho SecondChildCreateTimeTAG: 2013-01-08T01:40:55Z SecondChildTAG: Yes I do Juancho. If your job required an introductory course in electronics, or any of the other courses offered and edX, then yes a proctored certificate from Pearson VUE would be of value. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2013-01-08T04:52:18Z SecondChildTAG: This certificate can be viewed as an added benefit, all else equal chances. Including as the ability to self-improvement. But this is my opinion. SecondChildUserIdTAG: 402617 SecondChildUserNameTAG: Vl SecondChildCreateTimeTAG: 2013-01-08T05:48:53Z SecondChildTAG: $95 is so much costly for that because it won't contain your grade as EDX non-proctored.... SecondChildUserIdTAG: 108863 SecondChildUserNameTAG: shiviz SecondChildCreateTimeTAG: 2013-01-08T08:39:31Z IndexTAG: 1815 TitleTAG: Thanks You so much Thank you edx for Choosing a very large testing company like Pearson All the international test centers in my country are usually 200- 700 kilometers from my city ,But This time i have a test center in my own city FEELS Great Thanks UserIdTAG: 378150 UserNameTAG: GladIDoThis CreateTimeTAG: 2013-01-07T15:21:28Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1816 TitleTAG: Test centre locator in order to locate our center for Proctored Examination... which category i should choose in this website..[link][1] [1]: http://www.pearsonvue.com/vtclocator/ UserIdTAG: 114864 UserNameTAG: Vikaash CreateTimeTAG: 2013-01-07T12:39:48Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: *When* it becomes available later on this month, it will likely be under the Academic/Admissions section, followed by whatever they decide to call it. ie: edX or MITx etc. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2013-01-07T15:01:41Z IndexTAG: 1817 TitleTAG: proctored test benefits what are the benefits of taking the proctored test? UserIdTAG: 365225 UserNameTAG: doublegj526 CreateTimeTAG: 2013-01-06T18:09:26Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: -The exam is taken in a controlled environment under predetermined conditions. -Your identity will be verified. Example: You may be applying to a school or job that may not accept the Honor Code Certificate but would accept a certificate that was proctored by a reputable establishment (Pearson VUE), for the reasons I mentioned above. Essentially this gives the educational facility or employer added confidence in considering your qualifications. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2013-01-06T21:24:14Z IndexTAG: 1818 TitleTAG: Help with current sign How do you decide whether the current is leaving or entering the particular node? UserIdTAG: 702959 UserNameTAG: Sashkow CreateTimeTAG: 2013-01-06T16:27:48Z VoteTAG: 1 CoursewareTAG: Week 1 / Node method example CommentableIdTAG: 6002x_S2E5_Node_Method NumberOfReplyTAG: 3 FirstChildTAG: Asked and answered. FirstChildUserIdTAG: 90796 FirstChildUserNameTAG: ChaunceyGardiner FirstChildCreateTimeTAG: 2013-01-06T16:42:49Z FirstChildTAG: answered here : https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_L2Node2/threads/50e9a284c4d119270000000d FirstChildUserIdTAG: 236136 FirstChildUserNameTAG: Hamid-ch FirstChildCreateTimeTAG: 2013-01-06T21:14:37Z FirstChildTAG: Hi Sashkow If the above link is still confusing, here is a good example. You can assign current entering and leaving the node any way you want, as long as you are consistent. If you assign current entering the node as negative, you must assign current leaving the node as positive and vice versa. If you don't follow this convention or are not consistent, the sum of the currents will not add up to 0 and you will violate KCL laws. See below. ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13577167832550756.jpg FirstChildUserIdTAG: 132828 FirstChildUserNameTAG: rharris FirstChildCreateTimeTAG: 2013-01-09T07:43:23Z SecondChildTAG: Oh, and by the way, look at each equation above and solve for i1. In both cases i1 = i2 + i3. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2013-01-09T16:03:53Z IndexTAG: 1819 TitleTAG: Why no grade in proctored exam? I'm interested in proctored exam, but I hoped it could have a grade. More, I also hoped to have more time before the exam... UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2013-01-05T23:42:45Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: hope grading was involved...... FirstChildUserIdTAG: 91261 FirstChildUserNameTAG: mr_sophisticated FirstChildCreateTimeTAG: 2013-01-07T08:47:38Z IndexTAG: 1820 TitleTAG: To edX Team - Proctored Exam - Where? Hi! I have just received an e-mail from edX, announcing officialy the proctored exam option. I would really appreciate it if you could make a list of Countries and cities where a proctored exam can be taken. Thank you! UserIdTAG: 304587 UserNameTAG: Vasso CreateTimeTAG: 2013-01-04T20:52:08Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: http://www.pearsonvue.com/vtclocator/ FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2013-01-05T06:57:15Z SecondChildTAG: pls inform me about the date of commencements of exam and the duration. SecondChildUserIdTAG: 91261 SecondChildUserNameTAG: mr_sophisticated SecondChildCreateTimeTAG: 2013-01-08T15:55:19Z IndexTAG: 1821 TitleTAG: Staff: Much Thanks! Archiving Possibilities? Dear edX Staff, Thank you all so much for a challenging, illuminating, and inspiring experience with 6.0002x! I especially enjoyed Prof. Agarwal's clear, "aha moment"-producing lectures and the tutorials with Prof. Sussman and Dr. Mitros. (That real bread board was amazing to see!) It would be great if all the material from this course, including all the answers and explanations, could be archived along with that of the other edX courses -- possibly on some webpages with links to and from the MIT OCW pages (www.ocw.mit.edu/index.htm). This would allow for self-paced preview, study, and/or review (even by anyone who may not be registered for a current edX course). Thanks again! Happy New Year! Laureen UserIdTAG: 188609 UserNameTAG: Laureen CreateTimeTAG: 2013-01-04T05:05:48Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Well, I don't know what will happen here at edX, but you can find all the course material archived (except for the homework and exams unfortunately) on the previous site of 6.002x at http://6002x.mitx.mit.edu. FirstChildUserIdTAG: 129876 FirstChildUserNameTAG: msarabi95 FirstChildCreateTimeTAG: 2013-01-04T05:45:56Z SecondChildTAG: you can change the homework and final exam to pdf format easily and save them for ever. SecondChildUserIdTAG: 374393 SecondChildUserNameTAG: rmaleki SecondChildCreateTimeTAG: 2013-01-04T06:42:31Z SecondChildTAG: nice idea! SecondChildUserIdTAG: 361137 SecondChildUserNameTAG: Ericson SecondChildCreateTimeTAG: 2013-01-04T16:15:18Z IndexTAG: 1822 TitleTAG: Greetings from Kazakhstan! Thanks to all edX stuff! It was an amazing and unforgettable experience. Now I see why MIT holds its top positiion among all universities in the world. Great teaching, easy to follow the course syllabus, wonderful stuff... these are just a few examples of why MIT is the best. I had doubts at the beginning, but now I got inspired so I took two more courses from edX. My gratitude towards all the people who are involved in this initiative is endless! Again thanks, thanks, and thanks! I LIKE! :) UserIdTAG: 393625 UserNameTAG: makarlen CreateTimeTAG: 2013-01-04T03:15:28Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1823 TitleTAG: Certificate Validation Hi everyone :) congrats to all that get the certificate. I don't understand why the certificate should be certificated by a number on a software called Gpg4w. I could be great if someone can explain the importance about this and another cool, interesting or important information about. Thanks (Sorry, I only know a little of English) UserIdTAG: 110802 UserNameTAG: leoblack CreateTimeTAG: 2013-01-04T03:03:56Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50e5bb5bb3b7fe2b0000000f FirstChildUserIdTAG: 236136 FirstChildUserNameTAG: Hamid-ch FirstChildCreateTimeTAG: 2013-01-04T03:35:58Z SecondChildTAG: i have secured 100% marks however, i didnot get the certificate till date...its 4 jan now...will i be getting it in my email...?? SecondChildUserIdTAG: 460505 SecondChildUserNameTAG: QaiserIlyas SecondChildCreateTimeTAG: 2013-01-04T05:10:52Z SecondChildTAG: QaiserIlyas: Did u check on your dashboard..the option to sownload your certificate should be there..u wont get it on email.. SecondChildUserIdTAG: 173147 SecondChildUserNameTAG: cruiser_rahit SecondChildCreateTimeTAG: 2013-01-04T08:11:29Z SecondChildTAG: https://www.edx.org/dashboard SecondChildUserIdTAG: 173147 SecondChildUserNameTAG: cruiser_rahit SecondChildCreateTimeTAG: 2013-01-04T08:12:49Z IndexTAG: 1824 TitleTAG: Help Urgent How can I change my name on the certificate Help Urgent How can I change my name on the certificate UserIdTAG: 316899 UserNameTAG: elou CreateTimeTAG: 2013-01-03T22:25:21Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Unfortunately, the certificates were generated with the name that you had on your dashboard at the time of generation. If you change it now, future certificates should have the changed name. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2013-01-03T22:28:34Z IndexTAG: 1825 TitleTAG: To: staff. Where can I find my cerfiticate? Thanks you all. But I can not see my certificate. Where can I find it? Looking forward to hearing from you. Thanks again. UserIdTAG: 371617 UserNameTAG: rainbow12 CreateTimeTAG: 2013-01-03T22:13:21Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Have you checked your [dashboard][1]? [1]: https://www.edx.org/dashboard FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2013-01-03T22:21:14Z SecondChildTAG: ++ SecondChildUserIdTAG: 96452 SecondChildUserNameTAG: Lyla SecondChildCreateTimeTAG: 2013-01-03T22:26:09Z SecondChildTAG: I got certificate, but it is without my result. Will be certificate with result available? SecondChildUserIdTAG: 413784 SecondChildUserNameTAG: Dorotka SecondChildCreateTimeTAG: 2013-01-05T20:13:29Z SecondChildTAG: No grades on certificate. SecondChildUserIdTAG: 361137 SecondChildUserNameTAG: Ericson SecondChildCreateTimeTAG: 2013-01-06T05:25:28Z IndexTAG: 1826 TitleTAG: Certificate Congratulations to everyone who got their certificates. Why the Certificate does not contain the final grade?! UserIdTAG: 283909 UserNameTAG: MahmoudMosad CreateTimeTAG: 2013-01-03T21:49:27Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: ask to edx team FirstChildUserIdTAG: 174229 FirstChildUserNameTAG: fares27 FirstChildCreateTimeTAG: 2013-01-03T22:04:20Z SecondChildTAG: how? SecondChildUserIdTAG: 283909 SecondChildUserNameTAG: MahmoudMosad SecondChildCreateTimeTAG: 2013-01-03T22:22:16Z FirstChildTAG: The certificates are to show that you completed the course, and we didn't want to devalue the certificates of people who got C's. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2013-01-03T22:27:33Z SecondChildTAG: Grading system was from your side. Grading Letter grades will be based on the following weighting: homework 15%, labs 15%, midterm 30%, and final exam 40%. Each of the homework and labs carries equal weight. You will need to get a total mark of ***60% for a C, 70% for a B, and 87% for an A.*** **ATLEAST VALUE PEOPLE WHO GOT "A"**. i agree that learning and gaining knowlwdge is important,that is why hardwork of many students have got them A GRADE. SecondChildUserIdTAG: 353709 SecondChildUserNameTAG: anshu10750 SecondChildCreateTimeTAG: 2013-01-04T03:14:06Z SecondChildTAG: we all appreciate your hardwork.but also request you to consider our views,you cuold have given 2 certificates,one with grade and one without.i hope this the the best solution.it doesn't devalue the certificates of people who got C's. SecondChildUserIdTAG: 353709 SecondChildUserNameTAG: anshu10750 SecondChildCreateTimeTAG: 2013-01-04T03:19:31Z SecondChildTAG: Question for Staff i have got 100% marks in everything...and now i have received a certificate but there is nothing mentioned on it i mean no grade..no marks...nothing...its a suggestion that there should be some difference mentioned in the certificate for those who got 60% and for those who got 100%...thankyou for this course it helped us a lot in learning..God Bless SecondChildUserIdTAG: 460505 SecondChildUserNameTAG: QaiserIlyas SecondChildCreateTimeTAG: 2013-01-04T05:25:59Z SecondChildTAG: That`s true. Someone that get 60% maybe have not worked so hard as one that get 100%, i think that that the grades should appear on the certificate. SecondChildUserIdTAG: 110802 SecondChildUserNameTAG: leoblack SecondChildCreateTimeTAG: 2013-01-04T13:06:14Z SecondChildTAG: The whole point of this course is to learn electronics. How do you know that someone with 60% did not work as hard or harder then you? SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2013-01-04T13:22:08Z SecondChildTAG: this is a MAJOR FLAW in their evaluation!! THEY SHOULD HAVE PROVIDED DIFFERENT CERTIFICATES.... The fact that someone couldn't score higher or atleast 60% in this course itself displays the hardwork required and put in by those who scored higher..!! since they value feelings for ppl who scored less, they should equally value ppl who scored brilliantly!! what was the point of the grade system then and why have they switched over to certificates with no grade still remains a mystery!! a good explanation from staff is still awaited! SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2013-01-07T13:35:12Z IndexTAG: 1827 TitleTAG: Certificate Delay I am from Pakistan and certificate download button still not shown in my Dashboard. Why this delay? When will the certificates be available for download in Pakistan. UserIdTAG: 77046 UserNameTAG: Sarwar20 CreateTimeTAG: 2013-01-03T12:06:24Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I received mine. It's probably still distributing them. Be patient :D FirstChildUserIdTAG: 311022 FirstChildUserNameTAG: GordanS FirstChildCreateTimeTAG: 2013-01-03T12:09:28Z SecondChildTAG: Is the grade on it? Mine has no grade. SecondChildUserIdTAG: 359677 SecondChildUserNameTAG: guillermb SecondChildCreateTimeTAG: 2013-01-03T12:15:13Z FirstChildTAG: Shall I be informed though email when the certificate is available...? FirstChildUserIdTAG: 77046 FirstChildUserNameTAG: Sarwar20 FirstChildCreateTimeTAG: 2013-01-03T12:07:50Z SecondChildTAG: I feel the certificates are being distributed alphabetically .i am still waiting. SecondChildUserIdTAG: 99628 SecondChildUserNameTAG: Pranjal16 SecondChildCreateTimeTAG: 2013-01-03T12:14:40Z SecondChildTAG: hope so. i have got mine SecondChildUserIdTAG: 97359 SecondChildUserNameTAG: azeem179 SecondChildCreateTimeTAG: 2013-01-03T13:24:14Z IndexTAG: 1828 TitleTAG: certificate its third and i have not yet received the certificate..kindly tell when it will be available?? UserIdTAG: 433368 UserNameTAG: SurbhiMahajan CreateTimeTAG: 2013-01-03T11:48:06Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: hi surbhi!! u frm where? FirstChildUserIdTAG: 368712 FirstChildUserNameTAG: jmen FirstChildCreateTimeTAG: 2013-01-03T11:50:43Z SecondChildTAG: india SecondChildUserIdTAG: 433368 SecondChildUserNameTAG: SurbhiMahajan SecondChildCreateTimeTAG: 2013-01-03T11:53:40Z SecondChildTAG: its available surbhi...please check!!!! SecondChildUserIdTAG: 353709 SecondChildUserNameTAG: anshu10750 SecondChildCreateTimeTAG: 2013-01-03T14:14:19Z FirstChildTAG: Neither have I. Where am I supposed to find the certificate? Thank you... FirstChildUserIdTAG: 414462 FirstChildUserNameTAG: Pablo_C FirstChildCreateTimeTAG: 2013-01-03T11:58:51Z SecondChildTAG: me and my colleague are from Bulgaria and still do not have ours SecondChildUserIdTAG: 207728 SecondChildUserNameTAG: RossyFromBulgaria SecondChildCreateTimeTAG: 2013-01-03T12:03:23Z SecondChildTAG: i didn't get yet..now what should i do? SecondChildUserIdTAG: 219356 SecondChildUserNameTAG: Karan123 SecondChildCreateTimeTAG: 2013-01-03T12:03:31Z SecondChildTAG: You wil find it on the dashboard showing 'Download Your PDF Certificate'.Just wait for a little time. SecondChildUserIdTAG: 361137 SecondChildUserNameTAG: Ericson SecondChildCreateTimeTAG: 2013-01-03T15:15:20Z FirstChildTAG: I think that we just have to wait little more time ;) FirstChildUserIdTAG: 330357 FirstChildUserNameTAG: steryd FirstChildCreateTimeTAG: 2013-01-03T12:04:53Z SecondChildTAG: I got mine. SecondChildUserIdTAG: 359677 SecondChildUserNameTAG: guillermb SecondChildCreateTimeTAG: 2013-01-03T12:14:14Z SecondChildTAG: but this time it's good waiting, like the one on christmas eve or birthday, isn't it? SecondChildUserIdTAG: 207728 SecondChildUserNameTAG: RossyFromBulgaria SecondChildCreateTimeTAG: 2013-01-03T12:19:52Z FirstChildTAG: i got mine and hopefully u all will get yours :) FirstChildUserIdTAG: 433368 FirstChildUserNameTAG: SurbhiMahajan FirstChildCreateTimeTAG: 2013-01-04T14:31:35Z IndexTAG: 1829 TitleTAG: Use of Certificate I would love to hear how people are able to use the certificate....college or HS credit, perhaps continuing education credits, for a job, etc..... I'm trying to encourage advanced HS students to look into this course as a way of demonstrating to colleges their desire to study EE and/or general aptitude as they apply. Anyone have experience on this front? Thanks, Alan UserIdTAG: 214701 UserNameTAG: alanmatson CreateTimeTAG: 2013-01-02T16:26:03Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 6 FirstChildTAG: The certificate it important for me, since I get to show my family members something tangible, to convince them I'm not spending all my time on the computer doing useless things ;) This way I get some peace. FirstChildUserIdTAG: 4463 FirstChildUserNameTAG: pmj FirstChildCreateTimeTAG: 2013-01-02T16:57:10Z SecondChildTAG: What do you mean by "useless things"?:) SecondChildUserIdTAG: 252722 SecondChildUserNameTAG: vaboro SecondChildCreateTimeTAG: 2013-01-03T01:46:44Z SecondChildTAG: He means that education is not always appreciated for what it is, unless you earn something . In other words, pmj's family think that he does useless things standing in front of the computer all day long, so he needs the certificate to demonstrate that they were wrong. And he is not the only one in this situation . Sometimes my wife complains too . :)) SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2013-01-03T10:35:07Z SecondChildTAG: pmj very true. My mom has no idea and defintely thinks am doing useless courses without any value. They are some things you almost can't explain to those closest to you, you just have to pray they somehow understand. SecondChildUserIdTAG: 183507 SecondChildUserNameTAG: obiradaniel SecondChildCreateTimeTAG: 2013-01-04T22:30:16Z FirstChildTAG: In an inquire to a UK University, i asked if a certificate of such sort would be taken into consideration. They asked me to sent them the course structure and that they might consider it as APL. Furthermore, UK Universities will launch online platforms next year. So, i believe that this will be a new way of learning and that we are lucky to have found out about Edx. http://www.bbc.co.uk/news/education-20697392 FirstChildUserIdTAG: 304587 FirstChildUserNameTAG: Vasso FirstChildCreateTimeTAG: 2013-01-02T20:43:29Z SecondChildTAG: interesting! SecondChildUserIdTAG: 330357 SecondChildUserNameTAG: steryd SecondChildCreateTimeTAG: 2013-01-02T21:12:11Z SecondChildTAG: The program is called FutureLearn http://futurelearn.com SecondChildUserIdTAG: 252722 SecondChildUserNameTAG: vaboro SecondChildCreateTimeTAG: 2013-01-03T01:48:24Z FirstChildTAG: My employer provides educational benefit that pays the cost of professional development classes, provided that you get a B or better in the class. Since the edX course was free, I made a deal with my boss that I could take 4 hrs/week out of my schedule to work on the class. He agreed, but said I'd still have to get the B or he'd make me pay it back in (unpaid) overtime. So, I need the certificate (and my progress page) to show that I met my end of the bargain. Plus, since it's professional development it goes in my file. FirstChildUserIdTAG: 339668 FirstChildUserNameTAG: chickwebb FirstChildCreateTimeTAG: 2013-01-03T01:29:53Z SecondChildTAG: **Cool!** As time goes on and the popularity of edX and massive open-enrollment online classes grow, the number of students that had "the 6.002x experience" will also grow. Some of these former "alumni" will become (perhaps even already are) our managers and our bosses and the people that conduct job interviews in the corporate world. They will know the value and quality that stand behind an edX certificate, and respect the amount of time a student must put in in order to obtain the certificate, *as they were once enrolled themselves*! It's nice to hear that, even so ***early*** in the game, when edX is not yet a "household name," that 6.002x edX certificates are already being accepted by employers as a professional development credential. Its a sign that soon this will be the norm, and there won't have to be a special "deal with the boss," just fill out the standard professional development form, and if the classes cost the company next-to-nothing, you can work on the class on company time; its only fair! Better than some of these overpriced, dubious seminars, or boring, worthless community-college classes that are offered as professional development or continuing education. And better for those employees who only get reimbursed when the course(s) have something to do with their job function; the number of courses available will only grow; and hopefully those seeking (and especially those needing) highly specialized, rarely offered courses / training will find themselves with a class to take; without the long-distance travel or other special accommodation that are often necessary. SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2013-01-03T03:41:04Z FirstChildTAG: I took this class since I am extremely interested in the area of EE and so my high school let me take this as a class with credit. FirstChildUserIdTAG: 219780 FirstChildUserNameTAG: ReconIII FirstChildCreateTimeTAG: 2013-01-02T18:14:55Z SecondChildTAG: This is great! I agree. If you just finished the textbook you would get a tremendous advantage. So, finishing this course DOES mean something! It can't go unnoticed. SecondChildUserIdTAG: 252722 SecondChildUserNameTAG: vaboro SecondChildCreateTimeTAG: 2013-01-03T01:59:45Z FirstChildTAG: **Alan**,I do agree with your suggestion.Though the course needs some maths trick like differential and integration, complex number calculation,etc, the big 'aha' moments and many many surprises will be a nice introduction for high school students who develop or have a interest in EECS world. For me, learning this course is full of excitement and fun,though sometimes a little bit challenging :) .In my country , most HS students are buried in studying limited knowlege before going to colleges again and again preparing for big exam---which may have influence on their life. As a result, they have no idea about future directions when choosing a major because they do not develop an interest in some field and get confused. Introductory courses like 6.002x are of great help for HS students to develop a basic concept about future studying area, and may help them make a decision. And it couln't better if credits of the edx course or some other online courses are recognized by high school. FirstChildUserIdTAG: 361137 FirstChildUserNameTAG: Ericson FirstChildCreateTimeTAG: 2013-01-03T04:09:18Z FirstChildTAG: Got my certificate of completion! :) but i have got a query.. This certificate shows only our completion right? What about the grades? Suppose we add this certificate to our academic & job profile, what are our chances in accordance with this course completion & how can the interview panel or the institution authorities verify our grades? FirstChildUserIdTAG: 167618 FirstChildUserNameTAG: annvilla FirstChildCreateTimeTAG: 2013-01-03T08:52:39Z SecondChildTAG: No grades are shown in mine.But grades can be seen in the dashboard, and you may use screenshot though it is not formal. SecondChildUserIdTAG: 361137 SecondChildUserNameTAG: Ericson SecondChildCreateTimeTAG: 2013-01-03T15:05:40Z IndexTAG: 1830 TitleTAG: APPRECIATION FROM NIGERIA I am very delighted to have successfully completed this course, it basically fixed some major fault lines in my engineering foundation and I now have more professional confidence as an electrical engineer. My sincere gratitude to the whole EDX MITX 6002 team you've done a great job. To my fellow course mates who have issues with the certificate format I'll say though you deserve bragging rights for your success on an MIT COURSE , I think the focus should be on using this new found analytical knowledge to churn out innovative solutions which regular MIT grads are well known for and who knows a Nobel prize could be in the offing, "CERTIFICATES are not trophies they are an acknowledgement of capacity to DO."(Adetiba Ekundayo) UserIdTAG: 216600 UserNameTAG: Dayohigh CreateTimeTAG: 2013-01-02T08:06:39Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Nice thought. And I like your quote very much - Certificates are not trophies they are an acknowledgement of capacity to do. FirstChildUserIdTAG: 297370 FirstChildUserNameTAG: ayush3504 FirstChildCreateTimeTAG: 2013-01-03T00:44:35Z IndexTAG: 1831 TitleTAG: What name will you put on your certificate Your real name in your native language? Or translate it to English?...What if someone who doesn't have a English name.. Is there any other Chinese? 大家新年快乐 :) Happy New Year to everyone UserIdTAG: 136959 UserNameTAG: christerpher CreateTimeTAG: 2013-01-02T03:52:09Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: 同乐同乐 FirstChildUserIdTAG: 363456 FirstChildUserNameTAG: JiWentao FirstChildCreateTimeTAG: 2013-01-02T08:35:37Z FirstChildTAG: I guess you can always state a dual form of your name: i.e., Chinese version - English version (or viceversa). That should do. Best. Jordi FirstChildUserIdTAG: 366320 FirstChildUserNameTAG: Galli FirstChildCreateTimeTAG: 2013-01-02T10:26:19Z FirstChildTAG: Nice to meet you! I use my real name in Chinese. 新年快乐！ FirstChildUserIdTAG: 361137 FirstChildUserNameTAG: Ericson FirstChildCreateTimeTAG: 2013-01-02T11:32:06Z SecondChildTAG: Glad to meet you too:) SecondChildUserIdTAG: 136959 SecondChildUserNameTAG: christerpher SecondChildCreateTimeTAG: 2013-01-02T12:35:20Z SecondChildTAG: Hi Christerpher, how do you write your full name in English, I mean in the order of ''Xing Ming姓 名'' or the order of ''Ming Xing名 姓'' ? I am confused about that. SecondChildUserIdTAG: 361137 SecondChildUserNameTAG: Ericson SecondChildCreateTimeTAG: 2013-01-02T16:33:21Z SecondChildTAG: 我和你一样直接用汉字了。要是换英文的话还是“名姓”吧 SecondChildUserIdTAG: 136959 SecondChildUserNameTAG: christerpher SecondChildCreateTimeTAG: 2013-01-03T00:17:54Z SecondChildTAG: 非常感谢！ SecondChildUserIdTAG: 361137 SecondChildUserNameTAG: Ericson SecondChildCreateTimeTAG: 2013-01-03T03:30:02Z FirstChildTAG: Hi, but christerpher is not a Chinese name and not putonghua FirstChildUserIdTAG: 402617 FirstChildUserNameTAG: Vl FirstChildCreateTimeTAG: 2013-01-02T14:26:22Z SecondChildTAG: Yeah,it's my random typing...:) Ni Hao SecondChildUserIdTAG: 136959 SecondChildUserNameTAG: christerpher SecondChildCreateTimeTAG: 2013-01-02T14:34:12Z FirstChildTAG: We will use the name that you provided when you created your account. You can see it and change it from your dashboard. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2013-01-02T16:09:20Z IndexTAG: 1832 TitleTAG: Happy new year to all ! ok some questions now :o) Will there be a following to 6002X ? Do you think the certificate can have some utility ? ( in France. ) I need more maths and physics, are there courses at MIT ? Thank's !! UserIdTAG: 373752 UserNameTAG: Croak CreateTimeTAG: 2013-01-02T03:22:38Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: For math and physics you can always take a look at https://www.khanacademy.org/ FirstChildUserIdTAG: 288174 FirstChildUserNameTAG: matiasgrodriguez FirstChildCreateTimeTAG: 2013-01-02T13:57:41Z FirstChildTAG: I believe there will be following EECS courses soon. http://ocw.mit.edu/courses/find-by-department/ you will find more MIT courses here, maths and physics included. FirstChildUserIdTAG: 361137 FirstChildUserNameTAG: Ericson FirstChildCreateTimeTAG: 2013-01-02T11:48:52Z IndexTAG: 1833 TitleTAG: Certificates What should be final score to get a certificate ? UserIdTAG: 101902 UserNameTAG: DHEERAJK_VITS CreateTimeTAG: 2013-01-01T16:05:34Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I think greater than or equal to 60%. I'm not very sure because I don't remember where I had read this. According to the syllabus https://www.edx.org/static/content-mit-6002x/handouts/syllabus.a477535058a1.pdf 60% is minimum requirement to get a grade, so I think it would be reasonable to assume 60% as the minimum requirement for earning a certificate. FirstChildUserIdTAG: 297370 FirstChildUserNameTAG: ayush3504 FirstChildCreateTimeTAG: 2013-01-01T16:13:05Z IndexTAG: 1834 TitleTAG: To Myrimit Hola Myriam: Si me disculpas, sólo una última cuestión respecto al problema Q1 del Final Exam.Es que sigo sin ver el loop sobre el que se aplica KVL, para obtener las ecuaciones que dan en la explicación del problema; imagino que no tengo bien esquematizado el circuito.... Y quiero aprovechar esta oportunidad para darte las gracias por todo, y sobre todo, felicitarte por tu extraordinario trabajo que tanto nos ha ayudado. Te deseo un Feliz 2013. UserIdTAG: 706676 UserNameTAG: PabloVic CreateTimeTAG: 2012-12-31T19:00:50Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Gracias Pablo, Igualmente para ti, que tengas un hermoso 2013! Estoy subiendo un video explicativo de la Q1 a YouTube, pero va a demorar un poco, ya que está tardando mucho la subida del mismo. Espero que te sea de ayuda, en cuanto esté, adjunto el link. Sí, probablemente estés teniendo algún problema de signos...La ecuación no es tan fácil a simple vista, se deben plantear las ecuaciones correspondientes a cada nodo y luego se debe llegar a la expresión del voltaje de salida en función de rm, R, i1N e i2N...En el video esta explicado paso por paso. Espero que te sea de ayuda. Saludos, Myriam. ---- Thank you Pablo, I wish you to you too a nice 2013! I am uploading a video explanation about Q1 on YouTube, but it will take some minutes more, it will be might for the 2013 haha. But as soon as I have it uploaded it on my YouTube Account, I will Posst the link. Yes, it is possible that you are making some mistake with the signs...The equation is not easy, you can not get it at a first sight...You have to write the different node equations and then reach to the expression of the output voltage in terms of rm, R, iIN1 and iIN2...In the video explanation is explained step by step. I hope this can help you. Greetings, Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-31T23:40:47Z SecondChildTAG: [Q1- Video Explanation Link][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50e2579cd0df331f00000025 SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2013-01-01T03:30:22Z IndexTAG: 1835 TitleTAG: GRACIAS Gracias Lyla UserIdTAG: 264238 UserNameTAG: lean2209 CreateTimeTAG: 2012-12-31T18:49:26Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1836 TitleTAG: Happy New Year Happy New Year to everyone! from the island of the Philippines!... UserIdTAG: 349139 UserNameTAG: 1977ROYELMER CreateTimeTAG: 2012-12-31T15:40:43Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1837 TitleTAG: About textbook Hi! Please tell me how and where I can order and buy the textbook! Thanks in advance Happy New Year for everybody! UserIdTAG: 96873 UserNameTAG: Santyaga CreateTimeTAG: 2012-12-30T15:56:52Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: Places like Amazon.com etc. have them new and used. Example: http://www.amazon.com/Anant-Agarwal/e/B001K8QU2S May you have a great New Year too. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-12-30T16:02:30Z SecondChildTAG: If you are in India flipkart.com has a cheap copy. SecondChildUserIdTAG: 99628 SecondChildUserNameTAG: Pranjal16 SecondChildCreateTimeTAG: 2012-12-30T16:20:24Z SecondChildTAG: no you can download them on edx i will try to explain how you will download him SecondChildUserIdTAG: 174229 SecondChildUserNameTAG: fares27 SecondChildCreateTimeTAG: 2012-12-30T17:19:02Z SecondChildTAG: i will send him if you get me your facebook SecondChildUserIdTAG: 174229 SecondChildUserNameTAG: fares27 SecondChildCreateTimeTAG: 2012-12-30T17:30:16Z FirstChildTAG: [Here][1] - This was the Link provided in the Course Info. ![im][2] > The edX Circuits & Electronics Course (6.002x) uses the textbook > Foundations of Analog and Digital Electronic Circuits, by Anant > Agarwal and Jeffrey H. Lang. Save 40% when you buy print + electronic > titles together. Use code MIT40. Free shipping. *This offer is not > available in Australia. > > Save 25% on any print or electronic title. Use code MIT25. Free > shipping. ---------- Also you can find it in places as PennyPacker said, like Amazon ;) ---------- And, here it comes my Marketing again - I am taking seriously the comment of my friend Chauncey haha -, if you are willing to win a Textbook signed personally by Prof. Agarwal we are running a Contest, you have till January 14th 2013 to submit your video.[read here][3] also [here][4] Take care and Happy New Year, Myriam. [1]: http://store.elsevier.com/specialOffer.jsp?offerId=EST_MIT&utm_source=EMAP&utm_medium=text_link&utm_term=Circuits_Electronics&utm_campaign=edx#oid=1003_4 [2]: http://secure-ecsd.elsevier.com/est/20122517_B_MITx-Landing-Page-banner_034-317_1200_R2_FINAL.jpg [3]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50b2892629b7b1270000003b [4]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50c08a8f2196ec2300000022 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-30T16:45:06Z SecondChildTAG: Helpp!! Staff!!! I have a serious complaint: Today I noticed for the first time that your name was wrongly spelled on the course info page: Miramit and the community TA's have organized a student video competition. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-12-31T00:46:51Z FirstChildTAG: The are many available from several sellers on www.abebooks.com - search for isbn 1558607358. Be wary about international editions - in general, they aren't supposed to be sold in the USA and in some cases, may be inferior copies. Abebooks is my favorite source for used books if I strike out on amazon or amazon's pricing is too high to justify buying a new copy. Orin. FirstChildUserIdTAG: 141000 FirstChildUserNameTAG: OrinE FirstChildCreateTimeTAG: 2012-12-30T20:54:42Z FirstChildTAG: Oh thanks! I think i will choose amazon.com or if I`ll be lucky take part in competition FirstChildUserIdTAG: 96873 FirstChildUserNameTAG: Santyaga FirstChildCreateTimeTAG: 2012-12-31T10:53:09Z SecondChildTAG: Good luck on the competition! (Myrimit's link is actually little better, price wise.) SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-12-31T13:43:02Z IndexTAG: 1838 TitleTAG: thanxxx thank you edx.Over the course i have learnt a lot more than my university has taught me.Thank you mr. aggarwal,mr sussman ,mr piotr and everybody at edx who helped me with the course and have a better understanding of circuits and electronics. UserIdTAG: 386372 UserNameTAG: akashrao1991 CreateTimeTAG: 2012-12-29T16:59:30Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1839 TitleTAG: Is any Marine Engineer participating this course? In this modern world, Most of the machineries involve electronics control. The old version Electrical Officers are mostly unaware of basic electronics. Edx is providing a best opportunity to get the basic electronics knowlegde. I suggest all marine engineers and electrical officers must conduct this course. Best Regards UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-12-29T07:43:15Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1840 TitleTAG: Thank you Edx team. You guys are amazing. Thank you for your world wide vision. UserIdTAG: 371498 UserNameTAG: Engineer001 CreateTimeTAG: 2012-12-29T02:41:26Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 0 IndexTAG: 1841 TitleTAG: Doubt! Somebody knows how to request the "honor-code certificate"? Greetings! UserIdTAG: 377664 UserNameTAG: Bowen_Berty CreateTimeTAG: 2012-12-28T21:04:55Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50dad7f02fc042230000000f FirstChildUserIdTAG: 325197 FirstChildUserNameTAG: Vitali_Jerin FirstChildCreateTimeTAG: 2012-12-28T21:42:52Z SecondChildTAG: Thanks! SecondChildUserIdTAG: 377664 SecondChildUserNameTAG: Bowen_Berty SecondChildCreateTimeTAG: 2012-12-28T22:39:34Z IndexTAG: 1842 TitleTAG: EECS and the 'Five Senses' Digital or Analog, electronics still has a clearly demarcated boundary: the 5 senses that we humans have. You might argue that the Voyager sends signals in a frequency-band we can neither hear or see, or for that matter, a dog whistle that is not within our range of hearing. Whatever be the 'transients', the 'steady-state' always seem to boil down to our perceptive faculty. So far, though the gustatory (taste) and olfactory (smell) arena remain relatively virgin territory (for electronics' foray); I am certain that they are going to be deflowered soon. With the 'haptic' (touch) feedback of Android, the sense of touch has already made its way! UserIdTAG: 285616 UserNameTAG: AmiyaX CreateTimeTAG: 2012-12-28T17:42:45Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1843 TitleTAG: Final Exam - Q6 explanation wrong? I think there is an error in the explanation posted for the answer to Q6 - specifically for parts (c) and (d), which show the answer for Vmax as 1.0757 volts and Vmin as 2.9242 volts. On behalf of all of us who made similar dumb mistakes when entering the solution after getting the difficult parts of this (and other) questions right, it's good to know that we are in good company. Maybe we can call this a "duh" moment to go with our many "aha" ones??! Just to echo the sentiments of many other students, thanks to Prof Agarwal and all the MIT staff who made this course such a great experience, and to all those who posted in this forum and gave the rest of us encouragement that we were not struggling alone! Onward and upward to the next course, please! UserIdTAG: 145194 UserNameTAG: DerekH CreateTimeTAG: 2012-12-28T17:42:22Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Yes there are errors in the Q6 explanation. The answers are indeed reversed and there are minus signs added/missing. In addition, a $V_{min}$ should be $V_{max}$. This problem was easier to solve in terms of current; you just multiply by $R_L$ at the end to get the voltage. Also by judicious choice of origins, it can be analyzed as separate rising/falling exponentials from $t=0$ to $t=T$ with initial conditions of $V_{min}$ and $V_{max}$ - gets rid of the $t_0$ and $t_1$ in the official answer which cancel out anyway. I'll post my solution later. FirstChildUserIdTAG: 141000 FirstChildUserNameTAG: OrinE FirstChildCreateTimeTAG: 2012-12-28T20:48:34Z SecondChildTAG: Yep, i agree. Easier to solve in terms of current. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-12-28T23:44:56Z SecondChildTAG: OK, here is my solution. I'm going to do it in terms of the current through $R_L$, $i_R(t)$. The required voltages are obtained by multiplying by $R_L$. We are in steady state, so for each cycle, we start at a minimum value of current, $I_{min}$ and over $T$ seconds, the current rises to $I_{max}$, then over the next $T$ seconds, it falls to $I_{min}$. I'll call the time constant $\tau$ to simplify the equations. I'll deal with the second $T$ seconds first as it's easier. The current starts at $I_{max}$ and decays towards 0. After $T$ seconds, it will be $I_{min}$. (Ordinarily, I'd just write down the equation, but to make things clear, I'm choosing my time origin to be the point at which $v_S(t)$ switches from $V_0$ to 0. This stops time offsets cluttering up the equations as in the official answer.) This is a simple falling exponential: $I_{min} = I_{max}e^{-\frac{T}{\tau}}$ Now for the first $T$ seconds. It's a rising exponential, starting at $I_{min}$, rising towards a maximum current $I_0$ (which is given by $I_0=\frac{V_0}{R_w+R_L}$). Again, I'm choosing a convenient time origin (when $v_S(t)$ switches from 0 to $V_0$). So: $I_{max}=I_{min}+(I_0-I_{min})(1-e^{-\frac{T}{\tau}})$ (That might not be quite the way you remember it, but it's the way I do and it's the same once multiplied out.) Both these equations must be true in the steady state, so substituting for $I_{max}$ in the second equation, we get: $I_{min} = I_0\frac{1-e^{-\frac{T}{\tau}}}{e^{\frac{T}{\tau}}-e^{-\frac{T}{\tau}}}$ Now it's just a matter of substituting in the values for $I_0$ and $\tau$ and multiplying by $R_L$ to get $V_{min}$. Then use the first equation to get $I_{max}$ and hence $V_{max}$. SecondChildUserIdTAG: 141000 SecondChildUserNameTAG: OrinE SecondChildCreateTimeTAG: 2012-12-29T04:12:30Z IndexTAG: 1844 TitleTAG: certification when will we receive the certificate? UserIdTAG: 221736 UserNameTAG: samhita CreateTimeTAG: 2012-12-26T13:31:04Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Before New Year. FirstChildUserIdTAG: 183712 FirstChildUserNameTAG: colegiocientifico FirstChildCreateTimeTAG: 2012-12-26T13:35:02Z SecondChildTAG: Which new year?? SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-12-26T14:25:22Z SecondChildTAG: The Gregorian calendar. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-12-26T14:40:20Z SecondChildTAG: (Just the regular common calender, today is 2012-12-26) SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-12-26T14:42:37Z SecondChildTAG: Could also be Chinese new year ... SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-12-26T17:26:26Z SecondChildTAG: lol why? SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-12-26T17:40:46Z SecondChildTAG: Technically it will be before the Chinese New Year too I hope. lol SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-12-26T17:41:48Z SecondChildTAG: lol - if we run out of time, we will use the Chinese New Year as an excuse. No we meant Jan 1. SecondChildUserIdTAG: 96452 SecondChildUserNameTAG: Lyla SecondChildCreateTimeTAG: 2012-12-26T18:14:18Z SecondChildTAG: Or the next baktun, we've just started the new baktun ... SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-12-26T18:14:30Z SecondChildTAG: 2014 is a new year too :) SecondChildUserIdTAG: 454609 SecondChildUserNameTAG: Alwahsh SecondChildCreateTimeTAG: 2012-12-26T22:20:40Z SecondChildTAG: Exactly! SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-12-26T23:25:24Z SecondChildTAG: I'm not in a hurry. I remember from the previous course that there is a secret computer somewhere at mit, not connected to the internet, so no one can hack it, and that all the certificates had to be carried to that secret room, to save our results for eternity. All changes had to be made by hand. So in the far away future, when the creatures from the movie AI ever dig deep enough through the icy crust of earth, then my name will still be stored somewhere in the frozen memory depths of that computer. That knowledge gives me eternal peace! SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-12-26T23:41:20Z SecondChildTAG: those creatures are actually robots (evolved computers) :) SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-12-27T12:10:16Z SecondChildTAG: Salsero, you said no one could hack it; what about the insider enemies/Trojans? Just joking! Anyway, I feel it's time for a revision about how these characters look in the actual script: an "I" (capital i) vs a "l" (small L). Actually, I was first confused YOUR 'ai', Artificial Intelligence:) SecondChildUserIdTAG: 285616 SecondChildUserNameTAG: AmiyaX SecondChildCreateTimeTAG: 2012-12-27T16:10:52Z SecondChildTAG: One more thing, the 'follow' button on the top right hand corner shows (hover over text) follow, even after you've followed it, where actually, the hover/text should be 'unfollow'; in my opinion. SecondChildUserIdTAG: 285616 SecondChildUserNameTAG: AmiyaX SecondChildCreateTimeTAG: 2012-12-27T16:28:28Z SecondChildTAG: I can follow you. And about the capitals, that is a serious problem for me! I've a very bad eyesight and have often problems with distinguishing between l and I. If you edit this text in the textbox, then l=lowercase(L) and I=uppercase(i) appear different in the editbox, but appear as almost the same in the final text. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-12-27T17:10:43Z SecondChildTAG: I am sure they will take steps, salsero. SecondChildUserIdTAG: 285616 SecondChildUserNameTAG: AmiyaX SecondChildCreateTimeTAG: 2012-12-28T08:30:58Z FirstChildTAG: take a look here : https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50d343383c4ada2700000048 FirstChildUserIdTAG: 236136 FirstChildUserNameTAG: Hamid-ch FirstChildCreateTimeTAG: 2012-12-26T16:04:45Z IndexTAG: 1845 TitleTAG: soln for exam Hey does anybody know when will the answers of exam be put up !! Had done a mistake so curious to know what the correct answer could be :P UserIdTAG: 120850 UserNameTAG: Nitin1A CreateTimeTAG: 2012-12-26T13:02:29Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: What problem do you need solution to? FirstChildUserIdTAG: 252722 FirstChildUserNameTAG: vaboro FirstChildCreateTimeTAG: 2012-12-26T14:43:54Z SecondChildTAG: How to hack an ATM ... SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-12-26T18:15:34Z SecondChildTAG: Salsero is hilarious. + SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-12-26T20:18:46Z SecondChildTAG: :)) SecondChildUserIdTAG: 252722 SecondChildUserNameTAG: vaboro SecondChildCreateTimeTAG: 2012-12-27T10:18:53Z FirstChildTAG: Take it easy, relax. Enjoy The Christmas time! Staff is resting a bit :) FirstChildUserIdTAG: 325197 FirstChildUserNameTAG: Vitali_Jerin FirstChildCreateTimeTAG: 2012-12-26T13:07:55Z SecondChildTAG: Thanks for understanding. Solutions should be up. SecondChildUserIdTAG: 96452 SecondChildUserNameTAG: Lyla SecondChildCreateTimeTAG: 2012-12-26T18:15:33Z SecondChildTAG: :) ya sure.. SecondChildUserIdTAG: 120850 SecondChildUserNameTAG: Nitin1A SecondChildCreateTimeTAG: 2012-12-29T09:42:53Z FirstChildTAG: Should be up now. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-12-26T18:15:39Z SecondChildTAG: Q4. i wrote the exact answer for charghing capacitor fom 0 SecondChildUserIdTAG: 342221 SecondChildUserNameTAG: deepkar SecondChildCreateTimeTAG: 2012-12-27T03:01:21Z SecondChildTAG: :) :) SecondChildUserIdTAG: 120850 SecondChildUserNameTAG: Nitin1A SecondChildCreateTimeTAG: 2012-12-29T09:48:33Z IndexTAG: 1846 TitleTAG: Query about certificates I have completed 6.002 by 84%. But i haven't attempted practice session & video sequence... Will ,I get the certificate? UserIdTAG: 267220 UserNameTAG: ABHINAVSAXENA318619 CreateTimeTAG: 2012-12-26T03:51:02Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: Have a look at the Syllabus available in the Course Info page. ;) FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-12-26T04:10:11Z FirstChildTAG: yes ..there is nowhere mentioned that ,for getting certificates ,it is essential to watch the video lectures. FirstChildUserIdTAG: 267220 FirstChildUserNameTAG: ABHINAVSAXENA318619 FirstChildCreateTimeTAG: 2012-12-26T04:16:04Z SecondChildTAG: may be you seen this on some other site? SecondChildUserIdTAG: 623210 SecondChildUserNameTAG: preveen SecondChildCreateTimeTAG: 2012-12-26T04:50:52Z SecondChildTAG: Yes, on 6.00x you can earn points for the questions between the videos, but it's not required to answer them or to watch the videos. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-12-26T12:09:57Z FirstChildTAG: Yes you will. If your overal score is equal or more than 60% you will receive the Certificare . Exercises will not contribute to your grade.[read here][1] Interactive video sequences > Lecture style videos are presented in interactive video sequences (or > sequences for short), and are posted in the Courseware section of the > website. Each sequence includes a succession of short video clips and > online exercises, arranged in a logical progression. Two sequences > will be given each week; please take the time to watch each video and > each exercise in the sequence they are provided. Answer-check > mechanisms are provided in these exercises, but they will not > contribute towards your grade. [1]: https://www.edx.org/static/content-mit-6002x/handouts/syllabus.a477535058a1.pdf FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-26T12:03:49Z SecondChildTAG: hi myrimit. i believe you are the only reliable source of info here. How do i get my certificates? I tried going through the discussion, but couldnot find any? SecondChildUserIdTAG: 8897 SecondChildUserNameTAG: sam1202 SecondChildCreateTimeTAG: 2012-12-26T12:26:29Z SecondChildTAG: Hi sam1202, As far as I have read by a comment of a Staff, it will be available for downloading before New Year. Based on my experience, last Term happened the same, it took some days before we could see the button that allowed us to download our Certificate ;). I am sure that you will see it soon. Take care, Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-26T12:38:54Z SecondChildTAG: I guess nobody has received their certificates... They will be available for download before new year!! Happy new year!!! SecondChildUserIdTAG: 108929 SecondChildUserNameTAG: namit SecondChildCreateTimeTAG: 2012-12-26T12:39:15Z FirstChildTAG: "hi myrimit. i believe you are the only reliable source of info here. How do i get my certificates? I tried going through the discussion, but couldnot find any?" sam1202 You need to try harder, the certificate has been explained on an almost hourly basis. Please use the search function or look at a few of the threads to the left. Your question has been asked many, many times. It is not fair to call other members of the team unreliable, when in reality it is you who failed to make the effort. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-12-26T13:51:54Z SecondChildTAG: You are so right!! SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-12-26T14:28:08Z IndexTAG: 1847 TitleTAG: A fun problem with resistors Here is a fun problem I stumbled onto. Some of you have probably already seen this problem but I think a good portion of you haven't. What is the equivalent resistance of an infinite ladder of resistors connected as shown in the picture. ![infinite resistor ladder ][1] [1]: https://edxuploads.s3.amazonaws.com/13564701961343625.png Have fun! :) Edit: This problem was discussed in Week 1 tutorial. I wasn't aware of that before making this post. It's a fun problem nonetheless. UserIdTAG: 311022 UserNameTAG: GordanS CreateTimeTAG: 2012-12-25T21:18:31Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 6 FirstChildTAG: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_1/problem_1-2_part1/ FirstChildUserIdTAG: 623210 FirstChildUserNameTAG: preveen FirstChildCreateTimeTAG: 2012-12-26T17:23:44Z FirstChildTAG: Are you sure that all resistors are 1ohm? For example, with all resistors tied toghether (net B) 2Ohm and last resistor connected to this net 1Ohm, resistance of net AB will be 2Ohm. R-2R ladder is basic solution part for high speed AD converters, while it is VERY technological to make only two precision values instead alot ones.So this solution has some derivations too FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-12-25T21:42:02Z SecondChildTAG: Assume that all resistors are 1 Ohm, or R Ohms for that matter. SecondChildUserIdTAG: 311022 SecondChildUserNameTAG: GordanS SecondChildCreateTimeTAG: 2012-12-25T23:10:50Z FirstChildTAG: **GordanS** are you suggesting a **mathematical** solution, a **simulated** soln or **any of the 2 methods**. My guess is it **would approximate 1 ohm**, just guessing with the **increased parellism**. just approximating, unsure of the real soln. FirstChildUserIdTAG: 183507 FirstChildUserNameTAG: obiradaniel FirstChildCreateTimeTAG: 2012-12-25T21:48:01Z SecondChildTAG: I was aiming for a mathematical solution. It's not 1 ohm, in fact the solution is not even a rational number. :D SecondChildUserIdTAG: 311022 SecondChildUserNameTAG: GordanS SecondChildCreateTimeTAG: 2012-12-25T23:17:11Z FirstChildTAG: Who has passed the 6.002x knows that answer was given in Week 1 tutorials (video by Dr. Piotr Mitros) :) FirstChildUserIdTAG: 325197 FirstChildUserNameTAG: Vitali_Jerin FirstChildCreateTimeTAG: 2012-12-25T21:55:13Z SecondChildTAG: Haha, You are right. I didn't really watch tutorials so I wasn't aware that this problem was already discussed. A fun problem nonetheless. SecondChildUserIdTAG: 311022 SecondChildUserNameTAG: GordanS SecondChildCreateTimeTAG: 2012-12-25T23:08:07Z FirstChildTAG: (1+sqrt(5))/2 ohms The golden ratio !... FirstChildUserIdTAG: 373752 FirstChildUserNameTAG: Croak FirstChildCreateTimeTAG: 2012-12-26T01:29:03Z SecondChildTAG: right SecondChildUserIdTAG: 404188 SecondChildUserNameTAG: jishudasorissa SecondChildCreateTimeTAG: 2012-12-26T14:35:41Z FirstChildTAG: Since the ladder was solved, I suggest trying out an infinite grid instead: ![enter image description here][1] [1]: http://puu.sh/1FnhF Hint: The general problem is complicated, but for adjacent nodes, superposition and Theveninan simplifies it greatly. FirstChildUserIdTAG: 142857 FirstChildUserNameTAG: viking2 FirstChildCreateTimeTAG: 2012-12-26T04:07:22Z SecondChildTAG: You can complicate it more, by making it in 3D. Wonder what the solution will be in that case .. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-12-26T17:35:41Z SecondChildTAG: OK, so here is the solution. Think about it before you scroll down. . . . . . . . . . . . . . Assume the network is connected to ground at infinity. Now inject 1A into a node. Symmetry gives you 0.25 A in each adjacent resistor. Remove this source, and now, drain 1 A from the other node. Still 0.25 A in each adjacent resistor. Superposition of the two gives us 0.25+0.25=0.5 A in the connecting resistor, so the voltage across it must be 0.5V It also sums to 0 A at infinity, so this condition can be removed.(maybe add some rigor here) Thevenin's, with 0.5 V at 1 A gives us an equivalent resistance of 1/2 ohm. For the N-dimensional version, same argument with 2N edges at each node, so resistance should be 1/N ohms. SecondChildUserIdTAG: 142857 SecondChildUserNameTAG: viking2 SecondChildCreateTimeTAG: 2012-12-27T08:14:18Z IndexTAG: 1848 TitleTAG: 66 How I can turn the objects for 90 degrees UserIdTAG: 818353 UserNameTAG: strato CreateTimeTAG: 2012-12-25T17:16:17Z VoteTAG: 1 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 2 FirstChildTAG: if you have selected an object, just press 'r' and it will rotate FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-12-25T17:18:20Z SecondChildTAG: I have also noticed that sometimes you have to refresh the page before the object will rotate. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-12-25T19:36:34Z SecondChildTAG: I think You shuld use: "ctrl + r" SecondChildUserIdTAG: 1170067 SecondChildUserNameTAG: Zytkiewicz SecondChildCreateTimeTAG: 2013-02-09T10:09:14Z FirstChildTAG: Let's look at Myrimit's tutorial as well https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/MyrimitCircuitSimulatorTutorial/ https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/how-measure-current-sandbox/ FirstChildUserIdTAG: 325197 FirstChildUserNameTAG: Vitali_Jerin FirstChildCreateTimeTAG: 2012-12-26T10:30:22Z IndexTAG: 1849 TitleTAG: Honor of code:To Staff Sir, As we have completed our final exam when will we getting our Certificates for the course. And can we have more course related to electronics and communication engineering.? UserIdTAG: 182470 UserNameTAG: nitesh2703 CreateTimeTAG: 2012-12-25T15:29:23Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: The Certificate should be available before the New Year in the form of a PDF download, along with a verification link. If you log-out and visit the main edX page, there are some fairly new courses available, some starting within a month or so. Check back often, new and exciting things are being unveiled often. Have fun. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-12-25T15:48:51Z SecondChildTAG: Thanks Sir, But after checking the main page I found that most of the courses are either of Computer Science or general topics. But being a 3rd year student of Electronics branch I need some Electronics related courses that will help me in my area. The course that we completed just now was so helpful to me that I was the only one to get most of the answers in my class. Credit goes to MIT. Once again Thanks. SecondChildUserIdTAG: 182470 SecondChildUserNameTAG: nitesh2703 SecondChildCreateTimeTAG: 2012-12-27T04:57:43Z SecondChildTAG: try www.coursera.org for more course options. SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-12-27T12:15:33Z FirstChildTAG: for certificate look this post : https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50d343383c4ada2700000048 FirstChildUserIdTAG: 236136 FirstChildUserNameTAG: Hamid-ch FirstChildCreateTimeTAG: 2012-12-25T19:37:17Z IndexTAG: 1850 TitleTAG: proctored exam any further details of the proctored exam? UserIdTAG: 342221 UserNameTAG: deepkar CreateTimeTAG: 2012-12-25T11:32:51Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Not yet Sir. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-12-25T14:40:23Z SecondChildTAG: Will the proctored exam be held for the spring course? SecondChildUserIdTAG: 440331 SecondChildUserNameTAG: Shaun94 SecondChildCreateTimeTAG: 2012-12-30T12:50:36Z IndexTAG: 1851 TitleTAG: Answers for Final Exam TO STAFF Sir When will we get the answers for the final exam? UserIdTAG: 150120 UserNameTAG: krishna1993 CreateTimeTAG: 2012-12-25T04:30:41Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I wonder the same.... Thanks and Merry Christmas :)) Sandra FirstChildUserIdTAG: 342966 FirstChildUserNameTAG: SandraNavarro FirstChildCreateTimeTAG: 2012-12-25T09:44:11Z SecondChildTAG: Merry X'mas. I am sad that you dropped out of the course. Better luck next time and finish with a grand score! SecondChildUserIdTAG: 623210 SecondChildUserNameTAG: preveen SecondChildCreateTimeTAG: 2012-12-25T10:06:22Z SecondChildTAG: Thanks :)) SecondChildUserIdTAG: 342966 SecondChildUserNameTAG: SandraNavarro SecondChildCreateTimeTAG: 2012-12-25T18:17:14Z IndexTAG: 1852 TitleTAG: Thank You!! Greetings, i just want to say thank you to Prof. Agarwal and the wonderful people taking this fine course, especially to Myrimit and Skyhawk for helping me get a nice 99% :). I will totally recommend this course to my friends and classmates at Universidad Simón Bolívar in Venezuela, thank you and Merry Christmas to anyone who reads this ...(P.S.: i apologize for my funky english, not enough practice ha!). UserIdTAG: 293309 UserNameTAG: AlejoCCS CreateTimeTAG: 2012-12-25T02:24:24Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1853 TitleTAG: thanks thanks from algeria and thanks to sir agarwal and to the other staff and edx team UserIdTAG: 174229 UserNameTAG: fares27 CreateTimeTAG: 2012-12-24T23:56:40Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1854 TitleTAG: Any Dutch students here? So it's the end of this course, and I managed to get through it even with my very rusty math skills :) Two questions remain: am I the only Dutch student that took this course or are there more students from the Netherlands? Are there any numbers available of students per country and how many students started and how many actually finished it? UserIdTAG: 114881 UserNameTAG: xvink CreateTimeTAG: 2012-12-24T19:16:03Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1855 TitleTAG: Question in final：about the phase plot (Q5) ,urgent! After we get a transfer function H(w)=a+j*b ,what is the phase ? Is ''phase=arctan(b/a)'' right? Then,we plot the phase.What is the range of phase? I think as the period of function ''tan'' is pi(180º) ,so the range also should have a period of 180º.That is to say, range from 0º to -180º is the same as range from 180º to 0º. So,in Q5,U and Z , V and W are equivalent in my opinion. Is that right? Hope what I say is clear--My poor English:) Merry Christmas ! Best regards from Ericson. UserIdTAG: 361137 UserNameTAG: Ericson CreateTimeTAG: 2012-12-24T16:52:57Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: After we get a transfer function H(w)=a+j*b ,what is the phase ? Is ''phase=arctan(b/a)'' right? No, you have to add k*pi to it and test k for correctness, just as you have to test whether a root is positive or negative or true for both. Answer U and V tells me that there's no phase shift to a certain point, and Z and W tells me that there's a 180 degrees shift (or inverted polarity) to a certain point. The same tangent doesn't guarantee the same phase. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-12-24T17:14:55Z SecondChildTAG: Got it.Thank you vey much! I totally forgot to add k*pi and missed testing k. Best wishes to you! SecondChildUserIdTAG: 361137 SecondChildUserNameTAG: Ericson SecondChildCreateTimeTAG: 2012-12-24T17:30:00Z FirstChildTAG: - V: 0 degree shift for low frequencies and 180degree lead for high frequencies W: 180 degree lag for low frequencies and 0degree shift for high frequencies are they equivalent? FirstChildUserIdTAG: 623210 FirstChildUserNameTAG: preveen FirstChildCreateTimeTAG: 2012-12-24T17:29:06Z SecondChildTAG: No, see my answer above. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-12-24T17:35:08Z FirstChildTAG: The phase - frequency plot was poorly explained. It belongs to signals and systems domain, and 180 phase shifts were like a backstab to those who didn't had any background on bode diagrams. The poles and zeros method is very confusing sometimes, even to those who understand transfer functions. With every search you will find different explanations.I found this explanation to work every time : http://en.wikipedia.org/wiki/Bode_plot#Straight-line_phase_plot FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-12-25T12:12:54Z SecondChildTAG: yeah,I did some search about phase plot and become even more confused about this concept.Bode diagram is just one method,and ...so on. Anyway,thanks a lot! SecondChildUserIdTAG: 361137 SecondChildUserNameTAG: Ericson SecondChildCreateTimeTAG: 2012-12-26T05:29:59Z IndexTAG: 1856 TitleTAG: WHAT DOES YOUR PROGRESS LOOK LIKE? A very congratulations to all those who cleared the class!! So guys!!! what does your progress looks like?? Thanks edx and the whole workforce behind 6.002x!! Merry Christmas and a happy new year!! UserIdTAG: 368712 UserNameTAG: jmen CreateTimeTAG: 2012-12-24T16:21:32Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Looks like I progressed! FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-12-24T17:25:46Z FirstChildTAG: thanks to prof agarawal and to the team edx FirstChildUserIdTAG: 174229 FirstChildUserNameTAG: fares27 FirstChildCreateTimeTAG: 2012-12-24T18:32:54Z FirstChildTAG: My progress looks like an independent current source v-i characteristic :-) Great course! FirstChildUserIdTAG: 398594 FirstChildUserNameTAG: DaveyJC FirstChildCreateTimeTAG: 2012-12-24T23:00:08Z SecondChildTAG: Good one! SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-12-25T18:31:02Z IndexTAG: 1857 TitleTAG: i missed my finel.. i missed my final.... :::((( anyway this was a great experiance for us.. i would like to thank for all people who work for this.. happy xmas n happy new year UserIdTAG: 509517 UserNameTAG: randima CreateTimeTAG: 2012-12-24T15:53:02Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: sad to hear that $$Merry\ X'mas\ and\ Happy\ New\ Year!$$ FirstChildUserIdTAG: 623210 FirstChildUserNameTAG: preveen FirstChildCreateTimeTAG: 2012-12-24T16:13:43Z SecondChildTAG: T_T , I am so sorry for what happened to you randima.... Merry Christmas and Happy New Year. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-24T18:56:56Z FirstChildTAG: sorry dear n happy holiday FirstChildUserIdTAG: 285945 FirstChildUserNameTAG: lindalapiso FirstChildCreateTimeTAG: 2012-12-24T18:43:12Z IndexTAG: 1858 TitleTAG: Thanks Because of work I didn´t get to much time to participate on the discussion forums, but I really want to thank to Dr. Agarwal and the staff I really learned and re-learned a lot :) UserIdTAG: 199758 UserNameTAG: Varinia CreateTimeTAG: 2012-12-24T14:59:17Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1859 TitleTAG: Thank you very much! Thanks guys, especially Prof Agarwal for initiating edx and it's first course, Circuit and Electronics. It has been a truly wonderful learning experience. Finally, Merry Christmas to all of you! UserIdTAG: 143823 UserNameTAG: NathanNadeson CreateTimeTAG: 2012-12-24T09:17:45Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1860 TitleTAG: Thanks from Indian Student. Thank you for this awesome experience. I did my graduation recently in Electronics and this was the best course for revision. I really like the Lab section in this course and have really liked the way it was conducted. It was good and i got 100% in this course. A must take course to revise all your electronics concept. Just wanted to know when we will get the certificates and whether we will get a grade in certificate? UserIdTAG: 421726 UserNameTAG: Vikasb1701 CreateTimeTAG: 2012-12-24T09:13:20Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: congratulations and happy new year !!!!!!!! FirstChildUserIdTAG: 171199 FirstChildUserNameTAG: Dhananjay01 FirstChildCreateTimeTAG: 2013-01-01T09:32:40Z IndexTAG: 1861 TitleTAG: Thanks you very much from Thai student. Thank a lot for all 6.002x staff. UserIdTAG: 441436 UserNameTAG: Ooffy CreateTimeTAG: 2012-12-24T08:15:15Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1862 TitleTAG: Least i could do is to say thanks! Yes!! Thank you edx team for this wonderful course!! UserIdTAG: 423306 UserNameTAG: Puneeth_20 CreateTimeTAG: 2012-12-24T03:58:27Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1863 TitleTAG: 1st position of 6.002x course? who is overall topper of 6.002x course?? and do the topper get any prize?? UserIdTAG: 400909 UserNameTAG: WAHABAHMED CreateTimeTAG: 2012-12-23T12:20:33Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: i'm pretty sure there are more then a few people that got a perfect 100% the course wasn't that hard overall :) FirstChildUserIdTAG: 263693 FirstChildUserNameTAG: Coldberg FirstChildCreateTimeTAG: 2012-12-23T12:31:39Z SecondChildTAG: Not that hard? That depends on if/or how long ago one used the necessary math... %) Must say that I'm pleased with my final score and I surprised myself, but I'm also aware that I need A LOT more practice. SecondChildUserIdTAG: 114881 SecondChildUserNameTAG: xvink SecondChildCreateTimeTAG: 2012-12-23T13:16:25Z SecondChildTAG: I agree.Homeworks,labs and exams are not enough for us to understand the concepts of this course.I believe many ''aha'' moments will be found in textbook. SecondChildUserIdTAG: 361137 SecondChildUserNameTAG: Ericson SecondChildCreateTimeTAG: 2012-12-24T04:56:33Z FirstChildTAG: The exam is still ongoing so please refrain from posting results. FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-12-23T12:54:43Z FirstChildTAG: I did not get 100%, but I did get the best prize - a strong feeling of accomplishment! FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-12-23T14:56:50Z SecondChildTAG: yeah)) after solving another 2nd order differential equation i feel like being addicted to this things))) SecondChildUserIdTAG: 373498 SecondChildUserNameTAG: Cheblan SecondChildCreateTimeTAG: 2012-12-23T17:41:33Z SecondChildTAG: :-) SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-12-23T18:48:18Z FirstChildTAG: Materially talking, as far as I know, based on my experience in 6.002x Spring 2012, I have to tell you that there will not be a prize for the topper scores :p But not materially talking, the prize is the knowdlege and taking all this amazing experience forever XD. I guess this is the best award that one can receive. This is so amazing! My best wish to you WAHABAHMED, Myriam. P.D. But, if you want to win a prize - A Textbook signed personally by Prof. Agarwal, you can take a look at the Course Info and the Wiki, there is a Contest running for all of you made by Students to Students. I hope that you can participate. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-24T01:52:55Z SecondChildTAG: so true that we got knowledge and learned many things. if dead line of the contest is extended then it would be much better.. SecondChildUserIdTAG: 400909 SecondChildUserNameTAG: WAHABAHMED SecondChildCreateTimeTAG: 2012-12-24T07:28:11Z IndexTAG: 1864 TitleTAG: My dear classmates!! Im done with the course. :) I would like to know if my id along with all the videos will still be accessible next year, as i would love to revisit a few lectures for reference. UserIdTAG: 357292 UserNameTAG: RaghavAbboy CreateTimeTAG: 2012-12-23T10:41:42Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I just registered for this course next Spring, and both 6.002x courses show up on my dashboard, so it appears my ID is constant between them. Also, given that last Spring's materials are still available to those who registered, I have every reason to believe this semester's materials will remain available as well. FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-12-23T11:45:09Z IndexTAG: 1865 TitleTAG: Out of curiosity ! Hello, I was wondering what would be the number of students paasing out 6002X with a 100% score. Doing so requires great dedication on one's part. So just out of curiosity, anyone ? Stats from the course offered in Spring 2012 ? UserIdTAG: 724534 UserNameTAG: AnkitNegi CreateTimeTAG: 2012-12-23T10:04:44Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: I hope we will see **some** course statistics later. Im not sure that there will be number with 100% score, though. FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-12-23T10:11:52Z FirstChildTAG: Looks like last time (pilot course) there were around 340 students who aced the course. Am I right? May be somebody from previous course can answer. I'm guessing this time the number would be much higher than that. FirstChildUserIdTAG: 232157 FirstChildUserNameTAG: GayatriTR FirstChildCreateTimeTAG: 2012-12-24T08:42:28Z FirstChildTAG: I was one of them and I'm glad. Took me so much effort :D FirstChildUserIdTAG: 398594 FirstChildUserNameTAG: DaveyJC FirstChildCreateTimeTAG: 2013-01-22T09:59:07Z IndexTAG: 1866 TitleTAG: dependent sources How to create dependent sources in circuit sandbox..?? UserIdTAG: 101902 UserNameTAG: DHEERAJK_VITS CreateTimeTAG: 2012-12-23T05:52:21Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: See also [CircuitLab][1]. Though it does provide dependent sources, think very carefully about how you intend them to be controlled. An "independent" source with appropriate voltage or current values (even formulas!) may be what you really need. [1]: https://www.circuitlab.com FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-12-23T11:51:48Z FirstChildTAG: I suppose you could do that with a mosfet or an opamp. FirstChildUserIdTAG: 90796 FirstChildUserNameTAG: ChaunceyGardiner FirstChildCreateTimeTAG: 2012-12-23T06:39:40Z FirstChildTAG: It may be done easy. While we suppose that depended source is idealized, correct way is to use Op Amp. For the Voltage controlled dependent source you need to connect OA inputs to the specified nets. For the Current controlled dependent source you will need to add current sense resistor into specified net .You should understand that this resistor will cause some inaccuracy that may be compensated by relatively low value of the current sense resistor. Current sources may be build with an additional output stage.I hope it is clear. Good luck! FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-12-23T08:59:21Z IndexTAG: 1867 TitleTAG: grading?? my progress is 86% , so which grade will i get??? UserIdTAG: 404805 UserNameTAG: Aswinamar CreateTimeTAG: 2012-12-22T14:47:31Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: B FirstChildUserIdTAG: 173147 FirstChildUserNameTAG: cruiser_rahit FirstChildCreateTimeTAG: 2012-12-22T15:03:12Z SecondChildTAG: thanks a lot SecondChildUserIdTAG: 404805 SecondChildUserNameTAG: Aswinamar SecondChildCreateTimeTAG: 2012-12-22T15:14:03Z SecondChildTAG: 87% and u would have got A..I guess you missed it by a very small margin.. SecondChildUserIdTAG: 173147 SecondChildUserNameTAG: cruiser_rahit SecondChildCreateTimeTAG: 2012-12-22T16:58:50Z SecondChildTAG: I feel also sorry for your 1% difference, but it will be always tomorrow. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-12-22T19:28:51Z SecondChildTAG: b plus ...but sorry for 1 percent :) SecondChildUserIdTAG: 415376 SecondChildUserNameTAG: imab90 SecondChildCreateTimeTAG: 2012-12-23T16:53:14Z IndexTAG: 1868 TitleTAG: How many got a perfect score??? So how many got 100% score on final exam as well as overall??? UserIdTAG: 146770 UserNameTAG: solelonerajm CreateTimeTAG: 2012-12-21T22:25:07Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: 100% score on final exam --> contribut with 40% overall - **read section grade** : https://www.edx.org/static/content-mit-6002x/handouts/syllabus.a477535058a1.pdf FirstChildUserIdTAG: 236136 FirstChildUserNameTAG: Hamid-ch FirstChildCreateTimeTAG: 2012-12-22T00:47:26Z FirstChildTAG: 100 percent MAS ALLAH :) FirstChildUserIdTAG: 415376 FirstChildUserNameTAG: imab90 FirstChildCreateTimeTAG: 2012-12-22T13:41:45Z SecondChildTAG: Just 97% :-( Didn't get 2 question in final exam. SecondChildUserIdTAG: 345464 SecondChildUserNameTAG: Constantine_ru SecondChildCreateTimeTAG: 2012-12-22T14:08:36Z FirstChildTAG: Congrats perfect scoring folks but your work isn't done yet :) https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50d5c279ce1ccf2700000008 FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-12-22T14:25:44Z IndexTAG: 1869 TitleTAG: CERTIFICATE!! how would i get certificate ?? and is it possible that edx sends a hard copy of certificate at my home ?? UserIdTAG: 400909 UserNameTAG: WAHABAHMED CreateTimeTAG: 2012-12-21T19:30:35Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Have a look at this thread here: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50d343383c4ada2700000048 I myself, am not aware of the option for a hard copy certificate. (Aside from having your own printed.) FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-12-21T19:34:44Z SecondChildTAG: thanks for sharing thread. it would be greatif edx/MITx sends certificate's hard copy at our address SecondChildUserIdTAG: 400909 SecondChildUserNameTAG: WAHABAHMED SecondChildCreateTimeTAG: 2012-12-23T12:16:31Z FirstChildTAG: thanks for sharing thread. it would be greatif edx/MITx sends certificate's hard copy at our address FirstChildUserIdTAG: 400909 FirstChildUserNameTAG: WAHABAHMED FirstChildCreateTimeTAG: 2012-12-23T12:21:45Z IndexTAG: 1870 TitleTAG: FINAL: Extra attempts and time for questions 4 and 6 Hello, the original announcement thread about extra attempts is closed for posting but I still wanted to ask for some clarification. I did not start my final exam yet, I will have to do it sometime during weekend due to the family, job and all other obligations. I do not understand if questions Q4 and Q6 were either rephrased for a better understanding or not. And if I will not start exam in the next 24 hours, does it mean I will not get any extra attempts on those questions? Thank you Serge UserIdTAG: 16265 UserNameTAG: serge_korolev CreateTimeTAG: 2012-12-21T16:40:56Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I started the exam after the announcement, it appears that I have 6 submissions for questions 4 and 6, although I cannot confirm the time extension. I think it's safe to assume that you will get the time extension too. Good luck. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-12-21T16:56:48Z FirstChildTAG: Hi to everybody. Please, reload your final exam page and read VERY carefully Q4 and Q6 before [giving] your final answer. I finished the exam yesterday and got just [score removed] due to Q4 & Q6. I learned about the extra attempts when I came back home from work. I kept giving wrong answers for Q4 till I ran out of chances. And then I found out that I hadn't reloaded the page. I did, and just by reading carefully again, I found the way to solve Q4. So, please, read very carefully again and be sure you are reading what you have to. Good luck, Gonzalo Alcalá de Henares, Spain *Gonzalo, I removed the specific score you listed as receiving for the final. Posting exam scores is not allowed while the exam is still ongoing. Jersey Mark (Community TA).* FirstChildUserIdTAG: 260716 FirstChildUserNameTAG: gftp FirstChildCreateTimeTAG: 2012-12-21T19:11:52Z SecondChildTAG: Please do not comment on specifics of the Final Exam. SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-12-21T22:12:26Z IndexTAG: 1871 TitleTAG: FINALS yepee!..I made it, though not 100% but I got a passing grade..thanks MIT, this course helps me in my worst case analysis of certain circuits... UserIdTAG: 237941 UserNameTAG: per2x CreateTimeTAG: 2012-12-21T07:35:59Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1872 TitleTAG: Signal and system I knew a little about signal and system.. But what can it help in dealing with circuits?..What are its applications,or recommend a book? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-12-21T06:05:55Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: http://6003z.amolbhave.in/ FirstChildUserIdTAG: 136489 FirstChildUserNameTAG: ShahabSafa FirstChildCreateTimeTAG: 2012-12-21T12:27:10Z FirstChildTAG: Signals and systems(S&S) is the basis for Digital Signal Processing (DSP) It helps in the design of filters, reduction of noise & effective use of BandWidth(BW), etc. Refer OPPENHEIM textbook for S&S and DSP FirstChildUserIdTAG: 277808 FirstChildUserNameTAG: Hemanthmps FirstChildCreateTimeTAG: 2012-12-21T11:36:48Z SecondChildTAG: Prof. Oppenheim's book is one of the best I've read. I would suggest watching his lectures as well as you study it (link in my post below). It turned out very useful in my case because my intuition was pretty bad so self studying the book became a little hard. The lectures filled in for my lack of enough experience in math. SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-12-22T07:56:08Z FirstChildTAG: Signals and Systems by Oppenheim, Willsky and Nawab as Hemanthmps suggested. Additionally, there are lectures by Prof. Oppenheim which are quite good. You can check them out here: http://ocw.mit.edu/resources/res-6-007-signals-and-systems-spring-2011/ Also to answer your question, circuits are used to process signals. This changes their time domain and frequency domain characteristics (think about what a low pass filter does). So understanding how Signals and their properties along with system properties is is important. You'll also get a "tool chest" like we did in 6.002x which will make it easier to understand system properties and also design them. FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-12-21T15:54:12Z IndexTAG: 1873 TitleTAG: Amazing analogy Thanks for the string analogy, it was awesome. UserIdTAG: 378522 UserNameTAG: Alejo_Velasquez CreateTimeTAG: 2012-12-21T03:10:49Z VoteTAG: 1 CoursewareTAG: Week 14 / S28V5_What_s_Going_on_with_the_Double_Take_ CommentableIdTAG: 6002x_S28V5_What_s_Going_on_with_the_Double_Take_ NumberOfReplyTAG: 0 IndexTAG: 1874 TitleTAG: Fall 2008, Q5a The answer of question 5a of the fall 2008 exam puzzles me. Can someone please explain to me how one goes from: $\displaystyle\frac{V}{R+jwL}cos(wt) = Acos(wt)$ to $\displaystyle\frac{V}{\sqrt{R^2+(wL)^2}} = A $ I get that you want to loose the imaginary part, but as the denominator has been multiplied by $R-jwL$ I would expect the nominator to be $V*(R-jwL)$ and not just $V$ UserIdTAG: 114881 UserNameTAG: xvink CreateTimeTAG: 2012-12-20T20:33:06Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi xvink ! A is the modulus of that complex number.Multiply the denominator with it's complex conjugate and calculate it's modulus. In general you take the real part from the euler form V*e^(j*phi), where V is the modulus and phi is the phase. FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-12-20T22:54:15Z SecondChildTAG: Thanks! I had a hunch it would be something like that, but it's nice to have someone confirm it. I'm still struggling with the complex numbers / Euler equations, so I hope for the best for the final exam this weekend... SecondChildUserIdTAG: 114881 SecondChildUserNameTAG: xvink SecondChildCreateTimeTAG: 2012-12-21T08:36:14Z IndexTAG: 1875 TitleTAG: Anybody has Lecture Summaries? A month ago some very kind person posted here links for the lecture summaries. I've found the second part and link for it ( lectures 13-26) http://199.91.152.39/911pwsm9w0tg/bl53h4ndg3g3zbv/Lecture+Summaries+%2813-26small%29.updated.pdf Anybody has the link for the first part of this usefull summaries? UserIdTAG: 345464 UserNameTAG: Constantine_ru CreateTimeTAG: 2012-12-20T18:18:33Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Please see the [Course Wiki][1] under "Spring 2012 Student Contributions". [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/ FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-12-21T07:02:33Z FirstChildTAG: http://www.onrepare.com/midterm_summaries_small.pdf FirstChildUserIdTAG: 443358 FirstChildUserNameTAG: torkelh FirstChildCreateTimeTAG: 2012-12-21T08:45:22Z IndexTAG: 1876 TitleTAG: Capacitor and inductor small-signal model? What should be the small-signal model for capacitors and inductors and why? My doubts have arised due to a problem in the final review pdf (fall 2008 Q4). Thank you! UserIdTAG: 398594 UserNameTAG: DaveyJC CreateTimeTAG: 2012-12-20T13:55:35Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1877 TitleTAG: What will be the name signed on my certificate? I've asked to change my name..however it haven't changed yet.. I'm wondering what it will be...Can I change it? UserIdTAG: 136959 UserNameTAG: christerpher CreateTimeTAG: 2012-12-20T07:57:09Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: https://www.edx.org/dashboard FirstChildUserIdTAG: 623210 FirstChildUserNameTAG: preveen FirstChildCreateTimeTAG: 2012-12-20T10:10:52Z FirstChildTAG: i finished the cs188.1x ,the name in the certificate is the same as the name you subscribed with. - a lot of student have this problem , they wana to change the name in the certificate with their real name ,but......stil problem FirstChildUserIdTAG: 236136 FirstChildUserNameTAG: Hamid-ch FirstChildCreateTimeTAG: 2012-12-20T15:29:16Z SecondChildTAG: This is hilarious ! SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-12-20T17:14:25Z SecondChildTAG: the EDX will not ask you for your real full name, they put the name found in data base (the one you subscribed with), try to change it know and hope they will take your new name SecondChildUserIdTAG: 236136 SecondChildUserNameTAG: Hamid-ch SecondChildCreateTimeTAG: 2012-12-20T20:38:27Z FirstChildTAG: You have subscribed with a full name and a user name.Of course, they will put the full name. How would be to give a certificate to WixiePixie1234 ? :) You can always change your full name pressing the button from the up right corner that has your user name and the page with your courses will appear (if not already knew that). From the main page in the up left corner is n option to change your full name to the real one if you wish so. FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-12-20T22:37:59Z IndexTAG: 1878 TitleTAG: how to get power across R1 please help me out with it UserIdTAG: 513142 UserNameTAG: mahanteshvshetty CreateTimeTAG: 2012-12-18T05:23:19Z VoteTAG: 1 CoursewareTAG: Week 1 / KVL/KCL example CommentableIdTAG: 6002x_S2E3_Using_KVL_KCL_and_VI_constraints NumberOfReplyTAG: 1 FirstChildTAG: 1) Take node in the upper part and sum either current leaving or entering the node. Current entering the node give: (v-v2)/R1+I+(0-v2)/R2 = 0 which is a single equation in v2, rest are all known: (2-v2)/4+3+(0-v2)/5=0 => v2=70/9 or 7.777(7). Once you know v2 you can compute all rest. 2) Superposition: a)leave V source in and make I source null, then compute partial response; b)leave I source in and make V source null, then compute partial response; add a+b variables for total response. To null sources of current make it open and null source of voltage make it short. Because zero current passing = open while zero voltage difference = short. FirstChildUserIdTAG: 331483 FirstChildUserNameTAG: Doru FirstChildCreateTimeTAG: 2012-12-18T06:08:37Z IndexTAG: 1879 TitleTAG: Perpetually moving free electricity source in an engine I invented Through the syphening process only pure water will perpetually pull itself around in a circle of pvc piping. And only at the bottom of this circle of water and pvc pipes can you get the fan of an automobile alternator to turn as much as you need to generate this free electricity. Now you can use this electricity from these engines in vehicles, homes, or even businesses. I am trying to reach out to people to find the right crew to provide our planet with this luxury. So if anyone knows someone who fits that then contact me through responding to this message. Thank you. Yours truly, Mike Jones P.S. There is a way to strenghthen battery acids and battery cores themselves to hold more electricity at a reasonable strength around even babies. UserIdTAG: 886768 UserNameTAG: MIKJONZ CreateTimeTAG: 2012-12-18T00:38:00Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: This post clearly is worth of a few deductions: 1) First deduction is that world economy must be really bad if the scam industry is going so desperate. 2) Second deduction is that you can strengthen battery acids and battery cores themselves to hold more electricity at a reasonable strength around even babies, but clearly, you can NOT do it with an odd number of babies around. The number of babies necessarily must be even, for this delicate process to take place. Obviously. :) 3) The only way to get free electricity its the classic way of tapping into your neighbor power lines. FirstChildUserIdTAG: 331483 FirstChildUserNameTAG: Doru FirstChildCreateTimeTAG: 2012-12-18T04:35:32Z SecondChildTAG: Is there an MITx course on Thermodynamics we could point him to? ;-) SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-12-18T05:11:02Z SecondChildTAG: Anything related with free energy is a scam to some degree or another. Since it cant be anything else. He is gonna ask for some cash for postal costs for a dvd or whatever etc. Like that black-dyed money which come off with his own substance etc. But i guess free-energy stuff involves smaller sums of money and not so much in the spotlight of world greatest scams. But the thing really worth noticing about this is a scammer posting about free electricity here .... is either clueless or really really desperate. Cant be so clueless and do it for a living, so it has to be desperate. Now you can deduct how bad world economy must be going. :) SecondChildUserIdTAG: 331483 SecondChildUserNameTAG: Doru SecondChildCreateTimeTAG: 2012-12-18T05:54:04Z SecondChildTAG: I am just trying to take some of the burden off people with money, help mankind survive, and provide Africa and other poor countries with a livable environment. Please keep rude comments to yourself. And to prove I am a worthy person; this idea solved the perpetual motion question which is worthy of a Nobel Peace Prize. SecondChildUserIdTAG: 886768 SecondChildUserNameTAG: MIKJONZ SecondChildCreateTimeTAG: 2012-12-18T13:52:24Z SecondChildTAG: **for thermodynamics :** http://ocw.mit.edu/courses/chemistry/5-60-thermodynamics-kinetics-spring-2008/video-lectures/ SecondChildUserIdTAG: 236136 SecondChildUserNameTAG: Hamid-ch SecondChildCreateTimeTAG: 2012-12-18T20:03:45Z SecondChildTAG: I guess MIKJONZ thinks money is a burden, so taking some money from people would take some of the burden off. SecondChildUserIdTAG: 270011 SecondChildUserNameTAG: SnowmanZA SecondChildCreateTimeTAG: 2012-12-21T20:52:35Z IndexTAG: 1880 TitleTAG: final exam doubt I want to know how works exactly the time for the exam. If I've understood well, means that you have 24 hours to finish the exam and the time would start when you open the exam? is that wright? UserIdTAG: 433574 UserNameTAG: endika86 CreateTimeTAG: 2012-12-17T17:05:11Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Yes. FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-12-17T17:41:07Z FirstChildTAG: Right, with a small correction: if you start it very late, like on Sunday you'll have less than 24 hours. FirstChildUserIdTAG: 189680 FirstChildUserNameTAG: vnd FirstChildCreateTimeTAG: 2012-12-17T18:11:57Z FirstChildTAG: Yes you are right. t will go with the same conditions of time and check the button, as it was with the midterm exam, I understand. FirstChildUserIdTAG: 96873 FirstChildUserNameTAG: Santyaga FirstChildCreateTimeTAG: 2012-12-17T18:22:55Z IndexTAG: 1881 TitleTAG: From electronics to embedded programming What are good follow-up courses/books/resources for embedded/system programmer? E.g. someone who probably won't design hardware but rather needs a fairly good understanding of digital circuits (CPUs, buses, etc) mostly from interfacing point of view. UserIdTAG: 189680 UserNameTAG: vnd CreateTimeTAG: 2012-12-17T06:52:36Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: There is [Embedded][1] and its related magazines... [1]: http://www.embedded.com/ FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-12-17T07:47:36Z IndexTAG: 1882 TitleTAG: DIFF BETWEEN JFET AND MOSFET WHY DO WE INSIST ON STUDYING MOSFET INSTEAD OF JFET IN 6.002X IS MOSFET MORE ADVANTAGEOUS THAN JFET ? IF SO WHAT ARE THE ADVANTAGES OF USING MOSFET OVER JFET? UserIdTAG: 149154 UserNameTAG: santhosh1993 CreateTimeTAG: 2012-12-16T15:19:00Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: You have to start somewhere... FirstChildUserIdTAG: 9673 FirstChildUserNameTAG: xp42 FirstChildCreateTimeTAG: 2012-12-16T19:29:40Z FirstChildTAG: The family FET transistors have in common the isolation between the control terminal (gate) and the channel through which flows the current to be controlled. It is very important that control signal and high current power do not mix. Hence the isolation condition between the gate and the channel. In MOSFET this isolation is more efficient, and allows a much faster response rasoável. FirstChildUserIdTAG: 361823 FirstChildUserNameTAG: EliasOak FirstChildCreateTimeTAG: 2012-12-16T21:49:49Z FirstChildTAG: To expand on what EliasOak said, it's the "Metal Oxide Semiconductor"(MOS) in the MOSFET that allows the increased speed of a MOSFET vs a traditional FET. Although in reality most MOSFETS these days use Polysilicon for the gate material instead of Aluminum oxide. (The oxide is actually various dielectric materials) The MOSFET is much more relevant in both digital and analog circuits, then the BJT in 2012. Strategically, it's best to concentrate on the IGFET "Insulated-gate field effect transistor", otherwise known as the MOSFET. When we studied the Vacuum Tube earlier on, we were pseudo-studying the JFET as well. The Pentode and JFET are similar in operation. You can now expand your knowledge if you wish to include Legacy devices like BJT and Vacuum Tubes (Valves) etc. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-12-17T16:21:26Z FirstChildTAG: about [how transistors work][1] and about [MOSFET][2] [1]: http://www.youtube.com/watch?v=ZaBLiciesOU [2]: http://www.youtube.com/watch?NR=1&v=QO5FgM7MLGg&feature=endscreen FirstChildUserIdTAG: 96873 FirstChildUserNameTAG: Santyaga FirstChildCreateTimeTAG: 2012-12-17T12:25:22Z FirstChildTAG: How about IGBTs? Now in SMPS's up to ~500 khz or so, provide better switching performance than the mosfets. Or what about SCR, triacs ... etc. I think mosfets make sense as long the course is meant as to be part of the other courses you have at MIT. But if you consider it as a stand alone like most of the online ppls might take it, then the perspective needs some level of completeness. Not talking only about triacs, scr, igbs etc, but also alot more of other practical things. One can feel that course is old and everybody just added to it instead of rationalizing. Second order circuits stretched on too many chapters. Other stuff like reactive power left hanging in mid air for the followups which we cant take. Control theory also just touched, for one i would of liked to cut off or compress the useless derivative part on 2nd order and put more control theory instead, like nyquist, poles and zeroes, stability region etc. FirstChildUserIdTAG: 331483 FirstChildUserNameTAG: Doru FirstChildCreateTimeTAG: 2012-12-18T08:28:03Z IndexTAG: 1883 TitleTAG: it is explained toooooo well... Prof Agarwal is doing a great job: he is explaining all the basics of electronics and circuits very systematically, completely, in all the details and repeating often. But, I think, **it is completely too much and repeating too often**, not always but sometimes (@ opamp positive feedback, for example). I recognized that I'm not studying - I'm only listening. Ok, I know of the subject matter, but there is no doubt at all, I don't have any question, it's often lengthy ..... I don't know about the others, but I see that there is no discussion accompanying the lessons. Is this a good signal? (Please, do not get me wrong, I like the course...) UserIdTAG: 388524 UserNameTAG: dasbinich CreateTimeTAG: 2012-12-16T02:12:34Z VoteTAG: 1 CoursewareTAG: Week 13 / S25V12_Hysteresis_with_positive_feedback CommentableIdTAG: 6002x_S25V12_Hysteresis_with_positive_feedback NumberOfReplyTAG: 4 FirstChildTAG: I tend to agree. I would have delighted in this level of repetition a quarter century ago, when I felt the need to write down every word the professor said... Now, not so much. ;-) FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-12-16T04:32:14Z FirstChildTAG: The course is not suited to any one small group of students. Different people experience the lectures differently. If you find it too easy, you could just speed up the video. If you're already at the maximum speed, work out or read a book while you watch the lectures. Or realize that you have a luxury problem - anyone who can't fully grasp what is going on has a much bigger problem than you do. I found the number of repetitions just right for my level of understanding. Given the fact that a whopping 7,157 people got a passing grade the last time, they must have done *something* right. FirstChildUserIdTAG: 90796 FirstChildUserNameTAG: ChaunceyGardiner FirstChildCreateTimeTAG: 2012-12-16T06:38:26Z FirstChildTAG: AA understands the importance of getting the basics right. I'm sure the ideal situation is for every student to arrive at the same level of understanding of the materials that you have. AA is doing a great job. Repetition is repetitive only after you get it. Until then, it is invaluable. FirstChildUserIdTAG: 9673 FirstChildUserNameTAG: xp42 FirstChildCreateTimeTAG: 2012-12-16T19:34:25Z FirstChildTAG: I was new to this course (just finished final exam) so hopefully this can add something from a newbie's perspective. I noticed the repetitions, but it helped, on many levels, more often than bothered me. For one, it is an encouraging sign to follow an expert as he thinks along and struggles with issues that have often hurt my brain. Just like all the abstractions and Aha's, Prof AA makes the subject not only easy enough to grasp, but also fun, although i have no illusion at all about the field's (EE) complexity. Contrasting this with the sometimes terse home works explanations, according to some students, or the unexplained answers of the exercises, i think the lectures are a blessing, even if there are repetitions. BTW the seemingly absent comments after lecture sequences may not be a sign of lack of interest, because questions and comments are posted here on the discussion forum instead. Perhaps students think it's more appropriate, or more helpful, to help them posted here. Issues with the learning experience in a classroom environment are different altogether, and i won't get into that. In this one, as Chauncey mentioned, when i didn't feel the need to gear down to try to understand, i could just happily skip ahead, knowing, heavens be blessed, the lectures are still there for me to get back to if needed. FirstChildUserIdTAG: 184153 FirstChildUserNameTAG: tthngn FirstChildCreateTimeTAG: 2012-12-22T15:58:58Z IndexTAG: 1884 TitleTAG: refreshing im memory H8 Please help me understand why in expanation of problem 1 in homework 8 was told that charge wich will be transferred immediatly from the current impulse source to the capacitor mathematically, Q=∫(Iin)dt=1/R2? And will it be always right for similar surcuits with RC elements? UserIdTAG: 96873 UserNameTAG: Santyaga CreateTimeTAG: 2012-12-15T15:22:16Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1885 TitleTAG: Design of a Low pass Filter I have to design a low pass filter with a cutoff frequency of 10 MHz. Which circuit should i implement for a better output? What are the major benefits i"ll get if i use a pi section ? Thanks. UserIdTAG: 113365 UserNameTAG: satyabrata CreateTimeTAG: 2012-12-14T12:43:34Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Low pass "PI filters" are generally better at transmitting low frequencies and rejecting high frequencies. Conversely, a high pass "T filter" is generally better at transmitting high frequencies and rejecting low frequencies. Have fun. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-12-14T14:27:40Z FirstChildTAG: You should use more criteria for considerations , for example: what is a signal source and what is a signal receiver, and etc FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-12-14T15:28:16Z SecondChildTAG: Signal source is a low noise amplifier and output of LPF will go to ADC input. SecondChildUserIdTAG: 113365 SecondChildUserNameTAG: satyabrata SecondChildCreateTimeTAG: 2012-12-14T16:59:19Z SecondChildTAG: I suspected that it isnt radiotransmitter :) May you rebuild low noise amplifier? Do you know slope for this filter?I think you do need only Batterwort type, 2nd or 3rd order..I dont know what is real ADC resolution you do need, but may be good solution will be to use an additional OA. Try to check App Notes from AD,Linear and etc.. SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-14T20:00:04Z IndexTAG: 1886 TitleTAG: About Certificate I missed Homework 3, 8,10 and 11. I also missed Lab 1, 3, 10 and 11. My total percentage is 41%(including Homework, Lab, and Midterm Exam). Is it possible for me to get grade "B" in the certificate? Sombat UserIdTAG: 364129 UserNameTAG: Sombat CreateTimeTAG: 2012-12-14T09:49:55Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: You need $~\gt \frac{70-41}{40} = \frac{29}{40} = 72.5 \%$ in the final to get a B. FirstChildUserIdTAG: 9673 FirstChildUserNameTAG: xp42 FirstChildCreateTimeTAG: 2012-12-14T09:55:42Z SecondChildTAG: no you get b hhhhh SecondChildUserIdTAG: 880511 SecondChildUserNameTAG: THETRUE SecondChildCreateTimeTAG: 2012-12-14T20:32:53Z FirstChildTAG: Thaks a lot FirstChildUserIdTAG: 364129 FirstChildUserNameTAG: Sombat FirstChildCreateTimeTAG: 2012-12-15T04:06:32Z IndexTAG: 1887 TitleTAG: Digital to Analog Convertor in S24V12 Hello Friends have any body tried Digital to Analog Convertor by using circuit of problem 15.6 "page number 884 figure 15.74", if yes please share it with me because I could not understand it . UserIdTAG: 40209 UserNameTAG: sadaf78 CreateTimeTAG: 2012-12-12T08:46:38Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: I don't know which part you mean? Part a seems easy. Part b ,the two voltahe sources might reach 1 volt,so the output voltage might be 8 volts,which is within the first constraint.Now we need to make Ra and Rb small enough to make sure its current smaller than +/- 1 mA,so try to make Ra and Rb bigger.. Part c,just reverse v1 and v2.. FirstChildUserIdTAG: 136959 FirstChildUserNameTAG: christerpher FirstChildCreateTimeTAG: 2012-12-12T10:16:55Z SecondChildTAG: Im agree , there is wrong link to the task. By the meantime, I tried to solve problem 15.6.. Hmm I cant get $v_{out}$ similar to the solutions book :) SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-12T10:40:16Z FirstChildTAG: Have you watched last lecture of last sequence of week 12,in that lecture Dr.Anant Agarwal asked us to make Digital to analog convertor and gave hints by referring problem 15.6 circuit. plz tell me have anybody tried it? FirstChildUserIdTAG: 40209 FirstChildUserNameTAG: sadaf78 FirstChildCreateTimeTAG: 2012-12-12T11:32:09Z SecondChildTAG: It is simple, [link][1] [1]: http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/dac.html SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-12T12:20:21Z SecondChildTAG: Thank you Sergtronix SecondChildUserIdTAG: 40209 SecondChildUserNameTAG: sadaf78 SecondChildCreateTimeTAG: 2012-12-12T12:58:35Z FirstChildTAG: If you add another input v3, and weight the sum to be vO = 1*v1 + 2*v2 + 4*v3 you get a 3 bit AD-converter. All inputs high at 1V equals 7V out if the gain is set by Ra Rb to be 7. You can chose any voltage for input and output, so 5V at the inputs and 0--5 V out for 0--7 is also easy to get. FirstChildUserIdTAG: 151472 FirstChildUserNameTAG: Stensmed FirstChildCreateTimeTAG: 2012-12-12T23:01:14Z IndexTAG: 1888 TitleTAG: Final Exam (Weeks 1-12) I see that many posts in the discussion forum say that the syllabus for Final Exam is Weeks 1-12. Where is it mentioned officially? For Mid Term, the syllabus was mentioned in the "course info" page. I don't see a similar thing in Final Exam notification in the course info page. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-12-11T06:50:26Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: [https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50bf8166c67c571f00000031][1] By the way, posting as anonymous is NOT that nice. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50bf8166c67c571f00000031 FirstChildUserIdTAG: 376877 FirstChildUserNameTAG: AndBre FirstChildCreateTimeTAG: 2012-12-11T07:42:14Z SecondChildTAG: On the link above we have a post from staff RohanNagarkar saying that it is from 1 to 12 SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-12-11T13:56:49Z SecondChildTAG: But it was 1-13 last time. I wonder if they really did change it. SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-12-11T20:00:19Z SecondChildTAG: As you can see if you follow the link AndBre posted above, they did change it. It's a staff answer. SecondChildUserIdTAG: 90796 SecondChildUserNameTAG: ChaunceyGardiner SecondChildCreateTimeTAG: 2012-12-12T03:17:33Z FirstChildTAG: The way I look at it is: Suppose something covered in a lecture in Week 14 is what finally gets a bit of knowledge covered in Week 10 to "click". I'll be much better off having studied Week 14 than not, and there are no tutorials, exercises, homeworks or labs... There are tutorials and exercises for Week 13, but they seem pretty easy. So I am going all the way to the end. FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-12-11T20:07:59Z FirstChildTAG: Last time the course was run the final covered weeks 1-13. ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13552098241343672.png FirstChildUserIdTAG: 9673 FirstChildUserNameTAG: xp42 FirstChildCreateTimeTAG: 2012-12-11T07:11:00Z SecondChildTAG: They changed it. SecondChildUserIdTAG: 90796 SecondChildUserNameTAG: ChaunceyGardiner SecondChildCreateTimeTAG: 2012-12-12T03:19:12Z SecondChildTAG: They said the midterm was only up to week X last time but it really helped to do week x+1 first. Just saying... SecondChildUserIdTAG: 9673 SecondChildUserNameTAG: xp42 SecondChildCreateTimeTAG: 2012-12-12T23:51:41Z SecondChildTAG: Totally agree with [xp42][1]. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/users/9673 SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-12-13T05:31:50Z SecondChildTAG: I was pointing out that they changed it, as a response to the statement that *"Last time the course was run the final covered weeks 1-13."* Besides, they also said that the final covered weeks 1-13 and studying week 14 wouldn't have helped you one bit. SecondChildUserIdTAG: 90796 SecondChildUserNameTAG: ChaunceyGardiner SecondChildCreateTimeTAG: 2012-12-15T02:44:49Z IndexTAG: 1889 TitleTAG: h12p2 c I derivative vout with respect to vin and dvout = 0.2 d vin = 10 (20 - 10),but why I got Ro =99,was it rigt? Thanks UserIdTAG: 136959 UserNameTAG: christerpher CreateTimeTAG: 2012-12-11T02:49:40Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: There is no single 'correct' answer for $R_0$, but anything $ \gt 94 \Omega$ will work. If $R_0 = 94 \Omega$ then $V_{out}$ will range over $V_{out} = 5V \pm 0.1V$. If you choose $R_0$ greater than this, then the variation in $V_{out}$ will be smaller. For example, if $R_0 = 994 \Omega$, then $V_{out} = 5V \pm 0.01V$. Of course, for any chosen $R_0$ you will still have to find the $R_1$ and $R_2$ values to use. They also can be any valued, but the ratio of their values will be fixed for a given $R_0$ choice. FirstChildUserIdTAG: 9673 FirstChildUserNameTAG: xp42 FirstChildCreateTimeTAG: 2012-12-11T05:24:52Z SecondChildTAG: Thanks a lot SecondChildUserIdTAG: 136959 SecondChildUserNameTAG: christerpher SecondChildCreateTimeTAG: 2012-12-12T03:59:29Z SecondChildTAG: but why the ratio of their values will be fixed for a given R0 choice? Is not it right that R1/R2 = 1,then vout will be 5V? SecondChildUserIdTAG: 136959 SecondChildUserNameTAG: christerpher SecondChildCreateTimeTAG: 2012-12-12T04:03:16Z SecondChildTAG: The ratio is fixed because: R1 and R2 form a voltage divider, but while we don't have enough conditions to force unique values upon R1 and R2, we do know what the voltage out of the voltage divider must be; so lets suppose if R1 being twice the value of R2 say is a valid solution, then there are many, many (infinite) combinations of R1 and R2 values that will work to form the needed voltage divider. SecondChildUserIdTAG: 9673 SecondChildUserNameTAG: xp42 SecondChildCreateTimeTAG: 2012-12-14T04:55:34Z SecondChildTAG: have a look here also https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50c59f80a16468270000000c SecondChildUserIdTAG: 9673 SecondChildUserNameTAG: xp42 SecondChildCreateTimeTAG: 2012-12-14T05:08:20Z FirstChildTAG: The efficiency will depend upon the actual values chosen and used for R1 and R2. If they are large resistors (many $M \Omega$) there will be a small current flowing through them (especially compared to the small size of our load), and the power losses in the R1 and R2 resistors will be small (implies small losses in those resistors). If R1 and R2 are small values (say 10's of $\Omega$'s, then the current through them will be much higher, and the power losses in R1 and R2 will higher too. So, small R1 and R2 gives lower efficiency while larger R1 and R2 will give higher efficiency. Now, if we had been given a specific efficiency to satisfy (either exactly or better than), then that requirement would form an additional boundary condition which would allow us to find minimum or exact values for R1 and R2. FirstChildUserIdTAG: 9673 FirstChildUserNameTAG: xp42 FirstChildCreateTimeTAG: 2012-12-14T05:06:43Z FirstChildTAG: Thanks to you all! FirstChildUserIdTAG: 136959 FirstChildUserNameTAG: christerpher FirstChildCreateTimeTAG: 2012-12-16T05:17:25Z IndexTAG: 1890 TitleTAG: QUERY ABOUT GRADES I have a homework average of 91% and lab average of 89%. Is it possible for me to get an 'A' grade. Theoretically, it is still possible but how likely is it to get a perfect score in endterm? UserIdTAG: 449211 UserNameTAG: Deveshmonga CreateTimeTAG: 2012-12-10T11:50:13Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: So you have average homework/lab/midterm of 90% of 60% overall. = 54% so far. (Assuming 100% midterm) You need to get 82.5% minimum on **your** final exam to achieve an "A". You will not be able to get 100%, the best you can get is 94% (A), if you get 100% on your final exam. Good luck. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-12-10T15:03:40Z IndexTAG: 1891 TitleTAG: H12P2 - unable to submit answer Still unable to submit my answers to H12P2. Tried different browsers, clearing my cache. I've filled in all of the answer fields, but after pressing "Check" I don't get any feedback in the form of red crosses or green checks. I had originally submitted a formula with invalid variables for the first part, and now I can't seem to submit a new answer. When I load the homework from another browser, the answer box contains the first formula that I entered. UserIdTAG: 69075 UserNameTAG: ErikR CreateTimeTAG: 2012-12-10T03:28:56Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 9 FirstChildTAG: same problem FirstChildUserIdTAG: 214085 FirstChildUserNameTAG: shohin FirstChildCreateTimeTAG: 2012-12-10T07:30:08Z FirstChildTAG: same issue here as well, why isnt it taking any values??? FirstChildUserIdTAG: 128924 FirstChildUserNameTAG: arjun392 FirstChildCreateTimeTAG: 2012-12-10T07:24:15Z FirstChildTAG: An addendum... If I open the Javascript console, I see HTTP response codes of 500 (Internal Server Error) when I hit the "Check" button. FirstChildUserIdTAG: 69075 FirstChildUserNameTAG: ErikR FirstChildCreateTimeTAG: 2012-12-10T03:40:01Z FirstChildTAG: I am having the same issue. entering all answers ... but does not accept any FirstChildUserIdTAG: 714237 FirstChildUserNameTAG: PaxPolaris FirstChildCreateTimeTAG: 2012-12-10T07:15:53Z FirstChildTAG: yes same here. Any help from the staff? FirstChildUserIdTAG: 499671 FirstChildUserNameTAG: maliha266 FirstChildCreateTimeTAG: 2012-12-10T08:36:30Z SecondChildTAG: You have to have semi-valid answers for all the parts to get the grader to respond. Putting 0 for the last 4 entries won't cut it. Try 100 for $R_0$, $R_1$ and $R_2$ and 0.5 for the efficiency. These are wrong values but should get the grader to respond. For the first 2 equations, only use allowed symbols in each equation. vin is valid in the first, but not the second. vz is valid in the second, but not the first. So if you get messages that these are not allowed, make sure you are not using them in the equation where they are disallowed. SecondChildUserIdTAG: 9673 SecondChildUserNameTAG: xp42 SecondChildCreateTimeTAG: 2012-12-10T08:48:59Z FirstChildTAG: Also, you cannot use suffixes such as "470k" for a resistor value. You'll have to enter "470000" instead. FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-12-10T09:37:56Z SecondChildTAG: And you can *not* enter fractions like 1/2 either must enter 0.5 insted SecondChildUserIdTAG: 714237 SecondChildUserNameTAG: PaxPolaris SecondChildCreateTimeTAG: 2012-12-10T10:00:26Z FirstChildTAG: Same problem here also.... not able to submit the answer... trying for past 1 week... why edx staff is not taking any action on that??? FirstChildUserIdTAG: 258257 FirstChildUserNameTAG: Chandan358 FirstChildCreateTimeTAG: 2012-12-10T10:37:05Z FirstChildTAG: This episode seems to have ended much like it began. https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50aaa572cb8d0b250000006b FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-12-10T12:17:43Z SecondChildTAG: Same here... SecondChildUserIdTAG: 368981 SecondChildUserNameTAG: asimakhtar SecondChildCreateTimeTAG: 2012-12-10T19:03:41Z FirstChildTAG: same problem FirstChildUserIdTAG: 352647 FirstChildUserNameTAG: praveenjugge FirstChildCreateTimeTAG: 2012-12-11T07:18:51Z IndexTAG: 1892 TitleTAG: h12p2 i have analyzed the circuit and got ro according to .2 delta increasing =i*rz it gives me wrong answer and all the p2 gives me wrong may any one help me ? UserIdTAG: 221617 UserNameTAG: konan CreateTimeTAG: 2012-12-09T21:11:20Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Try the calculations for <0.05 Delta_V+ as explained [here][1]. See also: [Hints H12P1, H12P2, H12P3 requested by jmen][2] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50b8d3275516b62b00000010 [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50c3c129468bdc270000001c FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-12-09T22:30:12Z IndexTAG: 1893 TitleTAG: LAB12 cant get green mark i saw Myrimit post, so it is simple for R1 = 1620 R2 = 158000 C1=C2=1000p I cant get green mark, but the plot is right, calculated alpha and Q, they also meet req. idt what to do. UserIdTAG: 95280 UserNameTAG: Fortyq CreateTimeTAG: 2012-12-09T19:11:48Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: i found mistake, it was alpha, i forgot to divide on 2, and my Q was ~5 FirstChildUserIdTAG: 95280 FirstChildUserNameTAG: Fortyq FirstChildCreateTimeTAG: 2012-12-09T19:38:59Z FirstChildTAG: q is only 0.78588893148754 the criterion is q>5 FirstChildUserIdTAG: 623210 FirstChildUserNameTAG: preveen FirstChildCreateTimeTAG: 2012-12-09T19:23:46Z IndexTAG: 1894 TitleTAG: homework error my check box is not responding at the moment...i click on it and it gives me neither of the two ticks....does anyone have an idea about it? UserIdTAG: 413613 UserNameTAG: shaliesh CreateTimeTAG: 2012-12-09T17:07:51Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: Try to clean cash FirstChildUserIdTAG: 96873 FirstChildUserNameTAG: Santyaga FirstChildCreateTimeTAG: 2012-12-09T17:30:49Z SecondChildTAG: you mean bribe the system? SecondChildUserIdTAG: 623210 SecondChildUserNameTAG: preveen SecondChildCreateTimeTAG: 2012-12-09T19:04:23Z FirstChildTAG: I'm having the same problem with H12P2 only. Tried with Chrome and Safari already. FirstChildUserIdTAG: 110486 FirstChildUserNameTAG: Mirthyn FirstChildCreateTimeTAG: 2012-12-09T17:13:03Z SecondChildTAG: same here SecondChildUserIdTAG: 413613 SecondChildUserNameTAG: shaliesh SecondChildCreateTimeTAG: 2012-12-09T17:14:35Z SecondChildTAG: write message to STAFF SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-09T17:16:22Z SecondChildTAG: Sergtronix, How? SecondChildUserIdTAG: 110486 SecondChildUserNameTAG: Mirthyn SecondChildCreateTimeTAG: 2012-12-09T17:29:20Z SecondChildTAG: how can i message them? SecondChildUserIdTAG: 413613 SecondChildUserNameTAG: shaliesh SecondChildCreateTimeTAG: 2012-12-09T17:30:01Z SecondChildTAG: I found this: Santyaga 8 days ago Read careful task! Due to how these answer fields depend on each other, "H12P2 Linear Regulator" requires all 6 fields to contain answers for the "check" button to work. This means that writing the correct answer in the first two fields can yield a red X if the last part of the problem has not yet been answered. Also, as the last part of this problem is a design problem, the final four responses are related to each other; hence all four responses entered must agree with each other in order to receive a checkmark. Good Luck! SecondChildUserIdTAG: 110486 SecondChildUserNameTAG: Mirthyn SecondChildCreateTimeTAG: 2012-12-09T17:33:01Z SecondChildTAG: It works ;) SecondChildUserIdTAG: 110486 SecondChildUserNameTAG: Mirthyn SecondChildCreateTimeTAG: 2012-12-09T17:36:18Z SecondChildTAG: OOOPPS I thought that THESE, few times discussed, questions were read carefully already. Yes, yuo should FILL all required field. For H12P2 it is possible to place wrong values to R0, R1, R2 and efficiensy fields.This will help to check two first questions. Next, you should get required values for R0..R2 and efficiency. Only after these fields will be filled you may press CHECK button. You will see either red or green mark. Good luck! SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-09T17:57:55Z SecondChildTAG: sorry for typos :) SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-09T17:59:39Z FirstChildTAG: i tried with two different browsers, but still ended up with the same problem FirstChildUserIdTAG: 413613 FirstChildUserNameTAG: shaliesh FirstChildCreateTimeTAG: 2012-12-09T17:15:21Z FirstChildTAG: try to use different browser FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-12-09T17:10:39Z IndexTAG: 1895 TitleTAG: op-amp lab when i'm giving sine pulse to my source and click for ac analysis it show "ac analysis for unknown source"please get me rid of it...... UserIdTAG: 108863 UserNameTAG: shiviz CreateTimeTAG: 2012-12-09T05:00:04Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi Shiviz You may have got this by now but if you haven't, Select your sine source and type a name (I used Bob!) in the Name window. Then select ac at the top of the pane to run your simulation. In the simulation window you'll notice the "Name of V or I source for ac" option at the bottom. Simply enter the name that you chose for your sine source and the warning message goes away. smallblind FirstChildUserIdTAG: 259719 FirstChildUserNameTAG: Smallblindchris FirstChildCreateTimeTAG: 2012-12-09T12:31:30Z IndexTAG: 1896 TitleTAG: H12P2 c,d I have followed this steps to calculate the efficiency: efficiency = Pout/Pin, in which, Pin = Iin*Vin, Pout=Vout*Iout; That was straight forward and it is also easy to find Iout, which is Vout/RL. Here is the trick of the hard part: Iin = Iz+K*I (Iz is the current which goes through R0 and Rz and I is the BJT diode current). You can find I by knowing I = IL + Ir (IL is the current which goes through load resistor and Ir is the current which goes through R1 and R2). You also know that Vin = 20V and Vout is the value corresponding to that input. Be careful to compute Iz, remember you have to replace the zener diode by a voltage source + Rz. But I can't get green marks. I have tried with several values for R0, R1 and R2 (that I calculated with different methods): R0=1K R1=R2=100 ------- R0=10K R1=R2= 15k ------ R0=100 R1=90 R2=100 ------ R0=94 R1=90 R2=100 Someone can help me? Thank you very much!!! UserIdTAG: 558384 UserNameTAG: Josue9740 CreateTimeTAG: 2012-12-08T17:32:58Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Suppose you make the current through the zener VERY small, does vZ than change very much if Vin is changing? What happens to the power to that part of the circuit? Make the current through the branch with R1 VERY small, what happens to the power? FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-12-08T18:19:58Z SecondChildTAG: Ohhh my god! I'm stupid jajajaja I was calculating the current with K=0.5, when the exercise says K=0.95 Thank you!! SecondChildUserIdTAG: 558384 SecondChildUserNameTAG: Josue9740 SecondChildCreateTimeTAG: 2012-12-08T18:45:05Z IndexTAG: 1897 TitleTAG: H12P1 Please give your answer to within 0.1% of the exact answer. :o whats that mean??? UserIdTAG: 509517 UserNameTAG: randima CreateTimeTAG: 2012-12-08T13:34:25Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: pleas help me '-' FirstChildUserIdTAG: 509517 FirstChildUserNameTAG: randima FirstChildCreateTimeTAG: 2012-12-08T13:35:58Z FirstChildTAG: It means you have to include enough decimal places in the answer such that it is within .1% of the exact (infinite precision) answer. So if the correct answer is x, your answer has to be between x-.001*x and x+.001*x. FirstChildUserIdTAG: 397595 FirstChildUserNameTAG: mholin FirstChildCreateTimeTAG: 2012-12-08T13:57:20Z SecondChildTAG: thanx dear... but can yu tell me what is the format of this answer..should i write both parts?? SecondChildUserIdTAG: 509517 SecondChildUserNameTAG: randima SecondChildCreateTimeTAG: 2012-12-09T14:30:31Z IndexTAG: 1898 TitleTAG: why short vo? By superposition,we should not short dependent source,but he did.. Why? UserIdTAG: 136959 UserNameTAG: christerpher CreateTimeTAG: 2012-12-08T12:22:52Z VoteTAG: 1 CoursewareTAG: Week 12 / S23V18 Inverting amplifier analysis CommentableIdTAG: 6002x_S23V8_Inverting_amplifier_analysis NumberOfReplyTAG: 3 FirstChildTAG: I asked the same question. The book says you can do it but it might not work in all cases. The Prof must have known it would work for the V0 source. FirstChildUserIdTAG: 397595 FirstChildUserNameTAG: mholin FirstChildCreateTimeTAG: 2012-12-08T14:01:44Z FirstChildTAG: The dependend source is depending on something else in the circuit, so if Vd(ependend)=k*Vs(omethingelse) or k*Is, then if you differentiate: dVd/dVs=k or dVd/dIs=k and that is not zero. If you differentiate an independend voltage source then dV/di=0.. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-12-08T18:35:29Z FirstChildTAG: Shorting is not the issue, leaving it in or taking it out is, i think. FirstChildUserIdTAG: 184153 FirstChildUserNameTAG: tthngn FirstChildCreateTimeTAG: 2012-12-10T19:29:23Z IndexTAG: 1899 TitleTAG: Final Exam I just wanna know if all the weeks are in the exam... or just the second term (Week 7 -> Week 12)?? Thanks in advance. UserIdTAG: 365201 UserNameTAG: sirajmuhammad CreateTimeTAG: 2012-12-08T08:06:12Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50bf8166c67c571f00000031 FirstChildUserIdTAG: 288174 FirstChildUserNameTAG: matiasgrodriguez FirstChildCreateTimeTAG: 2012-12-08T13:38:08Z IndexTAG: 1900 TitleTAG: Fall time The fall is just the mirror image of the rise. UserIdTAG: 266912 UserNameTAG: pietvo CreateTimeTAG: 2012-12-07T19:49:14Z VoteTAG: 1 CoursewareTAG: Week 13 / S25V18_Op_Amp_oscillator_fall_time CommentableIdTAG: 6002x_S25V18_Op_Amp_oscillator_fall_time NumberOfReplyTAG: 0 IndexTAG: 1901 TitleTAG: Newbie here, please help. Hello there, I am honored to be among you passionate learners. However, I have some questions: I just registered a day ago, meaning the deadline for assignments and for the midterm is already reached, can I still complete them and will that still get me the certificate of completion of this course? Regards UserIdTAG: 856422 UserNameTAG: BilalDaya CreateTimeTAG: 2012-12-07T19:32:22Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: No, you cannot get a certificate: the course is almost over and a 60% score is needed for the certificate. As far as I know, this course will be offered again in Spring 2013. In the meantime, you can study the material you find here in self learning mode (without been graded). FirstChildUserIdTAG: 376877 FirstChildUserNameTAG: AndBre FirstChildCreateTimeTAG: 2012-12-07T20:02:26Z SecondChildTAG: Thank you so much that was very helpful, good luck in your final exam SecondChildUserIdTAG: 856422 SecondChildUserNameTAG: BilalDaya SecondChildCreateTimeTAG: 2012-12-07T20:08:57Z IndexTAG: 1902 TitleTAG: on the most common mistake Hate to admit it but I will still make this mistake if the devices are not marked in the diagram with +||-. Prof's explanation was based on knowing the markings so to me it is not definitive. My path to first principles is often obscured. These are the things I think are true. Please tell me if they don't qualify as first principles or are not precisely true. - Positive current moves (the arrow points in the direction) from the higher potential (voltage) toward the lower potential (voltage). - The end of the R/L/C that the current enters is always the positive end. - Unless the device is acting as a short circuit there is always a drop in voltage from the positive side to the negative side. If you put a meter from the + side to the negative it will read a positive voltage. - The voltage drop across a device is equal to the voltage on the '+' side minus the voltage on the '-' side Is this enough? Are there other 'first principles' that will keep me thinking straight? UserIdTAG: 440714 UserNameTAG: mcktim CreateTimeTAG: 2012-12-07T17:22:49Z VoteTAG: 1 CoursewareTAG: Week 12 / S24V6 Op amp integrator CommentableIdTAG: 6002x_S24V6_Op_amp_integrator NumberOfReplyTAG: 0 IndexTAG: 1903 TitleTAG: Problem about Homework 10,11and Lab10,11"staff" To Course Staff During EDX Website Maintenace I could not do my homework&Lab (10and 11). After EDX website comeback I expect to do my homework&Lab(10 and11) but I can not.[I means I do both Homework&Lab(10 and 11) and I hope to get marks. My bar graph(of Homework&Lab 10 and 11) on Course Progress page should show but it does not. I plan to get marks from Homework&Lab (10,11,and 12) to reach ten out of twelve homeworks&labs in order to avoid a grade penalty.Teacher please clearify my case and help me.Now I am worried about receiving a grade panalty although I do not intend to miss two homeworks and two labs(Homework&lab(10 and 11). It 's not my fault. Are there a way to avoid a grade penalty? Sincerely Sombat Singjoo( Bangkok, Thailand) My Email used to Log in to EDX is sombatsingjoo@gmail.com UserIdTAG: 364129 UserNameTAG: Sombat CreateTimeTAG: 2012-12-07T14:48:16Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1904 TitleTAG: Now here this. I hereby pledge my loyalty to the proposition that the virtual short method is well worthy of the title, rank and privilege of Aha moment standing. So it should be written. So it should be done. UserIdTAG: 226939 UserNameTAG: MarkJohnson CreateTimeTAG: 2012-12-07T04:53:14Z VoteTAG: 1 CoursewareTAG: Week 12 / S23V22 Inverting amplifier input resistance using virtual short method CommentableIdTAG: 6002x_S23V22_Inverting_amplifier_input_resistance_using_virtual_short_method NumberOfReplyTAG: 1 FirstChildTAG: Definitely Agree!! Working in electronics for 30 years with a much less rigorous understanding than I now have (Thanks so much Professor), the virtual short was THE way to understand OpAmp circuits. FirstChildUserIdTAG: 369519 FirstChildUserNameTAG: perpstud FirstChildCreateTimeTAG: 2012-12-13T04:34:56Z IndexTAG: 1905 TitleTAG: Red pencil sometimes irritates first I guess although it gets attention of the learner but it would be grate if the color would change in more intervals. just a suggestion.secondly I agree aha moment. UserIdTAG: 733429 UserNameTAG: EhsanMokhtari CreateTimeTAG: 2012-12-06T13:39:28Z VoteTAG: 1 CoursewareTAG: Week 12 / S23V22 Inverting amplifier input resistance using virtual short method CommentableIdTAG: 6002x_S23V22_Inverting_amplifier_input_resistance_using_virtual_short_method NumberOfReplyTAG: 0 IndexTAG: 1906 TitleTAG: 50 Watts Mosfet Amplifier I am trying to build a ultrasound generator to use it as dog repel device with a CD 4049 circuit ( http://dfi57.blogspot.ro/2011/02/un-montaj-care-ne-poate-apara-de-cainii.html ), but the signal generated it way too weak.I can hear it because i have switched the capacitor with a higher one, so that the generated sound would fall in 1-3 KHz audible range .I use a piezo ultrasound emitter but the signal is too weak. I tried to use a classic Mosfet amplifier circuit to boost it, but there is only a small change ( I have used a voltage divider with a variable resistor to bias the mosfet ).Should i use the power amplifier schematic with the output on the resistor ? Until then, i've found this schematic : http://www.circuitstoday.com/mosfet-amplifier-circuits .Why is there a resistor at the gate of Q5 ? Because there is no current entering the gate ? Thanks to anyone interested. UserIdTAG: 240487 UserNameTAG: AlexAlexandrescu CreateTimeTAG: 2012-12-05T21:38:31Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Question should be : what is D1 purpose? The resistor in the Q5 gate does limit current through D1. May be Im wrong, piezo ceramic require resonanse circuit for maximum output power, so you need some inductance. FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-12-05T22:22:59Z SecondChildTAG: That's what i thought. But on the same page there is a 18 watt mosfet amplifier, which does have gate resistors. Piezo speakers are not capacitors.When the excitation frequency reaches the crystal resonant harmonic frequency, it starts to vibrate. Here is an example with a piezo speaker and no inductor : http://www.electronicecircuits.com/electronic-circuits/cmos-piezo-transducer-buzzer-driver-circuit . Also here is a crystal oscillator without a inductor : http://www.electronicecircuits.com/electronic-circuits/ttl-crystal-oscillator-schematic .Besides, everybody hates inductors except radio amateurs because they don't have any choice. You cannot measure them properly, and very often you have to built then yourself as an amateur. Putting a 1k resistor in parallel with the speaker improved a bit, but i start to suspect that once my oscillator reaches ultrasound frequencies , it is louder. For once i wish i had dog ears. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-12-06T08:51:55Z SecondChildTAG: Im talking about **piezo ceramic**, not crystals, not completed speaker. For example, some ultrasonic cleaning bath has inductor connected to the ceramic, toghether with pretty high supply voltage. [One small example][1] [1]: http://www.fujicera.co.jp/product/e/03/image/ap_003.gif SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-06T09:38:12Z SecondChildTAG: I use a piezo ultrasound small speaker, the type that you use for emitting ultrasounds on car parking device.The one you showed is special, because it has 3 terminals and 8kv as source. For example, a small alarm clock , has only a small pcb , and the piezo speaker.No inductors, and it sounds very loud . Any thoughts why 18 watt mosfet amplifier from here http://www.circuitstoday.com/mosfet-amplifier-circuits has gate resistors ? Thanks ! SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-12-06T15:02:36Z SecondChildTAG: I'm sorry, but you are wrong about my link above (it is a just sample from hundreds), it is DC converter from 12V to +8kV and -8kV.This device is very simple, and it should not be hard to read this schematics. You may see LC circuit which give enough voltage for the next multiplying. Note where does voltage increases.Also [3 pins transducers][1] as well common as 2 pins.This third pin helps to simplify schematic.Ok. Do you want to build powerful **ultrasound** emitter with piezo speaker? Well, I tried to tell you that your enemy will be power source voltage and ceramic capasitance.Unfortunatelly piezo ceramic has a huge capasitance, [example here][2] . For example, check by oscope signals on the buffer gates inputs and their outputs.You may try to improve this situation by pnp-npn buffer. About gates resistors(18W amplifier).Power MOSFETs has pretty large $C_{GS}$ .In some cases it is necessary to eliminate the influence of such capasitance to MOSFET drivers.In this amplifier such resistors doing very important task :) PS Small buzzers have Resonanse freq within comfortable range for human ears.It may be 2.8 or 4kHz.Im not sure that these buzzers will be nice for your task. Good luck with your experiments! [1]: http://www.digikey.com/product-detail/en/CPE-3085/102-2199-ND/2364640 [2]: http://www.ctdco.com.tw/pdf4/LB-5020.pdf SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-06T17:01:40Z SecondChildTAG: Thank you Sergtronix ! That makes sense ! I will keep you posted if you're interested about my ultrasound emitter . Pnp npn buffer sounds pretty good . I will try that, but until then i will try to use another piezo speaker that has the frequency in the audible area. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-12-07T08:08:20Z SecondChildTAG: Hi Sergtronix ! You were right, i need an inductor to rise my voltage. Changing the piezo element make it sound louder, but not loud enough.As i only intend to supply max 9V to the circuit i need an inductor.Or a transformer. Now i have to make a inductor measure tool. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-12-10T19:02:56Z SecondChildTAG: I've also found the basics here to anyone interested : http://homemadecircuitsandschematics.blogspot.ro/2012/04/simplest-piezo-driver-circuit-explained.html SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-12-10T20:48:18Z IndexTAG: 1907 TitleTAG: Generating a pulse in the Sandbox Can anybody answer some questions I have regarding how to configure a voltage source to create a pulse in the Sandbox? What do they mean by "Time for first transition (sec):" and "Time for second transition (sec):"? Are these the rise times for the rising edge and falling edge of the pulse respectively? Also, why do they specify a period? Isn't it understood a pulse is half a period? UserIdTAG: 132828 UserNameTAG: rharris CreateTimeTAG: 2012-12-05T19:09:39Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Try to experiment setting different values for the parameters and check the resulting waveform using the voltage probe. For example you can replicate the following image. Besides, a pulse is not necessarily half a period, it could be whatever percentage: it's called [duty cycle][1]. ![enter image description here][2] [1]: http://en.wikipedia.org/wiki/Duty_cycle [2]: https://edxuploads.s3.amazonaws.com/1354782119134360.png FirstChildUserIdTAG: 376877 FirstChildUserNameTAG: AndBre FirstChildCreateTimeTAG: 2012-12-06T08:23:07Z SecondChildTAG: I made sure I tried every scenario I could before posting my question. Before creating the schematic below, I tried connecting the voltage probe directly to the voltage pulse generator like your suggestion, but no pulse was produced. Then I tried adding a load as shown, but still no pulse. This is why I was asking what was meant by the "time for first and second transition". The only thing that made sense to me is they are for rise and fall times. I understand what a duty cycle is, however why is a period required when a pulse typically only occurs once in a specified amount of time (it goes high, then low, done)? I would think that if a duty cycle came into play one would use a repeating waveform, such as a sine or square wave. I could see if the pulse functionality allowed a pulse train to be generated, but this doesn't seem to be the case. Anyway, it is just an observation and I'm curious what use it is in generating a single pulse. It is only one extra step to add the period to give the Sandbox what it wants. ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13548107221343689.jpg SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-12-06T17:01:11Z SecondChildTAG: The period value must be greater than the sum of the delay,first transition, second transition and width values; so in your example, try to increase the period value to 1.5m: you will see a pulse train. SecondChildUserIdTAG: 376877 SecondChildUserNameTAG: AndBre SecondChildCreateTimeTAG: 2012-12-06T20:01:19Z SecondChildTAG: Oh, OK I see, so it does create a pulse train. I thought all times were sequential, therefore I expected the pulse to begin at t = 1ms, rise for 1us, stay high for 0.098ms, fall for 1us then stop. Then I figured the period would just be twice the pulse width, figuring the tool always generated a duty cycle of 50%, which I accommodated. Thanks, it works now. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-12-07T04:26:57Z IndexTAG: 1908 TitleTAG: H12P1 Help with part C I have used V0 = A*VI / ( 1 + (A*R2) / (R1 + R2) ) = 9.32 V V0 / ( R1 + R2 ) = 0.0088 A What I'm doing Wrong? UserIdTAG: 433574 UserNameTAG: endika86 CreateTimeTAG: 2012-12-05T18:54:34Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: "Please give your answer to within 0.1% of the exact answer" 0.1% from ~8mA- how many is it? try to put more digits FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-12-05T19:36:31Z SecondChildTAG: Thanks. I thought that T put enough digits. SecondChildUserIdTAG: 433574 SecondChildUserNameTAG: endika86 SecondChildCreateTimeTAG: 2012-12-05T21:44:04Z IndexTAG: 1909 TitleTAG: Band stop filter? Wouldn't this filter stop low and high frequencies but allow values in the middle? Since there is high impedance for low frequency since the capacitor acts as an open and for high frequency the inductor acts as an open? Yet he is saying low values of omega and high values are allowed but values in the middle are not? UserIdTAG: 219780 UserNameTAG: ReconIII CreateTimeTAG: 2012-12-05T16:24:28Z VoteTAG: 1 CoursewareTAG: Week 11 / S21V6: Series RLC bandstop CommentableIdTAG: 6002x_S21V6_Series_RLC_bandstop NumberOfReplyTAG: 2 FirstChildTAG: It's a voltage divider. Look where he's taking the output. The impedance of that portion of the circuit is minimum at mid frequencies, so output is minimum. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-12-05T16:37:36Z SecondChildTAG: Ah that's right he is plotting the current across VLC not VR that makes more sense now, thanks. SecondChildUserIdTAG: 219780 SecondChildUserNameTAG: ReconIII SecondChildCreateTimeTAG: 2012-12-06T16:07:09Z FirstChildTAG: You can also make a notch filter with R's and C's, like this one. Run an AC-analysis from 10Hz to 100kHz. ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13547359601343612.png FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-12-05T19:33:18Z IndexTAG: 1910 TitleTAG: Inductor part labels? I was going to experiment with building an LC circuit, and I pulled this inductor out of some old electronics. It appears that the label is not a value in μH, but a part number? ![enter image description here][1] Is that common for inductors -- that you have to look up a part number, as opposed to reading off the value like with a resistor? How would you look it up? Or do you measure the inductance with a device? (Which I don't yet have.) [1]: https://edxuploads.s3.amazonaws.com/13547138698474434.jpg UserIdTAG: 468623 UserNameTAG: RobNik CreateTimeTAG: 2012-12-05T13:24:55Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Yes it is common. Aside from resistors and some capacitors, part numbers are usually used, some are different from brand to brand. Some inductors can be measured fairly easily, depending on what you have available to you. Have a look here, no affiliation. If you don't have a function generator you can download frequency tones you need and just use an iPod or stereo. http://daycounter.com/Articles/How-To-Measure-Inductance.phtml If the inductor is larger, you can measure the thickness of wire and estimate it's length by it's resistance. Use that length with the diameter of the coil to calculate inductance. Have fun. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-12-05T13:38:12Z SecondChildTAG: Thanks! SecondChildUserIdTAG: 468623 SecondChildUserNameTAG: RobNik SecondChildCreateTimeTAG: 2012-12-05T14:05:36Z SecondChildTAG: Hi Rob . Inductors are always tough to find to match your needs as you will soon discover.In an ideal world there are no damn inductors. Until then, here is a simple schematics used as an adapter to you multimeter to measure the inductance : http://dfi57.blogspot.ro/2011/01/adaptor-pentru-masurat-inductante.html .Hope it helps. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-12-06T09:00:00Z IndexTAG: 1911 TitleTAG: How Battery works I remember i read some where that electrons move from positive terminal to negative terminal inside a battery(NOT OUTSIDE BATTERY). Is it true and if yes can you please explain me why is it so? UserIdTAG: 132685 UserNameTAG: deepakmurali CreateTimeTAG: 2012-12-05T05:59:17Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: A battery has a Cathode (+) and an Anode (-) Chemical reactions in a battery cause a buildup of electrons on the Anode (-) terminal. The electrons would like to jump over to the Cathode (+), except the electrolyte stops this from happening. So they have to take the long way and travel out the Anode (-) terminal, through your circuit before they reach the Cathode (+). As the electrons start to balance out between the cathode and anode, the voltage starts to drop. At this point you will have to recharge or recycle the battery. Think of yourself as an electron. If you want to leave the store (a Battery), you will have to use the exit (-) terminal. Ideally there is little to no interaction of electrons inside the battery, the Cathode and Anode should be completely isolated from each other. If electrons are able to travel through the electrolyte at any significant rate, the battery will "drain" itself. Think of a battery as two balloons in one package, The Cathode (+) is an empty balloon and the Anode (-) is a full balloon. You let some air from the anode (-) balloon through your machine to do some work, eventually it exits the machine and starts to fill the cathode balloon. The machine will work until the two balloons are the same size. (Minus thermodynamic losses commonly associated with balloon powered machines.) So the labeling is semantics, depending on your point of view. Say you have two rooms connected by an door. Is it an exit or an entrance? It depends what you want to do. The confusion comes from devices needing batteries, They say "Insert AA Battery" and usually show a (+) and (-) sign. Well those are the signs used for orienting the battery the right way and not intended to be literal. I will add that I simplified the role of the electrolyte for the sake of my description. From a simplistic point of view, nothing flows in the battery unless the circuit has been completed through the cathode. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-12-05T13:10:19Z SecondChildTAG: thank u SecondChildUserIdTAG: 132685 SecondChildUserNameTAG: deepakmurali SecondChildCreateTimeTAG: 2012-12-07T09:06:36Z IndexTAG: 1912 TitleTAG: OK what I really mean is, what is "input resistance"? Resistance is a two-end measurement, like voltage (but unlike current). You measure resistance across an element, or many elements, not at a point. Where is the other end of the "resistor" we're working with for the "input resistance"? UserIdTAG: 91356 UserNameTAG: mnrsiat CreateTimeTAG: 2012-12-04T22:45:14Z VoteTAG: 1 CoursewareTAG: Week 12 / S23E3_L23AmplifierInputResistance CommentableIdTAG: 6002x_S23E3_L23AmplifierInputResistance NumberOfReplyTAG: 1 FirstChildTAG: Where is the other end of the input if you measure the input voltage ? Same thing, i.e. the ground node in the case at hand. FirstChildUserIdTAG: 90796 FirstChildUserNameTAG: ChaunceyGardiner FirstChildCreateTimeTAG: 2012-12-05T04:38:20Z IndexTAG: 1913 TitleTAG: How to write the equation of V* by intuition in S25V7? At t=0, V=K, and After a long time, it should be A(v+-v-). The time constant is RC. However, the "T" is associated with more than RC. How do I get the answer, especially the time constant by observation? UserIdTAG: 564606 UserNameTAG: ArequipaRadio CreateTimeTAG: 2012-12-04T15:42:57Z VoteTAG: 1 CoursewareTAG: Week 13 / S25V7_Dynamics_of_Op_Amp CommentableIdTAG: 6002x_S25V7_Dynamics_of_Op_Amp NumberOfReplyTAG: 0 IndexTAG: 1914 TitleTAG: H12P1 **Please give your answer to within 0.1% of the exact answer**. What does it mean?I calculated the current, but getting only red cross. Any help is highly appreciated. UserIdTAG: 156835 UserNameTAG: kphariprakash1968 CreateTimeTAG: 2012-12-04T06:48:22Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 6 FirstChildTAG: Same problem.I dont know where it is going wrong? iL=Vout/(Rl+R) Is it right? Or any problem in Vout calculation I got Vout={1+Rl/R}Vi FirstChildUserIdTAG: 156835 FirstChildUserNameTAG: kphariprakash1968 FirstChildCreateTimeTAG: 2012-12-04T12:07:18Z SecondChildTAG: You must involve A into the equation SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-04T13:22:28Z FirstChildTAG: I have the same problem. I have simulated the schema in the circuit sandbox and I am pretty sure that the current I am trying is correct but it keeps givind a red cross as result. FirstChildUserIdTAG: 35854 FirstChildUserNameTAG: dilandau FirstChildCreateTimeTAG: 2012-12-04T07:17:15Z FirstChildTAG: I got the green mark whith only one decimal of the answer. The exact answer didn't work. Try this. FirstChildUserIdTAG: 241012 FirstChildUserNameTAG: InakiL FirstChildCreateTimeTAG: 2012-12-04T08:38:17Z FirstChildTAG: At first you must find voltage considering A and then find current trough voltage and given resistance. But you must input your answer with high precision. FirstChildUserIdTAG: 96873 FirstChildUserNameTAG: Santyaga FirstChildCreateTimeTAG: 2012-12-04T10:53:06Z SecondChildTAG: If you have solved this try to check again.Will you have green check mark ? SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-04T11:09:26Z SecondChildTAG: Hmm... it is very strange and bad. I think it`s bug in this part of problem! SecondChildUserIdTAG: 96873 SecondChildUserNameTAG: Santyaga SecondChildCreateTimeTAG: 2012-12-04T13:12:56Z SecondChildTAG: try to press check again.It should works fine SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-04T19:20:03Z FirstChildTAG: It is pretty strange situation. I have solved H12P1 few days ago with success and all green marks. But today I tried to "check" this and got red cross on the last question. With the same as before values! What is happened with H12P1? It seems that staff made some corrections. Any suggestions? FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-12-04T11:11:43Z SecondChildTAG: WOW! It works again! My prevoius CORRECT value is correct again :) Im happy SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-04T19:19:08Z FirstChildTAG: Is the problem here basically with the sign of the answer? In my calculations, I believe I was correct with respect to conventions and arithmetic, yet received the red cross. Changing the sign got me the green check. The correct answer - whatever it is - would, in part, depend on what one considers the direction of positive current through the load resistor? What is the convention here? FirstChildUserIdTAG: 141108 FirstChildUserNameTAG: rl777 FirstChildCreateTimeTAG: 2012-12-04T19:10:37Z SecondChildTAG: In addition, of course, to including A in one's calculations. SecondChildUserIdTAG: 141108 SecondChildUserNameTAG: rl777 SecondChildCreateTimeTAG: 2012-12-04T19:13:18Z SecondChildTAG: 1. What is Vin for H12P1 part 2. 2. For this Vin compute Vout for new load resistance. 3. I=Vout/(R+Rload) SecondChildUserIdTAG: 297655 SecondChildUserNameTAG: Aljoska SecondChildCreateTimeTAG: 2012-12-05T14:23:04Z IndexTAG: 1915 TitleTAG: week 11's hw impossible?? is it me or was week 11's hw just impossible?? These last few weeks have been pretty hard, the first half of the semester was easy, finished all hw and labs and got 100 on midterm but it seems like 2nd ordered circuits that are in the second half of this class are difficult, especially with the sinusoidal inputs. I don't know if its because its the end of the year and im being lazy but man these last 2 homeworks have been tough! UserIdTAG: 435197 UserNameTAG: RichmondRichter CreateTimeTAG: 2012-12-04T04:36:25Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: It's not impossible. It's hard; I understand. I did not have the requisite math background to begin this course, and it's been a long brutal slog for me. And I'm feeling pretty burned out. But please, hang in there. Do not give up. Don't worry about your grades; you can always re-take the course. You're not alone. Just remember what an awesome opportunity this is, and try to relax and have fun. :-) FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-12-04T07:14:46Z SecondChildTAG: I'm burned out too... almost gave up on week 4 (small signal). But I have the filling that the second part was less stressful because of the help from the people of this forum and because we had a lot of holidays here in my country so i did have more time... :) SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-12-04T15:59:44Z SecondChildTAG: Also lacking some math skills here. Very frustrating when you know all the formula's and understand the concept but lack the skill to turn formula A into formula B... But I won't give up, NO WAY, not with less than three weeks to go. SecondChildUserIdTAG: 114881 SecondChildUserNameTAG: xvink SecondChildCreateTimeTAG: 2012-12-04T17:14:24Z SecondChildTAG: That is right, anyone here is not alone,and we can learn from each other.Just go ahead!For me,it is really a pleasant trip with the help of people here,though sometimes I was totally confused with the hard ,even seemingly impossible homework. SecondChildUserIdTAG: 361137 SecondChildUserNameTAG: Ericson SecondChildCreateTimeTAG: 2012-12-05T04:29:02Z IndexTAG: 1916 TitleTAG: To STAFF Hi There, Congratulations, for a great Job! I'd like to know if ,before final exam, are we going to have a week or two. I'd like to recheck all the videos from the beginning. Thanks again! UserIdTAG: 369339 UserNameTAG: vargaslen CreateTimeTAG: 2012-12-03T17:18:10Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1917 TitleTAG: single ended current source? All current sources presented are double-ended. Exists one (simple!) which is single-ended (where one terminal of the resistor 1k-10k is @ 0V)? UserIdTAG: 388524 UserNameTAG: dasbinich CreateTimeTAG: 2012-12-03T14:18:26Z VoteTAG: 1 CoursewareTAG: Week 12 / How to build a current source CommentableIdTAG: 6002x_How_to_build_a_current_source NumberOfReplyTAG: 0 IndexTAG: 1918 TitleTAG: wrong symbol for bandwidth bandwidth answers are not dw, but df. dw should be a factor of 2pi larger. UserIdTAG: 209655 UserNameTAG: typograph CreateTimeTAG: 2012-12-02T11:14:14Z VoteTAG: 1 CoursewareTAG: Week 11 / S21E4: AM_Radio_Tuning CommentableIdTAG: 6002x_S21V1_AM_Radio_Tuning NumberOfReplyTAG: 1 FirstChildTAG: Bandwidth can be described by Δw or Δf,in some way they are the same, but Δw=2*pi*Δf ,the unit is different.Actually,Δf is more used. Sorry about my english,have a look here: http://searchenterprisewan.techtarget.com/definition/bandwidth FirstChildUserIdTAG: 361137 FirstChildUserNameTAG: Ericson FirstChildCreateTimeTAG: 2012-12-02T12:23:35Z SecondChildTAG: Please, never apologize for your English! You communicate very well. I assure you, your English is far better than my [your native tongue]. ;-) SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-12-02T12:40:35Z SecondChildTAG: Haha,thanks for your kind inspiration! SecondChildUserIdTAG: 361137 SecondChildUserNameTAG: Ericson SecondChildCreateTimeTAG: 2012-12-02T16:15:46Z SecondChildTAG: Hmm, why in this exercise is Δw not Δf used in question? Δf would be more clear. SecondChildUserIdTAG: 413784 SecondChildUserNameTAG: Dorotka SecondChildCreateTimeTAG: 2012-12-02T19:16:14Z SecondChildTAG: Yeah,Δf is much better. SecondChildUserIdTAG: 361137 SecondChildUserNameTAG: Ericson SecondChildCreateTimeTAG: 2012-12-03T02:56:34Z IndexTAG: 1919 TitleTAG: H12P2 working late, still confused! Hello again guys, I've been reading some of the posts in this forum, and some are very confusing to me! for example, the use of case in the voltages and resistances make me think I'm missing something important here. Any way, I have come to this equation and before going any further, would appreciate your opinion of it. vIN*rz/(R0+rz)=vOUT*R2/(R1+R2) as you can see, I've applied voltage divider in both inputs of OP AMP and equated by virtue of "virtual short circuit" . thanks again guys for your assistance! UserIdTAG: 369339 UserNameTAG: vargaslen CreateTimeTAG: 2012-12-02T08:44:10Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hi, vargaslen! Could you explain, what do you want to describe by equation above? Youre lose $v_z$ at the left.. You may check [this][1] too [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50b8d3275516b62b00000010 FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-12-02T11:56:08Z SecondChildTAG: Thanks for answering! Sergtronix. I saw yor post before, and didn't get it! may be you can elaborate! thank you! SecondChildUserIdTAG: 369339 SecondChildUserNameTAG: vargaslen SecondChildCreateTimeTAG: 2012-12-02T13:52:16Z SecondChildTAG: My equation comes from trying to put vO in terms of vI. v- is vOUT*(R2/(R2+R1), since there is a "negative feedback" from a voltage divider. As v+ is more or less equal to v-, and from the graph there is a voltage divider there too! (rz is the zener resistance),then vin*rz/(R0+rz) is the voltage divider there. Please, tell me where am I wrong! Not sure if I'm getting this right! SecondChildUserIdTAG: 369339 SecondChildUserNameTAG: vargaslen SecondChildCreateTimeTAG: 2012-12-02T14:08:54Z SecondChildTAG: Yeah, I understand. I supposed that first two questions will not be hard. Ok. For the first question $V_{in}<{V_z}$.In this case you should remove Zener diode from the consideration.That mean $v_z$ and $r_z$ are excluded too.Final equation for this case will be very very simple - it is $v^+$ miltiplied by the Gain.I hope you will find $v^+$ easy.Remember that no current through + and - OA inputs, ie their input impedances are infinite. For the second question $V_{in}>{V_z}$ and $r_z=0$.Here you shouldnt worry about Zener current ( in the real life design flow is different) and do use $v_z$ as $v_z$ :). What will be your equation? Right, reference voltage multiplied by the Gain. Third question is a little bit more complex. You may do use my link above ( if it isnt edited yet :) ).In this case you must add $r_z$ into consideration.How to find $v^+$? What is Zener diode model here?You may use wide range R0 values ( again, we dont worry about Zener's current). Anyway, $v^+$ will be higher then $v_z$. Well, I hope you have solutions for 1, 2 and 3rd(a,b,c) questions. 3d is relation of $P_{load}$ to $P_{total}=SUM_{all current consumption nets}$. I hope I was helpful for you. SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-02T18:05:43Z SecondChildTAG: thanks! Sergtronix. SecondChildUserIdTAG: 369339 SecondChildUserNameTAG: vargaslen SecondChildCreateTimeTAG: 2012-12-02T19:47:06Z SecondChildTAG: SRY BUT CAN U EXPLAIN LAST PART OF H12P2 AGAIN SecondChildUserIdTAG: 519690 SecondChildUserNameTAG: vkjjw SecondChildCreateTimeTAG: 2012-12-03T10:52:08Z FirstChildTAG: Hola vargaslen :) Some tips: * The layout of the circuit is the non inverting amplifier that we saw on videos S23V7 and S23V13 (virtual short method). * When $Vin * When $Vin>Vz$ remember the nice properties of the zener and note that $Rz=0$ * To check if this correct put some values and ALL questions but do NOT put zero on the last part. Good luck FirstChildUserIdTAG: 288174 FirstChildUserNameTAG: matiasgrodriguez FirstChildCreateTimeTAG: 2012-12-02T14:29:08Z SecondChildTAG: * Vin < Vz * Vin > Vz SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-12-02T14:30:09Z SecondChildTAG: Gracias matiasgrodriguez! think I missundertand zener diode, I'll try again! SecondChildUserIdTAG: 369339 SecondChildUserNameTAG: vargaslen SecondChildCreateTimeTAG: 2012-12-02T15:52:11Z IndexTAG: 1920 TitleTAG: Can't open Lab 11...Staff Please Help! I am trying to open Lab 11 from the last night but couldn't open it instead the screen is going black... UserIdTAG: 122513 UserNameTAG: Mona77 CreateTimeTAG: 2012-12-02T05:07:26Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: What is your OS and browser? Versions? Having troubles with just the lab, or can you open other parts of Courseware? More details make for better error reports... FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-12-02T11:41:49Z SecondChildTAG: I am using Chrome and windows XP...I can open my course-ware but can't open the Lab SecondChildUserIdTAG: 122513 SecondChildUserNameTAG: Mona77 SecondChildCreateTimeTAG: 2012-12-02T12:33:37Z SecondChildTAG: Have you tried clearing your browser cache, restarting your browser, and reloading the page? SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-12-02T13:58:24Z SecondChildTAG: ok I'll try that...thanks :) SecondChildUserIdTAG: 122513 SecondChildUserNameTAG: Mona77 SecondChildCreateTimeTAG: 2012-12-02T14:35:15Z SecondChildTAG: I hope that worked! SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-12-02T15:58:19Z SecondChildTAG: mmm i've the same problem... and i can't open de progress page SecondChildUserIdTAG: 70519 SecondChildUserNameTAG: Fipe SecondChildCreateTimeTAG: 2012-12-03T01:21:54Z IndexTAG: 1921 TitleTAG: H11P2 last question I got the first 3 questions right, but I am struggling with the last one. I know that the transfer function of the system is $\frac{V_s}{V_p}=\frac{Z_s}{Z_s+Z_p}$ where $Z_s=\frac{R_s}{1+C_sR_ss}$ and $Z_p=\frac{R_p}{1+C_pR_ps}$ and $V_p=\frac{4}{s}$ (because 1/s represents a unit step function in the Laplace domain). So I solve for the time-domain response and $v_s(t=0)=2$ but this is not correct. UserIdTAG: 101750 UserNameTAG: caf960 CreateTimeTAG: 2012-12-01T22:08:42Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hi,try to consider about the characteristic of capacitor,maybe it will be much easier. FirstChildUserIdTAG: 361137 FirstChildUserNameTAG: Ericson FirstChildCreateTimeTAG: 2012-12-02T04:18:00Z FirstChildTAG: I never had to solve a differential equation for any part of H11P2. I suggest that you rethink your method. The "trick" is to choose Cp so that the transfer function is real. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-12-01T23:08:05Z SecondChildTAG: Have a look here: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50bbabcf4df6b51f0000004d SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-12-02T19:43:10Z IndexTAG: 1922 TitleTAG: check button not working in h12p2 ????????? UserIdTAG: 611666 UserNameTAG: Shaminder420 CreateTimeTAG: 2012-12-01T15:05:11Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Read careful task! Due to how these answer fields depend on each other, "H12P2 Linear Regulator" **requires all 6 fields to contain answers for the "check" button to work. This means that writing the correct answer in the first two fields can yield a red X if the last part of the problem has not yet been answered.** Also, as the **last part of this problem is a design problem, the final four responses are related to each other; hence all four responses entered must agree with each other in order to receive a checkmark**. Good Luck! FirstChildUserIdTAG: 96873 FirstChildUserNameTAG: Santyaga FirstChildCreateTimeTAG: 2012-12-01T16:27:19Z FirstChildTAG: Though they don't mention it: I think you also have to enter the resistance values in ohms! Don't use k(ilo) or, what I did: M(ega). FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-12-01T17:24:16Z IndexTAG: 1923 TitleTAG: H12 P1 Hi! Need some hints! For first and second qestion I got green stick, but can`t understand 3 qestion :-( according to the task we need to find current through the load and answer must be to within 0.1% of the exact answer. Now I want to understand! They say that "Using the value of R that you computed above, and assuming a load resistance of 500Ω"! ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13543572018739125.jpg Main qestion? Do we need find current through R, from load or through both of them R and load? UserIdTAG: 96873 UserNameTAG: Santyaga CreateTimeTAG: 2012-12-01T10:22:47Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: just find the output voltage..then i=Vout/(R+load)... FirstChildUserIdTAG: 114864 FirstChildUserNameTAG: Vikaash FirstChildCreateTimeTAG: 2012-12-01T11:10:34Z SecondChildTAG: can you tell me how to find the maximum load in part2.. SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-12-01T11:11:52Z SecondChildTAG: Thank you got it! There was one mistake. SecondChildUserIdTAG: 96873 SecondChildUserNameTAG: Santyaga SecondChildCreateTimeTAG: 2012-12-01T11:50:55Z FirstChildTAG: Ok, Vikaash here the answer - You have that Vs=X Volts Also you have R from first qestion. Then in task was told tham max current through circuit is 6mA. notise we have series resistance! Use Ohm law for Vs and max current and then find resistance difference with R That will be your answer! ;-) FirstChildUserIdTAG: 96873 FirstChildUserNameTAG: Santyaga FirstChildCreateTimeTAG: 2012-12-01T11:40:40Z SecondChildTAG: Yes but vout is equal to VS = 20? SecondChildUserIdTAG: 244225 SecondChildUserNameTAG: lerimock SecondChildCreateTimeTAG: 2012-12-04T22:58:53Z IndexTAG: 1924 TitleTAG: Low Noise Amplifier Design Hi Myrimit, I have to design a LNA meeting these specs. Bandwidth - DC to 12 MHz High Gain up to 80 dB Adjustable gain - 20 dB Noise down to 1 nV/Sqrt(Hz) Please help me. UserIdTAG: 113365 UserNameTAG: satyabrata CreateTimeTAG: 2012-12-01T10:00:28Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hi satyabrata, Do you have to design an Amplifier from zero? Using MOSFETS , BJT , resitors, etc ? or you can use OA? If it is the first, you have to calculate the components, you can make different stages in the design . Also you have to calculate the gain, etc ... If you can use a AO it will be more easy. But also, you will have to calculate the components in order to obtain that gain, etc ... Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-01T19:22:44Z SecondChildTAG: Hi Myrimit, I would go for Op-Amp. For such noise characteristics, gain and bandwidth consideration, will Instrument op-amp be a better option? SecondChildUserIdTAG: 113365 SecondChildUserNameTAG: satyabrata SecondChildCreateTimeTAG: 2012-12-02T04:14:00Z FirstChildTAG: Do you need to get 80dB@bandwidth12MHz? and noise within this range ? :) Should you design OA from discrete components? If not, you may use alot of OpAmps from well known vendors. For example, will LT1128 or OPA211 satisfy your task requirements? Check OA list [here][1]. [read it first][2] [1]: http://www.analog.com/en/high-speed-op-amps/low-noise-low-distortion-amplifiers/products/index.html [2]: http://www.ti.com/lit/an/slod006b/slod006b.pdf FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-12-01T12:33:46Z SecondChildTAG: Hi Sergtronix, Thanks for your reply. We need 80 dB gain(adjustable over 20 dB) at 3.5 MHz center frequency with 20% error.Currently we have 351 A-3 variable Gain DC Voltage Amplifier by Analog Modules Inc. I'm thinking to design an amplifier with the same specs with discrete components to reduce the cost of whole system. SecondChildUserIdTAG: 113365 SecondChildUserNameTAG: satyabrata SecondChildCreateTimeTAG: 2012-12-01T14:36:34Z SecondChildTAG: my post was deleted... SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-01T20:59:20Z IndexTAG: 1925 TitleTAG: VSM = Aha moment? in practical opamp applications the virtual **short** method **shorts** the time for calculating the circuits. In this sense, I think, it merits the Aha moment status. UserIdTAG: 388524 UserNameTAG: dasbinich CreateTimeTAG: 2012-11-30T22:59:36Z VoteTAG: 1 CoursewareTAG: Week 12 / S23V22 Inverting amplifier input resistance using virtual short method CommentableIdTAG: 6002x_S23V22_Inverting_amplifier_input_resistance_using_virtual_short_method NumberOfReplyTAG: 4 FirstChildTAG: Agree FirstChildUserIdTAG: 132828 FirstChildUserNameTAG: rharris FirstChildCreateTimeTAG: 2012-12-04T05:53:30Z FirstChildTAG: Definitely an Aha moment. FirstChildUserIdTAG: 393930 FirstChildUserNameTAG: tmciver FirstChildCreateTimeTAG: 2012-12-08T04:34:48Z FirstChildTAG: Agreed, as long as you always keep in the back of your head that VSM only works if: 1. A is huge, and 2. The op-amp is included in a negative feedback circuit. FirstChildUserIdTAG: 115880 FirstChildUserNameTAG: mikeaich FirstChildCreateTimeTAG: 2012-12-09T17:46:42Z FirstChildTAG: Definitely Agree!! Working in electronics for 30 years with a much less rigorous understanding than I now have (Thanks so much Professor), the virtual short was THE way to understand OpAmp circuits. FirstChildUserIdTAG: 369519 FirstChildUserNameTAG: perpstud FirstChildCreateTimeTAG: 2012-12-13T04:41:40Z IndexTAG: 1926 TitleTAG: What is the relation of 555 to clocks? The topic of clocks in the course book is explained in page 555 and the teacher said 555 is related to clocks.What do you think? UserIdTAG: 259238 UserNameTAG: omidsadeghi CreateTimeTAG: 2012-11-30T22:43:44Z VoteTAG: 1 CoursewareTAG: Week 13 / S25V19_Clocks_in_digital_systems CommentableIdTAG: 6002x_S25V19_Clocks_in_digital_systems NumberOfReplyTAG: 2 FirstChildTAG: The 555 "Timer" is the basis of a electronic clock. Components use this device as a reference to keep time, similar to an hour glass. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-11-30T22:53:41Z SecondChildTAG: The 555 Timer is associated to clocks which relates to page 555 in the text book. This forms the bedrock of clocks/clocking systems, then the advent of 556 which comprises of 2 clocks in a chip. SecondChildUserIdTAG: 329879 SecondChildUserNameTAG: Ugochukwu SecondChildCreateTimeTAG: 2013-01-09T08:11:05Z FirstChildTAG: Its one of the most popular named 555 IC's for producing astable and monostable RC timed pulses. FirstChildUserIdTAG: 168038 FirstChildUserNameTAG: smoosajee FirstChildCreateTimeTAG: 2012-12-02T06:18:23Z SecondChildTAG: And for unstable clocks. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-12-02T23:24:32Z IndexTAG: 1927 TitleTAG: A question about S21E3: THEVENIN TANK https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_11/Filters/ In the space provided below write an algebraic expression for the complex voltage-transfer ratio VoVi. (Remember, write w for ω.) correct (j*L*w)/((R-C*L*R*w^2)+j*L*w) How does this transfer ratio behave near ω=0? In the space provided below give an algebraic expression that best approximates this behavior. incorrect (j*L*w)/R------ **is this algebraic equal to 0?**while w=0? How does this transfer ratio behave for ω very large? In the space provided below give an algebraic expression that best approximates this behavior. incorrect-----same issue as above? UserIdTAG: 128276 UserNameTAG: ChengBin CreateTimeTAG: 2012-11-30T17:40:14Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You can use limits to solve this. It's not stated that w=0, but that w is GOING towards zero and GOING towards inf. It's important how you write you formulas. I use: s=jw and call C parallel to L: Z=sL/(1+LCs^2) and vo/vi=t=Z/(Z+R) and divide all the terms of t by Z so we will get: t1=1/(1+R/Z). I'll use the w->x as w GOING to x (not the same as w=x !!!) For (w->0)=>(s->0): You can see that in Z, that s^2 is going faster to ZERO then s, therefore write s^2=0 than the limit goes to Z=sL/(1+0)=sL and t1=1(1+R/sL) becomes t1=1/(R/sL)=sL/R=jwL/R because R/sL is increasing fast, so neglect the 1 in (1+R/sL). Now for (w->inf)=>(s->inf) we see that in this case the s^2 is INCREASING faster than s, so we neglect the 1 in (1+LCs^2) and we get: Z=sL/(LCs^2)=1/sC as a limit for w to inf. Now use the original t1 again and write Z=1/sC so we'll get: So for w->inf, t=1/(1+R/(1/sC))=1/sRC=1/jwRC (neglected the 1 again in (1+R/(1/sC)) Hope this will do. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-11-30T22:28:42Z SecondChildTAG: May be you are right. But in math. when w->0, the limit of jwl should be 0. So, I think the answer of o is right, too. Thank you. Salsero SecondChildUserIdTAG: 128276 SecondChildUserNameTAG: ChengBin SecondChildCreateTimeTAG: 2012-12-01T02:30:52Z SecondChildTAG: Here we don't need to solve the limit,it's just near w=0 and w very large(not inf),the key is to neglect smaller and keep the larger ones.We have to consider the size of 1, j*w*L, and 1/(j*w*C). You can get vo/vi=1/( 1+R*(Zl+Zc)/(Zl*Zc) ),Zl=j*w*L,Zc=1/(j*w*C), simplify the expression,compare the three elements,and then the aha moment! This is just one way,more views in the discussion part of S21E3 :) SecondChildUserIdTAG: 361137 SecondChildUserNameTAG: Ericson SecondChildCreateTimeTAG: 2012-12-01T04:05:05Z SecondChildTAG: Hi ChenBin, I agree: if they had ask for the limit, you are totally right. But they didn't, and in fact I didn't use it in all the terms. But I supposed that you wanted to know how you could deduce their answer. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-12-01T13:17:57Z IndexTAG: 1928 TitleTAG: Aha! moment or not **whether the virtual short trick rises to the level of an "Aha!" moment or not???.** ***if yes , vote it*** UserIdTAG: 114864 UserNameTAG: Vikaash CreateTimeTAG: 2012-11-30T11:53:03Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: for me..its yes.. FirstChildUserIdTAG: 114864 FirstChildUserNameTAG: Vikaash FirstChildCreateTimeTAG: 2012-11-30T11:55:36Z SecondChildTAG: for me, too SecondChildUserIdTAG: 374393 SecondChildUserNameTAG: rmaleki SecondChildCreateTimeTAG: 2012-12-14T04:32:11Z FirstChildTAG: Gonna vote no on this one. An "Aha!" moment doesn't just make a preceding calculation go a little faster. "Aha!" moments allow us to solve entirely new classes of problems based on preexisting knowledge, due to knew insight. For example: - The LMD allows us to use algebra to replace Maxwell's equations. - Thevenin and Norton equivalents allow us to abstract uninteresting portions of our circuits away so that we can "measure" the properties of new components. - The Impedance model allows us to use resistive network analysis to analyze RLC circuits. The op-amp abstraction itself counts as an "Aha!", and the virtual short trick is an interesting consequence of it. However, the trick doesn't give us anything new on its own. It belongs in the set of "useful patterns to look out for," along with voltage dividers and MOSFET logic gates. FirstChildUserIdTAG: 139646 FirstChildUserNameTAG: tmcnulty FirstChildCreateTimeTAG: 2012-11-30T21:54:41Z SecondChildTAG: It is just a...... op op,.. opam style ... SecondChildUserIdTAG: 282828 SecondChildUserNameTAG: Chibolator SecondChildCreateTimeTAG: 2012-12-01T20:45:45Z IndexTAG: 1929 TitleTAG: STAFF: Please upload Lecture Slides for Week 11 and onwards I've seen many requests for these in the forums but as yet we only have those through Week 10. Thanks in advance! UserIdTAG: 148542 UserNameTAG: planetscape CreateTimeTAG: 2012-11-30T09:43:01Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi,planetscape. There is one way to get the slides you need.You can find the slides of each week below the videos or exercises,where shows " Lecture Slides Handout [Clean] [Annotated] ".Help this will do. FirstChildUserIdTAG: 361137 FirstChildUserNameTAG: Ericson FirstChildCreateTimeTAG: 2012-12-01T04:13:10Z SecondChildTAG: Ah, good grief! I **AM** blind! THANKS!!! SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-12-01T07:58:58Z SecondChildTAG: Haha,you are welcome :) SecondChildUserIdTAG: 361137 SecondChildUserNameTAG: Ericson SecondChildCreateTimeTAG: 2012-12-01T12:01:23Z IndexTAG: 1930 TitleTAG: I could use a little more help on some problems (not homework related) Hello all, After many attempts I cannot find the second and third phases in S19E4, the last practice problem set of the Sinusoidal Steady State lecture. Could one of you kind folk be willing to give me a step by step of how to do solve it? I have looked at the posts regarding those problems but I still cannot figure out what I am doing wrong. Thanks in advance. UserIdTAG: 302713 UserNameTAG: EVega CreateTimeTAG: 2012-11-30T09:28:42Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: i worked out some formula [here][1] [1]: https://docs.google.com/open?id=0Bzyv_wJHcXnldHlDR2FPay1MTTA FirstChildUserIdTAG: 184153 FirstChildUserNameTAG: tthngn FirstChildCreateTimeTAG: 2012-11-30T20:03:03Z SecondChildTAG: Thank you very much this was very helpful. SecondChildUserIdTAG: 302713 SecondChildUserNameTAG: EVega SecondChildCreateTimeTAG: 2012-12-01T03:59:53Z FirstChildTAG: Hi - find the Real and Imaginary part of Vp/Vi. It is convenient to multiply numerator and denominator by (1-jwRC). You will get an expression in the form Re+jIm - the phase is then simply arctg(Im/Re) - remind that w = 2 x PI x f - the answer must be in radiant express the second phase with 3 significant digits and the third phase with 4 significant digits (digits after ".") Hoping to have been helpful! FirstChildUserIdTAG: 372321 FirstChildUserNameTAG: EnricoDona FirstChildCreateTimeTAG: 2012-11-30T11:15:51Z SecondChildTAG: Thank you for responding EnricoDona, I have: arctan((-w*2*pi)/(1/(R*C)) as my phase, is this correct? It worked to find the first phase but that must have been luck since it doesn't work for the rest. SecondChildUserIdTAG: 302713 SecondChildUserNameTAG: EVega SecondChildCreateTimeTAG: 2012-11-30T12:49:26Z SecondChildTAG: Vp/Vi = Re +j*Im = (wRC)²/[1+(wRC)²] +jwRC/[1+(wRC)²] phase = arctg(Im/Re) = arctg(1/wRC) Best regards Enrico SecondChildUserIdTAG: 372321 SecondChildUserNameTAG: EnricoDona SecondChildCreateTimeTAG: 2012-12-02T20:08:44Z IndexTAG: 1931 TitleTAG: Course Completion? I have joined the course just today(November 30th) and so I haven't completed my home-works, labs, midterm all other stuff. Can i cope up with others and can i complete my previous home-works and labs. What will be mine final score? UserIdTAG: 834454 UserNameTAG: Imthi CreateTimeTAG: 2012-11-30T02:08:13Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hi Imthi, Welcome to the 6.002x;) ! Unfortunately, you can not submit the Homeworks and Labs which its submission deadline have already passed... and late submitions in any format will not be accepted...[Click on here to read Syllabus - Part 5][1] And also you have missed the Midterm Exam... You can read [here][2] what I have said to another student that is in the same situation of you... Although, remember that you can follow this Course and that it will be offered again in the Spring ;) My best wish to you, Myriam. [1]: https://www.edx.org/static/content-mit-6002x/handouts/syllabus.a477535058a1.pdf [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50b7f4be048f0c250000000b FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-30T02:22:09Z FirstChildTAG: Each lecture of this course is a gem in itself. welcome to the world of learning. Forget worrying about your grades. FirstChildUserIdTAG: 623210 FirstChildUserNameTAG: preveen FirstChildCreateTimeTAG: 2012-12-01T11:26:08Z IndexTAG: 1932 TitleTAG: s^2 missing. (at 2:42) In the middle expression, sLC should be (s^2)LC. UserIdTAG: 282828 UserNameTAG: Chibolator CreateTimeTAG: 2012-11-30T00:23:35Z VoteTAG: 1 CoursewareTAG: Week 11 / S22V3: Q indicates peakiness CommentableIdTAG: 6002x_S22V3_Q_indicates_peakiness NumberOfReplyTAG: 1 FirstChildTAG: Yes, you are right :p ![im][1] [1]: https://edxuploads.s3.amazonaws.com/13542361621343636.png FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-30T00:43:55Z IndexTAG: 1933 TitleTAG: Student i want to join in this course from here only, is this possible. thank you UserIdTAG: 834289 UserNameTAG: Kool_Kaish CreateTimeTAG: 2012-11-29T23:50:22Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi Kool_Kaish :) Welcome to 6.002x! Yes, you can join. But unfortunately, you will not be able to get the Certificate as you have missed the Homeworks of the Previous Weeks and the Midtem Exam.... Although, you can take the Final Exam, you will not be able to reach to the minimun total score that you need to pass this Course (60% of the total score - remember that the total score will be based on Homeworks 15%, Labs 15%, Midterm Exam 30% and Final Exam 40% scores), even if you get 100% in the Final Exam, you will not reach to the minimun score to get the Certificate...[read here - Syllabus][1] But you can follow the Course. Also, it will be offered again next Spring (in March) :). I hope this can help you, Take care, Myriam. [1]: https://www.edx.org/static/content-mit-6002x/handouts/syllabus.a477535058a1.pdf FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-30T00:11:23Z IndexTAG: 1934 TitleTAG: Thanks profs for a blast! I need this humorous WBUL to attenuate all the BS (read band-stop) made in my calcs. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-11-29T20:18:14Z VoteTAG: 1 CoursewareTAG: Week 11 / S21V10: DEMO AM Radio receiver CommentableIdTAG: 6002x_S21V10_DEMO_AM_Radio_receiver NumberOfReplyTAG: 0 IndexTAG: 1935 TitleTAG: plzz help H11P1 how to proceed for bandwidth calc..? UserIdTAG: 183166 UserNameTAG: yogeshk CreateTimeTAG: 2012-11-29T14:18:01Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hi yogeshk, Can I help you? You can take a look at this Hints [read here][1] Hint: Bandwith = wo/Q Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50b54fce5fb7411f00000006 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-29T16:51:12Z SecondChildTAG: No need to calculate w0 and Q. Just read off answer. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-11-29T19:22:57Z SecondChildTAG: Yes, that is true @skyhawk... But is better to know why is that ;). If you know How is defined the Bandwith you will reach to the answer based on the characteristic equation coefficients that are in terms of the Circuit elements :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-29T19:44:50Z SecondChildTAG: I've watched the videos, read the book, and followed the math. It's not clear to me that saying bandwidth = w0/Q is any more of an explanation that saying that the bandwidth is given by the coefficient of s when the denominator is in canonical form. As the saying goes, "six of one; half a dozen of the other." SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-11-29T19:54:55Z SecondChildTAG: Hi @skyhawk, take a look at page 794 of the Textbook [read here][1] ;). I will copy and paste this, in order to be more easy to see this (no copyright infringement intended). Take a look at 14.48 and how they reach to that definition of Bandwith :) ![enter image description here][2] Take care, Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/818 [2]: https://edxuploads.s3.amazonaws.com/1354219301134369.bmp SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-29T20:05:29Z SecondChildTAG: Ah no! The bandwidth is defined on page 793 right below equation 14.46, and it is expressed in terms of alpha at the bottom of that page. On the page that you show it, is simply related to the previously defined Q. Note that in his video lectures on bandwidth the professor comments that an alternate definition of Q is Q = w0/BW, i.e. Q defined in terms of w0 and bandwidth not the other way around. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-11-30T00:42:51Z SecondChildTAG: I am agree that the bandwith is defined on page 793 as the width of the curve between the two 0.707 frecuencies... But can I ask you a question, I am tempted to ask you this haha, just for curious, without any intention of offense, you can or not answer me , are you always that stubborn ?:P See you, Myriam. P.D. Ah! Don´t forget to participate in the CECC 2 Contest @skyhawk ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-30T01:41:29Z FirstChildTAG: Compute the transfer function, put the denominator in canonical form, and read off the answer. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-29T15:23:19Z IndexTAG: 1936 TitleTAG: Elements used in input and feedback path Why is a series combination of Resistor and Capacitor used in input side and a parallel combination of Resistor and Capacitor used in Feedback path? Does the combination vary with type of filter required ,that is Band pass,High pass ,Low pass,Band Stop Filter ? UserIdTAG: 343388 UserNameTAG: bijojoseph CreateTimeTAG: 2012-11-29T05:50:14Z VoteTAG: 1 CoursewareTAG: Week 12 / S24V10 Op amps and impedances CommentableIdTAG: 6002x_S24V10_Op_amps_and_impedances NumberOfReplyTAG: 1 FirstChildTAG: You probably don't see a parallel RC in the input side because at high frequencies when the cap shorts out you would have a voltage source competing with a virtual ground which would not be good. You do see a series RC in the feedback path in something called an integral network used quite often in feedback control systems. FirstChildUserIdTAG: 397595 FirstChildUserNameTAG: mholin FirstChildCreateTimeTAG: 2012-11-29T12:41:46Z IndexTAG: 1937 TitleTAG: Plots appear to be wrong I tried sandbox too but i got different results. UserIdTAG: 477713 UserNameTAG: ikm104 CreateTimeTAG: 2012-11-28T18:27:24Z VoteTAG: 1 CoursewareTAG: Week 11 / Designing a frequency response CommentableIdTAG: 6002x_Designing_a_frequency_response NumberOfReplyTAG: 0 IndexTAG: 1938 TitleTAG: Lab 12 Hi all, How do you set up the equations of this problem to find the value of the resistors and condensers? Does f0=10KHz+/-100 means that the bandwidth is 200 Hz with f0 centered a 100KHz? I have the expressions for W0 and the Bandwidth, in terms of the resistors and condensers, but have not been able to move an inch for days. Any hint will be very welcomed. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-11-28T01:35:09Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: f0 = 10KHz +/- 100 means the center frequency is allowed to be in the range from 10000 - 100 Hz to 10000 + 100 Hz. They specified this because they ask you to use standard value resistors and capacitors so you need to find a combination that is close enough. The bandwidth you get from the given value of Q. FirstChildUserIdTAG: 397595 FirstChildUserNameTAG: mholin FirstChildCreateTimeTAG: 2012-11-28T02:02:28Z IndexTAG: 1939 TitleTAG: Where did that w^2 come from? The prof. says $(wRC)/sqrt((1-w^2LC)^2+(w^2RC)^2)$ but where did he get the $w^2$ in $(w^2RC)$ from? Shouldn't this be $(wRC)$ ? Edit: Right, Prof. made a mistake :) UserIdTAG: 114881 UserNameTAG: xvink CreateTimeTAG: 2012-11-27T19:23:42Z VoteTAG: 1 CoursewareTAG: Week 11 / S21V5: Series RLC bandpass CommentableIdTAG: 6002x_S21V5_Series_RLC_bandpass NumberOfReplyTAG: 1 FirstChildTAG: S21V5 : **8:17 min** ---------- Prof. Agarwal says: *Oh, wait a second here. How did I get that omega squared? If you had a live class going on, you would have caught me there. So this square should not have been there.* ---------- :) It should be: $(wRC)/sqrt((1-w^2LC)^2+(wRC)^2)$ FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-27T23:27:26Z IndexTAG: 1940 TitleTAG: Doubt in S21E3 I am even having difficult to understand why, in this circuit below, I have, in the first case, **(j*L*w)/R** to transfer function ratio (w=0). In the second case, I have **1/(R*C*j*w)** to transfer function ratio (w very large). In both cases the transfer fuction would be 0. Someone can explain for me, please? Thank you for something. ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/1354043418134365.png UserIdTAG: 366669 UserNameTAG: pedroramus CreateTimeTAG: 2012-11-27T19:15:35Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: see discussion right under S21E3 FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-11-27T21:01:03Z FirstChildTAG: This topology is an RLC bandpass filter, which attenuates gains at very high and very low frequencies (which is why the gain equated to zero at omega = infinity and omega = zero) but has a higher gain in the pass band. Check out the bode plot shown here: http://www.mathworks.com/help/control/ug/analyzing-the-response-of-an-rlc-circuit.html FirstChildUserIdTAG: 391929 FirstChildUserNameTAG: RohanNagarkar FirstChildCreateTimeTAG: 2012-11-27T21:21:57Z IndexTAG: 1941 TitleTAG: S21E3 Im stuck with part 1. Final equation should be $\frac{V_O}{V_I}$=$\frac{Z_{LC}}{R+Z_{LC}}$, where $Z_{LC}$=$\frac{Z_L*Z_C}{Z_L+Z_C}$=$\frac{j*w*L*\frac{1}{j*w*C}}{j*w*L+\frac{1}{j*w*C}}$=$\frac{\frac{L}{C}}{1-w^2LC}$. From this I have final equation $\frac{V_O}{V_I}$=$\frac{L}{L+R*C*(1-w^2*L*C)}$ Where am I wrong? UserIdTAG: 205311 UserNameTAG: Sergtronix CreateTimeTAG: 2012-11-27T15:56:24Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: ( (j*w*L) || (1/(j*w*C)) ) / (R + ( (j*w*L) || (1/(j*w*C)) )) --- It makes sense to leave it in this form so that you can see better what is going on. --- As $\omega$ nears $0$, the capacitor becomes irrelevant ($Z_C$ is very, very large) and can be ignored. The inductor and resistor form a voltage divider, $= {{Z_L} \over {R +Z_L}}$, but since $\omega$ is very small, $Z_L \ll R$. This means $Z_L$ can be removed from the denominator, since it is insignificant there wrt $R$. It can not be removed from the numerator, because it is the only term. Thus we are left with $= {{Z_L} \over {R}}$. As $\omega$ becomes very large, the inductor becomes irrelevant ($Z_L$ is very, very large) and can be ignored. The capacitor and resistor form a voltage divider, $= {{Z_C} \over {R +Z_C}}$, but since $\omega$ is very large, $Z_C \ll R$. This means $Z_C$ can be removed from the denominator, since it is insignificant there wrt $R$. It can not be removed from the numerator, because it is the only term. Thus we are left with $= {{Z_C} \over {R}}$. FirstChildUserIdTAG: 9673 FirstChildUserNameTAG: xp42 FirstChildCreateTimeTAG: 2012-11-27T16:13:33Z SecondChildTAG: xp42, thank you! But I do worry about difference between my equation and "reference" : (j*L*w)/((R-C*L*R*w^2)+j*L*w) SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-11-27T16:22:19Z SecondChildTAG: When you multiply $Z_{LC}$ by j*w*C (top and bottom) you should get $Z_{LC} = \frac {j \omega L}{1- {\omega}^2 C}$, not what you got. SecondChildUserIdTAG: 9673 SecondChildUserNameTAG: xp42 SecondChildCreateTimeTAG: 2012-11-27T16:36:20Z SecondChildTAG: oops, that should be $Z_{LC} = \frac {j \omega L}{1 - {\omega}^2 LC}$ SecondChildUserIdTAG: 9673 SecondChildUserNameTAG: xp42 SecondChildCreateTimeTAG: 2012-11-27T16:38:23Z SecondChildTAG: Ok, so my fault was to cancel jw here $\frac{j*w*L}{j*w*C}=\frac{L}{C}$ Thank you! Hmm But I MUST refresh this agebra part in my head.. SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-11-27T16:50:56Z IndexTAG: 1942 TitleTAG: H11P3 Graph Hi guys! I need help to obtain the fundamental frequency (w0)from the graph. The x axis is not clear, is it x 10^4 s or X10^-4 or x10^-1 on the horizontal axis? UserIdTAG: 143823 UserNameTAG: NathanNadeson CreateTimeTAG: 2012-11-27T15:13:48Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: It's x $10^{-4}$. (It's clear if you view the image at full size.) You shouldn't have to read $\omega_0$ from the graph, all you need can be read from the second graph. FirstChildUserIdTAG: 141000 FirstChildUserNameTAG: OrinE FirstChildCreateTimeTAG: 2012-11-27T19:06:40Z SecondChildTAG: thanks ! I just realised that the last problem can be solved without w0. SecondChildUserIdTAG: 143823 SecondChildUserNameTAG: NathanNadeson SecondChildCreateTimeTAG: 2012-11-27T20:59:35Z IndexTAG: 1943 TitleTAG: question to S21V5 Hi, I don't understand why the counter is unaccounted for partition in realpart and imaginärpart ![enter image description here][1] can anyone explain? [1]: https://edxuploads.s3.amazonaws.com/13540265081343672.jpg UserIdTAG: 389333 UserNameTAG: juergen CreateTimeTAG: 2012-11-27T14:27:25Z VoteTAG: 1 CoursewareTAG: Week 11 / S21V5: Series RLC bandpass CommentableIdTAG: 6002x_S21V5_Series_RLC_bandpass NumberOfReplyTAG: 2 FirstChildTAG: Hi! Could you please rephrase, because i didn't understand the question? FirstChildUserIdTAG: 304587 FirstChildUserNameTAG: Vasso FirstChildCreateTimeTAG: 2012-11-27T21:08:07Z SecondChildTAG: I will try... you can see in the counter w*R*C. To get the absolute value |Vo/Vi| you have to square the imaginary part and the real part. but what is with the counter? I guess there have to be a 1. SecondChildUserIdTAG: 389333 SecondChildUserNameTAG: juergen SecondChildCreateTimeTAG: 2012-11-27T21:44:38Z FirstChildTAG: Hi juergen, I didn´t undestand your question... Did you mean why the w*R*C is in the numerator ? The |Vo/Vi| is the module of Vo/Vi . You also know that the module of a complex number is the $\sqrt((realpart)^2 + (imaginarypart)^2)$ - In the numerator your complex number will be: 0 + j*w*R*C -So, your real part is 0 -and your imaginary part is w*R*C Now, what is your module in your numerator? Recall the definition of module: $\sqrt((realpart)^2 + (imaginarypart)^2)$ $\sqrt((0)^2 + (w*R*C)^2) = \sqrt((w*R*C)^2) = w*R*C$ **So, your module in your numerator will be w*R*C** ---------- - In the denominator your complex number will be: 1-w^2*L*C + j*w*R*C -So, your real part is 1-w^2*L*C -and your imaginary part is w*R*C Now, what is your module in your denominator? Recall the definition of module: $\sqrt((realpart)^2 + (imaginarypart)^2)$ $\sqrt((1-w^2*L*C)^2 + (w*R*C)^2)$ **So, your module in your denominator will be $\sqrt((1-w^2*L*C)^2 + (w*R*C)^2)$ ** ---- So, your |Vo/Vi| will be the division of the module of your numerator divided by your module of your denominator :). So, that is why you have in the numerator w*R*C nad not 1 ... Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-28T02:32:34Z IndexTAG: 1944 TitleTAG: misleaded Here I tried to applied voltage divider, but starting from vo-vI. ThenI realized, I can't make assumptions about vo(like vo being greater than vI. UserIdTAG: 369339 UserNameTAG: vargaslen CreateTimeTAG: 2012-11-27T09:31:32Z VoteTAG: 1 CoursewareTAG: Week 12 / S23V18 Inverting amplifier analysis CommentableIdTAG: 6002x_S23V8_Inverting_amplifier_analysis NumberOfReplyTAG: 1 FirstChildTAG: Good for you. FirstChildUserIdTAG: 90796 FirstChildUserNameTAG: ChaunceyGardiner FirstChildCreateTimeTAG: 2012-11-27T10:56:47Z SecondChildTAG: ChaunceyGardiner: if I don't recall bad, your alias is due to a character played by Peter Sellers in the movie "From the Garden", good for you! SecondChildUserIdTAG: 369339 SecondChildUserNameTAG: vargaslen SecondChildCreateTimeTAG: 2012-11-27T22:22:48Z SecondChildTAG: Close - it's from the Peter Sellers movie "Being there". ;o) SecondChildUserIdTAG: 90796 SecondChildUserNameTAG: ChaunceyGardiner SecondChildCreateTimeTAG: 2012-11-28T04:15:25Z SecondChildTAG: Haha :) @ChaunceyGardiner I always wondered why you haven´t participated more in the 6.002x Fall hahaha, you are smart, awesome and genuine person that I could meet in 6.002x Spring!- without any filters haha, but authentic haha- Take care, Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-28T04:45:40Z SecondChildTAG: Thank you, Myriam. I like you too. As for participating in the fall - I would have been here more if the forum was better. They made some improvements (the most important one was dropping the karma system), but they lost other important parts of what we had in the spring. I find it very hard to keep track of things in here now, and all the good discussions are gone before they get started. It seems that "I have a problem" - "Here you go" is the limit to how long discussions will get now. SecondChildUserIdTAG: 90796 SecondChildUserNameTAG: ChaunceyGardiner SecondChildCreateTimeTAG: 2012-11-29T05:03:14Z SecondChildTAG: ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-29T12:29:20Z SecondChildTAG: Oh, I should add that I think it's a major improvement that they now appoint community TAs. SecondChildUserIdTAG: 90796 SecondChildUserNameTAG: ChaunceyGardiner SecondChildCreateTimeTAG: 2012-11-29T21:50:03Z IndexTAG: 1945 TitleTAG: some one help!! can any one solve this: sin(z)=10 where z is a real number. can it have infinite number of complex solutons or no solution.... UserIdTAG: 563741 UserNameTAG: Mukundsingh89 CreateTimeTAG: 2012-11-26T16:17:05Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: If z is real: no solution. If z is complex: infinite solutions. You can check the following link for the calculations: http://www.suitcaseofdreams.net/Sine_Complex.htm FirstChildUserIdTAG: 376877 FirstChildUserNameTAG: AndBre FirstChildCreateTimeTAG: 2012-11-26T17:16:38Z IndexTAG: 1946 TitleTAG: Bridge Resonance: Millenium "wobbly" Bridge London The Millenium Bridge in London closed three days after opening, for safety reasons, as the initial irregular footsteps of walkers crossing the bridge forced it into resonance. Opened two years later after the detailed investigation and redesign with a spend of £5m on lateral dampers. Excellent BBC program "Science Shack - Millennium Bridge" on Bridge resonance, positive feedback (people as"tuned active exciters")damping and solutions, starts 2:41 minutes in and continues after the "unplanned programme interruption" http://www.youtube.com/watch?v=eiaM_LZUsqM Maths here: http://www2.eng.cam.ac.uk/~den/ICSV9_06.htm UserIdTAG: 273287 UserNameTAG: johnkh77 CreateTimeTAG: 2012-11-26T13:41:01Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1947 TitleTAG: Aha Given the beauty of negative feedback, it should definitely be an aha moment. UserIdTAG: 397595 UserNameTAG: mholin CreateTimeTAG: 2012-11-26T13:00:30Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I think the AHA should be when they show us how the Op Amp is built. FirstChildUserIdTAG: 288174 FirstChildUserNameTAG: matiasgrodriguez FirstChildCreateTimeTAG: 2012-11-26T15:38:11Z FirstChildTAG: An Op Amp is actually just a bunch of simple components assembled onto one die. Here is a schematic for the classic 741 Op Amp. The 741 Op Amp was the ubiquitous for a a long time, found in all sorts of electronics. It has since been subsided by more modern Op Amps such as the 4558, which are similar but of higher quality. Red: Current mirrors Blue: Differential amplifier Magenta: Class A gain stage Green: Voltage level shifter Cyan: Output stage ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13540192981343616.png Right away you will notice that it is just a bunch of transistors and resistors assembled as a mini amplifier. I even see a tiny capacitor on there. Typically, any capacitors needed to complete the amplifier are attached externally, due to large size and shape. You could literally make this "chip" on a piece of plywood using discrete (separate) transistors and resistors with the attached schematic. Of course you can take a single chip and combine it with another chip to make something more complex. Things like clock radios have specific chips made that not only incorporate a amplifier and radio tuner, but also contain the electronics for the clock and LED display. Some peripheral still need to be attached, like coils and potentiometers, but for the most part everything is contained on a single chip. http://www.wisc-online.com/objects/ViewObject.aspx?ID=SSE2903 FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-11-27T12:43:39Z IndexTAG: 1948 TitleTAG: Prof Agawar mentioned many time about the Appendix. How can I get it. I searched the whole textbook, but I still can't find where the appendix is. Who would like to help me? Thansk first. UserIdTAG: 128276 UserNameTAG: ChengBin CreateTimeTAG: 2012-11-26T05:04:25Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: You will find it in the last part of the textbook ,refer to Page 927,there are four parts A1,B,C,D.:) FirstChildUserIdTAG: 361137 FirstChildUserNameTAG: Ericson FirstChildCreateTimeTAG: 2012-11-26T05:38:54Z SecondChildTAG: Got it . thanks Ericson SecondChildUserIdTAG: 128276 SecondChildUserNameTAG: ChengBin SecondChildCreateTimeTAG: 2012-11-26T09:53:05Z FirstChildTAG: **Appendix A .** Maxwell´s Equations and the LMD[Click on here to read][1] **Appendix B.** Trigonometric Functions and Identities [Click on here to read][2] **Appendix C.** Complex Numbers[Click on here to read][3] **Appendix D.** Solving simultaneus Linear Equations[Click on here to read][4] :) [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/ [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/965 [3]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/971 [4]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/981 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-26T18:09:01Z IndexTAG: 1949 TitleTAG: what a great time! After couples of hours and hours of suffering from making kinds of small mistakes ,I finally got the right answer of homework 10 Part 3 Q1&Q2.Wow!It feels like I am reborn,I can have a long long sigh now!Thanks for all helpful hints on the discussion part.I really learned a lot from these posts.I am just too careless to spend so much time.Wish you good luck, my classmates!Go on to the following weeks! UserIdTAG: 361137 UserNameTAG: Ericson CreateTimeTAG: 2012-11-26T05:01:09Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1950 TitleTAG: :) the big "A-ha" moment! :) UserIdTAG: 385692 UserNameTAG: tpfslima CreateTimeTAG: 2012-11-26T01:42:40Z VoteTAG: 1 CoursewareTAG: Week 10 / RC Example Impedance Model Part 2 CommentableIdTAG: 6002x_RC_example_impedance_model_part_2 NumberOfReplyTAG: 0 IndexTAG: 1951 TitleTAG: Lab 10 How to convert decibels? I tried to convert magnitude which is equals 14.145 decibels. I tried this way: Vo/Vi=10^(-14.145/20). But I couldn't get right answer. What is wrong? UserIdTAG: 229266 UserNameTAG: Alexez CreateTimeTAG: 2012-11-25T16:01:14Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: The conversion from decibels to volts looks correct. Maybe the problem is somewhere else. FirstChildUserIdTAG: 288174 FirstChildUserNameTAG: matiasgrodriguez FirstChildCreateTimeTAG: 2012-11-25T16:15:24Z SecondChildTAG: @Alexez: the frequency on the magnitude graph is also on a log(Hz) scale! This might be the mistake :) SecondChildUserIdTAG: 252722 SecondChildUserNameTAG: vaboro SecondChildCreateTimeTAG: 2012-11-25T19:57:44Z FirstChildTAG: Right! But when I tried to go this direction I couldn't come anywhere, and this is why I used the $arctan$ or the $tan$-inverse way. Let's try to go your way together for I'm quite curious how to come to the write answer by following it. So, let me begin: $H_{RC}(j\omega) = \frac{V_o}{V_i}=\frac{1/RC}{j\omega + 1/RC}$ $H_{RL}(j\omega) = \frac{V_o}{V_i}=\frac{j\omega}{j\omega + R/L}$ I will first use the $RC$ formula, correct me if I'm wrong: $\frac {V_o}{V_i} = \frac{1}{j\omega RC + 1}$ $\frac {V_o}{V_i}\cdot (j\omega RC + 1) = 1$ $\frac {V_o}{V_i}\cdot j\omega RC + \frac {V_o}{V_i}-1 = 0$ $\left( \frac {V_o}{V_i}-1\right)+ j\cdot \left(\frac {V_o}{V_i}\cdot \omega RC \right) = 0$ And, unfortunately, I have no idea where to go next... FirstChildUserIdTAG: 252722 FirstChildUserNameTAG: vaboro FirstChildCreateTimeTAG: 2012-11-25T16:40:08Z SecondChildTAG: Actually you have to calculate the magnitude $|\frac{Vo}{Vi}|$. For the capacitor you will get a formula in terms of w, C and R and for the inductor you will get a formula in terms of w, L and R. From there is straightforward to look the given magnitude plot and solve the equations to get C and L. Just remember that you must convert from decibels to volts and from log(Hz) to rad/sec ... this also works for the last part! SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-25T17:19:53Z SecondChildTAG: Thanks vaboro! Your way is the best way))) SecondChildUserIdTAG: 229266 SecondChildUserNameTAG: Alexez SecondChildCreateTimeTAG: 2012-11-25T18:13:41Z SecondChildTAG: @matiasgrodriguez: I thought the magnitude $\lvert \frac {V_o}{V_i} \rvert$ can actually be found from the magnitude graph, not calculated. Yet, you may be right in that I should have transformed the transfer function differently: $\frac{1}{j\omega RC + 1}=\frac{j\omega RC - 1}{(j\omega RC + 1)(j\omega RC - 1)}=\frac{1-j\omega RC }{\omega^2 R^2C^2 + 1}= \frac{1}{\omega^2 R^2C^2 + 1}-j\cdot \frac{wRC}{\omega^2 R^2C^2 + 1}$ Now we can compute the magnitude of this complex number and equate it to the figure found in the graph: $\sqrt{\left(\frac{1}{\omega^2 R^2C^2 + 1}\right)^2+\left(\frac{-wRC}{\omega^2 R^2C^2 + 1}\right)^2}=\sqrt{\frac{1+w^2R^2C^2}{(1+w^2R^2C^2)^2}}=\frac{1}{\omega^2 R^2C^2 + 1}$ Now let's see if I can come to anything... SecondChildUserIdTAG: 252722 SecondChildUserNameTAG: vaboro SecondChildCreateTimeTAG: 2012-11-25T19:11:22Z SecondChildTAG: $\frac{1}{\sqrt{\omega^2 R^2C^2 + 1}}$ SecondChildUserIdTAG: 252722 SecondChildUserNameTAG: vaboro SecondChildCreateTimeTAG: 2012-11-25T19:34:55Z SecondChildTAG: Yep, the formula from your last comment (with the square root) is the formula $|\frac{Vo}{Vi}|$ for the circuit with the capacitor. Now look the given Magnitude plot. The vertical axis is $|\frac{Vo}{Vi}|$ in decibels. Alexez already told you how to convert from decibels to volts in the first post... the horizontal axis is $\omega$ but in log(Hz) ... just convert that to rad/sec. You are done: one formula, only one unknown. Find C. Do the same for L (find the magnitude formula, etc). Please note that the $|\frac{Vo}{Vi}|$ values on the magnitude plot are different for C and L. SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-25T19:48:14Z SecondChildTAG: @matiasgrodriguez: You're absolutely right about being careful about log(Hz)! :) I've been making this mistake again, again and again! :) haha SecondChildUserIdTAG: 252722 SecondChildUserNameTAG: vaboro SecondChildCreateTimeTAG: 2012-11-25T19:53:10Z SecondChildTAG: Yep... It took me some time to correctly convert from decibels and log(Hz). Did you get the correct answer? SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-25T20:04:21Z SecondChildTAG: @matiasgrodriguez:Yes, I got the correct answer a long time ago:) but I used the tan-inverse function and the phase graph. Now, I realize I could have used the transfer function and the magnitude graph if it hadn't been the mistake with log(Hz) frequency! :) SecondChildUserIdTAG: 252722 SecondChildUserNameTAG: vaboro SecondChildCreateTimeTAG: 2012-11-25T20:32:48Z SecondChildTAG: Cool :) You can build the circuit from the last problem very easily with this method... and no need to tweak values. SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-25T20:41:29Z IndexTAG: 1952 TitleTAG: lab 10... cant connect components hey there.... its kind of like a really silly problem that i am having but its been a real pain. I have solved all the problems. but the problem is i am unable to connect my components through wires. the wire just wont appear. please help me get through this. Thanx alot UserIdTAG: 117319 UserNameTAG: iqramalik CreateTimeTAG: 2012-11-25T10:43:53Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: how u have calculated break frequency plz help me i m not able to calculate :( :( FirstChildUserIdTAG: 158367 FirstChildUserNameTAG: brainyash1990 FirstChildCreateTimeTAG: 2012-11-25T11:03:58Z SecondChildTAG: in the answer area of 1st part i m writing 10^4.6*2*pi but's it's not showng me the green tick where m i wrong?? :( SecondChildUserIdTAG: 158367 SecondChildUserNameTAG: brainyash1990 SecondChildCreateTimeTAG: 2012-11-25T11:11:26Z SecondChildTAG: you have to write expression for break freq. ,not value SecondChildUserIdTAG: 64656 SecondChildUserNameTAG: AbuZar SecondChildCreateTimeTAG: 2012-11-25T11:29:12Z SecondChildTAG: attention, the showing value is not the break frequency, it is a random value. You have to estimate break frequency, look at the phase. SecondChildUserIdTAG: 389333 SecondChildUserNameTAG: juergen SecondChildCreateTimeTAG: 2012-11-25T14:52:09Z SecondChildTAG: Can I help you brainyash1990 ? SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-25T15:12:46Z FirstChildTAG: Hi iqramalik, Can I help you ? Try to Refresh with F5. If this does not work try to log Out and then Log In.... Can you Post a screen shot ? Myriam FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-25T14:58:42Z SecondChildTAG: thankyou myriam.... the suggestions you gave worked for me.... till now...:) SecondChildUserIdTAG: 117319 SecondChildUserNameTAG: iqramalik SecondChildCreateTimeTAG: 2012-12-09T18:07:43Z IndexTAG: 1953 TitleTAG: proctored final exam Need more information like test centre,fees etc UserIdTAG: 153869 UserNameTAG: rabindra CreateTimeTAG: 2012-11-25T09:57:38Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1954 TitleTAG: H11P1 Anyone,please? A question,please: I wonder if as long as I get the canonical form (characteristic equation for the denominator), doesn't matter what I have in the numerator? the rule always apply? thanks! UserIdTAG: 369339 UserNameTAG: vargaslen CreateTimeTAG: 2012-11-25T09:46:31Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi vargaslen, Yes, it doesn´t matter what you have in the numerator :). The characteristic equation is based on the denominator. You can also take a look at this Hints of H11P1 [here][1] Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50b54fce5fb7411f00000006 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-28T01:02:21Z IndexTAG: 1955 TitleTAG: Errata for text/pr15.3 Is there an errata for the text...tutorial and text ans for pr 15.3 seem to differ. UserIdTAG: 211150 UserNameTAG: Bernie1961 CreateTimeTAG: 2012-11-24T19:34:39Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 3 NumberOfReplyTAG: 2 FirstChildTAG: https://6002x.mitx.mit.edu/wiki/view/BookErrors FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-11-25T12:22:29Z FirstChildTAG: I've also got 15 volts as an answer before watching the tutorial.Even with the errata, there are a lot of other errors.But , it's a 1000 pages book , so it is understandable. FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-12-18T15:20:31Z IndexTAG: 1956 TitleTAG: H10P3 - last 4 questions The formula for the driving point impedance (Zdi) is clear. I'm having trouble with making it equal to R1 as the hint proposes. My idea on this is the following: as R1 is a clearly real value, this means that if Zdi = R1, then the real part of Zdi is equal to the value of R1, and the imaginary part of Zdi is equal to 0. So, by this method I have 2 equations for the 2 unknowns (Cmatch and Lmatch). The only problem is that this method seems way too difficult to solve. Can anyone suggest a maybe easier way, please? Is there a "trick" to dodge the above mentioned mess? UserIdTAG: 129309 UserNameTAG: Memphis823 CreateTimeTAG: 2012-11-24T18:05:27Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hi Memphis823, parts 3 and 4 it has been discussed here: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/undefined/threads/50a688f55b77f62300000035 two equations, two unknowns. FirstChildUserIdTAG: 288174 FirstChildUserNameTAG: matiasgrodriguez FirstChildCreateTimeTAG: 2012-11-24T18:15:41Z FirstChildTAG: It would be really messy that way since you would be taking the real and imaginary parts of a messy fraction. Instead, set Zdi=R1, multiply both sides by the denominator of the fraction, then your two equations come from the real part of the left hand side being equal to the real part of the right hand side; same for the imaginary parts. FirstChildUserIdTAG: 141000 FirstChildUserNameTAG: OrinE FirstChildCreateTimeTAG: 2012-11-24T18:56:05Z SecondChildTAG: Thanks very much, this way I could work it out! SecondChildUserIdTAG: 129309 SecondChildUserNameTAG: Memphis823 SecondChildCreateTimeTAG: 2012-11-25T09:50:32Z IndexTAG: 1957 TitleTAG: LAB 10 i am not getting the value of **L**..i have found the value of C by putting the value of w in the phase equation and doing the same procedure of L..but getting RED cross..can anyone help me? UserIdTAG: 114864 UserNameTAG: Vikaash CreateTimeTAG: 2012-11-24T17:57:20Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 6 FirstChildTAG: I am facing the same problem...the values works correctly even got the exact graph...but no green tick.. FirstChildUserIdTAG: 122513 FirstChildUserNameTAG: Mona77 FirstChildCreateTimeTAG: 2012-11-25T09:52:13Z FirstChildTAG: Hi Vikaash, Take a look at [here][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50b10a80244a1c2b0000002a FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-24T18:48:06Z FirstChildTAG: Try to simulate it at the sandbox FirstChildUserIdTAG: 113249 FirstChildUserNameTAG: EGuarch FirstChildCreateTimeTAG: 2012-11-24T18:42:24Z FirstChildTAG: I checked both my answers (for L and C) in the sand box and they work just fine yet the lab check indicates the answers are wrong! Also whenever I click on a link in this forum it goes nowhere! FirstChildUserIdTAG: 369519 FirstChildUserNameTAG: perpstud FirstChildCreateTimeTAG: 2012-11-25T02:07:55Z FirstChildTAG: I calculate L cnd C then simulate in sandbox just fine yet Lab Check marks it wrong! FirstChildUserIdTAG: 369519 FirstChildUserNameTAG: perpstud FirstChildCreateTimeTAG: 2012-11-25T02:15:52Z FirstChildTAG: even i have the same problem... the values work correctly in the simulator..but not in the check boxes... FirstChildUserIdTAG: 36121 FirstChildUserNameTAG: nikhilkumar FirstChildCreateTimeTAG: 2012-11-25T09:00:46Z IndexTAG: 1958 TitleTAG: How are RLC circuits "linear"? To double check my understanding of the word "linear"... In the impedance chapter it says we can use the impedance model because the circuits are "linear". Am I correct that these RLC circuits are only "linear" when we are looking at sinusoid inputs and complex numbers for the voltage and current outputs. But these circuits are **not** linear for arbitrary input signals, like the simple circuits in early chapters were linear -- the circuits that had only resistors and independent sources? Rob UserIdTAG: 468623 UserNameTAG: RobNik CreateTimeTAG: 2012-11-24T16:19:03Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: They are linear. The superposition principle applies. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-24T16:49:45Z FirstChildTAG: The circuits do not use "active" components (semi-conductors), cannot amplify or add power nor do they have sharp turn-on/turn-off points such as a diode. The inductor or capacitor cannot add power; but they can store power and return it to the circuit. FirstChildUserIdTAG: 375500 FirstChildUserNameTAG: Brej7665 FirstChildCreateTimeTAG: 2012-11-24T16:50:19Z IndexTAG: 1959 TitleTAG: H10P3 i am stuck at this problem..i am getting a huge expression and even gone through textbook but still getting wrong answer..can anyone help me?? UserIdTAG: 114864 UserNameTAG: Vikaash CreateTimeTAG: 2012-11-24T14:58:22Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: For part 1 simple numerator, rather long denominator. For part 2 complex expressions in numerator and denominator but neither is large. Where are you having difficulty? FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-24T16:18:38Z SecondChildTAG: finally got the correct answer... SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-11-25T08:57:25Z SecondChildTAG: Very good!! SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-11-25T20:27:30Z FirstChildTAG: look at page 724, there is an good example. Maybe you should 1/jwC and jwC replace at the end. Make steps This expression is not so big, but also not small FirstChildUserIdTAG: 389333 FirstChildUserNameTAG: juergen FirstChildCreateTimeTAG: 2012-11-24T16:00:04Z IndexTAG: 1960 TitleTAG: myrimit's heelp needed!! when ur hint's will be out because i m finding some difficulty in solving lab10 and H10q3 UserIdTAG: 158367 UserNameTAG: brainyash1990 CreateTimeTAG: 2012-11-24T14:32:33Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: Hi there, As to my experience with this lab,there are more than one way to build the requested filter. One way is derived from the two filters in Q1 (fig 1).Find L with first break frequency and C with the second break frequency (individually).then combine them one after the other(cascading).Here you start with the two resistances (R1,R1) at the same value, but you will leave one of them the same, and play with the other (hint: try to approach the original scenario, the one from which you got C and L individually, by reducing one of the two resistances).Good Luck! FirstChildUserIdTAG: 369339 FirstChildUserNameTAG: vargaslen FirstChildCreateTimeTAG: 2012-11-24T15:06:06Z FirstChildTAG: @vargasien i tried findind out the break frequency but failed :'( i m trying with 10^4.6*2*pi but my answer is not coming what might be the problem?? FirstChildUserIdTAG: 158367 FirstChildUserNameTAG: brainyash1990 FirstChildCreateTimeTAG: 2012-11-24T15:51:52Z SecondChildTAG: brainyash1990, break freq looks correct... how did you calculate the $|\frac{Vo}{Vi}|$? I mean, did you correctly converted from dB to Volts? SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-24T16:12:46Z FirstChildTAG: I will try to Post it. I were with Exams at my University. I hope that I can Post it...Today I have a Wedding :). Oh, my, I haven´t done my homework from 3.091x :P. Help GladIDothis haha! And also we have a Good News for you ! We will be announcing this soon with some students of 6.002x Spring 2012:) Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-24T15:49:44Z SecondChildTAG: Miriyam, is it YOUR WEDDING DAY??? SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-11-24T16:46:50Z SecondChildTAG: Oh no, haha! It is not my Wedding haha! Is my cousin Wedding, she is so happy :). She always dream with this day so I guess this must be a special day for her (Church at night + Party at night-morning ). I wish her the best. I hope to not cry as I am sensitive. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-24T18:13:49Z SecondChildTAG: it is so cool day..My best wishes to your cousin. And please do not cry, stay sensetive without an excessive tears :) Serge SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-11-24T18:20:20Z SecondChildTAG: Haha! :). Thank you Serge! By the way, here are the Lab10 Hints [here][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50b11357f6c1e02700000045 SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-24T18:49:52Z FirstChildTAG: brainyash1990, could you explain, what you cant do? I'll try to help you FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-11-24T17:02:19Z IndexTAG: 1961 TitleTAG: H11P1 Characteristic equation in transfer function A question,please: I wonder if as long as I get the canonical form (characteristic equation for the denominator), doesn't matter what I have in the numerator? the rule always apply? thanks! UserIdTAG: 369339 UserNameTAG: vargaslen CreateTimeTAG: 2012-11-24T14:25:50Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hi, vargaslen I have this doubt too. I would like to see if someone could answer this. From a practical view, it worked on the exercises... FirstChildUserIdTAG: 288174 FirstChildUserNameTAG: matiasgrodriguez FirstChildCreateTimeTAG: 2012-11-24T16:17:12Z SecondChildTAG: hey matiasgrodriguez, some times I wonder if someone is getting all of this clearly. Do you speak spanish? SecondChildUserIdTAG: 369339 SecondChildUserNameTAG: vargaslen SecondChildCreateTimeTAG: 2012-11-24T19:16:56Z SecondChildTAG: Yes, I do speak spanish :) SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-24T19:21:45Z SecondChildTAG: me too :P SecondChildUserIdTAG: 244706 SecondChildUserNameTAG: Miguel_Angel SecondChildCreateTimeTAG: 2012-11-26T18:44:32Z FirstChildTAG: Hola vargaslen, Sí, la ecuación característica se basa en lo que tienes en el denominador sin importar lo que haya en el numerador. Espero que esto te haya sido de ayuda. Si quieres también puedes leer estas Hints del HW 11 [Leer aquí][1]. También te invito a participar en el Concurso CECC 2 :). Puedes ser uno de los tres afortunados que ganen el Libro del Curso autografiado por el Prof. Agarwal! :). Saludos! Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50b54fce5fb7411f00000006 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-28T01:06:58Z SecondChildTAG: Sí Myrimit, ¿pero por qué la ecuación característica está asociada SÓLO al denominador?, ¿tiene que ver con que estamos usando Z en lugar de Vo/Vi? Gracias! SecondChildUserIdTAG: 398594 SecondChildUserNameTAG: DaveyJC SecondChildCreateTimeTAG: 2012-11-30T16:33:04Z IndexTAG: 1962 TitleTAG: H10P2: Zc I have got a correct Zc equation. But when I put *inf* or 0 in this equation, i get wrong answers. please guide me. remember that something divided by 0 is inf and something divided by inf is 0. similarly something plus inf is also inf. UserIdTAG: 64618 UserNameTAG: ashfaq2419 CreateTimeTAG: 2012-11-24T13:44:04Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: ashfaq2419 for this net "0" or "inf" answer isnt permissible. FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-11-24T16:58:01Z FirstChildTAG: look at [this][1] post or[here][2] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50af6aaa4684922b0000000c [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50af648e8ef9df230000000f FirstChildUserIdTAG: 96873 FirstChildUserNameTAG: Santyaga FirstChildCreateTimeTAG: 2012-11-24T18:00:11Z IndexTAG: 1963 TitleTAG: Lab 10 L,C values I can´t get the right values. This is what i'm doing to get the value of L: f=10^4.6 L=R/(2*pi*f)=3.99e-3 for C: C=1/(2*pi*f*R)=3.99e-9 What am I doing wrong? UserIdTAG: 137087 UserNameTAG: daxiel22 CreateTimeTAG: 2012-11-24T12:18:54Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Are you sure, that -14.5dB is a "break frequence"? ;-) Use the first formula Hrc=arctg(-w*R*C) and take Hrc from the graph (don't forget about radians-degrees conversion) FirstChildUserIdTAG: 345464 FirstChildUserNameTAG: Constantine_ru FirstChildCreateTimeTAG: 2012-11-24T12:32:43Z SecondChildTAG: Hrl=arctg(-w*L/R)= 78.684, -w*L/R=tan(78.684) L=R*(tan78.684)/2*pi*f,where f=10^4.6 Hz. But the answer is not correct? Why? What am I doing wrong? Any helps, thanks!! SecondChildUserIdTAG: 371617 SecondChildUserNameTAG: rainbow12 SecondChildCreateTimeTAG: 2012-11-24T15:33:01Z SecondChildTAG: Or use the magnitude 20log(1/(1+(w*L/R)^2)^(1/2)) = 14.148, the answer is not correct either.? why? Give me hints. thanks SecondChildUserIdTAG: 371617 SecondChildUserNameTAG: rainbow12 SecondChildCreateTimeTAG: 2012-11-24T15:36:57Z SecondChildTAG: Rainbow12, The correct would be L = R/(tan(angle)*w), where w = 2*pi*f, angle must be in rad/s. SecondChildUserIdTAG: 457034 SecondChildUserNameTAG: Ichihara SecondChildCreateTimeTAG: 2012-11-25T17:23:41Z FirstChildTAG: Ok, i could finally get the right values. All you need to do is take the frequency value and use it in the phase equations. Remember Angle(rad)=(Angleº*2*pi)/360 ;) FirstChildUserIdTAG: 137087 FirstChildUserNameTAG: daxiel22 FirstChildCreateTimeTAG: 2012-11-24T12:34:24Z SecondChildTAG: i have found the value of C but failed to find the value of L..i have tried a lot.. SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-11-24T16:36:23Z IndexTAG: 1964 TitleTAG: Please give me hint Lab10 last question ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13537445911343639.jpg Though the simulation gives the correct result I am not getting green tick any suggestion please UserIdTAG: 352647 UserNameTAG: praveenjugge CreateTimeTAG: 2012-11-24T08:11:17Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I've the same problem as you. I´ve tried to use pasaband filter with L-C-R in series configuration and R-L-R-C in cascade and exactly same problem magnitude perfect but phase oscillation behavior. Really strange. Any hint to solve this strange behavior? Million thanks! FirstChildUserIdTAG: 149058 FirstChildUserNameTAG: sotoroman FirstChildCreateTimeTAG: 2012-11-24T11:37:56Z SecondChildTAG: hi, I had that problem too. Did you notice that the axis of magnitude is not the same as shown. Just a small change in L and C and you'll get it right SecondChildUserIdTAG: 350842 SecondChildUserNameTAG: hasina SecondChildCreateTimeTAG: 2012-11-24T11:53:19Z SecondChildTAG: Thanks hasina, you are right. For all, just change + or - 0.1 in your C value, the phase curve does not change so much, but you get the green tick. SecondChildUserIdTAG: 296419 SecondChildUserNameTAG: oscman SecondChildCreateTimeTAG: 2012-11-24T15:16:17Z FirstChildTAG: I was getting what I thought was the correct output with the two breaks at "almost" -3dB, but apparently not close enough. I changed the value of one component in 0.1 steps, kept re-checking the break points until they were even closer to -3dB and finally got a green check. FirstChildUserIdTAG: 375500 FirstChildUserNameTAG: Brej7665 FirstChildCreateTimeTAG: 2012-11-24T16:37:26Z SecondChildTAG: 0.1 what? uF? nF? pF? I found 3 digits of precision to be enough for the L and C values. SecondChildUserIdTAG: 141000 SecondChildUserNameTAG: OrinE SecondChildCreateTimeTAG: 2012-11-24T18:45:04Z IndexTAG: 1965 TitleTAG: with this first equation you're able to get the job done V2=(v+r1×I)/(1+r1/r2) UserIdTAG: 805245 UserNameTAG: polson CreateTimeTAG: 2012-11-24T03:02:11Z VoteTAG: 1 CoursewareTAG: Week 1 / KVL/KCL example CommentableIdTAG: 6002x_S2E3_Using_KVL_KCL_and_VI_constraints NumberOfReplyTAG: 1 FirstChildTAG: how you write this equation??? FirstChildUserIdTAG: 553106 FirstChildUserNameTAG: ershibbu FirstChildCreateTimeTAG: 2012-11-27T05:07:55Z IndexTAG: 1966 TitleTAG: staff check button in H12P don(t work and thankx UserIdTAG: 331441 UserNameTAG: abdessamade CreateTimeTAG: 2012-11-23T20:25:26Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Me too FirstChildUserIdTAG: 357453 FirstChildUserNameTAG: BrunoCanoso FirstChildCreateTimeTAG: 2012-11-23T20:48:18Z SecondChildTAG: For me only H12P2 SecondChildUserIdTAG: 357453 SecondChildUserNameTAG: BrunoCanoso SecondChildCreateTimeTAG: 2012-11-23T21:15:55Z SecondChildTAG: You need all the boxes for part (c) and (d) filled in. I suspect a divide by zero in the checker, so use non-zero values and something between 0 and 1 for the efficiency. If it still doesn't work, clear your browser cache, reload the page and try again. SecondChildUserIdTAG: 141000 SecondChildUserNameTAG: OrinE SecondChildCreateTimeTAG: 2012-11-24T01:59:38Z FirstChildTAG: Look at my post about that problem, staff is looking into it: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_Troubleshooting/threads/50ae30a5a2185c2700000014 FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-11-24T13:57:37Z IndexTAG: 1967 TitleTAG: H10P3 L/C question Hi to all! I successfully got answer for the two first questions. Unfortunatelly I cant solve Q3/4 for C and L. At first I do use R1=Zf, where Zf is answer for Q2. But after this point I cant get correct answers.. Any hints, please? UserIdTAG: 205311 UserNameTAG: Sergtronix CreateTimeTAG: 2012-11-23T19:41:57Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hello Sergtronix, This link should do the trick! https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50a688f55b77f62300000035 Regards. FirstChildUserIdTAG: 153604 FirstChildUserNameTAG: autohost FirstChildCreateTimeTAG: 2012-11-23T20:24:12Z SecondChildTAG: Thank you ! Nevermind.. But I got it! thanks! SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-11-23T20:51:35Z IndexTAG: 1968 TitleTAG: lab10 i m nt able 2 calculate the values of L & C in 2nd part..... i have recognized the circuit well... i think i m making mistake in calculating w ...which is 39.81K...is tht rite... thnks..... UserIdTAG: 214277 UserNameTAG: yadsam CreateTimeTAG: 2012-11-23T14:23:16Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Your value for omega is correct, and you should be able to get the correct answer with it. FirstChildUserIdTAG: 390034 FirstChildUserNameTAG: kahlil FirstChildCreateTimeTAG: 2012-11-23T21:25:30Z SecondChildTAG: tan^-1(w*r*c) thts the formula rite...m still nt getting it..... SecondChildUserIdTAG: 214277 SecondChildUserNameTAG: yadsam SecondChildCreateTimeTAG: 2012-11-25T14:00:22Z FirstChildTAG: lost in this one aswell. how did you get w? how do i set up this type of circuit in sand box? i get AC analysis refers to unknown source is it natural log? FirstChildUserIdTAG: 388273 FirstChildUserNameTAG: kilmarta FirstChildCreateTimeTAG: 2012-11-25T21:56:48Z SecondChildTAG: ok got most of it, only thing tripping me was the log's. decible one was in base 2 and freq one was in base 10. was i meant to know that (honest question) SecondChildUserIdTAG: 388273 SecondChildUserNameTAG: kilmarta SecondChildCreateTimeTAG: 2012-11-25T22:58:46Z IndexTAG: 1969 TitleTAG: How is the answer for part-2 obtained in S21E1? For part-2, as w becomes really large, I thought the answer would be R1+jwL. Why is R1 eliminated? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-11-23T12:33:40Z VoteTAG: 1 CoursewareTAG: Week 11 / S21E1: Second-Order Impedance CommentableIdTAG: 6002x_S21E1_Second_Order_Impedance NumberOfReplyTAG: 1 FirstChildTAG: That is because the question states that w is very large (going to infinity) and therefore R1 value is insignificant compared to jwL. FirstChildUserIdTAG: 318577 FirstChildUserNameTAG: takeuchi FirstChildCreateTimeTAG: 2012-11-23T15:57:43Z SecondChildTAG: good answer SecondChildUserIdTAG: 226085 SecondChildUserNameTAG: Teto SecondChildCreateTimeTAG: 2012-12-01T03:18:22Z SecondChildTAG: thx SecondChildUserIdTAG: 226085 SecondChildUserNameTAG: Teto SecondChildCreateTimeTAG: 2012-12-01T03:18:28Z SecondChildTAG: I'm not sure, but it may be because w is so large that R1 is trivial. Just my guess. SecondChildUserIdTAG: 12667 SecondChildUserNameTAG: jwhall SecondChildCreateTimeTAG: 2012-12-02T20:51:55Z IndexTAG: 1970 TitleTAG: H12P2d Hint This problem can be done! A hint for part d is the expected answer is in terms of a fraction, not percent so do not multiply by 100. Also note the problem statement asks for the minimum efficiency so use the input voltage that produces the lowest number. UserIdTAG: 397595 UserNameTAG: mholin CreateTimeTAG: 2012-11-22T14:20:21Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: can you help me with the c part?please..... FirstChildUserIdTAG: 357747 FirstChildUserNameTAG: kishores FirstChildCreateTimeTAG: 2012-11-23T18:07:54Z FirstChildTAG: It is relatively straight forward. Set up two equations using the two input voltage extremes and the two desired output voltages and solve for the two unknowns R0 and the opamp noninverting gain. Model the zener diode as a voltage source in series with a resistor RZ. FirstChildUserIdTAG: 397595 FirstChildUserNameTAG: mholin FirstChildCreateTimeTAG: 2012-11-24T01:02:48Z IndexTAG: 1971 TitleTAG: df instaed of dw well, I got L, Cmin, Qmin and Qmax right. But, for the **bandwidths**, the right answers only result if I calculate dfmin and dfmax (instead of dwmin and dwmax). UserIdTAG: 388524 UserNameTAG: dasbinich CreateTimeTAG: 2012-11-22T14:05:33Z VoteTAG: 1 CoursewareTAG: Week 11 / S21E4: AM_Radio_Tuning CommentableIdTAG: 6002x_S21V1_AM_Radio_Tuning NumberOfReplyTAG: 3 FirstChildTAG: Yeah, that seems to work. Weird. FirstChildUserIdTAG: 85490 FirstChildUserNameTAG: yashsavani FirstChildCreateTimeTAG: 2012-11-22T14:30:11Z SecondChildTAG: The denominator (characteristic equation) should be in the canonic form: $\frac{something}{S^2+2\alpha S+\omega o^2}$ Now find the bandwidth is trivial because $\Delta\omega=2\alpha$ SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-24T20:56:58Z FirstChildTAG: How did u get things right? FirstChildUserIdTAG: 477713 FirstChildUserNameTAG: ikm104 FirstChildCreateTimeTAG: 2012-11-27T12:57:51Z SecondChildTAG: right the equation of the circuit first.... get it in standard (canonic) form... calculate the values of alpha n omega-not.... the rest is jst substitution of values in formulae....Q=w0/(2a) SecondChildUserIdTAG: 117319 SecondChildUserNameTAG: iqramalik SecondChildCreateTimeTAG: 2012-11-27T20:34:11Z SecondChildTAG: write*...:) SecondChildUserIdTAG: 117319 SecondChildUserNameTAG: iqramalik SecondChildCreateTimeTAG: 2012-11-27T20:34:25Z FirstChildTAG: Pay attention to what the question is. It asks the answer in kHz. So, find delta w and divide by 2pi. FirstChildUserIdTAG: 99238 FirstChildUserNameTAG: arthurltc FirstChildCreateTimeTAG: 2012-11-30T00:01:02Z SecondChildTAG: Thanks SecondChildUserIdTAG: 733429 SecondChildUserNameTAG: EhsanMokhtari SecondChildCreateTimeTAG: 2012-12-02T14:34:55Z IndexTAG: 1972 TitleTAG: Staff bug check of H12P2 not working and thankx UserIdTAG: 331441 UserNameTAG: abdessamade CreateTimeTAG: 2012-11-22T13:15:13Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Please staff, look at my problem too! I think it's related and I'm trying for days: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50ae30a5a2185c2700000014 FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-11-22T14:12:20Z IndexTAG: 1973 TitleTAG: H12P2 HELP!!!! need some hints.does vin < vz mean vin is negative? UserIdTAG: 357747 UserNameTAG: kishores CreateTimeTAG: 2012-11-21T16:57:46Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: No. It means vin is less than the zener reference voltage. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-21T17:02:36Z SecondChildTAG: zener reference voltage? SecondChildUserIdTAG: 357747 SecondChildUserNameTAG: kishores SecondChildCreateTimeTAG: 2012-11-21T17:10:27Z SecondChildTAG: If the vin voltage is less than vz, you should consider the voltage vin. Otherwisethe voltage should be vz. SecondChildUserIdTAG: 361823 SecondChildUserNameTAG: EliasOak SecondChildCreateTimeTAG: 2012-11-21T17:36:14Z SecondChildTAG: woudn't there be a drop across R0? SecondChildUserIdTAG: 357747 SecondChildUserNameTAG: kishores SecondChildCreateTimeTAG: 2012-11-21T18:01:31Z SecondChildTAG: **Can anyone help me regarding part c of this problem?** SecondChildUserIdTAG: 277808 SecondChildUserNameTAG: Hemanthmps SecondChildCreateTimeTAG: 2012-11-21T18:02:20Z SecondChildTAG: When vin is less than the zener voltage no current flows and no voltage drop across resistor. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-11-21T18:06:21Z SecondChildTAG: k thanks . SecondChildUserIdTAG: 357747 SecondChildUserNameTAG: kishores SecondChildCreateTimeTAG: 2012-11-23T18:10:13Z SecondChildTAG: Could be that the problem is a bit poorly specified?, or maybe I'm just not getting all of it? SecondChildUserIdTAG: 398594 SecondChildUserNameTAG: DaveyJC SecondChildCreateTimeTAG: 2012-12-07T22:34:14Z SecondChildTAG: Be very careful about your equation for $v^+$. Once you have that, you're golden; the rest is straightforward, and hints elsewhere, should you need them, will get you to the finish. What I finally did was put all the calculations in intermediary steps in Excel. That's when I caught my stupid math errors and finally solved H12P2. Good luck! SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-12-07T23:58:22Z IndexTAG: 1974 TitleTAG: About Rf amp and modulators.. Could you recommend me any book about Rf amp,modulator or oscillators... Really need some material .. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-11-21T10:00:44Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: The ARRL handbook is a very interesting book, used by radio-amateurs (hams) all over the world. Look at this: http://www.arrl.org/shop/ARRL-Handbook-2013-Softcover-Edition/?page=1 FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-11-21T10:13:36Z SecondChildTAG: Yes these are handy books. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-11-21T12:03:31Z SecondChildTAG: Also check out the application notes at: [Mini Circuits][1]. They may even have what is needed as a prebuilt module. Warning: there be dragons in RF land. You end up with 100s of lbs of test equipment and nothing is ever quite what it seems. [1]: http://www.minicircuits.com/applications/applications_notes.html SecondChildUserIdTAG: 141000 SecondChildUserNameTAG: OrinE SecondChildCreateTimeTAG: 2012-11-21T18:07:03Z FirstChildTAG: Thank you.This book indeed covers all of what I need,but it's just an introduction,not detailed enough.I want a book focusing on specific field,such as power amplifier and so on... I found mit's 6.061 textbook,any similar suggestion? FirstChildUserIdTAG: 136959 FirstChildUserNameTAG: christerpher FirstChildCreateTimeTAG: 2012-11-22T01:51:39Z IndexTAG: 1975 TitleTAG: capacitor small signal model So far ,we have derived the small signal models of MOSFET,BJT.What about the capacitor and inductor? Their constitutive law is a function containing time...Using the method differentiating the voltages and currents seems doesn't work...How could we do this ...Thanks UserIdTAG: 136959 UserNameTAG: christerpher CreateTimeTAG: 2012-11-21T00:45:17Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Ideal capacitors and inductors are linear devices, which means there is no need to linearize. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-21T01:04:58Z SecondChildTAG: does it mean we don't need to replace them? SecondChildUserIdTAG: 136959 SecondChildUserNameTAG: christerpher SecondChildCreateTimeTAG: 2012-11-21T01:15:21Z SecondChildTAG: And resistors are linear elements too...they still have a small signal model SecondChildUserIdTAG: 136959 SecondChildUserNameTAG: christerpher SecondChildCreateTimeTAG: 2012-11-21T01:20:16Z SecondChildTAG: Just like a resistor the large signal model and small signal model are the same. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-11-21T01:58:56Z SecondChildTAG: thank you SecondChildUserIdTAG: 136959 SecondChildUserNameTAG: christerpher SecondChildCreateTimeTAG: 2012-11-21T03:08:30Z SecondChildTAG: Awesome SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-21T10:32:36Z IndexTAG: 1976 TitleTAG: S19E3 complex numbers help I do not understand how to get the angle, specifically "The angle of the second factor is just the difference of the angle of the numerator and the angle of the denominator". Unfortunately I do not remember anything about complex numbers. I've watched the tutorials which has been of minimal help. UserIdTAG: 244115 UserNameTAG: Pedro1969 CreateTimeTAG: 2012-11-20T23:50:48Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: The [Khan Academy][1] has a section on complex math. [1]: http://www.khanacademy.org/ FirstChildUserIdTAG: 90796 FirstChildUserNameTAG: ChaunceyGardiner FirstChildCreateTimeTAG: 2012-11-21T00:45:41Z SecondChildTAG: Thank you and I've made some progress last night. I may be catching up but just a little confusion on notation: archtan(b,a) does that mean archtan(b/a)? SecondChildUserIdTAG: 244115 SecondChildUserNameTAG: Pedro1969 SecondChildCreateTimeTAG: 2012-11-21T12:57:29Z FirstChildTAG: Do either of these help you? [Complex Numbers][1] [Textbook Appendix C. Complex Numbers][2] [1]: https://www.edx.org/wiki/6.002x/complex-numbers [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/ FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-11-21T21:03:50Z IndexTAG: 1977 TitleTAG: Needing help with trig For this problem, I am having issues with the complex trig. I am still having difficuilty remembering all the tools that you are supposed to use, so I would appreciate some kind of resource, since the textbook doesnt seem to help me much at all. UserIdTAG: 248807 UserNameTAG: 1kingsman CreateTimeTAG: 2012-11-20T23:06:53Z VoteTAG: 1 CoursewareTAG: Week 10 / Magnitudes And Angles CommentableIdTAG: 6002x_Magnitudes_and_angles NumberOfReplyTAG: 1 FirstChildTAG: If you struggle with the math, the [Khan Academy][1] is your very best friend. They have a section on complex math as well as many, many other subjects. I wouldn't have made it through the first 6002x without them. [1]: http://www.khanacademy.org/ FirstChildUserIdTAG: 90796 FirstChildUserNameTAG: ChaunceyGardiner FirstChildCreateTimeTAG: 2012-11-20T23:45:31Z IndexTAG: 1978 TitleTAG: Course end... If the course ends will we still be able to watch the videos, do the work, and have access to the materials online? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-11-20T06:42:01Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: As far as I know, **yes**. FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-11-20T08:25:26Z IndexTAG: 1979 TitleTAG: H10P3 Part 1 just by reason...I need help to determin why it is correct As i say in the title...I came very close to the final solution of Vo/Vi, but the answer was wrong...after a little thought about what should happen to the circuit in DC, i found the right answer, but i am not sure what my (computed) error was... Can anyone help but without giving the equation..? UserIdTAG: 304587 UserNameTAG: Vasso CreateTimeTAG: 2012-11-19T21:35:50Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: the answer is taken as correct only without simplifications... i tried the thevenin method, solved right away, but you must leave the Zc (for example) equal to 1/jCw and do not make any simplifications such as multiply with jcw etc.. for example, if your thevenin equivalent is (R4+1/jCw)/(jLw+1/jCw) leave it as it is... FirstChildUserIdTAG: 301962 FirstChildUserNameTAG: labrinim FirstChildCreateTimeTAG: 2012-11-20T09:20:38Z SecondChildTAG: I don't understand why I need thevenin method. The reuslt should look like Vo/Vi=Z1/(R1+Z1). Unfortunately I don't get a green checkmark SecondChildUserIdTAG: 389333 SecondChildUserNameTAG: juergen SecondChildCreateTimeTAG: 2012-11-20T14:33:42Z SecondChildTAG: Vth equals to Vf(and has Vi inside). Rth includes all the embendances except R2. so, you final have just to solve the very simple circuit with Vf, Rth to find Vo and voila! SecondChildUserIdTAG: 301962 SecondChildUserNameTAG: labrinim SecondChildCreateTimeTAG: 2012-11-20T14:39:42Z IndexTAG: 1980 TitleTAG: S23V18 Superposition with dependent sources? I thought you could not use superposition with dependent sources. If that is true why did the analysis work? UserIdTAG: 397595 UserNameTAG: mholin CreateTimeTAG: 2012-11-19T17:03:17Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I don't think anyone said you cannot use superposition with dependent sources. Read pages 153-157, specifically section 3.5.1: Superposition Rules for Dependent Sources. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-11-19T20:44:00Z IndexTAG: 1981 TitleTAG: H9P1 - is this joke? ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13533428761343614.jpg UserIdTAG: 188778 UserNameTAG: xsaq CreateTimeTAG: 2012-11-19T16:34:54Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 9 FirstChildTAG: If it is, I'm guessing it's pretty funny. However, I think you are going to have to be more specific, because I can't tell what your issue is. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-11-19T16:57:53Z FirstChildTAG: I'm guessing that the grader was looking for 0. One value was small enough to pass. The other wasn't. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-19T17:01:57Z FirstChildTAG: haha I was caught here also. It wants 0 (zero) FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-11-19T17:11:27Z FirstChildTAG: Actually, -3.1 mV is not even close to zero. I'm wondering why he highlighted one of the variables in the equation instead of the answer. The two numbers are different by a factor of 1/100,000 (which is obvious by the variables in the equation). I go with JSChambers answer, we don't have enough information. FirstChildUserIdTAG: 132828 FirstChildUserNameTAG: rharris FirstChildCreateTimeTAG: 2012-11-19T17:26:52Z SecondChildTAG: click show answer - it wants ZERO :))) SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-11-19T17:56:18Z SecondChildTAG: I understand it is zero, that is what I got. All I'm saying is if one expects 0.0031 is to be interpreted as a zero, they are way off. This whole problem deals with equating floating point numbers when the computer program checks the answer. Here is a good lecture on the subject. [Computer Science lecture on floating point numbers][1] Also, it is explained here: [Five Tips for Floating Point Programming ][2] I happened across this when writing a Python program where I tried to compare the result of multiplying two floating point numbers together. You can't just say: if x == y in a program and expect them to match if x and y are two floating point numbers. The program has to figure out if the answer we give is close enough to the correct answer. There are numerous examples throughout this course where this happens. [1]: http://www.youtube.com/watch?v=Pfo7r6bjSqI&feature=edu&list=PL4C4720A6F225E074 [2]: http://www.codeproject.com/Articles/29637/Five-Tips-for-Floating-Point-Programming SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-11-19T18:24:37Z SecondChildTAG: The OP wasn't expecting that number to be interpreted as zero. He thought it to be the correct answer to the problem because he calculated w from the component values given and then computed cos(w*t) for the time of the impulse. The angle was intended to be 2*n*pi + pi/2 where n was an integer that varied from person to person. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-11-19T18:44:25Z FirstChildTAG: I believe that the OP's point is that the energy stored in the capacitor was calculated from the voltage across the capacitor. The energy was marked correct but the voltage wrong. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-19T17:37:37Z FirstChildTAG: I also had that problem, when I entered the exact value. Then I tried 0 and it was ok. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-11-19T17:08:18Z FirstChildTAG: Well I did have a problem with second answer (the one in which the OP has the red cross). If it wasn't for skyhawk's post in another thread, I wouldn't have got it right. We need to a approximate our answers so that we get an integer period. In my opinion, the exact values should be the ones which work. Otherwise this must be mentioned in the problem. FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-11-19T17:08:50Z FirstChildTAG: The grader was looking for 0 in the last three parts of H9P1. "-0.003106043..." must have been outside the range of error for the value 0, which is why the second to last problem was marked as incorrect. The "1.86e-8" term is considerably smaller, and within the range of error for the grader, which is why it was deemed correct, even though the correct answer was 0. FirstChildUserIdTAG: 391929 FirstChildUserNameTAG: RohanNagarkar FirstChildCreateTimeTAG: 2012-11-19T18:09:10Z SecondChildTAG: I guess the question is whether or not the grader **should** have been looking for zero or not. BTW, if your real name is Rohan, that is awesome. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-11-19T18:34:14Z FirstChildTAG: A small value is not accepted as 0. A small value squared and multiplied by another small value and divided by 2 *is* accepted as 0. FirstChildUserIdTAG: 90796 FirstChildUserNameTAG: ChaunceyGardiner FirstChildCreateTimeTAG: 2012-11-20T00:10:04Z IndexTAG: 1982 TitleTAG: a question regarding homework scores... I am in need of a small clarification. In the next two weeks, owing to semester exams in my college, i may find it hard to watch videos and complete homework/lab for week 11 and week 12... but i will be completing the first 10 weeks homework fully.... i ll resume watching the videos after my exams get over.... in that case, will i lose any mark/score from the homework/lab part, for not completing week 11 and week 12 homework/lab ??? please reply.... UserIdTAG: 316353 UserNameTAG: Vivekanand123 CreateTimeTAG: 2012-11-19T15:02:52Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Remember that you can skip two Labs and two Homeworks without penalty, that means, that it would not affect your final score. But, you should try to do them later, as might some of the homework ecxercises of week 11 and 12 can appearn in the Final Exam. I hope this can help you, Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-19T15:53:15Z SecondChildTAG: ok thanks SecondChildUserIdTAG: 316353 SecondChildUserNameTAG: Vivekanand123 SecondChildCreateTimeTAG: 2012-11-19T17:05:07Z IndexTAG: 1983 TitleTAG: Didn`t understand Can somebody explain me - why it didn`t oscillating with 2K Ohm Resistor and just with 50||2K? There was same loop, and same L and C parameteres... UserIdTAG: 131076 UserNameTAG: BoozZzilla CreateTimeTAG: 2012-11-19T12:43:37Z VoteTAG: 1 CoursewareTAG: Week 9 / S18V1 Review CommentableIdTAG: 6002x_S18V1_Review NumberOfReplyTAG: 1 FirstChildTAG: I would say that the 2K resistor provided enough damping to prevent oscilation from happening. 50 || 2K is arround 50 ohms and this resistor did not provide enough dampening to supress the oscilation. FirstChildUserIdTAG: 294766 FirstChildUserNameTAG: dejanst FirstChildCreateTimeTAG: 2012-11-19T23:29:09Z SecondChildTAG: Thank you. I did not think that way. I thought it doesn`t oscillating, but it does and damped... SecondChildUserIdTAG: 131076 SecondChildUserNameTAG: BoozZzilla SecondChildCreateTimeTAG: 2012-11-20T03:58:03Z IndexTAG: 1984 TitleTAG: H10P1: MAGNITUDE AND ANGLE trying hard but didn't get the success? any hint.. UserIdTAG: 114864 UserNameTAG: Vikaash CreateTimeTAG: 2012-11-19T09:01:23Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: You need to calculate how the magnitude and phase of $V_o/V_i$ vary with frequency for each circuit. Circuit A is done for you in video S19V14. The equations for the other three circuits can be derived fairly easily using the impedance method covered in lecture sequence S20. Using the given component values you can then calculate the break frequency for each circuit, and using this and the expected shape of the response, you should be able to identify which curve belongs to which circuit. Don't forget (like I did) that $\omega$ is in radians per second, but the frequency plots are in Hz. FirstChildUserIdTAG: 434869 FirstChildUserNameTAG: patey FirstChildCreateTimeTAG: 2012-11-19T10:30:22Z FirstChildTAG: As an example: circuits A and D will have transfer functions of the same functional form. For A: 1/(1+jwRC) and for D: 1/(1+jwL/R). Their magnitude and phase curves will have similar shapes. How do you distinguish them? ... by their break frequencies. There is a similar situation with circuits B and C. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-19T16:49:03Z SecondChildTAG: Hi,skyhawk, w = 2*pi*f = 6284 Till I know mod(1/(1+jwRC)) = 1/(sqrt(1+(jwRC)^2)) =1/sqrt(1+(6.8k*0.147n*6284)^2) and the angle is: arctan(-wRC) = arctan(-6284*6.8K*0.147n) what am doing wrong? Thanks SecondChildUserIdTAG: 299251 SecondChildUserNameTAG: TheRedBlackOne SecondChildCreateTimeTAG: 2012-11-19T17:29:03Z SecondChildTAG: and f = 1/sqrt(RC) SecondChildUserIdTAG: 299251 SecondChildUserNameTAG: TheRedBlackOne SecondChildCreateTimeTAG: 2012-11-19T17:32:07Z FirstChildTAG: RedBlack you don't need to do that math. Calculate the break frequency for the R-C and the R-L circuits. That will tell you which of the plots is for the R-C circuits and which for the R_L. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-19T17:43:46Z SecondChildTAG: yes, I know what they are, but I don't know what should I put in the boxes! SecondChildUserIdTAG: 299251 SecondChildUserNameTAG: TheRedBlackOne SecondChildCreateTimeTAG: 2012-11-19T17:57:58Z SecondChildTAG: It`says: A we want you to enter their product p∗u. !!! SecondChildUserIdTAG: 299251 SecondChildUserNameTAG: TheRedBlackOne SecondChildCreateTimeTAG: 2012-11-19T18:01:21Z SecondChildTAG: 1/sqrt(6.8k*0.147n) and then what? SecondChildUserIdTAG: 299251 SecondChildUserNameTAG: TheRedBlackOne SecondChildCreateTimeTAG: 2012-11-19T18:06:42Z SecondChildTAG: I'm sorry,I wanted to I mean that the break frequency is: 1/sqrt(2) and then what? SecondChildUserIdTAG: 299251 SecondChildUserNameTAG: TheRedBlackOne SecondChildCreateTimeTAG: 2012-11-19T18:13:58Z SecondChildTAG: Where did the square root come from? The break frequency (in rad/sec) for the R-C circuits solves wRC = 1. For the R_L circuits it solves wL/R = 1. All you need to know is the relative sizes and then you can pick which curves go with the R-C circuits and which with the R-L circuits. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-11-19T18:19:02Z SecondChildTAG: Ok, but what should I put in the boxes? a number? I'm feel desperate because I'm afraid that the answer is too easy SecondChildUserIdTAG: 299251 SecondChildUserNameTAG: TheRedBlackOne SecondChildCreateTimeTAG: 2012-11-19T18:31:24Z SecondChildTAG: Pick a curve from the magnitude plot and a curve from the phase plot and enter their labels separated by an *. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-11-19T18:35:35Z SecondChildTAG: skyhawk, Thank you very much for your time! I finally discover that the answer is a letter*letter! Brute force helps SecondChildUserIdTAG: 299251 SecondChildUserNameTAG: TheRedBlackOne SecondChildCreateTimeTAG: 2012-11-19T18:42:29Z SecondChildTAG: And sometimes finesse works! SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-11-19T18:48:33Z FirstChildTAG: You don't need to do detailed calculations. You simply need to identify which transfer functions go to one as w goes to zero and which go to one as w goes to infinity. Also which circuits have positive phase angles and which have negative phase angles. Finally you need to know which type circuit R-C or R-L have the largest break frequency. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-19T14:50:41Z FirstChildTAG: returned to edx after 2 days..i was just doing maths and then realized my mistake..finally got the answer..thanks @skyhawk... FirstChildUserIdTAG: 114864 FirstChildUserNameTAG: Vikaash FirstChildCreateTimeTAG: 2012-11-21T16:40:16Z IndexTAG: 1985 TitleTAG: Could some explain the last question? phi is 3*pi/2. I dont understand the result and why it is negative ϕ= (in radians −π<ϕ<π ) UserIdTAG: 11075 UserNameTAG: roncada CreateTimeTAG: 2012-11-19T01:06:53Z VoteTAG: 1 CoursewareTAG: Week 9 / An ILC circuit CommentableIdTAG: 6002x_an_ILC_circuit NumberOfReplyTAG: 1 FirstChildTAG: It's just a convention to express the phase as between $-\pi$ and $+\pi$ radians. With $2\pi$ radians per cycle, $\phi$ and $\phi \pm 2n\pi$ where $n$ is any integer are equivalent. FirstChildUserIdTAG: 141000 FirstChildUserNameTAG: OrinE FirstChildCreateTimeTAG: 2012-11-19T02:26:13Z SecondChildTAG: Also take a look on the polar coordinates of the tutorial https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_10/wk10_Complex/ SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-19T13:32:26Z IndexTAG: 1986 TitleTAG: H9P2 Hello guys, I'm having some problems in solving H9P2. I can't findo the undamped natural frequency, is it Wo? I found s²+2*alpha*s+wo^2 But I can't get the right question... And about the Quality Factor? Q= wo/2alpha? Any hint? Thank you in advance. UserIdTAG: 98262 UserNameTAG: rafaelmarques CreateTimeTAG: 2012-11-19T00:32:55Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Make sure you are expressing $\alpha$ and $\omega_0$ in terms of the components, L, R, and C. FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-11-19T00:35:25Z SecondChildTAG: Yes, I'm expressing right. The big hint is that the undamped natural frequency is in HZ. Thank you planescape. SecondChildUserIdTAG: 98262 SecondChildUserNameTAG: rafaelmarques SecondChildCreateTimeTAG: 2012-11-19T00:37:52Z SecondChildTAG: I already got the right answer. SecondChildUserIdTAG: 98262 SecondChildUserNameTAG: rafaelmarques SecondChildCreateTimeTAG: 2012-11-19T00:38:30Z SecondChildTAG: I'm glad. :-) SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-11-19T02:59:27Z SecondChildTAG: OMG, in Hz. Spent 10 mins to figure it out. Why it's not mentioned in text???? SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-11-19T03:48:31Z SecondChildTAG: I guess I assumed it was since the "natural frequency of oscillation" was given in [S17E5: An ILC Circuit][1] as $f=\frac{ω0}{2π}$. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_9/Undamped_Second-Order_Systems/ SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-11-19T04:12:37Z IndexTAG: 1987 TitleTAG: HW9 P2 Problem I had almost all green marks but the answer for alpha is not accepted. With Q (quality factor) and w (omega) and using the formula Q=omega/2*alpha, still the answer not accepted. Any hint? Regards UserIdTAG: 298547 UserNameTAG: AnthonyRF CreateTimeTAG: 2012-11-18T22:06:20Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You need equation for Q in terms of scheme components - R, L, C so what is your Omega_0 in terms of RLC and what is your Alfa? Here's Myrimit hints for HW9 https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50a7a289a0ff611f00000007 FirstChildUserIdTAG: 325197 FirstChildUserNameTAG: Vitali_Jerin FirstChildCreateTimeTAG: 2012-11-18T22:27:24Z SecondChildTAG: w=1/sqrt(LC), Q=(R*sqrt(LC))/L, and the "supposed" alpha= L/(2*R) SecondChildUserIdTAG: 298547 SecondChildUserNameTAG: AnthonyRF SecondChildCreateTimeTAG: 2012-11-18T22:55:18Z SecondChildTAG: AnthonyRF, I put a comment on https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50a903577dbff51f00000028 SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-18T23:10:59Z SecondChildTAG: Well, AnthonyRF's values for $\omega$ and Q would appear to give the right value for alpha if plugged into $Q=\frac{\omega _0}{2\alpha}$ . BTW, this circuit has the resistance parallel with the capacitor and you get a different alpha from a series LCR circuit. SecondChildUserIdTAG: 141000 SecondChildUserNameTAG: OrinE SecondChildCreateTimeTAG: 2012-11-18T23:45:17Z SecondChildTAG: Matias that was exactly what I did, the B term is L/R so 2α=L/R ==> α=L/2*R. Did you see now where my result is coming from? And that result is the same using the other relations like Q=ω/2α. OrinE I know it, please note that's different :) SecondChildUserIdTAG: 298547 SecondChildUserNameTAG: AnthonyRF SecondChildCreateTimeTAG: 2012-11-18T23:52:54Z SecondChildTAG: What is the term of $S^2$? Is it 1? SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-18T23:58:26Z IndexTAG: 1988 TitleTAG: Porcentaje de aprobación Hola con que porcentaje se aprueba el curso? Muchas gracias UserIdTAG: 42826 UserNameTAG: oscar_ed26 CreateTimeTAG: 2012-11-18T20:18:18Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: El porcentaje mínimo para obtener la certificación es el 60%. De nada hombre. FirstChildUserIdTAG: 314294 FirstChildUserNameTAG: victormp FirstChildCreateTimeTAG: 2012-11-18T20:38:00Z SecondChildTAG: Hola oscar_ed26, Tal y como te ha dicho @victormp, para obtener la certificación debes tener una puntuación total de 60% o más. Dicha ponderación se basará en la suma de la contribución de los HWs, Labs, Midterm Exam y Final Exam. Saludos, Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-18T20:59:04Z IndexTAG: 1989 TitleTAG: Error in H9P1 ? I think there is an error in H9P1. I've found iL(t) for t<9 (before the impulse), but my answer for iL(9-) seems to be wrong. I'm quite sure that my iL(t) is right because when I substitute it in the homogenous differential equation, I get an identity. Besides, I calculate vC(9-) as L*diL/dt(t=9) and I get the correct answer. This is strange: if my iL(t) is wrong, so should be vC(9-), which I calculate from iL(t). Can you please check the answers to H9P1? UserIdTAG: 84801 UserNameTAG: marcuspag CreateTimeTAG: 2012-11-18T17:31:59Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I find that a lot of the answer have to be rounded down to just 1 decimal place ... or even no decimal place if the numbers are big H9P1 FirstChildUserIdTAG: 714237 FirstChildUserNameTAG: PaxPolaris FirstChildCreateTimeTAG: 2012-11-18T17:35:57Z SecondChildTAG: guys, no math in this exercise.... just "SEE" it... SecondChildUserIdTAG: 301962 SecondChildUserNameTAG: labrinim SecondChildCreateTimeTAG: 2012-11-18T17:55:51Z SecondChildTAG: No math? How can you compute iL(9-) with no math? If you mean no big equations and so, I agree. It should be iL(9-)=cos(w0*9), or am I missing something? SecondChildUserIdTAG: 84801 SecondChildUserNameTAG: marcuspag SecondChildCreateTimeTAG: 2012-11-18T18:39:11Z FirstChildTAG: Make sure that your period of oscillation is an integer number of seconds then compare to 9 seconds to determine where you are in the cycle. No calculator needed for this part of the problem. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-18T19:07:29Z SecondChildTAG: Ok, I've caught your suggestion. No calculator needed. I've already been thinking about something like that, but I couldn't understand why they gave such "wrong" values for L and C. If you give me L=170.0H, with an approximation up to the first decimal position, I think I have to abide by the same level of precision. And so my period does not equate an integer number of seconds, and I start doing calculations... A better value of L should have been 170.3 H, so that T=4 s. Thanks SecondChildUserIdTAG: 84801 SecondChildUserNameTAG: marcuspag SecondChildCreateTimeTAG: 2012-11-18T19:42:07Z IndexTAG: 1990 TitleTAG: H9P2 - either i'm missing something or the grader is wrong My every calculation points to B being equal to C/R and alpha being 1/(2*R*L) And i did some tests in a commercial simulator to confirm , and it looks like they are correct , yet graded won't accept them. What could be wrong ??? UserIdTAG: 263693 UserNameTAG: Coldberg CreateTimeTAG: 2012-11-18T15:48:39Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You are missing something... The grader is right (I've solved it). How did you get the differential equation? FirstChildUserIdTAG: 84801 FirstChildUserNameTAG: marcuspag FirstChildCreateTimeTAG: 2012-11-18T16:32:40Z SecondChildTAG: oh nevermind i'v mixed up C and L it seems SecondChildUserIdTAG: 263693 SecondChildUserNameTAG: Coldberg SecondChildCreateTimeTAG: 2012-11-18T16:42:27Z SecondChildTAG: Do not feel bad - it's those stupid little things that get me EVERY TIME! All I can say is, double- and triple-check... Good luck! SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-11-18T17:04:13Z SecondChildTAG: same with me... SecondChildUserIdTAG: 316761 SecondChildUserNameTAG: gk_goel SecondChildCreateTimeTAG: 2012-11-18T17:46:10Z SecondChildTAG: I also cannot underestimate the importance of a good night's sleep - things that seemed impossible yesterday are falling readily to my sword today. ;-) SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-11-18T21:01:05Z SecondChildTAG: I'm having the same problem with HW9 P2. I had almost all green mark but the answer for alpha is not accepted. With Q (quality factor) and w (omega) and using the formula Q=omega/2*alpha, still the answer not accepted. Any hint? Regards SecondChildUserIdTAG: 298547 SecondChildUserNameTAG: AnthonyRF SecondChildCreateTimeTAG: 2012-11-18T22:04:14Z SecondChildTAG: I think there is no need to use quality factor formula here. Do the same professor did on the videos. Try to put your characteristic equation in the canonic form: $S^2+2\alpha S+\omega o^2 =0$ .... $\frac{d^2v}{dt^2}+a\frac{dv}{dt}+bv=0$ and relate $2\alpha$ with a. SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-18T23:03:26Z SecondChildTAG: Actually I think you can use the quality factor formula. The problem I believe is that you should put your equation in the canonic form. SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-18T23:52:07Z SecondChildTAG: Matias I did it like you're saying.. a=L/R so 2alpha=L/R that implies that alpha is equal to L/(2*R) Thanks for your time Gracias por tu tiempo (si eres habla-hispano como yo) SecondChildUserIdTAG: 298547 SecondChildUserNameTAG: AnthonyRF SecondChildCreateTimeTAG: 2012-11-18T23:58:48Z SecondChildTAG: The equation should be in terms of the components L, R, and C, and should be in Hertz. SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-11-19T04:15:48Z IndexTAG: 1991 TitleTAG: H9P1, I need a hint for using laplace transform here!! Hi, I want to try the Laplace approach also for finding the variables' value of this problem. May anyone give me a hint in this matter?? do I need to start from the ODE governing this circuit? I know that it is possible to solve the second order ODE by Laplace transform. Is it the case here or using Laplace here is for anything else??? UserIdTAG: 153707 UserNameTAG: masoud_np CreateTimeTAG: 2012-11-18T15:29:07Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Don't waste your time with Laplace transform. Most of the problem can be solved by inspection. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-18T19:10:44Z IndexTAG: 1992 TitleTAG: Help:@staff,What's wrong with the playlist? The playlist is inaccessible,as YouTube is blocked here,so I have to download every video one by one. I could get access to playlist before,but now it is wrong. Would you please help me? thanks:) UserIdTAG: 361137 UserNameTAG: Ericson CreateTimeTAG: 2012-11-18T13:22:40Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 1993 TitleTAG: y2 I'm getting y2 as -1.352?? Rest all were correct. Not understanding where I'm going wrong :( UserIdTAG: 504574 UserNameTAG: Abhyudha CreateTimeTAG: 2012-11-18T11:13:05Z VoteTAG: 1 CoursewareTAG: Week 2 / Superposition example CommentableIdTAG: 6002x_S3E3_superposition NumberOfReplyTAG: 1 FirstChildTAG: x1 = v1 * (r2 || r3) / (r1 + (r2 || r3)) x2 = v2 * (r1 || r3) / (r2 + (r1 || r3)) y1 = (x1 - v1) / r1 y2 = x2 / r1 The expressions for y1 and y2 are different because the two voltage sources are hooked up to R1 differently. Another way to put it would be that when you calculate x2, R1 is connected between x2 and ground (0 V). When you calculate x1, R1 is connected between x1 and v1. FirstChildUserIdTAG: 90796 FirstChildUserNameTAG: ChaunceyGardiner FirstChildCreateTimeTAG: 2012-11-19T03:48:56Z IndexTAG: 1994 TitleTAG: What's the output of the Oscillator? We want to get the triangle from the oscillator circuit. Therefore, the output port is of the capacitor? In the Video of DC-DC Oscillator and Control (6:29), it's strange that you connected the output of Op-amp to the comparator, although you have a right diagram in the breadboard. UserIdTAG: 564606 UserNameTAG: ArequipaRadio CreateTimeTAG: 2012-11-18T08:02:10Z VoteTAG: 1 CoursewareTAG: Week 9 / DC DC Converter CommentableIdTAG: 6002x_DC_DC_Converter NumberOfReplyTAG: 0 IndexTAG: 1995 TitleTAG: pls help me with this.... In the penultimate step, e^(jwdt)=coswdt+jsinwdt.... we can similarly write e^(-jwdt)=coswdt-jsinwdt.... and if we take only the real parts of both of these terms,and substitute in the RHS (in the place of those terms), will we not get the real part of RHS ?? In this case, the last term comes in cos(jwdt)? please help me with this.... UserIdTAG: 316353 UserNameTAG: Vivekanand123 CreateTimeTAG: 2012-11-18T07:38:48Z VoteTAG: 1 CoursewareTAG: Week 9 / RLC Dynamics -- Underdamped Case CommentableIdTAG: 6002x_LC_Dynamics_Underdamped_Case NumberOfReplyTAG: 1 FirstChildTAG: which video are you talking about? FirstChildUserIdTAG: 357747 FirstChildUserNameTAG: kishores FirstChildCreateTimeTAG: 2012-11-18T07:45:44Z IndexTAG: 1996 TitleTAG: I think it's wrong In the HW H9P2, I think there's an error. I've evaluated the natural frequency as 1/sqrt((L*C)), and it said I was wrong. Then I've evaluated the alpha as 1/(2*R*C), and it was right! Insisting the first one was right, I've evaluated the quality factor as (R*sqrt(L*C))/L, AND IT WAS RIGHT! So, I don't know wether my evaluation or the correction's wrong, because the quality factor is the natural frequency by 2 times alpha. So, if someone could give me a help... UserIdTAG: 378470 UserNameTAG: Iron CreateTimeTAG: 2012-11-18T02:10:17Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: It' been thrashed plenty. Here is but one example: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50a65d577a3ae92b0000001d BTW the grader wants the answer in Hz. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-18T02:20:59Z SecondChildTAG: skyhawk, I pointed the staff to our conversation, and they debating a change in the language. So all your good work is not in vain. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-11-18T03:33:35Z SecondChildTAG: Question updated. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-11-18T05:56:42Z SecondChildTAG: It's now clear that the undamped natural frequency is required, but still doesn't say it should be in Hz. I know that perhaps there is the argument that if they'd wanted angular frequency, they'd have said angular frequency, but I don't recall the term "angular frequency" being discussed and whenever $\omega _0$ and $\omega _d$ were discussed, they were simply referred to as frequency when in fact they were angular frequencies. It's no big deal for the homeworks where you essentially get infinite tries and can try both to see what is accepted, but for the final, I hope it will be clear which is required. SecondChildUserIdTAG: 141000 SecondChildUserNameTAG: OrinE SecondChildCreateTimeTAG: 2012-11-18T08:30:57Z FirstChildTAG: We need to be careful about the units of the frequency. If the desired answer is in hertz then the (w0) which is in Rad/second should be divided by 2*pi to arrive at the answer in hertz (unit: 1/s). 1 cycle or hertz contains 2*pi radians. FirstChildUserIdTAG: 150267 FirstChildUserNameTAG: vwsingh FirstChildCreateTimeTAG: 2012-11-18T07:32:45Z IndexTAG: 1997 TitleTAG: H10P3 Q2 Hi, I'm trying to calculate the driving point impedance in H10P3 but my answers seem to be all wrong. Isn't it ((sL+R2)||(1/sC))+R1? Where is my misunderstanding? Thanks! UserIdTAG: 210986 UserNameTAG: JorgeRmz CreateTimeTAG: 2012-11-18T00:03:14Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: R1 is part of the final amplifier and should not be included in the impedance. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-18T00:38:30Z SecondChildTAG: also can v use 's' instead of jw(omega)? SecondChildUserIdTAG: 356056 SecondChildUserNameTAG: Ganesh2810 SecondChildCreateTimeTAG: 2012-11-18T07:53:32Z SecondChildTAG: How you converted jw parts into real ones igot the answewr sqrt((1/(w^4*L^2*C^2))-(1/(w^4*C^2*R2^2))) but got the red mark. SecondChildUserIdTAG: 582521 SecondChildUserNameTAG: varunchabba SecondChildCreateTimeTAG: 2012-11-26T02:38:28Z FirstChildTAG: help me for h10p3 1st question FirstChildUserIdTAG: 356056 FirstChildUserNameTAG: Ganesh2810 FirstChildCreateTimeTAG: 2012-11-18T08:17:40Z SecondChildTAG: try to use voltage divider to find Vf/Vi and then V0/Vf=(Zc//(ZR2+ZL))...you will get V0/Vi... SecondChildUserIdTAG: 16263 SecondChildUserNameTAG: Noureddine SecondChildCreateTimeTAG: 2012-11-18T15:25:46Z FirstChildTAG: Hi, I still need help in this second question. What I am doing... Vf is the voltage divider relationship: $Vf=\frac{Zc\parallel(Zl+R2)}{Zc\parallel(Zl+R2)+R1}\cdot Vi$ Then I'm trying to solve using KCL on node Vf: $If=\frac{Vf}{Zc}+\frac{Vf}{R2+Zl}$ Then $\frac{Vf}{If}$, but this gives me a red mark :( Any hint? FirstChildUserIdTAG: 288174 FirstChildUserNameTAG: matiasgrodriguez FirstChildCreateTimeTAG: 2012-11-18T21:07:42Z SecondChildTAG: I know they say it's $\frac{V_f}{I_f}$, but you don't have to do it that way! The point of the impedance method is that you don't have to. You already calculated the requested impedance for the numerator of your voltage divider relationship, so there is no need to go any further. SecondChildUserIdTAG: 141000 SecondChildUserNameTAG: OrinE SecondChildCreateTimeTAG: 2012-11-18T21:41:09Z SecondChildTAG: Thank you very much OrinE!! Green mark, but still not very clear to me. Ok, I will try to understand this later. Regards. SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-18T22:03:32Z FirstChildTAG: R2/(R2+(R1-w^2*R1*C*L+j*w*L)/(1+j*w*R1*C)) This is my answer why is it wrong Please tell me Why?????????? FirstChildUserIdTAG: 106816 FirstChildUserNameTAG: Laith FirstChildCreateTimeTAG: 2012-11-22T07:55:06Z SecondChildTAG: Now I understand what OrinE said and what they are asking you... They want the impedance as seen by the driving-point. They put it as $\frac{Vf}{If}$ and that is simply $Rf$ (ohm's law). As stated above: $Zc\parallel(Zl+R2)$ Laith, if you are not getting a correct mark may there is some algebra mistake. Good luck. SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-23T16:03:51Z SecondChildTAG: How you converted jw parts into real ones igot the answewr sqrt((1/(w^4*L^2*C^2))-(1/(w^4*C^2*R2^2))) but got the red mark. SecondChildUserIdTAG: 582521 SecondChildUserNameTAG: varunchabba SecondChildCreateTimeTAG: 2012-11-26T02:38:07Z IndexTAG: 1998 TitleTAG: When will Week 12 be available????? When will Week 12 be available? I have done week 9,10 &11 My progress is 51% I wish to get to atleast 60% as soon as possible! UserIdTAG: 216684 UserNameTAG: Taimoor1017 CreateTimeTAG: 2012-11-17T19:34:47Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: as always on Monday, I guess... FirstChildUserIdTAG: 359107 FirstChildUserNameTAG: EliteTUM FirstChildCreateTimeTAG: 2012-11-17T19:36:21Z FirstChildTAG: Bad news! There is only one more week of new assignments. The most you can get from them is 3%. So you will have to wait until the final to get over 60%! FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-17T23:32:22Z SecondChildTAG: Yup! but the near you go the better it is.... :) SecondChildUserIdTAG: 216684 SecondChildUserNameTAG: Taimoor1017 SecondChildCreateTimeTAG: 2012-11-18T10:16:58Z FirstChildTAG: why don't you post your progress sheet screenshot to inspire all of us? FirstChildUserIdTAG: 623210 FirstChildUserNameTAG: preveen FirstChildCreateTimeTAG: 2012-11-18T11:51:31Z SecondChildTAG: ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13536552911343664.jpg SecondChildUserIdTAG: 216684 SecondChildUserNameTAG: Taimoor1017 SecondChildCreateTimeTAG: 2012-11-23T07:21:51Z SecondChildTAG: Praveen, it's for u. SecondChildUserIdTAG: 216684 SecondChildUserNameTAG: Taimoor1017 SecondChildCreateTimeTAG: 2012-11-23T07:22:33Z FirstChildTAG: i am at 53% - nothing I do at this point will move my grade other than the final exam. FirstChildUserIdTAG: 18073 FirstChildUserNameTAG: gburkhart FirstChildCreateTimeTAG: 2012-11-18T23:06:01Z IndexTAG: 1999 TitleTAG: Weeks 13 and 14. Do these weeks have no homework or labs? Will they be represented in the final? Just curious. UserIdTAG: 213386 UserNameTAG: dmascenik CreateTimeTAG: 2012-11-17T19:28:56Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: First Homeworks and Labs from Week 11 and 12 will have their deadline in Week 13 and 14. AFTER that, the final will be released. Simply check "Course Info" > Calendar. Everything explained there. FirstChildUserIdTAG: 359107 FirstChildUserNameTAG: EliteTUM FirstChildCreateTimeTAG: 2012-11-17T19:38:35Z SecondChildTAG: I already read that, but it doesn't answer what I was asking about whether the material would be on the final as far as I can see. SecondChildUserIdTAG: 213386 SecondChildUserNameTAG: dmascenik SecondChildCreateTimeTAG: 2012-11-19T13:03:31Z SecondChildTAG: If it is like last semester, the final will be all inclusive, from first week to last. You can also expect to see something you've never seen before, the idea being, if you are firmly grounded in the basics, you should be able to apply them to novel situations. SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-11-19T18:48:33Z FirstChildTAG: The final exam will deal with material from Week 1 to Week 12. -- Rohan FirstChildUserIdTAG: 391929 FirstChildUserNameTAG: RohanNagarkar FirstChildCreateTimeTAG: 2012-11-20T16:26:30Z IndexTAG: 2000 TitleTAG: impulse, t=0- and t=0+ Hi guys, can you give me some site or materials for more understanding the impulse and the response before and after the time it s happening. I already had problems with that in week 8 but now I think can't do with that anymore. Sorry for my english! UserIdTAG: 350842 UserNameTAG: hasina CreateTimeTAG: 2012-11-17T18:21:26Z VoteTAG: 1 CoursewareTAG: Week 8 / Response To Impulse Drug Delivery CommentableIdTAG: 6002x_Response_To_Impulse_Drug_Delivery NumberOfReplyTAG: 1 FirstChildTAG: HI, I had problems understanding impulses as well. Took me a good two weeks to get some understanding of it. The basic stuff to try and understand is that an impulse adds a certain quantity of either charge on a capacitor or flux on an inductor. The quantity of the whatever is delivered (charge/flux) is the value of (or area under) the impulse (this is normally given). I'm sure that there is material on the web to help you understand this further. Hope my little insight helps you in figuring impulses out. FirstChildUserIdTAG: 150267 FirstChildUserNameTAG: vwsingh FirstChildCreateTimeTAG: 2012-11-18T07:46:26Z SecondChildTAG: Also, in the case of the capacitor all that energy is instantaneously stored as a voltage increase (positive delta) if the direction of the impulse is up or a voltage decrease (negative delta) if the direction of the impulse is down. This delta in volts is Q/C (Q is the charge) SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-18T11:25:26Z IndexTAG: 2001 TitleTAG: Bandwidth I have problems with the bandwidth. Using delta omega=2alpha, I can not get the correct results. UserIdTAG: 147472 UserNameTAG: rojorpa CreateTimeTAG: 2012-11-17T13:12:22Z VoteTAG: 1 CoursewareTAG: Week 11 / S22E2: The filter is ringing CommentableIdTAG: 6002x_S22E2_The_filter_is_ringing NumberOfReplyTAG: 1 FirstChildTAG: Problem solved. I was using omega, and not frequencies in Hertz. FirstChildUserIdTAG: 147472 FirstChildUserNameTAG: rojorpa FirstChildCreateTimeTAG: 2012-11-17T13:18:54Z SecondChildTAG: Don't feel bad. Stupid math errors like the one you mentioned bite me ALL THE TIME! ARGH! You'd think eventually I'd learn... SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-11-30T02:45:00Z IndexTAG: 2002 TitleTAG: H9P2 How can I get α? I don't know how can I get α. Can somebody help me? Thanks. UserIdTAG: 558384 UserNameTAG: Josue9740 CreateTimeTAG: 2012-11-17T13:01:26Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Write node equation for the circuit, and get it into the same form as the diff equations given. Now you can simply read A, B, and C from your node equ. Just make sure the form is the same. If you are not sure, check last page here https://www.edx.org/static/content-mit-6002x/handouts/6002-L18-oei12-gaps-annotated.1114c3162825.pdf FirstChildUserIdTAG: 329254 FirstChildUserNameTAG: KGabor FirstChildCreateTimeTAG: 2012-11-17T13:09:32Z SecondChildTAG: just compare your characteristic eqn with s2+2as+w2=0.u will get the answer SecondChildUserIdTAG: 611666 SecondChildUserNameTAG: Shaminder420 SecondChildCreateTimeTAG: 2012-11-17T13:20:31Z SecondChildTAG: Ohhhhhh!!! Thank you very much for your help. I could solved the exercise. SecondChildUserIdTAG: 558384 SecondChildUserNameTAG: Josue9740 SecondChildCreateTimeTAG: 2012-11-17T13:27:50Z IndexTAG: 2003 TitleTAG: ϕ note that ϕ is in radian, not degree. UserIdTAG: 201818 UserNameTAG: ThreeHundred CreateTimeTAG: 2012-11-17T12:58:59Z VoteTAG: 1 CoursewareTAG: Week 9 / S18E3 Total Solution CommentableIdTAG: S18E3_Total_Solution NumberOfReplyTAG: 0 IndexTAG: 2004 TitleTAG: H10P1 anyone? Hey ,i need some hints on homework 10 part 1.desperately trying to get an answer but not a single tick. UserIdTAG: 357747 UserNameTAG: kishores CreateTimeTAG: 2012-11-17T10:49:18Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi, for each of the circuits perform the frequency response (Magnitude plot and Phase Plot). Look S19V14. Find what the magnitude looks like for low w and high w (first graph). Also look how phi behaves for low and high w (second graph). The inflection point is useful to compare the w (more to the left or right) between graphs. FirstChildUserIdTAG: 288174 FirstChildUserNameTAG: matiasgrodriguez FirstChildCreateTimeTAG: 2012-11-17T13:36:06Z SecondChildTAG: ok.thanks.Is the angle Vo/Vi in radians? SecondChildUserIdTAG: 357747 SecondChildUserNameTAG: kishores SecondChildCreateTimeTAG: 2012-11-17T14:49:38Z SecondChildTAG: Yes, it is in radians SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-17T19:04:37Z IndexTAG: 2005 TitleTAG: Lab 9 evaluation problem HI Everyone, I'm unable to evaluate Lab 9. The check button seems to have stopped working. Any hints on how to resolve this. Tried it on two OS - WIN7 and UBUNTU 12.04 with Firefox 15.0.1/16.0.2 & chrome 23.0 without luck. I won't be able to complete lab 9 before the deadline if this is not resolved. V. UserIdTAG: 150267 UserNameTAG: vwsingh CreateTimeTAG: 2012-11-17T10:16:00Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Well seems to have started working again...... V. FirstChildUserIdTAG: 150267 FirstChildUserNameTAG: vwsingh FirstChildCreateTimeTAG: 2012-11-17T10:20:26Z IndexTAG: 2006 TitleTAG: H9P1 Question 4 Hi, I have used the Laplace transform to come up with the equation for the current through the inductor as: $i_L(t) = cos(\omega_0t)$ When asked to solve for $i_L(5_-)$ I substituted $t$ with 5 giving $i_L(5_-) =cos\left(\sqrt{\frac{1}{LC}}\times 5\right)$ I used my calculator in radians mode to find the answer but I am still not getting it right. I have been breaking my head over it the whole day and am not able to find my error, any assistance on this problem would be greatly appreciated. Thank you in advance. UserIdTAG: 85490 UserNameTAG: yashsavani CreateTimeTAG: 2012-11-16T19:51:02Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Adjust your frequency so that your period is an **integer** number of seconds. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-16T19:58:04Z SecondChildTAG: Thank you so much. I was unaware that my answer was dependant on the frequency from the prior question. SecondChildUserIdTAG: 85490 SecondChildUserNameTAG: yashsavani SecondChildCreateTimeTAG: 2012-11-16T20:00:09Z SecondChildTAG: I cant get it ?my frequency is 0.25 i put cos(0.25*5) but the answere is wrong((( whY? SecondChildUserIdTAG: 338422 SecondChildUserNameTAG: Sl1mcO SecondChildCreateTimeTAG: 2012-11-17T12:24:40Z SecondChildTAG: Remember that it is cos(w*t) where w is 2*pi*frequency. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-11-17T12:30:34Z SecondChildTAG: I did it like that!But still i have no right answer(( SecondChildUserIdTAG: 338422 SecondChildUserNameTAG: Sl1mcO SecondChildCreateTimeTAG: 2012-11-17T13:09:23Z SecondChildTAG: make sure that unit of angle is radian in your calculator SecondChildUserIdTAG: 473933 SecondChildUserNameTAG: rahul_pradhan SecondChildCreateTimeTAG: 2012-11-17T13:24:35Z SecondChildTAG: Yep i got it thanks a lot!)) SecondChildUserIdTAG: 338422 SecondChildUserNameTAG: Sl1mcO SecondChildCreateTimeTAG: 2012-11-17T13:40:14Z SecondChildTAG: nice work SecondChildUserIdTAG: 307862 SecondChildUserNameTAG: HarshulNag SecondChildCreateTimeTAG: 2012-11-17T16:23:18Z FirstChildTAG: @skyhawk, why should the period be an integer? That did make a difference to the answer (and gave me a green tick) but I don't understand why :-/ FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-11-17T14:18:29Z SecondChildTAG: It was just a guess that the component values were chosen to give integer periods (In my case it was true to 3 significant digits). It just seemed the intent given how the remainder of the questions were expressed. I look or patterns. I don't blindly punch numbers into a calculator. In the end, just luck I suppose! SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-11-18T00:24:08Z SecondChildTAG: Hahahaha that was smart! Thanks! :-) SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-11-18T02:36:26Z FirstChildTAG: thanks guys! FirstChildUserIdTAG: 477713 FirstChildUserNameTAG: ikm104 FirstChildCreateTimeTAG: 2012-11-18T10:14:12Z IndexTAG: 2007 TitleTAG: H10P1: MAGNITUDE AND ANGLE , what should I do? Hi, I dont know what to do, if I should calculation with the grafics values or with equations with some w?, In this case what are those equations? Thanks! UserIdTAG: 299251 UserNameTAG: TheRedBlackOne CreateTimeTAG: 2012-11-16T18:20:42Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Are you given the break frequency? Look at the phase graphs. What phase value do you expect at break frequency? Use the phase values and find the values of L and C. Hope that helps. FirstChildUserIdTAG: 265795 FirstChildUserNameTAG: krishnakm FirstChildCreateTimeTAG: 2012-11-16T22:48:18Z IndexTAG: 2008 TitleTAG: Static D-Latch Vs D-Flip Flop? Just wondering which is better the static D-Latch which we learned as the 1 bit memory cell or a d-flip flop? The D-Flip flop uses nand gates and just seems like a smarter design, Here is a pic of it: ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13530808001343686.png I have used the d-flip flop many times and just was wondering for personal knowledge which is better, is there any advantage/difference using one or the other. UserIdTAG: 219780 UserNameTAG: ReconIII CreateTimeTAG: 2012-11-16T15:50:56Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 2009 TitleTAG: Lab 10: L and C values I have found the break frequencies for the RC and RL circuit and correctly identified the plots. However, I can't seem to get the correct values for L and C. I am not sure what I am doing wrong. I set the break frequency to 2*pi*f and solve for L and C. Am I missing something? Thanks. UserIdTAG: 317112 UserNameTAG: judy913 CreateTimeTAG: 2012-11-16T14:13:43Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Are you given the break frequency? Look at the phase plots. What value do you expect for phase at break frequency? You can use the phase value and find L and C very easily. I hope that helps. FirstChildUserIdTAG: 265795 FirstChildUserNameTAG: krishnakm FirstChildCreateTimeTAG: 2012-11-16T22:45:51Z SecondChildTAG: At the break frequency, w=R/L=1 L=R/w=R/(2*Pi*f), and f= 4.6Hz? Am I right? But the answer is wrong? What do I do incorrectly? Please give me a hint? Thanks! SecondChildUserIdTAG: 371617 SecondChildUserNameTAG: rainbow12 SecondChildCreateTimeTAG: 2012-11-22T02:47:02Z SecondChildTAG: The frequency in 4.6 on a log scale, which means it's 10^4.6 Hz. Then you can you the numbers on the phase plot and the angle formula. SecondChildUserIdTAG: 252722 SecondChildUserNameTAG: vaboro SecondChildCreateTimeTAG: 2012-11-22T05:04:48Z SecondChildTAG: Thanks SecondChildUserIdTAG: 371617 SecondChildUserNameTAG: rainbow12 SecondChildCreateTimeTAG: 2012-11-23T02:07:44Z FirstChildTAG: Krishnakm, is the any information about your hint in textbook? FirstChildUserIdTAG: 96873 FirstChildUserNameTAG: Santyaga FirstChildCreateTimeTAG: 2012-11-18T10:26:28Z SecondChildTAG: It took me a lot to get the green mark on these L and C values :( Maybe there is an easier way... but this is how I finally did it. The magnitude plot is in 20*log(Vo/Vi) x log(freq in Hz) scale so I converted the interesting values back to its normal form: log^-1((Vo/Vi)/20) and log^-1(break down freq)*2$\pi$ You should find the magnitude equation |Vo/Vi|in terms of w, C and R for the capacitor and w, L and R for the inductor. Just put the converted values that you found in the equations and find C and L. Note that log^-1 is the inverse of the log, is some calculator is the same as the 10^x button. Good luck SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-18T11:49:19Z FirstChildTAG: @judy913 I was stuck on this for a while as well and you do have the correct idea but are perhaps incorrectly estimating the break frequency and as it's on a log scale the grader is quite punishing. FirstChildUserIdTAG: 286002 FirstChildUserNameTAG: AndrewMacLachlan FirstChildCreateTimeTAG: 2012-11-20T15:46:37Z IndexTAG: 2010 TitleTAG: H10P1 tried everything but failed.the green tick remains elusive.besides,is the angle in radians? UserIdTAG: 357747 UserNameTAG: kishores CreateTimeTAG: 2012-11-16T13:37:55Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: that doesn't matters actually. it can be solved just by comparison FirstChildUserIdTAG: 446998 FirstChildUserNameTAG: kumar_vikash FirstChildCreateTimeTAG: 2012-11-17T18:27:57Z IndexTAG: 2011 TitleTAG: STAFF: Please see this thread re: "Math Review"! [Pointers Referenced in S18V2 Missing?][1] The "Math Review" section from the prototype course is missing from this version's "Course Info" section. Thanks! [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/509fb20c6d91002300000030 UserIdTAG: 148542 UserNameTAG: planetscape CreateTimeTAG: 2012-11-16T12:16:21Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: fixing.... FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-11-16T15:17:24Z SecondChildTAG: Adding the previous year's wiki to this year's. A link from course info is forthcoming, but for now, here is it https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/math-review/ SecondChildUserIdTAG: 96452 SecondChildUserNameTAG: Lyla SecondChildCreateTimeTAG: 2012-11-16T17:26:35Z SecondChildTAG: Thank you so much! SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-11-16T17:27:11Z IndexTAG: 2012 TitleTAG: h9p1 for calculating il at t=9- i got the eqn and the time period too but not able to get answer .can anyone help UserIdTAG: 611666 UserNameTAG: Shaminder420 CreateTimeTAG: 2012-11-16T07:16:38Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 2013 TitleTAG: Is S17E5 wrong? Im talking about parts 5 & 6. Could anyone explain where the differential equation in this problem differs from the example differential equation in the sequence of videos? Aren't them LC*iL''(t)+iL(t)=I0, and vL(t)=vC(t)=L*iL'(t)? Even initial conditions are the same. I think my theory is proved? by the following image. Current iL is the pink line and it follows iL[t] = I0 - I0*Cos[w0*t] equation. Voltage vC=vL is blue line and it follows vC[t] = L*I0*w0*Sin[w0*t] equation. The problem relates iL to sin function and vC to cosine function, when I think it's the opposite. ![enter image description here][1] Where am I wrong? Thank you! [1]: https://edxuploads.s3.amazonaws.com/13529347191343667.jpg UserIdTAG: 398594 UserNameTAG: DaveyJC CreateTimeTAG: 2012-11-14T23:33:27Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 2014 TitleTAG: [S18V21] - Error in the equation In writing the differential equation for the parallel RLC circuit shouldn't the last $i$ be divided by $C$ as well? UserIdTAG: 346056 UserNameTAG: fiatlux CreateTimeTAG: 2012-11-14T23:13:06Z VoteTAG: 1 CoursewareTAG: Week 9 / Driven_Parallel_RLC_Circuit CommentableIdTAG: 6002x_driven_parallel_RLC_circuit NumberOfReplyTAG: 0 IndexTAG: 2015 TitleTAG: help plz someone plz help me with homework 9 p1 all problems i didn't understood anything i saw hints by myrimit but still struggling plzz someone help!!!confused totally UserIdTAG: 285222 UserNameTAG: dhaval24 CreateTimeTAG: 2012-11-14T17:01:18Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: where did you see Myrimit's hints?..I didnt even get that..will have to search again.. FirstChildUserIdTAG: 173147 FirstChildUserNameTAG: cruiser_rahit FirstChildCreateTimeTAG: 2012-11-14T19:59:18Z SecondChildTAG: its better not to go with maths but go conceptually.. SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-11-17T07:32:11Z IndexTAG: 2016 TitleTAG: iCircuit App. Anyone used this? Has anyone tried the iCircuit app for mobile devices? Any reviews/thoughts on it? http://icircuitapp.com/ for those who don't know. I'm wondering if it's worth the $9.99 on the app store. UserIdTAG: 213386 UserNameTAG: dmascenik CreateTimeTAG: 2012-11-14T15:51:47Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: I have it for iPad. I wouldn't try using on a device smaller than an iPad mini. It is unusable on an iPod Touch. I like it. It's not perfect, but I have found it useful. It has it's own quirks, but works well enough. I think it's 10 dollars worth of fun. Circuitlab is also good, and LTSpice is good and free(although only for Windows). FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-11-14T16:16:18Z FirstChildTAG: I am very happy with [CircuitLab][1], and there is also the [NGSPICE Online simulator][2] for those inclined. As the site says, "Whether you are using a Windows, Linux, or Mac computer, a smart phone, or a tablet, you always should be able to run a SPICE simulation!" [1]: https://www.circuitlab.com/ [2]: http://www.ngspice.com/ FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-11-15T06:57:57Z FirstChildTAG: CircuitLab is also made by some MIT EECS grads, so they took this course, just like you and me. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-11-15T17:53:45Z IndexTAG: 2017 TitleTAG: Lab10 By using the circuit sandbox ,I finally managed to find C and L by trying various values. However I still can not calculate these values from the equations. It seems I am missing the correct way to read the frequency on the horizontal log axis.To my understanding point 4.6 on this axis means 6000 Hz.As the axis begins to 10^0 which corresponds to 1 ,4 on the axis corresponds to 10^3 .Is this right ? UserIdTAG: 38331 UserNameTAG: janadel CreateTimeTAG: 2012-11-14T14:40:33Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: If I remember correctly, I took the 4.6 to correspond to 10^4.6 or approximately 39,810. I seem to have answered the questions without difficulty. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-14T15:21:40Z SecondChildTAG: I have also taken the same thing. but for me its not coming correct SecondChildUserIdTAG: 365309 SecondChildUserNameTAG: chetnasinghaldas SecondChildCreateTimeTAG: 2012-11-23T05:13:46Z FirstChildTAG: rather than taking 4.6 take 4.49 as the value......really 0.1 difference is also causing error FirstChildUserIdTAG: 214277 FirstChildUserNameTAG: yadsam FirstChildCreateTimeTAG: 2012-11-25T13:54:43Z IndexTAG: 2018 TitleTAG: LC Circuit In week 9 ,under undamped 2nd order system,**in LC circuit we got max. voltage across the capacitor is twice of the input voltage..how thats possible?** can anyone explain it(**not mathematically but conceptually)?** may be i have missed the explanation in the lecture..a big thanks in advance UserIdTAG: 114864 UserNameTAG: Vikaash CreateTimeTAG: 2012-11-14T12:42:19Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: watch S17V14 :) FirstChildUserIdTAG: 357747 FirstChildUserNameTAG: kishores FirstChildCreateTimeTAG: 2012-11-14T15:21:23Z FirstChildTAG: Apparently even though the voltage accross the inductor is zero when the capacitor is fully charged, it still 'pumps' current, and the capacitor takes that current and raises it's voltage accordingly. FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-11-14T21:10:53Z SecondChildTAG: hazel1919, on a side note, you should make a tutorial about doing week 8 equations in Wolfram. You are the master of WA, show us how it's done. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-11-14T22:34:33Z SecondChildTAG: I second that! hazel1919, you're indispensable! SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-11-15T06:06:07Z IndexTAG: 2019 TitleTAG: mouse stick when i login , my mouse pointer just stick to the screen for 1-2min..is it normal?? UserIdTAG: 114864 UserNameTAG: Vikaash CreateTimeTAG: 2012-11-14T12:23:03Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I have the same experience. I believe (though I have not dug into it) that this is the result of the site initially loading some stuff like MathJax, etc., to facilitate the interactive experience. FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-11-14T12:35:18Z IndexTAG: 2020 TitleTAG: h9p2-d i am getting the answer for natural frequency as (1/(L*C)-(1/(2*R*C))^2)^(1/2). but it is showing as wrong. tell me where i am goin wrong UserIdTAG: 115703 UserNameTAG: munimunna CreateTimeTAG: 2012-11-14T11:30:49Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: You can just look up the formula for the natural frequency. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-11-14T14:23:12Z SecondChildTAG: w0/2*pi SecondChildUserIdTAG: 174229 SecondChildUserNameTAG: fares27 SecondChildCreateTimeTAG: 2012-11-15T17:28:05Z SecondChildTAG: So why does it say pi is not valid if I enter 1/(2*\pi*sqrt(L*C)) ? SecondChildUserIdTAG: 329254 SecondChildUserNameTAG: KGabor SecondChildCreateTimeTAG: 2012-11-17T13:26:02Z SecondChildTAG: well, it turns out that 1/(2pi) is != 0.159, but 0.1591549. SecondChildUserIdTAG: 329254 SecondChildUserNameTAG: KGabor SecondChildCreateTimeTAG: 2012-11-17T14:17:55Z FirstChildTAG: It wants the undamped natural frequency in Hz. FirstChildUserIdTAG: 141000 FirstChildUserNameTAG: OrinE FirstChildCreateTimeTAG: 2012-11-14T17:35:44Z SecondChildTAG: if Wo is the frequency in rad/sec. then Wo/(6.28) is the frequency in hz. am i right? SecondChildUserIdTAG: 115703 SecondChildUserNameTAG: munimunna SecondChildCreateTimeTAG: 2012-11-15T04:32:19Z SecondChildTAG: Same problem. I made circuit in sandbox, got correct results - the same from formulas and from AC analisys, but answer is shown as wrong! SecondChildUserIdTAG: 86420 SecondChildUserNameTAG: Aleksei_Katkov SecondChildCreateTimeTAG: 2012-11-15T07:28:59Z FirstChildTAG: This has got nothing to solve.. FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-11-15T06:24:16Z IndexTAG: 2021 TitleTAG: H10P3 parts d and e Can anyone please give some hints or help with C match and L match? Been working on this for a few days and getting nowhere. UserIdTAG: 18073 UserNameTAG: gburkhart CreateTimeTAG: 2012-11-14T04:22:06Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Equate the complex impedance to R1. Get equations for real and imaginary parts. Solve. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-15T03:45:16Z IndexTAG: 2022 TitleTAG: H9P1 part 7 This is the question: > At the time just after the impulse happens what is the voltage > vC(1.0+), in Volts, across the capacitor? I thought I would had to sum the charge sent from current source divided by the capacitance to the voltage at $v_C(1_{-})$. Apparently I'm wrong... Can somebody help me? --Excuse my english. UserIdTAG: 27938 UserNameTAG: pg1992 CreateTimeTAG: 2012-11-14T02:09:45Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Pay close attention to the sign of the charge. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-14T02:48:54Z SecondChildTAG: I use the same way with pg1992. Q/C+Vc(1-), the answer is 0.221443463147 if you use pi/2 but not 0.64, but it's still not right. SecondChildUserIdTAG: 564606 SecondChildUserNameTAG: ArequipaRadio SecondChildCreateTimeTAG: 2012-11-14T16:36:13Z SecondChildTAG: Hi ArequipaRadio, You are definitely on the right track, but the answer you give above is too big by three orders of magnitude. However, even typing in the smaller answer won't necessarily earn you a check mark. Consider that the problem is set out with a statement that the numbers are chosen to eliminate difficult math. Your method is correct, but perhaps you are using too much precision in your answer. How might you approximate a number that is a few orders of magnitude less than one? SecondChildUserIdTAG: 166031 SecondChildUserNameTAG: krebryna SecondChildCreateTimeTAG: 2012-11-15T03:34:13Z FirstChildTAG: Hi Skyhawk, Thanks for the reply. Actually I had the green check mark yet. I just wonder why the margin of error is so small here, therefore it should have some reason. If I used the right equation, the answer should be in the range. I don't think it's a good idea to round a number here(you just kill it :p), because it has a totally different meaning. I hope an Aha moment in the explanation. FirstChildUserIdTAG: 564606 FirstChildUserNameTAG: ArequipaRadio FirstChildCreateTimeTAG: 2012-11-15T13:38:18Z SecondChildTAG: Hey ArequipaRadio, I supose you are peruvian like me, it'd be glad to share answers or study togetter. About the problem, it's an ideal problem and you should all the time make asumptions to get the answer, pay atention to the capacitor's voltage just at t=9_ and sum it to voltage produced by impulse at t=9+, use rounded numbers and get the answer. SecondChildUserIdTAG: 148389 SecondChildUserNameTAG: chento SecondChildCreateTimeTAG: 2012-11-15T22:19:02Z FirstChildTAG: It looks like H9P1 has turned out to be a nightmare for a number of people. The components values were designed to give nice numbers. Unfortunately those who didn't round had trouble getting check marks. It will be interesting to see if panic sets in on Sunday evening. An impulse of 2/pi does produce the "correct" answer if the period of oscillation is rounded to an integer. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-15T14:57:52Z IndexTAG: 2023 TitleTAG: H9P2 Question F So if the quality factor is Q=w0/2a, and question d is the natural frequency, and question e is alpha, then why isn't the answer to question f: (answer to quesiton d)/2*(answer from question e)? Am I missing something? UserIdTAG: 21541 UserNameTAG: JSChambers CreateTimeTAG: 2012-11-13T18:48:09Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Answer to d in Hz, but omega naught in rad/sec? FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-13T19:11:55Z SecondChildTAG: oh right. Duh. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-11-13T19:17:18Z SecondChildTAG: Thanks skyhawk, sometimes my brain just isn't working. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-11-13T19:20:43Z FirstChildTAG: There is a number of discussions of H9P2 here. [This][1] may help. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/509812caad8cab230000008a FirstChildUserIdTAG: 329361 FirstChildUserNameTAG: ikrukov FirstChildCreateTimeTAG: 2012-11-13T19:15:35Z SecondChildTAG: I read your other post. I think they just want the answer in Hz instead of in radians. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-11-13T19:33:26Z IndexTAG: 2024 TitleTAG: final exam in final exam.. how many papers ...wether we have to give proctored exam or exam just like midterm pattern ? please enlight me on this topic!!! UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-11-13T11:03:13Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 2025 TitleTAG: Homeworks-Labs grading doubt Hello. In the last week I hadn't time to complete the week 8 Homeworks, but I completed all the past weeks. I can finish the Course then? I have read that you can miss 2 homeworks and 2 labs without a grade penalty. Is it true?. Thanks. UserIdTAG: 270769 UserNameTAG: 2214sanchez CreateTimeTAG: 2012-11-13T09:52:26Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Yes..only the 10 labs and homeworks with the best scores will be included in the grades..so as long as you do at least 3 of the 4 remaining ones, you have nothing to worry about..just make sure you learn the one that you missed..because it might come in handy during the final exam..:) FirstChildUserIdTAG: 173147 FirstChildUserNameTAG: cruiser_rahit FirstChildCreateTimeTAG: 2012-11-13T11:16:30Z FirstChildTAG: Yes, that is true, you can skip 2 Hw and 2 Labs without affecting your final grade. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-13T11:30:26Z FirstChildTAG: Thanks for response It's great FirstChildUserIdTAG: 270769 FirstChildUserNameTAG: 2214sanchez FirstChildCreateTimeTAG: 2012-11-13T18:24:59Z IndexTAG: 2026 TitleTAG: Lab 9 hints? Hi..can anyone put up hints to do lab 9?..Myrimit? UserIdTAG: 173147 UserNameTAG: cruiser_rahit CreateTimeTAG: 2012-11-13T05:33:45Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hi cruiser_rahit, I will try to post the Lab9 Hints :) FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-13T11:28:18Z SecondChildTAG: Thanks a lot.. :) SecondChildUserIdTAG: 173147 SecondChildUserNameTAG: cruiser_rahit SecondChildCreateTimeTAG: 2012-11-13T14:45:50Z SecondChildTAG: [Lab9 Hints][1] :). As I have promised to you, sorry for the delay. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50a78f0dce1ad22700000024 SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-17T13:52:04Z FirstChildTAG: Hi Myrimit, when can you post Lab 9 hints? FirstChildUserIdTAG: 536922 FirstChildUserNameTAG: arjshar FirstChildCreateTimeTAG: 2012-11-14T00:48:33Z SecondChildTAG: [Lab9 Hints][1] :) . Sorry for the delay. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50a78f0dce1ad22700000024 SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-17T13:52:32Z IndexTAG: 2027 TitleTAG: H8P3 Part 5 : the posted 'Answer: 31.7995722274' doesn't seem right. Is the posted Answer: 31.7995722274 correct? Two issues: 1) R = Ron + Ron = 2150+2150 = 4300, based on this R and (VOL/VOH)(0.5/3.5): t=29.286 2) If based on (VIL/VOH)(0.9/3.5): t=20.44 UserIdTAG: 370323 UserNameTAG: LeyonLee CreateTimeTAG: 2012-11-12T20:18:49Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Troubleshooting NumberOfReplyTAG: 2 FirstChildTAG: please, see "H8P3 - mistake in answer explanation?" FirstChildUserIdTAG: 188778 FirstChildUserNameTAG: xsaq FirstChildCreateTimeTAG: 2012-11-12T20:52:51Z SecondChildTAG: Thanks! SecondChildUserIdTAG: 370323 SecondChildUserNameTAG: LeyonLee SecondChildCreateTimeTAG: 2012-11-13T01:05:25Z FirstChildTAG: See this: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50a10d282dcf07230000004f and this: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50a13cfc28585d230000005e FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-11-12T21:45:44Z SecondChildTAG: Thanks! SecondChildUserIdTAG: 370323 SecondChildUserNameTAG: LeyonLee SecondChildCreateTimeTAG: 2012-11-13T01:06:02Z IndexTAG: 2028 TitleTAG: S18V18: Number of rings Isn't there an infinite number of rings? Each ring decreases in amplitude due to the decaying exponential, but the amplitude will never reach 0. So how can we say 'there are x number of rings'? What is the 'cutoff amplitude' from which we consider the rings to be negligible? UserIdTAG: 341293 UserNameTAG: Pietr CreateTimeTAG: 2012-11-12T19:56:55Z VoteTAG: 1 CoursewareTAG: Week 9 / Intuitive Analysis of Second Order Circuits CommentableIdTAG: 6002x_intuitive_analysis_of_second_order_circuits NumberOfReplyTAG: 0 IndexTAG: 2029 TitleTAG: Error in statement of H11P3 The third question in H11P3 should be Which branch voltage is labeled W? Enter R, L, or C in the space provided. and not Which branch voltage is labeled U? Enter R, L, or C in the space provided. The latter statement is the same as the first one. UserIdTAG: 153760 UserNameTAG: Kavka CreateTimeTAG: 2012-11-12T18:57:04Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Already noted. https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50a0ef1cc5c9722700000029 FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-12T19:44:21Z IndexTAG: 2030 TitleTAG: Making videos available in the wiki I think the lecture sequence videos should be made available in wiki for downloading. Youtube is still not operational in my country and the download links that appears below the videos are not helpful too. UserIdTAG: 162671 UserNameTAG: tuhin1991paul CreateTimeTAG: 2012-11-12T18:12:14Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 2031 TitleTAG: Staff - Typo H11P3 Third answer Question says branch U instead of branch W. UserIdTAG: 406202 UserNameTAG: skyhawk CreateTimeTAG: 2012-11-12T12:44:12Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Thank you for spotting that. Working on it now. - Rohan FirstChildUserIdTAG: 391929 FirstChildUserNameTAG: RohanNagarkar FirstChildCreateTimeTAG: 2012-11-13T13:15:10Z IndexTAG: 2032 TitleTAG: Staff:answer evaluation For H8P3 question 4, an answer first i got tick mark. After sometime i gave check with same answer, but it was wrong. pls clarify UserIdTAG: 624120 UserNameTAG: senthil1 CreateTimeTAG: 2012-11-12T10:54:33Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: maybe the values are changed. I wrote my values on paper to calculate, after some time I realized that my values are changed, so that for 2 hours I calculated with wrong values. FirstChildUserIdTAG: 389333 FirstChildUserNameTAG: juergen FirstChildCreateTimeTAG: 2012-11-12T22:14:40Z IndexTAG: 2033 TitleTAG: H9P1 wrong way? Hi all, I assumed for t<=9s d2iL(t)/d2t +iL(t)/LC = 0, which leads me to s^2 + wo^2 = 0 as the characteristic ecuation for this circuit ( before impulse) . Can anyone tell me what is wrong with this? thanks! UserIdTAG: 369339 UserNameTAG: vargaslen CreateTimeTAG: 2012-11-12T10:43:19Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Double-check this result with what you should have gotten as the characteristic equation for [S17E5: An ILC circuit][1] - they are the same circuit. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_9/Undamped_Second-Order_Systems/ FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-11-17T17:05:39Z IndexTAG: 2034 TitleTAG: Incorrect target location in textbook. Strangely the link for More information in the text is incorrect; it points to the chapter about Power Dissipation. Perhaps the correct one should be [this][1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/651 [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/651 UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-11-12T02:47:29Z VoteTAG: 1 CoursewareTAG: Week 9 / Characteristic Equation CommentableIdTAG: 6002x_characteristic_equation NumberOfReplyTAG: 0 IndexTAG: 2035 TitleTAG: Lab doesn t work for me correctly With chrome, the value is not always correctly updated for the calculation while onscreen correct values appear. Plz don t rely on graphic resolution else some student are gonna be really handicaped for returning the exams. thank you for your time UserIdTAG: 604236 UserNameTAG: philfrance CreateTimeTAG: 2012-11-11T23:34:53Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Troubleshooting NumberOfReplyTAG: 1 FirstChildTAG: Please, take screenshots and let us see the problem. Also, what browser, browser version, and OS you are on. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-11-12T01:26:56Z IndexTAG: 2036 TitleTAG: H8P3 - circuit doesn't satisfy VOL of static discipline? Hi, I must be forgetting something. If $R_{ON} = 1900\Omega$ and $R_{PU} = 14000\Omega$, doesn't that mean that when you calculate the low output by the voltage divider pattern, it will be 5.0 * 1900 / (1900+14000) = .59V, higher than the stated $V_{OL}$ of .5V? Rob UserIdTAG: 468623 UserNameTAG: RobNik CreateTimeTAG: 2012-11-11T22:31:44Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: It would appear to be so. That's what I concluded, and I reasoned that's why VOH was used in the problem. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-11T23:44:46Z IndexTAG: 2037 TitleTAG: Circuit sandbox CS doesn't work for a day or so. Yesterday night it has stopped correctly working, and it still cannot produce correct results. I have tried Chrome and IE too, with the same result. Do anyone knows about it something? THX - G UserIdTAG: 151942 UserNameTAG: GyuriK CreateTimeTAG: 2012-11-11T22:10:08Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: What OS are you on? Can you post a screenshot of the bad results? FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-11-12T01:27:29Z SecondChildTAG: Windows 8 with IE 10.00.9200.16384 and Chrome 22.0.1229.94 Thanks... ![Sandbox error][1] [1]: http://s1298.beta.photobucket.com/user/gyurik/media/error.png.html?sort=3&o=0 SecondChildUserIdTAG: 151942 SecondChildUserNameTAG: GyuriK SecondChildCreateTimeTAG: 2012-11-12T02:14:20Z SecondChildTAG: I can't see your screenshot, can you re-send? SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-11-12T04:59:43Z IndexTAG: 2038 TitleTAG: Lab 10 I have correctly found the formulas for the break frequency of both circuits.I have correctly determined which plot comes from which circuit.However I can not correctly determine L and C. I use the formulas I found for the the break frequency and equated with the break frequency noted on the plot but I don't get a correct answer. What am I doing wrong ? Thanks for your help. UserIdTAG: 38331 UserNameTAG: janadel CreateTimeTAG: 2012-11-11T17:53:01Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I had the same problem - I used the circuit sandbox to check and guess the value for C then used that to back calculate the correct place from the graph. I discovered that I was not being careful when I read the graph - I assumed that the break freq spot was on the vertical lines provided on the log plot but alas no. Once I figured out C I had no problem with L. FirstChildUserIdTAG: 18073 FirstChildUserNameTAG: gburkhart FirstChildCreateTimeTAG: 2012-11-12T03:01:30Z SecondChildTAG: Thanks for the tip. By using the circuit sandbox ,I finally managed to find C and L by trying various values .However I still can not calculate these values from the equations. It seems I am missing the correct way to read the frequency on the horizontal log axis.To my understanding point 4.6 on this axis means 6000 Hz.As the axis begins to 10^0 which corresponds to 1 ,4 on the axis is 10^3 .Is this right ? SecondChildUserIdTAG: 38331 SecondChildUserNameTAG: janadel SecondChildCreateTimeTAG: 2012-11-12T07:10:22Z SecondChildTAG: 10^4.6 SecondChildUserIdTAG: 206648 SecondChildUserNameTAG: TsvetanGeorgiev SecondChildCreateTimeTAG: 2012-11-23T18:22:51Z IndexTAG: 2039 TitleTAG: Pointers Referenced in S18V2 Missing? Question for staff. At the end of S18V2, Prof. A. mentions that. "...if you really want to understand how to solve these second-order differential equations and so on, **if you go to the course info section**, we give you some pointers to videos and so on that describe second-order differential equations and give you some background review material." I looked but did not see any. Am I missing something, or are the aforementioned pointers not there? Any information/help you can provide would be appreciated. Thanks. UserIdTAG: 339668 UserNameTAG: chickwebb CreateTimeTAG: 2012-11-11T14:11:24Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: They were part of the prototype version of this course, and IMHO, should also be linked here. I am not certain if you will be able to access these links if you do not have an account under the prototype course, but here they are, just in case: [Math Review][1] [First Order Differential Equations][2] [Second Order Differential Equations (Khan Academy)][3] [Complex Numbers][4] Non-rigorous treatments for first and second order equations - [wiki on first order circuits][5] - [wiki on second order circuits][6] For a different approach to both first and second order differential equations (just a brief review, no rigor), also see this [wiki article][7] Hope this helps! [1]: https://6002x.mitx.mit.edu/wiki/view/MathReview [2]: http://6002x.mitx.mit.edu/section/MathReview1/ [3]: http://www.khanacademy.org/math/differential-equations/v/2nd-order-linear-homogeneous-differential-equations-1 [4]: https://6002x.mitx.mit.edu/section/wk10_Complex/ [5]: https://6002x.mitx.mit.edu/wiki/view/CapacitorsandFirstOrderCircuits [6]: https://6002x.mitx.mit.edu/wiki/view/WorkedProblemsonTransientsinSecondOrderCircuits [7]: https://6002x.mitx.mit.edu/wiki/view/LinearOrdinaryDifferentialEquations FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-11-16T12:13:47Z FirstChildTAG: Adding the previous year's wiki to this year's. A link from course info is forthcoming, but for now, here is it https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/math-review/ FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-11-16T17:26:21Z IndexTAG: 2040 TitleTAG: Impulse simulation Has anybody tried to play with the impulse source in the sandbox? I think it's quite instructive to check the differences between the simulated/practical results and the theoretical ones (using the Dirac impulse): the results are only slightly different. You have to be careful with the area because in the sandbox the pulse is triangular shaped; it can also be useful, for example, to simulate the notorious homework 8 part 1. I get similar results using another simulator, gEDA, where a trapezoidal pulse is available; in fact, as prof. Agarwal says, it doesn't matter the pulse waveform, it's important that the pulse duration is much much smaller than the time constant. Have fun. UserIdTAG: 376877 UserNameTAG: AndBre CreateTimeTAG: 2012-11-11T11:48:07Z VoteTAG: 1 CoursewareTAG: Week 8 / Area CommentableIdTAG: 6002x_Area NumberOfReplyTAG: 0 IndexTAG: 2041 TitleTAG: Confusion about Q/T Could anybody explain, please, why for each curve $\frac{Q}{T}$ is equals. T for each curve is different, so $\frac{Q}{T}$ also should be different. I think it should be $\frac{Q}{T_1}$ for blue, $\frac{Q}{T}$ for red, $\frac{Q}{T_2}$ for green... In the video, voltage response curves looks different for each curve, but rising equation for each is equals: $\frac{Q}{T}R(1-e^{-\frac{t}{RC}})$. How does it possible? UserIdTAG: 322444 UserNameTAG: koluch CreateTimeTAG: 2012-11-11T11:35:55Z VoteTAG: 1 CoursewareTAG: Week 8 / Response To A Pulse As Pulse Gets Narrower Continued CommentableIdTAG: 6002x_Response_To_A_Pulse_As_Pulse_Gets_Narrower_Continued NumberOfReplyTAG: 0 IndexTAG: 2042 TitleTAG: Grading System What is the required percentage for A grade? UserIdTAG: 499671 UserNameTAG: maliha266 CreateTimeTAG: 2012-11-11T08:02:56Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: [atleast 87%..for more info..Look at here..][1] [1]: https://www.edx.org/static/content-mit-6002x/handouts/syllabus.a477535058a1.pdf FirstChildUserIdTAG: 114864 FirstChildUserNameTAG: Vikaash FirstChildCreateTimeTAG: 2012-11-11T08:06:56Z FirstChildTAG: Hi *maliha266*, [GRADING SYSTEM][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/509f6a7bb6f29f1f00000020 > **Regards:** asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-11-11T09:08:45Z IndexTAG: 2043 TitleTAG: H9P1 I solved first three question but unable to get any idea about other question of H9P1. Please Help UserIdTAG: 456732 UserNameTAG: anilkumar6745 CreateTimeTAG: 2012-11-11T05:17:35Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: This is an undamped LC circuit, so the initial energy is going to slosh back and forth between the inductor and capacitor. The current and voltage waveforms are going to be sinusoids. Hint for the 4th and 5th question: Since you've solved for the natural frequency and you are given initial conditions, you can work out where in the sinusoid for the current through the inductor the impulse occurs (draw it if necessary). You might be surprised... Once you have the current through the inductor, you can use the fact that energy is conserved to calculate the voltage across the capacitor. FirstChildUserIdTAG: 141000 FirstChildUserNameTAG: OrinE FirstChildCreateTimeTAG: 2012-11-11T06:44:02Z SecondChildTAG: Hi OrinE, It's wrong to assume d2iL(t)/d2t +iL(t)/LC = 0 befor impulse and work from there to find iL(9-)? SecondChildUserIdTAG: 369339 SecondChildUserNameTAG: vargaslen SecondChildCreateTimeTAG: 2012-11-12T10:49:30Z SecondChildTAG: How do you solve the second part of HW 9 PROBLEM 1? SecondChildUserIdTAG: 536922 SecondChildUserNameTAG: arjshar SecondChildCreateTimeTAG: 2012-11-14T01:00:05Z FirstChildTAG: Hi, [H9P1 Hints][1] [1]: http://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5098b2a1cd5fb82300000036 > **Regards:** asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-11-11T08:56:54Z SecondChildTAG: link doesnt seem to work.. SecondChildUserIdTAG: 173147 SecondChildUserNameTAG: cruiser_rahit SecondChildCreateTimeTAG: 2012-11-13T05:36:35Z FirstChildTAG: Start Now with the F=0.25 OK FirstChildUserIdTAG: 106816 FirstChildUserNameTAG: Laith FirstChildCreateTimeTAG: 2012-11-12T11:43:23Z SecondChildTAG: why? SecondChildUserIdTAG: 413869 SecondChildUserNameTAG: nlfigueiredo SecondChildCreateTimeTAG: 2012-11-12T22:03:19Z SecondChildTAG: F=2*pi*w0, and W0 is 1/(L*C) SecondChildUserIdTAG: 229979 SecondChildUserNameTAG: Hafezi SecondChildCreateTimeTAG: 2012-11-16T15:13:39Z SecondChildTAG: Sorry for the above Equation, it is wrong The Correct Equation is: W0=2*pi*f and w0=1/(L*C) SecondChildUserIdTAG: 229979 SecondChildUserNameTAG: Hafezi SecondChildCreateTimeTAG: 2012-11-16T15:16:49Z IndexTAG: 2044 TitleTAG: H8P1 b PLEASE~! TIME!! Regarding any length of time necessary to find the voltage at the first resistor? Относительно какого отрезка времени нужно находить напряжение на первом резисторе? UserIdTAG: 196404 UserNameTAG: Yel1owstone CreateTimeTAG: 2012-11-10T22:50:49Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Try to use: a) VR1 = R1 * delta(t)/L, where delta(t) for this instant is 1 b) VR1 = A *e^(-t/(R/L)) where A is the constant in time (0-) the part c) and d) is the same but with the apropiate time constant FirstChildUserIdTAG: 113249 FirstChildUserNameTAG: EGuarch FirstChildCreateTimeTAG: 2012-11-10T23:06:16Z SecondChildTAG: It's work. Thak you. But why? SecondChildUserIdTAG: 196404 SecondChildUserNameTAG: Yel1owstone SecondChildCreateTimeTAG: 2012-11-10T23:14:14Z SecondChildTAG: see Textbook,chapter 10 SecondChildUserIdTAG: 113249 SecondChildUserNameTAG: EGuarch SecondChildCreateTimeTAG: 2012-11-10T23:18:37Z SecondChildTAG: what page? SecondChildUserIdTAG: 196404 SecondChildUserNameTAG: Yel1owstone SecondChildCreateTimeTAG: 2012-11-10T23:23:17Z SecondChildTAG: Start at 504 and have a look. SecondChildUserIdTAG: 113249 SecondChildUserNameTAG: EGuarch SecondChildCreateTimeTAG: 2012-11-10T23:28:57Z SecondChildTAG: You can also see this post: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/509e68568e407d1f0000001b SecondChildUserIdTAG: 113249 SecondChildUserNameTAG: EGuarch SecondChildCreateTimeTAG: 2012-11-10T23:31:01Z SecondChildTAG: in part b R=R1?? SecondChildUserIdTAG: 316761 SecondChildUserNameTAG: gk_goel SecondChildCreateTimeTAG: 2012-11-11T09:15:07Z SecondChildTAG: how to solve e part? SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-11-11T10:34:39Z FirstChildTAG: Hi, [H8P1,H8P2, H8P3 Hints][1] > **Regards:** asadbhatti42 [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/509e68568e407d1f0000001b FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-11-11T08:54:24Z IndexTAG: 2045 TitleTAG: info needed can anyone let me know how many homeworks and labs we can miss at max in order to successfully complete the course? UserIdTAG: 320718 UserNameTAG: almas_rizvi CreateTimeTAG: 2012-11-10T19:46:42Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I have already missed 2 and a half homeworks...should I still continue with the course or not? FirstChildUserIdTAG: 320718 FirstChildUserNameTAG: almas_rizvi FirstChildCreateTimeTAG: 2012-11-10T19:55:58Z SecondChildTAG: By all means, yes! You can always retake the course in the spring (in which case I'll see you there). Besides missing 2 and a half homeworks isn't that bad! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-10T20:22:05Z SecondChildTAG: hazel1919 I will also see you there, I missed the midterm and honestly doubt I will make it this spring. almas_rizvi you are doing well if you all have missed is two homeworks since the worst 2 homeworks won't be counted as part of overall course assessment. SecondChildUserIdTAG: 183507 SecondChildUserNameTAG: obiradaniel SecondChildCreateTimeTAG: 2012-11-10T21:38:47Z SecondChildTAG: You should totally continue almas_rizvi :) Remember that you can skip two Homeworks without penalty [Syllabus][1]. [1]: https://www.edx.org/static/content-mit-6002x/handouts/syllabus.a477535058a1.pdf SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-11T00:48:06Z SecondChildTAG: That is to say, if you have, - 100 percent . HW1 - 100 percent . HW2 - 100 percent . HW3 - 100 percent . Hw4 - 100 percent . Hw5 - 100 percent . Hw6 - 100 percent . Hw7 - 100 percent . Hw8 - 100 percent . Hw9 - 100 percent . Hw10 - 100 percent . Hw11 - 100 percent . Hw12 You will get 100 percent of Hw. and your ponderation will be for the final score 30 percent. (100 percent/100 percent)*30 percent = 30 percent ---- Another example. - 100 percent . HW1 - 100 percent . HW2 - 100 percent . HW3 - 100 percent . Hw4 - **0 percent . Hw5** - **0 percent . Hw6** - 100 percent . Hw7 - 100 percent . Hw8 - 100 percent . Hw9 - 100 percent . Hw10 - 100 percent . Hw11 - 100 percent . Hw12 **The two lowest scores will not count.** You will obtain 100 percent. You will still get 100 percent of Hw. and your ponderation will be for the final score 30 percent. (100 percent/100 percent)*30 percent = 30 percent --- Another example. two Hw missed and done 50 percent of another. - 100 percent . HW1 - 100 percent . HW2 - 100 percent . HW3 - 100 percent . Hw4 - **0 percent . Hw5** - **0 percent . Hw6** - 100 percent . Hw7 - 50 percent . Hw8 - 100 percent . Hw9 - 100 percent . Hw10 - 100 percent . Hw11 - 100 percent . Hw12 So, here, you will have done from 10 Hw that counts for your grade,but you have done 50 percent of HW8. 100+100+100+100+100+100+100+100+100+50 = 950 (950/1000) * 100 percent = 95 percent HW and total contribution for final score (95/100)* 30 = 28.5 percent ---- Conclusion. You can skip 2 HW, but after the 2nd that you skip or you do it but not complete will affect your grade. I hope this can help you. So, my answer is yes, you should totally continue :). However, an advice, you should try to re-do the hw that you missed in order to practise for the Final Exam. See you, Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-11T01:10:41Z FirstChildTAG: Hi almas, You can score a zero on two homework's and two labs and still get 100 percent on the course, that means they drop the two lowest homework and lab scores and the end of the course. The lowest score for a pass is 60 percent so I can't tell you "how many homeworks and labs we can miss at max in order to successfully complete the course" because **that depends on what you got in the midterm and what you will get in the final**. FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-11-10T19:59:06Z IndexTAG: 2046 TitleTAG: LAB 8 unit Hi, I would like to double check the capacitor's value unit in LAB 8. It is farads, right? For example: 180pF = 0.00000000018F Thanks for help!! UserIdTAG: 282748 UserNameTAG: larryzhu CreateTimeTAG: 2012-11-10T17:33:48Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: Yes, in farads! FirstChildUserIdTAG: 239608 FirstChildUserNameTAG: guillegf84 FirstChildCreateTimeTAG: 2012-11-10T17:46:48Z FirstChildTAG: Right, farads. It accepts scientific notation, so I'd enter your example as 1.8e-10 rather than counting zeros after the decimal point. FirstChildUserIdTAG: 141000 FirstChildUserNameTAG: OrinE FirstChildCreateTimeTAG: 2012-11-10T17:46:03Z FirstChildTAG: You can use the submultiples of SI. For example, type exactly 180p for the number that you submitted. If you need the number of mycron notation, use the letter u (lowercase). FirstChildUserIdTAG: 361823 FirstChildUserNameTAG: EliasOak FirstChildCreateTimeTAG: 2012-11-10T18:11:39Z FirstChildTAG: Thanks for help! But I still can't get the green check. ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13525761279787253.jpg Then I use 180.605e-12. I made something wrong? Thanks!! FirstChildUserIdTAG: 282748 FirstChildUserNameTAG: larryzhu FirstChildCreateTimeTAG: 2012-11-10T19:38:27Z FirstChildTAG: sorry. I made a mistake. since the reading mark should be at 100ns not at 300ns. now everything is OK. FirstChildUserIdTAG: 282748 FirstChildUserNameTAG: larryzhu FirstChildCreateTimeTAG: 2012-11-10T20:55:55Z IndexTAG: 2047 TitleTAG: where is download link? where is download link? UserIdTAG: 341358 UserNameTAG: akbar13 CreateTimeTAG: 2012-11-10T17:09:57Z VoteTAG: 1 CoursewareTAG: Week 9 / LCOT CommentableIdTAG: 6002x_LCOT NumberOfReplyTAG: 1 FirstChildTAG: Staff, Kindly provide download link for this video. FirstChildUserIdTAG: 455950 FirstChildUserNameTAG: Wahabbaluch FirstChildCreateTimeTAG: 2012-11-13T07:08:05Z IndexTAG: 2048 TitleTAG: LAB 8 L-R AND C-R CKT ??? Can anyone give method to solve L-R ckt and C-R ckt. Is it different from R-L ckt?? UserIdTAG: 332321 UserNameTAG: akshayk CreateTimeTAG: 2012-11-10T13:49:38Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Look carefully at the associated graphs - they tell you what form ("Schick" or "Shook") the equations you are looking for will take. FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-11-10T16:25:03Z IndexTAG: 2049 TitleTAG: H8P1 Hi, I have problems to find vc. As in lekture S15V17 shown, vc is Q/C. For the inductor, I'm clear. But I don't get an green checkmark for the capacitor. Another question, the capacitor is in series with the resistor. To chanche to Norton, the resistor is equal. The pulse is with the energy of Q and doesn't change. So I can transform from thevnin to norton without changes the values. Is that right? UserIdTAG: 389333 UserNameTAG: juergen CreateTimeTAG: 2012-11-10T12:15:13Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Can you help me with the inductor? I'm clear with the capacitor. FirstChildUserIdTAG: 558384 FirstChildUserNameTAG: Josue9740 FirstChildCreateTimeTAG: 2012-11-10T17:26:10Z FirstChildTAG: Hint: $R_{TH} = R_N$, but $V_{TH} \neq I_N$. You can use the former equivalency to find either $V_{TH}$ or $I_N$ if you have the other two values. FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-11-10T16:28:39Z FirstChildTAG: Hi juerguen, Take a look at this Hints of HW8 [here][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/509e68568e407d1f0000001b FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-10T17:16:38Z IndexTAG: 2050 TitleTAG: h8p3 last question how to obtain the value of vth and rth for part 5.i have tried but didnt suceed. pls help UserIdTAG: 611666 UserNameTAG: Shaminder420 CreateTimeTAG: 2012-11-10T09:48:22Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Please use search feature of Discussion This question was clearly described Good luck! FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-11-10T11:14:23Z SecondChildTAG: thanks for the suggestion,now finally got the answer SecondChildUserIdTAG: 611666 SecondChildUserNameTAG: Shaminder420 SecondChildCreateTimeTAG: 2012-11-10T13:03:18Z SecondChildTAG: me too!!!am not getting the Rth value..can u please help? SecondChildUserIdTAG: 156585 SecondChildUserNameTAG: AdiRanga SecondChildCreateTimeTAG: 2012-11-10T18:53:54Z IndexTAG: 2051 TitleTAG: TO STAFF Sir, Can u tell us some important topics for final exam,as it would be easy for us to brush up those topics . UserIdTAG: 150120 UserNameTAG: krishna1993 CreateTimeTAG: 2012-11-10T06:37:55Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 2052 TitleTAG: Certificate What is the value of edx certificate UserIdTAG: 150120 UserNameTAG: krishna1993 CreateTimeTAG: 2012-11-10T06:17:05Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hi krishna, this might help https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/508446a8d210431f0000012c FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-11-10T11:04:57Z FirstChildTAG: Hi, [Check This][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/508446a8d210431f0000012c > **Regards:** asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-11-10T14:52:10Z IndexTAG: 2053 TitleTAG: Pulses with same charge, same voltage? If those pulses are all deposit the same charge $q$ on the capacitor, shouldn't all the voltage curves that he draws on the right be the same height, since $q = Cv$ for capacitors? Rob UserIdTAG: 468623 UserNameTAG: RobNik CreateTimeTAG: 2012-11-10T02:40:50Z VoteTAG: 1 CoursewareTAG: Week 8 / Response To A Pulse As Pulse Gets Narrower Continued CommentableIdTAG: 6002x_Response_To_A_Pulse_As_Pulse_Gets_Narrower_Continued NumberOfReplyTAG: 2 FirstChildTAG: It is not the same because $v$ is dependent of time so it is actually $v(t)$. FirstChildUserIdTAG: 304535 FirstChildUserNameTAG: hugo_h FirstChildCreateTimeTAG: 2012-11-10T05:28:35Z FirstChildTAG: Ah, I get now – I forgot that part of the charge $q$ goes through the resistor. So as you get narrower pulses, you get less charge going through the resistor, until at the limit, with pulse width = 0, all of $q$ goes to the capacitor, and $v = q/C$. But the wider the pulse, the lower the $v$ at the end of the pulse. Rob FirstChildUserIdTAG: 468623 FirstChildUserNameTAG: RobNik FirstChildCreateTimeTAG: 2012-11-10T17:18:20Z IndexTAG: 2054 TitleTAG: H8P1 - initial charge Hi everyone. Please can u help me understand how to get into this homework? It's supposed that the initial voltage in the capacitor is Q/T, but I can't find Q if I don't have T, and besides, when they mention the amount of time "1 ms" that time is in or out or the impulse??? If the impulse is 1 delta, I can assume that is 1V along 1 second or maybe 1mV along 1000 seconds too. Someone can help me? I'm really lost here... UserIdTAG: 376173 UserNameTAG: nacho110987 CreateTimeTAG: 2012-11-09T22:12:30Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi nacho110987, Take a look at here [Hints HW8][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/509e68568e407d1f0000001b FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-10T17:20:49Z IndexTAG: 2055 TitleTAG: H10 Impedance model In the impedance match, the circuit is constructed to maximize the power. This can not always be the best way to match a source to a load. Example: A car battery has internal resistance of 0.1ohm. A load with same impedance will maximize power in the load, but will also deposit half the power into the battery, so only 50% of the energy in the source is transferred to the load. It is possible to get much higher power transfer if the load resistance is much higher than the source resistance. For example when driving a windshield wiper motor with RL=10ohm, the power transfer efficiency is 99%. Of course impedance matching will avoid reflections, but must that be the best coupling between an amplifier and an antenna? How do we optimize power transfer? UserIdTAG: 142857 UserNameTAG: viking2 CreateTimeTAG: 2012-11-09T21:30:14Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 2056 TitleTAG: H8P1 b According to the Inductors and First Order Circuits Wiki: Inductors don't charge abruptly: iL(t=0+)=0 Ohmic drop across resistor is zero: vR(t=0+)=0 Vin appears entirely across L: vL(t=0+)=Vin If the current is zero then vR1(t) at t=0+ should be the same as the unit impulse or 1. That answer is not correct so I am missing something, but what? UserIdTAG: 231764 UserNameTAG: PaulP4881 CreateTimeTAG: 2012-11-09T20:04:38Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: For a finite value of voltage input the value of the current through an inductor does not change discontinuously. But an impulse represents an infinite pulse with an infinitesimal duration. In that case the current in an inductor can change discontinuously. In a like manner the charge on a capacitor can change discontinuously. A simple way to get the results you need is to write the differential equation for the R-L circuit like this: L*(di/dt) + R*i = Vin where Vin is a unit impulse. Then integrate from t = 0- to t = 0+, and the result that you want falls out immediately. You can do a similar thing for an R-C circuit. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-09T21:59:29Z SecondChildTAG: what should i put for vin for integration.? SecondChildUserIdTAG: 136490 SecondChildUserNameTAG: Ali_PU SecondChildCreateTimeTAG: 2012-11-10T17:44:48Z FirstChildTAG: How I solved it: In lecture S15V17, we learn the equation for current through an inductor in response to an impulse. iL(t) = (Lambda/L)*e^(-Rt/L) More specifically, the instantaneous current at time O+ is simply Lambda/L, where Lambda is the flux linkage ie the area of the impulse (given). Logically (using KCL, I suppose), this must be the same current that flows through the resistor R1, so now we go back to Ohm's Law, V=IR and from given value of R and the calculated value of I we trivially calculate V. Hope that helps. I don't like differential equations much and I love Ohm's Law, so that works for me. FirstChildUserIdTAG: 403503 FirstChildUserNameTAG: jmohrmann FirstChildCreateTimeTAG: 2012-11-10T08:59:37Z SecondChildTAG: thanks man.:) SecondChildUserIdTAG: 266739 SecondChildUserNameTAG: satvikchugh SecondChildCreateTimeTAG: 2012-11-10T09:14:12Z SecondChildTAG: still didnt get it..:( SecondChildUserIdTAG: 173147 SecondChildUserNameTAG: cruiser_rahit SecondChildCreateTimeTAG: 2012-11-10T18:57:10Z SecondChildTAG: thanks a lot.. SecondChildUserIdTAG: 169181 SecondChildUserNameTAG: NINI SecondChildCreateTimeTAG: 2012-11-11T11:47:17Z SecondChildTAG: thanks SecondChildUserIdTAG: 452559 SecondChildUserNameTAG: kuz1toro SecondChildCreateTimeTAG: 2012-11-11T18:29:31Z SecondChildTAG: I finally came up with the right answer. Thank you very much. SecondChildUserIdTAG: 231764 SecondChildUserNameTAG: PaulP4881 SecondChildCreateTimeTAG: 2012-11-11T20:05:39Z IndexTAG: 2057 TitleTAG: H8P3 - first question Hi! I've answered all the questions on HW8 correctly but on this one I just can't find the way to solve it! Can somebody please help me with some hints? Thank you! UserIdTAG: 342071 UserNameTAG: jmcalcada CreateTimeTAG: 2012-11-09T19:39:58Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: You are given the gate capacitance and the time to decay from $V_{OH}$ to $V_{IH}$ through the unknown parasitic resistance. You just have to find the unknown resistance. I know, that's just restating the question, so... You have to ignore Q3 for this question! There's no way to get the accepted answer if you try to use the given $R_{OFF}$. FirstChildUserIdTAG: 141000 FirstChildUserNameTAG: OrinE FirstChildCreateTimeTAG: 2012-11-09T22:42:08Z SecondChildTAG: Brilliant Effort ! SecondChildUserIdTAG: 477713 SecondChildUserNameTAG: ikm104 SecondChildCreateTimeTAG: 2012-11-10T05:39:27Z FirstChildTAG: Got it right!! VS was wrong :) Thank you FirstChildUserIdTAG: 342071 FirstChildUserNameTAG: jmcalcada FirstChildCreateTimeTAG: 2012-11-10T00:10:58Z FirstChildTAG: I used this formula: R = -t/(C*ln((VIH-VS)/(VOH-VS))) but it didn't work...Am I missing something? FirstChildUserIdTAG: 342071 FirstChildUserNameTAG: jmcalcada FirstChildCreateTimeTAG: 2012-11-09T23:52:36Z SecondChildTAG: Use discharging formula between VOH and VIH like Vc = A*E^(-t/Rp*C) form. SecondChildUserIdTAG: 149058 SecondChildUserNameTAG: sotoroman SecondChildCreateTimeTAG: 2012-11-10T22:33:06Z IndexTAG: 2058 TitleTAG: Value of ω From the previous examples I know that $\omega$ should be $1/ \sqrt{LC}$ , but is there a simple way to see that here (without going back and starting from $ v_c=Ae^{st} $) ? UserIdTAG: 434869 UserNameTAG: patey CreateTimeTAG: 2012-11-09T18:22:19Z VoteTAG: 1 CoursewareTAG: Week 9 / An LC Circuit CommentableIdTAG: 6002x_an_LC_Circuit NumberOfReplyTAG: 1 FirstChildTAG: Note that they say this is another way to attack a simple L-C circuit - you won't need to solve a second order differential equation. Put the 'trial' expressions for $v_C(t)$ and $i_L(t)$ into the two differential equations they give. Now you have two simultaneous equations with unknowns $\omega$, A and B. 'A' conveniently cancels out when solving for $\omega$. 'A' must be found by substituting the initial conditions into $v_C(t)= A\ cos({\omega}t)$ before you can solve for B. FirstChildUserIdTAG: 141000 FirstChildUserNameTAG: OrinE FirstChildCreateTimeTAG: 2012-11-09T22:33:05Z SecondChildTAG: Thanks very much. I knew there must be a way. I had managed to get A and B as you said, from the initial conditions in $v_C$ and then substituting into the expression for $i_L$, but I couldn't see how to get $\omega$. SecondChildUserIdTAG: 434869 SecondChildUserNameTAG: patey SecondChildCreateTimeTAG: 2012-11-09T23:35:42Z SecondChildTAG: Helpful SecondChildUserIdTAG: 442070 SecondChildUserNameTAG: MuhammadAsad SecondChildCreateTimeTAG: 2012-11-18T18:13:25Z IndexTAG: 2059 TitleTAG: Norton/Thevenin Equivalent Models Hi there, Does anyone understand why for the Norton equivalent model the resistor is in parallel to the current source and for the Thevenin model the resistor is in series with the voltage source? There is probably a simple answer but it has been bothering me for some time now. Thanks, Zak UserIdTAG: 278463 UserNameTAG: zniazi CreateTimeTAG: 2012-11-09T12:07:03Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Because to find the Norton/Thevenin resistor you must set to 0 the source and if you set to 0 a current source it is an open circuit so the resistor can only be in parallel to be viewed as a resistor from the port you are considering. FirstChildUserIdTAG: 376877 FirstChildUserNameTAG: AndBre FirstChildCreateTimeTAG: 2012-11-09T12:37:42Z IndexTAG: 2060 TitleTAG: fF or pF?? is there 5fF or pF...? fF = 10^-15 and pF = 10^-12.. if to take pF, so the answers are correct.. am i wrong with degrees?? UserIdTAG: 373498 UserNameTAG: Cheblan CreateTimeTAG: 2012-11-08T18:39:54Z VoteTAG: 1 CoursewareTAG: Week 8 / Charging And Discharging CommentableIdTAG: 6002x_Charging_And_Discharging NumberOfReplyTAG: 2 FirstChildTAG: In the question, it is given as 5 femtofarads(at least in my example), which as you say is 5*10^-15. If you are saying that the math only works if it's really picofarads, then go ahead and post your formula with those numbers and show it. It's only an exercise, so you are not violating the honor code. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-11-08T19:50:19Z FirstChildTAG: You get it correct with pF because you think that the answer is written in nanoseconds (10^-9) and not picoseconds (10^-12) You get XX,XX*10^(-12) if you use fF, and the answer would be XX,XX ps But if you use pF, you get XX,XX*10^(-9). If you think that the answer is in ns, you'd write XX,XX too, though you are wrong with the units. You are getting it right with a wrong reasoning :D FirstChildUserIdTAG: 193982 FirstChildUserNameTAG: Wfk FirstChildCreateTimeTAG: 2012-11-10T21:33:01Z IndexTAG: 2061 TitleTAG: Transformers When are we/are we even going to learn about the Transformer in 6.002x? In my opinion one of the first things we should have learned about. UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-11-08T17:57:03Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Why do you think it should be one of the first things we learned about? I'm not saying you are wrong or right, just wondering what your reasoning is. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-11-08T19:54:27Z SecondChildTAG: Hi JSChambers! My reasoning is that everything in an electronics needs power, and in allot of applications that power is 'transformed'. Transformers provide that power, that idea is so fundamental to electronics. We have been considering digital chips and MOS-FET, but we don't know how to power them (i.e. build a real world circuit like below). ![enter image description here][1] This circuit pattern should be very familiar as it's 'circuit pattern' is used in practically all devices with chips. But we cant know it (if we have kept solely to 6.002x), because there is this mysterious element (the transformer). Don't get me wrong, I'm not complaining, just asking :) [1]: https://edxuploads.s3.amazonaws.com/13524069381343662.bmp SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-08T20:35:48Z SecondChildTAG: There is a very small section on transformers in the book, pages 478-480. If I had to guess what they are thinking(always dangerous), maybe they consider them to be simple enough to model with dependent sources, so that they don't bother go into it. Not sure, really. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-11-08T23:45:44Z SecondChildTAG: Two pages on what seems to be such a widely used device? The Transformer is not complicated or particularly difficult to get a grasp of, that is, unless you go into the quantum mechanics of it. Just turns ratio, the effect of different cores, thickness of wire on current carrying capability, step up step down xformers etc... not that difficult. SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-09T08:11:54Z FirstChildTAG: I think transformers should be part of an intro course too. It's a great way to utilize diodes.But there's only so much that you can pack into a course. 6.002x opts for a bit of digital circuitry, at the expense of some analog stuff. Transformers are more complicated than inductors and capacitors, so I guess they had to leave them out. FirstChildUserIdTAG: 4463 FirstChildUserNameTAG: pmj FirstChildCreateTimeTAG: 2012-11-08T21:11:55Z SecondChildTAG: I see a basic theme in the course that they don't introduce something, apart from the basic abstractions, without having a theoretical/mathematical foundation. Note that at week 10, the term "Reactance" hasn't even been mentioned! But, you can solve the problems where you might use 'reactance' and have a better understanding of what's going on! Having a degree in Computing and Electronics from the early 80's and an amateur radio license for which the exams supposedly cover the same material, this course really brings it all together. They give you the basics, not a series of tricks (well, some might disagree about that ;)) which will serve you well in the future. Now, transformers... I suspect there simply wasn't time to cover them properly. As I recall, the assumption way back when I did my degree was that a voltage source would be a simple battery (or cell, there is a difference) and transformers wouldn't be involved. There is really no assumption either way with an ideal voltage source*. So, personally, I never noticed the lack of coverage. As for the 7806 circuit, it would probably like some capacitance on the output and a diode from Vo to Vi (see Figure 8 on the TI uA7800 series datasheet). Without the capacitor, it might oscillate with no load and without the diode, it might die prematurely. That's way too much complication for this course IMO. *yeah, I know, batteries/cells have internal resistance. SecondChildUserIdTAG: 141000 SecondChildUserNameTAG: OrinE SecondChildCreateTimeTAG: 2012-11-09T05:41:46Z FirstChildTAG: I will remind you that that this course has prerequisites, transformers would have been covered in the prerequisite courses. -6.01 Introduction to Electrical Engineering and Computer Science I -8.02 Electricity and Magnetism -18.03 Differential Equations While you can get by without the prerequisites, I would not consider 6.002x as an "into" course by any means. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-11-09T12:00:49Z SecondChildTAG: "6.01 Introduction to Electrical Engineering and Computer Science I" And the voltage source was not covered in an intro course on electronics but left to 6.002x? Sorry, I'm not trying to cause an issue, it's just how I feel. By the way, where can I find these courses you mentioned? They look really good. SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-09T15:53:14Z SecondChildTAG: "And the voltage source was not covered in an intro course on electronics but left to 6.002x? Sorry, I'm not trying to cause an issue, it's just how I feel." Are you just making assumptions? Anyways here are the courses, have fun. http://ocw.mit.edu/courses/mathematics/18-03sc-differential-equations-fall-2011/ http://ocw.mit.edu/courses/physics/8-02-electricity-and-magnetism-spring-2002/ http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/ SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-11-10T12:11:29Z SecondChildTAG: Thanks! Sorry about the assumptions! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-10T17:56:17Z IndexTAG: 2062 TitleTAG: H8P2 Help please!! I don't understand how can I calculate the R: I tried this ecuation: (Q/2C) = Q/C * e^(-14400/(R*70) UserIdTAG: 433574 UserNameTAG: endika86 CreateTimeTAG: 2012-11-08T16:07:52Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Q=400 mg C=70 kg FirstChildUserIdTAG: 297655 FirstChildUserNameTAG: Aljoska FirstChildCreateTimeTAG: 2012-11-08T17:28:42Z SecondChildTAG: 0.4/ 2 * 33e-9 = 0.4/33e-9 ^11400/(R*70) R = - 33.101 That is imposible. Any help??? SecondChildUserIdTAG: 433574 SecondChildUserNameTAG: endika86 SecondChildCreateTimeTAG: 2012-11-08T19:36:59Z SecondChildTAG: I dont understand were you found 33e-9, but it is easy to simplify "Yours" equation (note 1/2=0.5) : 0.5=exp(14400/(R*70)) from this point you might easily get correct answer SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-11-08T19:50:55Z FirstChildTAG: voltage across capactor is given by Vc=V(1-e^(-t/R*C)) At half time Vc=V/2 substituting this in aove one, we get e^(-t/R*C)=1/2 applying ln on oth sides we can get the value of resistence.... FirstChildUserIdTAG: 131558 FirstChildUserNameTAG: PAVAN6 FirstChildCreateTimeTAG: 2012-11-09T07:13:32Z FirstChildTAG: Hi endika86, Take a look at this Post [HW8 Hints][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/509e68568e407d1f0000001b FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-10T17:24:44Z IndexTAG: 2063 TitleTAG: H9P2 errors (d) The circuit would ring when excited with a step response Correct answer is appears in f and g parts, but cant be passed in (d) part. Is this a grader error? Also, in differential equation for this question we can see (dvo+o)/dt - typing mistake? UserIdTAG: 394836 UserNameTAG: v2g6ch4 CreateTimeTAG: 2012-11-08T15:05:38Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: It wants the undamped natural frequency in Hz. Yes, typing mistake, it should be $\frac{dv_o}{dt}$. FirstChildUserIdTAG: 141000 FirstChildUserNameTAG: OrinE FirstChildCreateTimeTAG: 2012-11-08T17:18:18Z IndexTAG: 2064 TitleTAG: HOMEWORK 8 plz plz someone give me a hint to solve problem 1 of h8 great difficulty in solving no idea of what is going on !! UserIdTAG: 285222 UserNameTAG: dhaval24 CreateTimeTAG: 2012-11-08T08:30:57Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Search the forum for H8P1, there are several tips. Regards. FirstChildUserIdTAG: 288174 FirstChildUserNameTAG: matiasgrodriguez FirstChildCreateTimeTAG: 2012-11-08T14:55:48Z FirstChildTAG: Hi dhaval24, Take a look at this Hints [here][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/509e68568e407d1f0000001b FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-10T17:26:06Z IndexTAG: 2065 TitleTAG: H9P1: value for impulse If you use .64 for the impulse, you can do this problem correctly and not obtain an acceptable answer. Instead, use the 2/pi instead of the approximate .64 All the best. UserIdTAG: 569199 UserNameTAG: neogerald CreateTimeTAG: 2012-11-07T21:05:49Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi, [H9P1 Hint][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5098b2a1cd5fb82300000036 > **Regards:** asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-11-08T16:18:43Z IndexTAG: 2066 TitleTAG: [H10P3] Need help I have stuck with calculating Cmatch and Lmatch. I have calculated Vf/If. For Cmatch and Lmatch I used this result and compared it to R1. Then I calculated C and L from equation. What I do wrong? Any hints? UserIdTAG: 146390 UserNameTAG: JGradzki CreateTimeTAG: 2012-11-07T19:37:05Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: I got the correct numerical value for Cmatch and Lmatch as well as the correct equation for Lmatch however my equation for Cmatch, which was used to obtain all of the above is marked as incorrect. FirstChildUserIdTAG: 397595 FirstChildUserNameTAG: mholin FirstChildCreateTimeTAG: 2012-11-07T23:19:48Z SecondChildTAG: I had trouble if I simplified the equation too much. I 'undid' the simplifications until it accepted my answer. SecondChildUserIdTAG: 141000 SecondChildUserNameTAG: OrinE SecondChildCreateTimeTAG: 2012-11-08T00:24:42Z FirstChildTAG: For $C_{match}$ and $L_{match}$, you have already calculated the impedance of the matching network plus antenna (no need to mess around with $v_F$ and $i_F$, the point of the impedance method is that you avoid calculating them) - set it equal to $R_1$**. The equation is probably of the form x/y = $R_1$ so multiply by y and divide by $R_1$ and collect all the terms on the left hand side, leaving the right hand side zero*. Now you can equate the real terms on the left hand side to zero and the imaginary terms on the left hand side to zero, giving a pair of simultaneous equations that you can solve for $C_{match}$ and $L_{match}$. *There's no need to collect everything on one side, but it makes it obvious to me that the imaginary terms sum to zero in that case. You could equate the imaginary terms on the left side with the imaginary terms on the right side and do the same for the real terms; you get the same result. **edit - actually, it's $R_1$+0j. We are just equating two impedances here. FirstChildUserIdTAG: 141000 FirstChildUserNameTAG: OrinE FirstChildCreateTimeTAG: 2012-11-08T00:22:50Z SecondChildTAG: Excellent explanation SecondChildUserIdTAG: 220837 SecondChildUserNameTAG: Calsomus SecondChildCreateTimeTAG: 2012-11-09T04:57:46Z SecondChildTAG: Excellent explanation [2] Thanks. SecondChildUserIdTAG: 39980 SecondChildUserNameTAG: MarinoJr SecondChildCreateTimeTAG: 2012-11-09T11:33:41Z SecondChildTAG: many many rays of goodness to you for the explanation :) SecondChildUserIdTAG: 352757 SecondChildUserNameTAG: Kerbyco SecondChildCreateTimeTAG: 2012-11-18T11:35:31Z SecondChildTAG: thank u very much! SecondChildUserIdTAG: 316353 SecondChildUserNameTAG: Vivekanand123 SecondChildCreateTimeTAG: 2012-11-19T17:21:32Z SecondChildTAG: a thousand thanks SecondChildUserIdTAG: 285945 SecondChildUserNameTAG: lindalapiso SecondChildCreateTimeTAG: 2012-11-22T12:28:26Z SecondChildTAG: Hi there! Thank you for the explanation. So, All I can get with my math is Cmatch=function(L,R1,R2,w), and Lmatch=function(C,R1,R2,w), which is useless I think. So, I wanna know, the expression for both Cmatch and Lmatch have which parameters? SecondChildUserIdTAG: 290028 SecondChildUserNameTAG: Fjuca SecondChildCreateTimeTAG: 2012-11-23T18:28:40Z SecondChildTAG: Ugh..I dont understand how to do.. OrinE, could you explain more precise? SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-11-23T20:35:23Z SecondChildTAG: @Fjuca Cmatch and Lmatch should be functions of only R1,R2&w. Use the two equations to find Cmatch and Lmatch independent of each other. @Sergtronix Use your answer from the previous question and equate it to R1. Cross multiply across the equal sign to form a equation. This equation will have some real terms and some imaginary terms. First equate the real terms to zero and then the imaginary terms to zero separately to obtain two equations. assuming C and L as unknowns use the pair of equations to find L and C. These are your values for Lmatch and Cmatch. SecondChildUserIdTAG: 315681 SecondChildUserNameTAG: nahuja1 SecondChildCreateTimeTAG: 2012-11-24T13:26:11Z SecondChildTAG: nahuja1, thank you. I still don't get it right, but i'll try harder later and maybe that will work. Thank you very much! SecondChildUserIdTAG: 290028 SecondChildUserNameTAG: Fjuca SecondChildCreateTimeTAG: 2012-11-24T16:35:21Z FirstChildTAG: Thanks for help! I was doing this in that way. Right now I just rewrite this equation and it works!! Thanks a lot! FirstChildUserIdTAG: 146390 FirstChildUserNameTAG: JGradzki FirstChildCreateTimeTAG: 2012-11-08T08:30:54Z FirstChildTAG: Hi, Is this the way? : 0 = R2*(1+C^2*R1^2*w^2)-(-R1+C*w^2*R1*L)+L*w^2*C*R1+j*(-L*w/(1+C^2*R1^2*w^2)-*C*R1*w) Re : {R2*(1+C^2*R1^2*w^2)-(-R1+C*w^2*R1*L)} Im : {j*(-L*w/(1+C^2*R1^2*w^2)-*C*R1*w)} if it's the case: What Should I do? Thanks! FirstChildUserIdTAG: 299251 FirstChildUserNameTAG: TheRedBlackOne FirstChildCreateTimeTAG: 2012-11-18T15:15:13Z SecondChildTAG: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/mathjax-tutorial-write-formula-forum/ please use it, its really a pain for the eye, and you will get more help if you formulate your question in a readable way. SecondChildUserIdTAG: 329254 SecondChildUserNameTAG: KGabor SecondChildCreateTimeTAG: 2012-11-23T11:16:06Z IndexTAG: 2067 TitleTAG: to staff about the proctored exam how can i take this one in my country can i take it after the course has been finished ? .... what is the value of it ? UserIdTAG: 221617 UserNameTAG: konan CreateTimeTAG: 2012-11-07T19:25:19Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I have asked a similar question in the past, perhaps it will be of help... https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/508446a8d210431f0000012c FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-11-07T21:25:23Z IndexTAG: 2068 TitleTAG: having problems with system I have tried everything that i can think of, my system is letting me access the labs and homework but not letting me enter anything in what do I do? Plus I keep losing my connection and get it back and still nothing want to keep going with class i enjoy it alot UserIdTAG: 160242 UserNameTAG: Mlevins35 CreateTimeTAG: 2012-11-07T02:41:39Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: What OS are you on? What browser are you using? What is the version number of the browser? Did you try another browser? If you did, did that work or not? Can you try another computer? If so, do you get the same result? FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-11-07T11:22:00Z SecondChildTAG: went on to google chrome was able to access class but not put anything on problem was with my computer. kept losing internet and would not let me post anything is there any kind of extra credit that i can make up for the week 7 lab and homework could not get computer fixed fast enough to get it turned in SecondChildUserIdTAG: 160242 SecondChildUserNameTAG: Mlevins35 SecondChildCreateTimeTAG: 2012-11-07T21:53:20Z IndexTAG: 2069 TitleTAG: problem in Lab i have problem in changing the component value and running a transient analysis in lab . UserIdTAG: 72786 UserNameTAG: Bhavesh CreateTimeTAG: 2012-11-06T19:15:52Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I am also having the same problem can not run transient analysis to finish lab any idea what to do need to turn lab in cant turn in if cant finish FirstChildUserIdTAG: 160242 FirstChildUserNameTAG: Mlevins35 FirstChildCreateTimeTAG: 2012-11-06T20:18:10Z SecondChildTAG: try running in another browser SecondChildUserIdTAG: 357453 SecondChildUserNameTAG: BrunoCanoso SecondChildCreateTimeTAG: 2012-11-06T21:04:35Z SecondChildTAG: The bug was out fault, and it should be fixed. Sorry for the inconvenience. SecondChildUserIdTAG: 96452 SecondChildUserNameTAG: Lyla SecondChildCreateTimeTAG: 2012-11-07T01:49:36Z SecondChildTAG: tried that even tried to reboot my system still not letting me SecondChildUserIdTAG: 160242 SecondChildUserNameTAG: Mlevins35 SecondChildCreateTimeTAG: 2012-11-07T02:19:03Z IndexTAG: 2070 TitleTAG: @ Myrimit - Step, Pulse, Impulse and Ramp in Spanish? In the field of electronics.... Hola, que tal? I already understand... but I would not know how to explain it to somedoby in my mother tongue, jajaja or just say this words in spanish :)) Gracias y un abrazo! Sandra UserIdTAG: 342966 UserNameTAG: SandraNavarro CreateTimeTAG: 2012-11-06T18:36:14Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Es un cambio de estado en la corriente o en el voltaje. Por ejemplo cuando tienes una señal cuadrada tienes un step o escalón... [Ejemplo de señal cuadrada][1] En cambio un impulso es cuando tienes un alza de corriente o tensión en un espacio muy reducido de tiempo. [Ejemplo de impulso (delta de Dirac)][2] y una rampa es un asenso de corriente que sube con una pendiente dada o sea como una recta del tipo y=ax+b [ejemplo de una señal rampa][3] esta puede subir y quedarse arriba con una continua, puede volver a bajar bruscamente o bajar con la misma pendiente con la que subió... hay muchas variaciones. Saludos, espero te sea útil! :D [1]: http://slack.codemaniacs.com/images/tutorial_sintesis/square.png [2]: http://2.bp.blogspot.com/_HcXX4owUqK8/SslUQbzBMZI/AAAAAAAAAL0/fXfgEhQh-14/s320/deltad.jpg [3]: http://facultad.bayamon.inter.edu/cgonzalezr/ELEN3312/Ejercicios%20de%20Practica_Filtros_files/image001.gif FirstChildUserIdTAG: 70519 FirstChildUserNameTAG: Fipe FirstChildCreateTimeTAG: 2012-11-06T19:37:50Z SecondChildTAG: Muchíssima gracias :)) claro seguro me sirve!!!! ahora me podré expresar con propiedad también en mi lengua materna, gracias :) un saludo!! SecondChildUserIdTAG: 342966 SecondChildUserNameTAG: SandraNavarro SecondChildCreateTimeTAG: 2012-11-06T21:34:52Z SecondChildTAG: :D pues pa' servirle señorita! SecondChildUserIdTAG: 70519 SecondChildUserNameTAG: Fipe SecondChildCreateTimeTAG: 2012-11-10T13:43:04Z FirstChildTAG: **Step = Escalón** En este caso, puedes observar la gráfica celeste. La misma, representa a un step o escalón, es decir, para un t menor a 0 será igual a cero y para un t mayor o igual a cero tendrá un valor constante V. El escalón suele llamarse \mu de te\ , simbólicamente, u(t). ![IM][1] ---- **Impulse = impulso o Delta de Dirac** El impulso o delta de dirac, se simboliza con $\delta(t)$ , se dice \delta de t\. Se representa con una flechita apuntando hacia arriba. Esto significa que su amplitud es casi muy pero muy grande, infinita, en un timpo muy pequeño. ![delta][2] ---- **Ramp = rampa** La rampa, se la suele llamar \ro de te\ , se simboliza como $\rho(t) $. Verás que para t menor a cero será igual a cero como has visto en la rampa y para, t mayor o igual a cero será una pendiente. ![enter image description here][3] ---- **Pulse = Pulso** El pulso es como la curva azul de abajo: ![pulso][4] Básicamente, si prestas atención es la suma de dos steps : uno que ya has visto u(t) más una rampa desplazada a ta (recuerda que si te desplazas a la derecha un tiempo ta, tu nueva rampa será u (t-ta); si en cambio, te desplazas a la izquierda, será u (t+ta)). Ok, pero tu rampa desplazada necesita estar invertida, para que sumada a tu otra rampa, te de un pulso, por eso escribes -u (t-ta). ![im][5] Un pulso podría decirse que es u (t) + -u (t-ta). ---- Espero haberte sido de ayuda, Sandra. Sí, a veces es confusa la nomenclatura de las cosas en inglés... espero que en el futuro haya un curso en español jaja! Un abrazo! Myriam. [1]: https://edxuploads.s3.amazonaws.com/1352246553134369.png [2]: https://edxuploads.s3.amazonaws.com/13522471781343652.png [3]: https://edxuploads.s3.amazonaws.com/13522488591343669.png [4]: https://edxuploads.s3.amazonaws.com/13522477001343615.png [5]: https://edxuploads.s3.amazonaws.com/13522483001343601.png FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-07T00:41:24Z SecondChildTAG: Muchas gracias Myriam!!!!! como no :)) Claro que me ha servido de ayuda, me lo guardo :)) Curso en español jaja :)) muy buena :)) Gracias de nuevo y un abrazo! Sandra SecondChildUserIdTAG: 342966 SecondChildUserNameTAG: SandraNavarro SecondChildCreateTimeTAG: 2012-11-07T20:46:38Z SecondChildTAG: Me alegro mucho que te haya sido de ayuda :) Abrazos! Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-07T22:05:01Z IndexTAG: 2071 TitleTAG: lack of hypothesis details I(t)=I0+Qδ(t)A where I0=10mA results to be Is at t=0, the dc steady component without the impulse. It doesnt actually say it was like that "long time" before, or that psu became on at either time. As a result the I0 could be 10mA at t=0 as it results from notation, and the student has to assume that it either psu started ON at t=0, or at t=-1ms or it was ON like that a "long time" before. Cant actually solve this without one assumption, either way. UserIdTAG: 331483 UserNameTAG: Doru CreateTimeTAG: 2012-11-06T18:08:14Z VoteTAG: 1 CoursewareTAG: Week 8 / Initial Conditions CommentableIdTAG: 6002x_Initial_Conditions NumberOfReplyTAG: 1 FirstChildTAG: The problem is 75% intuitive). don't look on the equation I(t)=I0+Qδ(t) as on the law of I(t) change in time - this is only a law for the impuls. Take a paper and graph how V changes from 3V to steady state an then turn on the impuls in the moment t=0.. FirstChildUserIdTAG: 373498 FirstChildUserNameTAG: Cheblan FirstChildCreateTimeTAG: 2012-11-07T08:17:32Z SecondChildTAG: I agree with you Doru that this question isn't very well posed. I don't know what the conditions are that we're meant to be considering. Some people seem to think it's very straightforward and obvious. They must have just made the same unstated assumption that the author of the question had in mind, because it's by no means explicitly stated what we're even meant to be considering here...as far as I can tell with my puny human brain. SecondChildUserIdTAG: 287805 SecondChildUserNameTAG: Beneficial SecondChildCreateTimeTAG: 2012-11-10T01:38:25Z IndexTAG: 2072 TitleTAG: H8P1 - VR1 I am struggling with finding VR1, which according to most of the posts should be very easy but for some reason I can not solve it. I appear to assume incorrectly that VR1 should equal the impulse voltage or should I be trying to find the inductor current and then calculate the voltage through R1? Does the inductor not act like a short? I've read the text and watched the lectures and for some reason none of week 8 is intuitive to me.... UserIdTAG: 244115 UserNameTAG: Pedro1969 CreateTimeTAG: 2012-11-06T13:01:54Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: Use Wikipedia. FirstChildUserIdTAG: 99628 FirstChildUserNameTAG: Pranjal16 FirstChildCreateTimeTAG: 2012-11-06T13:40:35Z FirstChildTAG: Hi Pedro 1969, Try using the fórmula that relates the magnetic flux in L (assuming this is one) and the current i (you have L) the rest is easy. Hang in there. These matters are not easy for me either. FirstChildUserIdTAG: 369339 FirstChildUserNameTAG: vargaslen FirstChildCreateTimeTAG: 2012-11-06T18:03:22Z SecondChildTAG: Doesn't the flux linkage V*T equal 1? And I would have thought that like Q/C, for this case the impulse at 0+ would have been (V*T)/L or 1/L but that is not quite right. SecondChildUserIdTAG: 244115 SecondChildUserNameTAG: Pedro1969 SecondChildCreateTimeTAG: 2012-11-07T14:43:32Z FirstChildTAG: Write KVL with impluse forcing. Integrate from 0- to 0+ to obtain i(0+). FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-06T19:40:12Z SecondChildTAG: Confused.....:0:0 HELP NEEDED SecondChildUserIdTAG: 127800 SecondChildUserNameTAG: MNB SecondChildCreateTimeTAG: 2012-11-08T03:19:43Z FirstChildTAG: Try it out in the sandbox. Should give you good intuition. FirstChildUserIdTAG: 142857 FirstChildUserNameTAG: viking2 FirstChildCreateTimeTAG: 2012-11-08T05:12:42Z IndexTAG: 2073 TitleTAG: Regarding the output voltage of a memory element While designing a memory element(buffer + decoupled refresh) sir has said that the resistance of the buffer or inverter is very high(in G ohm) . Now reading a value means whatever be the value stored in the capacitor m able to get it but because of high resistance of buffer due to voltage divider action the voltage appears across output resistance is very low (0) if the value stored in the capacitor is high(1). Then how is that it is a memory element?? please intimate in this regard UserIdTAG: 321556 UserNameTAG: sj31867 CreateTimeTAG: 2012-11-06T12:29:55Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: concerned authorities please reply. FirstChildUserIdTAG: 321556 FirstChildUserNameTAG: sj31867 FirstChildCreateTimeTAG: 2012-11-06T20:40:20Z SecondChildTAG: guys please help me out :( SecondChildUserIdTAG: 321556 SecondChildUserNameTAG: sj31867 SecondChildCreateTimeTAG: 2012-11-07T16:59:45Z SecondChildTAG: I don't quite understand what you are asking, could you reword your question perhaps? SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-07T21:45:40Z IndexTAG: 2074 TitleTAG: STAFF: I cannot log in on another computer! For some reason, when I try and login on a computer other than my home computer, MITx does not recognize me!! It will not even recognize my Email address. What could be the problem? UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-11-06T10:54:13Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Make sure your browser has cookies and Javascript enabled. What browser/version and operating system/version are you using on the other computer(s)? FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-11-07T13:21:51Z IndexTAG: 2075 TitleTAG: BYPASS CAPACITOR in the video of Week 7 Tutorials ... He applys voltage divider rule.. 1ohm/(1ohm+Kohm)...but it doesnt have to be 1Kohm/(1ohm+1Kohm) ? UserIdTAG: 60496 UserNameTAG: tomipiriyev CreateTimeTAG: 2012-11-06T09:59:08Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Could you provide a link to the video please? FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-11-06T10:13:17Z SecondChildTAG: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_7/Week_7_Tutorials/ ( time 1:15 ) SecondChildUserIdTAG: 60496 SecondChildUserNameTAG: tomipiriyev SecondChildCreateTimeTAG: 2012-11-06T18:40:37Z SecondChildTAG: Ok, thanks! Well I assume that is because we are not looking at the voltage drop from the power supply, but from the inverter, so the inverter is drawing a tiny bit of current on and off. SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-06T18:52:49Z IndexTAG: 2076 TitleTAG: H8P3 Dear friends, Please provide a hint for the 4th and 5th part of H8P3 regarding the duration of the pulse? For 4th part I think it is the charging time of the capacitor from VoL to VOH. Is it right? UserIdTAG: 156835 UserNameTAG: kphariprakash1968 CreateTimeTAG: 2012-11-06T06:20:36Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: yes i'd love some hints too , i's been driving me nuts for the past week FirstChildUserIdTAG: 263693 FirstChildUserNameTAG: Coldberg FirstChildCreateTimeTAG: 2012-11-06T11:47:12Z FirstChildTAG: check this out https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Time_To_Decay/threads/50964f5e988a091f0000001e FirstChildUserIdTAG: 525246 FirstChildUserNameTAG: Sujith92 FirstChildCreateTimeTAG: 2012-11-06T18:51:44Z IndexTAG: 2077 TitleTAG: Please change the syringe example Dear Prof.Anant, The injection example is not very pleasant. The picture with syringe on the leg also adds to the discomfort. Can you please change the example in future courses? I thoroughly enjoyed many of your earlier examples and illustrations and have found them extremely imaginative and artistic. Thanks and Regards. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-11-06T04:28:32Z VoteTAG: 1 CoursewareTAG: Week 8 / Motivation And Review Modeling Drug Delivery CommentableIdTAG: 6002x_Motivation_And_Review_Modeling_Drug_Delivery NumberOfReplyTAG: 1 FirstChildTAG: I think it's not a leg, it'a an arm. Anyway: discomfort, really? Come on. FirstChildUserIdTAG: 54453 FirstChildUserNameTAG: berzasnon FirstChildCreateTimeTAG: 2012-11-06T10:29:47Z SecondChildTAG: lololol!!! SecondChildUserIdTAG: 301962 SecondChildUserNameTAG: labrinim SecondChildCreateTimeTAG: 2012-11-06T13:29:07Z SecondChildTAG: We are all different! ;) SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-06T13:29:49Z IndexTAG: 2078 TitleTAG: Math Processing Error - can't load Web Fonts I'm having intermittent problems getting math formulae to display in the exercises er al. The web page tries to load the fonts, then after a short time it fails to load, flashes the URL (https://www.edx.org/static/js/vendor/mathjax-MathJax-c9db6ac/fonts/HTML-CSS/TeX/png/imagedata.js), and then changes all of the math formatting into [Math Processing Error]. I am using Chrome 22.0.1229.94 m on Windows 7, SP1, 64-bit. I get the same results on IE8 and Firefox. I accessed the URL directly and got a 404. The failure is not happening 100% of the time; I was able to complete HW 7 this weekend, but later in the weekend, I had to quickly copy the math text from Lab 7 and paste it into a text editor. And today, it's also broken, even for exercises where it used to work. Any ideas what might be going wrong, and how I might be able to work around it? UserIdTAG: 1501 UserNameTAG: pfagerburg CreateTimeTAG: 2012-11-06T03:22:40Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: That error is due to how your browser interacts with MathJax. MathJax is used by the server to present mathematical symbols on your screen. Here's how to fix it: 1) Place your cursor on the [Math Processing Error] message and right-click. 2) You will see a pull-down MathJax menu. 3) Select Math Settings, Math Renderer, and choose either MathML or SVG. I prefer the way MathML looks, but it drops some symbols when I print, so I have to use SVG. 4) You can also play with the zoom settings on the MathJax menu to adjust the zoom functionality. FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-11-06T03:29:14Z IndexTAG: 2079 TitleTAG: LAB 8 After doing the calculus for the first circuit the value I've got for C is 180.603 pico farads,tough the grader gives me a wrong answer.In the transient analysys, the values are very very close to those in the example ( with my C, the voltage for t 300n is 809.546 m andin the example is 809,545. What is wrong here? I have tried to adjust my C to obtain the exact values, but is a matter of guessing. UserIdTAG: 369339 UserNameTAG: vargaslen CreateTimeTAG: 2012-11-05T20:37:23Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: you must define how many the component on devise 1 and 2 and calcul the value of this component FirstChildUserIdTAG: 174229 FirstChildUserNameTAG: fares27 FirstChildCreateTimeTAG: 2012-11-05T21:11:27Z SecondChildTAG: I can tell you the calculations involved in this Lab are pretty clear and precise (as compared to H8P1 for example). If your answers are not accepted, your calc can only go wrong in 1 of 2 ways: you picked the wrong component (the other is given to be an R), or you put the components in the wrong order. I can't figure out how else (this is what i meant as clear, because, to me, H8P1 has way many more possible scenarios). I'd say don't worry about the values you might find in *your* circuit. If your estimates are not exact but close enough, you should see a matching graph signature. From there you can figure out which formula or equations to use (forget the differentials and integrals; this lab uses only exponentials). Then substitute in the only values that are important here, which are shown in the given graphs, e.g. 809.545m and 190.455u at time 100n for first circuit, to calculate the unknown components' values. SecondChildUserIdTAG: 184153 SecondChildUserNameTAG: tthngn SecondChildCreateTimeTAG: 2012-11-06T04:52:06Z SecondChildTAG: Thank you tthngn for your response. I'm sure to have picked the correct nature of the devices in the correct order, and the grader confirmed it. Is the value for C in the first circuit and for L in the second (I havn't done the other two circuits yet) that are not accepted for the grader,tough these values work pretty well in the sandbox. Any other idea will be very wellcome. SecondChildUserIdTAG: 369339 SecondChildUserNameTAG: vargaslen SecondChildCreateTimeTAG: 2012-11-06T09:53:18Z SecondChildTAG: Ok, I just realized that I was taking the wromg time for the values (did't notice the dash line). So, thanks pal, everything is clear now. SecondChildUserIdTAG: 369339 SecondChildUserNameTAG: vargaslen SecondChildCreateTimeTAG: 2012-11-06T10:18:07Z SecondChildTAG: @vargaslen...im having the same problem ...i dint quite understand what you did..what dash line..??please help SecondChildUserIdTAG: 285024 SecondChildUserNameTAG: nigelgomes SecondChildCreateTimeTAG: 2012-11-07T07:30:05Z SecondChildTAG: Same, I don't know how to put the answer the right way. How I should put in 1 pF for example? 0.000001? SecondChildUserIdTAG: 87808 SecondChildUserNameTAG: pumoneon SecondChildCreateTimeTAG: 2012-11-07T23:47:51Z IndexTAG: 2080 TitleTAG: H9P2 again I found the strange grader behaviour, which looks like an issue for me. The grader accepts my expressions for $$\alpha$$ and for Q, but rejects the expression for natural frequency. Do I properly understand, that "natural frequency" can be calculated by the following expression: $$\omega_{d} = \sqrt {\omega_{0}^2 - \alpha^2} $$ If yes, then probably we have an issue in grader. If no, please tell me where I'm wrong. Thanks. UserIdTAG: 329361 UserNameTAG: ikrukov CreateTimeTAG: 2012-11-05T19:26:02Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: It accepts the undamped natural frequency in terms of L and C in Hz. FirstChildUserIdTAG: 141000 FirstChildUserNameTAG: OrinE FirstChildCreateTimeTAG: 2012-11-05T20:11:17Z SecondChildTAG: Thanks! SecondChildUserIdTAG: 329361 SecondChildUserNameTAG: ikrukov SecondChildCreateTimeTAG: 2012-11-05T20:38:22Z FirstChildTAG: Very annoying! Last week I wasted a lot of time on that problem until in desperation I decided to try the undamped natural frequency. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-05T23:16:30Z IndexTAG: 2081 TitleTAG: S17E5 week 9 Can anyone explain how we get iL(0)=I0+Asin(ϕ) and vC(0)=ALω0cos(ϕ)? In my calculations i received opposite values for sin and cos. UserIdTAG: 352757 UserNameTAG: Kerbyco CreateTimeTAG: 2012-11-05T18:22:57Z VoteTAG: 1 CoursewareTAG: Week 9 / An ILC circuit CommentableIdTAG: 6002x_an_ILC_circuit NumberOfReplyTAG: 5 FirstChildTAG: See videos s17v11 and s17v12. Also valid for Norton equivalents. FirstChildUserIdTAG: 443358 FirstChildUserNameTAG: torkelh FirstChildCreateTimeTAG: 2012-11-07T15:22:46Z FirstChildTAG: I'm asking the same question. Could anyone explain where the differential equation in this problem differs from the example differential equation in the sequence of videos? Isn't it LC*iL''(t)+iL(t)=I0, and vL(t)=vC(t)=L*iL'(t)? Even initial conditions are the same. Where does it differ? Thank you very much! FirstChildUserIdTAG: 398594 FirstChildUserNameTAG: DaveyJC FirstChildCreateTimeTAG: 2012-11-14T22:54:56Z FirstChildTAG: I think my theory is proved? by the following image. Current iL is the pink line and it follows iL[t] = I0 - I0*Cos[w0*t] equation. Voltage vC=vL is blue line and it follows vC[t] = L*I0*w0*Sin[w0*t] equation. Where am I wrong? Thank you! ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13529347191343667.jpg FirstChildUserIdTAG: 398594 FirstChildUserNameTAG: DaveyJC FirstChildCreateTimeTAG: 2012-11-14T23:24:36Z SecondChildTAG: Your equations are right, of course. It was an arbitrary decision to continue using sine for currents, and cosine for voltage. But to use the sine function instead of cosine, it had to be offset by $\phi$, giving us: $ i_L\left(t\right) = I_0 + I_0\cdot \sin\left(\omega_0\cdot t+\phi\right)\\ $ If you look at the graphs, you can see that: $ -\cos(x)=\sin\left(x + \frac \pi 2 \right)$ SecondChildUserIdTAG: 714237 SecondChildUserNameTAG: PaxPolaris SecondChildCreateTimeTAG: 2012-11-18T03:10:21Z FirstChildTAG: could someone please help me understand the last question. if i'm not spoiling 'the thing' i would like to tell what i got till now. ok.... if we say that the step voltage started a bit late (from I(0+)=0) then we xpect a lag in capacitor's waveform. rite? now while solving the differential equation, this lagging factor is added up by involving 'phi' ..... could someone tell me how do we get to this..... from simple sin(wt) in the lectures to this sin(wt+phi).... thanx alot....:) FirstChildUserIdTAG: 117319 FirstChildUserNameTAG: iqramalik FirstChildCreateTimeTAG: 2012-11-16T22:05:26Z SecondChildTAG: another things troubling me like many others is that why is the current term having sin and voltage having cos in their equations. should'nt it be the other way round. as shown by DevayJC SecondChildUserIdTAG: 117319 SecondChildUserNameTAG: iqramalik SecondChildCreateTimeTAG: 2012-11-16T22:08:36Z FirstChildTAG: You can derive the 2nd order differential equation for $i_H(t)$ from the answer of the first question: $LC\cdot i_H''+i_H=0\\ \implies i_H=LC\cdot (-i_H'' )\\ $ Aft this you have to guess what $i_H$ looks like: If you guess $A\cdot e^{(st)}$ and simplify you will inevitably get a cosine in your answer. But note, you can also guess that $i_H(t)$ is some sin function ... since the second derivative of $\sin$ is $-\sin$. If you plug $i_H(t) = A\sin(\omega_0 t + \phi)$ in the 2nd order eq. above it should work Also, since the graphs of sin and cos are just translations of one another, they can easily replace one another. FirstChildUserIdTAG: 714237 FirstChildUserNameTAG: PaxPolaris FirstChildCreateTimeTAG: 2012-11-18T03:45:35Z SecondChildTAG: Thanks PaxPolaris! I hadn't thought about the equivalence between sines and cosines! Fool of me. SecondChildUserIdTAG: 398594 SecondChildUserNameTAG: DaveyJC SecondChildCreateTimeTAG: 2012-11-19T16:57:49Z IndexTAG: 2082 TitleTAG: Unable to understand how G and W are taken for case of inductance Can anyone explain physically how the area and length for inductance has been considered? From my understanding, I also need the depth/thickness of the power line to calculate the cross section area. So that, A = 2.5mm x Depth l = 10 cm L = ( u * N^2 * A ) / l (where N = 1) UserIdTAG: 297370 UserNameTAG: ayush3504 CreateTimeTAG: 2012-11-05T17:57:55Z VoteTAG: 1 CoursewareTAG: Week 7 / Geometry and Capacitance/Inductance CommentableIdTAG: 6002x_Geometry_Capacitance_Inductance NumberOfReplyTAG: 0 IndexTAG: 2083 TitleTAG: S15E1: REVIEW: A STEP UP problem to display exponential how to write exponential??? UserIdTAG: 316761 UserNameTAG: gk_goel CreateTimeTAG: 2012-11-05T16:34:32Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Your answer should look something like: $$x*(x \pm e ^{( \pm (x \pm x)*x/x)})$$ Use "^" for the exponential, as "3^3=9". Also, use parens liberally to insure terms are operated on as you intend. Hope this helps! FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-11-05T20:52:24Z IndexTAG: 2084 TitleTAG: Lab 7 P1, Q1 Integrating a variable that is the limit sorry but I just can't do Integration of sin(1000∗2π∗t) if the upper limit is t, Please help!!! -myrimit UserIdTAG: 231676 UserNameTAG: Roosemberth CreateTimeTAG: 2012-11-05T03:50:02Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: For those who care about the mathematical niceties, the following brief article is a good read: http://www.dam.brown.edu/people/anaray/classes/am35/materials/def_indef_int.pdf There is a real distinction between definite and indefinite integrals. It is the Fundamental Theorem that links the two. Importantly the variable that appears under the integral sign of a definite integral is a dummy variable and can be replaced by any other symbol. For those who exercise care in doing math the variable appearing in the upper or lower limit should not be the same as the dummy variable appearing under the integral sign. Here is another similar article: http://www.ms.uky.edu/~carl/ma123/kob98/kob98htm/chap18e1.html FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-05T14:07:34Z FirstChildTAG: Why not? Suppose in your head that t=12, but instead of 12 write everywhere t. Also, you may use wolfram alpha :) FirstChildUserIdTAG: 4076 FirstChildUserNameTAG: damians FirstChildCreateTimeTAG: 2012-11-05T04:23:07Z SecondChildTAG: Thank you, I had tried with microsoft mathematics but I get an error that by def the integrating variable couldn't be the upper limit Best Regards SecondChildUserIdTAG: 231676 SecondChildUserNameTAG: Roosemberth SecondChildCreateTimeTAG: 2012-11-05T05:29:41Z FirstChildTAG: Think of it as the integral of sin(1000*2*π*x) dx with an upper limit of t instead. Try inputting "integrate sin t dt from t=0 to t" on www.wolframalpha.com to see that it's OK for the upper limit to be t when integrating with respect to t. FirstChildUserIdTAG: 141000 FirstChildUserNameTAG: OrinE FirstChildCreateTimeTAG: 2012-11-05T04:37:18Z FirstChildTAG: Hi Roosemberth, Try to solve the integral in pencil and paper, Lets say that your result from your integral is, for example, $X^3 + 5$ and you have to evaluate it from 0 to -2 (remember the barrow rule, you replace the superior extreme in the formula minus the inferior extreme in the formula) $[(-2)^3 + 5 ] - [0^3 + 5] = -8 + 5 - 5 = -8$ ---------- Now, Lets say that your result from your integral is, for example, $X^3 + 5$ and you have to evaluate it from 0 to t (remember the barrow rule, you replace the superior extreme in the formula minus the inferior extreme in the formula) $[(t)^3 + 5 ] - [0^3 + 5] = (t)^3 + 5 - 5 = (t)^3$ Now is your turn try with the sin integral :) FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-05T11:52:38Z SecondChildTAG: You should become a professional teacher! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-05T12:51:38Z SecondChildTAG: Haha! I will consider your advice for the future ;). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-06T00:56:35Z FirstChildTAG: Hi Roosemberth! I definitely recommend Wolfram Alpha. Take a look here https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/508da86cd8a1b11f0000006c , this thread deals with exactly your problem! I hope this helps! Hazel. FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-11-05T12:50:58Z IndexTAG: 2085 TitleTAG: I have joined today Hello Every One! I have joined today How should I cover up with my first term work? Thank you. UserIdTAG: 762798 UserNameTAG: hshakir80 CreateTimeTAG: 2012-11-04T23:10:36Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Hi hshakir80! Welcome to 6.002x! :) Unfortunately is really late... You should take a look at this Response Post [here][1] Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Lab_0_Using_the_Tools/threads/5096bcb24ba7ce2b00000057 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-04T23:40:04Z FirstChildTAG: You happen to have joined at one of the lastest times when it is theoretically possible to still pass, but unless you already know that material and are doing the assignments for fun, it will probably make more sense to wait until next term to take the course. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-11-05T00:00:36Z FirstChildTAG: Hi hshakir80, If I understand your question correctly... start at week one in the courseware and *absorb the information thoroughly*. The deadlines for homework from weeks 1 to 6 have passed and homework 7 is due November the 6th, but I encourage you to attempt **all** the questions anyway. Think of this as a preparation for the spring, when the course will be offered again! Oh, and if you have any questions, please, just ask. There are loads of people that are more than willing to help on this forum. Hazel. FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-11-05T12:59:48Z IndexTAG: 2086 TitleTAG: Anybody Can anyone give some links for online electronic simulations or programms such as electronic work bench? UserIdTAG: 96873 UserNameTAG: Santyaga CreateTimeTAG: 2012-11-04T20:16:54Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: I really like [CircuitLab][1]. [1]: https://www.circuitlab.com/ FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-11-05T05:00:09Z FirstChildTAG: iCircuit can be useful. http://icircuitapp.com/ FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-11-05T15:09:31Z FirstChildTAG: Thank you for good advice!!! FirstChildUserIdTAG: 96873 FirstChildUserNameTAG: Santyaga FirstChildCreateTimeTAG: 2012-11-11T17:42:31Z IndexTAG: 2087 TitleTAG: LAB 7 how to evaluate for t=0.0005s can you help me with this? UserIdTAG: 413613 UserNameTAG: shaliesh CreateTimeTAG: 2012-11-04T20:02:49Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hi shaliesh, Can I help you? In which part are you lost? You can take a look at this hints [Lab 7 Hints][1] Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5091dd5d3fc09f2b0000018a FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-04T20:35:02Z FirstChildTAG: A hint: radians. FirstChildUserIdTAG: 54453 FirstChildUserNameTAG: berzasnon FirstChildCreateTimeTAG: 2012-11-05T00:17:39Z SecondChildTAG: Well now I feel stupid, but thank you. SecondChildUserIdTAG: 282209 SecondChildUserNameTAG: nklitzke SecondChildCreateTimeTAG: 2012-11-05T19:10:04Z IndexTAG: 2088 TitleTAG: pass margin what is the pass margin of this course UserIdTAG: 509517 UserNameTAG: randima CreateTimeTAG: 2012-11-04T16:43:24Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: 60% FirstChildUserIdTAG: 99628 FirstChildUserNameTAG: Pranjal16 FirstChildCreateTimeTAG: 2012-11-04T17:21:26Z SecondChildTAG: thanx pranjal... SecondChildUserIdTAG: 509517 SecondChildUserNameTAG: randima SecondChildCreateTimeTAG: 2012-11-11T12:15:10Z IndexTAG: 2089 TitleTAG: TextBook How can we download the text book. It is pretty tiresome to read it online UserIdTAG: 499671 UserNameTAG: maliha266 CreateTimeTAG: 2012-11-04T13:53:43Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Take snapshosts or save as webpage. FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-11-04T14:03:42Z FirstChildTAG: save the page that i used to do.. FirstChildUserIdTAG: 114864 FirstChildUserNameTAG: Vikaash FirstChildCreateTimeTAG: 2012-11-04T15:35:07Z FirstChildTAG: Downloading a Textbook that is not free is piracy... - You can buy it [here][1], also remember that edX Students have a discount. ![enter image description here][2] - Also, you can take a look at the Wiki, there is a Scrolling version of the Textbook, is more friendly for online reading, and you can go to different pages with "g" from the keyboard Wiki->[Scrolling Textbook Viewer][3] I hope this can help you, Myriam. [1]: http://store.elsevier.com/specialOffer.jsp?offerId=EST_MIT&utm_source=EMAP&utm_medium=text_link&utm_term=Circuits_Electronics&utm_campaign=edx#oid=1003_4 [2]: http://secure-ecsd.elsevier.com/est/20122517_B_MITx-Landing-Page-banner_034-317_1200_R2_FINAL.jpg [3]: https://6002x.mitx.mit.edu/static/contrib/xbook.html FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-04T15:55:08Z IndexTAG: 2090 TitleTAG: H8P3 can someone gimme a hint about H8P3 last question pls. wat should i consider the resistance value as? wat voltage does "drain of q1 is low" correspond to? should we use voltage divider rule or assume it to be 0? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-11-04T11:19:58Z VoteTAG: 1 CoursewareTAG: Week 8 / Time To Decay CommentableIdTAG: 6002x_Time_To_Decay NumberOfReplyTAG: 3 FirstChildTAG: You should consider that Q1 is ON, and that R_ON replaces the drain-source of Q1 and then use a voltage divider to find this low voltage which will not be zero. I have a question for you: Is there another way to express: "Now, suppose the drain of Q1 is high, as above, and the store line is held at the same voltage as the drain of Q1. What is the maximum voltage, in Volts, that the gate of Q2 can be charged to? Note, this value must be larger than VOH to satisfy the static discipline." in H8P3 I think that I'm misinterpreting the "and the store line is held at the same voltage as the drain of Q1" portion of the problem statement or that I'm missing the implication. FirstChildUserIdTAG: 569199 FirstChildUserNameTAG: neogerald FirstChildCreateTimeTAG: 2012-11-04T19:02:30Z SecondChildTAG: consider the voltage across the drain and source of transistor Q3. Store line is held at the voltage of drain of Q1 implies that VDS of Q1=VGS of Q3. note VDS>VGS-VT for Q3 to remain in saturation. this VDS is de value of the voltage to wch the gate of Q2 can be charged. SecondChildUserIdTAG: 525246 SecondChildUserNameTAG: Sujith92 SecondChildCreateTimeTAG: 2012-11-05T15:21:19Z FirstChildTAG: ok so i use the voltage divider rule and then use the formula VC=Vi+(Vo-Vi)e^(-t/RC) where Vi is the voltage obtained using voltage divider and Vo is VOH and VC is VIL. Is this correct? if yes what should be the value of R? pls help FirstChildUserIdTAG: 525246 FirstChildUserNameTAG: Sujith92 FirstChildCreateTimeTAG: 2012-11-05T15:47:56Z SecondChildTAG: **@Sujith92 I am not getting what is the use of Vi here....how to calculate it? Please let me know it.** SecondChildUserIdTAG: 156859 SecondChildUserNameTAG: PrithviMonangi SecondChildCreateTimeTAG: 2012-11-09T06:04:16Z FirstChildTAG: This R is the equivalent R that the "capacitor sees." This R is the sum of R_ON (from Q3) and the parallel combination of R_PU and R_ON (from Q1). Pls let me know if this helps. FirstChildUserIdTAG: 569199 FirstChildUserNameTAG: neogerald FirstChildCreateTimeTAG: 2012-11-06T15:50:58Z SecondChildTAG: Whoa! Thank u that's right! i guess i need to understand de concept of gate source capacitance better! did my hint to your question help? SecondChildUserIdTAG: 525246 SecondChildUserNameTAG: Sujith92 SecondChildCreateTimeTAG: 2012-11-06T18:49:04Z SecondChildTAG: Yes, your hint is helping me understand it better. Thank you! SecondChildUserIdTAG: 569199 SecondChildUserNameTAG: neogerald SecondChildCreateTimeTAG: 2012-11-07T14:06:52Z SecondChildTAG: For H8P3 Q5 I used the combination of Ron(Q3)+Ron(Q1)//Rpu and Voh as Vinitial and Vil as Vfinal and I can't get the green mark :( SecondChildUserIdTAG: 298547 SecondChildUserNameTAG: AnthonyRF SecondChildCreateTimeTAG: 2012-11-12T01:33:04Z IndexTAG: 2091 TitleTAG: Meaning of symbol. what is the evaluation of x||y? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-11-04T11:10:32Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hi, IT refers to x parallel to y 's equivalent and its mathematical evaluation is (x*y)/(x+y). FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-11-04T11:33:01Z SecondChildTAG: Also, 1/(1/x + 1/y) This form is easier to expand if you have more than two variables. SecondChildUserIdTAG: 90796 SecondChildUserNameTAG: ChaunceyGardiner SecondChildCreateTimeTAG: 2012-11-04T18:20:06Z FirstChildTAG: Just to echo the others: R1||R2 = (R1*R2)/(R1+R2) Two resistors in parallel. Now try R1||R2||R3 :) FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-11-04T18:44:12Z IndexTAG: 2092 TitleTAG: Writing the formula in the discussions Hi. everybody. Could anyone explain me who to write formula. Is it a tex syntax? Who can I use it? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-11-04T07:06:19Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13520988535744543.png FirstChildUserIdTAG: 9673 FirstChildUserNameTAG: xp42 FirstChildCreateTimeTAG: 2012-11-05T07:00:58Z SecondChildTAG: some links: http://www.onemathematicalcat.org/MathJaxDocumentation/TeXSyntax.htm http://www.mathjax.org/ http://amath.colorado.edu/documentation/LaTeX/Symbols.pdf SecondChildUserIdTAG: 9673 SecondChildUserNameTAG: xp42 SecondChildCreateTimeTAG: 2012-11-05T07:03:57Z FirstChildTAG: It's called mathjax, and in this forum it is broken. If they can ever fix it, that would be great. Go to the old mitx 6.002x forum and search on it. I'm not even gonna try to offer an explanation here because in all likelihood it would render here as nonsense. The math formulas would display ok, but everything would be all over the place. :-( Oh what the heck, let's see how it goes.... ---- **in line equations** *$* \\frac{ 1 }{ j \\omega C } *$* formats in-line as $ \frac{1}{j \omega C}$ --- **centered equations** *$$* \\frac{1}{j \\omega C}*$$* formats centered on a new line as $$ \frac{1}{j \omega C}$$ --- **enclosing braces** *$$* \\left [ \\frac{1}{j \\omega C} \\right ] *$$* $$ \left [ \frac{1}{j \omega C} \right ] $$ --- **exponents** *$$* \\left [ \\frac{1}{j \\omega C} \\right ]^6 *$$* $$ \left [ \frac{1}{j \omega C} \right ]^6 $$ --- **subscripts** *$$* R_1 C_2 Z_L *$$* $$ R_1 C_2 Z_L $$ --- **multiplication** *$$* \\omega = 2 \\pi \cdot f *$$* $$ \omega = 2 \pi \cdot f $$ --- **special characters** *$$* \\approx ~ \\rightarrow ~ \\mu ~ \\tau ~ \\infty ~ \\aleph_0 *$$* $$ \approx ~ \rightarrow ~ \mu ~ \tau ~ \infty ~ \aleph_0 $$ --- **spaces** *$$* (abc) = (~ab~c~) *$$* $$ (abc) = (~ab~c~) $$ --- **multi-character subscripts and superscripts** *$$* 10^{log_{10}(x)} = x *$$* $$ 10^{log_{10}(x)} = x $$ --- **text in a formula** *$$* \\text{normal font} ~~ equation font *$$* $$ \text{normal font} ~~ equation font $$ --- **square root** *$$* \sqrt{ \frac{R_L}{R1 + \frac{1}{j \omega C} } } *$$* $$ \sqrt{ \frac{R_L}{R1 + \frac{1}{j \omega C} } } $$ --- **more space** *$$*(a ~ b \quad c ~) *$$* $$(a ~ b \quad c ~) $$ --- **bigger** *$$* 12 {\large 34} {\Large 56} {\huge 78} {\Huge 90} 12 *$$* $$ 12 {\large 34} {\Large 56} {\huge 78} {\Huge 90} 12 $$ --- **integrals** *$$* \int_{- \infty}^{\infty} \frac{\sin(x)}{x^2+1} ~ dx *$$* $$\int_{-\infty}^{\infty}\frac{\sin(x)}{x^2+1}~dx$$ --- FirstChildUserIdTAG: 9673 FirstChildUserNameTAG: xp42 FirstChildCreateTimeTAG: 2012-11-04T09:12:10Z SecondChildTAG: Yep, still doing it's own thing here. Go check the mitx forum to see it displayed correctly. https://6002x.mitx.mit.edu/discussion/question/80142/how-do-you-enter-equations-in-posts#80323 SecondChildUserIdTAG: 9673 SecondChildUserNameTAG: xp42 SecondChildCreateTimeTAG: 2012-11-04T09:15:42Z SecondChildTAG: But i haven't the account for the old forum and i can't register to it. SecondChildUserIdTAG: 213452 SecondChildUserNameTAG: JoJosida SecondChildCreateTimeTAG: 2012-11-05T05:03:40Z SecondChildTAG: That sucks. It used to be the forum was open. i guess it's to keep answers out of the public domain. A lot of stuff is repeated from one class to another. SecondChildUserIdTAG: 9673 SecondChildUserNameTAG: xp42 SecondChildCreateTimeTAG: 2012-11-05T06:59:23Z FirstChildTAG: ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13520987811343691.png FirstChildUserIdTAG: 9673 FirstChildUserNameTAG: xp42 FirstChildCreateTimeTAG: 2012-11-05T06:59:53Z FirstChildTAG: ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13520988111343694.png FirstChildUserIdTAG: 9673 FirstChildUserNameTAG: xp42 FirstChildCreateTimeTAG: 2012-11-05T07:00:19Z FirstChildTAG: ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13520988361343697.png FirstChildUserIdTAG: 9673 FirstChildUserNameTAG: xp42 FirstChildCreateTimeTAG: 2012-11-05T07:00:42Z SecondChildTAG: Thank you. SecondChildUserIdTAG: 213452 SecondChildUserNameTAG: JoJosida SecondChildCreateTimeTAG: 2012-11-05T12:09:55Z IndexTAG: 2093 TitleTAG: S18V21 differentiation misses 'i' The differential equation developed in this lecture starts as: $C\displaystyle\frac{dv}{dt} + \frac{v}{R} - i + \frac{1}{L}\displaystyle\int^t_{-\infty} v\,dt = 0$ This is then differentiated and divided through by C to get: $\displaystyle\frac{d^2v}{d^2t} + \frac{1}{RC}\frac{dv}{dt} + \frac{v}{LC} = i$ 'i' got moved to the right hand side but didn't get differentiated nor divided by C! Surely, either the right hand side should be $\displaystyle\frac{1}{C}\frac{di}{dt}$ or, if i is assumed to be constant, 0. In either case, just i appears to be wrong. BTW, The textbook supplemental material for the parallel RLC circuit has $\displaystyle\frac{1}{C}\frac{di}{dt}$ on the right hand side. They mention the derivative on the right hand side as "an unnecessary complication" and get rid of it by substituting $\displaystyle v = L\frac{di}{dt}$ into the first equation above and solving for the current through the inductor instead of the voltage across it. Comments? UserIdTAG: 141000 UserNameTAG: OrinE CreateTimeTAG: 2012-11-04T05:52:35Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hello, If we think about this circuit as an anloguous case to the previously mentioned circuit, then we should find the solution such that we seek to find the current passing by the inductance. Let me give my proposition (I hope I didn't made any stupid mistake): ![my proposition of the initial 2nd Order DE][1] [1]: https://edxuploads.s3.amazonaws.com/13531144004409834.png We can notice easily that every term is analoguous to the previously mentioned circuit: 1. vC in the first circuit is analoguous to iL. 2. R in the first circuit is analoguous to 1/R. 3. L is analoguous to C. 4. C is analoguous to L. 5. V is analoguous to I. FirstChildUserIdTAG: 418182 FirstChildUserNameTAG: euldji2005 FirstChildCreateTimeTAG: 2012-11-17T01:17:26Z IndexTAG: 2094 TitleTAG: H7P3——Q3&Q4 I knew how to solve the questions and got the green tick,but I don't understand what is the meaning of the comparison between the two time duration.The results show that light across the 2cm wire lead takes less time than the output voltage reaching VO=2.5V(starts from 0V),so my question is from the results, what can we conclude? Is it related to the lumped matter discipline? please help! thx UserIdTAG: 361137 UserNameTAG: Ericson CreateTimeTAG: 2012-11-04T04:35:38Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Yes. In my opinion it demonstrates the delay that inductors impose in a particular circuit. While we like to assume electricity travels the loop in an "instant", you will now notice it takes four times as long with the inductor. (Atleast with the numbers I have) FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-11-04T11:35:35Z SecondChildTAG: Well,I got it.I find something on page 10 of the textbook,the third point of LMD,you are right,thanks a lot!　 SecondChildUserIdTAG: 361137 SecondChildUserNameTAG: Ericson SecondChildCreateTimeTAG: 2012-11-05T04:07:45Z SecondChildTAG: Even a broken clock is right twice a day! SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-11-05T11:32:05Z SecondChildTAG: Well,I'm not native English speaker,so I think i didn't understand this sentence,so ashamed of that:) SecondChildUserIdTAG: 361137 SecondChildUserNameTAG: Ericson SecondChildCreateTimeTAG: 2012-11-06T10:47:09Z IndexTAG: 2095 TitleTAG: Missed MidTerm Due to unavoidable circumstances I was not able to attempt Mid Term exam can I still make it looking at my LAB & Home work results. UserIdTAG: 272523 UserNameTAG: Jivraj CreateTimeTAG: 2012-11-03T21:09:01Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: yeah I missed midterm too,sorry Jivraj wa are in the same boat. am going to try to get all the homeworks and try getting a good score in the final exam. if I don't make it to al least 60% i will still have learnt alot and will re-offer the course the next time it is offered. FirstChildUserIdTAG: 183507 FirstChildUserNameTAG: obiradaniel FirstChildCreateTimeTAG: 2012-11-03T21:41:32Z SecondChildTAG: I NEED TO ENTER THE MIDTERM THAT I MIISED SecondChildUserIdTAG: 735825 SecondChildUserNameTAG: bolis SecondChildCreateTimeTAG: 2012-11-04T08:57:35Z SecondChildTAG: Same here bro..i would love to re-give it if possible..but yes whatever the case....learnt alot:-) SecondChildUserIdTAG: 478827 SecondChildUserNameTAG: Abbas1989 SecondChildCreateTimeTAG: 2012-11-04T16:54:15Z FirstChildTAG: Obiradaniel has the right idea. If you have 100% in your homework and 100% on your final exam you could still get a "B" grade. (These two represent 70% of the course) So basically you have to get within 10% of a perfect score for homework and final exam combined, from here on. As you may already know two of your lowest homework scores will not be counted, this helps a little too. Having said that, I do not know what your homework score is, so you will have to ultimately determine if it is possible. Either way, stick with it, you will either pass or end up with some advantage when you do take it again next time. Have fun. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-11-03T23:45:15Z SecondChildTAG: Thanks Pennypacker and Obiradaniel for positive feedback and I know I am not alone looking at my LAB 51% Home Work is 39%. so start doing LAB 7 and Home Work. Thanks SecondChildUserIdTAG: 272523 SecondChildUserNameTAG: Jivraj SecondChildCreateTimeTAG: 2012-11-04T07:25:40Z IndexTAG: 2096 TitleTAG: H9P1 Superconductors do exist in "high" temperatures - so far up to 130K UserIdTAG: 157610 UserNameTAG: mradziwo CreateTimeTAG: 2012-11-03T18:51:51Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: While I see your point, I would not consider ~ -225.67F / -143.15C to be considered a high temperature, from a practical point of view. :) FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-11-04T00:35:26Z IndexTAG: 2097 TitleTAG: "Survey closing soon" Please take it [here][1]! [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/508e0829ba8f10260000000c UserIdTAG: 374590 UserNameTAG: ahope CreateTimeTAG: 2012-11-03T18:27:42Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Do we get to see the results? FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-11-03T19:43:20Z IndexTAG: 2098 TitleTAG: MAY I CAN JOIN COURCE NOW OR ITS TOO LATE??? I JUST GOT INFO. ABOUT ONLINE COURSE SO IS IT WORTH TAKING CLASSES FROM NOW OR I SHOULD WAIT FOR NEXT TIME??? UserIdTAG: 757565 UserNameTAG: HIREN24 CreateTimeTAG: 2012-11-03T17:29:48Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: better wait for next time..you can still watch the lectures and solve the homeworks and all..the last dates for submission of the 2st 6 HWs and labs are over..you cannot manage a pass (60%) even if you solve all the labs and homeworks and the final exam (40+18=56)..so if you take it this time, it will only be for the knowledge..you can ofcourse learn this time and then take it again for the grade in spring.. FirstChildUserIdTAG: 173147 FirstChildUserNameTAG: cruiser_rahit FirstChildCreateTimeTAG: 2012-11-03T18:02:07Z SecondChildTAG: That was my doubt too, kinda sad, I saw some lectures and the course is fanstastic. But it's ok, I'll wait for next class too. Thanks SecondChildUserIdTAG: 700300 SecondChildUserNameTAG: Erikson SecondChildCreateTimeTAG: 2012-12-05T13:16:13Z FirstChildTAG: Hi HIREN24, You should take a look at Course Info - closing enrollment - [click on here][1] ... Remember that this Course will offered next spring... ---- **Closing Enrollment:** Although we usually like to give students the chance to do impressive and unlikely feats of learning, the due date for week 7 assignments marks the time when it is impossible to earn enough points for a certificate, and we will therefore not be accepting any new registrations for the course. Good luck to everyone who is still in the running! [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/info FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-03T20:56:27Z IndexTAG: 2099 TitleTAG: Week 7 Easiest of all? I almost completed week 7 lab and homework in about quarter of an hour, How do you guys feel about it? Was it the easiest of all? UserIdTAG: 148523 UserNameTAG: span993 CreateTimeTAG: 2012-11-03T12:05:54Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: true... FirstChildUserIdTAG: 582342 FirstChildUserNameTAG: achilees111 FirstChildCreateTimeTAG: 2012-11-03T13:47:38Z FirstChildTAG: Some they are a-easy, some they are a-hard. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-11-03T15:02:12Z SecondChildTAG: I agree SecondChildUserIdTAG: 342966 SecondChildUserNameTAG: SandraNavarro SecondChildCreateTimeTAG: 2012-11-04T17:21:32Z IndexTAG: 2100 TitleTAG: H7P3 part 3 Any hints will do. I try to approach it with the equation for iL, setting iL = vO/RO, II=V/RS, IO=0. UserIdTAG: 458621 UserNameTAG: GomesCesar CreateTimeTAG: 2012-11-02T18:57:09Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: Thank you both, I was able to get the answer because of your hints. FirstChildUserIdTAG: 458621 FirstChildUserNameTAG: GomesCesar FirstChildCreateTimeTAG: 2012-11-02T20:35:51Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-02T21:45:33Z FirstChildTAG: Hi GomesCesar, Can I help you? Let´s see this together: ![im][1] Hint 1: What current do you have in RO? Isn´t it the same of the one that shares the other component in the loop? Hint 2: Take a look at page 520 [read here][2]. How is the current in a RL circuit? What kind of Circuit do you have in this Problem ;). Hint 3: How did you calculate a voltage? Does it ring you a bell the Ohm´s Law? Do you have the current and the resistive element (recall Hint 1)? I hope this can help you. Myriam. [1]: https://www.edx.org/static/content-mit-6002x/images/circuits/lead-inductance.e6074fe36c87.gif [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/544 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-02T19:35:02Z SecondChildTAG: can u help me I am getting -L*ln(1-VO*((Ro+R)/(V*R)))/(RO+R) I am unable to get the correct answer SecondChildUserIdTAG: 185921 SecondChildUserNameTAG: Ravi_Teja SecondChildCreateTimeTAG: 2012-11-03T03:28:25Z SecondChildTAG: @Myriam on solving i got 0.29XXX but it says it is wrong can you help me where am i going wrong SecondChildUserIdTAG: 185921 SecondChildUserNameTAG: Ravi_Teja SecondChildCreateTimeTAG: 2012-11-03T12:00:15Z SecondChildTAG: @myriam...are Ro and Rs in the final formula...???i get an error that it cant parse Ro and Rs.:( SecondChildUserIdTAG: 285024 SecondChildUserNameTAG: nigelgomes SecondChildCreateTimeTAG: 2012-11-03T13:05:14Z SecondChildTAG: Hi @Ravi_Teja, Can I help you? you said that -L*ln(....) is a time unit, but L it is in Hy (Henry) unit, so Henry multiplied by a dimensionless number, will give Henry units and not seconds units... so, if you need time, is that correct? Hints: Take a look at page 520 of the Textbook, this is an example of an arbitrary RL (Resistive - inductive) series Circuit. After you have read that page, can you tell me the expression of iL (current in the inductor)? ;). Now, what voltage do we have in the output vO? How do you calculate a voltage in the Output? (Recall Ohm's Law) . What current do you have is the elements in a loop are in series? Do you have your output value? ;) do you have your output equation? ;). Is that equation expressed in terms of circuits elements an t? SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-03T13:38:32Z SecondChildTAG: Hi @nigelgomes, Be careful RS (R **uppercase** and S **uppercase**) **is not** the same as Rs (R **uppercase** and s **lowercase**). **You need RO and RS** and **not Rs and Ro**.... SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-03T13:42:34Z SecondChildTAG: Invalid input: Rs R not permitted in answer....is the error SecondChildUserIdTAG: 285024 SecondChildUserNameTAG: nigelgomes SecondChildCreateTimeTAG: 2012-11-03T14:12:17Z SecondChildTAG: this is for the part 1 of the question SecondChildUserIdTAG: 285024 SecondChildUserNameTAG: nigelgomes SecondChildCreateTimeTAG: 2012-11-03T14:13:44Z SecondChildTAG: Hi @nigelgomes, "Invalid input: Rs R not permitted in answer" Remember that you have to express your result in terms of the elements of your Circuit... So, Is R in your Circuit? You only have RS and RO and not R ... ![im][1] Also Rs is RS... Remember that in the Textbooks (page 520) it shows you an general RL (Resistive inductive) Circuit, so the R that they are using is the total R of the circuit ;). I hope this can help you. [1]: https://www.edx.org/static/content-mit-6002x/images/circuits/lead-inductance.e6074fe36c87.gif SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-03T15:00:24Z SecondChildTAG: Hi @Myrimit but i am dividing L with RS+RO u can see at the end of the equation i think this returns unit as seconds SecondChildUserIdTAG: 185921 SecondChildUserNameTAG: Ravi_Teja SecondChildCreateTimeTAG: 2012-11-03T16:47:59Z SecondChildTAG: Hi @Ravi_Teja, Sorry, I thoght that your R was inside the ln :P. Yes, you are right Henry = Ohm*s and divided by Ohm is s ... SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-03T21:05:55Z SecondChildTAG: Do you still having difficulties to find the t? ... Can I help you? SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-03T21:07:35Z SecondChildTAG: Hi Myrimit: I have got 0.29XXX too. Please help me. SecondChildUserIdTAG: 355852 SecondChildUserNameTAG: abuhossain SecondChildCreateTimeTAG: 2012-11-04T06:40:31Z SecondChildTAG: I have solved the probem, taking into account OHM's law i.e. V=IR. The I here is *iL* and *R=RO* and *V=VO*. *iL* is found using the equation given in Page 520 of text book, but we have two resistors in series and in the book example there is just one resistor. So for me I have done everything accordingly, but still my equation for *VO* is wrong. Second thing is that in the equation we have a *t*, but it says that *t* is wrong in the equation. How we can remove *t*? SecondChildUserIdTAG: 64618 SecondChildUserNameTAG: ashfaq2419 SecondChildCreateTimeTAG: 2012-11-04T10:22:30Z SecondChildTAG: okey by your idea, R will be the eq resistor Req=RS+RO=72ohms, taking that and picking the equations we have il=(Vs/Req)*e^-(Req/L)t and by that we have vO=RO*il okey let's do this by parts vO=1.65 from the exercise vs=3.3 so 1.65=50*((3.3/72)*e^-(72/(15.44*10^-9))t) let's solve this we get 1.65/50=0.033=iL so 0.033=(3.3/72)*(1-e^(blabla) we get (1-e^(blabla))=(3.3/72)/0.033=aprox: 1.3889 -(Req/L)*t=ln(1.3889-1) and t=ln((1.3889-1)/-Req/L) i have the value 2.0253e-10 what is wrong here? this answer is wrong i don't get to see where is the problem, and even if pass this value to nanoseconds getting 20.253e-9 I still get the x on the checkbox... help would be apreciated... SecondChildUserIdTAG: 317836 SecondChildUserNameTAG: VitorRodrigues SecondChildCreateTimeTAG: 2012-11-05T04:29:00Z SecondChildTAG: I'm sorry forgot the (1-e^blabla) the first time's i wroted the equation, but i used in the calculation so the value still's the same... SecondChildUserIdTAG: 317836 SecondChildUserNameTAG: VitorRodrigues SecondChildCreateTimeTAG: 2012-11-05T04:31:57Z FirstChildTAG: I had trouble with the same problem..but solved it.. the current iL was found following the same method given in Pages 518-520 of the text book..we get iL in terms of V, RS, RO, L and t.. this current multiplied with the load resistance gives output voltage..all variables except t are known..solving for t we get the solution.. FirstChildUserIdTAG: 173147 FirstChildUserNameTAG: cruiser_rahit FirstChildCreateTimeTAG: 2012-11-02T19:00:59Z SecondChildTAG: what do you mean by solving for t. as we have to find the *Vo* which is already on the left hand side of the equation? SecondChildUserIdTAG: 64618 SecondChildUserNameTAG: ashfaq2419 SecondChildCreateTimeTAG: 2012-11-04T10:05:48Z FirstChildTAG: is it V/X+X...I tried everything but couldn't get the answer. FirstChildUserIdTAG: 122513 FirstChildUserNameTAG: Mona77 FirstChildCreateTimeTAG: 2012-11-03T17:43:14Z SecondChildTAG: got it :) SecondChildUserIdTAG: 122513 SecondChildUserNameTAG: Mona77 SecondChildCreateTimeTAG: 2012-11-03T18:10:34Z SecondChildTAG: get the iL as in the example in page 517-520 (but remember you have here 2 resistance not one) ,then after you get iL ,Multipy it by RO then rearrange your equation to get "t" in one side and the rest stuff on the other side. SecondChildUserIdTAG: 185715 SecondChildUserNameTAG: amirengineer SecondChildCreateTimeTAG: 2012-11-03T18:15:45Z FirstChildTAG: Tried all the above suggested hints, still having a "Red Cross". Possibly a calculation error. Kindly do a clear explaination after due time. FirstChildUserIdTAG: 455950 FirstChildUserNameTAG: Wahabbaluch FirstChildCreateTimeTAG: 2012-11-04T03:57:02Z IndexTAG: 2101 TitleTAG: And what if we suppress the capacitor !? I think it works still ! UserIdTAG: 373752 UserNameTAG: Croak CreateTimeTAG: 2012-11-02T17:13:54Z VoteTAG: 1 CoursewareTAG: Week 8 / Building A Better Static Memory Element Continued CommentableIdTAG: 6002x_Building_A_Better_Static_Memory_Element_Continued NumberOfReplyTAG: 0 IndexTAG: 2102 TitleTAG: Deadline' Unofficial (November,4) Check [here][1] [1]: http://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/508e0829ba8f10260000000c UserIdTAG: 374590 UserNameTAG: ahope CreateTimeTAG: 2012-11-02T10:50:33Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 2103 TitleTAG: The number of the answer is 100 times than that I calculate. What is the approximate inductance (in Henrys) of a 100-turn solenoidal inductor wound on 1cm diameter cylindrical ferrite core 10cm long? My answer is "0.6310". But the answer is "0.0063". Why? UserIdTAG: 128276 UserNameTAG: ChengBin CreateTimeTAG: 2012-11-02T08:39:31Z VoteTAG: 1 CoursewareTAG: Week 7 / Scaling Inductors CommentableIdTAG: 6002x_Scaling_Inductors NumberOfReplyTAG: 1 FirstChildTAG: Did you use metres and not centimetres for A and l? FirstChildUserIdTAG: 434869 FirstChildUserNameTAG: patey FirstChildCreateTimeTAG: 2012-11-02T09:54:11Z SecondChildTAG: may be u r forget in calculation to balancing cm value check length unit SecondChildUserIdTAG: 575259 SecondChildUserNameTAG: rakeshgupta SecondChildCreateTimeTAG: 2012-11-03T21:39:52Z IndexTAG: 2104 TitleTAG: !!!!!! Very fast :):) UserIdTAG: 395257 UserNameTAG: blackguitar CreateTimeTAG: 2012-11-02T06:44:27Z VoteTAG: 1 CoursewareTAG: Week 7 / Geometry and Capacitance/Inductance CommentableIdTAG: 6002x_Geometry_Capacitance_Inductance NumberOfReplyTAG: 0 IndexTAG: 2105 TitleTAG: lab7 help help............ In Task 1 , my formula is correct for the voltage for question1, but whenever i am inserting the time value and calculating the voltage for 0.0005, it is showing incorrect. Have tried every possible way, still its showing incorrect. Please help... UserIdTAG: 231217 UserNameTAG: karmakargopal CreateTimeTAG: 2012-11-02T06:03:35Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Are you entering it with 6 digits of precision? I cant think of any other problem if your formula is really accepted. Just copy paste it in the calculator, and substitute t with 0.0005. Then copypaste the whole answer. FirstChildUserIdTAG: 329254 FirstChildUserNameTAG: KGabor FirstChildCreateTimeTAG: 2012-11-02T07:57:55Z SecondChildTAG: yes i have entered with 6 digit precision but still its showing incorrect......i am really helpless....just dis question only. SecondChildUserIdTAG: 231217 SecondChildUserNameTAG: karmakargopal SecondChildCreateTimeTAG: 2012-11-02T09:53:45Z FirstChildTAG: Hi karmakargopal, Remember that $\beta$ is in radians... Are you sure that are you calculating it correclty? Also, remember that it is 6 digits precision... FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-02T11:09:03Z SecondChildTAG: Hi Myrimit I solved Q2 but I am stuck to put the formula at Q1, any hint? Thanks. SecondChildUserIdTAG: 296419 SecondChildUserNameTAG: oscman SecondChildCreateTimeTAG: 2012-11-02T12:44:49Z SecondChildTAG: Hi oscman, Can I help you? I have Posted a Lab7 Hints (take a look at the Wiki-> Myrimit´s guide to 6.002x -> Myrimit´s Hints [here][1]). I hope this can help you :) [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/myrimits-hints-links/ SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-02T17:51:37Z SecondChildTAG: I did have the same situation as you, and Myrimit is right!Pay attention to cos(x),while x should multiply 180/PI in calculator,whose unit is degree. SecondChildUserIdTAG: 361137 SecondChildUserNameTAG: Ericson SecondChildCreateTimeTAG: 2012-11-05T04:33:09Z IndexTAG: 2106 TitleTAG: all this? are we going to be doing all this in this course? UserIdTAG: 123529 UserNameTAG: odarrk CreateTimeTAG: 2012-11-01T23:27:47Z VoteTAG: 1 CoursewareTAG: Week 1 / Course overview CommentableIdTAG: 6002x_S1V1toS1V4 NumberOfReplyTAG: 2 FirstChildTAG: yes FirstChildUserIdTAG: 90796 FirstChildUserNameTAG: ChaunceyGardiner FirstChildCreateTimeTAG: 2012-11-01T23:37:01Z FirstChildTAG: And more. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-11-02T04:57:18Z IndexTAG: 2107 TitleTAG: Some info http://en.wikipedia.org/wiki/Dirac_delta_function UserIdTAG: 95280 UserNameTAG: Fortyq CreateTimeTAG: 2012-11-01T15:15:06Z VoteTAG: 1 CoursewareTAG: Week 8 / Response To A Pulse As Pulse Gets Narrower CommentableIdTAG: 6002x_Response_To_A_Pulse_As_Pulse_Gets_Narrower NumberOfReplyTAG: 1 FirstChildTAG: thanks FirstChildUserIdTAG: 623210 FirstChildUserNameTAG: preveen FirstChildCreateTimeTAG: 2012-11-01T18:27:32Z IndexTAG: 2108 TitleTAG: To STAFF Will the final exams will be easy or it will be tough? UserIdTAG: 150120 UserNameTAG: krishna1993 CreateTimeTAG: 2012-11-01T11:51:53Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Can't it be just right? FirstChildUserIdTAG: 397804 FirstChildUserNameTAG: mdempsky FirstChildCreateTimeTAG: 2012-11-01T17:04:45Z FirstChildTAG: It will be tough and you will most likely make it. FirstChildUserIdTAG: 90796 FirstChildUserNameTAG: ChaunceyGardiner FirstChildCreateTimeTAG: 2012-11-01T23:07:11Z SecondChildTAG: Why I asked this question is,the midterm was so easy... SecondChildUserIdTAG: 150120 SecondChildUserNameTAG: krishna1993 SecondChildCreateTimeTAG: 2012-11-10T06:14:29Z IndexTAG: 2109 TitleTAG: Taylor series i didnt get the taylor series expansion of the function. i tried deriving it, but came up with a different answer than the one in the video. help pls... UserIdTAG: 285945 UserNameTAG: lindalapiso CreateTimeTAG: 2012-11-01T09:20:03Z VoteTAG: 1 CoursewareTAG: Week 8 / Response To Impulse Limit Case CommentableIdTAG: 6002x_Response_To_Impulse_Limit_Case NumberOfReplyTAG: 1 FirstChildTAG: Try this instead of :Lim (RQ/T)[1-e^(-T/RC)],when T->0 .You can use the L'Hospital Rule. FirstChildUserIdTAG: 282828 FirstChildUserNameTAG: Chibolator FirstChildCreateTimeTAG: 2012-11-04T06:16:50Z SecondChildTAG: Taylor expansion in this case is about the time zero. f[t] = e^(-t/(R*C)) f[0] = 1 f'[0] = -1/(R*C) So, f[0] + f'[0]*(t-0) = (1 - t/(R*C)) I think. SecondChildUserIdTAG: 398594 SecondChildUserNameTAG: DaveyJC SecondChildCreateTimeTAG: 2012-11-07T09:29:55Z SecondChildTAG: I meant: f[t] = e^(-t/(R*C)) f[0] = 1 f'[0] = -1/(R*C) So, f[0] + f'[0]*(t-0) = (1 - t/(R*C)) SecondChildUserIdTAG: 398594 SecondChildUserNameTAG: DaveyJC SecondChildCreateTimeTAG: 2012-11-07T09:31:14Z SecondChildTAG: thanks @DaveyJC So you are saying, if: $ V_C(t) = {RQ \over T} $ SecondChildUserIdTAG: 714237 SecondChildUserNameTAG: PaxPolaris SecondChildCreateTimeTAG: 2012-11-11T03:52:13Z SecondChildTAG: sorry... if: $ \large V_C(t) = {RQ \over T} \left(1-f[t]\right)$ then, you can replace $f[t]$ with its Taylor approximation $\large(1-{t \over RC})$ for small $t$. SecondChildUserIdTAG: 714237 SecondChildUserNameTAG: PaxPolaris SecondChildCreateTimeTAG: 2012-11-11T04:03:49Z IndexTAG: 2110 TitleTAG: LB8 Support request to Myrimit Dear Myrimit Thanks a lot for for your support I was just look at the lab 8 my eye comes out can you please give some clue what it means not stared studying the materials I am the first person in our class wishing you Happy "Depavali" ( "Depavali" is the festival of India ) Thanks MK.Prasanth UserIdTAG: 156694 UserNameTAG: mkprasanth CreateTimeTAG: 2012-11-01T07:35:15Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hi mkprasanth, Yes sure, I will make a Post with Hints for you soon :). Myriam. P.D: Happy "Depavali"! :). Sorry for my ignorance... but what do you celebrate in that day ? FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-01T14:05:09Z SecondChildTAG: Deepavali (Deewali) is festival of lights? happy Deevali to all of 6.002x/MITx/edX SecondChildUserIdTAG: 623210 SecondChildUserNameTAG: preveen SecondChildCreateTimeTAG: 2012-11-01T18:29:23Z SecondChildTAG: Thank you preveen! I didn't knew that :P Happy Deepavali to you too! I have now read that in the Wikipedia. It sounds so nice and curious! I wish to be there to see that :). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-01T19:36:20Z SecondChildTAG: Dear Myrimit , You are most welcome to India Thanks SecondChildUserIdTAG: 156694 SecondChildUserNameTAG: mkprasanth SecondChildCreateTimeTAG: 2012-11-02T00:16:41Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-02T11:47:02Z FirstChildTAG: OK, here are some things to think about until Myrimit gets her tutorial prepared. One of the devices is always a 1K resistor. The measurements are for device two. Voltage and current are proportional for a resistor (Ohm's Law). If device two is not a resistor then device one is the 1K resistor. Next think about the voltage-current relationships for capacitors and inductors. What device starts with a maximum current that decreases? What device starts with a zero current that increases? What is the voltage current relationship for two resistors in series? FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-01T21:58:51Z SecondChildTAG: Dear Mr skyhawk I got it Thanks SecondChildUserIdTAG: 156694 SecondChildUserNameTAG: mkprasanth SecondChildCreateTimeTAG: 2012-11-02T00:18:15Z SecondChildTAG: Thank you!skyhawk SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-02T11:47:28Z IndexTAG: 2111 TitleTAG: Bit of a minor annoyance Why do the labs keep dealing with subjects which correspond to readings which are assigned for NEXT week's materials? It's a bit annoying scheduling my week and then discovering that I actually have a section ore two more of the text to read towards the end of the week (or else if I don't do the surprise additional readings, as sometimes I don't, and just rely on my intuition to complete the labs then (maybe) I don't get as much out of the labs as I could, if I had first done the readings). UserIdTAG: 287805 UserNameTAG: Beneficial CreateTimeTAG: 2012-11-01T05:02:02Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: The labs, I believe, are designed so that you "self-discover" some of the concepts (even some of the homework assignments are like that). Personally, I feel it's a good thing because I understand things better when I struggle with it myself. Just hang in there. Towards week 11, you'll see that the past weeks have really built your intuition. Do not go through the next week's lectures for the labs. Also the labs are meant for you to play around with the circuit. You don't have to get the values exactly right before simulation. Iterate and refine the circuit. You'll get a really good feel of how various components affect the circuit. FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-11-01T07:53:36Z SecondChildTAG: I agree 100%. SecondChildUserIdTAG: 376877 SecondChildUserNameTAG: AndBre SecondChildCreateTimeTAG: 2012-11-01T08:31:21Z IndexTAG: 2112 TitleTAG: HW7P3Q4 I tried many things including .5=(1-e^t/(L/R)) I understand we are wondering how long for the current thru RO to get to half of 1/RO*[V*RO/(RO+RS)]. I need a pregnant hint UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-31T19:01:10Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: $$V_{out} = V_{in} * (resistive~~ voltage~~ divider) * (1 - e^{ -t * \frac {R}{L} } ) $$ At t=0, Vout = 0. At t = infinity, the inductor has no effect anymore, because the current has stopped flowing, and Vout is given by a simple resistive divider network. Whether it is R/L or L/R for the time constant?? Without the inductor, there is no time constant, and with a really big inductor, there will be a really big time constant (as it will oppose the current more strongly than a weak inductor would). Both these conditions are satisfied by R/L. FirstChildUserIdTAG: 9673 FirstChildUserNameTAG: xp42 FirstChildCreateTimeTAG: 2012-10-31T21:11:18Z SecondChildTAG: it's not Q4 sir SecondChildUserIdTAG: 395257 SecondChildUserNameTAG: blackguitar SecondChildCreateTimeTAG: 2012-11-02T14:00:33Z SecondChildTAG: Yeah, but Q4 is about the speed of light, and his description of the problem suggests Q3, so that is what I answered (I just ignored the subject). SecondChildUserIdTAG: 9673 SecondChildUserNameTAG: xp42 SecondChildCreateTimeTAG: 2012-11-02T19:41:36Z IndexTAG: 2113 TitleTAG: H8P3q5 need a hint ;) seems I miss something) I've got 31.471 picoseconds.. how far I'm from truth? I computed V as divider between RPU and RON and R for discharging as 2*RON hints please? UserIdTAG: 113561 UserNameTAG: Illogical CreateTimeTAG: 2012-10-31T14:53:45Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: The short answer: **There is probably a typo in the problem set.** I set up the problem in exactly the same way as you, and until just a second ago, I was banging my head against the wall. However, I'm glad I checked the forum, because your post gave me an idea! I decided to **use VIH instead of VOH**, and solve by your method. The checker marked my answer correct. It makes sense--doesn't it seem strange that we are computing the transition time between VOH and VIL, instead of two related quantities? And isn't it weird that none of the questions use VIH? FirstChildUserIdTAG: 139646 FirstChildUserNameTAG: tmcnulty FirstChildCreateTimeTAG: 2012-10-31T15:13:23Z SecondChildTAG: At first I trying to find error in my calculations ;-) SecondChildUserIdTAG: 113561 SecondChildUserNameTAG: Illogical SecondChildCreateTimeTAG: 2012-10-31T15:36:09Z FirstChildTAG: well, i got green mark after all, calculated another R for discharge.. 2*RON was wrong.. now I want someone smart to comment on it and convince me that I didn't just guessed it by playing with numbers :-( FirstChildUserIdTAG: 113561 FirstChildUserNameTAG: Illogical FirstChildCreateTimeTAG: 2012-10-31T15:34:41Z SecondChildTAG: Find the Thevenin equivalent resistance for the discharge circuit, then you should be ok. SecondChildUserIdTAG: 288323 SecondChildUserNameTAG: fatslow SecondChildCreateTimeTAG: 2012-10-31T15:39:18Z FirstChildTAG: Find out what is Vfinal. Use the voltage divider created. And it's not any typo or anything. PS: The answers are so tolerant, that anybody can find out the answer by mistake. FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-10-31T18:24:46Z IndexTAG: 2114 TitleTAG: RETAKE THE COURSE Will the course be offered again? UserIdTAG: 160677 UserNameTAG: ARMACK CreateTimeTAG: 2012-10-31T14:34:21Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: ya..in march..most probably.. FirstChildUserIdTAG: 114864 FirstChildUserNameTAG: Vikaash FirstChildCreateTimeTAG: 2012-10-31T15:03:51Z FirstChildTAG: yes it will be offered again..... most probably because it was also offered in spring semester..... FirstChildUserIdTAG: 51259 FirstChildUserNameTAG: Wardah FirstChildCreateTimeTAG: 2012-11-01T19:37:59Z FirstChildTAG: will other courses be offered again 'also ?? FirstChildUserIdTAG: 284515 FirstChildUserNameTAG: SMSMA FirstChildCreateTimeTAG: 2012-11-07T18:36:14Z IndexTAG: 2115 TitleTAG: Hello, I want to buy the textbook Does somebody have a link ? Thanks UserIdTAG: 373752 UserNameTAG: Croak CreateTimeTAG: 2012-10-31T02:33:52Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: http://www.amazon.com/Foundations-Electronic-Circuits-Kaufmann-Architecture/dp/1558607358/ FirstChildUserIdTAG: 373220 FirstChildUserNameTAG: oohall FirstChildCreateTimeTAG: 2012-10-31T04:05:07Z SecondChildTAG: At the bottom of the course info page they give the following link. Apparently you get a discount: http://store.elsevier.com/specialOffer.jsp?offerId=EST_MIT&utm_source=EMAP&utm_medium=text_link&utm_term=Circuits_Electronics&utm_campaign=edx#oid=1003_4 SecondChildUserIdTAG: 434869 SecondChildUserNameTAG: patey SecondChildCreateTimeTAG: 2012-11-01T18:40:29Z SecondChildTAG: Thanks, I had lost these pages! SecondChildUserIdTAG: 373752 SecondChildUserNameTAG: Croak SecondChildCreateTimeTAG: 2012-11-03T00:30:05Z FirstChildTAG: I have the book.... If you want then send me your email id so that i can share it with you.... FirstChildUserIdTAG: 51259 FirstChildUserNameTAG: Wardah FirstChildCreateTimeTAG: 2012-11-01T19:42:05Z SecondChildTAG: That's kind thank you, but i think I prefer buying it, because respect to M. Argawal ! And i like the paper ! SecondChildUserIdTAG: 373752 SecondChildUserNameTAG: Croak SecondChildCreateTimeTAG: 2012-11-03T00:29:22Z IndexTAG: 2116 TitleTAG: S18E3 RLC TOTAL SOLUTION Hi fellows, I've been bumping my head against the wall with fifth question, which is find "A" using v(0) = 0. I end up with VI + A*cos(ϕ) = 0. Can anyone give me a hand? It would be very much appreciated. UserIdTAG: 220837 UserNameTAG: Calsomus CreateTimeTAG: 2012-10-30T22:41:58Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: If you notice that cos(x)^2 = 1 / ( 1 + tan(x)^2) you will easily find the solution FirstChildUserIdTAG: 143593 FirstChildUserNameTAG: Tsybulkin FirstChildCreateTimeTAG: 2012-11-02T07:24:24Z FirstChildTAG: I think it's easiest to use the following formula: cos(arctan(x)) = 1/sqrt(1+x^2) Then you can easily find the answer using a^2 = w0^2 - wd^2. You can see more trigonometric identities on Wikipedia: [Relationships between trigonometric functions and inverse trigonometric functions][1] [1]: http://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Relationships_between_trigonometric_functions_and_inverse_trigonometric_functions "Relationships between trigonometric functions and inverse trigonometric functions" FirstChildUserIdTAG: 341293 FirstChildUserNameTAG: Pietr FirstChildCreateTimeTAG: 2012-11-12T18:55:23Z IndexTAG: 2117 TitleTAG: LAB 7 last part how to find the average current delivered by the bridge rectifier? can you explain it.. UserIdTAG: 114864 UserNameTAG: Vikaash CreateTimeTAG: 2012-10-30T14:03:34Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Average = (Area under the curve for one full cycle)/(length of one full cycle). This is the general formula for any average. Follow the "Hint" given in the Lab problem and apply that in this formula. FirstChildUserIdTAG: 232157 FirstChildUserNameTAG: GayatriTR FirstChildCreateTimeTAG: 2012-10-30T15:15:58Z SecondChildTAG: In my case this only works on second duty cycle (second of stabilized curve, of course) with three digits precision. On first cycle I got the answer different from second cycle by 0.002 and that was not correct. Please excuse my English. SecondChildUserIdTAG: 447663 SecondChildUserNameTAG: brigadakuznetsov SecondChildCreateTimeTAG: 2012-10-30T15:48:06Z SecondChildTAG: i have tried but not getting correct answer. SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-10-31T15:33:55Z SecondChildTAG: finally got it.. SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-10-31T15:40:00Z SecondChildTAG: i follow the hint for this part but i don't understand. SecondChildUserIdTAG: 314294 SecondChildUserNameTAG: victormp SecondChildCreateTimeTAG: 2012-11-04T12:47:55Z SecondChildTAG: Hint: read the last part of the lab very carefully :) SecondChildUserIdTAG: 114881 SecondChildUserNameTAG: xvink SecondChildCreateTimeTAG: 2012-11-04T18:52:02Z IndexTAG: 2118 TitleTAG: Semantics Isn't this like saying -1 person left the room when someone just entered? UserIdTAG: 740649 UserNameTAG: hqjb CreateTimeTAG: 2012-10-30T12:14:05Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 5_Simple_Power NumberOfReplyTAG: 3 FirstChildTAG: As long as it is understood that the room is always full, then in that case a person entering the room would equal the people leaving the room, providing they are all the same size. ;) FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-30T12:37:24Z FirstChildTAG: You completely lost me there. Where are we? FirstChildUserIdTAG: 623210 FirstChildUserNameTAG: preveen FirstChildCreateTimeTAG: 2012-10-30T12:39:34Z SecondChildTAG: I think he is in nodal analysis or dealing with current etc. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-30T12:55:39Z FirstChildTAG: Charge is conserved. I guess saying "-1 people left..." does rely on semantics. That's ok. And it also is a way of expressing some algebraic ideas and also a way of expressing some ideas algebraically. :) FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-10-30T16:40:48Z IndexTAG: 2119 TitleTAG: Categorized Certificate Will there be a better certificate for Grade-A than Grade-B or C? Or will grade be mentioned on the certificate? In which form do we have it, a soft copy v.i.a e-mail or a hard copy posted to our addresses????? UserIdTAG: 216684 UserNameTAG: Taimoor1017 CreateTimeTAG: 2012-10-30T11:25:35Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: I believe that they only provide the soft copy of the certificate. FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-10-30T12:01:02Z SecondChildTAG: what is minimum percentage we for the certificate SecondChildUserIdTAG: 106187 SecondChildUserNameTAG: veeranna SecondChildCreateTimeTAG: 2012-10-30T12:14:27Z SecondChildTAG: 60 or above, But dont worry yourself much about the percentage. My suggestion is that you should try to get maximum fun while learning and let not a certificate worry you. SecondChildUserIdTAG: 148523 SecondChildUserNameTAG: span993 SecondChildCreateTimeTAG: 2012-10-30T12:19:37Z FirstChildTAG: As far as I am aware, there will not be a grade on your Certificate of Mastery. You will be able to print your certificate from your home computer, or perhaps you have a local printing company that can print it out at a higher quality. You will also receive a link with which you can verify the completion of the course. If you want a certified grade, watch for the availability of a proctored exam by Pearson VUE, which does it's testing in 175 countries. Have fun. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-30T12:06:30Z SecondChildTAG: but they told it will be provided after completion of course SecondChildUserIdTAG: 342221 SecondChildUserNameTAG: deepkar SecondChildCreateTimeTAG: 2012-10-30T14:53:59Z SecondChildTAG: if the certificate is not graded..what difference would it show on a person getting A , B or C.....??? SecondChildUserIdTAG: 460505 SecondChildUserNameTAG: QaiserIlyas SecondChildCreateTimeTAG: 2012-10-30T17:42:54Z FirstChildTAG: Will this certificate be a bit helpful for job. FirstChildUserIdTAG: 216684 FirstChildUserNameTAG: Taimoor1017 FirstChildCreateTimeTAG: 2012-10-31T07:42:14Z SecondChildTAG: Sorry Taimoor1017, but I don't think so. If the certificate will not have a grade and your employer won't be able to validate it online, then, at least to the employer, it is as good as any other piece of paper... =( SecondChildUserIdTAG: 288637 SecondChildUserNameTAG: gotchi SecondChildCreateTimeTAG: 2012-10-31T11:43:44Z IndexTAG: 2120 TitleTAG: lab7 need help in the first question i got the formula but when i substitute in it i get a really really close number to the practical why its still wrong 6digits means 0.xxxxxx? or what? and in thelast question any idea how can i solve it? UserIdTAG: 84213 UserNameTAG: mohamed373 CreateTimeTAG: 2012-10-30T11:22:04Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: If your formula is correct, you just need to find out the voltage at 0.0005 using the formula. It doesnt matter if you put it as 0.xxxxxx or .xxxxxx FirstChildUserIdTAG: 171674 FirstChildUserNameTAG: wajahat1 FirstChildCreateTimeTAG: 2012-10-30T11:48:13Z SecondChildTAG: my formula is correct and the answer i get from it is soooooo close to the practical but still saying wrong answer dunnoi why SecondChildUserIdTAG: 84213 SecondChildUserNameTAG: mohamed373 SecondChildCreateTimeTAG: 2012-10-30T11:49:32Z SecondChildTAG: I had exactly the same issue in this lab. Then I clicked on Reset button, reinserted same formula and calculated voltage value and my answers were accepted. SecondChildUserIdTAG: 16265 SecondChildUserNameTAG: serge_korolev SecondChildCreateTimeTAG: 2012-10-30T12:05:27Z SecondChildTAG: i will try it SecondChildUserIdTAG: 84213 SecondChildUserNameTAG: mohamed373 SecondChildCreateTimeTAG: 2012-10-30T12:07:01Z SecondChildTAG: ffffff still getting it wrong they are to close teh change is from the forth digit i cant understand why he is saying so SecondChildUserIdTAG: 84213 SecondChildUserNameTAG: mohamed373 SecondChildCreateTimeTAG: 2012-10-30T12:10:27Z SecondChildTAG: my friend, did you change the radian unit into degress? SecondChildUserIdTAG: 349139 SecondChildUserNameTAG: 1977ROYELMER SecondChildCreateTimeTAG: 2012-10-31T03:19:59Z SecondChildTAG: agree with 1977ROYELMER . SecondChildUserIdTAG: 361137 SecondChildUserNameTAG: Ericson SecondChildCreateTimeTAG: 2012-11-05T04:37:48Z FirstChildTAG: Hi mohamed373! The answer has to be 6 digits precise... Try to not round $\pi$ and use all the possible decimals :) See you! Myriam. PD: Also you can see Lab7 Hints [here][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5091dd5d3fc09f2b0000018a FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-01T02:40:27Z FirstChildTAG: The trig functions can be evaluated exactly. A calculator is only needed for one final division! FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-01T13:47:50Z IndexTAG: 2121 TitleTAG: conversion problem even though i have convert the answer in ampere/ms the ans. is not matching...please help UserIdTAG: 153869 UserNameTAG: rabindra CreateTimeTAG: 2012-10-30T06:54:36Z VoteTAG: 1 CoursewareTAG: Week 7 / Thevenin Inductor Circuit CommentableIdTAG: 6002x_Thevenin_Inductor_Circuit NumberOfReplyTAG: 1 FirstChildTAG: Tip: Henry units are (V*seconds)/Amps FirstChildUserIdTAG: 398594 FirstChildUserNameTAG: DaveyJC FirstChildCreateTimeTAG: 2012-10-30T22:39:11Z IndexTAG: 2122 TitleTAG: HW7P3Q4 I am trying (time constant) tau*.693 (which is ln2) to no avail Thanks in advance :-) UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-30T05:45:41Z VoteTAG: 1 CoursewareTAG: Week 7 / Analyzing A First Order RL Circuit CommentableIdTAG: 6002x_Analyzing_A_First_Order_Circuit NumberOfReplyTAG: 2 FirstChildTAG: hmmmmmmm FirstChildUserIdTAG: 127800 FirstChildUserNameTAG: MNB FirstChildCreateTimeTAG: 2012-10-30T07:27:29Z SecondChildTAG: same probb SecondChildUserIdTAG: 127800 SecondChildUserNameTAG: MNB SecondChildCreateTimeTAG: 2012-10-30T07:27:46Z FirstChildTAG: No you don't have to calculate $\ln{2}$ for this problem. Think about the current required for the voltage. Just like Prof has done for the Inductor circuit do a comparison with the circuit you know and find out the formula. In this way you get to deepen you understanding of the circuit. Thanks to prof Anant for the comparison study. FirstChildUserIdTAG: 623210 FirstChildUserNameTAG: preveen FirstChildCreateTimeTAG: 2012-10-30T12:06:34Z IndexTAG: 2123 TitleTAG: A little bit faster if di/dt = 100A/s and dI/dt=400A/s ( I = total amps) Then it rests 300A/s for L1 L1=1V/300=0.0033H L2=1V/100=0.01H UserIdTAG: 373752 UserNameTAG: Croak CreateTimeTAG: 2012-10-30T04:00:37Z VoteTAG: 1 CoursewareTAG: Week 7 / Unknown Inductance CommentableIdTAG: 6002x_Unknown_Inductance NumberOfReplyTAG: 0 IndexTAG: 2124 TitleTAG: H7P3 I have been doing dis part 4 a long tym.....can't get the time ?? anybody ...hints plzz?? how will V relate 2 v0?? UserIdTAG: 127800 UserNameTAG: MNB CreateTimeTAG: 2012-10-30T02:06:26Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: You might find it useful to convert m/s to cm/s, now you know how many cm light travels in a second. Now you may also then want to convert seconds to nanoseconds. At this point you should be able to figure out how far light travels in a nanosecond, in cm. Now it should be fairly straight forward in determining how long it takes to travel for a given distance (cm) in nanoseconds. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-30T03:10:37Z SecondChildTAG: Which part are you having trouble with? I just realized you may not mean "Part 4" You will have to work on your delivery. Try not to use the number "4" in place of the word "for" to avoid confusion. Also it is "to" not "2", you need to make it easier for people to help you. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-30T03:33:28Z SecondChildTAG: simple s=vt just use 9th class formula SecondChildUserIdTAG: 164689 SecondChildUserNameTAG: muhammadfaizan SecondChildCreateTimeTAG: 2012-11-03T16:52:31Z FirstChildTAG: If you have a problem with the t-part, then I assume you already have a diff equation for $i_L$. Now simply $v_O=R_Oi_L$ where $i_L$ is an exponential function. Everything is known but t, so you solve it. Your equ is something like: $2.5 = (SolutionFromFirstQuestion)*(1-e^{-(t/SolutionFromSecondQuestion)})$ FirstChildUserIdTAG: 329254 FirstChildUserNameTAG: KGabor FirstChildCreateTimeTAG: 2012-10-30T15:45:46Z SecondChildTAG: my friend where did you get the value 2.5? please help. SecondChildUserIdTAG: 349139 SecondChildUserNameTAG: 1977ROYELMER SecondChildCreateTimeTAG: 2012-10-31T05:31:10Z SecondChildTAG: The 2.5 value is just $v_o$, and it differs among the homeworks. As an example, it is 0.5 in my case. Consider that you want to find the time when the value for $v_o$ is reached and the part on the right side of the equation gives you $v_o(t)$, so you have to equate both and solve for t. SecondChildUserIdTAG: 219823 SecondChildUserNameTAG: Karsroh SecondChildCreateTimeTAG: 2012-11-01T13:56:05Z SecondChildTAG: Did u get the answer using above equation i am totally stuck here even after using that equation can u help me my answer 0.294698662048 SecondChildUserIdTAG: 185921 SecondChildUserNameTAG: Ravi_Teja SecondChildCreateTimeTAG: 2012-11-01T18:50:12Z FirstChildTAG: still I am not able to understand how to solve part3 FirstChildUserIdTAG: 365309 FirstChildUserNameTAG: chetnasinghaldas FirstChildCreateTimeTAG: 2012-10-31T14:30:25Z FirstChildTAG: Read pages 518-520 and 524 from textbook. Q1- Solve voltaje divider at Ro, since L is a short circuit. Q2- Find Rth and then tau. Q3- Notice that iL=iR after some time, consider the current at Ro when Vo=0.5V, check page 324. Q4- Follow Pennypacker's post. FirstChildUserIdTAG: 296419 FirstChildUserNameTAG: oscman FirstChildCreateTimeTAG: 2012-10-31T15:48:35Z FirstChildTAG: I cant understand where am i going wrong!!! in the very last part of this section we are asked to calculate time for light to travel 2cm. It is dam easy II grade HW. but i cant get the answer i am doing the following calculation : time= distance/speed speed =299792458m/s distance =2cm =0.02m thus t=6.67*10^-11 sec Which is marked wrong!!! Please help me finish 100% HW FirstChildUserIdTAG: 232499 FirstChildUserNameTAG: Asim09 FirstChildCreateTimeTAG: 2012-11-01T04:48:31Z SecondChildTAG: You have to write the time in nanoseconds... maybe that was your fault? SecondChildUserIdTAG: 219823 SecondChildUserNameTAG: Karsroh SecondChildCreateTimeTAG: 2012-11-01T13:42:47Z SecondChildTAG: convert your answer in nanoseconds...and then enter your answer without 10^-9 sec...only enter the digits SecondChildUserIdTAG: 714369 SecondChildUserNameTAG: Manishrao SecondChildCreateTimeTAG: 2012-11-04T08:54:04Z IndexTAG: 2125 TitleTAG: 'Follow this post please!' Check [here][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/508e0829ba8f10260000000c UserIdTAG: 374590 UserNameTAG: ahope CreateTimeTAG: 2012-10-30T00:53:46Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 2126 TitleTAG: CANT GET ANSWER 3 IM WAY OFF AND TRIED GOING BACKWARDS FROM THE ANSWER AND STILL NOTHING. PLEASE COULD YOU JUST SHOW ME THE STEPS. THANKS UserIdTAG: 262875 UserNameTAG: GrantDennison CreateTimeTAG: 2012-10-29T22:37:33Z VoteTAG: 1 CoursewareTAG: Week 7 / Scaling Inductors CommentableIdTAG: 6002x_Scaling_Inductors NumberOfReplyTAG: 2 FirstChildTAG: 640*4*PI*0.0000001 * 100*100 * 0.005*0.005 *PI / 0.1 Dont forget to multiply by 640 to get permeability of manganese-zinc ferrite core material. permeability of air is 4 * pi * 10^(-7) permeability of core is 640 times that of air N = 100, so N^2 = 100 * 100 A = area of circle = pi * (radius)^2, units in meters. L = 10cm = 0.1 meters FirstChildUserIdTAG: 9673 FirstChildUserNameTAG: xp42 FirstChildCreateTimeTAG: 2012-10-29T22:52:21Z SecondChildTAG: Thanks was up late last night and was mixing up radius and diameter.thanks again that helped and you explained very well. SecondChildUserIdTAG: 262875 SecondChildUserNameTAG: GrantDennison SecondChildCreateTimeTAG: 2012-10-30T11:16:45Z FirstChildTAG: what do they the mean by width and depth of a solenoidal inductor on the first question? FirstChildUserIdTAG: 309255 FirstChildUserNameTAG: mamba747 FirstChildCreateTimeTAG: 2012-10-30T07:08:45Z SecondChildTAG: I'm guessing they mean the core is a rectangular prism of length l, and the area would be width * depth. SecondChildUserIdTAG: 373220 SecondChildUserNameTAG: oohall SecondChildCreateTimeTAG: 2012-10-30T07:27:33Z SecondChildTAG: Thanks for the idea.. SecondChildUserIdTAG: 309255 SecondChildUserNameTAG: mamba747 SecondChildCreateTimeTAG: 2012-10-30T18:41:19Z IndexTAG: 2127 TitleTAG: Dormsbee - Midterm problem 5 solution - formatting issues 1: **Problem 5.** 2: 3: **part one.** 4: 5: first consider DC only 6: 7: $$V_{OUT} = VS - R * i_{D} = V_S - f \cdot V_{GS}^3$$ 8: 9: substitute the values provided and the answer will pop right out. 10: 11: **part two.** 12: 13: Inject a small test signal, and measure the output change. 14: 15: First consider AC model 16: 17: $$v_{OUT} = VS - R * i_{D} = V_S - R \cdot f \cdot v_{GS}^3 ~~~~ (eqn 1)$$ 18: where 19: 20: $v_{GS} = V_{GS} + v_{gs}$ 21: 22: where $V_{GS} = V_{IN}$ because of circuit topology, 23: and $v_{gs} = v_{in}$ for the same reason. 24: $v_{in}$ is to be our small test input signal. 25: 26: and where 27: 28: $v_{OUT} = V_{OUT} + v_{out}$ 29: 30: where $v_{out}$ will be the response to our small test signal. ---- 31: So, we will inject $v_{in} = 0.0001 V $ say, 32: on top of the DC bias voltage $V_{IN}$. 33: 34: Substitute this into equation 1, and rearrange so we get 35: 36: $v_{out} = v_{OUT} - V_{OUT}$ 37: 38: where $V_{OUT}$ is the answer we obtained in part one, 39: and $v_{OUT}$ is the total signal we calculate using equation 1, 40: 41: and then the small signal gain is just $ \frac {v_{out}}{v_{in}}$. --- 42: Now you could use calculus also, and solve for partial derivatives if you 43: wished. But for an exam, this is faster, and for a simple circuit like 44: this, easier. Plus you would probably want to do an empirical test like 45: this test to check your math calcs anyway. 46: 47: And here to do the math to check the empirical answer, we solve 48: $$ \frac {\partial {v_{out}}} { \partial {v_{in}}} = \left [ \frac {\partial {v_{OUT}}} { \partial {v_{IN}}} \right ] _{ | v_{IN} = V_{IN} } = \left [ -R \cdot f \cdot v_{IN} \right ] _{| v_{IN} = {V_IN}} = -R \cdot 3 \cdot f \cdot {V_{IN}}^2$$ 49: 50: Now for me, f = 0.5, R = 6, $V_{IN} $ = 2, 51: so $ \frac {v_{out}}{v_{in} }$ = -36 UserIdTAG: 9673 UserNameTAG: xp42 CreateTimeTAG: 2012-10-29T18:38:34Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Wait. You typed this in the order 1-50 but it got mixed up by showing 18 - 40 first , then 1 - 17 and finally 42 - 48? FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-10-29T18:48:37Z SecondChildTAG: Yep, you got it. There are a lot of problems with mathjax. I have been helping them with this for over two weeks. I encounter problems almost every time I try to use a lot of equations. Dave says they have ideas what it may be, but a solution is not ready yet. Most annoyingly, stuff will usually render OK in the preview. It's later that it displays badly, after posting. This issue makes it very hard to contribute anything worthwhile to the forum. P.S. The last line "= -36" is actually part of line 51. SecondChildUserIdTAG: 9673 SecondChildUserNameTAG: xp42 SecondChildCreateTimeTAG: 2012-10-29T19:07:07Z SecondChildTAG: Yeah, if I hit the "edit" button on your post, the formulas look fine. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-29T20:16:10Z SecondChildTAG: Hmmm now it looks ok. Is it just me? SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-29T20:20:12Z SecondChildTAG: It's still mixed up for me. SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-10-29T20:35:35Z SecondChildTAG: So this mix up is caused by mathjax? Or is it due to long posts? The two solutions I posted for last term's midterms were pretty long too. Why didn't that have this problem? SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-10-29T20:37:55Z SecondChildTAG: Hmmm. After I hit the "Edit" button, I saw the correct text in the edit window. Then I hit the "Cancel" button and now the post displays correctly. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-29T20:38:48Z SecondChildTAG: Ok I reloaded the page about 30 times, and out of those 30 times it came up wrong twice, and the rest of the time it formatted correctly. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-29T20:47:50Z SecondChildTAG: Almost like the page stopped loading prematurely. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-29T20:48:34Z SecondChildTAG: look here https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5077bf337b5f442700000056 SecondChildUserIdTAG: 9673 SecondChildUserNameTAG: xp42 SecondChildCreateTimeTAG: 2012-10-29T21:06:47Z SecondChildTAG: and here for another example. https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Initial_Conditions/threads/508b555c0ec6c81f0000000d SecondChildUserIdTAG: 9673 SecondChildUserNameTAG: xp42 SecondChildCreateTimeTAG: 2012-10-29T21:08:16Z SecondChildTAG: Those all came out formatted correctly. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-29T22:39:03Z SecondChildTAG: You are missing the point. The order is all jumbled up. The math renders OK, but the resulting (jumbled output) is nonsense. SecondChildUserIdTAG: 9673 SecondChildUserNameTAG: xp42 SecondChildCreateTimeTAG: 2012-10-29T22:53:47Z SecondChildTAG: Oh. I was just looking at the formatting. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-29T23:14:58Z SecondChildTAG: Most likely it's a bad interaction between the Markdown renderer and the Mathjax renderer. I'm not working on this anymore and I've got other projects to work on unfortunately or I'd take a look :( SecondChildUserIdTAG: 323387 SecondChildUserNameTAG: ibrahimawwal SecondChildCreateTimeTAG: 2012-10-30T04:20:44Z FirstChildTAG: Now it's 1-17, 42-28, 18-41, 49-51 0n 12-25-12 FirstChildUserIdTAG: 9673 FirstChildUserNameTAG: xp42 FirstChildCreateTimeTAG: 2012-12-26T07:06:46Z IndexTAG: 2128 TitleTAG: RIPPLES FREE POWER SUPPLY IF I hAVE 220VAC 50hZ .AND I STEP DOWN IT USING TRANSFORMER RATING 1A AND 12Vac . AND I WANT A RIPPLES FREE 12Vdc POWER SUPPLY .hOW MANY FARAD CAPACITOR VALUE I WILL USE AND Which VOLTAGE RATING .IS I USE ELECTROLYTIC CAPACITOR OR NON ELECTROLYTIC CAPACITOR . UserIdTAG: 155008 UserNameTAG: sohailahmed CreateTimeTAG: 2012-10-29T18:16:58Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 9 FirstChildTAG: commonly speaking to get ideal ripples free voltage you need ideal infinite capacitor :) FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-29T20:34:08Z SecondChildTAG: Makes your diodes very happy when you switch the power on! SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-10-29T21:20:42Z SecondChildTAG: my ideal diodes always happy with my ideal current limiting resistors SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-29T21:48:07Z SecondChildTAG: On simulatorland only happiness prevails;-) SecondChildUserIdTAG: 623210 SecondChildUserNameTAG: preveen SecondChildCreateTimeTAG: 2012-10-30T12:47:55Z FirstChildTAG: Bridge rectification? You will need a lower voltage transformer if you want to achieve 12vDC, something in the 8v-9v range. You can use an electrolytic, as these are more cost effective and available in the sizes needed for filtering. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-29T18:43:36Z SecondChildTAG: Here is some useful software for designing your power supply. You will notice that in most cases, there is still a small amount of ripple, you have to decide what is good enough for your particular application. http://duncanamps.com/psud2/index.html SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-29T18:51:25Z SecondChildTAG: This one is also quite good =D https://www.circuitlab.com/ SecondChildUserIdTAG: 308853 SecondChildUserNameTAG: fmorato SecondChildCreateTimeTAG: 2012-10-30T00:01:05Z SecondChildTAG: Thanks Fmorato, I did not realize Circuitlab had this capability! SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-30T00:16:25Z SecondChildTAG: It would be great if one of the folks proposing simulation would build the circuit and test some component values. I would love to see the results! SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-30T00:42:12Z SecondChildTAG: See below. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-30T01:24:18Z FirstChildTAG: A voltage regulator should be used. A switching regulator is best, but a linear regulator can be used. Linear regulators typically have a dropout voltage of about 2 volts, but there are low dropout regulators available. So one would like an unregulated voltage with a minimum of 14 volts to the regulator. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-29T19:03:40Z FirstChildTAG: A 12V transformer at its rated current should output a 16.8V peak pulse, allowing for a two diode drop from the full bridge rectifier it will produce about 15.4V peak input to the filter capacitor. At 50Hz corresponding to a 100Hz ripple a 10,000 uF capacitor should minimize the ripple sufficiently for the regulator to work properly. A slightly higher voltage transformer will give a greater margin but will increase power dissipated in the linear regulator. If you use the 12V transformer I would use as a minimum a 25V rated capacitor, but better a 35V or 50V, which should cost little more and also not be much larger. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-29T19:34:08Z SecondChildTAG: why it should be 10000uF, not 6800uF or 22000uF ? SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-10-29T19:43:18Z SecondChildTAG: With a 100 Hz ripple and a 1 A draw, I calculated a 1V droop with 10,000 uF, which brings the input to the regulator down to 14.4V. That's a small margin considering the possibility of mains voltage variation. 6800 uF would be insufficient, 22000 uF would be better, but size and costs can be issues. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-29T19:52:48Z SecondChildTAG: You forgot to specify voltage pulsation value for rectifier output. SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-10-29T20:28:43Z SecondChildTAG: As for the regulator, I guess you can use a LM7812 =D SecondChildUserIdTAG: 308853 SecondChildUserNameTAG: fmorato SecondChildCreateTimeTAG: 2012-10-30T00:03:13Z SecondChildTAG: Yep! SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-30T00:37:16Z FirstChildTAG: Yes a voltage regulator would be suitable for this particular power supply. (I did not mention it due to the *slightly* added complexity, cost and energy consumption.) FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-29T20:01:40Z SecondChildTAG: He recommended 10000uf as a minimum. 20000uf would also work, while 6800uf may be insufficient. This value is dependent on the voltages you are working with. If you play around with PSUD, you will notice the requirement changes depend on the voltage you use. There may also be a limit on how much capacitance you use, as it will stress the rectifier(s) upon power-up. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-29T20:14:08Z SecondChildTAG: I'm not sure about the cost or complexity. A 12V, 1A regulator is a $.35 - $.40 item. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-29T21:47:49Z SecondChildTAG: If you don not know how to size a capacitor's voltage rating, then adding a regulator is complex. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-29T22:52:14Z SecondChildTAG: As for the regulator, I guess you can use a LM7812 =D There are capacitor values at it's data sheet, and how to use it. If you can't fully understand. SecondChildUserIdTAG: 308853 SecondChildUserNameTAG: fmorato SecondChildCreateTimeTAG: 2012-10-30T00:05:25Z SecondChildTAG: The capacitors specified in the data sheet are meant to stabilize the regulator and keep it from oscillating. They are not the specifications for the filter capacitor. http://www.futurlec.com/Linear/7812T.shtml SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-30T02:09:53Z FirstChildTAG: ...and why you are shouting? or lost caps lock key? FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-29T20:30:45Z FirstChildTAG: Warning: If you only use a rectifier and a VERY large capacitor, then it's possible that you blow up the rectifier when you switch the power on, because the currentpulses to load the empty capacitor, can exceed the maximum current rating of the diode(s), even when you don't connect a load to the output. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-10-29T21:24:24Z FirstChildTAG: ![PSUD][1] [1]: https://edxuploads.s3.amazonaws.com/13515593231343615.png FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-30T01:09:09Z SecondChildTAG: Here is the simulation for the very basic power supply that I originally suggested to Sohail. Feel free to do your own experiments. If you feel the need to compensate (drop) the voltage using a regulator as opposed to using a proper sized transformer to begin with, C'est la vie. The PSUD allows you to incorporate the load of your device into the test. Unloaded voltages will read higher. Obviously in some cases it is desirable to use a regulated power supply, however the OP did not state what kind of accuracy he was after. Even if you do regulate, you still need to design a power supply that is reasonable close to the regulated voltage you require, else you will burn out your regulator. I'm not even mentioning how inefficient regulators are. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-30T01:23:13Z SecondChildTAG: thanks SecondChildUserIdTAG: 623210 SecondChildUserNameTAG: preveen SecondChildCreateTimeTAG: 2012-10-30T12:51:00Z FirstChildTAG: I am fully aware of how to use these little guys. I use them all the time. Here is my original statement where I believe that I have dealt with these issues. >A 12V transformer at its rated current should output a 16.8V peak pulse, allowing for a two diode drop from the full bridge rectifier it will produce about 15.4V peak input to the filter capacitor. At 50Hz corresponding to a 100Hz ripple a 10,000 uF capacitor should minimize the ripple sufficiently for the regulator to work properly. **A slightly higher voltage transformer will give a greater margin but will increase power dissipated in the linear regulator.** But I was hoping that you would actually design a 12V, 1 amp power supply and demonstrate how you obtained low ripple (say .1v pp or better) with a capacitor only. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-30T02:05:00Z SecondChildTAG: With all do respect, I have tried to answer you, in the image above, which shows 1N4007 diode bridge rectifier, 10000uF cap and a load simulating resistor. The simulation shows approx 0.02V ripple. It even has a little schematic in the top left corner. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-30T02:30:35Z SecondChildTAG: I'm sorry, but I just don't have the resolution on my screen to make out the size of your load resistor. Sorry to be such a bother! SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-30T02:39:51Z SecondChildTAG: No worries. Load resistors from 10K to 1M have a nice low ripple. When you get down below 10K the ripple increases. (About 0.1V of ripple with a 1K load resistor, to ~0.2V with a 500ohm load resistor.) SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-30T02:52:57Z SecondChildTAG: OP asked about a 12V, 1 amp supply (at least that was his transformer spec). If I am not confused that corresponds to a 12 ohm load. That is the sort of load I have been designing for, and I think you will see that a low ripple supply with those specs is difficult with a capacitor alone. I am sorry for any trouble I caused! SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-30T09:23:16Z IndexTAG: 2129 TitleTAG: A flaw in Midterm's Q6 > B=1.5. Must be: B = 1.5 Ohm^(-1) I posted it twice already. But the first thread was closed immediately and the second, created today, was deleted altogether. But not only I think that dimensions **are important**, I am a rather persistent fellow too. UserIdTAG: 148226 UserNameTAG: Apprentice CreateTimeTAG: 2012-10-29T18:09:51Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: looks like some kind of mania :) FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-29T20:24:44Z SecondChildTAG: Ohm-ophilia. All those who study electricity have it. SecondChildUserIdTAG: 148226 SecondChildUserNameTAG: Apprentice SecondChildCreateTimeTAG: 2012-10-29T21:51:14Z SecondChildTAG: it's not true - I have an evidence :) SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-29T22:43:24Z FirstChildTAG: I also noticed B was missing its dimensions in that problem, but I think A/V would have been more intuitive than Ohm^-1. Unless you want to also argue the SuperT parameter should have been given as 1/(kOhm*V^2) instead of mA/V^3. ;) FirstChildUserIdTAG: 397804 FirstChildUserNameTAG: mdempsky FirstChildCreateTimeTAG: 2012-10-29T18:22:34Z SecondChildTAG: If i = f*v^3 f may only be in [a/(V^3)]. Or ma/(V^3). It does not metter from the dimensional point of view. [kOhm*V^2] is the same as [V^3/mA] as [A] = [V/Ohm] But if i = B*vb B must be in Ohm^(-1) for i to be in Ampers. Or mOhm^(-1). Or nOhm^(-1). That also does not metter from the dimensional point of view. What metters is **it ought to have the right dimention**. SecondChildUserIdTAG: 148226 SecondChildUserNameTAG: Apprentice SecondChildCreateTimeTAG: 2012-10-29T18:45:55Z SecondChildTAG: And I also notice that. It is important to me, to make work well. I have not made this question, becouse i did`n know or this is current or voltage independent source. Symbol and unit wasn`t fit together. SecondChildUserIdTAG: 413784 SecondChildUserNameTAG: Dorotka SecondChildCreateTimeTAG: 2012-10-29T18:51:53Z SecondChildTAG: Apprentice, I agree with you that B is missing a dimension. And I also agree the dimensions are the same. I was just pointing out that I think it would be more intuitive if the problem had provided "B = 1.5 A/V" instead of "B = 1.5 Ohm^-1". But I agree that either is better than "B = 1.5". I think you need to calm down though. You're behaving rather immaturely over what was a minor issue in the question. SecondChildUserIdTAG: 397804 SecondChildUserNameTAG: mdempsky SecondChildCreateTimeTAG: 2012-10-29T20:08:16Z SecondChildTAG: **mdempsky**, I am calm, I assure you. And after thinking for awhile I agree with you about the A / V. It would fit better. My insistence on Ohms^(-1) comes from the "3 laws of electricity" thinking. You know: I = V / R, V = I * R and R = V / I... :-) I just do not consider the lack of dimension in one of the problem's constants a "minor issue". Don't get me wrong. I am aware that mistakes are unavoidable. But I also think that mistakes need to be corrected. And absolutely can't see how my wish to correct some of them is immature. SecondChildUserIdTAG: 148226 SecondChildUserNameTAG: Apprentice SecondChildCreateTimeTAG: 2012-10-29T21:41:40Z SecondChildTAG: BTW, ohm^-1 is a siemens SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-29T22:36:49Z FirstChildTAG: I was confused by the missing units of B as well. For a moment I thought that maybe it was meant to be dimensionless, to give a dependent voltage source instead of a current source. Luckily I tried the solutions for missing units first. FirstChildUserIdTAG: 406420 FirstChildUserNameTAG: Picolo FirstChildCreateTimeTAG: 2012-10-31T17:30:44Z SecondChildTAG: I'm glad that I was not the only one. :) SecondChildUserIdTAG: 148226 SecondChildUserNameTAG: Apprentice SecondChildCreateTimeTAG: 2012-11-01T14:06:52Z IndexTAG: 2130 TitleTAG: Midterm Final Solutions Will the final solutions with explanations be posted for the midterm? If so, when and where? UserIdTAG: 77839 UserNameTAG: Rideathought CreateTimeTAG: 2012-10-29T17:14:21Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Yes. The showanswer button should be available before tomorrow. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-10-29T17:22:04Z SecondChildTAG: thanks lyla fisher SecondChildUserIdTAG: 623210 SecondChildUserNameTAG: preveen SecondChildCreateTimeTAG: 2012-10-29T17:37:52Z SecondChildTAG: Thanks! SecondChildUserIdTAG: 77839 SecondChildUserNameTAG: Rideathought SecondChildCreateTimeTAG: 2012-10-30T00:05:02Z SecondChildTAG: thanks SecondChildUserIdTAG: 115656 SecondChildUserNameTAG: rolando277 SecondChildCreateTimeTAG: 2012-10-30T00:38:28Z SecondChildTAG: Not available yet! SecondChildUserIdTAG: 216684 SecondChildUserNameTAG: Taimoor1017 SecondChildCreateTimeTAG: 2012-10-30T11:39:15Z FirstChildTAG: SOLUTIONS: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/508e8e5225f1ef2700000063 FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-10-30T12:09:11Z IndexTAG: 2131 TitleTAG: Midterm Exam Hi everyone! It seems need to be prepared more strongly to these courses. I understood that when the number of current courses grew from 1 to 4. But it is a good experience and we move on. By the way some one can tell is it possible go through the course when missed Midterm? UserIdTAG: 48745 UserNameTAG: viktar CreateTimeTAG: 2012-10-29T15:12:00Z VoteTAG: 1 CoursewareTAG: Week 7 / Ideal Linear Inductors CommentableIdTAG: 6002x_Ideal_Linear_Inductors NumberOfReplyTAG: 2 FirstChildTAG: Yes, it *may* be possible, depending on your homework scores form the first half. The final exam is worth 40%, plus you will need 20% minimum (out of a possible 30%) on your homework score. 60% is required for a passing "C". Good luck. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-29T15:34:43Z SecondChildTAG: :) I do not need even B. So for A you need extra 17%. How to get it? Any way B - lets do it! THANKS! SecondChildUserIdTAG: 48745 SecondChildUserNameTAG: viktar SecondChildCreateTimeTAG: 2012-10-29T19:12:48Z FirstChildTAG: Remember that this Course has: 12 Homeworks (but if you read syllabus here, they tell you that you can skip two without penality, that means that you will still have 100% of Homework score if you make only 10). If you get 100% in your Homework score, that homework score it will contribute your final score only a 15%. So, if you have done so far 50% of your homework, it will contribute with your final score with (50%*15%)/100% = 7.5%. 12 Labs (but if you read syllabus, they tell you that you can skip two without penality, that means that you will still have 100% of Labs score if you make only 10). 1 Midterm Exam. If you get 100% in the Midterm, it will contribute with your final score with 30%. 1 Final Exam.If you get 100% in the Final, it will contribute with your final score with 40%. So, it is not that late to get the certificate :), but you will have to do a lot of effort and work from now to the end of the Course... Also, as some students have said it to you, you can follow this Course and re-take it again, but that is always up to you :). See you! I hope this can help you. > **Regards:** asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-11-04T07:39:50Z IndexTAG: 2132 TitleTAG: [STAFF] When will be the mid-term solutions be out ? I wanted to know when you guys will be releasing mid-term solutions. UserIdTAG: 715231 UserNameTAG: jassi_jasraj CreateTimeTAG: 2012-10-29T11:44:08Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: In what problem's solution are you exactly interested in? FirstChildUserIdTAG: 252722 FirstChildUserNameTAG: vaboro FirstChildCreateTimeTAG: 2012-10-29T12:18:25Z SecondChildTAG: Q4, Q5 and Q6.... Will answers be out at some point? SecondChildUserIdTAG: 342966 SecondChildUserNameTAG: SandraNavarro SecondChildCreateTimeTAG: 2012-10-29T13:27:33Z SecondChildTAG: hey i have got them thank you SecondChildUserIdTAG: 318832 SecondChildUserNameTAG: sudhakarV SecondChildCreateTimeTAG: 2012-10-30T09:27:25Z FirstChildTAG: Have a look at my solution https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/508e8e5225f1ef2700000063 FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-10-29T14:20:03Z FirstChildTAG: Yes. They will be out by tomorrow. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-10-29T14:33:47Z IndexTAG: 2133 TitleTAG: Midterm Q6 last item Hi guys! I think it's already safe to ask questions about the midterm. I would like to know how to calculate the Thévenin's resistance in the last part of Question 6. I couldn't get it right. Thanks UserIdTAG: 288637 UserNameTAG: gotchi CreateTimeTAG: 2012-10-29T11:02:06Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: Standard notion: Calculate the thevenin Voltage... Calculate the Norton's Current... Divide Vth by In.... U will get Rth...... FirstChildUserIdTAG: 82597 FirstChildUserNameTAG: bondrajat FirstChildCreateTimeTAG: 2012-10-29T11:06:17Z SecondChildTAG: What about the Norton's current then? I tried this, but I was unlucky. SecondChildUserIdTAG: 288637 SecondChildUserNameTAG: gotchi SecondChildCreateTimeTAG: 2012-10-29T11:14:08Z SecondChildTAG: Short The Terminals Of the Ouptut...U will see that R3 will now be of no use... calculate the current flowing through that shorted wire... (couple of linear Equations will Do ur job) Where were u Stuck?? SecondChildUserIdTAG: 82597 SecondChildUserNameTAG: bondrajat SecondChildCreateTimeTAG: 2012-10-29T11:19:24Z SecondChildTAG: Aaaaaaaaaaaaaaaaaaaaa, after read your post, i knew why it is not correct when i tried to find Rth :| I forget remove R3 when short circuit at open port :( SecondChildUserIdTAG: 114913 SecondChildUserNameTAG: ngoctuan SecondChildCreateTimeTAG: 2012-10-29T11:38:54Z SecondChildTAG: By short-circuiting the output port I realized that no current passes throught $R_3$ because there isn't any difference of potencial between it's terms. While I was calculating the other items of this question, I found the value $V_a=20\,V$ and $I_b=3.75\,A$ to the dependent and current sources. Besides that, $i_1=5\,A$. So I considered $I_N=i_1-I_b=1.25\,A$ to find $$R_{TH}=\frac{V_{TH}}{I_N}=\frac{7.5}{1.25}=6.0\,\Omega.$$ But that was not the answer... SecondChildUserIdTAG: 288637 SecondChildUserNameTAG: gotchi SecondChildCreateTimeTAG: 2012-10-29T11:39:32Z SecondChildTAG: Just to correct myself: "because there isn't any difference of potencial between it's *terminals*." SecondChildUserIdTAG: 288637 SecondChildUserNameTAG: gotchi SecondChildCreateTimeTAG: 2012-10-29T11:42:32Z SecondChildTAG: $Va$ cannot be 20V as the $VIN$ is only 18V SecondChildUserIdTAG: 623210 SecondChildUserNameTAG: preveen SecondChildCreateTimeTAG: 2012-10-29T11:51:49Z SecondChildTAG: preveen, my $V_{in}$ was 45V. I think these values changes. By the way, my $R_1=1.5\,\Omega$, $R_2=2\,\Omega$, $R_3=6\,\Omega$, $A=2$ and $B=0.5$. SecondChildUserIdTAG: 288637 SecondChildUserNameTAG: gotchi SecondChildCreateTimeTAG: 2012-10-29T11:58:42Z SecondChildTAG: With these numbers, $R_{TH}=5 \Omega$. SecondChildUserIdTAG: 131214 SecondChildUserNameTAG: arnsfelt SecondChildCreateTimeTAG: 2012-10-29T12:07:28Z SecondChildTAG: You can't reuse the calculated quantaties after adding the short - they should be recomputed. SecondChildUserIdTAG: 131214 SecondChildUserNameTAG: arnsfelt SecondChildCreateTimeTAG: 2012-10-29T12:12:25Z SecondChildTAG: sorry about that, gotchi. I thought all were having the same questions. after shortcircuiting use KCL and KVL to find the unknowns. SecondChildUserIdTAG: 623210 SecondChildUserNameTAG: preveen SecondChildCreateTimeTAG: 2012-10-29T12:23:00Z SecondChildTAG: Thank you to all. It's now done. SecondChildUserIdTAG: 288637 SecondChildUserNameTAG: gotchi SecondChildCreateTimeTAG: 2012-10-29T15:23:54Z FirstChildTAG: 1) short out the independent voltage source 2) apply $V_{th}$ to the output port to obtain $i_N$ 3) find $R_{th}=\frac {V_{th}}{i_N}$ If you found $V_{th}$ correctly, which was 1.8V, and correctly calculated $i_N$, which was 0.5A, the value of $R_{th}$ should be 3.6 Ohms. FirstChildUserIdTAG: 252722 FirstChildUserNameTAG: vaboro FirstChildCreateTimeTAG: 2012-10-29T12:34:11Z SecondChildTAG: They use different numbers in people's tests, so your answers and other peoples will not line up. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-29T13:44:02Z SecondChildTAG: Aah... Alright! SecondChildUserIdTAG: 252722 SecondChildUserNameTAG: vaboro SecondChildCreateTimeTAG: 2012-10-29T13:58:25Z SecondChildTAG: However it can still be useful, just go back and put your initial conditions in your post so people can see how you got what you got. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-29T14:06:36Z FirstChildTAG: Solutions https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/508e8e5225f1ef2700000063 FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-10-29T14:58:23Z SecondChildTAG: Thank you span993. Certainly your solutions will be helpful! =D SecondChildUserIdTAG: 288637 SecondChildUserNameTAG: gotchi SecondChildCreateTimeTAG: 2012-10-29T15:24:40Z FirstChildTAG: For these conditions: Vin=27V, R1=5Ω, R2=7Ω, R3=12Ω, A=6 and B=0.1; I get a correct Vth of 2.7 volts and Rth of 16.2 but that didn't seem to be the correct Rth. Would someone post their formulae for this last question. Thanks. Jack. FirstChildUserIdTAG: 260299 FirstChildUserNameTAG: Jack1947 FirstChildCreateTimeTAG: 2012-10-29T15:07:08Z SecondChildTAG: jack1947, look at the link above: span993 solutions. SecondChildUserIdTAG: 288637 SecondChildUserNameTAG: gotchi SecondChildCreateTimeTAG: 2012-10-29T15:25:28Z SecondChildTAG: Those sound like the conditions I have. My answer for Rth was 10.8 ohms. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-29T15:26:49Z FirstChildTAG: **Step 1. Find VTh** VTh = Voc, the open circuit output voltage. Oops, they already had you calculate Vout, so just copy the answer from part 2 of the problem (from the answer box above). $$V_{Th} = V_{out}$$. **Step 2. Find RTh** There are two ways to do this. One is to find IN and use RTh = VTh / IN to get the desired result. Leave independent sources in for this approach. The other is to inject a test voltage Vtest at the output and measure (calculate) the current, Itest say, flowing through the test source, then use RTh = Vtest/Itest to get the desired result. But don't forget to remove the independent sources first. **step 2a. Find I_Norton** The Norton current, IN, is the current that flows through a short that has been applied so as to short-circuit the output. So redraw the circuit with a shorting wire placed parallel to R3, across the output. Now calculate the currents flowing through R1 and through the voltage controlled current source. There will be no current through R3, because it is parallel to a zero ohm shorting wire. I had R1 = 4 ohms, so using i1 for the current through R1, I can see that Va = i1 * 4. Now the total current through R1 splits, and some goes through the dependent current source (=BVa = B*i1*4), and the remainder goes through the short circuit across the output (=INorton = i1* (1-4*B) ). For me, B = 1/8, so the current splits evenly, and BVa = IN = 1/2 * i1. This description is just to convince you that there is current flowing through the short circuit we applied. You need to do KVL/KCL to find Va and thereby, i1. But this can probably be done by inspection since you have 2 resistors plus one dependent voltage source between Vs and the ground created by the short on the output. $$R_{Th} = \frac {V_{Th}}{I_N} = \frac {V_{out}}{i1*(1-R_1 \cdot B)} = \frac {V_{out}}{i1/2}$$ where the last equivalence only work for my R1 and B values. **step 2b. Find Vtest and Itest - remove independent sources first** I did this too to check my answers. After removing VS, the only independent source, I looked at the circuit and decided to inject a test voltage Vtest across Vout of 4V, but any voltage will do. By choosing 4V I didn't have to do a lot of involved KVL/KCL calculations because the ratios of components made it fairly straight forward to see what the currents in the different branches were. I first thought of inject 1A for Itest, and solving for Vtest, but with VS gone from the circuit, the polarity of Va will be reversed, so the dependent current source will actually be sending current upwards, against the direction of the arrow drawn in the problem diagram. What this is saying, it the value of Va is negative, and so will B*Va will be negative also. If not for the reversal, if I had 1A for Itest, (you see I had R3 = 7ohms) then i3 = 4V/7ohms, leaving 1A - 4/7A to go through the dependent current source and through the R1 circuit branch. And I know from above in part 2a that the current ratios would be 2:1. So 1A was going to be pretty close, but not exact since the ratio now (with VS removed) would be (-2):(-1), meaning the currents were flowing in the reverse direction and the sum of the currents at the top center node would be different. So I used Vtest = 4V for a nice round number, and the currents were easy to solve for, so I got $$R_{Th} = \frac {V_{test}}{I_{test}} = \frac {4V}{I_{test}}$$ and the value for RTh here matched that from step 2a, so all was ok. There is no need to make good/lucky choices for Vtest or Itest. Commonly you will choose Vtest = 1V, or Itest = 1A, and use KVL/KCL to solve for the other. If Itest = 1A, then RTh = Vtest/1, so the math at the end is easy. FirstChildUserIdTAG: 9673 FirstChildUserNameTAG: xp42 FirstChildCreateTimeTAG: 2012-10-29T16:56:55Z SecondChildTAG: Thank you xp42. This was a very detailed explanation. =) SecondChildUserIdTAG: 288637 SecondChildUserNameTAG: gotchi SecondChildCreateTimeTAG: 2012-10-29T18:08:37Z SecondChildTAG: Perhaps everyone has individual input data. I have, for instance R1=0.5, R2=1.5, R3=5, A=2, B=1.5. I'v written KVL and got: vb+A*va+va-Vin=0 va=In*R2 vb=In*R1 Solving the resulting system, i have In=3. Then i divide previously obtained Vth=3 by In=3, get RN=1 and is still unlucky. Where am i wrong? By the way how do we enter formula here? SecondChildUserIdTAG: 213452 SecondChildUserNameTAG: JoJosida SecondChildCreateTimeTAG: 2012-10-30T08:45:06Z SecondChildTAG: And i have a different schematic for Q6 then described by xp42.![Mine is look like this][1] [1]: https://edxuploads.s3.amazonaws.com/13515876021343637.png SecondChildUserIdTAG: 213452 SecondChildUserNameTAG: JoJosida SecondChildCreateTimeTAG: 2012-10-30T09:01:50Z IndexTAG: 2134 TitleTAG: end Midterm Midterm time is over? UserIdTAG: 51036 UserNameTAG: dj4vi CreateTimeTAG: 2012-10-29T04:18:22Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: don´t appear check and save buttons, what happened? FirstChildUserIdTAG: 51036 FirstChildUserNameTAG: dj4vi FirstChildCreateTimeTAG: 2012-10-29T04:24:12Z FirstChildTAG: monday already FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-29T04:29:46Z FirstChildTAG: Is there any way to get an extension on the mid-term. I need to fix that one last question. I joined the course late and every last point counts. Did any-one else get at *technical error* page when loading the mid-term. I was panicky there for a second. Refreshing the page did fix it though. :'( FirstChildUserIdTAG: 714237 FirstChildUserNameTAG: PaxPolaris FirstChildCreateTimeTAG: 2012-10-29T04:37:09Z SecondChildTAG: I do not think,(In my opinion) that extensions will be granted. You may decide to tough it out, or take the course again in the spring. You still have a chance at a decent grade this term, depending on how well you did so far. The final exam will be worth 40%. Also your two lowest (or incomplete) homework scores will be ignored. I hope this helps. Have fun. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-29T04:49:38Z SecondChildTAG: I suppose it is what it is... It is just frustrating after the *HOURS* I wasted today learning everything about capacitors, not even thinking of previewing the exam before I did that. And to have that "final check" button lock out just minute before they were working ... lol thanks for your reply. SecondChildUserIdTAG: 714237 SecondChildUserNameTAG: PaxPolaris SecondChildCreateTimeTAG: 2012-10-29T05:06:57Z IndexTAG: 2135 TitleTAG: The Curse of the Inductance Can anyone suggest the pages to be read for output voltage. Thanks in advance. UserIdTAG: 357381 UserNameTAG: krrish46 CreateTimeTAG: 2012-10-28T00:15:39Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 2136 TitleTAG: Practice Midterm - Q5 - Nonlinear Elements Hi, can anyone point out textbook sections or example problems to help with Q5? I looked over section 4.4 (Piecewise Linear Analysis), but didn't find it to be much help. Even a hint as to what methods would be appropriate to apply here would be much appreciated. Thanks. UserIdTAG: 194509 UserNameTAG: blackcompe CreateTimeTAG: 2012-10-27T23:31:28Z VoteTAG: 1 CoursewareTAG: Week 3 / Nonlinear Elements CommentableIdTAG: 6002x_nonlinear_elements NumberOfReplyTAG: 2 FirstChildTAG: I found it quite helpful to consider the two diodes (and voltage sources V1 and V2) in parallel as a single element, and to plot the i-v relation for this thing. (If you want to break it down a bit, then plot i-v for a diode and voltage source in series first.) Then if you also plot the load line for the rest of the circuit and look for where the two lines intersect, you'll get a pretty good idea of what is going on. FirstChildUserIdTAG: 288323 FirstChildUserNameTAG: fatslow FirstChildCreateTimeTAG: 2012-10-28T09:16:02Z FirstChildTAG: Hi blackcompe! Week 3 Tutorials videos are helpful :).[watch here][1] Also, Chapter 4 of the Textbook. See you, Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_3/Week_3_Tutorials/ FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-27T23:45:02Z IndexTAG: 2137 TitleTAG: the Midterm i finish my exam with 93% , is good?**strong text** UserIdTAG: 331441 UserNameTAG: abdessamade CreateTimeTAG: 2012-10-27T22:26:33Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I can understand your excitement, but will have to wait until everyone is finished before we can discuss it. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-27T22:31:52Z IndexTAG: 2138 TitleTAG: 100% on the Midterm Just celebrating... ;) UserIdTAG: 129876 UserNameTAG: msarabi95 CreateTimeTAG: 2012-10-27T21:32:30Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Congrats FirstChildUserIdTAG: 392100 FirstChildUserNameTAG: glenng FirstChildCreateTimeTAG: 2012-10-27T21:44:43Z FirstChildTAG: Congratulations msarabi95!:) But let's leave this Post Discussion for monday, after the deadline of the Midterm Exam ... Now you can go and relax! ^_^ See you! Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-27T22:08:02Z IndexTAG: 2139 TitleTAG: Why on Earth does the concept of Energy exist??? To know that is very crucial for me, because otherwise I can't make a good intuition on how does MOSFET (and, actually, all other staff) work... I understand, why was the concept of Impulse (or the amount of movement, as Newton called it), p=mv, developed. That's because of the conservation of that amount. We can predict lot's of mechanical interactions with that concept. Also, I understand, why do we need the concept of Force, F=dp/dt. It is a rate of change of impulse with time, so it's quite reasonable to use that concept in certain cases. But I can't understand the concept of energy and work, E=integral of F * ds. Why was it created? Why on earth did humanity need to develop it? What was the cause of defining that concept? Maybe there's some history of it's creation? Can anybody tell me? UserIdTAG: 326787 UserNameTAG: Angstrem CreateTimeTAG: 2012-10-27T20:17:55Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: Why on Earth? Simple: They invented Energy, so they could tax it! I think that the concept of E, makes life easier if you have to convert kinetic energy to electrical energy and vice versa. And energies can be added easily: it doesn't matter if you add kinetic energy and electric energy, they add up nicely. But how do I add the kinetic impulse p=mv to the electrical charge of a capacitor Q=CU? LI? FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-10-27T20:41:52Z SecondChildTAG: And why can we convert mechanical energy to electrical one? Why is mechanical energy defined as integral(F*ds) and the electrical one is defined as integral(F*dq)? SecondChildUserIdTAG: 326787 SecondChildUserNameTAG: Angstrem SecondChildCreateTimeTAG: 2012-10-28T08:47:15Z FirstChildTAG: My response would be similar to Salsero. It was and still is useful for farming. If you knew how much average work a horse could do in one day, you could figure out how many are needed to work a field of a certain size. If you did not have any horses, but you knew that a person was about ~0.15% horse's power, then it would take 6 or 7 men to do the work of one horse. In the mid 1700's the modern term for "Horsepower" was coined by James Watt, to compare the work of his machines to that of a horse. 1 HP = 745.7"Watts" FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-27T21:00:53Z SecondChildTAG: That's pretty interesting... 2 horses can move with different speeds both applying the same force, hence that's not correct to measure work done by a horse with an integral(F*dt), the change of impulse. But why did they decide to measure energy in integral(F*ds)? Wasn't there any other alternative? SecondChildUserIdTAG: 326787 SecondChildUserNameTAG: Angstrem SecondChildCreateTimeTAG: 2012-10-28T08:50:35Z FirstChildTAG: And what about the conservation rules? The energy is the invariant measure, remaining constant in any closed system, can take any forms (kinetic, potential, elastic, chemical etc..) allowing simple (relatively) calculations. Say, how would you calculate the velocity of an object falling from certain height? Taking integrals? In terms of energy it can be done with very simple arithmetics. FirstChildUserIdTAG: 194098 FirstChildUserNameTAG: ZhekaS FirstChildCreateTimeTAG: 2012-10-28T01:48:23Z SecondChildTAG: You're right, but that laws were definitely discovered *after* the energy was... SecondChildUserIdTAG: 326787 SecondChildUserNameTAG: Angstrem SecondChildCreateTimeTAG: 2012-10-28T08:56:08Z FirstChildTAG: I'm not a physicist but I can't resist chiming in on this thread. First, "What is energy?" The most concise answer I've ever heard is, "Energy is the capacity for doing work." And then the physicists chime in with their precise and calculable definitions of "work-energy equivalence." Rumford noticed the seemingly inexhaustible supply of heat energy while boring the barrels of cannons. From there, the supply of this mysterious 'energy' seems to be almost limitless. It seems to be an intrinsic property of the universe. But it can be 'lost' and 'dissipated' and that leads us to the notion of entropy...another pretty fundamental notion. I suspect this chase of reasoning goes on and on. Ultimately I don't think anyone really knows what energy 'is'...the ultimate reality of it all. But then we don't ultimately know the nature of Reality, anyway! :) ... and we're caught in some kind of mysterious, self-referencing puzzle. ('New Scientist' had a very nice series on just that topic, recently.) In a more practical sense, energy is that property of things which serves as a starting point for the construction of an understanding our world and our universe. Nothing all that wrong with that, is there? :) FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-10-29T00:58:54Z FirstChildTAG: If you are wondering about metaphysical concepts such as energy, matter, and time, both an introductory **physics** course and an introductory **philosophy** course would help you greatly. In EECS 6.002x we are not concerned with energy on the metaphysical level; just in it's use - hence engineering. FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-10-28T01:18:34Z SecondChildTAG: Thanks for the recommendations, although I have some hesitations about the philosophy course. Due to my university classes, it seems to me, that philosophy is quite not scientific discipline. It's not rigorous, hence it's insights may have errors. SecondChildUserIdTAG: 326787 SecondChildUserNameTAG: Angstrem SecondChildCreateTimeTAG: 2012-10-28T08:54:32Z IndexTAG: 2140 TitleTAG: Lab 7 helpp .. Q1 eq is is wrong UserIdTAG: 535546 UserNameTAG: mackallis006 CreateTimeTAG: 2012-10-27T14:20:36Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: What's your problem? Could you elaborate! FirstChildUserIdTAG: 277808 FirstChildUserNameTAG: Hemanthmps FirstChildCreateTimeTAG: 2012-10-27T15:07:25Z FirstChildTAG: Hi mackallis006, Can I help you? You can take a look here [Lab 7 - Hints][1] :) Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5091dd5d3fc09f2b0000018a FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-01T02:46:16Z IndexTAG: 2141 TitleTAG: About Administrivia How much must I have to get the certificate? In percent UserIdTAG: 70519 UserNameTAG: Fipe CreateTimeTAG: 2012-10-27T02:29:36Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: 60 percent. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-27T02:31:23Z IndexTAG: 2142 TitleTAG: Midterm exam bar gray ? does that mean it's not official yet ? or is it normal ? Midterm exam bar gray ? does that mean it's not official yet ? or is it normal ? Apretiate the staff's input on this :) UserIdTAG: 263693 UserNameTAG: Coldberg CreateTimeTAG: 2012-10-27T00:07:11Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: You mean on the progress page? I think that's normal. It will probably turn red when the deadline passes. Although I'm not on the staff. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-27T00:35:55Z SecondChildTAG: yes the progress page :) SecondChildUserIdTAG: 263693 SecondChildUserNameTAG: Coldberg SecondChildCreateTimeTAG: 2012-10-27T00:36:54Z SecondChildTAG: can you help me with the answers of MIDTERM? SecondChildUserIdTAG: 414461 SecondChildUserNameTAG: akashrandev SecondChildCreateTimeTAG: 2012-10-27T06:07:45Z FirstChildTAG: It will aways look like that. And the final will be dark grey. ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13512985451343614.png FirstChildUserIdTAG: 9673 FirstChildUserNameTAG: xp42 FirstChildCreateTimeTAG: 2012-10-27T00:44:15Z SecondChildTAG: thanks! SecondChildUserIdTAG: 263693 SecondChildUserNameTAG: Coldberg SecondChildCreateTimeTAG: 2012-10-27T00:57:08Z SecondChildTAG: BTW nice score :) SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-27T04:35:26Z SecondChildTAG: Classy SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-27T10:49:22Z SecondChildTAG: Is this from last course? SecondChildUserIdTAG: 623210 SecondChildUserNameTAG: preveen SecondChildCreateTimeTAG: 2012-10-27T16:11:17Z SecondChildTAG: I guess yes, it is his score @praveen :).xp42 did a great work in 6.002x (Spring)! In fact, he ended with the highest Karma Point of the Class! Karma Points were a system that was implemented in the Spring: Students on the forum get points based off of their activity in the forum. The main method of getting points is to ask good questions and give good answers. When a question or answer is upvoted, the user who posted them will gain "karma points". These points are used to determine how much moderation power a student has. If you like a question or answer, be sure to thank the contributor by up-voting their contribution. If you have enough karma, you will have the ability to down-vote bad contributions. I hope that he can back and help the new students, but it is up to him haha! See you! Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-27T22:33:11Z SecondChildTAG: **Myrimit:** I guess I must have gotten some "Karma" points with my questions lately, since I made "Community TA." I **didn't know** that "Karma" points count!!! If I did, I would have created many more new topics, instead of just replying in the "grey" area where no-one can up-vote you! **xp42:** I should say "**nice score**" as well! Although I am wondering - If I get 100% on all my homeworks up to HW11 & HW12, will I skip them (I can be lazy sometimes), or will I complete them 100% (just to show off) even though the two dropped grades don't count. SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-10-28T02:06:07Z SecondChildTAG: Hi @JerseyMark! haha! Congratulations for the Community TA :)! I guess that is based on your contribution. I am not sure if it is because your scores of the Posts haha, I can't confirm that. I become notice of the Community TA in the old forum (6.002x spring) days ago when kimt (staff) requested me if I would like to be a Community TA in 6.002x Fall. So, I am not sure if it is based on your scores of the Posts, I guess they choose students, but I am really not sure how they are based on... Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-28T16:02:55Z FirstChildTAG: Its just to differentiate your midterms contributions to your final score. red -homework dark red -lab grey-midterm dark grey -final exam Nothing to worry about :D FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-10-27T09:19:27Z IndexTAG: 2143 TitleTAG: Old midterm Q4 part 7 and 8 I don't get it, can someone help me? UserIdTAG: 392100 UserNameTAG: glenng CreateTimeTAG: 2012-10-26T23:39:27Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: How to do this is explained in lecture set 12 (the second set of lectures in week 6). According to the course info page where this years exam is described: "The exam will cover material through Small-Signal Analysis of Transistor Circuits (end of first lecture video sequence of Week 6)". In other words, don't panic if you don't understand this yet. FirstChildUserIdTAG: 174100 FirstChildUserNameTAG: markpolak FirstChildCreateTimeTAG: 2012-10-27T00:10:25Z SecondChildTAG: Actually I made a mistake. It's explained in the second lectures of week 7 (lecture set 14). SecondChildUserIdTAG: 174100 SecondChildUserNameTAG: markpolak SecondChildCreateTimeTAG: 2012-10-27T14:47:07Z SecondChildTAG: I wondered about that. I haven't watched that far yet. SecondChildUserIdTAG: 392100 SecondChildUserNameTAG: glenng SecondChildCreateTimeTAG: 2012-10-27T21:46:14Z FirstChildTAG: Have a look at my post here: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/508a8b09d8f4662b0000002e FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-10-27T15:08:27Z SecondChildTAG: Thanks SecondChildUserIdTAG: 392100 SecondChildUserNameTAG: glenng SecondChildCreateTimeTAG: 2012-10-27T21:45:18Z IndexTAG: 2144 TitleTAG: MTQ4 ANY HELP OR HINTS ON Q4 of midterm???? tottally stuck!! UserIdTAG: 127800 UserNameTAG: MNB CreateTimeTAG: 2012-10-26T19:24:38Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You are not allowed to discuss the mid-term until the deadline passes - Monday. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-26T19:26:22Z IndexTAG: 2145 TitleTAG: S3E2: Parser issue The parser doesn't seem to be working. I tried entering expressions like this: R1R2/(R1R2+R1R3+R2R3). Thanks a lot! UserIdTAG: 581004 UserNameTAG: minamora CreateTimeTAG: 2012-10-26T18:35:48Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 2 FirstChildTAG: What are you trying to say? FirstChildUserIdTAG: 277808 FirstChildUserNameTAG: Hemanthmps FirstChildCreateTimeTAG: 2012-10-26T18:39:27Z FirstChildTAG: You need to be using the "*" (Shift 8) for multiplication. Example: R1*R2 <-This means R1xR2 or R1R2. Hope this helps, have fun. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-26T19:00:11Z SecondChildTAG: > "*" (Shift 8) Well, it depends on the keyboard layout... SecondChildUserIdTAG: 376877 SecondChildUserNameTAG: AndBre SecondChildCreateTimeTAG: 2012-10-26T19:33:39Z SecondChildTAG: Agreed. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-26T19:35:23Z IndexTAG: 2146 TitleTAG: Current Sources in Series Hi, I would be grateful if someone could demystify the bizarre analysis of the circuit below. How can the current jump from 2 Amperes to 5 Amperes along an ideal wire? ![Current Sources in Series][1] [1]: https://edxuploads.s3.amazonaws.com/13512680167649838.png UserIdTAG: 85490 UserNameTAG: yashsavani CreateTimeTAG: 2012-10-26T16:15:42Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: A real current source will have an internal resistor in parallel. FirstChildUserIdTAG: 376877 FirstChildUserNameTAG: AndBre FirstChildCreateTimeTAG: 2012-10-26T16:34:32Z FirstChildTAG: You probably want to look at page 60 of the textbook, Fig. 2.13, for example, and related explanation in the text. Lab tool is based on KVL and KCL methods and lumped matter discipline, and since your circuit violates KCL and lumped matter discipline, you just can't simply go and solve this circuit. FirstChildUserIdTAG: 192703 FirstChildUserNameTAG: voffch FirstChildCreateTimeTAG: 2012-10-26T16:24:12Z FirstChildTAG: Impossible with ideal current sources. There will be some smoke . The same with ideal different voltages sources in parallel. FirstChildUserIdTAG: 373752 FirstChildUserNameTAG: Croak FirstChildCreateTimeTAG: 2012-10-27T08:48:55Z FirstChildTAG: If the current source has internal resistance then only it would be possible! Otherwise, it is wrong. Current suddenly can't change from 2A to 5A. FirstChildUserIdTAG: 277808 FirstChildUserNameTAG: Hemanthmps FirstChildCreateTimeTAG: 2012-10-26T18:28:49Z IndexTAG: 2147 TitleTAG: succes of midterm i have completed my midterm with a score of 100% good luck for those who will sit for the xm next. UserIdTAG: 154172 UserNameTAG: mofassair CreateTimeTAG: 2012-10-26T08:39:33Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Finaly! I've just understood how to solve the last problem of the last Q6! 100%! I wasn't so happy rather when I got my high school diploma :-) I like this courses, thanks a lot for an whole stuff! It's exiting! And I'd like to donate some money for edX-project. Where can I do it? FirstChildUserIdTAG: 345464 FirstChildUserNameTAG: Constantine_ru FirstChildCreateTimeTAG: 2012-10-26T10:11:07Z SecondChildTAG: Can You write to info@edx.org ? SecondChildUserIdTAG: 398733 SecondChildUserNameTAG: Smithamico SecondChildCreateTimeTAG: 2012-10-26T15:51:11Z SecondChildTAG: Yes the experience is great! SecondChildUserIdTAG: 623210 SecondChildUserNameTAG: preveen SecondChildCreateTimeTAG: 2012-10-27T16:25:07Z IndexTAG: 2148 TitleTAG: H8P3 In ap 3 of H8 I don't understand why Gate of Q2 can't be charged to màximum drain voltage of Q1. Please, help UserIdTAG: 160366 UserNameTAG: elcollar CreateTimeTAG: 2012-10-26T07:37:48Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: I was stuck on last question (part 5)and I put the result of part 4 and it was rigth! It was only luck, I wasn't able to solve this part numerically. I'm waiting for the right solution. Any hint for a real solution? Thanks! FirstChildUserIdTAG: 299251 FirstChildUserNameTAG: TheRedBlackOne FirstChildCreateTimeTAG: 2012-10-26T09:07:48Z SecondChildTAG: For the last part of Q5, draw the appropriate circuit with the capacitor when Q1 and Q3 are on, and include the pullup resistance for Q1 and the voltage source. Then one way to analyze this is to use the Thevenin equivalent circuit. SecondChildUserIdTAG: 288323 SecondChildUserNameTAG: fatslow SecondChildCreateTimeTAG: 2012-10-26T09:24:49Z SecondChildTAG: i don't even know where to start. surly i have to model the circuit first , but i don't know how to place in the 'store' gate? help plssss SecondChildUserIdTAG: 285945 SecondChildUserNameTAG: lindalapiso SecondChildCreateTimeTAG: 2012-11-03T11:27:16Z FirstChildTAG: For Q3, as for all our MOSFETS, the device conducts between drain and source when $V_{GS} \gt V_T$ and is OFF otherwise. The device still conducts a little when off, but the amount is negligible because $R_{OFF}$ is a massive resistance; ideally $ = \infty $. Anyway, for the problem at hand, the voltage on the gate of Q2 can only rise towards the value of the voltage on the drain of Q1 when the Q3 MOSFET is ON. Your task is to figure out at what point the Q3 MOSFET switches OFF when Q1 is charging the gate of Q2. Hint: They told you the drain of Q3 is the terminal at the higher voltage, and the source is at the lower voltage. Use this to figure out which two points $V_{GS}$ is measured between on Q3. FirstChildUserIdTAG: 9673 FirstChildUserNameTAG: xp42 FirstChildCreateTimeTAG: 2012-10-26T18:18:25Z SecondChildTAG: "The device still conducts a little when off, but the amount is negligible because $R_{OFF}$ is a massive resistance; ideally $ = \infty $" Yes, but in the question, they specify $R_{OFF}$ and if you were to actually use it for Q3, then the first part of the question doesn't make sense... SecondChildUserIdTAG: 141000 SecondChildUserNameTAG: OrinE SecondChildCreateTimeTAG: 2012-10-26T20:34:40Z SecondChildTAG: I don't recall saying how to use the $R_{OFF}$ value given in the problem statement to solve any part of H8P3, or saying when it is valid to do so. (For those who haven't done this problem yet, be warned you will use the given $R_{OFF}$ for modeling Q1 in some parts. You just won't use it for Q3). In the first part you have to use the ideal $R_{OFF} = \infty$ for Q3. Even if you didn't know this to be the case, when you calc the total leakage resistance $R_{leakage}$ you will get a number in the TeraOhms range, so clearly $R_{OFF}$ for Q3 must be greater than this since the total leakage resistance is $R_{OFF}$ of Q3 in parallel with the parasitic resistance of Q2 from its gate to its source. You can either calc $R_{leakage}$ with $R_{OFF}(Q3) = \infty$ and thus have $R_{leakage} = R_{parasitic}$, or you can calc $R_{leakage}$ and know in the back of your mind that $R_{leakage} = ( R_{OFF}(Q3) + R_{PU}(Q1) || R_{ON/OFF}(Q1) ) $ || $R_{parasitic}$. The fact they are asking for the parasitic resistance is the give-away that you should just ignore $R_{OFF}(Q3)$ by considering it to be ideal: i.e. $ R_{OFF}(Q3) = \infty$. SecondChildUserIdTAG: 9673 SecondChildUserNameTAG: xp42 SecondChildCreateTimeTAG: 2012-10-27T00:38:46Z SecondChildTAG: The fact they are asking for the parasitic resistance is that it has to do with Cgs. Ignore Roff of 95M, but take into consideration a Tohms resistor. I appreciate dear **xp42** that you are trying to help, but boy, sometimes you really get carried away. :P SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-10-27T17:54:42Z SecondChildTAG: Thanks a lot xp42 SecondChildUserIdTAG: 179140 SecondChildUserNameTAG: JoydeepSil SecondChildCreateTimeTAG: 2012-11-07T15:54:02Z FirstChildTAG: The correct answer for VGS2Max doesn't have any connection with the capacitor Cgs2 and it's an extremely simple problem. We know that the gate of Q3 is connected to the drain of Q1.If you draw it you will notice that you will have a diode connected mosfet, with VGS3 = VDS3. We know that VGS3 > VT, so VDS3 >=1 .If you write KVL you will get that VGS2 = Vs -VDS3 => VGS2max = Vs - VDS3min .So, VGS2max = Vs - 1. Vs is not really 5V due to Roff.It is 4.99947, but insignificant smaller . ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13513596791343676.gif FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-10-27T17:49:43Z SecondChildTAG: thanks SecondChildUserIdTAG: 366669 SecondChildUserNameTAG: pedroramus SecondChildCreateTimeTAG: 2012-11-11T23:55:52Z FirstChildTAG: Think carefully about what VGS is for Q3. FirstChildUserIdTAG: 288323 FirstChildUserNameTAG: fatslow FirstChildCreateTimeTAG: 2012-10-26T08:02:08Z FirstChildTAG: Hi I am able to solve part1,2,3 of H8p3 but unable to solve Part4 and 5 can any one give me hint to solve this problem FirstChildUserIdTAG: 352647 FirstChildUserNameTAG: praveenjugge FirstChildCreateTimeTAG: 2012-11-07T08:18:30Z IndexTAG: 2149 TitleTAG: PRACTICE Midterm - Q4, Part 6 Can anyone explain this? Thanks in advance. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-26T07:10:24Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: What? FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-10-26T07:22:20Z SecondChildTAG: It's all good, I solved it after reviewing the textbook. SecondChildUserIdTAG: 154685 SecondChildUserNameTAG: Kaichholz SecondChildCreateTimeTAG: 2012-10-26T08:03:54Z SecondChildTAG: Good for you. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-10-26T09:03:04Z FirstChildTAG: Can you specify which section of the book you used to help you? I'm stuck and from what i understood from the book i needed to treat it as three separate circuits and add the Power of each, unfortunately i get a wrong answer. I'm using the equation P=Vs^2/Rt where Rt is the total resistance in a particular circuit, the i add the three Powers and i get a wrong total. FirstChildUserIdTAG: 23555 FirstChildUserNameTAG: NickFrei FirstChildCreateTimeTAG: 2012-10-26T16:19:57Z SecondChildTAG: I looked in section 6.10 of the text book. From that example I was able to solve this by drawing the circuit as network of resistors for each value of A and B, and then using the V^2/R formula. It did take me more than 3 attempts though. Note: one of the mistakes I made was to incorrectly assume that power is consumed when C is a logical 1. SecondChildUserIdTAG: 373309 SecondChildUserNameTAG: njirving SecondChildCreateTimeTAG: 2012-10-26T21:08:12Z IndexTAG: 2150 TitleTAG: When will we be able to see the answers to the midterm questions..? Hello Everyone! I just completed my Midterm exams and scored 86%. I want to clarify my concepts of some of the questions in the midterm. when will the answers be available..???? UserIdTAG: 137686 UserNameTAG: Jaychandran CreateTimeTAG: 2012-10-26T03:47:13Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: **There is a show answer button once you have completed all the three check clicks. You can see there the answers! But the solutions will be available only after the completion of the MIDTERM.** [EDIT - The show answer button has been removed.] FirstChildUserIdTAG: 277808 FirstChildUserNameTAG: Hemanthmps FirstChildCreateTimeTAG: 2012-10-26T04:31:50Z SecondChildTAG: is the "Show Answer" button still available?I can't see it any more. SecondChildUserIdTAG: 222911 SecondChildUserNameTAG: bhaswardg SecondChildCreateTimeTAG: 2012-10-26T04:38:23Z SecondChildTAG: I'm not agree the show answer only appear once in some question and another not! SecondChildUserIdTAG: 299251 SecondChildUserNameTAG: TheRedBlackOne SecondChildCreateTimeTAG: 2012-10-26T05:06:33Z FirstChildTAG: Answers will be available after the deadline passes. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-26T14:21:29Z FirstChildTAG: There is no show answer button in the midterm. Solutions were available two days after the deadline. I'm pretty sure there will be tons of threads which will discuss the various ways of solving the problems much before that though :-) @Hemanthmps Just to be safe. Are you sure you attempted the midterm exam and NOT the Spring session's midterm which was offered for practice this time? The URL to the correct midterm is the following: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Midterm_Exam/MidtermFall2012/ FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-10-26T13:21:29Z SecondChildTAG: There was a Show Answer button in the mid-term, but it has been removed. We should discuss it after the mid-term is over. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-26T14:20:39Z IndexTAG: 2151 TitleTAG: Fantastic! That is brilliant! One of the things I was hoping for when I enrolled in 6002x was to learn what an oscillator was!! Why don't I see more of this practical circuit building in the main lecture sequences? UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-10-25T20:35:20Z VoteTAG: 1 CoursewareTAG: Week 6 / Ring Oscillator CommentableIdTAG: 6002x_Ring_Oscillator NumberOfReplyTAG: 0 IndexTAG: 2152 TitleTAG: Some help on the PRACTICE midterm-Q1(don't flag me) I'm probably just doing something insanely stupid-and it's pathetic that I'm having problems with something so basic.... but I'm not getting the voltage answers correct. Can somebody help me out? My KCL equations look like: el1-Vs/Rw1 = el1/RL1 + el1-el2/Rw2 el2-el1/Rw2 = eL2/RL2 God... I am so screwed. :( UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-25T20:09:49Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: Thanks all.I was getting confused in such a simple thing. FirstChildUserIdTAG: 281159 FirstChildUserNameTAG: ManasiS FirstChildCreateTimeTAG: 2012-10-26T00:06:28Z FirstChildTAG: You're nearly right, but don't forget that the currents leaving a node must _SUM_ to zero, so: el1-Vs/Rw1 + el1/RL1 + el1-el2/Rw2 = 0 el2-el1/Rw2 + eL2/RL2 = 0 FirstChildUserIdTAG: 502508 FirstChildUserNameTAG: amaher FirstChildCreateTimeTAG: 2012-10-25T20:34:06Z FirstChildTAG: You do not need to use them. You can turn the resistors R_w2, R_L1, R_L2 into one equivalent resistance. Then you have two resistors and a voltage to find between them... Then repeat for the next bit. FirstChildUserIdTAG: 261378 FirstChildUserNameTAG: es2377 FirstChildCreateTimeTAG: 2012-10-25T20:35:32Z FirstChildTAG: Every term you have written down is the current flowing out of a node, so you want equations where they sum to zero. So it looks to me like all the terms are correct, but your problem is just that you've got the = sign in the wrong place. FirstChildUserIdTAG: 288323 FirstChildUserNameTAG: fatslow FirstChildCreateTimeTAG: 2012-10-25T20:37:54Z FirstChildTAG: Now I really feel retarded. Thanks anyway! :) FirstChildUserIdTAG: 344331 FirstChildUserNameTAG: intellectualwanderer FirstChildCreateTimeTAG: 2012-10-26T02:47:09Z IndexTAG: 2153 TitleTAG: About the platform Firstly, I want to thank edX staff and community for edX as a really great platform for online education. Moreover, as far as I know, edX is developing as an open-source project. So the question is: where exactly can we look at what underlies edX (I mean, code), if it's really open-source? UserIdTAG: 192703 UserNameTAG: voffch CreateTimeTAG: 2012-10-25T18:35:43Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: We are still in the process of cleaning up the code so that when potential contributors look at the source, they do not run away screaming in terror. Releasing the code and constructing a good open source community is extremely high on our priority list, though. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-10-25T20:30:36Z SecondChildTAG: "Screaming in terror" eh? Many of us from the mitx class are already screaming. You don't need source code to convince us to run, run, run, far, far away. SecondChildUserIdTAG: 9673 SecondChildUserNameTAG: xp42 SecondChildCreateTimeTAG: 2012-10-26T04:30:24Z SecondChildTAG: And yet we stay, why do we stay?? SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-26T17:54:07Z SecondChildTAG: edx is fine platform. thanks for the experience SecondChildUserIdTAG: 623210 SecondChildUserNameTAG: preveen SecondChildCreateTimeTAG: 2012-10-27T16:32:05Z FirstChildTAG: I did not know that the edX platform is open-source. That's good to know. If enough people with the know-how get involved, 6.002x, which is already getting pretty good reviews, can become phenomenal. I'm wondering if the code will be available for the interactive stuff: applets such as the Circuit Simulator and the MOSFET Audio Amplifier Simulator, etc. With enough modification, tweaking and contribution, the Circuit Simulator, e.g. the "circuit sandbox", can evolve into a very useful stand-alone tool for technicians, hobbyists, and engineers, who now have to download bulky software from companies such as OrCAD just so they can build a circuit schematic and run a PSpice simulation on it. It's already pretty good as a "bare-bones" simulator to run a quick analysis of a basic circuit. Perhaps even a Logic Simulator can be built along these lines, similar to the Xilinx Foundations Series Student Edition that is bundled with some Logic Design textbooks; where you can place different gates along a schematic, wire them together, and even use higher-level ICs such as adders, shift registers, etc. FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-10-25T22:30:41Z SecondChildTAG: This thing could be revolutionary. SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-26T17:53:22Z SecondChildTAG: It would be great if the 6.002x students could collaborate on such a project. SecondChildUserIdTAG: 623210 SecondChildUserNameTAG: preveen SecondChildCreateTimeTAG: 2012-10-27T16:33:30Z IndexTAG: 2154 TitleTAG: Age of speaker in Two-stage MOSFET amplifier Tutorials week 6? What is the age of this person? https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_6/wk6_tiff1/ UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-10-25T18:18:28Z VoteTAG: 1 CoursewareTAG: Week 6 / Two-stage MOSFET amplifier CommentableIdTAG: 6002x_Two_Stage_MOSFET_amplifier NumberOfReplyTAG: 3 FirstChildTAG: Good one... FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-10-25T18:31:20Z SecondChildTAG: I don't get the joke.... SecondChildUserIdTAG: 96452 SecondChildUserNameTAG: Lyla SecondChildCreateTimeTAG: 2012-10-25T20:28:46Z SecondChildTAG: Neither do I... ? SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-25T20:30:03Z SecondChildTAG: I think hazel1919 was just asking how old the person doing the tutorial is. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-25T23:08:29Z SecondChildTAG: Never mind. Of course, why would someone would post such a question... SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-10-26T08:56:42Z FirstChildTAG: +1 FirstChildUserIdTAG: 454609 FirstChildUserNameTAG: Alwahsh FirstChildCreateTimeTAG: 2012-10-25T19:43:01Z FirstChildTAG: hahaha, I thought it was a woman the first time I saw the video. For some reason, now I think it's a small kid the one talking. FirstChildUserIdTAG: 392546 FirstChildUserNameTAG: JohnWayne FirstChildCreateTimeTAG: 2012-10-26T02:57:25Z SecondChildTAG: lol SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-10-26T09:02:33Z IndexTAG: 2155 TitleTAG: Graphing the curve. If you put the equations **e^-(t/tau)** and **1-e^-(t/tau)** And pick a arbitrary value for Tau (i.e 6) you can really see the curves!![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13511725577637814.png UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-10-25T13:42:48Z VoteTAG: 1 CoursewareTAG: Week 7 / Falling Delay Part 3 CommentableIdTAG: 6002x_Falling_Delay_Part_3 NumberOfReplyTAG: 1 FirstChildTAG: hazel1919, what program is that graph produced with? FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-25T15:03:45Z SecondChildTAG: http://www.graphcalc.com/ I recommend it, easy to download. All equations must have y= on the left. SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-25T15:07:13Z SecondChildTAG: google do the same, but no need to download... and not sure if it's possible to download google at all :) SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-26T00:04:06Z IndexTAG: 2156 TitleTAG: S1E7 SIR, HOW TO SOLVE S1E7 UserIdTAG: 563465 UserNameTAG: gbhatra CreateTimeTAG: 2012-10-25T13:12:42Z VoteTAG: 1 CoursewareTAG: Week 1 / KVL and KCL exercises CommentableIdTAG: 6002x_kvl_kcl_exercises NumberOfReplyTAG: 1 FirstChildTAG: ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13511783419005338.gif Using KCL : At node B: i5 = i3+i4 At node A: i1+i5+i2 = 0 Hope u got it! FirstChildUserIdTAG: 277808 FirstChildUserNameTAG: Hemanthmps FirstChildCreateTimeTAG: 2012-10-25T15:21:27Z IndexTAG: 2157 TitleTAG: Whats your Score? I ask all the members to share there score's!!! I congratulate all the people who have scored Cent %.... But according to me they should not get too much excited about it.... And the people scoring less should not get deprssed....!!! Coz there is still more to come... :) :) The best part is "Its really Fun"....!!! My personal score is 100%..... What's yours?? PS:(No details of the Midsem before the Deadline....!!! Please!!) UserIdTAG: 82597 UserNameTAG: bondrajat CreateTimeTAG: 2012-10-25T13:04:12Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: 100..:) FirstChildUserIdTAG: 266739 FirstChildUserNameTAG: satvikchugh FirstChildCreateTimeTAG: 2012-10-25T13:27:55Z SecondChildTAG: Great...!!! Many congratulations to you satvikchug!!! keep it up!! :) SecondChildUserIdTAG: 82597 SecondChildUserNameTAG: bondrajat SecondChildCreateTimeTAG: 2012-10-25T13:55:11Z FirstChildTAG: Lets leave this Post Discuss for monday , after the deadline, as this can might give some details of the Midterm Exam ... Congratulations!:) FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-25T13:46:35Z SecondChildTAG: Edited the post....!!! however it wasn't meant to do sumthing against 'Honor Code' or forum guidelines!!! SecondChildUserIdTAG: 82597 SecondChildUserNameTAG: bondrajat SecondChildCreateTimeTAG: 2012-10-25T13:57:20Z FirstChildTAG: Scored only 79%:(...paid the price for having a very slow internet connection repeatedly pressed check button and lost all submissions FirstChildUserIdTAG: 106395 FirstChildUserNameTAG: Arunmozhi FirstChildCreateTimeTAG: 2012-10-25T15:16:58Z SecondChildTAG: Ahhh!!! :( Thats O.K....!!!And Congrats!!! Try to Cover it up in the Finals!!! All the best!!! :) SecondChildUserIdTAG: 82597 SecondChildUserNameTAG: bondrajat SecondChildCreateTimeTAG: 2012-10-25T16:46:04Z FirstChildTAG: Did you guys not read Myrimit's post? Leave it until Monday. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-25T16:49:30Z IndexTAG: 2158 TitleTAG: Lastyear Midterm Exam Q2 HOW??? i don't understand. UserIdTAG: 196404 UserNameTAG: Yel1owstone CreateTimeTAG: 2012-10-25T12:37:31Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: *Black box* is linear, right? Then you can think of it as of it's Thevenin equivalent, a voltage source in series with the resistor, and resistive load is just a resistor. So you are left with *Vth* and two resistors forming a voltage divider, and you can write and solve two equations for voltage divider with different *RL*-s, where *Vth* and *Rth* are your unknowns. FirstChildUserIdTAG: 192703 FirstChildUserNameTAG: voffch FirstChildCreateTimeTAG: 2012-10-25T13:20:08Z FirstChildTAG: Or , you know that your black box have a Vth and a Rth, and you have V as output.You can model the output like a independent voltage source, and will give you the same results. FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-10-25T14:53:09Z FirstChildTAG: **Model BLACKBOX as voltage source (Vth) in series with resistor (Rth) and load as a single resistor (Rx). Then you'll end up with two equations from given two cases. Just solve them and find out the unknowns. Hope it has helped you.** FirstChildUserIdTAG: 277808 FirstChildUserNameTAG: Hemanthmps FirstChildCreateTimeTAG: 2012-10-25T15:44:30Z SecondChildTAG: You will have then Rx and Ry. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-10-25T16:07:08Z SecondChildTAG: First equation is for a resistor Rx.But the second is for another resistor Ry.I must admit that **voffch** has nail it the best possible way. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-10-25T16:09:24Z FirstChildTAG: Thank you. I solved. FirstChildUserIdTAG: 196404 FirstChildUserNameTAG: Yel1owstone FirstChildCreateTimeTAG: 2012-10-25T20:56:01Z FirstChildTAG: Just focus on the box so that u can involve the current from the source to get involved wid da device Z as well...!! its easyyyy.... FirstChildUserIdTAG: 127800 FirstChildUserNameTAG: MNB FirstChildCreateTimeTAG: 2012-10-26T03:36:48Z IndexTAG: 2159 TitleTAG: Midterm Confusion over time constraints ! 1. Midterm is open till 28th of October. 2. There is a 24 hours window which is ticking somewhere i can't see ! When does that 24 hours clock starts ticking ? I logged in and it straight-away made me land at the questionnaire page which i hadn't thought of. Is my clock ticking now ? Would anyone let me know about it ? Thanks UserIdTAG: 97340 UserNameTAG: AnkitRana CreateTimeTAG: 2012-10-25T09:21:52Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: The clock will only start ticking when you press the next page button at the cover page of the mid term. This means that once you have viewed the questions, the clock is start ticking. FirstChildUserIdTAG: 529515 FirstChildUserNameTAG: Low FirstChildCreateTimeTAG: 2012-10-25T10:04:05Z SecondChildTAG: I didn't quite get the 24 hour thing. When I start the exam, does that mean I have 24 hours to finish it or how much time do I have? SecondChildUserIdTAG: 259178 SecondChildUserNameTAG: bdrangova SecondChildCreateTimeTAG: 2012-10-25T10:53:51Z SecondChildTAG: Sorry, you don't have clock for that. You need to time yourself. SecondChildUserIdTAG: 529515 SecondChildUserNameTAG: Low SecondChildCreateTimeTAG: 2012-10-25T10:57:32Z SecondChildTAG: Okay ! I've read the whole thing now. Thanks a lot ! Got it SecondChildUserIdTAG: 97340 SecondChildUserNameTAG: AnkitRana SecondChildCreateTimeTAG: 2012-10-25T11:00:48Z FirstChildTAG: Hi. [Just Check It][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5088dbe5959a592b00000024 > **Regards:** asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-10-25T14:59:04Z IndexTAG: 2160 TitleTAG: MIDTERM EXAM Wow! I completed my midterm exam within 2 hours! The paper was way too easy for me. :) Is there anyone else who has completed the exam?? UserIdTAG: 277808 UserNameTAG: Hemanthmps CreateTimeTAG: 2012-10-25T08:30:32Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hi, Did you study from the textbook for the exam? FirstChildUserIdTAG: 85490 FirstChildUserNameTAG: yashsavani FirstChildCreateTimeTAG: 2012-10-25T09:07:42Z SecondChildTAG: Nope! SecondChildUserIdTAG: 277808 SecondChildUserNameTAG: Hemanthmps SecondChildCreateTimeTAG: 2012-10-25T10:32:33Z SecondChildTAG: @Hemanthmp: i did!! but it took me almost double ur time. i spent almost an hr on the last question :( @staff, the circuit parameters i calculated agrees with every physics law out there(i cross checked) and my answers were quite close to the official ones. i just dont get how i can be wrong???? help me understand aside from that, i enjoyed it!! i feel smart!(if u can do it @ MIT, u can do it anywhere)....THANK YOU im off to week 7 HW :( SecondChildUserIdTAG: 285945 SecondChildUserNameTAG: lindalapiso SecondChildCreateTimeTAG: 2012-10-25T11:14:19Z SecondChildTAG: The answers have a rounding error margin, usually 5%. If you are outside the margin then the system will check it as wrong. SecondChildUserIdTAG: 381746 SecondChildUserNameTAG: jelizon SecondChildCreateTimeTAG: 2012-10-25T12:32:44Z FirstChildTAG: Damn, we live in a crazy world ! FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-10-25T14:06:18Z IndexTAG: 2161 TitleTAG: where we have to click after completing the exam?? HI I have completed the exam. Where should I click for final mid-term submission?? Thanks UserIdTAG: 365309 UserNameTAG: chetnasinghaldas CreateTimeTAG: 2012-10-25T08:27:03Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You don't need to click anywhere. Just save your answers! FirstChildUserIdTAG: 277808 FirstChildUserNameTAG: Hemanthmps FirstChildCreateTimeTAG: 2012-10-25T08:31:32Z SecondChildTAG: ok.thanks. SecondChildUserIdTAG: 365309 SecondChildUserNameTAG: chetnasinghaldas SecondChildCreateTimeTAG: 2012-10-25T08:32:02Z SecondChildTAG: NO ! Just push Check ! If you save them, they will not be graded.Don't waste a submission for nothing ! SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-10-25T15:36:57Z SecondChildTAG: Just look at the course progress page and your score is listed there. SecondChildUserIdTAG: 623210 SecondChildUserNameTAG: preveen SecondChildCreateTimeTAG: 2012-10-27T16:41:25Z IndexTAG: 2162 TitleTAG: Mid term timer? Will we have any timer ticking down to show us how long we got? Or do we have to keep a check of that by ourselves? UserIdTAG: 108454 UserNameTAG: Raven7281 CreateTimeTAG: 2012-10-25T06:45:37Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You will get 24 hours from your starting time. No timer will be displayed. FirstChildUserIdTAG: 260154 FirstChildUserNameTAG: jjmattam FirstChildCreateTimeTAG: 2012-10-25T06:50:55Z IndexTAG: 2163 TitleTAG: Where would the link to midterm appear? Hi guys, Does anyone know _where_ would the link to the midterm exam appear? AFAIK, there is no link to the mock/practice midterm anywhere in Courseware section. The only way I can get to it is through links on the Forum or from Course Info page. So, is it possible the exam is up and there's just no UI link for it? PS Good luck everyone when the exam does go live :) UserIdTAG: 502508 UserNameTAG: amaher CreateTimeTAG: 2012-10-25T05:30:18Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: it here buddy go check out FirstChildUserIdTAG: 124534 FirstChildUserNameTAG: srihari46 FirstChildCreateTimeTAG: 2012-10-25T05:30:47Z SecondChildTAG: Awesome! Thanks SecondChildUserIdTAG: 502508 SecondChildUserNameTAG: amaher SecondChildCreateTimeTAG: 2012-10-25T05:31:31Z FirstChildTAG: Hi. Mid term available at **course ware** just under the **Week6**. > **Regards:** asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-10-25T14:23:42Z IndexTAG: 2164 TitleTAG: IT'S HERE IT'S HERE UserIdTAG: 342603 UserNameTAG: YakovO CreateTimeTAG: 2012-10-25T05:29:23Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: where FirstChildUserIdTAG: 144115 FirstChildUserNameTAG: mitangel FirstChildCreateTimeTAG: 2012-10-25T05:29:57Z SecondChildTAG: courseware after sixth week SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-25T05:31:08Z SecondChildTAG: Hi. Mid term available at **course ware** just under the **Week6**. > **Regards:** asadbhatti42 SecondChildUserIdTAG: 329051 SecondChildUserNameTAG: asadbhatti42 SecondChildCreateTimeTAG: 2012-10-25T14:23:29Z IndexTAG: 2165 TitleTAG: Midterm Start Hello. Help me, I dont see where start midterm. In courseware materials i don't see its. Moscow time is 9:15 am, that Boston time 1:15 am. If i correctly understand exam released. Where find link to start exam. UserIdTAG: 413517 UserNameTAG: mari_safonova CreateTimeTAG: 2012-10-25T05:28:20Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: уже FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-25T05:29:40Z SecondChildTAG: Да появилось) SecondChildUserIdTAG: 413517 SecondChildUserNameTAG: mari_safonova SecondChildCreateTimeTAG: 2012-10-25T05:32:26Z IndexTAG: 2166 TitleTAG: Great text book!! Great textbook... wish i could download for free but alas.....#sigh... lol.. Hope everybody is enjoying the course so far... couple of weeks left, keep at it people.. UserIdTAG: 14531 UserNameTAG: samutr3 CreateTimeTAG: 2012-10-25T01:55:15Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 2167 TitleTAG: TO STAFF: midterm graduation question Could you please clarify how will be calculated grade percentage of midterm - per problem (i.e. if even one question in problem was failed - whole problem counts failed), or per question basis? UserIdTAG: 342603 UserNameTAG: YakovO CreateTimeTAG: 2012-10-25T01:50:35Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi YakovO, **Each Problem will have a score:** **Q1:** Value_1 Score **Q2:** Value_2 Score **Q3:** Value_3 Score **Q4:** Value_4 Score **Q5:** Value_5 Score **Each sub_problem will have the following score:** **Q1:** - Q1.1: (Value_1/N1) Score - Q1.2: (Value_1/N1) Score - Q1.3: (Value_1/N1) Score - ... - ... - Q1.N1: (Value_1/N1) Score Where Q1 Score= Q1.1 Score + Q1.2 Score +Q1.3 Score +....+Q1.N1 Score **Q2:** - Q2.1: (Value_2/N2) Score - Q2.2: (Value_2/N2) Score - Q2.3: (Value_2/N2) Score - ... - ... - Q2.N2: (Value_2/N2) Score Where Q2 Score= Q2.1 Score + Q2.2 Score +Q2.3 Score +....+Q2.N2 Score And so on... It doesn't matter if you find your correct answer in the third attempt, second or first :). **Answering your question:** *(i.e. if even one question in problem was failed - **whole problem counts failed**?* **No.** *or per question basis?* **Yes :)** P.D: Good luck ! :) FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-25T02:36:21Z SecondChildTAG: thanks Myrimit SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-25T04:27:45Z SecondChildTAG: ***Where's is the midterm paper???*** SecondChildUserIdTAG: 277808 SecondChildUserNameTAG: Hemanthmps SecondChildCreateTimeTAG: 2012-10-25T04:55:34Z IndexTAG: 2168 TitleTAG: Duracion del Mid Term Buenas noches quisiera confirmar que la fecha para presentar el mid term ha sido extendida y formular una duda al respecto. Segun lo que entiendo, es que el Mid Term inicia el 25 de octubre y finaliza el 28. Eso quiere decir que puedo ver el examen el mismo 25 y tengo oportunidad de rellenar mis repuestas hasta el 28? o, si inicio el examen el 25 solo tendre 24 horas para dar respuesta? Me surge la duda porque debo organizar mi tiempo y asi cumplir con otras obligaciones academicas para mi universidad. Agradeceria mucho mas a quienes puedan contestarme en espanol. UserIdTAG: 110802 UserNameTAG: leoblack CreateTimeTAG: 2012-10-25T01:32:00Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hola leoblack :), Sí, la fecha del Midterm Exam ha sido extendida hasta el 28 de octubre. La respuesta a tu pregunta es no, es decir, no puedes ver el examen el 25 y rellenarlo el 28. Esto es porque, si bien tienes un rango de días para rendirlo, una vez que aceptas comenzarlo, a partir del momento en que has hecho clic en "aceptar realizar el Examen",sea cual sea el día dentro de aquellos permitidos, comenzará el conteo de las 24hs (recuerda que tú debes tomar tu propio tiempo, pues, no habrá ningún clock que revise cuántas horas te quesan para terminar tus 24hs). También, hay que mencionar, que si comienzas el 28 a las 22hs, como la fecha y hora límite es el 28 a las 23:59, sólo tendras 1 hora y 59 minutos para completarlo, pues, la hora de cierre de finalización del Examen es aquella, por más que no hayas tenido tus 24hs. Por ello, sugiero que calcules bien tus tiempos y revises el horario correspondiente al Boston Time, ya que el horario estará referenciado a dicho lugar y no a tu horario local... En resumen, eliges el día para rendir. Este será entre el 25 al 28 de Octubre. A partir del momento en el que ves el Examen comienza el conteo de las 24hs, es decir, sólo tendrás 24hs para responder tus preguntas (recuerda que está estipulado para que puedas realizar el Exámen en 2 horas, sin embargo, dan 24hs por si ocurriese algún eventual). Si comienzas muy tarde el Examen, como por ejemplo el domingo, recuerda que no tendrás las 24hs, ya que el cierre del mismo está estipulado que sea ese día a las 23:59 Boston Time. Espero que esto te haya sido de ayuda. Saludos y suerte! Myriam. P.D: Te sugiero practicar el modelo de Examen que figura en el Course Info. Aquél fue exactamente el que nos tomaron a nosotros en el Curso Prototipo. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-25T02:12:36Z SecondChildTAG: Muchas gracias Myriam! SecondChildUserIdTAG: 110802 SecondChildUserNameTAG: leoblack SecondChildCreateTimeTAG: 2012-10-25T02:31:27Z SecondChildTAG: Por nada :). Mucha suerte en el Examen! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-25T02:37:38Z IndexTAG: 2169 TitleTAG: midterm can we still access our previous assignments during the midterm exam? thanks UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-25T00:44:15Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: yes! FirstChildUserIdTAG: 414516 FirstChildUserNameTAG: randevakash FirstChildCreateTimeTAG: 2012-10-25T04:50:16Z IndexTAG: 2170 TitleTAG: Old Midterm Q2 I can't seem to figure out how to approach Q2 of last year's midterm. Suggestions? UserIdTAG: 327627 UserNameTAG: CharuN CreateTimeTAG: 2012-10-24T22:34:41Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hi! think there's a thevenin equivalent instead of the black box, remember you can model what's inside it. Also remember that equivalent determines a load line in a i-v graph, where the intersection with i-axis is Vth/Rth (also known as Isc, short circuit current) and the intersection with the v-axis is Vth (open circuit voltage). So, by finding that load line you get both answers. And the key for this is the two resistives loads, that gives you 2 points (intersections of both of the i-v relationships of the resistors, with the load line). Hope it helps, and best luck in mid-term! FirstChildUserIdTAG: 220837 FirstChildUserNameTAG: Calsomus FirstChildCreateTimeTAG: 2012-10-25T01:09:32Z FirstChildTAG: i = m*v + b Find your slope m , then find b which is your base current when v = 0. Then solve when i = 0. FirstChildUserIdTAG: 375824 FirstChildUserNameTAG: KMatariyeh FirstChildCreateTimeTAG: 2012-10-25T02:33:39Z SecondChildTAG: Ah, got it. Thanks and good luck to both of you as well! SecondChildUserIdTAG: 327627 SecondChildUserNameTAG: CharuN SecondChildCreateTimeTAG: 2012-10-26T02:56:27Z IndexTAG: 2171 TitleTAG: The future of high speed digital processing? The future of high speed digital processing? http://en.wikipedia.org/wiki/Memristor UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-10-24T20:33:14Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: yes, nice device - I dream of energy-independent RAM :) FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-25T02:00:22Z IndexTAG: 2172 TitleTAG: OLD MIDTERM, Q3, THE SMALL SIGNAL PART I FIND THE SMALL SIGNAL PART OF Q3 A BIT CHALLENGING. CAN SOMEONE PLEASE POST the step by step workout? surly it wont be a breach of agreement right? UserIdTAG: 285945 UserNameTAG: lindalapiso CreateTimeTAG: 2012-10-24T20:23:36Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Take a look at this thread https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/undefined/threads/508809cce965c0230000008a Towards the end there are a couple of posts that you might find useful. FirstChildUserIdTAG: 502508 FirstChildUserNameTAG: amaher FirstChildCreateTimeTAG: 2012-10-24T21:01:01Z FirstChildTAG: see textbook page 438 example 8.5 FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-25T01:56:31Z SecondChildTAG: Thanks! SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-25T05:34:43Z SecondChildTAG: thanks SecondChildUserIdTAG: 285945 SecondChildUserNameTAG: lindalapiso SecondChildCreateTimeTAG: 2012-10-25T11:59:21Z IndexTAG: 2173 TitleTAG: OLD MIDTERM EXAM Dear staff, Is it possible not to limit the number of time to check your answers on the Old Midterm Exam? I was checking almost each answer when I realized that the number of check is limited. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-24T19:34:49Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You can check your answer by clicking the show answer button. The goal was to simulate the midterm. It's best to keep the three checks as they are now so that the midterm is not the first time you see it. FirstChildUserIdTAG: 202032 FirstChildUserNameTAG: KeithD FirstChildCreateTimeTAG: 2012-10-24T19:42:39Z IndexTAG: 2174 TitleTAG: Review for Mid-Term: Question about H2P1 Hi, Could somebody pls. explain why in H2P1 the say the following: - "We are given an input voltage Vin=50.0V, and we need to provide an open-circuit output voltage of Vout≈12.5V" - BUT THEN, in the answer, they say: "For this solution, we assume that Vin=30V,Vout=7.5V" Isn't it contradictory? what do I miss?.... thanks in advance for a breve explanation :)) And good luck 4 the exam!! Suerte!! Drücke an alle die Daumen !! Regards, Sandra UserIdTAG: 342966 UserNameTAG: SandraNavarro CreateTimeTAG: 2012-10-24T19:23:43Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: I had different initial values, but explanation is same. I think it is answer for other H2P1 version , not ours. With best regards, Serge FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-10-24T20:23:44Z FirstChildTAG: To prevent plagiarism, student's initial values are randomized. Don't take the numbers as gospel; use the explanation provided, but plug in the numbers you were given. (For Midterm review, I created an Excel workbook with often-seen formulas; I plugged in both my numbers and the numbers for the sample Midterm, for example, side-by-side to confirm that the provided numbers and "my" numbers worked the same in the solution steps provided. They did.) Hope this helps! FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-10-25T05:39:35Z FirstChildTAG: also I am facing the same problem. can any one help? FirstChildUserIdTAG: 226085 FirstChildUserNameTAG: Teto FirstChildCreateTimeTAG: 2012-10-24T21:15:34Z IndexTAG: 2175 TitleTAG: Se evalua mi progreso igual for work and personal, I have not had time to solve the course would like to know if my assessment is now beginning the same, as I have several jobs without filing and assessments not given. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-24T16:51:52Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi anonymous! :) I didn't understand your question... ---- Hola anonymous! :) Hablas Español? Te puedo ayudar? No he comprendido muy bien lo que has querido preguntar... FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-24T20:39:33Z IndexTAG: 2176 TitleTAG: Is lectures and videos still there after session Hi all It seems that I cannot complete the course on time. Can I still do this after a couple of month not for certificate but just to complete it. UserIdTAG: 612226 UserNameTAG: Unknownxxxx CreateTimeTAG: 2012-10-24T14:31:10Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: *Hope so...* FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-10-24T15:10:39Z FirstChildTAG: y not? the course would be available in nxt spring.. u may do it then and ofcourse u may earn a certificate too if u completed it on time then..:p FirstChildUserIdTAG: 368712 FirstChildUserNameTAG: jmen FirstChildCreateTimeTAG: 2012-10-24T14:34:56Z SecondChildTAG: thanks jmen for the quick reply SecondChildUserIdTAG: 612226 SecondChildUserNameTAG: Unknownxxxx SecondChildCreateTimeTAG: 2012-10-24T14:41:12Z SecondChildTAG: aha,my pleasure!! SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-24T14:55:36Z FirstChildTAG: sir i have watched 4 weeks videos bt now rest 4 week videos ar not availble at direct download link so how can i download this FirstChildUserIdTAG: 531805 FirstChildUserNameTAG: abhishekpratapsingh FirstChildCreateTimeTAG: 2012-10-24T15:10:25Z SecondChildTAG: You can download the videos for week 5 and beyond from a link that is right next to the video player. SecondChildUserIdTAG: 96452 SecondChildUserNameTAG: Lyla SecondChildCreateTimeTAG: 2012-10-24T16:52:25Z IndexTAG: 2177 TitleTAG: Old midterm : ques4 (last 4 parts) Hi I tried solving the old mid term exam. I am not getting how to solve the last 4 parts of ques 4. Please help. UserIdTAG: 365309 UserNameTAG: chetnasinghaldas CreateTimeTAG: 2012-10-24T14:07:03Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Vout<=Vol, so you have to consider what is the worst case for Vout. Power = Vs^2/R so R must be minimized and you have Ron when the gate output is low: consider what happens when (A=1,B=0) and (A=0, B=1). For the time constants you have to find out the two circuit configurations for the transitions low to high and high to low at node X: again, you have Ron only when the output is low (that is when the MOSFET is in triode region). FirstChildUserIdTAG: 376877 FirstChildUserNameTAG: AndBre FirstChildCreateTimeTAG: 2012-10-24T15:09:53Z SecondChildTAG: the time constant portion etc is not for testing this mid term.it will be covered in end term SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-24T15:42:33Z IndexTAG: 2178 TitleTAG: mid term can anyone plz post which sections of the textbook should we read thoroughly to clear the mid term well UserIdTAG: 161899 UserNameTAG: harshvit12 CreateTimeTAG: 2012-10-24T13:54:36Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Check this. [MID Term Tip][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5086869b3e78a2270000007e > **Regards:** asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-10-24T15:09:57Z IndexTAG: 2179 TitleTAG: TAs available to walk through H4P3 part B? I was able to get the answer but I still do not have a good understanding for solving the Norton equivalent for H4P3 B. I have read through the many discussions on the forum which helped but if possible I would like to see it all in one spot like one of the tutorials. Thanks! UserIdTAG: 244115 UserNameTAG: Pedro1969 CreateTimeTAG: 2012-10-24T12:57:53Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: As the deadline for this homework has passed, you should be able to click on the now-active "Show Answer" button. Does this help you? FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-10-24T13:15:30Z SecondChildTAG: I did and was hoping that would help but it didn't. Plus is there an error in that solution? How can u=(R2/(R1+R2))⋅Vo. SecondChildUserIdTAG: 244115 SecondChildUserNameTAG: Pedro1969 SecondChildCreateTimeTAG: 2012-10-24T13:25:25Z IndexTAG: 2180 TitleTAG: Small signal doubt about amplification how we are increasing swing using small signal technique ? By small signal we are adding DC level in input but data is in swing and swing remain same in small signal .If noise is added then shape will disturb and there is no protection to data(swing) with amplifier . { if we choose large portion of VI versus VO plot for amplification then share got disturb ,Swing is not increase so how to increase swing because that is the actual information } UserIdTAG: 337713 UserNameTAG: ishan65 CreateTimeTAG: 2012-10-24T12:19:12Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 2181 TitleTAG: As things continue...it gets all complicated... I have my Mumbai University , semester 5 exams coming up for Electronics Engineering staring from 1st November , but still I will try giving my Mid term exam and the rest of it..... Damn !!! My exams had to come up now... What should i do ?? UserIdTAG: 381619 UserNameTAG: siddhantmishra007 CreateTimeTAG: 2012-10-24T11:18:32Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Prioritize. Pick the most important thing, and concentrate on that. Good luck! FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-10-24T11:52:05Z IndexTAG: 2182 TitleTAG: JFET AND MOSFET? The construction of amplifiers given in textbook is concerned with a JFET or a MOSFET? UserIdTAG: 368712 UserNameTAG: jmen CreateTimeTAG: 2012-10-23T17:23:16Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 2183 TitleTAG: HW6 Problem 2 part (f) Hi! I find that there is something unusual with the answer. Shouldn't the final answer be 1/(2*K*(VIN-VT)) instead of 1/(K*(VIN - VT)). This is because the second last line of the solution is Isc = -2*gm*vout. UserIdTAG: 143823 UserNameTAG: NathanNadeson CreateTimeTAG: 2012-10-23T15:19:19Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: and here also: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/508573b08669711f0000010c FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-24T01:13:33Z FirstChildTAG: look my comment there: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5085744f6f3be62300000126 FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-23T16:33:26Z IndexTAG: 2184 TitleTAG: i have written the truth answers but is showed me that is wrong. WHY????????? why??? UserIdTAG: 374512 UserNameTAG: arsen55591 CreateTimeTAG: 2012-10-23T08:49:11Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 2185 TitleTAG: Value of resistance of MOSFET in Triode region. Staff. I got $R_{on}$ (Resistance of MOSFET in triode region) to be $\frac{2}{(VGS-VT)\times K}$. As $V_{DS}$ of the MOSFET increases from zero to $V_{GS}-V_T$ , the MOSFET behaves like a resistor, right? So at the point when it starts to behave like a current source and the $I_{DS}$ Vs. $V_{DS}$ graph becomes horizontal: $V_{DS} = V_{GS}-V_T$ and $I_{DS} = (K/2)*(VGS-VT)^2 $ and $I_{DS}*R_{on} = V_{DS}$ Solving for Ron we get $R_{on} = \frac{2}{(VGS-VT)\times K}$ or $R_{on} = 2/g_m$ Where $g_m$ is the transconductance. What do you guys think? Is this correct? UserIdTAG: 154440 UserNameTAG: Aahlad CreateTimeTAG: 2012-10-22T16:43:10Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: your question is related to the last lab right? if it is, then I think you should try to get Ron from the simulation not the equations. FirstChildUserIdTAG: 397838 FirstChildUserNameTAG: Muhamad_Alaa FirstChildCreateTimeTAG: 2012-10-22T16:54:35Z SecondChildTAG: Nope, not related to the lab, just thinking of what it should be theoretically. SecondChildUserIdTAG: 154440 SecondChildUserNameTAG: Aahlad SecondChildCreateTimeTAG: 2012-10-22T16:59:44Z IndexTAG: 2186 TitleTAG: I think the answers of H6P2(f) missed a '2'.Staff In the answer of H6P2(f) the open circuit voltage is given by ${v_{oc}=-v_{out}}$, and the short-circuit current is given by ${i_{sc}=-2 \cdot g_{m} \cdot v_{out}}$, so the Thevenin equivalent resistance is ${R_{th}= \dfrac {v_{oc}} {i_{sc}}= \dfrac {- {v_{out}}} {{-2 \cdot g_{m} \cdot v_{out}}} = \dfrac 1 {2 \cdot K \cdot (V_{IN}-V_{T})}}$ So the given answer ${ \dfrac 1 {2 \cdot K \cdot (V_{IN}-V_{T})}}$ miss a ‘2’. Is that correct? Besides, in the small-signal model, one MOSFET can replaced by a resistantor which resistance is given by ${r_d=\dfrac 1 {g_m}} $,so in this case, the Thevenin equivalent resistance is equal to $ {r_{d1} \parallel r_{d2}= \dfrac 1 {2 \cdot K \cdot (V_{IN}-V_{T})}} $ which$ {r_{d1}}$、$ {r_{d1}}$ are small-signal resistances to Q1,Q2. Can I understand the question as above?Is that corret? UserIdTAG: 176939 UserNameTAG: electron625 CreateTimeTAG: 2012-10-22T16:29:03Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: you cant replace dependent source with resistor FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-22T17:13:53Z SecondChildTAG: i agree with you yakovo but the answer definitely misses a 2 there.. just go through the explanation and help us rectify the bug!! Maybe we r wrong! SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-23T12:21:54Z SecondChildTAG: It has correct answer. Equivalent resistance of Q2 for small signal is 1/gm and it's in parallel with equivalent dependent current source from Q1, so thevenin resistance will be same as Q2 equivalent resistance 1/gm SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-23T15:00:10Z SecondChildTAG: my records for questions 5 and 6 (two variants): ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/1351011619587294.png SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-23T17:00:40Z SecondChildTAG: Thank you very much,both jmen and YakovO. I think I'm wrong.In the saturation region ,the small-signal equivalent circuit of a MosFET is VCCS. Howerver,a MosFET with its drain and gate connected together can replaced by a resistor which resistance is equal to $r=K\cdot (V_{GS}-V_T)$. YakovO:In your records, the last diagram, you turn off the current source which is the small-signal model of Q1. It is regard as an independent current source, isn’t it? But why? SecondChildUserIdTAG: 176939 SecondChildUserNameTAG: electron625 SecondChildCreateTimeTAG: 2012-10-24T12:54:07Z SecondChildTAG: r=1/(K*(VGS-VT))=1/gm; VGS=VDS => r=1/(K*(VDS-VT)) Last case - I turned off small signal ids - after you convert schematic to small signal you can work with it the same way as with large signal because it becomes linear where KVL and KCL is working. But never forget that everything you will get here will be small signal equivalent - r is not just resistance - it's an incremental resistance! SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-25T01:32:04Z IndexTAG: 2187 TitleTAG: LAB 7 Last Question Hi, I would appreciatte any help with the last question of this Lab. I have taken some assumptions, like leaving out the first curve of the current graphic, because as time goes to infinity, the average is basically what the "triangles" give us. Also I consider,after calculating one "triangle" average, and given that this part is only 20% of time, the average of the current would be a factor of .2 of the triangle average. I'm assuming wrong?, please help. UserIdTAG: 369339 UserNameTAG: vargaslen CreateTimeTAG: 2012-10-22T15:19:52Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Both of your assumptions are completely correct, if I understand them correctly. Ignore the initial transient waveform and focus on the repeating triangles, and when you take the average, you have to account for the fact that your average current only takes up 0.2 of the total period. FirstChildUserIdTAG: 390034 FirstChildUserNameTAG: kahlil FirstChildCreateTimeTAG: 2012-10-22T23:46:55Z SecondChildTAG: Thank you kahlil. I'm glad to be correct on that. SecondChildUserIdTAG: 369339 SecondChildUserNameTAG: vargaslen SecondChildCreateTimeTAG: 2012-10-23T08:37:31Z SecondChildTAG: Hi, again. Tough I keep on checking my calculations, can't get to the correct answer.At first, calculated the integral of each half of the triangle,added them up and divided by 2 to calculate the triangle average. Then realized that the contribution of each half should be multiplied by the ratio of the base size of the half triangle respect to the total base of the triangle.May be I'm taking the wrong measurements. Any ideas? Deeply thankful for them. SecondChildUserIdTAG: 369339 SecondChildUserNameTAG: vargaslen SecondChildCreateTimeTAG: 2012-10-23T16:01:58Z SecondChildTAG: I get the correct answer by using the total area of the first triangule then divided the total time from where its zero until the end of the triangle. SecondChildUserIdTAG: 529515 SecondChildUserNameTAG: Low SecondChildCreateTimeTAG: 2012-10-24T03:30:18Z SecondChildTAG: Thank you Low, Appreciated your help. SecondChildUserIdTAG: 369339 SecondChildUserNameTAG: vargaslen SecondChildCreateTimeTAG: 2012-10-24T14:57:18Z SecondChildTAG: I still didn't got dis....:( SecondChildUserIdTAG: 127800 SecondChildUserNameTAG: MNB SecondChildCreateTimeTAG: 2012-11-04T09:40:42Z IndexTAG: 2188 TitleTAG: about midterm hey could u tell me how much time wil be awarded for the midtern test nd how much questn will be dere .,,actually on 26th my dream company coming to college for placement ,,,so i'll be preparing for that on 25th ,,,plz rply asap thnx in adv. UserIdTAG: 183166 UserNameTAG: yogeshk CreateTimeTAG: 2012-10-22T10:56:39Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Please see [this thread][1]. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5072e92588c9422200000097 FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-10-22T12:10:17Z FirstChildTAG: [Course Info][1] **Mid-Term Exam:** The mid-term exam will be released on October 25th at 00:01 am Boston time. The exam is designed to take 2 hours; however, in order to compensate for any Internet or power outages in your area you will have 24 hours to finish this exam. You can start the exam when it is convenient for you, but you must complete this exam by 11:59 pm (almost midnight) Boston time on October 27th, even if that is less than 24 hours after you started the exam. The exam will cover material through Small-Signal Analysis of Transistor Circuits (end of first lecture video sequence of Week 6). The exam must be completed online. Please understand that we cannot accept submissions in any format past the deadline, or delay the deadline for individual students for to any reason. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/info ---------- we have only 3 attempts to solve any query in **MIDTERM EXAM**, [Midterm Info of (MITx Spring FALL-2012)][1] [1]: https://6002x.mitx.mit.edu/wiki/view/MidtermInfo > Regards: **asadbhatti42** FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-10-22T12:55:15Z FirstChildTAG: Good news Guys..... The deadline to submit the midterm exam is extended to 23:59 (11:59 pm) on Sunday, October 28th, Boston time. The exam will still be released on Thursday, October 25th at 00:01 (12:01 am) Boston time. FirstChildUserIdTAG: 368712 FirstChildUserNameTAG: jmen FirstChildCreateTimeTAG: 2012-10-23T16:38:30Z IndexTAG: 2189 TitleTAG: Capacitance calculator http://www.daycounter.com/Calculators/Plate-Capacitor-Calculator.phtml UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-10-22T10:32:14Z VoteTAG: 1 CoursewareTAG: Week 6 / Capacitor scaling exercise CommentableIdTAG: 6002x_capacitor_scaling_exercise NumberOfReplyTAG: 1 FirstChildTAG: Hi, hazel1919! Can you tell us, or give some links, maybe instructions how to use wolframalpha, khanacademy and s.o.other than or in addition, that have already been announced? FirstChildUserIdTAG: 96873 FirstChildUserNameTAG: Santyaga FirstChildCreateTimeTAG: 2012-10-23T18:05:22Z IndexTAG: 2190 TitleTAG: I can't understand, what are the magnetic field and field interactions It's a shame to me, but I can't figure out, what is this field... OK, it's pretty clear for an electrical field: it's principle of work is similar to the one of gravitational field. The charge generates electrical field, while the mass generates the gravitational field. But what is the magnetic field? How does it appears? What does it depend? Please, someone, give me a piece of intuition for that. P.S.: Also, as a little offtopic, can anyone give me an intuition for how do these interactions look like? How does any charged particle "knows" that there's another charged particle nearby? I've heard that there's some "interaction particles", the gluons, the gravitons and so on. They are responsible for all that staff. But I still can't imagine, how is it possible for a particle to constantly emit those "interaction particles" and not to lose it's mass or energy? Also, if this interaction particle hits another particle, how does the latter "know", what it has to do? I know, that everything in P.S. sounds weird and is at most incorrect, but it's very, very interesting for me to know that staff. UserIdTAG: 326787 UserNameTAG: Angstrem CreateTimeTAG: 2012-10-22T09:41:40Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: When you first learn about them, magnetic fields can seem very weird. They exert forces on charged particles like electric fields, but the thing that makes them weird is this: a magnetic field will exert no force on a stationary charge; for its presence to even be felt in the first place, the charge has to be moving! This is described in what is called the Lorentz force law-- the force exerted on a charge q is **F** = q **v** x **B**, where the boldfaced letters are actually vectors and the "x" operation denotes the vector cross product-- something you may or may not have learned about yet in any class that deals with vectors. Furthermore, there are no sources of magnetic "charge," like with electric fields. Magnetic fields are created instead by moving electric charges-- otherwise known as current. Whenever you have a wire with current, it is actually generating a magnetic field around the wire. All of this if you haven't learned yet, you will when you take your first physics class in Electricity and Magnetism, so don't worry. Your second question, however, is a bit more complicated because it gets into more theoretical particle physics, which isn't exactly my specialty, so I'll let anyone who knows better comment on that. FirstChildUserIdTAG: 390034 FirstChildUserNameTAG: kahlil FirstChildCreateTimeTAG: 2012-10-22T11:39:04Z SecondChildTAG: Thanks for your explanation, now it seems a bit more clear for me. So, the magnetic fields appear only when the particles are moving. And they affect only those particles, who are moving. Regarding Electricity and Magnetism class: I haven't completely understood, will it be introduced in the program of this course as a part of some future weeks? SecondChildUserIdTAG: 326787 SecondChildUserNameTAG: Angstrem SecondChildCreateTimeTAG: 2012-10-22T23:03:12Z FirstChildTAG: Static magnetic fields are caused by electric currents, i.e. moving charges. Also, time varying magnetic fields give rise electric fields and time varying electric fields give rise to magnetic fields. The "interaction particle" for the electromagnetic is the photon. The photons are virtual in the case of static fields. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-22T11:44:33Z FirstChildTAG: Surprise, there is no magnetic field. It is only interactions between electric fields of charged particles subject to relativistic effects. The field is caused by photons. It looks like, smells like and calculates like a dipole field. If a positive charged particle is moving parallel to a wire with a current flowing through it, the Lorentz contraction will make the wire appear electrically charged (google Relativistic electromagnetism) for a fairly simple explanation. FirstChildUserIdTAG: 142857 FirstChildUserNameTAG: viking2 FirstChildCreateTimeTAG: 2012-10-23T00:22:58Z IndexTAG: 2191 TitleTAG: S14E1: Multiple correct answers? In S14E1 (RESPONSE TO STEP DOWN), why is only answer C correct? Isn't answer D a valid possibility as well? If VS = 5V, then answer C is correct, but if VS = 4V, then answer D is correct. We are given that A falls in a step from 5V to 0V, but that doesn't necessarily mean that VS = 5V, right? UserIdTAG: 341293 UserNameTAG: Pietr CreateTimeTAG: 2012-10-22T08:06:50Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 2192 TitleTAG: s13e1 There seems to be no dedicated discussion for this question. I dont seem to be getting the inductance right. (640*4*PI*10^-7)*(100)^2*(PI*(1*10^-2)^2)/(10*10^-2) UserIdTAG: 204745 UserNameTAG: allwynmendes CreateTimeTAG: 2012-10-22T07:58:41Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Check your formula for the area for the cross-section of the core. FirstChildUserIdTAG: 145420 FirstChildUserNameTAG: pmac12345 FirstChildCreateTimeTAG: 2012-10-22T09:11:51Z SecondChildTAG: AHHHH damn, thanks pmac12345 :) SecondChildUserIdTAG: 204745 SecondChildUserNameTAG: allwynmendes SecondChildCreateTimeTAG: 2012-10-22T10:44:20Z IndexTAG: 2193 TitleTAG: Home work 6 extended? I missed the deadline and completed homework 6 today. However, everything looks fine, I got green ticks and the progress page says HW6=100%.... Is that normal, or the deadline is extended? UserIdTAG: 254455 UserNameTAG: Ivelin_zhekov CreateTimeTAG: 2012-10-22T07:48:38Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 6 FirstChildTAG: ...Which is strange, because it's almost Tuesday in Far Siberia, but not yet Monday on the other side in Alaska! FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-10-22T07:55:14Z FirstChildTAG: Maybe you have the wrong due date? As long as it is **before** 12:00AM Monday *anywhere in the world*, Homework and Lab are accepted, which is good for late filers like us! Here in the U.S. on the East Coast (NYC) it's 3:50AM, but Homework is still being accepted, too. That's because it's **only 9:52PM in Hawaii**, one of the last time zones before the **International Date Line** (I believe only the Aleutian Islands of Alaska are further out), so at least another 2-3 hours to submit as of now (pretty late though!). FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-10-22T07:53:45Z SecondChildTAG: 3:50AM, Mark is really burning the midnight oil. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-22T13:23:57Z FirstChildTAG: I have observed that submissions close at 9AM MIT/Boston time (which is on EDT - Eastern Daylight Time). Here in Greece the above translates to 4PM which is absolutely great as I always submit the assignments late! FirstChildUserIdTAG: 145658 FirstChildUserNameTAG: Wh1t3w0lf FirstChildCreateTimeTAG: 2012-10-22T11:29:34Z FirstChildTAG: So depending on where you live in the world, you get a certain amount of "extra" hours past Midnight at the Date Line. For example, in New York, we get about 5-6 extra hours, Hawaii 1-2 extra hours, Europe gets about 12 extra hours, Japan gets at least 18, etc. I can't be exact because of things like "Daylight Savings Time" in the U.S. (I don't know how that applies, only that it expires Nov. 1st, and we have to set our clocks an hour back, including MIT in Boston), and I don't want to wait 'till the exact last minute to find out :D FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-10-22T08:06:11Z SecondChildTAG: I'm in Bulgaria...well, that's great! SecondChildUserIdTAG: 254455 SecondChildUserNameTAG: Ivelin_zhekov SecondChildCreateTimeTAG: 2012-10-22T09:04:54Z FirstChildTAG: This was new information to me. I thought the deadline was midnight local time. Because of this I had taken the attitude that by late Sunday night on the US east coast there wasn't much reason to try and help someone because there wasn't enough time to do any good. Live and learn. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-22T13:03:59Z FirstChildTAG: in india its usually 9.30am to 1:00 pm on monday...yippee! FirstChildUserIdTAG: 368712 FirstChildUserNameTAG: jmen FirstChildCreateTimeTAG: 2012-10-22T14:36:05Z IndexTAG: 2194 TitleTAG: How does this forum work? I heard there are something like 10,000 students active. Do all of them have access to this forum? Or are there separate forums? UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-10-22T07:29:36Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I'm pretty sure it's the same forum. FirstChildUserIdTAG: 154440 FirstChildUserNameTAG: Aahlad FirstChildCreateTimeTAG: 2012-10-22T11:45:47Z SecondChildTAG: Why don't we have more people commenting then? SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-22T11:52:10Z SecondChildTAG: My guess is that many can do the work without help, so they don't need the forum. And on the other hand, they don't feel the need to offer help to others. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-22T12:22:10Z SecondChildTAG: I think Skyhawk is right. SecondChildUserIdTAG: 154440 SecondChildUserNameTAG: Aahlad SecondChildCreateTimeTAG: 2012-10-22T16:53:02Z FirstChildTAG: I think separate forum for each course in **edX**. > **Regards:** asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-10-22T12:49:40Z IndexTAG: 2195 TitleTAG: Transcript The unit for capacitance is the farad. This can be expressed also as a "coulomb per volt", not a coulomb or volt. Volt is the unit of voltage. Coulomb is the unit of charge. Q = VC. Therefore C = Q/V. UserIdTAG: 164898 UserNameTAG: jbparkes CreateTimeTAG: 2012-10-21T21:30:00Z VoteTAG: 1 CoursewareTAG: Week 6 / Capacitor basics CommentableIdTAG: 6002x_capacitor_basics NumberOfReplyTAG: 1 FirstChildTAG: That is an important point. FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-10-22T09:11:55Z SecondChildTAG: Ah, the Prof says per, the translator says or. SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-22T09:20:35Z IndexTAG: 2196 TitleTAG: The Practical Side... **Please add your own experience to this thread...** Here are some things I have learned from the practical side... (1) - I am a big advocate of Weller TCP (Temperature controlled pencil) Soldering irons, they, for me, are singly the most important tool for physically building electronic prototype circuits. You can purchase one for cheap off Ebay if you look hard. Also, try and get a selection of tips, 9, 8 and 7 tips. Just use lead free solder and distilled water on your sponges, those tips can be expensive and you don't want mineral buildup! (2) For me, I use http://uk.farnell.com/ and http://uk.rs-online.com/web/ for most of my component orders and if they don't have it eBay will! (3) Eagle is by far the easiest PCB design software to download and use, just go here to get the full version: http://www.youtube.com/watch?v=PVbbJaGxmHQ (4) - If you are intending to do soldering (why wouldn't you be) build your self a solder fume extractor with a PC fan and some ventilation duct. Solder flux is really bad for you. Also, put a square of aluminum sheet over the part of table you are soldering on. (5) - Be well stocked with tools, needle nosed pliers - wire strippers - a good set of Drills - all sorts of screw drivers. Electronics is very hands on sometimes! (6) - Have a nice large work area, not easy for some, but it really does help!! (7) - Get your hands on an oscilloscope! I have a HAMEG 203-7, how about you? I'm sure you can add to the list... Hazel, UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-10-21T19:42:31Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Great speech Are you dutch ? What's your degree ? it seems that you are very interested in electronics FirstChildUserIdTAG: 382505 FirstChildUserNameTAG: AhmedGalal2 FirstChildCreateTimeTAG: 2012-10-21T21:22:56Z SecondChildTAG: No degree yet! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-22T13:52:46Z FirstChildTAG: If you buy common parts in bulk to stock your parts bin, then I suggest Futurlec. Their prices on a lot of common components are very good. Their shipping is slow, but I've made about 30 orders from them and never lost a one. I'm not yet into making my own circuit boards. I like to build on strip board. I've gotten quite good at laying out reasonably compact circuits on it. My interest is PIC micro controllers, but it has been a while since since I build anything. Right now I would like to find time to put together a motor controller for a simple robot ... H-bridge plus PWM speed control. PS I recently bought A Weller WP25 plus an extra ST7 tip. It replaces a junky old Radio Shack soldering iron that I nevertheless could make quite good joints with. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-21T23:37:53Z FirstChildTAG: Eagle is kind of complicated for beginners.You really have to know what you are looking for in the library, and it counter-intuitive. It is perfect though for PCBs. FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-11-22T07:52:40Z IndexTAG: 2197 TitleTAG: Regarding homework!! CHECKING WRONG ANSWERS Due to computational mistakes,,i checked wrong answers in homework and got them wrong!! But then i again 'checked' correct answers(before the deadline). In the progress page, it is showing 100% for that homework..what will they consider for grading? the wrong answer that i "checked" first? UserIdTAG: 489065 UserNameTAG: amiths CreateTimeTAG: 2012-10-21T18:24:55Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: no ... Final answers are counted :) FirstChildUserIdTAG: 454609 FirstChildUserNameTAG: Alwahsh FirstChildCreateTimeTAG: 2012-10-21T18:30:19Z FirstChildTAG: You can check however many times you want without affecting your grade as long as you finally got the right answer. FirstChildUserIdTAG: 154440 FirstChildUserNameTAG: Aahlad FirstChildCreateTimeTAG: 2012-10-21T18:30:54Z SecondChildTAG: and you can check wrong answers after you was graded for right answer - it will not change your summary - I use this feature to help ppl find where they wrong with formulas :) SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-21T18:47:29Z SecondChildTAG: Oh, I didn't know that. Useful trick. SecondChildUserIdTAG: 154440 SecondChildUserNameTAG: Aahlad SecondChildCreateTimeTAG: 2012-10-21T18:53:30Z FirstChildTAG: Thanks a lot:) but are u sure people? FirstChildUserIdTAG: 489065 FirstChildUserNameTAG: amiths FirstChildCreateTimeTAG: 2012-10-21T19:25:39Z IndexTAG: 2198 TitleTAG: h6p3......q5 and q6 help please? UserIdTAG: 462861 UserNameTAG: edxforme CreateTimeTAG: 2012-10-21T18:23:20Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Do you know the decay time constant for a normal network? (Normal=Thevenin) In case you don't its given by C*R where C is the capacitance and R is the Thevenin Resistance of the network. So here, find the Thevenin Resistance for each network and multiply with capacitance to get time constant. Hope this helps. :) Note: I'm not sure what exactly the honor code allows, could anyone tell me whether the answer I gave is allowed by the honor code? FirstChildUserIdTAG: 154440 FirstChildUserNameTAG: Aahlad FirstChildCreateTimeTAG: 2012-10-21T18:35:36Z SecondChildTAG: It's close, I suppose he still has to figure out the Thevenin equivalent and how to enter it correctly. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-21T18:41:48Z SecondChildTAG: Thanks. So what would be a perfect answer with regards to the honor code? Just trying to get the hang of it. P.S. How do you go to the next line? When typing I click enter but it ends up like the post above with all the sentences together. SecondChildUserIdTAG: 154440 SecondChildUserNameTAG: Aahlad SecondChildCreateTimeTAG: 2012-10-21T19:04:19Z SecondChildTAG: Try Hitting 3 spaces before enter SecondChildUserIdTAG: 442070 SecondChildUserNameTAG: MuhammadAsad SecondChildCreateTimeTAG: 2012-10-21T20:19:08Z SecondChildTAG: Lemme check and it works! Thanks :) SecondChildUserIdTAG: 154440 SecondChildUserNameTAG: Aahlad SecondChildCreateTimeTAG: 2012-10-22T11:25:11Z IndexTAG: 2199 TitleTAG: lab 6 . 3 HELP!!!!!!!!!!! -RTH*CGS*ln((VC-VTH)/(VS-VTH)),VC=VOL = 0.250V Why this formula dosen't work?((( I get X.XXXXX*10^-10 UserIdTAG: 192114 UserNameTAG: chuba CreateTimeTAG: 2012-10-21T17:27:09Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: express it in nano seconds FirstChildUserIdTAG: 489065 FirstChildUserNameTAG: amiths FirstChildCreateTimeTAG: 2012-10-21T18:26:28Z SecondChildTAG: Not Help SecondChildUserIdTAG: 192114 SecondChildUserNameTAG: chuba SecondChildCreateTimeTAG: 2012-10-21T18:32:40Z SecondChildTAG: The answer is in nanoseconds, not voltage as stated by Amiths . All you need is the TRANS analysis, no formula, just simple subtraction and division based off the graph. "Run a 50ns transient simulation on the ring oscillator and measure the period of oscillation." < (Time for peak to peak) "Divide the result by 9 to get an estimate....." < (Divide the peak to peak time by 9) Now you are done! SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-21T18:55:05Z FirstChildTAG: I was having the same trouble, turns out I had rounded VTH too much (though it accepted the answer in part 2a) and the resulting value was off just enough to not agree with the checker. Possibly check your VTH and input again. FirstChildUserIdTAG: 238627 FirstChildUserNameTAG: Dwilds FirstChildCreateTimeTAG: 2012-10-22T00:25:37Z FirstChildTAG: Make sure that CGS = 0.0002. FirstChildUserIdTAG: 457034 FirstChildUserNameTAG: Ichihara FirstChildCreateTimeTAG: 2012-10-22T00:44:43Z IndexTAG: 2200 TitleTAG: how to navigate in the book ? how to directly go to a certain page in the book ? UserIdTAG: 214085 UserNameTAG: shohin CreateTimeTAG: 2012-10-21T17:09:39Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I found the links given in some posts are usually about 25 pages more than the "page" number, probably due to the preface, TOC, etc. You can copy and paste a given link and change the last three numbers in the URL (accounting for the difference I mentioned above) to go to any page you want. You can also use the on-line reader, works like a Kindle..... FirstChildUserIdTAG: 375500 FirstChildUserNameTAG: Brej7665 FirstChildCreateTimeTAG: 2012-10-21T17:22:10Z FirstChildTAG: Hi shohin: You can go to the **Wiki-> Scrolling Textbook Viewer** (ashwith posted this, it is really useful). If you press "g", it will appear a window, where you can write the page and it will re-direct you to that page :). ![image][1] Also, you can go to the Textbook, and if you want a precise page, just sum 24 to that number, eg., if you want to read page 24, just write 48 like this: ![im2][2] I hope this can help you. See you, Myriam. [1]: https://edxuploads.s3.amazonaws.com/13508552631343658.png [2]: https://edxuploads.s3.amazonaws.com/13508556091343646.png FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-21T21:40:49Z IndexTAG: 2201 TitleTAG: Myrimit please help Hey Myrimit, Can you please give some hints for question 5 and 6 of problem 2 of week 6... My answer to the sixth question is (edited - please do not post actual answers for graded assignments). but the system is telling me it cannot parse it UserIdTAG: 308861 UserNameTAG: Ignaas CreateTimeTAG: 2012-10-21T17:04:48Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I just copied and pasted the formula above and it parsed correctly even though it wasn't the correct answer for the question I used as a test..... Usual problem seems to be missing multiplication (*) signs or missing (unmatched) parentheses.... FirstChildUserIdTAG: 375500 FirstChildUserNameTAG: Brej7665 FirstChildCreateTimeTAG: 2012-10-21T17:19:19Z SecondChildTAG: hmmm maybe i have a bug SecondChildUserIdTAG: 308861 SecondChildUserNameTAG: Ignaas SecondChildCreateTimeTAG: 2012-10-21T17:30:23Z FirstChildTAG: Hi Ignaas, Can I help you? Where are you lost? Ok, let me some minutes, I will post some hints and back to this post. Myriam. ---- Edit: **Part 4 - Hint:** What kind of circuit it is the one that you are viewing in H6P2? Isn't it an inverter? Hmmm... and what input do you have in the small signal? and if it is an inverter, what you will have in your output? What would happen with your input? What does an inverter? What happens with the signs? **Part 5 - Hint:** Take a look at page 417 [here][1]. What happens in the small signal with your Q2? Take a look at Figure 8.11, Figure 8.12, Figure 8.13. ;). [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/441 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-21T21:27:20Z SecondChildTAG: finding Rth and Vth.....for me vout=vin...my gm is K(VIN-VT) SecondChildUserIdTAG: 308861 SecondChildUserNameTAG: Ignaas SecondChildCreateTimeTAG: 2012-10-21T23:00:19Z SecondChildTAG: hmmm your hint on part 4 is a bit vague...where an i find the parts about inverters again?? SecondChildUserIdTAG: 308861 SecondChildUserNameTAG: Ignaas SecondChildCreateTimeTAG: 2012-10-22T00:47:27Z SecondChildTAG: thnx Myrimit....i have got it right.. SecondChildUserIdTAG: 308861 SecondChildUserNameTAG: Ignaas SecondChildCreateTimeTAG: 2012-10-22T01:08:26Z SecondChildTAG: Well done Ignaas! ;). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-22T01:22:03Z IndexTAG: 2202 TitleTAG: Hello Sri Lankans!! Is there anyone who follow this course from Sri lanka? UserIdTAG: 268444 UserNameTAG: Marlonabeykoon CreateTimeTAG: 2012-10-21T14:43:49Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 2203 TitleTAG: lab 6 plz give me hint in part 4.. UserIdTAG: 114864 UserNameTAG: Vikaash CreateTimeTAG: 2012-10-21T11:03:08Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: check the method given in the text at page 532 and 533. FirstChildUserIdTAG: 82651 FirstChildUserNameTAG: Ayazkazi FirstChildCreateTimeTAG: 2012-10-21T12:02:49Z FirstChildTAG: Hi Vikaash! Can I help you? You can take a look at Lab6 Hints [here][1] Visual Hint: Take a look at the third curve of figure 10.16 [here][2] . Can you find your tpd,0->1 ? ;) ![img][3] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Ring_Oscillator/threads/5082c6039e78031f000000c0 [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/550 [3]: https://edxuploads.s3.amazonaws.com/13508247331343602.png FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-21T13:06:33Z SecondChildTAG: @Myrimit i have understood the above concept from your previous post in the forum and got all the correct answers except 4 and 5 parts. i have taken help from textbook as mentioned and i got correct answer in part 3 but not in 4 and 5 part. and **plz explain 4th part **. i am very thankfull to you if you can help me SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-10-21T13:39:43Z SecondChildTAG: please tell me what is the value of VOUT SecondChildUserIdTAG: 550401 SecondChildUserNameTAG: revathisingh SecondChildCreateTimeTAG: 2012-10-21T17:27:22Z IndexTAG: 2204 TitleTAG: The algebraic expression always could not be sparsed Could not parse <....deleted....> as a formula. What happen to the expression? Any one could answer me ? UserIdTAG: 128276 UserNameTAG: ChengBin CreateTimeTAG: 2012-10-21T08:31:06Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: capital K FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-21T08:51:24Z SecondChildTAG: thanks Yakov0 SecondChildUserIdTAG: 128276 SecondChildUserNameTAG: ChengBin SecondChildCreateTimeTAG: 2012-10-21T09:01:24Z FirstChildTAG: use capital K.. FirstChildUserIdTAG: 114864 FirstChildUserNameTAG: Vikaash FirstChildCreateTimeTAG: 2012-10-21T08:51:27Z SecondChildTAG: Thanks SecondChildUserIdTAG: 128276 SecondChildUserNameTAG: ChengBin SecondChildCreateTimeTAG: 2012-10-21T09:01:36Z FirstChildTAG: Hi. **ChengBin** I requested to you that please avoid from uploading the exact answers of HW Questions.... Thnx.... FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-10-22T11:48:35Z IndexTAG: 2205 TitleTAG: Group Study Hi to all guys .. this is Fayzan Ahmed from Pakistan. kindly tell me are some of you doing "group study online" . If so then add me in too My id : fayzan_007@yahoo.com Skype id : fayzan_ahmed172 any individual can also add or send me his/her id so that we could do some group study. Best of luck to all .. :) Mitx Rockx UserIdTAG: 146930 UserNameTAG: fayzan_007 CreateTimeTAG: 2012-10-21T08:29:03Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi, *fayzan* me also from **PAKISTAN**.... FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-10-21T11:00:10Z IndexTAG: 2206 TitleTAG: Ready for the midterm Hi to all . .. Ready for the midterm . . "Best of Luck to All" :) UserIdTAG: 146930 UserNameTAG: fayzan_007 CreateTimeTAG: 2012-10-21T08:24:37Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi,i am new in this course, so not for midterm now then to study all lectures.please guys assist me to get along with you. FirstChildUserIdTAG: 469133 FirstChildUserNameTAG: olubori FirstChildCreateTimeTAG: 2012-10-21T13:05:28Z IndexTAG: 2207 TitleTAG: H6P2 Q5: VIN or vin invalid input I am unable to get my answer right for H6P2 Q5. No matter what I put (even an integer) in the field it says Vin is an invalid input. does any body have the same problem? UserIdTAG: 244670 UserNameTAG: ravikiran1201 CreateTimeTAG: 2012-10-21T00:14:23Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi ravikiran, The statement says "Write an algebraic expression for this vout in terms of the parameters **K** (K uppercase)and **VT** (V uppercase T uppercase), the bias voltage **VIN** (V uppercase IN uppercase), and the incremental input voltage **vin** (v lowercase in lowercase)" You said that "in the field it says **Vin** is an invalid input" and , if you see the statement they **ask you to write** the equation **in terms of** **K,VT,VIN** and **vin,** and your **Vin** (V **uppercase** in **lowercase**) it is **not a valid parameter** , because they are asking you to be in terms of vin, VIN, VT and K only ... - this 4, can appear or not in the answer ;) - FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-21T01:41:44Z SecondChildTAG: Thank you Myrimit. I totally agree. unfortunately i am getting this message for everything I enter in the field.If my answer was incorrect I guess should not say "Vin is an invalid input". SecondChildUserIdTAG: 244670 SecondChildUserNameTAG: ravikiran1201 SecondChildCreateTimeTAG: 2012-10-21T02:21:06Z SecondChildTAG: Hmmm..., that is strange... I had no problems with entering the answer in that question ... Do you have an updated version of Chrome or Firefox? Have you tried to Log Out and then Log In? or refresh the page with F5? If this persist you should contact to the edX Staff... SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-21T03:12:27Z SecondChildTAG: try to cleanup browser cache SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-21T09:04:48Z IndexTAG: 2208 TitleTAG: H6P2 Logic Behind Phase inverter Could someone please take the time to explain ... the small signal phase inverter analysis? I have got all the correct answers by reading the discussion, but I do not get the logic behind it, and it has no values to know the answers and not the logic. I'm missing it. Would it be possible, to do it , without violating the code of honor, pretty please?! UserIdTAG: 64512 UserNameTAG: OZSorescu CreateTimeTAG: 2012-10-20T23:03:09Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I think you might have to wait until the homework deadline has passed. I'm not sure it could be fully explained without revealing the answers. You have the correct answers, so there is no reason not to wait. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-21T02:06:35Z IndexTAG: 2209 TitleTAG: LAB6 Part 1 I don't get it that which color line is input and which color line is output to find RON ??? IF Blue line is input so for any value of input we get constant output voltage i.e VOL ... Am i thinking right ?? PLease help !!!! UserIdTAG: 145239 UserNameTAG: Maheenjd CreateTimeTAG: 2012-10-20T18:14:00Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi Maheenjd! If you see in the circuit, you will see different voltages probes: ![imag][1] - The $\color{blue}{blue}$ one, it is in the output, so the blue curve will correspond to the output. - The $\color{magenta}{magenta}$ one, it shows you the VOH level. So, the magenta line, It is only there in order that you can have see it in your plot and have some reference of where is that VOH level. - The $\color{cyan}{cyan}$ one, it shows you the VOL level. So, the cyan line, It is only there in order that you can have see it in your plot and have some reference of where is that VOL level. The input it is not plot there, but if you like you can add and extra voltage probe, just at the right of the Vin in sandbox and see it :). I hope this can help you . Myriam. [1]: https://edxuploads.s3.amazonaws.com/13507412791343667.png FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-20T18:43:54Z SecondChildTAG: Thanks! I was wondering about that! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-20T21:31:51Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-20T22:48:46Z IndexTAG: 2210 TitleTAG: H6P2 - solving for Vth, conflicting dependent current sources? Hi, I tried to draw a small signal circuit for this phase inverter and I get two dependent current sources in series, each with $i = g_m v$. The $v_o$ node (also $V_{th}$) is between them, but as I try to solve the circuit, I see that $v$ for one of them is tied to 0, but for the other it is the $v_i$ input signal. So the currents conflict, but they must be equal because of KCL. I don't get it... any hint on what I'm missing? thanks, Rob UserIdTAG: 468623 UserNameTAG: RobNik CreateTimeTAG: 2012-10-20T15:58:23Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: You just need to read pages 416 and 417 to solve the problem FirstChildUserIdTAG: 239608 FirstChildUserNameTAG: guillegf84 FirstChildCreateTimeTAG: 2012-10-20T19:17:39Z FirstChildTAG: Hi Rob! Hint 1: Are you sure that they are in series? What happens with a constant value of voltage in small signal? Isn't it a ground? So what happens with VDD in your circuit? Hint 2: So, based on Hint 1..., How are set Q1 and Q2 ? Edited: Can you make the small signal model including Q1 and Q2 in the same model? :p I hope this little hints can help you. I will be here if you need more help. See you! Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-20T17:10:10Z SecondChildTAG: Thanks, I see what you mean, they are not really in series. But I'm still confused. There is no small signal current in the upper MOSFET/VCCS, because it's gate/input is tied to ground. (Right?) And there is current in the lower one, because it's gate is $v_{i}$. They say to get the Thevenin voltage you take the open circuit voltage. But I still don't see how this circuit functions, because if there is a current through the lower fet/vccs, it has no where to come from -- in an open circuit it would be infinite voltage at $v_o$. ? Rob SecondChildUserIdTAG: 468623 SecondChildUserNameTAG: RobNik SecondChildCreateTimeTAG: 2012-10-20T17:35:47Z SecondChildTAG: Gate of Q2 isnt tied to the GND in the example schematic.But "tied" to gnd when you do try to find Thevenin's values. It is your mistake here. Resulting equation is crazy simple.You need to place exact sign and nothing more. SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-10-20T17:47:00Z SecondChildTAG: Doh. I see my brain misfire... just because Q2 gate is tied to ground doesn't mean $v_{gs} = 0$. I was thinking of a voltage as an absolute value, as opposed to it being a value "between" two points -- in this case the gate and source. Thanks everyone! SecondChildUserIdTAG: 468623 SecondChildUserNameTAG: RobNik SecondChildCreateTimeTAG: 2012-10-20T20:00:31Z FirstChildTAG: Q2 has Vgs=Vds,and gm1=gm2, id1=id2. You need write very simple equation and result will surpise you. I hope it will help you WBR FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-10-20T17:15:49Z IndexTAG: 2211 TitleTAG: H6P2 i have tried a lot in last two part of H6P2..i have read a lot in discussion forum but couldn't got it...please help.. UserIdTAG: 114864 UserNameTAG: Vikaash CreateTimeTAG: 2012-10-20T15:56:10Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hi Vikaash! :) How are you? Can I help you? Hint: remember that in the small signal a contast voltage source behaves like a ground, so How will behave your VDD? See Textbook page [here][1], Figure 8.10 (small signal equivalent models). Another Hint: Take a look at page 417 of the Textbook [read an example here][2]. So, How does your Q2 behaves? :). Another Hint: Can you make a small signal models for both, Q1 and Q2? [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/440 [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/441 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-20T16:09:33Z SecondChildTAG: I have read your comments and I got the right results, but do not quite understand the solution... SecondChildUserIdTAG: 239608 SecondChildUserNameTAG: guillegf84 SecondChildCreateTimeTAG: 2012-10-20T19:01:48Z FirstChildTAG: Sorry( But I got problem... For Q1 Ids=K*(Vdd-Vt)vds ------------------------ For Q2 Ids=K*(Vin-Vt)vin ------------------------ and I know that Idsq1=Idsq2 K*(Vin-Vt)vin=K*(Vdd-Vt)vds --------------------------- I have problem with Vdd. If I took min Vdd it incorrect.... PLz HELP FirstChildUserIdTAG: 192114 FirstChildUserNameTAG: chuba FirstChildCreateTimeTAG: 2012-10-21T17:52:01Z IndexTAG: 2212 TitleTAG: Lab 6 The second circuit connection is not responding when i click it. UserIdTAG: 137438 UserNameTAG: lummydee CreateTimeTAG: 2012-10-20T10:36:49Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi lummydee! Can I help you? Try with 50n TRAN time :) , also you can find Lab 6 hints [here][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Ring_Oscillator/threads/5082c6039e78031f000000c0 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-20T15:49:22Z IndexTAG: 2213 TitleTAG: H6P2: PHASE INVERTER how to find minimum vIN....want some hint...... UserIdTAG: 114864 UserNameTAG: Vikaash CreateTimeTAG: 2012-10-20T09:55:02Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You recall what the regions of operation of a MOSFET are, correct? For Q1 to stay in saturation, we must have vGS > VT satisfied at all times. For this case, our vGS is vIN. And if vIN has the potential to swing up or down by a known value Dv, what should we set it to (at a minimum) so we are in the saturation region at all times? FirstChildUserIdTAG: 390034 FirstChildUserNameTAG: kahlil FirstChildCreateTimeTAG: 2012-10-20T12:10:34Z SecondChildTAG: Thanks SecondChildUserIdTAG: 362299 SecondChildUserNameTAG: anandbaskaran SecondChildCreateTimeTAG: 2012-10-20T17:25:32Z IndexTAG: 2214 TitleTAG: Graph plotting? Can anyone suggest any graph plotting software to solve exponential equations faster? UserIdTAG: 108454 UserNameTAG: Raven7281 CreateTimeTAG: 2012-10-20T08:54:45Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: http://www.google.com/search?hl=en&q=y%3Dsin%28x%29 FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-20T09:36:59Z IndexTAG: 2215 TitleTAG: Where is 2*pi? Sorry, but I got confused with formulas for H6P3, where I have to write down expression for time constrains. All my previous teachers told me that (tau=2*pi*R*C), and it was valid recharge/discharge/charge time of RC-circuit, but in the HW, formula with 2*pi is invalid can someone tell me where is 2*pi? UserIdTAG: 210380 UserNameTAG: MurdocRus CreateTimeTAG: 2012-10-20T08:00:50Z VoteTAG: 1 CoursewareTAG: Week 6 / simple circuit with resistor and capacitor CommentableIdTAG: 6002x_Simple_RC_Circuit NumberOfReplyTAG: 2 FirstChildTAG: tau = R*C is the general rule. This is the time it takes to charge / discharge approx. 66%. FirstChildUserIdTAG: 270011 FirstChildUserNameTAG: SnowmanZA FirstChildCreateTimeTAG: 2012-10-20T08:39:55Z SecondChildTAG: 1-1/e=63.2% SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-20T08:57:10Z FirstChildTAG: omega=2*pi*f http://en.wikipedia.org/wiki/Angular_frequency probably you confused tau with cut of frequency for Hz frequency cut off: f=1/(2*pi*tau) for angular frequency: omega= 2*pi* 1/(2*pi*tau) = 1/tau FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-20T09:19:18Z SecondChildTAG: Yep, i know this, I'm graduating this year, as an engeneer:) Got confused, because i belive that in frequency domain the cutoff frequency is much more usefull than the frequency, where an edge of AMplitude-Frequency Characteristics starts SecondChildUserIdTAG: 210380 SecondChildUserNameTAG: MurdocRus SecondChildCreateTimeTAG: 2012-10-20T09:44:55Z SecondChildTAG: Not quite often you measure passband/stopband of filter with angular frequency:) SecondChildUserIdTAG: 210380 SecondChildUserNameTAG: MurdocRus SecondChildCreateTimeTAG: 2012-10-20T09:47:03Z SecondChildTAG: Anyway i believe that this is language trouble, in Russia we mean tau is 2piRC, so Fcut=1/tau, and in America it is a bit different, but have a same sence SecondChildUserIdTAG: 210380 SecondChildUserNameTAG: MurdocRus SecondChildCreateTimeTAG: 2012-10-20T09:55:24Z SecondChildTAG: такого не может быть - RC и 2piRC не инвариант - оно будет давать разные результаты при вычислениях. Я подозреваю что в формуле для частоты среза немного другая тау - просто исходят из того, что 2pi константа. SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-20T10:39:48Z SecondChildTAG: Яков все проще. Постоянная времени схемы при нулевых начальных условиях равна RC где сопротивление для А схемы r1+r2 для B как при паралельном. Умножил на С и получил ответы. Ты сделал H6P2? SecondChildUserIdTAG: 214324 SecondChildUserNameTAG: KostyaPunda SecondChildCreateTimeTAG: 2012-10-20T13:28:52Z SecondChildTAG: Вообщето мы не об этом. Например вот здесь утверждается что частота среза f=1/tau http://ru.wikipedia.org/wiki/%D0%9F%D0%BE%D1%81%D1%82%D0%BE%D1%8F%D0%BD%D0%BD%D0%B0%D1%8F_%D0%B2%D1%80%D0%B5%D0%BC%D0%B5%D0%BD%D0%B8 хотя на самом деле это формула для циклической частоты, если за тау принимать RC - в англоязычном варианте той же статьи все кошерно: f=1/(2*pi*tau) сделал SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-20T13:50:08Z SecondChildTAG: I can find the decay time IN H6P3.... Any one guide me SecondChildUserIdTAG: 37887 SecondChildUserNameTAG: Siddhu SecondChildCreateTimeTAG: 2012-10-20T14:54:16Z SecondChildTAG: просто логично что частота=1/время SecondChildUserIdTAG: 210380 SecondChildUserNameTAG: MurdocRus SecondChildCreateTimeTAG: 2012-10-25T07:22:15Z IndexTAG: 2216 TitleTAG: textbook hello i juste want to ask about the textbook , because i don't now how to choise the suitable pages for each week . thanks UserIdTAG: 651832 UserNameTAG: adnane235 CreateTimeTAG: 2012-10-20T00:31:49Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: https://www.edx.org/static/content-mit-6002x/handouts/at-a-glance.9674fe7f677e.pdf FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-20T01:12:23Z SecondChildTAG: can you explain this plan for me please , what we mean by HW ? Readings?? thanks SecondChildUserIdTAG: 651832 SecondChildUserNameTAG: adnane235 SecondChildCreateTimeTAG: 2012-10-20T23:57:21Z FirstChildTAG: HW = Homework exercise, the weekly homework assignment you must complete by the deadline. Lab = Laboratory exercise, the weekly lab assignment you must complete by the deadline. Readings = Sections of the textbook that you should read each week before attempting to work on the homework and labs. You should read these along with viewing the courseware videos and tutorials. The readings which are underlined in the syllabus are especially important, they contain key concepts and information that you will need often, they are intended to build your intuition about the material. I suggest reading material ahead at least one week of the actual course week. You will often find that the information you need to solve this week's homework and lab is in the following week's readings. FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-10-21T00:13:04Z SecondChildTAG: thank you SecondChildUserIdTAG: 651832 SecondChildUserNameTAG: adnane235 SecondChildCreateTimeTAG: 2012-10-21T15:18:10Z IndexTAG: 2217 TitleTAG: LAB 8 Circuit 3 I got the green answer, but I doubt. The circuit contains a non-linear R? UserIdTAG: 464744 UserNameTAG: attache CreateTimeTAG: 2012-10-19T22:56:53Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 2218 TitleTAG: H6P2 thevenin voltage and resistance Anybody please help me in this question. give some hints to solve. UserIdTAG: 242676 UserNameTAG: LinjharaSahil CreateTimeTAG: 2012-10-19T19:21:00Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: The key with this question is being able to draw the small-signal circuit model. In this problem, we use our typical small-signal model of the MOSFET, which assumes it acts like a voltage-controlled (by vgs) current source, with the small-signal current given by gm*vgs. So when we draw the small signal circuit, we do two things: replace the devices with their small signal models, and do the same for all the independent sources. That means, in addition to the MOSFETs, replace the vIN with its small-signal component, vin, and short out the DC power supply VDD to ground. Once we do this and end up with our equivalent small signal model, sub in for the "vgs" of both MOSFETs with what it should be based on the circuit's topology. From the simplified circuit, it should be much easier to solve for both the open circuit voltage (this is just solving for vout-- do you see why)? and the short circuit current (inject a test current i_test and solve for the ratio of vout to i_test) FirstChildUserIdTAG: 390034 FirstChildUserNameTAG: kahlil FirstChildCreateTimeTAG: 2012-10-19T19:38:01Z SecondChildTAG: Should I get the resistance of Q1 and Q2? And the value of VDD here is equal to minimum value? any hints.... SecondChildUserIdTAG: 182624 SecondChildUserNameTAG: Daoling SecondChildCreateTimeTAG: 2012-10-20T03:10:25Z SecondChildTAG: No, you do not have to somehow derive the small-signal resistances of the MOSFETs. Just leave them in as independent current sources-- you can still solve the circuit just fine and get the answers you need. And the value of VDD is zero here-- when we do small signal analysis we short out all DC sources. SecondChildUserIdTAG: 390034 SecondChildUserNameTAG: kahlil SecondChildCreateTimeTAG: 2012-10-20T12:18:09Z SecondChildTAG: ya got it.. thanks kahlil SecondChildUserIdTAG: 242676 SecondChildUserNameTAG: LinjharaSahil SecondChildCreateTimeTAG: 2012-10-20T13:53:42Z IndexTAG: 2219 TitleTAG: Correct way to parse equation When using exponential's, try parsing the equation similar to: (z*e^(((-t)/(x*y)))). There may be more or less parentheses, but when looking at the results (below the answer box), it looked correct and I got the correct answer (using the correct variables and proper equations, of course). Basically one has to enclose the exponent in parentheses and have the minus sign inside those parentheses. UserIdTAG: 132828 UserNameTAG: rharris CreateTimeTAG: 2012-10-19T16:21:06Z VoteTAG: 1 CoursewareTAG: Week 6 / First-order capacitor exercise CommentableIdTAG: 6002x_first-order_capacitor_exercise NumberOfReplyTAG: 0 IndexTAG: 2220 TitleTAG: mid term and discussion forum Hi everyone! During the mid terms, apart from the discussion forum, are their any other things that will be disabled.. I mean will we be able to access tutorials, lecture sequences, home-works, the online books, wikis etc during this period? please help! thanx! **!!!BTW GOOD LUCK EVERYONE!!!** UserIdTAG: 368712 UserNameTAG: jmen CreateTimeTAG: 2012-10-19T13:34:08Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Assuming that this class will be the same as the last class, everything will still be accessible during the mid-terms. That includes the wiki, the book, homework, lectures, etc. The forum will not be disabled, you just aren't allowed to use it to discuss the midterm. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-19T15:50:01Z SecondChildTAG: thanks ,,,n will there be limit on no. of checks during the xams.?? SecondChildUserIdTAG: 183166 SecondChildUserNameTAG: yogeshk SecondChildCreateTimeTAG: 2012-10-19T17:19:14Z SecondChildTAG: There is a limit of three checks you can use. The trick is that the three tries are for all the boxes in a question. So if there is a question that has 5 boxes to fill in, it will check all the boxes in that question when you hit the "check" button. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-19T17:24:57Z SecondChildTAG: thnx for the reply,, is there a save button along with check button in exam? SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-19T17:51:21Z SecondChildTAG: Yes, there is. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-20T16:08:30Z SecondChildTAG: Good news Guys..... The deadline to submit the midterm exam is extended to 23:59 (11:59 pm) on Sunday, October 28th, Boston time. The exam will still be released on Thursday, October 25th at 00:01 (12:01 am) Boston time. SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-23T16:32:29Z SecondChildTAG: it means that if i login to give exam on 25 october then the deadline to submit it is on sunday or 24 hours after the login? SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-10-24T09:43:38Z IndexTAG: 2221 TitleTAG: Why!!? in the 1st part of the question , why we didn't consider any incremental current ? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-19T11:32:23Z VoteTAG: 1 CoursewareTAG: Week 5 / Two Terminal Connection Exercise CommentableIdTAG: 6002x_2_term_con_e NumberOfReplyTAG: 1 FirstChildTAG: I believe it is because the first question is a not "small-signal" question. That comes in the second question, where it asks you to make a small-signal model of the first question. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-19T17:15:54Z IndexTAG: 2222 TitleTAG: Lab6 Q3 Hello. Can anyone help me find estimate tpd,0→1, So, I have estimate RTH,VTH, Vs and CGS I'm used following equation for tpd,0→1, tpd=-RTH*CGS*ln((1-VTH)/(Vs-VTH)) If it's equation is true my answer is 0.15944...nS In result i don't have a green check. Please help me!!! Where I'm wrong? UserIdTAG: 413517 UserNameTAG: mari_safonova CreateTimeTAG: 2012-10-19T11:24:15Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: VC don't equal 1V. Formula must be -RTH*CGS*ln((VC-VTH)/(VS-VTH)), where VC is a lowest voltage. In our case low voltage is equal VOL = 0.250V FirstChildUserIdTAG: 402617 FirstChildUserNameTAG: Vl FirstChildCreateTimeTAG: 2012-10-19T15:22:16Z SecondChildTAG: this problm also wd me SecondChildUserIdTAG: 228871 SecondChildUserNameTAG: ELINAKHAN SecondChildCreateTimeTAG: 2012-10-19T16:51:22Z SecondChildTAG: Thank you SecondChildUserIdTAG: 413517 SecondChildUserNameTAG: mari_safonova SecondChildCreateTimeTAG: 2012-10-19T17:34:52Z SecondChildTAG: How much you got for Rth? SecondChildUserIdTAG: 286954 SecondChildUserNameTAG: ododo SecondChildCreateTimeTAG: 2012-10-20T14:43:59Z SecondChildTAG: i'm using the exact eqution u show i have double checked the values and its not green checked, still wrong SecondChildUserIdTAG: 402850 SecondChildUserNameTAG: y1111 SecondChildCreateTimeTAG: 2012-10-20T21:45:53Z SecondChildTAG: oh , sorry . my result is right i was only writing in sec , and the answer ask for a value in nanosecond i just deleted nine zeros , and got my green check SecondChildUserIdTAG: 402850 SecondChildUserNameTAG: y1111 SecondChildCreateTimeTAG: 2012-10-20T21:49:34Z SecondChildTAG: thanks @Vl SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-10-21T10:17:08Z SecondChildTAG: i am doing the same thing but not found green tick for what the reason SecondChildUserIdTAG: 164689 SecondChildUserNameTAG: muhammadfaizan SecondChildCreateTimeTAG: 2012-10-21T14:13:24Z IndexTAG: 2223 TitleTAG: H6P2 Can anyone explain me what is D_v in phase inverter problem....pls help.... UserIdTAG: 308571 UserNameTAG: Sayantani CreateTimeTAG: 2012-10-19T07:17:16Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: It is just change in input voltage(like del V) FirstChildUserIdTAG: 132685 FirstChildUserNameTAG: deepakmurali FirstChildCreateTimeTAG: 2012-10-19T12:46:43Z SecondChildTAG: thanks.... SecondChildUserIdTAG: 308571 SecondChildUserNameTAG: Sayantani SecondChildCreateTimeTAG: 2012-10-19T16:41:44Z IndexTAG: 2224 TitleTAG: Finding the Transconductance gm It may be pretty obvious and simple thing that I am making harder than it is, but I am not comfortable unless I corrected my assumptions. For the first question to find the transconductance gm I set up an equation: iD = K(vIN-vOUT-VT)^2 (equation 1) and since we know that vOUT = iD*RS iD = K(vIN-iD*RS-VT)^2 (equation 2) or K(vIN-iD*RS-VT)^2-iD=0 The equation 2 is a implicit function with 2 variables iD and vIN, I found the differentials since that linearizes the equation, as suggested by the Professor and for d(iD/dvIN) which is negative of partial derivative of function w.r.t vIN over partial derivative of function w.r.t iD, I get something like he following: K(VIN-ID*RS-VT)/(1+K*RS(VIN-ID*RS-VT)) evaluated at operating points. ergo id = K(VIN-ID*RS-VT)/(1+K*RS(VIN-ID*RS-VT)) * vi which gives gm = K(VIN-ID*RS-VT)/(1+K*RS(VIN-ID*RS-VT)) but apparently that is not the correct solution. I would very much appreciate if somebody could help me find what assumption went wrong or why my math and the solution do not seem to agree with each other and how to set such system up. UserIdTAG: 22953 UserNameTAG: Linus CreateTimeTAG: 2012-10-19T05:20:07Z VoteTAG: 1 CoursewareTAG: Week 6 / Source-follower small-signal exercise CommentableIdTAG: 6002x_small_signal_source_follower NumberOfReplyTAG: 1 FirstChildTAG: I started with the definition from the lectures: slide 14 from [Lecture 10][1] and slide 4 from [Lecture 11][2]: $$g_m = K*(V_I-V_T)$$ Observe from the circuit that $$V_I = V_{GS}$$ and that $$V_{GS} = V_{IN} - V_{OUT}$$ So: $$gm = K*(V_{IN}-V_{OUT}-V_T)$$ We know that $$I_{DS}=\frac{V_{OUT}}{R_S}$$ therefore $$V_{OUT}=I_{DS}*R_S$$ And finally $$g_m=K*(V_{IN}-I_{DS}*R_S-V_T)$$ Hope that helps! [1]: https://www.edx.org/static/content-mit-6002x/handouts/6002-L10-oei12-gaps-annotated.1553fffa8ddc.pdf [2]: https://www.edx.org/static/content-mit-6002x/handouts/6002-L11-oei12-gaps-annotated.bcf50190b3f1.pdf FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-10-19T09:54:24Z SecondChildTAG: Thank you for your help but the thing is our model gm = K*(VIN-VT) was derived when the VGS=VIN, that is when the output was explicit function of VIN. But the problem has current which depends on VOUT or the current through RS also so I thought we can not use the gm = K*(VIN-VT). So going back to the more fundamental law that governs the device, we get a result that is pretty different. SecondChildUserIdTAG: 22953 SecondChildUserNameTAG: Linus SecondChildCreateTimeTAG: 2012-10-19T15:54:49Z SecondChildTAG: VIN=VGS+VO --> VO=VIN-VGS --> VGS=VIN-VO Those are the only things you need to change from the "standard" model we used in the lectures. Plus remember that the *generic* gm=k*(VGS-VT), and NOT k*(VIN-VT) ;) SecondChildUserIdTAG: 604636 SecondChildUserNameTAG: Klionheart SecondChildCreateTimeTAG: 2012-10-19T16:22:47Z SecondChildTAG: Thanks for that reminder, Klionheart. I should have started with equation 8.13 from the text, pp. 410. :-) SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-21T03:30:30Z IndexTAG: 2225 TitleTAG: HW6 Problem 2 Hey guys, I have no clue how to start working on the first question of problem 2...any hints?? UserIdTAG: 308861 UserNameTAG: Ignaas CreateTimeTAG: 2012-10-18T22:13:40Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Think about the regions of operation of a MOSFET. If we always want to keep Q1 out of cutoff, what is the constraint? VGS (in this case, the input bias VIN) has to be greater than VT. Now, we are told this VIN can potentially swing up or down by a given value, Dv. You should be able to figure out the minimum value of VIN such that the MOSFET always satisfies the constraint even in the worst-case scenario (what would this be?) FirstChildUserIdTAG: 390034 FirstChildUserNameTAG: kahlil FirstChildCreateTimeTAG: 2012-10-18T22:34:15Z SecondChildTAG: hmmm but there are no values given...I think 0 SecondChildUserIdTAG: 308861 SecondChildUserNameTAG: Ignaas SecondChildCreateTimeTAG: 2012-10-19T00:38:05Z IndexTAG: 2226 TitleTAG: Weird methods... I hope, we'll get an explanation of the meaning of this steps in the subsequent lectures... UserIdTAG: 326787 UserNameTAG: Angstrem CreateTimeTAG: 2012-10-18T19:29:41Z VoteTAG: 1 CoursewareTAG: Week 6 / First-order differential equation CommentableIdTAG: 6002x_first_order_differential_equation NumberOfReplyTAG: 1 FirstChildTAG: It's just the way they are solved. There is an equivalent way without any guessing at all. Suppose that you have an ODE dv/dt+v(t)/A=B(t), where A is a constant. In the first step you look for the solution of the homogeneous ODE, which is v = C*exp(-t/A), for some constant C. In the second step you assume that C(t) is a function and substitute back into your original ODE. If you've done it right, you'll have an equation for dC/dt. The general solution will be v = C(t)*exp(-t/A) In the third step you look for a particular solution which satisfies the boundary condition. FirstChildUserIdTAG: 365465 FirstChildUserNameTAG: Vzzzz FirstChildCreateTimeTAG: 2012-10-20T22:56:05Z SecondChildTAG: doubt it, I thought this course was going to be about electronics, after week 5 it's more about meaningless mathematical wanking with deliberately missing steps interspersed with comments about how easy it all is. SecondChildUserIdTAG: 61469 SecondChildUserNameTAG: Spacedog SecondChildCreateTimeTAG: 2012-10-21T13:53:59Z SecondChildTAG: Have faith. It will get easier as the course progresses. These concepts are important building blocks. SecondChildUserIdTAG: 19863 SecondChildUserNameTAG: Samir SecondChildCreateTimeTAG: 2012-10-27T06:50:21Z IndexTAG: 2227 TitleTAG: What's rout? I didn't understand how I get the rout. UserIdTAG: 49568 UserNameTAG: And_Sepulveda CreateTimeTAG: 2012-10-18T15:27:59Z VoteTAG: 1 CoursewareTAG: Week 6 / Source-follower small-signal exercise CommentableIdTAG: 6002x_small_signal_source_follower NumberOfReplyTAG: 1 FirstChildTAG: To find the output resistance, which is just the thevenin resistance looking in from the output port, we suppress all independent sources and find the equivalent resistance as seen through that output port. The key thing to realize with this question is, when you suppress the independent voltage source (in other words, short it out so it's just a wire), the voltage at the gate is zero (v_g = 0), and therefore v_gs = v_g - v_s = -v_s. But since v_s = v_out, v_gs = -v_out. Knowing this, to find the equivalent resistance, we can put an independent current source in and imagine injecting some test current i_test into the positive terminal of v_out, and the solve for what the voltage v_out would actually be. The ratio of our v_out to our i_test is the output resistance. If you put in a current source i_test and then solve the circuit equations for v_out in terms of i_test, their ratio should indeed give you the correct answer. FirstChildUserIdTAG: 390034 FirstChildUserNameTAG: kahlil FirstChildCreateTimeTAG: 2012-10-18T18:10:48Z SecondChildTAG: Ok! Now I got it. Thanks! =) SecondChildUserIdTAG: 49568 SecondChildUserNameTAG: And_Sepulveda SecondChildCreateTimeTAG: 2012-10-19T12:36:58Z SecondChildTAG: lol after using about one week and reading kahlil's comment and redo again, I found out i just miswrite a '-' with a '+' sign. SecondChildUserIdTAG: 157273 SecondChildUserNameTAG: ongchihang SecondChildCreateTimeTAG: 2012-10-19T14:43:00Z IndexTAG: 2228 TitleTAG: Invalid input: CF not permitted in answer capacitance of the capacitor is CF UserIdTAG: 194717 UserNameTAG: kaa CreateTimeTAG: 2012-10-18T12:04:35Z VoteTAG: 1 CoursewareTAG: Week 6 / First-order capacitor exercise CommentableIdTAG: 6002x_first-order_capacitor_exercise NumberOfReplyTAG: 1 FirstChildTAG: F is a measurement unit - farad FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-18T13:16:44Z IndexTAG: 2229 TitleTAG: What's wrong here!! L=100m W=100m d=10m er=(8.854*10^-12)*2.25 As C = (er*A)/d = 1.99215*10^-7 !! UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-18T11:47:54Z VoteTAG: 1 CoursewareTAG: Week 6 / Capacitor scaling exercise CommentableIdTAG: 6002x_capacitor_scaling_exercise NumberOfReplyTAG: 1 FirstChildTAG: is the answer right? 2.25 * 100m * 1cm / 0.1 mm = 2.25 * 100 m * 0.01 m / 0.0001 m = 22500 Please, clarify why the correct answer is 1.99215e-06 FirstChildUserIdTAG: 321152 FirstChildUserNameTAG: ser9ey FirstChildCreateTimeTAG: 2012-10-18T16:40:17Z SecondChildTAG: You missed: The permittivity of the vacuum is ϵ0≈8.854×10−12F/m. This is also called the electric constant. Also your distance is wrong. SecondChildUserIdTAG: 113561 SecondChildUserNameTAG: Illogical SecondChildCreateTimeTAG: 2012-10-18T23:02:19Z SecondChildTAG: I will transcribe the question *"The ratio of the permittivity of a material to the permittivity of the vacuum is called the "relative permittivity" or the "dielectric constant" of the material. For example, the dielectric constant of polyethylene is 2.25. For more information see. As a consequence of the very small permittivity of the vacuum, most common capacitors have a capacitance of only a tiny part of a Farad. For example, what is the capacitance (in Farads) of a parallel-plate capacitor with plates 100.0m long, 1.0cm wide, and separated by 0.01mm of polyethylene? "* So, we know______C= (e*A)/d ________which is translated C(F)= ("permittivity of the vacuum"*"relative permittivity" or the "dielectric constant"*Area)/distance so we have C=(8.854x10-12*2.25*1m)/0.01mm= 1.99215 microFarad :) success SecondChildUserIdTAG: 244706 SecondChildUserNameTAG: Miguel_Angel SecondChildCreateTimeTAG: 2012-10-19T23:56:58Z SecondChildTAG: I also came up with **1.99215*10^-7** using http://www.daycounter.com/Calculators/Plate-Capacitor-Calculator.phtml I converted all units into mm. That was where I made my mistake. Length = 100m = 100000mm Width = 10cm = 100mm A = 100*100000 = 1000000mm Depth = 0.01mm SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-22T10:37:10Z IndexTAG: 2230 TitleTAG: H6P1 Question 1 i have solved the equation but getting "Could not parse" after hitting "check". plz help. UserIdTAG: 114864 UserNameTAG: Vikaash CreateTimeTAG: 2012-10-18T10:47:35Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hi Vikaash! Can I help you? Are you sure that you are writting correctly the variables? Remember that VS it is not the same as Vs ... FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-19T03:14:24Z SecondChildTAG: Myrimit plz help me.i am is got stucked in the Homework6 question 1,i have solved the quadratic equation and got the ans like this Vout=+-X/[1+4*X*X*X(X-X)]^1/2 ,But not getting the green tick,plz help me SecondChildUserIdTAG: 97359 SecondChildUserNameTAG: azeem179 SecondChildCreateTimeTAG: 2012-10-19T15:43:02Z SecondChildTAG: i m also getting **vOUT= X/sqrt(1+4*X*(X-X)*X)** but getting RED cross..plx help SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-10-20T07:36:26Z SecondChildTAG: SORRY....its **vOUT= X/sqrt(1+4*X*X*(X-X)*X)** SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-10-20T09:46:00Z SecondChildTAG: @Myrimit PLZ HELP ME.... SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-10-20T09:53:27Z SecondChildTAG: Hi Vikaash! Sorry for the delay. Are you sure that your quadratic it is ok? remember that if you have a quadratic equation: $A*X^2+B*X+C$ Your possible roots are: $X_1= \frac {-B + \sqrt{B^2-4*A*C}}{2*A}$ $X_2= \frac {-B - \sqrt{B^2-4*A*C}}{2*A}$ So, if vOUT behaves like a quadratic equation, are you sure that your expression it is correct? why do you have the $\sqrt{....}$ in the denominator? ... SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-20T13:37:04Z SecondChildTAG: @Myrimit i got it.....please don't be sorry..i need your help in last two part of H6P2 i.e. thevenin vout and resistor. SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-10-20T13:51:37Z FirstChildTAG: I have got the same answer but with +ve sign still didn't shows green tick....... FirstChildUserIdTAG: 227508 FirstChildUserNameTAG: bhavyab FirstChildCreateTimeTAG: 2012-10-19T19:26:29Z SecondChildTAG: Hi bhavyab! Can I help you? Please read the comment that I have wrote it to Vikaash above. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-20T13:38:50Z IndexTAG: 2231 TitleTAG: Lab6 How come to this response? ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13505317901343637.gif thanks UserIdTAG: 327787 UserNameTAG: REINALDOPARANHOS CreateTimeTAG: 2012-10-18T03:43:50Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hi REINALDOPARANHOS! Can I help you? I didn't understand your question... Did you mean on how did they arrive to that result from the previous step of that example? If you go to the page 533 [here][1] You will see the equation 10.67, - you will see that 10000 it is the same as **10***10^3, - so, that is why 10000/11 it is the same as (10*10^3)/11 or (10/11)*10^3 - Then the 100*10^-15 it is the given value in that Texbook example of the Capacitor CGS2 , remember that f is femto and it is 10^-15 . Or Did you mean on how they get that formula? I hope this can help you. Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/557 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-19T03:31:29Z FirstChildTAG: the expression (In) is any value? FirstChildUserIdTAG: 327787 FirstChildUserNameTAG: REINALDOPARANHOS FirstChildCreateTimeTAG: 2012-10-21T11:47:04Z IndexTAG: 2232 TitleTAG: LAB 6 HEY GUYS, I need some hints on lab 6....can you give me some guidance UserIdTAG: 308861 UserNameTAG: Ignaas CreateTimeTAG: 2012-10-18T01:44:24Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Here's something to get you started: The first part is just a voltage divider problem. Measure the final voltage and calculate the other leg of the divider. The next part is about the Thevinin equivalent circuit. Vth is the measured final voltage and Rth is the resistance of the two legs of the divider in parallel. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-18T02:37:05Z FirstChildTAG: https://6002x.mitx.mit.edu/discussion/question/49385/lab-6-notes but I'm still confused about how it works.. why should we use a step source and these different capacitances? FirstChildUserIdTAG: 136959 FirstChildUserNameTAG: christerpher FirstChildCreateTimeTAG: 2012-10-18T13:09:50Z FirstChildTAG: Hi Ignaas! You can take a look at this [Lab 6 - Hints][1] If you have any doubt. I will be here :). See you! Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Ring_Oscillator/threads/5082c6039e78031f000000c0 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-20T16:24:10Z IndexTAG: 2233 TitleTAG: Joined Late... Sir , My name is Prafullkumar and I am From India, today morning only first time i visited this website (through search engine) and your class ,since long time i was searching for such a platform which can give online access to courses and certification ...I know its too late but is there any chance to join the course and get the edx certificate ..if given a chance i will complete the due assignments and lab work. UserIdTAG: 675836 UserNameTAG: kpkumar241 CreateTimeTAG: 2012-10-18T01:16:20Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: yes there is a chance if you perform well in mid term exam which is on 25th oct. and final exam. and one more thing you should complete your's homework and labs from today onwards . FirstChildUserIdTAG: 114864 FirstChildUserNameTAG: Vikaash FirstChildCreateTimeTAG: 2012-10-18T11:37:51Z SecondChildTAG: what is home work and lab work for today? SecondChildUserIdTAG: 681348 SecondChildUserNameTAG: alizarrar SecondChildCreateTimeTAG: 2012-10-18T16:48:51Z FirstChildTAG: To get the full benefit of the course, you should start at the beginning, and forget about the grade this semester. You should take it again for the grade another time. In the spring, this course will be offered again hopefully, and in the spring, there's a good chance **MITx** will offer a proctored exam certificate in addition to the free honor code certificate of completion that is currently offered. if you have a little bit interest in Computer then join **CS50x** to get good knowledge of Computer. Regards: asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-10-18T15:39:53Z FirstChildTAG: Hi kpkumar241, There are chances to get the certificate but you will not get the full grade, that is to say, you will not have chances to get a final score of 100% but yes to get an A with 91% (in the case that you do Midterm and Final Exam with 100% and also submit 100% of WEEK 6,7,8,9,10,11 and 12 assignments)... GENERAL > Remember that this Course has: - 12 Homeworks (but if you read [syllabus here][1], they tell you that you can skip two without penality, that means that you will still have 100% of Homework score if you make only 10). If you get 100% in your Homework score, that homework score it will contribute your final score only a 15%. So, if you have done so far 50% of your homework, it will contribute with your final score with (50%*15%)/100% = 7.5%. - 12 Labs (but if you read syllabus, they tell you that you can skip two without penality, that means that you will still have 100% of Labs score if you make only 10). - 1 Midterm Exam. If you get 100% in the Midterm, it will contribute with your final score with 30%. - 1 Final Exam.If you get 100% in the Final, it will contribute with your final score with 40%. ---- So, your CASE: - Tomorrow you it is the deadline of Week 6 asignments... Suppouse that you can do it and also, that you can do WEEK 7, 8,9,10 and 12 too. Your max score will be: Homework: (7*100%)/10 = 70% and final scoring contribution will be ---> (70%*15%)/100% = 10.5% Lab: (7*100%)/10 = 70% and final scoring contribution will be ---> (70%*15%)/100% = 10.5% - Remember that next week is the Midterm. Suppose that you get 100% in the Midterm, it will contribute to your final score with 30%. - Suppose that you get 100% in the Midterm. it will contribute to your final score with 40%. So, your max. score that you can get is 10.5% +10.5%+30%+40% = 91% . So, still have chances to get an A (but remember that this is the max. score that you can reach if you from now you do all to 100%). Remember that to get an certicate you have to get a minimun score of 60% (grade C). ---- So, it is not that late to get the certificate :), but you will have to do a lot of effort and work from now to the end of the Course... Also, as some students have said it to you, you can follow this Course and re-take it again, but that is always up to you :). See you! I hope this can help you. Sorry for my English. Myriam. [1]: https://www.edx.org/static/content-mit-6002x/handouts/syllabus.a477535058a1.pdf FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-20T16:59:05Z IndexTAG: 2234 TitleTAG: H5P1 wrong? Hello, I missed H5 and I'm doing it now. However I can't find my mistake in part (a) (haven't reached the other parts yet, I'm stuck here), even though I have the answer. The circuit is the following, with $V_{S+}=1.0V$, $V_{S−}=−1.0V$, $K=1 mA/V^2$ and $V_T=0.5V$: ![enter image description here][1] Question (a) says: *What is the minimum value of input voltage $v_{IN}$ in volts for the MOSFET to be operating in saturation region?* However, for a given $v_{DS}$ shouldn't there be an upper limit for $v_{IN}$ instead of a lower limit, to attain saturation? The answer says: $v_{IN}=-0.5\,V$ (MIN value?). However, my answer is $v_{IN}=+1.5\,V$ (MAX value). What am I doing wrong? Here's my reasoning: ![enter image description here][2] Thanks, Ignacio [1]: https://dl.dropbox.com/u/24096724/H5P1_Zero_Offset_Amplifier.png [2]: https://dl.dropbox.com/u/24096724/h5p1.jpg UserIdTAG: 276409 UserNameTAG: IgnacioUY CreateTimeTAG: 2012-10-17T16:17:29Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Anyone? FirstChildUserIdTAG: 276409 FirstChildUserNameTAG: IgnacioUY FirstChildCreateTimeTAG: 2012-10-18T01:21:36Z FirstChildTAG: There are two conditions to be satisfied for saturation, one on VGS and one on VDS. In the current circuit vIN only affects VGS, so that is the only condition we need to be concerned about. Because of Vs-, the source potential is -1.0V relative to ground. To be in the saturation region VGS > VT; therefore VGS > 0.5V, which means that relative to ground the gate potential > -0.5V. In this circuit, vIN is the gate potential relative to ground. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-18T02:24:44Z SecondChildTAG: Thank's skyhawk for your answer. I believe, however that VGS>VT alone, does not ensure saturation, and that VDS>=VGS-VT is a necessary condition which has to be taken into account as shown in the following figure: ![enter image description here][1] [1]: https://dl.dropbox.com/u/24096724/sat.png SecondChildUserIdTAG: 276409 SecondChildUserNameTAG: IgnacioUY SecondChildCreateTimeTAG: 2012-10-18T18:16:34Z SecondChildTAG: VDS decreases as iDS increases because of increasing VGS. This places a maximum on the value of vIN, which is the subject of question 3. Clearly the condition VDS >= VGS - VT can't place a lower limit on VGS. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-18T18:49:10Z SecondChildTAG: I understand what you say. However I feel the question statement is wrong.. It should be: *What is the minimum value of input voltage $v_{IN}$ in volts for the MOSFET to be **conducting**?* But anyway. Thanks skyhawk for your time SecondChildUserIdTAG: 276409 SecondChildUserNameTAG: IgnacioUY SecondChildCreateTimeTAG: 2012-10-18T19:18:53Z SecondChildTAG: But when the MOSFET starts conducting it is in the saturation region. When vIN rises to .093V the MOSFET leaves the saturation region, so -0.5V < vIN < .093V the MOSFET is conducting in the saturation region and iDS = (K/2)*(VGS-VT)^2. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-18T22:10:34Z SecondChildTAG: I got it now, thank you very much skyhawk! SecondChildUserIdTAG: 276409 SecondChildUserNameTAG: IgnacioUY SecondChildCreateTimeTAG: 2012-10-21T19:11:44Z SecondChildTAG: Well done :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-23T23:26:37Z IndexTAG: 2235 TitleTAG: Many correct answers for the first question... For the first question, you can solve for IDS in terms of the other parameters using techniques from the previous source follower exercise. It's a little annoying that the checker only accepts the overspecified answer in terms of all five parameters. UserIdTAG: 139646 UserNameTAG: tmcnulty CreateTimeTAG: 2012-10-17T15:39:01Z VoteTAG: 1 CoursewareTAG: Week 6 / Source-follower small-signal exercise CommentableIdTAG: 6002x_small_signal_source_follower NumberOfReplyTAG: 0 IndexTAG: 2236 TitleTAG: New here Hey guys, I just found out about edx and have registered for this course. I see that this course has started almost 6 weeks ago, and that that due dates for 5 HWs and Labs have passed. I am planning on submitting the HW 6 and Lab 6 but I wanted to know what happens of the previous HW and Labs. I am not sure how it works. Let me know what I can do. Thanks, Akhil UserIdTAG: 671533 UserNameTAG: AkhilKatkar CreateTimeTAG: 2012-10-17T15:28:58Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Akhil, Once the due date has passed for a homework assignment or a lab, you cannot go back and re-do it. I advise you to register for the next offering of 6.002x, which will be in the Spring semester. -- Rohan FirstChildUserIdTAG: 391929 FirstChildUserNameTAG: RohanNagarkar FirstChildCreateTimeTAG: 2012-10-17T18:25:59Z FirstChildTAG: To get the full benefit of the course, you should start at the beginning, and forget about the grade this semester. You should take it again for the grade another time. In the spring, this course will be offered again hopefully, and in the spring, there's a good chance **MITx** will offer a proctored exam certificate in addition to the free honor code certificate of completion that is currently offered. Regards: asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-10-18T15:33:45Z FirstChildTAG: Okay thanks. Will do that. FirstChildUserIdTAG: 671533 FirstChildUserNameTAG: AkhilKatkar FirstChildCreateTimeTAG: 2012-10-18T04:53:51Z IndexTAG: 2237 TitleTAG: Need help to get started on H6P2 Hey can anyone enlighten me on how to solve H6P2. I can't arrive to the correct answer on the first question and the succeeding questions are all related to the first question. Your help will be very much appreciated thanks :) UserIdTAG: 367072 UserNameTAG: jansenlopez CreateTimeTAG: 2012-10-17T10:58:47Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: vout = VDD minus the drop across Q2. What is the i-V relation for Q2 with the gate connected to the drain? FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-17T12:54:11Z SecondChildTAG: Q2 is a voltage drop right? hmmm i = gmVgs, gm = K(VDS-VT) for the small signal and I = K/2(VDS-VT)^2 for the large signal analysis. So I have to multiply it by Ron to have a voltage. But it doesn't make sense to me. And plus I know I have to use calculus here to get Dv which btw the derivative of I don't know. hahahah sorry dude I'm really lost at this point. Please help me!!!!! SecondChildUserIdTAG: 367072 SecondChildUserNameTAG: jansenlopez SecondChildCreateTimeTAG: 2012-10-18T07:17:50Z SecondChildTAG: The first two parts of the problem require no calculus or even complex math, just simple reasoning. Dv is some given value. You don't have to do anything to get it. To answer part 1 just think about it and write down the answer, no calculation required. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-18T12:07:12Z SecondChildTAG: VT+Dv SecondChildUserIdTAG: 285675 SecondChildUserNameTAG: Ascot SecondChildCreateTimeTAG: 2012-10-21T21:58:14Z FirstChildTAG: Check this link for some hints: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/507ec10ebc70a3681200000b FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-18T12:22:51Z IndexTAG: 2238 TitleTAG: Readings before Lectures Hi, my question is Which is better, read befores lectures, or attend lectures before reading? I had some problems with my schedule these last 2 weeks, and i looking for the fastest way to catch up. Thanks in advance. UserIdTAG: 43833 UserNameTAG: gmcentenom CreateTimeTAG: 2012-10-17T05:48:11Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: In my experience, it's always better to read before the lectures. That way, the lecturer will pretty much just confirm what you already know. The only downside is that it takes quite a bit of time. If you can afford that, go for it. You'll ace any subject. FirstChildUserIdTAG: 90796 FirstChildUserNameTAG: ChaunceyGardiner FirstChildCreateTimeTAG: 2012-10-17T06:10:46Z SecondChildTAG: ^^I agree with this^^ Have a look at the home-works and labs ahead of time too. This was recommended by someone on here earlier and I have started doing it that way, especially since the course has gotten progressively harder. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-17T13:00:11Z FirstChildTAG: Hi, I think it is better to watch the lectures first because you can grasp the concept faster and try solving problems. Afterwards you can read when you find time and skip the topics you have already understood. FirstChildUserIdTAG: 123982 FirstChildUserNameTAG: vasugupta9 FirstChildCreateTimeTAG: 2012-10-17T08:33:10Z IndexTAG: 2239 TitleTAG: H6P2 Why Q2 is in saturation? Hello everyone! Please explane why Q2 is in saturation? If for that is needed VDS>=VGS-VT, but VDS=VGS. Thanks. UserIdTAG: 387154 UserNameTAG: Junglist CreateTimeTAG: 2012-10-17T05:35:46Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: If VDS = VGS then surely VDS > VGS - VT ? FirstChildUserIdTAG: 90796 FirstChildUserNameTAG: ChaunceyGardiner FirstChildCreateTimeTAG: 2012-10-17T06:13:05Z SecondChildTAG: oops. absolutely:) SecondChildUserIdTAG: 387154 SecondChildUserNameTAG: Junglist SecondChildCreateTimeTAG: 2012-10-17T06:30:50Z IndexTAG: 2240 TitleTAG: Simulation error some simulation errors UserIdTAG: 657539 UserNameTAG: ajeeb24 CreateTimeTAG: 2012-10-17T02:23:38Z VoteTAG: 1 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: Hi ajeeb24! :) Can I help you? What kind of error are you referring about? From which Lab? FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-17T02:32:27Z SecondChildTAG: Circuit sandbox / Overview. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-17T03:08:29Z IndexTAG: 2241 TitleTAG: lab 7 put calculator in radians when solving the trig function UserIdTAG: 18073 UserNameTAG: gburkhart CreateTimeTAG: 2012-10-16T20:48:41Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 2242 TitleTAG: H6P2 Part 1 and 2 Hi.. I need some help in this exercise. Could someone help me? Thanks! UserIdTAG: 164572 UserNameTAG: Ranyeri_rocha CreateTimeTAG: 2012-10-16T17:56:21Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Its rather easy.... If u will write the saturation conditions for Q2,u will find that it will always be in saturation..... So all u have to worry for is the saturation for Q1... Write down its saturation conditions too.... From dere u will easily find Vin(min) and Vin(max) Part 1:all u have to do is Add Dv in Vin(min) Now find out the current equations of both the mosfets.... and EQUATE them as id in the circuit is same for both.... You will get a relation between vin, vout and vdd. Part 2:Relating it wid d conditions derived above (for Q1) u will easily get the answer!!! FirstChildUserIdTAG: 82597 FirstChildUserNameTAG: bondrajat FirstChildCreateTimeTAG: 2012-10-17T19:05:41Z IndexTAG: 2243 TitleTAG: For those who cannot watch Lecture Videos due to the blockage of you tube. Those Who can not watch videos due to the blockage of you tube download Hotspot sheild and see the magic :) UserIdTAG: 122513 UserNameTAG: Mona77 CreateTimeTAG: 2012-10-16T15:26:09Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: [JWplayer Video Link Week(1-6)][1] [1]: https://dl.dropbox.com/u/24096724/6002xMP4/menu.html# You can download videos with out using **Hotspot shield**. Regards: asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-10-18T15:53:13Z IndexTAG: 2244 TitleTAG: Help with problem HW6P1 what does this comment mean: Could not parse the formula.... UserIdTAG: 308861 UserNameTAG: Ignaas CreateTimeTAG: 2012-10-16T12:12:07Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: parenthesis are incorrect FirstChildUserIdTAG: 18073 FirstChildUserNameTAG: gburkhart FirstChildCreateTimeTAG: 2012-10-16T13:20:12Z SecondChildTAG: hmmm i don't think so..it was correct SecondChildUserIdTAG: 308861 SecondChildUserNameTAG: Ignaas SecondChildCreateTimeTAG: 2012-10-16T15:18:17Z SecondChildTAG: Myrimit you know sth about this SecondChildUserIdTAG: 308861 SecondChildUserNameTAG: Ignaas SecondChildCreateTimeTAG: 2012-10-16T15:18:42Z SecondChildTAG: the comment is telling you that you entered your formula in a way that does not make sense to the program. most common cause is parenthesis mistakes followed closely by a missing operation symbol such as writing K(VIN-VT) instead of K*(VIN-VT). SecondChildUserIdTAG: 18073 SecondChildUserNameTAG: gburkhart SecondChildCreateTimeTAG: 2012-10-16T23:21:16Z SecondChildTAG: Hi Ignaas! :) How are you? Yes, it might be because what gburkhart said in the previous comment. Are you sure that are you entering correctly the multiplications, parenthesis, etc? SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-17T02:16:37Z SecondChildTAG: Please check here Myrimit https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/507d85e59ba97423000001c1 SecondChildUserIdTAG: 308861 SecondChildUserNameTAG: Ignaas SecondChildCreateTimeTAG: 2012-10-17T02:27:09Z IndexTAG: 2245 TitleTAG: S1E1.5: SIMPLE POWER How did you find results? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-16T06:08:30Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 5_Simple_Power NumberOfReplyTAG: 1 FirstChildTAG: POWER=WORK PER UNIT TIME FirstChildUserIdTAG: 560501 FirstChildUserNameTAG: chandraprakash FirstChildCreateTimeTAG: 2012-10-16T09:28:25Z IndexTAG: 2246 TitleTAG: Question after lab 5 At the end of the lab,it says "Think about how your final choices for the amplitudes of Vbias and Vsignal relate to the input voltage range you reported above for linear operation." The answer was 0.3, but I mainly got it from just trying to plug different values in. The only relation I see was Taking the average of the two ends of the range (0.560, 1.175) which was around 0.6. IF you take half of this you get 0.3. Is this the relationship it was talking about? UserIdTAG: 381923 UserNameTAG: matteaton CreateTimeTAG: 2012-10-15T16:47:43Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Yes. Amplitude is peak-to-peak/2 FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-15T17:02:38Z FirstChildTAG: You had truncation for a input signal higher than 1.2V or lower than 0.6v (roughly) This, of course means that your optimal Vin must not exceed these two values at any time. So, you can take the Vbias as the average signal, and the oscillation is the peak of the small signal Vsignal. Thus, with the values found before, Vbias=0.9V and Vsignal has an amplitude of 0.3V. You'll get a signal that oscillates between 0.9+0.3=1.2V and 0.9-0.3=0.6V, and is exactly within the limits you found ;) FirstChildUserIdTAG: 604636 FirstChildUserNameTAG: Klionheart FirstChildCreateTimeTAG: 2012-10-15T17:20:24Z IndexTAG: 2247 TitleTAG: Question to staff ! i am unable to go through the lecture videos or do the home work as i am engaged with my academics, so i was wondering if i took up the same course next year would it hamper my grade when i take the course again or will there be any negative remarks if i am able to finish the course successfully next year without any discontinuities....??? UserIdTAG: 448671 UserNameTAG: anirudhsom3000 CreateTimeTAG: 2012-10-15T16:41:39Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: There is no penalty for dropping out. FirstChildUserIdTAG: 90796 FirstChildUserNameTAG: ChaunceyGardiner FirstChildCreateTimeTAG: 2012-10-15T23:41:32Z IndexTAG: 2248 TitleTAG: week 6 So week 6 starts....HW6 and lab 6.....ohw and the new solid state chemistry course.. good luck UserIdTAG: 308861 UserNameTAG: Ignaas CreateTimeTAG: 2012-10-15T13:57:12Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 2249 TitleTAG: H5P1 Explanation Mistakes Hi, I'm reading the explanation/answer and I think I found some minor mistakes, both in part C. First it says "The MOSFET leaves saturation when: $V_{DS}≥V_{GS}−V_T$". Shouldn't that inequality be reversed? Second, there are a series of simplifications for the equation for $V_{OUT}$, ending in "$V_{IN} = V_T$". Shouldn't that say "$V_{OUT} = V_{IN} - V_T$" ? This is a great course - thank you! Rob UserIdTAG: 468623 UserNameTAG: RobNik CreateTimeTAG: 2012-10-15T13:49:57Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 1 FirstChildTAG: Thank you. H5P1 solutions have been fixed :) FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-10-16T07:28:21Z IndexTAG: 2250 TitleTAG: Just enrolled and far behind.. what to do? hi all, i'm new to this course. i've just joined it , googling through online courses found it interesting and enrolled for it. But after joining, I came to know alot of it has been covered. Can I continue it from where it's or I've to do the rest first? will it be credited if I may take 2 weeks to accomplish the already delivered one of the course and get along for further. I need guidance in this aspect. UserIdTAG: 399502 UserNameTAG: Khadijah CreateTimeTAG: 2012-10-15T13:14:39Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: To get the full benefit of the course, you should start at the beginning, and forget about the grade this semester. You should take it again for the grade another time. In the spring, this course will be offered again, and in the spring, there's a good chance **MITx** will offer a proctored exam certificate in addition to the free honor code certificate of completion that is currently offered. Regards: asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-10-15T13:49:15Z SecondChildTAG: Profuse Thanx for such a helpful suggession! I was much worried about it. Ready to get started next morning. should I keep it doing with my own pace without considering the deadlines or hurrying to join upcoming lectures? SecondChildUserIdTAG: 399502 SecondChildUserNameTAG: Khadijah SecondChildCreateTimeTAG: 2012-10-15T16:38:14Z SecondChildTAG: Go at your own pace, learn the concepts thoroughly, it will pay off in the long-term for sure! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-15T16:42:49Z IndexTAG: 2251 TitleTAG: "No problem scores in this section"-in Progress what does this mean? i've seen all tutorial videos and still it shows in the progress i section UserIdTAG: 137509 UserNameTAG: rahulpark CreateTimeTAG: 2012-10-15T02:37:00Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: Help! I'm not scored for hw3 or lab3! FirstChildUserIdTAG: 311041 FirstChildUserNameTAG: sathyan FirstChildCreateTimeTAG: 2012-10-15T03:57:06Z SecondChildTAG: Because you are late to submit HW3 & LAB3.... SecondChildUserIdTAG: 329051 SecondChildUserNameTAG: asadbhatti42 SecondChildCreateTimeTAG: 2012-10-15T10:49:21Z FirstChildTAG: Share a screenshot with us. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-15T02:57:31Z SecondChildTAG: Hi JSChambers! :) What does it means the green Community TA? SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-15T11:50:30Z FirstChildTAG: You don't get a score for watching tutorials. You get that for completing homeworks and labs. FirstChildUserIdTAG: 90796 FirstChildUserNameTAG: ChaunceyGardiner FirstChildCreateTimeTAG: 2012-10-15T03:52:41Z FirstChildTAG: I completed the homework mostly correctly, and the lab correctly, and corrected my errors after clicking "Show Answers", but I am still not scored in HW3 or Lab3. The other homeworks were fine. FirstChildUserIdTAG: 311041 FirstChildUserNameTAG: sathyan FirstChildCreateTimeTAG: 2012-10-15T03:55:03Z SecondChildTAG: Because you are late to submit HW3 & LAB3.... SecondChildUserIdTAG: 329051 SecondChildUserNameTAG: asadbhatti42 SecondChildCreateTimeTAG: 2012-10-15T10:47:41Z IndexTAG: 2252 TitleTAG: possible to skip H5P2 q3? Hi all, is it possible to solve the remainder of H5 with just questions 1&2 of H5P2 solved? The quadratics I am deriving look ridiculous... by the way for the quadratic equation ax^2 + bx + c = 0 should I be aiming to solve for x = vout or ids? UserIdTAG: 335803 UserNameTAG: bobbanovski CreateTimeTAG: 2012-10-15T01:51:36Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Just go ahead,the root of the quadratic equation is necessary for the following questions,though the result would be seemingly complicated.Pay attention to the sign,which is case sensitive. FirstChildUserIdTAG: 361137 FirstChildUserNameTAG: Ericson FirstChildCreateTimeTAG: 2012-10-15T02:36:14Z FirstChildTAG: [Edited by staff -- please do not post answers to a graded part of the course] FirstChildUserIdTAG: 181081 FirstChildUserNameTAG: sant250784 FirstChildCreateTimeTAG: 2012-10-15T03:07:27Z SecondChildTAG: In the portion of the quadratic equation for sqrt(b^2-4*a*C), what happened to the -4*a*c in your equation above? SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-15T03:43:49Z SecondChildTAG: that ended up as 4*(K*R^2/2)*(K/2)*(VIN-VT)^2 SecondChildUserIdTAG: 335803 SecondChildUserNameTAG: bobbanovski SecondChildCreateTimeTAG: 2012-10-15T09:47:12Z FirstChildTAG: I had problems entering equations that would be accepted by the 6.002x checker, yet I was able to solve several of the numeric problems using those equations, because Wolfram Alpha was kind enough to correct some missing parens and such for me. If you can solve the numerics, DO SO. But then try to go back and get an equation that will be accepted. The point is, if you can continue despite difficulties, do so while you still have time before the deadline. Then go back for the troublesome bits. :-) Good luck! FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-10-15T03:28:01Z IndexTAG: 2253 TitleTAG: problem got an algrebraic expression for homework prob 3 but i keep getting the error message invalid entry cannot parse the equation what am i doing wrong UserIdTAG: 160242 UserNameTAG: Mlevins35 CreateTimeTAG: 2012-10-15T01:12:34Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: My first two guesses would be unbalanced parentheses or wrong case and a third would be missing * for multiply. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-15T01:46:05Z SecondChildTAG: Exactly right,I had the same situation.I tried many times,finally got the right tick.Pay much attention to the case,* for multiply is easily missing and the (). SecondChildUserIdTAG: 361137 SecondChildUserNameTAG: Ericson SecondChildCreateTimeTAG: 2012-10-15T02:54:31Z SecondChildTAG: Agreed, don't give up! I had several issues, missing parens, missing "*", case was a pain... Keep trying! Good luck! SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-15T03:29:34Z IndexTAG: 2254 TitleTAG: lab 5 Thanks to all that gave me hints for equations I was able to do lab 5 with a little problem but the only thing that got me was the measured gain. How do we figure the measured gain that is all that I have a problem with UserIdTAG: 160242 UserNameTAG: Mlevins35 CreateTimeTAG: 2012-10-14T23:27:02Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: measured gain is the ratio of Vp-p(out) and Vp-p(in) FirstChildUserIdTAG: 361137 FirstChildUserNameTAG: Ericson FirstChildCreateTimeTAG: 2012-10-15T02:29:11Z IndexTAG: 2255 TitleTAG: H5P3 question 1 The way my equation looked like for H5P2 for $v_{OUT}$ was: (Giant monster quadratic solution formula for $i_{DS}$) * $R_S$. Is this what I'm supposed to differentiate then multiply by $v_{in}$? Because it becomes an even more giant and more monster after the derivation. Which didn't really stop me, but the the formula didn't work. (Yes I typed it in correctly, * signs and all). Am I differentiating the correct equation? Did I just not do it right? UserIdTAG: 230226 UserNameTAG: zoliking CreateTimeTAG: 2012-10-14T23:10:22Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Yes, it's correct. In result many of members reduces, so it becomes even bit shorter. In discussion I found hint, that helped me to get correct formula hint: `d(sqrt(C1*x+C2))/dx = C1/(2*sqrt(C1*x+C2))` It works, but I can't understand this hint, can't find this rule of conversion. Could anyone explain how this hint conversion results from basic derivation rules? FirstChildUserIdTAG: 341573 FirstChildUserNameTAG: Exec FirstChildCreateTimeTAG: 2012-10-14T23:35:53Z SecondChildTAG: Let y = C1*x+C2, then d(sqrt(C1*x+C2))/dx = d(y^1/2)/dx Using chain rule and rules for differentiating powers: d(y^1/2)/dx = (1/2)(y^-1/2)(dy/dx) = C1/(2*sqrt(C1*x+C2)) SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-14T23:59:20Z FirstChildTAG: If we take the derivative of x^2, we get 2x. Same idea for taking derivative of sqrt(x) = x^(1/2). Derivative of x^(1/2) = (1/2)*x^(-1/2). In each case we're decrementing the exponent by exactly 1 and multiplying the original variable by what was the original exponent value. It's hard to explain in words. Derivative of x^3 = 3x^2. Derivative of x^2 = 2x^1 = 2x. Derivative of x^(1/2) = (1/2)*x^(-1/2). Then there's the chain rule as mentioned above. FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-10-15T01:26:45Z FirstChildTAG: Answer: vin*(((sqrt((2*K*RS*VIN-2*K*RS*VT+1)))-1)/(sqrt((2*K*RS*(VIN-VT)+1)))) FirstChildUserIdTAG: 447282 FirstChildUserNameTAG: Kuat FirstChildCreateTimeTAG: 2012-10-15T05:46:58Z IndexTAG: 2256 TitleTAG: i want answer from staff can i unregister from course now and join in next season? i cannot continue in course because of Military service begin 20/10 >> UserIdTAG: 292299 UserNameTAG: ammarsamir CreateTimeTAG: 2012-10-14T22:13:10Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: If you want to leave, just stop working on this course. Simple. FirstChildUserIdTAG: 359310 FirstChildUserNameTAG: AaronYeoh FirstChildCreateTimeTAG: 2012-10-14T23:11:27Z FirstChildTAG: Don't see why that should be a problem. FirstChildUserIdTAG: 434904 FirstChildUserNameTAG: mtrav FirstChildCreateTimeTAG: 2012-10-14T22:16:55Z FirstChildTAG: Hi, > ammarsamir I think u can unregister at any time of study.... FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-10-15T10:37:05Z IndexTAG: 2257 TitleTAG: LAB 5 LAST QUESTION somebody pease help me withthis, i dont know how can it have a particular answer UserIdTAG: 284628 UserNameTAG: jumana_mp CreateTimeTAG: 2012-10-14T20:12:49Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: It's actually simpler than it looks...think of the operational limits for Vin that you found in the first part...and how they apply to the combination of Vbias and Vsignal ;) FirstChildUserIdTAG: 604636 FirstChildUserNameTAG: Klionheart FirstChildCreateTimeTAG: 2012-10-14T21:23:38Z SecondChildTAG: hi Klionheart ! can you please help me in finding out the Input voltage upper end of linear operating range? is it not obtained by the expression: VT+(-1+sqrt(1+2*k*RL*VS))/(k*RL) ?? SecondChildUserIdTAG: 260272 SecondChildUserNameTAG: saikat24 SecondChildCreateTimeTAG: 2012-10-14T21:28:55Z SecondChildTAG: You don't need to deal with equations. As the lab stated, just play around with Vbias and Vsignal, until you get the greatest undistorted Vout possible. SecondChildUserIdTAG: 194509 SecondChildUserNameTAG: blackcompe SecondChildCreateTimeTAG: 2012-10-14T21:59:16Z IndexTAG: 2258 TitleTAG: Help with problems in course notes exercise 4.6(pg 232),problem 4.5(pg 234) and problem 4.7(pg 235)..? Can anyone suggest a method to solve the above problems. UserIdTAG: 273912 UserNameTAG: raj2691 CreateTimeTAG: 2012-10-14T19:13:44Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Nice question. I'll take a shot at 4.6. Thinking out loud (I'm thinking of text page 199): How about the Thevenin/Norton equivalent from the ends of the diode? Take it out, find the equivalent, and then put it back into the equivalent circuit? Maybe proceed graphically after that?? Intersection of the two characteristic i-v lines? FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-10-15T01:53:36Z FirstChildTAG: Problem 4.5, Pg. 234...here's my shot at this one: I'm reminded of one of our homework/lab assignments. The hint was to model the Zener as a voltage source in series with a resistor. The voltage source was at the point where the zener started conducting and the resistor was set by the slope of the zener i-v line during conduction. FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-10-15T01:57:14Z IndexTAG: 2259 TitleTAG: Help with H5P3!! At the 11th hour! So I took the equation for IDS in H5PT: **(XXX-XX)/XX+(1/(X*XX ^2) )*(1-sqrt(1+2*X*XX*(XXX-XX)))** (You know that long one ;) ) And modified it into the following form: ((**VIN+vin**-xx)/xx+(1/(x*xx ^2) )*(1-sqrt(1+2*x*xx*(**VIN+vin**-xx))))*16 So I replaced VIN with *VIN (the bias) and vin (the small signal component 0.003v)*. The whole multiplied by **16** to give me **VOUT+vout** (V=R*I). Setting VOUT to 0 I get the small signal out, and this shows a pretty shade of green for both the last questions of Homework 5. However... ...The equation: ((**VIN+vin**-xx)/xx+(1/(x*xx ^2) )*(1-sqrt(1+2*x*xx*(**VIN+vin**-xx))))*16 Which I must remind gave me pretty emeralds, is not the equation they are looking for (H5P3 Q1). I have given the equation in the correct terms (Derive an algebraic expression for the incremental output voltage vout in terms of parameters K, VT, and RS, the input bias voltage VIN, and the incremental input voltage vin). I have even tried simplifying the whole equation giving me something that looks like this: **16((1-sqrt(2*d*e(a+b-c)+1)/(e*d^2))+((a+b-c)/d))** a,b,c,d,e being the terms. Where am I going wrong?? Your help is much appreciated! UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-10-14T19:02:58Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I think that you don't need to substitue the values...and maybe your equation is too long, try to simplify it...mine worked at once FirstChildUserIdTAG: 308861 FirstChildUserNameTAG: Ignaas FirstChildCreateTimeTAG: 2012-10-14T19:16:16Z SecondChildTAG: Was this your equation? **(1+(K*RS*(vIN-VT))-sqrt(1+2*K*RS*(vIN-VT)^2))/(K*RS^2)** SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-14T19:38:39Z SecondChildTAG: Sorry, I mean: **(1+(K*RS*(VIN+vin-VT))-sqrt(1+2*K*RS*(VIN+vin-VT)^2))/(K*RS^2)** SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-14T19:39:29Z SecondChildTAG: all multiplied by 16 so: ((1+(K*RS*(VIN+vin-VT))-sqrt(1+2*K*RS*(VIN+vin-VT)^2))/(K*RS^2))*16 SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-14T19:40:12Z SecondChildTAG: no this not my equation....there are ways to simplify your equation...is your answer for question 3 of problem 2 correct?? SecondChildUserIdTAG: 308861 SecondChildUserNameTAG: Ignaas SecondChildCreateTimeTAG: 2012-10-15T00:55:27Z SecondChildTAG: you have to simplify the h5p2 vout => as short as you can. So, derive this equation and multiply by vin. That's it. It took me almost 4 hours(p2+p3)!!! But it's possible get succeeded! SecondChildUserIdTAG: 385692 SecondChildUserNameTAG: tpfslima SecondChildCreateTimeTAG: 2012-10-15T02:00:20Z IndexTAG: 2260 TitleTAG: DEFEATED Sorry but I just cannot get the equations part of this class I get the circuits and how they are put together and what makes them tick, but I just cannot get the equations part to back them up. I'm so afraid that I will not pass this class no matter how hard I try. No matter how hard I try and think that I get it right I get all the wrong numbers. I do not know what to do. I am not asking for someone to give me answers to the questions, I am just asking for someone to help me figure out how they get the equations to the circuits and how to get to that point. An easy way to come up with the equations to back up any circuits that I build in the future UserIdTAG: 160242 UserNameTAG: Mlevins35 CreateTimeTAG: 2012-10-14T19:02:16Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Wolfram alpha! FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-10-14T19:04:23Z SecondChildTAG: And JSChambers pointed to [Equation Solver][1] - invaluable! You can't possibly know less math than me, so DON'T GIVE UP! [1]: http://www.numberempire.com/equationsolver.php SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-15T03:45:03Z SecondChildTAG: Thank you I just got frustrated because I was having trouble figuring it out I had the right equations they were just put wrong on the page I forgot to put a * where it should have been SecondChildUserIdTAG: 160242 SecondChildUserNameTAG: Mlevins35 SecondChildCreateTimeTAG: 2012-10-15T22:51:38Z FirstChildTAG: Can you provide a specific example of your dificulty? Up to this point it's just been i-V relationships mainly for the MOSFET and ideal resistors (Ohm's Law). Until you have the basic relationship correct, you can't begin to manipulate whether manually or using software assistance. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-14T19:43:25Z FirstChildTAG: wolframalpha dot com, microsoft math (i think that's a free download). both are useful tools. don't hesitate to use them. There are also a lot of easier algebraic solutions we have to do, so don't be too discouraged. A couple of them are just large...no doubt about that. FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-10-14T23:40:57Z IndexTAG: 2261 TitleTAG: lab 5 plz anyone who can give me some hints?> UserIdTAG: 413613 UserNameTAG: shaliesh CreateTimeTAG: 2012-10-14T18:21:22Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi shaliesh, You have some Hints of Lab5 in this [Post][1] Also you can check in the wiki: Myrimit's guide to 6.002x -> [Myrimit's Hints][2] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/507628b2c4dd80250000004b [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/myrimits-hints-links/ FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-14T19:47:21Z SecondChildTAG: PAGE NOT FOUND for the POST SecondChildUserIdTAG: 110802 SecondChildUserNameTAG: leoblack SecondChildCreateTimeTAG: 2012-10-14T20:07:40Z SecondChildTAG: Hi leoblack, https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/507628b2c4dd80250000004b SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T20:57:09Z IndexTAG: 2262 TitleTAG: lab 5. last ques. How to find Largest input amplitude resulting in an undistorted output signal? UserIdTAG: 72647 UserNameTAG: NEEL11 CreateTimeTAG: 2012-10-14T16:00:02Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: from the first part you have max and min undistorted voltage - it's a peak-to-peak (some times called peak-to-peak amplitude). For symmetric waveform amplitude will be peak-to-peak/2 FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-14T18:48:58Z FirstChildTAG: Mosfet should operate in linear region so vds FirstChildUserIdTAG: 296945 FirstChildUserNameTAG: saket30 FirstChildCreateTimeTAG: 2012-10-14T19:20:36Z IndexTAG: 2263 TitleTAG: Part B How has the following result arrived? VT<=Vin<=VT+sqrt(2VinRK+1)/RK and sqrt(2VSRK+1)/RK-1<=VMID<=Vs. What do these roots imply? UserIdTAG: 343388 UserNameTAG: bijojoseph CreateTimeTAG: 2012-10-14T13:00:23Z VoteTAG: 1 CoursewareTAG: Week 6 / Two-stage MOSFET amplifier CommentableIdTAG: 6002x_Two_Stage_MOSFET_amplifier NumberOfReplyTAG: 0 IndexTAG: 2264 TitleTAG: H5P1 question 2 why do we keep Vs=1v and ignore the other value of Vs and again put Vs=2v in the third question??? can anybody help me with this? UserIdTAG: 413613 UserNameTAG: shaliesh CreateTimeTAG: 2012-10-14T12:10:24Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi shaliesh, You can take a look at this [Hints][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_mosfet_amp/threads/507a1d6dd894f12300000039 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-14T21:16:53Z IndexTAG: 2265 TitleTAG: the final answer is given in the text book. the answer is a little bit different in case of some sign and the root the answer on page 963 UserIdTAG: 226085 UserNameTAG: Teto CreateTimeTAG: 2012-10-14T12:05:43Z VoteTAG: 1 CoursewareTAG: Week 5 / MOSFET Amplifier with Source Degeneration CommentableIdTAG: 6002x_MOSFET_amp_w_source_degeneration_t NumberOfReplyTAG: 1 FirstChildTAG: also why does he write VGS= Vi-Vo our friend YakovO says that it is because of the common ground can any one help please. FirstChildUserIdTAG: 226085 FirstChildUserNameTAG: Teto FirstChildCreateTimeTAG: 2012-10-14T12:10:45Z SecondChildTAG: because GS means gate-source - voltage between gate and source node of the MOSFET - voltage on the gate is vI in respect to ground, voltage on the source is vO in respect to ground, so voltage between gate and source is vI-vO SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-14T18:34:06Z SecondChildTAG: ahaaaaa...!! now I understand thank you very very much. :) SecondChildUserIdTAG: 226085 SecondChildUserNameTAG: Teto SecondChildCreateTimeTAG: 2012-10-15T00:14:21Z IndexTAG: 2266 TitleTAG: S11V10 math How do we get `VA + VOUT + VB = 0` from `... + VA + ... + VOUT + ... + VB + ... = 0` ? UserIdTAG: 330638 UserNameTAG: systemfault CreateTimeTAG: 2012-10-14T12:04:14Z VoteTAG: 1 CoursewareTAG: Week 6 / Perspective on the small-signal circuit CommentableIdTAG: 6002x_small_signal_perspective NumberOfReplyTAG: 1 FirstChildTAG: I don't know. FirstChildUserIdTAG: 61469 FirstChildUserNameTAG: Spacedog FirstChildCreateTimeTAG: 2012-10-21T09:28:28Z IndexTAG: 2267 TitleTAG: question?? in this problem we need to find vpi and substitute it in the vout equation found from the current source and the RL in the right side of circuit. I found vpi from the voltage divider rule as follow: vpi=(rpi)/(rpi+Re)vin then I substituted by this equation in vout equation. the answer I get is wrong can any one help. UserIdTAG: 226085 UserNameTAG: Teto CreateTimeTAG: 2012-10-14T10:56:41Z VoteTAG: 1 CoursewareTAG: Week 5 / BJT Small Signal Model CommentableIdTAG: 6002x_bjt_small_signal_t NumberOfReplyTAG: 1 FirstChildTAG: Because the current through $R_E$ is not the same the current through $r_\pi$ (due to the presence of the current source), it is not the voltage divider formula that you posted. FirstChildUserIdTAG: 470934 FirstChildUserNameTAG: sonhx FirstChildCreateTimeTAG: 2012-10-14T11:22:20Z IndexTAG: 2268 TitleTAG: h5p2 how to find iDS in terms of K, vIN, vOUT and vT..? UserIdTAG: 114864 UserNameTAG: Vikaash CreateTimeTAG: 2012-10-14T10:27:58Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: firsttly write the formula of ids in terms of k,vgs and vt. now aplly kvl to get relation b/w vgs vin nd vout and subsitute that in one .u wl get the answer FirstChildUserIdTAG: 611666 FirstChildUserNameTAG: Shaminder420 FirstChildCreateTimeTAG: 2012-10-14T13:21:02Z FirstChildTAG: Watch tutorial problems from week 5. It really helped me. FirstChildUserIdTAG: 172304 FirstChildUserNameTAG: Demohunter FirstChildCreateTimeTAG: 2012-10-14T10:33:17Z SecondChildTAG: Please give a hint. How to find iDS in terms of K, vIN,vOUT,vT. In my formula of iDS, it includes RS. But it is not permitted. How can I cancel the RS in the iDS formula SecondChildUserIdTAG: 371617 SecondChildUserNameTAG: rainbow12 SecondChildCreateTimeTAG: 2012-10-14T13:11:12Z IndexTAG: 2269 TitleTAG: Lab5 answer to second part need help!! I have calculated all other values but this: Input voltage upper end of linear operating range: can some one please help!! do i need to see from the graph directly the value from where the curve starts to fall down linearly from 5V. Or its something else i am missing ! UserIdTAG: 232499 UserNameTAG: Asim09 CreateTimeTAG: 2012-10-14T09:19:02Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Remember it is the middle part of the graph where it starts to turn from the horizontal line to a slope and the slope line to a curve line. You only need to put your mouse cursor on the graph to read values. FirstChildUserIdTAG: 172304 FirstChildUserNameTAG: Demohunter FirstChildCreateTimeTAG: 2012-10-14T10:38:33Z FirstChildTAG: no...the first one is the upper point and the second one is the lower point...and your values are on the x-axis...i hope this helps FirstChildUserIdTAG: 308861 FirstChildUserNameTAG: Ignaas FirstChildCreateTimeTAG: 2012-10-14T17:57:32Z SecondChildTAG: how to calculate the measured gain (the second part of the lab)... SecondChildUserIdTAG: 429168 SecondChildUserNameTAG: arunprakashavm SecondChildCreateTimeTAG: 2012-10-14T18:17:15Z SecondChildTAG: still i was not able to solve the second part, my understanding is that it should be equal to Vs. please give a hint. SecondChildUserIdTAG: 141709 SecondChildUserNameTAG: charbelantonios SecondChildCreateTimeTAG: 2012-10-15T08:41:28Z IndexTAG: 2270 TitleTAG: Doomed I just started the course yesterday UserIdTAG: 619805 UserNameTAG: Fluxion CreateTimeTAG: 2012-10-14T07:59:52Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: To get the full benefit of the course, you should start at the beginning, and forget about the grade this semester. You should take it again for the grade another time. In the spring, this course will be offered again, and in the spring, there's a good chance MITx will offer a proctored exam certificate in addition to the free honor code certificate of completion that is currently offered. Regards: asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-10-14T12:23:49Z IndexTAG: 2271 TitleTAG: Remarkable!! That graph in the video is the **same as** the one that depicts the transfer function of an inverter gate made with a MOSFET. Thanks edX! UserIdTAG: 138769 UserNameTAG: ArturoPrado CreateTimeTAG: 2012-10-14T01:50:03Z VoteTAG: 1 CoursewareTAG: Week 5 / Triode Region Demo CommentableIdTAG: 6002x_triode_rgn_d NumberOfReplyTAG: 0 IndexTAG: 2272 TitleTAG: lab 5 last question...help i need a hint..i am changing the value between 0.1 and 1 V, but cant give the right answer...i have values between 0.43 and 0.48...but none is correct UserIdTAG: 308861 UserNameTAG: Ignaas CreateTimeTAG: 2012-10-14T01:42:40Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi Ignaas! Remember that the value is peak to peak (the double of amplitud of a wave)... FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-14T02:46:57Z SecondChildTAG: but we must answer the amplitud or the peak to peak value....because i measured the peak to peak value and then devided it by 2....is this wrong? SecondChildUserIdTAG: 308861 SecondChildUserNameTAG: Ignaas SecondChildCreateTimeTAG: 2012-10-14T02:51:59Z SecondChildTAG: just get the max and min values from input and output signals then (vmax-vmin)/2 on each it's the right amplitude, remember there are some offsets but if yo do like above they don't matter. SecondChildUserIdTAG: 50320 SecondChildUserNameTAG: marm496 SecondChildCreateTimeTAG: 2012-10-14T03:14:08Z SecondChildTAG: but for the answer we need to look only at the input wave right...yhey are asking for the amplitude of the input wave SecondChildUserIdTAG: 308861 SecondChildUserNameTAG: Ignaas SecondChildCreateTimeTAG: 2012-10-14T03:18:55Z SecondChildTAG: Remember to input answer in VOLTS, not millivolts! SecondChildUserIdTAG: 254686 SecondChildUserNameTAG: Hophmel SecondChildCreateTimeTAG: 2012-10-14T14:54:47Z SecondChildTAG: i also checked that...do i need tot take the blue wave or pink wave???.... SecondChildUserIdTAG: 308861 SecondChildUserNameTAG: Ignaas SecondChildCreateTimeTAG: 2012-10-14T18:20:46Z IndexTAG: 2273 TitleTAG: H5P2 question 6 please help!!!!!! Guys, I found the right answers for question 4 and question 5....but question 6 gives me a problem, because i found a negative value under my suareroot...please help!!! UserIdTAG: 308861 UserNameTAG: Ignaas CreateTimeTAG: 2012-10-14T00:56:12Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: See tutorial video of vaccum diode FirstChildUserIdTAG: 299251 FirstChildUserNameTAG: TheRedBlackOne FirstChildCreateTimeTAG: 2012-10-14T01:08:56Z SecondChildTAG: tutorial of which week??...i don't see it in week 5 SecondChildUserIdTAG: 308861 SecondChildUserNameTAG: Ignaas SecondChildCreateTimeTAG: 2012-10-14T01:37:47Z SecondChildTAG: I wonder how come you got right answers for question 4, as i am getting same "**negative value under my squareroot**" and thus unable to solve further .....need math help SecondChildUserIdTAG: 37464 SecondChildUserNameTAG: DoubleA SecondChildCreateTimeTAG: 2012-10-14T01:57:42Z FirstChildTAG: Myrimit....can you help us?? FirstChildUserIdTAG: 308861 FirstChildUserNameTAG: Ignaas FirstChildCreateTimeTAG: 2012-10-14T02:20:06Z SecondChildTAG: Please atleast can you produce a hint or two , How come you able to solve Q4 and Q5 I am getting that -ve square root even in that Q :( ???? SecondChildUserIdTAG: 37464 SecondChildUserNameTAG: DoubleA SecondChildCreateTimeTAG: 2012-10-14T02:29:13Z SecondChildTAG: Yes sure, tell me. In which part are you lost? SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T02:48:51Z SecondChildTAG: I am lost at question 6...I have found the correct answers for question 4 and 5....but for question 6 i get a negative value under my square root...i dont know how.... double a what values are you substituing...then i can help you maybe SecondChildUserIdTAG: 308861 SecondChildUserNameTAG: Ignaas SecondChildCreateTimeTAG: 2012-10-14T02:54:00Z SecondChildTAG: In the Square root part of Quadratic Equation : In **Square Root(B^2-4AC**) part Suppose I got **-4AC** like: **-4*(RS^2*K*VT**)*(K*vIN^2-2*vIN*VT*K+K*VT) [I am not writing the original equation due to Honour Code but this is Simmilar to that I am solving Then due to the Given RS value **16** the negative part become greater than B^2 part and thus unsovleable due to imaginary Iota Condition. Where am i wrong my Q3 is OK SecondChildUserIdTAG: 37464 SecondChildUserNameTAG: DoubleA SecondChildCreateTimeTAG: 2012-10-14T03:16:50Z SecondChildTAG: Hi Ignaas, That is strange.... So, let me know if I have understanded you: You have ok your iDS formula isn't it? (green check?) And then you calculated correctly the part 4 and 5 , isn't it? (green check?) But the part 6 gives you a negative root..., Can I ask you what values are giving to you in the statement? I need that you tell me from part 6: your given vIN value, your K, your VT and your RS ? It is odd that the sqrt() gives negative... I will check with my formula. But remember that I can not tell you the result, just to verifie if this gives negative... SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T03:41:50Z SecondChildTAG: Hi Myrimit I got iDS formula OK green check but i got that "negative value under my square root" problem in Q4 , i have tried to give my problem detail in this PAGE , can u find the cause of trouble Pre-Thanks SecondChildUserIdTAG: 37464 SecondChildUserNameTAG: DoubleA SecondChildCreateTimeTAG: 2012-10-14T03:58:14Z SecondChildTAG: Got it after working out for a long time Vout and vOUT are different , so each have different Protocols for Solving ;) SecondChildUserIdTAG: 37464 SecondChildUserNameTAG: DoubleA SecondChildCreateTimeTAG: 2012-10-14T05:39:31Z SecondChildTAG: Well done DoubleA! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T14:30:27Z SecondChildTAG: Myrimit My values are Vin=5,7V, Vt=T=2.0 V, Rs=16 ohm and K=2.0 A/V^2....please help me here SecondChildUserIdTAG: 308861 SecondChildUserNameTAG: Ignaas SecondChildCreateTimeTAG: 2012-10-14T18:24:43Z SecondChildTAG: Hi Ignaas, I have checked with your values in my formula and the value inside the root (with my formula), it isn't negative... so, you should have something wrong... Are you sure that you are replacing correctly the values ? SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T18:55:56Z SecondChildTAG: Hello, Do we set VIN=VT after we have our quadratic equation? This question seems to be quite difficult. :) Thanks SecondChildUserIdTAG: 322360 SecondChildUserNameTAG: jfree004 SecondChildCreateTimeTAG: 2012-10-14T22:08:54Z SecondChildTAG: Yes jfree004, But that is only to decide which equation will be the correct one. Remember that as iDS will be a quadratic equation, you will need to decide which expression of the two possibles will be the correct one, that means, that it has to verify the requirement that they give you in the statement, that is to say, that when vIN=VT iDS is zero. ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T22:41:57Z SecondChildTAG: Hmmm, but then shouldn't our answer consist of VIN? Because it says VIN is not permitted in the answer...Lol this is a good question :) It's going to feel good when I get it eventually :P SecondChildUserIdTAG: 322360 SecondChildUserNameTAG: jfree004 SecondChildCreateTimeTAG: 2012-10-14T23:08:14Z SecondChildTAG: Hi jfree004, Your answer it should consist on vIN ... The thing of doing vIN=VT is in order to verify if your chosen formula it is the correct one... but iDS it should be in terms of vIN ... be careful, **VIN** (V uppercase IN uppercase) it is not the same as **vIN** (v lowercase IN uppercase) - you need vIN, not VIN , so that is, why it shows you that "VIN is not permitted in the answer"-... SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T23:31:35Z SecondChildTAG: Yup, messed up lol, sorry :) and thank you! :D SecondChildUserIdTAG: 322360 SecondChildUserNameTAG: jfree004 SecondChildCreateTimeTAG: 2012-10-14T23:37:34Z SecondChildTAG: You are welcome :). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T23:44:18Z SecondChildTAG: Still lost so: 1.) plug in vOUT from part 2 for vOUT of part 1? 2.) 0 = (some values) - iDS which gives a quadratic form 3.) solve for iDS using quadratic formula 4.) determine root sign by using iDS = 0 for vIN=VT I obtained a very large answer for iDS, should that be? if so then I probably just have an algebraic error....if not, idk :P lol SecondChildUserIdTAG: 322360 SecondChildUserNameTAG: jfree004 SecondChildCreateTimeTAG: 2012-10-14T23:56:26Z SecondChildTAG: Try to get the minimun expression of it (factorize variables, agrupate variables, etc)... your expression should become a little bit more shorter... SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-15T00:05:30Z SecondChildTAG: I got it!! :'D just makes me want to cry lol, easy question, just so much algebra, and many mistakes to make thanks Myrimit :) <3 SecondChildUserIdTAG: 322360 SecondChildUserNameTAG: jfree004 SecondChildCreateTimeTAG: 2012-10-15T01:08:02Z SecondChildTAG: Well done jfree004! You are welcome! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-15T01:11:26Z IndexTAG: 2274 TitleTAG: What topics will the Midterm Exam cover? Hello everyone! So, how much weeks and how much textbook material should I know in order to complete the Midter Exam? UserIdTAG: 326787 UserNameTAG: Angstrem CreateTimeTAG: 2012-10-13T22:53:07Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: According to the post, it will cover everything through the first lecture sequence in the week six material. FirstChildUserIdTAG: 331664 FirstChildUserNameTAG: dwmnctrh3 FirstChildCreateTimeTAG: 2012-10-14T01:03:55Z SecondChildTAG: Yes, that it is true :) from week 1 to 6 SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T02:54:59Z SecondChildTAG: Thanks! Now I can feel a bit easier) SecondChildUserIdTAG: 326787 SecondChildUserNameTAG: Angstrem SecondChildCreateTimeTAG: 2012-10-14T07:45:29Z SecondChildTAG: Hey is'nt capacitors and first order circuits not there for the exam?Can someone please clarify? SecondChildUserIdTAG: 11538 SecondChildUserNameTAG: trishul SecondChildCreateTimeTAG: 2012-10-23T03:11:06Z IndexTAG: 2275 TitleTAG: HELP No matter how hard I try, I am having problems understanding how to figure out how to do the calculations. I was never good at math. I get lost in the book, it does make it sound easy to understand, I just cant get it. Does anyone have any suggestions? UserIdTAG: 160242 UserNameTAG: Mlevins35 CreateTimeTAG: 2012-10-13T22:34:14Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Where do you have problems? Is that quadratic equations, calculus from the last part or something else? Particularly, what equation, for example? FirstChildUserIdTAG: 326787 FirstChildUserNameTAG: Angstrem FirstChildCreateTimeTAG: 2012-10-13T22:43:44Z SecondChildTAG: just basically if the equation says vo = vi - vt. and it asks for a number i am confused on how to get the number for vi and vt. It looks easy in the book but I get totally lost SecondChildUserIdTAG: 160242 SecondChildUserNameTAG: Mlevins35 SecondChildCreateTimeTAG: 2012-10-14T02:17:05Z SecondChildTAG: Hi Mlevins35! Can I help you? As Angstrem said, where do you have problems? SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T02:59:12Z SecondChildTAG: trying to understand how to get the numbers for the equations how to figure out the solution for vi or vt I get lost when all I have is the equation like ids = vi -vt. I understand the equation but get lost on how to get the number for those. I stink at Math but have always been interested in electronics. I can figure out how to build circuits but when asked about the equations to back them up that is where I get lost Please help SecondChildUserIdTAG: 160242 SecondChildUserNameTAG: Mlevins35 SecondChildCreateTimeTAG: 2012-10-14T18:23:16Z SecondChildTAG: Hi Mlevins35, The important is that you have the concept, that is the first step, the math is a tool. So you can do it! :) The essential in math is to be tedious, one minus sign that you skip can be the tragic end of a problem haha. So, 1) Try to write in a paper step by step your procedure. This is a good practice. 2) In the case that you have to obtain an equation from a circuit, try to find all the possible equations that you can get it without caring if this will be useful. Try to use KVL, KCL. 3) Once you have all the equations, try to combine them in order to get your variable in function of other variables. 4) Be careful with signs, always you have to be consistent in your convention signs. For examble, choose if a current will be negative if it enters a node or choose to be it positive if it enters a node, no matter what convention you use, you have to use always that convention in all the problem. 5)Try to watch the week tutorials, they are really useful. Also, you can go to read the Textbook always :). 6) Read carefully the statement and review the conditions (regions of the MOSFET, etc...). 7) Ask as many times you want, remember that if you are asking you also are learning, and your doubts can be the doubts of others and so on. 8) You can take a look at Khan Academy, I have read in others Post that was helpful for the students that had issues with math [Khan Academy][1] And don't feel alone in this. Remember that there are a lot of TA's and students that can help you. I will be here if you need to review math or anything :). My best wish to you Mlevins35. Myriam. [1]: http://www.khanacademy.org/ SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T19:27:17Z SecondChildTAG: thank you for all the useful advice I just get frustrated and want to have them work out the first time. I guess that i am impatient and must learn patience first and the rest will come thank you for all of your hints they will help SecondChildUserIdTAG: 160242 SecondChildUserNameTAG: Mlevins35 SecondChildCreateTimeTAG: 2012-10-14T21:21:34Z SecondChildTAG: Hi Mlevins35, remember that I will be here if you need help . Be up! You can do it! ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T22:43:28Z SecondChildTAG: Thank you for all of your help. really good points I followed what you said was able to figure it out saw where I made my mistakes. Thank you so much. feel a lot better knowing that there is somewhere I can go to ask for help with this. I am not good at Math and am learning as I go along. I guess that when it gets to frustrating I need to just walk away from it and come back after a little bit. SecondChildUserIdTAG: 160242 SecondChildUserNameTAG: Mlevins35 SecondChildCreateTimeTAG: 2012-10-15T22:56:03Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-16T03:36:02Z IndexTAG: 2276 TitleTAG: How can I distinguish Drain and Source? Help me anyone please! This question have exploded my brain! I can not understand, how can I distinguish Source and Drain of the mosfet. So, how??? UserIdTAG: 326787 UserNameTAG: Angstrem CreateTimeTAG: 2012-10-13T21:01:47Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hi Angstrem! If you have a MOSFET: - The middle is the Gate. - The up is the Drain. - The down is the Source. ![image][1] The vDS(Drain Source) The vGS(Gate Source) ![IM2][2] See you! Myriam. [1]: https://edxuploads.s3.amazonaws.com/1350184167134366.png [2]: https://edxuploads.s3.amazonaws.com/13501843361343653.png FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-14T03:13:21Z SecondChildTAG: "If you have a MOSFET: The middle is the Gate. The up is the Drain. The down is the Source." You might want to say **top** is the drain and **bottom** is the source to avoid confusion with the direction of the arrows in the diagram. SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-14T03:46:30Z SecondChildTAG: Thank you, Myrimit! But I still can't figure out, how can I distinguish that two terminals in a particular circuit? For example, this one: ![enter image description here][1] It seems, that the source and the drain can appear anywhere, from any side of the MOSFET... [1]: https://edxuploads.s3.amazonaws.com/13502010331343635.png SecondChildUserIdTAG: 326787 SecondChildUserNameTAG: Angstrem SecondChildCreateTimeTAG: 2012-10-14T07:51:32Z FirstChildTAG: For NMOSFET, which is the only type covered so far in the course, the drain is more positive than the source for normal operation. Look to see which terminal is connected to the more positive potential. Actually, for the form of symbol being used the source and drain should be labelled with an S and D respectively to avoid any confusion. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-14T19:09:09Z SecondChildTAG: Thanks... It seems that I haven't managed to understand the essence of the mosfets. I'll have to work on it. SecondChildUserIdTAG: 326787 SecondChildUserNameTAG: Angstrem SecondChildCreateTimeTAG: 2012-10-14T21:58:08Z IndexTAG: 2277 TitleTAG: help with problems of week 5 I am stuck at the last part of problem 1....can't seem to find the right answer...and problem 2 looks also rather difficult....Minimith don't you have hints this week? UserIdTAG: 308861 UserNameTAG: Ignaas CreateTimeTAG: 2012-10-13T19:03:47Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: how u find vout in ist part of qustion 1?? FirstChildUserIdTAG: 228871 FirstChildUserNameTAG: ELINAKHAN FirstChildCreateTimeTAG: 2012-10-13T19:17:56Z FirstChildTAG: Hi Ignaas! :) Ok, I will write right now Hints of H5P1 . Give me some minutes to elaborate them. Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-13T19:58:16Z SecondChildTAG: Here what I have promised [Post H5P1][1] :) [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_mosfet_amp/threads/507a1d6dd894f12300000039 SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T02:36:20Z FirstChildTAG: Myrimit will you that also for problem 2....I am done wit problem 1 now FirstChildUserIdTAG: 308861 FirstChildUserNameTAG: Ignaas FirstChildCreateTimeTAG: 2012-10-13T20:32:15Z SecondChildTAG: Hi Ignaas for H5P2 take a look at this [post][1] :) [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5079c393be483a230000000e SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-13T21:02:33Z SecondChildTAG: thnx Myrimit....I tried it on my own and i got the first two questions right, but the third one is hard...actuallu it is easy but so much to write SecondChildUserIdTAG: 308861 SecondChildUserNameTAG: Ignaas SecondChildCreateTimeTAG: 2012-10-13T21:18:18Z SecondChildTAG: Myrimit, I have problems finding the correct answer of the first question of lab 5...i have the second question roght, but is that difficult to find the first one.i tried so many values which i found with my curson, but still not right SecondChildUserIdTAG: 308861 SecondChildUserNameTAG: Ignaas SecondChildCreateTimeTAG: 2012-10-13T21:53:05Z SecondChildTAG: Can I help you Ignaas? SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T02:37:08Z SecondChildTAG: Myrimit...thanks...i have found the answer..i was just reading the wrong axis... SecondChildUserIdTAG: 308861 SecondChildUserNameTAG: Ignaas SecondChildCreateTimeTAG: 2012-10-14T02:55:46Z SecondChildTAG: Well done Ignaas! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T03:43:07Z IndexTAG: 2278 TitleTAG: Problem with H5P1 in H5P1, in the second question it is asked that Vin=0 and then it is assumed that the MOSFET is in saturation!!! how is it possible. if Vin=0 it means Vgs will be -1 which is less than Vt and eventually the MOSFET will operate in cutoff. please explain me what is the problem. I have tried a lot but to no avail. UserIdTAG: 64618 UserNameTAG: ashfaq2419 CreateTimeTAG: 2012-10-13T18:33:52Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: No, Vgs will be +1. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-13T18:55:56Z SecondChildTAG: how +1? if Vg=Vin=0 and Source is connected to -1 V then 0-1=-1. isn't it? SecondChildUserIdTAG: 64618 SecondChildUserNameTAG: ashfaq2419 SecondChildCreateTimeTAG: 2012-10-13T19:15:45Z SecondChildTAG: 0 - (-1) SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-13T20:01:32Z SecondChildTAG: You can take a look at this [Hints][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_mosfet_amp/threads/507a1d6dd894f12300000039 SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T21:17:43Z IndexTAG: 2279 TitleTAG: H5P2 - setting up iDS equation I have no idea how to set this up. After reading almost all the posts regarding this question, I'm still not certain. A nudge in some direction would be really helpful. UserIdTAG: 141376 UserNameTAG: krtica CreateTimeTAG: 2012-10-13T18:17:32Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: iDS=K/2*(vIN-iDS*RS-VT)^2 put values..u will find qudrtc euation for iDs solve this...use iDS value that u will find from qudrtc eqution....aftr that calculte vout=iDS*RS.. ur problm will solve..do this..and telll me where u stuck..i will help FirstChildUserIdTAG: 228871 FirstChildUserNameTAG: ELINAKHAN FirstChildCreateTimeTAG: 2012-10-13T18:34:16Z SecondChildTAG: @Elinakhan: I'm totally blank for the 3rd one! I did put up the Ids value in eq 1 and solved but im getting something like 2/X*XX as the answer! Please help :( SecondChildUserIdTAG: 108454 SecondChildUserNameTAG: Raven7281 SecondChildCreateTimeTAG: 2012-10-13T19:36:20Z FirstChildTAG: Hi krtica! Take a look at this [Post][1] :). I hope this can help you. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5079c393be483a230000000e FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-13T20:23:25Z SecondChildTAG: Thanks! Your explanations are great. SecondChildUserIdTAG: 141376 SecondChildUserNameTAG: krtica SecondChildCreateTimeTAG: 2012-10-13T20:26:47Z SecondChildTAG: You are welcome :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-13T21:03:00Z IndexTAG: 2280 TitleTAG: Equation for MOSFET Is equation (K/2)*(Vgs-Vt)^2 is true for all types of real MOSFET transistors? And where I can find parametr K in datasheets (for example IRLML2502)? UserIdTAG: 104522 UserNameTAG: IgorNovice CreateTimeTAG: 2012-10-13T18:07:29Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: you can find transconductance, which related to K there is also channel-length modulation effect which is neglected http://en.wikipedia.org/wiki/Channel_length_modulation FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-13T18:35:06Z IndexTAG: 2281 TitleTAG: VT IS vt includes the answer? UserIdTAG: 395257 UserNameTAG: blackguitar CreateTimeTAG: 2012-10-13T17:31:22Z VoteTAG: 1 CoursewareTAG: Week 5 / MOSFET Amplifier Exercise 1 CommentableIdTAG: 6002x_mosfet_amp_e1 NumberOfReplyTAG: 1 FirstChildTAG: "show answer" bottom doesn't exist FirstChildUserIdTAG: 395257 FirstChildUserNameTAG: blackguitar FirstChildCreateTimeTAG: 2012-10-13T17:32:58Z SecondChildTAG: type " VT " instead of " vt" SecondChildUserIdTAG: 209930 SecondChildUserNameTAG: saikiraniitr SecondChildCreateTimeTAG: 2012-10-13T19:11:19Z SecondChildTAG: Hi blackguitar, I don't understand your question.... In which part are you lost? Can I help you? What saikiraniitr says it is true, write VT instead of vt. Remember that answers are case sensitive... SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-13T20:26:23Z SecondChildTAG: hi,Myrimit how are u? thank u for your offering to help me , i did it !!!!!!!!!.....after trying so many times ,and once the solution appears ,i get it and understand it,,, thank u again SecondChildUserIdTAG: 395257 SecondChildUserNameTAG: blackguitar SecondChildCreateTimeTAG: 2012-10-13T23:41:32Z SecondChildTAG: Well done blackguitar! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T00:57:11Z IndexTAG: 2282 TitleTAG: What is K in H5P2!!! What is K in H5P2!!! UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-13T17:19:20Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi anonymous! :) K is a constant having units of $\frac{A}{V^2}$ The value of K is related to the physical propieties of the MOSFET. In the statement they give you that value. See you! Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-13T20:20:51Z IndexTAG: 2283 TitleTAG: can anyone tell me how can i change my full name in my profile so that i can get certificate with that name?plz reply if anyone knows? can anyone tell me how can i change my full name in my profile so that i can get certificate with that name?plz reply if anyone knows? UserIdTAG: 274748 UserNameTAG: flynn_vijay CreateTimeTAG: 2012-10-13T17:04:20Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I am sure that when the time comes for you to get your certificate , MITx will require your full name. Further, it is an unfortunate fact, in our day and age, that it is not safe to give your full name on the internet. FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-10-13T17:11:51Z SecondChildTAG: trueeeeeey SecondChildUserIdTAG: 127800 SecondChildUserNameTAG: MNB SecondChildCreateTimeTAG: 2012-11-18T17:18:28Z SecondChildTAG: even i have a doubt regarding the way name appears on certificate,i want my name to appear as Xxxxxxxx Yyyyyyyy,but in my profile i have given my name as xxxxxxxx yyyyyyyy so should i change it??so actually it is not change of name just case(uuper case and lower case) change.someone please reply as i am about to complete my course,i have even forwarded the same query on info@edx.org,technical@edx.org. SecondChildUserIdTAG: 353709 SecondChildUserNameTAG: anshu10750 SecondChildCreateTimeTAG: 2012-12-22T08:07:11Z IndexTAG: 2284 TitleTAG: started course on october 13 i have log in to this course on October 13 2012.how can i submit the previous homework and lab work? UserIdTAG: 75873 UserNameTAG: priyaranjan8 CreateTimeTAG: 2012-10-13T16:48:12Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: no man..u can practice those out for ur mid term! by the way u bengali? FirstChildUserIdTAG: 128409 FirstChildUserNameTAG: arijitbme FirstChildCreateTimeTAG: 2012-10-13T16:52:46Z SecondChildTAG: mr. priyaranjan r u from Univ of maryland. SecondChildUserIdTAG: 232499 SecondChildUserNameTAG: Asim09 SecondChildCreateTimeTAG: 2012-10-13T17:47:07Z FirstChildTAG: To get the full benefit of the course, you should start at the beginning, and forget about the grade this semester. You should take it again for the grade another time. In the spring, this course will be offered again, and in the spring, there's a good chance **MITx** will offer a proctored exam certificate in addition to the free honor code certificate of completion that is currently offered. Regards: asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-10-13T17:20:24Z FirstChildTAG: My best suggestion is to understand as much as you can this time around, then take the course again when it becomes available. That way you will be extra prepared! FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-10-13T17:51:24Z IndexTAG: 2285 TitleTAG: HWP1 please tell me what value for RL we wil use?? i dnt able to understand UserIdTAG: 228871 UserNameTAG: ELINAKHAN CreateTimeTAG: 2012-10-13T14:57:06Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: We know solve: 0=Vs-(RL*K)VT^2/2. express and solve FirstChildUserIdTAG: 328920 FirstChildUserNameTAG: TisO FirstChildCreateTimeTAG: 2012-10-13T15:12:12Z SecondChildTAG: what value of vs will use? 2 or 1 SecondChildUserIdTAG: 228871 SecondChildUserNameTAG: ELINAKHAN SecondChildCreateTimeTAG: 2012-10-13T15:38:57Z FirstChildTAG: The value of K is in mV. I guess you would have not noticed it. FirstChildUserIdTAG: 362299 FirstChildUserNameTAG: anandbaskaran FirstChildCreateTimeTAG: 2012-10-13T16:09:33Z SecondChildTAG: (VS+) - (VS-) = K*VT*RL/2.. Substitute the known values and find RL. SecondChildUserIdTAG: 362299 SecondChildUserNameTAG: anandbaskaran SecondChildCreateTimeTAG: 2012-10-13T16:18:10Z SecondChildTAG: what value of k i will put?? tel me numeric value.. i stuck wd it SecondChildUserIdTAG: 228871 SecondChildUserNameTAG: ELINAKHAN SecondChildCreateTimeTAG: 2012-10-13T16:20:57Z SecondChildTAG: 8000 SecondChildUserIdTAG: 362299 SecondChildUserNameTAG: anandbaskaran SecondChildCreateTimeTAG: 2012-10-13T16:24:05Z SecondChildTAG: Help me in last question of Lab 5 SecondChildUserIdTAG: 362299 SecondChildUserNameTAG: anandbaskaran SecondChildCreateTimeTAG: 2012-10-13T16:24:21Z SecondChildTAG: the value of k still not workng :( SecondChildUserIdTAG: 228871 SecondChildUserNameTAG: ELINAKHAN SecondChildCreateTimeTAG: 2012-10-13T16:45:15Z SecondChildTAG: Hi, 'ELINAKHAN' value which 'anandbaskaran' mention is not worked till you find Minimum Vin. Regards: asadbhatti42 SecondChildUserIdTAG: 329051 SecondChildUserNameTAG: asadbhatti42 SecondChildCreateTimeTAG: 2012-10-13T17:26:03Z SecondChildTAG: not able to solve :( SecondChildUserIdTAG: 228871 SecondChildUserNameTAG: ELINAKHAN SecondChildCreateTimeTAG: 2012-10-13T17:52:09Z SecondChildTAG: K cant be in mV - it's A/V^2 so if you have mA/V^2 it means *10^-3 SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-13T18:48:46Z SecondChildTAG: now give me eqution for vout.... i tried many eqution but not wrkng SecondChildUserIdTAG: 228871 SecondChildUserNameTAG: ELINAKHAN SecondChildCreateTimeTAG: 2012-10-13T19:03:44Z SecondChildTAG: finaly find RL..thanks. May ALLAH gve u big succes.. now stuck wd vout :( SecondChildUserIdTAG: 228871 SecondChildUserNameTAG: ELINAKHAN SecondChildCreateTimeTAG: 2012-10-13T21:23:22Z IndexTAG: 2286 TitleTAG: H5P2 Vout HELP I have "I" Now how to find Vout? I Got the root. I have to dif. it. How UserIdTAG: 286954 UserNameTAG: ododo CreateTimeTAG: 2012-10-13T13:24:08Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: We have ids. Vout=ids*RS FirstChildUserIdTAG: 328920 FirstChildUserNameTAG: TisO FirstChildCreateTimeTAG: 2012-10-13T14:52:35Z SecondChildTAG: It's helped me SecondChildUserIdTAG: 328920 SecondChildUserNameTAG: TisO SecondChildCreateTimeTAG: 2012-10-13T14:53:39Z SecondChildTAG: good SecondChildUserIdTAG: 228871 SecondChildUserNameTAG: ELINAKHAN SecondChildCreateTimeTAG: 2012-10-13T14:58:30Z SecondChildTAG: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50798142cf35262700000083 please help me SecondChildUserIdTAG: 228871 SecondChildUserNameTAG: ELINAKHAN SecondChildCreateTimeTAG: 2012-10-13T14:59:03Z SecondChildTAG: oh, this page "not found"=( SecondChildUserIdTAG: 328920 SecondChildUserNameTAG: TisO SecondChildCreateTimeTAG: 2012-10-13T15:07:39Z SecondChildTAG: Can anybody help me with H5P2 2 question? what I did I put question 1 answer here Vout=ids*RS, I solve it got the roots. So I put one of the root as answer and its say wrong???? SecondChildUserIdTAG: 286954 SecondChildUserNameTAG: ododo SecondChildCreateTimeTAG: 2012-10-13T15:17:43Z SecondChildTAG: iDS*RS put ectly this.. its case sensitive SecondChildUserIdTAG: 228871 SecondChildUserNameTAG: ELINAKHAN SecondChildCreateTimeTAG: 2012-10-13T15:55:41Z SecondChildTAG: ELINAKHAN THANKS A LOT :))) SecondChildUserIdTAG: 286954 SecondChildUserNameTAG: ododo SecondChildCreateTimeTAG: 2012-10-13T17:11:17Z SecondChildTAG: ur welcome... help me in H5P1 :( ccnt able to solve SecondChildUserIdTAG: 228871 SecondChildUserNameTAG: ELINAKHAN SecondChildCreateTimeTAG: 2012-10-13T17:19:30Z FirstChildTAG: iDS=K/2*(vIN-iDS*RS-VT)^2_ iDS=K/2*(VT-iDS*RS-VT)^2 _ iDS=K/2*(-iDS*RS)^2 _1=K/2*-iDS*(RS)^2_ IDS=1/(K/2*RS^2) Why is it incorrect? FirstChildUserIdTAG: 192114 FirstChildUserNameTAG: chuba FirstChildCreateTimeTAG: 2012-10-13T15:24:43Z SecondChildTAG: tell me ur V1N value... i willl tel step by step SecondChildUserIdTAG: 228871 SecondChildUserNameTAG: ELINAKHAN SecondChildCreateTimeTAG: 2012-10-13T15:57:41Z SecondChildTAG: VIN = VT SecondChildUserIdTAG: 192114 SecondChildUserNameTAG: chuba SecondChildCreateTimeTAG: 2012-10-13T16:03:56Z SecondChildTAG: ?? SecondChildUserIdTAG: 374512 SecondChildUserNameTAG: arsen55591 SecondChildCreateTimeTAG: 2012-10-13T23:17:34Z FirstChildTAG: how u got ids FirstChildUserIdTAG: 611666 FirstChildUserNameTAG: Shaminder420 FirstChildCreateTimeTAG: 2012-10-13T13:28:36Z SecondChildTAG: it is very simple, but a little tedious > Solve the previous two results for iDS. >You will get a quadratic equation, and you will have to choose the correct root: make sure that iDS=0 when vIN=VT. You have to solve a quadratic equation.Be careful! SecondChildUserIdTAG: 328920 SecondChildUserNameTAG: TisO SecondChildCreateTimeTAG: 2012-10-13T14:49:35Z SecondChildTAG: iDS=K/2*(vIN-iDS*RS-VT)^2 this formula will give u iDS SecondChildUserIdTAG: 228871 SecondChildUserNameTAG: ELINAKHAN SecondChildCreateTimeTAG: 2012-10-13T14:54:55Z SecondChildTAG: Rs value??? SecondChildUserIdTAG: 127800 SecondChildUserNameTAG: MNB SecondChildCreateTimeTAG: 2012-10-14T08:09:58Z SecondChildTAG: But how to cancel Rs, it is not permitted in iDS formula? SecondChildUserIdTAG: 371617 SecondChildUserNameTAG: rainbow12 SecondChildCreateTimeTAG: 2012-10-14T13:14:42Z SecondChildTAG: it is permitted as RS SecondChildUserIdTAG: 324332 SecondChildUserNameTAG: petsol SecondChildCreateTimeTAG: 2012-10-14T13:21:12Z SecondChildTAG: But how to cancel IDS, it is not permitted in iDS formula? SecondChildUserIdTAG: 37708 SecondChildUserNameTAG: Jband SecondChildCreateTimeTAG: 2012-10-14T22:08:27Z IndexTAG: 2287 TitleTAG: what if the capacitor has a initial value Does it mean the capacitor is not a linear element anymore when it has a initial voltage ? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-13T12:47:22Z VoteTAG: 1 CoursewareTAG: Week 6 / Ideal linear capacitor CommentableIdTAG: 6002x_ideal_linear_capacitor NumberOfReplyTAG: 1 FirstChildTAG: until the capacitor is fully charged ,it acts as a closed circiut. we can always generate the condition u mentioned by focusing on the charging period. Say at half charged , take that as initial value, but that won't mean it is not linear FirstChildUserIdTAG: 168496 FirstChildUserNameTAG: poweltalwar FirstChildCreateTimeTAG: 2012-10-14T10:47:55Z SecondChildTAG: I know it's a linear element.But if it has a initial value,can we still treat it as a linear element? SecondChildUserIdTAG: 136959 SecondChildUserNameTAG: christerpher SecondChildCreateTimeTAG: 2012-10-15T07:52:47Z IndexTAG: 2288 TitleTAG: H5P1 B How do I go about finding RL? I have the equation VDS=K/2*RL*(VI-VT)^2, setting VDS and VI to 0 I get wrong answer! **EDIT:** I realise that VDS =/ VO so... ...I have also tried the following! **VO = VS - K/2(VI-VT)^2*RL** where **K = 1mA/VT^2** **K = 0.001/0.5^2 = 0.004** Substituting values I get: **0 = 1v - 0.004/2(0-(-0.5))^2*X** or **2k** Ohms which is wrong... Help would be much appreciated! UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-10-13T10:21:14Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: VI non equal VGS in your starting equation because Source isn't graund potential FirstChildUserIdTAG: 325197 FirstChildUserNameTAG: Vitali_Jerin FirstChildCreateTimeTAG: 2012-10-13T12:14:55Z SecondChildTAG: Thanks!! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-13T12:31:49Z SecondChildTAG: So from KVL VGS=VI-VS When I substitute this in VO = VS - K/2(VI-VS-VT)^2*RL Setting **VO to 0 VI to 0 K to 0.004 VS to 1 RL to x** and plugging the data into wolfram alpha I get 222.222 ohms I still get a wrong answer! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-13T13:28:48Z SecondChildTAG: I guess the offset must be applied to Vo also... But I am getting totally confused with signs, can anyone help. is the offset positive or negative...? SecondChildUserIdTAG: 137686 SecondChildUserNameTAG: Jaychandran SecondChildCreateTimeTAG: 2012-10-13T13:40:14Z SecondChildTAG: Try K = 0.001 SecondChildUserIdTAG: 192114 SecondChildUserNameTAG: chuba SecondChildCreateTimeTAG: 2012-10-13T13:50:32Z SecondChildTAG: I just figured that out!! How do we choose V^2 in the expression for K? Is it by convention Vs?? SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-13T14:01:38Z SecondChildTAG: A/V^2 in K=blahblahblah - it's a units of measurement, not a formula SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-13T17:24:32Z SecondChildTAG: So what do you put in place of k in the formula, say, for IDS? SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-13T17:36:08Z SecondChildTAG: for K=1mA/V^2 it's 0.001 and use capital K - some people already had problems when grader accepts 'k' instead of 'K' but don't give green mark SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-13T17:44:00Z SecondChildTAG: Thank you SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-13T17:45:39Z SecondChildTAG: I am still not getting it SecondChildUserIdTAG: 492496 SecondChildUserNameTAG: bikerider SecondChildCreateTimeTAG: 2012-10-14T11:13:23Z IndexTAG: 2289 TitleTAG: TOUBLESHOOT Invalid input: Could not parse ' (vIN-VT)/RS + (1/(K*RS ^2) )*(1-sqrt(1+2K*RS*(vIN-VT)))' as a formula i know my formula is correct as i am getting rest of the answers using this formula . UserIdTAG: 344720 UserNameTAG: Vaibhav_Kashyap CreateTimeTAG: 2012-10-13T09:27:27Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: 2K should be 2*K FirstChildUserIdTAG: 270011 FirstChildUserNameTAG: SnowmanZA FirstChildCreateTimeTAG: 2012-10-13T11:02:52Z IndexTAG: 2290 TitleTAG: HW due 11:59PM Boston Time? Hi, I'm a new member and I tried to search for this information but I couldn't find it. Is the HW due at 11:59PM Boston Time on Sunday? I'm trying to catch up on the materials. Thanks! TD UserIdTAG: 141199 UserNameTAG: dlanhtuan2114 CreateTimeTAG: 2012-10-13T03:12:48Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Yup Sir. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-13T03:32:01Z SecondChildTAG: thank you :D SecondChildUserIdTAG: 141199 SecondChildUserNameTAG: dlanhtuan2114 SecondChildCreateTimeTAG: 2012-10-13T05:25:46Z FirstChildTAG: no , local time FirstChildUserIdTAG: 382505 FirstChildUserNameTAG: AhmedGalal2 FirstChildCreateTimeTAG: 2012-10-13T21:28:23Z SecondChildTAG: please confirm. particular city's local time? or boston's 11:59pm? SecondChildUserIdTAG: 217094 SecondChildUserNameTAG: karanpanchal SecondChildCreateTimeTAG: 2012-10-14T06:14:31Z IndexTAG: 2291 TitleTAG: This seems circular... If the purpose of an AMP is to boost a signal, isn't it counter-productive to shrink the incoming signal? Can we boost it significantly above the incoming signal? Also, what if there's noise on the incoming voltage? Won't that completely mess with our signal? UserIdTAG: 164898 UserNameTAG: jbparkes CreateTimeTAG: 2012-10-13T02:06:40Z VoteTAG: 1 CoursewareTAG: Week 5 / Linear Amplification CommentableIdTAG: 6002x_linear_amp NumberOfReplyTAG: 2 FirstChildTAG: If I give you a boost to see over the fence, you are still short. If I zap you with my ray-gun and you become a 10' version of your former self, you have been amplified. That is the difference. If I have a AC signal that swings from +1 to -1, (2v peak to peak) and I add 7 volts of DC current I have now boosted the signal. It now swings from 6v to 8v, it is still 2v peak to peak. I have added 7 volts of DC bias. We shrink some signals to make the output more linear. We want to use as flat of a curve as possible, so sometimes you need to shrink the signal a bit to use a flatter part of the curve. Yes you lose a little bit of output in exchange for a more accurate waveform. Noise can mess up the signal, ideally the noise level is small enough not to bother it too much. There are wiring, shielding and grounding techniques that we can implement to minimise noise. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-13T03:18:30Z FirstChildTAG: let's say me shrink it 4 times and then amplify 40 = +20dB FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-13T04:39:31Z IndexTAG: 2292 TitleTAG: ¿Problems with the algebraic expressions? Don't use an equation complete expression ( for example A=B+C), use only the right side of the equation for each answer (for example B+C). :D UserIdTAG: 70519 UserNameTAG: Fipe CreateTimeTAG: 2012-10-13T00:36:03Z VoteTAG: 1 CoursewareTAG: Week 5 / MOSFET Amplifier Exercise 1 CommentableIdTAG: 6002x_mosfet_amp_e1 NumberOfReplyTAG: 2 FirstChildTAG: :) FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-13T03:42:18Z FirstChildTAG: thanks cap :) FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-13T04:25:06Z IndexTAG: 2293 TitleTAG: MOSFET Implementation of the Difference Amplifier, Page 433 This post discusses creating the "Difference-Mode Model" on page 433. At the beginning of the paragraph in the middle of the page, it states "Since gm and Ri are independent of each other, vs = 0." How is this determined? Is it due to the symmetry of gm1*Vgs1 and gm2*Vgs2, where their potentials cancel vs out? Is there someone who can explain this? UserIdTAG: 132828 UserNameTAG: rharris CreateTimeTAG: 2012-10-12T23:18:10Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 2294 TitleTAG: not finding direct video link As you know youtube is closed in some countries so i am unable to watch lecture video, as previous not finding direct link for week 5 and onward. UserIdTAG: 426051 UserNameTAG: sslohana CreateTimeTAG: 2012-10-12T18:34:55Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: http://www.ruudoleo.com/mitx/small/ FirstChildUserIdTAG: 79885 FirstChildUserNameTAG: ruudoleo FirstChildCreateTimeTAG: 2012-10-12T20:39:27Z FirstChildTAG: Did you tried to open edx in Google chrome browser ? FirstChildUserIdTAG: 227508 FirstChildUserNameTAG: bhavyab FirstChildCreateTimeTAG: 2012-10-12T19:28:01Z FirstChildTAG: Did you tried to open EDX in Google chrome browser? FirstChildUserIdTAG: 227508 FirstChildUserNameTAG: bhavyab FirstChildCreateTimeTAG: 2012-10-12T19:30:51Z FirstChildTAG: [Direct Video Link for Week (1-6)][1] [1]: https://dl.dropbox.com/u/24096724/6002xMP4/menu.html# FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-10-13T06:44:52Z IndexTAG: 2295 TitleTAG: H5P2 Q3 = sleepless nights I seem to be having difficulty with quadratic equations (what hope do I have of completing this course!). Now, explaining this without revealing answers could be tricky... I am using the equation I obtained in H5P2 Q1 and expanding this out... I then use the obtained values of "a", "b" and "c" to plug into the quadratic equation, however it seems that the value of "b^2" and "4ac" are the same and so cancel out, leaving "0" under the sqrt()... Has anyone else gone down this route and ended up with a similar problem? Is there something fundamental that I am doing wrong...? UserIdTAG: 277336 UserNameTAG: MarkPearce CreateTimeTAG: 2012-10-12T17:43:46Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: As I recall b was something like 1+x and 4ac was x^2 so the result was (1+x)^2-x^2, which simplified to 1 + 2x. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-12T18:21:36Z SecondChildTAG: Thanks, I was missing the "1", how silly!... no more sleepless nights :) SecondChildUserIdTAG: 277336 SecondChildUserNameTAG: MarkPearce SecondChildCreateTimeTAG: 2012-10-12T21:40:36Z FirstChildTAG: Hint: 1. you will find out what is vOUT . mean applying expression 1 to 2 2. make it simpler by: vIN - VT = VIT and G = K*RS/2 3. you will have something like: y = G*(VIT- y)^2 4. this is quadrat function: then you find y1, and y2. but must select only one of them is correct--> selection condion we have VIT =0 --> iDS must be zero. 5. but you know iDS = 0 only when vOUT = 0 mean when VIT = 0, one of y (or vout) must be =0. that is all FirstChildUserIdTAG: 114913 FirstChildUserNameTAG: ngoctuan FirstChildCreateTimeTAG: 2012-10-12T18:26:12Z SecondChildTAG: how you make the equation in step 3,i am much confused about that.plz help SecondChildUserIdTAG: 97359 SecondChildUserNameTAG: azeem179 SecondChildCreateTimeTAG: 2012-10-13T15:34:50Z FirstChildTAG: in hsp2 q1 m getting the error-couldn't parse this formulae,,,i've entered the exp of ids =k/2(vgs -vt)^2 &vgs as vin n vout,, FirstChildUserIdTAG: 183166 FirstChildUserNameTAG: yogeshk FirstChildCreateTimeTAG: 2012-10-12T19:07:55Z SecondChildTAG: are you remembering to use parentheses and the * symbol for multiplication? SecondChildUserIdTAG: 277336 SecondChildUserNameTAG: MarkPearce SecondChildCreateTimeTAG: 2012-10-12T19:15:32Z SecondChildTAG: even if i try that it keeps on saying couldn't parse as a formula vin*(1-1/sqrt(1+2K*RS*(VIN-VT)))) is this in the correct format SecondChildUserIdTAG: 424932 SecondChildUserNameTAG: aakash29 SecondChildCreateTimeTAG: 2012-10-13T07:02:49Z SecondChildTAG: oh wrong here: 1/sqrt(1+**2K** it must be 2*K SecondChildUserIdTAG: 114913 SecondChildUserNameTAG: ngoctuan SecondChildCreateTimeTAG: 2012-10-13T14:48:39Z FirstChildTAG: what worked for me was discovering how to put a radical sign in the box type sqrt before the term under the root instead of (blah)^.5 FirstChildUserIdTAG: 18073 FirstChildUserNameTAG: gburkhart FirstChildCreateTimeTAG: 2012-10-13T00:06:43Z IndexTAG: 2296 TitleTAG: week 5 lab I am tiring hard on the last part of week 5 lab. Still i am not done can some one tell how to do it? UserIdTAG: 151662 UserNameTAG: krishanunat6 CreateTimeTAG: 2012-10-12T16:54:16Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: what the problem ? do you mean the amplitube of signal? - amplitube mean you get the : max val - min val then get the mid val from this subtraction. max and min values you find at the first of lab. because the signal is swing between max and min. FirstChildUserIdTAG: 114913 FirstChildUserNameTAG: ngoctuan FirstChildCreateTimeTAG: 2012-10-12T18:57:43Z SecondChildTAG: last problem is still problematic for me SecondChildUserIdTAG: 227508 SecondChildUserNameTAG: bhavyab SecondChildCreateTimeTAG: 2012-10-12T22:33:57Z FirstChildTAG: I am submitting those values but it shows wrong. Also I calculated it from "interactive demonstration" but those values still shows wrong.. FirstChildUserIdTAG: 227508 FirstChildUserNameTAG: bhavyab FirstChildCreateTimeTAG: 2012-10-12T20:45:00Z FirstChildTAG: The first part just asks what the voltages are for the device to be in saturation. Click on TRANS and find the voltages for the upper and lower part of the saturation region. The second part is just a ratio of the input amplitude to the output amplitude. Hope this helps, Have fun. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-12T22:24:09Z IndexTAG: 2297 TitleTAG: exercise 3.9 i can't understand number c in the problem ? and what is the result of number b ? UserIdTAG: 221617 UserNameTAG: konan CreateTimeTAG: 2012-10-12T08:17:41Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 2298 TitleTAG: Wk 5 Tutorial Feedback I find all the tutorials very helpful, but this one where they pause between each equation along the way helps me think it through. I would like to see all the tutorials use this format. Thanks for listening! UserIdTAG: 175205 UserNameTAG: KT_MM CreateTimeTAG: 2012-10-12T01:48:33Z VoteTAG: 1 CoursewareTAG: Week 5 / BJT Small Signal Model CommentableIdTAG: 6002x_bjt_small_signal_t NumberOfReplyTAG: 0 IndexTAG: 2299 TitleTAG: Last equation: Should RL be squared? I don't think so; it looks like a typo to me. Anyone? UserIdTAG: 148542 UserNameTAG: planetscape CreateTimeTAG: 2012-10-11T20:32:01Z VoteTAG: 1 CoursewareTAG: Week 5 / Small Signal Mathematically Described CommentableIdTAG: 6002x_small_sig_math_des NumberOfReplyTAG: 1 FirstChildTAG: I think you are referring to 2:12 on S10V7. If you watch carefully, the "square" on the RL actually belongs to the RL*K/2 on the equation above the last equation. FirstChildUserIdTAG: 359310 FirstChildUserNameTAG: AaronYeoh FirstChildCreateTimeTAG: 2012-10-12T04:15:31Z SecondChildTAG: Ah, thank you! I believe you are correct. NOT having captions for this video proved too distracting for me, I think! Thanks again! SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-12T04:18:24Z IndexTAG: 2300 TitleTAG: ayuda myrimit hola espero estes bien aqui solicitando de tu ayuda estoy atorada el la tarea 5 problema 2 en la 3er pregunta en donde hay me tiene que salir una ecuacion cuadratica estoy confundida con el procedimiento me podrias ayudar gracias UserIdTAG: 320715 UserNameTAG: maraivette CreateTimeTAG: 2012-10-11T14:36:28Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Check this: http://www.purplemath.com/modules/quadform.htm FirstChildUserIdTAG: 127195 FirstChildUserNameTAG: princeofsudan FirstChildCreateTimeTAG: 2012-10-11T14:45:51Z SecondChildTAG: Thank you :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-13T19:44:56Z FirstChildTAG: Hola Maraivette! Si no hay problema, hoy más tarde, vuelvo a este Post y te ayudo a resolverlo ;). No te preocupes que te ayudaré. Un abrazo! Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-11T15:31:59Z SecondChildTAG: gracias te explico el problema por que al parecer no esta claro no se como sustituir una ecuacion en otra para poder llegar a la ec. cuadradita que mencionan gracias SecondChildUserIdTAG: 320715 SecondChildUserNameTAG: maraivette SecondChildCreateTimeTAG: 2012-10-11T20:17:23Z SecondChildTAG: en el siguiente problema tambien esta confuso hay que derivar una funcion pero cual? SecondChildUserIdTAG: 320715 SecondChildUserNameTAG: maraivette SecondChildCreateTimeTAG: 2012-10-11T22:20:32Z SecondChildTAG: Aquí te lo he respondido :) [Post][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_large_sig/threads/5079c393be483a230000000e SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-13T19:44:38Z IndexTAG: 2301 TitleTAG: Eh? Is this problem unlinear and thus hugely complicated, or am i not getting smth? vO>=vIN-VT -> vIN<=vO+VT vO=VS-K*RL/2 (vIN-VT)^2 vIN<=VS+VT-K*RL/2 (vIN-VT)^2 So we have vIN on one side and vIN^2 on the other. Although this can be solved, i feel like im doing it wrong. UserIdTAG: 380703 UserNameTAG: vargy CreateTimeTAG: 2012-10-11T12:44:09Z VoteTAG: 1 CoursewareTAG: Week 5 / MOSFET Amplifier Exercise 12 CommentableIdTAG: 6002x_mosfet_amp_e2 NumberOfReplyTAG: 1 FirstChildTAG: Watch the next lecture. S9V17, specifically. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-11T13:39:29Z IndexTAG: 2302 TitleTAG: Week5 video Please can some help me with the week 5 video. i cannot find it in wiki UserIdTAG: 341561 UserNameTAG: rameky CreateTimeTAG: 2012-10-11T08:50:52Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: click on "week 4" or any link you can see,then you may find FirstChildUserIdTAG: 361137 FirstChildUserNameTAG: Ericson FirstChildCreateTimeTAG: 2012-10-11T12:19:03Z SecondChildTAG: Thank you very much SecondChildUserIdTAG: 358539 SecondChildUserNameTAG: syed_abdullah SecondChildCreateTimeTAG: 2012-10-11T12:26:11Z SecondChildTAG: https://dl.dropbox.com/u/24096724/6002xMP4/menu.html SecondChildUserIdTAG: 122513 SecondChildUserNameTAG: Mona77 SecondChildCreateTimeTAG: 2012-10-12T06:44:04Z IndexTAG: 2303 TitleTAG: Be careful about the units! see, the unit of K is mA/V^2 ..... UserIdTAG: 240690 UserNameTAG: Rhapsowind CreateTimeTAG: 2012-10-11T07:23:08Z VoteTAG: 1 CoursewareTAG: Week 5 / Two Terminal Connection Exercise CommentableIdTAG: 6002x_2_term_con_e NumberOfReplyTAG: 1 FirstChildTAG: Sorry, but when I input 1m in formula it is still wrong response. I meen formula K/2*(VGS-VT0^2 FirstChildUserIdTAG: 96873 FirstChildUserNameTAG: Santyaga FirstChildCreateTimeTAG: 2012-10-11T18:38:48Z IndexTAG: 2304 TitleTAG: Paradox - Who made Who? Bad things will happen if you do not listen to Prof Agarwal. http://www.youtube.com/watch?v=_jvqPvDUEW8 Maybe this will be the 6.002x Theme song? ~Vote now!~ UserIdTAG: 292543 UserNameTAG: Pennypacker CreateTimeTAG: 2012-10-11T02:06:58Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Meh. They ought to dig up Bon Scott. I can't stand that new guy that can't carry a tune. FirstChildUserIdTAG: 90796 FirstChildUserNameTAG: ChaunceyGardiner FirstChildCreateTimeTAG: 2012-10-11T04:07:36Z SecondChildTAG: Fair enough. lol SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-11T04:26:26Z IndexTAG: 2305 TitleTAG: Midterm Exam I am a highschool student with a busy day and I am of the Jewish religion. Due to this I can't start the midterm on Friday or Saturday. Is it at all possible for the midterm to be released one day earlier. #staff UserIdTAG: 411504 UserNameTAG: taubrafi CreateTimeTAG: 2012-10-11T00:43:56Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: This was posted in the Course Info section recently. I doubt that they will release the exam any earlier. No harm in asking I suppose. Have fun. "Deadlines and extensions: We have received several forum posts and emails from students about missing homework or lab deadlines due to work or travel. While we would like to be as accommodating as possible when external circumstances arise, please note that it is not our policy to grant extensions to students on an individual basis because homework solutions become readily available past their official deadlines. In most cases, missing deadlines should not be an issue as long as you catch up on the learning afterwards. Only your best ten of your twelve homework scores and your best ten of your twelve lab scores will count towards your final grade, and each individual assignment only effects your grade by 1.5% each. This can add up if bad habits form, but it is generally not disasterous if something unavoidable forces you to miss a deadline." FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-11T00:56:14Z SecondChildTAG: I am trying to get them to do this beforehand "it is not our policy to grant extensions to students on an individual basis because homework solutions become readily available past their official deadlines." not an extra day afterward. But what's the worst that asking could do... SecondChildUserIdTAG: 411504 SecondChildUserNameTAG: taubrafi SecondChildCreateTimeTAG: 2012-10-11T04:34:05Z IndexTAG: 2306 TitleTAG: week 5 Videos not uploading yet?? Please upload week 5 videos as soon as possible few days are left behind in this week?? UserIdTAG: 315703 UserNameTAG: Sohaib786 CreateTimeTAG: 2012-10-10T20:19:57Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Troubleshooting NumberOfReplyTAG: 1 FirstChildTAG: I am yet to see week 5 videos. Please what do i do FirstChildUserIdTAG: 341561 FirstChildUserNameTAG: rameky FirstChildCreateTimeTAG: 2012-10-11T08:43:34Z IndexTAG: 2307 TitleTAG: H5P2 3rd First two part of the task is done, but now having some trouble with the language here, what does the 3rd question mean? Could someone rephrase it? "Solve the previous two results for iDS. You will get a quadratic equation, and you will have to choose the correct root: make sure that iDS=0 when vIN=VT. What is your solution for iDS?" UserIdTAG: 70211 UserNameTAG: Emberfej CreateTimeTAG: 2012-10-10T18:16:16Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: well that means:" rewrite iDS expression without vOUT" FirstChildUserIdTAG: 114913 FirstChildUserNameTAG: ngoctuan FirstChildCreateTimeTAG: 2012-10-10T18:29:15Z SecondChildTAG: and you have to choose correct root by checking the condition mentioned. SecondChildUserIdTAG: 170017 SecondChildUserNameTAG: Rajez SecondChildCreateTimeTAG: 2012-10-11T08:04:52Z SecondChildTAG: write expression for vOUT, iDS=vOUT/RS SecondChildUserIdTAG: 297655 SecondChildUserNameTAG: Aljoska SecondChildCreateTimeTAG: 2012-10-11T10:39:21Z SecondChildTAG: i have tried alot bt in vain although iDS=vIN-VT/RS is the correct answer bt x mark doesnt vanish SecondChildUserIdTAG: 282892 SecondChildUserNameTAG: wiky SecondChildCreateTimeTAG: 2012-10-11T16:41:39Z SecondChildTAG: Thanks, got it. Its a really long expression, takes all the space provided. SecondChildUserIdTAG: 70211 SecondChildUserNameTAG: Emberfej SecondChildCreateTimeTAG: 2012-10-12T10:36:21Z SecondChildTAG: my iDS equation is "[{RS*(vIN-VT)+2/K}-sqrt{4*RS/K*(vIN-VT)+(2/K)^2}]/RS^2" but it says invalid input. why? SecondChildUserIdTAG: 259693 SecondChildUserNameTAG: MehrozKhan SecondChildCreateTimeTAG: 2012-10-13T18:06:46Z SecondChildTAG: I only use () don't know if it is okay to use {}. Make sure not to forget the operators * SecondChildUserIdTAG: 394240 SecondChildUserNameTAG: ManU81 SecondChildCreateTimeTAG: 2012-10-14T10:20:42Z IndexTAG: 2308 TitleTAG: syllabus for mid term exam please tell me about which topic to be asked in mid term exam UserIdTAG: 181793 UserNameTAG: mitrahul CreateTimeTAG: 2012-10-10T17:38:47Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Lecture sequences week 1 through halfway of week 6. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-10T18:05:55Z SecondChildTAG: thanks SecondChildUserIdTAG: 181793 SecondChildUserNameTAG: mitrahul SecondChildCreateTimeTAG: 2012-10-11T05:12:23Z IndexTAG: 2309 TitleTAG: help H5p1 Im not able to get the min and the max value of Vin, I went through the book page 358 to find the vin MAx but im not getting the right answer. could anyone help me with this part, Im sort of stuck there and i dont see to get it. maybe a hint would be great, not asking for solution, just guidance. thank you UserIdTAG: 320871 UserNameTAG: seb1256 CreateTimeTAG: 2012-10-10T15:31:13Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: When I am stuck on something like this, I try alternate routes. Try building the circuit in a simulator and see what you can find out there. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-10T16:10:11Z SecondChildTAG: How do you set VT = 0.5 using the simulator? SecondChildUserIdTAG: 18073 SecondChildUserNameTAG: gburkhart SecondChildCreateTimeTAG: 2012-10-11T00:15:25Z SecondChildTAG: I guess you can't set VT in the 6.002x simulator. There are other simulators out there, though. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-11T00:55:18Z SecondChildTAG: links please? SecondChildUserIdTAG: 61469 SecondChildUserNameTAG: Spacedog SecondChildCreateTimeTAG: 2012-10-11T00:57:51Z SecondChildTAG: Try this one. If it does not help you now, it may be handy down the road. https://www.circuitlab.com/ SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-11T01:38:43Z FirstChildTAG: Consider these questions: How does Vin relate to VGS? How does max Vin relate to Vout and VDS? How are VGS and VDS related at the transition between the triode and saturated regions? FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-10T18:15:18Z FirstChildTAG: Hi Seb1256! You can take a look at this Hints [Post][1] :) [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_mosfet_amp/threads/507a1d6dd894f12300000039 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-14T04:16:00Z IndexTAG: 2310 TitleTAG: Misleading current sign I think the current sign (the arrow) is misleading. It should be below the junction as in the book (page 417). If iR is the current before the junction, there is a possibility some of the iR goes to G (gate). UserIdTAG: 201818 UserNameTAG: ThreeHundred CreateTimeTAG: 2012-10-10T13:10:46Z VoteTAG: 1 CoursewareTAG: Week 5 / Two Terminal Connection Exercise CommentableIdTAG: 6002x_2_term_con_e NumberOfReplyTAG: 1 FirstChildTAG: Well in S9V2 its clearly mentioned that for our purposes,we take IGS=0 i.e. the gate to source current is practically null for this entire course.Hence It doesn't matter if the arrow is above or below the terminal **here** as the whole current is **supposed** to flow through D to S. FirstChildUserIdTAG: 222809 FirstChildUserNameTAG: AshutoshTadkase FirstChildCreateTimeTAG: 2012-10-10T17:22:39Z IndexTAG: 2311 TitleTAG: Why is $v_{PK}$ also evaluated at its large signal bias value for the transconductance? Doesn't... $ \left . \cfrac {\partial i_p}{\partial v_{GK}} \right |_{v_{GK}=V_{GK}} = \cfrac {3}{2}K \mu (\mu\cdot V_{GK}+v_{PK})^{\frac{1}{2}}$ ? I'm trying to figure why $v_{PK}$ was also evaluated at its large signal bias in finding the transconductance. Perhaps I'm not clear on what exactly is going on; does anyone maybe have explanations or links to share? EDIT: This post is regarding the following video: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_6/Week_6_Tutorials/ UserIdTAG: 248902 UserNameTAG: Chanute CreateTimeTAG: 2012-10-10T09:48:05Z VoteTAG: 1 CoursewareTAG: Week 6 / Vacuum Triode Model CommentableIdTAG: 6002x_Vacuum_Triode_Model NumberOfReplyTAG: 1 FirstChildTAG: Because it was the mistake in the video. The second partial derivative should be evaluated at the bias point. The correct formula is: $i_p = v_{gk} \cdot\frac{\partial i_p}{\partial v_{GK}} \vert_{v_{GK = V_{GK}}; v_{PK} = V_{PK}} + v_{pk} \cdot\frac{\partial i_p}{\partial v_{PK}} \vert_{v_{GK = V_{GK}}; v_{PK} = V_{PK}}$ It is from the definition of the Taylor series for a function of several variables. FirstChildUserIdTAG: 345452 FirstChildUserNameTAG: ale-yoman FirstChildCreateTimeTAG: 2012-10-15T10:12:02Z IndexTAG: 2312 TitleTAG: H5P2 AND H5P3 equations??? Hi there! I have solved all the other questions of homework 5 except the H5P2's part 3 and H5P3's part 1. Please help me!!! It's driving me crazy. Also, for the H5P3, there's a hint saying that vout=vin⋅∂vOUT∂vIN|vIN=VIN. Has this got anything to do with the Taylor's series? Thank you in advance! Any help is deeply appreciated! UserIdTAG: 499268 UserNameTAG: ruinoah CreateTimeTAG: 2012-10-10T04:43:24Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: To H5P2's part 3 you can solve it with this: 1. find vOUT, solve this from the book: vOUT=K*(Vin - VT - vOUT) 2. Choose the right root and the put it in the iDS ec. (The answer is as long as the page) This is the way to solve it. Then to H5P3's part 1 you need preview vOUT: 1. vOUT = iDS*RS 2. Differentiate respect to Vin. (d(vOUT)/d(Vin)) 3. Replsase Vin with VIN 4. Multiply by vin. (this is an anwere more big than vOUT) So can you help me to me? Give me to me some hint to solve H5P2 part 7. and with H5P1 part 3: I have read [H5P1 Zero Offset Amplifier][1] but I don`t understand this hint! Thanks! [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50631b1199874e230000000e FirstChildUserIdTAG: 299251 FirstChildUserNameTAG: TheRedBlackOne FirstChildCreateTimeTAG: 2012-10-10T11:02:54Z SecondChildTAG: for H5P2 part 7: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5073bbb45da69a1f0000005a SecondChildUserIdTAG: 580037 SecondChildUserNameTAG: b25ch SecondChildCreateTimeTAG: 2012-10-10T13:25:12Z SecondChildTAG: Thanks but if you see that post I have asked in this! SecondChildUserIdTAG: 299251 SecondChildUserNameTAG: TheRedBlackOne SecondChildCreateTimeTAG: 2012-10-10T13:42:08Z SecondChildTAG: thanks this hint was really very helpful :) SecondChildUserIdTAG: 365309 SecondChildUserNameTAG: chetnasinghaldas SecondChildCreateTimeTAG: 2012-10-13T11:13:59Z IndexTAG: 2313 TitleTAG: late!!!! sorry I have problem with my course for to log in!!!!...Help UserIdTAG: 71157 UserNameTAG: llereador CreateTimeTAG: 2012-10-10T01:58:01Z VoteTAG: 1 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: If you can post here, you're logged in. FirstChildUserIdTAG: 90796 FirstChildUserNameTAG: ChaunceyGardiner FirstChildCreateTimeTAG: 2012-10-11T04:28:51Z IndexTAG: 2314 TitleTAG: Biasing and operating point? please tell me what is biasing and operating point?please explain it in very simple way with an example? UserIdTAG: 40209 UserNameTAG: sadaf78 CreateTimeTAG: 2012-10-09T22:29:08Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Say you have a signal that varies between -1 V and 1 V. You want to feed this into a component that can only deal with positive voltages. So, you add a 2 V voltage source in series with the signal (between ground and the signal). The combined signal will now vary between 1 and 3 V. It still has the same shape, but it has been offset to fit your requirements. The 2 V voltage source is your bias voltage, i.e. a voltage offset. The operating point is 2V, the operating range is 1V to 3V. Bias voltage, voltage offset and operating point are different phrases that mean the same thing. FirstChildUserIdTAG: 90796 FirstChildUserNameTAG: ChaunceyGardiner FirstChildCreateTimeTAG: 2012-10-09T23:56:58Z SecondChildTAG: Thank you ChaunceyGardiner you explained it very well. SecondChildUserIdTAG: 40209 SecondChildUserNameTAG: sadaf78 SecondChildCreateTimeTAG: 2012-10-10T16:25:06Z IndexTAG: 2315 TitleTAG: LAB 5 so, in lab 5, we don't have to do anything?...I mean we don't have to join extra stuff right?.....just take the readings,.....thought I would do it on the weekend.......but still...........it will be better if someone can just confirm for me now! UserIdTAG: 169784 UserNameTAG: thewiredbear CreateTimeTAG: 2012-10-09T21:33:41Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: No we don't have to add anything to the circuits. You get 100/100 by the measurements only FirstChildUserIdTAG: 304587 FirstChildUserNameTAG: Vasso FirstChildCreateTimeTAG: 2012-10-09T22:03:25Z SecondChildTAG: It was nice to have a lab that was easy for me after the week 4 beast SecondChildUserIdTAG: 18073 SecondChildUserNameTAG: gburkhart SecondChildCreateTimeTAG: 2012-10-10T02:46:21Z IndexTAG: 2316 TitleTAG: Dear TAs Hi, I just registered Circuits and Electronics today and want to do previous labs and assignments. Is it possible to do that ?? Thanks UserIdTAG: 601136 UserNameTAG: jeonghun CreateTimeTAG: 2012-10-09T19:58:29Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: It's not possible for you to get a grade on Homeworks 1-4 and Labs 1-4. The deadline for them has passed. You get to drop two of them, so you'd be down 2 homeworks/labs. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-09T20:38:54Z FirstChildTAG: The course will be offered again in Spring and you're welcome to give it another go then :) FirstChildUserIdTAG: 151427 FirstChildUserNameTAG: RichardZ FirstChildCreateTimeTAG: 2012-10-10T14:35:40Z IndexTAG: 2317 TitleTAG: homework scores and midterm exam hello to all 1) So far I thought that making mistakes at homework does not count to score as far as the mistakes are corrected and a green mark is shown. I usually do not pay attention and the answers I give initially may involve arithmetic mistakes which I correct when I see the red X and then move foreword) Is my assumption wrong? 2) Will the midterm exam give us the opportunity to check the answers real time and if we make mistakes to correct them or we will learn the mistakes afterward :-) thank you in advance UserIdTAG: 319453 UserNameTAG: leonidasGr CreateTimeTAG: 2012-10-09T12:52:57Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: three checks on answer available at the any of the exams FirstChildUserIdTAG: 206648 FirstChildUserNameTAG: TsvetanGeorgiev FirstChildCreateTimeTAG: 2012-10-09T13:34:45Z SecondChildTAG: to clarify further even for the midterm exam we will be given three shots to find the correct answers. is that right? SecondChildUserIdTAG: 319453 SecondChildUserNameTAG: leonidasGr SecondChildCreateTimeTAG: 2012-10-09T14:04:04Z SecondChildTAG: is that we can check the whole paper 3 times or each question 3 times SecondChildUserIdTAG: 388273 SecondChildUserNameTAG: kilmarta SecondChildCreateTimeTAG: 2012-10-12T16:43:21Z FirstChildTAG: Νόμιζα ότι ήμουν η μόνη απο Ελλάδα!! Καλή επιτυχία! :) FirstChildUserIdTAG: 304587 FirstChildUserNameTAG: Vasso FirstChildCreateTimeTAG: 2012-10-09T19:57:35Z SecondChildTAG: τελικά κατάλαβες τι θα παίξει με το διαγώνισμα?? θα μας δίνεται περιθώριο λάθους? SecondChildUserIdTAG: 319453 SecondChildUserNameTAG: leonidasGr SecondChildCreateTimeTAG: 2012-10-09T20:32:25Z SecondChildTAG: Απ' ότι κατάλαβα θα μπορούμε να κάνουμε 3 λάθη σε κάθε ενότητα του τεστ. πχ σαν να είχαμε περιθώριο εως 3 λάθη στο Η5P1 και 3 λάθη στο H5P2 κλπ. SecondChildUserIdTAG: 304587 SecondChildUserNameTAG: Vasso SecondChildCreateTimeTAG: 2012-10-09T20:53:59Z SecondChildTAG: ok γιατι είμαι λιγο στοκος ευχαριστώ SecondChildUserIdTAG: 319453 SecondChildUserNameTAG: leonidasGr SecondChildCreateTimeTAG: 2012-10-09T21:15:58Z SecondChildTAG: χαχαχα!!! Δε θα ήσουν εδω αν ησουν στόκος! :P Πόσο έχεις προχωρήσει? Σε ποιά εργασία βρίσκεσαι εννοώ? SecondChildUserIdTAG: 304587 SecondChildUserNameTAG: Vasso SecondChildCreateTimeTAG: 2012-10-09T22:09:11Z SecondChildTAG: Παιδια δεν ξερω για σας αλλα την τεταρτη εβδομαδα εγω πατωσα!! :( SecondChildUserIdTAG: 214275 SecondChildUserNameTAG: TheodoreGr SecondChildCreateTimeTAG: 2012-10-09T22:26:10Z SecondChildTAG: Είχα κάνει αυτό το post όταν τελείωσα την Τέταρτη εργασία https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5060cf634f44871f00000028 SecondChildUserIdTAG: 304587 SecondChildUserNameTAG: Vasso SecondChildCreateTimeTAG: 2012-10-09T22:41:12Z SecondChildTAG: Πρέπει να διαβάζουμε και τα επόμενα κεφάλαια δηλαδή και όχι μόνο όσα παρουσιάζονται. Που αν το σκεφτείς είναι σωστό, απλά για όσους έχουν το χρόνο :/ SecondChildUserIdTAG: 304587 SecondChildUserNameTAG: Vasso SecondChildCreateTimeTAG: 2012-10-09T22:43:29Z SecondChildTAG: 我们保持它在英语吗？ :) SecondChildUserIdTAG: 114881 SecondChildUserNameTAG: xvink SecondChildCreateTimeTAG: 2012-10-10T11:06:49Z SecondChildTAG: 我们没有做错任何事。 但我想我们可以阻止它，因为它是所有希腊给你 :) SecondChildUserIdTAG: 304587 SecondChildUserNameTAG: Vasso SecondChildCreateTimeTAG: 2012-10-10T22:01:21Z IndexTAG: 2318 TitleTAG: about links there are not available direct links of the lectures for download from 5th week.. UserIdTAG: 509517 UserNameTAG: randima CreateTimeTAG: 2012-10-09T07:57:11Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: [Week_5 Videos Playlist][1] [1]: http://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/LectureDownloadWeek5/ Check this Link.... FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-10-09T12:15:16Z SecondChildTAG: That's only a playlist. There are no direct links to the videos themselves for Weeks 5 and 6. SecondChildUserIdTAG: 94557 SecondChildUserNameTAG: g_hopper SecondChildCreateTimeTAG: 2012-10-09T16:12:58Z SecondChildTAG: yeah!! of course.. it is not just like other weeks.. SecondChildUserIdTAG: 509517 SecondChildUserNameTAG: randima SecondChildCreateTimeTAG: 2012-10-10T08:40:51Z SecondChildTAG: edX yet not produced direct link of these weeks videos...:) SecondChildUserIdTAG: 329051 SecondChildUserNameTAG: asadbhatti42 SecondChildCreateTimeTAG: 2012-10-13T17:15:54Z IndexTAG: 2319 TitleTAG: Voltage divider Vint is said to be 1/2Vout since it's a simple voltage divider, but voltage from vin that probably also affect Vint is ignored. Is this b/c vin is a small signal voltage that can be neglected? UserIdTAG: 189680 UserNameTAG: vnd CreateTimeTAG: 2012-10-09T06:35:32Z VoteTAG: 1 CoursewareTAG: Week 5 / Op Amp Small Signal Model CommentableIdTAG: 6002x_ap_amp_small_signal_t NumberOfReplyTAG: 1 FirstChildTAG: Same problem here. I don't understand how you can ignore vin for the vint node. Also if you do the node method for the vout node, you will get another expression for vout. Does anyone have a good explanation? FirstChildUserIdTAG: 329453 FirstChildUserNameTAG: Even83 FirstChildCreateTimeTAG: 2012-10-24T14:04:51Z IndexTAG: 2320 TitleTAG: math processing error Is their any guidance to rid oneself of "math processing error" showing up in this week's exercise's, problems, and lab? Thanks in advance for the help UserIdTAG: 242903 UserNameTAG: rcarpine CreateTimeTAG: 2012-10-08T23:40:00Z VoteTAG: 1 CoursewareTAG: Week 5 / MOSFET model exercise CommentableIdTAG: 6002x_mosfet_mod_e NumberOfReplyTAG: 1 FirstChildTAG: That error is due to how your browser interacts with MathJax. MathJax is used by the server to present mathematical symbols on your screen. Here's how to fix it: 1) Place your cursor on the [Math Processing Error] message and right-click. 2) You will see a pull-down MathJax menu. 3) Select Math Settings, Math Renderer, and choose either MathML or SVG. I prefer the way MathML looks, but it drops some symbols when I print, so I have to use SVG. 4) You can also play with the zoom settings on the MathJax menu to adjust the zoom functionality. FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-10-09T16:21:12Z SecondChildTAG: Thanks :) SecondChildUserIdTAG: 386004 SecondChildUserNameTAG: ravinarv SecondChildCreateTimeTAG: 2012-10-14T15:00:26Z IndexTAG: 2321 TitleTAG: Vch = 0 Why use e^x for Vch? RC*d0/dt + 0 = 0 also. UserIdTAG: 208296 UserNameTAG: OZ1 CreateTimeTAG: 2012-10-08T21:30:30Z VoteTAG: 1 CoursewareTAG: Week 6 / First-order differential equation CommentableIdTAG: 6002x_first_order_differential_equation NumberOfReplyTAG: 1 FirstChildTAG: Hi, OZ1! You can look at DE solution as a way to connect two points: the initial state of the circuit (when the VC was zero) and the final state (after a long period of time, when VC is stabilized). Zero function can connect only zero points, and can't handle any other situation, like real-world example. But e^x function can =) If you are interested in the math background... Your solution is called a trivial: http://en.wikipedia.org/wiki/Triviality_(mathematics) (there is exactly your example!) and it is not permitted in the searching of the general solution of ODE: http://en.wikipedia.org/wiki/Ordinary_differential_equation#General_definition_of_an_ODE. FirstChildUserIdTAG: 261503 FirstChildUserNameTAG: EugenyL FirstChildCreateTimeTAG: 2012-10-08T22:20:11Z IndexTAG: 2322 TitleTAG: Error at 3:18 of Part 2 of the Tutorial Example At 3:18 of Part 2, the simplified term should be $(v_o G_1 - v_i G_1)/(G_1+G_2)$. That is, the sign in the numerator should be $-$. UserIdTAG: 153760 UserNameTAG: Kavka CreateTimeTAG: 2012-10-08T21:02:01Z VoteTAG: 1 CoursewareTAG: Week 5 / Transistor Biasing with Feedback CommentableIdTAG: 6002x_transistor_biasing_t NumberOfReplyTAG: 1 FirstChildTAG: You are right! $G_1(v_o - v_i)/(G_1 + G_2)$ $\checkmark$ FirstChildUserIdTAG: 184827 FirstChildUserNameTAG: DiegoT FirstChildCreateTimeTAG: 2012-10-10T03:29:38Z IndexTAG: 2323 TitleTAG: S13E2 The exercise asks for the peak power, but having puzzled over it for awhile I couldn't figure it out. So I tried the average power...which worked. Is this correct? If so, why? It doesn't seem right to me. Thanks UserIdTAG: 145420 UserNameTAG: pmac12345 CreateTimeTAG: 2012-10-08T17:49:03Z VoteTAG: 1 CoursewareTAG: Week 7 / Constitutive Laws for Capacitors and Inductors CommentableIdTAG: 6002x_Constitutive_Laws_Caps_Inductors NumberOfReplyTAG: 1 FirstChildTAG: you have the peak energy from part 1. and you have been given the time. And we know that Power=Energy/Time. FirstChildUserIdTAG: 127195 FirstChildUserNameTAG: princeofsudan FirstChildCreateTimeTAG: 2012-10-11T14:33:48Z IndexTAG: 2324 TitleTAG: Just Started pls.i need to do my home work and Lab. work i just started yesterday thank you. UserIdTAG: 574840 UserNameTAG: shogs CreateTimeTAG: 2012-10-08T16:08:52Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: you have my permission FirstChildUserIdTAG: 18073 FirstChildUserNameTAG: gburkhart FirstChildCreateTimeTAG: 2012-10-09T02:31:18Z IndexTAG: 2325 TitleTAG: No Vs in the answer The question is a bit misleading because there is no Vs term in the answer. UserIdTAG: 150267 UserNameTAG: vwsingh CreateTimeTAG: 2012-10-08T13:54:13Z VoteTAG: 1 CoursewareTAG: Week 5 / Incremental Voltage Exercise CommentableIdTAG: 6002x_inc_volt_e NumberOfReplyTAG: 2 FirstChildTAG: Yeah, I think they did that on purpose. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-08T18:42:35Z FirstChildTAG: Hi is there any one know why the right answer doesn't have vi^2. my result is + vo = **(RL*K/2)*vi^2** - (RL*K*(VI-VT)*vi) + but answer is : + vo= - (RL*K*(VI-VT)*vi) + is **vi** very small then then quadrat of it is near zero and we can remove it, right? thanks FirstChildUserIdTAG: 114913 FirstChildUserNameTAG: ngoctuan FirstChildCreateTimeTAG: 2012-10-08T19:26:31Z SecondChildTAG: vi^2 is very small. For example if vi = 0.1, vi^2 =0.01 . SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-10-08T19:48:05Z SecondChildTAG: PS: I've noticed that small signal refers to a signal with an amplitude < 1. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-10-08T19:49:01Z IndexTAG: 2326 TitleTAG: mid term how many percent is the midterm worth UserIdTAG: 436749 UserNameTAG: waynebrown CreateTimeTAG: 2012-10-08T13:37:44Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Midterm is 30% of the total grade , homework and labs are 15% each , that's 60% , and the final is 40% and that makes it a complete 100% FirstChildUserIdTAG: 528897 FirstChildUserNameTAG: vepoo FirstChildCreateTimeTAG: 2012-10-08T13:47:03Z IndexTAG: 2327 TitleTAG: question??? what do we mean by dependent on the voltage or current?????? I mean practically if the voltage (or current) increase or decrease does this effect on the out put of the source. UserIdTAG: 226085 UserNameTAG: Teto CreateTimeTAG: 2012-10-08T11:07:05Z VoteTAG: 1 CoursewareTAG: Week 4 / Dependent Sources Example 2 CommentableIdTAG: 6002x_dep_src_ex_2 NumberOfReplyTAG: 1 FirstChildTAG: It's exactly what does it mean. The problem here that dependent source can influence itself by changing its own control parameter - as a result we have equilibrium equation for control parameter. FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-08T17:44:15Z IndexTAG: 2328 TitleTAG: H4P2: How can we from the graph take y=mx+b Many people are stating that it is possible to get from the Zenor diode graph the equation of iD with respect to vD, by using the formula y=mx+b. But I do not understand how to do it. Either you have two points in each sub-segment (that you don't have - you only have 1 point for rach sub-segment: (vD,iD): (0.6,0.0) and (-5.0,0.0)) or you must have 1 point and the slope. But the slope in the graph says that it is equal to 1A/V that I do not quite undertsnad its meaning - is it saying that is amperes over volts? That does not help too much... Can someone help me here? Thanks. UserIdTAG: 18880 UserNameTAG: pcarmo CreateTimeTAG: 2012-10-08T10:21:21Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: iD=vD+5 for -5 iD=0 -5 iD=vD-0.6 vD>0.6 FirstChildUserIdTAG: 134729 FirstChildUserNameTAG: zoranmrsa FirstChildCreateTimeTAG: 2012-10-08T11:19:51Z FirstChildTAG: The slope is 1/R, and that is why it is measured in A/V if i am not mistaken. ![As you see in this example, you must choose two points for which you know the parameters.][1] [1]: http://upload.wikimedia.org/wikipedia/commons/a/a9/Linear_function.png You already know the parameters of the (0.6,0) point that is marked on the given graph in Homework. You can find a , for 2 points in the same line e.g. (x1,y1) and (x2, y2) in the equation y=ax+b by: a = (g(x2)-g(x1))/(x2-x1) Furthermore, your b is the point in line where g(x)=0, so in our case we use the same point here. Sorry for my English, hope you got the idea :) FirstChildUserIdTAG: 304587 FirstChildUserNameTAG: Vasso FirstChildCreateTimeTAG: 2012-10-08T11:23:46Z SecondChildTAG: Your answer is brilliant and your English perfect! Thanks! SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-08T19:18:43Z IndexTAG: 2329 TitleTAG: LAB4 where is it wrong? plz help ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/1349688220134362.png UserIdTAG: 431224 UserNameTAG: erachopra10 CreateTimeTAG: 2012-10-08T09:23:56Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Flip the current meter FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-10-08T09:26:11Z SecondChildTAG: flipped it..still not working! SecondChildUserIdTAG: 431224 SecondChildUserNameTAG: erachopra10 SecondChildCreateTimeTAG: 2012-10-08T10:37:22Z SecondChildTAG: Voltage probe on the gate, triangle to the gate and 1V dc to the drain. SecondChildUserIdTAG: 416659 SecondChildUserNameTAG: Jonas3000 SecondChildCreateTimeTAG: 2012-10-08T11:22:34Z FirstChildTAG: Switch voltage sources. I think triangle goes to gate. Or am I wrong? Anyone? FirstChildUserIdTAG: 416659 FirstChildUserNameTAG: Jonas3000 FirstChildCreateTimeTAG: 2012-10-08T11:18:26Z SecondChildTAG: ...And then I measured voltage on the gate. SecondChildUserIdTAG: 416659 SecondChildUserNameTAG: Jonas3000 SecondChildCreateTimeTAG: 2012-10-08T11:20:58Z SecondChildTAG: here will help you: [LAB 4 sample][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5071cac722cc4a2700000015 SecondChildUserIdTAG: 114913 SecondChildUserNameTAG: ngoctuan SecondChildCreateTimeTAG: 2012-10-08T11:57:06Z FirstChildTAG: when you press tran do you reset time to 5ms? FirstChildUserIdTAG: 528897 FirstChildUserNameTAG: vepoo FirstChildCreateTimeTAG: 2012-10-08T13:58:50Z SecondChildTAG: of course, it should be, because when you choose 1 second it takes very long time to calculate and plot. SecondChildUserIdTAG: 114913 SecondChildUserNameTAG: ngoctuan SecondChildCreateTimeTAG: 2012-10-08T16:45:41Z IndexTAG: 2330 TitleTAG: Downloading videos week 5 and week 6 Hello, I´ve seen that the videos for week 5 and week 6 are already in the Wiki section but they are not available to download. It´s really really useful to have them downloaded in case yo don't have access to internet (this is something that happens to me very often). Could you please add the links to download the videos? Thank you so much, UserIdTAG: 255161 UserNameTAG: mlovelle CreateTimeTAG: 2012-10-08T06:35:10Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 2331 TitleTAG: H5P2 part A got part B but can't figure out how to write an expression that has vOUT, vIN AND VT in it. I put in the iDS equation from the lecture then I expanded the perfect square and did some substitution in order to get all three Voltages in one expression. It was a mess. I am pretty sure I am not making any mistakes with case sensitivity or parenthesis - any hints would be appreciated UserIdTAG: 18073 UserNameTAG: gburkhart CreateTimeTAG: 2012-10-08T02:03:36Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I put in the iDS equation from the lecture that has K vIN and VT in it and tried the version with the substitution for vOUT FirstChildUserIdTAG: 18073 FirstChildUserNameTAG: gburkhart FirstChildCreateTimeTAG: 2012-10-08T02:04:52Z FirstChildTAG: The difference between this problem and the one in the presentation's, is the place of the resistance. So in this case, your "VIN" is a little bit different from the one in the lecture. Think about what your input voltage will be when there is a resistance involved :) FirstChildUserIdTAG: 304587 FirstChildUserNameTAG: Vasso FirstChildCreateTimeTAG: 2012-10-08T07:31:57Z SecondChildTAG: that helped. dont know if this is right or if it is too much help but try grounding the bottom of the mosfet SecondChildUserIdTAG: 388273 SecondChildUserNameTAG: kilmarta SecondChildCreateTimeTAG: 2012-10-11T22:05:34Z IndexTAG: 2332 TitleTAG: LAB 4 Questions! Ayudenme por favor, queda poco tiempo y no entiendo para nada las preguntas del laboratorio!!! UserIdTAG: 331801 UserNameTAG: ingjesusortiz CreateTimeTAG: 2012-10-08T01:44:03Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: lograstes terminar el homework? FirstChildUserIdTAG: 337887 FirstChildUserNameTAG: kevil FirstChildCreateTimeTAG: 2012-10-08T03:47:59Z FirstChildTAG: i not understand lab 4 what can i do? FirstChildUserIdTAG: 369692 FirstChildUserNameTAG: andhale FirstChildCreateTimeTAG: 2012-10-08T05:46:47Z IndexTAG: 2333 TitleTAG: Calculation for K in Lab 4 3rd part I am not getting the correct answer for the value of constant K. Help please UserIdTAG: 369431 UserNameTAG: pkaistha CreateTimeTAG: 2012-10-07T23:18:08Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: thats weird...but check your power factor...because if you have your VT value, then it is just a question of substituting the correct values to obtain K FirstChildUserIdTAG: 308861 FirstChildUserNameTAG: Ignaas FirstChildCreateTimeTAG: 2012-10-08T00:41:03Z FirstChildTAG: I did use my home copy of Wolfram to get it. First run at it I plugged in the wrong numbers. FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-10-08T02:36:32Z FirstChildTAG: Watch your units! Review section 7.3-4 in the book. Check to make sure you've the right number of zeros in your answer. Hope this helps! FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-10-08T04:27:08Z IndexTAG: 2334 TitleTAG: experiment Very enjoyable! UserIdTAG: 352757 UserNameTAG: Kerbyco CreateTimeTAG: 2012-10-07T21:29:47Z VoteTAG: 1 CoursewareTAG: Week 5 / Music Amplifier Demonstration CommentableIdTAG: 6002x_snd_d NumberOfReplyTAG: 1 FirstChildTAG: Indeed! I have an old stereo with sliders, this makes me look at the equalizers in a whole new way. Anyway wanna play? FirstChildUserIdTAG: 175205 FirstChildUserNameTAG: KT_MM FirstChildCreateTimeTAG: 2012-10-12T01:15:16Z IndexTAG: 2335 TitleTAG: H4P3 Has anyone calculated Norton current(Part B) of the question? I still could not find the right answer.... UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-07T21:06:56Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You have to make sure that you are not repeating your equations and that they are carefully chosen. There are several ways to accomplish the problem. First, I came up with $R_{TH}$ via superposition; remember that independent voltage sources are replaced with a short, $=R_1+R_2 \parallel R_3$ (I may have mixed up the equation as I don't have the schematic in front of me). Then, I came up with equations for the loop current $(i)$ as it only takes one circular path (You can call the loop current whatever you want since it's not labeled, remember that current going into a (+) terminal means positive voltage). From there, I got $V_{TH}$ from $A \cdot u + i \cdot R_3$. Finally, $I_N = V_{TH} / R_{TH}$ : ) Now: dinner and bed. FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-10-07T21:57:29Z SecondChildTAG: Still not cleared with Rth...have been trying since 2 hours..Not Gettin ans correct SecondChildUserIdTAG: 390995 SecondChildUserNameTAG: coolnindz SecondChildCreateTimeTAG: 2012-10-07T23:28:48Z SecondChildTAG: Simulate the circuit in the circuit sandbox - that should give you enough info. Good luck! SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-08T04:22:03Z IndexTAG: 2336 TitleTAG: practical 4 Dear brother/sister please help in practical 4. what is "???" in tool. UserIdTAG: 342135 UserNameTAG: vikash902 CreateTimeTAG: 2012-10-07T19:08:03Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: That's the node label component. When you double click it, you can enter the name of the circuit node. In the lab it is used to name two 'vtest' nodes. FirstChildUserIdTAG: 502508 FirstChildUserNameTAG: amaher FirstChildCreateTimeTAG: 2012-10-07T19:14:19Z SecondChildTAG: Important note from lab description: _Nodes with the same name are considered to be electrically connected._ This is very useful in getting a clean circuit. SecondChildUserIdTAG: 502508 SecondChildUserNameTAG: amaher SecondChildCreateTimeTAG: 2012-10-07T19:15:53Z SecondChildTAG: what amaher says it is true :). The "???" it is a label, by default it is written in the box ??? so that is why you see that in the label . ![sand][1] So, this tool will be really useful when you want to design your circuit, like labeling inputs, nodes, etc. [1]: https://edxuploads.s3.amazonaws.com/13496377441431277.png SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T19:24:33Z FirstChildTAG: please send me friend request www.facebook.com/vikashverma9 FirstChildUserIdTAG: 342135 FirstChildUserNameTAG: vikash902 FirstChildCreateTimeTAG: 2012-10-07T19:18:07Z FirstChildTAG: If it's in the node label, than it's the name of the node. You can change it as you wish, and: > Nodes with the same name are considered to be electrically connected. FirstChildUserIdTAG: 192703 FirstChildUserNameTAG: voffch FirstChildCreateTimeTAG: 2012-10-07T19:13:28Z SecondChildTAG: What voffch says it is correct: Nodes with the same name are considered to be electrically connected :). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T19:25:52Z IndexTAG: 2337 TitleTAG: LAB4 curves I am not getting the curves . i have designed the circuit with voltage probe at drain , a dc value at vgs , triangular wave at vds and current probe b/w drain to source. where m i going wrong.? plz help! UserIdTAG: 431224 UserNameTAG: erachopra10 CreateTimeTAG: 2012-10-07T18:32:39Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: In the voltage probe use the option "x-axis" FirstChildUserIdTAG: 184827 FirstChildUserNameTAG: DiegoT FirstChildCreateTimeTAG: 2012-10-07T18:53:50Z FirstChildTAG: Also vary the color of your current probes. FirstChildUserIdTAG: 329015 FirstChildUserNameTAG: farah_sarwar FirstChildCreateTimeTAG: 2012-10-07T19:04:07Z FirstChildTAG: Your x-axis voltage probe needs to be at VDS and current probe at drain. FirstChildUserIdTAG: 502508 FirstChildUserNameTAG: amaher FirstChildCreateTimeTAG: 2012-10-07T19:00:43Z FirstChildTAG: You'd get a lot further posting a screen shot of your setup. It's likely you're overlooking something minor. If you did what the directions said verbatim, everything will work. And just to add to what's been said, for each MOSFET setup make sure you label the node before the current probe with "vtest." FirstChildUserIdTAG: 194509 FirstChildUserNameTAG: blackcompe FirstChildCreateTimeTAG: 2012-10-07T20:50:37Z SecondChildTAG: ![enter image description here][1] [1]: http:// SecondChildUserIdTAG: 431224 SecondChildUserNameTAG: erachopra10 SecondChildCreateTimeTAG: 2012-10-08T05:35:46Z SecondChildTAG: ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13496745844295505.png SecondChildUserIdTAG: 431224 SecondChildUserNameTAG: erachopra10 SecondChildCreateTimeTAG: 2012-10-08T05:36:33Z SecondChildTAG: where is this circuit going wrong.? plz tell SecondChildUserIdTAG: 431224 SecondChildUserNameTAG: erachopra10 SecondChildCreateTimeTAG: 2012-10-08T05:36:58Z SecondChildTAG: You should base on the example, then - replace the resistor in example with a mosfet, - plug the source to G gate of mosfet. - that is a circuit for a curve, for other curve just copy and paste the model then change the volt of source, 0.5, 1, 1.5...... to 3 V That is all SecondChildUserIdTAG: 114913 SecondChildUserNameTAG: ngoctuan SecondChildCreateTimeTAG: 2012-10-08T11:11:43Z SecondChildTAG: here is the example curve ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13496953954139501.jpg SecondChildUserIdTAG: 114913 SecondChildUserNameTAG: ngoctuan SecondChildCreateTimeTAG: 2012-10-08T11:23:20Z IndexTAG: 2338 TitleTAG: Equation https://www.dropbox.com/s/ss8gktvx1zebqtx/post.tiff Posted there is the equation from our professor. Why is it that for this problem we have to VI-VA My node equation is (VA-VI)/R - IA = 0 UserIdTAG: 230787 UserNameTAG: Gabriel007 CreateTimeTAG: 2012-10-07T18:22:16Z VoteTAG: 1 CoursewareTAG: Week 4 / Linearization Exercise 1 CommentableIdTAG: 6002x_linearization_ex_1 NumberOfReplyTAG: 2 FirstChildTAG: Probably because you're writing equation for currents entering the node, and he using currents leaving the node. Then you two have *the same equation* with the opposite signs before currents. Though it's not obvious what is VA and VI in your equation. FirstChildUserIdTAG: 192703 FirstChildUserNameTAG: voffch FirstChildCreateTimeTAG: 2012-10-07T19:08:20Z SecondChildTAG: Thanks for the response. But I am following his convention of exiting the node. VI is the battery and VA is e at the branch before the non-linear object. SecondChildUserIdTAG: 230787 SecondChildUserNameTAG: Gabriel007 SecondChildCreateTimeTAG: 2012-10-07T20:05:59Z FirstChildTAG: I got. I was subtracting... FirstChildUserIdTAG: 230787 FirstChildUserNameTAG: Gabriel007 FirstChildCreateTimeTAG: 2012-10-07T20:22:40Z IndexTAG: 2339 TitleTAG: STAFF - and general, about Homework and Labs Hi all, I have a question: Is there a way we can have the Homework and Labs solutions AFTER due date? For instance, now I've been trying for a long time to solve a homework exercise (IN for H4P3, PART B) and no way. Sure I still lack some knodledge. FOR SURE. But, I wonder, If I could have the answer and the path to the answer by tomorrow, for example, I would feel kind of releived If I see where I did the mistake/s. Thanks!! cheers, Sandra UserIdTAG: 342966 UserNameTAG: SandraNavarro CreateTimeTAG: 2012-10-07T16:50:26Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Sandra, After the deadline passes, go back to the homework. There will be a "Show Answer" button where the Check button was. That will let you see the answer. Of course, this is only after the deadline. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-07T16:56:51Z SecondChildTAG: Thanks! I did not know that! SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-07T17:01:03Z SecondChildTAG: You are right!! Thank you so much. Definitely I need more than 25 hrs a week to keep on with the course and still.... anyway, no matter if I drop, just thanks a lot!! SecondChildUserIdTAG: 342966 SecondChildUserNameTAG: SandraNavarro SecondChildCreateTimeTAG: 2012-10-07T21:36:58Z FirstChildTAG: I totally agree - in a "real" class, we'd be able to get feedback from the prof, teacher's assistants, tutors on homework assignments, etc... It would be a significant help for our learning process, especially since there's a MidTerm coming up. In the meantime, however, I suggest you search the Discussion Forum for "H4P3" and see if the many hints there help. If not, be sure to post a new (very specific!) question on where you're stuck, and I'll bet someone can provide you some hints. Good luck! FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-10-07T16:59:00Z SecondChildTAG: Maybe [this][1] would help; it was a sticking point for me. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5071a3f445001b2900000095 SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-07T17:00:48Z IndexTAG: 2340 TitleTAG: CALCULATIONS FOR LAB 4 I have tried to figure out what I am doing wrong on figuring out how to get the calculations for the problems. I think that I have it figured out and I am doing something wrong. I have looked at the figures in the textbook and still I cant figure it out. Can someone tell me a different way or easier way to get the figures UserIdTAG: 160242 UserNameTAG: Mlevins35 CreateTimeTAG: 2012-10-07T16:13:49Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: please use simulated graphs they are the best solution to your problem.. after you run transient analysis you can hover around your mouse an get the required current an voltage FirstChildUserIdTAG: 111320 FirstChildUserNameTAG: jhalife FirstChildCreateTimeTAG: 2012-10-07T16:48:39Z FirstChildTAG: **Mlevins35**, how difficult is it to set down a MOSFET, supply it with a given DC $V_{gs}=0$, a given 3V triangle wave $V_{ds}$, grounds, put plot markers at the $V_{ds}$, repeat this for $V_{gs}=0.5V, 1V, 1.5V ... 3V$, and run a transient analysis? Remember, you're not wiring the darn thing up on a breadboard, with a real MOSFET and real power supplies; it's just a sim that should take you about 30 minutes at most; faster if you just cut and paste the same circuit over. If you can't get it in a simulator, watch out if you're ever in real-life lab; you'll melt your proto-board, your MOSFET will catch fire, and the fuses in the power supplies will blow; that's when everyone wonders where the "burning" smell is coming from! FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-10-07T21:07:38Z SecondChildTAG: thats not the difficult part, I am not good at math I get the part of how to set a mosfet supply its just showing the equations to back it up. SecondChildUserIdTAG: 160242 SecondChildUserNameTAG: Mlevins35 SecondChildCreateTimeTAG: 2012-10-07T23:12:43Z IndexTAG: 2341 TitleTAG: h4p3>>>>>>>>>..partB i solved the norton resistance by guessing it .i didnot understand how can i find it.the first think that dependent source will replace with resistance with value=5 ohms...that give acorrect answer but why. UserIdTAG: 292299 UserNameTAG: ammarsamir CreateTimeTAG: 2012-10-07T15:41:10Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Rn = Vopen / In ---------- Vopen is voltage on port B without load. FirstChildUserIdTAG: 508733 FirstChildUserNameTAG: burdal1 FirstChildCreateTimeTAG: 2012-10-07T15:47:32Z SecondChildTAG: hint: For *u* (the driving voltage for controlled source), you can solve a equation for current in "main" loop (Vo, R1, R2, R3, Au). Then you can simply calculate A*u and then Vopen. ---------- I'm sorry for my primitive english. SecondChildUserIdTAG: 508733 SecondChildUserNameTAG: burdal1 SecondChildCreateTimeTAG: 2012-10-07T17:14:32Z FirstChildTAG: ammarsamir, Now that you have the answer, go back and try and figure out which equations could get you there. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-07T16:19:13Z IndexTAG: 2342 TitleTAG: H4P3. Could anyone provide me with a formula for finding In. Could anyone provide me with a formula for finding In. Clock is ticking... Urgent... UserIdTAG: 209363 UserNameTAG: phaneendraaa CreateTimeTAG: 2012-10-07T15:19:57Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Search the discussion forum for "H4P3". Good luck! FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-10-07T17:02:44Z FirstChildTAG: short the terminals across which u want to find u r circuit ..there is nothing like a simple formula.......... FirstChildUserIdTAG: 161899 FirstChildUserNameTAG: harshvit12 FirstChildCreateTimeTAG: 2012-10-07T17:47:29Z FirstChildTAG: find vth and rn and finally divide. FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-10-07T18:35:42Z IndexTAG: 2343 TitleTAG: SCRIPT ERROR LAB4 - Please Help :( I have a problem with lab 4, when i start the Transient Analysis appears that window about a problem with the script, what can i do? ![enter image description here][1] [1]: http://i839.photobucket.com/albums/zz311/CrisTamer07/scriptlab.jpg UserIdTAG: 440501 UserNameTAG: CGalindo CreateTimeTAG: 2012-10-07T14:52:02Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Hi Cgalindo! What time of TRAN are you setting? Might if you write a high value it colapses ;). Try with TRAN = 0.5m ... Myriam. P.D: Hablas en español? Necesitás ayuda? FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-07T18:02:12Z SecondChildTAG: Gracias, era problema del navegador, tuve que conseguir otro PC que tuviese IExplorer para poder hacerlo SecondChildUserIdTAG: 440501 SecondChildUserNameTAG: CGalindo SecondChildCreateTimeTAG: 2012-10-07T20:03:10Z SecondChildTAG: Me alegro que hayas podido solucionar tu inconveniente ;). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T20:09:19Z SecondChildTAG: alert(/xss/) SecondChildUserIdTAG: 678074 SecondChildUserNameTAG: pentest SecondChildCreateTimeTAG: 2012-10-18T07:39:44Z FirstChildTAG: Which browser/operating system are you using? And can you try another browser and see if you get the same results? FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-07T16:24:51Z SecondChildTAG: please use mozilla shall solve your problem and click on cintinue running script if still the pop up comes it comes in some pc due to antivirus SecondChildUserIdTAG: 111320 SecondChildUserNameTAG: jhalife SecondChildCreateTimeTAG: 2012-10-07T16:57:11Z SecondChildTAG: I had this in Mozilla. Try clicking continue. SecondChildUserIdTAG: 306244 SecondChildUserNameTAG: kevinsysum SecondChildCreateTimeTAG: 2012-10-07T17:46:31Z SecondChildTAG: Thanks! SecondChildUserIdTAG: 440501 SecondChildUserNameTAG: CGalindo SecondChildCreateTimeTAG: 2012-10-07T20:04:17Z SecondChildTAG: i'm using google chrome , and until now i have no problem . everithing is working fine. i have no doubt agout the post , i'm only sending the info maybe could help someone. SecondChildUserIdTAG: 402850 SecondChildUserNameTAG: y1111 SecondChildCreateTimeTAG: 2012-10-07T20:55:57Z FirstChildTAG: as mariam said , just use .0005 as stop time FirstChildUserIdTAG: 285675 FirstChildUserNameTAG: Ascot FirstChildCreateTimeTAG: 2012-10-07T18:20:28Z IndexTAG: 2344 TitleTAG: Lab 4 help needed!! i have completed the circuit and got the correct mark but my graph doesn't look like the given graph. so i'm unable to find iDS. since for triode region, vDS < vGS - VT, and given vGS=3 volt, vDS= 1 volt, i have vGS, vDS value. from here i found VT<2 now how to proceed?? please help. thanks. UserIdTAG: 204213 UserNameTAG: ratneshray CreateTimeTAG: 2012-10-07T14:51:02Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: please see if ur mosfet has w/l=1 and see that VDS is connected directly to drain then scroll through simualted graph i suppose correcting your graph shall help you. FirstChildUserIdTAG: 111320 FirstChildUserNameTAG: jhalife FirstChildCreateTimeTAG: 2012-10-07T16:54:00Z SecondChildTAG: @jhalife- my MOSFET has W/L=1 and VDS is connected to drain. SecondChildUserIdTAG: 204213 SecondChildUserNameTAG: ratneshray SecondChildCreateTimeTAG: 2012-10-07T17:37:56Z SecondChildTAG: I am also facing the same problem ... please help.![enter image description here][1] Here is my graph [1]: https://edxuploads.s3.amazonaws.com/13496340189944376.png SecondChildUserIdTAG: 413002 SecondChildUserNameTAG: Gauravjain88 SecondChildCreateTimeTAG: 2012-10-07T18:20:41Z SecondChildTAG: **Gauravjain88**: You have a few problems! I wish your schematic was larger, but it looks like you have your triangle waveform going into $V_{gs}$, if that's the case, that's incorrect; the triangle waveform goes into the top port, $V_{ds}$. Also, you have one too many probes. The x-axis probe should be at the TOP port, $V_{ds}$, also. Remember, the $V_{gs} =( 0, 0.5, 1, 1.5 ... 3)$ , (7) values in all, each goes into the GATE (the side port) and there is one separate circuit for EACH value in the "list" (yes, that means you'll need 7 MOSFETs total - you can take your basic circuit, copy and paste it, and change $V_{gs}$ in each, while keeping the triangle wave the same to all.) And try the trick where you use the "**label component**", it looks like a little wire, that you get to name as a node. See the first example with the resistor where they call it "vtest". For our purpose, name it "Vds" since the triangle wave $V_{ds}$ is common to ALL the circuits (0, 0.5, 1, ...3) you will draw. That way you only need to draw one triangle wave source $V_{ds}$, and use these "labels" wherever you need $V_{ds}$ in each of the (7) MOSFETS where it repeats. This helps the schematic from being a mess. SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-10-07T21:28:59Z SecondChildTAG: Have you gotten to the point yet where for the first question you measure Vds and Ids and use Ron = Vds/Ids? For some reason I'm not getting the right answer. I have the correct circuit as shown in the example, everything seems to be working correctly. The only thing I can think of is I got the units wrong on the current, however u means microamps, which is what I used. It is so simple, what can go wrong? SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-07T21:57:28Z IndexTAG: 2345 TitleTAG: buffer ? what does the buffer do exactly when its output is same as its input?? is it just an amplifier? UserIdTAG: 190441 UserNameTAG: JOHN92RAY CreateTimeTAG: 2012-10-07T13:37:06Z VoteTAG: 1 CoursewareTAG: Week 4 / Amplifiers CommentableIdTAG: 6002x_amplifiers NumberOfReplyTAG: 2 FirstChildTAG: Kinda. Ideally a buffer has a high input resistance and a low output resistance. The voltage in would equal voltage out, the difference being you would have more current available, without effecting the source much. Say you wanted to use a signal to drive two amplifiers, if you used a buffer or unity gain amplifier on the new branch, you could get a new duplicate signal without loading down (Taking current away from) the original one. If you did not use a buffer, the signal would have to drive both, the original amplifier and the new one. Neither amplifier would see the intended desired signal. They would have to share the current. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-07T15:09:50Z FirstChildTAG: Also a common use of a buffer is to keep a signal from degrading over distance. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-07T16:21:31Z IndexTAG: 2346 TitleTAG: homework after due time is there any penalty for doing homeworks after due time? UserIdTAG: 330357 UserNameTAG: steryd CreateTimeTAG: 2012-10-07T13:16:14Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Graded assignments are not counted if completed after the deadlines. Please read the Syllabus under the "Course Info" section. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-07T16:17:52Z FirstChildTAG: You will not be penalized (or "docked") for completing an assignment late; you simply will not receive credit for it. But there is no "punishment". FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-10-07T17:05:21Z IndexTAG: 2347 TitleTAG: Very interesting First of all, I would convey my sincere thanks to all out there making painstaking efforts in providing a quality education at my hand touch. This coure is really exciting and most of all the lab is mind boggling. Good piece of education. UserIdTAG: 145544 UserNameTAG: pandiya CreateTimeTAG: 2012-10-07T13:13:31Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 2348 TitleTAG: hw4p1 i am having serious trouble understand how we work out Ip if anyone can help me id be much aprechiated as i think im using the right way but am not getting the correct answers i am using IP = P*VPk^3/2 so 2.0*12^3/2 but that doesnt give me the correct answer am i going wrong somewhere UserIdTAG: 436749 UserNameTAG: waynebrown CreateTimeTAG: 2012-10-07T10:33:39Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Remember that the P value is in milliAmps. 10^-3. Everything else is fine. FirstChildUserIdTAG: 101329 FirstChildUserNameTAG: kellenproctor FirstChildCreateTimeTAG: 2012-10-07T10:45:04Z SecondChildTAG: where should it be 10 ^-3 please SecondChildUserIdTAG: 436749 SecondChildUserNameTAG: waynebrown SecondChildCreateTimeTAG: 2012-10-07T10:51:25Z SecondChildTAG: How did you find incremental resistance? SecondChildUserIdTAG: 360053 SecondChildUserNameTAG: Ohedul SecondChildCreateTimeTAG: 2012-10-07T12:24:42Z SecondChildTAG: @waynebrown Just multiply your equation by 10^-3 SecondChildUserIdTAG: 90075 SecondChildUserNameTAG: Samay SecondChildCreateTimeTAG: 2012-10-07T13:50:38Z SecondChildTAG: i have used the equation 2.0*12^3/2*10^-3 but it is still wrong im very confused where im going wrong SecondChildUserIdTAG: 436749 SecondChildUserNameTAG: waynebrown SecondChildCreateTimeTAG: 2012-10-07T16:44:11Z IndexTAG: 2349 TitleTAG: Pleasee help me with Lab 4 I got the ckt right and the plot but not able to calculate rOn, Where should i take vds and ids values(triode region)??? please help me calculate Ron.. UserIdTAG: 148523 UserNameTAG: span993 CreateTimeTAG: 2012-10-07T10:19:42Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Ok, just got it. Bad approximations Please help me with the second part thanks in advance. FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-10-07T10:33:06Z SecondChildTAG: Yippie! just completed Lab 4 SecondChildUserIdTAG: 148523 SecondChildUserNameTAG: span993 SecondChildCreateTimeTAG: 2012-10-07T10:44:07Z SecondChildTAG: some one please help me in lab 4 ...beacuse i m unable to c the video's beacuse of youtube blockage :( SecondChildUserIdTAG: 415376 SecondChildUserNameTAG: imab90 SecondChildCreateTimeTAG: 2012-10-07T10:45:43Z SecondChildTAG: U dont need the videoas just refer to sec 7.3 in text SecondChildUserIdTAG: 148523 SecondChildUserNameTAG: span993 SecondChildCreateTimeTAG: 2012-10-07T14:21:56Z IndexTAG: 2350 TitleTAG: Thank you mit Thanks a lot for providing this course in the best way and that too for free of cost. Expecting some other courses also like digital electronics or Digital signal processing etc UserIdTAG: 132685 UserNameTAG: deepakmurali CreateTimeTAG: 2012-10-07T09:21:48Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: This course is awesome. Thank you for it! FirstChildUserIdTAG: 326787 FirstChildUserNameTAG: Angstrem FirstChildCreateTimeTAG: 2012-10-07T09:36:23Z IndexTAG: 2351 TitleTAG: FORUM How do I search the forum? where do I have to click? I know about the schedule times. Thanks!! UserIdTAG: 342966 UserNameTAG: SandraNavarro CreateTimeTAG: 2012-10-07T07:13:50Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi SandraNavarro! How are you? You can go to the blue circle (look at the image) and then write in the box something that you want to find like: lab4. ![imagen][1] Also you can go to ALL and in the down arrow you will find some posts by topics of weeks. You can also see my Wiki hints ;) in Wiki-> Myrimit's guide to 6.002x -> [Myrimit's Hints][2] See you :), Myriam. [1]: https://edxuploads.s3.amazonaws.com/13495977101343662.png [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/myrimits-hints-links/ FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-07T08:19:54Z SecondChildTAG: Ok, then it is what I already new.... I was confused, thank you for clarifying!!!! I'll follow your wiki Hints.... it is amazing the support you're giving to all students. No words... Thank you!!!! Regards, cheers!! Sandra SecondChildUserIdTAG: 342966 SecondChildUserNameTAG: SandraNavarro SecondChildCreateTimeTAG: 2012-10-07T10:19:10Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T10:23:03Z SecondChildTAG: Gracias de nuevo!! (Thanks again!!) SecondChildUserIdTAG: 342966 SecondChildUserNameTAG: SandraNavarro SecondChildCreateTimeTAG: 2012-10-07T16:44:37Z IndexTAG: 2352 TitleTAG: how to change the mailing address and the password? plz help!! UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-07T03:40:27Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 2353 TitleTAG: CIRCUIT IN THE EXAMPLE **Someone can post the image from the circuit (input sound,led,amplifier) to do it in real??** UserIdTAG: 118093 UserNameTAG: Oktavious CreateTimeTAG: 2012-10-06T23:06:01Z VoteTAG: 1 CoursewareTAG: Week 4 / Incremental Analysis Demo CommentableIdTAG: 6002x_inc_analysis_demo NumberOfReplyTAG: 0 IndexTAG: 2354 TitleTAG: When exactly could we discuss H4P3 Part B to the bone ? Now I did get the magical green tick(never felt so good), but I feel that I just happened to stumble across the right answer. I got correct Rth using the combination of resistances, which I feel is not possible as there's a dependent source in series with R3, but it shows up as correct. And then using that and the Vth which I deduce using Node Analysis I got In, which was also right. Now my issue is with the Rth. Getting a green check makes me happy but its still killin me to know that it might just be blind luck. UserIdTAG: 204745 UserNameTAG: allwynmendes CreateTimeTAG: 2012-10-06T21:46:58Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Yes I know seems like some sort of black magic lol. We are missing something from those lectures. FirstChildUserIdTAG: 256543 FirstChildUserNameTAG: sidney23 FirstChildCreateTimeTAG: 2012-10-06T22:04:30Z SecondChildTAG: I can't get RN or the magical green tick- I have tried every combination of series and parallel addition SecondChildUserIdTAG: 18073 SecondChildUserNameTAG: gburkhart SecondChildCreateTimeTAG: 2012-10-06T23:39:07Z SecondChildTAG: pls help....... SecondChildUserIdTAG: 214277 SecondChildUserNameTAG: yadsam SecondChildCreateTimeTAG: 2012-10-07T06:45:49Z IndexTAG: 2355 TitleTAG: When time is over? Hi, I don't know which time exactly I have to finish my homework. I'm from Germany. In lab 4, there is no submit-button, is this correct? UserIdTAG: 389333 UserNameTAG: juergen CreateTimeTAG: 2012-10-06T20:58:32Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Is there a "Check" button there? FirstChildUserIdTAG: 326787 FirstChildUserNameTAG: Angstrem FirstChildCreateTimeTAG: 2012-10-06T21:00:41Z SecondChildTAG: yes, check-button is there, all check marks are green SecondChildUserIdTAG: 389333 SecondChildUserNameTAG: juergen SecondChildCreateTimeTAG: 2012-10-06T21:05:12Z SecondChildTAG: It seems that it's enough. Press the "Progress" button at the dashboard and you can see how good you are. It's reasonable to think, that if you've violated the deadline than the result's will not appear in the Progress section. SecondChildUserIdTAG: 326787 SecondChildUserNameTAG: Angstrem SecondChildCreateTimeTAG: 2012-10-06T21:14:05Z IndexTAG: 2356 TitleTAG: H4P3: did everybody have that big equations??? Hello everyone! I've just solved the 4th homework (in the last day! And lab4 is on queue =_=). In the process of solving I've had to solve enormous equations, like this: R = (R1*R2(R1+R2))/((R2+R3)(R1+R2)-(A*R2+R3)*R1) Maybe, I've missed something and there was a more simple and elegant way to proceed? UserIdTAG: 326787 UserNameTAG: Angstrem CreateTimeTAG: 2012-10-06T19:05:05Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: For parallel resistors, I prefer to use 1/Rp = 1/R1 + 1/R2 rather than Rp = (R1*R2)/(R1+R2) is a bit simpler and works or any number of parallel resistors. i.e. 1/Rp = 1/R1 + 1/R2 + 1/R3 + ..... + 1/Rx. FirstChildUserIdTAG: 270011 FirstChildUserNameTAG: SnowmanZA FirstChildCreateTimeTAG: 2012-10-06T20:15:54Z SecondChildTAG: Thanks for the advice. Actually, it's really more comfortable to use this form for parallel resistors, because if you use (R1*R2)/(R1+R2), you'll have problems with 3 or more parallel resistors. SecondChildUserIdTAG: 326787 SecondChildUserNameTAG: Angstrem SecondChildCreateTimeTAG: 2012-10-06T21:15:20Z FirstChildTAG: You need to plug in the values for these resistors, then combine as necessary as you solve the equations. You are right, it makes it enormous when you just use variables. FirstChildUserIdTAG: 132828 FirstChildUserNameTAG: rharris FirstChildCreateTimeTAG: 2012-10-06T20:16:27Z SecondChildTAG: That is, it's more compact if you operate digits, not variables? Well, it really is, but I've got used to operate with variables first. It's embarrassing for me to solve digital equations. SecondChildUserIdTAG: 326787 SecondChildUserNameTAG: Angstrem SecondChildCreateTimeTAG: 2012-10-06T21:17:10Z SecondChildTAG: I don't understand, why would you be embarrassed to solve digital equations? What is a digital equation? To me a digital equation would involve Boolean algebra. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-06T21:34:18Z SecondChildTAG: No, I've meant the usual decimal digits, not logical ones. It's simple a habit for me: first of all I find something in terms of variables and only then in terms of digits. I think, that's more comfortable, because you can then apply your equation-with-variables to multiple cases. SecondChildUserIdTAG: 326787 SecondChildUserNameTAG: Angstrem SecondChildCreateTimeTAG: 2012-10-06T21:45:26Z IndexTAG: 2357 TitleTAG: weekly solutions In the spring you provided weekly solutions after the due date of assignments. It was very helpful. UserIdTAG: 10512 UserNameTAG: asicok CreateTimeTAG: 2012-10-06T19:04:48Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: They still do post answers to assignments after the due date. Except now they are not in a download, they are in the homework themselves. Go back to the homework and hit the "Show Answer" button. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-06T19:19:17Z SecondChildTAG: Hi asicok! What JSChambers said it is true :). If you go to your Homework and Lab and click on "Show answer" it will display you the answer solved step by step. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T05:04:08Z IndexTAG: 2358 TitleTAG: Lab 6 help please I use the equ 10.67 from the textbook but my answer still wrong why please help and thankx UserIdTAG: 331441 UserNameTAG: abdessamade CreateTimeTAG: 2012-10-06T16:24:41Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: The transition is made from initial t=0. So you must check what is the value of V that you have there. You must not assume it is equal to VS. I believe that this is the error in your approach as it got me stuck for a while too. :) FirstChildUserIdTAG: 304587 FirstChildUserNameTAG: Vasso FirstChildCreateTimeTAG: 2012-10-07T20:12:26Z IndexTAG: 2359 TitleTAG: Lab 4 bug with checking ... It always gives me green check even with wrong circuit I tried to put a wrong circuit and pressed check ( don't ask why ) and it shows that my answer ( to the plotting only not the sections below ) as correct although I'm sure it was wrong . UserIdTAG: 454609 UserNameTAG: Alwahsh CreateTimeTAG: 2012-10-06T14:17:48Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi Alwahsh! Ok, I will not ask why you pressed check haha! ;) Yes, that is one bug of some Labs... Also it happened in the Prototype Course... Try to check if the answers of the questions are ok, you will corroborate your circuit if the answers are ok. Try to ignore this bug for the moment. See you, Myriam. P.D: In the case that you are needing some hints of lab4 you can see this [post][1]. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5070e1b72210d02400000055 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-07T05:42:44Z IndexTAG: 2360 TitleTAG: 40mV across zener diode when 40mV is applied across the zener diode it should behave as a open circuit isnt it? UserIdTAG: 257806 UserNameTAG: skoda CreateTimeTAG: 2012-10-06T12:06:52Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Same querry here :( FirstChildUserIdTAG: 97340 FirstChildUserNameTAG: AnkitRana FirstChildCreateTimeTAG: 2012-10-06T15:06:24Z FirstChildTAG: it should act like an open circuit as per its i-v characteristics but answers do not match with this condition. Also if it start acting like open circuit then value of $ v_0 $ should be same as in the previous circuit, but what will be its purpose then? We are definitely missing something. FirstChildUserIdTAG: 329015 FirstChildUserNameTAG: farah_sarwar FirstChildCreateTimeTAG: 2012-10-06T17:53:30Z FirstChildTAG: Follow this thread: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum#6002x_Fall2012_General/threads/50701d6df9e5af1f00000083 FirstChildUserIdTAG: 329015 FirstChildUserNameTAG: farah_sarwar FirstChildCreateTimeTAG: 2012-10-06T17:58:51Z SecondChildTAG: I get nothing when I click that link. SecondChildUserIdTAG: 339870 SecondChildUserNameTAG: rjlasota SecondChildCreateTimeTAG: 2012-10-06T23:45:17Z IndexTAG: 2361 TitleTAG: S8E0 why negative sign?? UserIdTAG: 353709 UserNameTAG: anshu10750 CreateTimeTAG: 2012-10-06T12:05:53Z VoteTAG: 1 CoursewareTAG: Week 4 / Dependent Sources Exercise 0 CommentableIdTAG: 6002x_dep_src_exsz_0 NumberOfReplyTAG: 2 FirstChildTAG: look at the way the current sign is facing. compared to VO FirstChildUserIdTAG: 262875 FirstChildUserNameTAG: GrantDennison FirstChildCreateTimeTAG: 2012-10-06T16:31:38Z FirstChildTAG: In addition to what GrantDennison said, KCL is V0 = RO*(-iD). FirstChildUserIdTAG: 194509 FirstChildUserNameTAG: blackcompe FirstChildCreateTimeTAG: 2012-10-07T02:04:30Z IndexTAG: 2362 TitleTAG: Lab4 help- how to plot current DS of Mosfet My transient analysis graph show current as 0, I checked my setup, I have DC voltage source on the gate, I have test voltage on the DS, the problem could be the wrong place to plot the current. I know the current is from drain to source, but how to represent it on the schematic sheet? Please help, I’ve spent so much time and no clue about it. Julie UserIdTAG: 240389 UserNameTAG: julieqpt CreateTimeTAG: 2012-10-06T11:57:06Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: so you're saying ,the current probe read $0A$ for different gate voltages? FirstChildUserIdTAG: 386004 FirstChildUserNameTAG: ravinarv FirstChildCreateTimeTAG: 2012-10-06T13:14:26Z SecondChildTAG: have the same problem kindly guide me thanks SecondChildUserIdTAG: 438395 SecondChildUserNameTAG: ali_PU1 SecondChildCreateTimeTAG: 2012-10-06T15:34:39Z SecondChildTAG: yes, changing DC voltage on the gate doesn't change anything, the current is the same 0, not sure which is missing here. SecondChildUserIdTAG: 240389 SecondChildUserNameTAG: julieqpt SecondChildCreateTimeTAG: 2012-10-06T16:43:00Z FirstChildTAG: Hi julieqpt! Can I help you? Are you sure that you are not missing the variable voltage source ;). P.D: in the case that you are needing hints of lab4 you can see [here][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5070e1b72210d02400000055 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-07T05:55:32Z IndexTAG: 2363 TitleTAG: H4P3 A I need help as to how to calculate ***i***. So far I have a node equation at ***Io***, ***R1*** and ***Zi*** that goes... **Io + i + Z*i = 0** Substituting I get , **2A + i + 2r*i = 0** This gives me *0.666A* = i which is wrong! UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-10-06T09:35:18Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Zi is the **voltage** across the current controlled voltage source (CCVS) not the current though it. The node equation at that node simply states that the current I0 splits, part goes through R1 and part goes through the CCVS. The part going through R1 is determined by the voltage at the node. In an open circuit the current through the CCVS goes through R2 and that current dependents on the output voltage. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-06T12:07:42Z SecondChildTAG: how are we going to get the voltage across R1? SecondChildUserIdTAG: 278792 SecondChildUserNameTAG: sali SecondChildCreateTimeTAG: 2012-10-07T01:03:52Z SecondChildTAG: Using the normal node method. if the node above R1 is e and node below R1 is ground, then the voltage across R1 is e/R1. SecondChildUserIdTAG: 366165 SecondChildUserNameTAG: silicon_ghost SecondChildCreateTimeTAG: 2012-10-07T02:08:03Z SecondChildTAG: thanx silicon_ghost ...actually i wasn't sure what to do with the dependent source ..but anyways i got the right answer :) SecondChildUserIdTAG: 278792 SecondChildUserNameTAG: sali SecondChildCreateTimeTAG: 2012-10-07T03:00:06Z IndexTAG: 2364 TitleTAG: staff : help with downloading week 5 & 6 & ... Staff My country has closed (filtered) You tube. I tried to use Proxifire, but it can not open the You tube. I enrolled in 6.002x Circuits and Electronics. Can some one provide direct links to download the lecture videos from week 5 to week 7? I do not want to miss this course. please help me. UserIdTAG: 374393 UserNameTAG: rmaleki CreateTimeTAG: 2012-10-06T09:11:11Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Now I have access to you tube with the **Hotspot Shield** VPN or **Spotflux** VPN. everyone can download them by free. I know that it works at least it works in Iran and Pakistan. FirstChildUserIdTAG: 374393 FirstChildUserNameTAG: rmaleki FirstChildCreateTimeTAG: 2012-10-06T19:50:28Z IndexTAG: 2365 TitleTAG: H4P1 I need some help please. Hi everyone, I have a problem calculating the incremental resistance, I'm using the equation: 1/(d(i(v))/dv) evaluated in the point V_PK but it says the answer is wrong... If any of you can help me, I'd be really thankful. UserIdTAG: 61423 UserNameTAG: libo654 CreateTimeTAG: 2012-10-06T03:04:45Z VoteTAG: 1 CoursewareTAG: Week 4 / Incremental Method Math CommentableIdTAG: 6002x_inc_method_math NumberOfReplyTAG: 5 FirstChildTAG: the equation u told is correct.... just try once more.. u will get the answer... FirstChildUserIdTAG: 477198 FirstChildUserNameTAG: Rajesh1993 FirstChildCreateTimeTAG: 2012-10-06T07:01:16Z FirstChildTAG: May I know the algebraic expression for your derivative? FirstChildUserIdTAG: 282828 FirstChildUserNameTAG: Chibolator FirstChildCreateTimeTAG: 2012-10-06T05:59:18Z FirstChildTAG: u just need to differentiate Ip with respect to V for the equation Ip=P⋅Vpk^(3/2). i.e. dIp/dVpk put the given values and then inverse it as incremental resistance is dV/dI. FirstChildUserIdTAG: 232499 FirstChildUserNameTAG: Asim09 FirstChildCreateTimeTAG: 2012-10-07T04:50:51Z SecondChildTAG: thanx SecondChildUserIdTAG: 162670 SecondChildUserNameTAG: charlesbabyt SecondChildCreateTimeTAG: 2012-10-07T18:44:12Z FirstChildTAG: I had the same problem - I just forgot to pick P = 2*10^(-3). I was counting for P = 2 ... FirstChildUserIdTAG: 413065 FirstChildUserNameTAG: anikey FirstChildCreateTimeTAG: 2012-10-07T08:56:57Z FirstChildTAG: OMG I already found my mistake, I didn't realize that P was in mA jaja. Thanks for your help :) FirstChildUserIdTAG: 61423 FirstChildUserNameTAG: libo654 FirstChildCreateTimeTAG: 2012-10-07T17:04:53Z IndexTAG: 2366 TitleTAG: Proof of linearity I would say that proof of linearity would be not in neglecting non-linear terms of series, but in they value estimation. Which would be wild to do in common. UserIdTAG: 66669 UserNameTAG: agarus CreateTimeTAG: 2012-10-05T22:49:34Z VoteTAG: 1 CoursewareTAG: Week 4 / Incremental Method Math CommentableIdTAG: 6002x_inc_method_math NumberOfReplyTAG: 0 IndexTAG: 2367 TitleTAG: What is transconductance then? If I understand it correctly transconductance shall be given by the following formula: $g_m=K\cdot(V_{GS}-V_T)$ Talking about either a buffer or an amplifier, if a large variable signal is supplied as input (not a DC bias and a small variable component), the transconductance will vary with the input signal. Am I correct? UserIdTAG: 252722 UserNameTAG: vaboro CreateTimeTAG: 2012-10-05T21:35:33Z VoteTAG: 1 CoursewareTAG: Week 6 / Perspective on the small-signal circuit CommentableIdTAG: 6002x_small_signal_perspective NumberOfReplyTAG: 1 FirstChildTAG: Yes. It's why we have distortion on large input signal amplitude - because transconductance will vary. For small signal it varies also but much much less and we can take it as a constant, so amplification will be more or less linear. FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-05T21:55:25Z IndexTAG: 2368 TitleTAG: calculation error in vid at video time 6:42: Calculation (2.45-2.39)/0.1 is said to be 0.573. In truth it is 0.6 UserIdTAG: 12905 UserNameTAG: Baer CreateTimeTAG: 2012-10-05T21:20:29Z VoteTAG: 1 CoursewareTAG: Week 4 / Small Signal Long Tutorial CommentableIdTAG: 6002x_small_sig_long_t NumberOfReplyTAG: 0 IndexTAG: 2369 TitleTAG: Some help with H4P1 Hopefully this will help with Q1 and Q3, but for Q2 and Q4 you will need to find the derivative of the equation, use wolfram alpha for this! But this is the starting point. The first step is to set up the equations that need to be solved using KVL and/or KCL. You have two unknowns, so you need two equations. KCL isn't going to do you any good, so use KVL to get one of them. Then the relationship betwee i_A and v_A is the other. Now combine the two equations to eliminate either i_A or v_A, whichever seems easier to do. You now have a single equation in the one that is left, say v_A for discussion's sake. Now solve it partially for v_A, meaning to get v_A alone on one side, even though you will still have v_A embedded in the other side, too. Now pick a trial value for v_A and plug it into the right hand side. If it is the correct value, it will yield that same value of v_A as a result. If not, then it will obviously either be greater than or less than the desired value. Examine the right hand side as see whether the result will increase or decrease if you increase v_A by a little bit (this is where knowing how to take derivatives is real handy, but you can do it without this). With this in mind, pick a new value of v_A and see what the result it. After a could iterations, you should get a feel for whether you are changing v_A by too big or too small a step. Often, a good starting point would be v_A = 0V. If that is negative (i.e., too small) then your immediate goal is to find a value of v_A that results in a value that is too large. Assuming that the nonlinear function is at least continuous, then you know that a solution exists between 0V and this last value of v_A. With that in hand, you can get to a solution very quickly by just using the value midway between them as your next guess. At each stage, decide if the correct value is in the lower or upper half of your range and then pick the midpoint of the new narrower range as the next trial value. You can set this up in a spread sheet very handily. UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-10-05T20:45:24Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: I think you are right amaher.We can use IP=f(VPK) FirstChildUserIdTAG: 282828 FirstChildUserNameTAG: Chibolator FirstChildCreateTimeTAG: 2012-10-06T06:48:30Z FirstChildTAG: Can't stress enough - for those like myself who lack the requisite math background - **USE WOLFRAM ALPHA** for the calculus! [Wolfram Alpha][1] You'll save yourself (and your furniture/walls) HOURS of head-banging! [1]: http://www.wolframalpha.com FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-10-06T18:17:52Z FirstChildTAG: Can we not just use $I_P = f(V_{PK}) = P \cdot V_{PK}^{3/2}$ which we get from the large signal circuit analysis? FirstChildUserIdTAG: 502508 FirstChildUserNameTAG: amaher FirstChildCreateTimeTAG: 2012-10-05T21:17:18Z SecondChildTAG: amaher This equation is exactly what you need for Q1 and Q3. SecondChildUserIdTAG: 345967 SecondChildUserNameTAG: Tsipis SecondChildCreateTimeTAG: 2012-10-07T14:06:17Z IndexTAG: 2370 TitleTAG: incremental resistance how can we get the incremental resistance UserIdTAG: 326507 UserNameTAG: mohamed200 CreateTimeTAG: 2012-10-05T20:13:01Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Incremental resistance is $\cfrac{1}{\cfrac{df(V_D)}{dV_D}}$ FirstChildUserIdTAG: 502508 FirstChildUserNameTAG: amaher FirstChildCreateTimeTAG: 2012-10-05T21:25:12Z SecondChildTAG: Using the main example from the lectures, where $i_D = f(v_D) = ae^{bv_D}$ Equation X: $I_D + i_d \approx f(V_D) + \cfrac{df(v_D)}{dv_D}|_{v_D = V_D} \cdot v_d$ From X: $I_D + i_d \approx ae^{bV_D} + ae^{bV_D}b \cdot v_d$ Here, $ae^{bV_D}b = \cfrac{df(V_D)}{dV_D}$ is the slope at $V_D, I_D$ and hence is equal to $\cfrac {1}{\Delta{R}}$ so incremental R, $\Delta{R} = \cfrac {1}{\cfrac{df(V_D)}{dV_D}} = \cfrac{1}{ae^{bV_D}b} = \cfrac{1}{I_D \cdot b}$ SecondChildUserIdTAG: 502508 SecondChildUserNameTAG: amaher SecondChildCreateTimeTAG: 2012-10-05T21:48:47Z SecondChildTAG: thanx SecondChildUserIdTAG: 326507 SecondChildUserNameTAG: mohamed200 SecondChildCreateTimeTAG: 2012-10-06T22:22:08Z IndexTAG: 2371 TitleTAG: Sinusoidal curves I don't understand what these sinusoidal curves denote. How can the current and Voltage be once +ve ,then -ve and back to +ve UserIdTAG: 284172 UserNameTAG: mayankg175 CreateTimeTAG: 2012-10-05T19:30:17Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Its like a pendulum swinging both directions alternatively. when we generate sin wave using a dynamo because of the circular motion of coil or magnet the current induced by magnetic field will be positive and negative alternatively. FirstChildUserIdTAG: 14726 FirstChildUserNameTAG: Sajilck FirstChildCreateTimeTAG: 2012-10-05T19:50:47Z SecondChildTAG: Thanks Sajilck ...but still I am confused that does the current coming out follows the same pattern ,just like what its graph denote. SecondChildUserIdTAG: 284172 SecondChildUserNameTAG: mayankg175 SecondChildCreateTimeTAG: 2012-10-06T05:00:04Z IndexTAG: 2372 TitleTAG: Keeping up with deadlines I'm currently a week behind the schedule and I'm increasingly finding it difficult to keep up with all the homework and reading(text) and understanding the whole thing. Especially I don't know which part of the text book I should read for a specific topic and whether I should read it before after the lecture and if its okay to leave out other parts which aren't asked by lecturer for reading. How should I plug the gap and once I'm done with that how should I divide my week for lectures, reading and homework? Thanks in advance, Dustinge UserIdTAG: 395953 UserNameTAG: dustinge CreateTimeTAG: 2012-10-05T17:13:02Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: yes, it would be nice if reference to textbook be provided - that links under video often misleading and chapters are mentioning in the lectures itself - it would be much better if there will be just a list of chapters of the textbook on every lecture's video web page FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-05T17:20:08Z SecondChildTAG: Pages are unresponsive too! I can't view the next page even after I hit the next button multiple times SecondChildUserIdTAG: 395953 SecondChildUserNameTAG: dustinge SecondChildCreateTimeTAG: 2012-10-05T18:11:58Z FirstChildTAG: I've found that reading the text first, and working hard to understand how the examples are done in particular, and then watching the lectures works very well. For me, anyway. The list of readings is here: [Link to Course-at-a-Glance Handout][1] [1]: https://www.edx.org/static/content-mit-6002x/handouts/at-a-glance.9674fe7f677e.pdf FirstChildUserIdTAG: 339668 FirstChildUserNameTAG: chickwebb FirstChildCreateTimeTAG: 2012-10-06T00:27:07Z IndexTAG: 2373 TitleTAG: mid term exam what is the time duration for the mid term exam? UserIdTAG: 114864 UserNameTAG: Vikaash CreateTimeTAG: 2012-10-05T13:49:08Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: 24 Hours. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-05T14:01:45Z SecondChildTAG: whats about syllabus???? SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-10-05T14:26:21Z SecondChildTAG: wow SecondChildUserIdTAG: 403660 SecondChildUserNameTAG: abuodeh SecondChildCreateTimeTAG: 2012-10-05T14:36:30Z SecondChildTAG: The syllabus is in the "Course Info" section. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-06T00:46:22Z IndexTAG: 2374 TitleTAG: H4P2 Hi, I'm having difficulty obtaining the correct values for vo in the second part of the problem (after the diode is connected). Because vi is smaller than 0.6 I believe the diode is in its region where it does not let current through, so it should be replaced with an open circuit. But in this case, the vo should be exactly the same value as in the first part of the problem, without the diode. I can't figure it out. Can anyone please help me? UserIdTAG: 363742 UserNameTAG: GAgnes CreateTimeTAG: 2012-10-05T10:50:51Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: I'm having the same problem here... Any suggestion?? Thanks. FirstChildUserIdTAG: 239608 FirstChildUserNameTAG: guillegf84 FirstChildCreateTimeTAG: 2012-10-05T13:51:58Z SecondChildTAG: No, you incorrectly identified the region of diode work. Try to find the points, where the load line intersects with x- and y-axes. And it is obviously from the question, that Zener diode must decrease v0 (noise level). If you replace the diode with open circuit so it has no any v0 correction effect. Sorry for my English. SecondChildUserIdTAG: 100840 SecondChildUserNameTAG: Gudin SecondChildCreateTimeTAG: 2012-10-05T14:26:22Z FirstChildTAG: Zener works on breakdown reverse voltage, and its operating point is behind negative breakdown. Please notice that IV you given is forward voltage - you need to convert it right way to get it ready for load line. http://en.wikipedia.org/wiki/Zener_diode FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-05T15:54:31Z FirstChildTAG: Like Gudin said, you incorrectly identified the region where the diode is operating. The region is determined by the total voltage across the diode - you can't break the signal into two components and have the diode act differently on each part. Also, as the hint says, be very careful about the signs! FirstChildUserIdTAG: 176592 FirstChildUserNameTAG: JLevitt FirstChildCreateTimeTAG: 2012-10-05T15:56:28Z SecondChildTAG: Thanks, my problem was to apply superposition. I forgot that the zener was working with Vi too. SecondChildUserIdTAG: 239608 SecondChildUserNameTAG: guillegf84 SecondChildCreateTimeTAG: 2012-10-05T23:07:40Z FirstChildTAG: me too not getting answer for that problem.... can any one give me a clue ? FirstChildUserIdTAG: 477198 FirstChildUserNameTAG: Rajesh1993 FirstChildCreateTimeTAG: 2012-10-06T07:42:44Z IndexTAG: 2375 TitleTAG: hp4 can anybody tell me how to find the answer for output noise in second part of second problem? UserIdTAG: 396893 UserNameTAG: Raajesh CreateTimeTAG: 2012-10-05T10:39:35Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hey pal! I have this problem too! H4P2 part 2 FirstChildUserIdTAG: 499268 FirstChildUserNameTAG: ruinoah FirstChildCreateTimeTAG: 2012-10-05T17:12:32Z IndexTAG: 2376 TitleTAG: certification I can't really understand whether the certification is free or not , and how are we getting it ? I'm from Egypt , will it be shipped or mailed? and how much would it cost ? UserIdTAG: 528897 UserNameTAG: vepoo CreateTimeTAG: 2012-10-05T08:30:58Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi vepoo! Based on my experience in the prototype course 6.002x, I can tell you that I got my certificate 100% free in the spring :) and I could download it, once the course was ended, in the place where you can see your progress (it will appear a buttom). The certificate had also a link in the .pdf where you can check and see that it is authentic (edx server). The certificate wasn´t delivered to your home in the 6.002x spring. See you! :) Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-05T12:27:24Z SecondChildTAG: thank you very much for illustration , but would that certification soft copy be trusted? as it can be easily imitated , I mean here by trusted from the point of view of any organization I would present that certification to as a part of my CV . SecondChildUserIdTAG: 528897 SecondChildUserNameTAG: vepoo SecondChildCreateTimeTAG: 2012-10-05T13:58:48Z SecondChildTAG: "The certificate had also a link in the .pdf where you can check and see that it is authentic (edx server)." SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-05T14:08:34Z SecondChildTAG: :) I'm sorry I didn't notice this . SecondChildUserIdTAG: 528897 SecondChildUserNameTAG: vepoo SecondChildCreateTimeTAG: 2012-10-05T14:10:42Z IndexTAG: 2377 TitleTAG: Is gm the derivative of Ids in the general case ? Hi ! Is gm the derivative of Ids for the mosfet amplifier ? Because a small change in the current with respect to a small change in Vin is the derivative.And it begins to make sense since the derivative of the current Id on a diode, was also a sort of conductance (1/Rd). And if this is true, we should have a clear path : gm is the derivative of the current, and Ro is the derivative of the output V0 , both with respect to VIN. Thanks ! UserIdTAG: 240487 UserNameTAG: AlexAlexandrescu CreateTimeTAG: 2012-10-04T19:37:10Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: The Week 6 tutorial video gives definitions for and shows exactly how to calculate gm and Ro. FirstChildUserIdTAG: 270011 FirstChildUserNameTAG: SnowmanZA FirstChildCreateTimeTAG: 2012-10-04T19:39:06Z SecondChildTAG: Yep, but that was a particular case.Suppose you don't have a load resistance.Suppose that the current expression is different, like in H6P1. I mean in the general case. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-10-05T08:41:46Z SecondChildTAG: The idea remains the same. i.e. The transconductance gm is del i / del Vin. So for a FET it would be del id / del Vgs. The conductance which is 1/Ro is del i / del Vs. So for a FET it would be del id / del Vds. SecondChildUserIdTAG: 270011 SecondChildUserNameTAG: SnowmanZA SecondChildCreateTimeTAG: 2012-10-05T14:58:26Z SecondChildTAG: Thank you! Nice to see another opinion. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-10-05T16:51:52Z IndexTAG: 2378 TitleTAG: Lab 4 Help Can anyone tell me how to build this circuit in Lab 4. I made it as it was in textbook 7.3 and connect probes to wire connected to Vds but I cannot get proper result. When I do transient analysis I see vertical line with voltage exactly the same as Vgs. UserIdTAG: 352142 UserNameTAG: DanteX CreateTimeTAG: 2012-10-04T16:08:47Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 2379 TitleTAG: Value for vi Hello all, In step 2 and 3 of this example we use a value $v_i$ as the small signal equivalent model. What´s its value? thnaks UserIdTAG: 35642 UserNameTAG: caled CreateTimeTAG: 2012-10-04T15:08:18Z VoteTAG: 1 CoursewareTAG: Week 4 / Small Signal Solution Example CommentableIdTAG: 6002x_small_signal_soln_ex NumberOfReplyTAG: 1 FirstChildTAG: VI=1V. See step 1 of the SS circuit solution (S7V15) FirstChildUserIdTAG: 166869 FirstChildUserNameTAG: Vasco FirstChildCreateTimeTAG: 2012-10-04T22:23:16Z SecondChildTAG: I don't understand one thing. Isn't the Input voltage source(VI) a DC source? And in the lecture, Mr. Agarwal said a DC Voltage source has to be replaced by a short circuit in its small signal model. Then why did he use vi as the small signal equivalent of VI? SecondChildUserIdTAG: 163264 SecondChildUserNameTAG: Aamir_edx SecondChildCreateTimeTAG: 2012-10-06T09:33:57Z SecondChildTAG: vi is the equivalent of the small signal component of the initial voltage source and in small signal model it stays as the AC voltage source. VI is the equivalent of the DC component of the initial voltage source and is replaced with short circuit in the SS model. SecondChildUserIdTAG: 380703 SecondChildUserNameTAG: vargy SecondChildCreateTimeTAG: 2012-10-06T10:18:14Z IndexTAG: 2380 TitleTAG: Midterm material As the midterm is approaching, I've been thinking about which material it is going to cover ? The videos, lectures and etc. from the first seven weeks ? Because on 15th of October there will be released the videos and the hw from week 8, so is it possible to have questions from the material from week 8, or only from the first seven weeks ? I'm asking just to know how to go through the material that week. Thanks. UserIdTAG: 206648 UserNameTAG: TsvetanGeorgiev CreateTimeTAG: 2012-10-04T13:01:14Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: The answer I recall given elsewhere is that the Midterm will be like last semester, and cover Weeks 1-6. FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-10-05T15:13:05Z SecondChildTAG: Thank you. It seems reasonable. SecondChildUserIdTAG: 206648 SecondChildUserNameTAG: TsvetanGeorgiev SecondChildCreateTimeTAG: 2012-10-05T15:34:11Z IndexTAG: 2381 TitleTAG: lab 4 I m getting the check sign but my transient analysis graph is not same as shown above in the question. I m getting 0(zero) Ids. why this is happening? please help......... UserIdTAG: 346003 UserNameTAG: anupamshakya CreateTimeTAG: 2012-10-04T11:31:53Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Did you put the probe to show x-axis? Remember that you need two sources to put the MOSFET in the work region: One is Vds (like the vtest in the example above) and the other is Vgs (dc voltage source). I hope that I help you Kelno, Recife-Brazil FirstChildUserIdTAG: 32164 FirstChildUserNameTAG: Kelno FirstChildCreateTimeTAG: 2012-10-04T13:43:40Z SecondChildTAG: could u please tell me how we have to use these probes?i mean what do they signify? SecondChildUserIdTAG: 137918 SecondChildUserNameTAG: jaisneha SecondChildCreateTimeTAG: 2012-10-04T15:22:53Z IndexTAG: 2382 TitleTAG: S7E3 Hi, can anybody help me how can i do this exercise? i tried for hours and i don't understand how i get those results Thanks UserIdTAG: 314294 UserNameTAG: victormp CreateTimeTAG: 2012-10-04T11:00:05Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 2383 TitleTAG: how to get these answers?? what formula can be used to solve these questions...? can anybody tell? i thought peak power is the maximum power..but my answer was wrong..also i couldn/t find the average power. UserIdTAG: 552187 UserNameTAG: ankita4237 CreateTimeTAG: 2012-10-04T10:04:40Z VoteTAG: 1 CoursewareTAG: Week 1 / AC power CommentableIdTAG: 6002x_S1E3_AC_power NumberOfReplyTAG: 1 FirstChildTAG: Peak power would be the peak voltage * the peak current. or Vpeak^2 / R. Remember the peak voltage is 1.414 * RMS voltage for a sinusoidal signal. The average power is the integral of the power over one cycle, or if you don't care to do integration it is Vrms^2/R. FirstChildUserIdTAG: 270011 FirstChildUserNameTAG: SnowmanZA FirstChildCreateTimeTAG: 2012-10-04T16:50:55Z SecondChildTAG: thanx..i get it.. SecondChildUserIdTAG: 552187 SecondChildUserNameTAG: ankita4237 SecondChildCreateTimeTAG: 2012-10-06T10:39:41Z SecondChildTAG: what is RMS? SecondChildUserIdTAG: 702959 SecondChildUserNameTAG: Sashkow SecondChildCreateTimeTAG: 2012-10-25T20:39:12Z IndexTAG: 2384 TitleTAG: How to watch the sequence naturely? Even if I use the Go-agnet which was provided by Google, I still can't watch the lecture sequence. I am a chinese. Is there anybody breaking it successfully? Though for now , I could watch some of the sequences through the download way,but as you know, it's not complete for us. Anybody would like to give me some kind suggestion to watch the sequence more naturely? UserIdTAG: 128276 UserNameTAG: ChengBin CreateTimeTAG: 2012-10-04T07:18:03Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hi ChengBin! How are you? I didn´t understand your question... Did you mean that how can you watch the video lectures in chinese? If it is that, you will notice in the video lectures (at the down right of the video window), that you will have a YouTube icon, if yo click on it, it will re-link you to YouTube where it is the original video. Ok, once you are there you can click on "cc" button and choose translate to chinese. Consecuently, you will watch the video lecture with the subtitle in chinese, so you can watch it more naturely... I hope this can help you :) Good luck! See you! Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-04T11:32:34Z FirstChildTAG: What a pity. My English is so poor that I can't express my self clearly. Judging from your name, you are not a Chinese. So, you completely misunderstand my question at all. In fact, I did not mean to the subtitle of the vedio. It's a advanced issue. For me, I even could not open the vedio, because of the firewall of the China's government. Can you understand that? I can't link to those webs, which were blocked by the firewall, such as 'Google.com', 'youtube.com', 'facebook.com', 'twintter.com', and so on and so forth. So,if I want to visit youtube, I have to use the proxy, which you might not ever use in your country. My question is that, even if I used the proxy,say, 'go-Agent', I still can't open the video.![enter image description here][1] Here attached a snapshot for discribing my issue. [1]: http:// FirstChildUserIdTAG: 128276 FirstChildUserNameTAG: ChengBin FirstChildCreateTimeTAG: 2012-10-05T05:22:28Z SecondChildTAG: Sorry,I don't know how to insert a snapshot. SecondChildUserIdTAG: 128276 SecondChildUserNameTAG: ChengBin SecondChildCreateTimeTAG: 2012-10-05T05:24:24Z SecondChildTAG: Any way! Thank you so much. Myrimit SecondChildUserIdTAG: 128276 SecondChildUserNameTAG: ChengBin SecondChildCreateTimeTAG: 2012-10-05T05:27:10Z SecondChildTAG: Hi ChengBin! You can download the video lectures here, in the [wiki - click here][1] :). If you want to insert a snapshot, you will se an icon 1) ![enter image description here][2] click on it, 2) select your image from your pc ![enter image description here][3] 3) You will see what it is in the red box, but you will see in the post your image :). ![enter image description here][4] See you! Myriam. P.D: Don´t worry haha, me too, I speak spanish :). I hope that this can be helpful. Can you post the snapshot? ;) [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/course_wiki [2]: https://edxuploads.s3.amazonaws.com/13494388678309194.png [3]: https://edxuploads.s3.amazonaws.com/13494389201343636.png [4]: https://edxuploads.s3.amazonaws.com/13494391079571274.png SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-05T12:13:41Z IndexTAG: 2385 TitleTAG: Problems with week 5 tutorials Hi everyone, First of all I'd like to congratulate Dr. argawal, Jerry and Peter as well as MIT for an excellent job which gives such a great opportunity to so many people around the world.Now I'd like to know if I'm the only one experiencing problems when trying to access the tutorials for the week 5. Thanks a lot and keep up with the great great job. Lenin Vargas, Caracas-Venezuela UserIdTAG: 369339 UserNameTAG: vargaslen CreateTimeTAG: 2012-10-03T22:37:49Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 2386 TitleTAG: Why we have to work with logs to facilitate convergence? Can anyone explain why trying values on equation va=-5*ln(va/20+15/20) converges to the ans and why not when trying values for the equation va=-15+20*e^(-va/5) UserIdTAG: 390678 UserNameTAG: pvlastaridis CreateTimeTAG: 2012-10-03T19:14:53Z VoteTAG: 1 CoursewareTAG: Week 4 / Linearization Exercise 1 CommentableIdTAG: 6002x_linearization_ex_1 NumberOfReplyTAG: 1 FirstChildTAG: Use the graphing calculator (fantastic tool) below and you will see why. Without logarithms your guesses could diverge pretty quickly, or converge much slower than with logs. http://my.hrw.com/math06_07/nsmedia/tools/Graph_Calculator/graphCalc.html FirstChildUserIdTAG: 132828 FirstChildUserNameTAG: rharris FirstChildCreateTimeTAG: 2012-10-03T20:38:18Z IndexTAG: 2387 TitleTAG: question 1 wrong answer? I don't get how question 1 is 0.787V. I get 0.837V Vo=Vs-v, For v, I took KCL at the note v/R-id=0, v/R-k/v^2=0 v=(R*K)^(1/3) Vo=Vs-(R*K)^(1/3)=0.837 What is wrong? UserIdTAG: 381386 UserNameTAG: MrMatt7k CreateTimeTAG: 2012-10-03T18:14:22Z VoteTAG: 1 CoursewareTAG: Week 4 / Dependent Sources Exercise 1 CommentableIdTAG: 6002x_dep_src_exsz_1 NumberOfReplyTAG: 2 FirstChildTAG: I'm not sure what you did, but your equation ( 5-(850*0.088)^(1/3) ) is correct. FirstChildUserIdTAG: 434869 FirstChildUserNameTAG: patey FirstChildCreateTimeTAG: 2012-10-03T18:37:55Z SecondChildTAG: the equation is v0=Vs-(R*K)/v^2 then v0=Vs-(R*K)/(R*K)^(2/3) SecondChildUserIdTAG: 236136 SecondChildUserNameTAG: Hamid-ch SecondChildCreateTimeTAG: 2012-10-05T04:19:16Z SecondChildTAG: Your equation is correct! Maybe you made a typo with numbers Vo = Vs - (R*K)^(1/3) = 5V - 4.213 = 0.7866 SecondChildUserIdTAG: 410033 SecondChildUserNameTAG: sagitta SecondChildCreateTimeTAG: 2012-10-05T12:37:16Z SecondChildTAG: oops SecondChildUserIdTAG: 410033 SecondChildUserNameTAG: sagitta SecondChildCreateTimeTAG: 2012-10-05T12:37:45Z SecondChildTAG: I'm a little confused. Can someone explain how did v^2 became equal to (R*K)^(2/3)? SecondChildUserIdTAG: 352373 SecondChildUserNameTAG: keyholder SecondChildCreateTimeTAG: 2012-10-07T11:25:52Z SecondChildTAG: I got it. The tutorial helps a lot. SecondChildUserIdTAG: 352373 SecondChildUserNameTAG: keyholder SecondChildCreateTimeTAG: 2012-10-07T11:34:45Z FirstChildTAG: Your equation is correct! Maybe you made a typo with numbers Vo = Vs - (R*K)^(1/3) = 5V - 4.213 = 0.7866 FirstChildUserIdTAG: 410033 FirstChildUserNameTAG: sagitta FirstChildCreateTimeTAG: 2012-10-05T12:37:57Z SecondChildTAG: How did you make only the sqr for (R*K)? Thanks SecondChildUserIdTAG: 166869 SecondChildUserNameTAG: Vasco SecondChildCreateTimeTAG: 2012-10-07T22:21:17Z IndexTAG: 2388 TitleTAG: What are the technical details of the setup? Great video. Can you please explain a couple questions about the setup: 1) How the nonlinear diode (id vs vd?) graph is displayed on the oscilloscope screen at the same time as the input and output signals (V vs time?). Don't all the signals have to be voltage, with respect to time, to display on the scope? 2) The CD player output is being sent from an auxiliary cable coming out of the headphone jack, and then...? Does it go into an input in a power supply, which adds the bias and then sends the combined signal to the scope? Thanks! UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-03T17:06:33Z VoteTAG: 1 CoursewareTAG: Week 4 / Incremental Analysis Demo CommentableIdTAG: 6002x_inc_analysis_demo NumberOfReplyTAG: 0 IndexTAG: 2389 TitleTAG: How tough will the mathematics get I wanted to thank MIT for giving me the opportunity to learn so much. I have acquired a profusion of knowledge thanks to this course. However as the weeks have gone by, I have noticed the rising difficulty in the level of mathematics required to solve the homework problems. For the week 6 homework we were asked to solve a multivariable equation. I would be much obliged if someone could let me know how much more rigorous the mathematics behind the electronics gets. Thank you in advance. UserIdTAG: 85490 UserNameTAG: yashsavani CreateTimeTAG: 2012-10-03T16:16:28Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: yup, I got the same question....someone help please! but still I suppose the maths would go as hard as calculus and not more than that.... but there maybe other complex stuff like multi-variable equations, but I don't think thats difficult but the part where you have to find the equations via the conditions, well that stings! though when it comes to solving, it is easy, though it may take a very very very very very very long time as it i once seriously tried 12 unknowns and 12 equations,.... believe that ain't no sunshine! FirstChildUserIdTAG: 169784 FirstChildUserNameTAG: thewiredbear FirstChildCreateTimeTAG: 2012-10-03T18:26:33Z FirstChildTAG: The math takes a step up when you start solving time dependent problems. Those first appear in week 6. You need to be able to work with differential equations. But you only use one standard trick. You really just need to know what derivatives and integrals are and how to differential an exponential function. If you are willing to be outside your comfort zone and take a few things on faith until you get more background, you can do the work. The truth is that you can take a purely operational approach, when ever you see a derivative of a function just replace it with s times the function, second derivative multiply by s^2. You just need to "turn the crank". FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-03T22:32:43Z SecondChildTAG: Thanks for the heads up. I am well acquainted with all calculus, single and multi variable, and have managed to complete and understand all the mathematics up till week 6, so I doubt that it will just be "turning the crank." I am glad to hear that the mathematics will be restricted to differential equations, though I am wondering whether we will be proceeding on to solving second order differential equations. SecondChildUserIdTAG: 85490 SecondChildUserNameTAG: yashsavani SecondChildCreateTimeTAG: 2012-10-04T12:10:22Z SecondChildTAG: Only linear second order differential equations with constant coefficients ... There is only one method used: assume a solution of the form A*exp(st) substitute into the equation and find what value of s that satisfies the equation. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-04T21:57:10Z IndexTAG: 2390 TitleTAG: [Math Processing Error] None of my math fonts are displaying so the homework problems are unreadable. UserIdTAG: 49861 UserNameTAG: codymartin CreateTimeTAG: 2012-10-03T16:10:08Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: It displays this message: [Math Processing Error] FirstChildUserIdTAG: 49861 FirstChildUserNameTAG: codymartin FirstChildCreateTimeTAG: 2012-10-03T18:20:07Z SecondChildTAG: I just started having the same problem an hour or so ago. Not sure what is going on. I printed my completed HW to study off of for the Mid-Term and I think it started happening after that. Not sure though. SecondChildUserIdTAG: 417864 SecondChildUserNameTAG: beauclark SecondChildCreateTimeTAG: 2012-10-03T18:48:13Z SecondChildTAG: Sorry to hear you are having the problem too but glad its not just me. Hopefully they fix it soon. Its putting me behind. SecondChildUserIdTAG: 49861 SecondChildUserNameTAG: codymartin SecondChildCreateTimeTAG: 2012-10-03T20:09:03Z SecondChildTAG: It happened to me previously, I don't know for sure what happened either. SecondChildUserIdTAG: 418182 SecondChildUserNameTAG: euldji2005 SecondChildCreateTimeTAG: 2012-10-03T22:22:56Z FirstChildTAG: For me this problem was my IE9 browser. I switched to Chrome and the fonts have been fine. FirstChildUserIdTAG: 397595 FirstChildUserNameTAG: mholin FirstChildCreateTimeTAG: 2012-10-03T21:54:49Z SecondChildTAG: Its failing for me in both IE9 and Chrome :( SecondChildUserIdTAG: 49861 SecondChildUserNameTAG: codymartin SecondChildCreateTimeTAG: 2012-10-04T00:07:55Z SecondChildTAG: I have been using only chrome and it is happening. I just tried IE9 just in case and it also has the same problem. I tried deleting all site data and restarting the browser. I guess I'll try restarting the entire machine. SecondChildUserIdTAG: 417864 SecondChildUserNameTAG: beauclark SecondChildCreateTimeTAG: 2012-10-04T17:21:38Z FirstChildTAG: So far a complete computer restart has seemed to fix the problem. That is usually the go-to answer but I honestly didn't think a font problem would need an entire restart. Anyway so far so good. I hope this helpful to others. FirstChildUserIdTAG: 417864 FirstChildUserNameTAG: beauclark FirstChildCreateTimeTAG: 2012-10-04T17:33:36Z SecondChildTAG: I've restarted since but will try again with fingers crossed. Thanks for the tip! SecondChildUserIdTAG: 49861 SecondChildUserNameTAG: codymartin SecondChildCreateTimeTAG: 2012-10-04T20:40:33Z SecondChildTAG: Its fixed now. It might have been the reboot after all. Thanks again. SecondChildUserIdTAG: 49861 SecondChildUserNameTAG: codymartin SecondChildCreateTimeTAG: 2012-10-05T13:58:08Z IndexTAG: 2391 TitleTAG: H4 P1 Please give some hints about chapters in textbook or videos to solve this problems by myself. It`s to hard and some abstract for me this nonlinear elements. UserIdTAG: 96873 UserNameTAG: Santyaga CreateTimeTAG: 2012-10-03T15:46:48Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Hi!! You could watch tutorials videos... First of all called "Naming Conventions" talk about non linear elements!! Link here!! [Tutorials W4][1] B.R. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_4/Week_4_Tutorials/ FirstChildUserIdTAG: 149058 FirstChildUserNameTAG: sotoroman FirstChildCreateTimeTAG: 2012-10-03T16:22:44Z FirstChildTAG: Diferentiate! FirstChildUserIdTAG: 299251 FirstChildUserNameTAG: TheRedBlackOne FirstChildCreateTimeTAG: 2012-10-03T16:50:54Z FirstChildTAG: Thank you got it! FirstChildUserIdTAG: 96873 FirstChildUserNameTAG: Santyaga FirstChildCreateTimeTAG: 2012-10-03T18:23:02Z IndexTAG: 2392 TitleTAG: Awesome concept This certainly was an "aha!" moment for me in all its literal meaning. One of the best demo's I've ever seen, or probably, will see in my life. Hats off to Prof. Anant Agarwal. UserIdTAG: 137686 UserNameTAG: Jaychandran CreateTimeTAG: 2012-10-03T15:18:52Z VoteTAG: 1 CoursewareTAG: Week 4 / Incremental Analysis Demo CommentableIdTAG: 6002x_inc_analysis_demo NumberOfReplyTAG: 0 IndexTAG: 2393 TitleTAG: Where to find this Sandbox simulator ? okay, this might sound stupid. But i am not able to locate this thing. plz help ! thanks UserIdTAG: 97340 UserNameTAG: AnkitRana CreateTimeTAG: 2012-10-03T13:58:03Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: HEY ANKIT ARE YOU FROM INDIA? FirstChildUserIdTAG: 368712 FirstChildUserNameTAG: jmen FirstChildCreateTimeTAG: 2012-10-03T14:22:43Z SecondChildTAG: Yes. Now that u've asked, i reckon you too ? :D Nice to meet you. Okay no more chatting ! ----> Honor code. SecondChildUserIdTAG: 97340 SecondChildUserNameTAG: AnkitRana SecondChildCreateTimeTAG: 2012-10-03T14:56:02Z SecondChildTAG: Honor code doesn't prevent students from enjoying and discussing..RYT?? are u from IIT ROPAR? :p SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-03T16:30:39Z SecondChildTAG: haha so funny AnkitRana! ;) Okay no more chatting ! ----> Honor code. haha. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T06:16:44Z FirstChildTAG: at the Overview section.. Click on progress, Overview and you'll see it. FirstChildUserIdTAG: 206648 FirstChildUserNameTAG: TsvetanGeorgiev FirstChildCreateTimeTAG: 2012-10-03T14:14:02Z SecondChildTAG: THANKS SecondChildUserIdTAG: 486919 SecondChildUserNameTAG: AEK48 SecondChildCreateTimeTAG: 2012-10-03T14:26:44Z SecondChildTAG: Thanks a lot :D SecondChildUserIdTAG: 97340 SecondChildUserNameTAG: AnkitRana SecondChildCreateTimeTAG: 2012-10-03T14:57:05Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T06:16:55Z IndexTAG: 2394 TitleTAG: H5P1 part 3 Hi guys, any leads on this part. I have obtained a quadratic equation for VIN and solved, but not the right answer UserIdTAG: 143823 UserNameTAG: NathanNadeson CreateTimeTAG: 2012-10-03T13:55:48Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Let VA=(-1+-sqrt(1+2kRLVs))/kRL Vo lies b/w VS and VA Vi lies b/w Vt and Vt+VA FirstChildUserIdTAG: 368712 FirstChildUserNameTAG: jmen FirstChildCreateTimeTAG: 2012-10-03T14:22:11Z SecondChildTAG: well frankly speaking i found this in the videos of aggarwal sir..I am using this concept but I am also stuck with this part..:( SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-03T14:35:34Z SecondChildTAG: yes, we can see this here: [enter link description here][1], what is the jedi mind trick? [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/MOSFETAmplifiersSmallsignalmodels/ SecondChildUserIdTAG: 299251 SecondChildUserNameTAG: TheRedBlackOne SecondChildCreateTimeTAG: 2012-10-03T14:58:21Z SecondChildTAG: to me it says Can't parse ! :/ Don't know what is being asked for ! SecondChildUserIdTAG: 97340 SecondChildUserNameTAG: AnkitRana SecondChildCreateTimeTAG: 2012-10-03T15:03:53Z SecondChildTAG: well TheRedBlackOne is correct... but has anyone tried it? I mean i am substituting values but my answer is not getting a tick here...Has anyone got one? SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-03T15:52:38Z SecondChildTAG: i'm also stuck here :( none of the calculations i tried work SecondChildUserIdTAG: 263693 SecondChildUserNameTAG: Coldberg SecondChildCreateTimeTAG: 2012-10-03T23:14:52Z SecondChildTAG: Infact the calculations i'v done seem to indicate the i never get into the saturation region with the voltages provided SecondChildUserIdTAG: 263693 SecondChildUserNameTAG: Coldberg SecondChildCreateTimeTAG: 2012-10-03T23:15:45Z SecondChildTAG: Bah my bad i missed the fact that K was in mA not A SecondChildUserIdTAG: 263693 SecondChildUserNameTAG: Coldberg SecondChildCreateTimeTAG: 2012-10-03T23:38:32Z SecondChildTAG: Speaking about saturation region you should understand that $V_o$ in the region's constraints is nothing but $V_{DS}$. In the task S is not grounded so come back to H5P1:3 and finish it B] SecondChildUserIdTAG: 190618 SecondChildUserNameTAG: Kirbabaev SecondChildCreateTimeTAG: 2012-10-06T22:48:53Z SecondChildTAG: i want to shout this from the roof tops, VDS is not Vo! SecondChildUserIdTAG: 285945 SecondChildUserNameTAG: lindalapiso SecondChildCreateTimeTAG: 2012-10-12T09:04:27Z SecondChildTAG: thank you Kirbabaev SecondChildUserIdTAG: 285945 SecondChildUserNameTAG: lindalapiso SecondChildCreateTimeTAG: 2012-10-12T09:05:21Z FirstChildTAG: This took me a long time to get correct, i finally had to write a few lines of code to get the correct answer. How did you guys solve for it? FirstChildUserIdTAG: 199421 FirstChildUserNameTAG: nandishratkal FirstChildCreateTimeTAG: 2012-10-07T12:01:44Z SecondChildTAG: I can't solve it ... SecondChildUserIdTAG: 297960 SecondChildUserNameTAG: sebseb95 SecondChildCreateTimeTAG: 2012-10-07T17:41:39Z SecondChildTAG: Mr. nandishratkal will you please help us with the equation SecondChildUserIdTAG: 92895 SecondChildUserNameTAG: shihab2555 SecondChildCreateTimeTAG: 2012-10-10T05:39:51Z IndexTAG: 2395 TitleTAG: How to find "Rn" with a external exitation Vtest or Itest when there are dependent sources? Can Anyone post any link that explain this method and examples? Thanks UserIdTAG: 299251 UserNameTAG: TheRedBlackOne CreateTimeTAG: 2012-10-03T13:44:32Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: The dependent sources can not be removed. You let them where they are aply an external excitation and compute V and I. Use ohm law for the rest. FirstChildUserIdTAG: 64512 FirstChildUserNameTAG: OZSorescu FirstChildCreateTimeTAG: 2012-10-03T16:19:37Z SecondChildTAG: I don`t understand this method!!!!!! an example please!!!! SecondChildUserIdTAG: 299251 SecondChildUserNameTAG: TheRedBlackOne SecondChildCreateTimeTAG: 2012-10-03T16:44:31Z FirstChildTAG: The same as Rth. The ratio Vn/In across that port is Rn. FirstChildUserIdTAG: 39980 FirstChildUserNameTAG: MarinoJr FirstChildCreateTimeTAG: 2012-10-03T15:32:13Z SecondChildTAG: But in this case I`m not able to find Vn. SecondChildUserIdTAG: 299251 SecondChildUserNameTAG: TheRedBlackOne SecondChildCreateTimeTAG: 2012-10-03T16:43:39Z FirstChildTAG: I wrote up an example a while ago about this, you can find it here: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5060f4a8a3f6d21f00000050 FirstChildUserIdTAG: 248902 FirstChildUserNameTAG: Chanute FirstChildCreateTimeTAG: 2012-10-04T01:48:40Z SecondChildTAG: It might be worth also noting that $R_{TH}=R_N$ always. SecondChildUserIdTAG: 248902 SecondChildUserNameTAG: Chanute SecondChildCreateTimeTAG: 2012-10-04T05:18:55Z IndexTAG: 2396 TitleTAG: Why can't we submit homework assignments late? What is the logic behind this? During most of my time in "real" university, I've been able to submit homework late, for a lowered score obviously or at least get my answers corrected/explained. But this is an online course anyway. Why is the Check Answer function disabled? It's really detrimental to the learning process. Just give us 0 credit but let us attempt the problems without having to look at the answers. What about people that joined late? Why not take advantage of what online courses allow us to do and let people do the course at their own pace? Don't give us the certificate, fine but do we really care about that? I just want to learn! UserIdTAG: 357225 UserNameTAG: MJBoa CreateTimeTAG: 2012-10-03T11:30:07Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: Hi MJBoa! That it seems a nice suggestion :) I would be nice to have two buttoms after the deadline of the assignment: "check without showing answer" and "show answer" FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-03T12:10:52Z FirstChildTAG: "Why not take advantage of what online courses allow us to do and let people do the course at their own pace? Don't give us the certificate, fine but do we really care about that? I just want to learn!" http://ocw.mit.edu/index.htm Yes we really care about the certificate. A lot actually. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-03T13:38:43Z SecondChildTAG: You're being willfully ignorant, you know very well that the Edx platform is much more powerful than OCW, which is just videos and PDFs. The certificate is irrelevant to my point anyway, just don't give it to us if we're late. Where's the problem? What was your post supposed to have contributed? SecondChildUserIdTAG: 357225 SecondChildUserNameTAG: MJBoa SecondChildCreateTimeTAG: 2012-10-04T14:37:47Z FirstChildTAG: I agree, I just started the course and I can not earn the points corresponding to four weeks... FirstChildUserIdTAG: 551838 FirstChildUserNameTAG: Fonciello FirstChildCreateTimeTAG: 2012-10-03T22:24:27Z FirstChildTAG: im having problem or dont know how this work....how do i submit my lab and home i did them but nothing or no button to submit. if i change the page all my work dissappears. please help me, thanks in advance FirstChildUserIdTAG: 462861 FirstChildUserNameTAG: edxforme FirstChildCreateTimeTAG: 2012-10-04T19:49:08Z IndexTAG: 2397 TitleTAG: Grounded If a chip contains millions of elements how they could be grounded individually in such small space ? UserIdTAG: 99628 UserNameTAG: Pranjal16 CreateTimeTAG: 2012-10-03T09:04:48Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: with multiple layers FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-03T10:31:32Z IndexTAG: 2398 TitleTAG: late joining I joined this course very late.so can anyone plz suggest me how to cover the earlier homeworks and tests?when will be the exams?becoz i am a engg student and having my final university exams from 10dec. UserIdTAG: 548400 UserNameTAG: Manitripathi CreateTimeTAG: 2012-10-03T06:36:31Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: same problem is mine,plz help... FirstChildUserIdTAG: 549249 FirstChildUserNameTAG: VIVEKSINGHBUNDELA FirstChildCreateTimeTAG: 2012-10-03T07:22:24Z SecondChildTAG: I HAVE SAME PROBLEM.... SecondChildUserIdTAG: 545379 SecondChildUserNameTAG: ASHSAN SecondChildCreateTimeTAG: 2012-10-03T07:43:06Z SecondChildTAG: **Same Problem bro** SecondChildUserIdTAG: 549768 SecondChildUserNameTAG: KamalSoni SecondChildCreateTimeTAG: 2012-10-03T08:15:24Z FirstChildTAG: Midterm exam is in three weeks, and final exam should be around December 20. More information will be posted as the date approaches. FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-10-03T13:59:30Z IndexTAG: 2399 TitleTAG: Beautiful Science That was just amazing. I graduated with a BA in Music and instead of an Aha moment I felt awe. + I love Sublime's music. hahah Thanks prof. You Rock! UserIdTAG: 230787 UserNameTAG: Gabriel007 CreateTimeTAG: 2012-10-03T04:29:12Z VoteTAG: 1 CoursewareTAG: Week 4 / Incremental Analysis Demo CommentableIdTAG: 6002x_inc_analysis_demo NumberOfReplyTAG: 0 IndexTAG: 2400 TitleTAG: Help? Why does it say 500e-6? How did you get that answer? By the way, what is a good way to learn the electricity laws and rules referenced in this class? I'm new to all of this. UserIdTAG: 542903 UserNameTAG: bwdmighall CreateTimeTAG: 2012-10-02T23:34:25Z VoteTAG: 1 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: Hi bwdmighall! Can I help you? Are you referring about Lab0 - Part 3 - meassure current? Ok, if you make the circuit that they give you in the statement: ![imasand][1] You will notice a button DC. If you click on that buttom, the simulator will show you the DC values in red. So, you can see that the current it is 500uA, so you have to write in the answer 500u or 500e-6 without units because they ask you the result value in Amperes, that is to say, without units :). Another way could be, the following: They give you this Law: V=I*R You know that the total R will be the sum of the 3 resistors because thay are in serial combination, so R=1kOhm+3kOhm+2kOhm=6kOhm. Ok, you know that V=3 Volt. So, your I= (V/R)=3 Volt/6kOhm = 500 *10^-6 Ampere = 500e-6 Ampere = 500u Ampere. Remember that u is micro, and micro it is 10^-6 that is the same as writting e-6 ;). That is why it is 500e-6. See you! Myriam. [1]: https://edxuploads.s3.amazonaws.com/13492237011673236.png FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-03T00:28:55Z SecondChildTAG: Very useful response to me too!Thanks! SecondChildUserIdTAG: 620830 SecondChildUserNameTAG: Lazaros1 SecondChildCreateTimeTAG: 2012-10-12T19:58:08Z SecondChildTAG: You are welcome Lazaros1! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T04:08:57Z IndexTAG: 2401 TitleTAG: KCL Mistake? In the first KCL the current through R2 is taken as going from the drain, vo, to the gate, VG. In the second it's taken as going the opposite way. Shouldn't these be consistent? By my reckoning the second KCL should be: (VG - vo) * G2 - (VG - vi) * G1 + iG = 0 UserIdTAG: 366083 UserNameTAG: smath CreateTimeTAG: 2012-10-02T21:40:10Z VoteTAG: 1 CoursewareTAG: Week 5 / Transistor Biasing with Feedback CommentableIdTAG: 6002x_transistor_biasing_t NumberOfReplyTAG: 1 FirstChildTAG: When one makes node equations there's a convention that the currents go either into the node or out of the node. Thus it is assumed that node potential is either lower than or higher than the potential of other corresponding nodes. The sign can only be changed in a node equation if the potential of the other corresponding nodes is known for sure. This is why both the lady with the pleasant voice in the tutorial and you are right! No polarity is mentioned for the source $V_I$, it could be that the $+$ is grounded, but remember that the source of the MOSFET is grounded. This is why in order for $i_{DS}$ to appear the gate's potential should be positive. In order for the gate's potential to be positive the 'minus' of the source $V_I$ should be grounded. FirstChildUserIdTAG: 252722 FirstChildUserNameTAG: vaboro FirstChildCreateTimeTAG: 2012-10-06T00:51:15Z IndexTAG: 2402 TitleTAG: Preview : LED Lighting Power Sources? Eventually I want to learn to build a circuit to drive higher power LEDs (1-10W) for lights in my apartment. I've heard that these circuits are basically current sources. Will I be able to build a basic version of this after this course? I know computer programming, so I'm hoping I can program a microcontroller to turn on different LEDs for different intensities, colors, times, etc. Can anyone give me of sneak preview of what principles and elements we will learn about that would go into a circuit like this? thanks, Rob UserIdTAG: 468623 UserNameTAG: RobNik CreateTimeTAG: 2012-10-02T21:17:33Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Yes We have touched on the transistor topic, these can be used in place of conventional relays in an LED current source. Since you already have TTL covered, the rest is fairly straight forward, involving a transistor(s) and a few resistors. I don't know for sure what is in store, but I am confident you will gain the tools needed for such projects. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-02T22:56:34Z FirstChildTAG: look at MOSFET in saturation region FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-03T01:23:42Z FirstChildTAG: I think the most interesting projects you could do, would be using an Arduino board. FirstChildUserIdTAG: 146377 FirstChildUserNameTAG: Lucas_Megassini FirstChildCreateTimeTAG: 2012-10-03T03:22:53Z IndexTAG: 2403 TitleTAG: Hp4P3 Help! Hi, I am stuck with HP4P3 Part B. The Norton resistance supposed to be (R1+R2)||R3 assuming short circuit to replace two voltage sources? where am I going wrong? Would appreciate a pointer to this problem. I have computed "u" to be 5.714V and open voltage to be 7.857v using node method. However, I am not sure of the accuracy. UserIdTAG: 246991 UserNameTAG: spatra CreateTimeTAG: 2012-10-02T21:14:03Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Perhaps my given numbers were different than yours but i found a different value for "u". Maybe you should check again the equation from which you derived that number. FirstChildUserIdTAG: 304587 FirstChildUserNameTAG: Vasso FirstChildCreateTimeTAG: 2012-10-02T21:41:48Z FirstChildTAG: Hi, The equations I used are as follows using node method: (u-V0)/R1 + (u-e)/R2 = 0 (e-u)/R2 + (e-Au)/R3 = 0 where u is node voltage between R1 and R2 and e is the node voltage between R2 and R3. I have grounded the bottom branch connecting V0 and Au. Is there any error in my equation? FirstChildUserIdTAG: 246991 FirstChildUserNameTAG: spatra FirstChildCreateTimeTAG: 2012-10-02T22:25:36Z SecondChildTAG: You must find IN so you must put that in your equation in order to find it. SecondChildUserIdTAG: 304587 SecondChildUserNameTAG: Vasso SecondChildCreateTimeTAG: 2012-10-03T10:34:02Z FirstChildTAG: I did not use IN in the second equation as I was computing open voltage e (Vth). The goal as to find Rth then IN = Vth/Rth. However, for Rth = (R1+R2)||R3 gives wrong answer. Any wrong in my logic? FirstChildUserIdTAG: 246991 FirstChildUserNameTAG: spatra FirstChildCreateTimeTAG: 2012-10-03T16:58:14Z IndexTAG: 2404 TitleTAG: Why Vth is not equal to Vs? Can someone please explain me why Vth= RP*VS/(RS+RP)? I tuoght it was equal to Vs Thanks... Amen... UserIdTAG: 385948 UserNameTAG: sedamen03 CreateTimeTAG: 2012-10-02T18:06:05Z VoteTAG: 1 CoursewareTAG: Week 3 / Isolation Via Thevenin CommentableIdTAG: 6002x_isolation_via_thevenin NumberOfReplyTAG: 1 FirstChildTAG: Hello, sedamen03! You can swap Non-Linear Element with Rp. Now it's simple to cut Non-Linear Element from the other network, which is a voltage divider, with $$ V_{TH} = V(R_P) = V_S * \frac{RP}{(RS+RP)} $$ Why can we swap the elements? Because element location doesn't matter, as far as connections remain the same... And they does =) ---------- You can also solve the KCL equation, and result will be the same, just don't forget about Rp when removing the NLE. FirstChildUserIdTAG: 261503 FirstChildUserNameTAG: EugenyL FirstChildCreateTimeTAG: 2012-10-02T18:51:14Z SecondChildTAG: Hi Eugenyl... Thanks for your reply. That was very helpful... SecondChildUserIdTAG: 385948 SecondChildUserNameTAG: sedamen03 SecondChildCreateTimeTAG: 2012-10-02T21:05:44Z IndexTAG: 2405 TitleTAG: Circuit Sandbox for download Hi guys! I was wondering if we can download the Circuit Sandbox to use it off-line. It is such a great tool that I'd like to keep it with me all the time! =D UserIdTAG: 288637 UserNameTAG: gotchi CreateTimeTAG: 2012-10-02T15:26:41Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I suspect that engine is on the server side. Try this: http://www.falstad.com/circuit/ FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-02T16:24:55Z SecondChildTAG: Thank you YakovO for the suggestion, but I'm willing to download edX's Circuit Sandbox because I'm learning about circuits using it. I think this way is a lot easier than to learn how to use another software. Besides this, Circuit Sandbox's way to nice to keep! =) SecondChildUserIdTAG: 288637 SecondChildUserNameTAG: gotchi SecondChildCreateTimeTAG: 2012-10-02T17:22:34Z SecondChildTAG: I would suggest learning SPICE if you want to get serious with circuits. SecondChildUserIdTAG: 323387 SecondChildUserNameTAG: ibrahimawwal SecondChildCreateTimeTAG: 2012-10-03T11:23:17Z SecondChildTAG: Thank you Ibrahim. I'll do this! =) SecondChildUserIdTAG: 288637 SecondChildUserNameTAG: gotchi SecondChildCreateTimeTAG: 2012-10-03T22:17:11Z IndexTAG: 2406 TitleTAG: getting trouble in playing video lectures video lectures are not playing,kindly tell me what to do or send me link for those video lectures! UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-02T14:58:00Z VoteTAG: 1 CoursewareTAG: Week 1 / KVL and KCL CommentableIdTAG: 6002x_kvl_and_kcl NumberOfReplyTAG: 1 FirstChildTAG: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/ Have fun! FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-02T23:03:20Z IndexTAG: 2407 TitleTAG: lab 4 help needed!! ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13491807059857381.png UserIdTAG: 156585 UserNameTAG: AdiRanga CreateTimeTAG: 2012-10-02T12:25:27Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: what is the mistake in my circuit? FirstChildUserIdTAG: 156585 FirstChildUserNameTAG: AdiRanga FirstChildCreateTimeTAG: 2012-10-02T12:26:06Z SecondChildTAG: you made a short circuit between the source and drain of the MOSFET SecondChildUserIdTAG: 252722 SecondChildUserNameTAG: vaboro SecondChildCreateTimeTAG: 2012-10-02T13:06:52Z SecondChildTAG: check the current coming into the drain, in other words, put the current meter above the drain of the MOSFET and connect the wires SecondChildUserIdTAG: 252722 SecondChildUserNameTAG: vaboro SecondChildCreateTimeTAG: 2012-10-02T13:08:43Z IndexTAG: 2408 TitleTAG: WEEK 4 :Norton Current I am so stuck with the Norton problem that not even the text and other forum discussions have helped me with it too the solution. I did the thevenin one in minutes and this has eaten two days at stretch. Kindly elaborate how to proceed with IN. I'd do the Rn part myself then. Thanks regards UserIdTAG: 97340 UserNameTAG: AnkitRana CreateTimeTAG: 2012-10-02T10:14:42Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: i did it like that: i found Rth first(which is the same with Rn), then i found Vth and then In=Vth/Rth FirstChildUserIdTAG: 301962 FirstChildUserNameTAG: labrinim FirstChildCreateTimeTAG: 2012-10-02T11:40:48Z SecondChildTAG: dont forget to calculate u first... SecondChildUserIdTAG: 301962 SecondChildUserNameTAG: labrinim SecondChildCreateTimeTAG: 2012-10-02T11:44:05Z SecondChildTAG: I have like 3 different values for Vth calculated. The problem being none of them works :p Guide me on how to find the correct one plz./ thanks SecondChildUserIdTAG: 97340 SecondChildUserNameTAG: AnkitRana SecondChildCreateTimeTAG: 2012-10-02T11:58:52Z SecondChildTAG: Vth=V(of node which has R3 and Au). so you need to find u the current of the particular node. SecondChildUserIdTAG: 301962 SecondChildUserNameTAG: labrinim SecondChildCreateTimeTAG: 2012-10-02T12:28:45Z SecondChildTAG: remember to ground the cirquit (i did it to the "-" of u) SecondChildUserIdTAG: 301962 SecondChildUserNameTAG: labrinim SecondChildCreateTimeTAG: 2012-10-02T12:30:22Z SecondChildTAG: Just use simple kvl method to find Vth.Than Vth/Rn=In SecondChildUserIdTAG: 92895 SecondChildUserNameTAG: shihab2555 SecondChildCreateTimeTAG: 2012-10-04T06:20:24Z IndexTAG: 2409 TitleTAG: plzz help regarding lab 4 transient analysis show no current through mos ,,,i hav taken vds at x axis n current on y & another marker at vgs stil no current ..wht to do plzz help UserIdTAG: 183166 UserNameTAG: yogeshk CreateTimeTAG: 2012-10-02T09:52:11Z VoteTAG: 1 CoursewareTAG: Week 4 / Amplifiers CommentableIdTAG: 6002x_amplifiers NumberOfReplyTAG: 2 FirstChildTAG: i had the same problem... Try to figure out the appropriate voltage source (Vds)!!! FirstChildUserIdTAG: 301962 FirstChildUserNameTAG: labrinim FirstChildCreateTimeTAG: 2012-10-02T11:54:50Z FirstChildTAG: Hi same problem here Yogesh :( Btw have you been able to find the Norton current in the last problem of week 4. plz help if so. FirstChildUserIdTAG: 97340 FirstChildUserNameTAG: AnkitRana FirstChildCreateTimeTAG: 2012-10-02T10:15:39Z SecondChildTAG: Same here!!!and for the norton equivalent,do exactly the same way as u did for the thevenin equivalent..in the end calculate norton current by doing Vth/Rth..ie norton current is same as Isc. SecondChildUserIdTAG: 156585 SecondChildUserNameTAG: AdiRanga SecondChildCreateTimeTAG: 2012-10-02T11:51:15Z IndexTAG: 2410 TitleTAG: Basics of ECE What are ACTIVE and PASIVE elements.? UserIdTAG: 182470 UserNameTAG: nitesh2703 CreateTimeTAG: 2012-10-02T09:50:41Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: A component may be classified as passive, active, or electromechanic. The strict physics definition treats **passive components as ones that cannot supply energy themselves**, **whereas a battery would be seen as an active component since it truly acts as a source of energy**. However, electronic engineers who perform circuit analysis use a more restrictive definition of **passivity**. When only concerned with the energy of signals, it is convenient to ignore the so-called DC circuit and pretend that the power supplying components such as transistors or integrated circuits is absent (as if each such component had its own battery built in), though it may in reality be supplied by the DC circuit. Then, the analysis only concerns the AC circuit, an abstraction that ignores DC voltages and currents (and the power associated with them) present in the real-life circuit. This fiction, for instance, lets us view an oscillator as "producing energy" even though in reality the oscillator consumes even more energy from a DC power supply, which we have chosen to ignore. Under that restriction, we define the terms as used in circuit analysis as: **Active components rely on a source of energy** (usually from the DC circuit, which we have chosen to ignore) and usually can inject power into a circuit, though this is not part of the definition.[1] **Active components include amplifying components such as transistors, triode vacuum tubes (valves), and tunnel diodes.** **Passive components can't introduce net energy into the circuit**. They also can't rely on a source of power, except for what is available from the (AC) circuit they are connected to. As a consequence they can't amplify (increase the power of a signal), although they may increase a voltage or current (such as is done by a transformer or resonant circuit). **Passive components include two-terminal components such as resistors, capacitors, inductors, and transformers.** FirstChildUserIdTAG: 304587 FirstChildUserNameTAG: Vasso FirstChildCreateTimeTAG: 2012-10-02T11:24:59Z IndexTAG: 2411 TitleTAG: HW question attempts How many attempts do you get to answer most homework problems? I am not seeing it in the syllabus. -Thx, MJ UserIdTAG: 61923 UserNameTAG: MikeJones CreateTimeTAG: 2012-10-02T09:16:12Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: There is no limit for the Homework questions, but i believe there will be one for the final test. FirstChildUserIdTAG: 304587 FirstChildUserNameTAG: Vasso FirstChildCreateTimeTAG: 2012-10-02T11:10:19Z SecondChildTAG: What Vasso says about the homework/lab quetios it is correct= unlimited check times. But not for the Exam = Limited check times . Also you can take a look [here][1] :) [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/505e2fd7e903b1230000001f SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-02T12:03:33Z IndexTAG: 2412 TitleTAG: I just registered today. Please how can i download previous lectures? the links to this purpose please. thanks! please reply ASAP UserIdTAG: 539343 UserNameTAG: Osinimu CreateTimeTAG: 2012-10-02T08:55:41Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Install IDM and it should be integrated to your browser. good luck :) FirstChildUserIdTAG: 97340 FirstChildUserNameTAG: AnkitRana FirstChildCreateTimeTAG: 2012-10-02T10:16:54Z SecondChildTAG: Hi Osinimu! Take a look in the [wiki][1], you can download the lectures :) [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/ SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-02T12:05:36Z IndexTAG: 2413 TitleTAG: what happens if my entire circuit contains non linear elements?? what happens if my entire circuit contains non linear elements?? UserIdTAG: 143597 UserNameTAG: aravindANV CreateTimeTAG: 2012-10-02T08:18:39Z VoteTAG: 1 CoursewareTAG: Week 3 / Analyzing Nonlinear Circuits CommentableIdTAG: 6002x_analyzing_nonlinear_circuits NumberOfReplyTAG: 0 IndexTAG: 2414 TitleTAG: Lab tool: Problem in changing colour of the voltage probe. In lab tool, when i place voltage probe to measure voltage value in transient analysis, I am not able to change colour of probe. I tried to double click it to open properties menu, but its not working. UserIdTAG: 533988 UserNameTAG: thakkerankit CreateTimeTAG: 2012-10-02T07:59:42Z VoteTAG: 1 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 2 FirstChildTAG: Also I am not able to name nodes. Please help! FirstChildUserIdTAG: 533988 FirstChildUserNameTAG: thakkerankit FirstChildCreateTimeTAG: 2012-10-02T08:00:51Z FirstChildTAG: Hi! Double click on the probe then you change it. Did you try to change the browser? I'am on Chrome and it works well! Sebastian FirstChildUserIdTAG: 252045 FirstChildUserNameTAG: JSeb FirstChildCreateTimeTAG: 2012-10-02T11:51:18Z IndexTAG: 2415 TitleTAG: WHO IS SENDING THE INPUT ? if it the sender who is sending the input then , why did the proffesor used the word VIH and VIL for INPUT to the gate ? remember that chart that he had drawn with VOH AND VOL on the sender side(input side) and VIH AND VIL on the receiver side . UserIdTAG: 242267 UserNameTAG: MOHITVASHISHTA CreateTimeTAG: 2012-10-02T06:22:59Z VoteTAG: 1 CoursewareTAG: Week 2 / Digital logic circuits CommentableIdTAG: 6002x_digital_logic NumberOfReplyTAG: 1 FirstChildTAG: I was confused by this a little too... If you are *SENDING* a signal *‘out’*: it is called **Output**. If you are *RECEIVING* a signal: that is the **Input**. There are stronger constraints on output than on input. FirstChildUserIdTAG: 714237 FirstChildUserNameTAG: PaxPolaris FirstChildCreateTimeTAG: 2012-10-26T18:03:55Z IndexTAG: 2416 TitleTAG: CONNECTING ALL SLIDES ... M I RIGHT GUYS !!! look this is what i can make out of this lecture ... 1. there is a sender , and a digital system whose input terminal is acting as receiver 2. noise is added to the signal while it is on its way to the input terminal through the wire or any other medium. 3. now the input terminal receives this corrupted signal but because of contract signed earlier it can understand the signal 4 now since the input was valid that is following the contract so output has to b valid . 5 now because of some system in the port the signal will b interpreted as better than VOH for a high and better than VOL for a low (although they were received as above VIH below VIL.. PLEASE TELL ME IF I M TOTALLY WRONG IN INTERPRETING THIS ... :) UserIdTAG: 242267 UserNameTAG: MOHITVASHISHTA CreateTimeTAG: 2012-10-02T05:43:24Z VoteTAG: 1 CoursewareTAG: Week 2 / Static Discipline CommentableIdTAG: 6002x_static_discipline NumberOfReplyTAG: 1 FirstChildTAG: I think you are correct, although I couldn't your fifth point... what do you mean with "better than VOH for a high and better than VOL for a low"? FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-10-02T14:09:41Z IndexTAG: 2417 TitleTAG: I got va = 1.43841036226, not 1.08825683594 Can anybody help I got va = 1.43841036226, not 1.08825683594 Can anybody help ? UserIdTAG: 229018 UserNameTAG: Changming CreateTimeTAG: 2012-10-02T02:50:46Z VoteTAG: 1 CoursewareTAG: Week 4 / Linearization Exercise 1 CommentableIdTAG: 6002x_linearization_ex_1 NumberOfReplyTAG: 1 FirstChildTAG: hi.. When you got VA Value, you can use this value to calculate a new VA Value. So, if you got va=1.4384. Now you can use this value in the expression. Many time after the VA value is converged to Va=1.088... Hope you understand!! FirstChildUserIdTAG: 164572 FirstChildUserNameTAG: Ranyeri_rocha FirstChildCreateTimeTAG: 2012-10-02T03:24:18Z SecondChildTAG: Just as a warning, you are giving out answers. The Edx staff doesn't like that. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-02T04:24:44Z SecondChildTAG: As far as I understand, the code doesn't allow giving precise answers for the coursework but, suggests that we collaborate for these examples... SecondChildUserIdTAG: 435193 SecondChildUserNameTAG: ManosP SecondChildCreateTimeTAG: 2012-10-06T20:06:33Z IndexTAG: 2418 TitleTAG: H2P1 Confused as to why we assume Vin to be 30 V and Vout to be 7.5 V, when the problem says Vin is 50 V and Vout is 17.5 V..thanks in advance UserIdTAG: 85261 UserNameTAG: ehanes7612 CreateTimeTAG: 2012-10-01T19:40:57Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: That's because the answers are unified to all students, but problems parameters differ from one to another... just to make it fair :) FirstChildUserIdTAG: 365201 FirstChildUserNameTAG: sirajmuhammad FirstChildCreateTimeTAG: 2012-10-01T20:03:48Z SecondChildTAG: thought it was something like that..thanks SecondChildUserIdTAG: 85261 SecondChildUserNameTAG: ehanes7612 SecondChildCreateTimeTAG: 2012-10-01T20:15:51Z SecondChildTAG: although i am not sure what you mean by unified..because i get a ratio of 1.8571: 1 instead of 3:1 when using 50 (Vin) and 17.5 (Vout) SecondChildUserIdTAG: 85261 SecondChildUserNameTAG: ehanes7612 SecondChildCreateTimeTAG: 2012-10-01T20:33:27Z SecondChildTAG: No, he means that the basic problem is the same for all students [=unified], but the values given for your personal problem, can differ from the values given to other students, to prevent that some students just copy an answer from someone else. However, the method given by the staff to find the solution, remains the same for all students, but the values that the staff uses, can differ from yours. So if you see the answer given by staff, you can check it for the given values in your problem. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-10-01T23:42:45Z SecondChildTAG: thanks SecondChildUserIdTAG: 85261 SecondChildUserNameTAG: ehanes7612 SecondChildCreateTimeTAG: 2012-10-02T03:30:21Z IndexTAG: 2419 TitleTAG: S7V6: DEMO - MUSIC OVER A LIGHT BEAM, DISTORTION AND NO DISTORTION I have two queries- 1. what is the physical or practical meaning of adding **DC OFFSET** in second part of demo i.e. MUSIC demostration ?? 2. How does music is added(or use as voltage source) to the circuit? UserIdTAG: 114864 UserNameTAG: Vikaash CreateTimeTAG: 2012-10-01T17:29:45Z VoteTAG: 1 CoursewareTAG: Week 4 / Incremental Analysis Demo CommentableIdTAG: 6002x_inc_analysis_demo NumberOfReplyTAG: 2 FirstChildTAG: Basically if you had a 5v DC source, in series with a AC source that varied from -1v to +1v, you would have a small signal that centers around 5v So instead of your small signal varying from -1v to +1v, it varies from +4v to +6v when you add 5v of "DC offset" FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-01T18:54:50Z SecondChildTAG: @pennypacker can u please elaborate it a little more. couldn't get Why the small signal would center around 5v? Thanks. SecondChildUserIdTAG: 204213 SecondChildUserNameTAG: ratneshray SecondChildCreateTimeTAG: 2012-10-02T19:42:42Z FirstChildTAG: Could any one please tell me the name the song that was played during the demo :D FirstChildUserIdTAG: 336001 FirstChildUserNameTAG: syd_buet12 FirstChildCreateTimeTAG: 2012-10-02T11:51:42Z SecondChildTAG: I got it its Santeria by Sublime :D SecondChildUserIdTAG: 336001 SecondChildUserNameTAG: syd_buet12 SecondChildCreateTimeTAG: 2012-10-02T11:55:48Z IndexTAG: 2420 TitleTAG: lab 3 unfair judgement in lab 3 my answer is marked wrong. when i consulted with my friend after the lab was over i was surprised to see that his answer was same as mine but was marked correct. i am not able to understand why it happend? so i would like to request the instructer to see in this matter and do a proper judgement as each lab is very imp in terms of marks. UserIdTAG: 477527 UserNameTAG: AASHISH_AMBER CreateTimeTAG: 2012-10-01T13:53:00Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Troubleshooting NumberOfReplyTAG: 1 FirstChildTAG: Did your friend have the same values for $V_{in}$ and $V_{out}$ as you? The numbers in the problem are not the same for everyone. FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-10-01T13:58:28Z SecondChildTAG: Yes it is ame fot both of us. SecondChildUserIdTAG: 477527 SecondChildUserNameTAG: AASHISH_AMBER SecondChildCreateTimeTAG: 2012-10-01T14:08:35Z SecondChildTAG: sorry its "same" not "ame"(typo error). SecondChildUserIdTAG: 477527 SecondChildUserNameTAG: AASHISH_AMBER SecondChildCreateTimeTAG: 2012-10-01T14:09:27Z SecondChildTAG: Hmmmm. Let's wait for a staff reply then. SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-10-01T14:15:17Z SecondChildTAG: Hi, ashwith I also have a problem with Lab3. Yesterday I tried to solve but I had not finished and I have not clicked the Check Button. Today, I successfully solved it, but the answer was marked as wrong, like AASHISH_AMBER's case. I believe the problem could be the use of comma instead of dot, because today I'm using a computer with a different keyboard layout. Probably that was my fault, but the judgement wasn't so just with me. Can you help me? SecondChildUserIdTAG: 175611 SecondChildUserNameTAG: Frango SecondChildCreateTimeTAG: 2012-10-01T16:28:39Z IndexTAG: 2421 TitleTAG: S8E2 Checker seems to be accepting incorrect results During the discussion on [this thread][1], I realized that the solution given for S8E2 could be incorrect. I'm getting a value of $400.9114713 \Omega$ for $R_{TH}$. The solution shows $399.314214464\Omega$. I do get the green tick for my answer but isn't the difference here too much to be due to a round off issue? Even $750||850$ works - which is definitely not the correct result. Is there something I have done wrong here (I get the same expression for $R_{TH}$ even without using superposition). My post contains all details of my solution. **Edit:** It looks like there is a mistake in my solution. However the checker is still accepting it as the right answer (along with the real right answer and 750|850). I now remember we saw this issue in the Spring run as well. That time, I entered the parallel combination of resistors and got it marked right. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/undefined/threads/506851eec64e0a2300000003 UserIdTAG: 14386 UserNameTAG: ashwith CreateTimeTAG: 2012-10-01T13:17:21Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 2 FirstChildTAG: If you look at the solution I gave to Ashwith in the thread, your answer is actually not correct. You have the term (IO+1) in the formula in the end and I don't have the 1 in it, so it should be (IO+0)=IO only. See my explanation to Ashwith in the thread, I also explain what you have to superimpose, because you or yakovo said it's not necessary, but it is. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-10-01T18:59:11Z SecondChildTAG: salsero, you are right. There is a mistake in my solution and I understand what it is. However this still means that the checker is accepting incorrect answers. My answer got the green tick. So did 750||850. SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-10-01T19:38:17Z SecondChildTAG: salsero, your solution is not only one, is not obvious and over complicated. In simplest solution you don't need superposition and differentiating at all. But I'm very interesting how to apply differentiating in this case. Could you please post it. Thank you. SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-01T21:49:45Z SecondChildTAG: I agree that my solution is not the only one leading to a correct answer. But you said you didn't see where you had to use superpostion. I show that by introducing an external CS you can convert it to a VS. I personally find it for myself more obvious to add a little deltaVS to it, and deltaVS is then imposed on Vs. So there you have the superposition. And about differentiating: In the lectures they use the taylor expansion to express a function as an example, but it can be any function. The point is, that you want to find the effects of the dependent source in the circuit, so for that reason, you introduce [in this problem] an external independent source and by introducing a small signal on that external source, you can find the effect of this small signal on the behavior of the circuit for small signal changes on the dependent source. If you only introduce an external source without a small signal, then you'll only find [if I'm correct] the bias point, or operating point, nothing more; by superimposing a small signal, you'll find the small signal behavior at that bias point, and then is differentiating the formula for calculating the biaspoint, for me the way to do it. In this problem, the vi-relation of the dependent source was linear, so this was relatively easy to solve. But if it's not linear, then it will become much harder. At the moment I can't scan anything, otherwise I would scan my way of solving it for you. I hope this will do for the moment. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-10-01T23:05:24Z SecondChildTAG: Got it! Did it your way :) Very unusual (using R2 as a part of tevenin) and clever solution :) It's complicated for linear dependent source, but will be very useful for nonlinear - great practice. Very appreciate you brought this solution out, thanks a lot :D SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-02T06:30:46Z FirstChildTAG: Hi Ashwith, The point is, that the checker allows some error margin, for HW, Labs but also for the midterm and final. Most of the time that is ok, but I discovered in the previous course, that it can also be a serious disadvantage, because serious mistakes, like yours in this case [and mine in others] stay unnoticed if you don't check for the reason of deviation. I call this mistake serious [in the way that the green check is misleading] , because if you hadn't tried to find out why your answer was deviating, then you would probably think that your way of solving was correct, and, when used to solve this kind of problem in the midterm or final with different values, then the possibility exists that your answer is just a little bit too much out of range for to get the green check and you'll never find out why .. Since I discovered this, I always check for the precise answer. To me it happened, in the previous course, that in one of the subquestion to calculate the Q-factor, I did the calculation totally wrong [didn't know it that time], so my answer had to be wrong, so I gave my answer and to my surprise all the following subquestions were marked red except the one I'm talking about. Took me hours to find out ... What had happened? When typing the answer in the box, I forgot one digit, right behind the dot, so by accident the anwer seemed correct ... Only when I checked over and over again why it deviated [I didn't notice the forgotten digit], I finally discovered the forgotten digit and the I found my wrong formula and since that formula was also needed to answer the rest of the question, everything stayed wrong ... And after the final I checked the anwers given by the staff. There was a problem where you could neglect [so stated in the problem] that you could neglect the Rp in that problem. So I did, and I got the green check. But after the final I noticed also a small deviation and I tried to find out why. So I discovered that in the real solution [the formula was not given in the answer from the final] that the Rp was not neglected. And I can give many of this kind of examples. So my tip: always check why your answer is deviating ... FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-10-01T22:17:42Z SecondChildTAG: In my case, my mistake is I don't use the show answer if I get the green tick. I should from now on. SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-10-02T15:35:04Z IndexTAG: 2422 TitleTAG: Observation about S7E2: choice of region; chart and reality This is more of an observation really which I think should help intuition. If you keep in mind our main goal - getting linear response from our non-linear device - the choice of (I,R) values is actually horrible, there're regions further along the curve that are more linear. So what you want is to move your load line by increasing changing I and R. Oh, and notice that the slope of the load line is unrealistic making the above picture a very crude approximation of what the chart should really look like. Obviously absent information about scale of both axis one cannot make any claims about the correctness of the chart, but I thought I'd put it out there in case there're other people out there who feel disturbed by it :) That said this particular exercise is more about making consistent mathematical statements and less about intuition, so all is well. After all how realistic 4A current really is. UserIdTAG: 5325 UserNameTAG: vkz CreateTimeTAG: 2012-10-01T13:00:15Z VoteTAG: 1 CoursewareTAG: Week 4 / Graphs Exercise CommentableIdTAG: 6002x_graphs_exercise NumberOfReplyTAG: 1 FirstChildTAG: Thanks for your observations, it's always interesting to hear others random thoughts in regards to these matters. I think I see what you are saying. I wonder though, if the graph is just working with what is available. ie: You may not have the voltage or current to operate the device in a different range, or the device may not be able to work properly with high or lower values. I also wonder what the slope does in relation to the output waveform. If you notice in examples that when the slope is around 45 degrees, the peak to peak "appears" to be the same size, less then 45 degrees it is smaller and would be larger with slopes over 45 degrees. Would it be ideal to have the slope as close to 45 degrees as possible, providing the linearity is acceptable? FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-01T14:57:55Z SecondChildTAG: Yes, 45 would be ideal. But! You do not get your linear output, you always get points on the curve - output of your non-linear device. Process of linearization is just approximation. You can't lineary approximate high curvatures. Recall the higher order elements that we said we could ignore in Tailor expansion? Well, you can in the segments of your curve that are close to straight lines, but they begin to be more important the "curlier" the V-I for your device is. Btw, I think it is ok to scarifies your peak-to-peak. You can always amplify the signal. SecondChildUserIdTAG: 5325 SecondChildUserNameTAG: vkz SecondChildCreateTimeTAG: 2012-10-01T15:36:28Z IndexTAG: 2423 TitleTAG: vi << VI? Its problematic, because, for neglect vi^2, it must be assumed that vi << (VI - VT) and not only VI - VT. But, in circuit which use (almost) all the dynamic range of the saturation range of a MOSFET, this is NOT the case! UserIdTAG: 388524 UserNameTAG: dasbinich CreateTimeTAG: 2012-10-01T12:10:16Z VoteTAG: 1 CoursewareTAG: Week 5 / Small Signal Mathematically Described CommentableIdTAG: 6002x_small_sig_math_des NumberOfReplyTAG: 2 FirstChildTAG: Can you elaborate? Thanks. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-01T15:03:27Z FirstChildTAG: I will give an numeric example: Lets VT = 1V and the bias input voltage VI = 1,01V. So: VI > VT --> we are in saturation region. Therefore, I can use the small signal model for the iDS = K/2 (vIN-VT)^2 formula with a small signal vi =, say, 0,01V which is << VI, ok?!?! But: 2(VI-VT)vi = 0,0002V and vi^2 = 0,0001V So: vi^2 is NOT NEGLECTABLE with respect to 2(VI-VT)vi!!! Ok, this case is very artificial and constructed; but matematically, vi << VI its not ok. It has to be: vi << (VI-VT). FirstChildUserIdTAG: 388524 FirstChildUserNameTAG: dasbinich FirstChildCreateTimeTAG: 2012-10-04T21:07:12Z IndexTAG: 2424 TitleTAG: Lab 4 I find it tough to use the node labels in lab 4...How to label them?As in ,if i double click,i am not getting the properties box..Also,am not getting connecting wires here.. UserIdTAG: 156585 UserNameTAG: AdiRanga CreateTimeTAG: 2012-10-01T11:27:09Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: clean your mouse and assign values and labels before you move the part into place. the "connections" between nodes with identical labels are invisible. FirstChildUserIdTAG: 331542 FirstChildUserNameTAG: james77901 FirstChildCreateTimeTAG: 2012-10-01T13:02:47Z IndexTAG: 2425 TitleTAG: Little Hint Remember that y2 is not i2 when v2 acting alone, y2 is i1 when v2 acting alone. Hope this helps. UserIdTAG: 393901 UserNameTAG: CamiloABarreto CreateTimeTAG: 2012-10-01T06:11:44Z VoteTAG: 1 CoursewareTAG: Week 2 / Superposition example CommentableIdTAG: 6002x_S3E3_superposition NumberOfReplyTAG: 1 FirstChildTAG: thanks FirstChildUserIdTAG: 367283 FirstChildUserNameTAG: Konstantin_Konov FirstChildCreateTimeTAG: 2012-10-13T14:42:05Z IndexTAG: 2426 TitleTAG: Problem on Lab3 ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13490666025654743.jpg Hey guys, I've got an issue on Lab3, apparently my answer is correct (the plot on the left), but the checker doesn't check it as a valid answer. I've calculated W/L and it seemed to be okay.. I would really appreciate if you could help me :) UserIdTAG: 197793 UserNameTAG: HazelLarack CreateTimeTAG: 2012-10-01T04:46:00Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I was able to tweak on my W/L (L/W?) value and the graph reflected no obvious changes. Try several values around your calculated value. I ended up using a value different from what I calculated and it was accepted by the grader. I think we're seeing some borderline glitch amplitudes. FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-10-01T05:00:21Z SecondChildTAG: Thanks MobiusTruth, I'll try it :) SecondChildUserIdTAG: 197793 SecondChildUserNameTAG: HazelLarack SecondChildCreateTimeTAG: 2012-10-01T05:04:52Z SecondChildTAG: It worked hahaha. I really appreciated your help, thanks a lot! SecondChildUserIdTAG: 197793 SecondChildUserNameTAG: HazelLarack SecondChildCreateTimeTAG: 2012-10-01T05:06:31Z SecondChildTAG: It would be nice to be able to select and expand a section of the display like we can with a decent scope. Set in a start time and a stop time and view just that time interval for the entire width of the display. Or maybe that would be easier said than done. SecondChildUserIdTAG: 92346 SecondChildUserNameTAG: MobiusTruth SecondChildCreateTimeTAG: 2012-10-01T05:16:33Z IndexTAG: 2427 TitleTAG: W/L is not 31.8?? I have found this value and still can not check lab 3???why??????help me!!!!!!! UserIdTAG: 308861 UserNameTAG: Ignaas CreateTimeTAG: 2012-10-01T02:27:42Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Does your waveform correspond? Do you know that W/L is just a lower limit, as explained in lecture? Trying higher values will only raise noise immunity, and should not affect the output waveform. FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-10-01T02:31:05Z SecondChildTAG: I calculated W/L (or was it L/W?) a few times and then tweaked it to watch how the waveform changed. I eventually settled on something other than the exact calculated value. Still got the Happy Emerald Checkmark, though. SecondChildUserIdTAG: 92346 SecondChildUserNameTAG: MobiusTruth SecondChildCreateTimeTAG: 2012-10-01T02:33:18Z SecondChildTAG: thnx guys...the above value is incorrect...the waveform corresponded...but it is antoher value..i got checked now...:))) SecondChildUserIdTAG: 308861 SecondChildUserNameTAG: Ignaas SecondChildCreateTimeTAG: 2012-10-01T02:35:16Z IndexTAG: 2428 TitleTAG: Lab3 - if all else fails, reset and draw again I could not figure out why I was not getting a check even though I was sure I had the right values and my waveform looked right. I reset, redrew my circuit with the same W/L values, got the exact same waveform, but this time I got my check. I thought I would throw that out there if anyone else is having the same problem. I can't figure out what the problem was with my original drawing. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-01T01:54:24Z VoteTAG: 1 CoursewareTAG: Week 3 / Logic Gates CommentableIdTAG: 6002x_logic_gates NumberOfReplyTAG: 2 FirstChildTAG: My big mistake in Week 1 lab was that I accidentally closed the waveform popup window, and then pressed 'check'. Of course I got a big red X, but I was stupid and it was the first week. I learned to make sure the transient analysis waveform window is open at the same time as I press 'check'! FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-10-01T02:32:43Z FirstChildTAG: I have to give up. i got the right wave form but the system did not 'check'. so had to start over again. did it twice! editing with a web based schematic editor is really awful. Instructors - Isnt there an alternative? Maybe the LTSPICE or some such free tool? The course is excellent but the schematic editor got me really frustrated. FirstChildUserIdTAG: 279379 FirstChildUserNameTAG: RajaSrinivasan FirstChildCreateTimeTAG: 2012-10-01T02:52:24Z SecondChildTAG: You can always 'tweak' your W/L value, as the calculated value is only a lower limit. You can raise the W/L value and after a certain point, it will have no effect on the waveform (unlike when the value is low). SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-10-01T03:38:11Z IndexTAG: 2429 TitleTAG: Question about Thevenin Equivalence. On the wiki: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/thevenin-norton/ The Rth obtained for the circuit between A and B was $4\Omega$. I don't understand how it was done. Can someone please explain how they got $4\Omega$? UserIdTAG: 213386 UserNameTAG: dmascenik CreateTimeTAG: 2012-10-01T01:49:08Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/1349059272459111.png Hi dmascenik, Let's name the resistors as in the picture above. The formula for calculating $R_{th}$ is: $R_{th}=((R_9+R_8)//((R_7//R_6)+R_5)+R_4)//R_3+R_2+R_1$ FirstChildUserIdTAG: 470934 FirstChildUserNameTAG: sonhx FirstChildCreateTimeTAG: 2012-10-01T02:54:11Z SecondChildTAG: ( ...............//R3) + r2 + r1 ( to be precise ! :o) 4ohms exact. SecondChildUserIdTAG: 373752 SecondChildUserNameTAG: Croak SecondChildCreateTimeTAG: 2012-10-02T00:30:02Z FirstChildTAG: R`a` = R`6`//R7 = (6*6)/(6+6) = 3... Add R`a` to R`5` and we get 3+3 = 6 = R`b`... Since R`8` and R`9` are in series, so we'll just add them up and get 3+3 = 6 = R`c`...; R`d` = R`b`//R`c` = (6*6)/(6+6) = 3... ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13490787386776292.png So the above (R`d`) results in 3 ohms. As it can be seen that R`d` is in series with R`4`, so we'll add them up to get R`e`= 7 Now this R`e` is parallel to R`3`, this gives us (7*3)/(7+3) = 2.1..., Simply add the above result to R`1` and R`2` (since they all are in series) and we'll get 2.1+1+1 = 4.1 I believe our answer isn't exactly 4, its 4.1; And lets hope I'm not wrong :) FirstChildUserIdTAG: 31529 FirstChildUserNameTAG: abbass FirstChildCreateTimeTAG: 2012-10-01T08:14:05Z SecondChildTAG: R4 is 4.. so, u r wrong. Rth is exactly 4 here. SecondChildUserIdTAG: 823 SecondChildUserNameTAG: naveenatmit SecondChildCreateTimeTAG: 2012-10-01T08:51:58Z SecondChildTAG: don't mess with R1 and R2 until you have reduced to one resistor between CD. Naveen is correct. SecondChildUserIdTAG: 331542 SecondChildUserNameTAG: james77901 SecondChildCreateTimeTAG: 2012-10-01T13:40:55Z SecondChildTAG: Rd is in series with R4, ..... YES so we'll add them up to get Re= 7 Nope : Re=6 ! SecondChildUserIdTAG: 373752 SecondChildUserNameTAG: Croak SecondChildCreateTimeTAG: 2012-10-02T01:07:14Z FirstChildTAG: R6 // R7 = 3 3+R5 = 6 R8+R9 = 6 6 // 6 = 3 3+R4 = 6 6 // R3 = 6//3= 18/9 = 2 2+R1+R2 = 4 ohms FirstChildUserIdTAG: 373752 FirstChildUserNameTAG: Croak FirstChildCreateTimeTAG: 2012-10-02T00:23:48Z SecondChildTAG: ( hint : if r1=r2, r1//r2= r1/2 ) SecondChildUserIdTAG: 373752 SecondChildUserNameTAG: Croak SecondChildCreateTimeTAG: 2012-10-02T00:33:20Z IndexTAG: 2430 TitleTAG: H3P4 Confused with Diodes I am kinda stuck on how to do H3P4. Diodes stop current from flowing in the opposite direction, correct? Also doesn't the center voltage sources with the diodes not supply current? Since D1 stops current flowing out of the positive terminal of V1 and D2 stops V2 from being grounded? Is that correct otherwise what is wrong and how would I do this problem? -Thanks for the help! :D UserIdTAG: 219780 UserNameTAG: ReconIII CreateTimeTAG: 2012-10-01T01:48:28Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I couldn´t do it either FirstChildUserIdTAG: 371220 FirstChildUserNameTAG: dan10 FirstChildCreateTimeTAG: 2012-10-01T02:04:19Z SecondChildTAG: Google "**clamper circuits**" and look up "**Clamper (electronics)**" in Wikipedia. This is a "general pattern" type circuit that appears again and again so it is essential to grasp the theory of it's operation. SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-10-01T02:05:25Z FirstChildTAG: I got the first half of it. The way to think about the first half is that current will flow through the diode when the voltage across it is positive. Since the diode acts like a short circuit, then the voltage across the terminals will be zero, which means that the voltage across the second resistor will also be zero. So you essentially have to figure out what the limits are when there won't be any current flowing through either diode. Second half, I'm not so sure. I thought I understood, but my answer appears to be wrong. FirstChildUserIdTAG: 95034 FirstChildUserNameTAG: mariogarcia FirstChildCreateTimeTAG: 2012-10-01T04:31:24Z IndexTAG: 2431 TitleTAG: late started Hi, im sorry, it took me a while to start, im a eng and i feel like i have dust in circuits and electronics, but i wanted to know, can i do it again in the future ?? how many time can i take this course?, also if i can take the lab and the homework that i havent done and its already past due time only for fun? i dont really care about grades right now. its only for fun and taking all that dust away from me, which by the way im not so happy with it. thank you, and this is a really cool course, UserIdTAG: 344660 UserNameTAG: issismilly CreateTimeTAG: 2012-10-01T00:12:07Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: 2 lowest scores for homework and lab will be dropped, so you can still continue FirstChildUserIdTAG: 340567 FirstChildUserNameTAG: sputnik1 FirstChildCreateTimeTAG: 2012-10-01T01:18:38Z FirstChildTAG: I believe you can take this course as many times as you wish. FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-10-01T02:36:59Z IndexTAG: 2432 TitleTAG: S6E1 My take https://www.dropbox.com/s/sdljc4wwscquhqj/S6E1.jpg UserIdTAG: 230787 UserNameTAG: Gabriel007 CreateTimeTAG: 2012-09-30T23:12:37Z VoteTAG: 1 CoursewareTAG: Week 3 / Analyzing Nonlinear Circuits Exercise CommentableIdTAG: 6002x_analyzing_nonlinear_circuits_exercise NumberOfReplyTAG: 0 IndexTAG: 2433 TitleTAG: LAB 3 HElP! Ok I get through all the hw fine but the labs are so confusing for me. I understand how to do it on paper but no the sand box thing. NOW the equation Z = complement(C(A+B)). I a NOR gate in series with a NAND gate the same as an OR in series with an AND complemented or do we some how have to implement an invert er somewhere? UserIdTAG: 435197 UserNameTAG: RichmondRichter CreateTimeTAG: 2012-09-30T21:53:27Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: What helped for me is instead of concentrating on making it a Nand gate or a Nor gate ect. what I did is just ignore those at first and just concentrate on building something that follows the truth table. Also this discussion should help: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50673e4c2193ea2300000009 FirstChildUserIdTAG: 219780 FirstChildUserNameTAG: ReconIII FirstChildCreateTimeTAG: 2012-09-30T22:54:00Z SecondChildTAG: **Richmond:** Follow **ReconIII**'s advice. Design the circuit using 3 MOSFETS as stated in the hint given. Do not attempt to design it using gates, otherwise you will get off track! **Another few ideas:** * When looking at your truth table, notice the patterns, and what you can extract. Especially look at **C**. * There should be one MOSFET connected to ground in your circuit. Think about which input will go to the gate of this MOSFET: **A, B, or C**, and why. * After you design for this constraint, you should figure out where to place the other two MOSFETS (e.g. in series or parallel to this). * Your output waveform will be incorrect unless the $W/L$ parameter of the MOSFET is correctly adjusted by double-clicking the MOSFET. Remember in the lecture that we were given $R_L >> Ron $ to meet the noise immunity threshold. Since 1. $Ron = 26500\cdot L/W$ and 2. $R_L=10000Ω$ are **given**, you can easily solve for $W/L$ with simple algebra if you choose $R_L/Ron>>1$ (i.e. = 50 or 100) * This should be an easy lab (unlike last week's). SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-10-01T01:37:35Z SecondChildTAG: Why won't my LaTeX work like in the preview? Oh...you cannot use the "much greater than" sign otherwise it screws with the HTML. Sorry! The first "box" should read: "$R_L$ $\gg$ $Ron$", and the second "box" should read "$ \frac{R_L}{Ron}$ $\gg$ 1". Sorry about the bad formatting! SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-10-01T01:50:06Z SecondChildTAG: And your $R_L$ and $Ron$ values may be different than mine (Homeworks and Labs are sometimes like that) but the equations still hold. SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-10-01T01:57:36Z SecondChildTAG: I don't see RL mentioned anywhere in the lab? SecondChildUserIdTAG: 270160 SecondChildUserNameTAG: Sethhhh SecondChildCreateTimeTAG: 2012-10-01T03:05:56Z SecondChildTAG: Sorry, I meant $Rpullup$ (insted of $R_L$). It's late here (sorry) and I forgot to substitute the exact resistor's name that appears in the lab's schematic. Thanks for catching that mistake of mine :D SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-10-01T03:42:47Z SecondChildTAG: And it would be nice if the system would let us edit our replies to comments (it allows us to edit comments themselves, though). I love Circuit Analysis BTW. A wrong subscript, a simple algebra mistake, or a wrong sign can mess up hours of work :D SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-10-01T03:46:08Z FirstChildTAG: I didn't bother with gates at all. I just went straight for the transistors and got the results that I wanted right off the bat. I felt so proud of myself until I saw the hint right in the directions that showed that it could be done with just three transistors. Now I've done the ol' plug and chug using the book's formula (pg. 305, whatever it says on the assignment), but can't figure out the W/L part. FirstChildUserIdTAG: 359097 FirstChildUserNameTAG: MaxAbramson FirstChildCreateTimeTAG: 2012-10-01T01:35:02Z IndexTAG: 2434 TitleTAG: Confused Hey guys can you give me a hand at understanding this? I can see the voltage divider question at the beginning. But, what is throwing me off is the Roff. We did not cover anything with that in lecture. I am failing at seeing the connection. Thanks UserIdTAG: 230787 UserNameTAG: Gabriel007 CreateTimeTAG: 2012-09-30T20:43:40Z VoteTAG: 1 CoursewareTAG: Week 3 / Switch Resister Model Exercise CommentableIdTAG: 6002x_switch_resistor_model_exercise NumberOfReplyTAG: 1 FirstChildTAG: I got the answers to all of the questions without using Roff...did I miss anything? FirstChildUserIdTAG: 230787 FirstChildUserNameTAG: Gabriel007 FirstChildCreateTimeTAG: 2012-09-30T20:50:17Z SecondChildTAG: Just like you made the on state calculation with RDS being equal to RON, make the off state calculation with RDS equal to ROFF. However ROFF is so big it probably does not affect the answer much which may explain why you got a check mark without even using it. SecondChildUserIdTAG: 397595 SecondChildUserNameTAG: mholin SecondChildCreateTimeTAG: 2012-09-30T21:49:04Z SecondChildTAG: The MOSFETs have two resistance values, $50 \Omega$ when on and $50 M\Omega$ when off. Rather than assuming the transistor is an open circuit when off (which would result in an output of exactly 3V) , you can use the voltage divider relationships to calculate the exact output voltage. If you look at the given answers, the voltages for the logical 1s are not quite 3V, due to the resistance of Roff. I think that in most cases Roff is so large that it can be ignored, but maybe someone will correct me there. Using the values of Ron to check that the pull-down circuitry will pull the output voltage sufficiently low for a logical zero seems much more important. SecondChildUserIdTAG: 434869 SecondChildUserNameTAG: patey SecondChildCreateTimeTAG: 2012-09-30T21:49:33Z SecondChildTAG: Let me do the math with it off at 50MOhms... thanks! SecondChildUserIdTAG: 230787 SecondChildUserNameTAG: Gabriel007 SecondChildCreateTimeTAG: 2012-10-01T00:00:11Z IndexTAG: 2435 TitleTAG: Question about homework and grading Since I just started following the course (I know that I am a few weeks late), and I am not able to finish the first three homeworks and labs in time, will this just affect my grade in the end or I will not receive a certificate at all? By my understanding from the syllabus it will only result in the grading. UserIdTAG: 259178 UserNameTAG: bdrangova CreateTimeTAG: 2012-09-30T19:32:42Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: MIT blindly overlooks 2 of your poorest scores in HWs and Labs and adds the others!........remember, if you want to pass this course, then you have a very small margin of error left now....... I wish I could give you all the answers, but I have left those roads now! Still I am always upto help! FirstChildUserIdTAG: 169784 FirstChildUserNameTAG: thewiredbear FirstChildCreateTimeTAG: 2012-09-30T20:08:39Z FirstChildTAG: You must get a grade of C (60%) or better to receive a certificate. You must do the first three homework assignments because they build on the rest of the course, but alas, you will receive a 0% for them, even if 100% correct. The first two 0% grades will be dropped at no penalty, but the third 0% grade will count towards your final grade, as shown on the 'Course Progress' page. Because each homework is worth 1.5% of your total grade, and each lab is worth 1.5% as well, and in your case Weeks #1 and #2 will be dropped at no penalty: If you get 100% on Homework #4 through #12, and a 100% on Labs # 4 through #12, and a 100% on both exams, your final score will be 97% at best. FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-10-01T02:49:32Z FirstChildTAG: Ok thanks for the answers. I plan to do all the homeworks from now on in time, as well as the first three ones (even if I don't get graded for them). FirstChildUserIdTAG: 259178 FirstChildUserNameTAG: bdrangova FirstChildCreateTimeTAG: 2012-10-01T11:42:21Z IndexTAG: 2436 TitleTAG: LAB 3: Formula to find out W/L: Remember the formula: **W/L > (R_on*(Vs-Vt))/Vt*R_pullup** , for a SR model of the MOSFET. Here you have to only calculate effective R_on... ...& remember the 2nd equation: **R_on= (L/W)Rn** UserIdTAG: 234636 UserNameTAG: KAYBEE CreateTimeTAG: 2012-09-30T19:11:14Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 2437 TitleTAG: Certificate After finishing the course at what name i will be getting the certificate. My edx full name is " Anirudh Muralidhar ". But my college certificate name is Anirudh.K.M , So please clarify me about this. UserIdTAG: 359968 UserNameTAG: Anirudh796 CreateTimeTAG: 2012-09-30T18:53:21Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Here's a [hint][1] of what's to come. [1]: http://6002x.blogspot.com/ FirstChildUserIdTAG: 132826 FirstChildUserNameTAG: fiat_veritas FirstChildCreateTimeTAG: 2012-09-30T19:39:46Z SecondChildTAG: i guess it will be delivered by hand at your address... as i have entered my address at the beginning of the course ---------- if there is something wrong about my information , please tell me SecondChildUserIdTAG: 395257 SecondChildUserNameTAG: blackguitar SecondChildCreateTimeTAG: 2012-10-01T00:56:27Z SecondChildTAG: But i don't remember me entering my address at the beginning of the course. SecondChildUserIdTAG: 359968 SecondChildUserNameTAG: Anirudh796 SecondChildCreateTimeTAG: 2012-10-01T06:19:03Z SecondChildTAG: that link posted by fiat_veritas says it does not exist. SecondChildUserIdTAG: 359968 SecondChildUserNameTAG: Anirudh796 SecondChildCreateTimeTAG: 2012-10-01T06:20:03Z SecondChildTAG: thanks SecondChildUserIdTAG: 156974 SecondChildUserNameTAG: ManojKumar SecondChildCreateTimeTAG: 2012-10-07T12:12:16Z IndexTAG: 2438 TitleTAG: I used the taylor's expansion as said in lecture but couldn't get right, Y? I solved as follows.. iA =10( 1 -e^(VI/5) )+( first derivative of iA at VA=VI )*VA substituting VI=5 volts and solving i get iA=6.3+3.6VA .....equation 1 From Ciruit applying KVL or KCL iA=(5-VA)/2 ....equation 2 On Solving equations 1 and 2.. I get VA= -0.9268 But this is a wrong answer... Please can any one help me ..... Thanks in advance Waiting !!!!!! UserIdTAG: 147459 UserNameTAG: Khanth CreateTimeTAG: 2012-09-30T18:52:29Z VoteTAG: 1 CoursewareTAG: Week 4 / Linearization Exercise 1 CommentableIdTAG: 6002x_linearization_ex_1 NumberOfReplyTAG: 2 FirstChildTAG: There is a step by step here: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_linearization_ex_1/threads/506842b98fda5c2700000005 FirstChildUserIdTAG: 308853 FirstChildUserNameTAG: fmorato FirstChildCreateTimeTAG: 2012-09-30T18:58:22Z FirstChildTAG: For the first equation you should not use Taylor, but the real equation, i.e. iA = 10*(1-e^(-vA/5)) = (vI-vA)/R and solve that using Newton's method. I use the program maxima to do all my calculations in this course and it makes it really easy. Maxima is a computer algebra like and Maple, but it is free (open source), and by the way, it is drived from the original Macsyma system developed at MIT. Here are my equations to solve it with maxima: VI:5$ R:2$ iA: 10*(1-%e^(-vA/5)); load(newton1); newton(iA-(VI-vA)/R, vA, VI, 0.001); and that gives the solution vA = 1.088... Then later I solve it again with VI=5.1. Maxima can also easily differentiate: slope: diff(iA, vA); 1/subst(1.088,vA,slope); gives you the incremental resistance. FirstChildUserIdTAG: 266912 FirstChildUserNameTAG: pietvo FirstChildCreateTimeTAG: 2012-10-02T02:04:04Z SecondChildTAG: Thanks Pietvo! But do we really need to take recourse to a program like "maxima' to solve "**15 - 20*e^(-Va/5) + Va = 0**"? are we really that weak that we can't solve it a more dignified manner? Pardon my naiveté, please! SecondChildUserIdTAG: 285616 SecondChildUserNameTAG: AmiyaX SecondChildCreateTimeTAG: 2012-10-03T21:53:16Z SecondChildTAG: can we solve the above eqn usin iteration method or newton-raphson method? SecondChildUserIdTAG: 337430 SecondChildUserNameTAG: aparna_b SecondChildCreateTimeTAG: 2012-10-04T16:49:47Z IndexTAG: 2439 TitleTAG: H3P1 help what could be the hints for the calculations of the maximum power consumed by the three gates? i tried it several times but couldnt get the proper ans. UserIdTAG: 413613 UserNameTAG: shaliesh CreateTimeTAG: 2012-09-30T16:31:06Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Express power in terms of V and I, V and R, or I and R. Use the formulas and the models for the MOSFETs. FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-09-30T16:44:00Z SecondChildTAG: Kindly tell the value for nand and nor and its formula SecondChildUserIdTAG: 414201 SecondChildUserNameTAG: uzaifakram SecondChildCreateTimeTAG: 2012-09-30T17:32:35Z FirstChildTAG: Have you attended the lectures? It uses a formula P=(Vs^2)/(Ron+Rl). Just put the values for these parameters to get the answer. FirstChildUserIdTAG: 310108 FirstChildUserNameTAG: Saakar FirstChildCreateTimeTAG: 2012-09-30T16:42:07Z SecondChildTAG: I have put the values and got the ans for inverter but didn't able to get power for nand and nor gate SecondChildUserIdTAG: 414201 SecondChildUserNameTAG: uzaifakram SecondChildCreateTimeTAG: 2012-09-30T17:32:01Z SecondChildTAG: I have got the ans. for nand also by using formula vs2/2Ron+Rpull but still dn't know how to find power for Nor SecondChildUserIdTAG: 414201 SecondChildUserNameTAG: uzaifakram SecondChildCreateTimeTAG: 2012-09-30T17:39:25Z IndexTAG: 2440 TitleTAG: LAB 3 Help needed I have got my Vol correctly but the cannot get the desired curve in my transient analysis. I don't know why i have connected the two mosfet in parallel and the 3rd in series well i get glitches in my transient analysis but not the same as given one. Can any one help me where i am wrong ??? UserIdTAG: 336001 UserNameTAG: syd_buet12 CreateTimeTAG: 2012-09-30T15:53:02Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi syd_buet12! Can I help you? Are you sure that you got the correct W/L? FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-30T16:01:56Z SecondChildTAG: try to connect A & B in the two parallel MOSFET SecondChildUserIdTAG: 383496 SecondChildUserNameTAG: winzard111 SecondChildCreateTimeTAG: 2012-09-30T16:26:40Z SecondChildTAG: The key point for this exercise is the W/L relation. After getting this value plug it into the three transistors and run the transient analysis. It should work. Best SecondChildUserIdTAG: 350458 SecondChildUserNameTAG: Aaron_PSX SecondChildCreateTimeTAG: 2012-09-30T16:40:26Z IndexTAG: 2441 TitleTAG: lab 3 ....please help !! with three MOSFETs ,setting the value of w/l to 31.8 produces the exact overall output waveform with those little spikes(almost exact) but still the circuit seems to be incorrect !! please help UserIdTAG: 260272 UserNameTAG: saikat24 CreateTimeTAG: 2012-09-30T15:50:01Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Hi saikat24! Can I help you? Hint: Are you sure that you are getting the maximun value of Rpulldown , the possible RON's, in order to find the worst condition of Vol? ;). FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-30T15:59:12Z SecondChildTAG: help , sure why not ! but i am unable to decode your hint....well, i am uploading the pic of the waveform that i obtained. SecondChildUserIdTAG: 260272 SecondChildUserNameTAG: saikat24 SecondChildCreateTimeTAG: 2012-09-30T16:08:17Z SecondChildTAG: I have answered down :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-30T16:40:16Z SecondChildTAG: hello Myrimit.. can you please help me in solving lab 3.PLEASE..i'm not getting anything.. expecting your reply soon SecondChildUserIdTAG: 137918 SecondChildUserNameTAG: jaisneha SecondChildCreateTimeTAG: 2012-09-30T18:28:10Z SecondChildTAG: Hi jaisneha! Yes, sure. Can I help you? In which part are you lost? SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-30T21:53:03Z FirstChildTAG: ![waveform shape][1] [1]: https://edxuploads.s3.amazonaws.com/13490213088950657.jpg FirstChildUserIdTAG: 260272 FirstChildUserNameTAG: saikat24 FirstChildCreateTimeTAG: 2012-09-30T16:09:01Z SecondChildTAG: w/l lies between 30 to 100 SecondChildUserIdTAG: 357747 SecondChildUserNameTAG: kishores SecondChildCreateTimeTAG: 2012-09-30T16:21:30Z SecondChildTAG: @kishores: i calculated it to be 31.8 ....for which i got the resultant waveform. SecondChildUserIdTAG: 260272 SecondChildUserNameTAG: saikat24 SecondChildCreateTimeTAG: 2012-09-30T16:24:06Z SecondChildTAG: Hi saikat24! I can see in your image that something it is wrong with your W/L ratio. I suggest two alternatives: 1- You can change the W/L experimentally (by +5,+10,+15,etc...), and see what happens. Don't forget to click on TRAN before check. 2- You can thing about what it is the maximun value of the Combinations of RON that you can have in the output, that is to say, the value of Rpulldown=RPD. for example, RPD=RON?, RPD= RON+ (RON//RON)?, RPD=...? Hint: Once you have your RPD you have to try to find what value will verify the Vol (use a voltage divider);). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-30T16:39:41Z FirstChildTAG: i got the result by first calculating a specific value for w/l ,but when it didn't match with the given waveform, i tried out different values for w/l (greater than that specific value) & then finally got it right. But is this the right procedure? is there any specific way to get that exact value? FirstChildUserIdTAG: 260272 FirstChildUserNameTAG: saikat24 FirstChildCreateTimeTAG: 2012-09-30T16:33:24Z SecondChildTAG: Hi saikat24! ;). Nice to see that you got it! Yes, there is a specific way to get the correct value of W/L by analyzing the Vol requirement (you can read what I have wrote you up)... SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-30T16:45:43Z SecondChildTAG: thank you :)) SecondChildUserIdTAG: 260272 SecondChildUserNameTAG: saikat24 SecondChildCreateTimeTAG: 2012-09-30T16:47:08Z SecondChildTAG: You are welcome saikat24! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-30T16:50:48Z SecondChildTAG: thanks SecondChildUserIdTAG: 162670 SecondChildUserNameTAG: charlesbabyt SecondChildCreateTimeTAG: 2012-10-01T10:56:31Z IndexTAG: 2442 TitleTAG: i'm late need to catch up. UserIdTAG: 514054 UserNameTAG: james14344 CreateTimeTAG: 2012-09-30T11:38:07Z VoteTAG: 1 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: Same here. We will have to work hard. FirstChildUserIdTAG: 533988 FirstChildUserNameTAG: thakkerankit FirstChildCreateTimeTAG: 2012-10-02T07:47:53Z SecondChildTAG: Oh God ! Same here. We really have to put the sky and earth together. SecondChildUserIdTAG: 526952 SecondChildUserNameTAG: DEVANKAR SecondChildCreateTimeTAG: 2012-10-03T08:28:05Z IndexTAG: 2443 TitleTAG: plz help why is i2=1.13723966428mA and V2=1.27447932857 the Wrong answer (when we assume that i>1mA)? UserIdTAG: 215271 UserNameTAG: JustStudent CreateTimeTAG: 2012-09-30T07:34:50Z VoteTAG: 1 CoursewareTAG: Week 3 / Analyzing Nonlinear Circuits Exercise CommentableIdTAG: 6002x_analyzing_nonlinear_circuits_exercise NumberOfReplyTAG: 1 FirstChildTAG: RTH=RS*RP/(RS+RP)=2987.59689922 ---------- VTH=VS*RP/(RS+RP)=3.17829457364 ---------- Vso=VTH*Rso/(RTH+Rso)-? ---------- Iso=Vso/Rso-? ---------- If Iso<1mA => Rso=Vso/Iso=1000 => Vso=0.79704510109 and Iso=0.00079704510109. and that is correct answer. ---------- But If Iso>1mA => Rso=Vso/Iso=2000 => Vso=1.27447932857 and Iso=Vso/2000+1/2000=1.13723966428mA. But it is wrong answer. ---------- Why? Smbdy plz explain me where is my mistake? :) FirstChildUserIdTAG: 215271 FirstChildUserNameTAG: JustStudent FirstChildCreateTimeTAG: 2012-09-30T07:45:34Z SecondChildTAG: with Iso=1.13723966428mA, if you calculate the voltage across Rth,it will be 3.3968164volts which is greater than Vth in the circuit SecondChildUserIdTAG: 274748 SecondChildUserNameTAG: flynn_vijay SecondChildCreateTimeTAG: 2012-09-30T08:57:14Z SecondChildTAG: it's a perpetuum mobile :) SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-30T09:29:57Z IndexTAG: 2444 TitleTAG: How to solve H3P4: diode limiter, plz help I am unable to understand how can i solve this. Please guide me in this regards and tell me the steps for solving it.. Thanks in advance UserIdTAG: 223378 UserNameTAG: hadi89 CreateTimeTAG: 2012-09-30T07:05:33Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hi, Case 1: Supply voltage > 0; D2 will be open, D1 will be on only when supply voltage is > 2.5V Case 2: Supply voltage < 0; D1 will be open, D2 will only switch on, when supply voltage < -2.5 V FirstChildUserIdTAG: 143823 FirstChildUserNameTAG: NathanNadeson FirstChildCreateTimeTAG: 2012-09-30T07:16:49Z SecondChildTAG: dear,NathanNadeson you said that Case 2: Supply voltage < 0; D1 will be open, D2 will only switch on, when supply voltage < -2.5 V ..... i guess it is "D2 will only switch on, when supply voltage > 3.25 V" .... AM I RIGHT? SecondChildUserIdTAG: 395257 SecondChildUserNameTAG: blackguitar SecondChildCreateTimeTAG: 2012-09-30T07:31:55Z SecondChildTAG: NathanNadeson, it's not correct. In both cases. blackguitar, it's not correct also. SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-30T09:18:00Z SecondChildTAG: dont miss - peak to peak voltage 20V SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-30T09:24:33Z SecondChildTAG: Guys don't forget that problem appears with different parameters for each student. @NathanNadeson may be right for the parameters appearing in his exercise. SecondChildUserIdTAG: 305491 SecondChildUserNameTAG: lefam SecondChildCreateTimeTAG: 2012-09-30T09:32:44Z SecondChildTAG: lefam, are you sure? SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-30T09:44:14Z SecondChildTAG: NathanNadeson, could you post your params please? SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-30T09:45:05Z SecondChildTAG: what?!!!!!!!!!!!!!!! SecondChildUserIdTAG: 395257 SecondChildUserNameTAG: blackguitar SecondChildCreateTimeTAG: 2012-09-30T11:11:09Z SecondChildTAG: Hopefully this will help you! https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5067499d2193ea230000000c Use KCL SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-09-30T11:14:09Z SecondChildTAG: You can play around with the sandbox if you want,you will definitely get the answer. SecondChildUserIdTAG: 294497 SecondChildUserNameTAG: Saurabhkotian SecondChildCreateTimeTAG: 2012-09-30T13:04:20Z SecondChildTAG: ppl, let's clarify question about different parameters in homework - post yours for H3P4 here SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-30T14:12:19Z FirstChildTAG: how to find out the max current flowing thru the diodes? FirstChildUserIdTAG: 401199 FirstChildUserNameTAG: ashritha FirstChildCreateTimeTAG: 2012-09-30T09:23:08Z SecondChildTAG: Yeah. even i want to know the currents passing through the diodes... SecondChildUserIdTAG: 309722 SecondChildUserNameTAG: kaushikraghavan1992 SecondChildCreateTimeTAG: 2012-09-30T09:41:21Z IndexTAG: 2445 TitleTAG: lab3 problem I get an output very similar to the one we have to get, with the voltage at 0.8V as low and with a w/l ratio of 4000, with the correct spikes at the correct positions. More than that(the w/l ratio) my browser freezes LOL. UserIdTAG: 204745 UserNameTAG: allwynmendes CreateTimeTAG: 2012-09-30T06:53:21Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: did you get the answer correct? FirstChildUserIdTAG: 8897 FirstChildUserNameTAG: sam1202 FirstChildCreateTimeTAG: 2012-09-30T08:14:54Z FirstChildTAG: I need help with the transistors. I don't know how to do it with just three transistors. FirstChildUserIdTAG: 244225 FirstChildUserNameTAG: lerimock FirstChildCreateTimeTAG: 2012-09-30T10:01:21Z SecondChildTAG: Look at the examples in Section 6.4 of the book. SecondChildUserIdTAG: 176592 SecondChildUserNameTAG: JLevitt SecondChildCreateTimeTAG: 2012-09-30T12:06:27Z SecondChildTAG: See Page 296 in the text book Fig 6.25 SecondChildUserIdTAG: 428658 SecondChildUserNameTAG: Moza SecondChildCreateTimeTAG: 2012-10-01T05:55:39Z FirstChildTAG: Hi allwynmendes! Are you sure that you are calculating ok the RON? I think that W/L=4000 it is really a huge value... FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-30T15:12:00Z SecondChildTAG: Haha I know, I resolved it. My circuitary was pretty insane and It was wrong. Got it corrected before the deadline. Thanks a lot SecondChildUserIdTAG: 204745 SecondChildUserNameTAG: allwynmendes SecondChildCreateTimeTAG: 2012-10-06T21:53:07Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T02:41:07Z IndexTAG: 2446 TitleTAG: H3P4 What is the maximum current (in Amperes) that can go through diode D2? any idea for getting the current UserIdTAG: 193792 UserNameTAG: TiTi89 CreateTimeTAG: 2012-09-30T01:50:07Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Do two KVL. FirstChildUserIdTAG: 197569 FirstChildUserNameTAG: darksamaro FirstChildCreateTimeTAG: 2012-09-30T04:51:38Z FirstChildTAG: KCL FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-09-30T03:20:40Z IndexTAG: 2447 TitleTAG: lab3 Does anybody can help me with the Lab3? I can't create the same graphic of trancient analysis as shown in example. I thought, I've done it all right but now I see that my results are not right. I don't understand, what should I do with this condition: "Please add the appropriate pulldown network of mosfet switches connected to node Z to implement the truth table above, with RON of the mosfets chosen so that Vol of the logic gate is less than 0.25V for any combination of inputs. In the schematic tool, the mosfet model has Vth=0.5V, so Vol<0.25V will ensure that when the output of the logic gate is 0, if it is used as the input to some other logic gate, the mosfet to which it connects will be off." UserIdTAG: 190057 UserNameTAG: Anastasia00435 CreateTimeTAG: 2012-09-29T21:41:45Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I think you should see Lab3 Hints by Myrimit. All the best FirstChildUserIdTAG: 281159 FirstChildUserNameTAG: ManasiS FirstChildCreateTimeTAG: 2012-09-29T21:46:28Z SecondChildTAG: I've seen. It didn't help me. Where do we use Vth from our task? SecondChildUserIdTAG: 190057 SecondChildUserNameTAG: Anastasia00435 SecondChildCreateTimeTAG: 2012-09-30T14:49:52Z FirstChildTAG: ok i can give you a hint first of all you will have to find the condition in order to get 0 low at output which is(ABC)=(111,101,011) and from this you will have to find the Ron for both case but since A and B are parraller this ould give you the same answer in case 101 and 011 so calculate and then in order to find the best value test you are answer in both case to see if you get the output Z<0.25 so from that i think you get the picture if not my skype name msamwelmollel FirstChildUserIdTAG: 475448 FirstChildUserNameTAG: msamwelmollel FirstChildCreateTimeTAG: 2012-09-29T22:38:40Z IndexTAG: 2448 TitleTAG: Lab 3 Need Help I'm trying to adjust W/L. All my efforts were to no avail. So I wanna ask whether my circuit is correct and here is mine: I have two MOSFETs in Paaralle which are in series with another MOSFET. Is that correct? UserIdTAG: 172304 UserNameTAG: Demohunter CreateTimeTAG: 2012-09-29T21:33:47Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hi Demohunter! Can I help you? I can not tell you if your circuit it is correct or incorrect.... But are you sure that when you are calculating ok your W/L? are you considering the maximun RPD (Rpulldown) in order to calculate RON? You can also take a look to this [Hints][1] ;) Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50673e4c2193ea2300000009 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-29T21:43:19Z SecondChildTAG: I did it. Thanks. I didn't see the Thevenin voltage. :) SecondChildUserIdTAG: 172304 SecondChildUserNameTAG: Demohunter SecondChildCreateTimeTAG: 2012-09-29T22:18:05Z SecondChildTAG: You are welcome! Congratulations! ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-29T22:21:26Z FirstChildTAG: ok i can give you a hint first of all you will have to find the condition in order to get 0 low at output which is(ABC)=(111,101,011) and from this you will have to find the Ron for both case but since A and B are parraller this ould give you the same answer in case 101 and 011 so calculate and then in order to find the best value test you are answer in both case to see if you get the output Z<0.25 so from that i think you get the picture if not my skype name msamwelmollel FirstChildUserIdTAG: 475448 FirstChildUserNameTAG: msamwelmollel FirstChildCreateTimeTAG: 2012-09-29T22:39:59Z IndexTAG: 2449 TitleTAG: How to think about parallel unequal voltage sources Here is a circuit: ![enter image description here][1] Am I right that depending on the value of R1, the current into the "battery" V2 can be either positive or negative - charging on discharging the battery? As it is now, with equal R1 and R2, the current in is 0. If I decrease R1, more current flows into that upper left hand corner node, so it has go somewhere, by KCL, and it charges the battery. I guess there is nothing logically wrong with this, but for some reason... the way I was imagining voltage sources and current flow... it seemed odd to me. [1]: https://edxuploads.s3.amazonaws.com/13489501991343611.png UserIdTAG: 468623 UserNameTAG: RobNik CreateTimeTAG: 2012-09-29T20:35:41Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I would try analyzing with superposition. Then use my results to view the relationships between the components and their quantities. Analysis solely by the "if this, then that" approach overwhelms me fairly quickly. FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-09-29T22:11:06Z IndexTAG: 2450 TitleTAG: STAFF Help needed, Lab 3 Homework Sir, I have built the schematic and have run the transient analysis and got the exact output wave form as given in the question. but when i checked the schematic, then it is checked as wrong. please guide. Thanks. Image attached:- ![enter image description here][1] [1]: http:// UserIdTAG: 204213 UserNameTAG: ratneshray CreateTimeTAG: 2012-09-29T20:35:18Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: I got the same problem FirstChildUserIdTAG: 219945 FirstChildUserNameTAG: Phobos9703 FirstChildCreateTimeTAG: 2012-09-29T22:03:35Z SecondChildTAG: The schematic may get right but the conditions is not satisfied for VOL<0.25. We have to choose RON such that VOL should be less than 0.25volts. Look for the 3rd week tutorial. hope it will help. SecondChildUserIdTAG: 204213 SecondChildUserNameTAG: ratneshray SecondChildCreateTimeTAG: 2012-09-30T11:27:53Z FirstChildTAG: unable to upload the image too. FirstChildUserIdTAG: 204213 FirstChildUserNameTAG: ratneshray FirstChildCreateTimeTAG: 2012-09-29T20:38:16Z FirstChildTAG: Try to check your Vol (that is, low voltage value, for logical 0). The logical 0 has to have a voltage < 0.25 (if I'm not mistaken). You can change Vol by changing MOSFETs Ron. FirstChildUserIdTAG: 326787 FirstChildUserNameTAG: Angstrem FirstChildCreateTimeTAG: 2012-09-29T20:40:04Z SecondChildTAG: take the equivalent resistances properly SecondChildUserIdTAG: 256221 SecondChildUserNameTAG: himanshu123 SecondChildCreateTimeTAG: 2012-09-29T20:48:08Z SecondChildTAG: thanks. got my answer correct. i was mistaking in calculation of Ron with respect to Vol. SecondChildUserIdTAG: 204213 SecondChildUserNameTAG: ratneshray SecondChildCreateTimeTAG: 2012-09-29T20:58:35Z IndexTAG: 2451 TitleTAG: The open circuit voltage I understand that when the circuit is open circuit we have this voltage Z = Vs because we ignore Rl as there's no current. But what is the point of this? As soon as any current flows then Rl's value is significant. Why do we even bother with this open circuit value? UserIdTAG: 357225 UserNameTAG: MJBoa CreateTimeTAG: 2012-09-29T20:09:28Z VoteTAG: 1 CoursewareTAG: Week 3 / Switch Model CommentableIdTAG: 6002x_switch_model NumberOfReplyTAG: 0 IndexTAG: 2452 TitleTAG: Is there a miscalculation in the power? Should not be (25 sqrt(V))/(13+2/3)? UserIdTAG: 155605 UserNameTAG: JaimeLopezCerezo CreateTimeTAG: 2012-09-29T19:45:59Z VoteTAG: 1 CoursewareTAG: Week 3 / Circuit Truth Table Tutorial CommentableIdTAG: 6002x_circuit_truth_table_tutorial NumberOfReplyTAG: 1 FirstChildTAG: I agree. If the question was "What is the power dissipation added by the on position resistors?" then I would agree with the answer. The total power should be power dissipated by Rl + power dissipated by Ron. FirstChildUserIdTAG: 230787 FirstChildUserNameTAG: Gabriel007 FirstChildCreateTimeTAG: 2012-09-30T21:04:46Z SecondChildTAG: there's no miscalculation,it asks power dissipated by the gate and gate here means the entire boolean logic that is realised by the combo.. the above circuit is actually a realization of a gate and so it will include both ron and rpull.. SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-19T11:54:24Z IndexTAG: 2453 TitleTAG: H5P3 SOURCE FOLLOWER SMALL SIGNAL What I should differentiate? This? RS*(K*(vIN-VT-vOUT)^2)/2 = vOUT Do I have expand the quadratic form?, solve and then differentiate? UserIdTAG: 299251 UserNameTAG: TheRedBlackOne CreateTimeTAG: 2012-09-29T19:41:36Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Yes, you need to find the solution to the quadratic equation then differentiate the solution. FirstChildUserIdTAG: 270011 FirstChildUserNameTAG: SnowmanZA FirstChildCreateTimeTAG: 2012-09-29T20:07:58Z SecondChildTAG: Are the roots a moster solutions? I get too big solutions. Are I on the right way? SecondChildUserIdTAG: 299251 SecondChildUserNameTAG: TheRedBlackOne SecondChildCreateTimeTAG: 2012-09-29T20:41:36Z SecondChildTAG: sorry Am I ... SecondChildUserIdTAG: 299251 SecondChildUserNameTAG: TheRedBlackOne SecondChildCreateTimeTAG: 2012-09-29T20:43:08Z SecondChildTAG: The quadratic equation gets quite complicated. SecondChildUserIdTAG: 270011 SecondChildUserNameTAG: SnowmanZA SecondChildCreateTimeTAG: 2012-09-29T23:10:37Z SecondChildTAG: The result can be simplified a bit before typing in. I realized that after putting in a more complex expression. Fortunately it was accepted. The final expression for gain isn't too bad. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-09-29T23:28:35Z SecondChildTAG: Okay, I solved the equation about vOUT. After that I differentiate in terms of vIN but still got vIN somewhere in my answer. As you know, vIN isn't permitted in the answer. Any clues ? Please, I'm stuck. SecondChildUserIdTAG: 206648 SecondChildUserNameTAG: TsvetanGeorgiev SecondChildCreateTimeTAG: 2012-10-04T11:48:57Z SecondChildTAG: The derivative should be evaluated at v_in = V_IN. So, it is perfectly fine that you're getting v_in in your answer. You just need to put the value of V_IN into the formula. SecondChildUserIdTAG: 252722 SecondChildUserNameTAG: vaboro SecondChildCreateTimeTAG: 2012-10-04T20:29:51Z FirstChildTAG: **When all else fail, watch the lectures ;)** Just a comment to above. If you drag with you all the parameters in the calculation you get a lot. Try to "loose" all the unnecessary parts *before* solving. This is shown clearly in the lectures. FirstChildUserIdTAG: 142857 FirstChildUserNameTAG: viking2 FirstChildCreateTimeTAG: 2012-10-01T03:00:53Z SecondChildTAG: If you're so kind which lecture/s SecondChildUserIdTAG: 299251 SecondChildUserNameTAG: TheRedBlackOne SecondChildCreateTimeTAG: 2012-10-03T17:34:00Z SecondChildTAG: S10V6: & S10V7: MATHEMATICAL VIEW OF SMALL SIGNAL SecondChildUserIdTAG: 149923 SecondChildUserNameTAG: Odessa SecondChildCreateTimeTAG: 2012-10-03T22:42:07Z IndexTAG: 2454 TitleTAG: Lab 2-How to Get Checked? I'm getting the correct waveform but the lab still marks me wrong. I'm using a voltage divider to cut down the sine wave, and the resulting voltage peaks are of the right size, but I still get marked wrong! UserIdTAG: 143345 UserNameTAG: jbeckett CreateTimeTAG: 2012-09-29T19:34:13Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Hi jbeckett! How are you? Can I help you? Hint: Are you sure that you are using the correct resistive network? You can also see this Post [here][1]. I also made a video tutorial for Lab2 but I can not upload it till the end of deadline... Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/505f41be015492230000001c FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-29T19:41:22Z FirstChildTAG: Are you sure it's the correct? You can check by moving your cursor over the peak and bottom and see the numbers. FirstChildUserIdTAG: 219945 FirstChildUserNameTAG: Phobos9703 FirstChildCreateTimeTAG: 2012-09-29T22:15:39Z FirstChildTAG: I'm pretty sure it's correct. When I run the transform function the result is identical to what we are shown in the lab, but when I press 'check' I get a big red X. Is there some special thing we're supposed to do before we press 'check'? I thought running the transform was enough. But thanks for answering. jbeckett FirstChildUserIdTAG: 143345 FirstChildUserNameTAG: jbeckett FirstChildCreateTimeTAG: 2012-09-30T12:20:35Z IndexTAG: 2455 TitleTAG: Discussion system I have a few questions about the discussion system. First of all, I can't see what posts I've made. Users need a list of all posts they've made. Easy, no questions. EDIT: I just realized the Following filter exists, which most likely includes posts I've made, but not always. Second, I can't search for "Lab 2" because 2 doesn't count as a word. The homework is called Lab 2 and that's what everyone titles their posts. Broken. Third, why is there nothing like sub forums? Then we'd post Lab questions under the corresponding lab subforum. Look at Coursera's forum system. It's wonderful. Why fix what isn't broken? Fourth, one that I just figured out when accidentally hitting backspace. Why must everything work through AJAX? Why is the back/forward button functionality broken? And why does it look like I'm using an iPhone theme on my laptop when I'm in the discussion system? Again, fixing what is not at all broken. Anyway, the actual Homework/Lecture system is awesome I think. Just perhaps the discission subsystem needs another look. UserIdTAG: 357225 UserNameTAG: MJBoa CreateTimeTAG: 2012-09-29T19:09:32Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: MJBoa, Thanks for the suggestions! Just a couple of clarifications... To see your own posts you can click on the button that says "All" on the top of the discussion forum subject list. If you select "Following" from the dropdown list you will see only the posts you create or that you decided to follow. You can also click on your own name, and that will take you to all the posts were you have replied. For the subtopics, you can also use the dropdown list. The sub-topics are organized by week, exercise, etc. Homeworks and Labs don't have sub-topics, but that was decided by the staff. About the search tool, that was noticed before and we are working to improve it. FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-29T19:44:17Z SecondChildTAG: Thank you jelizon! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-29T20:41:10Z IndexTAG: 2456 TitleTAG: Homework 3 part 2 I don´t quite understand the part two of Noise Margin... it says that NMh = NMl and there is the graph of Vout vs Vin Can someone explain me what do we have to do? Thanks! , Daniel UserIdTAG: 371220 UserNameTAG: dan10 CreateTimeTAG: 2012-09-29T17:59:03Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Watch the tutorials from Week 3. FirstChildUserIdTAG: 172304 FirstChildUserNameTAG: Demohunter FirstChildCreateTimeTAG: 2012-09-29T18:04:46Z SecondChildTAG: NMh = NMl is the constraint you're offered. Margin high = margin low or (Voh-Vih) = (Vil-Vol) SecondChildUserIdTAG: 5325 SecondChildUserNameTAG: vkz SecondChildCreateTimeTAG: 2012-09-29T18:07:47Z SecondChildTAG: Hint: Look at the graph and pick 2 points on one of the curves (or sloped lines), there are 3 of them. Then with the two points you picked, plug them into the equation NMh=NMl and if they are equal, the numbers you picked are your answers. Make sure the points you pick are only from one curve. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-09-29T22:04:13Z FirstChildTAG: Here some Hints for you dan10 ;): **H3P2: GRAPHICAL MODEL OF INVERTER** *“Recall that senders can output voltajes above VH and below 5V for the logical 1’s and voltages below VL and above 0V for logical 0’s. Receivers must correspondingly interpret output voltages above VH and below 5V for logical 1’s and voltages below VL as logical 0’s. A sender wishing to place a logical 0 on a wire can therefore output the voltage VL, which falls within the valid range for logical 0. Receivers observing the value VL transmitted on the wire wwill correctly interpret it as a logical 0. However, the presence of even the smallest amount of (positive) noise will force the voltage signal on the wire into the forbidden region, thereby causing the signal to become invalid…”* [Read here][1] ![enter image description here][2] **Part 1:** Hint: even the smallest amount of noise will force the voltage signal to the forbidden region.... So… what part of the curve it is more convenient to choose as a limit? The zone where the noise will vary much more or less more?? ;). So, can you find your VIL (in Volts) that will actually achieve the static discipline with the maximum positive noise margin? yes! **Part 2,3,4:** Hint: the same concept of 1) but for VIH, VOH and VOL ;). **Part 5:** Hint: Having your values from the parts above, can you find the noise margin? yes! [see this Post Text Links - Noise Margin][3] ;) I hope this can be helpful. See you! Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/272 [2]: https://www.edx.org/static/content-mit-6002x/images/circuits/inverter_graph.e109eb3e01bc.gif [3]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_logic_gates/threads/506616868032ee1f0000001e FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-29T22:20:10Z IndexTAG: 2457 TitleTAG: Week 3 - S5V16 MOSFET as SR Model Lecture At time 6:14mins - 6:17mins of S5V16 lecture, I noticed that Prof. said 10/11 gives 0.45 instead of 0.9090 or was there another value he divided by... (i.e did he divide by 2) somewhere? Anyone with an explanation should please post. Regards, Ugo UserIdTAG: 329879 UserNameTAG: Ugochukwu CreateTimeTAG: 2012-09-29T17:57:21Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: What I figured out here is that if we apply Vs = 5v it will give 0.45 as far as we still leave Rl/Ron = 10 and not Vs = 10v as seen on the right pane of the lecture. FirstChildUserIdTAG: 329879 FirstChildUserNameTAG: Ugochukwu FirstChildCreateTimeTAG: 2012-09-29T18:05:03Z IndexTAG: 2458 TitleTAG: Enable?? what is meant by "Enable"?? Why is it called so?? UserIdTAG: 316353 UserNameTAG: Vivekanand123 CreateTimeTAG: 2012-09-29T17:55:36Z VoteTAG: 1 CoursewareTAG: Week 3 / Circuit Truth Table Tutorial CommentableIdTAG: 6002x_circuit_truth_table_tutorial NumberOfReplyTAG: 1 FirstChildTAG: to enable/disable output - just like a light switch FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-09-29T18:26:23Z IndexTAG: 2459 TitleTAG: technical to administrator, sir i have been doing lab week-3,even though i am getting same waveform as shown in the figure in lab week -3 with numerical values correct it is giving wrong checkmate when submitted plz help!! UserIdTAG: 285222 UserNameTAG: dhaval24 CreateTimeTAG: 2012-09-29T17:34:00Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hi dhaval24! Can I help you? Have you clicked on TRAN before click on check? Have you chosed the correct value of W/L of the MOSFETS? Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-29T19:49:18Z FirstChildTAG: Dhaval...have you observed any of the "glitches" in your waveform? FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-09-30T02:04:51Z IndexTAG: 2460 TitleTAG: Lab 3 Homework I made the circuit for the logic sequence and simulated it to get the graph as shown in the input. The output obtained was similar to the output given. Still the answer seems to be wrong. I have been trying from the past few days but to no success. Is there anything I am missing here? And guidance will be appreciated the last date for submission is tomorrow as you may all know. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-29T16:48:12Z VoteTAG: 1 CoursewareTAG: Week 3 / Switch Resistor Model of a MOSFET CommentableIdTAG: 6002x_switch_resistor_model NumberOfReplyTAG: 1 FirstChildTAG: I suppose, you haven't worked on your Vol parameter. That is, I think, that your V, that is corresponding to the logical 0 is more than 0.25V (or what was the task value?). You can change it by changing the MOSFETs Ron. FirstChildUserIdTAG: 326787 FirstChildUserNameTAG: Angstrem FirstChildCreateTimeTAG: 2012-09-29T16:50:52Z SecondChildTAG: Thanks mate :D I did it! SecondChildUserIdTAG: 260713 SecondChildUserNameTAG: viktorfelix SecondChildCreateTimeTAG: 2012-09-29T17:29:22Z SecondChildTAG: setting the value of w/l to 31.8 produces those little spikes(almost exact) but still the circuit seems to be incorrect !! please help SecondChildUserIdTAG: 260272 SecondChildUserNameTAG: saikat24 SecondChildCreateTimeTAG: 2012-09-30T15:47:20Z IndexTAG: 2461 TitleTAG: lab 3 please tell me how to calculate w/l ratio UserIdTAG: 284628 UserNameTAG: jumana_mp CreateTimeTAG: 2012-09-29T15:23:34Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: its clearly given in the question itself.... jus read the 4th paragraph carefully.... all the best..:) FirstChildUserIdTAG: 367754 FirstChildUserNameTAG: srushti1211 FirstChildCreateTimeTAG: 2012-09-29T15:44:17Z SecondChildTAG: hey can i just registered for this, will u please tell me from where to start this course SecondChildUserIdTAG: 515619 SecondChildUserNameTAG: abdulhannan333 SecondChildCreateTimeTAG: 2012-09-29T15:49:39Z SecondChildTAG: @abdulhannan333 since we are allowed to miss two hw you must complete either one before the deadline.you may start by looking the exercises, if you are comfortable miss the lecture or just go through the slides and notes SecondChildUserIdTAG: 99628 SecondChildUserNameTAG: Pranjal16 SecondChildCreateTimeTAG: 2012-09-29T16:05:53Z SecondChildTAG: Hi juamana_mp! Take a look here [Post][1] there is an explanation with Sandbox. ;) [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50673e4c2193ea2300000009 SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-29T19:53:07Z IndexTAG: 2462 TitleTAG: lab 2 In which video is the mixer circuit explained? UserIdTAG: 100562 UserNameTAG: Extraordinary CreateTimeTAG: 2012-09-29T15:00:13Z VoteTAG: 1 CoursewareTAG: Week 2 / Linear combinations of source strenghts CommentableIdTAG: 6002x_S3E2_Linear_Combinations_of_Source_Strengths NumberOfReplyTAG: 2 FirstChildTAG: As mixer circuit uses voltage divider concept I remember it was not separately explained in video.you may use web for more details FirstChildUserIdTAG: 99628 FirstChildUserNameTAG: Pranjal16 FirstChildCreateTimeTAG: 2012-09-29T16:10:59Z FirstChildTAG: [Here][1] :) [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_S3E2_Linear_Combinations_of_Source_Strengths/threads/506a280d036af91f000000c9 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-01T23:35:51Z IndexTAG: 2463 TitleTAG: 2 last tasks i still can't solute it ..... I've read the old posts ... and i can't understand yet. please can any one explain the answer briefly ? and what's the relation between these tasks S6E1 with the Previous one S6E0...i see them as a same answer what's the difference? UserIdTAG: 395257 UserNameTAG: blackguitar CreateTimeTAG: 2012-09-29T14:24:55Z VoteTAG: 1 CoursewareTAG: Week 3 / Analyzing Nonlinear Circuits Exercise CommentableIdTAG: 6002x_analyzing_nonlinear_circuits_exercise NumberOfReplyTAG: 1 FirstChildTAG: In the last two questions, the trick is to determine in which region the device is operating so that it can be replaced by the appropriate model. One possibility to find the operating region is through trial and error either in the original circuit or (to make your life easier) using the Thevenin equivalent found in S6E0. Notice that the unknown device behaves as a resistor of one value if current is below 1 mA, and of a different value if above 1 mA. Just be sure to use the correct model (i.e. the correct resistor) and you can get the answer! I hope this helps FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-29T19:30:23Z SecondChildTAG: Dear friend, Look, you must satisfy two equations: By the thevenin method: Vx+2988Ix=3.18 (approx). And by the first graph, you have two option: Vx=1000*Ix (Ix<=1mA) or Vx=2000*Ix (Ix>=1mA). So, if you try the two options, you will get: If Vx=1000*Ix, Ix=0.797mA (SATISFY THE CONDICTION Ix<=1mA). If Vx=2000*Ix, Ix=0.637mA (DONT SATISFY THE CONDICTION Ix>=1mA). Therefore, you have just one answer!! Pick it up Best Regards SecondChildUserIdTAG: 263241 SecondChildUserNameTAG: yvesauad SecondChildCreateTimeTAG: 2012-09-30T03:45:06Z SecondChildTAG: thank you my friend SecondChildUserIdTAG: 395257 SecondChildUserNameTAG: blackguitar SecondChildCreateTimeTAG: 2012-10-01T01:10:59Z IndexTAG: 2464 TitleTAG: what % is required for getting certificate i am doing good in this course . what percentage is required for getting a certificate and also is it necessary to watch videos for getting certificate.? you tube is blocked in my country that's why i ask.. thanks UserIdTAG: 136490 UserNameTAG: Ali_PU CreateTimeTAG: 2012-09-29T13:33:56Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: You can download the lecture videos from the Wiki, personally I have watched most of them at least once. FirstChildUserIdTAG: 392546 FirstChildUserNameTAG: JohnWayne FirstChildCreateTimeTAG: 2012-09-29T14:30:31Z SecondChildTAG: yeah i download all the lectures but one thing more what % is required for certificate.? SecondChildUserIdTAG: 136490 SecondChildUserNameTAG: Ali_PU SecondChildCreateTimeTAG: 2012-09-29T14:36:19Z SecondChildTAG: I didn't found this information said explicitely anywhere. But my guess is that certificate will be granted to students with score >60% (mark C). SecondChildUserIdTAG: 104543 SecondChildUserNameTAG: sten SecondChildCreateTimeTAG: 2012-09-29T14:42:48Z SecondChildTAG: hi stern! Read what salsero answered . You need 60% of the Total Score to pass ;). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-29T20:24:19Z SecondChildTAG: *sten ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-29T20:43:01Z FirstChildTAG: You can find it in the Course Handouts. Look here at page 3 Grading: https://www.edx.org/static/content-mit-6002x/handouts/syllabus.a477535058a1.pdf If you get 60% you'll get a C. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-09-29T15:14:11Z SecondChildTAG: Hi Ali_PU! What salsero said it is correct ;). Also, you can find the videos posted in the wiki. See you! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-29T20:20:12Z IndexTAG: 2465 TitleTAG: H3P4: PART3 MAX Current through D1 !!!!!!!!!!!!!! UserIdTAG: 382505 UserNameTAG: AhmedGalal2 CreateTimeTAG: 2012-09-29T12:16:55Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Just take the maximum possible value of the source AC current (Vs), than all the terminals of D1 would have V1 voltage. Than you apply KCL. FirstChildUserIdTAG: 326787 FirstChildUserNameTAG: Angstrem FirstChildCreateTimeTAG: 2012-09-29T13:40:19Z SecondChildTAG: Thank you Sir, so much (: SecondChildUserIdTAG: 382505 SecondChildUserNameTAG: AhmedGalal2 SecondChildCreateTimeTAG: 2012-09-29T14:52:00Z SecondChildTAG: Yes but, if all terminals have the same voltage, there is not voltage drop, so there is not current. How is this possible? SecondChildUserIdTAG: 244225 SecondChildUserNameTAG: lerimock SecondChildCreateTimeTAG: 2012-09-29T15:02:24Z SecondChildTAG: AhmedGalal2, no problem =) lerimock, have a look here:![enter image description here][1] There's no voltage drop between e1 and e2 on this *ideal wire* piece, but still we have a current. How is that possible? ^_^ [1]: https://edxuploads.s3.amazonaws.com/1348932167985436.png SecondChildUserIdTAG: 326787 SecondChildUserNameTAG: Angstrem SecondChildCreateTimeTAG: 2012-09-29T15:24:03Z SecondChildTAG: Still don't understand..... please help SecondChildUserIdTAG: 232878 SecondChildUserNameTAG: Mjbora SecondChildCreateTimeTAG: 2012-09-29T21:50:06Z SecondChildTAG: Thanks for help :) SecondChildUserIdTAG: 365309 SecondChildUserNameTAG: chetnasinghaldas SecondChildCreateTimeTAG: 2012-09-30T10:42:51Z SecondChildTAG: understood SecondChildUserIdTAG: 391053 SecondChildUserNameTAG: Ahsanuddin SecondChildCreateTimeTAG: 2012-09-30T10:44:20Z FirstChildTAG: hi,AhmedGalal2.... where are you from Abo Galal :):) FirstChildUserIdTAG: 395257 FirstChildUserNameTAG: blackguitar FirstChildCreateTimeTAG: 2012-10-01T01:07:38Z IndexTAG: 2466 TitleTAG: Certificate worth to Universities I am currently studying in grade 12 and will be applying to the universities of my choice this year. Though I am doing this course solely because I am passionate about learning how to build cool things, I was wondering whether the certificate given to those students who pass this course has any worth to universities during the application procedure. I would be much obliged if someone could give me more information on this matter. UserIdTAG: 85490 UserNameTAG: yashsavani CreateTimeTAG: 2012-09-29T12:02:00Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: I've seen a lot of discussion about whether completing one of the many online courses currently on the internet is good for college admissions or to include on resumes. MOOC's are still pretty new and the final word is not out yet, but I have heard of people putting such courses on their resume and successfully get jobs. So although there isn't quite a final verdict yet, I believe completing one of these free online courses is an accomplishment worthy to include in your application. I can give you some ideas on how to include this course on your application, but ultimately ask the school's admissions office for advice on how to include this qualification on your application submission. No grades are offered at present so I don't see it going under GPA calculations. Although, if the grade record is self-completed you could consider taking a line under the "Official program of study" section and list the course there. Otherwise, you can at the very least definitely mention it in one of your application essays, or talk about it in the "extra considerations" section as well. But, like I said, no one really knows what your school is looking for unless you ask them directly, so give them a call or an e-mail regarding this matter. FirstChildUserIdTAG: 160489 FirstChildUserNameTAG: Spellstealz FirstChildCreateTimeTAG: 2012-09-29T12:39:31Z FirstChildTAG: Educators are aware of these efforts by edX, Coursera, et al. There is much discussion going on about them. When edX says its offerings are "Harvard hard" and "MIT hard," and a high school graduate can show that he or she has already performed credibly at that level, then I would expect their application for admission to be regarded more favorably. FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-09-29T17:19:00Z FirstChildTAG: is the grade (i.e A, B, C) written on the certificate or only "Pass"? FirstChildUserIdTAG: 263602 FirstChildUserNameTAG: elekov FirstChildCreateTimeTAG: 2012-09-29T20:37:46Z SecondChildTAG: The first 6.002x made both available for download...one with your grade and one without. I don't know if they will continue to do so. SecondChildUserIdTAG: 92346 SecondChildUserNameTAG: MobiusTruth SecondChildCreateTimeTAG: 2012-10-07T03:18:35Z IndexTAG: 2467 TitleTAG: practise scores what are practise scores that are mentioned in progress? i always get 0/4 , 0/5.how to get it done? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-29T11:57:11Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: same question FirstChildUserIdTAG: 136490 FirstChildUserNameTAG: Ali_PU FirstChildCreateTimeTAG: 2012-09-29T13:36:40Z SecondChildTAG: That are the scores of in-lecture-sequence exercises. They are not counted towards the final score. SecondChildUserIdTAG: 104543 SecondChildUserNameTAG: sten SecondChildCreateTimeTAG: 2012-09-29T14:47:23Z SecondChildTAG: what sten said it is true ;). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-29T20:27:39Z IndexTAG: 2468 TitleTAG: change password how to change password? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-29T11:51:50Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Logout of your account and in the log in screen click 'Forgot Password'. That will let you change the password. FirstChildUserIdTAG: 365201 FirstChildUserNameTAG: sirajmuhammad FirstChildCreateTimeTAG: 2012-09-29T15:39:51Z SecondChildTAG: ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-29T20:28:19Z IndexTAG: 2469 TitleTAG: Static discip valid range question textbook Hi There are some footnotes in the textbook, that I don't understand *Chapter 6 Page 308, note 19* "In fact, our inverter can satisfy a static discipl with a VOH as high as 5(uppercase minus)V" I don't understand this *uppercase minus* (is it some kind of node voltage)? UserIdTAG: 87106 UserNameTAG: Fabius CreateTimeTAG: 2012-09-29T10:53:32Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: They mean the range for VOH is going to [5 volt minus a little bit]. So the uppercase minus is just a notation for the VOH range not going precisely to 5 volt. You see the same notation in note 21 for VT uppercase + V, meaning for the VIH range going from VIH to not precisily VT, but to VT plus a little bit. In math you can approach a value of a function from the left or from the right, for example when taking a limit. this is in fact the same idea. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-09-29T15:50:26Z SecondChildTAG: thank you salsero :) SecondChildUserIdTAG: 87106 SecondChildUserNameTAG: Fabius SecondChildCreateTimeTAG: 2012-09-29T16:17:07Z SecondChildTAG: Thank you salsero ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-29T20:29:22Z IndexTAG: 2470 TitleTAG: Transient Analysis i cannot see the scope probes in the circuit sand box interactive tool; this means that i cannot correctly answer some of the questions that follow thereafter; HELP! UserIdTAG: 301849 UserNameTAG: agelu CreateTimeTAG: 2012-09-29T09:59:48Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi agelu! I can see them in Sandbox(voltages probes)... Do you have the last updated version of your Web Broser (Chrome or Firefox)? FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-29T20:32:18Z IndexTAG: 2471 TitleTAG: h2p1: no solution for me Seems like there is no solution with my given specs. See for yourself, you be the judge (there seems to be no solution even BEFORE considering the Thevenin voltage constraint): I am given: Vin = 30V, Vout = 9V (+/-10%). This means Vout/Vin = .3 +/- 10%. Therefore: 0.27 < Vout/Vin < 0.33 Transfer function for me is the voltage divider: R2/(R1+R2) = 1/(1+R1/R2) So: 0.27 < 1/(1+R1/R2) < .33 Or: 3.03 < 1 + R1/R2 < 3.704 Thus: 2.03 < R1/R2 < 2.704 Now, since both R1 and R2 can vary by +/-10%, R1/R2 has minimum value of (.9/1.1)R1/R2 and maximum value of (1.1/.9)R1/R2. Therefore R1/R2 must satisfy the inequalities 2.03 < (.9/1.1)R1/R2 and (1.1/.9)R1/R2 < 2.704 Therefore we have: R1/R2 > 2.481 and R1/R2 < 2.212, which have no solution(?). Anything wrong with my analysis, or is my problem bugged? UserIdTAG: 160489 UserNameTAG: Spellstealz CreateTimeTAG: 2012-09-29T09:29:50Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: The problem statement is **Assume first that the resistors have their nominal resistance.** Come up with resistors R1 and R2 such that the divider ratio Vout/Vin is within 10% of the requirement. Here you are trying to find the solution including resistor tolerance. I really appreciate your mathematical approach. Good work :) FirstChildUserIdTAG: 823 FirstChildUserNameTAG: naveenatmit FirstChildCreateTimeTAG: 2012-09-29T09:56:47Z SecondChildTAG: I think I see what you are saying, thank you for your kind explanation. :-) SecondChildUserIdTAG: 160489 SecondChildUserNameTAG: Spellstealz SecondChildCreateTimeTAG: 2012-09-29T12:17:43Z SecondChildTAG: you are welcome :) SecondChildUserIdTAG: 823 SecondChildUserNameTAG: naveenatmit SecondChildCreateTimeTAG: 2012-09-29T17:59:57Z SecondChildTAG: Thank you naveenatmit ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-29T20:33:52Z IndexTAG: 2472 TitleTAG: zener regulator "What is the minimum value of RL, in Ohms, that guarantees that the circuit will operate this way?" What does it mean? UserIdTAG: 343388 UserNameTAG: bijojoseph CreateTimeTAG: 2012-09-29T06:31:59Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Zener diode got a distinct operation region on the v-i graph where it actually operates, or "regulate" that is. So you need to calculate the value of RL that will bring it into this distinct region. Hint: use voltage divider equation. FirstChildUserIdTAG: 326409 FirstChildUserNameTAG: EugeneZ FirstChildCreateTimeTAG: 2012-09-29T09:27:33Z FirstChildTAG: If you were able to identify the operating region of the zener from the graph (that is made up of 3 piecewise segments), look for the maximum voltage across the zener to act as an open switch, and use that value to get RL (by using the voltage divider rule) FirstChildUserIdTAG: 143823 FirstChildUserNameTAG: NathanNadeson FirstChildCreateTimeTAG: 2012-09-30T07:29:47Z IndexTAG: 2473 TitleTAG: H3P3-solar cell problem hi, pl help me to solve H3P3 first problem.i got the answer for second one.How to find power when R=100ohms.this is the question. UserIdTAG: 219204 UserNameTAG: vsriram CreateTimeTAG: 2012-09-29T04:48:55Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Its surprising, when power is given - you have found the $R$ . Why can't you find the power when $R$ is given !! FirstChildUserIdTAG: 823 FirstChildUserNameTAG: naveenatmit FirstChildCreateTimeTAG: 2012-09-29T05:04:54Z FirstChildTAG: Just see a R = 100 ohm on power graph and that's the value. FirstChildUserIdTAG: 244225 FirstChildUserNameTAG: lerimock FirstChildCreateTimeTAG: 2012-09-29T14:48:48Z SecondChildTAG: I will make it simple. 1). P=I^2*R. I from graph 1 ,I is about 0.001 ,thus Power can be calculated. 2) This has nothing to do with part 1 of this question.From the second graph of the question it is very much understood that the maximum power is at .3 Volts and is about .00027 W. Thus from The relation P=V^2/R R can be easily calculated. SecondChildUserIdTAG: 232499 SecondChildUserNameTAG: Asim09 SecondChildCreateTimeTAG: 2012-09-29T19:29:43Z SecondChildTAG: that helped a lot :) SecondChildUserIdTAG: 433368 SecondChildUserNameTAG: SurbhiMahajan SecondChildCreateTimeTAG: 2012-09-30T07:53:29Z SecondChildTAG: some brain work and it solve! :-) SecondChildUserIdTAG: 96873 SecondChildUserNameTAG: Santyaga SecondChildCreateTimeTAG: 2012-09-30T14:58:35Z SecondChildTAG: tks SecondChildUserIdTAG: 150120 SecondChildUserNameTAG: krishna1993 SecondChildCreateTimeTAG: 2012-09-30T17:23:10Z IndexTAG: 2474 TitleTAG: plz help [staff] i have just noticed the start of course today. hence my 3 week home works and labs are due. What to do now?...plz help.. UserIdTAG: 170017 UserNameTAG: Rajez CreateTimeTAG: 2012-09-29T01:23:42Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Start doing homework 2 and 3 as they can still be submitted. FirstChildUserIdTAG: 42833 FirstChildUserNameTAG: KKA FirstChildCreateTimeTAG: 2012-09-29T02:21:29Z SecondChildTAG: you will miss only 3% of your score if you couldn't submit homework 3. Because, out of 12 homeworks and labs, only best 10 are selected. So, you can get 4 to 12 done, that gives you 27% of the score. SecondChildUserIdTAG: 823 SecondChildUserNameTAG: naveenatmit SecondChildCreateTimeTAG: 2012-09-29T05:10:24Z IndexTAG: 2475 TitleTAG: This works: Hint: Add an independent current source at the port 1) stop Io ( open circuit ) , let dependent source work 2) add a test source at the port ( i made it with a voltage source Vtest ) 3) compute the current through Vtest ( with respect to dependent source ) 4) rth = Vtest/Itest UserIdTAG: 373752 UserNameTAG: Croak CreateTimeTAG: 2012-09-28T17:04:49Z VoteTAG: 1 CoursewareTAG: Week 4 / Dependent Sources Exercise 1 CommentableIdTAG: 6002x_dep_src_exsz_1 NumberOfReplyTAG: 1 FirstChildTAG: can we use this test source to get vth FirstChildUserIdTAG: 432151 FirstChildUserNameTAG: rgdixit FirstChildCreateTimeTAG: 2012-09-29T17:52:45Z IndexTAG: 2476 TitleTAG: Lab5 The last exercise of Lab5 asks for "Largest input amplitude resulting in an undistorted output signal:" Is this vi or vI? Is the result meant to be expressed in Volts or miliVolts? Is a clipped signal distorted or vice-versa? What's the relationship between a clipped and distorted signal? I found the saturation operation range limits but I could not find the correct answer of this question. UserIdTAG: 39980 UserNameTAG: MarinoJr CreateTimeTAG: 2012-09-28T12:58:39Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: It is the biggest you can make vi and have the output still look like a sine wave. Yes a clipped signal is distorted. Any shape that does not look like a sine wave is distorted. You can play around with vI but vi is what you have to play with to find the bigest undistorted sine wave output. FirstChildUserIdTAG: 397595 FirstChildUserNameTAG: mholin FirstChildCreateTimeTAG: 2012-09-28T13:33:15Z SecondChildTAG: Is the answer expected to be expressed in Volts or miliVolts? Is this answer vi or vI? SecondChildUserIdTAG: 39980 SecondChildUserNameTAG: MarinoJr SecondChildCreateTimeTAG: 2012-10-01T20:09:43Z SecondChildTAG: In Volts. I got that answer. SecondChildUserIdTAG: 39980 SecondChildUserNameTAG: MarinoJr SecondChildCreateTimeTAG: 2012-10-01T20:21:18Z FirstChildTAG: Undistorted signal is generated only in the linear region. Try to obtain the limits of linear region from previous graph. FirstChildUserIdTAG: 157610 FirstChildUserNameTAG: mradziwo FirstChildCreateTimeTAG: 2012-09-28T14:58:03Z SecondChildTAG: this is the perfect help. thanks buddy :) SecondChildUserIdTAG: 438395 SecondChildUserNameTAG: ali_PU1 SecondChildCreateTimeTAG: 2012-10-14T15:55:25Z FirstChildTAG: Forget about linear distortion. ONLY UNCLIPPED ). FirstChildUserIdTAG: 394836 FirstChildUserNameTAG: v2g6ch4 FirstChildCreateTimeTAG: 2012-09-28T17:03:35Z IndexTAG: 2477 TitleTAG: What's the small signal model of attenuator circuit? I have watched the video of small signal analysis of attenuator but I cannot figure out what the small signal model is. To be concrete, in my opinion, the current source, although changeable, should be replaced by open circuit since it is a dependent current source when we do large signal analysis. By the way, I can understand most parts of large signal analysis but I'm not so clear about the relationship between large and small signal analysis. Can some one show me the relationship between large and small signal model by following the steps taught at lectures, i.e., find an operating point, develop small signal model, and do small signal analysis? UserIdTAG: 146640 UserNameTAG: CuiWenzhi CreateTimeTAG: 2012-09-28T12:50:12Z VoteTAG: 1 CoursewareTAG: Week 4 / Attenuator Tutorial CommentableIdTAG: 6002x_attenuator_t NumberOfReplyTAG: 1 FirstChildTAG: Hello, CuiWenzhi! You can treat 'attenuator' as a voltage divider. [Tutorial video, start from 1:00][1]. Our goal is to move diodes to the 'resistor' part of VA characteristic, and we use current source for it — and only for it. 1. If the current source is 0 A, the diode will be closed, and can't conduct current 2. If the current is too large, the diode voltage will not depend on current and will be constant 0.7 V 3. Some value of current will move diode VA to the small-linear part, and last video shows, how to find concrete value (with assumption: we already know the diode incremental resistance r = Vt/I0). That's idea of the small signal model, I believe. ---------- As for large signal... You can model it :D ![Large signal model][2] Watch to the red and magenta input-output signals. Left part: the diode is opened, the signal is same as input, minus some constant value. Right part: diode is closed, input voltage is the same as the output. And now pay attention to the central part, where input is near zero. The output is completely linear, diode acts as the resistor! We get the voltage divider, which reduces the input, that's our small signal area. ---------- And one final note: the current source is independent for this circuit. But if you want to change the attenuation ratio, you can change the current from outside. For example, by hands in your modeling sandbox ;) [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_4/wk4t2/2 [2]: https://edxuploads.s3.amazonaws.com/13488688481343639.png FirstChildUserIdTAG: 261503 FirstChildUserNameTAG: EugenyL FirstChildCreateTimeTAG: 2012-09-28T22:10:14Z IndexTAG: 2478 TitleTAG: expoDweep Thank you for that little thing! Makes me smile each time the professor says it :D Only why expodweep? Exponential... dweep?.. Does it stand for anything? :) UserIdTAG: 410033 UserNameTAG: sagitta CreateTimeTAG: 2012-09-28T12:45:16Z VoteTAG: 1 CoursewareTAG: Week 3 / Incremental Method Motivation CommentableIdTAG: 6002x_incremental_method_motivation NumberOfReplyTAG: 1 FirstChildTAG: yup.. very good name and very good transfer function FirstChildUserIdTAG: 257806 FirstChildUserNameTAG: skoda FirstChildCreateTimeTAG: 2012-09-29T03:04:42Z IndexTAG: 2479 TitleTAG: about the textbook how can i find the URL for the www in the textbook can you give us the URLs for the materiels that marked www in the textbook? UserIdTAG: 266386 UserNameTAG: zakzak200 CreateTimeTAG: 2012-09-28T11:30:37Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I knew it was there, but I forgot to bookmark the link after I had downloaded it. Look here for all Supplementary Sections and Examples. You can download all in one pdf or all chapters seperatly. http://www.elsevierdirect.com/v2/companion.jsp?ISBN=9781558607354 FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-09-28T16:29:15Z IndexTAG: 2480 TitleTAG: H3P1 I am not understanding how my math is wrong for RPUI - If Ron = 7000, Vs = 5.0, I would use equation Vt = 5.0(7000/7000+?) and I have tried 10500 and 21000 with no success. Please point me in a direction that works. Thank you. UserIdTAG: 298538 UserNameTAG: BrianBenedict CreateTimeTAG: 2012-09-28T02:28:26Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: are you solving for Vt ? its always the same if Vt "V threshold is 2V" the mosfet's on ,Ron = 15 k. i don't think you should have a Ron = 7000 anywhere.. if two 15k resistors are in parallel you will get a Ron of 7.5 k..... you need to solve for a Ron combination and R pullup " voltage divider " that will give you the minimum output voltage in the static discipline...... so think of it as the minimum voltage that you need to output from a transmitter " your mosfet circuit" that the receiver will be able to receive " detect" as a logical 0.... hope this helps... FirstChildUserIdTAG: 231206 FirstChildUserNameTAG: Bradley_B FirstChildCreateTimeTAG: 2012-09-28T02:53:10Z IndexTAG: 2481 TitleTAG: How to get correct signs in Nodal Alalysis or KCL/KVL. Foolproof method needed I'm still having a heck of a hard time getting my signs correct when I do nodal analysis (or KCL/KVL for that matter). I can follow through the examples and convince myself that the writer made the right choices, but I'm fooling myself because I can't always get the signs right when I do it from scratch. This comes up when I try to apply KCL in node analysis and there is a mixture of currents and voltages. As a concrete example, consider section 3.3.1, equation 3.19. Why is it (e1-e2)/R3 - e2/R4 + I = 0 and not (e1-e2)/R3 - e2/R4 -I = 0? BTW, I understand the basics. Current in = Current out. I can also say that when there are only voltages (eq 3.18 for example), it is quite easy. But whenever I have a mixure of voltages and currents as in 3.19, I have a 50/50 chance of messing it up. Can someone explain to me (again) how to go about looking at it? I would think that for the signs there is probably a "canned" method that will always give the correct results. Many thanks, Dave UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-28T01:41:03Z VoteTAG: 1 CoursewareTAG: Week 2 / Nodal Analysis CommentableIdTAG: 6002x_nodal_analysis_tutorial NumberOfReplyTAG: 2 FirstChildTAG: >Why is it (e1-e2)/R3 - e2/R4 + I = 0 and not (e1-e2)/R3 - e2/R4 -I = 0? BTW, I understand the basics. It's a little bit misleading in the book. It should be initially (e2-e1)/R3+e2/R4+(-I)=0. As you can see this equation written from the assumption that all currents flowing FROM the node. After transformation we have: (-(e1-e2)/R3) - (-e2/R4) + (-I) = 0 (-1) * (e1-e2/R3 - e2/R4 + I) = 0 e1-e2/R3 - e2/R4 + I = 0 FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-09-28T04:48:04Z FirstChildTAG: I think this will work: 1) establish conventions. 1a) this means choose directions for the quantities of interest consistently, and do your calculations. (For instance, currents OUT of a node treated as positive currents.) 2) If the answers come out +, then your choice of direction was ok. If the answers come out -, then the direction used in the calculation was reversed from the 'real' direction. EDIT: In Eqn. 3.19, they chose the opposite convention..current IN is positive. It still works. In your sample calculation, you chose current IN as positive for the first two terms and then current IN as negative for the last term (that was the current source). FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-09-28T18:56:04Z IndexTAG: 2482 TitleTAG: Tablets and our book Can we request that the website page where the book is displayed to be more mobile friendly? So that we can read the book in a more natural way on our e-readers and tablets? UserIdTAG: 230787 UserNameTAG: Gabriel007 CreateTimeTAG: 2012-09-27T22:56:38Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi Gabriel007! Take a look at the wiki, ashwith posted it in the wiki [Scrolling Textbook Viewer][1] :) [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/ FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-28T01:42:38Z IndexTAG: 2483 TitleTAG: Email Is edX supposed to email me when a discussion I am following has been replied? UserIdTAG: 230787 UserNameTAG: Gabriel007 CreateTimeTAG: 2012-09-27T22:54:31Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi Gabriel007! If you see in the left, you will see "**All"-> "Following"**. There you can see all the Posts that you are following. Also, I discovered that when the comment icon it is in $\color{blue}{light\ blue}$ color and not in gray, it is because **you have new comments :)** (I thought first that it was because the staff answered the post , because of the color light blue of staff when they post answers haha, but not, then I realized that was because of the new comments ). As far as I know, they don't e-mails you when someones replies to your Post. See you! Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-28T01:39:38Z SecondChildTAG: Email notifications and marking posts with staff responses are Coming Soon$^{TM}$ lol. Most of the work is done for both of those. SecondChildUserIdTAG: 323387 SecondChildUserNameTAG: ibrahimawwal SecondChildCreateTimeTAG: 2012-09-28T01:42:16Z SecondChildTAG: Cool! Thank you IbrahImawwal!! I hope that you are fine. Thank you for all your effort with edX I really apreciate it ! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-28T01:58:52Z SecondChildTAG: Haha no problem, I enjoy it. I was actually a TA for the equivalent class at Berkeley so sometimes I can't help jumping into the 6.002x forums, also just to look for feedback. I'm actually not working on this for much longer, but hopefully I can help square away some of these crucial things before I move on to other things. SecondChildUserIdTAG: 323387 SecondChildUserNameTAG: ibrahimawwal SecondChildCreateTimeTAG: 2012-09-28T02:02:28Z SecondChildTAG: Nice to meet you! and thank you for sharing your story ;). I hope that you can succeed with your future plans, I am sure you will! My best wish to you! Again, Thank You! P.D: Greetings to Berkeley's Students and Teachers! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-28T02:13:50Z SecondChildTAG: You should meet them in CS188.1x if you have time :) SecondChildUserIdTAG: 323387 SecondChildUserNameTAG: ibrahimawwal SecondChildCreateTimeTAG: 2012-09-28T02:30:41Z SecondChildTAG: Hi IbrahImawwal :). Totally yes! I would like to meet the People from Berkeley! I have heard that they have prestigious Scientists in Bekerley! a PhD in Physics who I work with told me that,;)! It has to be amazing there as MIT! I will register next year due my time: I am in the University,I also have a Undergraduate Fellowship in Research Solid State Laser and also I am Helping Ad-Honorem (free) in my University in a Investigation Group of Education since two years ago as a kind of thankful return for all that my University is giving me (Universidad Tecnológica Nacional), I really enjoy it doing that. So, next year, when I have more time, I will meet them in CS188.1x ;)! Totally sure! This fall I registered to 6.002x again because I wanted to help/colaborate here (Spanish and English), I really enjoyed 6.002x Prototype Course, I am really thankful, all the people from MITx were really kind - Piotr Mitros, Prof. Anant Agarwal, Lyla, Lauren, Kimberly, Peter, Dave, etc ;)- I felt like in my home, very comfortable! Also, I registered to 3.091x, that starts on October 15th :). Thank you for inviting me to the Course :)! I apreciate that gesture! See you! Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-28T03:16:38Z SecondChildTAG: Ibrahim: You're not going anywhere :| SecondChildUserIdTAG: 79361 SecondChildUserNameTAG: kimt SecondChildCreateTimeTAG: 2012-09-28T03:36:50Z SecondChildTAG: drat :P SecondChildUserIdTAG: 323387 SecondChildUserNameTAG: ibrahimawwal SecondChildCreateTimeTAG: 2012-09-28T05:34:53Z SecondChildTAG: Ahh! oK thanks SecondChildUserIdTAG: 230787 SecondChildUserNameTAG: Gabriel007 SecondChildCreateTimeTAG: 2012-09-30T00:19:22Z IndexTAG: 2484 TitleTAG: Why Use 3 Resitors for Lab2 !!! Help!!!! I have already submitted the lab and I would really appreciate if someone shares the scanned copy of the solution with explanation that how did they come up with answer. My email is tarana79@gmail.com UserIdTAG: 30155 UserNameTAG: electricalbest CreateTimeTAG: 2012-09-27T22:45:57Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: The lab due date hasn't passed yet. You can always continue to work on it until then. Giving solutions would not be appropriate before then. FirstChildUserIdTAG: 213386 FirstChildUserNameTAG: dmascenik FirstChildCreateTimeTAG: 2012-09-27T22:59:00Z FirstChildTAG: There is not much to explain - in case of two resistors they will form a voltage divider for each source, but with opposite proportions (that is, V1out=V1*R1/(R1+R2); V2out=V2*R2/(R1+R2) ), so Vout=a*V1+b*V2 where a+b=1. In therms of energy - all system's energy will be introduced on output node. For given proportions we have a+b=1/2+1/6=4/6 < 1 - which means we need a way to take away part of the energy from output node - and we can do it by adding a resistor between output node and ground. FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-09-27T23:17:55Z SecondChildTAG: Yakov0 - perfect mathematical explantion to the problem!! Good :) When i designed my circuit, i was writing equations to get this done. The problem would have been more interesting if they have asked Vout = 1/2V1 +1/4V2 , and Vout = 1/2V1+1/3V2. Because, once the equations are written, this is very easy to do so. And this way, we can force people to avoid trial and error method. SecondChildUserIdTAG: 823 SecondChildUserNameTAG: naveenatmit SecondChildCreateTimeTAG: 2012-09-28T06:27:09Z SecondChildTAG: What do you mean by: > all system's energy will be introduced on output node. The energy will rather be dissipated in R1 and R2. SecondChildUserIdTAG: 157610 SecondChildUserNameTAG: mradziwo SecondChildCreateTimeTAG: 2012-09-28T15:07:49Z SecondChildTAG: well, if you look at this from [Hydraulic analogy][1] point of view you will better understand what I mean - you need bypass to drop pressure at output node. [1]: http://en.wikipedia.org/wiki/Hydraulic_analogy SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-28T16:59:40Z SecondChildTAG: I totally agree with the sense that you need to use 3rd resistor and I appreciate your nice description of a+b=1. But I found energy (or rather power) analogy a bit misleading. SecondChildUserIdTAG: 157610 SecondChildUserNameTAG: mradziwo SecondChildCreateTimeTAG: 2012-09-29T09:53:00Z SecondChildTAG: It's not an analogy. Don't mix "power dissipation", "energy" and "power" - power is a rate of energy transferred through conductor. To lower this rate at output node we add bypass in order to transfer part of the energy directly to ground. SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-29T10:31:45Z FirstChildTAG: Hi electricalbest! You can see an explanation in this Post [Here][1]. :). [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_S3E2_Linear_Combinations_of_Source_Strengths/threads/506a280d036af91f000000c9 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-02T00:00:34Z IndexTAG: 2485 TitleTAG: H3P1 NOR problem i do the R_on in gate A and R_on in gate B on parallel with R_PuO too but still wrong why ? UserIdTAG: 331441 UserNameTAG: abdessamade CreateTimeTAG: 2012-09-27T20:30:59Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: *EDIT:* Post deleted by author cause he realized he was being silly :) FirstChildUserIdTAG: 324642 FirstChildUserNameTAG: StNas FirstChildCreateTimeTAG: 2012-09-27T20:45:24Z SecondChildTAG: A and B or what ? SecondChildUserIdTAG: 331441 SecondChildUserNameTAG: abdessamade SecondChildCreateTimeTAG: 2012-09-27T20:48:50Z SecondChildTAG: StNas please more explantion please SecondChildUserIdTAG: 331441 SecondChildUserNameTAG: abdessamade SecondChildCreateTimeTAG: 2012-09-27T20:49:24Z SecondChildTAG: I'm sorry! Disregard my previous answer! I didn't notice that you said "NOR". I thought you were doing the NAND! :) I'll check it and come back at you SecondChildUserIdTAG: 324642 SecondChildUserNameTAG: StNas SecondChildCreateTimeTAG: 2012-09-27T21:01:14Z SecondChildTAG: Ok. So you say you are calculating RonA and RonB in parallel and then the whole with RpuO in series. Think about this: What is the truth table for NOR? I mean when is the output of the NOR 0? SecondChildUserIdTAG: 324642 SecondChildUserNameTAG: StNas SecondChildCreateTimeTAG: 2012-09-27T21:05:44Z SecondChildTAG: if A=1 and B=1 => C=0 SecondChildUserIdTAG: 331441 SecondChildUserNameTAG: abdessamade SecondChildCreateTimeTAG: 2012-09-27T21:09:03Z SecondChildTAG: Not only. When else is the output 0? SecondChildUserIdTAG: 324642 SecondChildUserNameTAG: StNas SecondChildCreateTimeTAG: 2012-09-27T21:10:57Z SecondChildTAG: i use this formula v_ol/v_s=(n*R_on)/n*R_on + R_l in general but still wrong whith NOR gate SecondChildUserIdTAG: 331441 SecondChildUserNameTAG: abdessamade SecondChildCreateTimeTAG: 2012-09-27T21:15:40Z SecondChildTAG: Haha! Funny Edit, StNas! :). You are cool! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-27T21:16:55Z SecondChildTAG: hey Myrimit please help in my probelm thank you SecondChildUserIdTAG: 331441 SecondChildUserNameTAG: abdessamade SecondChildCreateTimeTAG: 2012-09-27T21:20:06Z SecondChildTAG: I would suggest drawing all the circuits that the NOR gate can turn into. SecondChildUserIdTAG: 213386 SecondChildUserNameTAG: dmascenik SecondChildCreateTimeTAG: 2012-09-27T21:21:42Z SecondChildTAG: The truth table for the NOR gate is: __A___B__+__C__ 0 0 | 1 0 1 | 0 1 0 | 0 1 1 | 0 So you see that you don't get 0 in the output only when both inputs are 1. You can also have 0 in the output when only one of the inputs is 1. Try to use this tip and see what equations can you come up with ;) I'm not sure I'm allowed to say more :) PS. Thanks Myrimit :D SecondChildUserIdTAG: 324642 SecondChildUserNameTAG: StNas SecondChildCreateTimeTAG: 2012-09-27T21:29:24Z SecondChildTAG: still give nothing SecondChildUserIdTAG: 331441 SecondChildUserNameTAG: abdessamade SecondChildCreateTimeTAG: 2012-09-27T21:43:58Z SecondChildTAG: Have you found RpuI and RpuA in the previous 2 questions? If yes, you will need to use a very similar thought process for this one too. When you have in either one of the inputs (A or B) a logic 1, the output will be logic 0 (according to the truth table). Please note that logic 0 in the output **is not the same** as 0V in the output. As dmascenik suggested, try drawing the equivalent circuit when A is logic 1 and B is logic 0. And calculate the Rpu0 for that circuit. Then draw the equivalent circuit when A is logic 0 and B is logic 1. Calculate the Rpu0 for that circuit too. Then draw the equivalent circuit when both A and B are logic 1. Then calculate the Rpu0 for this circuit too. Finally compare the 3 different values for Rpu0 that you calculated. The lowest value, is the right answer. SecondChildUserIdTAG: 324642 SecondChildUserNameTAG: StNas SecondChildCreateTimeTAG: 2012-09-27T22:04:07Z SecondChildTAG: i calculate RpuA and Rpul whith formula not the truth table SecondChildUserIdTAG: 331441 SecondChildUserNameTAG: abdessamade SecondChildCreateTimeTAG: 2012-09-27T22:22:50Z FirstChildTAG: for the case of the NOR gate resistor is only necessary to take a route through the combination of worst case 1-0 FirstChildUserIdTAG: 343291 FirstChildUserNameTAG: LuisEduardoH FirstChildCreateTimeTAG: 2012-09-29T23:16:56Z IndexTAG: 2486 TitleTAG: graphs can anyone explain why and how the voltage at the node (in black ) is going negative and the graphs in pink and black (points where the currents enter the diode), their starting and ending point UserIdTAG: 168496 UserNameTAG: poweltalwar CreateTimeTAG: 2012-09-27T18:44:52Z VoteTAG: 1 CoursewareTAG: Week 4 / Attenuator Tutorial CommentableIdTAG: 6002x_attenuator_t NumberOfReplyTAG: 0 IndexTAG: 2487 TitleTAG: fastrum Everything all right, today it´s my first day, i hope that i have enough time for finishing all the tasks- UserIdTAG: 508820 UserNameTAG: fastrum CreateTimeTAG: 2012-09-27T18:21:35Z VoteTAG: 1 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: :) You can do it @fastrum! FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-27T21:18:39Z IndexTAG: 2488 TitleTAG: Typo list for S6V13-S6V14 S6V13 At 0:49, "And the other one" should be "And the upper one". At 1:55, "I get those distortion" should be "I get this distortion". S6V14 At 0:21, "So what did I do?" should be "So what do I do?". At 2:19, "I will leave you "should be "I'm going to leave you". At 2:39, "be the next sequence" should be "be in the next sequence". At 2:56, "brilliant idea that it's a" should be "brilliant idea that is the". UserIdTAG: 280731 UserNameTAG: QuantumCaffeine CreateTimeTAG: 2012-09-27T17:05:40Z VoteTAG: 1 CoursewareTAG: Week 3 / Musical Demo CommentableIdTAG: 6002x_musical_demo NumberOfReplyTAG: 1 FirstChildTAG: Thank you very much! I have createad a [Post in the Wiki][1] : QuantumCaffeine's Typo lists I hope you don't mind :). See you! Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/ FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-27T21:43:48Z IndexTAG: 2489 TitleTAG: Homework 3 - Why is the current through the diodes not always 0? I don't want someone to tell me the answer - all I want is a pointer to a video that explains why the current sometimes changes. Thanks! UserIdTAG: 414306 UserNameTAG: YN300 CreateTimeTAG: 2012-09-27T14:56:03Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: please be more elaborate it thanks :) FirstChildUserIdTAG: 438395 FirstChildUserNameTAG: ali_PU1 FirstChildCreateTimeTAG: 2012-09-27T15:53:28Z FirstChildTAG: current appears when voltage across diode applied in forward dirrection FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-09-27T20:59:04Z IndexTAG: 2490 TitleTAG: I I have just join the course. But i can't find home assignments and lab work sheet. Can any one help me find where assigments and lab work sheet are? UserIdTAG: 506071 UserNameTAG: Sanamughal CreateTimeTAG: 2012-09-27T14:22:29Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Start here: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Overview/Welcome/ FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-09-27T14:49:36Z IndexTAG: 2491 TitleTAG: Will our answers to practice questions (not HOMEWORK) count during the grading process? Im talking about the questions which we will get in between lecture videos UserIdTAG: 489065 UserNameTAG: amiths CreateTimeTAG: 2012-09-27T13:22:57Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I don't think so. It's not mentioned, and it wasn't in the spring edition. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-09-27T13:31:46Z SecondChildTAG: What Myrimit said. Below. :) SecondChildUserIdTAG: 92346 SecondChildUserNameTAG: MobiusTruth SecondChildCreateTimeTAG: 2012-09-27T19:14:26Z FirstChildTAG: Hi amiths! Lecture exercises will not contribute to your final score. So, you can skip them. But it is recommendable to do them in order to get more practise :). **Interactive video sequences** Lecture style videos are presented in interactive video sequences (or sequences for short), and are posted in the Courseware section of the website. Each sequence includes a succession of short video clips and online exercises, arranged in a logical progression. Two sequences will be given each week; please take the time to watch each video and each exercise in the sequence they are provided. **Answer-check mechanisms are provided in these exercises, but they will not contribute towards your grade.** [More info read here][1] See you! Myriam. [1]: https://www.edx.org/static/content-mit-6002x/handouts/syllabus.a477535058a1.pdf FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-27T15:04:30Z IndexTAG: 2492 TitleTAG: Lab 4 tool set up Dear Sir , Please help me I found the LB4 I was shocked how to Change the X axis convert as Voltage I tried so many ways but I fails to do That i tried in sand box Please Help I am not completed the week 4 tutorials but I just look the Lab as usual Thanks Prasanth UserIdTAG: 156694 UserNameTAG: mkprasanth CreateTimeTAG: 2012-09-27T12:37:53Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi mkprasanth! Can I help you? If you double click on the Voltage Probe element you will see an option "Plot color" in Edit Propeties. If you click on the down arrow you will see an option "x-axis" :). See you! Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-27T12:50:33Z SecondChildTAG: Dear Sir , I was searching to you but I am not able to C in discussion Then I thought that you may busy any way I am trying hard to completed the course with my daily Job By the way this course again edx will do in feature ? and how much mark need for getting certificate I will I will try and get back I am sure if there is nay problem in my it will work based on your help I reach week 3 Thanks SecondChildUserIdTAG: 156694 SecondChildUserNameTAG: mkprasanth SecondChildCreateTimeTAG: 2012-09-27T12:58:18Z SecondChildTAG: Yes it is working thanks , now I have to complete the week 4 videos SecondChildUserIdTAG: 156694 SecondChildUserNameTAG: mkprasanth SecondChildCreateTimeTAG: 2012-09-27T13:11:54Z SecondChildTAG: :). You need 60% to Pass: Letter grades will be based on the following weighting: **homework 15%, labs 15%, midterm 30%, and final exam 40%.** Each of the homework and labs carries equal weight. You will need to get a total mark of 60% for a C, 70% for a B, and 87% for an A. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-27T14:52:53Z SecondChildTAG: thank you SecondChildUserIdTAG: 156694 SecondChildUserNameTAG: mkprasanth SecondChildCreateTimeTAG: 2012-09-27T16:27:23Z IndexTAG: 2493 TitleTAG: submission problem hey i'm not getting option of submit in my hw even though i have not submitted it even once UserIdTAG: 85790 UserNameTAG: vivektomar CreateTimeTAG: 2012-09-27T12:07:40Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: does it has to do something with date FirstChildUserIdTAG: 85790 FirstChildUserNameTAG: vivektomar FirstChildCreateTimeTAG: 2012-09-27T12:10:40Z SecondChildTAG: Hi vivektomar! Are you referring about Week1 assignments? If yes, remember that the deadline for this week1 assignment has already passed...and that is might you don´t see the check buttom...remember the Week2 assignments has been extended due to september 30th. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-27T12:58:47Z IndexTAG: 2494 TitleTAG: Progress Dear All, Please help me In my Progress menu Home work, Lab 11 & 12 Showing X mark what it means I am not understand Prasanth UserIdTAG: 156694 UserNameTAG: mkprasanth CreateTimeTAG: 2012-09-27T11:22:13Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Out of 12 best 10 will be counted. On other words you may skip 2 home works :) FirstChildUserIdTAG: 350697 FirstChildUserNameTAG: Khalil_Awan FirstChildCreateTimeTAG: 2012-09-27T11:36:17Z FirstChildTAG: Hi mkprasanth! How are you? I am preparing what you requested me about Week 3 in other Post! I will post it soon :). **Grading :** Letter grades will be based on the following weighting: homework 15%, labs 15%, midterm 30%, and final exam 40%. Each of the homework and labs carries equal weight. You will need to get a total mark of 60% for a C, 70% for a B, and 87% for an A. Homework and labs will be graded based on the best ten out of twelve individual grades. Therefore, **two homework assignments and two labs may be missed in total without a grade penalty.** (that is the two crosses that you see in the progress tab, by default they are in the last two, but they will shift to your low scored assignments, that then will not be considered for your final score). [You can read here for more info][1] See you! Myriam. [1]: https://www.edx.org/static/content-mit-6002x/handouts/syllabus.a477535058a1.pdf FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-27T12:25:08Z IndexTAG: 2495 TitleTAG: solution Simply omit the non linear element , and calculate the current that will flow in circuit, Imax = V/(Rs+Rp) Now, by VI graph, Imax < 1 mA so, Rstrange = 1kOhms, Reinsert the strange element and calculate current in strange element. UserIdTAG: 408534 UserNameTAG: kkashyap CreateTimeTAG: 2012-09-27T11:15:47Z VoteTAG: 1 CoursewareTAG: Week 3 / Analyzing Nonlinear Circuits Exercise CommentableIdTAG: 6002x_analyzing_nonlinear_circuits_exercise NumberOfReplyTAG: 3 FirstChildTAG: i am just joining the program yesterday and want to know if you guys will put me through. as in the past lectures. FirstChildUserIdTAG: 500023 FirstChildUserNameTAG: chukwueloka FirstChildCreateTimeTAG: 2012-09-27T13:55:34Z SecondChildTAG: I m sorry ... Why the I have to choose Imax<1 mA???/ SecondChildUserIdTAG: 115656 SecondChildUserNameTAG: rolando277 SecondChildCreateTimeTAG: 2012-10-16T23:38:10Z FirstChildTAG: perfect. thank you FirstChildUserIdTAG: 103719 FirstChildUserNameTAG: soha FirstChildCreateTimeTAG: 2012-09-28T20:18:04Z FirstChildTAG: I don't think this approach is working in every case. By omitting the strange element, you are changing the circuit completely. The current that flows in the circuit without the element can differ significantly from the current that flows with the element in it. Just imagine: By putting an element and its resistance in parallel to an existing resistor, you give the current more ways to flow through the circuit. So when there is a resistance added in parallel, the overall resistance decreases and this effect may have easily caused the whole current to increase over 1mA. In this case it worked because the current through the element was luckily 0.7mA at the end. But I don't think this way of calculating such questions works everytime. FirstChildUserIdTAG: 394599 FirstChildUserNameTAG: Albie FirstChildCreateTimeTAG: 2012-09-30T15:23:21Z IndexTAG: 2496 TitleTAG: Current Probe what is meant by plot offset of current probe?secondly how to point current probe in opposit direction? UserIdTAG: 99628 UserNameTAG: Pranjal16 CreateTimeTAG: 2012-09-27T09:05:44Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: push letter R twice FirstChildUserIdTAG: 402617 FirstChildUserNameTAG: Vl FirstChildCreateTimeTAG: 2012-09-27T11:24:08Z FirstChildTAG: Hi Pranjal16! - How to point current probe in opposit direction? Click on the element (current probe) and press "R" from keyboard. - What is meant by plot offset of current probe? It means that you will see the curve of your current plus a constant value of a current (the value that you write in the plot offset). So your curve will be shown in the graph displaced that value that you wrote as an offset, eg., suppouse that you have a current source of 5 A, you put a current probe with an offset of 5, so you will see in the graph 5A+5A=10A (curve goes up); eg, suppouse that you have a current source of 5 A, you put a current probe with an offset of -2, so you will see in the graph 5A-2A=3A (curve goes down);generally, we write 0 in the offset :). FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-27T11:52:51Z IndexTAG: 2497 TitleTAG: Lab 3 I have got the correct answer and been awarded marks (1/1) for my solution. However, the spike on my transient analysis is longer than the one suggested. Is this considered a full correct answer or a partial correct answer? Thanks. UserIdTAG: 157273 UserNameTAG: ongchihang CreateTimeTAG: 2012-09-27T07:37:09Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Some people say the spikes must be below Vout max. However, I've never had problems with the spikes, not in this lab nor the lab from the previous course. If your calculated solution for W/L is correct, then you won't have problems with the spikes. If you guess it, then it could be different .... If the grader gives the green ticks for all the questions, then it's ok. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-09-27T13:39:29Z SecondChildTAG: can u give me the hint about how u calculated W/L correctly? SecondChildUserIdTAG: 259693 SecondChildUserNameTAG: MehrozKhan SecondChildCreateTimeTAG: 2012-09-28T21:34:20Z IndexTAG: 2498 TitleTAG: (Rl/Ron)=5 if we apply (Rl/Ron)=5 --- Vout=0.833.... so,does it satisfy the conditions or not?and why?:):):) UserIdTAG: 395257 UserNameTAG: blackguitar CreateTimeTAG: 2012-09-27T06:09:18Z VoteTAG: 1 CoursewareTAG: Week 3 / Switch Resistor Model of a MOSFET CommentableIdTAG: 6002x_switch_resistor_model NumberOfReplyTAG: 1 FirstChildTAG: It will not satisfy the static discipline. By plotting the transfer curve you will see that it will violate the static discipline at Vin >= VIH. I mean when Vin is greater than or equal to VIH the graph enters the forbidden region (between VOL and VOH). In order to satisfy the discipline, the Vout should be less than or equal to VOL = 0.5V at (vin >= VIH) where VIH=4.1 FirstChildUserIdTAG: 305491 FirstChildUserNameTAG: lefam FirstChildCreateTimeTAG: 2012-09-27T18:27:03Z SecondChildTAG: thanks a lot SecondChildUserIdTAG: 395257 SecondChildUserNameTAG: blackguitar SecondChildCreateTimeTAG: 2012-09-28T01:40:04Z IndexTAG: 2499 TitleTAG: online textbook not loading The online textbook is currently refusing to load for me. It has worked fine up until now. Is it my system or is there a problem at edx. UserIdTAG: 226890 UserNameTAG: brax1961 CreateTimeTAG: 2012-09-27T06:09:00Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Thank you. FirstChildUserIdTAG: 226890 FirstChildUserNameTAG: brax1961 FirstChildCreateTimeTAG: 2012-09-27T12:23:19Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-27T12:46:43Z FirstChildTAG: Now it is working again :)! You can also see that ashwith was posted this in the wiki [scrolling Textbook Viewer][1]. [1]: https://6002x.mitx.mit.edu/static/contrib/xbook.html FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-27T11:20:38Z IndexTAG: 2500 TitleTAG: Small mistake Hi there. In the upper right corner, at about 0:39 (at 1.25x) the first eqn should be iD=... instead of vD... UserIdTAG: 378522 UserNameTAG: Alejo_Velasquez CreateTimeTAG: 2012-09-27T03:57:53Z VoteTAG: 1 CoursewareTAG: Week 4 / Small Signal Model CommentableIdTAG: 6002x_small_signal NumberOfReplyTAG: 0 IndexTAG: 2501 TitleTAG: queries about Q1 I dnt understand the question how to solve.................plz help me ... UserIdTAG: 282892 UserNameTAG: wiky CreateTimeTAG: 2012-09-26T23:04:48Z VoteTAG: 1 CoursewareTAG: Week 4 / Linearization Exercise 1 CommentableIdTAG: 6002x_linearization_ex_1 NumberOfReplyTAG: 1 FirstChildTAG: check: week 4 tutorials-> worked problems->small signal problem. problem 4.2 of text is being solved, which is very similar to this. watch it! it is very enlightening!!! FirstChildUserIdTAG: 301962 FirstChildUserNameTAG: labrinim FirstChildCreateTimeTAG: 2012-09-27T14:56:17Z SecondChildTAG: in numerical analysis exists a method called fixed point. This method helps us to solve non-linear equations. you have to put F(x) = x, like: by node method -> (Vi-Va)/R = ia = 20(1-e^(-va/5)) now you do f(x) = x and there is a hint using log. 1*(Va-Vi)/20= e^(-Va/5) applying ln ln(1+(Va-Vi)/20)=-Va/5 Va = -5ln(1+(Va-Vi)/20) -> Va = f(Va). Now choose one value for Va and iterate. This method converges if F'(X) < L < 1 and if your choice is close to the answer SecondChildUserIdTAG: 110214 SecondChildUserNameTAG: jeffsjunior SecondChildCreateTimeTAG: 2012-09-30T02:58:26Z SecondChildTAG: Thank you labrinim & jeffjunior. I am just adding one more point in addition to your comment. Consider Va=1, it coverges at 4th iteration. SecondChildUserIdTAG: 284006 SecondChildUserNameTAG: Alagan SecondChildCreateTimeTAG: 2012-09-30T05:16:54Z IndexTAG: 2502 TitleTAG: Lab 3 typo? In the Lab 3 problem description, it states "In the schematic tool, the mosfet model has Vth=0.5V,.." Is Vth a typo? I doubt it's referring to a Thevenin voltage. If not, what is it? What is this Vth? UserIdTAG: 141108 UserNameTAG: rl777 CreateTimeTAG: 2012-09-26T22:25:25Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: it's "threshold" FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-09-26T22:59:33Z SecondChildTAG: Ah, yes. Thanks! SecondChildUserIdTAG: 141108 SecondChildUserNameTAG: rl777 SecondChildCreateTimeTAG: 2012-09-26T23:02:39Z IndexTAG: 2503 TitleTAG: stuck with lab 3 i don't know how many R_on i need please help UserIdTAG: 331441 UserNameTAG: abdessamade CreateTimeTAG: 2012-09-26T20:40:35Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: just three R's u will need... just add one more R lyke the example given and calculate values according to what is needed using superposition FirstChildUserIdTAG: 171199 FirstChildUserNameTAG: Dhananjay01 FirstChildCreateTimeTAG: 2012-09-27T02:39:28Z FirstChildTAG: jst 3 gates ...... FirstChildUserIdTAG: 282892 FirstChildUserNameTAG: wiky FirstChildCreateTimeTAG: 2012-09-26T20:44:43Z SecondChildTAG: i connect just in Z SecondChildUserIdTAG: 331441 SecondChildUserNameTAG: abdessamade SecondChildCreateTimeTAG: 2012-09-26T20:45:31Z IndexTAG: 2504 TitleTAG: A couple of general information questions. I wondered whether the course, 6.002x, will be repeatable if it is offered again. Also, by way of suggestion, I think there's one tool, a graph generator, that might be a valuable addition to all the impressive course aids. Thanks to the edX staff always. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-26T20:10:22Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 1 FirstChildTAG: Yes, the course is repeatable! There are several people taking the course who completed it last Spring and are back helping and taking it again. Thanks for the suggestion, also. FirstChildUserIdTAG: 376895 FirstChildUserNameTAG: alecgg FirstChildCreateTimeTAG: 2012-09-27T06:02:14Z IndexTAG: 2505 TitleTAG: Why vista crashed Windows had no sense of the EECS playground back then, as they just really did not know 1.99 can easily jump above 2 by even bringing a lighter near the wire......vista guys, bunch of bozos, who made me spend a lot of money on their software, while I could have always used UBUNTU! FYI, I don't know how far I will last when calculus comes through, but seriously each and every second of this journey is just getting better than before! UserIdTAG: 169784 UserNameTAG: thewiredbear CreateTimeTAG: 2012-09-26T19:01:03Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 2506 TitleTAG: LAB 3 I couldn't do this part of the homeowrk. I can make A e B right, but C is wrong. I can make A+B, but I can't make (A+B)*C. How do I do that? UserIdTAG: 51667 UserNameTAG: Assuncao CreateTimeTAG: 2012-09-26T18:08:13Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: how many the pullup resistor you are used ? FirstChildUserIdTAG: 331441 FirstChildUserNameTAG: abdessamade FirstChildCreateTimeTAG: 2012-09-26T18:18:45Z SecondChildTAG: You dont need any additional pullup resistors. SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-26T18:34:02Z SecondChildTAG: i have make (A+B)*C but not getting correct output because of w/L ...! i cant understand its significance . RDS on should be very less then 10k resistor for proper voltage levels.. when we change w/l what thing is change and how it effects the circuit kindly guide me thanks :) SecondChildUserIdTAG: 438395 SecondChildUserNameTAG: ali_PU1 SecondChildCreateTimeTAG: 2012-09-26T19:34:03Z SecondChildTAG: Treat it as a divider to 26.5k of Ron SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-26T19:42:51Z SecondChildTAG: i don't how to connect R_on i need 1 R_on ?? SecondChildUserIdTAG: 331441 SecondChildUserNameTAG: abdessamade SecondChildCreateTimeTAG: 2012-09-26T19:50:31Z SecondChildTAG: Ron is the internal resistor from mosfet. Read the textbook about the sr-model and read chapter 6.7 about the W/L value SecondChildUserIdTAG: 389333 SecondChildUserNameTAG: juergen SecondChildCreateTimeTAG: 2012-09-26T19:54:53Z SecondChildTAG: Some guidance which might be useful: To find R_on for the 3 MOSFETs, I found an algebraic expression for the output in the 3 cases where Z = 0. The rest was algebra and logic... Hope this helps :) (and hope I'm not revealing too much haha) SecondChildUserIdTAG: 56851 SecondChildUserNameTAG: Alonso_sh SecondChildCreateTimeTAG: 2012-09-27T05:35:43Z FirstChildTAG: checkout the example in lectures...it is similar to that with little modofication FirstChildUserIdTAG: 171199 FirstChildUserNameTAG: Dhananjay01 FirstChildCreateTimeTAG: 2012-09-27T02:41:07Z SecondChildTAG: what number of lectures SecondChildUserIdTAG: 331441 SecondChildUserNameTAG: abdessamade SecondChildCreateTimeTAG: 2012-09-27T21:30:45Z SecondChildTAG: please SecondChildUserIdTAG: 331441 SecondChildUserNameTAG: abdessamade SecondChildCreateTimeTAG: 2012-09-27T21:30:57Z IndexTAG: 2507 TitleTAG: Glitches in Lab3 I don't understand why there are glitches. The signal transmit time for A and B is equal. Can anyone explain me? UserIdTAG: 389333 UserNameTAG: juergen CreateTimeTAG: 2012-09-26T16:32:37Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Because signal edge has slope. Signal edge has slope because MOSFET's gate has capacitance and it takes time to charge/discharge gate (and open/close transistor) FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-09-26T17:25:14Z SecondChildTAG: the figure is same and result is less than 0.25 but still got cross reason SecondChildUserIdTAG: 164689 SecondChildUserNameTAG: muhammadfaizan SecondChildCreateTimeTAG: 2012-09-26T17:52:40Z SecondChildTAG: how many Glitches use in the lab 3 please help SecondChildUserIdTAG: 331441 SecondChildUserNameTAG: abdessamade SecondChildCreateTimeTAG: 2012-09-26T17:59:38Z IndexTAG: 2508 TitleTAG: H3P4 Did anyone get the answer to the "current flowing through the diodes" question? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-26T13:05:52Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: i have same problem FirstChildUserIdTAG: 331441 FirstChildUserNameTAG: abdessamade FirstChildCreateTimeTAG: 2012-09-26T13:43:34Z SecondChildTAG: Did you solve H3P4 ???? Please can you help me???? SecondChildUserIdTAG: 106219 SecondChildUserNameTAG: veereshpatil SecondChildCreateTimeTAG: 2012-09-26T15:39:55Z FirstChildTAG: Apply a piecewise method. Diod is short circuit when its voltage is positive and open circuit when its voltage is negative. Dont forgot about additional voltage sources in series - they biasing zero point voltage. FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-09-26T15:47:22Z SecondChildTAG: still wrong :( SecondChildUserIdTAG: 331441 SecondChildUserNameTAG: abdessamade SecondChildCreateTimeTAG: 2012-09-26T16:15:53Z SecondChildTAG: Apply node method SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-26T17:05:16Z SecondChildTAG: still too wrong but thank you SecondChildUserIdTAG: 331441 SecondChildUserNameTAG: abdessamade SecondChildCreateTimeTAG: 2012-09-26T17:38:05Z SecondChildTAG: You should got 3 voltage intervals - two of them is your point of interest. SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-26T17:48:11Z SecondChildTAG: show me an example please for more understande SecondChildUserIdTAG: 331441 SecondChildUserNameTAG: abdessamade SecondChildCreateTimeTAG: 2012-09-26T19:46:18Z SecondChildTAG: On diod you have current flow when it has voltage difference in forward direction on its terminals. For each of the diods you have static voltage source for one terminal and sawtooth voltage generator with amplitude 10v for second terminal. SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-26T20:09:38Z SecondChildTAG: yes i do this id=(Vs-V1)/R but still wrong i d'ont understande why ? SecondChildUserIdTAG: 331441 SecondChildUserNameTAG: abdessamade SecondChildCreateTimeTAG: 2012-09-26T20:16:12Z SecondChildTAG: You forgot second resistor's branch. SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-26T20:37:55Z SecondChildTAG: still wrong id=2*(Vs-V1)/R please help SecondChildUserIdTAG: 331441 SecondChildUserNameTAG: abdessamade SecondChildCreateTimeTAG: 2012-09-26T20:57:03Z SecondChildTAG: It's not correct. Did you find subcircuit for every of voltage range? SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-26T21:07:41Z SecondChildTAG: Sorry, I cant write exact solution - it's against an honor code. SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-26T21:09:47Z SecondChildTAG: not write exact solution but give me just hint please thank you SecondChildUserIdTAG: 331441 SecondChildUserNameTAG: abdessamade SecondChildCreateTimeTAG: 2012-09-26T21:13:06Z SecondChildTAG: I gave you a hint - find 3 subcircuit corresponding to different voltage ranges. What voltage ranges did you find? SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-26T21:50:28Z SecondChildTAG: i have problem with diods so thank you i take more for your time im sorry YakovO SecondChildUserIdTAG: 331441 SecondChildUserNameTAG: abdessamade SecondChildCreateTimeTAG: 2012-09-26T21:54:51Z SecondChildTAG: try to use this circuit simulator: http://www.falstad.com/circuit/ SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-26T22:12:46Z SecondChildTAG: I have tried that the simulater I mean I found Id=0.01pA but no check SecondChildUserIdTAG: 331441 SecondChildUserNameTAG: abdessamade SecondChildCreateTimeTAG: 2012-09-27T15:27:52Z SecondChildTAG: it used not ideal diods by default to make it closer to ideal right click on diod, select edit, set parameter to minimal working value (if I remember correctly it's 150n) SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-27T19:45:53Z SecondChildTAG: link with working model: http://www.falstad.com/circuit/#%24+1+5.0E-6+10.20027730826997+67+5.0+50%0Ar+176+144+272+144+0+8200.0%0Ad+272+144+272+224+1+1.5000000000000002E-7%0Ad+352+224+352+144+1+1.5000000000000002E-7%0Av+272+272+272+224+0+0+40.0+2.75+0.0+0.0+0.5%0Av+352+224+352+272+0+0+40.0+2.25+0.0+0.0+0.5%0Ar+432+144+432+272+0+8200.0%0Aw+272+144+352+144+0%0Aw+352+144+432+144+0%0Aw+272+272+352+272+0%0Aw+352+272+432+272+0%0Av+176+144+176+272+0+3+40.0+10.0+0.0+0.0+0.5%0Aw+176+272+272+272+0%0Ag+176+272+176+304+0%0AO+432+144+496+144+0%0Ao+13+64+0+290+5.0+9.765625E-5+0+-1%0A SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-27T19:56:48Z SecondChildTAG: 150m not 150n SecondChildUserIdTAG: 331441 SecondChildUserNameTAG: abdessamade SecondChildCreateTimeTAG: 2012-09-27T19:57:37Z SecondChildTAG: YacovO is still wrong i gonna be crazy SecondChildUserIdTAG: 331441 SecondChildUserNameTAG: abdessamade SecondChildCreateTimeTAG: 2012-09-27T20:00:11Z SecondChildTAG: with scops: http://www.falstad.com/circuit/#%24+1+5.0E-6+10.20027730826997+67+5.0+50%0Ar+176+144+272+144+0+8200.0%0Ad+272+144+272+224+1+1.5000000000000002E-7%0Ad+352+224+352+144+1+1.5000000000000002E-7%0Av+272+272+272+224+0+0+40.0+2.75+0.0+0.0+0.5%0Av+352+224+352+272+0+0+40.0+2.25+0.0+0.0+0.5%0Ar+432+144+432+272+0+8200.0%0Aw+272+144+352+144+0%0Aw+352+144+432+144+0%0Aw+272+272+352+272+0%0Aw+352+272+432+272+0%0Av+176+144+176+272+0+3+40.0+10.0+0.0+0.0+0.5%0Aw+176+272+272+272+0%0Ag+176+272+176+304+0%0AO+432+144+496+144+0%0Ao+13+64+0+290+5.0+9.765625E-5+0+-1%0Ao+1+64+0+33+5.0+7.8125E-4+1+-1%0Ao+2+64+0+33+4.676805239458889+7.307508186654515E-4+2+-1%0A SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-27T20:01:33Z SecondChildTAG: YakovO i do all of this but still wrong i don't understand why ? SecondChildUserIdTAG: 331441 SecondChildUserNameTAG: abdessamade SecondChildCreateTimeTAG: 2012-09-27T20:09:44Z SecondChildTAG: i=0.01 pA SecondChildUserIdTAG: 331441 SecondChildUserNameTAG: abdessamade SecondChildCreateTimeTAG: 2012-09-27T20:11:03Z SecondChildTAG: thank you YakovO i understand now SecondChildUserIdTAG: 331441 SecondChildUserNameTAG: abdessamade SecondChildCreateTimeTAG: 2012-09-27T20:18:52Z SecondChildTAG: abdessamade, you need to understand how it works and find the analytic solution - else it will be useless SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-27T20:35:28Z SecondChildTAG: YakovO that is the problem i try many time to find it analytic solution but still wrong please if you want send me the analytic solution in e-mail please SecondChildUserIdTAG: 331441 SecondChildUserNameTAG: abdessamade SecondChildCreateTimeTAG: 2012-09-27T21:03:00Z SecondChildTAG: can't do it before due date, sorry SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-27T21:45:58Z SecondChildTAG: oki thank you after this sunday please if you want SecondChildUserIdTAG: 331441 SecondChildUserNameTAG: abdessamade SecondChildCreateTimeTAG: 2012-09-27T21:52:33Z SecondChildTAG: is this diod ideal SecondChildUserIdTAG: 374512 SecondChildUserNameTAG: arsen55591 SecondChildCreateTimeTAG: 2012-09-28T20:34:28Z SecondChildTAG: YakovO Have you simulated this diode limiter circuit in the circuit sandbox? I find it doesn't work, as the voltage across V (across the right most resistor) never gets below about 2.3V, which is a contradiction to the maximum negative voltage of -3V, required by the answer. I am using a 1KHz triangular wave, and simulate for 5ms. The diodes were set as ideal diodes. Thoughts? SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-09-29T16:23:59Z SecondChildTAG: Double click on diode and select ideal type. In case of nonideal diode it will work as a resistor. SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-29T17:33:25Z SecondChildTAG: ...and maximum negative voltage is not -3V SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-29T17:40:29Z SecondChildTAG: about rharris case , they change the values for each student , his max voltage could be -3v . about the simulation it must be right , because in the real world or in simulation sofwares even if i use a ideal device , it is a propriety of the diode to cause a litlle drop on the forward voltage , usually 0.7v , SecondChildUserIdTAG: 402850 SecondChildUserNameTAG: y1111 SecondChildCreateTimeTAG: 2012-09-30T19:00:59Z FirstChildTAG: KEEP IN MIND ABOUT RESISTANCE AND CURRENT THROUGH BOTH RESISTANCES FirstChildUserIdTAG: 230943 FirstChildUserNameTAG: samsung FirstChildCreateTimeTAG: 2012-09-26T17:09:00Z IndexTAG: 2509 TitleTAG: H5P2 parser fails Quite sure the answer is correct. Use the formula successfully to enter answers into subsequent questions! H5P2.1: No problem H5P2.2: No problem H5P2.3: Parsing errors, java crashes, math error H5P2.4: Using above formula gives correct answer! Checked my math with an algebraic solver, so it appears my answer is correct, including selecting the right root: iDS=0 when vIN=VT. Please correct Where can I post my answer to have someone check this issue? Solved: Tried multiple times with "Math parsing error". Then it was accepted UserIdTAG: 142857 UserNameTAG: viking2 CreateTimeTAG: 2012-09-26T11:25:28Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Just omit the left part of the equation with the "=" sign. I had the same problem and posted in https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50620fe3fb2bce2b00000040 FirstChildUserIdTAG: 39980 FirstChildUserNameTAG: MarinoJr FirstChildCreateTimeTAG: 2012-09-26T11:53:55Z SecondChildTAG: If this is not the case, it could also be the case that the formula was typed in incorrectly. Could you please send me a screenshot of the H5P2.3 error at kimth@mitx.mit.edu? SecondChildUserIdTAG: 79361 SecondChildUserNameTAG: kimt SecondChildCreateTimeTAG: 2012-09-26T13:46:57Z IndexTAG: 2510 TitleTAG: Linearity in the G-matrix Fig. 2.(a) Here you can check that if every resistor is increased in the same value (1, 2,..., 10) the voltages at all the points will remain constant, whereas the current from the source decrease proportionally. Fig. 2. (b) In a dual way if every resistor is increased equally the voltages increase in the same value, whereas the voltage in the source remain constant (Just check it to have some fun!) UserIdTAG: 155605 UserNameTAG: JaimeLopezCerezo CreateTimeTAG: 2012-09-26T10:10:30Z VoteTAG: 1 CoursewareTAG: Week 2 / Superposition experiment CommentableIdTAG: 6002x_Superposition_experiment NumberOfReplyTAG: 0 IndexTAG: 2511 TitleTAG: vA=-0.66 Using newtons iterative formula i got vA=-0.66 volts I can't understand what's wrong? UserIdTAG: 341358 UserNameTAG: akbar13 CreateTimeTAG: 2012-09-26T09:09:10Z VoteTAG: 1 CoursewareTAG: Week 4 / Linearization Exercise 1 CommentableIdTAG: 6002x_linearization_ex_1 NumberOfReplyTAG: 1 FirstChildTAG: That answer can't be right. If Va = -0.66 Volts, then the current ia = -1.41A. If you now calculate the voltage at the element using KVL or KCL you will find that it is Va = 7.82V, which is different from the original assumption! Check the equation you used in your iterative method, that might be the problem FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-26T13:35:25Z SecondChildTAG: for the iterative method I founded va= 5log20 - 5log ( va+15) va=14.98 - 5log ( va+15) by iteration va= 1.0890... 1.893... 1.0892 ...seems correct SecondChildUserIdTAG: 373752 SecondChildUserNameTAG: Croak SecondChildCreateTimeTAG: 2012-09-27T21:47:17Z SecondChildTAG: Pas clair!! va= 5log20 - 5log ( va+15) va=14.98 - 5log ( va+15) by iteration: va= 1.0890... 1.893... 1.0892 ...seems correct SecondChildUserIdTAG: 373752 SecondChildUserNameTAG: Croak SecondChildCreateTimeTAG: 2012-09-27T21:49:34Z SecondChildTAG: I used excel to converge the resulting equation using KVC (Vi-Va)/2=10(1-e^(-Va/5)) some math around ending in: Va=-5*ln(3/4 + Va/20) SecondChildUserIdTAG: 164471 SecondChildUserNameTAG: BillNieves SecondChildCreateTimeTAG: 2012-09-29T22:17:32Z SecondChildTAG: how come 5log20= 14.98 SecondChildUserIdTAG: 333877 SecondChildUserNameTAG: gaurab SecondChildCreateTimeTAG: 2012-10-02T11:45:03Z SecondChildTAG: in which calculator SecondChildUserIdTAG: 333877 SecondChildUserNameTAG: gaurab SecondChildCreateTimeTAG: 2012-10-02T11:45:19Z IndexTAG: 2512 TitleTAG: untill when are videos and resources available..?? Can anyone answer, untill when will the videos labs homeworks books etc would be available to us even after the course is over these are some really great resources and anyone would like to go through these quite a many times whenever needed to refresh their memory.... UserIdTAG: 210992 UserNameTAG: neerajnatu CreateTimeTAG: 2012-09-26T08:59:15Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 2513 TitleTAG: Lab 4 K value I'm having problems solving computen the K value in lab 4. I've a right value for RON and for Vt, and my transient analysis is quite similar to Figure 2. Nevertheless, when I compute the constant K from equation 7.8, I get a wrong value. How can I solve it? UserIdTAG: 314117 UserNameTAG: thuargon CreateTimeTAG: 2012-09-26T08:17:13Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: If you have the correct Vt, and you are using the appropriate pair of Vgs and Ids it should work. Make sure you are using the equation correctly, for example, don't miss the square in the $(v_{gs} - V_t)^2$ term. Another potential problems are the units, so make sure that you are using volts and amperes for all variables. FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-26T13:25:55Z FirstChildTAG: Thanks a lot jelizon The problem was the units of K. Maybe should be written somewhere in the exercise that it's in A/V^2 and I didn't find it Thanks again FirstChildUserIdTAG: 314117 FirstChildUserNameTAG: thuargon FirstChildCreateTimeTAG: 2012-09-26T14:05:36Z IndexTAG: 2514 TitleTAG: making week tutorial video available to download Hello everyone, i wanna thanks edx for making available the lecture videos to download . Due to world events in my country youtube is banned too. But my request to edx is that week tutorials videos are also very much important to complete homework. So please can you make tutorials video available to download directly as you did for lecture videos? Thank you UserIdTAG: 152551 UserNameTAG: sshakir CreateTimeTAG: 2012-09-26T06:52:54Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: They already are... See the link and message at the top of the home page FirstChildUserIdTAG: 385677 FirstChildUserNameTAG: Michaelc1 FirstChildCreateTimeTAG: 2012-09-26T07:43:35Z SecondChildTAG: this is great! SecondChildUserIdTAG: 239507 SecondChildUserNameTAG: murkennethie SecondChildCreateTimeTAG: 2012-09-26T08:37:58Z IndexTAG: 2515 TitleTAG: NEW STUDENT sorry, im just in this course, and i want to know if i can do the homework until the 3week? , or what can i do (sorry for the bad English) UserIdTAG: 496967 UserNameTAG: SSernaP CreateTimeTAG: 2012-09-26T02:12:12Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: It's too late to submit the Week 1 homework and lab as they were due on 16 September. However, the deadline for week 2 was extended, so week 2 and 3 are both due this Sunday (30 September) at midnight, which still leaves time to get them done. FirstChildUserIdTAG: 142402 FirstChildUserNameTAG: aaronrod FirstChildCreateTimeTAG: 2012-09-26T03:51:34Z FirstChildTAG: **SSernaP**: Hello and Welcome! Fortunately for you, **aaronrod** is correct in saying that Homework Week 2 and Lab Week 2 are extended and are now due on Sunday September 30th at midnight that Sunday (along with Homework Week 3 and Lab Week 3). Also fortunately for you, this course's grading policy includes a feature where your two lowest Homework grades and your two lowest Lab grades are dissmissed (e.g. they do not count towards your final average mark). So say for example, in the Homework category you get 0% for Week 1 because you miss it, and you only get 60% for Week 2 because you are catching up on the material, if you continue on to do better and get 100% for each of Weeks 3 through 12, you will still get 100% in the Homework category. If you miss **more than two** Homeworks, you will be penalized, however, so stick in there! FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-09-26T04:30:55Z SecondChildTAG: hi,i recently joined this,i missed two week's homework and labs,what to do? SecondChildUserIdTAG: 519516 SecondChildUserNameTAG: jasmine03 SecondChildCreateTimeTAG: 2012-09-30T15:50:12Z IndexTAG: 2516 TitleTAG: Need help on the last 2 problem in S6E1 I tried the trial and error method, but my range of voltage just keep expanding. I let the current through the strange object to be 1000*(voltage through the object(vD)) when 0 <= vD <= 1, and 2000*vD-1 when vD > 1. I shift the equation every time the value changes, but that doesn't seems to be the solution. Can anybody tell me how to solve this one? Is there a trick on the trial and error method? I think it's very important on how to choose your initial value. So that the value can narrow down to one single value. Any ideas what I should be looking out for, when I use this method? UserIdTAG: 98259 UserNameTAG: rogerloh0 CreateTimeTAG: 2012-09-26T00:00:25Z VoteTAG: 1 CoursewareTAG: Week 3 / Analyzing Nonlinear Circuits Exercise CommentableIdTAG: 6002x_analyzing_nonlinear_circuits_exercise NumberOfReplyTAG: 1 FirstChildTAG: I did the thevenin equivalent, which gave me a better sense of current so I could choose the right value FirstChildUserIdTAG: 440714 FirstChildUserNameTAG: mcktim FirstChildCreateTimeTAG: 2012-09-26T00:26:56Z SecondChildTAG: Oh, thank you. I think I got that sense of current too now. SecondChildUserIdTAG: 98259 SecondChildUserNameTAG: rogerloh0 SecondChildCreateTimeTAG: 2012-09-26T15:08:19Z SecondChildTAG: Let me add something...the thevenin helps when you think that RL could be zero, then you see that the max current in such circuit is 1 mA, from that assumption you can see the graph for the range 0 to 1 mA in this case the slope, meaning the resistance, will be 1000 ohms...and so on.....by the way thanks for your input help me a lot..... SecondChildUserIdTAG: 164471 SecondChildUserNameTAG: BillNieves SecondChildCreateTimeTAG: 2012-09-27T01:21:03Z IndexTAG: 2517 TitleTAG: Incremental change Not sure why I'm getting my answer wrong. I know that 5+2%=5.1 (delta vI) and now I'm getting for vA 1.112 (delta vA) so my ratio is 0.218 I don't know what's wrong since I have my first answer right (my ecuations should be right too). UserIdTAG: 23183 UserNameTAG: A1 CreateTimeTAG: 2012-09-25T23:33:02Z VoteTAG: 1 CoursewareTAG: Week 4 / Linearization Exercise 1 CommentableIdTAG: 6002x_linearization_ex_1 NumberOfReplyTAG: 2 FirstChildTAG: I was making the same mistake. So I looked at the Week 4 tutorials videos and see this: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_4/wk4t3/ In 6:39 I viewed what is wrong. Cause isn't 1.112/5.1. The correct is: (1.112 - Va initial)/(5.1 - Vi initial) FirstChildUserIdTAG: 300169 FirstChildUserNameTAG: castrogfx FirstChildCreateTimeTAG: 2012-09-27T22:17:53Z FirstChildTAG: my ratio is 0.237 not far ... with the slope= 1.61 = 1/ 0.621 dva=(5x0.02)x 0.621/2.621 = 0.0237 and 0.0237/0.1 =0.237 FirstChildUserIdTAG: 373752 FirstChildUserNameTAG: Croak FirstChildCreateTimeTAG: 2012-09-27T22:01:17Z IndexTAG: 2518 TitleTAG: formula to calculat v_min and v_max what is the formula please i did all the voltage divider and still nothing please a little help here and thank you everyone UserIdTAG: 331441 UserNameTAG: abdessamade CreateTimeTAG: 2012-09-25T22:00:21Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 2519 TitleTAG: Seriously, this is amazing Thanks, to EDx, I never thought being a high school student, I would get all this stuff....amazing work by everyone here, I am though already studying for the calculus part, which is soon to come.........because that might get tricky!...... #TNDO guys! UserIdTAG: 169784 UserNameTAG: thewiredbear CreateTimeTAG: 2012-09-25T20:44:03Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: If you are a high school student, and you get 70%-100% in this course, you should be especially proud of yourself. I'd say the calculus will become more important later on, and in later EECS courses. In basic circuit analysis and in digital combinatorial logic analysis, knowledge of algebra will suffice. When it comes to signal analysis that is when calculus plays a greater role. Also when we start AC analysis complex numbers will make an appearance and become important. By the time you get to analog communications system analysis, Bode plots, Fourier transforms, and the time vs. frequency domains, you will be pulling your hair out because there is so much calculus :) But these latter courses are junior-level university electrical engineering material, so you're getting a very good head-start and advantage over your peers! Mark in New Jersey USA FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-09-26T04:39:38Z SecondChildTAG: Thanks Mark, well I seriously hope I get till 70%-100%, but then again, I have an Olympiad, an international graphics championship, school exams, 3 other edx courses I have signed up for, all coming head on to me, I worry how will I study calculus, we have hardly done any in school till now! SecondChildUserIdTAG: 169784 SecondChildUserNameTAG: thewiredbear SecondChildCreateTimeTAG: 2012-09-27T21:23:01Z FirstChildTAG: All along i have been trying to get to you but there was a problem of internet cable cut due to the fact they are repairing Ugandan roads but i hope to be with you all the time. Missed the homeworrk and notes. Though it is the slowest network i hope to be around with you. Thank you once Again. Kenneth FirstChildUserIdTAG: 239507 FirstChildUserNameTAG: murkennethie FirstChildCreateTimeTAG: 2012-09-26T07:25:55Z SecondChildTAG: I am always here for your, brother!...... comment again, and I shall also give you my e-mail! SecondChildUserIdTAG: 169784 SecondChildUserNameTAG: thewiredbear SecondChildCreateTimeTAG: 2012-09-27T21:20:15Z IndexTAG: 2520 TitleTAG: H5P2 SOURCE FOLLOWER LARGE SIGNAL I'm experiencing troubles to check if the answer is correct. It shows, for instance, "Invalid input: iDS not permitted in answer" when I type "iDS=K/2*(vIN-vOUT-VT)^2" in the first text box of the exercise. I tried all the combinations using upper and lower cases and when I think it's solved, another error appears in vOUT and so on and so forth. Any clue? UserIdTAG: 39980 UserNameTAG: MarinoJr CreateTimeTAG: 2012-09-25T20:11:15Z VoteTAG: 1 CoursewareTAG: Week 5 / MOSFET Amplifier Exercise 1 CommentableIdTAG: 6002x_mosfet_amp_e1 NumberOfReplyTAG: 2 FirstChildTAG: Just omit the left part of the equation with the "=" sign. FirstChildUserIdTAG: 194098 FirstChildUserNameTAG: ZhekaS FirstChildCreateTimeTAG: 2012-09-25T20:34:37Z SecondChildTAG: I think that should be explicit in the exercise instructions. SecondChildUserIdTAG: 39980 SecondChildUserNameTAG: MarinoJr SecondChildCreateTimeTAG: 2012-09-26T11:55:49Z FirstChildTAG: remove iDS= from your answer FirstChildUserIdTAG: 412634 FirstChildUserNameTAG: galal FirstChildCreateTimeTAG: 2012-10-10T02:36:00Z IndexTAG: 2521 TitleTAG: lab 3 How can I use three mosfet switches ( as it was said in the hint ) when for the OR gate I need two, and for the NAND Gate two more.. can someone give a little explanation.. I'm stuck Thanks in advance. UserIdTAG: 206648 UserNameTAG: TsvetanGeorgiev CreateTimeTAG: 2012-09-25T16:51:57Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I did it.. They are equal as I thought :) FirstChildUserIdTAG: 206648 FirstChildUserNameTAG: TsvetanGeorgiev FirstChildCreateTimeTAG: 2012-09-25T19:43:25Z FirstChildTAG: When you create the NAND gate....you need two switches. But...look for a way to use the output of your other gate (which also uses two switches) as one switch of the NAND gate. FirstChildUserIdTAG: 214701 FirstChildUserNameTAG: alanmatson FirstChildCreateTimeTAG: 2012-09-25T19:36:08Z SecondChildTAG: I figured this out. Thanks. But I'm having trouble with the W/L ratio.. I did the algebra ( from pg 310 in the textbook ) and I come up with W/l>29.. so I put everywhere 30 butt got it wrong. Are the three mosfets with equal ratio ? SecondChildUserIdTAG: 206648 SecondChildUserNameTAG: TsvetanGeorgiev SecondChildCreateTimeTAG: 2012-09-25T19:39:40Z IndexTAG: 2522 TitleTAG: What is RL for? In a nand gate or inverter, ect. what is the purpose of the load resistor between the voltage supply and the mosfet? UserIdTAG: 219780 UserNameTAG: ReconIII CreateTimeTAG: 2012-09-25T14:46:15Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: The load resistor allows the supply voltage to actually drop so that we get inversion. Without it, Vout would simply be the supply voltage in our typical inverter setup. When Vin is high, the MOSFET is on and effectively has a small resistance, so most of the supply voltage is dropped over RL, and Vout is low. When Vin is low, the MOSFET is off and has a HUGE resistance, so most of the voltage drop occurs over the MOSFET and Vout is high. It's conceptually a voltage divider. FirstChildUserIdTAG: 376895 FirstChildUserNameTAG: alecgg FirstChildCreateTimeTAG: 2012-09-25T15:28:41Z SecondChildTAG: Ah ok, thanks for the info! SecondChildUserIdTAG: 219780 SecondChildUserNameTAG: ReconIII SecondChildCreateTimeTAG: 2012-09-25T15:34:39Z IndexTAG: 2523 TitleTAG: Error with "Check"? (Week 3 / IV Model of a MOSFET exercise) Hello Staff and everyone, 1. When I entered the Maximum possible Vt as just less than Vih and "Check", it gives me a "right" tick (obvious). To my understanding there can only be *one* maximum possible value. 2. When I tried a random value (e.g. 2.670) it still gives me a right tick. Is that correct? 3. And it does give me a "wrong" tick for a value like 2.0. But I am confused as I was assuming it to be a strictly single valued answer and not a range. Could somebody help me understand this? Regards. UserIdTAG: 372623 UserNameTAG: Amitraj CreateTimeTAG: 2012-09-25T11:51:35Z VoteTAG: 1 CoursewareTAG: Week 3 / IV Model of a MOSFET exercise CommentableIdTAG: 6002x_IV_mosfet_model_exercise NumberOfReplyTAG: 4 FirstChildTAG: Thanks for letting us know. This is okay behavior--a lot of the answers have a little wiggle room. FirstChildUserIdTAG: 376895 FirstChildUserNameTAG: alecgg FirstChildCreateTimeTAG: 2012-09-25T19:44:18Z SecondChildTAG: Thank you for answering my question. I still wonder why a range would be allowed, if there is to be only one answer in this particular case. Also, *I feel, if at all a range is allowed* its highest number should not be greater than 2.623V. The range allows me to enter even 2.7, while VIH itself is 2.624V. Dear team at MIT, I am re-raising my confusion herewith. Could we explore onto this? Regards. SecondChildUserIdTAG: 372623 SecondChildUserNameTAG: Amitraj SecondChildCreateTimeTAG: 2012-09-27T12:04:18Z FirstChildTAG: may be answer is allowing proper significant decimal places FirstChildUserIdTAG: 412703 FirstChildUserNameTAG: vijay15793 FirstChildCreateTimeTAG: 2012-09-25T17:02:31Z FirstChildTAG: so,what's the accurate answer? ,is this 2.623 for Vt FirstChildUserIdTAG: 395257 FirstChildUserNameTAG: blackguitar FirstChildCreateTimeTAG: 2012-09-27T04:54:41Z SecondChildTAG: Yes. That should be it, provided a maximum of 3 digits are allowed post decimal point. SecondChildUserIdTAG: 372623 SecondChildUserNameTAG: Amitraj SecondChildCreateTimeTAG: 2012-09-27T12:00:36Z FirstChildTAG: I believe that the maximum value should be 2.624 (as also shown when we hit show answer) but the minimum value should be .576 rather than .577. This is because, as mentioned in the slides. Vgs< Vt => open D-S Vgs>=Vt => Short D-S FirstChildUserIdTAG: 442070 FirstChildUserNameTAG: MuhammadAsad FirstChildCreateTimeTAG: 2012-09-27T17:58:21Z IndexTAG: 2524 TitleTAG: week 3 videos upload kindly upload week 3 videos .. youtube is still block in my country . or upload lectures to some other website. just 5 days are remaining i cant able to understand home work and lab , please make these lecturers available to us as soon as possible. thanks :) UserIdTAG: 438395 UserNameTAG: ali_PU1 CreateTimeTAG: 2012-09-25T11:37:10Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: On it. First of the Week 3 lectures are now up for download in the [Wiki][1]. Please let me know if you have any issues with the videos. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/LectureDownloadWeek3/ FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-25T13:30:22Z SecondChildTAG: Lab 3: finding W/L ! how did you that ? plz tell SecondChildUserIdTAG: 97340 SecondChildUserNameTAG: AnkitRana SecondChildCreateTimeTAG: 2012-09-25T15:02:22Z SecondChildTAG: Thank you!:) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-26T11:36:07Z FirstChildTAG: Thank you for providing alternative download link for the videos. Thankfully I don't have to worry about youtube being blocked, but my internet connection is unreliable. Downloading the videos is much more convenient. Thanks again! M. FirstChildUserIdTAG: 385677 FirstChildUserNameTAG: Michaelc1 FirstChildCreateTimeTAG: 2012-09-26T00:16:53Z IndexTAG: 2525 TitleTAG: Using the tools I had some problems but i solved it. It was a good practise. UserIdTAG: 491538 UserNameTAG: josemogollon CreateTimeTAG: 2012-09-25T05:13:42Z VoteTAG: 1 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: I have problem with using lab tools,please tell me how I make connections of circuit. because of this I have not submitted my lb work FirstChildUserIdTAG: 380404 FirstChildUserNameTAG: kakar FirstChildCreateTimeTAG: 2012-09-25T12:17:12Z IndexTAG: 2526 TitleTAG: S6V5 transcript error at ~2:39 "for analyzing long linear equations." *long* should be *non-* (Let me know if there is a better way to submit typo corrections like this.) UserIdTAG: 468623 UserNameTAG: RobNik CreateTimeTAG: 2012-09-25T02:41:51Z VoteTAG: 1 CoursewareTAG: Week 3 / Analyzing Nonlinear Circuits CommentableIdTAG: 6002x_analyzing_nonlinear_circuits NumberOfReplyTAG: 1 FirstChildTAG: Most likely, you should file your post under the category "Feedback". As always, I would like a Mod or Staff to confirm this, however. FirstChildUserIdTAG: 248807 FirstChildUserNameTAG: 1kingsman FirstChildCreateTimeTAG: 2012-09-25T03:25:47Z IndexTAG: 2527 TitleTAG: Muito Bom!!!! Excelente aula, excelente professor. Parabéns... UserIdTAG: 312043 UserNameTAG: Osmar17 CreateTimeTAG: 2012-09-25T02:27:14Z VoteTAG: 1 CoursewareTAG: Week 2 / Chain saw CommentableIdTAG: 6002x_chain_saw NumberOfReplyTAG: 0 IndexTAG: 2528 TitleTAG: dependent sources in s8 V3 we made a dependent current source such that I = k/v but v here is not the potential across the current source. what we did in earlier lectures was the graph of current source in which I was independent of V across it. if this is dependent shouldn't it be depending on voltage across itself??? UserIdTAG: 168496 UserNameTAG: poweltalwar CreateTimeTAG: 2012-09-24T21:30:03Z VoteTAG: 1 CoursewareTAG: Week 4 / Dependent Sources Example 1 CommentableIdTAG: 6002x_dep_src_ex_1 NumberOfReplyTAG: 1 FirstChildTAG: Don't be confused; the designer gets to pick what the dependent source is controlled by, it doesn't have to be anything near the diamond representation. FirstChildUserIdTAG: 248902 FirstChildUserNameTAG: Chanute FirstChildCreateTimeTAG: 2012-09-25T00:28:07Z IndexTAG: 2529 TitleTAG: Week 4 and week 5 lecture sequences for H4 Lab4 completion? I believe i wouldn't spent as much time on H4 and Lab4, if the sequence of videos was different. What i mean is that, the theory that was needed for homework completion in week 4, is uploaded now in week 5's lecture sequence. Maybe i rush too much through the weeks, but this is my point of view. UserIdTAG: 304587 UserNameTAG: Vasso CreateTimeTAG: 2012-09-24T21:23:47Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I kinda see what you are saying. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-24T22:13:22Z SecondChildTAG: At least you get to use what you have learned. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-24T22:14:18Z SecondChildTAG: Let's see! Hope this was a rewarding trick for those who rushed :P SecondChildUserIdTAG: 304587 SecondChildUserNameTAG: Vasso SecondChildCreateTimeTAG: 2012-09-24T22:16:41Z SecondChildTAG: Actually, it's just a common manner in which courses are taught at MIT. Often the problem comes out before the theory is formally introduced in lecture, so that students are forced to grapple with the material on their own a bit :) SecondChildUserIdTAG: 79361 SecondChildUserNameTAG: kimt SecondChildCreateTimeTAG: 2012-09-24T23:21:14Z SecondChildTAG: Sadly in my country you are not encouraged to try this hard for an engineering degree. You are given the tools and you go with the flow. The fact that i could cope with that well, even though i wasn't taught in this manner in my own degree, makes me happy!! :) SecondChildUserIdTAG: 304587 SecondChildUserNameTAG: Vasso SecondChildCreateTimeTAG: 2012-09-24T23:41:34Z SecondChildTAG: I enjoy that pedagogical trick actually. If you're really enjoying the material, it's fun to see if you come up with the method that's later presented in class, and it definitely reinforces the concept! SecondChildUserIdTAG: 248902 SecondChildUserNameTAG: Chanute SecondChildCreateTimeTAG: 2012-09-25T00:08:43Z IndexTAG: 2530 TitleTAG: Question about the resistor Is the resistor RL there just to prevent a short circuit on the terminals of the power supply? If so, its resistance should be quite high in order to drain as little current as possible, right? Cheers UserIdTAG: 274868 UserNameTAG: JPaquim CreateTimeTAG: 2012-09-24T21:07:06Z VoteTAG: 1 CoursewareTAG: Week 3 / Switch Model CommentableIdTAG: 6002x_switch_model NumberOfReplyTAG: 1 FirstChildTAG: As far as i know , the answer is yes . the text book have introduction about the usage of RL . You can refer to the testbook . FirstChildUserIdTAG: 232667 FirstChildUserNameTAG: KyleLiux FirstChildCreateTimeTAG: 2012-09-25T02:16:09Z IndexTAG: 2531 TitleTAG: About edx respected sir, Edx web is not properly opening so far. It's creating a problem to listen lectures and solve home work and labs . So, kindly solve the problem as soon as possible . Regards, UserIdTAG: 271448 UserNameTAG: UsmanRashid CreateTimeTAG: 2012-09-24T19:28:14Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: YouTube was recently blocked in several additional countries due to world events. We are now hosting a separate source of the lecture videos, https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/LectureDownloadWeek2/ Thanks for your patience. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-09-25T03:28:18Z IndexTAG: 2532 TitleTAG: Composite voltage sourse is linear or not? ![enter image description here][1] In week 2 tutorials instructor says that composite voltage sourse is not linear. Is it means that Thevenin equivalent is also not linear (Circuit of Thevenin equivalent is identical to composit voltage source in week 2 tutorial)? I'm little confused ...! ![enter image description here][2] What I'm doing wrong? [1]: https://edxuploads.s3.amazonaws.com/13485105623593943.jpg [2]: https://edxuploads.s3.amazonaws.com/13485110411343615.jpg UserIdTAG: 104522 UserNameTAG: IgorNovice CreateTimeTAG: 2012-09-24T18:25:04Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: The definition of a linear source is that the line formed by the source passes through the origin (0,0). Since the definition of a voltage source is a straight, vertical line (voltage stays the same, current varies), it cannot pass through the origin (unless voltage is 0, then it would no longer be a voltage source), and thus is Non-Linear. FirstChildUserIdTAG: 248807 FirstChildUserNameTAG: 1kingsman FirstChildCreateTimeTAG: 2012-09-24T23:50:49Z FirstChildTAG: Look at the explanation here: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_2/Week_2_Tutorials/8 The language is tricky here. Independent sources are technically not linear **elements**. But we have gone out of our way in the course to define linear **circuits** as circuits containing only linear elements **and** independent sources. FirstChildUserIdTAG: 248902 FirstChildUserNameTAG: Chanute FirstChildCreateTimeTAG: 2012-09-25T00:35:44Z SecondChildTAG: Ahh, so independent sources are put in their own category, and a linear circuit has a different definition than a linear element. SecondChildUserIdTAG: 248807 SecondChildUserNameTAG: 1kingsman SecondChildCreateTimeTAG: 2012-09-26T23:33:41Z IndexTAG: 2533 TitleTAG: Another solution When the equation at the end of the video is solved, -1 is also a solution (it's logical, with the current flowing in the opposite direction). UserIdTAG: 147472 UserNameTAG: rojorpa CreateTimeTAG: 2012-09-24T17:23:08Z VoteTAG: 1 CoursewareTAG: Week 4 / Dependent Sources Example 1 CommentableIdTAG: 6002x_dep_src_ex_1 NumberOfReplyTAG: 1 FirstChildTAG: we get v^2 = kr and what the answer actually is that v = +- ( |k|r )^-1/2 which gives both the answers FirstChildUserIdTAG: 168496 FirstChildUserNameTAG: poweltalwar FirstChildCreateTimeTAG: 2012-09-24T21:23:59Z IndexTAG: 2534 TitleTAG: problem watching videos i'm having some problem watching the videos. i use mozila firefox. i don't think it's a problem with my browser. because i was able to see videos earlier using the same. then what is the problem? any help...!! ? UserIdTAG: 163407 UserNameTAG: shuvajit CreateTimeTAG: 2012-09-24T16:50:12Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: same problem is here not listening lectures and not uploaded home work properly . FirstChildUserIdTAG: 271448 FirstChildUserNameTAG: UsmanRashid FirstChildCreateTimeTAG: 2012-09-24T19:32:11Z FirstChildTAG: i am not able to watch any videos for 2 days now. they hardly ever load. there surely is a problem with the site. does anyone have a word from the admin about it? i tried all the web explorers, changed internet connections and computers as well. FirstChildUserIdTAG: 337478 FirstChildUserNameTAG: Ghauri FirstChildCreateTimeTAG: 2012-09-24T20:00:35Z FirstChildTAG: that seems to be everyone here's problem. me too. :( FirstChildUserIdTAG: 484804 FirstChildUserNameTAG: thawtar FirstChildCreateTimeTAG: 2012-09-25T00:09:50Z IndexTAG: 2535 TitleTAG: H4P1 - Help finding Incremental Resistance All - I found the requested voltage by plugging and chugging; however, I cannot find the resistance. At first, I tried V/I, but that didn't work. After some research, I found that **incremental** resistance means dV/dI, so I manipulated the equation in terms of V=f(I), then found dV/dI. Unfortunately, the green checkmark still doesn't come. Any tips? UserIdTAG: 348141 UserNameTAG: TomTrieb CreateTimeTAG: 2012-09-24T15:33:52Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: The incremental resistance is 1/(dI/dV). I = f(V). FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-09-24T15:37:58Z SecondChildTAG: Thanks, Alex - That worked! SecondChildUserIdTAG: 348141 SecondChildUserNameTAG: TomTrieb SecondChildCreateTimeTAG: 2012-09-24T16:18:49Z SecondChildTAG: I'm confused on the units-am I supposed to raise it to the 1.5 power, or is it just V to the 1.5 power since that is the denominator of P, and I just use the v value? SecondChildUserIdTAG: 344331 SecondChildUserNameTAG: intellectualwanderer SecondChildCreateTimeTAG: 2012-09-24T19:29:46Z SecondChildTAG: It is Vpk raised to 1.5 . SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-09-24T20:16:05Z SecondChildTAG: Ive got a question for you guys, I found the incremental resistance for Vpk=8 and Vpk = 10 using the equation 1/(dI/dV). on the other hand they ask for the Ip for Vpk=8 which I found it looking at the graphic of 6al5 vacuum diode. I tried to do the same for Vpk=10v but the answer is wrong so i was wandering how i should find Ip. let me know guys if u have any idea. SecondChildUserIdTAG: 320871 SecondChildUserNameTAG: seb1256 SecondChildCreateTimeTAG: 2012-09-25T08:29:23Z SecondChildTAG: Thanks,Alex! It is simple way to understand what is "incremental resistance". SecondChildUserIdTAG: 367431 SecondChildUserNameTAG: VictorFedosenkov SecondChildCreateTimeTAG: 2012-09-25T12:08:15Z SecondChildTAG: by dI/dv..do u mean d(PV^3/2)...could you please explain me the working? SecondChildUserIdTAG: 122513 SecondChildUserNameTAG: Mona77 SecondChildCreateTimeTAG: 2012-10-04T19:51:19Z SecondChildTAG: still not getting the correct answer i know my answer is right but they don;t show green mark SecondChildUserIdTAG: 336001 SecondChildUserNameTAG: syd_buet12 SecondChildCreateTimeTAG: 2012-10-06T17:58:14Z IndexTAG: 2536 TitleTAG: about the Lectures sir here is a problem that im not able to see the lectures because youtube is blocked in our country thats why im not able to see the lectures what to do ....!! plz guide me UserIdTAG: 130290 UserNameTAG: RajkumarRughani CreateTimeTAG: 2012-09-24T14:50:44Z VoteTAG: 1 CoursewareTAG: Week 3 / Logic Gates CommentableIdTAG: 6002x_logic_gates NumberOfReplyTAG: 1 FirstChildTAG: You can see in the Wiki [Lecture Videos Download][1]:). Also see more info [here][2]: **Due to the YouTube outage in certain parts of the world, we have provided direct links to download the lecture videos, and extended the Week 2 homework and lab deadlines to the coming Sunday (September 30). Please note that Week 3 deadlines have not changed.** [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/ [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/info FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-26T12:01:15Z IndexTAG: 2537 TitleTAG: Lab5 > Now change the amplitude of Vsignal in Figure 2 from 0.1V to 1V and rerun the TRAN a nalysis. You should see significant distortion in the output signal, in this case clipping or truncation of the max and min signal values. Experiment with amplitudes of Vbias and Vsignal to find the largest amplitude for Vsignal for which amplifier produces an unclipped output. Largest input amplitude resulting in an **undistorted** output signal: We get unlinear distortion on 1xx mV. I think, the question must be like this: Largest input amplitude resulting in an **unclipped** output signal: UserIdTAG: 394836 UserNameTAG: v2g6ch4 CreateTimeTAG: 2012-09-24T13:35:41Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I got confused in this question and I did not get the solution yet. FirstChildUserIdTAG: 39980 FirstChildUserNameTAG: MarinoJr FirstChildCreateTimeTAG: 2012-09-27T18:53:48Z SecondChildTAG: You have to change the values of the DC value and the AC value to find the maximum value of the output signal without any kind of distorsion. Try changing a little bit only the DC value, and see the response. SecondChildUserIdTAG: 125880 SecondChildUserNameTAG: JoeUcvVzla SecondChildCreateTimeTAG: 2012-10-06T01:10:07Z SecondChildTAG: Are we also supposed to change the Vpower? Cause that will definitely change the maximum input we can have. SecondChildUserIdTAG: 324642 SecondChildUserNameTAG: StNas SecondChildCreateTimeTAG: 2012-10-10T09:04:49Z FirstChildTAG: Hi v2g6ch4! Take a look [here][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/507628b2c4dd80250000004b FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-11T02:12:52Z IndexTAG: 2538 TitleTAG: Please extend the time for Week 2 submissions As many of us are now the victims of YouTube ban in our countries. So I Request the edx Team to please extend the time for week 2 submissions so as week 3. I hope that you will provide immediate solution to the issue. Thanks UserIdTAG: 272417 UserNameTAG: MuKhan CreateTimeTAG: 2012-09-24T13:08:51Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You can see in the Wiki [Lecture Videos Download][1]:). Also see more info [here][2]: **Due to the YouTube outage in certain parts of the world, we have provided direct links to download the lecture videos, and extended the Week 2 homework and lab deadlines to the coming Sunday (September 30). Please note that Week 3 deadlines have not changed.** [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/ [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/info FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-26T11:44:12Z IndexTAG: 2539 TitleTAG: Lab4 Solving Quadratic Equation All - I found the following quadratic equation solver to be handy for question #2: http://www.mathsisfun.com/quadratic-equation-solver.html Quick question: the results from the solver both seemed accurate (in that they were both positive). Is there a way to check to see which of them is correct? (I simply plugged them into the solver until the correct one turned green, but in real life we won't have a solver . . . just curious.) Thanks. UserIdTAG: 348141 UserNameTAG: TomTrieb CreateTimeTAG: 2012-09-24T13:05:26Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You don't need a solver. I you use the second equation (ratio of $i_{DS1}/i_{DS2}$), do some manipulations to find $V_T$ and then use it in the first equation, you have everything you need. FirstChildUserIdTAG: 396446 FirstChildUserNameTAG: RousseauxS FirstChildCreateTimeTAG: 2012-09-24T17:50:03Z SecondChildTAG: vt punt attempt and that brand me wrong but SecondChildUserIdTAG: 320715 SecondChildUserNameTAG: maraivette SecondChildCreateTimeTAG: 2012-10-03T16:38:42Z SecondChildTAG: Just think physical - You get two solutions, but one of them is obviously too high to be the real value. SecondChildUserIdTAG: 329254 SecondChildUserNameTAG: KGabor SecondChildCreateTimeTAG: 2012-10-06T09:21:43Z IndexTAG: 2540 TitleTAG: Graphing Download GraphCalc http://www.graphcalc.com/download.shtml and use it! UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-09-24T12:32:38Z VoteTAG: 1 CoursewareTAG: Week 3 / IV Model of a MOSFET exercise CommentableIdTAG: 6002x_IV_mosfet_model_exercise NumberOfReplyTAG: 1 FirstChildTAG: GeoGebra is also nice : http://www.geogebra.org/cms/en FirstChildUserIdTAG: 396446 FirstChildUserNameTAG: RousseauxS FirstChildCreateTimeTAG: 2012-09-24T17:43:41Z IndexTAG: 2541 TitleTAG: Circuit c is a combination of two NOT gates and a NOR gate Z=[(X(bar)+Y)(bar)](bar)=X(bar)+Y`` UserIdTAG: 176939 UserNameTAG: electron625 CreateTimeTAG: 2012-09-24T12:05:21Z VoteTAG: 1 CoursewareTAG: Week 3 / Switch Model Exercise CommentableIdTAG: 6002x_switch_model_exercise NumberOfReplyTAG: 1 FirstChildTAG: Wont you say there are four gates in total? FirstChildUserIdTAG: 442070 FirstChildUserNameTAG: MuhammadAsad FirstChildCreateTimeTAG: 2012-09-27T17:11:20Z IndexTAG: 2542 TitleTAG: H3P1 can anyone help me with the formula for pullup resistor. I can't find it in the book UserIdTAG: 206648 UserNameTAG: TsvetanGeorgiev CreateTimeTAG: 2012-09-24T12:00:57Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: There is no such a "formula". You are asked to turn the switch ON under the SR-model so it behaves like a resistor of given resistance. So you replace the switch with a given resistance and apply the skills from week1 or even week0 FirstChildUserIdTAG: 190618 FirstChildUserNameTAG: Kirbabaev FirstChildCreateTimeTAG: 2012-09-24T12:13:18Z FirstChildTAG: It is quite simple. Vout must satisfy static discipline for "0" exit. So You can simply write expression, based on resistor divider for each circuit. FirstChildUserIdTAG: 195218 FirstChildUserNameTAG: Al_Incognito FirstChildCreateTimeTAG: 2012-09-24T12:13:35Z SecondChildTAG: Yeah! But how can u calculate Vout withou RL (I mean here Rpullup) because formula for that is Vo = Vs*Ron / (Rpullup+Ron). Am i right? please guide me if I am wrong,i am confused or I am not on the right way. SecondChildUserIdTAG: 281159 SecondChildUserNameTAG: ManasiS SecondChildCreateTimeTAG: 2012-09-28T20:35:07Z SecondChildTAG: My bad! Got it! SecondChildUserIdTAG: 281159 SecondChildUserNameTAG: ManasiS SecondChildCreateTimeTAG: 2012-09-28T20:39:51Z SecondChildTAG: Remember it is valid 0 to 1. SecondChildUserIdTAG: 143835 SecondChildUserNameTAG: sanpablc SecondChildCreateTimeTAG: 2012-09-29T21:46:53Z FirstChildTAG: Resistor divider.... replace each MOSFET with its respective Ron equivalent and then work you way from there so that static discipline is executed appropriately to satisfy given condition. FirstChildUserIdTAG: 14531 FirstChildUserNameTAG: samutr3 FirstChildCreateTimeTAG: 2012-09-30T10:21:42Z IndexTAG: 2543 TitleTAG: H3P1 Hi all, I cannot get the right result on the last question of H3P1: "What is the maximum power (in Watts) consumed by the NOR?" As far as I've understood, P = v^2 / R. The maximum R is calculated above (32000 ohms in my case), and v comes for the resistor divider (voltage fall on this resistor). But I do not get the correct result. Could somebody help me? UserIdTAG: 240876 UserNameTAG: Jecht CreateTimeTAG: 2012-09-24T08:28:52Z VoteTAG: 1 CoursewareTAG: Week 3 / MOSFET Power Tutorial CommentableIdTAG: 6002x_MOSFET_power_tutorial NumberOfReplyTAG: 4 FirstChildTAG: Why resistor divider? You need to calculate power of entire device, not only MOSFETS. FirstChildUserIdTAG: 195218 FirstChildUserNameTAG: Al_Incognito FirstChildCreateTimeTAG: 2012-09-24T08:45:19Z SecondChildTAG: And, by the way, You have strange R for NOR gate. Look carefully. SecondChildUserIdTAG: 195218 SecondChildUserNameTAG: Al_Incognito SecondChildCreateTimeTAG: 2012-09-24T08:46:50Z SecondChildTAG: I get confused writing. It was attending to the maximum V1l. I actually had good answer when I put on the pull-up resistor (previously) 32Kohm, like the not gate. Do you have another result? SecondChildUserIdTAG: 240876 SecondChildUserNameTAG: Jecht SecondChildCreateTimeTAG: 2012-09-24T12:00:14Z SecondChildTAG: Try writing it in the form X.XXXe-4 SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-09-25T13:27:58Z FirstChildTAG: Well, I've got a correct answer by interpreting "What is the maximum power (in Watts) consumed by the inverter? " as " What is the **minimum** power (in Watts) consumed by the inverter? ". And it seems, that they're asking about the power consuming by the ALL resistors, connected to the Vs. FirstChildUserIdTAG: 326787 FirstChildUserNameTAG: Angstrem FirstChildCreateTimeTAG: 2012-09-24T08:46:50Z FirstChildTAG: Hi everybody. My problem in H3P1 is that i couldn't calculate the low and high noise margins. Can anybody help me? FirstChildUserIdTAG: 213452 FirstChildUserNameTAG: JoJosida FirstChildCreateTimeTAG: 2012-09-24T11:18:22Z SecondChildTAG: Hi. If You will look in Wiki, You will see this: Static Discipline, a specification for digital devices, requires the devices to correctly interpret voltages that fall within the input thresholds (VIL and VIH). As long as valid inputs are provided to the devices, the discipline also requires the devices to produce valid output voltages that satisfy the output thresholds (VOL and VOH). **Noise margin is typically defined as the absolute value of the difference between the prescribed output voltage for a given logical value and the corresponding forbidden region voltage threshold for the receiver.** SecondChildUserIdTAG: 195218 SecondChildUserNameTAG: Al_Incognito SecondChildCreateTimeTAG: 2012-09-24T11:26:48Z SecondChildTAG: Also look S4V8. It is just about margins. SecondChildUserIdTAG: 195218 SecondChildUserNameTAG: Al_Incognito SecondChildCreateTimeTAG: 2012-09-24T11:29:30Z SecondChildTAG: Thanks, I have done it. SecondChildUserIdTAG: 213452 SecondChildUserNameTAG: JoJosida SecondChildCreateTimeTAG: 2012-09-24T11:30:30Z FirstChildTAG: > The maximum R is calculated above (32000 ohms in my case) Well you have two branches in parallel for NOR. Why do you only look at the case when the both branches are conductive? Try input 01 or 10 and you'll see it behaves the same as an invertor under input 1. For if you got 32K for input 11 and 64K for 01 and 10, what should be the minimal required resistance for ALL possible cases? FirstChildUserIdTAG: 190618 FirstChildUserNameTAG: Kirbabaev FirstChildCreateTimeTAG: 2012-09-24T12:05:02Z IndexTAG: 2544 TitleTAG: H3P3 Can't understand where is mistake Hello, everyone! I can't unserstand where is mistake in my calculation h3p3. 1 Step - If R=100Ω, what is the power drawn in Watts (W) from the solar cell? At graph (B) I take voltage V1 that correspond max power. Look up at chart (A) that shows the current vs voltage curve and find current I(V1)=0.0009amps. Then use P=0.0009^2*100Oms. But answer is incorrect. Thanks for help. UserIdTAG: 387154 UserNameTAG: Junglist CreateTimeTAG: 2012-09-24T06:45:34Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: It's got nothing to do with max power. This is a load line question. It might help to review S6V6 and S6V7 FirstChildUserIdTAG: 248902 FirstChildUserNameTAG: Chanute FirstChildCreateTimeTAG: 2012-09-24T08:55:32Z FirstChildTAG: Hmm. Why You take max power? If You will carefully read question #2 in H3P3 You will see, that the value of R can be not 100ohms to gain max power. FirstChildUserIdTAG: 195218 FirstChildUserNameTAG: Al_Incognito FirstChildCreateTimeTAG: 2012-09-24T07:24:33Z SecondChildTAG: I dont take max power, at graph (B) I take voltage (V=0.29V) that correspond max power. At graph (A) see I(V=0.29)=0.009Apms. SecondChildUserIdTAG: 387154 SecondChildUserNameTAG: Junglist SecondChildCreateTimeTAG: 2012-09-24T08:00:27Z SecondChildTAG: Ok, let's start from beginning. The question is to find Power, which drains if R=100. You have two graphs 1st is need to find working point of solar cell. Use Ohm law to find a point, where resistor i(v) graph crosses solar cell graph. After that You can see operating voltage and take power from graph 2. This will be Your answer. SecondChildUserIdTAG: 195218 SecondChildUserNameTAG: Al_Incognito SecondChildCreateTimeTAG: 2012-09-24T08:17:06Z SecondChildTAG: I get it! Thank You very much! SecondChildUserIdTAG: 387154 SecondChildUserNameTAG: Junglist SecondChildCreateTimeTAG: 2012-09-24T08:40:18Z SecondChildTAG: I think there's a problem. Shouldn't the problem statement say $R=100\,\Omega$? SecondChildUserIdTAG: 276409 SecondChildUserNameTAG: IgnacioUY SecondChildCreateTimeTAG: 2012-09-25T21:57:50Z SecondChildTAG: v=ri; v=100i (for i look 1graph); then look where v 1graph crosses 2graph SecondChildUserIdTAG: 440992 SecondChildUserNameTAG: calabrezcrv SecondChildCreateTimeTAG: 2012-09-29T14:38:55Z SecondChildTAG: From the second graph of the question it is very much understood that the maximum power is at .3 Volts and is about .00027 W. Thus from The relation P=V^2/R R can be easily calculated. SecondChildUserIdTAG: 232499 SecondChildUserNameTAG: Asim09 SecondChildCreateTimeTAG: 2012-09-29T19:23:08Z IndexTAG: 2545 TitleTAG: Lab2 I'm freaked out with this... Please would someone please just explain to me Lab, I'm tired of this... UserIdTAG: 317836 UserNameTAG: VitorRodrigues CreateTimeTAG: 2012-09-24T05:14:24Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hi VitorRodrigues! You can see an explanation in this Post [Here][1]. :). [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_S3E2_Linear_Combinations_of_Source_Strengths/threads/506a280d036af91f000000c9 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-02T00:01:34Z FirstChildTAG: Forget the square and sinusoidal waves for now. These could be any two waves! All you have to do is find appropriate resistors so that you could mix the signals. It's easier than you think! FirstChildUserIdTAG: 78396 FirstChildUserNameTAG: dharav FirstChildCreateTimeTAG: 2012-09-24T12:47:35Z IndexTAG: 2546 TitleTAG: H2P2 Took me quite awhile, days in fact, to correctly answer all parts of this question. I had two of the three answers days ago but I haven't done Calculus since 1987 so I have to use less analytical methods. This circuit should be treated just as it is shown in the assignment, do not lump together any values. You have a source, a load and losses in transmission. Remember the voltage drop across each resistor...... And there are multiple formulae for Power..... UserIdTAG: 375500 UserNameTAG: Brej7665 CreateTimeTAG: 2012-09-24T02:27:28Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: When they said "remember your calculus", they're just saying it's a max/min problem. If you remember how to take the derivative of a simple function, you've got all the calculus necessary to solve this analytically (i.e. whatever value of the dependent variable makes the derivative of the function = 0 will in turn make the function itself evaluate to a maximum or minimum value!) FirstChildUserIdTAG: 248902 FirstChildUserNameTAG: Chanute FirstChildCreateTimeTAG: 2012-09-24T02:41:10Z IndexTAG: 2547 TitleTAG: LAB2 Do the voltages of the sources matter? UserIdTAG: 127021 UserNameTAG: victorag CreateTimeTAG: 2012-09-24T01:38:48Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Not so far as I saw. All I did was make sure I had the right voltage divider values for the resistors. Hang tight! FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-09-24T02:07:49Z FirstChildTAG: Probably, but the question asks us not to change anything about the sources (voltage, amplitude, etc.). FirstChildUserIdTAG: 311920 FirstChildUserNameTAG: ghowell FirstChildCreateTimeTAG: 2012-09-24T06:04:53Z IndexTAG: 2548 TitleTAG: Help with Lab and Sandbox In the sandbox the instructions say that if you push the check button your results will be saved. I did a test circuit pushed the check button, but I don't know how to restore my circuit. If I do the lab in the sandbox can I copy it to the lab. Does the circuit have to be in the lab? I've already run the transient analysis and have numbers to work with but I lost my circuit due to computer problems. Do I have to reconstruct the circuit to get full credit for the lab? UserIdTAG: 406202 UserNameTAG: skyhawk CreateTimeTAG: 2012-09-23T23:05:56Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You can always reset the lab after checking. The circuit that will be saved will be your last one. FirstChildUserIdTAG: 130755 FirstChildUserNameTAG: sugoi100x FirstChildCreateTimeTAG: 2012-09-23T23:26:42Z SecondChildTAG: Yes, the circuit must be in the Lab to get credit. I recommend printing (at least screenshots) of your work. I've been burned too many times by computer glitches, and always take precautions to save my work so if necessary, I can easily reconstruct it in the text box or whatever web form. Good luck! SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-09-23T23:45:15Z SecondChildTAG: Thanks. Is there a way to copy from sandbox to lab? SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-09-24T00:04:30Z SecondChildTAG: I did my transient analysis in the circuit sandbox, copied the necessary data, did my calculations, and entered them into the answer boxes. I received full credit without doing the transient analysis in the lab. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-09-24T00:58:26Z SecondChildTAG: Before testing the circuit its always a good idea to click on 'check', as this will save a copy of your lab. In the event you lose part or all of your lab (as by hitting the reset button in error) simply close the lab page without saving (i.e. hitting 'check'). When you return to the lab page you'll find the copy of your pre-test circuit. SecondChildUserIdTAG: 150057 SecondChildUserNameTAG: Felinae SecondChildCreateTimeTAG: 2012-09-24T09:49:53Z IndexTAG: 2549 TitleTAG: H2P1 I dont understand this: Come up with resistors R1 and R2 such that the divider ratio Vout/Vin is within 10% of the requirement. the ratio between vout/vin is 0.25 but what about the 10% of the requirement UserIdTAG: 127021 UserNameTAG: victorag CreateTimeTAG: 2012-09-23T22:54:32Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You only need to use the tolerance in Vmax in Vmin. First calculate the nominal resitance(without the tolerance of 10%). Typically manafuctures have already a set o resistances for production. For example when you buy a resistor of 10ohm it has a tolerance of 10%, this means that your resistor can be 9ohm < R < 11ohhm. In some cases you don`t bother with this tolerance but when you need precision you need to consider tolerance in your project. FirstChildUserIdTAG: 130755 FirstChildUserNameTAG: sugoi100x FirstChildCreateTimeTAG: 2012-09-23T23:19:01Z IndexTAG: 2550 TitleTAG: H2P1 I am kind of confused... I've already chosen my resistances and i calculated the vmax and v min, the values of the resistances are wrong, but the the answers of the voltage questions are good... why?? and i still dont understand why my resistances are wrong of when i solve for Rth, the value is between the margin 10k-30k UserIdTAG: 127021 UserNameTAG: victorag CreateTimeTAG: 2012-09-23T22:49:42Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Are your resistances in the E12 list? for example 11.3k is not in the list, so you'll have to choose 10k or 12k FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-09-23T22:58:03Z SecondChildTAG: yes they are, thats why i dont know :/ SecondChildUserIdTAG: 127021 SecondChildUserNameTAG: victorag SecondChildCreateTimeTAG: 2012-09-23T23:01:25Z SecondChildTAG: Did you try the check or reset button for the circuit diagram again after changing something? SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-09-23T23:09:38Z SecondChildTAG: there is additional requirement - the real ratio Vin/Vout should be in 10% tolerance to the required also SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-24T00:47:12Z SecondChildTAG: * Vout/Vin SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-24T00:48:58Z SecondChildTAG: yeah, i got R values that complie with all the requirements but it still doesn't check... though my Vmax/min do check :S SecondChildUserIdTAG: 202345 SecondChildUserNameTAG: enio_383 SecondChildCreateTimeTAG: 2012-09-24T00:57:06Z SecondChildTAG: not R let's say R1 and R2 - resistances you picked from E12 list - make sure that Vout/Vin-10% <= (Vin*R2/(R1+R2))/Vin <= Vout/Vin+10% SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-24T01:11:02Z SecondChildTAG: maybe you made a error in typing, type the amount in Ohms, not kiloOhms SecondChildUserIdTAG: 82183 SecondChildUserNameTAG: 0Bojan SecondChildCreateTimeTAG: 2012-09-29T14:22:21Z IndexTAG: 2551 TitleTAG: Time zone....do i have time yet? the time here in Chile is 18:25 do i have time yet? right??? UserIdTAG: 149549 UserNameTAG: Java_Dido CreateTimeTAG: 2012-09-23T22:26:34Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Yes, you have until midnight at your local time FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-23T22:40:03Z SecondChildTAG: thanks you! SecondChildUserIdTAG: 149549 SecondChildUserNameTAG: Java_Dido SecondChildCreateTimeTAG: 2012-09-23T22:41:04Z SecondChildTAG: Isn`t it wrong? I thought it was due midnight in my local time zone. SecondChildUserIdTAG: 130755 SecondChildUserNameTAG: sugoi100x SecondChildCreateTimeTAG: 2012-09-23T23:04:07Z SecondChildTAG: Yes! sugoi100x thanks for pointing that out, the deadline is midnight. Sorry about that. The above post has been corrected SecondChildUserIdTAG: 381746 SecondChildUserNameTAG: jelizon SecondChildCreateTimeTAG: 2012-09-23T23:15:57Z SecondChildTAG: :D SecondChildUserIdTAG: 149549 SecondChildUserNameTAG: Java_Dido SecondChildCreateTimeTAG: 2012-09-23T23:25:37Z IndexTAG: 2552 TitleTAG: Feedback...MY H2P1 homework problem The spec calls for a 10% tolerance. I understand that to mean that if a Vo should be 16, then Vo_max should be no more than 17.6. Well - and I DID spend a LOT of time on this - the numbers I finally entered in order to meet tonight's deadline do NOT meet that spec but they were accepted by the automatic grader. (My Vo_max was the only value out of tolerance...too high.) I followed the advice of another post and ran a small Python thing to computer all possible outputs from the permitted resistor values and then selected what looked like the very best of the bunch. I repeat: so far as I can tell it did NOT meet the spec but the numbers were accepted by the grader. I find this a little annoying. :) UserIdTAG: 92346 UserNameTAG: MobiusTruth CreateTimeTAG: 2012-09-23T21:21:33Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 2 FirstChildTAG: Thanks a ton! I mean this was a simply logic! I mean really! I had spent my whole day and feeling bad for not able to solve this. But Vmax = Vo+(10%of Vo) and Vmin= Vo-(10% 0f Vo). and Vo is already given. Sometimes IT IS THAT SIMPLE! :) FirstChildUserIdTAG: 281159 FirstChildUserNameTAG: ManasiS FirstChildCreateTimeTAG: 2012-09-23T21:30:08Z SecondChildTAG: it's not correct you should calculate Vmax and Vmin from voltage divider's resistors applying 10% tolerance to resistors SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-23T22:04:58Z SecondChildTAG: A careful reading shows that we are given 10% resistors but also the ratio Vout/Vin must also meet a 10% tolerance. That means - as I interpret the problem - that worst-case resistors (at their full 10% variation from 'correct' values)must still allow a Vout/Vin ratio within 10% of the specified value. SecondChildUserIdTAG: 92346 SecondChildUserNameTAG: MobiusTruth SecondChildCreateTimeTAG: 2012-09-23T22:32:27Z SecondChildTAG: oh, yes, you are right. I just picked resistors with ratio closest to requirements - sorry, I'm pedant and get the right values without additional checking :) SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-24T00:37:57Z SecondChildTAG: Oh!then I may have thought different! SecondChildUserIdTAG: 281159 SecondChildUserNameTAG: ManasiS SecondChildCreateTimeTAG: 2012-09-24T19:43:01Z SecondChildTAG: Then why is that my Vmax is not working? It worked for Vmin though... SecondChildUserIdTAG: 21687 SecondChildUserNameTAG: Benadicta SecondChildCreateTimeTAG: 2012-09-28T20:22:11Z FirstChildTAG: I agree, it was a little confusing. I think the key is that the assignment asked you first to use the nominal resistances to get within 10% of the desired $\frac{V_{out}}{V_{in}}$ (i.e. ignoring that the resistors aren't perfect). I'm pretty sure that the next part, calculating the $V_{max}$ and $V_{min}$, was to demonstrate that... "dang, those gold and silver bands on the resistors can seriously affect your end result and make it go off-spec!" But the way I read the question (confusing as it may be), it doesn't matter if the non-nominal resistors go outside $\frac{V_{out}}{V_{in}} \pm 10\%$ for this problem. FirstChildUserIdTAG: 248902 FirstChildUserNameTAG: Chanute FirstChildCreateTimeTAG: 2012-09-24T01:57:09Z IndexTAG: 2553 TitleTAG: Bug -- I am unable to see comments to posts. I cannot find my posts or read comments associated with them. UserIdTAG: 298538 UserNameTAG: BrianBenedict CreateTimeTAG: 2012-09-23T21:16:14Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Troubleshooting NumberOfReplyTAG: 1 FirstChildTAG: You should post your browser, version, and relevant data about your OS, so if this is a browser-specific problem, it may be more quickly identified. Good luck! FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-09-23T23:51:49Z IndexTAG: 2554 TitleTAG: Operating Point Graph As mentioned previously by another student (sorry I forgot your name) you need to graph the load line using i = v^3 and the calculated resistive load line. The graph shown is not to scale. UserIdTAG: 341895 UserNameTAG: splash_dad CreateTimeTAG: 2012-09-23T21:05:03Z VoteTAG: 1 CoursewareTAG: Week 3 / Graphical Analysis of Nonlinear Circuits Exercise CommentableIdTAG: 6002x_graphical_analysis_of_nonlinear_circuits_exercise NumberOfReplyTAG: 0 IndexTAG: 2555 TitleTAG: YOUTUBE BLOCKED YOUTUBE IS BLOCKED IN MY REGION. I CANT SEE THE VIDEOS THAT ARE BEING UPLOADED. PLEASE HELP...... UserIdTAG: 173585 UserNameTAG: habib_akbar CreateTimeTAG: 2012-09-23T19:37:00Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: look up 'SSH proxy' and set up a linux box on amazon cloud, or ask a geek friend outside your country to crate you an SSH account on their linux computer. you can then divert your internet traffic through their connection. This is essential knowledge for circumventing censorship. FirstChildUserIdTAG: 61469 FirstChildUserNameTAG: Spacedog FirstChildCreateTimeTAG: 2012-09-23T21:38:30Z FirstChildTAG: Download "Super hide ip" from torrent, install, hide your ip and watch youtube. FirstChildUserIdTAG: 378096 FirstChildUserNameTAG: Jamshaid271 FirstChildCreateTimeTAG: 2012-09-23T19:52:09Z FirstChildTAG: Try TOR. Its not usually that fast as those your internets but you can pause videos while they are buffering so no problem FirstChildUserIdTAG: 190618 FirstChildUserNameTAG: Kirbabaev FirstChildCreateTimeTAG: 2012-09-24T08:45:37Z FirstChildTAG: You can see in the Wiki [Lecture Videos Download][1]:). Also see more info [here][2]: **Due to the YouTube outage in certain parts of the world, we have provided direct links to download the lecture videos, and extended the Week 2 homework and lab deadlines to the coming Sunday (September 30). Please note that Week 3 deadlines have not changed.** [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/ [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/info FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-26T11:45:33Z IndexTAG: 2556 TitleTAG: H2P2 How to calculate power delivered to the BEST load resistance? UserIdTAG: 353405 UserNameTAG: karthickks CreateTimeTAG: 2012-09-23T18:34:05Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: When the impedance is matched between the Thevenin equivalent power source and load, that would be ideal. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-23T18:41:52Z FirstChildTAG: No idea... Same question here... It should be something like infinite or absolutely nothing. Either of them... FirstChildUserIdTAG: 324332 FirstChildUserNameTAG: petsol FirstChildCreateTimeTAG: 2012-09-23T18:43:10Z SecondChildTAG: i have spend lot of time..pls help us SecondChildUserIdTAG: 353405 SecondChildUserNameTAG: karthickks SecondChildCreateTimeTAG: 2012-09-23T18:53:03Z SecondChildTAG: I assume u have found out the Rl for maximum power transfer , that is the optimum load resistance of previous question . After that its a simple question of finding the current flowing through it using the current divider method and the applying power P = i^2 * Rl . U can search for the current divider if u are unaware of it. SecondChildUserIdTAG: 321422 SecondChildUserNameTAG: navooooo SecondChildCreateTimeTAG: 2012-09-23T18:59:52Z SecondChildTAG: Power delivered expressed as a function of load. Now you have a maximum problem....That is why they say: " remember your calculus!"...I think. SecondChildUserIdTAG: 164171 SecondChildUserNameTAG: PacoJuarez SecondChildCreateTimeTAG: 2012-09-23T19:11:29Z SecondChildTAG: 1. calculate the power dissipated by arbitrary load resistance RL. You will obtain P(RL) function. 2. Calculate derivative dP(RL)/dRL of this function ("remember calculus";P) 3. Solve dP(RL)/dRL =0 for RL to find extremum of function 4. Calculate second derivative d^2P(RL)/dRL^2 and evaluate at found RL 5. If value is negative - we are happy - found extremum was maximum, other case we had no luck, because we identified a minimum SecondChildUserIdTAG: 157610 SecondChildUserNameTAG: mradziwo SecondChildCreateTimeTAG: 2012-09-23T20:26:38Z IndexTAG: 2557 TitleTAG: Can't load my HW page! I have problems loading my HW and Lab page properly. I'm taking a [Math processing error] message.It appears in the place where the numbers (values) of exercises should be. Any suggestions? UserIdTAG: 345967 UserNameTAG: Tsipis CreateTimeTAG: 2012-09-23T17:18:15Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Troubleshooting NumberOfReplyTAG: 2 FirstChildTAG: Hi Tsipis! What web broser are you using? Do you have the last updated version of Chrome or Firefox? [Chrome Download link][1] You can also, try to refresh the web page with F5... [1]: https://www.google.com/intl/us/chrome/browser/ FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-23T17:55:12Z FirstChildTAG: What Myrimit said. I've had similar problems, resolved by clearing my browser cache and reloading the page. Note it sometimes takes more than one clear+reload. I suspect server load is part of the problem. Good luck! FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-09-24T00:18:40Z IndexTAG: 2558 TitleTAG: lab: button check dissapears when I click on check button both buttons check and save dissapear and button reset appears which delete the whole lab and I've to start from the beginning How to check the lab again ? UserIdTAG: 341115 UserNameTAG: undsame CreateTimeTAG: 2012-09-23T17:14:12Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Do your lab in the Circuit Sandbox first. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-23T17:37:07Z IndexTAG: 2559 TitleTAG: h2p1 guys....this h2p1 is really horribl tried in al ways... can ny 1 help wit the way 2 proceed.. The resistances of commercially-available discrete resistors are restricted to particular sets. For example, the available values of resistors with 10% tolerance are selections from the E12 set multiplied by a power of ten from 100 through 105. The E12 set is: E12={10,12,15,18,22,27,33,39,47,56,68,82} Thus, you can buy 10% resistors with a nominal resistance of 330Ω or 33kΩ, but not 350Ω. Furthermore, the "tolerance" means that if you buy a 10% 390Ω resistor you can be sure that its resistance is between 351Ω and 429Ω. In this problem we need to choose 10% resistors to make a voltage divider that meets a given specification. We are given an input voltage Vin=40.0V, and we need to provide an open-circuit output voltage of Vout≈8.0V. An additional requirement is that the Thevenin resistance as seen from the output terminals is between 10kΩ and 30kΩ. Assume first that the resistors have their nominal resistance. Come up with resistors R1 and R2 such that the divider ratio Vout/Vin is within 10% of the requirement. Of course, the resistances you chose are just nominal. Given that they are only guaranteed to have resistances within 10% of the nominal value, what is the largest and smallest value that Vout may have? UserIdTAG: 214416 UserNameTAG: sivasai CreateTimeTAG: 2012-09-23T17:06:39Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: can sme 1help me in h.w 2 problem 1 FirstChildUserIdTAG: 381572 FirstChildUserNameTAG: jahanzaibsuleman FirstChildCreateTimeTAG: 2012-09-23T17:27:30Z SecondChildTAG: evn i hv d same problm in findin r1 n r2 som 1 help SecondChildUserIdTAG: 214416 SecondChildUserNameTAG: sivasai SecondChildCreateTimeTAG: 2012-09-23T17:35:13Z SecondChildTAG: hrd to imagin. shessh SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-23T17:39:50Z FirstChildTAG: Hi sivasai! Can I help you? In which part are you lost? You can see some hints [here][1] that I wrote it to sandeshacharya, I hope that it can help you... [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/505971f4dd2f4d1f00000004 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-23T17:44:51Z IndexTAG: 2560 TitleTAG: eqns for plotting H3P3 soln Somebody help me show me a way to deduce the eqns for the diode in the solar battery example. I got the second part right by following the graph trend and then calculating the resistance for power at the peak. On the first part I am stuck and I want to solve it graphically. Thanks anyone!!! UserIdTAG: 183507 UserNameTAG: obiradaniel CreateTimeTAG: 2012-09-23T17:00:03Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: obiradaniel, I don't understand your question, do you want to solve the problem algebraically? You can use the graphical method for part a and b. Keep in mind that you need to use both plots. FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-23T18:51:46Z SecondChildTAG: Exactly, how do you use the graphical method? You have to plot i and v of the resistor but I don't the exact diode eqn, i=kV^3 or anyother. jelizon my question is how do you plot the i-v relation for the 100 ohm resistor. I have no idea how. SecondChildUserIdTAG: 183507 SecondChildUserNameTAG: obiradaniel SecondChildCreateTimeTAG: 2012-09-23T20:53:55Z SecondChildTAG: The simplest way to get started solving graphically is probably to print 2 copies of the graph, overlay them, and find the operating point where the two graphs intersect. HTH! SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-09-29T04:52:02Z IndexTAG: 2561 TitleTAG: About the VL Will the VL be the simply: VL= IN * RL = 1.4035 * 10E-6 * 100*10E3 = 0.1404 V as IN has to be the current which flows to the load RL? Something goes wrong? UserIdTAG: 355666 UserNameTAG: bjdchwr CreateTimeTAG: 2012-09-23T16:45:53Z VoteTAG: 1 CoursewareTAG: Week 2 / Norton method example CommentableIdTAG: 6002x_S3E6_Norton_Model NumberOfReplyTAG: 1 FirstChildTAG: Notice that not all the current will go through RL because you also have the resistor RN in parallel. In fact, the current will split between RN and RL. Can you think of a way to calculate what fraction of the current goes through RL? Hint: Remember that voltage is the same when resistors are placed in parallel. FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-23T18:47:21Z IndexTAG: 2562 TitleTAG: H2P1 Help! I've been going over this exercise for way too long, and nothing I've tried worked so far. I've seen the Thevenin video and the example (S3V7). I've managed to get a couple sets of viable resistor values from the Vout formula, and managed to confirm that they are within 10 - 30 with the R1*R2/R1+R2 formula, but after that, I'm completely stuck, and none of the resistor values I've found are even accepted when I try and Check. I've gone through the discussions and read through most of what's been said on this exercise, but still can't get anywhere. Any further help would be greatly appreciated! UserIdTAG: 453009 UserNameTAG: LucasHughes CreateTimeTAG: 2012-09-23T15:47:03Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hello Lucas, Not sure if you went through this - https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/505f01c789b8882700000009 In all, the selected values should be from E12... that's the last condition we have to resolve for. Please let me know if further help is desired. FirstChildUserIdTAG: 372623 FirstChildUserNameTAG: Amitraj FirstChildCreateTimeTAG: 2012-09-23T15:55:24Z SecondChildTAG: ok, I finally managed to work it out, thank you very very much. SecondChildUserIdTAG: 453009 SecondChildUserNameTAG: LucasHughes SecondChildCreateTimeTAG: 2012-09-23T17:07:49Z IndexTAG: 2563 TitleTAG: In the transcription at 5:26 It says "It's not that important for [UNINTELLIGIBLE]." Prof. Argawal says that "It's not that important for 6.002 or 6.002x." UserIdTAG: 164898 UserNameTAG: jbparkes CreateTimeTAG: 2012-09-23T14:55:35Z VoteTAG: 1 CoursewareTAG: Week 3 / Switch Resistor Model of a MOSFET CommentableIdTAG: 6002x_switch_resistor_model NumberOfReplyTAG: 0 IndexTAG: 2564 TitleTAG: what is stativ discipline and what is its relation with vt ? what is stativ discipline and what is its relation with vt ? UserIdTAG: 171199 UserNameTAG: Dhananjay01 CreateTimeTAG: 2012-09-23T14:31:52Z VoteTAG: 1 CoursewareTAG: Week 3 / IV Model of a MOSFET exercise CommentableIdTAG: 6002x_IV_mosfet_model_exercise NumberOfReplyTAG: 1 FirstChildTAG: Thats exactly i am looking to.. i tried to equate current relation in saturation but again KW/L value is now known else Ids=KW/L(Vgs-Vt)2 would have been some meaningful FirstChildUserIdTAG: 270939 FirstChildUserNameTAG: lamsalritu FirstChildCreateTimeTAG: 2012-09-24T08:12:30Z SecondChildTAG: There is a good example in the TB: chapter 6 (the MOSFET switch), p.298 (under the Figure 6.28). SecondChildUserIdTAG: 308508 SecondChildUserNameTAG: Alex-Fion SecondChildCreateTimeTAG: 2012-09-24T09:51:41Z SecondChildTAG: Static discipline is where we choose to constrain ourselves (read: discipline) to a set of rules. I recommend that you watch S5V8-S5V13 once more, and, afterwards, read chapter 6, [p.285][1]-[p.300][2]. Alex-Fion had a good idea, but since you are having trouble in this area, I do not think it would be sufficient or a good idea to just read that one example, and those links are a part of the required reading for this section anyway. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/285/ [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/300/ SecondChildUserIdTAG: 248807 SecondChildUserNameTAG: 1kingsman SecondChildCreateTimeTAG: 2012-09-25T03:40:32Z SecondChildTAG: Thanks...to all SecondChildUserIdTAG: 171199 SecondChildUserNameTAG: Dhananjay01 SecondChildCreateTimeTAG: 2012-09-28T03:30:44Z IndexTAG: 2565 TitleTAG: small signal analysis Is small signal analysis similar to assuming ideal conditions for a device, like current source replaced by open ckt and voltage source replace by short ckt as mentioned above. UserIdTAG: 343388 UserNameTAG: bijojoseph CreateTimeTAG: 2012-09-23T14:18:07Z VoteTAG: 1 CoursewareTAG: Week 4 / Small Signal Model CommentableIdTAG: 6002x_small_signal NumberOfReplyTAG: 1 FirstChildTAG: Not really. It's actually approximating the device. If you look at the VI characteristics of these devices, the graph "looks" like a straight line at the central region. If you zoomed in, you would see that it is in-fact curved. However , if we zoom out to a scale which is about the same as our input signal, the straight line approximation is good enough for practical purposes. By practical purposes I mean - you can amply music with those tiny distortions and your ear won't realize it. You can modify a picture but your eyes won't perceive those distortions. So small signal analysis is really an approximation which makes the design easier. It introduces an error which we can live with. FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-09-23T16:36:38Z IndexTAG: 2566 TitleTAG: Just wondering, Do we have to read chapter 4 on non-linear circuits? ? UserIdTAG: 254325 UserNameTAG: bondablack CreateTimeTAG: 2012-09-23T12:13:58Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: At least read 4.1, 4.3 and 4.5. I don't think you "have" read anything if you don't want to. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-23T14:20:00Z IndexTAG: 2567 TitleTAG: midterm exam what is gng to be the pattern of the midterm exam?? i mean how much time i wud get to take my exam and do i have to do it on the same day? UserIdTAG: 145503 UserNameTAG: chirag3553 CreateTimeTAG: 2012-09-23T11:41:53Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 2568 TitleTAG: Equations: Dependent On page 58: "Summing equations 2.5 through .28 yields 0 = 0. This in turn shows that the four KCL equations are dependent. In fact, a circuit with N nodes will have only N-1 independent statements of KCL." I understand that if an identity is reached when solving simultaneous equations then the equations are dependent, however the last sentence doesn't make any sense to me, what does "N-1 independent statements of KCL" mean? Thanks UserIdTAG: 35758 UserNameTAG: willprice94 CreateTimeTAG: 2012-09-23T10:18:52Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 2569 TitleTAG: Error in transcript At the beginning of the tutorial the transcript reads "Circus" for circuits. UserIdTAG: 396673 UserNameTAG: anthonypraveen CreateTimeTAG: 2012-09-23T10:02:29Z VoteTAG: 1 CoursewareTAG: Week 2 / Is it linear? CommentableIdTAG: 6002x_is_it_linear NumberOfReplyTAG: 0 IndexTAG: 2570 TitleTAG: is there mistake? (Another load line example) When voltage VA>1 expression must be: iA=VA/R1+(VA-1V-1V)/R2 first 1V comes from voltage source, second 1V - from potential drop at open diode. UserIdTAG: 192653 UserNameTAG: Quaz CreateTimeTAG: 2012-09-23T08:52:00Z VoteTAG: 1 CoursewareTAG: Week 4 / Textbook Load Line Tutorial CommentableIdTAG: 6002x_txtbk_load_line_t NumberOfReplyTAG: 1 FirstChildTAG: The diode is a ideal one;so ,when VA>1 it is simply a wire,and therefore there is not voltage across it. FirstChildUserIdTAG: 282828 FirstChildUserNameTAG: Chibolator FirstChildCreateTimeTAG: 2012-10-05T16:42:45Z IndexTAG: 2571 TitleTAG: LAB2 Impossible After 3 days of unbearable pain I've solved the Lab 2, but can someone please explain me why is impossible to solve this Lab with the configuration shown in the example? Although I did the Lab, I still don't get it... why is impossible? What is the theoretical foundation? UserIdTAG: 376173 UserNameTAG: nacho110987 CreateTimeTAG: 2012-09-23T07:19:48Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: ok the point is it hard to generate the ration of 1/2V1 +1/6V2 by using two resistor cause the two resistor can only generate 1:2 ration FirstChildUserIdTAG: 475448 FirstChildUserNameTAG: msamwelmollel FirstChildCreateTimeTAG: 2012-09-23T07:45:44Z SecondChildTAG: It's not correct, msamwelmollel. With two resistors you can generate any ratio, but only proportional: a*V1+b*V2 where a+b=1 SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-23T07:51:08Z SecondChildTAG: so what ratio do we use? SecondChildUserIdTAG: 380743 SecondChildUserNameTAG: MANQ SecondChildCreateTimeTAG: 2012-09-23T08:46:48Z SecondChildTAG: ratio in the LAB 1/2 and 1/6 - as soon 1/2+1/6 != 1 it's impossible to solve it with two resistors SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-23T08:55:57Z SecondChildTAG: Probably could have been made clearer that figure 2 simply illustrates an example of simple signal mixing, and that the solution circuit might be a tad more complex as regards the resistor configuration. SecondChildUserIdTAG: 141108 SecondChildUserNameTAG: rl777 SecondChildCreateTimeTAG: 2012-09-23T09:38:54Z FirstChildTAG: Probably could have been made clearer that figure 2 simply illustrates an example of simple signal mixing, and that the solution circuit might be a tad more complex as regards the resistor configuration. FirstChildUserIdTAG: 141108 FirstChildUserNameTAG: rl777 FirstChildCreateTimeTAG: 2012-09-23T09:55:55Z FirstChildTAG: Using two resistors , you will get two different values of R1 in terms of R2(only i.e. no other variable) which is impossible.......so it can be solve using 3 resistors........ FirstChildUserIdTAG: 114864 FirstChildUserNameTAG: Vikaash FirstChildCreateTimeTAG: 2012-09-23T13:40:56Z FirstChildTAG: Hi nacho110987! You can see an explanation in this Post [Here][1]. :). [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_S3E2_Linear_Combinations_of_Source_Strengths/threads/506a280d036af91f000000c9 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-02T00:04:34Z IndexTAG: 2572 TitleTAG: again LAB2 This is what I did: when V1 is maximum (1V) and V2 is maximum (1V) the resulting voltage is 667V. when V1 is minimum (0V) and V2 is minimum (-1V) the resulting voltage is -167V. so I used KCL in the output node and I had 2 variables and 2 functions like so: (667-1)/r1 + (667-1)/r2 = 0 (-167-0)/r1 + (-167+1)/r2 = 0 but this doesn't have any solution, what am I doing wrong? UserIdTAG: 419186 UserNameTAG: DanielCuartas CreateTimeTAG: 2012-09-23T04:40:38Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Are you sure that it is 667 Volts and not 667mili Volts? ... You can read some [hints][1] here ;). Can I help you? - Edit: Now that the deadline has passed :). You can take a look [here][2] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/505e458499dd04270000001c [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_S3E2_Linear_Combinations_of_Source_Strengths/threads/506a280d036af91f000000c9 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-23T04:45:51Z SecondChildTAG: I will conect today but at afternoon (I am going to sleep now haha). But if you have any doubt I will be pleasured to help you. Please read my hints in that post. I will back later ;). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-23T04:49:51Z SecondChildTAG: i don't think that this post is a good advice to read about this lab. SecondChildUserIdTAG: 252475 SecondChildUserNameTAG: alexha SecondChildCreateTimeTAG: 2012-09-23T04:50:32Z SecondChildTAG: Hi alexha! of course there are a lot of posts more helpful about lab2 in search :). But there I wrote some comments to Jamshaid271 that might can help to DanielCuartas and I didn't want to re-typed it again ... SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-23T04:57:27Z SecondChildTAG: I will back later ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-23T05:00:12Z SecondChildTAG: use 3 resistor and appropriate value such that your graph wave is similar to the given....And one of resistor which is between Vout and Ground has value 0 SecondChildUserIdTAG: 393950 SecondChildUserNameTAG: RaoUmairTufailAnjum SecondChildCreateTimeTAG: 2012-09-23T05:10:40Z SecondChildTAG: Maybe you are doing it the wrong way, think about the resistive mixer circuit, what it does, and how its done, there you have a simple formula vOut=0.5V1+0.5V2, but now according to the description of the problem it has to be vOut=0.5V1+V2/6, then think, what can i change in the circuit in order to modify this?, remember too, the answers are always simple and easy. SecondChildUserIdTAG: 370343 SecondChildUserNameTAG: JJarquin SecondChildCreateTimeTAG: 2012-09-23T05:56:29Z SecondChildTAG: Hi DanielCuartas You can read an answer to you post request [here][1] ;) [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/505f41be015492230000001c SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-23T17:14:09Z IndexTAG: 2573 TitleTAG: date of homeworks when it is said homework is due 23rd does that mean we can suubmit the homework on or before 23rd or before 23rd UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-23T03:29:21Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: You can submit the homework on the 23rd. Good luck! FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-23T04:06:08Z SecondChildTAG: It's due before midnight your local time on the 23rd. SecondChildUserIdTAG: 303727 SecondChildUserNameTAG: skyking SecondChildCreateTimeTAG: 2012-09-23T04:55:21Z FirstChildTAG: i think it refers to submitting the homework before the start of 23rd GMT FirstChildUserIdTAG: 295303 FirstChildUserNameTAG: rickyzmkuo FirstChildCreateTimeTAG: 2012-09-23T07:25:57Z IndexTAG: 2574 TitleTAG: Lab 2 error? Transient analysis signal matches requirement but checking says i have wrong answer, help? ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/1348370396134361.jpg UserIdTAG: 185610 UserNameTAG: AdityaSR CreateTimeTAG: 2012-09-23T03:20:43Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Hi AdityaSR! Are you sure that matches the requirement? Remember that the signal wave form it is not the only condition. You have to verify that the value of the wave it it from some given Vomin value to some given Vomax value... does verify the Vomin too? If everything it is correct, try to press again the check buttom and then F5 to refresh ... If anything happens, Can I help you? might you are doing something wrong... Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-23T03:56:30Z FirstChildTAG: Did you by any chance delete the node named "output"? The checker will verify the voltage at the node named "output", so if you don't have it, it'll be incorrect. FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-23T03:56:59Z FirstChildTAG: use 3 resistor and appropriate value such that your graph wave is similar to the given....And one of resistor which is between Vout and Ground has value 0 FirstChildUserIdTAG: 393950 FirstChildUserNameTAG: RaoUmairTufailAnjum FirstChildCreateTimeTAG: 2012-09-23T05:14:15Z IndexTAG: 2575 TitleTAG: LAB2 anyone that has gotten lab2 right, please help me out, thanks UserIdTAG: 254607 UserNameTAG: werehenry CreateTimeTAG: 2012-09-23T00:02:14Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi werehenry! In which part of the Lab2 are you lost? Can I help you? :) Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-23T01:12:39Z SecondChildTAG: use 3 resistor and appropriate value such that your graph wave is similar to the given....And one of resistor which is between Vout and Ground has value 0 SecondChildUserIdTAG: 393950 SecondChildUserNameTAG: RaoUmairTufailAnjum SecondChildCreateTimeTAG: 2012-09-23T05:15:00Z SecondChildTAG: Hi werehenry you can see an answer to your post request [here][1] ;) [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/505f41be015492230000001c SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-23T17:15:27Z IndexTAG: 2576 TitleTAG: lab 2 week 2 i made the 3 resistors configuration and made the wave form exactly same as the given , but i still get that it is wrong , does it need a specific configuration of resistors or any config that give the same wave from with the same volts values , max and min please reply me UserIdTAG: 285675 UserNameTAG: Ascot CreateTimeTAG: 2012-09-22T21:52:57Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Even if the wave does look the same as in the example, are you sure the output is between 667 mV and -167 mV? FirstChildUserIdTAG: 101680 FirstChildUserNameTAG: 4o4error FirstChildCreateTimeTAG: 2012-09-22T22:12:47Z FirstChildTAG: finally i solved it , even i made like 50 try and error but the solution is that i just put the third resistors beside R1 , even i don't know what is the difference , and the third ressitor is got by Rth and get the new Vout from voltage driver between Rth and R3 FirstChildUserIdTAG: 285675 FirstChildUserNameTAG: Ascot FirstChildCreateTimeTAG: 2012-09-22T22:32:19Z SecondChildTAG: You could think of the third resistor as a volume control. The first two set the balance of R1 and R2, the third bleeds off excess power to ground. You could use 3 completely different resistor values then what you have, as long as the proportion between them is the same. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-22T23:14:07Z IndexTAG: 2577 TitleTAG: Awesome This guy is amazing. So smart, such a good teacher and puts so much effort in what he does. I'm not sure if i've ever actually met such a cool person. UserIdTAG: 405009 UserNameTAG: kubi CreateTimeTAG: 2012-09-22T21:30:54Z VoteTAG: 1 CoursewareTAG: Week 2 / Chain saw CommentableIdTAG: 6002x_chain_saw NumberOfReplyTAG: 0 IndexTAG: 2578 TitleTAG: How to read comments :( ? Guys I'm unable to read comments here ... i don't know how to open them .. can someone plz send me the procedure on my id sabasiddiqi098@hotmail.com UserIdTAG: 332941 UserNameTAG: SabaSiddiqi CreateTimeTAG: 2012-09-22T20:52:56Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: By any chance you are using Internet Explorer? The forum service has issues with IE. If so, could you try Chrome or Firefox? FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-22T21:56:16Z SecondChildTAG: thanks ... its working fine with chrome :) SecondChildUserIdTAG: 332941 SecondChildUserNameTAG: SabaSiddiqi SecondChildCreateTimeTAG: 2012-09-22T22:28:37Z IndexTAG: 2579 TitleTAG: lab 2 solution problem i calculate 1/6 of v2 by voltage divider then 1/2 of v1 again with voltage divider both have on resistance common but output is not same kindly guide me . i use R3 of 1k as common, 5k for v2 and 1k for v1 . what is the mistake . guide me kindly .thanks :) UserIdTAG: 438395 UserNameTAG: ali_PU1 CreateTimeTAG: 2012-09-22T20:20:22Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hint from the schematic, Vout=0.5(V1+V2) or, Vout=half of V1+V2 FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-22T20:32:18Z SecondChildTAG: comparing equations we have, (R2/(R1+R2))=1/2 & (R1/(R1+R2))= 1/6 which gives R1=R2 & 5R1=R2.. how is this possible? SecondChildUserIdTAG: 260272 SecondChildUserNameTAG: saikat24 SecondChildCreateTimeTAG: 2012-09-22T22:48:27Z SecondChildTAG: Those are 2 separate equations, the R1 and R2 on the first line have nothing to do with the R1 and R2 on the second line. I struggled for 90 minutes before I started breaking down the problem and watching how my attempts were shifting around. Hint: This can be solved with 3 resistors and it is 2 parallel circuits in combination. This is a neat application of "equivalent circuits" applied to a real-world problem. SecondChildUserIdTAG: 366165 SecondChildUserNameTAG: silicon_ghost SecondChildCreateTimeTAG: 2012-09-23T00:18:13Z FirstChildTAG: If you analyze your result, you will see that it doesn't give the correct result. For example, V1 is in series with R1 and R2//R3, which doesn't give the correct result. I suspect that you designed thinking that V1 was in series with R1 and R3 alone. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-09-22T23:13:13Z IndexTAG: 2580 TitleTAG: I need some info I just signed in the program, but I saw the characterics of the program, and I don't think that I can resolve the homeworks and activities for this sunday, how can i bring up to date? i want finish the course and i don't want to lose the credits for approving UserIdTAG: 479215 UserNameTAG: joseangelescobar CreateTimeTAG: 2012-09-22T19:25:54Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: If you miss tomorrows homework, that is OK, just be aware that you cannot miss any more without consequence. (You can miss two). Do your best to catch up when you can, you can still do very well in the course. Have fun. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-22T19:37:37Z SecondChildTAG: Thanks for the info and your time SecondChildUserIdTAG: 479215 SecondChildUserNameTAG: joseangelescobar SecondChildCreateTimeTAG: 2012-09-22T23:08:45Z IndexTAG: 2581 TitleTAG: Lab2 what am I missing Calculated the 1/2V1 and the 1/6V2 if connect one of the sources to the node that part seems fine, but when connecting both of them its way of. What am I missing?(used 3 resistors) UserIdTAG: 70211 UserNameTAG: Emberfej CreateTimeTAG: 2012-09-22T19:00:25Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: "Lab2 what am I missing" You are missing the search box. Type "Lab 2" into the search box, press enter. Voila! You'll have your answer and some left over for sandwiches. Have fun. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-22T19:25:25Z SecondChildTAG: Thanks but I am not a forum eater and didn't find what I am looking for. SecondChildUserIdTAG: 70211 SecondChildUserNameTAG: Emberfej SecondChildCreateTimeTAG: 2012-09-22T19:47:18Z SecondChildTAG: lol. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-22T19:55:17Z SecondChildTAG: Get your ratios right, then forget them. Use a resistor to shed the extra unwanted voltage to ground. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-22T19:56:45Z FirstChildTAG: i calculate 1/6 of v2 by voltage divider then 1/2 of v1 again with voltage divider both have on resistance common like in the above circuit but output id not same kindly guide me . i use R3 of 1k as common, 5k for v2 and 1k for v1 . what is the mistake . guide me kindly .thanks :) FirstChildUserIdTAG: 438395 FirstChildUserNameTAG: ali_PU1 FirstChildCreateTimeTAG: 2012-09-22T19:30:01Z FirstChildTAG: Hi darksamaro, i have the same configutarion. I´ve calculated this exercise over and over again, and starting to get frustated. This is as far as i am: 1. V1 acting alone, R1 = R2||R3 2. V2 acting alone is a problem, because i get 5R1 = R2||R3. I tried to solve for another value and i got R2 = 5R1R3 / (R3-5R1) 3. After having values for R1 and R2, I tried setting a value for R3 (ie R3=1), but when i do that the values for R1 and R2 resolve to 0!!! I am trying to stick to the first part, where I know that R1=R2||R3 and got close by trial and error, but its taking me too long and still no green check. Any clues? FirstChildUserIdTAG: 35642 FirstChildUserNameTAG: caled FirstChildCreateTimeTAG: 2012-09-22T19:24:47Z IndexTAG: 2582 TitleTAG: Calculus lab 0 I have to take on account the path of the current in each node to calculate the voltage value between that one, and the end KVL says that the sum of nodes is equal to the value of source. UserIdTAG: 472876 UserNameTAG: otakudany CreateTimeTAG: 2012-09-22T18:29:34Z VoteTAG: 1 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 0 IndexTAG: 2583 TitleTAG: Week 2 lab 2 Can we only use 2 resistors in this lab? Or can we use multiple resistors to create the different dividers required? UserIdTAG: 349840 UserNameTAG: Wilk CreateTimeTAG: 2012-09-22T18:18:59Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: You need 3 resistors. With two you only can divide voltage proportionally. FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-09-22T19:00:01Z FirstChildTAG: To solve this problem you will need only 2 resistors. You need to change the configurations of the sources. FirstChildUserIdTAG: 399292 FirstChildUserNameTAG: Qabali FirstChildCreateTimeTAG: 2012-09-22T18:32:36Z SecondChildTAG: I though the instruction said we can't change the sources. > Please do not modify the wiring or parameters of the voltage sources > -- your goal is to take the signals they generate and combine them, not to change what is generated. So with that in mind can you explain to me what you are saing exactly? SecondChildUserIdTAG: 349840 SecondChildUserNameTAG: Wilk SecondChildCreateTimeTAG: 2012-09-22T18:37:48Z SecondChildTAG: Attention: The instructions tell us to not modify the sources. It is not possible to solve the problem using just 2 resistors. If you try you will end up with a system of linear equations that will not produce valid solutions for our domain. Notice that adding more resistors in series, I mean trying 4 or more resistors in series will not help you again because the circuit could be simplified to an equivalent circuit with just 2 resistors in series and you would face the same problem I pointed up above. HINT: Try a different configuration of resistors. You could begin with 3 resistors. Eg. You could connect 3 resistors in parallel. SecondChildUserIdTAG: 305491 SecondChildUserNameTAG: lefam SecondChildCreateTimeTAG: 2012-09-22T19:02:52Z SecondChildTAG: Qabali, please don't change the configuration of the sources. SecondChildUserIdTAG: 381746 SecondChildUserNameTAG: jelizon SecondChildCreateTimeTAG: 2012-09-22T19:28:59Z FirstChildTAG: [please look through similar discussions before asking][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/505d5d527876112b00000029 FirstChildUserIdTAG: 5325 FirstChildUserNameTAG: vkz FirstChildCreateTimeTAG: 2012-09-22T19:08:49Z SecondChildTAG: I did read over the similar results before posting, but I have seen people who claim to have answers for networks ranging form 2-6 resistors. At this point I am sure you can't solve it with 2, but unsure of much else. SecondChildUserIdTAG: 349840 SecondChildUserNameTAG: Wilk SecondChildCreateTimeTAG: 2012-09-22T19:51:14Z FirstChildTAG: Simple answer is NO. If you try to use only 2 resistors, you'll end up in the very beginning of solving this problem with system of equations (for two voltage dividers) that contains a paradox, so it seems that you have to use (at least) 3 resistors. FirstChildUserIdTAG: 192703 FirstChildUserNameTAG: voffch FirstChildCreateTimeTAG: 2012-09-22T20:58:27Z FirstChildTAG: Attention: The instructions tell us to not modify the sources. It is not possible to solve the problem using just 2 resistors. If you try you will end up with a system of linear equations that will not produce valid solutions for our domain. Notice that adding more resistors in series, I mean trying 4 or more resistors in series will not help you again because the circuit could be simplified to an equivalent circuit with just 2 resistors in series and you would face the same problem I pointed up above. HINT: Try a different configuration of resistors. You could begin with 3 resistors. Eg. You could connect 3 resistors in parallel. FirstChildUserIdTAG: 305491 FirstChildUserNameTAG: lefam FirstChildCreateTimeTAG: 2012-09-22T19:03:44Z SecondChildTAG: I am working off of this idea now. but still coming up close, but not close enough. SecondChildUserIdTAG: 349840 SecondChildUserNameTAG: Wilk SecondChildCreateTimeTAG: 2012-09-22T19:52:37Z IndexTAG: 2584 TitleTAG: Urgent - Answers are not being checked Dear Staff, Could you help (a few of) us to resolve the error with Check button, on this page (S4E2)? I could enter 2 of the answers and get them checked. But for remaining two questions, the Check button does not do anything. I am using Firefox 15.0.01, on Win XP. Internet Explorer was also tried with, but it is worse there, as the "show Discussion" too, does not work. Request you to help us resolve this. Otherwise, things are just awesome here - the tools, the ease in accessing information and everything else, is just great. Thanks a lot for a pleasant experience. Warm regards, Amitraj UserIdTAG: 372623 UserNameTAG: Amitraj CreateTimeTAG: 2012-09-22T17:39:50Z VoteTAG: 1 CoursewareTAG: Week 2 / Boolean functions CommentableIdTAG: 6002x_S4E2_Boolean_Functions NumberOfReplyTAG: 1 FirstChildTAG: edX suggests that you use google chrome. FirstChildUserIdTAG: 254325 FirstChildUserNameTAG: bondablack FirstChildCreateTimeTAG: 2012-09-23T12:45:12Z IndexTAG: 2585 TitleTAG: Where is Discussion Forum Help? Is this where we get discussion forum help? UserIdTAG: 175706 UserNameTAG: rwskim CreateTimeTAG: 2012-09-22T17:36:03Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: YES FirstChildUserIdTAG: 40209 FirstChildUserNameTAG: sadaf78 FirstChildCreateTimeTAG: 2012-09-22T17:47:48Z FirstChildTAG: Why can't I see the comments in the discussions??? FirstChildUserIdTAG: 175706 FirstChildUserNameTAG: rwskim FirstChildCreateTimeTAG: 2012-09-22T18:04:20Z SecondChildTAG: Can you read this? SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-22T18:07:26Z SecondChildTAG: Are you using Internet Explorer? There is unfortunately a bug in our code when it runs on IE and we haven't figured it out yet. Unfortunately you probably won't be able to see this either... but I am narrowing the bug down and hope to find a fix this weekend. SecondChildUserIdTAG: 323387 SecondChildUserNameTAG: ibrahimawwal SecondChildCreateTimeTAG: 2012-09-23T02:34:53Z SecondChildTAG: Aha, I found it. IE should be fixed the next time the site is updated. SecondChildUserIdTAG: 323387 SecondChildUserNameTAG: ibrahimawwal SecondChildCreateTimeTAG: 2012-09-23T06:10:02Z FirstChildTAG: Hi rwskim! Might you can not see the comments because of this : ![enter image description here][1] If you see, the comments are in gray color... If you don't see that might it is because of what kimt said in that post ;). You can read that post [here][2]. See you! Myriam. [1]: https://edxuploads.s3.amazonaws.com/13483614815622154.png [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/505e25283451302300000021 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-23T00:53:49Z SecondChildTAG: Unfortunately he probably won't be able to see this either... Fortunately I have a spare Windows machine this weekend and can take a look at what the IE problem is. I will try and fix it. SecondChildUserIdTAG: 323387 SecondChildUserNameTAG: ibrahimawwal SecondChildCreateTimeTAG: 2012-09-23T02:35:51Z SecondChildTAG: Thank you IbrahImawwal ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-23T04:01:49Z FirstChildTAG: thanks everyone! FirstChildUserIdTAG: 175706 FirstChildUserNameTAG: rwskim FirstChildCreateTimeTAG: 2012-10-07T16:36:47Z SecondChildTAG: You are welcome rwskim ;). I am guessing that now you can see our comments haha. What were your issue? I am curious and how did you solve it? SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T17:55:59Z IndexTAG: 2586 TitleTAG: I NEED HELP IN LAB 2 BEFORE TOMORROW, PLEASE I could make circuit giving the required output voltage, I did lot of trials but not getting the same graph, somebody please tell me how to solve it than going for trial, I think I didnt get it how to approach the problem.Please help me out UserIdTAG: 284628 UserNameTAG: jumana_mp CreateTimeTAG: 2012-09-22T17:02:13Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: [Don't post answers in this forum] - Staff FirstChildUserIdTAG: 297163 FirstChildUserNameTAG: Suman005 FirstChildCreateTimeTAG: 2012-09-22T19:08:48Z FirstChildTAG: Hi jumana_mp! Yes, sure. In what can I help you? Do you undertand the objective of the Lab? Remember that you can not change the two voltage supply (not modify the parameters) and as this is a graded part I can only give you hints that you can solve this by yourself ;). Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-23T01:25:50Z IndexTAG: 2587 TitleTAG: Serious probleM Hi edx fellows. There's a serious problem . Im in Pakistan and here youtube is banned by the authorities. and it cant be opened on other proxy sites. So i cant watch the tutorial videos. now what to do .. please tell me tomorrow is the last date to submit the homework and labwork no. 2. UserIdTAG: 131113 UserNameTAG: MAKsci CreateTimeTAG: 2012-09-22T15:49:28Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: **yes you can easily do it by reading book plus you can go to MIT'S Open course site and download previous course videos but still watching tutorials is not possible without YouTube.** FirstChildUserIdTAG: 40209 FirstChildUserNameTAG: sadaf78 FirstChildCreateTimeTAG: 2012-09-22T17:25:49Z FirstChildTAG: Hi MAKsci, I'm from Indonesia, maybe you can try to solve it without having to see the vid, for example by reading the book and it is still OK because I do that FirstChildUserIdTAG: 452559 FirstChildUserNameTAG: kuz1toro FirstChildCreateTimeTAG: 2012-09-22T16:17:12Z SecondChildTAG: Even i need help in LAB 2 !! SecondChildUserIdTAG: 414516 SecondChildUserNameTAG: randevakash SecondChildCreateTimeTAG: 2012-09-22T17:02:13Z SecondChildTAG: what help tell us? SecondChildUserIdTAG: 40209 SecondChildUserNameTAG: sadaf78 SecondChildCreateTimeTAG: 2012-09-22T17:28:27Z FirstChildTAG: Hey Maksci, I know that books don't help nicely. So hide your ip address. You will able to switch on youtube. I am also from Pakistan and watching youtube. FirstChildUserIdTAG: 378096 FirstChildUserNameTAG: Jamshaid271 FirstChildCreateTimeTAG: 2012-09-22T23:53:36Z FirstChildTAG: How to hide IP FirstChildUserIdTAG: 358539 FirstChildUserNameTAG: syed_abdullah FirstChildCreateTimeTAG: 2012-09-23T18:04:34Z SecondChildTAG: Download "Super hide IP" from torrent and let things go slightly. SecondChildUserIdTAG: 378096 SecondChildUserNameTAG: Jamshaid271 SecondChildCreateTimeTAG: 2012-09-23T19:49:05Z IndexTAG: 2588 TitleTAG: How to get this, I have no clue Find the value for the incremental resistance of the nonlinear element N by linearizing the expression for iA about the operating point when vI=5.0V. UserIdTAG: 292360 UserNameTAG: Loai CreateTimeTAG: 2012-09-22T15:23:25Z VoteTAG: 1 CoursewareTAG: Week 4 / Linearization Exercise 1 CommentableIdTAG: 6002x_linearization_ex_1 NumberOfReplyTAG: 1 FirstChildTAG: Have you take a look on the videos S7V7, S7V8 and S7V9? You are suppose to use the incremental method based on the Taylor expansion of the function. The process and an example are given in the lectures. You can also go to your textbook (section 8.2) for a more detailed mathematical description of the small signal model. FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-22T19:11:41Z SecondChildTAG: Read section 4.5 of the textbook, that helped me to find the last answer, i hope that will help you too SecondChildUserIdTAG: 370343 SecondChildUserNameTAG: JJarquin SecondChildCreateTimeTAG: 2012-09-23T05:44:02Z SecondChildTAG: reading 4.5 helped me ...thanks! SecondChildUserIdTAG: 164471 SecondChildUserNameTAG: BillNieves SecondChildCreateTimeTAG: 2012-09-29T22:13:43Z IndexTAG: 2589 TitleTAG: How to find the RON for each MOSFET in Lab 3? My arrangement seems to be correct looking at the diagram. How to find the RON for each individual MOSFET? UserIdTAG: 295240 UserNameTAG: Jack_my CreateTimeTAG: 2012-09-22T14:12:54Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: You can make them all the same. And write down the eqn for a voltage divider with Ron as the unknown for the worst case VOL and then use the given formula to calculate L/W. The Rn is also given. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-09-22T14:50:38Z SecondChildTAG: 1)Found out (3/2)RON=909.09 taking V=0.25 2) Using the formula given, found out W/L but still it is showing wrong. 3) Any mistakes? SecondChildUserIdTAG: 295240 SecondChildUserNameTAG: Jack_my SecondChildCreateTimeTAG: 2012-09-22T15:13:16Z FirstChildTAG: 1)Found out (3/2)RON=909.09 taking V=0.25 2) Using the formula given, found out W/L but still it is showing wrong. 3) Any mistakes? FirstChildUserIdTAG: 295240 FirstChildUserNameTAG: Jack_my FirstChildCreateTimeTAG: 2012-09-22T15:15:26Z SecondChildTAG: I don't know if you made a mistake, because I don't know what circuit you use. But to be honest, you don't even have to calculate Ron itself, because Ron disappears when you equal the two formulas. If you have 2 fets parallel, then the worst case is when only 1 fet is on, and if you have 2 fets in series, then worst case is 2*Ron. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-09-22T20:08:47Z FirstChildTAG: You must pay attention on the spikes or glitches, take a look at it more clossely, it must below the VoL FirstChildUserIdTAG: 452559 FirstChildUserNameTAG: kuz1toro FirstChildCreateTimeTAG: 2012-09-22T16:09:32Z IndexTAG: 2590 TitleTAG: confused due to the "-7.2v" I think most of us including me are confused because of the V2 source. After solving this problem with KVL i found out that the phrase "*V2=-7.2v*" in the question is the value of V2 regardless of how it is connected and regardless of its polarity in the diagram! if we use the node method and solve while keeping that in mind, we get the correct answer ! UserIdTAG: 130810 UserNameTAG: moon_light CreateTimeTAG: 2012-09-22T13:30:21Z VoteTAG: 1 CoursewareTAG: Week 2 / Node equation review CommentableIdTAG: 6002x_S3E1_NodeEquationReview NumberOfReplyTAG: 2 FirstChildTAG: Actually it does matter how V2 is connected +/- or -/+. If you consider V1 -to+ is 5V, you should consider V2 -to+ is (-7.2V) as it given. And if you consider ground (which is V2 "+") is 0, then, considering V2 potential you should take V2 "-" equal to +7.2V. In case of opposite V2 polarity, you would took V2 "-" equal to (-7.2V). FirstChildUserIdTAG: 66669 FirstChildUserNameTAG: agarus FirstChildCreateTimeTAG: 2012-09-22T14:58:07Z FirstChildTAG: Just think about a voltage source as a device that creates a voltage difference "V" between two nodes. The voltage difference can be positive or negative. When you move from the node with labeled "-" to the one labeled "+", you have to add "V" to the potential. If "V" is negative, then you add a negative value, meaning that the "+" node is actually at a lower voltage than the "-" node, regardless of how the source is oriented (up, down, left, right or tilted). Setting V2 as negative was precisely meant to be confusing so that, by solving it correctly, you can be sure that negative voltages and/or currents won't surprise you later in this course! FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-22T18:58:53Z IndexTAG: 2591 TitleTAG: Vth Rth is R1 and R2 in parallel UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-22T12:58:51Z VoteTAG: 1 CoursewareTAG: Week 2 / Simple Thevenin CommentableIdTAG: 6002x_S3E4_Simple_Thevenin NumberOfReplyTAG: 1 FirstChildTAG: I am confused. For Rth, You have to remove all independent sources. i.e. Short Circuit Vs. When you do that, you are left with R1 and R2. How come they are parallel. They seem to be in series. See my description. ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13483874336515982.jpg FirstChildUserIdTAG: 229430 FirstChildUserNameTAG: MuhammadAli201 FirstChildCreateTimeTAG: 2012-09-23T08:04:26Z SecondChildTAG: I also see series... SecondChildUserIdTAG: 230787 SecondChildUserNameTAG: Gabriel007 SecondChildCreateTimeTAG: 2012-09-23T16:57:56Z SecondChildTAG: Hmm, ok I can see parallel now. I think you have to imagine some current is coming in through the top ---o and exiting on the bottom ---o, so the current must split into the edge where you find R2 and the edge where R1 resides. SecondChildUserIdTAG: 230787 SecondChildUserNameTAG: Gabriel007 SecondChildCreateTimeTAG: 2012-09-23T17:01:16Z IndexTAG: 2592 TitleTAG: H2P2 In homework 2 pt 2 my power through RL should be 0.024w, but that's wrong! Why? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-22T12:24:08Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I found the RL just with the sum of the other ones resistor at the source.. But I'm not quite sure that's simple like that. And also I can't find the power delivered to LOAD just making the equation for power using my resistor and current value = P = R*(I^2). Someone knows how to explain it? Tks. FirstChildUserIdTAG: 316902 FirstChildUserNameTAG: JesseTeixeira FirstChildCreateTimeTAG: 2012-09-22T13:20:36Z SecondChildTAG: i also have same problem. SecondChildUserIdTAG: 342135 SecondChildUserNameTAG: vikash902 SecondChildCreateTimeTAG: 2012-09-22T13:51:09Z SecondChildTAG: me too. I got the correct value for both resistors but can't figure out the power value! oO SecondChildUserIdTAG: 308853 SecondChildUserNameTAG: fmorato SecondChildCreateTimeTAG: 2012-09-22T14:06:58Z SecondChildTAG: here is some clue: 1. you must simplify the circuit using Thevenins theorem so there will be only Vth, Rth and RL in series 2. to find RL that give max power (Hint: remember your calculus!), you know that derivative stuff, in this case dP/dRL = 0 hope it usefull SecondChildUserIdTAG: 452559 SecondChildUserNameTAG: kuz1toro SecondChildCreateTimeTAG: 2012-09-22T14:35:31Z SecondChildTAG: Can someone give derivative example!! Really no idea bout this stuff. Thanks SecondChildUserIdTAG: 119160 SecondChildUserNameTAG: Thyaga SecondChildCreateTimeTAG: 2012-09-22T15:43:03Z SecondChildTAG: Hello all, accidentally enough, I came across this - In line with kuz1toro, if we get the Thevenin's equivalent of the circuit (again assuming that Rth and RL have been calculated) then you get to see a very well known pattern (as says Dr. Agrawal!) .. the voltage divider. That is surely a place, where one need not go to calculus. I could solve the problem without dealing with Calculus (though it is really underneath everywhere!) SecondChildUserIdTAG: 372623 SecondChildUserNameTAG: Amitraj SecondChildCreateTimeTAG: 2012-09-23T15:50:22Z FirstChildTAG: derivative have some theorems, I will give you some that related to the problem, to be more specific you need to read calculus book - the induction principle > let u = a.$x^n$+b where a and b are any constant and n are arbirary integer then $\frac{du}{dx}$=a.n.$x^{(n-1)}$. > example : $u=4 x^5+25 $ then $\cfrac{du}{dx}=4 \cdot 5 \cdot x^{(5-1)}+ 0=20x^4$ - Power rule theorem > let v = f(x) then $\cfrac{d(v^n)}{dx}$= $n\cdot v^{(n-1)}\cdot\cfrac{dv}{dx}$ > example : $v=5 x^2 $ then $\cfrac{d(v^4)}{dx}= 4\cdot 5 {x^2}^{(4-1)}\cdot 10x=200x^7$ - Sum rule > $\cfrac{d(u+v)}{dx}= \cfrac{du}{dx}+\cfrac{dv}{dx}$ CMIIW FirstChildUserIdTAG: 452559 FirstChildUserNameTAG: kuz1toro FirstChildCreateTimeTAG: 2012-09-22T20:53:54Z IndexTAG: 2593 TitleTAG: Lab-2 In Lab 2 I used voltage divider resistor just for Sin wave power source, do we need to add voltage divider circuit for Square wave power source as well? UserIdTAG: 441835 UserNameTAG: KAMBIZAMEEN CreateTimeTAG: 2012-09-22T11:41:12Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Can you rephrase that pls? Not clear what you mean. FirstChildUserIdTAG: 5325 FirstChildUserNameTAG: vkz FirstChildCreateTimeTAG: 2012-09-22T13:43:52Z FirstChildTAG: Resistor networks akts in the same way for DC, sin or square waves, they are linear elements. With two signal sources they are added to the output with the constants (fractions) times the original value. The constants you get from the equations. Variations in the signal don't influence the fraction of that signal that passes the resistor network. FirstChildUserIdTAG: 151472 FirstChildUserNameTAG: Stensmed FirstChildCreateTimeTAG: 2012-09-22T13:45:24Z IndexTAG: 2594 TitleTAG: vout Vout can be between 4 to 5 volt, too. if we write the voltage divider equation, the Rpu should be between 0 and 0.75 .So the minimum value of Rpu is 0. it contradicts the above solution. Please help me to understand the correct solution?![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13483125041343646.jpg UserIdTAG: 374393 UserNameTAG: rmaleki CreateTimeTAG: 2012-09-22T11:16:28Z VoteTAG: 1 CoursewareTAG: Week 3 / Circuit Truth Table Tutorial CommentableIdTAG: 6002x_circuit_truth_table_tutorial NumberOfReplyTAG: 1 FirstChildTAG: **Vout= (5v * Ron)/(Rp+Ron) thus: Rp = [((Vs*Ron)/ Vout) - Ron] Vout = 1V since the required is the minimum Rp** FirstChildUserIdTAG: 399292 FirstChildUserNameTAG: Qabali FirstChildCreateTimeTAG: 2012-09-22T13:03:57Z SecondChildTAG: The formula is correct. Just suppose that Ron is again 3K and vout is 4: Rp=5*3/4-3=0.75 that is smaller than 12k.!!!! SecondChildUserIdTAG: 374393 SecondChildUserNameTAG: rmaleki SecondChildCreateTimeTAG: 2012-09-22T15:08:54Z IndexTAG: 2595 TitleTAG: h3p3 &h3p4 with regard to h3p4, i dont even know how to go abt it. i am not even sure i get what is asked(i read it a million times :( ) i am terribly lost, please show me the way. h3p3 is relatively ok, but i still hv some issues. again, please help me understand.... UserIdTAG: 285945 UserNameTAG: lindalapiso CreateTimeTAG: 2012-09-22T10:58:27Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: ok, first you need to understand few things 1. logic gate are circuit (digital circuit) that can only have an input value 0 or 1 and also only can produce output value of 0 or 1. The number of the terminal for the input is from 1 to 2 but theirs only one terminal for the output, however we can combine several logic gate to form a more complex circuit or gate that can take more input terminal larger than two 2. there are 3 basic type of logic gate plus 2 type of combination gate, which is: > a. the AND gate, it is a gate that give an output 1, if and only if the two input value are 1, others will give a 0; this gate is like multiplication of the input value, that is why the arithmetic symbol for this gate are dot. > > b. the OR gate, it is a gate that give and output 0, if and only if the two input value are 0, others will give a 1. this gate represent in arithmetic symbol as + > > c. the NOT gate, it is an inverting gate that only have one input terminal and also one output terminal. it have the ability to inverse the input, so if the input is 1 then the output is 0, if the input is 0 then the output is 1 > > d. the NAND gate, it is a combination of the NOT and AND gate, the result is always the inverse version of the AND gate. for example, if there are two input at state 1 then the output for the AND gate are 1 but for the NAND gate, the output are 0, and so on for the other state > > e. the NOR gate, it is a combination of the NOT and OR gate so the result are always the inverse of the OR gate. 3. you must know the circuit symbol for this 5 gate, look for it at the lecture note or the books 4. the H2P3 and H2P4 problems are circuit that used combination of this five gate, try to solve it one gate at a time, try from the closest gate with the input terminal and move on until the farthest gate and of course one condition at a time I hope you found your "AHA moment" :) and I'm sorry for my English if it isn't clear FirstChildUserIdTAG: 452559 FirstChildUserNameTAG: kuz1toro FirstChildCreateTimeTAG: 2012-09-22T13:03:34Z SecondChildTAG: thank you for taking the time to help me out,kuz1toro. but question H3P4(diode limiter)of week 3 is about non-linear devices not gates. i think im supposed to implement piecewise linear analysis technique. intuitively, i do understand for the first half cycle of 'Vs' diode one is on(short) while the second diode is off(open) and vise-versa for the next half cycle. how do i prove this analytically n what comes next(im stuck from here on out). what am i missing? SecondChildUserIdTAG: 285945 SecondChildUserNameTAG: lindalapiso SecondChildCreateTimeTAG: 2012-09-22T13:21:51Z SecondChildTAG: man, I'm sorry, what a embarrassing, anyway for H3P4, you are right that for the firs half cycle of Vs, D1 is short and D2 is open. Ok stop here and draw that circuit (dont forget to replace the short diode with wire because it is in short state and vice versa for the open diode), from there you can calculate the V and I Now for the second half cycle of Vs, draw the circuit again and look and calculated and AHA so you analysis it by breaking it in two state, first in high state Vs and the second in low state Vs SecondChildUserIdTAG: 452559 SecondChildUserNameTAG: kuz1toro SecondChildCreateTimeTAG: 2012-09-22T15:00:27Z SecondChildTAG: thank you dear, i got it right! i owe u one SecondChildUserIdTAG: 285945 SecondChildUserNameTAG: lindalapiso SecondChildCreateTimeTAG: 2012-09-23T13:27:09Z SecondChildTAG: thanks a lot for the help :) SecondChildUserIdTAG: 157273 SecondChildUserNameTAG: ongchihang SecondChildCreateTimeTAG: 2012-09-25T10:11:53Z IndexTAG: 2596 TitleTAG: H2P1 could someone help me in understanding this problem..... the question says thevenin resistance is between 10 and 30.... is it total or defines a range.... UserIdTAG: 276450 UserNameTAG: ritece CreateTimeTAG: 2012-09-22T06:23:20Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Its just a range. You need complete an equation with 2 parallel resistor and given range FirstChildUserIdTAG: 444204 FirstChildUserNameTAG: hekan FirstChildCreateTimeTAG: 2012-09-22T06:30:41Z FirstChildTAG: thanks for your response..... i entered r1 as 82k and r2 as 18k. the thevenin resistance is well within the range defined and the last two answers which i found using these two values are being displayed green... but the resistance values are red... where am i making a mistake??? could i be helped please??? FirstChildUserIdTAG: 276450 FirstChildUserNameTAG: ritece FirstChildCreateTimeTAG: 2012-09-22T07:09:36Z SecondChildTAG: yes me too, my R1 and R2 values are also well within the range, r1 = 82k, r2 = 33/39k , but its still hating my answers uggg SecondChildUserIdTAG: 204745 SecondChildUserNameTAG: allwynmendes SecondChildCreateTimeTAG: 2012-09-22T09:07:37Z SecondChildTAG: I've got the same problem, my Vmin and Vmax are good, also my thevenin resitance is within range, but my two first answers are wong. I don't see where i'm wrong ;/ SecondChildUserIdTAG: 330357 SecondChildUserNameTAG: steryd SecondChildCreateTimeTAG: 2012-09-22T10:45:54Z SecondChildTAG: there are two condition that has to fulfill, the first is that Thevenins range and the second is the voltage divider condition,,,your answer must fulfill this two,,,, note that the resistor value are not fixed but rather its has range, so the output voltage also has range SecondChildUserIdTAG: 452559 SecondChildUserNameTAG: kuz1toro SecondChildCreateTimeTAG: 2012-09-22T11:18:01Z SecondChildTAG: Hi, keep in mind that for resistors in parallel the result is less than the least resistance is to say that if you need a resistor between 10 and 30 put a resistor 20 and calculate the otherone, thank SecondChildUserIdTAG: 343291 SecondChildUserNameTAG: LuisEduardoH SecondChildCreateTimeTAG: 2012-09-22T20:37:16Z FirstChildTAG: thank you everyone for your exciting response...... finally i have successfully completed week 2 problems.... FirstChildUserIdTAG: 276450 FirstChildUserNameTAG: ritece FirstChildCreateTimeTAG: 2012-09-22T23:50:01Z IndexTAG: 2597 TitleTAG: Labs Don't Like Fractions? While completing lab 2 I hit a wall for 4 hours. The problem was writing a resistance as a fraction instead of a decimal. (4/5 instead of .8 type of thing). I get that the inputs on the Labs aren't meant to be calculators, but I didn't realize how little arithmetic they could handle. Just a heads up, use decimals in lab, sorry if this has already been posted. UserIdTAG: 334952 UserNameTAG: ciaranfrancismoloney CreateTimeTAG: 2012-09-22T05:30:21Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I don't like fractions either :0-x FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-23T01:29:15Z IndexTAG: 2598 TitleTAG: i2? i used voltage divider and it should be okey for this but it gives me an error... Let's see, deactivating V2, we find R2//R3, I made this equal to Rx so Rx=1.875ohm, then the voltage in Rx, VRx=(Rx/(R1+Rx))*V1=0.4225, this is the value for x1. Forward i1=VR1/R1 so VR1=(R1/(R1+Rx))*V1=1.5775V finally i1=VR1/R1=-0.2254A. So far so good. Deactivating V1, we find (V2 in series with R2)//R1//R3 so I made R1//R3 equal to Ry, Ry=2.917, then x2=VRy so x2=(Ry/(R2+Ry))*V2=3.944V, so far so good. Now I was hoping to make the same trick with i1 for i2, unless something went wrong here I made i2=VR2/R2 VR2=(R2/(R2+Ry))*V2=4.056V by this result my causing result was -1.352A for i2.... :S the answer is wrong, what do i have to do? please someone there, that might would have patience to tell me.... Appreciated... UserIdTAG: 317836 UserNameTAG: VitorRodrigues CreateTimeTAG: 2012-09-21T22:42:20Z VoteTAG: 1 CoursewareTAG: Week 2 / Superposition example CommentableIdTAG: 6002x_S3E3_superposition NumberOfReplyTAG: 1 FirstChildTAG: hahahahaha .. u fell in the same fault as me :D !! you went and solved for i2 !!! you should not solve for i2 :D .. you should find the current in R1 which is i1 (y2) not R2 !! FirstChildUserIdTAG: 130810 FirstChildUserNameTAG: moon_light FirstChildCreateTimeTAG: 2012-09-22T14:43:21Z SecondChildTAG: im sprry dude how did you get 1.875ohms... SecondChildUserIdTAG: 133084 SecondChildUserNameTAG: Warrensiggs SecondChildCreateTimeTAG: 2012-09-23T13:21:10Z SecondChildTAG: nvm i got it....hmm.. your method for finding x2 is long this is how i got x2...R1//R3=2.917=(Rx2)... Vx2=Rx2(V2)/R2+Rx2... just exactly like how u did the first one... SecondChildUserIdTAG: 133084 SecondChildUserNameTAG: Warrensiggs SecondChildCreateTimeTAG: 2012-09-23T15:28:51Z SecondChildTAG: oh.. sorry... its exactly what u did.. lol... SecondChildUserIdTAG: 133084 SecondChildUserNameTAG: Warrensiggs SecondChildCreateTimeTAG: 2012-09-23T15:32:50Z IndexTAG: 2599 TitleTAG: H4P3 In the first part of H4P3 I was given I0=2A,Z=2Ω,R1=2 and,R2=4Ω Now if we find the thevenin resistance its supposed to be the one in connection with the port A right? As the resistance R1 will get disconnected when we disconnect the sources? But why is it not so? I know the answer but i'm hoping to get the correct reasoning for it! UserIdTAG: 108454 UserNameTAG: Raven7281 CreateTimeTAG: 2012-09-21T18:44:58Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: R1 will not be disconnected as only the *independent* sources are to be disconnected. Dependent sources are to remain connected. Also, remember to take care that you short circuit the voltage sources and the current sources become open circuits. FirstChildUserIdTAG: 331664 FirstChildUserNameTAG: dwmnctrh3 FirstChildCreateTimeTAG: 2012-09-21T18:57:12Z FirstChildTAG: If you look at the derivation of the Thevenin Theorem, you will see that it uses an independent current source I connected to the output of the circuit. The effects on Vout due to the original independent sources is given by Vth and the effect on Vout due to I with the independent voltage sources shorted and independent current sources open is given by Rth*I. For the homework problem you open the original independent current source and connect another current source to the output and calculate Vout. Then Rth = Vout/I. Z appears in the expression for Rth as thought the current dependent voltage source were a resistor with resistance Z. At this point I don't know a general principle. I wish I did! FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-09-21T22:49:45Z SecondChildTAG: I didn't need an independent current source connected to the output, I just came up with equations for both nodes on either side of the dependent source, then found other relationships throughout the circuit to come up with 2 equations with 2 unknowns. Also, when calculating Rth, I set the independent current source to zero, and the dependent voltage source to Z, since, in the problem statement it indicated Z = 2 ohms. This worked. This was similar to S8E2, but alpha was so small compared to R1 and R2, it didn't matter. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-06T19:43:26Z IndexTAG: 2600 TitleTAG: H3P4 Diode limiter Hi everybody, Could someone tell me if I am doing anything wrong with the last question of H3P4: What is the maximum current (in Amperes) that can go through diode D2? So, D1 ideally becomes an open circuit, so gone. D2 is short so we have -|V2|. My opinion is that Id2 will be max when Vs is negative max. Doing so, Id2 =|Ivs| + |Iv| = |(Vs-Vd2)/R| + |Vd2/R|. I do not want to wirte down values in here, but I think I am doing it correctly but does not accept the answer... THanks! UserIdTAG: 331379 UserNameTAG: leoxweb CreateTimeTAG: 2012-09-21T18:30:27Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You can use the node method for that one. Beware for signals! FirstChildUserIdTAG: 308853 FirstChildUserNameTAG: fmorato FirstChildCreateTimeTAG: 2012-09-21T18:49:22Z IndexTAG: 2601 TitleTAG: Question to Staff about Font contrast Is it possible to use a Font with more contrast on Edx? I find it often very difficult to read the text on screen and even when I print HW or Lab, the letters are grey and hard to read. Problem is, that when I choose a larger fontsize in Firefox or Chrome, then, if printed, the pages are a mess. I don't have this contrast problem normally with other sites or documents. UserIdTAG: 83276 UserNameTAG: salsero CreateTimeTAG: 2012-09-21T16:18:16Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Maybe it's a setting on your computer, my letters are black and the font is easy on the eyes. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-21T17:04:51Z SecondChildTAG: I've tried a lot, with Display properties/Appearance/AdvancedAppearance/Window all black. That works OK for the normal windows of Windows itself, and for the text on the menubar and addressbar in FF and Chrome, but not for the onlinetext itself inside the windows in Firefox and Chrome. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-09-21T17:37:53Z SecondChildTAG: I tried several fonts in FF and now I use Arial Black. It's much better now, but only if I disable the option to let the pages choose their own font. If I enable that option, then the text is gray again, so it seems to me that the font used on edx is the main cause. Thanks for answering. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-09-21T17:59:34Z SecondChildTAG: Actually, I have a monitor at edX where the contrast shows up terribly. I will bug the team again about it SecondChildUserIdTAG: 79361 SecondChildUserNameTAG: kimt SecondChildCreateTimeTAG: 2012-09-21T18:01:04Z SecondChildTAG: Thank you staff. Because nowI discovered a new problem: If I disable the Let the page choose its own font, then the HW pages will have problems, because they want to load one or another font, and that's not possible when disabled, and then my answers cause an js.script error .... And disabling the colors choosen by the page, makes my green checks invisible and some of the drawings, normally blue, become then black and white ... SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-09-21T18:20:08Z IndexTAG: 2602 TitleTAG: incorrect result interpretation In math -x-y=-(x+y) In the b1 and c2 answer this rule is not working. only one of these two parts of equation is correct for answer to get green mark UserIdTAG: 191392 UserNameTAG: erboreus CreateTimeTAG: 2012-09-21T16:08:53Z VoteTAG: 1 CoursewareTAG: Week 2 / Linear combinations of source strenghts CommentableIdTAG: 6002x_S3E2_Linear_Combinations_of_Source_Strengths NumberOfReplyTAG: 1 FirstChildTAG: So you tried (-R2-R3)/(R1*R2+R1*R3+R2*R3) and it didn't work for you? FirstChildUserIdTAG: 11 FirstChildUserNameTAG: dormsbee FirstChildCreateTimeTAG: 2012-09-21T16:24:06Z SecondChildTAG: YUP... SecondChildUserIdTAG: 364086 SecondChildUserNameTAG: jordan3110 SecondChildCreateTimeTAG: 2012-09-21T17:06:36Z SecondChildTAG: and i jst realised wat went wrong.#feelingDumb. SecondChildUserIdTAG: 364086 SecondChildUserNameTAG: jordan3110 SecondChildCreateTimeTAG: 2012-09-21T17:13:59Z SecondChildTAG: No worries, glad you're all set. SecondChildUserIdTAG: 11 SecondChildUserNameTAG: dormsbee SecondChildCreateTimeTAG: 2012-09-21T22:53:04Z IndexTAG: 2603 TitleTAG: Lab2 Disappeared Hi My Lab 2 work area has disappeared !! UserIdTAG: 377602 UserNameTAG: Goby CreateTimeTAG: 2012-09-21T15:08:16Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I had today the same problem with Lab3. Yesterday it worked fine with FF15.0.1 and I printed it, to work it out. But when I looked today, the Lab3 editor didn't appear in FF. When I switched later today to Chrome, it did appear and I finished the lab3 and got the green mark. However, when I switched again to FF, the Lab3 circuit didn't appear again, but the green mark was visible .... FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-09-21T15:33:01Z FirstChildTAG: Hmmm ! Just downloaded chrome as you did and it works FirstChildUserIdTAG: 377602 FirstChildUserNameTAG: Goby FirstChildCreateTimeTAG: 2012-09-21T15:58:51Z IndexTAG: 2604 TitleTAG: correction (To Course staff) v in example 2.14 (page 72 of the online version of the textbook: Foundations of Analog and digital electronics) must be 2 volts(not 0.5 volt). UserIdTAG: 364129 UserNameTAG: Sombat CreateTimeTAG: 2012-09-21T14:45:37Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Look at the Wiki for the Errata: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/Book/ FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-09-21T15:36:09Z IndexTAG: 2605 TitleTAG: week-2 problem-1 the last 2 parts of question 1 of week 2, no matter what answer i give it says "not a valid input" please help UserIdTAG: 330833 UserNameTAG: lavakumar18 CreateTimeTAG: 2012-09-21T12:19:16Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: can you please specify your question? FirstChildUserIdTAG: 252722 FirstChildUserNameTAG: vaboro FirstChildCreateTimeTAG: 2012-09-21T14:57:06Z IndexTAG: 2606 TitleTAG: Effect on devices when signal speed increases speed of light If LMD 3rd rule gets violated what are affects on circuit & what are precautions fro that. Like now uP with GHZ speed are available so do they take some extra precaution for avoiding LMD 3 rd rule UserIdTAG: 137709 UserNameTAG: DeepakBansal CreateTimeTAG: 2012-09-21T10:22:14Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Element laws would become more complicated since time and space need to be taken into account. FirstChildUserIdTAG: 252722 FirstChildUserNameTAG: vaboro FirstChildCreateTimeTAG: 2012-09-21T15:51:20Z IndexTAG: 2607 TitleTAG: lab 1 detailed answer (step to step, with graph) light bulb: 1.5V, 0.5A → R=1.5/0.5=3(ohm) ![enter image description here][1] equs.: (R2||3)/(R1+R2/||3)×6=1.5 R2/(R1+R2)×6=2 ans.: R1=3(ohm) R2=1.5(0hm) ![enter image description here][2] [1]: https://edxuploads.s3.amazonaws.com/134821565248235.bmp [2]: https://edxuploads.s3.amazonaws.com/134821599948253.bmp UserIdTAG: 138445 UserNameTAG: jih43711 CreateTimeTAG: 2012-09-21T08:27:40Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 2608 TitleTAG: Lab 4 I don't understand how i'm supposed to connect the circuit? I built a circuit with a Vgs and a Vds source and grounded the source, where am i supposed to connect the probes? I am not getting any Current to Voltage graph anywhere! :( UserIdTAG: 108454 UserNameTAG: Raven7281 CreateTimeTAG: 2012-09-21T07:38:10Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hint1. Use more than one transistor. Hint2. You can choose "x-axis" in probe parameter to make it's data x axis on graph. Hint3. You need graph of ids depends of Vds. Hope it will be helpful. FirstChildUserIdTAG: 195218 FirstChildUserNameTAG: Al_Incognito FirstChildCreateTimeTAG: 2012-09-21T07:51:47Z SecondChildTAG: More than one transistor???? you mean to get the multiple v-i curves? But i'm not even getting one singe V-I curve in the first place! :( i did take x-axis as the probe parameter and i connected the voltage probe at the Drain and The current probe along the wire from Vds to the drain![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13482146424954455.png SecondChildUserIdTAG: 108454 SecondChildUserNameTAG: Raven7281 SecondChildCreateTimeTAG: 2012-09-21T08:04:21Z SecondChildTAG: Look at figure 1 in lab. You have a little, but very importand difference in one of voltage sources. SecondChildUserIdTAG: 195218 SecondChildUserNameTAG: Al_Incognito SecondChildCreateTimeTAG: 2012-09-21T08:08:16Z SecondChildTAG: Wait! Are we supposed to use a sin wave generator for the Vds source? :D SecondChildUserIdTAG: 108454 SecondChildUserNameTAG: Raven7281 SecondChildCreateTimeTAG: 2012-09-21T08:08:17Z SecondChildTAG: Look more carefully))) You had to measure, how ids changes, during linear Vds change (in lab craph from 0 to 3v). SecondChildUserIdTAG: 195218 SecondChildUserNameTAG: Al_Incognito SecondChildCreateTimeTAG: 2012-09-21T08:19:29Z SecondChildTAG: *graph SecondChildUserIdTAG: 195218 SecondChildUserNameTAG: Al_Incognito SecondChildCreateTimeTAG: 2012-09-21T08:20:19Z SecondChildTAG: Traingle? And then we just take the current readings with respect to our given voltage in the graph and then calculate the required values? SecondChildUserIdTAG: 108454 SecondChildUserNameTAG: Raven7281 SecondChildCreateTimeTAG: 2012-09-21T08:23:27Z SecondChildTAG: Try to build propper graph (as shown on fig.2) All other You will get from graph and simple equations. You think in right direction. SecondChildUserIdTAG: 195218 SecondChildUserNameTAG: Al_Incognito SecondChildCreateTimeTAG: 2012-09-21T08:30:51Z SecondChildTAG: I got them all :) Thanks a lot for your help! SecondChildUserIdTAG: 108454 SecondChildUserNameTAG: Raven7281 SecondChildCreateTimeTAG: 2012-09-21T08:34:53Z IndexTAG: 2609 TitleTAG: 除了看视频，教程是不是也是需要看的啊 ！ 除了看视频，教程是不是也是需要看的啊 ！ UserIdTAG: 471183 UserNameTAG: snow_yxx CreateTimeTAG: 2012-09-21T06:10:04Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: From what I understood using Google translate, you want to know if it is compulsory to watch the tutorial videos. No it is not compulsory to watch tutorials. These are there to supplement lectures. I would however recommend watching them because they do provide a lot of important guidance in understanding concepts better, solving assignments and the exams. Many of them are also a lot of fun! :-) FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-09-21T07:41:46Z IndexTAG: 2610 TitleTAG: Problem 2.I4, text book, is the answer correct and how? Page 72 of the text book.![enter image description here][1] The problem is on conservation of energy and the final answer is given as v = 0.5V where as it should be v = 2V. [1]: https://edxuploads.s3.amazonaws.com/13481932371969287.jpg UserIdTAG: 211715 UserNameTAG: pitankar CreateTimeTAG: 2012-09-21T02:07:42Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: looks like an error in the book :) FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-09-21T06:29:02Z FirstChildTAG: Look at the wiki for the errata: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/Book/ FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-09-21T15:43:30Z IndexTAG: 2611 TitleTAG: confusion about the power supply labeling I don't want to sound petty or pedantic, I'm just hoping to get it straight. The ground point in these diagrams is being labeled "S" for 'source', and the other pole of the triode "D" for 'drain', but the power supply voltage is referred to as $V_S$ instead of $V_D$. Why is that? UserIdTAG: 280220 UserNameTAG: pilgrimCycle CreateTimeTAG: 2012-09-21T02:07:16Z VoteTAG: 1 CoursewareTAG: Week 3 / MOSFET CommentableIdTAG: 6002x_mosfet_model NumberOfReplyTAG: 1 FirstChildTAG: @pilgrimCycle :its just a notation..u can write it as Vd ,vcc or anything else which suits u better...ter is no relation like its source thats why u shuld rite vs .just remove this frm ur mind...it is just a simple notation... FirstChildUserIdTAG: 99583 FirstChildUserNameTAG: sanjeevkm0912 FirstChildCreateTimeTAG: 2012-09-23T17:34:48Z SecondChildTAG: I can certainly do that. Thank you. SecondChildUserIdTAG: 280220 SecondChildUserNameTAG: pilgrimCycle SecondChildCreateTimeTAG: 2012-09-25T01:16:16Z IndexTAG: 2612 TitleTAG: Diode is active or passive What are active & passive devices. Diode is active or passive UserIdTAG: 137709 UserNameTAG: DeepakBansal CreateTimeTAG: 2012-09-21T01:44:54Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: Diode is an active component because it only works when is a current through it. Thanks to this property you can control how it works. Diode is also a nonlinear component. A resistor is a passive and linear component. Take a look at this document from [TI][1]. Is very ilustrative. [1]: http://www.ti.com/lit/an/sloa026a/sloa026a.pdf FirstChildUserIdTAG: 184827 FirstChildUserNameTAG: DiegoT FirstChildCreateTimeTAG: 2012-09-21T02:18:17Z FirstChildTAG: But diode don't have any gain. Or its like ideally its gain is 1 in forward biased its gain is 0 in reverse direction. FirstChildUserIdTAG: 137709 FirstChildUserNameTAG: DeepakBansal FirstChildCreateTimeTAG: 2012-09-21T10:19:00Z FirstChildTAG: I agreed with Diego that a diode is an active device. It does not have a gate, but it could be used as a switch, with an on and off state, depending on the forward current. However in this case the power supply is the signal. The more I think about it, the more I am unsure, it certainly has passive qualities. I can see the confusion. I think because it does have active qualities, you could consider it active, or both. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-21T15:33:18Z FirstChildTAG: I read on Babylon, that Active device is one which has gain or can switch the flow of current. If that is the case them diode clearly falls in active. Is it right defination FirstChildUserIdTAG: 137709 FirstChildUserNameTAG: DeepakBansal FirstChildCreateTimeTAG: 2012-09-22T10:18:12Z IndexTAG: 2613 TitleTAG: Nice course this will be amazing course i just bought the book UserIdTAG: 470657 UserNameTAG: migrator CreateTimeTAG: 2012-09-21T00:00:57Z VoteTAG: 1 CoursewareTAG: Week 1 / Course overview CommentableIdTAG: 6002x_S1V1toS1V4 NumberOfReplyTAG: 0 IndexTAG: 2614 TitleTAG: H2P1 help please I have r2/(r1+r2)=27/90=0.3 And Rth I think is r1r2/(r1+r2), I do not know, what to do then. I am trying to solve this already 4 day please help UserIdTAG: 286954 UserNameTAG: ododo CreateTimeTAG: 2012-09-20T23:13:05Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Note the Rth is in K ohm and that the resistors come in multiples of powers of 10. FirstChildUserIdTAG: 66101 FirstChildUserNameTAG: bluefin FirstChildCreateTimeTAG: 2012-09-20T23:21:50Z SecondChildTAG: so how to get them? SecondChildUserIdTAG: 286954 SecondChildUserNameTAG: ododo SecondChildCreateTimeTAG: 2012-09-20T23:33:14Z SecondChildTAG: hit n try.. SecondChildUserIdTAG: 266739 SecondChildUserNameTAG: satvikchugh SecondChildCreateTimeTAG: 2012-09-21T03:40:35Z FirstChildTAG: ododo equate $R_th$ to a value between 10k and 30k..and solve for $R_1$ and $R_2$ FirstChildUserIdTAG: 11538 FirstChildUserNameTAG: trishul FirstChildCreateTimeTAG: 2012-09-23T11:13:50Z IndexTAG: 2615 TitleTAG: answer my question please why y1 is negaive UserIdTAG: 284184 UserNameTAG: Rodrigo10 CreateTimeTAG: 2012-09-20T19:26:10Z VoteTAG: 1 CoursewareTAG: Week 2 / Superposition example CommentableIdTAG: 6002x_S3E3_superposition NumberOfReplyTAG: 1 FirstChildTAG: because i1 direction is going into the source "remember--->sources deliver not consuming" FirstChildUserIdTAG: 185715 FirstChildUserNameTAG: amirengineer FirstChildCreateTimeTAG: 2012-09-20T20:20:18Z SecondChildTAG: In the original circuit the positive direction of I1 is shown by an arrow, and this positive direction is when I1 is going into V1 source. The y1 is the component of I1 when only V1 source is taking into account in the circuit. In this case, the only voltage source in the circuit is V1. When you have only one voltage source in a resistive circuit the current always going out from this source. This direction is opposite to the direction of the arrow of I1, so y1 is negative. SecondChildUserIdTAG: 447011 SecondChildUserNameTAG: Higinio SecondChildCreateTimeTAG: 2012-09-20T20:31:05Z SecondChildTAG: its the same case with y2 also...but why y2 is positive??? SecondChildUserIdTAG: 132396 SecondChildUserNameTAG: pandu1993 SecondChildCreateTimeTAG: 2012-09-29T14:07:09Z IndexTAG: 2616 TitleTAG: Should I go through the Text??! Is it Ok if I Go thru the the Lecture slides?? Cos I dont have sufficient time to go through the text.. Does the video slides cover everything?? UserIdTAG: 148523 UserNameTAG: span993 CreateTimeTAG: 2012-09-20T19:22:33Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I think you should read the text and make the exercises FirstChildUserIdTAG: 184827 FirstChildUserNameTAG: DiegoT FirstChildCreateTimeTAG: 2012-09-20T21:47:47Z IndexTAG: 2617 TitleTAG: NMH=NML Does the "NM" stand for NOISE MARGIN? UserIdTAG: 209041 UserNameTAG: SmartEngine CreateTimeTAG: 2012-09-20T18:32:25Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Yes! **N**oise **M**argin **H**igh and **L**ow. FirstChildUserIdTAG: 396446 FirstChildUserNameTAG: RousseauxS FirstChildCreateTimeTAG: 2012-09-20T18:46:02Z IndexTAG: 2618 TitleTAG: Did the Prof really break that mouse? I'm super curious. Does a mouse break so nicely to reveal the circuit inside? I cant try that on mine right away as i just have only one. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-20T17:39:43Z VoteTAG: 1 CoursewareTAG: Week 3 / Inside a Mouse CommentableIdTAG: 6002x_inside_a_mouse NumberOfReplyTAG: 1 FirstChildTAG: He has practice with chainsaws and hammers, you can tell! xD Nice! FirstChildUserIdTAG: 398594 FirstChildUserNameTAG: DaveyJC FirstChildCreateTimeTAG: 2012-09-26T20:05:26Z IndexTAG: 2619 TitleTAG: dimension by a dimensional constant of 1 v^-2 UserIdTAG: 160366 UserNameTAG: elcollar CreateTimeTAG: 2012-09-20T17:23:02Z VoteTAG: 1 CoursewareTAG: Week 4 / Graphs Exercise CommentableIdTAG: 6002x_graphs_exercise NumberOfReplyTAG: 0 IndexTAG: 2620 TitleTAG: Any hints for lab 3 please?? 1)I have connected two MOSFET in series and another MOSFET in parallel to the two resistors 2)How do I find the RON for each MOSFET? UserIdTAG: 295240 UserNameTAG: Jack_my CreateTimeTAG: 2012-09-20T15:28:14Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Check your mosfets arrangement first. Apart from three mosfet and propper wiring, you do not need to add any other elements to the circuit FirstChildUserIdTAG: 210954 FirstChildUserNameTAG: Shahrouz FirstChildCreateTimeTAG: 2012-09-20T20:48:15Z SecondChildTAG: Rearranged the MOSFETs, two in parallel and both connected to one MOSFET. All looks good including the graph. But still getting wrong. Is it because of the spike? How do I find the RON for each MOSFETS? SecondChildUserIdTAG: 295240 SecondChildUserNameTAG: Jack_my SecondChildCreateTimeTAG: 2012-09-21T04:27:09Z FirstChildTAG: Rearranged the MOSFETs, two in parallel and both connected to one MOSFET. All looks good including the graph. But still getting wrong. Is it because of the spike? How do I find the RON for each MOSFETS? FirstChildUserIdTAG: 295240 FirstChildUserNameTAG: Jack_my FirstChildCreateTimeTAG: 2012-09-21T04:27:13Z IndexTAG: 2621 TitleTAG: Doubt in Text Book Chapter 3 Dear Sir , I have a doubt while I was going through Chapter of Book ( page 154) Source acting alone Vo1=(1/100)v1*1kohm =5v in that this 100ohm fro where it is coming I am not getting please explain Thanks MK.Prasanth (India) UserIdTAG: 156694 UserNameTAG: mkprasanth CreateTimeTAG: 2012-09-20T15:28:09Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: The dependant current source it is that up arrow symbol. It says that this current source will have a value of (1/100 Ohm)*v . Why it is dependant? because it depends on the value of v. If you see, v it is taken from the point up to the voltage source 2V and the resistor 1kOhm. Then, the problem analyze the behaviour per each independant voltage source. That is to say: the functionality of the circuit when v2=0 and v1 not; and when v1=0 and v2 not. 1-V acting alone: you willl have to do v2=0. so you will only have v1=1V. Now, what will be our output? In the output we will have I*1kOhm. But our I, it depends on v. So, Vout=I*1kOhm=(1/100 Ohm)*v*1kOhm. Our, v, in this case it is v=1V*(1kOm/1kOhm+1kOhm)= 0.5 Volt (go to voltage devider formula in Textboook). So, that its why v=(1/100 Ohm)*0.5Volt*1kOhm=5Volt. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-20T17:02:50Z IndexTAG: 2622 TitleTAG: Joining in the middle of a course Hello, can someone please let me know whether I can join in the middle of a course. 6.002x started on Sep 5th. Would it be ok to join now? UserIdTAG: 468193 UserNameTAG: DeeVar CreateTimeTAG: 2012-09-20T14:39:18Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: hi, I also just joined this week, your not the only one but we need to speed it up or we might left behind FirstChildUserIdTAG: 452559 FirstChildUserNameTAG: kuz1toro FirstChildCreateTimeTAG: 2012-09-20T16:47:21Z FirstChildTAG: Yes, it's only the second week. If you are writing in here you must already be joined. You can miss up to two homeworks, as they only grade 10 out of 12. Relax and don't forget to use the search function if you have any problems, many good posts already exist. Have fun. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-20T14:48:00Z SecondChildTAG: Thanks a bunch Pennypacker for this reply. SecondChildUserIdTAG: 468193 SecondChildUserNameTAG: DeeVar SecondChildCreateTimeTAG: 2012-09-20T15:12:43Z IndexTAG: 2623 TitleTAG: a small doubt related to this video... if the light bulb is analogous to resistor, why isnt the graph linear?? does it have anything to do with lumped matter discipline?? UserIdTAG: 316353 UserNameTAG: Vivekanand123 CreateTimeTAG: 2012-09-20T14:20:27Z VoteTAG: 1 CoursewareTAG: Week 2 / Lightbulb model CommentableIdTAG: 6002x_lightbulb_model NumberOfReplyTAG: 1 FirstChildTAG: The graph is not linear because the light bulb is not linear. The straight line they drew with the ruler represents a linear model they could use to design that light bulb into a circuit. In this case the model would not be an exact representation of the light bulb, but it may be close enough for some circuits. A light bulb is a temperature dependent resistor. As it gets hotter, the resistance increases, as you can see by the curve in the graph. If you imagine the apex of the curve from the straight line and express that as a percentage, that would be the tolerance of that resistor. Ex: If the curve was 10% away from your straight line, you could say you have a resistor of a certain value within 10% tolerance. This particular resistor would be more accurate at the lower and higher voltages, as shown by the graph, but would be off by 10% with voltages in the middle range. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-20T14:37:13Z SecondChildTAG: It's not correct about tolerance. Tolerance is a permissible variation, so if resistor has a tolerance 10% it just means that factual resistance can be more or less in 10% limit from nominal - but IV must be linear in nominal power range. All other explanations about bulb is correct. SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-21T06:16:24Z IndexTAG: 2624 TitleTAG: Brilliant Learning That's sooooooooooo funny!! UserIdTAG: 285241 UserNameTAG: Kazemi CreateTimeTAG: 2012-09-20T14:12:30Z VoteTAG: 1 CoursewareTAG: Week 2 / Chain saw CommentableIdTAG: 6002x_chain_saw NumberOfReplyTAG: 0 IndexTAG: 2625 TitleTAG: Resistor in parallel versus in series Does anyone understand why the resistance in parallel with the current for the Norton method versus in series for the Thevenin method? UserIdTAG: 278463 UserNameTAG: zniazi CreateTimeTAG: 2012-09-20T06:15:24Z VoteTAG: 1 CoursewareTAG: Week 2 / Norton method CommentableIdTAG: 6002x_norton NumberOfReplyTAG: 1 FirstChildTAG: In the Thevenin method, the resistance is in series with a voltage source, not a current source. They are both handy, but not necessarily related to each other, despite their similarities. The Mayer-Norton theorem was introduced in 1926. Thevenin's theorem was discovered earlier in 1853 by Herman von Helmholtz. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-20T14:02:52Z IndexTAG: 2626 TitleTAG: Problem with lab 2 I'm having a problem coming up with the resistances for lab 2. Now, the equation I need to replicate is Vo = 1/2*V1 + 1/6*V2 .....(1) From the lecture, I understand where he gets the equation: Vo = V1*R2/(R1+R2) + V2*R1/(R1+R2) ......(2) If we equated the two equations, we'd get: 1/2 = R2/(R1+R2); 1/6 = R1/(R1+R2) Solving simultaneously: 3R1 = R2 ......(3) Now, here's the bit that gets me stumped, substituting this back into equation 2 gives me: Vo = V1(3*R1/(4*R1)) + V2(R1/(*4R1)) = V1*3/4 + V2(1/4) which is not the equation I want. What am I doing wrong? :( UserIdTAG: 141053 UserNameTAG: Albert0ng CreateTimeTAG: 2012-09-20T06:01:18Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: If you use the Hint and add 1 resistor extra then you can solve this. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-09-20T14:08:36Z FirstChildTAG: can anyone help me doing lab 2 FirstChildUserIdTAG: 162670 FirstChildUserNameTAG: charlesbabyt FirstChildCreateTimeTAG: 2012-09-20T08:40:03Z SecondChildTAG: If you go back to the post from a day and a half ago with the title "I am stucked with LAB #2" there is a long discussion on how to solve the problem. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-09-20T09:10:35Z SecondChildTAG: Thank you SecondChildUserIdTAG: 328336 SecondChildUserNameTAG: prince1 SecondChildCreateTimeTAG: 2012-09-20T17:19:21Z FirstChildTAG: your method are right by comparing equation 1 and 2 but I think you must add one more resistor on it, which placed between vo and ground and to analysis it, you must use the Thevenin theorems and then you equated the equations to find the relation between R1, R2 and R3 hope it usefull FirstChildUserIdTAG: 452559 FirstChildUserNameTAG: kuz1toro FirstChildCreateTimeTAG: 2012-09-20T19:46:46Z FirstChildTAG: Anyone with a proper clue to lab 2 should help me out.I have tried several times but only get a close waveform which is wrong. FirstChildUserIdTAG: 429851 FirstChildUserNameTAG: ssembajjwe FirstChildCreateTimeTAG: 2012-09-21T08:01:42Z FirstChildTAG: Thanks guys got it finally after a lot of work. FirstChildUserIdTAG: 429851 FirstChildUserNameTAG: ssembajjwe FirstChildCreateTimeTAG: 2012-09-21T08:30:28Z IndexTAG: 2627 TitleTAG: RE: How to measure current in Sandbox? (praveenjugge) Hi praveenjugge! Yes, you can measure currents in Sandbox. I have made a Wiki Post [here][1] about your request [here][2]. I hope this can be useful ;). See you! Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/how-measure-current-sandbox/ [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5059e3225aa28a230000000d UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2012-09-20T04:53:58Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: **STEP 1:** - Let's choose an arbitrary Circuit. ![image1][1] **STEP 2:** - Go to the Toolbar and choose the following symbol. ![image2][2] **STEP 3:** - Select the element and drag it to the work place. ![image3][3] **STEP 4:** - Carry it (with the mouse, without releasing it) to the point of the Circuitwhere you want to measure the current. ![image4][4] **STEP 5:** - Click on DC and that is all! You will have the current measured in that point! ![image5][5] [1]:https://edxuploads.s3.amazonaws.com/13481126378183067.png [2]:https://edxuploads.s3.amazonaws.com/13481128761343615.png [3]:https://edxuploads.s3.amazonaws.com/1348113222234286.png [4]:https://edxuploads.s3.amazonaws.com/13481137561343664.png [5]:https://edxuploads.s3.amazonaws.com/13481140431343659.png FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-21T02:34:45Z IndexTAG: 2628 TitleTAG: RE: Problem with Courseware and Discussion (TeTAn) This is a reply to TeTan post... I had the same problem of not being able to see the forum replies when I was using Internet Explorer. As soon as I started using Chrome everything seem to be working great. Let me know if this helps. UserIdTAG: 303762 UserNameTAG: leocarrasco CreateTimeTAG: 2012-09-20T03:20:48Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: If this is a reply to TetAn's post, why not use the reply function for this post? I'm not admonishing you, I just wondered why you did it this way? Something to do with how the forum works, or ? FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-09-24T16:50:33Z IndexTAG: 2629 TitleTAG: Chévere ¡Muy bacano! So cool! UserIdTAG: 184827 UserNameTAG: DiegoT CreateTimeTAG: 2012-09-20T01:23:15Z VoteTAG: 1 CoursewareTAG: Week 2 / Chain saw CommentableIdTAG: 6002x_chain_saw NumberOfReplyTAG: 1 FirstChildTAG: Hi DiegoT! Cómo estás? ( How are you?) Yes, it was a funny educative way of teaching and learning! Those students were really fortunate to see it live and direct haha! ;) FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-20T02:17:20Z SecondChildTAG: I ♥ it! It would be very strange to see these things in Spain. I am learning more than electronics in this course $:)$ SecondChildUserIdTAG: 243424 SecondChildUserNameTAG: FernandoSaez SecondChildCreateTimeTAG: 2012-09-20T10:09:10Z SecondChildTAG: Que bueno que estés disfrutando de este Curso! :) (Nice to see that you are enjoying this Course!) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-21T02:37:27Z IndexTAG: 2630 TitleTAG: How can I find VTH by superposition method? Please can someone help me? I found VTH 1,2162 by node method, but I'd like to know how I can find it by superpostion method. Thanks! UserIdTAG: 177243 UserNameTAG: rdrigoskt CreateTimeTAG: 2012-09-20T01:22:43Z VoteTAG: 1 CoursewareTAG: Week 2 / Simple Thevenin CommentableIdTAG: 6002x_S3E4_Simple_Thevenin NumberOfReplyTAG: 4 FirstChildTAG: How are you planning to apply superposition? It seems that you only have one source in the circuit. FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-20T13:54:13Z SecondChildTAG: There is only a V source! SecondChildUserIdTAG: 289132 SecondChildUserNameTAG: Nestor_Escala SecondChildCreateTimeTAG: 2012-09-23T01:07:01Z SecondChildTAG: Rth = R1 AND R2 IN PARALLEL WHICH IS EQUAL TO (56000*18000)/(56000+18000)=13621.621621 OHMS SecondChildUserIdTAG: 446984 SecondChildUserNameTAG: Msobki SecondChildCreateTimeTAG: 2012-12-18T12:36:25Z FirstChildTAG: Don't you think Vth = Voltage across R2 ? FirstChildUserIdTAG: 254346 FirstChildUserNameTAG: moijes12 FirstChildCreateTimeTAG: 2012-09-20T18:30:41Z SecondChildTAG: use voltage divider rule to find Vth coz r1 n r2 are in series.. SecondChildUserIdTAG: 337430 SecondChildUserNameTAG: aparna_b SecondChildCreateTimeTAG: 2012-09-22T08:50:07Z FirstChildTAG: use voltage divider rule to find Vth coz r1 n r2 are in series.. FirstChildUserIdTAG: 337430 FirstChildUserNameTAG: aparna_b FirstChildCreateTimeTAG: 2012-09-22T08:50:38Z SecondChildTAG: r1 and r2 r not in series as the is divided at the node in to resistore SecondChildUserIdTAG: 463557 SecondChildUserNameTAG: n1manish SecondChildCreateTimeTAG: 2012-10-07T07:42:07Z FirstChildTAG: u cannot use superposition to this circuit because it has only one source or we can say we don't need to use superposition to solve the circuit. FirstChildUserIdTAG: 231749 FirstChildUserNameTAG: utshau FirstChildCreateTimeTAG: 2012-09-24T05:56:20Z IndexTAG: 2631 TitleTAG: ? I think (I´m not sure), is the different configurations you can have, for example with three signals you can have A*B*C, and AND gate with 8 different arragements (000,101, etc), or you can have A+B+C, and OR gate, with eight ways again, or A*B+C, and you can have the inverse of them also, ..., it could be that UserIdTAG: 450547 UserNameTAG: erikita CreateTimeTAG: 2012-09-20T00:36:15Z VoteTAG: 1 CoursewareTAG: Week 2 / Boolean functions CommentableIdTAG: 6002x_S4E2_Boolean_Functions NumberOfReplyTAG: 0 IndexTAG: 2632 TitleTAG: h2p2 5 hours in cant wait until solutions getting it a few diff ways keep getting formula of order 1/RL. learned a lot while failing at this one though, always next week UserIdTAG: 388273 UserNameTAG: kilmarta CreateTimeTAG: 2012-09-20T00:10:11Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: i guess 5 hours and 10mins was the key, hint to all. look up Maximum Power Transfer Theorem FirstChildUserIdTAG: 388273 FirstChildUserNameTAG: kilmarta FirstChildCreateTimeTAG: 2012-09-20T00:19:15Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-20T03:05:16Z IndexTAG: 2633 TitleTAG: H2P1 R1, R2 values are not correct in my homework H2P1, Vmax and Vmin correct. However R1, R2 values are not correct. Vout value is similar to 8V and R1, R2 value are included E12 set. I don't know R1, R2 result are wrong answer? UserIdTAG: 345399 UserNameTAG: mybrand CreateTimeTAG: 2012-09-19T23:57:22Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: You have two constraints in your circuit: Voltage and Thevenin resistance. Have you check that the equivalent Thevenin resistance is within the required values? FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-20T00:31:58Z SecondChildTAG: I had the same problem, and I was pretty sure my pair had met both constraints, so I must have some misunderstanding about the Thevenin resistance here. Is the Thevenin resistance only the sum of resistances we encounter in the path from one terminal to the other $(R_1)$, or am I wrong to think the resistance of the entire circuit $(R_1+R_2)$ needs to be in that range? SecondChildUserIdTAG: 267702 SecondChildUserNameTAG: curio_city SecondChildCreateTimeTAG: 2012-09-20T00:54:53Z SecondChildTAG: Ah, I see, finally! Current due to the voltage source sees resistors in series, but current injected from the port would see resistors in parallel! SecondChildUserIdTAG: 267702 SecondChildUserNameTAG: curio_city SecondChildCreateTimeTAG: 2012-09-20T01:44:45Z SecondChildTAG: what value are u getting for Vmax and Vmin? SecondChildUserIdTAG: 414516 SecondChildUserNameTAG: randevakash SecondChildCreateTimeTAG: 2012-09-20T09:15:48Z SecondChildTAG: what is the vaue for vmax and vmin SecondChildUserIdTAG: 276808 SecondChildUserNameTAG: DEBASMITAMAJUMDER SecondChildCreateTimeTAG: 2012-09-20T12:15:43Z FirstChildTAG: I'm having the same issue. I can't work out why. :/ FirstChildUserIdTAG: 243871 FirstChildUserNameTAG: Vanilly FirstChildCreateTimeTAG: 2012-09-20T00:10:07Z SecondChildTAG: you need to make sure that R1 and R2 lie within the Thevenin Resistance range. SecondChildUserIdTAG: 141053 SecondChildUserNameTAG: Albert0ng SecondChildCreateTimeTAG: 2012-09-20T05:24:34Z IndexTAG: 2634 TitleTAG: S5E1 : Logic with switches Hi! Anyone can explain why x = 1 and y = 1 -> Z = 1 ? I think I understand others, but not this. Thanks! UserIdTAG: 300169 UserNameTAG: castrogfx CreateTimeTAG: 2012-09-19T20:01:31Z VoteTAG: 1 CoursewareTAG: Week 3 / Switch Model Exercise CommentableIdTAG: 6002x_switch_model_exercise NumberOfReplyTAG: 2 FirstChildTAG: Hi, Castrogfx It is more easy to build a ligical function Z=F(X,Y) and then comupite the truth table rather then computing the truth table by substituting all possible values of X and Y into the circuts. In my case the function was Z = Y + ... - that is why when Y is 1 Z will be always 1 too. Thanks, Sergestus FirstChildUserIdTAG: 375724 FirstChildUserNameTAG: sergestus FirstChildCreateTimeTAG: 2012-09-20T00:56:51Z SecondChildTAG: Two switches recieve X and Y directly, but two of them recieve output from switch X and the other one receive output from inverter(x) and y switch. Look my picture: https://edxuploads.s3.amazonaws.com/13485441005063727.jpg Sorry, my english is not very well. SecondChildUserIdTAG: 268104 SecondChildUserNameTAG: SaulCardenasG SecondChildCreateTimeTAG: 2012-09-25T20:07:12Z SecondChildTAG: When u/p of X from first switch, it is inverted; should it not be again inverted as it passes through the second (left to right). I see it as 1> X' 2> X'' = X 3> X+Y' 4> (X+Y')' = Z The results are opposite as Z' instead of Z. Please comment. SecondChildUserIdTAG: 442070 SecondChildUserNameTAG: MuhammadAsad SecondChildCreateTimeTAG: 2012-09-27T17:05:59Z SecondChildTAG: Forget the last line! SecondChildUserIdTAG: 442070 SecondChildUserNameTAG: MuhammadAsad SecondChildCreateTimeTAG: 2012-09-27T17:18:43Z SecondChildTAG: The entire circuit is equivalent to NOT(X)+Y. That's why you get a 1 when Y=1 OR X=0. SecondChildUserIdTAG: 9100 SecondChildUserNameTAG: dmgongora SecondChildCreateTimeTAG: 2012-10-02T03:02:59Z FirstChildTAG: z = x&y FirstChildUserIdTAG: 372453 FirstChildUserNameTAG: mkimo FirstChildCreateTimeTAG: 2012-09-19T20:11:55Z SecondChildTAG: But X =0 and Y = 0, Z = 1. x = 0, y = 1, z =1 x=1, y=0, z = 0 SecondChildUserIdTAG: 300169 SecondChildUserNameTAG: castrogfx SecondChildCreateTimeTAG: 2012-09-19T22:25:49Z SecondChildTAG: castrogfx is incorrect. On the middle resistor if either the second switch (x switch's output) or the third switch (the y switch) is shorted, then the output of that sub-gate is 0, which leaves the fourth gate open, meaning that Z has voltage Vs across it. SecondChildUserIdTAG: 164898 SecondChildUserNameTAG: jbparkes SecondChildCreateTimeTAG: 2012-09-20T02:02:59Z SecondChildTAG: Thanks jbparkes! SecondChildUserIdTAG: 300169 SecondChildUserNameTAG: castrogfx SecondChildCreateTimeTAG: 2012-09-20T14:03:47Z IndexTAG: 2635 TitleTAG: Something wrong with the 3rd box !!!??? shouldn't the current be -8.98 ??? an "e-06" appear in the answer wht is tht !! UserIdTAG: 40417 UserNameTAG: jorj_khalil CreateTimeTAG: 2012-09-19T18:54:56Z VoteTAG: 1 CoursewareTAG: Week 3 / Piecewise Linear Exercise CommentableIdTAG: 6002x_piecewise_linear_exercise NumberOfReplyTAG: 5 FirstChildTAG: 8.98/1000000 = 8.98e-06 = 0.00000898 FirstChildUserIdTAG: 180092 FirstChildUserNameTAG: MrBondBoca FirstChildCreateTimeTAG: 2012-09-19T21:16:43Z SecondChildTAG: -8.98u SecondChildUserIdTAG: 371000 SecondChildUserNameTAG: Joseph090892 SecondChildCreateTimeTAG: 2012-09-20T21:58:32Z FirstChildTAG: If you use the correct resistance, the answer is correct ; ). FirstChildUserIdTAG: 396446 FirstChildUserNameTAG: RousseauxS FirstChildCreateTimeTAG: 2012-09-19T19:11:31Z SecondChildTAG: -9/(1e6+1200)=8.98921294447e-06 SecondChildUserIdTAG: 374393 SecondChildUserNameTAG: rmaleki SecondChildCreateTimeTAG: 2012-09-21T05:22:22Z SecondChildTAG: -8.98921294447e-06 SecondChildUserIdTAG: 264320 SecondChildUserNameTAG: israel05 SecondChildCreateTimeTAG: 2012-09-22T23:00:49Z FirstChildTAG: To say what is "e-06" it is a scientific notation for power of 10 in "e" form. So e-06 stays for $10^{-6}$ = 0.000001 and -8.98e-06 = -8.98 * $10^{-6}$ = -0.00000898 FirstChildUserIdTAG: 190618 FirstChildUserNameTAG: Kirbabaev FirstChildCreateTimeTAG: 2012-09-23T20:49:15Z FirstChildTAG: For the first time I haven't noticed the "e-06" at the end of the answer. Well, I was REALLY wondered by getting SUCH a big number as a result. But when I saw "e-06" I was laughing really loud ^_^ FirstChildUserIdTAG: 326787 FirstChildUserNameTAG: Angstrem FirstChildCreateTimeTAG: 2012-09-23T21:26:49Z FirstChildTAG: e-06 means 10^-6 which is micro Amp FirstChildUserIdTAG: 272523 FirstChildUserNameTAG: Jivraj FirstChildCreateTimeTAG: 2012-09-28T19:57:06Z IndexTAG: 2636 TitleTAG: H2P1 Confusion. -R1+R2 error? Hi all, I'm having a little bit of a problem with this part of homework 2. Whilst the rest of the homework has been fine, I've been struggling to complete this bit. I appear to have both Vmax and Vmin correct, but my R1 and R2 selections (are from the E12 set) keep getting marked as incorrect. I'm confused as to why that could be, and could someone please explain how I'm going wrong? Thanks :) UserIdTAG: 243871 UserNameTAG: Vanilly CreateTimeTAG: 2012-09-19T18:47:37Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Wot r ur vals of R1 and R2? FirstChildUserIdTAG: 375305 FirstChildUserNameTAG: sandy07 FirstChildCreateTimeTAG: 2012-09-19T19:58:42Z FirstChildTAG: wot r ur vals of R1 and R2? FirstChildUserIdTAG: 375305 FirstChildUserNameTAG: sandy07 FirstChildCreateTimeTAG: 2012-09-19T20:07:05Z SecondChildTAG: My values were 47 and 15. SecondChildUserIdTAG: 243871 SecondChildUserNameTAG: Vanilly SecondChildCreateTimeTAG: 2012-09-19T20:35:18Z FirstChildTAG: E12 set is given for the nominal values..the actual resistance can be any value in this set multiplied by a power of 10..Important point to note here is that thevenin resistance is between 10k and 30k..so choose R1 and R1 as resistances from E12 set multiplied by power of 10 such that this condition is satisfied..and enter the value in ohm and not kohm.. for example, if its 22k write as 22000.. FirstChildUserIdTAG: 173147 FirstChildUserNameTAG: cruiser_rahit FirstChildCreateTimeTAG: 2012-09-19T19:57:47Z SecondChildTAG: I tried that, and it still wouldn't accept them as right. SecondChildUserIdTAG: 243871 SecondChildUserNameTAG: Vanilly SecondChildCreateTimeTAG: 2012-09-19T20:36:14Z SecondChildTAG: same for me. the values do not work. SecondChildUserIdTAG: 5345 SecondChildUserNameTAG: caspik SecondChildCreateTimeTAG: 2012-09-20T04:05:40Z SecondChildTAG: By voltage divider rule : Vout = Vin *(R2/(R1+R2)) Substituting Vin = 10 and Vout = 40 and solving You can get relation between R1 and R2 (as in R1= kR2)..and then choose two values from E12 such that R1 = K R2.. SecondChildUserIdTAG: 173147 SecondChildUserNameTAG: cruiser_rahit SecondChildCreateTimeTAG: 2012-09-20T18:22:13Z SecondChildTAG: same here. It is not accepting values for R1 and R2. Banging mu head since the last 10 minutes. Such a simple problem and it is not accepting the values. Help pls SecondChildUserIdTAG: 332360 SecondChildUserNameTAG: srinivasav SecondChildCreateTimeTAG: 2012-09-23T06:05:11Z SecondChildTAG: Got is guys, i was using k for kilo. write in terms of ohm only. if its 1k, write it as 1000. SecondChildUserIdTAG: 332360 SecondChildUserNameTAG: srinivasav SecondChildCreateTimeTAG: 2012-09-23T06:08:17Z IndexTAG: 2637 TitleTAG: S7E2 Hi! I didn't figure out how to solve this by using load curve method(solved with numerical one). When I try to measure operating point current and voltage from graph I get values V=28.35V and I=0.51A. What do I do wrong? ![Graph from course][1] [1]: https://edxuploads.s3.amazonaws.com/13480782809599065.png UserIdTAG: 192653 UserNameTAG: Quaz CreateTimeTAG: 2012-09-19T18:12:39Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: The graph does not have any unit of measure on it, so you cant use that lines to measure anything. For example you dont know if the scales are linear or logarithmic, so the numbers you have up there mean nothing. Its only a drawing and cant be used for anything else. Its way much easier to get the formulas. We have a current source that provides and I known. That current splits in two, through the resistor and then through the device. What we want to do is to take the current through device and write it in two different ways: 1) our device being a "load", the current that it can get from this circuit is I minus the current passing through resistor. This is linear and shows what the circuit is able to "give" to an unknown load. The circuit says "this is what i can offer, i want to operate on that line". I guess hence the term "load line". 2) at the same time, the same thing as above, the current through the device can be written according to its internal elemental law; this represents what the device is willing to "take". Element says "this is what i can accept - i want to operate on this other (curved) line". Like into a negotiation, when put together, the circuit operates where the two sides meet. If we say 1) = 2) then we find the intersection and where the circuit operates. E.g. operating point has to satisfy both. Its what we see on graph as green and red. Solving for the intersection gives us the current in device. Once we know that we can easily compute all rest. At this point we know every value in the circuit. *=*=*=*=*=*=*=*=*=*=*=*=*=* The 3rd question its hypothetical. Its about the blue line. Think of x = voltage and y = current. If we change x just a little (e.g. incremental), how much would the y change. Its the (x-minuscule change in x) / (y-minuscule change in y) all this near the operating point. Can be done 2 ways: 1) compute the derivative function for the device. The derivative by definition, its about the same thing: the derivative is the limit (minuscule) difference in a function output / the change in input. Derivative definition is limit (for h->0) out of : [f(x+h) - f(x)] / [h] . Note this is generally speaking, limit of the (change in y) / (change in x). That is amps / voltage, so if we talk in terms of "resistance" it has to be the other way around, hence the 1/~. 2) compute that fraction directly: we have x and y at operating point. We establish a new x different with 0.001% either in plus or minus, doesnt matter much. We compute x*1.001 into the device law and we get a new y to pair with it. We make the fraction like so: (x-x*1.001) / (y-y_new) and we got it. I guess its the incremental "resistance" because its a fraction of voltage over current, so it doesnt need 1/~ like the differential. Of course we can change the upper term with he low term and have it the other way. This should compute roughly to the same value as to what derivative method gives out. FirstChildUserIdTAG: 331483 FirstChildUserNameTAG: Doru FirstChildCreateTimeTAG: 2012-09-23T07:08:54Z IndexTAG: 2638 TitleTAG: S4E2 EXERCISE I do not understand very well how to calculate all possible boolean valued functions if there are two or three boolean-valued signals. Can anyone help me please? Thanks!!! UserIdTAG: 352637 UserNameTAG: Juaniba CreateTimeTAG: 2012-09-19T17:13:15Z VoteTAG: 1 CoursewareTAG: Week 2 / Boolean functions CommentableIdTAG: 6002x_S4E2_Boolean_Functions NumberOfReplyTAG: 1 FirstChildTAG: If we represent boolean function as y=f(x), than 1st question means "How many different arguments may we get". Ex: for 2 boolean valued signal y=f(00), f(01), f(10), f(11) N=4 or N=2^n for n boolean valued signal. Truth table will consist of 4 strings. A|B| C 0|0|0 or 1 (2 possible values) 0|1|0 or 1 ... 1|0|0 or 1 ... 1|1|0 or 1 ... Second question I understood as "How many different truth tables we can make" N=2*2*2*2 or N=2^(number of strings in truth table) FirstChildUserIdTAG: 192653 FirstChildUserNameTAG: Quaz FirstChildCreateTimeTAG: 2012-09-19T19:01:18Z SecondChildTAG: Thanks Quaz.. your explanation was very helpful SecondChildUserIdTAG: 335370 SecondChildUserNameTAG: SumitN SecondChildCreateTimeTAG: 2012-09-20T15:26:39Z SecondChildTAG: Yes, this is helpful, thank you! SecondChildUserIdTAG: 148389 SecondChildUserNameTAG: chento SecondChildCreateTimeTAG: 2012-09-20T21:01:30Z SecondChildTAG: good explanation, thanks SecondChildUserIdTAG: 171378 SecondChildUserNameTAG: ABHISHEKFROMINDIA SecondChildCreateTimeTAG: 2012-09-21T12:59:09Z IndexTAG: 2639 TitleTAG: Not able to watch the videos Hi, I am a student from India. When I tried to see the recorded videos of the lecture, it says "This is video is unavailable now" . I am not sure why this is happening. Please do help me watch the videos. Thank you !! UserIdTAG: 335980 UserNameTAG: thendral CreateTimeTAG: 2012-09-19T17:10:13Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 2640 TitleTAG: S3E3 Problem with calculating x1 and x2 I tried using the voltage divider formula and got x1 = V1 * R3/(R1+R3) = 0.8333 and x2 = V2 * R3/(R2+R3) = 5. What did I do wrong? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-19T16:14:52Z VoteTAG: 1 CoursewareTAG: Week 2 / Superposition example CommentableIdTAG: 6002x_S3E3_superposition NumberOfReplyTAG: 2 FirstChildTAG: You can use the voltage divider pattern, ¡but with carefull! When V2 is off, you have R2 and R3 in parallel, so you have to calculate an equivalent value of this two resistors, for example you can named it as R23. As you know, when you have two resistors in parallel, the equivalent resistor is R23 = (R2 * R3)/(R2 + R3). Then, you can apply the voltage divider pattern to R1 and R23. Similar actions you have to do in order to calculate the other case, when V1 is off. in this other case R1 and R3 are in paralell, you have to calculate their equivalent value, and then apply the voltage divider patter to R2 and R13. P.S. Sorry for mi english written. I'm spanish native and my level of english is no so good as you can read. FirstChildUserIdTAG: 447011 FirstChildUserNameTAG: Higinio FirstChildCreateTimeTAG: 2012-09-20T19:08:56Z SecondChildTAG: As you will turn off V2 you can make a simplification to the resistors & get i1 in terms of V1. SecondChildUserIdTAG: 185715 SecondChildUserNameTAG: amirengineer SecondChildCreateTimeTAG: 2012-09-20T20:24:21Z SecondChildTAG: sorry , I mixed between X1 & y1.... in X1, you will do a v.divider after applying parallel series technique 1st. SecondChildUserIdTAG: 185715 SecondChildUserNameTAG: amirengineer SecondChildCreateTimeTAG: 2012-09-20T20:31:20Z FirstChildTAG: You used the voltage divider pattern, but that pattern it works with two resistors, and we have three here. Maybe we can look for the pattern in this case, or otherwise, applied the node method. That is what i did FirstChildUserIdTAG: 149549 FirstChildUserNameTAG: Java_Dido FirstChildCreateTimeTAG: 2012-09-19T16:27:21Z SecondChildTAG: When using superposition there are only two resistors in each calculation. The question is asking for the component voltages in each of those situations. If V2 is open, its resistor(R2) has no effect. The answers don't make sense. Anonymous did it correctly. SecondChildUserIdTAG: 167370 SecondChildUserNameTAG: alphathreethree SecondChildCreateTimeTAG: 2012-09-20T02:35:33Z SecondChildTAG: Be careful when you tern off a voltage source ,you replace it with a short circuit "not open" so the branch that contains R2 is still there. SecondChildUserIdTAG: 185715 SecondChildUserNameTAG: amirengineer SecondChildCreateTimeTAG: 2012-09-20T20:26:29Z IndexTAG: 2641 TitleTAG: Troubles with R_TH I have troubles with understanding how to compute $R_{TH}$. I got $V_{TH}$ correctly, but now if I try to get current through the short cut, I cannot understand how to write equations for the scheme. UserIdTAG: 59903 UserNameTAG: Lumag CreateTimeTAG: 2012-09-19T14:36:06Z VoteTAG: 1 CoursewareTAG: Week 4 / Dependent Sources Exercise 1 CommentableIdTAG: 6002x_dep_src_exsz_1 NumberOfReplyTAG: 5 FirstChildTAG: Also you can compute ITH in similar way than VTH and then obtain RTH as relation VTH and ITH. FirstChildUserIdTAG: 143268 FirstChildUserNameTAG: jborrego FirstChildCreateTimeTAG: 2012-09-20T18:12:04Z SecondChildTAG: ITH = ISC IF you short circuit the output, the voltage across R1 will be the negative of the output of the dependent voltage source. R can't be negative alpha. I'm stuck. Please help. Thanks! SecondChildUserIdTAG: 245291 SecondChildUserNameTAG: rlicas SecondChildCreateTimeTAG: 2012-09-20T23:41:41Z SecondChildTAG: I suggest not short-circuiting the output then. Pick something arbitrary. I went with a 1A independent current source because that makes the math easy! See my comment on nevical's response to this post for more details SecondChildUserIdTAG: 248902 SecondChildUserNameTAG: Chanute SecondChildCreateTimeTAG: 2012-09-21T09:26:26Z FirstChildTAG: ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13483404842712824.jpg FirstChildUserIdTAG: 197569 FirstChildUserNameTAG: darksamaro FirstChildCreateTimeTAG: 2012-09-22T19:01:34Z SecondChildTAG: Thank you ! SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-09-23T10:29:43Z SecondChildTAG: Sorry to say, but short circuiting and expecting that the potential e to be the same doesn't seem right to me. By example ,if you have a resistor in paralel with a voltage source, by short circuiting, the potential isn't the same anymore. Node potential is a lot more complicated in this case than KVL. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-09-23T11:08:16Z SecondChildTAG: And you wrote the same equation for the node e, twice.Once by kcl, and once with node potential. Results that -alpha/R1=1 and -0,004/850 != 1 . SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-09-23T11:39:40Z SecondChildTAG: @AlexAlexandrescu: I don't think darksamaro meant for the two sets of equations to be interchangeable, except for $V_{TH}$ and $I_{SC}$ (i.e. $e_{OC~case} \neq e_{SC~case}$ and $i_{OC~case}\neq i_{SC~case}$). It looks like the open-cicuited terminal case has 3 eq's and 3 unknowns, and the short-circuited terminal case has the same; then there is one additional equation to determine $R_{TH}$. SecondChildUserIdTAG: 248902 SecondChildUserNameTAG: Chanute SecondChildCreateTimeTAG: 2012-09-24T01:39:04Z SecondChildTAG: Oh, i got it .Thanks ! SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-09-24T20:30:24Z FirstChildTAG: After doing some reading, I think I figured out what my problem was! It's all got to do with the equation $ V_{terminal} = V_{TH}+I_{excitation} \cdot R_{TH} $ and what happens when you do different things to the circuit. If all you do is leave the terminal as an open circuit, then $I_{excitation}=0$ and $ V_{terminal}=V_{TH} $. On the other hand, if all you do is short-circuit the terminal, then $V_{terminal}=0$ and $V_{TH}=I_{excitation} \cdot R_{TH}$ where $I_{excitation} = I_{SC}$. I know it should be 'negative', but that's just a matter of which direction you take to be the positive reference direction. What effect does zero-ing all the independent sources have on this equation? As far as I have been able to determine that sets $V_{TH}=0$. Which is great for determining the resistance in a purely resistive system, but less elucidating when there are dependent sources in the mix. Adding an $I_{excitation}$ seems to solve this problem nicely, because then you have $V_{terminal}=I_{excitation} \cdot R_{TH}$ and if you use $I_{excitation}=1A$, $R_{TH}=V_{terminal}$ (I'm omitting the unit conversion for the sake of the simplicity of the equation). If anybody sees something wrong with this explanation please let me know! FirstChildUserIdTAG: 248902 FirstChildUserNameTAG: Chanute FirstChildCreateTimeTAG: 2012-09-24T01:30:13Z SecondChildTAG: It is ok from my part, as i discovered from S3V7 .Choosing the external source of 1A was the easiest way, at least for me. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-09-24T16:24:47Z FirstChildTAG: short-circuit all current sources, and open-circuit all voltage sources. now assume current is flowing from the output of the network into the network and watch how the current is dividing, then you know which resistors are in series or parallel FirstChildUserIdTAG: 203899 FirstChildUserNameTAG: nevical FirstChildCreateTimeTAG: 2012-09-19T16:37:14Z SecondChildTAG: Wouldn't it be open-circuiting independent current sources, and short circuiting independent voltage sources (i.e. setting all the values to 0)? Then if you apply an arbitrary current you can see how the system reacts and you can use the equation $ V_{out} = V_{TH}+I_{arbitrary} \cdot R_{TH} $ where $ V_{out}$ is the voltage you find across the port with your arbitrary current applied and all independent sources zero-d (and $V_{TH} $ found as before). SecondChildUserIdTAG: 248902 SecondChildUserNameTAG: Chanute SecondChildCreateTimeTAG: 2012-09-21T09:24:07Z SecondChildTAG: ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/134834027989342.jpg SecondChildUserIdTAG: 197569 SecondChildUserNameTAG: darksamaro SecondChildCreateTimeTAG: 2012-09-22T18:58:14Z SecondChildTAG: I don't follow that all the way darksamaro. Why is $I_{R_2} = 0$ in the second circuit? SecondChildUserIdTAG: 248902 SecondChildUserNameTAG: Chanute SecondChildCreateTimeTAG: 2012-09-23T08:09:12Z SecondChildTAG: When two resistors are in parallel connection, current through resistor with lower value is bigger. Short-circuited resistor looks like parallel connection with R=0, and all current flows through R=0. SecondChildUserIdTAG: 192653 SecondChildUserNameTAG: Quaz SecondChildCreateTimeTAG: 2012-09-23T09:25:00Z SecondChildTAG: Thank you chanute for the explanations ! SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-09-23T10:23:57Z SecondChildTAG: IR2 = 0, because the resistor R2 itself it's practically short-circuited. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-09-23T10:41:12Z SecondChildTAG: No problem, and thank you for the explanation as well. But I think more to the point what I'm having trouble with is figuring out when to leave independent sources in, when to set them to zero, and when to add an arbitrary current and when not to. My difficulties were made very evident when trying to solve the last part of week 4's homework. SecondChildUserIdTAG: 248902 SecondChildUserNameTAG: Chanute SecondChildCreateTimeTAG: 2012-09-23T11:07:45Z SecondChildTAG: Hi, I resolved the exercise, but I didn't use another current source at the output port. I started like darksamaro: -IO + e/850 + Vth/750 = 0 e = -4*i + Vth, so -IO +(-4*i + Vth)/850 + Vth/750 = 0 By earranging Vth I finally got: Vth = 1.5936 + 1.8725*i The missing part is i. We know that Vth = e + 4i, and e = 850*i so Vth = 854*i. The final substitution gives Vth a value of 1.5971 V. What do you think of this? SecondChildUserIdTAG: 220837 SecondChildUserNameTAG: Calsomus SecondChildCreateTimeTAG: 2012-09-23T22:25:47Z SecondChildTAG: RTH can also be calculated by turning off all sources and solving for R (two resistors in parallel) SecondChildUserIdTAG: 348141 SecondChildUserNameTAG: TomTrieb SecondChildCreateTimeTAG: 2012-09-24T14:29:03Z SecondChildTAG: @TomTrieb: I'm pretty sure that will give you an incorrect value in the case where you have a dependent source in the mix SecondChildUserIdTAG: 248902 SecondChildUserNameTAG: Chanute SecondChildCreateTimeTAG: 2012-09-25T00:16:38Z SecondChildTAG: I think that wont be a problem, because in the process of making Io Zero, we are anyway making i (the current responsible for alpha*i term) zero. So, this way answer works.... SecondChildUserIdTAG: 367886 SecondChildUserNameTAG: KashwinKohli SecondChildCreateTimeTAG: 2012-09-30T08:22:19Z SecondChildTAG: While it is true that removing the independent sources also brings the $i$ term to 0, calculating the resistance in this way will neglect the effect the dependent source has on the effective resistance of the circuit and give you an incorrect answer for $R_{TH}$. Take a super simple example as a check: an independent voltage source ($V_I$), a dependent voltage source ($V_D=0.5 \cdot V_R$) and a resistor ($R$) all in series. By that logic $V_{TH}=V_I$ and $R_{TH}=R$. But we know by short-circuiting the device that $I_{N}= \frac {2 \cdot V_I}{R}$, so $R_{TH}=\frac {V_{TH}}{I_N}=\frac {R}{2}$ SecondChildUserIdTAG: 248902 SecondChildUserNameTAG: Chanute SecondChildCreateTimeTAG: 2012-10-04T05:05:39Z FirstChildTAG: It's very simple. The resistence of dependent source is α. ![dependent source as resistance ][1] [1]: https://edxuploads.s3.amazonaws.com/1349447241850045.jpg FirstChildUserIdTAG: 325246 FirstChildUserNameTAG: efys FirstChildCreateTimeTAG: 2012-10-05T14:26:51Z SecondChildTAG: Ahhh! I can't believe I didn't see that sooner! I don't suppose all CCVS's will be that simple, but that was just something that was so obvious I completely overlooked it. Nice catch! SecondChildUserIdTAG: 248902 SecondChildUserNameTAG: Chanute SecondChildCreateTimeTAG: 2012-10-09T12:03:09Z IndexTAG: 2642 TitleTAG: Question about the discussion forum I was wondering how we get to know about changes in the posts we follow. Right now what I do is find a post I'm following by browsing through the list, then go to my profile. Then I have to check each post listed there to see if there are any updates. Is there a shorter way to do this? Thanks in advance. UserIdTAG: 154440 UserNameTAG: Aahlad CreateTimeTAG: 2012-09-19T14:09:10Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Very soon there will be tracking of which posts have read/unread content as well as an entry in the dropdown menu to view all your followed threads. Hopefully today, but I can't guarantee that. FirstChildUserIdTAG: 323387 FirstChildUserNameTAG: ibrahimawwal FirstChildCreateTimeTAG: 2012-09-19T14:14:35Z SecondChildTAG: Thanks for the info! SecondChildUserIdTAG: 154440 SecondChildUserNameTAG: Aahlad SecondChildCreateTimeTAG: 2012-09-19T15:17:41Z SecondChildTAG: Should be live in a few minutes :) SecondChildUserIdTAG: 323387 SecondChildUserNameTAG: ibrahimawwal SecondChildCreateTimeTAG: 2012-09-21T13:11:34Z IndexTAG: 2643 TitleTAG: help in lab 2 hai... if anyone can give me some hints in lab 2 i will be helped a lot... UserIdTAG: 447302 UserNameTAG: phanisrikar CreateTimeTAG: 2012-09-19T13:26:20Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Try to get the proportions right first. 1/2 of one 1/6th of the other, combine them. Then figure out how to end up end up with 0.5 of this combined voltage. (Your actual numbers may be different.) FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-19T13:53:33Z SecondChildTAG: what about the minimun and maximum voltages?/ how are they used in solution? SecondChildUserIdTAG: 447302 SecondChildUserNameTAG: phanisrikar SecondChildCreateTimeTAG: 2012-09-19T16:49:29Z IndexTAG: 2644 TitleTAG: Omega value Can someone explain for me why answers are not depended on Omega value? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-19T10:40:24Z VoteTAG: 1 CoursewareTAG: Week 4 / Linearization Exercise 1 CommentableIdTAG: 6002x_linearization_ex_1 NumberOfReplyTAG: 1 FirstChildTAG: By Omega you mean the symbol next to the "2" in the resistor? That is only the units of the resistance, not a constant multiplying it. FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-19T13:34:52Z SecondChildTAG: Great thanks! It was my huge miss! SecondChildUserIdTAG: 329361 SecondChildUserNameTAG: ikrukov SecondChildCreateTimeTAG: 2012-09-19T13:59:40Z SecondChildTAG: this question will be solved by numerical method known as newton raphson method but first do check the convergence of this id equation google it and find this equation to solve SecondChildUserIdTAG: 294370 SecondChildUserNameTAG: Naif1125 SecondChildCreateTimeTAG: 2012-10-09T06:14:51Z IndexTAG: 2645 TitleTAG: STAFF While reading about MOSFET , I was introduced to BJT(Bipolar junction Transistor)in a book. I read that Ic=(beeta)*Ib (i.e) Collector current = beeta * base current in Common emitter(CE) configuration. My doubt is what makes Collector current to depend only on base current. Why that relation holds good for BJT? On what factors beeta will depend? Will it depend on size of base, size of contact of Base and collector, amount of doping etc.Please please explain me. UserIdTAG: 132685 UserNameTAG: deepakmurali CreateTimeTAG: 2012-09-19T10:16:28Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 2646 TitleTAG: hello every one what is noise margin UserIdTAG: 133844 UserNameTAG: RSSK133 CreateTimeTAG: 2012-09-19T10:08:40Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: http://en.wikipedia.org/wiki/Noise_margin FirstChildUserIdTAG: 436749 FirstChildUserNameTAG: waynebrown FirstChildCreateTimeTAG: 2012-09-19T10:15:26Z FirstChildTAG: [Read here ][1]and [here][2] ,Textbook. **Noise margin:** The absolute value of the difference between the prescribed output voltage for a given logical value and the corresponding forbidden region votage threshold for the receiver is called the noise margin for that logical value. **Noise Margin for logical 0:** NM0=VIL-VOL **Noise margin for logical 1:** NM1=VOH-VIH [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/274 [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/275 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-19T12:59:57Z IndexTAG: 2647 TitleTAG: Bummer... Well, I'm making an interesting observation this time around. I did the inaugural 6.002x course and did fairly well, so I thought I'd go through it again and firm up some of the ideas (my day-to-day does not offer an opportunity to use the info so much). Well, I did better the first time than I'm doing this time. I guess I was more single-minded about it all that first time. It remains to be seen if I will remain determined...at the moment I am somewhat bummed out. It may pass. :) EDIT...18 days later: Well, it was my hope to add some depth, this time, to what I learned last time. I am currently optimistic it's working out that way. Yay! Stay tuned! :) UserIdTAG: 92346 UserNameTAG: MobiusTruth CreateTimeTAG: 2012-09-19T03:44:28Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Hang in there. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-19T04:02:19Z FirstChildTAG: Hi MobiusTruth! I am sure that is because of the saying:) "As much as you learn as much that you feel that you don´t know". You did a great work in the spring! So be up! I am sure that this fall you will improve your concepts and will have an excellent performance XD See you! Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-19T13:34:23Z FirstChildTAG: I also did the first course and achieved 59.9%, so am attempting a pass this time round. Stick with it...it is easier second time round (so far). Melbur FirstChildUserIdTAG: 95673 FirstChildUserNameTAG: melbur FirstChildCreateTimeTAG: 2012-09-19T15:31:05Z IndexTAG: 2648 TitleTAG: everything oki! good test to familiarize withe the tool UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-18T22:36:33Z VoteTAG: 1 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: what is noise margin??? FirstChildUserIdTAG: 133844 FirstChildUserNameTAG: RSSK133 FirstChildCreateTimeTAG: 2012-09-19T10:06:15Z IndexTAG: 2649 TitleTAG: Fractional resistance Can someone help and explain to me how to interpret fractional resistances, for example, R=0.5 Ohms? The interactive tool chest accepts fractional resistances. Is this the OK thing? UserIdTAG: 252722 UserNameTAG: vaboro CreateTimeTAG: 2012-09-18T21:05:59Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: .. strange question, 0.5 Ohms is OK, and 0.00056701 Ohms is OK too as 157.875 Ohms. Could you describe your question more widely ? FirstChildUserIdTAG: 277168 FirstChildUserNameTAG: AndrewKiselev FirstChildCreateTimeTAG: 2012-09-18T21:18:58Z SecondChildTAG: There's no particular question. I just couldn't imagine a 0.5 Ohms resistor. I was particularly confused by the formula I = V/R. With a resistor less than 1 Ohm one can actually increase the current with the same voltage. I was just thinking how might the power be affected. SecondChildUserIdTAG: 252722 SecondChildUserNameTAG: vaboro SecondChildCreateTimeTAG: 2012-09-19T11:50:30Z SecondChildTAG: ...I think I got it: the power will increase; yet, power is a derivative of the amount of energy, and when resistance is less than one energy source will be depleted faster I think, right? SecondChildUserIdTAG: 252722 SecondChildUserNameTAG: vaboro SecondChildCreateTimeTAG: 2012-09-19T11:56:21Z SecondChildTAG: ...however, what about superconductive materials with almost no resistance... I need to clarify this mess somehow...:) SecondChildUserIdTAG: 252722 SecondChildUserNameTAG: vaboro SecondChildCreateTimeTAG: 2012-09-19T11:58:17Z SecondChildTAG: ...alright, the lump circuit representation of an ideal wire is a superconductive wire with no resistance ...now this is more or less clear SecondChildUserIdTAG: 252722 SecondChildUserNameTAG: vaboro SecondChildCreateTimeTAG: 2012-09-19T19:49:39Z FirstChildTAG: 0.5 ohms = 500 milliohms - 1,000,000 ohms = 1 megaohm - 100,000 ohms = 100 kilohms - 10,000 ohms = 10 kilohms - 1,000 ohms = 1 kilohm - 1 ohm = 1 ohm - 0.100 = 100 milliohms - 0.010 = 10 milliohms - 0.001 = 1 milliom - 0.000100 = 100 microohms - 0.000010 = 10 microohms - 0.000001 =1 microohm FirstChildUserIdTAG: 454075 FirstChildUserNameTAG: neme2987 FirstChildCreateTimeTAG: 2012-09-18T22:14:18Z IndexTAG: 2650 TitleTAG: H2P1 help I can not get the right answer of H2P1. Since Rth = R1*R2 / (R1+ R2) and Rth is between 10 k and 30 k, also R2 / (R1 + R2) = 7/20, I can get R1 between 28.6 k and 85.71 k and R2 between 15.4 and 46k. How to proceed further? UserIdTAG: 240389 UserNameTAG: julieqpt CreateTimeTAG: 2012-09-18T20:02:44Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: An assumption should be made about nominal values of resistors. As a result either Vmax or Vmin can be higher or lower than the tolerance of Vout/Vin +/- 10% Just make sure that the tolerance of Vout/Vin +/- 10% holds for the nominal values of the resistors. FirstChildUserIdTAG: 252722 FirstChildUserNameTAG: vaboro FirstChildCreateTimeTAG: 2012-09-18T20:28:13Z SecondChildTAG: I still don't understand. When it says Vout/Vin, is that putting 4/20? SecondChildUserIdTAG: 187400 SecondChildUserNameTAG: Charizard SecondChildCreateTimeTAG: 2012-09-19T04:38:19Z IndexTAG: 2651 TitleTAG: a good principle Be conservative in what you send and liberal in what you accept. UserIdTAG: 107219 UserNameTAG: AlexandreZ CreateTimeTAG: 2012-09-18T16:08:20Z VoteTAG: 1 CoursewareTAG: Week 2 / Static Discipline CommentableIdTAG: 6002x_static_discipline NumberOfReplyTAG: 0 IndexTAG: 2652 TitleTAG: Unable to access courses I have a problem with my vision after a minor accident last week. It has blurred my sight, hence could not start the course as per schedule on the 5 th of september. my qsn is : Am i allowed to continue the course despite missing the DEADLINE ? i would very much want to ! at present my friend is reading out the material to me ....hence my progress is slow............. - Thanks a lot , for your time and patience UserIdTAG: 274093 UserNameTAG: Martandatilaka CreateTimeTAG: 2012-09-18T16:03:00Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Troubleshooting NumberOfReplyTAG: 2 FirstChildTAG: You won't be able to get grades for Week 1's assignment. However, you can still get full credit for the course because they count the best 10 out of the 12 assignments. So focus on the remaining homework and labs. FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-09-18T16:23:14Z FirstChildTAG: The course grading drops the two lowest grades for homeworks and labs. So you are not in any trouble until you either miss more than two of each, or get less than 100% on any of the remaining 10 labs and homeworks that will constitute part of your grade (the remaining part of your grade will consist of midterm and final exam grades). I suggest you review the course syllabus https://www.edx.org/static/content-mit-6002x/handouts/syllabus.a477535058a1.pdf and other materials on the 'Course Info' page. The Course Syllabus explains how the course is graded, among other things. Course grades are computed as follows: Homeworks - 15% Labs - 15% Midterm - 30% Final - 40% Grades are distributed as follows: A - 87% B - 70% C - 60% You need a minimum of a C to receive a certificate in this course. Each of the homeworks and labs has equal weight. Therefore, since the two lowest grades of homeworks and labs are dropped, only ten of each count toward your overall grade. That means that each homework and lab can contribute a maximum of 1.5 points toward your overall grade (assuming you get all the questions correct on each lab or homework). FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-09-18T23:12:18Z IndexTAG: 2653 TitleTAG: Does it possible to open next weeks lectures? Hello, EDX team. I want to study this course few weeks forward, because next month I won't have enough time. Is it possible open next week lectures, when student finishes current week HW and Labs. With Best wishes! UserIdTAG: 192653 UserNameTAG: Quaz CreateTimeTAG: 2012-09-18T15:40:54Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I am just a student here, but yes you can, many are working on week 4 now. I highly recommend getting ahead, because if you have problems, it's best to ask when the work first becomes available. That seems to be when the more experienced people respond. If you wait too long to ask for help, you are just asking questions that have already been asked, and are less likely to get a response. Also it gives you time to work on your weaknesses once identified. If you can't ask questions when the work is fresh then use the search feature, you will find similar questions asked and can get your answer quickly. If you can't find your answer then, ask away. Have fun. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-19T15:53:22Z IndexTAG: 2654 TitleTAG: help plzz!! how to find Distinct boolean valued function is that by 16^(n-1) or 2^(2^n)????????? UserIdTAG: 177128 UserNameTAG: MARYAAM CreateTimeTAG: 2012-09-18T12:15:49Z VoteTAG: 1 CoursewareTAG: Week 2 / Boolean functions CommentableIdTAG: 6002x_S4E2_Boolean_Functions NumberOfReplyTAG: 1 FirstChildTAG: Hi, pls. look at the post[Difference between "Distinct value"and "Distinct boolean valued function".][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_S4E2_Boolean_Functions/threads/505626a378dbb71f0000002c FirstChildUserIdTAG: 277168 FirstChildUserNameTAG: AndrewKiselev FirstChildCreateTimeTAG: 2012-09-18T12:27:25Z IndexTAG: 2655 TitleTAG: Few questions about this course i enter this course now 18-sep-2012. meaning that a week has passed where i wasn't here. so a h.w. has been submitted that i couldn't do. will this affect my grade? UserIdTAG: 456939 UserNameTAG: Ice_Shadow CreateTimeTAG: 2012-09-18T11:16:42Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Check out your progress timeline. In HW11 and HW12 it says "The lowest 2 Homework are dropped". FirstChildUserIdTAG: 311022 FirstChildUserNameTAG: GordanS FirstChildCreateTimeTAG: 2012-09-18T11:46:58Z FirstChildTAG: I posted a reply to a similiar question here: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5058581a5dc8a52300000002 The course grading drops the two lowest grades for homeworks and labs. So you are not in any trouble until you either miss more than two of each, or get less than 100% on any of the remaining 10 labs and homeworks that will constitute part of your grade (the remaining part of your grade will consist of midterm and final exam grades). I suggest you review the course syllabus https://www.edx.org/static/content-mit-6002x/handouts/syllabus.a477535058a1.pdf and other materials on the 'Course Info' page. The Course Syllabus explains how the course is graded, among other things. Course grades are computed as follows: Homeworks - 15% Labs - 15% Midterm - 30% Final - 40% Grades are distributed as follows: A - 87% B - 70% C - 60% You need a minimum of a C to receive a certificate in this course. Each of the homeworks and labs has equal weight. Therefore, since the two lowest grades of homeworks and labs are dropped, only ten of each count toward your overall grade. That means that each homework and lab can contribute a maximum of 1.5 points toward your overall grade (assuming you get all the questions correct on each lab or homework). FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-09-18T13:12:55Z IndexTAG: 2656 TitleTAG: CIRCUIT CATEGORIES How do we know that a circuit is linear without solving for circuit parameters.What other types of circuits are there apart from linear circuits? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-18T09:53:42Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: A circuit that contains only resistors, capacitors, inductors, dependent and independent voltage sources is always linear. We will see non-linear circuits from Week 3 onward (actually, digital circuits of week 2 are also non-linear). FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-09-18T10:54:51Z SecondChildTAG: Great answer, but just to clarify, dependent voltage or current sources can also be non-linear as we will study in the next weeks. SecondChildUserIdTAG: 381746 SecondChildUserNameTAG: jelizon SecondChildCreateTimeTAG: 2012-09-18T14:06:33Z SecondChildTAG: Ah yes I should have added that. Thanks jelizon :-) SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-09-18T14:20:07Z SecondChildTAG: but how do i make out which dependent sources make the circuit non linear? SecondChildUserIdTAG: 344720 SecondChildUserNameTAG: Vaibhav_Kashyap SecondChildCreateTimeTAG: 2012-09-19T16:05:35Z SecondChildTAG: The dependent source is linear if the equation describing it is linear, i.e. of the form $y = m*x + b$. You will see that some dependent sources will have an equation such as $i = k*V^2$, and so the equation is not linear but quadratic. SecondChildUserIdTAG: 381746 SecondChildUserNameTAG: jelizon SecondChildCreateTimeTAG: 2012-09-19T21:14:23Z IndexTAG: 2657 TitleTAG: H2P1 - Answer correct ... but why? Actually, the question is : why not 33kOhm and 22kOhm? They fit the Rth (13.2kOhms) and Vth(20V) and they are part of the E set ... anybody care to explain why? I can't explain myself. They even fit to the 10% threshold easily - IF i am calculating correctly. Any help appreciated. UserIdTAG: 19774 UserNameTAG: Gor131313 CreateTimeTAG: 2012-09-18T08:57:20Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hi, 1.What about: "open-circuit output voltage of Vout≈12.0V." if R1=33K and R2=22K ? 2.What about: the same thing in case R1=33K+10% and R2=22K-10% (and vice versa ) ? FirstChildUserIdTAG: 277168 FirstChildUserNameTAG: AndrewKiselev FirstChildCreateTimeTAG: 2012-09-18T11:38:02Z FirstChildTAG: Your calculations seem to be correct but the Vout=20V is outside of the 17.5 + 10% tolerance (19.25V). Your Vout has to be between 15.75V and 19.25V. FirstChildUserIdTAG: 208488 FirstChildUserNameTAG: moelarrycheese FirstChildCreateTimeTAG: 2012-09-19T01:39:00Z IndexTAG: 2658 TitleTAG: What is that mean' boolean-valued'? How many distinct boolean-valued functions are there of two boolean-valued signals? I can't understand this sentence. why the answer is 16. what is it want me to calculate? UserIdTAG: 128276 UserNameTAG: ChengBin CreateTimeTAG: 2012-09-18T05:53:47Z VoteTAG: 1 CoursewareTAG: Week 2 / Boolean functions CommentableIdTAG: 6002x_S4E2_Boolean_Functions NumberOfReplyTAG: 3 FirstChildTAG: i think its 16^(n-1)where n=2 here.... i need confirmation hope its right .. FirstChildUserIdTAG: 177128 FirstChildUserNameTAG: MARYAAM FirstChildCreateTimeTAG: 2012-09-18T12:02:32Z FirstChildTAG: Hi, look at this post:[Difference between "Distinct value"and "Distinct boolean valued function".][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_S4E2_Boolean_Functions/threads/505626a378dbb71f0000002c FirstChildUserIdTAG: 277168 FirstChildUserNameTAG: AndrewKiselev FirstChildCreateTimeTAG: 2012-09-18T12:05:13Z FirstChildTAG: You can think about how many different outputs can I have? If you have N input values (which is the number of distinct values possible), then you'd have 2 possible output values for each of the N values (0 or 1). Then, you simply have 2^N possible outputs, where N is the number of input values of your truth table. Thus, as N = 2^n (where n is the number of boolean-valued signals), you have 2^(2^n) as the answer. FirstChildUserIdTAG: 99238 FirstChildUserNameTAG: arthurltc FirstChildCreateTimeTAG: 2012-09-18T16:49:25Z IndexTAG: 2659 TitleTAG: REQUEST FOR APOLOGIZE AND DOUBT SIR..,I HAD EXAMS LAST WEEK SO I COULDN'T COMPLETE 1ST WEEK ASSIGNMENT.., PLEASE TELL ME WHAT TO DO AND I HAD SEEN ALL VIDEOS OF 1ST WEEK ..,ALSO I HAD DOUBT IN INDEPENDENT AND REDUNDANT EQUATIONS AFTER APPLYING KCL AND KVL METHOD PLEASE HELP ME..I.E HELP ME IN DECIDING CRITERIA FOR REDUNDANT EQNS UserIdTAG: 293926 UserNameTAG: YATHISH121 CreateTimeTAG: 2012-09-18T02:45:55Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: let think about 2nd assignment :) FirstChildUserIdTAG: 415376 FirstChildUserNameTAG: imab90 FirstChildCreateTimeTAG: 2012-09-18T17:31:05Z FirstChildTAG: You can do 10 out of 12 assignments. FirstChildUserIdTAG: 138769 FirstChildUserNameTAG: ArturoPrado FirstChildCreateTimeTAG: 2012-09-18T03:32:46Z FirstChildTAG: I posted a reply to a similiar question here: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5058581a5dc8a52300000002 The course grading drops the two lowest grades for homeworks and labs. So you are not in any trouble until you either miss more than two of each, or get less than 100% on any of the remaining 10 labs and homeworks that will constitute part of your grade (the remaining part of your grade will consist of midterm and final exam grades). I suggest you review the course syllabus https://www.edx.org/static/content-mit-6002x/handouts/syllabus.a477535058a1.pdf and other materials on the 'Course Info' page. The Course Syllabus explains how the course is graded, among other things. Course grades are computed as follows: Homeworks - 15% Labs - 15% Midterm - 30% Final - 40% Grades are distributed as follows: A - 87% B - 70% C - 60% You need a minimum of a C to receive a certificate in this course. Each of the homeworks and labs has equal weight. Therefore, since the two lowest grades of homeworks and labs are dropped, only ten of each count toward your overall grade. That means that each homework and lab can contribute a maximum of 1.5 points toward your overall grade (assuming you get all the questions correct on each lab or homework). FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-09-18T13:08:43Z IndexTAG: 2660 TitleTAG: Time Zone !!! Which time zone one should follow to complete all the homework and lab. work within due date and time? Please help with some relevant information... UserIdTAG: 295278 UserNameTAG: souravfn7 CreateTimeTAG: 2012-09-18T01:36:36Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: If you complete the HW and Lab by the (ending) midnight of Sunday of your local time, your submission will be accepted. FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-18T01:43:32Z FirstChildTAG: i am new applicant for this program, which time zone i follow for submitting homework FirstChildUserIdTAG: 456614 FirstChildUserNameTAG: atulnarayan786 FirstChildCreateTimeTAG: 2012-09-18T10:40:35Z IndexTAG: 2661 TitleTAG: Lab 2: It looked right to me! still "X" I don't know how the demands of the grader can be so exacting! My answer looked right, on the transient analysis, but I earned the "X" anyhow. Has anyone else experienced this? UserIdTAG: 280220 UserNameTAG: pilgrimCycle CreateTimeTAG: 2012-09-17T20:09:06Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: Wonderful! I got it! Hint: Superposition problem. FirstChildUserIdTAG: 385692 FirstChildUserNameTAG: tpfslima FirstChildCreateTimeTAG: 2012-09-18T02:29:48Z FirstChildTAG: Yes, i did. My actual output signal had a similar shape as the one required, except that the amplitude was a little lower. I submitted it and got an X. I reset my circuit, made new calculations, and I finally got it right!... it's a tricky one (I checked it again and got the check mark and credit for it) FirstChildUserIdTAG: 322021 FirstChildUserNameTAG: mastropiero FirstChildCreateTimeTAG: 2012-09-17T21:41:25Z SecondChildTAG: Thanks for the encouragement. I shall keep trying. SecondChildUserIdTAG: 280220 SecondChildUserNameTAG: pilgrimCycle SecondChildCreateTimeTAG: 2012-09-17T23:35:26Z FirstChildTAG: I have a hunch that the 'problem' is in the simulation using non-ideal models for the plain wire, giving it overall a slightly increased resistance. Does anyone know for a fact whether this is so? FirstChildUserIdTAG: 280220 FirstChildUserNameTAG: pilgrimCycle FirstChildCreateTimeTAG: 2012-09-18T00:29:57Z SecondChildTAG: Nope, the circuit simulator wire is an ideal wire. SecondChildUserIdTAG: 79361 SecondChildUserNameTAG: kimt SecondChildCreateTimeTAG: 2012-09-18T00:30:26Z SecondChildTAG: Does the simulation put variances in the resistor values, like the variance for non-ideal resistor tolerances? I'd just like to know why the calculation yields an exact proportion, but it's not within the tolerance needed to earn the 'check.' Is this an engineering lesson in throwing the algebra out the window and resorting to trial and error after all? SecondChildUserIdTAG: 280220 SecondChildUserNameTAG: pilgrimCycle SecondChildCreateTimeTAG: 2012-09-18T12:03:04Z FirstChildTAG: Same problem here. I designed simple mixer but it does not matter how I try, I just cannot balance voltage dividers to exactly 1/2 and 1/6. FirstChildUserIdTAG: 167413 FirstChildUserNameTAG: TeTAn FirstChildCreateTimeTAG: 2012-09-18T02:39:19Z SecondChildTAG: Tetan, Can you post a screenshot of your lab circuit? SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-09-18T05:31:48Z SecondChildTAG: It just dawned on me, I think I know why we are having trouble. We're dealing with sinusoidal and square wave voltages, which change based on time. For instance when V1 = .667, V2 = sin(Pi/2) and other multiples of t. Therefore, we're trying to solve the problem considering the voltages are constant (which, so far probably 95% of the problems we have solved are with constant voltages and we have not had a lot of experience with varying voltage sources.), which is not the case. I am going to approach the problem with this knowledge and see what happens. Stay tuned. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-09-19T03:49:30Z SecondChildTAG: I made a mistake, let me modify my statement. When Vout = .667V, V1 = 1 and V2 = sin(Pi/2). I'm pretty sure this is correct, but as I get further into the problem I may discover differently. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-09-19T04:21:55Z FirstChildTAG: It seems the consensus is to use more than two resistors and I'm assuming trying different combinations until an answer is found, starting with the sample circuit given in Figure 2. Superposition also seems to be the best method to use. What would happen if the circuit was more complicated? There would be hundreds of different resistor combinations and it would take forever to find the right one. Shouldn't there be a way to solve this problem analytically, or some other way besides by trial and error, similar to Lab1 where we used 2 equations with 2 unknowns to solve the problem? Perhaps if we used all the data given in the problem and somehow set it up to solve for a solution. Data available is a square wave with the given parameters and a sin wave given by V = sin(pi*10^4*t). Also given is a maximum output value of 667mV and a minimum output value of -167mV. Could't these be set equal to Vout, coming up with 2 equations, such as 667mV = (1/2)V1 + (1/6)V2 and -167mV = (1/2)V1 + (1/6)V2? Has anyone used a method similar to this, without having to use trial and error starting with a similar circuit? FirstChildUserIdTAG: 132828 FirstChildUserNameTAG: rharris FirstChildCreateTimeTAG: 2012-09-18T07:22:27Z IndexTAG: 2662 TitleTAG: Maybe helpful After solving for a1 and a2, I was stuck for a bit. I used KCL and used the voltages divided by their resistances to get an equation in terms of V1 and V2 to solve for v3. Using KCL again, and leaving i1 alone, I again converted i2 and i3 into the respective V/R. This left me with an equation with V2 and v3. Since we already solved for v3, just plug that equation in. The denominator is quite large (R1*R2+R1*R3+R2*R3), so I called that x. This helped me keep my algebra straight. Solving for the coefficient of V1 was simple, since certain resistances cancelled (R2*R3/R2*x)=(R3/x). Solving for V2 was much more difficult. There were many V2 terms, so getting a nice compact answer required finding the common denominator. This led to an equation similar to x/(R2*x) + (R1*R3)/(R2*x). This worked out nicely, since x contains terms such as R1*R3, so these cancelled out, leaving just once simple coefficient. I know this was long winded, but perhaps someone will find this useful. I stopped for a while thinking I had screwed up the answer just to find that I needed to keep going. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-17T19:54:53Z VoteTAG: 1 CoursewareTAG: Week 2 / Linear combinations of source strenghts CommentableIdTAG: 6002x_S3E2_Linear_Combinations_of_Source_Strengths NumberOfReplyTAG: 0 IndexTAG: 2663 TitleTAG: Comments are not visible In the previous layout comments, which are actually answers to the asked problem, were visible. But now I cannot find those comments, only number of comments are visible. Kindly let them visible again. UserIdTAG: 329015 UserNameTAG: farah_sarwar CreateTimeTAG: 2012-09-17T19:49:29Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Can you provide a screenshot of the issue? I'm not sure I understand. FirstChildUserIdTAG: 323387 FirstChildUserNameTAG: ibrahimawwal FirstChildCreateTimeTAG: 2012-09-17T22:27:13Z FirstChildTAG: I was going to ask about this too! I'm considering putting my screen-name in each title so I can search for it and find it again. Does anyone know how to get the list of 'followed' threads for one's own account? I noticed if you click on someone else's username, it brings up their thread activity on the right. FirstChildUserIdTAG: 280220 FirstChildUserNameTAG: pilgrimCycle FirstChildCreateTimeTAG: 2012-09-17T20:34:06Z FirstChildTAG: After I'd posted that, I found that you can click your own user-name in the Discussion forum and find a list of threads... very useful! FirstChildUserIdTAG: 280220 FirstChildUserNameTAG: pilgrimCycle FirstChildCreateTimeTAG: 2012-09-17T20:44:40Z IndexTAG: 2664 TitleTAG: there might be some confusion here The speaker says "V low is zero volts"; whereas the transcript reads "V low is two volts" UserIdTAG: 252722 UserNameTAG: vaboro CreateTimeTAG: 2012-09-17T19:43:17Z VoteTAG: 1 CoursewareTAG: Week 2 / Static Discipline CommentableIdTAG: 6002x_static_discipline NumberOfReplyTAG: 0 IndexTAG: 2665 TitleTAG: another question about the signs on y1, y2 I see Kerbyco's question, and I understand AndrewKiselev's answer, about the different signs for y1 and y2. My question concerns why it is y1 that has the negative sign and y2 the positive sign, instead of y1 positive and y2 negative. The reason I wonder is i1 is directed into the voltage source V1 at the "+" end, and I thought this was our convention, that the current entering at the "+" end is positive current. Can anyone help clarify for me? UserIdTAG: 280220 UserNameTAG: pilgrimCycle CreateTimeTAG: 2012-09-17T18:23:09Z VoteTAG: 1 CoursewareTAG: Week 2 / Superposition example CommentableIdTAG: 6002x_S3E3_superposition NumberOfReplyTAG: 0 IndexTAG: 2666 TitleTAG: Math processing error Hello Sir, Each time I load the the homework/worked example page I get "Math processing error" on all the numeric values. My firefox is unable to load some mathjax image format. I already sent mail to bugs@edx.org . No reply. Please help if anybody facing the same issues. Thanks in advance. UserIdTAG: 407691 UserNameTAG: shettyma CreateTimeTAG: 2012-09-17T17:49:42Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Dear Shettyma, Kindly use Google chrome as it has proven to be the best for our results. Regards, Ugo FirstChildUserIdTAG: 329879 FirstChildUserNameTAG: Ugochukwu FirstChildCreateTimeTAG: 2012-09-17T18:15:13Z FirstChildTAG: I've answered this question in several threads (hint: Use the 'Search' function to find answers before posting your question). Here's the answer again: That error is due to how your browser interacts with MathJax. MathJax is used by the server to present mathematical symbols on your screen. Here's how to fix it: 1) Place your cursor on the *[Math Processing Error]* message and right-click. 2) You will see a pull-down MathJax menu. 3) Select Math Settings, Math Renderer, and choose either MathML or SVG. I prefer the way MathML looks, but it drops some symbols when I print, so I have to use SVG. 4) You can also play with the zoom settings on the MathJax menu to adjust the zoom functionality. FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-09-18T02:01:53Z IndexTAG: 2667 TitleTAG: S3E1 I've used the sandbox to confirm the node potential, labeled e. It's minus 1.69V. UserIdTAG: 260994 UserNameTAG: capvl CreateTimeTAG: 2012-09-17T16:24:41Z VoteTAG: 1 CoursewareTAG: Week 2 / Node equation review CommentableIdTAG: 6002x_S3E1_NodeEquationReview NumberOfReplyTAG: 1 FirstChildTAG: becareful with the sign placement of -7.2 voltage source FirstChildUserIdTAG: 201508 FirstChildUserNameTAG: datle FirstChildCreateTimeTAG: 2012-09-18T09:16:28Z IndexTAG: 2668 TitleTAG: How to change my user name I want to change my user name how can I change it? Thanks UserIdTAG: 136959 UserNameTAG: christerpher CreateTimeTAG: 2012-09-17T14:47:11Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 2669 TitleTAG: Lab 2 HELP needed i just cant figure out what kind of orientation/circuit should i do with my 3 resistors to solve this lab problem, can someone help me?, just an idea how can connect the resistors even without telling me the value.., please.., UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-17T14:42:07Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Look at what Vout is... a fraction of V1 and V2 added together. To get some fractional value of each of these input voltages, you need a voltage divider, and you have to add both voltages together. I suggest you focus on getting one of the input voltages to the correct value, then figure out how to get the other one fractionated and added to get the correct waveform. Experiment in the Circuit Sandbox (Lab 0) instead of the actual lab itself, it's easier. FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-09-18T02:30:25Z SecondChildTAG: I was only able to solve this in the circuit sandbox. First I made sure I could get the correct outputs for each input on its own. Then I played around with resistors with both sources and made a matrix showing how each effected the output amplitude. From there it was just a matter of fine tuning. Still trying to solve it on paper but at least I have a correct answer! SecondChildUserIdTAG: 244115 SecondChildUserNameTAG: Pedro1969 SecondChildCreateTimeTAG: 2012-09-19T13:16:53Z IndexTAG: 2670 TitleTAG: Bug - Mistake at 5.59? Hi, Current I have the right direction form e. I think it should be left direction - to the e. Node equation is ok for the left direction. Sorry for my English. UserIdTAG: 413784 UserNameTAG: Dorotka CreateTimeTAG: 2012-09-17T10:46:24Z VoteTAG: 1 CoursewareTAG: Week 2 / Linearity CommentableIdTAG: 6002x_linearity NumberOfReplyTAG: 1 FirstChildTAG: Hi Dorotka, I don't fully understand your question. Are you confused about the arrow direction at 4:40? I agree that current is actually going to the left but notice how he mentioned that current going to the right is -I. By using a negative sign he is implying that current is actually going in the opposite direction to the one shown by the arrow. You may also consider the arrow pointing towards "e". The equation is not affected because then you have to subtract the current going into the node (namely +I). I hope this explanation helps! FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-17T13:13:53Z SecondChildTAG: Hi, tkanks for your answer. I get it. The arrow have right direction, but current is -I. I drew arrow in my notice with left direction, but in equation I wrote -I becouse current going to e ( out +; in -). It is the same. Thanks for your attention. SecondChildUserIdTAG: 413784 SecondChildUserNameTAG: Dorotka SecondChildCreateTimeTAG: 2012-09-17T14:18:38Z IndexTAG: 2671 TitleTAG: Homework 2 (H2P1) Hi Guys, I'm having lots of fun in this course, and thankful for it.I'm kind of stuck in H2P1. Even when I recognize a familiar circuit pattenrn here and came up with some values for R1 and R2 (within restrictions impossed), can't get a green mark when I press the check button. Any hint?, please. UserIdTAG: 369339 UserNameTAG: vargaslen CreateTimeTAG: 2012-09-17T07:57:23Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: same as you mate, Breezed through h2p2 and h2p3 but cant figure out p1. FirstChildUserIdTAG: 436749 FirstChildUserNameTAG: waynebrown FirstChildCreateTimeTAG: 2012-09-17T09:57:39Z SecondChildTAG: One possible method: 1. Create an expression for Vout 2. Express R2 as a function of R1 using desired value for Vout 3. Create an expression for Rth as a function of R1 using 2. 4. Select closest R1 from E12 series based on Rth expression in 3. 5. Use 2. to select closest value in E12 series based on R1 value selected in 4. 6. Check $\frac{V_{out}}{V_{in}}$ with nominal values of R1 and R2 selected. If everything checked out then you have it. Otherwise try again. SecondChildUserIdTAG: 392100 SecondChildUserNameTAG: glenng SecondChildCreateTimeTAG: 2012-09-17T21:49:20Z FirstChildTAG: I figured that no matter what you take the best result for Vout/Vin is 15% in a worst case (r1 is +10% and r2 is -10% for example), even if you will take the perfect match! It's something terrible. I tryed every combination in my model and still I do not get how to achieve the requirements. The Vout/Vin simply will not be 10% no matter what you do!!! Any clues? FirstChildUserIdTAG: 446597 FirstChildUserNameTAG: garry_crannon FirstChildCreateTimeTAG: 2012-09-17T10:46:55Z SecondChildTAG: You only need to use nominal values of resistances R1 and R2 to meet the output voltage constraint of 10% SecondChildUserIdTAG: 381746 SecondChildUserNameTAG: jelizon SecondChildCreateTimeTAG: 2012-09-17T13:04:31Z FirstChildTAG: I do not know why but I got all four green marks. But my Vout ecxeeds the 10% criteria! I just tryed to see what the combinations within given range of components will give a lowest difference between an ideal calculated nominals and a real parts, taking the larger component as a basis. It needs a bit of calculation so I just use electronic tables. Good luck guys, I think it's the proper way. FirstChildUserIdTAG: 446597 FirstChildUserNameTAG: garry_crannon FirstChildCreateTimeTAG: 2012-09-17T11:49:54Z FirstChildTAG: There are two constraints in the problem: 1) Output voltage 2) Thevenin resistance. Are you sure you are considering both for selecting R1 and R2? Also, notice that to meet the constraints you need to use the nominal value of the resistances only. FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-17T13:03:32Z IndexTAG: 2672 TitleTAG: lab 2 help can anyone help me for LAB 2???plzzz UserIdTAG: 197539 UserNameTAG: irfansultan94 CreateTimeTAG: 2012-09-17T07:09:19Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: please some 1 help.. FirstChildUserIdTAG: 171170 FirstChildUserNameTAG: rinutituschakkattil FirstChildCreateTimeTAG: 2012-09-17T09:46:36Z SecondChildTAG: I have got resistor values with T network. The problem on paper gives proper results, but not in simulation. Kindly help... SecondChildUserIdTAG: 100418 SecondChildUserNameTAG: balashyamala SecondChildCreateTimeTAG: 2012-09-17T11:47:41Z FirstChildTAG: I posted a list of hints for Lab 2 some days ago. The post name is "Lab2 hints" FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-17T17:19:46Z IndexTAG: 2673 TitleTAG: What is the load line? What does load line represent? UserIdTAG: 229896 UserNameTAG: Svilen CreateTimeTAG: 2012-09-17T05:40:32Z VoteTAG: 1 CoursewareTAG: Week 3 / Graphical Analysis of Nonlinear Circuits Exercise CommentableIdTAG: 6002x_graphical_analysis_of_nonlinear_circuits_exercise NumberOfReplyTAG: 2 FirstChildTAG: http://en.wikipedia.org/wiki/Load_line_%28electronics%29 FirstChildUserIdTAG: 40209 FirstChildUserNameTAG: sadaf78 FirstChildCreateTimeTAG: 2012-09-17T08:12:58Z FirstChildTAG: The load line representing the Thevenin equivalent circuit at the terminals of the nonlinear element. In this case, the equation for the load line is $i = \dfrac{1}{R_{th}} v + I$ FirstChildUserIdTAG: 129288 FirstChildUserNameTAG: Tinchito FirstChildCreateTimeTAG: 2012-09-18T13:30:06Z IndexTAG: 2674 TitleTAG: Using the schematic layout or circuit tool - rotating resistor to horizontal I completed lab 1 successfully. However, my diagrams looked horrible. I realize that the software just checks to see if I have the write values for the resistors by looking at the voltage at reference A, but still, I had trouble reading my diagrams. One of the biggest problems is that when I pull a resistor onto the grid, it stays vertical; I can't rotate it. Any way to rotate? I am on a MacBook Air. If there is some notes on this, please point me to them. Thanks. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-17T04:44:27Z VoteTAG: 1 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: "R" for rotate. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-17T04:49:51Z SecondChildTAG: could you post the lab 1 ans??? SecondChildUserIdTAG: 400239 SecondChildUserNameTAG: LincyG SecondChildCreateTimeTAG: 2012-09-19T17:26:05Z IndexTAG: 2675 TitleTAG: Anyone else having trouble with the forum date sorting? When I click sort by "votes" and then back to "date" the forum doesn't sort by dates properly, and now I can't get it working properly again. Anyone else having this trouble? EDIT: The devs told me this: the workaround for now is to hit "load more" at the bottom of the list and then refresh Which did work for me. UserIdTAG: 21541 UserNameTAG: JSChambers CreateTimeTAG: 2012-09-17T03:55:59Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: I can't tell what is happening because I don't see dates. I really, really, ***REALLY*** dislike this new format. Why are the edX folks trying to reinvent the wheel here? What was wrong with Askbot? It worked fine in the pilot, all it needed was top-level topics so everything wasn't dumped in one random mess. Neither the first edX forum nor this mess can hold a candle to the Askbot forum for clarity and ease of use. This thing doesn't have tags, the sidebar is useless, and there's no easy way to figure out what is going on. Boo hiss. And of course, there's no karma. Despite some of the overzealous students who abused the karma system, overall it worked pretty well for self-policing and organizing the discussion. FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-09-17T04:17:36Z SecondChildTAG: I'm not speaking for other EdX devs here, but for one thing, I don't think Askbot scales particularly well. And most of the features you're requesting are on the way/partially implemented. Why do you think the sidebar is useless? We felt that the categorization with specific courseware content made tags less useful, but we are open to suggestions of course. Tags are actually in the system but disabled because we didn't have a good UI for them. Some sort of reputation system is forthcoming as well. SecondChildUserIdTAG: 323387 SecondChildUserNameTAG: ibrahimawwal SecondChildCreateTimeTAG: 2012-09-17T22:19:22Z FirstChildTAG: g_hopper, I'm with ya, but from what I've heard they can't go back to Askbot, so we are going to have to work on making this one better. I know it's harder for us because we had the other forum, and they kinda knocked it out of the park the first time. My understanding is that the forum was oriented for discussions around videos, or something like that. How does that saying go? All your plans go out the window when the shooting starts. Plus now you have three universities, and that means three sets of ideas. The sidebar can be made useful, by adding dates and other stuff. I think it can be saved. I know you just want to say "get the old system" which is what I said. But I would just make detail lists of the problems and post them. It's what the developers have said they want. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-09-17T05:11:05Z FirstChildTAG: Ibrahimawwal - I find the sidebar useless because all I can see are the first two or three words of each subject title, which is not enough to understand what the post is about. I have to click on the post to see if it's of interest or germane to me. Under Askbot, the subject title, tags, and the first few lines were visible, which made it very easy to determine if the post was something I would be interested in. Also, the sidebar columns can't be expanded/contracted, nor can the sidebar itself. Lastly, all these sliders (one on the post, the other on the sidebar) interact, so while I am attempting to scroll through the post, the subject lines visible in the sidebar change. FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-09-18T03:03:25Z FirstChildTAG: I was, but it appears to be OK now. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-17T04:02:21Z IndexTAG: 2676 TitleTAG: It took me a while to see where I messed up... ...on the polarity signs next to the voltage sources, they switch positions from v1 to v3 - but I had thought that the voltage source's poles switched as well! (DUH! lol)... so, am I to assume that the switching pole marks at the farthest left of each schematic is from the point of view of a multimeter? Seems to make worlds more sense that way... as I mentioned before, elsewhere, I'm about as sharp as a bag of wet mice (my brain is *WAY* too slow lol) ahahahaha UserIdTAG: 340856 UserNameTAG: RanmaSaotome CreateTimeTAG: 2012-09-17T02:12:56Z VoteTAG: 1 CoursewareTAG: Week 1 / Associated reference directions CommentableIdTAG: 6002x_S2E2_Associated_Reference_Directions NumberOfReplyTAG: 1 FirstChildTAG: You can do it RanmaSaotome ;)! We always learn a new interesting thing each day! Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-17T02:57:36Z IndexTAG: 2677 TitleTAG: lab 1, still not working I still can't figure it out, what am I doing wrong here? Here's my solution: http://i.imgur.com/VDh0T.jpg I have calculated this several times, can't find an error. Even inside the toolbox the numbers for the first one are perfectly fine(?), the bulb gets its 1.5V but it is still showing up as "wrong" and I can't save the project. The second one gives silly numbers inside the toolbox... yet when I draw that circuit normally and calculate it with a simple voltage divider it should work perfectly fine: $Vs=6V*(1/(1+2.25))=24/13$ Is that incorrect somehow? Seriously, I think my calculations are correct, so maybe I'm making mistakes when trying to handle the program? UserIdTAG: 228979 UserNameTAG: boysen CreateTimeTAG: 2012-09-17T00:55:45Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Your voltage divider is not dividing the voltage in the right place. You need to rearrange the wires. See p73 in the book. Your calculation also needs a double-check. I did this by a node analysis writing KCL at A. FirstChildUserIdTAG: 419029 FirstChildUserNameTAG: resonate FirstChildCreateTimeTAG: 2012-09-17T01:15:17Z SecondChildTAG: Sorry, my mistake, your voltage divider works with that arrangement it's just the resistances that are wrong. The only other thing is that putting the voltage divider with A at the centre makes it easier to see what's going on, but it's not necessary. Did you do the node analysis (in your case at the lower-right node)? Write KCL for your circuit at the lower-right node. SecondChildUserIdTAG: 419029 SecondChildUserNameTAG: resonate SecondChildCreateTimeTAG: 2012-09-17T01:24:17Z SecondChildTAG: PS: in mine the voltage without the bulb had to be *equal to the limit*, even though the question asked for *less than the limit*. This may in fact be the problem. SecondChildUserIdTAG: 419029 SecondChildUserNameTAG: resonate SecondChildCreateTimeTAG: 2012-09-17T01:28:13Z SecondChildTAG: After all that, when I go back I see that indeed your voltage arrangement, although correct in practical terms, doesn't do exactly what the question asks for. *The system marks the voltage measured at node A.* SecondChildUserIdTAG: 419029 SecondChildUserNameTAG: resonate SecondChildCreateTimeTAG: 2012-09-17T01:31:51Z FirstChildTAG: Hi boysen! I saw your image. Have you Checked the current value? I don't know how many ampers do you have in your statement... You have a 6-volt battery (assumed ideal) and **a 1.5-volt flashlight bulb, which is known to draw 0.5A when the bulb voltage is 1.5V** (see figure below). Design a network of resistors to go between the battery and the bulb to give vs=1.5V when the bulb is connected, yet ensures that vs does not rise above 2V when the bulb is disconnected. In your schematic the current it is 2 Ampers , I have simulated in the sandbox... Can I help you? Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-17T01:29:39Z SecondChildTAG: Can you help me? I put one resistor r1 with 4.5 ohm before de bulb and another resitor r2 with 4.5 ohm after the bulb... and they are in series. what's wrong? SecondChildUserIdTAG: 116489 SecondChildUserNameTAG: matheusribeiro SecondChildCreateTimeTAG: 2012-09-17T01:58:04Z SecondChildTAG: Hi matheusribeiro! Yes, sure. I can. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-17T02:06:39Z SecondChildTAG: Here's the photo: [Lab1_First_Circuit][1] [1]: https://edxuploads.s3.amazonaws.com/1347847801257715.png SecondChildUserIdTAG: 116489 SecondChildUserNameTAG: matheusribeiro SecondChildCreateTimeTAG: 2012-09-17T02:10:41Z SecondChildTAG: The wrong thing it is that when you disconnect the bulb, if the three resistors are in series, you will have something like this : ![enter image description here][1] Will will have an open circuit an in the output you will have a voltage of V (the same as the voltage source)... and you need a low voltage (in my statement 2 V)... So, That is why it is not correct... don't verify the second condition... [1]: https://edxuploads.s3.amazonaws.com/13478478823113196.png SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-17T02:13:48Z SecondChildTAG: Hint: [Take a Look to 2.3.4 of the Textbook][1] ;). [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/97 SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-17T02:21:11Z SecondChildTAG: I was thinking that the first circuit was for JUST when the bulb is connected... because of this , here's the photo:![LAB1][1] [1]: https://edxuploads.s3.amazonaws.com/1347848572257728.png SecondChildUserIdTAG: 116489 SecondChildUserNameTAG: matheusribeiro SecondChildCreateTimeTAG: 2012-09-17T02:23:37Z SecondChildTAG: The Circuit has to be the same. The only difference is that in the Up circuit you have connected the bulb and in the other one not... I see in your images that you are changing the Circuit... In the second circuit (without the bulb), you are adding a new resistence of 4.5 Ohms... and you shouldn't...because the statement ask you not to change anything of the circuit when the bulb it is not connected... SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-17T02:29:02Z SecondChildTAG: Now it's cleared to me! Thank you a lot Myriam. SecondChildUserIdTAG: 116489 SecondChildUserNameTAG: matheusribeiro SecondChildCreateTimeTAG: 2012-09-17T02:30:55Z SecondChildTAG: Hint 2: Try to think about a Circuit similar of your second circuit ;), that can verifies the two conditions: with the bulb connected (...Voltage and ... Current) and when is disconnected (... Voltage). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-17T02:35:16Z SecondChildTAG: You are welcome matheusribeiro ;)! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-17T02:39:43Z SecondChildTAG: I did it! Thank a lot. SecondChildUserIdTAG: 116489 SecondChildUserNameTAG: matheusribeiro SecondChildCreateTimeTAG: 2012-09-17T02:48:15Z SecondChildTAG: So, happy for you ;). Congratulations matheusribeiro! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-17T02:58:41Z IndexTAG: 2678 TitleTAG: Math Processing Error I have been getting a "[Math Processing Error]" for every variable in the homework and the lab. I updated Java and still getting the same problem. Can anyone tell me how to fix this since everything is due tonight. Thanks! UserIdTAG: 152890 UserNameTAG: jeanmarc CreateTimeTAG: 2012-09-17T00:55:16Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Here is the problem... Homework 1 - http://i.imgur.com/oKhbC.gif Homework 2 - http://i.imgur.com/wmLRC.gif Lab 1 - http://i.imgur.com/3Kar9.gif Lab 2 - http://i.imgur.com/r7H5Z.gif Halp! FirstChildUserIdTAG: 152890 FirstChildUserNameTAG: jeanmarc FirstChildCreateTimeTAG: 2012-09-17T01:02:33Z SecondChildTAG: Please try Chrome or Firefox! SecondChildUserIdTAG: 79361 SecondChildUserNameTAG: kimt SecondChildCreateTimeTAG: 2012-09-17T01:30:50Z FirstChildTAG: I've posted the fix for this in a few places, here is one of them: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/undefined/threads/504a0c61afe5cd2300000010 That error is due to how your browser interacts with MathJax. MathJax is used by the server to present mathematical symbols on your screen. Here's how to fix it: 1) Place your cursor on the *[Math Processing Error]* message and right-click. 2) You will see a pull-down MathJax menu. 3) Select Math Settings, Math Renderer, and choose either MathML or SVG. I prefer the way MathML looks, but it drops some symbols when I print, so I have to use SVG. 4) You can also play with the zoom settings on the MathJax menu to adjust the zoom functionality. FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-09-17T01:31:40Z FirstChildTAG: I just refresh my browser and that does the trick. :) FirstChildUserIdTAG: 175205 FirstChildUserNameTAG: KT_MM FirstChildCreateTimeTAG: 2012-09-17T01:48:51Z IndexTAG: 2679 TitleTAG: Question on Node Voltage page 120 of textbook I am confused by page 120 of the textbook and their introductory remarks about Node voltage. Look at figure 3.1. The authors argue that e(a) or potential between a and c is 2 Volts. That's fine. It was given in figure 3.1 that it was 2 volts. But where are they coming up with e(b) or potential between b and c being 1.5 volts? 1.5 volts wasn't given in the original figure. It just appears in figure 3.2 and is introduced in the discussion as if it was there all along. It's not obviousl to me that the drop across resistor of 1 ohms is 1.5 volts. Should it be? It seems that one would have had to have already solved the circuit using the methods of chapter 2 to make the claim that the difference is 1.5 volts. I most be missing something obvious. Please explain. Thanks. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-17T00:05:40Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: There is nothing obvious you missed; they just omit the math and equations. FirstChildUserIdTAG: 390034 FirstChildUserNameTAG: kahlil FirstChildCreateTimeTAG: 2012-09-17T00:14:34Z SecondChildTAG: Thanks for your quick response. I'm not sure I understand your response. What good is the Node method if I must first solve the entire circuit simply to get the Node voltages? In this case, it seems I must do that to get the voltage of 1.5 volts. This relates by the way to one of the homework questions. However, I have now solved it by simply using KVL, KCL and I=IR equations. It would be nice to solve it using this "node method"; the homework instructions indicate that it should be easier. SecondChildUserIdTAG: 94009 SecondChildUserNameTAG: David1956 SecondChildCreateTimeTAG: 2012-09-17T00:54:55Z IndexTAG: 2680 TitleTAG: lab1 it checked my answer to be correct but wont let me send them. tried mutliple times now. very close to deadline UserIdTAG: 205581 UserNameTAG: kkn CreateTimeTAG: 2012-09-16T23:52:50Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: If it's checked then you're done! Take a look at the "Progress" page FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-16T23:58:12Z IndexTAG: 2681 TitleTAG: entering answers Trying to enter, in algebraic terms, answer to H1P1 with little luck. What is the format? UserIdTAG: 234797 UserNameTAG: voxac30 CreateTimeTAG: 2012-09-16T23:04:52Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: The system will not accept $\frac{xR}{y}$ so you must either type $(x*R)/y$ or $(x/y)*R$. Basically the $*$ multiplication sign is required, not optional. FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-09-16T23:08:30Z SecondChildTAG: it is not accepting my simple equation... R+R+R SecondChildUserIdTAG: 280332 SecondChildUserNameTAG: gamble89 SecondChildCreateTimeTAG: 2012-09-17T01:10:54Z SecondChildTAG: Hi gamble89! Try to simplify your expression to the minimun expression, that is to say, if you have a+a, the minumun expression will be 2*a (don't forget the asteric). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-17T02:01:36Z IndexTAG: 2682 TitleTAG: H1P3 Hi, I hope I am not repeating a question already asked (tried searching but could not find anything similar). I am finding it difficult to work out the individual heater resistances for this question, would anyone be able to kindly refer me to an appropriate section of the content that might give me some guidance, or drop any hints? Any assistance would be most welcome! Many thanks, Wil1e UserIdTAG: 103161 UserNameTAG: Wil1e CreateTimeTAG: 2012-09-16T22:30:02Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: There is a relationship between voltage, current, resistance and power. See, for example, page 27 of the textbook. Don't get confused with the AC nature of the problem. If you use RMS values for voltage, it is the same as if it was in DC (as stated in the problem itself). FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-16T22:40:06Z SecondChildTAG: Many thanks, much appreciated! SecondChildUserIdTAG: 103161 SecondChildUserNameTAG: Wil1e SecondChildCreateTimeTAG: 2012-09-16T22:41:48Z SecondChildTAG: AC with Resistance we solve it like DC and you can use P = IV = I^2 * R = V^2/R to get R and then use it to get Volt and Curr and use Ohm eqn to solve the 2nd part of Question SecondChildUserIdTAG: 284864 SecondChildUserNameTAG: H0ssam SecondChildCreateTimeTAG: 2012-09-16T23:42:26Z FirstChildTAG: AC with Resistance we solve it like DC and you can use P = IV = I^2 * R = V^2/R to get R and then use it to get Volt and Curr and use Ohm eqn to solve the 2nd part of Question FirstChildUserIdTAG: 284864 FirstChildUserNameTAG: H0ssam FirstChildCreateTimeTAG: 2012-09-16T23:42:40Z FirstChildTAG: Practically, we do not need the value of the individual heater resistances. What we need is the individual current flowing through each of them. In the "normal" case I've divided the expected current to be drawn by the 3 heaters connected in parallel and divided it by 3. Then, in the "sleepy Joe" case I use this value as a reference. Let's say, if the first heater is supposed to work under 240V with 5.8A, now that the voltage between its terminals (V1-V2) is not 240V (because it is in series with the supply V1=240V and V2=the voltage fall due to the heater resistance) I could make a proportional calculation between currents and voltages and I get a new current value for the first heater, this value multiplied by (V1-V2), gives me the power this heater consumes. FirstChildUserIdTAG: 59536 FirstChildUserNameTAG: kepler2007 FirstChildCreateTimeTAG: 2012-09-17T01:14:01Z FirstChildTAG: Ok. in this Quastion they give you the Voltage source, the Power for each heater and the total power of all three heaters. Q1. the current flow in this circuit when all heater are ON? Ans. you can find the current from P=V*I that very simple Q2. If the system will supply by 120 V find the current ? Ans. as we say befor from P=V*I Q3. What power was being dissipated in H1? Ans. First you should be find the resistance of H1 which is H2 and H3 are the same by P=V(squar)/R. after that find the total current of this circuit I. then you can find the v1 on the H1 v1=I * R1 then the power dissipated in H1 = v1 * I Q4. What power was being dissipated in H2 (or in H3)? Ans. Same Q3 but you should be know you have Node so the current will be seperated we will divided the current by 2 because H2 and H3 have same resistance then I2=I/2. after that v2 = I2 * R2. then the Power P2= v2 *I2 Q5. So the total heating power in Joe's shop was? Ans. you can find this by your self it's very simple # I Hope that be useful to you and I'm sory if I have mistake in my write FirstChildUserIdTAG: 318037 FirstChildUserNameTAG: Maher-84 FirstChildCreateTimeTAG: 2012-09-17T04:07:57Z IndexTAG: 2683 TitleTAG: what is current source? what is current source and how it is dual of voltage source? any real example ? UserIdTAG: 40209 UserNameTAG: sadaf78 CreateTimeTAG: 2012-09-16T22:22:07Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: why we use current source or what is its benefits over voltage source. FirstChildUserIdTAG: 40209 FirstChildUserNameTAG: sadaf78 FirstChildCreateTimeTAG: 2012-09-16T22:36:48Z FirstChildTAG: A current source provides a constant current in a particular branch. It is usually consider "dual" of a voltage source because it has similar properties with currents and voltages interchanged. For example, if you short the two terminals of a voltage source then you will have infinite current (in reality you will burn you source), and if you open the two terminals of a current source, then you will have infinite voltage across it. Several similar examples exist. A current source is usually constructed by adding a very large inductor in series with a battery or any voltage source. You will probably understand why later in the course. Also, some devices actually behave more or less as a current source, for example, a solar panel. FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-16T22:29:41Z SecondChildTAG: understand but not 100 percent SecondChildUserIdTAG: 40209 SecondChildUserNameTAG: sadaf78 SecondChildCreateTimeTAG: 2012-09-16T22:33:56Z FirstChildTAG: Also, the reason why we would use a current source over a voltage source is if we want to model a real-world device **that behaves like** an ideal current source. In later EECS lessons we study a model of the *bipolar junction transistor* (BJT). This is the standard transistor that is used for amplification purposes; often in audio devices. When we model this device to look at how it behaves, the model contains a current source, not a voltage source. That is because of the physics of the device. But current is what drives audio waves, and most speakers operate at a constant voltage, so the current must vary to produce the sound you hear. If you supply too much current though, that which exceeds the speaker's rating, you will end up ruining it. That is why almost all speakers are labeled with a voltage and a maximum wattage: $P(max)=v \cdot i(max)$ FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-09-16T22:51:55Z SecondChildTAG: very nice explained THANK YOU VERY VERY MUCH. SecondChildUserIdTAG: 40209 SecondChildUserNameTAG: sadaf78 SecondChildCreateTimeTAG: 2012-09-17T00:32:16Z FirstChildTAG: Dear sir thank you very much for this nice explanation actually I am getting confused between voltage and current source what I understand is current source CANNOT produce itself we use voltage source to produce current source means we need potential difference to get current source, you gave example of battery (DC VOLTAGE) which is itself voltage source and produce current(DC CURRENT). I got that what ever alternating voltage is power supply control the current and give the constant current required for the particular circuit element. FirstChildUserIdTAG: 40209 FirstChildUserNameTAG: sadaf78 FirstChildCreateTimeTAG: 2012-09-16T23:12:04Z FirstChildTAG: EECS concerns itself with abstractions, so it talks about **ideal** current sources and ideal voltage sources for the most part. It is possible to model real-world current sources, such as batteries, but they are not ideal, and hence they do not obey the EECS abstractions if not modeled correctly. In EECS, an ideal current source puts out a **constant** current *without regard to external conditions*. Take the example of a battery that has been shorted out (e.g. the positive terminal is connected directly to the negative without a load, just a bare wire.) That battery is **not** an ideal current source because it will deliver maximum current for only a few seconds before a) it overheats, or causes the wire to overheat, and thus open the connection, and/or b) it dies because all of the chemical energy stored inside has been depleted. The closest real-world devices to a current source, if you ever have been in an electronics lab, are power supplies that can be adjusted to deliver a constant current irregardless of the voltage and/or resistance across the power supply's terminals. **But** even a power supply like this cannot exceed it's rated current. That is the constraint. You can buy a $300 (U.S. dollar) model with a rated output of 3 to 5 amps, and if you want to spend more money, you can get supplies rated even higher. Note that if you exceed the rated current, the more-expensive supplies have an emergency shut-off or regulator that warns you of this condition, before your experimental circuit lets it overheat and fail. FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-09-16T22:41:13Z IndexTAG: 2684 TitleTAG: capital V, small v...a tad confusing. "The properties of the element are described by its V-I relation, and this is given by capital V." "So the small v in this equation is related to the V-I for the element, while capital V is the voltage supplied by the element" UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-16T21:57:31Z VoteTAG: 1 CoursewareTAG: Week 1 / Lumped Element Abstraction CommentableIdTAG: 6002x_lumped_element_abstraction NumberOfReplyTAG: 1 FirstChildTAG: A big "V" would be a parameter of the part in question. Ex: 1.5V Battery. A small "v" would be a variable, at say a node. Ex: The voltage after this particular resistor is 1v. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-16T23:57:33Z IndexTAG: 2685 TitleTAG: S2E5: Node Method ?????? We choose the top terminal of the current source as the ground , where it connects to the two resistors. So now e2=0 so in part five how v2 is 8.75????? UserIdTAG: 40209 UserNameTAG: sadaf78 CreateTimeTAG: 2012-09-16T21:07:02Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: Actually when I tried schematic voltage across R2 is 0 FirstChildUserIdTAG: 40209 FirstChildUserNameTAG: sadaf78 FirstChildCreateTimeTAG: 2012-09-16T22:12:49Z FirstChildTAG: Because $e3 = -8.75V$, so $v2 = e2 - e3 = 0V - -8.75V = 8.75V$. FirstChildUserIdTAG: 397804 FirstChildUserNameTAG: mdempsky FirstChildCreateTimeTAG: 2012-09-16T21:41:43Z FirstChildTAG: Notice how the polarity was defined for v2. The positive terminal is e2 and the negative terminal is e3, so as mdempsky mentioned above $v_2 = e_2 - e_3$ FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-16T22:08:42Z FirstChildTAG: **thanx jelizon and mdempsky NOW I understand v2 is the potential difference of e2and e3** FirstChildUserIdTAG: 40209 FirstChildUserNameTAG: sadaf78 FirstChildCreateTimeTAG: 2012-09-16T22:18:27Z IndexTAG: 2686 TitleTAG: Contact Teachers The time slots are given when the teachers stay online in the Course Info. But how & where can we communicate with them in those given time slots? UserIdTAG: 317445 UserNameTAG: arghya33 CreateTimeTAG: 2012-09-16T20:10:18Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: his any one there? FirstChildUserIdTAG: 442102 FirstChildUserNameTAG: fridexcool FirstChildCreateTimeTAG: 2012-09-16T20:36:43Z SecondChildTAG: in my progress part, what does HWavg indicate? SecondChildUserIdTAG: 200391 SecondChildUserNameTAG: subinabraham SecondChildCreateTimeTAG: 2012-09-16T20:38:04Z FirstChildTAG: what happens if the tasks are not delivered on time?. I start the course for 3 days and I will not have enough time to finish FirstChildUserIdTAG: 435253 FirstChildUserNameTAG: renevalderrama FirstChildCreateTimeTAG: 2012-09-16T20:49:33Z SecondChildTAG: You can drop two problem sets and two labs without penalty. Just make sure that you know the material and keep chugging along with the course. SecondChildUserIdTAG: 96452 SecondChildUserNameTAG: Lyla SecondChildCreateTimeTAG: 2012-09-16T20:53:24Z FirstChildTAG: They are simply going to be particularly active on the forums then. You will see more posts with the handy dandy little staff flag. :P FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-09-16T20:52:52Z FirstChildTAG: We're all here but there are a lot of threads to go through! FirstChildUserIdTAG: 151427 FirstChildUserNameTAG: RichardZ FirstChildCreateTimeTAG: 2012-09-16T23:51:58Z FirstChildTAG: I think right here. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-16T20:17:14Z SecondChildTAG: I didn't find them in the discussion forum. SecondChildUserIdTAG: 317445 SecondChildUserNameTAG: arghya33 SecondChildCreateTimeTAG: 2012-09-16T20:29:03Z SecondChildTAG: Maybe it's a different timezone then where you are. I would imagine it is -5 or -4 DST. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-16T20:36:19Z SecondChildTAG: We're online at the posted times in Boston time. If you make a post during these times there's a very good chance that we'll read it and reply to it. SecondChildUserIdTAG: 151427 SecondChildUserNameTAG: RichardZ SecondChildCreateTimeTAG: 2012-09-16T23:53:33Z IndexTAG: 2687 TitleTAG: violation of rule ? some 1 from the staff please informed me with my progress ? u have suppose to be an alert email for those students who violate the rules ..!!!!!! UserIdTAG: 415376 UserNameTAG: imab90 CreateTimeTAG: 2012-09-16T18:56:42Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 0 IndexTAG: 2688 TitleTAG: how I may solve this problem? power in watts ( W ) can the smallest-valued composite resistor dissipate before burnning up? UserIdTAG: 227508 UserNameTAG: bhavyab CreateTimeTAG: 2012-09-16T17:07:51Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: * There is no diagram/schematic for this problem, but you should create your own. * You know that $I^2 \cdot R$ $= 1$ for *each* resistor, but **not** the *composite* resistor (i.e. the equivalent resistance you get by combining the three, which was the subject of the previous question in the homework.) * Work out the equation I gave you above to get $I_1$, $I_2$, and $I_3$ for *each* resistor, where $R_1 = R_2 = 4$, and $R_3 = 6$. You will have three equations, two of which will be the same, since the resistances are the same. * Hint: The equation for the first 4-ohm resistor should look like: $4 \cdot I^2_1 =1$. Repeat for the other two. * Next, apply KCL where $I(max) = I_1 + I_2 + I_3$ * Finally, apply the power equation, $P(max) = I(max)^2 \cdot R(equivalent)$, where $R(equivalent)$ is the answer you got from the previous problem. FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-09-16T17:36:17Z FirstChildTAG: Basically you need to figure out what kind of voltage is being delivered to the three resistors. If you know what happens to voltage when it is shared across parallel devices, you will know what kind of voltage each resistor sees. Then you can figure out how much power each resistor is dissipating. Once you figure out which resistor can handle the least amount of power, that will dictate what the other two resistors will be dissipating at the same time. Have fun. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-16T17:39:42Z IndexTAG: 2689 TitleTAG: somebody plzz. help in getting the last answer. plzz how to get the voltage across r4 UserIdTAG: 163748 UserNameTAG: abhhisheak1 CreateTimeTAG: 2012-09-16T15:17:51Z VoteTAG: 1 CoursewareTAG: Week 1 / Series and parallel example CommentableIdTAG: 6002x_S2E4_Series_and_Parallel NumberOfReplyTAG: 2 FirstChildTAG: ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13478098901343622.png FirstChildUserIdTAG: 374590 FirstChildUserNameTAG: ahope FirstChildCreateTimeTAG: 2012-09-16T15:38:21Z FirstChildTAG: how to find the voltage across R4 can any one explain me?? FirstChildUserIdTAG: 644283 FirstChildUserNameTAG: ragav78 FirstChildCreateTimeTAG: 2012-10-15T17:15:06Z IndexTAG: 2690 TitleTAG: technical problems!! hey the ansers which are marked as correct to my friend are being shown incorrect to me.....can anybody help plz UserIdTAG: 161899 UserNameTAG: harshvit12 CreateTimeTAG: 2012-09-16T14:26:19Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Many of the problems are randomized per student, so the same answers won't necessarily be correct. FirstChildUserIdTAG: 11 FirstChildUserNameTAG: dormsbee FirstChildCreateTimeTAG: 2012-09-16T15:01:47Z FirstChildTAG: Are you saying you are getting the answers from your friend? FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-09-17T03:28:49Z SecondChildTAG: hey !!!!it is not like that SecondChildUserIdTAG: 161899 SecondChildUserNameTAG: harshvit12 SecondChildCreateTimeTAG: 2012-09-17T13:27:32Z IndexTAG: 2691 TitleTAG: Lab session How to connect wires between two components? I am unable to rotate the component any solution? UserIdTAG: 342717 UserNameTAG: kale_anurag CreateTimeTAG: 2012-09-16T14:22:12Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: press R FirstChildUserIdTAG: 204571 FirstChildUserNameTAG: JFKlepac FirstChildCreateTimeTAG: 2012-09-16T14:22:53Z SecondChildTAG: To rotate a component Click in a component, when it becomes green then type the letter "r" on the keyboard. SecondChildUserIdTAG: 183712 SecondChildUserNameTAG: colegiocientifico SecondChildCreateTimeTAG: 2012-09-16T15:40:50Z FirstChildTAG: See this [wiki tutorial][1] on connecting wires [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/MyrimitCircuitSimulatorTutorial/ FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-16T14:30:51Z SecondChildTAG: I saw it nd tried the same.... i used the google chrome as well as IE8 but both show the same SecondChildUserIdTAG: 342717 SecondChildUserNameTAG: kale_anurag SecondChildCreateTimeTAG: 2012-09-16T14:40:51Z FirstChildTAG: i have seen wiki tutorial nd implemented the same in chrome nd IE8 yet facing the same problem when i click on the end of the wire & drag it a box appears not wire please help Homework is pending FirstChildUserIdTAG: 342717 FirstChildUserNameTAG: kale_anurag FirstChildCreateTimeTAG: 2012-09-16T14:47:32Z FirstChildTAG: To rotate a component Click in a component, when it becomes green then type the letter "r" on the keyboard. FirstChildUserIdTAG: 183712 FirstChildUserNameTAG: colegiocientifico FirstChildCreateTimeTAG: 2012-09-16T15:39:58Z IndexTAG: 2692 TitleTAG: (V0-e1)/R1 I wrote the answer like (V0-e1)/R1 ,as my understanding. What is the wrong in it? UserIdTAG: 147577 UserNameTAG: rithi3 CreateTimeTAG: 2012-09-16T13:49:50Z VoteTAG: 1 CoursewareTAG: Week 1 / Node analysis practice, part 2 CommentableIdTAG: 6002x_L2Node1 NumberOfReplyTAG: 1 FirstChildTAG: I will explain. The current is going up (as it is stated in the question), so according the associated variables convention the positive terminal is located at the bottom of R1 and the negative terminal - at the top (current flows from + to -). In order to compute the value of the voltage across the resistor you need to subtract the potential of the negative terminal from the potential of positive. So, V1=e1-V0. The current is V1/R1. FirstChildUserIdTAG: 370573 FirstChildUserNameTAG: pberkovich FirstChildCreateTimeTAG: 2012-09-16T16:39:13Z SecondChildTAG: THANKS SecondChildUserIdTAG: 116107 SecondChildUserNameTAG: samisna SecondChildCreateTimeTAG: 2012-09-20T12:14:31Z IndexTAG: 2693 TitleTAG: submission of assignment Friends,plz tell me how to submit home work assignment?? UserIdTAG: 214319 UserNameTAG: ramangupta CreateTimeTAG: 2012-09-16T12:56:29Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: same question heRe.. FirstChildUserIdTAG: 294630 FirstChildUserNameTAG: babarumt FirstChildCreateTimeTAG: 2012-09-16T13:37:11Z SecondChildTAG: Hi babarumt! Please read what I have answered ramangupta ;). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-16T15:13:19Z FirstChildTAG: you must put in the values of the exercise and than click on the button "check", when your answer is ok,then has the exercise completed. FirstChildUserIdTAG: 274475 FirstChildUserNameTAG: Konredus FirstChildCreateTimeTAG: 2012-09-16T13:37:18Z FirstChildTAG: Hi Ramangupta! With the **Check buttom you submit** your Homework and Labs. Remember that this buttom does the following **three things** at the **same time**: 1-Submit. 2- Save. 3- Says if your answer it is correct or incorrect. ---- You **can corroborate** that was submited correctly in your [Progress Tab][1]. You will see if you have done ok the respective Homework and Lab, that it will show you 100%. If not, something it is wrong (not submited well or with incorrect answers). I hope this can help you. Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/progress FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-16T15:11:55Z IndexTAG: 2694 TitleTAG: feedback how will we get the answers to the wrong questions?? UserIdTAG: 443968 UserNameTAG: shobhitsajwan CreateTimeTAG: 2012-09-16T12:45:42Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: do the exercise again and then recheck it. FirstChildUserIdTAG: 274475 FirstChildUserNameTAG: Konredus FirstChildCreateTimeTAG: 2012-09-16T13:39:21Z FirstChildTAG: Hi shobhitsajwan! I don't understand your question. Do you mean that when we will have the results of the wrong questions? Based on my experience in the Prototype Course 6.002x , after the deadline of the Homework and Lab, the solved steps of (Hw, lab and exercises) were released in the Course Info ( Course Handouts) in a .pdf. Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-16T15:23:00Z IndexTAG: 2695 TitleTAG: Discussion Im having some problems searching the discussions, it doesnt seem to load any thing. UserIdTAG: 47733 UserNameTAG: Solmaz CreateTimeTAG: 2012-09-16T11:12:28Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 2696 TitleTAG: doubt my progress in practice scores of administivia and circuit elements is 0/3 0/2 0/2 0/3 .....and all of them show ZERO!how to fill it? UserIdTAG: 431563 UserNameTAG: vamsi45 CreateTimeTAG: 2012-09-16T10:37:20Z VoteTAG: 1 CoursewareTAG: Week 1 / Course overview CommentableIdTAG: 6002x_S1V1toS1V4 NumberOfReplyTAG: 1 FirstChildTAG: those are the practice questions which come in between the lecture videos ...go attend them FirstChildUserIdTAG: 102529 FirstChildUserNameTAG: laksh FirstChildCreateTimeTAG: 2012-09-16T10:55:01Z IndexTAG: 2697 TitleTAG: Resistors in parallel or series tutorials Is there really no simpler way to do resistors in parallel, like grouping both into (2 sets)of 2 different unit, as with Resistors in series. Just wondering if any one knows any way not so equation intensive. Or is that simply the point of the exercise . That it is and there is no way around it. Refer to combinations of resistors in parallel and series of course. UserIdTAG: 359302 UserNameTAG: GaryG CreateTimeTAG: 2012-09-16T10:02:52Z VoteTAG: 1 CoursewareTAG: Week 1 / Resistor Voltages CommentableIdTAG: 6002x_resistor_voltages NumberOfReplyTAG: 0 IndexTAG: 2698 TitleTAG: One way to pass this course Okay, now we all know that all of us are from different backgrounds some are adults, some are school students, etc., also all of us, love electronics, that's why we joined this course, but sadly, all of us don't have enough per-requisite knowledge to go through this course easily, I mean it was a near nightmare for me to solve the homework and lab 1 and I saw most people saying that they can't solve the 1st homework itself, which i suppose is the easiest, now I don't people to lose hope and continue in the course, I have decided that I will create another website where all the important per-requisite concepts will be shows, every practice question will be solved(by me or by other members), and a lot of open discussion can be done, I will start it after my MID-Year exams, they will end at 24th and I will start after that, till that I want everyone to hold on, I am ready to help them in any way possible, till the site comes up......now, the site would help you as you will learn easier methods and what not, what I want you to do is to UP or comment this post as many times you can so other people see it and I get to know that If I make a website as such....how many of you will join... P.S. till, then mail at battlewiredbear@gmail.com, for anything(homework help,etc., remember till then I will be able to help only with my power, but when the site starts, the whole community is welcome there....) Also to the STAFF.... i am not breaking the rules but bending them, by continuing open discussion of this website... AND REMEMBER, NO HOMEWORK OR LAB HELP WILL BE GIVEN AFTER THE WEBSITE IS UP AND RUNNING, I WANT TO KEEP THE SANCTITY OF MITX ALIVE BUT THE WEBSITE WILL BE A GREAT HELP UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-16T09:26:59Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: LETS DO IT! FirstChildUserIdTAG: 169784 FirstChildUserNameTAG: thewiredbear FirstChildCreateTimeTAG: 2012-09-16T09:27:27Z FirstChildTAG: hey i dont think there is any need to do so.we have already got mit ocw and discussion forum in edx which every one knows FirstChildUserIdTAG: 397298 FirstChildUserNameTAG: ZA90 FirstChildCreateTimeTAG: 2012-09-16T09:53:32Z FirstChildTAG: Yes. We can discuss everything under the 'discussions' section, like what we are doing now. FirstChildUserIdTAG: 199839 FirstChildUserNameTAG: vijaykrishnankk FirstChildCreateTimeTAG: 2012-09-16T10:48:40Z SecondChildTAG: right SecondChildUserIdTAG: 169784 SecondChildUserNameTAG: thewiredbear SecondChildCreateTimeTAG: 2012-09-16T10:57:27Z SecondChildTAG: and for your sarcasm, I mentioned a lot of other things SecondChildUserIdTAG: 169784 SecondChildUserNameTAG: thewiredbear SecondChildCreateTimeTAG: 2012-09-16T10:58:04Z FirstChildTAG: I want to say that your spirit of cooperation is TRULY COMMENDABLE! I SINCERELY believe that everyone is better served if your idea is develop within this venue. You can always use the wiki. FirstChildUserIdTAG: 81830 FirstChildUserNameTAG: FFoulenIT FirstChildCreateTimeTAG: 2012-09-16T13:45:54Z SecondChildTAG: Hi FFoulenIT! How are you? Nice to see that you are here! What happened to you? ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-16T15:45:40Z IndexTAG: 2699 TitleTAG: Circuit please I donot know the circuit i am trying on lab1 due today. UserIdTAG: 396372 UserNameTAG: vsnarendran CreateTimeTAG: 2012-09-16T09:11:16Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: finally after long hrs of messing, i gor 100% correct ans .happy to help u all.just try r1 in series with source and r2 in parallel with bulb. in both case same combination.now in 1st case source voltage 6,and suppy volt-1.5.and in 2nd case source volt-6 and supply volt-2. u have to satisfy both eqtion simultaneously for r1 and r2. just try r1 in integer values 2 to 5 and r2 values b/w 1.5 10 3. u will get exact ans. your welcome FirstChildUserIdTAG: 169784 FirstChildUserNameTAG: thewiredbear FirstChildCreateTimeTAG: 2012-09-16T09:48:48Z IndexTAG: 2700 TitleTAG: home work 2 how to solve H2P2?? if i give values as 8.2k ohms and 2.7k ohms ans is incorrect??? UserIdTAG: 334613 UserNameTAG: aswinshankar CreateTimeTAG: 2012-09-16T09:06:29Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: it is the solar power problem isnt it?? Try to find the Thevenin resistance seen from the load resistance (node). That should help you find the optimal resistance... FirstChildUserIdTAG: 257806 FirstChildUserNameTAG: skoda FirstChildCreateTimeTAG: 2012-09-16T09:17:48Z FirstChildTAG: sorry it is H2P1 voltage divider problem... FirstChildUserIdTAG: 334613 FirstChildUserNameTAG: aswinshankar FirstChildCreateTimeTAG: 2012-09-16T15:08:32Z FirstChildTAG: One possible method: 1. Create an expression for Vout 2. Express R2 as a function of R1 using desired value for Vout 3. Create an expression for Rth as a function of R1 using 2. 4. Select closest R1 from E12 series based on Rth expression in 3. 5. Use 2. to select closest value in E12 series based on R1 value selected in 4. 6. Check $\frac{Vout}{Vin}$ with nominal values of R1 and R2 selected. If everything checked out then you have it. Otherwise try again. FirstChildUserIdTAG: 392100 FirstChildUserNameTAG: glenng FirstChildCreateTimeTAG: 2012-09-18T00:39:03Z IndexTAG: 2701 TitleTAG: lab help please how to change the value of resistor in lab1 UserIdTAG: 396372 UserNameTAG: vsnarendran CreateTimeTAG: 2012-09-16T08:42:16Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: double click on the resistor, then change value.. it is found in the lab tutorial. :) FirstChildUserIdTAG: 209082 FirstChildUserNameTAG: Nikemurton FirstChildCreateTimeTAG: 2012-09-16T08:46:08Z SecondChildTAG: ok thanks SecondChildUserIdTAG: 396372 SecondChildUserNameTAG: vsnarendran SecondChildCreateTimeTAG: 2012-09-16T08:59:16Z SecondChildTAG: i didnt finish lab1 yet. SecondChildUserIdTAG: 396372 SecondChildUserNameTAG: vsnarendran SecondChildCreateTimeTAG: 2012-09-16T09:04:22Z FirstChildTAG: Double click on the resistor and there will appear a box where you can give a name and value to the resistor. You can use prefixes such k and M for kilo and Mega ohms. FirstChildUserIdTAG: 441835 FirstChildUserNameTAG: KAMBIZAMEEN FirstChildCreateTimeTAG: 2012-09-16T09:05:38Z IndexTAG: 2702 TitleTAG: i wanna make some club on fb. (check plz) I wanna make some club on fb. just like small talk and discussion. :) in ma case, im university student. if ur guys wanna join this club, can u write sth on this comment? As a matter of fact, i joined another club (6002x or mitx) but, their semester was end. and it looks like huge stone that had no talking.... so... UserIdTAG: 208923 UserNameTAG: Kylekim CreateTimeTAG: 2012-09-16T07:51:28Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: Yes i would like to join it. FirstChildUserIdTAG: 441835 FirstChildUserNameTAG: KAMBIZAMEEN FirstChildCreateTimeTAG: 2012-09-16T08:14:48Z SecondChildTAG: http://www.facebook.com/groups/mitxfallsemester plz join this group :) SecondChildUserIdTAG: 208923 SecondChildUserNameTAG: Kylekim SecondChildCreateTimeTAG: 2012-09-16T09:07:24Z FirstChildTAG: amazing, I wanted this too........im open, create it and send an email to battlewiredbear@gmail.com, I will join it, or friend request me at spaceparth@gmail.com and I am from school, you have place for me? FirstChildUserIdTAG: 169784 FirstChildUserNameTAG: thewiredbear FirstChildCreateTimeTAG: 2012-09-16T07:58:38Z SecondChildTAG: group name ?? SecondChildUserIdTAG: 430030 SecondChildUserNameTAG: imali SecondChildCreateTimeTAG: 2012-09-16T08:06:18Z SecondChildTAG: I can't think of something catchy.... SecondChildUserIdTAG: 169784 SecondChildUserNameTAG: thewiredbear SecondChildCreateTimeTAG: 2012-09-16T08:09:22Z SecondChildTAG: http://www.facebook.com/groups/mitxfallsemester plz join this group :) SecondChildUserIdTAG: 208923 SecondChildUserNameTAG: Kylekim SecondChildCreateTimeTAG: 2012-09-16T09:07:33Z FirstChildTAG: I would like to join too :) FirstChildUserIdTAG: 442358 FirstChildUserNameTAG: mauddy FirstChildCreateTimeTAG: 2012-09-16T08:28:11Z SecondChildTAG: http://www.facebook.com/groups/mitxfallsemester plz join this group :) SecondChildUserIdTAG: 208923 SecondChildUserNameTAG: Kylekim SecondChildCreateTimeTAG: 2012-09-16T09:07:39Z FirstChildTAG: me too, if it's another place for me:P. thx FirstChildUserIdTAG: 443442 FirstChildUserNameTAG: Nelutu FirstChildCreateTimeTAG: 2012-09-16T09:06:47Z SecondChildTAG: http://www.facebook.com/groups/mitxfallsemester plz join this group :) SecondChildUserIdTAG: 208923 SecondChildUserNameTAG: Kylekim SecondChildCreateTimeTAG: 2012-09-16T09:07:44Z FirstChildTAG: I also joined this group..... https://www.facebook.com/groups/mitxfallsemester/ FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-09-16T09:53:07Z IndexTAG: 2703 TitleTAG: Please provide practical circuits IN h1p2 (6.002X) Here the currents i1,i2 and i3 current flow is impossible . With respect to the direction of arrow heads in h1p2 UserIdTAG: 396372 UserNameTAG: vsnarendran CreateTimeTAG: 2012-09-16T07:24:43Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: all because of difference current and voltage source.... don't worry if you get stuck somewhere, I will give you all the answers FirstChildUserIdTAG: 169784 FirstChildUserNameTAG: thewiredbear FirstChildCreateTimeTAG: 2012-09-16T07:29:03Z SecondChildTAG: just tell me SecondChildUserIdTAG: 169784 SecondChildUserNameTAG: thewiredbear SecondChildCreateTimeTAG: 2012-09-16T07:29:17Z SecondChildTAG: ok thewiredbear. Only do help. Dont tell answers. SecondChildUserIdTAG: 396372 SecondChildUserNameTAG: vsnarendran SecondChildCreateTimeTAG: 2012-09-16T08:40:02Z IndexTAG: 2704 TitleTAG: fee payment for the last exam How much is the fees to be paid for writing the exam? UserIdTAG: 396372 UserNameTAG: vsnarendran CreateTimeTAG: 2012-09-16T07:00:08Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: It's all FREE OF COST dear:-) FirstChildUserIdTAG: 364473 FirstChildUserNameTAG: vibgyor FirstChildCreateTimeTAG: 2012-09-16T07:24:11Z FirstChildTAG: Hi vsnarendran! Go to Overview-> edX Tutorial -> CHECKING AND SUBMITTING AN ANSWER ---------- LIMITED NUMBER OF CHECKS, PART 1 Sometimes, you will have a limited number of times that you can check your answer. If you would like to input an answer and keep it there until you come back to the same question, you can use the "save" button. This will not use up any of your checks for the question. **How much does it cost to take an edX course? Enter the number of dollars.** ---------- You will find your answer here ;) ---------- ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13478117828474421.png FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-16T16:10:16Z IndexTAG: 2705 TitleTAG: How is this possible? how is 7v possible from a 4v source, in H1P2 question 1 http://troll.me?p=429659 ![how???][1] [1]: http://www.troll.me/images/jackie-chan-whut/4volts-from-source-7-volts-in-branchjust-messed-up-all-my-concepts.jpg UserIdTAG: 169784 UserNameTAG: thewiredbear CreateTimeTAG: 2012-09-16T06:56:26Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: what was the current through that branch for you??? think there is some problem with that. FirstChildUserIdTAG: 210146 FirstChildUserNameTAG: Divyakumari FirstChildCreateTimeTAG: 2012-09-16T07:16:32Z SecondChildTAG: 7v SecondChildUserIdTAG: 169784 SecondChildUserNameTAG: thewiredbear SecondChildCreateTimeTAG: 2012-09-16T07:27:28Z SecondChildTAG: m asking for the current.. not the volgtage SecondChildUserIdTAG: 210146 SecondChildUserNameTAG: Divyakumari SecondChildCreateTimeTAG: 2012-09-16T07:30:55Z SecondChildTAG: 8 SecondChildUserIdTAG: 169784 SecondChildUserNameTAG: thewiredbear SecondChildCreateTimeTAG: 2012-09-16T07:32:18Z SecondChildTAG: how come 8?? isnt that jst 1?? SecondChildUserIdTAG: 210146 SecondChildUserNameTAG: Divyakumari SecondChildCreateTimeTAG: 2012-09-16T07:39:12Z SecondChildTAG: no it v1 is right next to the current source SecondChildUserIdTAG: 169784 SecondChildUserNameTAG: thewiredbear SecondChildCreateTimeTAG: 2012-09-16T07:59:57Z SecondChildTAG: you are considering only voltage source that's connected on the left there is also a current source that's driving that node due to which you get a higher voltage. regards Neeraj SecondChildUserIdTAG: 210992 SecondChildUserNameTAG: neerajnatu SecondChildCreateTimeTAG: 2012-09-16T08:04:29Z SecondChildTAG: hmm.......these things won't get into the mind of a 11th standard boy, still current source explanation gives me some clearance...thanks a lot! SecondChildUserIdTAG: 169784 SecondChildUserNameTAG: thewiredbear SecondChildCreateTimeTAG: 2012-09-16T08:10:41Z SecondChildTAG: Take note that there is a current source contributing to the cct. At the right hand side where the voltage source is, an amount of current must pass there making the power across the 3ohms 1 watt where I=1amp, then ask yourself, where is my remaining 7amps.. that 7amps is what is dissipated through the 1ohm resistor R2. so I*R becomes 7*1=7V. Regards. Ugochukwu. SecondChildUserIdTAG: 329879 SecondChildUserNameTAG: Ugochukwu SecondChildCreateTimeTAG: 2012-09-16T09:01:51Z FirstChildTAG: Adding a 3v in parrallel with same polarity FirstChildUserIdTAG: 396372 FirstChildUserNameTAG: vsnarendran FirstChildCreateTimeTAG: 2012-09-16T08:40:53Z SecondChildTAG: lol SecondChildUserIdTAG: 169784 SecondChildUserNameTAG: thewiredbear SecondChildCreateTimeTAG: 2012-09-16T08:42:29Z IndexTAG: 2706 TitleTAG: H1P2 how do you find v2, i mean by node analysis, you don't have a resistor between e2(left of voltage source, on the lower side) and V source....or you can see the branch connected to voltage source without a resistor UserIdTAG: 169784 UserNameTAG: thewiredbear CreateTimeTAG: 2012-09-16T06:09:05Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: probably you still have not considered one of the nodes to be the reference node.And another point is that the ends of a component( by which i mean a resistor alone in this case) can be considered to be a node. FirstChildUserIdTAG: 210146 FirstChildUserNameTAG: Divyakumari FirstChildCreateTimeTAG: 2012-09-16T06:32:16Z SecondChildTAG: I considered the point below the voltage source as a ground node, but then there is no resistor in that branch, so how will you equate? SecondChildUserIdTAG: 169784 SecondChildUserNameTAG: thewiredbear SecondChildCreateTimeTAG: 2012-09-16T06:37:46Z SecondChildTAG: and i think considering the resistor's end as a node won't be helpful because then the parallel connection loses its identity ( i hope, don't know what am i talking) SecondChildUserIdTAG: 169784 SecondChildUserNameTAG: thewiredbear SecondChildCreateTimeTAG: 2012-09-16T06:41:20Z SecondChildTAG: i just have to ask you something..have you got the total number of nodes right? SecondChildUserIdTAG: 210146 SecondChildUserNameTAG: Divyakumari SecondChildCreateTimeTAG: 2012-09-16T06:48:56Z SecondChildTAG: and the ground node is actually common for the resistor 1ohm, the current source and the voltage source. SecondChildUserIdTAG: 210146 SecondChildUserNameTAG: Divyakumari SecondChildCreateTimeTAG: 2012-09-16T06:51:02Z SecondChildTAG: just sorted that.... I got that just before this post..... but still how can there be 7v in the branch with 4v from source!!!! http://troll.me?p=429659 SecondChildUserIdTAG: 169784 SecondChildUserNameTAG: thewiredbear SecondChildCreateTimeTAG: 2012-09-16T06:58:43Z SecondChildTAG: still thanks a lot SecondChildUserIdTAG: 169784 SecondChildUserNameTAG: thewiredbear SecondChildCreateTimeTAG: 2012-09-16T06:59:20Z SecondChildTAG: thanks divya for your wounder answer.... SecondChildUserIdTAG: 204017 SecondChildUserNameTAG: KOMMURI SecondChildCreateTimeTAG: 2012-09-16T07:17:06Z FirstChildTAG: Hey man, remember that you have a potential different between the current source terminal. So, you can try put the ground at the lowest branch and use KCL for 2 nodes (for the nodes which has at least 3 wires coming out). So you determine v2. Best Regards FirstChildUserIdTAG: 263241 FirstChildUserNameTAG: yvesauad FirstChildCreateTimeTAG: 2012-09-16T07:05:13Z IndexTAG: 2707 TitleTAG: Power supplied by current source. Hey! can any body tell me how to find power supplied by current source in above problem.I am not able to convince with solution given above. UserIdTAG: 46119 UserNameTAG: sasidhar_vajrala CreateTimeTAG: 2012-09-16T04:52:20Z VoteTAG: 1 CoursewareTAG: Week 1 / KVL/KCL example CommentableIdTAG: 6002x_S2E3_Using_KVL_KCL_and_VI_constraints NumberOfReplyTAG: 3 FirstChildTAG: P=VI P=(V^2)/R ,with V given P=(I^2)*R, with I given mind the direction of the currents and the signs. FirstChildUserIdTAG: 209082 FirstChildUserNameTAG: Nikemurton FirstChildCreateTimeTAG: 2012-09-16T05:05:10Z SecondChildTAG: 56 watts is the answer. SecondChildUserIdTAG: 401259 SecondChildUserNameTAG: amangodne SecondChildCreateTimeTAG: 2012-09-16T07:47:02Z SecondChildTAG: For the voltage source, how come the answer isn't 4/9? Since v = 2 and R = 9? SecondChildUserIdTAG: 381923 SecondChildUserNameTAG: matteaton SecondChildCreateTimeTAG: 2012-09-17T23:42:30Z FirstChildTAG: always remember that power is V*I so as in the current source you have got I that is the current you know that specific amount of power is needed so you just have to find the voltage plugging in the values..(in the heater problem) regards Neeraj FirstChildUserIdTAG: 210992 FirstChildUserNameTAG: neerajnatu FirstChildCreateTimeTAG: 2012-09-16T05:15:18Z FirstChildTAG: whats your question FirstChildUserIdTAG: 233221 FirstChildUserNameTAG: JPShukla FirstChildCreateTimeTAG: 2012-09-16T06:22:21Z IndexTAG: 2708 TitleTAG: Voltage and Current source question Hello, I am confused by current source being separate from voltage source, what I understood so far was that voltage (v) and current (I) are from the same source. How can you have a current source without a potential difference? Can someone please help. UserIdTAG: 385224 UserNameTAG: shilpiji991 CreateTimeTAG: 2012-09-16T04:31:17Z VoteTAG: 1 CoursewareTAG: Week 1 / Node analysis CommentableIdTAG: 6002x_node_analysis NumberOfReplyTAG: 2 FirstChildTAG: ideally you can, but in reality you will always have an internal resistance associated with the current source. That's the idea of ​​exercise, you see that in reality there is always associated resistance. FirstChildUserIdTAG: 274475 FirstChildUserNameTAG: Konredus FirstChildCreateTimeTAG: 2012-09-16T04:54:37Z FirstChildTAG: its not that current source doesn't have a potential difference you will actually come across a question i don't remember where,where you have to find it. its just that current source has a FIXED current output no matter whats the potential difference across it, on the other hand voltage source has a FIXED voltage output no matter whats the current across it. regards Neeraj FirstChildUserIdTAG: 210992 FirstChildUserNameTAG: neerajnatu FirstChildCreateTimeTAG: 2012-09-16T05:20:14Z SecondChildTAG: thnx neeraj SecondChildUserIdTAG: 156225 SecondChildUserNameTAG: sanjana_m SecondChildCreateTimeTAG: 2012-09-23T18:31:29Z IndexTAG: 2709 TitleTAG: help with lab I've drawn the circuit on paper but i'm having difficulty mobving elements in the sandbox, when i left click and draw a green box over the element and try to drag the element, nothing happens, please any help would be welcome UserIdTAG: 224975 UserNameTAG: Croby CreateTimeTAG: 2012-09-16T03:57:11Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Left click and hold onto the object to move it around. You can also rotate a highlighted object by pressing "R". Press "Delete" to remove a highlighted object. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-16T04:08:22Z SecondChildTAG: The elements are drawn using lines, right. You have draw a box and click on the lines of the circuit, so as to select it properly. SecondChildUserIdTAG: 199839 SecondChildUserNameTAG: vijaykrishnankk SecondChildCreateTimeTAG: 2012-09-16T04:11:36Z SecondChildTAG: thanks, but how do you edit values of the components SecondChildUserIdTAG: 224975 SecondChildUserNameTAG: Croby SecondChildCreateTimeTAG: 2012-09-16T05:03:56Z SecondChildTAG: Double click the component! SecondChildUserIdTAG: 79361 SecondChildUserNameTAG: kimt SecondChildCreateTimeTAG: 2012-09-16T05:13:05Z FirstChildTAG: Hi Croby! I have answered your Resquest [Here][1] ;) [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50554d1778dbb71f00000004 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-16T05:29:39Z IndexTAG: 2710 TitleTAG: Lab 1 Can anyone help me solve the lab 1 assignment...plz i need help urgently UserIdTAG: 128409 UserNameTAG: arijitbme CreateTimeTAG: 2012-09-16T03:52:55Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Hi arijitbme! How are you? What are your difficulties? Can I help you? Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-16T03:57:39Z SecondChildTAG: i'm having problems moving the circuit elements in the lab, could you give me a few pointers Croby SecondChildUserIdTAG: 224975 SecondChildUserNameTAG: Croby SecondChildCreateTimeTAG: 2012-09-16T04:44:45Z SecondChildTAG: Hi Croby! Sure, I will be pleasured to help you ;). 1) You will see in the Simulator a Toolbox (I will use Sandbox to explain this) ![enter image description here][1] 2) Suppouse that we want to make a simple Resistor Divider Circuit. We will need a Voltage Source, a Ground and two Resistors: ![enter image description here][2] 3) Now click on the element that you want and **don't release the mouse!** and move it to the work place. ![enter image description here][3] 4) You will see something like this: ![enter image description here][4] 5) Let's try with the other resistor! ![enter image description here][5] 6) Ok, now, in order to rotate the component press "r" (keyboard). ![enter image description here][6] 7) wire it with the resistor one. Just go to the extreme of the resistor and don't release the mouse till reach to the other resistor. ![enter image description here][7] 8)Try it! Now you should be able to do the same with the source and the ground. ![enter image description here][8] 9) Now , you can wire them. ![enter image description here][9] 10)Ready! You have your Circuit! Now you can Play with the Parameters: changing the values of the elements, meassuring , etc! Have fun Croby!;) I hope this can help you! See you! Myriam. [1]: https://edxuploads.s3.amazonaws.com/13477720491700108.png [2]: https://edxuploads.s3.amazonaws.com/13477722771343605.png [3]: https://edxuploads.s3.amazonaws.com/1347772554104507.png [4]: https://edxuploads.s3.amazonaws.com/1347772644554092.png [5]: https://edxuploads.s3.amazonaws.com/13477727422265003.png [6]: https://edxuploads.s3.amazonaws.com/1347772832861453.png [7]: https://edxuploads.s3.amazonaws.com/13477729324011119.png [8]: https://edxuploads.s3.amazonaws.com/13477730701343607.png [9]: https://edxuploads.s3.amazonaws.com/13477731298474461.png SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-16T05:27:26Z FirstChildTAG: Hi, in which circuit you have problems? Konrad. FirstChildUserIdTAG: 274475 FirstChildUserNameTAG: Konredus FirstChildCreateTimeTAG: 2012-09-16T04:05:37Z FirstChildTAG: Hint: You can use 3 extra resistors so as to produce such conditions. Check, only after a dc simulation is done, without closing the dc simulation. FirstChildUserIdTAG: 199839 FirstChildUserNameTAG: vijaykrishnankk FirstChildCreateTimeTAG: 2012-09-16T04:10:18Z SecondChildTAG: In the lab1 you only need 2 resistors, and the Hint is: use a two-resistor voltage divider to create the voltage for node A. You'll have two unknowns (R1 and R2) which can be determined by solving the two equations for vs derived from the constraints above: one involving R1, R2 and Rbulb where vs=1.5, and one involving R1 and R2 where vs=2. I do not understand why you want to use 3 resistors SecondChildUserIdTAG: 274475 SecondChildUserNameTAG: Konredus SecondChildCreateTimeTAG: 2012-09-16T04:19:25Z IndexTAG: 2711 TitleTAG: What does short-circuit mean ?? short-circuit what does it mean in Now, suppose we short-circuit the compound battery. (This is very dangerous. NEVER do this to a large battery, such as a lead-acid battery in a car, or to a lithium-ion battery from your laptop. You MAY live to regret it, but you may not.) What is the current (in Amperes) you should expect to go through the short circuit? UserIdTAG: 290966 UserNameTAG: AhmedImam CreateTimeTAG: 2012-09-16T02:47:50Z VoteTAG: 1 CoursewareTAG: Week 1 / Battery model CommentableIdTAG: 6002x_S1E9_Battery_Model NumberOfReplyTAG: 1 FirstChildTAG: To connect the positive terminal directly to the negative terminal would be a "short circuit" Sometimes the term "short" is used to bypass an individual component. Ex: To short out a resistor would mean to attach a wire to both sides of the resistor, effectively cancelling out the resistor. The resistor would then be "shorted" FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-16T03:04:03Z SecondChildTAG: thanks very much :) SecondChildUserIdTAG: 290966 SecondChildUserNameTAG: AhmedImam SecondChildCreateTimeTAG: 2012-09-16T03:21:48Z SecondChildTAG: Acordding to the textbook: **Open-circuit:** "if we choose a voltage, then because the intercept is by definition at **zero current**" **Short-circuit:** "if the intercept is specified in terms of current (...) by definition the **voltage is zero** at that point" SecondChildUserIdTAG: 38538 SecondChildUserNameTAG: Alienista SecondChildCreateTimeTAG: 2012-09-16T03:22:23Z SecondChildTAG: If you are using an ideal conductor, V/R = I = infinite. If you are using a very good conductor(eg: copper) and assume R = 1ohms, then V/R = V, which is actually a very large current. Not even the power supply can withstand that current. If you use a fuse in between, then it may not be a problem(I haven't tried short circuiting any device). SecondChildUserIdTAG: 199839 SecondChildUserNameTAG: vijaykrishnankk SecondChildCreateTimeTAG: 2012-09-16T04:07:40Z IndexTAG: 2712 TitleTAG: circuit sandbox pointer on rotating elements' icon to horizontal I'm surprised at myself not having noticed from the video, but it seems that the elements sometimes appear horizontally too, and I don't know how to rotate them. Please advise. UserIdTAG: 280220 UserNameTAG: pilgrimCycle CreateTimeTAG: 2012-09-16T02:15:00Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Press "R" to rotate Highlight and press "Delete" to remove. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-16T02:25:19Z SecondChildTAG: jajaja,very well, I do not know how to rotate, thanks SecondChildUserIdTAG: 274475 SecondChildUserNameTAG: Konredus SecondChildCreateTimeTAG: 2012-09-16T03:47:44Z FirstChildTAG: Hi pilgrimCycle! Yes, Elements can be rotated. 1) Suppouse that you have choose an element like a Resistor. Just select the resistive element and dragg it to your work place window. ![uno][1] 2) Once you did step 1), click on it. You will notice that the element it is selected because the element will have a change color (from blue to green). ![dos][2] 3)Let's see how the element will be rotated as we press "r" from the keyboard. Once, the element it is selected, you will notice that when you press "r" or "r""r" or "r""r""r" .... the element will change of direction. So, this will enable you to design your circuit as the way you like ;). ![tres][3] ---- Hola pilgrimCycle! Sí, los elementos se pueden rotar. 1) Supongamos que deseamos elegir un elemento tal y como lo es un Resistor. Pues, si así lo deseamos, debemos hacer un clic sobre el elemento en la barra de herramientas localizada a la derecha y arrastrarlo a nuestra ventana de trabajo. ![][1] 2) Una vez que has hecho el paso 1), debes seleccionar el elemento. Para ello, debes hacer clic sobre él. Notarás que su aspecto cambiará del color azul al verde. Esto quiere decir, que el elemento se encuentra seleccionado. ![][2] 3) Ahora veamos como el elemento se puede rotar al teclear con la "r" del teclado de tu PC siempre y cuando hayas seleccionado previamente el elemento. Verás que, cada vez que presionas la "r" el elemento rotará. Notarás, que si presionas dos veces la "r", girará 90 grados, si lo haces 3 veces, rotará 135 grados, y así sucesivamente. Esto, te permitirá a ti, crear el diseño circuital a tu gusto ;). ![][3] [1]: https://edxuploads.s3.amazonaws.com/13477666306001938.png [2]: https://edxuploads.s3.amazonaws.com/13477666565619128.png [3]: https://edxuploads.s3.amazonaws.com/13477666685619116.png FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-16T03:54:19Z IndexTAG: 2713 TitleTAG: Could someone post a sample certificate Hello, Could one of the staff members post an image of a sample certificate in the "Course Info" section. Someone asked about it and I realized it would be great to have something to point people to. UserIdTAG: 21541 UserNameTAG: JSChambers CreateTimeTAG: 2012-09-16T00:57:29Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 2 FirstChildTAG: hi try to see here https://www.facebook.com/EdxOnline?ref=ts you can find here are posted FirstChildUserIdTAG: 235727 FirstChildUserNameTAG: amor FirstChildCreateTimeTAG: 2012-09-16T01:05:29Z FirstChildTAG: ![enter image description here][1] [1]: http://gilbertojunqueira.com/wp-content/uploads/2012/06/cert.png FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-16T01:07:14Z SecondChildTAG: wow I am very anxious to get it ^_^ SecondChildUserIdTAG: 358712 SecondChildUserNameTAG: emansabah SecondChildCreateTimeTAG: 2012-09-16T05:56:17Z SecondChildTAG: me toooooooooo SecondChildUserIdTAG: 276808 SecondChildUserNameTAG: DEBASMITAMAJUMDER SecondChildCreateTimeTAG: 2012-09-16T06:08:01Z SecondChildTAG: how this certificate can be delivered to us SecondChildUserIdTAG: 326507 SecondChildUserNameTAG: mohamed200 SecondChildCreateTimeTAG: 2012-09-18T22:06:05Z SecondChildTAG: And can i get a hard copy of it and how much will it cost? SecondChildUserIdTAG: 213452 SecondChildUserNameTAG: JoJosida SecondChildCreateTimeTAG: 2012-09-19T05:46:23Z IndexTAG: 2714 TitleTAG: H1P2: How can I get V1 ? Please help, How can I get V1 ? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-15T23:54:16Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: V1 is in parallel with the 1ohm resistor, that means the current sourse have the equal voltage at the resistor. FirstChildUserIdTAG: 274475 FirstChildUserNameTAG: Konredus FirstChildCreateTimeTAG: 2012-09-16T03:45:03Z FirstChildTAG: Find the voltage across the 1 ohm resistor.Since the current source is in parallel with the 1 ohm resistor v1=v2.Elements connected is parallel have their voltages same and any elements connect in series have their currents same. FirstChildUserIdTAG: 1825 FirstChildUserNameTAG: shaik1990jaffer FirstChildCreateTimeTAG: 2012-09-16T04:20:51Z SecondChildTAG: I thought R1 and R3 are in serie with V4 so R1 = 1 Volt and R3 = 3Volt. Apparently this is to simple. SecondChildUserIdTAG: 410943 SecondChildUserNameTAG: JPalaciosRoman SecondChildCreateTimeTAG: 2012-09-16T06:25:17Z IndexTAG: 2715 TitleTAG: R .... constant? why is the resistant of a material a constant ? UserIdTAG: 351107 UserNameTAG: sanket383 CreateTimeTAG: 2012-09-15T22:54:03Z VoteTAG: 1 CoursewareTAG: Week 1 / Lumped Element Abstraction CommentableIdTAG: 6002x_lumped_element_abstraction NumberOfReplyTAG: 2 FirstChildTAG: the resistant of a material depend of many external factors like temperature, presure, size and even shape, so the resistant of a material is constant only under certain conditions, if that conditions change the resistant also change. FirstChildUserIdTAG: 73837 FirstChildUserNameTAG: ricardo87 FirstChildCreateTimeTAG: 2012-09-15T23:15:09Z SecondChildTAG: But in LMD we suppose that this material is ideal, so resistant is constant SecondChildUserIdTAG: 230187 SecondChildUserNameTAG: sormon SecondChildCreateTimeTAG: 2012-09-15T23:16:48Z SecondChildTAG: yes, but only in LMD, so if you have to solve problems in real life you have to consider this behaivor in some cases. SecondChildUserIdTAG: 73837 SecondChildUserNameTAG: ricardo87 SecondChildCreateTimeTAG: 2012-09-15T23:24:26Z FirstChildTAG: If a device is to be used under the lumped matter discipline, in needs to have a constant R. Not all materials have Static resistance, some are very non-linear and are known as Differential resistance. $Rdiff = dv/di$ In future chapters you will be taught how to deal with nonlinear devices within the EECS playground. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-15T23:21:00Z IndexTAG: 2716 TitleTAG: HW1 P1 Hi, I know the answer for: "What is the equivalent resistance as an algebraic expression (in terms of R) of network as C viewed from its port?" should be R+R+(1/(1/R+1/R)) but it keeps saying wrong??? Is it wrong? Also can someone help me with how to solve the last question (I dont want answer, but just method). And this is the question: Given that each individual resistor can dissipate up to 1 watt of power before burning up, how much total power in watts ( W ) can the smallest-valued composite resistor dissipate before burning up? Thanks alot! UserIdTAG: 275798 UserNameTAG: Basemjs CreateTimeTAG: 2012-09-15T22:01:45Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 8 FirstChildTAG: I THINK 3*R/5 FirstChildUserIdTAG: 215287 FirstChildUserNameTAG: yomi FirstChildCreateTimeTAG: 2012-09-16T01:56:37Z FirstChildTAG: hi...ans is 3/5*R FirstChildUserIdTAG: 307088 FirstChildUserNameTAG: PRAKHAR2012 FirstChildCreateTimeTAG: 2012-09-16T04:49:50Z FirstChildTAG: that is Rp = R2(R3+R4)/R2+ (R3+R4) or Rp+ R1 + Rp but i dont know the keyword for this it says invalid keywork but its so unfair for me coz that all question never found in my homework so i can't get a perfect score how to answer that i never found it FirstChildUserIdTAG: 235727 FirstChildUserNameTAG: amor FirstChildCreateTimeTAG: 2012-09-15T22:15:24Z SecondChildTAG: Amor, when you enter things like "R3", you will notice an error shows up near the "check" box. So try to input answers based upon that. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-15T22:26:00Z SecondChildTAG: Also as others have mentioned on previous occasions, you must use "*" for multiplication. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-15T22:33:36Z FirstChildTAG: Network c ((R + R)||R)+R FirstChildUserIdTAG: 402510 FirstChildUserNameTAG: Masumul FirstChildCreateTimeTAG: 2012-09-15T22:30:23Z FirstChildTAG: Could someone help me with the HW1P1 answers or did anybody finish HW1P1 please help me with the answers. FirstChildUserIdTAG: 243273 FirstChildUserNameTAG: verma1129 FirstChildCreateTimeTAG: 2012-09-15T22:50:45Z FirstChildTAG: Hello man, how are you? Actually, the right answer is 1/(1/(R+R)+1/R)+R which can be reescripeted in the easier form: ( (R+R)^{-1} + (R)^{-1} )^{-1} + R. If you develop this expression you´ll get 5/3*R FirstChildUserIdTAG: 263241 FirstChildUserNameTAG: yvesauad FirstChildCreateTimeTAG: 2012-09-15T23:03:23Z FirstChildTAG: (R+R)=2R; which is in parallel with R;therefore (2R*R)/(2R+R) which in last is in series with the terminal R. hence equivalent resistor will be 5R/3 FirstChildUserIdTAG: 181793 FirstChildUserNameTAG: mitrahul FirstChildCreateTimeTAG: 2012-09-16T02:58:33Z FirstChildTAG: THE ANSWER IS 5/3*R FirstChildUserIdTAG: 441979 FirstChildUserNameTAG: shrutimehrotra FirstChildCreateTimeTAG: 2012-09-16T13:49:08Z IndexTAG: 2717 TitleTAG: H3P4, maximum current-I swear I'm doing this right..... Also, H3P1. OK, so say I want to solve for D1. I can ignore D2, because it is an open circuit. The one with D1 acts as a short. So, I have the D1, and the resistor. At that point I just use KCL: (V-Vs) = (V-V1)R + V I get something like 3 A, it says it is wrong. What am I not getting? And for that matter, I'm also having problems with the NOR gate on H3P1, and finding the power of all of them. So if anyone could give me some advice, I'd appreciate it. I think you'd just use the geometry, so I divided the R value that I had in the first one by 2, but that's wrong. UserIdTAG: 344331 UserNameTAG: intellectualwanderer CreateTimeTAG: 2012-09-15T21:47:53Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: If you look at the NOR, you will find that it is actually similar to the inverter. Once you have the RpuI, the others can be found in a similar fashion, the difference being, they have a higher "on state resistance" then the inverter by itself. I am not sure I understand the first part of your question, are you referring to H3P3? Have fun. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-15T22:19:56Z SecondChildTAG: H3P4. SecondChildUserIdTAG: 344331 SecondChildUserNameTAG: intellectualwanderer SecondChildCreateTimeTAG: 2012-09-16T02:11:58Z SecondChildTAG: I use the value of the first in parallel, and solve using the voltage divider equation, but no go. SecondChildUserIdTAG: 344331 SecondChildUserNameTAG: intellectualwanderer SecondChildCreateTimeTAG: 2012-09-16T02:36:41Z SecondChildTAG: I got H3P4 done. Comments on H3P1 still WELCOME. SecondChildUserIdTAG: 344331 SecondChildUserNameTAG: intellectualwanderer SecondChildCreateTimeTAG: 2012-09-16T03:13:30Z SecondChildTAG: For H3P1 go to this link. You will get it. https://6002x.mitx.mit.edu/section/week3_gjs/ SecondChildUserIdTAG: 328404 SecondChildUserNameTAG: NIDHIN_RS SecondChildCreateTimeTAG: 2012-09-16T13:53:12Z SecondChildTAG: Thank you! SecondChildUserIdTAG: 344331 SecondChildUserNameTAG: intellectualwanderer SecondChildCreateTimeTAG: 2012-09-16T17:01:51Z SecondChildTAG: @intellectualwanderer What did you do for H3P4 and the max currents? Using KCL about each Diode, I got the following: `Positive Voltage: iD1 + Vs/R + V1/R = 0` `Negative Voltage: iD2 + Vs/R + V2/R = 0` but solving for each iD yields incorrect values. Thanks. SecondChildUserIdTAG: 348141 SecondChildUserNameTAG: TomTrieb SecondChildCreateTimeTAG: 2012-09-21T15:45:57Z IndexTAG: 2718 TitleTAG: Working out resistances The network of resistors can be classed as one big resistor (lumping them all together). Start the easy way. We know all the resistance values, so we can add R3 to R4 to get R(3,4), which we can consider as one resistor. Then R2 and R(3,4) can be classed as one resistor (parallel relation), say R(2,3,4). Then that is in series with R1, so the problem of i1 boils down to one R with one V. For an explanation of current polarity, please see my post in S2E3. UserIdTAG: 261378 UserNameTAG: es2377 CreateTimeTAG: 2012-09-15T21:37:35Z VoteTAG: 1 CoursewareTAG: Week 1 / Series and parallel example CommentableIdTAG: 6002x_S2E4_Series_and_Parallel NumberOfReplyTAG: 1 FirstChildTAG: I don't think I have seen a (simple solution) for combinations of parallel and series Resistors. If I have missed something please tell me where. FirstChildUserIdTAG: 359302 FirstChildUserNameTAG: GaryG FirstChildCreateTimeTAG: 2012-09-16T10:39:35Z IndexTAG: 2719 TitleTAG: problem with h1p1 You are given three resistors: two 4Ω resistors and one 6Ω resistor. What is the value in Ohms ( Ω ) of the smallest-valued resistor that can be fabricated by combining these three resistors? can any one tell me what will be the answer of this UserIdTAG: 376345 UserNameTAG: Hassan-Ahmed CreateTimeTAG: 2012-09-15T20:11:29Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 6 FirstChildTAG: the smallest-valued resistor that can be fabricated by combining these three resistors - parallel connection of all ;o) http://www.1728.org/resistrs.htm FirstChildUserIdTAG: 325197 FirstChildUserNameTAG: Vitali_Jerin FirstChildCreateTimeTAG: 2012-09-15T20:29:56Z SecondChildTAG: the value is 3/2 SecondChildUserIdTAG: 445425 SecondChildUserNameTAG: Rafaykhan SecondChildCreateTimeTAG: 2012-09-15T20:58:01Z FirstChildTAG: it will be as they connecting in parallel way so we can say : ( 4*4*6)/ (4+4+6) FirstChildUserIdTAG: 284144 FirstChildUserNameTAG: reem28 FirstChildCreateTimeTAG: 2012-09-15T20:32:33Z SecondChildTAG: This works only for 2 resistors in paralleled. For more the 2 resistors paralleled use 1/RV = 1/R1+1/R2+1/R3 and so on. In the example: 1/4+1/4+1/6 = 3/12+3/12+2/12 ==>> 12/8=1.5 ohm SecondChildUserIdTAG: 410943 SecondChildUserNameTAG: JPalaciosRoman SecondChildCreateTimeTAG: 2012-09-15T20:47:49Z FirstChildTAG: Did you not read the code of conduct? FirstChildUserIdTAG: 233700 FirstChildUserNameTAG: ANDREW77 FirstChildCreateTimeTAG: 2012-09-15T21:28:12Z FirstChildTAG: cuando haces un arreglo de resistencias por decir 2 resistencias de 1 ohm en serie se suman y te dara una R(total)= 2 OHM en paralelo te dara una R(total)=0.5 OHM con este ejemplo puedes deducir que arreglo necesitas para obtener una resistencia de valor pequeño,espero te sirva,suerte FirstChildUserIdTAG: 299697 FirstChildUserNameTAG: ramirito FirstChildCreateTimeTAG: 2012-09-15T20:31:32Z FirstChildTAG: You cannot directly ask for an answer. It would be faster and easier for you to use the search function, this question has been asked many, many times. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-15T20:53:09Z FirstChildTAG: Join them in parallel, and get total value using 1/(Total Resistance) = 1/R1 + 1/R2 + 1/R3 FirstChildUserIdTAG: 57012 FirstChildUserNameTAG: sam747 FirstChildCreateTimeTAG: 2012-09-15T21:16:39Z IndexTAG: 2720 TitleTAG: Diodes and LambertW function ![enter image description here][1] ![enter image description here][2] [1]: https://edxuploads.s3.amazonaws.com/13477359895963964.jpg [2]: https://edxuploads.s3.amazonaws.com/13477366042767063.jpg UserIdTAG: 27399 UserNameTAG: juancho CreateTimeTAG: 2012-09-15T19:17:20Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hola Juan! Cómo estás? Hace mucho que no veía tus Posts. Gracias! :) ---------- Hi Juan! How are you? Since a long time that I haven't seen your Posts! Thank you! :) FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-16T05:59:21Z SecondChildTAG: Hi Myriam, many thanks, forever you will be my queen, all the best. SecondChildUserIdTAG: 27399 SecondChildUserNameTAG: juancho SecondChildCreateTimeTAG: 2012-09-16T13:53:32Z IndexTAG: 2721 TitleTAG: Current Source, Current Directions, Nodes Now, I know that normally, the current is going out of the positive terminal of the current source .. ok ?! So what should I do if the given network has the current moving in the opposite direction ?! .. should I calculate it in negative values when I substitute in the equations of the KCL ?! .. well, our convention is to make the currents going out of the node (and that what the figure show to me) .. but if we go with the normal situation where the current should go out of the +ve terminal of the current source, then it should be going into the node .. and that means I should substitute with negative values. Which sign should I use ? negative one because the current source must push the current out of it? or positive because this is the current direction in the figure ?!! Another question, we know that voltage sources have constant voltage difference between their terminals, and a current equals the voltage supplied divided by the total resistance of the network. O, this is good, but what about current sources? what is the voltage difference between their terminals ?! Do they have resistance of some kind to calculate the voltage ?! So, again .. my basic question is .. should I make the direction of the current of my own and ignore the directions given in the figure? or should I stick with what is given ?! And thanks UserIdTAG: 289472 UserNameTAG: OsamaAdel CreateTimeTAG: 2012-09-15T19:08:05Z VoteTAG: 1 CoursewareTAG: Week 1 / Node analysis CommentableIdTAG: 6002x_node_analysis NumberOfReplyTAG: 2 FirstChildTAG: If you are calculating currents leaving a node, a current entering the node would be negative, as it is going the opposite direction as those leaving the node. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-15T19:56:57Z FirstChildTAG: ya < i find this a big problem with me .... FirstChildUserIdTAG: 284144 FirstChildUserNameTAG: reem28 FirstChildCreateTimeTAG: 2012-09-15T19:27:09Z IndexTAG: 2722 TitleTAG: Mixing of signals ... LAb 2 ??? some 1 please explain that whether we choose value of V1 and v2 by hit and trial method ??? or by any logic ... UserIdTAG: 430030 UserNameTAG: imali CreateTimeTAG: 2012-09-15T18:41:40Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: V1 and V2 are already indicated. Our task was to use resistors to design the rest of the circuit such that the voltage measured at the output equaled 1/2 V1 + 1/6 V2. It's similar to the simple resistive mixer shown but with a little something added to it. FirstChildUserIdTAG: 331664 FirstChildUserNameTAG: dwmnctrh3 FirstChildCreateTimeTAG: 2012-09-15T18:55:22Z SecondChildTAG: it is a impossible thing to reduce a voltage source to half of its magnitude at the o/p terminals.... give me some hint about the component which is necessary ........ lol thnx SecondChildUserIdTAG: 395966 SecondChildUserNameTAG: sambo007 SecondChildCreateTimeTAG: 2012-09-15T19:23:20Z SecondChildTAG: ..like turning a volume control to half? SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-15T19:59:25Z IndexTAG: 2723 TitleTAG: LAB 1 Difficulties HI everyone it would be great if smone could help me do the lab1 i use potential divider method and i completed the circuit but they are both wrong... plz someone help !! thnx UserIdTAG: 151205 UserNameTAG: jesher777 CreateTimeTAG: 2012-09-15T18:04:23Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: i am also stuck at this... can somebody help ? FirstChildUserIdTAG: 129030 FirstChildUserNameTAG: vjbhatt786 FirstChildCreateTimeTAG: 2012-09-15T18:18:04Z SecondChildTAG: Ya I hav performed the dc analysis as well...n the answers are correct...But It is displaying it as wrong...R u facing the same problem??? SecondChildUserIdTAG: 375695 SecondChildUserNameTAG: MsnShiva SecondChildCreateTimeTAG: 2012-09-15T18:20:02Z SecondChildTAG: yeah the same!! :S but i'm not totally sure that it may be the right answer but mathematically it is correct... but it shows wrong their!! I may have disregard the bulb resistance(3ohms) but the voltage is the same if i consider it to be paralleled to the other resistor! the second circuit may be the wrong one too!! i don't understand how to make the second circuit!! is it an open circuit?? we don't need to put any resistor their? isn't it? i'm a bit confused!! SecondChildUserIdTAG: 151205 SecondChildUserNameTAG: jesher777 SecondChildCreateTimeTAG: 2012-09-15T18:50:32Z SecondChildTAG: hey this post could possibly help ...read the 3rd comment from Chaunt ..i finally got the correct answer ..give it a try https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50515dd092ce11210000000d SecondChildUserIdTAG: 129030 SecondChildUserNameTAG: vjbhatt786 SecondChildCreateTimeTAG: 2012-09-15T19:31:32Z FirstChildTAG: hey this post could possibly help ...read the 3rd comment from Chaunt ..i finally got the correct answer ..give it a try https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50515dd092ce11210000000d FirstChildUserIdTAG: 129030 FirstChildUserNameTAG: vjbhatt786 FirstChildCreateTimeTAG: 2012-09-15T19:43:04Z FirstChildTAG: Hi I think first you should figure out where you should have the resistors and why! in this case if you haven’t a resistor paralleled to the bulb then in the second model your circuit is not completed, and you also need a resistor in series to the bulb in order to change the voltage by changing its R! ![suggestion formation for the lab 1 (the R1 and R2 are not calculated)][1] [1]: https://edxuploads.s3.amazonaws.com/1347778979113529.png now you may use the hint in the lab page (hard way) or you can perform a easy way! First you should select the R2 resistant’s or its current and find the other one! By considering that the bulb, needs 0.5A to consume 1.5v! And by writing the KCL in the B (B is the same as A) you can figure out the total current in the circuit. Then, by calculating the total parametric resistant (R1 is the only parameter in that equation) of the circuit you can find the R1 resistant’s. By playing whit values of R1 and R2 you will soon find out that by increasing current of the R2 the voltage differences of the point A in the first and second model decreased. For instance if you select .1 Ohms for R2 the voltage of point A in second model will be 1.54V. I hoop this helps you and good luck! FirstChildUserIdTAG: 420339 FirstChildUserNameTAG: AliJenabi FirstChildCreateTimeTAG: 2012-09-16T07:23:22Z IndexTAG: 2724 TitleTAG: online text book help on online book section page no. 10 2nd paragraph says that such a selection of element boundaries make "(del phi B / del t =0)". so my question is " how the selection of element boundaries make (del phi B / del t =0)" ? UserIdTAG: 419552 UserNameTAG: saurabhkrchauhan CreateTimeTAG: 2012-09-15T17:53:06Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 2725 TitleTAG: confused how to finish lab 2 i am much confused and i dont know how to do lab 2 UserIdTAG: 149154 UserNameTAG: santhosh1993 CreateTimeTAG: 2012-09-15T17:52:27Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hello! Think about using a resistive devider being connected to one of the voltage sources. Hint: one resistive devider will be enough, so you will need to calculate 4 resistors. To find their values write down the equation for Vout using variables V1, V2, R1,...,R4. To get a system of 4 equations consider 4 different reference values of V1 and V2: 1) V1 = 1V and V2 = 1V 2) V1 = 1V and V2 = 0V 3) V1 = 1V and V2 = -1V 4) V1 = 0V and V2 = 1V By doing so you'll get V1 and V2 as knowns and four equations with 4 unknown variables R1,...,R4. Good luck! FirstChildUserIdTAG: 209041 FirstChildUserNameTAG: SmartEngine FirstChildCreateTimeTAG: 2012-09-16T16:16:37Z IndexTAG: 2726 TitleTAG: Any hints on lab 2? Hi, I have no idea on what to do with lab 2? Any hints? Thanks. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-15T17:31:43Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: 3 days ago I posted some hints about Lab 2. You can use the search tool of the discussion forum to look for them! The name of the post is "Lab 2 hints" FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-15T17:40:38Z SecondChildTAG: 1)By superposition, I found out that v=R2V1/(R1+R2)+R1V2/(R1+R2) 2)Since Vout=V1/2+V2/6, by comparison, I got R2/(R1+R2)=1/2 --> 1st equation and R1/(R1+R2)=1/6 -->2nd equation 3) Solving it I obtained, for the first equation R1=R2 and second equation 5R1=R2. No idea what to do after this of whether this is correct? SecondChildUserIdTAG: 295240 SecondChildUserNameTAG: Jack_my SecondChildCreateTimeTAG: 2012-09-16T15:43:38Z SecondChildTAG: The two equations of your third point have the unique solution R1 = R2 = 0. Then the lab is not solvable by using only two resistors. What else can you do? Can you think of a different configuration of resistors that will allow you to have coefficients that doesn't add to 1? SecondChildUserIdTAG: 381746 SecondChildUserNameTAG: jelizon SecondChildCreateTimeTAG: 2012-09-16T16:54:22Z SecondChildTAG: Progress made so far 1) Put in a third resistor connected to ground as what I read in some other post. 2) Ended up with R2R3/R1R2+R1R3+R2R3)=1/2 AND R1R3/(R1R2+R1R3+R2R3)=1/6 3) Solving both of it, got R1R3:R2R3:R1R2 = 1:3:2 4) Hence, R1/R2=1/3 AND R3/R2 = 1/2 Is this correct? SecondChildUserIdTAG: 295240 SecondChildUserNameTAG: Jack_my SecondChildCreateTimeTAG: 2012-09-17T14:16:35Z FirstChildTAG: Progress made so far 1) Put in a third resistor connected to ground as what I read in some other post. 2) Ended up with R2R3/R1R2+R1R3+R2R3)=1/2 AND R1R3/(R1R2+R1R3+R2R3)=1/6 3) Solving both of it, got R1R3:R2R3:R1R2 = 1:3:2 4) Hence, R1/R2=1/3 AND R3/R2 = 1/2 Is this correct? FirstChildUserIdTAG: 295240 FirstChildUserNameTAG: Jack_my FirstChildCreateTimeTAG: 2012-09-18T06:59:38Z IndexTAG: 2727 TitleTAG: ifrah i want to ask that is edx offering any course in bio-medical engineering or not?? UserIdTAG: 380132 UserNameTAG: IFRAHJAFFRI CreateTimeTAG: 2012-09-15T17:23:33Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Here is a list of the current offerings from edX: https://www.edx.org/courses FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-15T17:43:44Z SecondChildTAG: ok SecondChildUserIdTAG: 380132 SecondChildUserNameTAG: IFRAHJAFFRI SecondChildCreateTimeTAG: 2012-09-21T18:14:23Z IndexTAG: 2728 TitleTAG: what is the use of a giga voltage what is the use of the high voltage? and I calculated the $R_{th}$ as the following ***$R_{th} = 1k + (1/1k) + (1/1k) = 1000.002ohm$*** what I did to get the wrong answer? please help UserIdTAG: 292360 UserNameTAG: Loai CreateTimeTAG: 2012-09-15T16:31:44Z VoteTAG: 1 CoursewareTAG: Week 2 / Thevenin Example CommentableIdTAG: 6002x_thevenin_example_1 NumberOfReplyTAG: 1 FirstChildTAG: Your expression for the parallel resistance is not right. Recall that $R_1$ and $R_2$ in parallel has an equivalent resistance $R_\mathrm{eq}$ given by: $\frac{1}{R_\mathrm{eq}} = \frac{1}{R_1}+\frac{1}{R_2}$ FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-15T16:55:14Z SecondChildTAG: You might also like to try a parallel resistor calculator, like this one: http://www.sengpielaudio.com/calculator-paralresist.htm There are a bunch of them on the web. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-09-15T17:33:30Z IndexTAG: 2729 TitleTAG: What do you think of the new forum? Just wondering what people think about the new forum. UserIdTAG: 21541 UserNameTAG: JSChambers CreateTimeTAG: 2012-09-15T16:28:51Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I like it and look forward to the final incarnation! I am also impressed with how quickly the staff can resolve problems and implement changes. Cheers. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-15T17:06:32Z IndexTAG: 2730 TitleTAG: doubt in finding v2 in 1st question got t answer to be (8.75-0)/5=1.75....can someone pls help us to know why it is same as the nodal voltage?? UserIdTAG: 414118 UserNameTAG: raje93 CreateTimeTAG: 2012-09-15T15:13:57Z VoteTAG: 1 CoursewareTAG: Week 1 / Node method example CommentableIdTAG: 6002x_S2E5_Node_Method NumberOfReplyTAG: 1 FirstChildTAG: They're looking for the voltage at node e1, not the voltage difference between e1 and e2 (which would be the voltage across R1). Hope this helps. FirstChildUserIdTAG: 207204 FirstChildUserNameTAG: hgustavii FirstChildCreateTimeTAG: 2012-09-15T21:28:39Z SecondChildTAG: @raje93- its like voltage difference /resistance.So this will be a current;Not an voltage. To get a voltage you have to multiply by corresponding R(R2=5).So you will get 8.75V. Is that Ok? SecondChildUserIdTAG: 147577 SecondChildUserNameTAG: rithi3 SecondChildCreateTimeTAG: 2012-09-16T14:31:34Z IndexTAG: 2731 TitleTAG: Lab 1 Question How do you get the voltage of node A to show up right next to node A? "Because we will be checking the voltage at node A, you should have an assigned voltage at node A after the DC analysis; otherwise, your submission will be deemed incorrect." Not sure how to do this. Thanks! UserIdTAG: 213386 UserNameTAG: dmascenik CreateTimeTAG: 2012-09-15T13:50:17Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Just click on DC analysis before clicking check.. :) FirstChildUserIdTAG: 257806 FirstChildUserNameTAG: skoda FirstChildCreateTimeTAG: 2012-09-15T13:51:57Z SecondChildTAG: I forgot to mention that I did that. The only voltages that show up when I do that are the voltages across the resistors, but not at the node. SecondChildUserIdTAG: 213386 SecondChildUserNameTAG: dmascenik SecondChildCreateTimeTAG: 2012-09-15T14:27:10Z SecondChildTAG: If node A is connected by a wire to the point where a voltage value is shown, then you should be good. Give it a try! (You can submit an infinite number of times.) SecondChildUserIdTAG: 79361 SecondChildUserNameTAG: kimt SecondChildCreateTimeTAG: 2012-09-15T14:34:22Z IndexTAG: 2732 TitleTAG: lab1 help i couldnt understand what to do in lab1.plz help UserIdTAG: 276808 UserNameTAG: DEBASMITAMAJUMDER CreateTimeTAG: 2012-09-15T13:33:41Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Basically, you have to draw two different circuits, to satisfy the two different requirements in the question. When you think you have the right circuit drawn, do a "DC analysis" to see if you have the required voltages in each circuit. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-15T14:30:56Z SecondChildTAG: is it really two *different* circuits or should be just one? SecondChildUserIdTAG: 330569 SecondChildUserNameTAG: cuckoo_clock SecondChildCreateTimeTAG: 2012-09-15T16:25:24Z SecondChildTAG: I'v been at it for 6 hours and still can't see it... I get everything so far, my brain is just melting away looking at this question though... I'm so peeved i can't see something I'm sure is simple I can tell you 1000 ways how not to solve this question... and 1 way to cheat and get a pass lol... SecondChildUserIdTAG: 329964 SecondChildUserNameTAG: SmartMike SecondChildCreateTimeTAG: 2012-09-15T16:44:04Z SecondChildTAG: If there is way to cheat and it really is cheating, you should email the admins and let them know. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-09-15T17:14:17Z IndexTAG: 2733 TitleTAG: Help with lab 2 I have tried many resistor combinations but am not able to get the solution. Can anyone help me by giving some ideas!!! Thank you!!! UserIdTAG: 100418 UserNameTAG: balashyamala CreateTimeTAG: 2012-09-15T13:08:45Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Hints:you only need 3 resistors.. and I used " trials and errors" method .. Don't expect the resistances of them to be integers FirstChildUserIdTAG: 136959 FirstChildUserNameTAG: christerpher FirstChildCreateTimeTAG: 2012-09-15T13:58:54Z SecondChildTAG: 3 resitors? i tried only 2 resistors thats what im mistaken :( SecondChildUserIdTAG: 235727 SecondChildUserNameTAG: amor SecondChildCreateTimeTAG: 2012-09-15T15:38:50Z FirstChildTAG: Start with a 'T' kind of network... You can use KVL and KCL to arrive at the values precisely... No need for trail and error.. but of course, tat will also help... :) FirstChildUserIdTAG: 257806 FirstChildUserNameTAG: skoda FirstChildCreateTimeTAG: 2012-09-15T14:18:59Z FirstChildTAG: It might help to redraw the circuit(on a piece of paper) as a straight line. Gives you a more visually immediate overview of what is going on. Also, remember that the grounds are all tied together. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-09-15T16:58:21Z IndexTAG: 2734 TitleTAG: home work 2 H2p1 can any one help me understanding homework 2 part one plz I put R1=18000 and calculated R2 through votage devider and it was R2=10000 but it is still wrong i can't understand what he really wants UserIdTAG: 206794 UserNameTAG: Abdofarrag CreateTimeTAG: 2012-09-15T10:32:17Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: > An additional requirement is that the Thevenin resistance as seen from the output terminals is between 10kΩ and 30kΩ. R2 = 10K which means that the thevinin resistance is lesser than 10K... tat is why it is wrong... :) FirstChildUserIdTAG: 257806 FirstChildUserNameTAG: skoda FirstChildCreateTimeTAG: 2012-09-15T14:21:58Z SecondChildTAG: how to calculate R1 SecondChildUserIdTAG: 197422 SecondChildUserNameTAG: sheryar SecondChildCreateTimeTAG: 2012-09-15T14:47:55Z SecondChildTAG: you can make calculates through voltage devider but i dont know how to calculate Rth SecondChildUserIdTAG: 206794 SecondChildUserNameTAG: Abdofarrag SecondChildCreateTimeTAG: 2012-09-16T01:49:14Z IndexTAG: 2735 TitleTAG: where is mine Older posts ???? please some 1 from Staff help me in finding my older posts thanks !!!! UserIdTAG: 415376 UserNameTAG: imab90 CreateTimeTAG: 2012-09-15T07:55:31Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You can get to your old posts by clicking on your username: [imab90][1]. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/users/415376 FirstChildUserIdTAG: 12663 FirstChildUserNameTAG: arjun FirstChildCreateTimeTAG: 2012-09-15T08:40:15Z IndexTAG: 2736 TitleTAG: How to follow posts? Is there anything which can let us know if our questions have been answerd or sombody replied to our comments or posts? UserIdTAG: 314624 UserNameTAG: Owais001 CreateTimeTAG: 2012-09-15T06:32:08Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: sort of notifications??? FirstChildUserIdTAG: 314624 FirstChildUserNameTAG: Owais001 FirstChildCreateTimeTAG: 2012-09-15T06:33:47Z FirstChildTAG: In the top right there is a star, click on it to follow. FirstChildUserIdTAG: 312035 FirstChildUserNameTAG: philhiggins FirstChildCreateTimeTAG: 2012-09-15T08:10:41Z SecondChildTAG: Thanks for the help. SecondChildUserIdTAG: 222234 SecondChildUserNameTAG: ivsprasad SecondChildCreateTimeTAG: 2012-09-15T08:22:33Z IndexTAG: 2737 TitleTAG: Is there a list showing which sections of the text book to read? It would be helpful for me if new how much of the text book to read before I watch the Lectures and attempt the problems. Is there a list showing which text book section are relevant to each week of the course? Thanks UserIdTAG: 147178 UserNameTAG: gtissington CreateTimeTAG: 2012-09-15T05:04:14Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: yes u can go to the course info & download 6.002x At-A-Glance & u will find what sections to read every weak under the column " Readings " FirstChildUserIdTAG: 138871 FirstChildUserNameTAG: corabict FirstChildCreateTimeTAG: 2012-09-15T05:58:57Z IndexTAG: 2738 TitleTAG: how to count the number of loops what is the exact technique to be followed while counting the number of loops? UserIdTAG: 401199 UserNameTAG: ashritha CreateTimeTAG: 2012-09-15T04:56:21Z VoteTAG: 1 CoursewareTAG: Week 1 / Circuit topology CommentableIdTAG: 6002x_S2E1_Circuit_Topology NumberOfReplyTAG: 1 FirstChildTAG: "Circuits **Loops** are defined to be closed paths along its branches" "The connections between the nodes are referred to as **branches** of a Circuit" "The junction points which at which the terminals at which the terminals of two or more elements are connected are referred as the **nodes** of a circuit " [More Info in the Textbook][1] Also you can watch [S2V3][2] and also [S2V4][3] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/78 [2]: https://www.youtube.com/watch?v=NAlQWlfiB_Y&feature=player_embedded [3]: https://www.youtube.com/watch?v=wlcwr1VIQ7Y&feature=player_embedded FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-15T05:10:01Z IndexTAG: 2739 TitleTAG: -ve power can -ve power be possible???? if yes den in what other cases and conditions?? UserIdTAG: 430318 UserNameTAG: abhi24 CreateTimeTAG: 2012-09-15T04:34:22Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 5_Simple_Power NumberOfReplyTAG: 2 FirstChildTAG: is power loss considered to be -ve here.... FirstChildUserIdTAG: 430318 FirstChildUserNameTAG: abhi24 FirstChildCreateTimeTAG: 2012-09-15T04:36:07Z SecondChildTAG: power is the rate at which **energy** is transferred, used, or transformed.............as we know that energy can be negative and positive........... SecondChildUserIdTAG: 123036 SecondChildUserNameTAG: ABHINAV_5 SecondChildCreateTimeTAG: 2012-09-15T10:10:14Z SecondChildTAG: negative energy is not posible conceptually but it is possible mathematically SecondChildUserIdTAG: 524754 SecondChildUserNameTAG: raj001 SecondChildCreateTimeTAG: 2012-10-01T13:04:14Z FirstChildTAG: Direction of power flow depends on how you are looking at it. To a device that is normally supplying power to a load, it would probably see that as positive power. If however you connected something to it that forced power to flow back into it's 'output' terminals, then it would probably measure that as negative power. FirstChildUserIdTAG: 270011 FirstChildUserNameTAG: SnowmanZA FirstChildCreateTimeTAG: 2012-09-15T07:14:17Z IndexTAG: 2740 TitleTAG: Lab 1 IRL: Smoke, no fire, everything fine as long as you're not a resistor So I just tried to replicate the Lab 1 circuit IRL thinking to myself: "What could possibly go wrong?" Turns out I burned a resistor within seconds. (I used a bunch of resistors I picked up from Sparkfun) I don't think I failed to replicate the correct circuit. I just think my resistor really didn't like what I was shoving through it. What I find interesting is that I didn't have the correct resistor for "R1", so I instead assembled R1 from 2 equal resistors in series. However, only the first resistor in the series burned out before I managed to kill the voltage source. I'm sort of wondering why that would happen. From what we've learned so far, the current and voltage would be identical in R1a and R1b. So wouldn't they both burn out simultaneously? Also, does anyone know what's the best way to figure out what the operating parameters of my resistors are? This was kind of fun once, but I'd rather not do it again. Those things are cheap, but not free. UserIdTAG: 107219 UserNameTAG: AlexandreZ CreateTimeTAG: 2012-09-15T04:19:36Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: > So wouldn't they both burn out simultaneously? In the real world, no two components are *exactly* alike. > Also, does anyone know what's the best way to figure out what the operating parameters of my resistors are? Common resistors are either 1/4 watt I think, ie. they can dissipate about 1/4 of a watt before letting out the magic smoke. Looking at Lab 1's circuit - yeah, they would have gone fast. FirstChildUserIdTAG: 312035 FirstChildUserNameTAG: philhiggins FirstChildCreateTimeTAG: 2012-09-15T08:14:15Z FirstChildTAG: If you can find what kind of resistor specifically you have, you can look up its spec sheet online. That should tell you the maximum amount of power it can dissipate. Most electronics retailers and the manufacturing company will have references to the spec sheets. Also - awesome post. :) FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-09-15T13:45:04Z IndexTAG: 2741 TitleTAG: Getting [Math Processing Error] While working the homework, I'm getting a [Math Processing Error], so can't determine what the question is asking. I've been using this same computer to these same problems before, but something has changed and it now does not work. I an error on the page says: Webpage error details User Agent: Mozilla/4.0 (compatible; MSIE 8.0; Windows NT 6.1; Trident/4.0; SLCC2; .NET CLR 2.0.50727; .NET CLR 3.5.30729; .NET CLR 3.0.30729; Media Center PC 6.0; .NET CLR 1.1.4322; .NET4.0C; .NET4.0E; ATT) Timestamp: Fri, 14 Sep 2012 21:50:32 UTC Message: 'HTMLCanvasElement' is undefined Line: 4252 Char: 2 Code: 0 URI: https://www.edx.org/static/js/lms-modules.9b7377108c8f.js Message: 'console' is undefined Line: 848 Char: 9 Code: 0 URI: https://www.edx.org/static/js/discussion.145352719a17.js Hope this helps in solving the problem. UserIdTAG: 29275 UserNameTAG: Mbarsalou CreateTimeTAG: 2012-09-14T21:51:31Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 2742 TitleTAG: lab 2 in lab two i would like to know how to get rms value of the sin wave i know that sin(2*pi*f*t) by root 2 is rms value but i cant figure out what t value should be also can i use rms value as fixed voltage values to represent them as v sources?? [enter link description here][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_2/Mixing_Two_Signals/ UserIdTAG: 375082 UserNameTAG: satya1889 CreateTimeTAG: 2012-09-14T21:12:13Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 2743 TitleTAG: S1E7 help ! Plz help about S1E7... how to measure i5 and i2? UserIdTAG: 152551 UserNameTAG: sshakir CreateTimeTAG: 2012-09-14T19:47:10Z VoteTAG: 1 CoursewareTAG: Week 1 / KVL and KCL exercises CommentableIdTAG: 6002x_kvl_kcl_exercises NumberOfReplyTAG: 2 FirstChildTAG: at first i did i1+i2+i3+i4=0 after putting the value the answer was right!! i mean -3.6 . but i think it's wrong. FirstChildUserIdTAG: 152551 FirstChildUserNameTAG: sshakir FirstChildCreateTimeTAG: 2012-09-14T19:52:05Z SecondChildTAG: Hello I am Colombian not speak good English. The correct answer is -3.6 due to the direction of the current We have i2 = -i1 -i5 = -(-0.7A) - 4.3A -3.6 A SecondChildUserIdTAG: 163395 SecondChildUserNameTAG: jasonlll88 SecondChildCreateTimeTAG: 2012-09-15T02:47:13Z FirstChildTAG: Hi sshakir! How are you? In S1E7 you have the violet node and the green node. Ok, let's use KCL! ![enter image description here][1] In KCL you will have that the sum of currents in a node it is zero. Use a convention of sign, eg., Go to node violet, you will see that i5 enters the node so it will rest in the equation. However, i3 and i4 gets out of the violet node, so they will sum. The same happens with node green, in this case, all the currents goes out, so all will sum in that node equation. Ok, Now, lets write the node equation of the violet node: -i5+i3+i4=0 i5=i3+i4 You know i4, i3 so you can get i5 Now, write the green node equation. It will be a equation expressed in i1,i2 and i5. I will let this to you, you will see that having i5 from previous equation and having i1, you can easily obtain i2 ;) . Try it! You can do it! Myriam. ----- Hola sshakir! Cómo estás? En el Ejercicio S1E7 verás que tendrás dos nodos. Puedes ver en el Circuito de abajo, te he marcado dos nodos: uno violeta y uno azul. Para hallar i2 e i5, debemos utilizar el KCL! ![enter image description here][1] Recordando qué era KCL: "La suma de todas las corrientes pertenecientes a un nodo es igual a cero", estamos listos para hallar i2 e i5. Para plantear las ecuaciones, debes utilizar una convención de signos, por ejemplo: Si miras el nodo violeta, verás que la corriente i5 ingresa al nodo (aquí utilizaremos la convención que dice que a las corrientes ingresantes se las considerará de signo negativo, es decir, restarán en la ecuación);en cambio, como las corrientes i4 e i3, salen del nodo, diremos que al ser corrientes salientes, sumarán en la ecuación. Puedes observar que en el nodo verde, las corrientes son todas salientes, es decir, todas sumarán en la ecuación del nodo. Bien, ahora te ayudaré con la primera ecuación de nodo (violeta): -i5+i3+i4=0 i5=i3+i4 Al tener i3 e i4 como datos, será posible obtener i5 :). Ahora escribe la ecuación del nodo verde. Será una ecuación expresada en función de i1, i2 e i5. Te dejaré esto para ti, verás que teniendo i5 de la ecuación del nodo violeta, podrás fácilmente obtener el valor de i2 ;). Inténtalo! Tú puedes! Myriam. [1]: http://s16.postimage.org/ujvu04y7p/nodos.png FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-14T20:47:02Z SecondChildTAG: Thanks for your nice explanation Myriam. I was just confused about sum of currents. I thought sum of currents in a circuit is zero. But it will be the sum of currents in a node is zero. right ? But one think i didn't understand. In another discussion it was said that when current enters in a node then it's positive and if it leaves a node then it's negative. But in your equation you wrote i5+i3+i4=0 which measn i5 enters a node and it's positive. I'm just bit confused ! SecondChildUserIdTAG: 152551 SecondChildUserNameTAG: sshakir SecondChildCreateTimeTAG: 2012-09-15T04:50:33Z SecondChildTAG: Hi sshakir! For the convention of signs, you can use any convention, that is to say: **Convention 1:** Negative when enter to a node; Positive when you get out a node. **Convention 2:** Positive when you enter a node; Negative when you get out a node. You can use any of this two conventions, but, you have to use one while you are solving a problem, you can not mix the two conventions at the same time... So , both are valids. You can choose any of them. The Final result will be the same ;). SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-15T05:50:07Z IndexTAG: 2744 TitleTAG: h1 last part for people who are asking about h1 last part, i think these concepts might be helpful: P=V^2/R (play with the formula to get you R) another formula is P=I^2*R parallel resistors (same voltage across each of them ) series resistors (same current passing through them) just remember these and everything will be fine UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-14T19:34:48Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: The power dissipated should effectively be the sum of addition of power dissipated in each individual elements irrespective of circuit configuration. So adding power dissipated till burnout will be equal to 3. But when I submit it shows answer as wrong. FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-14T22:23:04Z FirstChildTAG: Also the smallest valued composite resistor will be a parallel connection. FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-14T22:24:26Z IndexTAG: 2745 TitleTAG: enquiry about the basic approach hello, i just wanted to know if my approach to the problem was right.what i did was i considered the flow of current to be from the voltage source to the resistor to be positive, in that case i1 and i4 become negative and that is how i solved the problem, that way even the power P1 has a negative sign..i just wanted to make sure if i got the concept right..please help UserIdTAG: 401199 UserNameTAG: ashritha CreateTimeTAG: 2012-09-14T17:17:54Z VoteTAG: 1 CoursewareTAG: Week 1 / Associated reference directions CommentableIdTAG: 6002x_S2E2_Associated_Reference_Directions NumberOfReplyTAG: 1 FirstChildTAG: yep...you got the concept right ... if power has negative sign that implies that the circuit is supplying power to the source ,i.e the source is consuming power! FirstChildUserIdTAG: 367639 FirstChildUserNameTAG: paikartik FirstChildCreateTimeTAG: 2012-09-14T17:22:51Z SecondChildTAG: thanks a ton! SecondChildUserIdTAG: 401199 SecondChildUserNameTAG: ashritha SecondChildCreateTimeTAG: 2012-09-15T04:57:03Z SecondChildTAG: power is entering/giving to the source - is negative power is dissipated/consuming by the source - is positive SecondChildUserIdTAG: 572704 SecondChildUserNameTAG: mitali1994 SecondChildCreateTimeTAG: 2012-10-21T04:31:16Z IndexTAG: 2746 TitleTAG: linearity I always thought that when the I-V characteristic is a straight line then the component is linear no mater if the line's slope is zero negative or positive and no matter whether it intersects point (0,0). I also have heard that systems are considered linear when they are described by linear differential equations. In that essence we are interested in "incremental" resistance dV/dI. Am I totally wrong? What do you think? UserIdTAG: 319453 UserNameTAG: leonidasGr CreateTimeTAG: 2012-09-14T16:47:55Z VoteTAG: 1 CoursewareTAG: Week 2 / Is it linear? CommentableIdTAG: 6002x_is_it_linear NumberOfReplyTAG: 1 FirstChildTAG: You're on the right track. The reason why we're interested in linearity is because linear systems have two very nice characteristics: 1. They obey superposition. You can solve for each portion separately and combine them together. 2. They can be scaled up or down. You can multiply all voltages and currents by 10, and everything should still be conserved. Anything that satisfies these two criteria is a linear device. So generally resistors and capacitors are very linear. Inductors can be fairly linear within some operating range. Diodes and transistors are non-linear, although incremental models can be constructed to "linearize" them within some operating range. Digital circuits are generally non-linear. FirstChildUserIdTAG: 151427 FirstChildUserNameTAG: RichardZ FirstChildCreateTimeTAG: 2012-09-14T17:15:07Z SecondChildTAG: Thanks for adding this explanation. Without it, linearity means different things to different people. SecondChildUserIdTAG: 12905 SecondChildUserNameTAG: Baer SecondChildCreateTimeTAG: 2012-09-15T00:00:40Z SecondChildTAG: additivity+homgeneity=linearity. SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-17T16:13:29Z IndexTAG: 2747 TitleTAG: Help . .Regading to Homework and exam To Staff Please Clear it . . and i would also like to ask the same thing from edx or MITX officials. Kindly Tell me that .. Plzz plzzz Question : In how much time i have to complete the Homework and Lab work. If some of my answer went wrong will i be allowed to submit another answer?? Question : Same question about Midterm Exam and Final Exam. . ? UserIdTAG: 146930 UserNameTAG: fayzan_007 CreateTimeTAG: 2012-09-14T14:04:34Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: As far as I know,you can submit your homework as many times as you would like.It seems that the exam submission have a limitation . FirstChildUserIdTAG: 136959 FirstChildUserNameTAG: christerpher FirstChildCreateTimeTAG: 2012-09-14T14:16:22Z SecondChildTAG: What about the time limit for the homework. if i started the homework or lab homework by submitting only one answer do i have to complete all answers within 24 hours?? SecondChildUserIdTAG: 146930 SecondChildUserNameTAG: fayzan_007 SecondChildCreateTimeTAG: 2012-09-14T19:08:40Z SecondChildTAG: Thank you soo much christerpher SecondChildUserIdTAG: 146930 SecondChildUserNameTAG: fayzan_007 SecondChildCreateTimeTAG: 2012-09-15T05:37:30Z SecondChildTAG: you're welcome SecondChildUserIdTAG: 136959 SecondChildUserNameTAG: christerpher SecondChildCreateTimeTAG: 2012-09-26T10:02:29Z FirstChildTAG: in homework and labs you can submit as many times you like before deadlines. And in midterm and final you will have 3 chances!!! FirstChildUserIdTAG: 385662 FirstChildUserNameTAG: droidme FirstChildCreateTimeTAG: 2012-09-14T14:37:20Z SecondChildTAG: have you submitted homework ? that's why you know this SecondChildUserIdTAG: 146930 SecondChildUserNameTAG: fayzan_007 SecondChildCreateTimeTAG: 2012-09-14T18:54:10Z IndexTAG: 2748 TitleTAG: Re: Exactly 2.5 V, threshold The fact is, that the value for a false should be _lower_ and for a true it should be _higher_ than the threshold. Equality is not allowed, so in the given example the 2.5 Volt is an _illegal_ value. UserIdTAG: 151942 UserNameTAG: GyuriK CreateTimeTAG: 2012-09-14T12:35:29Z VoteTAG: 1 CoursewareTAG: Week 2 / Static Discipline CommentableIdTAG: 6002x_static_discipline NumberOfReplyTAG: 0 IndexTAG: 2749 TitleTAG: meaninig of [Clean] [Annotated] can any one please tell me what does this two terms means? [Clean] [Annotated] UserIdTAG: 360137 UserNameTAG: nivish CreateTimeTAG: 2012-09-14T11:09:29Z VoteTAG: 1 CoursewareTAG: Week 1 / Course overview CommentableIdTAG: 6002x_S1V1toS1V4 NumberOfReplyTAG: 1 FirstChildTAG: I guess it's about the instructor's slides shown in the presentations. If you press 'Clean' you can see them without the instructor's notes on them, and if you press 'annotated' you can see them with notes. FirstChildUserIdTAG: 304587 FirstChildUserNameTAG: Vasso FirstChildCreateTimeTAG: 2012-09-14T11:58:11Z SecondChildTAG: Yes, that is true ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-14T12:00:49Z IndexTAG: 2750 TitleTAG: What is the power (in Watts) supplied by the voltage source? In this question: What is the power (in Watts) supplied by the voltage source? The value should be positive, as the voltage source is consuming energy, and not providing power. So the power should be positive. Also, if you sum up the watts with the solutions provided, the result is 40W and it should be 0W. Regards! UserIdTAG: 65274 UserNameTAG: ratonxi CreateTimeTAG: 2012-09-14T09:50:47Z VoteTAG: 1 CoursewareTAG: Week 1 / KVL/KCL example CommentableIdTAG: 6002x_S2E3_Using_KVL_KCL_and_VI_constraints NumberOfReplyTAG: 1 FirstChildTAG: Your assumption is incorrect. If the voltage source is consuming power, it's power should be negative. FirstChildUserIdTAG: 270011 FirstChildUserNameTAG: SnowmanZA FirstChildCreateTimeTAG: 2012-09-14T17:28:35Z SecondChildTAG: but if we take power consumed to be negative, then power dissipated by resistor should also be negative. SecondChildUserIdTAG: 168180 SecondChildUserNameTAG: deep20jain SecondChildCreateTimeTAG: 2012-09-16T05:23:44Z IndexTAG: 2751 TitleTAG: lab 1 please i found it difficult to edit my resistor during my lab work if click it, it just pop up but can't change anything what do i do. UserIdTAG: 95877 UserNameTAG: iscofield CreateTimeTAG: 2012-09-14T04:45:07Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 2752 TitleTAG: Maxwell's Equations Will it be necessary for us to use or apply Maxwell's Equations (on assignments or tests) in this course, or is the professor simply using them to demonstrate how we arrive at simplified abstractions so we may use those abstractions instead? UserIdTAG: 405370 UserNameTAG: dPiccolo CreateTimeTAG: 2012-09-14T03:33:59Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: I believe intends the latter. FirstChildUserIdTAG: 310007 FirstChildUserNameTAG: ErinF FirstChildCreateTimeTAG: 2012-09-14T04:41:58Z FirstChildTAG: Hi dPicollo, the treatment of Maxwell's equations in 6.002x is simply to give (cursorily) the physics background that underlies circuit analysis. We will not be revisiting them in assignments or tests in 6.002x. FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-14T14:54:55Z IndexTAG: 2753 TitleTAG: KCL (e1+6)/3 + (e1-e2)/2 - 3 = 0. This all makes perfect sense... but why the heck is e1-e2 divided by 2? where did the 2 come from? I assumed that the first component being divided by 3 was due to the 3amp current source... is it in fact due to the resistor? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-14T03:02:13Z VoteTAG: 1 CoursewareTAG: Week 1 / Nodal analysis example CommentableIdTAG: 6002x_NodalAnalysisExample NumberOfReplyTAG: 1 FirstChildTAG: The divide by 3 is indeed the resistor. The - 3 at the end is the current source. The divide by 2 is the other resistor. FirstChildUserIdTAG: 312035 FirstChildUserNameTAG: philhiggins FirstChildCreateTimeTAG: 2012-09-14T03:53:11Z IndexTAG: 2754 TitleTAG: Technical problem with textbook ! The new way of turning the pages of the book is very annoying when reading. Please change it or make the book available for download as a pdf. Also turning the pages of the book is too slow which can be frustrating at times. UserIdTAG: 154541 UserNameTAG: AhmedMedhat CreateTimeTAG: 2012-09-13T19:03:25Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: This way its free. If you want a pdf of a book...your gonna have to buy it. (Which you can...) FirstChildUserIdTAG: 25084 FirstChildUserNameTAG: waf102 FirstChildCreateTimeTAG: 2012-09-13T19:45:31Z FirstChildTAG: One of the students in the Spring 2012 version of the course wrote a scrolling textbook viewer for 6.002x: https://6002x.mitx.mit.edu/static/contrib/xbook.html We'll try to make it at least as good :) FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-13T20:50:55Z IndexTAG: 2755 TitleTAG: unable to open this video s1v8 why i'm unable to play this video UserIdTAG: 374805 UserNameTAG: LAKSHMIKANTH3900 CreateTimeTAG: 2012-09-13T17:35:05Z VoteTAG: 1 CoursewareTAG: Week 1 / Lumped Element Abstraction CommentableIdTAG: 6002x_lumped_element_abstraction NumberOfReplyTAG: 1 FirstChildTAG: Hello I'm having problems loading the videos. Even videos I've watched are no longer available to be play. FirstChildUserIdTAG: 377998 FirstChildUserNameTAG: AlineFerreira FirstChildCreateTimeTAG: 2012-09-13T18:33:34Z SecondChildTAG: Please use google chrome browser thanks SecondChildUserIdTAG: 382532 SecondChildUserNameTAG: emmanuelpeace SecondChildCreateTimeTAG: 2012-09-16T10:59:58Z IndexTAG: 2756 TitleTAG: H2P2 : Solar Power woes. I don't think I am understanding question 2 correctly. -"What is the power (in Watts) that is delivered to this best load resistance?" I do know the optimum load resistance and the Thevenin equivalent resistance of the source as seen by the load resistance. (Questions 1 and 3) I can't seem to figure this one out, any help or hint would be appreciated, I keep coming up with the same 3 or 4 wrong answers. Thanks! UserIdTAG: 292543 UserNameTAG: Pennypacker CreateTimeTAG: 2012-09-13T15:09:11Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Hmm, if you have 1 and 3, the answer should be straightforward. You know the value of the best resistor (maximum power transfered) and the Thevenin equivalent of the system so you can find the voltage across this resistance (or the current) and calculate the power. FirstChildUserIdTAG: 396446 FirstChildUserNameTAG: RousseauxS FirstChildCreateTimeTAG: 2012-09-13T15:33:02Z SecondChildTAG: I agree it should be straightforward. I know the Thevenin equivalent as seen by the source, maybe I need to be using the Thevenin equivalent of the complete system? SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-13T15:39:17Z SecondChildTAG: you should be able to get your thevenin equivalent ckt. Then afterwards you connect in series the RL you computed with the VTH and RTH. Use ohm's law to get the current passing through RL. The last step is use the I^2*RL power formula to get the answer. SecondChildUserIdTAG: 371000 SecondChildUserNameTAG: Joseph090892 SecondChildCreateTimeTAG: 2012-09-13T16:36:58Z SecondChildTAG: Hi, I dont know the Max. Power formula, please help me. SecondChildUserIdTAG: 171674 SecondChildUserNameTAG: wajahat1 SecondChildCreateTimeTAG: 2012-09-15T18:40:21Z SecondChildTAG: @RousseauxS How can I figure out the VTH of the Thevenin equivalent???? SecondChildUserIdTAG: 296511 SecondChildUserNameTAG: sky0917 SecondChildCreateTimeTAG: 2012-09-21T16:20:50Z SecondChildTAG: ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13482446227525379.png so I just know the I,Rth and RL .How can I work out the question-'What is the power (in Watts) that is delivered to this best load resistance?' I did this: P=(I/2)^2*RL but it is wrong!! SecondChildUserIdTAG: 296511 SecondChildUserNameTAG: sky0917 SecondChildCreateTimeTAG: 2012-09-21T16:29:02Z IndexTAG: 2757 TitleTAG: Transcript on the right side of video disappeard - Youtube issue? Hello, Could someone explain how to recover transcript on the right side of video? Is it possible? UserIdTAG: 412458 UserNameTAG: maar CreateTimeTAG: 2012-09-13T08:21:49Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: YouTube has changed things and broken it; they're trying to fix it. See https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/MITx_6_002x_2012_Fall/threads/5051442739934a2b0000001a FirstChildUserIdTAG: 312035 FirstChildUserNameTAG: philhiggins FirstChildCreateTimeTAG: 2012-09-13T08:32:13Z SecondChildTAG: Thanks! SecondChildUserIdTAG: 412458 SecondChildUserNameTAG: maar SecondChildCreateTimeTAG: 2012-09-13T08:49:06Z IndexTAG: 2758 TitleTAG: BUG IN LECTURE VIDEO IS THERE IS A BUG IN THE VIDEO LECTURER THERE IS NO BUTTON IT IS NOW JUST LIKE THE YOUTUBE FORMULA CAN FIX IT, UserIdTAG: 266386 UserNameTAG: zakzak200 CreateTimeTAG: 2012-09-13T08:03:51Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Please see [Dave's (edX developer) post][1]. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/MITx_6_002x_2012_Fall/threads/5051442739934a2b0000001a FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-13T12:45:16Z IndexTAG: 2759 TitleTAG: What is the prove for the negative - 2 Please can we prove the negative -2 since by logic the device is loosing power. should be equal to the power in dissipated in the resistor so I think mathematical 2 should be acceptable but minus -2 should just be noted that power is consumed.hence -2 or 2 should be acceptable... Please I stand for correction UserIdTAG: 382532 UserNameTAG: emmanuelpeace CreateTimeTAG: 2012-09-13T07:29:29Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 5_Simple_Power NumberOfReplyTAG: 1 FirstChildTAG: If we reverse the direction of power(or current or voltage etc), then the sign of the value of the power(or current or voltage etc) reverses. FirstChildUserIdTAG: 347274 FirstChildUserNameTAG: Mahesh87 FirstChildCreateTimeTAG: 2012-09-14T18:43:11Z IndexTAG: 2760 TitleTAG: h2p1 hi guys! I have been answering this question for a long time already. Can someone please explain to me what does this part of the question mean? "Assume first that the resistors have their nominal resistance. Come up with resistors R1 and R2 such that the divider ratio Vout/Vin is within 10% of the requirement." Thanks in advance! UserIdTAG: 371000 UserNameTAG: Joseph090892 CreateTimeTAG: 2012-09-13T03:08:08Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: The question is telling you how to proceed. When is says to assume they have their nominal resistance, it means to ignore that the resistance values may vary by ±10%. FirstChildUserIdTAG: 248902 FirstChildUserNameTAG: Chanute FirstChildCreateTimeTAG: 2012-09-13T04:09:37Z SecondChildTAG: I understood the nominal resistance part but what i can't really comprehend is the one that involves the "divider ratio Vout/Vin is within 10% of the requirement". Thanks for the response! SecondChildUserIdTAG: 371000 SecondChildUserNameTAG: Joseph090892 SecondChildCreateTimeTAG: 2012-09-13T04:20:00Z IndexTAG: 2761 TitleTAG: lab 2 Im trying to solve lab2. My idea is to solve applying superposition method and i obtain that Vout=V1*r2/(r1+r2)+v2*r1/(r1+r2). Solving that for the numerical values i get 5r1=r2 and r1=r2. Then when im running a transient analysis, assigning a random value to let's say r1, i dont get the right answer. What am i doing wrong? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-13T02:07:38Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 6 FirstChildTAG: Don't forget that this is an exercise in superposition. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-09-15T23:53:42Z FirstChildTAG: "5r1=r2 and r1=r2" You need to find a way of changing the circuit so both of those can be true at the same time. FirstChildUserIdTAG: 312035 FirstChildUserNameTAG: philhiggins FirstChildCreateTimeTAG: 2012-09-13T02:46:39Z SecondChildTAG: you are right. It was impossible to do it this way, as the sum of voltage coeff of the exercise is less than 1, so one needs to modify the 2 scheme resistors. Thanks! SecondChildUserIdTAG: 146446 SecondChildUserNameTAG: adrianp SecondChildCreateTimeTAG: 2012-09-13T03:03:29Z FirstChildTAG: Hi, What values are you substituting for r1 and r2 ? You have to find values for r1 and r2 which satisfy not only one, but both the equations. Id est the point where both the equations converge. The only possible solution to these two equations is zero, so I suggest you to recheck your circuit design. Hope this helps! FirstChildUserIdTAG: 330269 FirstChildUserNameTAG: Jaikr FirstChildCreateTimeTAG: 2012-09-13T03:06:57Z SecondChildTAG: consider the two voltage as two. for v1 fully neglect v2 part(like superposition).then make it to give value of v1/2 by adding appropriate resistors(by voltage divider rule). now go to v2, make it to give v2/6. by solving these two as separate, then make connection between them. observe the output graph. again adjust the resistors for small change in value. remember you will get answer in between 3ohm to 1ohm range including 3 and 1. SecondChildUserIdTAG: 378967 SecondChildUserNameTAG: krish2012 SecondChildCreateTimeTAG: 2012-09-19T19:00:36Z FirstChildTAG: Your equation is correct. But I dont think that you get right answer using just **two** resistors. Try to build a bit more complicated circuit than given as example on fig.2 FirstChildUserIdTAG: 258500 FirstChildUserNameTAG: karas FirstChildCreateTimeTAG: 2012-09-13T05:26:59Z SecondChildTAG: That worked for me. I originally thought that we could only build the example in Figure 2. Once I realized I could add to, but not modify the Figure 2 example, I was able to get the correct answer. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-13T13:47:45Z FirstChildTAG: Hi Pennypack, I tried to put an extra resistor. The only choice could be to shunt a resistor at Vout. But when I do the calculation, it still does not work. any hint? Thanks! ![enter image description here][1] ![enter image description here][2] [1]: http://www.flickr.com/photos/87235783@N05/7989889813/in/photostream [2]: http://www.flickr.com/photos/87235783@N05/7989889813/in/photostream/lightbox/ FirstChildUserIdTAG: 282748 FirstChildUserNameTAG: larryzhu FirstChildCreateTimeTAG: 2012-09-15T22:28:43Z SecondChildTAG: Keep trying, make sure you have the proportions of the other two resistors right. 1/2 of one, 1/6 of the other, which could be viewed another way. ;) SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-15T22:46:54Z SecondChildTAG: ![enter image description here][1] [1]: http://www.flickr.com/photos/87235783@N05/7989889813/in/photostream/lightbox/ SecondChildUserIdTAG: 282748 SecondChildUserNameTAG: larryzhu SecondChildCreateTimeTAG: 2012-09-15T23:00:01Z FirstChildTAG: Finally I find what mistake I made.I should first "save" and then click "check". Otherwise, it always fails. FirstChildUserIdTAG: 282748 FirstChildUserNameTAG: larryzhu FirstChildCreateTimeTAG: 2012-09-16T03:37:23Z IndexTAG: 2762 TitleTAG: Video Lecture videos are stopping after a few minutes. I can't even review the ones that I could previously without this problem. UserIdTAG: 265795 UserNameTAG: krishnakm CreateTimeTAG: 2012-09-13T01:29:45Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: YouTube made a recent change to their Flash player that broke our video embeds. As a temporary workaround while we figure things out, you can use YouTube's HTML5 player. To do so, please go to YouTube's [Join the HTML5 Trial][1] page and click "Join the HTML5 Trial" at the bottom. Then reload our site in your browser. Thank you for your patience as we work on this. FirstChildUserIdTAG: 11 FirstChildUserNameTAG: dormsbee FirstChildCreateTimeTAG: 2012-09-13T02:26:32Z IndexTAG: 2763 TitleTAG: videos stop after 50" Hello everybody From today, all the videos stop after approximately 50s . youtube bug ? or something else ? ( I'm in France ) UserIdTAG: 373752 UserNameTAG: Croak CreateTimeTAG: 2012-09-13T01:21:32Z VoteTAG: 1 CoursewareTAG: Week 3 / Weeks 1-3 Review CommentableIdTAG: 6002x_weeks_1_3_review NumberOfReplyTAG: 3 FirstChildTAG: YouTube made a recent change to their Flash player that broke our video embeds. As a temporary workaround while we figure things out, you can use YouTube's HTML5 player. To do so, please go to YouTube's [Join the HTML5 Trial][1] page and click "Join the HTML5 Trial" at the bottom. Then reload our site in your browser. Thank you for your patience as we work on this. FirstChildUserIdTAG: 11 FirstChildUserNameTAG: dormsbee FirstChildCreateTimeTAG: 2012-09-13T02:26:59Z FirstChildTAG: Please see [Dave's (edX developer) post][1]. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/MITx_6_002x_2012_Fall/threads/5051442739934a2b0000001a FirstChildUserIdTAG: 208296 FirstChildUserNameTAG: OZ1 FirstChildCreateTimeTAG: 2012-09-13T19:51:46Z FirstChildTAG: Which part of France? FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-10-22T16:24:57Z IndexTAG: 2764 TitleTAG: BUG This video CONSTANTLY stops without any reason and in order to continue watching the video I have to start it all over again because when I press play nothing happens. I have to take the purple circle of the time line back to the start, I'm getting mad with that. My system is windows 7 ultimate 64 bits (SP1), and my browser is mozilla firefox 15.0.1 HELP! UserIdTAG: 181432 UserNameTAG: enriqueferreralcala CreateTimeTAG: 2012-09-12T22:25:43Z VoteTAG: 1 CoursewareTAG: Week 1 / Nodal analysis with floating voltage CommentableIdTAG: 6002x_FloatingVoltage NumberOfReplyTAG: 5 FirstChildTAG: My video is doing this as well. Freezes first 1:09 into lecture (happened on several lectures), then at shorter intervals throughout. Text seems to stop scrolling just before this happens. Running Firefox (latest) and Vista 64-bit. Updated all the addons/plugins, same. Tried it on IE, same symptoms. Video from other sources streams ok. FirstChildUserIdTAG: 21592 FirstChildUserNameTAG: wallingjl FirstChildCreateTimeTAG: 2012-09-12T22:53:01Z FirstChildTAG: For now, watch it on youtube instead (click on bottom right corner of the video). FirstChildUserIdTAG: 27851 FirstChildUserNameTAG: Arman FirstChildCreateTimeTAG: 2012-09-13T00:05:54Z FirstChildTAG: YouTube made a recent change to their Flash player that broke our video embeds. As a temporary workaround while we figure things out, you can use YouTube's HTML5 player. To do so, please go to YouTube's [Join the HTML5 Trial][1] page and click "Join the HTML5 Trial" at the bottom. Then reload our site in your browser. Thank you for your patience as we work on this. FirstChildUserIdTAG: 11 FirstChildUserNameTAG: dormsbee FirstChildCreateTimeTAG: 2012-09-13T02:27:57Z FirstChildTAG: I am using windows XP and firefox 15.0.1 it is working kindly clear the history and tem internet file and check FirstChildUserIdTAG: 156694 FirstChildUserNameTAG: mkprasanth FirstChildCreateTimeTAG: 2012-09-15T02:06:00Z FirstChildTAG: Thanks for the bug report. We are working on it now. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-09-13T01:18:35Z IndexTAG: 2765 TitleTAG: [BUG] wk 1 Tutorials circuit simulater through e-text Wk. 1 e-text through A/c analysis all stopping short in videos. I think this is new. please advise me. Thanks UserIdTAG: 359302 UserNameTAG: GaryG CreateTimeTAG: 2012-09-12T21:50:41Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Please see [this thread][1] for a quick workaround. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/MITx_6_002x_2012_Fall/threads/5051442739934a2b0000001a FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-13T02:39:10Z IndexTAG: 2766 TitleTAG: Missing something basic I'm trying to calculate x1, but I'm missing something basic. I got .4, but that shows as incorrect. What am I missing? UserIdTAG: 29275 UserNameTAG: Mbarsalou CreateTimeTAG: 2012-09-12T21:32:37Z VoteTAG: 1 CoursewareTAG: Week 2 / Superposition example CommentableIdTAG: 6002x_S3E3_superposition NumberOfReplyTAG: 1 FirstChildTAG: Too much rounding! From trial and error I've been able to determine that the answers will be accepted as correct if they're ±5% of the exact answer. FirstChildUserIdTAG: 248902 FirstChildUserNameTAG: Chanute FirstChildCreateTimeTAG: 2012-09-13T00:36:37Z IndexTAG: 2767 TitleTAG: Confused with signs Can somebody explain me why in previous videos, he used for i1, e1-V0, and in this video, he uses the opposite, V0-e1... it isn't the first time that it gets me confused... Thanks UserIdTAG: 265133 UserNameTAG: Elienai CreateTimeTAG: 2012-09-12T21:20:30Z VoteTAG: 1 CoursewareTAG: Week 1 / Node analysis CommentableIdTAG: 6002x_node_analysis NumberOfReplyTAG: 3 FirstChildTAG: It's dependent on which node is the "node of interest" for which he's writing the equations. The convention he's using is to always sum the currents *leaving* the node. The current leaving the node through any branch can be expressed by: [(voltage at node of interest)-(voltage at the node opposite the branch)]/(resistance of that branch) Any currents *entering* the node would end up being expressed as negative (since the negative signs indicate they're going in the direction opposite whatever you've defined them to be) and any currents *leaving* the node would end up being positive. FirstChildUserIdTAG: 248902 FirstChildUserNameTAG: Chanute FirstChildCreateTimeTAG: 2012-09-13T00:47:31Z SecondChildTAG: > `![Blockquote][1]` [1]: https://edxuploads.s3.amazonaws.com/13478076549390567.jpg SecondChildUserIdTAG: 254325 SecondChildUserNameTAG: bondablack SecondChildCreateTimeTAG: 2012-09-16T15:01:05Z SecondChildTAG: **ha ha ha ha haa h ** SecondChildUserIdTAG: 254325 SecondChildUserNameTAG: bondablack SecondChildCreateTimeTAG: 2012-09-16T15:01:28Z FirstChildTAG: Dear sir I think nothing need to confuse just how we practically measuring in the real life FirstChildUserIdTAG: 156694 FirstChildUserNameTAG: mkprasanth FirstChildCreateTimeTAG: 2012-09-13T05:12:22Z FirstChildTAG: Thank you too much Chanute! FirstChildUserIdTAG: 265133 FirstChildUserNameTAG: Elienai FirstChildCreateTimeTAG: 2012-09-13T16:12:25Z IndexTAG: 2768 TitleTAG: Scrolling text freezes. I was just watching some Week 2 Tutorials,(the first three so far), the scrolling text has been freezing between 57 and 59 seconds into the video. Windows 7, Firefox 15.0.1 UserIdTAG: 292543 UserNameTAG: Pennypacker CreateTimeTAG: 2012-09-12T21:19:45Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Please see [this thread][1] for a quick workaround. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/MITx_6_002x_2012_Fall/threads/5051442739934a2b0000001a FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-13T02:39:26Z IndexTAG: 2769 TitleTAG: Small and big letters when we use lower or bigger case letters for v, i r? UserIdTAG: 408796 UserNameTAG: GrzegorzPietrzak CreateTimeTAG: 2012-09-12T20:14:39Z VoteTAG: 1 CoursewareTAG: Week 1 / Lumped Element Abstraction CommentableIdTAG: 6002x_lumped_element_abstraction NumberOfReplyTAG: 3 FirstChildTAG: From Agarwal et Lang: "The variables v and i (lowercase) are called the terminal variables for the element" (1.5.3) Use uppercase letters to refer to the parameters of the element. So, R for the resistor or V for the voltage source. FirstChildUserIdTAG: 178447 FirstChildUserNameTAG: mmart FirstChildCreateTimeTAG: 2012-09-12T20:43:39Z FirstChildTAG: BIG **V** is a property of voltage source Or you can say it is component value So if it's a battery, it might supply a voltage **V** of 1.5 volts.while small v and i are the branch variables.it means when we put that voltage source into a circuit,it's going to draw some current i and there will be some voltage across it,this current and voltage is branch current and voltage and denoted by small **v** and small **i**.we have not used small r yet in the course we denote resistor with letter R. FirstChildUserIdTAG: 40209 FirstChildUserNameTAG: sadaf78 FirstChildCreateTimeTAG: 2012-09-12T20:59:00Z FirstChildTAG: Apart this material, big letter(U,I) we use for constant signals in time domain and small letter(u(t),i(t)) we use for variable signals. FirstChildUserIdTAG: 406360 FirstChildUserNameTAG: TN_PL FirstChildCreateTimeTAG: 2012-09-12T21:56:38Z IndexTAG: 2770 TitleTAG: important explanation needed please wht i know about loop analysis is that : this circuit consists of 7 loops , but we should not use them all because we choos a specific number of loops equals ( b-n+1 ) not all the 7 loops . and we have ( b = 6 ) and ( n = 4 ) . so i think that we should have just 3 equations not 7 or 4 because ( 6 - 4 +1 = 3 ) . UserIdTAG: 293070 UserNameTAG: Makary CreateTimeTAG: 2012-09-12T16:19:28Z VoteTAG: 1 CoursewareTAG: Week 1 / KVL and KCL CommentableIdTAG: 6002x_kvl_and_kcl NumberOfReplyTAG: 1 FirstChildTAG: only 3 of the equations are of interest to us. the others come up as some linear combinations of the 3 and as a result are redundant. FirstChildUserIdTAG: 152516 FirstChildUserNameTAG: shuvo915 FirstChildCreateTimeTAG: 2012-09-12T16:34:05Z IndexTAG: 2771 TitleTAG: difference b/w lecture sequence and week tutorials What is difference b/w lecture sequence and week tutorials? UserIdTAG: 332159 UserNameTAG: muhammadtahir CreateTimeTAG: 2012-09-12T15:56:31Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Lecture sequences are the primary way for essential teaching material to be communicated to you students. It's basically your "class." Tutorials can be thought of as "recitations-" They either reinforce the critical material taught in lectures or give you additional practice or insight on material related to what you've learned. FirstChildUserIdTAG: 390034 FirstChildUserNameTAG: kahlil FirstChildCreateTimeTAG: 2012-09-12T16:04:40Z IndexTAG: 2772 TitleTAG: Delta shape Why the delta shaped circuit is not valid?And how did he calculate the number of combinations? UserIdTAG: 373262 UserNameTAG: nwtn_pro CreateTimeTAG: 2012-09-12T15:18:11Z VoteTAG: 1 CoursewareTAG: Week 1 / Black-Box Resistor Network CommentableIdTAG: 6002x_black_box_resistor_network NumberOfReplyTAG: 1 FirstChildTAG: It should be easy to see why the delta shaped circuit isn't valid- what is the resistance between A and B? There's a direct path in between and only one resistor in the way, so it's just one ohm. That doesn't satisfy the very first constraint. And when he was talking about the number of combinations, he just was referring to the number of ways you could combine either one or two resistors. You could just have one resistor alone, or connect two in series or parallel- either 1 ohm, 2 ohms, or half and ohm. FirstChildUserIdTAG: 390034 FirstChildUserNameTAG: kahlil FirstChildCreateTimeTAG: 2012-09-12T15:58:01Z SecondChildTAG: how did he know that it was a y network? what if i use his 3 resistance theory, calculate the three unknown resistance and find that my constraints don't match? i don't think thats a correct way to approach this problem. its sheer hit and trial.. SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-17T14:11:52Z IndexTAG: 2773 TitleTAG: How? How are you expected to solve this? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-12T15:06:41Z VoteTAG: 1 CoursewareTAG: Week 1 / KVL/KCL example CommentableIdTAG: 6002x_S2E3_Using_KVL_KCL_and_VI_constraints NumberOfReplyTAG: 1 FirstChildTAG: You should start with writing down the KVL loop equations and KCL node equations of the circuit. There are four unknowns- v1, v2, i1, and i2. With your equations along with the v = ir relations for each element you should be able to get the answer. FirstChildUserIdTAG: 390034 FirstChildUserNameTAG: kahlil FirstChildCreateTimeTAG: 2012-09-12T15:55:28Z IndexTAG: 2774 TitleTAG: S3E5 result is different with answer Hi is there any one explain for me, why i calculate Vth = 2.62262 V is still correct when i push check button?? because the answer is 2.574V UserIdTAG: 114913 UserNameTAG: ngoctuan CreateTimeTAG: 2012-09-12T11:32:12Z VoteTAG: 1 CoursewareTAG: Week 2 / Thevenin model CommentableIdTAG: 6002x_S3E5_Thevenin_Model NumberOfReplyTAG: 5 FirstChildTAG: Did you forget about some resistors? ) FirstChildUserIdTAG: 394836 FirstChildUserNameTAG: v2g6ch4 FirstChildCreateTimeTAG: 2012-09-12T13:48:21Z FirstChildTAG: Usually the answers have some margin to account for rounding errors. However, in this case, with the given numbers getting the answer Vth = 2.574 should be direct, so you shouldn't have any rounding errors. How are you getting the Vth = 2.62262? FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-12T13:09:30Z SecondChildTAG: He's forget about resistor's.... SecondChildUserIdTAG: 394836 SecondChildUserNameTAG: v2g6ch4 SecondChildCreateTimeTAG: 2012-09-12T14:12:41Z SecondChildTAG: I guess that he think that Vth=I*(Rp+Rs+Rs)=2.62262. SecondChildUserIdTAG: 333731 SecondChildUserNameTAG: Timophei_NhaTrang SecondChildCreateTimeTAG: 2012-09-13T15:41:09Z SecondChildTAG: Simply convert the source and u will get the answer. bcoz resistances in series will not effect the open circuit voltage. SecondChildUserIdTAG: 279136 SecondChildUserNameTAG: rudra_ece SecondChildCreateTimeTAG: 2012-09-18T11:25:02Z SecondChildTAG: this is very misleading SecondChildUserIdTAG: 270284 SecondChildUserNameTAG: nkukushkin SecondChildCreateTimeTAG: 2012-09-20T18:11:50Z SecondChildTAG: VL = 0.139 is also accepted as right answer. SecondChildUserIdTAG: 254346 SecondChildUserNameTAG: moijes12 SecondChildCreateTimeTAG: 2012-09-20T18:57:59Z SecondChildTAG: Hey! How to find voltage 'VL' of last figure? SecondChildUserIdTAG: 398903 SecondChildUserNameTAG: PriteshAmrelia SecondChildCreateTimeTAG: 2012-09-22T09:57:51Z FirstChildTAG: 2.57V is correct under open circuit conditions, i.e., there is no voltage drop at the 17k contacts; then Vth = 1.8M * 1.43u = 2.57V FirstChildUserIdTAG: 12905 FirstChildUserNameTAG: Baer FirstChildCreateTimeTAG: 2012-09-13T06:08:27Z FirstChildTAG: I think that ngoctuan means that there may be a bug. If you put in 2.62262 as the answer for the first question and click check, you get the green tick instead of the red cross. Is this because of the rounding tolerance? FirstChildUserIdTAG: 142402 FirstChildUserNameTAG: aaronrod FirstChildCreateTimeTAG: 2012-09-13T06:21:05Z FirstChildTAG: This will be a bug, I agree with Baer´s post, there are no voltaje drops at resistors Rs, because the ends of them are in open circuit. FirstChildUserIdTAG: 296419 FirstChildUserNameTAG: oscman FirstChildCreateTimeTAG: 2012-09-16T04:20:22Z IndexTAG: 2775 TitleTAG: Challenges I am having probs using the interface. It keeps telling me invalid input. Please what do I do. I am seriously behind in my assignments. Pls pls.. I need support to enable me perform well. UserIdTAG: 346235 UserNameTAG: teexlim CreateTimeTAG: 2012-09-12T07:19:00Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Be more specific about your problems and I'll try to help. It's tough to give you advice when I don't know what the problem is. FirstChildUserIdTAG: 248902 FirstChildUserNameTAG: Chanute FirstChildCreateTimeTAG: 2012-09-12T07:49:04Z SecondChildTAG: I tried doing the homework 1. Most of the answers gives me error palse, invalid input, and so on. How do i solve the assignments? SecondChildUserIdTAG: 346235 SecondChildUserNameTAG: teexlim SecondChildCreateTimeTAG: 2012-09-12T14:36:19Z SecondChildTAG: Need the sample. For explame: How I must write answer at 1 exercise 1 HW, if I got, suppose, 999? SecondChildUserIdTAG: 199530 SecondChildUserNameTAG: Vasily_K SecondChildCreateTimeTAG: 2012-09-12T15:02:07Z IndexTAG: 2776 TitleTAG: Is this a simplification for 2nd part of this question?(s1E3)??? Hi. I believe someone mentioned Integrating some...thing, but does this equation work or help with this 2nd part to S1E3? Average power =(1/r) x max.voltage^2 UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-12T05:20:26Z VoteTAG: 1 CoursewareTAG: Week 1 / AC power CommentableIdTAG: 6002x_S1E3_AC_power NumberOfReplyTAG: 1 FirstChildTAG: Yes there's integration involved. See my answer below. FirstChildUserIdTAG: 194509 FirstChildUserNameTAG: blackcompe FirstChildCreateTimeTAG: 2012-09-12T05:31:52Z IndexTAG: 2777 TitleTAG: What are "Practice Scores"? What is meant by "Practice scores" for the Lecture Sequences? unable to understand this? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-12T05:08:23Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 2778 TitleTAG: problem in getting y2 Hi! I was able to get the value of y1 correctly, but when I did the same procedure in getting y2, I arrived at a wrong answer which is 1.35. What have I done wrong and why do they have different signs? Thanks! UserIdTAG: 371000 UserNameTAG: Joseph090892 CreateTimeTAG: 2012-09-12T04:00:46Z VoteTAG: 1 CoursewareTAG: Week 2 / Superposition example CommentableIdTAG: 6002x_S3E3_superposition NumberOfReplyTAG: 5 FirstChildTAG: Please note that y2 is the current flowing through R1 when V2 is acting alone, so the value of y2 will be X2/R1. Thanks. FirstChildUserIdTAG: 355972 FirstChildUserNameTAG: gagiks FirstChildCreateTimeTAG: 2012-09-12T08:48:47Z SecondChildTAG: Joseph, I've mistaken as you first time. You mustn't do "the same" procedure for y2, because we are analysing R1(!!!) resistor. So for y1 we've got sequential R1 and R2||R3, and for y1 - sequential R2 + R1||R3. Remember we are analysing R1! SecondChildUserIdTAG: 325199 SecondChildUserNameTAG: korifey SecondChildCreateTimeTAG: 2012-09-15T15:00:28Z FirstChildTAG: 1.35A is the current through R2 not R1 ; ). Once you have this value, you can use a "current divider" or simply use $x_2/R_1$ as said before. FirstChildUserIdTAG: 396446 FirstChildUserNameTAG: RousseauxS FirstChildCreateTimeTAG: 2012-09-12T11:17:25Z FirstChildTAG: I also spend the same, what happens is that y2 is corrinete in resistor R1 so the equation is y2 = x2/R1. FirstChildUserIdTAG: 51696 FirstChildUserNameTAG: blasco23 FirstChildCreateTimeTAG: 2012-09-12T15:42:04Z FirstChildTAG: I had the same problem computing y2. The i2 mark on the picture had confused me. It seems to me it is redundant :) FirstChildUserIdTAG: 375724 FirstChildUserNameTAG: sergestus FirstChildCreateTimeTAG: 2012-09-14T08:47:51Z FirstChildTAG: It is slyly question: "Similarly, the current i1 into the resistor R1 can be expressed as the sum of the current y1 due to V1 acting alone and the current y2 due to V2 acting alone". As result equations will not be similar. y1=V1/(R1+R23), where R23=R2*R3/(R2+R3); y2=x2/R1. FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-09-14T18:47:40Z IndexTAG: 2779 TitleTAG: look at the graph Hi, there. I hope I have this part correct when I say, "Look at the graph." The graph shows time from 0.0 to 0.1 seconds in intervals of 0.01 seconds. Furthermore, it seems there are 6 complete cycles in 0.10 seconds. There are 10 such intervals in 1 seconds. 6(from cycles per 0.10 seconds) x 10(because there are 10, 0.10 intervals every second) so, 10x6 = 60 cycles per seconds. I hope that helps you. Now, it is my time to be confused about another problem...wish us luck, pref. good luck. -me in nowhere, or nearly so... UserIdTAG: 119440 UserNameTAG: WG CreateTimeTAG: 2012-09-12T03:19:08Z VoteTAG: 1 CoursewareTAG: Week 1 / AC power CommentableIdTAG: 6002x_S1E3_AC_power NumberOfReplyTAG: 1 FirstChildTAG: Or if you just look at the description "120V 60Hz AC," the 60 Hz refers to the frequency of the circuit. (So here's more mumbo jumbo for heuristics!) FirstChildUserIdTAG: 333348 FirstChildUserNameTAG: easherly FirstChildCreateTimeTAG: 2012-09-12T23:49:33Z IndexTAG: 2780 TitleTAG: pos and neg?? what do we mean by ''pos and neg''?? UserIdTAG: 226085 UserNameTAG: Teto CreateTimeTAG: 2012-09-12T02:14:11Z VoteTAG: 1 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 2 FirstChildTAG: positive and negative terminals of the source FirstChildUserIdTAG: 304082 FirstChildUserNameTAG: frjeremias FirstChildCreateTimeTAG: 2012-09-12T04:56:45Z FirstChildTAG: "pos" means, "positive". "neg" means, "negative" FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-14T20:53:04Z IndexTAG: 2781 TitleTAG: Help On Homework1 Good evening: I'm sorry professors, but I was checking the Homework1 and I have a doubt, let's see the last exercise has three devices dissipating 2460W each one of them 820W, all connected in parallel, if p=vi, and total p=120(rms)*i, assuming that we're using ideal devices and the rms value of the source, the result shall be 2460/120=i that's equal to 20.5A, the check button gives an error on this question and I don't understand why even if i use the minus simbol, please if you could help me I'd be appreciated. (I've done the math and got round value for resistance of each device as 17.57ohm, I've made the simulation in the Circuit Sandbox and the value for the current was closely the same, I tried this value for my answer and it didn't accept.). With no further questions by now, and my thanks for the given attention: Vitor Rodrigues UserIdTAG: 317836 UserNameTAG: VitorRodrigues CreateTimeTAG: 2012-09-12T00:36:47Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Your approach and numbers seem correct to me, actually. The only thing is, when I see that question, I see a 240V AC source, not 120V- maybe that could be it? FirstChildUserIdTAG: 390034 FirstChildUserNameTAG: kahlil FirstChildCreateTimeTAG: 2012-09-12T00:48:44Z SecondChildTAG: Yes i've noticed it now... i'm sorry... SecondChildUserIdTAG: 317836 SecondChildUserNameTAG: VitorRodrigues SecondChildCreateTimeTAG: 2012-09-12T02:03:03Z IndexTAG: 2782 TitleTAG: the voltage across R4 about the voltage across the R4 :if we take a loop between R2 ,R3 and R4 then -v2+v3+v4=0 but V3=i3*r3 and i3 is half of i1 because the equivalent res of r3 and r4 is 4 and its equal to r2 so the current will be divided into two equal values so v3= 2*0.1667 and the loop become -0.666+(2*0.1667)+v4=0 and v4=0.333 an easier method is v4 = r4*i4 = r4*i3= 2*0.16667=0.333 . UserIdTAG: 403660 UserNameTAG: abuodeh CreateTimeTAG: 2012-09-11T18:28:00Z VoteTAG: 1 CoursewareTAG: Week 1 / Series and parallel example CommentableIdTAG: 6002x_S2E4_Series_and_Parallel NumberOfReplyTAG: 1 FirstChildTAG: Can I solve this problem if I mean that R3 and R4 is a symmetric voltage dividers of R2, and then v3=R4/(R3+R4)*v2=2/(2+2)*0.666=0.333. or this is a mistake? FirstChildUserIdTAG: 269368 FirstChildUserNameTAG: StAlex FirstChildCreateTimeTAG: 2012-09-14T19:11:25Z IndexTAG: 2783 TitleTAG: Problem in finding total power I'm facing problem in calculating total power in last part of H1P1. I've tried but it seems as if I'm doing a mistake in computing it. Can anyone guide me through it? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-11T18:19:38Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: help me out... FirstChildUserIdTAG: 125998 FirstChildUserNameTAG: omnath92 FirstChildCreateTimeTAG: 2012-09-15T10:20:05Z IndexTAG: 2784 TitleTAG: Issue with the Captions list in the video player The video player has the captions on the right which let us jump to a particular part of the video. The problem is this. Suppose I scroll down to a point much later in the captions list, the player automatically jumps back up after some time. This happens when the video has reached the end of the current set of captions. This becomes a problem when I have to review a video to recap something or to understand questions asked on the forum. It would be nice if the player doesn't automatically jump back up unless there has been no mouse activity over that area for a while. UserIdTAG: 14386 UserNameTAG: ashwith CreateTimeTAG: 2012-09-11T15:28:43Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 2785 TitleTAG: Practise questions Is it necessary to finish the practice tests in order to complete the course? Or will the homeworks suffice UserIdTAG: 108454 UserNameTAG: Raven7281 CreateTimeTAG: 2012-09-11T13:13:05Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: As far as I know, only the homeworks, labs, midterm exam and final exam will count to evaluate your progress. So, it is not needed to do the practise tests. FirstChildUserIdTAG: 9806 FirstChildUserNameTAG: AneHCraM FirstChildCreateTimeTAG: 2012-09-11T15:22:03Z FirstChildTAG: By practice tests do you mean the exercises which are in between the lecture sequences? If yes, these questions don't count towards your grade. They are there to help you check if you've understood the lecture up to that point as well as apply what you've learned immediately. FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-09-11T14:49:02Z IndexTAG: 2786 TitleTAG: Lab 1 - Can not Check or Reset circuit Hi, I am trying to pass lab1. I believe that i invented right divider and then I typed save. Now If i type "check" nothing happens. Reset key does not appear too. In the progress I see that lab1 is not completed. I tried IE, chrome, firefox from the different machines with different versions of the java but nothing helps. Seems that i need help to pass it. BTW lab1 has got symmetric solution if voltage on A is 6V but voltage drop vs is expected. This solution is considered as wrong. Thanks Alexander UserIdTAG: 125219 UserNameTAG: shulzr CreateTimeTAG: 2012-09-11T09:27:06Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 2787 TitleTAG: Here is the link which explain all about the AC concepts http://www.allaboutcircuits.com/vol_2/chpt_1/3.html UserIdTAG: 329866 UserNameTAG: Balaji_Gopal CreateTimeTAG: 2012-09-11T02:24:47Z VoteTAG: 1 CoursewareTAG: Week 1 / AC power CommentableIdTAG: 6002x_S1E3_AC_power NumberOfReplyTAG: 2 FirstChildTAG: Thanks I didn't really have a good reference sight yet for AC in my folder. FirstChildUserIdTAG: 359302 FirstChildUserNameTAG: GaryG FirstChildCreateTimeTAG: 2012-09-11T22:26:47Z FirstChildTAG: I think RMS of the blurred signal in the link can be measured using Fourier transform method. FirstChildUserIdTAG: 378967 FirstChildUserNameTAG: krish2012 FirstChildCreateTimeTAG: 2012-09-11T07:01:16Z IndexTAG: 2788 TitleTAG: differences Why in current IN I write 1.43 and is correct, but in answer is 1.40348964013 ? UserIdTAG: 136519 UserNameTAG: damiS CreateTimeTAG: 2012-09-11T01:57:03Z VoteTAG: 1 CoursewareTAG: Week 2 / Norton method example CommentableIdTAG: 6002x_S3E6_Norton_Model NumberOfReplyTAG: 4 FirstChildTAG: I(n) should be V(th)/R(th)...so you can verify your answer by check the previous exercises. FirstChildUserIdTAG: 397247 FirstChildUserNameTAG: pcbolt FirstChildCreateTimeTAG: 2012-09-11T02:03:03Z FirstChildTAG: Usually when they have to compare right answer and given answers it might happen that because the student rounded some values the final answer is a bit different from the correct one, despite everything being made without errors. So the checking system compares whether your answer is in a certain interval, rather than discrete value, an example interval could be <1.40348*(1-2%);1.40348*(1+2%)>. So 1.403+2%*1.403=1.43106, because 1.43<1.43106 system accepted your answer FirstChildUserIdTAG: 404284 FirstChildUserNameTAG: tomdrifmach FirstChildCreateTimeTAG: 2012-09-11T06:36:22Z FirstChildTAG: The answer is i = I*Rp/(Rp+2Rs) .If you keep 5 digits after the point, it gives precisely 1.40348 .It is not I. FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-09-14T21:22:27Z FirstChildTAG: VL= (Vth/(Rth+RL))*RL FirstChildUserIdTAG: 143835 FirstChildUserNameTAG: sanpablc FirstChildCreateTimeTAG: 2012-09-19T16:55:02Z IndexTAG: 2789 TitleTAG: Lab2 Could anybody give me a hint about how to do the Lab 2? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-11T01:37:49Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 4 FirstChildTAG: I'm confused about how did you get the correct resistance ? by error and trial? Or some computation? FirstChildUserIdTAG: 136959 FirstChildUserNameTAG: christerpher FirstChildCreateTimeTAG: 2012-09-13T00:33:18Z SecondChildTAG: consider the two voltage as two. for v1 fully neglect v2 part(like superposition).then make it to give value of v1/2 by adding appropriate resistors(by voltage divider rule). now go to v2, make it to give v2/6. by solving these two as separate, then make connection between them. observe the output graph. again adjust the resistors for small change in value. remember you will get answer in between 3ohm to 1ohm range including 3 and 1. SecondChildUserIdTAG: 378967 SecondChildUserNameTAG: krish2012 SecondChildCreateTimeTAG: 2012-09-19T19:00:46Z FirstChildTAG: I'll try. I hope this helps. If you look at figure 2, you can derive the equation for Vout using superposition. Now, can you write the expression for the desired Vout of the problem in a form that looks like the one in figure 2 and then use superposition to find a circuit that gives that expression? FirstChildUserIdTAG: 184529 FirstChildUserNameTAG: tfors FirstChildCreateTimeTAG: 2012-09-11T02:57:53Z SecondChildTAG: crap. SecondChildUserIdTAG: undefined SecondChildUserNameTAG: undefined SecondChildCreateTimeTAG: 2012-09-11T05:00:16Z SecondChildTAG: I tried to do this way, but I arrived in an inconsistent expression: R2 = 5R1 and R1=R2 where, R1 is next the V1 and R2 is next the V2 SecondChildUserIdTAG: 219752 SecondChildUserNameTAG: Nagata SecondChildCreateTimeTAG: 2012-09-11T12:21:35Z SecondChildTAG: But I dont Know whats the need of 3rd resistor and if its required then what should be its value. SecondChildUserIdTAG: 182470 SecondChildUserNameTAG: nitesh2703 SecondChildCreateTimeTAG: 2012-09-19T09:45:54Z FirstChildTAG: Another way to look at is that you have to divide both voltages down. With two resistors you cab only get one division, so you will need mote than that. Then give one of the resistors a value and solve for the other two. There are other threads on this topic. If you get the equations right and insert the values the correct plot will appear, no need to fiddle the values. FirstChildUserIdTAG: 66101 FirstChildUserNameTAG: bluefin FirstChildCreateTimeTAG: 2012-09-11T03:50:09Z SecondChildTAG: Sorry, but didn't understand why I need more than 2 resistors SecondChildUserIdTAG: 219752 SecondChildUserNameTAG: Nagata SecondChildCreateTimeTAG: 2012-09-11T12:24:26Z FirstChildTAG: after trying for 3 hours I finally did it.Thank you all. FirstChildUserIdTAG: 99628 FirstChildUserNameTAG: Pranjal16 FirstChildCreateTimeTAG: 2012-09-13T15:46:45Z SecondChildTAG: using how many resistors ? I do this few hours but don't understand how to solve . SecondChildUserIdTAG: undefined SecondChildUserNameTAG: undefined SecondChildCreateTimeTAG: 2012-09-14T03:13:24Z SecondChildTAG: damn this lab SecondChildUserIdTAG: 206794 SecondChildUserNameTAG: Abdofarrag SecondChildCreateTimeTAG: 2012-09-15T08:34:15Z SecondChildTAG: stilll i didn't get the correct answer.. :( SecondChildUserIdTAG: 171170 SecondChildUserNameTAG: rinutituschakkattil SecondChildCreateTimeTAG: 2012-09-17T15:42:46Z SecondChildTAG: @pranjal16 If you got it can you please try to help us by telling the answer.Its a humble request. SecondChildUserIdTAG: 182470 SecondChildUserNameTAG: nitesh2703 SecondChildCreateTimeTAG: 2012-09-19T09:47:40Z IndexTAG: 2790 TitleTAG: thermal to current analogy GND symbol and presentation Related to video "Speakers Heatsinks and Houses". Just iron out the thermal-current analogy. In the presentation i am looking at the gnd simbol which is at the bottom, then its the thermal source, then its resistor going upwards, like a pullup. Gnd position should connect to the upper side, making upper side low. Thermal flow is a given by entropy law saying two different states of energy with contact to each other will tend to combine to equal each other out, applied to electricity or thermal alike. I'd rather think that ambient is the low side (gnd). So lets say gnd is the ambiental air, and radiator is the big surface metal to contact to air in the room. IMHO nicer analogy to medium voltage safety gnd which is also a big metal surface 1 meter burried in soil. Like this, thermal energy can go from high potential (energy) to low side level - gnd. Through the resistor, so gnd needs to be after resistor. UserIdTAG: 331483 UserNameTAG: Doru CreateTimeTAG: 2012-09-11T01:08:52Z VoteTAG: 1 CoursewareTAG: Week 2 / Speakers CommentableIdTAG: 6002x_speakers_tutorial NumberOfReplyTAG: 1 FirstChildTAG: I really liked this video, it was a cool expansion on the material and is nice to see how useful these tools actually are. Thanks! FirstChildUserIdTAG: 274263 FirstChildUserNameTAG: CoreyO FirstChildCreateTimeTAG: 2012-09-15T16:00:34Z IndexTAG: 2791 TitleTAG: what is alpha? what is alpha and beta? UserIdTAG: 349662 UserNameTAG: gaaral CreateTimeTAG: 2012-09-10T23:24:17Z VoteTAG: 1 CoursewareTAG: Week 2 / Thevenin method CommentableIdTAG: 6002x_thevenin NumberOfReplyTAG: 1 FirstChildTAG: If you go back to S3V1, towards the end of the video the professor talks about a general form for voltage equations: e = a1V1 + a2V2 + ... + b1I1 + b2I2 + ... a (alpha) and b (beta) represent constants determined by the arrangement of resistors. Check out S3V1 again, it summarizes how they are calculated using the node method. FirstChildUserIdTAG: 142402 FirstChildUserNameTAG: aaronrod FirstChildCreateTimeTAG: 2012-09-12T12:36:54Z IndexTAG: 2792 TitleTAG: Help I use the equation: (e-5)/6800 + (e-(-7.2))/5600 = 0 And I got -1.69. Why is this wrong? Then I used the same equation but with the -7.2 (e-5)/6800 + (e-7.2))/5600 = 0 And I got the 6.20 volts. Do you just ignore the - on the 7.2 of V2 or not? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-10T22:48:05Z VoteTAG: 1 CoursewareTAG: Week 2 / Node equation review CommentableIdTAG: 6002x_S3E1_NodeEquationReview NumberOfReplyTAG: 2 FirstChildTAG: Your method and result are correct. Take a step back and look at the circuit. The two voltages at each side of the resistor string, +5V and -7.2V, would be at an average (5-7.2)/2 or -1.1 in the middle if the two resistors were equal. You would expect that even though one resistor is approximately 20% larger than the other the centre voltage would be around the value of -1.6 not +6.2 Also, where would you get +6.2 volts with respect to ground when that positive level doesn't even exist in the network? FirstChildUserIdTAG: 345958 FirstChildUserNameTAG: PWilson123 FirstChildCreateTimeTAG: 2012-09-10T23:03:30Z FirstChildTAG: No matter how many times I do my math, I get between 1.6 and 1.8 depending on how many decimal places I use. I don't get where this 6.2 number comes along. FirstChildUserIdTAG: 426266 FirstChildUserNameTAG: TheNetImp FirstChildCreateTimeTAG: 2012-09-15T10:36:10Z SecondChildTAG: you are making the same mistake which even I made the first time. Check the circuit diagram. The Voltage V2 has reverse potential. Hence in the KCL equation you will get (e+V2)/R2 for the current. Now if you substitute -7.2 for V2 you will get the answer. SecondChildUserIdTAG: 335370 SecondChildUserNameTAG: SumitN SecondChildCreateTimeTAG: 2012-09-15T21:06:17Z SecondChildTAG: it's so amazing because i computed and found the same answer but not the answer given above. SecondChildUserIdTAG: 273039 SecondChildUserNameTAG: SOCRATE23 SecondChildCreateTimeTAG: 2012-09-18T08:59:22Z IndexTAG: 2793 TitleTAG: Question 3 I didn't understand the first question. Total resistance between pos and neg terminals (ohms): 6000 ohms??? isn't it 6ohms? UserIdTAG: 357906 UserNameTAG: Debora_Oliveira CreateTimeTAG: 2012-09-10T22:45:27Z VoteTAG: 1 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 6 FirstChildTAG: The 'k' in the values of the resistors in Fig. 1 is an abbreviation for "thousands", so writing 3k Ohm is equivalent to writing 3000 Ohm. FirstChildUserIdTAG: 122737 FirstChildUserNameTAG: MakerDyne FirstChildCreateTimeTAG: 2012-09-10T23:06:38Z FirstChildTAG: Dear all values in KOhms. FirstChildUserIdTAG: 117588 FirstChildUserNameTAG: Kashif13377 FirstChildCreateTimeTAG: 2012-09-11T09:33:32Z FirstChildTAG: you have to sum the each of the R1, R2 and R3 resistances. Remember that each resistance was expressed in Kilo ohms which has to be converted to ohms by multiplying by 1000 FirstChildUserIdTAG: 374226 FirstChildUserNameTAG: ngozi FirstChildCreateTimeTAG: 2012-09-11T16:26:21Z FirstChildTAG: as it is in series it will be sum of all three 1K+2k+3k=6k(k=1000) FirstChildUserIdTAG: 374805 FirstChildUserNameTAG: LAKSHMIKANTH3900 FirstChildCreateTimeTAG: 2012-09-11T19:47:49Z FirstChildTAG: Me too confused over it, And i have getting 666.484 in terminal c in last question! FirstChildUserIdTAG: 216235 FirstChildUserNameTAG: newinnov FirstChildCreateTimeTAG: 2012-09-16T16:03:00Z FirstChildTAG: Hi! I had the same error, until I saw your post. Here's the really annoying part: the circuit diagram had a check underneath it, even though I had the wrong values on my resistors. If I have the wrong value resistors, shouldn't that be a "X"? Anyway, easy enough once the kohms are taken into account. FirstChildUserIdTAG: 23698 FirstChildUserNameTAG: mtmentat FirstChildCreateTimeTAG: 2012-09-16T19:54:26Z IndexTAG: 2794 TitleTAG: Negative currents doesn't exist We have a negative current but it really doesn't exist so the real answer is a positive current in the oposite direction, but they asked for a current flowing from the resistor to the network so it has to be a negative current to answer correctly, with that we are saying that the current really flows in the oposite direction. (I know that my english is not good enought but I try to do my best). UserIdTAG: 340568 UserNameTAG: witedonkey CreateTimeTAG: 2012-09-10T20:11:34Z VoteTAG: 1 CoursewareTAG: Week 1 / Power CommentableIdTAG: 6002x_S1E2_Power NumberOfReplyTAG: 1 FirstChildTAG: Kirchoff"s current law: abstract of a resistor is that Sum of the current flowing into a node with current flowing out of the node being negative. as stated in the Wiki FirstChildUserIdTAG: 359302 FirstChildUserNameTAG: GaryG FirstChildCreateTimeTAG: 2012-09-10T23:49:12Z IndexTAG: 2795 TitleTAG: Homework I read another post that says that the HWK is submitted when we press check. Does that mean that we only get one shot at the problem? Or, is it similar to the problems we worked on during the lectures? UserIdTAG: 230787 UserNameTAG: Gabriel007 CreateTimeTAG: 2012-09-10T19:16:28Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 3 FirstChildTAG: They're like problems in lectures, only you have a due date. FirstChildUserIdTAG: 267993 FirstChildUserNameTAG: mavlijas FirstChildCreateTimeTAG: 2012-09-10T19:40:42Z FirstChildTAG: You can check as many times as you like, until the due date. (Sept 16th for week 1) Correct answers are submitted as you go along and can be viewed under the "Progress" tab. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-10T19:43:21Z FirstChildTAG: Thanks for your answers! FirstChildUserIdTAG: 230787 FirstChildUserNameTAG: Gabriel007 FirstChildCreateTimeTAG: 2012-09-12T05:03:59Z IndexTAG: 2796 TitleTAG: Is answer saved? I select the Check button. The diagram I drew seems gone. The window is blank. Is it supposed to be saved for later visit. Or I have some setting error. Please help. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-10T18:11:28Z VoteTAG: 1 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: I suspect that you are using Firefox. The circuit simulator has this bug with Firefox. Can you try out Google Chrome? FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-10T18:12:54Z SecondChildTAG: Thanks. You are right. I retried in Chrome and the answer is saved. However the previous answer input in Firefox is gone. I have to reenter it in chrome. SecondChildUserIdTAG: undefined SecondChildUserNameTAG: undefined SecondChildCreateTimeTAG: 2012-09-10T18:43:39Z SecondChildTAG: Which version(s) of Firefox have this bug? All of them? SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-09-11T11:01:56Z IndexTAG: 2797 TitleTAG: S3E3: y1 and y2 signs Please explain why y1 and y2 with different signs? Where V1 or V2 are alone - the circuits are same. Only value changes. Thank you. UserIdTAG: 352757 UserNameTAG: Kerbyco CreateTimeTAG: 2012-09-10T17:58:11Z VoteTAG: 1 CoursewareTAG: Week 2 / Superposition example CommentableIdTAG: 6002x_S3E3_superposition NumberOfReplyTAG: 1 FirstChildTAG: Hi, It because in the first case the current (real) moves from + V1 source up via the R1 resistor to R2 and R3 resistors. But in the second case the current (real) moves from + V2 source up via the R2 resistor and further down via R1 and R3 resistors. And in both cases the direction of movement of a current (real) should be compare with the direction of a current of i1 . (through R1 from top to a bottom). The both scheme are NOT identical, in the first case a bottom of R1 is connected to + V1, but in the second case the bottom of R1 is connected to - of V2 source. (V1 looks like a wire in second case) FirstChildUserIdTAG: 277168 FirstChildUserNameTAG: AndrewKiselev FirstChildCreateTimeTAG: 2012-09-10T18:29:09Z SecondChildTAG: Oh. I see. Just mixed up in my head y2 and i2. This is actually so easy. Thank you. SecondChildUserIdTAG: 352757 SecondChildUserNameTAG: Kerbyco SecondChildCreateTimeTAG: 2012-09-10T19:57:22Z IndexTAG: 2798 TitleTAG: Help in S1E9 Hi, guys! I'm having some problems answering the last question in this sheet: "Now, suppose that the voltages of the two component batteries are not quite the same. For example, suppose that V 2 =1.6 . Then when we hook the two batteries together current will flow and the higher voltage battery will charge the lower voltage one. What is the current (in Amperes) that will flow? ". Can someone tell what technic to use? Thanks! UserIdTAG: 304905 UserNameTAG: SergioSilva CreateTimeTAG: 2012-09-10T16:05:30Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: First, look at the potential difference (Volts) between both batteries. Then look at the total resistance connecting the two batteries (Ohms). With V = RI you can calculate the current. FirstChildUserIdTAG: 144694 FirstChildUserNameTAG: johndoe31415 FirstChildCreateTimeTAG: 2012-09-10T16:31:35Z SecondChildTAG: Thank you! sorted out! :) SecondChildUserIdTAG: 304905 SecondChildUserNameTAG: SergioSilva SecondChildCreateTimeTAG: 2012-09-10T19:03:47Z IndexTAG: 2799 TitleTAG: Negative Voltage? When a battery supplies voltage(energy) to a load, is that voltage to be considered as negative?? Because the battery is loosing its energy....... UserIdTAG: 375233 UserNameTAG: Rosaac CreateTimeTAG: 2012-09-10T15:35:33Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 5_Simple_Power NumberOfReplyTAG: 1 FirstChildTAG: yes! FirstChildUserIdTAG: 117661 FirstChildUserNameTAG: BUNDAY FirstChildCreateTimeTAG: 2012-09-10T16:19:31Z IndexTAG: 2800 TitleTAG: THE CONCEPT OF WAVE~LENGTH OF LIGHT IN THE 1.3. The Lumped Matter Discipline OF the Textbook How do we calculate a wavelength equivalent to that of the speed of light wavelength of a circuit when given the distance in kilometeres/centimeters and a frequency in Hertz... So as to compare the Wavelength of the circuit and that of the speed of light?!!! Need your help guys/gals. Huh! UserIdTAG: 117661 UserNameTAG: BUNDAY CreateTimeTAG: 2012-09-10T12:02:01Z VoteTAG: 1 CoursewareTAG: Week 1 / Lumped Element Abstraction CommentableIdTAG: 6002x_lumped_element_abstraction NumberOfReplyTAG: 0 IndexTAG: 2801 TitleTAG: Textbook error The textbook seems to be in error in Example 2.14. I believe that the correct result is v=2 volts rather than v=0.5 volts. UserIdTAG: 147367 UserNameTAG: dtjohnsonis CreateTimeTAG: 2012-09-10T06:02:39Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Agree. FirstChildUserIdTAG: 331926 FirstChildUserNameTAG: KS99 FirstChildCreateTimeTAG: 2012-09-12T05:21:39Z IndexTAG: 2802 TitleTAG: S3E3: How to determine x2 I get 3.94V for the potential across R2 and V2 together. Why is it that x2 = 3.94V. i.e. the potential difference across R2 with V1 shorted out? UserIdTAG: 310007 UserNameTAG: ErinF CreateTimeTAG: 2012-09-10T04:03:34Z VoteTAG: 1 CoursewareTAG: Week 2 / Superposition example CommentableIdTAG: 6002x_S3E3_superposition NumberOfReplyTAG: 2 FirstChildTAG: Silly me. I misread the question and thought it was asking for the voltage across R2, instead of what it actually is asking for: the voltage cross R3. Thanks for the help. FirstChildUserIdTAG: 310007 FirstChildUserNameTAG: ErinF FirstChildCreateTimeTAG: 2012-09-14T04:43:09Z FirstChildTAG: V1 when shorted behaves as a simple wire.now r1,r2,r3 are in parellal and they have same voltage across them. FirstChildUserIdTAG: 99628 FirstChildUserNameTAG: Pranjal16 FirstChildCreateTimeTAG: 2012-09-10T14:36:35Z SecondChildTAG: But R1, R2, and R3 are not all in parallel when V1 is shorted-- R1&R3 are in parallel but in series with R2 and V2. So the resistors do not all have the same delta V across them. SecondChildUserIdTAG: 310007 SecondChildUserNameTAG: ErinF SecondChildCreateTimeTAG: 2012-09-11T04:59:10Z IndexTAG: 2803 TitleTAG: How to submit the Lab1? How can I submit the Lab 1?, I have already solve it, but there are only two bottons: Check and Save, if I hit the Check one the exercise is good, but everything gets cleared up, and nothing happens when I hit the Save one. Can anybody help me please? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-10T01:05:16Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: Once you give 'check' option, it submits your results and shows up if its correct. And to get to know about your submission status, goto 'Progess' tab on your top navigation bar. FirstChildUserIdTAG: 364166 FirstChildUserNameTAG: vaamarnath FirstChildCreateTimeTAG: 2012-09-10T01:09:52Z SecondChildTAG: Thanks SecondChildUserIdTAG: undefined SecondChildUserNameTAG: undefined SecondChildCreateTimeTAG: 2012-09-10T01:15:10Z FirstChildTAG: The 'Check' button both saves and submits your answers. If you get a green tick checkmark, your answer is correct. If you get a red X cross, your answer is incorrect. The 'Save button only saves your answers, it does not submit them. For homeworks and labs, you have unlimited attempts to use the 'Check' button. On the midterm and final exams, you will be limited to three attempts to submit your answers. This is a good reason to practice using the 'Save' button on homeworks and labs. During the pilot 6.002x course, some students were so accustomed to the unlimited 'Check' button that they ran out of attempts to submit their answers on the exams. After you use the 'Check' button, a 'Reset' button will appear. If you click on it, it will delete all your answers, and reset the progress to zero on the 'Progress' tab. If you have some correct and incorrect answers on your homework or lab, you shouldn't click on 'Reset' unless you really intend to start over from scratch. It's better to reenter only the answers for the questions which you incorrectly answered, and click 'Check' again. You can repeat this until you either give up or get them all correct. Beware, the green tick marks are addictive, you will find yourself working very hard to get them, sort of like a videogame. After the deadline for the homework and lab has passed, the 'Check' and 'Save' buttons will disappear and be replaced by a 'Show Answers' button. Click on it and the correct answers will be displayed next to each answer box. Also note that while students all receive the same general questions, the component values and other parameters in their questions are different. For example, in Lab 1, the voltage and resistance values that are in my lab are probably different from yours. Same thing on exams. FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-09-10T04:52:57Z IndexTAG: 2804 TitleTAG: Problem with power signs I find a contradiction between this problem and problem stated in "S1E1.5: SIMPLE POWER", where the sum of the power dissipated by the resistor (positive by convention), and the the power generated by the voltage power supply (negative by convention) add up to zero, as I think it should be. In this case however, it seems (according to the "correct answers"), that the power dissipated by the resistors (+12.09W and +8.34W), plus the power of the current source (+23.33W), and the power of the voltage source (-2.88W) do not add up to zero. According to my convention analysis, the power of the current source should be -23.33W, and the power of the voltage source should be +2.88. Only in this way the total power of the circuit adds up to zero. By the way, according to my analysis, all the currents, voltages, and resistor powers match the "correct answers", the only discrepancies I have are in the power signs of the voltage and current sources.. Is it correct to say that the sum of the powers on a circuit should be add up to zero as it does in problem S1E1.5 ??? UserIdTAG: 322021 UserNameTAG: mastropiero CreateTimeTAG: 2012-09-09T23:50:39Z VoteTAG: 1 CoursewareTAG: Week 1 / KVL/KCL example CommentableIdTAG: 6002x_S2E3_Using_KVL_KCL_and_VI_constraints NumberOfReplyTAG: 1 FirstChildTAG: Hi, even in this case the power adds up to 0. The total power dissipated by resistors = (12.09+8.34)= 20.43W. And total power supplied by the sources = (23.33-2.88)= 20.45W. And by convention, power supplied by sources = (-ve) of the power dissipated by the sources. That is, power dissipation at sources is -20.45W. Hence, total power of circuit = 20.43 - 20.45 ~ 0W. FirstChildUserIdTAG: 364166 FirstChildUserNameTAG: vaamarnath FirstChildCreateTimeTAG: 2012-09-10T01:14:47Z SecondChildTAG: Thanks for replying, But I still don't see how saying that 12.09 + 8.34 -23.33 +2.88 = 0 is not valid. This equation implies that the power of the resistors is positive, the power of the voltage source is positive, and the power of the current source is negative, and the sum of all equals zero. All these results obtained by applying the convention already. Why to apply the convention a second time to make the results +23.33, and -2.88? I still see the results in problem S1E1.5 where the power of the resistance was +2W, the power of the voltage supply was -2W, and therefore +2 - 2 = 0. All these results applying the same convention once. SecondChildUserIdTAG: 322021 SecondChildUserNameTAG: mastropiero SecondChildCreateTimeTAG: 2012-09-10T01:44:15Z IndexTAG: 2805 TitleTAG: "BUG"LAB1 , LAB2 , VIDEOS In the Lab1 and Lab2 schematic just one component,i.e,resistor is available in the right side of the worksheet,how is it possible to connect the other elements such as bulb, battery & ground. when I play video it says " this video is currently unavailable". please help me and fix the problem as soon as possible. UserIdTAG: 40209 UserNameTAG: sadaf78 CreateTimeTAG: 2012-09-09T22:30:57Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 3 FirstChildTAG: The resistor is all you need. Click on the resistor and drag it onto the design field. You can then rotate the position using the r key. Connect it to the circuit by clicking on a node and then dragging a line to another node. FirstChildUserIdTAG: 66101 FirstChildUserNameTAG: bluefin FirstChildCreateTimeTAG: 2012-09-09T22:47:46Z SecondChildTAG: dear bluefin unfortunately there is no circuit available in the sheet, sheet is just blank and there is just one resister in tool box.plz tell me what should i do then. SecondChildUserIdTAG: 40209 SecondChildUserNameTAG: sadaf78 SecondChildCreateTimeTAG: 2012-09-09T22:57:29Z FirstChildTAG: i have both Firefox and chrome and problem is in both browsers even in chrome i can not wire the components. FirstChildUserIdTAG: 40209 FirstChildUserNameTAG: sadaf78 FirstChildCreateTimeTAG: 2012-09-09T23:45:49Z FirstChildTAG: I had the same problem. I made a circuit from resistors and clicked "check". Off course it was wrong, but the Reset button appeared in the bottom. And after reseting, there were schematic diagrams with voltage sources. FirstChildUserIdTAG: 365465 FirstChildUserNameTAG: Vzzzz FirstChildCreateTimeTAG: 2012-09-11T09:58:27Z IndexTAG: 2806 TitleTAG: I'm confused about how to get the voltage in R2 AND R4 There is something I don't understand. When I calculate the voltage for R2 doing this.... V`2` = i`1` * R`2` I got the result as 1.32, but when I apply KVL for the first loop I got V`2` = 0.68 that according to the solution is the correct result. But then, when I apply KVL for the second loop I got the wrong answer using 0.68, but I get the good one if I use 1.32 I'm in a mess, could anybody explain me this. UserIdTAG: 45307 UserNameTAG: josejimenez2 CreateTimeTAG: 2012-09-09T21:22:29Z VoteTAG: 1 CoursewareTAG: Week 1 / Series and parallel example CommentableIdTAG: 6002x_S2E4_Series_and_Parallel NumberOfReplyTAG: 1 FirstChildTAG: The first method you use (i.e. V2 = i1*R2) is not correct because a fraction of i1 will go to the second branch (the one with R3 and R4). Try working again the calculations because you should be able to get the correct answer when using V2 = 0.68V. Remember that i1 is not going entirely through the second branch either. That might be the problem. FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-09T22:41:51Z IndexTAG: 2807 TitleTAG: dependent sources How to enter dependent sources in the simulator? UserIdTAG: 318210 UserNameTAG: Alvin12 CreateTimeTAG: 2012-09-09T19:58:20Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 2808 TitleTAG: 5th question how to calculate current in 5th task? UserIdTAG: 219082 UserNameTAG: uzair1993 CreateTimeTAG: 2012-09-09T19:00:33Z VoteTAG: 1 CoursewareTAG: Week 1 / Battery model CommentableIdTAG: 6002x_S1E9_Battery_Model NumberOfReplyTAG: 1 FirstChildTAG: You have 2 voltage sources and two resistors in a single loop. Are the resistors in series or in parallel? Can you calculate the voltage drop in them? (Try to combine them in an equivalent resistor first) FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-09T20:47:41Z IndexTAG: 2809 TitleTAG: Lab 1 Voltage divider Ok here goes.... I understand how a voltage divider works and I know where the resistors need to be placed in the circuit to achieve this, what I can't for the life of me fathom out is the equations needed to find values for R1 and R2. Could someone kindly point me in the right direction for the equations UserIdTAG: 377602 UserNameTAG: Goby CreateTimeTAG: 2012-09-09T18:44:34Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 3 FirstChildTAG: You need more than two resistors. Sorry, this was not a good response, Please ignore it. FirstChildUserIdTAG: 66101 FirstChildUserNameTAG: bluefin FirstChildCreateTimeTAG: 2012-09-09T18:50:56Z SecondChildTAG: Well yes if you include the resistor for the bulb, if this is what you are referring to ? Or am I missing something ? SecondChildUserIdTAG: 377602 SecondChildUserNameTAG: Goby SecondChildCreateTimeTAG: 2012-09-09T18:54:09Z FirstChildTAG: For the bulb off write the divide equation of 6 volts to 1.9 volts using R1 and R2. For the bulb on write the divide equation of 6 volts to 1.5 with R1 and R2+bulb in parallel. That gives you two equations and two unknowns. The general divide equation is V(divided) = V * (R2/R1+R2). FirstChildUserIdTAG: 66101 FirstChildUserNameTAG: bluefin FirstChildCreateTimeTAG: 2012-09-09T19:45:15Z FirstChildTAG: With the bulb connected R1 plus R2 and the bulb must divide to the required voltage. Went the bulb is not connected R1 and R2 divide to the second voltage. When the bulb is not connected there is a current flow in the circuit. FirstChildUserIdTAG: 66101 FirstChildUserNameTAG: bluefin FirstChildCreateTimeTAG: 2012-09-09T19:12:56Z SecondChildTAG: Indeed I understand what you say, but I can't see the equations needed to calculate the values of R1 and R2. SecondChildUserIdTAG: 377602 SecondChildUserNameTAG: Goby SecondChildCreateTimeTAG: 2012-09-09T19:33:19Z IndexTAG: 2810 TitleTAG: Current flow not negative ...Read question carefully Although the diagnostics shows -500uA the question asks what the current flow into the circuit is; which is always going to be a positive value of current as the (-)sign indicates directional flow of the current in the diagram and is not a value indicator as much as it is a directional flow indicator. The **amount** or **value** of the current flow cannot be less then zero or there would be no flow of current to measure. The current flow is the absolute value of -500uA. UserIdTAG: 404685 UserNameTAG: SamGS CreateTimeTAG: 2012-09-09T18:39:46Z VoteTAG: 1 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: so what is your question ? FirstChildUserIdTAG: 275256 FirstChildUserNameTAG: Riyansh FirstChildCreateTimeTAG: 2012-09-11T13:28:24Z IndexTAG: 2811 TitleTAG: lab1 ckt tools query In the Lab1 schematic diagram sheets,just one ckt component,i.e,resistor is available in my toolbox,on right side of the worksheet. then how is it possible to connect the other elements such as bulb, battery & ground? & where is node A?? it is not given in the question diagram nor in the worksheet! is my worksheet missing any additional displays?? plz help. . UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-09T16:41:50Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Bulb is the 3 ohm resistor,its already given and no need to connect it .Neglect node A .Resistors are enough to represent the circuit. FirstChildUserIdTAG: 120850 FirstChildUserNameTAG: Nitin1A FirstChildCreateTimeTAG: 2012-09-09T18:42:28Z SecondChildTAG: open the lab work page.wait for a while .If the bulb and voltage source (it may take few seconds) do not appear then reload the page... SecondChildUserIdTAG: undefined SecondChildUserNameTAG: undefined SecondChildCreateTimeTAG: 2012-09-11T09:17:27Z IndexTAG: 2812 TitleTAG: lab1 after building the resistor network, when i clicked DC button, it asked me to add ground(inverted T symbol). But there is no ground in it. there is only resistor in it. And how to do DC analysis without having voltage source. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-09T14:43:26Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: The ground node should be in the original diagram. If it's not there anymore, try reloading the page or using the "Reset" button at the bottom of the lab. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-09-09T16:00:45Z SecondChildTAG: where is reset button?cudnt find any. .& reloading the page is not making any difference. . my schematic diagram worksheet is empty & the toolbox on the right side has got just a resistor element! what could be done? SecondChildUserIdTAG: undefined SecondChildUserNameTAG: undefined SecondChildCreateTimeTAG: 2012-09-09T17:03:44Z IndexTAG: 2813 TitleTAG: no reset button in my lab exercises i am not able to reset the lab i just have a save and a check button.. i accidently delete the sources now i dont know what to do?? any suggestions UserIdTAG: 78647 UserNameTAG: shiva24 CreateTimeTAG: 2012-09-09T14:33:30Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: After you "check" it, the reset button should appear. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-09T14:45:29Z SecondChildTAG: i am not getting the reset button SecondChildUserIdTAG: 78647 SecondChildUserNameTAG: shiva24 SecondChildCreateTimeTAG: 2012-09-11T15:53:06Z IndexTAG: 2814 TitleTAG: AC or DC? Do we have to consider 240V as DC constant value, or as the peak value of AC at 60Hz? In this second case, P=V^2/(2*R)... UserIdTAG: 341020 UserNameTAG: franjescribano CreateTimeTAG: 2012-09-09T12:36:19Z VoteTAG: 1 CoursewareTAG: Week 1 / Power delivery modeling CommentableIdTAG: 6002x_S2E6_Modeling NumberOfReplyTAG: 2 FirstChildTAG: Just consider it as 240V DC. FirstChildUserIdTAG: 310007 FirstChildUserNameTAG: ErinF FirstChildCreateTimeTAG: 2012-09-10T04:30:26Z SecondChildTAG: Thanks a lot! I now understand it ;-) I didn't realize it was the rms value! SecondChildUserIdTAG: 341020 SecondChildUserNameTAG: franjescribano SecondChildCreateTimeTAG: 2012-09-10T23:27:53Z FirstChildTAG: 240is the rms value.peak value=240*(2)^.5 FirstChildUserIdTAG: 55345 FirstChildUserNameTAG: sarmaji FirstChildCreateTimeTAG: 2012-09-09T12:59:08Z IndexTAG: 2815 TitleTAG: New here - startup problems Hi, I'm new to the EDX site and the courseware, ... Some suggestions ... if I may ... I'm experiencing problems with the circuit simulation application - like many of us, it seems - and the discussion forum does not really help. I tried different computers home and at work and different OS's and browsers. I can't get the TRAN result graph on the screen ... This problem should be solved asap since I ant my home work and labs to be entered on time. Maybe a FAQ could be helpful to give solutions to common problems? It seems that some people are experiencing some kind of bugs - why not make a bug submission utility, to get a structured view on problem(s) and solutions? Lab 0 starts with exercises where the theory was not explained before having to answer questions - maybe the exercise could be preceded with some kind of introduction lesson ? Thank you for this great opportunity to learn on line. J.M. UserIdTAG: 389424 UserNameTAG: jmlietaer CreateTimeTAG: 2012-09-09T12:19:57Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: Hi, I have used the same circuit simulation application, and it works flawlessly. I have used it with Chrome, Firefox, and on windows 7 64bit as well as a laptop with win 7 32bit. No problems at all. The transient popups work perfect. You should probably check your setup. FirstChildUserIdTAG: 142857 FirstChildUserNameTAG: viking2 FirstChildCreateTimeTAG: 2012-09-12T11:49:17Z FirstChildTAG: Hi jmlietaer, What OS/browsers have you tried so far? If you are on Chrome, try logging out from the website, clearing the cache, then logging back in. Does the DC functionality work? The first graded lab only requires an ability to run DC analysis. I agree with you that a brief theoretical introduction prior to the lab exercise would be very helpful. FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-09T12:58:16Z SecondChildTAG: Hello kimt, I tried WinXP64 with IE8, latest FF, Safari and Chrome and with Mac OS X Lion with latest Chrome, FF and Safari. Same result in every configuration... Can't see the Transient analysis result graph. Clearing cache, history, ... and/or logging off does not work. DC functionality is OK. Some people indicate server-side issues? Currently working on Mac OS X Lion with latest version of Chrome 21.0.1180.82 and with latest download of Firefox 15.0.1 SecondChildUserIdTAG: 389424 SecondChildUserNameTAG: jmlietaer SecondChildCreateTimeTAG: 2012-09-09T14:38:53Z IndexTAG: 2816 TitleTAG: current negative explanation: the reason is that the current is not taken in conventional direction, instead electronic flow is considerd in this example UserIdTAG: 328820 UserNameTAG: saifee CreateTimeTAG: 2012-09-09T09:33:37Z VoteTAG: 1 CoursewareTAG: Week 1 / Power CommentableIdTAG: 6002x_S1E2_Power NumberOfReplyTAG: 0 IndexTAG: 2817 TitleTAG: Powers do we have to follow the given formula for power only?..or we can substitute v=iR in all power,,if so then power will be positive(if we square the -ve currents)... UserIdTAG: 219442 UserNameTAG: aki31 CreateTimeTAG: 2012-09-09T06:46:35Z VoteTAG: 1 CoursewareTAG: Week 1 / Associated reference directions CommentableIdTAG: 6002x_S2E2_Associated_Reference_Directions NumberOfReplyTAG: 1 FirstChildTAG: Power will always be positive. If the current through a device is "negative" (Remember that this depends only on how you measure it), then the voltage across that device will also be negative if you use the same convention for measuring the voltage as for the current. That means that power will be positive if you multiply negative current with negative voltage. So if you use P = VI or P = I^2*R or P = V^2/R it doesn't matter, you will get the same answer. FirstChildUserIdTAG: 270011 FirstChildUserNameTAG: SnowmanZA FirstChildCreateTimeTAG: 2012-09-09T08:58:02Z SecondChildTAG: negative or positive sign is decided by whether the power is consumed or supply to the conductor SecondChildUserIdTAG: 572704 SecondChildUserNameTAG: mitali1994 SecondChildCreateTimeTAG: 2012-10-21T04:35:51Z IndexTAG: 2818 TitleTAG: negative i1 analytically i1 comes out to be negative..hence P1..is it right? UserIdTAG: 219442 UserNameTAG: aki31 CreateTimeTAG: 2012-09-09T06:38:40Z VoteTAG: 1 CoursewareTAG: Week 1 / Associated reference directions CommentableIdTAG: 6002x_S2E2_Associated_Reference_Directions NumberOfReplyTAG: 1 FirstChildTAG: According to the sign criteria which has been used, yes, it is right. However, it is a good practise choosing a criteria which makes the outgoing power positive. So, this is not the best election. And, actually, they have chosen it to make us think. FirstChildUserIdTAG: 9806 FirstChildUserNameTAG: AneHCraM FirstChildCreateTimeTAG: 2012-09-11T15:54:19Z IndexTAG: 2819 TitleTAG: Equation 2.109 from textbook? On page 84 of the textbook there is a reference to equation 2.109. I cannot find equation 2.109 in fact the numbered equation sequence appears to jump from 2.100 to 2.110. Am i just lost and equation 2.109 does in fact exist, or is this a published oversight? Disregard... i found the notation in the beginning of the book where the equation is to be found on the internet. As usual it was me who was lost. If anyone else knows the whereabouts of equation 2.109 i would appreciate some guidance. UserIdTAG: 226890 UserNameTAG: brax1961 CreateTimeTAG: 2012-09-09T03:48:50Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: It's in the supplemental sections you can see here: http://www.mkp.com/companions/1558607358 FirstChildUserIdTAG: 312035 FirstChildUserNameTAG: philhiggins FirstChildCreateTimeTAG: 2012-09-09T04:08:12Z SecondChildTAG: Note that the supplemental sections of the textbook are freely downloadable as a PDF. The actual textbook is ***NOT*** freely downloadable. Please don't post links to pirated copies of the textbook. SecondChildUserIdTAG: 94557 SecondChildUserNameTAG: g_hopper SecondChildCreateTimeTAG: 2012-09-09T04:36:10Z SecondChildTAG: Also note that the ***www*** symbol preceding equation 2.109 in the text indicates that the material is available for download on the web. I suggest reading page xxiii in the Preface which explains this and provides the link to the downloadable material. SecondChildUserIdTAG: 94557 SecondChildUserNameTAG: g_hopper SecondChildCreateTimeTAG: 2012-09-09T04:39:46Z IndexTAG: 2820 TitleTAG: How Does V2 = 7.777? I am stuck on how v2 equals 7.777V when I calculated it to be 17.5V. I first used I=v/r to find I1 which equals 0.5A. Then using KCL at the top node to get -0.5A-3A+I3=0 (I3 being the current flowing though R2) which gives me that I3=3.5A then using V=IR so V2=3.5*5 so V2=17.5v. UserIdTAG: 219780 UserNameTAG: ReconIII CreateTimeTAG: 2012-09-09T02:01:39Z VoteTAG: 1 CoursewareTAG: Week 1 / KVL/KCL example CommentableIdTAG: 6002x_S2E3_Using_KVL_KCL_and_VI_constraints NumberOfReplyTAG: 2 FirstChildTAG: R1 lies in a series sckt so I1 cannot be found out through I=V/r you have to apply voltage diver first to get the voltage on R2 throug voltage source then you can calculate the voltage through current source on R2 and by current divider or any other metho you know then sum up both the results.. the answer is V2 FirstChildUserIdTAG: 285548 FirstChildUserNameTAG: Faizan89 FirstChildCreateTimeTAG: 2012-09-09T09:06:37Z FirstChildTAG: I got that first too, then i read "You should observe that the sum of the power supplied by the sources is the sum of the power dissipated by the resistors. If this is not true you have done something wrong." At the end, so according to this v1+v2 = V -> v1 = 2 -v2 i1 = (2 - v2)/4 = 1/2 - v2/4 The you use KCL at the node top of R2 and get this i1 + I - i2 = 0 -> i2 = i1 + I = = 1/2 - v2/4 + 3 = v2/5 ( because i = v/R ) multiply by 20 and you get 70 - 5*v2 = 4*v2 from here it's simple math FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-09T12:27:03Z IndexTAG: 2821 TitleTAG: Cross- referencing lectures to textbook I would like to see a more complete cross-referencing from the lectures to the book. What we have now is ok, but it could be better...more complete, precise. More than the occasional comment in lecture and more precise than the single link now given at the bottom of some of the pages. Just a thought. UserIdTAG: 92346 UserNameTAG: MobiusTruth CreateTimeTAG: 2012-09-08T23:50:48Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 3 FirstChildTAG: Just an additional thought.... I wish i could see both pages of the text at once. At many points in the text book, it references diagrams that are on the next page. I find it cumbersome to go back and forth. FirstChildUserIdTAG: 175205 FirstChildUserNameTAG: KT_MM FirstChildCreateTimeTAG: 2012-09-09T01:56:37Z FirstChildTAG: In another thread @blakcompe points out that "CourseInfo/6002x at a Glance" gives this information. FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-09-11T03:02:47Z FirstChildTAG: Touche. At both RPI and UNH, my professors were very much on top of things about keeping the page or section number in the text referenced on almost every lecture. Sometimes, they would even give the Chapter and Section numbers in the assignment or practice test. FirstChildUserIdTAG: 359097 FirstChildUserNameTAG: MaxAbramson FirstChildCreateTimeTAG: 2012-09-09T01:09:39Z IndexTAG: 2822 TitleTAG: In H1P1: strict input requirements for first three Q's Question requires answer "in terms of R." Placed just the multiple of R that the answers are and that was marked wrong. Placed [correct multiple number]R and that was also marked wrong. When I changed it to [number]*R it became correct. Is marking on problem sets based on first trial or final result? If first trial, is it possible this could this be fixed retroactively? If not whatever, because what's being given here for free is already fantastic :) UserIdTAG: 340606 UserNameTAG: cgranfield CreateTimeTAG: 2012-09-08T20:43:01Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: You have an infinity number of chances to submit -- your score is only based on the final result! FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-08T20:46:22Z SecondChildTAG: it is very easy but your have to use maths for example if you have two resistor on parallel r and r the answer is (1/R+1/R)*(R^2)/4and for the three resistor use maths good luck. SecondChildUserIdTAG: 266386 SecondChildUserNameTAG: zakzak200 SecondChildCreateTimeTAG: 2012-09-08T21:33:58Z IndexTAG: 2823 TitleTAG: Two wires! There are two wires on the transmission line! Be careful! UserIdTAG: 208296 UserNameTAG: OZ1 CreateTimeTAG: 2012-09-08T19:15:52Z VoteTAG: 1 CoursewareTAG: Week 1 / Power delivery modeling CommentableIdTAG: 6002x_S2E6_Modeling NumberOfReplyTAG: 0 IndexTAG: 2824 TitleTAG: Problem with equation editor for A*xˆ2 I use a MacBook Pro and when I type A*xˆ2+sqrt(y) the equation is not recognized but when I type A*x*x+sqrt(y) the equation is accepted. Through trial and error I have determined it does not accept the "ˆ" symbol. Any suggestions? UserIdTAG: 183663 UserNameTAG: mathorburn CreateTimeTAG: 2012-09-08T18:31:48Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 4 FirstChildTAG: Try typing "^" and then press the SPACE bar, works without no problems with my MacBook-Pro FirstChildUserIdTAG: 278353 FirstChildUserNameTAG: vgmariucci FirstChildCreateTimeTAG: 2012-09-08T18:53:56Z FirstChildTAG: Hi mathorburn! I have not Problems entering A*xˆ2+sqrt(y) : ![enter image description here][1] Do you have the lasted updated version of Chrome or Firefox? ---- Hola mathorburn! Yo no tengo inconvenientes al entrar la fórmula A*xˆ2+sqrt(y) en el ejercicio de prueba... ![enter image description here][1] Tienes la última versión de Chrome o Firefox en tu Navegador? [1]: http://s7.postimage.org/hyg3u0mwb/problema.jpg FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-08T18:55:49Z FirstChildTAG: Thank you all. I have tried both Firefox and Safari to no avail. I tried my iPad and it seemed to work so there is some problem with a setting on my MacBook Pro. The error message always says "Invalid input: Could not parse 'xˆ2' as a formula. If someone does know the answer - I would be grateful. Regardless I will make do. MT FirstChildUserIdTAG: 183663 FirstChildUserNameTAG: mathorburn FirstChildCreateTimeTAG: 2012-09-09T01:37:30Z SecondChildTAG: Your caret ˆ is not the same character as the "true" caret ^. On US keyboards, the "true" caret is obtained by SHIFT+6. SecondChildUserIdTAG: 79361 SecondChildUserNameTAG: kimt SecondChildCreateTimeTAG: 2012-09-09T03:46:46Z FirstChildTAG: Gracias. My problem was solved. I use an international keyboard and as kimt pointed out, the ^ is not the same as ^. The first is to create characters such as â. The later to write equations like x^2. FirstChildUserIdTAG: 183663 FirstChildUserNameTAG: mathorburn FirstChildCreateTimeTAG: 2012-09-10T00:29:19Z IndexTAG: 2825 TitleTAG: Thank God for the tool-chest!! The tutorials in the tool-chest are great! UserIdTAG: 230787 UserNameTAG: Gabriel007 CreateTimeTAG: 2012-09-08T17:39:42Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 2826 TitleTAG: "BUG" Actually this is not exactly a bug but just for the sake of better understanding and classification, it would be better if the questions in homework can be classified as **1-a,1-b, 3-c or 1.1.2, 3.2** etc. I mean numbering and sub numbering. It helps better understand that which question relates to which values for calculation. Also it would be easier to discuss and identify problems. Regards UserIdTAG: 232499 UserNameTAG: Asim09 CreateTimeTAG: 2012-09-08T17:30:52Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 2827 TitleTAG: The sign Why the voltage V4 and the current i4 is negative??? UserIdTAG: 336001 UserNameTAG: syd_buet12 CreateTimeTAG: 2012-09-08T17:27:37Z VoteTAG: 1 CoursewareTAG: Week 1 / Associated reference directions CommentableIdTAG: 6002x_S2E2_Associated_Reference_Directions NumberOfReplyTAG: 2 FirstChildTAG: Remember that current and voltage can be positive or negative depending on how you define them in your circuit. A negative current just means that the direction of current is opposite to the one that the arrow is showing. A negative voltage just means that the node labeled "-" has a larger voltage than the node labeled "+". It is very common to find negative voltages and currents in circuit analysis, specially in complex circuits when you don't know beforehand the direction of all currents and polarity of all voltages FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-08T18:24:07Z FirstChildTAG: This is as far as I understand a matter of convention. The associated variables discipline postulates that current flows from plus to minus. This is the 'positive' current. Look at the battery. The associated variable i3 goes from plus to minus, thus its direction corresponds with positive direction according to the convention. i4 on the other hand goes in the opposite direction, therefore, it's negative. I am not so sure about the voltages but I think in the following way: a resistor should dissipate power, so, the product of variables associated with the resistor should be positive. Since i is negative, v should be negative as well. Since the resistor dissipates power, the power should be supplied by the voltage source, therefore, the product of variables associated with the voltage source should be negative and since i is positive, the voltage is negative. Besides, the negative voltage on the voltage source is obvious from the circuit abstraction. FirstChildUserIdTAG: 252722 FirstChildUserNameTAG: vaboro FirstChildCreateTimeTAG: 2012-09-12T08:37:30Z IndexTAG: 2828 TitleTAG: Cant understand the question: Cant understand the question: Given that each individual resistor can dissipate up to 1 watt of power before burning up, how much total power in watts ( W ) can the smallest-valued composite resistor dissipate before burning up? can any body help me understand what are they trying to ask?? Thanks UserIdTAG: 232499 UserNameTAG: Asim09 CreateTimeTAG: 2012-09-08T17:25:27Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 3 FirstChildTAG: Think composite (all resistors parallel for example.) If one burns up at 1 watt...It means one can stand 0.? before burning up. How much can 3 of them withstand :) 0.? x 3. I guess its a spoiler but moderators are free to remove my comment. FirstChildUserIdTAG: 238946 FirstChildUserNameTAG: AwesomeO FirstChildCreateTimeTAG: 2012-09-08T19:27:32Z FirstChildTAG: The smallest composite resistor is combination of two 4ohm and one 6ohm resistors in parallel. find then current in the 4 ohm resistor when it dissipates 1w of power. with this condition on find voltage across parallel combination of resistors and u will subsequently find current through 6 ohm resistor. add all the currents and find total power dissipated in this situation. FirstChildUserIdTAG: 180402 FirstChildUserNameTAG: smeiraj FirstChildCreateTimeTAG: 2012-09-08T17:46:32Z FirstChildTAG: The question is asking how much power can be dissipated by all three resistors when connected to make the smallest-valued composite resistor (i.e. when connected in the configuration you found in the question before this one). It might seem like you can dissipate 3W (each resistor dissipating 1W), but that is not possible because when connected the resistors share the same current (if in series) or the same voltage (if in parallel). FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-08T18:18:04Z IndexTAG: 2829 TitleTAG: TYPES OF GROUND In most of the text books three different type of symbols are used to represent the ground. What's the difference between these symbols? UserIdTAG: 192216 UserNameTAG: endplayer CreateTimeTAG: 2012-09-08T17:13:08Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Hi endplayer, From what I have seen, they are often used to diffirienciate between analog and digital grounds. Also in cases where within the same circuit you want to have separete grounds. I'm sure someone has a more complete answer, but I hope this helps FirstChildUserIdTAG: 303762 FirstChildUserNameTAG: leocarrasco FirstChildCreateTimeTAG: 2012-09-08T17:18:56Z SecondChildTAG: To add to this, in practical electronic circuits, it can be desirable to have separate analog and digital grounds, because digital grounds often contain excess "noise" which 6.002x will cover in the coming weeks. It's also often common practice to have a separate ground symbol for "earth". SecondChildUserIdTAG: 79361 SecondChildUserNameTAG: kimt SecondChildCreateTimeTAG: 2012-09-08T17:27:22Z IndexTAG: 2830 TitleTAG: Image upload I cannot upload images to the course server directly. I need to use another image host. Can you guys please fix this? -- Kind Regards as always UserIdTAG: 193033 UserNameTAG: NIslam CreateTimeTAG: 2012-09-08T16:36:59Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: I am having the same problem. I can not view any of the lectures or videos. At this point I'm thinking their is no way I'm going to be able to keep up with turning stuff in on time. FirstChildUserIdTAG: 321603 FirstChildUserNameTAG: gabanetti FirstChildCreateTimeTAG: 2012-09-10T12:25:38Z SecondChildTAG: Never mind, I moved from IE to google chrome. It appears to fix this. I will also try FireFox as others have suggested in other posts. SecondChildUserIdTAG: 321603 SecondChildUserNameTAG: gabanetti SecondChildCreateTimeTAG: 2012-09-10T13:15:21Z IndexTAG: 2831 TitleTAG: Node method Can a circuit have more than one ground nodes? UserIdTAG: 99628 UserNameTAG: Pranjal16 CreateTimeTAG: 2012-09-08T16:18:31Z VoteTAG: 1 CoursewareTAG: Week 1 / Node analysis practice, part 2 CommentableIdTAG: 6002x_L2Node1 NumberOfReplyTAG: 2 FirstChildTAG: no. because the neutral wire is considered only one in a circuit. FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-08T16:33:39Z FirstChildTAG: surely, ground nodes are kinda imaginary, but 2 grounds, thats no FirstChildUserIdTAG: 254607 FirstChildUserNameTAG: werehenry FirstChildCreateTimeTAG: 2012-09-11T12:53:48Z IndexTAG: 2832 TitleTAG: What time will lectures commence (in GMT) for Week 2_September 10? Good day, To whom it may concern: I am Ugochukwu who is offering this course from Nigeria. I am very delighted to see a new way of learning with the concept of getting to educate people worldwide in disciplines of interest. I earnestly belief this will do what it was aimed for. But, it would be appreciated if students outside the United States know the time for the lectures in GMT so that we can stream in to listen to lectures at such times in order to feel the live interactions. I missed the first lecture because I did not know the time the lecture commenced but thank you for making it available on our course-ware portals. This I am presently viewing. I have come, am seeing and it is my wish to conquer - my value in this course. Till I get a response, please accept my best wishes and warm regards. Ugo. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-08T13:10:14Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: To my knowledge, the lectures for week 2 have already been posted. Just look under the Week 1 segment to find week 2. FirstChildUserIdTAG: 193033 FirstChildUserNameTAG: NIslam FirstChildCreateTimeTAG: 2012-09-08T14:58:40Z IndexTAG: 2833 TitleTAG: Fractional resistances I just spent the better part of an hour trying to figure out why my solution in lab2 was wrong. Turns out that I shouldn't have used fractional resistor values. I understand that this is my fault, I wasn't through enough in reading the tutorial. Plus I didn't look the problem up in the discussion board. Still I feel like it would be a good Idea to put a reminder -maybe in italics- about that. Please don't get me wrong, I understand that this is a minor issue, but I bet there are more people who don't "read the manual" just like me! UserIdTAG: 24953 UserNameTAG: Dennis_Georgiadis CreateTimeTAG: 2012-09-08T12:36:42Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 4 FirstChildTAG: I think problem is not tat.tat arrangement of two resistors might not work.hint:try more than two resistor FirstChildUserIdTAG: 55345 FirstChildUserNameTAG: sarmaji FirstChildCreateTimeTAG: 2012-09-08T15:22:48Z SecondChildTAG: consider the two voltage as two. for v1 fully neglect v2 part(like superposition).then make it to give value of v1/2 by adding appropriate resistors(by voltage divider rule). now go to v2, make it to give v2/6. by solving these two as separate, then make connection between them. observe the output graph. again adjust the resistors for small change in value. remember you will get answer in between 3ohm to 1ohm range including 3 and 1. SecondChildUserIdTAG: 378967 SecondChildUserNameTAG: krish2012 SecondChildCreateTimeTAG: 2012-09-19T19:00:59Z FirstChildTAG: I have had no problems in using fractional resistor values. FirstChildUserIdTAG: 66101 FirstChildUserNameTAG: bluefin FirstChildCreateTimeTAG: 2012-09-09T18:46:49Z SecondChildTAG: But how to proceed about it ? Kindly guide ! SecondChildUserIdTAG: 97340 SecondChildUserNameTAG: AnkitRana SecondChildCreateTimeTAG: 2012-09-10T10:09:35Z FirstChildTAG: I had trouble doing Lab1, too. I entered a fractional resistor value (i.e. 7/2) and the schematic did not do proper DC analysis. When I changed it to a decimal (i.e. 3.5), everything worked fine, This seems like a bug! FirstChildUserIdTAG: 175706 FirstChildUserNameTAG: rwskim FirstChildCreateTimeTAG: 2012-09-14T17:25:12Z FirstChildTAG: I can confirm this, twice. Once using Iceweasel 11.0 on Crunchbang Squeeze, once on Firefox 15 on Win7 x64. In the schematic I had integer value and fractional value resistors. The analysis really is simple for this lab, so I was completely baffled why the transient analysis showed me an error. I tried modeling the same circuit with [Falstads simulator][1], boom, looks good. Finally tried changing the frantional resistor value to a 4-place approximation and the transient analysis looked good. Lesson learned. [1]: http://www.falstad.com/circuit/ FirstChildUserIdTAG: 257171 FirstChildUserNameTAG: arcticSearcher FirstChildCreateTimeTAG: 2012-09-14T19:51:25Z IndexTAG: 2834 TitleTAG: regarding practice problems is it compulsory to solve practice problems for grades or they can just be left?i m asking this because in progress report they are also showing practice problems score UserIdTAG: 285222 UserNameTAG: dhaval24 CreateTimeTAG: 2012-09-08T12:25:29Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Hi dhaval24! How are you ? Yes, you can left the Practice Problems. They would not affect your grade. You can read [here - syllabus][1]. So , you can skip them, but if you can do them it is better. I hope this information can help you. : "3 - Interactive video sequences Lecture style videos are presented in interactive video sequences (or sequences for short), and are posted in the Courseware section of the website. Each sequence includes a succession of short video clips and online exercises, arranged in a logical progression. Two sequences will be given each week; please take the time to watch each video and each exercise in the sequence they are provided. Answer-check mechanisms are provided in these exercises, but they will not contribute towards your grade." See you! Myriam. ---- Hola dhaval24! Cómo estás? Sí, puedes no realizar los Problemas de la Práctica. El hacerlos o no, no afectan tu calificación. Puedes [leer - syllabus][1] .Es decir, puedes evitarlos sin problemas, pero es recomendable que los hagas, ya que son de mucha ayuda en el el proceso de aprendizaje de los Alumnos. Espero que esta información te sea de utilidad. Nos vemos! Myriam. [1]: https://www.edx.org/static/content-mit-6002x/handouts/syllabus.a477535058a1.pdf FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-08T19:23:01Z SecondChildTAG: thanks for the info it helped really!! SecondChildUserIdTAG: 285222 SecondChildUserNameTAG: dhaval24 SecondChildCreateTimeTAG: 2012-09-09T09:31:31Z IndexTAG: 2835 TitleTAG: an error i entered the answer for 3rd part as"b"...and when i clicked "check" a red cross appeared on it...saying that the correct answer is "b"...which is same as i entered..what to do?? http://postimage.org/image/8qu99dyy3/ UserIdTAG: 126705 UserNameTAG: arpitchugh CreateTimeTAG: 2012-09-08T11:04:08Z VoteTAG: 1 CoursewareTAG: Week 1 / Various V-I characteristics CommentableIdTAG: 6002x_S1E1_Various_V-I_characteristics NumberOfReplyTAG: 6 FirstChildTAG: delete spacebar symbol in begin of answer FirstChildUserIdTAG: 394836 FirstChildUserNameTAG: v2g6ch4 FirstChildCreateTimeTAG: 2012-09-08T11:19:24Z SecondChildTAG: thanx buddy :-) SecondChildUserIdTAG: 126705 SecondChildUserNameTAG: arpitchugh SecondChildCreateTimeTAG: 2012-09-08T11:35:30Z FirstChildTAG: also, we need to contact with developers. Spacebars must be ex-teeer-mi-na-ted ) FirstChildUserIdTAG: 394836 FirstChildUserNameTAG: v2g6ch4 FirstChildCreateTimeTAG: 2012-09-08T11:21:07Z SecondChildTAG: yeah... Ure right! SecondChildUserIdTAG: 117661 SecondChildUserNameTAG: BUNDAY SecondChildCreateTimeTAG: 2012-09-10T15:38:26Z FirstChildTAG: ![enter image description here][1] [1]: http://postimage.org/image/8qu99dyy3/ FirstChildUserIdTAG: 126705 FirstChildUserNameTAG: arpitchugh FirstChildCreateTimeTAG: 2012-09-08T11:05:38Z FirstChildTAG: http://postimage.org/image/8qu99dyy3/ FirstChildUserIdTAG: 126705 FirstChildUserNameTAG: arpitchugh FirstChildCreateTimeTAG: 2012-09-08T11:06:44Z FirstChildTAG: click on that link to see the snapshot FirstChildUserIdTAG: 126705 FirstChildUserNameTAG: arpitchugh FirstChildCreateTimeTAG: 2012-09-08T11:07:32Z FirstChildTAG: I had no problem like you. FirstChildUserIdTAG: 186782 FirstChildUserNameTAG: Edgardmaua FirstChildCreateTimeTAG: 2012-10-09T00:32:03Z IndexTAG: 2836 TitleTAG: Possible bad solution S2E4 In this circuit I think that it is not possible that current in R2 were bigger than current i1. Could someone confirm this fact? UserIdTAG: 143268 UserNameTAG: jborrego CreateTimeTAG: 2012-09-08T09:57:21Z VoteTAG: 1 CoursewareTAG: Week 1 / Series and parallel example CommentableIdTAG: 6002x_S2E4_Series_and_Parallel NumberOfReplyTAG: 2 FirstChildTAG: Ok, I see the problem. I read currents instead of voltages. Sorry. FirstChildUserIdTAG: 143268 FirstChildUserNameTAG: jborrego FirstChildCreateTimeTAG: 2012-09-08T12:31:14Z FirstChildTAG: Yes, I1 is greater than I2 and I3. Please link to you scheme, maybe it's different.. FirstChildUserIdTAG: 394836 FirstChildUserNameTAG: v2g6ch4 FirstChildCreateTimeTAG: 2012-09-08T10:57:09Z SecondChildTAG: ىخ SecondChildUserIdTAG: 206794 SecondChildUserNameTAG: Abdofarrag SecondChildCreateTimeTAG: 2012-09-08T12:12:39Z IndexTAG: 2837 TitleTAG: Node analysis practice, part 3 Mistake? Why I1 is leaving the node with potential e2? As I see I1 is coming to this node UserIdTAG: 94978 UserNameTAG: Skyer CreateTimeTAG: 2012-09-08T07:44:20Z VoteTAG: 1 CoursewareTAG: Week 1 / Node analysis practice, part 3 CommentableIdTAG: 6002x_L2Node2 NumberOfReplyTAG: 1 FirstChildTAG: Yes, I1 is entering the node, so -I1 is leaving the node. FirstChildUserIdTAG: 376895 FirstChildUserNameTAG: alecgg FirstChildCreateTimeTAG: 2012-09-08T15:29:32Z IndexTAG: 2838 TitleTAG: subscript in homework equations When i check home work on H1P1 i get Invalid input: R1 R2 R3 not permitted in answer I read: > Subscripts: When writing by hand, variable names often have subscripts. (for example A1, A2, A3….) However, in order to enter symbolic answers into the course answer boxes, the subscripts on those variables should be written with normal text (for example A1, A2, A3) Also having difficulty understanding how to handle denominators in equations in homework forms. Any help out of this state of confusion would be greatly appreciated. UserIdTAG: 226890 UserNameTAG: brax1961 CreateTimeTAG: 2012-09-08T07:19:59Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: Umm... in the first homework problem the resistors don't have subscripts... they're all the same. FirstChildUserIdTAG: 267993 FirstChildUserNameTAG: mavlijas FirstChildCreateTimeTAG: 2012-09-08T08:21:10Z FirstChildTAG: to write equations you have to know that (3r)=not (3*R) FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-10T09:11:34Z IndexTAG: 2839 TitleTAG: Text Book The textbook is not getting loaded at all. can any suggest some thing about how to rectify it and getting that done plz UserIdTAG: 358302 UserNameTAG: girishgowtham CreateTimeTAG: 2012-09-08T05:54:11Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 2840 TitleTAG: 3rd constraint in Lumped circuit abstraction I am in great confusion. Can anybody please explain me the 3rd constriant. I can't understand clearly these points: 1.Time scales of intrest are much longer than the propagation delay of electromagnetic waves. 2.The signal speeds of intrest are much lower than the electromagnetic waves speeds. And what is the relation between these two points? UserIdTAG: 188460 UserNameTAG: Lokesh17 CreateTimeTAG: 2012-09-08T03:31:02Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: The speed of an electromagnetic signal in vacuum is c, the speed of light. It is reduced when travelling in a substance like a wire. It doesn’t matter in slowly changing systems, the ends of a conductor have the same voltage. For very very fast frequencies the starting signal level may have changed before it reaches the other end of the wire a short time later. You can then have different signals at each end of the same conductor in a device such as a logic chip or microprocessor. In this case the rules like ohms law cannot apply. To make the rules apply we must have the third constraint that the signal changes are slow enough that this does not happen. FirstChildUserIdTAG: 345958 FirstChildUserNameTAG: PWilson123 FirstChildCreateTimeTAG: 2012-09-08T04:18:34Z SecondChildTAG: That's right. But this is the problem with the frequency. The frequency of the signal should be low. If the signal speed is high then the system will be faster. Then we can write in this way "The signal speed should be high and the frequency should be low". But in the book it is stated that the speed of the signal should be much lower than the speed of electromagnetic waves. SecondChildUserIdTAG: 188460 SecondChildUserNameTAG: Lokesh17 SecondChildCreateTimeTAG: 2012-09-08T05:35:06Z SecondChildTAG: the signal speed depends on the mobility of electrons in the wires and elements. How can we make the speed of the signals much slower than the electromagnetic waves speed?
If the propagation delay is much smaller, signal speeds will always be comparable to EM Wave speeds. SecondChildUserIdTAG: 188460 SecondChildUserNameTAG: Lokesh17 SecondChildCreateTimeTAG: 2012-09-08T12:23:12Z SecondChildTAG: I'm still a little confused. How does the signal changes? Does it have anything to do with electromagnetic waves? And what will change if the signal changes? Will the value for the voltage change? SecondChildUserIdTAG: 98259 SecondChildUserNameTAG: rogerloh0 SecondChildCreateTimeTAG: 2012-09-09T14:29:10Z IndexTAG: 2841 TitleTAG: Current sign Ok What is with the -ve sign in the current? nd that.I am not able to understand that.I calculated and i have both the opposite sign than the answer!If v1= +6v then i1= V1/R =6/18= +0.3333 why -ve current value for i1 ?? UserIdTAG: 281159 UserNameTAG: ManasiS CreateTimeTAG: 2012-09-07T19:55:36Z VoteTAG: 1 CoursewareTAG: Week 1 / Associated reference directions CommentableIdTAG: 6002x_S2E2_Associated_Reference_Directions NumberOfReplyTAG: 1 FirstChildTAG: Because the current i1 (consisting of electrons) is shown as if it's entering the voltage source from the positive to the negative terminal. Conventionally, current flows from the negative to the positive terminal from the voltage source, which is why i1 is negative and i2 (opposite direction of i1) is positive. FirstChildUserIdTAG: 176605 FirstChildUserNameTAG: sdnath FirstChildCreateTimeTAG: 2012-09-07T23:00:49Z IndexTAG: 2842 TitleTAG: lab 0 voltage of node C in transient analysis I didn't understand at 1.25ms in analysis window it shows 666.504m but in answer it is 0.665. why????? UserIdTAG: 40209 UserNameTAG: sadaf78 CreateTimeTAG: 2012-09-07T19:08:50Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: notice the "m" after 666.504, this means that the actual value is in "milivolts"´or "miliamps" or any other unit, "mili" translates numerically to e-3, so 666.504m = 666.504e-3 = 666.504*10^-3 = 0.666 FirstChildUserIdTAG: 156608 FirstChildUserNameTAG: luiseramos FirstChildCreateTimeTAG: 2012-09-07T19:29:26Z SecondChildTAG: thank you luiseramos SecondChildUserIdTAG: 40209 SecondChildUserNameTAG: sadaf78 SecondChildCreateTimeTAG: 2012-09-07T19:50:27Z SecondChildTAG: ok another thing node a and b are in what?. In millivolt or?? please solve this little confusion . SecondChildUserIdTAG: 40209 SecondChildUserNameTAG: sadaf78 SecondChildCreateTimeTAG: 2012-09-07T19:53:14Z IndexTAG: 2843 TitleTAG: Voltage can anyone tell that v1 and v2 have positive sign and v3 and v4 have negative sign..?? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-07T18:42:18Z VoteTAG: 1 CoursewareTAG: Week 1 / Associated reference directions CommentableIdTAG: 6002x_S2E2_Associated_Reference_Directions NumberOfReplyTAG: 2 FirstChildTAG: if you measure with a voltmeter a battery with your probe backwards(red in - and black in +) you read a negative voltage. Is the same thing: you have a diferent convention of signs to analyze a circuit. PD:sorry inglish is not my native lenguage FirstChildUserIdTAG: 265815 FirstChildUserNameTAG: ivan_mzr FirstChildCreateTimeTAG: 2012-09-08T02:24:30Z FirstChildTAG: v1 and v2 are positive because of the polarity of the battery and the measuring probe the same apply to v3 and v4 being negative FirstChildUserIdTAG: 336466 FirstChildUserNameTAG: kenstan FirstChildCreateTimeTAG: 2012-09-09T15:19:56Z IndexTAG: 2844 TitleTAG: H1P1 For the power of the smallest combined value resistor, What I did is that I find out the current in each parallel branch using P=(I^2)R and then sum up the current (I know the smallest valued resistor is just put all three individual resistors in parallel). After that, I use P=(I^2)R again to find the power but the answer comes up wrong. Need help with that. Thank you! UserIdTAG: 168210 UserNameTAG: tronicNuk CreateTimeTAG: 2012-09-07T18:31:55Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: Use the equation $$p_{{\max}}=R_{{1}}P\left( {R_{{1}}}^{-1}+{R_{{2}}}^{-1}+{R_{{3} }}^{-1} \right)$$ FirstChildUserIdTAG: 27399 FirstChildUserNameTAG: juancho FirstChildCreateTimeTAG: 2012-09-07T18:50:18Z SecondChildTAG: Thanks ... SecondChildUserIdTAG: 51599 SecondChildUserNameTAG: beatlecal SecondChildCreateTimeTAG: 2012-09-09T14:47:03Z SecondChildTAG: thanks SecondChildUserIdTAG: undefined SecondChildUserNameTAG: undefined SecondChildCreateTimeTAG: 2012-09-13T12:04:21Z FirstChildTAG: You have some circular logic going on there - you assume the power in each resistor is the maximum allowed (which isn't true) and then go calculate the power again. Parallels all have the same voltage, right? So the higher resistance resistor will be limited by the lower resistance resistors. What you need to do is find the maximum voltage allowed for the 4ohm resistors, and use that voltage to calculate the power used in the 6ohm resistor. FirstChildUserIdTAG: 267993 FirstChildUserNameTAG: mavlijas FirstChildCreateTimeTAG: 2012-09-07T18:49:17Z SecondChildTAG: Thanks @mavlijas for the info provided. I too, was completely stuck at that point. But the intuition you provided has given me some confidence. I'll try again. Thanks a lot. SecondChildUserIdTAG: 137686 SecondChildUserNameTAG: Jaychandran SecondChildCreateTimeTAG: 2012-09-08T15:02:48Z SecondChildTAG: Thanks @mavlijas fot providing this concept SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-09-15T11:53:52Z SecondChildTAG: sorry, its 'for' in place of 'fot' above. SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-09-15T11:56:40Z IndexTAG: 2845 TitleTAG: H1P1 No se cómo puedo introducir la fórmula de la resistencia equivalente en paralelo. No me la da por buena. ¿Alguien me ayuda? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-07T17:48:35Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Hola, para introducir las expreciones debes tener en cuenta que debes ingresar los datos respetando mayúsculas y minúsculas, y utilizando asterisco para la multiplicación y la barra / para la división. Te daré un ejemplo arbitrario: Si supones 2 resistencias R en paralelo y necesitas hallar la resistencia total en paralelo: deberás escribir en el casillero algo así como: R*R/(R+R) Puedes también ver el Textbook en esta página [Leer aquí-Resistores en Paralelo][2] ---- Hi, remember that you have to introduce the formulas using asteric for the product and / for the division. Also remember that is case sensitive "r" is different from "R". I will give you an arbitrary example: Suppouse to have two resistors R in paralell. They ask you the total paralell resistance. So you have to write in the box this as an answer: R*R/(R+R) you can also read in the Textbook more info about paralell resistances [here][2] [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/106 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-07T18:36:42Z SecondChildTAG: Gracias por la ayuda pero no consigo solucionarlo. La última R de la multiplicación me la sube al dividendo y las otras dos me las deja fuera. Me da como incorrecta todo el tiempo. SecondChildUserIdTAG: undefined SecondChildUserNameTAG: undefined SecondChildCreateTimeTAG: 2012-09-07T19:08:04Z IndexTAG: 2846 TitleTAG: Significance of negative power What is the physical significance of the negative power supplied by the voltage source? UserIdTAG: 279379 UserNameTAG: RajaSrinivasan CreateTimeTAG: 2012-09-07T16:52:15Z VoteTAG: 1 CoursewareTAG: Week 1 / KVL/KCL example CommentableIdTAG: 6002x_S2E3_Using_KVL_KCL_and_VI_constraints NumberOfReplyTAG: 3 FirstChildTAG: i think if the voltage supplies negative power, it's actually getting charged.... i think so.... FirstChildUserIdTAG: 316353 FirstChildUserNameTAG: Vivekanand123 FirstChildCreateTimeTAG: 2012-09-07T17:16:57Z FirstChildTAG: . FirstChildUserIdTAG: 244352 FirstChildUserNameTAG: foo1226 FirstChildCreateTimeTAG: 2012-09-07T21:00:08Z FirstChildTAG: On my opinion there are two types of elements. First are the sources. They give energy to the circuit. They make the charged particles (electrons) to move. How does this happen is not important. It might be result of chemical reaction (battery) or result of energy transformation (solar panel for example). There are two types of sources - voltage and current source. But both of them give energy to the circuit. The second type of elements are the consumers. They turn the energy of the moving electrons into heat. Once turned to heat the energy doesn't return back to the circuit. There is one situation when source can turn to a consumer. It's charging. It is to pump energy back to the source - some external force pushes the electrons backwards. So the sign of the power is only to show the direction of the energy. It is not important which sign you use for the sources and which for the consumers. The only important thing is to use different signs. The rule adopted in the lecture was to use minus for the sources and plus for the consumers. This means that for the energy to be conserved, the amount of the sourced power must be equal to the amount of the dissipated power. This is unbreakable (so far ;-) ) law of physics. So if you sum all the power in your circuit you should always have zero result. The most simple example is voltage source connected to resistor. The same voltage ant the same current for both elements. The same amount of power. The only difference is the sign. Sum them and you get zero. What puzzled me here are the answers. The sum of the power supplied by the sources is a positive number equal to the sum of the dissipated by R1 and R2 power. They have same sign. They are positive and this means as far as I understand that both the sources and the resistors dissipate power. Hmmm..... Where did I go wrong? FirstChildUserIdTAG: 229896 FirstChildUserNameTAG: Svilen FirstChildCreateTimeTAG: 2012-09-07T19:34:58Z SecondChildTAG: Power is positive when supplied by a source or dissipated in a resistor. SecondChildUserIdTAG: 267993 SecondChildUserNameTAG: mavlijas SecondChildCreateTimeTAG: 2012-09-07T21:40:46Z IndexTAG: 2847 TitleTAG: H1P2 - green ticks but wrong answers Dear 6.002x Community! This is the second time I'm taking this course - in order to move the information deeper inside my brain, I've challenged myself not to make any notes, not to write down anything - but fight the problems using only my brain and a handy calculator ;) It's good to see this course starting again! I hope that more and more students will join - it's fun! And now, to the point - I've finished H1P2 with all green ticks, however the power budget does not add up. I'm certain that the calculations are correct. Is there some kind of bug in the problem? Is power being dissipated on the wires ? :) UserIdTAG: 7114 UserNameTAG: piotroxp CreateTimeTAG: 2012-09-07T16:17:38Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: They add up for me, and no power is being dissipated on the wires. Are you sure you understand the notation? Power coming out of the voltage source is negative, which means the source is consuming power. If you take the power used in the voltage source and the two resistors it should add up to the power provided by the current source. FirstChildUserIdTAG: 267993 FirstChildUserNameTAG: mavlijas FirstChildCreateTimeTAG: 2012-09-07T16:27:09Z SecondChildTAG: May I ask what value are your resistors? SecondChildUserIdTAG: 7114 SecondChildUserNameTAG: piotroxp SecondChildCreateTimeTAG: 2012-09-07T21:15:42Z FirstChildTAG: Same here, the system doesn't take the right numbers, but it will mark green any wrong answer. FirstChildUserIdTAG: 132866 FirstChildUserNameTAG: evgkib FirstChildCreateTimeTAG: 2012-09-07T16:37:25Z SecondChildTAG: I have assigned the same values and have green ticks, but if I sum up the power it is not zero. The signs of the powers accepted as right (green tick) are: Power supply 8A = + Power supply 4V = - 1 ohm resistor = + 3 ohm resistor = + I think the real answers would be: Power supply 8A = - Power supply 4V = + 1 ohm resistor = + 3 ohm resistor = + Please provide any feedback. SecondChildUserIdTAG: 3702 SecondChildUserNameTAG: Pedro SecondChildCreateTimeTAG: 2012-09-14T18:43:21Z IndexTAG: 2848 TitleTAG: Reading Assignments I am having trouble finding a list of readings from the text that correspond to the course lectures and laboratory assignments. Also, I find the Discussion to be very low bandwidth. Hard to find appropriate posts. My suggestion is to list just the topic headings, and let me drill-down if I want more information. That way, I can survey past discussions more quickly. - Subsequently found this material. I think the cases are: 1. open the At-A-Glance page 2. wait until an Administrivia where you are told that the At-A-Glance page contains the reading assignments. At-A-Glance seems a strange place to put the reading assignments; not intuitively obvious. UserIdTAG: 194450 UserNameTAG: wrbuckley CreateTimeTAG: 2012-09-07T15:49:29Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: Course info -> 6.002x At-A-Glance FirstChildUserIdTAG: 267993 FirstChildUserNameTAG: mavlijas FirstChildCreateTimeTAG: 2012-09-07T15:57:30Z FirstChildTAG: I did not find this post until after I created a new post suggesting that reading assignments be embedded in the Courseware Progress. FirstChildUserIdTAG: 330688 FirstChildUserNameTAG: chota300 FirstChildCreateTimeTAG: 2012-09-11T00:29:58Z IndexTAG: 2849 TitleTAG: 6.002x Circuits and Electronics Prototype Course is there any difference in the video lectures from the prototype course to this course???? UserIdTAG: 82651 UserNameTAG: Ayazkazi CreateTimeTAG: 2012-09-07T15:15:46Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Most of the lectures are the same. This term, we did a lot of platform changes, reorganized the tutorial videos, and we are staffing the discussion forum more thoroughly. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-09-07T21:50:29Z IndexTAG: 2850 TitleTAG: [Math Processing Error] I can't see variable in questions, only [Math Processing Error] !!! UserIdTAG: 259086 UserNameTAG: Acan86 CreateTimeTAG: 2012-09-07T15:01:53Z VoteTAG: 1 CoursewareTAG: Week 1 / Associated reference directions CommentableIdTAG: 6002x_S2E2_Associated_Reference_Directions NumberOfReplyTAG: 1 FirstChildTAG: That error is due to how your browser interacts with MathJax. MathJax is used by the server to present mathematical symbols on your screen. Here's how to fix it: 1) Place your cursor on the *[Math Processing Error]* message and right-click. 2) You will see a pull-down MathJax menu. 3) Select Math Settings, Math Renderer, and choose either MathML or SVG. I prefer the way MathML looks, but it drops some symbols when I print, so I have to use SVG. 4) You can also play with the zoom settings on the MathJax menu to adjust the zoom functionality. I hope this helps. FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-09-09T03:25:44Z IndexTAG: 2851 TitleTAG: MP3s of video lectures Is it possible to download mp3s of the video lectures. It would definitely not be equivalent in anyway to the videos but those of us in some parts of Africa really have bandwidth problems and ready internet availability. Is there anyway they can be processed for this platform. UserIdTAG: 183507 UserNameTAG: obiradaniel CreateTimeTAG: 2012-09-07T12:34:19Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Using mp3s alone would be difficult as you would not be able to follow along with the class. If yo u want to download the video however, you can download the video from YouTube using Free Download Manager (FDM). http://www.informationweek.com/security/attacks/java-still-not-safe-security-experts-say/240006876 But if you are still certain that the mp3 file is enough, then you can use this website: http://www.youtube-mp3.org/ I apologize in advance if I have broken any laws by posting links. Please do inform me if I have, and I will not repeat it again. :) FirstChildUserIdTAG: 193033 FirstChildUserNameTAG: NIslam FirstChildCreateTimeTAG: 2012-09-07T12:45:31Z IndexTAG: 2852 TitleTAG: internal resistance of the batteries got all the answers when i ignored the internal resistance of the batteries. why do these not come into play? UserIdTAG: 388273 UserNameTAG: kilmarta CreateTimeTAG: 2012-09-07T12:18:24Z VoteTAG: 1 CoursewareTAG: Week 1 / Battery model CommentableIdTAG: 6002x_S1E9_Battery_Model NumberOfReplyTAG: 4 FirstChildTAG: You simply considered the battery to have no losses. Which means that the resistance of the battery, "r" ,is equal to 0. FirstChildUserIdTAG: 173585 FirstChildUserNameTAG: habib_akbar FirstChildCreateTimeTAG: 2012-09-07T16:08:38Z SecondChildTAG: but we are told that r is not 0 SecondChildUserIdTAG: 388273 SecondChildUserNameTAG: kilmarta SecondChildCreateTimeTAG: 2012-09-09T20:54:10Z SecondChildTAG: well,its really simple guys..R1 AND R2 are themselves the internal resistances of the batteries...0.2 ohm is for general case..Here we r given the exact value n these internal resistances are in series to the battery.. SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-17T14:40:38Z FirstChildTAG: same here...i don't get it..why should we neglect the internal resistance of the batteries? FirstChildUserIdTAG: 163407 FirstChildUserNameTAG: shuvajit FirstChildCreateTimeTAG: 2012-09-08T19:15:18Z SecondChildTAG: because internal resistance is .25 and .32 , .2 is considered as general case. SecondChildUserIdTAG: 231749 SecondChildUserNameTAG: utshau SecondChildCreateTimeTAG: 2012-09-09T05:35:06Z FirstChildTAG: You must use the internal resistance which is $0.32\Omega$ for one battery and $0.25\Omega$ for the second battery. The $0.2\Omega$ is an example of an average value but here we get the exact values for both batteries. FirstChildUserIdTAG: 396446 FirstChildUserNameTAG: RousseauxS FirstChildCreateTimeTAG: 2012-09-10T13:00:44Z SecondChildTAG: but i didnt use these numbers at all, yet it told me my answers were correct SecondChildUserIdTAG: 388273 SecondChildUserNameTAG: kilmarta SecondChildCreateTimeTAG: 2012-09-10T15:44:21Z SecondChildTAG: ---------- Hei! Internal resistance it's matter only when current flow, if not dont think about it, because is not there then. When you measure voltage on open circuit, current dosn't flow. Hope it help :-) SecondChildUserIdTAG: 97792 SecondChildUserNameTAG: SzymonJozef SecondChildCreateTimeTAG: 2012-09-16T11:42:53Z FirstChildTAG: Perhaps the wording of the problem's conditions is a little misled, and you thought that each battery along with an internal resistance (0,2 Ω) is further connected to an external resistor (0,25Ω and 0,32Ω). In fact, **these resistors (R1 and R2) — are the internal resistance of each battery**, as they are inside it (as you can see, each battery is indicated by a dotted line). To avoid such confusion, apparently, the better way to formulate a condition: "*Let's assume that both component batteries have the same voltage V1 = V2 = 1.5. The internal resistances **(R1 and R2)** of small batteries are about 0.2Ω, but they vary a bit* .... " FirstChildUserIdTAG: 343633 FirstChildUserNameTAG: lexder FirstChildCreateTimeTAG: 2012-09-15T15:57:35Z IndexTAG: 2853 TitleTAG: What is a "signal"? An abstraction? Section 1.4 of Chapter 1 of the textbook discusses the limitations of the lumped circuit abstraction and, in particular, its third postulate about signal timescales and propagation delays. On page 14 it is said: "If we are interested in signal speeds that are comparable to the speed of electromagnetic waves..." As far as I understand we have a mention of a physical phenomenon here - the electromagnetic wave, and this physical phenomenon is used to transmit a "signal". I infer that "signal" in this context is an abstraction of some type. What type of abstraction is it? P.S. In a "discussion forum schedules" handout there are mentioned the times when the TA's are available for asking questions. How exactly can contact with a TA be established to discuss this "signal" abstraction if it is? UserIdTAG: 252722 UserNameTAG: vaboro CreateTimeTAG: 2012-09-07T11:51:46Z VoteTAG: 1 CoursewareTAG: Week 1 / Lumped Element Abstraction CommentableIdTAG: 6002x_lumped_element_abstraction NumberOfReplyTAG: 1 FirstChildTAG: signal is a waveform.LMD requires that speed of this signal should be much less than speed of electromagnetic waves,that is,less than 'c'. FirstChildUserIdTAG: 55345 FirstChildUserNameTAG: sarmaji FirstChildCreateTimeTAG: 2012-09-07T11:55:58Z SecondChildTAG: Alright, so, the relationship between frequency and wave length is $\frac{1}{t} = \frac{c}{\lambda}$, where ${\lambda}$ is the wave length, and if signal speed is less than the speed of light, how do we measure signal speed? And by the way c is the speed of light in a vacuum, whereas in other matter speed of electromagnetic waves is lower. SecondChildUserIdTAG: 252722 SecondChildUserNameTAG: vaboro SecondChildCreateTimeTAG: 2012-09-07T12:22:45Z IndexTAG: 2854 TitleTAG: Measure current - explain please ! How to find the right answer: 6000 ohm = 1k + 3k + 2k (add all R's in serie) 500uA = 3 V / 6000 ohm (Ohm's Law: I = U / R) UserIdTAG: 389424 UserNameTAG: jmlietaer CreateTimeTAG: 2012-09-07T08:43:55Z VoteTAG: 1 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: Voltage source has an internal resistance. FirstChildUserIdTAG: 395953 FirstChildUserNameTAG: dustinge FirstChildCreateTimeTAG: 2012-09-08T17:53:46Z IndexTAG: 2855 TitleTAG: Transient analysis Nothing happens after starting the Transient analysis with correct parameters No graph showing UserIdTAG: 389424 UserNameTAG: jmlietaer CreateTimeTAG: 2012-09-07T08:38:23Z VoteTAG: 1 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 3 FirstChildTAG: https://6002x.mitx.mit.edu/courseware/6.002_Spring_2012/Overview/Circuit_Sandbox/ it helped to me. FirstChildUserIdTAG: 295103 FirstChildUserNameTAG: Syavick FirstChildCreateTimeTAG: 2012-09-11T08:15:56Z FirstChildTAG: hi buddy :) , you will get the graph if you set the DC to AC and add offset voltage if you want . FirstChildUserIdTAG: 362198 FirstChildUserNameTAG: Sampathram FirstChildCreateTimeTAG: 2012-09-07T12:35:20Z FirstChildTAG: befor staring you have to connect probe throug leads... probe is the last component shown in ckt elements tool box FirstChildUserIdTAG: 285548 FirstChildUserNameTAG: Faizan89 FirstChildCreateTimeTAG: 2012-09-07T12:58:22Z SecondChildTAG: This is frustrating, they should definately make an instruction video for this tool! SecondChildUserIdTAG: 355887 SecondChildUserNameTAG: martinliland SecondChildCreateTimeTAG: 2012-09-07T14:02:07Z IndexTAG: 2856 TitleTAG: Transient analysis I couldn't get this - Now run a transient analysis for 5ms. Move the mouse over the plot until the marker (a vertical dashed line that follows the mouse when it's over the plot) is at approximately 1.25ms. Please report the measured voltages for nodes A, B and C. after this graph doesn't appear i.e. the result of analysisi doesn't occurin my plot what to do UserIdTAG: 293719 UserNameTAG: jaiprakash1 CreateTimeTAG: 2012-09-07T07:26:34Z VoteTAG: 1 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 4 FirstChildTAG: Watch tutorial video in Courseware->edX tutorial video number 7 "EDX TUTORIAL: CIRCUIT SIMULATOR FOR 6.002X" This video explains how to use sandbox FirstChildUserIdTAG: 332449 FirstChildUserNameTAG: 4a4ik FirstChildCreateTimeTAG: 2012-09-07T07:35:50Z SecondChildTAG: In my console, the **AC button** on the toolbar is missing. Thus, AC transient function does not work. Also, the *second column* on the "parts bin" (FETs, diodes etc) is **absent** (though the probe is present in the single-columned parts-bin. SecondChildUserIdTAG: 285616 SecondChildUserNameTAG: AmiyaX SecondChildCreateTimeTAG: 2012-09-07T11:56:44Z FirstChildTAG: go to the tutorial for the simulator. the transient analysis is the continuous time analysis, which shows how a circuit responds to a time varying signal over the time. click the voltage source and edit the voltage, offset voltage, and the frequency . put the probe at the required places. then click on tran button then you will see the graph. FirstChildUserIdTAG: 114721 FirstChildUserNameTAG: vinayv FirstChildCreateTimeTAG: 2012-09-07T13:05:19Z FirstChildTAG: I am having the same issue. I have a probe connected to all three nodes, and I have a sin generator set for an amplitude of 1 and a freq of either 1000, or 1K. I set my transient time in seconds to either .005 or 5m and when I click on "ok" I do not get a graph to appear. I use firefox, is this a brower issue? FirstChildUserIdTAG: 147626 FirstChildUserNameTAG: bizzol FirstChildCreateTimeTAG: 2012-09-08T00:43:38Z FirstChildTAG: in firefox under Tools menu option, go to options. click on "Content" in Block pop-up, add exceptions as edx.edu FirstChildUserIdTAG: 385224 FirstChildUserNameTAG: shilpiji991 FirstChildCreateTimeTAG: 2012-09-15T16:23:48Z IndexTAG: 2857 TitleTAG: About solution......... are they going to show solutions for questions? UserIdTAG: 225946 UserNameTAG: vishal22 CreateTimeTAG: 2012-09-07T07:00:52Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: Only for questions between videos.not for homeworks.answers to homework will be available after deadline FirstChildUserIdTAG: 55345 FirstChildUserNameTAG: sarmaji FirstChildCreateTimeTAG: 2012-09-07T07:02:43Z FirstChildTAG: Hi vishal! Based my experience in the 6.002x Prototype Course, yes. But after the due of the Homeworks and Labs. For example, once that you have finished the Homework 1 and the due date has passed (for the fist Homework you have till september 16th to submit it), you will have a "show" buttom where you can see the answers of each question. Also they will upload a general solved .pdf with the steps of the exercise of Hw and Lab. Remember that the students has different values in their Homework´s/Lab´s Problems. ---- Hola vishal! Basandome en mi experiencia en el Curso Prototipo 6.002x, te puedo decir que sí. Pero esto será así cuando haya pasado el tiempo de entrega, es decir, para el Homework 1 será después del 16 de septiembre. Aparecerá un botón "Show", que si haces clic sobre él te enseñará los resultados correctos al lado de cada respuesta. También se subirá un .pdf resuelto de dicho Homework y Lab. Recuerda que cada Alumno en particular posee un valor distinto en los ejercicios. Esto es para evitar copias en las respuestas. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-07T13:37:42Z IndexTAG: 2858 TitleTAG: about resistor values ... how can i connect 2k ohms ,3k ohms resistors . the tool box always gives only 1k ohm resistor | UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-07T05:47:52Z VoteTAG: 1 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 5 FirstChildTAG: Double click on each resistor and change the value. FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-07T06:06:59Z FirstChildTAG: Double click to access and manipulate features. FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-07T05:58:29Z FirstChildTAG: just double click on the resistor (which only shows 1K at fist). There comes a edit option so you can edit the numeric values as you wish. FirstChildUserIdTAG: 170475 FirstChildUserNameTAG: sandeshacharya FirstChildCreateTimeTAG: 2012-09-07T15:18:54Z FirstChildTAG: I recommend you to see the tutorial videos, they explain how to use the tool. FirstChildUserIdTAG: 86810 FirstChildUserNameTAG: Ariel12 FirstChildCreateTimeTAG: 2012-09-07T05:57:32Z FirstChildTAG: you have to edit the properties of each individual resistor by left clicking on it, once set up. Then just type 2k (use the letter k) instead of the 1 in the properties of the resistor. FirstChildUserIdTAG: 231554 FirstChildUserNameTAG: MChaudhary FirstChildCreateTimeTAG: 2012-09-07T09:13:54Z SecondChildTAG: infact double click on the resistor and change the setting where it says r: this should read as 2k for a 2kohm resistor and 3k for a 3k ohm resistor SecondChildUserIdTAG: 231554 SecondChildUserNameTAG: MChaudhary SecondChildCreateTimeTAG: 2012-09-07T09:14:57Z IndexTAG: 2859 TitleTAG: Can`t watch videos. Help Could someone help me please!!!! I am from Turkmenistan. I can`t watch any videos on youtube and it it very hard to view the textbook as well. I am passionate about taking this course but internet connection does not let me to fulfill all my aspirations. Please, help me. UserIdTAG: 206658 UserNameTAG: Madinasabirova CreateTimeTAG: 2012-09-07T05:44:25Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 3 FirstChildTAG: Please try with modem portable max speed 14.7mbs firstly and feel your connection it's still slow are more fast. or You can try to looking for a free wireless network in many place side you. Because I using free wireless network in my place here and it all work fine. Thanks. FirstChildUserIdTAG: 367558 FirstChildUserNameTAG: lifnur FirstChildCreateTimeTAG: 2012-09-07T05:53:55Z SecondChildTAG: it seems your government or ISP blocked youtube SecondChildUserIdTAG: undefined SecondChildUserNameTAG: undefined SecondChildCreateTimeTAG: 2012-09-07T06:46:37Z FirstChildTAG: Friend,go to internet centre or anyone near u having gud connection,download videos from ocw spring 2007 course 6.002x.download lecture slides also.i think questions and homework can be seen (without much speed in conncection also),gud luck FirstChildUserIdTAG: 55345 FirstChildUserNameTAG: sarmaji FirstChildCreateTimeTAG: 2012-09-07T06:54:59Z SecondChildTAG: http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-electronics-spring-2007/ SecondChildUserIdTAG: 55345 SecondChildUserNameTAG: sarmaji SecondChildCreateTimeTAG: 2012-09-07T06:56:07Z FirstChildTAG: You can see in the Wiki [Lecture Videos Download][1]:). Also see more info [here][2]: **Due to the YouTube outage in certain parts of the world, we have provided direct links to download the lecture videos, and extended the Week 2 homework and lab deadlines to the coming Sunday (September 30). Please note that Week 3 deadlines have not changed.** [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/ [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/info FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-26T11:52:14Z IndexTAG: 2860 TitleTAG: DC TRAN I Didn't get any dc and tran at top of the tool box UserIdTAG: 218442 UserNameTAG: srinath231 CreateTimeTAG: 2012-09-07T05:40:02Z VoteTAG: 1 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: Same here. Can anyone please help? FirstChildUserIdTAG: 131963 FirstChildUserNameTAG: FahimSyed FirstChildCreateTimeTAG: 2012-09-07T09:06:36Z SecondChildTAG: Yeah theres missing tools in the box =( No bueno. Is this a browser issue? Im using firefox. What are you guys using? SecondChildUserIdTAG: 355887 SecondChildUserNameTAG: martinliland SecondChildCreateTimeTAG: 2012-09-07T13:31:48Z IndexTAG: 2861 TitleTAG: "the rule of thumb" which helps you get the plus or minus signs right when applying the loop rule -When a resistor is traversed in the same direction as the current, the change in potential is -IR -When a resistor is traversed in the direction opposite to the current, the change in potential is +IR. -When a voltage sources is traversed from – to + (the same direction it moves positive charge), the change in potential is +V. -When a voltage sources is traversed from + to – (opposite to the direction it moves positive charge), the change in potential is -V. UserIdTAG: 323963 UserNameTAG: gtrf CreateTimeTAG: 2012-09-07T05:19:49Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 2862 TitleTAG: Confusing question The following question is confusing: "The strength of the source V is given. Also, there are two known branch voltages: we know v1 and v2. In terms of the known voltages, write an algebraic expression for the branch voltage v3." Usually when a voltage is known, a number such as 1 or 2 is given. v1 and v2 are variables, which by definition are unknown. UserIdTAG: 132828 UserNameTAG: rharris CreateTimeTAG: 2012-09-07T05:10:51Z VoteTAG: 1 CoursewareTAG: Week 1 / KVL and KCL exercises CommentableIdTAG: 6002x_kvl_kcl_exercises NumberOfReplyTAG: 1 FirstChildTAG: The reason you do not see any numbers, is because its an excercise in algebra, not in elementary maths. Better get used to it. So the goal is to express sth unknown with known variables, and do not care about the values. If you have ur expression, you can then set v1=1V, v2=50kV, and calculate. FirstChildUserIdTAG: 329254 FirstChildUserNameTAG: KGabor FirstChildCreateTimeTAG: 2012-09-07T11:35:37Z IndexTAG: 2863 TitleTAG: Negative power??? Can somebody explain an idea of negative power. Does it mean that AC current(sine wave) flowing thru resistive load will dissipate 0 power ? UserIdTAG: 188219 UserNameTAG: GenadiOs CreateTimeTAG: 2012-09-07T02:56:00Z VoteTAG: 1 CoursewareTAG: Week 1 / Associated reference directions CommentableIdTAG: 6002x_S2E2_Associated_Reference_Directions NumberOfReplyTAG: 3 FirstChildTAG: Power can either be absorbed or supplied. Sources normally supply power while components like resistors always absorb power. Therefore, a voltage/current source will absorb "negative power" implying that it is actually supplying power. If a source absorbs positive power, it means it is getting charged. FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-09-07T03:24:06Z FirstChildTAG: Check out the examples in the text, starting about Pgs. 26-29. Depending on the device (resistor, battery) the sign indicates if the device is supplying power to or dissipating power from the circuit. FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-09-07T03:41:28Z FirstChildTAG: Well, as far as I understand it, it was agreed that if the positive current flows into the positive terminal of an element, then the power dissipated (or entering) the element, was positive. Then, however, the power supplied by the voltage source must be negative, as dissipation is the "negative" process of supplying. FirstChildUserIdTAG: 179862 FirstChildUserNameTAG: MieszkoM FirstChildCreateTimeTAG: 2012-09-09T12:22:19Z IndexTAG: 2864 TitleTAG: lab1 bug? I saved the circuits for lab1 and later I clicked on the check button and it graded them as wrong, but when I reseted the lab, and clicked the check button first (without saving), it graded them as right. UserIdTAG: 130045 UserNameTAG: JRRNTR CreateTimeTAG: 2012-09-07T01:40:08Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: smae problem for me.. plz tell me how to reset FirstChildUserIdTAG: 102370 FirstChildUserNameTAG: aninda FirstChildCreateTimeTAG: 2012-09-09T14:25:42Z IndexTAG: 2865 TitleTAG: S1E8 KCL I have a question! Can we take another node for finding the value of i4? that would consider i4,i2,i1 . any suggestion ? I tried and so thee answer was different.What if we solve taking that node? UserIdTAG: 281159 UserNameTAG: ManasiS CreateTimeTAG: 2012-09-07T00:06:29Z VoteTAG: 1 CoursewareTAG: Week 1 / KVL and KCL exercises CommentableIdTAG: 6002x_kvl_kcl_exercises NumberOfReplyTAG: 3 FirstChildTAG: Answer should be same.remember,i4 should be written in terms of i1,i2,i3 FirstChildUserIdTAG: 55345 FirstChildUserNameTAG: sarmaji FirstChildCreateTimeTAG: 2012-09-07T01:31:44Z FirstChildTAG: In order to do that you would need to consider i1, i2, i3 and i4. Remember, what goes in must come out. I like to think of these KCL problems as cars navigating a highway, or water flowing through a pipe network. FirstChildUserIdTAG: 151427 FirstChildUserNameTAG: RichardZ FirstChildCreateTimeTAG: 2012-09-07T00:53:52Z FirstChildTAG: yes you are r8...the answer is different ! If i write i4 in terms of i1,i2 and i3 then in the answer where is the i3? There might be some rules about selecting node ! i dont know...someone plx explain... FirstChildUserIdTAG: 152551 FirstChildUserNameTAG: sshakir FirstChildCreateTimeTAG: 2012-09-14T20:29:02Z IndexTAG: 2866 TitleTAG: Offline viewing/study? Is there a way to view and work with the coursework offline? Sometimes I'm away where there is limited internet access or not enough mobile broadband bandwidth. Any help in the matter appreciated. UserIdTAG: 207204 UserNameTAG: hgustavii CreateTimeTAG: 2012-09-06T22:42:53Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: If you go to the Course Info tab you will find the lecture slides available in PDF form. Other than that you can buy the textbook and download the lectures off youtube, but the exercises, labs and homework must be completed online. FirstChildUserIdTAG: 151427 FirstChildUserNameTAG: RichardZ FirstChildCreateTimeTAG: 2012-09-07T00:56:44Z SecondChildTAG: Ok, great, thanks! That's all I need. SecondChildUserIdTAG: 207204 SecondChildUserNameTAG: hgustavii SecondChildCreateTimeTAG: 2012-09-07T11:52:44Z SecondChildTAG: I too find it tough to find 12 hours a week online and would love a better way to study offline. The only way that i can find to download the lectures for viewing offline is to open each lecture series in the courseware, open each lecture segment, click on the youtube icon in the video and from there download it - this is almost as time consuming as just watching them online! This looks like a really great course, but i don't think i will be able to spend 12 hours online each week participating. I'm happy to go online to participate in exercises, labs and homework etc, but just don't have the ability to spend 12 hours a week connected to the internet outside work. Ideally (for me at least) would be a bulk download pack of lectures, notes, slide etc This way i could transfer the videos to my phone, print out the slides, and work through the lectures during my lunch break and during transit. This would greatly reduce wi-fi or g3 dependance and would make it much easier to study on the go. Thanks SecondChildUserIdTAG: 429032 SecondChildUserNameTAG: EMW SecondChildCreateTimeTAG: 2012-09-18T01:58:26Z IndexTAG: 2867 TitleTAG: number of loops and independent KVL 7 loops and B-N+1=3 independent applications of KVL (page 63 of textbook) UserIdTAG: 206669 UserNameTAG: dimitrios66 CreateTimeTAG: 2012-09-06T21:43:01Z VoteTAG: 1 CoursewareTAG: Week 1 / Circuit topology CommentableIdTAG: 6002x_S2E1_Circuit_Topology NumberOfReplyTAG: 2 FirstChildTAG: dimitrios, LOOP 1: Vca Vad Vdc LOOP 2: Vca Vab Vbc LOOP 3: Vad Vdb Vba LOOP 4: Vbd Vdc Vcb LOOP 5: Vad Vdc Vcv Vba LOOP 6: Vab Vbd Vdc Vca LOOP 7: Vad Vdb Vbc Vca ;) FirstChildUserIdTAG: 385692 FirstChildUserNameTAG: tpfslima FirstChildCreateTimeTAG: 2012-09-06T23:07:19Z SecondChildTAG: hmm. may be 14? if you add reverse loop? Reverse loop1: Vac Vcd Vda Reverse loop2: Vac Vcb Vba ....etc ;) SecondChildUserIdTAG: 324957 SecondChildUserNameTAG: mazzay SecondChildCreateTimeTAG: 2012-09-07T09:35:12Z FirstChildTAG: Page 63 of the textbook hints it´s just 4 loops, but only 3 independent ones. FirstChildUserIdTAG: 296965 FirstChildUserNameTAG: LGMailhos FirstChildCreateTimeTAG: 2012-09-08T04:38:48Z IndexTAG: 2868 TitleTAG: Number of loops In the lecture S2V4 professor say there are 4 loops: 1. V-a-b-c 2. a-b-d 3. b-c-d 4. V-a-d-c UserIdTAG: 327746 UserNameTAG: sciamannalucio CreateTimeTAG: 2012-09-06T21:31:17Z VoteTAG: 1 CoursewareTAG: Week 1 / Circuit topology CommentableIdTAG: 6002x_S2E1_Circuit_Topology NumberOfReplyTAG: 1 FirstChildTAG: I agree, there are 7 branches and 4 loops; a to b via R1 is not a loop FirstChildUserIdTAG: 306110 FirstChildUserNameTAG: AEJ FirstChildCreateTimeTAG: 2012-09-11T02:11:38Z SecondChildTAG: yeah i also agree.. There are four loops not 7. can someone please explain SecondChildUserIdTAG: 235917 SecondChildUserNameTAG: Joebenny SecondChildCreateTimeTAG: 2012-09-16T14:29:48Z IndexTAG: 2869 TitleTAG: phi What is phi? (phi-B from the explanation of when V is defined) UserIdTAG: 331345 UserNameTAG: LieuweR CreateTimeTAG: 2012-09-06T20:39:06Z VoteTAG: 1 CoursewareTAG: Week 1 / Lumped Element Abstraction CommentableIdTAG: 6002x_lumped_element_abstraction NumberOfReplyTAG: 1 FirstChildTAG: Phi is usually the symbol for the magnetic flux. So, from Faraday's Law, a time changing magnetic field (or the magnetic flux) creates an electromotive force. FirstChildUserIdTAG: 183947 FirstChildUserNameTAG: gfenton06 FirstChildCreateTimeTAG: 2012-09-06T20:45:33Z IndexTAG: 2870 TitleTAG: Sign Convention Suppose we have a simple circuit with only one load on it, and two terminals of wire say 'A' and 'B' connected with any battery or voltage source where A is +ve and B is -ve terminal. It does not matter what is the value of load or resistor, voltage and current. For this always remember a sign convention: 1.If the current enters the terminal labeled with +ve voltage of an element, it is receiving energy. For example Resistors, Capacitors(usually). 2.If the current enters the terminal labeled with the -ve voltage, it is supplying energy. For example Voltage source, Current source, Inductors, Capacitors(sometimes). 3.When an element is supplying power or energy, its sign is -ve. Negative Power = Power being Supplied. 4.When an element is receiving power or energy, its sign is +ve. Positive Power = Power being Absorbed. So to get exact sign of power we have to take care of current direction. UserIdTAG: 378096 UserNameTAG: Jamshaid271 CreateTimeTAG: 2012-09-06T19:27:07Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Here is a very simple way to understand this. Just remember because we are using the conventional flow of current (holes moving) that at a location with a high voltage with respect to another location that is at a lower voltage, means that holes will then move from the high voltage down to the low voltage. On the holes journey if they happen to pass through a resistor then the resistor will dissipate energy due to the movement of this current. Hence a current flow from high voltage to low voltage through a resistor is supplying energy to the resistor which dissipates it as heat. Just remember the above and the passive sign convention all falls out naturally and is obvious. All you need to remember is that when we have a location of high voltage to one of low voltage this is the direction in which current flows (always high to low. Never low to high.). FirstChildUserIdTAG: 15344 FirstChildUserNameTAG: kob FirstChildCreateTimeTAG: 2012-09-07T01:38:29Z IndexTAG: 2871 TitleTAG: S2E5: NODE METHOD "There are three node potentials..." There are only 2 nodes, e1 is a point of circuit, not a node. UserIdTAG: 326135 UserNameTAG: Uhimovich CreateTimeTAG: 2012-09-06T19:14:43Z VoteTAG: 1 CoursewareTAG: Week 1 / Node method example CommentableIdTAG: 6002x_S2E5_Node_Method NumberOfReplyTAG: 2 FirstChildTAG: It's confusing...By definition, node is a place of the circuit where 3 or more branches are connected. But here the connection between 2 elements is also node. FirstChildUserIdTAG: 196353 FirstChildUserNameTAG: Semen95 FirstChildCreateTimeTAG: 2012-09-07T18:12:59Z SecondChildTAG: A node is defined as a "junction point at which the terminals of *two or more* elements are connected." SecondChildUserIdTAG: 183711 SecondChildUserNameTAG: MMatheson SecondChildCreateTimeTAG: 2012-09-08T16:29:13Z FirstChildTAG: Its a node! Its the node connecting the supply to the resistor. FirstChildUserIdTAG: 80920 FirstChildUserNameTAG: MichaelSturgess FirstChildCreateTimeTAG: 2012-09-06T19:19:20Z IndexTAG: 2872 TitleTAG: Info for H1P1 With Reference to the question : Given that each individual resistor can dissipate up to 1 watt of power before burning up, how much total power in watts ( W ) can the smallest-valued composite resistor dissipate before burning up? Can some1 tell me whats smallest-valued composite resistor mean??? Thanks. UserIdTAG: 364884 UserNameTAG: Ambli CreateTimeTAG: 2012-09-06T19:13:03Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: Hey can u help me solve this question...? I m not able to solve this FirstChildUserIdTAG: 108929 FirstChildUserNameTAG: namit FirstChildCreateTimeTAG: 2012-09-07T11:07:50Z FirstChildTAG: Composite means made up of parts, so the smallest-valued composite resistor would be the smallest value resistor made up of the three resistors they gave you earlier in the question. FirstChildUserIdTAG: 267993 FirstChildUserNameTAG: mavlijas FirstChildCreateTimeTAG: 2012-09-06T19:27:44Z IndexTAG: 2873 TitleTAG: What is the current through R2? If the Current Supply is 3A, and we use the Current divider formula, the current through R2 would be 1,333A, and through R1 would be 1.6667 amps. But using the volt given in the solutions, the current would need to be 1.55 on R2, how is this possible? UserIdTAG: 315841 UserNameTAG: JoaoFerreira1337 CreateTimeTAG: 2012-09-06T19:04:55Z VoteTAG: 1 CoursewareTAG: Week 1 / KVL/KCL example CommentableIdTAG: 6002x_S2E3_Using_KVL_KCL_and_VI_constraints NumberOfReplyTAG: 1 FirstChildTAG: Nevermind, found it out: So, writing the KCL: I2 = I1 + I <=> I2 = I1 + 3 KVL: -V + V1 + V2 = 0 -2 + R1I1 + R2I2 = 0 Doing all the math: I1 = -1.44A I2 = 1.56 A FirstChildUserIdTAG: 315841 FirstChildUserNameTAG: JoaoFerreira1337 FirstChildCreateTimeTAG: 2012-09-06T19:20:27Z SecondChildTAG: But u2 = 7.777777, so i2 = u2 / R2 = 7.77777 / 5 = 1.555555, what about this? SecondChildUserIdTAG: 328831 SecondChildUserNameTAG: Fandir SecondChildCreateTimeTAG: 2012-09-07T13:49:31Z IndexTAG: 2874 TitleTAG: S1E7: KCL-0 why in problem S1E7: KCL-0 the answer is 4.3 instead of -4.3 for i5. UserIdTAG: 155029 UserNameTAG: cohete2006 CreateTimeTAG: 2012-09-06T18:32:14Z VoteTAG: 1 CoursewareTAG: Week 1 / KVL and KCL exercises CommentableIdTAG: 6002x_kvl_kcl_exercises NumberOfReplyTAG: 1 FirstChildTAG: I think it is because of the node between i3, i5 and i4 current i5 becomes -ve in the equation so the equation is now i4-i5+i3=0 FirstChildUserIdTAG: 281159 FirstChildUserNameTAG: ManasiS FirstChildCreateTimeTAG: 2012-09-07T00:09:59Z SecondChildTAG: IT IS 4.3 NOT MINUS 4.3 REMEMBER CURRENT COMING TO NODE IS POSITIVE AND LEAVING THE NODE IS NEGATIVE. hence ** ** i5-i4-i3=0 ========== ** ---------- so **i5=i4+i3 = 4.3** ============== hope you understand now. SecondChildUserIdTAG: 40209 SecondChildUserNameTAG: sadaf78 SecondChildCreateTimeTAG: 2012-09-09T12:50:21Z IndexTAG: 2875 TitleTAG: How to rotate components in the simulator I don't think my simulator is broken as I can place and connect components, run analyses, etc. I would like to be able to place some components horizontally sometimes and it seems there should be a way to rotate them but I can't find it. Is it possible to rotate a component, before or after placing it in the diagram? UserIdTAG: 91356 UserNameTAG: mnrsiat CreateTimeTAG: 2012-09-06T17:39:16Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 3 FirstChildTAG: Well immediately after posting this, I found the answer, in the "Using the Interactive Laboratory" handout in the course info tab. "Click to select a component in the diagram (it will turn green) and then type the letter "r" on the keyboard. The component will be rotated 90 degrees. Additional rotations will move the component through its eight possible orientations (4 compass points and their reflections)." FirstChildUserIdTAG: 91356 FirstChildUserNameTAG: mnrsiat FirstChildCreateTimeTAG: 2012-09-06T17:45:01Z FirstChildTAG: Press R key on keyboard to rotate your highlight components. FirstChildUserIdTAG: 324774 FirstChildUserNameTAG: eduard_k FirstChildCreateTimeTAG: 2012-09-06T17:46:37Z FirstChildTAG: ITS NOT WOrking on my laptop..i m working in google chrome...OS-ubuntu 12.04 lts FirstChildUserIdTAG: 126705 FirstChildUserNameTAG: arpitchugh FirstChildCreateTimeTAG: 2012-09-08T19:59:20Z IndexTAG: 2876 TitleTAG: answer why so strange answer? I1 goes to the node, but not leave it. And answer expression must be equal zero. UserIdTAG: 376258 UserNameTAG: dostt CreateTimeTAG: 2012-09-06T17:11:05Z VoteTAG: 1 CoursewareTAG: Week 1 / Node analysis practice, part 3 CommentableIdTAG: 6002x_L2Node2 NumberOfReplyTAG: 2 FirstChildTAG: the opposite of leaving is going into .... FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-09-06T17:21:00Z FirstChildTAG: misleading when the question says not to include conductance. FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-12T02:24:02Z IndexTAG: 2877 TitleTAG: Homework submission Should I answer all questions and then **check** or can I answer some of them, then **check** and answer others **at another time**?? UserIdTAG: 255408 UserNameTAG: Omer9 CreateTimeTAG: 2012-09-06T14:52:09Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: You can answer them one at a time. There is no limit on the number of attempts. They've done this because in many cases a question depends on the answer of the previous so you can work through the problem in steps. FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-09-06T14:54:18Z FirstChildTAG: ashwith is correct. But beware if you get into the habit of doing this that the midterm and final will probably be different, and there you will have limited submissions, so you should not submit until you have answered all the parts you intend to answer FirstChildUserIdTAG: 9147 FirstChildUserNameTAG: SueP FirstChildCreateTimeTAG: 2012-09-06T15:43:34Z SecondChildTAG: yes absolutely +1 for both ashwith & Sue SecondChildUserIdTAG: 82571 SecondChildUserNameTAG: Nidhi3 SecondChildCreateTimeTAG: 2012-09-06T16:05:41Z IndexTAG: 2878 TitleTAG: Hotkeys for Videos Not sure if this is the place for this, but is it possible to add a hotkey that enables to continue to the next video sequence by pressing the arrow keys? UserIdTAG: 181745 UserNameTAG: Yoavrott CreateTimeTAG: 2012-09-06T14:50:29Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 2879 TitleTAG: Homework submission Invalid input: Could not parse '3R' as a formula... What to do in this case?? UserIdTAG: 255408 UserNameTAG: Omer9 CreateTimeTAG: 2012-09-06T14:21:18Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Hi Omer9, remember that the Formulas has to be wrote like 5*R and not 5R... Hola Omer9, recuerda que las fórmulas tienen que ser ingresadas como 5*R y no como 5R... FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-06T14:24:38Z SecondChildTAG: Thanks alot. SecondChildUserIdTAG: 255408 SecondChildUserNameTAG: Omer9 SecondChildCreateTimeTAG: 2012-09-06T14:26:20Z IndexTAG: 2880 TitleTAG: Homework submission By clicking on **Check**; I **save** my homework or **submit** it?? UserIdTAG: 255408 UserNameTAG: Omer9 CreateTimeTAG: 2012-09-06T13:44:25Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: both FirstChildUserIdTAG: 82571 FirstChildUserNameTAG: Nidhi3 FirstChildCreateTimeTAG: 2012-09-06T13:47:07Z FirstChildTAG: The Check Buttom has 3 functions: one is to save your Homework; at the same time, they also tell you if your answer is correct or incorrect; Finally, this allows you to submit your Homework. So, when you click on Check happens 3 things at the same time: save+review/check the answer+send/submit your answer.Remember that you have unlimited opportunities to click on Check. But not in the Midterm or Final Exam (you only have 3 possibilities). ---- El Botón Check tiene tres funciones: la de guardar lo que has escrito; a su vez, revisarlo y finalmente, de enviarlo. Es decir, cuando presionas el Botón de Check suceden 3 cosas: guarda+corrige+envía tu respuesta.Pero no en el Midterm ni en el Examen Final (solo tienes 3 oportunidades). FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-06T14:52:06Z SecondChildTAG: Can anyone from the staff confirm that please (unlimited checks for homework/lab and only 3 checks allowed for midterm/final exam)? Thanks in advance! SecondChildUserIdTAG: 322552 SecondChildUserNameTAG: riv SecondChildCreateTimeTAG: 2012-09-10T17:47:06Z IndexTAG: 2881 TitleTAG: confused by (dq/dt) isn't dq/dt=I and by making it equals to zero, we desided that there is no current inside of elements? UserIdTAG: 210380 UserNameTAG: MurdocRus CreateTimeTAG: 2012-09-06T13:32:16Z VoteTAG: 1 CoursewareTAG: Week 1 / Lumped Element Abstraction CommentableIdTAG: 6002x_lumped_element_abstraction NumberOfReplyTAG: 4 FirstChildTAG: dq/dt is the rate of change of total charge within the element, not the charge passing through the element. If a current is flowing through it, then electrons are coming in SA and leaving SB, so the total charge stays the same. FirstChildUserIdTAG: 256880 FirstChildUserNameTAG: Jetbeard FirstChildCreateTimeTAG: 2012-09-06T13:39:18Z FirstChildTAG: dq/dt=0 means total charge is constant, non zero. There's a constant current flow and this will not increase or decrease over time. FirstChildUserIdTAG: 268104 FirstChildUserNameTAG: SaulCardenasG FirstChildCreateTimeTAG: 2012-09-07T05:39:52Z FirstChildTAG: Could this be explained like this? dq/dt means "the change in charge over some change in time" or "is there a change in charge over time?", so if there is NO CHANGE, or ZERO CHANGE we say it must be equal to 0. Therefore: dq/dt = 0 So, all the charge (Ia) that is going in is all the charge (Ib) going out, no charge is left or no charge is added, so the current that goes in is the same that goes out: Ia=Ib FirstChildUserIdTAG: 155619 FirstChildUserNameTAG: ECamou FirstChildCreateTimeTAG: 2012-09-14T05:38:34Z FirstChildTAG: I is not defined as dq/dt .. its integral of J.dS FirstChildUserIdTAG: 351107 FirstChildUserNameTAG: sanket383 FirstChildCreateTimeTAG: 2012-09-16T01:38:21Z IndexTAG: 2882 TitleTAG: Captions at 7:32 I believe "time [? keeping ?] signals" should be "time-domain signals". UserIdTAG: 256880 UserNameTAG: Jetbeard CreateTimeTAG: 2012-09-06T11:12:33Z VoteTAG: 1 CoursewareTAG: Week 1 / Course overview CommentableIdTAG: 6002x_S1V1toS1V4 NumberOfReplyTAG: 1 FirstChildTAG: Yeah it is time domain or time varying signals FirstChildUserIdTAG: 151751 FirstChildUserNameTAG: Senade FirstChildCreateTimeTAG: 2012-09-06T15:48:24Z IndexTAG: 2883 TitleTAG: I AM FACING PROBLEMS IN LAB 0 I am unable to connect the components with wires, when I Click on a connection point to start a wire whole component gets green and starts moving with cursor and there is no any wire appear,please tell me what is the problem and how long will it take to be fixed. UserIdTAG: 40209 UserNameTAG: sadaf78 CreateTimeTAG: 2012-09-06T11:06:58Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Click on the terminal of component and **DRAG** it. you shall get wire. FirstChildUserIdTAG: 908 FirstChildUserNameTAG: m_umair72001 FirstChildCreateTimeTAG: 2012-09-06T13:28:03Z SecondChildTAG: I think there were some problem in Google chrome because in fire fox it is working better. SecondChildUserIdTAG: 40209 SecondChildUserNameTAG: sadaf78 SecondChildCreateTimeTAG: 2012-09-06T21:01:04Z IndexTAG: 2884 TitleTAG: all vectors exit or entering a node S1E8 KCL some nodes have all vector currents exiting from all sides and no current entering. Interesting nodes. Anybody knows where i can buy a node like that to hack it into my mains power lines so i can have my free energy forever? Apparently all the youtube videos with free energy arent so bogus. Which reminds me i gonna look into the three legged chickens right away. ;) UserIdTAG: 331483 UserNameTAG: Doru CreateTimeTAG: 2012-09-06T09:50:17Z VoteTAG: 1 CoursewareTAG: Week 1 / KVL and KCL exercises CommentableIdTAG: 6002x_kvl_kcl_exercises NumberOfReplyTAG: 2 FirstChildTAG: Forgot to explain that 3 legged chickens have more chances turning true. Now thinking again at the exercise in question if arrows represent the sign of current and a 3 way node shows 3 vectors leaving the node, and if sum of currents going in = sum of going out = zero, and all 3 have same sign, then only solution is all currents are zero. And if all currents are zero thus are all equal to each other, correct answers is to put whatever other i in answer box and it will be true. Correct current flows must be sorted before asking answers otherwise single solution is that all i1-6=0 and then equal to each other FirstChildUserIdTAG: 331483 FirstChildUserNameTAG: Doru FirstChildCreateTimeTAG: 2012-09-06T09:58:30Z SecondChildTAG: Doru, remember that currents can also be negative meaning that the direction is actually opposite to the one the arrow is showing. You can either reverse your arrow and use the current as positive, or you can just leave the arrow as it is and consider the current as negative. Both options should give the same answer. In circuit analysis it´s very common to find negative currents in your circuits, specially if you don´t know in advance the direction in which current is going! SecondChildUserIdTAG: 381746 SecondChildUserNameTAG: jelizon SecondChildCreateTimeTAG: 2012-09-06T13:25:05Z FirstChildTAG: Is it possible if values were given for the variables, one of the arrows could change directions? If it turns out that say, i5 is negative, the arrow would reverse and current would be flowing into the node. Food for thought. FirstChildUserIdTAG: 132828 FirstChildUserNameTAG: rharris FirstChildCreateTimeTAG: 2012-09-07T05:55:44Z IndexTAG: 2885 TitleTAG: UI tweek I think order of weeks in the left sidebar should be other way round. In 10 weeks it would be annoying to scroll down to find current week. And its nice to go bottom up, as knowledge rise and etc, etc. UserIdTAG: 95280 UserNameTAG: Fortyq CreateTimeTAG: 2012-09-06T09:35:04Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 2886 TitleTAG: S1E9 #4 #4 cut off right side, and treat left side as a loop. the by KVL find voltage that drops on equivalent resistor. and then you know current that will flow through V1. is my explanation correct? UserIdTAG: 95280 UserNameTAG: Fortyq CreateTimeTAG: 2012-09-06T09:23:18Z VoteTAG: 1 CoursewareTAG: Week 1 / Battery model CommentableIdTAG: 6002x_S1E9_Battery_Model NumberOfReplyTAG: 1 FirstChildTAG: yes FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-09-06T16:07:36Z IndexTAG: 2887 TitleTAG: Answer test is just a little too picky! I had problems with the system trying to enter in my answers until I realized that the answers are without V and A at the end. The programming could be better to make them optional. And currently your teaching people bad habits when it comes to writing numbers without identifying what the number is for. UserIdTAG: 323442 UserNameTAG: DanDegagne CreateTimeTAG: 2012-09-06T07:41:07Z VoteTAG: 1 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 2 FirstChildTAG: Me, too, I found it odd to write numerical values without units. Also f'ed up the amp question 'cause I didn't have the correct notation for milliamps. 500e-6 and then no "amps" or A? Weird. FirstChildUserIdTAG: 253337 FirstChildUserNameTAG: Remguy FirstChildCreateTimeTAG: 2012-09-06T14:26:23Z FirstChildTAG: Thanks, You solved one problem I was having on several questions. The system did not accept almost anything I put in. FirstChildUserIdTAG: 359302 FirstChildUserNameTAG: GaryG FirstChildCreateTimeTAG: 2012-09-07T11:58:59Z IndexTAG: 2888 TitleTAG: lab 1 how to get the inverted T symbol since dc analysis is not possible without it and it is not appearing at the right side in tool section(inverted T). Thanks in advance UserIdTAG: 230943 UserNameTAG: samsung CreateTimeTAG: 2012-09-06T07:30:31Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: For lab1, u dnt need any T symbol, as it is already there, u jst have to choose the appropriate valused of the resistors. dts all. jst reread the statement once again. May be u wud get sm hint. also read the paragraph given in the last. i thnk it shud help. FirstChildUserIdTAG: 82571 FirstChildUserNameTAG: Nidhi3 FirstChildCreateTimeTAG: 2012-09-06T07:38:27Z FirstChildTAG: It's needed to rotate the t Symbol, typing "R". Regards FirstChildUserIdTAG: 43833 FirstChildUserNameTAG: gmcentenom FirstChildCreateTimeTAG: 2012-09-06T09:06:41Z SecondChildTAG: Its not needed, as they have given d complete circuit, u jst have 2 insert 2 resistors with appropriate values over there, n dts all. SecondChildUserIdTAG: 82571 SecondChildUserNameTAG: Nidhi3 SecondChildCreateTimeTAG: 2012-09-06T13:09:29Z IndexTAG: 2889 TitleTAG: Facing the problem my tool board is not showing the DC,AC analysis....I'm using the latest updated mozilla firefox....How to overcome this? UserIdTAG: 226563 UserNameTAG: SAIKUMAR9 CreateTimeTAG: 2012-09-06T06:51:41Z VoteTAG: 1 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 4 FirstChildTAG: Same situation here, no AC/DC/TRANS buttons available. Mozilla Firefox 15.0 on Win32. FirstChildUserIdTAG: 113632 FirstChildUserNameTAG: Quinta FirstChildCreateTimeTAG: 2012-09-06T07:01:05Z SecondChildTAG: I submitted random answers to the remaining questions. Pushed "Check" and eventually the DC and TRAN labels appeared. SecondChildUserIdTAG: 113632 SecondChildUserNameTAG: Quinta SecondChildCreateTimeTAG: 2012-09-06T07:03:40Z FirstChildTAG: Why are people down voting this? There is a real problem many of us are seeing with one of their platforms (Obviously the platform/server "all who are having this isssue" are unlucky enough to be on). Please do not down vote posts that are hi-liting the bugs in the system. FirstChildUserIdTAG: 15344 FirstChildUserNameTAG: kob FirstChildCreateTimeTAG: 2012-09-06T10:48:41Z SecondChildTAG: YEAH. Its a problem though but the temporary solution is to click on check and viola, they all appear SecondChildUserIdTAG: 300532 SecondChildUserNameTAG: mjb247 SecondChildCreateTimeTAG: 2012-09-06T11:45:56Z FirstChildTAG: y dnt u search d forum instead of asking d same question again & again. anywyz try using chrome if u r using windows. FirstChildUserIdTAG: 82571 FirstChildUserNameTAG: Nidhi3 FirstChildCreateTimeTAG: 2012-09-06T08:07:05Z SecondChildTAG: This is nothing to do with any browser. If you are not seeing the problem, then your edx account is running from a stable server. Those of us who are seeing this, need the flaky platform server we are allocated to, fixed or patched. I have tested this under all platforms and browsers. Same bugs on all combinations. It is a server issue not a client issue. And on the contrary, I would suggest people do keep asking the same question, until either it gets fixed or a member of staff posts an update. At moment we have heard nothing. SecondChildUserIdTAG: 15344 SecondChildUserNameTAG: kob SecondChildCreateTimeTAG: 2012-09-06T10:54:42Z FirstChildTAG: Just press "Check" below. FirstChildUserIdTAG: 191609 FirstChildUserNameTAG: AlexPoloz FirstChildCreateTimeTAG: 2012-09-06T07:43:24Z IndexTAG: 2890 TitleTAG: Problem 4 Why voltage changes across R1, R2, R3 should be negative? Is it because current is flowing into the reference nodes of these resisters, and these nodes are treated as '+' node? UserIdTAG: 176234 UserNameTAG: hejinjie CreateTimeTAG: 2012-09-06T06:42:55Z VoteTAG: 1 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 6 FirstChildTAG: i am also having the same doubt FirstChildUserIdTAG: 382930 FirstChildUserNameTAG: vikku FirstChildCreateTimeTAG: 2012-09-06T08:07:02Z SecondChildTAG: voltage change from node A to node B is B-A. SecondChildUserIdTAG: 176234 SecondChildUserNameTAG: hejinjie SecondChildCreateTimeTAG: 2012-09-07T01:54:36Z FirstChildTAG: The change in voltage is negative for the KVL equation: Vsupply - V1 - V2 - V3 = 0 FirstChildUserIdTAG: 385796 FirstChildUserNameTAG: occidental FirstChildCreateTimeTAG: 2012-09-06T09:40:21Z SecondChildTAG: we are about to prove KVL, maight not use it, don't we? SecondChildUserIdTAG: 176234 SecondChildUserNameTAG: hejinjie SecondChildCreateTimeTAG: 2012-09-07T02:01:49Z FirstChildTAG: As we starting from the point A, When we reach the point B then the drop of voltage 3V to 2.5 volt shows a drop of .5V. Thus it should be -.5V across the resister R1, in KVL and so on. However, once we reach to the DC voltage source, its value increases from 0V to 3V, so it is became +3V as its value, in KVL. FirstChildUserIdTAG: 224620 FirstChildUserNameTAG: TomVO FirstChildCreateTimeTAG: 2012-09-06T10:13:04Z SecondChildTAG: much clear now, thanks! :) SecondChildUserIdTAG: 176234 SecondChildUserNameTAG: hejinjie SecondChildCreateTimeTAG: 2012-09-07T02:00:05Z FirstChildTAG: I had the same problem,and now I realize that I have misunderstood 'change'.When we define a change from A to B, then the change value should be 'B-A'.So when the voltage drop is positive,the change should be negative. FirstChildUserIdTAG: 361137 FirstChildUserNameTAG: Ericson FirstChildCreateTimeTAG: 2012-09-06T12:28:01Z SecondChildTAG: That helps a lot. Thanks very much! SecondChildUserIdTAG: 176234 SecondChildUserNameTAG: hejinjie SecondChildCreateTimeTAG: 2012-09-07T01:50:17Z FirstChildTAG: http://www.facstaff.bucknell.edu/mastascu/elessonsHTML/Basic/Basic5Kv.html FirstChildUserIdTAG: 252722 FirstChildUserNameTAG: vaboro FirstChildCreateTimeTAG: 2012-09-06T18:42:20Z SecondChildTAG: that helps, thanks! SecondChildUserIdTAG: 176234 SecondChildUserNameTAG: hejinjie SecondChildCreateTimeTAG: 2012-09-07T02:04:39Z FirstChildTAG: It is negative because there is a voltage drop. From one node of the resistor to the other node in clock wise direction, there is a drop. FirstChildUserIdTAG: 355726 FirstChildUserNameTAG: shishirbs FirstChildCreateTimeTAG: 2012-09-06T09:29:02Z SecondChildTAG: now i know voltage drop is a negetive change in clock wise direction, thanks! :) SecondChildUserIdTAG: 176234 SecondChildUserNameTAG: hejinjie SecondChildCreateTimeTAG: 2012-09-07T02:04:00Z IndexTAG: 2891 TitleTAG: component how to change the value of component? UserIdTAG: 206513 UserNameTAG: undhad CreateTimeTAG: 2012-09-06T06:05:25Z VoteTAG: 1 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: Double click, I think FirstChildUserIdTAG: 176234 FirstChildUserNameTAG: hejinjie FirstChildCreateTimeTAG: 2012-09-06T06:31:28Z IndexTAG: 2892 TitleTAG: impulse signals when the signal is impulse, time in nanoseconds UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-06T03:55:35Z VoteTAG: 1 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: sounds like one of the problems with the new supercomputers too much power and speed too little area of chip. since power time should be instantaneous til you get too much power or speed try checking settings on that. either power setting or delay FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-06T14:19:55Z IndexTAG: 2893 TitleTAG: Hope I can help It´s minus 2 because the input power plus the output power has to be 0, so, 2-2=0 UserIdTAG: 244109 UserNameTAG: MichFer CreateTimeTAG: 2012-09-06T03:45:42Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: 5_Simple_Power NumberOfReplyTAG: 2 FirstChildTAG: hi , u are true and there is another explanation ==> the direction of the current which entering (the current enter the voltage source to the negative terminal >>> the current (-0.2)* the voltage (10) which = -2 watt correct me if there is some thing wrong FirstChildUserIdTAG: 193445 FirstChildUserNameTAG: eslammagdy FirstChildCreateTimeTAG: 2012-09-06T16:57:12Z FirstChildTAG: thanks, now i realize that i know this! FirstChildUserIdTAG: 369511 FirstChildUserNameTAG: Year FirstChildCreateTimeTAG: 2012-09-06T23:38:48Z IndexTAG: 2894 TitleTAG: Tools This is great, i like... UserIdTAG: 359257 UserNameTAG: Rony0691 CreateTimeTAG: 2012-09-06T03:28:42Z VoteTAG: 1 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 0 IndexTAG: 2895 TitleTAG: Simulator This is great!! UserIdTAG: 304605 UserNameTAG: George_cba CreateTimeTAG: 2012-09-06T01:16:29Z VoteTAG: 1 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 0 IndexTAG: 2896 TitleTAG: video issue Any one else having issues loading the pages? I can see some problems but not the videos. Using windows 7. Thanks! UserIdTAG: 244115 UserNameTAG: Pedro1969 CreateTimeTAG: 2012-09-05T22:19:53Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Hi Pedro1969! How are you? I have no Problems uploading the pages... What Web broser are you using? Try to use the last Updated version of Chrome or Firefox. I think that can be because of that... ;) Hola Pedro1969! Cómo estás? Yo no tengo este tipo de problemas. Puedo ver perfectamente las páginas del Curso como así también los videos... Puedo preguntarte algo? Qué Navegador estás utilizando? Por experiencia personal por haber cursado el Curso Prototipo de Circuits and Electronics puedo intuir que se deba por el Navegador. Trata de utilizar la última versión de Chrome or Firefox (que son los Navegadores recomendados) y verás que no tendrás inconvenientes. ;) FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-05T22:34:05Z SecondChildTAG: I am also facing same problem SecondChildUserIdTAG: 40209 SecondChildUserNameTAG: sadaf78 SecondChildCreateTimeTAG: 2012-09-05T22:49:47Z SecondChildTAG: When posting questions about browser or video problems, it is very helpful to specify what browser and version, and what operating system and version you are using, so we can assist. SecondChildUserIdTAG: 94557 SecondChildUserNameTAG: g_hopper SecondChildCreateTimeTAG: 2012-09-06T00:57:51Z IndexTAG: 2897 TitleTAG: How to get a copy of the book Hi all, my name´s Marcelo Nascimento Barreto. I´m from manaus, Amazon - Brazil. I am descended from Spanish by my father and my mother is Brazilian. I´m an electrical engineer, 38 years old, married, 3 kids/boys(12 years and 6 years). last friday(August 31) my third son has borned. I´ove been graduaded on november 2008 by University of State of Amazon. but unfortunately I´m not working. So I hope guys that doing this course with U, help me to open oportunities for getting a job. However, I´d like to know how to get a copy of this book: FOUNDATIONS OF ANALOG AND DIGITAL CIRCUITS. Finally, my facebook address is : facebook/marcelo nascimento barreto and my email is: marcelo_nbarreto@yahoo.com.br Its a pleasure to be part of this course! its a dream come true! Believe me! Tnks MIT ! UserIdTAG: 384870 UserNameTAG: marcelo_nbarreto CreateTimeTAG: 2012-09-05T22:00:50Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 4 FirstChildTAG: Hello Marcelo. My name is Rodrigo and I´m from Brazil too. I live in Salto, Sao Paulo, and I´m a software enginner. I saw in Syllabus that the book is available in Amazon.com and I supposed that is possible buy this in digital version (to tablet like iPad) or printed (this option is available). I will add you in my facebook. FirstChildUserIdTAG: 7209 FirstChildUserNameTAG: rodrigoferrazazevedo FirstChildCreateTimeTAG: 2012-09-05T22:11:28Z SecondChildTAG: Hello Rodrigo! How are U? Nice to meet U too ! As I told you before, I´m electrical engineer, graduated in 2008 by Universidade do Estado do Amazonas - UEA(visit its web site as a suggestion). So Does Salto stays near campinas/sp ? Tnks a lot for your comments and you are welcome Rodrigo! SecondChildUserIdTAG: 384870 SecondChildUserNameTAG: marcelo_nbarreto SecondChildCreateTimeTAG: 2012-09-06T01:25:43Z FirstChildTAG: E ai Marcelo, meu nome é Elson, sou de Fortaleza e estou nessa também juntamente com um amigo de faculdade. Estamos cursando Engenharia de Controle e Automação (Engenharia Mecatrônica) na Universidade de Fortaleza (Unifor). Vamos ficar nos falando aqui e tirando nossas dúvidas. Abraço. Hey Marcelo, my name's Elson, I'm from Fortaleza e we're together here. I'm doing it with a friend of mine, we're studying Mechatronics Engineering at the University of Fortaleza (Unifor). Let's keep talking here to talk about the questions that will appear ok? Hugs from a friend! FirstChildUserIdTAG: 156864 FirstChildUserNameTAG: Elson_mpf FirstChildCreateTimeTAG: 2012-09-05T23:08:24Z SecondChildTAG: Olá Elson...acabei de ver seu post. Obrigado pela mensagem. desejo sucesso pra vc´s na unifor. vc agora tem meu contato de face e email acima, qualquer duvida nos ajudamos mutuamente. um abraço! SecondChildUserIdTAG: 384870 SecondChildUserNameTAG: marcelo_nbarreto SecondChildCreateTimeTAG: 2012-09-06T01:14:34Z FirstChildTAG: Hello all. I think this course is a great opportunity for all of us, specially for you, Marcelo. Hope you find a good job soon. I'm from Brazil too: live in Goiânia. Have you (all of you) got Google+? FirstChildUserIdTAG: 198988 FirstChildUserNameTAG: brunoconterato FirstChildCreateTimeTAG: 2012-09-10T11:04:11Z FirstChildTAG: Oi Marcelo! I am from Argentina, it is really near Brazil ;). I hope you can find a Job soon. I am sure that you will! You can take a look to edX opportunities [here][1] You can freely read online the Textbook in this Web Page's Course [here][2] . Also, you can buy it [here][3] Please read the Course Info where it says that "If you would like to buy a print version of the book, a discounted copy is available directly from the publisher" [Course Info][4]. Nice to meet you! ----- Hola Marcelo! Yo soy de Argentina! Estamos muy cerca! ;). Espero que puedas encontrar un trabajo pronto! Seguro que sí! Creo que edX tiene Oportunidades Laborales, si quieres puedes ver [Aquí][5] En cuanto al libro del Texto, lo puedes leer en forma online y gratuita desde esta página web del Curso [Aquí te dejo el Link][6]. Además, puedes comprar el libro [aquí][7]. Si quieres puedes ver en donde dice [Course Info][4] en donde se menciona que "Si quieres obtener una versión del libro en papel, existe un descuento a través del Publicador Directo". Encantada de Conocerte! [1]: https://www.edx.org/jobs [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/ [3]: http://%20http://store.elsevier.com/specialOffer.jsp?offerId=EST_MIT&utm_source=EMAP&utm_medium=text_link&utm_term=Circuits_Electronics&utm_campaign=edx#oid=1003_4 [4]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/info [5]: https://www.edx.org/jobs [6]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/ [7]: http://%20http://store.elsevier.com/specialOffer.jsp?offerId=EST_MIT&utm_source=EMAP&utm_medium=text_link&utm_term=Circuits_Electronics&utm_campaign=edx#oid=1003_4 [8]: http:// FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-05T22:57:01Z SecondChildTAG: Hola que tal! Como miraste arriba yo soy engeniero electrico, vivo en manaus/brasil. tengo 38 años, pero usted sabes, no estoy trabajando, creo por la crise economica mundial. Muchas gracias por los "links" que me enviaste y la ayuda tambien. Cómo te llamas? es engieniera tbn? mi facebook: facebook/marcelo nascimento barreto Saludos a usted y argentina ! buenas noches...como diria mi papa...salud y suerte ! SecondChildUserIdTAG: 384870 SecondChildUserNameTAG: marcelo_nbarreto SecondChildCreateTimeTAG: 2012-09-06T01:21:17Z IndexTAG: 2898 TitleTAG: Problem in Lab 1 The interactive tool is just showing resistance, nothing else. Then how to solve this and submit the result. Also I clicked check(button): after which it showed me the DC analysis options, but doining this gave a cross (wrong) on my window indicating a wrong answer. So will this be counted as submitted with wrong answer. please help. UserIdTAG: 315378 UserNameTAG: gbp1984 CreateTimeTAG: 2012-09-05T21:13:27Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: I cleared browser cache, it helped FirstChildUserIdTAG: 279867 FirstChildUserNameTAG: Lisenik FirstChildCreateTimeTAG: 2012-09-05T21:22:05Z FirstChildTAG: Clearing the cache won't make any difference. (those not seeing these problem are running from stable servers. I suspect that some of the servers (Maybe its just one) with user accounts on it, needs patching. These problem are nothing to do with the browser (client side code). The issue is software glitch on the faulty platform we've been unlucky enough to be allocated to. Everyone else will not see these problems and they are always the ones that assume its just the browser ha ha! FirstChildUserIdTAG: 15344 FirstChildUserNameTAG: kob FirstChildCreateTimeTAG: 2012-09-06T11:27:26Z IndexTAG: 2899 TitleTAG: its been a while i forgot the resistors were in 1,000 range...ugh UserIdTAG: 252746 UserNameTAG: tommyd CreateTimeTAG: 2012-09-05T21:05:32Z VoteTAG: 1 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 2 FirstChildTAG: Same here FirstChildUserIdTAG: 241495 FirstChildUserNameTAG: NickT02LS FirstChildCreateTimeTAG: 2012-09-06T02:50:23Z FirstChildTAG: Hi tommyd! If this help you, You can also write in the box (simulation window) 1k instead of 1000 ;). It is less confusing like that. ---- Hola tommyd! Si esto te es de ayuda, hay algo interestante. En la ventana en donde escribes los parámetros de los elementos que arrojas a la ventana del sandbox o de simulación, puedes escribir 1k en vez de 1000 ;). Esto puede ayudarte a que sea menos confuso. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-05T23:32:31Z IndexTAG: 2900 TitleTAG: Current dilemma My only error was the current question :P Also, how is current negative? I haven't heard much bout current flowing from negative to positive terminal. UserIdTAG: 200391 UserNameTAG: subinabraham CreateTimeTAG: 2012-09-05T20:50:20Z VoteTAG: 1 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 2 FirstChildTAG: see the current direction indicated in the circuit, the reference is opposite to the actual :-) FirstChildUserIdTAG: 185815 FirstChildUserNameTAG: adilmeersha FirstChildCreateTimeTAG: 2012-09-05T21:01:03Z FirstChildTAG: hi..the simple answer is that if you take the direction of current clockwise (positive to negative) the sign will positive..and vice versa FirstChildUserIdTAG: 282892 FirstChildUserNameTAG: wiky FirstChildCreateTimeTAG: 2012-09-05T21:22:38Z IndexTAG: 2901 TitleTAG: unable to view course material I am using internet explorer and am unable to view labs, tutorials etc Can any help explain why... UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-05T20:08:52Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: It seems that tools don't work equally with all browsers. Today I tried Firefox, IE, and GChrome, and at this time Chrome behaves better than the other two. Maybe they will fix this later. FirstChildUserIdTAG: 164171 FirstChildUserNameTAG: PacoJuarez FirstChildCreateTimeTAG: 2012-09-05T20:18:49Z FirstChildTAG: Use Google Chrome. FirstChildUserIdTAG: 42833 FirstChildUserNameTAG: KKA FirstChildCreateTimeTAG: 2012-09-05T20:53:15Z IndexTAG: 2902 TitleTAG: Voltage component missing in lab1 Managed to see the DC tool in lab 1 now. But only resisitor component showing. Anyone figured out how this works yet? Looks like its now working. A reset button has now appeared and that is able to show the correct compoents on the page. Hooray! UserIdTAG: 15344 UserNameTAG: kob CreateTimeTAG: 2012-09-05T20:07:41Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 3 FirstChildTAG: I have exact same issue. All visible in sandbox so must be a code issue on platform itself. Be good if a staff member could post an update. FirstChildUserIdTAG: 15344 FirstChildUserNameTAG: kob FirstChildCreateTimeTAG: 2012-09-05T20:51:07Z SecondChildTAG: I had same problem. Clearing browser cashe helped. SecondChildUserIdTAG: 279867 SecondChildUserNameTAG: Lisenik SecondChildCreateTimeTAG: 2012-09-05T20:59:32Z FirstChildTAG: You should only need to add resistors to the circuits in Lab 1 in order to complete the lab. FirstChildUserIdTAG: 376895 FirstChildUserNameTAG: alecgg FirstChildCreateTimeTAG: 2012-09-05T21:43:48Z FirstChildTAG: Why are people down voting bug reports. If you are not having the problem many of us are seeing then you are running on a stable platform at the server. Users are allocated to different servers and it seems many of us have been unlucky and are running on one that has problems. So quit with down voting and just do your course! Idiot! FirstChildUserIdTAG: 15344 FirstChildUserNameTAG: kob FirstChildCreateTimeTAG: 2012-09-06T11:31:11Z IndexTAG: 2903 TitleTAG: using comma here a linear resistor is both a and f.so i used a comma in between them.my answer is shown wrong.but why?i didn't understand. UserIdTAG: 138095 UserNameTAG: zobzyer CreateTimeTAG: 2012-09-05T19:59:34Z VoteTAG: 1 CoursewareTAG: Week 1 / Various V-I characteristics CommentableIdTAG: 6002x_S1E1_Various_V-I_characteristics NumberOfReplyTAG: 10 FirstChildTAG: I used whitespace to seperate the answer and it was also shown wrong:( FirstChildUserIdTAG: 190711 FirstChildUserNameTAG: heliumhgy FirstChildCreateTimeTAG: 2012-09-14T02:42:53Z FirstChildTAG: that is the reason edx asked us to enter only one correct answer FirstChildUserIdTAG: 156713 FirstChildUserNameTAG: HARITEJAREDDYP FirstChildCreateTimeTAG: 2012-09-06T13:41:56Z FirstChildTAG: you must choose on answer not 2 FirstChildUserIdTAG: 236136 FirstChildUserNameTAG: Hamid-ch FirstChildCreateTimeTAG: 2012-09-05T20:52:17Z FirstChildTAG: Note, there may be more than one correct answer to each question, but please only provide one answer. FirstChildUserIdTAG: 367260 FirstChildUserNameTAG: Jesusrb FirstChildCreateTimeTAG: 2012-09-06T04:14:33Z FirstChildTAG: in the instruction it is clearly mentoned that..... Note, there may be more than one correct answer to each question, but please only provide one answer......... FirstChildUserIdTAG: 346913 FirstChildUserNameTAG: nashid FirstChildCreateTimeTAG: 2012-09-07T14:41:11Z FirstChildTAG: "Note, there may be more than one correct answer to each question, **but please only provide one answer**". This is what the instruction says. FirstChildUserIdTAG: 214990 FirstChildUserNameTAG: Shuaibu FirstChildCreateTimeTAG: 2012-09-08T13:44:22Z FirstChildTAG: Perhaps it confused the software... as it states, there *may* be more than one correct answer to a problem, but they still ask that you choose *only one* correct answer. FirstChildUserIdTAG: 340856 FirstChildUserNameTAG: RanmaSaotome FirstChildCreateTimeTAG: 2012-09-09T05:59:25Z FirstChildTAG: please read question carefully......we have to give only one answer..........so we can put either a or f but one at a time FirstChildUserIdTAG: 123036 FirstChildUserNameTAG: ABHINAV_5 FirstChildCreateTimeTAG: 2012-09-15T10:03:57Z FirstChildTAG: try using "or" instead of comma FirstChildUserIdTAG: 186808 FirstChildUserNameTAG: Adot FirstChildCreateTimeTAG: 2012-09-08T19:48:41Z FirstChildTAG: well perhaps its because they mentioned to give only one answer, so try either a or f FirstChildUserIdTAG: 191292 FirstChildUserNameTAG: hitu271 FirstChildCreateTimeTAG: 2012-09-05T20:27:11Z IndexTAG: 2904 TitleTAG: Checking simulator for the first time. Hello everyone, after some minutes, I checked my errors, in KVL forgot the sign for voltages, and when calculating forgot that if we want to know the KVL for a resistor, in a circuit in series, we must make the sum of all values of the other resistors divided by the value of resistor of witch we want to know the KVL, and multiply this by the value of the source. In transient analysis missed the offset value. But it's all ok now and I understood my errors. Good luck. UserIdTAG: 317836 UserNameTAG: VitorRodrigues CreateTimeTAG: 2012-09-05T19:34:52Z VoteTAG: 1 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 2 FirstChildTAG: KVL is Kirchoff Voltages Law, and it is calculated in a loop, not in a single component. So I don't know what you meant when you said about the "KVL for a resistor". FirstChildUserIdTAG: 188586 FirstChildUserNameTAG: FLara FirstChildCreateTimeTAG: 2012-09-05T20:44:47Z SecondChildTAG: I was misunderstanding KVL and KCL, don't worry I've read some pages and watched the videos. I understand it now. SecondChildUserIdTAG: 317836 SecondChildUserNameTAG: VitorRodrigues SecondChildCreateTimeTAG: 2012-09-12T01:02:03Z FirstChildTAG: I recommend you to check page 73 of the textbook, about voltage dividers. FirstChildUserIdTAG: 188586 FirstChildUserNameTAG: FLara FirstChildCreateTimeTAG: 2012-09-05T20:50:15Z SecondChildTAG: I still need a help, on the little exercise, if we have two resistors of 4 ohms and another of 6ohms, connected in parallel making a smaller resistor of 1.5 ohms, and each one of them can dissipate 1W before burning, how much will dissipate the one of 1.5ohms before burning, according to exercise of the heaters i should have an amount of 3W dissipating in the three resistors,(because they're in parallel) so if i have 3w and 1.5ohms p=vi, should be those 3w... wrong answer? SecondChildUserIdTAG: 317836 SecondChildUserNameTAG: VitorRodrigues SecondChildCreateTimeTAG: 2012-09-12T02:46:34Z IndexTAG: 2905 TitleTAG: Fractions vs decimals Really? You guys put all this work into these courses, but couldn't do something as simple as make your answer check recognize that '.333*R' and 'R/3' are the same thing? Disappointing. What is going to happen during an "exam" when someone doesn't do their answer exactly the way your answer check is expecting? UserIdTAG: 357979 UserNameTAG: 3rdto1st CreateTimeTAG: 2012-09-05T19:11:17Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: I understand your frustration, but perhaps this isn't worded in the best way? FirstChildUserIdTAG: 25252 FirstChildUserNameTAG: cwm9 FirstChildCreateTimeTAG: 2012-09-05T19:18:53Z SecondChildTAG: He's incorrect anyway SecondChildUserIdTAG: 80920 SecondChildUserNameTAG: MichaelSturgess SecondChildCreateTimeTAG: 2012-09-05T19:21:15Z FirstChildTAG: Maybe you shouldn't be so abrupt with your questioning. If you had a 10M resistor and divided it by three you wouldn't get the same as 10M*0.333 would you! Try adding further 3's to your accuracy and you'll find it does accept it. They put all this effort in so you could be more precise with your answer. FirstChildUserIdTAG: 80920 FirstChildUserNameTAG: MichaelSturgess FirstChildCreateTimeTAG: 2012-09-05T19:19:34Z SecondChildTAG: "precision, precision, precision.". :) SecondChildUserIdTAG: 92346 SecondChildUserNameTAG: MobiusTruth SecondChildCreateTimeTAG: 2012-09-06T02:22:07Z IndexTAG: 2906 TitleTAG: Power supplied by individual Sources I did not follow how to calculate the power supplied by individual sources UserIdTAG: 135072 UserNameTAG: Shriniwas CreateTimeTAG: 2012-09-05T19:03:31Z VoteTAG: 1 CoursewareTAG: Week 1 / KVL/KCL example CommentableIdTAG: 6002x_S2E3_Using_KVL_KCL_and_VI_constraints NumberOfReplyTAG: 1 FirstChildTAG: The voltage of the current sourse is the same as the R2 resistor. So you can multiply it buy current of this source. V2 = 7.778V, so 7.778*3 = 23.33W. FirstChildUserIdTAG: 200319 FirstChildUserNameTAG: Virviil FirstChildCreateTimeTAG: 2012-09-05T20:09:40Z SecondChildTAG: hi, by associated reference directions, the voltage across the current source is equal to V2=7.778 in that case, aren`t the reference direction current through Source is -3, rather than positive 3? thus, the power is negative. i`m really confused here... would explain this to me, thanks in advance. ^0^ SecondChildUserIdTAG: 386237 SecondChildUserNameTAG: forlsy SecondChildCreateTimeTAG: 2012-09-07T01:23:36Z IndexTAG: 2907 TitleTAG: textbook hey can we download the book from here UserIdTAG: 350334 UserNameTAG: nikhilpadho CreateTimeTAG: 2012-09-05T18:41:56Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: The online book is not available for downloading, but you can either purchase the book directly from Elsevier using the discount as shown on the Course Info page, or from another bookseller. If you are in India, you can buy an inexpensive legal copy of the text from Flipkart. If you have an eBay account, you can purchase an international version, they are legal to import for personal use in the US, I don't know about the laws in other countries. FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-09-09T03:43:05Z IndexTAG: 2908 TitleTAG: suggestion for an interactive video before this task,a small video explaining all the elements used in the tools could have been made to make things much clearer.. UserIdTAG: 347775 UserNameTAG: praveen_yvs CreateTimeTAG: 2012-09-05T17:56:35Z VoteTAG: 1 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: In fact, there is already a video which explains the different elements. You can find it in the **edX Tutorial** section. FirstChildUserIdTAG: 158985 FirstChildUserNameTAG: n1xx FirstChildCreateTimeTAG: 2012-09-05T18:17:16Z IndexTAG: 2909 TitleTAG: Difficulties and problems faced in first lab Sir i was unable to change the values of resistance like figure 1. i wrote the answer for 3rd question as 6k and -.0005 which are showing wrong ? I could not understand the question 5 .. UserIdTAG: 18782 UserNameTAG: tyagi CreateTimeTAG: 2012-09-05T16:58:51Z VoteTAG: 1 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 2 FirstChildTAG: double click on resistance after placing it in circuit....there you can change the value. FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-05T17:18:35Z FirstChildTAG: The answer boxes appear to not accept "k" and other notations FirstChildUserIdTAG: 383637 FirstChildUserNameTAG: IanCull FirstChildCreateTimeTAG: 2012-09-05T17:31:31Z SecondChildTAG: 6k is my answer, and it's ok SecondChildUserIdTAG: 210380 SecondChildUserNameTAG: MurdocRus SecondChildCreateTimeTAG: 2012-09-05T18:51:45Z IndexTAG: 2910 TitleTAG: Naming the nodes How do I name the nodes A, B, C and D in my schematic? UserIdTAG: 140184 UserNameTAG: nagudaku CreateTimeTAG: 2012-09-05T16:45:07Z VoteTAG: 1 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 2 FirstChildTAG: With node labeling tool FirstChildUserIdTAG: 219182 FirstChildUserNameTAG: Glider23 FirstChildCreateTimeTAG: 2012-09-05T16:51:02Z SecondChildTAG: thanks, somehow i missed it. SecondChildUserIdTAG: 140184 SecondChildUserNameTAG: nagudaku SecondChildCreateTimeTAG: 2012-09-05T16:52:53Z FirstChildTAG: You are not supposed to name the nodes with A, B, C and D. In the image those letters appear only to identify the nodes so that in the next question you know which node the question refers to. FirstChildUserIdTAG: 300827 FirstChildUserNameTAG: PFonseca FirstChildCreateTimeTAG: 2012-09-05T16:48:36Z SecondChildTAG: this being a tutorial the point is to get a hang of the tools so that we can use them effectively when we build larger schematics later in the course (i presume). SecondChildUserIdTAG: 140184 SecondChildUserNameTAG: nagudaku SecondChildCreateTimeTAG: 2012-09-05T16:56:00Z IndexTAG: 2911 TitleTAG: power supplied by current source i have a question for calculating power supplied by the current source .if i or current source equals i0. haven't in our case power v*I should be -v2*I ? UserIdTAG: 277593 UserNameTAG: aleks83 CreateTimeTAG: 2012-09-05T16:12:09Z VoteTAG: 1 CoursewareTAG: Week 1 / KVL/KCL example CommentableIdTAG: 6002x_S2E3_Using_KVL_KCL_and_VI_constraints NumberOfReplyTAG: 3 FirstChildTAG: I think, that the key word is "supply". Thats why Current Source gives huge power with "+", and Voltage Source - with "-". It looks like in this curcuit voltage source is cunsumer, thats why it's "supplied" power is "-". Otherwize - current source. But as for me, it's also a bit hard to understand, what sign is asked... FirstChildUserIdTAG: 200319 FirstChildUserNameTAG: Virviil FirstChildCreateTimeTAG: 2012-09-05T16:44:58Z FirstChildTAG: The power supplied by current source is the voltage drop across R2 * current source, V2 * 3 = 23.3 Amps. FirstChildUserIdTAG: 183947 FirstChildUserNameTAG: gfenton06 FirstChildCreateTimeTAG: 2012-09-05T23:05:25Z SecondChildTAG: Nice! Thanks SecondChildUserIdTAG: 230787 SecondChildUserNameTAG: Gabriel007 SecondChildCreateTimeTAG: 2012-09-08T18:34:12Z FirstChildTAG: True. Thanks for the clue. :) FirstChildUserIdTAG: 136093 FirstChildUserNameTAG: KrishYash FirstChildCreateTimeTAG: 2012-09-06T13:49:22Z IndexTAG: 2912 TitleTAG: How i place wire connections? I am feeling issue in placing wire connections. whenever i try to connect wires , respective components becomes selected instead of making wire connection.Can any one guide me in this regards? Thanks UserIdTAG: 347789 UserNameTAG: engr_zohaib CreateTimeTAG: 2012-09-05T15:53:21Z VoteTAG: 1 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 2 FirstChildTAG: What web browser are you using? I had the same problem a few minutes ago. I change my browser to Mozilla and it solved the problem. Please try other web browsers FirstChildUserIdTAG: 132229 FirstChildUserNameTAG: ahlo FirstChildCreateTimeTAG: 2012-09-05T16:05:13Z SecondChildTAG: Dear i am using Google chrome . Well i try it on Mozilla as well.Thanks buddy .... SecondChildUserIdTAG: 347789 SecondChildUserNameTAG: engr_zohaib SecondChildCreateTimeTAG: 2012-09-05T16:26:14Z SecondChildTAG: Thank you very much SecondChildUserIdTAG: undefined SecondChildUserNameTAG: undefined SecondChildCreateTimeTAG: 2012-09-05T17:57:22Z FirstChildTAG: OK. So If I'm on Firefox I can't see the DC tool, if I switch to Chrome, I can't draw wires! FirstChildUserIdTAG: 326402 FirstChildUserNameTAG: Net79 FirstChildCreateTimeTAG: 2012-09-05T18:24:26Z SecondChildTAG: Sorry Net79, but in both browsers DC tool is present. SecondChildUserIdTAG: 132229 SecondChildUserNameTAG: ahlo SecondChildCreateTimeTAG: 2012-09-05T23:03:51Z IndexTAG: 2913 TitleTAG: progress actually when i go to progress tool bar , i am getting zero marks in welcome to 6.002x and circuit sandbox ? kindly provide me information about it ! UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-05T15:07:22Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 3 FirstChildTAG: Those sections are not graded: 0/0 :) FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-05T15:21:44Z FirstChildTAG: they are not graded. FirstChildUserIdTAG: 320303 FirstChildUserNameTAG: yoshimi FirstChildCreateTimeTAG: 2012-09-05T15:23:16Z FirstChildTAG: thanks for providing info . thanks alot :) FirstChildUserIdTAG: 158348 FirstChildUserNameTAG: mudz FirstChildCreateTimeTAG: 2012-09-05T17:42:15Z IndexTAG: 2914 TitleTAG: will there be any reminders about course activity Hi. I was wondering if the edx system is going to remind about missing home work or unwatched lectures automatically ? I can easily see myself forget to sign into the site and check the calendar. UserIdTAG: 271081 UserNameTAG: YonJah CreateTimeTAG: 2012-09-05T14:50:10Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 2915 TitleTAG: Transient Analysis is done with scope referenced to ground I don't think the transient analysis is correct. The 1 VOlt offset would be split like a voltage divider, and the 3 signals would NOT be referenced to ground. Then again, the scope probe is referenced to ground on all 3 measurements. So now I see it. If the scope probe was not referenced to ground, but placed across the resistors, would we see the voltage division of the DC offset voltage? UserIdTAG: 145676 UserNameTAG: CathyPK CreateTimeTAG: 2012-09-05T14:41:52Z VoteTAG: 1 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 0 IndexTAG: 2916 TitleTAG: Number Of Loops i can't understand how the number of loops is 7. i am getting only 4 loops. could someone please explain? thanks... UserIdTAG: 222911 UserNameTAG: bhaswardg CreateTimeTAG: 2012-09-05T14:20:05Z VoteTAG: 1 CoursewareTAG: Week 1 / Circuit topology CommentableIdTAG: 6002x_S2E1_Circuit_Topology NumberOfReplyTAG: 4 FirstChildTAG: let V- = e, V+ = f so loops are 1.fabcef 2.adba 3.bdcb 4.fabdcef 5.fadbcef 6.adcba 7.fadcef FirstChildUserIdTAG: 326083 FirstChildUserNameTAG: kaiwalya8 FirstChildCreateTimeTAG: 2012-09-05T14:51:07Z FirstChildTAG: I can see 6, I guessed 2, because I thought it was talking comeplete loops; ie 2 connections back to the source, but looking at the answer it's going through the circuit and wrapping back around to the source. But, I'm still having trouble finding the last one. FirstChildUserIdTAG: 306208 FirstChildUserNameTAG: Creamsaw FirstChildCreateTimeTAG: 2012-09-05T14:24:10Z SecondChildTAG: I guess the straight line of v to a to b to d counts as a loop? I'm still confused SecondChildUserIdTAG: 306208 SecondChildUserNameTAG: Creamsaw SecondChildCreateTimeTAG: 2012-09-05T14:29:00Z SecondChildTAG: Loop 1 - the whole thing, all the way around the outside Loop 2 - the big one on the left Loops 3 and 4 - the smaller ones on the right Loop 5 - the two smaller ones together Loop 6 - the big one and the top smaller one Loop 7 - the big one and the bottom smaller one SecondChildUserIdTAG: 196693 SecondChildUserNameTAG: danbshaw SecondChildCreateTimeTAG: 2012-09-06T17:48:45Z FirstChildTAG: 1 Loop is the simply closed circuit, and in this bridge network you can form 7 different closed circuits. For example VaR1bR2c one of them. VaR4dR5c other. FirstChildUserIdTAG: 325634 FirstChildUserNameTAG: Aleksey_Stepanov FirstChildCreateTimeTAG: 2012-09-05T14:34:25Z FirstChildTAG: Think of all the possible paths you can traverse the circuit - i.e., start at one point and get back to the same point - each distinct closed path is a loop. FirstChildUserIdTAG: 42833 FirstChildUserNameTAG: KKA FirstChildCreateTimeTAG: 2012-09-05T14:47:38Z IndexTAG: 2917 TitleTAG: summary SO basically we always start out our circuits from voltage and when we hit (-) the current is negative and if we hit (+) the current is positive in writing equation? UserIdTAG: 277593 UserNameTAG: aleks83 CreateTimeTAG: 2012-09-05T13:52:48Z VoteTAG: 1 CoursewareTAG: Week 1 / Associated reference directions CommentableIdTAG: 6002x_S2E2_Associated_Reference_Directions NumberOfReplyTAG: 2 FirstChildTAG: Yeah, I felt like a retard, I had all the right answers on the second half just on the wrong side of the spectrum... FirstChildUserIdTAG: 306208 FirstChildUserNameTAG: Creamsaw FirstChildCreateTimeTAG: 2012-09-05T15:13:18Z FirstChildTAG: No in electricity a current is positive when the electrons goes from the negative point to the positive one!! FirstChildUserIdTAG: 151444 FirstChildUserNameTAG: mael FirstChildCreateTimeTAG: 2012-09-06T06:56:46Z IndexTAG: 2918 TitleTAG: Can a ground symbol be added to the Lab 1 circuit elements column? Could someone from MITx add the ability to draw a new ground connection point (the inverted T), to Lab 1. If for whatever reason it wasn't there, or it gets deleted, there is no way of completing Lab 1 which is really annoying. If I cannot complete the first lab simply because of this feature, I'm going to struggle to motivate myself to continue with the rest of the course. UserIdTAG: 80920 UserNameTAG: MichaelSturgess CreateTimeTAG: 2012-09-05T13:41:49Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 2919 TitleTAG: Lab 1 impossible - unless I'm missing something? Has anyone attempted to complete Lab 1 - the inverted T symbol is not available and so the DC analysis cannot be completed. I have tried on Google Chrome, Firefox and IE. Any ideas folks? UserIdTAG: 80920 UserNameTAG: MichaelSturgess CreateTimeTAG: 2012-09-05T13:11:07Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Did you delete the inverted T by mistake? It should be there by default. In case you did delete it, you can copy it from the second circuit. Select the ground (inverted T), click copy in the toolbar. Then click paste in the first circuit sandbox. If you've deleted both T's I'm not sure what to do :-( FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-09-05T13:15:51Z SecondChildTAG: I don't think I've deleted them, there's normally the simple to add a ground connection on the right hand side along with the other circuit elements. SecondChildUserIdTAG: 80920 SecondChildUserNameTAG: MichaelSturgess SecondChildCreateTimeTAG: 2012-09-05T13:18:48Z IndexTAG: 2920 TitleTAG: I've just noticed that the old Profile link for the discussion board is missing. Does this mean we won't have to worry about KARMA EGO striking again? :-) (If you find this baffling, see https://6002x.mitx.mit.edu/discussion/question/62686/has-anyone-else-noticed-that-karma-ego-is-haunting .) UserIdTAG: 79337 UserNameTAG: AppliedImagination CreateTimeTAG: 2012-09-05T12:45:31Z VoteTAG: 1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: So does that mean there is no Karma anymore? FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-09-05T20:21:13Z SecondChildTAG: On a cosmic scale I'm not sure... :-) As far as this board goes, I'm puzzled: I've just given you what I think is an upvote, but I don't know what good it does or how you see what you've accumulated, etc. SecondChildUserIdTAG: 79337 SecondChildUserNameTAG: AppliedImagination SecondChildCreateTimeTAG: 2012-09-06T13:48:12Z IndexTAG: 2921 TitleTAG: Problems **Many problems in our country is about the system of education so i use edX to help the new génération in our country and in our city** UserIdTAG: 1555513 UserNameTAG: GENIUSSI666 CreateTimeTAG: 2013-04-08T12:59:48Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 2922 TitleTAG: problem ? e2/R5 + (e2-e1)/R3 + (e2-V0)/R4 - I1=0 and when I press the check button i get "Error: Invalid input: Could not parse 'e2/R5 + (e2-e1)/R3 + (e2-V0)/R4 - I1=0' as a formula " !!!!! UserIdTAG: 1515242 UserNameTAG: Muradalfgeeh CreateTimeTAG: 2013-04-06T08:01:10Z VoteTAG: 0 CoursewareTAG: Week 1 / Node analysis practice, part 3 CommentableIdTAG: 6002x_L2Node2 NumberOfReplyTAG: 1 FirstChildTAG: thnx I get the answer we should write the LFS means without " =0 " FirstChildUserIdTAG: 1515242 FirstChildUserNameTAG: Muradalfgeeh FirstChildCreateTimeTAG: 2013-04-06T08:04:37Z IndexTAG: 2923 TitleTAG: EEC What's the EEC ? UserIdTAG: 1515242 UserNameTAG: Muradalfgeeh CreateTimeTAG: 2013-04-05T08:32:09Z VoteTAG: 0 CoursewareTAG: Week 1 / Lumped Element Abstraction CommentableIdTAG: 6002x_lumped_element_abstraction NumberOfReplyTAG: 0 IndexTAG: 2924 TitleTAG: K constant in real life As we have seen in the mosfet Characteristics, we have an equation for iD in terms of "k", I was wondering how can We Determinate this constant "K" according to the mosfet datasheet?, what about BJT Transistors?, what is their formula?, as we saw in Week 5 Sideloads BJT Signal Model, the equation is given as -(gm)(RL)(Vpi)=Vout but these are not realistic parameters. Best Regards UserIdTAG: 231676 UserNameTAG: Roosemberth CreateTimeTAG: 2013-04-05T03:57:13Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 2925 TitleTAG: To develop a electrical circuit for a gas sensor. I have very few knowledge about electrical circuit designing as I am not from this field. But now I have to develop a circuit for my gas sensor material which I synthesized by myself. As far my knowledge a sensor has three terminals such as Vcc- to power up the sensor, GND- to provide a fixed negative reference and a output. But I don't know how to integrate these three terminals to develop my own sensor. Moreover I don't have any idea about the significance of each terminals. Please help me in this regard. Thanking you. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2013-04-01T17:37:23Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi! I would advise you to post your question in the current session (Spring 2013) of this course. FirstChildUserIdTAG: 297370 FirstChildUserNameTAG: ayush3504 FirstChildCreateTimeTAG: 2013-04-01T20:45:22Z IndexTAG: 2926 TitleTAG: What is the value during the transition from 0-1 or 1-0? In the digital circuit, at the time of transition from 0-1 or from 1-0, what is the value of the circuit? It is neither a 1 nor a 0. So what is it? UserIdTAG: 1499380 UserNameTAG: Priyankamg CreateTimeTAG: 2013-03-30T16:38:33Z VoteTAG: 0 CoursewareTAG: Week 2 / Digital Abstraction CommentableIdTAG: 6002x_digital NumberOfReplyTAG: 1 FirstChildTAG: Hi Priyankamg, Can I help you? This is the 6.002x Course of the Fall 2012... There is running a new Course of 6.002x Spring 2013. https://www.edx.org/courses/MITx/6.002x/2013_Spring/about Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2013-03-31T15:39:43Z IndexTAG: 2927 TitleTAG: At 2.5 V, is it considered as logical 1 or logical 0? Above 2.5 V, it is considered as 1. Below 2.5 V, it is considered as 0. What is the value at 2.5 V? UserIdTAG: 1499380 UserNameTAG: Priyankamg CreateTimeTAG: 2013-03-30T16:32:46Z VoteTAG: 0 CoursewareTAG: Week 2 / Digital Abstraction CommentableIdTAG: 6002x_digital NumberOfReplyTAG: 1 FirstChildTAG: Hi Priyankamg, Can I help you? This is the 6.002x Course of the Fall 2012... There is running a new Course of 6.002x Spring 2013. https://www.edx.org/courses/MITx/6.002x/2013_Spring/about Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2013-03-31T15:38:57Z IndexTAG: 2928 TitleTAG: Homework1 and Lab1 I cannot find any "Check" button so how can I put my answers? Instead of "Check" button, I can only find "Show Answer" button and Explanations. Or is "check" button appeared on the day of 24th March(submission date)? I accidentally posted this question in General tag. I am really worried that staffs or course-mates won't see my post so I posted it in Troubleshooting tag again. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2013-03-23T14:55:47Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Troubleshooting NumberOfReplyTAG: 1 FirstChildTAG: Hi anonymous, This Course of the fall have already end... You can take a look at here for the new 6.002x Course (Spring) [here][1]... [1]: https://www.edx.org/courses/MITx/6.002x/2013_Spring/about FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2013-03-25T02:24:54Z IndexTAG: 2929 TitleTAG: This is not the forum for the Spring class This forum is not for the Spring 2013 class that just started. This is the forum for the Fall 2012 class. You can find the Spring class here: https://www.edx.org/courses/MITx/6.002x/2013_Spring/courseware/Week_1/ UserIdTAG: 21541 UserNameTAG: JSChambers CreateTimeTAG: 2013-03-14T00:00:32Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 2930 TitleTAG: - I1? The sum of the currents "leaving" the node shouldn't include I1 since it is entering the node. Do the answer should be (e2-e1)/R3 + (e2-V0)/R4 + e2/R5 , without I1. Isn't it? UserIdTAG: 136742 UserNameTAG: giovanni_83 CreateTimeTAG: 2013-03-12T14:07:05Z VoteTAG: 0 CoursewareTAG: Week 1 / Node analysis practice, part 3 CommentableIdTAG: 6002x_L2Node2 NumberOfReplyTAG: 1 FirstChildTAG: You realize that this forum is not for the Spring 2013 class that just started, but for the class that has already ended, right? You can find the Spring class here: https://www.edx.org/courses/MITx/6.002x/2013_Spring/courseware/Week_1/ FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2013-03-13T21:32:05Z IndexTAG: 2931 TitleTAG: sign problem HI I cant understand how to do that "Choose the sign of the change for a component using the first terminal you come to in the clockwise traversal as the reference node" the sign confused me it said start from node a and go clockwise why the voltage difference on resistance is negative? thank you UserIdTAG: 1402055 UserNameTAG: SHREIF CreateTimeTAG: 2013-03-11T13:06:57Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 0 IndexTAG: 2932 TitleTAG: Thanks !! great intuition, it was really confusing to me and my professor couldn't answer me when I asked him about it- although he's paid around $16 000 a month !!! UserIdTAG: 154541 UserNameTAG: AhmedMedhat CreateTimeTAG: 2013-02-28T20:46:08Z VoteTAG: 0 CoursewareTAG: Week 13 / S25V2_Negative_vs._positive_feedback CommentableIdTAG: 6002x_S25V2_Negative_vs._positive_feedback NumberOfReplyTAG: 0 IndexTAG: 2933 TitleTAG: how to use lab tools...pls help.... i feel great difficulty in using the lab tools ,especially in making connection wires.. what can i do?... UserIdTAG: 538070 UserNameTAG: newuser CreateTimeTAG: 2013-02-27T10:34:11Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 2 FirstChildTAG: first select a component; resistor or something click with the left mousebutton on the end of the wire of the component,hold the button pressed and start drawing a straight line, when you release the line is drawed and you can go on drawing the next piece from there.If you end a drawed line onto a existing line a connection will be made between the two lines (black dot). FirstChildUserIdTAG: 79885 FirstChildUserNameTAG: ruudoleo FirstChildCreateTimeTAG: 2013-02-28T20:28:00Z FirstChildTAG: Take a look here https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/MyrimitCircuitSimulatorTutorial/ Thanks to Myrimit FirstChildUserIdTAG: 325197 FirstChildUserNameTAG: Vitali_Jerin FirstChildCreateTimeTAG: 2013-03-02T07:22:55Z IndexTAG: 2934 TitleTAG: Gnuplot For the gnu linux and derivatives one can use gnuplot. To run gnuplot simply type gnuplot in the command prompt on a xterminal (X-Windows needs to be running). Once in gnuplot type the following to define the window scale aspect: gnuplot> set xrange [-5:35] gnuplot> set yrange [-30:250] then to plot the graph enter the following: gnuplot> plot x**3 title 'Device Constrains', 4-x/8.2 title 'Thevenin Equivalent Constrains' Note: you can always tweak the range to better visualize certain aspects of the graph UserIdTAG: 968923 UserNameTAG: tk3000 CreateTimeTAG: 2013-02-26T10:08:33Z VoteTAG: 0 CoursewareTAG: Week 3 / Graphical Analysis of Nonlinear Circuits Exercise CommentableIdTAG: 6002x_graphical_analysis_of_nonlinear_circuits_exercise NumberOfReplyTAG: 0 IndexTAG: 2935 TitleTAG: Good Very good this tool. UserIdTAG: 1243345 UserNameTAG: Jocrisdias CreateTimeTAG: 2013-02-25T23:00:19Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: yeah it's very good FirstChildUserIdTAG: 1515242 FirstChildUserNameTAG: Muradalfgeeh FirstChildCreateTimeTAG: 2013-04-02T08:04:07Z IndexTAG: 2936 TitleTAG: Old exams to go over before the exam . where can i find them ? UserIdTAG: 1308469 UserNameTAG: Nati88 CreateTimeTAG: 2013-02-25T16:39:25Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: hope this helps https://www.edx.org/static/content-mit-6002x~2012_Fall/handouts/6002x-FinalReview-S2012-clean.pdf and with solutions https://www.edx.org/static/content-mit-6002x~2012_Fall/handouts/6002x-FinalReview-S2012.pdf FirstChildUserIdTAG: 325197 FirstChildUserNameTAG: Vitali_Jerin FirstChildCreateTimeTAG: 2013-03-02T07:32:03Z IndexTAG: 2937 TitleTAG: BINGO THANK YOU for giving the overview on simulation tools... UserIdTAG: 1255566 UserNameTAG: Deepak95 CreateTimeTAG: 2013-02-24T02:51:17Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 0 IndexTAG: 2938 TitleTAG: Good luck with the exam everyone For those taking the proctored exam, I hope it went/goes well for you. Like many others (I suspect) I am waiting to hear all about it. UserIdTAG: 9673 UserNameTAG: xp42 CreateTimeTAG: 2013-02-13T18:38:49Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Good luck from my part also, but what's all the fuss with this proctored exam ? Does this help someone to get a job ? This is just an introductory course and can serve only as a prerequisite in the best case scenario. Does it have university credit value ? Because except it is from mighty mit it can also help a lot of people to brag. FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2013-02-14T08:51:27Z SecondChildTAG: Well, there is nothing wrong with bragging rights. It also might be very helpful in job interviews, even if it's just to show that you can do things like this. If I thought I could pass it in the time given, I would have taken it. I don't think I could have. Especially without wolfram alpha, as my math skills are not the best. Personally, I don't see why you can't use WA, because if you don't understand how circuits really work, a math assistant will not save you. But whatever. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2013-02-14T16:26:00Z IndexTAG: 2939 TitleTAG: iphone/ipad app development?? hello 6.002x...i have a rather irrevelant question...does anyone know if there is an online app development course available for iphone/ipad? im a newbie in this field so any kind of help would be welcomed best regards :) UserIdTAG: 278792 UserNameTAG: sali CreateTimeTAG: 2013-02-12T04:45:59Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: You could start with the youtube lessons at : [ipad/iphone lessons][1] or follow the free stanford university course at: [iPad and iPhone App Development (Fall 2011)][2] or one of the three other Iphone app courses at: [stanford][3] [1]: http://www.youtube.com/user/RubyA1234/videos [2]: https://itunes.apple.com/us/course/ipad-iphone-app-development/id495052415 [3]: http://www.openculture.com/computer_science_free_courses FirstChildUserIdTAG: 79885 FirstChildUserNameTAG: ruudoleo FirstChildCreateTimeTAG: 2013-02-12T08:25:24Z FirstChildTAG: if you are interested in android..then this course will help you.. [link][1] ...hope there will be more such course in future.. [1]: https://www.coursera.org/course/androidapps101 FirstChildUserIdTAG: 114864 FirstChildUserNameTAG: Vikaash FirstChildCreateTimeTAG: 2013-02-12T12:55:45Z FirstChildTAG: thanks friends :) FirstChildUserIdTAG: 278792 FirstChildUserNameTAG: sali FirstChildCreateTimeTAG: 2013-02-19T04:05:08Z IndexTAG: 2940 TitleTAG: The importance of a period Grammar Teacher: Do you know the importance of a period? Kid: Yeah, once my sister said she has missed one, my mom fainted, dad got a heart attack & our driver ran away I found this funny until I realized a grammar teacher is not a math teacher and wouldn't ask for the importance of the period of a sinusoidal signal. Looks like I've become a ‘narrow-minded specialist’ . UserIdTAG: 79885 UserNameTAG: ruudoleo CreateTimeTAG: 2013-02-11T21:03:44Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 2941 TitleTAG: about second question i am nor able to find the solution... UserIdTAG: 1069402 UserNameTAG: Kamleshpri CreateTimeTAG: 2013-01-29T16:49:25Z VoteTAG: 0 CoursewareTAG: Week 1 / Power delivery modeling CommentableIdTAG: 6002x_S2E6_Modeling NumberOfReplyTAG: 2 FirstChildTAG: what is your question ? FirstChildUserIdTAG: 236136 FirstChildUserNameTAG: Hamid-ch FirstChildCreateTimeTAG: 2013-01-29T21:29:31Z FirstChildTAG: Kamleshpri, are you sure you are in the right forum. This course has ended. You should subscribe to the new Spring version in Edx if you want to restart the course. FirstChildUserIdTAG: 79885 FirstChildUserNameTAG: ruudoleo FirstChildCreateTimeTAG: 2013-01-29T22:54:27Z IndexTAG: 2942 TitleTAG: Pearson exam registration. Hi, i have found the option for Pearson exam registration in my dashboard.But how can i complete my registration and attend for Pearson exam?How can i pay $95?Please give me details information. UserIdTAG: 360053 UserNameTAG: Ohedul CreateTimeTAG: 2013-01-29T07:48:39Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Sorry about lack of coordination with the availability of the registration button and full communication about exam logistics. https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5107e1ff6732881f0000001b FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2013-01-29T14:52:51Z IndexTAG: 2943 TitleTAG: peareson registeration they said in the announcement of the proctored exam that the last eligible appointment is 13th feb? does that mean that the last day for taking the exam is in 13th feb? UserIdTAG: 221617 UserNameTAG: konan CreateTimeTAG: 2013-01-29T07:44:39Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: No, the only date available for taking the exam is Feb 13th. We can't be as flexible with times as the Pearson exam seats are a much more limited resource than server time. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2013-01-29T14:55:14Z SecondChildTAG: thanks SecondChildUserIdTAG: 221617 SecondChildUserNameTAG: konan SecondChildCreateTimeTAG: 2013-01-29T15:04:49Z IndexTAG: 2944 TitleTAG: Hi ,how many of you will register for this Pearson exam ? How much is it? And what's the value of it?...Hesitating UserIdTAG: 136959 UserNameTAG: christerpher CreateTimeTAG: 2013-01-29T07:05:56Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 2945 TitleTAG: Unable to find the centre for the proctored I have not been able to find the center for the proctored examination in Nepal. Can you help me with it? UserIdTAG: 413613 UserNameTAG: shaliesh CreateTimeTAG: 2013-01-29T05:11:27Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: that mean they are no avalible centers for edx exam in your area ; FirstChildUserIdTAG: 236136 FirstChildUserNameTAG: Hamid-ch FirstChildCreateTimeTAG: 2013-01-29T08:09:11Z SecondChildTAG: it means there are no available centers for the exam in the whole country??.... i tried to search for it and would always get the nearest center as New Delhi(India). i cant reach there to give the exam and i want to really give it. SecondChildUserIdTAG: 413613 SecondChildUserNameTAG: shaliesh SecondChildCreateTimeTAG: 2013-01-30T03:34:43Z SecondChildTAG: how can i ask the staffs at edx about it?? is it in this discussion forum SecondChildUserIdTAG: 413613 SecondChildUserNameTAG: shaliesh SecondChildCreateTimeTAG: 2013-01-30T03:36:10Z IndexTAG: 2946 TitleTAG: typying error?? I have write down the expression (e2-e1)/R3+(e2-V0)/R4+e2/R5+I1=0?? where i am making mistake in typing?? UserIdTAG: 1069402 UserNameTAG: Kamleshpri CreateTimeTAG: 2013-01-28T18:41:14Z VoteTAG: 0 CoursewareTAG: Week 1 / Node analysis practice, part 3 CommentableIdTAG: 6002x_L2Node2 NumberOfReplyTAG: 1 FirstChildTAG: The sum of all currents leaving a node is zero. You are summing three currents leaving the node with one current entering the node. (I1 is entering the e2 node, not leaving it). FirstChildUserIdTAG: 90796 FirstChildUserNameTAG: ChaunceyGardiner FirstChildCreateTimeTAG: 2013-01-28T21:58:11Z IndexTAG: 2947 TitleTAG: typeing problem... Invalid input: a e1 r1 a e2 r4 not permitted in answer how to remove this problem? UserIdTAG: 1069402 UserNameTAG: Kamleshpri CreateTimeTAG: 2013-01-28T18:14:47Z VoteTAG: 0 CoursewareTAG: Week 1 / Node analysis practice, part 1 CommentableIdTAG: 6002x_L2Node0 NumberOfReplyTAG: 0 IndexTAG: 2948 TitleTAG: Duality explained Haven't done the Fall version of 6002x but in the Spring version we got a complimentary surprise dessert by Piotr and Gerry. I don't think it was posted this year as only 144 people looked at it in youtube. [duality explained][1] [1]: http://www.youtube.com/watch?v=vgGDm-rdsrQ UserIdTAG: 79885 UserNameTAG: ruudoleo CreateTimeTAG: 2013-01-28T13:39:17Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 2949 TitleTAG: 6003z already started? Hi, ammubhave I have seen week 2 is already posted. Can you inform a bit about the planning as I am just starting (and sweating already). UserIdTAG: 79885 UserNameTAG: ruudoleo CreateTimeTAG: 2013-01-25T23:18:56Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_6003z NumberOfReplyTAG: 1 FirstChildTAG: Hi, no no it hasn't started just yet. We are right now developing a tentative calender. I was just transferring data from the old server. FirstChildUserIdTAG: 118131 FirstChildUserNameTAG: ammubhave FirstChildCreateTimeTAG: 2013-01-26T14:17:30Z IndexTAG: 2950 TitleTAG: Guys, I do really need your help Guys, can you give me some materials of references for Single-axis Accelerometer. This is my topic for my thesis. I will make an electronic single-axis accelerometer tester. Could you please help me? It is very hard to get some sources. I am hoping for your support, suggestion and help. Thanks. UserIdTAG: 340272 UserNameTAG: ezekielbrizuela CreateTimeTAG: 2013-01-22T18:38:27Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: What do you really need? For example try [this][1]. I dont see any problems with accelerometers.. [1]: http://www.analog.com/en/mems-sensors/mems-inertial-sensors/adxl103/products/product.html FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2013-01-22T21:56:24Z FirstChildTAG: The Arduino board could offer a simple and low-cost way to control an accelerometer. Give a look at www.arduino.cc and you will find plenty of discussions about using accelerometers with this board. best regards FirstChildUserIdTAG: 372321 FirstChildUserNameTAG: EnricoDona FirstChildCreateTimeTAG: 2013-01-23T10:18:07Z FirstChildTAG: Maybe you need this: http://www.analog.com/static/imported-files/application_notes/AN-1057.pdf FirstChildUserIdTAG: 297370 FirstChildUserNameTAG: ayush3504 FirstChildCreateTimeTAG: 2013-01-27T09:12:22Z IndexTAG: 2951 TitleTAG: certificate name change I want to change the name in my certificate. It is printed wrong..to what id can I mail the name change request.. UserIdTAG: 394124 UserNameTAG: ManeeshaSatya CreateTimeTAG: 2013-01-21T15:04:16Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: discussion about that here : https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50e59e1886b3ec1f00000007 FirstChildUserIdTAG: 236136 FirstChildUserNameTAG: Hamid-ch FirstChildCreateTimeTAG: 2013-01-21T15:31:35Z FirstChildTAG: Hi ManeeshaSatya, Unfortunately, as far as I have read, *the Certificate will have the complete name as it was in your dashboard when the Certificate it was generated*... You can edit your name in your dashboard *but this new name will be applied for the future Certificates* (this was wrote by a Staff in previous Posts)... Myriam. P.D. As a student,by hart, if I were in your same situation, I would try to write them explaining what have happened and if it is any possibility of editing my name( mit-6002x@edx.org )... FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2013-01-21T21:00:27Z IndexTAG: 2952 TitleTAG: to staff what is the deadline for the proctored exam ? UserIdTAG: 221617 UserNameTAG: konan CreateTimeTAG: 2013-01-17T15:18:13Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: As far as I am aware, a date has not been announced yet. Check back here soon! FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2013-01-18T00:52:38Z IndexTAG: 2953 TitleTAG: Sandbox I'm not sure if we lose access to the course material in the next few weeks, but it would sure be nice if we could maintain access to the sandbox. :-) Thanks again for an excellent course! UserIdTAG: 194122 UserNameTAG: J_Dennis CreateTimeTAG: 2013-01-16T16:38:27Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: You will continue to have access to the Circuit Sandbox as well as other course materials. Have fun. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2013-01-16T16:44:43Z FirstChildTAG: or if you want you can always download for free LTspice that is similar to the Sandbox! FirstChildUserIdTAG: 99441 FirstChildUserNameTAG: coyarce FirstChildCreateTimeTAG: 2013-01-17T01:28:34Z FirstChildTAG: circuitlab.com is almost identical to the sandbox and can be used on-line without any installation. FirstChildUserIdTAG: 372321 FirstChildUserNameTAG: EnricoDona FirstChildCreateTimeTAG: 2013-01-18T08:44:41Z IndexTAG: 2954 TitleTAG: certificate till now i dont get the certificate UserIdTAG: 414688 UserNameTAG: rajkumarpettikkal CreateTimeTAG: 2013-01-15T20:12:48Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: did you get more than 60% ? FirstChildUserIdTAG: 454609 FirstChildUserNameTAG: Alwahsh FirstChildCreateTimeTAG: 2013-01-15T20:20:37Z FirstChildTAG: Hi rajkumarpettikkal, Can I help you? Have you checked your dashboard? if your final score is equal or more than 60 %, you will see a blue button that says: download the certificate... https://www.edx.org/dashboard If you still having problems, please let me know... I hope this can help you. Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2013-01-15T21:28:57Z IndexTAG: 2955 TitleTAG: Circuit's SandBox trick hey guys, first post (thx for any redirection to a better sub-forum/wiki or any answer at all) Like you, i enjoy key like ctrl+z/y to undo or redo ctrl+x/c/v/ to cut, copy or paste But we can't do that in the SandBox. It also make me sad. Will there be more sandbox later on the course? If not, which program do EECS guys use to design electronical schema? EDIT for clarity: *The sandbox before week1; (I'm on week 1 here) UserIdTAG: 1000511 UserNameTAG: georgesLurk CreateTimeTAG: 2013-01-14T07:30:40Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: LTspice IV is absolutely free http://www.linear.com/designtools/software/ FirstChildUserIdTAG: 325197 FirstChildUserNameTAG: Vitali_Jerin FirstChildCreateTimeTAG: 2013-01-14T14:24:15Z SecondChildTAG: I said it's free of charge but it's not free software. http://en.wikipedia.org/wiki/Free_software SecondChildUserIdTAG: 376877 SecondChildUserNameTAG: AndBre SecondChildCreateTimeTAG: 2013-01-14T14:56:28Z SecondChildTAG: Now is clear. Thanks for explanation. SecondChildUserIdTAG: 325197 SecondChildUserNameTAG: Vitali_Jerin SecondChildCreateTimeTAG: 2013-01-14T16:28:28Z SecondChildTAG: I use the same...for simple circuits it is fine, but form advanced ones you need to use Cadence or similar SecondChildUserIdTAG: 99441 SecondChildUserNameTAG: coyarce SecondChildCreateTimeTAG: 2013-01-14T17:56:35Z FirstChildTAG: In the sandbox you can't do ctrl-z/y but you can do ctrl-x/c/v (if you can't do that, try to install an updated browser and/or clear the browser cache). About circuits simulation, personally I use, besides the edX sandbox (it is quite good actually, especially if you need to draw a schema quickly), "LTspice" for Windows systems (you can download it free of charge, but it's not free software) and the "geda" suite for unix-like systems (if you prefer free software). FirstChildUserIdTAG: 376877 FirstChildUserNameTAG: AndBre FirstChildCreateTimeTAG: 2013-01-14T09:03:41Z FirstChildTAG: edX Staff is also very accommodating and always looking for your feedback. If you have any suggestions for new features or fixes, they would like to hear about it. You can list such threads under "feedback" when creating them. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2013-01-14T14:30:52Z IndexTAG: 2956 TitleTAG: Excellent! This is an excellent tool. It takes a little getting used to but it works great. It would help if we can re-size the scope window. UserIdTAG: 621482 UserNameTAG: jfarhat CreateTimeTAG: 2013-01-13T16:13:14Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 0 IndexTAG: 2957 TitleTAG: S8E1: what is little v from the formula for i? is it vs? UserIdTAG: 795417 UserNameTAG: rnepxk CreateTimeTAG: 2013-01-11T05:42:40Z VoteTAG: 0 CoursewareTAG: Week 4 / Dependent Sources Exercise 1 CommentableIdTAG: 6002x_dep_src_exsz_1 NumberOfReplyTAG: 1 FirstChildTAG: is the same v across R FirstChildUserIdTAG: 236136 FirstChildUserNameTAG: Hamid-ch FirstChildCreateTimeTAG: 2013-01-11T06:58:59Z IndexTAG: 2958 TitleTAG: to ashwith rego(CCEC) to ashwith rego(CCEC) this is my name which i use it in this edx forum- NAME-aditya jayant mhatre email-adjmhatre@gmail.com UserIdTAG: 533341 UserNameTAG: adjmhatre CreateTimeTAG: 2013-01-08T11:47:02Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Got it. Thanks :-) FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2013-01-08T12:41:01Z IndexTAG: 2959 TitleTAG: Certificate Hi, Could you please change my name in certificate from "akbar" to "Akbar Shadakov" Pleeeease.... Thanks UserIdTAG: 341358 UserNameTAG: akbar13 CreateTimeTAG: 2013-01-07T11:22:05Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi akbar13, As far as I have read, the certificate will show the name that was in your dashboard at the moment that the Certificate was being generated... so, unfortunately, this certificate will display the name akbar... But, they said that you can change-edit your name in your dashboard with Akbar Shadakov for the furture Certificates... But, as student, in my point of view, if I were in your position, I would try to not close any door and I would e-mail them ( mit-6002x@edx.org )if there is any change to change it... I am not sure that if they will accept your petition but you can always try... My best wish to you, Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2013-01-07T13:27:41Z IndexTAG: 2960 TitleTAG: about proctored exam pls inform about the Pearson VUE test center locations (an edX partner)available in india, esp in Maharashtra. also the date and duration of exam. UserIdTAG: 91261 UserNameTAG: mr_sophisticated CreateTimeTAG: 2013-01-07T08:40:55Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Please do use Discussion forum! [link][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50e740f8ca362e1f0000000d FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2013-01-07T09:29:06Z SecondChildTAG: thnks for the link. but whn is the exam dated and whts the duration ??? SecondChildUserIdTAG: 91261 SecondChildUserNameTAG: mr_sophisticated SecondChildCreateTimeTAG: 2013-01-08T15:50:52Z SecondChildTAG: It is yet to be announced, it should be later on this month of January. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2013-01-09T01:10:35Z SecondChildTAG: whn will the registeration start ??? SecondChildUserIdTAG: 91261 SecondChildUserNameTAG: mr_sophisticated SecondChildCreateTimeTAG: 2013-01-11T07:15:27Z IndexTAG: 2961 TitleTAG: examination fee for proctored exam before or after do we have to pay before giving the exam or after if we clear it & want the certificate? UserIdTAG: 267220 UserNameTAG: ABHINAVSAXENA318619 CreateTimeTAG: 2013-01-07T04:37:09Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I think you have to pay BEFORE exam. FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2013-01-07T08:15:19Z IndexTAG: 2962 TitleTAG: proctored exam hi I'm from egypt but i can't find my country on pareson site UserIdTAG: 221617 UserNameTAG: konan CreateTimeTAG: 2013-01-05T07:56:53Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: السلام عليكم ازيك يا عمي الحج و مبروك الشهادة FirstChildUserIdTAG: 156535 FirstChildUserNameTAG: badawiIiIi FirstChildCreateTimeTAG: 2013-01-05T08:48:59Z SecondChildTAG: السلام عليكم :) SecondChildUserIdTAG: 283909 SecondChildUserNameTAG: MahmoudMosad SecondChildCreateTimeTAG: 2013-01-05T17:20:19Z SecondChildTAG: و عليكم السلام و رحمة الله و بركاتة SecondChildUserIdTAG: 156535 SecondChildUserNameTAG: badawiIiIi SecondChildCreateTimeTAG: 2013-01-05T22:30:19Z SecondChildTAG: ممكن تعطوني فكره عن pearson vue في اش ممكن استفيد منه SecondChildUserIdTAG: 156535 SecondChildUserNameTAG: badawiIiIi SecondChildCreateTimeTAG: 2013-01-05T22:31:39Z FirstChildTAG: دول بعض مراكز الاختبار زي ماشفتهم في الموقع Test Center: ORASCOM TRAINING AND TECHNOLOGY Address: 13 Demeshk Street Orascom Co Roxy Heliopolis Cairo, 20 Egypt Map MapQuest directions are sometimes inaccurate. Be sure to read the test center directions below. Telephone: +20222563322 --------------------------------------------------------------------------------------------------------------- Test Center: N.E.N (National Education Network) Address: 66 main mehwar, 5th district Vodafone Square 6th October, 12573 Egypt Map MapQuest directions are sometimes inaccurate. Be sure to read the test center directions below. Telephone: +201068827096/7 Directions: "6October Branch" Building # 66, Main Mehwar, 5th District, 6october city, (Vodafone Square) MAP: https://maps.google.com/maps/ms?msid=204748542318851200028.0004c4026349450f884dd&msa=0&ie=UTF8&t=m&z=10&vpsrc=0&iwloc=0004c402838ed1bbbe6ab You are welcome any time between 11:00 am and 10:00 pm Sat to Wed. Pearson VUE: Room no (1) Prometric: Room no (2) ETS TOEFl-IBT Room no (5) ICDL: Room no (3) CertiPort: Room no (6) *** please check the online schedule *** Contacts US: Tel.: +20 (2) 3830 8960/1 Fax: +20 (2) 3830 3773 D.S: +20106 882 7096/7 Testing Team: TCA@nen-global.org http://www.nen-global.org/nen_contactus.htm Test Center: Raya Academy for IT & Management Address: 26th July Street Touristic Zone Behind Dar El Fouad hospital 6th october, 12568 Egypt Map MapQuest directions are sometimes inaccurate. Be sure to read the test center directions below. Telephone: 202 38276091 Directions: After Juhayna Square, behind Dar El Fouad You are welcome any time between 09:00 am and 2:00 pm Sundays to Thursdays. Test Center: New Vision Address: 220 Saad Ben Ebada 7th zone- 6 October, 12573 Egypt Map MapQuest directions are sometimes inaccurate. Be sure to read the test center directions below. Telephone: 0020201229906081 Test Center: New Vision Address: 90 D Ahmed Oraby St El Mohandseen Floor 11 , Flat 5 Giza, 12411 Egypt Map MapQuest directions are sometimes inaccurate. Be sure to read the test center directions below. Telephone: 20233025353 Test Center: YAT Education Center Address: 77 Syria st. Crossing of Sudan St. Mohandessen Giza, 12311 Egypt Map MapQuest directions are sometimes inaccurate. Be sure to read the test center directions below. Telephone: +202-33057716 Test Center: AasyeaTech Company Address: 22A Salah Salem RD Emarat El Obour, 15th Floor - Apt. 3 Heliopolis Cairo, Egypt Map MapQuest directions are sometimes inaccurate. Be sure to read the test center directions below. Telephone: 24033141/01221723484 Test Center: Harvest Egypt Address: Al Galaa St, Above Toty Center, Harvest Training Center, Mansoura, 35111 Egypt Map MapQuest directions are sometimes inaccurate. Be sure to read the test center directions below. Telephone: 502220370 Directions: In front of Mowafy Restaurante Cross of Al-Galaa St and Al-Gammal St Phone: 502220370 502220371 Mob : 116662516 Web: http://www.harvesteg.com Test Center: Information Technology Engineering Consultancy Address: Ahmed Maher St. on the corner of Elzareaf St. before Panda Mall Mansoura, 35111 Egypt Map MapQuest directions are sometimes inaccurate. Be sure to read the test center directions below. Telephone: 002 050 223 2331 Test Center: YAT Education - Mansoura Address: 95 El-Gomhoreya Street in front of the Univeristy Hospital 3rd Floor Mansoura, 35111 Egypt Map MapQuest directions are sometimes inaccurate. Be sure to read the test center directions below. Telephone: +20502223064 Test Center: Out Box Address: 50 Al Gomhoria St. Ops to Children Hospital Mansoura, 35111 Egypt Map MapQuest directions are sometimes inaccurate. Be sure to read the test center directions below. Telephone: 0020502212232 Test Center: Hi-Q Academy Address: 16 Botros street above El-Salam Market Hi-Q Academy Tanta, 31515 Egypt Map MapQuest directions are sometimes inaccurate. Be sure to read the test center directions below. Telephone: 20 40 3314283 Directions: Above El-Salam supermarket Test Center: YAT Education - Tanta Address: 9 Moaweya St. from El Geish St. Tanta, 31121 Egypt Map MapQuest directions are sometimes inaccurate. Be sure to read the test center directions below. Telephone: +2 040 3299035 Directions: 9 Moawya St. from El Bahr ST. (Formaly El Geish), EL Khashab building, above El Delta Engoneerong Co for printers and copiers,. 1st floor, YAT Education Center. FirstChildUserIdTAG: 528897 FirstChildUserNameTAG: vepoo FirstChildCreateTimeTAG: 2013-01-05T08:53:09Z FirstChildTAG: http://www.pearsonvue.com/servlet/vue.web2.core.Dispatcher?webContext=CandidateSite&webApp=TestCenterLocator&requestedAction=register&cid=346 ده لينك تحديد الاماكن حسب البلد FirstChildUserIdTAG: 528897 FirstChildUserNameTAG: vepoo FirstChildCreateTimeTAG: 2013-01-05T09:02:06Z FirstChildTAG: بالله ممكن تعطوني فكره عن هذا الامتحان و في اش ممكن استفيد منة FirstChildUserIdTAG: 156535 FirstChildUserNameTAG: badawiIiIi FirstChildCreateTimeTAG: 2013-01-05T22:32:56Z IndexTAG: 2963 TitleTAG: Grade To: Admin My progress states that I have a 84% final grade, but on my dash board it shows only, that I got 81%, why it comes like this how it is done and computed? thanks! UserIdTAG: 349139 UserNameTAG: 1977ROYELMER CreateTimeTAG: 2013-01-04T10:55:00Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: The readjusted the scoring to account for some misunderstanding in the wording of question 6 I believe. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2013-01-04T13:29:51Z SecondChildTAG: In certificate Grade is not mention why ??? SecondChildUserIdTAG: 106219 SecondChildUserNameTAG: veereshpatil SecondChildCreateTimeTAG: 2013-01-07T07:28:57Z IndexTAG: 2964 TitleTAG: video lectures download I have a suggestion for the edX team. They should provide a download link that don't use YOUTUBE. Because people like download this lecture in the company and study at home. And in our company YOUTUBE is blocked.the coursera website provide download link that don't use YOUTUBE.If edX use same kind of thing then it will be really easy for people like us to download the video lectures. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2013-01-04T09:48:20Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 3 FirstChildTAG: Below each lecture, the option to download is given..that works even in places where youtube is blocked.. FirstChildUserIdTAG: 173147 FirstChildUserNameTAG: cruiser_rahit FirstChildCreateTimeTAG: 2013-01-04T10:42:20Z FirstChildTAG: Yes and changing service providers will not resolve the issues between your government and the internet. If you mean "Company" instead of "Country", I don't think edX is interested in helping you circumvent your companies rules or internet policy. If you ask your bosses permission, perhaps he/she will enable you use of the companies internet for your own persoanal educational use. If your company finds out about Coursera, they will likely block that too. You may find the use of a Proxy server helpful. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2013-01-04T13:24:14Z FirstChildTAG: http://www.ruudoleo.com/mitx/small/ to download small outside youtube use videolan videoplayer(http://www.videolan.org/) to see the subtitles, put srt file in same directory as the video file if your internet speed is average you can start viewing the video in your browser. FirstChildUserIdTAG: 79885 FirstChildUserNameTAG: ruudoleo FirstChildCreateTimeTAG: 2013-01-07T06:50:11Z IndexTAG: 2965 TitleTAG: grade The progress tab on my window shows 81% but the final grade on my dashboard shows 79%.Just wanted to know why is that?Thankyou everyone for your support. Regards Jay UserIdTAG: 358198 UserNameTAG: Jay1492 CreateTimeTAG: 2013-01-04T06:23:56Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 2966 TitleTAG: Thankyou Thankyou edx Team 4 organizing such a wonderful course. Learned alot. Just saw my certificate....!! Feels GREAT :D UserIdTAG: 127800 UserNameTAG: MNB CreateTimeTAG: 2013-01-03T18:20:58Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 2967 TitleTAG: certificat comment je peu ajoute mon prenom sur le certificte UserIdTAG: 316899 UserNameTAG: elou CreateTimeTAG: 2013-01-03T13:15:42Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: je croix que c'est deja trop tard , mais vous pouvez essayer de changer votre nom sur https://www.edx.org/dashboard et d'essayer de contacter EDX par mail. il y a une deuxieme methode qui ne va pas changer votre nom sur le certificat mais plutot de l'enregistrer sur le code de verification si je comprend bien);y a un lien en bas de certificat cliquez sur ce lien et essayez de suivre les instruction ou bien faites une petite recherche sur wikipedia sur les signatures numerique FirstChildUserIdTAG: 236136 FirstChildUserNameTAG: Hamid-ch FirstChildCreateTimeTAG: 2013-01-03T14:09:53Z SecondChildTAG: Je ne crois pas que ca soit encore possible, voir : https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50e59e1886b3ec1f00000007 Ici, on ne parle pas de changement complet de nom mais juste d'ajout de prenom, mais ca me semble un peu tard... --------------- I don't think it's still possible, see : https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50e59e1886b3ec1f00000007 In your case, we don't talk about a full name modification with only the first name to be added but that seems a bit late... SecondChildUserIdTAG: 396446 SecondChildUserNameTAG: RousseauxS SecondChildCreateTimeTAG: 2013-01-03T17:09:38Z SecondChildTAG: - un autre moyen c'est l'examen surveillé "proctored exame", cette option n'est pas disponible en ce moment - ou bien réinscrire dans la nouvelle session SecondChildUserIdTAG: 236136 SecondChildUserNameTAG: Hamid-ch SecondChildCreateTimeTAG: 2013-01-03T22:53:24Z IndexTAG: 2968 TitleTAG: Homeworks, mid, final downloads How to download homeworks, mid, final and practice questions? I tried to open the save final, but it didnt reopen. So any suggestions!! UserIdTAG: 254355 UserNameTAG: waqasbinabbas CreateTimeTAG: 2013-01-03T12:53:42Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I printed them. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2013-01-03T13:25:55Z SecondChildTAG: There is no link to download them but a USEFULL tool for that (as I did) is to install in my MOZILLA FIREFOX browser the application called "Print Edit". It is a nice tool developed for printing websites. Furthermore, if you install "PDF Creator" (a free software) you can create a virtual print fo your docs. Hope my advice helps you! SecondChildUserIdTAG: 99441 SecondChildUserNameTAG: coyarce SecondChildCreateTimeTAG: 2013-01-03T16:55:23Z SecondChildTAG: I didn't know that there was an app for that. Thanks for the info. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2013-01-03T17:12:27Z SecondChildTAG: google chrome works for printing too, and pdf redirect (free software) is good too SecondChildUserIdTAG: 207728 SecondChildUserNameTAG: RossyFromBulgaria SecondChildCreateTimeTAG: 2013-01-03T17:13:22Z SecondChildTAG: you are welcome SecondChildUserIdTAG: 99441 SecondChildUserNameTAG: coyarce SecondChildCreateTimeTAG: 2013-01-03T17:46:52Z SecondChildTAG: Thanks for the suggestions SecondChildUserIdTAG: 254355 SecondChildUserNameTAG: waqasbinabbas SecondChildCreateTimeTAG: 2013-01-04T09:47:10Z IndexTAG: 2969 TitleTAG: certificate Problem When will we get the certificate?i didnt get it still! UserIdTAG: 150120 UserNameTAG: krishna1993 CreateTimeTAG: 2013-01-03T12:33:23Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: same here... i didn't get yet. FirstChildUserIdTAG: 219356 FirstChildUserNameTAG: Karan123 FirstChildCreateTimeTAG: 2013-01-03T13:03:06Z IndexTAG: 2970 TitleTAG: Are certificates given alphabetically ? I feel it is so looking at the usernames of persons who got it and the others who are still waiting. UserIdTAG: 99628 UserNameTAG: Pranjal16 CreateTimeTAG: 2013-01-03T12:16:46Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: If so, lucky you!! I am a "V", you will propably get it before me. FirstChildUserIdTAG: 304587 FirstChildUserNameTAG: Vasso FirstChildCreateTimeTAG: 2013-01-03T12:31:39Z SecondChildTAG: )) SecondChildUserIdTAG: 394836 SecondChildUserNameTAG: v2g6ch4 SecondChildCreateTimeTAG: 2013-01-03T12:37:36Z FirstChildTAG: I don't think so. Friend of mine is with J and still hasn't got his certificate. But I'm curious to know what the principe is. Anybody knows? FirstChildUserIdTAG: 207728 FirstChildUserNameTAG: RossyFromBulgaria FirstChildCreateTimeTAG: 2013-01-03T12:43:00Z SecondChildTAG: me too frm J and not received yet! they have distributed till 'go' now.. 'h', 'i' ,'j' ....... :( SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2013-01-03T12:45:42Z IndexTAG: 2971 TitleTAG: changin name in Certificate i forgot to chnage my name to full one .. so if u can change it in the certificate !! plzz can anyone heeelp UserIdTAG: 397578 UserNameTAG: bensiamar CreateTimeTAG: 2013-01-03T11:23:42Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi bensiamar, You can edit your name in your dashboard. https://www.edx.org/dashboard I hope this can help you. Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2013-01-03T11:30:19Z SecondChildTAG: I changed the name to my full name but it is showing only the previous one in the certificate. How to solve it. SecondChildUserIdTAG: 369051 SecondChildUserNameTAG: praveencbe0 SecondChildCreateTimeTAG: 2013-01-03T14:56:10Z SecondChildTAG: It is possible that the system have not accepted the s edit of your name because it was after your Certificate generation... I remember that last term happened the same of you with some students -they got their Certificates but after some days more after their request by email of changing...-...You should e-mail them to mit-6002x@edx.org I hope this can help you... Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2013-01-03T17:16:27Z IndexTAG: 2972 TitleTAG: Certificate I'm very happy. Good new year to EDX TEAM. Thanks. UserIdTAG: 260994 UserNameTAG: capvl CreateTimeTAG: 2013-01-03T11:19:42Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 2973 TitleTAG: my calender keep on telling me today is 3rd n u got nothing.... lollllllllll........:D UserIdTAG: 108863 UserNameTAG: shiviz CreateTimeTAG: 2013-01-03T10:17:02Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 2974 TitleTAG: What about statistics ? At the spring Course about 17,000 were certified from 150,000. How's things now? UserIdTAG: 323963 UserNameTAG: gtrf CreateTimeTAG: 2013-01-03T09:23:13Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: For clarification in the Spring course the official figures were 7,157 people (not 17,000) earned a certificate from 154,763 registrants. FirstChildUserIdTAG: 3646 FirstChildUserNameTAG: freespirit FirstChildCreateTimeTAG: 2013-01-03T10:18:15Z IndexTAG: 2975 TitleTAG: Above 90% but no certificate. Yet Please help. UserIdTAG: 99628 UserNameTAG: Pranjal16 CreateTimeTAG: 2013-01-03T07:59:03Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: same case here..! FirstChildUserIdTAG: 219356 FirstChildUserNameTAG: Karan123 FirstChildCreateTimeTAG: 2013-01-03T08:01:31Z SecondChildTAG: Check https://www.edx.org/dashboard and u can see " Download your PDF certificate" button. SecondChildUserIdTAG: 230076 SecondChildUserNameTAG: ANANDAM SecondChildCreateTimeTAG: 2013-01-03T08:06:30Z SecondChildTAG: Nothing there SecondChildUserIdTAG: 529515 SecondChildUserNameTAG: Low SecondChildCreateTimeTAG: 2013-01-03T08:09:47Z SecondChildTAG: Maybe they give in alphabetical order. SecondChildUserIdTAG: 99628 SecondChildUserNameTAG: Pranjal16 SecondChildCreateTimeTAG: 2013-01-03T08:14:56Z SecondChildTAG: still waiting for certificate...anybody get from here? SecondChildUserIdTAG: 219356 SecondChildUserNameTAG: Karan123 SecondChildCreateTimeTAG: 2013-01-03T11:21:03Z FirstChildTAG: Same problem. Maybe certificates for A+ will be some different? FirstChildUserIdTAG: 394836 FirstChildUserNameTAG: v2g6ch4 FirstChildCreateTimeTAG: 2013-01-03T08:28:06Z SecondChildTAG: yeah, and every day is christmas :) SecondChildUserIdTAG: 207728 SecondChildUserNameTAG: RossyFromBulgaria SecondChildCreateTimeTAG: 2013-01-03T09:07:40Z SecondChildTAG: I got my certificate. I got a grade of 99% but nothing appear about the grade on it. SecondChildUserIdTAG: 359677 SecondChildUserNameTAG: guillermb SecondChildCreateTimeTAG: 2013-01-03T11:39:19Z SecondChildTAG: wait for grade letter on e-mail SecondChildUserIdTAG: 394836 SecondChildUserNameTAG: v2g6ch4 SecondChildCreateTimeTAG: 2013-01-03T11:43:27Z SecondChildTAG: what is grade letter? SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2013-01-03T11:48:47Z IndexTAG: 2976 TitleTAG: HELP i didn't get certificate yet...my progerss above 60%... UserIdTAG: 219356 UserNameTAG: Karan123 CreateTimeTAG: 2013-01-03T07:46:04Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I didn't receive the certificate yet too, maybe they are making it availabe for download in alphabetical order FirstChildUserIdTAG: 229979 FirstChildUserNameTAG: Hafezi FirstChildCreateTimeTAG: 2013-01-03T08:36:50Z SecondChildTAG: hope so.. SecondChildUserIdTAG: 219356 SecondChildUserNameTAG: Karan123 SecondChildCreateTimeTAG: 2013-01-03T08:41:10Z SecondChildTAG: I'm 88% and still nothing, too SecondChildUserIdTAG: 207728 SecondChildUserNameTAG: RossyFromBulgaria SecondChildCreateTimeTAG: 2013-01-03T08:52:14Z SecondChildTAG: I got mine but the grade doesn't appear on it. SecondChildUserIdTAG: 359677 SecondChildUserNameTAG: guillermb SecondChildCreateTimeTAG: 2013-01-03T11:37:42Z SecondChildTAG: Just wait some time and soon it will come out on your dashboard. SecondChildUserIdTAG: 361137 SecondChildUserNameTAG: Ericson SecondChildCreateTimeTAG: 2013-01-03T15:08:17Z IndexTAG: 2977 TitleTAG: certificate what is the min. Percentage marks required for a certificate? I got 60 ,but no certi yet :( UserIdTAG: 260272 UserNameTAG: saikat24 CreateTimeTAG: 2013-01-03T07:33:45Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: **9 Grading** Letter grades will be based on the following weighting: homework 15%, labs 15%, midterm 30%, and final exam 40%. Each of the homework and labs carries equal weight. You will need to get a total mark of 60% for a C, 70% for a B, and 87% for an A. https://www.edx.org/static/content-mit-6002x/handouts/syllabus.a477535058a1.pdf FirstChildUserIdTAG: 325197 FirstChildUserNameTAG: Vitali_Jerin FirstChildCreateTimeTAG: 2013-01-03T07:36:47Z IndexTAG: 2978 TitleTAG: Certificate delay? I have got a score of 68% but haven't got any certificate yet! I will get a certificate right? I worked very hard throughout the course but screwed up the finals as I was outta town :( I really need this certificate! UserIdTAG: 108454 UserNameTAG: Raven7281 CreateTimeTAG: 2013-01-03T03:35:01Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: It should be available some time tomorrow January 3rd. Congratulations! FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2013-01-03T04:03:53Z SecondChildTAG: just got it...im so happy.Certificate is absolutely awesome ! SecondChildUserIdTAG: 230076 SecondChildUserNameTAG: ANANDAM SecondChildCreateTimeTAG: 2013-01-03T06:03:09Z SecondChildTAG: anandam, u r frm where?? SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2013-01-03T11:19:08Z IndexTAG: 2979 TitleTAG: What Address the certificate will be sent to? I don't remember whether the address was requested at the beggining of the course or not, but I didn't manage to find anywhere to change or insert it. Does anyone know where I can put or change it? UserIdTAG: 378470 UserNameTAG: Iron CreateTimeTAG: 2013-01-02T16:02:17Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You can change your email adress here: https://www.edx.org/dashboard FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2013-01-02T16:19:05Z SecondChildTAG: Hmmm... So, the certificate's just virtual? SecondChildUserIdTAG: 378470 SecondChildUserNameTAG: Iron SecondChildCreateTimeTAG: 2013-01-02T16:29:45Z SecondChildTAG: yes, it's only virtual SecondChildUserIdTAG: 207728 SecondChildUserNameTAG: RossyFromBulgaria SecondChildCreateTimeTAG: 2013-01-02T19:38:12Z IndexTAG: 2980 TitleTAG: Video of Taipei 101 New Year http://www.youtube.com/watch?feature=player_detailpage&v=S28CUTy1600 UserIdTAG: 183712 UserNameTAG: colegiocientifico CreateTimeTAG: 2013-01-01T16:59:32Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 2981 TitleTAG: lab0 good but i didn't understand the transient analysis with the sinusoid UserIdTAG: 938093 UserNameTAG: Mtraore CreateTimeTAG: 2012-12-30T21:49:26Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: the roadmap we see: - Change the waveform produced by the voltage source from a DC value of 3V to a sinusoid with a 1V amplitude, an offset voltage of 1V, and a frequency of 1kHz. - Add scope probes to nodes A, B and C and edit their properties so that the plots will be different colors. - Now run a transient analysis for 5ms. Just do it this way. what was the particular non-clear? You cann't repeat step by step procedure above? FirstChildUserIdTAG: 325197 FirstChildUserNameTAG: Vitali_Jerin FirstChildCreateTimeTAG: 2012-12-31T07:11:48Z IndexTAG: 2982 TitleTAG: question about telecom course at edx i want to ask the question which might be out of these topics.i am very curiosity about the telecom and ii wish one day to become expert on that fields.is there any possibility to learn telecom from edx platform UserIdTAG: 475448 UserNameTAG: msamwelmollel CreateTimeTAG: 2012-12-30T20:16:59Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 2983 TitleTAG: The textbook word in honor of The Feynman Lectures on Physics books which help me pass this course UserIdTAG: 323963 UserNameTAG: gtrf CreateTimeTAG: 2012-12-30T10:32:36Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: There are two important points in Physics: Physics to develop and teach physics to new generations. Richard Feynman was able to achieve these two points with much geniality. FirstChildUserIdTAG: 361823 FirstChildUserNameTAG: EliasOak FirstChildCreateTimeTAG: 2012-12-30T10:42:56Z SecondChildTAG: An interesting detail is that Feynman graduated from MIT. SecondChildUserIdTAG: 361823 SecondChildUserNameTAG: EliasOak SecondChildCreateTimeTAG: 2012-12-30T11:05:55Z IndexTAG: 2984 TitleTAG: Question to staff regarding certificate!!! Its been a week since we have finished our final exam and still no certificate... How long we have to wait for the certificate??? UserIdTAG: 108929 UserNameTAG: namit CreateTimeTAG: 2012-12-30T06:25:05Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: **namit**: We, the Community TAs, would love to answer questions regarding electronics, circuits, and even have philosophical discussions regarding the direction of edX. Unfortunately, we are bombarded constantly by students asking the same question over and over again; who, without doing a simple search in the discussion section (hint: click the magnifying glass icon); encourage other students to continue to do the same, setting forth a vicious cycle. Never mind that such a search that should occur before any question is asked, because its likely that simple questions have been answered before, and because that is course policy - in order to avoid wasting resources and wasting Staff time. I keep asking myself, why, especially university-level students, studying material from one of the world's most prestigious universities, would basically act in the manner of children; being so impatient, especially when something has been explained and answered over and over. I cannot find the answer; and the stereotypical answers of age or culture are meaningless when a global internet community is being discussed. Perhaps people just expect things on the 'net to be instantaneous, similar to being able to download a library's worth of music and movies at any hour of the day. Anyways, Lyla said that certificates would be released before the New Year. She did not specify if she meant Chinese New Year's, though she was being humorous. FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-12-30T06:55:31Z SecondChildTAG: This answer was very rude. You could just post the last sentence and avoid wasting your precious time. SecondChildUserIdTAG: 73251 SecondChildUserNameTAG: vhvidall SecondChildCreateTimeTAG: 2012-12-30T07:10:55Z SecondChildTAG: ..but then, everyone could also avoid wasting their and others time simply by not asking about certificates ? For me, I think this is a fine answer. It shows people care about educating you, considering we pay nothing here. SecondChildUserIdTAG: 201818 SecondChildUserNameTAG: ThreeHundred SecondChildCreateTimeTAG: 2012-12-30T09:13:27Z SecondChildTAG: not rude at all, I think that this answer could have been posted weeks ago. SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-12-30T11:12:15Z SecondChildTAG: @JerseyMark: Outstanding answer! In fact I suggest an entry level test to these courses, where reading and searching is tested. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-12-30T12:37:58Z SecondChildTAG: Outstanding answer. It should be considered positively. SecondChildUserIdTAG: 455950 SecondChildUserNameTAG: Wahabbaluch SecondChildCreateTimeTAG: 2012-12-30T15:45:58Z SecondChildTAG: **vhvidall:** I deliberately avoided being rude in the response: Do I threaten the respondent? Do I personally demean him/her? Did I make it personal? **No.** If I simply told the student, "Find out yourself," that would have been rude on my part as a Community TA. I was simply being thoughtful. Note that the post **immediately below** this one asks the same exact question, and an answer is provided. There is a difference between *constructive criticism* (a mix of satire and wit was what I was aiming for, but I wrote it late last night when I should have been sleeping) and *rudeness*. As for the "wasting your time" you point out, we Community TA's "waste our time" each time a repetitious, elsewhere-answered, easy-to-research-on-the -Discussion-boards question is asked! Ultimately, the answer is provided, **and** hopefully the poster learned how to solve problems for himself/herself, one of the most important lessons one can learn! Jersey Mark SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-12-30T17:33:45Z SecondChildTAG: No rudeness in response. There have been too many repetitious posts on certificates. The post is good advice. SecondChildUserIdTAG: 190670 SecondChildUserNameTAG: chauhan1955 SecondChildCreateTimeTAG: 2012-12-31T03:31:42Z FirstChildTAG: I'm afraid that even if staff puts the answer clearly in a post, you will never find it. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-12-30T13:30:20Z SecondChildTAG: I concur. Hence my earlier rant for some sort of minimum test of competence, and a test of ability to function on the internet properly. A ***simple multiple choice test*** on the course's syllabus and Honor Code before being able to enroll. We would cut in half the number of irrelevant questions that clearly do not belong on the Discussions forum. **Half!** Think of the extra time for Community TAs and Staff to assist other students really in need! SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-12-30T17:38:45Z SecondChildTAG: Well, that is how it works... We have tons of the same question (certificate) every day that you have answered with a "before new year". The question itself and the community TA answers are a waste of time for all. Edx should post all the info over Course Info Section and problem resolved, other students can reply to the ones that not see it there. Personally, I think that your first response was not rude, but looks like you are complaining about your wasted time with the students and not with your superior or coordinator or whatever. Also, the behavior you seen about not searching more deeply is because is not worth of time searching on previous days that obviously have a negative response about the certificate. The course is free and the people is trying to learn, so your comment about a test before being able to enroll is very rude by your attitude, because you are trying to cut in half the number of irrelevant questions on the forums and not waste your time, when you should be thinking about the people that want to learn from the course, that is the objective of Edx. I agree that are a lot of people that is very annoying asking over and over again, but that is because they not have the information they need and not wasting your time replying over and over again is very simple as I previosly said, Edx need to post relevant information about how to get the certificate on course info section. I hope I haven't wasted your time with my response. As a note, any spelling or gramatical error is because english is not my primary lenguage. SecondChildUserIdTAG: 265027 SecondChildUserNameTAG: edumm1 SecondChildCreateTimeTAG: 2012-12-30T19:33:26Z SecondChildTAG: **JerseyMark**, please take a look at cs50's discussion forum! That's how the repeated discussion would be cut to the minimum! Nevertheless, I believe the amount of certificate questioning that bothers you, were asked because most people would thought their efforts are inconclusive or ineffective. Please accept my season’s greetings. SecondChildUserIdTAG: 420339 SecondChildUserNameTAG: AliJenabi SecondChildCreateTimeTAG: 2012-12-31T12:00:26Z IndexTAG: 2985 TitleTAG: Certificado No puedo bajar el certificado esta disponible? de ser a si donde lo bajo? UserIdTAG: 264238 UserNameTAG: lean2209 CreateTimeTAG: 2012-12-29T21:39:37Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Se lo puedes pedir a los Reyes ;) FirstChildUserIdTAG: 244706 FirstChildUserNameTAG: Miguel_Angel FirstChildCreateTimeTAG: 2012-12-29T23:05:47Z FirstChildTAG: Layla, coordinadora de Edx dijo que iba a ser lanzado antes de año nuevo, es decir hoy lo van a poner a disposición de quienes aprobaron el curso. FirstChildUserIdTAG: 183712 FirstChildUserNameTAG: colegiocientifico FirstChildCreateTimeTAG: 2012-12-31T12:35:37Z IndexTAG: 2986 TitleTAG: Help with onscreen calculator I am struggling to use the onscreen calculator with the sqrt function, I seem to get a "syntax error" each time. Could someone please post a few examples of mathmatical problems and how they are entered on the calcultor. Regards John UserIdTAG: 873955 UserNameTAG: Wendelspanswick CreateTimeTAG: 2012-12-29T13:41:02Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: sqrt(4) result must be 2.0 just try [Copy]=>[Paste] into calculator FirstChildUserIdTAG: 325197 FirstChildUserNameTAG: Vitali_Jerin FirstChildCreateTimeTAG: 2012-12-29T13:49:16Z IndexTAG: 2987 TitleTAG: certificate when are we going to get certificates ?? UserIdTAG: 65255 UserNameTAG: jaideepjalla CreateTimeTAG: 2012-12-29T06:37:39Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: before new year FirstChildUserIdTAG: 114864 FirstChildUserNameTAG: Vikaash FirstChildCreateTimeTAG: 2012-12-29T08:16:06Z SecondChildTAG: LOL... that means tomorrow ~~ SecondChildUserIdTAG: 415375 SecondChildUserNameTAG: ZWX SecondChildCreateTimeTAG: 2012-12-30T21:31:41Z SecondChildTAG: 14 more hours (Central Time) until Certificates!!! SecondChildUserIdTAG: 45933 SecondChildUserNameTAG: Qubit SecondChildCreateTimeTAG: 2012-12-31T15:41:47Z IndexTAG: 2988 TitleTAG: Certificate Searching Problem hello frens, i have been having problem with searching my certificate....where can i get it?? UserIdTAG: 413613 UserNameTAG: shaliesh CreateTimeTAG: 2012-12-29T01:31:42Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hi shaliesh, You can not see your Certificate because it is not available yet ... As far as I have read in the Forum Discussion, it is possible that it will be available for downloading it before New Year :). Last term, some days after the final Exam, appeared a button were we could download our Certificate in the Progress Tab. This Fall,in my point of view, I suspect, that it is possible to be available for downloading it in our Dashboard :p I hope this can help you, Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-29T02:20:31Z FirstChildTAG: take a look here (example with image): https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50d343383c4ada2700000048 FirstChildUserIdTAG: 236136 FirstChildUserNameTAG: Hamid-ch FirstChildCreateTimeTAG: 2012-12-29T02:26:32Z IndexTAG: 2989 TitleTAG: certificado gente cuando estaran los certificados!!! saludos desde buenos aires-Argentina UserIdTAG: 264238 UserNameTAG: lean2209 CreateTimeTAG: 2012-12-28T23:21:17Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hola lean2209, Los Certificados estarán muy pronto, probablemente para antes de Año Nuevo. Esta información es en base a un comentario que ha hecho el Staff con anterioridad. Saludos, Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-29T01:11:00Z SecondChildTAG: What are you talking about? SecondChildUserIdTAG: 128276 SecondChildUserNameTAG: ChengBin SecondChildCreateTimeTAG: 2012-12-29T07:17:38Z SecondChildTAG: Hi ChengBin, lean2209 requested to know, in Spanish, when will be available the Certificates. I have answered to him or her that it is possible to be available before New Year :) Take care, Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-29T15:59:10Z SecondChildTAG: today SecondChildUserIdTAG: 183712 SecondChildUserNameTAG: colegiocientifico SecondChildCreateTimeTAG: 2012-12-31T12:24:37Z SecondChildTAG: when? SecondChildUserIdTAG: 342221 SecondChildUserNameTAG: deepkar SecondChildCreateTimeTAG: 2012-12-31T12:47:19Z IndexTAG: 2990 TitleTAG: op amp I forgot where's circuit of op amp(what is inside it). Can anyone remember where I can find the circuit? UserIdTAG: 342144 UserNameTAG: Orest02 CreateTimeTAG: 2012-12-28T18:51:25Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: text book from page 382 FirstChildUserIdTAG: 325197 FirstChildUserNameTAG: Vitali_Jerin FirstChildCreateTimeTAG: 2012-12-28T19:20:43Z FirstChildTAG: Here is national 741 Op amp ![enter image description here][1] [1]: http://puu.sh/1GLvS FirstChildUserIdTAG: 142857 FirstChildUserNameTAG: viking2 FirstChildCreateTimeTAG: 2012-12-29T13:09:35Z SecondChildTAG: With a circuit description ![enter image description here][1] [1]: http://puu.sh/1GLwK SecondChildUserIdTAG: 142857 SecondChildUserNameTAG: viking2 SecondChildCreateTimeTAG: 2012-12-29T13:11:03Z IndexTAG: 2991 TitleTAG: certificado ¿Cuándo vamos a obtener la certificación? ......... por favor díganos UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-12-28T15:18:08Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: Scroll a bit down or use search-engine. You will discover answer in 30 seconds. Cheers! FirstChildUserIdTAG: 325197 FirstChildUserNameTAG: Vitali_Jerin FirstChildCreateTimeTAG: 2012-12-28T16:48:13Z FirstChildTAG: Hola, En base a lo que he leído hasta el momento, el Staff ha dicho que esperan tener los Certificados para antes de Año Nuevo :). Por experiencia personal, he hecho este mismo Curso pero en el semestre pasado, entregaron los Certificados días después del Examen Final - el Certificado es en formato .pdf y se puede descargar una vez que ellos habilitan un botón, probablemente esta vez se encuentre en tu dashboard, la vez pasada estaba en el Progress Tab. Saludos, Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-28T16:57:31Z FirstChildTAG: Sorry to intrude like that, but i have noticed on several comments in spanish that a reverse question mark is used at the beginning of a question .Is that an error, or is it the way to formulate a question in spanish ? Just curios.Thank you ! FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-12-28T23:49:33Z SecondChildTAG: Hi AlexAlexandrescu, Yes, sure, no problem. If you ask you learn :p I will answer to your concern. In Spanish when you ask a question, your question must be between that two symbols. ¿? **¿** .....Here you ask your question, it must be between this symbols.... **?** I sometimes, skip the first symbol when I am writting fast, but when you are writting in Spanish you must use those two symbols when you ask a question, is a rule :p I hope this could answer to your concern. Take care, Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-29T01:25:17Z SecondChildTAG: Very interesting~ SecondChildUserIdTAG: 361137 SecondChildUserNameTAG: Ericson SecondChildCreateTimeTAG: 2012-12-29T02:12:17Z SecondChildTAG: Thank you very much! I wasn't aware of this. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-12-29T20:20:13Z FirstChildTAG: Muchísimas gracias por todo Myriam. Supongo, que en estos, los últimos del 2012, sí que vendrá específicado el grado en el Certificado; entiendo que es en los del 2013 donde no vendrá. Por otra parte, reiterarte de nuevo mi más sincera enhorabuena por tu trabajo, y desearte, que pronto veas cumplido tu sueño de obtener una beca para ir al MIT como decías en otro post; no dudo de que una persona de tu talento no tendrá problemas para ello. Un saludo desde España. FirstChildUserIdTAG: 329444 FirstChildUserNameTAG: albmartin FirstChildCreateTimeTAG: 2012-12-28T17:45:00Z SecondChildTAG: Por nada albmartin :) Un abrazo, Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-12-29T01:27:00Z IndexTAG: 2992 TitleTAG: Grade in certificate I understand to be a reference only. UserIdTAG: 260994 UserNameTAG: capvl CreateTimeTAG: 2012-12-28T12:40:56Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: There are certain procedures that can be used for your degree to be recognized by institutions and organizations. One is the apostille, method consisting of a notary public faith that saw progress on your computer and the level you reached in your course, then in a printout of the certificate he can authenticate you reached a especific note and a degree in that course. After that a diplomatic representation can attest that the course was taught by MITX indeed, and that the signatures on the certificate are true and they put an international stamp wich is valid in any country that is adhered to the Hague Convention of The Apostille. This course can be recognized in universities and organizations. Here in Costa Rica that procedure with certificates issued by other institucciones have already been endorsed by the Constitutional Court, admitting that education is a fundamental human right as also the right to information is too, so I think those would be some good steps if you want to have all the credibility of his information. http://www.state.gov/m/a/auth/c16921.htm Hay ciertos procedimientos que se pueden utilizar para que su grado pueda ser reconocido por instituciones y organizaciones. Uno es el apostillado,método que consiste en que un notario público de fé pública de que el vió en su computadora el progreso y el grado que usted alcanzó en su curso,para que luego una vez impreso el certificado el pueda autenticar que usted alcanzó un a nota y un determinado grado en ese curso. Ese certificado se lleva a una representación diplomática que puede dar fé que el curso efectivamente fue impartido por Mitx, y que las firmas en el certificado son auténticas y le ponen un sello internacional que es válido en cualquier país que esté adherido al convenio de la Haya sobre el apostillado. Así su curso puede ser reconocido en universidades y organizaciones. Aquí en Costa Rica ese procedimiento con certificaciones emitidas por otras instituciones ya ha sido avalado por el Tribunal Constitucional, admitiendo que la educación es un derecho humano fundamental así como el derecho a la información también lo es, por lo que creo que esos serían unos buenos pasos a seguir si usted quiere tener toda la credibilidad de su información. FirstChildUserIdTAG: 183712 FirstChildUserNameTAG: colegiocientifico FirstChildCreateTimeTAG: 2012-12-28T13:27:54Z SecondChildTAG: While the edX certificate itself carries very little weight in getting credit at a U.S. university, it will help in the admissions process, and a good score (85%+) on the Final Exam will virtually guarantee that you will pass your university's "course credit by examination" exam. The exam itself is given by the Department of Electrical and Computer Engineering (or similar) in which you plan on enrolling, and depending on the University, should give you credit for that university's equivalent for 6.002x (i.e. the introductory electrical engineering course, often called Fundamentals of Circuit Analysis, Introduction to Circuits and Electronics, etc.) as long as you are qualified. Such an exam is similar to the Final Exam. Search for my earlier post about "course credit by examination." This is standard procedure for U.S. universities, which I am personally very familiar with. SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-12-28T19:00:10Z IndexTAG: 2993 TitleTAG: Same Video Lectures? Are the video lectures same as those used in the spring 2012 version of this course? UserIdTAG: 623210 UserNameTAG: preveen CreateTimeTAG: 2012-12-28T10:56:00Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Yes, besides overview, week 1 to week 14 videos are the same as those in spring 2012 course. FirstChildUserIdTAG: 361137 FirstChildUserNameTAG: Ericson FirstChildCreateTimeTAG: 2012-12-28T13:06:31Z IndexTAG: 2994 TitleTAG: admission how to register for spring UserIdTAG: 550401 UserNameTAG: revathisingh CreateTimeTAG: 2012-12-28T08:40:33Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: simple click on register button appear at the below link :) https://www.edx.org/courses/MITx/6.002x/2013_Spring/about FirstChildUserIdTAG: 108863 FirstChildUserNameTAG: shiviz FirstChildCreateTimeTAG: 2012-12-28T08:59:52Z IndexTAG: 2995 TitleTAG: passing marks? what are the minimum passing marks UserIdTAG: 158367 UserNameTAG: brainyash1990 CreateTimeTAG: 2012-12-27T16:28:02Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: 60% for a completion certificate. take a look at the progress page. Hope you are eligible. FirstChildUserIdTAG: 623210 FirstChildUserNameTAG: preveen FirstChildCreateTimeTAG: 2012-12-27T16:33:09Z IndexTAG: 2996 TitleTAG: analysis available? is there any official analysis available regarding the number of students who have completed the course successfully and going to receive certificate? i have scored 92% overall. is it good in terms of percentile or ranking as lot of students got 100% or it looks good as a number only? kindly reply to this query so that i can get to know where i stand in this global platform.. UserIdTAG: 114864 UserNameTAG: Vikaash CreateTimeTAG: 2012-12-27T13:39:35Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: vikaash .. where are you from? FirstChildUserIdTAG: 342221 FirstChildUserNameTAG: deepkar FirstChildCreateTimeTAG: 2012-12-27T13:53:34Z SecondChildTAG: i am from INDIA..i guess you too belong to the same SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-12-27T14:29:50Z SecondChildTAG: yes SecondChildUserIdTAG: 342221 SecondChildUserNameTAG: deepkar SecondChildCreateTimeTAG: 2012-12-27T14:35:08Z IndexTAG: 2997 TitleTAG: to staff when will be staff back from their vacation? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-12-27T12:59:27Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: The staff was back working hard yesterday. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-12-27T13:03:07Z FirstChildTAG: they will answer that after coming back from vacation;-) FirstChildUserIdTAG: 623210 FirstChildUserNameTAG: preveen FirstChildCreateTimeTAG: 2012-12-27T13:11:06Z IndexTAG: 2998 TitleTAG: When will be receiving our certifictes? When will be receiving our certifictes? UserIdTAG: 126911 UserNameTAG: sidhant7 CreateTimeTAG: 2012-12-27T11:03:21Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: please use search engine left up side of the screen. Searching by key-word "certificate" Good luck! FirstChildUserIdTAG: 325197 FirstChildUserNameTAG: Vitali_Jerin FirstChildCreateTimeTAG: 2012-12-27T11:30:32Z SecondChildTAG: Or just scroll down a bit... :) SecondChildUserIdTAG: 345502 SecondChildUserNameTAG: ElectroVova SecondChildCreateTimeTAG: 2012-12-27T11:43:50Z IndexTAG: 2999 TitleTAG: Something is wrong when I am using firefox as a browser Hello everybody, Firefox isn't allowing me to visit the courseware and the discussion forum. Furthermore, in the progress section, the finger diagram isn't showing properly. I already mentioned this remark three days ago to the staff. But for some reason, they are avoiding me ! at least send me a confirmation message. Yesterday, I could finally access the previous material via the chrome browser. Thank you for your precious help. UserIdTAG: 418182 UserNameTAG: euldji2005 CreateTimeTAG: 2012-12-26T19:43:27Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I dont have any problems with Firefox under Win XP.But on the same laptop Chrome doesnt work correctly as well as IE.Other laptop with Win7 has problems with connecting to the edx.org with Chrome and Firefox and no any propblems with another sites.. FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-12-26T19:49:26Z SecondChildTAG: emm... But this is something new, I already used firefox as the default browser when I took this course. Since three days ago, I have not been able to access freely all the previously mentioned sections. Thank you for your reply. SecondChildUserIdTAG: 418182 SecondChildUserNameTAG: euldji2005 SecondChildCreateTimeTAG: 2012-12-26T20:09:15Z FirstChildTAG: I have not been able to log in at all using firefox for a few days now, however IE works fine. FirstChildUserIdTAG: 270011 FirstChildUserNameTAG: SnowmanZA FirstChildCreateTimeTAG: 2012-12-26T19:50:08Z SecondChildTAG: Then, I am not the only one with this problem. Thank you for your reply. The version of firefox I am installing right now is 17.0.1. SecondChildUserIdTAG: 418182 SecondChildUserNameTAG: euldji2005 SecondChildCreateTimeTAG: 2012-12-26T20:06:14Z SecondChildTAG: Firefox 17.0.1 works for me. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-12-26T21:01:02Z SecondChildTAG: My current FF version is same , problems been started ~2 weeks ago with the FF v17.0.0 SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-26T21:02:22Z SecondChildTAG: I have the 17.0.1 version. I was using firefox as my primary browser when I access 6.00x and 6.002x and CS50. You can see what's happening when you look at the associated images. I can browse the web sites that I usually use without problem (at least that's what I think). I cleaned the cache, but It's still not working :( Anyway, It's not a problem. I will try to use chrome, at least to pass the next PSET, and then I will try to revise this problem. Thank you so much my friend. I appreciate your help very much. ![enter image description here][1] ![enter image description here][2] [1]: https://edxuploads.s3.amazonaws.com/1356564155847449.png [2]: https://edxuploads.s3.amazonaws.com/13565641851343644.png SecondChildUserIdTAG: 418182 SecondChildUserNameTAG: euldji2005 SecondChildCreateTimeTAG: 2012-12-26T23:26:29Z SecondChildTAG: do you accept requests to the amazon cloud (cloudfront.net and amazonaws.com)? If I don't allow it in the addon "request policy", I get the same result as you do. SecondChildUserIdTAG: 406420 SecondChildUserNameTAG: Picolo SecondChildCreateTimeTAG: 2012-12-27T22:30:24Z SecondChildTAG: That sounds like it's worth a shot. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-12-28T12:07:36Z SecondChildTAG: No, I have never had such requests. Maybe the firefox via its popup blocker is blocking them in the first place SecondChildUserIdTAG: 418182 SecondChildUserNameTAG: euldji2005 SecondChildCreateTimeTAG: 2012-12-28T19:57:17Z IndexTAG: 3000 TitleTAG: videos can we have a link so that i can download all the lectures together.? UserIdTAG: 266739 UserNameTAG: satvikchugh CreateTimeTAG: 2012-12-26T14:41:28Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Weeks 1-6 can be downloaded in the Wiki as you may already know Also there were some students that had compressed versions available as well. If I find the post, I will let you know here. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-12-26T14:48:34Z SecondChildTAG: sure..:) Thanks in advance.;) SecondChildUserIdTAG: 266739 SecondChildUserNameTAG: satvikchugh SecondChildCreateTimeTAG: 2012-12-27T07:40:31Z IndexTAG: 3001 TitleTAG: Brazilian students Do you know how many Brazilian people take a this course ? UserIdTAG: 260994 UserNameTAG: capvl CreateTimeTAG: 2012-12-26T12:55:59Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: Take a look here https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50d914bbef2ecd2b0000002a Seems You are the first marked on map :) FirstChildUserIdTAG: 325197 FirstChildUserNameTAG: Vitali_Jerin FirstChildCreateTimeTAG: 2012-12-26T13:04:45Z FirstChildTAG: OK! Thank you. FirstChildUserIdTAG: 260994 FirstChildUserNameTAG: capvl FirstChildCreateTimeTAG: 2012-12-26T13:34:15Z FirstChildTAG: Shout: "Corinthians!!", and you may have many returns. FirstChildUserIdTAG: 182664 FirstChildUserNameTAG: Glaucus FirstChildCreateTimeTAG: 2012-12-26T15:22:57Z FirstChildTAG: Vai Corinthians. I am from São Paulo. FirstChildUserIdTAG: 361823 FirstChildUserNameTAG: EliasOak FirstChildCreateTimeTAG: 2012-12-26T21:10:46Z FirstChildTAG: I'm brazilian too.... From Minas Gerais! Happy new Year for Everyone! See you! FirstChildUserIdTAG: 164572 FirstChildUserNameTAG: Ranyeri_rocha FirstChildCreateTimeTAG: 2012-12-27T12:15:55Z IndexTAG: 3002 TitleTAG: Certificate Enquire Will we get the certificates by post or they will be electronically mailed to us? UserIdTAG: 396903 UserNameTAG: vasudoegar CreateTimeTAG: 2012-12-26T09:42:08Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50d343383c4ada2700000048 To Staff: seems that better to add some explanation about certificate getting into the FAQ chapter. :) FirstChildUserIdTAG: 325197 FirstChildUserNameTAG: Vitali_Jerin FirstChildCreateTimeTAG: 2012-12-26T09:51:50Z SecondChildTAG: duely noted. We'll change it once we get back from the holidays. SecondChildUserIdTAG: 96452 SecondChildUserNameTAG: Lyla SecondChildCreateTimeTAG: 2012-12-26T18:13:25Z FirstChildTAG: I hope we'll get the certificate in january, because than it will be new. If we get it before, then in january it will be from last year! I want a new one! FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-12-26T12:30:42Z IndexTAG: 3003 TitleTAG: Video Lectures !!! Are we able to have access to the video lectures and electronic textbook after the end of this courses. UserIdTAG: 239736 UserNameTAG: khatal CreateTimeTAG: 2012-12-26T02:40:29Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Yes sir. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-12-26T02:44:17Z IndexTAG: 3004 TitleTAG: Final Exam Explanation - as a Christmas gift? I actually haven't been all that good this year. However, such a gift would spur me to good deeds in the coming year UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-12-26T00:40:09Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: Patience, Grasshopper. FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-12-26T01:50:19Z FirstChildTAG: I AM WAITING FOR THE SAME. FirstChildUserIdTAG: 449185 FirstChildUserNameTAG: akhilprasad1993 FirstChildCreateTimeTAG: 2012-12-26T05:01:17Z SecondChildTAG: waiting for the answers or methods? I guess the questions are too easy for those who took the pain to go through the homework. SecondChildUserIdTAG: 623210 SecondChildUserNameTAG: preveen SecondChildCreateTimeTAG: 2012-12-26T05:45:16Z FirstChildTAG: I have some questions on the graph question...but the other questions I know where my mistakes were...anyway, I'll try to upload my versions of the solutions to some of the questions in a day or too.. FirstChildUserIdTAG: 143823 FirstChildUserNameTAG: NathanNadeson FirstChildCreateTimeTAG: 2012-12-26T15:05:09Z FirstChildTAG: Should be up now. Thanks for your patience. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-12-26T18:12:35Z SecondChildTAG: Lyla, What was up with the Final Exam? I retook the course after I didn't do well (well enough, but not well) on the final last time; but this final seemed like it covered only half as much material and at 60% of the difficulty level. In fact, although for various reasons, I was unable to review any of the material I didn't understand last time, I still got 94% on this final. My higher "achievement" in the course this time is meaningless as I know less now than I did when I took last semester's final. Regards, SecondChildUserIdTAG: 126825 SecondChildUserNameTAG: aphoenix SecondChildCreateTimeTAG: 2012-12-27T14:45:10Z SecondChildTAG: LOL. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-12-27T20:48:30Z IndexTAG: 3005 TitleTAG: When can we get the certificate Is there any one Knows the time of receiving the certificates? Any guess? UserIdTAG: 463816 UserNameTAG: theKnight CreateTimeTAG: 2012-12-25T23:20:24Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Hi theKnight, As far as I have read, they will be before New Year. Last Term, a few days after the Final Exam appeared a button in the progress tab were you could download the certificate. Take care, Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-25T23:31:08Z SecondChildTAG: but they aren't giving it in progress tab... SecondChildUserIdTAG: 342221 SecondChildUserNameTAG: deepkar SecondChildCreateTimeTAG: 2012-12-26T06:31:24Z SecondChildTAG: But, will the on-line storage of certificates be provided so you can use links in the summary for example. SecondChildUserIdTAG: 213452 SecondChildUserNameTAG: JoJosida SecondChildCreateTimeTAG: 2012-12-26T09:13:44Z SecondChildTAG: thank you all SecondChildUserIdTAG: 463816 SecondChildUserNameTAG: theKnight SecondChildCreateTimeTAG: 2012-12-26T17:58:48Z FirstChildTAG: I hope we'll get the certificate in january, because than it will be new. If we get it before, then in january it will be from last year! I want a new one! FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-12-26T12:14:55Z SecondChildTAG: thanks SecondChildUserIdTAG: 463816 SecondChildUserNameTAG: theKnight SecondChildCreateTimeTAG: 2012-12-26T17:58:18Z FirstChildTAG: look this post : https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50d343383c4ada2700000048 FirstChildUserIdTAG: 236136 FirstChildUserNameTAG: Hamid-ch FirstChildCreateTimeTAG: 2012-12-26T15:54:18Z SecondChildTAG: Thank you very much SecondChildUserIdTAG: 463816 SecondChildUserNameTAG: theKnight SecondChildCreateTimeTAG: 2012-12-26T17:57:53Z IndexTAG: 3006 TitleTAG: Certificate Myrimit, would you please let me know what will be the certificate procedure? And how we will get it? UserIdTAG: 131426 UserNameTAG: M_Inam_Ul_Haq CreateTimeTAG: 2012-12-25T15:20:16Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Please read some of the previous posts. This question has already been answered several times. The link to print out your certificate will appear before the start of the New Year. Jersey Mark FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-12-25T18:12:45Z FirstChildTAG: here : https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50d343383c4ada2700000048 FirstChildUserIdTAG: 236136 FirstChildUserNameTAG: Hamid-ch FirstChildCreateTimeTAG: 2012-12-25T19:34:20Z FirstChildTAG: Hi M_Inam_Ul_Haq, As far as I have read they will be available before New Year. Based on my experience, last term, it appeared in the progress tab a button where you could download your certificate. This Certificate had my complete name and also a link - in the .pdf, where it directed you to the edx servers showing the authenticity of your Certificate (the servers displayed your names too). Take care, Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-25T23:38:19Z SecondChildTAG: Thanks Myriam, that helps really well. ^_^ Sincerely, Inam SecondChildUserIdTAG: 131426 SecondChildUserNameTAG: M_Inam_Ul_Haq SecondChildCreateTimeTAG: 2012-12-26T15:21:38Z IndexTAG: 3007 TitleTAG: Certificates and answers From when can we get the answers for Final exam qwestions and what is the procedure to get certificate UserIdTAG: 132685 UserNameTAG: deepakmurali CreateTimeTAG: 2012-12-25T06:24:29Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50d729451f858b2700000011 FirstChildUserIdTAG: 325197 FirstChildUserNameTAG: Vitali_Jerin FirstChildCreateTimeTAG: 2012-12-25T08:48:39Z FirstChildTAG: an example here (certificate): https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50d343383c4ada2700000048 FirstChildUserIdTAG: 236136 FirstChildUserNameTAG: Hamid-ch FirstChildCreateTimeTAG: 2012-12-25T13:43:19Z IndexTAG: 3008 TitleTAG: Certificates Sir I have completed the 6.002x course with A grade.What's the procedure for obtaining the certificates? Thank you UserIdTAG: 431237 UserNameTAG: rohinraj CreateTimeTAG: 2012-12-25T04:29:11Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50d729451f858b2700000011 FirstChildUserIdTAG: 325197 FirstChildUserNameTAG: Vitali_Jerin FirstChildCreateTimeTAG: 2012-12-25T09:03:35Z FirstChildTAG: an example here : https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50d343383c4ada2700000048 FirstChildUserIdTAG: 236136 FirstChildUserNameTAG: Hamid-ch FirstChildCreateTimeTAG: 2012-12-25T13:42:40Z IndexTAG: 3009 TitleTAG: certificate its about a week i have completed the awesome course of circuits and electronics. i was just curious to know when certificate will be give . I hope today ? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-12-24T14:58:22Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: It should be available before the New Year. Congratulations. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-12-24T15:08:25Z FirstChildTAG: can any one tell me what is the passing grade for a certificate? FirstChildUserIdTAG: 404188 FirstChildUserNameTAG: jishudasorissa FirstChildCreateTimeTAG: 2012-12-24T15:31:24Z SecondChildTAG: 60% Overall. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-12-24T15:45:15Z FirstChildTAG: How certificate will be made available i.e. either it will be sent to our email or it will be available in this account or something else. FirstChildUserIdTAG: 404188 FirstChildUserNameTAG: jishudasorissa FirstChildCreateTimeTAG: 2012-12-24T15:27:22Z SecondChildTAG: A PDF Download will be made available along with a verification link. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-12-24T15:45:13Z IndexTAG: 3010 TitleTAG: Qestion to stuff about sandbox. What is the limit of circuit complexity of the Sandbox? I'm really enjoyed it works and find it quite useful in small cases modelling for my tasks. And also, could we access it a little bit? After the course is off.... It really nice... Thanks in advance!!! UserIdTAG: 345502 UserNameTAG: ElectroVova CreateTimeTAG: 2012-12-24T11:09:16Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: It is handy! AS far as I am aware everything will be available to you. However if you ever find it is not available there are other alternatives like https://www.circuitlab.com/ which you do not have to download. Have fun! FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-12-24T13:32:58Z SecondChildTAG: Is circuit sandbox same as PSPICE? SecondChildUserIdTAG: 449211 SecondChildUserNameTAG: Deveshmonga SecondChildCreateTimeTAG: 2012-12-24T13:39:23Z SecondChildTAG: Thanks for reply and advice! Have a nice Holidays! MC! SecondChildUserIdTAG: 345502 SecondChildUserNameTAG: ElectroVova SecondChildCreateTimeTAG: 2012-12-24T13:40:18Z SecondChildTAG: Circuit-Lab is not spice as far as I know. It is Spice-like though not as complex as Spice. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-12-24T14:14:32Z SecondChildTAG: For colleagues, ho may interested, here some soft, that designed in our country. Not much peoples knows about it, but it really powerful. Especially, topologic router for PCB. http://eda.eremex.com/products/topor/ I haven't seen better... Take a look: http://www.eremex.com/ P.S. It is not the Ad, just an advice... :) Feel free do delete this post, if some laws are broken by me... SecondChildUserIdTAG: 345502 SecondChildUserNameTAG: ElectroVova SecondChildCreateTimeTAG: 2012-12-24T14:25:02Z SecondChildTAG: Hey thanks for the circuitlab link.... :-) SecondChildUserIdTAG: 259719 SecondChildUserNameTAG: Smallblindchris SecondChildCreateTimeTAG: 2012-12-24T14:38:30Z FirstChildTAG: I love the Sandbox. One feature I would have liked would have been **a way to save/store circuits** which one constructed which could be retrieved at a later time. Not simply the single save which is provided now so that a current sandbox project is available after leaving and logging back in to the web site, but the ability to save multiple circuits and selectively retrieve them from one's personal library of multiple circuits. Even a store and retrieve capability of ten circuits would be extremely useful. There would only be one active circuit able to be worked on at any time. But one could immediately save that and call another from the library. In any case, the sandbox is a fantastic tool and adds a significant dimension to an already amazing experience - 6.002x. FirstChildUserIdTAG: 141108 FirstChildUserNameTAG: rl777 FirstChildCreateTimeTAG: 2012-12-24T13:58:18Z SecondChildTAG: Even if you could store the files on your computer, that would be handy. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-12-24T14:15:51Z IndexTAG: 3011 TitleTAG: Thank you from Chilean student Thank a lot for all 6.002x staff!! Henry A. Estrada UserIdTAG: 175734 UserNameTAG: hestrada CreateTimeTAG: 2012-12-24T11:00:44Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Saludos de otro chileno :) FirstChildUserIdTAG: 378294 FirstChildUserNameTAG: Zaybort FirstChildCreateTimeTAG: 2012-12-24T14:43:58Z FirstChildTAG: Hola Henry, Soy de los Estados Unidos. Yo fui a Chile en 1978 como estudiante intercambio. Vivi por diez semanas con una familia en Vitacura, Las Condes, Santiago. Quiero decirte que Chile es un pais muy lindo, y toda la gente fue muy simpatico conmigo. Espero que un dia puedo visitar tu pais de nuevo. FirstChildUserIdTAG: 428560 FirstChildUserNameTAG: MikeDayton FirstChildCreateTimeTAG: 2012-12-24T12:44:22Z SecondChildTAG: Vivi en Arauco, Chile por un año y medio. Me diverti muchisimo. Saludos!! SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-12-24T13:16:54Z IndexTAG: 3012 TitleTAG: problems about submit I've finished my final exam on 21. Now I return to verify my number of submits but there is an error because the final check still appears but the worse is that it says I've done 7 of 3 tries. I'm really worried about because I don't want a bad degree for that error. Thanks to the staff that can reply. (sorry, I only know a little of English) UserIdTAG: 110802 UserNameTAG: leoblack CreateTimeTAG: 2012-12-24T03:24:35Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Troubleshooting NumberOfReplyTAG: 1 FirstChildTAG: Hi leoblack, Have you checked your Progress tab? Does it shows you what you supposed to have as score in the Final Exam ? If not, I will report that to the Staff. If yes, remember that was a confusion in one question of the Exam and the Staff gave some extra attempts only for a short time, 24hs, then it dissapeared that chance of extra attempts. So, might is why you see 6 of 3... Please let me know if your Progress is not showing what it should be. Myriam. Edit. https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50d466e76766661f00000005 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-24T03:59:17Z SecondChildTAG: I did not see that attempts SecondChildUserIdTAG: 156535 SecondChildUserNameTAG: badawiIiIi SecondChildCreateTimeTAG: 2012-12-24T04:28:10Z SecondChildTAG: Muchas gracias Myrimit. He revisado mi progreso y tengo la puntuacion que esperaba, sin embargo, lo que me causaba preocupacion es que en el numero de intentos aparecian 7 y no el maximo que suponia 6. Feliz Navidad. SecondChildUserIdTAG: 110802 SecondChildUserNameTAG: leoblack SecondChildCreateTimeTAG: 2012-12-24T17:46:17Z IndexTAG: 3013 TitleTAG: final exam question no.6 I felt that there is some error in the answer to the last two parts of question no.6. Could you please verify it as soon as possible so that I could improve my percentage. UserIdTAG: 107674 UserNameTAG: liya CreateTimeTAG: 2012-12-24T01:24:05Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3014 TitleTAG: Brazilian students Where are you ? UserIdTAG: 260994 UserNameTAG: capvl CreateTimeTAG: 2012-12-23T20:32:05Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 12 FirstChildTAG: Mais um aqui FirstChildUserIdTAG: 156411 FirstChildUserNameTAG: Mtebaldi FirstChildCreateTimeTAG: 2012-12-23T22:11:39Z FirstChildTAG: Here FirstChildUserIdTAG: 11075 FirstChildUserNameTAG: roncada FirstChildCreateTimeTAG: 2012-12-23T20:36:07Z FirstChildTAG: Hay, I am from São Paulo. Vai Corinthians!!! FirstChildUserIdTAG: 361823 FirstChildUserNameTAG: EliasOak FirstChildCreateTimeTAG: 2012-12-23T21:14:21Z FirstChildTAG: Será que só tem "nóis treis"???? FirstChildUserIdTAG: 361823 FirstChildUserNameTAG: EliasOak FirstChildCreateTimeTAG: 2012-12-23T21:59:44Z FirstChildTAG: Hi, I'm from Belém - Pará :) FirstChildUserIdTAG: 355503 FirstChildUserNameTAG: fabriciogs FirstChildCreateTimeTAG: 2012-12-23T21:39:10Z FirstChildTAG: Ronaldo Salvador-BA FirstChildUserIdTAG: 58613 FirstChildUserNameTAG: ronaldocavalcante FirstChildCreateTimeTAG: 2012-12-24T01:02:57Z FirstChildTAG: Hi from São Paulo!!! FirstChildUserIdTAG: 385692 FirstChildUserNameTAG: tpfslima FirstChildCreateTimeTAG: 2012-12-24T00:54:21Z FirstChildTAG: Pedro Guilherme - DF FirstChildUserIdTAG: 27938 FirstChildUserNameTAG: pg1992 FirstChildCreateTimeTAG: 2012-12-24T02:00:48Z FirstChildTAG: Gabriel Takeuchi - PR FirstChildUserIdTAG: 318577 FirstChildUserNameTAG: takeuchi FirstChildCreateTimeTAG: 2012-12-24T02:21:35Z FirstChildTAG: curitiba, PR FirstChildUserIdTAG: 288174 FirstChildUserNameTAG: matiasgrodriguez FirstChildCreateTimeTAG: 2012-12-24T02:24:27Z FirstChildTAG: Não sou brasileiro, sou cubano, mais moro no Brasil, em São Carlos, SP. Um grande abraço a todos e tudo de bom para o novo ano. Karel Negrín. FirstChildUserIdTAG: 138934 FirstChildUserNameTAG: KarelNN FirstChildCreateTimeTAG: 2012-12-24T03:25:48Z FirstChildTAG: Aqui tá paid'egua. Belém Pará FirstChildUserIdTAG: 366669 FirstChildUserNameTAG: pedroramus FirstChildCreateTimeTAG: 2012-12-24T14:27:53Z IndexTAG: 3015 TitleTAG: Link to print certificate Hello there, I have completed my exam and my aggregate has been finalized, when (or where if I have failed to notice) will I find a link to my certificate. Also can we use our indian version of naming by using initials in our names or do we have to expand the initials to surnames. And by the way, a million thanks to Dr. Agarwal and the entire 6.002X team for this wonderful journey. Cheers, Raghu UserIdTAG: 333806 UserNameTAG: raghu47 CreateTimeTAG: 2012-12-23T15:54:45Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: It should available on your Dashboard page before the New Year. As far as I'm aware, you can use whatever name, however you wish, which can also be edited on the Dashboard page. Cheers FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-12-23T16:14:08Z SecondChildTAG: thanks ...... SecondChildUserIdTAG: 430030 SecondChildUserNameTAG: imali SecondChildCreateTimeTAG: 2012-12-23T16:24:04Z SecondChildTAG: when before new year? SecondChildUserIdTAG: 342221 SecondChildUserNameTAG: deepkar SecondChildCreateTimeTAG: 2012-12-25T11:34:20Z IndexTAG: 3016 TitleTAG: Finish the final exam Hi, I am very excited my final exam. I get XX%. I appreciate the Professor professor Agarwal sir, and other staff and those who study this program friends and MIT,further more, and America. Thank you all offer many people, like me, such a great chance to learn electronics acknowledge. I can't express my feeling using words. Only say, thanks to you all. UserIdTAG: 371617 UserNameTAG: rainbow12 CreateTimeTAG: 2012-12-23T15:05:31Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3017 TitleTAG: Finished Finally finished it with [*score removed*].............. Thanx to Almighty ALLAH. *I edited out your score; the exam is still ongoing for many of us and we're not supposed to discuss our scores until it is officially over (same policy as the midterm). I hope you understand - Jersey Mark (Community TA)* UserIdTAG: 216684 UserNameTAG: Taimoor1017 CreateTimeTAG: 2012-12-23T13:56:10Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Congratulations, let's try not to discuss our scores until the exam is over. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-12-23T13:58:59Z IndexTAG: 3018 TitleTAG: edx staff; reference homework, labs, practice exercise, and lecture does on CD or on anything. UserIdTAG: 462861 UserNameTAG: edxforme CreateTimeTAG: 2012-12-23T01:04:03Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3019 TitleTAG: edx staff: personal info how does i add my address and/or date of birth to my account? does date of birth and home address needed in regard to the certificate? UserIdTAG: 462861 UserNameTAG: edxforme CreateTimeTAG: 2012-12-23T00:58:05Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Your home address is not required in order to download and print your certificate. You will also receive a link to reference your completion, should you need it in the future. Just make sure you have your "Full Name" correct which can be edited on the Dashboard page. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-12-23T01:19:32Z FirstChildTAG: I was also going to ask this too. Because I didn't fill my address when I registered, for privacy maybe. If it's not needed, why do they ask for it? FirstChildUserIdTAG: 398594 FirstChildUserNameTAG: DaveyJC FirstChildCreateTimeTAG: 2012-12-23T13:48:24Z SecondChildTAG: I don't remember filling out my address, I guess I must have. lol If they do want your address I do not think it is unreasonable, it may be used to distinguish between you and someone else with the same name. etc. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-12-23T14:02:53Z IndexTAG: 3020 TitleTAG: WHERE DID THE 3 EXTRA ATTEMPTS OF QUESTION 6 GO ? Hello, I noticed that the 6th Question got 6 attempts last night. I decided to stop the exam for a while and go to sleep since I still have 20 legal hours in my possession. When I re-established the exam, I noticed that the extra 3 attempts have gone ! I decided to pass the 6th question without them but I didn't get the green marks I desired. I request the staff to offer me those extra 3 attempts if this doesn't include any problem. I still have 6 hours. Thank you in advance UserIdTAG: 418182 UserNameTAG: euldji2005 CreateTimeTAG: 2012-12-22T16:23:56Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I believe the "extra" attempts were for those who, like myself, began the exam when the wording was "flawed" or "confusing." Since the wording has subsequently been changed, fairness seems to dictate that the number of attempts revert to 3. FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-12-22T16:43:25Z SecondChildTAG: hmmm... I don't know. My bother is with the final sub-question of Q6. I guess I gave correct answer but the grader refused my first and second attempt. My third attempt was a gamble that I lost. Anyway, if the staff decided to add me those extra three attempts I will be thankful, otherwise no problem SecondChildUserIdTAG: 418182 SecondChildUserNameTAG: euldji2005 SecondChildCreateTimeTAG: 2012-12-22T16:48:26Z SecondChildTAG: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50d466e76766661f00000005 time is up SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-22T16:58:00Z SecondChildTAG: emmmmmmm........ No problem. Thank you Sergtronix. SecondChildUserIdTAG: 418182 SecondChildUserNameTAG: euldji2005 SecondChildCreateTimeTAG: 2012-12-22T17:01:48Z SecondChildTAG: I wish I didn't go to sleep. ha ha ha ! SecondChildUserIdTAG: 418182 SecondChildUserNameTAG: euldji2005 SecondChildCreateTimeTAG: 2012-12-22T17:04:03Z IndexTAG: 3021 TitleTAG: Rodolphe Hi! dear Staff, Just to know when should my certificate will be available? Thank to all the Staff for this course and I wish you start another course as PIC programming. Kind regard. UserIdTAG: 373916 UserNameTAG: larou CreateTimeTAG: 2012-12-22T16:21:19Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: They are hoping to distribute certificates before the end of the year to students who passed the course. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-12-22T17:04:25Z FirstChildTAG: I think it would be more usefull to learn ARM-architecture or PowerPC. 8bit microcontrollers it's last days. FirstChildUserIdTAG: 345464 FirstChildUserNameTAG: Constantine_ru FirstChildCreateTimeTAG: 2012-12-22T20:52:50Z IndexTAG: 3022 TitleTAG: Can we ask for advice for Q1? Can we? UserIdTAG: 244225 UserNameTAG: lerimock CreateTimeTAG: 2012-12-22T14:01:04Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I think it is not allowed.. FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-12-22T14:02:38Z IndexTAG: 3023 TitleTAG: fair play thanks for the Q6 extra tries, it's fair play UserIdTAG: 323963 UserNameTAG: gtrf CreateTimeTAG: 2012-12-22T12:12:09Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Q6 is clearly explained... FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-12-22T12:25:08Z SecondChildTAG: yes, honestly after I opened the exam I could not see any issues with Q4 and Q6. So, if there was any confusion with those question definitions it's clearly gone. SecondChildUserIdTAG: 16265 SecondChildUserNameTAG: serge_korolev SecondChildCreateTimeTAG: 2012-12-22T18:09:56Z SecondChildTAG: There weren't any issues from my part also . But carrots are at high demand these days. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-12-22T19:26:55Z IndexTAG: 3024 TitleTAG: change of name somebody please tell me how to change my name as it would appear in certificate .... UserIdTAG: 285222 UserNameTAG: dhaval24 CreateTimeTAG: 2012-12-22T05:26:53Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: even i have a doubt regarding the way name appears on certificate,i want my name to appear as Xxxxxxxx Yyyyyyyy,but in my profile i have given my name as xxxxxxxx yyyyyyyy so should i change it??so actually it is not change of name just case(uuper case and lower case) change.someone please reply as i am about to complete my course,i have even forwarded the same query on info@edx.org,technical@edx.org. FirstChildUserIdTAG: 353709 FirstChildUserNameTAG: anshu10750 FirstChildCreateTimeTAG: 2012-12-22T06:11:20Z FirstChildTAG: In your dashboard , to the up left corner it's an option to edit your full name.The dashboard contains all the courses you have enrolled.To access the dashboard, simply click on the top right corner of this page. FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-12-22T07:03:32Z SecondChildTAG: thanx got it SecondChildUserIdTAG: 285222 SecondChildUserNameTAG: dhaval24 SecondChildCreateTimeTAG: 2012-12-23T16:08:39Z IndexTAG: 3025 TitleTAG: content download First of all, thanks to Mr. Anant Agarwal and team for such a wonderful experiense. and i want to know whether the contents will be as it is or will be removed ,because i want these for further reference. Please reply ASAP. Again Thanks to all of you!! UserIdTAG: 408534 UserNameTAG: kkashyap CreateTimeTAG: 2012-12-22T04:47:33Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Just take what you can, and make notes ASAP .Besides there is MIT Open CourseWare...The spring session is still available, so i guess this will be too. FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-12-22T07:07:46Z FirstChildTAG: Yes. I would also like to know, if we will have access to course material, after the exam has ended. And for how long. Staf??! FirstChildUserIdTAG: 400567 FirstChildUserNameTAG: JoseMartins FirstChildCreateTimeTAG: 2012-12-22T19:26:19Z IndexTAG: 3026 TitleTAG: problem course slides Hi freinds I try to download the course slides but i find empty pages in some of them. is anyone else have the same problem??? UserIdTAG: 324041 UserNameTAG: sadrab CreateTimeTAG: 2012-12-21T22:00:42Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3027 TitleTAG: REQUIREMENT FOR CERTIFICATE Mine total score is 67% can i get certificate.If yes how? UserIdTAG: 342135 UserNameTAG: vikash902 CreateTimeTAG: 2012-12-21T20:25:47Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: The current information about the certificate can be found here: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50d343383c4ada2700000048 67% would be high enough to earn a certificate, congratulations! FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-12-21T20:39:44Z SecondChildTAG: thank you SecondChildUserIdTAG: 342135 SecondChildUserNameTAG: vikash902 SecondChildCreateTimeTAG: 2012-12-21T21:25:29Z IndexTAG: 3028 TitleTAG: Final Progress I did get 100% at final exam, but at Total final, appears "Final Exam = 40.0% of possible 40%", what's that mean? Isn't enough for to get the certificate? UserIdTAG: 514105 UserNameTAG: rubensousa CreateTimeTAG: 2012-12-21T11:01:44Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: Was the final exam timed?? Because I had power cuts in my locality and now i am unable to submit my answers... Pls reply... FirstChildUserIdTAG: 128924 FirstChildUserNameTAG: arjun392 FirstChildCreateTimeTAG: 2012-12-21T14:02:06Z SecondChildTAG: You have 24 hours to complete once you have started, good luck! SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-12-21T14:05:01Z FirstChildTAG: Final Exam = 40.0% of possible 40% means you got 100% on the exam ... the exam is only 40% of the total marks. you need to get at least 60% to get a certificate ( I think ) ... If you've got more than 60% at the Total ( which is at the very right at the progress tab ) then you should get a certificate. That means you need to have done some home works and labs or the mid term ( you got 20% from any other source ) FirstChildUserIdTAG: 454609 FirstChildUserNameTAG: Alwahsh FirstChildCreateTimeTAG: 2012-12-21T11:25:19Z FirstChildTAG: Hi rubensousa, That is because the final exam contributes to a 40 % if you get 100%: eg. if you get 50% in the final exam, your result will be 20%. You can read more info in syllabus [here][1] In order to receive the Certificate, you must sum 60% (based on Hw, Labs, Midterm and Final), remember that if you get 100% in Hw will contribute with an 15%, if you get 100% in Lab will contribute with an 15%, 100% in Midterm Exam will contribute with an 30%, 100% in Final Exam will contribute with an 40%...You must sum 60% in your final score in order to receive the Certificate... Take care, Myriam. [1]: https://www.edx.org/static/content-mit-6002x/handouts/syllabus.a477535058a1.pdf FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-21T11:28:45Z SecondChildTAG: ^ yeah that's exactly what I wanted to say xD ... SecondChildUserIdTAG: 454609 SecondChildUserNameTAG: Alwahsh SecondChildCreateTimeTAG: 2012-12-21T11:39:29Z FirstChildTAG: The final exam contributes 40% to your final grade. If you got 100% on the exam, that adds a full 40% to your final grade for the course; a 50% on the final exam would have added 20% to your final grade. FirstChildUserIdTAG: 340370 FirstChildUserNameTAG: Alaric7 FirstChildCreateTimeTAG: 2012-12-21T14:27:12Z FirstChildTAG: Man... and women :) I would like to see a video lesson from Professor Anant Agarwal because if he is able to teach us non trivial electronics he also will be able to teaching us how to read instructions on how to sum our scores. [... i'm just being ironic :) don't take me too seriously...] FirstChildUserIdTAG: 288174 FirstChildUserNameTAG: matiasgrodriguez FirstChildCreateTimeTAG: 2012-12-21T15:01:38Z IndexTAG: 3029 TitleTAG: question to Staff will we be getting a certificate after successfully finishing this course?? if yes, when will we receive our certificate.?? hope for an early response UserIdTAG: 108929 UserNameTAG: namit CreateTimeTAG: 2012-12-21T06:52:59Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: We hope to distribute certificates before the new year. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-12-21T13:35:15Z FirstChildTAG: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50d343383c4ada2700000048 FirstChildUserIdTAG: 236136 FirstChildUserNameTAG: Hamid-ch FirstChildCreateTimeTAG: 2012-12-21T13:41:39Z IndexTAG: 3030 TitleTAG: certificate?? will we get any certificate for doing this course and if yes, when will we get it?? UserIdTAG: 108929 UserNameTAG: namit CreateTimeTAG: 2012-12-21T06:33:49Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: We hope to distribute certificates before the new year. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-12-21T13:35:26Z FirstChildTAG: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50d343383c4ada2700000048 FirstChildUserIdTAG: 236136 FirstChildUserNameTAG: Hamid-ch FirstChildCreateTimeTAG: 2012-12-21T13:40:53Z IndexTAG: 3031 TitleTAG: Was it grounded in the previous video? Definitely not grounded here, I think. In the diagram in the previous video, it seems that you are putting a resistance between VE and ground. Here in this demo, it is pretty clear that you are not grounding it, you just are plugging a nice resistor in to the end, leaving it open and ungrounded. Am I misinterpreting the diagram or is that a standard way of doing it? UserIdTAG: 318779 UserNameTAG: jbell9 CreateTimeTAG: 2012-12-20T20:43:26Z VoteTAG: 0 CoursewareTAG: Week 14 / S28V7_Parallel_Termination_DEMO CommentableIdTAG: 6002x_S28V7_Parallel_Termination_DEMO NumberOfReplyTAG: 1 FirstChildTAG: In the demo, he has a long piece of coax. The resistor goes between the center conductor and the shield. It is connected to 'ground', but via the coax shield. I suppose in the diagram, it would help if the ground line that the resistor is connected to were dotted as well. FirstChildUserIdTAG: 141000 FirstChildUserNameTAG: OrinE FirstChildCreateTimeTAG: 2012-12-20T20:59:58Z IndexTAG: 3032 TitleTAG: What's happened? I was just dropped from the final exam. For a while I couldn't open anything on the site, but finally can. When I now open the exam, I still see green checks on Q1 through Q3 but the course progress tab isn't giving me any credit on the final. Should I try finishing the exam now or are there system glitches making this problematic? Will the progress tab eventually self-correct? UserIdTAG: 260299 UserNameTAG: Jack1947 CreateTimeTAG: 2012-12-20T20:28:09Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I see now the credit is appearing. FirstChildUserIdTAG: 260299 FirstChildUserNameTAG: Jack1947 FirstChildCreateTimeTAG: 2012-12-20T20:30:00Z IndexTAG: 3033 TitleTAG: Proctored Exams? I was just looking for a confirmation that there is no news as regards to *if* and *when* proctored exams for this course might be offered ... ? Are any other edX courses offering poctored finals this fall semester? UserIdTAG: 714237 UserNameTAG: PaxPolaris CreateTimeTAG: 2012-12-20T19:20:20Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Not that I am aware of. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-12-20T19:30:12Z IndexTAG: 3034 TitleTAG: QUERY ABOUT FINAL EXAM:TO STAFF HI.. To secure a grade of A in the 6.002x course, is it sufficient to get above 87% in the overall stats, or should the mid-sems, finals, homework overalls, lab overalls must all individually be above 87% as well.. i ask this because, my overall stat is above 87% but my midsem is less than 87%.. UserIdTAG: 457713 UserNameTAG: ASHWIN_DT CreateTimeTAG: 2012-12-20T17:35:53Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi ASHWIN ! Overall should be above 87%.... not midsem!!! FirstChildUserIdTAG: 277808 FirstChildUserNameTAG: Hemanthmps FirstChildCreateTimeTAG: 2012-12-20T17:55:59Z IndexTAG: 3035 TitleTAG: Total Energie Hallo Who is given the Energie? Isn't it the Voltage source? At T2, the Energie of the charged capacitor was given from the Voltage source at T1. Am I wrong if I think that the total Energie is given only at T1, from the Voltage source? and isn't the total Energie consumed by the 2 Resistant's? UserIdTAG: 219473 UserNameTAG: jsavvidis CreateTimeTAG: 2012-12-19T16:54:31Z VoteTAG: 0 CoursewareTAG: Week 13 / S26V7_Total_energy_dissipated_in_T1_and_T2 CommentableIdTAG: 6002x_S26V7_Total_energy_dissipated_in_T1_and_T2 NumberOfReplyTAG: 1 FirstChildTAG: the totale energy is given by the voltage source, but we looke for the energy disipated then: - 1 part is consumed at T1 - 2part is consumed at T2 - another thing to consider is the power,(we consider the energy disipated),this energy is disipated in 2 times T1 and T2(if you just make T1 thats wrong) FirstChildUserIdTAG: 236136 FirstChildUserNameTAG: Hamid-ch FirstChildCreateTimeTAG: 2012-12-19T17:25:55Z IndexTAG: 3036 TitleTAG: characteristic impedance of a wire I never understand this: 50 Ohms means 50 Ohms per unit length? And what is the unit length - it is never mentioned? UserIdTAG: 388524 UserNameTAG: dasbinich CreateTimeTAG: 2012-12-19T14:24:17Z VoteTAG: 0 CoursewareTAG: Week 14 / S28V5_What_s_Going_on_with_the_Double_Take_ CommentableIdTAG: 6002x_S28V5_What_s_Going_on_with_the_Double_Take_ NumberOfReplyTAG: 3 FirstChildTAG: It means just 50Ohms charaterictic impendance, nothing related to the resistance per unit length. It means that each signal transmittion line has own characterisic impendance, but in some cases we cant use this for consideration.In another cases we must remember about this parameter: It means that if we want to get undistorted signal we must terminate transmittion line by the resistor. Where you can meet this problem? Almost in the all hi frequency devices:cell phones, TVsets , labs tools and etc and etc. Even more..your laptop or desktop computer has alot of same nets. FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-12-19T14:43:35Z SecondChildTAG: so, a cable with 50 Ohms cha. imp., for ex., and a length of 1mm, for. ex., has also this char. imp.? And with 0,1mm of length??? Mr Prof said it in lesson: it is the impedance seen by a step transition of signal. And we know these impedances: sL and 1/sC. But it depends on the length of the cable!???!?!?!?! SecondChildUserIdTAG: 388524 SecondChildUserNameTAG: dasbinich SecondChildCreateTimeTAG: 2012-12-19T15:29:04Z SecondChildTAG: It depends from the few phisical conditions, as well as transmittion lines not only cable. Let you do have coaxial cable with 50Ohm impedance. But what is a phisical sense to use 0.1 or 1.0 mm of this cable? Right, it has a sense only at and from some length. But you should remember, that first question is How to get Undistorted signal at the receiver end?It is very important meaning, because, as I mentioned above, you can easy find matched nets in your laptop, with length only 10mm. Second important parameter for transmittion lines is signal attenuation. This parameter may be given as some attenuation in dB at the defined Frequency per one meter. This parameter directly depended from the length. SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-19T15:58:58Z SecondChildTAG: is week 14 cmng? SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-12-19T16:51:53Z SecondChildTAG: Sergtronix, thank you, but it is complex... SecondChildUserIdTAG: 388524 SecondChildUserNameTAG: dasbinich SecondChildCreateTimeTAG: 2012-12-19T21:34:46Z FirstChildTAG: See wikipedia: http://en.wikipedia.org/wiki/Characteristic_impedance FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-12-19T17:44:28Z SecondChildTAG: salsero, good, the link, I am studying. Thank you. SecondChildUserIdTAG: 388524 SecondChildUserNameTAG: dasbinich SecondChildCreateTimeTAG: 2012-12-19T21:35:14Z FirstChildTAG: Just to illustrate: The impedance of empty space is 376.73031.. ohms The impedance of a superconducting coax made to RG-59/U spec is 75 ohms. All this means if that you can terminate the medium with a 75 ohm resistor, and there will be no reflections. It has nothing to do with a the resistive properties of the medium. FirstChildUserIdTAG: 595503 FirstChildUserNameTAG: epi FirstChildCreateTimeTAG: 2012-12-20T08:53:35Z IndexTAG: 3037 TitleTAG: Themes in final exam I'd like to know, if the final exam includes mostly themes from the 7-12 weeks, or entire material from the first week to the 12th? And would it contains questions from the last two weeks, whicj didn't have homework? UserIdTAG: 345464 UserNameTAG: Constantine_ru CreateTimeTAG: 2012-12-19T07:25:10Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: The final covers material from weeks 1-12. FirstChildUserIdTAG: 90796 FirstChildUserNameTAG: ChaunceyGardiner FirstChildCreateTimeTAG: 2012-12-19T07:56:33Z SecondChildTAG: Thank you! And how many excersies would it there? Edx-team told that it would a few longer, then midterm, but my friends, who passed MITx6.002x in last spring, told that final-exam was a several times longer and more complicated. SecondChildUserIdTAG: 345464 SecondChildUserNameTAG: Constantine_ru SecondChildCreateTimeTAG: 2012-12-19T08:19:35Z SecondChildTAG: it was 10 tasks , 47 questions last final exam SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-19T09:42:51Z SecondChildTAG: Yes, as Sergtronix says it was 10 tasks with a total of 47 questions. To me, the final was much harder and took much longer than the midterm to complete but it was certainly doable. I had a very lacking math background, and I was still able to get 45 out of 47 questions right on the final. I was in it to do my very best, though. If I was just looking for a passing grade, I would have been in and out in no time at all. SecondChildUserIdTAG: 90796 SecondChildUserNameTAG: ChaunceyGardiner SecondChildCreateTimeTAG: 2012-12-19T10:13:40Z IndexTAG: 3038 TitleTAG: Source termination Because of R,we call it source termination? Or because of measuring the voltage of the end ,we call it source termination And still confused about why ramps are better than step sometimes.. UserIdTAG: 136959 UserNameTAG: christerpher CreateTimeTAG: 2012-12-19T01:13:07Z VoteTAG: 0 CoursewareTAG: Week 14 / S28V6_So_why_did_our_Circuits_Work_ CommentableIdTAG: 6002x_S28V6_So_why_did_our_Circuits_Work_ NumberOfReplyTAG: 0 IndexTAG: 3039 TitleTAG: S27V5: "...never has a path from power to ground"????? what is when vIN is at transition from 0V and 5V or from 5V to 0V? When vIN is, for example, = 3V: ==> VGSN = 3V > VTN and VGSP = -2V < VTP ==> os dois MOSFETs conduct and there will be a huge current between power and ground.... UserIdTAG: 388524 UserNameTAG: dasbinich CreateTimeTAG: 2012-12-19T00:41:45Z VoteTAG: 0 CoursewareTAG: Week 14 / S27V5_The_PFET_and_CMOS_Logic CommentableIdTAG: 6002x_S27V5_The_PFET_and_CMOS_Logic NumberOfReplyTAG: 0 IndexTAG: 3040 TitleTAG: thanks MIT for this cool example! very helpful! By the way, does anyone know what song is the Jazz one? I like it a lot UserIdTAG: 663109 UserNameTAG: JaviJap CreateTimeTAG: 2012-12-18T20:18:23Z VoteTAG: 0 CoursewareTAG: Week 5 / Music Amplifier Demonstration CommentableIdTAG: 6002x_snd_d NumberOfReplyTAG: 0 IndexTAG: 3041 TitleTAG: Log scale Could someone explain me how in the "Final exam soultions" page 14 is it possible to get bandwidth ~4e5 rad/sec? I see first point ~8.5e5 rad/sec and second point approximatelly 10.5e6 rad/sec so difference is about 2e5 rad/sec .. UserIdTAG: 205311 UserNameTAG: Sergtronix CreateTimeTAG: 2012-12-18T08:54:32Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3042 TitleTAG: s_stavrev Hello, I tried to build this comparator in the cirquit sand box. I put two 9000 ohm resistors in the positive feedback divider. Output connected to + input and + input to ground through resistors. Amplification of the op amp is A=3000. To the - input I connected triangle voltage source with 3V amplitute. Also tried with sin voltage with 0 offset. The same result - the cirquit behaves not like comparator, but like x2 amplifier with negative feedback. Please check this situation. Seems like + and - inputs may act as their names changed. UserIdTAG: 229896 UserNameTAG: Svilen CreateTimeTAG: 2012-12-18T08:16:03Z VoteTAG: 0 CoursewareTAG: Week 13 / S25V11_Hysteresis_with_positive_feedback CommentableIdTAG: 6002x_S25V11_Hysteresis_with_positive_feedback NumberOfReplyTAG: 1 FirstChildTAG: For the Opamp in the circuit sandbox, you cannot set the power supplies so, I think, you cannot simulate the positive feedback. FirstChildUserIdTAG: 376877 FirstChildUserNameTAG: AndBre FirstChildCreateTimeTAG: 2012-12-18T12:37:24Z IndexTAG: 3043 TitleTAG: i+ =0, i- =0? I am a little confused on how the current into the op amp is 0? How does that work? UserIdTAG: 219780 UserNameTAG: ReconIII CreateTimeTAG: 2012-12-17T15:33:21Z VoteTAG: 0 CoursewareTAG: Week 12 / S23V14 Virtual short method CommentableIdTAG: 6002x_S23V14_Virtual_short_method NumberOfReplyTAG: 1 FirstChildTAG: We do suppose that an ideal OpAmp has infinite input resistance. FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-12-17T16:05:16Z SecondChildTAG: Well with nothing going to the op amp in a real circuit how would it even know what V+ or V- is? SecondChildUserIdTAG: 219780 SecondChildUserNameTAG: ReconIII SecondChildCreateTimeTAG: 2012-12-17T16:25:41Z SecondChildTAG: Real OPAmps have different input currents: as example LM258 with pnp input stage has ~100nA input current, OPA2134 with FET stage has ~5pA input current (but some "current feedback" OAs may have significant input current, for example 5uA). In the most cases you can forget about input current while doing schematic analyse.But do not forget when working under real project- you need to choose correct resistors values, may be to place resistors with the same values to $V^+$ and $V^-$ related nets.. I have posted link to the AD or TI Op Amp handbook in this forum. SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-17T17:47:06Z SecondChildTAG: I've searches everywhere for the handbook.Where is it more exactly please? Thanks ! SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-12-19T00:55:36Z IndexTAG: 3044 TitleTAG: for busy person i couldn't complete this great course and i intend to join at next year can i or not UserIdTAG: 322638 UserNameTAG: khalid1988 CreateTimeTAG: 2012-12-16T11:09:30Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Yes you can, see you there. FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-12-16T15:49:31Z SecondChildTAG: thank you SecondChildUserIdTAG: 322638 SecondChildUserNameTAG: khalid1988 SecondChildCreateTimeTAG: 2012-12-17T07:23:18Z FirstChildTAG: Hello, When does the next course starts - i mean this one, again. FirstChildUserIdTAG: 372623 FirstChildUserNameTAG: Amitraj FirstChildCreateTimeTAG: 2012-12-16T18:09:49Z IndexTAG: 3045 TitleTAG: 6.002x for next semester For the class next semester, would the same homework sets for this semester be applied for the next time the class is offered. I understand is late to register but I would like to start on the right set of homework assignments. UserIdTAG: 883007 UserNameTAG: JuanSancen CreateTimeTAG: 2012-12-16T05:25:21Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I registered for mitx,some sets might be the same,but the certain values might be changed. FirstChildUserIdTAG: 136959 FirstChildUserNameTAG: christerpher FirstChildCreateTimeTAG: 2012-12-16T06:55:35Z IndexTAG: 3046 TitleTAG: About clocks So what is the coincidence with 555? UserIdTAG: 342144 UserNameTAG: Orest02 CreateTimeTAG: 2012-12-15T15:07:22Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: May be this is an answer: http://en.wikipedia.org/wiki/555_timer_IC FirstChildUserIdTAG: 329361 FirstChildUserNameTAG: ikrukov FirstChildCreateTimeTAG: 2012-12-15T18:49:45Z SecondChildTAG: I actually used the 555 timer chip in a project at work about 15 years ago, automating a certain process in a manufacturing plant. SecondChildUserIdTAG: 428560 SecondChildUserNameTAG: MikeDayton SecondChildCreateTimeTAG: 2012-12-23T05:34:13Z SecondChildTAG: I built a circuit to simulate a keystroke from a VT520 Vax keaboard, so that the operator would not have to make that keystroke 100 times/hour during an 8 or 12 hour shift. I used the 555 as the clock. SecondChildUserIdTAG: 428560 SecondChildUserNameTAG: MikeDayton SecondChildCreateTimeTAG: 2012-12-23T05:36:53Z SecondChildTAG: keyboard...duh SecondChildUserIdTAG: 428560 SecondChildUserNameTAG: MikeDayton SecondChildCreateTimeTAG: 2012-12-23T05:37:43Z IndexTAG: 3047 TitleTAG: Node method for v+ Hi, guys. I'm wondering why when I solve equation for v+ using node method (i.e. equation that I get is: (vIN-vP)/R1 = (v0-vP)/R2 where vP = v+) I don't get correct result afterwards when I substitute vP, that I got on previous step, into v0=A*(vP-vM). UserIdTAG: 413065 UserNameTAG: anikey CreateTimeTAG: 2012-12-11T16:03:50Z VoteTAG: 0 CoursewareTAG: Week 13 / S25E1_Positive_Feedback_Gain CommentableIdTAG: 6002x_S25E1_Positive_Feedback_Gain NumberOfReplyTAG: 2 FirstChildTAG: The expression for $v^+$, derived using a voltage divider is: $ v^+=v_{IN}+\frac{R_1(v_{OUT}-v_{IN})}{R_1+R_2}$ but also $ v_{OUT}=Av^+ $ so $v^+= \frac{v_{OUT}}{A}$ After equating expressions for $v^+$ $ \frac{v_{OUT}}{A}=v_{IN}+\frac{R_1(v_{OUT}-v_{IN})}{R_1+R_2}$ Rearranging yields $ \frac{v_{OUT}}{v_{IN}} = \frac{A \cdot R_2}{R_1 + R_2 - A \cdot R_1}$ as $A \rightarrow \pm\infty$ this becomes $ \frac{v_{OUT}}{v_{IN}} \rightarrow \frac{A \cdot R_2}{ - A \cdot R_1}$ which is solved using L'Hopital's Rule to give $ \frac{v_{OUT}}{v_{IN}} \rightarrow \frac{ - R_2}{ R_1}$ FirstChildUserIdTAG: 9673 FirstChildUserNameTAG: xp42 FirstChildCreateTimeTAG: 2012-12-14T10:09:12Z FirstChildTAG: anikey, (vIN-vP)/R1 = (v0-vP)/R2 is false because you applied a wrong KCL! Your equation for v+ should be: (vP-vIN)/R1 = (v0-vP)/R2 Now you will get the correct results. FirstChildUserIdTAG: 388524 FirstChildUserNameTAG: dasbinich FirstChildCreateTimeTAG: 2012-12-15T00:45:21Z IndexTAG: 3048 TitleTAG: S25V19 coincidence? In the video about Clocks in Digital Systems we are referred to page 555 in the text, with a comment about something being curious about that page number. The 555 timer Integrated Circuit was introduced in 1972 and is still in use today, a fitting tribute to a 40 year-old design. UserIdTAG: 231764 UserNameTAG: PaulP4881 CreateTimeTAG: 2012-12-11T02:53:45Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3049 TitleTAG: If we want a short circuit between Vs and 0v, we must have A=1 AND A=0 (! ) ( and also b=1) or B=1 and B=0 ( and also a=1) That is never true ! No static power... Good system ! UserIdTAG: 373752 UserNameTAG: Croak CreateTimeTAG: 2012-12-11T02:07:29Z VoteTAG: 0 CoursewareTAG: Week 14 / S27V10_CMOS_Logic_Gate_Design CommentableIdTAG: 6002x_S27V10_CMOS_Logic_Gate_Design NumberOfReplyTAG: 0 IndexTAG: 3050 TitleTAG: A suggestion for better videos There is a software called 'CursorFx', that gives various options for cursors. The cursors it provide are more visible and good looking and also can provide special effects. If those are used, the videos will be most effective. Regards UserIdTAG: 131426 UserNameTAG: M_Inam_Ul_Haq CreateTimeTAG: 2012-12-10T11:36:13Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 3 FirstChildTAG: Using the mouse's cursor to enhance the video is a great feature. It will be still better if the cursor looked different from the standard one, right now it may be a bit confusing and may trigger security concerns. A bigger, colored cursor, would be ideal IMHO. FirstChildUserIdTAG: 296965 FirstChildUserNameTAG: LGMailhos FirstChildCreateTimeTAG: 2012-12-10T16:37:45Z SecondChildTAG: We need one like this, so when it moves the eye can follow it. ![enter image description here][1] It is transparent so text can be seen behind it. The problem I had, is the pointer used now [1]: https://edxuploads.s3.amazonaws.com/13552720291343603.jpg SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-12-12T00:28:16Z SecondChildTAG: Oops, forgot to finish my sentence. The pointer used now gets lost if no marks big enough are made for the eye to see. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-12-12T00:31:43Z FirstChildTAG: These are some of the snapshots of optional cursors it provide! FirstChildUserIdTAG: 131426 FirstChildUserNameTAG: M_Inam_Ul_Haq FirstChildCreateTimeTAG: 2012-12-15T06:35:28Z FirstChildTAG: ![CursorFx showing some of the available cursors options][1] [1]: https://edxuploads.s3.amazonaws.com/13555539911343619.png FirstChildUserIdTAG: 131426 FirstChildUserNameTAG: M_Inam_Ul_Haq FirstChildCreateTimeTAG: 2012-12-15T06:47:39Z IndexTAG: 3051 TitleTAG: h12p2 - not seeing the role of current source at top of diagram Looking at this voltage regulator circuit, it seems like the output voltage is determined by $v_-$, which is between the R1 and R2 resistors. That should be equal to $v_+$, and isn't $v_+$ determined by $V_{in}$ and R0 and the zener diode? I don't see what that current controlled current source is doing at the top of the circuit diagram. Can anyone offer a hint? Thanks. Rob UserIdTAG: 468623 UserNameTAG: RobNik CreateTimeTAG: 2012-12-10T00:06:16Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Do you mean BJT transistor? If so it is just emitter follower, nothing else. This example is very simplified. FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-12-10T00:24:43Z SecondChildTAG: Yes. SecondChildUserIdTAG: 468623 SecondChildUserNameTAG: RobNik SecondChildCreateTimeTAG: 2012-12-10T00:27:01Z FirstChildTAG: You are correct about what is determining Vout. The op amp is controlling The BJT. Most of the current supplied to the load does not pass through the op amp at all. Although there are power op amps the typical op amp is a low current device. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-12-10T00:23:30Z SecondChildTAG: Right, the BJT is supplying 95% of the current to the load. That makes it a lot easier on the Op Amp and allows appropriate choice of BJT do deliver the desired amount of power to the load, which could be very high. BJTs have very high transconductance and low $R_{DS_{on}}$ and so can deliver a lot of power compared to an Op Amp or MOSFET. SecondChildUserIdTAG: 339668 SecondChildUserNameTAG: chickwebb SecondChildCreateTimeTAG: 2012-12-11T00:01:34Z IndexTAG: 3052 TitleTAG: Change is nice I love this video , it was different and new! UserIdTAG: 733429 UserNameTAG: EhsanMokhtari CreateTimeTAG: 2012-12-09T15:46:38Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: _positive_feedback NumberOfReplyTAG: 0 IndexTAG: 3053 TitleTAG: about the response curve in the middle of the response curve we r getting a notch... according to the mathematical derivation, the notch hits zero at resonant frequency. but in the graph it's still having some positive value, is this due to the resistor?? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-12-09T04:58:39Z VoteTAG: 0 CoursewareTAG: Week 11 / S21V6: Series RLC bandstop CommentableIdTAG: 6002x_S21V6_Series_RLC_bandstop NumberOfReplyTAG: 0 IndexTAG: 3054 TitleTAG: Lab 12 help required I choose Q as 10 and w0 as 200 hz and C=C1=C2 as 10 nF ... for these values i found R2 as 10M ohm and R1 as 25000 ohm my question is how should i write R1 in the box ??? Means i dont know how to read that Resistors table ... So please guide me and sorry for my english !!! UserIdTAG: 145239 UserNameTAG: Maheenjd CreateTimeTAG: 2012-12-08T19:39:18Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: You must put straight 25000 or another value using only digits. I used this way. FirstChildUserIdTAG: 96873 FirstChildUserNameTAG: Santyaga FirstChildCreateTimeTAG: 2012-12-08T19:44:31Z FirstChildTAG: Hi Maheenjd, Can I help you? Take a look at this Hints of Lab12 [here][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50c3b82f1f09202b0000002a FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-12-09T04:17:01Z SecondChildTAG: ohh thank you so much Myrimit :) SecondChildUserIdTAG: 145239 SecondChildUserNameTAG: Maheenjd SecondChildCreateTimeTAG: 2012-12-09T16:58:38Z IndexTAG: 3055 TitleTAG: S23V19 Connecting "+" is inverting too, why we use "-" ? If we change places of amplifiers - and + , then this method would give us the same answer, thus we would get inverting amplifier again. Why did we connected - not + as before? UserIdTAG: 66669 UserNameTAG: agarus CreateTimeTAG: 2012-12-08T09:29:35Z VoteTAG: 0 CoursewareTAG: Week 12 / S23V19 Inverting amplifier analysis using virtual short method CommentableIdTAG: 6002x_S23V19_Inverting_amplifier_analysis_using_virtual_short_method NumberOfReplyTAG: 0 IndexTAG: 3056 TitleTAG: If additionaly consider v0=A((vi+) - (vi-)), then v0=vi=0. if additionally consider v0=A((vi+) - (vi-)), then v0=vi=0. UserIdTAG: 66669 UserNameTAG: agarus CreateTimeTAG: 2012-12-07T18:43:10Z VoteTAG: 0 CoursewareTAG: Week 12 / S23V15 Buffer circuit CommentableIdTAG: 6002x_S23V15_Buffer_circuit NumberOfReplyTAG: 0 IndexTAG: 3057 TitleTAG: Completely lost in H12P2 Hi!! I'm completely lost in this exercise, can one give me any hint?? Th UserIdTAG: 250473 UserNameTAG: pajaropica CreateTimeTAG: 2012-12-07T18:10:24Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Try to search, keyword H12P2..You will find alot explanation ( my too ) and discussions. FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-12-07T18:19:24Z FirstChildTAG: Here's one link: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50b8d3275516b62b00000010 FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-12-07T19:18:35Z IndexTAG: 3058 TitleTAG: H12P12 - c and d Hi everyone, First of all, I want to say thanks for all your help during this course. About H12P2 question, I found the answer for a) and b), but I'm having some trouble to find the values of R1, R2 and the efficience Thanks to vargaslen and SkyHawk posts, I use the formula above to find R0: delta_v+ = 10*Rz/(R0+Rz) (I use delta_v+ = 0.01 to calculate R0) Now, how can I find the R1, R2 and efficience values? Please, do not send me to similar topics. I read all of them, but none of them give me a satisfactory answer. Best regards UserIdTAG: 310147 UserNameTAG: ildomarcarvalho CreateTimeTAG: 2012-12-07T17:41:19Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: H12P2 is pretty simple. At first please read [here][1].You may find there other link to my post. Ok.Before some explanation I want note that it is possible to calculate R0 by the two or more ways(In the real design you will need to check $I_{min}$ and $I_{max}$ for the Zener.Here you shouldnt worry about Zeners current value).Anyway, you should understand THAT smaller R0 means larger $delta$ $v^+$ and vice versa. It is possible to use other way to define R0: from the change of output voltage.It is my preferable way here.So, what do you need?You need to understand gain $G$ value, it is clear, and suppose that you want to design MORE precise voltage regulator.This will help you to understand relativities between delta $v_O$ and R0. Next, let you have $v^+=mmm...2.554V$.May you get $G$ ? How to get R1 and R2 from this $G$? I think that even with $G=2$ you will have correct $v_O$ value. Ok, How to calculate efficiency? It is $P_{output}$/$P_{total}$...What you must involve into this calculation?Correct, it is current through :Zener, BJT (as function from $I_O$), and divider R1/R2.Please dont forget to add $I_O$ for $P_{total}$ calculation. Ok, you have all values for consumptioning nets. Now you may try to press $CHECK$ button. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50bb14dad5bfda270000007a FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-12-07T19:55:58Z FirstChildTAG: Olá Idiomar, lembra de mim. Sou do Brasil também. Nós já nos conhecemos. Quanto a esse problema, eu também encontrei muita dificuldade. Existe um macetizinho nele que creio, você não tentou. Para aparecer as marcas verdinhas você deve responder simultaneamente os itens c e d. Em seguida mandar checar. Não tente responder só a letra c e tentar. Coloque, por exemplo, valores de resistências iguais nos três resistores e um ganho menor que 0,5. Tente e veja se dá certo. Um abraço: Elias. FirstChildUserIdTAG: 361823 FirstChildUserNameTAG: EliasOak FirstChildCreateTimeTAG: 2012-12-07T23:10:09Z IndexTAG: 3059 TitleTAG: Problems with calculating Vo I got e1=e2=v1*R4/(R3+R4), now my logic is that vo = - (v2-e2)*R2/R1 i.e. the current going through R2. Then the final expression is vo=R2/R1*(v1*R4/(R3+R4)-v2). Why that is wrong? UserIdTAG: 415375 UserNameTAG: ZWX CreateTimeTAG: 2012-12-07T08:45:21Z VoteTAG: 0 CoursewareTAG: Week 12 / S24E2 Difference Amplifier CommentableIdTAG: 6002x_S24E2_Difference_Amplifier NumberOfReplyTAG: 1 FirstChildTAG: In addition I got the right result using the superposition method, and calculating vo from the equation e1=e2, where e1=v1*R4/(R3+R4), and e2=(v2-vo)*R2/(R1+R2)+vo. So looks like my first attempt is wrong, because it is taking into account only that e1 is affecting e2, but not that e2 is affecting e1 either. What confused me was that the value for e1 is the same as the value for e2 i.e. R4/(R3+R4)*v1, and I still don't understand why actually that is wrong... FirstChildUserIdTAG: 415375 FirstChildUserNameTAG: ZWX FirstChildCreateTimeTAG: 2012-12-07T09:18:23Z SecondChildTAG: here: 1. e1=(v1*R4/(R3+R4)) 2. e2=e1 3. Current through R2 i=(v2-i2)/R1 4. vo=e2-(v2-e2)*R2/R1 Then you have to replace e2 and you'll get vo. SecondChildUserIdTAG: 220304 SecondChildUserNameTAG: sergei_m SecondChildCreateTimeTAG: 2012-12-07T20:20:28Z IndexTAG: 3060 TitleTAG: current through R if the voltage v+ is 7.5V in the start wont the current through R be (vo-7.5)/R ? and as the capacitor charges the voltage vc keeps increasing and so the (vo-vc)/R must keep decreasing ...is this rite or am i missing something ? UserIdTAG: 307137 UserNameTAG: kishen CreateTimeTAG: 2012-12-07T06:37:03Z VoteTAG: 0 CoursewareTAG: Week 13 / S25V16_Op_Amp_oscillator CommentableIdTAG: 6002x_S25V16_Op_Amp_oscillator NumberOfReplyTAG: 0 IndexTAG: 3061 TitleTAG: LAB12 Cannot get the right value Hello, I cannot get the right values for this circuit. If my analysis is correct, then w0 value is 1/sqrt(R1*R2*C1*C2) and 2*alpha is (C1+C2/(R2*C1*C2)). I am able to choose the components according to those values but cannot get the green checks and sandbox AC analysis doesn't seem right. I am starting wondering if my KCLs are correct, I assumed V3 voltage is zero volt. Can you tell me if I messed up the analysis or is it the values calculation that I am doing wrong? Thank you UserIdTAG: 16265 UserNameTAG: serge_korolev CreateTimeTAG: 2012-12-06T23:52:52Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Привет, тезка! Hi, youre right.Except that $w_0$=2*$\pi$*f . You can get R1 from $f$ and $Q$, next R2. After that you should choose C1 and C2. (My value is C=C1=C2=10nF;while it is just a LAB we do use only $f$ and $Q$ values). I think all next steps will not be hard for you. Good luck! :) FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-12-07T00:25:12Z SecondChildTAG: привет!-)))) ok, so those are my calculations that didn't go right, though of course I remembered to convert the frequency value. Thanks for the response, I think I'll go easy with this particular lab and the second part of H12P2 too (as I've already earned 30% for doing my homework in time) and will concentrate on the next two weeks material and the final exam preparations. Good luck to you too SecondChildUserIdTAG: 16265 SecondChildUserNameTAG: serge_korolev SecondChildCreateTimeTAG: 2012-12-07T14:20:38Z FirstChildTAG: You wrote the $2\alpha$ wrong, you left out some parentheses. Maybe in your calculations you did it right but I want to point you to the error anyway. $2\alpha = \frac{C_1+C_2}{R_2*C_1*C_2}$ FirstChildUserIdTAG: 266912 FirstChildUserNameTAG: pietvo FirstChildCreateTimeTAG: 2012-12-07T20:20:55Z SecondChildTAG: Yeah, it is possible SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-07T21:53:10Z SecondChildTAG: no, the formula was as you wrote it, I typed it badly in the message. Anyhow I straighten my numeric calculations out and got the green marks. Thanks anyway))) SecondChildUserIdTAG: 16265 SecondChildUserNameTAG: serge_korolev SecondChildCreateTimeTAG: 2012-12-09T12:28:45Z FirstChildTAG: I think you need to correct your equations for α and frequency from textbook on page 865. And try to find right values using excel worksheet. Maybe it will help you. FirstChildUserIdTAG: 96873 FirstChildUserNameTAG: Santyaga FirstChildCreateTimeTAG: 2012-12-08T11:28:16Z IndexTAG: 3062 TitleTAG: pls guide me on my method!! I've solved the above case using my own consideration of a non ideal opamp ii.e. having gain 'A' Now since opamp is a differential amp it shud lead to the result I obtained... PS-my result perfectly matches R2 when A approaches infinity.... The equations used are the kcl at negative terminal of opamp and another one is for the A(input)=output one.... I just need to know why my answers are not accepted using this technique...I guess I've solved correctly :-[ Here's a screenshot![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13547424701343662.jpg UserIdTAG: 367886 UserNameTAG: KashwinKohli CreateTimeTAG: 2012-12-05T21:25:26Z VoteTAG: 0 CoursewareTAG: Week 12 / S23V21_Inverting_amplifier_input_resistance_-_2 CommentableIdTAG: 6002x_S23V21_Inverting_amplifier_input_resistance_-_2 NumberOfReplyTAG: 1 FirstChildTAG: You have on the third line of equation $v_{in} = \frac{e (R_1+R_2)}{R1} - \frac{v_o}{R_1} $ The last term should be $\frac{v_o R_2}{R_1}$, however. You forgot the $R_2$. You can see that the dimensions are not right. The first part is Volts, the second is Volts/Ohms. FirstChildUserIdTAG: 266912 FirstChildUserNameTAG: pietvo FirstChildCreateTimeTAG: 2012-12-06T15:49:14Z IndexTAG: 3063 TitleTAG: Someone expirienced, please help There is an interesting questing about H11P1 with now answer. Could you please take a look at it: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50bba11a916b05270000004a UserIdTAG: 66669 UserNameTAG: agarus CreateTimeTAG: 2012-12-05T16:03:28Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Assuming you are saying 'no answer' and not 'now answer', it seems that the question has answers. What part didn't get answered? FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-12-05T23:47:04Z SecondChildTAG: Yes, you are right it's "no answer", sorry. I'm having the same confusion as RobNik in that question. The standard form of denominator in H11P1 is like s^2+A*s+B/(CL), where (if I'm looking in right drafts) A=(RL/L+1/C*RC) and B= RL/RC. System green-checked the answer for 2*alpha=A, but it says omega^2=1/CL but not omega^2=B/CL. So the question is: why is it so? Why omega^2 is not B/CL? In answers Doru and skyhawk assume that equation isn't in standard form (but it is in standard form). Ericson shares our curiosity. And salsero says that some times we can use simplified form w=1/CL, but right answer comes from standard form. But system says its wrong. So I'm still curious for answer: is it right that omega^2=B/CL is right answer and omega^2=1/CL is just simplified one? SecondChildUserIdTAG: 66669 SecondChildUserNameTAG: agarus SecondChildCreateTimeTAG: 2012-12-06T13:34:36Z SecondChildTAG: I did not say that the equation wasn't in standard form. What I said in my previous post was: **The constant term in the denominator may, in fact, not be equal to 1/LC when the coefficient of S^2 is one. The constant term is, nevertheless, equal to w0^2, where w0 is the resonant (angular) frequency.** I see the equation in standard form when the coefficient of s^2 is one. The constant term is not in this case 1/(L*C) but (1 + RL/RC)/(L*C). It is not (RL/RC)/(L*C) as you indicate. For the data given in the problem RC >> RL; therefore, RL/RC << 1 and to a good approximation w0^2 = 1/(L*C). Your expression gives a far different numerical value. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-12-06T14:18:13Z SecondChildTAG: The constant term is, nevertheless, nevertheless, nevertheless... sorry, I'm stupid! Couldn't catch that speech expression. Thank for the answer!!! JSChambers thank you too. SecondChildUserIdTAG: 66669 SecondChildUserNameTAG: agarus SecondChildCreateTimeTAG: 2012-12-07T17:46:27Z SecondChildTAG: Thank you for explanation (RL/RC<<1) of answer simplification reasons either! SecondChildUserIdTAG: 66669 SecondChildUserNameTAG: agarus SecondChildCreateTimeTAG: 2012-12-07T17:49:21Z FirstChildTAG: I didn't mean to sound harsh. No, you are not stupid. I just wanted anyone reading this thread to understand what I meant. I hope things are clear now. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-12-07T22:25:29Z IndexTAG: 3064 TitleTAG: H12P2 R0 R1 R2 correct values, but red cross?????????? Hi. I have calculated R0=94, R1 = 90, R2 = 100 and i check the result with those values and i obtain for Vin= 10 v, Vout = 4.9V, and for Vin = 20 V, Vout = 5.1v, but i have red cross. Why????????????????????????????????????? UserIdTAG: 244225 UserNameTAG: lerimock CreateTimeTAG: 2012-12-05T07:16:54Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: I think you have to specify zenor voltage so that i will be able to debug your values. I assumed Vz=2.5Volt and used your values of R0,R1 and R2 for which when 20 volt is supplied as input, 4.75035372Volt is output which is wrong FirstChildUserIdTAG: 132685 FirstChildUserNameTAG: deepakmurali FirstChildCreateTimeTAG: 2012-12-05T08:10:14Z SecondChildTAG: If i use V+ = (Vin-Vz)*Rz/(R0+Rz) +Vz and Vout = V+*(R1+R2)/R2 i obtain the correct values. SecondChildUserIdTAG: 244225 SecondChildUserNameTAG: lerimock SecondChildCreateTimeTAG: 2012-12-05T08:50:55Z SecondChildTAG: if H12P2(c & d) correct = green cross SecondChildUserIdTAG: 297655 SecondChildUserNameTAG: Aljoska SecondChildCreateTimeTAG: 2012-12-05T12:02:04Z SecondChildTAG: What? SecondChildUserIdTAG: 244225 SecondChildUserNameTAG: lerimock SecondChildCreateTimeTAG: 2012-12-05T12:50:21Z SecondChildTAG: Without correct H12P2 part d, you not get green cross for H12P2 part c. SecondChildUserIdTAG: 297655 SecondChildUserNameTAG: Aljoska SecondChildCreateTimeTAG: 2012-12-05T14:11:31Z FirstChildTAG: 1) Look again at the values you have computed for the resistors. What's the gain of the circuit, required to give a nominal 5 volts out with a 2.5v zener voltage reference? Will your quoted R1/R2 values give you a virtual ground between the +- inputs of the amplifier? . Its a voltage divider, work it out for your values (I worked it out to be 1.9 using your values unless I'm having another mental aberration. Is 1.9 OK?). Do you get the same voltage 2.5v at the two terminals with this gain/ feedback resistor network? 2. You are required to work out the efficiency of this circuit, Power Out divided by Power In. An element of the power in into this circuit is caused by quiescent power loss within the resistors - they are drawing current all the time. Power is V^2/R, the lower the resistor values, the higher the power loss. Can you choose higher valued resistors for R1 and R2, to significantly reduce the power loss currently associated with your quoted values but still meet the gain requirement? 3. This question has lots of posts on the discussion board read them all and you will get all the help required. 4. As has been stated on previous posts the final value of RO, R1, R2 you achieve can have some variation , I suspect my values will be different from many other students values, what's important is the efficiency calculation that you are asked to produce it must be correct based upon the resistor values you have chosen. I ended up with quite a low efficiency but in agreement with the chosen values No doubt other students got better efficiency figures than I did also correct for their values, The efficiency will be low (it asks for minimum efficiency) which you will get using the maximum input voltage. The question suggests you model the zener as a voltage source and a resistor. Given that the output can swing +- 0.1 volts about a nominal 5V, for the quoted input supply variation, +10 to +20V, the reference voltage will have to change . Its possible to arrive at two equations using "some" voltage source for the zener with a varying voltage drop across the 1 ohm resistor to give you R0, the current through R0 passes through the zener. Hope this helps. FirstChildUserIdTAG: 273287 FirstChildUserNameTAG: johnkh77 FirstChildCreateTimeTAG: 2012-12-05T13:57:15Z SecondChildTAG: Those values work perfectly for a Vz of 2.5V giving $v_{OUT}$ of exactly 4.9V for $v_{IN}$ of 10V and 5.1V for 20V. Efficiency is horrible and I couldn't get a 'correct' answer that way. I don't know what the checker is really looking for as I think most people are using large values for R0, R1 and R2 such that the power used in them is insignificant in the efficiency calculation i.e. in the order of $10^{-4}$ when the answer is in the order of $10^{-1}$. SecondChildUserIdTAG: 141000 SecondChildUserNameTAG: OrinE SecondChildCreateTimeTAG: 2012-12-06T00:35:10Z SecondChildTAG: Still can't get it. looks impossible. If R0 is higher i obtain two different values for (R1+R2)/R2. I don't know what to do. SecondChildUserIdTAG: 244225 SecondChildUserNameTAG: lerimock SecondChildCreateTimeTAG: 2012-12-06T16:46:03Z FirstChildTAG: Still can't get it. looks impossible. If R0 is higher i obtain two different values for (R1+R2)/R2. I don't know what to do. FirstChildUserIdTAG: 244225 FirstChildUserNameTAG: lerimock FirstChildCreateTimeTAG: 2012-12-06T16:46:09Z FirstChildTAG: I'll tell you the process that I did to achieve full marks for the question - no doubt there are alternative methods and maybe better methods. 1. parts a and b require you to produce generic equations for the output voltage in terms of vin, and vz I assume there are no problems there. Parts b and c: Work out R1 and R2 for a zener reference voltage of 2.5v, working back from an output of 5V to give a virtual ground between the OP amp input terminals. **Ignore** the specified input voltage variation for this calculation of R1, R2, just focus on the use of a 2.5v reference. These values of R1 and R2 are now fixed. To work out Ro. As stated above, you model the zener as a voltage source in series with a resistor of 1 ohms the voltage source is referenced to ground and connected to one side of the resistor the other side of the resistor is connected to the appropriate op amp terminal and Ro. **Do not** fix the zener voltage source model at 2.5V, (the output voltage is allowed to swing between 4.9v and 5.1v and that swing occurs because of variation in the reference voltage value, changing with the input voltage values, i.e. if the output voltage rises or falls it does so because it follows the reference voltage changes, which results from the changes in input voltage. Use your output voltages, previous posts suggested using lower swings than given, about the nominal +5v with the corresponding input(power supply) voltages to determine the resulting virtual ground voltages and hence R0. You should have two equations from which you can determine Ro. My efficiency was not in the 10^-4 range rather in the 10^-1 ranges. FirstChildUserIdTAG: 273287 FirstChildUserNameTAG: johnkh77 FirstChildCreateTimeTAG: 2012-12-06T17:29:55Z SecondChildTAG: I obtain (R1+R2)/R2=2 for Vout=5volts, but i can't continue because the input in V- only con grow up, so it's impossible to obtain Vout=4.9 volts. SecondChildUserIdTAG: 244225 SecondChildUserNameTAG: lerimock SecondChildCreateTimeTAG: 2012-12-07T12:08:37Z FirstChildTAG: Try the ideas in my latest post here: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50b8d3275516b62b00000010 FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-12-06T18:39:03Z IndexTAG: 3065 TitleTAG: H12P1 part 3 If I use my previously calculated value for R, which is over 500 ohms, and insert 500 ohms as my load resistor, I will get a gain of 1 or slightly less for the circuit. Yet I am also given A=100. What am I missing? UserIdTAG: 231764 UserNameTAG: PaulP4881 CreateTimeTAG: 2012-12-05T02:00:46Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: 1. What is Vin for H12P1 part 2. 2. For this Vin compute Vout for new load resistance. 3. I=Vout/(R+Rload) FirstChildUserIdTAG: 297655 FirstChildUserNameTAG: Aljoska FirstChildCreateTimeTAG: 2012-12-05T14:30:01Z IndexTAG: 3066 TitleTAG: How is the resistance ever not equal to R2? For this circuit, as the voltage source and current are marked, how can V/I=R not be equal to R2? Either the voltage or the current has to change, if the resistance changes. The current is marked as what passes through R2. Therefore how can the resistance of that current combined with the (fixed) voltage source be anything other than R2? UserIdTAG: 91356 UserNameTAG: mnrsiat CreateTimeTAG: 2012-12-04T22:36:34Z VoteTAG: 0 CoursewareTAG: Week 12 / S23E3_L23AmplifierInputResistance CommentableIdTAG: 6002x_S23E3_L23AmplifierInputResistance NumberOfReplyTAG: 1 FirstChildTAG: If A were low (as in the example, but not in real op-amps), the voltage on the other side of R2, the $v^-$, might not be zero. So, yes you have a fixed $v_I$, but you don't have a fixed $v^-$, and so you don't have a fixed drop across R2. FirstChildUserIdTAG: 468623 FirstChildUserNameTAG: RobNik FirstChildCreateTimeTAG: 2012-12-05T13:44:21Z IndexTAG: 3067 TitleTAG: H12P2 - the efficiency question By which method should I calculate the minimum efficiency in general? My idea is that it means the proportion of the power consumed by the 100 Ohm load RL (let’s call this Pout) and the sum of the input power provided by the 2 sources (Pin). So the efficiency is Pout / Pin. To calculate Pout, I shall simply use (vOUT squared) divided by RL (for the case when vOUT is 5.1 V). But how to calculate Pin? I guess it shall be the sum of the input powers of the 2 sources. To get the input power of the independent voltage source, I believe I shall take the product of vIN and the input current iIN, which is the sum of iR0 and K*i (according to the node method). To get the input power of the dependent current source, I believe I shall take the product of its current K*i and the voltage dropped through it, which is vIN – vOUT (for the case when vIN is 20 V and vOUT is 5.1 V). Are these assumpions good, or is there something I see wrong? Thanks for any help! UserIdTAG: 129309 UserNameTAG: Memphis823 CreateTimeTAG: 2012-12-04T16:48:27Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi check this post: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50bd76e645ea1f1f0000007b Take care the efficiency has to be negative. FirstChildUserIdTAG: 296419 FirstChildUserNameTAG: oscman FirstChildCreateTimeTAG: 2012-12-04T17:36:41Z IndexTAG: 3068 TitleTAG: to SkyHawk! Hi Dear friend, In the past all your advises have bee very useful to me!, and you push people to think (which is good).This time I'm going to ask you to be more flexible.I'm still struggling with H12P2, and is my very last pending assignment. Really, I don't know what is wrong with my logic!For instance, tough I applied "virtual short circuit" at the OP AMP (because of negative feedback response),my answer doesn't match to what other fellow mates have done.My answer to question b is so simple as vout=vz*(R2+R1)/R2. When they say vIN < or > vz, can I assume that there is or not current thru zener, even tough there is no direct relation betwen vIN and vz ?(because of R0 voltage drop).I'm really concern in validating my concepts here. Would you care to help me? UserIdTAG: 369339 UserNameTAG: vargaslen CreateTimeTAG: 2012-12-04T15:45:01Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: That is what I got for the answer two weeks ago, and it was marked correct. I know that there was discussion with the staff, and the wording of the problem was changed, but in my mind that is the correct answer. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-12-04T15:59:02Z SecondChildTAG: So, what you're saying is that this is the right answer for question b?, What about vOUT=vIN*(R2+R1)/R2 for question a? SecondChildUserIdTAG: 369339 SecondChildUserNameTAG: vargaslen SecondChildCreateTimeTAG: 2012-12-04T16:49:25Z SecondChildTAG: Yes, that is what I have. With the current wording that should be correct. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-12-04T16:55:38Z SecondChildTAG: ok, thanks buddy. SecondChildUserIdTAG: 369339 SecondChildUserNameTAG: vargaslen SecondChildCreateTimeTAG: 2012-12-04T17:16:59Z SecondChildTAG: by the way, what do you mean with current wording? SecondChildUserIdTAG: 369339 SecondChildUserNameTAG: vargaslen SecondChildCreateTimeTAG: 2012-12-04T17:22:23Z SecondChildTAG: There was a lengthy discussion over a week ago, and as a result the staff changed the wording of the problem. Please see here: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50b3b5967ebf302500000029 SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-12-04T17:56:26Z SecondChildTAG: ok! I saw that discussion , and must confess, it was heavy!. My problem now is trying to get R0,R1 and R2. Based in a and b (which I'm now certain), as far as I've got is R2=R1 and 7.5 SecondChildUserIdTAG: 369339 SecondChildUserNameTAG: vargaslen SecondChildCreateTimeTAG: 2012-12-04T18:20:43Z SecondChildTAG: my bet is on R0=7.5,for trying to keep vIN above Vz, an current thru zener = 1. Am I lost here?Thanks again! SecondChildUserIdTAG: 369339 SecondChildUserNameTAG: vargaslen SecondChildCreateTimeTAG: 2012-12-04T18:23:00Z FirstChildTAG: Ro has to be large enough so that the reference voltage created by the Zener diode does not vary too much. The output voltage is allowed to vary .2V, i.e. 5V +/ .1V. Since the gain of the op amp will be set to approximately 2 then the reference voltage cannot vary more than .1V. But it doesn't have to vary that much. Setting R0 to allow a variation of .01V is fine and will make the rest of the design easy. The resistor branches of the circuit represent losses, so the smaller the current through those branches the better the efficiency, although unless you choose unrealistically small resistors their effect on efficiency is small. The ratio of R1 and R2 is chosen to give the gain that you need. All my resistors were in the 10K to 20K range. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-12-04T18:48:49Z SecondChildTAG: Thanks a lot SkyHawk! SecondChildUserIdTAG: 369339 SecondChildUserNameTAG: vargaslen SecondChildCreateTimeTAG: 2012-12-04T20:45:55Z SecondChildTAG: Great hint SkyHawk :) SecondChildUserIdTAG: 47372 SecondChildUserNameTAG: Digius SecondChildCreateTimeTAG: 2012-12-08T13:27:26Z IndexTAG: 3069 TitleTAG: H12P2 Hej guys, I have a problem with the values of $R_0$, $R_1$, and $R_2$. I found the equation for $v_{out}(v_{in}, v_z)$ and used the form $v_{out}=\frac{a·b}{1+b}·v_z+\frac{a}{1+a}·v_{in}$. Then I solved for (15, 5) and $\frac{\delta v_{out}}{\delta v_{in}}<0.02$. I checked the results in excel, and they are ok. I cannot get the green tick. Is the BJT affecting the equation and I just missed it? What's the problem? I assumed, $v^+\approx v^-$, no A taken into account. UserIdTAG: 7057 UserNameTAG: raiabril CreateTimeTAG: 2012-12-04T11:58:22Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Try [this][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50bb14dad5bfda270000007a FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-12-04T13:25:59Z SecondChildTAG: Hi raiabril. Assuming that you got the correct resistors and vout function I read a comment that said to solve for vin=20... It worked for me (they are asking for the minimum efficiency, so you have to maximize the denominator of the efficiency calculation). SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-12-04T15:47:45Z IndexTAG: 3070 TitleTAG: lost in H12P2 and H12P3 i have solved H12P1 but stuck in problem H12P2 and H12P3...can any one give me hints.. UserIdTAG: 114864 UserNameTAG: Vikaash CreateTimeTAG: 2012-12-04T11:32:24Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3071 TitleTAG: To STAFF: Missing videos in Sequence 27 (Week 14) I noticed that the last three videos in S27 cannot be accessed. They can be viewed on the old site, though. https://6002x.mitx.mit.edu/courseware/6.002_Spring_2012/Week_14/Energy_and_CMOS_Design/ UserIdTAG: 94751 UserNameTAG: ivigorth CreateTimeTAG: 2012-12-04T08:26:06Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 1 FirstChildTAG: The staff are working on it. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-12-04T17:00:40Z IndexTAG: 3072 TitleTAG: Link error in week 14 handout Hi, there is a link error for the week 14 handout. Could anyone please update it? Thanks. UserIdTAG: 529515 UserNameTAG: Low CreateTimeTAG: 2012-12-04T01:31:21Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3073 TitleTAG: regarding final exam during the final exam ,,,does the lecture sequence be available to us or they may be locked?? UserIdTAG: 183166 UserNameTAG: yogeshk CreateTimeTAG: 2012-12-03T19:11:40Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3074 TitleTAG: Differential pair, week 14 Can't quite see how the current is sourced to the NFET ( i.e. when its ON) when its load (PFET)is switched off. Does the NFET obtain its source drain current from the load attached to Vo? UserIdTAG: 273287 UserNameTAG: johnkh77 CreateTimeTAG: 2012-12-03T16:14:20Z VoteTAG: 0 CoursewareTAG: Week 14 / S27V1_Review CommentableIdTAG: 6002x_S27V1_Review NumberOfReplyTAG: 1 FirstChildTAG: Correction,posting refers to S27V6 Power dissipation in CMOS logic FirstChildUserIdTAG: 273287 FirstChildUserNameTAG: johnkh77 FirstChildCreateTimeTAG: 2012-12-03T16:18:14Z IndexTAG: 3075 TitleTAG: LAB 11 -Resonance frequency given as w0; For the Lab 11, the resonance frequency is given as **w0=10 KHz** instead of f0=10 KHz. I have just checked the answer by clicking "show answer" and it has been taken as **f0=10 KHz** and calculated the values for L and C. I got answer wrong since I used w0=10 KHz. the Lab question : **Consider the series RLC circuit shown in Figure 1. Leaving the resistance fixed at 10Ω your task is choose values of L and C such that the undamped resonance frequency, ω0, is 10kHz, with Q=10.** UserIdTAG: 356391 UserNameTAG: anoopkrishnan CreateTimeTAG: 2012-12-03T13:14:24Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Whenever you derive the values of $\omega, \omega_0, \omega_d, \Delta \omega$ or $\alpha$ using the circuit components ... eg. $\Delta \omega = \frac RL$, for this course you're supposed to think of their units as ***radians/second*** Whenever you are given something in ***Hz*** think ***Hz = cycles/second***. You have to first convert it to *radians/second* before using. $ 1\text{ cycle} = 2\pi\text{ radians}$. I know it can be confusing as we don't usually think of either *radians* or *cycles* as units. You may be right that it might be less confusing if $f_0$ was used. But, since it is the same quantity with a slightly different unit, we can be expected to know convert them even if the label is the same. FirstChildUserIdTAG: 714237 FirstChildUserNameTAG: PaxPolaris FirstChildCreateTimeTAG: 2012-12-03T16:45:58Z SecondChildTAG: damn, I forgot 2PI SecondChildUserIdTAG: 413869 SecondChildUserNameTAG: nlfigueiredo SecondChildCreateTimeTAG: 2012-12-04T13:40:39Z FirstChildTAG: Thanks a lot !!! FirstChildUserIdTAG: 356391 FirstChildUserNameTAG: anoopkrishnan FirstChildCreateTimeTAG: 2012-12-07T14:38:33Z IndexTAG: 3076 TitleTAG: H12P2 PLEASE HELP ME IN SOLVING C AND D PART OF THIS QUESTION UserIdTAG: 519690 UserNameTAG: vkjjw CreateTimeTAG: 2012-12-03T12:49:46Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: check this link https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50bb14dad5bfda270000007a FirstChildUserIdTAG: 357747 FirstChildUserNameTAG: kishores FirstChildCreateTimeTAG: 2012-12-03T14:16:44Z IndexTAG: 3077 TitleTAG: final exam Hi, how much time do we have to take the final exam.. 24 hours like the midterm ? UserIdTAG: 206648 UserNameTAG: TsvetanGeorgiev CreateTimeTAG: 2012-12-03T08:46:05Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: yes same time as the mid term... FirstChildUserIdTAG: 513738 FirstChildUserNameTAG: RPSINGHELE FirstChildCreateTimeTAG: 2012-12-03T11:07:06Z SecondChildTAG: thx mate :) SecondChildUserIdTAG: 206648 SecondChildUserNameTAG: TsvetanGeorgiev SecondChildCreateTimeTAG: 2012-12-04T08:18:56Z IndexTAG: 3078 TitleTAG: Transfer function! reduce it up to where? Hi again! This time I'd like to raise a question that is making my life miserable lately,and is this: up to what point should be the transfer function reduced? Any toughs will be very welcomed? UserIdTAG: 369339 UserNameTAG: vargaslen CreateTimeTAG: 2012-12-02T22:47:32Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Your question is a little vague however this may help. The important issue, in this course is to end up with a **denominator** in the transfer function of the form of a polynomial in s^2 and S plus some constant i.e. the canonical form which it is referred to in the text, it is acceptable to have a numerator in the transfer function as a polynomial in s - I believe the rule is that the degree of the denominator polynomial must be higher than that of the numerator polynomial. so if you have s+1/ (s^2 + 6s+7) this is acceptable and you can effectively ignore the numerator (s+1 )and, in this course to answer the questions, concentrate on the denominator ( s^2+6s+7) . In control engineering and indeed much of electronic engineering designers use the transfer function to determine the poles and zeros of the circuit which can tell you a lot about the circuit performance. The zeros effectively come from the numerator i.e. some value of s that effectively makes the numerator 0, in the above example transfer function, when s=-1, (s+1) =0. In this situation the transfer function effectively becomes 0. Of more interest to designers are the poles. The poles occur at the roots of the denominator ie finding the values of s that make (s^2 +6s+ 7) = 0. These roots may be real or imaginary, or a mixture of both - again in this course you are normally asked to find these roots. If these roots are used in the transfer function then the numerator is effectively being divided by 0 giving- simplistically -infinite gain thus we call these roots, poles. If you look at a 3D plot of the transfer function in the s plane , the transfer function rises dramatically in the regions of the poles. http://www.chem.mtu.edu/~tbco/cm416/PolesAndZeros.html. A health warning: my description above has some simplifications and a Google search on Poles and zeros will uncover some heated debates on the meaning of the transfer function and poles and zeros - such is life! Hope I've answered your question and it helps FirstChildUserIdTAG: 273287 FirstChildUserNameTAG: johnkh77 FirstChildCreateTimeTAG: 2012-12-03T12:43:31Z SecondChildTAG: Thanks a lot Johnkh77! that is the answer I was looking for! SecondChildUserIdTAG: 369339 SecondChildUserNameTAG: vargaslen SecondChildCreateTimeTAG: 2012-12-03T17:14:08Z SecondChildTAG: Nice link! SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-12-03T17:34:58Z SecondChildTAG: Very nice explanation! SecondChildUserIdTAG: 361137 SecondChildUserNameTAG: Ericson SecondChildCreateTimeTAG: 2012-12-04T02:20:38Z FirstChildTAG: A correction to my original posting where I suggested that the degree of the numerator polynomial should be less than that of the denominator, - not true "Physically realizable control systems must have a **number of poles greater than or equal to the number of zeros**. Systems that satisfy this relationship are called proper." source Wikibooks. Or otherwise its not permitted to have more holes than poles. (Poles arise from the denominator, zeros from the numerator), the number of the actual poles and zeros arise from the degree (s, s^2 etc) of the polynomials in the numerator and denominator It would thus be acceptable to have **both** numerator and denominator polynomials in s^2 or s^3 etc i.e. of equal degrees, **but not** for the numerator to have a higher degree polynomial say s^2, than the denominator say s. This would give a non realizable transfer function, or more holes than poles. All that said I believe my original statements about this course hold given the material we are studying i.e. get the denominator in the canonical form, which of course may mean reducing the numerator to its simplest form to achieve. FirstChildUserIdTAG: 273287 FirstChildUserNameTAG: johnkh77 FirstChildCreateTimeTAG: 2012-12-04T17:17:33Z IndexTAG: 3079 TitleTAG: Why my expression doesn't accepted as right answer? $1/(1+R*(1-w^2*L*C)/j*w*L)$ UserIdTAG: 95521 UserNameTAG: Gatling CreateTimeTAG: 2012-12-02T20:47:41Z VoteTAG: 0 CoursewareTAG: Week 11 / S21E3: Thevenin Tank CommentableIdTAG: 6002x_Thevenin_Tank NumberOfReplyTAG: 1 FirstChildTAG: You need parentheses around j*w*L. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-12-02T22:05:13Z SecondChildTAG: You are right. Thanks!!! SecondChildUserIdTAG: 95521 SecondChildUserNameTAG: Gatling SecondChildCreateTimeTAG: 2012-12-03T09:01:11Z IndexTAG: 3080 TitleTAG: Lab 12 completed! Hi guys, thanks to your help, finally did complete Lab 12, tough, would like to get things crystal clear. So , I assumed Bandwidth 200 Hz, assumed C1=C2=1 microFarad, got R2=1590 ohms (so far ok), but then when calculated R1 I've got a value under 10 ohms. Isn't 10 ohms the least value from those available? (Tough I got green check marks all over). Am I misunderstanding the resistance table?Thanks again guys! UserIdTAG: 369339 UserNameTAG: vargaslen CreateTimeTAG: 2012-12-02T20:17:33Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: Thanks Sergtronix and skyhawk and all for your suggestions finally I could solve LAB12 Thanks again FirstChildUserIdTAG: 352647 FirstChildUserNameTAG: praveenjugge FirstChildCreateTimeTAG: 2012-12-06T10:41:36Z FirstChildTAG: Hmm....Interesting.. What is BASIC problem in this LAB. You can ONLY use E96 resistors and !!!10% capasitors and arnt allowed to use trim pots or/and Caps bank and etc. Thats why $w_0$ specified as 10kHz+-100Hz. Ok, next , my C1=C2 and in the nF range.It allowed me to get standart and correct resistors values. FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-12-02T21:14:02Z FirstChildTAG: Resistors are made in the single digit values and even fractions, just not as commonly used. Choosing C to be smaller would have resulted in larger resistances. I used C = 2nF, certainly a common value for ceramic capacitors. With that choice, my smaller resistance was in the 100-1000 ohm decade. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-12-02T21:53:15Z FirstChildTAG: As I understand selected values ​​of the elements must satisfy specific criteria Q and w0=2*pi*"resonant frequency" must be the same as w0 wich can be calculated from equation from section 15.6? FirstChildUserIdTAG: 96873 FirstChildUserNameTAG: Santyaga FirstChildCreateTimeTAG: 2012-12-03T19:22:55Z SecondChildTAG: Yes.You have two equations and may easily get required values: 1)you have w0, Q and E12/E96 values requirement. 2)so let you already made your choise for capasitor values. 3)apply some algebra for both equations and voila - you got R1 and R2! SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-03T22:31:16Z IndexTAG: 3081 TitleTAG: Lab 12,huge equations from node method! Hi guys, I'd like to know if I'm in the right way here. From the node method, I've got rid of v2 (assuming v3=0 from the op amp circuit), but the equations I get are huge,even when haven't come to solve for s yet! UserIdTAG: 369339 UserNameTAG: vargaslen CreateTimeTAG: 2012-12-01T20:46:35Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: Have a look in the textbook. It's worked out there two ways starting page 863. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-12-01T21:05:33Z SecondChildTAG: Skyhawk is correct as always :) BTW LAB12 is pretty simple, and with my lab version requirements I got very cool values for both resistors and capasitors. SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-12-01T21:10:30Z SecondChildTAG: thanks! Skyhawk, think I'll have to learn more about crammer's rule (no wonder my equations were so big) SecondChildUserIdTAG: 369339 SecondChildUserNameTAG: vargaslen SecondChildCreateTimeTAG: 2012-12-02T16:40:34Z SecondChildTAG: There are just two equations but four unknowns, so there are extra degrees of freedom. A simplification suggested by OrinE is to let C1 = C2 = C. Then you still have one free variable to work with. There is still additional freedom since the design requirement for Q is an inequality. You can choose Q to make the design easier, if that will help. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-12-02T19:09:20Z FirstChildTAG: Thanks Guys! FirstChildUserIdTAG: 369339 FirstChildUserNameTAG: vargaslen FirstChildCreateTimeTAG: 2012-12-02T01:49:35Z FirstChildTAG: Hi vargaslen, I did not understand what you mean by get "rid of v2". As the the Lab itself suggests you can take a look at section 15.6 to see if you are going through the right path. Regards. FirstChildUserIdTAG: 288174 FirstChildUserNameTAG: matiasgrodriguez FirstChildCreateTimeTAG: 2012-12-01T21:08:51Z FirstChildTAG: Hi vargaslen, Just to answer you question whether you're taking the right path, I believe you are. Ok, you can get the equations from the textbook, but if you are like me, you will want take on the math. I can tell you the equations are huge, indeed, but the math is not that hard. And the nice thing about having the equation in the textbook is that you can check whether you got the right equations. Regards FirstChildUserIdTAG: 318577 FirstChildUserNameTAG: takeuchi FirstChildCreateTimeTAG: 2012-12-02T01:47:31Z FirstChildTAG: Just wanted to add link to an earlier thread with good information. https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50b41469d5463f1f00000076 FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-12-02T19:49:55Z IndexTAG: 3082 TitleTAG: h11p1 Got the correct expression of bandwidth and the capacitance values are also correct but system not accepting bandwidth value.why? UserIdTAG: 611666 UserNameTAG: Shaminder420 CreateTimeTAG: 2012-12-01T13:24:03Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: same problem here. I could't figure it out so far. FirstChildUserIdTAG: 319324 FirstChildUserNameTAG: SIZUSKA FirstChildCreateTimeTAG: 2012-12-01T13:46:56Z SecondChildTAG: divide your bandwidth answer by 2pi SecondChildUserIdTAG: 611666 SecondChildUserNameTAG: Shaminder420 SecondChildCreateTimeTAG: 2012-12-01T13:58:47Z SecondChildTAG: oh, that's obvious. Thank you ;) SecondChildUserIdTAG: 319324 SecondChildUserNameTAG: SIZUSKA SecondChildCreateTimeTAG: 2012-12-01T14:41:58Z SecondChildTAG: That didn't help me :( Formula correct, C correct but cant get the answer to both bandwidth questions. (divided by 2*pi*1000 for frequency in khz) SecondChildUserIdTAG: 360675 SecondChildUserNameTAG: MichielT SecondChildCreateTimeTAG: 2012-12-01T22:10:24Z IndexTAG: 3083 TitleTAG: AHA MOMENT I think it certainly deserves an aha moment since it saves quite a lot of time UserIdTAG: 206669 UserNameTAG: dimitrios66 CreateTimeTAG: 2012-12-01T12:04:44Z VoteTAG: 0 CoursewareTAG: Week 12 / S23V22 Inverting amplifier input resistance using virtual short method CommentableIdTAG: 6002x_S23V22_Inverting_amplifier_input_resistance_using_virtual_short_method NumberOfReplyTAG: 2 FirstChildTAG: I adhere. My vote goes to the AHA moment. By the way, will you post a summary of AHA moments? Thanks a lot from Spain. FirstChildUserIdTAG: 260716 FirstChildUserNameTAG: gftp FirstChildCreateTimeTAG: 2012-12-02T18:18:27Z FirstChildTAG: Definitely Agree!! Working in electronics for 30 years with a much less rigorous understanding than I now have (Thanks so much Professor), the virtual short was THE way to understand OpAmp circuits. FirstChildUserIdTAG: 369519 FirstChildUserNameTAG: perpstud FirstChildCreateTimeTAG: 2012-12-13T04:39:36Z IndexTAG: 3084 TitleTAG: S23V18 Is superposition justified? Good morning everyone, congratulations for the amazing course. Well, i have a question concerning the S23V18. During the video the superposition method is used in order that the v- is found. Since we have a VCVS and one independent, is it ok to use it? From what i have understood so far a dependent source cannot stand on its own. However by using the node method we come to the same result. That means that the result is correct. But the method was the right one to use? UserIdTAG: 206669 UserNameTAG: dimitrios66 CreateTimeTAG: 2012-12-01T11:14:32Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3085 TitleTAG: H11P1 What precisely was this supposed to mean? "So it is apparent that this is not the only circuit in the radio that selects the desired station from stations on adjacent channels." UserIdTAG: 287805 UserNameTAG: Beneficial CreateTimeTAG: 2012-12-01T03:18:08Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Stations may be spaced 10 kHz apart, but the bandwidth of the tuner is greater than 10 kHz; therefore, the tuner by itself does not have sufficient selectivity. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-12-01T03:50:54Z SecondChildTAG: but my stations were about 400khz apart and the respective bandwidths were only about 10 and 20 khz. SecondChildUserIdTAG: 287805 SecondChildUserNameTAG: Beneficial SecondChildCreateTimeTAG: 2012-12-01T06:21:07Z SecondChildTAG: I do not believe you can infer how far apart stations are spaced by your problem definition. Rather, in [S21E4][1], we are told, "Stations are allocated 10kHz channels, so in the worst case two stations may be separated by as little as 10kHz," per [these docs][2]. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_11/Filters/ [2]: http://www.ntia.doc.gov/files/ntia/publications/spectrum_wall_chart_aug2011.pdf SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-12-01T07:52:50Z SecondChildTAG: ahh, I see. Thank you. SecondChildUserIdTAG: 287805 SecondChildUserNameTAG: Beneficial SecondChildCreateTimeTAG: 2012-12-01T08:26:05Z SecondChildTAG: Glad I could help! :-) SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-12-01T13:19:08Z IndexTAG: 3086 TitleTAG: H11P1 I'm trying to figure out the impedance of the circuit, but I've got stuck. I found Z= RC*(s*L+RL)/((RC*C*s+1)*(s*L+RL)+RC), but I didn't manage to put it in the canonic form. If someone could give me a help, I'll appreciate... UserIdTAG: 378470 UserNameTAG: Iron CreateTimeTAG: 2012-11-30T23:53:38Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Admitancia. FirstChildUserIdTAG: 197569 FirstChildUserNameTAG: darksamaro FirstChildCreateTimeTAG: 2012-12-01T01:22:36Z SecondChildTAG: The things got better but there is a term RL/RC which doesn't let me finish it. I've got LCs²+...s+1+(RL/RC). If that last term doesn't exist, it'll be fine... Another tip? SecondChildUserIdTAG: 378470 SecondChildUserNameTAG: Iron SecondChildCreateTimeTAG: 2012-12-01T01:34:38Z SecondChildTAG: What's wrong with the last term? It's just a constant (a very small constant) that's no different than the 1. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-12-01T03:22:28Z SecondChildTAG: In this case, if there was no last term, I'd simply divide all this thing by LC, then I'd have the canonic form (s²+(.../LC)s+(1/LC)). But I have this last term, and I don't know what to do with it. SecondChildUserIdTAG: 378470 SecondChildUserNameTAG: Iron SecondChildCreateTimeTAG: 2012-12-01T05:59:29Z SecondChildTAG: Add the "last term" and the 1 together so that after you divide by LC you have s^2 + .... s + (1+RL/RC)/LC. Remember you are looking for a form S^2+a*s+b, in this case b = (1+RL/RC)/LC. When you plug in numbers you will find that RL/RC is very small compared to 1. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-12-01T13:03:50Z SecondChildTAG: The term b doesn't must be always 1/LC? SecondChildUserIdTAG: 378470 SecondChildUserNameTAG: Iron SecondChildCreateTimeTAG: 2012-12-01T14:13:13Z SecondChildTAG: No, it does't have to be 1/LC, but it is w0^2. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-12-01T15:59:56Z FirstChildTAG: Hi,Iron,to put Z in the canonic form, the form of denominator should be like this: s^2+a*s+b, we don't care about the numerator. You can get this form by mutiplying or division.Hope this can help you :) FirstChildUserIdTAG: 361137 FirstChildUserNameTAG: Ericson FirstChildCreateTimeTAG: 2012-12-01T03:17:03Z FirstChildTAG: Hello I don't understand where I am wrong. ![I find Z][1] Ok. Denominator in canonic form and Q ![Denominator][2] I have a green tick for value capacitors, but I haven't green tick at othet questions. Please help me find mistake. Sorry for my english [1]: https://edxuploads.s3.amazonaws.com/13543551992330899.jpg [2]: https://edxuploads.s3.amazonaws.com/13543558041343651.jpg FirstChildUserIdTAG: 413517 FirstChildUserNameTAG: mari_safonova FirstChildCreateTimeTAG: 2012-12-01T10:08:55Z SecondChildTAG: your expressions are correct. Do you remember that w0=2*pi*f0? SecondChildUserIdTAG: 177560 SecondChildUserNameTAG: ss7loginov SecondChildCreateTimeTAG: 2012-12-01T11:09:49Z SecondChildTAG: yes, but when enter sqrt((1+RL/RC)*1/(L*C))/(1/(C*RC)+RL/L) for answer first question I don't have green tick. SecondChildUserIdTAG: 413517 SecondChildUserNameTAG: mari_safonova SecondChildCreateTimeTAG: 2012-12-01T11:44:38Z SecondChildTAG: Sorry I'm found my mistake SecondChildUserIdTAG: 413517 SecondChildUserNameTAG: mari_safonova SecondChildCreateTimeTAG: 2012-12-01T11:46:53Z SecondChildTAG: ss7loginov is right, pay attention to that,I made the mistake too. SecondChildUserIdTAG: 361137 SecondChildUserNameTAG: Ericson SecondChildCreateTimeTAG: 2012-12-01T11:55:20Z SecondChildTAG: I've got the same equation, and also trying to put in sqrt((1+RL/RC)*1/(L*C))/(1/(C*RC)+RL/L) And it's still wrong. This result is in radians, so I tryed to divide it by 2pi, but it's still wrong. What should I stare at? And why do we ignore numerator? SecondChildUserIdTAG: 345464 SecondChildUserNameTAG: Constantine_ru SecondChildCreateTimeTAG: 2012-12-01T17:15:12Z SecondChildTAG: Ok, I got it, I should stare at the text of question :-) SecondChildUserIdTAG: 345464 SecondChildUserNameTAG: Constantine_ru SecondChildCreateTimeTAG: 2012-12-01T17:18:41Z SecondChildTAG: Wow,you got it,congratulations! About ignoring numerator,because after simplification we get the canomic form, and the characteristic equation s^2+2α*s+w0^2 is the denominator,during the process of simplifying, we can ignore the form of numerator,that is, just focus on the denominator, not on the whole Z,which is not a necessary step. SecondChildUserIdTAG: 361137 SecondChildUserNameTAG: Ericson SecondChildCreateTimeTAG: 2012-12-01T17:52:21Z IndexTAG: 3087 TitleTAG: Proctored Exam for the final . Is the proctored option available for the final exam that commences next month and if it is can someone kindly provide a link as to where to apply for it and centers where one can do so. thanks :) UserIdTAG: 321422 UserNameTAG: navooooo CreateTimeTAG: 2012-11-30T14:31:56Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I don't have an answer to your first question. You can however have a look around http://www.pearsonvue.com/ to see where the closest location is should the proctored option become available. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-12-01T00:25:16Z FirstChildTAG: Prof Agarwal talks about edX certificates in this article [report about MOOC][1] [1]: http://cacm.acm.org/magazines/2012/12/157884-in-the-year-of-disruptive-education/fulltext FirstChildUserIdTAG: 623210 FirstChildUserNameTAG: preveen FirstChildCreateTimeTAG: 2012-12-01T14:40:09Z IndexTAG: 3088 TitleTAG: H12P2 I have correctly found Vout as a function of a Vin,Vz,R0,R1 & R2 for both Vin>Vz and Vin UserIdTAG: 38331 UserNameTAG: janadel CreateTimeTAG: 2012-11-30T05:24:41Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Do a discussion search on "H12P2" and you will find the answers to most all your questions. FirstChildUserIdTAG: 397595 FirstChildUserNameTAG: mholin FirstChildCreateTimeTAG: 2012-11-30T10:55:19Z FirstChildTAG: Efficiency: Power Out/ Power In. Power Out is the power supplied to the load, via the transistor, and Power In is the power delivered to the circuit: (**quiescent**) plus power supplied to the load by the Power supply. Be careful with your calculation of the latter for the power supply, there is a dependent source transistor in the feedback path of the OP amp, what current flows through the dependent source, and what current flows through the load? It requires **minimum** efficiency. Given that the efficiency is the ratio of two numbers, maximise the denominator value, the input power. Power is VI, which term dominates in this relationship for the power supply options specified and what impact does this value have on the output voltage value? Note it tells you you can ignore the power used by the amplifier - they tell you that for a reason it impacts upon your answer. If they give you numerical values in the question you need to use them all. FirstChildUserIdTAG: 273287 FirstChildUserNameTAG: johnkh77 FirstChildCreateTimeTAG: 2012-11-30T15:23:39Z IndexTAG: 3089 TitleTAG: why w0 The period also could be w,if we use another supply. Why he use w0 instead? Thanks UserIdTAG: 136959 UserNameTAG: christerpher CreateTimeTAG: 2012-11-30T02:35:27Z VoteTAG: 0 CoursewareTAG: Week 11 / S21V17: Another_way_of_looking_at_Q CommentableIdTAG: 6002x_S21V17_Another_way_of_looking_at_Q NumberOfReplyTAG: 1 FirstChildTAG: w can be any frequency. w0 is one specific frequency, i.e. the resonant frequency. FirstChildUserIdTAG: 90796 FirstChildUserNameTAG: ChaunceyGardiner FirstChildCreateTimeTAG: 2012-11-30T06:33:05Z IndexTAG: 3090 TitleTAG: Current mirror ! It's very strange because we had to do an exercise like this one ( I don't remember if it was homework or other ). I said the currents must be the same, but I was wrong ! (I search where it was ... ) UserIdTAG: 373752 UserNameTAG: Croak CreateTimeTAG: 2012-11-28T16:16:39Z VoteTAG: 0 CoursewareTAG: Week 12 / How to build a current source CommentableIdTAG: 6002x_How_to_build_a_current_source NumberOfReplyTAG: 1 FirstChildTAG: It was a tutorial on week 6. It is this tut : https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_6/wk6_bCS/ FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-12-09T10:37:16Z IndexTAG: 3091 TitleTAG: Bug report, staff. Hello, I dunno if anyone's still here :). I want to report that videos S6V6 and S6V7 are the same. I hope that by "tagging" post You meant typing those words in the title. Best regards from Krakow, Poland! What You do is Remarkable. UserIdTAG: 778186 UserNameTAG: Pabl CreateTimeTAG: 2012-11-28T15:20:30Z VoteTAG: 0 CoursewareTAG: Week 3 / Graphical Analysis of Nonlinear Circuits CommentableIdTAG: 6002x_graphical_analysis_of_nonlinear_circuits NumberOfReplyTAG: 2 FirstChildTAG: Yes they are ) FirstChildUserIdTAG: 66669 FirstChildUserNameTAG: agarus FirstChildCreateTimeTAG: 2012-11-28T16:49:29Z FirstChildTAG: Actually they are both the same only if played at normal(1.0x) speed. At other speeds, they are different. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-11-28T20:34:11Z SecondChildTAG: What difference it makes with speed? Please specify. Thanks in advance! SecondChildUserIdTAG: 455950 SecondChildUserNameTAG: Wahabbaluch SecondChildCreateTimeTAG: 2012-11-29T10:51:45Z SecondChildTAG: yes u're correct!! i thought i was going **nuts** when every1 pointed that they r same and i couldn't see it! actually i was on 1.5x speed. btw 1.5x is pretty slow considering that our sir repeats things so many times! SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-12-19T16:15:40Z IndexTAG: 3092 TitleTAG: S17V10 why are we adding two roots? Why don't we consider them 2 separate roots? Do we have to add all the roots we find of homogeneous and all roots of particular solutions while analyzing any circuit? UserIdTAG: 66669 UserNameTAG: agarus CreateTimeTAG: 2012-11-28T14:29:44Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3093 TitleTAG: How to find L? how can I find L? f is not w0? UserIdTAG: 329532 UserNameTAG: Dinko CreateTimeTAG: 2012-11-28T13:07:05Z VoteTAG: 0 CoursewareTAG: Week 11 / S21E4: AM_Radio_Tuning CommentableIdTAG: 6002x_S21V1_AM_Radio_Tuning NumberOfReplyTAG: 1 FirstChildTAG: 2*pi*f = w FirstChildUserIdTAG: 219473 FirstChildUserNameTAG: jsavvidis FirstChildCreateTimeTAG: 2012-11-28T16:49:24Z IndexTAG: 3094 TitleTAG: Why not a Inductor? I tought of a Inductor instead of a Resistor at first! Would this work? At t=0 the inductor behaves as a open circuit and vI gets dropped across it.But the I tought, at t=0 i would be 0 and start to build up! so as t goes to infinity i starts to look like the one I'm looking for. It's that it? (I'm trying to build up some intuition here). UserIdTAG: 369339 UserNameTAG: vargaslen CreateTimeTAG: 2012-11-28T12:04:44Z VoteTAG: 0 CoursewareTAG: Week 12 / S24V5 First try to build integrator CommentableIdTAG: 6002x_S24V5_First_try_to_build_integrator NumberOfReplyTAG: 1 FirstChildTAG: The differential equation for a vI - L - C - circuit is: LC*d2vO/(dt)2 + vO = vI We know that this oscillates, but doesn´t integrate. FirstChildUserIdTAG: 388524 FirstChildUserNameTAG: dasbinich FirstChildCreateTimeTAG: 2012-12-03T00:13:51Z IndexTAG: 3095 TitleTAG: some one tell which filter is it?? Low pass filter with lower cutoff fc=20hz is cascaded with the high pass filter with cutoff fc=30 hz. The resultant system is- (a) All stop filter (b) band stop filter i will be thank ful to you UserIdTAG: 563741 UserNameTAG: Mukundsingh89 CreateTimeTAG: 2012-11-28T10:16:08Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: The low pass filter cut away the the high part that the high pass filter is intended to let pass. The result will be a "all stop filter" but the frequencies are so close, so the result will be a band pass filter with a damping factor in the pass band. FirstChildUserIdTAG: 151472 FirstChildUserNameTAG: Stensmed FirstChildCreateTimeTAG: 2012-11-28T16:48:54Z FirstChildTAG: low pass filter cuts everything that exceeds 20Hz High pass filter cuts everything under 30hz ... answear (a) !! FirstChildUserIdTAG: 373752 FirstChildUserNameTAG: Croak FirstChildCreateTimeTAG: 2012-11-28T17:34:09Z IndexTAG: 3096 TitleTAG: So what is it about? According to the best exp I found : for low w -> jwL/(R+jwL) -> jwL/jwL(R/jwL+1) -> 1/(R/jwL+1) here I can ignore the 1 because R/jwL for small w we can assume it will be much larger than 1, so I would rather stick to 1/(R/jwL)=jwL/R. For high w -> 1/(1+jwRC), 1 is ignored, as we assume (wRC)^2 is large enough -> 1/jwRC :))) Not bad at all ^^^ :) UserIdTAG: 415375 UserNameTAG: ZWX CreateTimeTAG: 2012-11-27T21:27:54Z VoteTAG: 0 CoursewareTAG: Week 11 / S21E3: Thevenin Tank CommentableIdTAG: 6002x_Thevenin_Tank NumberOfReplyTAG: 0 IndexTAG: 3097 TitleTAG: Real MOFSTETs What model of real MOSFET behaves like our MOSFET model that we learnde? I meen is it P-channel or n-channel or something else? Thanx. UserIdTAG: 500965 UserNameTAG: barka0 CreateTimeTAG: 2012-11-27T21:20:35Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: p and n channel, in the grand scheme of things they both behave the same. FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-11-27T22:10:20Z IndexTAG: 3098 TitleTAG: Double check me please - Book error On page 607 of the text book, equation (11.23) The second part of the equation contains an RL^2 term. Shouldn't that be RON^2??? most of subsequent equations follow this one. UserIdTAG: 127195 UserNameTAG: princeofsudan CreateTimeTAG: 2012-11-27T21:06:56Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3099 TitleTAG: How you calculated Q? How you calculated Q of this circuit? There are actually three Q factors due to each element. Thanks in advance UserIdTAG: 341358 UserNameTAG: akbar13 CreateTimeTAG: 2012-11-27T13:24:24Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: output NumberOfReplyTAG: 1 FirstChildTAG: Which Q? Where?? Which circuit??? FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-11-27T14:21:05Z SecondChildTAG: Hi akbar13, Can I help you? Which circuit are you talking about? :P SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-27T23:18:01Z SecondChildTAG: It's the S22E1 circuit. It seems that the Q only applies to the voltage spike seen on the capacitor and inductor. SecondChildUserIdTAG: 359310 SecondChildUserNameTAG: AaronYeoh SecondChildCreateTimeTAG: 2012-11-28T05:38:01Z SecondChildTAG: I just looked at the graph and guessed it. Not sure if that is the right way to find out Q. If we have to do it using the formula, w0 can be calculated, as f is 1e06 Hz. But how do we find out R and L? SecondChildUserIdTAG: 232157 SecondChildUserNameTAG: GayatriTR SecondChildCreateTimeTAG: 2012-11-28T14:33:22Z SecondChildTAG: Look at page 817 in the text book, which discusses Q in a resonant circuit. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-11-29T01:16:52Z SecondChildTAG: I think you can't exactly calculate it, or at least I can't find the way. By the given resonant frequency w0, you can guess a pair of L*C, but we need R too. Another way would be measuring the bandwidth in the Y circuit (voltage taken at Vr), so we know w0 and delta w (~0.2MHz), but it would only be an approximation. How to find the exact 4.176 value of Q with just these parameters? Thank you! SecondChildUserIdTAG: 398594 SecondChildUserNameTAG: DaveyJC SecondChildCreateTimeTAG: 2012-11-29T11:02:34Z SecondChildTAG: the answer is that of this question:how many times Vi is Vo?see S21V13 SecondChildUserIdTAG: 439986 SecondChildUserNameTAG: DanPop SecondChildCreateTimeTAG: 2012-11-29T18:45:40Z SecondChildTAG: Yeah, S21V13, but how do you calculate its EXACT value in this case? SecondChildUserIdTAG: 398594 SecondChildUserNameTAG: DaveyJC SecondChildCreateTimeTAG: 2012-11-29T19:01:24Z SecondChildTAG: You can not calculate the Q here as you don't have all the parameter values. You do have a freq. plot from which you can read the Q value (+/-, about, not very accurate). SecondChildUserIdTAG: 114881 SecondChildUserNameTAG: xvink SecondChildCreateTimeTAG: 2012-11-29T19:40:19Z SecondChildTAG: In the textbook is said that **abs(Vout)=Vin*Q**,and so the Q is just the ratio of magnitude of Vout and Vin SecondChildUserIdTAG: 342144 SecondChildUserNameTAG: Orest02 SecondChildCreateTimeTAG: 2012-12-01T14:57:29Z SecondChildTAG: By visual inspection, the width at 1/sqrt(2) is about 1/4 the w0 frequency. SecondChildUserIdTAG: 381386 SecondChildUserNameTAG: MrMatt7k SecondChildCreateTimeTAG: 2012-12-01T18:15:04Z IndexTAG: 3100 TitleTAG: LAB 12 I have the expressions for w0 and the bandwidth in terms of the resistors and condensers. How do you go from here to find the value of each resistor or condenser? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-11-27T01:16:25Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Same Problem here.We have two equations with four unknowns.Where should we start ? FirstChildUserIdTAG: 38331 FirstChildUserNameTAG: janadel FirstChildCreateTimeTAG: 2012-11-27T14:03:10Z FirstChildTAG: Make the capacitors equal to get rid of one unknown. Then trial and error with the two resistors. I put the equations into an excel spreadsheet and played with the values until I got acceptable results. If you have access to the book "Active Filter Cookbook" by Don Lancaster, the circuit is in it (actually, it's in the textbook too if you just want to check your math). Here's a hint: make the feedback resistor $4Q^2$ times the input resistor. FirstChildUserIdTAG: 141000 FirstChildUserNameTAG: OrinE FirstChildCreateTimeTAG: 2012-11-27T18:14:28Z SecondChildTAG: Hahaa, moment :) SecondChildUserIdTAG: 244706 SecondChildUserNameTAG: Miguel_Angel SecondChildCreateTimeTAG: 2012-11-28T00:19:51Z SecondChildTAG: $4Q^2$, very valuable tip. Thank again OrinE! SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-29T00:06:30Z SecondChildTAG: that was nice.. OrinE :-) SecondChildUserIdTAG: 185815 SecondChildUserNameTAG: adilmeersha SecondChildCreateTimeTAG: 2012-11-29T07:16:11Z IndexTAG: 3101 TitleTAG: S26V8 stops at 1:32 Video S26V8 freezes at time 1:32. Moving the time cursor forward just brings it back to 1:32. UserIdTAG: 397595 UserNameTAG: mholin CreateTimeTAG: 2012-11-26T21:50:33Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: It's happened to me several times in different videos. If you start running another video and come back to the same, you can then drag the cursor to 1:32 and it'll resume normaly. FirstChildUserIdTAG: 381024 FirstChildUserNameTAG: QZ FirstChildCreateTimeTAG: 2012-11-27T14:12:19Z IndexTAG: 3102 TitleTAG: h12p3 help i have tried for both break voltage=1/2vi and sqrt(2)/2 vi and 1khz gain=2 vi and sqrt2 vi but it is wrong. I got 3 equations from the transfer function, 2 break frequencies and 1khz. Here it is http://www.wolframalpha.com/input/?i=solve+system+of+equations&a=*C.solve+system+of+equations-_*Calculator.dflt-&a=FSelect_**SolveSystemOf3EquationsCalculator-.SolveSystemOf4EquationsCalculator.SolveSystemOf2EquationsCalculator-&f3=%281250000*a*13*pi*%28c%2B20000%29%29%2F%28sqrt%28%28%28a*13*pi%29^2%2B25*10^10%29*%28121*%28b*13*pi%29^2%2B625*10^18%29%29%29%3Dsqrt%282%29%2F2&f=SolveSystemOf3EquationsCalculator.equation1_%281250000*a*13*pi*%28c%2B20000%29%29%2F%28sqrt%28%28%28a*13*pi%29^2%2B25*10^10%29*%28121*%28b*13*pi%29^2%2B625*10^18%29%29%29%3Dsqrt%282%29%2F2&f4=%281250000*a*17000*pi*%28c%2B20000%29%29%2F%28sqrt%28%28%28a*17000*pi%29^2%2B25*10^10%29*%28121*%28b*17000*pi%29^2%2B625*10^18%29%29%29%3Dsqrt%282%29%2F2&f=SolveSystemOf3EquationsCalculator.equation2_%281250000*a*17000*pi*%28c%2B20000%29%29%2F%28sqrt%28%28%28a*17000*pi%29^2%2B25*10^10%29*%28121*%28b*17000*pi%29^2%2B625*10^18%29%29%29%3Dsqrt%282%29%2F2&f5=%281250000*a*1000*pi*%28c%2B20000%29%29%2F%28sqrt%28%28%28a*1000*pi%29^2%2B25*10^10%29*%28121*%28b*1000*pi%29^2%2B625*10^18%29%29%29%3Dsqrt%282%29&x=1&y=8&f=SolveSystemOf3EquationsCalculator.equation3_%281250000*a*1000*pi*%28c%2B20000%29%29%2F%28sqrt%28%28%28a*1000*pi%29^2%2B25*10^10%29*%28121*%28b*1000*pi%29^2%2B625*10^18%29%29%29%3Dsqrt%282%29 what is wrong? thanks UserIdTAG: 138709 UserNameTAG: tom_smith CreateTimeTAG: 2012-11-26T11:37:18Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: No need for simultaneous equations. The three conditions can be satisfied independently. Calculate resistance with R = 1/(C*2*pi*f) where f is the break frequency. RF computed to give required gain of a non-inverting amplifier. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-26T12:53:09Z SecondChildTAG: Hi skyhawk, Can you provide some hints on how you got the equation for the resistance? I know the equation works, but when I try to derive it I end up in a tangle of algebra with coupling between R3 and R4 that is nasty. A few folks have been commenting that it's really easy but I don't yet see how to do it. I know I can just use the formula you posted but I want to understand why. Any help would be appreciated. Thanks SecondChildUserIdTAG: 367878 SecondChildUserNameTAG: spoida SecondChildCreateTimeTAG: 2012-12-04T04:39:33Z SecondChildTAG: For the input network H = R1/(R1+1/(j*w*C1)) = j*w*R1*C1/(1+j*w*R1*C1) The break frequency is where the real part and imaginary part have the same magnitude, so 1 = w*R1*C1 or R1 = 1/(w*C1) = 1/(2*pi*f*C1). A **similar** analysis for the output network gives R2 = 1/(2*pi*f*C2) SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-12-07T14:52:43Z FirstChildTAG: I think you are trying way too hard. Think in terms of just break frequencies. Maybe the prof didn't spend much time on it but where in terms of omega is the response down 1/sqrt(2). Is just 1/(RC) or R/L then don't forget to convert to Hz from radians/sec if the problem calls for Hz. omega is 2*pi*f. FirstChildUserIdTAG: 397595 FirstChildUserNameTAG: mholin FirstChildCreateTimeTAG: 2012-11-26T12:54:25Z FirstChildTAG: Hi guys finally got it. very simple indeed. thanks a bunch FirstChildUserIdTAG: 138709 FirstChildUserNameTAG: tom_smith FirstChildCreateTimeTAG: 2012-11-26T14:25:54Z IndexTAG: 3103 TitleTAG: Homework 10 part 3 question 2 the question is to find VI/II the complex input voltage over the complex input current...What exactly are we trying to find, is II the same as If on the circuit? which current is it talking about and where is it located? UserIdTAG: 435197 UserNameTAG: RichmondRichter CreateTimeTAG: 2012-11-26T04:35:32Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi,RichmondRichter,the voltage is Vf(across the capacitor),the current is If(through R1).We need to find a relation between Vf and If applying KCL. Myrimit' post is helpful. https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50b12405ca9a44220000002f FirstChildUserIdTAG: 361137 FirstChildUserNameTAG: Ericson FirstChildCreateTimeTAG: 2012-11-26T05:49:18Z SecondChildTAG: Ok thank you, so thats an f not an I, I can bearly see it lol. I thought it was VI and II not Vf and If, makes sense now SecondChildUserIdTAG: 435197 SecondChildUserNameTAG: RichmondRichter SecondChildCreateTimeTAG: 2012-11-26T08:38:04Z SecondChildTAG: That's great!Come on! SecondChildUserIdTAG: 361137 SecondChildUserNameTAG: Ericson SecondChildCreateTimeTAG: 2012-11-26T11:25:18Z IndexTAG: 3104 TitleTAG: Mechanical impedance model analogy Hello 6.002x people, As a mechanical engineering student, I try to find as much analogies between the two subjects. Mechanical systems and electrical systems can be seen as analogs, and solved with the same equations. In mechanical engineering, the sinusoidal steady state response is usually, as far as i know, solved with Laplace transformations. Is there a reason the element impedance model is not widely used in mechanical system analysis? Or is the element impedance model actually an simplified version of the laplace transformation? Greetings! UserIdTAG: 556514 UserNameTAG: RvanErp CreateTimeTAG: 2012-11-26T00:57:20Z VoteTAG: 0 CoursewareTAG: Week 10 / Element impedance models CommentableIdTAG: 6002x_Element_impedance_models NumberOfReplyTAG: 2 FirstChildTAG: Both the laplace transform and complex impedance methods are completely valid ways of approaching and solving mechanical and electrical problems, which are at their heart, solving differential equations. There is no one inherently better method-- the emphasis on one over the other often comes down to mere convention, or practicality. For instance, we can successfully model a harmonic oscillator as an LC circuit, but that requires an extra level of abstraction, first thinking of your spring as a capacitor, and furthermore thinking of your capacitor as a resistor. But both methods are completely fine. FirstChildUserIdTAG: 390034 FirstChildUserNameTAG: kahlil FirstChildCreateTimeTAG: 2012-11-26T01:43:06Z FirstChildTAG: whatever your method, always make it a point to have the "Aha moment" FirstChildUserIdTAG: 623210 FirstChildUserNameTAG: preveen FirstChildCreateTimeTAG: 2012-11-26T10:10:12Z SecondChildTAG: Aha! That's the secret! SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-11-26T12:10:01Z IndexTAG: 3105 TitleTAG: help for S21E4 i have some problem to calculate Q. Can you give me some hint? UserIdTAG: 206669 UserNameTAG: dimitrios66 CreateTimeTAG: 2012-11-25T18:07:13Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: $Q=\frac{\omega o}{2\alpha} $ If you put you characteristic equation in the canonical form then you have all this terms. FirstChildUserIdTAG: 288174 FirstChildUserNameTAG: matiasgrodriguez FirstChildCreateTimeTAG: 2012-11-25T18:16:34Z FirstChildTAG: Get 2a for the circuit in exercise (parallel L, C). It is not R/L, as in the course. FirstChildUserIdTAG: 357924 FirstChildUserNameTAG: DanI20 FirstChildCreateTimeTAG: 2012-11-25T18:16:40Z SecondChildTAG: thanks for the hints. It worked fine. However i used only the denominator (in canonical form)of the transfer function. Do we care for the numerator as well or our purpose is to locate either the numerator or the denominator in canonical form? SecondChildUserIdTAG: 206669 SecondChildUserNameTAG: dimitrios66 SecondChildCreateTimeTAG: 2012-11-25T18:52:26Z SecondChildTAG: I did all the simplifications needed to get the numerator equal to 1 and the denominator in canonical form. SecondChildUserIdTAG: 357924 SecondChildUserNameTAG: DanI20 SecondChildCreateTimeTAG: 2012-11-25T19:23:56Z SecondChildTAG: I did the simplifications so the denominator was on the canonical form. The numerator did not got 1. SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-25T19:29:32Z SecondChildTAG: There are some question from yesterday and today about whether it is always correct to make the denominator on the canonical form and don't care about the numerator but nobody answered. SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-25T19:32:35Z SecondChildTAG: Well it seems that we have to care for the denominator only. I suppose this has to do with the stability of the system and perhaps is outside of the scope of the current course. In any case if there is some answer from the staff, it will be highly apreciated SecondChildUserIdTAG: 206669 SecondChildUserNameTAG: dimitrios66 SecondChildCreateTimeTAG: 2012-11-25T20:15:55Z IndexTAG: 3106 TitleTAG: help regarding problem https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_1/Basic_Circuit_Analysis/ [H1P2] A KVL equation can be written around each of these loops. How many of the KVL equations are independent? can anyone explain the reason for this answer? UserIdTAG: 810332 UserNameTAG: jsmita CreateTimeTAG: 2012-11-25T16:56:19Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: take a look at page 63 and page 69 of textbook FirstChildUserIdTAG: 325197 FirstChildUserNameTAG: Vitali_Jerin FirstChildCreateTimeTAG: 2012-11-25T19:56:52Z IndexTAG: 3107 TitleTAG: help regarding problem I have just started following this course. i am not getting the true voltages at node A & B in lab 0. so please anyone could help me out of this situation? UserIdTAG: 810332 UserNameTAG: jsmita CreateTimeTAG: 2012-11-25T16:14:24Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Node A must be the same voltage as the power supply, 1V, since there is no component between the power supply and the node. We assume conductors are perfect. Node B is 0.83V, must apply the "voltage divisor" formula. In this case, 6K ohms are related to 1V, so 1K -> 0.17, 3K -> 0.5, 2K -> 0.33. Now, node A is 1V, node B is 1-0.17=0.83V, node C is 1-0.17-0.5=0.33 and node D is zero (ground). FirstChildUserIdTAG: 296965 FirstChildUserNameTAG: LGMailhos FirstChildCreateTimeTAG: 2012-11-25T16:36:34Z SecondChildTAG: thanks a lot... SecondChildUserIdTAG: 810332 SecondChildUserNameTAG: jsmita SecondChildCreateTimeTAG: 2012-11-25T17:03:37Z IndexTAG: 3108 TitleTAG: Solution to second order differential equation in the textbook On page 630 of the textbook the following is written: $v(t)=A_1e^{+jw_0t}+A_2e^{-jw_0t}$ And then based on the above the formula 12.11 is provided: $v(t)=K_1cos(w_0t)+K_2sin(w_0t)$ But I wonder where is $j$? I mean that according to Euler's equation: $v(t)=A_1(cos(w_0t)+j\cdot sin(w_0t))+A_2(cos(w_0t)-j\cdot sin(w_0t))=$ $=(A_1+A_2)cos(w_0t)+j\cdot (A_1-A_2)sin(w_0t)$ Is it legitimate to assign $j\cdot (A_1-A_2)$ to a new variable $K_2$? UserIdTAG: 252722 UserNameTAG: vaboro CreateTimeTAG: 2012-11-25T16:07:31Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3109 TitleTAG: another formula checker bug arctan(-R*C*w)+w*t is not accepted as valid angle, although it is the same as w*t + arctan(-w*R*C) UserIdTAG: 209655 UserNameTAG: typograph CreateTimeTAG: 2012-11-25T14:15:56Z VoteTAG: 0 CoursewareTAG: Week 10 / Complex Numbers CommentableIdTAG: 6002x_Complex_numbers NumberOfReplyTAG: 2 FirstChildTAG: In lab 0, i am not getting correct voltages at node B & C. i have just started following this course. so can anyone please help me out solve that problem? FirstChildUserIdTAG: 810332 FirstChildUserNameTAG: jsmita FirstChildCreateTimeTAG: 2012-11-25T15:29:21Z SecondChildTAG: Hi jsmita! It's a wee bit late to start out on this course because the course is finishing soon. Have you read the [Couse Info][1] page? > Enrollment: Newly registered students should be aware that it is impossible to pass the class at this late stage. Although we have re-opened enrollment for technical reasons, we encourage you to enroll in next semester's offering instead of this one. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/info "Course Info" SecondChildUserIdTAG: 359310 SecondChildUserNameTAG: AaronYeoh SecondChildCreateTimeTAG: 2012-11-26T06:13:53Z FirstChildTAG: @jsmita, are you using voltage probes to measure the voltages for you? ![enter image description here][1] Once you have probes in place, use the DC button (at the top of the circuit tool) to run a simulation. [1]: https://edxuploads.s3.amazonaws.com/13538846048754385.png FirstChildUserIdTAG: 9673 FirstChildUserNameTAG: xp42 FirstChildCreateTimeTAG: 2012-11-25T23:06:54Z IndexTAG: 3110 TitleTAG: Some Advice Needed I have 60% after week 10 and will be having my exams from 5th to 21 so I do not wish to do Hw and Labs of week 11,and 12.How can I cover these topics ? Is textbook sufficient(it takes less time to read than watch )or watching lectures is still better.Secondly does anyone know any good review strategy for midterm ,I guess it will be sometime between my exams. UserIdTAG: 99628 UserNameTAG: Pranjal16 CreateTimeTAG: 2012-11-25T03:10:31Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Read through the pdf's of the lecture slides. If anything is unclear just drag the slider through the video to get to where you need to hear an explanation. You can find the lecture slides on the RHS of the 'Course Info' page FirstChildUserIdTAG: 9673 FirstChildUserNameTAG: xp42 FirstChildCreateTimeTAG: 2012-11-25T07:47:52Z IndexTAG: 3111 TitleTAG: not sure where i am going wrong i getting magnitude = wRC / sqrt(1+(wRC)^2) and phase = 90 - arctan(wRC) UserIdTAG: 388273 UserNameTAG: kilmarta CreateTimeTAG: 2012-11-24T22:06:19Z VoteTAG: 0 CoursewareTAG: Week 10 / Magnitudes And Angles CommentableIdTAG: 6002x_Magnitudes_and_angles NumberOfReplyTAG: 2 FirstChildTAG: is it something about the freq? is w not = 170 in first one? am i going to feel stupid when i realise my mistake? FirstChildUserIdTAG: 388273 FirstChildUserNameTAG: kilmarta FirstChildCreateTimeTAG: 2012-11-24T22:14:07Z FirstChildTAG: ah i never converted freq to radians, move along FirstChildUserIdTAG: 388273 FirstChildUserNameTAG: kilmarta FirstChildCreateTimeTAG: 2012-11-24T22:15:43Z IndexTAG: 3112 TitleTAG: I'm stumped on H12P3 part 2 I've started this problem over about 5 times and keep getting the same answer, but the grader doesn't like it. I've even simulated my answer in the circuit sandbox, performed an AC analysis, and confirmed I have -3dB gain at both cutoff frequencies. It looks right to me, but the only thing marked correct are the capacitor values, C3 and C4. I've re-read the problem to see if there is some requirement I might have missed, but I'm at a loss. UserIdTAG: 184529 UserNameTAG: tfors CreateTimeTAG: 2012-11-24T21:15:50Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Maybe you are trying too hard, the problem is super easy. What is the only difference between Part 1 and Part 2? FirstChildUserIdTAG: 397595 FirstChildUserNameTAG: mholin FirstChildCreateTimeTAG: 2012-11-24T22:15:58Z FirstChildTAG: You're right. I should have looked back to the page on which I worked out the transfer function for Part 1. Your clue was most helpful. Thank you. FirstChildUserIdTAG: 184529 FirstChildUserNameTAG: tfors FirstChildCreateTimeTAG: 2012-11-25T00:28:32Z IndexTAG: 3113 TitleTAG: Problem with checkmark Lab 11 In Lab 11, I calculated the values of L and C as *mH and *uf. I changed the values of the components in the circuit and put the exact values in the boxes below. However, on checking I'm getting a green checkmark for the circuit and two big red crosses for the values written in the boxes. Can someone explain why is that so, even when the values are same. UserIdTAG: 499671 UserNameTAG: maliha266 CreateTimeTAG: 2012-11-24T19:38:37Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3114 TitleTAG: Cursor vs Pen I like a pen more than a cursor. Cursor obscures some area behind it, making content harder to read. The location of the pen is not a problem because usually you expect it to be somewhere near the current writing and its size is sufficient to distinguish it. The trouble may be with rapid movements when you lose it out of sight for a couple of seconds. Visual feedback may be in help there. Like circles when you press Ctrl in Windows. Or some noticeable transparent border around the pen. UserIdTAG: 208296 UserNameTAG: OZ1 CreateTimeTAG: 2012-11-24T19:13:50Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: The best cursor to use is the pen highlighted with a yellow circle ([Highlighted yellow cursor][1]). The yellow circle is transparent so it doesn't' cover up the material behind it. This way the professor doesn't have to switch between cursors. [1]: http://www.youtube.com/watch?v=Nv2lV-geBbs FirstChildUserIdTAG: 132828 FirstChildUserNameTAG: rharris FirstChildCreateTimeTAG: 2012-12-04T16:50:18Z SecondChildTAG: agree SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-12-04T19:25:04Z IndexTAG: 3115 TitleTAG: Lab10 first part the break frequency for an RC-lowpass is 1/(2*pi*R*C). Why I get a red cross?! UserIdTAG: 389333 UserNameTAG: juergen CreateTimeTAG: 2012-11-24T18:45:25Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: because question is: Break frequency in **radians/sec** for RC circuit (formula using R and C): FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-11-24T18:48:24Z SecondChildTAG: 2*pi equates radiants, or not? SecondChildUserIdTAG: 389333 SecondChildUserNameTAG: juergen SecondChildCreateTimeTAG: 2012-11-24T19:12:34Z SecondChildTAG: okay, my fault. because this declaration is unusual SecondChildUserIdTAG: 389333 SecondChildUserNameTAG: juergen SecondChildCreateTimeTAG: 2012-11-24T19:16:21Z FirstChildTAG: You should verify which equations you are using. You should use either > tan^-1( argument ) = +- 45° -> argument = +- 1 or > |H(jw)| = 1/sqrt(2) FirstChildUserIdTAG: 147694 FirstChildUserNameTAG: FilhoJoel FirstChildCreateTimeTAG: 2012-11-25T03:19:41Z IndexTAG: 3116 TitleTAG: force the acceleration will be F/m . ther's one equation F=ma UserIdTAG: 821000 UserNameTAG: chandan88 CreateTimeTAG: 2012-11-24T18:12:07Z VoteTAG: 0 CoursewareTAG: Week 1 / Lumped Element Abstraction CommentableIdTAG: 6002x_lumped_element_abstraction NumberOfReplyTAG: 0 IndexTAG: 3117 TitleTAG: H10P2 I can't get the expressions for circuit B and C correct. NO matter how hard I try. I'm pretty sure that I've used the right formula? Isn't it resistors parallel with cap and inductance in B and in C ,,resistor in series with cap parallel with resistor in series with an inductance? I tried this and I get a long expression which is wrong according to edX. any hint please? UserIdTAG: 8897 UserNameTAG: sam1202 CreateTimeTAG: 2012-11-24T17:35:16Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: These questions are pretty simple. You may try to solve them in the some ways. For example, to use admittance or impendance. With admittance:Zb=1/Yb; Yb=Yr+Yc+Yl, where Yx is admittance of entire branch. and etc FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-11-24T18:10:54Z FirstChildTAG: Look at this [post][1] or [here][2] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50af648e8ef9df230000000f [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50af6aaa4684922b0000000c FirstChildUserIdTAG: 96873 FirstChildUserNameTAG: Santyaga FirstChildCreateTimeTAG: 2012-11-24T17:55:31Z FirstChildTAG: Hi, I tried with the admittance method. It simplified the algebra a lot, but still having the wrong answer. The value calculated by my for Zb = (R1+R2) - j (w*L) / (w^2*L*C+1) Can some one help us a little bit? FirstChildUserIdTAG: 87808 FirstChildUserNameTAG: pumoneon FirstChildCreateTimeTAG: 2012-11-24T23:56:16Z IndexTAG: 3118 TitleTAG: doubt about S19E4 I don't understand this problem. We can say that: abs(z1/z2) = abs(z1)/abs(z2) In our problem the magnitude of the numerator is w*R*C, an the magnitude of the denominator is sqrt(1+(w*R*C)^2). Therefore, if w*R*C is big, the result is alwais 1. Why I am wrong? UserIdTAG: 252135 UserNameTAG: dgv CreateTimeTAG: 2012-11-24T17:32:59Z VoteTAG: 0 CoursewareTAG: Week 10 / Magnitudes And Angles CommentableIdTAG: 6002x_Magnitudes_and_angles NumberOfReplyTAG: 1 FirstChildTAG: What happens if wRC is small? FirstChildUserIdTAG: 261378 FirstChildUserNameTAG: es2377 FirstChildCreateTimeTAG: 2012-11-24T21:09:43Z IndexTAG: 3119 TitleTAG: Submissions I am unable to submit my answers for home work 10 and following it. Please help. Is is because of the maintenance work?? But i don't face any such problems with the lab. Please help. UserIdTAG: 332360 UserNameTAG: srinivasav CreateTimeTAG: 2012-11-24T15:08:19Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: try to use different browser. For example, I do have problems with EDX with Chrome on XP, no problem with Firefox there. On Win7 and Chrome- no any problem FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-11-24T17:09:25Z IndexTAG: 3120 TitleTAG: H10P3 Q1 I am stuck in part 1 while i think it's right.... i entered Vo/Vf * Vf/Vi.... but not getting the right answer..... UserIdTAG: 51259 UserNameTAG: Wardah CreateTimeTAG: 2012-11-24T12:15:36Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 7 FirstChildTAG: Isn't it R2/(R2+ZL)*Zc/(R1+Zc)??? FirstChildUserIdTAG: 51259 FirstChildUserNameTAG: Wardah FirstChildCreateTimeTAG: 2012-11-24T12:42:59Z FirstChildTAG: same problem, I think Z1/(R+Z1) with Z1=R2+L||C is the right answer, but I'm getting a red cross. Maybe I don't understand the question correct. I'm helpless FirstChildUserIdTAG: 389333 FirstChildUserNameTAG: juergen FirstChildCreateTimeTAG: 2012-11-24T12:24:56Z FirstChildTAG: Finally done...... Thanx matiasgrodriguez........ i got the right answer...... was missing a resistor in the answer.... FirstChildUserIdTAG: 51259 FirstChildUserNameTAG: Wardah FirstChildCreateTimeTAG: 2012-11-24T13:01:50Z SecondChildTAG: You're welcome! SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-24T14:09:37Z FirstChildTAG: IN HI0 P3 Q1 MY ANSWER OF V0/VI CAME (R2*((j*w*L)+R2+R1+1/(j*w*C)))/((j*w*C)*((j*w*L)+R2+1/(j*w*C))*((j*w*L)+R2)*(R1+1/(j*w*C))).I WAS PRETTY SURE THAT IT WAS CORRECT I DID IT MANY TIMES BUT STILL A RED MARK:(..PLESE ANYBODY COULD HELP ME IN DETERMINING MY MISTAKE....ADVANCE THANKS:) FirstChildUserIdTAG: 276808 FirstChildUserNameTAG: DEBASMITAMAJUMDER FirstChildCreateTimeTAG: 2012-11-24T13:03:56Z SecondChildTAG: Hi DEBASMITAMAJUMDER. If you are trying to solve as in the exercise from page 724 of the text book, then there must be an algebra mistake. Good luck. SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-24T14:12:48Z FirstChildTAG: Hi Wardah. two voltage dividers... take a look at exercise from page 724 of the text book. FirstChildUserIdTAG: 288174 FirstChildUserNameTAG: matiasgrodriguez FirstChildCreateTimeTAG: 2012-11-24T12:26:49Z SecondChildTAG: Thanks for the hint. Not one voltage divider, there are a double divider. SecondChildUserIdTAG: 389333 SecondChildUserNameTAG: juergen SecondChildCreateTimeTAG: 2012-11-24T15:56:29Z FirstChildTAG: Is there a problem with the official answer for the first question for the H10P3? The official solutions is: $\frac{V_o}{V_i} = \frac{\frac{R_2}{j*\omega*C}}{R_1 * (\frac{1}{j*\omega*C} + j*\omega*L + R_2) + \frac{1}{j*\omega*C} * (j*\omega*L + R_2)}$ But I get: $\frac{V_o}{V_i} = \frac{\frac{(j*\omega*L + R_2)}{j*\omega*C}}{R_1 * (\frac{1}{j*\omega*C} + j*\omega*L + R_2) + \frac{1}{j*\omega*C} * (j*\omega*L + R_2)}$ Is there a problem with the official answer or do I have a mistake somewhere? FirstChildUserIdTAG: 294766 FirstChildUserNameTAG: dejanst FirstChildCreateTimeTAG: 2012-12-02T22:15:25Z FirstChildTAG: Apparently I made a mistake :) My result was the result of the first voltage divider: $V_{intermediate} = Vi * \frac{Z_C || (Z_L + Z_{R2})}{Z_{R1} + (Z_C ||(Z_L + Z_{R2}))}$ For the correct solution you need to solve the second voltage divider: $V_o = V_{intermediate} * \frac{Z_{R2}}{Z_{R2} + Z_L}$ After a log of fiddling arround with equation, you should get the solution that is presented in the official answer. Happy calculus ;) FirstChildUserIdTAG: 294766 FirstChildUserNameTAG: dejanst FirstChildCreateTimeTAG: 2012-12-03T21:58:24Z IndexTAG: 3121 TitleTAG: H10 P2 Part a Hi everyone! I' ve got this problem on getting the rigth answers on the H10 P2 PartA for w=0 and infinity.I wrote the right Za and I' m sure my answers after that are all right, I mean, it's an easy one. but when I check my answers out, they say I'm wrong. So, is any of you got the same problem as me or am I actually wrong? For W=0, I have the Zc and for W=infinity, ZL... UserIdTAG: 350842 UserNameTAG: hasina CreateTimeTAG: 2012-11-24T08:48:55Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: can you give me the link of online player.? FirstChildUserIdTAG: 136490 FirstChildUserNameTAG: Ali_PU FirstChildCreateTimeTAG: 2012-11-24T09:24:45Z SecondChildTAG: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_10/Homework_1478/ SecondChildUserIdTAG: 350842 SecondChildUserNameTAG: hasina SecondChildCreateTimeTAG: 2012-11-24T09:36:28Z SecondChildTAG: Hope that what's you meant ^^ SecondChildUserIdTAG: 350842 SecondChildUserNameTAG: hasina SecondChildCreateTimeTAG: 2012-11-24T09:37:41Z FirstChildTAG: Hi Ali_PU. When w->0, Zc prevails but just that it is not the answer.... because w is part of Zc: 1/(j*w*C), so what gives you a 1/(an extremely small number)? The same for w->infinity... the Zl part prevails but you have j*w*L ... and with w reaching an extremely big number you will get... Hope this helps. Regards. FirstChildUserIdTAG: 288174 FirstChildUserNameTAG: matiasgrodriguez FirstChildCreateTimeTAG: 2012-11-24T12:43:24Z SecondChildTAG: Hi, I understand that, but, how do I actually write it? SecondChildUserIdTAG: 350842 SecondChildUserNameTAG: hasina SecondChildCreateTimeTAG: 2012-11-24T16:53:47Z SecondChildTAG: Hi, I didn t know I had to write infinity= inf. Thanks SecondChildUserIdTAG: 350842 SecondChildUserNameTAG: hasina SecondChildCreateTimeTAG: 2012-11-24T17:22:29Z SecondChildTAG: I'm glad you did it. SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-24T18:27:32Z IndexTAG: 3122 TitleTAG: H10 P1 IN HI0 P1 MY ANSWER OF V0/VI CAME (R2*((j*w*L)+R2+R1+1/(j*w*C)))/((j*w*C)*((j*w*L)+R2+1/(j*w*C))*((j*w*L)+R2)*(R1+1/(j*w*C))).I WAS PRETTY SURE THAT IT WAS CORRECT I DID IT MANY TIMES BUT STILL A RED MARK:(..PLESE ANYBODY COULD HELP ME IN DETERMINING MY MISTAKE....ADVANCE THANKS:) UserIdTAG: 276808 UserNameTAG: DEBASMITAMAJUMDER CreateTimeTAG: 2012-11-24T06:09:41Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: can you give me the link of online player of videos.? FirstChildUserIdTAG: 136490 FirstChildUserNameTAG: Ali_PU FirstChildCreateTimeTAG: 2012-11-24T06:14:32Z IndexTAG: 3123 TitleTAG: link for online player of video lectures please admins mentioned this link on some suitable place where it is easy to find.. give me the link of online player to watch video lectures UserIdTAG: 136490 UserNameTAG: Ali_PU CreateTimeTAG: 2012-11-24T06:00:29Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/ FirstChildUserIdTAG: 136959 FirstChildUserNameTAG: christerpher FirstChildCreateTimeTAG: 2012-11-25T11:02:33Z SecondChildTAG: i want online player link... which inculde dropbox .. on which videos are stored and played online.. thanks SecondChildUserIdTAG: 136490 SecondChildUserNameTAG: Ali_PU SecondChildCreateTimeTAG: 2012-11-27T12:10:02Z FirstChildTAG: It seems that video download links for the recent sequences (week 7 onwards) are not being posted to wiki. You could go to any of the sequence pages with a video and click on the link shown in the "download video here". Hope this helps. FirstChildUserIdTAG: 359310 FirstChildUserNameTAG: AaronYeoh FirstChildCreateTimeTAG: 2012-11-26T06:08:49Z IndexTAG: 3124 TitleTAG: NOW instead of NOT This should NOW be one of the most fun parts of the course! UserIdTAG: 339870 UserNameTAG: rjlasota CreateTimeTAG: 2012-11-24T05:30:47Z VoteTAG: 0 CoursewareTAG: Week 10 / Is there a simpler way to get Vp CommentableIdTAG: 6002x_Is_there_a_simpler_way_to_get_Vp NumberOfReplyTAG: 0 IndexTAG: 3125 TitleTAG: CHECKER DID NOT WORKING FOR THE WHOLE DAY CHECK BUTTON HAS NOT BEEN WORKING FOR THE WHOLE DAY. UserIdTAG: 216684 UserNameTAG: Taimoor1017 CreateTimeTAG: 2012-11-23T18:06:34Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3126 TitleTAG: H10P2 .. i have found **Z** B...but not getting green tick. UserIdTAG: 114864 UserNameTAG: Vikaash CreateTimeTAG: 2012-11-23T17:09:48Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: me too , don't understand why? FirstChildUserIdTAG: 227508 FirstChildUserNameTAG: bhavyab FirstChildCreateTimeTAG: 2012-11-24T05:29:30Z SecondChildTAG: I also don't get green on 1st part h10p2 SecondChildUserIdTAG: 227508 SecondChildUserNameTAG: bhavyab SecondChildCreateTimeTAG: 2012-11-24T05:56:45Z SecondChildTAG: me too SecondChildUserIdTAG: 350842 SecondChildUserNameTAG: hasina SecondChildCreateTimeTAG: 2012-11-24T08:37:19Z SecondChildTAG: I think the reader has an issue. I know I have the right answer and got the next two questions correct but cant get a green tick on the impedence expression. SecondChildUserIdTAG: 287613 SecondChildUserNameTAG: gbaral SecondChildCreateTimeTAG: 2012-11-25T18:42:53Z IndexTAG: 3127 TitleTAG: H10P31 i am getting a red cross but i believe that i have adopted the right method for h10p3 part 1 UserIdTAG: 153584 UserNameTAG: SalmanShahzad CreateTimeTAG: 2012-11-23T15:54:16Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: What method you tried? It will be easier to help if spot the problem if you state your method. FirstChildUserIdTAG: 318577 FirstChildUserNameTAG: takeuchi FirstChildCreateTimeTAG: 2012-11-23T16:15:20Z FirstChildTAG: same is the case with me......... i entered Vo/Vf * Vf/Vi.......... isn't it right????? FirstChildUserIdTAG: 51259 FirstChildUserNameTAG: Wardah FirstChildCreateTimeTAG: 2012-11-24T12:18:44Z FirstChildTAG: another method i used is thevinine equivalent circuit but it's not working FirstChildUserIdTAG: 51259 FirstChildUserNameTAG: Wardah FirstChildCreateTimeTAG: 2012-11-24T12:26:41Z SecondChildTAG: thevinin equivalent works . jus remove R2 n find out thevinin equivalent n finally place R2 in thevinin equinalent circuit SecondChildUserIdTAG: 337430 SecondChildUserNameTAG: aparna_b SecondChildCreateTimeTAG: 2012-11-24T14:40:48Z IndexTAG: 3128 TitleTAG: small and large "w" I cannot find expressions for small and large "w" Below i gave a correct expression for magnitude of asked ratio. (sqrt((wRC)^4+(wRC-w^3RLC^2)^2))/(1-2w^2LC+w^4L^2C^2+(wRC)^2) UserIdTAG: 341358 UserNameTAG: akbar13 CreateTimeTAG: 2012-11-23T15:20:02Z VoteTAG: 0 CoursewareTAG: Week 10 / LCR Voltage Divider Frequency Limits CommentableIdTAG: 6002x_LCR_voltage_divider_frequency_limits NumberOfReplyTAG: 1 FirstChildTAG: To solve the problem, you must look at the scale. For example, let't analize for small w: The lower part is 1+jωRC−ω^2LC Since w is very small, 1 is large compared to anything that has w multiplying. Therefore, the lower part is very close to 1, so you can consider it 1. The upper part is what remains, just take is magnitude and voila. That is, you have to think what is considered insignificant compared to the other terms involved in the problem. Another example, if you have: (Aw^2+w)/A and w is very large, then w is insignificant compared to w^2. So, the expression is close to Aw^2/A, and simplifying, you get w^2. So, you don't get accurate values, you just get an expression that will give you an approximation of the real equation. FirstChildUserIdTAG: 318577 FirstChildUserNameTAG: takeuchi FirstChildCreateTimeTAG: 2012-11-23T16:10:19Z IndexTAG: 3129 TitleTAG: S22V3: max|Vc/Vi| is not = Q in the series RLC-circuit?!?! Vc/Vi = w02 / (-w2 + j2aw + w02) with 2a=R/L and w02=1/(LC) as shown in S22V3. so: |Vc/Vi| = w02 / sqrt((w02-w2)2 + (2aw)2), ok?! now: derivation of |Vc/Vi| with respect of w and setting = 0 yields: w^2 = w02 - 2a2 so: w^2 = w02 - 2a2 is a (local) extremum if sqrt(2)*a < w0 (+-underdamped case). But, I expected w2 = w02 ?!??!! Furthermore, at this point: |Vc/Vi| = Q * w0/sqrt(w02-a2) > Q !!!! What did I make wrong? (At the parallel RLC the same calculation gives the expected results). UserIdTAG: 388524 UserNameTAG: dasbinich CreateTimeTAG: 2012-11-23T12:42:37Z VoteTAG: 0 CoursewareTAG: Week 11 / S22V3: Q indicates peakiness CommentableIdTAG: 6002x_S22V3_Q_indicates_peakiness NumberOfReplyTAG: 0 IndexTAG: 3130 TitleTAG: Op Amp Power Hi, I was designing a non-inverting op Amp circuit, but all I have to power the entire circuit is one 9V battery. I needed to increase my Vin by a factor of 2 so I used the Non-inverting amplifier model with two $1M\Omega$ resistors to prevent power wastage. But the Op amp seems to need to be powered by both a positive Voltage and a negative Voltage. I would be grateful if anybody could explain to me how I could go about powering the Op amp with my 9V cell if it is possible. Thank you in advance UserIdTAG: 85490 UserNameTAG: yashsavani CreateTimeTAG: 2012-11-23T09:33:50Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: can't you ground the negative and connect 9V to the positive? FirstChildUserIdTAG: 357747 FirstChildUserNameTAG: kishores FirstChildCreateTimeTAG: 2012-11-23T10:30:24Z SecondChildTAG: I thought that prof. Anant said that there has to be a positive supply connected to the positive power rail and a negative power supply connected to the negative power rail. The ground would be in between them. By connecting 9V to the positive power rail and ground to the negative power rail wouldn't your ground be at 4.5V? SecondChildUserIdTAG: 85490 SecondChildUserNameTAG: yashsavani SecondChildCreateTimeTAG: 2012-11-23T10:53:58Z SecondChildTAG: Well, it all depends on the op-amp in question and the range of input and output voltages required. You can get "single supply" op-amps where the negative supply to the op-amp can be ground, but obviously, you can't get a negative output from it and in fact, may not be able to get a 0V output from it. You have to read the datasheets. Some op-amps will get upset if you put an input into them less than their negative supply or above their positive supply (see common mode input voltage range on the datasheet). Few will manage to produce an output as low as their negative or as high as their positive supplies (look for "rail-to-rail" in the specifications for ones that do). So in this case, it's fine to define ground as 4.5V using a circuit like that in the link that Pennypacker posted for AC signals. You just have to remember that neither power rail would be at ground potential. It's not going to work if the circuit is producing a DC output relative to the 4.5V ground if any current is required since the current would have to go through one of the resistors used as a voltage divider to provide the 4.5V ground and the 4.5V ground wouldn't be 4.5V any more. There are other ways of deriving a 4.5V ground which would avoid this problem, but that's getting a little too complicated to get into here. [Virtual Ground Circuits][1] is a good resource for solving this problem. [1]: http://tangentsoft.net/elec/vgrounds.html SecondChildUserIdTAG: 141000 SecondChildUserNameTAG: OrinE SecondChildCreateTimeTAG: 2012-11-24T01:50:52Z FirstChildTAG: Have a look at this power supply from the Cmoy headphone amplifier. You will notice that it uses two 9v batteries to achieve +/- 9v, in your case just use one 9v battery and you will have +/- 4.5v. It's a neat little trick that does have limitations, but it is handy from time to time. Also known as a "virtual ground" or a "bipolar power supply". Have fun. http://tangentsoft.net/audio/cmoy-tutorial/misc/cmoy-tangent-sch.pdf FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-11-23T11:27:57Z SecondChildTAG: Thank you very much. I think I understood the concept. SecondChildUserIdTAG: 85490 SecondChildUserNameTAG: yashsavani SecondChildCreateTimeTAG: 2012-11-23T11:32:15Z SecondChildTAG: I just had one doubt, Why do they use the two 220uF capacitances? I seem to see capacitances being used in almost every circuit, however don't really understand their purpose. SecondChildUserIdTAG: 85490 SecondChildUserNameTAG: yashsavani SecondChildCreateTimeTAG: 2012-11-23T11:35:05Z SecondChildTAG: The 9V batteries have a high internal resistance, therefor have a hard time delivering large amounts of current when needed. So the capacitors are used in conjunction with the battery to deliver instant bursts of power when needed. (This improves the bass response and transient sounds as they require large and fast amounts of current.) That is one function of the capacitors. Another purpose is in power supplies using an AC source such as a wall outlet The capacitors act as a filter to "smooth" out and ripples that are left behind after the rectification stage. (Rectification is converting AC into DC) They are also used as fast-acting power reserves just as in the little cMoy amp. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-11-23T12:25:48Z SecondChildTAG: Correction: '"smooth" out ***any***ripples that are...' SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-11-23T12:28:00Z FirstChildTAG: You could use the op-amp to build a non-inverting amplifier. If you're not sure how, I think Google will help you find examples. I have done this several times, when I wanted to use a single supply to power my op-amp for simplicity. Be careful though that most op-amps will have a minimum of about 0.7V on the output when used this way. If you need the output to go to ground, you would need a rail to rail op-amp, which are generally more expensive than standard types. FirstChildUserIdTAG: 270011 FirstChildUserNameTAG: SnowmanZA FirstChildCreateTimeTAG: 2012-11-24T21:35:34Z IndexTAG: 3131 TitleTAG: S21E3 Part1 (R+j*w*L)/(R-L*C*w^2+j*w*L) Please explain　R mystery disappears!! UserIdTAG: 89440 UserNameTAG: yuk CreateTimeTAG: 2012-11-23T07:55:26Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3132 TitleTAG: Lab 11 AC analysis Hi, I get green check mark for L and C values but when I run the AC analysis. With the right values. I dont get the correct graphs. My values gives w0=1/sqrt(L*C)=62831 so f0=10000Hz as requested. The graphs shows 5.5=log(316227Hz) Thanks UserIdTAG: 392699 UserNameTAG: lu2adw CreateTimeTAG: 2012-11-22T20:55:10Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: My result for w0 is about 62893, so f0=10014.8 Maybe it's a matter of accuracy. FirstChildUserIdTAG: 304587 FirstChildUserNameTAG: Vasso FirstChildCreateTimeTAG: 2012-11-22T23:07:53Z IndexTAG: 3133 TitleTAG: staff...... I think there is a mistake . mr.Anant Agarwal give us a sinusoidal wave for the input (Vi cos wt) and he draw it for us. I think it is suppose to be (Vi sin wt). UserIdTAG: 226085 UserNameTAG: Teto CreateTimeTAG: 2012-11-22T20:30:21Z VoteTAG: 0 CoursewareTAG: Week 10 / Approach To Solving Sinusoidal Problems CommentableIdTAG: 6002x_Approach_to_solving_sinusoidal_problems NumberOfReplyTAG: 1 FirstChildTAG: doesn't make a difference.notice there are no cosinusoids(coined a new word :p).you can always treat the cosine input as a sinusoid lagging or leading by a phase of pi/2 radians. FirstChildUserIdTAG: 357747 FirstChildUserNameTAG: kishores FirstChildCreateTimeTAG: 2012-11-23T04:02:35Z SecondChildTAG: aha O.K. thanks SecondChildUserIdTAG: 226085 SecondChildUserNameTAG: Teto SecondChildCreateTimeTAG: 2012-11-24T01:42:50Z IndexTAG: 3134 TitleTAG: Lab 12 AC Response Anybody else having problems getting the plot to look right? (got check marks for all values but plot shows HPF response instead of BPF) UserIdTAG: 397595 UserNameTAG: mholin CreateTimeTAG: 2012-11-22T16:45:41Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: AC Response doesn't work for me either. FirstChildUserIdTAG: 4463 FirstChildUserNameTAG: pmj FirstChildCreateTimeTAG: 2012-12-03T18:18:15Z IndexTAG: 3135 TitleTAG: H12P3 I need some help in the very last question. I cant figure out mid band gain. To start with what is mid band gain? Can anybody please give me a clue? UserIdTAG: 162671 UserNameTAG: tuhin1991paul CreateTimeTAG: 2012-11-22T11:48:47Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: You need to calculate Vo/Vi :) FirstChildUserIdTAG: 304587 FirstChildUserNameTAG: Vasso FirstChildCreateTimeTAG: 2012-11-22T12:25:51Z SecondChildTAG: thank you....:) i calculated Vo/Vi for midband but the answer is still wrong SecondChildUserIdTAG: 162671 SecondChildUserNameTAG: tuhin1991paul SecondChildCreateTimeTAG: 2012-11-22T14:23:35Z SecondChildTAG: Are you sure you were careful with signs? SecondChildUserIdTAG: 304587 SecondChildUserNameTAG: Vasso SecondChildCreateTimeTAG: 2012-11-22T22:27:57Z FirstChildTAG: For th midband gain , It actually is the H(w) =Vo/Vi . It's difficult to calcutale by compose algebra equations . The most efficient way is to draw the circuit in Sandbox and run a AC analysis , you can find the Magnitude chart . Base on this equation: DB=20*lg H(w) . You will know Db through Magnitude chart , then you can find the H(w) FirstChildUserIdTAG: 232667 FirstChildUserNameTAG: KyleLiux FirstChildCreateTimeTAG: 2012-12-06T06:55:31Z IndexTAG: 3136 TitleTAG: LAB10 I calculated the unknown values for L and C and same checked in the circuit sandbox.It shows the -3dB point exactly at the breakfrequency given.But when it is entered it is showing the red cross...... Any idea will be highly appreciated UserIdTAG: 156835 UserNameTAG: kphariprakash1968 CreateTimeTAG: 2012-11-22T10:15:40Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Are the flank slopes right too? Have you tried to round-down the values? Drop any decimal value. FirstChildUserIdTAG: 372321 FirstChildUserNameTAG: EnricoDona FirstChildCreateTimeTAG: 2012-11-22T11:01:00Z IndexTAG: 3137 TitleTAG: H10P3 Q1 b/(b+j*w*(a*b*x+y)-w^2*a*x*y) I cant understand the mistake in this equation.Can anybody help? Thanks UserIdTAG: 156835 UserNameTAG: kphariprakash1968 CreateTimeTAG: 2012-11-22T07:19:04Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Make a consistency check Vo/Vi must be a pure number. To solve the problem just substitute L and C with their impedances (i.e., jwL and 1/jwC) and solve the resulting resistive ladder. - L and R2 are connected in series - (L+R2) is connected in parallel with C -(L+R2)||C is connected in series with R1 FirstChildUserIdTAG: 372321 FirstChildUserNameTAG: EnricoDona FirstChildCreateTimeTAG: 2012-11-22T07:49:53Z IndexTAG: 3138 TitleTAG: 5 missing result : 50sqrt(2)*exp(j*95) a 5 is missing in the video UserIdTAG: 373752 UserNameTAG: Croak CreateTimeTAG: 2012-11-22T04:42:43Z VoteTAG: 0 CoursewareTAG: Week 10 / Complex Multiplication CommentableIdTAG: 6002x_Complex_Multiplication NumberOfReplyTAG: 0 IndexTAG: 3139 TitleTAG: About the Lab 10 C &L value w=R/L, L=R/w=R/(2*Pi*f), f=4.6Hz. Is it right? But the answer is not right? Why? Can a friend give me hints? Thanks! UserIdTAG: 371617 UserNameTAG: rainbow12 CreateTimeTAG: 2012-11-22T02:38:59Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: The frequency is 4.6 on a log scale which means it's 10^4.6Hz. Then use the phase plot and the angle formulas to find C and L. Remember the two reciprocal functions tan-1 and tan. I basically gave you the answer :-) FirstChildUserIdTAG: 252722 FirstChildUserNameTAG: vaboro FirstChildCreateTimeTAG: 2012-11-22T05:06:58Z IndexTAG: 3140 TitleTAG: Decibels Hi, I would be much obliged if someone could explain to me the concept of decibels, in the electronic context, or guide me to a page where it is properly explained. Thank you in advance UserIdTAG: 85490 UserNameTAG: yashsavani CreateTimeTAG: 2012-11-21T20:03:51Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Did you try Google ? The first hit for decibel is [this Wikipedia article][1]. [1]: http://en.wikipedia.org/wiki/Decibel FirstChildUserIdTAG: 90796 FirstChildUserNameTAG: ChaunceyGardiner FirstChildCreateTimeTAG: 2012-11-22T00:17:59Z IndexTAG: 3141 TitleTAG: How to make a band pass filter Hi, In Lab 10 we were asked to make a band pass filter. I was able to successfully do so, however while completing the task I was confronted by a problem; the value of the resistor. I began with an arbitrary value of $100\Omega$ which worked fine for the RL sub-circuit but for the RC sub-circuit there seemed to be an issue with the decay rate. I would be much obliged if someone could explain to me why it was necessary to change the resistance of the RC sub-circuit to $10k\Omega$. Thank you in advance. UserIdTAG: 85490 UserNameTAG: yashsavani CreateTimeTAG: 2012-11-21T20:00:53Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Did you have the RL first? If you cascade an RL (first) with an RC (second), you have to use both resistors in series as the R for the RC section (and vice-versa). Setting the R for the RC circuit to $10k\Omega$ would make the $100\Omega$ contribution from the RL circuit insignificant. I used equal resistors and used twice the value when calculating the second time constant/component value. FirstChildUserIdTAG: 141000 FirstChildUserNameTAG: OrinE FirstChildCreateTimeTAG: 2012-11-21T21:10:19Z IndexTAG: 3142 TitleTAG: Can anyone please help me on H10P2 I can't seem to get the right equation when w->0 and w-) infinity please help me thanks :) UserIdTAG: 367072 UserNameTAG: jansenlopez CreateTimeTAG: 2012-11-21T09:02:01Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi JansenLopez! Remember that: Capacitor will acts as SHORT circuit for w -> INFINITY & acts as OPEN circuit for w -> 0(ZERO) Inductor will acts as OPEN circuit for w -> INFINITY & acts as SHORT circuit for w -> 0(ZERO) And any element parallel to SHORT circuit can be neglected. FirstChildUserIdTAG: 277808 FirstChildUserNameTAG: Hemanthmps FirstChildCreateTimeTAG: 2012-11-21T12:59:48Z IndexTAG: 3143 TitleTAG: H10P3-Q1 Dear All, I calculated the equation for Vo/Vi,but getting only red cross.I entered the equation in different formats ; but no scope. My equation is in the following format. b/(b+jwaby+jwx-w^2axy) Expecting reply UserIdTAG: 156835 UserNameTAG: kphariprakash1968 CreateTimeTAG: 2012-11-20T15:08:35Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Dear kphariprakash1968, I suspect you've forgotten multiplication signs in between variables. Try: b/(b+j*w*a*b*y+j*w*..........) --Rohan FirstChildUserIdTAG: 391929 FirstChildUserNameTAG: RohanNagarkar FirstChildCreateTimeTAG: 2012-11-20T18:43:16Z SecondChildTAG: Mr.Rohan, I put the sign....I am asking the given eqn is correct or not...... SecondChildUserIdTAG: 156835 SecondChildUserNameTAG: kphariprakash1968 SecondChildCreateTimeTAG: 2012-11-20T19:37:55Z SecondChildTAG: Unfortunately I'm not allowed to help you on this as it is a homework problem due in 5 days time. I suggest first reviewing your current answer to find if it's a mistake in the the way you've entered it into the formula box. If that proves to be fruitless, then you'd likely be best off trying it again later. SecondChildUserIdTAG: 391929 SecondChildUserNameTAG: RohanNagarkar SecondChildCreateTimeTAG: 2012-11-20T20:33:25Z FirstChildTAG: Dear kphariprakash1968, Try to read https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/ on page 724 it is a rather similar situation, redo you calculations ;). And as stated here https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50aaa6368eb5a7290000007a do not do simplify it toooooo much with doing multiplications with j and w it took me a bit to figure it out also :P. Good luck :) FirstChildUserIdTAG: 157714 FirstChildUserNameTAG: Arkas FirstChildCreateTimeTAG: 2012-11-22T21:39:40Z IndexTAG: 3144 TitleTAG: Characteristic equation in Impedance analysis Hi, I was just wondering why after analyzing the circuit using impedances do we get the characteristic equation in the denominator. For example in the series RLC circuit: $V_r=\frac{V_i \cdot R \cdot \frac{s}{L}}{s^2+\frac{R \cdot s}{L}+\frac{1}{LC}}$ The denominator $s^2+\frac{R \cdot s}{L}+\frac{1}{LC}$ Is the characteristic equation for the circuit. Why is this so? I would be much obliged if someone could help clear my doubt. Thank you in advance UserIdTAG: 85490 UserNameTAG: yashsavani CreateTimeTAG: 2012-11-20T13:57:59Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: From what I know, that's the definition. The characteristic equation is defined the polynomial in the denominator. The reason why it's important is because it gives you a lot of information about a system. You can get a feel of the system's speed and stability. The location of poles (the roots of the characteristic equation) allow us to study these aspects. This is usually dealt with in a course on Control Systems (can't wait for edX to offer that and Signals & Systems). I was studying stability just now (will be moving on to steady state errors in a bit) so if you want any more info I'll be happy to help. It's still fresh in my mind :) The numerator is important in system design as well. The roots of the numerator are called the zeros of the system. FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-11-20T18:45:51Z IndexTAG: 3145 TitleTAG: Lab 10 C&L values I am unable to find the values of C and L I used the formulae ∠HRL(jω)=tan−1(R/Lω), plugged in the corresponding values from the graph but the resultant is showing up incorrect(red tick). I also plugged in w=2*pi*f what did I do wrong? Pls help... UserIdTAG: 425976 UserNameTAG: P_V_Rohith CreateTimeTAG: 2012-11-20T12:12:21Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: See if this helps https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/undefined/threads/50a64a17615d3b2a00000018 FirstChildUserIdTAG: 288174 FirstChildUserNameTAG: matiasgrodriguez FirstChildCreateTimeTAG: 2012-11-20T16:32:04Z IndexTAG: 3146 TitleTAG: What is Vp ? 1st) the real part of complex Vps 2nd) the "complex amplitude" of Vps Is that the same ...? I'm a little confused.. UserIdTAG: 373752 UserNameTAG: Croak CreateTimeTAG: 2012-11-20T05:27:10Z VoteTAG: 0 CoursewareTAG: Week 10 / Particular Solution To Cosine Input Part 2 CommentableIdTAG: 6002x_Particular_solution_to_cosine_input_part_2 NumberOfReplyTAG: 1 FirstChildTAG: Complex amplitude of Vps looks like "a+jb". So, the real part of Vps is "a", imagine part is "jb". IMHO. FirstChildUserIdTAG: 323416 FirstChildUserNameTAG: Alexxkr FirstChildCreateTimeTAG: 2012-11-20T05:36:02Z SecondChildTAG: Vp=Re(Vp*e(jwt)) That is strange ... ---------- SecondChildUserIdTAG: 373752 SecondChildUserNameTAG: Croak SecondChildCreateTimeTAG: 2012-11-20T17:07:51Z SecondChildTAG: OK i got it There is Vp and vp ( small v ) Vp is complex ( amplitude ) vp is real SecondChildUserIdTAG: 373752 SecondChildUserNameTAG: Croak SecondChildCreateTimeTAG: 2012-11-20T17:29:07Z IndexTAG: 3147 TitleTAG: LAB 10 Issue I can not get values of L and C based on the information in the plots. For example, according my plot for RL (red), breaking point is 5200 Hz. w=R/L (correct); w=2*pi*f so, L=R/(2*pi*f) Inserting values: R=1000 ohms, f=5200 Hz Calculating: L=0.0306067198254 henries. Answer is incorrect. I tried to model the circuit using AC analisys. I got the same plot with same breaking point in parameters R=1000 ohms and L=9.5e-07 henries, but this answer is incorrect too. What is wrong? UserIdTAG: 323416 UserNameTAG: Alexxkr CreateTimeTAG: 2012-11-20T04:11:51Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Are you sure that the cursors are set precisely at breaking points? In my case they were at some diffrent frequency. FirstChildUserIdTAG: 272136 FirstChildUserNameTAG: przemgaw FirstChildCreateTimeTAG: 2012-11-20T19:25:24Z SecondChildTAG: 1. I tried different frequencies near breaking point in my model. What accuracy is required? 2. What is wrong in my calculations? my calculated aswer is more than modeled answer in 3000 times! SecondChildUserIdTAG: 323416 SecondChildUserNameTAG: Alexxkr SecondChildCreateTimeTAG: 2012-11-21T00:45:59Z SecondChildTAG: Hi Alex. Did you notice that the plot is in **log(Frequency in Hertz)**? Take a look at my comment on https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/undefined/threads/50a64a17615d3b2a00000018 I hope this helps SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-21T02:21:05Z SecondChildTAG: To matiasgrodriguez Thank you! I seemed that it was frequensy in log scale... SecondChildUserIdTAG: 323416 SecondChildUserNameTAG: Alexxkr SecondChildCreateTimeTAG: 2012-11-21T08:58:32Z SecondChildTAG: You're welcome! SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-21T11:34:59Z IndexTAG: 3148 TitleTAG: H10P3 I think it is easy, but I have problems with this task. I have voltage divider, so vo/vi I have an expression like this A/(B+A). antenna is include. Same problem like in Task 2?! UserIdTAG: 389333 UserNameTAG: juergen CreateTimeTAG: 2012-11-19T21:29:30Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I got it FirstChildUserIdTAG: 389333 FirstChildUserNameTAG: juergen FirstChildCreateTimeTAG: 2012-11-24T17:37:39Z IndexTAG: 3149 TitleTAG: S18V19 I think that v(0) starting value needs to be higher than Vi, so the current would flow in reverse through the inductor, and go from a higher potential to a lower one. I am missing something about this? Thanks. UserIdTAG: 331483 UserNameTAG: Doru CreateTimeTAG: 2012-11-18T19:56:19Z VoteTAG: 0 CoursewareTAG: Week 9 / Intuitive Analysis of Second Order Circuits Continued CommentableIdTAG: 6002x_intuitive_analysis_of_second_order_circuits_continued NumberOfReplyTAG: 0 IndexTAG: 3150 TitleTAG: Lab 11 some mistake > Consider the series RLC circuit shown in Figure 1. Leaving the > resistance fixed at 10Ω your task is choose values of L and C such > that the undamped resonance frequency, ω0, is 10kHz, with Q=10. Hey, this circuit is already have valid values! That's not fair! ) UserIdTAG: 394836 UserNameTAG: v2g6ch4 CreateTimeTAG: 2012-11-18T14:40:35Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: it seems to give a green check mark for a variety of values not just the correct values. FirstChildUserIdTAG: 18073 FirstChildUserNameTAG: gburkhart FirstChildCreateTimeTAG: 2012-11-18T23:03:30Z IndexTAG: 3151 TitleTAG: H10Q2P1 - possible fault in grader? Hello there. Im looking at the impedance, Z, for the first circuit. We have 2 resistors, one cap and one inductor in series. As they are in series, we simply sum the impedances to get Z. Impedance of a resistor is R, a cap is 1/wC and an inductor is Lw. The equation follows on from that but when I type that into the grader for the first part I get the red x of doom. Is this my fault or the grader's? Cheers! Ben UserIdTAG: 56024 UserNameTAG: onidaito CreateTimeTAG: 2012-11-18T13:23:50Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Impedance of a resistor is Zr=R, a cap Zc=1/jcw and an inductor is Zl=jlw.. FirstChildUserIdTAG: 16263 FirstChildUserNameTAG: Noureddine FirstChildCreateTimeTAG: 2012-11-18T15:18:02Z SecondChildTAG: Then that question is misleading as it says in terms of R1,R2,C,L and w; j is not mentioned as a variable (and indeed, being imaginary, isn't too useful :S) SecondChildUserIdTAG: 56024 SecondChildUserNameTAG: onidaito SecondChildCreateTimeTAG: 2012-11-18T15:51:36Z SecondChildTAG: j is a mathematical operator, not a variable and is required in some answers, even if they don't mention it. SecondChildUserIdTAG: 141000 SecondChildUserNameTAG: OrinE SecondChildCreateTimeTAG: 2012-11-18T18:21:18Z IndexTAG: 3152 TitleTAG: Help with phase of Vp/Vi I understand I am asking a trivial question for many, however I would appreciate some help:). To calculate the phase of Vp/Vi, I would take the angle(Vp)-angle(Vi). I know that the way to calculate the angle would be: arctan(Im/Re). So the angle(Vi)=arctan(wRC/1). However, if I try angle(Vp) which has real part=0, then I would have angle(Vp)=arctan(wRC/0), in other words I would have division by 0, which doesnt seem right. What am I doing wrong? How do I calculate the angle of a complex number which has real part=0. THanks a lot for any help. GT UserIdTAG: 350183 UserNameTAG: gt94539 CreateTimeTAG: 2012-11-18T12:03:35Z VoteTAG: 0 CoursewareTAG: Week 10 / Magnitudes And Angles CommentableIdTAG: 6002x_Magnitudes_and_angles NumberOfReplyTAG: 2 FirstChildTAG: Remember your trig. Tan(pi/2) is + inf and tan(-pi/2) is - inf. A complex impedance with zero real part is going to have a phase of pi/2 or -pi/2 depending on the sign of the imaginary part. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-18T14:09:44Z SecondChildTAG: Yes, I had forgotten some of that:), thank you for your help. regards, GT SecondChildUserIdTAG: 350183 SecondChildUserNameTAG: gt94539 SecondChildCreateTimeTAG: 2012-11-19T09:12:47Z FirstChildTAG: The answer can also be found there: http://en.wikipedia.org/wiki/Complex_number - search for keyword "phase". FirstChildUserIdTAG: 413065 FirstChildUserNameTAG: anikey FirstChildCreateTimeTAG: 2012-11-21T20:24:57Z IndexTAG: 3153 TitleTAG: damping frequency for overdamped??? is it possible to compute the damping frequency for overdamped case??? it turns out to be complex???? can it be complex?? [ by the equation: Wd=sqrt(Wo^2-alpha^2) ] UserIdTAG: 316353 UserNameTAG: Vivekanand123 CreateTimeTAG: 2012-11-18T09:43:51Z VoteTAG: 0 CoursewareTAG: Week 9 / Driven_Parallel_RLC_Circuit CommentableIdTAG: 6002x_driven_parallel_RLC_circuit NumberOfReplyTAG: 2 FirstChildTAG: Hi, If you keep on trying like I was, with wd, you'll never get there! Notice there is an input voltage step u(t),so, It's really damped? FirstChildUserIdTAG: 369339 FirstChildUserNameTAG: vargaslen FirstChildCreateTimeTAG: 2012-11-18T11:37:36Z FirstChildTAG: by definition when over-damped: $\alpha$ > $\omega_0$ so, $\omega_d=\sqrt{\omega_0^2-\alpha^2}$ has to be imaginary. you should NOT have to use $\omega_d$ in any analysis of over-damped systems. FirstChildUserIdTAG: 714237 FirstChildUserNameTAG: PaxPolaris FirstChildCreateTimeTAG: 2012-11-19T01:54:13Z IndexTAG: 3154 TitleTAG: H10P3 I hav solvd the circuit in frequency domain using fourier transforms...i could not get the ratio Vo/Vi...evn though got the crct eqn grader shows my ans as wrong...pls helpp UserIdTAG: 356056 UserNameTAG: Ganesh2810 CreateTimeTAG: 2012-11-18T08:16:51Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: try to find Vf/Vi depending of (R1 and C), it's easy. Then V0/Vf depending of (C//(R2+L)). You will get V0/Vi... FirstChildUserIdTAG: 16263 FirstChildUserNameTAG: Noureddine FirstChildCreateTimeTAG: 2012-11-18T15:08:12Z IndexTAG: 3155 TitleTAG: LAB 10 - BPF ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13532262961343674.png I am getting the AC analysis has shown in the figure..... It is similar to the given AC analysis.... Still i am not getting green tick.... Can anyone tell me what i am doing wrong...? UserIdTAG: 477198 UserNameTAG: Rajesh1993 CreateTimeTAG: 2012-11-18T08:12:02Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: It looks OK. I used 3 significant digits for my component values and entered the capacitor in pF and the inductor in H. The magnitude should be very close to -3dB at 3 and 6 on the log(Frequency in Hz) axis. My magnitude graph however only went to -60dB so I guess your graph is dipping a little below -60dB at 1e9 Hz. I'd try calculating more accurate values for the inductor and capacitor and leave the resistor values the same as the first part of the lab. FirstChildUserIdTAG: 141000 FirstChildUserNameTAG: OrinE FirstChildCreateTimeTAG: 2012-11-18T08:46:18Z SecondChildTAG: Thanks for the comment.... But i am still not getting the green tick... SecondChildUserIdTAG: 477198 SecondChildUserNameTAG: Rajesh1993 SecondChildCreateTimeTAG: 2012-11-18T09:33:03Z SecondChildTAG: try 1 ohm for resistor value and accordingly calculate L and C. SecondChildUserIdTAG: 357747 SecondChildUserNameTAG: kishores SecondChildCreateTimeTAG: 2012-11-18T10:29:58Z SecondChildTAG: Make sure you have EXACTLY -3dB at 1KHz and 1MHz (-3.1 or -2.9 won't give you the tick) SecondChildUserIdTAG: 277787 SecondChildUserNameTAG: kirilaska SecondChildCreateTimeTAG: 2012-11-18T15:25:58Z IndexTAG: 3156 TitleTAG: great video This is a great video - both educational and funny. UserIdTAG: 12905 UserNameTAG: Baer CreateTimeTAG: 2012-11-18T04:58:03Z VoteTAG: 0 CoursewareTAG: Week 9 / LCOT CommentableIdTAG: 6002x_LCOT NumberOfReplyTAG: 0 IndexTAG: 3157 TitleTAG: well done! who would have thought that this problem can be solved simply by KVL and KCL! UserIdTAG: 12905 UserNameTAG: Baer CreateTimeTAG: 2012-11-18T01:41:49Z VoteTAG: 0 CoursewareTAG: Week 9 / A&L Problem 12.3 CommentableIdTAG: 6002x_AL_P_12_3 NumberOfReplyTAG: 0 IndexTAG: 3158 TitleTAG: Possible Typo in H11P2 I think that the variable capacitor must be the one that is between 12 and 25pF. So perhaps there is a typo at 12 < **Cp** < 25pF? UserIdTAG: 304587 UserNameTAG: Vasso CreateTimeTAG: 2012-11-17T23:28:57Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: They are correct. Cs is the input capacitance of the oscilloscope and is usually some value between 12 and 25pF (the specs for an oscilloscope will tell you what it is). The variable capacitor, Cp, is in the probe and in this problem, they don't specify the range of values it could be FirstChildUserIdTAG: 141000 FirstChildUserNameTAG: OrinE FirstChildCreateTimeTAG: 2012-11-17T23:49:44Z SecondChildTAG: Thank you very much! I will try to figure it out :) SecondChildUserIdTAG: 304587 SecondChildUserNameTAG: Vasso SecondChildCreateTimeTAG: 2012-11-18T00:19:45Z IndexTAG: 3159 TitleTAG: S17E4 Solving for B How do I solve for B? Substituting into the given, assumed equations, the sin term goes to zero and B could be any constant whatsoever. If I take my complete solution for vc(t) and substitute it into one of the two given state equations and THEN try to solve for B, I end up with B = C*V/sqrt(L*C) where am I going wrong? UserIdTAG: 287805 UserNameTAG: Beneficial CreateTimeTAG: 2012-11-17T18:30:16Z VoteTAG: 0 CoursewareTAG: Week 9 / An LC Circuit CommentableIdTAG: 6002x_an_LC_Circuit NumberOfReplyTAG: 2 FirstChildTAG: You do everything right. Now just simplify your expression for B. FirstChildUserIdTAG: 146246 FirstChildUserNameTAG: Gromo3eka FirstChildCreateTimeTAG: 2012-11-17T21:49:21Z FirstChildTAG: I substituted $i_L(t) = B\sin(\omega t)$ into $L \frac{d i_L(t)}{d t} = v_C(t)$. $\frac{d i_L(t)}{d t}$ becomes $B\omega\cos(\omega t)$ Now you have $cos(0)$ rather than $sin(0)$ and you know $\omega$ and $v_C(t) = V$ at $t = 0$, so a little rearranging and you have the answer you had above. FirstChildUserIdTAG: 141000 FirstChildUserNameTAG: OrinE FirstChildCreateTimeTAG: 2012-11-17T23:14:20Z IndexTAG: 3160 TitleTAG: Lab 10 frequency confirm please. They sad "Recall that 1 Hz=2*pi radians/sec." but formula for the frequency is w=2*pi*f than 1Hz must be 1/(2*pi)radians/sec. Is that bug or my mistake? UserIdTAG: 352757 UserNameTAG: Kerbyco CreateTimeTAG: 2012-11-17T18:17:04Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: hi, yes, it's confusing! What we have to do? FirstChildUserIdTAG: 299251 FirstChildUserNameTAG: TheRedBlackOne FirstChildCreateTimeTAG: 2012-11-18T06:29:51Z FirstChildTAG: It's ur mistake.... Just try to understand the formula... `w=2*pi*f` FirstChildUserIdTAG: 477198 FirstChildUserNameTAG: Rajesh1993 FirstChildCreateTimeTAG: 2012-11-18T09:36:14Z IndexTAG: 3161 TitleTAG: about third time constant Third time constant is not time constant, it's frequency (omega 0). Third time constant is just square root of LC. Frequency constant (omega 0) is 1 divided by time constant. UserIdTAG: 451397 UserNameTAG: Tertei CreateTimeTAG: 2012-11-17T18:05:32Z VoteTAG: 0 CoursewareTAG: Week 9 / Preview of RLC Circuits CommentableIdTAG: 6002x_preview_of_RLC_circuits NumberOfReplyTAG: 0 IndexTAG: 3162 TitleTAG: Finally This was the most intense thriller ever. Why did the wave look like a damped sin wave. Season 17 Episode 18 phew. UserIdTAG: 204745 UserNameTAG: allwynmendes CreateTimeTAG: 2012-11-17T17:55:05Z VoteTAG: 0 CoursewareTAG: Week 9 / Preview of RLC Circuits CommentableIdTAG: 6002x_preview_of_RLC_circuits NumberOfReplyTAG: 0 IndexTAG: 3163 TitleTAG: H10P3 Q1 Hello! I can't seem to get the right answer to question 1 form the third problem, even though I thing I have solved it right. I managed to get green for all the rest of the questions in P3. In order to determine the algebraic expression I found VF in terms of Vi and then I got the expression of Vo in terms of VF, and by replacing VF from the first relation in the second one I get something of the form: b/(b-w^2*a*x*y+j*w*(a*b*x+y)) where a,b,x,y are elements in the circuit. Isn't this the final form? Can anyone tell me where I went wrong? UserIdTAG: 319221 UserNameTAG: stefan31i CreateTimeTAG: 2012-11-17T16:19:15Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: ONE **R1** TERM IS MISSING IN THE DENOMINATOR. CHECK YOUR CALCULATION AGAIN. FirstChildUserIdTAG: 446998 FirstChildUserNameTAG: kumar_vikash FirstChildCreateTimeTAG: 2012-11-17T17:47:13Z SecondChildTAG: Thanks a lot. I've just redone the calculation and R1 is exactly what is missing. It always helps to take a day off and come back later to redo the calculation. SecondChildUserIdTAG: 319221 SecondChildUserNameTAG: stefan31i SecondChildCreateTimeTAG: 2012-11-18T13:21:27Z SecondChildTAG: I've gone trough the same calculations over and over and I still get the result just the same as stefan31i posted above. As kumar stated, there's a missing R1 there, but I can't find it. If there's anyway to help, that's not answer given, thank you very much +) SecondChildUserIdTAG: 290028 SecondChildUserNameTAG: Fjuca SecondChildCreateTimeTAG: 2012-11-23T05:12:33Z SecondChildTAG: I did the same exact thing as stefan31 and got the same answer, but I don't understand why it is wrong and why an R1 might be missing. SecondChildUserIdTAG: 101750 SecondChildUserNameTAG: caf960 SecondChildCreateTimeTAG: 2012-11-23T21:16:50Z SecondChildTAG: Yeah, it's really buzzing me all day. SecondChildUserIdTAG: 290028 SecondChildUserNameTAG: Fjuca SecondChildCreateTimeTAG: 2012-11-23T21:36:22Z IndexTAG: 3164 TitleTAG: Lab 10 i used the break freqency formulae, which were calculated earlier, to obtain the L and C for the last question( Assumed R to be 1K in both cases).. I'm still not getting the right magnitude plot.. can anyone please tell me why?? UserIdTAG: 457713 UserNameTAG: ASHWIN_DT CreateTimeTAG: 2012-11-17T13:56:24Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: use very high value of resistance(>=40000ohm) in case of RC filter. Use corresponding value for capacitor. FirstChildUserIdTAG: 446998 FirstChildUserNameTAG: kumar_vikash FirstChildCreateTimeTAG: 2012-11-17T18:24:44Z SecondChildTAG: Great.... Very nice. Now I get a green mark. But, could you explain me why?. Why other lower values (near to the RL value) are not so good to get the response we are looking for? Thank you very much SecondChildUserIdTAG: 261081 SecondChildUserNameTAG: DanielSP SecondChildCreateTimeTAG: 2012-11-23T10:46:41Z IndexTAG: 3165 TitleTAG: week 9 completed Feeling a bit relaxed now......... UserIdTAG: 611666 UserNameTAG: Shaminder420 CreateTimeTAG: 2012-11-17T12:50:56Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Can you help me to get α?¿ I don't know how to do it. Thanks. FirstChildUserIdTAG: 558384 FirstChildUserNameTAG: Josue9740 FirstChildCreateTimeTAG: 2012-11-17T13:02:10Z SecondChildTAG: help me in H9P1 7 SecondChildUserIdTAG: 473933 SecondChildUserNameTAG: rahul_pradhan SecondChildCreateTimeTAG: 2012-11-17T13:13:13Z SecondChildTAG: just compare your characteristic eqn with s2+2as+w2=0.u will get the answer SecondChildUserIdTAG: 611666 SecondChildUserNameTAG: Shaminder420 SecondChildCreateTimeTAG: 2012-11-17T13:20:54Z SecondChildTAG: @rahul q/c+v(t-) SecondChildUserIdTAG: 611666 SecondChildUserNameTAG: Shaminder420 SecondChildCreateTimeTAG: 2012-11-17T13:21:28Z SecondChildTAG: i got it SecondChildUserIdTAG: 473933 SecondChildUserNameTAG: rahul_pradhan SecondChildCreateTimeTAG: 2012-11-17T17:34:04Z FirstChildTAG: pls help me doing lab 9 last part FirstChildUserIdTAG: 220425 FirstChildUserNameTAG: rishienggju FirstChildCreateTimeTAG: 2012-11-18T12:02:10Z IndexTAG: 3166 TitleTAG: H9P1 part 5 voltage across capacitor I'm stuck on this point. I have no clue how to attack this without actually doing the math. I have the current through inductor, but trying to work this problem keeps resulting in 0. UserIdTAG: 248807 UserNameTAG: 1kingsman CreateTimeTAG: 2012-11-17T01:22:12Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: me too have same problem...... FirstChildUserIdTAG: 368981 FirstChildUserNameTAG: asimakhtar FirstChildCreateTimeTAG: 2012-11-17T01:38:21Z FirstChildTAG: You know what the total energy stored in the circuit is, so you can work out the energy stored in the capacitor as total energy - energy stored in the inductor. Once you know the energy stored in the capacitor, it's easy to calculate the voltage across it. Just be careful of the sign. FirstChildUserIdTAG: 141000 FirstChildUserNameTAG: OrinE FirstChildCreateTimeTAG: 2012-11-17T01:58:15Z SecondChildTAG: Thanks OrinE, this sign was bothering me SecondChildUserIdTAG: 296419 SecondChildUserNameTAG: oscman SecondChildCreateTimeTAG: 2012-11-17T23:34:43Z FirstChildTAG: just apply your voltage equation of before impulse and fill w0 in terms of time period.u wll get the answer FirstChildUserIdTAG: 611666 FirstChildUserNameTAG: Shaminder420 FirstChildCreateTimeTAG: 2012-11-17T07:43:59Z FirstChildTAG: I'm stuck here too. FirstChildUserIdTAG: 364126 FirstChildUserNameTAG: shunyi FirstChildCreateTimeTAG: 2012-11-17T10:14:54Z IndexTAG: 3167 TitleTAG: S19E4 Phase Angles I'm not feeling it; guess my trig is a little weak. Can somebody give me a hint as to how to come up with the arctan of that transfer function? Thanks in advance. UserIdTAG: 339668 UserNameTAG: chickwebb CreateTimeTAG: 2012-11-17T00:50:12Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3168 TitleTAG: Charge Pump in Ex 12.4 of textbook - how can periods be same length? I'm looking at this example in the textbook. I see how the first part of the cycle works: S1 is closed, and S2 is open, so we have nothing but the voltage source and the inductor, and current ramps up. But in the next phase when the inductor is switched from the voltage source to the capacitor... I don't see how the time to zero current is the same each cycle. In the first cycle the cap will start at zero volts, but the next cycle it will be charged some and push back harder on the inductor, no? Shouldn't the inductor than go to i=0 faster with each cycle? Rob UserIdTAG: 468623 UserNameTAG: RobNik CreateTimeTAG: 2012-11-16T21:33:53Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Ah, I just found that week 9 tutorials discuss the same circuit. Maybe I will get it after I watch those... FirstChildUserIdTAG: 468623 FirstChildUserNameTAG: RobNik FirstChildCreateTimeTAG: 2012-11-16T21:43:09Z IndexTAG: 3169 TitleTAG: Revision resources in Electricity and Magnetism I am revising my Physics (electricity & magnetism). Does anyone know of really simple resources, other than those from mit opencourseware. UserIdTAG: 5252 UserNameTAG: rohit1971 CreateTimeTAG: 2012-11-16T21:00:22Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: http://ocw.mit.edu/courses/physics/8-02-electricity-and-magnetism-spring-2002/video-lectures/ FirstChildUserIdTAG: 236136 FirstChildUserNameTAG: Hamid-ch FirstChildCreateTimeTAG: 2012-11-16T22:09:37Z SecondChildTAG: these two r major power sources SecondChildUserIdTAG: 443544 SecondChildUserNameTAG: vinoth123 SecondChildCreateTimeTAG: 2012-11-17T00:37:21Z IndexTAG: 3170 TitleTAG: Lab Week 9 Hi! Anyone with any hints on how to get the value of L in the 3rd part? UserIdTAG: 143823 UserNameTAG: NathanNadeson CreateTimeTAG: 2012-11-16T19:07:35Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Just change the value of the capacitor first, then the inductor and finally the duty cycle. Play around with the circuit using guess and check. That's how I solved it. Best of luck. FirstChildUserIdTAG: 85490 FirstChildUserNameTAG: yashsavani FirstChildCreateTimeTAG: 2012-11-16T19:45:24Z FirstChildTAG: do not touch!!!!!!!!! :P FirstChildUserIdTAG: 244706 FirstChildUserNameTAG: Miguel_Angel FirstChildCreateTimeTAG: 2012-11-16T19:46:25Z IndexTAG: 3171 TitleTAG: H9P2(d) any trick in there? I'm for sure my alpha and wo are correct (I've got a green mark in every other response I gave). Then, there is this question about wd ( I think there is only one formula), but can get the correct answer? What is going on? UserIdTAG: 369339 UserNameTAG: vargaslen CreateTimeTAG: 2012-11-16T15:35:51Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 6 FirstChildTAG: H9P2(d) is the w0, but you said you got the w0 correctly. So I'm confused about what your question is. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-11-16T15:49:24Z SecondChildTAG: hello, they ask about wd which is supposed to be derived from sqrt(wo^2 + alpha^2), but the formula doesn't work here. SecondChildUserIdTAG: 369339 SecondChildUserNameTAG: vargaslen SecondChildCreateTimeTAG: 2012-11-16T16:40:00Z SecondChildTAG: My comment from last evening: The question is badly stated. They want the **undamped** frequency in Hz. Wasted lots of time there!!! SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-11-16T17:00:20Z SecondChildTAG: I don't know, I never thought that they were looking for the damped frequency. But maybe that's just where my head was at the time. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-11-16T17:35:37Z SecondChildTAG: I don't get it! Isn't the wd (natural oscillation frequency for LCR circuit) the same in either case? ## SecondChildUserIdTAG: 369339 SecondChildUserNameTAG: vargaslen SecondChildCreateTimeTAG: 2012-11-16T18:00:15Z SecondChildTAG: no, wd is the damped frequency, and w0 is the undamped frequency. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-11-16T19:15:43Z SecondChildTAG: I didn't bother with wd. Just make sure your w0 is in Hertz for d), and only d). SecondChildUserIdTAG: 184153 SecondChildUserNameTAG: tthngn SecondChildCreateTimeTAG: 2012-11-16T19:34:28Z SecondChildTAG: Ok guys, That worked! thinking I'll have to check the videos again! SecondChildUserIdTAG: 369339 SecondChildUserNameTAG: vargaslen SecondChildCreateTimeTAG: 2012-11-16T20:47:43Z SecondChildTAG: No need to check the videos. The question is just incorrectly stated. The system naturally oscillates at the damped frequency, but that is not what the grader accepts for the correct answer. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-11-16T22:08:35Z SecondChildTAG: Thanks guys, you are great. The second time this evening I have spent a lot of time looking for my fault, and in the end I was right all along. Really stupid that they want w0 but ask for the natural frequency. In S18E3 they explicitely call wd the natural frequency. SecondChildUserIdTAG: 406420 SecondChildUserNameTAG: Picolo SecondChildCreateTimeTAG: 2012-11-17T00:20:17Z SecondChildTAG: I wasted a great deal of time on this question. This was the last answer in the homework set that I got correct. It was only after I was so desperate that I was willing to try anything, no matter how silly, that I stumbled on the "correct" answer. I am not a happy camper. I worry that they will do something similar on the final where there is neither the time nor attempts available to happen onto their desired answer! SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-11-17T00:53:24Z SecondChildTAG: Rethinking about the undamped frequency, I think I'd missed the fact of a step u(t) input. That is why vo will never fade out (undamped),so instead of sqrt(wo^2 - alpha^2) (I wroted wrongly upstairs), I should have used the wo (undamped oscillation frequency), Thanks again guys! I think my brain boiled out with all that complex algebra. SecondChildUserIdTAG: 369339 SecondChildUserNameTAG: vargaslen SecondChildCreateTimeTAG: 2012-11-17T01:11:14Z FirstChildTAG: Rethinking about the undamped frequency, I think I've missed the fact of a step u(t) input. That is why vo will never fade out (undamped),so instead of sqrt(wo^2 - alpha^2) (I wroted wrongly upstairs), I should have used the wo (undamped oscillation frequency), Thanks again guys! I think my brain boiled out with all that complex algebra. FirstChildUserIdTAG: 369339 FirstChildUserNameTAG: vargaslen FirstChildCreateTimeTAG: 2012-11-16T21:21:20Z SecondChildTAG: No, the ringing dies out. The damped frequency is the correct frequency. The problem statement is just wrong. In my opinion this was a horrid problem set. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-11-17T01:00:39Z SecondChildTAG: skyhawk, I'm not making a judgement either way as to whether they made a bad problem or not, but if you think the problem is stated inaccurately, you need to lay it all out and tell us why. The whole thing, from the beginning, with equations. They are engineers and if you lay it out like that and you are right, they will change it. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-11-17T03:02:35Z SecondChildTAG: Here is why I think the problem is correct. The damped resonance frequency wd=sqrt(wo^2 - alpha^2). So wd is composed of the natural frequency and the damping factor. Because you have to have a natural frequency to get a damped frequency, the question is correct. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-11-17T03:29:53Z SecondChildTAG: Both are called natural frequencies (please see comment below), but the one that is actually observed or measured when excited by a step is the damped natural frequency. Of course, in a driven circuit w0 will be observed as a resonance frequency. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-11-17T04:06:09Z SecondChildTAG: I understand what you are saying, which is that the "natural frequency" that would be observed in this circuit is the damped frequency. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-11-17T04:38:59Z FirstChildTAG: Over two weeks ago when I did this set, I should have been more vocal. However, at the time, I was happy to get the check mark and watch my score build toward the magic 60%. It is only in the past couple days that I have realized the grief and confusion this answer is causing. In addition, parts of problem 1 although not technically wrong are causing pain. It is now bedtime for me on the US east coast. Tomorrow will be another day. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-17T03:40:42Z FirstChildTAG: The question is: **The circuit would ring when excited with a step response vi=u(t). What is the natural frequency of this circuit, in terms of the component values?** The circuit rings at the frequency that appears in the trig function, and that frequency is the damped natural frequency. Have a look at the last paragraph on page 646 of the text where the term damped natural frequency is introduced and explained. In the preceding sentence, w0 is called undamped natural frequency. P.S. Before commenting I went back to the text to make sure that the term natural frequency was not reserved for the undamped case alone. I only started questioning the wording of the problem when I was sure that damped natural frequency was a term that was used in **this course**. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-17T04:00:55Z SecondChildTAG: Hmmm. Maybe the question is stated badly. Because without values, the possibility exists that R=0, therefore alpha=0, in which case the answer would be the undamped natural frequency. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-11-17T04:47:07Z SecondChildTAG: Please look at the circuit. If R = 0 then the capacitor is shorted out and the circuit consists of an ideal inductor alone. Also note that the B coefficient is undefined. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-11-17T04:57:13Z SecondChildTAG: (Y) skyhawk SecondChildUserIdTAG: 162671 SecondChildUserNameTAG: tuhin1991paul SecondChildCreateTimeTAG: 2012-11-17T06:14:23Z SecondChildTAG: Oh yeah, I didn't think of that. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-11-17T16:31:21Z FirstChildTAG: Ok, just to sum up: skyhawk's argument is that if you were observing this circuit, what you would see as the "natural frequency" would be the damped natural frequency, so the accepted answer should be wd. My argument is that since wd is composed of wo and alpha, you need the undamped natural frequency to obtain the damped frequency, so the question is correct. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-11-17T16:43:53Z SecondChildTAG: I assumed what was meant was the natural frequency in radians per second. I tried that. It was marked wrong. I tried the damped frequency. It was marked wrong. I tried each of them over 2pi and they were marked wrong. I tried each of them times 2pi just in case physics had suddenly changed since I last checked. Every single thing I have tried has been marked wrong. There aren't that many possibilities and anyway, the answer should be the natural frequency because that is what the question asks for. What the heck is going on? SecondChildUserIdTAG: 126825 SecondChildUserNameTAG: aphoenix SecondChildCreateTimeTAG: 2012-11-18T00:51:01Z FirstChildTAG: finally any1 pls say wts the ans for the frequency wich the grader accepts!! giv the expression for it FirstChildUserIdTAG: 356056 FirstChildUserNameTAG: Ganesh2810 FirstChildCreateTimeTAG: 2012-11-18T07:51:42Z IndexTAG: 3172 TitleTAG: H9P1 Hi, I have calculated the equation $i_L(t) = cos(\omega _0t)$ But when I substitute $t=5$ in the equation I do not get the correct answer. I have got the answer for $v_c(5_-)$ by using $v_c = L\frac{di_L(t)}{dt}$ and found the same answer for $i_L(5_-)$ using the energy equation: $\frac{C(v_c(5_-))^2}{2}+\frac{L(i_L(5_-))^2}{2} = Total\ Energy$ and calculating for $i_L$. However I am still not getting the green tick. Any help would be much appreciated. Thank You, in advance UserIdTAG: 85490 UserNameTAG: yashsavani CreateTimeTAG: 2012-11-16T10:16:17Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: was your calculator in radians?.. FirstChildUserIdTAG: 173147 FirstChildUserNameTAG: cruiser_rahit FirstChildCreateTimeTAG: 2012-11-16T16:03:40Z SecondChildTAG: Yes, I made sure that it was. SecondChildUserIdTAG: 85490 SecondChildUserNameTAG: yashsavani SecondChildCreateTimeTAG: 2012-11-16T19:42:57Z FirstChildTAG: Half of the questions, including the last 3, can be solved intuitively: - C is like short circuit, and - L like open circuit, for abrupt transitions (t+). Once v (i) is known for C (L, reps.) then power/energy is easy (as in no calculation). FirstChildUserIdTAG: 184153 FirstChildUserNameTAG: tthngn FirstChildCreateTimeTAG: 2012-11-16T19:51:43Z FirstChildTAG: Adjust the frequency slightly so that the period in an **integer** number of seconds. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-16T20:01:55Z IndexTAG: 3173 TitleTAG: Question Week 9 - S18V15 Hello all, I'd like to how I can prove the identity: ![enter image description here][1] I have this curiosity. XD Thank you for any help. [1]: https://edxuploads.s3.amazonaws.com/13530606031903454.png UserIdTAG: 366669 UserNameTAG: pedroramus CreateTimeTAG: 2012-11-16T09:55:44Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: $A_1cos(\theta) + A_2sin(\theta) = R\cdot cos(\theta + \alpha)$ $\therefore A_1cos(\theta) + A_2sin(\theta) = R(cos(\theta)cos(\alpha)-sin(\theta)sin(\alpha))$ $\therefore A_1cos(\theta) + A_2sin(\theta) = (Rcos(\alpha))cos(\theta)-(Rsin(\alpha))sin(\theta)$ $\therefore A_1= Rcos(\alpha)$ and $\therefore A_2= Rsin(\alpha)$ Since $cos^2(\alpha) + sin^2(\alpha) = 1$ $\therefore (A_1)^2 + (A_2)^2 = R^2(cos^2(\alpha) + sin^2(\alpha)) = R^2$ $\therefore \sqrt{(A_1)^2 + (A_2)^2} = R$ Since $tan(\alpha) = \frac{sin(\alpha)}{cos(\alpha)}$ $\therefore \frac{A_1}{A_2} = \frac{Rsin(\alpha)}{Rcos(\alpha)} = tan(\alpha)$ $\therefore \alpha = arctan\left(\frac{A_1}{A_2}\right) = tan^{-1}\left(\frac{A_1}{A_2}\right)$ $\therefore A_1cos(\theta) + A_2sin(\theta) = \sqrt{(A_1)^2 + (A_2)^2}\cdot cos\left(\theta + arctan\left(\frac{A_1}{A_2}\right)\right)$ FirstChildUserIdTAG: 85490 FirstChildUserNameTAG: yashsavani FirstChildCreateTimeTAG: 2012-11-16T10:29:22Z SecondChildTAG: Thank you very much!!!! SecondChildUserIdTAG: 366669 SecondChildUserNameTAG: pedroramus SecondChildCreateTimeTAG: 2012-11-16T22:53:38Z IndexTAG: 3174 TitleTAG: Finding Phase Hint please. I'm not getting it. Thank you UserIdTAG: 108959 UserNameTAG: cpfaivre CreateTimeTAG: 2012-11-16T02:27:11Z VoteTAG: 0 CoursewareTAG: Week 10 / Magnitudes And Angles CommentableIdTAG: 6002x_Magnitudes_and_angles NumberOfReplyTAG: 1 FirstChildTAG: Found my problem. I was not taking into account the 90 degree component FirstChildUserIdTAG: 108959 FirstChildUserNameTAG: cpfaivre FirstChildCreateTimeTAG: 2012-11-16T04:39:23Z SecondChildTAG: What 90 degrees component? SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-11-24T00:38:14Z IndexTAG: 3175 TitleTAG: It should be 1/(G1 + G2 + G3 + ... + Gn) and NOT as shown in this video! (0:58) I seriously almost smacked my head to a wall while trying to figure out where I went wrong in the parallel resistors question on HW1.. please correct this section. UserIdTAG: 699728 UserNameTAG: yuvall CreateTimeTAG: 2012-11-15T22:30:52Z VoteTAG: 0 CoursewareTAG: Week 1 / Element combination rules CommentableIdTAG: 6002x_S2V5_Method_2_-_Element_combination_rules NumberOfReplyTAG: 1 FirstChildTAG: Dear yuvall, Conductances in parallel can be added together to create an equivalent conductance value. 1/(G1+G2+G3+...+GN) will give you the **resistance** of a network of parallel **conductances**. The video is not wrong, it is simply stating that the equivalent **conductance** 'Geq' of a network of parallel conductances is the sum of the conductances. Rohan FirstChildUserIdTAG: 391929 FirstChildUserNameTAG: RohanNagarkar FirstChildCreateTimeTAG: 2012-11-16T16:12:52Z IndexTAG: 3176 TitleTAG: S17E5, 3rd question does not make sense. The question says: "Unfortunately, our particular solution cannot be the solution to our problem because it does not match the initial conditions. So we must add in an appropriate amount of solution of the homogeneous equation to match the initial conditions." However, I came up with the a characteristic equation without adding anything to the homogeneous equation, in fact it came out exactly the same as the characteristic equation in the lecture S17V10, except in this case it is current and not voltage. Does anybody have any theory why? UserIdTAG: 132828 UserNameTAG: rharris CreateTimeTAG: 2012-11-15T21:47:34Z VoteTAG: 0 CoursewareTAG: Week 9 / An ILC circuit CommentableIdTAG: 6002x_an_ILC_circuit NumberOfReplyTAG: 2 FirstChildTAG: Also, in the last question, anybody have any idea why they wanted the negative value of ϕ, which corresponded to -A, instead of the +A? FirstChildUserIdTAG: 132828 FirstChildUserNameTAG: rharris FirstChildCreateTimeTAG: 2012-11-15T23:01:39Z FirstChildTAG: > Unfortunately, our particular solution cannot be the solution to our > problem because it does not match the initial conditions. So we must > add in an appropriate amount of solution of the homogeneous equation > to match the initial conditions. You get the general solution by adding the particular solution and the homogeneous solution. $i_L(t)=i_{LP} + i_{LH}(t)$ If your particular solution already matches the initial conditions ... you don't need a homogeneous solution at all i.e. it must be 0. That is all the question is saying. The characteristic equation is used to derive the frequency constant $\omega_0$. As we have seen before time/frequency constants only depend on R\L\C values and not the power source. So, it shouldn't be too surprising that you get the same characteristic equation as before. FirstChildUserIdTAG: 714237 FirstChildUserNameTAG: PaxPolaris FirstChildCreateTimeTAG: 2012-11-18T04:01:51Z IndexTAG: 3177 TitleTAG: Still needing assistance with H9P1 Almost There, but not quite!. Í understand this is a circuit with energy bouncing between capacitor and inductor (no doubt about it). Just before impulse, all the energy is in the capacitor, and I've got the right energy in the circuit in joules (again,no doubt about it). So applying the equation for energy in equilibrium between L and C, knowing the total energy and knowing that all that energy is in the capacitor, I should be able to get the voltage in the capacitor from (1/2)Cvc^2, then, Why doesn't this work? Please help me to clarify! UserIdTAG: 369339 UserNameTAG: vargaslen CreateTimeTAG: 2012-11-15T20:26:55Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Have you checked the sign of your answer? FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-15T20:40:11Z SecondChildTAG: Thanks a lot! that is it! by the way in H9P2 I've got right alpha and wo, (I've right A,B and C and characteristic equation),but when calculating the natural frequency (wd=sqrt(wo^2 + alpha^2) ) can't get the green mark! Any sugestions? and thanks again! SecondChildUserIdTAG: 369339 SecondChildUserNameTAG: vargaslen SecondChildCreateTimeTAG: 2012-11-15T20:57:05Z SecondChildTAG: The question is badly stated. They want the **undamped** frequency in Hz. Wasted lots of time there!!! SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-11-15T21:00:02Z SecondChildTAG: why the capacitor voltage has to be negative? I don't understand!. SecondChildUserIdTAG: 369339 SecondChildUserNameTAG: vargaslen SecondChildCreateTimeTAG: 2012-11-15T21:02:58Z SecondChildTAG: The capacitor charges thru the negative plate, it's that it? SecondChildUserIdTAG: 369339 SecondChildUserNameTAG: vargaslen SecondChildCreateTimeTAG: 2012-11-15T21:18:08Z SecondChildTAG: Thanks again , you've helped me a lot! SecondChildUserIdTAG: 369339 SecondChildUserNameTAG: vargaslen SecondChildCreateTimeTAG: 2012-11-15T21:37:26Z SecondChildTAG: Glad you got it. If you connect the + lead of a voltmeter to the + node and the ground lead to the - node, the meter will read a negative voltage because the + lead is at a lower potential than the ground lead. Half a cycle earlier the voltmeter would have read positive with the same magnitude. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-11-15T22:33:17Z IndexTAG: 3178 TitleTAG: Typo in Example 12.7 at the textbook Result that had shown for w0 = 62017 obtained if assume C = 13nF, but not for 13pF. This example is analog 12.6 UserIdTAG: 188778 UserNameTAG: xsaq CreateTimeTAG: 2012-11-15T16:33:45Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3179 TitleTAG: please give some hint to solve p7 in H9P1 please give some hint to solve p7 in H9P1 I am using q/c + V(5-)but it is not accepting any hint where i am going wrong Thanks UserIdTAG: 352647 UserNameTAG: praveenjugge CreateTimeTAG: 2012-11-15T10:55:29Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: Same question. In our case q=A? FirstChildUserIdTAG: 500965 FirstChildUserNameTAG: barka0 FirstChildCreateTimeTAG: 2012-11-15T12:30:55Z FirstChildTAG: How did ypu find vc(-5)? FirstChildUserIdTAG: 96873 FirstChildUserNameTAG: Santyaga FirstChildCreateTimeTAG: 2012-11-16T19:14:10Z FirstChildTAG: Use basic capacitor equation that relates charge, voltage and capacitance. Also take into account existing capacitor voltage at the moment of impulse. FirstChildUserIdTAG: 189680 FirstChildUserNameTAG: vnd FirstChildCreateTimeTAG: 2012-11-15T15:26:59Z SecondChildTAG: Check the sign of your answer. SecondChildUserIdTAG: 99628 SecondChildUserNameTAG: Pranjal16 SecondChildCreateTimeTAG: 2012-11-16T04:33:09Z SecondChildTAG: Hi! How do we find the value of the instantaneous current iL after the impulse is applied? Thanks! SecondChildUserIdTAG: 143823 SecondChildUserNameTAG: NathanNadeson SecondChildCreateTimeTAG: 2012-11-16T09:55:53Z FirstChildTAG: Hi I got correct voltage for vc(5-) and now we are calculating vc(5+) after an impulse of .64 coulombs charge capacitance is given so **vc(5+) = charge/capacitance+ vc(5-)** still i am not getting green check mark any hint where am i going wrong FirstChildUserIdTAG: 352647 FirstChildUserNameTAG: praveenjugge FirstChildCreateTimeTAG: 2012-11-16T10:22:25Z IndexTAG: 3180 TitleTAG: Help with Week 11 homeworkP1 problem 1 of week 11 is complicated. i dont knoe how to start i.e which definition to use UserIdTAG: 137509 UserNameTAG: rahulpark CreateTimeTAG: 2012-11-15T07:25:52Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Try putting the denominator of the impedance in standard form and use the definition of Q. FirstChildUserIdTAG: 397595 FirstChildUserNameTAG: mholin FirstChildCreateTimeTAG: 2012-11-15T14:27:24Z SecondChildTAG: which one? SecondChildUserIdTAG: 137509 SecondChildUserNameTAG: rahulpark SecondChildCreateTimeTAG: 2012-11-17T14:40:46Z SecondChildTAG: Q=(w/del(w)) SecondChildUserIdTAG: 137509 SecondChildUserNameTAG: rahulpark SecondChildCreateTimeTAG: 2012-11-17T14:41:23Z SecondChildTAG: i'm having problem with complex analysis and root finding as the equation gets complex in the standard form SecondChildUserIdTAG: 137509 SecondChildUserNameTAG: rahulpark SecondChildCreateTimeTAG: 2012-11-17T14:42:51Z FirstChildTAG: You don't really have to solve any equation; just look at the impedance of the circuit in standard form (s=jw) and look for the bandwidth definition. It easier than you think; you can also check an example in the cap 13 of the textbook. FirstChildUserIdTAG: 277787 FirstChildUserNameTAG: kirilaska FirstChildCreateTimeTAG: 2012-11-18T04:57:30Z SecondChildTAG: thnx.... SecondChildUserIdTAG: 137509 SecondChildUserNameTAG: rahulpark SecondChildCreateTimeTAG: 2012-11-18T11:08:47Z IndexTAG: 3181 TitleTAG: Not finding text box to put answer for boost converter LAB9 Dear sir I am Not finding text box to put answer for boost converter LAB9 please do something UserIdTAG: 352647 UserNameTAG: praveenjugge CreateTimeTAG: 2012-11-15T06:52:00Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: you dont have a text box for that..I think you need to run a trans analysis and then check you answer..like we had in 1st few labs.. FirstChildUserIdTAG: 173147 FirstChildUserNameTAG: cruiser_rahit FirstChildCreateTimeTAG: 2012-11-15T07:53:25Z SecondChildTAG: You should made that they want. When all required values will be set, run transient analysis and press Check. SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-11-16T11:02:59Z SecondChildTAG: try Thomas Backman's (exscape) Notes in wiki SecondChildUserIdTAG: 357747 SecondChildUserNameTAG: kishores SecondChildCreateTimeTAG: 2012-11-16T14:35:51Z IndexTAG: 3182 TitleTAG: resistors in series or in parallel ?? The last two resistors at the right are in series, are they not? UserIdTAG: 134637 UserNameTAG: runsbarefeet CreateTimeTAG: 2012-11-14T20:44:35Z VoteTAG: 0 CoursewareTAG: Week 1 / Long Resistor Chains CommentableIdTAG: 6002x_long_resistor_chains NumberOfReplyTAG: 0 IndexTAG: 3183 TitleTAG: H11P1 Hello there! How can I calculate bandwidth Δω of the impedance Z ? I can calculate Z, but its bandwidth I am not really sure. Any tip? UserIdTAG: 39980 UserNameTAG: MarinoJr CreateTimeTAG: 2012-11-14T18:06:43Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Try putting the denominator of the impedance in standard form and use the definition of Q. FirstChildUserIdTAG: 397595 FirstChildUserNameTAG: mholin FirstChildCreateTimeTAG: 2012-11-15T14:29:23Z SecondChildTAG: Why the denominator?, and not the whole expresion as before? Gotta be because this is Z and not Vo/Vi, but I don't see why! Please! SecondChildUserIdTAG: 398594 SecondChildUserNameTAG: DaveyJC SecondChildCreateTimeTAG: 2012-11-30T16:09:26Z IndexTAG: 3184 TitleTAG: S21E4: AM RADIO TUNING S21E4: question c asks for the Q of the circuit. I just calculated L, Cmin and Cmax is given. Therefore, Q = w0*L/R = 1/sqrt(L*Cmax)*L/R The result is not matching. Any tip? UserIdTAG: 39980 UserNameTAG: MarinoJr CreateTimeTAG: 2012-11-14T15:49:29Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: It's a parallel LC circuit so $\alpha = \frac{1}{2RC}$ . (Then don't forget the $2\pi$ when converting from frequency to $\omega$ like I did.) FirstChildUserIdTAG: 141000 FirstChildUserNameTAG: OrinE FirstChildCreateTimeTAG: 2012-11-14T17:32:06Z SecondChildTAG: Thanks. It worked. SecondChildUserIdTAG: 39980 SecondChildUserNameTAG: MarinoJr SecondChildCreateTimeTAG: 2012-11-14T18:01:31Z FirstChildTAG: I found Q=w0/(2*alfa)=w0*R*C. Then question e asks for bandwidth delta_w.. I calculated delta_w=w0/Q=w0/(w0*R*C)=1/(R*C) but the result again does not match. Any tip? FirstChildUserIdTAG: 39980 FirstChildUserNameTAG: MarinoJr FirstChildCreateTimeTAG: 2012-11-14T18:03:48Z SecondChildTAG: You must divide by 2*pi. I would have call it delta_f, but they don't. I've found more than once during the course a confusion in notation for w and f. They call w both of them, and use their respective units of measurement (rad/s or Hertz) to tell them apart. Here you're asked to express the solution in kHz, so you must divide by 2*pi. SecondChildUserIdTAG: 84801 SecondChildUserNameTAG: marcuspag SecondChildCreateTimeTAG: 2012-11-28T11:07:48Z IndexTAG: 3185 TitleTAG: video S6V6 = video S6V7 Yes, as some people have already noticed videos S6V6 and S6V7 are identical.... UserIdTAG: 710567 UserNameTAG: oleg_a CreateTimeTAG: 2012-11-14T15:29:45Z VoteTAG: 0 CoursewareTAG: Week 3 / Graphical Analysis of Nonlinear Circuits CommentableIdTAG: 6002x_graphical_analysis_of_nonlinear_circuits NumberOfReplyTAG: 0 IndexTAG: 3186 TitleTAG: help me hi friends how to do this course UserIdTAG: 793223 UserNameTAG: kichukrishna CreateTimeTAG: 2012-11-14T06:57:19Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: How long you have been following this ? FirstChildUserIdTAG: 99628 FirstChildUserNameTAG: Pranjal16 FirstChildCreateTimeTAG: 2012-11-14T10:33:38Z IndexTAG: 3187 TitleTAG: electronics how to design voltage to frequency converter.please explain step wise,..if possible with lab tools using animation UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-11-14T05:13:22Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3188 TitleTAG: H11P1 Anybody have a hint for the expression for the delta omega of the tank impedance? UserIdTAG: 397595 UserNameTAG: mholin CreateTimeTAG: 2012-11-14T00:40:26Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Try putting the impedance denominator in canonical form. FirstChildUserIdTAG: 397247 FirstChildUserNameTAG: pcbolt FirstChildCreateTimeTAG: 2012-11-14T06:57:56Z SecondChildTAG: and the numerator must be equal to one SecondChildUserIdTAG: 18073 SecondChildUserNameTAG: gburkhart SecondChildCreateTimeTAG: 2012-11-15T02:03:09Z SecondChildTAG: Is it really must be? i got right answer with numenator s/(Rc*C)+RL/(L*C) which is not equal to one. SecondChildUserIdTAG: 352757 SecondChildUserNameTAG: Kerbyco SecondChildCreateTimeTAG: 2012-11-20T14:32:24Z SecondChildTAG: same here! SecondChildUserIdTAG: 369339 SecondChildUserNameTAG: vargaslen SecondChildCreateTimeTAG: 2012-11-23T23:12:11Z SecondChildTAG: The numerator **does not matter**. It will change depending on the circuit, but the denominator will not change from the canonical equation. SecondChildUserIdTAG: 248807 SecondChildUserNameTAG: 1kingsman SecondChildCreateTimeTAG: 2012-12-02T03:55:39Z FirstChildTAG: Why the denominator?, and not the whole expresion as before? Gotta be because this is Z and not Vo/Vi, but I don't see why! Please! FirstChildUserIdTAG: 398594 FirstChildUserNameTAG: DaveyJC FirstChildCreateTimeTAG: 2012-11-30T16:09:33Z IndexTAG: 3189 TitleTAG: h9p1 help please q..5 im having difficulties with it UserIdTAG: 462861 UserNameTAG: edxforme CreateTimeTAG: 2012-11-14T00:21:00Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Check your sign. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-14T02:53:15Z IndexTAG: 3190 TitleTAG: H9P2 e) Don't understand the question I don't understand what does it mean with "The ringing will be damped by the factor e−αt. What is the expression for α in terms of the component values?" UserIdTAG: 244225 UserNameTAG: lerimock CreateTimeTAG: 2012-11-13T22:28:50Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Please elaborate on what you are having problems with. The information you need is all in Chapter 12, so you might want to skim through that. If it's ringing you don't understand, there is a great picture of it on page 626. It's that unwanted oscillation. There is also the wikipedia article on ringing: http://en.wikipedia.org/wiki/Ringing_(signal) If it's the damping, then page 649 has a good picture of it. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-11-14T02:14:21Z SecondChildTAG: It is 0.1591549*sqrt(1/(L*C)) SecondChildUserIdTAG: 106816 SecondChildUserNameTAG: Laith SecondChildCreateTimeTAG: 2012-11-18T07:24:54Z IndexTAG: 3191 TitleTAG: Undamped second-order system: how to get rid of j? For undamped second-order systems we apply Eulers formula `e^(j*t) = cos(t) + j*sin(t)` to the solution of the form `e^(s*t)` where `s = j*w0` or `-j*w0`. But then somehow general solution appears to be of the form `A1*cos(w0*t) + A2*sin(w0*t)`. What I don't get is disappearance of j in this solution since there's still j before `sin(t)` in Eulers formula. The only explanation I can come up with is that A1 and A2 are complex but that doesn't seem to be right.. UserIdTAG: 189680 UserNameTAG: vnd CreateTimeTAG: 2012-11-13T20:14:53Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Here's the approach I learned when taking D.E. You want the solution to be a fundamental set of 2 solutions and guess that y will be in some form $e^{rx}$ where $r=\lambda \pm i\mu$ So $Y_1=A1*e^{(\lambda+i\mu)x}$ and $Y_2=A2*e^{(\lambda-i\mu)x}$ $Y=A1*e^{(\lambda+i\mu)x}+A2*e^{(\lambda-i\mu)x}$ $A1*e^{(\lambda+i\mu)x} = A1*e^{\lambda x}*e^{i\mu x} = A1*e^{\lambda x}[cos(\mu x)+i*sin(\mu x)] = Y_1$ $A2*e^{(\lambda-i\mu)x} = A2*e^{\lambda x}*e^{-i\mu x} = A1*e^{\lambda x}[cos(\mu x)-i*sin(\mu x)] = Y_2$ To simplify things we can substitute $A1=A2=1$ So from here $Y_1 + Y_2 = 2e^{\lambda x}cos(\mu x)$ divide by 2 to get $e^{\lambda x}cos(\mu x)$ There's one solution. The other solution is $Y_1 - Y_2 = 2*i*e^{\lambda x}*sin(\mu x)$ divide by 2i to get $e^{\lambda x}sin(\mu x)$ Now you have the solution $Y = A1*e^{\lambda x}cos(\mu x) + A2*e^{\lambda x}sin(\mu x)$ Which is a similar form. FirstChildUserIdTAG: 213386 FirstChildUserNameTAG: dmascenik FirstChildCreateTimeTAG: 2012-11-13T21:46:59Z SecondChildTAG: Thanks, makes sense but still leaves question: original fundamental set covered much larger class of functions since coefficients were complex and our final set covers only real valued functions so it feels like we cheated somewhere along the way.. I guess for applied field like electronics complex coefficients don't make sense so they are silently discarded. SecondChildUserIdTAG: 189680 SecondChildUserNameTAG: vnd SecondChildCreateTimeTAG: 2012-11-14T08:26:37Z IndexTAG: 3192 TitleTAG: Stuck in H9P1 Hi, I'm still stuck in H9P1.Maybe someone can help me. I've got for iL iL=cos(w0*t), then tried to get iL for t=9. Even when my answer for w0 is correct, my answer for iL is not, tough I've tried w0 and w0*2pi (w0 is in hertz). I'm missing something? UserIdTAG: 369339 UserNameTAG: vargaslen CreateTimeTAG: 2012-11-13T18:14:33Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: If you got the answer for part 1 correct, then you should be able to calculate the current at t = 9- by inspection! FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-13T19:57:45Z FirstChildTAG: Can anybody help me with HW 9 problem 1 part 2 FirstChildUserIdTAG: 536922 FirstChildUserNameTAG: arjshar FirstChildCreateTimeTAG: 2012-11-14T01:02:34Z IndexTAG: 3193 TitleTAG: Help H9P2 B coefficient!!!!!!!!! Hello. I don't know what's the coefficient for B. I'm not sure if it is a differential issue or just R, L C or something, and i'm not sure how to obtain it. Please, could anybody give me a track? Thanks. UserIdTAG: 244225 UserNameTAG: lerimock CreateTimeTAG: 2012-11-13T16:32:44Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Write node equation involving r,l,c. Differentiate it to get a second order differential eqn. compare the coefficients to find A,B,C valkue. FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-11-15T08:35:22Z FirstChildTAG: Basically the key was working out the voltage over the inductor. Everything else then followed FirstChildUserIdTAG: 373309 FirstChildUserNameTAG: njirving FirstChildCreateTimeTAG: 2012-11-16T23:22:14Z IndexTAG: 3194 TitleTAG: Buck converter, or similar? Are these covered anywhere in the text or courseware? I want to build a circuit that can throttle current (about 1A) to a 3W LED. It sounds like inductors and capacitors are key to stepping down the voltage without dissipating as much energy, but I don't understand how [this circuit][1] would work. Rob [1]: http://en.wikipedia.org/w/index.php?title=File:Buck_circuit_diagram.svg&page=1 UserIdTAG: 468623 UserNameTAG: RobNik CreateTimeTAG: 2012-11-13T15:20:26Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: It sounds like you need a constant current source rather than a buck voltage regulator. If you still are interested in the buck regulator, here is an interesting circuit that I have been studying: http://romanblack.com/smps/smps.htm This circuit may work for you: http://romanblack.com/smps/a05.htm FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-13T16:23:52Z SecondChildTAG: My first simple circuit idea was to use a MOSFET as a constant current source, as we studied. But if I start with a 5V PC power supply, and the LED only drops 2.5V, then there is alot of V*I power at the MOSFET or a resistor in series with it, so it dissipates a lot of energy, not through the LED. I'll try to study those circuits you linked but they appear complicated so I may have problems without some stepping stones in 6.002. SecondChildUserIdTAG: 468623 SecondChildUserNameTAG: RobNik SecondChildCreateTimeTAG: 2012-11-13T17:01:23Z FirstChildTAG: From 110 V to 3 Volt, you need a transformer, We won´t see them in the course, but, think of Cell phone charger, it normally provide 110 to 5 V DC conversion. So it reduces and converts from AC to DC. Make sure the rating of the charger is at least 5 Watts. A more practical option is the charger of a laptop, but the output is 19 Volts DC. In both cases the next step is to use a resistor divider or a resistor in series with the LED. If interested in the subject, I will be glad to work with you a more detailed solution. FirstChildUserIdTAG: 403987 FirstChildUserNameTAG: electricon FirstChildCreateTimeTAG: 2012-11-13T17:16:54Z SecondChildTAG: Of course, the usual solution for controlling the LED current is a series resistor, but I believe that in this case the OP is looking for a more efficient method that minimizes I-R losses. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-11-13T17:53:58Z SecondChildTAG: Yes, I can't use a resistive divider -- too much power loss. I'm going to end up powering more than one of these, for an indoor garden. Here's the [LED][1]. And no, I'm not growing marijuana. [1]: http://www.mouser.com/Search/ProductDetail.aspx?R=LZ1-10R200virtualkey62410000virtualkey897-LZ110R200 SecondChildUserIdTAG: 468623 SecondChildUserNameTAG: RobNik SecondChildCreateTimeTAG: 2012-11-13T22:50:28Z IndexTAG: 3195 TitleTAG: S17V3-Fast case I have a doubt regarding the fast case in which RL is shunted with 50ohm, Till now we have encountered with RON of MOSFET in 2K ohm range, hence by considering the same, wont the output at Node B be always high instead of it being inverted, but in the demo it doesnt seem so, maybe RON of MOSFET is quite low?? Can somebody tell the reason for the output to be inverted! Thanks in advance. UserIdTAG: 149428 UserNameTAG: BLogical CreateTimeTAG: 2012-11-13T13:54:13Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3196 TitleTAG: error in lab 11 in lab 11 i entered the values of l & c in the circuit and also in boxes provided in ckt is is giving green while in boxes it is showing red UserIdTAG: 285222 UserNameTAG: dhaval24 CreateTimeTAG: 2012-11-13T09:49:30Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: yeah same problem for me too!!! FirstChildUserIdTAG: 156585 FirstChildUserNameTAG: AdiRanga FirstChildCreateTimeTAG: 2012-11-13T11:17:13Z FirstChildTAG: the values are not graded in the circuit..that green mark will appear even if you dont make any change..probably just an indicator for the circuit being there..:)..but only if you get the right values you will get a green in the boxes for the answers..and these are the ones that are graded.. FirstChildUserIdTAG: 173147 FirstChildUserNameTAG: cruiser_rahit FirstChildCreateTimeTAG: 2012-11-13T11:18:27Z SecondChildTAG: no but it is showing some percentage of work done in progress section SecondChildUserIdTAG: 285222 SecondChildUserNameTAG: dhaval24 SecondChildCreateTimeTAG: 2012-11-13T16:24:56Z IndexTAG: 3197 TitleTAG: H9P2 I really need some help with the first part of this question (a, b, c). I must be missing something easy here but I can't seem to understand the question. The remaining parts I was able to answer. Can anyone help me get started? I sure others could use the help also. Thanks UserIdTAG: 108959 UserNameTAG: cpfaivre CreateTimeTAG: 2012-11-13T02:05:07Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: i = CdVo/dt, VL = Ldi/dt... FirstChildUserIdTAG: 244225 FirstChildUserNameTAG: lerimock FirstChildCreateTimeTAG: 2012-11-13T07:13:05Z FirstChildTAG: All I could say is to establish differential equations by node method and compare the coefficients . FirstChildUserIdTAG: 136959 FirstChildUserNameTAG: christerpher FirstChildCreateTimeTAG: 2012-11-13T10:12:51Z SecondChildTAG: Yes but, for C, the differential equation gives i, not v. SecondChildUserIdTAG: 244225 SecondChildUserNameTAG: lerimock SecondChildCreateTimeTAG: 2012-11-13T12:49:12Z FirstChildTAG: I got it. Thanks for the help. A good night sleep also helps! FirstChildUserIdTAG: 108959 FirstChildUserNameTAG: cpfaivre FirstChildCreateTimeTAG: 2012-11-13T17:35:48Z IndexTAG: 3198 TitleTAG: LED Power Circuit? Hi, Would the circuit below work for driving a high power (3W) LED? Here is the [part info][1]. The LED wants about 1A of current. I'm thinking I can use the variable resistor to set Vgs on the MOSFET so that it allows 1A to flow to the LED. For the 5V supply I have a computer power supply, so I could actually use other wires and get 3, or 12 V. Rob ![enter image description here][2] [1]: http://www.mouser.com/Search/ProductDetail.aspx?R=LZ1-10R200virtualkey62410000virtualkey897-LZ110R200 [2]: https://edxuploads.s3.amazonaws.com/13527602061343626.png UserIdTAG: 468623 UserNameTAG: RobNik CreateTimeTAG: 2012-11-12T22:50:09Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: any information about the MOSFET? FirstChildUserIdTAG: 136959 FirstChildUserNameTAG: christerpher FirstChildCreateTimeTAG: 2012-11-13T10:30:44Z SecondChildTAG: No, not yet. I figured I'd buy one with high enough power rating. But now I want to abandon this circuit and try to use a "buck converter" or something like that will not waste as much energy. SecondChildUserIdTAG: 468623 SecondChildUserNameTAG: RobNik SecondChildCreateTimeTAG: 2012-11-13T15:08:06Z SecondChildTAG: What about an IGBT SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-11-13T17:09:14Z IndexTAG: 3199 TitleTAG: Help with q4 Can anybody explain q4? Thanks! UserIdTAG: 220304 UserNameTAG: sergei_m CreateTimeTAG: 2012-11-12T21:08:29Z VoteTAG: 0 CoursewareTAG: Week 9 / S18E3 Total Solution CommentableIdTAG: S18E3_Total_Solution NumberOfReplyTAG: 1 FirstChildTAG: Taking the derivative of v(t) and setting t=0 will give to a sum with two terms involving a cosine and a sine. Their ratio is ... = tan(x) = sin(x)/cos(x). Then take inverse tangent to find x = arctan(...) FirstChildUserIdTAG: 153760 FirstChildUserNameTAG: Kavka FirstChildCreateTimeTAG: 2012-11-13T05:21:11Z IndexTAG: 3200 TitleTAG: o<>k "Show answer" gives identical answers both for entry 'o' and entry 'k' = 2.99970003 V. This can not be true. It has to be 2.99880... V for parallel MOSFETs. UserIdTAG: 710567 UserNameTAG: oleg_a CreateTimeTAG: 2012-11-12T17:17:15Z VoteTAG: 0 CoursewareTAG: Week 3 / Switch Resister Model Exercise CommentableIdTAG: 6002x_switch_resistor_model_exercise NumberOfReplyTAG: 0 IndexTAG: 3201 TitleTAG: H9P1 I'm I in the wrong way? Hi all, I assumed that the ecuation given for this circuit dynamics, reduces to d2iL(t)/dt2 + iL/LC=0 for t<=9s which gives me s^2 + w0^2=0 as the characteristic ecuation. Is this incorrect? UserIdTAG: 369339 UserNameTAG: vargaslen CreateTimeTAG: 2012-11-12T10:30:37Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I did the same, but I can't work out why my frequency isn't just 1/sqrt(LC)..... FirstChildUserIdTAG: 395588 FirstChildUserNameTAG: GDibb FirstChildCreateTimeTAG: 2012-11-12T14:36:22Z SecondChildTAG: Well, remember that in harmonic/sinusoidal motion, ω = 2πf SecondChildUserIdTAG: 219461 SecondChildUserNameTAG: shuncobra SecondChildCreateTimeTAG: 2012-11-13T12:20:37Z FirstChildTAG: No you are in the right way keep going S1=+jw & S2=-jw FirstChildUserIdTAG: 106816 FirstChildUserNameTAG: Laith FirstChildCreateTimeTAG: 2012-11-12T11:41:02Z SecondChildTAG: yes, that is what I've got!thanks! SecondChildUserIdTAG: 369339 SecondChildUserNameTAG: vargaslen SecondChildCreateTimeTAG: 2012-11-12T16:20:08Z SecondChildTAG: Hi, I'm still stuck.Maybe you can give me some help. I've got for iL iL=cos(w0*t), then tried to get iL for t=9. Even when my answer for w0 is correct, my answer for iL is not, tough I've tried w0 and w0*2pi (w0 is in hertz). I'm missing something? SecondChildUserIdTAG: 369339 SecondChildUserNameTAG: vargaslen SecondChildCreateTimeTAG: 2012-11-13T16:24:41Z IndexTAG: 3202 TitleTAG: H9 Hello Myriam and Skyhawk.. Waiting for your hints on HW9 :) UserIdTAG: 499671 UserNameTAG: maliha266 CreateTimeTAG: 2012-11-12T05:07:04Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: OrinE has already given an excellent hint for H9P1 here: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/509f34ef6d91001f0000000c For H9P2, question 4 is looking for the undamped frequency in Hz. Otherwise, I think the homework set is straightforward. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-12T13:01:17Z IndexTAG: 3203 TitleTAG: Vc transition from low to high v Wont the same thing happen to Vc as in Vb(sinusoid output) when Vc moves from low to high state UserIdTAG: 112682 UserNameTAG: Pratik94 CreateTimeTAG: 2012-11-12T04:43:53Z VoteTAG: 0 CoursewareTAG: Week 9 / Motivation Whats Going On With The Fast Case CommentableIdTAG: 6002x_whats_going_on_with_fast_case NumberOfReplyTAG: 0 IndexTAG: 3204 TitleTAG: To the course Staff: I am sorry to bother you but I have some questions about the Boost Converter lab(lab 9). I have studied that circuit thoroughly. But, I am not sure if I actually can get zero inductor current for any frequency. What exactly am I supposed to achieve? Hope you can explain to me what the final goal is. Thanks UserIdTAG: 265795 UserNameTAG: krishnakm CreateTimeTAG: 2012-11-11T22:01:24Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3205 TitleTAG: Bug? I'm not sure if it's a bug but, won't be the complex current and voltage through the resistor Ir and Vr instead of Il and Vl? UserIdTAG: 414462 UserNameTAG: Pablo_C CreateTimeTAG: 2012-11-11T17:59:32Z VoteTAG: 0 CoursewareTAG: Week 10 / Element impedance models CommentableIdTAG: 6002x_Element_impedance_models NumberOfReplyTAG: 1 FirstChildTAG: I guess its a bug too. FirstChildUserIdTAG: 204745 FirstChildUserNameTAG: allwynmendes FirstChildCreateTimeTAG: 2012-12-09T14:06:53Z IndexTAG: 3206 TitleTAG: H8P2 last part I am stuck in this one..can anyone help? UserIdTAG: 173147 UserNameTAG: cruiser_rahit CreateTimeTAG: 2012-11-11T17:52:18Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: V(initial/4)=V(2inital/3)*e^-t/RC i use this equation for this question 3rd part but git wrong answer can you tell me my mistake thanks FirstChildUserIdTAG: 136490 FirstChildUserNameTAG: Ali_PU FirstChildCreateTimeTAG: 2012-11-11T17:53:21Z SecondChildTAG: i am still solving question no 1 SecondChildUserIdTAG: 342135 SecondChildUserNameTAG: vikash902 SecondChildCreateTimeTAG: 2012-11-11T17:59:40Z SecondChildTAG: for 3rd part..the half time is given..so after half time it will be Vinitial/2..so time taken for this to again reduce to half (Vinitial/4)..is double the half time? SecondChildUserIdTAG: 173147 SecondChildUserNameTAG: cruiser_rahit SecondChildCreateTimeTAG: 2012-11-11T18:58:20Z SecondChildTAG: V(initial)/4=V(inital)*e^(-t/RC)..R from part 2..C from question..u can find t..u get in seconds..convert to hours(divide by (60*60)).. SecondChildUserIdTAG: 173147 SecondChildUserNameTAG: cruiser_rahit SecondChildCreateTimeTAG: 2012-11-11T19:33:48Z IndexTAG: 3207 TitleTAG: H8P3 Part3 confusion capacitor discharge equation id V(initial/4)=V(2inital/3)*e^-t/RC is use this equation to calculate t but no green tick where i am wrong plz guide me.. thanks UserIdTAG: 136490 UserNameTAG: Ali_PU CreateTimeTAG: 2012-11-11T16:34:45Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: right side..it will start at V(initial)..not V(2inital/3)... FirstChildUserIdTAG: 173147 FirstChildUserNameTAG: cruiser_rahit FirstChildCreateTimeTAG: 2012-11-11T19:31:39Z SecondChildTAG: and its H8P2 right? SecondChildUserIdTAG: 173147 SecondChildUserNameTAG: cruiser_rahit SecondChildCreateTimeTAG: 2012-11-11T19:53:33Z IndexTAG: 3208 TitleTAG: H8P3 memory confusion I was able to eventually get my green check marks on all these questions, but I'm not sure that answers are truly correct. For example, the first question leads me to calculate the 190 T-ohm parasitic resistance, so that I get proper RC time. But it seems to me that there is a much easier path to discharge the capacitor through the "off" Q3 transistor, which has a million times more conductance than the parasitic resistance we're asked to calculate. Similarly, in the third question, I think the capacitor would eventually charge up to 4.9993V through the "off" Q3. Do I have things wrong?? Would appreciate comments. Tim p.s. the course is wonderful! UserIdTAG: 410846 UserNameTAG: tbrunner CreateTimeTAG: 2012-11-11T16:31:42Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3209 TitleTAG: H8P3: MEMORY H8P3 (Last question)...please guide me how to solve it. I have done all the remaining assignment but the last question have put me into deep trouble. UserIdTAG: 64618 UserNameTAG: ashfaq2419 CreateTimeTAG: 2012-11-11T16:22:40Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Keyword is H8P3 [first link][1] [second link][2] and etc [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/509e68568e407d1f0000001b [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/509a18aedb6fed1f0000000d FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-11-11T16:32:49Z SecondChildTAG: dear i am finding Vth using Vs.(Ron)/(Ron+Rpu) and it is 2.66v. But there is a negative value on left hand side and when i am going to take ln of it, it is undefined. whats wrong i am doing? SecondChildUserIdTAG: 64618 SecondChildUserNameTAG: ashfaq2419 SecondChildCreateTimeTAG: 2012-11-11T16:52:37Z SecondChildTAG: Calculate Vth (from your comments it looks correct), and Rth. Draw the VCgsQ2(t) x t graph to help you find your equation in terms of VIL, VTH and VOH and e^(-t/RthC). If you are getting an undeffined ln, there should be an error in your equation. Good luck SecondChildUserIdTAG: 288174 SecondChildUserNameTAG: matiasgrodriguez SecondChildCreateTimeTAG: 2012-11-11T18:13:09Z IndexTAG: 3210 TitleTAG: ans to first ques at t=-1ms Vc=3v. the 10 mA current and 100ohm resistance can change maximum of 1v, which is less than 3 volt, so capacitor will discharge with equation, Vc=1-(1-3)e^(-t/R*C) pot t=1ms. then Vc=1.7357V UserIdTAG: 309863 UserNameTAG: jithugeorge CreateTimeTAG: 2012-11-11T15:23:43Z VoteTAG: 0 CoursewareTAG: Week 8 / Initial Conditions CommentableIdTAG: 6002x_Initial_Conditions NumberOfReplyTAG: 1 FirstChildTAG: the voltage across the capacitor is 3v, the maximum voltage that can be across R is 1v , how that ? FirstChildUserIdTAG: 236136 FirstChildUserNameTAG: Hamid-ch FirstChildCreateTimeTAG: 2012-11-12T01:30:07Z SecondChildTAG: I believe that we assume that there's 3v charged capacitor and we connect I=I0+delta(t) at t=-1.0 ms . Just after t=-1.0 + , 1A current source is connected to R and capacitor in parallel. So cap tries to discharge from 3v to 1v(You can just use thevenin equivalent I0(10mA) x R (100 ohm) = 1v Voltage soruce, which is connected to R and Cap in series). And suddenly , there's 2v dropout at t=0. After impulse is gone, the circuit just like 1A current source is connected to R and cap in parallel. So the cap tries to change from {1+2*e^(-1)} v to 1v . I just attach an image file, which shows what I believe. Because it's not easy to write formula, I do not write down the formula for -0.26424 to 1v transition.![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13533741681343601.jpg SecondChildUserIdTAG: 10762 SecondChildUserNameTAG: Junior SecondChildCreateTimeTAG: 2012-11-20T01:16:38Z IndexTAG: 3211 TitleTAG: For staff: problem with Course Progress bars Hi everyone, congratulations for the amazing course!! There is a slight problem with the total bar in the Course Progress. My total should have been 57% instead of 56% (the numbers are not rounded). thanks in advance UserIdTAG: 206669 UserNameTAG: dimitrios66 CreateTimeTAG: 2012-11-11T14:37:28Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3212 TitleTAG: Problem with Chrome Chrome works smoothly in my iPad for most of the edx website, but for some reason I keep getting this message when I try to open the lecture notes from i.e. the link below one of the videos. ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/1352636743211117.jpg There is no way to ignore this warning message and get the notes, uless I use Safari, that seems to be a bit less picky. Not a big problem, but it could probably be easily solved. (I don't think the desktop version of Chrome shows this problem - I guess you can force the browser to proceed anyway, but I haven't check it.) Thanks and best, Jordi UserIdTAG: 366320 UserNameTAG: Galli CreateTimeTAG: 2012-11-11T12:33:13Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Troubleshooting NumberOfReplyTAG: 0 IndexTAG: 3213 TitleTAG: LAB 10 For finding the value of L and C i used fc=1/(2piRC) to find C what is wrong in the above expression UserIdTAG: 224318 UserNameTAG: ashwin12312 CreateTimeTAG: 2012-11-11T10:59:02Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi **ashwin12312**: Read the Question Carefully, - *Break frequency in radians/sec for RC circuit (formula using R and C)* Above phrase demanding algebraic expression with the exemption of term **radians**. I hope it will help you. > **Regards:** asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-11-11T11:17:02Z IndexTAG: 3214 TitleTAG: Around 1:20 "So make sure that you realize that [the initial voltage] has changed because the capacitance here has also changed" Isn't it instead because the resistors are in a voltage divider configuration ? UserIdTAG: 351871 UserNameTAG: Lamarque CreateTimeTAG: 2012-11-11T10:57:40Z VoteTAG: 0 CoursewareTAG: Week 8 / Exercise 10.19 CommentableIdTAG: 6002x_Ex10_19 NumberOfReplyTAG: 0 IndexTAG: 3215 TitleTAG: Lab8 i m using eqn vc=v0(1-e^(-t/rc)) for capacitor charging v0 will be vstep i.e 1v.is my approach correct.help me UserIdTAG: 611666 UserNameTAG: Shaminder420 CreateTimeTAG: 2012-11-11T10:18:06Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: yep and use the required values from the graph and you'll get the value for C. FirstChildUserIdTAG: 8897 FirstChildUserNameTAG: sam1202 FirstChildCreateTimeTAG: 2012-11-11T15:50:00Z IndexTAG: 3216 TitleTAG: h8p2 last part Last question as: "What dose, in milligrams, will be required at that time to get the concentration back to 2/3 of the initial concentration?"..how to find it? UserIdTAG: 114864 UserNameTAG: Vikaash CreateTimeTAG: 2012-11-11T06:15:00Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi, [H8P1,H8P2, H8P3 Hints][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/509e68568e407d1f0000001b > **Regards:** asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-11-11T08:59:01Z SecondChildTAG: i have already gone through it but couldn't get it.. SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-11-11T10:01:26Z SecondChildTAG: It took me a long time and then it struck me that the answer/solution is very simple. You know what dose your at (1/4) and where you need to get to (2/3). You already have a simple formula for Q.... SecondChildUserIdTAG: 244115 SecondChildUserNameTAG: Pedro1969 SecondChildCreateTimeTAG: 2012-11-11T12:45:42Z IndexTAG: 3217 TitleTAG: H9P2 question d Hi there, I have a problem with this question. I assume that the natural frequency of the circuit was sqrt(1/(L*C)) (according to the characteristic polynomial), but it says it's a wrong answer. I assume that expression of the natural frequency for the questions e,f,g and, surprisingly, i had the check marks. So, what am i doing wrong? UserIdTAG: 464522 UserNameTAG: Josbrach CreateTimeTAG: 2012-11-11T03:47:38Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: They want the answer in Hz. FirstChildUserIdTAG: 141000 FirstChildUserNameTAG: OrinE FirstChildCreateTimeTAG: 2012-11-11T06:28:08Z SecondChildTAG: Ok, i have the check mark, thanks for the help!! SecondChildUserIdTAG: 464522 SecondChildUserNameTAG: Josbrach SecondChildCreateTimeTAG: 2012-11-11T16:54:04Z IndexTAG: 3218 TitleTAG: says short instead of open at 1:02 The current source gets replaced by an open circuit and he draws an open circuit, but he says short circuit. UserIdTAG: 339870 UserNameTAG: rjlasota CreateTimeTAG: 2012-11-11T02:01:44Z VoteTAG: 0 CoursewareTAG: Week 8 / Exercise 10.1 CommentableIdTAG: 6002x_Ex10_1 NumberOfReplyTAG: 0 IndexTAG: 3219 TitleTAG: H8P3Q4 Because of your wonderful tips, I've solved everything for week 8 -- except H8P3Q4. 3.5=.5*(1-e^(-t/RC)) R is Ron//Rpu(from Q1) + Ron(from Q2)+ Ron(from Q3) C = 3.5fF Spent a long time on this to no avail, so need real help. Charging problems are harder for me than discharging ones UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-11-10T23:21:05Z VoteTAG: 0 CoursewareTAG: Week 8 / Digital Memory Array CommentableIdTAG: 6002x_Digital_Memory_Array NumberOfReplyTAG: 2 FirstChildTAG: The ec. of charging is not correct, the value would reach for infinity t would be 5 V; Calculate the time it takes to reach that 0.5 and 3.5 and is mecesita for the difference. , FirstChildUserIdTAG: 95638 FirstChildUserNameTAG: cperezf FirstChildCreateTimeTAG: 2012-11-10T23:53:45Z FirstChildTAG: Hi, [H8P1,H8P2, H8P3 Hints][1] [1]: http://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/509e68568e407d1f0000001b > **Regards:** asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-11-11T08:55:08Z IndexTAG: 3220 TitleTAG: H8P1 please help me in Vr(0-) UserIdTAG: 155008 UserNameTAG: sohailahmed CreateTimeTAG: 2012-11-10T19:45:01Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: This is prior to the impulse, so the initial conditions apply (Read the problem statement carefully!), so Vr(0-) = R1*i(0-) FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-10T20:26:33Z FirstChildTAG: Hi, [H8P1,Hints][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/509e68568e407d1f0000001b > **Regards:** asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-11-11T08:52:56Z IndexTAG: 3221 TitleTAG: Can someone tell me what I'm neglecting? strange answers to me....how can a cap in decaying mode go from -0.2v to + 0.5 v...can anyone tell me what I'm missing? UserIdTAG: 256543 UserNameTAG: sidney23 CreateTimeTAG: 2012-11-10T18:41:08Z VoteTAG: 0 CoursewareTAG: Week 8 / Initial Conditions CommentableIdTAG: 6002x_Initial_Conditions NumberOfReplyTAG: 1 FirstChildTAG: You are probably missing the fact that the current source is supplying current $I_0$ at all times. Replace the current source and R with their Thevenin equivalent and you'll see that the capacitor is actually charging. FirstChildUserIdTAG: 141000 FirstChildUserNameTAG: OrinE FirstChildCreateTimeTAG: 2012-11-10T19:47:41Z IndexTAG: 3222 TitleTAG: About v_s Could we take for granted that dv_s/dt=0? Nothing is said about how v_s is. The result in terms of the natural frequencies would be the same (we only need the homogeneous equation), but it would make quite a difference if we want to calculate currents and voltages. I think we should keep dv_s/dt in the equation not to create any confusion. Take care! UserIdTAG: 341020 UserNameTAG: franjescribano CreateTimeTAG: 2012-11-10T08:05:02Z VoteTAG: 0 CoursewareTAG: Week 9 / A&L Problem 12.8 CommentableIdTAG: 6002x_AL_P_12_8 NumberOfReplyTAG: 0 IndexTAG: 3223 TitleTAG: capacitor voltage I really couldn't understand if the circuit is decaying then there should be an initial condition on the circuit. initial conditions are zero ,then he equates the voltage across the r2 to be that across capacitor then plugs it into the circuit,couldn't really get as to how? UserIdTAG: 133068 UserNameTAG: swarn99 CreateTimeTAG: 2012-11-09T14:28:20Z VoteTAG: 0 CoursewareTAG: Week 8 / Example 10.16 CommentableIdTAG: 6002x_Ex10_16 NumberOfReplyTAG: 0 IndexTAG: 3224 TitleTAG: Taylor expansion constant term Hi, I was just going over the incremental analysis lecture when I came across a doubt which I could not resolve. The Taylor Series expansion of f(x) is $f(x) = \sum_{n=0} ^ {\infty } \frac {f^{(n)}(a)}{n!}(x-a)^{n}$ However when we are doing linear analysis we always neglect the $a$ term in $(x-a)^n$ $i_D\approx f(V_D)+\frac{df(v_D)}{dv_D}|_{v_D=V_D}\cdot v_d$ What happened to the constant? Shouldn't it be: $i_D\approx f(V_D)+\frac{df(v_D)}{dv_D}|_{v_D=V_D}\cdot (v_d-V_D)$ Any help would be much appreciated. Thank You in advance UserIdTAG: 85490 UserNameTAG: yashsavani CreateTimeTAG: 2012-11-09T11:07:16Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: The constant part is the operating point and the small signal is the variation around this operating point: if vD (small v, capital D) is the total signal and VD (capital V, capital D) is the operating point, the small signal vd (small v, small d) is vd = vD - VD. That's why it seems we neglected the constant part, but we don't. FirstChildUserIdTAG: 376877 FirstChildUserNameTAG: AndBre FirstChildCreateTimeTAG: 2012-11-09T12:14:33Z IndexTAG: 3225 TitleTAG: Lab 7 Part1, problem Hello I get the right result, when I double my value. Where is my mistake? *Answer removed* UserIdTAG: 389333 UserNameTAG: juergen CreateTimeTAG: 2012-11-09T10:57:33Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: What is the integration interval you are using? FirstChildUserIdTAG: 376877 FirstChildUserNameTAG: AndBre FirstChildCreateTimeTAG: 2012-11-09T11:27:24Z SecondChildTAG: 0... 0.0005 SecondChildUserIdTAG: 389333 SecondChildUserNameTAG: juergen SecondChildCreateTimeTAG: 2012-11-09T12:53:00Z SecondChildTAG: Have you solved the integral manually? Using wolframalpha with the correct integration interval gives the correct result. Even manually it's very simple if you know how to integrate functions. SecondChildUserIdTAG: 376877 SecondChildUserNameTAG: AndBre SecondChildCreateTimeTAG: 2012-11-09T14:16:47Z SecondChildTAG: -(cos(2000*pi*t))/(2000*pi*C) SecondChildUserIdTAG: 389333 SecondChildUserNameTAG: juergen SecondChildCreateTimeTAG: 2012-11-09T16:08:29Z SecondChildTAG: Definite integrals have contributions from both the upper and lower limits. The lower limit is t = 0 and cos is equal to 1 at that point. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-11-09T16:25:25Z SecondChildTAG: The first value for the integration interval is 0 and cos(0) is not 0... You know how to solve a DEFINITE integral, don't you? SecondChildUserIdTAG: 376877 SecondChildUserNameTAG: AndBre SecondChildCreateTimeTAG: 2012-11-09T16:31:48Z SecondChildTAG: I remember. Thanks for the hint SecondChildUserIdTAG: 389333 SecondChildUserNameTAG: juergen SecondChildCreateTimeTAG: 2012-11-09T17:52:52Z FirstChildTAG: you tried to put the result with 6 digits of precision? Note that there is a point to be considered. FirstChildUserIdTAG: 361823 FirstChildUserNameTAG: EliasOak FirstChildCreateTimeTAG: 2012-11-09T11:06:07Z FirstChildTAG: Hi juergen, Can I help you? FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-09T18:32:04Z SecondChildTAG: Thanks a lot. I got it. Next questions will be come. SecondChildUserIdTAG: 389333 SecondChildUserNameTAG: juergen SecondChildCreateTimeTAG: 2012-11-09T20:02:14Z SecondChildTAG: Well done! ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-11-10T17:23:27Z FirstChildTAG: Hi, Check this, [Lab7 Hint][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5091dd5d3fc09f2b0000018a > **Regards:** asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-11-10T14:50:43Z IndexTAG: 3226 TitleTAG: Lab 2 Part 1 When i computed the voltage, my value is half the amount. Where is my mistake? *Please do not post answers in the forum.* UserIdTAG: 389333 UserNameTAG: juergen CreateTimeTAG: 2012-11-09T10:16:47Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3227 TitleTAG: H9P1 Q4 Problem I'm pretty sure I'm writing the correct answer to this question, but for some reason it doesn't get accepted as the correct answer... can someone please double check? > At the time just before the impulse happens what is the current iL(1.0−), in Amperes, through the inductor? By the way, I'm getting Q5 correct, and I've done all sorts of check like adding up the energy for C and L and getting the required energy prior to the impulse. UserIdTAG: 269325 UserNameTAG: Fed_Ang CreateTimeTAG: 2012-11-09T08:31:02Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: okay, I figured it out... they approximated the value FirstChildUserIdTAG: 269325 FirstChildUserNameTAG: Fed_Ang FirstChildCreateTimeTAG: 2012-11-09T10:36:07Z FirstChildTAG: Hi, Check this, [H9P1 Hint][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5098b2a1cd5fb82300000036 > **Regards:** asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-11-10T14:48:49Z IndexTAG: 3228 TitleTAG: mistake found the equation of the output of capacitor for a falling input is wrong for the blue pulse. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-11-09T06:20:24Z VoteTAG: 0 CoursewareTAG: Week 8 / Response To A Pulse As Pulse Gets Narrower Continued CommentableIdTAG: 6002x_Response_To_A_Pulse_As_Pulse_Gets_Narrower_Continued NumberOfReplyTAG: 0 IndexTAG: 3229 TitleTAG: krishnan I do not understand the step 2A - precisely the value of Vh = A * e power st UserIdTAG: 101278 UserNameTAG: Krishnan CreateTimeTAG: 2012-11-08T15:54:32Z VoteTAG: 0 CoursewareTAG: Week 9 / Homogeneous Solution and Characteristic Equation CommentableIdTAG: 6002x_homogeneous_solution_and_characteristic_equation NumberOfReplyTAG: 1 FirstChildTAG: if the above expression is selected,then the derivative is same as the given expression,hence the above expression is selected FirstChildUserIdTAG: 374351 FirstChildUserNameTAG: vindhya123 FirstChildCreateTimeTAG: 2012-11-15T17:04:41Z IndexTAG: 3230 TitleTAG: proctored exam what is proctored exam? which courses will be having this exam? can anyone tell me in details? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-11-08T15:22:18Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: hi, [Check This][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/508446a8d210431f0000012c > **Regards:** asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-11-08T16:24:07Z IndexTAG: 3231 TitleTAG: H8P2 question Last question sounds as: "What dose, in milligrams, will be required at that time to get the concentration back to 2/3 of the initial concentration?" What is meaning for "at that time" - as time calculated before ( in the previous quuestion)? Thanks alot! плиз UserIdTAG: 205311 UserNameTAG: Sergtronix CreateTimeTAG: 2012-11-08T15:15:58Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: It means a time,when concentration is 1/4 of initial concentration. FirstChildUserIdTAG: 367431 FirstChildUserNameTAG: VictorFedosenkov FirstChildCreateTimeTAG: 2012-11-08T16:48:02Z SecondChildTAG: Q from 1/4 to 2/3 SecondChildUserIdTAG: 297655 SecondChildUserNameTAG: Aljoska SecondChildCreateTimeTAG: 2012-11-08T17:31:09Z SecondChildTAG: Ok, I got time T alowed to keep desired concentration between 1/4 and 2/3 from the initial value. Next I should define **I** (models the rate of injection of the drug into the body) to meet requirement "to get the concentration back to 2/3 of the initial concentration". By the other words, I should get "I" using time T and 1/4, 2/3 values too. Is this correct ? SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-11-08T18:06:56Z SecondChildTAG: I calculated it by first calculating the change in concentration and placing the value in the standard equation of the capacitor (that defines charge on a capacitor for a voltage input) SecondChildUserIdTAG: 315681 SecondChildUserNameTAG: nahuja1 SecondChildCreateTimeTAG: 2012-11-10T13:12:27Z IndexTAG: 3232 TitleTAG: h8p1 hi friends!!! please help me with h8p1 b,c,d,e parts UserIdTAG: 108929 UserNameTAG: namit CreateTimeTAG: 2012-11-08T13:47:23Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: me too FirstChildUserIdTAG: 37708 FirstChildUserNameTAG: Jband FirstChildCreateTimeTAG: 2012-11-08T18:48:55Z FirstChildTAG: careful thinking is needed. FirstChildUserIdTAG: 4463 FirstChildUserNameTAG: pmj FirstChildCreateTimeTAG: 2012-11-08T21:17:26Z FirstChildTAG: Seek help from Textbook. FirstChildUserIdTAG: 477713 FirstChildUserNameTAG: ikm104 FirstChildCreateTimeTAG: 2012-11-09T07:09:23Z IndexTAG: 3233 TitleTAG: Placement of the parasitic inductance Why do stuck the inductance between the drain of Q1 and the gate of Q2? I think it should be stuck in the outer parts of the loop, where the length of a wire is large. UserIdTAG: 213452 UserNameTAG: JoJosida CreateTimeTAG: 2012-11-08T04:12:52Z VoteTAG: 0 CoursewareTAG: Week 9 / Whats Going On With The Fast Case Continued CommentableIdTAG: 6002x_whats_going_on_with_fast_case_continued NumberOfReplyTAG: 1 FirstChildTAG: It doesn't really matter where the inductor goes. The KVL equation is going to be the same wherever you put it - see slide 9 of the annotated lecture slide handout for the Damped Second-Order Systems lecture sequence. You should look at the Walter Lewin's lecture in the week 9 tutorials for another point of view. FirstChildUserIdTAG: 141000 FirstChildUserNameTAG: OrinE FirstChildCreateTimeTAG: 2012-11-08T21:03:38Z IndexTAG: 3234 TitleTAG: liya cud anyone just help in plotting the graphs of input current and output capacitor voltage?? UserIdTAG: 107674 UserNameTAG: liya CreateTimeTAG: 2012-11-08T01:50:36Z VoteTAG: 0 CoursewareTAG: Week 8 / Initial Conditions CommentableIdTAG: 6002x_Initial_Conditions NumberOfReplyTAG: 1 FirstChildTAG: There are 2 output voltages, one is the function before the impulse and another is function after impulse. **Before:** V = VI + (VO - VI) e ^-(t/tau) Here the initial voltage is 3V, Final Voltage is 1V given by I0 = 10mA (initial condition) So, V function before impulse is *V = 1 + 2 e^-(t/tau)* **After:** because of the impulse the voltage will jump to -2V (Q/C) from t=0−, therefore t=0+ is 1.7357-2 = -0.2643V, it will now increase from -0.2643V at t=0+ to 1V, due to I0 =10mA (initial condition), V function after impulse is *V = 1 - 1.26424 e^-(t/tau)* where tau = R*C FirstChildUserIdTAG: 544644 FirstChildUserNameTAG: Ethanaung FirstChildCreateTimeTAG: 2012-11-13T02:41:44Z IndexTAG: 3235 TitleTAG: H9P2 ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13523355141343695.bmp what is mean by o in eq. UserIdTAG: 155008 UserNameTAG: sohailahmed CreateTimeTAG: 2012-11-08T00:47:19Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: It should be ... $B.\displaystyle\frac{dv_o}{dt}$ ... FirstChildUserIdTAG: 141000 FirstChildUserNameTAG: OrinE FirstChildCreateTimeTAG: 2012-11-08T01:36:42Z SecondChildTAG: thanks. fixing. SecondChildUserIdTAG: 96452 SecondChildUserNameTAG: Lyla SecondChildCreateTimeTAG: 2012-11-08T13:15:49Z SecondChildTAG: thanks SecondChildUserIdTAG: 155008 SecondChildUserNameTAG: sohailahmed SecondChildCreateTimeTAG: 2012-11-08T13:23:40Z IndexTAG: 3236 TitleTAG: S12E5: Solving the differential equation I am having a real hard time with solving the differential equation for the exercise S12E5. So I hope somebody can help me out with this ;) This is what I have got so far. We are solving the problem for this circuit: ![Circuit we need to solve][1] If I attack this with the node method, I get this equation: $\cfrac {v_c-V_s}{R} + \cfrac {v_c}{R_s} + i_c = 0$ If I fiddle arround with the equation, I get: $\cfrac {v_c-V_s}{R} + \cfrac {v_c}{R_s} + C\cdot \cfrac {dv_c}{dt} = 0$ $v_c \cdot (R_s+R) + R\cdot R_s\cdot C\cdot \cfrac {dv_c}{dt} = R_s\cdot V_s$ This is as far as I got for now. Does somebody have a good tip on how to continue to solve the differential equation? [1]: https://edxuploads.s3.amazonaws.com/13523289981343634.bmp UserIdTAG: 294766 UserNameTAG: dejanst CreateTimeTAG: 2012-11-07T23:07:58Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Divide the initial equation by $dv_c$, then move $\cfrac{C}{dt}$ to the right side, make common denominator in the left side, you will get the following: $\cfrac{( v_c -V_s) \cdot R_S + v_c \cdot R}{R \cdot R_S \cdot dv_c} = -\cfrac{C}{dt}$ Then invert both sides so you will have $dv_c$ in numerator on the left side and $dt_c$ in numerator on the right side. Now integrate left side by $dv_c$ and right side by $dt_c$. Don't forget about constant value, which defines initial conditions. FirstChildUserIdTAG: 363868 FirstChildUserNameTAG: vslapik FirstChildCreateTimeTAG: 2012-11-07T23:56:15Z FirstChildTAG: It's much easier if you replace $V_s$, R and $R_s$ with their Thevenin equivalent... FirstChildUserIdTAG: 141000 FirstChildUserNameTAG: OrinE FirstChildCreateTimeTAG: 2012-11-08T03:45:24Z IndexTAG: 3237 TitleTAG: S14V3 Confusion with the intuitive approach to RC circuits In S14V3 it is said that in RC circuits the relationship between $t$ and $v_C$ can either be of the form $(1-e^\frac{-t}{\tau}$) or the form $e^\frac{-t}{\tau}$ and this results in either the equation $v_C=V_0+(V_I-V_0)(1-e^\frac{-t}{\tau})$ or the equation $v_C=V_I+(V_0-V_I)\cdot e^\frac{-t}{\tau}$. But these equations are identical. The second equation can be obtained by solving the first equation. I think it would be much clearer if said that either $V_0$ or $V_I$ could be larger depending on whether the capacitor is charging or discharging, which results in different forms of the $t-v_C(t)$ curves. UserIdTAG: 252722 UserNameTAG: vaboro CreateTimeTAG: 2012-11-07T22:07:39Z VoteTAG: 0 CoursewareTAG: Week 7 / An Intuitive Approach Continued CommentableIdTAG: 6002x_An_Intuitive_Approach_Continued NumberOfReplyTAG: 0 IndexTAG: 3238 TitleTAG: Inverter Delays For the inverter delay between two inverters denoted as "B" in the lectures is the delay of it caused from the mosfet of the first inverter? Or is it casued from the mosfet it is going into? Thanks :D UserIdTAG: 219780 UserNameTAG: ReconIII CreateTimeTAG: 2012-11-07T16:19:44Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3239 TitleTAG: S20E4 Question c asks 'What is the approximate form of the magnitude of this ratio for large ω?' Any hint in how to find that? What approximation can I use when w->inf? UserIdTAG: 39980 UserNameTAG: MarinoJr CreateTimeTAG: 2012-11-07T16:10:23Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I realized. You must consider just the term -w^2*L*C in denominator. FirstChildUserIdTAG: 39980 FirstChildUserNameTAG: MarinoJr FirstChildCreateTimeTAG: 2012-11-07T16:36:16Z IndexTAG: 3240 TitleTAG: problem downloading week 8 videos is any else besides me facing this problem of downloading the videos. my net speed is quite fine but the videos taking so long to get downloaded and ultimatly the error occurs and the video is still not downloaded. I have also tried switching from firefox to google chrome, still no success.... could some help me with this.? thanx UserIdTAG: 117319 UserNameTAG: iqramalik CreateTimeTAG: 2012-11-07T16:03:16Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3241 TitleTAG: problem with watching the videos the youtube videos aren't working, what should i do? they were working until three days before.... i have no idea what is going on UserIdTAG: 413613 UserNameTAG: shaliesh CreateTimeTAG: 2012-11-07T15:38:56Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3242 TitleTAG: Lab 8 i have done almost all question in lab 8 but couldn't find value of C and L. i make equation from graph for Vc and putting the value of Vc and corresponding time in the equation, i am trying to find out C..but that doesn't works..plz help UserIdTAG: 114864 UserNameTAG: Vikaash CreateTimeTAG: 2012-11-07T11:09:25Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 6 FirstChildTAG: You should use the equation Vc(t) = Vc(0)e^(-t/(R*C))...i got stuck with that one too.Hope this helps FirstChildUserIdTAG: 143823 FirstChildUserNameTAG: NathanNadeson FirstChildCreateTimeTAG: 2012-11-07T16:02:28Z SecondChildTAG: can u tell the eqn for inducator voltage SecondChildUserIdTAG: 611666 SecondChildUserNameTAG: Shaminder420 SecondChildCreateTimeTAG: 2012-11-11T10:51:22Z FirstChildTAG: Hi, [Check This][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5099c01db386582b00000005 > **Regards:** asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-11-08T16:13:27Z FirstChildTAG: Hola Vikaash could you please help me with the part H8P1 IMPULSE FirstChildUserIdTAG: 37708 FirstChildUserNameTAG: Jband FirstChildCreateTimeTAG: 2012-11-08T18:38:06Z FirstChildTAG: Try using [WolframAlpha][1]. Plug in something like the equation given above, and append " solve for C". That should give you an equation you can work with to find your numbers. Similarly with L. Also, don't forget there are two forms: "schick" and "shook". ;-) [1]: http://www.wolframalpha.com FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-11-09T00:32:25Z FirstChildTAG: Yes, you can solve all this using math... or use the sandbox with a bit of intuition. Both methods worked well for me. FirstChildUserIdTAG: 114881 FirstChildUserNameTAG: xvink FirstChildCreateTimeTAG: 2012-11-10T16:47:37Z FirstChildTAG: finally did it..i was just making some wrong interpretation which i had corrected..thax to all for responding my question. FirstChildUserIdTAG: 114864 FirstChildUserNameTAG: Vikaash FirstChildCreateTimeTAG: 2012-11-11T10:32:19Z IndexTAG: 3243 TitleTAG: to staff link given in lab8 for chapter 10 is not working. UserIdTAG: 114864 UserNameTAG: Vikaash CreateTimeTAG: 2012-11-07T10:17:00Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I notified the staff. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-11-07T11:40:41Z IndexTAG: 3244 TitleTAG: capacitor voltage at 2:29 min, vc=v2..why? UserIdTAG: 114864 UserNameTAG: Vikaash CreateTimeTAG: 2012-11-07T07:45:17Z VoteTAG: 0 CoursewareTAG: Week 8 / Example 10.16 CommentableIdTAG: 6002x_Ex10_16 NumberOfReplyTAG: 1 FirstChildTAG: In an RC circuit the voltages across the elements (resistor or capacitor doesn't matter) can be exponential functions going down or going up (e^-t/tau or 1-e^-t/tau). Once you have figured out the initial and final voltage for a resistor, you can apply the differential equation solution found for the capacitor voltage. FirstChildUserIdTAG: 376877 FirstChildUserNameTAG: AndBre FirstChildCreateTimeTAG: 2012-11-07T16:39:28Z IndexTAG: 3245 TitleTAG: Staff [bugs in Lab 10] AC in last quaqtion not worked ? UserIdTAG: 331441 UserNameTAG: abdessamade CreateTimeTAG: 2012-11-06T19:59:39Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Should be fixed. Please let us know if it is not. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-11-07T01:49:01Z SecondChildTAG: is work thnkx lyla SecondChildUserIdTAG: 331441 SecondChildUserNameTAG: abdessamade SecondChildCreateTimeTAG: 2012-11-08T14:50:01Z IndexTAG: 3246 TitleTAG: lab10 is it broken i cant double click to change the numbers :S UserIdTAG: 436749 UserNameTAG: waynebrown CreateTimeTAG: 2012-11-06T18:55:53Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: i have also problem in lab FirstChildUserIdTAG: 72786 FirstChildUserNameTAG: Bhavesh FirstChildCreateTimeTAG: 2012-11-06T19:16:22Z SecondChildTAG: email to bugs@edx.org SecondChildUserIdTAG: 70519 SecondChildUserNameTAG: Fipe SecondChildCreateTimeTAG: 2012-11-06T20:06:19Z FirstChildTAG: should be fixed. Thanks for the report. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-11-07T01:49:54Z IndexTAG: 3247 TitleTAG: Lab 10 technical issue The circuit builder will not let me edit the properties of the lumped elements or let me run an AC analysis. UserIdTAG: 331664 UserNameTAG: dwmnctrh3 CreateTimeTAG: 2012-11-06T17:50:29Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: why is sand box not responding to any input ? FirstChildUserIdTAG: 526952 FirstChildUserNameTAG: DEVANKAR FirstChildCreateTimeTAG: 2012-11-06T17:58:23Z SecondChildTAG: the same problem SecondChildUserIdTAG: 189282 SecondChildUserNameTAG: akopyan SecondChildCreateTimeTAG: 2012-11-06T18:16:13Z FirstChildTAG: same problem here FirstChildUserIdTAG: 72786 FirstChildUserNameTAG: Bhavesh FirstChildCreateTimeTAG: 2012-11-06T19:23:59Z FirstChildTAG: Thanks for the reports. This should be fixed now. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-11-07T01:50:14Z IndexTAG: 3248 TitleTAG: LAB 7 task 2: Isn't the Average current equal to peak current devided by period time Average current supplied by bridge rectifier, in amps: Isn't the Ave= peak I / period time The factors above should 2.0(Amp), 8.3(ms) respectively, as measured from the plot. but this number is not the right one. why? UserIdTAG: 128276 UserNameTAG: ChengBin CreateTimeTAG: 2012-11-06T15:59:22Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3249 TitleTAG: Lab 10 last problem Are you sure that everything ok with last problem in Lab10? I've made a scheme which is seemly correct, but it doesn't work (the magnitude graph the same as on Figure 3 and break frequencies 1kHz and 1MHz) UserIdTAG: 189282 UserNameTAG: akopyan CreateTimeTAG: 2012-11-06T15:05:29Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: it worked for me & got the green tick FirstChildUserIdTAG: 124534 FirstChildUserNameTAG: srihari46 FirstChildCreateTimeTAG: 2012-11-06T15:19:00Z SecondChildTAG: Me too SecondChildUserIdTAG: 357453 SecondChildUserNameTAG: BrunoCanoso SecondChildCreateTimeTAG: 2012-11-06T21:46:24Z FirstChildTAG: It seems to be quite sensitive to component values FirstChildUserIdTAG: 142857 FirstChildUserNameTAG: viking2 FirstChildCreateTimeTAG: 2012-11-09T06:38:04Z FirstChildTAG: Same problem here.Did you find any solution ? FirstChildUserIdTAG: 38331 FirstChildUserNameTAG: janadel FirstChildCreateTimeTAG: 2012-11-11T19:00:47Z IndexTAG: 3250 TitleTAG: Help needed with lab 8! Just a simple question: Do we need to make adjustment to the component values we find? I mean, I found my values from calculations, then went to the transient analysys. Tough the values I got for t=400n (first case) and t=800n (second case) are pretty close to the voltages and current at those times,as far as to the decimal point ,they don't match exactly. UserIdTAG: 369339 UserNameTAG: vargaslen CreateTimeTAG: 2012-11-06T00:08:40Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I think the *general* shapes of the curves corresponding to different combinations of RCL do not require exact numbers. Once we know the unknown components and their order, their *exact* values can be calculated with the appropriate formula wherein the values shown in the graph are substituted. In short, my answer to your question is No. I love the Circuit Sandbox, but there have been many instances where what i see there do not match the values given in labs and homework's, and i'm not clear what are its limitations or in what circumstances it can be close enough (unless clearly instructed in the labs). FirstChildUserIdTAG: 184153 FirstChildUserNameTAG: tthngn FirstChildCreateTimeTAG: 2012-11-06T04:35:39Z IndexTAG: 3251 TitleTAG: H8P2: Half-Life I've found nice hint about half-life in circuits. Just look at the following article: http://en.wikipedia.org/wiki/Half-life It becomes more clear. UserIdTAG: 81712 UserNameTAG: dzhon CreateTimeTAG: 2012-11-05T22:57:23Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Thank you, this really helped! FirstChildUserIdTAG: 176173 FirstChildUserNameTAG: SDdad FirstChildCreateTimeTAG: 2012-11-12T06:50:24Z IndexTAG: 3252 TitleTAG: answer to lab 7 Hello, I was unable to solve the lab. How can I get the solution like the homework? UserIdTAG: 389333 UserNameTAG: juergen CreateTimeTAG: 2012-11-05T21:58:04Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You need to wait until the deadline is over. FirstChildUserIdTAG: 131214 FirstChildUserNameTAG: arnsfelt FirstChildCreateTimeTAG: 2012-11-06T12:09:26Z IndexTAG: 3253 TitleTAG: H7P3 [Edited- Asking for answers on the forum is prohibited.] UserIdTAG: 374512 UserNameTAG: arsen55591 CreateTimeTAG: 2012-11-05T17:39:51Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3254 TitleTAG: How to express cosine in formula? For LAB 7 I want to write a formula like below: 1/0.001*(((-COS(2000*PI*T)+1)/2000*PI)) but it prompt that 'Invalid input: Could not parse '1/0.01*(((-cos(2000*PI*T))/(2000*PI)+(1/2000*PI))' as a formula'. What happen? UserIdTAG: 128276 UserNameTAG: ChengBin CreateTimeTAG: 2012-11-05T15:56:58Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: I believe that the parser is expecting lower case "t". FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-05T16:01:16Z FirstChildTAG: two possible mistakes. Try 't' instead of 'T' and check the parenthesis. You have more of ')'. FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-11-05T17:18:50Z FirstChildTAG: 1/2000*PI this should be like 1/(2000*PI) FirstChildUserIdTAG: 60496 FirstChildUserNameTAG: tomipiriyev FirstChildCreateTimeTAG: 2012-11-06T08:37:46Z FirstChildTAG: Also cos not COS FirstChildUserIdTAG: 375500 FirstChildUserNameTAG: Brej7665 FirstChildCreateTimeTAG: 2012-11-06T00:12:48Z IndexTAG: 3255 TitleTAG: H9P1: Solve by inspection To everyone who stuck in H9P1. The hint is in the task description itself. Forget all those nasty math! Think in terms of periods, energy and charge. It is so easy then! UserIdTAG: 208296 UserNameTAG: OZ1 CreateTimeTAG: 2012-11-05T15:44:54Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I can get all "green answers" without math, only by "hmm it should work like this", but its not a goal. I need math ground under my feet for that question. And i dont understand anything after second question. FirstChildUserIdTAG: 352757 FirstChildUserNameTAG: Kerbyco FirstChildCreateTimeTAG: 2012-11-05T16:18:12Z SecondChildTAG: even i have also solved all the part without any maths..can you help me in part 5.. SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-11-16T14:52:48Z FirstChildTAG: I hope it will satisfy you, [CHECK THIS POST][1] > **Regards:** asadbhatti42 [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5098b2a1cd5fb82300000036 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-11-06T06:49:46Z SecondChildTAG: PAGE NOT FOUND SecondChildUserIdTAG: 352757 SecondChildUserNameTAG: Kerbyco SecondChildCreateTimeTAG: 2012-11-06T14:34:45Z IndexTAG: 3256 TitleTAG: LAB7 task 1 and 2 Help me please UserIdTAG: 374512 UserNameTAG: arsen55591 CreateTimeTAG: 2012-11-05T15:38:58Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Take a look here: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/508da86cd8a1b11f0000006c FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-11-05T16:21:39Z SecondChildTAG: thanks SecondChildUserIdTAG: 374512 SecondChildUserNameTAG: arsen55591 SecondChildCreateTimeTAG: 2012-11-05T17:40:29Z FirstChildTAG: I hope it will satisfy you, [Take a look here][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5091dd5d3fc09f2b0000018a > **Regards:** asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-11-06T06:35:38Z IndexTAG: 3257 TitleTAG: Correction I think you meaned "time in seconds", because 4e-10 of 1 nanosecond it's 4E-19 seconds, and this time is smaller, then takes to light beam to across hydrogen atom :-) UserIdTAG: 345464 UserNameTAG: Constantine_ru CreateTimeTAG: 2012-11-05T07:02:01Z VoteTAG: 0 CoursewareTAG: Week 7 / Response To A Step Down CommentableIdTAG: 6002x_Response_To_A_Step_Down NumberOfReplyTAG: 0 IndexTAG: 3258 TitleTAG: HW8P2 I got the vC(0+) correct but not the R associated with a half life of 4 hours(4x3600s =14400s). At the half life point, our vC will be half of vC(0+). Ln of 1/2 = .69. My thinking is half life = 14400 = 0.69RC = .69R(400mg) and solve for R and get 52. I assume that is 52seconds/mg in a 70L person. But neither dividing nor multiplying by 70 produces green check. Ideas? Thanks for your time UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-11-04T23:15:11Z VoteTAG: 0 CoursewareTAG: Week 8 / Time To Decay CommentableIdTAG: 6002x_Time_To_Decay NumberOfReplyTAG: 2 FirstChildTAG: You are using the wrong value for C. The question defines it as the size of the body in L. FirstChildUserIdTAG: 141000 FirstChildUserNameTAG: OrinE FirstChildCreateTimeTAG: 2012-11-05T00:33:16Z FirstChildTAG: ## Hint HW8P2 ## **Q1.** Since we're modeling the injection as a pulse function, then knowing that vC(0-) = 0 **Q2.** Dynamics of vC(t) after the pulse is exponential decay with time constant of τ = RC and initial value of vC(0+). vC(t) = vC(0+) * exp(-t/τ) **Q3.** vC(t)/vC(0+) = 1/4 exp(-t/τ) = 1/4 **Q4.** vC(T-) = 1/4 * vC(0+), where T is the time of second injection. We would like to have vC(T+) = 2/3 * vC(0+) And knowing due to the effect of injection pulse, vC(T+) = vC(T-) + Q'/C > **Regards:** asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-11-05T13:13:22Z SecondChildTAG: asadbhatti I am using same formula which u have mentioned for Q.2 but not able to get green tick SecondChildUserIdTAG: 413002 SecondChildUserNameTAG: Gauravjain88 SecondChildCreateTimeTAG: 2012-11-05T15:40:01Z IndexTAG: 3259 TitleTAG: Relationship between battery voltage and capacity I know that the capacity left in a battery can be estimated by measuring voltage across it with a multimeter, but I don't know why this is so. Anyone? Thanks! UserIdTAG: 398594 UserNameTAG: DaveyJC CreateTimeTAG: 2012-11-04T22:40:07Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: When the battery gets depleted, its internal resistance changes (think how). Measuring the voltage across terminals is in fact measuring voltage drop on the internal resistance of the meter (making voltage divider with internal resistance of the battery). FirstChildUserIdTAG: 4076 FirstChildUserNameTAG: damians FirstChildCreateTimeTAG: 2012-11-05T16:19:53Z IndexTAG: 3260 TitleTAG: power entering of the source that means that an element that "provides" power, that power delivery is negative. Otherwise, if an element "consumes" power, that power is determinated as positive. UserIdTAG: 737053 UserNameTAG: jhonrodriguez92 CreateTimeTAG: 2012-11-04T21:31:28Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 5_Simple_Power NumberOfReplyTAG: 2 FirstChildTAG: When an element is positive it dissipates energy. When an element is negative energy it provides. FirstChildUserIdTAG: 361823 FirstChildUserNameTAG: EliasOak FirstChildCreateTimeTAG: 2012-11-04T22:30:37Z FirstChildTAG: The power of a voltage source is considered positive when it gives power to the circuit, and negative when it receives power from the circuit. That's equivalent to saying that the power of a voltage source is positive when the current exits the source by its + terminal, and negative when the current enters the source by its + terminal. For example, if you connect a voltage source (let's say V=10) to a resistor of 5 Ohm there will be a current of 2 A going out from the + terminal of the voltage source, so you can say that the voltage source is providing a power of 20 W (the power going out of the source is 20 W). You can say also that the power *entering* the source is -20 W (obviously the power entering the source is the same as the power going out of the source, but with the opposite sign). Now think of two voltage sources with their + terminals connected with a resistor between them. The current will go out by the + terminal of the voltage source with the highest voltage, and will enter on the + terminal of the source with the lowest, so the first one will be giving power to the circuit (its power will be possitive) and the second will be receiving power from the curcuit (its power will be negative); you can say also that the power *entering* the firt source (the one with the higher value) is negative, and the power entering the second one is positive. The same reasoning works with the current sources. FirstChildUserIdTAG: 54453 FirstChildUserNameTAG: berzasnon FirstChildCreateTimeTAG: 2012-11-04T23:08:56Z IndexTAG: 3261 TitleTAG: VS=VI ?? Since capacitor us essentially connected in series with RL and Voltage divides in series resistances, howcome capacitor sstill charges to VS. UserIdTAG: 442070 UserNameTAG: MuhammadAsad CreateTimeTAG: 2012-11-04T17:14:55Z VoteTAG: 0 CoursewareTAG: Week 7 / Rise Time CommentableIdTAG: 6002x_Rise_Time NumberOfReplyTAG: 2 FirstChildTAG: If you want to think about it like a voltage divider, just pretend the capacitor is a variable resistor that starts out with a low resistance that keeps growing. As it becomes larger, less and less voltage is dropped across the resistor and more is dropped across the capacitor. FirstChildUserIdTAG: 90796 FirstChildUserNameTAG: ChaunceyGardiner FirstChildCreateTimeTAG: 2012-11-04T18:12:02Z FirstChildTAG: Looks like problem is appealing to S14E1 pic. After A is off, Vs go through Resistor (called RL in this problem) right to Capacitor CGS (not to the ground through transistor) and charges it to VS value. FirstChildUserIdTAG: 66669 FirstChildUserNameTAG: agarus FirstChildCreateTimeTAG: 2012-11-04T18:15:55Z IndexTAG: 3262 TitleTAG: H7P3 Q4 ow much time, in nanoseconds, does it take for the output voltage to reach vO=2.5V? please help me UserIdTAG: 374512 UserNameTAG: arsen55591 CreateTimeTAG: 2012-11-04T12:40:56Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/509664dc1a46ae2700000039 see this post FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-11-04T12:52:36Z IndexTAG: 3263 TitleTAG: Hint for H7P3 Q2 I don't know what is wanted. Is it time invariant, there is no induction, so I have two resistors in series UserIdTAG: 389333 UserNameTAG: juergen CreateTimeTAG: 2012-11-04T10:05:18Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: time constant = L/Rth .where Rth = thevenin equivalent resistor FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-11-04T10:56:39Z FirstChildTAG: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/509664dc1a46ae2700000039 check this post. FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-11-04T12:53:03Z FirstChildTAG: Thanks, that helps FirstChildUserIdTAG: 389333 FirstChildUserNameTAG: juergen FirstChildCreateTimeTAG: 2012-11-04T13:31:58Z SecondChildTAG: :) SecondChildUserIdTAG: 148523 SecondChildUserNameTAG: span993 SecondChildCreateTimeTAG: 2012-11-04T14:02:34Z IndexTAG: 3264 TitleTAG: Hint for H7P3 Please give me some hint about H7P3. I am trying to solve it for a long time but to no avail. Actually what I am doing is to apply KVL on the circuit i.e. V=iL.Rs+L.(diL/dt)+vo. From here onwards, I am unable to solve it. Please guide me. UserIdTAG: 64618 UserNameTAG: ashfaq2419 CreateTimeTAG: 2012-11-04T09:22:36Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/509664dc1a46ae2700000039. check this post. FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-11-04T12:53:21Z FirstChildTAG: Check out the initial response of a inductor in a circuit and thus calculate initial current. FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-11-04T10:58:14Z IndexTAG: 3265 TitleTAG: lab 7 formula checker problem I came up with this formula in part 1 of the lab <.... deleted , please don't post graded answers ....> if I enter t=0.0005 I get the right result but for some reason but for some reason the checker finds this formula wrong. any ideas ? P.S sorry if this is a double post I was sure I posted it before but couldn't find my old post UserIdTAG: 271081 UserNameTAG: YonJah CreateTimeTAG: 2012-11-03T15:59:49Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: try use \cos and \pi FirstChildUserIdTAG: 189766 FirstChildUserNameTAG: shuhuan84 FirstChildCreateTimeTAG: 2012-11-03T16:19:07Z FirstChildTAG: Take a careful look at your "denominator". FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-11-03T16:53:51Z FirstChildTAG: re-check the formula. something might be missing FirstChildUserIdTAG: 405473 FirstChildUserNameTAG: Avisec FirstChildCreateTimeTAG: 2012-11-03T16:42:34Z FirstChildTAG: For Sure your post might be deleted as it contains answer to the HW which violates Honor code and regarding your quest, They asked us to answer upto 6 digit precision. I suggest you to take pi value appropriately coz a minor change in pi could make huge difference in answer (with 6 digit precision) FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-11-03T19:20:48Z FirstChildTAG: Try to put your denominator inside 2 prentheses & seperate the 2000 like this 2*1000 FirstChildUserIdTAG: 185715 FirstChildUserNameTAG: amirengineer FirstChildCreateTimeTAG: 2012-11-03T19:51:16Z SecondChildTAG: thank you for pointing this out SecondChildUserIdTAG: 271081 SecondChildUserNameTAG: YonJah SecondChildCreateTimeTAG: 2012-11-03T20:21:11Z IndexTAG: 3266 TitleTAG: Reg HW and LABS The last four weeks(ie....week 7,8,9,10) and the labs account for how many percent? UserIdTAG: 150120 UserNameTAG: krishna1993 CreateTimeTAG: 2012-11-03T13:09:32Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: week10? FirstChildUserIdTAG: 501454 FirstChildUserNameTAG: gasnikovkv FirstChildCreateTimeTAG: 2012-11-03T13:37:54Z FirstChildTAG: Total 10 HWs & 10 labs include a total of 30% . 10 HWs-- 15% 10 LABs--15% FirstChildUserIdTAG: 277808 FirstChildUserNameTAG: Hemanthmps FirstChildCreateTimeTAG: 2012-11-03T14:12:40Z IndexTAG: 3267 TitleTAG: TO STAFF i'm interested in analog ic design could you guide me to the track i should take and the tools i should learn to become adesigner and will you offer any course in this field ? UserIdTAG: 221617 UserNameTAG: konan CreateTimeTAG: 2012-11-03T09:30:37Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3268 TitleTAG: huh? a,b, and d are identical graphs. How can they be considered different by the presenter? UserIdTAG: 12905 UserNameTAG: Baer CreateTimeTAG: 2012-11-03T02:07:13Z VoteTAG: 0 CoursewareTAG: Week 7 / Constitutive Laws for Capacitors and Inductors CommentableIdTAG: 6002x_Constitutive_Laws_Caps_Inductors NumberOfReplyTAG: 1 FirstChildTAG: Please notice that the big black arrow is the [Dirac delta function][1] , so graph d is different from graph a and b; as far as graph a is concerned, it is wrong: the correct graph is without the constant part (that is it should be zero), so the correct graph consists of the Dirac delta function only. [1]: http://en.wikipedia.org/wiki/Dirac_delta_function FirstChildUserIdTAG: 376877 FirstChildUserNameTAG: AndBre FirstChildCreateTimeTAG: 2012-11-03T08:52:45Z IndexTAG: 3269 TitleTAG: LAB 7 part1 what are the limits of the integral in the question ? UserIdTAG: 382505 UserNameTAG: AhmedGalal2 CreateTimeTAG: 2012-11-02T22:53:50Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: since the current is zero for t<=0. Your integral limits are 0 and t. FirstChildUserIdTAG: 127195 FirstChildUserNameTAG: princeofsudan FirstChildCreateTimeTAG: 2012-11-02T23:03:25Z SecondChildTAG: thanks from egypt SecondChildUserIdTAG: 382505 SecondChildUserNameTAG: AhmedGalal2 SecondChildCreateTimeTAG: 2012-11-02T23:41:56Z IndexTAG: 3270 TitleTAG: A new spring session i wanted to ask if there is a spring session by the new year ? UserIdTAG: 196718 UserNameTAG: HaAr CreateTimeTAG: 2012-11-02T22:31:51Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3271 TitleTAG: Clues/steps for solving all HW 8 prob. plzzz !! Can any1 plz provide clues or steps to follow to find the solutions to HW8 all questions ? UserIdTAG: 332321 UserNameTAG: akshayk CreateTimeTAG: 2012-11-02T16:04:49Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3272 TitleTAG: h7p3 part 4 i'm gonna be crazy i tried sooooo many times with that question plz provide me with ( HINTS ) UserIdTAG: 395257 UserNameTAG: blackguitar CreateTimeTAG: 2012-11-02T14:08:34Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: So to travel x amount of cm with a speed of 299792458m/s, you get: t=( x cm )/ (29979245800 cm/s) This gives you the answer in sec. We want nanosec. 1ns=1e-9s So t from above in nanosec is: t/1e-9 Hope this helps. FirstChildUserIdTAG: 329453 FirstChildUserNameTAG: Even83 FirstChildCreateTimeTAG: 2012-11-02T14:27:44Z SecondChildTAG: gee , thank you very much SecondChildUserIdTAG: 395257 SecondChildUserNameTAG: blackguitar SecondChildCreateTimeTAG: 2012-11-02T14:47:05Z SecondChildTAG: im still not getting it!! i've tried a million times SecondChildUserIdTAG: 338685 SecondChildUserNameTAG: varshaD SecondChildCreateTimeTAG: 2012-11-04T17:03:10Z IndexTAG: 3273 TitleTAG: What is the peak power (in Watts) of this transaction? I think it should be 4.4 miliWatts instead of 4.4 Watts, because the inductor is miliHerry. UserIdTAG: 128276 UserNameTAG: ChengBin CreateTimeTAG: 2012-11-02T10:09:54Z VoteTAG: 0 CoursewareTAG: Week 7 / Inductors Store Energy CommentableIdTAG: 6002x_Inductors_Store_Energy NumberOfReplyTAG: 1 FirstChildTAG: the inductor is 2.2 $\mu H$, not $mH$. This leads to an energy of 1.1 micro joules. Since this is released over 1/4 microsecond, the answer is 4.4 W FirstChildUserIdTAG: 434869 FirstChildUserNameTAG: patey FirstChildCreateTimeTAG: 2012-11-02T11:37:33Z IndexTAG: 3274 TitleTAG: What does it mean? length, width and depth? Suppose we double the length, width, and depth of a solenoidal inductor, by what multiple does the inductance increase? UserIdTAG: 128276 UserNameTAG: ChengBin CreateTimeTAG: 2012-11-02T08:36:31Z VoteTAG: 0 CoursewareTAG: Week 7 / Scaling Inductors CommentableIdTAG: 6002x_Scaling_Inductors NumberOfReplyTAG: 2 FirstChildTAG: Here's a handy calculator for just that: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/indsol.html FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-11-02T14:02:07Z SecondChildTAG: very helpful SecondChildUserIdTAG: 397636 SecondChildUserNameTAG: victordeman SecondChildCreateTimeTAG: 2012-11-03T04:15:53Z FirstChildTAG: I used google translate ! i thought depth was somethig else ... FirstChildUserIdTAG: 373752 FirstChildUserNameTAG: Croak FirstChildCreateTimeTAG: 2012-11-03T00:21:46Z IndexTAG: 3275 TitleTAG: HW7 P3 Q3 OK with what was show in the lecture slides we always solved for the voltage across the inductor. Vout is across the Load resistor and is throwing me completly of. Can someone tell me how our RL equation changes or can I not create my equation of inspection and actually have to solve the differential equation for the answer. Please and thank you UserIdTAG: 435197 UserNameTAG: RichmondRichter CreateTimeTAG: 2012-11-02T07:52:57Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Try to find an expression for the current first, then you'll get the expression for $v_0$ FirstChildUserIdTAG: 470934 FirstChildUserNameTAG: sonhx FirstChildCreateTimeTAG: 2012-11-02T08:41:55Z FirstChildTAG: The key is in the question "What is the final voltage across the load?" What variable determines when the **final** voltage is reached? Don't let the initial explanation of inductance of a wire throw you off. FirstChildUserIdTAG: 132828 FirstChildUserNameTAG: rharris FirstChildCreateTimeTAG: 2012-11-02T14:59:29Z IndexTAG: 3276 TitleTAG: Videos not loading with Ubuntu 12.10 Hi, if I try to watch the videos with Firefox on Windows 7 everything is ok. When I use Ubutnu 12.10 and Firefox the youtube videos are not loading. I tried to install chromium-browser but I got the same result. Is anyone experiencing the same problem? UserIdTAG: 55480 UserNameTAG: gattosilvestro CreateTimeTAG: 2012-11-02T07:30:45Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I'm not sure if it has changed recently, but youtube always used to require a flash player. Ubuntu doesn't come with patent restricted audio/video codecs or Adobe flash player installed by default - you have to do this yourself. See, for example, https://help.ubuntu.com/community/RestrictedFormats/Flash FirstChildUserIdTAG: 434869 FirstChildUserNameTAG: patey FirstChildCreateTimeTAG: 2012-11-02T09:50:25Z SecondChildTAG: Sounds about right, in the past I recall having to add Flash in the past to Ubuntu. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-11-02T12:36:44Z FirstChildTAG: I've already installed flash player. If I visit youtube website I can watch the videos correctly, but it seams that edx is not able to load the youtube videos. I need also to use a VPN to access youtube sinsce I'm in China but again, I can watch youtube videos if I go directly to www.youtube.com. I don't know if it's a problem of Ubuntu 12.10 or some setting on my pc. Is anybody using Ubuntu 12.10 without problems? FirstChildUserIdTAG: 55480 FirstChildUserNameTAG: gattosilvestro FirstChildCreateTimeTAG: 2012-11-02T14:34:47Z SecondChildTAG: You could try downloading the videos using the download link and then viewing directly.. SecondChildUserIdTAG: 173147 SecondChildUserNameTAG: cruiser_rahit SecondChildCreateTimeTAG: 2012-11-02T15:48:39Z IndexTAG: 3277 TitleTAG: Download links for Weeks 6+ Are there links posted anywhere to download videos for Weeks 6-9? UserIdTAG: 264596 UserNameTAG: Nuru CreateTimeTAG: 2012-11-02T04:27:45Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Ah, nevermind, I did not know they were below the videos now. FirstChildUserIdTAG: 264596 FirstChildUserNameTAG: Nuru FirstChildCreateTimeTAG: 2012-11-02T04:36:26Z SecondChildTAG: Yes they were moved to a more convenient location. :) SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-11-02T12:38:36Z IndexTAG: 3278 TitleTAG: It's as easy as going Shick or Shook ...am I right? UserIdTAG: 287805 UserNameTAG: Beneficial CreateTimeTAG: 2012-11-02T02:36:58Z VoteTAG: 0 CoursewareTAG: Week 7 / An Intuitive Approach CommentableIdTAG: 6002x_An_Intuitive_Approach NumberOfReplyTAG: 0 IndexTAG: 3279 TitleTAG: H9P1 Can anyone help me? I have found w - the natural frequency, E - the total energy, iL(5.0_) and vC(5.0_). But inductor current is wrong. And I can't understand why? I have computed vC(5.0_) using iL(5.0_), but vC is correct and iL is wrong. At first I supposed that iL(t)=iL(0)*cos(w*t), after I try to find iL using (L*iL(t)^2)/2+(C*uC(t)^2)/2=E and I had the same answer. My wrong answer is -0.0047. Who can explain me what I have done incorrectly? UserIdTAG: 391745 UserNameTAG: we_told CreateTimeTAG: 2012-11-02T00:50:30Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I suspect your answer should be exactly zero. The values given for the inductor and capacitor result in a natural frequency very close to a common fraction. It's close enough for the grader to mark big numbers (like the energy and voltage) as correct, but not close enough for small numbers, like the current through the inductor at 1 second. FirstChildUserIdTAG: 366083 FirstChildUserNameTAG: smath FirstChildCreateTimeTAG: 2012-11-02T08:58:52Z IndexTAG: 3280 TitleTAG: A comment about the first part of H8P3 The "correct" response to the first question encountered in the homework problem H8P3 apparently requires that the OFF resistance of Q3 must be infinite [not 115M]. This seems unlikely in real world. UserIdTAG: 286880 UserNameTAG: herbsteiner CreateTimeTAG: 2012-11-01T18:30:37Z VoteTAG: 0 CoursewareTAG: Week 8 / Time To Decay CommentableIdTAG: 6002x_Time_To_Decay NumberOfReplyTAG: 1 FirstChildTAG: actually, no. I used my Roff value (it really does vary the voltage VERY slightly) - and everything was ok FirstChildUserIdTAG: 196898 FirstChildUserNameTAG: Undead FirstChildCreateTimeTAG: 2012-11-11T23:37:49Z IndexTAG: 3281 TitleTAG: Lab7 Template Hi all, Could any person help me? my LAB7 template doesn't show anything. I see only blank paper. MAY UserIdTAG: 373916 UserNameTAG: larou CreateTimeTAG: 2012-11-01T12:59:48Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3282 TitleTAG: lecture 7, 8 download links hey there.... could some one please tell me where can i find download links for lecture 7,8 and also the upcoming lectures so that i wont bother u again ....:) thnakyou..... UserIdTAG: 117319 UserNameTAG: iqramalik CreateTimeTAG: 2012-10-31T18:40:15Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3283 TitleTAG: Midterm Q3 question about the Thevenin equivalent circuit I am not clear about the equivalent circut for Q3 in this year's midterm. In the midterm solutions, only the current source and the paralel resistor was transfered to equivalent Thevenin circuit. Why isn't the Thevenin equivalent made for the whole "left side of the" circuit (current source and both resistors)? This is my view of the Thevenin equivalent circuit for this problem. ![Circuit for Q3][1] R_th - resistance measured on open connectors: Since we have open connectors, the current through R_2 equals 0, so R_th equals R_1. `R_th = 1 Ohm`. V_th - voltage measured on open connectors: Since we have open connectors, the current through R_2 equals 0, so V_th equals voltage measured on R_1. `V_th = I*R_1 = 1A*1Ohm = 1V` [1]: https://edxuploads.s3.amazonaws.com/13517010531343655.bmp UserIdTAG: 294766 UserNameTAG: dejanst CreateTimeTAG: 2012-10-31T16:40:38Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: $R_{th}$ is resistance seen from the output with independent sources replaced by their thevinin equivalent. FirstChildUserIdTAG: 623210 FirstChildUserNameTAG: preveen FirstChildCreateTimeTAG: 2012-10-31T17:41:05Z FirstChildTAG: Two ways of calculating Rth: First is the resistance at the open connectors *with the sources turned off*, so it's simply R_1 + R_2, i.e. 2 ohm. The second way is the open circuit voltage divided by the short circuit current. With the connectors shorted, the current is shared equally between the two resistors as they are the same value, so the short circuit current is I/2 amps. Open circuit voltage is I*R_1, or just 'I' volts. I/(I/2) is therefore 2 ohms. In this problem however, since we are asked for the voltage across R_1, i.e. that at the left terminal of R2, (picking the negative terminal of the current source as a ground node), it would be easier to replace just the current source and R_1 by their Thevenin equivalent (Vth = I*R_1, Rth = R_1). FirstChildUserIdTAG: 141000 FirstChildUserNameTAG: OrinE FirstChildCreateTimeTAG: 2012-10-31T17:03:33Z FirstChildTAG: Ah, I understand now. Voltage source turned off - replace with short circuit Current source turned off - replace with open connectors This is true for Norton and Thevenin method. If I replace the current source with open connectors, I see 2 resistors connected in series from the open output connectors. So the R_th is R_1 + R_2. ![Circuit for finding T_th][1] The mistake that I was making was replacing the current source with a short circuit. Thanks for helping me to clear this up. [1]: https://edxuploads.s3.amazonaws.com/13517104291343625.bmp FirstChildUserIdTAG: 294766 FirstChildUserNameTAG: dejanst FirstChildCreateTimeTAG: 2012-10-31T19:09:15Z IndexTAG: 3284 TitleTAG: Fourth attempt? If the output is completely decoupled from the load, how we can use the voltage data stored in the capacitor then? UserIdTAG: 201818 UserNameTAG: ThreeHundred CreateTimeTAG: 2012-10-31T15:55:41Z VoteTAG: 0 CoursewareTAG: Week 8 / Building A Better Static Memory Element Continued CommentableIdTAG: 6002x_Building_A_Better_Static_Memory_Element_Continued NumberOfReplyTAG: 1 FirstChildTAG: When he says that it completely decoupled, he means that the external circuit connected to the output cannot influence the charge in the capacitor. The value after the buffer can still be read by the external circuit though. FirstChildUserIdTAG: 270011 FirstChildUserNameTAG: SnowmanZA FirstChildCreateTimeTAG: 2012-11-01T17:07:50Z IndexTAG: 3285 TitleTAG: Time in nanoseconds or seconds? Is the rise time really around 5e-10 nanoseconds (5e-19 seconds)? UserIdTAG: 105523 UserNameTAG: eero CreateTimeTAG: 2012-10-31T13:46:34Z VoteTAG: 0 CoursewareTAG: Week 7 / Response To A Step Down CommentableIdTAG: 6002x_Response_To_A_Step_Down NumberOfReplyTAG: 0 IndexTAG: 3286 TitleTAG: Progress : Midterm's bar is still grey Is it normal? at the moment it should be red? Please answer me... UserIdTAG: 204303 UserNameTAG: fabnelli CreateTimeTAG: 2012-10-31T06:20:47Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: should it change to red? FirstChildUserIdTAG: 201818 FirstChildUserNameTAG: ThreeHundred FirstChildCreateTimeTAG: 2012-10-31T06:34:39Z FirstChildTAG: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/508b25af09bc342300000004 FirstChildUserIdTAG: 9673 FirstChildUserNameTAG: xp42 FirstChildCreateTimeTAG: 2012-10-31T06:47:14Z FirstChildTAG: Nothing to worry about. It is grey for everyone. I guess they use different colors in progress indicating Homework, lab, midterm and finals. FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-10-31T08:26:51Z FirstChildTAG: Yes. Thanks. I have seen that the percentage from total in grey is the midterm contribution. FirstChildUserIdTAG: 204303 FirstChildUserNameTAG: fabnelli FirstChildCreateTimeTAG: 2012-10-31T13:55:17Z IndexTAG: 3287 TitleTAG: Erratum re. initial condition? At time mark 0:37 / 5:19: I guess "So since the **input voltage is at I** for a long period of time..." should be understood as: "So since the **input current is at I** for a long period of time..." UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-30T22:57:36Z VoteTAG: 0 CoursewareTAG: Week 8 / RC Response To A Falling Step Continued CommentableIdTAG: 6002x_RC_Response_To_Falling_Step_Continued NumberOfReplyTAG: 0 IndexTAG: 3288 TitleTAG: Case sensitive... Nice to know that the answer checker is case sensitive. A|B|C|D != a|b|c|d UserIdTAG: 114881 UserNameTAG: xvink CreateTimeTAG: 2012-10-30T20:17:02Z VoteTAG: 0 CoursewareTAG: Week 7 / Response To A Step Down CommentableIdTAG: 6002x_Response_To_A_Step_Down NumberOfReplyTAG: 1 FirstChildTAG: I got burned by the same thing. I had the correct answer the first time and it said I was wrong :). FirstChildUserIdTAG: 366165 FirstChildUserNameTAG: silicon_ghost FirstChildCreateTimeTAG: 2012-11-02T03:35:58Z IndexTAG: 3289 TitleTAG: finished with homework 9 lab 9 and only one week till all labs and hw will be finished as 11 and 12 are not needed if you got 100% on all the homeworks :) when is the final exam if anyone knows UserIdTAG: 436749 UserNameTAG: waynebrown CreateTimeTAG: 2012-10-30T14:27:33Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: what is your current progress percentage? FirstChildUserIdTAG: 623210 FirstChildUserNameTAG: preveen FirstChildCreateTimeTAG: 2012-10-30T14:54:15Z SecondChildTAG: 56% SecondChildUserIdTAG: 357453 SecondChildUserNameTAG: BrunoCanoso SecondChildCreateTimeTAG: 2012-10-30T18:49:13Z FirstChildTAG: Would be in mid december.. FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-10-30T15:14:20Z FirstChildTAG: The calendar shows Dec 20 as the date for the final. Based on the midterm, I believe that is the release date with the due date several days later, but I could be wrong. Since there is nothing scheduled the previous week, I would prefer moving the the exam to an earlier date. I would like to get the course completed well before Christmas. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-30T16:02:06Z FirstChildTAG: with 100% in midterm 100% in all HW and Labs (Include 11 and 12), your score will 60%. FirstChildUserIdTAG: 357453 FirstChildUserNameTAG: BrunoCanoso FirstChildCreateTimeTAG: 2012-10-30T18:52:10Z IndexTAG: 3290 TitleTAG: No graded certificate in this course?? i recently visited the page of certification where it is written that certificate of merit will be given. No specific grade will be provided...what is that?? they had mentioned initially that graded certificate will be provided. UserIdTAG: 321556 UserNameTAG: sj31867 CreateTimeTAG: 2012-10-30T14:17:35Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: where you get that they won't provide graded certificate? FirstChildUserIdTAG: 342221 FirstChildUserNameTAG: deepkar FirstChildCreateTimeTAG: 2012-10-30T14:52:06Z SecondChildTAG: https://www.edx.org/courses/MITx/6.002x/2012_Fall/about SecondChildUserIdTAG: 321556 SecondChildUserNameTAG: sj31867 SecondChildCreateTimeTAG: 2012-10-30T16:35:34Z SecondChildTAG: Certificate on this page has a grade http://lh5.ggpht.com/-hpIcOErEk90/T_eLh24i-9I/AAAAAAAAA5Q/hWwHk6dV_eE/Certificate%25255B13%25255D.jpg SecondChildUserIdTAG: 171674 SecondChildUserNameTAG: wajahat1 SecondChildCreateTimeTAG: 2012-10-30T18:24:26Z SecondChildTAG: Yeah as the grade is mentioned in above certificate i want to ask the concern authorities that whether such grade is provided in this course or not? SecondChildUserIdTAG: 321556 SecondChildUserNameTAG: sj31867 SecondChildCreateTimeTAG: 2012-10-30T20:07:02Z FirstChildTAG: I do not think a graded certificate was ever offered. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-30T16:05:55Z SecondChildTAG: Before joining this course it was written on 'CERTIFICATION' section that a graded certification will be provided. In recent past they change their statement. SecondChildUserIdTAG: 321556 SecondChildUserNameTAG: sj31867 SecondChildCreateTimeTAG: 2012-10-30T16:34:32Z SecondChildTAG: if a graded certificate is not provided then it is unfair to many of the top scorer of this course as they worked consistently on this course. Also how will one differentiate whether a particular student was a A grader(>87%) or a C grader?? Everyone then comes down to to the same level.. SecondChildUserIdTAG: 321556 SecondChildUserNameTAG: sj31867 SecondChildCreateTimeTAG: 2012-10-30T16:39:50Z SecondChildTAG: It seems that there is confusion in the graduation word. This word is applied to those who form a university. This cusro of EDX can not graduate. He will just leveled in categories A, B and C. SecondChildUserIdTAG: 361823 SecondChildUserNameTAG: EliasOak SecondChildCreateTimeTAG: 2012-10-30T16:54:14Z FirstChildTAG: It's said in the section "About this course": Those who succesfully earn enough points ( more than 60%)will receive an honor code certificate from MITx. FirstChildUserIdTAG: 329444 FirstChildUserNameTAG: albmartin FirstChildCreateTimeTAG: 2012-10-30T17:48:08Z FirstChildTAG: we can only do our best and hope MITx admin will not disappoint us.. FirstChildUserIdTAG: 114864 FirstChildUserNameTAG: Vikaash FirstChildCreateTimeTAG: 2012-11-01T15:17:07Z IndexTAG: 3291 TitleTAG: lab7 Q1 Integral help By calculating the integral manually or with wolfram i get: http://www.wolframalpha.com/input/?i=1%2F0.001+*+integrate+sin%28+2000*pi*x+%29+dx This is wrong, could anyone tell me why? I guess this integral from -inf to t is the same as from 0-t, where the I(0) tag is 0 at the final evaluation. Somehow I get the real solution / 2, but I cannot understand why. UserIdTAG: 329254 UserNameTAG: KGabor CreateTimeTAG: 2012-10-30T12:49:13Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: You need a set of parentheses to properly specify the denominator of your constant. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-30T21:34:45Z SecondChildTAG: Thanks for advice skyhawk, but this was already solved yesterday. Those comments were somehow deleted, because I do not see them anymore. The problem was actually by evaluating the definite integral between I(t) - I(0), where I thought I(0) = 0, but there I should have made t=0... SecondChildUserIdTAG: 329254 SecondChildUserNameTAG: KGabor SecondChildCreateTimeTAG: 2012-10-31T08:08:31Z SecondChildTAG: "Those comments were somehow deleted, because I do not see them anymore." Answers were removed from the discussion forum. I cannot edit replies in grey,(Like mine here for example), so if answers are posted as replies and not new posts,the whole post may be removed. Thanks for your understanding. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-31T12:07:56Z SecondChildTAG: I'am get a half value in answer too. When may be error? I have the same integral, but I can't understant why answer is half of right. SecondChildUserIdTAG: 269368 SecondChildUserNameTAG: StAlex SecondChildCreateTimeTAG: 2012-10-31T13:50:51Z SecondChildTAG: I don't think you are integrating properly. Remember you need to integrate from 0 to t. Take your first value - the second. SecondChildUserIdTAG: 146347 SecondChildUserNameTAG: vmcpherron SecondChildCreateTimeTAG: 2012-10-31T18:47:36Z FirstChildTAG: Hi KGabor, Take a look at this [explanation Post][1]. Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5091dd5d3fc09f2b0000018a FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-11-01T02:36:59Z IndexTAG: 3292 TitleTAG: lab#9 i need help with the boost converter..kindly guide for the same. UserIdTAG: 433368 UserNameTAG: SurbhiMahajan CreateTimeTAG: 2012-10-30T07:40:19Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3293 TitleTAG: 4µC is a quantity of electrons given by the source isn't it ? so how is it possible to have 4µC in C1 + 4µC in C2 if the source gived no more than 4 ? UserIdTAG: 373752 UserNameTAG: Croak CreateTimeTAG: 2012-10-30T03:41:37Z VoteTAG: 0 CoursewareTAG: Week 7 / Unknown Capacitance CommentableIdTAG: 6002x_Unknown_Capacitance NumberOfReplyTAG: 1 FirstChildTAG: The spike in current at $t=0$ delivers a charge $q = 4 \ \mu C$ to both capacitors. This must be true since charge, $q$ is equal to the integral of the current over time, and both capacitors have equal current flow through them. Note the units for charge here, $C$ are coulombs (or amps-seconds) and should not be confused with the C for capacitance, measured in Fahrads (or coulombs-per-volt). Since voltage, $v$, is related to charge, $q$, and the capacitance of the capacitor, C, by: $v = \frac{q}{C}$, the voltage across the two capacitors can be different even though they both contain the same amount of charge. FirstChildUserIdTAG: 434869 FirstChildUserNameTAG: patey FirstChildCreateTimeTAG: 2012-10-30T16:50:00Z SecondChildTAG: OK OK means i must reconsider my intuitive notion of "charge" !! thanks SecondChildUserIdTAG: 373752 SecondChildUserNameTAG: Croak SecondChildCreateTimeTAG: 2012-11-01T17:52:29Z SecondChildTAG: I think that your original statement was pretty much correct: "$4\mu C $ is a quantity of electrons". Effectively it is the amount of charge contained on the electrons delivered by a 1A current in 4 microseconds. With two capacitors connected in series, the amount of charge pushed into the first capacitor is the same as the amount of charge "pulled" out of the second. Therefore the amount of charge stored in each capacitor is the same. SecondChildUserIdTAG: 434869 SecondChildUserNameTAG: patey SecondChildCreateTimeTAG: 2012-11-02T14:48:52Z IndexTAG: 3294 TitleTAG: S15E1 May some one help for the "Invalid input: Is not permitted in answer" I am submitting XX*(1-e^(-(R/X)*t)) and similarly other. What I have to do? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-29T23:54:32Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3295 TitleTAG: Wrong graphics? Initially I thought, how can the current be still higher if the voltage is zero or dropped down? Shouldn't the current drop after the voltage? But, indeed after I solved the integral it's really like that, the new current will be VT/L... At least in the ideal case... UserIdTAG: 415375 UserNameTAG: ZWX CreateTimeTAG: 2012-10-29T22:31:24Z VoteTAG: 0 CoursewareTAG: Week 7 / Inductor and a Voltage Source CommentableIdTAG: 6002x_Inductor_And_A_Voltage_Source NumberOfReplyTAG: 0 IndexTAG: 3296 TitleTAG: mistake at end? If the time constant is 1/3 ms, shouldn't the final exponent be e^(-3000t)? UserIdTAG: 434869 UserNameTAG: patey CreateTimeTAG: 2012-10-29T22:02:13Z VoteTAG: 0 CoursewareTAG: Week 8 / Exercise 10.1 CommentableIdTAG: 6002x_Ex10_1 NumberOfReplyTAG: 1 FirstChildTAG: The value of 't' in this expression is entered in milli seconds. So if time is 1 second I have to enter 1000 as a value of 't', this takes care of 1000 multiplying factor. Using your expression, which is also correct, for time of 1 second I will enter t=1. FirstChildUserIdTAG: 19863 FirstChildUserNameTAG: Samir FirstChildCreateTimeTAG: 2012-11-04T02:10:57Z IndexTAG: 3297 TitleTAG: Precision What does 6 digits of precision mean ?? My answer is right but still its not accepting :( Kindly explain it with the help of an example !! UserIdTAG: 145239 UserNameTAG: Maheenjd CreateTimeTAG: 2012-10-29T21:10:54Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Have a look here: http://en.wikipedia.org/wiki/Arithmetic_precision FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-29T21:13:45Z IndexTAG: 3298 TitleTAG: Text book wrong link Hi, I think that link on text book should direct to pg. 525, 10th. chapter - "Propagation Delay and Digital Abstraction": Currently text book link direct to pg. 433 - that is end of Chapter 8. And it is looks like the same problem for video: S12V1-S12V5 UserIdTAG: 410845 UserNameTAG: rvpilipen CreateTimeTAG: 2012-10-29T19:07:42Z VoteTAG: 0 CoursewareTAG: Week 6 / Motivation for capacitor CommentableIdTAG: 6002x_motivation_for_capacitor NumberOfReplyTAG: 0 IndexTAG: 3299 TitleTAG: Why Why 0.00099 and not 0.0099? UserIdTAG: 326787 UserNameTAG: Angstrem CreateTimeTAG: 2012-10-29T17:08:21Z VoteTAG: 0 CoursewareTAG: Week 7 / Fall Time Constant CommentableIdTAG: 6002x_Fall_Time_Constant NumberOfReplyTAG: 1 FirstChildTAG: If I hit 9.9*0.1e-12 in the calculator it gives me 9.9e-13. now multiply it with 1e9 as you want your answer in nanosec it gives 0.00099 FirstChildUserIdTAG: 329254 FirstChildUserNameTAG: KGabor FirstChildCreateTimeTAG: 2012-10-29T19:12:09Z IndexTAG: 3300 TitleTAG: Lab week 9 why is there an option to show answer to lab 9?? is the lab not graded?? if it is not graded then will the top 10 lab scores out of 11 labs only be considered for grading? UserIdTAG: 525246 UserNameTAG: Sujith92 CreateTimeTAG: 2012-10-29T16:23:49Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Sujith92, the staff have been notified. We will let you know the outcome when it becomes available to us. Thanks. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-29T16:41:54Z SecondChildTAG: glad i cud help..the option is gone now :) SecondChildUserIdTAG: 525246 SecondChildUserNameTAG: Sujith92 SecondChildCreateTimeTAG: 2012-10-29T17:42:23Z IndexTAG: 3301 TitleTAG: Capacitor usage Sir Can we use a high value capacitor as an emergency power source? UserIdTAG: 125478 UserNameTAG: saket_sharad CreateTimeTAG: 2012-10-29T08:15:04Z VoteTAG: 0 CoursewareTAG: Week 7 / Types of Capacitors CommentableIdTAG: 6002x_Types_of_Capacitors NumberOfReplyTAG: 2 FirstChildTAG: http://en.wikipedia.org/wiki/Electric_double-layer_capacitor FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-29T08:35:05Z SecondChildTAG: thanks SecondChildUserIdTAG: 623210 SecondChildUserNameTAG: preveen SecondChildCreateTimeTAG: 2012-10-29T10:23:43Z FirstChildTAG: Super Cap: http://goo.gl/3Y8b4 - 3000 Farads - 3020W Usable Power - 147A Maximum Continuous Current - 2170 Maximum Peak Current (1 second) - 3.04Wh Available Energy FirstChildUserIdTAG: 142857 FirstChildUserNameTAG: viking2 FirstChildCreateTimeTAG: 2012-10-29T16:21:32Z IndexTAG: 3302 TitleTAG: edit resistor value how to edit resistor value? UserIdTAG: 153803 UserNameTAG: bhargavpokala CreateTimeTAG: 2012-10-29T06:25:26Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Please visit https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Overview/Lab0_Using_the_tools/ Change a component's properties Double click on the component. This will bring up an Edit Properties window that has input fields for each of the component's properties. Click OK to change the values. Click CANCEL or the window's close button to abort the changes. Numeric values can be entered using engineering HTH FirstChildUserIdTAG: 623210 FirstChildUserNameTAG: preveen FirstChildCreateTimeTAG: 2012-10-29T06:55:31Z IndexTAG: 3303 TitleTAG: is there new lectures and homework this week as it says UserIdTAG: 436749 UserNameTAG: waynebrown CreateTimeTAG: 2012-10-29T05:20:52Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Yes, check back after 8:00 AM Boston time, this the 29th of October. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-29T05:27:15Z IndexTAG: 3304 TitleTAG: So close.... I was just finishing the midterm, but when I hit the check button there was no response. I realize that I ran out of time, but I have a screenshot at 9:26pm PDT with my answers to the final question. I know there's no way I'll get any points, but is there any way I can check my answers? I had %100 up until the final question, and I'm dying to know if I got the last question right as well. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-29T04:39:29Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: At some point they will put a "Show Answer" button in the mid-term just like the homework. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-29T04:58:41Z FirstChildTAG: You should still get points if you hit the "check" button for Questions 1-5. You will only miss points for Question 6 if you did not hit the "check" button only for that question itself. So if you had 100% up to that point, your Midterm score should be 5/6 = 83%, a 'B', not bad. **Hint:** I always leave the "Progress" tab open in a separate window and continually refresh it after every Homework or Exam Question I complete just to make sure my credit is recorded properly. Once it appears in the "Progress" window, I believe it cannot be taken away. FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-10-29T06:56:04Z FirstChildTAG: Hint: Answers to final question are so related to one another;0) FirstChildUserIdTAG: 623210 FirstChildUserNameTAG: preveen FirstChildCreateTimeTAG: 2012-10-29T06:59:18Z SecondChildTAG: tell them SecondChildUserIdTAG: 127800 SecondChildUserNameTAG: MNB SecondChildCreateTimeTAG: 2012-10-30T04:37:25Z IndexTAG: 3305 TitleTAG: midterm ends Dear all, just finished midterm with 93%(due to silly mistakes,that i can't even think of), not quite happy with my efforts. Hoping final would get better kick . Thanks a lot to EDX UserIdTAG: 408534 UserNameTAG: kkashyap CreateTimeTAG: 2012-10-29T04:28:35Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Congratulations Kkashyap! I too, would like to thank edX for this truly remarkable opportunity. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-29T04:38:08Z IndexTAG: 3306 TitleTAG: aaha MIDTERM OVER UserIdTAG: 156835 UserNameTAG: kphariprakash1968 CreateTimeTAG: 2012-10-29T04:17:57Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: don´t appear check and save buttons, what happened? midterm time over? FirstChildUserIdTAG: 51036 FirstChildUserNameTAG: dj4vi FirstChildCreateTimeTAG: 2012-10-29T04:24:59Z SecondChildTAG: It appears to be over. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-29T04:33:52Z IndexTAG: 3307 TitleTAG: Midterm exam time is over? UserIdTAG: 51036 UserNameTAG: dj4vi CreateTimeTAG: 2012-10-29T04:16:09Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Should be. FirstChildUserIdTAG: 127195 FirstChildUserNameTAG: princeofsudan FirstChildCreateTimeTAG: 2012-10-29T04:18:45Z SecondChildTAG: but no yet 23.59 Boston time SecondChildUserIdTAG: 51036 SecondChildUserNameTAG: dj4vi SecondChildCreateTimeTAG: 2012-10-29T04:20:05Z SecondChildTAG: Its 12:21 AM Boston time SecondChildUserIdTAG: 127195 SecondChildUserNameTAG: princeofsudan SecondChildCreateTimeTAG: 2012-10-29T04:22:30Z SecondChildTAG: =( SecondChildUserIdTAG: 51036 SecondChildUserNameTAG: dj4vi SecondChildCreateTimeTAG: 2012-10-29T04:25:38Z SecondChildTAG: I was over 79% of the midterm SecondChildUserIdTAG: 51036 SecondChildUserNameTAG: dj4vi SecondChildCreateTimeTAG: 2012-10-29T04:27:15Z SecondChildTAG: I was over 93% of the midterm SecondChildUserIdTAG: 475577 SecondChildUserNameTAG: kibens SecondChildCreateTimeTAG: 2012-10-29T06:38:30Z SecondChildTAG: 87% could have done more if not for power outtage. SecondChildUserIdTAG: 42560 SecondChildUserNameTAG: circuitron SecondChildCreateTimeTAG: 2012-10-29T07:05:40Z SecondChildTAG: I was over 93% of the midterm too. SecondChildUserIdTAG: 364126 SecondChildUserNameTAG: shunyi SecondChildCreateTimeTAG: 2012-10-29T09:54:25Z IndexTAG: 3308 TitleTAG: Midterm it is biased as above UserIdTAG: 51036 UserNameTAG: dj4vi CreateTimeTAG: 2012-10-29T04:15:14Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3309 TitleTAG: WEEK 9 When will week 9 show up?? Did anyone that? UserIdTAG: 277808 UserNameTAG: Hemanthmps CreateTimeTAG: 2012-10-29T04:01:29Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: It should be available sometime today, the 29th of October. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-29T04:32:16Z FirstChildTAG: New weeks are released (and weekly assignments are due) on the change of day at the international date line, ie 8am Boston time. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-10-29T04:50:16Z IndexTAG: 3310 TitleTAG: Yupyyyyyyyy ^_^ I have just finished the Midterm with a nice shining brand new 100% ^_^ UserIdTAG: 365201 UserNameTAG: sirajmuhammad CreateTimeTAG: 2012-10-28T23:30:45Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Good for you Siraj! Try to contain your excitement until next week. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-28T23:39:53Z IndexTAG: 3311 TitleTAG: H8P1 HELP SOCORRO please someone help me with the formula for h8p1 UserIdTAG: 327787 UserNameTAG: REINALDOPARANHOS CreateTimeTAG: 2012-10-28T22:20:11Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: H8P1 is a graded homework. Someone can't just give you the formula. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-29T01:43:00Z IndexTAG: 3312 TitleTAG: Wrong Answer the answer is supposed to be 0.16094379n They forgot the nano abb. UserIdTAG: 451917 UserNameTAG: tamtam CreateTimeTAG: 2012-10-28T11:30:25Z VoteTAG: 0 CoursewareTAG: Week 7 / Rise Time CommentableIdTAG: 6002x_Rise_Time NumberOfReplyTAG: 1 FirstChildTAG: They didn't forget it: "What is the time tr, in nanoseconds"... FirstChildUserIdTAG: 443358 FirstChildUserNameTAG: torkelh FirstChildCreateTimeTAG: 2012-10-28T14:04:47Z IndexTAG: 3313 TitleTAG: Please help me how to plot current in lab? Please help me how to plot current in lab? UserIdTAG: 114473 UserNameTAG: Veerabasanagouda CreateTimeTAG: 2012-10-28T05:40:18Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hi, first you need define a Voltage Probe as x-axis and then placeed the current probe where you want to messure the current. Have fun. FirstChildUserIdTAG: 274475 FirstChildUserNameTAG: Konredus FirstChildCreateTimeTAG: 2012-10-28T06:03:54Z FirstChildTAG: Hi Veerabasanagouda, Take a look at the Tool of Sandbox (it is on the right), you will see a dash with a magenta arrow, if you want to meassure a current in a specific part of a circuit, just drag it to that place [for more info read this Tutorial that I have made in the Wiki][1]. I hope this can help you, See you! Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/how-measure-current-sandbox/ FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-28T06:16:10Z IndexTAG: 3314 TitleTAG: Supernode Review Hi guys, I wanted to do a quick review of supernode trick before starting in on the exam. Can't seem to find where that method was described. Does anyone recall which lecture video that appeared in? UserIdTAG: 287805 UserNameTAG: Beneficial CreateTimeTAG: 2012-10-27T16:46:08Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Great review item! FirstChildUserIdTAG: 188609 FirstChildUserNameTAG: Laureen FirstChildCreateTimeTAG: 2012-10-27T17:21:02Z FirstChildTAG: [Week 1 Tutorials][1] -> Nodal Analysis with Floating Voltage Source (6th video in the sequence) [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_1/Week_1_Tutorials/ FirstChildUserIdTAG: 405936 FirstChildUserNameTAG: Whible FirstChildCreateTimeTAG: 2012-10-27T16:56:24Z SecondChildTAG: thanks SecondChildUserIdTAG: 287805 SecondChildUserNameTAG: Beneficial SecondChildCreateTimeTAG: 2012-10-27T16:58:42Z IndexTAG: 3315 TitleTAG: Integrating Factor Method Hi, In week 6 the we were taught how to solve the first order differential equation related to the first order circuit. However since I am not accustomed to solving first order differential equations in such a manner, I solved it using the Integrating Factor method. $RC\frac{dV_c}{dt} + V_c = V_I$ $\frac{dV_c}{dt} + \frac{V_c}{RC} = \frac{V_I}{RC}$ $\frac{dV_c}{dt} + V_c\cdot P(t) = Q(t)$ Integrating Factor = $e^{\int P(t) dt}$ $\therefore$ Integrating Factor = $e^{\int \frac{1}{RC} dt}$ $\therefore$ Integrating Factor = $e^{\frac{t}{RC}}$ $\therefore V_c \cdot e^{\frac{t}{RC}} = \int \frac{V_I}{RC} \cdot e^{\frac{t}{RC}} dt$ $\therefore V_c \cdot e^{\frac{t}{RC}} = \frac{V_I}{RC} \int e^{\frac{t}{RC}} dt$ $\therefore V_c \cdot e^{\frac{t}{RC}} = \frac{V_I}{RC} \cdot RC \cdot e^{\frac{t}{RC}} + c$ $\therefore V_c \cdot e^{\frac{t}{RC}} = V_I \cdot e^{\frac{t}{RC}} + c$ $\therefore V_c = V_I + c \cdot e^{-\frac{t}{RC}}$ $c = V_O - V_I$ $\therefore V_c = V_I + (V_O-V_I) \cdot e^{-\frac{t}{RC}}$ QED My only doubt is the constant "$c$". I was able to get the constant from the final equation. But can anybody explain to my why $c = V_O - V_I$? Thank you in advance UserIdTAG: 85490 UserNameTAG: yashsavani CreateTimeTAG: 2012-10-27T14:31:57Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Your math didn't come out right. Is this what you meant? $RC\frac{dV_c}{dt} + V_c = V_I$ $\frac{dV_c}{dt} + \frac{V_c}{RC} = \frac{V_I}{RC}$ <=> $\frac{dV_c}{dt} + V_c\cdot P(t) = Q(t)$ Integrating Factor = $e^{\int P(t) dt}$ $\therefore$ Integrating Factor = $e^{\int \frac{1}{RC} dt}$ $\therefore$ Integrating Factor = $e^{\frac{t}{RC}}$ $\therefore V_c \cdot e^{\frac{t}{RC}} = \int \frac{V_I}{RC} \cdot e^{\frac{t}{RC}} dt$ $\therefore V_c \cdot e^{\frac{t}{RC}} = \frac{V_I}{RC} \int e^{\frac{t}{RC}} dt$ $\therefore V_c \cdot e^{\frac{t}{RC}} = \frac{V_I}{RC} \cdot RC \cdot e^{\frac{t}{RC}} + c$ $\therefore V_c \cdot e^{\frac{t}{RC}} = V_I \cdot e^{\frac{t}{RC}} + c$ $\therefore V_c = V_I + c \cdot e^{-\frac{t}{RC}}$ $c = V_O - V_I$ $\therefore V_c = V_I + (V_O-V_I) \cdot e^{-\frac{t}{RC}}$ QED FirstChildUserIdTAG: 141000 FirstChildUserNameTAG: OrinE FirstChildCreateTimeTAG: 2012-10-27T17:02:36Z SecondChildTAG: Hi, sorry but I don't see the difference between the two proofs, I would be grateful if you could elaborate. My question was why $c = V_O-V_I$? SecondChildUserIdTAG: 85490 SecondChildUserNameTAG: yashsavani SecondChildCreateTimeTAG: 2012-10-27T17:29:30Z SecondChildTAG: can you explain how do we enter math thanks SecondChildUserIdTAG: 623210 SecondChildUserNameTAG: preveen SecondChildCreateTimeTAG: 2012-10-27T18:09:02Z SecondChildTAG: I should have said your math didn't display right - the $ signs were missing. You get c by substituting the initial values in - you know Vc is Vo at t = 0. SecondChildUserIdTAG: 141000 SecondChildUserNameTAG: OrinE SecondChildCreateTimeTAG: 2012-10-27T18:18:14Z SecondChildTAG: preveen https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/mathjax-tutorial-write-formula-forum/ SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-27T19:02:50Z SecondChildTAG: Thank you SecondChildUserIdTAG: 85490 SecondChildUserNameTAG: yashsavani SecondChildCreateTimeTAG: 2012-10-27T19:12:09Z IndexTAG: 3316 TitleTAG: What does VT depend on? (about MOSFETs) Currently I'm trying to build some digital circuits in our Circuit Sandbox. I want to know what does VT, the threshold voltage of the mosfets, depend on? I thought it depends on W/L, but it turns out that it's not true. UserIdTAG: 326787 UserNameTAG: Angstrem CreateTimeTAG: 2012-10-27T14:16:40Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: If you are just playing around the VT is usually defined by the datasheet for the particular device. The manufacturer find the VT by the sum of the flat-band voltage, twice the bulk potential and the voltage across the oxide due to the depletion layer charge. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-27T14:30:36Z SecondChildTAG: And what about our circuit sandbox? It seems, that VT equals 0.75V there. SecondChildUserIdTAG: 326787 SecondChildUserNameTAG: Angstrem SecondChildCreateTimeTAG: 2012-10-27T15:02:40Z SecondChildTAG: Congrats on the TA badge Pennypacker! :-) SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-10-27T16:12:28Z SecondChildTAG: Thanks! SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-27T16:33:46Z SecondChildTAG: Angstrem, in our circuit sandbox VT is 0.5V as you should know after [Lab 4][1]. It seems there's no way to change its value, you can only change K playing with the ratio L / W. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_4/Curve_Tracer/ SecondChildUserIdTAG: 376877 SecondChildUserNameTAG: AndBre SecondChildCreateTimeTAG: 2012-10-27T17:23:43Z SecondChildTAG: Congrats Pennypacker! Welcome to the Community TA! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-27T22:36:57Z SecondChildTAG: Thank you too, Myrimit! SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-27T23:10:54Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-27T23:20:07Z FirstChildTAG: Threshold voltage depends on manufacturing process and materials used: http://en.wikipedia.org/wiki/Threshold_voltage W and L influence K: http://en.wikipedia.org/wiki/MOSFET K=$\mu_n$$C_{ox}$$\frac{W}{L}$ where $\mu_n$ - the charge-carrier effective mobility, $C_{ox}$ - the gate oxide capacitance per unit area, W - the gate width, L - the gate length $\mu_n$ and $C_{ox}$ are constants for given material and manufacturing process FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-27T18:59:12Z IndexTAG: 3317 TitleTAG: I made two different midterm-exams !? Yesterday, i was working on the first one and when I came back in the evening, it was replaced by another one ! ( this actually on line ) !!! [Edited - Do not discuss questions on the mid-term] UserIdTAG: 373752 UserNameTAG: Croak CreateTimeTAG: 2012-10-27T08:44:36Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I guess you were in last year's midterm and then you found this year's midterm ?! FirstChildUserIdTAG: 454609 FirstChildUserNameTAG: Alwahsh FirstChildCreateTimeTAG: 2012-10-27T08:53:19Z FirstChildTAG: I'm not sure which one you are talking about so I'm editing this. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-27T13:37:54Z IndexTAG: 3318 TitleTAG: Can anyone help me? My name is Fernando The system says: "Invalid input: R1R2 R1R3 R2R3 not permitted in answer" because I wrote a1 like:"R2R3/(R1R2+R1R3+R2R3)". Is there anyting wrong with my type of fonts? UserIdTAG: 335463 UserNameTAG: chinon98 CreateTimeTAG: 2012-10-26T23:10:19Z VoteTAG: 0 CoursewareTAG: Week 2 / Linear combinations of source strenghts CommentableIdTAG: 6002x_S3E2_Linear_Combinations_of_Source_Strengths NumberOfReplyTAG: 2 FirstChildTAG: Fernando, você é do Brasil? FirstChildUserIdTAG: 361823 FirstChildUserNameTAG: EliasOak FirstChildCreateTimeTAG: 2012-10-26T23:33:15Z FirstChildTAG: Hello, Fernando Try writing "R2*R3/(R1*R2+R1*R3+R2*R3)" instead. The system probably can't tell where variable name ends so it needs all multiplications to be denoted explicitly. FirstChildUserIdTAG: 405936 FirstChildUserNameTAG: Whible FirstChildCreateTimeTAG: 2012-10-26T23:28:12Z SecondChildTAG: Yes, that would be the answer. It worked for me too. Thank you SecondChildUserIdTAG: 221050 SecondChildUserNameTAG: jucapini SecondChildCreateTimeTAG: 2012-10-28T03:12:53Z IndexTAG: 3319 TitleTAG: Mid term problem 6 Hi, I would appreciate if the TA will publish the result after closure of mid-term exam. I would like to know where did I go wrong. Regards, Susant UserIdTAG: 246991 UserNameTAG: spatra CreateTimeTAG: 2012-10-26T20:29:05Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: We should try to avoid discussing this topic at this time. I am confident the staff will enable you to review this topic at a later date. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-26T20:49:42Z SecondChildTAG: Pennypacker, welcome to the Community TA party. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-26T22:58:37Z IndexTAG: 3320 TitleTAG: Hey I'm new here Hi I am new here and can'I start doing all the homeworks ! and then passing the exam? please help me ! UserIdTAG: 718305 UserNameTAG: daidi CreateTimeTAG: 2012-10-26T15:16:12Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Try taking the Midterm exam. See if you could answer it, and remember it is an open book exam . As of now it might be difficult for you to go through the whole lecture sequences/ material form week 1, so I suggest you to take a quick peek at the exam and take a look at those concepts in the text/lecture slides. FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-10-26T16:15:06Z FirstChildTAG: Quite a few homework deadlines have passed already and the midterm is going on. You could however stick around and view the material at your own pace. The course will be offered again during Spring next year. You'll need to enroll again that time. FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-10-26T15:20:49Z SecondChildTAG: I dont think that course wil be free... SecondChildUserIdTAG: 151762 SecondChildUserNameTAG: AADHITHYA SecondChildCreateTimeTAG: 2012-10-28T10:00:50Z FirstChildTAG: Hi, Sorry Dude... > **As Myrimit Said:** There are chances to get the certificate but you will not get the full grade, that is to say, you will not have chances to get a final score of 100% but yes to get an A with 91% (in the case that you do Midterm and Final Exam with 100% and also submit 100% of WEEK 6,7,8,9,10,11 and 12 assignments)... GENERAL > Remember that this Course has: 12 Homeworks (but if you read syllabus here, they tell you that you can skip two without penality, that means that you will still have 100% of Homework score if you make only 10). If you get 100% in your Homework score, that homework score it will contribute your final score only a 15%. So, if you have done so far 50% of your homework, it will contribute with your final score with (50%*15%)/100% = 7.5%. 12 Labs (but if you read syllabus, they tell you that you can skip two without penality, that means that you will still have 100% of Labs score if you make only 10). 1 Midterm Exam. If you get 100% in the Midterm, it will contribute with your final score with 30%. 1 Final Exam.If you get 100% in the Final, it will contribute with your final score with 40%. So, it is not that late to get the certificate :), but you will have to do a lot of effort and work from now to the end of the Course... Also, as some students have said it to you, you can follow this Course and re-take it again, but that is always up to you :). See you! I hope this can help you. > **Regards:** asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-11-04T07:10:17Z IndexTAG: 3321 TitleTAG: wasnt midterm too easy ? i was expecting something more challenging. I hope this is not how midsems are like in the real MIT course :P UserIdTAG: 214085 UserNameTAG: shohin CreateTimeTAG: 2012-10-26T13:49:52Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: In the real exam you only have 2 h, you don't know if your answer is right or wrong, and probably you didn't sleep too well with a night before because you were worried sick. But of course ,nothing that a guy like you can't handle. FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-10-26T14:06:06Z FirstChildTAG: Let's leave discussion of the midterm until after the deadline. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-26T14:10:50Z IndexTAG: 3322 TitleTAG: Dimensionless B in Middterm's Q6 [Edited - Do NOT discuss questions on the midterm] UserIdTAG: 148226 UserNameTAG: Apprentice CreateTimeTAG: 2012-10-26T13:28:25Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You are not allowed to post questions/considerations about midterm. FirstChildUserIdTAG: 376877 FirstChildUserNameTAG: AndBre FirstChildCreateTimeTAG: 2012-10-26T13:46:01Z IndexTAG: 3323 TitleTAG: i dint get it i wrote capocks for the resister and voltage values but i dint get the right answer? can someone put some light on it....??? UserIdTAG: 541174 UserNameTAG: HASITAKAJA CreateTimeTAG: 2012-10-26T11:05:08Z VoteTAG: 0 CoursewareTAG: Week 3 / Isolation Via Thevenin CommentableIdTAG: 6002x_isolation_via_thevenin NumberOfReplyTAG: 1 FirstChildTAG: you can get Vth by calculating the value of V on Rp, as shown : I=VS/(RS+RP) current in the circuit then V=I*RP=(VS*RP)/(RS+RP) about Rth make the independent voltage source SC and if it was independent current source make it OC. and calculate Rth you will see that RS and RP are parallel, so Rth=(RS*RP)/(RS+RP) FirstChildUserIdTAG: 397838 FirstChildUserNameTAG: Muhamad_Alaa FirstChildCreateTimeTAG: 2012-10-26T13:02:19Z IndexTAG: 3324 TitleTAG: ¡¡¡Help!!! Problems are changing wording values When i entered several times to solve the problems i have found that the wording is changing and i've commited serveral mistakes with no return. Please, could you help me? UserIdTAG: 244225 UserNameTAG: lerimock CreateTimeTAG: 2012-10-26T06:32:00Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: What are you talking about? Please be more specific. FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-10-26T07:22:02Z IndexTAG: 3325 TitleTAG: Multiplying Inner Req with Outer Req I'm confused about how we can take the inner Req and multiply with the outer Req. The inner Req being the Req to the nth-1 power and outer Req being the R nth solution. When trying to solve, we have a quadratic resultant. We take the denominator and multiply with the outer Req to have two quadratics equal one another. Is this allowed? UserIdTAG: 62354 UserNameTAG: graysilver CreateTimeTAG: 2012-10-26T04:28:16Z VoteTAG: 0 CoursewareTAG: Week 1 / Long Resistor Chains CommentableIdTAG: 6002x_long_resistor_chains NumberOfReplyTAG: 1 FirstChildTAG: Hello, graysilver You have an infinite ladder of resistors. If you look at the first step of that ladder, what is it connected to? It's connected to exactly the same infinite ladder of resistors. So infiniteladder = 1step :: infiniteladder. You write that down and solve it, and that is done in the video. That is the magic of 'infinite'. If this clears your doubts - stop reading this post! You can take a different approach. Start with 1 step ladder, add 2nd step, add 3rd step and so on and so forth and look at what happens if you keep going. So what happens? Resistance of 1 step ladder is R1 = R+R. For two steps it is R2 = R + 1/(1/R + 1/(R+R)), going further: R3 = R + 1/(1/R + 1/(R + 1/(1/R + 1/(R+R)))) R4 = R + 1/(1/R + 1/(R + 1/(1/R + 1/(R + 1/(1/R + 1/(R+R)))))) ... Now let me take R out of brackets: R1 = R*(1+1/1) R2 = R*(1+1/(1+1/(1+1/1))) R3 = R*(1+1/(1+1/(1+1/(1+1/(1+1/1))))) R4 = R*(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/1))))))) ... Where 1 + 1/1 is another form of 1 + 1. I don't know how to make pretty formulae in forums, so let me post a picture from [wikipedia][1] instead. ![enter image description here][2] Do you see where am i going here? Using wikipedia (not only wikipedia) notation we get Rinfinite = R*[1,1,1,1,1,1,1,1...]. For example [here][3] you can read that [1,1,1,1,1,1,1,1...] = (1+sqrt(5))/2 and it's exactly the same result as in the video. In the end you still calculate [1,1,1,1,1,1,1,1...] by solving x = 1 + 1/x so this approach isn't that different from one in the video. I hope this helps to clarify something and that my English wasn't too terrible. [1]: http://en.wikipedia.org/wiki/Continued_fraction [2]: http://upload.wikimedia.org/math/1/1/d/11dc14afeeb64dad18b916638aa287d7.png [3]: http://neilbickford.com/picf.htm FirstChildUserIdTAG: 405936 FirstChildUserNameTAG: Whible FirstChildCreateTimeTAG: 2012-10-26T11:15:16Z IndexTAG: 3326 TitleTAG: Input format for currents in S1E15 My answer to the last question reads like: -(2(i1) + i3 + i2) However, the checking tool can't parse it. Can anyone help me out with this? What is the correct format? UserIdTAG: 78396 UserNameTAG: dharav CreateTimeTAG: 2012-10-26T03:15:06Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL and KCL exercises CommentableIdTAG: 6002x_kvl_kcl_exercises NumberOfReplyTAG: 1 FirstChildTAG: Try putting a * between 2 and ( -(2*(i1) + i3 + i2) Does it work? FirstChildUserIdTAG: 303762 FirstChildUserNameTAG: leocarrasco FirstChildCreateTimeTAG: 2012-10-26T03:36:59Z IndexTAG: 3327 TitleTAG: Doubt Q5 - Mid-term - staff a) don't talk about the midterm while it is still open b) Your question didn't seem to actually have anything to do with the midterm? At the very least I couldn't understand it. It just looked like it was mid-term-ish, thus the editing. UserIdTAG: 211715 UserNameTAG: pitankar CreateTimeTAG: 2012-10-26T02:08:21Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3328 TitleTAG: Miss the first half of this course for mailbox filter reason How to catch up or take a new session? UserIdTAG: 303197 UserNameTAG: tengw CreateTimeTAG: 2012-10-26T01:34:47Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: 6.002x will be offered again in the spring. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-10-26T02:07:35Z SecondChildTAG: Will it be for free again? SecondChildUserIdTAG: 277808 SecondChildUserNameTAG: Hemanthmps SecondChildCreateTimeTAG: 2012-10-26T04:35:24Z SecondChildTAG: edx is always free SecondChildUserIdTAG: 397838 SecondChildUserNameTAG: Muhamad_Alaa SecondChildCreateTimeTAG: 2012-10-26T13:21:58Z IndexTAG: 3329 TitleTAG: Midterm Q4 problem!!! Hello [Edited - Do not discuss mid-term questions] UserIdTAG: 71413 UserNameTAG: BorSusMos CreateTimeTAG: 2012-10-25T22:51:25Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3330 TitleTAG: First Voltage V(t=0-) edit: I had a mistake in the unit conversion and figured my mistake. If you have a different result yourself, you can figure the result like that (spoiler alert): I figured the steady-state voltage to be 1V so I need to have an exponential decay from 3V to 1V: vC (t) = 3V - 2V * (1-exp(-(t-0.001s)/RC) = 3V - 2V * ( 1 - exp( -(t-0.001s)/1ms) and: vC (t=0-) = 3V - 2V * (1-exp(-1)) = 1.73576V UserIdTAG: 388844 UserNameTAG: AnAppleADay CreateTimeTAG: 2012-10-25T22:32:23Z VoteTAG: 0 CoursewareTAG: Week 8 / Initial Conditions CommentableIdTAG: 6002x_Initial_Conditions NumberOfReplyTAG: 1 FirstChildTAG: the problem says that vc=3 at time 1 ms,,, why you replace 1 ms in that equation and the result is 1.73....??? and the problem says that at that time the vc is 3 v?? plase help FirstChildUserIdTAG: 115656 FirstChildUserNameTAG: rolando277 FirstChildCreateTimeTAG: 2012-11-15T00:20:08Z IndexTAG: 3331 TitleTAG: Download link missing Please provide download link for this video. Thanks. UserIdTAG: 120983 UserNameTAG: mykisaacs CreateTimeTAG: 2012-10-25T19:02:25Z VoteTAG: 0 CoursewareTAG: Week 8 / Response To A Pulse As Pulse Gets Narrower CommentableIdTAG: 6002x_Response_To_A_Pulse_As_Pulse_Gets_Narrower NumberOfReplyTAG: 1 FirstChildTAG: http://www.youtube.com/watch?gl=IN&feature=player_embedded&v=hCdWt5ainuw FirstChildUserIdTAG: 277808 FirstChildUserNameTAG: Hemanthmps FirstChildCreateTimeTAG: 2012-10-26T04:39:46Z IndexTAG: 3332 TitleTAG: midterm exam feeling good after finishing the midterm exam.........:) almost scored 100% if not made a silly mistake......... UserIdTAG: 162671 UserNameTAG: tuhin1991paul CreateTimeTAG: 2012-10-25T18:25:48Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: cant find the mid-term exam in my account here,pls help me out,joshua from nigeria FirstChildUserIdTAG: 693792 FirstChildUserNameTAG: Jodapson FirstChildCreateTimeTAG: 2012-10-25T19:21:35Z SecondChildTAG: Click on "Courseware", located in the top left of your screen. You will then notice the "Midterm Exam" located between weeks 6 and 7 on the left hand side of your screen. Good luck! SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-25T19:41:42Z FirstChildTAG: Congratulations Paul. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-25T19:42:27Z IndexTAG: 3333 TitleTAG: Why? Why 0.00099 and not 0.001? UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-10-25T17:24:02Z VoteTAG: 0 CoursewareTAG: Week 7 / Fall Time Constant CommentableIdTAG: 6002x_Fall_Time_Constant NumberOfReplyTAG: 1 FirstChildTAG: I gess you mean ...*why not 0.001* FirstChildUserIdTAG: 414462 FirstChildUserNameTAG: Pablo_C FirstChildCreateTimeTAG: 2012-10-25T17:45:49Z SecondChildTAG: Zing ;) SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-25T17:49:33Z SecondChildTAG: Actually you have 2 resistors RL and RON for the first transistor of the pattern from S14E1.I've made the Thevenin equivalent of the first mosfet (wich is a voltage divider composed by RL and RON) and Rth = (RL*RON)/(RL+RON) = 9.9 Ohms.Tau = RTH*Cgs = 0.1*10^-12*9.9 = 0.00099 ns . SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-10-25T18:40:06Z IndexTAG: 3334 TitleTAG: instruction i want to know about the procedure of lab work, UserIdTAG: 621018 UserNameTAG: abhishek0506 CreateTimeTAG: 2012-10-25T16:22:32Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Troubleshooting NumberOfReplyTAG: 1 FirstChildTAG: Which lab are you referring to? FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-25T19:45:19Z IndexTAG: 3335 TitleTAG: midterm i just went to "courseware" and says i was most recently in midterm even tho i wuzn't. time hasn't started for me yet rite? UserIdTAG: 365225 UserNameTAG: doublegj526 CreateTimeTAG: 2012-10-25T13:55:35Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: No, the page is showing that you were there last - simply to advertise that you can take the test from now. When you click into the midterm, you have a page of text to read - on that page it will say that if you go to the next page you have started and have 24 hours to complete the exam. FirstChildUserIdTAG: 243871 FirstChildUserNameTAG: Vanilly FirstChildCreateTimeTAG: 2012-10-25T14:25:42Z IndexTAG: 3336 TitleTAG: Be true to your MITx: 6.002x ! I'll take the stage to talk about it. UserIdTAG: 323963 UserNameTAG: gtrf CreateTimeTAG: 2012-10-25T12:40:35Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3337 TitleTAG: Midterm Deadline? So the midterm we need to complete by Sunday however if I start it right now which is Thursday is there still that 24 hour time line so I would need it done by Friday? UserIdTAG: 219780 UserNameTAG: ReconIII CreateTimeTAG: 2012-10-25T12:32:33Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: you have to finish it before 24 hours since you opened it or by sunday night. Whichever comes first. ( btw: I am not staff but I've seen this somewhere ) FirstChildUserIdTAG: 454609 FirstChildUserNameTAG: Alwahsh FirstChildCreateTimeTAG: 2012-10-25T12:56:24Z SecondChildTAG: The Sunday night being referred to is **11:59pm BOSTON TIME**, not your local time. See the Course Info tab above. SecondChildUserIdTAG: 9673 SecondChildUserNameTAG: xp42 SecondChildCreateTimeTAG: 2012-10-25T15:08:12Z FirstChildTAG: Hi. [Just Check It][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5088dbe5959a592b00000024 > **Regards:** asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-10-25T15:00:06Z IndexTAG: 3338 TitleTAG: Midterm Q- (content edited) UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-25T11:21:01Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Please do not discuss any of the questions on the exam. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-10-25T13:57:33Z IndexTAG: 3339 TitleTAG: midSeM what is end time of exam in INDIA. Can i save answer without checking it. UserIdTAG: 342135 UserNameTAG: vikash902 CreateTimeTAG: 2012-10-25T10:13:54Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: We in INDIA are +9:30 from Boston. So, the end time of exam in India should be 09:29 am on MONDAY morning. However, once you start the test you need to answer it within 24hrs of starting it. There is no clock with the test, keep track of time yourself. Yes, there is a save button that allows you to save your answer without checking. BEST OF LUCK!! FirstChildUserIdTAG: 315681 FirstChildUserNameTAG: nahuja1 FirstChildCreateTimeTAG: 2012-10-25T10:26:00Z SecondChildTAG: thank you SecondChildUserIdTAG: 342135 SecondChildUserNameTAG: vikash902 SecondChildCreateTimeTAG: 2012-10-25T13:28:23Z IndexTAG: 3340 TitleTAG: Mistake in Textbook Page No.195,Example 4.1,In the last part the calculation is wrongly shown.Given that iDS=4mA,but in the calculation it is taken as 8mA. UserIdTAG: 156835 UserNameTAG: kphariprakash1968 CreateTimeTAG: 2012-10-25T08:43:50Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3341 TitleTAG: midterm deadline hi they said that the dead line on 28 oct 11:59 is this mean that sunday will be with us in the exam ? UserIdTAG: 221617 UserNameTAG: konan CreateTimeTAG: 2012-10-25T08:20:45Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: konan, yes the exam will be open on Sunday FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-10-25T12:30:10Z FirstChildTAG: Hi. [Just Check It][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5088dbe5959a592b00000024 > **Regards:** asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-10-25T14:57:43Z IndexTAG: 3342 TitleTAG: Is there a bug in Midterm Q3 first problem? [Edited- Do NOT discuss mid-term questions] UserIdTAG: 18877 UserNameTAG: johnkhiangte CreateTimeTAG: 2012-10-25T07:33:48Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: Should you be posting even as much info as the title of a question on the Midterm? I thought that was expressly forbidden. If I were you, I'd edit my post ASAP. FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-10-25T08:03:45Z FirstChildTAG: no,, i didnt face any problem FirstChildUserIdTAG: 489065 FirstChildUserNameTAG: amiths FirstChildCreateTimeTAG: 2012-10-25T07:50:46Z FirstChildTAG: i didnt face any prob FirstChildUserIdTAG: 365309 FirstChildUserNameTAG: chetnasinghaldas FirstChildCreateTimeTAG: 2012-10-25T08:00:02Z FirstChildTAG: Thanks guys. it must have been the network. everything is okay now. FirstChildUserIdTAG: 18877 FirstChildUserNameTAG: johnkhiangte FirstChildCreateTimeTAG: 2012-10-25T08:12:04Z FirstChildTAG: Due to the Honor Code restrictions on Discussion during the Mid-term exam, please refrain from even posting Exam Questions (for example Q1) as they appear in the exam, parts of Questions, or even *the topic they deal with* (for example, "I'm having a problem with Q1, where it talks about...." ) in Discussion, **because many users have not even opened their exams yet!** FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-10-25T16:11:46Z SecondChildTAG: JerseyMark, go ahead and edit questions and close posts. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-25T17:08:07Z IndexTAG: 3343 TitleTAG: Shouldn't be the C in Farads? Here C was expressed in MiliFarads! to find the Tao. UserIdTAG: 369339 UserNameTAG: vargaslen CreateTimeTAG: 2012-10-24T18:26:58Z VoteTAG: 0 CoursewareTAG: Week 6 / simple circuit with resistor and capacitor CommentableIdTAG: 6002x_Simple_RC_Circuit NumberOfReplyTAG: 1 FirstChildTAG: micro FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-24T20:50:42Z IndexTAG: 3344 TitleTAG: Error in calculator? LAB 7 q2 I think there's an error on the cos() function. I suspect it because of this: cos(2000*PI*0.0005)=0.9984971499 according to my beloved Casio calculator. But. cos(2000*PI*0.0005)=-1.0 according to the edx calculator. Even after getting the formula right, I couldn't answer the next question, v(0.0005), because it keeps rejecting my results. It was when I got it from the Edx calculator when I got it right. This error drives the calculation to a very confusing result. I don't know whether it is my computer's fault, or if there's an error on the calculators code. UserIdTAG: 482646 UserNameTAG: elgambitero CreateTimeTAG: 2012-10-24T17:42:40Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 5 FirstChildTAG: Measuring the value on the transient analysis gives a very similar number... I'm confused. FirstChildUserIdTAG: 482646 FirstChildUserNameTAG: elgambitero FirstChildCreateTimeTAG: 2012-10-24T17:45:34Z FirstChildTAG: Please, please, please. cos(Pi) is one of those values we memorize. It's **-1**. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-24T18:23:49Z SecondChildTAG: THANK YOU skyhawk!!! SecondChildUserIdTAG: 285945 SecondChildUserNameTAG: lindalapiso SecondChildCreateTimeTAG: 2012-10-30T08:58:38Z FirstChildTAG: Casio calculator in degree mode. Mystery solved. I'll pay more attention next time... FirstChildUserIdTAG: 482646 FirstChildUserNameTAG: elgambitero FirstChildCreateTimeTAG: 2012-10-25T10:00:34Z SecondChildTAG: ditto! SecondChildUserIdTAG: 285945 SecondChildUserNameTAG: lindalapiso SecondChildCreateTimeTAG: 2012-10-30T08:59:15Z FirstChildTAG: Bet your calculator was in degree mode! FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-24T18:25:41Z FirstChildTAG: Your Casio calculator has 3 operation modes: degree, radian and grado. This calculation must be done in radians. FirstChildUserIdTAG: 361823 FirstChildUserNameTAG: EliasOak FirstChildCreateTimeTAG: 2012-10-24T18:34:16Z SecondChildTAG: Press key DRG to change mode RAD. SecondChildUserIdTAG: 464744 SecondChildUserNameTAG: attache SecondChildCreateTimeTAG: 2012-10-24T20:20:45Z IndexTAG: 3345 TitleTAG: A missing term(Too many jumps!) I think there is a missing term here: What happened with 1/RK (minute 4)? UserIdTAG: 369339 UserNameTAG: vargaslen CreateTimeTAG: 2012-10-24T16:01:46Z VoteTAG: 0 CoursewareTAG: Week 6 / Two-stage MOSFET amplifier CommentableIdTAG: 6002x_Two_Stage_MOSFET_amplifier NumberOfReplyTAG: 0 IndexTAG: 3346 TitleTAG: 2nd part I can't understand how the answer of the second question is 1.0. Could someone please explain? Thanks. UserIdTAG: 222911 UserNameTAG: bhaswardg CreateTimeTAG: 2012-10-24T15:15:38Z VoteTAG: 0 CoursewareTAG: Week 6 / Norton circuit capacitor exercise CommentableIdTAG: 6002x_norton_capacitor_circuit_exercise NumberOfReplyTAG: 1 FirstChildTAG: you are probably getting 1000 as the answer. The answer is required in volts per millisecond. So divide by 1000. FirstChildUserIdTAG: 19863 FirstChildUserNameTAG: Samir FirstChildCreateTimeTAG: 2012-10-27T05:57:15Z SecondChildTAG: thanks! :) SecondChildUserIdTAG: 222911 SecondChildUserNameTAG: bhaswardg SecondChildCreateTimeTAG: 2012-10-27T06:24:12Z IndexTAG: 3347 TitleTAG: how can i get any specific page in textbook????????? how can i get any specific page in textbook????????? UserIdTAG: 346003 UserNameTAG: anupamshakya CreateTimeTAG: 2012-10-23T14:57:01Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: see here https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5077436048cffb1f00000019 FirstChildUserIdTAG: 9673 FirstChildUserNameTAG: xp42 FirstChildCreateTimeTAG: 2012-10-23T15:36:33Z IndexTAG: 3348 TitleTAG: textbook i want to download textbook..how can I dwnld it??? UserIdTAG: 346003 UserNameTAG: anupamshakya CreateTimeTAG: 2012-10-23T14:55:22Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You have to buy it. http://www.amazon.com/Anant-Agarwal/e/B001K8QU2S FirstChildUserIdTAG: 9673 FirstChildUserNameTAG: xp42 FirstChildCreateTimeTAG: 2012-10-23T15:38:18Z SecondChildTAG: Hi xp42! Welcome back! Let us know if you plan on being as active in the forum this term as you were last term. SecondChildUserIdTAG: 96452 SecondChildUserNameTAG: Lyla SecondChildCreateTimeTAG: 2012-10-23T15:45:49Z SecondChildTAG: I'm currently waiting for Dave to **FIX** mathjax. And I'm hoping a bunch of really annoying bugs in the interface will go away soon too. I have little interest in fighting with the tools. I attempted to offer hints for weeks 5 and 6, but stopped when matjax failed. SecondChildUserIdTAG: 9673 SecondChildUserNameTAG: xp42 SecondChildCreateTimeTAG: 2012-10-23T16:20:02Z SecondChildTAG: Buy it you will not regrate it. It will be handy reference for circuit design SecondChildUserIdTAG: 272523 SecondChildUserNameTAG: Jivraj SecondChildCreateTimeTAG: 2012-10-23T21:33:19Z SecondChildTAG: you can download it , it isnt a problem SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-10-24T12:27:58Z IndexTAG: 3349 TitleTAG: When will the faculty be avialiable for discussion i saw in the intro videos that faculty will be available in the discussion.i heard that timings will be specified for it.can someone help me and tell where will the timings be specified UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-23T14:29:32Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3350 TitleTAG: resistor value hi,i understood the concept of replacing the bulb in order to find current but what should be the value of resistor that we take. should it be any random value or from where do we get the value of resistor? UserIdTAG: 624563 UserNameTAG: Gypsymaitu CreateTimeTAG: 2012-10-23T13:54:44Z VoteTAG: 0 CoursewareTAG: Week 1 / Lumped Element Abstraction CommentableIdTAG: 6002x_lumped_element_abstraction NumberOfReplyTAG: 1 FirstChildTAG: Hi! If you want to simulate a real bulb you must to know some more about it. Take a look closer on it and you can see a mark, which tell to us that “Dissipated power is 100Watts at 220V(in my country)”. So, using a dissipated power formula for a resistor P=U^2/R, we can calculate it. So, R=U^2/R. But!!! 1. This value of resistance is approximate! To have more accuracy you must remember, that dissipated power consist of at least 2 parts: heat and light. In bulb, approximately 80% of power spent for a heat (Light is just 20% of power. Engineers were sure that it was made for a light? ;) ). So to be more accurate we must to use this coefficient because it is just for a 80% a resistor in such case R=0.8*U^2/R 2. Calculated value of “R” approximately true just @220V(written on the bulb). At another voltage you can’t be sure, that the resistance is the same, because it is NOT a linear device. You can see it by yourself trying to measure current through a bulb at different voltage and calculating resistance using an Ohm law. And this is the second way to get it’s resistance. Sorry for my English, I am not a native speaker. FirstChildUserIdTAG: 94834 FirstChildUserNameTAG: Gelmut FirstChildCreateTimeTAG: 2012-10-24T07:50:43Z IndexTAG: 3351 TitleTAG: Lab 7 Hi please help me to calculate the integral for Vc (t) and thankx UserIdTAG: 331441 UserNameTAG: abdessamade CreateTimeTAG: 2012-10-23T12:22:25Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: There's a list of integrals [here][1]. If you don't want to do the substitution of values yourself, you can put the given formula for I(t) into Wolfram Alpha (with "integral(...)"), or into the [Number Empire's Integral Calculator][2]. Be warned that the correct answer is more than just the formula that you get from any of the above methods (cf. my other post [here][3]). [1]: http://en.wikipedia.org/wiki/Lists_of_integrals#Tables_of_integrals [2]: http://www.numberempire.com/integralcalculator.php [3]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5086890dda70411f00000099 FirstChildUserIdTAG: 184153 FirstChildUserNameTAG: tthngn FirstChildCreateTimeTAG: 2012-10-24T17:44:39Z SecondChildTAG: Not sure if someone who doesn't know the integral of sin(x) (It's -cos(x)) will know how to handle sin(k*x) or what to do with the limits on a definite integral! Or realize that they should be working in radians. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-24T19:34:09Z IndexTAG: 3352 TitleTAG: Any IITians (Indian institute of technology) here ? I am from IIT G. Do you guys find it easy or hard ? UserIdTAG: 214085 UserNameTAG: shohin CreateTimeTAG: 2012-10-23T09:59:33Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: i am from NSIT,DELHI..The course seems pretty easy.. What about you? FirstChildUserIdTAG: 368712 FirstChildUserNameTAG: jmen FirstChildCreateTimeTAG: 2012-10-23T11:35:52Z SecondChildTAG: yep same here. SecondChildUserIdTAG: 214085 SecondChildUserNameTAG: shohin SecondChildCreateTimeTAG: 2012-10-23T16:47:35Z SecondChildTAG: which branch shohin? and year? SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-23T17:19:40Z SecondChildTAG: mechanical major, electronics minor. 2nd yr. u ? SecondChildUserIdTAG: 214085 SecondChildUserNameTAG: shohin SecondChildCreateTimeTAG: 2012-10-25T04:09:02Z IndexTAG: 3353 TitleTAG: Thevenin Voltage expression Good morning Sir! I noticed that about H6P2 phase Inverter, you asked us to express Thevenin voltage with K, VIN, VT and vin. I tried, but found only (-vin) instead, but your system didn't recognize that expression because of none existence of VIN, VT and K. How should I do for those marks I lost? Rodolphe MAY UserIdTAG: 373916 UserNameTAG: larou CreateTimeTAG: 2012-10-23T09:54:55Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3354 TitleTAG: Problem in ac analysis hello sir, i m unable to do ac analysis bcoz the ac analysis tool is missing in ckt simulator . plz help me to sort out this. UserIdTAG: 403456 UserNameTAG: SandeepKumarKhichar786 CreateTimeTAG: 2012-10-23T06:34:26Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: are you just joined the course like me? FirstChildUserIdTAG: 733429 FirstChildUserNameTAG: EhsanMokhtari FirstChildCreateTimeTAG: 2012-11-01T18:00:03Z IndexTAG: 3355 TitleTAG: Problem in getting the answers being accepted Sir, while I was attempting to enter the answers into the space provided for homework 6, it was not accepting the answers showing errors that the parameters in which the term needed to be expressed thereby mentioned with are not allowed & now when the answers are shown after the given time, many of them are exactly the same I was attempting to enter.So, many of my answers were not accepted. UserIdTAG: 435644 UserNameTAG: abhinav92 CreateTimeTAG: 2012-10-23T04:54:14Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3356 TitleTAG: H6P2 part D Although the answer has been released, I still don't get part D of H6P2. How can you express gm2 (transconductance of Q2) in terms of VIN? Q2's gate is connected to its drain (=VDD) and not not VIN, so VIN is not variable in Q2s equation. I expected gm2 to be in terms of VDD, vOUT and VT. UserIdTAG: 114881 UserNameTAG: xvink CreateTimeTAG: 2012-10-22T15:16:44Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Since the same current passes through both MOSFETs VGS must be the same for MOSFETs. This establishes a relationship between VIN, VDD, and VOUT. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-22T16:37:24Z SecondChildTAG: Is this because they used equal MOSFETs? SecondChildUserIdTAG: 114881 SecondChildUserNameTAG: xvink SecondChildCreateTimeTAG: 2012-10-22T16:41:58Z SecondChildTAG: Yes, if the K's are the same then they cancel out. But even if they are not equal, one can still write an equation connecting VIN, VDD, and VOUT. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-22T16:47:16Z IndexTAG: 3357 TitleTAG: Error on algebric formula for HW6 problem 1 & 2 I have a lot of issue on posting my answer.The system keep telling me either "the term gm1 is not allowed here" or "the system cannot parse the formula...". Does anybody face the same problem and fix it? UserIdTAG: 36838 UserNameTAG: Atomoly CreateTimeTAG: 2012-10-22T09:08:26Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: 1. for H6P2 there should not be gm1 or gm2 - you asked to write formulas for them 2. check parentheses FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-22T15:50:26Z IndexTAG: 3358 TitleTAG: lab6 i did not get the right answer while taking vout=2.5 what is the value of ron UserIdTAG: 550401 UserNameTAG: revathisingh CreateTimeTAG: 2012-10-22T07:57:16Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50843aed24ad21270000013d FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-10-22T08:21:15Z FirstChildTAG: **Don't** use the IDEAL value when making the calculation for $V_{OL} = V_S \cdot R_{ON} / (R_{ON} +R_L )$, use the MEASURED value from the Transient plot. It will be lower. Measure the bottom of the blue curve. Not much time left though... FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-10-22T08:42:36Z IndexTAG: 3359 TitleTAG: HomeWork 6 Problem 2 Can someone please help me with the thevinin resistance for part 2. Isn't the resistance of the small signal of the mosfet just the inverse of the transconductance? UserIdTAG: 435197 UserNameTAG: RichmondRichter CreateTimeTAG: 2012-10-22T02:54:45Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: You're finding the thevenin resistance of the entire phase inverter through the vout terminals, not just a single mosfet device. Do you have the small-signal equivalent circuit? All you have to do is find the thevenin resistance of this circuit as seen through vout-- and do this by injecting a "test" current i_test into the vout terminal and solving for the ratio of vout to i_test. FirstChildUserIdTAG: 390034 FirstChildUserNameTAG: kahlil FirstChildCreateTimeTAG: 2012-10-22T03:21:12Z SecondChildTAG: I got the answer but it still doesnt make sense. The inverse of gm1 and gm2 are the resistors correct? so the why isn't rth both of those in parallel looking in from vout? and Also can you give me a hint on the thevinin voltage? which is vth I'm presuming. SecondChildUserIdTAG: 435197 SecondChildUserNameTAG: RichmondRichter SecondChildCreateTimeTAG: 2012-10-22T04:45:21Z SecondChildTAG: Did you do **Exercise S11E2**? If so, that should almost exactly lead you in the correct direction. Not much time left though... SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-10-22T08:16:45Z FirstChildTAG: Current through the resistor depends on the voltage across that same resistor. Current through the VCCS (being small signal model for mosfet) depends on the voltage across anywhere in the circuit (hence the prefix "trans"). Thats why you cant just use the inverse of the transconductance. But in case of Q2 mosfet its control terminal is set so current is controlled by just the voltage across Q2. And so it does indeed behaves like a normal resistor. But that is not the case for Q1. FirstChildUserIdTAG: 405936 FirstChildUserNameTAG: Whible FirstChildCreateTimeTAG: 2012-10-22T03:22:16Z SecondChildTAG: Ok I got it I think, so when you looking for rth you have to short circuit VIN, which makes Q1 and open circuit..living gm2 and rth is the inverse of that.... any help with VTH? SecondChildUserIdTAG: 435197 SecondChildUserNameTAG: RichmondRichter SecondChildCreateTimeTAG: 2012-10-22T04:58:53Z FirstChildTAG: Try it and see. If you haven't already, I would suggest you draw the small signal circuit and short all independent sources to find Rth, just like you would do normally. You will find something unique, but not unexpected, about how one of the dependent current sources is open circuited, but not for reasons why an independent source would. FirstChildUserIdTAG: 132828 FirstChildUserNameTAG: rharris FirstChildCreateTimeTAG: 2012-10-22T03:33:11Z SecondChildTAG: for the small circuit model what happens to VDD, since its also an input does it have a small signal voltage as its small signal model? I know VIN become vin, and both mosfets become gm*vgs but I have no idea what happens to VDD SecondChildUserIdTAG: 435197 SecondChildUserNameTAG: RichmondRichter SecondChildCreateTimeTAG: 2012-10-22T04:47:55Z FirstChildTAG: You can read chapter 8 from A&L or a little hint, you have to take the small signal current, and this about, what this really is, it's units for example, and you'll find it straight ahead FirstChildUserIdTAG: 231676 FirstChildUserNameTAG: Roosemberth FirstChildCreateTimeTAG: 2012-10-22T05:38:43Z IndexTAG: 3360 TitleTAG: Source of the noise? I'm confused about the source of the noise. Is it due to capacitance between the long power wire and other wires, as in the previous videos? UserIdTAG: 203941 UserNameTAG: JacobN CreateTimeTAG: 2012-10-22T01:54:56Z VoteTAG: 0 CoursewareTAG: Week 7 / Bypass Capacitor CommentableIdTAG: 6002x_Bypass_Capacitor NumberOfReplyTAG: 1 FirstChildTAG: The noise in the circuit is from the 100MHz input at the MOSFET. So they use the capacitor to filter or reduce the noise. FirstChildUserIdTAG: 359310 FirstChildUserNameTAG: AaronYeoh FirstChildCreateTimeTAG: 2012-10-22T02:40:48Z SecondChildTAG: yes and this input imho causes the MOSFET turning on and off and as a consequence a pulsed current goes through 1Ω resistor so we have a pulsed voltage drop and a pulsed voltage power rail. SecondChildUserIdTAG: 319453 SecondChildUserNameTAG: leonidasGr SecondChildCreateTimeTAG: 2012-10-22T16:33:55Z SecondChildTAG: I THINK they both are themselves a source of noise. they just try to cut across the problem and one can not understand a word they say. i wsh prof. aggarwal explain such tutorials.. SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-12-17T10:53:18Z IndexTAG: 3361 TitleTAG: H6P2: Finding Rth. With Itest. I got Vth, but now I'm stuck on getting Rth. I short the output port and find Itest, but it equates to zero due to gm1 being equal to gm2 and the answer I got for Vth. I'm following kahlil's directions at [H6P2][1] to find Vth and using the results to find Rth. It seems so simple, what am I doing wrong? [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/508282c564146d1f000000a4 UserIdTAG: 132828 UserNameTAG: rharris CreateTimeTAG: 2012-10-21T23:14:08Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: "It seems so simple, what am I doing wrong?" You are shorting the output port. You should short the independent source *vin* and place some test current source *Itest* at the output port. Then you calculate voltage across the test current source *Vtest* and do the division. Btw Q2 behaves exactly like a resistor since it has gate and drain shorted*, and if you short *vin* Q1 behaves like an open circuit. *for small signal model of course FirstChildUserIdTAG: 405936 FirstChildUserNameTAG: Whible FirstChildCreateTimeTAG: 2012-10-21T23:24:52Z SecondChildTAG: OK, I did short VIN for Q1, but I forgot to set VIN = 0 in the equation for gm1, which it should be K*(-VT). Is this correct? From what I read from kahlil (staff member), he said we didn't need to use the the resistance from Q2, just use Itest. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-22T00:08:09Z SecondChildTAG: No, no, no. Once u got gm1 and gm2 from large signal circuit u don't change them. VIN threated as some constant for small signal model. Have u drawn small signal model circuit? SecondChildUserIdTAG: 405936 SecondChildUserNameTAG: Whible SecondChildCreateTimeTAG: 2012-10-22T00:12:31Z SecondChildTAG: I mean once u get all vGS from large signal circuit u know all gm for small signal circuit. And that gms are parameters of elements u are using in small signal circuit. SecondChildUserIdTAG: 405936 SecondChildUserNameTAG: Whible SecondChildCreateTimeTAG: 2012-10-22T00:24:34Z SecondChildTAG: I have drawn the small signal circuit, that is where I I got vgs = -VOUT for Q2. I will re-draw it and see if something needs to be different. Due to the way this class uses upper and lower case variables for large and small signal circuits confuses me, and I have a hard time telling which variable is used when. Now that I know gm and VIN are generated from the large signal analysis and used in the small signal analysis, it will help me. Thanks. I read your first note again and It sounds like you meant connect a voltage source (Vtest) to the output port, then calculate Itest generated by Vtest. Ahah, I was shorting the output port like we do to find the Norton equivalent current. I'll try Vtest then calculate Itest. Thank you. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-22T02:16:42Z SecondChildTAG: Holy be-jeebers, I never knew that shorting out VIN in a small signal circuit for finding Rth open circuited the transistor? Being it is a dependent current source I thought you couldn't do that. But now I understand, it is not open circuited like an independent current source, it is open circuited because the gate (VIN) is short circuited and set to zero. It now all makes sense, when you set the gate to zero it goes into cutoff. How about that? Whible, you're a genius. Thank you so much. You have a good day/evening, whichever applies. As the other genius in this course would say, Dr. Agarwal, I had an aha moment. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-22T02:51:43Z SecondChildTAG: I'm glad I could help. I will just add that in small signal model you don't have cutoff regions and any other regions because you replace the mosfet with a *linear* voltage controlled current source. But it is indeed open circuited when u short circuit its control terminal. SecondChildUserIdTAG: 405936 SecondChildUserNameTAG: Whible SecondChildCreateTimeTAG: 2012-10-22T03:12:33Z IndexTAG: 3362 TitleTAG: HELP My name is Javier I had problems with the platform. I tried to make laboratories. but it saves me the answers. Question: previous labs can not be made? 2. in some circuits I could turn some elements? 3. Exposible the overtake or not let me by closing labs? any help UserIdTAG: 395337 UserNameTAG: lujafo CreateTimeTAG: 2012-10-21T22:20:21Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 0 IndexTAG: 3363 TitleTAG: Please Help, how do you get gm??? in H6P1 I don't know how to get the transconductance (gm), I know that it is d-iout/d-vin so, if my i=k(vin-vt)^2 so the derivative should be 2k(vin-vt) but I still don´t get where should I use vds! UserIdTAG: 231676 UserNameTAG: Roosemberth CreateTimeTAG: 2012-10-21T20:40:17Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 7 FirstChildTAG: Used I1+I2 and you understand where you should use vds FirstChildUserIdTAG: 192114 FirstChildUserNameTAG: chuba FirstChildCreateTimeTAG: 2012-10-21T20:44:49Z FirstChildTAG: hi Roosemberth... the expression you are to use is this i=K/2(vgs-vT)^2... now take the derivative.... ul get your answer.... FirstChildUserIdTAG: 117319 FirstChildUserNameTAG: iqramalik FirstChildCreateTimeTAG: 2012-10-21T20:46:39Z FirstChildTAG: gm =di/dvgs when vds is constant FirstChildUserIdTAG: 95638 FirstChildUserNameTAG: cperezf FirstChildCreateTimeTAG: 2012-10-21T23:01:34Z FirstChildTAG: I put the derivative and it's still wrong. I check it in the book and are the same :/ FirstChildUserIdTAG: 298547 FirstChildUserNameTAG: AnthonyRF FirstChildCreateTimeTAG: 2012-10-21T23:09:42Z FirstChildTAG: In H6P1 you work with I = K\*(vin-VT)\*vDS^2, not I = k(vin-vt)^2 and not even I = (k/2)\*(vin-vt)^2. If u derive that by d/dvin u get vDS somewhere in the answer since it is treated as a constant for that purpose. FirstChildUserIdTAG: 405936 FirstChildUserNameTAG: Whible FirstChildCreateTimeTAG: 2012-10-21T23:48:21Z FirstChildTAG: So that's why xD, Tkank you! FirstChildUserIdTAG: 231676 FirstChildUserNameTAG: Roosemberth FirstChildCreateTimeTAG: 2012-10-22T01:18:05Z FirstChildTAG: Please my question is: Is (VDS = VS-Vout)? Thank you. FirstChildUserIdTAG: 329879 FirstChildUserNameTAG: Ugochukwu FirstChildCreateTimeTAG: 2012-10-22T07:57:14Z IndexTAG: 3364 TitleTAG: H6P2 Q5 HELPPPPPPPPPPPPPPPPP Hi! I read a lot posts her. And I have some problem. 1) gm1=gm2 2) I=gm *v ## _________ ## 3) the voltage at the vout node is also the voltage at the source of Q2 4) I1=I2 K(Vin-VT)*vin=K(Vout-VT)*vout -------------------------------------- if I used 3) it will be K(Vin-VT)*vin=K(Vin-VT)*vout => vin=vout What next? HELP PLZ.... UserIdTAG: 192114 UserNameTAG: chuba CreateTimeTAG: 2012-10-21T20:24:17Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Chuba can you help me.... VOUT=VIN...but how do i find Vth then? FirstChildUserIdTAG: 308861 FirstChildUserNameTAG: Ignaas FirstChildCreateTimeTAG: 2012-10-21T22:43:02Z SecondChildTAG: and how about Rth? SecondChildUserIdTAG: 308861 SecondChildUserNameTAG: Ignaas SecondChildCreateTimeTAG: 2012-10-21T22:46:26Z SecondChildTAG: Are you sure your signs are correct. Does VOUT=VIN? SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-21T22:53:47Z SecondChildTAG: no..vout=-vin SecondChildUserIdTAG: 308861 SecondChildUserNameTAG: Ignaas SecondChildCreateTimeTAG: 2012-10-21T23:04:46Z SecondChildTAG: vout=K(vIN-VT)vin SecondChildUserIdTAG: 308861 SecondChildUserNameTAG: Ignaas SecondChildCreateTimeTAG: 2012-10-21T23:06:59Z FirstChildTAG: Check your signs on the variables. Also, don't expect the answer to have the same variables asked for in the question. FirstChildUserIdTAG: 132828 FirstChildUserNameTAG: rharris FirstChildCreateTimeTAG: 2012-10-21T20:32:47Z SecondChildTAG: 1)Check your signs on the variables ----------------------------------- problem with the signs her K(VOUT-VT)*VOUT. It must be K(-VOUT-VT)*VOUT? 2) Also, don't expect the answer to have the same variables asked for in the question. ------------------------------------------------------------------------ You meant that the answer will be with out K VIN .... ? SecondChildUserIdTAG: 192114 SecondChildUserNameTAG: chuba SecondChildCreateTimeTAG: 2012-10-21T20:38:44Z SecondChildTAG: I found) Thanks) And now I have a problem How to get R q6 SecondChildUserIdTAG: 192114 SecondChildUserNameTAG: chuba SecondChildCreateTimeTAG: 2012-10-21T20:47:19Z SecondChildTAG: It's all) Thanks) good night) SecondChildUserIdTAG: 192114 SecondChildUserNameTAG: chuba SecondChildCreateTimeTAG: 2012-10-21T20:49:44Z FirstChildTAG: hey there... could you please tell me why is it necessary for gm1 be equal to gm2.... when KCL can be satisfied by the products gm1*vgs1 = gm2*vgs2??? FirstChildUserIdTAG: 117319 FirstChildUserNameTAG: iqramalik FirstChildCreateTimeTAG: 2012-10-21T20:53:39Z SecondChildTAG: gm1 = gm2 because both transistors are the same. The key is to assigning what vgs equals, they are different for each transistor. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-21T21:12:58Z SecondChildTAG: for q1 vgs=Vin.... for q2 Vgs= Vdd- Vout correct? SecondChildUserIdTAG: 117319 SecondChildUserNameTAG: iqramalik SecondChildCreateTimeTAG: 2012-10-21T21:19:23Z SecondChildTAG: You are getting there, but not quite. The answer to your question is no, regarding Q2. Check out [the input from staff][1]. Do exactly what he says with gm1 and gm2, but with the correct Vgs for Q2. Don't let the signs of vgs mix you up. Look at page 417 in the text book at fig. 8.13. You don't need r. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/508282c564146d1f000000a4 SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-21T23:03:36Z SecondChildTAG: the small-signal circuit vgs1=vin; vgs2=-vout how gm1*vgs1= gm2*vgs2 so ---> SecondChildUserIdTAG: 95638 SecondChildUserNameTAG: cperezf SecondChildCreateTimeTAG: 2012-10-21T23:12:41Z SecondChildTAG: Hay cperezf, any success on finding Rth? I found Vth, but I'm having trouble finding Rth. - Thanks SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-21T23:19:00Z SecondChildTAG: @ rharris.... if u have found Vth correctly.... the only thing left is to take the ratio rth= Vth/ids where ids would be the small signal ids value....i hope u got that.... SecondChildUserIdTAG: 117319 SecondChildUserNameTAG: iqramalik SecondChildCreateTimeTAG: 2012-10-22T02:49:34Z SecondChildTAG: Making fool of myself by solving for so many hrs... Really easy ..... It is simply Phase inverter :D SecondChildUserIdTAG: 413002 SecondChildUserNameTAG: Gauravjain88 SecondChildCreateTimeTAG: 2012-10-22T12:27:19Z IndexTAG: 3365 TitleTAG: Can someone post an explanation for the second part please? I don't quite get how to get the power in Watts. UserIdTAG: 42787 UserNameTAG: rafuk CreateTimeTAG: 2012-10-21T20:04:06Z VoteTAG: 0 CoursewareTAG: Week 6 / Capacitor energy storage exercise CommentableIdTAG: 6002x_capacitor_energy_storage_exercise NumberOfReplyTAG: 1 FirstChildTAG: Hi rafuk, We know Energy=Power*time. So Power=Energy/time. Plug in the values and you will get the answer(make sure you use the correct units). :) FirstChildUserIdTAG: 222911 FirstChildUserNameTAG: bhaswardg FirstChildCreateTimeTAG: 2012-10-24T14:29:49Z IndexTAG: 3366 TitleTAG: HELP THIS ONE What is the minimum value that the power supply voltage VDD must be to ensure that both transistors are operating in the saturated region? Write an algebraic expression involving VIN and VT for this minimum value of VDD in the space provided below: UserIdTAG: 475448 UserNameTAG: msamwelmollel CreateTimeTAG: 2012-10-21T20:03:32Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: THIS IS H6 HELP AM STUCKED FirstChildUserIdTAG: 475448 FirstChildUserNameTAG: msamwelmollel FirstChildCreateTimeTAG: 2012-10-21T20:04:21Z IndexTAG: 3367 TitleTAG: problem in question of homework 7 This question is from homework 7:- Now, let's get down to numbers. Let V=1.8V, RS=22.0Ω, RO=50.0Ω, and L=15.44nH. How much time, in nanoseconds, does it take for the output voltage to reach vO=0.9V? Answer should be 0.2946 nano Seconds but website is showing that it's not the correct answer. UserIdTAG: 379749 UserNameTAG: deepaknangru CreateTimeTAG: 2012-10-21T19:31:27Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: it's not correct FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-21T19:37:47Z SecondChildTAG: what is not correct?? MY answer or their answer?? SecondChildUserIdTAG: 379749 SecondChildUserNameTAG: deepaknangru SecondChildCreateTimeTAG: 2012-10-22T16:59:42Z SecondChildTAG: Your answer is not correct. Check the values of the resistances again. SecondChildUserIdTAG: 219461 SecondChildUserNameTAG: shuncobra SecondChildCreateTimeTAG: 2012-10-23T14:41:20Z IndexTAG: 3368 TitleTAG: Awesome HW6 and Lab6! Thanks for that! Dear staff, thanks for that amazing HW6 and Lab6! As for me, the main feature of these HWs and Labs is that you discover something new while solving them. You'll never solve the tasks only knowing the appropriate formulas - the understanding is essential. I've spent about 6 hours on week 6 tasks and finished them at something about 3-4 AM (according to my local time) - and that was wonderful time. Thanks! UserIdTAG: 326787 UserNameTAG: Angstrem CreateTimeTAG: 2012-10-21T17:42:49Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: ok am still stuck with lab 6 on the estimated time how to solve that FirstChildUserIdTAG: 475448 FirstChildUserNameTAG: msamwelmollel FirstChildCreateTimeTAG: 2012-10-21T18:12:51Z IndexTAG: 3369 TitleTAG: Consider rearranging a video I think that the video of week 6 (S11V11) should be inserted here (in week 5). The reason is that the first question of homework (of week 5) has Vgs not equal to Vi, and it becomes really confusing when no words of caution are there that Vgs can be something else than Vi. I wonder if it is deliberately done to initiate thought process, but if that is the case, then you sure have succeeded cause it had given me terrible time (and I assume others as well). UserIdTAG: 213503 UserNameTAG: jyotech CreateTimeTAG: 2012-10-21T16:59:11Z VoteTAG: 0 CoursewareTAG: Week 5 / Small Signal Amplifier Demo CommentableIdTAG: 6002x_small_sig_amp_d NumberOfReplyTAG: 0 IndexTAG: 3370 TitleTAG: Dear EDX Staff: Issue with my browser. Unable to answer H6P2 Q5 question Dear EDX Staff, I am unable to get my answer right for H6P2 Q5. No matter what I answer (even an answer that doesn't include VIN or vin) in the field it says Vin is an invalid input. can you please advise. I have tried to reach out for users support to see if its a problem anyone else faced but it seem to be just me. Need help!! Thanks, Ravi UserIdTAG: 244670 UserNameTAG: ravikiran1201 CreateTimeTAG: 2012-10-21T16:41:24Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 6 FirstChildTAG: Post a screen shot please FirstChildUserIdTAG: 623210 FirstChildUserNameTAG: preveen FirstChildCreateTimeTAG: 2012-10-21T16:44:53Z FirstChildTAG: having the same problem with question 6 FirstChildUserIdTAG: 308861 FirstChildUserNameTAG: Ignaas FirstChildCreateTimeTAG: 2012-10-21T17:07:21Z FirstChildTAG: You may need to record (write down) all your correct answers to the other problems, click on reset on the bottom of the page and re-enter all your answers, sort of clear your browser cache. You may even need to re-boot. My Mozilla and Google Chrome both stopped displaying all the Mathjax characters and I had to remove Mozilla and Chrome . I then re-installed Mozilla it before I could proceed with the course (haven't re-installed Chrome yet as Mozilla is working....) FirstChildUserIdTAG: 375500 FirstChildUserNameTAG: Brej7665 FirstChildCreateTimeTAG: 2012-10-21T17:14:03Z FirstChildTAG: Since I cannot see what you are seeing I cannot say for sure, but check out if you have written Vin for any other answer for H6P2. Because the autograder checks for issues in the whole part, so for example if you are checking Q2 and you have an invalid parameter in Q4 then it won't work, rather give an error message. Hope it helps. FirstChildUserIdTAG: 251544 FirstChildUserNameTAG: kilersins93 FirstChildCreateTimeTAG: 2012-10-21T17:15:51Z FirstChildTAG: ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13508427561343613.jpg FirstChildUserIdTAG: 244670 FirstChildUserNameTAG: ravikiran1201 FirstChildCreateTimeTAG: 2012-10-21T18:06:04Z SecondChildTAG: just to show an example i have answered the question as K.. there is no VIN in the answer but it still says VIN not permitted SecondChildUserIdTAG: 244670 SecondChildUserNameTAG: ravikiran1201 SecondChildCreateTimeTAG: 2012-10-21T18:06:25Z SecondChildTAG: I think you are using vin in the answer of other parts of same question. Just remove vin from other parts of this question. I hope it will work. SecondChildUserIdTAG: 154979 SecondChildUserNameTAG: Heetesh SecondChildCreateTimeTAG: 2012-10-22T06:49:11Z FirstChildTAG: try to cleanup your browser's cache - looks like browser showing cached page instead of send an answer and get a new one FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-21T18:59:21Z SecondChildTAG: weird thing.. this is happening on every computer I try this on. SecondChildUserIdTAG: 244670 SecondChildUserNameTAG: ravikiran1201 SecondChildCreateTimeTAG: 2012-10-22T02:24:38Z SecondChildTAG: do you connect through proxy? SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-22T14:29:40Z IndexTAG: 3371 TitleTAG: H6P1:The NewFET device. Im don't understand??? Iid=K* (vGS-vT)* vDS^2; [A]=([A]/ [V]^2)* [V]* [V]^2=[A]* [V]; [A] - Amper;Volt. It's not correctly??? Or Im dont understand. Thanks for the My answer. UserIdTAG: 350249 UserNameTAG: 1s_BricK CreateTimeTAG: 2012-10-21T16:25:11Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: hazel1919 has a very nice hint post here: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50818210ca4eff27000000b8 You can find it and more hints by searching the posts for "H6P1". FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-21T16:36:53Z FirstChildTAG: I suspect K for this device is A/V^3 FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-21T18:36:22Z IndexTAG: 3372 TitleTAG: H6P2, 5 and 6 Can anyone help with the circuit? I think i'm doing it wrong! Do we get a resistance of some unknown RL in series with Q2's small signal model ( resistor) I just am not able to figure it out! I'm too saturated :( UserIdTAG: 108454 UserNameTAG: Raven7281 CreateTimeTAG: 2012-10-21T16:24:06Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Just because the question asks for the answer in certain variables, it doesn't mean all those variables need to be in the answer. See the explanation [from a staff member][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/508282c564146d1f000000a4 FirstChildUserIdTAG: 132828 FirstChildUserNameTAG: rharris FirstChildCreateTimeTAG: 2012-10-21T19:13:51Z IndexTAG: 3373 TitleTAG: S15V14 Approximation Why first order approximation of "**e^-T/RC**" is "**1 - T/RC**". Where does this "**+1**" come from? Why it is not just "**-T/RC**"? UserIdTAG: 208296 UserNameTAG: OZ1 CreateTimeTAG: 2012-10-21T16:11:25Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Troubleshooting NumberOfReplyTAG: 0 IndexTAG: 3374 TitleTAG: LAB6 1 question how to find the final output of the inverter..plz help.. isn't it the value of output at v=VOH? UserIdTAG: 431224 UserNameTAG: erachopra10 CreateTimeTAG: 2012-10-21T15:56:28Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Myrimit did an excellent write-up about lab 6. It's is here: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Ring_Oscillator/threads/5082c6039e78031f000000c0 FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-21T16:00:54Z SecondChildTAG: sorry but i am still not able to get the final output voltage..plz help! SecondChildUserIdTAG: 431224 SecondChildUserNameTAG: erachopra10 SecondChildCreateTimeTAG: 2012-10-21T16:50:13Z SecondChildTAG: i m not to get ron pleasse help someone SecondChildUserIdTAG: 550401 SecondChildUserNameTAG: revathisingh SecondChildCreateTimeTAG: 2012-10-21T17:05:40Z SecondChildTAG: same problem... can somebody help. Not sure how to interpret the graph as the language is a bit confusing. SecondChildUserIdTAG: 251544 SecondChildUserNameTAG: kilersins93 SecondChildCreateTimeTAG: 2012-10-21T17:58:20Z SecondChildTAG: got it!.. check for the value when output curve reaches its final value..(i.e when it tends to a stable value after some time.) SecondChildUserIdTAG: 431224 SecondChildUserNameTAG: erachopra10 SecondChildCreateTimeTAG: 2012-10-22T06:56:10Z IndexTAG: 3375 TitleTAG: mid term exam i have following queries regarding the mid term: **1. is there any chances that exam shall postponed to sat-sunday?** 2. i am just going through all lectures and tutorials and doing homeworks and labs. is it just enough for mid term.?? 3. is there any portion of **lab** in mid term? UserIdTAG: 114864 UserNameTAG: Vikaash CreateTimeTAG: 2012-10-21T15:05:43Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Unless there is some technical glitch with the website, the chances of the exam schedule being altered are very close to zero. Doing the lectures, tutorials, homeworks and labs is the class, there isn't much more you can do. If you have extra time, I would suggest doing problems from the book. If you mean "are they going to directly reproduce a question from the labs?", I have no idea. But the concepts you learn in the labs will certainly be represented in both the mid-term and final. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-21T15:35:51Z FirstChildTAG: Good news Vikaash..... The deadline to submit the midterm exam is extended to 23:59 (11:59 pm) on Sunday, October 28th, Boston time. The exam will still be released on Thursday, October 25th at 00:01 (12:01 am) Boston time. FirstChildUserIdTAG: 368712 FirstChildUserNameTAG: jmen FirstChildCreateTimeTAG: 2012-10-23T16:40:37Z SecondChildTAG: it means we have more than 24 hours to submit our answer!!! m i right.. SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-10-24T08:22:50Z SecondChildTAG: no,it means that you can give it on sunday too but within 24 hours of the time you first open it. SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-24T15:45:30Z SecondChildTAG: ok..thanks SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-10-24T16:42:38Z IndexTAG: 3376 TitleTAG: plzz help h6p2 vout ,,,,i hav to replace diode connected tx with resistor n another with small signal n hav to apply node method ,,but i m not getting it right UserIdTAG: 183166 UserNameTAG: yogeshk CreateTimeTAG: 2012-10-21T14:19:51Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: refer texbook page number 417 FirstChildUserIdTAG: 432171 FirstChildUserNameTAG: manasa100 FirstChildCreateTimeTAG: 2012-10-21T15:09:47Z IndexTAG: 3377 TitleTAG: Current direction in Second question In the second question, why the result is positive? Isn't the current direction to the batteries? UserIdTAG: 270856 UserNameTAG: Cris_11 CreateTimeTAG: 2012-10-21T13:59:14Z VoteTAG: 0 CoursewareTAG: Week 1 / Battery model CommentableIdTAG: 6002x_S1E9_Battery_Model NumberOfReplyTAG: 1 FirstChildTAG: No. The question is "What is the current (in Amperes) you should expect to go through the short circuit?". Shor circuit can be represented as resistor with resistace close to 0 Ohms. So the current goes from the batteries through this resistor. FirstChildUserIdTAG: 1254417 FirstChildUserNameTAG: egor_g FirstChildCreateTimeTAG: 2013-02-23T09:00:09Z IndexTAG: 3378 TitleTAG: help me in H6P2: I use the condition: VGS>= VT, VDS>= VGS-VT. And for Q2: VDD = VGS; VGS>=VT+VOUT; For Q1: VGS>=VT, VDS>=VGS-VT, and VDS =VOUT. But the answer is not correct. Why, what mistake do I make here? UserIdTAG: 371617 UserNameTAG: rainbow12 CreateTimeTAG: 2012-10-21T13:59:08Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: [Check this][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5082ba3a7021a8230000009d FirstChildUserIdTAG: 268444 FirstChildUserNameTAG: Marlonabeykoon FirstChildCreateTimeTAG: 2012-10-21T14:40:44Z FirstChildTAG: mistake in Q2: VDD = VGS; VDD=VG2 only VS2=VD1 FirstChildUserIdTAG: 113249 FirstChildUserNameTAG: EGuarch FirstChildCreateTimeTAG: 2012-10-21T22:40:37Z IndexTAG: 3379 TitleTAG: H6P3 How to find the time constant of A and B?What is the formula we have to use UserIdTAG: 150120 UserNameTAG: krishna1993 CreateTimeTAG: 2012-10-21T13:56:48Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: time constant is R(equivalent)*C FirstChildUserIdTAG: 432171 FirstChildUserNameTAG: manasa100 FirstChildCreateTimeTAG: 2012-10-21T14:14:14Z IndexTAG: 3380 TitleTAG: BASIC WHY IS VS=VI IN SMALL SIGNAL ANALYSIS? UserIdTAG: 342135 UserNameTAG: vikash902 CreateTimeTAG: 2012-10-21T12:38:26Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3381 TitleTAG: H6p2 VTH prob i tried to replace mosfet by its small signal model.still im unable to get vth. UserIdTAG: 337430 UserNameTAG: aparna_b CreateTimeTAG: 2012-10-21T10:49:23Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: [see this][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5082ba3a7021a8230000009d FirstChildUserIdTAG: 268444 FirstChildUserNameTAG: Marlonabeykoon FirstChildCreateTimeTAG: 2012-10-21T14:36:43Z IndexTAG: 3382 TitleTAG: h6p2 Q5 and Q6 I have tried alot but not able to get green tick. For thevenin resistance the Q1 and Q2 are parallel so it must be gm1+gm2....am i right? and for thevenin voltage its ids*(gm1+gm2).....i.e, ids=K(VIN-VT)^2/2 Please tell me if i am going the right way.? UserIdTAG: 182470 UserNameTAG: nitesh2703 CreateTimeTAG: 2012-10-21T07:05:33Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Try to separate the MOSFETs. Replace them with the small signal model. Check the relations you can find from each of them (One will behave as current source, One will behave as a resistor). FirstChildUserIdTAG: 107306 FirstChildUserNameTAG: Amir54 FirstChildCreateTimeTAG: 2012-10-21T07:33:19Z FirstChildTAG: [Check this][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5082ba3a7021a8230000009d FirstChildUserIdTAG: 268444 FirstChildUserNameTAG: Marlonabeykoon FirstChildCreateTimeTAG: 2012-10-21T14:38:10Z IndexTAG: 3383 TitleTAG: help me out with H6P2 Q2 according to me minimum value should be VIN but answer is wrong help me out UserIdTAG: 321556 UserNameTAG: sj31867 CreateTimeTAG: 2012-10-21T06:47:22Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Using the conditions for keeping the MOSFET out of cutoff, what do you have to do to keep Q1 out of cutoff. You have to maintain the relationships: VGS >= VT, and VDS >= VGS-VT. FirstChildUserIdTAG: 132828 FirstChildUserNameTAG: rharris FirstChildCreateTimeTAG: 2012-10-21T07:18:31Z SecondChildTAG: There is one more condition that needs to be satisfied. VGS of Q1 needs to be equal to VGS of Q2, because the current in both these have to be same to satisfy KCL. SecondChildUserIdTAG: 329544 SecondChildUserNameTAG: Jasjot SecondChildCreateTimeTAG: 2012-10-21T15:50:44Z SecondChildTAG: Thanx for the hint! SecondChildUserIdTAG: 316353 SecondChildUserNameTAG: Vivekanand123 SecondChildCreateTimeTAG: 2012-10-21T18:52:17Z SecondChildTAG: Thanks! SecondChildUserIdTAG: 192653 SecondChildUserNameTAG: Quaz SecondChildCreateTimeTAG: 2012-10-22T10:49:53Z IndexTAG: 3384 TitleTAG: H6p2 Q5 i need some hint to find vth. UserIdTAG: 337430 UserNameTAG: aparna_b CreateTimeTAG: 2012-10-21T06:30:58Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: [check this][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5082ba3a7021a8230000009d FirstChildUserIdTAG: 268444 FirstChildUserNameTAG: Marlonabeykoon FirstChildCreateTimeTAG: 2012-10-21T14:47:10Z IndexTAG: 3385 TitleTAG: I dont get the 2nd part Why multiply 1/2*C*v^2 with 10^3....isent milli 10^-3 UserIdTAG: 204745 UserNameTAG: allwynmendes CreateTimeTAG: 2012-10-21T05:42:01Z VoteTAG: 0 CoursewareTAG: Week 6 / Capacitor energy storage exercise CommentableIdTAG: 6002x_capacitor_energy_storage_exercise NumberOfReplyTAG: 2 FirstChildTAG: 10^-6 FirstChildUserIdTAG: 192114 FirstChildUserNameTAG: chuba FirstChildCreateTimeTAG: 2012-10-21T15:51:04Z SecondChildTAG: Exactumundo SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-22T11:37:53Z FirstChildTAG: Hi allwynmendes, Yes milli is 10^-3. Here we are using the formula Power=Energy/time. So it is getting multiplied by 10^3. :) FirstChildUserIdTAG: 222911 FirstChildUserNameTAG: bhaswardg FirstChildCreateTimeTAG: 2012-10-24T14:33:17Z IndexTAG: 3386 TitleTAG: H6P2 Rth Hi, I have got some trouble with the last question. Both FETs have the same gm. gm I have solved, for thevenin both resistors are parallel and have the same value. 1/gm=R --> Rth=2/gm where is my error in reasoning? UserIdTAG: 389333 UserNameTAG: juergen CreateTimeTAG: 2012-10-20T22:59:43Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Resistors in parallel add inversely-- not directly! You should have the resistance, not double it. FirstChildUserIdTAG: 390034 FirstChildUserNameTAG: kahlil FirstChildCreateTimeTAG: 2012-10-20T23:05:19Z SecondChildTAG: hmmmm what does it mean? "Resistors in parallel add inversely-- not directly!" SecondChildUserIdTAG: 325197 SecondChildUserNameTAG: Vitali_Jerin SecondChildCreateTimeTAG: 2012-10-20T23:14:31Z SecondChildTAG: I think he means "half" the resistance, not "have" the resistance. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-21T00:10:40Z FirstChildTAG: ... Think simple... how many resistors do you have in the circuit? FirstChildUserIdTAG: 64512 FirstChildUserNameTAG: OZSorescu FirstChildCreateTimeTAG: 2012-10-20T23:09:57Z SecondChildTAG: It seems to be one resistor, but I don't know why. Is the reason, that Q2 is a fixed courrentsource? SecondChildUserIdTAG: 389333 SecondChildUserNameTAG: juergen SecondChildCreateTimeTAG: 2012-10-20T23:16:54Z SecondChildTAG: That is also my problem ... I do not understand why! SecondChildUserIdTAG: 64512 SecondChildUserNameTAG: OZSorescu SecondChildCreateTimeTAG: 2012-10-20T23:26:20Z SecondChildTAG: Q1 dependenet current source of vin, but vin is independent source, so you should turn off it. Then current through Q1 equal to 0 and you left with 1 resistor SecondChildUserIdTAG: 198822 SecondChildUserNameTAG: Gennady SecondChildCreateTimeTAG: 2012-10-21T18:01:27Z FirstChildTAG: Two resistors, R1 and R2 in parallel, the net resistance, Rp = 1/(1/R1 + 1/R2) (add the inverse), OR Rp = (R1 + R2)/(R1 + R2) but this formula works only for two resistors (or two at a time). IF the two resistors are equal in value, the total parallel resistance is equal to 1/2 the value of either one. Note, the net parallel resistance will ALWAYS be less than the value of the smallest resistor,,,, FirstChildUserIdTAG: 375500 FirstChildUserNameTAG: Brej7665 FirstChildCreateTimeTAG: 2012-10-21T02:47:59Z IndexTAG: 3387 TitleTAG: Problem 2 : find Vth .week 6 Hello Can someone help me with this part I have read alot of the posted comments and I still cant solve it UserIdTAG: 246143 UserNameTAG: edx90 CreateTimeTAG: 2012-10-20T21:49:19Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Having the same trouble..:( FirstChildUserIdTAG: 173147 FirstChildUserNameTAG: cruiser_rahit FirstChildCreateTimeTAG: 2012-10-20T22:32:34Z FirstChildTAG: Read the discussion, post by post... you'll find it ... even if you will not understand why ... trust me! FirstChildUserIdTAG: 64512 FirstChildUserNameTAG: OZSorescu FirstChildCreateTimeTAG: 2012-10-20T23:13:11Z IndexTAG: 3388 TitleTAG: Internal resistance of multimeter Could someone help me draw the circuit diagram for measuring the internal resistance r of a multimeter? I'm using a dry-cell battery, high-ohm resistor (of resistance R), and the multimeter. My guess is that the battery should power up the high-ohm resistor and that the meter should be connected across the high-ohm resistor. The voltage measured by the meter is then equal to that across the internal resistance r and the high-ohm resistor R, right? Also, wouldn't the resistor R carry a voltage equal to the battery voltage? The meter voltage should then be divided amongst resistors r and R, and the fraction of the meter voltage across R is then R/(R+r), which should be equal to the battery voltage, I think: V_batt = V_meter * R/(R+r) I am confused because this would imply that the meter voltage is higher than the battery voltage, which, according to several sources I've consulted online, is not correct (but maybe it is?) Is this the correct reasoning? UserIdTAG: 581004 UserNameTAG: minamora CreateTimeTAG: 2012-10-20T18:14:52Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: ... I do not believe that the meter is acting as a power source. If you want to measure voltages(You connect the voltmeter in parallel!!!!), than your meter has a very high resistance, so that it does not influence the measurement (~1MOhm). If you want to measure currents, the resistance is very low, next to nothing, so that it does not oppose the the flow of current(you connect the ampere-meter in series with the resistance). Also ... you are connecting the voltmeter in parallel thus having the same voltage on the resistor and on the meter... this is not a voltage divider(connected in series)... as it comes out of expression: V_batt = V_meter * R/(R+r) FirstChildUserIdTAG: 64512 FirstChildUserNameTAG: OZSorescu FirstChildCreateTimeTAG: 2012-10-20T23:22:42Z IndexTAG: 3389 TitleTAG: Minimum VDD!!! I am following this approach to find minimum VDD: i1 + i2 = 0 K(VGS1 - VT).VDS1^2 + K(VGS2 - VT).VDS2^2 = 0 after taking VGS2 = VDD - Vout and then Vout = Vin-VT I have finally got a very large equation for VDD of order 3. Please tell me whats wrong in my approach? UserIdTAG: 64618 UserNameTAG: ashfaq2419 CreateTimeTAG: 2012-10-20T14:50:32Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: I1 es igual a I2 FirstChildUserIdTAG: 613206 FirstChildUserNameTAG: grubio FirstChildCreateTimeTAG: 2012-10-20T14:55:16Z SecondChildTAG: but how? i have applied KCL at the output node and added up the currents leaving form the node. isn't it like this? SecondChildUserIdTAG: 64618 SecondChildUserNameTAG: ashfaq2419 SecondChildCreateTimeTAG: 2012-10-20T15:07:11Z SecondChildTAG: Clearly the current that flows through Q2 must also flow through Q1. If not where would it go? Therefore, I1 = I2, which is key to solving the problem. **No long equations to solve!!!** SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-20T16:07:31Z FirstChildTAG: You are using equations of the NewFET device from H6P1, but you have to solve H6P2: in this part of the homework there are simple and traditional MOSFETs. FirstChildUserIdTAG: 376877 FirstChildUserNameTAG: AndBre FirstChildCreateTimeTAG: 2012-10-20T17:13:25Z FirstChildTAG: Hi, Any hint for Q2 (min VDD) solving please? FirstChildUserIdTAG: 325197 FirstChildUserNameTAG: Vitali_Jerin FirstChildCreateTimeTAG: 2012-10-20T23:29:38Z IndexTAG: 3390 TitleTAG: H6P1 i have solved all parts of H6P1..but still getting a **red cross** on 1st part i.e. finding equation of vOUT. i am getting answer after applying L-hosp. is of the form **X/sqrt(1+4*X*(X-X)*X)** where X are different parameters. plz help. a big thanks in advance. UserIdTAG: 114864 UserNameTAG: Vikaash CreateTimeTAG: 2012-10-20T07:51:49Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: First of all, your final answer will be what you get before applying L'hospital's rule, not after; L'hospital's rule is only for you to decide what root to take. And you should be able to simplify what you get after applying L-hospital's rule to an extremely simple looking expression- much simpler than what you have. A word of help- did you combine your expression for vOUT into one term with a common denominator? You might get something that looks a little hairy but it simplifies nicely when taking derivatives. FirstChildUserIdTAG: 390034 FirstChildUserNameTAG: kahlil FirstChildCreateTimeTAG: 2012-10-20T12:01:04Z SecondChildTAG: thax for the help @kahlil.....i got **vOUT=(-1+sqrt(1+4*X*X*(X-X)*X))/(2*X*X*(X-X))** SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-10-20T12:21:39Z SecondChildTAG: but getting "could not parse" SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-10-20T12:22:46Z SecondChildTAG: The form of the answer is completely correct. Try re-entering and checking for mismatched parenthesis maybe? But it looks like you have it. SecondChildUserIdTAG: 390034 SecondChildUserNameTAG: kahlil SecondChildCreateTimeTAG: 2012-10-20T12:56:43Z SecondChildTAG: finally got it...MITx should make equation checker more efficient...hope they will do.. SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-10-20T13:07:18Z SecondChildTAG: vOUT=(-1+sqrt(1+4*X*X*(X-X)*X))/(2*X*X*(X-X)) i am writing my answer in in above format and fed up of parse error!!!! nvalid input: Could not parse"......" please help..... SecondChildUserIdTAG: 353709 SecondChildUserNameTAG: anshu10750 SecondChildCreateTimeTAG: 2012-10-20T17:47:38Z SecondChildTAG: try -2*X*X etc... SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-20T21:34:32Z SecondChildTAG: @HAZEL-even tried -2*....still parse error.even tried both the roots(+/-). SecondChildUserIdTAG: 353709 SecondChildUserNameTAG: anshu10750 SecondChildCreateTimeTAG: 2012-10-21T05:48:02Z FirstChildTAG: My entries such as VOUT entered as given in the exercise are invalid. Please tell whats wrong? thanx FirstChildUserIdTAG: 498458 FirstChildUserNameTAG: jefwa FirstChildCreateTimeTAG: 2012-10-20T13:22:37Z SecondChildTAG: may be you entering some wrong pararmeter..for eg. in place of VS, you may be entering Vs.. SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-10-20T13:29:44Z IndexTAG: 3391 TitleTAG: H6P2: Phase Inverter I can't understand the figure given in the problem in relevance with MOSFETs please help me out. UserIdTAG: 140842 UserNameTAG: srvz39 CreateTimeTAG: 2012-10-20T04:20:23Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Troubleshooting NumberOfReplyTAG: 2 FirstChildTAG: same here. FirstChildUserIdTAG: 114864 FirstChildUserNameTAG: Vikaash FirstChildCreateTimeTAG: 2012-10-20T08:17:48Z FirstChildTAG: This circuit is simply two MOSFETs stacked on top of each other. The top one, Q2, has its gate and drain tied together, so they're at the same voltage (VDD) at all times. Then, its source is tied to the drain of the bottom MOSFET, Q1. We also apply our input voltage vIN to the gate of Q1. Each MOSFET still individually obeys the equations and laws for its region of operation that we already learned. So for both to be in saturation, for example, Q1 has to satisfy vGS = vIN > VT and Q2 has to satisfy vGS = VDD-vOUT > VT. FirstChildUserIdTAG: 390034 FirstChildUserNameTAG: kahlil FirstChildCreateTimeTAG: 2012-10-20T12:15:43Z IndexTAG: 3392 TitleTAG: H6P2 minimum VDD Currently, I've gotten to the point where I have $V_{DD}= V_T-2/K$ or $V_{DD}=2/K-V_T$ I've used KCL (the current through the two MOSFETs are equal), KVL, and the MOSFET characteristic $V_{VSI}=v_{IN}-V_T$. My question becomes, what am I missing? UserIdTAG: 248807 UserNameTAG: 1kingsman CreateTimeTAG: 2012-10-19T23:25:21Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: if you equate the currents of both the MOSFETs you will get an equation having VDD,vOUT,vIN in this equation substitute your equation vOUT=vIN-VT find out VDD...n get your aha moment..:P FirstChildUserIdTAG: 266739 FirstChildUserNameTAG: satvikchugh FirstChildCreateTimeTAG: 2012-10-20T08:59:38Z SecondChildTAG: I can't understand, how to solve it :-( I'm using the rule that $$Vds>=Vgs-VT$$ For Q1 it's always true. Let's solve it for Q2 $$Vout>Vin-V_T$$ And now we are using KCL method $$K/2*(Vdd-Vout-V_T)^2=K/2*(Vout-Vt)^2$$ Here we get, that $$Vout=Vdd/2$$ and further $$Vdd>=2*(Vin-VT)$$ But it's not a correct answer :-( Where is my mistake? SecondChildUserIdTAG: 345464 SecondChildUserNameTAG: Constantine_ru SecondChildCreateTimeTAG: 2012-10-20T09:21:14Z SecondChildTAG: For Q1 $$Vout>=Vin-Vt$$ And my attempt to get answer is $$Vdd>2*(Vin-Vt)$$ SecondChildUserIdTAG: 345464 SecondChildUserNameTAG: Constantine_ru SecondChildCreateTimeTAG: 2012-10-20T09:24:53Z SecondChildTAG: Oh, I've found my mistake) Парсер лох! SecondChildUserIdTAG: 345464 SecondChildUserNameTAG: Constantine_ru SecondChildCreateTimeTAG: 2012-10-20T09:29:50Z SecondChildTAG: :) SecondChildUserIdTAG: 266739 SecondChildUserNameTAG: satvikchugh SecondChildCreateTimeTAG: 2012-10-20T09:36:07Z SecondChildTAG: that worked for me too, thanks ;) SecondChildUserIdTAG: 47372 SecondChildUserNameTAG: Digius SecondChildCreateTimeTAG: 2012-10-20T12:59:30Z SecondChildTAG: Could you please let us know where was the problem? SecondChildUserIdTAG: 64512 SecondChildUserNameTAG: OZSorescu SecondChildCreateTimeTAG: 2012-10-20T21:43:17Z SecondChildTAG: And what an aha moment it was! Thanks everybody. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-21T03:32:02Z SecondChildTAG: aha moment is that between VDD and Vout exist VT. Why ? SecondChildUserIdTAG: 333731 SecondChildUserNameTAG: Timophei_NhaTrang SecondChildCreateTimeTAG: 2012-10-21T11:43:44Z FirstChildTAG: aha :D Thanks to satvikchugh FirstChildUserIdTAG: 131726 FirstChildUserNameTAG: chemiboy11 FirstChildCreateTimeTAG: 2012-10-21T15:02:37Z SecondChildTAG: :) SecondChildUserIdTAG: 266739 SecondChildUserNameTAG: satvikchugh SecondChildCreateTimeTAG: 2012-10-22T02:51:23Z IndexTAG: 3393 TitleTAG: How do we get dvc/dt? I cant understand this, smbd pls help me UserIdTAG: 214584 UserNameTAG: Askeroff CreateTimeTAG: 2012-10-19T19:02:56Z VoteTAG: 0 CoursewareTAG: Week 6 / Norton circuit capacitor exercise CommentableIdTAG: 6002x_norton_capacitor_circuit_exercise NumberOfReplyTAG: 1 FirstChildTAG: The current that flows through a capacitor is proportional to the derivative of the voltage across it-- this relation is: i = C*dv_C/dt Given this, and also given that the voltage across the resistor is also v_C (do you see this)? We simply write a KCL equation at the node to solve for dv_c/dt. FirstChildUserIdTAG: 390034 FirstChildUserNameTAG: kahlil FirstChildCreateTimeTAG: 2012-10-19T19:44:21Z SecondChildTAG: I see that the potential right before the resistor is the same as right before the capacitor. But how can we assume that the voltage drop across the resistor and the capacitor is the same? If the capacitor were replaced with a resistor of a different resistance than R, you could not assume that the voltage drop is the same. I'm having a hard time wrapping my mind around it. SecondChildUserIdTAG: 311920 SecondChildUserNameTAG: ghowell SecondChildCreateTimeTAG: 2012-10-21T00:29:15Z IndexTAG: 3394 TitleTAG: lecture 5 download link???? hey there guys...when are you posting download links for lecture 5.... UserIdTAG: 117319 UserNameTAG: iqramalik CreateTimeTAG: 2012-10-19T17:52:58Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: [JWplayer Video Link Week(1-6)][1] [1]: https://dl.dropbox.com/u/24096724/6002xMP4/menu.html# Regards: asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-10-19T18:08:14Z SecondChildTAG: thanx dude... apppreciate it... SecondChildUserIdTAG: 117319 SecondChildUserNameTAG: iqramalik SecondChildCreateTimeTAG: 2012-10-19T18:21:09Z IndexTAG: 3395 TitleTAG: h6p2 vout i hav to replace diode connected tx with resistor n another with small signal n hav to apply node method ,,but i m not getting it right UserIdTAG: 183166 UserNameTAG: yogeshk CreateTimeTAG: 2012-10-19T16:58:16Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3396 TitleTAG: H6P1 syntax problem while entering this is what i calculated for vout but it said cannot parse ..? where i am wrong kindly guide me thanks UserIdTAG: 136490 UserNameTAG: Ali_PU CreateTimeTAG: 2012-10-19T16:37:47Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Ali, You posted this: (1-sqrt(1+4*X*X*(X-X)*X))/(2*X*X*(X-X) A- You are missing a ) on the end. B- It's not the right root. Oh and next time I ask you to replace all the variables with X's, please replace ALL the variables with X's. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-19T17:02:29Z IndexTAG: 3397 TitleTAG: LAB 6 Please anybody! Give me hints how to find Ron for 1st question! It`s awful but i`m confused in this question UserIdTAG: 96873 UserNameTAG: Santyaga CreateTimeTAG: 2012-10-19T15:57:31Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: refer chap 10 of text book for formulas FirstChildUserIdTAG: 357747 FirstChildUserNameTAG: kishores FirstChildCreateTimeTAG: 2012-10-19T16:33:46Z SecondChildTAG: I read it, but still haven`t got any idea :-( SecondChildUserIdTAG: 96873 SecondChildUserNameTAG: Santyaga SecondChildCreateTimeTAG: 2012-10-19T17:50:50Z SecondChildTAG: measure final voltage from simulation and apply voltage divider SecondChildUserIdTAG: 340640 SecondChildUserNameTAG: Hariprathin SecondChildCreateTimeTAG: 2012-10-20T02:47:41Z FirstChildTAG: Once the circuit has reached equilibrium you have a simple divider circuit. You know the voltages and one of the resistances, just solve for the other resistance. FirstChildUserIdTAG: 174100 FirstChildUserNameTAG: markpolak FirstChildCreateTimeTAG: 2012-10-19T17:58:05Z SecondChildTAG: thanks! Got it! SecondChildUserIdTAG: 96873 SecondChildUserNameTAG: Santyaga SecondChildCreateTimeTAG: 2012-10-19T19:42:01Z IndexTAG: 3398 TitleTAG: Hard Rock I really like the connection prof. Agarwal made with Rock Music and distortion. I would like to see, however, next time he plays something from the east culture like from India, China or Japan. UserIdTAG: 415375 UserNameTAG: ZWX CreateTimeTAG: 2012-10-19T14:32:22Z VoteTAG: 0 CoursewareTAG: Week 6 / Amplifier operating point CommentableIdTAG: 6002x_amplifier_operating_point NumberOfReplyTAG: 0 IndexTAG: 3399 TitleTAG: H8p1 (b) How, will you find a solution? I have been seeking solutions! Vth,Rth? Q/C=1*(1-e^(-t/(33*10^-5))) t=?V=1? In me, still a mystery without your help. UserIdTAG: 89440 UserNameTAG: yuk CreateTimeTAG: 2012-10-19T05:22:20Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 8 FirstChildTAG: The problem is not very difficult. I analysed the RL and RC circuits individually, since the resistance between them (R1+R2) is quite large, they are effectively uncoupled. They give a hint that the integral of the unit impulse is 1. That is the unit impulse is 1 volt-second. That could be 10000V for 100us, 1000000V for 1us or even 100000000V for 10ns. As long as you choose a time period that is much smaller than the time constant of the RL or RC circuit you are analysing. (Ideally a unit impulse has infinite voltage for infinitely small time). FirstChildUserIdTAG: 270011 FirstChildUserNameTAG: SnowmanZA FirstChildCreateTimeTAG: 2012-10-19T14:43:02Z SecondChildTAG: Thanks SnowmanZA ! SecondChildUserIdTAG: 299251 SecondChildUserNameTAG: TheRedBlackOne SecondChildCreateTimeTAG: 2012-10-21T00:48:14Z SecondChildTAG: Thank you man ! This is finally explaining a lot of things with the impulse.You really nailed it ! SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-10-21T15:23:22Z FirstChildTAG: couldt do this, I solve the parts which I can, so that I have atleast some part of the homework complete and then solve the difficult parts. FirstChildUserIdTAG: 171674 FirstChildUserNameTAG: wajahat1 FirstChildCreateTimeTAG: 2012-10-19T16:10:55Z FirstChildTAG: Any hints for the last two parts? FirstChildUserIdTAG: 35192 FirstChildUserNameTAG: JoaoVMC FirstChildCreateTimeTAG: 2012-10-19T18:31:19Z SecondChildTAG: Part 2 should be fairly simple. There is a reason they draw it as an electrical circuit. If you ignore the units of seconds/liter for example, you can model an RC circuit with the same time-constant and then calculate the answer from there. In Part 3 they give a hint to use the RS model for the MOSFETs to simplify things. So model an ON FET as a resistor and an OFF FET as an open circuit and you should get most of the answers fairly easily. SecondChildUserIdTAG: 270011 SecondChildUserNameTAG: SnowmanZA SecondChildCreateTimeTAG: 2012-10-19T19:39:03Z FirstChildTAG: No need to make pulse with V=1e12V being active for T=1e-12s Just separate them. For LR see S15V17. RC is different. Just find the Q delivered to capacitor and do the same as in S15V17. FirstChildUserIdTAG: 302188 FirstChildUserNameTAG: Pepek FirstChildCreateTimeTAG: 2012-10-20T17:41:25Z SecondChildTAG: Thank you. The origin of 1e-12? SecondChildUserIdTAG: 89440 SecondChildUserNameTAG: yuk SecondChildCreateTimeTAG: 2012-10-21T05:49:12Z SecondChildTAG: I was able to get a good luck in your favor! SecondChildUserIdTAG: 89440 SecondChildUserNameTAG: yuk SecondChildCreateTimeTAG: 2012-10-22T07:37:01Z SecondChildTAG: How do i find Q.Please help. SecondChildUserIdTAG: 92895 SecondChildUserNameTAG: shihab2555 SecondChildCreateTimeTAG: 2012-10-23T09:49:42Z FirstChildTAG: Q/C=1*(e^(1-t/(33*10^-5))) v=1? t=? thanks in advance for your comment! FirstChildUserIdTAG: 89440 FirstChildUserNameTAG: yuk FirstChildCreateTimeTAG: 2012-10-21T05:23:33Z SecondChildTAG: v*t = 1, so just choose a t that is much shorter than the time constant of the circuit you are analysing. eg. v = 1e6, t = 1e-6 or v = 1e10, t = 1e-10. SecondChildUserIdTAG: 270011 SecondChildUserNameTAG: SnowmanZA SecondChildCreateTimeTAG: 2012-10-27T17:05:59Z FirstChildTAG: I don't understand this exercise and the information that is see in this discussion don't me help much. Please give me any hint for this problem. FirstChildUserIdTAG: 331159 FirstChildUserNameTAG: milanosanta FirstChildCreateTimeTAG: 2012-10-21T01:01:37Z FirstChildTAG: The truth is that without a transfer function, you can't predict a impulse response.Q/C is just an approximation and it's not the only one.So, that's why the answer is WHEREVER between 2879 and 3189. To a impulse problem? You just can fool around with the time constant and find the answer by mistake . So please give a break with " Oh, there it is, you just have to find Q and bla bla .Hint : BLA".If someone has something to say, please form a sentence so we can argue the method. The integral is 1, which means that the area A of the impulse is 1.From the textbook, VC(t) = A/RC*e^(-t/RC).VC(0+) = A/RC =1/(R2*C) . FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-10-26T07:34:38Z SecondChildTAG: Thanks a lot for your help!!!!!!!! SecondChildUserIdTAG: 433574 SecondChildUserNameTAG: endika86 SecondChildCreateTimeTAG: 2012-11-08T15:10:09Z SecondChildTAG: Thank you for, as you say, "forming a sentence." The "it's so easy!" followed by an aimless hint comments are getting really tiresome. SecondChildUserIdTAG: 287805 SecondChildUserNameTAG: Beneficial SecondChildCreateTimeTAG: 2012-11-10T05:15:53Z SecondChildTAG: Thank you! All I needed was a reminder that A was the area of the impulse. That hint alone would have been enough but thanks for making it clear. SecondChildUserIdTAG: 467169 SecondChildUserNameTAG: Eyowzitgoin SecondChildCreateTimeTAG: 2012-11-11T06:25:11Z FirstChildTAG: Please give me some hint on part 3 .. FirstChildUserIdTAG: 413002 FirstChildUserNameTAG: Gauravjain88 FirstChildCreateTimeTAG: 2012-11-04T12:51:53Z IndexTAG: 3400 TitleTAG: H6P1 Hello. I need help finding r0 in small-signal-model. I just lost and don't know where to start. I know that it's value same as R. But have no idea how to derive it from K, VT, VGS and VDS. Please help. Any hints? UserIdTAG: 113561 UserNameTAG: Illogical CreateTimeTAG: 2012-10-18T21:32:39Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: please check this video....it is very helpful..good luck https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_6/Week_6_Tutorials/ FirstChildUserIdTAG: 308861 FirstChildUserNameTAG: Ignaas FirstChildCreateTimeTAG: 2012-10-18T22:04:49Z SecondChildTAG: Seen 'em all ;-) nice vids, btw done lab 6 and hw 6, except r0 question :-( SecondChildUserIdTAG: 113561 SecondChildUserNameTAG: Illogical SecondChildCreateTimeTAG: 2012-10-18T22:10:54Z SecondChildTAG: See the previous thread. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-18T22:14:41Z FirstChildTAG: Thanks! I also learn a lot from kahlil's explaination. Thanks again! ![enter image description here][1] [1]: http:// FirstChildUserIdTAG: 182624 FirstChildUserNameTAG: Daoling FirstChildCreateTimeTAG: 2012-10-19T03:12:18Z FirstChildTAG: Finding the output resistance is a lot like finding the transconductance, gm; in fact, we proceed in nearly an identical manner. Looking at the small-signal model, do you know why we even have the current source and the resistor there? They both represent the response of the newFET to small-signal variations in voltage, but they differ in what voltage we're actually varying. As you already calculated, the transconductance is the derivative of iD with respect to vGS evaluated at the DC vGS = VGS, holding everything else constant. But what if we took the derivative with respect to vDS instead-- and held VGS constant? That's where we get the output resistance term from. Well, almost- because we did d(iD)/d(vDs), what we actually end up with is a conductance-- you need to take the inverse of whatever you got to get the output resistance. FirstChildUserIdTAG: 390034 FirstChildUserNameTAG: kahlil FirstChildCreateTimeTAG: 2012-10-18T22:12:18Z SecondChildTAG: Ouch.. thanx.. it worked. It's seems my luck of attention, as I thought that on Fig3 was drawn SSM of amp built on NewFET, which actually was SSM of lonely NewFET. ;-) SecondChildUserIdTAG: 113561 SecondChildUserNameTAG: Illogical SecondChildCreateTimeTAG: 2012-10-18T22:33:53Z SecondChildTAG: Wow. Thanks a lot. Now i undestand. It works SecondChildUserIdTAG: 204303 SecondChildUserNameTAG: fabnelli SecondChildCreateTimeTAG: 2012-10-19T13:29:24Z FirstChildTAG: plz help me.i am is got stucked in the Homework6 question 1,i have solved the quadratic equation and got the ans like this Vout=+-X/[1+4*X*X*X(X-X)]^1/2 ,But not getting the green tick,plz help me FirstChildUserIdTAG: 97359 FirstChildUserNameTAG: azeem179 FirstChildCreateTimeTAG: 2012-10-19T20:10:41Z IndexTAG: 3401 TitleTAG: HOMEWORK 6 Part 1 Can anyone please help me out on finding gm for the small signal model for the newFet. I know the gm for the mosfet is takeing the derivative of the current function with respect to vIN and evaluated at VIN, but the equationg for the newFet is throwing me of because there is a vOUT term in there. Also I dont understand the drawing of the small signal model. HELP PLEASE!! UserIdTAG: 435197 UserNameTAG: RichmondRichter CreateTimeTAG: 2012-10-18T20:30:46Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: The gm for the mosfet is taking the derivative of the current function with respect to vGS, and treating everything else as a constant. Treat vDS as a constant and take the derivative, and it should be very simple. FirstChildUserIdTAG: 390034 FirstChildUserNameTAG: kahlil FirstChildCreateTimeTAG: 2012-10-18T20:41:17Z SecondChildTAG: we have to take gm=id*ro-vDS/(r0*VGS) SecondChildUserIdTAG: 550401 SecondChildUserNameTAG: revathisingh SecondChildCreateTimeTAG: 2012-10-18T20:51:28Z SecondChildTAG: WHETHER IT IS CORRECT SecondChildUserIdTAG: 550401 SecondChildUserNameTAG: revathisingh SecondChildCreateTimeTAG: 2012-10-18T20:52:00Z SecondChildTAG: thnks, that was exactly what I was doing just realized I was taking the derivative wrong..now in terms of ro I did KCL and got an equation involving small vgs and vds but those are not permitted in the answer. any help for this one? SecondChildUserIdTAG: 435197 SecondChildUserNameTAG: RichmondRichter SecondChildCreateTimeTAG: 2012-10-18T21:48:59Z SecondChildTAG: The output resistance is calculated in a similar manner, except we take the derivative with respect to vDS (instead of vGS), treating everything else as a constant. Well, to be exact, that will give you 1/ro. The multiplicative inverse of what you get for the derivative is the output resistance. SecondChildUserIdTAG: 390034 SecondChildUserNameTAG: kahlil SecondChildCreateTimeTAG: 2012-10-18T22:05:15Z SecondChildTAG: alright thans...Would you be so kind to explain why though? SecondChildUserIdTAG: 435197 SecondChildUserNameTAG: RichmondRichter SecondChildCreateTimeTAG: 2012-10-18T22:06:20Z SecondChildTAG: Well, what is the resistance of an element defined as, anyway? It's the ratio of the voltage across the element to the current through it. When we take the derivative of iD with respect to vDS, we're doing exactly that- finding, for small signal variations around an operating point, what is the approximately constant ratio of current to voltage for our element-- AKA, its equivalent resistance. Because the voltage across our element corresponds to the output voltage, its resistance is the output resistance. SecondChildUserIdTAG: 390034 SecondChildUserNameTAG: kahlil SecondChildCreateTimeTAG: 2012-10-18T22:28:37Z SecondChildTAG: gotcha! but why exactly is the resitor parralel with the current source... SecondChildUserIdTAG: 435197 SecondChildUserNameTAG: RichmondRichter SecondChildCreateTimeTAG: 2012-10-19T01:08:09Z SecondChildTAG: Thanks alot kahlil, very well explained. SecondChildUserIdTAG: 183507 SecondChildUserNameTAG: obiradaniel SecondChildCreateTimeTAG: 2012-10-19T09:05:52Z SecondChildTAG: Thanks a lot.... SecondChildUserIdTAG: 308571 SecondChildUserNameTAG: Sayantani SecondChildCreateTimeTAG: 2012-10-19T14:26:16Z SecondChildTAG: The resistor is in parallel, because both are components of the small-signal drain current, id, and you can think of adding them both together to get the "total" current, and two parallel current sources will add. SecondChildUserIdTAG: 390034 SecondChildUserNameTAG: kahlil SecondChildCreateTimeTAG: 2012-10-19T17:52:01Z IndexTAG: 3402 TitleTAG: H6P2 HELP HOW TO FIND MIN VDD PLZZ ,,I KNW Q2 WILL ALWAYZ BE IN SATURATION,,,I THINK VDD MAY BE INPUT VOLTAGE PLUS THE THRESHOLD VOLTAGE..? UserIdTAG: 183166 UserNameTAG: yogeshk CreateTimeTAG: 2012-10-18T17:12:58Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Look here for some hints: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/507ec10ebc70a3681200000b FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-18T17:22:53Z SecondChildTAG: thanks ,,,could u help me out to find gm2 SecondChildUserIdTAG: 183166 SecondChildUserNameTAG: yogeshk SecondChildCreateTimeTAG: 2012-10-18T17:54:20Z SecondChildTAG: Just like gm1 then use the fact that VGS1 = VGS2 to express the way requested. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-18T18:09:42Z IndexTAG: 3403 TitleTAG: HW6 Problem 2 Hi! Anyone here with any leads on how to solve the RTH part? Thanks UserIdTAG: 143823 UserNameTAG: NathanNadeson CreateTimeTAG: 2012-10-18T11:37:14Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Inject a current increment at the output and calculate the voltage change. Then divide the voltage change by the current increment. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-18T11:51:17Z IndexTAG: 3404 TitleTAG: H6P1 : solution Solved the quadratic equation ,the root i got ,which is correct anyway,is being infinity at vIN = VT. is it possible ? UserIdTAG: 408534 UserNameTAG: kkashyap CreateTimeTAG: 2012-10-18T05:56:01Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: apply l'hospital. FirstChildUserIdTAG: 114864 FirstChildUserNameTAG: Vikaash FirstChildCreateTimeTAG: 2012-10-18T11:51:04Z SecondChildTAG: but the result is not acceptable. SecondChildUserIdTAG: 408534 SecondChildUserNameTAG: kkashyap SecondChildCreateTimeTAG: 2012-10-19T04:39:29Z IndexTAG: 3405 TitleTAG: H6P2 Rth Hi, I've got the answer for the Rth, but unfortunately I got it with the inverted sign in my analysis. Please can someone explain me where does it come from? Just to see why I have the opposite sign. Thanks UserIdTAG: 376173 UserNameTAG: nacho110987 CreateTimeTAG: 2012-10-18T03:45:48Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You are good. I don't know where to start with the problem myself. FirstChildUserIdTAG: 143823 FirstChildUserNameTAG: NathanNadeson FirstChildCreateTimeTAG: 2012-10-18T11:36:18Z IndexTAG: 3406 TitleTAG: Processor Speeds I recently built a pc and the most popular gaming chip is the Intel i53750k. http://www.microcenter.com/product/388577/Core_i5_3570K_34GHz_LGA_1155_Processor I foolishly bought this on newegg. When I watch this video on my television and use another monitor at the same time it clocks at 1.6 MHz, but its stock maximum is 3.4. But these are recommendations, the limit on the GHz is heat dissipation, and many purchase a 3rd party cooler to allow 'overclocking'. UserIdTAG: 334952 UserNameTAG: ciaranfrancismoloney CreateTimeTAG: 2012-10-17T23:33:58Z VoteTAG: 0 CoursewareTAG: Week 6 / Motivation for capacitor CommentableIdTAG: 6002x_motivation_for_capacitor NumberOfReplyTAG: 2 FirstChildTAG: The heat dissipation is a side effect - you need to increase voltage in order to make it work, and you need to increase voltage in order to make signal raise steeper in order to make period of sync signal (which is reciprocal of the frequency) shorter. So the real limit is signal's raising time. FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-18T00:07:20Z FirstChildTAG: Most newer CPU's run at lower speeds when they are not being used heavily, this reduces power consumption. Run a processor intensive application, such as one designed to do burn-in tests of CPU's and check your clock speed while it is running, you will probably see that it will be running at 3.4GHz. FirstChildUserIdTAG: 270011 FirstChildUserNameTAG: SnowmanZA FirstChildCreateTimeTAG: 2012-10-18T16:42:05Z IndexTAG: 3407 TitleTAG: H6P2 Minimum value of Vin Hello from Greece, although i have finished the second question i am stuck on the first one. It is frustrating because i cannot find many questions in the forum about the first one which leads me to believe it is easy to be solved. Can somebody quide me? Thanks in advance. UserIdTAG: 214275 UserNameTAG: TheodoreGr CreateTimeTAG: 2012-10-17T19:24:10Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: It is the DC bias that when combined with the maximum negative signal causes the MOSFET to just reach cutoff. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-17T19:41:26Z SecondChildTAG: Thanks a million! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-17T20:30:17Z SecondChildTAG: Thanks a lot. I got the point.... SecondChildUserIdTAG: 381797 SecondChildUserNameTAG: Saira180 SecondChildCreateTimeTAG: 2012-10-20T11:08:14Z FirstChildTAG: 1stly realise what should be the Vin(min) to keep Q1 in Saturation!!! Then add Dv so as to boost it to the operation region of minimum Bias!!! Hope it helps!!! FirstChildUserIdTAG: 82597 FirstChildUserNameTAG: bondrajat FirstChildCreateTimeTAG: 2012-10-17T19:39:17Z SecondChildTAG: Thanks a lot both of you. Problem solved. Some times when you read too much you are stuck to the simpler problems. SecondChildUserIdTAG: 214275 SecondChildUserNameTAG: TheodoreGr SecondChildCreateTimeTAG: 2012-10-17T19:47:22Z IndexTAG: 3408 TitleTAG: Lab 7 - Task 1 I think the result of the integral is not the result the question is looking for. Since the current also goes negative, the voltage across the capacitor can go negative, which in the graph (and in the answer formula) does not happen, as it is biased by a certain value. Moreover, the links to the equations do not work. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-17T17:42:24Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: The negative current discharges the capacitor and returns its voltage to zero. The voltage across the capacitor never goes negative. The value of the integral is the answer to the question. Edited to correct typo. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-17T17:53:15Z IndexTAG: 3409 TitleTAG: there's a glitch!!! well in the question it is stated that fill only 1 answer for a question..so when i put just "a" for the 1st question,i got it wrong and when i checked it, the answer was "a or f".Please do rectify this mistake... UserIdTAG: 625118 UserNameTAG: alex_003 CreateTimeTAG: 2012-10-17T17:16:20Z VoteTAG: 0 CoursewareTAG: Week 1 / Various V-I characteristics CommentableIdTAG: 6002x_S1E1_Various_V-I_characteristics NumberOfReplyTAG: 1 FirstChildTAG: It seems to be working for me. FirstChildUserIdTAG: 90796 FirstChildUserNameTAG: ChaunceyGardiner FirstChildCreateTimeTAG: 2012-10-17T21:43:41Z IndexTAG: 3410 TitleTAG: invalid input Invalid input: Could not parse '(­ 1+ sqrt( 1+4*R*K*VS*( vIN-VT) ) )/( 2*R*K*( vIN -­ VT ) )' as a formula UserIdTAG: 219204 UserNameTAG: vsriram CreateTimeTAG: 2012-10-17T17:00:24Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I've looked at it quite awhile, and it looks OK to me. At the risk of giving away the answer it is equivalent to mine, which got a check. The only difference from mine is the order of the factors in the square root. I wish I could be more help. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-17T17:23:46Z IndexTAG: 3411 TitleTAG: GUYS help me Invalid input: Could not parse '1/(2K*VDS*(VGS-VT)' as a formula. i'm getting this what should i do UserIdTAG: 219204 UserNameTAG: vsriram CreateTimeTAG: 2012-10-17T16:29:35Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Make sure parentheses balance and use * for **all** multiplication. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-17T16:31:23Z SecondChildTAG: also please help me: Invalid input: Could not parse '(­ 1+ sqrt( 1+4*R*K*VS*( vIN-VT) ) )/( 2*R*K*( vIN -­ VT ) )' as a formula SecondChildUserIdTAG: 219204 SecondChildUserNameTAG: vsriram SecondChildCreateTimeTAG: 2012-10-17T16:47:23Z SecondChildTAG: all the things are balanced here i suppose SecondChildUserIdTAG: 219204 SecondChildUserNameTAG: vsriram SecondChildCreateTimeTAG: 2012-10-17T16:47:56Z SecondChildTAG: if u hav copied it and pasted in the box, it wont work!! try re writing it using * for all multiplications, balanced ( and ) and omit any unnecessary spaces.. SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-17T17:30:35Z SecondChildTAG: The "2K" should be 2*K SecondChildUserIdTAG: 391929 SecondChildUserNameTAG: RohanNagarkar SecondChildCreateTimeTAG: 2012-10-17T18:19:56Z IndexTAG: 3412 TitleTAG: >>1 or <<1 is gm*RL meant to be much greater than 1 or much less than 1? i get vo = vin / ( 1 + gm*RL ) so gm*RL << 1 would give vo = vin UserIdTAG: 388273 UserNameTAG: kilmarta CreateTimeTAG: 2012-10-17T16:20:08Z VoteTAG: 0 CoursewareTAG: Week 6 / Perspective on the small-signal circuit CommentableIdTAG: 6002x_small_signal_perspective NumberOfReplyTAG: 0 IndexTAG: 3413 TitleTAG: bug shouldn't he write f(2i) instead of 2f(i) for voltage source case? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-17T16:06:27Z VoteTAG: 0 CoursewareTAG: Week 2 / Is it linear? CommentableIdTAG: 6002x_is_it_linear NumberOfReplyTAG: 1 FirstChildTAG: yes u r ryt! FirstChildUserIdTAG: 368712 FirstChildUserNameTAG: jmen FirstChildCreateTimeTAG: 2012-12-19T13:57:32Z IndexTAG: 3414 TitleTAG: The homwork system has a little fault. The homework system has a little fault. It can't realize this algebra K/2(vIN-vOUT-VT)^2 as a correct one, while the answer given by the system is (k/2)*((vIN-vOUT-VT)^2 ). UserIdTAG: 128276 UserNameTAG: ChengBin CreateTimeTAG: 2012-10-17T15:31:16Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Or you could have just as easily written K/2*(vIN-vOUT-VT)^2 The checker requires a * sign between the parentheses and the variable. Just a fact of life :) FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-10-17T15:37:59Z SecondChildTAG: Case matters. It's not too great a hurdle to overcome, is it? SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-18T23:51:46Z IndexTAG: 3415 TitleTAG: Why is the result of the Differential Vout/Vin equal to that one in H5p3 VOUT=(KRs(Vin−Vt)+1−sqrt(2KRs(Vin−Vt)+1))/KRS (Vout/Vin)' should be =1- 1/(2KRs*sqrt(2KRs(Vin-Vt)+1)) ,but not 1-1/sqrt(2KRs(Vin-Vt)+1). Is there anybody could understand what I mean? Who is wrong? Me or the answer? UserIdTAG: 128276 UserNameTAG: ChengBin CreateTimeTAG: 2012-10-17T15:22:11Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3416 TitleTAG: Multi-Gate Transistors Industry is going to a multi-gate transistor technology, such as Tri-Gate transistors, where one gate controls 3 transistors. Similar technologies include Flexfet, Gate-all-around FETs and Planar double-gate transistors. It will be interesting on how these new transistors will affect our analysis of MOSFET transistors, as we're doing now. I'm sure MIT has done plenty of research in this area. Hopefully our current study of MOSFET's will be a good foundation for understanding this new technology, when it is our turn design IC's with it. Has anybody had any experience comparing our current methods for analyzing MOSFET's to that of analyzing multi-gate transistors? UserIdTAG: 132828 UserNameTAG: rharris CreateTimeTAG: 2012-10-17T15:10:39Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: MOSFET just an example - we are given principles - with this knowledge you can calculate any device FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-17T16:14:43Z IndexTAG: 3417 TitleTAG: How i did it? **p=v*i r=v*i** so *11=v*i* so *8=v/i* **v=i*8 + v=11/i =** 2v=(11/i)+(i*8) 2v=11+8 v=9.5 UserIdTAG: 657539 UserNameTAG: ajeeb24 CreateTimeTAG: 2012-10-17T04:10:57Z VoteTAG: 0 CoursewareTAG: Week 1 / Power CommentableIdTAG: 6002x_S1E2_Power NumberOfReplyTAG: 1 FirstChildTAG: Your answer doesn't make sense at all. I can't make out what you are doing. FirstChildUserIdTAG: 622419 FirstChildUserNameTAG: npham048 FirstChildCreateTimeTAG: 2012-10-18T07:56:39Z IndexTAG: 3418 TitleTAG: H6P1 Question 1 I can't see where am i going wrong. I find my equation $i_D$ $i_D = K \cdot (v_{IN}- v_T) \cdot {v_{OUT}}^2$ And then i use the Node method to get $v_{OUT}$ in terms of $i_D$ $v_{OUT} = \frac{V_S}{R} - i_D \cdot R$ Combining both and solving the quadratic for $v_{OUT}$ i get: $\displaystyle v_{OUT} = \frac{1 \pm \sqrt{ 1 + 4 \cdot K \cdot V_S \cdot (v_{IN} - V_T)}}{-2 \cdot K \cdot R \cdot (v_{IN}-V_T)}$ By checking ($v_{IN}=V_T$), one of the roots become indeterminate ($\frac{0}{0}$), so i apply L'hopital till i get a concrete value. When i input the root that i was able to check ($v_{IN}=V_T$) in the answer box, it gives me wrong answer. Please help UserIdTAG: 40865 UserNameTAG: guilima10 CreateTimeTAG: 2012-10-17T03:56:47Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Vs / R gives you a current. How can that be a part of a voltage ? FirstChildUserIdTAG: 90796 FirstChildUserNameTAG: ChaunceyGardiner FirstChildCreateTimeTAG: 2012-10-17T06:15:45Z SecondChildTAG: Thanks, now i see where i was going in the wrong way SecondChildUserIdTAG: 40865 SecondChildUserNameTAG: guilima10 SecondChildCreateTimeTAG: 2012-10-19T14:09:32Z FirstChildTAG: You don't need to use l'Hopital's rule! Just use your answer as it is. FirstChildUserIdTAG: 359310 FirstChildUserNameTAG: AaronYeoh FirstChildCreateTimeTAG: 2012-10-18T11:32:16Z FirstChildTAG: I have the same problem. I got that formula but i don't know why the check button gives me this notice... "Invalid input: Could not parse '(1-sqrt(4*K*R*VS*(vIN-VT)+1))/(2*K*R*(vIN-VT)' as a formula" FirstChildUserIdTAG: 314294 FirstChildUserNameTAG: victormp FirstChildCreateTimeTAG: 2012-10-18T10:44:17Z SecondChildTAG: Look at your formula carefully. You are *almost* there! (((Check your parentheses (brackets!)) SecondChildUserIdTAG: 359310 SecondChildUserNameTAG: AaronYeoh SecondChildCreateTimeTAG: 2012-10-18T11:29:46Z SecondChildTAG: Thanks ;) SecondChildUserIdTAG: 314294 SecondChildUserNameTAG: victormp SecondChildCreateTimeTAG: 2012-10-18T14:11:18Z SecondChildTAG: i have the same problem where parentheses are wrong.? SecondChildUserIdTAG: 136490 SecondChildUserNameTAG: Ali_PU SecondChildCreateTimeTAG: 2012-10-19T16:20:09Z SecondChildTAG: I have the same problem where is the mistake!!!!!?????I cant fix it///so stupid??? SecondChildUserIdTAG: 338422 SecondChildUserNameTAG: Sl1mcO SecondChildCreateTimeTAG: 2012-10-20T16:55:51Z SecondChildTAG: cant get the same expression correct....!! SecondChildUserIdTAG: 127800 SecondChildUserNameTAG: MNB SecondChildCreateTimeTAG: 2012-10-21T02:54:57Z IndexTAG: 3419 TitleTAG: number it please number it please. UserIdTAG: 657539 UserNameTAG: ajeeb24 CreateTimeTAG: 2012-10-17T03:39:04Z VoteTAG: 0 CoursewareTAG: Week 1 / Various V-I characteristics CommentableIdTAG: 6002x_S1E1_Various_V-I_characteristics NumberOfReplyTAG: 0 IndexTAG: 3420 TitleTAG: Help with H6P1 Q1 please I am struggling with **H6P1 Q1**. So, I have the equation for **Vout=VS+K*R*(VIN-VT)*VDS^2**. Then, to substitute **VDS** in terms of **Vout** and **VS** doing **KVL** for the outer circuit loop I get: **VDS=2VS-VOUT** ![enter image description here][1] I am unsure what to do from here! Or even if I am correct thus far! Your help will be kindly appreciated. Hazel. [1]: https://edxuploads.s3.amazonaws.com/13504226941343647.bmp UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-10-16T21:25:04Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You have a sign wrong (It is physically impossible for VOUT to be greater than VS!) and your expression for VDS is incorrect. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-16T21:35:21Z SecondChildTAG: Thanks SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-19T17:56:39Z IndexTAG: 3421 TitleTAG: lab 7 Computed value for v(c) in volts: I have the correct formula and have calculated a value that is very close to the one measured with the simulation but am getting the hated red x. Any hints or help would be appreciated UserIdTAG: 18073 UserNameTAG: gburkhart CreateTimeTAG: 2012-10-16T20:38:29Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: put your calculator in RADIANS not degrees. FirstChildUserIdTAG: 18073 FirstChildUserNameTAG: gburkhart FirstChildCreateTimeTAG: 2012-10-16T20:48:03Z IndexTAG: 3422 TitleTAG: Kindly ,expalin what happened here ![this example is in page 343][1] [1]: http:// UserIdTAG: 185715 UserNameTAG: amirengineer CreateTimeTAG: 2012-10-16T18:46:04Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi amirengineer, Can I help you? :) What are you referring about with that question? Do you refer about an specific problem or about the edX Plataform? FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-16T18:52:43Z SecondChildTAG: Thank You Myrlmlt for your response actually i though i posted a photo about a Problem, I'll post it again "in case it didn't appears it's about Example 7.6 in the text book,Page 343" , i didn't get what Prof. Agarwal means by this "Since we are told that both MOSFETs operate in the saturation region, and since iDS for both MOSFETs is the same, their respective gate-to-source voltages must also be equal."!!? Also how did he solved for Vo ? ![7.6][1] [1]: http:// allow me to ask another question, in Example 7.10 "page 361", I'm Confused about the answer of Vo at Vi(min.) which = 0 , I'm confused about distinguishing between when to use the fact that at the edge of transition between cutoff & Saturation Vo=Vs but at the same time the Equation says that Vo at the Edge is Vo=Vin - Vt which will equals Zero as mentioned above? SecondChildUserIdTAG: 185715 SecondChildUserNameTAG: amirengineer SecondChildCreateTimeTAG: 2012-10-17T20:05:25Z IndexTAG: 3423 TitleTAG: H6P2 QS 5 CAN sum1 pls explain the correct aproach to find the vout for this case step by step??I couldnt solve it correctly..please help someone? UserIdTAG: 368712 UserNameTAG: jmen CreateTimeTAG: 2012-10-16T16:20:14Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: VDD minus the drop across Q2. What is the i-V relation for Q2 with the gate connected to the drain? FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-16T16:35:18Z SecondChildTAG: well q2 acts like a resistance whose value i knw.. moreover VS2=VD2=VGS1=VIN. SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-16T16:40:13Z SecondChildTAG: now what to do? SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-16T16:40:25Z SecondChildTAG: oops i mean VGS2=VDS2=VGS1=VIN SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-16T16:42:24Z SecondChildTAG: If you know VDS2 then you have everything you need. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-16T17:33:03Z SecondChildTAG: I understand VGS2=VDS2, no problem there but how does VDS2=VGS1=VIN? SecondChildUserIdTAG: 244115 SecondChildUserNameTAG: Pedro1969 SecondChildCreateTimeTAG: 2012-10-17T12:36:32Z SecondChildTAG: iDS2 = iDS1 therefore VGS2 = VGS1 SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-17T12:53:23Z SecondChildTAG: how do i find resistance across Q2? SecondChildUserIdTAG: 357747 SecondChildUserNameTAG: kishores SecondChildCreateTimeTAG: 2012-10-17T13:33:24Z SecondChildTAG: Took a little bit but I got it. Thank you very much! These MOSFET problems have been a real pain for me. SecondChildUserIdTAG: 244115 SecondChildUserNameTAG: Pedro1969 SecondChildCreateTimeTAG: 2012-10-17T13:44:52Z FirstChildTAG: **H6P2** how do i find resistance across Q2? FirstChildUserIdTAG: 357747 FirstChildUserNameTAG: kishores FirstChildCreateTimeTAG: 2012-10-17T13:25:02Z SecondChildTAG: Use i-V relation for MOSFET. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-17T18:20:44Z SecondChildTAG: kishores: see page 417, figure 8.13 :) SecondChildUserIdTAG: 278792 SecondChildUserNameTAG: sali SecondChildCreateTimeTAG: 2012-10-20T09:40:18Z IndexTAG: 3424 TitleTAG: I need some help with the first problem of week 6 All, this is my answer for the first question sqrt((vIN-VS)/(-K*R*(vIN-VT) is this incorrect???, because i cannto get the green tick....i get a comment that the system can't parse it.... UserIdTAG: 308861 UserNameTAG: Ignaas CreateTimeTAG: 2012-10-16T16:05:57Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Skyhawk, I have the following equations: Vin=Vs-idR Vin=Vs-KR(vin-VT)vout^2... I have taken vgs=vin and vds=vout. FirstChildUserIdTAG: 308861 FirstChildUserNameTAG: Ignaas FirstChildCreateTimeTAG: 2012-10-17T00:42:32Z SecondChildTAG: I believe that you meant vout instead Vin on the left hand side. With that you have a quadratic equation for vout. Solve it. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-17T03:00:50Z FirstChildTAG: It is not correct. In addition, as written the parentheses are not balanced. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-16T16:16:24Z SecondChildTAG: Can u help me with h6p2 qs 5 sky? SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-16T16:39:00Z SecondChildTAG: hmmmm....can you give me a hint?? SecondChildUserIdTAG: 308861 SecondChildUserNameTAG: Ignaas SecondChildCreateTimeTAG: 2012-10-16T17:39:08Z SecondChildTAG: What is your starting point? We can go step by step from there. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-16T18:31:36Z SecondChildTAG: Skyhawk, I have the following equations: Vin=Vs-idR Vin=Vs-KR(vin-VT)vout^2... I have taken vgs=vin and vds=vout. SecondChildUserIdTAG: 308861 SecondChildUserNameTAG: Ignaas SecondChildCreateTimeTAG: 2012-10-17T02:14:30Z FirstChildTAG: Myrimit, Do you have any suggestions?? FirstChildUserIdTAG: 308861 FirstChildUserNameTAG: Ignaas FirstChildCreateTimeTAG: 2012-10-17T02:14:52Z SecondChildTAG: Take a look at this [Hints][1] :) [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/507e218af4d89e230000009d SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-17T03:14:51Z IndexTAG: 3425 TitleTAG: S11V3 i haven't understood about input swing. please tell me, what is input swing ? UserIdTAG: 114864 UserNameTAG: Vikaash CreateTimeTAG: 2012-10-16T12:39:57Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3426 TitleTAG: H6P2 What is Dv value in H6P2? UserIdTAG: 407765 UserNameTAG: Igor14 CreateTimeTAG: 2012-10-16T09:13:43Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: it is the small change in voltage D stands for delta FirstChildUserIdTAG: 18073 FirstChildUserNameTAG: gburkhart FirstChildCreateTimeTAG: 2012-10-16T13:21:15Z IndexTAG: 3427 TitleTAG: confused how to submit all there is no progress while homework can be done.....how submit it please show me UserIdTAG: 561013 UserNameTAG: sohilfaruqui CreateTimeTAG: 2012-10-16T05:20:47Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: a dc source is of battery, then how it can begave as open circuit? FirstChildUserIdTAG: 561769 FirstChildUserNameTAG: Ranjankumar251 FirstChildCreateTimeTAG: 2012-10-16T06:18:57Z SecondChildTAG: *behave SecondChildUserIdTAG: 561769 SecondChildUserNameTAG: Ranjankumar251 SecondChildCreateTimeTAG: 2012-10-16T06:19:09Z IndexTAG: 3428 TitleTAG: Why is a DC current source considered a open circuit? "And so DC voltage sources behave like short circuit small signals." Okay, I understand the reason. "And similarly, DC current sources behave as open circuit small signals." I had a look in the textbook but it also gives no reason for why a current source is treated as an open circuit. Please refresh my memory here, I remember this in one of the past videos somewhere but searching through videos is difficult at best. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-16T04:13:51Z VoteTAG: 0 CoursewareTAG: Week 6 / Small-signal circuit model CommentableIdTAG: 6002x_small_signal_ckt_model NumberOfReplyTAG: 3 FirstChildTAG: In small signal model you suppose, that all large signal active elements are turned off. Turning off means that current through the branch with current source=0, no circuit - open circuit. So turned off current source is equivalent to open circuit. FirstChildUserIdTAG: 345464 FirstChildUserNameTAG: Constantine_ru FirstChildCreateTimeTAG: 2012-10-16T12:51:53Z FirstChildTAG: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/SmallSignalCircuitModels/ FirstChildUserIdTAG: 361137 FirstChildUserNameTAG: Ericson FirstChildCreateTimeTAG: 2012-10-17T16:17:07Z FirstChildTAG: You differentiate the large signal model device equations to get the small signal ones.So if you differentiate the equation of Voltage source Vs which is Vs=Const. ,you get vs=0. Hence if you differentiate equation for current source, i.e. Is=Const, you get is=0 , which is represented by a open circuit(as no current flows through it.) Remember that both sources are independent and don't depend on any changes whatsoever. FirstChildUserIdTAG: 222809 FirstChildUserNameTAG: AshutoshTadkase FirstChildCreateTimeTAG: 2012-10-25T17:15:36Z IndexTAG: 3429 TitleTAG: Trade-off between volume and dynamic range in music I think there's also a trade-off in modern music between volume and dynamic range in all the masterized versions of old albums we're getting in the last years. In sound engineering forums is a common topic. It's usually said that old vinyl versions of 50-70's albums had a much higher dynamic range (signal amplitude, peak to peak, I understand) than nowadays masterized versions of the same albums; i.e., actual music is thought and engineered to sound on radios, speakers without mid-range, cars, etc. I think amplifiers have to do with this as well, right? UserIdTAG: 398594 UserNameTAG: DaveyJC CreateTimeTAG: 2012-10-15T20:02:14Z VoteTAG: 0 CoursewareTAG: Week 6 / Amplifier operating point CommentableIdTAG: 6002x_amplifier_operating_point NumberOfReplyTAG: 2 FirstChildTAG: warm tube sound? ;) FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-15T20:15:14Z FirstChildTAG: Most records released these days have been compressed to death. They sacrifice everything to make the music sound (and look) louder. It wasn't all better in the old days, though. For instance, there is a physical limit to how much bass you can put on vinyl records. FirstChildUserIdTAG: 90796 FirstChildUserNameTAG: ChaunceyGardiner FirstChildCreateTimeTAG: 2012-10-15T23:38:32Z IndexTAG: 3430 TitleTAG: doubt e1=e3+V how and why ?? explain in detail . UserIdTAG: 640228 UserNameTAG: vishal119 CreateTimeTAG: 2012-10-15T19:47:24Z VoteTAG: 0 CoursewareTAG: Week 1 / Node method example CommentableIdTAG: 6002x_S2E5_Node_Method NumberOfReplyTAG: 1 FirstChildTAG: explain what? V is connected to e3 by its negative terminal, so voltage on its positive terminal, which is e1, will be obviously e3+V FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-15T20:04:38Z IndexTAG: 3431 TitleTAG: doubt what is nodal voltage ? diff. between nodal voltage and supply voltage ? UserIdTAG: 640228 UserNameTAG: vishal119 CreateTimeTAG: 2012-10-15T19:45:13Z VoteTAG: 0 CoursewareTAG: Week 1 / Node method example CommentableIdTAG: 6002x_S2E5_Node_Method NumberOfReplyTAG: 1 FirstChildTAG: if second node not specified then between node and common ground FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-15T20:00:18Z IndexTAG: 3432 TitleTAG: H5p2 In H5P2 problem in the 4th part my answer was 3.90591528528 and i was given a wrong answer. Now the right answer is 3.906 ?? UserIdTAG: 309933 UserNameTAG: miramar CreateTimeTAG: 2012-10-15T17:04:29Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3433 TitleTAG: Lab 8 problem In lab 8 i check the answer i give and it says "Invalid input: dv dt v V not permitted in answer" but i haven't put any of these characters in the text boxes, only in the beginning, but after i erased them and put another answer without any illegal character. How can i fix it because it is not allow to me to check any answer i give.. thanks UserIdTAG: 20608 UserNameTAG: jimz CreateTimeTAG: 2012-10-15T16:43:31Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: OK i solved it, after logouts and perhaps session timeouts(?) FirstChildUserIdTAG: 20608 FirstChildUserNameTAG: jimz FirstChildCreateTimeTAG: 2012-10-15T21:20:54Z IndexTAG: 3434 TitleTAG: Gain related to -K(VI-VT)*RL? How is "**gain**" related to gm*RL?? UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-10-15T16:18:52Z VoteTAG: 0 CoursewareTAG: Week 6 / Amplifier operating point CommentableIdTAG: 6002x_amplifier_operating_point NumberOfReplyTAG: 1 FirstChildTAG: vo= -K(VI-VT)*RL*vi. but gm= -K(VI-VT) Thus, vo=-gm*RL*vi vo/vi= -gm*RL Recall, Gain=Vo/VI Hence, Gain =-gm*RL Bearing in mind that gain is negative, then we can say: Gain = gm*RL FirstChildUserIdTAG: 332954 FirstChildUserNameTAG: Folashade FirstChildCreateTimeTAG: 2012-10-21T09:55:19Z SecondChildTAG: Все правильно, но только посчитай на конце уравнения VI-маленького сигнала. Ответ простой будет SecondChildUserIdTAG: 399289 SecondChildUserNameTAG: Minasyan SecondChildCreateTimeTAG: 2012-10-22T09:11:55Z IndexTAG: 3435 TitleTAG: Where am I doing the mistake? (S11E3 last q) Where do I go wrong? Step1: I get this circuit by shorting out the independent source: ![enter image description here][1] Step2: Applying KCL above RS gives me: $g_mv_{out} -\frac{V{out}}{R_S}-1 = 0$ which results in $v_{out}=\frac{R_s}{g_mR_S-1}$ Now this one is wrong, but I am not able to see why? It must be something superobvious I fear... [1]: https://edxuploads.s3.amazonaws.com/13503114611343647.png UserIdTAG: 329254 UserNameTAG: KGabor CreateTimeTAG: 2012-10-15T14:35:04Z VoteTAG: 0 CoursewareTAG: Week 6 / Source-follower small-signal exercise CommentableIdTAG: 6002x_small_signal_source_follower NumberOfReplyTAG: 3 FirstChildTAG: What is the ground node when you set vin=0? FirstChildUserIdTAG: 376877 FirstChildUserNameTAG: AndBre FirstChildCreateTimeTAG: 2012-10-15T15:22:12Z SecondChildTAG: It is said to dd an independent current source to inject a current INTO !!! the output port. Change direction of independent current source). And vgs = -vout. SecondChildUserIdTAG: 337855 SecondChildUserNameTAG: flexo SecondChildCreateTimeTAG: 2012-10-16T17:18:42Z FirstChildTAG: well... you should decide what you consider as a positive current - incoming into or outgoing from the node - now it's total mess FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-15T15:40:10Z SecondChildTAG: Well as I thought ground node is under Rs, the current of the dependent source comes in, the current of the 1A source goes out, and through Rs it also leaves. Thats how i figured the formula with 2 negatives. SecondChildUserIdTAG: 329254 SecondChildUserNameTAG: KGabor SecondChildCreateTimeTAG: 2012-10-15T18:48:45Z SecondChildTAG: So, you considered that leaving current is negative. In this case current from dependent source is incorrect. Note, that schematics you provided is not clear - you said you shorted out independent source - it's not shorted on you schematics. Think what vgs is. SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-15T19:54:10Z SecondChildTAG: I think vgs=-vout at that case. SecondChildUserIdTAG: 213452 SecondChildUserNameTAG: JoJosida SecondChildCreateTimeTAG: 2012-10-16T04:52:26Z SecondChildTAG: vgs is the neg vout...but shouldn't the answer be RS/(RS(1+gm)) SecondChildUserIdTAG: 256543 SecondChildUserNameTAG: sidney23 SecondChildCreateTimeTAG: 2012-10-16T15:15:47Z SecondChildTAG: JoJosida is right. sidney23 is wrong. SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-16T15:18:36Z SecondChildTAG: JoJosida is correct.... but can you figure out why? I recommend that you use the convention taught in the lectures, that is, "the currents leaving the node must sum to zero", and then solve that equation for $v_{out}$ SecondChildUserIdTAG: 248807 SecondChildUserNameTAG: 1kingsman SecondChildCreateTimeTAG: 2012-10-16T22:54:14Z SecondChildTAG: In addition, you do not show that your dependent source is shorted, but rather left open. Try to fix that. SecondChildUserIdTAG: 248807 SecondChildUserNameTAG: 1kingsman SecondChildCreateTimeTAG: 2012-10-16T22:55:43Z SecondChildTAG: Actually, i am getting resistance as -RS/1+gm*RS. And since -resistace don't have physical meaning that's why it's RS/1+gm*RS. Am i making sense. SecondChildUserIdTAG: 408534 SecondChildUserNameTAG: kkashyap SecondChildCreateTimeTAG: 2012-10-17T05:10:36Z SecondChildTAG: Absolutely not. Resistance is defined as v/i when i enters the positive terminal. If you get a negative resistance, that means you chose the wrong sign for i or v. You cannot invert the sign of a solution simply because the negative doesn't make sense: you can do that only when you have multiple solutions (like from a quadratic equation) and you discard those that don't make sense. SecondChildUserIdTAG: 376877 SecondChildUserNameTAG: AndBre SecondChildCreateTimeTAG: 2012-10-17T08:53:03Z SecondChildTAG: kkashyap, first equation is correct, second is not. It's not a question of "negative resistance" - it's a question how you choose direction of currents. It's why it's better to follow convention for node method that outgoing current is positive - less mess. SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-17T16:43:34Z SecondChildTAG: YakovO, it's not a question how you choose directions of currents (positive outgoing or positive ingoing doesn't make difference), it's a question of combination of signs between voltage and currents: it's called "associated variables discipline". Check this out: S2V3 at 3:58. SecondChildUserIdTAG: 376877 SecondChildUserNameTAG: AndBre SecondChildCreateTimeTAG: 2012-10-17T20:22:26Z SecondChildTAG: AndBre > it's a question of combination of signs between voltage and currents it's question of combination of voltage signs and current direction in common case. In this particular case voltage signs remained usual, but current direction chosen opposite - so it's a question of choice of direction of current (sorry for many 'of's :) - consider, it would be pretty weird if he choose opposite voltage signs for node method :) SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-17T23:52:30Z SecondChildTAG: YakovO, try to solve this, taking ingoing currents as positive (or negative if you like). Stick to the associated variables discipline and you'll be fine. That's it. ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13505508801343618.png SecondChildUserIdTAG: 376877 SecondChildUserNameTAG: AndBre SecondChildCreateTimeTAG: 2012-10-18T09:03:55Z SecondChildTAG: AndBre, did you read what I wrote about switching voltage? Did you realize that KGabor left voltage usual, but considered that incoming current is positive - instead of outgoing - WHICH IS question of current direction? Would you object that (-Vout/RS) means negative current? "Discipline" means "social contract" - do you realize, that our "social contract" about current direction in metal conductors is totally wrong and accepted by historical mistake? Open your mind and think about it :) That's it. SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-18T13:48:47Z SecondChildTAG: probably there is a little misunderstanding - under "current direction" I mean how to consider outgoing current - positive or negative - there is nothing about discipline that current incoming into positive terminal - it remains intact of course SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-18T13:54:59Z SecondChildTAG: Thanks all for your comments, it worked out for at the end, I learned from this. SecondChildUserIdTAG: 329254 SecondChildUserNameTAG: KGabor SecondChildCreateTimeTAG: 2012-10-30T15:53:03Z FirstChildTAG: why vgs=-vout? FirstChildUserIdTAG: 500965 FirstChildUserNameTAG: barka0 FirstChildCreateTimeTAG: 2012-10-17T20:28:10Z SecondChildTAG: vgs = vin - vout, and vin = 0. Therefore, vgs = - vout. I happened to set the bottom wire to ground in my node analysis, so that made vin = 0, but they also said to evaluate the output resistance with vin = 0. SecondChildUserIdTAG: 106229 SecondChildUserNameTAG: jc_lounge SecondChildCreateTimeTAG: 2012-10-21T05:43:16Z IndexTAG: 3436 TitleTAG: homework 8 help iv finieshed all but the first section part 1 can sumone suggest helpull places to read up to equire the formula in wich to appl sorry for my english UserIdTAG: 436749 UserNameTAG: waynebrown CreateTimeTAG: 2012-10-15T14:10:19Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Just realise what should be the initial conditions to satrt up with!!! Hint:( No current no Voltage whatsoever :) ) I wish it may help!! FirstChildUserIdTAG: 82597 FirstChildUserNameTAG: bondrajat FirstChildCreateTimeTAG: 2012-10-17T19:29:48Z IndexTAG: 3437 TitleTAG: Thanks to staff Hello...I just took a look at the 'show answer' displays for a couple problems that I messed up. Thank you. It should be a big help. UserIdTAG: 92346 UserNameTAG: MobiusTruth CreateTimeTAG: 2012-10-15T14:06:20Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3438 TitleTAG: H6P3 How to find Decay Time constant In Network A UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-15T13:19:51Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi, just find RTH , remember your decaying rate RC. FirstChildUserIdTAG: 369339 FirstChildUserNameTAG: vargaslen FirstChildCreateTimeTAG: 2012-10-15T14:38:22Z SecondChildTAG: I can not get it SecondChildUserIdTAG: 37887 SecondChildUserNameTAG: Siddhu SecondChildCreateTimeTAG: 2012-10-17T09:32:18Z SecondChildTAG: Hi. Ignore the capacitance first, and look at the circuit from its place. Whats the resistant you see? Thats your RTH. Now, the answer you are looking for is simply C*RTH , as it is your time constant. Enter your answer algebraically. SecondChildUserIdTAG: 329254 SecondChildUserNameTAG: KGabor SecondChildCreateTimeTAG: 2012-10-18T15:43:19Z SecondChildTAG: why wouldn't the decaying be $\frac{1}{R\cdot C}$?? SecondChildUserIdTAG: 40865 SecondChildUserNameTAG: guilima10 SecondChildCreateTimeTAG: 2012-10-19T14:53:02Z SecondChildTAG: RC is the time constant it has dimensions of time . SecondChildUserIdTAG: 266739 SecondChildUserNameTAG: satvikchugh SecondChildCreateTimeTAG: 2012-10-20T11:15:09Z SecondChildTAG: I try to put my answer wrong SecondChildUserIdTAG: 37887 SecondChildUserNameTAG: Siddhu SecondChildCreateTimeTAG: 2012-10-20T14:56:04Z IndexTAG: 3439 TitleTAG: H5P2 iDS Hello, I have been attempting to input the accurate expression for iDS in H5P2 (since yesterday before the deadline for HW5) and when I check the website always returns that it is incorrect. I suspect for whatever reason, it is not accepting the correct answer, for when I refresh the page it always returns to the last incorrect answer I had typed in. Has anyone encountered this problem before? Any assistance is appreciated. Thank You Soniya UserIdTAG: 275806 UserNameTAG: soniyatn CreateTimeTAG: 2012-10-15T13:06:17Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Troubleshooting NumberOfReplyTAG: 0 IndexTAG: 3440 TitleTAG: sufix f in capacitor value hi i just want to ask why the parser doesn't understand **200fF = 200*10^(-15)F** as other sufixes : 1m= 10^(-3), 1n=10^(-6), 1k=1000 Tuan UserIdTAG: 114913 UserNameTAG: ngoctuan CreateTimeTAG: 2012-10-15T11:35:31Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: femto is not implemented in the parser. They never explained why, in spite of many posts about this during the first 6002x last spring. For your example expression you can use this notation, though: **200e-15** FirstChildUserIdTAG: 90796 FirstChildUserNameTAG: ChaunceyGardiner FirstChildCreateTimeTAG: 2012-10-15T23:53:26Z IndexTAG: 3441 TitleTAG: download lecture Hi Dear i cannot download lectures for week 5,6,7 if it possible prepare download link . UserIdTAG: 153075 UserNameTAG: rezaei CreateTimeTAG: 2012-10-15T08:19:27Z VoteTAG: 0 CoursewareTAG: Week 5 / Dependent Sources Review CommentableIdTAG: 6002x_dep_src_rvw NumberOfReplyTAG: 1 FirstChildTAG: [JWplayer Video Link week(1-6)][1] [1]: https://dl.dropbox.com/u/24096724/6002xMP4/menu.html# Regards: asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-10-18T15:21:12Z IndexTAG: 3442 TitleTAG: H5P3.1 where are these equations wrong.. plzz help.. vin*(K*RS-(K*RS)/sqrt(2*K*RS*VIN-2*K*RS*VT+1))/(K*RS^2) OR vin*((sqrt((2*K*RS*VIN-2*K*RS*VT+1)))-1)/(RS*sqrt(2*K*RS*(VIN-VT)+1)) UserIdTAG: 431224 UserNameTAG: erachopra10 CreateTimeTAG: 2012-10-15T06:49:54Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: 1. Use the $i_{DS}$ equation from H5P2, but multiply by $R_S$ to get $V_{OUT}$ 2. Plug it into [www.derivative-calculator.net][1] making sure $R_S$=R, $V_{IN}$=V, etc. 3. Multiply the derivative by $v_{in}$. It seems you correctly did Step 3, but the equation is off. Try cancelling and combining terms. A hint: $(1 - \frac{1}{ \sqrt (...................)}) \cdot v_{in}$ [1]: http://www.derivative-calculator.net/ FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-10-15T07:05:43Z SecondChildTAG: vin*(1-1/sqrt(1+2*K*RS*(VIN-VT))) SecondChildUserIdTAG: 244517 SecondChildUserNameTAG: JoseRodrigues SecondChildCreateTimeTAG: 2012-10-15T10:42:46Z IndexTAG: 3443 TitleTAG: H5P2: Setting iDS = 0 I have my quadratic equation, but I can't seem to get any further. When I make sure that iDS=0 when vIN=VT, I have been setting all (vIN-VT) to 0 throughout the whole quadratic equation. This doesn't seem to be working. Can anybody tell me what I'm doing wrong? UserIdTAG: 132828 UserNameTAG: rharris CreateTimeTAG: 2012-10-15T06:19:04Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3444 TitleTAG: week 5 not showing answer week 5 hw is not showing answer after deadline UserIdTAG: 256221 UserNameTAG: himanshu123 CreateTimeTAG: 2012-10-15T06:15:34Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: It's still not the deadline yet...almost. It's 2:48 AM EST (Boston time), but Homework is still being accepted. It is almost midnight in California, but still 8:50PM in Hawaii, USA in the Pacific Ocean; they are close to the International Date Line, which this course goes by (i.e. as long as it's before Monday **anywhere** in the world!) FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-10-15T06:53:02Z SecondChildTAG: You wouldn't happen to have an answer to this one, do you? I have my quadratic equation, but I can't seem to get any further. When I make sure that iDS=0 when vIN=VT, I have been setting all (vIN-VT) to 0 throughout the whole quadratic equation. This doesn't seem to be working. Can anybody tell me what I'm doing wrong? SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-15T07:06:28Z IndexTAG: 3445 TitleTAG: help me with H5P3 I need a help with H5P3 UserIdTAG: 37708 UserNameTAG: Jband CreateTimeTAG: 2012-10-14T23:14:33Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: [Edited by staff -- please do not post answers to a graded part of the course] FirstChildUserIdTAG: 181081 FirstChildUserNameTAG: sant250784 FirstChildCreateTimeTAG: 2012-10-15T03:07:31Z SecondChildTAG: Thnks sant250784, but i have a problem with VIN SecondChildUserIdTAG: 37708 SecondChildUserNameTAG: Jband SecondChildCreateTimeTAG: 2012-10-15T03:34:42Z IndexTAG: 3446 TitleTAG: Lab 5 Post question Hey I finished lab 5, and was wondering if the largest input amplitude is related to the linear voltage operating range by (V(High)-V(Low))/2 It looks like that's the case, but want to make sure! UserIdTAG: 381923 UserNameTAG: matteaton CreateTimeTAG: 2012-10-14T20:03:13Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Hi matteaton Please, can you help me with the lab? I'm trying to find the Input voltage at lower end of linear operating range but every value gives me error. thanks FirstChildUserIdTAG: 110802 FirstChildUserNameTAG: leoblack FirstChildCreateTimeTAG: 2012-10-14T20:06:09Z SecondChildTAG: matteaton, can u please help me out in getting the value for V(High) ?? SecondChildUserIdTAG: 260272 SecondChildUserNameTAG: saikat24 SecondChildCreateTimeTAG: 2012-10-14T20:30:50Z FirstChildTAG: Hi matteaton. Please, can you help me with the lab? I'm trying to find the Input voltage at lower end of linear operating range but every value gives me error. I need a hint. thnaks. FirstChildUserIdTAG: 110802 FirstChildUserNameTAG: leoblack FirstChildCreateTimeTAG: 2012-10-14T20:09:00Z SecondChildTAG: Please click on TRAN in Figure 1 to generate the plot and then use the plot to determine the approximate range of input voltages during the amplifier will operate in its linear region. The lower input is when vout start to decrease SecondChildUserIdTAG: 95638 SecondChildUserNameTAG: cperezf SecondChildCreateTimeTAG: 2012-10-14T22:47:52Z FirstChildTAG: In general, the maximum output from an amplifier is limited by the supply voltage. You can't get more out than is available from the power supply, unless you have discovered how to create energy..... In this case, there is no output transformer so the max Vs is a limiting factor. If you have an output transformer, then you have to think in terms of max Power, not voltage.... FirstChildUserIdTAG: 375500 FirstChildUserNameTAG: Brej7665 FirstChildCreateTimeTAG: 2012-10-15T02:36:09Z IndexTAG: 3447 TitleTAG: H5P2 could someone help me solve the last part of H5P2, the Vdd? UserIdTAG: 214085 UserNameTAG: shohin CreateTimeTAG: 2012-10-14T19:36:20Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: check this out https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_large_sig/threads/5079c393be483a230000000e FirstChildUserIdTAG: 110802 FirstChildUserNameTAG: leoblack FirstChildCreateTimeTAG: 2012-10-14T20:03:29Z SecondChildTAG: I need help with the solution for iDS SecondChildUserIdTAG: 37708 SecondChildUserNameTAG: Jband SecondChildCreateTimeTAG: 2012-10-14T23:17:29Z IndexTAG: 3448 TitleTAG: How did you choose... How did you choose negative or positive value of the sqrt? And, what this means? UserIdTAG: 155605 UserNameTAG: JaimeLopezCerezo CreateTimeTAG: 2012-10-14T19:29:51Z VoteTAG: 0 CoursewareTAG: Week 4 / Dependent Sources Example 1 CommentableIdTAG: 6002x_dep_src_ex_1 NumberOfReplyTAG: 0 IndexTAG: 3449 TitleTAG: please help / point out where m i going wrong.. Lab 5 to get the input voltage upper end of linear operating range, m i not supposed to use the expression, VT+(-1+sqrt(1+2*k*RL*VS))/(k*RL) ?? with the values ,k=100mA/V^2 RL=10kohm VS=5v & VT which is calculated/ observed from the 1st ques, equal to .487 help can any one point me out where m i conceptually getting diverted?? UserIdTAG: 260272 UserNameTAG: saikat24 CreateTimeTAG: 2012-10-14T19:27:06Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hey saikat24 are you doing lab5 right? the answer is obtained by graphical analysis of the graph vO vs. vI FirstChildUserIdTAG: 110802 FirstChildUserNameTAG: leoblack FirstChildCreateTimeTAG: 2012-10-14T19:52:23Z SecondChildTAG: yup... thnx leoblack for the info. graphical analysis is fine , but, to get the precise value should i not calculate it? or merely inferring from the graph will do? SecondChildUserIdTAG: 260272 SecondChildUserNameTAG: saikat24 SecondChildCreateTimeTAG: 2012-10-14T20:10:28Z SecondChildTAG: Saikat, No precise value is necessary...infer the answer from the graph SecondChildUserIdTAG: 308861 SecondChildUserNameTAG: Ignaas SecondChildCreateTimeTAG: 2012-10-15T00:51:27Z FirstChildTAG: @saikat24...I don't see the K=100mA/V^2 as part of the Lab 5 information. I do see a W/L ratio of 100 and that will affect the value of K. Maybe that's skewing your numbers? Just a thought. FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-10-15T01:35:58Z IndexTAG: 3450 TitleTAG: common collector can anybody explain me the working of common collector amplifier? UserIdTAG: 633981 UserNameTAG: Areeb CreateTimeTAG: 2012-10-14T18:56:41Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: common collector is an emitter follower - it doesn't amplify voltage amplitude but amplifies power by lowering output impedance => higher load current => higher output power http://en.wikipedia.org/wiki/Common_collector FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-14T22:21:28Z IndexTAG: 3451 TitleTAG: need help for H5P3 hey can anyone tell me how to solve for vout ?? got stuck here for an hour .. UserIdTAG: 120850 UserNameTAG: Nitin1A CreateTimeTAG: 2012-10-14T18:48:43Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: you need use the hint and substitute the values. take the vout from the 2 nd problem and differentiate it with respect to vi as suggested in the hint. the remainder problems are just substitution of values into the resultant equation. FirstChildUserIdTAG: 273912 FirstChildUserNameTAG: raj2691 FirstChildCreateTimeTAG: 2012-10-14T19:22:08Z SecondChildTAG: still i am unable to get it right ,i got vout as (1+(2/(K*RS*sqrt(1+2*K*RS*(VIN-VT)))))*vin) can you tell me where i have gone wrong!! SecondChildUserIdTAG: 120850 SecondChildUserNameTAG: Nitin1A SecondChildCreateTimeTAG: 2012-10-14T20:01:19Z SecondChildTAG: It's vo = D(VO)*vi, where D is the derivative. You didn't differentiate VO. I suggest using Wolfram to do it quickly. SecondChildUserIdTAG: 194509 SecondChildUserNameTAG: blackcompe SecondChildCreateTimeTAG: 2012-10-14T20:20:59Z IndexTAG: 3452 TitleTAG: Electrical meaning Therefore, the slope is the inverse of a linear resistance. UserIdTAG: 155605 UserNameTAG: JaimeLopezCerezo CreateTimeTAG: 2012-10-14T18:41:14Z VoteTAG: 0 CoursewareTAG: Week 4 / Incremental Method Math CommentableIdTAG: 6002x_inc_method_math NumberOfReplyTAG: 0 IndexTAG: 3453 TitleTAG: H5P1 question 1 Here is my thought process: * For the circuit to operate in the saturation region $V_{GS}$ > $V_T$ and $V_{DS}$ > $V_{GS}$ - $V_T$ must be true. * $V_{GS}$ = $V_{IN}$ - $V_S$ and $V_{DS}$ = $V_o$ - $V_S$ * Since I'm looking for the minimum value of $V_{IN}$ I make the ineqalities into equasions and solve for $V_{IN}$. The problem is that since $V_{DS}$ is expressed in terms of $V_o$, I'd have to find the value of $V_o$ first, which depends on RL a number that I'm not given. Obviously there is a problem in my way of thinking, but I can't seem to figure out another method. Any ideas? Thanks, zoliking UserIdTAG: 230226 UserNameTAG: zoliking CreateTimeTAG: 2012-10-14T18:29:09Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I suggest the p.345 of the textbook. FirstChildUserIdTAG: 469757 FirstChildUserNameTAG: Kanta FirstChildCreateTimeTAG: 2012-10-14T21:17:34Z FirstChildTAG: I have the same problem zoliking. Someone please help as early as possible. FirstChildUserIdTAG: 514878 FirstChildUserNameTAG: Gaurav006 FirstChildCreateTimeTAG: 2012-10-14T18:48:35Z SecondChildTAG: You can take a look at this post [here][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_mosfet_amp/threads/507a1d6dd894f12300000039 SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T21:15:09Z IndexTAG: 3454 TitleTAG: Tutorials w7 - A&L Pb9.2 - whats up with vt is it 1v or 0.3v? If in problem its stated that v(t)=0.3v u(t), then why in the solution it appears as 1.0v u(t)? Anybody can pls enlighten me on this? UserIdTAG: 331483 UserNameTAG: Doru CreateTimeTAG: 2012-10-14T18:24:08Z VoteTAG: 0 CoursewareTAG: Week 7 / Unknown Inductance CommentableIdTAG: 6002x_Unknown_Inductance NumberOfReplyTAG: 0 IndexTAG: 3455 TitleTAG: h5 p2 how to write the algebraic expression ,please can any one help me ? it always tells me invalid inputs. i don't know how to write iDS correctly ! or any other parameter UserIdTAG: 215974 UserNameTAG: Esmail CreateTimeTAG: 2012-10-14T18:20:42Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: The interpreter (the script that runs when you hit "CHECK") is very picky. If you have mismatched parentheses, it will flag that as an invalid input that cannot be parsed, **even though it will appear in the equation box and appear largely correct**. I wasted a couple of hours on this myself yesterday. One way to check is to replace your constants with letters (a for VT, b for RS, etc) and input the equation in Wolfram Alpha. It will return an error "An attempt was made to fix mismatched delimiters" if you are missing a parentheses. This will also allow you to solve the equation - your still have to go back in and re-substitute your variables (VT for a, RS for b etc.) Also, keep in mind that for these problems, the case of each letter has meaning. There is a difference to the interpreter between vIN and VIN, for instance. If you use a variable that is not in the solution it will not accept your answer - it will then give you an answer like "Invalid input: VDD not permitted in answer" where the variable VDD is not part of the answer for instance, even if maybe vDD is part of the answer (note lower case "v"). Additionally, I find that the interpreter does not seem to play well with my version of Internet Explorer. Google Chrome seems to work pretty well though. Also, may be better to write the thing in a word processor and paste it - I also lost time yesterday due to the plug in crashing. Hope this helps FirstChildUserIdTAG: 342181 FirstChildUserNameTAG: JUSTIN_W FirstChildCreateTimeTAG: 2012-10-14T19:03:10Z IndexTAG: 3456 TitleTAG: Transcript Access in the transcript should read "excess", I think UserIdTAG: 266912 UserNameTAG: pietvo CreateTimeTAG: 2012-10-14T17:23:23Z VoteTAG: 0 CoursewareTAG: Week 6 / Capacitor basics CommentableIdTAG: 6002x_capacitor_basics NumberOfReplyTAG: 0 IndexTAG: 3457 TitleTAG: WEEK 5----- WEEK 6 --- course videos download link --- please upload I request the administration to please add the liks to download WEEK 5 and WEEK 6 videos of the lectures... as mid-term exam is so near.... Thank you in advance.... :) UserIdTAG: 209930 UserNameTAG: saikiraniitr CreateTimeTAG: 2012-10-14T16:54:33Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: [Week 5 Video Playlist][1] [Week 6 Video Playlist][2] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/LectureDownloadWeek5/ [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/LectureDownloadWeek6/ FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-10-15T10:30:37Z SecondChildTAG: how to download videos, as i would have sporadic access to internet.It will help me if i can download videos and go through them whenever i want. Thanks. SecondChildUserIdTAG: 364884 SecondChildUserNameTAG: Ambli SecondChildCreateTimeTAG: 2012-10-15T15:43:18Z IndexTAG: 3458 TitleTAG: PROBLEM IN H5P3 WHAT IS THE EXPRESSION OF IDS?? CAN SOMEONE EXPLAIN ME THE WAY???? UserIdTAG: 296945 UserNameTAG: saket30 CreateTimeTAG: 2012-10-14T15:57:02Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: i have the same problem have you got expression for H5P2 first one FirstChildUserIdTAG: 342135 FirstChildUserNameTAG: vikash902 FirstChildCreateTimeTAG: 2012-10-14T17:22:20Z IndexTAG: 3459 TitleTAG: About the iDS express formula. Use iDS=(K*(vIN-vT- iDS*Rs)^2)/2, get the iDS formula. But it includes the RS parameter. RS is not permitted as requirement. How to cancel RS from iDS formula? UserIdTAG: 371617 UserNameTAG: rainbow12 CreateTimeTAG: 2012-10-14T13:18:06Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Write RS in caps FirstChildUserIdTAG: 429149 FirstChildUserNameTAG: divyakavi11 FirstChildCreateTimeTAG: 2012-10-14T14:12:16Z SecondChildTAG: I also have the same problem. how can I cancel RS from the formula. I have written the resistence as RS. SecondChildUserIdTAG: 279303 SecondChildUserNameTAG: Raja91 SecondChildCreateTimeTAG: 2012-10-14T20:39:19Z IndexTAG: 3460 TitleTAG: H5P3 Ιn the first question asked the algebraic expression of the Vout, the answer I give was not accepted. Using the above formula for the rest of my questions the answers are received. UserIdTAG: 345779 UserNameTAG: tpsonis CreateTimeTAG: 2012-10-14T12:09:15Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: vin*(1-2/(K*RS*sqrt((2/(K*RS)*(VIN-VT)+1/(K^2*RS^2))))) is this ans is right? FirstChildUserIdTAG: 328998 FirstChildUserNameTAG: vipinmsp FirstChildCreateTimeTAG: 2012-10-14T14:47:09Z FirstChildTAG: You might not be putting * sign whenever required.for example 2K must be written as 2*K,vin(......) should be written as vin*(....). FirstChildUserIdTAG: 99628 FirstChildUserNameTAG: Pranjal16 FirstChildCreateTimeTAG: 2012-10-14T12:11:35Z SecondChildTAG: I've written it, as you say but continue to consider it my mistake vin*(1-1/(K*RS*sqrt((2/(K*RS)*(VIN-VT)+1/(K*RS^2))))) SecondChildUserIdTAG: 345779 SecondChildUserNameTAG: tpsonis SecondChildCreateTimeTAG: 2012-10-14T12:34:54Z SecondChildTAG: please use dimensional analysis to correct your expression... SecondChildUserIdTAG: 316353 SecondChildUserNameTAG: Vivekanand123 SecondChildCreateTimeTAG: 2012-10-14T15:11:05Z IndexTAG: 3461 TitleTAG: H5P2 SOURCE FOLLOWER LARGE SIGNAL i am getting a long answer for iDS and it satisfy the condition when vIN=VT then iDS=0. but still getting "red cross". please help!!! UserIdTAG: 114864 UserNameTAG: Vikaash CreateTimeTAG: 2012-10-14T11:37:48Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: [Edited by staff -- please do not post answers to a graded part of the course] FirstChildUserIdTAG: 429149 FirstChildUserNameTAG: divyakavi11 FirstChildCreateTimeTAG: 2012-10-14T15:49:12Z SecondChildTAG: is it correct ?? SecondChildUserIdTAG: 469547 SecondChildUserNameTAG: Vedantastro SecondChildCreateTimeTAG: 2012-10-14T18:24:55Z FirstChildTAG: have you solved lab 5 last problem .. I need hint.. FirstChildUserIdTAG: 227508 FirstChildUserNameTAG: bhavyab FirstChildCreateTimeTAG: 2012-10-14T11:59:46Z FirstChildTAG: I think you have got a quadratic equation in terms of Vin VT K Vout etc. You just have to make sure that you are writing in the form as RS,vIN,1/K,VT... you will get the green tick.. FirstChildUserIdTAG: 227508 FirstChildUserNameTAG: bhavyab FirstChildCreateTimeTAG: 2012-10-14T11:57:36Z SecondChildTAG: How to get iDS formula with vIN,vOUT, K.VT, but does not RS. I can not to do. How to cancel the RS? SecondChildUserIdTAG: 371617 SecondChildUserNameTAG: rainbow12 SecondChildCreateTimeTAG: 2012-10-14T13:00:25Z SecondChildTAG: my is in terms of RS,vIN, k,VT.......infact i am getting the correct answer for vOUT for different values given in the question.. SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-10-14T13:12:02Z SecondChildTAG: then check that r u applying * sign in every multipcation .this worked for me SecondChildUserIdTAG: 611666 SecondChildUserNameTAG: Shaminder420 SecondChildCreateTimeTAG: 2012-10-14T13:22:55Z SecondChildTAG: yes SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-10-14T13:30:00Z SecondChildTAG: but still getting wrong. SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-10-14T13:30:22Z SecondChildTAG: Can anyone tell me the correct expression for this question... SecondChildUserIdTAG: 429149 SecondChildUserNameTAG: divyakavi11 SecondChildCreateTimeTAG: 2012-10-14T14:05:27Z SecondChildTAG: Got the answer....:-) SecondChildUserIdTAG: 429149 SecondChildUserNameTAG: divyakavi11 SecondChildCreateTimeTAG: 2012-10-14T14:11:50Z SecondChildTAG: Can anyone tell me the correct expression for this question SecondChildUserIdTAG: 429168 SecondChildUserNameTAG: arunprakashavm SecondChildCreateTimeTAG: 2012-10-14T19:18:01Z IndexTAG: 3462 TitleTAG: lab 5 anybody may help me for lab 5th last problem ," Largest input amplitude" UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-14T11:02:43Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3463 TitleTAG: charging of a capacitor why is their a straight line between v and i in graph UserIdTAG: 168496 UserNameTAG: poweltalwar CreateTimeTAG: 2012-10-14T10:49:02Z VoteTAG: 0 CoursewareTAG: Week 6 / Ideal linear capacitor CommentableIdTAG: 6002x_ideal_linear_capacitor NumberOfReplyTAG: 0 IndexTAG: 3464 TitleTAG: LAB 5 i've got the the lower end input voltage,but it is showing wrong for every possible value i've entered for higher end,,m i missing something..plz help UserIdTAG: 219442 UserNameTAG: aki31 CreateTimeTAG: 2012-10-14T10:35:48Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: X axis will give the values of input voltage... look at the point where the curve is changing....... FirstChildUserIdTAG: 227508 FirstChildUserNameTAG: bhavyab FirstChildCreateTimeTAG: 2012-10-14T11:00:07Z SecondChildTAG: aki31 How did you get the lower end input voltage? thanks SecondChildUserIdTAG: 110802 SecondChildUserNameTAG: leoblack SecondChildCreateTimeTAG: 2012-10-14T19:31:47Z FirstChildTAG: aki31 How did you get the lower end input voltage? thanks FirstChildUserIdTAG: 110802 FirstChildUserNameTAG: leoblack FirstChildCreateTimeTAG: 2012-10-14T19:32:04Z IndexTAG: 3465 TitleTAG: H5P1 third sub question i have tried all formulas for calculating maximum vin,, quadratic in which vin=-1+sqrt(1+2vs*Rl*K)/Rl*K +VT i use vs=2 vt=0.7 R=8000 and k =1mA/v^2 but not getting right answer please guide me thanks UserIdTAG: 438395 UserNameTAG: ali_PU1 CreateTimeTAG: 2012-10-14T07:19:52Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: which condition applied for H5P1 third sub question FirstChildUserIdTAG: 526734 FirstChildUserNameTAG: darkknight90 FirstChildCreateTimeTAG: 2012-10-14T08:47:35Z SecondChildTAG: try 0.093 SecondChildUserIdTAG: 429149 SecondChildUserNameTAG: divyakavi11 SecondChildCreateTimeTAG: 2012-10-14T13:42:09Z SecondChildTAG: thanks SecondChildUserIdTAG: 403660 SecondChildUserNameTAG: abuodeh SecondChildCreateTimeTAG: 2012-10-14T18:53:48Z SecondChildTAG: how to find RL?? SecondChildUserIdTAG: 429168 SecondChildUserNameTAG: arunprakashavm SecondChildCreateTimeTAG: 2012-10-14T19:16:42Z FirstChildTAG: VT = 0.5 v FirstChildUserIdTAG: 217094 FirstChildUserNameTAG: karanpanchal FirstChildCreateTimeTAG: 2012-10-14T08:17:26Z FirstChildTAG: use formula vt+(-1+sqrt(1+2vs.R.K))/(Rl.k) and get the answer. after that subtract vt from answer to get the final answer FirstChildUserIdTAG: 611666 FirstChildUserNameTAG: Shaminder420 FirstChildCreateTimeTAG: 2012-10-14T09:06:57Z SecondChildTAG: i use this formula put vt=0.5 vs=2 , R=8000 ,K=1mA but still no green check :( if i need to subtract vt then i can subtract vt from this equation before putting values isnt it.? SecondChildUserIdTAG: 438395 SecondChildUserNameTAG: ali_PU1 SecondChildCreateTimeTAG: 2012-10-14T11:56:36Z SecondChildTAG: You have to subtract Vs not Vt. SecondChildUserIdTAG: 99628 SecondChildUserNameTAG: Pranjal16 SecondChildCreateTimeTAG: 2012-10-14T12:16:22Z SecondChildTAG: srry my mistake bro subtract vs =1 v frm ans SecondChildUserIdTAG: 611666 SecondChildUserNameTAG: Shaminder420 SecondChildCreateTimeTAG: 2012-10-14T12:37:18Z SecondChildTAG: How did you get the formula? What's the logic behind $\frac{-1+\sqrt{(1+2VsR_LK)}}{R_LK}$ It looks like a solution to the quadratic equation $0.5R_LKV_i^2 + V_i - V_S$, but how did you get this equation? SecondChildUserIdTAG: 230226 SecondChildUserNameTAG: zoliking SecondChildCreateTimeTAG: 2012-10-14T19:52:54Z IndexTAG: 3466 TitleTAG: H5P2 part 3 hello everyone...can someone please help me with H5P2 part 3, where we have to find ids...i have completed rest of homework 5 +lab5...just stuck with this part...im getting such a long equation that its giving me an error "Invalid input: Could not parse xxxxxxx as a formula :(...im tired of this part now :( UserIdTAG: 278792 UserNameTAG: sali CreateTimeTAG: 2012-10-14T07:17:26Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: use the right parantheses and be sure u enter the simplified result FirstChildUserIdTAG: 611666 FirstChildUserNameTAG: Shaminder420 FirstChildCreateTimeTAG: 2012-10-14T07:18:54Z SecondChildTAG: thanks Shaminder420 :) ummm does anyone know how to simplify your result in matlab? if that even exists :S SecondChildUserIdTAG: 278792 SecondChildUserNameTAG: sali SecondChildCreateTimeTAG: 2012-10-14T07:55:05Z FirstChildTAG: sali, i think that with xxxxxxx u mean vout , right ? in matlab ,i got vout = ((49/10-((vin-Vt-vout)^2)^(1/2)-vin-Vt)^2)^(1/2)+89/10 and i get the final expression vout = 1/2*vout+49/20 (with vIN=VT) . so, it still have the vout on the same sides.(so cannot substitute vout on the ids equation) i was thinking in make a simple .m file with a short numerical method to solve the equation(maybe its your idea too) but i guess that its not the way to do it. it's a valid approach but i guess the lesson is trying to teach us other way to do it. i will take a look on the book , recommend the same to u , if u figure it out give us a hint, i ll do t same too. FirstChildUserIdTAG: 402850 FirstChildUserNameTAG: y1111 FirstChildCreateTimeTAG: 2012-10-14T08:30:27Z SecondChildTAG: y1111 yupp sure i'll do :) no actually xxxxxx is root of ids that i got from solving the quadratic equation in matlab but is huge !!!! i put alot of effort in making it compact but no use ....n yesss ur right they want us to solve it so i guess i'll put some more time in it :( SecondChildUserIdTAG: 278792 SecondChildUserNameTAG: sali SecondChildCreateTimeTAG: 2012-10-14T15:29:45Z SecondChildTAG: i already solded it , i feel like a dumb because i did not sleep trying to solve , i only sleep i bit on the mornig and i guess i had dream about trying to solve it thinking that i'm awake. well... some ours ago i realize that the thing is not make a equation isolating the vout. is easily noticeable that with tha the **vouts whe got a quadratic equation**. so the thing is to **get the roots of this equation(root of vout) and substitute each root on the ids equation** . now u make vIN=VT and substitute it on each one of the 2 ids eq that u have one of them will result 0 (zero) so is this one that u use , and is the root that u used in it that you assume as the vout equation. SecondChildUserIdTAG: 402850 SecondChildUserNameTAG: y1111 SecondChildCreateTimeTAG: 2012-10-14T17:19:20Z IndexTAG: 3467 TitleTAG: h5p3 help please i have solved all the answers with my equation for ids and they are correct and i got a green check but still getting wrong in ids checkbox .please help UserIdTAG: 611666 UserNameTAG: Shaminder420 CreateTimeTAG: 2012-10-14T06:36:35Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Try to. Enter it in quadratic formula form without taking out anything common. FirstChildUserIdTAG: 99628 FirstChildUserNameTAG: Pranjal16 FirstChildCreateTimeTAG: 2012-10-14T06:41:07Z SecondChildTAG: did so but still gettng could not parse it as formula SecondChildUserIdTAG: 611666 SecondChildUserNameTAG: Shaminder420 SecondChildCreateTimeTAG: 2012-10-14T06:48:55Z SecondChildTAG: at last got solved thanks for suggestion SecondChildUserIdTAG: 611666 SecondChildUserNameTAG: Shaminder420 SecondChildCreateTimeTAG: 2012-10-14T06:55:36Z SecondChildTAG: how you solve.? i waste hours on it but no solution :( SecondChildUserIdTAG: 438395 SecondChildUserNameTAG: ali_PU1 SecondChildCreateTimeTAG: 2012-10-14T06:59:28Z SecondChildTAG: use the two equations to get the quadratic equqtion and then select the right root and answer is long but is easy SecondChildUserIdTAG: 611666 SecondChildUserNameTAG: Shaminder420 SecondChildCreateTimeTAG: 2012-10-14T07:10:56Z IndexTAG: 3468 TitleTAG: H4P2 Quadratic Eq solution has a bug! The part to solve for iDS and find the correct root, I solved it, I got the perfect answer but somehow the site says its wrong! I was checking the discussions and found someone using the same formula i got to solve for the next part, I just casually copy pasted the formula and voila! Its correct! It was the same exact answer that i got! For sure! Edx people, please rectify this problem. Many people are wasting their time solving for an answer they already have! UserIdTAG: 108454 UserNameTAG: Raven7281 CreateTimeTAG: 2012-10-14T06:34:48Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3469 TitleTAG: Lab 6 #3 Equation 10.66 vc(t) = ... what is V(c) equal to - a hint for how to determine Vc(t) the voltage across the capacitor. I have tried using 2.4 - finding the voltage using the voltage divider and setting it to zero thanks! I really enjoy the labs. Really hate all the quadratic formulas in the HW :> UserIdTAG: 18073 UserNameTAG: gburkhart CreateTimeTAG: 2012-10-14T04:43:02Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Not sure what your question is. vc(t) is the voltage across a capacitor as a function of time when driven by a Thevinin equivalent circuit. It can be inverted to find the time that it takes to reach a specified voltage. The parameters for the Thevinin circuit were computed in the previous parts of the lab. P.S. Quadratic equations aren't a problem; typing **any formula** into a text parser is a pain!!! FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-14T13:35:54Z FirstChildTAG: I figured it out. I did the next question first and then guessed the tau that I should have calculated first using the hints in the problem. then I back calculated the Vc and at that point I had a "duh" moment and knew how to get the answer FirstChildUserIdTAG: 18073 FirstChildUserNameTAG: gburkhart FirstChildCreateTimeTAG: 2012-10-15T01:58:27Z IndexTAG: 3470 TitleTAG: H7P3 THE CURSE OF LEAD INDUCTANCE, what's the diff. Ec. To solve? We know that the carge of a self is: I(t) = Io-Io*e^(-t*R/L) And V(0) = 0, with the given parameters the ec. Looks like this: 0.9/50=1.8/77-1.8/77*e^(-77*x/(15.44*10^(-9)) But it's wrong! Any idea? UserIdTAG: 299251 UserNameTAG: TheRedBlackOne CreateTimeTAG: 2012-10-14T00:53:52Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3471 TitleTAG: Lab 5 question 1 Hey guys, I have a problem...I can't get the right answer of the first question of lab 5...I have the second question right, but somehow my cursor doesn't give me the right value....can i please get a hind in what range the right answer lies...i have tried so many values... UserIdTAG: 308861 UserNameTAG: Ignaas CreateTimeTAG: 2012-10-13T22:39:06Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hint: the values over some axis are given in MILIvolts! FirstChildUserIdTAG: 326787 FirstChildUserNameTAG: Angstrem FirstChildCreateTimeTAG: 2012-10-13T22:41:56Z SecondChildTAG: thnx man...i was clearly looking at the wrong axis SecondChildUserIdTAG: 308861 SecondChildUserNameTAG: Ignaas SecondChildCreateTimeTAG: 2012-10-14T00:33:00Z IndexTAG: 3472 TitleTAG: H2P5 part 3 help please i have this expresiion which should fulfil the requirments (1+(K*RS*(vIN-VT))-sqrt(1+2*K*RS*(vIN-VT)^2))/(K*RS^2) what did i do wrong , i used the 2 equations in parts 1 , 2 , whic i got green tick for them , means they are absolutely right , UserIdTAG: 285675 UserNameTAG: Ascot CreateTimeTAG: 2012-10-13T21:50:45Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: u have done it exactly right except that u have wrongly sqaured this - 2*K*RS*(vIN-VT)^2) It would be sqrt(1+2*K*RS*(vIN-VT)) :D FirstChildUserIdTAG: 214085 FirstChildUserNameTAG: shohin FirstChildCreateTimeTAG: 2012-10-14T08:38:13Z IndexTAG: 3473 TitleTAG: Upload Week 5 Videos for Download..???? we are unable to download week 5 videos plz upload week 5 videos thanks! UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-13T21:25:08Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Troubleshooting NumberOfReplyTAG: 1 FirstChildTAG: [Week 5 Lecture Playlist][1] Above Link for those who can't view lecture due to 'youtube issue'. Regards: asadbhatti42 [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/LectureDownloadWeek5/ FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-10-14T11:55:13Z IndexTAG: 3474 TitleTAG: second question which method is used to solve this vr UserIdTAG: 550401 UserNameTAG: revathisingh CreateTimeTAG: 2012-10-13T19:05:20Z VoteTAG: 0 CoursewareTAG: Week 5 / Two Terminal Connection Exercise CommentableIdTAG: 6002x_2_term_con_e NumberOfReplyTAG: 0 IndexTAG: 3475 TitleTAG: Help with LAB 5 How do I go about finding the formula that governs the "Largest input amplitude resulting in an undistorted output signal"? UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-10-13T18:53:41Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: spend some time to watch the first videos of 6th week... it clears out! FirstChildUserIdTAG: 301962 FirstChildUserNameTAG: labrinim FirstChildCreateTimeTAG: 2012-10-13T19:28:07Z SecondChildTAG: I hope so! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-13T19:32:31Z SecondChildTAG: "To obtain maximum useful input signal range, we might choose the input bias voltage VIN to be at the center of the valid range of input voltages for the amplifier, as illustrated in Figure 7.42.", from page 366. SecondChildUserIdTAG: 9100 SecondChildUserNameTAG: dmgongora SecondChildCreateTimeTAG: 2012-10-14T00:07:23Z SecondChildTAG: how to choose approximate Vsignal amplitude? SecondChildUserIdTAG: 182653 SecondChildUserNameTAG: monkeyfoahead SecondChildCreateTimeTAG: 2012-10-14T00:56:27Z FirstChildTAG: Recall what the definition of amplitude is. Myself i found a review of H1 -the AC signals- helped. Then use the figures you got from the graphs for the first coupla questions. Remember too the Shrink part from the lectures and demos last week: we'd want to keep the whole input signals amplitude, peak to peak (what is the peak amplitude for an AC sig?) within the saturation region. Good luck, FirstChildUserIdTAG: 184153 FirstChildUserNameTAG: tthngn FirstChildCreateTimeTAG: 2012-10-13T19:54:51Z IndexTAG: 3476 TitleTAG: Lab 5 Can anyone plz help me solve the problem asking gain in figure 2??actually i am calcutating the highest peak value of Vout and dividing it by the highest peak value of Vin....is it the right way??if not plz can u guys help me out? UserIdTAG: 128409 UserNameTAG: arijitbme CreateTimeTAG: 2012-10-13T16:51:33Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You have to divide the differences between highest point and lowest points of both sinusoids. FirstChildUserIdTAG: 352142 FirstChildUserNameTAG: DanteX FirstChildCreateTimeTAG: 2012-10-13T17:22:27Z SecondChildTAG: hey, i have got the answer but can you tell me why we are not taking **-ve sign** even after we are getting the phase change of 180 degree between signal of vo and vi? SecondChildUserIdTAG: 242676 SecondChildUserNameTAG: LinjharaSahil SecondChildCreateTimeTAG: 2012-10-13T20:02:53Z SecondChildTAG: will the output pk 2 pk come in numerator?? and vice versa?? SecondChildUserIdTAG: 127800 SecondChildUserNameTAG: MNB SecondChildCreateTimeTAG: 2012-10-14T19:25:21Z IndexTAG: 3477 TitleTAG: lecture video I still can not get the lecture video course - because youtube is banned here, please help UserIdTAG: 153317 UserNameTAG: hdjt CreateTimeTAG: 2012-10-13T16:07:23Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: This was a major issue for me too.. FirstChildUserIdTAG: 362299 FirstChildUserNameTAG: anandbaskaran FirstChildCreateTimeTAG: 2012-10-13T16:25:16Z SecondChildTAG: i hve also this problm SecondChildUserIdTAG: 228871 SecondChildUserNameTAG: ELINAKHAN SecondChildCreateTimeTAG: 2012-10-13T17:23:13Z FirstChildTAG: [6.002x Lecture Direct Videos Link][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/ Above Link for those who can't view lecture due to 'youtube issue'. Regards: asadbhatti42 FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-10-13T17:18:28Z SecondChildTAG: this link doesn't have week 5 and onward videos, please provide link for week 5 videos ASAP because tomorrow is last date of week 5 assignment & Lab. SecondChildUserIdTAG: 426051 SecondChildUserNameTAG: sslohana SecondChildCreateTimeTAG: 2012-10-13T19:59:07Z SecondChildTAG: [JWplayer Video Link(1-6)][1] [1]: https://dl.dropbox.com/u/24096724/6002xMP4/menu.html# Regards: asadbhatti42 SecondChildUserIdTAG: 329051 SecondChildUserNameTAG: asadbhatti42 SecondChildCreateTimeTAG: 2012-10-18T15:19:45Z SecondChildTAG: WK7?? SecondChildUserIdTAG: 127800 SecondChildUserNameTAG: MNB SecondChildCreateTimeTAG: 2012-10-28T11:32:05Z IndexTAG: 3478 TitleTAG: lab 5 Hey all, Is sequence 10 important for the work in lab 5? UserIdTAG: 308861 UserNameTAG: Ignaas CreateTimeTAG: 2012-10-13T15:17:49Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: No. Its not necessary but its better if u finish seeing Sequence 10 too. Because it will help in solving Homework. FirstChildUserIdTAG: 362299 FirstChildUserNameTAG: anandbaskaran FirstChildCreateTimeTAG: 2012-10-13T15:58:54Z FirstChildTAG: ok thank you...i'be just finished sequence 9...so i can start wit lab 5? FirstChildUserIdTAG: 308861 FirstChildUserNameTAG: Ignaas FirstChildCreateTimeTAG: 2012-10-13T16:07:00Z SecondChildTAG: Better finish the next sequence too. So that you can spend ample amount of time in solving HW and LAB. SecondChildUserIdTAG: 362299 SecondChildUserNameTAG: anandbaskaran SecondChildCreateTimeTAG: 2012-10-13T16:28:12Z IndexTAG: 3479 TitleTAG: hsp1 in hsp1 q1 the min vin required to work tx in saturation region is VT na ..but it is showing red x mrk plz help if m wrong UserIdTAG: 183166 UserNameTAG: yogeshk CreateTimeTAG: 2012-10-13T15:17:21Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: There is a voltage source Vs- kept ad the Source. So the answer will be in negative. FirstChildUserIdTAG: 362299 FirstChildUserNameTAG: anandbaskaran FirstChildCreateTimeTAG: 2012-10-13T16:08:40Z IndexTAG: 3480 TitleTAG: How K is derived from the TN0205A By reading the spec for the TN0205A MOSFET how can we derive that the K value is 0.4A/(V^2)? UserIdTAG: 18880 UserNameTAG: pcarmo CreateTimeTAG: 2012-10-13T12:05:01Z VoteTAG: 0 CoursewareTAG: Week 5 / MOSFET model exercise CommentableIdTAG: 6002x_mosfet_mod_e NumberOfReplyTAG: 2 FirstChildTAG: Well... I discovered the answer... By looking to the 1st graphic of the MOSFET spec, we can see that with vGS= 2V, we have a iD of 0.25A. By using the SCS model we know that K= 2*iDS / (vGS-VT)^2. From the text of the question we know that VT is approximately = 1 (even though the specs says 0.9V). By replacing on the equation we have: K= 2*0.25 / (2-0.9)^2= 0.4 A/V^2... Nice... FirstChildUserIdTAG: 18880 FirstChildUserNameTAG: pcarmo FirstChildCreateTimeTAG: 2012-10-13T16:04:53Z FirstChildTAG: for amplifiers MOSFET you can find transconductance g for specific GS voltage, so g=K*(VGS-VT) K=g/(VGS-VT) FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-13T17:39:10Z IndexTAG: 3481 TitleTAG: BJT voltages. Hello. I consider example 7.15 from the textbook. ![enter image description here][1] I can't understand one thing: why is $v_{BC} = v_{BE} - v_{CE}$ (1)? I think, it should be $v_{CE} = v_{BE} + v_{BC}$, thus $v_{BC} = v_{CE} - v_{BE}$. But it is wrong. Can somebody explain, please, how to obtain (1) equation? [1]: https://edxuploads.s3.amazonaws.com/1350128694134363.jpg UserIdTAG: 311784 UserNameTAG: ilya07 CreateTimeTAG: 2012-10-13T11:48:56Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I feel it is wrong FirstChildUserIdTAG: 132685 FirstChildUserNameTAG: deepakmurali FirstChildCreateTimeTAG: 2012-10-13T18:03:59Z SecondChildTAG: vBC=vBE−vCE is right. Since vBC is the voltage of node B with respect to node C. These nodes B & C can have any voltage and hence can be referenced to any independent node (In this case, it is node E). vCE=vBE+vBC is not correct. vCE=vCB+ vBE SecondChildUserIdTAG: 364851 SecondChildUserNameTAG: Varindra SecondChildCreateTimeTAG: 2012-10-13T19:50:18Z IndexTAG: 3482 TitleTAG: not linear ? But how we can say that a circuit is linear and using superposition, thevenin and norton methods while the circuit has a non-linear element that is a voltage source ?? UserIdTAG: 402183 UserNameTAG: Threepwood CreateTimeTAG: 2012-10-13T10:51:00Z VoteTAG: 0 CoursewareTAG: Week 2 / Is it linear? CommentableIdTAG: 6002x_is_it_linear NumberOfReplyTAG: 1 FirstChildTAG: peicewise linear? FirstChildUserIdTAG: 368712 FirstChildUserNameTAG: jmen FirstChildCreateTimeTAG: 2012-10-17T16:06:51Z IndexTAG: 3483 TitleTAG: lab5 last part , what to vary and what not to vary,simultaneously or one at a time?in what intervals.pls help. UserIdTAG: 269641 UserNameTAG: BAUWA CreateTimeTAG: 2012-10-13T08:27:29Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3484 TitleTAG: H6P1 l'Hospital's rule Correct. Hi guys! I got this part correct by trial and error of the roots. Anyone out there knows why l'Hospital's rule applies here? Any help will be deeply appreciated! Cheers! UserIdTAG: 499268 UserNameTAG: ruinoah CreateTimeTAG: 2012-10-13T05:37:35Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi, ruinoah! If we will substitute vIN=VT, we will get (0/0) indeterminate form for one root. Now we have to solve it, and most useful way is the l'Hospital's rule. If we get the concrete number, we will be able to check if it corresponds to our circuit. FirstChildUserIdTAG: 261503 FirstChildUserNameTAG: EugenyL FirstChildCreateTimeTAG: 2012-10-13T07:45:14Z SecondChildTAG: l'hopital rule applies only limit calculation. SecondChildUserIdTAG: 361823 SecondChildUserNameTAG: EliasOak SecondChildCreateTimeTAG: 2012-10-13T12:41:31Z SecondChildTAG: You are right, EliasOak. But when we have zero denominator, we are already unable to calculate actual value. All we can do is to find the limit of expression, and hello, l'Hospital =) However, we can keep limits in mind and don't put them on the paper for simple practical exercises. SecondChildUserIdTAG: 261503 SecondChildUserNameTAG: EugenyL SecondChildCreateTimeTAG: 2012-10-13T18:17:49Z IndexTAG: 3485 TitleTAG: [H5P2] Help with Q7 I see a lot of topics showing tips how to solve the Q7 of H5P2, but I still can't solve! This condition should be checked: **vDS >= vGS - VT** As I'm looking for the MINIMUN value, I have: **vDS = vGS - VT** As **vDS = VDD - vOUT** and **vGS = vIN-vOUT**: **VDD - vOUT = vIN - vOUT - vT**, simplifying: **VDD = vIN - vT** Can somebody help me? UserIdTAG: 310147 UserNameTAG: ildomarcarvalho CreateTimeTAG: 2012-10-13T05:11:31Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Yes, you are right! Now just put the MAXIMUM VIN value and get answer in volts :-) FirstChildUserIdTAG: 345464 FirstChildUserNameTAG: Constantine_ru FirstChildCreateTimeTAG: 2012-10-13T07:04:32Z SecondChildTAG: But how I calculate maximum vIN? To calculate max vIN I have the formula: **vIN = VT + vOUT** VT is given but to find vOUT I have to use the formula: **vOUT = (-1 + sqrt(1+2*K*RS*VS))/(K*RL)** I have the value for all variables, except the VS, because he is the VDD! If I put all of them together, I get the formula: **VDD = (-1+sqrt(1+(2*K*RS*VDD)))/(K*RS)** Solving this equation I get 0, and that is not the right answer. SecondChildUserIdTAG: 310147 SecondChildUserNameTAG: ildomarcarvalho SecondChildCreateTimeTAG: 2012-10-13T22:04:26Z SecondChildTAG: Oh, now I get it! The MAXIMUM vIN is given by the exercise :D SecondChildUserIdTAG: 310147 SecondChildUserNameTAG: ildomarcarvalho SecondChildCreateTimeTAG: 2012-10-14T04:37:38Z IndexTAG: 3486 TitleTAG: h6p1 need help in finding ro. i m writing ro=1/(2*K*VDS*(VGS-VT)) but it doesn't parse it as formula UserIdTAG: 582342 UserNameTAG: achilees111 CreateTimeTAG: 2012-10-13T04:55:30Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: write as following way then u'll get ur answer: .5*K^-1*(VGS-VT)^-1*VDS^-1 FirstChildUserIdTAG: 189650 FirstChildUserNameTAG: schn345 FirstChildCreateTimeTAG: 2012-10-13T05:35:16Z SecondChildTAG: THNX SecondChildUserIdTAG: 582342 SecondChildUserNameTAG: achilees111 SecondChildCreateTimeTAG: 2012-10-13T07:13:12Z SecondChildTAG: achilles111 YOUR ANSWER IS SO RIGHT. O.K. SecondChildUserIdTAG: 106816 SecondChildUserNameTAG: Laith SecondChildCreateTimeTAG: 2012-10-14T08:33:42Z FirstChildTAG: This probably won't help you if you are not good at maths. I have however simply used the two-dimensional Taylor formula to get gm and ro in this context. EDIT: Just seen this method was also discussed in the Week 6 tutorials. FirstChildUserIdTAG: 375382 FirstChildUserNameTAG: apatriarca FirstChildCreateTimeTAG: 2012-10-13T14:28:56Z IndexTAG: 3487 TitleTAG: saad When we apply small signal to MOSFET,then why output is in inverted form? UserIdTAG: 442829 UserNameTAG: saadzulfiqar CreateTimeTAG: 2012-10-13T03:42:36Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: No reason to be sad, there are other types of FETs and ways to hook them up so they don't invert. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-13T04:06:46Z IndexTAG: 3488 TitleTAG: I started too late Hello, first let me apolozige for my bad english, fell free to correct me. My friend introduce me this new online course too late. I really wanna do it, really wanna learn it. But, i joined too late and i too far away. 7weak´s late, the midle exam is coming soon... and i think i cannot study all this in 2 weaks. After all this, i have some questions? -When course end,will site still available with all resource? I wanna learn even if i don´t get the certificated. -Next year, this course will mantain free? If not,how much i have to pay to get it? -There´s any chances to get certificate even if i started today? Sorry again for my bad english,i know i make too many mistakes... Appreciate any answer you can give me. And thanks too staff to make this possible, it´s just so simple and so much information to learn... Respects, Diogo Machdo UserIdTAG: 607132 UserNameTAG: DiogoMachado CreateTimeTAG: 2012-10-13T02:21:45Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You are almost 5 weeks late not 7. Check out the calender to show you all the due dates. The course ends in late December. You can follow along and learn without certification, or hustle and get the certification. I don't know if the course will be free next year. You might have to pay for the certificate next year at the very least. Yes you can get the certificate even though you started today, in fact it is still possible for you to get 97%. Use the search function to get good answers to questions already asked in the earlier weeks. Ask in here if you don't find a suitable answer. Have fun. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-13T03:39:11Z SecondChildTAG: Calender here. > https://www.edx.org/static/content-mit-6002x/handouts/calendar.c7b2799155f8.pdf SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-13T03:40:03Z IndexTAG: 3489 TitleTAG: V0 is not equal to VGS. VI = VGS At 2:00 he says V0 = VGS. Surely he made a mistake? It is really bad for beginners like me who are already confused! UserIdTAG: 409867 UserNameTAG: shazmiah CreateTimeTAG: 2012-10-12T21:44:15Z VoteTAG: 0 CoursewareTAG: Week 5 / MOSFET Amplifier Graphically Described CommentableIdTAG: 6002x_mosfet_amp_graph NumberOfReplyTAG: 0 IndexTAG: 3490 TitleTAG: example 5.1 in example 5.1 the yehaa adders interpret the level between 3&3.5 is high and that for desco iterpret that under level 3.5 as un defined why was yehaa accepted ? UserIdTAG: 221617 UserNameTAG: konan CreateTimeTAG: 2012-10-12T19:52:21Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3491 TitleTAG: H6P1, hint needed !!!!!!!!!!!!!!!!!!!!!!!!!!!!! Stucked in to find "ro". I'm not looking for a solution from you, but at least a one hint. Thanks for support in advance UserIdTAG: 269412 UserNameTAG: andrei836 CreateTimeTAG: 2012-10-12T18:31:08Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: view tutorial of week 6, that helps to solve exactly the problem. FirstChildUserIdTAG: 114913 FirstChildUserNameTAG: ngoctuan FirstChildCreateTimeTAG: 2012-10-12T19:09:52Z SecondChildTAG: vaccum diode SecondChildUserIdTAG: 299251 SecondChildUserNameTAG: TheRedBlackOne SecondChildCreateTimeTAG: 2012-10-12T19:12:02Z IndexTAG: 3492 TitleTAG: a small doubt what's the actual meaning of Pull up and pull down resistances? UserIdTAG: 316353 UserNameTAG: Vivekanand123 CreateTimeTAG: 2012-10-12T18:05:44Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: pull up mean connect to V+, pull down mean connect to grnd FirstChildUserIdTAG: 114913 FirstChildUserNameTAG: ngoctuan FirstChildCreateTimeTAG: 2012-10-12T19:12:47Z IndexTAG: 3493 TitleTAG: h5p2 Q7 struggling with this one, firstly the what it means, am i looking for the minimun VDD such that the previous 3 questions are in the saturation zone or the min VDD so that there is a saturation zone at all? is min VDD 0 for 2nd option? i got a VDS vs VGS relation ship. where i have equated VDS and VGS to Vout, Vin and VT. from that i can get a range of VDS that relies on VDD.and from there i can relate VDD to the range of Vout, but when i solve that for range = 0 i get VDD = 0 range = 'the range i get from the previous questions' the answer i get for VDD is wrong does anything i am doing sound right, if infact you can understand my babbling. UserIdTAG: 388273 UserNameTAG: kilmarta CreateTimeTAG: 2012-10-12T16:41:19Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: 1. Be careful the condition for saturation of mosfet are: 2. VGS >= VT 3. VDS >= VGS + VT 4. VGS, VDS does not means VIN or VOUT in many case. just look to the diagram and you will notice that VDS == VDD - [some volt]. and wat is VGS == [some volt with vin and vout] only apply the VDS condition with both VDS , and VGS then finish FirstChildUserIdTAG: 114913 FirstChildUserNameTAG: ngoctuan FirstChildCreateTimeTAG: 2012-10-12T18:35:30Z SecondChildTAG: in hsp2 q1 m getting the error-couldn't parse this formulae,,,i've entered the exp of ids =k/2(vgs -vt)^2 &vgs as vin n vout,, SecondChildUserIdTAG: 183166 SecondChildUserNameTAG: yogeshk SecondChildCreateTimeTAG: 2012-10-12T18:40:26Z IndexTAG: 3494 TitleTAG: saad Week 5 direct videos download links are not available in course info and also youtube is not working, then how can i get the week 5 videos? UserIdTAG: 442829 UserNameTAG: saadzulfiqar CreateTimeTAG: 2012-10-12T16:30:56Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/LectureDownloadWeek5/ FirstChildUserIdTAG: 122513 FirstChildUserNameTAG: Mona77 FirstChildCreateTimeTAG: 2012-10-12T18:47:58Z IndexTAG: 3495 TitleTAG: h5p3 after computation i m getting this. can ny1 tell me what's d problem in this, differenciation error or xtra term. wasted more than half an hour on this question. vout===VIN/(K*RS)*(K*RS+(.5/sqrt((2*(K*RS*(VIN-VT)+1))^2-4*K*RS*RS*K*(VIN-VT)^2)*(4*RS*K*(K*RS*(VIN-VT)+1)-4*RS*K*K*RS(VIN-VT)))) UserIdTAG: 72647 UserNameTAG: NEEL11 CreateTimeTAG: 2012-10-12T16:24:14Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: i think this is help to make the differenciation for dvout/dvin - Hint for math y = (ax+b)^(1/2) - y^2 = ax+b - 2y.y' = a - y' = (a/2y) with y !=0 - y' is dvout/dvin and y = vout FirstChildUserIdTAG: 114913 FirstChildUserNameTAG: ngoctuan FirstChildCreateTimeTAG: 2012-10-12T18:54:32Z FirstChildTAG: Oh... May be you will try to calculate derivative without using "mathematica", Matlab and others? You will see, that it would me more easer! Hint: (Vin)'-(1/Rs*)sqrt(1/K^2+2Rs/K*(Vin-VT)))', Anothet hint d(sqrt(C1*x+C2))/dx=C1/(2*sqrt(C1*x+C2)) Try to calulate d' of this equation honesty , and then you will get very simlpe and beatifull formula :-) FirstChildUserIdTAG: 345464 FirstChildUserNameTAG: Constantine_ru FirstChildCreateTimeTAG: 2012-10-13T07:27:18Z FirstChildTAG: As Constantine_ru said you con use these tools, or you can use http://www.wolframalpha.com/ FirstChildUserIdTAG: 310147 FirstChildUserNameTAG: ildomarcarvalho FirstChildCreateTimeTAG: 2012-10-14T04:08:35Z IndexTAG: 3496 TitleTAG: "right offset"? What does "right offset" means? Thank you. UserIdTAG: 166869 UserNameTAG: Vasco CreateTimeTAG: 2012-10-12T15:01:16Z VoteTAG: 0 CoursewareTAG: Week 5 / Amplifier Distortion Demo CommentableIdTAG: 6002x_amp_dist_demo NumberOfReplyTAG: 0 IndexTAG: 3497 TitleTAG: V=IR true for Non-linear elements? because for the "D" element the equation is totally different. Though it worked in this example. UserIdTAG: 138239 UserNameTAG: LaFolle CreateTimeTAG: 2012-10-12T14:35:41Z VoteTAG: 0 CoursewareTAG: Week 3 / Analyzing Nonlinear Circuits Exercise CommentableIdTAG: 6002x_analyzing_nonlinear_circuits_exercise NumberOfReplyTAG: 2 FirstChildTAG: Though the element said to be non-linear, it can be considered linear piecewise. Break its characteristic into two pieces and the element will act as a resistor on each piece (with different values), thus $V=I*R$. FirstChildUserIdTAG: 470934 FirstChildUserNameTAG: sonhx FirstChildCreateTimeTAG: 2012-10-12T15:51:42Z FirstChildTAG: nope FirstChildUserIdTAG: 334613 FirstChildUserNameTAG: aswinshankar FirstChildCreateTimeTAG: 2012-10-12T18:01:10Z IndexTAG: 3498 TitleTAG: MOSFET: Large signals Hi 6.002x! In the lecture sequence "MOSFET: Large signals" after few lectures we start to use VI and VO instead of VGS and VDS. But when I start do homework, I understand that for problems in HW this is not right. So my question is: -all formulas that we use to find VI and VO in lectures at real are formulas for VGS and VDS? -to find expression of VGS and VDS for example at HW5P2, we say that VDD=VDS+Vo and VGS=VIN? Sorry for my english. UserIdTAG: 500965 UserNameTAG: barka0 CreateTimeTAG: 2012-10-12T11:41:30Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: yea what ur are saying is completly true in the sequence he used a special case were the source is grounded in cascaded amplifiers or as the circuit in hm we should use vds and vgs FirstChildUserIdTAG: 84213 FirstChildUserNameTAG: mohamed373 FirstChildCreateTimeTAG: 2012-10-12T14:08:50Z FirstChildTAG: Yes he simply wants us to stay on our toes and not get lax with our understanding of the principles that make the mosfet function. FirstChildUserIdTAG: 256543 FirstChildUserNameTAG: sidney23 FirstChildCreateTimeTAG: 2012-10-12T14:52:40Z IndexTAG: 3499 TitleTAG: I keep getting "this video is currently unavailable" Anyone else have this? Thanks ALex UserIdTAG: 358806 UserNameTAG: alexB1 CreateTimeTAG: 2012-10-12T11:15:05Z VoteTAG: 0 CoursewareTAG: Week 5 / Saturation Region Model Demo CommentableIdTAG: 6002x_sat_rgn_mdl_demo NumberOfReplyTAG: 0 IndexTAG: 3500 TitleTAG: Week 4-INCREMENTAl ANALYSIS (S7E3-LINEARIZATION how can the last two questions be solved...i have some idea but can't get the proper ans....plz help with yours....... UserIdTAG: 413613 UserNameTAG: shaliesh CreateTimeTAG: 2012-10-12T10:06:34Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3501 TitleTAG: H5P3 this is what i have tried, i changed the whole circuit to its small signal equivalent(mosfet to VCS(where id=K*(VI-VT)*vi), VDD to 0(shorted),VIN to vin and RS remains the same). apply simple ohms law, where vo=id*RS= this is perfectly logical(i think), so y am i getting the 'X' sign UserIdTAG: 285945 UserNameTAG: lindalapiso CreateTimeTAG: 2012-10-12T09:14:06Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: try to figure out the equation in question number 1 in h5p3 then directly substitute in it its really really easier FirstChildUserIdTAG: 84213 FirstChildUserNameTAG: mohamed373 FirstChildCreateTimeTAG: 2012-10-12T09:42:10Z SecondChildTAG: i forgot to tell u something if u r going to use the modeling u should first do the dc analysis u have 2 equations one id as a funtion of (k VGS AND VT) the other is kvl in the input loop u will have a quadratic equations solve it and choose s suitable value of VGS hint VGS IS calculated from dc analysis (large signal) the apply the small signal model u will find that vgs is equal to i vout and solve this is the looogic answer from electronis and circuits point of view SecondChildUserIdTAG: 84213 SecondChildUserNameTAG: mohamed373 SecondChildCreateTimeTAG: 2012-10-12T09:55:48Z SecondChildTAG: corrected down SecondChildUserIdTAG: 84213 SecondChildUserNameTAG: mohamed373 SecondChildCreateTimeTAG: 2012-10-12T10:00:07Z SecondChildTAG: MOhamed 3afac help H5P3 SecondChildUserIdTAG: 331441 SecondChildUserNameTAG: abdessamade SecondChildCreateTimeTAG: 2012-10-12T13:05:31Z FirstChildTAG: i forgot to tell u something if u r going to use the modeling u should first do the dc analysis u have 2 equations one id as a funtion of (k VGS AND VT) the other is kvl in the input loop u will have a quadratic equations solve it and choose s suitable value of VGS use it in finding the gm hint gmIS calculated from dc analysis (large signal) the apply the small signal model u will find that vgs is equal to i vout and solve this is the looogic answer from electronis and circuits point of view FirstChildUserIdTAG: 84213 FirstChildUserNameTAG: mohamed373 FirstChildCreateTimeTAG: 2012-10-12T09:56:02Z FirstChildTAG: - You must finish H5P2 where you have vOUT expression. - vout = vin* d(vOUT)/d(vIN) with vIN = VIN - that means you need to calculate d(vOUT)/d(vIN) with vIN = VIN - apply result d(vOUT)/d(vIN) with vIN = VIN to vout - ---------------------- Hint for math y = (ax+b)^(1/2) - y^2 = ax+b - 2y.y' = a - y' = (a/2y) with y !=0 FirstChildUserIdTAG: 114913 FirstChildUserNameTAG: ngoctuan FirstChildCreateTimeTAG: 2012-10-12T18:46:01Z FirstChildTAG: Thank God for wolfram alpha - finally got that dang derivative done - calculus was in 1992 - a little rusty FirstChildUserIdTAG: 18073 FirstChildUserNameTAG: gburkhart FirstChildCreateTimeTAG: 2012-10-13T01:41:36Z IndexTAG: 3502 TitleTAG: very clear excellent explanation of a highly technical topic. nevver knew transistors had their own internal capacitance, had always just abstracted them as switches that achieved amplification somehow..:p UserIdTAG: 132135 UserNameTAG: mizzlemax CreateTimeTAG: 2012-10-11T21:58:39Z VoteTAG: 0 CoursewareTAG: Week 6 / Capacitor basics CommentableIdTAG: 6002x_capacitor_basics NumberOfReplyTAG: 0 IndexTAG: 3503 TitleTAG: homework 1 I started the course last week and i am trying to catch up the homeworks , i did the homework 1 and lab 1 , but it does not seems to be submitted , is still possible to do it or not? UserIdTAG: 514732 UserNameTAG: meliyka CreateTimeTAG: 2012-10-11T21:54:24Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Troubleshooting NumberOfReplyTAG: 2 FirstChildTAG: It is too late for homework 1,2,3,4. You can miss two of them, the other two have cost you 3% of your final grade. You can still do really good if you get caught up. Week 5 is due on Sunday evening. Good luck, have fun. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-11T22:42:42Z FirstChildTAG: I AM UNABLE TO FIND MY HOMEWORK PAGE.PLEASE GUIDE ME HOW AND WHERE I CAN FIND IT.PLEASE SEND ME THE LINK ON MY EMAIL ID. THANKS. FirstChildUserIdTAG: 478475 FirstChildUserNameTAG: AliFarhan FirstChildCreateTimeTAG: 2012-10-12T14:46:01Z IndexTAG: 3504 TitleTAG: H5P1 vIN If $v_{in}$ is negative, then how can $v_{IN}= V_T$, with VT being positive? If $v_{IN}$ is not equal or greater than $V_T$, then the mosfet is not on, so it cannot be in saturation. UserIdTAG: 248807 UserNameTAG: 1kingsman CreateTimeTAG: 2012-10-11T21:23:53Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Because vIN is not equal to VGS in this circuit. It is VGS that controls the current. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-11T21:54:04Z SecondChildTAG: But how is VGS not equal to VI? What is VGS? SecondChildUserIdTAG: 248807 SecondChildUserNameTAG: 1kingsman SecondChildCreateTimeTAG: 2012-10-11T22:18:48Z SecondChildTAG: VGS is gate-source voltage. vIN is applied between the gate and ground, but the source is not at ground because of VS-. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-11T22:35:50Z SecondChildTAG: i hope i can help u in the sequence he took a special case wehre the source is grounded he assumed that id=k/2 *(VIN-VT)^2 WHICH IS A SPEEECIAL case as the source is grounded vin in this equation should be replaced by vgs which is equal to vg-vs the two terminals of the mosfet it self u can find it using kvl SecondChildUserIdTAG: 84213 SecondChildUserNameTAG: mohamed373 SecondChildCreateTimeTAG: 2012-10-12T09:38:45Z SecondChildTAG: You can take a look at this [post][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_mosfet_amp/threads/507a1d6dd894f12300000039 SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T04:13:50Z IndexTAG: 3505 TitleTAG: What is V in V^2? What is V in V^2? VT?? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-11T19:23:50Z VoteTAG: 0 CoursewareTAG: Week 5 / MOSFET model exercise CommentableIdTAG: 6002x_mosfet_mod_e NumberOfReplyTAG: 2 FirstChildTAG: Do you mean K≈0.4A/V^2? It's unit of measurement, ampere/(volt squared). FirstChildUserIdTAG: 261503 FirstChildUserNameTAG: EugenyL FirstChildCreateTimeTAG: 2012-10-11T20:40:20Z FirstChildTAG: To calculate, you should take Vgs. It means: ids = K*(Vgs-1)^2 Probably, i'm wrong... I mean, mb you should take ids = K*(Vgs-Vt)^2 FirstChildUserIdTAG: 200319 FirstChildUserNameTAG: Virviil FirstChildCreateTimeTAG: 2012-10-14T18:34:18Z IndexTAG: 3506 TitleTAG: Late!!! Sorry got little late... hope i will cover it.. looking for your support ahead...This is my first distant learning eduction..its fun and cool..Thanks to whole team for such a nice effort :) UserIdTAG: 164840 UserNameTAG: dwijaybane CreateTimeTAG: 2012-10-11T18:54:05Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 0 IndexTAG: 3507 TitleTAG: lack of english language i have got errors in my calculations due to lack in english language comprehension , i have made a banal mistakes, i will try be careful next time. UserIdTAG: 488779 UserNameTAG: nadir22 CreateTimeTAG: 2012-10-11T17:11:38Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 0 IndexTAG: 3508 TitleTAG: H5P2 iDS BUG!!! Hi all, in the third part of h5p2 where we have to find the equation for iDS i have got this ridiculously long equation but i know it is correct as it satisfies the condition that iDS=0 when vIN=VT and also i got the tick mark in the very next question which proves that the equation i got is correct. However i am constantly getting the wrong mark for the third part though the equation looks correct in the math format after i wrote it in the text box. Could someone please help me out with this? Thanks!!! UserIdTAG: 222911 UserNameTAG: bhaswardg CreateTimeTAG: 2012-10-11T16:34:50Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 6 FirstChildTAG: There is a quadratic equation. Which root did you use? FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-11T17:44:13Z FirstChildTAG: Ok here's the deal: If you think it's really a bug you should send an email to bugs@edx.org. You need to include all your evidence along with specific references to what the problem is. Make sure you include all your equations. If it's really a bug they will fix it. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-11T16:40:41Z SecondChildTAG: I have the same problem as bhaswardg SecondChildUserIdTAG: 400832 SecondChildUserNameTAG: panosbr SecondChildCreateTimeTAG: 2012-10-13T13:40:12Z SecondChildTAG: the answer to this question should be in this format (xx-xx)/xx + 1/(xx^x*x)*(1-sqrt(1+x*x*xx*(xx-xx))) finally after hours of head banging and lots if note book pages i was sure that my answer is correct & i finally got the green holy tick!! Just make sure that you put a * for every multiplication!!! This was the mistake that I was doing SecondChildUserIdTAG: 232499 SecondChildUserNameTAG: Asim09 SecondChildCreateTimeTAG: 2012-10-13T15:50:08Z SecondChildTAG: i got the tick too. :) thanks anyway. SecondChildUserIdTAG: 222911 SecondChildUserNameTAG: bhaswardg SecondChildCreateTimeTAG: 2012-10-14T14:42:22Z FirstChildTAG: same i get a messy-ish equation that i use to answer following questions correctly so the eqn works although there is prob a clearer version. the error i get is Invalid input: Could not parse 'equation i used' as a formula will i post what i used here? prob just typing it in wrong, although i have tried multi times FirstChildUserIdTAG: 388273 FirstChildUserNameTAG: kilmarta FirstChildCreateTimeTAG: 2012-10-11T22:54:09Z SecondChildTAG: rewrote it longer and got the tick, could too many brackets screw it up?? SecondChildUserIdTAG: 388273 SecondChildUserNameTAG: kilmarta SecondChildCreateTimeTAG: 2012-10-11T23:22:42Z SecondChildTAG: i am just getting the wrong mark, nothing as invalid input. What do you mean by 'rewrote it longer'? SecondChildUserIdTAG: 222911 SecondChildUserNameTAG: bhaswardg SecondChildCreateTimeTAG: 2012-10-12T05:10:41Z SecondChildTAG: duh! still not getting the tick mark. As you said i 'rewrote it longer' by expanding the brackets, but no tick mark. Please help. SecondChildUserIdTAG: 222911 SecondChildUserNameTAG: bhaswardg SecondChildCreateTimeTAG: 2012-10-12T05:30:49Z SecondChildTAG: You assumed that got right equation and it's a bug of the grader. According to my experience, as many people got it right, most likely it's time to search something wrong in your solution. As I asked before - there is a quadratic equation with two roots - which root did you use? SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-12T06:56:02Z SecondChildTAG: the one with the negative sign. SecondChildUserIdTAG: 222911 SecondChildUserNameTAG: bhaswardg SecondChildCreateTimeTAG: 2012-10-12T09:47:52Z SecondChildTAG: It's correct. If you want contact me on skype and give me your equation - I'll check it. My skype ieee802.1 SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-12T14:00:22Z SecondChildTAG: instead of skype could you give me your email id? i wll send you the equation. :) SecondChildUserIdTAG: 222911 SecondChildUserNameTAG: bhaswardg SecondChildCreateTimeTAG: 2012-10-12T15:43:43Z SecondChildTAG: tmpbox(at)hotmail.com replace (at) with @ SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-12T18:50:23Z SecondChildTAG: sent you the mail YakovO. SecondChildUserIdTAG: 222911 SecondChildUserNameTAG: bhaswardg SecondChildCreateTimeTAG: 2012-10-13T04:43:43Z SecondChildTAG: i am sure my equation is correct but it doesn't give the holy green tick!! it is some thing like this (xx-xx)/xx + (xx^xx/xx)*(1-sqrt(xx*xx*(xx-xx)+1)) i hope i am not violating the honor code!! SecondChildUserIdTAG: 232499 SecondChildUserNameTAG: Asim09 SecondChildCreateTimeTAG: 2012-10-13T09:18:13Z SecondChildTAG: finally!!! the answer to this question should be in this format (xx-xx)/xx + (1/(xx^x*x)*(1-sqrt(1+x*x*xx*(xx-xx))) finally after hours of head banging and lots if note book pages i was sure that my answer is correct & i finally got the green holy tick!! Just make sure that you put a * for every multiplication!!! This was the mistake that I was doing SecondChildUserIdTAG: 232499 SecondChildUserNameTAG: Asim09 SecondChildCreateTimeTAG: 2012-10-13T15:50:58Z FirstChildTAG: DO IT STEP BY STEP>and in every time check answer and make sure that give you wrong only< not : Could not parse 'equation i used' as a formula> > and after write formula will be right isa. FirstChildUserIdTAG: 292299 FirstChildUserNameTAG: ammarsamir FirstChildCreateTimeTAG: 2012-10-12T01:31:30Z FirstChildTAG: I am not getting any idea how to come with equation. Any hint !!!! FirstChildUserIdTAG: 181935 FirstChildUserNameTAG: vishsahare FirstChildCreateTimeTAG: 2012-10-12T01:59:20Z SecondChildTAG: use ohm's law to get the relation between vOUT, iDS and RS. Then substitute the value of vOUT in the first answer you got for iDS. :) SecondChildUserIdTAG: 222911 SecondChildUserNameTAG: bhaswardg SecondChildCreateTimeTAG: 2012-10-12T05:06:52Z SecondChildTAG: whats about rs SecondChildUserIdTAG: 164689 SecondChildUserNameTAG: muhammadfaizan SecondChildCreateTimeTAG: 2012-10-12T13:16:59Z SecondChildTAG: I have tried replacing what I got for vOUT in the equation for iDS, however I can't get it right... First thing that pops up is that RS is not permitted, can someone give some insight on this?? SecondChildUserIdTAG: 30599 SecondChildUserNameTAG: Novelo SecondChildCreateTimeTAG: 2012-10-13T03:10:41Z SecondChildTAG: the answer to this question should be in this format (xx-xx)/xx + (1/(xx^x*x)*(1-sqrt(1+x*x*xx*(xx-xx))) finally after hours of head banging and lots if note book pages i was sure that my answer is correct & i finally got the green holy tick!! Just make sure that you put a * for every multiplication!!! This was the mistake that I was doing SecondChildUserIdTAG: 232499 SecondChildUserNameTAG: Asim09 SecondChildCreateTimeTAG: 2012-10-13T15:49:01Z SecondChildTAG: Novelo, make sure you are writing RS..all in caps. SecondChildUserIdTAG: 222911 SecondChildUserNameTAG: bhaswardg SecondChildCreateTimeTAG: 2012-10-14T14:44:15Z FirstChildTAG: Don't forget to put all need parenthesis. I stay stucked with this question 2 hours because the Wolfram calculator don't generated all the parenthesis :D FirstChildUserIdTAG: 310147 FirstChildUserNameTAG: ildomarcarvalho FirstChildCreateTimeTAG: 2012-10-13T03:36:12Z IndexTAG: 3509 TitleTAG: lab5 last part tried everything still not able to get the answer plz help asap.... tried different values of sin wave my answer is coming around 1XX mv UserIdTAG: 126911 UserNameTAG: sidhant7 CreateTimeTAG: 2012-10-11T13:46:58Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Measure peak-to-peak for each signal. Divide peak-to-peak by 2 to get the reference. Subtract peak value measured from reference to get the real peak value. FirstChildUserIdTAG: 127195 FirstChildUserNameTAG: princeofsudan FirstChildCreateTimeTAG: 2012-10-11T14:05:16Z SecondChildTAG: I vary voltages then find Vpeak_peak values for Input and Output...what next? SecondChildUserIdTAG: 122513 SecondChildUserNameTAG: Mona77 SecondChildCreateTimeTAG: 2012-10-13T11:18:17Z FirstChildTAG: yes mine also...but it won't accept it...hmmmmm FirstChildUserIdTAG: 256543 FirstChildUserNameTAG: sidney23 FirstChildCreateTimeTAG: 2012-10-14T17:41:04Z SecondChildTAG: they want the peak to peak input SecondChildUserIdTAG: 256543 SecondChildUserNameTAG: sidney23 SecondChildCreateTimeTAG: 2012-10-14T17:49:23Z IndexTAG: 3510 TitleTAG: Lab 5 last question Since 1.15 - 0.6 = 0.55, and that's the peak to peak amplitude, the amplitude is 0.55/2= 0.275, so Vbias is (1.15+0.6)=0.875 volts. Ok, but i don't get the green tick, still a red cross. WHY?!!!!!!!!! UserIdTAG: 244225 UserNameTAG: lerimock CreateTimeTAG: 2012-10-11T13:40:03Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Measure peak-to-peak for each signal. Divide peak-to-peak by 2 to get the reference. Subtract peak value measured from reference to get the real peak value. Check your measured values, they don't seem right, at least one of them. FirstChildUserIdTAG: 127195 FirstChildUserNameTAG: princeofsudan FirstChildCreateTimeTAG: 2012-10-11T14:06:12Z FirstChildTAG: 0.875-1.15=-0275 red cross. Something happens. It's not just graphs. It's calculus. I think graphs is not necessary. with the results of the previous parts is enough. FirstChildUserIdTAG: 244225 FirstChildUserNameTAG: lerimock FirstChildCreateTimeTAG: 2012-10-11T14:17:40Z SecondChildTAG: 0.875-0.6=0.275 red cross. SecondChildUserIdTAG: 244225 SecondChildUserNameTAG: lerimock SecondChildCreateTimeTAG: 2012-10-11T14:18:11Z SecondChildTAG: To measure peak-to-peak you add the absolute value of the positive peak to the absolute value of the negative peak. No Calculus is involved here. SecondChildUserIdTAG: 127195 SecondChildUserNameTAG: princeofsudan SecondChildCreateTimeTAG: 2012-10-11T14:23:50Z SecondChildTAG: Yes. 1.15+0.6=1.75-->1.75/2=0.875, that is what i wrote above. SecondChildUserIdTAG: 244225 SecondChildUserNameTAG: lerimock SecondChildCreateTimeTAG: 2012-10-11T17:14:41Z SecondChildTAG: I wonder if it is the Vbias which is asked? I thought you just had to find the amplitude. SecondChildUserIdTAG: 222809 SecondChildUserNameTAG: AshutoshTadkase SecondChildCreateTimeTAG: 2012-10-11T17:36:28Z SecondChildTAG: Vbias = 0.875 volts; vin = 0.275 volts. Answer?????????? SecondChildUserIdTAG: 244225 SecondChildUserNameTAG: lerimock SecondChildCreateTimeTAG: 2012-10-11T18:44:13Z SecondChildTAG: Finally... SecondChildUserIdTAG: 244225 SecondChildUserNameTAG: lerimock SecondChildCreateTimeTAG: 2012-10-11T19:12:14Z SecondChildTAG: ...and... How did you do it? SecondChildUserIdTAG: 413869 SecondChildUserNameTAG: nlfigueiredo SecondChildCreateTimeTAG: 2012-10-11T22:21:25Z SecondChildTAG: never mind, I get it. I just "round" too much. SecondChildUserIdTAG: 413869 SecondChildUserNameTAG: nlfigueiredo SecondChildCreateTimeTAG: 2012-10-11T22:24:08Z IndexTAG: 3511 TitleTAG: mid term exam can we get a chance to check our answer during exam? UserIdTAG: 114864 UserNameTAG: Vikaash CreateTimeTAG: 2012-10-11T13:08:25Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You will have 3 tries. FirstChildUserIdTAG: 127195 FirstChildUserNameTAG: princeofsudan FirstChildCreateTimeTAG: 2012-10-11T13:19:43Z IndexTAG: 3512 TitleTAG: Suggestion... Perhaps the course makers could clear up mess like this in the future... it is hard enough to understand the concepts! :) ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/1349958336757923.jpg UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-11T12:27:11Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: there are lecture notes without scribblings. FirstChildUserIdTAG: 8897 FirstChildUserNameTAG: sam1202 FirstChildCreateTimeTAG: 2012-10-11T13:31:05Z IndexTAG: 3513 TitleTAG: i_{DS} i couldn't solve it , i need a help UserIdTAG: 459680 UserNameTAG: jejy CreateTimeTAG: 2012-10-11T11:57:17Z VoteTAG: 0 CoursewareTAG: Week 5 / MOSFET model exercise CommentableIdTAG: 6002x_mosfet_mod_e NumberOfReplyTAG: 1 FirstChildTAG: Use the given formula: Ids = K/2 * (Vgs-Vt)^2 where everything is given. Remember to answer in milliampers FirstChildUserIdTAG: 329254 FirstChildUserNameTAG: KGabor FirstChildCreateTimeTAG: 2012-10-11T14:47:36Z SecondChildTAG: thanks SecondChildUserIdTAG: 459680 SecondChildUserNameTAG: jejy SecondChildCreateTimeTAG: 2012-10-13T11:48:24Z IndexTAG: 3514 TitleTAG: question?? I totally lost after 3:18 ....... what does Mr. Agarwal mean by iDS <= K/2 vO^2 I am a little bit confuse. can any one help. UserIdTAG: 226085 UserNameTAG: Teto CreateTimeTAG: 2012-10-11T11:39:20Z VoteTAG: 0 CoursewareTAG: Week 5 / MOSFET Amplifier Graphically Described CommentableIdTAG: 6002x_mosfet_amp_graph NumberOfReplyTAG: 1 FirstChildTAG: from the minute 3:18......to......4:58 FirstChildUserIdTAG: 226085 FirstChildUserNameTAG: Teto FirstChildCreateTimeTAG: 2012-10-11T11:42:28Z SecondChildTAG: as v0>= (Vin-Vt), for saturation mode so in eq. Ids=K/2(Vin-Vt)^2 you can replace Vin-Vt with v0 and if v0>=Vin-Vt, for saturation mode, then Ids should be less than (K/2)*v0^2, yo stay in triode mode. anyway that means that the border between saturation mode and triode mode is looks like half of parabola SecondChildUserIdTAG: 210380 SecondChildUserNameTAG: MurdocRus SecondChildCreateTimeTAG: 2012-10-11T12:20:27Z SecondChildTAG: You explained it better than the video. Thanks. SecondChildUserIdTAG: 398594 SecondChildUserNameTAG: DaveyJC SecondChildCreateTimeTAG: 2012-12-14T09:22:02Z IndexTAG: 3515 TitleTAG: S7E3 I tried to post my solution to S7E3 giving in Latex-code in order to be nicely formatted. It seemed to work, although images and references to equations were not found. Can I post my thing in PDF format? I eagerly want to take part in the discussions, but I even eagerly want my solutions to be nicely formatted and readable. Can someone help me? Thanks in advance. UserIdTAG: 330170 UserNameTAG: H_Litzroth CreateTimeTAG: 2012-10-11T11:20:21Z VoteTAG: 0 CoursewareTAG: Week 4 / Linearization Exercise 2 CommentableIdTAG: 6002x_linearization_ex_2 NumberOfReplyTAG: 0 IndexTAG: 3516 TitleTAG: invalid input error why it says "couldn't parse (some expression) as formula". What should i've to do to avoid that? UserIdTAG: 368150 UserNameTAG: vasanthan3886 CreateTimeTAG: 2012-10-11T11:20:18Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Are you getting this error for S9E3(MOSFET Amplifier 2)? I got it for this exercise and I see that the formula i typed seems right. FirstChildUserIdTAG: 135674 FirstChildUserNameTAG: sowmyas FirstChildCreateTimeTAG: 2012-10-11T12:01:26Z FirstChildTAG: It would be helpful if you posted a screenshot of what you did. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-11T13:40:51Z FirstChildTAG: I usually got that when I tried 2K instead of 2*K and so on. FirstChildUserIdTAG: 222809 FirstChildUserNameTAG: AshutoshTadkase FirstChildCreateTimeTAG: 2012-10-11T17:21:24Z SecondChildTAG: thank you. I just saw that i've committed the same mistake. SecondChildUserIdTAG: 368150 SecondChildUserNameTAG: vasanthan3886 SecondChildCreateTimeTAG: 2012-10-12T04:53:41Z IndexTAG: 3517 TitleTAG: entering answers hey ...how to enter the answers of algebraic expressions we get in problems .. UserIdTAG: 161899 UserNameTAG: harshvit12 CreateTimeTAG: 2012-10-11T10:28:48Z VoteTAG: 0 CoursewareTAG: Week 4 / Dependent Sources Exercise 0 CommentableIdTAG: 6002x_dep_src_exsz_0 NumberOfReplyTAG: 0 IndexTAG: 3518 TitleTAG: New and troubled Hello everyone. I just discovered this site and I'm happy about it. I want to be able to finish the course even though I'm way too late already. Can I have some tips on catching up on everything I missed? Can I still submit homeworks? UserIdTAG: 605039 UserNameTAG: danicaguinid CreateTimeTAG: 2012-10-11T08:29:03Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: The deadline for weeks 1-4 homework and labs has already passed. You get to drop two homework and two lab assignments, so you would have to take zeros on two. The deadline for week 5 is Sunday, October 14th. So you have three days. Mid-terms start on October 25th, two weeks from today. The only way to catch up is to start at the beginning and move forward. If you have really excellent algebra and calculus skills, and pick things up fast, you might make it. The midterm is 30% of your grade and the final is 40%, so if you can nail those, you can pass the class. Otherwise, you might want to view this semester as a practice for the next one. It will be offered in the Spring. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-11T14:30:02Z FirstChildTAG: [For New Joining][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50759a15c702ae2b00000099 It is best for you to join this course in next session.... FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-10-11T14:31:12Z IndexTAG: 3519 TitleTAG: Two compound batteries While calculating the equivalent resistance of two bigger compound batteries,connected to the above circuit.I get Req=.2412ohm if i consider internal resistance along with external resistance,but correct answer is Req=.1403ohm.So plz clarify,So should I consider internal Resistance or not?If not why? UserIdTAG: 607857 UserNameTAG: AnikethEP CreateTimeTAG: 2012-10-11T07:53:06Z VoteTAG: 0 CoursewareTAG: Week 1 / Battery model CommentableIdTAG: 6002x_S1E9_Battery_Model NumberOfReplyTAG: 0 IndexTAG: 3520 TitleTAG: Homework 7- 2nd last problem Now, let's get down to numbers. Let V=3.3V, RS=22.0Ω, RO=50.0Ω, and L=15.44nH. How much time, in nanoseconds, does it take for the output voltage to reach vO=1.65V? could anyone help me to proceed this problem..? UserIdTAG: 189650 UserNameTAG: schn345 CreateTimeTAG: 2012-10-11T06:16:26Z VoteTAG: 0 CoursewareTAG: Week 6 / Vacuum Triode Model CommentableIdTAG: 6002x_Vacuum_Triode_Model NumberOfReplyTAG: 2 FirstChildTAG: You have final voltage from first section and the time constant from the second section... you can use those to write the equation of VO (Intuitively)and solve for t. Hope that would help. FirstChildUserIdTAG: 127195 FirstChildUserNameTAG: princeofsudan FirstChildCreateTimeTAG: 2012-10-11T11:31:15Z SecondChildTAG: same i did but i could'nt get the answer....btw thnx for help... SecondChildUserIdTAG: 189650 SecondChildUserNameTAG: schn345 SecondChildCreateTimeTAG: 2012-10-12T11:44:03Z FirstChildTAG: You already have vo(t)=1.65. Your initial voltage is going to be 0 at time 0, so you already have that. The final voltage is from question 1, your time constant is question 2, and your unknown is time. Just sub all that into the complete response equation and you are done. You can do it all in one long equation but it's easier to break it into pieces. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-11T13:31:17Z SecondChildTAG: same i did but i could'nt get the answer....btw thnx for help... SecondChildUserIdTAG: 189650 SecondChildUserNameTAG: schn345 SecondChildCreateTimeTAG: 2012-10-12T11:43:48Z IndexTAG: 3521 TitleTAG: reversed polarity I find the drawing confusing because the polarity of Vout is reversed and the connections are unclear. Vin is connected to where? UserIdTAG: 12905 UserNameTAG: Baer CreateTimeTAG: 2012-10-11T04:09:03Z VoteTAG: 0 CoursewareTAG: Week 5 / Op Amp Small Signal Model CommentableIdTAG: 6002x_ap_amp_small_signal_t NumberOfReplyTAG: 0 IndexTAG: 3522 TitleTAG: H5P2 I am struggling with Vout. I solve it 2 times same think. I do not want to give answer. it looks like this (ab+1+-sqrt(1+2ab))/a. Any idea? And which root we take one with - sign right? UserIdTAG: 286954 UserNameTAG: ododo CreateTimeTAG: 2012-10-11T01:10:39Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Has to go to zero when vIN = VT, which probably corresponds to b going to zero. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-11T01:53:53Z SecondChildTAG: it gets o for one of the root so whats wrong? SecondChildUserIdTAG: 286954 SecondChildUserNameTAG: ododo SecondChildCreateTimeTAG: 2012-10-11T13:00:38Z SecondChildTAG: I got wrong mark although i have full fill all the conditions i.e vIN=VT but x sign does not vanish. SecondChildUserIdTAG: 282892 SecondChildUserNameTAG: wiky SecondChildCreateTimeTAG: 2012-10-11T14:32:10Z SecondChildTAG: (vIN-VT)/(RS)^1/2 is int wrong answer SecondChildUserIdTAG: 282892 SecondChildUserNameTAG: wiky SecondChildCreateTimeTAG: 2012-10-11T14:32:59Z SecondChildTAG: do we have to differentiate the root? SecondChildUserIdTAG: 286954 SecondChildUserNameTAG: ododo SecondChildCreateTimeTAG: 2012-10-12T14:20:14Z SecondChildTAG: i think when it said iDS=0 when vIN=VT. , it means that the term ( vIN-VT) must be in the solution , so when vin = vt , ids turns to zero , but this won't happen untill the whole formula turn to zero , i have a formula that fulfil all the requirments but still give red X , my formula is (1+(K*RS*(vIN-VT))-sqrt(1+2*K*RS*(vIN-VT)^2))/(K*RS^2) i don't know what is wrong with it , if any one know , tell me please SecondChildUserIdTAG: 285675 SecondChildUserNameTAG: Ascot SecondChildCreateTimeTAG: 2012-10-13T21:49:12Z IndexTAG: 3523 TitleTAG: H6P1: THE NEWFET DEVICE, part 3 and part 4. Please some hint to solve those parts, I have read all posts that says "H6P1" and I don`t found something interesting. UserIdTAG: 299251 UserNameTAG: TheRedBlackOne CreateTimeTAG: 2012-10-10T22:12:20Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_6/Week_6_Tutorials/ VACUUM TRIODE MODEL video FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-10T22:26:54Z SecondChildTAG: Thanks SecondChildUserIdTAG: 299251 SecondChildUserNameTAG: TheRedBlackOne SecondChildCreateTimeTAG: 2012-10-10T22:59:59Z SecondChildTAG: the only tutorial i skipped.... fff.... SecondChildUserIdTAG: 301962 SecondChildUserNameTAG: labrinim SecondChildCreateTimeTAG: 2012-10-15T13:20:46Z IndexTAG: 3524 TitleTAG: Largest amplitude in Lab 5 I can't get the "largest amplitude" to be accepted. How exact does it have to be? The graph shows no clipping at my value, which is V_bias - V_lower_end, which should be the largest amplitude, but the answer is not accepted. What am I doing wrong? UserIdTAG: 209655 UserNameTAG: typograph CreateTimeTAG: 2012-10-10T21:47:48Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: make sure that you understand amplitude amplitude=peak-to-peak/2 FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-10T22:06:04Z SecondChildTAG: Is it? Hmmmm I thought amplitude was just Peak High minus Peak Low. Can you point me to a page number or something? SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-11T00:47:29Z SecondChildTAG: Peak to peak/2 is the same as peak high minus peak low if we are talking sinusoid. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-11T01:01:48Z SecondChildTAG: Disregard my post above. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-11T01:03:21Z SecondChildTAG: Basically it is measured from one extreme to the equilibrium and if we are talking sinusoid, that would be "peak to peak/2" SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-11T01:05:10Z SecondChildTAG: >I thought amplitude was just Peak High minus Peak Low. It's not quite right. Symmetric periodic signal can be biased in respect to zero - so peak max and peak min will have different absolute values. So amplitude will be (peak_max-peak_min)/2, and (peak_max-peak_min) is peak-to-peak. And don't mix it with "peak-to-peak amplitude" - it's the same as just peak-to-peak, but "amplitude" without any other specification for symmetric periodic signal is what I said above. SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-11T05:14:22Z IndexTAG: 3525 TitleTAG: question? how do we know the value of VA that suppose to be reach to make the diode start working or change it's behavior ? I mean why VA > 1 ??? UserIdTAG: 226085 UserNameTAG: Teto CreateTimeTAG: 2012-10-10T19:08:27Z VoteTAG: 0 CoursewareTAG: Week 4 / Textbook Load Line Tutorial CommentableIdTAG: 6002x_txtbk_load_line_t NumberOfReplyTAG: 1 FirstChildTAG: now I think I get it. I think it's because of the voltage source. FirstChildUserIdTAG: 226085 FirstChildUserNameTAG: Teto FirstChildCreateTimeTAG: 2012-10-10T19:14:11Z IndexTAG: 3526 TitleTAG: H5P1 formula hey guys whats the formula for this problem thank you. In this circuit, VS+=1.0V, VS−=−1.0V, and the MOSFET parameters are K=1 mAV2 and VT=0.5V. What is the VIN in volts for the MOSFET to be operating in saturation region? UserIdTAG: 462861 UserNameTAG: edxforme CreateTimeTAG: 2012-10-10T18:59:37Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Posting formulas is against the code of conduct, unless you have proven that you are completely stuck. FirstChildUserIdTAG: 248807 FirstChildUserNameTAG: 1kingsman FirstChildCreateTimeTAG: 2012-10-11T20:39:18Z SecondChildTAG: It's against the honor code whether you are completely stuck or not. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-11T22:37:11Z SecondChildTAG: Oh, misunderstood his question. I thought he was posting his own formula, not someone else giving him the formula. SecondChildUserIdTAG: 248807 SecondChildUserNameTAG: 1kingsman SecondChildCreateTimeTAG: 2012-10-15T23:58:42Z IndexTAG: 3527 TitleTAG: date for homework and lab 7 the date for hw and lab 7 is 4 of nov or 7 of nov? I assume 4 is the correct date UserIdTAG: 206669 UserNameTAG: dimitrios66 CreateTimeTAG: 2012-10-10T18:27:07Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Why would you assume that? The date is given as Nov. 7th. in the "Coursewares" tab. ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13499073072550785.png FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-10T22:15:16Z SecondChildTAG: INDEED SO. BUT IN COURSE INFO, CALENDAR AND 6002X AT A GLANCE, IT SAYS NOV 4 SecondChildUserIdTAG: 206669 SecondChildUserNameTAG: dimitrios66 SecondChildCreateTimeTAG: 2012-10-10T22:55:09Z SecondChildTAG: As a matter of fact it seems likely that nov 7 came from hom 7 by mistake SecondChildUserIdTAG: 206669 SecondChildUserNameTAG: dimitrios66 SecondChildCreateTimeTAG: 2012-10-10T22:56:41Z SecondChildTAG: Hmmmm I think you are right. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-11T00:28:39Z FirstChildTAG: I emailed someone on the staff about it. I think it should be Nov 4th because Nov 7th is a Wends. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-11T00:44:06Z SecondChildTAG: ok, thanks for that SecondChildUserIdTAG: 206669 SecondChildUserNameTAG: dimitrios66 SecondChildCreateTimeTAG: 2012-10-11T07:09:30Z IndexTAG: 3528 TitleTAG: new joining i joined my course today i will be able to manage my text works but how will i cope up with my lab n tutorial works please help me out if anyone can i am very tensed.... Akanksha UserIdTAG: 606726 UserNameTAG: Akanksha071992 CreateTimeTAG: 2012-10-10T15:53:57Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Well you are past the deadline for the Homework and Labs for Weeks 1-4. The two lowest scores in the HM and Labs are dropped, so you will have zeros for two homework and labs. I would start at the beginning of the videos and plow through. Good luck. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-10T16:08:41Z SecondChildTAG: **Akanksha071992**: To get the full benefit of the course, you should start at the beginning, and forget about the grade this semester. You should take it again for the grade another time. In the spring, this course will be offered **again**, and in the spring, there's a good chance MITx will offer a **proctored exam** certificate in additon to the free honor code certificate of completion that is currently offered. SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-10-10T19:53:00Z IndexTAG: 3529 TitleTAG: TA Forum Schedules, and Answers? Hi, In the course info there is a schedule of when TAs will be online to answer questions. Does that mean simply that they are browsing the discussion boards and answering as many questions as they can, during that time? Is there a way to tell which answers are coming from TAs, so I know which ones I can trust? :) thanks, Rob UserIdTAG: 468623 UserNameTAG: RobNik CreateTimeTAG: 2012-10-10T14:59:45Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: See the thread: "Access to student assistants". jelizon, a staff member, answers this question for me. FirstChildUserIdTAG: 132828 FirstChildUserNameTAG: rharris FirstChildCreateTimeTAG: 2012-10-10T18:23:26Z SecondChildTAG: Perfect, thanks. SecondChildUserIdTAG: 468623 SecondChildUserNameTAG: RobNik SecondChildCreateTimeTAG: 2012-10-10T20:48:53Z IndexTAG: 3530 TitleTAG: Getting confused with the current source I still don't get it...how do we know every time if the voltage source is supplying power to the circuit or it's charging ? I don't get it UserIdTAG: 402183 UserNameTAG: Threepwood CreateTimeTAG: 2012-10-10T14:57:17Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL/KCL example CommentableIdTAG: 6002x_S2E3_Using_KVL_KCL_and_VI_constraints NumberOfReplyTAG: 2 FirstChildTAG: I think you'll have to explain more clearly what you mean. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-10T16:32:40Z FirstChildTAG: Is the current entering the Vs greater then it's stated value? If so, it is "charging". FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-10T16:41:18Z IndexTAG: 3531 TitleTAG: VT?? it the transition voltage? how do we get it in this case, tried to look it up, anyone point me to a lecture seq i should look over? UserIdTAG: 388273 UserNameTAG: kilmarta CreateTimeTAG: 2012-10-10T14:49:45Z VoteTAG: 0 CoursewareTAG: Week 5 / Saturation Region CommentableIdTAG: 6002x_sat_rgn NumberOfReplyTAG: 1 FirstChildTAG: VT(in this course) is the MOSFET threshold voltage. It's a physical characteristic of the MOSFET. Think of the MOSFET as a switch. It's off or on state is controlled by the gate to source voltage(VGS). The VT is the voltage level that must be reached by VGS to change the state of the switch. This is a very simplistic way to describe it, though. In general, VT is the point when the current(iDS) goes from 0 to VDS(drain to source voltage)/RON. There are three regions of operation for n-Channel MOSFETs. Cutoff (or sub-threshold) where VGSVT and VDS<(VGS-VT) Saturation (or Active) where VGS>VT and VDS>(VGS-VT) Page 304 of the text has what you are looking for, although I suggest you read pages 301-305 which is all about the physical structure of the MOSFET. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-10T15:45:56Z SecondChildTAG: so we define VT for a particular mosfet, do we do that through the W/L imput? will give those parts a read thanks SecondChildUserIdTAG: 388273 SecondChildUserNameTAG: kilmarta SecondChildCreateTimeTAG: 2012-10-10T15:54:50Z SecondChildTAG: No, the W/L input controls RON. VT is controlled by something called the Body Effect. It's outside my scope of knowledge. You can read more about it here: http://en.wikipedia.org/wiki/MOSFET#Body_effect SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-10T16:05:23Z IndexTAG: 3532 TitleTAG: no flatline at saturation point when i set up this circuit in the sandbox, instead of a flatline when it becomes saturated i get a sloped line, that normal? UserIdTAG: 388273 UserNameTAG: kilmarta CreateTimeTAG: 2012-10-10T14:36:48Z VoteTAG: 0 CoursewareTAG: Week 5 / Saturation Region Model Demo CommentableIdTAG: 6002x_sat_rgn_mdl_demo NumberOfReplyTAG: 2 FirstChildTAG: ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13498810041343667.jpg FirstChildUserIdTAG: 388273 FirstChildUserNameTAG: kilmarta FirstChildCreateTimeTAG: 2012-10-10T14:57:15Z FirstChildTAG: it's a channel-length modulation effect: http://en.wikipedia.org/wiki/Channel_length_modulation the real formula for MOSFET is I=K/2*(VGS-VT)^2*(1+alpha*VDS) where alpha - channel-length modulation parameter FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-10T16:36:44Z IndexTAG: 3533 TitleTAG: transient analysis in sandbox not working works fine in a lab. but when i try set up a simple one or even an exact copy of a lab in the sand box, it doesnt work. i click tran, imput the time interval and then nothing happens. tried in chrome and firefox UserIdTAG: 388273 UserNameTAG: kilmarta CreateTimeTAG: 2012-10-10T14:18:22Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Troubleshooting NumberOfReplyTAG: 1 FirstChildTAG: simple problem in the end, the tran wont run if you have check clicked, need to reset it FirstChildUserIdTAG: 388273 FirstChildUserNameTAG: kilmarta FirstChildCreateTimeTAG: 2012-10-10T14:24:51Z SecondChildTAG: Thanks for the update! SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-10T20:51:20Z IndexTAG: 3534 TitleTAG: Homeworks Is it possible to pass the course without the homeworks, as I only started the course this week and cannot submit the homeworks. UserIdTAG: 534164 UserNameTAG: orlandong CreateTimeTAG: 2012-10-10T09:33:17Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Homework and Labs are the 30% of the total score. You may not do 2 out of 12 and still get a chance of getting that 30%. It is advisable that you complete them as they help you with your study for the Midterm and the final exam. If you start this weeks Homework and Lab, having completed all of the homework and labs to come 100% correct, you can add about 24% to your final score. But this means that your maximum score would be 94% in this case. What i mean if i understand correctly is ithat you only lost 6% of the final score, but you don't have the chance of missing another asignment anymore. :) FirstChildUserIdTAG: 304587 FirstChildUserNameTAG: Vasso FirstChildCreateTimeTAG: 2012-10-10T10:56:12Z IndexTAG: 3535 TitleTAG: Suggestion I think, that a drawing will help a lot in the explanations. ABR UserIdTAG: 309359 UserNameTAG: abarea10 CreateTimeTAG: 2012-10-10T08:59:29Z VoteTAG: 0 CoursewareTAG: Week 6 / Two-stage MOSFET amplifier CommentableIdTAG: 6002x_Two_Stage_MOSFET_amplifier NumberOfReplyTAG: 0 IndexTAG: 3536 TitleTAG: OTHER WAY? for the last two questions ( operating voltage and operating current). i solved the equation Vd = V - Vd^3 * R Vd = voltage across the strange device as i got R= 8.2 and V = 32.8 .. i plugged them in the above eqn and using my scientific calculator i solved this cubic equation ..and i got 3 roots ... i got the right answer. but i was wondering if there's any other way to solve this ????? sry for my bad english UserIdTAG: 140065 UserNameTAG: Madhav92 CreateTimeTAG: 2012-10-10T06:55:19Z VoteTAG: 0 CoursewareTAG: Week 3 / Graphical Analysis of Nonlinear Circuits Exercise CommentableIdTAG: 6002x_graphical_analysis_of_nonlinear_circuits_exercise NumberOfReplyTAG: 0 IndexTAG: 3537 TitleTAG: Help Somebody help me!!! I can't get this.... :( UserIdTAG: 268104 UserNameTAG: SaulCardenasG CreateTimeTAG: 2012-10-10T06:15:37Z VoteTAG: 0 CoursewareTAG: Week 4 / Dependent Sources Exercise 0 CommentableIdTAG: 6002x_dep_src_exsz_0 NumberOfReplyTAG: 1 FirstChildTAG: first question: vB=VI so iD=K1*VI . now be carefull: for a resistor we accept current to be positive IF it goes from higher potential (+) to lower potential (-) and here this IS NOT the case. so we take ohms low with negative iD second question we need to calculate iB first. use ohms low for that then do the same as above. hth FirstChildUserIdTAG: 319453 FirstChildUserNameTAG: leonidasGr FirstChildCreateTimeTAG: 2012-10-10T13:02:29Z SecondChildTAG: thanks!!! SecondChildUserIdTAG: 268104 SecondChildUserNameTAG: SaulCardenasG SecondChildCreateTimeTAG: 2012-10-10T14:24:19Z IndexTAG: 3538 TitleTAG: Axis How can i get "Voltage" on x-axis and "Current" on y-axis in TRAN analysis? Is it possible? Picture in Lab4 show Voltage vs Current. UserIdTAG: 268104 UserNameTAG: SaulCardenasG CreateTimeTAG: 2012-10-10T06:10:05Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Did you read the text in that lab exercise ? *"2. The voltage probe on the vtest node has its "Plot color" set to "x-axis", which asks the tool to use the values sampled by the probe as the x-coordinate when plotting. So the horizontal axis of the plot showing the transient results will be the voltage vtest instead of time"* FirstChildUserIdTAG: 90796 FirstChildUserNameTAG: ChaunceyGardiner FirstChildCreateTimeTAG: 2012-10-11T04:27:31Z SecondChildTAG: Oops thanks SecondChildUserIdTAG: 268104 SecondChildUserNameTAG: SaulCardenasG SecondChildCreateTimeTAG: 2012-10-11T19:57:57Z IndexTAG: 3539 TitleTAG: STAFF: Are the directions correct in question in S10E1. Hello In S10E1, the directions state "Express your answer in terms of the device parameters K, VT, RL, the bias voltages VS and VI, and the incremental input voltage vi". Is VS supposed to be in the equation for the answer? Thanks. UserIdTAG: 132828 UserNameTAG: rharris CreateTimeTAG: 2012-10-10T05:10:21Z VoteTAG: 0 CoursewareTAG: Week 5 / Incremental Voltage Exercise CommentableIdTAG: 6002x_inc_volt_e NumberOfReplyTAG: 1 FirstChildTAG: In some of the exercises/homeworks/labs, one or more of the parameters they specify aren't actually required in the answer. I suspect it is to try to give you one of those "Aha" moments when you find the result is independent of one or more of the parameters. However, you can be pretty sure you got it wrong if you get an answer with a parameter they didn't specify! FirstChildUserIdTAG: 141000 FirstChildUserNameTAG: OrinE FirstChildCreateTimeTAG: 2012-10-11T02:58:55Z IndexTAG: 3540 TitleTAG: Lab 5 Input voltage range I'm having trouble finding the lower end value. Probably it is my interpretation. I found the input voltage upper end linear operating range, the voltage at lower end means that this value i'm looking for is smaller or greater than the one I found ? UserIdTAG: 92699 UserNameTAG: LuKiNhA CreateTimeTAG: 2012-10-10T01:34:35Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: It's smaller than the upper end value. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-10T16:37:30Z SecondChildTAG: Thank you. SecondChildUserIdTAG: 92699 SecondChildUserNameTAG: LuKiNhA SecondChildCreateTimeTAG: 2012-10-10T18:19:45Z SecondChildTAG: hix im opposite with you, i can only find lower voltage, but can not find out upper voltage :(. help SecondChildUserIdTAG: 114913 SecondChildUserNameTAG: ngoctuan SecondChildCreateTimeTAG: 2012-10-10T18:30:54Z SecondChildTAG: Be sure you are looking at the input axis. SecondChildUserIdTAG: 393045 SecondChildUserNameTAG: SrChasJC SecondChildCreateTimeTAG: 2012-10-13T02:29:08Z IndexTAG: 3541 TitleTAG: Using small signal defeating the purpose of the amplifier If we have to reduce the signal for it fit in a linear range of the mosfet, doesn't that defeat the goal of the amplifier circuit? as we are re-amplifying the signal to it's original size. UserIdTAG: 61469 UserNameTAG: Spacedog CreateTimeTAG: 2012-10-09T23:34:04Z VoteTAG: 0 CoursewareTAG: Week 5 / Small Signal Graphically Described CommentableIdTAG: 6002x_small_sig_graph_des NumberOfReplyTAG: 1 FirstChildTAG: In a real-world applications we fit the amplifier for the signal, not the signal for the amplifier (well, sometimes we have to shift it's DC level). FirstChildUserIdTAG: 194098 FirstChildUserNameTAG: ZhekaS FirstChildCreateTimeTAG: 2012-10-10T00:36:36Z SecondChildTAG: fair enough, many thanks. SecondChildUserIdTAG: 61469 SecondChildUserNameTAG: Spacedog SecondChildCreateTimeTAG: 2012-10-10T19:16:23Z IndexTAG: 3542 TitleTAG: S9E2 Can someone tell me what the answer is for this exercise? I typed in the conditions for saturation which are listed bottom of textbook page 344, top of 345 but for some reason I am not getting a check. Vin >= Vt Vout >= Vin - Vt I am asking for the answer for a mere exercise(not a HW or Lab) Thanks so much UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-09T22:08:31Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You have the right answers, but you are not putting them in the form that the checker wants. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-09T22:53:05Z SecondChildTAG: yeah, I know that SecondChildUserIdTAG: 191220 SecondChildUserNameTAG: Breffni SecondChildCreateTimeTAG: 2012-10-10T01:34:41Z IndexTAG: 3543 TitleTAG: H5P2 Hello! My problem is that despite the fact that i have calculated correctly the algebraic expression for iDS and vOUT i cannot calculate the iDS and the vOUT solution. I have even used the Microsoft Mathematics software to solve the equations with the given numbers and the vOUT is found equal to ... something which is obviously wrong. Any idea? UserIdTAG: 214275 UserNameTAG: TheodoreGr CreateTimeTAG: 2012-10-09T21:43:08Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: [Edited by staff -- Please do not directly share answers to graded portions of the course!] FirstChildUserIdTAG: 297655 FirstChildUserNameTAG: Aljoska FirstChildCreateTimeTAG: 2012-10-09T21:56:00Z SecondChildTAG: Thanks for the immediate reply. Yes Aljoska, this is vOUT. Problem solved. VOUT was correctly calculated but a bug prevented the check button to present the solution as "correct". After several refreshes the problem was solved. SecondChildUserIdTAG: 214275 SecondChildUserNameTAG: TheodoreGr SecondChildCreateTimeTAG: 2012-10-09T22:39:55Z SecondChildTAG: You are not supposed to post answers to homework problems. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-10-09T23:01:40Z SecondChildTAG: Find solution for Vout. Next step ids=Vout/RS and its OK. SecondChildUserIdTAG: 297655 SecondChildUserNameTAG: Aljoska SecondChildCreateTimeTAG: 2012-10-09T23:05:33Z IndexTAG: 3544 TitleTAG: Lab 7 link broken The Link of equation 9.18 in Lab 7 is broken. I managed to find it but i thought of reporting that. UserIdTAG: 304587 UserNameTAG: Vasso CreateTimeTAG: 2012-10-09T20:33:21Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3545 TitleTAG: midsem can nybody tell me the syllabus of mid term exams UserIdTAG: 582342 UserNameTAG: achilees111 CreateTimeTAG: 2012-10-09T19:29:41Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: This should be in the announcement. Everything up to week 6 but no capacitors. FirstChildUserIdTAG: 151427 FirstChildUserNameTAG: RichardZ FirstChildCreateTimeTAG: 2012-10-10T14:36:24Z IndexTAG: 3546 TitleTAG: week four tutorial doubt 6002-L8-oei12-8_100b(week 4) in this tutorial video at (3.30 time) how come the Vout becomes 10 volts for Vi=0v; that must be 5V is in it??? UserIdTAG: 334613 UserNameTAG: aswinshankar CreateTimeTAG: 2012-10-09T18:59:33Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3547 TitleTAG: H5P2 - What is your solution for iDS ??? Hi, is there any one explain for me the meaning here, because we already have the expression for iDS what should we make with this again ?? ---------- UserIdTAG: 114913 UserNameTAG: ngoctuan CreateTimeTAG: 2012-10-09T18:31:33Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: yes but they want it without the vO FirstChildUserIdTAG: 310474 FirstChildUserNameTAG: aldaris565 FirstChildCreateTimeTAG: 2012-10-10T02:18:50Z SecondChildTAG: some times we have use oms law SecondChildUserIdTAG: 352647 SecondChildUserNameTAG: praveenjugge SecondChildCreateTimeTAG: 2012-10-10T17:02:42Z SecondChildTAG: thanks guys i did it, yeah a bunch of thing when we replace vOUT with VIN,RS,K and VT. SecondChildUserIdTAG: 114913 SecondChildUserNameTAG: ngoctuan SecondChildCreateTimeTAG: 2012-10-11T09:00:30Z FirstChildTAG: use ohms law FirstChildUserIdTAG: 352647 FirstChildUserNameTAG: praveenjugge FirstChildCreateTimeTAG: 2012-10-10T17:03:27Z IndexTAG: 3548 TitleTAG: H5P1 Hello everyone.I have stucked in h5p1 both 1st & 3rd ques.I couldn't find Max & min Vi. Though i have able to find Rl with the equation Vo=Vs-k/2*Rl*(Vi-Vt)^2.Where both vi and vo is 0.But how do i find vi.I have tried the equation vi-vt=Vs-k/2*Rl*(Vi-Vt)^2 But it seems wrong.Please help. UserIdTAG: 92895 UserNameTAG: shihab2555 CreateTimeTAG: 2012-10-09T17:26:17Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Part 1: VIN-VT-(VS-)>=0 Remember: (VS-) is a negative value FirstChildUserIdTAG: 66440 FirstChildUserNameTAG: lilly FirstChildCreateTimeTAG: 2012-10-09T18:36:01Z SecondChildTAG: Thanx a lot.I got it.But How do I find it in 3rd part SecondChildUserIdTAG: 92895 SecondChildUserNameTAG: shihab2555 SecondChildCreateTimeTAG: 2012-10-09T18:51:27Z SecondChildTAG: Most of us are making same mistake.We are taking Vgs=Vi & Vds=Vo which is not true due to present of offset.So we should take Vi=Vgs+Vs(-) & Vo=Vds+Vs(-) instead.Hope this will help. SecondChildUserIdTAG: 92895 SecondChildUserNameTAG: shihab2555 SecondChildCreateTimeTAG: 2012-10-11T16:04:32Z FirstChildTAG: lilly how VIN-VT-(VS-)>=0? FirstChildUserIdTAG: 278792 FirstChildUserNameTAG: sali FirstChildCreateTimeTAG: 2012-10-10T05:28:51Z SecondChildTAG: Because VGS = Vin-Vs- and mofset must be in saturation region! SecondChildUserIdTAG: 149058 SecondChildUserNameTAG: sotoroman SecondChildCreateTimeTAG: 2012-10-10T17:13:08Z SecondChildTAG: Where is it written that VGS=Vin-Vs-? SecondChildUserIdTAG: 373585 SecondChildUserNameTAG: radami1 SecondChildCreateTimeTAG: 2012-10-11T13:26:27Z FirstChildTAG: Lilly.........part 3?? FirstChildUserIdTAG: 92895 FirstChildUserNameTAG: shihab2555 FirstChildCreateTimeTAG: 2012-10-10T08:09:05Z SecondChildTAG: pls help part 3? SecondChildUserIdTAG: 276808 SecondChildUserNameTAG: DEBASMITAMAJUMDER SecondChildCreateTimeTAG: 2012-10-11T15:14:03Z SecondChildTAG: Most of us are making same mistake.We are taking Vgs=Vi & Vds=Vo which is not true due to present of offset.So we should take Vi=Vgs+Vs(-) & Vo=Vds+Vs(-) instead.Hope this will help. SecondChildUserIdTAG: 92895 SecondChildUserNameTAG: shihab2555 SecondChildCreateTimeTAG: 2012-10-11T16:05:13Z IndexTAG: 3549 TitleTAG: h5p2 what's the use of input signal variation 0.5v in this question. UserIdTAG: 582342 UserNameTAG: achilees111 CreateTimeTAG: 2012-10-09T17:17:48Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: vdd max must be vin(max)-threshold voltage thus ans shud be 5.6v but ans given is 5.1v which is when u neglect the uppr swing of .5v in input signal. FirstChildUserIdTAG: 582342 FirstChildUserNameTAG: achilees111 FirstChildCreateTimeTAG: 2012-10-09T17:29:24Z IndexTAG: 3550 TitleTAG: do best how to do gr8 in this subject give me some suggesion UserIdTAG: 563034 UserNameTAG: shubh279 CreateTimeTAG: 2012-10-09T15:28:39Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Start looking at your homework at the beginning of the week. FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-10-09T15:41:31Z IndexTAG: 3551 TitleTAG: H5P2 min. VDD Need help in finding min. VDD? Any suggestions.... UserIdTAG: 425842 UserNameTAG: rish1992 CreateTimeTAG: 2012-10-09T13:09:20Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Mosfet is in saturation. You know the equations for this condition. Also the current can be derived easily. FirstChildUserIdTAG: 380242 FirstChildUserNameTAG: a_niyath FirstChildCreateTimeTAG: 2012-10-09T14:27:54Z SecondChildTAG: but still ans is not cming can u pls tell me right ans SecondChildUserIdTAG: 582342 SecondChildUserNameTAG: achilees111 SecondChildCreateTimeTAG: 2012-10-09T17:09:55Z SecondChildTAG: vdd max must be vin(max)-threshold voltage thus ans shud be 5.6v but ans given is 5.1v which is when u neglect the uppr swing of .5v in input signal. SecondChildUserIdTAG: 582342 SecondChildUserNameTAG: achilees111 SecondChildCreateTimeTAG: 2012-10-09T17:29:03Z SecondChildTAG: I had 0.6 volts signal variation so my answer was 5 - but that doesn't make sense to me at all :( Why is that so? SecondChildUserIdTAG: 413065 SecondChildUserNameTAG: anikey SecondChildCreateTimeTAG: 2012-10-11T20:57:35Z SecondChildTAG: vdd max must be vin max. could you point me to text book to read more about that SecondChildUserIdTAG: 388273 SecondChildUserNameTAG: kilmarta SecondChildCreateTimeTAG: 2012-10-11T23:18:19Z IndexTAG: 3552 TitleTAG: How to downlod tutorial videos Is anyone able to download the tutorial videos?could someone guide me on how to download those videos?? pls do help UserIdTAG: 119534 UserNameTAG: altis CreateTimeTAG: 2012-10-09T12:48:35Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: If you use firefox, DownloadHelper will help you. In Opera you can find videos in the cache. I don't know about IE or Chrome. p.s.: sorry for my engrish. FirstChildUserIdTAG: 311784 FirstChildUserNameTAG: ilya07 FirstChildCreateTimeTAG: 2012-10-09T13:28:50Z IndexTAG: 3553 TitleTAG: H5P2 Rs It tells me that Rs is not permitted in answer. How is it possible? It does not tell me nothing about parse...etc. UserIdTAG: 244225 UserNameTAG: lerimock CreateTimeTAG: 2012-10-09T12:27:49Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi, Use RS instead. Even for Vt, you need to use VT. Equation parsing is case sensitive -- Abhi FirstChildUserIdTAG: 380242 FirstChildUserNameTAG: a_niyath FirstChildCreateTimeTAG: 2012-10-09T14:22:50Z SecondChildTAG: It's still not accepting RS. SecondChildUserIdTAG: 141376 SecondChildUserNameTAG: krtica SecondChildCreateTimeTAG: 2012-10-13T17:42:45Z IndexTAG: 3554 TitleTAG: Beside the suggested parts,should I read textbook thoroughly ? I am a college student so I don't have much time spent on the course, but suggested reading is not missed every week, which is not enough for my deep understanding of the course sometimes.Is it necessary to read the whole chapter? That means taking me a lot of time . UserIdTAG: 361137 UserNameTAG: Ericson CreateTimeTAG: 2012-10-09T11:02:27Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I don't read textbook at all, doing ok so far (done with first 5 weeks). FirstChildUserIdTAG: 189680 FirstChildUserNameTAG: vnd FirstChildCreateTimeTAG: 2012-10-09T15:01:49Z SecondChildTAG: Thank you :) SecondChildUserIdTAG: 361137 SecondChildUserNameTAG: Ericson SecondChildCreateTimeTAG: 2012-10-15T06:19:10Z FirstChildTAG: Reading and comprehending the book is a personal issue. The book is very well written and targeted to the audience with no or little experience in circuits. The introduction to the book and the first couple of lectures are extremely important because they explain the idea behind engineering in general and circuit engineering in particular. I would say that the book contains fundamental ideas about circuitry. In other words, hardly anything can be extracted from the book without loss of value to the subject covered. The above leads to the conclusion that covering suggested reading only and reading whole chapters would result in different levels of understanding of the subject. FirstChildUserIdTAG: 252722 FirstChildUserNameTAG: vaboro FirstChildCreateTimeTAG: 2012-10-09T15:45:24Z SecondChildTAG: Very helpful!Thank you very much! SecondChildUserIdTAG: 361137 SecondChildUserNameTAG: Ericson SecondChildCreateTimeTAG: 2012-10-15T06:21:54Z IndexTAG: 3555 TitleTAG: About Homework H5P2 Hi, Please tell me how to write good formula. I tried all my best, but the feedback after checking are: Can not parse.... as a formula or ....does not appear in the formula. Help me R. May UserIdTAG: 373916 UserNameTAG: larou CreateTimeTAG: 2012-10-09T08:48:33Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3556 TitleTAG: Mosfet I-V charectiristics Dear sir in the demo of mosfet characteristics can you please let me know how can we see different parellel lines (curves) of the same device. Or have you connected three different mosfets with different VgS values UserIdTAG: 352647 UserNameTAG: praveenjugge CreateTimeTAG: 2012-10-09T06:37:33Z VoteTAG: 0 CoursewareTAG: Week 5 / Saturation Region CommentableIdTAG: 6002x_sat_rgn NumberOfReplyTAG: 2 FirstChildTAG: http://www.falstad.com/circuit/#%24+1+5.0E-6+9.78399845368213+47+5.0+50%0Ag+528+160+528+208+0%0AR+400+64+336+64+0+3+1000.0+6.0+4.0+0.0+0.5%0Af+336+144+400+144+0+1.5%0Aw+400+64+400+128+0%0Av+144+368+144+304+0+2+100.0+1.0+1.0+0.0+0.9%0Av+144+304+144+240+0+2+100.0+0.5+0.5+0.0+0.8%0Ag+144+368+144+384+0%0Av+144+112+144+48+0+2+100.0+0.5+0.5+0.0+0.5%0Av+144+176+144+112+0+2+100.0+0.5+0.5+0.0+0.6%0Av+144+240+144+176+0+2+100.0+0.5+0.5+0.0+0.7000000000000001%0Aw+144+48+240+48+0%0Aw+240+48+240+144+0%0Aw+240+144+336+144+0%0Aw+400+160+528+160+0%0Ao+2+64+0+98+15.05123000691892+0.30102460013837856+0+-1%0Ao+10+64+0+35+9.353610478917778+9.765625E-55+1+-1%0A FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-09T07:51:06Z FirstChildTAG: i believe you get those when we change vgs FirstChildUserIdTAG: 388273 FirstChildUserNameTAG: kilmarta FirstChildCreateTimeTAG: 2012-10-10T14:51:06Z IndexTAG: 3557 TitleTAG: h5p2, how to calculate Vdd hey everyone, what must i consider to find Vdd? i tried Vds=Vdd-Vout>Vgs-Vt, where Vgs is Vin-Vout... it got me nowhere, help pls UserIdTAG: 285945 UserNameTAG: lindalapiso CreateTimeTAG: 2012-10-09T05:52:52Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi lindalapiso, In order to find minimum VDD, one should check both conditions of saturation: 1. vDS ≥ vGS-VT 2. VDD-vOUT ≥ vIN-vOUT-VT FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-10-09T12:12:32Z SecondChildTAG: so simple... SecondChildUserIdTAG: 301962 SecondChildUserNameTAG: labrinim SecondChildCreateTimeTAG: 2012-10-09T15:26:56Z SecondChildTAG: I not able to see the hint! SecondChildUserIdTAG: 299251 SecondChildUserNameTAG: TheRedBlackOne SecondChildCreateTimeTAG: 2012-10-10T11:35:30Z SecondChildTAG: LOL it was indeed so simple! Thanks SecondChildUserIdTAG: 299251 SecondChildUserNameTAG: TheRedBlackOne SecondChildCreateTimeTAG: 2012-10-10T14:12:18Z SecondChildTAG: I am also thinking the same way. But by using the second equation, I got vdd>=vIN-vT since vOUT can be cancelled and got vdd>=3.8 which is wrong. I need help please.... SecondChildUserIdTAG: 157273 SecondChildUserNameTAG: ongchihang SecondChildCreateTimeTAG: 2012-10-11T15:16:21Z SecondChildTAG: I cannot understand your hint. I am solving this problem for about two hour. Give me some help. T_T SecondChildUserIdTAG: 134925 SecondChildUserNameTAG: Hanbyul SecondChildCreateTimeTAG: 2012-10-11T15:57:17Z SecondChildTAG: I got the answer! T_T You have to think with previous little problems... then it's easy. I differentiated useless equation continuously... SecondChildUserIdTAG: 134925 SecondChildUserNameTAG: Hanbyul SecondChildCreateTimeTAG: 2012-10-11T16:40:42Z SecondChildTAG: How I can get needed value of VIN? SecondChildUserIdTAG: 269368 SecondChildUserNameTAG: StAlex SecondChildCreateTimeTAG: 2012-10-13T11:09:56Z IndexTAG: 3558 TitleTAG: H5P2 homework DOESN'T CHECK When I click check for H5P2 it issues a message Invalid Input VIN VIN not permitted, if I click again it says Invalid Input VIN VOUT not permitted and once more come Invalid Input IDS not permitted in answer. Does anyone have the same problem? UserIdTAG: 11075 UserNameTAG: roncada CreateTimeTAG: 2012-10-09T05:51:55Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: is vIN and vOUT ... FirstChildUserIdTAG: 310474 FirstChildUserNameTAG: aldaris565 FirstChildCreateTimeTAG: 2012-10-09T17:05:02Z FirstChildTAG: The answers are case sensitive. ![enter image description here][1] That's iDS, K, vIN, vOUT, and VT. [1]: https://edxuploads.s3.amazonaws.com/13498037281343635.png FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-09T17:29:28Z IndexTAG: 3559 TitleTAG: Gain? What it 'gain' a measure of? increase of signal voltage? UserIdTAG: 61469 UserNameTAG: Spacedog CreateTimeTAG: 2012-10-09T00:42:07Z VoteTAG: 0 CoursewareTAG: Week 5 / MOSFET Amplifier CommentableIdTAG: 6002x_mosfet_amp NumberOfReplyTAG: 1 FirstChildTAG: Gain could be the relation between the output and input of a system, let's say if you have a black box and you apply two volts at the input and you obtain three volts at the output, that means the gain is 1.5 (output/input -> 3/2). The gain could be unitless or a combination (volts/amps, etc) FirstChildUserIdTAG: 55932 FirstChildUserNameTAG: otgonz FirstChildCreateTimeTAG: 2012-10-09T01:35:03Z SecondChildTAG: thanks for clarifying this :D SecondChildUserIdTAG: 61469 SecondChildUserNameTAG: Spacedog SecondChildCreateTimeTAG: 2012-10-15T00:01:48Z IndexTAG: 3560 TitleTAG: Really informative Lab 7 I've put together numerous power supplies like the one in the second part of lab 7. It never occurred to me how large the peak current is relative to the average current. This is definitely something I will think about in the future. UserIdTAG: 406202 UserNameTAG: skyhawk CreateTimeTAG: 2012-10-09T00:09:46Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: The power company must hate it as more and more of the load in their network probably looks like this. FirstChildUserIdTAG: 142857 FirstChildUserNameTAG: viking2 FirstChildCreateTimeTAG: 2012-10-09T01:17:05Z IndexTAG: 3561 TitleTAG: H5P2 question 3, what equation should I solve? From the page 349 I get vOUT = RS*(K*(vIN-VT-vOUT)^2)/2 at the beginning I try to solve this by hand but finaly I gone crazy an I try it with a software and I got the same result. What I`m doing wrong? Please a hint or some advise!!! UserIdTAG: 299251 UserNameTAG: TheRedBlackOne CreateTimeTAG: 2012-10-08T21:06:55Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: Do not forget that you find "Vout". To find the "Ids" to make another small step. What should be a step? See the answer to the second question H5P2. FirstChildUserIdTAG: 365162 FirstChildUserNameTAG: AndrewKhvalko FirstChildCreateTimeTAG: 2012-10-08T23:10:22Z FirstChildTAG: Hello, TheRedBlackOne! It looks like you are doing all right :) However, here are two tips: 1. You can simplify equation with $$v_{IN}-V_T=V$$ $$\frac{1}{K*R_S}=T$$ and substitute them after you have solved the equation. This might minimize chance of algebraic errors. 2. You are asked about iDS, and if you would prefer to find vOUT first, then don't forget to make a final step! FirstChildUserIdTAG: 261503 FirstChildUserNameTAG: EugenyL FirstChildCreateTimeTAG: 2012-10-08T21:55:48Z SecondChildTAG: Thank you so much! Yes, my answere was right but I forgot that the question was answering by iDS not vOUT, It was the final step. SecondChildUserIdTAG: 299251 SecondChildUserNameTAG: TheRedBlackOne SecondChildCreateTimeTAG: 2012-10-08T23:05:31Z SecondChildTAG: @TheRedBlackOne @EugenyL May I know if the answers are ridiculously long and complicated? Because mine is! Please do advise on how I should go about this question. I had both parts 1 and 2 correct. Thank you very much! SecondChildUserIdTAG: 499268 SecondChildUserNameTAG: ruinoah SecondChildCreateTimeTAG: 2012-10-09T10:10:35Z SecondChildTAG: Hi, ruinoah! You are right, the answer even may exceed the textbox length. If you have wrote the correct quadratic equation and solved it, but the system reports error, maybe you can try a symbolic math programm, for example, WxMaxima. SecondChildUserIdTAG: 261503 SecondChildUserNameTAG: EugenyL SecondChildCreateTimeTAG: 2012-10-09T13:16:24Z SecondChildTAG: Thank you, EugenyL. These step's really simplificates calculations! SecondChildUserIdTAG: 269368 SecondChildUserNameTAG: StAlex SecondChildCreateTimeTAG: 2012-10-11T12:56:36Z SecondChildTAG: hi all! i have got a ridiculously long equation and i know it is correct as it satisfies the condition that iDS=0 when vIN=VT and it also looks ok in the math format after i wrote it out in the text box. But still getting the wrong mark. Could someone please help me out with this? Thanks! SecondChildUserIdTAG: 222911 SecondChildUserNameTAG: bhaswardg SecondChildCreateTimeTAG: 2012-10-11T16:03:12Z FirstChildTAG: i have got the right expression for iDS in H5P2,but it is not accepting my answer Invalid input: Could not parse eqn as a formula !! help!! FirstChildUserIdTAG: 502784 FirstChildUserNameTAG: milindm FirstChildCreateTimeTAG: 2012-10-09T02:10:40Z SecondChildTAG: mistake in parenthesis SecondChildUserIdTAG: 18073 SecondChildUserNameTAG: gburkhart SecondChildCreateTimeTAG: 2012-10-09T02:32:24Z SecondChildTAG: With some complex equations I had problems as well where I couldn't spot the mistake. It might help to rebuild the equation from scratch in small steps and press "check" several times in between to verify that you don't have parsing errors. Or copy the equation into a text editor that does parentheses highlighting (e.g. the scratch buffer in emacs). SecondChildUserIdTAG: 148015 SecondChildUserNameTAG: ralfh SecondChildCreateTimeTAG: 2012-10-09T07:55:38Z FirstChildTAG: It tells me that Rs is not permitted in answer. How is it possible? It does not tell me nothing about parse...etc. FirstChildUserIdTAG: 244225 FirstChildUserNameTAG: lerimock FirstChildCreateTimeTAG: 2012-10-09T11:15:39Z SecondChildTAG: Because is RS!! SecondChildUserIdTAG: 299251 SecondChildUserNameTAG: TheRedBlackOne SecondChildCreateTimeTAG: 2012-10-10T10:28:43Z FirstChildTAG: since vout = Id * Rs, now put this value in vout to find the answer FirstChildUserIdTAG: 126911 FirstChildUserNameTAG: sidhant7 FirstChildCreateTimeTAG: 2012-10-12T18:37:19Z IndexTAG: 3562 TitleTAG: H4P2 mismatch with simulation Hello, I tried to solve VO for RL=2K with multisim. But the result is different. Right value should be 5.007488. Any suggestions? ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13497283571343674.jpg UserIdTAG: 389333 UserNameTAG: juergen CreateTimeTAG: 2012-10-08T20:34:27Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3563 TitleTAG: Transconductance? Okey, I'm totally lost now. I don't get how they got Transconductance. What is **BIAS** current IDS? I was thinking we needed to follow on the node equation in the end of a previous lecture, vo/RS-iDS=0, so we substitute iDS for K((VI-VO)-VT)*(vi-vo) and solve but in the end you end up with unknown large signal model VI-VO. Can somebody explain step by step how to get Transconductance? Thank you very much in advance! UserIdTAG: 326409 UserNameTAG: EugeneZ CreateTimeTAG: 2012-10-08T18:53:44Z VoteTAG: 0 CoursewareTAG: Week 6 / Source-follower small-signal exercise CommentableIdTAG: 6002x_small_signal_source_follower NumberOfReplyTAG: 1 FirstChildTAG: $gm = K(V_{GS}-V_T) = K(v_{IN}-v_{OUT}-V_T)$ (1) $I_{DS} = v_{out}/R_S \Rightarrow v_{out}=I_{DS}R_S$ (2) Inject (2) into (1) and u'r done FirstChildUserIdTAG: 190618 FirstChildUserNameTAG: Kirbabaev FirstChildCreateTimeTAG: 2012-10-09T11:48:15Z IndexTAG: 3564 TitleTAG: tutorial videos? I don't have access to weekly tutorials due to youtube outreach and I didn't find direct links to download them. how can I have them? UserIdTAG: 355370 UserNameTAG: omaransari CreateTimeTAG: 2012-10-08T17:49:50Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: brother, just try using tor browser.......a lot solves from that! FirstChildUserIdTAG: 169784 FirstChildUserNameTAG: thewiredbear FirstChildCreateTimeTAG: 2012-10-08T18:58:38Z SecondChildTAG: Just want to remind - if something blocked by government authorities, bypassing this blockage can be unlawful, and TOR itself can be banned in some countries as a result of government's totalitarian stupid paranoia. Be careful. SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-08T23:25:19Z SecondChildTAG: thanks a lot bro.. SecondChildUserIdTAG: 355370 SecondChildUserNameTAG: omaransari SecondChildCreateTimeTAG: 2012-10-09T19:35:49Z IndexTAG: 3565 TitleTAG: H5P2 I write the correct expression for iDS, with capital letters where it must, but iDS is not accepted by the page. How do i have to write the letters of iDS? Thanks a lot. UserIdTAG: 244225 UserNameTAG: lerimock CreateTimeTAG: 2012-10-08T15:33:04Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: enter in quadratic formula form. FirstChildUserIdTAG: 99628 FirstChildUserNameTAG: Pranjal16 FirstChildCreateTimeTAG: 2012-10-08T18:38:45Z SecondChildTAG: I know. What i'm asking is about the way to write "iDS", because i had problems with vOUT and the rest, and i tried with capital letters and several ways to write, and finally got it but not with iDS. SecondChildUserIdTAG: 244225 SecondChildUserNameTAG: lerimock SecondChildCreateTimeTAG: 2012-10-08T18:55:16Z FirstChildTAG: I write this iDS = (K(vIN-vOUT-VT)^2)/2 and says that iDS is not permitted in answer, and i don't understand, because what he writes below what i write is perfectly correct. ¡¡¡¡HEEEEEEEEEEEELP!!!! FirstChildUserIdTAG: 244225 FirstChildUserNameTAG: lerimock FirstChildCreateTimeTAG: 2012-10-08T22:03:58Z SecondChildTAG: leave off the "iDS =" part SecondChildUserIdTAG: 18073 SecondChildUserNameTAG: gburkhart SecondChildCreateTimeTAG: 2012-10-08T23:01:45Z SecondChildTAG: I have the same problem, and deleting iDS is not a solution, because then it says: "Invalid input: Could not parse 'K/2 (vIN-vOUT-VT)^2' as a formula". SecondChildUserIdTAG: 241379 SecondChildUserNameTAG: CrisAzc SecondChildCreateTimeTAG: 2012-10-10T20:38:06Z SecondChildTAG: same problem here SecondChildUserIdTAG: 319905 SecondChildUserNameTAG: andrea490 SecondChildCreateTimeTAG: 2012-10-12T14:16:29Z SecondChildTAG: ok i got it, i wasn't entering capital K SecondChildUserIdTAG: 319905 SecondChildUserNameTAG: andrea490 SecondChildCreateTimeTAG: 2012-10-12T14:17:45Z IndexTAG: 3566 TitleTAG: h5p2 question 3 im slightly confused at how we make the equations can someone tell me what pages to read thanks UserIdTAG: 436749 UserNameTAG: waynebrown CreateTimeTAG: 2012-10-08T11:35:53Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: page 349 for H5P2! FirstChildUserIdTAG: 122513 FirstChildUserNameTAG: Mona77 FirstChildCreateTimeTAG: 2012-10-10T10:30:10Z SecondChildTAG: We should write an equation in terms of ***V***out? SecondChildUserIdTAG: 269368 SecondChildUserNameTAG: StAlex SecondChildCreateTimeTAG: 2012-10-10T20:11:38Z IndexTAG: 3567 TitleTAG: My homework check buttons are still active Even in Massachusetts it's October 8th and my check buttons for HW4 are still active. What's going on? UserIdTAG: 416659 UserNameTAG: Jonas3000 CreateTimeTAG: 2012-10-08T11:33:53Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Try refreshing your screen, the check buttons should change to 'Show Answer'. FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-10-08T15:14:39Z IndexTAG: 3568 TitleTAG: ampere amperes is spelled with a small 'a' and not capital 'A' (like Ampere). However, when you write in abbreviated form, you use 'A', for instance, 1A UserIdTAG: 497378 UserNameTAG: zaherva CreateTimeTAG: 2012-10-08T11:28:25Z VoteTAG: 0 CoursewareTAG: Week 2 / Is it linear? CommentableIdTAG: 6002x_is_it_linear NumberOfReplyTAG: 1 FirstChildTAG: it depends what you mean - [André-Marie Ampère][1] or [ampere][2] ;D [1]: http://en.wikipedia.org/wiki/Andr%C3%A9-Marie_Amp%C3%A8re [2]: http://en.wikipedia.org/wiki/Ampere FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-08T17:06:44Z IndexTAG: 3569 TitleTAG: lab 4 characteristic problem ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13496747051343672.png UserIdTAG: 431224 UserNameTAG: erachopra10 CreateTimeTAG: 2012-10-08T05:38:31Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: where is this circuit going wrong..plzz tell. FirstChildUserIdTAG: 431224 FirstChildUserNameTAG: erachopra10 FirstChildCreateTimeTAG: 2012-10-08T05:39:07Z SecondChildTAG: current sense going to locate between source-ground SecondChildUserIdTAG: 151026 SecondChildUserNameTAG: atelco2008 SecondChildCreateTimeTAG: 2012-10-08T05:57:02Z SecondChildTAG: You want to measure current going *into* the MOSFET. Flip the chevron. SecondChildUserIdTAG: 194509 SecondChildUserNameTAG: blackcompe SecondChildCreateTimeTAG: 2012-10-08T06:08:54Z IndexTAG: 3570 TitleTAG: Lab 4 Vt i) My circuit got green checkmark and found Ron. But for determining Vt I found values around .000071 and .000046 for Ids at Vgs 3 and 2.5, respectively. Used my ti-89 solver and get 2.7 for Vt ii) for the above 71 and 46, I took those values at the end of the scale (3) which should be well into saturation but my saturation region was not very flat I know this is simple so there must be some blind spot Ideas? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-08T02:44:41Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You can change your offset value a bit so that you can have your scale a little bit grater. :) FirstChildUserIdTAG: 304587 FirstChildUserNameTAG: Vasso FirstChildCreateTimeTAG: 2012-10-08T07:42:18Z IndexTAG: 3571 TitleTAG: HW 4 Missing something in both H4P2 and H4P3, Part B. Graduated college 32 years ago, never used calculus since (we didn't have PCs then either). I've determined Vo for both loads and the minimum RL for the circuit with the Zener but cannot figure out how to draw the load line or get vo at all. I also calculated RN for part B, Dependent Source, but cannot get IN. What the heck is A? A constant? What is u? I short Port B as stated in the book, which mean Au and R3 are shorted but my answer for IN is still incorrect. PS - Haven't understood any of the practice problems either. UserIdTAG: 375500 UserNameTAG: Brej7665 CreateTimeTAG: 2012-10-08T00:02:32Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: u is the voltage at the node between R1 and R2 relative to ground. Here's what to do: short the output and calculate the current through R1 and R2. Call that i1. Next calculate u. Now you can solve for the current through R3. Call it i3. Add i1 and i3, and you have Norton current. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-08T01:46:24Z FirstChildTAG: Hello Brej, Glad you're still into learning dude. Sorry if this response is coming a little late but I will try my best to help you out. Let's begin with the load line in H4P2. 1st: Find the Thev. equivalent of the rest of the circuit from the perspective of the diode. (in other words, replace the diode with a port and find the voltage across the port and the resistance seen from the port.) 2nd: Find the slope and intercept for the load line. The Thev. values we just obtained are going to be used to find this slope and intercept. To see how, click 'Week 4 Tutorials' and then watch 'Load Line Experimental Demo'. 3rd: write the equation of the load line. We are going to be writing the current i as a function of voltage, v. It will look something like... i = (?)v + (?). 4th: Check your signs. Note that in the problem the zener diode is 'backwards', i.e. vd = -vo. If you did not take this into account, no worries. Just note that the load 'i' you calculated = -i for the actual value. 5th: Find the intersection between the i-v relation of the zener and the load line. Take your load line equation and set it equal to the appropriate linear region on the zener i-v graph. To find the appopriate region, graph your load line and see which part of the zener i-v curve it seems like it will actually intersect with. Solving for the intersection will give you the operating point. But again, remember that vd = -vo. (if you are confused why this works, think of it this way. The circuit is setting some limits on what the voltage across the element can be (the load line). But the diode is also setting limits on what its voltage can be (the i-v charachteristic). So the intersection gives us where the voltage satisfies both parameters, a place we call 'the operating point'). As well, note that wolframalpha.com is your friend for solving and graphing. I hope this will be somewhat benefical! Regards. FirstChildUserIdTAG: 153604 FirstChildUserNameTAG: autohost FirstChildCreateTimeTAG: 2012-10-08T02:05:59Z SecondChildTAG: for any input i.e VI or VI+Vi, the output voltage would be the same.hence we should get null Vo ,but its giving the wrong ans...pls help me... SecondChildUserIdTAG: 443660 SecondChildUserNameTAG: itsme71 SecondChildCreateTimeTAG: 2012-10-08T04:25:08Z SecondChildTAG: Hello itsme, Sorry for the inconvenience, but I am afraid I am having a bit of difficulty interpreting your question. Would you mind elaborating on what question part you are stuck on? SecondChildUserIdTAG: 153604 SecondChildUserNameTAG: autohost SecondChildCreateTimeTAG: 2012-10-09T00:59:28Z FirstChildTAG: Using the tips above, I found IN but I'm still working the Zener.... If the Zener becomes a port, I get RTH = RIN//RL and VTH is the same as Vo from the circuit before the Zener was added. This gives me a slope very close to Zero .... FirstChildUserIdTAG: 375500 FirstChildUserNameTAG: Brej7665 FirstChildCreateTimeTAG: 2012-10-08T12:26:24Z FirstChildTAG: Glad to see you are back in education!! Hope autohost's post helped you! :) FirstChildUserIdTAG: 304587 FirstChildUserNameTAG: Vasso FirstChildCreateTimeTAG: 2012-10-08T09:13:48Z FirstChildTAG: Vasso, Autohost & Skyhawk; Just getting moving this morning as I have the day off, Columbus Day is a Federal Holiday in the US. Thanks for the responses. I didn't have time last night to work on the problems anymore but your efforts weren't wasted. I still like to know what I missed, how to do the problem correctly. I work as a Power Engineer for the US Government and almost never use any of the knowledge from 30+ years ago anymore. I've actually understood much of the work up to this point and even remembered how to take the 1st derivative! But I've never analysed circuits like this, even when I was in college. Out of all the work in Lesson 4, I was unable to answer only 3 of the questions, not bad for an old man! I got all of Lesson's 1 & 2 but missed some on 3. I did finally see the "u" in the circuit late last night before your responses were posted (I can be a little slow or dense, my wife says "Stubborn"). I then tried a few more calculations but was still unsuccessful. I then had to stop as getting to bed at Midnight, getting up at 5 AM, working full-time and taking care of my family leaves little time for extra pleasures such as studying and learning.... Drew Brejda FirstChildUserIdTAG: 375500 FirstChildUserNameTAG: Brej7665 FirstChildCreateTimeTAG: 2012-10-08T11:51:41Z SecondChildTAG: Hello Mr. Brejda, Glad to be of service. Please feel free to let me know if you'd care for additional help. I will try my best to oblige. Regards. SecondChildUserIdTAG: 153604 SecondChildUserNameTAG: autohost SecondChildCreateTimeTAG: 2012-10-09T01:04:17Z SecondChildTAG: Mr. Brejda, You are serving as a source of encouragement to me. For someone like you who graduated 30yrs ago to still solve these problems, is something I hold dear. I missed out solvings for HW4 & Lab4 due to illness but thanks to people like you, am back on track as the best 10 are what will be used. Going through the solutions worked by edx team, brings ack my Algebra alive. All the best friend. Ugo SecondChildUserIdTAG: 329879 SecondChildUserNameTAG: Ugochukwu SecondChildCreateTimeTAG: 2012-10-09T09:29:58Z IndexTAG: 3572 TitleTAG: S12V2: there seems to be more?!?! The rising edge of signal B is determined by a capacitor, ok. Now i don`t understand: 1. and the falling slope; it seems that there don`t act any capacitor!?!? 2. why is there no capacitor acting at signal C - principally at its raising edge? UserIdTAG: 388524 UserNameTAG: dasbinich CreateTimeTAG: 2012-10-07T23:18:37Z VoteTAG: 0 CoursewareTAG: Week 6 / Motivation for capacitor CommentableIdTAG: 6002x_motivation_for_capacitor NumberOfReplyTAG: 2 FirstChildTAG: 1. it depends on the pulldown resister - smaller resistance - faster gate discharge - you just don't see the curve because it's very steep 2. there could be different reasons - transistor B could have much bigger GS capacitance, or power supplied to gate much lesser - probably there is an additional series resistor from output A to input C - for education purpose ;) If you want I can try to reproduce this case in this circuit simulator: http://www.falstad.com/circuit/ FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-08T01:32:22Z SecondChildTAG: http://www.falstad.com/circuit/#%24+1+5.0E-6+0.32112705431535615+40+5.0+50%0Af+512+464+560+464+0+1.5%0Af+688+464+736+464+0+1.5%0AR+512+464+464+464+0+2+500.0+2.5+2.5+0.0+0.5%0Ar+560+448+560+368+0+2000.0%0Ar+736+448+736+368+0+2000.0%0AR+736+368+736+288+0+0+40.0+5.0+0.0+0.0+0.5%0Aw+560+368+736+368+0%0Ag+560+544+560+608+0%0Ag+736+528+736+592+0%0Aw+688+448+688+464+0%0AO+736+448+800+448+0%0Aw+736+480+736+528+0%0Ar+560+480+560+544+0+200.0%0Ac+736+528+688+528+0+2.0E-8+-4.99989857485769%0Aw+688+464+688+528+0%0Ar+560+448+640+448+0+1000.0%0Aw+640+448+688+448+0%0Ao+2+1+0+38+5.0+9.765625E-5+0+-1%0Ao+16+1+0+38+5.0+0.003125+0+-1%0Ao+10+1+0+38+5.0+9.765625E-5+0+-1%0A play with 1k and 200 resistors and capacitor SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-08T02:39:24Z FirstChildTAG: YakovO, thanks for the simulation, now I understand: 1. the pull-up resistor is much higher than the Ron of the MOSFET: so it takes time to **charge** the GS-capacitor of the right MOSFET via pullup-resistor of the left MOSFET; but the **discharge** of the GS-capacitor of the right MOSFET via Ron of the left MOSFET is much faster. 2. at the output of the right MOSFET isn´t any capacitor (only a little scope´s capacitor) --> charge and discharge are very quickly. FirstChildUserIdTAG: 388524 FirstChildUserNameTAG: dasbinich FirstChildCreateTimeTAG: 2012-10-09T01:03:30Z SecondChildTAG: in that circuit simulator MOSFET represents as an ideal model - without any capacitance between terminals, so I added it to represent more physical model for this case - of course the real capacitance much smaller (ones-hundreds picofarads) - I increased it because of restriction of simulator's time resolution to show effect on low frequencies. There is no capacitor on first transistor because it directly connected to ideal voltage source (no internal resistance), so it will change nothing - there is no voltage drop from ideal voltage source even to charge capacitor - connect one and try. SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-09T02:43:09Z IndexTAG: 3573 TitleTAG: Week 5 HW what does this mean? Invalid input: Could not parse ... as a formula this is what I get when I enter a formula into Week 5 HW UserIdTAG: 18073 UserNameTAG: gburkhart CreateTimeTAG: 2012-10-07T22:27:07Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: This happens when one opens parenthesis but forgets to close it or vice versa closes parenthesis that was not opened. Happened to me many times:) FirstChildUserIdTAG: 252722 FirstChildUserNameTAG: vaboro FirstChildCreateTimeTAG: 2012-10-07T23:36:44Z SecondChildTAG: OK - thanks SecondChildUserIdTAG: 18073 SecondChildUserNameTAG: gburkhart SecondChildCreateTimeTAG: 2012-10-08T01:19:43Z SecondChildTAG: It may also happen if one forgets to mention operation, for example, the multiplication sign. Just happened to me recently. SecondChildUserIdTAG: 252722 SecondChildUserNameTAG: vaboro SecondChildCreateTimeTAG: 2012-10-08T03:43:39Z IndexTAG: 3574 TitleTAG: Dont understand the third question I understand now how to get the first and second question, but I dont understand the third question, can someone help me ?? UserIdTAG: 154373 UserNameTAG: JReyes87 CreateTimeTAG: 2012-10-07T22:06:16Z VoteTAG: 0 CoursewareTAG: Week 4 / Linearization Exercise 1 CommentableIdTAG: 6002x_linearization_ex_1 NumberOfReplyTAG: 1 FirstChildTAG: The incremental resistance is the ratio of the change in voltage due to an infinitesimal change in current. Mathematically, this can be expressed as: $r_d = \frac{1}{\frac{di}{dt}}$. From a circuits point of view, it is the resistance that will best model the non-linear component when deviation from the DC operating point is small. FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-10-07T22:19:55Z IndexTAG: 3575 TitleTAG: I need to find vd to make the 2nd and 3rd questions, is there anybody that can help me? I substituted the value for I0 and VT, has VT=0.026V, i made the node equation for a VI fo 1V, but i tried the values for vd, has 1, 0.01, 0.99, and even -1.99 and -0.99 and they diverge a lot to make a check and guess on this... Is there any other way??? Help me please... UserIdTAG: 317836 UserNameTAG: VitorRodrigues CreateTimeTAG: 2012-10-07T21:45:52Z VoteTAG: 0 CoursewareTAG: Week 4 / Linearization Exercise 2 CommentableIdTAG: 6002x_linearization_ex_2 NumberOfReplyTAG: 1 FirstChildTAG: ups missed me something here, forgot the R valuea, i'll try to do it again... FirstChildUserIdTAG: 317836 FirstChildUserNameTAG: VitorRodrigues FirstChildCreateTimeTAG: 2012-10-07T21:49:29Z IndexTAG: 3576 TitleTAG: H4P3b and node method When I'm trying to use node method to solve H4P3b, I get for node "u" the following equation: (u-V0)/R1 + (u-Au)/(R2+R3) = 0, which gives u=5.71V and completely incorrect results. So I cannot understand what am I doing wrong here. Similar approach worked perfectly for H4P3a. Any hints? UserIdTAG: 192703 UserNameTAG: voffch CreateTimeTAG: 2012-10-07T18:04:31Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: 5.71 V is correct value. (with no load on port B). FirstChildUserIdTAG: 508733 FirstChildUserNameTAG: burdal1 FirstChildCreateTimeTAG: 2012-10-07T18:23:54Z SecondChildTAG: burdal1 is right, 5.71V is correct, but it's not the end of your calculation. If, say, you wanted to get I as V/R, it's the node adjacent to u you need to calculate (i.e. one above R3 and Au) SecondChildUserIdTAG: 502508 SecondChildUserNameTAG: amaher SecondChildCreateTimeTAG: 2012-10-07T18:31:23Z SecondChildTAG: OK, thank you all. I see that it's a valid way of solving this problem, if I found the proper value for u. Nevertheless, when I wanted to calculate the current flowing into R1 as IN (because it's the same for all elements in the loop, right?), so I took (V0 - 5.71) as a voltage across R1, divided it by value of R1 and... couldn't get a right answer. SecondChildUserIdTAG: 192703 SecondChildUserNameTAG: voffch SecondChildCreateTimeTAG: 2012-10-07T18:43:21Z SecondChildTAG: Current is _not_ the same for all elements, because you calculate I_N for a closed circuit network and your current splits over components in parallel. SecondChildUserIdTAG: 502508 SecondChildUserNameTAG: amaher SecondChildCreateTimeTAG: 2012-10-07T18:49:20Z SecondChildTAG: ok, it's not the end. Then you can calculate (with same method) voltage on port B and current trought shorted port B. SecondChildUserIdTAG: 508733 SecondChildUserNameTAG: burdal1 SecondChildCreateTimeTAG: 2012-10-07T18:50:49Z SecondChildTAG: current flowing into R1 is NOT IN. IN is current trough shorted port B SecondChildUserIdTAG: 508733 SecondChildUserNameTAG: burdal1 SecondChildCreateTimeTAG: 2012-10-07T18:56:15Z SecondChildTAG: Aha! That was my problem. Solved successfully by calculating equivalent voltage and dividing it by R_N. Thanks! SecondChildUserIdTAG: 192703 SecondChildUserNameTAG: voffch SecondChildCreateTimeTAG: 2012-10-07T18:57:20Z SecondChildTAG: How did u get R_N? I have tried several times, but it's always incorrect... SecondChildUserIdTAG: 61423 SecondChildUserNameTAG: libo654 SecondChildCreateTimeTAG: 2012-10-07T20:38:14Z SecondChildTAG: To get R_N, short the independent sources and connect a voltage source, V, across the port, and thus the node voltage at the + terminal is V. Find I going into that node and you got R (i.e. R_N) across the port. You can give V a concrete value, although it's not absolutely necessary. Once you have R_N, compute V_TH using Nodal Analysis, and you got I_N. SecondChildUserIdTAG: 194509 SecondChildUserNameTAG: blackcompe SecondChildCreateTimeTAG: 2012-10-07T20:46:33Z SecondChildTAG: R_N not clear how to solve with V dependent source in the circuit SecondChildUserIdTAG: 251287 SecondChildUserNameTAG: JBusquets SecondChildCreateTimeTAG: 2012-10-07T22:35:11Z IndexTAG: 3577 TitleTAG: H4P2 b and H4P3 b (SOLVED) (SOLVED) 1. H4P2 b : the main problem is: we dont understand what the graph of zener meaning: ---> we have an expression between Id and Vd here: something like Id = slopeVd + b. the slope and b can take from graph. Apply this expression with node and we have only Vd in --> have Vd. hint V0 is opposite Vd.(KVL for loop) 1. H4P3 actually it is really easy: IN = contanst*U then we find U by other node Apply u for IN. That all. Wat two days :(( ----------------------------------------- Help me!! after take hours almost 2 days reading in Discussion threads, it is not clear how to solve it right: 1. H4p2 b how to find Id because we dont have Rd. of course i know Vd = - V0 2. H4p3 b i calculate u = 5/4 that means Au = 5/2 but after make a short circuit in open port to calculate IN the result is false. Help plz Tuan UserIdTAG: 114913 UserNameTAG: ngoctuan CreateTimeTAG: 2012-10-07T17:34:48Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I'm having the same problem - I can't solve part B FirstChildUserIdTAG: 306244 FirstChildUserNameTAG: kevinsysum FirstChildCreateTimeTAG: 2012-10-07T17:44:05Z SecondChildTAG: Please can you help me solve H4P3 part A Rth and Vth SecondChildUserIdTAG: 178115 SecondChildUserNameTAG: nehamakhija SecondChildCreateTimeTAG: 2012-10-07T17:47:40Z SecondChildTAG: please can you help me SecondChildUserIdTAG: 178115 SecondChildUserNameTAG: nehamakhija SecondChildCreateTimeTAG: 2012-10-07T17:50:01Z SecondChildTAG: For H4P3, Check to verify that the values for your variables are correct. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-07T17:50:43Z SecondChildTAG: Give us some scenarios, tell us what you are thinking, how far have you gotten? SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-07T17:54:12Z SecondChildTAG: Kevin Please can you help me SecondChildUserIdTAG: 178115 SecondChildUserNameTAG: nehamakhija SecondChildCreateTimeTAG: 2012-10-07T18:00:31Z SecondChildTAG: So I Turn off the current source , leaving the dependent source, I don't know how to proceed after this SecondChildUserIdTAG: 178115 SecondChildUserNameTAG: nehamakhija SecondChildCreateTimeTAG: 2012-10-07T18:08:16Z SecondChildTAG: And I compute Rth =(R1//R2)+Z but the answer is wrong SecondChildUserIdTAG: 178115 SecondChildUserNameTAG: nehamakhija SecondChildCreateTimeTAG: 2012-10-07T18:10:18Z SecondChildTAG: rharris , did you understand my scenario. SecondChildUserIdTAG: 178115 SecondChildUserNameTAG: nehamakhija SecondChildCreateTimeTAG: 2012-10-07T18:19:50Z SecondChildTAG: please help i have tried for 2 hours SecondChildUserIdTAG: 178115 SecondChildUserNameTAG: nehamakhija SecondChildCreateTimeTAG: 2012-10-07T18:22:21Z SecondChildTAG: Where is R3? SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-07T18:24:32Z SecondChildTAG: Oh, which problem are you working on? SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-07T18:25:19Z SecondChildTAG: h4p3 part a, r3 is resistance of the dependent source SecondChildUserIdTAG: 178115 SecondChildUserNameTAG: nehamakhija SecondChildCreateTimeTAG: 2012-10-07T18:28:49Z SecondChildTAG: i tried 2 days :((!! help SecondChildUserIdTAG: 114913 SecondChildUserNameTAG: ngoctuan SecondChildCreateTimeTAG: 2012-10-07T18:29:33Z SecondChildTAG: Looking at your equation, I believe you are working on H4P3, part A SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-07T18:31:24Z SecondChildTAG: yes, please guide me SecondChildUserIdTAG: 178115 SecondChildUserNameTAG: nehamakhija SecondChildCreateTimeTAG: 2012-10-07T18:34:16Z SecondChildTAG: Your equation for Rth isn't quite right. Look at where Z is positioned with respect to R1 and R2, with the 2 amp source set to zero. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-07T18:34:31Z SecondChildTAG: to find Rth and Vth SecondChildUserIdTAG: 178115 SecondChildUserNameTAG: nehamakhija SecondChildCreateTimeTAG: 2012-10-07T18:34:39Z SecondChildTAG: I am a bit confused SecondChildUserIdTAG: 178115 SecondChildUserNameTAG: nehamakhija SecondChildCreateTimeTAG: 2012-10-07T18:36:44Z SecondChildTAG: can you please elaborate SecondChildUserIdTAG: 178115 SecondChildUserNameTAG: nehamakhija SecondChildCreateTimeTAG: 2012-10-07T18:37:58Z SecondChildTAG: Look at your equation "Rth =(R1//R2)+Z". Is it set up right? SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-07T18:39:05Z SecondChildTAG: ok I got Rth , how do i compute Vth SecondChildUserIdTAG: 178115 SecondChildUserNameTAG: nehamakhija SecondChildCreateTimeTAG: 2012-10-07T18:40:13Z SecondChildTAG: Cool. You have to find Voc at port A, which is Vth, using the node method. Realize, Voc is 2i volts above the node to the left of the dependent source, with 2i being the voltage across the dependent source. Also, remember that i is the current through the R1. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-07T18:47:23Z FirstChildTAG: **H4P2** : check for the voltage drop across RL after removing the zener diode i.e. make the zener diode port open circuit. Then reconnect it. HINT: we know, that zener conducts if the voltage across it is equal to or greater than breakdown voltage. you can find the breakdown voltage from graph. If the voltage across it is lesser then zener will be "OFF". the voltage at which the zener will be "ON" is the VD. Hope it helps. FirstChildUserIdTAG: 204213 FirstChildUserNameTAG: ratneshray FirstChildCreateTimeTAG: 2012-10-07T17:50:46Z SecondChildTAG: uhm, and what about v0 ? the small signal ? SecondChildUserIdTAG: 271772 SecondChildUserNameTAG: Ondrej SecondChildCreateTimeTAG: 2012-10-07T18:17:46Z SecondChildTAG: the voltage without zener diod is 7.33 v, that mean zener will conduct when we attach it to circuit again right? how to find b in load line y = mx +b with m = 1 A/V and x = 7.33 ? do i think correct way? SecondChildUserIdTAG: 114913 SecondChildUserNameTAG: ngoctuan SecondChildCreateTimeTAG: 2012-10-07T18:23:47Z SecondChildTAG: Ondrej: If you have solved the other part then you have changed the zener with a voltage source and a resistor, right. For the small signal remember to turn off all the DC sources including the one you added and calculate the response to the small signal voltage source. SecondChildUserIdTAG: 392100 SecondChildUserNameTAG: glenng SecondChildCreateTimeTAG: 2012-10-07T19:59:34Z IndexTAG: 3578 TitleTAG: H4P2 minimum load resistance in Zener case? Hi..I have completed everything except this one question..can someone please help me with this?.. UserIdTAG: 173147 UserNameTAG: cruiser_rahit CreateTimeTAG: 2012-10-07T17:07:57Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: can be claculated by thinking it as a volatge divider whose output voltage should be 5v an rl is unknown FirstChildUserIdTAG: 111320 FirstChildUserNameTAG: jhalife FirstChildCreateTimeTAG: 2012-10-07T17:12:06Z FirstChildTAG: *Stabilization effect* can't work when voltage on zener diode is smaller than the *Zener voltage* (5 V) FirstChildUserIdTAG: 508733 FirstChildUserNameTAG: burdal1 FirstChildCreateTimeTAG: 2012-10-07T17:27:09Z FirstChildTAG: How did you calculate the rest. I solved only the first and the third one.. please help me.. FirstChildUserIdTAG: 309722 FirstChildUserNameTAG: kaushikraghavan1992 FirstChildCreateTimeTAG: 2012-10-07T17:42:29Z SecondChildTAG: Consider only the 50mV small signal source..now since zener is in reverse bias it can be replaced with resistance whose value can be obtained from graph (1 A/V)..not just solve for output voltage..the answers seems right.. SecondChildUserIdTAG: 173147 SecondChildUserNameTAG: cruiser_rahit SecondChildCreateTimeTAG: 2012-10-07T18:23:58Z FirstChildTAG: You have to calculate the maximun current for Rin. And then supose that this current will pass throught RL for a zener voltage. Then you obtain the RL value FirstChildUserIdTAG: 113249 FirstChildUserNameTAG: EGuarch FirstChildCreateTimeTAG: 2012-10-07T18:18:00Z IndexTAG: 3579 TitleTAG: Typo list for S7V1 - S7V2 S7V1 0:12 : "been learning about" --> "be learning about" 0:55 : "a lump circuit" --> "the lumped circuit" 1:57 : "Within a nonlinear" --> "Within the nonlinear" 2:42 : "network of the sort" --> "network of this sort" 2:51 : "within a nonlinear" --> "within the nonlinear" 3:27 : "what you've done" --> "what we've done" 3:28 : "at the analytical" --> "at analytical" 3:36 : "piecewise linear methods" --> "piecewise linear method" S7V2 0:08 "why small signal" --> "why the small signal" 2:16 : "is related to" --> "related to" 2:21 : "and it can find" --> "and I can find" 2:44 : "verses" --> "versus" 3:09 : "given and input" --> "given an input" 4:30 : "to time equal" --> "to time t equal" 4:56 : "look at how the voltage" --> "look at how, as the voltage" 5:17 : "along to be vD curve" --> "along the vD curve" UserIdTAG: 280731 UserNameTAG: QuantumCaffeine CreateTimeTAG: 2012-10-07T16:36:31Z VoteTAG: 0 CoursewareTAG: Week 4 / Introduction to Incremental Analysis CommentableIdTAG: 6002x_intro_to_inc_analysis NumberOfReplyTAG: 0 IndexTAG: 3580 TitleTAG: norton resistance how to find Norton resistance in week 4 homework problem?.plz reply UserIdTAG: 219204 UserNameTAG: vsriram CreateTimeTAG: 2012-10-07T16:16:49Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: think about input as only U FirstChildUserIdTAG: 111320 FirstChildUserNameTAG: jhalife FirstChildCreateTimeTAG: 2012-10-07T17:13:22Z SecondChildTAG: exclude R1 and one more resistance fin out which SecondChildUserIdTAG: 111320 SecondChildUserNameTAG: jhalife SecondChildCreateTimeTAG: 2012-10-07T17:13:44Z FirstChildTAG: I found Rno but unable to find Ino.. How should i solve? FirstChildUserIdTAG: 309722 FirstChildUserNameTAG: kaushikraghavan1992 FirstChildCreateTimeTAG: 2012-10-07T17:47:06Z IndexTAG: 3581 TitleTAG: varying dc source i want to ask can we also vary the dc voltage UserIdTAG: 181793 UserNameTAG: mitrahul CreateTimeTAG: 2012-10-07T16:16:10Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You realize that since you make no mention of what Lab, Homework or exercise you are referring to, we have no idea what you are talking about, right? FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-07T16:40:03Z SecondChildTAG: which dc are you talking about? where you wanna change? SecondChildUserIdTAG: 111320 SecondChildUserNameTAG: jhalife SecondChildCreateTimeTAG: 2012-10-07T16:46:31Z SecondChildTAG: i mean that dc mean a constant voltage ,but in incremental analysis we are saying varying dc,that the problem SecondChildUserIdTAG: 181793 SecondChildUserNameTAG: mitrahul SecondChildCreateTimeTAG: 2012-10-08T12:19:07Z IndexTAG: 3582 TitleTAG: H4P2 SECOND PART I am totally lost in the second part of the zaner diode ............ help me plss UserIdTAG: 395966 UserNameTAG: sambo007 CreateTimeTAG: 2012-10-07T15:49:25Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: its pretty simple.. for small signal consider it as resistance parallel to Rl and solve.. for large dc voltage consider it as constant voltage source of its voltage equal to breakdown & go ahead FirstChildUserIdTAG: 111320 FirstChildUserNameTAG: jhalife FirstChildCreateTimeTAG: 2012-10-07T16:38:04Z SecondChildTAG: refer zener as shunt regulator on google :-) SecondChildUserIdTAG: 111320 SecondChildUserNameTAG: jhalife SecondChildCreateTimeTAG: 2012-10-07T16:49:50Z FirstChildTAG: I think this will help you :) good luck https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5070be20fabaf62b0000004d FirstChildUserIdTAG: 45307 FirstChildUserNameTAG: josejimenez2 FirstChildCreateTimeTAG: 2012-10-07T16:46:08Z FirstChildTAG: If i consider it as a resistance. then what kinda values should i assign to it?? FirstChildUserIdTAG: 309722 FirstChildUserNameTAG: kaushikraghavan1992 FirstChildCreateTimeTAG: 2012-10-07T18:25:06Z IndexTAG: 3583 TitleTAG: Lab 4 I have designed the circuit correctly and getting the i-v transient response plot. The problem is when I calculate the resistance value (Ron) (by the given formula vds/ids) by noting the values of current for a particular voltage from the graph in the triode region, it displays an in correct answer. Next, I have correctly calculated the value of VT, but when i plugged back the values for calculating K in the given equation, it displays an incorrect answer. I want to know is there something that i am doing wrong and if yes then how to correct it. UserIdTAG: 254355 UserNameTAG: waqasbinabbas CreateTimeTAG: 2012-10-07T14:53:43Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Problem solved: Actually i forgot to change the value of W/L from 2 to 1. FirstChildUserIdTAG: 254355 FirstChildUserNameTAG: waqasbinabbas FirstChildCreateTimeTAG: 2012-10-07T15:23:33Z FirstChildTAG: we have to find RON at Vgs =3v and vds=1V also Ids is in microamps! how do u got Vt, i m getting it wrong plz help FirstChildUserIdTAG: 357089 FirstChildUserNameTAG: KBM FirstChildCreateTimeTAG: 2012-10-07T15:14:54Z FirstChildTAG: KBM, To find VT you have to use the IDS1/IDS2 equation provided. Just measure the IDS1 and IDS2 in saturation region for VGS= 3 and 2.5 V and the substitute those value in the IDS ratio equation. You should be able to solve for VT FirstChildUserIdTAG: 175706 FirstChildUserNameTAG: rwskim FirstChildCreateTimeTAG: 2012-10-07T16:00:58Z IndexTAG: 3584 TitleTAG: As i =V^3/1 i consider from the above Equation that R "of the Device" = 1 hence, solve with current divider, it gives me a very closed answer but still not the same. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-07T14:27:08Z VoteTAG: 0 CoursewareTAG: Week 4 / Graphs Exercise CommentableIdTAG: 6002x_graphs_exercise NumberOfReplyTAG: 1 FirstChildTAG: If the device is non-linear then there isn't a single resistance to model the device. You can calculate an incremental resistance that will depend on the operation point, and it won't be constant. That means, for example, that if the device is operating at 2V, the resistance to model it (if variations are small) will be different from the resistance to model it if operating at 3V FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-10-07T22:32:57Z IndexTAG: 3585 TitleTAG: H4P3 Can anyone give the solution for this problem ?? UserIdTAG: 336001 UserNameTAG: syd_buet12 CreateTimeTAG: 2012-10-07T14:24:57Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: If they gave you the solution, they would be violating the honor code. However, there have been some excellent posts on this topic. Here's a good recent one: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/507191da7722361f000000a4 FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-07T16:25:14Z IndexTAG: 3586 TitleTAG: H4p3 part b pls Help help?? I have tried the various methods suggested in the discussions but couldnt get the answer ?? pls elobrate .. UserIdTAG: 346960 UserNameTAG: priya91 CreateTimeTAG: 2012-10-07T13:37:51Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: somebody pls help FirstChildUserIdTAG: 346960 FirstChildUserNameTAG: priya91 FirstChildCreateTimeTAG: 2012-10-07T15:25:02Z FirstChildTAG: Find this H4P3 B HELp In FirstChildUserIdTAG: 192114 FirstChildUserNameTAG: chuba FirstChildCreateTimeTAG: 2012-10-07T14:14:13Z SecondChildTAG: but i am not getting correct ans ..help?? SecondChildUserIdTAG: 346960 SecondChildUserNameTAG: priya91 SecondChildCreateTimeTAG: 2012-10-07T14:59:38Z SecondChildTAG: I too have the same problem... Could anybody enlighten me in finding In.. plz.. SecondChildUserIdTAG: 209363 SecondChildUserNameTAG: phaneendraaa SecondChildCreateTimeTAG: 2012-10-07T15:01:06Z SecondChildTAG: I spent 5 hours on this problem yesterday and I finally got the answers. In the last 3 circuits involving dependent sources in which I have analyzed, the resistance of the dependent source has played a role in calculating Rth and Rn. It is either given in the problem statement, as with the last 2 problems we have solved, or it has to be calculated with the voltage across and the current through the dependent device. In solving this problem, everything I have calculated for u and Voc at port B were correct, so when I considered the resistance of the dependent source I got the answer. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-07T15:34:07Z SecondChildTAG: what is the resistance of dependent source?? is it 'A' which is given in the problem SecondChildUserIdTAG: 346960 SecondChildUserNameTAG: priya91 SecondChildCreateTimeTAG: 2012-10-07T15:59:47Z SecondChildTAG: No, I tried that. It doesn't have the units stated in ohms. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-07T16:22:47Z IndexTAG: 3587 TitleTAG: H5P1 Zero-Offset Amplifier Hi guys, I was able to find the minimum value of input voltage VIN in volts for the MOSFET to be operating in saturation region (first part) And I tried evything to find the Rl value and the VIN value in volts such that the MOSFET will remain in the saturation region!!!!!!! HELPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP UserIdTAG: 297960 UserNameTAG: sebseb95 CreateTimeTAG: 2012-10-07T13:04:34Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I guess you found your VIN value by writing KVL for -VIN+VGS+VS-=0. Write another KVL for VDS value. Use those values in MOSFET IDS formula to solve the equation for RL by zeroing VIN and VO as written in the requirements. Once you have RL value, solve the same equation to find VIN. FirstChildUserIdTAG: 16265 FirstChildUserNameTAG: serge_korolev FirstChildCreateTimeTAG: 2012-10-07T15:21:28Z SecondChildTAG: Used all of this. Found right answers for first and second questions, but still wrong third. SecondChildUserIdTAG: 352757 SecondChildUserNameTAG: Kerbyco SecondChildCreateTimeTAG: 2012-10-07T17:17:28Z SecondChildTAG: I still can't calculate the RL and VIN .... It will make me crazy.... SecondChildUserIdTAG: 297960 SecondChildUserNameTAG: sebseb95 SecondChildCreateTimeTAG: 2012-10-07T17:40:53Z SecondChildTAG: re: Kerbyco ditto I followed the procedure described in the lecture to find VI but can't get a green check mark SecondChildUserIdTAG: 18073 SecondChildUserNameTAG: gburkhart SecondChildCreateTimeTAG: 2012-10-07T22:16:22Z SecondChildTAG: A got the Rl, but I also can't calculate VIN SecondChildUserIdTAG: 190057 SecondChildUserNameTAG: Anastasia00435 SecondChildCreateTimeTAG: 2012-10-08T19:07:27Z SecondChildTAG: You can take a look at this [Post][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_mosfet_amp/threads/507a1d6dd894f12300000039 SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-14T21:19:37Z IndexTAG: 3588 TitleTAG: lab4 i have designed the circuit. but to get the VI characteristic for different value of VGS in a single graph what should i do? should i make more no.circuit? i have read the lab section for this but can't understood. please help. UserIdTAG: 114864 UserNameTAG: Vikaash CreateTimeTAG: 2012-10-07T13:04:19Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You just duplicate your first one with different constant DC values on the same simulator. FirstChildUserIdTAG: 172304 FirstChildUserNameTAG: Demohunter FirstChildCreateTimeTAG: 2012-10-07T13:11:44Z SecondChildTAG: how? SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-10-07T13:12:17Z SecondChildTAG: Ctrl + C and Ctrl + V? And change DC values on the source. SecondChildUserIdTAG: 172304 SecondChildUserNameTAG: Demohunter SecondChildCreateTimeTAG: 2012-10-07T13:13:03Z SecondChildTAG: I asked a question in another topic. Please, answer my question if you know the answer to my question. SecondChildUserIdTAG: 172304 SecondChildUserNameTAG: Demohunter SecondChildCreateTimeTAG: 2012-10-07T13:14:06Z SecondChildTAG: thanx @demohunter and in which topic u have asked question? SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-10-07T13:56:14Z SecondChildTAG: I got it now. Thanks anyway. @Vikaash. SecondChildUserIdTAG: 172304 SecondChildUserNameTAG: Demohunter SecondChildCreateTimeTAG: 2012-10-07T14:04:08Z IndexTAG: 3589 TitleTAG: H4P2 HELLPP common guys UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-07T12:43:44Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3590 TitleTAG: H4P1 How can i find incremental resistance? UserIdTAG: 360053 UserNameTAG: Ohedul CreateTimeTAG: 2012-10-07T12:42:39Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Just look here http://forum.allaboutcircuits.com/showthread.php?t=75112 Q2 and Q4 require differential calculus. To get the required knowledge just check out the Khan academy http://www.youtube.com/watch?v=ANyVpMS3HL4&feature=BFa&list=EC19E79A0638C8D449 These videos will do you well. Remember, incremental resistance is the "derivative" of the slope of the curve at the point 6V. Just punch the equation of the weird element into wolfram alpha and your derivative is your incremental resistance! FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-10-07T13:04:04Z IndexTAG: 3591 TitleTAG: R is not permited (e1-V0)/R1 + e1/R2 + (e1-e2)/R3 = 0 what is wrong UserIdTAG: 580208 UserNameTAG: shivamyadav10 CreateTimeTAG: 2012-10-07T11:35:50Z VoteTAG: 0 CoursewareTAG: Week 1 / Node analysis practice, part 3 CommentableIdTAG: 6002x_L2Node2 NumberOfReplyTAG: 1 FirstChildTAG: You are calculating at node e1, but the problem asks you to calculate the currents at e2 FirstChildUserIdTAG: 600559 FirstChildUserNameTAG: SergioFP FirstChildCreateTimeTAG: 2012-10-15T17:21:47Z IndexTAG: 3592 TitleTAG: submission of previous homeworks hey frns, i've recently registered on the course and i,ve missed the deadlines for submission of previous homeworks. plz tell me if ther is ny way of submitting them. UserIdTAG: 582342 UserNameTAG: achilees111 CreateTimeTAG: 2012-10-07T11:25:31Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: nope..derez no way that you can submit the previous homeworks. FirstChildUserIdTAG: 433368 FirstChildUserNameTAG: SurbhiMahajan FirstChildCreateTimeTAG: 2012-10-07T12:48:00Z IndexTAG: 3593 TitleTAG: currents!!! i got all answers right except my directions of currents. UserIdTAG: 541174 UserNameTAG: HASITAKAJA CreateTimeTAG: 2012-10-07T11:06:35Z VoteTAG: 0 CoursewareTAG: Week 1 / Associated reference directions CommentableIdTAG: 6002x_S2E2_Associated_Reference_Directions NumberOfReplyTAG: 0 IndexTAG: 3594 TitleTAG: H4P3 B HElp In I have trouble with H4P3 B . I used KVL metod to find I(CURRENT) which I used to find U=Vo-R1*I KVL Vo=IR1+IR2+IR3+aU=IR1+IR2+IR3+a(Vo-R1*I)? and I get i=-1. What next step I must do? UserIdTAG: 192114 UserNameTAG: chuba CreateTimeTAG: 2012-10-07T10:59:11Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: I found Rn. How can Ifind In? FirstChildUserIdTAG: 192114 FirstChildUserNameTAG: chuba FirstChildCreateTimeTAG: 2012-10-07T11:07:09Z SecondChildTAG: How did you found Rn SecondChildUserIdTAG: 106816 SecondChildUserNameTAG: Laith SecondChildCreateTimeTAG: 2012-10-07T11:15:16Z SecondChildTAG: when you find au. You can used I=u/r. And you find R of dependent source. Next step will like in A part. SecondChildUserIdTAG: 192114 SecondChildUserNameTAG: chuba SecondChildCreateTimeTAG: 2012-10-07T11:20:05Z SecondChildTAG: but isn't it with minus sign SecondChildUserIdTAG: 106816 SecondChildUserNameTAG: Laith SecondChildCreateTimeTAG: 2012-10-07T11:40:12Z SecondChildTAG: what with mines? SecondChildUserIdTAG: 192114 SecondChildUserNameTAG: chuba SecondChildCreateTimeTAG: 2012-10-07T12:07:42Z SecondChildTAG: Hi friends RN can be found by considering input as only U.. while calculating RN forget VO & R1 as U is the input SecondChildUserIdTAG: 111320 SecondChildUserNameTAG: jhalife SecondChildCreateTimeTAG: 2012-10-07T12:07:53Z SecondChildTAG: Do you know how find IN???????? SecondChildUserIdTAG: 192114 SecondChildUserNameTAG: chuba SecondChildCreateTimeTAG: 2012-10-07T12:36:08Z SecondChildTAG: Calculate the short circuit current. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-07T13:06:47Z SecondChildTAG: i have calculated In how to find Rn SecondChildUserIdTAG: 219204 SecondChildUserNameTAG: vsriram SecondChildCreateTimeTAG: 2012-10-07T16:19:27Z FirstChildTAG: ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/134961618110428.jpg Something like this? FirstChildUserIdTAG: 192114 FirstChildUserNameTAG: chuba FirstChildCreateTimeTAG: 2012-10-07T13:23:43Z SecondChildTAG: For a short circuit v out is zero. Calculate the current through R1 and R2. Next calculate u, then find current through R3. Add and you are done. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-07T13:43:02Z SecondChildTAG: ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13496176411343631.jpg something like this? SecondChildUserIdTAG: 192114 SecondChildUserNameTAG: chuba SecondChildCreateTimeTAG: 2012-10-07T13:47:46Z SecondChildTAG: What U next calculate? SecondChildUserIdTAG: 192114 SecondChildUserNameTAG: chuba SecondChildCreateTimeTAG: 2012-10-07T13:52:42Z SecondChildTAG: That's step 1. Then calculate u. Finally find current through R3. Add SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-07T13:54:16Z SecondChildTAG: if right u we calculate U=V-R1*I(short circuit ) yes? SecondChildUserIdTAG: 192114 SecondChildUserNameTAG: chuba SecondChildCreateTimeTAG: 2012-10-07T14:00:39Z SecondChildTAG: ДЯкую_) thanks) I found it))))) SecondChildUserIdTAG: 192114 SecondChildUserNameTAG: chuba SecondChildCreateTimeTAG: 2012-10-07T14:05:51Z SecondChildTAG: how do i find current thru r3 SecondChildUserIdTAG: 375565 SecondChildUserNameTAG: SAILESHSMART SecondChildCreateTimeTAG: 2012-10-07T15:10:44Z SecondChildTAG: pls guys pls time is running i got rth value but i am not getting In so pls tell me how to find out In? SecondChildUserIdTAG: 372330 SecondChildUserNameTAG: masters91 SecondChildCreateTimeTAG: 2012-10-07T18:13:00Z SecondChildTAG: ok i got it thank u SecondChildUserIdTAG: 372330 SecondChildUserNameTAG: masters91 SecondChildCreateTimeTAG: 2012-10-07T18:23:36Z FirstChildTAG: I used here : (V0-V)/R1+(V0-Vp)/R2=0 (Vp-V0)/R2+(Vp-aU)/R3+In=0 U=V-IR1=V-(V0-V0) am I right ? FirstChildUserIdTAG: 192114 FirstChildUserNameTAG: chuba FirstChildCreateTimeTAG: 2012-10-07T13:29:54Z IndexTAG: 3595 TitleTAG: H4P3: DEPENDENT SOURCE CIRCUIT Part b i am trying to find IN but not getting the correct answer. i am making the port short and applying KVL in both mesh and also using relation u=5-R1(current in mesh 1) for dependent source but still not getting the correct one. Is i am doing wrong?? UserIdTAG: 114864 UserNameTAG: Vikaash CreateTimeTAG: 2012-10-07T10:51:36Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Did you get Rn? FirstChildUserIdTAG: 192114 FirstChildUserNameTAG: chuba FirstChildCreateTimeTAG: 2012-10-07T11:13:28Z SecondChildTAG: I have Rn but I am lost on how to find In SecondChildUserIdTAG: 238005 SecondChildUserNameTAG: isisbocardo SecondChildCreateTimeTAG: 2012-10-07T23:51:55Z IndexTAG: 3596 TitleTAG: H4P2 I am trying to find out the minimum value of RL by VIN * RL/(RL+RIN) ≥ 5V where vi = 25mA and Rin = 1K and get a value of -1005.025 ohm and i get wrong answer why??? UserIdTAG: 336001 UserNameTAG: syd_buet12 CreateTimeTAG: 2012-10-07T10:10:47Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: ur formula is correct but ur values are wrong. FirstChildUserIdTAG: 433368 FirstChildUserNameTAG: SurbhiMahajan FirstChildCreateTimeTAG: 2012-10-07T11:00:38Z SecondChildTAG: Vi=0.035 Try this value I think Its in the example too clear SecondChildUserIdTAG: 106816 SecondChildUserNameTAG: Laith SecondChildCreateTimeTAG: 2012-10-07T11:23:52Z SecondChildTAG: I tried but still cant get it right dont know why can you tell me where my values are wrong SecondChildUserIdTAG: 336001 SecondChildUserNameTAG: syd_buet12 SecondChildCreateTimeTAG: 2012-10-07T12:23:34Z IndexTAG: 3597 TitleTAG: Transiant analysis how to change transiant analysis to voltage on x-axis and current on y-axis for lab 4? UserIdTAG: 459071 UserNameTAG: akshaykalaskar CreateTimeTAG: 2012-10-07T09:27:17Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Just double-click the voltage probe and set it's colour to "x-axis". Than double-click the current probe and sat it's colour to whatever you want. Than you perform the analysis. FirstChildUserIdTAG: 326787 FirstChildUserNameTAG: Angstrem FirstChildCreateTimeTAG: 2012-10-07T09:35:49Z SecondChildTAG: thanks ... SecondChildUserIdTAG: 459071 SecondChildUserNameTAG: akshaykalaskar SecondChildCreateTimeTAG: 2012-10-07T09:40:54Z IndexTAG: 3598 TitleTAG: lab 5 the last question asks us to find the largest input amplitude..does it refer to the amplitude of the small signal sinusoidal waveform or the total input?if total input,then we have already found the answer in the 2nd question of lab 5 ie input voltage upper end of linear operating range.. what exactly should i do?someone please help me!!! UserIdTAG: 156585 UserNameTAG: AdiRanga CreateTimeTAG: 2012-10-07T09:17:06Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: you have linear operating range (from first part) - peak-to-peak voltage of small signal should be in this range. Hint: amplitude=peak-to-peak/2 ;) FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-07T09:32:09Z IndexTAG: 3599 TitleTAG: cant solve h4p2 at all:(:(:(.....feels like quitting:( cant solve h4p4 at all...my head is bursting now after sitting over for 3 days:(:(....feels like quitting..i have never been so much depressed:( UserIdTAG: 276808 UserNameTAG: DEBASMITAMAJUMDER CreateTimeTAG: 2012-10-07T08:22:38Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: h4p2 hints https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5070be20fabaf62b0000004d FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-07T08:42:27Z FirstChildTAG: If you're referring to problem 4 part b, you can get the Norton Resistance by shorting the independent voltage source and adding a voltage source (V_0) across the port. R_N = V_0/I_0. Using Node Analysis find I_0. Once you get the R_N, come back to the original circuit and calculate V_TH using Node Analysis. Then you have I_N = V_TH/R_N. I ended up with 3 KCL equations in my analysis. FirstChildUserIdTAG: 194509 FirstChildUserNameTAG: blackcompe FirstChildCreateTimeTAG: 2012-10-07T09:38:34Z SecondChildTAG: if you short out output port the I_N through this will be obvious - no need for KCL ;) SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-07T10:04:39Z IndexTAG: 3600 TitleTAG: Hard algebraic expression I can write the algebraic, but I don't know how to solve the algebraic. Use KVL, 5=2*10*(1-e^(-VA/5)+VA. Could somebody give some tools (software, etc) to me? Huge thanks! UserIdTAG: 146995 UserNameTAG: ZYJ CreateTimeTAG: 2012-10-07T08:05:34Z VoteTAG: 0 CoursewareTAG: Week 4 / Linearization Exercise 1 CommentableIdTAG: 6002x_linearization_ex_1 NumberOfReplyTAG: 3 FirstChildTAG: No additional software is needed. All you have to do is linearise the exponential function in the expression. You just write the equation for the tangential to the operation point (2 first terms of Tailor's series) and substitute that linear equation to the KCL. FirstChildUserIdTAG: 326787 FirstChildUserNameTAG: Angstrem FirstChildCreateTimeTAG: 2012-10-07T08:22:07Z FirstChildTAG: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_linearization_ex_1/threads/50710dee3749032b00000052 or just google it: [y=2*10*(1-e^(-x/5))+x-5][1] [1]: http://www.google.com/search?hl=en&site=&source=hp&q=y=2*10*%281-e%5E%28-x/5%29%29%2bx-5 FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-07T09:07:52Z FirstChildTAG: [Wolfram][1] will give you the roots. Look under the section "Real Root." [1]: http://www.wolframalpha.com/input/?i=2*10*%281-e%5E%28-x/5%29%29%2bx-5 FirstChildUserIdTAG: 194509 FirstChildUserNameTAG: blackcompe FirstChildCreateTimeTAG: 2012-10-07T09:22:24Z IndexTAG: 3601 TitleTAG: finding the resistance linearizing this question, well if i have 1.088Volts for Va at VI=5.0V and and increment of 0.2376 volts va, if i sum those voltages va+va'=1.088+0.2376=1.3256V, and using the expression above, ia=10(1-e^(-1.3256/5))=2.3288A well 2.3288*2%=0.046577, it is wrong why???? UserIdTAG: 317836 UserNameTAG: VitorRodrigues CreateTimeTAG: 2012-10-07T06:40:31Z VoteTAG: 0 CoursewareTAG: Week 4 / Linearization Exercise 1 CommentableIdTAG: 6002x_linearization_ex_1 NumberOfReplyTAG: 1 FirstChildTAG: you should use this IV to find constant, for slope you need derivative FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-07T09:25:11Z IndexTAG: 3602 TitleTAG: Homework4 problem 2 Can someone help me with Homework 4 problem 3 the Zener diode, I not sure how to approach it. PLEASE AND THAN YOU!! UserIdTAG: 435197 UserNameTAG: RichmondRichter CreateTimeTAG: 2012-10-07T06:34:16Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: how to do home work? and how to learn on this website?? FirstChildUserIdTAG: 577267 FirstChildUserNameTAG: jaimish FirstChildCreateTimeTAG: 2012-10-07T06:40:37Z SecondChildTAG: no i finished homework 4 the only thing I didn't do is the Zener diode problem... Any help? SecondChildUserIdTAG: 435197 SecondChildUserNameTAG: RichmondRichter SecondChildCreateTimeTAG: 2012-10-07T07:12:05Z FirstChildTAG: @RichmondRichter ur post is confusing . your problem is where ?? to understand the problem or to solve the problem FirstChildUserIdTAG: 395966 FirstChildUserNameTAG: sambo007 FirstChildCreateTimeTAG: 2012-10-07T06:54:41Z SecondChildTAG: I need help with problem 3 the zener Diode? sorry for the confusion, Im just typing fast due to frustration. SecondChildUserIdTAG: 435197 SecondChildUserNameTAG: RichmondRichter SecondChildCreateTimeTAG: 2012-10-07T07:48:21Z IndexTAG: 3603 TitleTAG: lab-4 lost in lab-4. coldnt make out what they are trying to say.... should i connect dc supply accross MOSFET?or ac supply?? UserIdTAG: 395966 UserNameTAG: sambo007 CreateTimeTAG: 2012-10-07T06:25:34Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: hint: both ;) FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-07T08:11:37Z FirstChildTAG: You connect DC to the gate in order to provide it with a constant VGS. And you connect AC, triangular wave as VDS voltage in order to observe vi relationship. FirstChildUserIdTAG: 326787 FirstChildUserNameTAG: Angstrem FirstChildCreateTimeTAG: 2012-10-07T08:25:32Z SecondChildTAG: I did the same, but the graphic is not correct. Why? And others? How to use the diode? SecondChildUserIdTAG: 371617 SecondChildUserNameTAG: rainbow12 SecondChildCreateTimeTAG: 2012-10-07T11:01:05Z SecondChildTAG: there is no need to use the diode SecondChildUserIdTAG: 131726 SecondChildUserNameTAG: chemiboy11 SecondChildCreateTimeTAG: 2012-10-07T13:49:46Z FirstChildTAG: I did it as you said. But the graphic is not correct. How to use the diode? FirstChildUserIdTAG: 371617 FirstChildUserNameTAG: rainbow12 FirstChildCreateTimeTAG: 2012-10-07T10:35:49Z SecondChildTAG: there is no need to use the diode SecondChildUserIdTAG: 131726 SecondChildUserNameTAG: chemiboy11 SecondChildCreateTimeTAG: 2012-10-07T13:50:10Z IndexTAG: 3604 TitleTAG: h4p3 help me with part A.NOT GETTING THE ANSWER UserIdTAG: 433368 UserNameTAG: SurbhiMahajan CreateTimeTAG: 2012-10-07T05:37:47Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Search the forum, click on the magnifying glass and type h4p3. FirstChildUserIdTAG: 392100 FirstChildUserNameTAG: glenng FirstChildCreateTimeTAG: 2012-10-07T05:52:37Z SecondChildTAG: Got it..:-) SecondChildUserIdTAG: 433368 SecondChildUserNameTAG: SurbhiMahajan SecondChildCreateTimeTAG: 2012-10-07T06:03:09Z IndexTAG: 3605 TitleTAG: Slope?? In the Taylor expansion How??? d(f(vD)) / d(vD) |vD=VD gives the slope of VD and ID UserIdTAG: 87035 UserNameTAG: Ritu CreateTimeTAG: 2012-10-07T04:24:34Z VoteTAG: 0 CoursewareTAG: Week 4 / Incremental Method Math CommentableIdTAG: 6002x_inc_method_math NumberOfReplyTAG: 1 FirstChildTAG: It's a small signal part. you forgot vD at the end. iD = (d(f(vD)) / d(vD)|vD=VD) * vD A slop is 1/RD, R = v/i, RD = vD/iD and slope = 1/RD = iD/vD correspondingly slope = d(f(vD)) / d(vD)|vD=VD RD = 1 / slope FirstChildUserIdTAG: 240170 FirstChildUserNameTAG: seca FirstChildCreateTimeTAG: 2012-10-07T04:37:57Z IndexTAG: 3606 TitleTAG: H4P3 finding In Can someone please give me a hint how to find the norton current? I found Rn but have problems finding In. Please help me! UserIdTAG: 263602 UserNameTAG: elekov CreateTimeTAG: 2012-10-07T00:52:26Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: How you find Rn? FirstChildUserIdTAG: 196404 FirstChildUserNameTAG: Yel1owstone FirstChildCreateTimeTAG: 2012-10-07T01:22:39Z SecondChildTAG: To find the norton current you short out the port terminals. First up find the voltage input for your dependent sources Now short out any interdependent voltage sources then the current In is the current going through the short circuited port terminals. I can't work out the Rn though, what happens to a dependent supply when working out Rn for a network? SecondChildUserIdTAG: 329964 SecondChildUserNameTAG: SmartMike SecondChildCreateTimeTAG: 2012-10-07T01:34:30Z SecondChildTAG: The VCCS is a VOLTAGE controlled current source. Think of it as a function of some other voltage in the circuit. ;-) Hope this helps! SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-10-07T05:33:10Z SecondChildTAG: Isn't this dependent source a VCVS? Also, I saw someone state that i = A*u. Do you have any input on this? - Thanks SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-07T05:53:15Z IndexTAG: 3607 TitleTAG: H4P2 very last question I am taking the slope of my function i(v) and setting it equal to .014/.6 and solving for RL, but I do not get the correct answer. Can anyone please help me? Thanks. UserIdTAG: 101750 UserNameTAG: caf960 CreateTimeTAG: 2012-10-06T23:29:49Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: You can model the zener as a resistor in series with a DC voltage source. If you check the graph for the zener you will see that the breakdown region is in the range of the -5V and the slope 1A/V. Use this values in your equation and you'll find the answer for the second part quickly... FirstChildUserIdTAG: 350458 FirstChildUserNameTAG: Aaron_PSX FirstChildCreateTimeTAG: 2012-10-07T07:29:28Z FirstChildTAG: use these formula V0 = Vin (Rl/Rl+Ri) Vin total = dc offset + noise FirstChildUserIdTAG: 382505 FirstChildUserNameTAG: AhmedGalal2 FirstChildCreateTimeTAG: 2012-10-07T03:59:33Z SecondChildTAG: Is this the hint for the first question of the second part of H4P2 ,if so ,can you please tell what is the value of DC offset and noise as i cant get the answer. SecondChildUserIdTAG: 232499 SecondChildUserNameTAG: Asim09 SecondChildCreateTimeTAG: 2012-10-07T07:12:03Z IndexTAG: 3608 TitleTAG: h4p2 problem solving part 2 h4p2 I need some help or any clear hint to solve part 2 of h4p2... I tried a lot of things but I couldn't get it. Please help me UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-06T23:21:02Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Hope I can help. 1. Start with replacing the circuit with its Thevenin equivalent on the left and the Zener diode on the right. Once you get $V_{TH}$ and $R_{TH}$, you can use the $v = V_{TH} + R_{TH}\cdot i$ to plot the load line (just remember to rearrange so that $i$ is on LHS). Plot line will show you which region the diode is operating in. 2. Now you have your region and you can get equation for $i_D$ from the graph. 3. Calculate $V_O$ using the node method over the Large Signal circuit. This is where you need $i_D$ equation from 2 to plug into KCL. 4. Calculate $v_O$ using Small Signal analysis. Just remember that in SSC, diode acts as a resistor and its resistance value can be inferred from the diagram (slope). 5. Minimum value for $R_L$: You can see the limits for the diode's $V_D$ in the diagram. Use the limit for your operating region in conjunction with KVL & KCL to solve for $R_L$ FirstChildUserIdTAG: 502508 FirstChildUserNameTAG: amaher FirstChildCreateTimeTAG: 2012-10-06T23:45:52Z SecondChildTAG: just remember to rearrange so that i is on LHS) what is it mean ? SecondChildUserIdTAG: 192114 SecondChildUserNameTAG: chuba SecondChildCreateTimeTAG: 2012-10-07T12:21:35Z SecondChildTAG: LHS: left hand side...he means that keep in one side just your current and in the other side of your = sign all other terms SecondChildUserIdTAG: 99441 SecondChildUserNameTAG: coyarce SecondChildCreateTimeTAG: 2012-10-07T13:02:36Z SecondChildTAG: Solved everything but the min value of RL part..can u help with that please? SecondChildUserIdTAG: 173147 SecondChildUserNameTAG: cruiser_rahit SecondChildCreateTimeTAG: 2012-10-07T17:06:25Z SecondChildTAG: cruiser_rahit, I've just seen your question now and you've probably solved it already, but here it goes just in case. Look at the loop on the right side of the circuit. You have only 2 voltages $V_D$ and $V_O$ for KVL in that loop. You know the limit for $V_D$ is -5V from the vi graph, so you also know the value for $V_O$ (from KVL). Now, you've already solved for $V_O$. Use that same solution, but now you know $V_O$ (from above) and you're solving for $R_L$. SecondChildUserIdTAG: 502508 SecondChildUserNameTAG: amaher SecondChildCreateTimeTAG: 2012-10-07T18:56:46Z FirstChildTAG: Try modeling the circuit in the circuit sandbox and see if that helps. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-06T23:48:41Z FirstChildTAG: the following link was helpful : http://www.allaboutcircuits.com/vol_3/chpt_3/11.html FirstChildUserIdTAG: 278792 FirstChildUserNameTAG: sali FirstChildCreateTimeTAG: 2012-10-07T00:46:50Z IndexTAG: 3609 TitleTAG: YEAH!! ajas! ajas! thanks Dr!!! this is very interesant!!!! tss UserIdTAG: 118093 UserNameTAG: Oktavious CreateTimeTAG: 2012-10-06T23:02:07Z VoteTAG: 0 CoursewareTAG: Week 4 / Incremental Analysis Demo CommentableIdTAG: 6002x_inc_analysis_demo NumberOfReplyTAG: 0 IndexTAG: 3610 TitleTAG: CHILD - LANGMUIR LAW Could someone tell me where I can find more information about this law I dont really understand it? UserIdTAG: 160242 UserNameTAG: Mlevins35 CreateTimeTAG: 2012-10-06T22:02:46Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You might try Wikipedia. FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-10-06T22:20:07Z IndexTAG: 3611 TitleTAG: Any one from Rajkot - Gujrat India Please let me know if you are from Rajkot Gujrat. UserIdTAG: 272523 UserNameTAG: Jivraj CreateTimeTAG: 2012-10-06T21:39:55Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3612 TitleTAG: H4P3 Part A and B For Part A, I place a short across port B, so then all the current would bypass the leg with R3 and dependent voltage source. Therefore, the the current is just 5V /(R1+R2), except this doesn't give me the right answer. Anyone see what I'm doing wrong? For Part B, I place a 1A current source across port B, and I place a short across the independent voltage source. I solve for the voltage across port B and divide this voltage by 1A to get the resistance. This doesn't work either. Can anyone help me? Thanks. UserIdTAG: 101750 UserNameTAG: caf960 CreateTimeTAG: 2012-10-06T21:36:57Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Nevermind about part A, I got it, but I still need help with part B. Thanks! FirstChildUserIdTAG: 101750 FirstChildUserNameTAG: caf960 FirstChildCreateTimeTAG: 2012-10-06T21:43:26Z SecondChildTAG: Your method is correct..I did it in the same way..I assumed I=9A and then just applied nodal analysis to find the voltage across port b..dependent source was kept as it is and found by finding u..and that divided by 9 gave the answer..but I am having trouble with norton current..how did u do that?.. SecondChildUserIdTAG: 173147 SecondChildUserNameTAG: cruiser_rahit SecondChildCreateTimeTAG: 2012-10-06T22:21:08Z SecondChildTAG: Ok..solved that one too.. SecondChildUserIdTAG: 173147 SecondChildUserNameTAG: cruiser_rahit SecondChildCreateTimeTAG: 2012-10-06T22:30:32Z SecondChildTAG: Actually, it doesn't matter what I is. You may assume I=10^(10^1000) amps and nothing change. In the right equations for Rth there's a point where you have to divide I by I, so there's no I left in the resulting equations. SecondChildUserIdTAG: 326787 SecondChildUserNameTAG: Angstrem SecondChildCreateTimeTAG: 2012-10-06T22:31:50Z FirstChildTAG: I also tried the short circuit method - why isn't this right? How did you end up solving for the current? "For Part A, I place a short across port B, so then all the current would bypass the leg with R3 and dependent voltage source. Therefore, the the current is just 5V /(R1+R2), except this doesn't give me the right answer." FirstChildUserIdTAG: 216374 FirstChildUserNameTAG: OliverBarham FirstChildCreateTimeTAG: 2012-10-07T22:57:38Z IndexTAG: 3613 TitleTAG: H4P2 - Part 2 Finding VO? I know this should probably be the easy part of the question, but how do I find VO in part 2 of this question? I've tried a lot of things so far. Is the diode supposed to be an open circuit? I think I've found the operating point (in the flat region), and I've been using that as my bias point. If we are supposed to model the diode as voltage source in series with a resistor, do we simply add (or subtract) that voltage from VO in Part 1? UserIdTAG: 193625 UserNameTAG: coreykruger CreateTimeTAG: 2012-10-06T21:09:37Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi coreykruger, I'm not sure about your version of HW (different students get different values for V's, I's, etc so that results can't be plagiarized), but I got correct results with operating point being in the negative region (i.e. VD <= -5). So, I got my equation for $i_D$ from the graph, noting that for $i_D = 0$, $V_D = -5$ and $R=1$, and used it with the node analysis in the Large Signal circuit. FirstChildUserIdTAG: 502508 FirstChildUserNameTAG: amaher FirstChildCreateTimeTAG: 2012-10-06T21:33:44Z SecondChildTAG: Hmmm...still no luck. Did you arrange the diode voltage source in front of or behind its 1ohm resistor? SecondChildUserIdTAG: 193625 SecondChildUserNameTAG: coreykruger SecondChildCreateTimeTAG: 2012-10-06T21:56:02Z SecondChildTAG: I didn't need to worry about the diode voltage source, because I have the equation for $i_D$ (from the graph), so using node analysis: $ \cfrac{V_O - V_I}{R_{IN}} + i_D + \cfrac{V_O}{R_L} = 0$ SecondChildUserIdTAG: 502508 SecondChildUserNameTAG: amaher SecondChildCreateTimeTAG: 2012-10-06T22:19:12Z SecondChildTAG: Okay, I actually solved the first question a few minutes before you posted that equation- but thanks for showing it. I think my main problem is finding the proper load line. I guess I didn't understand what the load line was as well as I thought I did. Is there a process you followed to find yours? SecondChildUserIdTAG: 193625 SecondChildUserNameTAG: coreykruger SecondChildCreateTimeTAG: 2012-10-06T22:31:14Z SecondChildTAG: I got my plot line from the Thevenin equivalent values. The equation is: $v = V_{TH} + R_{TH}\cdot i$, so if you substitute your $V_{TH}$ and $R_{TH}$ and rearrange so that $i$ is on LHS, you get your plot line equation. SecondChildUserIdTAG: 502508 SecondChildUserNameTAG: amaher SecondChildCreateTimeTAG: 2012-10-06T22:58:00Z SecondChildTAG: $i = \cfrac {v - V_{TH}}{R_{TH}}$ SecondChildUserIdTAG: 502508 SecondChildUserNameTAG: amaher SecondChildCreateTimeTAG: 2012-10-06T23:00:42Z IndexTAG: 3614 TitleTAG: lab 5 last question what do they mean by '' clipping or truncation of the max and min signal values''? am unable to get the answer as i dont understand the question.. UserIdTAG: 156585 UserNameTAG: AdiRanga CreateTimeTAG: 2012-10-06T19:57:58Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You just try different values and look for values at which the resulting output waveform still has the same form as the input waveform (magnified). The distorted ones have flatter tops or bottoms, so you should recognize them. And then you just squiggle around a bit and see which value is the last one without that distortion. You might have to modify your answer slightly and input it several times but even my answer was accepted by the grader... Good luck! FirstChildUserIdTAG: 186276 FirstChildUserNameTAG: OliverMun FirstChildCreateTimeTAG: 2012-10-06T20:18:40Z SecondChildTAG: till how many decimal points did u modify? SecondChildUserIdTAG: 156585 SecondChildUserNameTAG: AdiRanga SecondChildCreateTimeTAG: 2012-10-06T20:33:25Z IndexTAG: 3615 TitleTAG: Lab 4 Hi, I have an Problem to find VT. RdsON is right. For Vt I have created a circuit with VGS=2.5V and one with VGS=3V VDS is rising up to 3V with triangle. Now I measure the current in both circuits at VDS=3V ( I put the slider on the right side) Then I put the result in the formula and point 2 and get 2 values. unfortunately I'm wrong UserIdTAG: 389333 UserNameTAG: juergen CreateTimeTAG: 2012-10-06T19:49:49Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hi, I am having the same problem. The first question i got it right, but when i put the values of Vgs = 2V and Vgs = 3V with their respect currents,it gives me a value that is not right. FirstChildUserIdTAG: 40865 FirstChildUserNameTAG: guilima10 FirstChildCreateTimeTAG: 2012-10-06T20:48:45Z SecondChildTAG: I don't know what exactly was the problem. I got it, the procedure I reported, is right. Try VGS=2.5 and measure the current at VDS=3V SecondChildUserIdTAG: 389333 SecondChildUserNameTAG: juergen SecondChildCreateTimeTAG: 2012-10-06T20:54:06Z SecondChildTAG: rounding is important! SecondChildUserIdTAG: 122513 SecondChildUserNameTAG: Mona77 SecondChildCreateTimeTAG: 2012-10-06T21:06:56Z FirstChildTAG: Watch out for correct rounding! I've had VERY big and painful problems on that place, because of I've rounded 71.8 to 71 and some other value. FirstChildUserIdTAG: 326787 FirstChildUserNameTAG: Angstrem FirstChildCreateTimeTAG: 2012-10-06T21:02:42Z IndexTAG: 3616 TitleTAG: control port voltage is "different" than V across resistor? I'm having trouble with how the control port is hooked into the circuit. About halfway through the lecture, it is implied by the full drawing that the control port is connected across the resistor, while the output port is connected in series with the resistor as in the shorthand notation circuit. (Is there always going to be a resistor there to connect the control port to? If not, and why should there be, how can we depend on this shorthand notation where the control port is just left off?) However, since the positive node of the resistor is wired to the output of the dependent source, that produces a dependent source where the control port and the output port are the same terminals, just hooked up with opposite polarities. This would produce no current, or infinite current? Where is the voltage difference coming from? Anyway a bit later on he says, "My dependent current source has a relationship K divide by V where V is some other voltage in the circuit." And later still, he analyzes this circuit assuming that the "other" voltage V is the same as the voltage V across the resistor. I would love to hear any thoughts or explanations for this that anyone has. UserIdTAG: 91356 UserNameTAG: mnrsiat CreateTimeTAG: 2012-10-06T19:28:40Z VoteTAG: 0 CoursewareTAG: Week 4 / Dependent Sources Example 1 CommentableIdTAG: 6002x_dep_src_ex_1 NumberOfReplyTAG: 0 IndexTAG: 3617 TitleTAG: Misprint at Homework 6 The text rightbefore **Figure 3** is "With the NewFET biased into its active region, its small-signal model is as shown in **Figure 4.**". I think it should be "With the NewFET biased into its active region, its small-signal model is as shown in **Figure 3.**" UserIdTAG: 376572 UserNameTAG: sergy_kan CreateTimeTAG: 2012-10-06T17:41:08Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 0 IndexTAG: 3618 TitleTAG: S8E2 Why can we add the independent current source at the port? Wouldn't that alter the circuit? UserIdTAG: 380703 UserNameTAG: vargy CreateTimeTAG: 2012-10-06T17:03:25Z VoteTAG: 0 CoursewareTAG: Week 4 / Dependent Sources Exercise 1 CommentableIdTAG: 6002x_dep_src_exsz_1 NumberOfReplyTAG: 1 FirstChildTAG: we use independent source to find Rth FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-06T18:19:19Z SecondChildTAG: I set the dependent source to its Z value, then calculated Rth. It worked for this problem and S8E1. Will this come back to haunt me later? SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-06T20:25:52Z SecondChildTAG: R1||R2 is not correct even it was graded green due to grader tolerance. For S8E1 you don't need Rth at all. Though for S8E2 you also can find Rth without prove current source, because current through shorted output is obvious. SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-06T22:00:07Z SecondChildTAG: *prove current = probe current SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-06T22:01:25Z IndexTAG: 3619 TitleTAG: lab 4 resistor there's no resistor...don't we need it at the o/p?? pls help UserIdTAG: 214277 UserNameTAG: yadsam CreateTimeTAG: 2012-10-06T16:44:57Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3620 TitleTAG: h4p3-a i m not able to find vth ...can ne 1 help me...i gt Rth right...pls help me thts the only 1 left UserIdTAG: 214277 UserNameTAG: yadsam CreateTimeTAG: 2012-10-06T16:30:05Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: How did you find Rth? FirstChildUserIdTAG: 305491 FirstChildUserNameTAG: lefam FirstChildCreateTimeTAG: 2012-10-06T17:39:04Z FirstChildTAG: ohme law FirstChildUserIdTAG: 331441 FirstChildUserNameTAG: abdessamade FirstChildCreateTimeTAG: 2012-10-06T16:58:36Z SecondChildTAG: during thevenin's finding vth do we have to remove R2 or nt.... or any load in parallel with the general port.... jst help me thnks SecondChildUserIdTAG: 214277 SecondChildUserNameTAG: yadsam SecondChildCreateTimeTAG: 2012-10-06T17:16:47Z SecondChildTAG: don't overthink. it's really simple. SecondChildUserIdTAG: 172304 SecondChildUserNameTAG: Demohunter SecondChildCreateTimeTAG: 2012-10-06T19:20:35Z IndexTAG: 3621 TitleTAG: lab 4 tool using problem kindly check this picture and tell me what i am doing wrong that..![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13495393731343631.png UserIdTAG: 438395 UserNameTAG: ali_PU1 CreateTimeTAG: 2012-10-06T15:48:25Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hi, your wrong is that Vds = V triangle, and Vgs = V dc Excuse me my bad english, Bye. FirstChildUserIdTAG: 87950 FirstChildUserNameTAG: neoacademic FirstChildCreateTimeTAG: 2012-10-06T18:13:29Z SecondChildTAG: again m not getting any plot ...i used vgs of 0,0.5,1,1.5,2,2.5,3 nut mothing at the output :( SecondChildUserIdTAG: 438395 SecondChildUserNameTAG: ali_PU1 SecondChildCreateTimeTAG: 2012-10-07T04:37:23Z FirstChildTAG: Hi ali_PU1, Some hints for you: 1) Make sure that you are using the W/L given in the statement. 2) Which is the voltage source that has to be variable the one of the gate-source or the one of the drain -source? Which is the voltage source that has to be constant the one of the gate-source or the one of the drain -source? 3) What voltage do you need to plot in the x-axis? are you sure that is the vGS? 4) You can take a look at this [Hints of Lab4][1] See you! P.D: A suggestion, you should delet your snapshop from this post :P . Remember that you can not show partial solutions or they will delet it or erase it... [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5070e1b72210d02400000055 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-07T05:34:15Z IndexTAG: 3622 TitleTAG: i cant know the right use of probe to plot curves correctly what should be the Vds of the mosfet in lab.? m not getting the graph like it was in the shown there problem is that i dnt know the use of currrent probe peopert editing. kindly guide me thanks UserIdTAG: 438395 UserNameTAG: ali_PU1 CreateTimeTAG: 2012-10-06T15:17:18Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3623 TitleTAG: ans They assume it..usually 1 V . UserIdTAG: 266739 UserNameTAG: satvikchugh CreateTimeTAG: 2012-10-06T15:02:46Z VoteTAG: 0 CoursewareTAG: Week 4 / Load Line Tutorial CommentableIdTAG: 6002x_load_line_t NumberOfReplyTAG: 0 IndexTAG: 3624 TitleTAG: Diode Voltage Threshold How do we calculate the diode voltage threshold? UserIdTAG: 230787 UserNameTAG: Gabriel007 CreateTimeTAG: 2012-10-06T14:40:04Z VoteTAG: 0 CoursewareTAG: Week 4 / Load Line Tutorial CommentableIdTAG: 6002x_load_line_t NumberOfReplyTAG: 1 FirstChildTAG: diode voltage threshold doesn't need to be calculated. its given in the diode datasheet and provided by the manufacturer! FirstChildUserIdTAG: 368712 FirstChildUserNameTAG: jmen FirstChildCreateTimeTAG: 2012-10-19T14:28:35Z IndexTAG: 3625 TitleTAG: About the Lab4 What is diode for? UserIdTAG: 371617 UserNameTAG: rainbow12 CreateTimeTAG: 2012-10-06T14:32:04Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: No use. FirstChildUserIdTAG: 97340 FirstChildUserNameTAG: AnkitRana FirstChildCreateTimeTAG: 2012-10-06T15:02:35Z SecondChildTAG: i can draw circuit there ... how to edit the properties to make graph like it shows there.. SecondChildUserIdTAG: 438395 SecondChildUserNameTAG: ali_PU1 SecondChildCreateTimeTAG: 2012-10-06T15:13:17Z SecondChildTAG: do not need any diode. about the graph make sure that you have edited W/L to 1 on the mosfets because the default value is 2 SecondChildUserIdTAG: 402850 SecondChildUserNameTAG: y1111 SecondChildCreateTimeTAG: 2012-10-07T20:39:42Z IndexTAG: 3626 TitleTAG: Amplifier produces drop, but how does that amplify? @kingsman::for a small change in vi we have a large change in vo.that is amplification.:) UserIdTAG: 266739 UserNameTAG: satvikchugh CreateTimeTAG: 2012-10-06T14:27:26Z VoteTAG: 0 CoursewareTAG: Week 4 / Amplifier with Dependent Source CommentableIdTAG: 6002x_amplifier_w_dep_src NumberOfReplyTAG: 0 IndexTAG: 3627 TitleTAG: Error in subscript S11V1? There may be an error in lecture video S11V1 subscript @ 5:31. The text is "times vi minus gm **auto**-transconductance". But in video it is heard "times vi minus gm **or the**-transconductance". UserIdTAG: 376572 UserNameTAG: sergy_kan CreateTimeTAG: 2012-10-06T13:39:04Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 0 IndexTAG: 3628 TitleTAG: lab4 Why is that i get only one gate voltage at the x axis although i use all combinations specified UserIdTAG: 357381 UserNameTAG: krrish46 CreateTimeTAG: 2012-10-06T13:18:10Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Did you add the current probe ? FirstChildUserIdTAG: 454609 FirstChildUserNameTAG: Alwahsh FirstChildCreateTimeTAG: 2012-10-06T13:50:42Z SecondChildTAG: yes SecondChildUserIdTAG: 357381 SecondChildUserNameTAG: krrish46 SecondChildCreateTimeTAG: 2012-10-06T23:37:49Z FirstChildTAG: Hi krrish46! Can I help you? Are you sure that you are connecting in the correct place the different voltage sources? Are you sure that you are placing correctly the variable voltage source and the constant voltage source? Have you put the current probe in the correct position? See you! Myriam. P.D: if you are needing hints of lab4 you can read [here][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5070e1b72210d02400000055 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-07T05:52:09Z SecondChildTAG: thanks myrimit SecondChildUserIdTAG: 357381 SecondChildUserNameTAG: krrish46 SecondChildCreateTimeTAG: 2012-10-07T13:04:12Z IndexTAG: 3629 TitleTAG: How about a sledge hammer method for solving these... Hope I don't get in trouble for posting this but if I did cross the line then I apologize, I most profusely do! After checking in the work I started reading other posts and folks going to online calculators and what not. Well, talk about building new gates but why not write a little program to let the existing gates do the number crunching for us. Right here! For those on Windows platform, here's a little VBS script. ******************************************************************************************************************** Option Explicit Dim R, VD, RD, ID, objFSO, objFile, strFile Const I0 = 8E-14 Const VT = 26E-3 Const EE = 2.71828 strFile = "C:\Users\ADMIN\Desktop\Results.txt" Set objFSO = CreateObject("Scripting.FileSystemObject") Set objFile = objFSO.CreateTextFile(strFile, True) For VD = 0.5 to 0.6 step 0.0005 ID = I0 * EE^(VD/VT) RD = 1 / ((I0/VT) * EE^(VD/VT)) objFile.Write (VD & "V" & vbTab & ID & "A" & vbTab & RD & "OHM" & vbLf) Next objFile.Close Set objFile = Nothing ******************************************************************************************************************** After it has computed the numbers, just walk down the list and find the ID's where it matches the required RD's. Add the voltage drop on R and it should do it. Or is this approach faulty? UserIdTAG: 559256 UserNameTAG: BrianLind CreateTimeTAG: 2012-10-06T12:34:55Z VoteTAG: 0 CoursewareTAG: Week 4 / Linearization Exercise 2 CommentableIdTAG: 6002x_linearization_ex_2 NumberOfReplyTAG: 1 FirstChildTAG: spreadsheets? :) FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-06T18:05:17Z IndexTAG: 3630 TitleTAG: S8E2 Can anybody explain, why is my KCL is wrong. Assume, that node upper R1 has potential V1 and upper R2 is V2=VTH. So we can write KCL I0+V2/R2=V1/R1 And equation for depended source V2-V1=a*V1/R But it gives me wrong answer. Where is my mistake? UserIdTAG: 345464 UserNameTAG: Constantine_ru CreateTimeTAG: 2012-10-06T12:25:14Z VoteTAG: 0 CoursewareTAG: Week 4 / Dependent Sources Exercise 1 CommentableIdTAG: 6002x_dep_src_exsz_1 NumberOfReplyTAG: 1 FirstChildTAG: или V2-V1=-a*V1/R1 или I0-V2/R2=V1/R1 FirstChildUserIdTAG: 324957 FirstChildUserNameTAG: mazzay FirstChildCreateTimeTAG: 2012-10-06T13:28:39Z IndexTAG: 3631 TitleTAG: Regarding download link Hello staff, It would help us if you can upload week5 week6 download links soon. Thanks. UserIdTAG: 364884 UserNameTAG: Ambli CreateTimeTAG: 2012-10-06T11:51:06Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3632 TitleTAG: Lab 4 help please i plot circuit as shown in fig. 7.7 in page 336. i don't understand how much VDS value use in the circuit? i am not getting any plot in my transient analysis? UserIdTAG: 171103 UserNameTAG: hemanth463 CreateTimeTAG: 2012-10-06T11:37:54Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Vds will vary with time (for example it will be a triangular wave) and a probe attached to it will be set as x-axis. All other probes (included the current probe) will be drawn respect to Vds. To draw different plots corresponding to different Vgs values you have to replicate the same circuit making some appropriate changes. FirstChildUserIdTAG: 376877 FirstChildUserNameTAG: AndBre FirstChildCreateTimeTAG: 2012-10-06T12:11:41Z SecondChildTAG: thank so much. i got it now. SecondChildUserIdTAG: 171103 SecondChildUserNameTAG: hemanth463 SecondChildCreateTimeTAG: 2012-10-06T12:19:07Z IndexTAG: 3633 TitleTAG: lab4 need help how can i draw them together? UserIdTAG: 84213 UserNameTAG: mohamed373 CreateTimeTAG: 2012-10-06T10:15:15Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Draw one circuit, select all the components and use ctrl-c ctrl-v (or the icons in the toolbar)to replicate it. Make the due changes to each replica and run the simulation. FirstChildUserIdTAG: 376877 FirstChildUserNameTAG: AndBre FirstChildCreateTimeTAG: 2012-10-06T10:45:48Z SecondChildTAG: when i substitute by vt in the equation i didnt get a correct answer for k why? SecondChildUserIdTAG: 84213 SecondChildUserNameTAG: mohamed373 SecondChildCreateTimeTAG: 2012-10-06T11:03:31Z SecondChildTAG: If you solved the quadratic equation correctly, you should get 2 values for Vt: only one of them is correct. SecondChildUserIdTAG: 376877 SecondChildUserNameTAG: AndBre SecondChildCreateTimeTAG: 2012-10-06T11:58:02Z SecondChildTAG: Hi mohamed373! Remember that you have to work in the saturated region ;). In the case that you are needing some hints of Lab4 you can [read here][1]. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5070e1b72210d02400000055 SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T05:58:36Z FirstChildTAG: I had the same problem, when I use my Vt in equation the answer was wrong, that's because I forgot that current is in uA FirstChildUserIdTAG: 140867 FirstChildUserNameTAG: DarkWishMaster FirstChildCreateTimeTAG: 2012-10-06T14:10:34Z IndexTAG: 3634 TitleTAG: Awesome explanation! Thank you so much :) UserIdTAG: 405009 UserNameTAG: kubi CreateTimeTAG: 2012-10-06T10:11:30Z VoteTAG: 0 CoursewareTAG: Week 6 / Capacitor basics CommentableIdTAG: 6002x_capacitor_basics NumberOfReplyTAG: 0 IndexTAG: 3635 TitleTAG: H4P2 Dear all, How do I find the operating point of the zener diode for vo? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-06T09:42:28Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I am also having some problems with that. I can find it for the first value of RL but not the second value. So if you follow the instruction on page 222 in the textbook, the first step is to set all small signal sources to zero. Then you solve this circuit to find the Vo, operating point. I used the Node method: -(Vo-VI)/Rin + ID + Vo/RL = 0 (a) And ID is by reading from the graph: ID = Vo + 5 (b) Put (a) into (b) and solve for Vo. However, I cannot get the right answer for the 2nd resistance (in my case 4kOhms) but I get the correct voltage for 2kOhms. Can anyone spot an error with my setup? FirstChildUserIdTAG: 329453 FirstChildUserNameTAG: Even83 FirstChildCreateTimeTAG: 2012-10-06T10:22:52Z SecondChildTAG: I just got it! You need to model the Zener as a 1 ohm resistor and be crazy accurate with your answer (e-5). SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-10-06T10:27:54Z SecondChildTAG: Yes, that is for vo, not Vo=operating point. As I understood from the original poster he had problems finding the operating point, not the small signal amplitude. SecondChildUserIdTAG: 329453 SecondChildUserNameTAG: Even83 SecondChildCreateTimeTAG: 2012-10-06T11:14:40Z SecondChildTAG: kindly guide me how to find the output voltage of small signal when zener is attached.? i have done all the parts but lost in these 2 parts ... kindly guide me how i can figure it out.? thanks SecondChildUserIdTAG: 438395 SecondChildUserNameTAG: ali_PU1 SecondChildCreateTimeTAG: 2012-10-06T12:06:00Z IndexTAG: 3636 TitleTAG: when we isolat the non-linear element what we put instead of it in the circut befor we do thevenin method > UserIdTAG: 437659 UserNameTAG: ahmed3000 CreateTimeTAG: 2012-10-06T08:57:07Z VoteTAG: 0 CoursewareTAG: Week 3 / Analyzing Nonlinear Circuits CommentableIdTAG: 6002x_analyzing_nonlinear_circuits NumberOfReplyTAG: 1 FirstChildTAG: Nothing - we exclude non linear element in order to find thevenin equivalent for rest of the circuit - the nodes which element was connected to will be port of thevenin equivalent. Some later you will be introduced to load line and operating point - it what you need this abstraction for. FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-06T09:32:46Z IndexTAG: 3637 TitleTAG: difference between the speed of the video, and text there was a difference between the speed of the video, and text. can anyone tell me what to do? UserIdTAG: 343092 UserNameTAG: Rustam84 CreateTimeTAG: 2012-10-06T08:37:05Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Troubleshooting NumberOfReplyTAG: 1 FirstChildTAG: It's a bug. Simply put, post a report of the video, and how far into the video you were, and at what speed. Post it twice, once in the discussion pane under the video that bugged, and once in the feedback category here. FirstChildUserIdTAG: 248807 FirstChildUserNameTAG: 1kingsman FirstChildCreateTimeTAG: 2012-10-06T08:57:22Z IndexTAG: 3638 TitleTAG: thevenin parameter in lab in lab why do we get the terminla voltage when connecting some known resistance to the voltage source ? UserIdTAG: 221617 UserNameTAG: konan CreateTimeTAG: 2012-10-06T07:49:39Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3639 TitleTAG: Op amp small signal model analysis I had the same result with a slight different path: taking the node equation for node "vint" we get: (vint-0)/1k - (vint-vout)/1k = 0 combining with the observations that: vint=vin-vop and vop=vout/a and substituting these to the node equation we get to the same result. UserIdTAG: 319453 UserNameTAG: leonidasGr CreateTimeTAG: 2012-10-06T07:15:15Z VoteTAG: 0 CoursewareTAG: Week 5 / Op Amp Small Signal Model CommentableIdTAG: 6002x_ap_amp_small_signal_t NumberOfReplyTAG: 1 FirstChildTAG: Yep, this would be the sane method as far as i am concern. FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-10-06T09:03:48Z IndexTAG: 3640 TitleTAG: norton example in example 3.26 when getting RN why don't we take the resistor of(1K) in our calculation ? UserIdTAG: 221617 UserNameTAG: konan CreateTimeTAG: 2012-10-06T07:03:29Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3641 TitleTAG: norton derivation how did we drive norton equiv for the circuit? UserIdTAG: 221617 UserNameTAG: konan CreateTimeTAG: 2012-10-06T05:48:56Z VoteTAG: 0 CoursewareTAG: Week 2 / Norton method CommentableIdTAG: 6002x_norton NumberOfReplyTAG: 0 IndexTAG: 3642 TitleTAG: h4p2 partb hello everyone...could anyone please help me with h4p2 partb where we have to find v0 when RL=2K...i found parta quite easily but i dont know what im doing wrong in part b UserIdTAG: 278792 UserNameTAG: sali CreateTimeTAG: 2012-10-06T04:39:05Z VoteTAG: 0 CoursewareTAG: Week 4 / Small Signal Model CommentableIdTAG: 6002x_small_signal NumberOfReplyTAG: 1 FirstChildTAG: Are you following the textbook systematic procedure -page222 ?. FirstChildUserIdTAG: 282828 FirstChildUserNameTAG: Chibolator FirstChildCreateTimeTAG: 2012-10-06T05:37:46Z SecondChildTAG: just got my answer right :) SecondChildUserIdTAG: 278792 SecondChildUserNameTAG: sali SecondChildCreateTimeTAG: 2012-10-06T05:45:59Z SecondChildTAG: i have the same problem but i don´t get solution yet SecondChildUserIdTAG: 314294 SecondChildUserNameTAG: victormp SecondChildCreateTimeTAG: 2012-10-06T10:30:13Z SecondChildTAG: I think I'm confused about the concepts because I can't figure it out SecondChildUserIdTAG: 45307 SecondChildUserNameTAG: josejimenez2 SecondChildCreateTimeTAG: 2012-10-06T22:57:48Z SecondChildTAG: Take a look at [h4p2 hints][1] victormp and josejimenez2 [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5070be20fabaf62b0000004d SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T06:03:24Z IndexTAG: 3643 TitleTAG: due sunday is that mean the work rejected that day or the day before 4 ex HW due 7 Oct dead time 7oct at 12 Pm or all the day UserIdTAG: 382505 UserNameTAG: AhmedGalal2 CreateTimeTAG: 2012-10-06T04:38:21Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: 11:59pm FirstChildUserIdTAG: 269641 FirstChildUserNameTAG: BAUWA FirstChildCreateTimeTAG: 2012-10-06T07:11:29Z SecondChildTAG: And does it for UTC(0) Time Zone, or for MIT time zone? SecondChildUserIdTAG: 345464 SecondChildUserNameTAG: Constantine_ru SecondChildCreateTimeTAG: 2012-10-06T08:46:37Z SecondChildTAG: Hi Constantine_ru! It is based on you local time of your country ;), you have till your 11:59 pm of your local time of october 7th. See you! Myriam. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-07T06:06:20Z IndexTAG: 3644 TitleTAG: mosfet amplifier What is the difference between MOSFET and BJT amplifier?? UserIdTAG: 567227 UserNameTAG: Ravi015 CreateTimeTAG: 2012-10-06T04:23:28Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: MOSFET is a voltage controlled device while BJT is a current controlled device.. MOSFET are smaller and thus easy to accomodate on chips..Moreover they are much more immune to distortion than BJT FirstChildUserIdTAG: 368712 FirstChildUserNameTAG: jmen FirstChildCreateTimeTAG: 2012-10-06T04:46:13Z SecondChildTAG: but BJT has better gain and lower noise :) SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-06T05:52:46Z SecondChildTAG: does this imply that BJT is more appropriate for accurate analog signal processing? SecondChildUserIdTAG: 252722 SecondChildUserNameTAG: vaboro SecondChildCreateTimeTAG: 2012-10-09T20:15:32Z FirstChildTAG: Description of BJT in example 7.13 mentions exact threshold values of $v_{CE}$ and $v_{BE}$ for a BJT to operate in its active region: 1) $v_{CE} \gt 0.2V$ 2) $v_{BE} \lt v_{CE}+0.4V$ Where do these thresholds come from? For example, for a MOSFET we have a threshold $V_T$ which is a variable that depends on MOSFET design. However, for a BJT these thresholds seem to be fixed. Why is that? Is this really this way or the thresholds mentioned in the book are simply there for pedagogical reasons? FirstChildUserIdTAG: 252722 FirstChildUserNameTAG: vaboro FirstChildCreateTimeTAG: 2012-10-09T20:30:22Z IndexTAG: 3645 TitleTAG: LATE COMER FEELING VERY FRUSTRATED CAUSE I HAVE JUST STARTED THE COURSE . WILL IT AFFECT MY GRADES? I AM THINKING FOR SPRING COURSE INSTEAD OF THIS ONE! PLEASE HELP ME! UserIdTAG: 569477 UserNameTAG: laxmikantmundkar CreateTimeTAG: 2012-10-06T04:16:24Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: A few tips.... 1. Search the discussion boards. It's the little magnifying glass in the left pane. 2. Search the wiki (granted, this might not help you too much.) 3. If for some reason you couldn't find it by searching, your lowest 2 scores in your homeworks and labs are ignored (will not effect your grade) In other words, you will not get credit for the first 3 weeks (you got a zero on them, because you didn't complete them on time), 2 of those are thrown out, and so you would start with a penalty of 1 failed week. Since I doubt you will complete the 4th week in 2 days, make that 2 weeks failed. Still can get a passing grade, since 55% or better is a passing grade. So do your best. FirstChildUserIdTAG: 248807 FirstChildUserNameTAG: 1kingsman FirstChildCreateTimeTAG: 2012-10-06T08:20:14Z SecondChildTAG: since i have not attended any of the assessments ..... its not workin for me so kindly can u say when will the next session begin for the same course so that i will be aware of it early itself and work on time. SecondChildUserIdTAG: 486092 SecondChildUserNameTAG: vimpana SecondChildCreateTimeTAG: 2012-10-06T10:06:34Z SecondChildTAG: it will start in the spring season generally in jan-feb. be alert ! SecondChildUserIdTAG: 569477 SecondChildUserNameTAG: laxmikantmundkar SecondChildCreateTimeTAG: 2012-10-09T07:59:12Z SecondChildTAG: Hey, like you I've just started the course and so far I like it! I don't really care about the exam or degrees. I'll just enjoy the experience, the content and if I have enough time I will follow it 'till the end. For me is not about the destination is about the journey. And if that proves worthy then it's a time well spent. SecondChildUserIdTAG: 612047 SecondChildUserNameTAG: mandreamanuel SecondChildCreateTimeTAG: 2012-10-11T14:18:34Z IndexTAG: 3646 TitleTAG: Lab 4 quadratic equation? To solve question 2 I resorted to using the quadratic equation to find VT, resulting in two positive solutions, one of which was correct. How would I work out which one was correct without trial and error? UserIdTAG: 335803 UserNameTAG: bobbanovski CreateTimeTAG: 2012-10-06T02:00:00Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: vGS >= VT FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-06T02:58:39Z SecondChildTAG: I solve it to but I get k wrong, any idea? SecondChildUserIdTAG: 286954 SecondChildUserNameTAG: ododo SecondChildCreateTimeTAG: 2012-10-07T00:05:37Z FirstChildTAG: Hi bobbanovski! Take a look at [here][1] :) [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5070e1b72210d02400000055 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-07T06:08:27Z IndexTAG: 3647 TitleTAG: Lab5 gain I got the right answer on the gain of the amplifier question, but I can't understand why that gain is positive if there is a sign change. Thanks. UserIdTAG: 125880 UserNameTAG: JoeUcvVzla CreateTimeTAG: 2012-10-06T01:47:04Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: i think sign change is about phase shift, not about gain. gain is - absolute value. and sign change: 180 deg phase shift FirstChildUserIdTAG: 95280 FirstChildUserNameTAG: Fortyq FirstChildCreateTimeTAG: 2012-10-06T08:55:33Z SecondChildTAG: what do they mean by '' clipping or truncation of the max and min signal values''? am unable to get the answer as i dont understand the question.. SecondChildUserIdTAG: 156585 SecondChildUserNameTAG: AdiRanga SecondChildCreateTimeTAG: 2012-10-06T19:59:34Z FirstChildTAG: what do they mean by '' clipping or truncation of the max and min signal values''? am unable to get the answer as i dont understand the question.. FirstChildUserIdTAG: 156585 FirstChildUserNameTAG: AdiRanga FirstChildCreateTimeTAG: 2012-10-06T19:59:59Z SecondChildTAG: Hi AdiRanga Take a look at [here Lab5 Hints][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/507628b2c4dd80250000004b SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-11T02:09:08Z IndexTAG: 3648 TitleTAG: What happens if we have problem reading the test? Sometimes the mathmatical expression are shown as \cfrac{1}{\cfrac{df(V_D)}{dV_D}} in my internet browser. If this happens during test I know there is not much I can do. UserIdTAG: 175706 UserNameTAG: rwskim CreateTimeTAG: 2012-10-05T23:32:34Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: disable the translator of google chrome and see if the defect is resolved. FirstChildUserIdTAG: 361823 FirstChildUserNameTAG: EliasOak FirstChildCreateTimeTAG: 2012-10-06T00:05:26Z FirstChildTAG: looks like browser failed to download some scripts - try to refresh page FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-05T23:39:52Z SecondChildTAG: If refreshing the page doesn't work, try A) a different browser, and/or B) making sure you have all the required fonts (not so much for this issue, but it's helped with others - google MathML fonts) SecondChildUserIdTAG: 138690 SecondChildUserNameTAG: Alyxi SecondChildCreateTimeTAG: 2012-10-05T23:51:17Z IndexTAG: 3649 TitleTAG: First question Hi, For the first question I¨m getting as an answer 1/((I0*VD/VT)*e^(VD/VT)). Can somebody say? UserIdTAG: 342966 UserNameTAG: SandraNavarro CreateTimeTAG: 2012-10-05T23:12:11Z VoteTAG: 0 CoursewareTAG: Week 4 / Linearization Exercise 2 CommentableIdTAG: 6002x_linearization_ex_2 NumberOfReplyTAG: 2 FirstChildTAG: the derivative of I0 in relation to voltage VD is the conductance. The conductance is the inverse of resistance. FirstChildUserIdTAG: 361823 FirstChildUserNameTAG: EliasOak FirstChildCreateTimeTAG: 2012-10-06T00:24:03Z SecondChildTAG: Thanks!! I "forgot" about that, will try again. SecondChildUserIdTAG: 342966 SecondChildUserNameTAG: SandraNavarro SecondChildCreateTimeTAG: 2012-10-06T05:24:59Z FirstChildTAG: you mistake with derivative of e^(ax) FirstChildUserIdTAG: 324219 FirstChildUserNameTAG: Dmitry79 FirstChildCreateTimeTAG: 2012-10-06T05:15:00Z IndexTAG: 3650 TitleTAG: What does "short out" mean? This "short out" expression confused me. On the one hand independent voltage sources are replaced with short circuits when analyzing circuits with dependent sources. On the other hand there is this "out" in the expression, which kind of implies to me something of a "break up". I am not a native speaker and this is why I am confused. What does this expression exactly mean? UserIdTAG: 252722 UserNameTAG: vaboro CreateTimeTAG: 2012-10-05T22:43:27Z VoteTAG: 0 CoursewareTAG: Week 6 / Source-follower small-signal exercise CommentableIdTAG: 6002x_small_signal_source_follower NumberOfReplyTAG: 4 FirstChildTAG: All right, I figured it out. "Short out" - replacing the voltage source with a short circuit. FirstChildUserIdTAG: 252722 FirstChildUserNameTAG: vaboro FirstChildCreateTimeTAG: 2012-10-05T23:30:07Z FirstChildTAG: We "short out" two points when we connect them with zero resistance. Once those two points have been "shorted out," there exists a "short circuit" between them. Hope that works. :) FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-10-07T01:03:04Z SecondChildTAG: Right! Thank you! SecondChildUserIdTAG: 252722 SecondChildUserNameTAG: vaboro SecondChildCreateTimeTAG: 2012-10-09T15:52:36Z FirstChildTAG: see page 437 of the textbook FirstChildUserIdTAG: 225977 FirstChildUserNameTAG: diego_ruiz FirstChildCreateTimeTAG: 2012-10-15T02:21:41Z FirstChildTAG: "short circuit" - is a noon, "short out" - is a verb means make a short circuit FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-05T22:55:15Z IndexTAG: 3651 TitleTAG: telll meeeee plz tell me ki how to solve first question... :(:(:( UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-05T20:19:15Z VoteTAG: 0 CoursewareTAG: Week 4 / Dependent Sources Exercise 1 CommentableIdTAG: 6002x_dep_src_exsz_1 NumberOfReplyTAG: 2 FirstChildTAG: You definitely need to use both KCL and KVL. FirstChildUserIdTAG: 172304 FirstChildUserNameTAG: Demohunter FirstChildCreateTimeTAG: 2012-10-05T22:31:45Z FirstChildTAG: Simply use the given formula for iP. FirstChildUserIdTAG: 282828 FirstChildUserNameTAG: Chibolator FirstChildCreateTimeTAG: 2012-10-06T06:53:30Z IndexTAG: 3652 TitleTAG: Final Current I don't fully understand how the current of 0.39mA is obtained at the end. Have I missed something? UserIdTAG: 278602 UserNameTAG: Timothyking2011 CreateTimeTAG: 2012-10-05T19:36:13Z VoteTAG: 0 CoursewareTAG: Week 4 / Attenuator Tutorial CommentableIdTAG: 6002x_attenuator_t NumberOfReplyTAG: 2 FirstChildTAG: rd = VT / ID. Since rd = 67 Om, we have ID = 0.39 mA. We are using here following condition: VT = 0.026 FirstChildUserIdTAG: 337855 FirstChildUserNameTAG: flexo FirstChildCreateTimeTAG: 2012-10-08T16:14:28Z FirstChildTAG: Just to add to this, the thermal voltage for a diode is given by VT=kT/q, where k is Boltzmann's constant, T is the temperature in Kelvin, and q is the electron charge. At T=300K, this does indeed give VT=0.026V. Take a look at section 16.2 of the textbook for more on this. FirstChildUserIdTAG: 288323 FirstChildUserNameTAG: fatslow FirstChildCreateTimeTAG: 2012-10-15T16:04:14Z IndexTAG: 3653 TitleTAG: youtube the videos are not working and i cant solve the home work in time i wish if the dead lines are postponed UserIdTAG: 326507 UserNameTAG: mohamed200 CreateTimeTAG: 2012-10-05T18:45:27Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3654 TitleTAG: Alternative to videos Since I cannot view any of the tutorial videos, Can anyone tell of any possible alternative? Like downloadable PDFs for lectures etc UserIdTAG: 442070 UserNameTAG: MuhammadAsad CreateTimeTAG: 2012-10-05T18:37:37Z VoteTAG: 0 CoursewareTAG: Week 4 / Small Signal Long Tutorial CommentableIdTAG: 6002x_small_sig_long_t NumberOfReplyTAG: 1 FirstChildTAG: yes. you can get those under course info >>>> lecture slides FirstChildUserIdTAG: 387026 FirstChildUserNameTAG: Adefilaedward FirstChildCreateTimeTAG: 2012-10-05T18:51:34Z IndexTAG: 3655 TitleTAG: difference what is difference between vd and VD? UserIdTAG: 177968 UserNameTAG: manmeet143 CreateTimeTAG: 2012-10-05T18:18:29Z VoteTAG: 0 CoursewareTAG: Week 4 / Incremental Method Math CommentableIdTAG: 6002x_inc_method_math NumberOfReplyTAG: 1 FirstChildTAG: VD is the DC Value "Operating DC", vd is the small signal value "the change around the DC", vD is the Sum of the 2 signals "DC & AC" FirstChildUserIdTAG: 185715 FirstChildUserNameTAG: amirengineer FirstChildCreateTimeTAG: 2012-10-07T07:35:14Z IndexTAG: 3656 TitleTAG: Well, I think that's right I just found the voltage on port which is Vth by standart node method, and then total resistance seen from this port is just Vth/I0, voltage divide by total current true the circuit. UserIdTAG: 140867 UserNameTAG: DarkWishMaster CreateTimeTAG: 2012-10-05T18:08:34Z VoteTAG: 0 CoursewareTAG: Week 4 / Dependent Sources Exercise 1 CommentableIdTAG: 6002x_dep_src_exsz_1 NumberOfReplyTAG: 0 IndexTAG: 3657 TitleTAG: how to perform labs Please upload any video which shows how to use labs? UserIdTAG: 520394 UserNameTAG: cmayur CreateTimeTAG: 2012-10-05T17:52:58Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Go to Overview->Edx tutorial FirstChildUserIdTAG: 395953 FirstChildUserNameTAG: dustinge FirstChildCreateTimeTAG: 2012-10-05T18:26:34Z IndexTAG: 3658 TitleTAG: h4p2 plz i need help i cant draw the iv characteristics curve of the zener diode how come the slope is 1 UserIdTAG: 84213 UserNameTAG: mohamed373 CreateTimeTAG: 2012-10-05T17:07:03Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: slope 1 is 1ohm incremental resistance FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-05T17:14:45Z SecondChildTAG: didnt get it yet SecondChildUserIdTAG: 84213 SecondChildUserNameTAG: mohamed373 SecondChildCreateTimeTAG: 2012-10-05T17:17:56Z SecondChildTAG: what exactly you didn't get? there is an IV of zener - the ziner's operating zone is negative breakdown voltage - it has slope 1A/V - which means 1ohm incremental resistance in this operating zone. Note that IV given for forward voltage - the working schematic uses it reversed. SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-05T17:30:01Z SecondChildTAG: read wiki about zener to understand it's property and uses: http://en.wikipedia.org/wiki/Zener_diode SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-05T17:32:35Z SecondChildTAG: how we can calculate small vo or vo from ac source plzz help SecondChildUserIdTAG: 164689 SecondChildUserNameTAG: muhammadfaizan SecondChildCreateTimeTAG: 2012-10-05T17:36:39Z SecondChildTAG: first i need to get the bias voltage which is vo i can get it by getting the intersection point between the loadline and the iv characteristics curve all i want to know is how this curve is drawn in need 1 point on it SecondChildUserIdTAG: 84213 SecondChildUserNameTAG: mohamed373 SecondChildCreateTimeTAG: 2012-10-05T17:49:04Z SecondChildTAG: which curve? load line? for load line consider zener as an device you examine and a rest of the circuit as a source this device is connected to - so to find load line you need replace this part of the circuit with thevenin or norton equivalent. And again - pay attention to IV of zener - it's given as for forward voltage across it - in your schematic it's reversed, so you need to covert this IV to correct one to get a correct intersection. SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-05T18:12:29Z SecondChildTAG: i found that dc bias voltage is a factor multiplied by the vI SO why when i multiply this factorby 50mv i cant get the correct answer vo why? SecondChildUserIdTAG: 84213 SecondChildUserNameTAG: mohamed373 SecondChildCreateTimeTAG: 2012-10-05T18:48:48Z SecondChildTAG: For part without zener it should be correct. For part with zener it will be different - it's what the purpose of zener. SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-05T21:24:56Z SecondChildTAG: when you calculate vo - it's a small signal analysis - zener will act as a resistor 1ohm SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-05T21:32:52Z SecondChildTAG: finallllly i ca solve it thanks you all for your help yesterday i was sick my brain was about to explode :D thnx SecondChildUserIdTAG: 84213 SecondChildUserNameTAG: mohamed373 SecondChildCreateTimeTAG: 2012-10-06T08:20:34Z SecondChildTAG: Thanks YakovO for your input.... SecondChildUserIdTAG: 428658 SecondChildUserNameTAG: Moza SecondChildCreateTimeTAG: 2012-10-07T09:14:44Z IndexTAG: 3659 TitleTAG: Textbook Exercise 4.3 Hello all, Can anyone please help me in solving Ex. 4.3 on page 231 of the textbook? How do we find the relation of such a curve? UserIdTAG: 441979 UserNameTAG: shrutimehrotra CreateTimeTAG: 2012-10-05T15:47:09Z VoteTAG: 0 CoursewareTAG: Week 4 / Linearization Exercise 2 CommentableIdTAG: 6002x_linearization_ex_2 NumberOfReplyTAG: 2 FirstChildTAG: looks like a cubic function to me FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-05T16:17:53Z SecondChildTAG: this one looks close :) y=((x-5)^3/26.32)+4.75 SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-05T16:40:34Z SecondChildTAG: I think they want you to do it graphically. SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-05T16:41:35Z FirstChildTAG: As the v-i relationship(as an equation) for the NLD is not given, you're left with the graphical approach alone. Superimpose the load-line for the circuit onto the graph given. The intersection of the load line and the v-i curve gives you the solution. FirstChildUserIdTAG: 95362 FirstChildUserNameTAG: asvictor FirstChildCreateTimeTAG: 2012-10-06T04:34:33Z IndexTAG: 3660 TitleTAG: Crazy capital letter I'm feel confused about the capital letter of the answer. Could administrator write the note under this lecture? Everybody can input right answer. UserIdTAG: 146995 UserNameTAG: ZYJ CreateTimeTAG: 2012-10-05T14:40:22Z VoteTAG: 0 CoursewareTAG: Week 3 / Isolation Via Thevenin CommentableIdTAG: 6002x_isolation_via_thevenin NumberOfReplyTAG: 1 FirstChildTAG: I'm not sure what you are asking. If you are asking why are they using capital letters, that just the convention in the exercise. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-05T16:55:08Z IndexTAG: 3661 TitleTAG: The circuit sandbox not working well with me Hello guys, I'm just wondering if anybody has the same problem with me in using circuit sandbox. Whenever I double click the tools (for example the transistor; to change Ohm), they move upward. This make it hard to make a neat connection and I always need to delete it and doing it all over again. And it is really hard for me to make connection between the tools. The wiring just do not work well. And instead of the making connection, the tools, again, moving upward. Is there any particular reason for this to happen? And what should I do with it? Thanks :) UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-05T13:53:30Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Try a different browser. Firefox has been working pretty good for me. Sometimes I have a problem "selecting" small things other then that it works perfect. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-05T14:14:48Z IndexTAG: 3662 TitleTAG: Amplifier ports At 2:06 the amplifier abstraction on the right totally looks like a spaceship. I just couldn't resist to write it. UserIdTAG: 410033 UserNameTAG: sagitta CreateTimeTAG: 2012-10-05T13:20:32Z VoteTAG: 0 CoursewareTAG: Week 4 / Amplifier with Dependent Source CommentableIdTAG: 6002x_amplifier_w_dep_src NumberOfReplyTAG: 0 IndexTAG: 3663 TitleTAG: Static discipline I'm getting confused with the **Static Discipline** here... Maybe I already forgot what we've learned :) Voh and Vol are the parameters for our **sender**, right? And Vih, Vil are for the **receiver**. We made large margins, so that even when the value sent by the sender jumps lower then Voh because of the noise, it will still not affect the signal reading. So the sender is supposed to have **small range** of values allowed, so then at which point **amplification** is useful? UserIdTAG: 410033 UserNameTAG: sagitta CreateTimeTAG: 2012-10-05T13:09:43Z VoteTAG: 0 CoursewareTAG: Week 4 / Amplifiers CommentableIdTAG: 6002x_amplifiers NumberOfReplyTAG: 2 FirstChildTAG: "Voh and Vol are the parameters for our sender, right? And Vih, Vil are for the receiver. We made large margins, so that even when the value sent by the sender jumps lower then Voh because of the noise, it will still not affect the signal reading." Yes. I don't think amplification is directly tied to Static Discipline. Use amplification if needed, but it is not "necessary". If you have suitable voltages for Vol and Voh, then you don't need to amplify anything. I hope I understood your question correctly, if not just ignore my response. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-05T14:20:42Z SecondChildTAG: Oh, I got this! In this video he uses different notations: V **i** H and V **i** L are the **sender** (**input**, logically) and V **o** H, V **o** L are the **reciever** (**output**). So since the output has higher values we need to amplify to get the correct result. In the **S4 Static Discipline** it's the other way around! ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13494604236025332.png SecondChildUserIdTAG: 410033 SecondChildUserNameTAG: sagitta SecondChildCreateTimeTAG: 2012-10-05T18:07:53Z FirstChildTAG: Maybe what he is referring to is the concept of the buffer. A buffer is used to keep the voltages from degrading to ranges outside of the static discipline. It's a logical input=output device. It's not necessary if you already have the correct voltages. You might want to check out pages 314-319 of the text. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-05T17:28:45Z IndexTAG: 3664 TitleTAG: H5P3 not solved hey i successfully solved H5P2 but i m not able to get the answer for H5P3. Vout=IDS*RS. My IDS is correct.. Now when i multiply it by RS and differentiate it with Vin and then multiply the whole thing by Vin and finally substitute Vin= VIN,my answer doesn't matches.. Please help!! I am not that bad at calculus...! Someone please help!! Is my approach wrong? I guess not.. Then what is the problem m not getting it..:( UserIdTAG: 368712 UserNameTAG: jmen CreateTimeTAG: 2012-10-05T13:05:51Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I am also doing it the same way but it is showing could not parse (.......) as a formula. FirstChildUserIdTAG: 99628 FirstChildUserNameTAG: Pranjal16 FirstChildCreateTimeTAG: 2012-10-05T13:23:21Z SecondChildTAG: Hey i solved the next 2 parts easily..SEE they have given vin separately. This means that the vin given before derivative of Vout w.r.t Vin has not to be replaced by VIN .Only the Vin in derivative of Vout w.r.t Vin has to be replaced by VIN. SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-05T13:25:37Z SecondChildTAG: Did you got the answer? This means that if we do not replace Vin by VIN in the term before the derivative term, we'll get correct answer..So i hav found the correct derivative. Just that now i need to write Vin in terms of VIN..Do you know how to do that? SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-05T13:28:18Z SecondChildTAG: has anyone got the answer for H5P3 part 1? I am not able to understand what mistake i hav done?? My other questions hav got a check! SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-06T02:55:09Z SecondChildTAG: I solved it by taking the function for ids from H5P2 and multiplying by Rs to get Vout. Then taking the derivative of that function with respect to vIN. Then simply multiply by vin. SecondChildUserIdTAG: 270011 SecondChildUserNameTAG: SnowmanZA SecondChildCreateTimeTAG: 2012-10-06T07:39:31Z SecondChildTAG: OOPS my bad!! I multiplied by Vin instead of vin...Got it now! Thanx man! SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-06T11:10:32Z FirstChildTAG: how did you work out part 2 3rd question FirstChildUserIdTAG: 436749 FirstChildUserNameTAG: waynebrown FirstChildCreateTimeTAG: 2012-10-08T12:12:53Z SecondChildTAG: Take the answer from the first question in Part 2 and substitute the answer for the second question in place of vOUT. Then solve for iDS. You will get a quadratic equation which you have to solve. SecondChildUserIdTAG: 270011 SecondChildUserNameTAG: SnowmanZA SecondChildCreateTimeTAG: 2012-10-08T16:10:15Z IndexTAG: 3665 TitleTAG: H4P1 what is incremental resistance? is there any formula to find it out???? UserIdTAG: 131558 UserNameTAG: PAVAN6 CreateTimeTAG: 2012-10-05T09:21:43Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: dR is the incremental resistance dR=dV/dI FirstChildUserIdTAG: 396893 FirstChildUserNameTAG: Raajesh FirstChildCreateTimeTAG: 2012-10-05T10:30:47Z SecondChildTAG: Raajesh, but how to differentiate this expression? Please explain if you can. Maybe you will give some hints? SecondChildUserIdTAG: 96873 SecondChildUserNameTAG: Santyaga SecondChildCreateTimeTAG: 2012-10-05T11:07:55Z SecondChildTAG: Yep, i also cant get it(( SecondChildUserIdTAG: 338422 SecondChildUserNameTAG: Sl1mcO SecondChildCreateTimeTAG: 2012-10-05T11:43:28Z SecondChildTAG: differentiate the given current expression with respect to v so u get dI/dV is some expression . reciprocal of that is the incremental resistance SecondChildUserIdTAG: 396893 SecondChildUserNameTAG: Raajesh SecondChildCreateTimeTAG: 2012-10-05T11:50:13Z SecondChildTAG: Thats the thing i did and its wrong(((I am sure thate I did it correctly cuz its easy(( SecondChildUserIdTAG: 338422 SecondChildUserNameTAG: Sl1mcO SecondChildCreateTimeTAG: 2012-10-05T12:29:22Z SecondChildTAG: dude it's easy.... differentiate the given expression and reciprocal it ... simple SecondChildUserIdTAG: 477198 SecondChildUserNameTAG: Rajesh1993 SecondChildCreateTimeTAG: 2012-10-05T14:51:41Z SecondChildTAG: guys, read wiki Derivative: http://en.wikipedia.org/wiki/Derivative and understand it's graphical meaning SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-05T15:43:08Z SecondChildTAG: Guyz its simple to differentiate Ip=(3/2)*p*Vpk^0.5 SecondChildUserIdTAG: 414201 SecondChildUserNameTAG: uzaifakram SecondChildCreateTimeTAG: 2012-10-05T16:47:04Z SecondChildTAG: the above formula is dIp. after this take 1/Ip put all values u will get the answer SecondChildUserIdTAG: 414201 SecondChildUserNameTAG: uzaifakram SecondChildCreateTimeTAG: 2012-10-05T16:48:09Z SecondChildTAG: How did you get to this formula? Please explain the steps, as the formula works, I just don't understand how you got there. Thanks! SecondChildUserIdTAG: 207204 SecondChildUserNameTAG: hgustavii SecondChildCreateTimeTAG: 2012-10-08T12:51:00Z SecondChildTAG: This was directed to uzaifakram, by the way. SecondChildUserIdTAG: 207204 SecondChildUserNameTAG: hgustavii SecondChildCreateTimeTAG: 2012-10-08T12:52:26Z IndexTAG: 3666 TitleTAG: lab4 in the lab4,do we have to connect vds? ok i never knew it was an 'aha'moment... UserIdTAG: 55345 UserNameTAG: sarmaji CreateTimeTAG: 2012-10-05T08:42:13Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3667 TitleTAG: Lab 4 Problems! I tried to create the circuit but somehow I am getting no values for Id. What am I doing wrong? I have a simple circuit just like figure 7.7 in the textbook. then I assign different DC values to Vgs but I get no Ids ploted and I am not sure why. I tried triangle wave for Vgs but the waveform looks inverted. Ids stays close to 0 A up to close to Vds = 1 V then it's up to 80 uA for Vds = 3 V. Oddly enough, my graph has voltages for x axis but has both voltages (left) and currents (right) for y axis. Can anyone help me? UserIdTAG: 175706 UserNameTAG: rwskim CreateTimeTAG: 2012-10-05T08:41:16Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: there is a sample circuit in lab4 (remember the resistor??). just try to replicate the circuit; but this time change the resistor with the MOS and you will be needing two power supplies. the one going to the Drain of your MOS has to be ramping (triangular; same as the state in the resistor sample). the second supply has to be pure DC goung into the Gate of your MOS. all supplies has to have a common ground. FirstChildUserIdTAG: 236810 FirstChildUserNameTAG: jmendego FirstChildCreateTimeTAG: 2012-10-05T09:24:34Z SecondChildTAG: Thanks for the help. I actually had wrong set up for the voltage source and the probe. I had connected Vtest to the gate. SecondChildUserIdTAG: 175706 SecondChildUserNameTAG: rwskim SecondChildCreateTimeTAG: 2012-10-05T21:40:09Z IndexTAG: 3668 TitleTAG: Discrete Resistor....?? can anyone tell me why a reistor is said to be discrete one..?? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-05T05:49:28Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: http://en.wikipedia.org/wiki/Electronic_component FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-05T06:34:32Z IndexTAG: 3669 TitleTAG: H4 I have trouble finding incremental resistance and operating point of Zener diode..I will appreciate if someone help me out with the steps of doing the calculations.. Thanks in advance UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-05T05:49:27Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: 1. Thevenise the circuit and get an equation. You will get a straight line equation. 2. Finding the short circuit current (when the zener is replaced by a short) and open circuit voltage across the zener (disconnect the zener). Sketch the line on the i - v curve for the diode. Get intersect point = operating point. 3. The incremental resistance is the inverse of the slope of the i - v characteristics of the zener at the operating point. Hope this helps! FirstChildUserIdTAG: 143823 FirstChildUserNameTAG: NathanNadeson FirstChildCreateTimeTAG: 2012-10-05T09:02:19Z FirstChildTAG: Incremental resistance is dR=dV/dI FirstChildUserIdTAG: 396893 FirstChildUserNameTAG: Raajesh FirstChildCreateTimeTAG: 2012-10-05T10:29:52Z IndexTAG: 3670 TitleTAG: H4P3 Part B "Norton Analysis" good day everyone, i just wanna ask for some help regarding on how to get the value of IN needed on the circuit. I already got my RN, and based on the ohm's law I=V/R where V=VTH=Vin and R=RN=RTH, but whenever i use this equation IN=Vin/RN it gives me a wrong answer >.< any idea how to solved this IN?.., thanks... UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-05T05:41:00Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I am stuck too. I have been trying on computing 'u' first and then finding Vth. Have failed like 8 times with diiferent values. :/ FirstChildUserIdTAG: 97340 FirstChildUserNameTAG: AnkitRana FirstChildCreateTimeTAG: 2012-10-06T15:17:28Z FirstChildTAG: I was stuck on this problem for the longest time. https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5060f4a8a3f6d21f00000050 This discussion really helped. I used this to find Norton Current and also the voltage 'u'. To find the RN, I just simply used that 'u' I calculated to find the current through the original circuit by using node analysis on R1 which allowed me to calculate the resistance of the dependent voltage source. Finally, just sum up those resistances to get the RN! FirstChildUserIdTAG: 281693 FirstChildUserNameTAG: aznayc FirstChildCreateTimeTAG: 2012-10-07T07:41:51Z IndexTAG: 3671 TitleTAG: How can I show the subtitles when I watch the sequence I downloaded? How can I show the subtitles when I watch the sequence I downloaded? I find that the subtitles disappeared when I watch the vedio I downloaded, while I could see it on-line. What happened? UserIdTAG: 128276 UserNameTAG: ChengBin CreateTimeTAG: 2012-10-05T04:35:12Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3672 TitleTAG: Difficulty with lab4 tools I'm facing problems in dealing with lab 4. I don't know why but I'm not able to select the components in the grid to change their properties. It is not responding properly. The component is not getting selected even on double clicking on it several times.Also I'm not able to draw the connecting lines. I don't know what to do. Can anyone please help me out with this? UserIdTAG: 381971 UserNameTAG: Suhas0992 CreateTimeTAG: 2012-10-05T04:09:31Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Troubleshooting NumberOfReplyTAG: 1 FirstChildTAG: I had the same problem before. This was because of the browser. Try Google Chrome browser. FirstChildUserIdTAG: 343092 FirstChildUserNameTAG: Rustam84 FirstChildCreateTimeTAG: 2012-10-06T17:34:04Z IndexTAG: 3673 TitleTAG: Can some one please help with the correct syntaxes that have to use in the boxes its quite annoying when you know the answer and you can't put the values in right just because..its not in correct format, on the top of which you don't even know the correct format or syntax.In S8E2 i faced this problem you need to do just simple things, but i keep on getting invalid input every time i type in the values. can someone please help with it. thanks in advance. UserIdTAG: 210992 UserNameTAG: neerajnatu CreateTimeTAG: 2012-10-05T04:05:27Z VoteTAG: 0 CoursewareTAG: Week 4 / Dependent Sources Exercise 0 CommentableIdTAG: 6002x_dep_src_exsz_0 NumberOfReplyTAG: 1 FirstChildTAG: K1, K2, VI, RO, RI, iB, iD, vB, vO FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-05T04:33:55Z SecondChildTAG: thanks for the answer but still, it says iB not permitted when i put in the values. what do i do? SecondChildUserIdTAG: 210992 SecondChildUserNameTAG: neerajnatu SecondChildCreateTimeTAG: 2012-10-05T04:43:54Z SecondChildTAG: hey, thanks for these i just manipulated my answers and other variables works just fine. I got it done now. SecondChildUserIdTAG: 210992 SecondChildUserNameTAG: neerajnatu SecondChildCreateTimeTAG: 2012-10-05T04:55:14Z SecondChildTAG: It's just a little undercover hint - it accepts only terms you need ;) SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-05T04:57:47Z SecondChildTAG: yeah, thats true, thats what i did i wasn't using the correct terms when i used the correct one got the answer right. thank you for the help. not being from any computers background i thought that, that might be causing the problems but that wasn't the issue. SecondChildUserIdTAG: 210992 SecondChildUserNameTAG: neerajnatu SecondChildCreateTimeTAG: 2012-10-05T05:15:59Z SecondChildTAG: We can only use resistances (RI and RO), constants (K1 and K2) and VI. SecondChildUserIdTAG: 296965 SecondChildUserNameTAG: LGMailhos SecondChildCreateTimeTAG: 2012-10-06T13:45:43Z IndexTAG: 3674 TitleTAG: Typo @ 0:58 in captioning "So I'm going to show you a circuit technique that uses circuit analysis to directly write these down almost by **inspection**," not expansion. UserIdTAG: 164898 UserNameTAG: jbparkes CreateTimeTAG: 2012-10-05T00:16:40Z VoteTAG: 0 CoursewareTAG: Week 4 / Small Signal Model CommentableIdTAG: 6002x_small_signal NumberOfReplyTAG: 0 IndexTAG: 3675 TitleTAG: S8E2 - Absolutely stalled I have no sense of a starting point for this problem. I don't even know how to pick a node or anything. Can I have a little guidance about attacking this please? For example, how does the node voltage on the current source side of the dependent voltage source related to the node voltage on the other side of the dependent current source? Is it something like: v0/R1 + (v0 + alpha i)/R2 -I0 = 0 I'm ready to put a fist through the wall in frustration. Thanks for any pointers. UserIdTAG: 166031 UserNameTAG: krebryna CreateTimeTAG: 2012-10-04T21:39:28Z VoteTAG: 0 CoursewareTAG: Week 4 / Types of Dependent Sources CommentableIdTAG: 6002x_types_of_dep_src NumberOfReplyTAG: 1 FirstChildTAG: My tip: ignore their hint. It adds so much work, and is harder to solve. For this problem, due to the dependent source, it is (in my opinion) harder to solve using the node method, so just solve it directly. 1. Write down your laws ($R_1$=, $R_2$=, etc.) 2. Write down KCL and KVL *for both loops*. 3. Solve for $V_{th}$, which just happens to be the voltage across $R_1$. 4. Having a good graphic calc, or at least an online one that can handle calculus level stuff, can't hurt (not for graphing, just solving if the math throws your for a loop.) FirstChildUserIdTAG: 248807 FirstChildUserNameTAG: 1kingsman FirstChildCreateTimeTAG: 2012-10-05T03:40:35Z IndexTAG: 3676 TitleTAG: S12E5: NEON RELAXATION OSCILLATOR , traduction to spanish or another English definiton , please!!! I dont understand the third part of the problem: What is the duty cycle of the lamp? (The duty cycle is the ratio of the time it is lit to the total time.) What does it mean? Thanks UserIdTAG: 299251 UserNameTAG: TheRedBlackOne CreateTimeTAG: 2012-10-04T21:01:34Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: http://en.wikipedia.org/wiki/Duty_cycle FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-04T22:06:28Z FirstChildTAG: Hello, TheRedBlackOne! If you turn on your table-lamp for 1H per day, then it will have duty cycle of 1H/24H=0.04. This process is periodic, and you calculate ratio for one period (one day). Now the same for the quiz: you have calculated charge and discharge time. One charge and one discharge forms one period; the lamp lits while it is discharging, and you can calculate the duty cycle. FirstChildUserIdTAG: 261503 FirstChildUserNameTAG: EugenyL FirstChildCreateTimeTAG: 2012-10-04T22:12:43Z FirstChildTAG: Thanks a lot both of you!!! Now I undertand it was very easy answer2/answer1 FirstChildUserIdTAG: 299251 FirstChildUserNameTAG: TheRedBlackOne FirstChildCreateTimeTAG: 2012-10-05T07:23:41Z IndexTAG: 3677 TitleTAG: Lagging behind in the course! I've joined this course quite late and right now I'm way behind.will this hamper my grades as I am lagging behind? Should I continue my course? Please suggest. UserIdTAG: 481452 UserNameTAG: TheAshutoshGupta CreateTimeTAG: 2012-10-04T20:00:32Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: How much lag?..if u have missed two homeworks and labs..its ok as long as u catch up and perform well in all others..only the top ten out of the total 12 are taken into account..max u might have missed till 3rd hw..u can still catch up..4 th hw and lab are due this weekend..try and complete that..and then put in a little extra effort..and in a week or two u can be as good as anyone here..and just go along.. FirstChildUserIdTAG: 173147 FirstChildUserNameTAG: cruiser_rahit FirstChildCreateTimeTAG: 2012-10-04T20:23:57Z IndexTAG: 3678 TitleTAG: question 2 I dont understand the second question at all. Can someone elaborate on it please? UserIdTAG: 435197 UserNameTAG: RichmondRichter CreateTimeTAG: 2012-10-04T19:55:53Z VoteTAG: 0 CoursewareTAG: Week 4 / Linearization Exercise 1 CommentableIdTAG: 6002x_linearization_ex_1 NumberOfReplyTAG: 1 FirstChildTAG: Hello, RichmondRichter! The 'brute force' way is to increase VI by 2%: $$V_I =5*1.02=5.1$$ then solve non-linear circuit for VA once again, and finally calculate (Va2-Va1)/(5.1-5) But it's not a required method, as I think :) ---------- Another method is to check: aha, 2% is a small change, I can use calculus! First of all, write the formula for VA: $$ V_A=V_I-2*i_A=V_I-2*10*(1-e^{-V_A/5}) $$ Now we can move to deltas with partial derivatives: $$ 1 * \Delta V_A + 0 *\Delta V_I = 1*\Delta V_I+(-20* (-1) * (-1/5) *e^{-V_A/5}$$ Calculate the numbers... $$ \Delta V_A = \Delta V_I -4 * e^{-V_A/5} \Delta V_A $$ And move all delta Va to the left: $$ (1 + 4 * e^{-V_A/5}) \Delta V_A = \Delta V_I $$ Now, if we substitute VA from the first question, we will calculate (delta VA)/(delta VI) and get the answer. You can ask: this formula has nothing about 2% of VI! Where is it? Wel, it only used to launch this method, because increase is small. No matter, if it 2%, or 1%, or 0.001%, the answer is the same. However, if growth was 50%, we can't go this way, and have to use the first method. FirstChildUserIdTAG: 261503 FirstChildUserNameTAG: EugenyL FirstChildCreateTimeTAG: 2012-10-04T21:59:16Z SecondChildTAG: Hello EugenyL, I didn't understand very well how you did the partial derivatives. What is the reason of partial derivatives? What is the principle of doing so? thanks in advance SecondChildUserIdTAG: 125388 SecondChildUserNameTAG: Glauber SecondChildCreateTimeTAG: 2012-10-11T15:28:56Z SecondChildTAG: Hello, Glauber! It's easy to use regular derivatives in this topic: $$V_A = V_I−20∗(1−e^{−V_A/5})$$ Replace variables with x and y. $$x=y−20∗(1−e^{−x/5})$$ Move all x to the right, and swap left and right $$y= x+20∗(1−e^{−x/5})$$ And go to the deltas: $$\Delta y = (derivativeYByX) = (1 + 4 * e^{-x/5}) * \Delta x$$ **But sometimes it's hard or even impossible to separate y from x variables... Or someone is too lazy to do so** =) In these cases the partial derivatives will make life easier: we do nothing with the initial (may be non-linear!) equation. We take partial derivatives, and then we solve simple linear equation for deltaX and deltaY. SecondChildUserIdTAG: 261503 SecondChildUserNameTAG: EugenyL SecondChildCreateTimeTAG: 2012-10-11T23:49:26Z SecondChildTAG: I got that in this method u had neglected the 2% change in the vi and take derivative to get the require ratio but if if the change is not important than you can't say that it's small signal method because in that we get some constant operating point on which change occurred. In this case we have operating point is Va 1.1v and Ia 1.95Amp and with the change in VI there should be change in this bias point than we get linearity that is all about in incremental method.and for that exercise is give. SecondChildUserIdTAG: 78064 SecondChildUserNameTAG: BHAVIN SecondChildCreateTimeTAG: 2012-12-23T07:07:59Z IndexTAG: 3679 TitleTAG: lads and homework im having problem or dont know how this work....how do i submit my lab and home i did them but nothing or no button to submit. if i change the page all my work dissappears. please help me, thanks in advance UserIdTAG: 462861 UserNameTAG: edxforme CreateTimeTAG: 2012-10-04T19:32:22Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3680 TitleTAG: H6P2 Question Why VDS_of_Q2 = VIN ? UserIdTAG: 39980 UserNameTAG: MarinoJr CreateTimeTAG: 2012-10-04T18:14:48Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: It's rather VGS = VDS. And VIN = VGS for Q2 since it's one and the same.The gate beeing connected to the drain, has the same potential.Watch the diode connected tutorial from week 6 for more. PS: The book and the lectures insist on using VIN notations instead of VGS. The correct form of IDS = K/2(VGS - VT)^2. At the end of the part 1 of the lectures from week 6 is a kind of advice regarding this confusion.( After 2 weeks of confusion it's about time , hehehe...) FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-10-04T19:28:28Z SecondChildTAG: Yes. I saw that one. Thank you. SecondChildUserIdTAG: 39980 SecondChildUserNameTAG: MarinoJr SecondChildCreateTimeTAG: 2012-10-04T19:48:19Z IndexTAG: 3681 TitleTAG: using determinant instead of matrix if we use the determinant does our answer going to be wrong in some cases? I use it and it give same answer. UserIdTAG: 226085 UserNameTAG: Teto CreateTimeTAG: 2012-10-04T18:11:33Z VoteTAG: 0 CoursewareTAG: Week 2 / Nodal Analysis CommentableIdTAG: 6002x_nodal_analysis_tutorial NumberOfReplyTAG: 1 FirstChildTAG: Are you using analytical software like SAGE or Mathematica? FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-10-05T15:11:42Z IndexTAG: 3682 TitleTAG: progress error ive completed the whole course of the first week..and a part of the second weeks..but when i check my progress it shows 0 ..any idea whats wrong ?! UserIdTAG: 380750 UserNameTAG: Aditya1990 CreateTimeTAG: 2012-10-04T16:42:18Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: The deadline might be over. FirstChildUserIdTAG: 99628 FirstChildUserNameTAG: Pranjal16 FirstChildCreateTimeTAG: 2012-10-05T13:40:41Z FirstChildTAG: Yeah, if you scroll down on the progress page, it shows the deadline for the homework and lab assignments. The deadline for Week 1 was Sept. 16, and for Week 2 it was Sept. 30th. Week 3 was also Sept 30th. Week 4 is due Oct. 7th. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-05T14:12:54Z IndexTAG: 3683 TitleTAG: problem in using lab tool yet i am new here facing many problem in lab tutorials....don't know what to do??? UserIdTAG: 546344 UserNameTAG: ravijat89 CreateTimeTAG: 2012-10-04T14:02:37Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: You need to say what your problems are, exactly. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-05T14:13:34Z IndexTAG: 3684 TitleTAG: h4p1 serious trouble help im having trouble understanding everything to do with p1 can sumone explain it for me UserIdTAG: 436749 UserNameTAG: waynebrown CreateTimeTAG: 2012-10-04T13:56:07Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: For Ip just plug in given values to the ip equation, R is the derivative of the f(ip). FirstChildUserIdTAG: 73809 FirstChildUserNameTAG: rhyssouth FirstChildCreateTimeTAG: 2012-10-04T14:16:04Z SecondChildTAG: So R is dV/dIp???? SecondChildUserIdTAG: 338422 SecondChildUserNameTAG: Sl1mcO SecondChildCreateTimeTAG: 2012-10-05T11:35:53Z SecondChildTAG: And if so, how to differantiate it??? SecondChildUserIdTAG: 338422 SecondChildUserNameTAG: Sl1mcO SecondChildCreateTimeTAG: 2012-10-05T11:45:50Z FirstChildTAG: u just need to differentiate Ip with respect to V for the equation Ip=P⋅Vpk^(3/2). i.e. dIp/dVpk put the given values and then inverse it as incremental resistance is dV/dI. FirstChildUserIdTAG: 232499 FirstChildUserNameTAG: Asim09 FirstChildCreateTimeTAG: 2012-10-07T04:49:46Z IndexTAG: 3685 TitleTAG: H6P2 part 2 I am having problems to find the minimum value of VDD. I know that in Q2 VGS=VDS. Then I need to use VGS>VT to find the minimum value. If VGS=VDD-VOUT, then I need to find the maximum value of VOUT, but I don't know how to do it. Could anyone give me a hint? Is my approach wrong? UserIdTAG: 370769 UserNameTAG: PabloFCid CreateTimeTAG: 2012-10-04T09:45:04Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: Hello, PabloFCid! Don't forget: both transistors operate in saturated region. Recall the textbook, "MOSFET Amplifier", p 338: the condition for saturated mode is $$ V_{DS} \ge V_{GS} - V_T$$ Now you can calculate Vout, I believe. FirstChildUserIdTAG: 261503 FirstChildUserNameTAG: EugenyL FirstChildCreateTimeTAG: 2012-10-04T10:52:16Z SecondChildTAG: Hello EugnyL, thanks for your reply, I have tried to solve the problem with that equation but it is still not working. I think I need the maximum value for VOUT, and with that equation I get the minimum value of VOUT. I need the maximum value because the bigger VOUT is, the bigger minimum value of VDD is. Do you know if the rest of my approach is correct?? Or do I need to rethink the problem? SecondChildUserIdTAG: 370769 SecondChildUserNameTAG: PabloFCid SecondChildCreateTimeTAG: 2012-10-04T12:22:58Z SecondChildTAG: Well, if you assume min VGS(Q2)=VT, it will not help. Q2 is not standalone, it is connected to another MOSFET. We directly set up operating point for the Q1 with Vin... But Q2 also biased (with the Q1 current), and we can't say that VGS=VT is the minimum voltage (because it is true only for zero current). If you need a hint, look at MarinoJr comment, or read below. ---------- Recall the simple (MOSFET+resistor) amplifier. We calculate the current, defined by the MOSFET input. Now we know the current of resistor, and can calculate it voltage. And all other voltages in the circut, if required. Can you use the same operations to calculate the homework circuit? SecondChildUserIdTAG: 261503 SecondChildUserNameTAG: EugenyL SecondChildCreateTimeTAG: 2012-10-04T13:25:36Z SecondChildTAG: EugenyL , i know that you saw that tutorial about diode connected mosfet. Like all the tutorials made by that guy , that is very confusing. If you know the current, you have to know the resistor value . In saturation, mosfets are nonlinear for large signals. They behave like a resistor only in triode region and for small signals. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-10-06T09:16:50Z SecondChildTAG: Hi, AlexAlexandrescu! I've tried to tell about sequence of operations: first find the current, then use it to get voltage for pull-up element, and finally calculate voltages for all nodes. That is similar to the simple amplifier. You are right, we can't replace MOSFET with resistor, especially when we know it's in saturation ;) However, we know the current for this element. It is non-linear? No matter! We still can calculate voltage, even if it requires some math instead of simple multiplication by R. SecondChildUserIdTAG: 261503 SecondChildUserNameTAG: EugenyL SecondChildCreateTimeTAG: 2012-10-06T15:39:20Z FirstChildTAG: PabloFCid, You may see that VDDminimum = VGS_in_Q2 + VDS_in_Q1. For your MOSFET Q1 operate in saturation, you need that VDS >= VGS-VT. MOSFET Q2 is already in saturation (2-terminal mosfet). Now you just need to realize what VGS_in_Q2 is equal. Remember that IDS of both transistors is the same. FirstChildUserIdTAG: 39980 FirstChildUserNameTAG: MarinoJr FirstChildCreateTimeTAG: 2012-10-04T12:28:59Z SecondChildTAG: IDS of both transistors is the same ? even though they have different VGS Voltages? SecondChildUserIdTAG: 149844 SecondChildUserNameTAG: amitaratnayake SecondChildCreateTimeTAG: 2012-10-04T14:41:03Z SecondChildTAG: Of course Ids is the same, both FET's are in series. SecondChildUserIdTAG: 270011 SecondChildUserNameTAG: SnowmanZA SecondChildCreateTimeTAG: 2012-10-04T16:52:57Z SecondChildTAG: i dont know why, i find Vddmin = VGS2 + VDS1. But after apply iDS for both FETs => VGS2 = VGS1, but it is wrong when apply it, an ideas SecondChildUserIdTAG: 114913 SecondChildUserNameTAG: ngoctuan SecondChildCreateTimeTAG: 2012-10-12T13:10:14Z SecondChildTAG: solved :| write wrong sign for VT T_T SecondChildUserIdTAG: 114913 SecondChildUserNameTAG: ngoctuan SecondChildCreateTimeTAG: 2012-10-12T13:23:07Z FirstChildTAG: Thanks for all your replies. I finally understood the problem and I got the correct solution. FirstChildUserIdTAG: 370769 FirstChildUserNameTAG: PabloFCid FirstChildCreateTimeTAG: 2012-10-04T13:56:02Z FirstChildTAG: Sorry, but after reading all this I still can't understand. Is the voltage across Q2 not equal to that across Q1? I thought it should be since the currents are the same. When I try to enter twice what I think the minimum vDS1 should be for Q1 it tells me I'm wrong. Can anyone tell me what I'm missing? FirstChildUserIdTAG: 434869 FirstChildUserNameTAG: patey FirstChildCreateTimeTAG: 2012-10-14T01:49:56Z SecondChildTAG: OK, I have finally understood. The DS voltages across the two devices cannot be the same (or vOUT wouldn't vary with vIN!). But the voltage vDS2=vGS2 and this is fixed by the current (which is fixed by Q1). It is quite simple when you realise! SecondChildUserIdTAG: 434869 SecondChildUserNameTAG: patey SecondChildCreateTimeTAG: 2012-10-14T11:53:06Z IndexTAG: 3686 TitleTAG: homework how can i submit my homework and lab? i'm so late i write answers in 1st week's homework and lab but when i logout answers will be destroyed UserIdTAG: 547566 UserNameTAG: nikhil_patel509 CreateTimeTAG: 2012-10-04T09:25:58Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 3 FirstChildTAG: i missed my 3rd weeks HW & lab , how can i overcome that loss ??? FirstChildUserIdTAG: 91261 FirstChildUserNameTAG: mr_sophisticated FirstChildCreateTimeTAG: 2012-10-04T10:47:03Z FirstChildTAG: First decide if you need to worry - the lowest two homework scores will be dropped; likewise the lowest two labs. FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-10-05T15:17:11Z FirstChildTAG: facing problem in submitting homework and labs. can anyone help me out? FirstChildUserIdTAG: 588785 FirstChildUserNameTAG: ahsan810 FirstChildCreateTimeTAG: 2012-10-15T14:42:13Z IndexTAG: 3687 TitleTAG: H4P3 Part B I have tried to find In many times.But i couldn't.Than i tried to find the thevenin voltage.Now in respect to port B thevenin voltage should be same as Vo.Am i doing anything wrong here? Again i tried to use KVL method in this way: V0=iR1 + u v0=i(r1+r2+r3)+au But according to this equation i is negative.What should i do? UserIdTAG: 92895 UserNameTAG: shihab2555 CreateTimeTAG: 2012-10-04T06:08:38Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Right, Vth is the same as the voltage at the node of the output port B. Your current is negative with respect to the loop direction you are using. Just because you assume the current, for example, goes in a clock wise loop; it doesn't mean it's true. Your negative sign means that it is going the opposite direction of your loop. In this case, this probably means that your voltage dependent voltage source is at a higher potential than the 5volt source; lazy electrons (moving electrons = current) are always seeking ground, just as water flows downhill. BTW, where is the independent (5V) voltage source in your loop equation. Also, follow KVL method, and create the whole equation for the loop; always setting it to zero. You can always check that equation with the current you solve for; making sure it equals zero. example: look at the following circuit representation. <--(-+)10V---^^^^5Ohms-----100Ohms----(+-)03V---> ------------------------------------------------> clock wise loop direction <----------------------------------------------- KVL eq.) 10V - i(50) - i(100) - 30V = 0 -i(50 + 100) = +30 - 20V -i(150) = +20 -i = 20/150, or 2/15 (reduced) THEREFORE i= -2/15. So the current is actually going counter clockwise; or opposite the direction I am referencing. Hope this helps. -MJ FirstChildUserIdTAG: 61923 FirstChildUserNameTAG: MikeJones FirstChildCreateTimeTAG: 2012-10-04T08:57:20Z SecondChildTAG: how to calculate the value of u? SecondChildUserIdTAG: 122513 SecondChildUserNameTAG: Mona77 SecondChildCreateTimeTAG: 2012-10-05T20:22:10Z IndexTAG: 3688 TitleTAG: H4P3 I have being able to find norton resistance in part B.Than the norton current should be Vo/Rn.But my answer is not correct.Am i doing anything wrong? UserIdTAG: 92895 UserNameTAG: shihab2555 CreateTimeTAG: 2012-10-04T05:52:10Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Which method did you use to solve for the Resistance, and did you account for the current of the branch of the dependent voltage source. Hint. short the output terminals. Vth/Isc = Rth = Rn. In=Vth/Rn Remember, you are finding the current coming out of the circuit; use KCL method. All currents leaving a node sum to 0. Branch1 current + Branch2 current + Isc = 0.; etc. BTW, what is the voltage at ground. Remember you are essentially attaching a bare wire from the output node to ground; this makes it easy to find the voltage of that node (i.e. Gnd, or 0V) Good Luck. MJ FirstChildUserIdTAG: 61923 FirstChildUserNameTAG: MikeJones FirstChildCreateTimeTAG: 2012-10-04T09:05:12Z IndexTAG: 3689 TitleTAG: [S8E0] For problem 1 Invalid Input I have $X*Y*-Z$, and it says invalid input. It would appear that it only likes negative symbols at the beginning. UserIdTAG: 248807 UserNameTAG: 1kingsman CreateTimeTAG: 2012-10-04T04:41:54Z VoteTAG: 0 CoursewareTAG: Week 4 / Dependent Sources Exercise 0 CommentableIdTAG: 6002x_dep_src_exsz_0 NumberOfReplyTAG: 1 FirstChildTAG: that mean that the variables used are invalid (dont use vB and iB) substitut them FirstChildUserIdTAG: 236136 FirstChildUserNameTAG: Hamid-ch FirstChildCreateTimeTAG: 2012-10-05T02:18:49Z SecondChildTAG: Not this time. The variables I used were correct (I chose to remove them from this post, as to follow the honor code), just the problem couldn't understand $*-$ together. SecondChildUserIdTAG: 248807 SecondChildUserNameTAG: 1kingsman SecondChildCreateTimeTAG: 2012-10-05T03:42:00Z IndexTAG: 3690 TitleTAG: Lab 4 V-i curves Hello, the online tool doesnt give me any curve for V-i, it was V-t.Is there a way to set the axis to V instead of t. The circuit is marked correct. UserIdTAG: 342243 UserNameTAG: drrahulbasu CreateTimeTAG: 2012-10-04T00:59:01Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Click on the voltage probe. When the dialog box opens select x-axis. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-04T02:05:08Z FirstChildTAG: I'm missing something. When I click on the probe, all I can change is the color and the offset. How do I get any other options? FirstChildUserIdTAG: 375500 FirstChildUserNameTAG: Brej7665 FirstChildCreateTimeTAG: 2012-10-05T10:10:36Z SecondChildTAG: When you click on the probe click the drop down box for the color options and the last one is x-axis, select it. SecondChildUserIdTAG: 282642 SecondChildUserNameTAG: snakeman SecondChildCreateTimeTAG: 2012-10-05T10:51:56Z IndexTAG: 3691 TitleTAG: week 4 hi any body may hint me for solving this problem.. " Find the value for the incremental resistance of the nonlinear element N by linearizing the expression for iA about the operating point when vI=5.0V." please.. UserIdTAG: 227508 UserNameTAG: bhavyab CreateTimeTAG: 2012-10-04T00:03:14Z VoteTAG: 0 CoursewareTAG: Week 4 / Linearization Exercise 1 CommentableIdTAG: 6002x_linearization_ex_1 NumberOfReplyTAG: 1 FirstChildTAG: Just find dI(U)/dU, and then find dI(Ua), when V1 is 5V. And Rl is 1/(dI(Ua)/dt) ---------- dI/dU=(10)'-(10*exp(-0.2*U))'=0-(-0.2*exp(-0.2*U))=5*exp(-0.2U); dI'(Ua)=dI(1.088)=... R=1/(dI'(Ua))=.... FirstChildUserIdTAG: 345464 FirstChildUserNameTAG: Constantine_ru FirstChildCreateTimeTAG: 2012-10-04T07:32:57Z SecondChildTAG: ТЫ не правильно взял производную, Константин. (10*exp(-U/5))'=10exp(..)*(-1/5)=2*exp(..). м? SecondChildUserIdTAG: 215271 SecondChildUserNameTAG: JustStudent SecondChildCreateTimeTAG: 2012-10-05T10:15:40Z IndexTAG: 3692 TitleTAG: Stuck with the resistor in HW4P3-A and B Hello, Though I used the same method to find the value of the Thevenin resistor been indicated in S8E2, but the answer is always wrong. Can anyone please explain to me what happened ?! UserIdTAG: 418182 UserNameTAG: euldji2005 CreateTimeTAG: 2012-10-03T22:29:46Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Connect a current source to the output, short **independent** voltage sources and open **independent** current sources, compute output voltage and divide by I. In Part A you will discover that the current controlled voltage source behaves like a resistor when calculating Rth. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-04T00:00:56Z SecondChildTAG: Thank you so much for your reply. I appreciate it very much SecondChildUserIdTAG: 418182 SecondChildUserNameTAG: euldji2005 SecondChildCreateTimeTAG: 2012-10-05T14:31:38Z IndexTAG: 3693 TitleTAG: H6P2 The small-signal model of this circuit is 2 VCCS in series in only 1 loop and vin providing the voltage that controls the current, but outside this loop. How can I find Rth and Vth across one of the VCCS, therefore? It's related to part 5 and 6 of the exercise. UserIdTAG: 39980 UserNameTAG: MarinoJr CreateTimeTAG: 2012-10-03T20:11:07Z VoteTAG: 0 CoursewareTAG: Week 6 / Small-signal circuit analysis of amplifier CommentableIdTAG: 6002x_small_signal_amplifier NumberOfReplyTAG: 4 FirstChildTAG: A more physical way of looking at the problem is the following. With the output open the current through Q1 and Q2 is the same. If you connect an incremental current source at the output the two currents will no longer be the same. The current through Q1 is determined by VIN and will not change, so the current through Q2 must decrease. That current is determined by VDD-VOUT, which must decrease. Since VDD fixed this requires VOUT to increase. Calculate the incremental increase in VOUT due to the incremental current source at the output and proceed from there. Note that this approach will also work for H6P1. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-03T23:30:25Z FirstChildTAG: Thanks for the hint, skyhawk. I found the current in dependent current sources as: for Q2: -K*(VDD-VOUT-VT)* vout and for Q1: K*(VIN-VT)*vin. Applying current law in the node (after connecting an independent current source of 1A), I calculated the expression for vout, but in terms of, among others permitted variables (as VT), VDD and VOUT, which are not permitted in the answer. FirstChildUserIdTAG: 39980 FirstChildUserNameTAG: MarinoJr FirstChildCreateTimeTAG: 2012-10-04T14:19:05Z FirstChildTAG: The large signal Vout of the circuit is VS-Vin. The small signal vout should therefore be -vin. Create the small signal model of the circuit by using one VCCS and letting this current source develop the necessary voltage over a resistor. FirstChildUserIdTAG: 145420 FirstChildUserNameTAG: pmac12345 FirstChildCreateTimeTAG: 2012-10-03T20:57:34Z SecondChildTAG: When you say VS, you mean VDD? Even so, I can't see that relationship. What is the resistor that you are talking about? I guess we are not observing the same circuit. SecondChildUserIdTAG: 39980 SecondChildUserNameTAG: MarinoJr SecondChildCreateTimeTAG: 2012-10-04T14:25:47Z SecondChildTAG: The Resistance of Q2 --> See Textbook page 417 Figure 8.13 SecondChildUserIdTAG: 149844 SecondChildUserNameTAG: amitaratnayake SecondChildCreateTimeTAG: 2012-10-04T15:38:23Z SecondChildTAG: amitaratnayake, now I realize. I was using another VCSS instead of a resistor. But the result is the same. SecondChildUserIdTAG: 39980 SecondChildUserNameTAG: MarinoJr SecondChildCreateTimeTAG: 2012-10-04T17:39:03Z SecondChildTAG: First: draw the small signal Circit on paper have a look and point out what voltage you are looking for... Then: On the lefthand side of Figure 8.13 .... there is a formular for v[ds] Have a look there and also think about the Titel of H6P2... I'am not 100% sure if this is the right way but my solution got accepted :) SecondChildUserIdTAG: 149844 SecondChildUserNameTAG: amitaratnayake SecondChildCreateTimeTAG: 2012-10-04T18:33:36Z SecondChildTAG: Thanks amitaratnayake. I got that. SecondChildUserIdTAG: 39980 SecondChildUserNameTAG: MarinoJr SecondChildCreateTimeTAG: 2012-10-04T19:46:49Z SecondChildTAG: can sum1 please help me find out vout in this case? SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-16T16:24:24Z SecondChildTAG: I spent a while on this today and then somewhere in these posts it was mentioned to think of what a MOSFET is. Thinking in the simplest terms back to when we first started with inverters etc. I got the answer. SecondChildUserIdTAG: 244115 SecondChildUserNameTAG: Pedro1969 SecondChildCreateTimeTAG: 2012-10-17T23:30:29Z FirstChildTAG: With the output open the currents through Q1 and Q2 are the same; therefore, VIN-VT = VDD-VOUT-VT since the constants for Q1 and Q2 are the same, which allows you to express VDD-VOUT in terms of VIN. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-04T21:49:44Z SecondChildTAG: To find RTH , both Q1 and Q2 will be like a resistor with resistance as shown on page 417. Using that i calculatd the value of VDS for both and then applied the parallel resistance formulae. But my answer is not correct. Can anyone help? SecondChildUserIdTAG: 365309 SecondChildUserNameTAG: chetnasinghaldas SecondChildCreateTimeTAG: 2012-10-15T12:27:52Z SecondChildTAG: chetnasinghaldas : how can u put Q1 equal to resistor shown in figure 8.13...it is the small signal model for MOSFET whose G=D SecondChildUserIdTAG: 278792 SecondChildUserNameTAG: sali SecondChildCreateTimeTAG: 2012-10-20T09:36:08Z SecondChildTAG: any hint to find VTH ? SecondChildUserIdTAG: 337430 SecondChildUserNameTAG: aparna_b SecondChildCreateTimeTAG: 2012-10-21T04:56:53Z IndexTAG: 3694 TitleTAG: H4P2 I can't understand what difference the zener makes. Why can't the the answer be the V1 less the v of the zener and everithing else like the other part of the question? UserIdTAG: 51667 UserNameTAG: Assuncao CreateTimeTAG: 2012-10-03T20:04:53Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I struggled with this problem and finally got it right. I found this site to be very helpful: http://www.allaboutcircuits.com/vol_3/chpt_3/11.html FirstChildUserIdTAG: 244115 FirstChildUserNameTAG: Pedro1969 FirstChildCreateTimeTAG: 2012-10-03T20:17:08Z SecondChildTAG: This website was very useful. Thanks a lot dear Pedro1969. I read it carefully and then I solved the homework in half hour. for getting Vo we should use 13.035 in our equations and then subtract the answer from VO (the first part of the question or 13v substitution in equations). The difference is the answer. SecondChildUserIdTAG: 374393 SecondChildUserNameTAG: rmaleki SecondChildCreateTimeTAG: 2012-10-04T20:51:17Z SecondChildTAG: Awesome glad it helped! SecondChildUserIdTAG: 244115 SecondChildUserNameTAG: Pedro1969 SecondChildCreateTimeTAG: 2012-10-04T23:13:23Z SecondChildTAG: how u get the value 13.035? SecondChildUserIdTAG: 122513 SecondChildUserNameTAG: Mona77 SecondChildCreateTimeTAG: 2012-10-05T05:39:44Z SecondChildTAG: Pedro1969, I would be taking you out for a beer this instant if you happened to live in Calgary. Thank you thank you thank you for that link. SecondChildUserIdTAG: 166031 SecondChildUserNameTAG: krebryna SecondChildCreateTimeTAG: 2012-10-06T01:11:33Z SecondChildTAG: Pedro1969 u r d man,ssly u hav made my day!!! Cheers SecondChildUserIdTAG: 375305 SecondChildUserNameTAG: sandy07 SecondChildCreateTimeTAG: 2012-10-07T19:34:41Z IndexTAG: 3695 TitleTAG: what about neglect of second differentialand higher order? it is explaining as I'm sorry, because delta vD is small, I'm going to neglect can u give me an example UserIdTAG: 295983 UserNameTAG: qassam CreateTimeTAG: 2012-10-03T19:43:26Z VoteTAG: 0 CoursewareTAG: Week 4 / Incremental Method Math CommentableIdTAG: 6002x_inc_method_math NumberOfReplyTAG: 0 IndexTAG: 3696 TitleTAG: Source Follower amp. Is Vs and Vout same for source follower amplifier??? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-03T18:39:41Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Almost, but not exactly. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-03T22:34:17Z SecondChildTAG: not at all SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-04T02:44:12Z IndexTAG: 3697 TitleTAG: offset hi friends ,can anyone explain what does dc offset(Id offset and Vd offset) practically mean? i can't get it as explained in s7v4 video. thanx UserIdTAG: 269641 UserNameTAG: BAUWA CreateTimeTAG: 2012-10-03T17:52:35Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3698 TitleTAG: H5P3 wrong input I wished to find vout by differentiating iDS*Rs with respect to vIN and then finally multiplying the answer by vin.it gave warning Invalid input: Could not parse (......) as a solution.I have tried it as a simplified one taking few common terms out.second way i entered it raw(without simplifying) then I tried to enter it in form of individual terms but the same warning reappeared. Please help UserIdTAG: 99628 UserNameTAG: Pranjal16 CreateTimeTAG: 2012-10-03T17:12:49Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: We need find the roots of vOUT = (RS*K*(vIN-VT-vOUT)^2)/2 choose the correct an then diferentiate! see page 349 of the book Please tell me if you wes able! FirstChildUserIdTAG: 299251 FirstChildUserNameTAG: TheRedBlackOne FirstChildCreateTimeTAG: 2012-10-03T17:23:32Z FirstChildTAG: i am putting (vin)*(((sqrt((2*K*RS(VIN-VT)+1)))-1)/(sqrt((2*K*RS(VIN-VT)+1)))) but it is still showing wrong input. FirstChildUserIdTAG: 99628 FirstChildUserNameTAG: Pranjal16 FirstChildCreateTimeTAG: 2012-10-04T08:20:37Z SecondChildTAG: Could you use MathJax to rewrite your formula? SecondChildUserIdTAG: 252722 SecondChildUserNameTAG: vaboro SecondChildCreateTimeTAG: 2012-10-04T20:26:49Z SecondChildTAG: I TRIED D SAME THING. ANSWER IS COMING WRONG.. SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-05T12:26:54Z SecondChildTAG: pranjal it should be VIN instead of Vin in your expression..Rest is fine.. I am also not getting it..Have you got the answer.? SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-05T13:08:12Z IndexTAG: 3699 TitleTAG: H4P2 HELP! Hi guys! I have been trying the second part involving the zener diode hours and I can't seem to arrive at the answer! First, i converted the circuit into its thevenin equivalent. Meaning 2k||1k. 1.Then, for vth, i feel kinda lost. Do i find vth for both AC and DC or just merely DC? 2. I also read where the resistance of the zener diode is 1ohms. Is this true? Can anyone kindly post a step by step solution for this question? Thank you so much! UserIdTAG: 499268 UserNameTAG: ruinoah CreateTimeTAG: 2012-10-03T16:18:58Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Remember: Differentiate!!!!! FirstChildUserIdTAG: 299251 FirstChildUserNameTAG: TheRedBlackOne FirstChildCreateTimeTAG: 2012-10-03T16:49:10Z SecondChildTAG: Sorry I think that was part 1, what is specifically your problem, which part? SecondChildUserIdTAG: 299251 SecondChildUserNameTAG: TheRedBlackOne SecondChildCreateTimeTAG: 2012-10-03T16:56:55Z SecondChildTAG: Sorry I`m on the moon, Vo and vo are voltage dividers, this cover the points 1,2,3,4. For the points 5,6,7,8 you need to solve this circuits:![Circuit to solve][1] an for the final you need to find a value in the worst case at least VI+vi [1]: https://edxuploads.s3.amazonaws.com/13492843501343663.png SecondChildUserIdTAG: 299251 SecondChildUserNameTAG: TheRedBlackOne SecondChildCreateTimeTAG: 2012-10-03T17:14:06Z SecondChildTAG: I think that i haven´t to post an answere specificaly but think (1V)/(1A) what is it? SecondChildUserIdTAG: 299251 SecondChildUserNameTAG: TheRedBlackOne SecondChildCreateTimeTAG: 2012-10-03T17:16:44Z SecondChildTAG: the value in worst case must be VI+vi?we must adjust the 1Ω properly? SecondChildUserIdTAG: 400832 SecondChildUserNameTAG: panosbr SecondChildCreateTimeTAG: 2012-10-03T19:34:54Z SecondChildTAG: yes the worst case is VI+Vi and VI is where the resitence starts. As you can see you need to find RS (value). SecondChildUserIdTAG: 299251 SecondChildUserNameTAG: TheRedBlackOne SecondChildCreateTimeTAG: 2012-10-03T20:48:45Z IndexTAG: 3700 TitleTAG: Lab4 curves Do your TRAN curves look like the ones shown in the picture? UserIdTAG: 64512 UserNameTAG: OZSorescu CreateTimeTAG: 2012-10-03T16:16:57Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: Yes, but I put diferent colours!!! FirstChildUserIdTAG: 299251 FirstChildUserNameTAG: TheRedBlackOne FirstChildCreateTimeTAG: 2012-10-03T16:49:47Z FirstChildTAG: Yes, they do. You just have to be careful assigning the values for the mosfets and the sources FirstChildUserIdTAG: 210561 FirstChildUserNameTAG: Alejovillapar FirstChildCreateTimeTAG: 2012-10-03T16:54:10Z SecondChildTAG: the default W/L value is 2. so make sure you change to what the question give. SecondChildUserIdTAG: 402850 SecondChildUserNameTAG: y1111 SecondChildCreateTimeTAG: 2012-10-07T20:58:21Z FirstChildTAG: Figured it out in the End. Pays to read carefully the whole description. Thank you for the answers. FirstChildUserIdTAG: 64512 FirstChildUserNameTAG: OZSorescu FirstChildCreateTimeTAG: 2012-10-03T17:00:30Z FirstChildTAG: I get voltage vs time. How do I put current and voltage on the axes, there is no option in the input to set the x, y axes? FirstChildUserIdTAG: 342243 FirstChildUserNameTAG: drrahulbasu FirstChildCreateTimeTAG: 2012-10-03T17:11:36Z SecondChildTAG: you have... on the probe, from where you set the color you can ... choose x axis SecondChildUserIdTAG: 64512 SecondChildUserNameTAG: OZSorescu SecondChildCreateTimeTAG: 2012-10-03T17:38:17Z IndexTAG: 3701 TitleTAG: better Explanation of (d) I do not understand the tutorial for (d). The incremental resistance is given by the inverse of the diA/dvA? Was this taught in the teorethical sequnces? And the next doubt is the following: the inverse of a exp is a ln, am i right? So, the inverse of 2*exp(-vA/5) should be -5*ln(vA/2) that with a vA equal to 2.39V gives -0.8907??? Unless here the inverse is just 1/x... in which case we would have 1/1.24 what would result in 0.806 and not 0.807. Sorry for my current bad maths... I am working to re-learn it... Thanks for the clarification. UserIdTAG: 18880 UserNameTAG: pcarmo CreateTimeTAG: 2012-10-03T15:52:08Z VoteTAG: 0 CoursewareTAG: Week 4 / Small Signal Long Tutorial CommentableIdTAG: 6002x_small_sig_long_t NumberOfReplyTAG: 4 FirstChildTAG: Also i have a problem with point d, i have done the derivative of iA, but what is the derivative of vA? FirstChildUserIdTAG: 216115 FirstChildUserNameTAG: Orsi FirstChildCreateTimeTAG: 2012-10-03T16:07:34Z FirstChildTAG: Conductance $G=\frac{I}{V}$ Resistance $R=\frac{V}{I}$ FirstChildUserIdTAG: 184827 FirstChildUserNameTAG: DiegoT FirstChildCreateTimeTAG: 2012-10-04T05:52:09Z FirstChildTAG: $(2e^(-x/5))^-1$ = $\frac{e^(x/5)}{2}$ In this case x = $v_A$ = 2.393 r = 0.8069 FirstChildUserIdTAG: 184827 FirstChildUserNameTAG: DiegoT FirstChildCreateTimeTAG: 2012-10-04T06:02:29Z FirstChildTAG: what is mathematica?? anyone? have never used it? FirstChildUserIdTAG: 368712 FirstChildUserNameTAG: jmen FirstChildCreateTimeTAG: 2012-10-19T15:16:31Z IndexTAG: 3702 TitleTAG: H6P2 part 5 and 6 How to find the Thevenin equivalent open-circuit voltage at the output of the small-signal circuit if the circuit is 2 VCCS in series? The same question for Rth. UserIdTAG: 39980 UserNameTAG: MarinoJr CreateTimeTAG: 2012-10-03T15:29:01Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: just make the small signal equivalent.. FirstChildUserIdTAG: 310474 FirstChildUserNameTAG: aldaris565 FirstChildCreateTimeTAG: 2012-10-03T16:28:07Z SecondChildTAG: 2 VCCS in series SecondChildUserIdTAG: 39980 SecondChildUserNameTAG: MarinoJr SecondChildCreateTimeTAG: 2012-10-03T17:10:48Z FirstChildTAG: It's important to note that the voltage controlled current source representing Q2 is controlled by the output voltage (with a minus sign), while the voltage controlled current source representing Q1 is controlled by the input voltage. Write the small signal equivalent and the answers are obvious. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-04T01:58:27Z FirstChildTAG: What is the relationship between VDD, VIN and VOUT? Differentiate to get the small-signal gain and you know why H6P2 is called a "Phase Inverter" :) FirstChildUserIdTAG: 259653 FirstChildUserNameTAG: CKW FirstChildCreateTimeTAG: 2012-10-04T13:54:10Z SecondChildTAG: I'm stucked in that question, also. The small-signal gain of Q2 relates to VDD and VOUT. SecondChildUserIdTAG: 39980 SecondChildUserNameTAG: MarinoJr SecondChildCreateTimeTAG: 2012-10-04T14:32:02Z SecondChildTAG: or stuck. SecondChildUserIdTAG: 39980 SecondChildUserNameTAG: MarinoJr SecondChildCreateTimeTAG: 2012-10-04T14:33:08Z FirstChildTAG: Now I realize. The small-signal equivalent of a 2-terminal MOSFET is a Resistor (page 417 of the Textbook). And VDS_in_Q2 = VIN. With this, you solve both parts. FirstChildUserIdTAG: 39980 FirstChildUserNameTAG: MarinoJr FirstChildCreateTimeTAG: 2012-10-04T18:06:09Z SecondChildTAG: thats what i said ;) SecondChildUserIdTAG: 149844 SecondChildUserNameTAG: amitaratnayake SecondChildCreateTimeTAG: 2012-10-04T19:02:22Z SecondChildTAG: Sorry for insisting, I got Vth, but I'm stuck in rth, if I inject a current 'i' then I get that my current should be 0 from voltage inversion. I'm using equivalent resistance for 2-terminal Mosfet, any hint is welcome! SecondChildUserIdTAG: 148389 SecondChildUserNameTAG: chento SecondChildCreateTimeTAG: 2012-10-17T00:01:07Z IndexTAG: 3703 TitleTAG: superpositioning with nonlinear sources? I think, that superposition works only for linear depended sources. But in S8E1, the depended source is not linear. Where is my error in reasoning. UserIdTAG: 389333 UserNameTAG: juergen CreateTimeTAG: 2012-10-03T13:15:34Z VoteTAG: 0 CoursewareTAG: Week 4 / Dependent Sources Exercise 1 CommentableIdTAG: 6002x_dep_src_exsz_1 NumberOfReplyTAG: 2 FirstChildTAG: if the dependent source has a linear expression, it will also work. If you read the text book in the superposition chapter there is a mention regarding this issue. FirstChildUserIdTAG: 64512 FirstChildUserNameTAG: OZSorescu FirstChildCreateTimeTAG: 2012-10-03T13:52:48Z SecondChildTAG: the problem ist, that the expression of f(v) is not linear. I'm confused SecondChildUserIdTAG: 389333 SecondChildUserNameTAG: juergen SecondChildCreateTimeTAG: 2012-10-03T14:30:43Z FirstChildTAG: Hi juergen, How are you trying to use superposition in that exercise? I think it is easier to solve either with KCL (node method) or KVL (loop method), and those work for linear and non-linear circuits FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-10-03T13:55:29Z SecondChildTAG: The hint advises it! It's totally misleading. I solved it with node method and then spent bunch of time trying understand how it's possible to apply superposition - I didn't get clear answer. SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-06T18:18:32Z SecondChildTAG: oh, sorry, it was about E1. Could you please clarify how to use superposition in E2 - I didn't find the way. SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-06T18:24:09Z IndexTAG: 3704 TitleTAG: S11E3: SOURCE FOLLOWER, how to find the solution? I can´t see the solution to "a" and "rout", can anyone explain it? UserIdTAG: 299251 UserNameTAG: TheRedBlackOne CreateTimeTAG: 2012-10-03T11:24:20Z VoteTAG: 0 CoursewareTAG: Week 6 / Source-follower small-signal exercise CommentableIdTAG: 6002x_small_signal_source_follower NumberOfReplyTAG: 1 FirstChildTAG: To find "a", in ids=gm*vgs, you may find vout/vin using vgs=vin-vout and ids=vout/RS To find "rout", use the small-signal model of the source-follower and add a 1A Current Source in parallel with RS, and find the voltage (Vth) across it. Hence, rout=Vth/1 FirstChildUserIdTAG: 39980 FirstChildUserNameTAG: MarinoJr FirstChildCreateTimeTAG: 2012-10-03T12:40:04Z SecondChildTAG: for source follower, just go through the text book.There is a solved example there.all your queries will get answered there...TRY IT ONCE! SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-04T03:18:17Z SecondChildTAG: The source follow model in the second figure where the MOSFET is replaced with a current source is liner. This is why Ohm's law $V=I \cdot R$ applies directly to it. SecondChildUserIdTAG: 252722 SecondChildUserNameTAG: vaboro SecondChildCreateTimeTAG: 2012-10-05T22:01:46Z SecondChildTAG: @ jmen - how about a page number? SecondChildUserIdTAG: 18073 SecondChildUserNameTAG: gburkhart SecondChildCreateTimeTAG: 2012-10-13T02:52:10Z SecondChildTAG: This was hard to do. SecondChildUserIdTAG: 359310 SecondChildUserNameTAG: AaronYeoh SecondChildCreateTimeTAG: 2012-10-15T04:10:08Z SecondChildTAG: textbook 437 SecondChildUserIdTAG: 329254 SecondChildUserNameTAG: KGabor SecondChildCreateTimeTAG: 2012-10-15T13:56:15Z SecondChildTAG: Well, not exactly "hard", just took a bit of fiddling. If every time you use a 1 A independent current source, and find vout, then you just use the node method to find vout, since the node you should pick has a voltage of vout. SecondChildUserIdTAG: 248807 SecondChildUserNameTAG: 1kingsman SecondChildCreateTimeTAG: 2012-10-16T22:58:22Z SecondChildTAG: that helped a lot SecondChildUserIdTAG: 111540 SecondChildUserNameTAG: tksanthosh SecondChildCreateTimeTAG: 2012-12-02T11:47:23Z IndexTAG: 3705 TitleTAG: I give up I got to week 4, and it was hard, but manageable. At this point it is above my power of comprehension. I bow to those able to complete the series. UserIdTAG: 64512 UserNameTAG: OZSorescu CreateTimeTAG: 2012-10-03T09:27:52Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Stick around, it's all good FirstChildUserIdTAG: 18123 FirstChildUserNameTAG: femiSmile FirstChildCreateTimeTAG: 2012-10-03T09:39:26Z SecondChildTAG: If you are only watching the lectures and hoping to make sense of it all then I'm not surprised. Try reading the textbook, it actually explains things in a logical manner with many examples of each topic. Maybe i was just expecting too much for the lectures, but i'v found them more confusing then helpful for actually learning.... SecondChildUserIdTAG: 329964 SecondChildUserNameTAG: SmartMike SecondChildCreateTimeTAG: 2012-10-03T12:08:57Z FirstChildTAG: Please do not give in, everyone in this class is so nice and supportive. I had a bad week last week and was going to throw in the towel, and everyone said that it is ok to have a bad week just dont let it get you down keep trying and you will get it and guess what after trying I found out my mistakes and kept going FirstChildUserIdTAG: 160242 FirstChildUserNameTAG: Mlevins35 FirstChildCreateTimeTAG: 2012-10-07T01:55:19Z IndexTAG: 3706 TitleTAG: Bug around minute 5 There is a bug around minute 5. It does not affect the final result. UserIdTAG: 244841 UserNameTAG: eyubero CreateTimeTAG: 2012-10-03T09:02:47Z VoteTAG: 0 CoursewareTAG: Week 6 / Two-stage MOSFET amplifier CommentableIdTAG: 6002x_Two_Stage_MOSFET_amplifier NumberOfReplyTAG: 1 FirstChildTAG: I got confused there too. Should the correct equation at minute 5 be: $\frac{2}{RK}v_s+\frac{1}{(RK)^2}=(v_{MID}+\frac{1}{RK})^2$ ? FirstChildUserIdTAG: 138131 FirstChildUserNameTAG: fliang FirstChildCreateTimeTAG: 2012-10-08T19:32:34Z SecondChildTAG: Yes, that makes more sense. VS+(1/RK)^2 doesn't make sense as VS has units Volt and 1/(RK)^2 is Volts squared. SecondChildUserIdTAG: 266912 SecondChildUserNameTAG: pietvo SecondChildCreateTimeTAG: 2012-10-16T16:23:21Z IndexTAG: 3707 TitleTAG: Didn't know abt this course untill now. Joined late ! What losses will I be facing ? I joined the course at about the end of the 4th week of this course. Will I be able to submit my home works for the previous weeks? How am I going to be affected by this delay in my certification? What should I do ? Please help! UserIdTAG: 526952 UserNameTAG: DEVANKAR CreateTimeTAG: 2012-10-03T08:42:55Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: You can find the answers to those questions here on [Course syllabus][1], particularly #5 and #9 [1]: https://www.edx.org/static/content-mit-6002x/handouts/syllabus.a477535058a1.pdf FirstChildUserIdTAG: 470934 FirstChildUserNameTAG: sonhx FirstChildCreateTimeTAG: 2012-10-03T09:52:39Z IndexTAG: 3708 TitleTAG: for late submission of homework dear sir, i have not submit my homework and my due dates are also gone how can i submit it now. UserIdTAG: 202082 UserNameTAG: navin_gsd3 CreateTimeTAG: 2012-10-03T08:29:40Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: You cannot submit them for grading now, but you can carry on with the rest, try not to miss anymore. Have fun. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-03T14:13:54Z FirstChildTAG: im having problem or dont know how this work....how do i submit my lab and home FirstChildUserIdTAG: 462861 FirstChildUserNameTAG: edxforme FirstChildCreateTimeTAG: 2012-10-04T19:27:02Z SecondChildTAG: i did them but nothing or no button to submit. if i change the page all my work dissappears SecondChildUserIdTAG: 462861 SecondChildUserNameTAG: edxforme SecondChildCreateTimeTAG: 2012-10-04T19:28:49Z IndexTAG: 3709 TitleTAG: About exam again As we know der is a deadline for every week's homework..likewise does midterm and final have any deadlines?,reason behind m asking this is because i have my regular course engineering exam starting from nov 22 to dec 8('university' exam), and this course ends on (Dec 20, 2012),,so will it be fine if i will give midterm and final after my regular exams,,of course (excluding week's homework). UserIdTAG: 219442 UserNameTAG: aki31 CreateTimeTAG: 2012-10-03T06:29:31Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: plz help!.. FirstChildUserIdTAG: 219442 FirstChildUserNameTAG: aki31 FirstChildCreateTimeTAG: 2012-10-03T06:44:34Z SecondChildTAG: There is a deadline. They will give you 3-4 days to write the exam. However, once you begin the exam you will have 24 hours to complete it. These were the rules last time. I'm not sure if it will change. SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-10-03T08:30:35Z SecondChildTAG: is there any deadline for homework?.where and how it submitted can any one tell me the procedure as there is no guideline to submit it. SecondChildUserIdTAG: 202082 SecondChildUserNameTAG: navin_gsd3 SecondChildCreateTimeTAG: 2012-10-03T08:49:28Z IndexTAG: 3710 TitleTAG: Lab 3 - explanation mistake - Staff please check / correct ... Thank you [update: solved/corrected by Staff]. Hi, Consecuently of Gabriel007 [issue][1] with and excel of Lab 3, we have noticed while seeing the excel that the explanation of Lab 3 has a mistake in the explanation: RPU it is not Rn value ... RPU it should be Rpullup value... If you use RPU=Rn, W/L it is 22 and not as you are showing in the explanation (58.3)... Please correct, this can be confusing for the students when they calculate W/L. Thank you! Myriam. P.D: All credits goes to Gabriel007! ---- EDIT: solved/corrected by Staff. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/506bb3669be1831f000000a1# UserIdTAG: 58095 UserNameTAG: Myrimit CreateTimeTAG: 2012-10-03T05:19:55Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Thanks for pointing this out, we are fixing it now. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-10-03T12:24:06Z SecondChildTAG: :) Thank you Lyla! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-03T12:25:34Z IndexTAG: 3711 TitleTAG: K Factor Hi, I went through some mosfet datasheets looking for the different factors in the formula: ids=k/2 * (vgs-vt)^2. Found vt and, obviously, vds is a voltage you apply in your circuit. I just didn't find k. Does anybody know how to calculate it from any other value given in a mosfet datasheet? Or maybe there's an alternate nomenclature for it? How can I know what the k factor is for a determined mosfet? As an example, I took mosfet datasheets from On Semiconductor page: http://www.onsemi.com/pub_link/Collateral/NTNS3A91PZ-D.PDF UserIdTAG: 210986 UserNameTAG: JorgeRmz CreateTimeTAG: 2012-10-03T04:14:22Z VoteTAG: 0 CoursewareTAG: Week 5 / MOSFET Amplifier CommentableIdTAG: 6002x_mosfet_amp NumberOfReplyTAG: 2 FirstChildTAG: For a MOSFET K is the product of several terms. $K = \mu_nC_{ox}\frac{W}{L}$ $\mu_n$ is the effective mobility of electrons, $C_{ox}$ is the capacitance per unit area of the insulating gate oxide, $W$ is the transistor's gate width and $L$ is the transistors gate length FirstChildUserIdTAG: 370247 FirstChildUserNameTAG: Dijkstra FirstChildCreateTimeTAG: 2012-10-03T21:21:45Z SecondChildTAG: don't think you will find these parameters in the specs of industrial MOSFETs :) SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-04T03:44:05Z FirstChildTAG: This is an analog switch MOSFET and not intended to work in saturation regime as amplifier - I think it's why there is no any parameter related to K-factor For amplifying MOSFET you can figure out K from transconductance (gm) at given VDS and threshold voltage (VT): gm=(VDS-VT)*K K=gm/(VDS-VT) so, for http://www.vishay.com/docs/71065/71065.pdf it will be: VT=(VTmax+VTmin)/2=(1.5+0.6)/2=1.05 K=8/(10-1.05)=0.894 FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-10-03T19:12:51Z SecondChildTAG: Actually I think it's possible to calculate K from the table on the right side of the first page... SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-03T19:48:20Z IndexTAG: 3712 TitleTAG: Disappearing Math As I am trying to solve the problems given to us during the lectures, ex:S3E3, I am given my information and then it all disappears and it reads in red math error. This is hurting me a lot. How can I solve this problem? UserIdTAG: 478114 UserNameTAG: TheDoodleFiend CreateTimeTAG: 2012-10-03T01:26:15Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: It's your browser. Chrome works for me. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-10-03T01:49:52Z FirstChildTAG: how to repaire my tv its no on FirstChildUserIdTAG: 545956 FirstChildUserNameTAG: chawand FirstChildCreateTimeTAG: 2012-10-03T03:23:21Z IndexTAG: 3713 TitleTAG: Which method to use? With so many different methods to use (KVl, KCL, node etc) I find it very confusing to decide which one to use for a particular problem. Is there a convention for this or is the node method the best one to use for everything? Save me before my head explodes! UserIdTAG: 195916 UserNameTAG: pflynn CreateTimeTAG: 2012-10-02T20:52:22Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: When in doubt, use the Node Method. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-10-02T22:17:12Z IndexTAG: 3714 TitleTAG: H4P2 Problem I've gotten the 1st part of H4P2 done and the parts of the second half where we're supposed to calculate V0. I figured out how to model the zener as a equivalent source+resistor. But using the same procedure I can't get the v0 parts done. I think there too we need to model the diode in the same way but obviously I'm going wrong sompeplace. Could someone point that out? Or give me a hint to get going? I hope this doesn't violate the CoC in anyway. UserIdTAG: 278301 UserNameTAG: prateektaneja CreateTimeTAG: 2012-10-02T19:44:35Z VoteTAG: 0 CoursewareTAG: Week 4 / Small Signal Model CommentableIdTAG: 6002x_small_signal NumberOfReplyTAG: 1 FirstChildTAG: Assuming you figured out the op point, you would have known the Zener's resistance RZ by now. You only have to calculate v0 based on vi (as if it's fixed and not variable) and the given config of the 3 resistors. In my case, I had to enter very precisely the result, and as a hint, it's smaller than 10e-3 (a value derived from the sandbox's analysis wasn't good enough for me). I say this because i wasted a lot of time trying to figure out how good is enough for precision and it seems for this problem, unlike previous weeks, close is not enough, even though we might get the point about the utility of Zener from the Transient graph. Oh, and don't worry about RMS. FirstChildUserIdTAG: 184153 FirstChildUserNameTAG: tthngn FirstChildCreateTimeTAG: 2012-10-04T20:08:15Z IndexTAG: 3715 TitleTAG: Simulator diode Please explain the function of the Area parameter for a diode. I'd also like to know how to substitute an ON normal diode with a resistor, i.e. what would be the resistance? UserIdTAG: 184153 UserNameTAG: tthngn CreateTimeTAG: 2012-10-02T18:20:14Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3716 TitleTAG: There is a problem in downloading week 4 videos?? plz help us to download the videos!! UserIdTAG: 315703 UserNameTAG: Sohaib786 CreateTimeTAG: 2012-10-02T14:54:56Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Troubleshooting NumberOfReplyTAG: 0 IndexTAG: 3717 TitleTAG: Week 6 annotated and clean lecture slides When will the week 6 lecture slides be posted? I would like to be able to print them out before I watch the videos. UserIdTAG: 64151 UserNameTAG: guba11ro CreateTimeTAG: 2012-10-02T14:38:26Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3718 TitleTAG: Lab 4 I am getting my Ids as zero always..where do you connect the current probe?and what should Vds be? UserIdTAG: 156585 UserNameTAG: AdiRanga CreateTimeTAG: 2012-10-02T11:37:29Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: read the first question, just after the sandbox, and u will figure it out(i was stuck too...)! FirstChildUserIdTAG: 301962 FirstChildUserNameTAG: labrinim FirstChildCreateTimeTAG: 2012-10-02T11:48:42Z SecondChildTAG: I dint get u.... can u please elaborate!!! SecondChildUserIdTAG: 156585 SecondChildUserNameTAG: AdiRanga SecondChildCreateTimeTAG: 2012-10-02T11:55:29Z SecondChildTAG: it says: "report vDS/iDS" so, the current probe and the voltage probe should be connected to the same node. and if you still don't get a result, check that all the sources are grounded. and if you still don't get a result, the you haven't used the appropriate DS source... SecondChildUserIdTAG: 301962 SecondChildUserNameTAG: labrinim SecondChildCreateTimeTAG: 2012-10-02T12:23:24Z SecondChildTAG: what DS source should you use?i used the same one that they used for the resistor example!!!ie the triangular waveform SecondChildUserIdTAG: 156585 SecondChildUserNameTAG: AdiRanga SecondChildCreateTimeTAG: 2012-10-02T12:32:00Z SecondChildTAG: thats right! SecondChildUserIdTAG: 301962 SecondChildUserNameTAG: labrinim SecondChildCreateTimeTAG: 2012-10-02T12:34:20Z SecondChildTAG: we must use only one mosfet? SecondChildUserIdTAG: 400832 SecondChildUserNameTAG: panosbr SecondChildCreateTimeTAG: 2012-10-03T20:26:54Z FirstChildTAG: I think current probe and voltage probe are not connected to the same node. Textbook says fet's output is a function of input alone but separate. What we want to do is measure the output current iDS (which depends on vGS). For vDS the instruction says 1V. So answer to first question is 1/iDS whatever iDS you can find by TransAn at vGS=3V, assuming your circuit is right. Mine is accepted (from simu to measurements used in answers), but the curves produced look quite different than the illustration in the instructions, or those in the textbook. I spent a few hours trying to bend the curves to how they're supposed to look, including using diodes (actually the diodes are for a different purpose in this lab, as mentioned in Food for thought at the bottom) since there is no resistor available. Anyone knows how please comment. FirstChildUserIdTAG: 184153 FirstChildUserNameTAG: tthngn FirstChildCreateTimeTAG: 2012-10-02T18:33:21Z IndexTAG: 3719 TitleTAG: MID TERM EXAM when is the mid term exam and how are we going to attend it? UserIdTAG: 413613 UserNameTAG: shaliesh CreateTimeTAG: 2012-10-02T08:54:01Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: This link shows the dates of the lectures, homework and lab releases and the midterm and final exam as well. https://www.edx.org/static/content-mit-6002x/handouts/at-a-glance.9674fe7f677e.pdf There are some dates announced when you can take your exam in the same way that you took the homework and Lab assignments, but you will only have 24 hours to complete it. This means that once you open your exam you must complete it in a day (choosing from dates 22-25 if i am not mistaken of October) FirstChildUserIdTAG: 304587 FirstChildUserNameTAG: Vasso FirstChildCreateTimeTAG: 2012-10-02T11:18:20Z SecondChildTAG: Is the Check button (not Show, obviously) available at exams? It helps in so many ways. Thanks. SecondChildUserIdTAG: 184153 SecondChildUserNameTAG: tthngn SecondChildCreateTimeTAG: 2012-10-02T20:49:19Z SecondChildTAG: Never mind, i should have searched the forum first. Thanks. SecondChildUserIdTAG: 184153 SecondChildUserNameTAG: tthngn SecondChildCreateTimeTAG: 2012-10-02T20:56:46Z SecondChildTAG: you will get only 3 checks per qs...:( SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-04T03:21:04Z FirstChildTAG: please tell me about grading FirstChildUserIdTAG: 94196 FirstChildUserNameTAG: mishra_prakhar FirstChildCreateTimeTAG: 2012-10-02T15:36:14Z IndexTAG: 3720 TitleTAG: Difficulty in taking Lectures Hi As most of the peoples are affected by the YouTube Ban in there countries. We are really affected by it. However Edx has provided a link to download videos and then watch them, it is really time consuming as most of us have only 1 MB data connection. It takes a lot time to First download and then watch videos. As Edx initiative is education for all. It should also make it user friendly to make sure every one can get benefit from it in his premises. I request Edx to embed these videos in it own site for example like the way other sites ( e.g Cousera.org ) do.. I believe this is the most effective way. I hope you will consider my request and provide as with the best. Thanks UserIdTAG: 272417 UserNameTAG: MuKhan CreateTimeTAG: 2012-10-02T08:34:59Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3721 TitleTAG: New Registration - Late I have registered myself on October 2nd. Can I still go ahead with the course? Will it affect my final scores? UserIdTAG: 539018 UserNameTAG: Aditya94 CreateTimeTAG: 2012-10-02T07:42:21Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: It won't affect much.Your 3 weeks homeworks and labs exercices are lost,bt for final grading only top 10 out of the total of 12 exercises wud be considered.anyways u can watch the video lectures anythime. FirstChildUserIdTAG: 334777 FirstChildUserNameTAG: AAYUSH007 FirstChildCreateTimeTAG: 2012-10-02T08:02:09Z SecondChildTAG: You've only missed out on 3% of the final mark if you complete the rest of the course. SecondChildUserIdTAG: 99941 SecondChildUserNameTAG: Nexus1974 SecondChildCreateTimeTAG: 2012-10-02T08:06:15Z SecondChildTAG: Sorry was meant to reply to the OP SecondChildUserIdTAG: 99941 SecondChildUserNameTAG: Nexus1974 SecondChildCreateTimeTAG: 2012-10-02T08:10:51Z SecondChildTAG: I also have the same challenge. please how can I download the previous videos? SecondChildUserIdTAG: 539343 SecondChildUserNameTAG: Osinimu SecondChildCreateTimeTAG: 2012-10-02T08:51:22Z FirstChildTAG: You've only missed out on 3% of the final mark if you complete the rest of the course. FirstChildUserIdTAG: 99941 FirstChildUserNameTAG: Nexus1974 FirstChildCreateTimeTAG: 2012-10-02T08:10:25Z IndexTAG: 3722 TitleTAG: H2_P1 Solution The requirements for the problem states that Vout/Vin must be within %10 of the required (nominal) value. This is the way I understood the problem. For example: The worked solution Vo/Vin (nominal) = 7.5/30 = .25; (+/-)%10 [.275,.225]. However, the solution's max, and min ratios' are .282,.202 (outside the %10 requirement). I guess I was thrown by the wording of the problem. I couldn't find a resistor combo for my problem's values to meet the apparent %10 deviation from nominal Vo/Vi ratio constraint. BTW, I am referring to the following statement in the problem. "Come up with resistors R1 and R2 such that the divider ratio Vout/Vin is within 10% of the requirement". The requirement being desired Vo/Vin(source). Thanks, M.J UserIdTAG: 61923 UserNameTAG: MikeJones CreateTimeTAG: 2012-10-02T06:59:42Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: M.J., notice that the problem state in the previous sentence that you should assume that the resistors have their nominal value first. That is, without considering the +/- 10%. Anyway, I agree with you that the problem description could be improved to avoid this confusion. FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-10-02T14:17:47Z IndexTAG: 3723 TitleTAG: registering to the edx.org Hello sir, my friend wants to join the MITx:6.002 course now. Please help him to join and to learn more. I thought he can follow the weeks. What is the solution for it? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-02T06:12:53Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: https://www.edx.org I don't think it is too late, send him to the link, you can help him sign up. Have fun. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-02T23:15:57Z IndexTAG: 3724 TitleTAG: Lab5 I have a doubt regarding Q3 of Lab5. It seems to be a very easy by just clicking on tran and measuring max output voltage and max input voltage and dividing it by each other with a negative sign for the 180deg phase shift. But still it is saying wrong answer.Am i going wrong anywhere. UserIdTAG: 363482 UserNameTAG: preeteshs2012 CreateTimeTAG: 2012-10-02T05:19:33Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I am stuck too. :( The same thought here ! FirstChildUserIdTAG: 97340 FirstChildUserNameTAG: AnkitRana FirstChildCreateTimeTAG: 2012-10-02T10:20:26Z SecondChildTAG: I can't find the values too.. SecondChildUserIdTAG: 122513 SecondChildUserNameTAG: Mona77 SecondChildCreateTimeTAG: 2012-10-13T11:22:19Z FirstChildTAG: Hello friends. The question says something about amplitudes, not max values. FirstChildUserIdTAG: 309359 FirstChildUserNameTAG: abarea10 FirstChildCreateTimeTAG: 2012-10-02T19:11:43Z SecondChildTAG: Could u please be a little more clear about what u want to say SecondChildUserIdTAG: 363482 SecondChildUserNameTAG: preeteshs2012 SecondChildCreateTimeTAG: 2012-10-03T15:05:18Z SecondChildTAG: ur max and minimum voltages are not same for the o/p waveform. neither are they same for the input waveform.. so first find the span of o/p voltage=max voltage achieved-minimum voltage achieved .do d same fr input and then find the gain. SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-04T03:24:42Z SecondChildTAG: That's the answer. I told you only a hint becouse, when you notice where the trick is, it becomes very intuitive. SecondChildUserIdTAG: 309359 SecondChildUserNameTAG: abarea10 SecondChildCreateTimeTAG: 2012-10-04T19:04:20Z SecondChildTAG: You can take a look [here][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/507628b2c4dd80250000004b SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-10-11T02:11:21Z IndexTAG: 3725 TitleTAG: Begining of the course I've just started the course and I don´t know if I could complete it satisfactorily. I would like if I can continue with it without problems at this point, overtake the topics viewed and recieve the certificate without problems. Thanks UserIdTAG: 537549 UserNameTAG: davidleot7 CreateTimeTAG: 2012-10-02T04:56:25Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: davidleot7 welcome to 6.002x! Good luck with the next assignments. The first three homeworks and labs were already submitted but remember that 2 of them won't be counted towards the final score, so you still have chances of receiving the certificate! FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-10-02T14:06:18Z IndexTAG: 3726 TitleTAG: Anybody knows how to change the e-mail adress in edx? Please, I need change mine UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-01T22:23:22Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3727 TitleTAG: MathJax bug report S7E1 won't display on ie8 but ok on ie9 running internet explorer 8 on win7HE/sp1 32bit all recommended Winups (except ie9) shows a bunch of [Math Processing Error] or the proper formulae etc but in the markup language rather than superscripted and readable. side by side, ie9 on windows7sp1 64bit works fine to display and calc on Incremental Analysis Lecture, S7E1 LINERAIZATION. UserIdTAG: 331542 UserNameTAG: james77901 CreateTimeTAG: 2012-10-01T21:51:09Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Troubleshooting NumberOfReplyTAG: 0 IndexTAG: 3728 TitleTAG: H2P2 Zene Regulator I lost some time thinking that before Vd = -5 and after Vd = 0.6 I could substitute for a simple resistor. However the relation isn't linear. i.e: before Vd = -5 (id = vd - 5). And with this and the node method I could solve the problem. Remember Vd=-V0 UserIdTAG: 110214 UserNameTAG: jeffsjunior CreateTimeTAG: 2012-10-01T18:25:46Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You could also substitute the zener by its equivalent model in a particular region. In the region Vd < -5V, the model is an independent voltage source in series with a resistor, can you see why? FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-10-02T13:57:34Z SecondChildTAG: Because if You look on i(V) graph, You can see, that for each branch (V<0 and V>0) graph seems like a resistor with some offset from "0" point. So You can model zener diode like a resistor and offset voltage source. SecondChildUserIdTAG: 195218 SecondChildUserNameTAG: Al_Incognito SecondChildCreateTimeTAG: 2012-10-02T16:07:31Z SecondChildTAG: Exactly, that is the reason. I think that it is easier to solve the problem by replacing with the equivalent model, because then it is a linear circuit and all the techniques studied so far can be used. SecondChildUserIdTAG: 381746 SecondChildUserNameTAG: jelizon SecondChildCreateTimeTAG: 2012-10-02T16:44:59Z IndexTAG: 3729 TitleTAG: question regarding to "full name" i have written my full name as "fayzan ahmed" . .. . but is should be "Fayzan Ahmed" so what will appear on certificate and how can i chage my full name to "Fayzan Ahmed" here. . . any suggestions please... UserIdTAG: 146930 UserNameTAG: fayzan_007 CreateTimeTAG: 2012-10-01T17:32:48Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3730 TitleTAG: Ooops... Piece of cake, except second question... Can`t understand, how 3 volts devided by 2 mA equals 2k ????? Could somebody explain? UserIdTAG: 345502 UserNameTAG: ElectroVova CreateTimeTAG: 2012-10-01T17:24:40Z VoteTAG: 0 CoursewareTAG: Week 3 / Analyzing Nonlinear Circuits Exercise CommentableIdTAG: 6002x_analyzing_nonlinear_circuits_exercise NumberOfReplyTAG: 1 FirstChildTAG: its easy. R=delta_V/delta_i= (V2-V1)/(i2-i1)= (3V-1V)/(2mA-1mA)=2V/1mA= 2kOhm FirstChildUserIdTAG: 325197 FirstChildUserNameTAG: Vitali_Jerin FirstChildCreateTimeTAG: 2012-10-01T21:24:25Z SecondChildTAG: IT'S BECAUSE IT ISN'T 3V DIVIDED BY 2mA IT SHOULD BE: (3-1) VOLTS DIVIDED BY 2mA SecondChildUserIdTAG: 65651 SecondChildUserNameTAG: Arnaldo_Martell SecondChildCreateTimeTAG: 2012-10-02T23:06:36Z SecondChildTAG: But why so? SecondChildUserIdTAG: 500965 SecondChildUserNameTAG: barka0 SecondChildCreateTimeTAG: 2012-10-03T09:58:51Z SecondChildTAG: ![enter image description here][1] Because when you divide 3v by 2mA you get slope for the red line in the graph i posted. You need to find the slope for the blue line. As you remember slope of the VI relation graph gives R [1]: https://edxuploads.s3.amazonaws.com/13495230051610395.png SecondChildUserIdTAG: 364894 SecondChildUserNameTAG: tremo SecondChildCreateTimeTAG: 2012-10-06T11:30:34Z IndexTAG: 3731 TitleTAG: Problem with Question 5 of Lab 6 Question 5 of Lab 6 says: "Finally, change the $Vin$ voltage source so that input is making a $1→0$ transition (just interchange the initial and plateau voltages)..." However, I am not able to edit the voltage source values in the first schematic provided for the lab. UserIdTAG: 153760 UserNameTAG: Kavka CreateTimeTAG: 2012-10-01T16:46:44Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: How are you getting to Lab 7? I only see 6 weeks of courseware. FirstChildUserIdTAG: 401592 FirstChildUserNameTAG: netghost FirstChildCreateTimeTAG: 2012-10-01T16:56:53Z SecondChildTAG: Because he lives in a parallel universe, where his current time is our future SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-10-02T13:16:49Z SecondChildTAG: Sorry, I meant Lab 6. I've updated the post. SecondChildUserIdTAG: 153760 SecondChildUserNameTAG: Kavka SecondChildCreateTimeTAG: 2012-10-02T17:42:57Z IndexTAG: 3732 TitleTAG: Staff: Contact for a personal query please. Hi, I need to speak or email a staff member about a personal matter. Could you please send me some details to my registered email address? Thank you, Alejandro01 UserIdTAG: 217971 UserNameTAG: Alejandro01 CreateTimeTAG: 2012-10-01T15:02:23Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3733 TitleTAG: no. of h/ws can we choose any 2 h/ws of our choice to leave? UserIdTAG: 85790 UserNameTAG: vivektomar CreateTimeTAG: 2012-10-01T14:13:29Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: My recommendation is NOT to choose to leave any of them undone. They are all useful. Don't underestimate the value of working these problems. Just a friendly thought. FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-10-01T14:39:03Z FirstChildTAG: yes FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-10-01T14:15:53Z IndexTAG: 3734 TitleTAG: H3P4 I don't under stand why when you're trying to find the voltage across the resistor, its just the voltage of V1 when Vs is positive and V2 when Vs is negative. Why isn't the voltage drop after the first resistor Vs/2 like it for 2 like resistors, so that when you apply the superposition principle, shouldnt the voltage be Vs/2 + V1 when positve and Vs/2 + V2 when negative? UserIdTAG: 381923 UserNameTAG: matteaton CreateTimeTAG: 2012-10-01T14:12:38Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Mmm... no, in fact it shouldn't! Let's discuss the circuit together and find out the voltage across the resistor. Here is the circuit: ![enter image description here][1] Assuming that V1 is 3v and V2 is 3v and Vs has this wave: ![enter image description here][2] Now, we want to find the maximum "positive" voltage across the resistor, that means when Vs is at its positive peak, which means Vs = 7.5v (as noticed in the problem) Great, now we may substitute Vs with another DC voltage source taking into account the polarity is left untouched! Like this: ![enter image description here][3] But why?!! Well, that's because we are dealing with the positive part of the wave. So, if you took the negative part, you should flip the polarity. Good, now we have Vs = 7.5v , V1 = 3v , V2 = 3v but wait a minute! Look at the diodes how are they connected. Let's discuss D2... its positive terminal is connected to the negative of V2 and its negative terminal is connected to the positive of Vs (of course through the resistor R); okay what does that mean? That means D2 is reverse-biased and represents an open-circuit condition. Great now we got rid of that branch. What about D1? Both of its terminals are connected to positive terminals of the sources!! But let's take a closer look, Vs is higher than V1, in other words we can consider that the negative terminal of the diode is connected to a NEGATIVE voltage! Because it is NOT about signs, it's about "Voltage difference across an element". Well, Vs (7.5v) is connected to the positive terminal and V1 (3v) is connected to the negative terminal of D1 (Here we should take the voltage drop in resistor R into account, but it's only 6.8 kOhm so there won't be big drop). So, D1 represents a short-circuit condition. And the resulting circuit is: ![enter image description here][4] Obviously the voltage across the resistor is V1 ☺ In the same way you can figure out that negative part of Vs wave would lead you to v = V2 I Hope that was useful ^_^ [1]: https://edxuploads.s3.amazonaws.com/13491187361343685.jpg [2]: https://edxuploads.s3.amazonaws.com/13491189081343638.jpg [3]: https://edxuploads.s3.amazonaws.com/13491195161343677.jpg [4]: https://edxuploads.s3.amazonaws.com/13491208301343662.jpg FirstChildUserIdTAG: 365201 FirstChildUserNameTAG: sirajmuhammad FirstChildCreateTimeTAG: 2012-10-01T19:56:01Z SecondChildTAG: I still don't understand how Vs doesn't affect the voltage across the resistor. Why isn't superposition applied? Is the remaining voltage from Vs dissipated by the diode in the "short circuit"? SecondChildUserIdTAG: 381923 SecondChildUserNameTAG: matteaton SecondChildCreateTimeTAG: 2012-10-02T14:11:51Z SecondChildTAG: No, it is a short-circuit there is no dissipation here. $V_1$ and the resistor are in parallel so they have the same voltage across them. Now, the remaining voltage from $V_S$ is drawn by resistor $R$ in series with $V_S$. You can find out that by writing KVL in the left loop of the last circuit: > $V_S - V_1 - V_R = 0$ > $V_R = V_S - V_1$ > $V_R = 7.5 - 3 = 4.5v$ SecondChildUserIdTAG: 365201 SecondChildUserNameTAG: sirajmuhammad SecondChildCreateTimeTAG: 2012-10-05T11:56:26Z IndexTAG: 3735 TitleTAG: OA Small signal problem Thank you for the tutorial, but i don't understand where is the port for Vin on the operational amplifier, since the report is between Vout a Vin and not between Vout and Vop like you have illustrated in the video.Thank you! UserIdTAG: 240487 UserNameTAG: AlexAlexandrescu CreateTimeTAG: 2012-10-01T13:48:37Z VoteTAG: 0 CoursewareTAG: Week 5 / Op Amp Small Signal Model CommentableIdTAG: 6002x_ap_amp_small_signal_t NumberOfReplyTAG: 0 IndexTAG: 3736 TitleTAG: help LAB 4 how to get different values of vgs simultaneously UserIdTAG: 183166 UserNameTAG: yogeshk CreateTimeTAG: 2012-10-01T12:28:16Z VoteTAG: 0 CoursewareTAG: Week 4 / Amplifiers CommentableIdTAG: 6002x_amplifiers NumberOfReplyTAG: 1 FirstChildTAG: Use scissors and clip board in upper left corner of the circuits sandbox. Click and drag in the sandbox creates a green box around stuff that can be sent to the clip board then pasted repeatedly. "To get multiple v-i curves at the same time, simply replicate the mosfet test setup (i.e., the current probe, mosfet and gate voltage source), chosing different values for gate voltage source and changing the plot color for the current probe" FirstChildUserIdTAG: 331542 FirstChildUserNameTAG: james77901 FirstChildCreateTimeTAG: 2012-10-01T12:46:04Z SecondChildTAG: I wish I had though of that. lol Sometimes I have a hard time manipulating those little current taps. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-01T15:02:08Z SecondChildTAG: Me too! I sure could have saved a lot of time if I had known that. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-10-01T21:33:10Z SecondChildTAG: how do i even set the value of Vds? SecondChildUserIdTAG: 156585 SecondChildUserNameTAG: AdiRanga SecondChildCreateTimeTAG: 2012-10-02T11:05:34Z IndexTAG: 3737 TitleTAG: Videos on tablet Hi all, I bought myself a tablet so I can watch the course videos wherever I want. Now on my PC I set the videoplayer to 1.5x speed, as it saves some precious time. When I watch it on my tablet however, I'm not able to set the speed. I tried different browsers (Explorer, Chrome, Firefox and Opera) and different O/S' (Android 4.0, MeeGoo, Windows 8), but at best I can open the speed menu, but not select anything :( Also tried Damians userscript to fine tune the player speed in Chrome, but that also didn't work. [edit]The problem seems to be caused by the 'window' of the speed settings getting closed after releasing the speed button; so no way to select the speed except when one manages to pause it by opening the context menu of the browser :P [/edit] Is this a known problem? And if so, is there a solution? Or any suggestions? UserIdTAG: 114881 UserNameTAG: xvink CreateTimeTAG: 2012-10-01T11:46:35Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3738 TitleTAG: LAB 3 In lab 3 experiment the system is not allowing me to change the value of L/W property of MOSFET. can any one help me in this matter..? thanks in advance.. UserIdTAG: 509592 UserNameTAG: chinmaya24aug CreateTimeTAG: 2012-10-01T11:39:20Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: double click on the MOSFET .. FirstChildUserIdTAG: 464142 FirstChildUserNameTAG: wedadMA FirstChildCreateTimeTAG: 2012-10-01T12:29:25Z IndexTAG: 3739 TitleTAG: week 4 dear, is it possible to load week 4 videos as was done for the previous weeks? from where i work i cannot access these videos directly online. i can only download mp4 thanks! UserIdTAG: 141709 UserNameTAG: charbelantonios CreateTimeTAG: 2012-10-01T11:36:47Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I agree totally. It´s really useful to have the possibility to watch the videos without being connected to Internet. Thank you. FirstChildUserIdTAG: 255161 FirstChildUserNameTAG: mlovelle FirstChildCreateTimeTAG: 2012-10-02T10:22:24Z IndexTAG: 3740 TitleTAG: h2p1 shows red mark it shows red mark for right answer. Please help UserIdTAG: 508330 UserNameTAG: adheedhan CreateTimeTAG: 2012-10-01T11:27:34Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Most likely is the answer is not correct according to the automated checker. However I have found by trying again different bugs can be found. Same input to the courseware can give " arithmetic calc error" or red or green. I never found it gives green when answer actually wrong. and when I got red USUALLY I was wrong. FirstChildUserIdTAG: 331542 FirstChildUserNameTAG: james77901 FirstChildCreateTimeTAG: 2012-10-01T12:57:59Z IndexTAG: 3741 TitleTAG: lab grading Hey guys, I don't know how you guys think about this, but I find it a little odd that we receive 100% for the lab only...what I mean its like a multiple choice question, you either have it right or wrong....and in my perspective this not good UserIdTAG: 308861 UserNameTAG: Ignaas CreateTimeTAG: 2012-10-01T11:06:26Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Lab 3 or Lab 4? If you are referring to Lab 3, it seems fine. Digital either works or it doesn't.(To a point) Not much of a gray area there. If you sent a Digital signal that was 95% perfect, it would not work at all. Analog is a little more forgiving, things will still work with a less then perfect signal, just at a degraded quality. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-10-01T15:10:29Z IndexTAG: 3742 TitleTAG: LAB 3 Please help. I am at my wits end. Why do I keep getting this plot ? http://db.tt/heiKa41Z UserIdTAG: 134080 UserNameTAG: singhn9 CreateTimeTAG: 2012-10-01T09:46:03Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You need to adjust the l/w parameter. FirstChildUserIdTAG: 299251 FirstChildUserNameTAG: TheRedBlackOne FirstChildCreateTimeTAG: 2012-10-01T10:30:59Z SecondChildTAG: How do I calculate RON as it requires l/w ? Am I missing something ? Both l/w and RON are unknown. SecondChildUserIdTAG: 134080 SecondChildUserNameTAG: singhn9 SecondChildCreateTimeTAG: 2012-10-01T10:44:47Z SecondChildTAG: Rn is given in the statement so we can find RON out. SecondChildUserIdTAG: 409143 SecondChildUserNameTAG: spider023 SecondChildCreateTimeTAG: 2012-10-01T11:06:42Z IndexTAG: 3743 TitleTAG: S1E7 how to find i2 in the 4th element UserIdTAG: 530763 UserNameTAG: khizar35 CreateTimeTAG: 2012-10-01T09:27:39Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL and KCL exercises CommentableIdTAG: 6002x_kvl_kcl_exercises NumberOfReplyTAG: 1 FirstChildTAG: So it is based on kirschoff's law of all current coming to a node = 0. The first node between i5 and i4 will be called node a. As you can see the currents also have the directions in which they go. so at node a you have i4 current as -1.3 coming towards the node. Notice the difference in the sign of the number. i3 is also -3.0 because again the sign changes because of the direction of the current. Thus the current for i4 in the direction towards the node is 4.3 . The other node ( b) in this case will have now a -4.3 current coming from i4 because of the change in direction and a -(-).7 from i1 making the current coming in from i2 towards node b to be 3.6 so that the full sum of all the numbers is 0. Then looking at the question it says what is the current going into the fourth element and since now we are looking at the current in the opposite direction it will be -3.6 . Hopefully this makes sense. FirstChildUserIdTAG: 234626 FirstChildUserNameTAG: kpsingh1 FirstChildCreateTimeTAG: 2012-10-03T07:35:11Z IndexTAG: 3744 TitleTAG: Good afternoon every one Actually i submitted lab2 answer and it is correct but in progress report they did not show. UserIdTAG: 456614 UserNameTAG: atulnarayan786 CreateTimeTAG: 2012-10-01T08:28:16Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: When did u submit it ?.due date was 30th midnight. FirstChildUserIdTAG: 357747 FirstChildUserNameTAG: kishores FirstChildCreateTimeTAG: 2012-10-01T09:36:06Z IndexTAG: 3745 TitleTAG: So ready to give up on lab 3 Any hints on selecting the w/l? I am getting Ron=909.09 Am I in the right track? UserIdTAG: 230787 UserNameTAG: Gabriel007 CreateTimeTAG: 2012-10-01T06:04:37Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I had been entering 40-44 for the past 3 hours!!! WHAT!? I saw some post to round up to the next 10th...that worked. How is this explained? If I am dying with labs and hwk I am toast on the test. Any help? FirstChildUserIdTAG: 230787 FirstChildUserNameTAG: Gabriel007 FirstChildCreateTimeTAG: 2012-10-01T06:11:43Z SecondChildTAG: i entered 100, it works SecondChildUserIdTAG: 324219 SecondChildUserNameTAG: Dmitry79 SecondChildCreateTimeTAG: 2012-10-01T08:28:31Z SecondChildTAG: Is your waveform that you simulated lower than .25V for a logic low level? If not, your Ron is incorrect. Selecting W/L is just a formality once you find Ron. As you probably already know, the formula for Ron is Ron = Rn(L/W), with Rn = 26.5kOhms, W/L = Rn/Ron. Please read the discussion forum about giving out direct answers. It could be fatal. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-01T12:00:34Z SecondChildTAG: Thanks rharris. I did not think it was giving an answer because the problem requires you to place three switches in a specific order before you even get to the l/w setting. SecondChildUserIdTAG: 230787 SecondChildUserNameTAG: Gabriel007 SecondChildCreateTimeTAG: 2012-10-01T13:45:47Z SecondChildTAG: It wasn't you that gave the answer, it was a reply to your question. I didn't want to bring any more attention to it than needed. Did you figure out the correct answer yet? Think Vth and what Voh needs to be then find the resistance. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-01T14:30:24Z IndexTAG: 3746 TitleTAG: H3P3 nad H3P4 Homework need a little help Hello everyone, i need some help, I can't think in way of solving the third problem in week 3 homework, and beyond that i can't find the result for the current in the last question, the simulation gives me a -10fA, this is, -10*10^-30A, which gives a wrong answer, I tried some calculation, but i didn't got the result... UserIdTAG: 317836 UserNameTAG: VitorRodrigues CreateTimeTAG: 2012-10-01T05:30:05Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: H3P3...check out tutorial, wk.3 on load lines. FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-10-01T05:38:10Z SecondChildTAG: :D thanks for the tip! I´m already done with hw3 thanks to you SecondChildUserIdTAG: 79854 SecondChildUserNameTAG: AlbertoGT SecondChildCreateTimeTAG: 2012-10-01T07:51:49Z IndexTAG: 3747 TitleTAG: LAB4 Hello all! Does anybody know how to configure the probe (Lab 4) to get the plot with voltage on horizontal axis and current on vertical one? I connected the probe to the gate and put the current sense (a small arrow) inside the drain circuit, but my plot shows me time axis. UserIdTAG: 209041 UserNameTAG: SmartEngine CreateTimeTAG: 2012-10-01T05:29:51Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Position your voltage probe and then use the drop-down box to select x-axis. FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-10-01T05:35:13Z SecondChildTAG: Thank you so much! SecondChildUserIdTAG: 209041 SecondChildUserNameTAG: SmartEngine SecondChildCreateTimeTAG: 2012-10-01T10:22:11Z SecondChildTAG: ... but how to set different colors for a number of traces without reset x-axis mode? SecondChildUserIdTAG: 209041 SecondChildUserNameTAG: SmartEngine SecondChildCreateTimeTAG: 2012-10-01T10:46:26Z SecondChildTAG: Each device will have it's own current probe, you select individual colours that way. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-10-01T15:25:10Z IndexTAG: 3748 TitleTAG: H4P2, i think there is an error... Could someone please post their V-I graph for Zener diode in H4P2? I think there's an error in mine. There need to be a mA/V not A/V. If i have "1 A/V" here then resistance of diode will be 1ohm, and this is weird. there is my graph: https://www.edx.org/static/content-mit-6002x/images/circuits/Zener-diode.a7b4f06d4768.gif PS: sorry for my english, it is not my native language. :) UserIdTAG: 323703 UserNameTAG: kkosh CreateTimeTAG: 2012-10-01T04:12:04Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: There is no problem. Mine was 1A/V too. It's more about the solution than the values. FirstChildUserIdTAG: 270011 FirstChildUserNameTAG: SnowmanZA FirstChildCreateTimeTAG: 2012-10-01T04:49:44Z SecondChildTAG: Oh, i got it. Diode is not a resistor, its equivalent is resistor + dc source. Thank you. SecondChildUserIdTAG: 323703 SecondChildUserNameTAG: kkosh SecondChildCreateTimeTAG: 2012-10-01T05:10:12Z FirstChildTAG: Hi kkosh, I have been having trouble with this problem for the past 4 hrs.... If it is of any help the "1A/V" is correct. But remember that Resistance is "V/A" Let me know if this helps. Regards FirstChildUserIdTAG: 303762 FirstChildUserNameTAG: leocarrasco FirstChildCreateTimeTAG: 2012-10-01T05:16:58Z SecondChildTAG: Understanding that the marking is correct is helpful in itself. In fact, I made an error finding the bias point, which led me to absurd results. The realization of this fact dawn upon me in the restroom - a true "place for contemplation", indeed: struggling with the task for 6 hours only to have a eurica moment in the loo! SecondChildUserIdTAG: 323703 SecondChildUserNameTAG: kkosh SecondChildCreateTimeTAG: 2012-10-01T05:38:22Z SecondChildTAG: hehehe!! I will do this when i'm frustrated too if it helps! :P SecondChildUserIdTAG: 304587 SecondChildUserNameTAG: Vasso SecondChildCreateTimeTAG: 2012-10-01T07:37:11Z SecondChildTAG: Feed the zener enough current and it will reach it's breakdown voltage. Datasheets always supply what this current is (Iz - I believe) SecondChildUserIdTAG: 61923 SecondChildUserNameTAG: MikeJones SecondChildCreateTimeTAG: 2012-10-02T09:09:33Z IndexTAG: 3749 TitleTAG: When is homework due? I'm on the West Coast and had to wait until this evening to start on HW3. When I started I wanted to find out how long I would have to finish so i decided to try to find out if the homework is due by the end of Sept 30 Eastern time or Pacific Time. I've spent over an hour on the site and on google doing nothing but trying to find out whether its due EST or PST and I've come up empty. I suppose I'll find out in a few minutes at 9pm PST if I try to submit a homework, but I'm pretty amazed that its so hard to find out the answer to such a fundamental question. On coursera the time is denoted as PST but I can find no such designation anywhere on this site. I'll probably have to punt on HW3 if the due time is EST but really, somebody should make it easy to find out. I'd rather use the hour doing HW than just trying to find out when its due. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-01T03:58:13Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I have seen it in the comments before that the homework is due at midnight in your own timezone. FirstChildUserIdTAG: 336061 FirstChildUserNameTAG: mcgeheejb FirstChildCreateTimeTAG: 2012-10-01T04:40:09Z IndexTAG: 3750 TitleTAG: Lab 3 WILL NOT CHECK Created circuit, calculated W/L, ran transient and my output looks IDENTICAL to the sample. But I cannot get a Check! It's 11:36 PM and I have to get up at 5:00 AM, I'm going to bed! Haven't got a clue how to solve the Diode problem, gave up on that one a long time ago. Seems to me it should be a simple resistor voltage divider and a +/- DC offset to the signal, depending on which diode is conducting, but none of my anwers will check. UserIdTAG: 375500 UserNameTAG: Brej7665 CreateTimeTAG: 2012-10-01T03:39:25Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Go and check myrimit's hint. You can do it! :) FirstChildUserIdTAG: 281159 FirstChildUserNameTAG: ManasiS FirstChildCreateTimeTAG: 2012-10-01T04:13:07Z IndexTAG: 3751 TitleTAG: digital memory guys, i cant access the digital memory subject when mouse clicked on this through getting inside the text book UserIdTAG: 525327 UserNameTAG: kinzangbhutan CreateTimeTAG: 2012-10-01T03:39:11Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: page thru to 10.7... click on 10.7 digital memory bookmark works fine for me on courseware and directly in the PDF. I run Windows7SP1+ie9. FirstChildUserIdTAG: 331542 FirstChildUserNameTAG: james77901 FirstChildCreateTimeTAG: 2012-10-01T13:18:31Z IndexTAG: 3752 TitleTAG: S3E3: Circuit Variables are Superpositions of values due to each source separately. I have a question about one of the lecture exercises. Specifically S3E3. I am having a hard time getting the correct answer for the first part which is "get the voltage v3 due to just V1". Can someone tell me if I'm on the right track? I'm using Nodal analysis and have 3 equations from KVL for loop 1 Equation 1) 2 - i1*R1 + i2*R2 = 0 from KVL for loop 2 Equation 2) v3 + i2*R2 = 0 from KCL Equation 3) i1 + i2 - v3/R3 = 0. I solved the equations and got v3 = 2, which is not correct. (We are given the answer and it is .422535211268. Like in all the problems involving nodal analysis or KVL/KCL, I strongly suspect I made a sign error somewhere. So my specific questions are: A) Are my 3 equations correct (except possibly for signs)? B) Assumming my 3 eqations are correct, can you give me ideas for this problem and in general, on how to get the signs correct? Thanks, Dave P.S. I believe this problem is somewhat related to HW2 - Lab. Not sure, but it seems worth going through in any case. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-10-01T00:00:43Z VoteTAG: 0 CoursewareTAG: Week 2 / Linear combinations of source strenghts CommentableIdTAG: 6002x_S3E2_Linear_Combinations_of_Source_Strengths NumberOfReplyTAG: 1 FirstChildTAG: Ok, I think I got the first part. And the equations posted above were what I used. Notice there are 2 KVL equations and 1 KCL equation. I don't recall that in the lectures/homework. For the second part, "v3 due to V2", I have these three equations and things are not working out. Can anyone see an error (including sign error?) 1) v2 - i2*R2 + i1*R1 = 0 This is KVL loop1 2) v2 - i2*R2 + v3 = 0 KVL loop 2 3) i1 + i2 - v3 /R3 = 0 KCL Thanks FirstChildUserIdTAG: 94009 FirstChildUserNameTAG: David1956 FirstChildCreateTimeTAG: 2012-10-01T01:09:09Z IndexTAG: 3753 TitleTAG: Aussie Mates Are there any other Aussies out there? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-30T23:08:34Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Yup - living on the West coast. You? FirstChildUserIdTAG: 281858 FirstChildUserNameTAG: aspringv FirstChildCreateTimeTAG: 2012-10-01T00:46:34Z SecondChildTAG: Mildura, VIC SecondChildUserIdTAG: 345671 SecondChildUserNameTAG: cbjerregaard SecondChildCreateTimeTAG: 2012-10-03T10:45:11Z FirstChildTAG: Yep, in Sydney what about you? FirstChildUserIdTAG: 217971 FirstChildUserNameTAG: Alejandro01 FirstChildCreateTimeTAG: 2012-10-01T06:58:03Z SecondChildTAG: Mildura VIC SecondChildUserIdTAG: 345671 SecondChildUserNameTAG: cbjerregaard SecondChildCreateTimeTAG: 2012-10-03T10:45:33Z IndexTAG: 3754 TitleTAG: Course Book? Just a real quick question. The book for this course, even though it is here free, is called "Foundations of analog and digital electronics circuits" right? I prefer a book you can hold and keep vs the online copy and just wanted to be sure. UserIdTAG: 219780 UserNameTAG: ReconIII CreateTimeTAG: 2012-09-30T22:45:48Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: by Agarwal and Lang, try Amazon FirstChildUserIdTAG: 345671 FirstChildUserNameTAG: cbjerregaard FirstChildCreateTimeTAG: 2012-09-30T23:17:00Z SecondChildTAG: Thanks, and actually ebay has it for around $50 less SecondChildUserIdTAG: 219780 SecondChildUserNameTAG: ReconIII SecondChildCreateTimeTAG: 2012-09-30T23:35:22Z SecondChildTAG: You want soft copy , like in pdf ? SecondChildUserIdTAG: 37464 SecondChildUserNameTAG: DoubleA SecondChildCreateTimeTAG: 2012-09-30T23:52:33Z SecondChildTAG: A paperback copy, like the actual book. I prefer to have it so I can write notes in it and always have it. I ended up finding the book on ebay for only $35 usd shipped which really isn't bad. SecondChildUserIdTAG: 219780 SecondChildUserNameTAG: ReconIII SecondChildCreateTimeTAG: 2012-10-01T00:31:39Z SecondChildTAG: Seems Good :) SecondChildUserIdTAG: 37464 SecondChildUserNameTAG: DoubleA SecondChildCreateTimeTAG: 2012-10-01T00:50:48Z IndexTAG: 3755 TitleTAG: How to solve DIODE LIMITER H3P4? i am stuck in finding the currents through D1 and D2 diodes, please help me understand how can i find them.. UserIdTAG: 223378 UserNameTAG: hadi89 CreateTimeTAG: 2012-09-30T21:12:36Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: there is already an excellent post about this, just look for it and your problem will be solved FirstChildUserIdTAG: 203899 FirstChildUserNameTAG: nevical FirstChildCreateTimeTAG: 2012-09-30T21:29:02Z FirstChildTAG: I think thIs will help you to solve H3P4 https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5067499d2193ea230000000c FirstChildUserIdTAG: 45307 FirstChildUserNameTAG: josejimenez2 FirstChildCreateTimeTAG: 2012-09-30T22:04:11Z FirstChildTAG: I just did a dc-analysis in CIRCUIT SANDBOX and got right answers for currents. Thought not yet understood them) FirstChildUserIdTAG: 66669 FirstChildUserNameTAG: agarus FirstChildCreateTimeTAG: 2012-09-30T23:31:14Z SecondChildTAG: Google "**clamper circuits**" and look up "**Clamper (electronics)**" in Wikipedia. This is a "general pattern" type circuit that appears again and again so it is essential to grasp the theory of it's operation. SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-10-01T02:04:06Z IndexTAG: 3756 TitleTAG: Why is Coefficient V sometimes = 2? I can only get a green check-mark for j, n, p, q, r if I use 2 as the coefficient of V. UserIdTAG: 150057 UserNameTAG: Felinae CreateTimeTAG: 2012-09-30T20:37:56Z VoteTAG: 0 CoursewareTAG: Week 3 / Switch Resister Model Exercise CommentableIdTAG: 6002x_switch_resistor_model_exercise NumberOfReplyTAG: 0 IndexTAG: 3757 TitleTAG: lab 4 please give me some hint about the circuit also i m not able to get how to place probe for current. what's the need of diode in the cicuit? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-30T20:29:42Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I don't believe that it's a diode. I believe it is an arrow indicating the direction of current flow. The color you pick for the arrow is the color of the line in the output plot for that device. You need a current probe at the drain of each of your MOSFETs. You need 7 MOSFETs each with a different gate voltage and all driven by the same drain-source voltage. Does that get you started? I thought creating the circuit was quite tedious! FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-09-30T23:32:00Z SecondChildTAG: I keep setting up the circuit but cannot get any graphs when I do the analysis. I think I am hooking it up properly but it is like their is an open somewhere and I can't figure it out SecondChildUserIdTAG: 18073 SecondChildUserNameTAG: gburkhart SecondChildCreateTimeTAG: 2012-10-01T03:38:04Z SecondChildTAG: check if you have applied the ground node. SecondChildUserIdTAG: 357747 SecondChildUserNameTAG: kishores SecondChildCreateTimeTAG: 2012-10-01T09:43:43Z SecondChildTAG: same problem, i did ground my circuit but still no graph. how can that be? here i thought this was a very basic lab exercise :( SecondChildUserIdTAG: 285945 SecondChildUserNameTAG: lindalapiso SecondChildCreateTimeTAG: 2012-10-01T18:39:22Z SecondChildTAG: i am also not getting any graph :( SecondChildUserIdTAG: 97340 SecondChildUserNameTAG: AnkitRana SecondChildCreateTimeTAG: 2012-10-02T09:58:23Z IndexTAG: 3758 TitleTAG: lab 3 pls help.. my output graph is exactly the same as shown.. bt it is showing wrong UserIdTAG: 509366 UserNameTAG: shikhagoel003 CreateTimeTAG: 2012-09-30T19:36:28Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: pls anyone help...i m getting exactly the same output.. FirstChildUserIdTAG: 509366 FirstChildUserNameTAG: shikhagoel003 FirstChildCreateTimeTAG: 2012-09-30T19:45:43Z SecondChildTAG: A few people already had this issue, and a few people already posted helpful answers here: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50689296d4e4002b0000001f SecondChildUserIdTAG: 287805 SecondChildUserNameTAG: Beneficial SecondChildCreateTimeTAG: 2012-09-30T19:56:39Z SecondChildTAG: ![enter image description here][1] [1]: http:// SecondChildUserIdTAG: 509366 SecondChildUserNameTAG: shikhagoel003 SecondChildCreateTimeTAG: 2012-09-30T19:58:05Z SecondChildTAG: still not working SecondChildUserIdTAG: 509366 SecondChildUserNameTAG: shikhagoel003 SecondChildCreateTimeTAG: 2012-09-30T20:04:18Z SecondChildTAG: You've described a ***very*** general difficulty. People have provided several possible resolutions to the difficulty you might be having. If none of those worked. You will ***have*** to be more specific than "still not working" in order for people to provide you with useful help, within the scope of the honor code (or honour code, as I'm from Canada, where our English is chiefly british and where men of honour always possess 'U's of the most respected sort). SecondChildUserIdTAG: 287805 SecondChildUserNameTAG: Beneficial SecondChildCreateTimeTAG: 2012-09-30T20:49:01Z FirstChildTAG: Because your graph looks similar, you probably have the MOSFETs in the correct order, but you have to adjust the W/L value of your MOSFET. The equation given is RON=Rn*(L/W) while the input for the MOSFET is W/L, or the inverse. So,to find the lowest possible RON to satisfy your requirements, you have to use the VS*(R1/(R1 + R2+...Rn))=VOL. FirstChildUserIdTAG: 371588 FirstChildUserNameTAG: DifulTakveh FirstChildCreateTimeTAG: 2012-09-30T22:19:43Z IndexTAG: 3759 TitleTAG: H4P2 - Zener Regulator I'm having issues at the last part, where we're supposed to recalculate vo with the Zener diode in place. I've identified the correct region, but I'm not sure how I'm supposed to model the diode. I presume it should have the breakdown voltage identified through a source, but how am I supposed to give the resistance values. Is it just based on the graph (because it's very very low based on it)? Thanks. UserIdTAG: 466665 UserNameTAG: raresmihaipopa CreateTimeTAG: 2012-09-30T19:20:21Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Yes, the resistor value should be based on the I-V characteristic graph FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-30T19:34:13Z IndexTAG: 3760 TitleTAG: i cant solve lab 3 with only 3 switches! please help! every which way i apply and/or follow the logic to assemble the gates i need 4 or more switches to get the desired output at Z. a NAND requires 2 mosfet swtiches. a NOR requires 2 switches an inverter/negator requires 1 switch the negation of C AND (A OR B) leads to an OR gate and a NAND gate (5 switches), applying demorgan's laws NOT C OR NOT(A OR B) leads to 2 NOR gates I believe (4 switches)... What combination of 3 switches can possibly lead to Z? Thanks everyone for your help! UserIdTAG: 74193 UserNameTAG: SpaceAgeRobot CreateTimeTAG: 2012-09-30T19:11:43Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: read page 295 there something similar. FirstChildUserIdTAG: 152106 FirstChildUserNameTAG: JoaoBR FirstChildCreateTimeTAG: 2012-09-30T19:16:26Z FirstChildTAG: Build your result incrementally. 1) How would I implement an OR gate? 2) How would I implement an AND gate? 3) How would I implement the combination that this question is asking for? That was exactly my process for this problem. Never give up! :) One more thing: this may sound odd, but I get the impression you're trying to solve this "horizontally." Notice that the examples in the text are more easily understood "vertically," because they are showing us the actual circuits instead of just the logic gates. FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-09-30T19:25:33Z FirstChildTAG: What combination you ask? Take a good look at Fig. 6.24 on p. 295 of the textbook. Note that it required only four MOSFETs to implement NOT((A+B)CD). The logic that the lab asks you to implement is simpler, suggesting that the solution circuit is simpler also with respect to the number of logic elements (MOSFETS) required. Hope this helps. FirstChildUserIdTAG: 141108 FirstChildUserNameTAG: rl777 FirstChildCreateTimeTAG: 2012-09-30T20:00:39Z FirstChildTAG: The lab can be done with just 3 mosfets. Pay particular attention to the negation in the equation. Try writing it out by hand using pen and paper, it might make things a little clearer. Don't think in terms of OR, AND, think in terms of NOR, NAND - hope this helps. FirstChildUserIdTAG: 385677 FirstChildUserNameTAG: Michaelc1 FirstChildCreateTimeTAG: 2012-09-30T20:25:33Z FirstChildTAG: Look at the truth table. The gate pulls the output low if C is high AND either A is high or B is high. You can do a lot of boolean math, or you can just think of what it takes to pull that resistor low. How low? Down to a maximum of 0.25 V. You need to choose a Ron for the MOSFETs that produces 0.25 V when the MOSFETs are providing the maximum voltage drop. (Some combinations of inputs will provide less voltage drop.) Then you need to choose W/L for the MOSFETS to get the Ron you desire. So then the question becomes how to calculate W/L. Oddly, they give you the value of Rn and the equation in terms of L/W. Think of W/L that you need to calculate as a variable that is the inverse of L/W. (It's simply the ratio of the length to the width of the silicon area.) It's a straightforward calculation from the given values and formula to calculate L/W; W/L is just its inverse 1/(L/W). FirstChildUserIdTAG: 131970 FirstChildUserNameTAG: elucify FirstChildCreateTimeTAG: 2012-10-01T04:27:54Z IndexTAG: 3761 TitleTAG: How do i find IN in h4p3? How do i find IN in h4p3? UserIdTAG: 92895 UserNameTAG: shihab2555 CreateTimeTAG: 2012-09-30T19:04:16Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Take a look at your textbook section 3.6 for a description of Thevenin and Norton equivalent models. The fact that there is a dependent source won't affect the procedure for calculating IN, although calculating RN can be somewhat different from the method for circuits with independent sources only FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-30T19:37:37Z SecondChildTAG: I thought that the IN was simply placing a short across the port B terminals and calculating the current (Vo/(R1+R2), but that is not the right answer...why? SecondChildUserIdTAG: 244115 SecondChildUserNameTAG: Pedro1969 SecondChildCreateTimeTAG: 2012-10-02T18:39:05Z SecondChildTAG: I have done it in the same way.But i steel don't know why it's incorrect SecondChildUserIdTAG: 92895 SecondChildUserNameTAG: shihab2555 SecondChildCreateTimeTAG: 2012-10-04T04:23:29Z IndexTAG: 3762 TitleTAG: I m late !!help !! :( i have just started this course.. its been just a week till now and i havent finished week 1 yet :( i m trying to finish the week 3 by another 48 hours.. will this late finish affect my course at the end when i receive this certificate ??? pls help :( :( UserIdTAG: 89535 UserNameTAG: Rudhra CreateTimeTAG: 2012-09-30T18:25:19Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: From what I understood from the syllabus it should only affect the grade, but you'll still receive a certificate. I just started the course as well and wondering the same. FirstChildUserIdTAG: 259178 FirstChildUserNameTAG: bdrangova FirstChildCreateTimeTAG: 2012-09-30T19:35:31Z SecondChildTAG: the least 2 scores of lab and homework will not be considered for grading!! so chill SecondChildUserIdTAG: 489065 SecondChildUserNameTAG: amiths SecondChildCreateTimeTAG: 2012-09-30T19:54:16Z SecondChildTAG: what does that least of 2 scores wont be considered ??i don't understand the concept behind it amiths...!! can you pls explain in detail about it ??? SecondChildUserIdTAG: 89535 SecondChildUserNameTAG: Rudhra SecondChildCreateTimeTAG: 2012-10-02T06:24:06Z FirstChildTAG: do week 2 or 3 now, as in the end MIT wont consider the least two HWs and Labs, so just do one week now, i say do week 3, the first 2 problems are still doable and you get to a 70% score like me and that is still commendable, get going right now!....im here to help you within the realms of the honor code..... FirstChildUserIdTAG: 169784 FirstChildUserNameTAG: thewiredbear FirstChildCreateTimeTAG: 2012-09-30T20:04:48Z IndexTAG: 3763 TitleTAG: about lab 3 can any one help me how to draw the circuit diagram plzz UserIdTAG: 130290 UserNameTAG: RajkumarRughani CreateTimeTAG: 2012-09-30T18:24:32Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I don't think we can help you draw the circuit diagram. That's half the answer right there, man! Didn't you watch the videos? This week's lab is pretty easy: How many ways are there to arrange **three MOSFETS** with respect to ground? There are only a limited amount of combinations! Try each out; but if you compare the truth table and think a little, one combination should be obvious! Use the search function, e.g. click the magnifying glass icon and search for "lab 3" and I bet you there's plenty of hints out there. One student posted a whole guide on how to approach that lab. There is no excuse for laziness, though, and no such thing as a free lunch! FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-10-01T02:59:24Z IndexTAG: 3764 TitleTAG: regarding hw/lab submission hi every one i wanted to know the clock of edx database & my laptop are synchronized ?? as i uaually complete my hw/lab on last day ie sunday.. UserIdTAG: 183166 UserNameTAG: yogeshk CreateTimeTAG: 2012-09-30T18:22:11Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: if there is still time to mid-night in any part of the world on that date ( local for every place), then you got time to submit! FirstChildUserIdTAG: 169784 FirstChildUserNameTAG: thewiredbear FirstChildCreateTimeTAG: 2012-09-30T20:10:48Z FirstChildTAG: There is some flexibility on the submission deadline time to account for that, but always try to submit the homeworks and labs before midnight at your local time. You should be ok if you do that FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-30T19:42:49Z IndexTAG: 3765 TitleTAG: How does the 2k * 3 work? Hey Guys, How does the splitting of the 1k work? UserIdTAG: 230787 UserNameTAG: Gabriel007 CreateTimeTAG: 2012-09-30T17:54:08Z VoteTAG: 0 CoursewareTAG: Week 3 / Circuit Truth Table Tutorial CommentableIdTAG: 6002x_circuit_truth_table_tutorial NumberOfReplyTAG: 1 FirstChildTAG: Hi, It's just a trick to calculate the parallel value easily. 1k//2k gives no straightforward answer unless you have some experience (or a calculator : )). But if you have the same value for each component, the answer is simply the value of the resistor divided by the number of resistors. For example, with 2 resistors you get $(\frac{1}{R_1}+\frac{1}{R_1})^{-1}=\frac{R_1\cdot R_1}{2R_1}=\frac{R_1}{2}$ So here we can replace (for the calculation) the 1k resistor with 2x 2k in parallel (the same value). Since we have 3 times 2k in parallel, the value of the total parallel resistance is 2k/3. FirstChildUserIdTAG: 396446 FirstChildUserNameTAG: RousseauxS FirstChildCreateTimeTAG: 2012-09-30T18:58:25Z SecondChildTAG: Ahh, got it SecondChildUserIdTAG: 230787 SecondChildUserNameTAG: Gabriel007 SecondChildCreateTimeTAG: 2012-09-30T20:21:43Z IndexTAG: 3766 TitleTAG: LAB 3 how can we know the value of Rn and whats really this(Ron=Rn*L/W)? UserIdTAG: 413613 UserNameTAG: shaliesh CreateTimeTAG: 2012-09-30T17:47:43Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Rn is fixed to 26.5kOhms FirstChildUserIdTAG: 99419 FirstChildUserNameTAG: Texada FirstChildCreateTimeTAG: 2012-09-30T19:05:42Z SecondChildTAG: check the text book page 305 equation 6.4 SecondChildUserIdTAG: 285024 SecondChildUserNameTAG: nigelgomes SecondChildCreateTimeTAG: 2012-09-30T19:56:38Z IndexTAG: 3767 TitleTAG: H3P4 current part i need help in last question of week 3 assessment to find the diode currents... UserIdTAG: 304875 UserNameTAG: Mian-A CreateTimeTAG: 2012-09-30T17:30:23Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: The current through each is to calculated by assuming the other branch open circuited. The current through each diode is the current from the AC source minus the current going through resistor R. You can use the usual node method to calculate the current. Hope this helps. Cheers FirstChildUserIdTAG: 310108 FirstChildUserNameTAG: Saakar FirstChildCreateTimeTAG: 2012-09-30T17:44:59Z FirstChildTAG: ok if i`m not wrong then current will bee 0.803 and so on. But it is give me deadline FirstChildUserIdTAG: 96873 FirstChildUserNameTAG: Santyaga FirstChildCreateTimeTAG: 2012-09-30T19:13:21Z IndexTAG: 3768 TitleTAG: Lab 3 and the interactive circuit tool Everything appeared to be OK with my circuit and values, yet I was not getting anything like the expected output. When I added additional voltage probes right at the gate inputs, no voltages were showing at these inputs, despite the desired/specified voltages being shown on the default probes. This led me to the discovery that apparently good wire connections I'd added between the voltage sources and the mosfets were, despite appearances, actually open circuits. I deleted these connections, made new connections and, lo and behold, the desired voltage signals appeared at the gates, and the desired solution output was obtained. Note that at no time did I change the original configuration. I think my problems arose by my picking off my gate connections from the A/B/C points. Despite appearing to be solid connections, these were points of disconnection. (Was I connecting to a label and not a node??) With this tool, which I love, all is not always as it appears to be connection-wise. Always double-check. UserIdTAG: 141108 UserNameTAG: rl777 CreateTimeTAG: 2012-09-30T17:22:15Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Interactive Laboratory Usage Summary: Wires start at connection points, the open circles that appear at the terminals of components or the ends of wires. Click on a connection point to start a wire -- a green wire will appear with one end anchored at the starting point. Drag the mouse and release the mouse button when the other end of the wire is positioned as you wish. Once a wire has been added to the diagram it can be manipulated like any other component. FirstChildUserIdTAG: 141108 FirstChildUserNameTAG: rl777 FirstChildCreateTimeTAG: 2012-09-30T18:24:43Z IndexTAG: 3769 TitleTAG: I strongly believe these are right answ but they are showing its worng [Don't post answers in this forum] - Staff UserIdTAG: 383699 UserNameTAG: Gogineni CreateTimeTAG: 2012-09-30T17:19:06Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: [Don't post answers in this forum] - Staff FirstChildUserIdTAG: 443057 FirstChildUserNameTAG: punisher_2012 FirstChildCreateTimeTAG: 2012-09-30T18:03:05Z IndexTAG: 3770 TitleTAG: Help in Lab 3 What exactly is this W/L factor ??? Im getting the wave shape correct, but the spikes are bit different (in size) with that of the sample output. what should i do?? UserIdTAG: 288344 UserNameTAG: nkgoutham CreateTimeTAG: 2012-09-30T17:16:23Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: W/L is just some parameter, that you need to provide in transistor configuration. So you need to calculate Ron values, and than, use formula from description Ron = Rn * L/W , find out value W/L. FirstChildUserIdTAG: 132996 FirstChildUserNameTAG: re_test FirstChildCreateTimeTAG: 2012-09-30T18:47:13Z IndexTAG: 3771 TitleTAG: lab 3 check problems When I am pressing the TRAN button, i don't see my curves?why is that??and wthat is W/L?? UserIdTAG: 308861 UserNameTAG: Ignaas CreateTimeTAG: 2012-09-30T16:55:33Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Text, pg. 305. FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-09-30T19:52:08Z IndexTAG: 3772 TitleTAG: Lab4 Hi, In Lab4 I get the TRAN analysis right but all other answers wrong (even after various hours of reviewing them). To compute VT I thought it would give a better result using Vgs=1V instead of 2.5V and it gives me the result I could expect from the curve I have in the TRANS analysis, but still marked wrong. I have also drawn with Excel the curve ids=f(vds) for ids=(K(Vgs-Vt)^2)/2 and I get something correct for vds=3V (comparing with the TRANS analysis). It seems that assuming the TRANS analysis is right, the other results should also be right: I am confused... Thanks for your help, Frédéric UserIdTAG: 384914 UserNameTAG: fbarnel CreateTimeTAG: 2012-09-30T16:52:37Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Check [this][1] thread. It may help. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/505ed4c14140e02300000003 FirstChildUserIdTAG: 329361 FirstChildUserNameTAG: ikrukov FirstChildCreateTimeTAG: 2012-09-30T17:00:01Z SecondChildTAG: Thank you ikrukov!! I also forgot some connections to the ground. SecondChildUserIdTAG: 384914 SecondChildUserNameTAG: fbarnel SecondChildCreateTimeTAG: 2012-10-04T20:46:44Z FirstChildTAG: I followed instructions and got a check, but I agree that using a lower Vgs seems reasonable. Did you check your units? My answer turned out rather small. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-09-30T18:50:25Z FirstChildTAG: hola myrimit una pregunta es que en la pregunta 3 soy atorada ya que no se que valores de vgs e ids hay que tomar para el calculo de k me puedes ayudar gracias FirstChildUserIdTAG: 320715 FirstChildUserNameTAG: maraivette FirstChildCreateTimeTAG: 2012-10-05T02:53:40Z IndexTAG: 3773 TitleTAG: urgent i cant access lecture videos. it started this Saturday(sept 29th) and i don't know why. i was watching 4th week lecture sequence and it suddenly stopped. help me pls. i'm from ethiopia UserIdTAG: 285945 UserNameTAG: lindalapiso CreateTimeTAG: 2012-09-30T16:05:21Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi lindalapiso! You can download the videos in the Wiki. Take a look[here][1]. ;) [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/ FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-30T16:49:11Z SecondChildTAG: Many thanks Myrimit, ill check it out SecondChildUserIdTAG: 285945 SecondChildUserNameTAG: lindalapiso SecondChildCreateTimeTAG: 2012-09-30T16:56:41Z SecondChildTAG: You are welcome lindalapiso! SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-30T16:57:30Z IndexTAG: 3774 TitleTAG: H5P2-Yet another parsing error.... I'm not putting iDS = in the equation, what else can go wrong? I've checked obsessively, but I didn't forget a subscript or do something in lower case... I've confirmed that the formula I have is correct elsewhere. Help? I really want to move on from this problem. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-30T15:56:20Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: 1. Double check the K and VT is in uppercase and vIN abd vOUT is in lower case. 2. Check that you have put explicit multiplication sign between expression's members. Omitting '*'-sign can cause an error. FirstChildUserIdTAG: 329361 FirstChildUserNameTAG: ikrukov FirstChildCreateTimeTAG: 2012-09-30T16:56:16Z SecondChildTAG: Also, make sure parentheses balance. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-09-30T17:50:18Z SecondChildTAG: I am in the same situation... SecondChildUserIdTAG: 240876 SecondChildUserNameTAG: Jecht SecondChildCreateTimeTAG: 2012-09-30T21:20:45Z SecondChildTAG: Do not forget to put a capital K SecondChildUserIdTAG: 457034 SecondChildUserNameTAG: Ichihara SecondChildCreateTimeTAG: 2012-10-11T00:26:17Z SecondChildTAG: I had the same problem, but after putting capital K my answer was accepted SecondChildUserIdTAG: 457034 SecondChildUserNameTAG: Ichihara SecondChildCreateTimeTAG: 2012-10-11T00:28:06Z FirstChildTAG: Had same problem. Just edit the whole string in a seperate word processor, and paste it all in. Also, if you get parsing errors, just exit the browser and reload. Make sure you have last version of jave. I use Chrome btw. FirstChildUserIdTAG: 142857 FirstChildUserNameTAG: viking2 FirstChildCreateTimeTAG: 2012-10-01T02:51:39Z IndexTAG: 3775 TitleTAG: A simple doubt what if the voltage received by the receiver exeeds 5v or goes below 0v? It is as much possible since the noise margin could affect the input voltage when it is exactly 5v and the voltage at the receiver's end could go beyond 5v.... Do plz answer me. UserIdTAG: 380324 UserNameTAG: y_a_s_h CreateTimeTAG: 2012-09-30T15:49:18Z VoteTAG: 0 CoursewareTAG: Week 2 / Static Discipline CommentableIdTAG: 6002x_static_discipline NumberOfReplyTAG: 1 FirstChildTAG: Greater than 5 would be a "stronger" logic 1, and less than 0 would be a stronger logic 0, wouldn't it? Of course, too much in either direction would destroy the device, but we have yet to consider that. FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-09-30T15:55:28Z SecondChildTAG: Practical devices **may** have diode clamps to protect against these out of range conditions. I am familiar with PIC l microcontrollers that usually have such protection. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-09-30T19:11:07Z IndexTAG: 3776 TitleTAG: does the number of checks matter? i was just wondering, does our percentage increase if the answer's right on the first try? UserIdTAG: 34786 UserNameTAG: ss93 CreateTimeTAG: 2012-09-30T15:42:13Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: no.. FirstChildUserIdTAG: 206648 FirstChildUserNameTAG: TsvetanGeorgiev FirstChildCreateTimeTAG: 2012-09-30T15:48:16Z FirstChildTAG: ss93, in the homeworks and labs the percentage won't be affected if you use multiple tries to get the answer. However, keep in mind that in the midterm and final exams you only get 3 opportunities. FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-30T23:22:21Z IndexTAG: 3777 TitleTAG: H3P3 please help to solve the problem.Waht is the power when R=100ohms UserIdTAG: 219204 UserNameTAG: vsriram CreateTimeTAG: 2012-09-30T15:22:23Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Well, the graph shows the v-i relationship and the Power-voltage relationship(where P=-v*i). In both graphs you have a set of values that will help to find the power when R=100 ohm. According to ohm's law you know that V=R*I, thus R=V/I Knowing that R=100 ohm and that R=V/I and also having all those values in the graphs you can find the power. Tell me if you don't understand, then I'll try to give you more hints that could help you find the right answer. FirstChildUserIdTAG: 45307 FirstChildUserNameTAG: josejimenez2 FirstChildCreateTimeTAG: 2012-09-30T21:47:10Z IndexTAG: 3778 TitleTAG: Lab 3 - Way to draw crossing wires? For lab 3, I'd like to draw wires that cross, but are not connected. Is there a way to do that? Or perhaps I don't need to do that at all for the question... UserIdTAG: 160439 UserNameTAG: tiffguo CreateTimeTAG: 2012-09-30T14:18:12Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Just join the terminals you want to be joined without worrying about the crossing. New nodes will not be formed until you actually drag a wire exactly on top of the other. In the lab here, wire crossover is not represented by a bridging symbol. So any two wires crossing over without a node(a black dot) between them are actually not connected. Hope this helps. Cheers FirstChildUserIdTAG: 310108 FirstChildUserNameTAG: Saakar FirstChildCreateTimeTAG: 2012-09-30T14:34:23Z IndexTAG: 3779 TitleTAG: How to write the answer? im confused. which one should i take?, just mention variables like iB or IB or Ib?? reply ASAP UserIdTAG: 440453 UserNameTAG: Asandyz CreateTimeTAG: 2012-09-30T14:08:39Z VoteTAG: 0 CoursewareTAG: Week 4 / Dependent Sources Exercise 0 CommentableIdTAG: 6002x_dep_src_exsz_0 NumberOfReplyTAG: 1 FirstChildTAG: IB is unknown.so is VB .ur answer should not contain them. FirstChildUserIdTAG: 357747 FirstChildUserNameTAG: kishores FirstChildCreateTimeTAG: 2012-09-30T16:25:31Z SecondChildTAG: Why does that? It is said "Write the algebraic expression required for each question below". And "If iD=K1⋅vB what is vO? If iD=K2⋅iB what is vO?" So iB and vB are given values. SecondChildUserIdTAG: 213452 SecondChildUserNameTAG: JoJosida SecondChildCreateTimeTAG: 2012-10-02T06:59:12Z SecondChildTAG: I think if your answer is rejected in this type of question, just put the answer in the other terms so the system will accept it. As long this doesn't happen when there is only one set of terms available, we should be fine. Also, as long as this type of unacceptable answer doesn't count as one of our 3 tries on the midterm, we should be fine. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-03T15:51:00Z SecondChildTAG: You must write the algebraic expression in terms of circuit parameters, since these aren't unknown. SecondChildUserIdTAG: 129288 SecondChildUserNameTAG: Tinchito SecondChildCreateTimeTAG: 2012-10-05T10:42:20Z IndexTAG: 3780 TitleTAG: LAB 3 TRIED EVERYTHING NOT UNDERSTANDING AT ALL I have tried to solve the lab 3 and have tried everthing that i can nothing works. Can someone point me in the right direction? I have no idea what i am doing wrong HELP UserIdTAG: 160242 UserNameTAG: Mlevins35 CreateTimeTAG: 2012-09-30T13:58:11Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: MOSFETs in parallel make an OR gate while MOSFETS in series make an AND gate. Taking the output from the top most DRAIN tied to a resistor to supply voltage makes an inverter. That is all you need to impliment the equation they are asking for. FirstChildUserIdTAG: 397595 FirstChildUserNameTAG: mholin FirstChildCreateTimeTAG: 2012-09-30T14:13:30Z FirstChildTAG: Apart from the response above, keep in mind that the W/L ratio has to be calculated to change the R.on according to the problem or you will not get the correct response. FirstChildUserIdTAG: 310108 FirstChildUserNameTAG: Saakar FirstChildCreateTimeTAG: 2012-09-30T14:36:42Z FirstChildTAG: Have a look at these hints which Myrimit has put up. https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50673e4c2193ea2300000009 Steps to check. 1. Does you circuit, theoretically, implement the boolean function which they ask? 2. Have you calculated the $R_{ON}$ of the MOSFETs and changed them in the simulator (i.e. changed the W/L ratio from the default value)? FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-09-30T15:02:57Z FirstChildTAG: Regarding the W/L ratio you can in fact note that the bigger the ratio the smaller the resistance is when the MOSFET is ON. If you model the MOSFET as a switch you probably prefer the resistance to be the smallest possible when the switch is closed. So you can just try to write a very big number and it will work almost surely FirstChildUserIdTAG: 346056 FirstChildUserNameTAG: fiatlux FirstChildCreateTimeTAG: 2012-09-30T16:39:07Z FirstChildTAG: I have tried to make the connections and get one-half of the equation right but I can not get the rest. I get the glitch on one end but the one end that the example shows between 2 and 3, thats the part that I can not get. I have tried everything , I just dont think that I can get this. What am I doing wrong and why cant I get it to work? FirstChildUserIdTAG: 160242 FirstChildUserNameTAG: Mlevins35 FirstChildCreateTimeTAG: 2012-09-30T19:42:08Z SecondChildTAG: I want to be able to get it turned in on time am a little worried about that. SecondChildUserIdTAG: 160242 SecondChildUserNameTAG: Mlevins35 SecondChildCreateTimeTAG: 2012-09-30T19:42:36Z IndexTAG: 3781 TitleTAG: Lab2, needing help Hi, I have just started the course today, and I'm having problems to solve the lab2. I think I get the right equations, but when I check the answer it is wrong. I use three resistors. I've read that one of the resistors have to be less than 1 Ohm. Why? UserIdTAG: 241379 UserNameTAG: CrisAzc CreateTimeTAG: 2012-09-30T13:00:04Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: ... Reading problems :) I didn't do the tran analysis before hiting the check button. FirstChildUserIdTAG: 241379 FirstChildUserNameTAG: CrisAzc FirstChildCreateTimeTAG: 2012-09-30T13:02:41Z IndexTAG: 3782 TitleTAG: pls help- lab 3 unable to get a check mark in spite of getting transient analysis correctly.. what to do?? pls help.... UserIdTAG: 128924 UserNameTAG: arjun392 CreateTimeTAG: 2012-09-30T12:14:48Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: look at the small spikes in your output waveform.. do they meet the requirement for the static discipline " assuming this is your problem " ...you have to design your circuit that gives the required output voltage FirstChildUserIdTAG: 231206 FirstChildUserNameTAG: Bradley_B FirstChildCreateTimeTAG: 2012-09-30T12:44:31Z IndexTAG: 3783 TitleTAG: Help required on H3P1 Kindly please anyone help me on H3P1 q4-7 UserIdTAG: 414201 UserNameTAG: uzaifakram CreateTimeTAG: 2012-09-30T11:43:58Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: kindly plz help as soon as possible FirstChildUserIdTAG: 414201 FirstChildUserNameTAG: uzaifakram FirstChildCreateTimeTAG: 2012-09-30T11:54:15Z SecondChildTAG: I have seen myrit hints but didn't get it SecondChildUserIdTAG: 414201 SecondChildUserNameTAG: uzaifakram SecondChildCreateTimeTAG: 2012-09-30T11:55:16Z SecondChildTAG: Hi uzaifakram! Can I help you? SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-30T15:48:27Z FirstChildTAG: Let's look at q4. We want to find the minimum value of R_PuI. That means, when it receives a "low" input, A < 1.5 V, it must output a "high" output, B > 4.5 V, and when it receives a "high" input, A > 4.0 V, it must output a "low" output, B < 1.0 V. Also, according to the SR model, when the mosfet is on, it behaves like a resistor with resistance R_on, which we are told is 8000 Ohm. When A is low, the mosfet is off, so B = V_s = 5.0 V, which satisfies the static discipline. When A is high, the mosfet is on, so B depends on V_s, R_on, and R_PuI. You can write an equation for B and set B < V_OL. Solving this will give you a range of values for R_PuI. If you need a more complete explanation of how this all works, I recommend reading through Chapter 6. FirstChildUserIdTAG: 176592 FirstChildUserNameTAG: JLevitt FirstChildCreateTimeTAG: 2012-09-30T12:00:36Z SecondChildTAG: boss i am not getting that how can i will calculate Rpull I SecondChildUserIdTAG: 414201 SecondChildUserNameTAG: uzaifakram SecondChildCreateTimeTAG: 2012-09-30T12:38:37Z SecondChildTAG: Once you replace the mosfet with a resistor with resistance R_On, calculate the voltage at B. Then find R_PuI so that B < V_OL. If you do not understand the reasoning for this, watch lectures and/or read the book. SecondChildUserIdTAG: 176592 SecondChildUserNameTAG: JLevitt SecondChildCreateTimeTAG: 2012-09-30T12:44:46Z SecondChildTAG: I have dne it for nand and inverter but not able to do for NOR can kindly tell me the formula for VOut for nor implementation SecondChildUserIdTAG: 414201 SecondChildUserNameTAG: uzaifakram SecondChildCreateTimeTAG: 2012-09-30T13:45:15Z SecondChildTAG: thx man, nice explication. SecondChildUserIdTAG: 322923 SecondChildUserNameTAG: Serhii_S SecondChildCreateTimeTAG: 2012-09-30T15:20:14Z SecondChildTAG: I have got the ans. for nand also by using formula vs2/2Ron+Rpull but still dn't know how to find power for Nor SecondChildUserIdTAG: 414201 SecondChildUserNameTAG: uzaifakram SecondChildCreateTimeTAG: 2012-09-30T17:43:21Z SecondChildTAG: Kindly plz help me urgently SecondChildUserIdTAG: 414201 SecondChildUserNameTAG: uzaifakram SecondChildCreateTimeTAG: 2012-09-30T17:43:39Z IndexTAG: 3784 TitleTAG: me talking a lot of Jealousy come to me ... when talking about undergraduate students achievements ..... is the digital abstraction easy for human to deal with ... or it's the optimal way when it's come to computation ? it's look like OOP and the procedural programming UserIdTAG: 364253 UserNameTAG: goon CreateTimeTAG: 2012-09-30T11:27:31Z VoteTAG: 0 CoursewareTAG: Week 3 / Logic Gates CommentableIdTAG: 6002x_logic_gates NumberOfReplyTAG: 1 FirstChildTAG: I believe that the digital abstraction is of great help when dealing with digital circuits. With it, one does not have to deal with the details of the MOSFETs, resistors, etc used to construct them. Although, of course, you should be aware that behind the digital abstraction there are analog components and for detailed calculations one has to take them into account. For example, we will study the effect of the MOSFET gate capacitance in week 7 FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-30T23:17:07Z IndexTAG: 3785 TitleTAG: lab3-w/l how to find w/l ,is Ron same for 3 mosfet.i have designed several times,but still can't find correct solution.whats the right combinatin.,plase help. UserIdTAG: 269641 UserNameTAG: BAUWA CreateTimeTAG: 2012-09-30T11:22:37Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: R_on is not necessarily the same for the 3 mosfets. You want to have values of w/l so that V_ol < 0.25 V. So, calculate the output voltage V_ol when the output is "low." You should find the output voltage is a function of the on resistances R_on of the mosfets. Then you can adjust the R_on by changing the w/l values as described in the problem. You can solve the problem just by calculating V_in in terms of R_on and choosing w/l accordingly. But if you want more information about how mosfets are modeled with R_on resistances, see Section 6.6: The SR Model of the Mosfet, and if you want to know how the w/l factor changes R_on, see section 6.7: Physical Structure of the Mosfet and Example 1.8 on p. 20. FirstChildUserIdTAG: 176592 FirstChildUserNameTAG: JLevitt FirstChildCreateTimeTAG: 2012-09-30T11:49:39Z SecondChildTAG: @JLevitt-thanx well i got correct ans,but still i need to understand the logic behind o output and truthtable. SecondChildUserIdTAG: 269641 SecondChildUserNameTAG: BAUWA SecondChildCreateTimeTAG: 2012-09-30T12:21:02Z SecondChildTAG: well i got Ron equal for all 3 MOSFETS....and its correct....!!! :) THIS IS POSSIBLE SecondChildUserIdTAG: 127800 SecondChildUserNameTAG: MNB SecondChildCreateTimeTAG: 2012-09-30T15:41:01Z SecondChildTAG: mnb cud u plzz help me with the lab 3 SecondChildUserIdTAG: 365792 SecondChildUserNameTAG: pabhijeet SecondChildCreateTimeTAG: 2012-09-30T18:30:13Z FirstChildTAG: Lab2: Have been trying to solve lab2 but still having problems with it,don't know how exaclty to go about it. 1. The mixer cicuit cannot be give us d required out of 1/2V1 + 1/6V2. 2. I think this can only be done using OP-Amp. Do any body has any idea on lab2,please? FirstChildUserIdTAG: 433119 FirstChildUserNameTAG: SHOMUYIWA FirstChildCreateTimeTAG: 2012-09-30T12:26:21Z FirstChildTAG: you are given an approximate Rn for a certain voltage , in this case 3 V.... look on page 295 of the text for an example... your truth table should match your circuit, you can test it by plugging in CBA from you truth table the output should follow. FirstChildUserIdTAG: 231206 FirstChildUserNameTAG: Bradley_B FirstChildCreateTimeTAG: 2012-09-30T12:37:04Z FirstChildTAG: Hi, I made up the Lab3 circuit and the Vol=0.22V, Trans figure is almost same to the given graphic's figure, but it is still error. Why? I set the W/L = 40. FirstChildUserIdTAG: 371617 FirstChildUserNameTAG: rainbow12 FirstChildCreateTimeTAG: 2012-09-30T13:40:09Z SecondChildTAG: you are close to w/l.try higher range. SecondChildUserIdTAG: 357747 SecondChildUserNameTAG: kishores SecondChildCreateTimeTAG: 2012-09-30T16:29:13Z IndexTAG: 3786 TitleTAG: H3P4 Currents I am confused applying node method for currentd. If I assume when Vs is positive D1 is on(short) and D2 is Open, then the currents coming out of node is e-Vs/R + e/R + id = 0. Then I have to assume e=2.5 V because it is ideal diode and there is no voltage drop across diode. However, the answer I get is wrong. What am I doing wrong? Help UserIdTAG: 175706 UserNameTAG: rwskim CreateTimeTAG: 2012-09-30T11:20:24Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I don't think you can make that assumption. The behavior of D1 and D2 depend on their characteristics. Examining values of e based on the characteristic of the ideal diode is a good idea. But e = 2.5 V isn't the one to look at. Check your node values again and try values of e that will make D1 or D2 short circuit. Also note which values of e are possible and which are not. FirstChildUserIdTAG: 176592 FirstChildUserNameTAG: JLevitt FirstChildCreateTimeTAG: 2012-09-30T11:34:52Z FirstChildTAG: NOTE THAT both the diodes are both reverse biased to the DC supplies. When the ac supplys polarity is positive D1 will conduct current and D2 will not. When the ac supply is negitive D2 will conduct current and D1 will not. I just managed to complete it :D FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-09-30T14:35:57Z IndexTAG: 3787 TitleTAG: general question We all know that according to ohm's law 'v' is directly proportinal to 'i' at constant temperature. All i want to know is does constant tempeature means oom temperature or temperature in resistor. "staff" UserIdTAG: 516123 UserNameTAG: PrabhakarX CreateTimeTAG: 2012-09-30T11:18:15Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: It's temperature in the resistor - that's what affects the properties of the material. Change in room temperature may change the resistance of the device but that's because the heat will get transferred to (of from, if the room is cooler) the device. This will cause the temperature of the device to change. This change in the device's temperature will cause the resistance to change. So it's always the temperature of the resistor that needs to be constant. FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-09-30T14:57:52Z IndexTAG: 3788 TitleTAG: H3P1 The on resistance for NOR gate should be calculated by calculating the equivalent parallel resistance of both the resistances, right? Which should give 7000 ohms, and then the minimum pull up resistance should be 4 times, which makes 28000. Tell me where did I go wrong here? Please help UserIdTAG: 499671 UserNameTAG: maliha266 CreateTimeTAG: 2012-09-30T10:34:09Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: The question is asking for minimum value of the pullup resistor R. Think about when R will have minimum value! (try to think of different combinations of input that requires the minimum value of the pullup resistor R. FirstChildUserIdTAG: 175706 FirstChildUserNameTAG: rwskim FirstChildCreateTimeTAG: 2012-09-30T10:52:18Z SecondChildTAG: ok but we are not given the value for off resistance SecondChildUserIdTAG: 499671 SecondChildUserNameTAG: maliha266 SecondChildCreateTimeTAG: 2012-09-30T10:57:04Z SecondChildTAG: @maliha: The values mentioned in your question might not match that for others. Hence please try to provide us with the process with which you are proceeding instead of outright values. SecondChildUserIdTAG: 314654 SecondChildUserNameTAG: Mandar001 SecondChildCreateTimeTAG: 2012-09-30T13:17:05Z SecondChildTAG: maliha266, Off resistance is a open circuit so when A < Vt the output B will be Vs SecondChildUserIdTAG: 385224 SecondChildUserNameTAG: shilpiji991 SecondChildCreateTimeTAG: 2012-09-30T15:36:16Z FirstChildTAG: I'm not seeing where you got 7000 Ohms. A||B = (8000^-1 + 8000^-1)^-1 = 4000 Ohms. This approach may give you R_PuO for when both A and B are on. But in order to satisfy the static discipline, we need C < V_OL for ANY inputs for which the output is low... FirstChildUserIdTAG: 176592 FirstChildUserNameTAG: JLevitt FirstChildCreateTimeTAG: 2012-09-30T12:29:54Z FirstChildTAG: Thanx everyone I got it right :) FirstChildUserIdTAG: 499671 FirstChildUserNameTAG: maliha266 FirstChildCreateTimeTAG: 2012-09-30T13:13:10Z IndexTAG: 3789 TitleTAG: H3P2 GRAPHICAL MODEL OF INVERTER how to start this problem?? please help... UserIdTAG: 114864 UserNameTAG: Vikaash CreateTimeTAG: 2012-09-30T10:28:52Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: you can refer to tutorial 3 so as to get an insight to this problem. FirstChildUserIdTAG: 137918 FirstChildUserNameTAG: jaisneha FirstChildCreateTimeTAG: 2012-09-30T10:32:00Z SecondChildTAG: i have already gone through it.....but it help me in solving H3P1.... SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-09-30T10:35:53Z FirstChildTAG: I would recommend looking at Section 5.1: Voltage levels and the static discipline. It made my understand why we want certain voltage levels. For this problem, we are given an input/output transfer characteristic. This plots the output Vout versus the input Vin of an inverter. Also we are given that the noise margins are symmetric. Say we choose some value for VIL. All inputs below VIL will be interpreted as "low" by the inverter. From the characteristic, we can see what voltages will be output for our chosen VIL, and we can determine VOH. Now, we want to choose VIH < VOH. And with VIH, we can use the characteristic to find VOL. Now, we can compare this VOL with the original VIL we chose to ensure that VIL > VOL. Now we can do it again and try choosing different values until we have some that have maximal symmetric noise margins. Maybe try to come up with an argument for why your voltage levels are optimal. (I think it has something to do with the slope of the characteristic...) If you didn't understand any of this, I would really recommend reading the book. FirstChildUserIdTAG: 176592 FirstChildUserNameTAG: JLevitt FirstChildCreateTimeTAG: 2012-09-30T11:25:32Z IndexTAG: 3790 TitleTAG: "H3P4" HOW TO CALCULATE CURRENT THROUGH "DIODE D2"?i've open circuited D1 branch and then applying KVL,but answer is coming wrong.someone please help me as soon as possible. UserIdTAG: 137918 UserNameTAG: jaisneha CreateTimeTAG: 2012-09-30T10:15:06Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: have u changed the polarity of Vs????? FirstChildUserIdTAG: 114864 FirstChildUserNameTAG: Vikaash FirstChildCreateTimeTAG: 2012-09-30T10:30:35Z SecondChildTAG: YES..STILL M NOT GETTING THE CORRECT ANSWER SecondChildUserIdTAG: 137918 SecondChildUserNameTAG: jaisneha SecondChildCreateTimeTAG: 2012-09-30T10:35:05Z SecondChildTAG: u have to apply KVL in both loop and obtained current through diode in each case......and finally the current through diode is the difference of current obtained in each case. SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-09-30T10:39:43Z SecondChildTAG: i've taken -ve value of Vs(RMS) for -ve half cyle and then applied KVL..still not getting the answer SecondChildUserIdTAG: 137918 SecondChildUserNameTAG: jaisneha SecondChildCreateTimeTAG: 2012-09-30T10:42:17Z SecondChildTAG: then what i think answer should come...if you are not getting the answer...go for superposition theorem SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-09-30T10:47:59Z SecondChildTAG: Hopefully this will help! https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5067499d2193ea230000000c SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-09-30T11:13:29Z IndexTAG: 3791 TitleTAG: H3P4 I got the voltages across the resistor. But i dont know how to get the current passing through the diodes. Help me... please UserIdTAG: 309722 UserNameTAG: kaushikraghavan1992 CreateTimeTAG: 2012-09-30T09:40:02Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5067499d2193ea230000000c FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-09-30T09:46:12Z IndexTAG: 3792 TitleTAG: Music on the voltage source Please Explain how to apply Music to the voltage source? UserIdTAG: 215271 UserNameTAG: JustStudent CreateTimeTAG: 2012-09-30T09:35:55Z VoteTAG: 0 CoursewareTAG: Week 3 / Incremental Method Motivation CommentableIdTAG: 6002x_incremental_method_motivation NumberOfReplyTAG: 1 FirstChildTAG: Output jack on the player already a voltage source. It's just too weak to drive LED, so you need an amplifier. FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-09-30T09:50:15Z IndexTAG: 3793 TitleTAG: h3p4 !!!!!!!!!! can any one explain to me how to get the currents? UserIdTAG: 395257 UserNameTAG: blackguitar CreateTimeTAG: 2012-09-30T09:11:37Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: apply superposition method FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-09-30T09:39:02Z SecondChildTAG: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5067499d2193ea230000000c SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-09-30T09:46:03Z SecondChildTAG: i saw your explanation hazel1919 before,I'm sorry i can't got it with this explanation SecondChildUserIdTAG: 395257 SecondChildUserNameTAG: blackguitar SecondChildCreateTimeTAG: 2012-09-30T10:29:01Z SecondChildTAG: What exactly is the part you are having trouble with, I know I got confused with the signs when applying KCL to the -9v circuit. SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-09-30T11:09:03Z SecondChildTAG: ok ,i got it ,,,, the mistake that i have the Ron=8.2 (ohms) .... but finally i realized it was (K ohms):):):) SecondChildUserIdTAG: 395257 SecondChildUserNameTAG: blackguitar SecondChildCreateTimeTAG: 2012-10-01T01:04:59Z IndexTAG: 3794 TitleTAG: S2E3 SHINJI 01 Do you someone have the expression and video link of the answer of S2E3 ? Shinji UserIdTAG: 158876 UserNameTAG: SHINJI01 CreateTimeTAG: 2012-09-30T08:12:37Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3795 TitleTAG: homework due time what time is the homework due exactly on sunday? UserIdTAG: 365225 UserNameTAG: doublegj526 CreateTimeTAG: 2012-09-30T06:11:57Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Midnight, your time zone FirstChildUserIdTAG: 118266 FirstChildUserNameTAG: Nisheet FirstChildCreateTimeTAG: 2012-09-30T06:31:15Z IndexTAG: 3796 TitleTAG: mosfet w/l value explanation Is the lab wording wrong or am i missing something? Ron=Rn*L/W Ron = total resistance of mosfet Rn = resistance per square inch LW = lengh / width = what we can set Cutouts from the lab ** the mosfet model has Rn≈26.5kΩ ** setting a mosfet's W/L to 10 would result in RON=2.65kΩ. 26.5k * 10 = 2.65k? in what universe.... UserIdTAG: 329964 UserNameTAG: SmartMike CreateTimeTAG: 2012-09-30T06:01:58Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: setting ratio W/L to 10 sets the reciprocal L/W to 0.1 therefore Ron = Rn * 1/(W/L) = 26.5k * 0.1 = 2.65k that got me too, for like a half hour :( FirstChildUserIdTAG: 74193 FirstChildUserNameTAG: SpaceAgeRobot FirstChildCreateTimeTAG: 2012-09-30T06:27:20Z IndexTAG: 3797 TitleTAG: Lab 2 hack Hey folks, just in case you're still stuck with Lab 2, here's a little hack to help you solve it ;) You've been doing the practice excercises in the lectures, haven't you? Well then, return to [S3E2][1], incline your head to the left and *look closely*. ![enter image description here][2] If you have the excercise done, all the equations are already there. It only requires to express the resistor values via one of the resistors out there to solve it. Good luck! [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_2/Linearity_and_Superposition/3 [2]: https://www.edx.org/static/content-mit-6002x/images/circuits/superposition-v.0dc7d284480b.gif UserIdTAG: 378267 UserNameTAG: artfwo CreateTimeTAG: 2012-09-30T06:00:50Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3798 TitleTAG: h5p1.MaximaumVin H5P1 Part3 Why? I want to know the minimum value of Vout in Saturation region! ids=K/2{vout)^2 1=Vout+RL*ids Thank you in advance. UserIdTAG: 89440 UserNameTAG: yuk CreateTimeTAG: 2012-09-30T05:01:02Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3799 TitleTAG: BUG : S5V6 SWITCH MODEL Here for the case when C1 and C2 are both 1, it is told as both as zero at time around 0:50. but in the board it is written correctly. It is told when c1 and c2 are 0, we have short circuits there which is wrong. UserIdTAG: 177329 UserNameTAG: nanduharshan CreateTimeTAG: 2012-09-30T03:11:17Z VoteTAG: 0 CoursewareTAG: Week 3 / Switch Model CommentableIdTAG: 6002x_switch_model NumberOfReplyTAG: 1 FirstChildTAG: yeah it sounds a little confusing..... what Dr. Anant is saying .... is you have a short circuit "0 volts " when the switch is closed... the switch is the mosfet.. so a logical one at the input "C1 or C2" to the mosfet turns on the mosfet " swicth" when the switch turns on it shorts " 0 Volts " allowing maximum current to flow with O Volts " ideal case". so in short think of mosfets as switches, you apply a logical one to the input " as seen in the truth table" the switch turns on ansdshort circuits, if both mosfets shorts to ground " 0 volts" then the output is 0 volts.. hopes this helps.. FirstChildUserIdTAG: 231206 FirstChildUserNameTAG: Bradley_B FirstChildCreateTimeTAG: 2012-09-30T03:32:45Z IndexTAG: 3800 TitleTAG: hw 2 how to find the resistor values for the h2p1 in hw 2 UserIdTAG: 383641 UserNameTAG: bountyhunter CreateTimeTAG: 2012-09-30T02:53:49Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi bountyhunter! Can I help you? You can read this Post with hints of H2P1 [here][1] also you can visit the wiki [read this Post][2];) Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/505971f4dd2f4d1f00000004 [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5067b3ab2193ea2300000022 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-30T03:11:34Z IndexTAG: 3801 TitleTAG: LAB 2 Guys, I am stuck with lab 2 homework....do we need 3 resistors...and are 1, 2 and 3 ohm then te correct results for those 3 resistors???...i dont know how to incorporate the third resistor... please help UserIdTAG: 308861 UserNameTAG: Ignaas CreateTimeTAG: 2012-09-30T02:53:39Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Hi Ignass! How are you? Can I help you? In which part are you lost? You can see this Post of Lab2 [Here][1] also you can visit the Wiki [read this Post][2] ;). Myriam. P.D: I have made a video Tutorial of Lab2 but I can not upload it till the deadline of Lab 2 ... [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/505f41be015492230000001c [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5067b3ab2193ea2300000022 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-30T03:09:00Z SecondChildTAG: i've started this week with the course...so now i've so much to do...i don't know how to combine the three resistors...i know that i need to satisfy the equation that is given...i've punt one resistor in serie with V1 and another resister in serie with V2.....but now i need to plase the third resistor SecondChildUserIdTAG: 308861 SecondChildUserNameTAG: Ignaas SecondChildCreateTimeTAG: 2012-09-30T03:22:04Z SecondChildTAG: Hi Ignaas! Welcome to 6.002x! Ok, here some hints: Hint 1: Try to think of the voltage dividers formula (You will see that the output voltage will be the the voltage input by (R2/R2+1)) . This formula it is general and you can use it in Lab 2 ;). You can take a look at page 75 of the Textbook [here][1] Voutput=Vinput*(R2/R2+R1) ![Divisor][2] Hint 2: knowing the Hint 1, can you in some way relate your resistors in a similar way? I hope this two hints can help you ;) [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/99 [2]: https://edxuploads.s3.amazonaws.com/13489758508135765.png SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-30T03:37:25Z SecondChildTAG: thnx...i hope so now SecondChildUserIdTAG: 308861 SecondChildUserNameTAG: Ignaas SecondChildCreateTimeTAG: 2012-09-30T11:39:06Z FirstChildTAG: hi lgnass... there has been alot of suggestions , maybe search the forum for lab 2... other than that take each circuit one at a time.... you need a third resistor... so basically short one source at a time and re draw each circuit then use superposition... the trick is the fine the resistor combination that will give you the ratio you want... what is true so far the resistor combination to get 1/2 V1 means that R combination has to equal each other as seen in the lecture FirstChildUserIdTAG: 231206 FirstChildUserNameTAG: Bradley_B FirstChildCreateTimeTAG: 2012-09-30T03:12:06Z IndexTAG: 3802 TitleTAG: MOS GATES how do i plot transfr chara for the nor gate using a mosfet and how do i check the static discipline wkth the given threshold UserIdTAG: 383641 UserNameTAG: bountyhunter CreateTimeTAG: 2012-09-30T02:38:56Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3803 TitleTAG: Lab 3 setup I Know that switch A and B needs to be in parallel, and the in series with switch C but I don't see how to do it in the sand box. I am not sure how to connect the ports. Can someone please help me thank you UserIdTAG: 435197 UserNameTAG: RichmondRichter CreateTimeTAG: 2012-09-30T01:10:10Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: look on page 295 of textbook for an example FirstChildUserIdTAG: 231206 FirstChildUserNameTAG: Bradley_B FirstChildCreateTimeTAG: 2012-09-30T01:23:29Z FirstChildTAG: Hi RichmondRichter! Can I help you? I answered here some way to conect wires without overlapping [read the post here][1]. Also you can see some help of Lab 3 [here][2] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50670ed25336b42700000004 [2]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/50673e4c2193ea2300000009 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-30T01:17:26Z IndexTAG: 3804 TitleTAG: H3P3 Solar Cell Hi everyone, I don't understand how to get the power drawn from the solar cell given a resistor. I have to find any number, but it doesn't say the voltage or current in order to calculate it, it only says that you have to use the graphical technique. I'd be really thankful if you can help me out. UserIdTAG: 61423 UserNameTAG: libo654 CreateTimeTAG: 2012-09-30T00:13:11Z VoteTAG: 0 CoursewareTAG: Week 3 / Nonlinear Elements CommentableIdTAG: 6002x_nonlinear_elements NumberOfReplyTAG: 3 FirstChildTAG: Fig. 4.20 in the text (pg. 204) shows an example of a graphical solution of a linear circuit (Fig. 4.16) with a non-linear element. Basically, each element has its own characteristic i-v curve. We plot them both and see where they intersect. FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-09-30T01:38:31Z SecondChildTAG: Thank you, I already understand that part :) SecondChildUserIdTAG: 61423 SecondChildUserNameTAG: libo654 SecondChildCreateTimeTAG: 2012-09-30T02:23:16Z FirstChildTAG: you really need to look at the graphs and understand what you are seeing...start with ohms law, rearrange your equation in terms of Resistance, that will tell you what graph to use. so if you solve for R, whats the relation that gives you R... if you still don't get it start plugging in values from the graph into your equation... hope this helps FirstChildUserIdTAG: 231206 FirstChildUserNameTAG: Bradley_B FirstChildCreateTimeTAG: 2012-09-30T00:31:12Z FirstChildTAG: There is a part of the both graphs that the slope is constant.. meaning if you take the derivative you will get a constant. The part of the graph is your I which you will pull out from graph A. In other words I is constant for a long period of time in graph A, thats your I, you also are given R. That should be helpful enough FirstChildUserIdTAG: 435197 FirstChildUserNameTAG: RichmondRichter FirstChildCreateTimeTAG: 2012-09-30T01:07:52Z IndexTAG: 3805 TitleTAG: about solution i get answer in one case only at ix less than 1mA...... vx=ix*r.........r=1k> but if i > 1mA............. not correct answer UserIdTAG: 292299 UserNameTAG: ammarsamir CreateTimeTAG: 2012-09-29T22:49:51Z VoteTAG: 0 CoursewareTAG: Week 3 / Analyzing Nonlinear Circuits Exercise CommentableIdTAG: 6002x_analyzing_nonlinear_circuits_exercise NumberOfReplyTAG: 1 FirstChildTAG: Dear friend, Look, you must satisfy two equations: By the thevenin method: Vx+2988Ix=3.18 (approx). And by the first graph, you have two option: Vx=1000*Ix (Ix<=1mA) or Vx=2000*Ix (Ix>=1mA). So, if you try the two options, you will get: If Vx=1000*Ix, Ix=0.797mA (SATISFY THE CONDICTION Ix<=1mA). If Vx=2000*Ix, Ix=0.637mA (DONT SATISFY THE CONDICTION Ix>=1mA). Therefore, you have just one answer!! Pick it up Best Regards FirstChildUserIdTAG: 263241 FirstChildUserNameTAG: yvesauad FirstChildCreateTimeTAG: 2012-09-30T03:43:57Z SecondChildTAG: OK I made the same mistake. The problem is that one tries to figure out the Ix in the 2nd case from the IV curve, whereas the current flowing through is not determined by the IV curve but the joint resistance of Rth and Rx, where Rx can be of two values. (I guess...) Now the use of the IV-curve comes as the Ix value must fit on the curve... (thats how I guess to solve it) The results you get with the first resistance (and with the wrong order of methods) is still good because it crosses the origin... but the logic is wrong... (I guess again...) SecondChildUserIdTAG: 324332 SecondChildUserNameTAG: petsol SecondChildCreateTimeTAG: 2012-09-30T16:29:25Z IndexTAG: 3806 TitleTAG: Ports, MOSFETs... As I've understood so far, a port is a pair of terminals: + and - terminal. The dependent current sources have 4 terminals: 2 for output and 2 for control, hence they are 2 port devices. Resistors and independent current/voltage sources have only 2 terminals - they are 1 port devices. And how about MOSFET with it's 3 terminals? UserIdTAG: 326787 UserNameTAG: Angstrem CreateTimeTAG: 2012-09-29T20:55:14Z VoteTAG: 0 CoursewareTAG: Week 4 / Introduction to Dependent Sources CommentableIdTAG: 6002x_int_dep_src NumberOfReplyTAG: 1 FirstChildTAG: MOSFET inputs and outputs use a common ground, even if there happens to be a resistor or other component in the path(s) to ground. I vote for "2-Port." FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-09-29T22:56:34Z IndexTAG: 3807 TitleTAG: H3P1 I think one of the answers is wrong ! The one before the last one : What is the maximum power (in Watts) consumed by the NAND? I think I could find a power that is more than the one in the answer. Could you please verify your answer is right ? UserIdTAG: 454609 UserNameTAG: Alwahsh CreateTimeTAG: 2012-09-29T20:03:28Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: FWIW, I did that one and got it right. I don't think there is a problem. (I'm a student) FirstChildUserIdTAG: 468623 FirstChildUserNameTAG: RobNik FirstChildCreateTimeTAG: 2012-09-29T20:09:20Z SecondChildTAG: Let us discuss it after the deadline ... may be I got something wrong SecondChildUserIdTAG: 454609 SecondChildUserNameTAG: Alwahsh SecondChildCreateTimeTAG: 2012-09-29T20:20:44Z SecondChildTAG: yes i got it tooo.......... jst check ur answer once again (iam astudent 4m india) SecondChildUserIdTAG: 383641 SecondChildUserNameTAG: bountyhunter SecondChildCreateTimeTAG: 2012-09-30T02:56:29Z FirstChildTAG: I got that problem right, too FirstChildUserIdTAG: 371220 FirstChildUserNameTAG: dan10 FirstChildCreateTimeTAG: 2012-09-29T20:32:48Z FirstChildTAG: for the power use the resistor value and use the equation= (Vs^2)/Rpull+Ron FirstChildUserIdTAG: 343291 FirstChildUserNameTAG: LuisEduardoH FirstChildCreateTimeTAG: 2012-09-29T23:20:00Z FirstChildTAG: as LuisEduardoH stated above, this is power dissipation formula, to find maximum dissipation: you need to find the minimum Ron equivalent, in order to obtain the minum total resistence and then the maximum power dissipation. FirstChildUserIdTAG: 50320 FirstChildUserNameTAG: marm496 FirstChildCreateTimeTAG: 2012-09-30T04:57:51Z FirstChildTAG: Did you get it right? I didn't get the correct answer at the begining and the reason was I didn't understand properly "...with the minimum pullup resistors that you have already calculated", in that case with the one calculated for NAND. I was using the minimum resistor in general. FirstChildUserIdTAG: 401263 FirstChildUserNameTAG: VicI FirstChildCreateTimeTAG: 2012-09-30T17:39:13Z IndexTAG: 3808 TitleTAG: Resistor model for the amp How an amplifier can be represented as a resistor..I can't get it Thanks UserIdTAG: 302050 UserNameTAG: Talaat CreateTimeTAG: 2012-09-29T19:55:25Z VoteTAG: 0 CoursewareTAG: Week 2 / Thevenin model CommentableIdTAG: 6002x_S3E5_Thevenin_Model NumberOfReplyTAG: 2 FirstChildTAG: I assume you're referring to the RS model for the MOSFET. It's just a model. It is approximately correct for a useful range of values. For instance, the S model alone was ok for us at first. Then the RS model was introduced. Another model will be soon be required. FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-09-30T01:51:46Z FirstChildTAG: Well, you aren't actually modelling an amplifier as a resistor, what you are doing is looking into the amplifier and seeing the input resistance, that is you are abstracting the amplifier. We do this so we can easily find out how much of the original voltage is going to be sent to the load (amplifier). On a tangent, R = V/I, and R has a unit of ohms, or volts/amp, if you consider volts/amp to be an output/input, then the resistor seems like some kind of amplifier that inputs current and outputs voltage. This isn't related to the problem though, just a side note. FirstChildUserIdTAG: 3571 FirstChildUserNameTAG: ifoughtsharks FirstChildCreateTimeTAG: 2012-10-03T19:44:44Z IndexTAG: 3809 TitleTAG: Solved using superposition method I solved it by applying the superposition method. I guess my solution is correct only this time since the dependant voltage source is linear. Forgot something, It is obvious to use the superposition method since the objective of this problem is to find VTH and RTH which is correct only if the circuit is linear. I want to know how to solve this problem with the node method only. UserIdTAG: 418182 UserNameTAG: euldji2005 CreateTimeTAG: 2012-09-29T19:08:41Z VoteTAG: 0 CoursewareTAG: Week 4 / Dependent Sources Exercise 1 CommentableIdTAG: 6002x_dep_src_exsz_1 NumberOfReplyTAG: 1 FirstChildTAG: Could you please clarify where did you apply superposition? I don't see where it can be applied. FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-09-30T19:56:42Z SecondChildTAG: You should find first the expression of the current i which is the sum of currents resulted from the contribution of each source at a time in the circuit. Then, find the thevenin voltage source by the same procedure. Since the dependant source is linear, then you can apply superposition by keeping a single source at a time in order to calculate what you want. SecondChildUserIdTAG: 418182 SecondChildUserNameTAG: euldji2005 SecondChildCreateTimeTAG: 2012-10-02T22:23:12Z SecondChildTAG: You don't need an additional source to find Vth. You need a probe source to find Rth - without Io. There is no place for superposition. SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-03T06:39:13Z SecondChildTAG: Could you please post in formulas. Probably I dont understand something. SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-03T06:42:13Z SecondChildTAG: After reading so many comments, I found out that the soltion I found was totaly a very special case. You can see the post of ashwith to understand the soltion SecondChildUserIdTAG: 418182 SecondChildUserNameTAG: euldji2005 SecondChildCreateTimeTAG: 2012-10-05T03:26:18Z SecondChildTAG: you can see my answer to his solution - his solution is not correct in part of use of superposition SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-06T18:53:35Z SecondChildTAG: actually looks like he just forgot to add Io, but it totally unnecessary - probe current source is absolutelly enough to get Rth SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-10-06T18:56:43Z IndexTAG: 3810 TitleTAG: Possibility of doing a weeks in advance I have a very important exam coming up in November from school. As a result I doubt that for two weeks, I will be able to complete the Homework or Lab Assignments. I have done all the Lecture Series, Homework and Lab Assignments till now (including Week 5). Is there a way to do the requirements for subsequent weeks now, when I have more time, to cover up for the weeks I will be missing due to my exams? UserIdTAG: 85490 UserNameTAG: yashsavani CreateTimeTAG: 2012-09-29T19:07:33Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: If you look on Week 5 Homework you can see the due date is 15 days from now, so if you miss the next two weeks, you should still have time to do them after you come back. Or even better, take a loot at the [6.002x At-A-Glance][1] and plan accordingly. Notice the mid term exam :) [1]: https://www.edx.org/static/content-mit-6002x/handouts/at-a-glance.9674fe7f677e.pdf FirstChildUserIdTAG: 138981 FirstChildUserNameTAG: Pr0bability FirstChildCreateTimeTAG: 2012-09-29T19:23:18Z IndexTAG: 3811 TitleTAG: HINT i din't get the hint even though i solved it using some other method(and i know the method is right one).....still i wanna understand the hint...plz help UserIdTAG: 432151 UserNameTAG: rgdixit CreateTimeTAG: 2012-09-29T17:51:58Z VoteTAG: 0 CoursewareTAG: Week 4 / Dependent Sources Exercise 1 CommentableIdTAG: 6002x_dep_src_exsz_1 NumberOfReplyTAG: 1 FirstChildTAG: i meanhow do we get Vth and Rth using that additional source. FirstChildUserIdTAG: 432151 FirstChildUserNameTAG: rgdixit FirstChildCreateTimeTAG: 2012-09-29T18:00:24Z SecondChildTAG: Let me clearly define what is meant by Rth. Rth is the equivalent resistance seen by a source(voltage source or current source) if it is added at the port where we calculate Rth. Let me know if this clue is not sufficient for you. SecondChildUserIdTAG: 823 SecondChildUserNameTAG: naveenatmit SecondChildCreateTimeTAG: 2012-09-29T18:22:51Z SecondChildTAG: chech this: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5060f4a8a3f6d21f00000050 SecondChildUserIdTAG: 374393 SecondChildUserNameTAG: rmaleki SecondChildCreateTimeTAG: 2012-09-30T10:42:10Z IndexTAG: 3812 TitleTAG: Finding a fixed point in S6V5 Hello, There is a mistake in the explanations on how to find a fixed point: it starts on 3m27s with "So I plug 0.32 volts on the left hand side, I get the 0.65" When you put Vd = 1.0V on the right side, you get 0.32V on the left side. In order to continue, you have to plug in 0.32V on the *right* side again. Because if you plug it in on the left side you will just get the original 1.0V on the right. UserIdTAG: 188680 UserNameTAG: corsairnv CreateTimeTAG: 2012-09-29T17:34:29Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 0 IndexTAG: 3813 TitleTAG: voltage and current sources in thevenin n norton's.... Hi.. Can someone explain me why the voltage source in thevenin's theorem must be placed in series and the norton's current source in parallel with the thevenin's equivalent resistor? UserIdTAG: 390803 UserNameTAG: Spurty CreateTimeTAG: 2012-09-29T17:20:52Z VoteTAG: 0 CoursewareTAG: Week 2 / Thevenin method CommentableIdTAG: 6002x_thevenin NumberOfReplyTAG: 3 FirstChildTAG: For the way to arrive at both equivalents look it up in the textbook. It'll do a far superior job than anyone here. If however you just want to convince yourself that both N and Th are equivalent connect each to the same load. Simple resistor would be easiest. Now that you have two circuits calculate the V across and the current through the resistor in each network. They should be the same for each circuit, which proves that both N and Th are indeed equivalent. FirstChildUserIdTAG: 5325 FirstChildUserNameTAG: vkz FirstChildCreateTimeTAG: 2012-09-29T18:16:51Z FirstChildTAG: Thevenin equivalent models the network as a voltage source whereas the Norton equivalent models it as Current source. How do you model the current source and voltage source ? is the concept that you to have learn from this question. Learn the VI characteristics of voltage source(not the ideal one) and current source(not the ideal one), you will get to know the concept. FirstChildUserIdTAG: 823 FirstChildUserNameTAG: naveenatmit FirstChildCreateTimeTAG: 2012-09-29T18:31:47Z FirstChildTAG: We get Vth in series with Rth because the circuit being modeled will have its own resistance which decreases the voltage available to a load. Similarly, for the parallel current source, the internal resistance (Rth) will model the fact that the circuit being modeled is also in parallel with the circuit external to the circuit being modeled and requires some current to operate. A careful reading of Section 3.6 of the text will show that both the Thevenin and Norton models construct the i-v characteristic of the circuit which is being modeled. Consider an i-v graph (i, vertical axis and v, horizontal axis). Open circuit voltage will be the x-axis intercept (max voltage and zero current). Short circuit current will be the y-axis intercept (max current and zero voltage). FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-09-30T02:23:42Z IndexTAG: 3814 TitleTAG: Homework 4 Dependant Sources Hello, I am wondering if anyone please clear me how to calculate the thevenin resistance? UserIdTAG: 131426 UserNameTAG: M_Inam_Ul_Haq CreateTimeTAG: 2012-09-29T16:44:18Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: You could do it by calculating norton current then thevinin resistance would be r=Vth/In FirstChildUserIdTAG: 99628 FirstChildUserNameTAG: Pranjal16 FirstChildCreateTimeTAG: 2012-09-29T17:25:55Z FirstChildTAG: See Example 3.23 in the text (Thevenin model of ckt with dependent source). FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-09-30T02:29:56Z IndexTAG: 3815 TitleTAG: MOSFET AS LOGICAL GATE how to analyze the nand gate meets the static discipline with the given threshold UserIdTAG: 383641 UserNameTAG: bountyhunter CreateTimeTAG: 2012-09-29T16:24:31Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You simply have 2*Ron in your circuit. With given threash hold calculate Rl. The procedure is exactly the same as that of inverter except that you have 2*Ron in place of Ron. FirstChildUserIdTAG: 131426 FirstChildUserNameTAG: M_Inam_Ul_Haq FirstChildCreateTimeTAG: 2012-09-29T16:50:19Z SecondChildTAG: please for NOR gate thankx help SecondChildUserIdTAG: 331441 SecondChildUserNameTAG: abdessamade SecondChildCreateTimeTAG: 2012-09-29T18:22:36Z IndexTAG: 3816 TitleTAG: mosfet as a logical gate (HW 3) how do i plot the transfer characteristics of mos "nand gate" ? UserIdTAG: 383641 UserNameTAG: bountyhunter CreateTimeTAG: 2012-09-29T16:22:01Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: In the S-model they are absolutely the same. In he SR-model notice that when input is low effectively you have an inverter, so plot your high output. When both gates are ON (input high in both) you have two pull-down resistors Ron in series, so the only difference is that your OH will be higher than in the inverter chart. FirstChildUserIdTAG: 5325 FirstChildUserNameTAG: vkz FirstChildCreateTimeTAG: 2012-09-29T18:25:01Z IndexTAG: 3817 TitleTAG: circuit sandbox not listing all relevant values I'm doing circuit simulations of various constructions and finding that many relevant values are not displayed. I assume this is some sort of a bug. Any recommendations? I'm looking to download circuit simulation software and also to hopefully get this course's browser tool circuit sandbox working well. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-29T14:58:17Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: try this one: http://www.falstad.com/circuit/ FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-09-29T18:16:14Z IndexTAG: 3818 TitleTAG: Lab 4 I did the scheme and get the green tick but I can't get the true graphics. In fact, the only thing i see after the tran analysis ( 0.5us ) is the x and y-axes. Can anyone help me with this ? How can I get green tick, if I did wrong scheme ? I use regular DC source, some vtest labels and my mosfet switch exactly the same as in the example. Please help me. This is drving me crazy.. UserIdTAG: 206648 UserNameTAG: TsvetanGeorgiev CreateTimeTAG: 2012-09-29T14:15:42Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: The stop time for transient analysis should be 0.5**m**s. Also check your probes' connections. FirstChildUserIdTAG: 104543 FirstChildUserNameTAG: sten FirstChildCreateTimeTAG: 2012-09-29T14:34:42Z SecondChildTAG: hello. could you please help me? all i am getting is a vertical line in the graph. SecondChildUserIdTAG: 222911 SecondChildUserNameTAG: bhaswardg SecondChildCreateTimeTAG: 2012-09-30T05:43:02Z IndexTAG: 3819 TitleTAG: S5V13: MOSFET INVERTER Why do we write Voh=5.1V in this video? I think if Vol is 0.2V it means that Voh will be 4.8V. Am I right or not? UserIdTAG: 190057 UserNameTAG: Anastasia00435 CreateTimeTAG: 2012-09-29T14:11:20Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3820 TitleTAG: week tutorial lectures . . any downlaod link for these the lectures of "tutorials" can't be seen due to blockage of Youtube in my country. are there any links for these vidoes??? UserIdTAG: 146930 UserNameTAG: fayzan_007 CreateTimeTAG: 2012-09-29T11:58:08Z VoteTAG: 0 CoursewareTAG: Week 3 / Circuit Truth Table Tutorial CommentableIdTAG: 6002x_circuit_truth_table_tutorial NumberOfReplyTAG: 1 FirstChildTAG: download it from wiki. direct link to download is available. FirstChildUserIdTAG: 136490 FirstChildUserNameTAG: Ali_PU FirstChildCreateTimeTAG: 2012-09-29T14:37:01Z SecondChildTAG: "tutorial videos" are also there .. .. i have seen that there is only lecture sequences of each week but not the tutorial lectures SecondChildUserIdTAG: 146930 SecondChildUserNameTAG: fayzan_007 SecondChildCreateTimeTAG: 2012-09-30T18:17:10Z IndexTAG: 3821 TitleTAG: correction 0:22 Thevenin not feminine thx UserIdTAG: 445656 UserNameTAG: MHMTMU CreateTimeTAG: 2012-09-29T10:27:32Z VoteTAG: 0 CoursewareTAG: Week 4 / Amplifiers CommentableIdTAG: 6002x_amplifiers NumberOfReplyTAG: 0 IndexTAG: 3822 TitleTAG: Local time, checking. Hi guys! Can you explane me how do the site programm now my local time? I did not write my country or time zone in my profile. So, that meens that i can cheat by changing time on my computer? :) Also, how does the site programm see that i realy done lab or homework in time? and made it right? UserIdTAG: 500965 UserNameTAG: barka0 CreateTimeTAG: 2012-09-29T08:19:58Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3823 TitleTAG: Problem in HW3 pls help me to find max power consumed by the Nor gate in H3P2 also in H3P4 i m unable to find the maximum current in diode D1 and D2..... UserIdTAG: 308248 UserNameTAG: dpc11 CreateTimeTAG: 2012-09-29T08:11:07Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Do you complete the FET power calculation FirstChildUserIdTAG: 156694 FirstChildUserNameTAG: mkprasanth FirstChildCreateTimeTAG: 2012-09-29T08:37:54Z IndexTAG: 3824 TitleTAG: technically what exactly happened in the demo i would like to know the basic description of how the demo was performed...as in what exactly were the connections made.. UserIdTAG: 401199 UserNameTAG: ashritha CreateTimeTAG: 2012-09-29T07:54:40Z VoteTAG: 0 CoursewareTAG: Week 3 / Musical Demo CommentableIdTAG: 6002x_musical_demo NumberOfReplyTAG: 0 IndexTAG: 3825 TitleTAG: distance between the led and the receiver what is the max possible distance b/w the LED and the receiver for it to work? UserIdTAG: 401199 UserNameTAG: ashritha CreateTimeTAG: 2012-09-29T07:51:04Z VoteTAG: 0 CoursewareTAG: Week 3 / Musical Demo CommentableIdTAG: 6002x_musical_demo NumberOfReplyTAG: 3 FirstChildTAG: how to find diode Max current in Diode D1 and D2 in problem H3P4???? FirstChildUserIdTAG: 308248 FirstChildUserNameTAG: dpc11 FirstChildCreateTimeTAG: 2012-09-29T07:58:12Z SecondChildTAG: also i m unable to find maximum power consumed by the NOR in problem H3P1..plss help me....... SecondChildUserIdTAG: 308248 SecondChildUserNameTAG: dpc11 SecondChildCreateTimeTAG: 2012-09-29T07:59:57Z SecondChildTAG: Dear Ashritha Sir/Mam Please note there are LED and Photo diode / Transistors couplers avilable or you can tray with Photo couplers available, S erch any laser printers and VCR for such couplers SecondChildUserIdTAG: 156694 SecondChildUserNameTAG: mkprasanth SecondChildCreateTimeTAG: 2012-09-29T13:40:04Z FirstChildTAG: also i m unable to find maximum power consumed by the NOR in problem H3P1..plss help me....... FirstChildUserIdTAG: 308248 FirstChildUserNameTAG: dpc11 FirstChildCreateTimeTAG: 2012-09-29T08:00:18Z FirstChildTAG: for LED it will be very bad actually: http://en.wikipedia.org/wiki/Inverse-square_law FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-09-29T09:54:00Z IndexTAG: 3826 TitleTAG: Homework 3 question 3 How are we suppose to find the power with the given resistor, I've solved everything else that is the only thing I cant figure out UserIdTAG: 435197 UserNameTAG: RichmondRichter CreateTimeTAG: 2012-09-29T07:29:48Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Power chart for the cell defines the constraint on the Power it generates with respect to the voltage across. Power generated must be consumed. What part of the circuit dissipates energy? Can you produce the P,v relationship for it? If you could then you'd have another constraint on P,v. Your solution would obviously have to satisfy both. Try it. FirstChildUserIdTAG: 5325 FirstChildUserNameTAG: vkz FirstChildCreateTimeTAG: 2012-09-29T15:53:57Z SecondChildTAG: Ok so the constraint on the cell is -v*i and the has to be equivalent to the power out consumed by the resistor which is v^2/R. Im guessing I set these equal to each other and pick a random voltage, find i then use that to solve for the power? SecondChildUserIdTAG: 435197 SecondChildUserNameTAG: RichmondRichter SecondChildCreateTimeTAG: 2012-09-29T21:41:32Z FirstChildTAG: Chart B shows the relation between p and v. If you differentiate that relation and determine dp/dv then you can link that outcome to chart A. Ask yourself: is dp/dv constant or not? That implies something ... FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-09-29T20:06:40Z SecondChildTAG: dp/dv is constant from 0 to about .25 volts and di/dv for graph is is also the same...I need one more hint. SecondChildUserIdTAG: 435197 SecondChildUserNameTAG: RichmondRichter SecondChildCreateTimeTAG: 2012-09-29T21:50:37Z SecondChildTAG: You are close now. If di/dv is constant in A, is it zero or not? If it is 0, is it a CS or VS? Convert to norton or thevenin? SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-09-30T12:49:33Z IndexTAG: 3827 TitleTAG: why does it have to always be a resistor? what if the curve is slightly inclined to either the x or y axes.what if its not linear or just a straight line which represents a voltage or a current source? what do we do then? UserIdTAG: 401199 UserNameTAG: ashritha CreateTimeTAG: 2012-09-29T06:36:47Z VoteTAG: 0 CoursewareTAG: Week 3 / Piecewise Linear Analysis CommentableIdTAG: 6002x_piecewise_linear_analysis NumberOfReplyTAG: 2 FirstChildTAG: then we look for thevenin or norton equivalent circuit with IDEAL source and series/shunt resistance FirstChildUserIdTAG: 157610 FirstChildUserNameTAG: mradziwo FirstChildCreateTimeTAG: 2012-09-29T09:45:22Z FirstChildTAG: For straight line in terms of incremental resistance it's short circuit for voltage-like and open circuit for current-like graphs. Equivalent circuits for these curves are ideal voltage source and ideal current source, respectively. Commonly speaking any linear graph can be introduced by voltage or current source with, respectively, series or parallel resistor - source corresponds to shift from origin and resistor corresponds to tangent. for nonlinear see small signal FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-09-29T09:46:05Z IndexTAG: 3828 TitleTAG: confirmation the load line is nothing but the representation of the thevinins equivalent circuit..right? UserIdTAG: 401199 UserNameTAG: ashritha CreateTimeTAG: 2012-09-29T06:15:40Z VoteTAG: 0 CoursewareTAG: Week 3 / Graphical Analysis of Nonlinear Circuits Exercise CommentableIdTAG: 6002x_graphical_analysis_of_nonlinear_circuits_exercise NumberOfReplyTAG: 1 FirstChildTAG: right but also Norton's FirstChildUserIdTAG: 157610 FirstChildUserNameTAG: mradziwo FirstChildCreateTimeTAG: 2012-09-29T09:46:05Z SecondChildTAG: It's already set up as a Norton equivalent circuit the way it's drawn... right? SecondChildUserIdTAG: 318779 SecondChildUserNameTAG: jbell9 SecondChildCreateTimeTAG: 2012-09-29T23:48:24Z SecondChildTAG: Yes. SecondChildUserIdTAG: 372623 SecondChildUserNameTAG: Amitraj SecondChildCreateTimeTAG: 2012-09-30T07:40:05Z IndexTAG: 3829 TitleTAG: juan carlos a1=(R2*R3)/(R1*R2+R2*R3+R1*R3) a2=(R1*R3)/(R1*R2+R2*R3+R1*R3) b1=(-R2-R3)/(R1*R2+R2*R3+R1*R3) b2=(R3)/(R1*R2+R2*R3+R1*R3) c1=(R3)/(R1*R2+R2*R3+R1*R3) c2=(-R1-R3)/(R1*R2+R2*R3+R1*R3) UserIdTAG: 471282 UserNameTAG: juancarlosxd CreateTimeTAG: 2012-09-29T05:13:20Z VoteTAG: 0 CoursewareTAG: Week 2 / Linear combinations of source strenghts CommentableIdTAG: 6002x_S3E2_Linear_Combinations_of_Source_Strengths NumberOfReplyTAG: 0 IndexTAG: 3830 TitleTAG: how do i download the lecture sequences..?? please could anyone show how to download the videos of lectures...??? UserIdTAG: 210992 UserNameTAG: neerajnatu CreateTimeTAG: 2012-09-29T04:42:23Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: install internet download manager when you play video pop up window will open and you will be able to download video FirstChildUserIdTAG: 342135 FirstChildUserNameTAG: vikash902 FirstChildCreateTimeTAG: 2012-09-29T06:48:59Z IndexTAG: 3831 TitleTAG: "that range" what should be understood by that range in prob no1? UserIdTAG: 401199 UserNameTAG: ashritha CreateTimeTAG: 2012-09-29T04:31:37Z VoteTAG: 0 CoursewareTAG: Week 3 / Analyzing Nonlinear Circuits Exercise CommentableIdTAG: 6002x_analyzing_nonlinear_circuits_exercise NumberOfReplyTAG: 1 FirstChildTAG: consider the case when current is greater than 1mA. FirstChildUserIdTAG: 12144 FirstChildUserNameTAG: Sonam FirstChildCreateTimeTAG: 2012-09-29T05:08:29Z SecondChildTAG: ok got it!!! thanks! SecondChildUserIdTAG: 401199 SecondChildUserNameTAG: ashritha SecondChildCreateTimeTAG: 2012-09-29T06:16:33Z IndexTAG: 3832 TitleTAG: juan carlos the response is 6.206 UserIdTAG: 471282 UserNameTAG: juancarlosxd CreateTimeTAG: 2012-09-29T04:28:29Z VoteTAG: 0 CoursewareTAG: Week 2 / Node equation review CommentableIdTAG: 6002x_S3E1_NodeEquationReview NumberOfReplyTAG: 2 FirstChildTAG: orly? FirstChildUserIdTAG: 116304 FirstChildUserNameTAG: Wegrzyn FirstChildCreateTimeTAG: 2012-10-12T17:50:48Z FirstChildTAG: that's right FirstChildUserIdTAG: 411696 FirstChildUserNameTAG: mendez FirstChildCreateTimeTAG: 2012-10-12T17:48:02Z IndexTAG: 3833 TitleTAG: R_ON = 5kohms?? That big? Did I misunderstand, or Prof. Argawal said that the ON-state resistance of the MOSFET was 5 thousand ohms? Isn't it too much? UserIdTAG: 109327 UserNameTAG: Felipe_Martins CreateTimeTAG: 2012-09-29T04:06:28Z VoteTAG: 0 CoursewareTAG: Week 3 / Switch Resistor Model of a MOSFET CommentableIdTAG: 6002x_switch_resistor_model NumberOfReplyTAG: 1 FirstChildTAG: As he is mentioning many many times... it is ABSTRACTION. It is a big value for a mosfet to be used as switch, but there are some applications, where such resistance would be desired. Nevertheless it is only for training purposes and in our case it could be 1mOhm or 1MOhm as well... just to exaggerate some effects to make us aware of them. FirstChildUserIdTAG: 157610 FirstChildUserNameTAG: mradziwo FirstChildCreateTimeTAG: 2012-09-29T09:49:55Z IndexTAG: 3834 TitleTAG: help with the lab 3 where am i wromg?? [1]: https://edxuploads.s3.amazonaws.com/13488851304954474.png UserIdTAG: 161899 UserNameTAG: harshvit12 CreateTimeTAG: 2012-09-29T02:19:13Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: What is your problem? FirstChildUserIdTAG: 70325 FirstChildUserNameTAG: Zotyaka FirstChildCreateTimeTAG: 2012-09-29T09:15:23Z IndexTAG: 3835 TitleTAG: static discipline when we say output high and output low and vout such sort of things whether are we talking of the things at the recievers end or senders end??iam very much confused plz help UserIdTAG: 161899 UserNameTAG: harshvit12 CreateTimeTAG: 2012-09-29T01:18:19Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I think both.. actually both means the same. you connect the output of one transistor to the input of another one on the belief that what we see as the output is fed as input to the next guy. Hope it explains.. :) FirstChildUserIdTAG: 257806 FirstChildUserNameTAG: skoda FirstChildCreateTimeTAG: 2012-09-29T02:58:55Z IndexTAG: 3836 TitleTAG: Just trying out something. It is not ok to take someone else’s solution and simply copy the answers from their solution into your checkboxes. • It is not ok to take someone else’s formula and plug in your own numbers to get the final answer. • It is not ok to post answers to lecture and problem set problems before the submission deadline. • It is not ok to look at a full step-by-step solution to a problem before the submission deadline. UserIdTAG: 418095 UserNameTAG: Ezeiyoke CreateTimeTAG: 2012-09-29T00:58:50Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3837 TitleTAG: Chapter 4, formula 4.1 What is the meaning or value of the letter e in formula 4.1? UserIdTAG: 231764 UserNameTAG: PaulP4881 CreateTimeTAG: 2012-09-29T00:40:21Z VoteTAG: 0 CoursewareTAG: Week 3 / Inside a Mouse CommentableIdTAG: 6002x_inside_a_mouse NumberOfReplyTAG: 1 FirstChildTAG: It's a constant. More info here: http://en.wikipedia.org/wiki/E_(mathematical_constant) FirstChildUserIdTAG: 213386 FirstChildUserNameTAG: dmascenik FirstChildCreateTimeTAG: 2012-09-29T03:43:22Z IndexTAG: 3838 TitleTAG: I am lost and need a guidance. I am lost. How do l follow your day to day progress? I don't know what to do, is it just watching video clips? Where are the materials to read? I saw an announcement on deadline for homework but have n't seen the homework nor the material to read before attending to the homework. I need guidance. UserIdTAG: 418095 UserNameTAG: Ezeiyoke CreateTimeTAG: 2012-09-28T23:52:04Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Homework and Labs are under courseware - after each week's lectures and tutorials. The book is under the textbook tab. FirstChildUserIdTAG: 42833 FirstChildUserNameTAG: KKA FirstChildCreateTimeTAG: 2012-09-29T02:24:29Z IndexTAG: 3839 TitleTAG: Boston magazine about EDX [link to boston gazette article september][1] [1]: http://www.bostonmagazine.com/articles/2012/08/edx-online-classes-schools-out-forever/4/ UserIdTAG: 79885 UserNameTAG: ruudoleo CreateTimeTAG: 2012-09-28T22:56:51Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Very good FirstChildUserIdTAG: 361823 FirstChildUserNameTAG: EliasOak FirstChildCreateTimeTAG: 2012-09-28T23:54:29Z IndexTAG: 3840 TitleTAG: 2nd in 2nd one i had entered v but it showed invalid input v UserIdTAG: 516164 UserNameTAG: RamadeviEdamakanti CreateTimeTAG: 2012-09-28T18:26:29Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL and KCL exercises CommentableIdTAG: 6002x_kvl_kcl_exercises NumberOfReplyTAG: 0 IndexTAG: 3841 TitleTAG: Official answers I would like to know where we get official answers once the deadline is over. UserIdTAG: 99628 UserNameTAG: Pranjal16 CreateTimeTAG: 2012-09-28T17:00:01Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: I would like to know how is this online court work? FirstChildUserIdTAG: 496272 FirstChildUserNameTAG: ammekakuzukera2004 FirstChildCreateTimeTAG: 2012-09-28T17:09:19Z SecondChildTAG: I don't think if they provide SecondChildUserIdTAG: 131426 SecondChildUserNameTAG: M_Inam_Ul_Haq SecondChildCreateTimeTAG: 2012-09-28T18:11:22Z FirstChildTAG: If you mean the correct answers for homework, there will be a "show answer" button after the due date has passed. FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-09-28T18:49:03Z FirstChildTAG: I am lost. How do l follow your day to day progress? Is it just watching video clips? Where are the materials to read? I saw an announcement on deadline for homework but have n't seen the homework and the material to read before attending to the homework. I need guidance. FirstChildUserIdTAG: 418095 FirstChildUserNameTAG: Ezeiyoke FirstChildCreateTimeTAG: 2012-09-28T23:45:33Z SecondChildTAG: Try avery menu you find in this page, is very intuitive. SecondChildUserIdTAG: 410943 SecondChildUserNameTAG: JPalaciosRoman SecondChildCreateTimeTAG: 2012-09-29T17:28:57Z IndexTAG: 3842 TitleTAG: Homework 3 Help! I would like to know what is low noise margin and high noise margin. UserIdTAG: 252983 UserNameTAG: IgorCastelo CreateTimeTAG: 2012-09-28T16:44:14Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: i would like to know where all the assignment is located at? FirstChildUserIdTAG: 496272 FirstChildUserNameTAG: ammekakuzukera2004 FirstChildCreateTimeTAG: 2012-09-28T17:09:40Z SecondChildTAG: That has nothing to do with my problem but the answer is in courseware ¬¬ SecondChildUserIdTAG: 252983 SecondChildUserNameTAG: IgorCastelo SecondChildCreateTimeTAG: 2012-09-28T17:15:02Z SecondChildTAG: Read section 5.1 of your textbook. https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/269 SecondChildUserIdTAG: 248807 SecondChildUserNameTAG: 1kingsman SecondChildCreateTimeTAG: 2012-09-28T21:23:21Z IndexTAG: 3843 TitleTAG: More practice exercises Does anyone know where I can find more practice exercise similar to H3P4? UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-09-28T16:02:13Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3844 TitleTAG: terminals for battery drawn i ve observed that the battery terminal he drew in the video is wrong ..what do you say guys?? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-28T15:38:06Z VoteTAG: 0 CoursewareTAG: Week 2 / Is it linear? CommentableIdTAG: 6002x_is_it_linear NumberOfReplyTAG: 0 IndexTAG: 3845 TitleTAG: what is discount code for Foundations of Analog and Digital Electronic Circuits textbook? what is discount code for Foundations of Analog and Digital Electronic Circuits textbook? for edx students UserIdTAG: 170636 UserNameTAG: shanthasuresh29 CreateTimeTAG: 2012-09-28T14:28:30Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: From http://store.elsevier.com/specialOffer.jsp?offerId=EST_MIT "Save 40% when you buy print + electronic titles together. Use code MIT40. Free shipping. *This offer is not available in Australia. Save 25% on any print or electronic title. Use code MIT25. Free shipping." FirstChildUserIdTAG: 213386 FirstChildUserNameTAG: dmascenik FirstChildCreateTimeTAG: 2012-09-28T15:14:20Z SecondChildTAG: Thank you dmascenik SecondChildUserIdTAG: 170636 SecondChildUserNameTAG: shanthasuresh29 SecondChildCreateTimeTAG: 2012-09-28T17:26:36Z FirstChildTAG: you can get free from www.4shared.com FirstChildUserIdTAG: 131426 FirstChildUserNameTAG: M_Inam_Ul_Haq FirstChildCreateTimeTAG: 2012-09-28T18:13:27Z SecondChildTAG: Piracy is not encouraged here. SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-09-28T18:38:51Z SecondChildTAG: @ M_Inam_Ul_Haq thanz a lot SecondChildUserIdTAG: 170636 SecondChildUserNameTAG: shanthasuresh29 SecondChildCreateTimeTAG: 2012-09-30T14:02:32Z FirstChildTAG: In case you're from India, you can buy the Indian edition here: http://www.flipkart.com/foundations-analog-digital-electronic-circuits-8131200892/p/itmdytr5dbf4tdth?pid=9788131200896&ref=8deb85e1-b597-4f6b-8ccf-317287d1fb39&srno=s_1&otracker=from-search It's about 1/10th the price of the international edition. FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-09-28T15:44:06Z SecondChildTAG: @ashwith I found most books at flipkart printed in a newspaper like quality.WHAT ABOUT THIS ? SecondChildUserIdTAG: 99628 SecondChildUserNameTAG: Pranjal16 SecondChildCreateTimeTAG: 2012-09-28T17:14:49Z SecondChildTAG: ashwith thanx a lot :-). I am from Indian SecondChildUserIdTAG: 170636 SecondChildUserNameTAG: shanthasuresh29 SecondChildCreateTimeTAG: 2012-09-28T17:24:05Z SecondChildTAG: @Pranjal16 that depends on the publisher not Flipkart. The international edition costs about 5k while this one is about 600. So I don't think we can comment on that. Having said that, I'm happy with the quality of the book although I've rarely bothered about it. SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-09-28T18:37:44Z IndexTAG: 3846 TitleTAG: Battery mnemonics The longer line corresponds to the higher potential (+), the shorter one to the lower potential (-). I think it's always better when mnemonics reflect the physical meaning. UserIdTAG: 410033 UserNameTAG: sagitta CreateTimeTAG: 2012-09-28T14:05:31Z VoteTAG: 0 CoursewareTAG: Week 3 / Load Line Tutorial CommentableIdTAG: 6002x_load_line_tutorial NumberOfReplyTAG: 1 FirstChildTAG: yes.. you need longer line to draw + than to draw - ;p therefore the rule:) FirstChildUserIdTAG: 157610 FirstChildUserNameTAG: mradziwo FirstChildCreateTimeTAG: 2012-09-28T14:55:26Z SecondChildTAG: well, that works, but it just plain memorizing without understanding what the + and - actually stand for :) SecondChildUserIdTAG: 410033 SecondChildUserNameTAG: sagitta SecondChildCreateTimeTAG: 2012-10-04T09:56:19Z IndexTAG: 3847 TitleTAG: Last two tasks Somebody can say How to find correct answer to last two tasks ? UserIdTAG: 192114 UserNameTAG: chuba CreateTimeTAG: 2012-09-28T13:52:43Z VoteTAG: 0 CoursewareTAG: Week 3 / Analyzing Nonlinear Circuits Exercise CommentableIdTAG: 6002x_analyzing_nonlinear_circuits_exercise NumberOfReplyTAG: 1 FirstChildTAG: find the thevenin equivalent. Lets say the thevenin resistance and voltage is "Rth" and "Vth" respectively. Notice that the nonlinear element will then be in series with the "Rth". Now in the thevenin equivalent ckt assume that only Rth and Vth exists in this case the current "In"(i.e Norton Current) will be Vth/Rth. In this question "In = .000106A". Now plug in the device and note that it is a non linear resistor in series with "Rth". Since the device and the "Rth" are in series the current will decrease and therefore current through the device is definitely going to be less than 1mA. Bingo!! Then from the previous two question we found out that the resistance of the device is 1k when the current is less than 1mA. Now it is simple!! FirstChildUserIdTAG: 12144 FirstChildUserNameTAG: Sonam FirstChildCreateTimeTAG: 2012-09-29T05:20:47Z SecondChildTAG: Thanks Sonam, your answer actually warped all the good previous posts up in some short sentences with best resolution for the tasks. SecondChildUserIdTAG: 153707 SecondChildUserNameTAG: masoud_np SecondChildCreateTimeTAG: 2012-09-29T19:17:04Z IndexTAG: 3848 TitleTAG: why distortion please i want in clearly way why there is distortion i - due to non linearity 2- clipping in negative signal UserIdTAG: 295983 UserNameTAG: qassam CreateTimeTAG: 2012-09-28T12:57:24Z VoteTAG: 0 CoursewareTAG: Week 3 / Musical Demo CommentableIdTAG: 6002x_musical_demo NumberOfReplyTAG: 1 FirstChildTAG: Clipping is a non-linearity and all non-linearity is a distortion. FirstChildUserIdTAG: 157610 FirstChildUserNameTAG: mradziwo FirstChildCreateTimeTAG: 2012-09-28T14:59:11Z SecondChildTAG: yeah SecondChildUserIdTAG: 131426 SecondChildUserNameTAG: M_Inam_Ul_Haq SecondChildCreateTimeTAG: 2012-09-28T18:14:47Z IndexTAG: 3849 TitleTAG: Official Answers to Homework 2 and Lab not yet available? When will the official answers to past-due homeworks and labs be made available? In particular, the official answers for Week 2 were available at some point, but they are gone now. UserIdTAG: 153760 UserNameTAG: Kavka CreateTimeTAG: 2012-09-28T12:49:17Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: deadline for week 2 was extended, so most likely solutions will appear next week FirstChildUserIdTAG: 157610 FirstChildUserNameTAG: mradziwo FirstChildCreateTimeTAG: 2012-09-28T15:00:04Z IndexTAG: 3850 TitleTAG: Notes Could you explain some progression to answer the five questions of the expression ? Measure current. The DC analysis annotates each voltage source with the current it is providing to the rest of the circuit. Please report the signed current I flowing from the source into the circuit, where the current is considered to be positive when it is flowing out of the positive terminal of the source, through the series of resistors, then back into the negative terminal. Next, report the total resistance R of the path connecting the positive and negative terminals of the source. Using the voltage value V of the source, verify that Ohm's law V=IR is obeyed. Current flowing from source into the circuit (amps, with correct sign): ※ 500e-6 Total resistance between pos and neg terminals (ohms): ※ 6000 Transient analysis. Change the waveform produced by the voltage source from a DC value of 3V to a sinusoid with a 1V amplitude, an offset voltage of 1V, and a frequency of 1kHz. Add scope probes to nodes A, B and C and edit their properties so that the plots will be different colors. Now run a transient analysis for 5ms. Move the mouse over the plot until the marker (a vertical dashed line that follows the mouse when it's over the plot) is at approximately 1.25ms. Please report the measured voltages for nodes A, B and C. Voltage for node A (volts): ※incorrect 1.995 Voltage for node B (volts): ※incorrect 1.662 Voltage for node C (volts): ※incorrect .665 UserIdTAG: 158876 UserNameTAG: SHINJI01 CreateTimeTAG: 2012-09-28T11:07:07Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3851 TitleTAG: H4P3 part B Notron current I have tried all the methods but not getting the correct answer for norton current.i am afraid I might give up this now.please help UserIdTAG: 99628 UserNameTAG: Pranjal16 CreateTimeTAG: 2012-09-28T08:25:13Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 6 FirstChildTAG: I am afraid, you might miss the happiness that everyone of us enjoy in solving circuits if you give up now. I am very much experienced in solving circuits even then it took some time for me to solve this Norton current. This problem is little tricky. Take it as a challenge, change your approach every time when you see a REDx instead of GREENtick. then you can really have an EUREKA moment. :) If you have come this far, I believe, you can solve it yourself, have patience!! FirstChildUserIdTAG: 823 FirstChildUserNameTAG: naveenatmit FirstChildCreateTimeTAG: 2012-09-28T08:44:37Z FirstChildTAG: I got stuck on this one for quite a while. I watched the courseware sequence videos again couple of times, I read section 3.5.1 of the textbook, and still could not figure it out. Then I started searching online for additional info on Thevenin/Norton equivalents of circuits with both independent and dependent sources, and ran into the following page: http://en.wikibooks.org/wiki/Electronics/Thevenin/Norton_Equivalents. This helped me A LOT! I was finally able to figure out that Rth==Rn = Voc / Isc. You basically need to separately calculate V on open circuit port B (or Vth) and I on short circuit port B (this would be your resulting In). Then to get Rn you just have to divide Voc by Isc. Hope this helps. I wish this was explained better in the sequence material though... FirstChildUserIdTAG: 401592 FirstChildUserNameTAG: netghost FirstChildCreateTimeTAG: 2012-09-28T09:05:37Z FirstChildTAG: (R2//R3)+ R1 FirstChildUserIdTAG: 316899 FirstChildUserNameTAG: elou FirstChildCreateTimeTAG: 2012-09-29T09:23:30Z SecondChildTAG: how? SecondChildUserIdTAG: 89663 SecondChildUserNameTAG: farazali316 SecondChildCreateTimeTAG: 2012-09-30T10:51:06Z SecondChildTAG: I'm asking the same question - how? SecondChildUserIdTAG: 7114 SecondChildUserNameTAG: piotroxp SecondChildCreateTimeTAG: 2012-10-01T19:02:25Z SecondChildTAG: how is it possible? SecondChildUserIdTAG: 259693 SecondChildUserNameTAG: MehrozKhan SecondChildCreateTimeTAG: 2012-10-07T09:52:50Z FirstChildTAG: After understanding the textbook example and going through the links mentioned by netghost I am happy to do it.I feel I am now enjoying the same happiness that naveenatmit mentioned. FirstChildUserIdTAG: 99628 FirstChildUserNameTAG: Pranjal16 FirstChildCreateTimeTAG: 2012-09-29T15:55:41Z FirstChildTAG: I had the same issue that pranjal60 had, I tried all the methods that I had and anyone seem to work. I felt dissapointed and angry. First I tried to find IN, doing a shortcircuit at terminals A and B. In that way, I guess that the shortcircuit makes the dependent source and the 5 ohm resistor dissapear (because they are in parallel to a short circuit), I tried calculating that short circuit current because that should be IN. It didnt work. I tried every form to calculate IN, and when I thought I was going to expote, I just surrender to the exercise, and applied the convertion equation that helps you get IN in function of VTH and RTH. So basically I calculated RTH (which is the same as RN) and then VTH. Then I just divided VTH/RTH and got IN. That answer worked. Still I dont know why the other method (shortcircuit between A and B) never worked. If someone knows, please let me know. And sorry for my English, it is not my native language. FirstChildUserIdTAG: 181432 FirstChildUserNameTAG: enriqueferreralcala FirstChildCreateTimeTAG: 2012-10-01T23:02:57Z SecondChildTAG: I'm hoping my ah-ha moment comes before the work is due....I would love to know why shorting across the terminals does not work to solve for IN. That is how it's supposed to work, yes?? SecondChildUserIdTAG: 244115 SecondChildUserNameTAG: Pedro1969 SecondChildCreateTimeTAG: 2012-10-02T22:28:16Z SecondChildTAG: Please help me. I applied your method, but cant get the answer. SecondChildUserIdTAG: 171674 SecondChildUserNameTAG: wajahat1 SecondChildCreateTimeTAG: 2012-10-05T07:20:11Z FirstChildTAG: Frustration here too :( Mahn i have like 20 odd pages working this problem out. I know i have this Vth value wrong. But can't seem to figure out how to get the correct one. Any help would be appreciated ! :) FirstChildUserIdTAG: 97340 FirstChildUserNameTAG: AnkitRana FirstChildCreateTimeTAG: 2012-10-03T15:07:04Z IndexTAG: 3852 TitleTAG: H3P4: How to deal with diodes? I've got a problem with diodes. It seems that I don't know how to analyse circuits with diodes... For example, I have no idea, how to figure out voltages in all the nodes (except the ground ones) in this circuit: ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13488157698390905.png Please, can anybody show me, how to analyse such circuits? UserIdTAG: 326787 UserNameTAG: Angstrem CreateTimeTAG: 2012-09-28T07:03:21Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: V(e1) should be greater than V(e2) for the diode to act in forward biased region. e1 node is the anode of the diode d1. e2 node is the cathode of the diode d1. V(anode) > V(cathode), diode is forward biased. else, reverse biased. FirstChildUserIdTAG: 823 FirstChildUserNameTAG: naveenatmit FirstChildCreateTimeTAG: 2012-09-28T07:20:08Z SecondChildTAG: Thank you, but I know, how do diodes work. I want to know, how to calculate e1, e2, e3 when knowing only V1, V2, V3. SecondChildUserIdTAG: 326787 SecondChildUserNameTAG: Angstrem SecondChildCreateTimeTAG: 2012-09-28T07:26:54Z SecondChildTAG: Treat a diode as a short circuit when voltage across it is positive (respectively to diode polarity) and as a open circuit when voltage across it is negative. So voltage across upper diode is e1-e2=V1-V2, and across lower diode e3-e1=-V3-V1. Please, notice signs in these equations - it's a key point to understanding when diode behaves as a short or open circuit. SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-28T07:43:13Z SecondChildTAG: Thanks a lot, but why e1 = V1? Isn't it affected by V2 and V3? SecondChildUserIdTAG: 326787 SecondChildUserNameTAG: Angstrem SecondChildCreateTimeTAG: 2012-09-28T07:53:55Z SecondChildTAG: In general, equation goes likes this. V1=V(across Diode)+V2. If we assume diode as a ideal diode, V(across Diode) will be zero, so, above equation becomes, V1 = V2. This is purely, mathematical point of viewing the situation. In the real scenario, the series resistance of V1 and V2 battery, diode's Vbi and series resistance comes into picture. So, in this case, If V1 and V2 are different, and V1>V2, you cannot assume IDEAL diode. All the physical consequences of connecting two battery's in parallel will occur if the Diode is assumed a ideal one. Are you able to understand and see the situation here? I recommend you to solve more circuits involving diodes. Keep in mind, in forward biased mode, diode can conduct currents in mA range limited by power capability whereas in reverse biased mode, reverse current of uA/nA range will flow through the diode. try to understand the physical meaning of VI characteristics of diode. SecondChildUserIdTAG: 823 SecondChildUserNameTAG: naveenatmit SecondChildCreateTimeTAG: 2012-09-28T08:16:10Z SecondChildTAG: >Thanks a lot, but why e1 = V1? Isn't it affected by V2 and V3? It's a property of ideal independent voltage source - at its terminal it's always its voltage. Actually your schematics, if every element is ideal, as naveenatmit noticed, is meaningless because, let's say for upper diode, if voltage across it is positive (in respect to diode's polarity) it will act as a short circuit, and V1 and V2 will be connected directly and voltage at this node should be V1 and V2 at same time - it's a contradiction. Physically it means that current at this node will be infinite. SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-28T08:44:22Z SecondChildTAG: you can better understand the voltage on the source terminals if you replace diodes with resistors and write down nodes' equations SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-28T08:46:17Z FirstChildTAG: Thank you all very much for your explanations! It seems that now I understand how to analyse diode circuits. Ideal diodes are basically almost the same thing as pieces of ideal wire, except the point that they allow current to flow in only one direction. And hence they behave like ideal wires, we cannot have positive voltage on their terminals. This is just like we can't have any potential difference between 2 nodes taken on a single piece of ideal wire. FirstChildUserIdTAG: 326787 FirstChildUserNameTAG: Angstrem FirstChildCreateTimeTAG: 2012-09-28T16:53:29Z IndexTAG: 3853 TitleTAG: What does it this mean ? I didn't understand that question"How many **distinct boolean-valued functions** are there of two boolean-valued signals?" Specially this part "distinct boolean-valued functions" , can any one explain it please. UserIdTAG: 185715 UserNameTAG: amirengineer CreateTimeTAG: 2012-09-28T06:29:09Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: - signals can take any value. Boolean valued signals can take either 0 or 1. - functions are dependent variables. In this case, it depends on signals. - Boolean-valued functions are functions which are dependent on Boolean-valued signals. - distinct Boolean-valued functions : unique functions. - Ex: $\operatorname{func1} A$ = $A$ and $\operatorname{func2} A$ = $A$ are not distinct. They are same. Not unique. So, here the question is to find the all possible different distinct Boolean-valued function definitons possible for two variables. FirstChildUserIdTAG: 823 FirstChildUserNameTAG: naveenatmit FirstChildCreateTimeTAG: 2012-09-28T09:59:11Z IndexTAG: 3854 TitleTAG: H2P1 Hi guys! I have read some of the posts and many of which kept saying that Rth is the parallel combination of R1 and R2. Could anyone please shed light on this? Thanks in advance! Cheers! UserIdTAG: 499268 UserNameTAG: ruinoah CreateTimeTAG: 2012-09-28T06:22:04Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Rth is the equivalent resistance seen between two terminals when all the independent sources are removed. That is, by replacing the Voltage source by a short circuit and by opening the current source. Rth is not the parallel combination of R1 and R2 in all circuits. If you are finding an equivalent resistance between two terminals, try to travel from one end of the terminal and draw the resistance that you see in the path to the other end of the terminal. If R1 and R2 are parallel in the path between two terminals,then Rth is R1||R2. If R1 and R2 are in series, Rth is R1 + R2. FirstChildUserIdTAG: 823 FirstChildUserNameTAG: naveenatmit FirstChildCreateTimeTAG: 2012-09-28T06:49:11Z FirstChildTAG: this is very importend FirstChildUserIdTAG: 285450 FirstChildUserNameTAG: hussain2010 FirstChildCreateTimeTAG: 2012-09-28T12:46:51Z IndexTAG: 3855 TitleTAG: S6E1: A Nonlinear Element problem I used node analysis theorem to first find the current. I used the non linear resistor value 2k and got the correct answer for current "0.00079". But when I calculate voltages with the same value of resistor, I got "1.58", which is double of what given in answer. Can anyone help what is the real scenario? UserIdTAG: 378096 UserNameTAG: Jamshaid271 CreateTimeTAG: 2012-09-28T05:21:01Z VoteTAG: 0 CoursewareTAG: Week 3 / Analyzing Nonlinear Circuits Exercise CommentableIdTAG: 6002x_analyzing_nonlinear_circuits_exercise NumberOfReplyTAG: 1 FirstChildTAG: notice we do not know the voltage,current or resistance across the unknown device branch.even in nodal analysis you would need to know the current across this branch.you could assume the node potential as 'V' but you would still do not know the resistance across this branch.therefore assume R as 2k.solve for V.check in the graph if values of V and I are satisfied. FirstChildUserIdTAG: 357747 FirstChildUserNameTAG: kishores FirstChildCreateTimeTAG: 2012-09-28T13:03:18Z IndexTAG: 3856 TitleTAG: Value of e in Chapter 4, paragraph 1 What does e represent in the formula in paragraph 4.1? UserIdTAG: 231764 UserNameTAG: PaulP4881 CreateTimeTAG: 2012-09-28T05:19:55Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3857 TitleTAG: Sandbox: Simulating the circuit of H3P4. It doesn't work. Once I finished H3P4, I put the circuit in Sandbox, however It doesn't work correctly. I put a probe on the node in question (above the resistor with "v" across it), a probe above Vs and programmed Vs to be a triangular waveform of 1KHz (I'm assuming for Vs, the 8.5V peak voltage is the voltage from 0V to 8.5V, making the peak-to-peak voltage 17V). I simulate for 1ms (one period of Vs) and I see Vs vary from peak-to-peak correctly, but the node voltage, v stays at 2.504V. According to the homework, this node should at one point reach -3V. Has anybody else tried simulating this circuit, and if so any idea what is going on? UserIdTAG: 132828 UserNameTAG: rharris CreateTimeTAG: 2012-09-28T02:00:57Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: hii rharris I thought ur Vs waveform creating problem because according to you , u take peak2peak right but are u sure that is ur Vs waveform goes below the 0 level.if this not so , diode on-off disturbed as there condition depend on voltages at their terminals so u have to simulate vs waveform correctly i.e. upper & lower peak should be at 9.5 & -9.5. FirstChildUserIdTAG: 419552 FirstChildUserNameTAG: saurabhkrchauhan FirstChildCreateTimeTAG: 2012-09-29T17:37:27Z SecondChildTAG: HI Saurabhkrchauhan Yes, my waveform goes from -8.49 to 8.49, close to the 8.5 peak voltage required. How come your's is -9.5 to 9.5, which I see no problem of working? Oh, by the way, I haven't been able to solve and get the -3V either. I guessed and got lucky. With the positive terminal of the 3V independent voltage source connected directly to the node between the resisitors (the diode is shorted due to it being turned on), I don't understand how the node can deviate from 3V, either higher or lower. ?? - Thanks SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-09-29T21:34:13Z SecondChildTAG: Guess what the problem was? I had the polarities on the 3V source backwards. No wonder the voltage wouldn't go to -3V, the node was held at 3V. I started to explain the problem to someone and while looking at the 3V source I noticed the polarity change. Thanks SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-10-01T18:07:57Z IndexTAG: 3858 TitleTAG: Lab 3 question. Ok, I have watched, and re-watched the lectures over, and over. I cannot find a way to sum to that answer in a pure digital scheme. So how, or why does Ron, and W/L have anything to do with the timing of the waveform instead of just the voltage levels at the output? Where was this explained in the lectures or textbook? UserIdTAG: 349840 UserNameTAG: Wilk CreateTimeTAG: 2012-09-28T00:03:53Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Yes, in the textbook appears AND, NOT and OR gates with MOSFET (page 296 and more) FirstChildUserIdTAG: 459662 FirstChildUserNameTAG: FRANCISCOG FirstChildCreateTimeTAG: 2012-09-28T02:36:45Z SecondChildTAG: Thank You. SecondChildUserIdTAG: 283808 SecondChildUserNameTAG: Hanboushy SecondChildCreateTimeTAG: 2012-09-28T14:06:29Z IndexTAG: 3859 TitleTAG: Explanation Can somebody xplain me the question given in H3P3? UserIdTAG: 89170 UserNameTAG: frost CreateTimeTAG: 2012-09-27T22:59:01Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You would get a better response posting as H3P3 explanation, but here it goes; use the two curves and ohm's law to solve for the two situations presented. What values of voltage and current result in the desired resistance, then use iv=p, 2nd part basically working backwards start at peak power, what i, and v does that give, then ohms law. FirstChildUserIdTAG: 73809 FirstChildUserNameTAG: rhyssouth FirstChildCreateTimeTAG: 2012-09-28T01:36:12Z SecondChildTAG: Also, it would help if he posted this in the right topic instead of under General. SecondChildUserIdTAG: 323387 SecondChildUserNameTAG: ibrahimawwal SecondChildCreateTimeTAG: 2012-09-28T01:49:29Z SecondChildTAG: thnx rhyssouth!!!! it helped a lot SecondChildUserIdTAG: 89170 SecondChildUserNameTAG: frost SecondChildCreateTimeTAG: 2012-09-28T04:50:31Z SecondChildTAG: I will make it simple. 1). P=I^2*R. I from graph 1 ,I is about 0.001 ,thus Power can be calculated. 2) This has nothing to do with part 1 of this question.From the second graph of the question it is very much understood that the maximum power is at .3 Volts and is about .00027 W. Thus from The relation P=V^2/R R can be easily calculated. SecondChildUserIdTAG: 232499 SecondChildUserNameTAG: Asim09 SecondChildCreateTimeTAG: 2012-09-29T19:32:13Z SecondChildTAG: thank u rhyssouth. your post really helped SecondChildUserIdTAG: 369431 SecondChildUserNameTAG: pkaistha SecondChildCreateTimeTAG: 2012-09-30T20:54:59Z IndexTAG: 3860 TitleTAG: Please help Why is that, in the thevenin equivalent, the current source makes voltage only with the resistor in parallel, and not further resistors joined in series..... Please see S3E5 UserIdTAG: 169784 UserNameTAG: thewiredbear CreateTimeTAG: 2012-09-27T21:56:22Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: Please explain if your answer is going to be "ohm's Law" FirstChildUserIdTAG: 169784 FirstChildUserNameTAG: thewiredbear FirstChildCreateTimeTAG: 2012-09-27T21:57:00Z SecondChildTAG: In the future, please refrain from responding to yourself unless you are actually answering your own quesiton, and instead edit your thread to include messages like this. SecondChildUserIdTAG: 323387 SecondChildUserNameTAG: ibrahimawwal SecondChildCreateTimeTAG: 2012-09-28T01:48:56Z SecondChildTAG: Current will flow through a path/branch only if it can come back to the source again. I mean, there should be a closed path for the current to return. In this case, Rs resistances are open, current cannot flow through this path as the resistance of the open path is Infinity. SecondChildUserIdTAG: 823 SecondChildUserNameTAG: naveenatmit SecondChildCreateTimeTAG: 2012-09-28T07:11:20Z FirstChildTAG: because VI of any linear circuit is linear - so it can be substitute by any other linear circuit with same VI, and as soon to reproduce any linear VI you need only voltage or current source and one resistor - Thevenin and Norton are minimal linear circuits. FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-09-27T22:31:32Z FirstChildTAG: When finding the Thevenin equivalent, you want to find the open circuit voltage across the terminals. When the circuit is open, as in S3E5, no current can flow through the branch of the circuit that those other resistors are on, because that branch is open. It may help to consider an open circuit as a resistor with value $\infty$. In that case, if you analyze the circuit as a current divider circuit, current will choose the path of least resistance (heh) and all the current will flow through the path with resistance $R_P$ and none of it will flow through the open circuit. Intuitively current can't flow through an open circuit, but perhaps it may help you to consider the full analysis when intuition is not so intuitive. FirstChildUserIdTAG: 323387 FirstChildUserNameTAG: ibrahimawwal FirstChildCreateTimeTAG: 2012-09-28T01:48:15Z FirstChildTAG: Hi! Each element, including sources, have it's own equivalent resistance. The ideal V source has r = 0, the ideal I source has r = ∞ Now let's connect resistor R in parallel and in series to our sources. | source type | r | r parallel R | r sequental R | | voltage source | 0 | 0*R/(0+R) = 0 | 0 + R = R | | current source | ∞ | R * ∞ /(R+∞) -> R | ∞ + R -> ∞ | The parallel resistor does nothing to the V source (no matter, what we add in parallel to the wire, this area is still short circuit). The series resistor does nothing to the I source (no matter, what we add in series to the open circuit, it still open circuit) As you can see, the V-source allows only series connections, and I-source allows only parallel elements. ---------- And another approach. The Thevenin/Norton resistance represents energy losses in the circuit, if the load is not ideal. The V-source will loss voltage (obviously!), and the load will get only part of the maximum value. So we add a series resistor to emulate the voltage divider. The I-source will loss current, and we add a parallel resistor to get the current divider. FirstChildUserIdTAG: 261503 FirstChildUserNameTAG: EugenyL FirstChildCreateTimeTAG: 2012-09-28T06:33:36Z FirstChildTAG: There are no current sources in Thevenin equivalent FirstChildUserIdTAG: 157610 FirstChildUserNameTAG: mradziwo FirstChildCreateTimeTAG: 2012-09-28T15:09:49Z IndexTAG: 3861 TitleTAG: Methods to pick a threshold Vt I've been thinking on this issue for a several days, even because I do agree that with lower value than .576v will produce a "0" sing in the receiver part of the MOS. Thus, to address the first question any value below .576v would satisfy it. But not, the best way of thinking in Vt is including it on the state "1" (not in state "0"). Indeed, when the voltage achieves Vt, the gate will be open and Vs will be zero. Similar analysis can be done to address the second question but having in mind the forbidden zone. I'd extend some concerns to the fact that this is not a continuos function. Ones can assume threshold on one side, others may assume to the other. UserIdTAG: 139622 UserNameTAG: farasipe CreateTimeTAG: 2012-09-27T21:42:24Z VoteTAG: 0 CoursewareTAG: Week 3 / IV Model of a MOSFET exercise CommentableIdTAG: 6002x_IV_mosfet_model_exercise NumberOfReplyTAG: 0 IndexTAG: 3862 TitleTAG: Maximum current D2 H3P4 I have a question about H3P4 diode limiter. To get the maximun current through diode D2 (last question) I have an open circuit in D1 and find the Thevenin equivalent of the circuit, so I have Vs/2 in series with R/2 and the D2 branch (diode + voltage source). To get the current I use Ohm's Law and it doesnt work for D2, but it does for D1. What am I doing wrong? UserIdTAG: 255016 UserNameTAG: AngelCasaseca CreateTimeTAG: 2012-09-27T21:40:00Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hello! You are using the correct method. As far as I can see, the only possible error is the sign of the Thevenin voltage. FirstChildUserIdTAG: 261503 FirstChildUserNameTAG: EugenyL FirstChildCreateTimeTAG: 2012-09-27T23:44:23Z SecondChildTAG: Yeah, it was the sign of Vth. Thank you!!! SecondChildUserIdTAG: 255016 SecondChildUserNameTAG: AngelCasaseca SecondChildCreateTimeTAG: 2012-09-28T05:49:28Z SecondChildTAG: you could have used nodel analysis to solve this easily. try solving it through nodal analysis as well. :) it will be simple. u can solve by looking at the circuit rather than writing down the thevenin equivalent. SecondChildUserIdTAG: 823 SecondChildUserNameTAG: naveenatmit SecondChildCreateTimeTAG: 2012-09-28T07:05:19Z IndexTAG: 3863 TitleTAG: Trial and Error Method The hint states that you should use trial and error to solve this problem. I have a nice graphing calculator, but I am not attempting to try to solve with that. I am attempting to solve it by the trial and error method only. When I attempt to solve it, I end up getting 2 constant and unchanging values for $v$ and as such, my attempt at T&E doesn't anywhere. What am I supposed to do in order to actually find the fixed point? UserIdTAG: 248807 UserNameTAG: 1kingsman CreateTimeTAG: 2012-09-27T21:22:46Z VoteTAG: 0 CoursewareTAG: Week 3 / Graphical Analysis of Nonlinear Circuits Exercise CommentableIdTAG: 6002x_graphical_analysis_of_nonlinear_circuits_exercise NumberOfReplyTAG: 1 FirstChildTAG: Well, I get 2 values for $v$, and one of them was the correct answer. If I substituted it back into the equation, I would get the guess I started with. Strange? FirstChildUserIdTAG: 248807 FirstChildUserNameTAG: 1kingsman FirstChildCreateTimeTAG: 2012-09-27T21:40:52Z IndexTAG: 3864 TitleTAG: [HELP] S6E1 Q3 and Q4 Again, sorry for the poor English :) We can reduce the circuit using the Thevenin's method as seen on the S6E0 exercise, so: **Vth = (Vs * Rp)/(Rp + Rs)**; Vth = 3.18 **Rth = (Rs * Rp)/(Rs + Rp)**; Rth = 2987.60 **In = Vth / Rth**; In = 0.00106 As seen on the last video, the formula to find the Vx is: **Vx = V - R * Ix** and: **Ix = Vx / Rx**; so the formula is: **Vx = V - R * (Vx / Rx)**; To find the Vx value I replace the Rx with 1 ohm and 2 ohm (the two possible values for Rx) but both answers I got are wrong! Can someone help me? UserIdTAG: 310147 UserNameTAG: ildomarcarvalho CreateTimeTAG: 2012-09-27T19:59:21Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hello! You are doing all right. But check the graphics: current units are mili Amperes (not the Amperes!), and resistance Rx=1-2KOhm instead of 1-2Ohms. FirstChildUserIdTAG: 261503 FirstChildUserNameTAG: EugenyL FirstChildCreateTimeTAG: 2012-09-27T21:49:33Z SecondChildTAG: Oh right, that was a stupid error =) Now it works! Thank you very much. SecondChildUserIdTAG: 310147 SecondChildUserNameTAG: ildomarcarvalho SecondChildCreateTimeTAG: 2012-09-28T00:42:10Z IndexTAG: 3865 TitleTAG: large signal analysis of Attenuator tutorial 4 can anyone explain why and how the voltage at the node (in black ) is going negative and the graphs in pink and black (points where the currents enter the diode), their starting and ending point UserIdTAG: 168496 UserNameTAG: poweltalwar CreateTimeTAG: 2012-09-27T18:52:09Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3866 TitleTAG: how and why?! according to KVL, the sum of voltages around the loops are supposed to be zero, then why v1 and v2 (and v3, v4) are of the same signs? shouldn't they be 6V and -6V respectively? similarly, why are i1 and power to v1 negative, while those entering the resistor are positive? it is opposite in case of i3 and i4...please explain the how the current is flowing in both circuits. Thank you! UserIdTAG: 214221 UserNameTAG: UZMA1234 CreateTimeTAG: 2012-09-27T18:40:54Z VoteTAG: 0 CoursewareTAG: Week 1 / Associated reference directions CommentableIdTAG: 6002x_S2E2_Associated_Reference_Directions NumberOfReplyTAG: 2 FirstChildTAG: they are of the same sign so that as we go around the loop we get v1 - v2 = 0 FirstChildUserIdTAG: 168496 FirstChildUserNameTAG: poweltalwar FirstChildCreateTimeTAG: 2012-09-27T19:11:22Z FirstChildTAG: these are not of same sign we have to assume the sign convention for them as i take positive to negative +v(voltage) FirstChildUserIdTAG: 572704 FirstChildUserNameTAG: mitali1994 FirstChildCreateTimeTAG: 2012-10-21T04:27:26Z IndexTAG: 3867 TitleTAG: In the case when C=1. I know current flows through the path which offers less resistance to it but why does output get shorted to the ground? Also is the 0 at Vz in this case a logical zero? ( I don't think so since I see 0V written!). UserIdTAG: 254325 UserNameTAG: bondablack CreateTimeTAG: 2012-09-27T17:38:57Z VoteTAG: 0 CoursewareTAG: Week 3 / Switch Model CommentableIdTAG: 6002x_switch_model NumberOfReplyTAG: 0 IndexTAG: 3868 TitleTAG: H2P1 I am still not getting the answer I have read through all the discussion and tried n number of methods to solve but still i am not getting the answer right...when i chose a specific pair of resistors ( for eg 39 & 15) i get Vmax and Vmin right( 12.5 and 9.67) but the resistors shows wrong.And if a give a pair of resistors alone it always comes wrong ..i am really stuck and i cant get through this.. UserIdTAG: 294497 UserNameTAG: Saurabhkotian CreateTimeTAG: 2012-09-27T17:27:49Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: use different combination which gives output closest to the required. FirstChildUserIdTAG: 438395 FirstChildUserNameTAG: ali_PU1 FirstChildCreateTimeTAG: 2012-09-27T17:38:21Z SecondChildTAG: i have used a combination of 33 and 15 which is closest to the output of 12v but the weird part is that in that i am getting Vmax and Vmin as the same value of 12.5v. SecondChildUserIdTAG: 294497 SecondChildUserNameTAG: Saurabhkotian SecondChildCreateTimeTAG: 2012-09-27T17:54:44Z FirstChildTAG: solve it like a normal sum of resistors without the tolerances.find a relation between R1 and R2 using input and output voltage conditions that comes out to be R1=kR2.assume r2 as 15 .you wud get R2. FirstChildUserIdTAG: 357747 FirstChildUserNameTAG: kishores FirstChildCreateTimeTAG: 2012-09-27T17:56:13Z IndexTAG: 3869 TitleTAG: week 3 2nd question i have solve graphical question AL HUM DO LILAH .. now only one question remain which is 2nd question.tell me how i get values from graph for 2nd question.. guide me thanks :) i am confused in it :( UserIdTAG: 438395 UserNameTAG: ali_PU1 CreateTimeTAG: 2012-09-27T17:27:07Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: hey ali you want to calculate value ou R or what ?? FirstChildUserIdTAG: 331441 FirstChildUserNameTAG: abdessamade FirstChildCreateTimeTAG: 2012-09-27T18:57:02Z IndexTAG: 3870 TitleTAG: problem week 3 question of solar cell and 2nd question hello every one i am completely lost in week three solar cell and 2nd question kindly guide me thanks :) i cant understand how to calculate things form graph and also i can understand the meaning of NMh=NML ( i dnt remember what it is but i think you people understand which thing i want to point out) UserIdTAG: 438395 UserNameTAG: ali_PU1 CreateTimeTAG: 2012-09-27T15:04:11Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: NMH = noise margin high, NML = noise margin low.Please look in the text book at page 250.Regarding the graph, you will have to eyeball some values. FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-09-27T15:40:56Z SecondChildTAG: i have solve graphical question AL HUM DO LILAH .. now only one question remain which is 2nd question.tell me how i get values from graph for 2nd question.. guide me thanks :) SecondChildUserIdTAG: 438395 SecondChildUserNameTAG: ali_PU1 SecondChildCreateTimeTAG: 2012-09-27T15:52:27Z SecondChildTAG: you can eyeball the coordinates for v and p max, and you deduce i.Then you just use the other formula for power on a resistor i^2*R. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-09-27T19:07:05Z SecondChildTAG: Dear Ali, From the second graph of the question it is very much understood that the maximum power is at .3 Volts and is about .00027 W. Thus from The relation P=V^2/R R can be easily calculated. SecondChildUserIdTAG: 232499 SecondChildUserNameTAG: Asim09 SecondChildCreateTimeTAG: 2012-09-29T19:20:17Z IndexTAG: 3871 TitleTAG: doubts what is meant by independent KCL & KVL equations in a circuit? and how are 7 loops present in the circuit diagram given above? please explain. Thank you! UserIdTAG: 214221 UserNameTAG: UZMA1234 CreateTimeTAG: 2012-09-27T13:59:42Z VoteTAG: 0 CoursewareTAG: Week 1 / Circuit topology CommentableIdTAG: 6002x_S2E1_Circuit_Topology NumberOfReplyTAG: 1 FirstChildTAG: it means that the equations are only dependent on the internal system parameters (in our case resistors). that means that the equations are linearly independent (see linear algebra, professor Strang's book gives a very good idea of linear independency). As for the loops you to find the highest possible number of closed routes tha include circuit parameters including nodes resistors, capacitors and so forth.For a example if you go through the voltage source to node to node d to node b to node c and then back to the voltage source that is another loop.So is the loop than goes from node a to node d to node c to node b and then back to node a forming again a closed route. if you try all possible routes you end up with 7. However only three of them are fundamental loops and so they are also independent FirstChildUserIdTAG: 529743 FirstChildUserNameTAG: Anteo FirstChildCreateTimeTAG: 2012-10-02T07:03:17Z IndexTAG: 3872 TitleTAG: H3P1 hi everyone.. please tell me how to calculate width of forbidden region? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-27T13:57:41Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: hi go to textbook chapter five the digital abstraction FirstChildUserIdTAG: 331441 FirstChildUserNameTAG: abdessamade FirstChildCreateTimeTAG: 2012-09-27T14:39:50Z IndexTAG: 3873 TitleTAG: S9E3 has anybody got the same error as below? **Invalid input: Could not parse 'MY ANSWER' as a formula** UserIdTAG: 210954 UserNameTAG: Shahrouz CreateTimeTAG: 2012-09-27T13:28:09Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: try upper case/lower case :) FirstChildUserIdTAG: 357747 FirstChildUserNameTAG: kishores FirstChildCreateTimeTAG: 2012-09-27T15:56:30Z IndexTAG: 3874 TitleTAG: H4P3 - Part B – how to calculate RN? Hello everyone, Let’s suppose that in the exercise H4P3/B I already have the correct result for the current going through the dependent source (so i = A*u is already calculated) and also for the voltage dropping across the dependent source (let’s call it v). Is it OK if – for calculating RN – I consider the dependent source as a resistance, the value of which is RD = v/i (using the variables explained in the previous sentence)? If so, do I calculate RN as the paralell combination of the R1-R2 line and the R3-RD line, so as: RN = (R1+R2)*(R3+RD)/(R1+R2+R3+RD)? My problem is that I used this method to calculate RN but it seems to be giving an incorrect answer. If this assumption is not correct, can somebody explain the correct method for calculating RN? UserIdTAG: 129309 UserNameTAG: Memphis823 CreateTimeTAG: 2012-09-27T13:14:38Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I have used the same trick with the independent current source at the exit port and i didn't shut down anything. RN = RTH FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-09-27T15:29:02Z FirstChildTAG: Memphis823 I thought A*u was a voltage for the vcvs? Does i actually = A*u? If this is the case, no wonder I can't get the right answer. FirstChildUserIdTAG: 132828 FirstChildUserNameTAG: rharris FirstChildCreateTimeTAG: 2012-10-07T04:56:06Z IndexTAG: 3875 TitleTAG: Please explain current divider in H3P4 Hello. I have a trouble to understand a current dividers relations in H3P4. I find it with circuit simulator, and I've got right answers. I understand the relations between currents. But I cannot understand by what rule current divide between diod voltage source and right resistor. I would be very grateful if someone could explain it. Sorry if my english's not fine. UserIdTAG: 269368 UserNameTAG: StAlex CreateTimeTAG: 2012-09-27T12:51:41Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi StAlex. For a particular time when you know that the diode is on, replace it with a short circuit and then apply the circuit analysis techniques that we have learn so far such as node or loop method. You should be able to find currents easily that way, the trick is to get the right circuit model. FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-27T15:22:37Z SecondChildTAG: Oh, thank you! I understand, it's so simple. I was initially confused and thinking in the wrong direction. SecondChildUserIdTAG: 269368 SecondChildUserNameTAG: StAlex SecondChildCreateTimeTAG: 2012-09-27T16:56:52Z IndexTAG: 3876 TitleTAG: USING TOOLS Could you explain some progression to answer the five questions of the expression ? Measure current. The DC analysis annotates each voltage source with the current it is providing to the rest of the circuit. Please report the signed current I flowing from the source into the circuit, where the current is considered to be positive when it is flowing out of the positive terminal of the source, through the series of resistors, then back into the negative terminal. Next, report the total resistance R of the path connecting the positive and negative terminals of the source. Using the voltage value V of the source, verify that Ohm's law V=IR is obeyed. Current flowing from source into the circuit (amps, with correct sign): ※ 500e-6 Total resistance between pos and neg terminals (ohms): ※ 6000 Transient analysis. Change the waveform produced by the voltage source from a DC value of 3V to a sinusoid with a 1V amplitude, an offset voltage of 1V, and a frequency of 1kHz. Add scope probes to nodes A, B and C and edit their properties so that the plots will be different colors. Now run a transient analysis for 5ms. Move the mouse over the plot until the marker (a vertical dashed line that follows the mouse when it's over the plot) is at approximately 1.25ms. Please report the measured voltages for nodes A, B and C. Voltage for node A (volts): ※incorrect 1.995 Voltage for node B (volts): ※incorrect 1.662 Voltage for node C (volts): ※incorrect .665 UserIdTAG: 158876 UserNameTAG: SHINJI01 CreateTimeTAG: 2012-09-27T11:47:58Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I don't understand your question. It seems like you have all the correct answers. Are they checked as incorrect?? Make sure you are typing them correctly, that might be the problem. FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-27T15:06:37Z SecondChildTAG: Thank you vrey much for your answer. The answer I have gotten is from the edX's answer system. So I would like to know how do you guide into thiese answers. SecondChildUserIdTAG: 158876 SecondChildUserNameTAG: SHINJI01 SecondChildCreateTimeTAG: 2012-09-28T11:06:29Z IndexTAG: 3877 TitleTAG: Please explain for the answer. Please explain bellow answer's progressing ?I have five missing answers to understand. Measure current. The DC analysis annotates each voltage source with the current it is providing to the rest of the circuit. Please report the signed current I flowing from the source into the circuit, where the current is considered to be positive when it is flowing out of the positive terminal of the source, through the series of resistors, then back into the negative terminal. Next, report the total resistance R of the path connecting the positive and negative terminals of the source. Using the voltage value V of the source, verify that Ohm's law V=IR is obeyed. Current flowing from source into the circuit (amps, with correct sign): ※incorrect 500e-6 Total resistance between pos and neg terminals (ohms): ※incorrect 6000 Verify KVL. The four components in the schematic form a large loop -- let's verify that Kirchoff's Voltage Law holds by summing the voltage changes across the devices. Starting with node A, travel clockwise around the loop of components, entering the voltage change across each component. Choose the sign of the change for a component using the first terminal you come to in the clockwise traversal as the reference node. Verify that these four values sum to 0. ※Voltage for node A (volts): incorrect 1.995 ※Voltage for node B (volts): incorrect 1.662 Voltage for node C (volts): ※incorrect .665 UserIdTAG: 158876 UserNameTAG: SHINJI01 CreateTimeTAG: 2012-09-27T11:44:26Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3878 TitleTAG: Negative current In the first question i find the answer in negative value. VS/R+İD=0 what is wrong with my solution? EDIT: ok i got it UserIdTAG: 182653 UserNameTAG: monkeyfoahead CreateTimeTAG: 2012-09-27T09:33:20Z VoteTAG: 0 CoursewareTAG: Week 3 / Piecewise Linear Exercise CommentableIdTAG: 6002x_piecewise_linear_exercise NumberOfReplyTAG: 0 IndexTAG: 3879 TitleTAG: Control port current is zero Is it obvious or not, but there should be said that you mean $i_I=0$ (as it is explicitly said in the textbook page 333 figure 7.5). Otherwise you have to put $i_I$ into the node equation UserIdTAG: 190618 UserNameTAG: Kirbabaev CreateTimeTAG: 2012-09-27T08:56:44Z VoteTAG: 0 CoursewareTAG: Week 4 / Dependent Sources Example 1 CommentableIdTAG: 6002x_dep_src_ex_1 NumberOfReplyTAG: 0 IndexTAG: 3880 TitleTAG: Lab3 Help I've been trying to solve this for the last 3 days... finally yesterday I got what seemed to be the rigth answer, the output signal is the same as shown in the graph under the workspace, and by saying the same I mean that values change at the same points (I satisfy the conditions of the problem getting by 1's and 0's when I am supossed to get them), and the high value and low value are equal to the ones shown in the graph, but still I don't get the magic green tick... and I don't know what else I can do... UserIdTAG: 456610 UserNameTAG: Estebanterrero CreateTimeTAG: 2012-09-27T08:18:44Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: look at the textbook pg305 to 310 and adjust your W/l ratio according to what's written in the textbook :) You'll get your magic green tick FirstChildUserIdTAG: 206648 FirstChildUserNameTAG: TsvetanGeorgiev FirstChildCreateTimeTAG: 2012-09-27T10:56:17Z FirstChildTAG: Hi, This helped me alot, is part of the [textbook][1] [1]: http://www.elsevierdirect.com/companions/9781558607354/casestudies/06~Chapter_6/Section_6_11.pdf FirstChildUserIdTAG: 184827 FirstChildUserNameTAG: DiegoT FirstChildCreateTimeTAG: 2012-09-28T23:15:37Z IndexTAG: 3881 TitleTAG: Lab 4 How do we change the direction of the -> key as given in the example circuit. UserIdTAG: 99628 UserNameTAG: Pranjal16 CreateTimeTAG: 2012-09-27T08:05:41Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: With R FirstChildUserIdTAG: 244841 FirstChildUserNameTAG: eyubero FirstChildCreateTimeTAG: 2012-09-27T08:10:29Z FirstChildTAG: mark it as active component and press "R" key FirstChildUserIdTAG: 157610 FirstChildUserNameTAG: mradziwo FirstChildCreateTimeTAG: 2012-09-29T09:54:04Z IndexTAG: 3882 TitleTAG: H2P1 it's confusing...i've calculated everything right still Vmax and Vmin are the only quantities m getting right and the resistor pair with which i've calculated this(Vmax and Vmin)are wrong i've tried every other combination but still it's showing Vmax and Vmin with 'green tick' and any other resistor pair 'wrong'..actually the value of Vamx and Vmin are 'resistors pair dependent' then how it is showing right irrespective of any resistor pair... please help! UserIdTAG: 219442 UserNameTAG: aki31 CreateTimeTAG: 2012-09-27T05:43:05Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You can get Vmax and Vmin right with the wrong resistors. Remember the resistor need to satisfy two conditions: 1) Provide the proper output voltage. 2) Have the proper Thevenin equivalent resistance as seen from the output port. From previous posts, the most common problem was that Rth was calculated incorrectly, so make sure you are doing it right (hint: Rth is **NOT** R1+R2) I hope this helps! FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-27T14:53:22Z SecondChildTAG: thanx jelizon for you help.I know resistors are in parallel,the pair i've selected does satisfy the 2nd condition but still m getting it wrong would u please give any further explanation abt 1st condition.. SecondChildUserIdTAG: 219442 SecondChildUserNameTAG: aki31 SecondChildCreateTimeTAG: 2012-09-27T18:19:28Z SecondChildTAG: Hi Jelizon In the Text Book there is solved problem similar to the HW go and read it. SecondChildUserIdTAG: 272523 SecondChildUserNameTAG: Jivraj SecondChildCreateTimeTAG: 2012-09-27T20:19:51Z IndexTAG: 3883 TitleTAG: Typo/slip-of-the-tongue in "S6V5: Method 1 - Analytical Method" At 2:38 @1.0x speed, the speaker states that " you're not responsible for understanding and learning about sophisticated numerical techniques for analyzing long **linear** equations." The equation in question was defined to be a non-linear equation, so by definition, it cannot be linear. UserIdTAG: 248807 UserNameTAG: 1kingsman CreateTimeTAG: 2012-09-27T01:26:35Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 0 IndexTAG: 3884 TitleTAG: Bug in dictation for "S6V4: Method 1 - Analytical Method" At 1:34 @1.0x speed, the dictation says "with a **node** method"; however, it should read "with a **Norton** method". UserIdTAG: 248807 UserNameTAG: 1kingsman CreateTimeTAG: 2012-09-27T00:46:21Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 1 FirstChildTAG: unfortunately there are many bugs like that. It seems that typewriter was not following the course :p. Request to staff.. maybe it would be good idea to make "report bug" button somewhere next to video player. FirstChildUserIdTAG: 157610 FirstChildUserNameTAG: mradziwo FirstChildCreateTimeTAG: 2012-09-29T09:56:15Z IndexTAG: 3885 TitleTAG: Bug in dictation for "S6V1: Review of the Course So Far" At 1:07 at 1.0x speed, the dictation says, "put them into a **but** hopper," but it should be "put them into a **big** hopper". UserIdTAG: 248807 UserNameTAG: 1kingsman CreateTimeTAG: 2012-09-26T23:39:30Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 1 FirstChildTAG: In addition, the PPT slide at 3:59 has a really brief flash of red art on the lower-left corner of the slide. FirstChildUserIdTAG: 248807 FirstChildUserNameTAG: 1kingsman FirstChildCreateTimeTAG: 2012-09-26T23:51:12Z IndexTAG: 3886 TitleTAG: significance of W/L and how it effect the output wave form i have make (A+B)*C but not getting correct output because of w/L ...! i cant understand its significance . RDS on should be very less then 10k resistor for proper voltage levels.. when we change w/l what thing is change and how it effects the circuit kindly guide me thanks :) UserIdTAG: 438395 UserNameTAG: ali_PU1 CreateTimeTAG: 2012-09-26T19:42:11Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: read chapter 6.7 in your textbook. FirstChildUserIdTAG: 389333 FirstChildUserNameTAG: juergen FirstChildCreateTimeTAG: 2012-09-26T19:45:47Z FirstChildTAG: It's a divider for 26.5k Ron FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-09-26T19:46:52Z IndexTAG: 3887 TitleTAG: h1p1 I can't understand what are the V's of this problem. Could anyone explain to me the about theses voltages. UserIdTAG: 51667 UserNameTAG: Assuncao CreateTimeTAG: 2012-09-26T18:09:56Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Sorry, can't find in H1P1 any voltages. It's only about resistor combinations. FirstChildUserIdTAG: 131747 FirstChildUserNameTAG: hasi FirstChildCreateTimeTAG: 2012-09-26T19:36:53Z IndexTAG: 3888 TitleTAG: plz help... i don't know what's wrong..i can't watch any of the videos for the last couple of days..some of my friends have also taken the course..they are also facing the same problem...plz help me out..otherwise i'm afraid i won't be able to continue the course..i want to finish the course badly..plz help... UserIdTAG: 163407 UserNameTAG: shuvajit CreateTimeTAG: 2012-09-26T18:06:12Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: try using TOR browser...... you may have to install flash again, but then, everything works! FirstChildUserIdTAG: 169784 FirstChildUserNameTAG: thewiredbear FirstChildCreateTimeTAG: 2012-09-26T18:58:31Z FirstChildTAG: The edX Team announced "Due to the YouTube outage in certain parts of the world, we have provided direct links to download the lecture videos, and extended the Week 2 homework and lab deadlines to the coming Sunday (September 30). Please note that Week 3 deadlines have not changed." The link where you will find the lectures is this [https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/ FirstChildUserIdTAG: 304587 FirstChildUserNameTAG: Vasso FirstChildCreateTimeTAG: 2012-09-26T18:52:13Z IndexTAG: 3889 TitleTAG: Vibrational analysis and the electrical model Could we apply the electrical model to vibrational analysis? I guess that mass, inertia or momentum is analog to inductance. Damping is represented by resistors. And stiffness is equivalent to capacity. UserIdTAG: 155605 UserNameTAG: JaimeLopezCerezo CreateTimeTAG: 2012-09-26T17:48:01Z VoteTAG: 0 CoursewareTAG: Week 2 / Speakers CommentableIdTAG: 6002x_speakers_tutorial NumberOfReplyTAG: 1 FirstChildTAG: That's right, there is a huge analogy betwwen these two domains. But I am not sure of the analogy of each element. FirstChildUserIdTAG: 418182 FirstChildUserNameTAG: euldji2005 FirstChildCreateTimeTAG: 2012-09-27T00:06:52Z IndexTAG: 3890 TitleTAG: last part of the exercise Hi,i am a bit confused about the solution of the very last question. What is the value of vC (in Volts) at entry r?I can see that the answer is 0.0370.Can somebody explain to me how i get this number? UserIdTAG: 400832 UserNameTAG: panosbr CreateTimeTAG: 2012-09-26T17:17:32Z VoteTAG: 0 CoursewareTAG: Week 3 / Switch Resister Model Exercise CommentableIdTAG: 6002x_switch_resistor_model_exercise NumberOfReplyTAG: 2 FirstChildTAG: 1 Look you get both resistances of mosfet opened(working) 2 they are in parallel 3 find quivalent resistance 4 do the last part FirstChildUserIdTAG: 378150 FirstChildUserNameTAG: GladIDoThis FirstChildCreateTimeTAG: 2012-09-26T22:57:07Z FirstChildTAG: Hi! When digital MOSFET is in static mode (on or off), then it can be replaced with resistor. The value of resistor is given at beginning of page: $$ R_{ON}=50\Omega, R_{OFF}=10M\Omega$$ In your case both transistors are opened, and they are replaced with 50Ω resistors in parallel, with overall resistance of 25Ω. Now look at the circuit: it is voltage divider, with 2kΩ and 25Ω resistors. $$ V_R = 3V \times \frac{25}{25 + 2000} $$ FirstChildUserIdTAG: 261503 FirstChildUserNameTAG: EugenyL FirstChildCreateTimeTAG: 2012-09-26T22:58:52Z IndexTAG: 3891 TitleTAG: H3P3 SOLAR CELL Hi there!! I have a little problem to find the power drawn in Watts (W) from the solar cell, with R=100 Ohm, I dont know how to do it... I need some explanation, idea or something... Wait your answer, greetings!! UserIdTAG: 118022 UserNameTAG: E_Dorf CreateTimeTAG: 2012-09-26T15:52:56Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: you work with graph 1 and 2 FirstChildUserIdTAG: 331441 FirstChildUserNameTAG: abdessamade FirstChildCreateTimeTAG: 2012-09-26T16:04:40Z SecondChildTAG: Of course, but I dont know how to relate the plot of V vr I with R=100 SecondChildUserIdTAG: 118022 SecondChildUserNameTAG: E_Dorf SecondChildCreateTimeTAG: 2012-09-26T16:18:40Z SecondChildTAG: If we consider the next question to this part, we determine the resistance R from the relation v=ir. So we can safely imagine that 'v' and 'i' follow linear relation. But the graph shows otherwise. So for R=100, we cannot take a linear characteristics. Supposedly its an exponential relation like i= a*e^bv. Can anyone please tell me how to find resistance from such non linear expression? SecondChildUserIdTAG: 467386 SecondChildUserNameTAG: gp12saheb SecondChildCreateTimeTAG: 2012-09-29T10:11:43Z SecondChildTAG: I will make it simple. 1). P=I^2*R. I from graph 1 ,I is about 0.001 ,thus Power can be calculated. 2) This has nothing to do with part 1 of this question.From the second graph of the question it is very much understood that the maximum power is at .3 Volts and is about .00027 W. Thus from The relation P=V^2/R R can be easily calculated. SecondChildUserIdTAG: 232499 SecondChildUserNameTAG: Asim09 SecondChildCreateTimeTAG: 2012-09-29T19:28:28Z SecondChildTAG: Thanks Asim09 SecondChildUserIdTAG: 378096 SecondChildUserNameTAG: Jamshaid271 SecondChildCreateTimeTAG: 2012-09-30T07:53:21Z IndexTAG: 3892 TitleTAG: one doubt in H3P4: DIODE LIMITER. Hello friends, I have one doubt in H3P4: DIODE LIMITER. **What is the maximum positive voltage (in Volts) that can appear at v?** In above question "v" means which node it is??????? Please answer to my doubt. Waiting for your reply...... Thank you, UserIdTAG: 106219 UserNameTAG: veereshpatil CreateTimeTAG: 2012-09-26T15:36:31Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: you try one of the voltage that you have FirstChildUserIdTAG: 331441 FirstChildUserNameTAG: abdessamade FirstChildCreateTimeTAG: 2012-09-26T15:38:34Z FirstChildTAG: "v" is the voltage across the rightmost resistor. FirstChildUserIdTAG: 378267 FirstChildUserNameTAG: artfwo FirstChildCreateTimeTAG: 2012-09-27T02:21:38Z IndexTAG: 3893 TitleTAG: Download of videos May i know how could i download week1 ,week4 lectures and also week 2 course non linear part. It would be a pleasure knowing this. UserIdTAG: 359968 UserNameTAG: Anirudh796 CreateTimeTAG: 2012-09-26T15:32:59Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Nonlinear part is coming up right now FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-26T15:41:21Z SecondChildTAG: My pleasure. SecondChildUserIdTAG: 359968 SecondChildUserNameTAG: Anirudh796 SecondChildCreateTimeTAG: 2012-09-26T16:10:19Z FirstChildTAG: You can download the videos here: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/ Or use the playlists I've made here, for online viewing of the videos: Week 2 https://dl.dropbox.com/u/24096724/6002xMP4/S3.html https://dl.dropbox.com/u/24096724/6002xMP4/S4.html Week 3 https://dl.dropbox.com/u/24096724/6002xMP4/S5.html FirstChildUserIdTAG: 276409 FirstChildUserNameTAG: IgnacioUY FirstChildCreateTimeTAG: 2012-09-26T15:46:00Z SecondChildTAG: but what about week 1 and week 4 week 5 and all.Only online viewing is available. whrer could i download those.Plz help me. SecondChildUserIdTAG: 359968 SecondChildUserNameTAG: Anirudh796 SecondChildCreateTimeTAG: 2012-09-26T16:11:13Z SecondChildTAG: Week 1 will be ready in about an hour. You'll be able to access it here: https://dl.dropbox.com/u/24096724/6002xMP4/S1.html SecondChildUserIdTAG: 276409 SecondChildUserNameTAG: IgnacioUY SecondChildCreateTimeTAG: 2012-09-26T18:51:57Z SecondChildTAG: Thankk you.My pleasure. SecondChildUserIdTAG: 359968 SecondChildUserNameTAG: Anirudh796 SecondChildCreateTimeTAG: 2012-09-27T03:17:30Z IndexTAG: 3894 TitleTAG: H5P2 Bug The formula checker for iDS does not seem to work correctly. I have all the correct numerical answers but it says the formula is wrong. UserIdTAG: 397595 UserNameTAG: mholin CreateTimeTAG: 2012-09-26T14:10:46Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: then it is wrong.. check it again FirstChildUserIdTAG: 342221 FirstChildUserNameTAG: deepkar FirstChildCreateTimeTAG: 2012-09-26T16:56:49Z FirstChildTAG: Be sure not to include the "ids =" in the formula FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-27T14:39:29Z SecondChildTAG: Here is my answer: (sqrt(2*K*RS*(vIN-VT)+1)-K*RS*(vIN-VT)-1)/(K*RS^2) It evaluates the correct numerical values but the equation is still marked with an x. I believe this is a genuine bug. Thanks, Mark SecondChildUserIdTAG: 397595 SecondChildUserNameTAG: mholin SecondChildCreateTimeTAG: 2012-09-28T14:30:00Z SecondChildTAG: Don't post your answer(s) please... (and check your signs) SecondChildUserIdTAG: 396446 SecondChildUserNameTAG: RousseauxS SecondChildCreateTimeTAG: 2012-09-29T19:39:05Z FirstChildTAG: My all other answers are correct using the same solution of quadratic equation but this third part is again showing wrong only. Can anyone help? FirstChildUserIdTAG: 365309 FirstChildUserNameTAG: chetnasinghaldas FirstChildCreateTimeTAG: 2012-10-08T15:11:10Z IndexTAG: 3895 TitleTAG: H4P3 Hello there. I'm trying the final task of Homework 4, the Norton equivalent network for a circuit with a voltage dependent voltage source. By applying the node method, I got to a point where I don't know what to do. Anyone could push me in the right direction? (So far I've applied the node method to node U and the node between R2 and R3, which leaves me with a single equation with 2 unknowns...) Thanks. UserIdTAG: 138981 UserNameTAG: Pr0bability CreateTimeTAG: 2012-09-26T14:03:55Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Applying Norton, the node between R2 and R3 gives you an equation with 2 unknowns (u and IN). In order to solve u, you should solve the node between R1 and R2. If it helps you, I put the ground in the node under the dependent voltage source. Good luck FirstChildUserIdTAG: 314117 FirstChildUserNameTAG: thuargon FirstChildCreateTimeTAG: 2012-09-26T14:19:37Z FirstChildTAG: Isn't Rth=(R1+R2)*R3/(R1+R2+R3)? But it doesn't work(( FirstChildUserIdTAG: 349662 FirstChildUserNameTAG: gaaral FirstChildCreateTimeTAG: 2012-09-26T21:06:34Z SecondChildTAG: Apply an arbitrary voltage to the dead (V0 shorted) network at port B then find u and A*u in terms of the arbitrary voltage. You could then find the current through R3 and the VCVS in terms of the arbitrary voltage and determine the effective resistance of R3 in series with the VCVS, but the answer is rather obvious at this point (think voltage divider). SecondChildUserIdTAG: 141000 SecondChildUserNameTAG: OrinE SecondChildCreateTimeTAG: 2012-09-27T00:26:43Z SecondChildTAG: I got the answer by calculating Voc, then using your method to obtain Rth, then substituting to find Isc or the Norton current (Rth = Voc/Isc). But the question seems to suggest that we can calculate Isc first independently before we calculate Rth. How does that work? I shorted the circuit, then added the current flowing from the independent source and the dependent source across the short, but it seems to be incorrect. =[ help please SecondChildUserIdTAG: 132521 SecondChildUserNameTAG: jimmynyge SecondChildCreateTimeTAG: 2012-09-30T14:23:13Z SecondChildTAG: how did you get V0c? SecondChildUserIdTAG: 278792 SecondChildUserNameTAG: sali SecondChildCreateTimeTAG: 2012-10-07T04:31:13Z IndexTAG: 3896 TitleTAG: Week 5 MOSFET Amplifier Experiment Does not work with Win7/IE9. What is the recommended browser for Win7? Thanks. UserIdTAG: 397595 UserNameTAG: mholin CreateTimeTAG: 2012-09-26T10:21:47Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Google Chrome works fine for me. FirstChildUserIdTAG: 157273 FirstChildUserNameTAG: ongchihang FirstChildCreateTimeTAG: 2012-09-26T13:43:13Z IndexTAG: 3897 TitleTAG: Lab 3 Hello I am working on Lab 3, and i have gotten the correct transient response with RON less than 0.606060K and W/L = 43.725.And also my Vol is less that o.25 Volts. But the checker still considers it wrong. I do not understand what is going wrong. Please answer. UserIdTAG: 423976 UserNameTAG: Sirbrevis CreateTimeTAG: 2012-09-26T08:37:46Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: check the glitches, they must be smaller than Vol FirstChildUserIdTAG: 389333 FirstChildUserNameTAG: juergen FirstChildCreateTimeTAG: 2012-09-26T16:12:25Z SecondChildTAG: how many the pullup resistor you are used ? SecondChildUserIdTAG: 331441 SecondChildUserNameTAG: abdessamade SecondChildCreateTimeTAG: 2012-09-26T18:19:28Z SecondChildTAG: i have make (A+B)*C but not getting correct output because of w/L ...! i cant understand its significance . RDS on should be very less then 10k resistor for proper voltage levels.. when we change w/l what thing is change and how it effects the circuit kindly guide me thanks :) SecondChildUserIdTAG: 438395 SecondChildUserNameTAG: ali_PU1 SecondChildCreateTimeTAG: 2012-09-26T19:45:23Z IndexTAG: 3898 TitleTAG: A simple and precise definition for abstraction? . UserIdTAG: 1580 UserNameTAG: user CreateTimeTAG: 2012-09-26T04:36:49Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3899 TitleTAG: H5P1 ZERO-OFFSET AMPLIFIER hi, I find the bug in the ask 3, because my ans is true but don't show, help me please ( I don't speak english fine) UserIdTAG: 58618 UserNameTAG: ingeniero13 CreateTimeTAG: 2012-09-26T02:46:38Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Seems fine to me. I received a check mark. I does require a bit more computation than the first two parts of the problem. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-09-26T02:56:39Z IndexTAG: 3900 TitleTAG: H4P3 THE NORTON EQUIVALENT RESISTANCE AND CURRENT ARE REQUIRED. The resistance should be easy to calculate (equals to Thevenin resistance). However i get wrong answer. Any suggestions? UserIdTAG: 206669 UserNameTAG: dimitrios66 CreateTimeTAG: 2012-09-25T23:57:50Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Connect a current source to the output and short out the independent voltage source. Calculate Vout and divide by I. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-09-26T02:05:07Z SecondChildTAG: correct @skyhawk. u'll get the correct resistance by this method. actually if a circuit contains only dependent sources then one has to assume a voltage or current source of magnitude 1 unit at the terminal and then calculate v/i to get resistance. however how will we calculate the norton current? can we find out the thevenin voltage and divide it by norton resistance to get norton current? or is there any other simple way? SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-09-26T06:43:29Z SecondChildTAG: Short the output and see how much current the circuit delivers. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-09-26T09:53:06Z IndexTAG: 3901 TitleTAG: HW4P2 Issues/Comments I had a hard time with HW4P2 - specifically the Zener diode section. Previously I rounded many of my intermediate calcs to about 2 decimal places and didn't have a problem, but here one of the answers required a very small number, that I called it zero, which of course didn't pass the check - could EdX post some guidance on how accurate we need to be with our answers? If it's a percentage and we're dealing with very small numbers it becomes a bigger issue (no pun intended!). Also, the last part of the question is very vague - "What is the minimum value of RL, in Ohms, that guarantees that the circuit will operate this way?" The "this way" is quite open. Without giving away the answer, I won't say too much, but my initial approach was to check against the static discipline output voltage level requirements, but that failed - could this question be a little more clarified for everyone, I spent way too much time trying to figure out what value the answer was based on rather than understanding what was going on in the circuit. Thanks for EdX staff's hard work! UserIdTAG: 323230 UserNameTAG: Xango5346 CreateTimeTAG: 2012-09-25T23:31:32Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: "This way" means with the Zener diode functioning to reduce the noise as in the previous part of the problem. I took that approach and got a check mark. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-09-26T02:21:19Z IndexTAG: 3902 TitleTAG: Problem browsers Hi, the problem with videos of classes only happens with google chrome or with all browsers, await return. Thank you. UserIdTAG: 375517 UserNameTAG: CristianeTeburcio CreateTimeTAG: 2012-09-25T21:25:52Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: The latest versions of Google Chrome and Firefox appear to work well. I'd recommend them. I've only tested them on my Mac, however. More or less, if you can use Youtube with your browser, you should be okay. FirstChildUserIdTAG: 376895 FirstChildUserNameTAG: alecgg FirstChildCreateTimeTAG: 2012-09-25T22:48:43Z IndexTAG: 3903 TitleTAG: MITX 6002x Hi there. I was following this very same course this spring. Unfortunately I had to stop halfway through (~week 8). My question is now, if it was possible to catch up on this course running now, and jump in at week 8? Just discovered right now that it was going again. Is this possible, or do I have to start all over again (Which is too late now, since week 5 already). Regards Jesper UserIdTAG: 85799 UserNameTAG: jesperhn CreateTimeTAG: 2012-09-25T20:50:38Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Where can i access the assignments, Lab and Homework? I need to catch up in Uganda we had a problem of Internet problem. FirstChildUserIdTAG: 239507 FirstChildUserNameTAG: murkennethie FirstChildCreateTimeTAG: 2012-09-26T07:29:41Z FirstChildTAG: Hi Jesper, Sorry, I don't think it's possible to jump in at week 8. But there's hope! The due dates for Lab/HW 2 & 3 are September 30th, so you do have time to complete them if you wanted to take the course again. You would miss Week 1 assignments, however. Edit: Your lowest two HWs and your lowest two Labs are also dropped, so you could still get 100%. And week 1's homework and lab are only ~3% of your grade. FirstChildUserIdTAG: 376895 FirstChildUserNameTAG: alecgg FirstChildCreateTimeTAG: 2012-09-25T22:45:46Z IndexTAG: 3904 TitleTAG: i registered for this course today!! what to do??plz help Can i still submit the assignments of first 3 weeks? Can i still get a certificate? UserIdTAG: 489065 UserNameTAG: amiths CreateTimeTAG: 2012-09-25T19:38:33Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Perhaps you cant submit the assignments, but as for learning you can always follow the lectures and solve the problems and the lab. you can perhaps still get an honor certificate if you do well in the comings weeks. and yes you have only missed 1st week assignment yet, you can submit the 2nd and 3rd weeks assignment upto 30th sep. it is still not very late! FirstChildUserIdTAG: 314624 FirstChildUserNameTAG: Owais001 FirstChildCreateTimeTAG: 2012-09-25T19:52:50Z IndexTAG: 3905 TitleTAG: got some wrong answers !! i had some problem with calculation in it !! don't know.. applied correctly and solved but still got some wrong answers... !! is there anything we should learn more than this to solve this ??? but had such fun in solving it !! UserIdTAG: 89535 UserNameTAG: Rudhra CreateTimeTAG: 2012-09-25T18:08:44Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: Kindly specify the exact problems where you are having a problem. Not everyone is "on the same page." You must realize that since this is an online course, some students have just signed up and have already missed the Homework Week 1 deadline (but they still can complete the course without penalty as the grading policy 'throws out' your two lowest Homework scores) and some of the more advanced students are working on Homework Week 5! (The calender shows that we are in Week 3 if you are new to the course.) FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-09-26T04:45:01Z IndexTAG: 3906 TitleTAG: uddan its was fun doing it.Got some answers wrong.. UserIdTAG: 492993 UserNameTAG: uddan CreateTimeTAG: 2012-09-25T17:18:17Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 0 IndexTAG: 3907 TitleTAG: how can i get i =0? I'm confused. Is it possible that i can get a corrent of 0? How? because if i have a voltage source there is a corrent from this source. or ther is not? UserIdTAG: 309952 UserNameTAG: alexjose CreateTimeTAG: 2012-09-25T17:12:14Z VoteTAG: 0 CoursewareTAG: Week 3 / MOSFET CommentableIdTAG: 6002x_mosfet_model NumberOfReplyTAG: 2 FirstChildTAG: An ideal voltage source can supply any current necessary to maintain the voltage difference at its terminals. So, yes! It's possible to have a current of 0. Also, think about an open circuit. It can have a voltage difference across it and not allow any current to flow between its terminals. FirstChildUserIdTAG: 376895 FirstChildUserNameTAG: alecgg FirstChildCreateTimeTAG: 2012-09-25T19:38:11Z SecondChildTAG: Yes, you are right =) SecondChildUserIdTAG: 309952 SecondChildUserNameTAG: alexjose SecondChildCreateTimeTAG: 2012-09-25T20:40:15Z FirstChildTAG: it is possible, I=0 mean there is no connection on that line or it is open circuit (actually in off state, the drain to source resistant in MOSFET as switch is high enough so that the current are near zero), MOSFET as switch operate like a device called [Relay][1] (an electromagnet operated switch). commonly relay has 4 terminal, 2 of which are forming switch circuit and the other 2 are the electromagnetic circuit (consists of a coil of wire wrapped around an iron core). this relay device is design in such a way so that if we put voltage across the electromagnetic circuit terminal then it would create magnetic force that can pull the switch so it change state from the open state to close state, so the Drain Source circuit in Mosfet is like a switch circuit in relay that controlled by the relay electromagnetic circuit which is like a gate source circuit in Mosfet, and to be noted this analogy only valid for Mosfet as swich. and of course actual MOSFET didn't work base on electromagnetic but base on electric field and Relay is just a switch but MOSFET is more than a switch CMIIW [1]: http://en.wikipedia.org/wiki/Relay FirstChildUserIdTAG: 452559 FirstChildUserNameTAG: kuz1toro FirstChildCreateTimeTAG: 2012-09-25T19:57:52Z SecondChildTAG: I know the relay. I Understand de way of MOSFET function. My question was because acording to the V-I graphics Ids =0, but Vds is variable. or not? .. . ahhh!!!.. i understand. Ids is 0 when the resistance doesnt exist!. Its obviously. jejeje. But in a wire, if we suppose that doesn't have a resistance, the voltage is 0 ? amh... V=IR, if R=0, then V=0. Ohhh!!. You are a genius!. Thank you =). Then if R = oo, I=V/R = 0. Wau! SecondChildUserIdTAG: 309952 SecondChildUserNameTAG: alexjose SecondChildCreateTimeTAG: 2012-09-25T20:39:50Z IndexTAG: 3908 TitleTAG: S3E1 Attention Staff: Can you please confirm this question is correct? If you build this cct in the sandbox so that V2 outputs -7.2V into the cct (which appears to be the intent of the question), you will not get 6.2. However if you configure V2 so that it outputs -7.2v from its positive terminal, you will get 6.2. In practice, I have never seen a voltage source put out a - value from its + terminal. Staff: can you please clarify this question? UserIdTAG: 217971 UserNameTAG: Alejandro01 CreateTimeTAG: 2012-09-25T16:36:51Z VoteTAG: 0 CoursewareTAG: Week 2 / Node equation review CommentableIdTAG: 6002x_S3E1_NodeEquationReview NumberOfReplyTAG: 2 FirstChildTAG: The positive terminal (V+) and negative terminal (V-) of a voltage source and the voltage source's value (Vs) are related like this: V+ - V- = Vs. So if the positive terminal is grounded (V+ = 0) and the voltage source's value is -7.2V (V=-7.2V), then the voltage at the negative terminal (V-) is 7.2V. FirstChildUserIdTAG: 376895 FirstChildUserNameTAG: alecgg FirstChildCreateTimeTAG: 2012-09-25T17:03:10Z SecondChildTAG: alecgg you are right, this should be the approch to solve this kinda questions. But how does positive voltage , to a negative terminal make sense? SecondChildUserIdTAG: 226838 SecondChildUserNameTAG: SaSSer SecondChildCreateTimeTAG: 2012-09-28T04:06:33Z FirstChildTAG: Alecgg, Thanks for your prompt response. What you have described above is confusing and does not refelct a practical circuit. A DC voltage source such as a battery or even a lab power supply does not behave like you are describing in your response. If you placed a battery in a circuit as per the question and per your description, you would not be getting +7.2V from its negative terminal, or vice versa. I have two Lab power supplies. Both have have a '+' terminal, a 'COM' (-) terminal and a 'Ground' terminal. One of these has a -12v terminal and a +5V terminal. None of these outputs an inverse voltage. The diagram in the question has the source inverted. As it is drawn and defined, it is explicitly communicating an output of -7.2V from its '-' terminal. Is it not appropriate for the questions to reflect that which could be experienced in practice?, perhaps it would have been better understood with a signal generator outputting a -V? This could have been better understood and still provide the same educative effect? Can you please also confirm the intent of such a question. It seems to be un-necessarily confusing? FirstChildUserIdTAG: 217971 FirstChildUserNameTAG: Alejandro01 FirstChildCreateTimeTAG: 2012-09-26T13:12:48Z IndexTAG: 3909 TitleTAG: Too behind I have only finished first lecture of week three till now.Am I much behind others? UserIdTAG: 99628 UserNameTAG: Pranjal16 CreateTimeTAG: 2012-09-25T16:02:55Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Same here, lots of work, but don't get discouraged. We can do it! FirstChildUserIdTAG: 88556 FirstChildUserNameTAG: Luxedrina FirstChildCreateTimeTAG: 2012-09-25T16:21:29Z SecondChildTAG: Get it! If you all work hard, you can definitely do it :) SecondChildUserIdTAG: 376895 SecondChildUserNameTAG: alecgg SecondChildCreateTimeTAG: 2012-09-25T17:05:20Z IndexTAG: 3910 TitleTAG: LAB 3 Troubleshoot I am not getting the save option for Lab 3 , i am only getting the check option . My solution is correct but how do i save it . HELP ME UserIdTAG: 344720 UserNameTAG: Vaibhav_Kashyap CreateTimeTAG: 2012-09-25T15:34:54Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Troubleshooting NumberOfReplyTAG: 1 FirstChildTAG: I got it . i figured out my mistake . FirstChildUserIdTAG: 344720 FirstChildUserNameTAG: Vaibhav_Kashyap FirstChildCreateTimeTAG: 2012-09-25T15:38:54Z SecondChildTAG: help me for Lab 3 please SecondChildUserIdTAG: 331441 SecondChildUserNameTAG: abdessamade SecondChildCreateTimeTAG: 2012-09-25T16:55:08Z SecondChildTAG: abdessamade: can you maybe please be a bit more specific? SecondChildUserIdTAG: 131747 SecondChildUserNameTAG: hasi SecondChildCreateTimeTAG: 2012-09-25T17:07:46Z SecondChildTAG: I dont understand very well the problem set help please and thank you SecondChildUserIdTAG: 331441 SecondChildUserNameTAG: abdessamade SecondChildCreateTimeTAG: 2012-09-25T19:05:23Z IndexTAG: 3911 TitleTAG: Textbook Is there anyway to download the textbook that has been put up??.. It would be a great pleasure if that's possible...:) UserIdTAG: 359968 UserNameTAG: Anirudh796 CreateTimeTAG: 2012-09-25T15:27:28Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You can see the parts of the textbook you'll need in the Textbook tab. You could buy an eBook version. We have a discount for it at this site (you have to enter a code): http://store.elsevier.com/product.jsp?locale=en_US&isbn=9781558607354 or perhaps you could find it cheaper elsewhere. FirstChildUserIdTAG: 376895 FirstChildUserNameTAG: alecgg FirstChildCreateTimeTAG: 2012-09-25T19:40:50Z SecondChildTAG: hey ok thanks for your response. pleasure. SecondChildUserIdTAG: 359968 SecondChildUserNameTAG: Anirudh796 SecondChildCreateTimeTAG: 2012-09-26T08:39:41Z SecondChildTAG: Dear Anirudh Sir , If you are in India it available in Local book store i got it From Bangalore swapna book stall just below 500 INR SecondChildUserIdTAG: 156694 SecondChildUserNameTAG: mkprasanth SecondChildCreateTimeTAG: 2012-09-26T16:49:45Z IndexTAG: 3912 TitleTAG: At 4:16 At that time in the lecture u have mention that , A valid input is one that adheres to the VIL, VIH constraints. but that's wrong.. it has to be above Voh and Vol UserIdTAG: 296303 UserNameTAG: DEEPatXUniv CreateTimeTAG: 2012-09-25T10:31:05Z VoteTAG: 0 CoursewareTAG: Week 2 / Static Discipline CommentableIdTAG: 6002x_static_discipline NumberOfReplyTAG: 1 FirstChildTAG: If you are *SENDING* a signal *‘out’*: it is called **Output**. If you are *RECEIVING* a signal: that is the **Input**. There are stronger constraints on output than on input. FirstChildUserIdTAG: 714237 FirstChildUserNameTAG: PaxPolaris FirstChildCreateTimeTAG: 2012-10-26T18:25:10Z IndexTAG: 3913 TitleTAG: S9V16 There is a typo in the formula for iDS on 8th minute of S9V16 video sequence. There should be minus sign in 'iDS=Vs/RL+V0/RL' instead of '+'. UserIdTAG: 143593 UserNameTAG: Tsybulkin CreateTimeTAG: 2012-09-25T08:20:37Z VoteTAG: 0 CoursewareTAG: Week 5 / Large Signal Analysis of MOSFET Amplifier CommentableIdTAG: 6002x_large_sig_MOSFET_amp_t NumberOfReplyTAG: 0 IndexTAG: 3914 TitleTAG: logic with switches S5E1 FROM CIRCUIT "C" YOU CAN WRIGHT Z=((X(bar)+Y)(bar))(bar) UserIdTAG: 370257 UserNameTAG: hassanano CreateTimeTAG: 2012-09-25T02:11:27Z VoteTAG: 0 CoursewareTAG: Week 3 / Switch Model Exercise CommentableIdTAG: 6002x_switch_model_exercise NumberOfReplyTAG: 0 IndexTAG: 3915 TitleTAG: Assistance on HP31 NOR net resistance As far as I understand, we can use conventional Linear Methods for the internal MOSFET circuitry. If true, then the net resistance for the NOR circuit as stated in H3P1 has A and B in parallel with each other. That means the net resistance is (Ron*Ron)/(Ron+Ron) = Ron/2, right (which I'm guessing is where I'm wrong)? Now if so, using this I tried to deduce for Rpuo and I'm getting it wrong. I did get the right resistances for the NOT and AND given, using the exact same methodology, I aint getting it with NOR though. UserIdTAG: 204745 UserNameTAG: allwynmendes CreateTimeTAG: 2012-09-25T00:47:43Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: hello allwynmendes,In case of NOR gate the two switches are in parallel so Vout would be same for A as for the B. using your mentioned formula you will easily evaluate the R for NOR gate.. FirstChildUserIdTAG: 282892 FirstChildUserNameTAG: wiky FirstChildCreateTimeTAG: 2012-09-25T01:31:53Z SecondChildTAG: With your advice I could answer correctly the question, but still doesn't make any sense to me... On my point of view, allwynmendes method is correct SecondChildUserIdTAG: 310147 SecondChildUserNameTAG: ildomarcarvalho SecondChildCreateTimeTAG: 2012-09-25T03:20:26Z SecondChildTAG: forget it, now I see why should be your method :D SecondChildUserIdTAG: 310147 SecondChildUserNameTAG: ildomarcarvalho SecondChildCreateTimeTAG: 2012-09-25T13:22:07Z FirstChildTAG: You have to calculate for the worst case condition. For parallel part: If one fet is open and the other closed, then there is only one Ron in series with Rpuo. What gives a higher Vout: Ron or Ron/2? FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-09-25T09:55:20Z SecondChildTAG: appreciated reply SecondChildUserIdTAG: 419552 SecondChildUserNameTAG: saurabhkrchauhan SecondChildCreateTimeTAG: 2012-09-26T19:07:48Z IndexTAG: 3916 TitleTAG: Following topics What good does following topics do with the new forum system? There is no easy way to see topics that ive posted in, and/or see what topics I've followed, so what purpose does it serve? UserIdTAG: 248807 UserNameTAG: 1kingsman CreateTimeTAG: 2012-09-25T00:17:00Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 1 FirstChildTAG: Found my answer. In the dropdown menu in the top left sidebar, you can select "Following" to view what topics you have followed. FirstChildUserIdTAG: 248807 FirstChildUserNameTAG: 1kingsman FirstChildCreateTimeTAG: 2012-09-25T00:22:27Z SecondChildTAG: If you want to see topics you've posted on but haven't necessarily followed you can also click on your hyperlinked name! SecondChildUserIdTAG: 248902 SecondChildUserNameTAG: Chanute SecondChildCreateTimeTAG: 2012-09-25T00:25:23Z SecondChildTAG: That is true. I'll have to make a suggestion, however, that the there will be a "profile page" or some similar function, in order to allow me to look at my posts. As for now, I have chosen to bookmark my page, so I just input my username in the address bar of my browser, and go to the link that appears (I use Firefox). SecondChildUserIdTAG: 248807 SecondChildUserNameTAG: 1kingsman SecondChildCreateTimeTAG: 2012-09-25T01:30:33Z IndexTAG: 3917 TitleTAG: Mosfet gate internal inherently iverting The various combinational gate designs we are shown are all inherently inverting, which lead us from the basic inverter,to the NAND and NOR circuits. Fair enough. I'm just wondering if there is some other beneficial reason for using this pattern. It would appear to me, just as easy to take the output from the source terminal, and attach RL to the ground. i.e. make RL a pull down resistor instead of a pull up. The is would give allow a more intuitive AND and OR gate configurations, but obviously we would still have to use the original inverter design. Is there any reason for preferring to use RL as pull up? Is there power saving this way? By inverting the o/p of either the NAND or NOR gates, you end up using an extra switch (for the inversion) which wouldn't be necessary if you used the pull down config. Just curious. Thanks. M. UserIdTAG: 385677 UserNameTAG: Michaelc1 CreateTimeTAG: 2012-09-24T23:13:33Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: The inverters inject energy into circuit, right from the V+ source, and they are used as buffers. You can connect 1000 of the AND-NOT devices, and this chain will work. The pull-down elements will drain the power, and at some moment it will be not enough to handle the next Gate input. FirstChildUserIdTAG: 261503 FirstChildUserNameTAG: EugenyL FirstChildCreateTimeTAG: 2012-09-24T23:24:36Z SecondChildTAG: Thanks! I think see what you mean, although I would think that using the normal inverting pattern (with a pull up RL) would still drain power from the circuit is a similar way when the mosfet was ON, and eventually lead to the same problem - but I'll assume not a quickly.. SecondChildUserIdTAG: 385677 SecondChildUserNameTAG: Michaelc1 SecondChildCreateTimeTAG: 2012-09-24T23:37:50Z FirstChildTAG: [Not 6.002x staff, but edx developer with some background in digital circuits, among other things] I think this is in Week 3's lectures, but as you probably know an NMOS device has a minimum $V_T$ needed across gate and source to "turn on." Thus, when using the NMOS as pull up device, you will not be able to pull up all the way to $V_{supply}$ because once the output reaches $V_{supply}-V_T$ your transistor will shut off. You would have to use a PMOS to get the desired behavior, and I think all the examples just showed NMOS which is why you never saw a MOSFET in the pull up role. However, in digital circuits you will generally see Complementary MOSFETs used (aka CMOS) where the pull up network is made of PMOS devices and the pull down network is made of NMOS devices. This gives the interesting property that your circuit will not burn any static power because there is never a path from $V_{supply}$ to ground (ignoring leakage/crossbar current, which is actually a big deal these days at the small process nodes that modern CPUs are fabbed at). FirstChildUserIdTAG: 323387 FirstChildUserNameTAG: ibrahimawwal FirstChildCreateTimeTAG: 2012-09-25T00:28:22Z IndexTAG: 3918 TitleTAG: dependent current source in s8 V3 we made a dependent current source such that I = k/v but v here is not the potential across the current source. what we did in earlier lectures was the graph of current source in which I was independent of V across it. if this is dependent shouldn't it be depending on voltage across itself??? UserIdTAG: 168496 UserNameTAG: poweltalwar CreateTimeTAG: 2012-09-24T21:28:39Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I don't know quite what you are asking, bu the aspects of a dependent source are that it depends on some other portion of the circuit, which is typically not the voltage across the dependent source itself, but rather for somewhere else in the circuit. Once we get into how dependent sources are built, it should be more clear regarding the mechanism of dependence and how the dependent source is analyzed in the circuit (one part of the circuit that the dependent source derives its value from, and the part of the circuit which has the dependent source itself and the voltage/current/resistance aspects that the dependent source changes in the circuit) FirstChildUserIdTAG: 323230 FirstChildUserNameTAG: Xango5346 FirstChildCreateTimeTAG: 2012-09-25T23:37:35Z IndexTAG: 3919 TitleTAG: Could you help me please Something that might be pretty simple for understanding, but I am just not being able to. So the question is: Shouldn't the VIL,VIH (that is inputs values) be in the left side of the figure (the box), whereas the VOH, VOL (that is output values) be in the right side of the box? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-24T20:39:30Z VoteTAG: 0 CoursewareTAG: Week 2 / Static Discipline CommentableIdTAG: 6002x_static_discipline NumberOfReplyTAG: 2 FirstChildTAG: I'm not sure. Maybe it means that the output of a system will be transmitted to the input of another system, through a communication channel. So that´s why we're receiving outputs at one side and feeding it to the input of another system. I just know that he was using this standard all along then at the end of this video he switched averything in his explanations and annotations, althought the presentation it self still mantains Voh and Vol as the constraints for the input (of the communication channel). So there was a mistake there. FirstChildUserIdTAG: 349595 FirstChildUserNameTAG: danvisk FirstChildCreateTimeTAG: 2012-09-24T23:32:16Z FirstChildTAG: I never thought about it that way, but that's true. What danvisk said is pretty much true. Prof A is trying to demonstrate what can happen to the signal between the output of one logic circuit and the input of another due to noise, so his box there can basically be interpreted as the transmission lines. FirstChildUserIdTAG: 248902 FirstChildUserNameTAG: Chanute FirstChildCreateTimeTAG: 2012-09-25T00:32:44Z IndexTAG: 3920 TitleTAG: Please explain? H2P1 The specs for the problem say that Vin is 50 v. The answer says that we assume Vin is 30v. How do we go from 50 v to 30 v? UserIdTAG: 373585 UserNameTAG: radami1 CreateTimeTAG: 2012-09-24T19:21:32Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: There are different set of data for the homeworks. Your input voltage was 50V, mine 70V and the problem is solved for 30V. So the answer states "Let's assume" to cover the different input voltage (or output). FirstChildUserIdTAG: 396446 FirstChildUserNameTAG: RousseauxS FirstChildCreateTimeTAG: 2012-09-24T19:29:46Z FirstChildTAG: Because the students get different values in their HW en Lab, the values in the solutions given by the staff can differ from the values in your personal solution. However, the way to solve the problems remains the same. The reason is simple: if the values given in the HW/Lab for each student are the same, then some people copy the values from other students. This system was also used for the midterm and final. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-09-25T10:04:45Z IndexTAG: 3921 TitleTAG: Thanks! Thanks for the explanation of linearity!! UserIdTAG: 230787 UserNameTAG: Gabriel007 CreateTimeTAG: 2012-09-24T18:25:37Z VoteTAG: 0 CoursewareTAG: Week 2 / Is it linear? CommentableIdTAG: 6002x_is_it_linear NumberOfReplyTAG: 0 IndexTAG: 3922 TitleTAG: lab 0 Everything is good, it was funny to do it UserIdTAG: 230573 UserNameTAG: aichaamma CreateTimeTAG: 2012-09-24T16:30:54Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: Thanks for letting us know. FirstChildUserIdTAG: 189680 FirstChildUserNameTAG: vnd FirstChildCreateTimeTAG: 2012-09-24T16:32:12Z IndexTAG: 3923 TitleTAG: error in website the internet explorer is showing some kind of error, especially "math processing error" because of which i could not complete my homework. it is not responding to the answers given. please help me out!!!! UserIdTAG: 89612 UserNameTAG: himanshu29 CreateTimeTAG: 2012-09-24T16:00:59Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: problem is about your browser,use chrome,firefox FirstChildUserIdTAG: 112406 FirstChildUserNameTAG: MKG FirstChildCreateTimeTAG: 2012-09-24T17:55:23Z SecondChildTAG: I agree with MKG, use a real browser, use Mozilla Firefox or Google Chrome. Internet Explorer since version 6 has been an outdated and full of bugs web browser. Microsoft Internet Explorer is the most used web browser because people just use what is in front of their eyes and not search a good application. SecondChildUserIdTAG: 265027 SecondChildUserNameTAG: edumm1 SecondChildCreateTimeTAG: 2012-09-24T18:08:36Z IndexTAG: 3924 TitleTAG: Video not working... (Youtube not banned) Where is only white area instead of video. I used another computer to watch before and all was fine. Youtube works well separately. UserIdTAG: 189808 UserNameTAG: AlexanderZ CreateTimeTAG: 2012-09-24T15:56:25Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Troubleshooting NumberOfReplyTAG: 1 FirstChildTAG: Use another browser maybe? FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-09-24T18:22:17Z IndexTAG: 3925 TitleTAG: homework 2 hi I want to ask that in homework 2 why we assume that " Vin=30V,Vout=7.5V,10kΩ≤RTH≤30kΩ." please tell me any one. UserIdTAG: 227508 UserNameTAG: bhavyab CreateTimeTAG: 2012-09-24T15:37:27Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Because, you could chose one real resistor from E12 between 10 and 30 kOhm. Your RTH moved into a real one. FirstChildUserIdTAG: 394042 FirstChildUserNameTAG: Gurgussum FirstChildCreateTimeTAG: 2012-09-24T16:23:20Z FirstChildTAG: The solution is taking the wrong values. The values of input and output are 10V and 2V, not 30V and 7.5V. How do we get in contact with the people managing the course? Thanks! FirstChildUserIdTAG: 349595 FirstChildUserNameTAG: danvisk FirstChildCreateTimeTAG: 2012-09-24T18:19:15Z FirstChildTAG: Because the students get different values in their HW en Lab, the values in the solutions given by the staff can differ from the values in your personal solution. However, the way to solve the problems remains the same. The reason is simple: if the values given in the HW/Lab for each student are the same, then some people copy the values from other students. This system was also used for the midterm and final. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-09-24T19:00:58Z IndexTAG: 3926 TitleTAG: ExpoDweeb I don't understand the equation iD=a*e^b*vD, can someone help me? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-24T14:41:42Z VoteTAG: 0 CoursewareTAG: Week 3 / Analyzing Nonlinear Circuits CommentableIdTAG: 6002x_analyzing_nonlinear_circuits NumberOfReplyTAG: 1 FirstChildTAG: I wouldn't too caught up in the particulars of the actual equation. It's not important. What is important is that it's not linear. That's the important point, that you have this one non linear component,in what is otherwise a linear circuit. Any non linear equation could have been used. It may look strange, but using e is a really good way to make a simple non linear equation. Basically this equation is just e^x where x = Vd. the a and b are just constants. So as in the example. When Vd =0, then b * Vd is also zero, (because zero time anything is zero). That leaves us with e^0. By definition anything to the power of zero =1. That leave the equation now as iD = a * 1. So, when Vd is zero, the current at that point is a. FirstChildUserIdTAG: 385677 FirstChildUserNameTAG: Michaelc1 FirstChildCreateTimeTAG: 2012-09-24T22:49:45Z IndexTAG: 3927 TitleTAG: iDS why iDS is positive which is flowing drain to source? UserIdTAG: 67648 UserNameTAG: mgm CreateTimeTAG: 2012-09-24T13:34:52Z VoteTAG: 0 CoursewareTAG: Week 3 / MOSFET CommentableIdTAG: 6002x_mosfet_model NumberOfReplyTAG: 1 FirstChildTAG: That is the typical convention. In fact, that is why is it call $i_{ds}$. Placing the "d" first and "s" second on the current name means that the current is positive when going from Drain to Source. FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-25T13:19:58Z IndexTAG: 3928 TitleTAG: S6E1: Current flow once the equations from 0 to 1mA and 2 - 3 mA are created? Once the different functions are created, V=1000i-0.999 from 0-1mA and V=2000i-1 from 1-3mA, how do you calculate the current when I connect the function back up the the Thevenin equivalent circuit? I would think one would find Vth and Rth using Rs=4700 and Rp=8200 from the Thevenin equivalent, then find the Norton equivalent current due to these values. If this current is above 1mA, use the formula: V=2000i-1. If it was below 1mA, use V=1000i-0.999. With this method I found i=1.06mA, therefore to calculate the voltage I used V=2000i-1, however my answers were wrong. If I have to consider current from 0 to 3mA, how do I deal with the different functions? It is like one function with a discontinuity in the middle of it. UserIdTAG: 132828 UserNameTAG: rharris CreateTimeTAG: 2012-09-24T13:20:37Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3929 TitleTAG: KCL and KVL method Please Can't we solve the H2P1 - week 2 assignment with kcl and kvl methods? since I am certain that KCL and KVL can be use to solve all elctrical problems Thanks UserIdTAG: 382532 UserNameTAG: emmanuelpeace CreateTimeTAG: 2012-09-24T13:16:01Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Yes, you can always apply KVL and KCL to set up your equations. In this case the circuit is a simple divider, so the equation relating output and input voltage is very simple, but certainly you can use KVL to find it. FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-25T13:17:20Z IndexTAG: 3930 TitleTAG: Voltages Is there not a voltage accross RL so that Vout is not equal to VS but rather VS-VL? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-24T11:48:06Z VoteTAG: 0 CoursewareTAG: Week 3 / MOSFET CommentableIdTAG: 6002x_mosfet_model NumberOfReplyTAG: 1 FirstChildTAG: If you speak about the S-model, there is either $V_s$ or 0 at output FirstChildUserIdTAG: 190618 FirstChildUserNameTAG: Kirbabaev FirstChildCreateTimeTAG: 2012-09-24T12:35:52Z IndexTAG: 3931 TitleTAG: H2P1 Resistances. Please help. i Already pick good values from E12 by formula Vout/Vin-10% <= (Vin*R2/(R1+R2))/Vin <= Vout/Vin+10% but answer in resistances sections does not accept. Please help, i don't know how much time left to answer! UserIdTAG: 205077 UserNameTAG: Altleonidas CreateTimeTAG: 2012-09-24T10:51:48Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Please please FirstChildUserIdTAG: 205077 FirstChildUserNameTAG: Altleonidas FirstChildCreateTimeTAG: 2012-09-24T11:06:41Z FirstChildTAG: And what about **An additional requirement is that the Thevenin resistance as seen from the output terminals is between 10kΩ and 30kΩ.** Are Your resistors satisfy this requirement? FirstChildUserIdTAG: 195218 FirstChildUserNameTAG: Al_Incognito FirstChildCreateTimeTAG: 2012-09-24T11:58:43Z SecondChildTAG: By fact You can gather unlimited anount of resistor pairs to satisfy Vin/Vout range, but this requirement limit Your choise for several pairs. Choose the best of them. SecondChildUserIdTAG: 195218 SecondChildUserNameTAG: Al_Incognito SecondChildCreateTimeTAG: 2012-09-24T12:03:58Z SecondChildTAG: Yes, they are. I already recalculate 3 pairs of resistances, with the correct answers, i spent about 5 hrs, trying to figure out this question. I am quet not shure what this question asks. I forgot to mension - Vmax and Vmin have been accepted all 3 times. SecondChildUserIdTAG: 205077 SecondChildUserNameTAG: Altleonidas SecondChildCreateTimeTAG: 2012-09-24T12:09:08Z SecondChildTAG: Or maybe not. A little hand of help? SecondChildUserIdTAG: 205077 SecondChildUserNameTAG: Altleonidas SecondChildCreateTimeTAG: 2012-09-24T12:09:29Z SecondChildTAG: Ok. 1st find a formula likr R1=x*R2. According to Your 1st post You already find one. 2nd Find a formula for Thevenin resistance (shortcut voltage source and look from terminal). Then Try to combine Rs from table to satisfy both requirements. Due to some peculiarity of E12 row, one combination will be better, according to Vin/Vout. Try to calculate several pairs and You will see dependence. SecondChildUserIdTAG: 195218 SecondChildUserNameTAG: Al_Incognito SecondChildCreateTimeTAG: 2012-09-24T12:23:04Z SecondChildTAG: thank you. SecondChildUserIdTAG: 205077 SecondChildUserNameTAG: Altleonidas SecondChildCreateTimeTAG: 2012-09-24T12:29:17Z SecondChildTAG: Ok, now after deadline i am disappointed. I already pick 56 and 18 resistors, and calculate correct Voltage, but it still was wrong. Why so??? Why the system didn't accept my the very first pair I've been trying to submit?? SecondChildUserIdTAG: 205077 SecondChildUserNameTAG: Altleonidas SecondChildCreateTimeTAG: 2012-10-01T14:48:11Z IndexTAG: 3932 TitleTAG: lab 3 my output graph is coming perfectly. but still server saying incorrect. plz help. UserIdTAG: 72647 UserNameTAG: NEEL11 CreateTimeTAG: 2012-09-24T09:39:17Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: pay attention on the spikes or glitches, take a look at it more clossely, it must below the VoL FirstChildUserIdTAG: 452559 FirstChildUserNameTAG: kuz1toro FirstChildCreateTimeTAG: 2012-09-24T10:52:57Z SecondChildTAG: Dear Neel 11 , iam not getting the lab 3 can you just give some clue SecondChildUserIdTAG: 156694 SecondChildUserNameTAG: mkprasanth SecondChildCreateTimeTAG: 2012-09-25T07:57:04Z SecondChildTAG: i can't find the right W/L ratio :( SecondChildUserIdTAG: 97340 SecondChildUserNameTAG: AnkitRana SecondChildCreateTimeTAG: 2012-09-25T15:01:44Z SecondChildTAG: how many MOSFET do I need and just connected to Z ?? SecondChildUserIdTAG: 331441 SecondChildUserNameTAG: abdessamade SecondChildCreateTimeTAG: 2012-09-25T21:54:30Z IndexTAG: 3933 TitleTAG: Lab 2: How I solved it I did a quick check on http://www.timeanddate.com/worldclock/custom.html?sort=1 to see if it was still Sunday everywhere. It looks like it's Monday all around, so I wanted to post how I went about this to see if I made any incorrect leaps. $V_{out} = \frac{1}{2}V_1 + \frac{1}{6}V_2$ What we know about linearity means: $\frac{1}{2} = \frac{R_1}{R_2+R_1}$ and $\frac{1}{6} = \frac{R_2}{R_1+R_2}$ Solving the second equation I got $R_2 = 3R_1$. Since when it comes to resistors it is the ratio we care about I chose 2 as an arbitrary $R_1$ and then that meant $R_2$ was 6. So now $R_1 = 2$, $R_2 = 6$, so $R_{th} = 2||6$, or 1.5. The max amplitude of the wave for $V_{out}$ is supposed to be .667 which is $\frac{2}{3}$ the value of the current amplitude peak of 1. Using this value I was able to solve $R_3$ (my resistor to ground) with $\frac{2}{3} = \frac{R_2}{1.5 + R_2}$, or $R_3 = 3$. I ended up with a valid solution, but I'm not sure whether two assumptions I made were correct: 1) I obtained the $R_1$ and $R_2$ ratio via only using one of the pair of equations. It works, but was it right? 2) My choice of using $\frac{2}{3}$ as a voltage proportion to calculate $R_3$. Was there a more "correct" way to do this? UserIdTAG: 264596 UserNameTAG: Nuru CreateTimeTAG: 2012-09-24T08:41:43Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: You described a two step process to design the mixer. From the design point of view those weren't assumptions but a statement of how the design worked. First you designed a simple mixer with two resistors with 3/4 and 1/4 weights for the signals. The proportions were right but the signal was too strong. The two voltage sources and two resistors are equivalent to a Thevenin voltage and Thevenin resistance in series. By choosing the proper load resistor across the circuit any fraction of Vth can be taken as output. Choosing the load resistance to produce an output of 2/3 Vth gave the desired weights. In my opinion an important part of the design process was realizing that the design could not be accomplish with only two resistors because the weights did not sum to one. After posting I began to wonder whether it is still Sunday in Hawaii or some other Pacific islands not on that world clock. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-09-24T09:19:15Z SecondChildTAG: It is S3E2, the simplest way, and already calculated if you made that seminar. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-09-24T20:23:27Z FirstChildTAG: Hold on... if R1 = 2 and R2 = 6 then R1/(R1 + R2) is not coming out to be 1/2 which was assumed before. How? or am I doing something wrong? FirstChildUserIdTAG: 358307 FirstChildUserNameTAG: Avneesh92 FirstChildCreateTimeTAG: 2012-09-27T09:54:14Z IndexTAG: 3934 TitleTAG: Way to cheat - Staff Help https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/505c404c2d40882b00000014 The above link shows a discussion. Here a comment by 2214sanchez shows a link which redirects to the MITx page where we can view the course as a guest. I found that we could view all the Lab problems as well as the answers to those problems. Some students might just find a way to cheat here by copying the answers. So i request the staff members to look into this. UserIdTAG: 222911 UserNameTAG: bhaswardg CreateTimeTAG: 2012-09-24T07:36:30Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 1 FirstChildTAG: I don't think the answers of the labs can be seen there. The Show Answer button never worked for me during the Spring session of the course. Lab solutions were posted on the Course Info page after the deadline. Does the Show Answer button work for you? I just checked. It doesn't in my case. FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-09-24T10:47:56Z SecondChildTAG: Yes the Show Answer button is working for me. That's the reason i posted it here. SecondChildUserIdTAG: 222911 SecondChildUserNameTAG: bhaswardg SecondChildCreateTimeTAG: 2012-09-24T13:33:39Z SecondChildTAG: I tried this link, but the show answer button didnt work for me either. I'm not sure why it worked in his case (I tried it w/ lab 1). SecondChildUserIdTAG: 248807 SecondChildUserNameTAG: 1kingsman SecondChildCreateTimeTAG: 2012-09-25T00:20:59Z SecondChildTAG: Not working for me either. Are you a previous student from MITx and you forgot you have logged in using that account? SecondChildUserIdTAG: 157273 SecondChildUserNameTAG: ongchihang SecondChildCreateTimeTAG: 2012-09-25T14:20:13Z SecondChildTAG: That button didn't work at all in the Spring session. In case of labs, the answers were posted on the Course Info page. SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-09-25T19:54:20Z SecondChildTAG: no i am taking this course for the first time. SecondChildUserIdTAG: 222911 SecondChildUserNameTAG: bhaswardg SecondChildCreateTimeTAG: 2012-09-28T19:03:08Z IndexTAG: 3935 TitleTAG: HOW MANY DIGIT Hi, can anyone told me how many digit or number i put in my answer after putting the point... as a example i have an answer of a problem 6.333333333333333 now how many 3`s i also used in my answer. thank you. UserIdTAG: 352582 UserNameTAG: Sumanghosal CreateTimeTAG: 2012-09-24T05:13:20Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: If you have three place functions you only have three place accuracy. Do not imply false accuracy unless you have it. Are your measurements corrected and traceable to the National Bureau of Standards? Is this an "IDEAL" problem? Mel FirstChildUserIdTAG: 146284 FirstChildUserNameTAG: Melvyn FirstChildCreateTimeTAG: 2012-09-24T05:27:37Z IndexTAG: 3936 TitleTAG: Lab2 HELP!!! can someone please help me with lab 2...I feel like I have tried every single possibilty. I have 30 mins left. Please help me UserIdTAG: 435197 UserNameTAG: RichmondRichter CreateTimeTAG: 2012-09-24T04:23:33Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: did you figure it out FirstChildUserIdTAG: 233700 FirstChildUserNameTAG: ANDREW77 FirstChildCreateTimeTAG: 2012-09-24T04:47:59Z FirstChildTAG: try using a r3 in betwen the node and gnd to get a better graph and usig resitor values betwen .5 to 1 resitors FirstChildUserIdTAG: 156060 FirstChildUserNameTAG: radeon9550 FirstChildCreateTimeTAG: 2012-09-24T04:29:57Z SecondChildTAG: and for the r2 use values betwen 1.5 to 2 resistors SecondChildUserIdTAG: 156060 SecondChildUserNameTAG: radeon9550 SecondChildCreateTimeTAG: 2012-09-24T04:33:11Z SecondChildTAG: so basically trial and error. I have about 15 mins left. I know you have to use 2 voltage dividers but the equations get messy. you have 3 unknowns and 2 equations and when u plug in values for one it give u negative in return SecondChildUserIdTAG: 435197 SecondChildUserNameTAG: RichmondRichter SecondChildCreateTimeTAG: 2012-09-24T04:47:57Z FirstChildTAG: Use three resistors. R1 between v1 and node, R2 between v2 and node and R3 between node and ground int he following ratio:- 1.5 R1 = R3, 3 R1 = R2, 2R3 = R2 These equations can be made by looking at equivalent resistance for each input. I hope this will help FirstChildUserIdTAG: 338752 FirstChildUserNameTAG: SaadatAli FirstChildCreateTimeTAG: 2012-09-24T04:49:14Z SecondChildTAG: thanks for your help bro but unfortunately Im gonna have to take the L for this lab lol...is there partial points for late hw SecondChildUserIdTAG: 435197 SecondChildUserNameTAG: RichmondRichter SecondChildCreateTimeTAG: 2012-09-24T05:01:51Z SecondChildTAG: just submit it and you will get marks. where are you right now? because different locations have different submission times SecondChildUserIdTAG: 338752 SecondChildUserNameTAG: SaadatAli SecondChildCreateTimeTAG: 2012-09-24T05:23:24Z SecondChildTAG: and i hope you have found the answers SecondChildUserIdTAG: 338752 SecondChildUserNameTAG: SaadatAli SecondChildCreateTimeTAG: 2012-09-24T05:23:39Z SecondChildTAG: Awesome! Thank you so much! SecondChildUserIdTAG: 205077 SecondChildUserNameTAG: Altleonidas SecondChildCreateTimeTAG: 2012-09-24T08:43:41Z SecondChildTAG: bro i am still not getting the right answer please help SecondChildUserIdTAG: 217045 SecondChildUserNameTAG: ANURAG1993 SecondChildCreateTimeTAG: 2012-09-24T11:04:15Z SecondChildTAG: now i have got the answer by your manner can u please tell how this idea germinated in your mind please do tell SecondChildUserIdTAG: 217045 SecondChildUserNameTAG: ANURAG1993 SecondChildCreateTimeTAG: 2012-09-24T11:27:10Z SecondChildTAG: pleas a bit explanation about the relation between resistence SecondChildUserIdTAG: 115656 SecondChildUserNameTAG: rolando277 SecondChildCreateTimeTAG: 2012-10-01T12:03:44Z FirstChildTAG: try trial and error with variables..use 1 as a reference and change others FirstChildUserIdTAG: 443809 FirstChildUserNameTAG: Duvindu FirstChildCreateTimeTAG: 2012-09-24T04:51:05Z IndexTAG: 3937 TitleTAG: Homework 2 p1 why in hwp1 i put in the r1= 67000 and r2= 18000 are wrong but the values of vmax= 2.5 and vmin 1.8 are correct.???? can anyone tellme why please.??? thanks.! UserIdTAG: 156060 UserNameTAG: radeon9550 CreateTimeTAG: 2012-09-24T04:20:57Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: thanks for the advice FirstChildUserIdTAG: 156060 FirstChildUserNameTAG: radeon9550 FirstChildCreateTimeTAG: 2012-09-24T04:38:56Z FirstChildTAG: the values are correct because they base it on the values you put in for r1 and r2...their saying you calculated the correct but wrong max and mins. FirstChildUserIdTAG: 435197 FirstChildUserNameTAG: RichmondRichter FirstChildCreateTimeTAG: 2012-09-24T04:24:39Z FirstChildTAG: I have the same question. I took correct R1 and R2 resistances, figured out the voltage, but system does not accept resistances. I did something wrong? Maybe wright it in another format? Please HELP FirstChildUserIdTAG: 205077 FirstChildUserNameTAG: Altleonidas FirstChildCreateTimeTAG: 2012-09-24T10:16:25Z IndexTAG: 3938 TitleTAG: Lab 1 - why R1=2*R2? For the circuit without ligthbulb we have following equations (R1+R2)*I=6 R2*I=2 How did we get R1=2*R2 from these equations? UserIdTAG: 269639 UserNameTAG: ishalyapin CreateTimeTAG: 2012-09-24T03:30:05Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: For the circuit without the lightbulb, we have: $ I_{R_1}=I_{R_2}=I$ and yes, $ (R_1 + R_2) \cdot I = 6 ~~~ \boxed {eq 1}$ $ R_2 \cdot I = 2 $ $ I = \cfrac {2}{R_2} ~~~ \boxed {eq 2} $ substituting 2 into 1 $ (R_1 + R_2) \cdot \cfrac {2}{R_2} = 6$ $ 2 \cdot R_1 + 2 \cdot R_2 = 6 \cdot R_2 $ $ 2 \cdot R_1 = 4 \cdot R_2 $ $ \boxed {R_1 = 2 \cdot R_2} $ QED FirstChildUserIdTAG: 248902 FirstChildUserNameTAG: Chanute FirstChildCreateTimeTAG: 2012-09-24T06:11:23Z SecondChildTAG: Thank you!!! Perfect explanation! SecondChildUserIdTAG: 269639 SecondChildUserNameTAG: ishalyapin SecondChildCreateTimeTAG: 2012-09-24T08:04:09Z IndexTAG: 3939 TitleTAG: This question is not fair! Well, what I mean by that is: If you had seen 2 batteries instead of the 2 symbols then you would surely commit a mistake? Don't you agree with me? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-23T22:23:22Z VoteTAG: 0 CoursewareTAG: Week 2 / Node equation review CommentableIdTAG: 6002x_S3E1_NodeEquationReview NumberOfReplyTAG: 2 FirstChildTAG: no FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-09-23T22:24:37Z FirstChildTAG: This question has caused a lot of problems. In my opinion, the way it is written it is very convenient from an educational point of view. One of the key concepts of circuit analysis is understanding what a voltage difference between two nodes means, independent on how the source is oriented and which is its sign. When you start designing your own circuits, you will need reversed sources, negative voltages, etc, quite often! FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-23T22:29:49Z SecondChildTAG: I agree with you, it is logical and mathematical real SecondChildUserIdTAG: 327787 SecondChildUserNameTAG: REINALDOPARANHOS SecondChildCreateTimeTAG: 2012-09-24T02:12:02Z IndexTAG: 3940 TitleTAG: Math error on H2P1 I think I have a bug on my exercise. My specs are: 50V in and 10V out, Rth between 10k and 30k, tolerances for resistors are 10% and same percent for Vout. If you go on with the two voltage divisor inequations, according with the voltage tolerances: 1.1*R2/(0.9*R1+1.1*R2)<=1.1/5 0.9*R2/(1.1*R1+0.9*R2)>=0.9/5 you get two thresholds of R2, in terms of R1, like this: R2<=0.2307*R1 R2>=0.268*R1 This leaves the problem without any mathematical solution. I'm confused. In the exercise, I've left the values that fit better with the voltage divisor spec, and calculated the Vmax and Vmin (system says it's correct), being Vmin out of range. UserIdTAG: 482646 UserNameTAG: elgambitero CreateTimeTAG: 2012-09-23T21:59:59Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 3 FirstChildTAG: You could use: Vout = Vin R2 / (R1+R2) FirstChildUserIdTAG: 197569 FirstChildUserNameTAG: darksamaro FirstChildCreateTimeTAG: 2012-09-23T22:21:22Z FirstChildTAG: why you apply resistors' tolerance to ratio? forget tolerance, write equation for resistors ratio - it will be first equation. Then calculate Rth and apply 10k-30k spec to it - it will be the second equation. Solve it, pick resistors, apply tolerance to them - you will have R1min, R1max, R2min and R2max - calculate Vmax and Vmin. FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-09-23T22:13:58Z SecondChildTAG: Why it doesn't work: Vout max: Vin*R2max/(R1min+R2max) Vout min: Vin*R2min/(R1max+R2min) ???? SecondChildUserIdTAG: 95521 SecondChildUserNameTAG: Gatling SecondChildCreateTimeTAG: 2012-09-23T23:10:08Z SecondChildTAG: It's correct. Should work SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-24T00:29:06Z SecondChildTAG: I just lost one more requirement - the real Vin/Vout ratio should be also not more then 10% from required, so just write down all possible variants (there only 5 of them) and pick the best one SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-24T00:40:45Z SecondChildTAG: * Vout/Vin SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-24T00:49:32Z FirstChildTAG: Hmmm, now I doubt if the problem asked the Vout being in a 10% tolerance too. What I say is, if resistors have 10% tolerance, it is not mathematically possible to make a voltage divisor that places the output voltage with a 10% tolerance. FirstChildUserIdTAG: 482646 FirstChildUserNameTAG: elgambitero FirstChildCreateTimeTAG: 2012-09-24T15:06:25Z IndexTAG: 3941 TitleTAG: Kirchoff's Law I learn the Kirchoff's Law with different sign convention The equation 0.5(1)+0.5(3)+0.5(2)-3 = 0 UserIdTAG: 483990 UserNameTAG: RobertoGea CreateTimeTAG: 2012-09-23T21:47:24Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 0 IndexTAG: 3942 TitleTAG: practice very good practice. I had problems at the beginning of the simulation but the fix aid review UserIdTAG: 395337 UserNameTAG: lujafo CreateTimeTAG: 2012-09-23T21:03:29Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 0 IndexTAG: 3943 TitleTAG: Please Explain Question one and two for the week 2 Home work que 1 part and b Please the question 1 of H2P1 is not clear to me I am trying to come out with the solution.the concept is not clear to me The thevenin problem is also given me a though time. I need an immediate help Thanks UserIdTAG: 382532 UserNameTAG: emmanuelpeace CreateTimeTAG: 2012-09-23T20:44:49Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Select your R values from the list and use a V divider to ensure that the Rth value is within the required range. Once the R values are found you can work out the tolerances fro the second part. FirstChildUserIdTAG: 195916 FirstChildUserNameTAG: pflynn FirstChildCreateTimeTAG: 2012-09-23T21:22:05Z IndexTAG: 3944 TitleTAG: H2P1 Help please! I read all the comments and all the posts on the discussion forum, and I did all what I could but I can't arive to the answer. I think I'm having a mistake, but I don't know where. Maybe I'm not choosing the correct resistors, because those meet all the requierements but the system says I'm still wrong. Let's say I've got R1= 22222 and R2=200000 (Those are not my real answers) is it right if I choose R1=22000 and R2= 220000? UserIdTAG: 45307 UserNameTAG: josejimenez2 CreateTimeTAG: 2012-09-23T20:42:10Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: You can only chose the resistance values that are provided to you, i.e. 10,12,15,18,22,27,33,39,47,56,68,82. So it would have to be 22000 or 22k. So you have to chose the numbers that give the closest ratio (to a 10% deviation). FirstChildUserIdTAG: 207204 FirstChildUserNameTAG: hgustavii FirstChildCreateTimeTAG: 2012-09-23T20:51:10Z FirstChildTAG: There is an extra condition on $R_{TH}$ in that task. Maybe thats your trouble FirstChildUserIdTAG: 190618 FirstChildUserNameTAG: Kirbabaev FirstChildCreateTimeTAG: 2012-09-23T22:04:38Z IndexTAG: 3945 TitleTAG: H3P4 Help with Current portion of Question All - I got the voltages correct for H3P4, but am having problems with the current. Methodology: Let Triangle wave = positive peak; therefore, D1 passes voltage and D2 is open circuit Using KCL method: iD1 + VS/R + V1/R = 0; solve for iD1 Let Triangle wave = negative peak; therefore, D2 passes voltage and D1 is open circuit Using KCL method: iD2 + VS/R + V2/R = 0; solve for iD2 What am I missing? Thanks. UserIdTAG: 348141 UserNameTAG: TomTrieb CreateTimeTAG: 2012-09-23T19:57:12Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I think the problem is a bit more logical, sometimes you dont need to solve equations you just need to figure out what is the answer. Think logically then solve it mathematically :P The ansers are easy and simple (sorry bad english xD) FirstChildUserIdTAG: 370343 FirstChildUserNameTAG: JJarquin FirstChildCreateTimeTAG: 2012-09-23T20:22:15Z SecondChildTAG: Well, I set up the circuit in the sandbox and did a current analysis to brute force the answer and it worked . . . but I don't understand it. I look forward to reading the explanation after next week. SecondChildUserIdTAG: 348141 SecondChildUserNameTAG: TomTrieb SecondChildCreateTimeTAG: 2012-09-24T03:04:17Z SecondChildTAG: it is so easy just use simple use adding two source formula and use ohms law SecondChildUserIdTAG: 164689 SecondChildUserNameTAG: muhammadfaizan SecondChildCreateTimeTAG: 2012-09-26T15:43:56Z SecondChildTAG: What is sandbox? a simulator? where can we downloaded if so ? SecondChildUserIdTAG: 410943 SecondChildUserNameTAG: JPalaciosRoman SecondChildCreateTimeTAG: 2012-09-29T11:35:55Z SecondChildTAG: @JPalaciosRoman Sandbox is a simulator available in the overview section of your courseware in 6.002x itself. SecondChildUserIdTAG: 315681 SecondChildUserNameTAG: nahuja1 SecondChildCreateTimeTAG: 2012-09-29T13:47:32Z IndexTAG: 3946 TitleTAG: Pls find me my mistake By nodal analysis, as current entering a node is equal to the current leaving the node. we have, ((V1-e)/R1)=((e-V2)/R2) =>((5-e)/6800)=((e-(-7.2))/5600) Dividing denominator of both sides with 100 we get, ((5-e)/68)=((e-(-7.2))/56) =>280-56e=68e+489.6 =>-209.6=124e =>e=-1.69 UserIdTAG: 134454 UserNameTAG: Anvesh_Maddula CreateTimeTAG: 2012-09-23T19:54:09Z VoteTAG: 0 CoursewareTAG: Week 2 / Node equation review CommentableIdTAG: 6002x_S3E1_NodeEquationReview NumberOfReplyTAG: 1 FirstChildTAG: You should to change sign of V2 to "plus" in the right part of equation, because the voltage source V2 on the figure has negative polarity (positive terminal is conected to the ground). The right equation is ((V1-e)/R1)=((e+V2)/R2) FirstChildUserIdTAG: 278853 FirstChildUserNameTAG: MaratGafarov FirstChildCreateTimeTAG: 2012-09-24T04:44:35Z IndexTAG: 3947 TitleTAG: It works if I consider the difference between z and ground For some of these cases, I find Vs going to ground among resistors in parallel. In these situations, z would be equal to 1/3*Vs, assuming all three resistors are the same size. Even if the resistors are not equal in size, wouldn't this violate the idea that we want our output voltage to be as clean if not cleaner than our input voltage? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-23T19:50:19Z VoteTAG: 0 CoursewareTAG: Week 3 / Switch Model Exercise CommentableIdTAG: 6002x_switch_model_exercise NumberOfReplyTAG: 1 FirstChildTAG: Z can have only 2 values Z = Vs or Z = 0 try to use KVL ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13484348622384249.jpeg FirstChildUserIdTAG: 236136 FirstChildUserNameTAG: Hamid-ch FirstChildCreateTimeTAG: 2012-09-23T21:15:17Z IndexTAG: 3948 TitleTAG: HW2 p1 @__@ 3 hours now, i really can't figure this out I even went through matlab to narrow down the numbers but it's just not working can anyone please help me check if i'm understanding the conditions right? I think Rth= R1+R2 and has to be 10k-30k R2/R1+R2 has to be about 1/4 because we're using V. divider there to get around 20V? (80/20 >> 1/4) just with that much am i getting things right? UserIdTAG: 151030 UserNameTAG: Eskamo CreateTimeTAG: 2012-09-23T19:48:35Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Rth = R1||R2 = R1*R2/R1+R2 ("||" means parallel) FirstChildUserIdTAG: 207204 FirstChildUserNameTAG: hgustavii FirstChildCreateTimeTAG: 2012-09-23T20:54:57Z FirstChildTAG: Rth is wrong , Vin is short circuited, calculate Rth as seen from vout. Sorry for my english :p FirstChildUserIdTAG: 370343 FirstChildUserNameTAG: JJarquin FirstChildCreateTimeTAG: 2012-09-23T20:47:48Z FirstChildTAG: Eskamo, as the posts above mentioned, your Rth calculation is wrong. That is probably the problem you are having. FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-23T22:31:45Z IndexTAG: 3949 TitleTAG: LAB 2 - showing blank page Hi, LAB 2 just shows a blank page, I am using Opera, but tried on Firefox and chrome. Other lab pages are working correctly but LAB 2 is due in the next few hours, please help!!! (also sent ths message to bugs@edx) UserIdTAG: 61469 UserNameTAG: Spacedog CreateTimeTAG: 2012-09-23T19:40:08Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: ok, I have less than an hour left... still nothing, no response to e-mail. it has been like this for says, I've waited for it to resolve itself, but now I'm stuck :( someone please reply. FirstChildUserIdTAG: 61469 FirstChildUserNameTAG: Spacedog FirstChildCreateTimeTAG: 2012-09-23T21:09:57Z SecondChildTAG: I've had a few problems similar to this. I use Firefox, and resolved by clearing my browser cache and reloading the page. (Sometimes, it took more than one clear/reload.) SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-09-24T00:02:31Z SecondChildTAG: Also, you might see if there is a "Reset" button toward the bottom of the page. SecondChildUserIdTAG: 148542 SecondChildUserNameTAG: planetscape SecondChildCreateTimeTAG: 2012-09-24T00:03:29Z SecondChildTAG: page is blank on all browsers, only for this lab. I was progressing well on it a few days back, but made a mistake, I tried to reload the page to reset the simulator and it was blank ever since. deadline is passed now and I'm disappointed I missed a lab credit and have no official response for this issue. SecondChildUserIdTAG: 61469 SecondChildUserNameTAG: Spacedog SecondChildCreateTimeTAG: 2012-09-24T00:40:54Z IndexTAG: 3950 TitleTAG: vO sign Why expressions are negative? Is it because vO is turn upside down with respect to current source? UserIdTAG: 208296 UserNameTAG: OZ1 CreateTimeTAG: 2012-09-23T19:35:21Z VoteTAG: 0 CoursewareTAG: Week 4 / Dependent Sources Exercise 0 CommentableIdTAG: 6002x_dep_src_exsz_0 NumberOfReplyTAG: 1 FirstChildTAG: Yes, your idea is right. Notice that the dependent current source is pointing downwards, and therefore current through the output resistor will cause a voltage drop that is negative with respect to its defined polarity. FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-23T22:35:54Z SecondChildTAG: Thanks for the explanation. I also have overlooked it ( SecondChildUserIdTAG: 484464 SecondChildUserNameTAG: IreneK SecondChildCreateTimeTAG: 2012-10-04T00:44:02Z IndexTAG: 3951 TitleTAG: H2P1 - Rth value between 10k and 30k?! Hiya, I've been reading loads of posts about getting the correct value for R1 and R2, but nowhere seems to answer what it actually means with having "...Thevenin resistance as seen from the output terminals is between 10kΩ and 30kΩ". Does it actually mean that R2/(R1+R2)=Rth=? How does that work when numbers tend more to be 0.2 or 0.25 etc? With Vin=60 and Vout=15 I have put numbers R1=27k and R2=10k (to comply with the resistor sets). Obviously this is wrong, since I read that R1 and R2 have to be within 10k to 30k. Can anyone shed a light on this matter? Cheers! UserIdTAG: 207204 UserNameTAG: hgustavii CreateTimeTAG: 2012-09-23T19:22:43Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 7 FirstChildTAG: i thought Rth was R1+R2 FirstChildUserIdTAG: 151030 FirstChildUserNameTAG: Eskamo FirstChildCreateTimeTAG: 2012-09-23T19:51:16Z SecondChildTAG: If that's the case, then I've missed something... By the way, what does the || mean in say 1||2? SecondChildUserIdTAG: 207204 SecondChildUserNameTAG: hgustavii SecondChildCreateTimeTAG: 2012-09-23T20:09:29Z SecondChildTAG: || means Parallel combination SecondChildUserIdTAG: 489237 SecondChildUserNameTAG: asdasDasd SecondChildCreateTimeTAG: 2012-09-26T15:34:17Z FirstChildTAG: Finally, I got all answers correct. FirstChildUserIdTAG: 378096 FirstChildUserNameTAG: Jamshaid271 FirstChildCreateTimeTAG: 2012-09-23T19:42:58Z SecondChildTAG: Ok... SecondChildUserIdTAG: 207204 SecondChildUserNameTAG: hgustavii SecondChildCreateTimeTAG: 2012-09-23T19:49:44Z SecondChildTAG: You have Rth, and i assume that you know a formula that relates R1, R2 and Rth, then you can assume a value for R1 and calculate R2 but remember that it has to satisfy the constraint of Vout. I solved it first matematically to find the relation between R1,R2,Rth,Vin,Vout, all in a simple formula. The answers are simple and easy. SecondChildUserIdTAG: 370343 SecondChildUserNameTAG: JJarquin SecondChildCreateTimeTAG: 2012-09-23T19:52:48Z SecondChildTAG: In that case it is the maths that gets me (my weakness in all this), I just can't see it. Where do I have "Rth" by the way? SecondChildUserIdTAG: 207204 SecondChildUserNameTAG: hgustavii SecondChildCreateTimeTAG: 2012-09-23T19:55:57Z SecondChildTAG: ...and I thought the formula was Rth=R2/(R1+R2)... So I'm still confused... SecondChildUserIdTAG: 207204 SecondChildUserNameTAG: hgustavii SecondChildCreateTimeTAG: 2012-09-23T19:57:10Z FirstChildTAG: Just know that R1 and R2 are in parallel, when we are going to find Rth. By which you will get between 10k to 30k. I got 23k. And Vin/Vout ratio is in between 3.5 to 4.5. That's all. FirstChildUserIdTAG: 378096 FirstChildUserNameTAG: Jamshaid271 FirstChildCreateTimeTAG: 2012-09-23T20:06:39Z SecondChildTAG: I thought it was the Vout/Vin ratio we were looking for? SecondChildUserIdTAG: 207204 SecondChildUserNameTAG: hgustavii SecondChildCreateTimeTAG: 2012-09-23T20:15:25Z SecondChildTAG: Thanks for the parallel part! I finally get that part! :) SecondChildUserIdTAG: 207204 SecondChildUserNameTAG: hgustavii SecondChildCreateTimeTAG: 2012-09-23T20:16:28Z SecondChildTAG: Hi Jamshaid271, When I tried with 10k & 30k for Rth, the answer seems not correct. But with 23k, the answer is correct.On what basis did you take the value of Rth as 23k?? Can you pls explain? SecondChildUserIdTAG: 127352 SecondChildUserNameTAG: Kathir SecondChildCreateTimeTAG: 2012-09-23T21:22:51Z SecondChildTAG: Rth should be between 10k and 30k, so 23k is fine :) SecondChildUserIdTAG: 207204 SecondChildUserNameTAG: hgustavii SecondChildCreateTimeTAG: 2012-09-23T21:31:56Z FirstChildTAG: Thanks everyone! I finally got it! ("only" took like half a day). Thanks again for your patience, it was really appreciated. Now on to the next question... FirstChildUserIdTAG: 207204 FirstChildUserNameTAG: hgustavii FirstChildCreateTimeTAG: 2012-09-23T20:28:09Z SecondChildTAG: hahahaha SecondChildUserIdTAG: 378096 SecondChildUserNameTAG: Jamshaid271 SecondChildCreateTimeTAG: 2012-09-23T20:48:57Z FirstChildTAG: I took this course last time as well and still facing problem to find the answers... FirstChildUserIdTAG: 21687 FirstChildUserNameTAG: Benadicta FirstChildCreateTimeTAG: 2012-09-24T00:37:27Z FirstChildTAG: Your Vin and Vout is wrong!! FirstChildUserIdTAG: 233700 FirstChildUserNameTAG: ANDREW77 FirstChildCreateTimeTAG: 2012-09-24T04:56:37Z FirstChildTAG: muchas gracias Miriam eres una gran ayuda para nosotros y no te preocupes por la tardanza se que tambien tienes cosas que hacer y te das un tiempecito para poder ayudarnos y eso es lo que importa gracias de verdad FirstChildUserIdTAG: 320715 FirstChildUserNameTAG: maraivette FirstChildCreateTimeTAG: 2012-10-08T02:51:22Z IndexTAG: 3952 TitleTAG: Here, here and over here That does not explain much! UserIdTAG: 146329 UserNameTAG: soniaroa CreateTimeTAG: 2012-09-23T19:06:25Z VoteTAG: 0 CoursewareTAG: Week 2 / Thevenin Example CommentableIdTAG: 6002x_thevenin_example_1 NumberOfReplyTAG: 0 IndexTAG: 3953 TitleTAG: h2p1 Vin/Vout ratio out of range I can't get correct resistance values satisfying the condition Vin/Vout<=10. The closest ratio i got was something around 0.15. Here's what i did: given V_in=80V and V_out=24(approx) Vin/Vout=R2/(R1+R2) 0.3=R2/(R1+R2) R1=7*R2/3 Next, Rth=R1*R2/(R1+R2) Rth=0.7*R2(OR)0.3*R1 And Rth lies between 10k and 30k ohms 6 values of R2(15,18,22,27,33,39) and 5 values of R1(39,47,56,68,82) satisfy the above condition. When i try to substitute different values of R1 and R2 in: Vin/Vout=R2/(R1+R2), the closest i can get to is 15/(15+82)=0.154. Any help? I spent an entire day on this; first trying to understand the question and then not getting the correct answer. UserIdTAG: 395953 UserNameTAG: dustinge CreateTimeTAG: 2012-09-23T19:02:45Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: dustinge, notice that you can also multiply the resistors by multiples of 10. Have you tried that? I recommend you to first find an appropriate pair for getting the output voltage, and then make the necessary adjustments to get the correct Rth (hint: For the second part, keep in mind the possibility of multiplying by 10, 100, 1000, etc.) FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-23T22:54:51Z SecondChildTAG: never mind, too late for that! SecondChildUserIdTAG: 395953 SecondChildUserNameTAG: dustinge SecondChildCreateTimeTAG: 2012-09-24T17:18:29Z IndexTAG: 3954 TitleTAG: Please Help me with H3P4 Please give me a tip to solve H3P4.Whatever i try is wrong. Thank you. UserIdTAG: 38331 UserNameTAG: janadel CreateTimeTAG: 2012-09-23T18:55:53Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Observe the characteristics of the diode and think, if i apply certain input (lets say the positive semicycle), how is the behavior of the diode?(What equation is satisfied), observe the graph of the diode again ¿Can you replace the diode for another thing?? (resistor, open switch, conductor, etc), i hope that will help you. Remember that the answers are easy and simple. FirstChildUserIdTAG: 370343 FirstChildUserNameTAG: JJarquin FirstChildCreateTimeTAG: 2012-09-23T20:02:45Z FirstChildTAG: Thanks .You helped a lot. FirstChildUserIdTAG: 38331 FirstChildUserNameTAG: janadel FirstChildCreateTimeTAG: 2012-09-23T21:30:38Z IndexTAG: 3955 TitleTAG: Please Help me out with Lab 3 I am having problems in designing a circuit which can perform the fifth line of the truth table i.e. 1 0 0 | 1 function. I can however obtain the same transient response when the output is zero 0 for this the same combination of inputs 1 0 0. C B A | Z ========= 0 0 0 | 1 0 0 1 | 1 0 1 0 | 1 0 1 1 | 1 1 0 0 | 1 1 0 1 | 0 1 1 0 | 0 1 1 1 | 0 Thanks for responding UserIdTAG: 423976 UserNameTAG: Sirbrevis CreateTimeTAG: 2012-09-23T18:40:26Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3956 TitleTAG: Clarification request Every grey rectangle is a switch? We only got x and y but 4 switches in the circuit? Why are the circuit a and b here? UserIdTAG: 388555 UserNameTAG: 4lk4tr43 CreateTimeTAG: 2012-09-23T18:13:59Z VoteTAG: 0 CoursewareTAG: Week 3 / Switch Model Exercise CommentableIdTAG: 6002x_switch_model_exercise NumberOfReplyTAG: 2 FirstChildTAG: Every grey rectangle is a switch? - Yes We only got x and y but 4 switches in the circuit? - Yes Why are the circuit a and b here? - Maybe for example? Or tip ) FirstChildUserIdTAG: 394836 FirstChildUserNameTAG: v2g6ch4 FirstChildCreateTimeTAG: 2012-09-23T18:38:32Z SecondChildTAG: Ah I see its a tip, I used it and dint even notice. So the circuit c can basically be written as !(!((!X)||Y)) or !X||Y SecondChildUserIdTAG: 388555 SecondChildUserNameTAG: 4lk4tr43 SecondChildCreateTimeTAG: 2012-09-23T20:15:01Z FirstChildTAG: yes you can see that the input of the second switch is th output of the first switch FirstChildUserIdTAG: 236136 FirstChildUserNameTAG: Hamid-ch FirstChildCreateTimeTAG: 2012-09-23T21:04:04Z IndexTAG: 3957 TitleTAG: S6V5 Week 3 Could some kind person please explain why at 3mins 9secs into this video where on earth 0.32 comes from the equation. UserIdTAG: 377602 UserNameTAG: Goby CreateTimeTAG: 2012-09-23T17:57:01Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: It is the result of $1V-1\Omega\cdot\frac{1}{4}Ae^{(1V\cdot V^-1)}$. He tries to solve the equation by guessing $v_D$. He uses $1V$ as a first guess (it could have been anything) and solve the right side of the equation. He then finds $\text{left} \neq \text{right}$, it is not the good value of $v_d$. He continues the process until the values match. FirstChildUserIdTAG: 396446 FirstChildUserNameTAG: RousseauxS FirstChildCreateTimeTAG: 2012-09-23T18:03:57Z SecondChildTAG: Its the actual equation that has me stumped...I can't get 0.32 from it, maybe I need to go to sleep and wake up fresh tomorrow ! SecondChildUserIdTAG: 377602 SecondChildUserNameTAG: Goby SecondChildCreateTimeTAG: 2012-09-23T18:58:58Z SecondChildTAG: 1-e^(1v)/4 = 0.3204295428852 SecondChildUserIdTAG: 370343 SecondChildUserNameTAG: JJarquin SecondChildCreateTimeTAG: 2012-09-23T20:17:59Z SecondChildTAG: And thats it right there...the "e"....what is that ? what is its value ? Sorry...my algebra isn't that good in my old age :( SecondChildUserIdTAG: 377602 SecondChildUserNameTAG: Goby SecondChildCreateTimeTAG: 2012-09-23T21:03:02Z SecondChildTAG: e = Euler's number $\cong$ 2.718 SecondChildUserIdTAG: 366165 SecondChildUserNameTAG: silicon_ghost SecondChildCreateTimeTAG: 2012-09-24T04:22:54Z SecondChildTAG: AHA !!! :) SecondChildUserIdTAG: 377602 SecondChildUserNameTAG: Goby SecondChildCreateTimeTAG: 2012-09-24T06:29:25Z SecondChildTAG: Thank you very very much SecondChildUserIdTAG: 377602 SecondChildUserNameTAG: Goby SecondChildCreateTimeTAG: 2012-09-24T06:30:00Z IndexTAG: 3958 TitleTAG: how does the noise effect voltage in circuit I have trouble understanding how can longitude pressure wave i.e. sound effect voltage at node which depends on electromagnetic wave? Can anybody explain it? UserIdTAG: 230632 UserNameTAG: shubhajeet CreateTimeTAG: 2012-09-23T17:46:38Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: shubhajeet, I don't understand your question, are you trying to see the effect of sound to voltage? Usually when we say "noise" affecting a signal we refer to electromagnetic noise, not sound. FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-23T22:14:57Z IndexTAG: 3959 TitleTAG: homework2 found the values of r1 and r2 that satisfy all the conditions and in E12 set but apparently its incorrect.when my friends checked it with the same value its correct for them..HOW??? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-23T17:18:41Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: They give different questions to you, to keep you from sharing your answers with your friends. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-23T17:35:12Z SecondChildTAG: it is possible .notice R1=3R2 is required condition.there are many such values in E12 set. SecondChildUserIdTAG: 357747 SecondChildUserNameTAG: kishores SecondChildCreateTimeTAG: 2012-09-23T17:56:34Z SecondChildTAG: how do you get this condition?? the relation that I get is R1=(7/3)R2... and next I searched the values in the E12 set. I get a correct answer.. SecondChildUserIdTAG: 86179 SecondChildUserNameTAG: aldajo92 SecondChildCreateTimeTAG: 2012-09-23T18:56:59Z IndexTAG: 3960 TitleTAG: using lab tools m not able to get the connecting wires even after clicking on the connection points..plz help me as soon as possible...plzzz UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-23T17:10:01Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 0 IndexTAG: 3961 TitleTAG: lab problem m not able to get the connecting wires even after clicking on the connection points.please help me to solve this problem as soon as possible. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-23T17:06:48Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Same problem here!!! FirstChildUserIdTAG: 355420 FirstChildUserNameTAG: parkavi FirstChildCreateTimeTAG: 2012-09-23T17:12:20Z SecondChildTAG: TRY..USING ANOTHER BROWSER SecondChildUserIdTAG: 137918 SecondChildUserNameTAG: jaisneha SecondChildCreateTimeTAG: 2012-09-23T18:47:22Z SecondChildTAG: Try and hold the mouse key not on the connector but little bit above or side ways and mouse (make sure component is not turned green)and move the mouse to desired direction you will get a green line attached to the connector and drag it to the next connector. SecondChildUserIdTAG: 272523 SecondChildUserNameTAG: Jivraj SecondChildCreateTimeTAG: 2012-09-23T20:58:17Z IndexTAG: 3962 TitleTAG: Other signal systems than digital ones I've just watched "S4V4: Why Digital" right now and I'm curious: why don't we use in the real world other signals then digital ones? For instance, I don't see any problem in considering a voltage source that emits 0V, 5V, 10V and 15V signals and, in presence of noise, read them as "0" for [-2.5V, 2.5V], "1" for (2.5V, 7.5V], "2" for (7.5V, 12.5V] and "3" for (12.5V, 17.5V]. Where's the flaw? Thanks in advance! =) UserIdTAG: 288637 UserNameTAG: gotchi CreateTimeTAG: 2012-09-23T17:05:54Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: real world talk in analog language which is continue and digital world talk in binary which is discreet. this two has well develop right now and to create something new on base we need to be sure the advantage of it compare to existing one if there isn't or the same then we are taking step back down. the flaw, which I can think, it need higher volt which mean need more power supply. digital can be as low as 5 volt, FirstChildUserIdTAG: 452559 FirstChildUserNameTAG: kuz1toro FirstChildCreateTimeTAG: 2012-09-23T19:34:06Z SecondChildTAG: Hi kuz1toro! It isn't a problem to find a use to such a "machine". Think about nucleobases: Adenine (A), Cytosine (C), Guanine (G) and Thymine (T). We could sequence a DNA sample using it! In fact, I'm quite sure there is some kind of DNA sequencer that does this, but using light and some kind of chemical reagent instead of electrical signals. Besides that, I'm not sure if a change from 5.0V to 15.0V voltage source to a machine is a great problem, once my pretty standard notebook works at 19.5V. I would appreciate some more opinions. Thanks. SecondChildUserIdTAG: 288637 SecondChildUserNameTAG: gotchi SecondChildCreateTimeTAG: 2012-09-24T01:31:46Z SecondChildTAG: I'm a platform developer, not course staff (just want to clarify since I have the Staff tag), but I have taken several courses on analog and digital electronics at Berkeley and was in fact a TA for a similar course there, so sometimes I can't resist jumping in here :) One major issue is that now you've got 3 transition points, and thus many more noise margins to consider. I think you have covered noise margins in 6.002x so that should give you a starting point to understand. First and foremost, as a CPU designer you need to make sure your circuit will function as desired, and it becomes harder to guarantee that when you have to worry about the noise margin when transitioning from 0->5, 5->0, 5->10, etc... Consider that a company like Intel fabricates millions of CPUs, and they need to make sure that a high percentage of them work as designed. More room for error is generally not a good idea in engineering, especially in the CPU business where startup costs are extremely high and marginal costs are nearly nonexistant (i.e. to make one CPU from scratch will cost you millions of dollars, but to make 100,000 the cost would not be much more). Also, while your notebook itself may run at 19.5V, the CPU inside your notebook likely runs on something between 1.5V-2V. Power consumed is roughly speaking $$\alpha fC_{L}V_{DD}^2$$ where $\alpha$ is the probability of switching, $f$ is the clock frequency, $C_L$ is the load capacitance of your circuit (proportional to how many transistors you have and how big they are, roughly speaking), and $V_{DD}$ is the supply voltage (see Chapter 11 of your book for more, although the terminology there is a bit different). As you can see the voltage term is squared in the power estimate, so a twofold increase in voltage results in a 4-fold increase in power consumed. Thus, you can see why raising the supply voltage is generally a bad idea. If you're curious about how supply voltages have changed over time, [wikipedia](http://en.wikipedia.org/wiki/Logic_family#Lowering_the_power_supply_voltage) has a brief overview of it. These days with computer performance being more or less sufficient for 90% of users, the main area of competition is really power consumption, and in the case of mobile devices, battery life. As such the general trend has been to push down supply voltages and use dynamic voltage and frequency scaling techniques to save battery life as much as possible. And with smaller and smaller supply voltages, noise margins become tighter and tighter. Your laptop may take in 19.5V from the power supply, but that voltage is then stepped down in many parts of your laptop depending on what voltage that specific part needs. And if different circuits use different voltages, there may be level shifters to allow communication between different circuits with different logic levels. There is really no one voltage that your laptop uses. This is a very common question that a lot of students ask, and you're very clever for thinking of it, but at the end of the day, it's not very practical and offers little real benefit. SecondChildUserIdTAG: 323387 SecondChildUserNameTAG: ibrahimawwal SecondChildCreateTimeTAG: 2012-09-24T06:11:45Z SecondChildTAG: Thanks a lot, Ibrahim. That was really a complete answer to my question! SecondChildUserIdTAG: 288637 SecondChildUserNameTAG: gotchi SecondChildCreateTimeTAG: 2012-09-24T11:31:54Z IndexTAG: 3963 TitleTAG: lab 2 problem can anybody just help me with lab 2 .submission time is near but i was nt able to solve it UserIdTAG: 391854 UserNameTAG: ashishsmvdu CreateTimeTAG: 2012-09-23T16:13:51Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi Ashish, Could you explore the problem? FirstChildUserIdTAG: 372623 FirstChildUserNameTAG: Amitraj FirstChildCreateTimeTAG: 2012-09-23T16:19:20Z SecondChildTAG: Not sure, where you are stuck, but following thread seems to be resolving for people - https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/505ec30cc162402b0000003f All the best. SecondChildUserIdTAG: 372623 SecondChildUserNameTAG: Amitraj SecondChildCreateTimeTAG: 2012-09-23T16:21:23Z SecondChildTAG: thanks amitraj......at last got the concept SecondChildUserIdTAG: 391854 SecondChildUserNameTAG: ashishsmvdu SecondChildCreateTimeTAG: 2012-09-23T16:33:56Z SecondChildTAG: that realy works.. SecondChildUserIdTAG: 118756 SecondChildUserNameTAG: abdulrazzaq789 SecondChildCreateTimeTAG: 2012-09-23T16:55:29Z SecondChildTAG: Just compare the 2 equations given and get a relationship between R1 and R2.Then consider a 3rd resistor to divide the voltage between V0 and ground.Hope that will help you work. SecondChildUserIdTAG: 429851 SecondChildUserNameTAG: ssembajjwe SecondChildCreateTimeTAG: 2012-09-23T17:05:33Z IndexTAG: 3964 TitleTAG: help for homework4-zener diod. everybody:please say me,how calculate noise voltage in output terminals? UserIdTAG: 153317 UserNameTAG: hdjt CreateTimeTAG: 2012-09-23T16:09:43Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: It is not too difficult. The graph of the Zener response gives a V/A ratio in the conducting area. Calculate the current that the noise voltage will generate through the Zener, and then you can calculate the voltage at the output terminals. FirstChildUserIdTAG: 270011 FirstChildUserNameTAG: SnowmanZA FirstChildCreateTimeTAG: 2012-09-23T20:24:45Z IndexTAG: 3965 TitleTAG: having a bit of a problem finding i1(y1) i1=VR1/R1... VR1= R1/((R1+Rx))V1.... is this the right formula for finding the ampere? UserIdTAG: 133084 UserNameTAG: Warrensiggs CreateTimeTAG: 2012-09-23T16:02:35Z VoteTAG: 0 CoursewareTAG: Week 2 / Superposition example CommentableIdTAG: 6002x_S3E3_superposition NumberOfReplyTAG: 1 FirstChildTAG: Yes! However make sure you are looking at the current direction of your answer (is it positive or negative?).. although you can define your positive current direction 2 different ways, the site will only accept one standard as an answer. Also, you can save yourself by time if you use your answers with the voltages. If you know V1, and you know x1 [V(R2//R3 as caused by V1)], then you can very simply find VR1 as caused by V1 simply by.. VR1 = V1 - x1 This works because the sum VR1+x1 = V1, if considering V2 a short cct. FirstChildUserIdTAG: 3571 FirstChildUserNameTAG: ifoughtsharks FirstChildCreateTimeTAG: 2012-10-03T19:25:57Z IndexTAG: 3966 TitleTAG: Confused on example 3.19 of textbook (or perhaps, "What is a dependent source") I am thoroughly confused on example 3.19, figure 3.53 of textbook. Why does setting v2=0 imply i2=0? I suspect that the root of my problem is I don't understand what is a "dependent source" - in this case a "dependent current source". I recall independent current source from week 1. So my questions: 1. What is dependent source (textbook or lecture reference?) 2. Why does v2=0 imply i2=0 in example 3.19? Dave P.S. When is ability to follow posts coming back? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-23T15:59:01Z VoteTAG: 0 CoursewareTAG: Week 2 / Linearity CommentableIdTAG: 6002x_linearity NumberOfReplyTAG: 1 FirstChildTAG: it is related to component like transistor or mosfet, this creature have that behaviour, the voltage or current that cross or flow on two of their terminal is controlled by or function other terminal. in example 3,19, i2 is function of v2, i2 = g.v2, you will get the big picture as you proceed... wright now you just need to understand the technique to analysis it FirstChildUserIdTAG: 452559 FirstChildUserNameTAG: kuz1toro FirstChildCreateTimeTAG: 2012-09-23T18:40:19Z IndexTAG: 3967 TitleTAG: h2p1 cant solve by any one , ha i am still have a problem , i see all discussion , but all have confuse ,what we do no, plz any one have easy method ,so till us ,dont make more confussion ,for god sake UserIdTAG: 133891 UserNameTAG: junaidkarim CreateTimeTAG: 2012-09-23T15:58:33Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi Junaid, Just responded to your earlier post - https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/505ef9500dabf42300000011 It is surely a tricky question.... please go through the steps enlsited in the above post. Take care and all the very best .. :) FirstChildUserIdTAG: 372623 FirstChildUserNameTAG: Amitraj FirstChildCreateTimeTAG: 2012-09-23T16:12:13Z IndexTAG: 3968 TitleTAG: STAFF HELP NEEDED i tried to solve home work 3...i solved some problems...but i think the way of the following question which i solved is right..but i got the answer is wrong...may you please check whether the answer declared by you is right? the ques is **How about the NOR gate of this family. What is the minimum value of the pullup resistor RPuO (in Ohms) for which this inverter can obey the required static discipline?** UserIdTAG: 118611 UserNameTAG: mitianhari CreateTimeTAG: 2012-09-23T15:24:05Z VoteTAG: 0 CoursewareTAG: Week 3 / Circuit Truth Table Tutorial CommentableIdTAG: 6002x_circuit_truth_table_tutorial NumberOfReplyTAG: 0 IndexTAG: 3969 TitleTAG: Is a resistor linear? Yes UserIdTAG: 384271 UserNameTAG: GEORGEE CreateTimeTAG: 2012-09-23T15:08:09Z VoteTAG: 0 CoursewareTAG: Week 2 / Is it linear? CommentableIdTAG: 6002x_is_it_linear NumberOfReplyTAG: 3 FirstChildTAG: don't overlook the video in week 1 covering this topic *glowingvegetables* ;) FirstChildUserIdTAG: 131747 FirstChildUserNameTAG: hasi FirstChildCreateTimeTAG: 2012-09-23T15:15:49Z FirstChildTAG: linearity is always a myth... it is only within the play ground and with the set of rules... that the device is really linear... :) FirstChildUserIdTAG: 257806 FirstChildUserNameTAG: skoda FirstChildCreateTimeTAG: 2012-09-23T15:28:38Z FirstChildTAG: Real resistors, used in electronic devices is not linear (nearly linear, but not full), resistors in EECS playground are linear. FirstChildUserIdTAG: 104522 FirstChildUserNameTAG: IgorNovice FirstChildCreateTimeTAG: 2012-09-23T16:03:04Z IndexTAG: 3970 TitleTAG: How to rotate a device in LAB2 I don't know which key can make a device (i.e. resistance) rotate, can anyone help me out? Thanks. UserIdTAG: 298314 UserNameTAG: js93082011 CreateTimeTAG: 2012-09-23T14:56:03Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: it's simple 'r' on your keyboard. FirstChildUserIdTAG: 131747 FirstChildUserNameTAG: hasi FirstChildCreateTimeTAG: 2012-09-23T15:19:20Z SecondChildTAG: thanks vary good SecondChildUserIdTAG: 153317 SecondChildUserNameTAG: hdjt SecondChildCreateTimeTAG: 2012-09-23T16:12:46Z FirstChildTAG: just press R after u have selected the device FirstChildUserIdTAG: 479321 FirstChildUserNameTAG: TarunVarma FirstChildCreateTimeTAG: 2012-09-23T15:19:20Z FirstChildTAG: Thanks a lot, got it. FirstChildUserIdTAG: 298314 FirstChildUserNameTAG: js93082011 FirstChildCreateTimeTAG: 2012-09-23T15:32:37Z IndexTAG: 3971 TitleTAG: question 2 and 4 I have my solve for those question like this, but i don't know whether it were true or not. Suppose that we have 2 signal A, B. There are 4 distinct value that **(1*1)** **(1*0)** **(0*1)** **(0*0)**, which * may be AND or OR.... We have 4 value, change the place of A and B: **(A*B)--> (B*A)** we have 8 value, but there are **2 function AND and OR** so we have **16** boolean-valued function as answer. In case three signal A,B and C. Let D=(A*B) then D*C because D=0 or 1 we have the answer 8 for question 3, and with the same techniques, we have **16*16=256** for the question 4. Do I have any mistakes? UserIdTAG: 471408 UserNameTAG: SplendorVN CreateTimeTAG: 2012-09-23T14:23:39Z VoteTAG: 0 CoursewareTAG: Week 2 / Boolean functions CommentableIdTAG: 6002x_S4E2_Boolean_Functions NumberOfReplyTAG: 1 FirstChildTAG: why haven't you considered NOT gate? FirstChildUserIdTAG: 357747 FirstChildUserNameTAG: kishores FirstChildCreateTimeTAG: 2012-09-23T16:02:51Z IndexTAG: 3972 TitleTAG: Arithmetic signsmi Please follow the equation as suggested by a respondent (read through), be MINDFULL about your signs ie variable and actual value. Thanks UserIdTAG: 320661 UserNameTAG: AbbaHanna CreateTimeTAG: 2012-09-23T14:04:12Z VoteTAG: 0 CoursewareTAG: Week 2 / Node equation review CommentableIdTAG: 6002x_S3E1_NodeEquationReview NumberOfReplyTAG: 0 IndexTAG: 3973 TitleTAG: H2P1 I'm not able to find the right value of resistances for first que..can anybody help me?? UserIdTAG: 230891 UserNameTAG: abhigupta CreateTimeTAG: 2012-09-23T14:02:44Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hello Abhi, You can follow this thread - https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/505f01c789b8882700000009 FirstChildUserIdTAG: 372623 FirstChildUserNameTAG: Amitraj FirstChildCreateTimeTAG: 2012-09-23T15:40:54Z IndexTAG: 3974 TitleTAG: lab 3 I deleted the resistance of the lab3, what do I do? UserIdTAG: 327787 UserNameTAG: REINALDOPARANHOS CreateTimeTAG: 2012-09-23T13:58:39Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: try resetting Lab 3 FirstChildUserIdTAG: 357747 FirstChildUserNameTAG: kishores FirstChildCreateTimeTAG: 2012-09-23T15:40:03Z SecondChildTAG: It is 10kiloohms by default. But do help me if you get to know how to find the correct W/L SecondChildUserIdTAG: 97340 SecondChildUserNameTAG: AnkitRana SecondChildCreateTimeTAG: 2012-09-25T15:01:02Z SecondChildTAG: need help in lab3 SecondChildUserIdTAG: 403493 SecondChildUserNameTAG: aiai SecondChildCreateTimeTAG: 2012-09-25T17:32:15Z SecondChildTAG: i can't see lectures as youtube is blocked so i need your help plz SecondChildUserIdTAG: 403493 SecondChildUserNameTAG: aiai SecondChildCreateTimeTAG: 2012-09-25T17:33:21Z SecondChildTAG: Check [this][1] [1]: http://www.elsevierdirect.com/companions/9781558607354/casestudies/06~Chapter_6/Section_6_11.pdf SecondChildUserIdTAG: 184827 SecondChildUserNameTAG: DiegoT SecondChildCreateTimeTAG: 2012-09-28T23:16:49Z IndexTAG: 3975 TitleTAG: Can anyone help me? iam unable to solve this problem.... UserIdTAG: 190150 UserNameTAG: deepusandy CreateTimeTAG: 2012-09-23T13:54:10Z VoteTAG: 0 CoursewareTAG: Week 2 / Norton method example CommentableIdTAG: 6002x_S3E6_Norton_Model NumberOfReplyTAG: 1 FirstChildTAG: FOR 1.Rn open circuit the current source and measure resistance across the two required terminals. 2.In short the ends of required terminals.Using nodal analysis find voltage across the two terminals and then find current across the short. FirstChildUserIdTAG: 357747 FirstChildUserNameTAG: kishores FirstChildCreateTimeTAG: 2012-09-23T15:48:54Z IndexTAG: 3976 TitleTAG: Can you get Rth by measuring the current? Hi, In this video, he says that to measure $R_{th}$ you turn off the voltage sources and measure the resistance of the network. Could you also measure the short circuit current coming out and going back into the Thevenin equivalent network? Then you could get $R_{th}$ from the measured $i$, $V_{th}$, and Ohm's law, correct? thanks, Rob UserIdTAG: 468623 UserNameTAG: RobNik CreateTimeTAG: 2012-09-23T13:41:49Z VoteTAG: 0 CoursewareTAG: Week 2 / Thevenin method CommentableIdTAG: 6002x_thevenin NumberOfReplyTAG: 0 IndexTAG: 3977 TitleTAG: jaymin current sign convention is opposite here?/ UserIdTAG: 367877 UserNameTAG: jaymin1328 CreateTimeTAG: 2012-09-23T13:37:05Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: your voltage source is an AC source.I think the current direction is at a particular instant. FirstChildUserIdTAG: 357747 FirstChildUserNameTAG: kishores FirstChildCreateTimeTAG: 2012-09-23T16:00:13Z IndexTAG: 3978 TitleTAG: H2P2: SOLAR POWER dead end ? Hi there. I'm trying to get the resistance RL by Thevenin method, but apparently i don't understand something. Always ends that I need the value of current which is with unknown RL. I enclose a picture. Any ideas ? ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13484065721708326.jpg UserIdTAG: 427347 UserNameTAG: Aminds CreateTimeTAG: 2012-09-23T13:24:50Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: @Aminds you are going in right direction....find the power dissipated by RL(load resistor) using i^2*r........you know all values except RL....... for max. power, use maxima which we use in calculus.......... go ahead..... all the best FirstChildUserIdTAG: 114864 FirstChildUserNameTAG: Vikaash FirstChildCreateTimeTAG: 2012-09-23T13:32:50Z SecondChildTAG: I can't get it right :P. Maximum power wile be VTH^2/RTH ? SecondChildUserIdTAG: 427347 SecondChildUserNameTAG: Aminds SecondChildCreateTimeTAG: 2012-09-23T14:09:42Z SecondChildTAG: Sorry i get You wrong, You was saying that i can calculate i ? But how ? SecondChildUserIdTAG: 427347 SecondChildUserNameTAG: Aminds SecondChildCreateTimeTAG: 2012-09-23T14:16:57Z SecondChildTAG: OK i have it thanks for Your help :) SecondChildUserIdTAG: 427347 SecondChildUserNameTAG: Aminds SecondChildCreateTimeTAG: 2012-09-23T14:30:26Z SecondChildTAG: could u pls explain a bit how you get it? SecondChildUserIdTAG: 329525 SecondChildUserNameTAG: udhayaraj12 SecondChildCreateTimeTAG: 2012-09-23T14:40:49Z SecondChildTAG: Hint - Think of Rp and the two Rs as ONE resistor - Thevenin Resistance, RL is another resistor. You got two resistors in parallel. What is the power relationship in parallel resistors? SecondChildUserIdTAG: 27898 SecondChildUserNameTAG: adebon SecondChildCreateTimeTAG: 2012-09-23T15:32:21Z FirstChildTAG: There is a maximum power transfer theorem, which you can probably use.. it states that maximum power is transferred from source to load when source resistance and load resistance are the same... so the Thevenin resistance should be your RL.. FirstChildUserIdTAG: 257806 FirstChildUserNameTAG: skoda FirstChildCreateTimeTAG: 2012-09-23T15:33:50Z IndexTAG: 3979 TitleTAG: about week 2 homework plz can you help me with the H2P1 and H2P2 of week one. i am having great difficulty in doing them UserIdTAG: 413613 UserNameTAG: shaliesh CreateTimeTAG: 2012-09-23T12:34:15Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: Hello there, Let me enlist the 'Givens' here (so just exploring the question )- 1. Regarding the Rth of the ckt. (which is a voltage divider) - 10KOhm <= Rth <= 30KOhm 2. Vopen = 15V (approx) 3. Vin = 60V 4. Vin/Vout = 4 (approx) but acc/to the question this approximation should be between +- 10% of 15. That gives us the min and max of the Vout we are allowed to get using our values of R1 and R2. 5. R1 and R2 are selected with 10% tolerance. Question is to calculate R1 and R2 and then max Vout and min Vout. With this, following clues can be used - 1. First - Realise what is Rth for a voltage divider. 2. Once Rth is understood you get the relationship between R1 and R2. 3. Once R1 and R2 are calculated (which I suppose is not going to be same amongst us all), using the 10% tolerance, calculate the maximum and minimum Vout (for extremes of R1 and R2). Just tried to simplify the question (which I myself took much time to understand). All the best. FirstChildUserIdTAG: 372623 FirstChildUserNameTAG: Amitraj FirstChildCreateTimeTAG: 2012-09-23T13:05:02Z SecondChildTAG: I have a similar issue but, here I have Vin=20 & Vout=6 Simply I think Vin\Vout=R2\(R2+R1) & Rth=(R2*R1)\(R2+R1) But I never Got the answer. SecondChildUserIdTAG: 106816 SecondChildUserNameTAG: Laith SecondChildCreateTimeTAG: 2012-09-23T13:52:24Z SecondChildTAG: I think that Rth=(R1*R2)/(R1+Rb). Let me know if you ever figure it out; I'm as confused as you. SecondChildUserIdTAG: 358858 SecondChildUserNameTAG: alice144 SecondChildCreateTimeTAG: 2012-09-23T14:14:20Z SecondChildTAG: Is it correct to consider an arbitrary value for Rth (comprehended in the class of value from 10k to 30k) for calculating the value of R1 and R2, knowing the relations: Rth=R1//R2 Vout/Vin=R2/(R1+R2)?? SecondChildUserIdTAG: 105018 SecondChildUserNameTAG: ema_cuo SecondChildCreateTimeTAG: 2012-09-23T14:57:40Z FirstChildTAG: Guys, I think we all have the same equations but I am not sure how to solve towards them. The problem comes because Rth doesn't have a specific value but 10k<=Rth<=30k. I am not sure how to solve for it... FirstChildUserIdTAG: 435193 FirstChildUserNameTAG: ManosP FirstChildCreateTimeTAG: 2012-09-23T15:08:43Z SecondChildTAG: I agree with you... If I'll find a solution I'll explain it SecondChildUserIdTAG: 105018 SecondChildUserNameTAG: ema_cuo SecondChildCreateTimeTAG: 2012-09-23T15:10:08Z SecondChildTAG: Ok friends, I've found the values of R1 and R2 just regarding the medium value of the range in which could vary the Rth. Than, I've just selected the nearest values to those obtained by calculation, among the values commercially available. SecondChildUserIdTAG: 105018 SecondChildUserNameTAG: ema_cuo SecondChildCreateTimeTAG: 2012-09-23T15:20:07Z FirstChildTAG: Dear all, Just resolved the problem. A few points below - In addition to all the assumptions and conclusions, we have to keep in mind to use values from E12 set, strictly, given at the very start of question. For example, I came across atleast 2 different pairs of R1 and R2 values, which had solved the problem in all respects, except that these values were not from the E12 set. Hence, sometimes, even if Vmax and Vmin get resolved, even if Rth is within the expected range, the software does not assume it to be a correct answer. My statements above are based on experience first, verified by the question itself. Please let me know if we still need to explore this further. All the best. FirstChildUserIdTAG: 372623 FirstChildUserNameTAG: Amitraj FirstChildCreateTimeTAG: 2012-09-23T15:22:17Z SecondChildTAG: What you put the value of vout/vin. I am putting 8/20=0.4 and sloved using above method but my answer is not accepted although calculations meet the requirements given in questions are being met. SecondChildUserIdTAG: 4620 SecondChildUserNameTAG: mabdullah169 SecondChildCreateTimeTAG: 2012-09-23T16:03:43Z SecondChildTAG: Are the values from E12 set, given at the start of question? Even if your values satisfy all the requirements, **they should be present in the E12 set**. Please confirm once. Please let me know if any help needed further. All the best. SecondChildUserIdTAG: 372623 SecondChildUserNameTAG: Amitraj SecondChildCreateTimeTAG: 2012-09-23T16:14:50Z SecondChildTAG: Thanks for the help. I was doing a silly mistake and the problem is solved now. SecondChildUserIdTAG: 4620 SecondChildUserNameTAG: mabdullah169 SecondChildCreateTimeTAG: 2012-09-23T16:27:10Z FirstChildTAG: How do you get R1 and R2? FirstChildUserIdTAG: 150641 FirstChildUserNameTAG: DungTienTran FirstChildCreateTimeTAG: 2012-09-23T15:39:00Z SecondChildTAG: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/505ef9500dabf42300000011 SecondChildUserIdTAG: 372623 SecondChildUserNameTAG: Amitraj SecondChildCreateTimeTAG: 2012-09-23T16:16:47Z FirstChildTAG: One last comment.... You will need to input both values of R1 and R2 or it will not accept a single answer as right even if it is so. I thought of saving myself some time and not calculate R2 until I had a valid R1...Very bad move! Eventually I got I included both values and it passed FirstChildUserIdTAG: 435193 FirstChildUserNameTAG: ManosP FirstChildCreateTimeTAG: 2012-09-23T15:56:02Z IndexTAG: 3980 TitleTAG: problem in reference point the posted answer is correct only if the reference point is taken in between 'R2' and voltage source 'V2'. as we can see if e=6.2 then through R1 the current must be in anticlockwise direction, but at the same time current through R2 be in clockwise... which is itself contradictory. UserIdTAG: 480957 UserNameTAG: rishikantsharan CreateTimeTAG: 2012-09-23T12:23:03Z VoteTAG: 0 CoursewareTAG: Week 2 / Node equation review CommentableIdTAG: 6002x_S3E1_NodeEquationReview NumberOfReplyTAG: 1 FirstChildTAG: I think in both cases the current will be anti-clockwise. Notice that V2 is negative and it is also the positive terminal is pointing downwards. FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-23T23:13:48Z SecondChildTAG: That's for pointing that out. Made a world a difference! SecondChildUserIdTAG: 194509 SecondChildUserNameTAG: blackcompe SecondChildCreateTimeTAG: 2012-09-24T00:13:26Z SecondChildTAG: Thanks* SecondChildUserIdTAG: 194509 SecondChildUserNameTAG: blackcompe SecondChildCreateTimeTAG: 2012-09-24T00:13:38Z IndexTAG: 3981 TitleTAG: h2p1 help me urgent plz ha i see all discussion , but i cant understand what u say , plz be clear explain ,what my first step, and after that my next step, plz give me little clue ,so i can do it UserIdTAG: 133891 UserNameTAG: junaidkarim CreateTimeTAG: 2012-09-23T11:58:08Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Hello there, Let me enlist the 'Givens' here (*so just exploring the question* )- **1.** Regarding the Rth of the ckt. (which is a voltage divider) - 10KOhm <= Rth <= 30KOhm **2.** Vopen = 15V (approx) **3.** Vin = 60V **4.** Vin/Vout = 4 (approx) but acc/to the question this approximation should be between +- 10% of 15. That gives us the min and max of the Vout we are allowed to get using our values of R1 and R2. **5.** R1 and R2 are selected with 10% tolerance. **Question is to calculate R1 and R2 and then max Vout and min Vout.** With this, following clues can be used - **1.** First - Realise what is Rth for a voltage divider. **2.** Once Rth is understood you get the relationship between R1 and R2. **3.** Once R1 and R2 are calculated (which I suppose is not going to be same amongst us all), using the 10% tolerance, calculate the maximum and minimum Vout (for extremes of R1 and R2). Just tried to simplify the question (which I myself took much time to understand). All the best. FirstChildUserIdTAG: 372623 FirstChildUserNameTAG: Amitraj FirstChildCreateTimeTAG: 2012-09-23T13:01:46Z SecondChildTAG: plz till me how i found r1 ,and r2, to calculated ,what shold i do, simple explain me plz SecondChildUserIdTAG: 133891 SecondChildUserNameTAG: junaidkarim SecondChildCreateTimeTAG: 2012-09-23T15:50:39Z FirstChildTAG: how do you get R1 and R2 FirstChildUserIdTAG: 150641 FirstChildUserNameTAG: DungTienTran FirstChildCreateTimeTAG: 2012-09-23T15:37:46Z SecondChildTAG: that a main problem ,Vout/Vin= 0.4, max is 0.44, mim is 0.36, after that what i do now , Rth is 10k, so its vdr, what then, SecondChildUserIdTAG: 133891 SecondChildUserNameTAG: junaidkarim SecondChildCreateTimeTAG: 2012-09-23T15:45:53Z SecondChildTAG: Hi. Try following - 1. For a while, forget about the tolerances and approximations. 2. The circuit is clearly a Voltage Divider. Hence, write down the relationship between Vin and Vout (for a voltage divider). 3. We already know Vin and Vout. Hence using the equation and these known values, we know the relationship between R1 and R2. 4. Select a value for R1 or R2 from E12 (at the start of question). 5. Using the relation found in 3 above, find R2. 6. Now verify if these values satisfy other constraints in the Question. If not, try another value for R1 or R2 from the table. **An observation -** It seems, once you select the correct value (from E12) for R1 (or R2) the other value automatically comes to be from the E12 set. So we, perhaps, need no select both R1 and R2. One has to select only one of those. Please follow E12... that is also one of the conditions. All the best. SecondChildUserIdTAG: 372623 SecondChildUserNameTAG: Amitraj SecondChildCreateTimeTAG: 2012-09-23T16:08:48Z SecondChildTAG: the relation obtained by me was R1=4R2. plzz correct me if i am wrong. SecondChildUserIdTAG: 365792 SecondChildUserNameTAG: pabhijeet SecondChildCreateTimeTAG: 2012-09-23T17:07:17Z SecondChildTAG: I found a combination by writing a simple brute force python program, it does the following: 1.) Find all combinations of resistors up to 10^5. 2.) Filter results by Vout > 9.5 and < 10.5 with tolerance check 3.) Filter those by Rth > 10000 and < 30000. Although this may not be the most elegant way to solve for R1 and R2, but it works. SecondChildUserIdTAG: 355773 SecondChildUserNameTAG: Albright4edx SecondChildCreateTimeTAG: 2012-09-23T17:18:07Z FirstChildTAG: it is given that Vout = 7 volt, Vin = 20 volt and vot/vin < 10% then how you are writing Vout/Vin =.4 ; Vin = 60 and Vout= 15? please tell me FirstChildUserIdTAG: 227508 FirstChildUserNameTAG: bhavyab FirstChildCreateTimeTAG: 2012-09-23T21:16:30Z IndexTAG: 3982 TitleTAG: H2P2.i m struck :( HOW TO CALCULATE THE MAXIMUM POWER IN THIS PROBLEM..I HAVE GOT THE RTH VALUE BUT POWER IS NOT COMING OUT CORRECT,PLEASE HELP In this system we have I=0.3A, Rp=2.0Ω, and Rs=1.8Ω. UserIdTAG: 433368 UserNameTAG: SurbhiMahajan CreateTimeTAG: 2012-09-23T11:18:20Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: This is the only question I couldn't answer. Finding this power looks to be difficult for calculation. Any help will be welcome. FirstChildUserIdTAG: 221793 FirstChildUserNameTAG: Manchev FirstChildCreateTimeTAG: 2012-09-23T14:12:26Z SecondChildTAG: Hi SurbhiMahajan and Manchev, [here][1] is a link that maybe can help you but if this still not clear enough try to search the forum [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/undefined/threads/505dade89f8fb01f00000002 SecondChildUserIdTAG: 452559 SecondChildUserNameTAG: kuz1toro SecondChildCreateTimeTAG: 2012-09-23T15:38:50Z FirstChildTAG: Hi, If we present the ckt in Thevenin's equivalent form, and if the 1st and the 3rd questions are resolved, we can clearly see that the Th. equivalent ckt. is a voltage divider!! I was shocked to realise it! All the best. FirstChildUserIdTAG: 372623 FirstChildUserNameTAG: Amitraj FirstChildCreateTimeTAG: 2012-09-23T15:39:06Z FirstChildTAG: if you have found the Rth then you are probably (very) close to the answer for the optimal load resistance... Read about maximum power transfer theorem for more info regarding this.. it states that the source impedance (Rth) must be same as the load impedance for maximum power transfer.. FirstChildUserIdTAG: 257806 FirstChildUserNameTAG: skoda FirstChildCreateTimeTAG: 2012-09-23T15:56:06Z FirstChildTAG: i did it..thanks to all..:-) FirstChildUserIdTAG: 433368 FirstChildUserNameTAG: SurbhiMahajan FirstChildCreateTimeTAG: 2012-09-29T11:44:03Z IndexTAG: 3983 TitleTAG: Alternate Standards? Why cant a "common ground" be chosen and be assigned to either "0" or "1"? In such a case the sender doesnt have to restrict himself to "tougher" standards. - Eg: 0 < V <= 3 => logic 0 3 < V <= 5 => logic 1 UserIdTAG: 151751 UserNameTAG: Senade CreateTimeTAG: 2012-09-23T09:23:28Z VoteTAG: 0 CoursewareTAG: Week 2 / Static Discipline CommentableIdTAG: 6002x_static_discipline NumberOfReplyTAG: 1 FirstChildTAG: Because in reality there's exist a noise that could interface with the line/wire/circuit, even in common ground. let say you have 2 circuit/device that have their own power and you try to connect the ground by wire it up. fortunately you have a precision voltage measurement and you measure the ground voltage between the circuit/device, you will see that it isn't exactly 0.0000000000000000 Volt, the longer the wire and or the higher the noise would give this difference higher. that's why it need a margin or static discipline to compensate this problem. so its present not to make things tougher instead it make things work FirstChildUserIdTAG: 452559 FirstChildUserNameTAG: kuz1toro FirstChildCreateTimeTAG: 2012-09-23T16:16:35Z IndexTAG: 3984 TitleTAG: Week 3, Homework, H3P1 [Solved] We have two questions in the section H3P1 "What is the minimum value of the pullup resistor RPuI (in Ohms) for which this inverter can obey the required static discipline?" and "Now, consider the NAND gate of this family. What is the minimum value of the pullup resistor RPuA (in Ohms) for which this inverter can obey the required static discipline?" I don't understand why the answers must be different for both the questions since the pullup resistor value for either of the circuits is determined using only VOL and RON. Am I missing some important concept here? UserIdTAG: 383316 UserNameTAG: Srujana_K CreateTimeTAG: 2012-09-23T09:18:58Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: How many FETs in series in a NAND gate vs an inverter? FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-09-23T13:46:30Z SecondChildTAG: Got it. Thank you :) SecondChildUserIdTAG: 383316 SecondChildUserNameTAG: Srujana_K SecondChildCreateTimeTAG: 2012-09-23T15:47:24Z IndexTAG: 3985 TitleTAG: textbook. Can i get the pdf of the textbook. UserIdTAG: 295511 UserNameTAG: bishu CreateTimeTAG: 2012-09-23T09:04:44Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3986 TitleTAG: I have a confusion ! I did not notice the difference between calculating (x1,x2) and (y1,y2) :( UserIdTAG: 393116 UserNameTAG: RANDOUCH CreateTimeTAG: 2012-09-23T08:45:30Z VoteTAG: 0 CoursewareTAG: Week 2 / Superposition example CommentableIdTAG: 6002x_S3E3_superposition NumberOfReplyTAG: 1 FirstChildTAG: IN CALCULATING X1,X1 U HAVE TO FIND ACROSS RESISTOR R3 WHEREAS IN CAL. Y1 AND Y2 U HAVE TO FIND IN THE RESISTOR R1 FirstChildUserIdTAG: 231749 FirstChildUserNameTAG: utshau FirstChildCreateTimeTAG: 2012-09-24T05:47:03Z IndexTAG: 3987 TitleTAG: Bug report in homework In H2P2: Solar Power question TWO "What is the power (in Watts) that is delivered to this best load resistance?" The answer is ***CLEARLY*** *Exactly 0.024W* And yet the answer is wrong! What is happening?! UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-23T08:23:26Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: it's obvious - your answer is not correct calculate current through RL, then P=I^2*RL don't forget that Rp and Rs+RL+Rs - current divider FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-09-23T08:31:20Z IndexTAG: 3988 TitleTAG: edx player hey.. i m from pakistan and at my side edx player is not running so i m unable to see the luctures... please kindly help me out UserIdTAG: 60845 UserNameTAG: unk589 CreateTimeTAG: 2012-09-23T08:18:28Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You can view them here: https://6002x.mitx.mit.edu/wiki/view/YoutubeLinks FirstChildUserIdTAG: 383316 FirstChildUserNameTAG: Srujana_K FirstChildCreateTimeTAG: 2012-09-23T09:23:05Z IndexTAG: 3989 TitleTAG: H2P1 Guys,how to find vmax and vmin UserIdTAG: 150120 UserNameTAG: krishna1993 CreateTimeTAG: 2012-09-23T08:14:45Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: i'm still confused .. but thx about this decleration FirstChildUserIdTAG: 193445 FirstChildUserNameTAG: eslammagdy FirstChildCreateTimeTAG: 2012-09-23T18:08:53Z FirstChildTAG: Ok krishna, so this is the simple part! Once you have found your resistors from the E12 set that allow approximately 9V Vout you need to factor in the 10 percent resistance tolerance. So say you have R1 = 39K and R2 = 15K, Then what is ten percent of 39K and 10 percent of 15K? The answers are 3.9K and 1.5K respectively. (For harder percentages just use a calculator http://www.percentagecalculator.net/ ) Therefore to calculate ***Vmin*** we add 3.9k (10 percent) to the 39k resistor and likewise subtract 1.5k (also 10 percent) from the 15k resistor. This gives use the values of R1 and R2 where the voltage is at it's minimum if you happen to have resistors on the limit of their tolerance. For ***Vmax***, you just have to do the opposite, 39k - 3.9k = 35.1k = R1 and 15k + 1.5k = 16.5k = R2. Plug the info into our voltage divider equation: R2*Vin/R1+R2=Vout And ***Viola!*** FirstChildUserIdTAG: 365551 FirstChildUserNameTAG: hazel1919 FirstChildCreateTimeTAG: 2012-09-23T08:47:53Z SecondChildTAG: Hey I am finding it difficult to calculate R1 an R2 please help SecondChildUserIdTAG: 178115 SecondChildUserNameTAG: nehamakhija SecondChildCreateTimeTAG: 2012-09-23T09:16:02Z SecondChildTAG: it worked but we should be careful while choosing the resistances satisfying the relation we got. thanks SecondChildUserIdTAG: 176149 SecondChildUserNameTAG: gkc143 SecondChildCreateTimeTAG: 2012-09-23T11:17:09Z SecondChildTAG: gkc143, Can you help me find R1 and R2 SecondChildUserIdTAG: 178115 SecondChildUserNameTAG: nehamakhija SecondChildCreateTimeTAG: 2012-09-23T11:35:10Z SecondChildTAG: hey m mot getting how u got r1 and r2.. help.. SecondChildUserIdTAG: 219950 SecondChildUserNameTAG: sumit_010 SecondChildCreateTimeTAG: 2012-09-23T11:48:28Z SecondChildTAG: pls help me to get the value of R1 & R2???can u ellaborate the relation which it should satisfy?? SecondChildUserIdTAG: 414118 SecondChildUserNameTAG: raje93 SecondChildCreateTimeTAG: 2012-09-23T12:01:50Z SecondChildTAG: if u have not got ur r1 and r2 values then ur vmax and vmin also are wrong but due to software calculation problem it will show as correct after finding ur r1 and r2 by the process mentioned above u have to calculate vmax and vmin u should choose r1 and r2 fron the equation vout=vin*r2/r1+r2. suppose if vin =80 and vout=24 then u will get the relation 3r1=7r2 so u should choose r1 and r2 which satisy the above condition not exact values there can be some variations . SecondChildUserIdTAG: 176149 SecondChildUserNameTAG: gkc143 SecondChildCreateTimeTAG: 2012-09-23T12:55:00Z FirstChildTAG: Let say you have two resistors R1 and R2 with 10% tolerance, so the min value of R1 will be R1-R1/10, and max value R1+R1/10, and the same for R2. As soon output voltage is proportional to resistors ratio it's obvious that V maximum will be when ratio is maximum, and minimum when ratio is minimum FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-09-23T08:24:56Z IndexTAG: 3990 TitleTAG: Error in caption at 0:03 At 0:03, the caption is "how a wave *sound* might look like in this scenario with the". It should be "how a wave *form* might look like in this scenario with the". UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-23T08:13:46Z VoteTAG: 0 CoursewareTAG: Week 2 / Static Discipline CommentableIdTAG: 6002x_static_discipline NumberOfReplyTAG: 0 IndexTAG: 3991 TitleTAG: Vd? the math is trivial, but how do we choose operating point Vd? Just from observing plot? UserIdTAG: 95280 UserNameTAG: Fortyq CreateTimeTAG: 2012-09-23T08:02:08Z VoteTAG: 0 CoursewareTAG: Week 4 / Small Signal Model CommentableIdTAG: 6002x_small_signal NumberOfReplyTAG: 3 FirstChildTAG: I think you find a spot on the curve that is the most linear, or flat and that also works in your given application. Most devices have an optimum operating point that is a compromise of effectiveness vs. durability. Ex: An LED's lifespan will be reduced if used with too high of voltage. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-23T11:18:50Z FirstChildTAG: I would think the best point to choose is the one where the slope is as close to 1:1 as possible as that will result in a minimum distortion of the signal. Or am I wrong here? FirstChildUserIdTAG: 114881 FirstChildUserNameTAG: xvink FirstChildCreateTimeTAG: 2012-10-03T17:53:37Z FirstChildTAG: Actually **remarkable**, "the device behaves as a resistor for small signal" FirstChildUserIdTAG: 369431 FirstChildUserNameTAG: pkaistha FirstChildCreateTimeTAG: 2012-10-06T13:45:22Z IndexTAG: 3992 TitleTAG: H2P1 What is the value of R1 and R2.I am not able to solve at all.Please help me... UserIdTAG: 150120 UserNameTAG: krishna1993 CreateTimeTAG: 2012-09-23T07:15:14Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: There isn't an only answer, there are a lot of valid arrangements of values of resistors that you can use according to the problem specifications. As a hint, have in mind that Vin/Vout has to be in the range from 3.6 to 4.4 as you calculate it with the nominal values of the resistors, after that the division can be out of the range. I hope it can be helpful for you. FirstChildUserIdTAG: 376173 FirstChildUserNameTAG: nacho110987 FirstChildCreateTimeTAG: 2012-09-23T07:27:05Z SecondChildTAG: can u help me out of this problem SecondChildUserIdTAG: 150120 SecondChildUserNameTAG: krishna1993 SecondChildCreateTimeTAG: 2012-09-23T07:59:00Z SecondChildTAG: guys with ur help i have found r1 and r2 .please give me a clue to find vmax and vin SecondChildUserIdTAG: 150120 SecondChildUserNameTAG: krishna1993 SecondChildCreateTimeTAG: 2012-09-23T08:09:38Z SecondChildTAG: for vmax the value the value of R2 shud be max ie R2+10%of R2 and R1 shud be minimum ie R1-10%of R1 similarly, vmin will be for the case when R2 is min and R1 is maximum.... SecondChildUserIdTAG: 89170 SecondChildUserNameTAG: frost SecondChildCreateTimeTAG: 2012-09-23T11:30:06Z SecondChildTAG: guys if i choose two values for my R1 and R2 i am getting my my Vmax and min correct but the values of R1 & R2 are wrong???pls help me out... SecondChildUserIdTAG: 414118 SecondChildUserNameTAG: raje93 SecondChildCreateTimeTAG: 2012-09-23T11:55:38Z FirstChildTAG: ya .. there are so many ans... u can calculate using VDR fix one resistor then find other... FirstChildUserIdTAG: 93123 FirstChildUserNameTAG: Bilawal FirstChildCreateTimeTAG: 2012-09-23T07:36:50Z FirstChildTAG: Guys i have found out Vmax and Vmin. But R1 and R2 values are wrong. Is there any way to find out R1 and R2 from Vman and Vmin? FirstChildUserIdTAG: 309722 FirstChildUserNameTAG: kaushikraghavan1992 FirstChildCreateTimeTAG: 2012-09-23T08:32:57Z SecondChildTAG: Vmax and Vmin only LOOKS LIKE correct - just because engine calculates it the same way from YOUR R1 and R2. You have a special requirement: "An additional requirement is that the Thevenin resistance as seen from the output terminals is between 10kΩ and 30kΩ." SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-23T08:46:06Z SecondChildTAG: keep those are in kilo values SecondChildUserIdTAG: 114473 SecondChildUserNameTAG: Veerabasanagouda SecondChildCreateTimeTAG: 2012-09-23T10:15:14Z SecondChildTAG: R1 and R2 values are wrong...pls help SecondChildUserIdTAG: 414118 SecondChildUserNameTAG: raje93 SecondChildCreateTimeTAG: 2012-09-23T11:58:03Z IndexTAG: 3993 TitleTAG: Confused **actually i am confused in number of nodes present in this circuit... nodal analysis is applied when there are 2 principal nodes .but i dont see any principal node ..rether there is a refrence node making me confuse.. ** UserIdTAG: 342886 UserNameTAG: hthimz CreateTimeTAG: 2012-09-23T07:11:39Z VoteTAG: 0 CoursewareTAG: Week 2 / Node equation review CommentableIdTAG: 6002x_S3E1_NodeEquationReview NumberOfReplyTAG: 1 FirstChildTAG: Start with marking the points at same potential with the same label. For example, in the above question, one end of the voltage source and one end of the resistor R1 are at same potential V1..so mark them as V1 The other end of resistor R1 and one end of resistor R2 are at same potential, say, e The other end of R2, V2 are at same potential -V2 And the remaining ends at potential 0 So number of nodes = 4 (V1,e,-V2,0) You got your nodes and their voltages, now using nodal analysis is simple. FirstChildUserIdTAG: 383316 FirstChildUserNameTAG: Srujana_K FirstChildCreateTimeTAG: 2012-09-23T09:30:57Z IndexTAG: 3994 TitleTAG: Answers to Lab 2 Again i am doing my home work at the 11th hour and am sorry for the late post but i hope it will help other solve the lab 2 here is a quick hint: 1. we need only voltage divider circuits ,no series parallel keep it simple its not that difficult 2. for V1 R1=R2 ratio is 1:1 3. for V2 ratio of resistance is like 6:1 try values form 0 to 3 including 3. 4. You also have to use ground some where for V2 There can be no simpler hint than this! if you still cant do it. forget it .its not for you!! I hope it helps! UserIdTAG: 232499 UserNameTAG: Asim09 CreateTimeTAG: 2012-09-23T06:45:38Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Well actualy it can be done with 3 very small resistors, iv've got my answer correct with ratio of R1/R2 for va such as 2/3 and R3 for V2 ratio 5/6. As everybody can see there is tons of options to do that. I believe looking in coursebook at page about 76 can help you. Good luck :) FirstChildUserIdTAG: 405305 FirstChildUserNameTAG: Tomaszeek FirstChildCreateTimeTAG: 2012-09-23T06:52:44Z SecondChildTAG: by diagram as example we can place the 2 resistors and 3rd one as load am i right SecondChildUserIdTAG: 164689 SecondChildUserNameTAG: muhammadfaizan SecondChildCreateTimeTAG: 2012-09-23T14:00:50Z FirstChildTAG: Finally, I have solved this dreaded lab2 with the hints that you gave . thank you very much. FirstChildUserIdTAG: 340270 FirstChildUserNameTAG: revathi_srinivasa FirstChildCreateTimeTAG: 2012-09-23T11:01:25Z IndexTAG: 3995 TitleTAG: lab2 i am unable to connect the resistors in the lab 2 session as in lab 1.. and i couldn't join them with wires too.. does anyone know the way to sort it out?? UserIdTAG: 279312 UserNameTAG: shravanipalla CreateTimeTAG: 2012-09-23T06:30:27Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Are you using a IPAD? Because I wasn't able to do the lab on it. But either a Mac with OSX either a PC with Windows 7 Home did the job... FirstChildUserIdTAG: 348965 FirstChildUserNameTAG: patruclo FirstChildCreateTimeTAG: 2012-09-23T06:37:05Z IndexTAG: 3996 TitleTAG: Additional Buttons My first thanks go to the designers and those who mounted this course However I have some suggestions to make if it can fit into the design Please I will like you to show solutions with answers in the show answer button or possibly create additional buttons to show soutions hint to questions if possibly, This should be applied to both practice questions and Homeworks even labs should be given detail solution . Thanks UserIdTAG: 382532 UserNameTAG: emmanuelpeace CreateTimeTAG: 2012-09-23T06:21:19Z VoteTAG: 0 CoursewareTAG: Week 2 / Digital Abstraction CommentableIdTAG: 6002x_digital NumberOfReplyTAG: 0 IndexTAG: 3997 TitleTAG: Feedback...pages not iPad friendly. Just an item to be considered...the Discussion pages are not iPad friendly. I'll guess it's all that dynamic resizing the iPad does. EDIT: I'm using Chrome browser on the iPad. UserIdTAG: 92346 UserNameTAG: MobiusTruth CreateTimeTAG: 2012-09-23T05:09:35Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: haha iPad :P FirstChildUserIdTAG: 156974 FirstChildUserNameTAG: ManojKumar FirstChildCreateTimeTAG: 2012-09-23T05:14:49Z IndexTAG: 3998 TitleTAG: H2P1 I'm stuck at homework 2 problem 1 and 2. How should i proceed for both the problems. Please help me... UserIdTAG: 309722 UserNameTAG: kaushikraghavan1992 CreateTimeTAG: 2012-09-23T05:02:47Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 3999 TitleTAG: LAB 2 HELP I've watched the lectures, skimmed through text, done the homework, and read through multiple threads about Lab 2 all week. I've tried grounding the excess voltage (resistance) through a resistance 1/3 the total resistance of two resistors in series in the circuit. I grounded that resistor between my R1 resistor, which is its arbitrary value, and my R2 resistor, which is 3*R1. Someone please lecture me further on what to do. I am not getting the answer. I understand the concept of superposition. I don't understand the schematic to the correct answer. UserIdTAG: 357929 UserNameTAG: chips67 CreateTimeTAG: 2012-09-23T05:02:37Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: use VDR ... and satisfy the given equation.... FirstChildUserIdTAG: 93123 FirstChildUserNameTAG: Bilawal FirstChildCreateTimeTAG: 2012-09-23T07:38:19Z FirstChildTAG: use 3 resistor and appropriate value such that your graph wave is similar to the given....And one of resistor which is between Vout and Ground has value 0 FirstChildUserIdTAG: 393950 FirstChildUserNameTAG: RaoUmairTufailAnjum FirstChildCreateTimeTAG: 2012-09-23T05:13:05Z IndexTAG: 4000 TitleTAG: Week 2 LAB 2 : I give up just can't crack it I am trying to solve LAB 2 since 5 hours and i just don't get the values. Equations from the circuit simply don't help determine the resistance values. UserIdTAG: 94464 UserNameTAG: sarkar_abhijoy CreateTimeTAG: 2012-09-23T04:20:46Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: Take a break, sleep on it. Come back tomorrow and read some "lab 2" threads. You will get it. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-23T04:25:43Z SecondChildTAG: use 3 resistor and appropriate value such that your graph wave is similar to the given....And one of resistor which is between Vout and Ground has value 0 SecondChildUserIdTAG: 393950 SecondChildUserNameTAG: RaoUmairTufailAnjum SecondChildCreateTimeTAG: 2012-09-23T05:12:29Z FirstChildTAG: oh well I spent much more trying to solve it.... FirstChildUserIdTAG: 252475 FirstChildUserNameTAG: alexha FirstChildCreateTimeTAG: 2012-09-23T04:48:07Z SecondChildTAG: Hi alexha, if anything I can do to help you please tell me. I will conect here later... SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-23T05:03:08Z SecondChildTAG: use 3 resistor and appropriate value such that your graph wave is similar to the given....And one of resistor which is between Vout and Ground has value 0 SecondChildUserIdTAG: 393950 SecondChildUserNameTAG: RaoUmairTufailAnjum SecondChildCreateTimeTAG: 2012-09-23T05:12:47Z SecondChildTAG: I really really really don't understand why RaoUmairTufailAnjum always recommends 0 SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-23T08:03:12Z SecondChildTAG: Hi sakar_abjiyoy and alexha read here [hints][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/505f41be015492230000001c SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-23T17:11:37Z FirstChildTAG: its sooo simple ...dear friendzzz u have to just satisfy the given equation...using Voltage division rule.. FirstChildUserIdTAG: 93123 FirstChildUserNameTAG: Bilawal FirstChildCreateTimeTAG: 2012-09-23T07:39:58Z FirstChildTAG: aahhh ..finally got the right answer after spending 4hrs :) FirstChildUserIdTAG: 219950 FirstChildUserNameTAG: sumit_010 FirstChildCreateTimeTAG: 2012-09-23T11:40:34Z FirstChildTAG: lab 2 was one of the toughest labs till week 5....its great we all could solve it with much hardwork!! CHEERS TO ALL! FirstChildUserIdTAG: 368712 FirstChildUserNameTAG: jmen FirstChildCreateTimeTAG: 2012-09-26T07:06:41Z IndexTAG: 4001 TitleTAG: LAB 2 PROBLEM how can i solve lab-2 using resistors?Please give some hints. UserIdTAG: 256102 UserNameTAG: sadique CreateTimeTAG: 2012-09-23T03:14:42Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: hint: using 3 resistors and node method FirstChildUserIdTAG: 338466 FirstChildUserNameTAG: thcnhan FirstChildCreateTimeTAG: 2012-09-23T04:07:20Z FirstChildTAG: use VDR .. and satisfy the given equation.. FirstChildUserIdTAG: 422358 FirstChildUserNameTAG: SajidHussain FirstChildCreateTimeTAG: 2012-09-23T07:32:59Z IndexTAG: 4002 TitleTAG: H2P1 Problem After filling the all text boxes in H2P1 when I click on "check", it showed that Vmax and Vmin are true and resistances are false. I didn't understand what was wrong. Anybody could explain it? UserIdTAG: 378096 UserNameTAG: Jamshaid271 CreateTimeTAG: 2012-09-23T03:05:39Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 6 FirstChildTAG: How did you calculate Vmax and Vmin?? I just don't understand the problem... Please help.. FirstChildUserIdTAG: 309722 FirstChildUserNameTAG: kaushikraghavan1992 FirstChildCreateTimeTAG: 2012-09-23T03:53:46Z SecondChildTAG: Vmax will occur when R2 is max (ie R2+tolerence) and when R1 is minimum.. smlly Vmin can be found out for R2 min and R1 max SecondChildUserIdTAG: 332360 SecondChildUserNameTAG: srinivasav SecondChildCreateTimeTAG: 2012-09-23T04:45:07Z SecondChildTAG: you people are doing mistake in calculating thevenin resistance SecondChildUserIdTAG: 373281 SecondChildUserNameTAG: Gaurav025 SecondChildCreateTimeTAG: 2012-09-23T05:05:30Z FirstChildTAG: Hi Jamshaid271, I believe you need to enter the nearest standard values. FirstChildUserIdTAG: 11538 FirstChildUserNameTAG: trishul FirstChildCreateTimeTAG: 2012-09-23T04:48:11Z SecondChildTAG: could you plz explain? SecondChildUserIdTAG: 378096 SecondChildUserNameTAG: Jamshaid271 SecondChildCreateTimeTAG: 2012-09-23T04:55:22Z SecondChildTAG: choose the values from the provided set SecondChildUserIdTAG: 83746 SecondChildUserNameTAG: Mamoona SecondChildCreateTimeTAG: 2012-09-23T04:56:31Z SecondChildTAG: After I have chosen the values I got Vmax and Vmin. But still not believing that R1 and R2 is wrong and Vmin and Vmax is correct SecondChildUserIdTAG: 378096 SecondChildUserNameTAG: Jamshaid271 SecondChildCreateTimeTAG: 2012-09-23T04:59:45Z FirstChildTAG: I have no idea what could be your mistake. Have you checked the resistors´Potencies???Because it will get the right result in Vmax and Vmin if you just get the same potency of R1 and R2, but not the right value of those... FirstChildUserIdTAG: 116471 FirstChildUserNameTAG: haravez FirstChildCreateTimeTAG: 2012-09-23T05:19:54Z FirstChildTAG: hey i got somethng..answers in last 2 boxes are depending on answer in 1st 2 boxes.. FirstChildUserIdTAG: 219950 FirstChildUserNameTAG: sumit_010 FirstChildCreateTimeTAG: 2012-09-23T12:07:19Z FirstChildTAG: The same happens to me. You could find the answer? FirstChildUserIdTAG: 440992 FirstChildUserNameTAG: calabrezcrv FirstChildCreateTimeTAG: 2012-09-23T20:34:10Z FirstChildTAG: Can any one explain H2P1 (Vmax and Vmin) FirstChildUserIdTAG: 150120 FirstChildUserNameTAG: krishna1993 FirstChildCreateTimeTAG: 2012-09-30T17:48:29Z IndexTAG: 4003 TitleTAG: Why is it still not right .. UserIdTAG: 229018 UserNameTAG: Changming CreateTimeTAG: 2012-09-23T03:03:48Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: please ask more specific questions. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-09-23T03:14:39Z IndexTAG: 4004 TitleTAG: H2P1 - Vmax - accepting wrong values The system is accepting wrong values. It is possible to get a greater Vmax mathematically on the basis of R1 and R2 values. I will post a more elaborate comment after the deadline has passed. UserIdTAG: 370538 UserNameTAG: samcore2804 CreateTimeTAG: 2012-09-23T00:31:50Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: More then 10%, are you saying that the two resistors maximum tolerances compound? (+-20%) FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-23T00:42:47Z SecondChildTAG: you have to see which error, + 10% or -10%, you have to use in order to have a Vmax SecondChildUserIdTAG: 344814 SecondChildUserNameTAG: clerio SecondChildCreateTimeTAG: 2012-09-23T04:03:49Z IndexTAG: 4005 TitleTAG: The Node Method & algebra Hi!! Could someone please explain the reasoning behind formula 3.10-3.15 pp 127-128 in the textbook? There are some basic algebraic steps from (3.8) to (3.10) - but I just dont get it :( (3.8) e=(1/(G1+G2))*I+(G1/(G1+G2))*V (3.10) V-e= -(1/(G1+G2))*I+(G2/(G1+G2))*V UserIdTAG: 87106 UserNameTAG: Fabius CreateTimeTAG: 2012-09-22T21:12:26Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi, you have to remember that G1 is the conductante, which is the inverse of resistance. Sometimes circuit algebra is clearer if you write conductance terms, rather than resistances. (3.6) e((1/R1)+(1/R2)) = I + (V/R1) (3.7) here resistances are replaced by conductances G1 and G2 (3.8) both sides of equation are divided by G1 + G2, and terms are slightly rearranged (3.9) it's the same 3.8 equation, but written using resistances (3.10) v1 = V - e. So make this: v1 = V - (((1/(G1 + G2))*I) + (G1/(G1 + G2))*V. Arrange terms containing V: v1 = V*(1 - G1/(G1 + G2)) - (1/(G1 + G2))*I v1 = V*((G1 + G2 - G1)/(G1 + G2)) - (1/(G1 + G2))*I you simplify G1 and (-G1), obviously they cancel mutually, and finally you get 3.10. Hope it helps FirstChildUserIdTAG: 220837 FirstChildUserNameTAG: Calsomus FirstChildCreateTimeTAG: 2012-09-22T23:34:07Z SecondChildTAG: now it's all clear in my head - thanks for your explanation :) SecondChildUserIdTAG: 87106 SecondChildUserNameTAG: Fabius SecondChildCreateTimeTAG: 2012-09-23T00:56:31Z IndexTAG: 4006 TitleTAG: Was that system linear ?! In the previous tutorials, we knew that a linear system must intersect with the origin. In this very example, the graph did not ! we know as well that Thevenin and Norton can only be applied on linear systems ! so, was that graph only an assumption ?! or we actually can use Thevenin and Norton on such graphs ?! UserIdTAG: 130810 UserNameTAG: moon_light CreateTimeTAG: 2012-09-22T21:05:31Z VoteTAG: 0 CoursewareTAG: Week 2 / Thevenin Example CommentableIdTAG: 6002x_thevenin_example_1 NumberOfReplyTAG: 1 FirstChildTAG: I dont think linear system must intersect with origin. The simple form of linear equation is $y=a \cdot x+b$ where a and b are any constant, let say a and b are not 0, so if $x=0$ then $y=b$, which mean not crossing the origin FirstChildUserIdTAG: 452559 FirstChildUserNameTAG: kuz1toro FirstChildCreateTimeTAG: 2012-09-23T16:48:25Z IndexTAG: 4007 TitleTAG: 2 days of working, Still I have no clue about H2P1. Please help. I know the Vout = (R2/R1+R2)*Vin. The Rth = (1/R1)+(1/R2)(I have consider Rth=20k as per the range given). I get R2 as 25k and R1 as 100k after mapping to E12 is again wrong?? "Come up with resistors R1 and R2 such that the divider ratio Vout/Vin is within 10% of the requirement." which requirement is the question speaking about. Please let me know a proper way of understanding the problem. UserIdTAG: 156974 UserNameTAG: ManojKumar CreateTimeTAG: 2012-09-22T21:03:08Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: In this problem, our available resistor values are constrained to real world conditions. The only values we can use are listed in the problem (at the top). I found that the easiest way to solve it would be to use a spreadsheet. This way you can calculate the tolerances of the resistors and the maximum and minimum outputs automatically. Keep in mind the Thevinen resistance requirement and use that to narrow down the resistance range you need. Good Luck! FirstChildUserIdTAG: 227432 FirstChildUserNameTAG: JasonTraud FirstChildCreateTimeTAG: 2012-09-22T21:22:39Z FirstChildTAG: [here][1] is a link that may can help you out [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/undefined/threads/505aa3fae87fdd2800000001 FirstChildUserIdTAG: 452559 FirstChildUserNameTAG: kuz1toro FirstChildCreateTimeTAG: 2012-09-22T21:34:49Z FirstChildTAG: insure that your resistor selections are in harmony with the E12 set requirements FirstChildUserIdTAG: 153068 FirstChildUserNameTAG: tobyone FirstChildCreateTimeTAG: 2012-09-23T00:03:31Z IndexTAG: 4008 TitleTAG: Week Two Home work I cant understand the first problem. Can someoen please help me... Thanks UserIdTAG: 309722 UserNameTAG: kaushikraghavan1992 CreateTimeTAG: 2012-09-22T20:11:00Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 4009 TitleTAG: external excitation? What is meant by external excitation? UserIdTAG: 254325 UserNameTAG: bondablack CreateTimeTAG: 2012-09-22T20:04:43Z VoteTAG: 0 CoursewareTAG: Week 2 / Thevenin method CommentableIdTAG: 6002x_thevenin NumberOfReplyTAG: 1 FirstChildTAG: okay, Got it! FirstChildUserIdTAG: 254325 FirstChildUserNameTAG: bondablack FirstChildCreateTimeTAG: 2012-09-22T20:05:29Z SecondChildTAG: :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-23T00:45:24Z IndexTAG: 4010 TitleTAG: how to submit homeworks when i go to homework section i can only find check option but not the submit option so how can i submit my homeworks?can someone here pleasae help me out on this?? UserIdTAG: 479321 UserNameTAG: TarunVarma CreateTimeTAG: 2012-09-22T19:16:14Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I think that when you are checking your homework, you are also submiting it. There is no submit button. FirstChildUserIdTAG: 45307 FirstChildUserNameTAG: josejimenez2 FirstChildCreateTimeTAG: 2012-09-22T19:23:13Z SecondChildTAG: yes... SecondChildUserIdTAG: 393950 SecondChildUserNameTAG: RaoUmairTufailAnjum SecondChildCreateTimeTAG: 2012-09-22T19:28:15Z SecondChildTAG: i don't know either but i wish if you know help me please SecondChildUserIdTAG: 475448 SecondChildUserNameTAG: msamwelmollel SecondChildCreateTimeTAG: 2012-09-22T19:36:08Z IndexTAG: 4011 TitleTAG: Typo at 1:02 The text in the caption at 1:02 is "So VS is my normally input". It should be "So VS is my normal input." UserIdTAG: 254346 UserNameTAG: moijes12 CreateTimeTAG: 2012-09-22T18:49:58Z VoteTAG: 0 CoursewareTAG: Week 2 / Digital Abstraction CommentableIdTAG: 6002x_digital NumberOfReplyTAG: 0 IndexTAG: 4012 TitleTAG: i need help hi.i have joined edx just on 23rd sep.but now what about the homeworks,online lab etc for which due dates are over.will i be loosing any points for that.will i be given any grace period to submit them.im very new to this edx and its concept so i will be very greatful to any one who will help on this issue.i am all confused UserIdTAG: 479321 UserNameTAG: TarunVarma CreateTimeTAG: 2012-09-22T18:39:13Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: hmm well you are absolutely right but i just asked about the marks for the sake of general knowledge and thanks.but one more question.if we complete the course successfully will they send certificates direct to our home.and also for online mid exams how can mit verify on computer if the student is not cheating and answering.i will be very happy if u clear this doubt of mine too FirstChildUserIdTAG: 479321 FirstChildUserNameTAG: TarunVarma FirstChildCreateTimeTAG: 2012-09-22T19:00:54Z SecondChildTAG: They don't check whether you are cheating or not. Everything is on the honor code, so successfully passing the exams assumes you are following the honor code. It is outlined here: https://www.edx.org/honor SecondChildUserIdTAG: 213386 SecondChildUserNameTAG: dmascenik SecondChildCreateTimeTAG: 2012-09-22T20:12:53Z FirstChildTAG: if I recall correctly two worst homeworks will be dropped for the sake of your final score. So you can come two homeworks late, get full marks for the rest of the assignments and get 100% as final score. Do you really care though? Automatic checker will still work past the deadline so you'll know if your giving correct answers. Sure marks and deadlines keep you focused but at the end of the day you're here to learn, not to get marks. Knowledge first, everything else inessential. Good reason to keep up though is so you can participate in the discussions. I'm getting the feeling these are extremely valuable. FirstChildUserIdTAG: 5325 FirstChildUserNameTAG: vkz FirstChildCreateTimeTAG: 2012-09-22T18:53:27Z IndexTAG: 4013 TitleTAG: HP2P: solar Power Can I get a hint of where to look in the text book regarding this question? I am completely lost in what are asking. This appears to be a basic transmission line so RL needs to equal RP for the most efficient transmission, but somehow that is not a correct answer? Are each one of the RS labels 1.9 ohms, or the path as a whole 1.9 ohms? I have to be missing something simple. UserIdTAG: 349840 UserNameTAG: Wilk CreateTimeTAG: 2012-09-22T18:27:22Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: please give me some hint on where to look up in the book regarding this formula for electric power. FirstChildUserIdTAG: 263602 FirstChildUserNameTAG: elekov FirstChildCreateTimeTAG: 2012-09-22T20:18:32Z SecondChildTAG: Find I with Vth and Rth. After you have to find voltage across RL. And finally - P=I*V. SecondChildUserIdTAG: 295103 SecondChildUserNameTAG: Syavick SecondChildCreateTimeTAG: 2012-09-22T23:11:06Z FirstChildTAG: Yes those formulas are right and everything but... well i find best load resistance, thevenin resistance find out the voltage throught formula V = RI then use any of power formulas and... i lost... i can't find the value of power. and i tried almost eveything... almost becouse it still don't work and i know i should :) FirstChildUserIdTAG: 405305 FirstChildUserNameTAG: Tomaszeek FirstChildCreateTimeTAG: 2012-09-23T07:03:46Z FirstChildTAG: RL = Rp+2*Rs FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-09-22T18:57:00Z SecondChildTAG: which formula do you use for computing the power? I tried with P=(V^2)/R but it doesn't work. I am using RL for R and the voltage across RL. Can you please tell me where is my mistake? Thanks! SecondChildUserIdTAG: 263602 SecondChildUserNameTAG: elekov SecondChildCreateTimeTAG: 2012-09-22T20:06:57Z SecondChildTAG: Thanks SecondChildUserIdTAG: 349840 SecondChildUserNameTAG: Wilk SecondChildCreateTimeTAG: 2012-09-22T20:08:40Z SecondChildTAG: yeah its 1.7 not 1.9 SecondChildUserIdTAG: 378150 SecondChildUserNameTAG: GladIDoThis SecondChildCreateTimeTAG: 2012-09-22T20:57:28Z SecondChildTAG: P=(V^2)/R works good - as good as P=(I^2)*R If you want use voltage anyway - just remember how to calculate voltage for Thevenin equivalent and replace Norton with Thevenin. But I don't see a purpose for such complication :) SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-23T02:17:39Z SecondChildTAG: How is it that the RL = Rp+2*Rs? In other words, the resistance of the load is equal to the resistance of the source? SecondChildUserIdTAG: 101329 SecondChildUserNameTAG: kellenproctor SecondChildCreateTimeTAG: 2012-09-23T13:47:06Z SecondChildTAG: Yes, maximum power dissipated by load when Rload=Rsource. Consider this schematics: ---------- ![][1] ---------- where Vsource=10V and Rsource=100ohm and Rload - unknown. Let's calculate power dissipated by Rload: ---------- Pload=I^2*Rload I=Vsource/(Rsource+Rload) Pload=(Vsource/(Rsource+Rload))^2*Rload Pload=y; Rload=x; Vsource=10; Rsource=100 y=(10/(100+x))^2*x now you can find maximum for y with derivative or just graph it: ---------- ![enter image description here][2] ---------- as you can see y is max when x=100, so Rload=Rsource [1]: https://edxuploads.s3.amazonaws.com/13484184631343683.jpg [2]: https://edxuploads.s3.amazonaws.com/1348419088134365.jpg SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-23T16:53:28Z IndexTAG: 4014 TitleTAG: Help please I see all of this partial information on when to be able to discuss issues with the instructors, but I can't find anything as to the actual link to the forums? Can anyone provide me with a link to ask a question to an instructor? UserIdTAG: 349840 UserNameTAG: Wilk CreateTimeTAG: 2012-09-22T18:14:41Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: I want to know also how contact with a instructor, please help us FirstChildUserIdTAG: 479215 FirstChildUserNameTAG: joseangelescobar FirstChildCreateTimeTAG: 2012-09-22T18:23:54Z FirstChildTAG: Go to course info tab and you'll find that : "The 6.002x teaching staff will be online to answer questions at the following times. Please note that these times are listed with respect to Boston. If you need help understand how these times translate into your local time, find your nearest city here and compare that to the time in Boston: Monday-Friday: 9:00-9:30am, 5:30-6pm, 9:00-9:30pm Saturday: 10:30-11:00am, 1:30-2:00pm, 4:30pm-5:00pm Sunday: 12:00-12:30pm, 3:00pm-3:30pm, 4:30-5:00pm, 6:00-8:00pm" That's all I know !!! I hope it helps FirstChildUserIdTAG: 211036 FirstChildUserNameTAG: Mohammad91 FirstChildCreateTimeTAG: 2012-09-22T19:05:09Z SecondChildTAG: Online as in watching the forums? or somewhere else SecondChildUserIdTAG: 349840 SecondChildUserNameTAG: Wilk SecondChildCreateTimeTAG: 2012-09-22T19:43:11Z SecondChildTAG: Yes right here. The staff will have a blue bar, like the staff member that posted below. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-22T19:58:24Z FirstChildTAG: Please be aware that there are many questions to be answered and the staff can't go through all of them. Using the search tool of the discussion forum to find similar questions that were answered before might be the most efficient way to clear up a doubt. FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-22T19:28:02Z SecondChildTAG: I understand, thanks SecondChildUserIdTAG: 349840 SecondChildUserNameTAG: Wilk SecondChildCreateTimeTAG: 2012-09-22T20:09:39Z IndexTAG: 4015 TitleTAG: error and a point to note ! during the current flow diagram , i think the direction of currents are marked wrongly...(am i right?) and another interesting point to note is that at the last, the variables a1,a2,a3... and b1,b2,b3.... are all constant coefficients having different units (a's dont have any unit, while b's have Ohm !) !...i think its a interesting point that is missing in the lecture ! UserIdTAG: 479160 UserNameTAG: chankii CreateTimeTAG: 2012-09-22T18:01:14Z VoteTAG: 0 CoursewareTAG: Week 2 / Linearity CommentableIdTAG: 6002x_linearity NumberOfReplyTAG: 1 FirstChildTAG: a's don't have any unit, and b's have Ohn, so when b's multiply with I's you get Volts. So on the left and right sides of the equation we get the same dimension (Volt). FirstChildUserIdTAG: 391687 FirstChildUserNameTAG: Gadzhi FirstChildCreateTimeTAG: 2012-09-22T18:29:56Z IndexTAG: 4016 TitleTAG: A Suggestion to staffs! Hi, I have a suggestion for improving the efficiency of this course. I think it can be more helpful for students that after students struggled with the labs and home works, you the staffs, show theory and reasons that support the labs, after each lab is finished, and also for the homework the best procedure that you as experts can suggest. I think in the labs, getting the results by experiment is more common to using the right theory and looking the matter from theory aspect, and if you can connect the results of our experiments to the theories, we can learn much more. UserIdTAG: 153707 UserNameTAG: masoud_np CreateTimeTAG: 2012-09-22T17:20:37Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: How do you feel about the solutions that we release after the deadline is up? FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-09-23T02:58:15Z SecondChildTAG: ya! that wud indeed be very helpful! SecondChildUserIdTAG: 357292 SecondChildUserNameTAG: RaghavAbboy SecondChildCreateTimeTAG: 2012-09-23T08:00:43Z SecondChildTAG: im sorry i mean it is indeed very helpful! SecondChildUserIdTAG: 357292 SecondChildUserNameTAG: RaghavAbboy SecondChildCreateTimeTAG: 2012-09-23T08:05:47Z SecondChildTAG: I feel it can be so helpful, especially after the deadline is up and after students struggle with assignments. Actually I think it can help lots of students, for both Labs and home works. The explanation about some of possible ways of doing a lab (and maybe homeworks) and getting the result correctly, like a supervised lab, can illuminate some shady parts. SecondChildUserIdTAG: 153707 SecondChildUserNameTAG: masoud_np SecondChildCreateTimeTAG: 2012-09-23T11:58:55Z IndexTAG: 4017 TitleTAG: For Lab 3 (in Chrome browser) there are no resistors in the sandbox menu Doing Lab 3 I need to insert pulldown resistors, but there are no resistors in the sandbox menu. Can we get some resistors? UserIdTAG: 228637 UserNameTAG: DCounsell CreateTimeTAG: 2012-09-22T16:48:31Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 2 FirstChildTAG: You shouldn't need resistors to solve Lab 3 -- MOSFETs are sufficient. FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-22T16:54:23Z FirstChildTAG: ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13483445492550784.jpg FirstChildUserIdTAG: 197569 FirstChildUserNameTAG: darksamaro FirstChildCreateTimeTAG: 2012-09-22T20:09:28Z SecondChildTAG: Almost right .... you need the worst case for Vout. Then rearrange so that Ron is a function of RL,Vs and Vout and set it equal to the given formula. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-09-22T20:51:22Z SecondChildTAG: Thanks for the hint.I was having problem placing mosfets.This hint helped me out. SecondChildUserIdTAG: 182653 SecondChildUserNameTAG: monkeyfoahead SecondChildCreateTimeTAG: 2012-09-27T21:19:34Z IndexTAG: 4018 TitleTAG: TUT1: Nodal Analysis: KCL question When summing the current at (e1) leaving the node at the top. The current going down the vertical branch is written as (e1-e2)/2 why is it that as opposed to (e2-e1)/2 ? UserIdTAG: 35758 UserNameTAG: willprice94 CreateTimeTAG: 2012-09-22T15:24:16Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: because we calculate all voltage differences relative to the node of interest - so in the equation for this node we always subtract from its voltage FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-09-22T15:57:20Z IndexTAG: 4019 TitleTAG: problem in lab 2 in the box at the top of the page its given that vout = 1/2v1+ 1/6 v2 but in the sample circuit it is gigen vout=1/2(v1+v2). sombody please explain this to me........ UserIdTAG: 168396 UserNameTAG: rik2008 CreateTimeTAG: 2012-09-22T14:41:45Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: You must change given circuit with vout=1/2(v1+v2) somehow to get $vout = 1/2v1+ 1/6u2$ FirstChildUserIdTAG: 95521 FirstChildUserNameTAG: Gatling FirstChildCreateTimeTAG: 2012-09-22T14:47:19Z SecondChildTAG: ok thanks SecondChildUserIdTAG: 168396 SecondChildUserNameTAG: rik2008 SecondChildCreateTimeTAG: 2012-09-22T14:51:37Z SecondChildTAG: $1/2(v1+v2)$ SecondChildUserIdTAG: 391687 SecondChildUserNameTAG: Gadzhi SecondChildCreateTimeTAG: 2012-09-22T18:07:24Z SecondChildTAG: use $ before and after formula :-) SecondChildUserIdTAG: 391687 SecondChildUserNameTAG: Gadzhi SecondChildCreateTimeTAG: 2012-09-22T18:07:47Z FirstChildTAG: it's a hint. Just one case of a relationship V = a*V1 + b*V2. Hopefully this should tell you something about R1 and R2 relationship. It's there to point you to the right direction, but otherwise unrelated to the problem FirstChildUserIdTAG: 5325 FirstChildUserNameTAG: vkz FirstChildCreateTimeTAG: 2012-09-22T14:53:01Z SecondChildTAG: actually example only gives V=a*V1+(1-a)*V2 the trick is - how to make it V=a*V1+b*V2 :) SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-22T15:48:52Z FirstChildTAG: the sample circuit is just a guide , but you have to design your circuit to give Vout = 1/2 V1+ 1/6 V2 . FirstChildUserIdTAG: 204336 FirstChildUserNameTAG: Amin86 FirstChildCreateTimeTAG: 2012-09-22T16:23:07Z FirstChildTAG: so my question is i got the relationship between r1 and r2 but what is the real is depend on how you choose the value or what? FirstChildUserIdTAG: 475448 FirstChildUserNameTAG: msamwelmollel FirstChildCreateTimeTAG: 2012-09-22T18:25:23Z IndexTAG: 4020 TitleTAG: H2P2 How come when you try to find the voltage to use in finding the power you multiply the given current by the parallel resistor. I thought Vth = RNo(INo) UserIdTAG: 381923 UserNameTAG: matteaton CreateTimeTAG: 2012-09-22T14:26:44Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: If you see the CS and the Rp as a norton circuit, without the rest of the circuit connected, then the conversion to thevenin is Vth= I*Rp when not loaded. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-09-22T21:01:23Z IndexTAG: 4021 TitleTAG: STATIC DISCIPLINE Can anyone tell me that why is it V(Output high & Output low) are on sender's side while V(Input high & Input low) on receiver's side .Since intuitively it appears to be opposite. Thanks in advance UserIdTAG: 230943 UserNameTAG: samsung CreateTimeTAG: 2012-09-22T13:19:50Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Suppose I send you a signal. Since you receive it, you'll have a port open for input. Hence the name. Since the signals you receive go through wires they are subjected to distortion so you'll need to be forgiving to correctly interpret them. How forgiving? Static discipline for inputs tells you that. This is your margin for error. If you then want to pass this signal to somebody else, that is output the signal, you'll have to amplify it - make it strong enough - just as you static discipline dictates for output. FirstChildUserIdTAG: 5325 FirstChildUserNameTAG: vkz FirstChildCreateTimeTAG: 2012-09-22T14:03:26Z FirstChildTAG: Its not really like that. Every unity has both of them. The mosfet for exemple, has input and output. So the value of the gate, which sets if it is open or not is the V(input). And the output as you probably have seen, is the V(output). That is why in a family of devices (in a group of devices that comunicates between themselves) they must have the same static discipline, the same values for input and output. So as that the output of one will be correctly interpreted by the input of another. Hope you can understand what i meant and it helps.. =D FirstChildUserIdTAG: 308853 FirstChildUserNameTAG: fmorato FirstChildCreateTimeTAG: 2012-09-22T14:05:55Z IndexTAG: 4022 TitleTAG: Using tablet I'm using Google Nexus 7 when traveling. Site works fine except for one little feature - I can not drug elements into "sand box" on touch screen. Yesterday I tried half a dozen of different tablets in local computer store and all of them have same issue - can not drug-and-drop elements. Is there a way to copy and paste elements into "sand box"? UserIdTAG: 167413 UserNameTAG: TeTAn CreateTimeTAG: 2012-09-22T12:45:53Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I dont know about your issue but the word you mean is DRAG. drug is something usually illegal stuff to smoke. FirstChildUserIdTAG: 308853 FirstChildUserNameTAG: fmorato FirstChildCreateTimeTAG: 2012-09-22T14:10:52Z SecondChildTAG: We will learn about DRUG DELIVERY is this course. I promise that :) SecondChildUserIdTAG: 95521 SecondChildUserNameTAG: Gatling SecondChildCreateTimeTAG: 2012-09-22T14:51:41Z SecondChildTAG: Autocorrect feature on mobile devices occasionally has mind of its own. :-) Yes, you are right, I meant DRAG. SecondChildUserIdTAG: 167413 SecondChildUserNameTAG: TeTAn SecondChildCreateTimeTAG: 2012-09-22T14:53:05Z SecondChildTAG: psst, I heard there was drugs in here? SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-22T19:48:56Z FirstChildTAG: In my tablet PC with wacom touch screen "sand box" work fine except I can't use "R" button. FirstChildUserIdTAG: 95521 FirstChildUserNameTAG: Gatling FirstChildCreateTimeTAG: 2012-09-22T14:54:52Z SecondChildTAG: Do you just drag elements on touch screen with your finger or you use external Bluetooth mouse? SecondChildUserIdTAG: 167413 SecondChildUserNameTAG: TeTAn SecondChildCreateTimeTAG: 2012-09-22T15:07:19Z SecondChildTAG: The Wacom pen has a hover function, so I imagine that's how it works. SecondChildUserIdTAG: 323387 SecondChildUserNameTAG: ibrahimawwal SecondChildCreateTimeTAG: 2012-09-23T02:37:09Z SecondChildTAG: IMHO adding copy-paste feature to already existing drag-and-drop one should not be complicated but it will make edx run properly on most of current mobile phones and tablets. SecondChildUserIdTAG: 167413 SecondChildUserNameTAG: TeTAn SecondChildCreateTimeTAG: 2012-09-23T05:00:11Z IndexTAG: 4023 TitleTAG: staff help-math processing error h2 althogh i got correct ans with green tick mark but its displaying math processing error everywhere on this page. whats this. sir or one kindly suggest whether ans are correct or what to do. UserIdTAG: 269641 UserNameTAG: BAUWA CreateTimeTAG: 2012-09-22T11:32:18Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 4024 TitleTAG: doubt in homework H2P1 What should i do with that E12 set given? I couldn't understand the problem completely. Can anyone help me out? UserIdTAG: 431422 UserNameTAG: vivek2609 CreateTimeTAG: 2012-09-22T10:44:40Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: this H2P1 problem are actual problem that we might encounter in real life. let say I want to make a circuit as described and after I compute it, I found the resistors value but I confuse where can I get a 1758.84 ohm resistor, there is no resistor like that in the market. there is only that E12 set resistor value in the market. and also the reality is not as good as in theory, real resistor have certain range value or tolerance so the problem is want you to choose the right resistor value from the set and also it has to fulfill the requirement of the Thevenins and the divider ratio (Vout/Vin) requirement FirstChildUserIdTAG: 452559 FirstChildUserNameTAG: kuz1toro FirstChildCreateTimeTAG: 2012-09-22T11:34:49Z SecondChildTAG: In what format should I enter the resistance values? SecondChildUserIdTAG: 431422 SecondChildUserNameTAG: vivek2609 SecondChildCreateTimeTAG: 2012-09-22T17:03:27Z SecondChildTAG: Ohms SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-22T17:41:17Z SecondChildTAG: for example if R1 is 1000 ohm should i enter 1000 or 1K? SecondChildUserIdTAG: 431422 SecondChildUserNameTAG: vivek2609 SecondChildCreateTimeTAG: 2012-09-22T19:10:32Z SecondChildTAG: 1000 SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-22T19:46:47Z IndexTAG: 4025 TitleTAG: unable to draw a line i am unable to draw a line in between the ckt elements..pls suggest me what shud i do ? UserIdTAG: 154314 UserNameTAG: raghav3190 CreateTimeTAG: 2012-09-22T10:37:05Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: try in internet explorer FirstChildUserIdTAG: 364991 FirstChildUserNameTAG: coolvishal FirstChildCreateTimeTAG: 2012-09-22T11:04:00Z IndexTAG: 4026 TitleTAG: Lab Correctness question Does an earned check mark always mean that the lab was successfully solved? I ask this because I got the right transient analysis on lab 2 but with a little discrepancy on the summation ratio of the voltages (e= +- .005). EDIT: Is there a human reviewing the answers? Greetz and have a nice weekend UserIdTAG: 388555 UserNameTAG: 4lk4tr43 CreateTimeTAG: 2012-09-22T09:58:10Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: yap. I experience the same thing on lab 3, the plot of the result seems same with the answer but when i press the check button, it is wrong FirstChildUserIdTAG: 452559 FirstChildUserNameTAG: kuz1toro FirstChildCreateTimeTAG: 2012-09-22T11:23:06Z SecondChildTAG: That is possible when your W/L is not correct, so Ron can turn out to be to high. Then it's possible that the value of VOL exceeds it's max value. SecondChildUserIdTAG: 83276 SecondChildUserNameTAG: salsero SecondChildCreateTimeTAG: 2012-09-22T21:18:37Z IndexTAG: 4027 TitleTAG: homework problem didnt understood :-What is the Thevenin equivalent resistance (in Ohms) of the power source as seen by the load resistance?plz help UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-22T09:35:07Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 4028 TitleTAG: Regarding week 2 homwork plz make me understand what H2P1 is exactly and also which two variable would be such that v1/(v1+v2)=1/2 and v2/(v1+v2)=1/6 lab 2 UserIdTAG: 85797 UserNameTAG: Nilayam CreateTimeTAG: 2012-09-22T09:33:58Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 4029 TitleTAG: plz give proper hint of H2P1 and Lab 2. plz give proper hint of H2P1 and Lab 2. UserIdTAG: 393950 UserNameTAG: RaoUmairTufailAnjum CreateTimeTAG: 2012-09-22T09:12:46Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: The problem can be viewed as a Thevenin problem with a voltage divider. You're given $V_{in}$ and $V_{out}$ already, so you can use what you know to find a relationship between $R_1$ and $R_2$. With that you should be able to select a pair of $E_{12}$ resistors that fit within the constraints on the Ohms it asks you. It's a much simpler question than it seems; you have most of the information provided to you. FirstChildUserIdTAG: 264596 FirstChildUserNameTAG: Nuru FirstChildCreateTimeTAG: 2012-09-22T09:31:59Z FirstChildTAG: Hi, I dont know this correct way to solve .But i got green marks. First solve for Vin/Vout.You will get ratio of R1,R2. From that choose R1, R2 values such that Rth will be in between 10k & 30k . Then calculate for min ,max voltages. This is called **sensitivity analysis**,which will be checked for output voltage variation with tolerance values of resistors. FirstChildUserIdTAG: 118839 FirstChildUserNameTAG: sateeshtal FirstChildCreateTimeTAG: 2012-09-22T09:43:04Z SecondChildTAG: @sateeshtal. I think I understood and solved the problem but the checker gave me red X. If you are correct for Vmin and Vmax, how much significant figures did you use? So that I may try it again. SecondChildUserIdTAG: 143616 SecondChildUserNameTAG: mekusha SecondChildCreateTimeTAG: 2012-09-22T12:59:53Z IndexTAG: 4030 TitleTAG: Help with H2P2 You are to determine the load resistance, RL , for which the maximum power is transferred to the load. (Hint: remember your calculus!) What's optimum load for maximum power? UserIdTAG: 175706 UserNameTAG: rwskim CreateTimeTAG: 2012-09-22T07:02:58Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: I did this by differentiating power Il^2*RL with respect to RL where I is current through RL which I got from current divider Rp/(2Rs+Rl)*I FirstChildUserIdTAG: 230632 FirstChildUserNameTAG: shubhajeet FirstChildCreateTimeTAG: 2012-09-22T08:04:11Z FirstChildTAG: for maximum power transfer,we have the formula dp/dRl=vt[(rt+rl)^2-2*(rt+rl)/(rt+rl)^2]=0 where p=i^2*rl.this is the hint. i think this would help. i solved the problem like this. FirstChildUserIdTAG: 340270 FirstChildUserNameTAG: revathi_srinivasa FirstChildCreateTimeTAG: 2012-09-22T13:15:30Z FirstChildTAG: optimum load for maximum power is the nearest closed loop sum of resistors. In this problem Rl max= 2Rs+Rp FirstChildUserIdTAG: 399292 FirstChildUserNameTAG: Qabali FirstChildCreateTimeTAG: 2012-09-22T07:29:57Z IndexTAG: 4031 TitleTAG: Help with Lab 2! I can't figure out what I am doing wrong in this lab. Using superposition, the Vo shoukd depend on voltage dividers. With V1 off, vo should be R1*V2/(R1+R2) and with V2 off vo should be equal to R2*V1/(R1+R2). But there are no R1 and R2 that could satisfy 1/2 anf 1/6 voltage divider. What am I doing wrong? I tried puting in different values of resistance to change output waveform but can't change amplitude of the output waveform. Please help. One problem with this discussion is that Ican't see comments. I seen posts asking about lab 2 but was not able to see the comments. This seems very useless. UserIdTAG: 175706 UserNameTAG: rwskim CreateTimeTAG: 2012-09-22T06:40:18Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 8 FirstChildTAG: I have done the lab. However, it is misleading because it says you can not make changes to wiring and I did not realize you could add another resistor. I however do not know how to figure out the resistor values using equations. FirstChildUserIdTAG: 175706 FirstChildUserNameTAG: rwskim FirstChildCreateTimeTAG: 2012-09-22T06:58:59Z SecondChildTAG: Check the exercise S3E3 out SecondChildUserIdTAG: 99441 SecondChildUserNameTAG: coyarce SecondChildCreateTimeTAG: 2012-09-24T04:02:47Z FirstChildTAG: You need more resistances. Think in an audio mixer, for example, and how you could make something like that, in the lab. That worked for me. I used 6 resistances. FirstChildUserIdTAG: 220837 FirstChildUserNameTAG: Calsomus FirstChildCreateTimeTAG: 2012-09-22T06:59:33Z FirstChildTAG: I'm going crazy too. I can't figure out the lab, I've tried to solve it with equations but the only result I get is that there's no solution, at least with only two resistances... I'm trying to solve it with 3 resistances but... strange things happen again... FirstChildUserIdTAG: 456610 FirstChildUserNameTAG: Estebanterrero FirstChildCreateTimeTAG: 2012-09-22T08:37:02Z FirstChildTAG: Do a search. There has been much discussion of this problem including a very long thread from about four days ago. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-09-22T11:43:07Z SecondChildTAG: do you remember the title of that thread? there are so many posts, it's going to be difficult to find the right one SecondChildUserIdTAG: 267702 SecondChildUserNameTAG: curio_city SecondChildCreateTimeTAG: 2012-09-22T17:58:58Z FirstChildTAG: see..i solved lab2 after a lot of hardwork... what i found was..when i ws using resistors in the range of hundreds..i was getting wrong answer..when itried in megaohms or more...i got right!!! i dont know how..but may be u can give it a shot too!! We are simply interested in getting ratios right... FirstChildUserIdTAG: 126705 FirstChildUserNameTAG: arpitchugh FirstChildCreateTimeTAG: 2012-09-22T13:04:49Z SecondChildTAG: moreover i used 4 resistors! SecondChildUserIdTAG: 126705 SecondChildUserNameTAG: arpitchugh SecondChildCreateTimeTAG: 2012-09-22T13:05:36Z SecondChildTAG: 5* SecondChildUserIdTAG: 126705 SecondChildUserNameTAG: arpitchugh SecondChildCreateTimeTAG: 2012-09-22T13:05:58Z SecondChildTAG: 3 is enough. And it's not a question of resistors values but ratios of resistors values SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-22T19:19:37Z SecondChildTAG: In fact, more than 2 resistors are needed. I solved this Lab assuming if I use n resistors, then n-1 of them should be written as a function of the last one. So, once u have all your n-1 resistors written as functions of your last one, give it a value (for example from the E12 list) and..done! SecondChildUserIdTAG: 99441 SecondChildUserNameTAG: coyarce SecondChildCreateTimeTAG: 2012-09-24T03:59:10Z FirstChildTAG: If you think about the problem correctly, it is not hard. The number are so easy to work with that you don't need a calculator to do the arithmetic. Since there are no other constrains given the absolute value of the resistors doesn't matter, only the ratios. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-09-22T13:20:17Z FirstChildTAG: As other people mentioned you'll need to add something. With any one source turned off you always get a voltage divider network. Say, turn the V2 off, you now have a divider which gives you the coefficient for V1 (r2/(r1+r2)) just like you said yourself. There's only two things you can do: a) add some resistance in series b) add some in parallel. Adding in series doesn't really change the picture, does it? Say, you added r3 in series to r1. Now rename r1+r3 to be your new r1 and you're back where you started. So it must be the other option. You'll need to add resistors in parallel. Do you need to add more than one? Probably not, since any resistors in parallel can be replaced with one, which resistance you can calculate easily. Now you can write your V = a*V1 + b*V2, where a and b are functions of r1, r2 and the resistance you added r3. Since you have a = 1/2 and b = 1/6 you seem to be one equation short. Or are you? Look very carefully at the coefficients to V1 and V2 in your original network. Do you really need to know the actual values of r1 and r2? Or their relationship will be enough? (Hint: there's infinite number of networks that satisfy V = 1/2*V1 + 1/6*V2) FirstChildUserIdTAG: 5325 FirstChildUserNameTAG: vkz FirstChildCreateTimeTAG: 2012-09-22T13:41:09Z SecondChildTAG: Which one? SecondChildUserIdTAG: 298867 SecondChildUserNameTAG: JNPH SecondChildCreateTimeTAG: 2012-09-22T15:51:57Z SecondChildTAG: I add R3 in parallel and Get the values of R1 ,R2,R3 By using superposition and assume one value of R1 or R2 or R3 thne , i Reach to the relation V = a*V1 + b*V2, a = 1/2 and b = 1/6 , But the final waveform is not the same of the Required one , where My problem now SecondChildUserIdTAG: 448321 SecondChildUserNameTAG: Keroo SecondChildCreateTimeTAG: 2012-09-22T16:26:06Z SecondChildTAG: then you'll need to check your solution. Without looking at your particular numbers I won't help. As long as you understand the intuition behind and careful with your arithmetic, you'll get there. You should arrive at very simple expressions: R2 = k*R1 R3 = m*R2 = m*k*R1 SecondChildUserIdTAG: 5325 SecondChildUserNameTAG: vkz SecondChildCreateTimeTAG: 2012-09-22T19:01:40Z FirstChildTAG: I add R3 in parallel and Get the values of R1 ,R2,R3 By using superposition and assume one value of R1 or R2 or R3 thne , i Reach to the relation V = a*V1 + b*V2, a = 1/2 and b = 1/6 , But the final waveform is not the same of the Required one , where My problem now FirstChildUserIdTAG: 448321 FirstChildUserNameTAG: Keroo FirstChildCreateTimeTAG: 2012-09-22T16:55:33Z SecondChildTAG: You can satisfy V = 1/2 V1 + 1/6 V2 without having the right waveform? I used a 3rd resistor connected to ground (because its easy to calculate) to bring the amplitude down. SecondChildUserIdTAG: 388555 SecondChildUserNameTAG: 4lk4tr43 SecondChildCreateTimeTAG: 2012-09-22T18:46:42Z SecondChildTAG: You can't. The important feature of linear circuits is that they preserve the waveform. If your R1, R2, R3 indeed satisfy V = 1/2V1 + 1/6V2 you get the waveform automatically. Basically thats exactly what this equation says: Vout is superposition of scaled signals. SecondChildUserIdTAG: 5325 SecondChildUserNameTAG: vkz SecondChildCreateTimeTAG: 2012-09-22T19:05:56Z IndexTAG: 4032 TitleTAG: what is operating point and volt graphical method what is operating point and volt graphical method and how to calculate that?? UserIdTAG: 118611 UserNameTAG: mitianhari CreateTimeTAG: 2012-09-22T06:35:00Z VoteTAG: 0 CoursewareTAG: Week 3 / Incremental Method Motivation CommentableIdTAG: 6002x_incremental_method_motivation NumberOfReplyTAG: 0 IndexTAG: 4033 TitleTAG: HW2 P1 Guys, i solve the equation, and found in which range my values are correct. But mitx's checker that values are incorrect! i substitude all values from array, but each oh them are incorrect. I try to use another formats(68*10^3, 68000, 68), but they are also incorrect. Where i mistaked? UserIdTAG: 444204 UserNameTAG: hekan CreateTimeTAG: 2012-09-22T06:34:55Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Same problem is there FirstChildUserIdTAG: 393950 FirstChildUserNameTAG: RaoUmairTufailAnjum FirstChildCreateTimeTAG: 2012-09-22T09:15:39Z SecondChildTAG: try making a spreadsheet with various combinations and using that. it helps. SecondChildUserIdTAG: 150267 SecondChildUserNameTAG: vwsingh SecondChildCreateTimeTAG: 2012-09-22T09:30:02Z SecondChildTAG: Yes, use spreadsheet. Works for me!! SecondChildUserIdTAG: 119160 SecondChildUserNameTAG: Thyaga SecondChildCreateTimeTAG: 2012-09-22T11:14:35Z IndexTAG: 4034 TitleTAG: stuck with hw 2 problem 1 help I have been trying to complete hw 2 problem but have been stuck. Now I am really confused whether i have not grasped the question or have not understood course material. Here is my way of thinking plz correct me. **So from volatge divider** ---------- - 14= R1*70/(R1+R2) - R1/(R1+R2) = 14/70 = 1/5 ---------- - Rth = R1 || R2 - 10*10^3 < R1*R2/(R1+R2) < 30*10^3 - from voltge divider eqn - 10*10^3 < R2/5 < 30*10^3 - 50*10^3 < R2 < 150*10^3 ---------- possible values - r2 = 56 kohm, 68 kohm, 82 kohm, 100 kohm, 120 kohm, 120 kohm 150 kohm ---------- simplifying voltage divider eqn we get - r2=4r1 - r1= 14 kohm, 17kohm , 20.5kohm , 25kohm, 30kohm, 37.5 kohm - possible nearest values: - r1= 15 kohm, 18kohm , 22kohm , 27kohm, 33kohm, 39 kohm - using above value we get output voltage - vout=14.78, 14.65, 14.8, 14.8, 15, 14.4 - choosing the best answer: - *150, 39 kohm* **this doesnot seem to be the answer.help!!!** - also i calculated maxvolt by increasing r1 by 10 and decreasing r2 by 10 = 16.88 - also i calculated mvoinlt by decreasing r1 by 10 and increasing r2 by 10 = 12.27 UserIdTAG: 230632 UserNameTAG: shubhajeet CreateTimeTAG: 2012-09-22T06:27:46Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 4035 TitleTAG: c2 wrong solution?! The value of c2 should be similar in form to b1. So the numerator should be (R1 + R3) instead of R1*R3. Kindly look into this. Thanks UserIdTAG: 129669 UserNameTAG: nano_vlsi CreateTimeTAG: 2012-09-22T06:26:36Z VoteTAG: 0 CoursewareTAG: Week 2 / Linear combinations of source strenghts CommentableIdTAG: 6002x_S3E2_Linear_Combinations_of_Source_Strengths NumberOfReplyTAG: 1 FirstChildTAG: hello, c2 and b1 are indeed similar, numerators being (-R1-R3) AND (-R2-R3) respectively. jus check ur KCL at the top node again :) FirstChildUserIdTAG: 357292 FirstChildUserNameTAG: RaghavAbboy FirstChildCreateTimeTAG: 2012-09-22T12:28:51Z IndexTAG: 4036 TitleTAG: H2P1 I got my resistors value correct but unable to find Vmax and Vmin correct.Pls tell me how to proceed. UserIdTAG: 401259 UserNameTAG: amangodne CreateTimeTAG: 2012-09-22T06:19:11Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Think in the 10 % tolerance of the resistors and how it affects their values. If they change their value, Vout won't be what you calculated, right? Resistances can be 10 % higher or lower than their nominal values. Hope it helps. FirstChildUserIdTAG: 220837 FirstChildUserNameTAG: Calsomus FirstChildCreateTimeTAG: 2012-09-22T06:55:01Z FirstChildTAG: how did u find the value of R1 and R2? FirstChildUserIdTAG: 393950 FirstChildUserNameTAG: RaoUmairTufailAnjum FirstChildCreateTimeTAG: 2012-09-22T09:18:05Z FirstChildTAG: The same problem the value of resistors its OK but I could not find the value of Vmax and Vmin (R1=39k and R2=22k). Any ideas please. FirstChildUserIdTAG: 319792 FirstChildUserNameTAG: chou25alg FirstChildCreateTimeTAG: 2012-09-22T14:17:01Z SecondChildTAG: i found the values vmax and Vmin..do u need the steps? or have you got it? :) SecondChildUserIdTAG: 338685 SecondChildUserNameTAG: varshaD SecondChildCreateTimeTAG: 2012-09-22T15:27:02Z IndexTAG: 4037 TitleTAG: Debate: Does God Exist? Atheists vs. Theists Religion is faith driven. What I find is that this faith gives theists freedom to circumvent all logical arguments against religion. All theists have to do is believe and everything they do is justified. But when atheists present clear and concise evidence, everything they say is automatically wrong regardless of the evidence given to support their standpoint. Atheists have plate tectonics, the big bang theory, evolution, relativity, and more scientific approaches that support the justification that god does not exist. In contrast, theists have nothing to support their assertions. Everyone has the right to believe in god. However, when you use your beliefs to justify acts that violates ethics we have the right to question those acts. Theists have always asked us to respect their decision, but what about our decisions? We can condone what you do, but then we'd have to grant the ban of abortion, blood transfusions, stem-cell research, and more. That's when we have to step in. Theists show no respect. Well then, we show no respect either. That's my perspective on religion. Anybody have objections? UserIdTAG: 282221 UserNameTAG: TylerVo CreateTimeTAG: 2012-09-22T06:16:44Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: İrrevelant... FirstChildUserIdTAG: 182653 FirstChildUserNameTAG: monkeyfoahead FirstChildCreateTimeTAG: 2012-09-22T09:40:08Z FirstChildTAG: Why are you posting this here? FirstChildUserIdTAG: 264596 FirstChildUserNameTAG: Nuru FirstChildCreateTimeTAG: 2012-09-22T06:21:36Z SecondChildTAG: Is there a particular reason why I can't? I already know all about circuits. Hm. I wish to know why the world exists. SecondChildUserIdTAG: 282221 SecondChildUserNameTAG: TylerVo SecondChildCreateTimeTAG: 2012-09-22T06:28:38Z SecondChildTAG: Yes. This is a discussion forum for the class at hand. SecondChildUserIdTAG: 264596 SecondChildUserNameTAG: Nuru SecondChildCreateTimeTAG: 2012-09-22T09:09:00Z FirstChildTAG: So...uhh, *nobody* wants to debate? FirstChildUserIdTAG: 282221 FirstChildUserNameTAG: TylerVo FirstChildCreateTimeTAG: 2012-09-22T06:37:34Z SecondChildTAG: there are theists who are close minded people and atheists who are close minded people and both types are jerks, not get out of here please SecondChildUserIdTAG: 219945 SecondChildUserNameTAG: Phobos9703 SecondChildCreateTimeTAG: 2012-09-22T16:02:23Z SecondChildTAG: now* SecondChildUserIdTAG: 219945 SecondChildUserNameTAG: Phobos9703 SecondChildCreateTimeTAG: 2012-09-22T16:06:19Z FirstChildTAG: I object to having this on a forum dedicated to learning about circuits. Please find somewhere else to discuss this. FirstChildUserIdTAG: 248902 FirstChildUserNameTAG: Chanute FirstChildCreateTimeTAG: 2012-09-22T06:22:00Z FirstChildTAG: Hi TylerVo, Thank you for sharing your thoughts, you are right to express your point of view. But I think that as this is a Forum Discussion of Circuits and Electronics that are visited from many people worldwide with different cultures, religions and sexual orientation, this post-debate can offend or hurt to someone’s belief … and debating this will become inappropiate … I hope that this comment does not offend you…and if yes, my apologies. My best wish to you, Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-22T16:00:52Z IndexTAG: 4038 TitleTAG: H2P1 Resistors table While choosing the resistance from the values in the table (we "buy" the resistors) can I **add** some of them to form R1 or R2? e.g. R1 = 40kOhm which is **4x**(10kOhm) resistors together. Because that seems to be the only way of solving this problem, if I just choose single values I don't get the Thevenin resistance in the required range UserIdTAG: 410033 UserNameTAG: sagitta CreateTimeTAG: 2012-09-22T05:36:21Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I believe you are supposed to use exactly two resistors. FirstChildUserIdTAG: 264596 FirstChildUserNameTAG: Nuru FirstChildCreateTimeTAG: 2012-09-22T05:40:00Z SecondChildTAG: Well, probably yes. I'll stick to that. It's just that the combination of resistors as a solution would give an exact solution for nominal values (without the tolerance) :) SecondChildUserIdTAG: 410033 SecondChildUserNameTAG: sagitta SecondChildCreateTimeTAG: 2012-09-22T06:33:58Z SecondChildTAG: it's solvable. automate - use a spreadsheet using various resistor values (highs and lows - include and reduce tolerances) and use that to calculate various Vth using the voltage divider formulas. you'll get the right answers in no time. SecondChildUserIdTAG: 150267 SecondChildUserNameTAG: vwsingh SecondChildCreateTimeTAG: 2012-09-22T09:35:09Z IndexTAG: 4039 TitleTAG: multiple +/- voltages, when to add and subtract Say i have two power sources, especially with one revered and a resistor in series, i'm getting totally lost with what's pos and neg and whether i should be adding or subtracting voltages. I get all the analyses methods and maths i'm just getting wrong answers half the time as i mix these up. I just cant get it to sit in my head right. UserIdTAG: 329964 UserNameTAG: SmartMike CreateTimeTAG: 2012-09-22T04:14:07Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi, first of all you've got to asign polarities to resistors in the circuit. After that, when you travel around the loop writing down KVL, you write down the FIRST sign encountered when you "arrive" to the element. Hope it helps. FirstChildUserIdTAG: 220837 FirstChildUserNameTAG: Calsomus FirstChildCreateTimeTAG: 2012-09-22T04:23:38Z IndexTAG: 4040 TitleTAG: H2P2 I got the Thevenin equivalent finally as 2Rs + Rp... but I dont get how Rp is in series with Rs and not in parallel! I mean it definitely looks like its in parallel UserIdTAG: 381923 UserNameTAG: matteaton CreateTimeTAG: 2012-09-22T01:08:57Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi, imagine the current source and RL are gone and you put a multimeter at the port where RL was connected to. What equivalent resistance you find? RS + RP + RS. RP is in series in that loop. FirstChildUserIdTAG: 220837 FirstChildUserNameTAG: Calsomus FirstChildCreateTimeTAG: 2012-09-22T02:25:24Z SecondChildTAG: Look at the load side, you would find that the current source will be "open". Then you will be left with the three resistances in series. Add up and then you have your Rth. SecondChildUserIdTAG: 209082 SecondChildUserNameTAG: Nikemurton SecondChildCreateTimeTAG: 2012-09-22T02:35:51Z IndexTAG: 4041 TitleTAG: how i get Vth in S3E5? please explain UserIdTAG: 290966 UserNameTAG: AhmedImam CreateTimeTAG: 2012-09-22T00:19:14Z VoteTAG: 0 CoursewareTAG: Week 2 / Thevenin model CommentableIdTAG: 6002x_S3E5_Thevenin_Model NumberOfReplyTAG: 2 FirstChildTAG: You've got a current source in parallel with the resistor Rp. By ohm's law, that flow of current will produce a voltage across Rp only. That is VTH in this case. Remember that you have no current flowing through the output port (i), so there's no drop of voltage across both resistors Rs and Vth appears "untouched" at the output port. FirstChildUserIdTAG: 220837 FirstChildUserNameTAG: Calsomus FirstChildCreateTimeTAG: 2012-09-22T02:33:51Z SecondChildTAG: thanks :) SecondChildUserIdTAG: 290966 SecondChildUserNameTAG: AhmedImam SecondChildCreateTimeTAG: 2012-09-23T05:38:15Z FirstChildTAG: its simple Vth is voltage across v which is product of series combination of(rs,rp,rs) and current source I. FirstChildUserIdTAG: 231749 FirstChildUserNameTAG: utshau FirstChildCreateTimeTAG: 2012-09-24T06:16:02Z IndexTAG: 4042 TitleTAG: Scratching my head I spent 30 minutes on finding x1 because for some dumb reason I though we needed current.hahhahaha UserIdTAG: 230787 UserNameTAG: Gabriel007 CreateTimeTAG: 2012-09-21T23:18:01Z VoteTAG: 0 CoursewareTAG: Week 2 / Superposition example CommentableIdTAG: 6002x_S3E3_superposition NumberOfReplyTAG: 0 IndexTAG: 4043 TitleTAG: H2P1 catastroph Vth=Vin*(R2/..+..) : this is the first equation . R1//R2=10 to 30 Kohm :this is the second equation vin=30, vout=6 , so why i dont find right answer (82 -22 ) please help UserIdTAG: 231318 UserNameTAG: moutasem CreateTimeTAG: 2012-09-21T22:19:39Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi, you could start with Vout = Vin * R2 / (R1 + R2), so you know that desired Vout is about 24.5 V, and Vin is 70 V, so make Vout/Vin = R2 / (R1 + R2) and that gaves you the relationship between R1 and R2 values. For me that was the starting point. By the way, you wrote vout = 6?? FirstChildUserIdTAG: 220837 FirstChildUserNameTAG: Calsomus FirstChildCreateTimeTAG: 2012-09-21T22:45:33Z SecondChildTAG: Hey, I was wondering if you could help me with this: I've got my resistor values, but even after trying for hours, I'm not able to get the correct Vmax and Vmin, any idea ? SecondChildUserIdTAG: 358606 SecondChildUserNameTAG: st1046 SecondChildCreateTimeTAG: 2012-09-22T04:22:36Z SecondChildTAG: thank you , but how deduced that desired Vout= 24,5 , and Vin=70 . for your question i wrote vout = 6 v as they demanded in the case SecondChildUserIdTAG: 231318 SecondChildUserNameTAG: moutasem SecondChildCreateTimeTAG: 2012-10-17T08:51:56Z IndexTAG: 4044 TitleTAG: please clarify "We are given an input voltage Vin=80.0V, and we need to provide an open-circuit output voltage of Vout≈24.0V. An additional requirement is that the Thevenin resistance as seen from the output terminals is between 10kΩ and 30kΩ. Assume first that the resistors have their nominal resistance. Come up with resistors R1 and R2 such that the divider ratio Vout/Vin is within 10% of the requirement." can someone please clarify what this last statement means?-within 10% of the requirement stands for which requirement? UserIdTAG: 218653 UserNameTAG: jaz07 CreateTimeTAG: 2012-09-21T21:54:40Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: It's asking for $ \cfrac {V_{out}}{V_{in}} \cdot 90 \% \le \cfrac {V_{out~actual}}{V_{in}} \le \cfrac {V_{out}}{V_{in}} \cdot 110 \% $ according to your numbers, $ \cfrac {V_{out}}{V_{in}} = \cfrac {24.0}{80.0}$ FirstChildUserIdTAG: 248902 FirstChildUserNameTAG: Chanute FirstChildCreateTimeTAG: 2012-09-21T23:31:58Z SecondChildTAG: thanxs Chanute...that was indeed very helpful!!:) SecondChildUserIdTAG: 218653 SecondChildUserNameTAG: jaz07 SecondChildCreateTimeTAG: 2012-09-27T19:22:57Z FirstChildTAG: if you buy a resistor with a value, for example 1000 ohm 10%, the real value can be from 900 to 1100(+-10%) If the value of resistor varies, also the value of voltage varies FirstChildUserIdTAG: 466996 FirstChildUserNameTAG: tntvlad FirstChildCreateTimeTAG: 2012-09-21T23:35:10Z IndexTAG: 4045 TitleTAG: wow that's spectacular ! UserIdTAG: 158348 UserNameTAG: mudz CreateTimeTAG: 2012-09-21T21:29:17Z VoteTAG: 0 CoursewareTAG: Week 2 / Chain saw CommentableIdTAG: 6002x_chain_saw NumberOfReplyTAG: 0 IndexTAG: 4046 TitleTAG: week2 homework1 error no matter what values i provide to the answer box.it shows a cross X mark.how is it possible..it have to have an answer......is it that.once i give a wrong answer i am not allowed to change it??? please reply ASAP...as 23 is the last day of submission. UserIdTAG: 137813 UserNameTAG: aSRamOS CreateTimeTAG: 2012-09-21T21:26:49Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: It can be changed. try to get the correct answer. FirstChildUserIdTAG: 131726 FirstChildUserNameTAG: chemiboy11 FirstChildCreateTimeTAG: 2012-09-21T21:33:58Z FirstChildTAG: yes you are allowed to change your answer before submission dates..u might get the wrong result thats why its shows a cross mark FirstChildUserIdTAG: 321556 FirstChildUserNameTAG: sj31867 FirstChildCreateTimeTAG: 2012-09-21T21:42:43Z FirstChildTAG: yes you can change the answer before last date FirstChildUserIdTAG: 321556 FirstChildUserNameTAG: sj31867 FirstChildCreateTimeTAG: 2012-09-21T21:43:41Z FirstChildTAG: This is not reasonable. I must pick from a limited set of resistors and no matter which I put in, it's wrong. There are only 12 resistors in the set. Am I on the wrong track? FirstChildUserIdTAG: 373585 FirstChildUserNameTAG: radami1 FirstChildCreateTimeTAG: 2012-09-22T04:18:53Z SecondChildTAG: there are not only 12 , that twelve are multipied by the powers off 10 , from 10e+0 until 10e+5 , so 6 powers that multiply each basic value (those 12 ) , so 12 values x 6 powers = 72 values .........................."the available values of resistors with 10% tolerance are selections from the E12 set multiplied by a power of ten from 100 through 105. The E12 set is:E12={10,12,15,18,22,27,33,39,47,56,68,82}" ; "Thus, you can buy 10% resistors with a nominal resistance of 330Ω or 33kΩ, but not 350Ω" SecondChildUserIdTAG: 402850 SecondChildUserNameTAG: y1111 SecondChildCreateTimeTAG: 2012-09-22T04:34:29Z SecondChildTAG: OK. R1 is the thevinin resistance as seen from Vout and the problem states that the thevinin resistance from Vout is between 10k and 30k. So that narrows the R1 value to 6 possible resistors {10k,12k,15k,18k,22k,27k} SecondChildUserIdTAG: 373585 SecondChildUserNameTAG: radami1 SecondChildCreateTimeTAG: 2012-09-22T11:49:28Z IndexTAG: 4047 TitleTAG: synaxis for HW2 last time in Lab I made mistake by wrote incorrect decimal devider, Now I ask to help by pointing how I must insert kOhm as 000 or k= f.e. 15k is correct or 15000? Thanks in advance UserIdTAG: 202249 UserNameTAG: ABZ CreateTimeTAG: 2012-09-21T21:22:07Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: the box usually ask for an answer and give the unit in parenthesis. if in the parants is "(Ohm)" then u write 15000 , if the parantss is "(kOhm)" then u write just 15 ,. but look , i just check in the S3E4 that it recognizes 15k too , so both modes will work FirstChildUserIdTAG: 402850 FirstChildUserNameTAG: y1111 FirstChildCreateTimeTAG: 2012-09-22T04:41:40Z IndexTAG: 4048 TitleTAG: V0 I do not understand UserIdTAG: 459680 UserNameTAG: jejy CreateTimeTAG: 2012-09-21T21:17:37Z VoteTAG: 0 CoursewareTAG: Week 1 / Node analysis practice, part 1 CommentableIdTAG: 6002x_L2Node0 NumberOfReplyTAG: 2 FirstChildTAG: The voltage at that node is measured with respect to(w.r.t) ground node that is zero. Now the voltage source of the rating V0 is providing the voltage to all the elements present in the circuit.Now at node 'a' we find that it is directly connected to the voltage source that is there is no voltage consuming element like a resistor between the voltage source and node 'a', so there is no voltage drop between 'a' and the voltage source.Thus w.r.t the ground node 'a' has V0 - 0 = V0 potential. FirstChildUserIdTAG: 156225 FirstChildUserNameTAG: sanjana_m FirstChildCreateTimeTAG: 2012-09-23T18:19:38Z FirstChildTAG: simply apply kvl around the node which contain the voltage source you'll find that Va to ground =V0 but v at ground equal 0 so Va-Vg=V0 so Va=V0 FirstChildUserIdTAG: 221617 FirstChildUserNameTAG: konan FirstChildCreateTimeTAG: 2012-09-28T11:53:20Z IndexTAG: 4049 TitleTAG: S8E1 Can't solve it. It should solve as: (Vo-Vs)/R = K/(Vo^2) Am I right? UserIdTAG: 378778 UserNameTAG: happylife CreateTimeTAG: 2012-09-21T21:01:58Z VoteTAG: 0 CoursewareTAG: Week 4 / Dependent Sources Exercise 1 CommentableIdTAG: 6002x_dep_src_exsz_1 NumberOfReplyTAG: 2 FirstChildTAG: sorry. understood. S8V3 FirstChildUserIdTAG: 378778 FirstChildUserNameTAG: happylife FirstChildCreateTimeTAG: 2012-09-21T21:19:42Z SecondChildTAG: S8E1 (vo-VS)/R = -K/(VS-vo)^2 SecondChildUserIdTAG: 197569 SecondChildUserNameTAG: darksamaro SecondChildCreateTimeTAG: 2012-09-22T15:50:35Z SecondChildTAG: use the node method to find first v of R and than use the kvl law to find v0 good luck SecondChildUserIdTAG: 266386 SecondChildUserNameTAG: zakzak200 SecondChildCreateTimeTAG: 2012-09-30T22:38:45Z FirstChildTAG: (vo-VS)/R = -K/(VS-vo)^2 FirstChildUserIdTAG: 197569 FirstChildUserNameTAG: darksamaro FirstChildCreateTimeTAG: 2012-09-22T15:51:06Z SecondChildTAG: Good luck with that.This is solved in a very mathematical whird way in one of the tutorials.V0 = VS-cubicrooot(K*R) , AND it is the simplest way. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-09-23T09:21:21Z SecondChildTAG: PS : the dependent source depends of v, the voltage across the rezistor. SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-09-23T09:24:28Z IndexTAG: 4050 TitleTAG: Incorrect source data in H2P1: VOLTAGE-DIVIDER DESIGN Hello. I have some problems with Homework 2 in H2P1. > We are given an input voltage Vin=50.0V, and we need to provide an > open-circuit output voltage of Vout≈10.0V. An additional requirement > is that the Thevenin resistance as seen from the output terminals is > between 10kΩ and 30kΩ. Assume first that the resistors have their > nominal resistance. Come up with resistors R1 and R2 such that the > divider ratio Vout/Vin is within 10% of the requirement. I tried to solve this problem, but I failed. I used Nodal Analysis: (e-Vin)/R1+e/R2=0 => R1=4R2. But "An additional requirement is that the Thevenin resistance as seen from the output terminals is between 10kΩ and 30kΩ". Maybe, I did mistake, maybe, data is incorrect.. Please, help me! UserIdTAG: 328920 UserNameTAG: TisO CreateTimeTAG: 2012-09-21T21:00:55Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: HI!!! what you are supposed to do is to make two equations one for Vout/Vin and another one for thevenin resistance for(10k ohm) then solve for R1 and R2, and again take thevenin resistance to be 30k ohm and again find the value for R1 and R2..now it is given that only E12 set of resistance elements are possible so you have to take a nearby permissible values for your resistance..now you got 2 pair of permissible values of R1 and R2..now you have to check which pair is abiding the given conditions like(thevenin resitance between 10-30k-ohm and Vout/Vin within 10%of requirement)the pair which may qualify for this relation is R1 and R2...now using these values find Vmax and Vmin FirstChildUserIdTAG: 321556 FirstChildUserNameTAG: sj31867 FirstChildCreateTimeTAG: 2012-09-21T21:35:42Z SecondChildTAG: What does Vout/Vin is within 10% of the requirement? That Vout/Vin should be 10% (0.1) ??? If that's the case then Vout should not be 10V cause 10v/50v = 0.20 so if Vout/Vin should be 0.1 then Vout should not be 10V but 5V Am I making any sense? SecondChildUserIdTAG: 29736 SecondChildUserNameTAG: Euly SecondChildCreateTimeTAG: 2012-09-22T03:45:48Z SecondChildTAG: sj31867 , i was doing like TisO , only my values are different but on the same logic, and i got stuck the same way. did u make in this way that u explain and it worked ? SecondChildUserIdTAG: 402850 SecondChildUserNameTAG: y1111 SecondChildCreateTimeTAG: 2012-09-22T04:47:03Z SecondChildTAG: I think I got the correct resistor Values. For me Vin = 90V and Vout = 18V. Rth should be between 10K and 30K. So after doing some calculations I got this: ![Resistor Values for a Perfect Vout of 18V][1] Now since I need to pick from the E12 resistor list. I use the approximate nominal values of: R1=22k and R2=5600 and I still get a Vout that is 10% within what is required: ![Values of Resistor from E12 set][2] aaaaaaaaaand yet I input this values as my answer and it is wrong... So I'm wondering what does the Vout/Vin really means... Cause 18/90 = 0.20 or 20% and not 10% as the problem request... so in the end... are they asking me to have 18V on Vout or 10% of Vin... I'm confused... [1]: https://edxuploads.s3.amazonaws.com/13482914034779373.jpg [2]: https://edxuploads.s3.amazonaws.com/134829153151111.jpg SecondChildUserIdTAG: 29736 SecondChildUserNameTAG: Euly SecondChildCreateTimeTAG: 2012-09-22T05:27:13Z SecondChildTAG: and I'm confused too, because my friend solved this problem like me and he got the right answer ... Mystic.. SecondChildUserIdTAG: 328920 SecondChildUserNameTAG: TisO SecondChildCreateTimeTAG: 2012-09-22T09:34:57Z IndexTAG: 4051 TitleTAG: Video is not working The video tutorials of week 3 are not working. UserIdTAG: 79024 UserNameTAG: WesllyMorais CreateTimeTAG: 2012-09-21T19:07:01Z VoteTAG: 0 CoursewareTAG: Week 3 / Load Line Tutorial CommentableIdTAG: 6002x_load_line_tutorial NumberOfReplyTAG: 1 FirstChildTAG: Try to pause and unpause the video. FirstChildUserIdTAG: 376877 FirstChildUserNameTAG: AndBre FirstChildCreateTimeTAG: 2012-09-22T08:07:16Z IndexTAG: 4052 TitleTAG: Staff: In Pakistan youtube is banned how can I see the lectures kindly plz reply fast UserIdTAG: 414201 UserNameTAG: uzaifakram CreateTimeTAG: 2012-09-21T19:01:03Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You can used proxy software, like tor: https://www.torproject.org/ FirstChildUserIdTAG: 308853 FirstChildUserNameTAG: fmorato FirstChildCreateTimeTAG: 2012-09-21T19:26:55Z IndexTAG: 4053 TitleTAG: H3P1 - Mosfet SR model The question asks for the minimum value of the pullup resistor RPuI. I guess it has to be so that Vout is less than the specified for Vol. Wouldn't that be 1, for the resistor? UserIdTAG: 308853 UserNameTAG: fmorato CreateTimeTAG: 2012-09-21T18:57:18Z VoteTAG: 0 CoursewareTAG: Week 3 / Switch Resistor Model of a MOSFET CommentableIdTAG: 6002x_switch_resistor_model NumberOfReplyTAG: 3 FirstChildTAG: Vout = Vol so use Vout= 1 if Vol = 1 FirstChildUserIdTAG: 73809 FirstChildUserNameTAG: rhyssouth FirstChildCreateTimeTAG: 2012-09-21T19:50:50Z SecondChildTAG: Use Vol=1, because it's asking for the minimum Rp witch will be in the minimum output. SecondChildUserIdTAG: 399292 SecondChildUserNameTAG: Qabali SecondChildCreateTimeTAG: 2012-09-22T06:58:20Z FirstChildTAG: Use Vol=1, because it's asking for the minimum Rp witch will be in the minimum output. FirstChildUserIdTAG: 399292 FirstChildUserNameTAG: Qabali FirstChildCreateTimeTAG: 2012-09-22T06:58:42Z FirstChildTAG: Ok. I understood. Thank you guys. If the resistence is greater than when the voltage is 1V, the voltage in that resistor will decrease. Though I'm not getting the right value. Looks like I have a basic math mistake, as usual. FirstChildUserIdTAG: 308853 FirstChildUserNameTAG: fmorato FirstChildCreateTimeTAG: 2012-09-22T14:17:43Z IndexTAG: 4054 TitleTAG: Errors on screen I can not see any lectures. The labs and homework Show "Math Processing Error" after a few seconds of real numbers. Help Raymond R Sutton UserIdTAG: 425297 UserNameTAG: raymondrsutton CreateTimeTAG: 2012-09-21T18:42:04Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I also cann't proceed labs and homework. the cheek button doesn't work. FirstChildUserIdTAG: 197641 FirstChildUserNameTAG: mtanju FirstChildCreateTimeTAG: 2012-09-21T19:03:50Z FirstChildTAG: Yes me too FirstChildUserIdTAG: 49861 FirstChildUserNameTAG: codymartin FirstChildCreateTimeTAG: 2012-10-03T23:48:09Z IndexTAG: 4055 TitleTAG: del phi B del t What is del over del t stands for? UserIdTAG: 139448 UserNameTAG: meno CreateTimeTAG: 2012-09-21T18:39:30Z VoteTAG: 0 CoursewareTAG: Week 1 / Lumped Element Abstraction CommentableIdTAG: 6002x_lumped_element_abstraction NumberOfReplyTAG: 1 FirstChildTAG: This comes from electromagnetic theory. ${\phi}_B$ stands for magnetic flux. $\frac{\partial\phi_B}{\partial t}$ is the rate of change of magnetic flux with time. Michael Faraday showed that a changing flux through a conductor induces a voltage which causes a current to flow. The voltage is given by $V = -\frac{\partial\phi_B}{\partial t}$ That's what Prof. Agarwal is talking about. The point he is trying to make is, as we go around a loop in the circuit (in our EECS playground), there is no external changing magnetic field. Hence the only sources of voltage and current are the ones you see in the circuit. If that term was not zero, you would have to include an extra voltage term while using KVL. Hope this helps! :-) FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-09-21T20:00:52Z IndexTAG: 4056 TitleTAG: Help with Lab Need help with lab 2!!!!! I have been breaking my head with Lab 2 for a while but haven't been able to figure it out yet! just got the equations "r1=r2" n "5r2=r1"... ![][1] [1]: https://edxuploads.s3.amazonaws.com/1348252175713744.png Wanted to know why my circuit is wrong, is it cuz it is finding difference between the potentials of the two points at the voltage dividers instead of adding them? if so how do i correct it??? UserIdTAG: 468052 UserNameTAG: Charan94 CreateTimeTAG: 2012-09-21T18:32:10Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Try with different arrangements of resistors. and make the calculation. that will give the right values. FirstChildUserIdTAG: 308853 FirstChildUserNameTAG: fmorato FirstChildCreateTimeTAG: 2012-09-21T18:59:46Z FirstChildTAG: The Hint and figure 2 is just a hint. If you look at the formula at the hint, then it applies only to figure 2. Your circuit is almost ok, if you remove one of the resistors to ground [in fact your resistors to ground are parallel, so you can replace them by 1 resistor] My extra hint: the resistors all have different values. You'll need [how I solved it] Thevenin and superposition for this problem FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-09-21T19:09:44Z IndexTAG: 4057 TitleTAG: help really i need an explanation in this part? UserIdTAG: 139448 UserNameTAG: meno CreateTimeTAG: 2012-09-21T17:58:11Z VoteTAG: 0 CoursewareTAG: Week 1 / AC power CommentableIdTAG: 6002x_S1E3_AC_power NumberOfReplyTAG: 1 FirstChildTAG: witch part? FirstChildUserIdTAG: 399292 FirstChildUserNameTAG: Qabali FirstChildCreateTimeTAG: 2012-09-22T06:59:20Z IndexTAG: 4058 TitleTAG: Lab 3 help? Could anyone just help me get started on this problem? I can build an inverter using a MOSFET but how would I implement the C.(A+B) function? As in how would I know where to put a MOSFET in series or in parallel? Any help appreciated. Thanks! UserIdTAG: 278301 UserNameTAG: prateektaneja CreateTimeTAG: 2012-09-21T17:54:06Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi. Look for S5E3. Is the truth tables of devices in example tell You anything? Try to analyse them and use in Your circuit. Hope it will be usefull. FirstChildUserIdTAG: 195218 FirstChildUserNameTAG: Al_Incognito FirstChildCreateTimeTAG: 2012-09-21T18:15:30Z IndexTAG: 4059 TitleTAG: home work 2 question 1 i use 10% resistance of 47k and 15k . now for max voltage we add 10% of 15k in original resistance and we get 15k+1500 resistance and for minimum voltage we subtract 15k-1500 is that correct.? kindly guide me thanks :) UserIdTAG: 438395 UserNameTAG: ali_PU1 CreateTimeTAG: 2012-09-21T16:51:52Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: The principle is correct FirstChildUserIdTAG: 361634 FirstChildUserNameTAG: Z400 FirstChildCreateTimeTAG: 2012-09-21T17:24:57Z SecondChildTAG: Nice way of thinking .. but think of this question in a different manner First 1. Only the resistance from the set we can take. 2. Now write down the conditions given in the question a. First R1 and R2 should satisfy in getting the value of Rth within 10k to 30k. So try the values from 10k that will be easy for u. 3. the value Vout/Vin= 9/30= 0.3 ... approximatly from the diagram we can get Vout/Vin= R2/(R1+R2).. 4. then using the above ideas try to work it out... u did not get then i will tell the solution SecondChildUserIdTAG: 296303 SecondChildUserNameTAG: DEEPatXUniv SecondChildCreateTimeTAG: 2012-09-21T17:44:17Z SecondChildTAG: I choose 27 ohms for R1 and 18 ohms for R2. Please, help us! SecondChildUserIdTAG: 265133 SecondChildUserNameTAG: Elienai SecondChildCreateTimeTAG: 2012-09-22T02:35:18Z FirstChildTAG: The values, of your resistors can be within 10% of the answer. So once you find an answer that works, find the closest value resistors from the E12 chart. As long as they are within 10% of your answer, it will be correct. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-21T17:01:18Z SecondChildTAG: In order to find the maximum voltage, you need to think about R2 being 10% over the stated value, and R1 being 10% under - not just one of them. Similarly for the minimum voltage. SecondChildUserIdTAG: 145194 SecondChildUserNameTAG: DerekH SecondChildCreateTimeTAG: 2012-09-21T17:09:47Z SecondChildTAG: I suppose that i have to find a pair of resistors which values could divide the voltage for Vout, then I found a relationship between Vout and Vin, which resulted in 0.4. Therefore I looked for a couple of resistors to accomplish that, and I choose R1=33 and R2=22 ohms. Summing up, for minimum voltage, R1 is supposed to be the highest value, and R2 the lowest value. So, R1h=36.3 and R2l=18. What am I doing wrong??? Why can't I make it right? Please, give me a hand!!! SecondChildUserIdTAG: 265133 SecondChildUserNameTAG: Elienai SecondChildCreateTimeTAG: 2012-09-21T19:33:34Z FirstChildTAG: my solution was R1=68 K , R2=22 and it gives me wrong also !! FirstChildUserIdTAG: 204336 FirstChildUserNameTAG: Amin86 FirstChildCreateTimeTAG: 2012-09-21T19:52:53Z IndexTAG: 4060 TitleTAG: Homework 2 Problem 2 I'm trying to calculate the Thevenin Equivalent by saying the parallel resistor is in parallel with the two transmission resistors (which are in series) but apparently thats wrong. For Rs = 1.3 and Rp = 6, I get 1.81ohms (2.6)(6.0)/(2.6+6.0) UserIdTAG: 381923 UserNameTAG: matteaton CreateTimeTAG: 2012-09-21T16:51:50Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Look at the port. Consider the port as terminal.The rest as circuit. Apply element combinations and you would be able to figure out. FirstChildUserIdTAG: 361634 FirstChildUserNameTAG: Z400 FirstChildCreateTimeTAG: 2012-09-21T17:26:21Z FirstChildTAG: its better to use Norton equivalent circuit ..i did it . FirstChildUserIdTAG: 204336 FirstChildUserNameTAG: Amin86 FirstChildCreateTimeTAG: 2012-09-21T20:33:32Z SecondChildTAG: Yeah, when I do that I get the same answer since Rth = RN SecondChildUserIdTAG: 381923 SecondChildUserNameTAG: matteaton SecondChildCreateTimeTAG: 2012-09-22T00:55:50Z FirstChildTAG: I used the parallel concept on resistors RP = 3 and rs = 1.9 and found out 3.95 as RTH, but the system did not consider it... I really don't understand why not, we have an example at the classes (S3E5) very similar and it works there FirstChildUserIdTAG: 316902 FirstChildUserNameTAG: JesseTeixeira FirstChildCreateTimeTAG: 2012-09-21T20:38:53Z SecondChildTAG: You are making mistake as how to sum up the resistances. Remember that to find Rth you do not consider the power source. For voltage sources it is a short circuit and for current ones a open circuit! Try looking at the resistances as the load would, from the open terminals of the thevenin circuit! Hope it helps. SecondChildUserIdTAG: 308853 SecondChildUserNameTAG: fmorato SecondChildCreateTimeTAG: 2012-09-22T14:21:41Z IndexTAG: 4061 TitleTAG: H2P1 V_out = 1/2 V1 + 1/6 V2 where V_out = (R2/R1+R2)* V1 + (R1/R1+R2)*V2 (by superposition). V_out Maximum = 667m v. when V1=1 and V2=1 V_out Minmmum =-167m v. when V1=0 and V2 = -1 I'm try every thing and don't have any correct solve what can I do? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-21T16:50:35Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: For the lab problem you have to realize that you need another resistor R3 that goes from Vout to ground. So you can put that into your circuit and calculate again. Then you can choose R3 = 1 and you have two equations to solve with two unknowns. FirstChildUserIdTAG: 392100 FirstChildUserNameTAG: glenng FirstChildCreateTimeTAG: 2012-09-21T17:20:31Z FirstChildTAG: Vout=+-10%Vin Vout max = 1.10*Vin Vout min = 0.90*Vin FirstChildUserIdTAG: 143835 FirstChildUserNameTAG: sanpablc FirstChildCreateTimeTAG: 2012-09-22T13:03:54Z IndexTAG: 4062 TitleTAG: Current flowing from source into the circuit? Current flowing from source into the circuit?Can any one help for getting the answer of this question? UserIdTAG: 139448 UserNameTAG: meno CreateTimeTAG: 2012-09-21T16:14:45Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: 500e-6 FirstChildUserIdTAG: 391687 FirstChildUserNameTAG: Gadzhi FirstChildCreateTimeTAG: 2012-09-21T20:20:09Z IndexTAG: 4063 TitleTAG: h2p1 - help me regain integrity V_out is approx. 4V and V_in = 20V. Here are the equations I used: V_out/V_in = R_2/(R_1 + R_2) {using the node method} R_thevenin (R_th) = R_2*R_1/(R_1+R_2) And given the conditions => 10k < R_th < 30k Manipulating this inequality I found that 50 < R_1 < 150, which left me with three possible values for R_1 (56, 68, 82). Using the first equation, I put in the numbers, which gave me 0.2=R_2/(R_2+56) and solved for R_2 with every value I put in for R_1 (56, 68, 82). R_2 values were 14, 17, and 20, respectively, which came to equate to 15, 18, and 22 when compared to the original set of resistors and the 10% condition. I tried putting in the values of these resistors I used, because all gave me the right ratio of voltages, but they're incorrect. Pardon how nuanced I may seem, it's been too long since I've done any physics.. UserIdTAG: 141376 UserNameTAG: krtica CreateTimeTAG: 2012-09-21T15:50:58Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Edit: My possible values for R_1 are {56k, 68k, 82k, 100k, 120k, 150k} FirstChildUserIdTAG: 141376 FirstChildUserNameTAG: krtica FirstChildCreateTimeTAG: 2012-09-21T16:12:00Z SecondChildTAG: u find all the values R1,R2,Vmax, and Vmin and then put all the vales in respective boxes and click check...in this question u have to fill all the answer boxxes then only u will get the the answer correct....understood???? SecondChildUserIdTAG: 99583 SecondChildUserNameTAG: sanjeevkm0912 SecondChildCreateTimeTAG: 2012-09-21T19:00:44Z SecondChildTAG: Thanks! SecondChildUserIdTAG: 141376 SecondChildUserNameTAG: krtica SecondChildCreateTimeTAG: 2012-09-23T05:06:55Z FirstChildTAG: You could simplify your equations more, they are correct but you could try to get this: R2 = 0.25*R1 R1 = RTH/0,25 If you use these then it should be pretty easy to find a good value R1 and from the R1 you pick you should be able to find a reasonably good R2. Edit: All of your values seem ok. Are you sure you are writing it correctly in the boxes. You have to write eg. R2 = 15k or R2 = 15000. FirstChildUserIdTAG: 392100 FirstChildUserNameTAG: glenng FirstChildCreateTimeTAG: 2012-09-21T17:00:21Z SecondChildTAG: i didnt understand it yet how to find r1 and r2 can u give me clear expressions please SecondChildUserIdTAG: 444708 SecondChildUserNameTAG: jagadeeshv16 SecondChildCreateTimeTAG: 2012-09-22T15:30:04Z SecondChildTAG: I am; thanks for your help! SecondChildUserIdTAG: 141376 SecondChildUserNameTAG: krtica SecondChildCreateTimeTAG: 2012-09-23T05:07:13Z SecondChildTAG: So I chose R1=15k and R2=56k, then Vout=4.23, so my min would be 3.807 and max would be 4.653 (plus/minus 10% of 4.23). It's incorrect. SecondChildUserIdTAG: 141376 SecondChildUserNameTAG: krtica SecondChildCreateTimeTAG: 2012-09-23T05:58:43Z IndexTAG: 4064 TitleTAG: FAO EdX staff - Homework answers not accepted When I click 'Check' in H2P1, there is no response, and my answer fields are cleared if I leave the page. H2P2 and H2P3 submitted without problems. UserIdTAG: 261378 UserNameTAG: es2377 CreateTimeTAG: 2012-09-21T15:02:39Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Actually that may be the problem with the browser same happened to me in firefox now working fine in chrome FirstChildUserIdTAG: 444708 FirstChildUserNameTAG: jagadeeshv16 FirstChildCreateTimeTAG: 2012-09-21T15:24:39Z IndexTAG: 4065 TitleTAG: Voltage drop question confusing The resistance from the house to the barn is 0.179444, the wire that closes the circuit back to the house is plus 0.179444. So the voltage drop caused by the wire that goes to the barn is 0.17944 * I = 0.74... . I think you see my confusion. I believe the question would be better framed like this: Whats the total voltage drop caused by the wires? UserIdTAG: 236220 UserNameTAG: diegoizidoro CreateTimeTAG: 2012-09-21T14:39:20Z VoteTAG: 0 CoursewareTAG: Week 1 / Power delivery modeling CommentableIdTAG: 6002x_S2E6_Modeling NumberOfReplyTAG: 1 FirstChildTAG: You have to use a voltage divider, taking into account the load as "the other" resistance. FirstChildUserIdTAG: 9100 FirstChildUserNameTAG: dmgongora FirstChildCreateTimeTAG: 2012-09-26T04:16:28Z IndexTAG: 4066 TitleTAG: I have a problem with Week 2 tutorials Hello, I have a problem with week 2 tutorials, tutorials videos cannot open. Kindly please help me. Thank You UserIdTAG: 291506 UserNameTAG: usmanahmad87 CreateTimeTAG: 2012-09-21T14:14:47Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: youtube must be blocked...!! FirstChildUserIdTAG: 127800 FirstChildUserNameTAG: MNB FirstChildCreateTimeTAG: 2012-09-21T19:20:35Z IndexTAG: 4067 TitleTAG: node method and KC L and KVL method Please which of these two will you recommend the we stick to that comparing the node method and KCL and KVL method. Again, please you explain the node method to me with some examples since it is not very clear to me. I fully understand the KCL and KVL approach to linear circuit Thanks UserIdTAG: 382532 UserNameTAG: emmanuelpeace CreateTimeTAG: 2012-09-21T11:23:23Z VoteTAG: 0 CoursewareTAG: Week 2 / Superposition CommentableIdTAG: 6002x_superposition NumberOfReplyTAG: 0 IndexTAG: 4068 TitleTAG: H3P1 Where is the error? Hello. For Nor Mosfet I have this: Vout = VOL <= Vs(Ron||Ron)/((Ron||Ron)+Rpuo) 1 <= 5*4000/(4000+Rpuo) -> Rpuo >= 16000 ohm But the correct answer is Rpuo=32000 ohm Where is the error? UserIdTAG: 270769 UserNameTAG: 2214sanchez CreateTimeTAG: 2012-09-21T10:24:12Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: thanks to NIDHIN_RS for the web https://6002x.mitx.mit.edu/section/week3_gjs/ –posted 5 days ago by NIDHIN_RS This resolved my dude. FirstChildUserIdTAG: 270769 FirstChildUserNameTAG: 2214sanchez FirstChildCreateTimeTAG: 2012-09-21T13:01:47Z IndexTAG: 4069 TitleTAG: Reverse Power to electrolytic cap If I give reverse voltage connection to elec cap will it burst. Like +ve voltage on -ve plate of cap and -ve voltage on +ve plate of cap UserIdTAG: 137709 UserNameTAG: DeepakBansal CreateTimeTAG: 2012-09-21T10:20:23Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Yup. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-21T13:58:45Z SecondChildTAG: There is the odd occasion where an electrolytic capacitor "appears" to be backwards. If you are feeding the capacitor negative voltage, then the positive terminal will be connected to ground. Ground is a higher voltage then a negative voltage. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-21T14:52:32Z IndexTAG: 4070 TitleTAG: about the batteries why does the mutimeter not show exact readings?why is there an error? UserIdTAG: 401199 UserNameTAG: ashritha CreateTimeTAG: 2012-09-21T04:45:26Z VoteTAG: 0 CoursewareTAG: Week 2 / Lightbulb model CommentableIdTAG: 6002x_lightbulb_model NumberOfReplyTAG: 2 FirstChildTAG: 我也有这样的问题 ，为什么mutimeter不显示精确的读数，为何有错误吗？ FirstChildUserIdTAG: 471183 FirstChildUserNameTAG: snow_yxx FirstChildCreateTimeTAG: 2012-09-21T06:03:05Z FirstChildTAG: because battery itself has resistance (as you have to know already :) - with lamp (and multimeter's shunt in parallel) it creates a voltage divider, so on the lamp you will detect Vsrc*(Rlamp||Rshunt)/(Rlamp||Rshunt+Rint) voltage, where (Vsrc) - battery voltage, (Rlamp) - lamp resistance, (Rshunt) - multimeter's shunt resistance, (Rint) - battery internal resistance. FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-09-21T06:05:28Z IndexTAG: 4071 TitleTAG: new discussion format I found the discussion format in the spring very helpful - the current format I find in-tracktable. UserIdTAG: 10512 UserNameTAG: asicok CreateTimeTAG: 2012-09-21T01:59:06Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Could you elaborate? FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-09-21T02:14:53Z IndexTAG: 4072 TitleTAG: Parallel mean less resistance? Hi, For the resisters connected in series, it intuitively make sense that the resulting element have more resistance than the individual resister. Assume from this video tute that R1=R2=R, then the resulting resistance for the element would be R/2, which is less than R. My question is why connecting in parallel result in less total resistance? Thanks, Ravuth, UserIdTAG: 455668 UserNameTAG: Ravuth CreateTimeTAG: 2012-09-21T01:44:18Z VoteTAG: 0 CoursewareTAG: Week 1 / Three parallel resistors CommentableIdTAG: 6002x_parallel_resistors2 NumberOfReplyTAG: 3 FirstChildTAG: disclaimer: I know this is not an accurate physics or flow dynamics example, and is used only to describe a way of thinking. Image two set of hoses. Connect them together, and you should see that (with no additional fall being implied) that two hoses move water more slowly than just one hose. The resistance is "more". Now image two hoses both emptying the same bucket. The flow would be more. The resistance is "less". FirstChildUserIdTAG: 88550 FirstChildUserNameTAG: Neil_S_Berry FirstChildCreateTimeTAG: 2012-09-21T01:54:03Z FirstChildTAG: In fact there is an analogy between a basic electric circuit and the water in a hose. Here's a [link][1] in spanish. [1]: http://www.windows2universe.org/physical_science/physics/electricity/circuit_analogy_water_pipes.html&lang=sp FirstChildUserIdTAG: 184827 FirstChildUserNameTAG: DiegoT FirstChildCreateTimeTAG: 2012-09-21T02:11:46Z FirstChildTAG: Just think of conductances FirstChildUserIdTAG: 190618 FirstChildUserNameTAG: Kirbabaev FirstChildCreateTimeTAG: 2012-09-24T19:40:49Z IndexTAG: 4073 TitleTAG: Trigger Function Oscillosope What is the purpose of trigger function in oscilloscope. Like if set it as edge trigger at 2V amp , then CRO will not show the waveform until some rising edge appears with amplitude crossing 2V. UserIdTAG: 137709 UserNameTAG: DeepakBansal CreateTimeTAG: 2012-09-21T01:43:53Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: The first purpose - crossing triggering voltage will be always on the Y axis - so for periodic signal you will have steady picture. The second - if you want to detect some special event you can use single snapshot mode - in this case crossing triggering voltage will start snapshot capture and after that oscilloscope stops and shows captured signal. FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-09-21T06:53:12Z IndexTAG: 4074 TitleTAG: Resistors Does someone knows how to turn the resistor into horizontal position? UserIdTAG: 43205 UserNameTAG: filipe25 CreateTimeTAG: 2012-09-21T01:37:51Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 2 FirstChildTAG: Use R key FirstChildUserIdTAG: 184827 FirstChildUserNameTAG: DiegoT FirstChildCreateTimeTAG: 2012-09-21T02:01:29Z SecondChildTAG: which program ? SecondChildUserIdTAG: 471575 SecondChildUserNameTAG: Hamodeh SecondChildCreateTimeTAG: 2012-09-21T06:12:38Z FirstChildTAG: **Thanks!** FirstChildUserIdTAG: 298577 FirstChildUserNameTAG: msolis FirstChildCreateTimeTAG: 2012-09-23T17:34:47Z IndexTAG: 4075 TitleTAG: S5E3 - Very last question Vc I've worked out correct answer for the others, but for this question, I am getting 0.00748, which is approximately twice as much as the answer given. My equation is: 3 * (10/2)/(2000 + 10/2) = 0.00748...... Am I missing something? UserIdTAG: 142402 UserNameTAG: aaronrod CreateTimeTAG: 2012-09-21T01:03:47Z VoteTAG: 0 CoursewareTAG: Week 3 / Switch Resister Model Exercise CommentableIdTAG: 6002x_switch_resistor_model_exercise NumberOfReplyTAG: 1 FirstChildTAG: Never mind, just saw it. Ron is 50, not 10. I mixed up the R values, missed some decimal places... I think it's time for a break! FirstChildUserIdTAG: 142402 FirstChildUserNameTAG: aaronrod FirstChildCreateTimeTAG: 2012-09-21T01:06:00Z IndexTAG: 4076 TitleTAG: Oh Snap! I think I have to go over there and teach you guys a couple of moves. haha! Entertaining. UserIdTAG: 378733 UserNameTAG: rlagomar CreateTimeTAG: 2012-09-20T20:54:06Z VoteTAG: 0 CoursewareTAG: Week 2 / Chain saw CommentableIdTAG: 6002x_chain_saw NumberOfReplyTAG: 0 IndexTAG: 4077 TitleTAG: Help with homework I have solved all the other Homework 2 questions except the "H2P1: VOLTAGE-DIVIDER DESIGN" I dont understand how to solve it... all i know is Vo=Vin*R2/(R1+R2) and when he says "An additional requirement is that the Thevenin resistance as seen from the output terminals is between 10kΩ and 30kΩ" Does that mean Rth=R1+R2( which lies between 10k n 30k?) UserIdTAG: 468052 UserNameTAG: Charan94 CreateTimeTAG: 2012-09-20T20:44:55Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: You've got to remember the $R_{TH}$ is found by replacing the supply voltage with a short circuit. Then you'll see that the two resistors are in parallel, not in series. $R_1+R_2$ of course is the series sum. I hope I'm telling you right . . . I'm not the expert here either. FirstChildUserIdTAG: 280220 FirstChildUserNameTAG: pilgrimCycle FirstChildCreateTimeTAG: 2012-09-20T21:17:00Z SecondChildTAG: Hi, I'll give you a clue: Look were are the output teminals, is evident that R2 = Rth. V2 = Vth = 20 (remember, the exercise says that Vout is like 20V). Now choose a nominal value for Vth betwen 10-30 kohms. Use your formula and find R1 and then the other answers. Can you help me with H2P2. I don't whats wrong but I only have found Rth SecondChildUserIdTAG: 184827 SecondChildUserNameTAG: DiegoT SecondChildCreateTimeTAG: 2012-09-20T21:34:32Z SecondChildTAG: i can help u with h2p2 but i still didnt understand h2p1 SecondChildUserIdTAG: 444708 SecondChildUserNameTAG: jagadeeshv16 SecondChildCreateTimeTAG: 2012-09-22T16:07:45Z FirstChildTAG: pilgrimCycle is correct, (R)th = R1 || R2 Replace the supply voltage with a short circuit. Look at the top terminal of R1. Instead of connecting to the + terminal of Vin, it now connects to the bottom terminal of R1. From where you're taking your output voltage, the two resistors are in parallel. If you were taking your output voltage from the bottom of R2 to the "top" of R1, they would be in series. FirstChildUserIdTAG: 208488 FirstChildUserNameTAG: moelarrycheese FirstChildCreateTimeTAG: 2012-09-20T22:27:27Z FirstChildTAG: Hi, I have some clue in [this post][1] hope it useful [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/505aa3fae87fdd2800000001 FirstChildUserIdTAG: 452559 FirstChildUserNameTAG: kuz1toro FirstChildCreateTimeTAG: 2012-09-20T23:12:45Z FirstChildTAG: Thankx for the help guys!! Silly me.... Got the answer!! :) FirstChildUserIdTAG: 468052 FirstChildUserNameTAG: Charan94 FirstChildCreateTimeTAG: 2012-09-21T17:18:03Z SecondChildTAG: hw u got the answer please do explain me SecondChildUserIdTAG: 444708 SecondChildUserNameTAG: jagadeeshv16 SecondChildCreateTimeTAG: 2012-09-22T16:11:02Z IndexTAG: 4078 TitleTAG: counting Hi I have a problem with the counting of the voltages in this circuit. I increment and decrement the resistors, but i can't figure out how is the voltages change in the branch-s after resistorts. I use this formula V=IR Maybe this is a stupid a question, but i will be really happy if anyone can help me :) UserIdTAG: 413714 UserNameTAG: Banszki_Istvan CreateTimeTAG: 2012-09-20T20:37:07Z VoteTAG: 0 CoursewareTAG: Week 2 / Superposition experiment CommentableIdTAG: 6002x_Superposition_experiment NumberOfReplyTAG: 1 FirstChildTAG: You can solve it using node method FirstChildUserIdTAG: 184827 FirstChildUserNameTAG: DiegoT FirstChildCreateTimeTAG: 2012-09-20T21:42:42Z IndexTAG: 4079 TitleTAG: Lab 2 solved but not sure how! Hi every one! I managed to solve Lab 2 but I'm not sure how and why. :P Could someone enlighten me please? Here's the deal. At first I tried doing it on my own, but couldn't grasp the concept so I searched the discussions. So, someone said that you have to use 3 resistors in total and that you should solve 2 separate circuits as voltage dividers, and find the resistor values and then combine them into one. So I did. And I found some results in the form: R1 = R3 R2 = 5*R3 With R1 being the resistor connected to V1, R2 connected to V2 and R3 the "common" one. So I enter the values in the schematic and the output plot looks quite good but with less amplitude, ie max ~540 and min ~ -130. Then I started experimenting with the values of the resistors and voila! i got it right. But when comparing the new values of the resistors, I got something in the form: R1 = R3/3 R2 = 3*R3 [These are random example numbers and not the ones I actually found, so I don't give out the solution] Bottom line is that the final results were completely different from the ones found in the calculations. So what is the right procedure to do it after all? UserIdTAG: 324642 UserNameTAG: StNas CreateTimeTAG: 2012-09-20T20:30:18Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: If by R3 being the "common" one you mean it goes from the node formed formed by the other two and ground, then the ratios given in your second paragraph are wrong. If you analyze that circuit you can see immediately without doing any calculations that those ratios will not exactly produce the specified result. Yes, it will come "close". By using any of the circuit analysis techniques taught in the course you can generate two equation that can be solved to give exactly what is requested. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-09-20T22:17:44Z FirstChildTAG: But still i am unable to solve it :( Do i have to apply Delta-Wye transformation??? FirstChildUserIdTAG: 336001 FirstChildUserNameTAG: syd_buet12 FirstChildCreateTimeTAG: 2012-09-21T08:58:44Z IndexTAG: 4080 TitleTAG: Question about Switch Model I think I'm having a conceptual problem with the resistor in this model, or possibly resistors in general. When the switch is closed, the output is connected to ground and so the voltage at the output is zero. But when the switch is open, we have Vs connected to a resistor Rl. The lecture says that VS appears at the output. But, doesn't the resistor cause a voltage drop? UserIdTAG: 62503 UserNameTAG: Elizabot CreateTimeTAG: 2012-09-20T20:10:33Z VoteTAG: 0 CoursewareTAG: Week 3 / Switch Model CommentableIdTAG: 6002x_switch_model NumberOfReplyTAG: 1 FirstChildTAG: I was thinking the same thing, but something in my mind told me perhaps the reason the call the apparent voltage $V_S$ is because it's an open circuit, and therefore the resistance is effectively infinite. That makes the finite value of $R_L$ insignificant. Tell me if you get a better answer... FirstChildUserIdTAG: 280220 FirstChildUserNameTAG: pilgrimCycle FirstChildCreateTimeTAG: 2012-09-20T21:11:11Z SecondChildTAG: Right, for a voltage drop to occur across a resistor there must be a current ($V=IR$). Since the circuit is open, no current can flow through the resistor. SecondChildUserIdTAG: 323387 SecondChildUserNameTAG: ibrahimawwal SecondChildCreateTimeTAG: 2012-09-21T00:10:54Z SecondChildTAG: If there is no current then why is there a need for a resistor Rl? If the circuit wasn't open and if we assume the wires to have negligible resistance, then the current in the wires would be infinite( current inversely proportional to resistance). Therefore, we add a resistance Rl ( with a small value??) to limit the current flowing through them. CORRECT ME IF I'M WRONG! SecondChildUserIdTAG: 254325 SecondChildUserNameTAG: bondablack SecondChildCreateTimeTAG: 2012-09-27T17:46:42Z SecondChildTAG: In response to ibrahimawwal's comment, what is the reason we even look at this voltage? How can we have this voltage at Z if as soon as we connect something to Z, then the current WILL flow through Rl and then it's value becomes very significant. Isn't it meaningless to ignore Rl since it's there? SecondChildUserIdTAG: 357225 SecondChildUserNameTAG: MJBoa SecondChildCreateTimeTAG: 2012-09-29T20:07:48Z IndexTAG: 4081 TitleTAG: Not shown my progress Not shown my progress, anyone know I have to do it then appears? UserIdTAG: 435253 UserNameTAG: renevalderrama CreateTimeTAG: 2012-09-20T19:29:07Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I beleive you do not need to do anything special. Progree will be calculated and posted automatically when you watched movies and get the all exercises done properly. the same applies to homework and labs. FirstChildUserIdTAG: 210954 FirstChildUserNameTAG: Shahrouz FirstChildCreateTimeTAG: 2012-09-20T20:51:39Z IndexTAG: 4082 TitleTAG: Current measurement I cant measure current can any one help me UserIdTAG: 457331 UserNameTAG: deepansh91 CreateTimeTAG: 2012-09-20T17:25:00Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 4 FirstChildTAG: Hi, Once you click the DC analysis tool the simulator will plot the current next to the source. Because the Resistors are in series, if you want to know the current in a particular element you can use Ohm's law V=IR. The voltage you need to use to make the calculation is the one that the element drops. If you need more help just make another post. Regards, Luis FirstChildUserIdTAG: 355859 FirstChildUserNameTAG: LuisZorrillaEE FirstChildCreateTimeTAG: 2012-09-20T18:21:03Z FirstChildTAG: To measure current through an element, add a voltage source in series with the element. Set the value of the voltage to 0. Next run a DC analysis. You'll see the current which enters the source displayed. Since the 0V source is in series with your circuit element, this is also the current through that element. I don't remember why it behaves like that (SPICE behaves like this as well). FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-09-20T18:47:00Z FirstChildTAG: to measure the current you need to wire properly to input and output port. do not just lay the current measure device on an existing line. FirstChildUserIdTAG: 210954 FirstChildUserNameTAG: Shahrouz FirstChildCreateTimeTAG: 2012-09-20T20:44:12Z FirstChildTAG: As a general rule. when you have all the elements connected in series, then the current is constant through the circuit. or Current is constant through elements connected in series. FirstChildUserIdTAG: 458519 FirstChildUserNameTAG: hooba21 FirstChildCreateTimeTAG: 2012-09-21T12:14:56Z IndexTAG: 4083 TitleTAG: cannot view the video I cannot view the video in my browser (ie or firefox or chrome) as shown in the following image. ![image][1] Is there possible to download the lecture videos? [1]: https://edxuploads.s3.amazonaws.com/13481595431343661.png UserIdTAG: 452617 UserNameTAG: tuqiu CreateTimeTAG: 2012-09-20T16:45:57Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 4084 TitleTAG: peak power we can get the value of peak power by putting cosx=1. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-20T15:37:41Z VoteTAG: 0 CoursewareTAG: Week 1 / AC power CommentableIdTAG: 6002x_S1E3_AC_power NumberOfReplyTAG: 0 IndexTAG: 4085 TitleTAG: Simple question about superposition (min 2 +o-) Why when the voltages and currents are set to 0, the Professor write Vm-1 and In-1. Why -1?? ¿It is because the "-1" is the current or voltage that don't set to 0? thank you folks! UserIdTAG: 149549 UserNameTAG: Java_Dido CreateTimeTAG: 2012-09-20T15:21:07Z VoteTAG: 0 CoursewareTAG: Week 2 / Thevenin method CommentableIdTAG: 6002x_thevenin NumberOfReplyTAG: 1 FirstChildTAG: He said that we exclude all the voltage sources from V1 to Vm-1 and left only one source - Vm FirstChildUserIdTAG: 342603 FirstChildUserNameTAG: YakovO FirstChildCreateTimeTAG: 2012-09-20T15:35:18Z SecondChildTAG: Okey I get it. The voltage acting alone that we will use, it will be Vm, so the voltages V1 to V(m-1) are set to 0. I mean, the "-1" is acting only over the m, not on the "V". Thanks YakovO, you made me think a little bit more about this issue. and sorry by my english, maybe it's not the best. SecondChildUserIdTAG: 149549 SecondChildUserNameTAG: Java_Dido SecondChildCreateTimeTAG: 2012-09-20T17:06:53Z IndexTAG: 4086 TitleTAG: Internal battery resistance? What happened to internal battery resistance in questions 2-4? UserIdTAG: 251793 UserNameTAG: chellaton CreateTimeTAG: 2012-09-20T14:49:59Z VoteTAG: 0 CoursewareTAG: Week 1 / Battery model CommentableIdTAG: 6002x_S1E9_Battery_Model NumberOfReplyTAG: 1 FirstChildTAG: hey 0.2 ohm is just an approx value of int. reisitance. In our case we have to use internal resis for battery 1 as 0.25 and other as 0.32 ohm respectively. FirstChildUserIdTAG: 456457 FirstChildUserNameTAG: userdivya FirstChildCreateTimeTAG: 2012-09-22T09:10:16Z SecondChildTAG: sorry i m bit late to join this course..bt i hv a doubt.. as the internal resistance is specified in the que..so it shouldn't be included in the calculation?? if no then plzz correct me.. SecondChildUserIdTAG: 545481 SecondChildUserNameTAG: BRAJESH1 SecondChildCreateTimeTAG: 2012-10-05T03:09:51Z IndexTAG: 4087 TitleTAG: strength of the noise signal what if the magnitude of the noise signal is high enough to change the magnitude of the output frm 1 to 0 or vice versa? UserIdTAG: 401199 UserNameTAG: ashritha CreateTimeTAG: 2012-09-20T13:35:27Z VoteTAG: 0 CoursewareTAG: Week 2 / Digital Abstraction CommentableIdTAG: 6002x_digital NumberOfReplyTAG: 1 FirstChildTAG: yes that's a problem in digital systems too. but less pronounced as in analog systems and the sender will send some extra cross check information appended at the end of the actual information which is to be sent to the receiver, so that receiver can cross check whether the received information is right or wrong. [ see [parity bits][1] for some info on your doubt [1]: http://en.wikipedia.org/wiki/Parity_bit FirstChildUserIdTAG: 440453 FirstChildUserNameTAG: Asandyz FirstChildCreateTimeTAG: 2012-09-20T20:07:57Z SecondChildTAG: thanku!! SecondChildUserIdTAG: 401199 SecondChildUserNameTAG: ashritha SecondChildCreateTimeTAG: 2012-09-21T04:47:17Z IndexTAG: 4088 TitleTAG: confused with H2P1!!!! i got really confused with the types of resistors to be chosen with problem H2P1. can anyone please help me with this problem!! UserIdTAG: 170475 UserNameTAG: sandeshacharya CreateTimeTAG: 2012-09-20T13:04:08Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: for H2P2, the system is not picking my answer and I feel it is correct. FirstChildUserIdTAG: 125652 FirstChildUserNameTAG: Adedayo FirstChildCreateTimeTAG: 2012-09-20T16:35:04Z FirstChildTAG: The best way is to start with choosing Rth seen from Vout in the middle of the given range for Rth. Write the expression for Vout, this is also Vth. Write the expression for Ith. Then you can calculate Ith and from that, R1 and R2. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-09-20T13:35:14Z FirstChildTAG: Hi sandeshacharya! I have answered you here ;)[Post][1] If you have any doubt please tell me. I will do my best to help you. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/505971f4dd2f4d1f00000004 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-20T15:38:07Z IndexTAG: 4089 TitleTAG: H3P4: Diode Limiter I have issues with last two questions on H3P4: Diode Limiter. My understanding is that the maximum D1 current must be the (max Vs - max V) / R but it seems it is not right. Any hint? UserIdTAG: 210954 UserNameTAG: Shahrouz CreateTimeTAG: 2012-09-20T12:30:35Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: I have made the circuit on the Sandbox and as I have seen the Max V will be three but when you calculate it it is different. Any idea what is wrong? ![Sandbox Circuit][1] ![Result][2] [1]: http:// [2]: http://Result FirstChildUserIdTAG: 210954 FirstChildUserNameTAG: Shahrouz FirstChildCreateTimeTAG: 2012-09-20T12:59:33Z FirstChildTAG: The circuit behaves different from calculation when you make it in Sandbox. Any idea why is that so? ![Sandbox1][1] ![Result][2] [1]: http://Sandbox%201 [2]: http://Result FirstChildUserIdTAG: 210954 FirstChildUserNameTAG: Shahrouz FirstChildCreateTimeTAG: 2012-09-20T13:10:30Z FirstChildTAG: when the D1 or D2 conducts, the voltages are pegged to V1 or V2 respectively. FirstChildUserIdTAG: 252126 FirstChildUserNameTAG: sudhakarvenkatesh FirstChildCreateTimeTAG: 2012-09-24T00:56:08Z IndexTAG: 4090 TitleTAG: direction of current in voltage source A voltage source is supposed to supply voltage and current to the circuit. If that is the case then the current should be flowing away from the source. Why is it flowing towards the voltage source? UserIdTAG: 111540 UserNameTAG: tksanthosh CreateTimeTAG: 2012-09-20T12:17:23Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL and KCL CommentableIdTAG: 6002x_kvl_and_kcl NumberOfReplyTAG: 1 FirstChildTAG: Hello tksanthosh, thanks 4 raising this question. Okay, It's true that a voltage source supplies voltage but whether we CONSIDER the corresponding current as positive or negative depends ENTIRELY on us. Its as we make it according to our convenience and ease of understanding. Either of the two is correct but by chance WE ourselves get confused its better we stick to one DISCIPLINE or RULE .say if we OURSELVES assume that current entering a node or say any LUMPED element as positive, we must keep on assigning a '+' sign to all the currents entering each of the lumped element as '+' inorder to calculate correctly during our analysis of the circuit. Prof Agarwal in the lecture ,for example, has followed the ASSOCIATED VARIABLE DISCIPLINE where he has ASSUMED that any current entering any positive terminal of the branch voltages ( or +ve terminal of the LUMPED ELEMENTS in each of the branches, which consume some part of the voltage from the source,hence they are in short referred as branch voltages) as positive. This is just a convention followed to ease up the calculations and make it error-free. It is a convention and has got nothing to do with the Physical meaning of voltage source and flow of currents. I will quote what Prof Agarwal said in the video : "...And notice here that in this example, I have used the Associated Variable Discipline, that I have assigned all my currents flowing into the positive terminal of each of the assigned branch voltages." I hope you can now see the point why current is flowing towards the voltage source....Just because we have "ASSUMED" it and thus "ASSIGNED" it to ease our calculations following a single DISCIPLINE. FirstChildUserIdTAG: 156225 FirstChildUserNameTAG: sanjana_m FirstChildCreateTimeTAG: 2012-09-23T06:51:07Z IndexTAG: 4091 TitleTAG: Can't see the answers or open the videos I'm not able to compute any of the questions which are for homework as the numbers are displayed on my screen for barely one second and is replaced by 'math computation error'. I'm not able to use the 'show answers' option and am not able to access the video tutorial as well. please help. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-20T12:09:05Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: i think you need to update your java FirstChildUserIdTAG: 363894 FirstChildUserNameTAG: vishak31292 FirstChildCreateTimeTAG: 2012-09-20T13:51:02Z FirstChildTAG: I had similar problems and I realized that for the 'math computation error' I had to sign out of EdX and sign back in to fix it. Also make sure that you only have one EdX webpage open. All of this happened when I was using Internet Explorer, but as soon I changed to Chrome, everything seem to be working fine for me. Hope this helps FirstChildUserIdTAG: 303762 FirstChildUserNameTAG: leocarrasco FirstChildCreateTimeTAG: 2012-09-20T16:32:05Z IndexTAG: 4092 TitleTAG: what's refer to in the textbook ? hi , ever one .... i want to know these numbers what refer to in the Course-at-a-Glance like these ( 3.5, 3.6 5.1-5.4, 5.6, 5.7 ) i actually know refer to the main point in the chapter ex: (5.4) refer to : chapter five and headline number four but this sequence above dosent match with the text book like 5.7 in the textbook is the summery >>> thx to all UserIdTAG: 193445 UserNameTAG: eslammagdy CreateTimeTAG: 2012-09-20T12:07:02Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: They are suggested reading from Professor Agarwal's book. The lecture numbers do not match the chapters in the book, some might by coincidence. Professor Agarwal suggests for you reading particular chapters during the video lectures, those are the chapters he suggests you read. Make a note during the videos of what chapter is related to the video. It is a large book, not all of the books content is covered in this course. That is why the numbers do not match. Have fun. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-20T13:50:23Z IndexTAG: 4093 TitleTAG: lab 2 can someone help me out with lab 2? i'm totally stuck on this... can't just figure out how to generate the desired output by only using resistors? help needed..thanks in advance. UserIdTAG: 163407 UserNameTAG: shuvajit CreateTimeTAG: 2012-09-20T10:26:44Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Try the search feature. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-20T13:39:33Z FirstChildTAG: If you use the Hint and add 1 resistor extra then you can solve this. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-09-20T13:43:16Z SecondChildTAG: you say that this lab solves with 3 resistors? SecondChildUserIdTAG: 302803 SecondChildUserNameTAG: eiyawii SecondChildCreateTimeTAG: 2012-09-20T15:14:13Z SecondChildTAG: yes two resistors can only divide voltage proportionally SecondChildUserIdTAG: 342603 SecondChildUserNameTAG: YakovO SecondChildCreateTimeTAG: 2012-09-20T15:23:21Z SecondChildTAG: what method can i use to this problem? superposition and nodes analyse? SecondChildUserIdTAG: 302803 SecondChildUserNameTAG: eiyawii SecondChildCreateTimeTAG: 2012-09-20T17:38:09Z SecondChildTAG: Those methods will work. SecondChildUserIdTAG: 406202 SecondChildUserNameTAG: skyhawk SecondChildCreateTimeTAG: 2012-09-20T22:55:23Z SecondChildTAG: Guys i am totally lost on this one.Any help. SecondChildUserIdTAG: 429851 SecondChildUserNameTAG: ssembajjwe SecondChildCreateTimeTAG: 2012-09-21T07:51:09Z IndexTAG: 4094 TitleTAG: textbook pg. 300 constraint for SR-model of MOSFET I'm confused by the equation given as a constraint for the MOSFET's SR-model. The text says, "$v_{DS}$ << $v_{GS} -V_T$." Unless I'm misunderstanding something, $v_{DS}$ must always be greater than $v_{GS}$, let alone $v_{GS}-V_T$. Is it the intention of the textbook to say that since the current source will be called the "negative" voltage potential, then $v_{DS}$, being the greatest potential, will have the greater negative evaluation? Thank you as always for your patience and time. UserIdTAG: 280220 UserNameTAG: pilgrimCycle CreateTimeTAG: 2012-09-20T10:00:42Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 4095 TitleTAG: How to submit homework?? Once the homework/lab work is completed, how to submit it?? UserIdTAG: 224940 UserNameTAG: Shan1987 CreateTimeTAG: 2012-09-20T06:50:43Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I knew solution to first part of my wk1 home , could not input it cause I didn't know format of input field.Please help with formats of input for homeworks FirstChildUserIdTAG: 326390 FirstChildUserNameTAG: Aishah FirstChildCreateTimeTAG: 2012-09-20T08:31:19Z SecondChildTAG: In the edX Tutorial (Overview in Courseware) you can find information about how to input the equations. SecondChildUserIdTAG: 370769 SecondChildUserNameTAG: PabloFCid SecondChildCreateTimeTAG: 2012-09-20T12:22:25Z FirstChildTAG: You only need to check to submit your homework. FirstChildUserIdTAG: 370769 FirstChildUserNameTAG: PabloFCid FirstChildCreateTimeTAG: 2012-09-20T12:18:10Z IndexTAG: 4096 TitleTAG: my Email to tech support hi I really feel pain that my system is not loading edx videos which it did before and i submitted week 1 assignment and lab. after that i am not seeing videos. examples, book and all written material does load in it. youtube is blocked in our country after that controvercial video, may be this is why videos are not loaded but one update in edx said that thy have loaded their video player. i am regretting that i am getting late in the course and i would be very ashamed to quit this course. when i play video there is no error on screen but the bottom left corner status shows SENDING REQUEST and i keep on looking at that. i have no option but to close the window. so kindly i would appreciate any suggestions to help me out and it will worth morth more than a million bux.......... Regards: marshad.mit@gmail.com UserIdTAG: 279621 UserNameTAG: MuhammadArshadshamsher CreateTimeTAG: 2012-09-20T06:44:28Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 4097 TitleTAG: Problem in loading Courseware From last few days I am getting problem in loading Courseware. It takes a lot of time to load. any of you getting the same problem?? UserIdTAG: 908 UserNameTAG: m_umair72001 CreateTimeTAG: 2012-09-20T05:02:34Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: i'm getting the problem with this discussion board. Took forever to load the discussion board in chrome. But, the rest of the site works fine for me. What browser are you using? Do you have the right plug-ins installed? FirstChildUserIdTAG: 141053 FirstChildUserNameTAG: Albert0ng FirstChildCreateTimeTAG: 2012-09-20T05:21:39Z FirstChildTAG: Had some answers for homework, could not get school website on several attempts till date for submission expired.It is painful,Please help on what to do,who to contact.Thank u. FirstChildUserIdTAG: 326390 FirstChildUserNameTAG: Aishah FirstChildCreateTimeTAG: 2012-09-20T08:37:34Z IndexTAG: 4098 TitleTAG: week 2 video not loading it says vid not available ............... UserIdTAG: 152471 UserNameTAG: amrith CreateTimeTAG: 2012-09-20T04:36:42Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Must be a problem in your adobe flash player. Update it and restart your computer. FirstChildUserIdTAG: 908 FirstChildUserNameTAG: m_umair72001 FirstChildCreateTimeTAG: 2012-09-20T04:49:43Z SecondChildTAG: same here SecondChildUserIdTAG: 403493 SecondChildUserNameTAG: aiai SecondChildCreateTimeTAG: 2012-09-20T09:18:41Z IndexTAG: 4099 TitleTAG: H3P1: NOR 'staff help required' Dear sir can you please help me to solve maximum power at resistor at NOR gate give some clue UserIdTAG: 352647 UserNameTAG: praveenjugge CreateTimeTAG: 2012-09-20T04:30:41Z VoteTAG: 0 CoursewareTAG: Week 3 / Static Power of a MOSFET CommentableIdTAG: 6002x_mosfet_static_power NumberOfReplyTAG: 1 FirstChildTAG: Use the SR model of the MOSFET. Lectures from first sequence of week 3 explain the SR model. Substituting the MOSFET with the appropriate model your allow you to calculate the maximum power in a straightforward way because the circuit will be only voltage sources and resistors. FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-20T14:07:11Z IndexTAG: 4100 TitleTAG: What is happening here? Hi All, I have attempted to solve the system with the liner equation: - (v1-e)/r1+(e+v2)/r2=0 - (5-e)/6800+(e+7.2)/5600=0 - Find LCM of 95200 - 6800 goes into 95200 14 times and 5600 goes into 95200 17 times so: - (5*14)-(e*14) + (e*17)+(7.2*17)=0 - 70-14e+17e+122.4=0 - Join like terms - 192.4+3e=0 - 192.4=-3e - 192.4/-3=e - -64.1333 =e obviously my answer is wrong but I'm not sure why. Can someone tell me what I did wrong. Thanks, Scorliss UserIdTAG: 267537 UserNameTAG: scorliss CreateTimeTAG: 2012-09-20T03:43:44Z VoteTAG: 0 CoursewareTAG: Week 2 / Node equation review CommentableIdTAG: 6002x_S3E1_NodeEquationReview NumberOfReplyTAG: 2 FirstChildTAG: when you apply nod analysis you always start with the voltage of the node that you want to find: (e-5)/6.8K+(e-7.2)/5.6k=0 note: as the Right source's polarity is reversed (- +) & because it's value is -ve , so you can solved as a +ve value, not reversed polarity (+ -) source . FirstChildUserIdTAG: 185715 FirstChildUserNameTAG: amirengineer FirstChildCreateTimeTAG: 2012-09-20T04:44:55Z SecondChildTAG: good tip. Thanks engineer SecondChildUserIdTAG: 267537 SecondChildUserNameTAG: scorliss SecondChildCreateTimeTAG: 2012-09-20T04:50:19Z SecondChildTAG: Hello Amir, I think it should be (e-5)/6.8K+(e+7.2)/5.6k=0. Because if we start freom "e" point current flows from there to ground (assumption). but as voltage is neagtive, so current is in opposite direction. I am confused still.. Can you explain bit more.. Thanks SecondChildUserIdTAG: 181368 SecondChildUserNameTAG: Hanumanta SecondChildCreateTimeTAG: 2012-09-20T09:20:04Z SecondChildTAG: There are many ways to apply the node method. I will recommend assuming the currents directions and then assigning the voltage polarities after that. Also, don't get confused by the sources signs. For that, I always recommend working the problem with variables and substituting the values at the end. From what you post, I can see that you have a sign error in your first equation. Try again and be careful with the signs! SecondChildUserIdTAG: 381746 SecondChildUserNameTAG: jelizon SecondChildCreateTimeTAG: 2012-09-20T14:03:51Z SecondChildTAG: Hello Jelizon, “staff” Thanks for your inputs. As per your recommendations, I did the following steps: 1. Sum of currents leaving the node =0 Therefore (e-v1)/r1+(e-v2)/r2=0 2. After this e(1/r1+1/r2)=v1/r1+v2/r2 3. I have to substitute the values in place of variables v1,v2,r1 and r2 to find "e". Now do we have to substitute v2 = -7.2 ? I did the same and getting -1.69 Please let me know where I went wrong.. Thanks SecondChildUserIdTAG: 181368 SecondChildUserNameTAG: Hanumanta SecondChildCreateTimeTAG: 2012-09-23T10:38:10Z FirstChildTAG: I have solved it with the answer 6.8 FirstChildUserIdTAG: 107465 FirstChildUserNameTAG: Abasit FirstChildCreateTimeTAG: 2012-09-22T15:13:46Z IndexTAG: 4101 TitleTAG: Supplementary condition Please show me! iD=δf(VD)/δvD*⊿vd I do not know the meaning of vD=VD (Supplementary condition)　why? UserIdTAG: 89440 UserNameTAG: yuk CreateTimeTAG: 2012-09-20T02:48:51Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Can you be more specific on your question? Where in the courseware did you see that equation? FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-20T13:58:40Z SecondChildTAG: Thank you! SecondChildUserIdTAG: 89440 SecondChildUserNameTAG: yuk SecondChildCreateTimeTAG: 2012-09-23T08:02:29Z IndexTAG: 4102 TitleTAG: Problem with Courseware and Discussion Problem1: I can only see initial post on forum. No replies can be seen (even my own!). Problem2: Today I tried to continue with Lab2 and cannot see formulas anymore! In the box where I use to be able to see formula now says" [Math Processing Error]. Now I get this error message everywhere where formulas supposed to be! I do not what to do. Even if somebody knows what happened and is going to post it here how to fix this bug, I'll not be able to read solution because I do not see any replies to initial tread. eDx team - PLEASE HELP! UserIdTAG: 167413 UserNameTAG: TeTAn CreateTimeTAG: 2012-09-20T02:32:08Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Troubleshooting NumberOfReplyTAG: 1 FirstChildTAG: Can you tell us what OS/browser you are using? FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-20T03:06:14Z SecondChildTAG: Same problem is here kimt...I'm using Windows 7 Ultimate version/Google Chrome 21.0.1180.89 m SecondChildUserIdTAG: 401259 SecondChildUserNameTAG: amangodne SecondChildCreateTimeTAG: 2012-09-20T04:59:32Z SecondChildTAG: i had the problem with mozilla , but it worked fine with chrome SecondChildUserIdTAG: 330881 SecondChildUserNameTAG: mmmishra SecondChildCreateTimeTAG: 2012-09-23T14:52:02Z SecondChildTAG: Windows 7 with all updates installed. Tried chrome And Firefox - nothing works. Ended up using my 5 years old lenovo T60 with XP. It work like a charm. SecondChildUserIdTAG: 167413 SecondChildUserNameTAG: TeTAn SecondChildCreateTimeTAG: 2012-09-23T22:40:35Z IndexTAG: 4103 TitleTAG: Title numbering Hi... just to inform you that this chapter should be S5V10 instead of S5V9! Regards. P.D. Excellent course UserIdTAG: 378522 UserNameTAG: Alejo_Velasquez CreateTimeTAG: 2012-09-20T01:38:32Z VoteTAG: 0 CoursewareTAG: Week 3 / MOSFET CommentableIdTAG: 6002x_mosfet_model NumberOfReplyTAG: 0 IndexTAG: 4104 TitleTAG: Solving the first S3E1 Hi... My name is Julian, i from colombia, and i have a question... solving the first laboratory, i have that in the e node a KVL, (v1-e)/r1=(e+v2)/R2, that is correct?. thank for you time UserIdTAG: 67238 UserNameTAG: guisaojulian CreateTimeTAG: 2012-09-20T00:48:48Z VoteTAG: 0 CoursewareTAG: Week 2 / Node equation review CommentableIdTAG: 6002x_S3E1_NodeEquationReview NumberOfReplyTAG: 1 FirstChildTAG: Hi Julian, Your equation looks correct. Are you getting the correct answer? FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-20T13:52:43Z SecondChildTAG: good! SecondChildUserIdTAG: 71157 SecondChildUserNameTAG: llereador SecondChildCreateTimeTAG: 2012-10-11T04:52:12Z IndexTAG: 4105 TitleTAG: Homework 2 is not opening i tried to open my courseware but it is just googling i mean it shows busy icon and my homework is not opening???? UserIdTAG: 198382 UserNameTAG: Rizi2k5 CreateTimeTAG: 2012-09-19T23:52:07Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 4106 TitleTAG: I can't understand at all! Hello; I watched the video at least five times and I can't understand what its talking about. I think is because I didn't understand the arbitrary network, but... Somebody could present the Thevenin method with an easier way? I really try, but I don't understand nothing!! Thanks for the attention. UserIdTAG: 291362 UserNameTAG: Gudson CreateTimeTAG: 2012-09-19T23:02:40Z VoteTAG: 0 CoursewareTAG: Week 2 / Thevenin method CommentableIdTAG: 6002x_thevenin NumberOfReplyTAG: 1 FirstChildTAG: Thevenin's theorem says that you can replace any **linear** (only linear) network with a simpler network - one that contains just a voltage source and a series resistance. In what way is this circuit equivalent? It is equivalent in it's terminal characteristics. Let me use an example. Consider the following circuit. ![Example Ciruit][1] We want to get the Thevenin equivalent of this network. This equivalent will have the same terminal characteristics. To understand what this means, lets connect a load resistance to this network across the terminals (AB) and then find the voltage and current. ![Terminal Characteristics][2] In the above figure, you can see that the voltage across the $1\Omega$ load is $0.33V$. Also, the current through this resistance is $0.33A$ (I also connected a voltage source so that the sandbox shows the current as well. The source has been set to zero so that it doesn't affect the characteristics of the network). Now our Thevenin equivalent network should behave the same way. This means that if I connect a $1\Omega$ resistance to it, the voltage across it must be $0.33V$ and current must be $0.33A$. Thevenin's theorem says that it is possible to do this for a Linear network. It says that the simplified circuit consists of a voltage source and a series resistance like this: ![General Thevenin Equivalent][3] So if we connect $1\Omega$ to that, we'll get the same voltage and current. How do we figure out what value of voltage and resistance to use? Well Thevenin's theorem tells us how. It says that in order to find out the equivalent voltage, remove the load resistance (i.e. the $1\Omega$) and measure or calculate the voltage across the terminals. Since the load has been removed, the terminals are open-circuited. So we call this the open circuit voltage (denoted by $V_{OC}$). Doing that is simple in the sandbox. Just run a DC analysis on the original network. This is what you'll see. ![Open Circuit Voltage][4] The voltage across AB is $0.57V$. Therefore $V_{OC} = 0.57V$. This is our Thevenin equivalent voltage. That's half of the job. Next we need the resistance. How do we find that? In order to find the Thevenin resistance, Thevenin's theorem says that we have to set all the independent sources to $0$. So voltage sources will be replaced by a short circuit as the voltage across a wire is $0V$ and current sources will be replaced by an open circuit as the current through an open circuit is $0A$. We then calculate the equivalent resistance of the network. For our example, this is what we get: ![Thevenin Resistance][5] Now, we calculate the equivalent resistance of the above network. This should be quite easy so I'll skip the details. Using our understanding of series and parallel resistances, we get an equivalent resistance of $\frac{5}{7}\Omega$ And that's it! We've found the Thevenin equivalent network. Lets now construct it on the sandbox, connect the $1\Omega$ load (along with the $0V$ source for the reason I mentioned above) and see if we get the same voltage and current. ![Final Result!][6] You can see we have exactly the same voltage and current as our original circuit. So as far as the load resistance is concerned, both circuits are equivalent. The internal details of the original circuit can be ignored and we can use a simple representation like this. Hope this helps. Do let me know if something is still not clear and I'll be happy to explain better. :-) Also, note that our procedure doesn't directly work if there are dependent sources. The textbook explains what to do in such a case. [1]: https://edxuploads.s3.amazonaws.com/13481098492550727.png [2]: https://edxuploads.s3.amazonaws.com/13481100114954492.png [3]: https://edxuploads.s3.amazonaws.com/13481104542550767.png [4]: https://edxuploads.s3.amazonaws.com/13481106756745443.png [5]: https://edxuploads.s3.amazonaws.com/13481110058787291.png [6]: https://edxuploads.s3.amazonaws.com/13481119504954448.png FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-09-20T03:35:48Z SecondChildTAG: Thanks! Really, thanks! I understood now all the explanation and I'm sure that I will can do the exercises. Your text was absolutely helpful, to me and (I'm sure) to everybody that still are doubted. SecondChildUserIdTAG: 291362 SecondChildUserNameTAG: Gudson SecondChildCreateTimeTAG: 2012-09-20T10:13:05Z SecondChildTAG: My pleasure! :-) I'd like to remind you that things are a little different for circuits with dependent sources so do read the textbook (In fact, any book on network analysis will tell you how to handle circuits with dependent sources) SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-09-20T10:41:22Z SecondChildTAG: Thank you ashwith! :) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-21T02:39:16Z SecondChildTAG: Haha. Using screenshots like this is your idea :-P SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-09-21T04:18:35Z SecondChildTAG: I'v started looking more at these great posts before i watch the lectures, as most the time u have to already understand what he is talking about to make any sense of it... SecondChildUserIdTAG: 329964 SecondChildUserNameTAG: SmartMike SecondChildCreateTimeTAG: 2012-09-21T13:29:14Z SecondChildTAG: Watch the lectures first. We learned from Prof. Agarwal, Prof Gerry and Piotr in the Spring run of this course and they are super awesome teachers! :-) Maybe the struggle to figure out something yourself will actually help you a lot more. SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-09-21T14:28:44Z SecondChildTAG: Oh and the brilliant TAs too! There were a lot of homework problem I could have never been able to solve if it wasn't for their solved examples. I wish there was a video featuring an introduction by all the TAs. SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-09-21T14:35:03Z IndexTAG: 4107 TitleTAG: H3P4 Can anybody help me to solved this problem, I don't know what exactly I have to do with the diodes, if the voltage source bellow them is the one that tell that the diode is on or if I only have to take into account the sinusoidal source UserIdTAG: 307811 UserNameTAG: AlejandraBlanco CreateTimeTAG: 2012-09-19T21:33:03Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Positive wave remove from circuit V2 D2, Negative wave remove V1 D1. FirstChildUserIdTAG: 297655 FirstChildUserNameTAG: Aljoska FirstChildCreateTimeTAG: 2012-09-19T22:49:26Z SecondChildTAG: Why do you think so? It doesn't seem to be correct, you can even try to use a simulator (for ex 1V voltage) to check. SecondChildUserIdTAG: 189680 SecondChildUserNameTAG: vnd SecondChildCreateTimeTAG: 2012-09-22T19:02:04Z IndexTAG: 4108 TitleTAG: exercise 2.3 in problem num d can't find resistance UserIdTAG: 221617 UserNameTAG: konan CreateTimeTAG: 2012-09-19T19:36:51Z VoteTAG: 0 CoursewareTAG: Week 1 / Series and parallel example CommentableIdTAG: 6002x_S2E4_Series_and_Parallel NumberOfReplyTAG: 1 FirstChildTAG: When finding resistances it might be easier to use mathematical notation, where there are the following definitions: Series Resistance -> R1 + R2 + ... + Rn Parallel Resistance -> R1//R2//...//Rn And using the following definition then c) is asking for R1 + (R2//(R3+R4)). In English that is R1 in series with R2 in parallel with R3 and R4 in series. Sometimes language can be confusing. FirstChildUserIdTAG: 3571 FirstChildUserNameTAG: ifoughtsharks FirstChildCreateTimeTAG: 2012-09-28T19:01:32Z IndexTAG: 4109 TitleTAG: good test,,, but the last answer appeared to be 666.5 not 665,,, why ! UserIdTAG: 464142 UserNameTAG: wedadMA CreateTimeTAG: 2012-09-19T19:21:33Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: It's because the analysis tool listed the answer in millivolts. If you look closely at the graph you'll notice it says 666.504m, the "m" meaning milli or thousandths of volts. FirstChildUserIdTAG: 123628 FirstChildUserNameTAG: stickybeats FirstChildCreateTimeTAG: 2012-09-23T00:54:39Z IndexTAG: 4110 TitleTAG: oh yaa...great oh yaa...great UserIdTAG: 162670 UserNameTAG: charlesbabyt CreateTimeTAG: 2012-09-19T18:35:08Z VoteTAG: 0 CoursewareTAG: Week 2 / Chain saw CommentableIdTAG: 6002x_chain_saw NumberOfReplyTAG: 0 IndexTAG: 4111 TitleTAG: how to complete lab work.. i cant understand how to complete lab work.... in lab 1 my answer is same which required but after chect that was wrong......in lab work how to rotate equipment...... UserIdTAG: 413500 UserNameTAG: falconbest_555 CreateTimeTAG: 2012-09-19T18:29:34Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: rotate - press R key while item selected(green) FirstChildUserIdTAG: 192653 FirstChildUserNameTAG: Quaz FirstChildCreateTimeTAG: 2012-09-19T18:37:02Z SecondChildTAG: maybe answer still wrong, because time to submit this lab is over. SecondChildUserIdTAG: 192653 SecondChildUserNameTAG: Quaz SecondChildCreateTimeTAG: 2012-09-19T18:39:40Z IndexTAG: 4112 TitleTAG: H2P2 :What does best load resistance means? does best load means minimum resistance or optimum resistance? UserIdTAG: 123484 UserNameTAG: hudkmr CreateTimeTAG: 2012-09-19T17:59:19Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: best load resistance here means the resistance at which maximum power tranfer occurs.. FirstChildUserIdTAG: 173147 FirstChildUserNameTAG: cruiser_rahit FirstChildCreateTimeTAG: 2012-09-19T20:18:35Z SecondChildTAG: best load resistance is the sum of the nearest closed loop resistance which will be connected to the load. (as I know) SecondChildUserIdTAG: 399292 SecondChildUserNameTAG: Qabali SecondChildCreateTimeTAG: 2012-09-19T20:31:01Z FirstChildTAG: The answer is 0.11 FirstChildUserIdTAG: 425325 FirstChildUserNameTAG: IranagoudaN FirstChildCreateTimeTAG: 2012-09-19T20:34:36Z SecondChildTAG: show me hw SecondChildUserIdTAG: 254607 SecondChildUserNameTAG: werehenry SecondChildCreateTimeTAG: 2012-09-20T03:53:44Z IndexTAG: 4113 TitleTAG: S5E3: SR MODEL Last diagram: on question what is the value of vc at entry o? my answer 2.99880048 (also acceptable) differs from 2.99970003. Is there some error there or this is the accurate number? UserIdTAG: 206669 UserNameTAG: dimitrios66 CreateTimeTAG: 2012-09-19T17:10:28Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 4114 TitleTAG: H3P1 - possible bug in the assignment regarding NOR When calculating the NOR, I've stumbled upon a strange thing - the system does not take into account the situation, when the NOR is turned on "twice" - "11". When the nor is in such a state, the equvalent resistance of two FETs in such a configuration is half of the resistance of a single one. Thus, when calculating, the nor will consume maximum power. However, the grader accepts an invalid (as far as I understand) answer - which is in fact the same resistance as for the inverter. Thoughts? UserIdTAG: 7114 UserNameTAG: piotroxp CreateTimeTAG: 2012-09-19T16:31:15Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: i think that we can safely say that the max consumption occurs when R is minimum (due to p=v^2/R). If some other answer is acceptable, perhaps that happens because of the error margin paused by the system. FirstChildUserIdTAG: 206669 FirstChildUserNameTAG: dimitrios66 FirstChildCreateTimeTAG: 2012-09-19T17:04:09Z IndexTAG: 4115 TitleTAG: H2P1 - No such value of resistances available Using the E12 series + ratio of 0.4 and other constraints, I was unable to properly design the voltage divider. As far as I understand, I am supposed to choose an arbitrary resistance from E12 and shuffle around an additional E12 resistance in order to satisfy the required constraints (ratios, the thev equiv resistance and +-10%). Is it possible that the task is impossible to take on? UserIdTAG: 7114 UserNameTAG: piotroxp CreateTimeTAG: 2012-09-19T16:17:13Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 6 FirstChildTAG: I'm facing the same problem and have tried in many ways. FirstChildUserIdTAG: 329015 FirstChildUserNameTAG: farah_sarwar FirstChildCreateTimeTAG: 2012-09-19T18:49:38Z FirstChildTAG: you should do is make your problem with two equations which leave the VTH and RTH respectively and the values ​​you already know Vin and Vout and RTH which is between 10k and 30k, I suggest you choose 20k, you find the two unknowns (R1 and R2 ) and these values ​​are approaching depues nominal values, and the test results so that Vout / Vin are above or below 10%! 0% of the ideal value in this case is 21V FirstChildUserIdTAG: 441072 FirstChildUserNameTAG: jonathan726 FirstChildCreateTimeTAG: 2012-09-19T18:49:03Z FirstChildTAG: i am also facing the same problem...... FirstChildUserIdTAG: 99583 FirstChildUserNameTAG: sanjeevkm0912 FirstChildCreateTimeTAG: 2012-09-19T19:07:13Z FirstChildTAG: same here .... is there a bug ? maybe only for people having 80V as Vin? FirstChildUserIdTAG: 297960 FirstChildUserNameTAG: sebseb95 FirstChildCreateTimeTAG: 2012-09-19T19:29:09Z FirstChildTAG: I am not getting that question can some body explain the question?? FirstChildUserIdTAG: 425325 FirstChildUserNameTAG: IranagoudaN FirstChildCreateTimeTAG: 2012-09-19T20:31:29Z FirstChildTAG: Look this post and see if it helps. It helped me to realize that there is more than one correct answer. https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/505a0b9a859a82270000000d FirstChildUserIdTAG: 288174 FirstChildUserNameTAG: matiasgrodriguez FirstChildCreateTimeTAG: 2012-09-20T00:08:57Z IndexTAG: 4116 TitleTAG: Current I1 Am i wright or wrong, the total current I1 will not be coming to the node e1.I1 divides at e2 node so how come we represent I1 in the KCL at e2????? UserIdTAG: 374566 UserNameTAG: OmkarVijay CreateTimeTAG: 2012-09-19T14:53:09Z VoteTAG: 0 CoursewareTAG: Week 1 / Node analysis practice, part 3 CommentableIdTAG: 6002x_L2Node2 NumberOfReplyTAG: 1 FirstChildTAG: **see..i didnt get it ...the way u represent the value eg:- |1 and e1.|1 .so i am bit confused** FirstChildUserIdTAG: 342886 FirstChildUserNameTAG: hthimz FirstChildCreateTimeTAG: 2012-09-19T14:56:02Z SecondChildTAG: correction : Am i wright or wrong, the total current I1 will not be coming to the node e1 but I1 divides at e2 node so how come we represent I1 in the KCL at e2????? SecondChildUserIdTAG: 374566 SecondChildUserNameTAG: OmkarVijay SecondChildCreateTimeTAG: 2012-09-19T15:42:39Z IndexTAG: 4117 TitleTAG: H2P2: SOLAR POWER HELP!!! I have problem with 2 ex. I found RL= 3.8 . I found I=0.078 and U=0.6 P=i*u=0.047 why is it incorrect ? PLS HELP!! UserIdTAG: 192114 UserNameTAG: chuba CreateTimeTAG: 2012-09-19T14:44:38Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: power calculation through P=i^R ..... FirstChildUserIdTAG: 415376 FirstChildUserNameTAG: imab90 FirstChildCreateTimeTAG: 2012-09-19T15:27:35Z SecondChildTAG: $P=I^2 * R$ SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-19T16:02:42Z SecondChildTAG: Thank's ) SecondChildUserIdTAG: 192114 SecondChildUserNameTAG: chuba SecondChildCreateTimeTAG: 2012-09-19T19:28:20Z FirstChildTAG: You have to find the best load resistance and its a theorem that maximum power can be provided only if load matches the circuit impedance. I hope that it'll give some clue. FirstChildUserIdTAG: 329015 FirstChildUserNameTAG: farah_sarwar FirstChildCreateTimeTAG: 2012-09-19T18:48:28Z SecondChildTAG: Thevenin or Norton Resistance must be equal to Load Resistance for maximum power transfer. SecondChildUserIdTAG: 413002 SecondChildUserNameTAG: Gauravjain88 SecondChildCreateTimeTAG: 2012-09-19T18:55:18Z SecondChildTAG: Thank's ) SecondChildUserIdTAG: 192114 SecondChildUserNameTAG: chuba SecondChildCreateTimeTAG: 2012-09-19T19:28:25Z IndexTAG: 4118 TitleTAG: Meetup. Let there be Meetup like thing for EdX as well so that students may meet and discuss about the Courses as well. UserIdTAG: 2956 UserNameTAG: akshayb CreateTimeTAG: 2012-09-19T14:28:17Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 0 IndexTAG: 4119 TitleTAG: Circuit Voltages and Currents are Linear Combinations of Source Strengths Please, my s3v1: Introduction to linearity is not showing due to bad network. can you please help me out on how to go about this. thanks UserIdTAG: 442257 UserNameTAG: KingAdex CreateTimeTAG: 2012-09-19T13:24:57Z VoteTAG: 0 CoursewareTAG: Week 2 / Linear combinations of source strenghts CommentableIdTAG: 6002x_S3E2_Linear_Combinations_of_Source_Strengths NumberOfReplyTAG: 2 FirstChildTAG: Here are the slides. Both "clean" and "annotated". (With Professor's notes) https://www.edx.org/static/content-mit-6002x/handouts/6002-L3-oei12-gaps.2bb9987ac7b9.pdf https://www.edx.org/static/content-mit-6002x/handouts/6002-L3-oei12-gaps-annotated.d5b91fd3778f.pdf Hope this helps in the meantime. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-19T14:05:23Z FirstChildTAG: Every Video is actually a YouTube video. You can get the YouTube link by right clicking. Maybe you can use some YouTube download software then so that you can watch the video offline.... FirstChildUserIdTAG: 297279 FirstChildUserNameTAG: ZlatkoF FirstChildCreateTimeTAG: 2012-09-23T10:40:45Z IndexTAG: 4120 TitleTAG: Are the signs of y1 and y2 wrong? I solved the exercice, but I got these solutions y1 = 0.22A and y2 = -0.56A. y1 is positive because the current moves from V1 to the others branches and y2 is negative because the current moves from V2 to R1. However the solution says that my signs are wrong. Can someone help me to tell me where I'm mistaken? Thanks! UserIdTAG: 301442 UserNameTAG: Casama CreateTimeTAG: 2012-09-19T13:13:43Z VoteTAG: 0 CoursewareTAG: Week 2 / Superposition example CommentableIdTAG: 6002x_S3E3_superposition NumberOfReplyTAG: 1 FirstChildTAG: The solutions on the exercise are correct. Did you notice that the arrow of i1 is pointing towards the positive terminal of the source V1? When you apply superposition and use only V1 you will find the y1 is leaving the positive terminal of V1, i.e. it is in the direction opposite to that of i1... what does it mean to have i1 and y1 in opposite directions? When you use only V2, then you find y2 going in the same direction of i1... what does that mean? FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-19T13:58:51Z SecondChildTAG: guys try to draw your own circuit on a paper (1).. when V1 is connected x1=0.421 (i.e. node potential) so ... y1= (x1 - V1)/R1 = (0.421 - 2)/7 = -0.22 (2)..when V2 is connected... then look carefully R1 is connected to -ve terminal of V2 so .. y2= (V2 - x2)/R1 = (8 - 4.056)/7 = 0.56 depending upon the direction of current and the resistor across which the value is to be calculated SecondChildUserIdTAG: 389831 SecondChildUserNameTAG: saahilaa SecondChildCreateTimeTAG: 2012-09-22T07:53:15Z IndexTAG: 4121 TitleTAG: regarding v5 there is no connection between v5 why do we take that for calculation? and how can this have a voltage in between? UserIdTAG: 134904 UserNameTAG: ASHWIN07 CreateTimeTAG: 2012-09-19T13:06:13Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL and KCL exercises CommentableIdTAG: 6002x_kvl_kcl_exercises NumberOfReplyTAG: 1 FirstChildTAG: A connection is not needed for a voltage difference to exist. You can consider v5 to be part of a loop, for example, V - v1 - v5 - v2. Just be careful with the signs. FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-19T13:51:31Z SecondChildTAG: If there was a connection -using an ideal wire- we would have a short circuit scenario (no voltage whatsoever). Measuring the voltage, for instance, in a 9V battery is an example of what jelizon said (a connection is not needed). SecondChildUserIdTAG: 9100 SecondChildUserNameTAG: dmgongora SecondChildCreateTimeTAG: 2012-09-24T04:22:45Z IndexTAG: 4122 TitleTAG: Access to student assistants This is the first time I have been able to be online when teachers assistants are available. With the hundreds or even thousands of students taking the class, how can you handle so much activity? Also, how does it work? Do you just join the conversation or do you actually chat with students somehow etc.? UserIdTAG: 132828 UserNameTAG: rharris CreateTimeTAG: 2012-09-19T13:02:46Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: It is only via the discussion forum. We try to answer as many questions as we can, but with so many posts, many don't get answered. It is more likely for you question to be answer if you post it during the times when a TA is online (like now). You can find the hours in "Course Info". We have noticed, however, that fellow students answer questions very well and we also want promote that. So, if you feel you understand a topic correctly, feel free to help others! Just don't give away answers for homeworks and labs FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-19T13:48:57Z SecondChildTAG: Thanks jelizon. I got up this morning specifically to ask this question, knowing I have to be online at 6:00 A.M. CA time. I agree with you not to give out answers, or take answers from someone else. I am here to learn to do it myself, and copying from someone else gives me nothing. This course is excellent, thank you for your participation. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-09-19T14:35:27Z IndexTAG: 4123 TitleTAG: good test of skills i failed at some of the math but that is because i have been up over 24 hours UserIdTAG: 461815 UserNameTAG: wile2u CreateTimeTAG: 2012-09-19T10:30:03Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 0 IndexTAG: 4124 TitleTAG: S8E2 I don't understand hint method. I did it with Thevenin (cut of dependent source and get VTH and RTH) and then I got current flowing by dependent source. Then I get VTH for port A with is voltage on R2. But this method is not working in homework... UserIdTAG: 302188 UserNameTAG: Pepek CreateTimeTAG: 2012-09-19T09:50:52Z VoteTAG: 0 CoursewareTAG: Week 4 / Dependent Sources Exercise 1 CommentableIdTAG: 6002x_dep_src_exsz_1 NumberOfReplyTAG: 2 FirstChildTAG: Do not cut anything.Simply add a currrent source of 1A at the port and solve using KCL and KVL.Watch S3V7 again, and will be a lot clearer.At least it worked for me, and i am not a clever guy. FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-09-24T16:29:09Z FirstChildTAG: I solved it by applying the superposition method. I guess my solution is correct only this time since the dependant voltage source is linear. FirstChildUserIdTAG: 418182 FirstChildUserNameTAG: euldji2005 FirstChildCreateTimeTAG: 2012-09-29T18:32:33Z SecondChildTAG: Forgot something, It is obvious to use the superposition method since the objective of this problem is to find VTH and RTH which is correct only if the circuit is linear. I want to know how to solve this problem with the node method only. SecondChildUserIdTAG: 418182 SecondChildUserNameTAG: euldji2005 SecondChildCreateTimeTAG: 2012-09-29T18:58:20Z SecondChildTAG: since between two nodes there is one voltage source, apply supernode method SecondChildUserIdTAG: 373559 SecondChildUserNameTAG: aditya14 SecondChildCreateTimeTAG: 2012-10-06T09:09:05Z IndexTAG: 4125 TitleTAG: late course start Can i start taking this circuit and electronics course on September 19? Thanks. UserIdTAG: 461593 UserNameTAG: yishagerew CreateTimeTAG: 2012-09-19T09:44:27Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I think you can but you will have missed one homework and lab. However I understand that you can drop 2 of each over the whole course so you could still catch up. I'm not staff so you may want to check anything I say with them. FirstChildUserIdTAG: 3646 FirstChildUserNameTAG: freespirit FirstChildCreateTimeTAG: 2012-09-19T11:02:36Z IndexTAG: 4126 TitleTAG: What song is that? I'm just wondering. Thank you in advance. UserIdTAG: 280220 UserNameTAG: pilgrimCycle CreateTimeTAG: 2012-09-19T09:24:44Z VoteTAG: 0 CoursewareTAG: Week 2 / Chain saw CommentableIdTAG: 6002x_chain_saw NumberOfReplyTAG: 1 FirstChildTAG: http://www.youtube.com/watch?v=Z2uRLq50Z8g FirstChildUserIdTAG: 436749 FirstChildUserNameTAG: waynebrown FirstChildCreateTimeTAG: 2012-09-19T10:25:23Z SecondChildTAG: Thanks :D SecondChildUserIdTAG: 280220 SecondChildUserNameTAG: pilgrimCycle SecondChildCreateTimeTAG: 2012-09-25T01:18:40Z IndexTAG: 4127 TitleTAG: Lab 01 assesment I have a small problem. I completed my Lab 01 before the due date, run the DC analysis and saved it successfully but I see that it has not been assessed and my progress shows 0 out of 2 marks in my lab 01. I do not know why this has happened. I request the course regulator to please review the matter. UserIdTAG: 131726 UserNameTAG: chemiboy11 CreateTimeTAG: 2012-09-19T08:44:37Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Troubleshooting NumberOfReplyTAG: 1 FirstChildTAG: You Must click At Check Button To Get the Evaluation FirstChildUserIdTAG: 341818 FirstChildUserNameTAG: Elshazly FirstChildCreateTimeTAG: 2012-09-19T12:03:26Z SecondChildTAG: There is no check button there now. SecondChildUserIdTAG: 131726 SecondChildUserNameTAG: chemiboy11 SecondChildCreateTimeTAG: 2012-09-19T17:18:27Z IndexTAG: 4128 TitleTAG: h2p1 help please i dont know were iv gone wrong i have the right vin vout but its not accepting my resistors am i supposed to put k or ohms after it or what its the only thing i need to complete then iv done everything so can someone help UserIdTAG: 436749 UserNameTAG: waynebrown CreateTimeTAG: 2012-09-19T07:31:56Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Since the Thevenin resistance as seen from the output terminals is between 10kΩ and 30kΩ, it is better to take R1 and R2 in K-Ohms... To enter values in KOhms simply multiply a value by 10^3 or the value followed by k.. For e.g. 22*10^3 or 22k FirstChildUserIdTAG: 285089 FirstChildUserNameTAG: jobysam7 FirstChildCreateTimeTAG: 2012-09-19T08:10:48Z FirstChildTAG: i have and to my knowledge done it correctly but it does not work only my out and in gets the tick so iv referd it to the bugs department as im not the only one who has had this happen but thank you for the advice :) FirstChildUserIdTAG: 436749 FirstChildUserNameTAG: waynebrown FirstChildCreateTimeTAG: 2012-09-19T08:27:48Z IndexTAG: 4129 TitleTAG: H2P1 Any 1 who have done question H2P1..............kINdly help me UserIdTAG: 382186 UserNameTAG: noorsyani CreateTimeTAG: 2012-09-19T07:30:23Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: H2P1: VOLTAGE – DIVIDER DESIGN The resistances of commercially-available discrete resistors are restricted to particular sets. For example, the available values of resistors with 10% tolerance are selections from the E12 set multiplied by a power of ten from 10^0 through 10^5. The E12 set is: E12={10,12,15,18,22,27,33,39,47,56,68,82} Ok, What are that numbers inside the {}? They are standard values of resistors, values that you can buy comercially. Imagine if there were not this values available and any manufacturer could make any arbitrary value depending on their available materials in stock … it would be a caos! Fortunately, exists standards values ;) What they mean when they say that they have a 10% of a tolerance??? That means that if you buy a resistor of 22 Ohm, if you meassure that resistor it values can be 10% more or less: the resistor can be R= 22+2=24 Ohm or R=22-2=20 Ohm. What about that range that they give (10^0 to 10^5)? The values inside the {} are: 10 * 10^0 Ohm= 10 Ohm 10 * 10^1 Ohm = 100 Ohm 10* 10^2=1000 Ohm = 1kOhm 10* 10^3=10000Ohm = 10kOhm 10*10^4=100000 Ohm = 100kOhm 10*10^5=1000000 Ohm = 1 MOhm . . . 47 * 10^0 Ohm= 47 Ohm 47* 10^1 Ohm = 470 Ohm 47* 10^2=4700 Ohm = 4.7kOhm 47* 10^3=47000Ohm = 47kOhm 47*10^4=470000 Ohm = 470kOhm 47*10^5=4700000 Ohm = 4.7 MOhm And so on with all the values inside the {}… The statement says : In this problem we need to choose 10% resistors to make a voltage divider that meets a given specification. We are given an input voltage Vin=70.0V, and we need to provide an open-circuit output voltage of Vout≈24.5V. An additional requirement is that the Thevenin resistance as seen from the output terminals is between 10kΩ and 30kΩ. Assume first that the resistors have their nominal resistance. Come up with resistors R1 and R2 such that the divider ratio Vout/Vin is within 10% of the requirement. Of course, the resistances you chose are just nominal. Given that they are only guaranteed to have resistances within 10% of the nominal value, what is the largest and smallest value that Vout may have? Here my first Hint to you: review this part of the Textbook Read Here. The statement it is a Little confusing. Ok, let’s try to understand what are they asking in order that you can later solve this by your own: Hint PART 1 and PART 2: They give you Vin and Vout. You know how to relate Vout in parameters of R1, R2 and Vin (have you read the first hint? - Textbook Chapter). Can you find R1 and R2? … No, we need another equation more... (because we have 2 incognits and to solve the system we need the same number of equations as variables that we have)… ... But they tell you: An additional requirement is that the thevenin resistance as seen from the output terminals is between 10kΩ and 30kΩ. (Here it is your Aha moment!). (Read Thevenin in the Textbook) Can you find R1 and R2 now? Yes!!!! .... Ok, but wait a minute, what does it mean: “Come up with resistors R1 and R2such that the divider ratio Vout/Vin is within 10% of the requirement” This is, that once you have your values, with the values that you have chosen, that voutput/vinput has to be less-equal than a 10%. It is to verify your chosen values! If it is less or equal it is correct if not, it is wrong and you have to re-calculate again. Hint PART 4 and 5: If you Know that the resistances that you have chosen has an Rmáx value and Rmín value (Remember that this is because they have a 10% tolerance), So, If you have chosen R1 and R2, what combination of the possibles values gives me the V output máx and the Voutput mín of the Circuit? FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-09-19T08:03:53Z SecondChildTAG: thanks for such detailed explanation. SecondChildUserIdTAG: 5345 SecondChildUserNameTAG: caspik SecondChildCreateTimeTAG: 2012-09-20T04:11:35Z IndexTAG: 4130 TitleTAG: pages of the book? which the pages of the book to study and solve the two laboratory ? UserIdTAG: 327787 UserNameTAG: REINALDOPARANHOS CreateTimeTAG: 2012-09-19T00:46:08Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Take a look at "Course Info" (top of page) and click on "6002x at a Glance." It gives the chapters and sections of the book for each week. FirstChildUserIdTAG: 92346 FirstChildUserNameTAG: MobiusTruth FirstChildCreateTimeTAG: 2012-09-19T04:11:34Z IndexTAG: 4131 TitleTAG: Schematic Can you post schematic for the circuit? I'd really appreciate it. Cheers! UserIdTAG: 322552 UserNameTAG: riv CreateTimeTAG: 2012-09-18T22:56:42Z VoteTAG: 0 CoursewareTAG: Week 2 / Cardiac Experiment CommentableIdTAG: 6002x_cardiac NumberOfReplyTAG: 1 FirstChildTAG: I would rather not. This really is a "Don't try this at home." Aside from circuit design, there are a lot of details to doing this experiment safely. For instance, both the scope and the power supply are isolated -- neither is connected directly to 120VAC or ground. The scope is battery-powered. I do not recall whether we ran the circuit off of batteries, or used a power supply+isolation transformer, but regardless, there was no direct path to ground. There are probably a half-dozen other things you have to do like this, and I'm not confident I could give them all, or figure out all the other ways things could go wrong. A lot of this is just built into my brain at this point. This is really not the sort of experiment you can do safely just by copying what we did. You have to be able to understand the circuit, and understand how to use the circuit. Being able to design the circuit is a minimum, but not sufficient level of knowledge for this. A secondary reason is that I no longer have the schematic. The circuit is sitting in a box in edX, and the only way to figure it out would be to trace it out. FirstChildUserIdTAG: 2 FirstChildUserNameTAG: pm FirstChildCreateTimeTAG: 2012-09-19T03:36:53Z SecondChildTAG: Ok, no problem. The reason for the question was to know another way to build EMG sensor than this one: http://www.instructables.com/id/Muscle-EMG-Sensor-for-a-Microcontroller/?ALLSTEPS . I know that the output of of these sensors (yours and from instructables) are different but if I understand correctly the latter one contains some parts from the former one. Another reason was that the one from instructables uses quite pricey and hard to find on some countries differential amplifier - INA106 (~$10) so I was hoping to find a decent replacement for it. Thanks for the response anyway. Cheers! SecondChildUserIdTAG: 322552 SecondChildUserNameTAG: riv SecondChildCreateTimeTAG: 2012-09-19T19:04:46Z IndexTAG: 4132 TitleTAG: Tell me if I did it right. I'm confused by the instructions. For y1 and y2, I did the superposition. After I solved for x1 & x2, I simply took the value of the part that was divided by R1 from each equation. In the 1st equation I took the(x1-V1)/R1 part. For the 2nd equation it was x2/R1 and simply got the answer. Is this correct? UserIdTAG: 378733 UserNameTAG: rlagomar CreateTimeTAG: 2012-09-18T21:41:22Z VoteTAG: 0 CoursewareTAG: Week 2 / Superposition example CommentableIdTAG: 6002x_S3E3_superposition NumberOfReplyTAG: 1 FirstChildTAG: Yes, for the **Y1** it works but in the case of **Y2** even I am stucked. lol ! FirstChildUserIdTAG: 211164 FirstChildUserNameTAG: Hamoodi FirstChildCreateTimeTAG: 2012-09-20T16:19:25Z IndexTAG: 4133 TitleTAG: average power hi, guys. who can explain me how to calculate the average power in "S1E3: AC POWER" diagramme? UserIdTAG: 187207 UserNameTAG: sundog CreateTimeTAG: 2012-09-18T20:37:50Z VoteTAG: 0 CoursewareTAG: Week 1 / AC power CommentableIdTAG: 6002x_S1E3_AC_power NumberOfReplyTAG: 1 FirstChildTAG: average power = average voltage/resistance. In this example average voltage is given: 120V. And peak voltage is 120V*sqrt(2)~168V. So, peak power = peak voltage/resistance. FirstChildUserIdTAG: 367431 FirstChildUserNameTAG: VictorFedosenkov FirstChildCreateTimeTAG: 2012-09-19T07:20:31Z SecondChildTAG: average power is simply defined as: p(avg)={V(rms)}^2/(resistance) here V(rms)=120v,so we can get easily the value of average power. SecondChildUserIdTAG: 451865 SecondChildUserNameTAG: kumar51 SecondChildCreateTimeTAG: 2012-09-20T15:42:28Z SecondChildTAG: $P(avg)={V^2(rms)/R}$ SecondChildUserIdTAG: 539003 SecondChildUserNameTAG: amitgupta25121993 SecondChildCreateTimeTAG: 2012-10-04T20:08:43Z IndexTAG: 4134 TitleTAG: Urgent about course text I found a the course's text book as a pdf on the website It claims it's a legal copy under creative commons license. So is that true? Can I freely use it (the online version is very slow in turning pages)? UserIdTAG: 154541 UserNameTAG: AhmedMedhat CreateTimeTAG: 2012-09-18T20:02:20Z VoteTAG: 0 CoursewareTAG: Week 2 / Is it linear? CommentableIdTAG: 6002x_is_it_linear NumberOfReplyTAG: 1 FirstChildTAG: No, the textbook has never been released under creative commons. We got permission from the publisher to offer png's of the pages, but that is a limited agreement. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-09-18T22:47:05Z IndexTAG: 4135 TitleTAG: Replacing the excitation current by a resistor or voltage Since Thevenin theorem will be used as a generalization, how would we prove Thevenin theorem if the voltage v we intend to measure is across a resistor 'r' in place of a current source 'i'? UserIdTAG: 335370 UserNameTAG: SumitN CreateTimeTAG: 2012-09-18T18:46:09Z VoteTAG: 0 CoursewareTAG: Week 2 / Thevenin method CommentableIdTAG: 6002x_thevenin NumberOfReplyTAG: 1 FirstChildTAG: Anybody? FirstChildUserIdTAG: 335370 FirstChildUserNameTAG: SumitN FirstChildCreateTimeTAG: 2012-09-24T00:25:49Z IndexTAG: 4136 TitleTAG: trouBle in LAB-2, i m trying to solve it but, the answer in not right, Any HInts?? ........ UserIdTAG: 210639 UserNameTAG: ictscholariiu CreateTimeTAG: 2012-09-18T18:24:47Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: i dont know what i am doing wrong i have tried to do this in every way but nothing works we are not supposed to change anything on the resistors are we because it does not work on 1 FirstChildUserIdTAG: 160242 FirstChildUserNameTAG: Mlevins35 FirstChildCreateTimeTAG: 2012-09-18T22:24:39Z FirstChildTAG: Hi, 1.First try to get result Vout= 1/2*V1+1/2*V2 . (like in Figure 2. Simple resistive mixer) 2.Independently from point 1 try to get Vout_=1/3*V2 (to do this you can use 2 resistors Ra and Rb with relationship Ra/Rb=2/1). 3.Try to combine V1 and 1/3 of V2 in this way like in point 1. (Take into account that you'll need to harmonize the resistance of all resistors in the diagram at step 3.) FirstChildUserIdTAG: 277168 FirstChildUserNameTAG: AndrewKiselev FirstChildCreateTimeTAG: 2012-09-18T18:58:05Z SecondChildTAG: explain first step please ..mean we put only V1 and V2 value in this equation ??? and also explain the relationship ...we suppose to use 2 resistors ?? SecondChildUserIdTAG: 415376 SecondChildUserNameTAG: imab90 SecondChildCreateTimeTAG: 2012-09-18T20:11:44Z SecondChildTAG: For step 1 - yes, just 2 resistors and 2 voltage sources. (Reproduce the diagram which you can see in "Figure 2. Simple resistive mixer".) SecondChildUserIdTAG: 277168 SecondChildUserNameTAG: AndrewKiselev SecondChildCreateTimeTAG: 2012-09-18T20:53:18Z SecondChildTAG: "we put only V1 and V2 value in this equation " -You should better not to use any equation (for step 1), but do it in lab simulator!! equations you can use later to harmonize the resistance of all resistors in the diagram at step 3. The simulator will give you sense how the diagram behave yourself and equations will give you certainty in the result. SecondChildUserIdTAG: 277168 SecondChildUserNameTAG: AndrewKiselev SecondChildCreateTimeTAG: 2012-09-18T21:09:49Z IndexTAG: 4137 TitleTAG: Exercise wrong Y2 does not accept the correct answer, please correct it. I will not waste my time with this exercise. UserIdTAG: 412912 UserNameTAG: Terleira CreateTimeTAG: 2012-09-18T17:01:27Z VoteTAG: 0 CoursewareTAG: Week 2 / Superposition example CommentableIdTAG: 6002x_S3E3_superposition NumberOfReplyTAG: 1 FirstChildTAG: Hello, Terleira. Try to read carefully WHAT y2 IS. Maybe it's will help you: y2 is not current over R2. FirstChildUserIdTAG: 277168 FirstChildUserNameTAG: AndrewKiselev FirstChildCreateTimeTAG: 2012-09-18T17:50:22Z SecondChildTAG: can u pls tell in detail how to proceed from the beginning.....please...am totally confused...like how do they get x1 and x2 atleast?? thanks SecondChildUserIdTAG: 414118 SecondChildUserNameTAG: raje93 SecondChildCreateTimeTAG: 2012-09-18T18:00:09Z SecondChildTAG: You can read the comment which was maded by staff in the post [What am I missing?][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_S3E3_superposition/threads/50585c9c2c49d42a00000001 SecondChildUserIdTAG: 277168 SecondChildUserNameTAG: AndrewKiselev SecondChildCreateTimeTAG: 2012-09-18T18:14:49Z SecondChildTAG: Since the scheme is LINEAR therefore we can to use a superposition method. For a finding x1 (voltage on resistance R3, caused by action of source V1) we will replace V2 with a wire piece!! In this case x1=V1/R_common; R_common=R1+R2 || R3. (R2 || R3 means resistance of parallel R2 and R3). SecondChildUserIdTAG: 277168 SecondChildUserNameTAG: AndrewKiselev SecondChildCreateTimeTAG: 2012-09-18T18:37:12Z SecondChildTAG: Thanks Andrej, you help me a lot! Rgds SecondChildUserIdTAG: 380287 SecondChildUserNameTAG: gayetan SecondChildCreateTimeTAG: 2012-09-18T18:38:40Z SecondChildTAG: :) SecondChildUserIdTAG: 277168 SecondChildUserNameTAG: AndrewKiselev SecondChildCreateTimeTAG: 2012-09-18T19:10:21Z SecondChildTAG: I made mistake in my comments x1=(V1/R_common)*R1 !! Sorry! V1/R_common will just a current over R1 in case when V2 replaced by a wire. SecondChildUserIdTAG: 277168 SecondChildUserNameTAG: AndrewKiselev SecondChildCreateTimeTAG: 2012-09-18T19:27:33Z SecondChildTAG: Y2 is not correct. The expression: y2 = R3/(R1R2+R2R3+R1R3) means y2 = 5/71, which is 0.0704, far from the given answer. SecondChildUserIdTAG: 484464 SecondChildUserNameTAG: IreneK SecondChildCreateTimeTAG: 2012-09-28T23:29:39Z IndexTAG: 4138 TitleTAG: Lab 3 hi friends i was solving lab 3 and i got the desired graph as shown in the lab 3. but i m not awarded any grade.. So, please help me out... thanx UserIdTAG: 108929 UserNameTAG: namit CreateTimeTAG: 2012-09-18T16:55:59Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Did you click 'tran' before you 'check'? FirstChildUserIdTAG: 95673 FirstChildUserNameTAG: melbur FirstChildCreateTimeTAG: 2012-09-19T15:41:46Z SecondChildTAG: yeah i clicked 'tran' function and i got the desired plot but after checking it marked me with a cross. SecondChildUserIdTAG: 108929 SecondChildUserNameTAG: namit SecondChildCreateTimeTAG: 2012-09-19T17:55:31Z IndexTAG: 4139 TitleTAG: No signal or null value The convention between sender and receiver should have an additional item to cover the case when there is no signal, such as when the connection is broken. For instance if the receiver receives 0V during a time period exceeding tx then the connection has been lost. Perhaps that condition is outside the playground of Static Discipline for now. UserIdTAG: 184153 UserNameTAG: tthngn CreateTimeTAG: 2012-09-18T16:39:06Z VoteTAG: 0 CoursewareTAG: Week 2 / Static Discipline CommentableIdTAG: 6002x_static_discipline NumberOfReplyTAG: 1 FirstChildTAG: Well you can create any discipline you like, as long as your devices adhere to that discipline. Essentially what you are proposing is an additional forbidden zone say in the 0v-1v range? (Possibly requiring additional noise margin near 1v), and anything that lands in that voltage range would be a "no signal"? If the Tx still had power during a "no signal" event, maybe it could be setup to send a signal in the 2.5 volt range.(In the already present forbidden zone.) FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-19T04:21:48Z IndexTAG: 4140 TitleTAG: problem - unable to view textbooks i am unable to view any of the textbooks available in this course. could you please look into it ASAP..... UserIdTAG: 448671 UserNameTAG: anirudhsom3000 CreateTimeTAG: 2012-09-18T15:44:43Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: What browser are you using and what shows up when you click on the Textbook tab? FirstChildUserIdTAG: 11 FirstChildUserNameTAG: dormsbee FirstChildCreateTimeTAG: 2012-09-18T18:01:57Z SecondChildTAG: i am using google chrome, and when i press the textbook tab above all i can see is the content list on the left and rest of the page is blank.... SecondChildUserIdTAG: 448671 SecondChildUserNameTAG: anirudhsom3000 SecondChildCreateTimeTAG: 2012-09-18T18:12:49Z SecondChildTAG: here is an image for u to have an idea what my page looks like after selecting the textbook's tab...![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13479937681343688.png SecondChildUserIdTAG: 448671 SecondChildUserNameTAG: anirudhsom3000 SecondChildCreateTimeTAG: 2012-09-18T18:42:58Z SecondChildTAG: It looks like some javascript problem, check if you have javascript enabled or try with another browser. SecondChildUserIdTAG: 276409 SecondChildUserNameTAG: IgnacioUY SecondChildCreateTimeTAG: 2012-09-18T20:24:00Z IndexTAG: 4141 TitleTAG: superposition rule Last lesson practice the superposition. The rule can be use more than or equal to two sources schematic. Keep the one source to analysis in schematic. Final you can superposition the result, you will get the correct results. Attention: the minus and plus of current and voltage. UserIdTAG: 146995 UserNameTAG: ZYJ CreateTimeTAG: 2012-09-18T15:29:15Z VoteTAG: 0 CoursewareTAG: Week 2 / Superposition example CommentableIdTAG: 6002x_S3E3_superposition NumberOfReplyTAG: 0 IndexTAG: 4142 TitleTAG: Impressive schematic Adjust 4 resistors value, the voltage of node will be go up and down. UserIdTAG: 146995 UserNameTAG: ZYJ CreateTimeTAG: 2012-09-18T14:30:29Z VoteTAG: 0 CoursewareTAG: Week 2 / Superposition experiment CommentableIdTAG: 6002x_Superposition_experiment NumberOfReplyTAG: 0 IndexTAG: 4143 TitleTAG: Norton resistance H4P3 What is the Norton resistance in Ohms? Vth=7/11*5=3.181818 Rth=4*7/(4+7)...? Rth=Norton resistance UserIdTAG: 89440 UserNameTAG: yuk CreateTimeTAG: 2012-09-18T14:21:48Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: I'm having difficulty with this one, as well. By no means should anyone just give me the answer but I'm not sure how to begin. No matter what I do, I can't find the Norton equivalent current. I'm obviously doing something wrong and need someone to point me in the right direction. Thanks. FirstChildUserIdTAG: 331664 FirstChildUserNameTAG: dwmnctrh3 FirstChildCreateTimeTAG: 2012-09-19T00:55:04Z FirstChildTAG: I'm not sure how much I can say within the honor code and so my answer is cryptic. The first thing is that it can be done i.e. there isn't a mistake which was my first thought when my first attempt was wrong. The second thing is that I didn't solve it the same way as I would normally solve these problems with independent sources - the challenge is of course how to allow correctly for the dependent source which I don't think was covered in the videos. I didn't check out the book on how to handle dependent sources so I'm not sure if that would help. I couldn't solve either part a or b initially and when I sorted out how to solve one part the second followed more easily. The 'aha' moment for me came from one of the additional tutorials. It is the realisation that if 2 circuits are equivalent (i.e original circuit and it's Norton or Thevanin equivalent) then if you add something to the output of one and see a response you should see the same response if you add the same thing to the output of the other. Sorry I don't feel I can be more explicit due to the honor code. I hope it helps. FirstChildUserIdTAG: 3646 FirstChildUserNameTAG: freespirit FirstChildCreateTimeTAG: 2012-09-19T10:59:34Z FirstChildTAG: r2 et r3 en paralle et r1 en serie avec r2//r3 FirstChildUserIdTAG: 316899 FirstChildUserNameTAG: elou FirstChildCreateTimeTAG: 2012-09-24T12:01:54Z SecondChildTAG: merci mec ;) SecondChildUserIdTAG: 297960 SecondChildUserNameTAG: sebseb95 SecondChildCreateTimeTAG: 2012-09-29T18:42:08Z SecondChildTAG: You are violating the honor code by providing such an explicit answer. SecondChildUserIdTAG: 129876 SecondChildUserNameTAG: msarabi95 SecondChildCreateTimeTAG: 2012-10-07T17:36:32Z IndexTAG: 4144 TitleTAG: doubt why we are applying this concept , we can also use it another way like , if we are applying 2 or more inputs then it is obvious that we get the output equals to the sum of the both the inputs applied . sir please calrify my doubt . UserIdTAG: 146625 UserNameTAG: ece CreateTimeTAG: 2012-09-18T14:04:04Z VoteTAG: 0 CoursewareTAG: Week 2 / Linearity CommentableIdTAG: 6002x_linearity NumberOfReplyTAG: 1 FirstChildTAG: It is not always true that if you sum the inputs then obviously we will have the summed outputs ! that is why he emphasizing on the concept and relate to the linear systems only! As you can see in the picture, the summed inputs does not give the summed outputs! ![Linear Systems vs Non-Linear Systems][1] [1]: https://edxuploads.s3.amazonaws.com/134832307972693.gif FirstChildUserIdTAG: 130810 FirstChildUserNameTAG: moon_light FirstChildCreateTimeTAG: 2012-09-22T14:11:37Z IndexTAG: 4145 TitleTAG: doubt please explain it once why v2 source is rotated . UserIdTAG: 146625 UserNameTAG: ece CreateTimeTAG: 2012-09-18T13:48:35Z VoteTAG: 0 CoursewareTAG: Week 2 / Node equation review CommentableIdTAG: 6002x_S3E1_NodeEquationReview NumberOfReplyTAG: 1 FirstChildTAG: The exercise shows that even though the battery might appear like it would lower the voltage of e, it actually increases the voltage. e is a is a voltage suspended between one potential and a higher potential. Even though V2 is "backwards", it's voltage is negative, so it acts the same as a battery with positive voltage, rotated the other way. Give it a try in the sandbox. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-18T15:12:02Z IndexTAG: 4146 TitleTAG: Lab 3 Can somebody please help me with lab 3? UserIdTAG: 162671 UserNameTAG: tuhin1991paul CreateTimeTAG: 2012-09-18T13:22:02Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: combine 3 MOSFET such as two MOSFET give nor gate operation and one give the nand operation. hope this will help you FirstChildUserIdTAG: 5070 FirstChildUserNameTAG: wajid2786 FirstChildCreateTimeTAG: 2012-09-18T13:42:13Z SecondChildTAG: i did that.....but my output curve (4th one) is not the expected one. I does not resemble the one which was given after 3ms..thank u....can u do a little more help? SecondChildUserIdTAG: 162671 SecondChildUserNameTAG: tuhin1991paul SecondChildCreateTimeTAG: 2012-09-18T15:42:18Z IndexTAG: 4147 TitleTAG: solution KVL and KCL combination can get the answer UserIdTAG: 146995 UserNameTAG: ZYJ CreateTimeTAG: 2012-09-18T13:00:45Z VoteTAG: 0 CoursewareTAG: Week 2 / Linear combinations of source strenghts CommentableIdTAG: 6002x_S3E2_Linear_Combinations_of_Source_Strengths NumberOfReplyTAG: 0 IndexTAG: 4148 TitleTAG: Asking About logic outputs A B Z This is the NAND Gate Truth Table 0 0 1 0 1 0 1 0 0 1 1 1 What You Always Saying Is That The Truth Table Of NAND Gate is: A B Z 0 0 1 0 1 1 1 1 1 1 1 0 As Far As I Know This is Wrong. Please Response & Advise. UserIdTAG: 106816 UserNameTAG: Laith CreateTimeTAG: 2012-09-18T12:07:03Z VoteTAG: 0 CoursewareTAG: Week 3 / Switch Model CommentableIdTAG: 6002x_switch_model NumberOfReplyTAG: 1 FirstChildTAG: The truth table of a NAND gate is: A B **Z** - 0 0 **1** - 0 1 **1** - 1 0 **1** - 1 1 **0** FirstChildUserIdTAG: 159031 FirstChildUserNameTAG: GabrielNeves FirstChildCreateTimeTAG: 2012-09-19T19:03:25Z IndexTAG: 4149 TitleTAG: problem reg direction of flow of current **strong text**hi........ this is my first time digram is correct but proving kvl law i am fail.but it is interesting. UserIdTAG: 456614 UserNameTAG: atulnarayan786 CreateTimeTAG: 2012-09-18T12:00:00Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 0 IndexTAG: 4150 TitleTAG: What am I missing? If -----> y1 = (x1 - V1)/R1, then shouldn't ------> y2 = (x2 - V2)/R2? UserIdTAG: 237421 UserNameTAG: Lmotloch CreateTimeTAG: 2012-09-18T11:35:56Z VoteTAG: 0 CoursewareTAG: Week 2 / Superposition example CommentableIdTAG: 6002x_S3E3_superposition NumberOfReplyTAG: 1 FirstChildTAG: Hi, y2 is not a curent over R2 !! Pls. read carefully what y2 is. FirstChildUserIdTAG: 277168 FirstChildUserNameTAG: AndrewKiselev FirstChildCreateTimeTAG: 2012-09-18T12:22:01Z SecondChildTAG: You should be taking the following steps: - Set $V_2$ to zero, leaving only $V_1=2V$. - Solve for the current into R1, call this value $y_1$. - Set $V_1$ to zero, leaving only $V_2=8V$. - Solve for the current into R1, call this value $y_2$. - Sum $y_1$ and $y_2$ to get $i_1$, the total current through resistor 1 due to both sources. If you do that, you should be able to see why the expressions for y1 and y2 aren't similar. SecondChildUserIdTAG: 151427 SecondChildUserNameTAG: RichardZ SecondChildCreateTimeTAG: 2012-09-18T13:43:26Z SecondChildTAG: Thanks for the response, I just misread what the question was looking for as pointed out by AndrewKiselev. SecondChildUserIdTAG: 237421 SecondChildUserNameTAG: Lmotloch SecondChildCreateTimeTAG: 2012-09-18T19:04:18Z IndexTAG: 4151 TitleTAG: How did you get Rth? How did you get Rth? I get 7.34126984127e-05 by summing the resistors in paralel UserIdTAG: 231676 UserNameTAG: Roosemberth CreateTimeTAG: 2012-09-18T11:14:33Z VoteTAG: 0 CoursewareTAG: Week 2 / Simple Thevenin CommentableIdTAG: 6002x_S3E4_Simple_Thevenin NumberOfReplyTAG: 3 FirstChildTAG: In general you can compute the equivalent value of resistors connected in parallel using: $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + ... +\frac{1}{R_n}$ So in the case where you have only 2 resistors in parallel this can also be written as: $R_{eq} = \frac{R_{1}R_{2}}{R_1 +R_2}$ Your value is the reciprocal of the correct value, so I assume you just switched the places of numerator and denominator. FirstChildUserIdTAG: 311022 FirstChildUserNameTAG: GordanS FirstChildCreateTimeTAG: 2012-09-18T11:31:09Z SecondChildTAG: I just made the calculus of the resistors in parallel mode.... SecondChildUserIdTAG: 316902 SecondChildUserNameTAG: JesseTeixeira SecondChildCreateTimeTAG: 2012-09-20T20:31:22Z FirstChildTAG: I made this mistake as well. When you have resistors in parallel, you can add their conductances to get the equivalent conductance: 1/R1 + 1/R2 ~ 7.34126984127e-05 But I forgot that this is a conductance value, you have to invert it to get the resistance: 1 / (1/R1 + 1/R2) = 1 / 7.34126984127e-05 ~ 13621.62 FirstChildUserIdTAG: 104102 FirstChildUserNameTAG: ZeusFET FirstChildCreateTimeTAG: 2012-09-21T23:48:23Z FirstChildTAG: at first you just turn off the voltage source so now your vth is the equivalent resistor across v, which is parallel combination of r1 and r2. FirstChildUserIdTAG: 231749 FirstChildUserNameTAG: utshau FirstChildCreateTimeTAG: 2012-09-24T05:58:40Z IndexTAG: 4152 TitleTAG: y2 please i obtain y2=1.35 as i take (R1&R3) parallel and current will be v2/r2+(r1 parallel r3) where the error UserIdTAG: 295983 UserNameTAG: qassam CreateTimeTAG: 2012-09-18T09:15:40Z VoteTAG: 0 CoursewareTAG: Week 2 / Superposition example CommentableIdTAG: 6002x_S3E3_superposition NumberOfReplyTAG: 1 FirstChildTAG: y2=i2*R3/(R1+R3) and i2 = V2*(R1+R3)/(R1R2+R2R3+R3R1) FirstChildUserIdTAG: 416042 FirstChildUserNameTAG: Venturl FirstChildCreateTimeTAG: 2012-09-18T10:45:02Z IndexTAG: 4153 TitleTAG: Bug Help! I can't see what other students post (or even my own posts). UserIdTAG: 369339 UserNameTAG: vargaslen CreateTimeTAG: 2012-09-18T08:52:41Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 4154 TitleTAG: strange made on mistake in positive or negative change the answer !!!!1 UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-18T06:07:26Z VoteTAG: 0 CoursewareTAG: Week 2 / Node equation review CommentableIdTAG: 6002x_S3E1_NodeEquationReview NumberOfReplyTAG: 0 IndexTAG: 4155 TitleTAG: unknown potential e ((e-V1)/R1)-((e+V2)/R2)=0 UserIdTAG: 50320 UserNameTAG: marm496 CreateTimeTAG: 2012-09-18T05:47:24Z VoteTAG: 0 CoursewareTAG: Week 2 / Node equation review CommentableIdTAG: 6002x_S3E1_NodeEquationReview NumberOfReplyTAG: 1 FirstChildTAG: I suppose it should be ((e-V1)/R1)+((e+V2)/R2)=0 FirstChildUserIdTAG: 111320 FirstChildUserNameTAG: jhalife FirstChildCreateTimeTAG: 2012-09-18T11:42:48Z IndexTAG: 4156 TitleTAG: Strange e=6.20 doesn't make sense, it would if V2 has plus at top UserIdTAG: 298314 UserNameTAG: js93082011 CreateTimeTAG: 2012-09-18T01:37:16Z VoteTAG: 0 CoursewareTAG: Week 2 / Node equation review CommentableIdTAG: 6002x_S3E1_NodeEquationReview NumberOfReplyTAG: 1 FirstChildTAG: Take a look at this: [**Discussion - Double Trouble**][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_S3E1_NodeEquationReview/threads/5057c8ac1aef61250000000f FirstChildUserIdTAG: 138769 FirstChildUserNameTAG: ArturoPrado FirstChildCreateTimeTAG: 2012-09-18T03:46:42Z IndexTAG: 4157 TitleTAG: I don't fully understand how Ei was obtained ! What Dr.Agarwal did was to calculate the resultant resistance of the parallel resistors R1 and R2 and then get the voltage by multiplying this by the current ! I don't understand the directions of the currents and how would he do that using the KCl at the node e1 ? Any help will be appreciated UserIdTAG: 289472 UserNameTAG: OsamaAdel CreateTimeTAG: 2012-09-17T21:37:18Z VoteTAG: 0 CoursewareTAG: Week 2 / Superposition CommentableIdTAG: 6002x_superposition NumberOfReplyTAG: 1 FirstChildTAG: Using KCL: We consider currents coming out of a node negative and the sum of them is 0. - -i1 -i2 -ix = 0 - ix = -I ![][1] Using a little bit of math: - -i1 -i2 -(-I) = 0 - -i1 -i2 +I = 0 Then we end up with this equation: - I = +i1 +i2 Now we can replace R1 and R2 with an equivalent resistor called R3 (parallel R1||R2). ![][2] Note that node e1 is in the same position and its value is the voltage across resistor R3. Then we have this equation for e1: e1 = R3 (i1 + i2) but i1 + i2 = I and R3 = (R1*R2)/(R1+R2) So we just replace R3 and i1+i2 into the formula for e1. And it'll look something like this: **e1 = I * ( (R1*R2)/(R1+R2) )** Don't forget that this is e1 when current source ***I is acting alone***. You need to find out e1 when voltage source ***V is acting alone*** as well and finally add those two e1 values and you'll get your final answer. I hope this explanation helps. I'll try to buy a pen mouse to make more dynamic explanations. Circuits & Electronics works!! [1]: https://edxuploads.s3.amazonaws.com/13479255242550762.png [2]: https://edxuploads.s3.amazonaws.com/13479274781252395.png FirstChildUserIdTAG: 138769 FirstChildUserNameTAG: ArturoPrado FirstChildCreateTimeTAG: 2012-09-18T00:51:55Z SecondChildTAG: Thanks a lot! I was not perceiving that R1 and R2 was in a parallel circuit, but I still don't know how find the answer when V is acting alone. SecondChildUserIdTAG: 291362 SecondChildUserNameTAG: Gudson SecondChildCreateTimeTAG: 2012-09-18T13:55:47Z IndexTAG: 4158 TitleTAG: certification i wonder if we will actually have acertificate that is equivalent to that the original students in mit have taken UserIdTAG: 326507 UserNameTAG: mohamed200 CreateTimeTAG: 2012-09-17T21:35:35Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: No, we will probably get a statement of accomplishment, but it won't be endorsed by MIT. FirstChildUserIdTAG: 395349 FirstChildUserNameTAG: bgr FirstChildCreateTimeTAG: 2012-09-17T21:45:24Z SecondChildTAG: actually we need an answer about this from the staff SecondChildUserIdTAG: 395257 SecondChildUserNameTAG: blackguitar SecondChildCreateTimeTAG: 2012-09-18T00:12:18Z SecondChildTAG: blackguitar, you can see a certificate here: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/undefined/threads/505523f9fe12522b00000018 SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-09-18T05:09:09Z SecondChildTAG: JSChambers,thanks a lot SecondChildUserIdTAG: 395257 SecondChildUserNameTAG: blackguitar SecondChildCreateTimeTAG: 2012-09-19T00:56:56Z FirstChildTAG: For me these courses from edX are much more important than attending any university in my country. People in my country study engineering just because they need to make money and not because they want to reach the edge of the knowledge. Teachers don't care whether their students learned or not, they also want to make money to pass by. I ♥ edX. FirstChildUserIdTAG: 138769 FirstChildUserNameTAG: ArturoPrado FirstChildCreateTimeTAG: 2012-09-18T01:29:13Z IndexTAG: 4159 TitleTAG: Applying KVL Hi, How do you apply KVL to solve b1, b2 and c1, c2? A detailed answer would be much appreciated. Many thanks in advance. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-17T20:09:42Z VoteTAG: 0 CoursewareTAG: Week 2 / Linear combinations of source strenghts CommentableIdTAG: 6002x_S3E2_Linear_Combinations_of_Source_Strengths NumberOfReplyTAG: 1 FirstChildTAG: I think you should use and remember your algebra course. That problem is all about KVL and KCL but try to use additional independent equations. It take me some time but not more than 10 minutes. It's not really a tough problem you'll be factoring out a lot and doing other things. I hope this helps you. FirstChildUserIdTAG: 138769 FirstChildUserNameTAG: ArturoPrado FirstChildCreateTimeTAG: 2012-09-18T01:45:13Z IndexTAG: 4160 TitleTAG: Problem regarding proving KVL As it is mentioned that you need to travel clockwise then the change of voltage across each resister comes out to be positive not negative. but the when you click on "show answer", it shows you negative sign of voltage change across each resistor. According to me, this should be true only when we traverse anti-clockwise. Kindly help me understand this! Thank you in advance Regards Hadi UserIdTAG: 223378 UserNameTAG: hadi89 CreateTimeTAG: 2012-09-17T19:58:04Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: I think... Taking into consideration that current is going from a positive terminal to a less positive terminal we can interpret the change in voltage as the difference between the final voltage(B) and the initial voltage(A). Whereas in the case of the voltage source the current is going from a positive terminal(I don't know how to consider 0 volts) to a more positive terminal. That means final voltage(A) and initial voltage(D). I hope this helps you. Circuits & Electronics works!! FirstChildUserIdTAG: 138769 FirstChildUserNameTAG: ArturoPrado FirstChildCreateTimeTAG: 2012-09-18T01:18:10Z IndexTAG: 4161 TitleTAG: Lab 1 how save ? How save the answer ON lab 1? because I did the DC analysis, but in "my progress" no appear as done. UserIdTAG: 298395 UserNameTAG: smolder CreateTimeTAG: 2012-09-17T19:36:24Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: You need to click the Check Button at the bottom (scroll to the bottom of the page). Lab1's deadline is over though so you cannot submit that for grading (and it looks like they remove the check button too after the deadline). FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-09-17T19:39:47Z SecondChildTAG: THE PROBLEM IS THAT BUTTON NOT APPEAR HERE. SecondChildUserIdTAG: 298395 SecondChildUserNameTAG: smolder SecondChildCreateTimeTAG: 2012-09-17T19:44:49Z SecondChildTAG: That's because week 1's deadline is over. You can still score full grade however because two assignments can be dropped. Concentrate on doing well in the remaining assignments. SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-09-18T02:25:54Z FirstChildTAG: It usually get saved automatically when we click on check. Try again. FirstChildUserIdTAG: 329015 FirstChildUserNameTAG: farah_sarwar FirstChildCreateTimeTAG: 2012-09-17T19:41:13Z SecondChildTAG: I TRIED, BUT THE BUTTON NOT APPEAR SecondChildUserIdTAG: 298395 SecondChildUserNameTAG: smolder SecondChildCreateTimeTAG: 2012-09-17T19:49:20Z SecondChildTAG: cox u r late :( SecondChildUserIdTAG: 415376 SecondChildUserNameTAG: imab90 SecondChildCreateTimeTAG: 2012-09-17T20:19:42Z IndexTAG: 4162 TitleTAG: H2P1 I can't understand what difference the 10% makes. Why the answer can't be 16000 e 24000 ohms? UserIdTAG: 51667 UserNameTAG: Assuncao CreateTimeTAG: 2012-09-17T18:51:28Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: This is what I thought at first, but it is not correct. Rth is based on the resistance of R1 in parallel with R2. This is not the same as R1 and/or R2 themselves having to be between 10kOhms and 30kOhms. FirstChildUserIdTAG: 377339 FirstChildUserNameTAG: nickd717 FirstChildCreateTimeTAG: 2012-09-18T01:20:04Z IndexTAG: 4163 TitleTAG: Lab 2 I can't make the voltage be less than 1 volt of maximum. How can I do that??? The output looks like the picture UserIdTAG: 51667 UserNameTAG: Assuncao CreateTimeTAG: 2012-09-17T18:50:16Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 4164 TitleTAG: This is crazy For the past four days I am making all the same mistakes every time! Every time this god damn current or voltage have reversed sign! I see I am not alone here. I am not consider my self as a dummy so I think maybe the reason is somewhere else. English is not my native language and maybe this is the issue. Maybe the explanation of this signs should be more clearer? At this point I still have to spend 2 minutes thinking about this +/- and at the end I pick the wrong one! Please... (for the record, I did write the formula for "e" to the calculator below and reversing random signs I finally got the right answer, but I am not proud of this method). UserIdTAG: 427245 UserNameTAG: Yakuzza CreateTimeTAG: 2012-09-17T18:42:06Z VoteTAG: 0 CoursewareTAG: Week 2 / Node equation review CommentableIdTAG: 6002x_S3E1_NodeEquationReview NumberOfReplyTAG: 2 FirstChildTAG: Sign mistakes will happen plenty of times. I agree it does get very frustrating. I would suggest practicing more problems from the textbook. Do them slowly and carefully even if you are totally confident about the solution. That way you'll get better at being more careful and can gradually speed up. FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-09-17T18:45:48Z SecondChildTAG: if sum of outgoing currents from e is 0, then -(e-e1)/6800=(e-e2)/5600=>5600e-28000=-6800e+48960=>12400e=76960. Why are the signs thus? We see from the figure that V1 goes -+ and the plus will meet + at the resistor R1. So the other side of R1 is -. Opposite is the other side of the circuit. So e works out to have + on the left and - on the right. The currents flowing out of e are - on the left and + on the right since the left arrow encounters "-" first and on the right "+" first. SecondChildUserIdTAG: 373585 SecondChildUserNameTAG: radami1 SecondChildCreateTimeTAG: 2012-09-18T01:41:13Z FirstChildTAG: Yakuzza, having trouble with the signs is very common. I remember struggling with that myself several years ago when I initially learnt about circuits. My recommendation is to start the problem by defining polarities of the voltage in each element and the direction of all currents, and then stick to what you defined. If you get any negative voltage or current, don't change your polarities because that will make it more confusing. Just keep working with the negative numbers, but keep in mind what they mean. I am sure that with practice you will start solving circuits more naturally and without the sign mistakes. FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-18T00:32:10Z IndexTAG: 4165 TitleTAG: e its a node is important to consider current entering in the point e according to R1 or R2; natural; they are equal. UserIdTAG: 241050 UserNameTAG: eddietoni CreateTimeTAG: 2012-09-17T18:36:38Z VoteTAG: 0 CoursewareTAG: Week 2 / Node equation review CommentableIdTAG: 6002x_S3E1_NodeEquationReview NumberOfReplyTAG: 0 IndexTAG: 4166 TitleTAG: Please Provide the textbook in pdf to download Please Provide the textbook in pdf to download UserIdTAG: 163045 UserNameTAG: premprajan CreateTimeTAG: 2012-09-17T18:29:29Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Dear premprajan, The Professor and his team have tried so much in making this course a reality at no cost and also provided portions of the textbook under the subhead **"textbook"**. It is best we appreciate their efforts. So asking for it to be in pdf, to me will bw asking of too much like Oliver Twist. I will suggest you visit Amason's website as I just did to know how much it cost.Here is the cost to save you time: ***"Foundations of Analog and Digital Electronic Circuits (The Morgan Kaufmann Series in Computer Architecture and Design) by Anant Agarwal and Jeffrey Lang (Jul 15, 2005)"*** Rent is $25.75 Buy is $79.99 New is $45.94 Used is $45.94 Order in the next 3 hours to get it by Tuesday, Sep 18. Eligible for FREE Super Saver Shipping and 1 more promotion Cheers. FirstChildUserIdTAG: 329879 FirstChildUserNameTAG: Ugochukwu FirstChildCreateTimeTAG: 2012-09-17T20:24:29Z IndexTAG: 4167 TitleTAG: H2P2: question to the staff Hello, with my friend we were working separately on the homework and have found several ways to determine the optimal load resistance RL. Here they are: 1. By selection 2. By using derivative 3. By using maximum power transfer theorem (we have found this way in a post here). The hint for this homework was: remember your calculus! We don't understand how the ways to solve the task connect to the calculus or is there another method? Thank you in advance, Sergestus UserIdTAG: 375724 UserNameTAG: sergestus CreateTimeTAG: 2012-09-17T18:08:04Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Derivatives is a part of calculus. And the maximum power transfer theorem is derived by using derivatives. FirstChildUserIdTAG: 392100 FirstChildUserNameTAG: glenng FirstChildCreateTimeTAG: 2012-09-18T00:32:44Z FirstChildTAG: Right, derivatives are a part of calculus. Remember in your calculus class taking the derivative to find the maximum and minimum of a function? To also help jog your memory, the second derivative gave the concavity of the function, which changed at the inflection points. This was all a method to describe functions, both analytically and graphically. I had to pull out my calculus book to refresh my memory on how to take the derivative of a quotient, hint, hint. FirstChildUserIdTAG: 132828 FirstChildUserNameTAG: rharris FirstChildCreateTimeTAG: 2012-09-18T14:32:48Z IndexTAG: 4168 TitleTAG: Marks Why "b" starts from 70? UserIdTAG: 444204 UserNameTAG: hekan CreateTimeTAG: 2012-09-17T17:56:36Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 4169 TitleTAG: Independent KCL/KVL What is an independent KCL/KVL equation? I assumed that if a circuit has n nodes there must be (n-1) KCL independent equations. Well this is true in this case. I also assumed that if a circuit has n loops there must be (n-1) independent KVL equations which failed in this case. So what is that INDEPENDENT KCL/KVL equation means? Help me with this. UserIdTAG: 87309 UserNameTAG: Vijayenthiran CreateTimeTAG: 2012-09-17T17:37:38Z VoteTAG: 0 CoursewareTAG: Week 1 / Circuit topology CommentableIdTAG: 6002x_S2E1_Circuit_Topology NumberOfReplyTAG: 2 FirstChildTAG: As an example, 3 equations are independent when you cannot get an equation by adding two other equations. Here is a specific example. Consider the following three equations. 3x + 2y + 5z = 10 -------- (1) x + y + z = 1 ------------ (2) 11x + 8y + 17z = 22 ------ (3) If you study these three equations carefully, you can see that we are able to obtain the third equation by multiplying Eq(1) by 3, Eq(2) by 2 and adding the two. Here are the steps in more detail 3x + 2y + 5z = 10 -------- (1) x 3 9x + 6y + 15z = 20 x + y + z = 1 ------------ (2) x 2 2x + 2y + 2z = 2 Now add the two equations and you'll end up with Eq(3) 9x + 6y + 15z = 20 2x + 2y + 2z = 2 \------------------- 11x + 8y + 17z = 22 What you just saw is called **Linear dependence**. This means that the equations depend on each other - we can obtain one of the equations if we know the other two. If you look through linear algebra, you will learn that if you want your equations to have unique solutions, you need to have **Linearly Independent** equations. Clearly, the equations I gave above are **dependent** so we cannot get unique values of x, y and z. So how does this relate to circuits? KVL and KCL translate into Linear equations like the equations above. The only difference is that the variables will be voltages and currents instead of x, y and z. So to get unique values of currents and voltages, you'll need the minimum number of independent equations. Once you have enough of these equations, it doesn't make sense to look for more (and end up spending more time) because any new equation will be dependent on the equations you've already written. Does this help? (I hope I haven't confused you more :-(. Please do let me know if I haven't been clear) FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-09-17T18:21:20Z SecondChildTAG: Number of independent KVL equations is given by B-(N-1) where B is number of branches and N is number of nodes. SecondChildUserIdTAG: 154440 SecondChildUserNameTAG: Aahlad SecondChildCreateTimeTAG: 2012-09-19T11:45:27Z SecondChildTAG: And number of independent KCL equations *is* N-1 SecondChildUserIdTAG: 154440 SecondChildUserNameTAG: Aahlad SecondChildCreateTimeTAG: 2012-09-19T11:46:19Z SecondChildTAG: Sorry, I meant to post the above as a response to the original post. SecondChildUserIdTAG: 154440 SecondChildUserNameTAG: Aahlad SecondChildCreateTimeTAG: 2012-09-19T11:47:36Z SecondChildTAG: how to prove that we will have **B-(n-1)** kvl equations ? SecondChildUserIdTAG: 345541 SecondChildUserNameTAG: brishiedx SecondChildCreateTimeTAG: 2012-09-24T14:28:02Z SecondChildTAG: Excellent answer, Aswith. I would like to clarify that in the context of KVL and KCL, what you have to do is to find the minimum number of equation, with which you can obtain all the remaining equations in the circuit. You will notice there are three such equations for both tthe KVL and KCL equation. You can take any three equations as long as they are not dependent on each other,i.e. one of these equations cant be obtained by adding or subtracting the other two. Once you get these 3 equations, it will be possible to obtain all the other equations by adding/subtracting each other. SecondChildUserIdTAG: 332470 SecondChildUserNameTAG: Nithinsp SecondChildCreateTimeTAG: 2012-10-20T06:39:07Z FirstChildTAG: Dear colleague, I understand your explanation, but you are mistaken in your example. I'm referring to this passage 3x + 2y + 5z = 10 -------- (1) x 3 9x + 6y + 10z = 20 The correct thing 9x +6 y +15 z = 30 FirstChildUserIdTAG: 457034 FirstChildUserNameTAG: Ichihara FirstChildCreateTimeTAG: 2012-09-19T12:12:49Z SecondChildTAG: Whoops! Good catch. I'll go ahead and fix it. Thanks! :-) SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-09-20T18:39:44Z IndexTAG: 4170 TitleTAG: H1P1- help You are given three resistors: two 4Ω resistors and one 6Ω resistor. Given that each individual resistor can dissipate up to 1 watt of power before burning up, how much total power in watts ( W ) can the smallest-valued composite resistor dissipate before burning up? UserIdTAG: 287613 UserNameTAG: gbaral CreateTimeTAG: 2012-09-17T17:14:52Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: It's a bit tricky, but you can solve it by using a simple relationship: Since you know that the smalles resistor in this circuit is 4 ohms, it uses 1 amp, then you want to know how much does a total resistance consume (1.5 ohms)Or the value of the composite resistance. 4 ohms --------------> 1 watt 1.5 ohms ------------> ? Watts FirstChildUserIdTAG: 355438 FirstChildUserNameTAG: Alexgta FirstChildCreateTimeTAG: 2012-09-17T17:31:00Z IndexTAG: 4171 TitleTAG: Here is the trick Think of the voltage values given for each voltage source as being relative to the positive side. If V1 is 5v then it is +5v relative to ground node. If the + side of V2 is -7.2v and your ground reference is the ground node then the - side of V2 will be +7.2v. After thinking about it like this I got the right answer. UserIdTAG: 132086 UserNameTAG: parbro CreateTimeTAG: 2012-09-17T15:57:26Z VoteTAG: 0 CoursewareTAG: Week 2 / Node equation review CommentableIdTAG: 6002x_S3E1_NodeEquationReview NumberOfReplyTAG: 1 FirstChildTAG: I used superposition method e=eV1+eV2 eV1=R2*V1/(R1+R2) eV2=-R2*-V2/(R1+R2) Then summed the two. FirstChildUserIdTAG: 424002 FirstChildUserNameTAG: KinuthiaMugi FirstChildCreateTimeTAG: 2012-09-22T13:04:30Z IndexTAG: 4172 TitleTAG: HW and LAB I NEED HELP ON WHERE TO FIND THE HOMEWORK AND LABS UserIdTAG: 97511 UserNameTAG: ORASIO CreateTimeTAG: 2012-09-17T15:13:49Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Click on "Courseware" (Upper left), then click on whatever week you want. You will see two lecture sequences, a Homework and a Lab. It's best not to use CAPS LOCK if you want to get any serious responses in the future. Have fun FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-17T15:27:32Z SecondChildTAG: Thanks SecondChildUserIdTAG: 97511 SecondChildUserNameTAG: ORASIO SecondChildCreateTimeTAG: 2012-09-17T16:18:59Z FirstChildTAG: The reason why Pennypacker said you should avoid the use of all-round capital letters when typing is because, all Caps means you are shouting. Take care. Ugo. FirstChildUserIdTAG: 329879 FirstChildUserNameTAG: Ugochukwu FirstChildCreateTimeTAG: 2012-09-17T18:36:21Z IndexTAG: 4173 TitleTAG: Correction There are things wrong in description of the systematic way. With two 1k resistors, when we put them in parallel and work that out we will see that this is 1k/2(is not 1/2k). With three 1k resistors, when we put 1k resistor parallei to 2k resistor we will get 2k/3(will not get 2/3k). With three 1k resistors, when we take 1k resistor and put an additional one parallel.1,2,3. This gives us k/3(doesn't give us 1/3k). With three 1k resistors,1k plus 1k/2 is 3k/2(is not 3/2k). With four 1k resistors,we can put four in series and we will get 4k. With four 1k resistors,we can put four in parallel and we will get 1k/4(won't get 1/4k). With four 1k resistors,three 1k(in series) parallel to 1k gives us 3k/4(750 ohms) That doesn't give us 3/4k(0.00075 ohm).And if we take three in parallel,add one in series we get 4k/3(doesn't get 4/3k) UserIdTAG: 364129 UserNameTAG: Sombat CreateTimeTAG: 2012-09-17T14:07:12Z VoteTAG: 0 CoursewareTAG: Week 1 / Equivalent Resistances CommentableIdTAG: 6002x_equivalent_resistances NumberOfReplyTAG: 0 IndexTAG: 4174 TitleTAG: results where & how i can see my total result of my lab individually? UserIdTAG: 276803 UserNameTAG: srinivasaram CreateTimeTAG: 2012-09-17T13:56:45Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: you can see it in the option 'progress'.. FirstChildUserIdTAG: 171170 FirstChildUserNameTAG: rinutituschakkattil FirstChildCreateTimeTAG: 2012-09-17T15:29:14Z IndexTAG: 4175 TitleTAG: what about the internal resistances? My answers came out wrong because i added the internal resistances also to the resistance in the branch. Is that not required? UserIdTAG: 180498 UserNameTAG: iturhs CreateTimeTAG: 2012-09-17T13:48:16Z VoteTAG: 0 CoursewareTAG: Week 1 / Battery model CommentableIdTAG: 6002x_S1E9_Battery_Model NumberOfReplyTAG: 1 FirstChildTAG: hi ,, the internal resistance its not added in ur calculations until tell u in the problem and give its value .... here .. in week 1/ battery model .. u will calculate the current out from both batteries and take in ur attention the internal resistance by using ohm's law and as the both of them connected in parallel get the equivalent by the following (R1*R2)/(R+R2)... finally want to know the current in the loop as following : V2-V1=0.1V and the resistance in this loop is 0.75 ohm then I=V/R.......... i hope u get it >> FirstChildUserIdTAG: 193445 FirstChildUserNameTAG: eslammagdy FirstChildCreateTimeTAG: 2012-09-17T14:12:10Z IndexTAG: 4176 TitleTAG: no of attempts how many maximum attempts can i make in answering my lab or hw..?? i mean how many times i can check my answers UserIdTAG: 353479 UserNameTAG: dimple92 CreateTimeTAG: 2012-09-17T13:14:33Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Unlimited. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-17T13:33:19Z IndexTAG: 4177 TitleTAG: Stuck! I have found v3 in terms of V1, V2, R1 ,R2 and R3. v3 = ((V1*R2) + (V2*R1)) / ((R2*R3) + (R1*R3) + R1*R2)) Where do I go from here??? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-17T13:08:32Z VoteTAG: 0 CoursewareTAG: Week 2 / Linear combinations of source strenghts CommentableIdTAG: 6002x_S3E2_Linear_Combinations_of_Source_Strengths NumberOfReplyTAG: 1 FirstChildTAG: That correct. Just rearrange it you will get it FirstChildUserIdTAG: 441436 FirstChildUserNameTAG: Ooffy FirstChildCreateTimeTAG: 2012-09-17T15:55:15Z SecondChildTAG: How do I solve for b1, b2, c1 and c2? SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-09-17T19:00:53Z SecondChildTAG: Use KVL method for the loop on the left. If you follow in a clockwise manner you will get ---------- i1*R1+V1-V2-i2*R2=0 ---------- And you already know the voltage for the resistence $R1$ ---> ($V3-V1$) and for the resistence $R2$ ---> ($V3-V2$), then you replace into the equation and obtain $i1$ and $i2$ in terms of $V1$ and $V2$ SecondChildUserIdTAG: 136253 SecondChildUserNameTAG: chemeng SecondChildCreateTimeTAG: 2012-09-17T19:42:04Z SecondChildTAG: Thank you very much! That is the exact help I needed! Perfect! SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-09-17T19:54:19Z SecondChildTAG: I have my answers but it's not parsing the formula in the box.cudn't even process my answers. can you help me please! SecondChildUserIdTAG: 433119 SecondChildUserNameTAG: SHOMUYIWA SecondChildCreateTimeTAG: 2012-09-27T10:46:39Z IndexTAG: 4178 TitleTAG: First circuit is not a choice The first statement when A=0, B=0 and C=0 will be 1 only in right circuit and it is not true in left circuit because A+B is simply 0 and C NOT is 1 and AND operator between them will result with 0 (A+B).C NOT = 0 Therefore I think the answer to the first question must be 1 not 2 because only one of the circits can satisfy the first statement. UserIdTAG: 210954 UserNameTAG: Shahrouz CreateTimeTAG: 2012-09-17T13:05:04Z VoteTAG: 0 CoursewareTAG: Week 2 / Truth table CommentableIdTAG: 6002x_S4E3_Truth_Table NumberOfReplyTAG: 2 FirstChildTAG: To solve, work back from the output Z. Consider Circuit 2, for the first entry in the truth table, Z =1, since Z is the output of an OR gate if **either**, or **both** of the OR gate inputs are set to **1** then Z will be **1**, else Z will be ***0*** if **both** inputs to the OR gate are ***0***. One input of the OR gate is fed from an inverter, the inverter output can only be 1 if C, its input, is set to 0, thus when ever C is zero, the inverter output will be 1 producing an output (Z) from the OR gate, =1 ( which is confirmed from the first line entry in the truth table). The other input into the OR gate is from an AND gate which processes the A and B inputs. this can only produce a 1 output when **both** A and B are set to 1. The 1 output from the AND gate feeds into the OR gate and produces a 1 at the OR gate output (Z=1). Thus Z will only be 1 when either A and B =1 or C =0,( or A and B = 1 and C =0) verify in the truth table. With circuit 1: the Z output is from an AND gate, which can only give a 1 at its output when **both** of its inputs are set to 1. One of these inputs is connected to the output of an OR gate, which processes input A and input B. For the first line in the truth table, input A =0 and input B =0 therefore the OR gate output will be 0 which would force the output of the AND gate (Z) to 0 which violates the first line of the truth table ( it should be 1) Circuit 2 meets the requirements of the given truth table FirstChildUserIdTAG: 273287 FirstChildUserNameTAG: johnkh77 FirstChildCreateTimeTAG: 2012-09-17T15:56:00Z FirstChildTAG: You explained the circuit right, but, how come you end up with "Circuit 1" is the answer?! Your explanation meets "Circuit 2". FirstChildUserIdTAG: 365201 FirstChildUserNameTAG: sirajmuhammad FirstChildCreateTimeTAG: 2012-09-28T17:53:18Z SecondChildTAG: I can only guess that he read the question as "Enter the number of circuits that satisfy the table" instead of "Enter the number corresponding to the circuit which satisfies the table". Hopefully he has figured it out by now, but just in case: the second one is right ;) SecondChildUserIdTAG: 604636 SecondChildUserNameTAG: Klionheart SecondChildCreateTimeTAG: 2012-10-11T20:40:12Z IndexTAG: 4179 TitleTAG: STAFF Is the question given under section "H3P2 GRAPHICAL MODEL OF INVERTER" complete? I feel that it is incomplete. UserIdTAG: 132685 UserNameTAG: deepakmurali CreateTimeTAG: 2012-09-17T12:53:30Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: It is complete. FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-17T13:02:40Z SecondChildTAG: Will answers of how to solve home work questions be posted after the due date by MITx: 6.002x Circuits and Electronics team? SecondChildUserIdTAG: 132685 SecondChildUserNameTAG: deepakmurali SecondChildCreateTimeTAG: 2012-09-17T15:07:42Z IndexTAG: 4180 TitleTAG: general doubt.. guys have a doubt.... if one is not able to make up the course in his first attempt... what will be done with him?? will he be allowed to take it once more for free?? or what options will he have then?? and by the way,.. when is this site going to charge money?? right after this semister?? any idea?? UserIdTAG: 447302 UserNameTAG: phanisrikar CreateTimeTAG: 2012-09-17T12:33:47Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I have read the U.S. news reports regarding EdX, and all I have heard so far is "free" for the time being. So if the course is offered next semester (which I think it will), you will likely be able to make it up. **But do not drop this course yet!** Note that it coves in one semester in what most major U.S. universities accomplish in two: In most other universities, this course has the "analog circuit analysis" component in one semester (what we have been doing so far) and the "digital logic and design" component (which will start in the second half of week 2) as a complete other course in the second semester. So this is really a one-year course "crammed" or "distilled" into half-a-year. You have to be fast and accurate with complex mathematics to be able to even do the homeworks! *This means that maybe you "do not get" the analog stuff yet, but when the course gets to the digital stuff, you may excel at that portion!* If you are just **starting** grade 9, then you should not be expected to take this course once and pass, except if you are a "brain". At this level, students have other concerns to worry about, like "fitting in" with their peers, and dating, and having fun. If you are in **not yet** in U.S. high school "honors" or "advanced placement math and physics" (or it's U.K equivalent in A-levels) then this course will be very, very challenging, and you should definitely take it the first time just to learn, and the second time for the grades. That's the beauty of EdX over "real" university, where you can only take a class only once or twice. However, **if you are finished** with U.S. advanced placement (AP) or U.K. (and I think India) A-level math and physics, this course should be **expected of you** if you are to proceed in the engineering field. If you are having "major" problems, not just stupid simple algebric and sign mistakes, then you should reconsider engineering as a career, because this is one of the "**weed out**" courses in U.S. engineering colleges that seperates the students who "can" from those who "cannot". FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-09-17T13:08:22Z SecondChildTAG: 'That's the beauty of EdX over "real" university, where you can only take a class only once or twice.' what is meant by this line?? SecondChildUserIdTAG: 447302 SecondChildUserNameTAG: phanisrikar SecondChildCreateTimeTAG: 2012-09-17T16:21:14Z SecondChildTAG: It's meant, I believe in the context, that in university, generally you do not get the option to take a course or module again should you not do well in it, whereas with this course, you could potentially take it twice (this time round and when it becomes next available through the Edx page). It's been an interesting course so far, very enjoyable, considering I haven't studied maths or physics since my GCSEs. SecondChildUserIdTAG: 243871 SecondChildUserNameTAG: Vanilly SecondChildCreateTimeTAG: 2012-09-18T00:05:54Z IndexTAG: 4181 TitleTAG: lab 1 isn't added to my progress i did my lab1 yesterday successfully with both of them correct.but i didn't got that in my progress. pls any one help me.. UserIdTAG: 353405 UserNameTAG: karthickks CreateTimeTAG: 2012-09-17T10:49:10Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Troubleshooting NumberOfReplyTAG: 3 FirstChildTAG: Any more information? If you go back to the Lab 1 page, are the checkmarks still there? Did you press the reset button in the meantime? FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-17T10:58:45Z SecondChildTAG: i reset that.and again put the same circuit->save->check.it displays"Your answers have been saved but not graded. Hit 'Check' to grade them." and i also press check.but not upload in my progress. pls help me!! SecondChildUserIdTAG: 353405 SecondChildUserNameTAG: karthickks SecondChildCreateTimeTAG: 2012-09-17T11:29:19Z SecondChildTAG: Do you get a checkmark when you press 'Check'? SecondChildUserIdTAG: 79361 SecondChildUserNameTAG: kimt SecondChildCreateTimeTAG: 2012-09-17T12:52:04Z FirstChildTAG: I have the same problem. I had 100% progress until I pressed Reset and as the result my whole work was reseted. How can I change, pls, help me out!!! FirstChildUserIdTAG: 396141 FirstChildUserNameTAG: nuidem FirstChildCreateTimeTAG: 2012-09-17T14:35:59Z SecondChildTAG: Unfortunately we cannot allow submissions to be made after the deadline. (You should not be able to press reset after the deadline.) Please keep in mind that out of a total of 12 labs, only 10 are required for your final grade. SecondChildUserIdTAG: 79361 SecondChildUserNameTAG: kimt SecondChildCreateTimeTAG: 2012-09-17T17:26:25Z FirstChildTAG: i got the progress!!Thank u. FirstChildUserIdTAG: 353405 FirstChildUserNameTAG: karthickks FirstChildCreateTimeTAG: 2012-09-18T18:50:15Z IndexTAG: 4182 TitleTAG: proctored final exam Which edX course has the option of proctored final exam this Fall? UserIdTAG: 30000 UserNameTAG: gopakumar CreateTimeTAG: 2012-09-17T07:08:18Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I have read about the option of a proctored final exam in the university newspaper of Harvard, the "Crimson". Someone posted an excerpt somewhere on this site but I forgot where. So I would wager that one of the Harvardx courses being offered would have this option, over one of the MITx courses. Note that **edX** is just in it's beginning, trial stages, so not everything is "set in stone" yet. While the courses are "free" right now, and I think they will be free for at least a few more years, there was a controversial article in the *New York Times Magazine on Sunday Sept. 16th, where a University of Virginia president was forced out by the school's board$^1$ for **not** offering "free" online courses. These online courses are also very "big" for the conservative$^2$ newspaper *The Wall Street Journal*. * Footnote **1**: The University of Virginia's board is composed of people selected by the governor of that state, mainly as a reward for political contributions, and it consists of venture capitalists and businessmen/women, not professional educators, who's job it is is to choose the president. They do not lead the university day-to-day, similar to a corporate board. That's the university president's job, and they disagreed with her leadership. * Footnote **2**: In the U.S., conservatives are those who support free-market, *lazes-faire* capitalism, with minimal government intrusion, as opposed to those that lean socialist. Hence, most millionaires, billionaires, the "top 1%", business owners, those with a financial interest, etc. lean towards the "purported" party of fiscal conservatives, the Republicans. Note that this party also espouses social conservatism (i.e. religion in government and the schools). So as a U.S. engineering student taking this course for the second time for mastery, the first time in a major public university many years ago for my degree (I am **not** staff **or** affiliated with edX **or** any company or news organization) and reading these articles and rumors, **on one hand**, it seems like the *venture capitalists* and Wall Street will want to charge money for these online courses, or have them raise revenue through advertising or proctored exams, etc. because U.S. capitalists care about *profit* above all else. **On the other hand**, MIT and Harvard, while private schools, are funded with public U.S. government research grants. They also have very large endowments, in the billions of dollars already. So maybe these courses will be free forever, because MIT and Harvard don't really need the money and are perhaps being truly altruistic, and serving the world public by this service. That's the controversy for now. If an "insider" has a tip, maybe we will be lucky enough to hear about the "true" direction of edX. FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-09-17T13:40:13Z IndexTAG: 4183 TitleTAG: H1P1 algebra. I am going crazy( So, for figure B i had: (3*R)/(R+R+R), (R*R*R)/(R+R+R), 1/(1/R+1/R+1/R), 1/R+1/R+1/R I don't know what to do, please, i want to go sleep( UserIdTAG: 205077 UserNameTAG: Altleonidas CreateTimeTAG: 2012-09-17T07:02:42Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 1 FirstChildTAG: Page 83, formulas 2.99 and 2.100 are what you're looking for. FirstChildUserIdTAG: 264596 FirstChildUserNameTAG: Nuru FirstChildCreateTimeTAG: 2012-09-17T07:06:41Z SecondChildTAG: Thank you so much! SecondChildUserIdTAG: 205077 SecondChildUserNameTAG: Altleonidas SecondChildCreateTimeTAG: 2012-09-17T07:10:33Z IndexTAG: 4184 TitleTAG: Problems with my progress Hello. I have a problem with my progress. I made the Homework1 and lab1, and i check my progress immediately and i saw that my grades for hm1 and lab1 were 100%, but after i few hours later i check again my progress and appears that i dont make the hw1 and lab1, What can i do with this problem? UserIdTAG: 262044 UserNameTAG: AdrianSNV CreateTimeTAG: 2012-09-17T07:01:28Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: i had that problem in lab1. pls help to both of us!! FirstChildUserIdTAG: 353405 FirstChildUserNameTAG: karthickks FirstChildCreateTimeTAG: 2012-09-17T11:31:10Z IndexTAG: 4185 TitleTAG: Is time up for lab 1??? Is time up in India here its Monday morning. UserIdTAG: 351425 UserNameTAG: brain2boom CreateTimeTAG: 2012-09-17T06:44:17Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: yes FirstChildUserIdTAG: 364254 FirstChildUserNameTAG: ANSHULAG FirstChildCreateTimeTAG: 2012-09-17T11:22:18Z IndexTAG: 4186 TitleTAG: Lab 1 assumptions If this was a simple series divider it would be easy, but when it becomes a resistor leading into a set of parallel resistors I'm a bit lost, especially if I need to deal with one resistor just disappearing sometimes. I can solve for if I have $R_1$ and $R_2$, or $R_1, R_2, R_{bulb}$, but if I change the resistors to handle the bulb being there won't they need to change for it not being there? Or is there a sweet spot where 2V plus another 1.5 ohm resistor in parallel drops us to 1.5V? By base, I have basically $V = 6$ $V_{bulb} = 1.5$ $V_1 = 6 - V_{bulb} = 4.5$ (using KVL) $V_2 = V_{bulb}$ (also using KVL on the inner loop) $I_{bulb} = 0.5$ $I_1 = I_2 + I_{bulb} = I_2 + 0.5$ I know for voltage I have $V_1 = \frac{R_1}{R_1 + R_2}V$ and $V_2 = \frac{R_2}{R_2+R_1}V$, and then for resistance the general parallel formula $R_s = \frac{R_1R_2}{R_1+R_2}$. But I am not sure how to use them with the information I have above. Do I simply need to solve for more elements to continue KVL/KCL? UserIdTAG: 264596 UserNameTAG: Nuru CreateTimeTAG: 2012-09-17T05:55:11Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: > "...but when it becomes a resistor leading into a set of parallel resistors I'm a bit lost" You know Rb, the resistance for the bulb, and it is in parallel with -let's name it- R2; so you have Re = Rb||R2. Now you have a voltage divider, and the required voltage for the node common to R1 and Re. FirstChildUserIdTAG: 164171 FirstChildUserNameTAG: PacoJuarez FirstChildCreateTimeTAG: 2012-09-17T06:21:48Z SecondChildTAG: Thanks, PacoJuarez. If I knew $R_{bulb}$ would always be there it would make sense to me that I could do a voltage divider formula with that setup, but with it gone in another case wouldn't that alter what we're calling $R_2$'s required value? Or maybe I need to try to solve with $R_{bulb}$ for both 2v and 1.5v? SecondChildUserIdTAG: 264596 SecondChildUserNameTAG: Nuru SecondChildCreateTimeTAG: 2012-09-17T06:28:25Z SecondChildTAG: "solve with $R_e$" I mean SecondChildUserIdTAG: 264596 SecondChildUserNameTAG: Nuru SecondChildCreateTimeTAG: 2012-09-17T06:29:51Z FirstChildTAG: Looks like I got it, though I didn't expect what I tried to work. I'll have to go over it again to make sense of it. FirstChildUserIdTAG: 264596 FirstChildUserNameTAG: Nuru FirstChildCreateTimeTAG: 2012-09-17T06:59:40Z IndexTAG: 4187 TitleTAG: Current direction for the voltage source I just don't understand why current is entering the positive sign of the voltage source. If you have a simple circuit with a battery and a resistor, why the battery and the resistor have different current directions? UserIdTAG: 374 UserNameTAG: mturilin CreateTimeTAG: 2012-09-17T05:40:58Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL and KCL CommentableIdTAG: 6002x_kvl_and_kcl NumberOfReplyTAG: 2 FirstChildTAG: It is just a **convention** we use. If you have an electronic component (resistor, voltage source, ...) with 2 terminals, you assign a **+** terminal and **-** terminal for the **voltage** **v** and define the **current** **i** as positive when it enters the + terminal. Remember that the **v-i variables** you assign to each component may be **positive** or **negative**. In case of the current variable **i** : if you calculate all branch variables for your network, the value of i may be positive or negative, if it is positive, this means the actual current will flow into the + terminal. This convention assures us that when a component dissipates power p = v*i will be **positive**, if it is **negative** (so v*i<0), it means the device delivers power. This will be the case for the voltage source. It is important that you differentiate between the v-i variables (parameters) you assign to a component and for instance the given voltage of a voltage source, which is an internal property of the component. As for the **currents** for the resistor and voltage source in our simple circuit, if you do the calculations, one will be positve and one will be negative, and both will have the same magnitude. FirstChildUserIdTAG: 107802 FirstChildUserNameTAG: protonman71 FirstChildCreateTimeTAG: 2012-09-17T10:37:47Z FirstChildTAG: Hello Mturilin, well i agree with PROTONMAN71 that its a convention. We are assuming it to get an error-free calculation. During the course of solving the problem ,it could have happened that we confuse the rule we adopted. say we might once assign a +ve sign to a current entering a + terminal of an element and maybe sometime later ,again we assign a "-" sign to the current entering the "+" terminal of an element....this will lead to confusion and error both.To avoid this we ,right from the starting, will stick to a rule or a discipline. Here, Prof makes use of the ASSOCIATED VARIABLE DISCILINE where it is assigned that current will flow TOWARDS the +ve terminal of a branch voltage. FirstChildUserIdTAG: 156225 FirstChildUserNameTAG: sanjana_m FirstChildCreateTimeTAG: 2012-09-23T07:04:49Z IndexTAG: 4188 TitleTAG: I think we have mistake in Tutorial Week 1 In Tutorial of Week 1 at OCW Problem 1-1/Textbook 2.7 - I think there are mistake in the Equation whene the teacher say V=I*R and then say I = V2 / R3 -----{1} then he replace V2 in equation {1} by V * (R2//R3) /(R1+R2//R3) I think you must write R3 in Epua. {1} at the top of equa. with V2 then the equation {1} will be I = V * R3 * (R2//R3) / (R1+R2//R3) # If I'm rong then I'm soory for this confuse UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-17T05:24:00Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hey I don't know if this an error or not, but I just wanted to say that there is no reason you can't post this under your own username. Certainly you can if you want to, but there isn't going to be any kind of retribution if you point out an error. I can tell you that there are errors. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-09-17T06:42:49Z SecondChildTAG: Definitely. If you help us catch mistakes, that would be excellent :) SecondChildUserIdTAG: 79361 SecondChildUserNameTAG: kimt SecondChildCreateTimeTAG: 2012-09-17T10:42:30Z IndexTAG: 4189 TitleTAG: Lab tool very unstable..... :-( I hope we get an extension on the deadline... So far I've re-built my lab about five times, re-inserted resistors about 20 times (when I change the resistor values, sometimes deletes the element from the page). Still going.... UserIdTAG: 217971 UserNameTAG: Alejandro01 CreateTimeTAG: 2012-09-17T05:23:59Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 2 FirstChildTAG: I tried lab tool on Firefox and Internet Explorer and it worked just fine. I think you should read Myriamit tutorial: https://www.edx.org/wiki/6.002x/MyrimitCircuitSimulatorTutorial FirstChildUserIdTAG: 164171 FirstChildUserNameTAG: PacoJuarez FirstChildCreateTimeTAG: 2012-09-17T06:43:04Z FirstChildTAG: I tried IE10, Firefox and Chrome. Had no problems only with Chrome FirstChildUserIdTAG: 372987 FirstChildUserNameTAG: ulman FirstChildCreateTimeTAG: 2012-09-17T07:08:06Z IndexTAG: 4190 TitleTAG: H2P2 H2P2 was a challenge, but I finished it. The calculus part of it was easy, the algebra to solve for RL was a wild ride. - Good Luck UserIdTAG: 132828 UserNameTAG: rharris CreateTimeTAG: 2012-09-17T05:11:14Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: please pass the answer for first 3 FirstChildUserIdTAG: 364254 FirstChildUserNameTAG: ANSHULAG FirstChildCreateTimeTAG: 2012-09-17T11:23:13Z SecondChildTAG: part 2 and 3 are easy its part one thats giving me trouble do you know a good refrence page to see where im going wrong SecondChildUserIdTAG: 436749 SecondChildUserNameTAG: waynebrown SecondChildCreateTimeTAG: 2012-09-17T12:05:51Z IndexTAG: 4191 TitleTAG: lab 2 I designed a circuit with voltage source v2 only. In my circuit there were 2 resistors of resistances 2R and R in series. The problem is when I do transient analysis by palcing the probe between 2R and R, I still get peak voltage of v2 while I think I should be getting v2/3.So where is the problem?? anyone help plz.. UserIdTAG: 201414 UserNameTAG: ArcherJeffrey CreateTimeTAG: 2012-09-17T04:36:37Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Don´t do transient analysis; do dc analysis, no need to add probes either. FirstChildUserIdTAG: 403987 FirstChildUserNameTAG: electricon FirstChildCreateTimeTAG: 2012-09-17T05:13:39Z SecondChildTAG: three resistors is the best way to go with lab2 SecondChildUserIdTAG: 436749 SecondChildUserNameTAG: waynebrown SecondChildCreateTimeTAG: 2012-09-17T09:59:48Z FirstChildTAG: if u r doing lab 2 use both sources and use three resistors FirstChildUserIdTAG: 294370 FirstChildUserNameTAG: Naif1125 FirstChildCreateTimeTAG: 2012-09-17T09:48:09Z SecondChildTAG: can anyone help me with the schematic.. SecondChildUserIdTAG: 206648 SecondChildUserNameTAG: TsvetanGeorgiev SecondChildCreateTimeTAG: 2012-09-17T14:23:49Z IndexTAG: 4192 TitleTAG: about the deadline any one have faced any changes in his homework?. according to deadline UserIdTAG: 395257 UserNameTAG: blackguitar CreateTimeTAG: 2012-09-17T04:24:19Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: and How do I know that the homework had been accepted FirstChildUserIdTAG: 395257 FirstChildUserNameTAG: blackguitar FirstChildCreateTimeTAG: 2012-09-17T04:30:15Z SecondChildTAG: Check the Progress tab which shows your score SecondChildUserIdTAG: 157967 SecondChildUserNameTAG: Jithinraj SecondChildCreateTimeTAG: 2012-09-17T04:36:33Z SecondChildTAG: i finished my lab.but i didn't got that in my progress.why?? SecondChildUserIdTAG: 353405 SecondChildUserNameTAG: karthickks SecondChildCreateTimeTAG: 2012-09-17T10:59:33Z SecondChildTAG: **blackguitar:** As to being a challenge versus the deadline: I think it depends on your grade level. This is probably the most challenging Circuits/Electronics course offered at any major university. When I went to a major Public university in the U.S. about 10 years ago, the material that this *one* course covers was split into *two* courses: Circuit Analysis for one semester, and Digital Logic and Design for the second semester. Most "lower level" U.S. *high-school* students would **not** even *attempt* this course, they would be struggling with basic algebra, and even in some of the poorer schools even English and reading! But most "honors", or "advanced placement" (AP) U.S. high school students would have **no problem** doing the math, or with the rigors and demands of the course, because first-year major U.S. university is the same. This would also go for those U.K. (and I think India) students that took Physics and Math A-levels. (Sorry I don't know the equivalents in the German or East Asian school systems if you are from there). It should take such students 1-3 days depending on mastery and experience. Here, all that work is "crammed" into one semester, so to succeed, you have to be able to think *fast*, and complex mathematical calculations need to be done quickly and accurately. Like when I went to university and took this course the first time, you would get "partial credit" if your equations were correct, but you made a stupid algebraic or sign mistake in your answer. Here, you have to be 100% correct to get the green "check mark". (As for me, *having taken the course many years before*, the Week 1 homework questions took about two hours, and that's only because of stupid mistakes in forgetting basic algebra or a sign.) SecondChildUserIdTAG: 292546 SecondChildUserNameTAG: JerseyMark SecondChildCreateTimeTAG: 2012-09-17T12:49:44Z IndexTAG: 4193 TitleTAG: Linear and Non linear Are capacitive and inductive networks linear? UserIdTAG: 309863 UserNameTAG: jithugeorge CreateTimeTAG: 2012-09-17T04:18:37Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Yes, although in some real world applications they may not be perfect. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-17T04:41:11Z FirstChildTAG: No FirstChildUserIdTAG: 294370 FirstChildUserNameTAG: Naif1125 FirstChildCreateTimeTAG: 2012-09-17T09:46:48Z FirstChildTAG: Yes they are, as you will see later, differentiating and integrating are linear operators. FirstChildUserIdTAG: 107802 FirstChildUserNameTAG: protonman71 FirstChildCreateTimeTAG: 2012-09-17T10:10:19Z IndexTAG: 4194 TitleTAG: Lab Check/Reset For Lab 1 if I hit check once it will check both, but then I from then on only have the Reset button. Do I have to rework my circuit each time? I probably should be using the sandbox come to think of it? UserIdTAG: 264596 UserNameTAG: Nuru CreateTimeTAG: 2012-09-17T03:49:19Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Yes, I agree the circuit lab would be a little easier. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-17T04:04:32Z SecondChildTAG: I intend to have this feature for you all soon, the ability to check without resetting. SecondChildUserIdTAG: 79361 SecondChildUserNameTAG: kimt SecondChildCreateTimeTAG: 2012-09-17T04:07:42Z SecondChildTAG: How's that for service? SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-17T04:25:17Z SecondChildTAG: Good stuff. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-17T04:25:47Z IndexTAG: 4195 TitleTAG: Recip of [1/R1+1/R2]... In this video, the professor used (R1+R2)/(R1R2) HOW would you find the reciprocal of more than 2 terms? UserIdTAG: 365894 UserNameTAG: jGeorge CreateTimeTAG: 2012-09-17T03:07:52Z VoteTAG: 0 CoursewareTAG: Week 2 / Linearity CommentableIdTAG: 6002x_linearity NumberOfReplyTAG: 1 FirstChildTAG: http://en.wikipedia.org/wiki/Series_and_parallel_circuits If you have two resistors, the parallel combination resistance is R = 1 / ((1 / R1) + (1 / R2)) equivalent to (R1+R2)/(R1R1). That's a shortcut equation. If you have more, it's just R = 1 / ((1 / R1) + (1 / R2) + ... + (1 / Rn)) Makes for some pretty hairy algebra if you have a lot of resistors. :-) FirstChildUserIdTAG: 131970 FirstChildUserNameTAG: elucify FirstChildCreateTimeTAG: 2012-09-17T03:19:02Z SecondChildTAG: The hairy algebra is why we often use conductance (G) in the equations, it keeps all those fractions out of the picture. Conductance is the inverse of resistance, 1/R = G. SecondChildUserIdTAG: 94557 SecondChildUserNameTAG: g_hopper SecondChildCreateTimeTAG: 2012-09-17T04:23:37Z IndexTAG: 4196 TitleTAG: H1P3 how to solve that problem any tricks r there to solve it in few steps UserIdTAG: 82846 UserNameTAG: suraj2794 CreateTimeTAG: 2012-09-17T03:00:32Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Some tricks: * Keep in mind that all three elements (when connected in parallel) consume the same amount of power; the problem tells you that up front. So the current through each is the same. Or, looked at it another way, the total current divides into three, 1/3 going through each heating element. The voltage, of course, across all three is (in the first diagram) the same: 240VAC. * You can treat those heating elements as resistors. Then you have 3 resistors in parallel, which you already know how to solve. (Or see http://en.wikipedia.org/wiki/Series_and_parallel_circuits) * The power versions of Ohm's Law are P = (V^2)/R and P = (I^2) * R. So if you know (for example) the power and the voltage, you can get the equivalent resistance Rp for the three heating elements connected in parallel. * If you know the resistance of the three in parallel, then you can calculate the resistance the individual resistances (that is, the heaters). Since they are identical, they all have the same resistance R. So the equivalent parallel resistance Rp = R / 3, where R is the resistance of one single element. * In the second part of the problem, since you know R, then that incorrectly-wired circuit is just a voltage divider, with R at the top, and (R || R) on the bottom (R||R means "R in parallel with R"). FirstChildUserIdTAG: 131970 FirstChildUserNameTAG: elucify FirstChildCreateTimeTAG: 2012-09-17T03:29:57Z SecondChildTAG: GIVE ME THE ANSWERS SecondChildUserIdTAG: 157972 SecondChildUserNameTAG: devanar SecondChildCreateTimeTAG: 2012-09-17T08:38:28Z IndexTAG: 4197 TitleTAG: Voltage and Current variables I have always been taught that DC circuits should have upper case letters for voltages and currents such as V1 and I1 and AC circuits should have lower case letters such as v1 and i1, but you don't follow that convention. I hope I don't carry this over into real life. UserIdTAG: 247286 UserNameTAG: ronald_borunda CreateTimeTAG: 2012-09-17T02:55:14Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 4198 TitleTAG: circuit sandbox i cannot rotate the resisters and other elements in circuit sandbox plz give mouse shortcut UserIdTAG: 82846 UserNameTAG: suraj2794 CreateTimeTAG: 2012-09-17T02:42:00Z VoteTAG: 0 CoursewareTAG: Week 1 / A light bulb circuit CommentableIdTAG: 6002x_RealLightBulbCircuit NumberOfReplyTAG: 2 FirstChildTAG: I cannot rotate also. I use anly vertical resistors. FirstChildUserIdTAG: 289132 FirstChildUserNameTAG: Nestor_Escala FirstChildCreateTimeTAG: 2012-09-17T02:50:04Z SecondChildTAG: use litter "R" on keyboard SecondChildUserIdTAG: 395257 SecondChildUserNameTAG: blackguitar SecondChildCreateTimeTAG: 2012-09-17T03:00:47Z SecondChildTAG: thank u very much coz using vertical resisters ;my circuit did not look neat and clean SecondChildUserIdTAG: 82846 SecondChildUserNameTAG: suraj2794 SecondChildCreateTimeTAG: 2012-09-17T03:05:01Z SecondChildTAG: I too used vertical resistors only. But suddenly the resistor rotated horizontally at one time. So resistors can be rotated, but I donot know how. SecondChildUserIdTAG: 222234 SecondChildUserNameTAG: ivsprasad SecondChildCreateTimeTAG: 2012-09-17T03:21:53Z SecondChildTAG: Click on the resistor and hit "r" (for "rotate") on your keyboard. SecondChildUserIdTAG: 131970 SecondChildUserNameTAG: elucify SecondChildCreateTimeTAG: 2012-09-17T03:31:03Z FirstChildTAG: Use the letter R. Cheers! FirstChildUserIdTAG: 153986 FirstChildUserNameTAG: johnhess FirstChildCreateTimeTAG: 2012-10-12T17:10:47Z SecondChildTAG: thanks! SecondChildUserIdTAG: 261944 SecondChildUserNameTAG: BSharkey SecondChildCreateTimeTAG: 2012-10-12T17:12:26Z IndexTAG: 4199 TitleTAG: H1P1 6th question Hey, I need some help with the 6th question of H1P1 where the cumulative power dissipated from the smallest valued composite resistor is asked. The smallest valued composite resistor network would be in parallel combo, r8? If each of the resistor dissipate 1W, should they add up in the total power dissipated? UserIdTAG: 298958 UserNameTAG: tuks CreateTimeTAG: 2012-09-17T02:14:45Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: The set of resistors is limited by the weakest link, when a resistor has reached its burnout limit, are all the other resistors dissipating 1W? Hopefully that helps. FirstChildUserIdTAG: 291544 FirstChildUserNameTAG: omdaniel FirstChildCreateTimeTAG: 2012-09-17T02:35:00Z FirstChildTAG: Try this: * Draw the circuit for the smallest composite resistor. * You have two 4 ohm and one 6 ohm resistor. The question implies that either the 6 ohm or one (or both) of the 4 ohm resistors will be dissipating 1 W (since you are running at least one of them up to its maximum power.) * So figure out the current through one of the resistors, then calculate the power through the others. * You have the right answer when one (or more) of them is dissipating 1W, and the all the remaining ones are dissipating less than 1W. * If you get an answer where one of them is dissipating > 1W, pick another and try again. FirstChildUserIdTAG: 131970 FirstChildUserNameTAG: elucify FirstChildCreateTimeTAG: 2012-09-17T03:37:17Z SecondChildTAG: thanks! this helped me a lot SecondChildUserIdTAG: 199008 SecondChildUserNameTAG: matuko SecondChildCreateTimeTAG: 2012-09-17T04:12:01Z IndexTAG: 4200 TitleTAG: PROBLEMS WITH h1p1 In Exercises h1p1 appears that we put everything in terms of c. I placed but I get the error of resistors in parallel and combined. confirmeme please if are these. According to the text because these are 1 / R +1 / R + +1 / R for parallel R + R +1 / R +1 / R Combined UserIdTAG: 45303 UserNameTAG: brenda CreateTimeTAG: 2012-09-17T01:24:54Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Another method for expressing Parallel is A||B. I have yet to find an appropriate way to express the combined. FirstChildUserIdTAG: 267537 FirstChildUserNameTAG: scorliss FirstChildCreateTimeTAG: 2012-09-17T01:41:27Z SecondChildTAG: A+B SecondChildUserIdTAG: 131970 SecondChildUserNameTAG: elucify SecondChildCreateTimeTAG: 2012-09-17T03:38:43Z FirstChildTAG: You forgot to use 1 to divide the expression again. FirstChildUserIdTAG: 350214 FirstChildUserNameTAG: HeatDynasty FirstChildCreateTimeTAG: 2012-09-17T02:47:03Z FirstChildTAG: where you get number 1 the circuit dosen't get any number so you juste write algebraic expression and don't forget tne Brackets ({}) to be clear FirstChildUserIdTAG: 318037 FirstChildUserNameTAG: Maher-84 FirstChildCreateTimeTAG: 2012-09-17T03:09:17Z IndexTAG: 4201 TitleTAG: direction of current in a battery In S2E2, the direction of the current is flowing from negative to positive ?. what is the convention ? UserIdTAG: 45441 UserNameTAG: devnieL CreateTimeTAG: 2012-09-17T01:22:37Z VoteTAG: 0 CoursewareTAG: Week 1 / Associated reference directions CommentableIdTAG: 6002x_S2E2_Associated_Reference_Directions NumberOfReplyTAG: 2 FirstChildTAG: The convention is that positive current flows out of the positive terminal of a voltage source and into the positive terminal of a resistor. If the value for the current is negative, it's actually flowing the other way. FirstChildUserIdTAG: 390034 FirstChildUserNameTAG: kahlil FirstChildCreateTimeTAG: 2012-09-17T01:30:01Z SecondChildTAG: I think my dyslexia gene just had a hernia lol SecondChildUserIdTAG: 340856 SecondChildUserNameTAG: RanmaSaotome SecondChildCreateTimeTAG: 2012-09-17T01:34:21Z FirstChildTAG: Electrons are negatively charged, however "conventional" science states that it flows from the positive to the negative. This is a carry over from very early days. However in reality, electrons flow from negative into positive. It's best to consider the flow within a given context. To avoid trouble. Just do whatever the book tells you. Lightning also strikes from the ground up. So you have "conventional" flow and "Electron" flow, in opposite directions. I hope you are as confused as I am on the matter. Have fun. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-17T01:44:21Z SecondChildTAG: Lightning also strikes from the ground up.....Love it! SecondChildUserIdTAG: 267537 SecondChildUserNameTAG: scorliss SecondChildCreateTimeTAG: 2012-09-17T01:58:19Z SecondChildTAG: Here is a cool video demonstrating this exchange using high speed cameras. The last few moments of the video are extraordinary. http://www.youtube.com/watch?v=6NZ7BollRo4 SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-17T03:42:37Z SecondChildTAG: @Pennypacker ~Thnx 4 d AWESOME video!!! and i liked the last few words u used..."i hope u r as confused as i am on the matter.have fun" SecondChildUserIdTAG: 156225 SecondChildUserNameTAG: sanjana_m SecondChildCreateTimeTAG: 2012-09-23T10:05:07Z IndexTAG: 4202 TitleTAG: help with H1P1 sixth question Can anyone give any hint on how to resolve it? thanks in advance! UserIdTAG: 79854 UserNameTAG: AlbertoGT CreateTimeTAG: 2012-09-17T01:08:12Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 9 FirstChildTAG: lol I got it! thanks to all of you :D FirstChildUserIdTAG: 79854 FirstChildUserNameTAG: AlbertoGT FirstChildCreateTimeTAG: 2012-09-17T02:57:21Z FirstChildTAG: got it too! thanks FirstChildUserIdTAG: 237583 FirstChildUserNameTAG: bresendez3 FirstChildCreateTimeTAG: 2012-09-17T03:47:55Z FirstChildTAG: See also [a more detailed guide][1] (see comments) [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/5056879524c6bd2b00000028 "a more detailed guide" FirstChildUserIdTAG: 131970 FirstChildUserNameTAG: elucify FirstChildCreateTimeTAG: 2012-09-17T03:42:18Z FirstChildTAG: so am i ,i need it FirstChildUserIdTAG: 395257 FirstChildUserNameTAG: blackguitar FirstChildCreateTimeTAG: 2012-09-17T01:15:51Z FirstChildTAG: What happens to the voltage across parallel resistors? When you find the voltage, find out what resistor would burn up first. When that resistor is maxed out at 1W, what is going on with the other two resistors? Then combine the total wattage of all three. Have fun FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-17T01:23:44Z FirstChildTAG: I know how to deal with the fact that we´re making the assumption of each individual resistor being capable of taking 1 watt before burning up, according to the resistor arrangement used in the previous question (H1P1 - 5) , I guess I should not explain it, because I´d need to tell the answer of of the previous question :) FirstChildUserIdTAG: 79854 FirstChildUserNameTAG: AlbertoGT FirstChildCreateTimeTAG: 2012-09-17T01:34:47Z SecondChildTAG: :) SecondChildUserIdTAG: 395257 SecondChildUserNameTAG: blackguitar SecondChildCreateTimeTAG: 2012-09-17T01:38:36Z SecondChildTAG: please, I need to know how to solve SecondChildUserIdTAG: 449365 SecondChildUserNameTAG: RafaelR SecondChildCreateTimeTAG: 2012-09-17T02:58:10Z FirstChildTAG: Hi AlbertoGT! Can I help you? FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-17T01:45:38Z FirstChildTAG: I can not figure this out for some reason. FirstChildUserIdTAG: 237583 FirstChildUserNameTAG: bresendez3 FirstChildCreateTimeTAG: 2012-09-17T02:08:51Z FirstChildTAG: Hi, Hint: Work out the element voltage for each individual resitor in question. With the 1W value and each resistor's value given in H1P1, (6ohms & 4ohms) you should be able to work out 'V' and/or 'I' for each element. The rest is fairly straight forward regarding the smallest-valued composite resistor (which you would have answered in H1P1 fifth question). Good luck !!! FirstChildUserIdTAG: 217971 FirstChildUserNameTAG: Alejandro01 FirstChildCreateTimeTAG: 2012-09-17T03:18:50Z SecondChildTAG: Thanks a lot! :D SecondChildUserIdTAG: 80283 SecondChildUserNameTAG: secastanomu SecondChildCreateTimeTAG: 2012-09-17T03:41:06Z IndexTAG: 4203 TitleTAG: the deadline of H1 is the deadline of H1 finished? if yes so , is there answers? and the "clock sign" didn't change ( that beside H1 and lab1) "there is no changes" UserIdTAG: 395257 UserNameTAG: blackguitar CreateTimeTAG: 2012-09-17T01:04:33Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: more explanation please FirstChildUserIdTAG: 395257 FirstChildUserNameTAG: blackguitar FirstChildCreateTimeTAG: 2012-09-17T02:56:40Z FirstChildTAG: is it accept any more solutions? FirstChildUserIdTAG: 395257 FirstChildUserNameTAG: blackguitar FirstChildCreateTimeTAG: 2012-09-17T01:05:53Z SecondChildTAG: Probably a few more hours to go. It's only 9:15 here. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-17T01:14:10Z SecondChildTAG: here in Egypt 3.20 AM and now GMT is 1.20 how? SecondChildUserIdTAG: 395257 SecondChildUserNameTAG: blackguitar SecondChildCreateTimeTAG: 2012-09-17T01:21:10Z SecondChildTAG: It's 9:28pm in Massachusetts, I guess they are not using GMT, they are using GMT -4(DST). SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-17T01:28:45Z SecondChildTAG: yea, it's quite possible SecondChildUserIdTAG: 395257 SecondChildUserNameTAG: blackguitar SecondChildCreateTimeTAG: 2012-09-17T01:32:27Z SecondChildTAG: The instructions for the course say that homeworks and labs are due at midnight your local time. So they aren't using MIT time, they're using your time. SecondChildUserIdTAG: 131970 SecondChildUserNameTAG: elucify SecondChildCreateTimeTAG: 2012-09-17T03:20:49Z SecondChildTAG: but there is no changes happen to me until now h still can edit and re edit my homework till now what's going on? SecondChildUserIdTAG: 395257 SecondChildUserNameTAG: blackguitar SecondChildCreateTimeTAG: 2012-09-17T04:10:55Z SecondChildTAG: Now it is 9:45 am ,17th September, but still all coursework options are as they were ..... so, what is going on now??? What is the time deadline? SecondChildUserIdTAG: 295278 SecondChildUserNameTAG: souravfn7 SecondChildCreateTimeTAG: 2012-09-17T04:29:38Z SecondChildTAG: If you will follow Boston time...according to midnight there, it should be closed on 9 am,17th Sept. in Indian time Zone(IST) But even at 10am in India,all the options are remaining same.... SecondChildUserIdTAG: 295278 SecondChildUserNameTAG: souravfn7 SecondChildCreateTimeTAG: 2012-09-17T04:33:46Z FirstChildTAG: Look at [Course info][1]: > The first problem set and lab are due this Sunday (at midnight of your local time) [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/info FirstChildUserIdTAG: 366786 FirstChildUserNameTAG: LuisFernandoMurguia FirstChildCreateTimeTAG: 2012-09-17T02:14:11Z FirstChildTAG: no, the time with respect to Boston FirstChildUserIdTAG: 395257 FirstChildUserNameTAG: blackguitar FirstChildCreateTimeTAG: 2012-09-17T03:21:05Z IndexTAG: 4204 TitleTAG: silly mistake I just kept getting the wrong answer because instead of seeing 18ohms I kept seeing 8ohms, then I went back and I felt very silly :P. UserIdTAG: 309501 UserNameTAG: SalT CreateTimeTAG: 2012-09-17T00:52:33Z VoteTAG: 0 CoursewareTAG: Week 1 / Associated reference directions CommentableIdTAG: 6002x_S2E2_Associated_Reference_Directions NumberOfReplyTAG: 0 IndexTAG: 4205 TitleTAG: Some Tricks Some tricks and correctly solving algebraic expressions. V3 expressed in terms of v1 and v2. When you have a 1 / R replace it by its conductance G. It is easier to score. When you have the first expressions (a1, a2) Use them to find the other variables. UserIdTAG: 80601 UserNameTAG: MiltonM CreateTimeTAG: 2012-09-17T00:43:59Z VoteTAG: 0 CoursewareTAG: Week 2 / Linear combinations of source strenghts CommentableIdTAG: 6002x_S3E2_Linear_Combinations_of_Source_Strengths NumberOfReplyTAG: 0 IndexTAG: 4206 TitleTAG: H1P2 How do I calculate the power coming out of the current source and the voltage source? How do I find out the power dissipated in the 1ohm and 3 ohm resistor? UserIdTAG: 283740 UserNameTAG: sujathad CreateTimeTAG: 2012-09-17T00:30:04Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: P=IV FirstChildUserIdTAG: 272702 FirstChildUserNameTAG: mindo FirstChildCreateTimeTAG: 2012-09-17T00:44:29Z SecondChildTAG: i need current i1,i2,i3. SecondChildUserIdTAG: 449365 SecondChildUserNameTAG: RafaelR SecondChildCreateTimeTAG: 2012-09-17T03:22:44Z FirstChildTAG: Also P = I^2 * R and P = (V ^ 2) / R FirstChildUserIdTAG: 131970 FirstChildUserNameTAG: elucify FirstChildCreateTimeTAG: 2012-09-17T03:45:01Z IndexTAG: 4207 TitleTAG: H1P1 Last resistor Combo Hi All, I have been banging my head against the wall here trying to figure out how to express two sets of resistors in series that are parallel to each other....Any clues? (R+R*R+R)/((R+R)+(R+R)) does not seem to work UserIdTAG: 267537 UserNameTAG: scorliss CreateTimeTAG: 2012-09-16T22:47:25Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Hint: $R+R=2R$ and $R*R=R^2$. Simplify your equation until only one R remains. FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-09-16T22:54:48Z SecondChildTAG: THX Jersey. SecondChildUserIdTAG: 267537 SecondChildUserNameTAG: scorliss SecondChildCreateTimeTAG: 2012-09-16T22:55:49Z FirstChildTAG: First group 2 parallel resistor into a new one. Then the new resistor and the last remaining one. 2 parallel resistors: R1*R2 / (R1+R2) FirstChildUserIdTAG: 388555 FirstChildUserNameTAG: 4lk4tr43 FirstChildCreateTimeTAG: 2012-09-16T22:58:26Z FirstChildTAG: Thanks for all of your help. I am still at a loss for expressing this Network. I have used (2R^2)/(2R+2R) which can be simplified down to R/2 but it says that it is wrong. I don't doubt that it is wrong but I'm sure that I'm missing something here. FirstChildUserIdTAG: 267537 FirstChildUserNameTAG: scorliss FirstChildCreateTimeTAG: 2012-09-17T01:45:39Z SecondChildTAG: I'm in the same. I still have not given my task. :( SecondChildUserIdTAG: 449365 SecondChildUserNameTAG: RafaelR SecondChildCreateTimeTAG: 2012-09-17T02:21:03Z IndexTAG: 4208 TitleTAG: i m verrrrry confused !!! hey , i'm very confused about where i concentrate plz any one tell me how he study (is write some notes , depend only the videos , tell me some steps to get and make all of this videos easy to collect because in the mid term i didnt know where i should be start >> advanced thx for all UserIdTAG: 193445 UserNameTAG: eslammagdy CreateTimeTAG: 2012-09-16T22:44:51Z VoteTAG: 0 CoursewareTAG: Week 2 / Linearity CommentableIdTAG: 6002x_linearity NumberOfReplyTAG: 3 FirstChildTAG: 1)watch videos + do exercise with videos 2)if stuck don't be hesitate to ask in discussion forum 3)watch tutorials 4)reading selected or suggested sections in the book 5)if stuck in any section in the book don't get shy to ask of course in discussion forum. 6)try to solve home work and lab on time 7)before watching new video revise what have you learned and if you have time practice the previous lab and exercises. I hope at the end you will have no problem. FirstChildUserIdTAG: 40209 FirstChildUserNameTAG: sadaf78 FirstChildCreateTimeTAG: 2012-09-16T23:24:04Z SecondChildTAG: thx alot for ur interest and ur explanation SecondChildUserIdTAG: 193445 SecondChildUserNameTAG: eslammagdy SecondChildCreateTimeTAG: 2012-09-16T23:33:14Z FirstChildTAG: Watch all videos and lectures do all exercises (writting) and see discussions, are sometimes very helpful. And Textbook, also very helpful. FirstChildUserIdTAG: 342966 FirstChildUserNameTAG: SandraNavarro FirstChildCreateTimeTAG: 2012-09-16T22:48:45Z SecondChildTAG: firstly ,, thx for ur interest i actualy do this except for reading textbook well i will do and i hope work because i feel i didnt understand any thing ,, but i do the exercises and labs good thx again sandra SecondChildUserIdTAG: 193445 SecondChildUserNameTAG: eslammagdy SecondChildCreateTimeTAG: 2012-09-16T23:06:18Z FirstChildTAG: hi beside to watch a videos here just go to the course info then in the right side find the lecture slide so u can print those lectures and also theirs a textbook free to read or u can also to print the textbook if ever still so hard to you to understand just search google or youtube or maybe u want to buy textbook u can order here for i know theirs 40% discount and the last u ask anybody who knows electronics..hope this advice can help u FirstChildUserIdTAG: 235727 FirstChildUserNameTAG: amor FirstChildCreateTimeTAG: 2012-09-17T00:20:37Z IndexTAG: 4209 TitleTAG: Example 2.14, page 72 clarification I'm a bit stuck on a leap this problem makes. I understand the goal is to obtain voltage so we want our equations to be in the form of V. Power supplied by the source is vi, i = .002 A so Ps = (.002)v Power into the resistor is Ps = vi, i = v/r, r = 1K so Pr = v * (v / 1000) = .001v^2 Assuming I understood that so far, then the power supplied and power dissipated should be the same so we can equate .002v = .001v^2. The book says the solution is V = .5, but that equation results in: .002/.001 = v, so v = 2. What am I missing here? If I go from another approach wouldn't just be P = vi, v = ir, so P = i^2r = (.002)^2 * 1000. So Pr = .004 and Pr = v(.002) so v = 2 again. Either way I'm getting v = 2. Any help would be appreciated. UserIdTAG: 264596 UserNameTAG: Nuru CreateTimeTAG: 2012-09-16T21:58:26Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Nuru, your answer v = 2V is correct. The book has a mistake. You can alse see that the book is incorrect directly by Ohm's law V = I*R = (0.002 A)*(1000 ohm) = 2V. FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-16T22:22:37Z SecondChildTAG: Thanks Jelizon. Is there a pdf or listing of corrections in the book exercises? A cursory google search didn't net me much. SecondChildUserIdTAG: 264596 SecondChildUserNameTAG: Nuru SecondChildCreateTimeTAG: 2012-09-16T22:24:51Z SecondChildTAG: Book eratta https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/Book/ SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-16T23:54:58Z IndexTAG: 4210 TitleTAG: H1P2 How do you get V1 UserIdTAG: 348251 UserNameTAG: PaulchenPanther CreateTimeTAG: 2012-09-16T21:50:55Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: v1 = v2 (its a loop). Now you can use the node method, I used the grounding node below the v2 resistor. You will get one very simple equation for the node potential that is missing. If you know that potential you know the v2 and basically v1. EDIT(corrected polarity according to the staff post, I'm always messing them up) FirstChildUserIdTAG: 388555 FirstChildUserNameTAG: 4lk4tr43 FirstChildCreateTimeTAG: 2012-09-16T22:07:29Z SecondChildTAG: Be careful with you loop equations! The one that you wrote (i.e. -v1 = v2) is incorrect. Check the polarities of v1 and v2 SecondChildUserIdTAG: 381746 SecondChildUserNameTAG: jelizon SecondChildCreateTimeTAG: 2012-09-16T22:18:00Z FirstChildTAG: is i2=2v3 FirstChildUserIdTAG: 215287 FirstChildUserNameTAG: yomi FirstChildCreateTimeTAG: 2012-09-16T23:58:10Z IndexTAG: 4211 TitleTAG: anyone else working on lab 2 yet? Obviously I'm doing something wrong here, but for the life of me, I cannot get the maximum value of the output to go below 1V, although I'm somehow supposed to get it to 667mV. Having no trouble with getting the minimum right around -167mV though. Hmm :/ Any help would be appreciated. UserIdTAG: 275839 UserNameTAG: Mel9826 CreateTimeTAG: 2012-09-16T21:34:17Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Once you have the proportions right, you need to get rid of some of the voltage. ;) FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-16T21:48:02Z FirstChildTAG: Choose two resistors to get the 3:1 ratio, then add a third resistor from their intersection to ground to get the proper reduction. FirstChildUserIdTAG: 406202 FirstChildUserNameTAG: skyhawk FirstChildCreateTimeTAG: 2012-09-17T00:06:42Z SecondChildTAG: Thanks for your help! But can you explain why 3:1 ratio? How did you come to this value? I couldn't get calculate this for a long time. SecondChildUserIdTAG: 251637 SecondChildUserNameTAG: rus_ivan SecondChildCreateTimeTAG: 2012-09-18T08:14:40Z SecondChildTAG: how will the voltage be divided by putting a third resistor from the intersection of the other two to ground? won't this third resistor be parallel with the other two, thus unaltering the voltage? Please explain... SecondChildUserIdTAG: 457713 SecondChildUserNameTAG: ASHWIN_DT SecondChildCreateTimeTAG: 2012-09-22T06:17:19Z IndexTAG: 4212 TitleTAG: H1P3 Can anyone help with H1P3? I thought p=Vrms^2/R where p= 4350w and Vrms= 240*sqrt(2)= 340. R should be equal to Vrms^2/power But the answer i got seems to be wrong? Help! UserIdTAG: 175706 UserNameTAG: rwskim CreateTimeTAG: 2012-09-16T21:31:40Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Actuall 240 is the rms voltage. Multiplying by square root of 2 will give you the peak voltage. FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-16T22:15:22Z FirstChildTAG: i'm also stuck in this section since last 3 days... my brain is going to burst out with the last 2 problems c and e. Let me know if you get the hints. FirstChildUserIdTAG: 130118 FirstChildUserNameTAG: manis FirstChildCreateTimeTAG: 2012-09-16T21:49:31Z SecondChildTAG: Thanks for the help! Let me try when i get back SecondChildUserIdTAG: 175706 SecondChildUserNameTAG: rwskim SecondChildCreateTimeTAG: 2012-09-16T22:50:20Z SecondChildTAG: For e you can combine all resistors, then you'll have v and i. (p=vi) SecondChildUserIdTAG: 388555 SecondChildUserNameTAG: 4lk4tr43 SecondChildCreateTimeTAG: 2012-09-16T23:08:03Z IndexTAG: 4213 TitleTAG: but mathematical is linear dependent UserIdTAG: 337902 UserNameTAG: Dan_Pop CreateTimeTAG: 2012-09-16T21:16:43Z VoteTAG: 0 CoursewareTAG: Week 2 / Is it linear? CommentableIdTAG: 6002x_is_it_linear NumberOfReplyTAG: 0 IndexTAG: 4214 TitleTAG: Local Time & Account info How is the local time for submission determined? (It isn't the time on my laptop, is it?) I can only see my full name and my email address when I click on the home button on the top right, is that normal? (Edit: I thought there should be an addresse) UserIdTAG: 388555 UserNameTAG: 4lk4tr43 CreateTimeTAG: 2012-09-16T20:27:00Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: compare your local time with GMT FirstChildUserIdTAG: 80970 FirstChildUserNameTAG: cruse FirstChildCreateTimeTAG: 2012-09-16T20:32:09Z SecondChildTAG: Thank you, I just got confused by "The first problem set and lab are due this Sunday (at midnight of your local time)". Not that it would matter much, but who knows what happens on week 5+ SecondChildUserIdTAG: 388555 SecondChildUserNameTAG: 4lk4tr43 SecondChildCreateTimeTAG: 2012-09-16T20:51:23Z IndexTAG: 4215 TitleTAG: H1p2 Kindly anyone help me on last questions of H1p2 UserIdTAG: 414201 UserNameTAG: uzaifakram CreateTimeTAG: 2012-09-16T20:12:07Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: you need i3 and v3. For v3 you can create a loop -v1+v3+v4=0. i3 can be calculated using the relationship i=v/R. I hope this helps (EDIT please check the loop equation yourself, I always mess up with the polarities) FirstChildUserIdTAG: 388555 FirstChildUserNameTAG: 4lk4tr43 FirstChildCreateTimeTAG: 2012-09-16T21:06:13Z FirstChildTAG: A question. some of the worked examples use "e1, e2, etc" The diagram uses "v1, v2,..." What is going on with nomenclature? FirstChildUserIdTAG: 118098 FirstChildUserNameTAG: Sandhog FirstChildCreateTimeTAG: 2012-09-17T03:18:36Z IndexTAG: 4216 TitleTAG: solutions to hw and labs... will the solutions to the home work qs n lab sessions be made available???and if so when? UserIdTAG: 390803 UserNameTAG: Spurty CreateTimeTAG: 2012-09-16T19:57:43Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Yes, usually the day after the deadline, if they go by what happened last spring. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-09-16T20:13:30Z SecondChildTAG: Yep, although instead of releasing PDF's, we added functionality to show-answer button, so it can show full explanations as well as final answers. SecondChildUserIdTAG: 96452 SecondChildUserNameTAG: Lyla SecondChildCreateTimeTAG: 2012-09-16T20:54:23Z IndexTAG: 4217 TitleTAG: practise sessions.. do the practise questions carry any marks? UserIdTAG: 390803 UserNameTAG: Spurty CreateTimeTAG: 2012-09-16T19:56:07Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: No. Only Homeworks, Labs, Midterm, and the Final count towards your marks (See the Progress tab). Your two lowest homework grades AND your two lowest lab grades get thrown out, so if you have a bad week, or a busy one, you can miss an assignment or two and not be penalized. I would rather get a 100% on the first two homeworks and the first two labs, since they are usually the easiest, and then take it from there. FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-09-16T20:28:55Z SecondChildTAG: great! thanks for the Info. SecondChildUserIdTAG: 237948 SecondChildUserNameTAG: cyberjayar SecondChildCreateTimeTAG: 2012-09-16T20:33:33Z IndexTAG: 4218 TitleTAG: Lab 1 The value of the resisence in Lab 1, i tried with normaly valuos, example 1,2,3,4. But its wrong i read in he forum about 10% someone could explain me how to put them? Thanks UserIdTAG: 348251 UserNameTAG: PaulchenPanther CreateTimeTAG: 2012-09-16T19:42:30Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Please tell your question clearly in proper English. I didn't understand your problem. FirstChildUserIdTAG: 317445 FirstChildUserNameTAG: arghya33 FirstChildCreateTimeTAG: 2012-09-16T19:55:28Z SecondChildTAG: Thanks, look i tried with differents valuos in Lab one. I use two resistence with the bulb three. But the valous that i use (6 and 3) dont macht. How do you resolve the Lab 1? SecondChildUserIdTAG: 348251 SecondChildUserNameTAG: PaulchenPanther SecondChildCreateTimeTAG: 2012-09-16T20:10:15Z SecondChildTAG: he means that he tried to write the answer of the H.W , but there is something wrong in the forming of the answer, how can we write the answer & be accepted by the system ? SecondChildUserIdTAG: 284395 SecondChildUserNameTAG: AhmedAbuZeid SecondChildCreateTimeTAG: 2012-09-16T20:10:25Z FirstChildTAG: Try using the hint at the bottom of the Lab. Don't use random values for the resistances, but actually set up the equations and solve them. FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-16T21:23:20Z IndexTAG: 4219 TitleTAG: password how to change the password for edx UserIdTAG: 326964 UserNameTAG: vinothraj CreateTimeTAG: 2012-09-16T19:31:00Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: I> First log out. II> Load edx.org and click on the sign in button III> Click on Forgot password option & enter your email id. You will receive a link in your email inbox. IV> Click on the link & reset your password. Done!!! FirstChildUserIdTAG: 317445 FirstChildUserNameTAG: arghya33 FirstChildCreateTimeTAG: 2012-09-16T19:53:44Z FirstChildTAG: just click on forgot password FirstChildUserIdTAG: 80970 FirstChildUserNameTAG: cruse FirstChildCreateTimeTAG: 2012-09-16T20:00:26Z IndexTAG: 4220 TitleTAG: Sorry but can anyone explain how to do the last question?? I don't really understand how to do the last one... UserIdTAG: 225436 UserNameTAG: ElmerTsai CreateTimeTAG: 2012-09-16T19:13:10Z VoteTAG: 0 CoursewareTAG: Week 1 / Battery model CommentableIdTAG: 6002x_S1E9_Battery_Model NumberOfReplyTAG: 2 FirstChildTAG: Hai,i'm thouheed india. We have 3 heaters,1st calculate eqlnt resistance and try to find out p=i^r,p=v^2/r,p=v*i and apply potential divider rule FirstChildUserIdTAG: 319286 FirstChildUserNameTAG: thouheed FirstChildCreateTimeTAG: 2012-09-16T19:30:24Z SecondChildTAG: i still cudnt understand SecondChildUserIdTAG: 444708 SecondChildUserNameTAG: jagadeeshv16 SecondChildCreateTimeTAG: 2012-09-16T19:32:36Z SecondChildTAG: The resistance must be calculated because the material of the heater has some internal resistance in it. check the material its internal resistance and then apply the power equation. SecondChildUserIdTAG: 365612 SecondChildUserNameTAG: ibetojumbo SecondChildCreateTimeTAG: 2012-09-16T19:44:07Z SecondChildTAG: I couldn't understand either ........ SecondChildUserIdTAG: 457331 SecondChildUserNameTAG: deepansh91 SecondChildCreateTimeTAG: 2012-09-23T12:10:16Z SecondChildTAG: from where in the world did the heater pop from? :O SecondChildUserIdTAG: 486510 SecondChildUserNameTAG: RamanRJN SecondChildCreateTimeTAG: 2012-09-27T20:44:06Z FirstChildTAG: the difference in voltages is 0.1 V so that is the voltage that will charge the lower voltage source. take that and divide with summed resistance (0.57 ohm) and you get the answer FirstChildUserIdTAG: 281752 FirstChildUserNameTAG: amilinovic FirstChildCreateTimeTAG: 2012-09-17T00:17:03Z IndexTAG: 4221 TitleTAG: write... how can i write the answer in the answer block in the home work like when 3 resisters R are in series , then their equivalent resistance is 3R. so how can i write this answer in answer block. when i write 3R it shows wronge answer???? anyone know how to write??? UserIdTAG: 80970 UserNameTAG: cruse CreateTimeTAG: 2012-09-16T19:11:57Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: 3*R FirstChildUserIdTAG: 382791 FirstChildUserNameTAG: 6814tav FirstChildCreateTimeTAG: 2012-09-16T19:25:24Z SecondChildTAG: indeed and the sum of resistances and not their product. SecondChildUserIdTAG: 375517 SecondChildUserNameTAG: CristianeTeburcio SecondChildCreateTimeTAG: 2012-09-16T19:39:54Z SecondChildTAG: thankshttps://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/users/375517 SecondChildUserIdTAG: 80970 SecondChildUserNameTAG: cruse SecondChildCreateTimeTAG: 2012-09-16T20:28:28Z IndexTAG: 4222 TitleTAG: Value I can not see the values of the problem , [Math Processing Error] is writing instead of values.i am using chrome or explorer , this problem occur again and again. UserIdTAG: 132475 UserNameTAG: mekazanc CreateTimeTAG: 2012-09-16T19:07:30Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Can you take a screenshot of this? FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-16T20:58:16Z IndexTAG: 4223 TitleTAG: h1p3 Joe wants to heat his 12'X20' workshop with electric heat. He has hired the HACME electrician company to build the system. They propose to use three 1490.0W 240V baseboard heaters to provide a total heating capacity of 4470.0W how to calculate 3 4 and 5 questions please help me UserIdTAG: 444708 UserNameTAG: jagadeeshv16 CreateTimeTAG: 2012-09-16T19:02:26Z VoteTAG: 0 CoursewareTAG: Week 1 / AC power CommentableIdTAG: 6002x_S1E3_AC_power NumberOfReplyTAG: 1 FirstChildTAG: You can calculate the total R cose you have know values of each resistor (or can calculate easyly, next calculate total current for serie - paralel then you can calculate power for each resistor, serie with total current and paralel with half of it, the just add to total hope help take care of http://2.bp.blogspot.com/_IGPyEGm6pT0/TEr7980y-9I/AAAAAAAAAh4/Via21LTOjgM/s1600/OhmsWattsLaws.png ohms law do a lot FirstChildUserIdTAG: 298612 FirstChildUserNameTAG: fabianh FirstChildCreateTimeTAG: 2012-09-16T19:58:30Z SecondChildTAG: thats the problem i face i am getting various values of R i couldnt find R SecondChildUserIdTAG: 444708 SecondChildUserNameTAG: jagadeeshv16 SecondChildCreateTimeTAG: 2012-09-16T20:17:25Z SecondChildTAG: Why are you getting different values of R? Notice that the power of each resistor is 1490W and you also know it's voltage to be 240V. You should be able to derive it from there... SecondChildUserIdTAG: 381746 SecondChildUserNameTAG: jelizon SecondChildCreateTimeTAG: 2012-09-16T21:07:44Z SecondChildTAG: Thank you jelizon i figured it out later SecondChildUserIdTAG: 444708 SecondChildUserNameTAG: jagadeeshv16 SecondChildCreateTimeTAG: 2012-09-21T15:26:55Z IndexTAG: 4224 TitleTAG: What is the right way? I'm feeling bewildered. Until now I just used the KVL to get the simply equations, and all my attempts were wrong, because I dunno how to convert the variables to resistances in this exercise. Somebody can help me? :) UserIdTAG: 291362 UserNameTAG: Gudson CreateTimeTAG: 2012-09-16T18:55:12Z VoteTAG: 0 CoursewareTAG: Week 2 / Linear combinations of source strenghts CommentableIdTAG: 6002x_S3E2_Linear_Combinations_of_Source_Strengths NumberOfReplyTAG: 1 FirstChildTAG: Gudson, I think the bit you are missing happens at 7:10 and onward in the video previous to the exercise. EDIT: Maybe I misunderstood what you are asking. First step: Write node equations. 2nd: Create conductance matrix 3rd: Find coefficients It's only algebra but it does get confusing with the all the R's. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-09-16T19:15:32Z SecondChildTAG: Aw, thanks... I didn't understand correctly the video previous this exercise, about conductance matrix and I'll review it now. SecondChildUserIdTAG: 291362 SecondChildUserNameTAG: Gudson SecondChildCreateTimeTAG: 2012-09-16T20:30:17Z IndexTAG: 4225 TitleTAG: Problem getting polarities right using KVL When using KVL to analyze a circuit it seems that we can arbitrarily assign polarities to circuit elements and then use a rule such as "use the sign that is encountered first when traversing the loop." If you look at Figure 2.17 on page 62 of the text, the positive side of both circuit elements is at the top and the resulting calculation shows that v1 = v2. But what if I reversed the polarity of *one* of the elements? Then would not v1 = -v2? How do we reconcile this? UserIdTAG: 393930 UserNameTAG: tmciver CreateTimeTAG: 2012-09-16T18:25:07Z VoteTAG: 0 CoursewareTAG: Week 1 / Associated reference directions CommentableIdTAG: 6002x_S2E2_Associated_Reference_Directions NumberOfReplyTAG: 2 FirstChildTAG: any reversal has be done over the whole circuit perhaps? FirstChildUserIdTAG: 241170 FirstChildUserNameTAG: Mauette FirstChildCreateTimeTAG: 2012-09-16T18:48:11Z FirstChildTAG: Yes, you can define the polarities anyway you want. If you encounter that the value of a voltage is negative, that just means that the terminal you labeled as "+" has a lower voltage than the terminal labeled "-". If you find v1 = -v2, that just mean that with that polarity, the sign of v1 and v2 will be opposite. I hope this clarify things! FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-16T21:02:39Z SecondChildTAG: Jelizon: yes, I've been thinking about this and your explanation makes sense. Thanks! SecondChildUserIdTAG: 393930 SecondChildUserNameTAG: tmciver SecondChildCreateTimeTAG: 2012-09-16T23:03:14Z IndexTAG: 4226 TitleTAG: Lab 1 : I don't understand what is being asked Hi, I just spent an hour on this and it just feels like I'm misunderstanding something. Is the model for both have to be the same? even the R values? Cause I did it with the R values different and it told me I'm all good. but i understood that the whole circuit had to have the same values @___@ I'm just really confused. Please help me out. UserIdTAG: 151030 UserNameTAG: Eskamo CreateTimeTAG: 2012-09-16T18:07:17Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: The model should have same resistor values in the two schematics. FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-16T18:08:18Z SecondChildTAG: Thank you very much. Could you also please explain what "two-resistor voltage divider" means? Will there be a total of 3 or 2 resistors including the bulb? SecondChildUserIdTAG: 151030 SecondChildUserNameTAG: Eskamo SecondChildCreateTimeTAG: 2012-09-16T18:17:09Z SecondChildTAG: This had to be written somewhere in the beginning, and not in the *Hint* section below the Lab. Especially if putting different R values in the schematics can get a green checkmark after a **Check** SecondChildUserIdTAG: 187709 SecondChildUserNameTAG: ieroglif SecondChildCreateTimeTAG: 2012-09-16T18:19:25Z SecondChildTAG: "Two resistor voltage divider" refers to the two resistors R1 and R2. In the case of the bulb, there would be three total resistors. SecondChildUserIdTAG: 79361 SecondChildUserNameTAG: kimt SecondChildCreateTimeTAG: 2012-09-16T18:20:07Z SecondChildTAG: Thank you ^^ SecondChildUserIdTAG: 151030 SecondChildUserNameTAG: Eskamo SecondChildCreateTimeTAG: 2012-09-16T18:31:15Z SecondChildTAG: can you really get the circuit done ... i dont know how to connect :( SecondChildUserIdTAG: 133307 SecondChildUserNameTAG: Bharath_Kumar SecondChildCreateTimeTAG: 2012-09-16T18:32:34Z SecondChildTAG: stll trying to figure out lab 1 SecondChildUserIdTAG: 424965 SecondChildUserNameTAG: inteldana SecondChildCreateTimeTAG: 2012-09-16T23:32:47Z IndexTAG: 4227 TitleTAG: Symbol Mistake In tutorial IR1, IR2, IR3 are used for representation of voltage across resistors, so this mistake should be corrected... UserIdTAG: 426051 UserNameTAG: sslohana CreateTimeTAG: 2012-09-16T18:04:52Z VoteTAG: 0 CoursewareTAG: Week 1 / Three series resistors CommentableIdTAG: 6002x_three_series_resistors NumberOfReplyTAG: 1 FirstChildTAG: Remember that by Ohm's law a voltage across a resistor can be calculated as $V = I*R$. This equation is being used in the tutorial to represent the voltage across each resistor. FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-16T20:57:56Z SecondChildTAG: by prior knowledge, this could be wrong but (I == Input) (R == Resistance) (V == Volt) , hence the equation above *V* = *I* * *R* SecondChildUserIdTAG: 222572 SecondChildUserNameTAG: EmmanuelAppiah SecondChildCreateTimeTAG: 2012-09-17T02:01:57Z IndexTAG: 4228 TitleTAG: value of resistance R1 and R2 ????? Please any one like to clear me that value of resistance R1 and R2 should be from E12={10,12,15,18,22,27,33,39,47,56,68,82} ..... ? Em i getting right? UserIdTAG: 347789 UserNameTAG: engr_zohaib CreateTimeTAG: 2012-09-16T17:59:34Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Yep, values of the resistors must be from the set E12. FirstChildUserIdTAG: 95058 FirstChildUserNameTAG: KeshawP FirstChildCreateTimeTAG: 2012-09-16T18:10:33Z SecondChildTAG: Thanks KeshawP SecondChildUserIdTAG: 347789 SecondChildUserNameTAG: engr_zohaib SecondChildCreateTimeTAG: 2012-09-16T18:28:53Z FirstChildTAG: "For example, the available values of resistors with 10% tolerance are selections from the E12 set multiplied by a power of ten from $10^0$ through $10^5$." So it may be a multiple of the E12 type resistors listed {10,12,15,18,22,27,33,39,47,56,68,82} Find a resistor that is closest to your answer, within 10%. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-16T18:11:06Z SecondChildTAG: It means i can select even 10K 18K or 100K as well. ????? SecondChildUserIdTAG: 347789 SecondChildUserNameTAG: engr_zohaib SecondChildCreateTimeTAG: 2012-09-16T18:29:41Z SecondChildTAG: Yep. SecondChildUserIdTAG: 396446 SecondChildUserNameTAG: RousseauxS SecondChildCreateTimeTAG: 2012-09-16T18:48:14Z SecondChildTAG: Thanks RousseauxS ... SecondChildUserIdTAG: 347789 SecondChildUserNameTAG: engr_zohaib SecondChildCreateTimeTAG: 2012-09-16T18:52:38Z IndexTAG: 4229 TitleTAG: lab 2 plzz help me in lab 2 how do i get v1/2 & v2/6 UserIdTAG: 221736 UserNameTAG: samhita CreateTimeTAG: 2012-09-16T17:50:41Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I posted a series of hints for Lab 2 a couple of days ago. You can look for the post using the search tool of the discussion forum. The post name is: "Lab2 hints" FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-16T20:50:22Z IndexTAG: 4230 TitleTAG: The check button? I know the check button is used to submit your homework to be graded. My question is on week one homework there are 3 check buttons , one at the end of each section - does hitting anyone of these submit all the homework or just the section it is located in? UserIdTAG: 241170 UserNameTAG: Mauette CreateTimeTAG: 2012-09-16T17:41:36Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Just the section located within, as far as I am aware. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-16T17:43:45Z SecondChildTAG: well I guess I won't submit until I have answered all the questions on a piece of paper and then carefully entered them using * instead of x for multiplication and the like. Looks like its designed to be a "1" or "o" action. SecondChildUserIdTAG: 241170 SecondChildUserNameTAG: Mauette SecondChildCreateTimeTAG: 2012-09-16T17:46:15Z SecondChildTAG: Have no fear, you can check your homework as many times as you like. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-16T17:47:18Z SecondChildTAG: Many answers are simple integers between 1 to 10. If one can check any number of times then instead of actually solving students can arrive at correct solutions of many problems by inserting integers iteratively. That seems like a loophole for many problems. I wonder if that issue should be addressed by edx authorities. SecondChildUserIdTAG: 168495 SecondChildUserNameTAG: blueorb SecondChildCreateTimeTAG: 2012-09-16T18:53:01Z SecondChildTAG: I don't think it is enough to worry about. There really are not too many questions with answers between 1 and 10, in the first 3 weeks anyways. I don't think you can completely avoid low whole numbers unless they draw circuits with 52 nodes etc. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-16T18:58:51Z IndexTAG: 4231 TitleTAG: LATE? I wonder how strict the deadline is for the Assignments and Labs... cause while its already 1:46am where I am (I finished Lab1 at around 1:30am?), its only afternoon on the other side of the world. (My progress bar states that I have 100% in both labs and assignments even though the course info states that it should be done by midnight local time) UserIdTAG: 126153 UserNameTAG: JohannesL CreateTimeTAG: 2012-09-16T17:34:29Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I have heard that it is due by the end of the 16th in your local time. I have also heard that assignments will be accepted as long as it's 16th of September somewhere on Earth. I am leaning towards the latter. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-16T17:46:01Z SecondChildTAG: Deadline is according to boston. SecondChildUserIdTAG: 145503 SecondChildUserNameTAG: chirag3553 SecondChildCreateTimeTAG: 2012-09-16T18:36:19Z IndexTAG: 4232 TitleTAG: add a wire , ha i have problem of add a wire , i cleck on wire , but cant darg it ,what i do , please help me ,quickly , UserIdTAG: 133891 UserNameTAG: junaidkarim CreateTimeTAG: 2012-09-16T17:20:51Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Make sure you are clicking on the ends of wires or parts. If that does not work, try a different browser as the programers may or may not get your issue fixed in time, if in fact there is an issue, before your homework is due. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-16T17:26:21Z FirstChildTAG: Does it apply to all your wires? Cause I think some wires cant be dragged cause they're probably locked to allow the system a reference point to grade your answers. Nevertheless, if the reason is cause you have not enough space, you should be able to work around it easily by creating your extra circuit outside of the main drawing and connect them together by wires? (btw.. If in the event that you are working on LAB 1, there's more than enough space :) FirstChildUserIdTAG: 126153 FirstChildUserNameTAG: JohannesL FirstChildCreateTimeTAG: 2012-09-16T17:29:52Z SecondChildTAG: ha i doing lab1 , i but resistor , and i cleck end of wire too, but i cant drang it ,i am stuck what i do, SecondChildUserIdTAG: 133891 SecondChildUserNameTAG: junaidkarim SecondChildCreateTimeTAG: 2012-09-16T17:34:16Z SecondChildTAG: Can you place resistors, rotate elements(etc resistors), and create wires by clicking on the end of wires/parts? If so you will be able to do lab1, there is no need to shift any wires around. If you accidentally draw extra wires and are unable to move/delete them, try refreshing your browser? SecondChildUserIdTAG: 126153 SecondChildUserNameTAG: JohannesL SecondChildCreateTimeTAG: 2012-09-16T17:40:45Z SecondChildTAG: i know the ans, but i cont do it ,bez i cand work , plz come on yahoo, junaid_ali156@yahoo.com, SecondChildUserIdTAG: 133891 SecondChildUserNameTAG: junaidkarim SecondChildCreateTimeTAG: 2012-09-16T18:04:40Z IndexTAG: 4233 TitleTAG: Rotate resistor How can i rotate resistor to make it horizontal in sand tool? UserIdTAG: 309863 UserNameTAG: jithugeorge CreateTimeTAG: 2012-09-16T16:55:21Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Yes, after pulling the resistor in the working area press the key '**R**' on the keyboard to rotate it. FirstChildUserIdTAG: 95058 FirstChildUserNameTAG: KeshawP FirstChildCreateTimeTAG: 2012-09-16T16:58:53Z IndexTAG: 4234 TitleTAG: Lab 1 - submit Cannot see if my answer is correct to wrong. Nothing happens after pressing "CHECK" UserIdTAG: 132294 UserNameTAG: Clemelyn CreateTimeTAG: 2012-09-16T16:55:18Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Before submitting the result run the **DC button** on left corner and then go for **CHECK button**. FirstChildUserIdTAG: 95058 FirstChildUserNameTAG: KeshawP FirstChildCreateTimeTAG: 2012-09-16T17:05:11Z SecondChildTAG: Actually, I already finished the DC analysis and Voltage readings in red font appeared on their respective node but I still cannot see result after hitting the "CHECK" button SecondChildUserIdTAG: 132294 SecondChildUserNameTAG: Clemelyn SecondChildCreateTimeTAG: 2012-09-16T17:15:13Z FirstChildTAG: I ran the DC simulation and the voltages are correct - after I pressed **Check** both schematics had green checkmarks. But the **Save** button is gone now - is it OK this way? Or do I need to press **Save** to submit my lab? FirstChildUserIdTAG: 187709 FirstChildUserNameTAG: ieroglif FirstChildCreateTimeTAG: 2012-09-16T17:53:51Z SecondChildTAG: You are complete. You only need to save a work in progress. If you hit "check" and got a green check-mark, it is submitted. You can verify that it has been accepted by clicking on the "Progress" tab at the top of the page. Look for a bar above the particular Homework or Lab in question. Have fun. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-16T17:57:29Z IndexTAG: 4235 TitleTAG: plz help me , in lab 1 ha i have a problem of making connection connect , but i cant connect , ido now wht the reason , so plz if any on online hear , mgs me , junaid_ali156@yahoo.com , hear of my yahoo id , UserIdTAG: 133891 UserNameTAG: junaidkarim CreateTimeTAG: 2012-09-16T16:47:37Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Try a different browser. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-16T17:54:26Z IndexTAG: 4236 TitleTAG: Cannot see the numbers in the homework problems I see the following: You are given three resistors: two [Math Processing Error] resistors and one [Math Processing Error] resistor. What is the value in Ohms ( [Math Processing Error] ) of the largest-valued resistor that can be fabricated by combining these three resistors? I am using IE Version 8.0 UserIdTAG: 217722 UserNameTAG: prasar55 CreateTimeTAG: 2012-09-16T16:39:46Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Could you please try Chrome or Firefox? FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-16T17:19:23Z FirstChildTAG: Right click on the [Math Processing Error] and select a different format... It worked for me. FirstChildUserIdTAG: 386223 FirstChildUserNameTAG: mtburgler FirstChildCreateTimeTAG: 2012-09-16T17:55:38Z IndexTAG: 4237 TitleTAG: download video Please how can I download the videos lectures? UserIdTAG: 176407 UserNameTAG: DrHajara CreateTimeTAG: 2012-09-16T16:35:01Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Install a youtube downloader on your computer and download it from youtube. I cannot recommend a good program, you will have to find a suitable one for yourself. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-16T16:37:49Z SecondChildTAG: ha u online of not u are their please help me i have some problem , mgs me junaid_ali156@yahoo.com SecondChildUserIdTAG: 133891 SecondChildUserNameTAG: junaidkarim SecondChildCreateTimeTAG: 2012-09-16T17:06:34Z SecondChildTAG: and till response me SecondChildUserIdTAG: 133891 SecondChildUserNameTAG: junaidkarim SecondChildCreateTimeTAG: 2012-09-16T17:06:55Z SecondChildTAG: As I said above, I cannot recommend a program for downloading youtube videos, I have never used one before. It looks like KeshawP linked you a program below this post, you can try that out. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-16T17:49:40Z FirstChildTAG: Use **Internet Download Manager** (IDM) form the link given bellow.... http://www.internetdownloadmanager.com/download.html FirstChildUserIdTAG: 95058 FirstChildUserNameTAG: KeshawP FirstChildCreateTimeTAG: 2012-09-16T17:26:50Z FirstChildTAG: If this lectures are the same, as at mitx 6.002 spring course, I can download all videos and upload as torrent. Is anybody interested in it? FirstChildUserIdTAG: 95521 FirstChildUserNameTAG: Gatling FirstChildCreateTimeTAG: 2012-09-16T18:15:52Z SecondChildTAG: surely man...please do it.. SecondChildUserIdTAG: 343159 SecondChildUserNameTAG: prem_1101 SecondChildCreateTimeTAG: 2012-09-16T18:20:47Z SecondChildTAG: I have some troubles. Original jar file was removed from https://github.com/terriblybored/MITx-6002x-Video-Downloader Can anyone compile source code for me? SecondChildUserIdTAG: 95521 SecondChildUserNameTAG: Gatling SecondChildCreateTimeTAG: 2012-09-17T10:42:05Z SecondChildTAG: Sorry, it doesn't work for some reason... SecondChildUserIdTAG: 95521 SecondChildUserNameTAG: Gatling SecondChildCreateTimeTAG: 2012-09-17T17:13:46Z SecondChildTAG: A torrent would be great! Thanks! Even better, could staff provide torrents of the lectures, perhaps week-by-week? SecondChildUserIdTAG: 86632 SecondChildUserNameTAG: LCL SecondChildCreateTimeTAG: 2012-09-22T19:55:02Z IndexTAG: 4238 TitleTAG: LAB 1 I can't see voltage source! How do I solve this problem :( UserIdTAG: 271425 UserNameTAG: rinzo CreateTimeTAG: 2012-09-16T16:29:11Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Click on "DC" for a DC analysis. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-16T16:35:16Z SecondChildTAG: Or try a different web browser. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-16T16:36:12Z IndexTAG: 4239 TitleTAG: Confirmation E-mail Problem I am already in week 2 of the course work and doing my coursework,and I have started in the fall,2012.which was started 5th September as normal, I got the news from some of my friends that edx had sent a confirmation email to all just before starting of the course. But I did not get any such email yet.. Is that thing will create any problem to get grade or certificate in near future or hamper any important stuff during the course? Please help me with some relevant information about it.... UserIdTAG: 295278 UserNameTAG: souravfn7 CreateTimeTAG: 2012-09-16T16:20:40Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 1 FirstChildTAG: Check your junk folder. I think it was just a courtesy reminder that the course is starting. I do not think it required any action. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-16T16:33:12Z IndexTAG: 4240 TitleTAG: Calculating i1 with superposition I know that by principle the current y2 will come as x2/R1 but I am making some mistake if I try to calculate y2 from scratch after shorting V1. Since, V1 is shorted we have R1||R3 so the current i2 in this scenario comes as i2 = V2*(R1+R3)/(R1R2+R2R3+R3R1) so by applying current divider for y2 the equation should be y2= i2*R1/(R1+R3) According to this the answer I get is 0.789Amps which is not correct. Can someone explain where I am making the mistake? UserIdTAG: 335370 UserNameTAG: SumitN CreateTimeTAG: 2012-09-16T15:59:32Z VoteTAG: 0 CoursewareTAG: Week 2 / Superposition example CommentableIdTAG: 6002x_S3E3_superposition NumberOfReplyTAG: 2 FirstChildTAG: You have calculated i2 right. But when you are calculating y2,you have written wrong expression. The expression you have written will calculate current through R3 register but in question it is asked to find current through R1 register when V2 acting alone.. So the correct expression is y2=i2*R3/(R1+R3)=0.563, Try it.. FirstChildUserIdTAG: 369153 FirstChildUserNameTAG: supratip09 FirstChildCreateTimeTAG: 2012-09-16T20:22:41Z FirstChildTAG: This task uses the results of S3E2 and if you calculated S3E2 - you can beat this task even without calculator. I typed y2=(1/7)*(1/3)/(1/7+1/3+1/5)*8 and it just fit all right. FirstChildUserIdTAG: 446597 FirstChildUserNameTAG: garry_crannon FirstChildCreateTimeTAG: 2012-09-17T08:53:20Z IndexTAG: 4241 TitleTAG: LAb 1 issue how to make resistor horizontal?? I mean if I wanna put resistor horizontally, how to do it?? UserIdTAG: 351425 UserNameTAG: brain2boom CreateTimeTAG: 2012-09-16T15:55:19Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Type "r" : https://www.edx.org/static/content-mit-6002x/handouts/schematic_tutorial.ba422f80d72b.pdf FirstChildUserIdTAG: 396446 FirstChildUserNameTAG: RousseauxS FirstChildCreateTimeTAG: 2012-09-16T16:11:35Z SecondChildTAG: just press R to change horizontal SecondChildUserIdTAG: 286165 SecondChildUserNameTAG: nchibwe SecondChildCreateTimeTAG: 2012-09-16T16:23:54Z FirstChildTAG: Press R on your keyboard to rotate FirstChildUserIdTAG: 374010 FirstChildUserNameTAG: franckpager23 FirstChildCreateTimeTAG: 2012-09-16T16:12:01Z IndexTAG: 4242 TitleTAG: How to rotate a device in lab experiment? I want to put a resistance horizontal instead of vertical, how can I do? Thanks UserIdTAG: 298314 UserNameTAG: js93082011 CreateTimeTAG: 2012-09-16T15:46:16Z VoteTAG: 0 CoursewareTAG: Week 1 / Node analysis CommentableIdTAG: 6002x_node_analysis NumberOfReplyTAG: 1 FirstChildTAG: click R button FirstChildUserIdTAG: 150641 FirstChildUserNameTAG: DungTienTran FirstChildCreateTimeTAG: 2012-09-16T15:48:14Z SecondChildTAG: thanks ........ alot SecondChildUserIdTAG: 415376 SecondChildUserNameTAG: imab90 SecondChildCreateTimeTAG: 2012-09-16T16:33:08Z SecondChildTAG: Hi, I tried to press R key in upper case and in lower case after I move some element to working area but it does not rotate. SecondChildUserIdTAG: 376572 SecondChildUserNameTAG: sergy_kan SecondChildCreateTimeTAG: 2012-10-02T00:34:08Z IndexTAG: 4243 TitleTAG: What is the peak power (in Watts) dissipated by the resistor? the answer is 262V. How?? UserIdTAG: 315278 UserNameTAG: Dexon CreateTimeTAG: 2012-09-16T15:44:18Z VoteTAG: 0 CoursewareTAG: Week 1 / AC power CommentableIdTAG: 6002x_S1E3_AC_power NumberOfReplyTAG: 4 FirstChildTAG: I don't know either. FirstChildUserIdTAG: 150641 FirstChildUserNameTAG: DungTienTran FirstChildCreateTimeTAG: 2012-09-16T16:10:02Z SecondChildTAG: Strange thing. I just computed it with a formula P=V^2/R and I got 259.64 But probably it is an incorrect way to computing. SecondChildUserIdTAG: 397712 SecondChildUserNameTAG: JohnBartko SecondChildCreateTimeTAG: 2012-09-16T16:47:02Z SecondChildTAG: We're trying to find peak power (P) in watts which will be at peak voltage. I = V/R. P = VI. So P = (V^2)/R. V = 120*sqrt(2)*cos(2pi60t) take t=0, for example. (that's one of the times when voltage peaks) so then cos(2pi60t) = cos(2pi60(0)) = cos(0) = 1. So V = 120*sqrt(2)*1. P = (V^2)/R, R = 110.0 P = ((120*sqrt(2))^2)/110.0 = 261.8 watts. I hope that makes sense. SecondChildUserIdTAG: 135369 SecondChildUserNameTAG: pskagen SecondChildCreateTimeTAG: 2012-09-16T20:34:03Z SecondChildTAG: pskagen, great explanation. Thanks SecondChildUserIdTAG: 8840 SecondChildUserNameTAG: MWaddell SecondChildCreateTimeTAG: 2012-09-19T14:40:31Z SecondChildTAG: for the peak value,we have to put the maximum value of cos,which is 1.putting the value of cos in v(t),after squaring it and dividing by resistance we get the result SecondChildUserIdTAG: 451865 SecondChildUserNameTAG: kumar51 SecondChildCreateTimeTAG: 2012-09-20T15:46:12Z SecondChildTAG: Mannnn! Thanks a lot! SecondChildUserIdTAG: 464647 SecondChildUserNameTAG: Carlosrolando11 SecondChildCreateTimeTAG: 2012-09-25T00:00:50Z FirstChildTAG: Power is a V^2/R, so power would be maximum, when V would be maximum. V is maximum, when cos() in V equation is maximum. So, maximum power would be, for example, when t = 0. FirstChildUserIdTAG: 322444 FirstChildUserNameTAG: koluch FirstChildCreateTimeTAG: 2012-09-16T16:49:31Z FirstChildTAG: The specific numbers in a students homework or labs may be different from person to person. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-09-16T17:10:12Z SecondChildTAG: I first calculated the power dissipated by resistors...130.9 From there I did this equation: 130.9 X 1.414 (AVERAGE power times 1.414 equals peak power)X 0.707(which equals average peak time)x 2 = 261.720936 SecondChildUserIdTAG: 229419 SecondChildUserNameTAG: NNieves SecondChildCreateTimeTAG: 2012-09-16T17:49:21Z FirstChildTAG: I first calculated the power dissipated by resistors...130.9 From there I did this equation: 130.9 X 1.414 (AVERAGE power times 1.414 equals peak power)X 0.707(which equals average peak time)x 2 = 261.720936 FirstChildUserIdTAG: 229419 FirstChildUserNameTAG: NNieves FirstChildCreateTimeTAG: 2012-09-16T17:49:39Z IndexTAG: 4244 TitleTAG: Confused question I feel confused about the source and battery, i will read the txt book again to understand. Could somebody can explain clearly? UserIdTAG: 146995 UserNameTAG: ZYJ CreateTimeTAG: 2012-09-16T15:40:05Z VoteTAG: 0 CoursewareTAG: Week 1 / Battery model CommentableIdTAG: 6002x_S1E9_Battery_Model NumberOfReplyTAG: 1 FirstChildTAG: what do you specific want to know? FirstChildUserIdTAG: 274475 FirstChildUserNameTAG: Konredus FirstChildCreateTimeTAG: 2012-09-16T16:12:58Z IndexTAG: 4245 TitleTAG: H1P3 How to count H1 and H2.I'm using Pt=V^2/Rt. and i got Rt=102.42.? im stuck at here? anyone can help? UserIdTAG: 396953 UserNameTAG: raisfikri18 CreateTimeTAG: 2012-09-16T15:00:47Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: check the sign. FirstChildUserIdTAG: 274475 FirstChildUserNameTAG: Konredus FirstChildCreateTimeTAG: 2012-09-16T16:15:42Z SecondChildTAG: ok so total resistance is 18.11 , there fore if H1=H2=H3=H then H =18.11/3= 6.03 ohms. P=I2R , my answer is still wrong . Please can some one help. SecondChildUserIdTAG: 178115 SecondChildUserNameTAG: nehamakhija SecondChildCreateTimeTAG: 2012-09-16T18:06:47Z SecondChildTAG: please help me SecondChildUserIdTAG: 178115 SecondChildUserNameTAG: nehamakhija SecondChildCreateTimeTAG: 2012-09-16T18:06:56Z IndexTAG: 4246 TitleTAG: H1P3 Hello, I have calculated the total resistance to be only 18.64R based on 240^2/3090W. This works ok for working out the currents of part 1 and 2 but not for part 3 (working out H1 resistance, then power) based on this. I have calculated parallel resistance and power for part 4, but am stuck for part 3. Anyone know where I'm going wrong? UserIdTAG: 195916 UserNameTAG: pflynn CreateTimeTAG: 2012-09-16T14:54:32Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: check the sign. FirstChildUserIdTAG: 274475 FirstChildUserNameTAG: Konredus FirstChildCreateTimeTAG: 2012-09-16T16:16:16Z SecondChildTAG: I can't give you the answer but hope my logic will help.(it helped me after 3 hours struggling at 3,4&5th question). If you have done the 4th part and found out the parallel resistance,compare it with the original 1040 W walue, what do you see? How many times H(parallel) is smaller than 1040W? Hint: Note that the value of H(p) is in parallel and has 2 R's...and is at a certain rate higher than the original value (1040).Find the connection and after one quick multiplication you're done. H1>H2(or H3) SecondChildUserIdTAG: 376101 SecondChildUserNameTAG: MirceaMihai SecondChildCreateTimeTAG: 2012-09-16T17:01:23Z SecondChildTAG: ok so total resistance is 18.11 , there fore if H1=H2=H3=H then H =18.11/3= 6.03 ohms. P=I2R , my answer is still wrong . Please can some one help. SecondChildUserIdTAG: 178115 SecondChildUserNameTAG: nehamakhija SecondChildCreateTimeTAG: 2012-09-16T18:07:19Z IndexTAG: 4247 TitleTAG: About these all probems. Guys can anybody explain me about all these problems, serially. Seems like i'm confused more and more. I'm keep getting wrong answers often. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-16T14:42:36Z VoteTAG: 0 CoursewareTAG: Week 1 / Battery model CommentableIdTAG: 6002x_S1E9_Battery_Model NumberOfReplyTAG: 1 FirstChildTAG: me too :( FirstChildUserIdTAG: 315278 FirstChildUserNameTAG: Dexon FirstChildCreateTimeTAG: 2012-09-16T16:51:59Z SecondChildTAG: calculate the voltage across each resistor (.32+.2) and (.25+.2) from each voltage source with the other shorted. then add the two voltages. that will give your the first solution. for 2-3, the answer expected takes off the internal .2ohms - which some of us thought odd. you can assume all current from each source goes through only the resistor in series because of the external short. prob 3: - is the "network resistance" viewed from the terminal. prob 4: - again kirchoff's rule applies - assume open circuit SecondChildUserIdTAG: 251793 SecondChildUserNameTAG: chellaton SecondChildCreateTimeTAG: 2012-09-20T14:54:46Z IndexTAG: 4248 TitleTAG: Error How to send my homework because the save button is not appear on the website ? UserIdTAG: 319792 UserNameTAG: chou25alg CreateTimeTAG: 2012-09-16T14:38:41Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Just hit "Check", your homework is then saved. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-16T15:37:42Z SecondChildTAG: thanks SecondChildUserIdTAG: 404572 SecondChildUserNameTAG: raj_ie24 SecondChildCreateTimeTAG: 2012-09-16T15:38:24Z SecondChildTAG: if i edit my answers after checking it once does it saves the changes made??? SecondChildUserIdTAG: 404572 SecondChildUserNameTAG: raj_ie24 SecondChildCreateTimeTAG: 2012-09-16T15:39:12Z SecondChildTAG: yes ,when you click on the check button again. SecondChildUserIdTAG: 274475 SecondChildUserNameTAG: Konredus SecondChildCreateTimeTAG: 2012-09-16T16:18:50Z SecondChildTAG: thanks once again... SecondChildUserIdTAG: 404572 SecondChildUserNameTAG: raj_ie24 SecondChildCreateTimeTAG: 2012-09-16T16:30:40Z SecondChildTAG: thank you very much. SecondChildUserIdTAG: 319792 SecondChildUserNameTAG: chou25alg SecondChildCreateTimeTAG: 2012-09-16T17:13:54Z IndexTAG: 4249 TitleTAG: Lab1 Problem. if the bulb is not connected then the voltage at the output will same as input due to the open circuit terminology,, then how can we get less than 2 volts if the battery remains same...??? can anY one ... https://edxuploads.s3.amazonaws.com/1347806157250728.jpg UserIdTAG: 294630 UserNameTAG: babarumt CreateTimeTAG: 2012-09-16T14:36:23Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: read hint at the foot of the Lab page FirstChildUserIdTAG: 188160 FirstChildUserNameTAG: lofe FirstChildCreateTimeTAG: 2012-09-16T15:38:14Z FirstChildTAG: The problem asks you to do a voltage divider, and the tip on the exercise is to use a resistor in parallel(R2) with the Bulb resistor (Rbulb) so when you disconect the bulb from the circuit, R2 won't let the circuit open. Hope that it answered your question Sorry for the bad english. FirstChildUserIdTAG: 122637 FirstChildUserNameTAG: Caetano FirstChildCreateTimeTAG: 2012-09-16T15:41:16Z FirstChildTAG: The voltage across the bulb decreases because initially there was no current.But when the bulb is connected then a current flows.But current flows only when there is a voltage drop(water dropped from a height is good example-difference in potential energy).Hence as the current flows the voltage drops.(mechanical analogy-potential energy is converted to kinetic energy so potential energy is decreased because energy is fixed. FirstChildUserIdTAG: 1825 FirstChildUserNameTAG: shaik1990jaffer FirstChildCreateTimeTAG: 2012-09-16T15:52:15Z IndexTAG: 4250 TitleTAG: Error in transcript In the video, at 2:05 he says "node a", but the transcript has "node eight".Please correct. UserIdTAG: 86810 UserNameTAG: Ariel12 CreateTimeTAG: 2012-09-16T14:20:56Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL and KCL CommentableIdTAG: 6002x_kvl_and_kcl NumberOfReplyTAG: 0 IndexTAG: 4251 TitleTAG: :) tnx UserIdTAG: 407512 UserNameTAG: bjrebollos27 CreateTimeTAG: 2012-09-16T14:15:56Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL and KCL exercises CommentableIdTAG: 6002x_kvl_kcl_exercises NumberOfReplyTAG: 0 IndexTAG: 4252 TitleTAG: independent Kcl !! someone plz explain to me what independent kcl means ?!! UserIdTAG: 123616 UserNameTAG: fettah901 CreateTimeTAG: 2012-09-16T14:15:11Z VoteTAG: 0 CoursewareTAG: Week 1 / Circuit topology CommentableIdTAG: 6002x_S2E1_Circuit_Topology NumberOfReplyTAG: 2 FirstChildTAG: Ok, this is what did it for me from a conceptual standpoint (I don't "think" in math very well). Let's start with the loops. Look at the circuit. Your first thought when asked the question, "How many loops are there?", was probably 3 -- abc, adb, and bdc. (That was my first thought, anyway.) Well, although there are other loops, you only need three loops to have "touched" all the branches in the circuit, so you only need three KVL equations to have voltage equations for all the branches in the circuit. You can do the same thing with the currents. Look at each node. Make a list of the branches in the circuit and, one-by-one, check off the branches touched by each node. By the time you've gotten to the last node, you'll have found that all the branches have already been touched by the other nodes. So, you don't need to use the equation for every single node to have an equation for every current. I can add pics of a marked-up circuit if that would help. FirstChildUserIdTAG: 154526 FirstChildUserNameTAG: silentquasar FirstChildCreateTimeTAG: 2012-09-17T02:36:24Z SecondChildTAG: That's a useful way to look at the problem and get the result. Thanks! SecondChildUserIdTAG: 434948 SecondChildUserNameTAG: StefanoDB SecondChildCreateTimeTAG: 2012-09-17T19:52:22Z FirstChildTAG: A set of independent equations is that, in which any of the equations cannot be derived from the others. FirstChildUserIdTAG: 370573 FirstChildUserNameTAG: pberkovich FirstChildCreateTimeTAG: 2012-09-16T14:29:39Z SecondChildTAG: One way to check this is to set the equations in a matrix and do row reduction. I'm not sure if there is a more clever way to determine this with physical intuition, but if the reduced row echelon form has a row of zeroes, the equation corresponding to that row can be thrown out. SecondChildUserIdTAG: 267702 SecondChildUserNameTAG: curio_city SecondChildCreateTimeTAG: 2012-09-16T16:01:44Z IndexTAG: 4253 TitleTAG: madness i can't believe, i was one hour trying to resolve this. First of all a basic error with least common multiple!!. and then, the voltage source and its sign, such a damn trick. UserIdTAG: 149549 UserNameTAG: Java_Dido CreateTimeTAG: 2012-09-16T14:07:55Z VoteTAG: 0 CoursewareTAG: Week 2 / Node equation review CommentableIdTAG: 6002x_S3E1_NodeEquationReview NumberOfReplyTAG: 0 IndexTAG: 4254 TitleTAG: Lab 1 I'm wondering for lab 1 part 2 when the bulb is disconnected is it treated as an open circuit or a short cicruit? Thanks UserIdTAG: 151991 UserNameTAG: dgpking CreateTimeTAG: 2012-09-16T14:03:17Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: When the bulb is removed, it is treated as an open circuit. FirstChildUserIdTAG: 331664 FirstChildUserNameTAG: dwmnctrh3 FirstChildCreateTimeTAG: 2012-09-16T14:04:43Z FirstChildTAG: It is treated as an open circuit FirstChildUserIdTAG: 374477 FirstChildUserNameTAG: Nauman12345 FirstChildCreateTimeTAG: 2012-09-23T19:42:29Z IndexTAG: 4255 TitleTAG: cube form let us assume that we are having cube and resistors are arranged in such manner that on every side of the cube we are having resistance R, find equivalent resistance across corners. UserIdTAG: 73754 UserNameTAG: fadi99 CreateTimeTAG: 2012-09-16T13:57:05Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: May be 5*R/6 FirstChildUserIdTAG: 402617 FirstChildUserNameTAG: Vl FirstChildCreateTimeTAG: 2012-09-16T14:19:50Z IndexTAG: 4256 TitleTAG: Hi! Can anybody help me with explanation on how I can get to practice each sequences? UserIdTAG: 430895 UserNameTAG: Mahmud41 CreateTimeTAG: 2012-09-16T13:15:13Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Can you be more specific? The sequences do have practice exercises mixed into them. Since they are practice exercises, completion is optional. Have fun. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-16T14:44:10Z IndexTAG: 4257 TitleTAG: Invalid input: Could not parse '1.666R' as a formula What Does this Mean,,,>>?? UserIdTAG: 294630 UserNameTAG: babarumt CreateTimeTAG: 2012-09-16T12:46:33Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: try this: (5/3)*R FirstChildUserIdTAG: 443968 FirstChildUserNameTAG: shobhitsajwan FirstChildCreateTimeTAG: 2012-09-16T12:51:26Z FirstChildTAG: Just write the whole formula, not an answer FirstChildUserIdTAG: 444204 FirstChildUserNameTAG: hekan FirstChildCreateTimeTAG: 2012-09-16T12:58:51Z SecondChildTAG: YuP,,, I have tried the whole formula,,, and also (5/3)R... But same error again and again and again and,,,, SecondChildUserIdTAG: 294630 SecondChildUserNameTAG: babarumt SecondChildCreateTimeTAG: 2012-09-16T13:33:08Z SecondChildTAG: JUST WRITE 5R/3 SecondChildUserIdTAG: 441979 SecondChildUserNameTAG: shrutimehrotra SecondChildCreateTimeTAG: 2012-09-16T13:35:15Z SecondChildTAG: You're missing an asterisk for the multiplication. Instead of (5/3)R, type (5/3)*R SecondChildUserIdTAG: 79361 SecondChildUserNameTAG: kimt SecondChildCreateTimeTAG: 2012-09-16T13:36:16Z FirstChildTAG: In the button "syntax" of this flash's application, [http://www.colegio cientifico.con/funciones.swf][1] you can find the rules to use math equations [1]: http://www.colegiocientifico.com/funciones.swf FirstChildUserIdTAG: 183712 FirstChildUserNameTAG: colegiocientifico FirstChildCreateTimeTAG: 2012-09-16T14:26:00Z IndexTAG: 4258 TitleTAG: fb group created frns!!! i created a open group in fb "MITX12-CIRCUITS".al r welcome.... UserIdTAG: 356056 UserNameTAG: Ganesh2810 CreateTimeTAG: 2012-09-16T12:45:48Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: hey iam not able to find plz provide the link.. FirstChildUserIdTAG: 161899 FirstChildUserNameTAG: harshvit12 FirstChildCreateTimeTAG: 2012-09-16T13:00:09Z SecondChildTAG: me too cannot find any! SecondChildUserIdTAG: 151205 SecondChildUserNameTAG: jesher777 SecondChildCreateTimeTAG: 2012-09-16T13:03:21Z FirstChildTAG: https://www.facebook.com/groups/mitxfallsemester join this one for 6.002x help...... FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-09-16T16:00:58Z IndexTAG: 4259 TitleTAG: About the LAb1? I have figured out the lab1. But I don't modify the circuit of lab1? How can change the resistors' label? Please help me. UserIdTAG: 371617 UserNameTAG: rainbow12 CreateTimeTAG: 2012-09-16T12:38:39Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL and KCL CommentableIdTAG: 6002x_kvl_and_kcl NumberOfReplyTAG: 1 FirstChildTAG: double click on the resistor. A small dialog box will open for changing values FirstChildUserIdTAG: 443968 FirstChildUserNameTAG: shobhitsajwan FirstChildCreateTimeTAG: 2012-09-16T12:44:34Z IndexTAG: 4260 TitleTAG: about oscillator, need help from instructors or any body who knows? is it possible to make a high power high frequency (in MHz) oscillator (inverter)?i mean Is switching device is available that can lead me to MHz with high power? please reply asap. UserIdTAG: 107972 UserNameTAG: nice CreateTimeTAG: 2012-09-16T12:26:55Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Just use your stereo with a function generator feeding MHz signal with reverse polarity. I quite often use my stereo as an impromptu signal generator for measuring the inductance of air core coils. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-16T14:51:26Z IndexTAG: 4261 TitleTAG: About the lab? How can I change the resistors' label? UserIdTAG: 371617 UserNameTAG: rainbow12 CreateTimeTAG: 2012-09-16T12:26:46Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Click mouse on resistor icon and in windows on upper string insert/update new label. FirstChildUserIdTAG: 402617 FirstChildUserNameTAG: Vl FirstChildCreateTimeTAG: 2012-09-16T12:38:12Z SecondChildTAG: DOUBLE CLICK THE RESISTOR A LABEL(BOX) WILL APPEAR WHERE U INPUT YOUR VALUE. SecondChildUserIdTAG: 338327 SecondChildUserNameTAG: abbasufi SecondChildCreateTimeTAG: 2012-09-16T14:16:35Z IndexTAG: 4262 TitleTAG: Electronics Internship ! Hello ! Dis is Channpreet frm India ! Well am a 2nd yr student of Electronics n Communication! Lookin for an internship job for 2 months in Singapore/Malysia/Hong Kong/china/Korea after my 2nd year ! Thnx :) UserIdTAG: 119311 UserNameTAG: channpreet_695 CreateTimeTAG: 2012-09-16T12:11:57Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 4263 TitleTAG: H2P2 - Please help 2nd question... Power delivered to the ideal load? Please help me.. How to do that? Thanks!! UserIdTAG: 199839 UserNameTAG: vijaykrishnankk CreateTimeTAG: 2012-09-16T12:11:32Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: I tried many times, its showing error. FirstChildUserIdTAG: 199839 FirstChildUserNameTAG: vijaykrishnankk FirstChildCreateTimeTAG: 2012-09-16T12:11:56Z FirstChildTAG: Power delivered to the ideal load is equal power on thevenin resistor. In this case, load resistor must be equal Thevenin resistor. FirstChildUserIdTAG: 402617 FirstChildUserNameTAG: Vl FirstChildCreateTimeTAG: 2012-09-16T12:44:13Z SecondChildTAG: I am having the same problem. I tried what you said but it is still wrong SecondChildUserIdTAG: 304587 SecondChildUserNameTAG: Vasso SecondChildCreateTimeTAG: 2012-09-16T13:05:15Z SecondChildTAG: R-OTH = RP + 2*RS V-OTH = I * RP P-MAX = V-OTH ^ 2 / 4 / R-OTH I calculate and write above, my answer was correcting SecondChildUserIdTAG: 402617 SecondChildUserNameTAG: Vl SecondChildCreateTimeTAG: 2012-09-16T13:20:47Z SecondChildTAG: R-OTH = RP + 2*RS; ---------- V-OTH = I * RP; ---------- P-MAX = V-OTH ^ 2 / 4 / R-OTH; ---------- Correctly formated SecondChildUserIdTAG: 402617 SecondChildUserNameTAG: Vl SecondChildCreateTimeTAG: 2012-09-16T13:23:09Z SecondChildTAG: Hi VI I have already tried what you have presented, however, it does not give the right answer as it is clearly indicated by error X. Have you tried something else? did you get right signal though? let me know please. SecondChildUserIdTAG: 241062 SecondChildUserNameTAG: Al-Temimy SecondChildCreateTimeTAG: 2012-09-16T16:36:44Z SecondChildTAG: It was a matter of accuracy in digits for me thanx :) SecondChildUserIdTAG: 304587 SecondChildUserNameTAG: Vasso SecondChildCreateTimeTAG: 2012-09-16T20:29:19Z SecondChildTAG: Hello ! Please clarify in details for me : Why P-MAX = V-OTH ^ 2 / 4 / R-OTH ? Regards, SecondChildUserIdTAG: 194717 SecondChildUserNameTAG: kaa SecondChildCreateTimeTAG: 2012-09-19T12:43:37Z FirstChildTAG: Thanks. Thank you so much. I got the answer. I think it was a matter of precision. FirstChildUserIdTAG: 199839 FirstChildUserNameTAG: vijaykrishnankk FirstChildCreateTimeTAG: 2012-09-17T05:21:47Z IndexTAG: 4264 TitleTAG: Last question H1P1 Hi I'm trying to solve this question to complete, at least, H1P1 and i didn't find the way. I'm having serious troubles to find equations and tools to solve this question and previous, so i would appreciate any help. Thanks UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-16T11:58:36Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Me too. It sounds simple but I can't get my head around it. Anyone? FirstChildUserIdTAG: 195916 FirstChildUserNameTAG: pflynn FirstChildCreateTimeTAG: 2012-09-16T12:49:26Z SecondChildTAG: hi....it is just the series,parallel resistive networks....for series its like R1+R2+...+Rn,n for parallel R1R2...Rn/R1+R2+...+Rn SecondChildUserIdTAG: 414118 SecondChildUserNameTAG: raje93 SecondChildCreateTimeTAG: 2012-09-16T13:46:16Z SecondChildTAG: Given that each individual resistor can dissipate up to 1 watt of power before burning up, how much total power in watts ( W ) can the smallest-valued composite resistor dissipate before burning up? heip plz..... SecondChildUserIdTAG: 286106 SecondChildUserNameTAG: Saqib367 SecondChildCreateTimeTAG: 2012-09-16T15:16:39Z FirstChildTAG: Hi! Its all about P=V^2/R. So its given each resistor dissipates 1W power, so using above equation we can get the value of voltage through each resistors. so that we can find out separate & total current through that circuit. That"s all we can calculate the total power of that circuit using P=IV equation. FirstChildUserIdTAG: 389433 FirstChildUserNameTAG: Deepica FirstChildCreateTimeTAG: 2012-09-16T14:08:14Z SecondChildTAG: Thanks for your answer, I have calculated separately each of the three currents and voltages. But I am unable to find total voltages and currents to solve the problem. SecondChildUserIdTAG: 334890 SecondChildUserNameTAG: DAntequera SecondChildCreateTimeTAG: 2012-09-16T15:48:46Z FirstChildTAG: Given that each individual resistor can dissipate up to 1 watt of power before burning up, how much total power in watts ( W ) can the smallest-valued composite resistor dissipate before burning up? FirstChildUserIdTAG: 286106 FirstChildUserNameTAG: Saqib367 FirstChildCreateTimeTAG: 2012-09-16T15:16:53Z IndexTAG: 4265 TitleTAG: mine full name is coming wrong how can i make it correct how can i make my full name correct.. and put my correct mailling adress UserIdTAG: 392009 UserNameTAG: Umaishmed CreateTimeTAG: 2012-09-16T11:46:02Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi Umaishmed! I suggest to e-mail the edX Staff for your concern: ---------- HELP EMAIL System-related questions: technical@edx.org Content-related questions: content@edx.org Bug reports: bugs@edx.org Suggestions: suggestions@edx.org ---------- FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-16T15:26:51Z IndexTAG: 4266 TitleTAG: H1P3 Last 3 questions OK I have done my Calculations. I calculated that H1 dissipates 160W of power and the H2 and H3 dissipate 80W of power each which really does explain why Joe was getting cold however the system says that my answers are wrong. Why? UserIdTAG: 256356 UserNameTAG: WesleyHanes CreateTimeTAG: 2012-09-16T11:42:52Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 7 FirstChildTAG: I am also not getting Ra,,, Please help me if you got it. FirstChildUserIdTAG: 106219 FirstChildUserNameTAG: veereshpatil FirstChildCreateTimeTAG: 2012-09-16T12:19:31Z SecondChildTAG: i dont have any idea like how to proceed with the last 3 questions of H1P3....can anybody help me out...??? thanks. SecondChildUserIdTAG: 414118 SecondChildUserNameTAG: raje93 SecondChildCreateTimeTAG: 2012-09-16T12:35:20Z SecondChildTAG: H1 ,H2 and H3 urz all are wrong...try n solve it again SecondChildUserIdTAG: 130834 SecondChildUserNameTAG: Shawn07 SecondChildCreateTimeTAG: 2012-09-16T15:27:00Z SecondChildTAG: ok so total resistance is 18.11 , there fore if H1=H2=H3=H then H =18.11/3= 6.03 ohms. P=I2R , my answer is still wrong . Please can some one help. SecondChildUserIdTAG: 178115 SecondChildUserNameTAG: nehamakhija SecondChildCreateTimeTAG: 2012-09-16T18:07:30Z FirstChildTAG: i dont have any idea like how to proceed with the last 3 questions of H1P3....can anybody help me out...??? thanks. FirstChildUserIdTAG: 414118 FirstChildUserNameTAG: raje93 FirstChildCreateTimeTAG: 2012-09-16T12:39:20Z FirstChildTAG: I calculated the parallel combo of H2 and H3, then used the power equation to give a correct answer. I can't however calculate the power output of H1! FirstChildUserIdTAG: 195916 FirstChildUserNameTAG: pflynn FirstChildCreateTimeTAG: 2012-09-16T12:54:29Z SecondChildTAG: hi...can u please tell tat how did u go about to find the power across H2?? thanks SecondChildUserIdTAG: 414118 SecondChildUserNameTAG: raje93 SecondChildCreateTimeTAG: 2012-09-16T13:51:07Z SecondChildTAG: H2*4=H1 dissipated power SecondChildUserIdTAG: 376101 SecondChildUserNameTAG: MirceaMihai SecondChildCreateTimeTAG: 2012-09-16T21:08:08Z FirstChildTAG: Well your calculations are not correct,thats why! if you have done the first two questions correctly it doesn't seem you have a problem with the concept.just revisit your calculations section.cheers FirstChildUserIdTAG: 218653 FirstChildUserNameTAG: jaz07 FirstChildCreateTimeTAG: 2012-09-16T13:02:03Z SecondChildTAG: wow thanks that is a really helpful answer :/ SecondChildUserIdTAG: 256356 SecondChildUserNameTAG: WesleyHanes SecondChildCreateTimeTAG: 2012-09-16T13:54:38Z SecondChildTAG: OK I see where I went wrong :) SecondChildUserIdTAG: 256356 SecondChildUserNameTAG: WesleyHanes SecondChildCreateTimeTAG: 2012-09-16T14:32:27Z FirstChildTAG: Hi, I'm stuck with this problem, could somebody help me ?? FirstChildUserIdTAG: 443442 FirstChildUserNameTAG: Nelutu FirstChildCreateTimeTAG: 2012-09-16T14:38:13Z FirstChildTAG: I got it , I have the right answers, thx. FirstChildUserIdTAG: 443442 FirstChildUserNameTAG: Nelutu FirstChildCreateTimeTAG: 2012-09-16T15:22:31Z SecondChildTAG: Hey can you help me out with the last three questions SecondChildUserIdTAG: 178115 SecondChildUserNameTAG: nehamakhija SecondChildCreateTimeTAG: 2012-09-16T16:41:59Z SecondChildTAG: hello, hint me about hw to find power through h1 h2 h3 SecondChildUserIdTAG: 444708 SecondChildUserNameTAG: jagadeeshv16 SecondChildCreateTimeTAG: 2012-09-16T17:37:01Z SecondChildTAG: ok so total resistance is 18.11 , there fore if H1=H2=H3=H then H =18.11/3= 6.03 ohms. P=I2R , my answer is still wrong . Please can some one help. SecondChildUserIdTAG: 178115 SecondChildUserNameTAG: nehamakhija SecondChildCreateTimeTAG: 2012-09-16T18:08:30Z SecondChildTAG: I just have 30 mins please help me SecondChildUserIdTAG: 178115 SecondChildUserNameTAG: nehamakhija SecondChildCreateTimeTAG: 2012-09-16T18:09:17Z FirstChildTAG: Because your results are wrong. H1 dissipated 537.824 w H2 or H3 dissipated 134.456 w the total power is 806.7376 w. You have to review your calculations. FirstChildUserIdTAG: 8345 FirstChildUserNameTAG: geronimo1 FirstChildCreateTimeTAG: 2012-09-17T22:43:30Z IndexTAG: 4267 TitleTAG: how to study?????? do we need to go through the whole weeks videos and understand the subject and solve the homework...?? or will there be everyday classes where we can ask doubts on the spot to the faculty... ?? i m confused... i m new to the course... kindly help me.... how do u guys clear your doubts and how do you interact with faculty and how frequently?? how to clear my doubts while studying?? UserIdTAG: 447302 UserNameTAG: phanisrikar CreateTimeTAG: 2012-09-16T11:39:22Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: join the group as soon as you can FirstChildUserIdTAG: 169784 FirstChildUserNameTAG: thewiredbear FirstChildCreateTimeTAG: 2012-09-16T13:16:27Z FirstChildTAG: http://www.facebook.com/groups/mitxfallsemester/ study group on FACEBOOK FirstChildUserIdTAG: 208923 FirstChildUserNameTAG: Kylekim FirstChildCreateTimeTAG: 2012-09-16T11:44:15Z FirstChildTAG: You really need to watch the videos, they don't take that long. If you have doubts, watch it again or read the book. This forum is open 24 hours a day to answer questions. Teaching staff from MIT is also on here everyday, check the schedual. There are a total of 7 teachers or teachers assistants for this particular course. I would also note that these facebook "study" groups seem a little suspect and I will refer you to the collaboration guidelines before you join such a group. https://www.edx.org/static/content-mit-6002x/handouts/Collaboration_guidelines.65d4e81ed157.pdf FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-16T15:16:48Z SecondChildTAG: Totally agree with Pennypacker ;) SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-16T15:30:26Z IndexTAG: 4268 TitleTAG: difficulty in lab 1 Hey all, Can any one plz elaborate step by step to solve lab 1 question..plzz any one help..i have got struck on it..and cant solve it Thanks Arijit Ghosh UserIdTAG: 128409 UserNameTAG: arijitbme CreateTimeTAG: 2012-09-16T11:38:54Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: me toooooooooooooooooooo pls helpppppppppppppp.... FirstChildUserIdTAG: 276808 FirstChildUserNameTAG: DEBASMITAMAJUMDER FirstChildCreateTimeTAG: 2012-09-16T12:19:38Z FirstChildTAG: R2 ,which is 1.5 Ohom,is parallel to bulb, and R1 is connected to the main circuit. R1=3 Ohom. That's it. FirstChildUserIdTAG: 371617 FirstChildUserNameTAG: rainbow12 FirstChildCreateTimeTAG: 2012-09-16T12:30:33Z SecondChildTAG: lab 2 please SecondChildUserIdTAG: 396372 SecondChildUserNameTAG: vsnarendran SecondChildCreateTimeTAG: 2012-09-16T12:41:01Z FirstChildTAG: why am i still not correct even if got the correct resistance?![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13478101363705861.jpg FirstChildUserIdTAG: 236270 FirstChildUserNameTAG: kevinkjd FirstChildCreateTimeTAG: 2012-09-16T15:42:27Z SecondChildTAG: Not aplicable here also . . SecondChildUserIdTAG: 286106 SecondChildUserNameTAG: Saqib367 SecondChildCreateTimeTAG: 2012-09-16T16:11:05Z FirstChildTAG: give some clue . . FirstChildUserIdTAG: 286106 FirstChildUserNameTAG: Saqib367 FirstChildCreateTimeTAG: 2012-09-16T16:11:49Z IndexTAG: 4269 TitleTAG: Problem with the Answer CHECK ???? @adim hi how are you? There is a problem with the answer check button here !!!!! Kindly do fix this ASAP... thanks as always..;) regards Hassan UserIdTAG: 107038 UserNameTAG: Hassankhan CreateTimeTAG: 2012-09-16T11:01:53Z VoteTAG: 0 CoursewareTAG: Week 2 / Linear combinations of source strenghts CommentableIdTAG: 6002x_S3E2_Linear_Combinations_of_Source_Strengths NumberOfReplyTAG: 1 FirstChildTAG: I used it like 15 min ago and it worked (in my computer) cheers FirstChildUserIdTAG: 342966 FirstChildUserNameTAG: SandraNavarro FirstChildCreateTimeTAG: 2012-09-16T11:20:17Z IndexTAG: 4270 TitleTAG: If you all don't know the unofficial page is now online, please join it!....also has solutions for HW1 https://www.facebook.com/groups/mitxfallsemester/ UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-16T10:56:39Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 4271 TitleTAG: H1P1 Hi. I cannot submit my answers for the homework. I keep getting issues with my answers. is it just numbers or how do I write them? UserIdTAG: 313509 UserNameTAG: Manu18 CreateTimeTAG: 2012-09-16T10:55:51Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: you should write a formula with "R" letter, according to task. Just calculate the total resistor FirstChildUserIdTAG: 444204 FirstChildUserNameTAG: hekan FirstChildCreateTimeTAG: 2012-09-16T12:10:43Z IndexTAG: 4272 TitleTAG: How can we determine the direction of ac current? The direction of ac current changes with the time and we can assume that if the waveform rises on the left hand side, it's positive. So if we measure 3 current signals simultaneously, we have a referenced signal and the other can at the same phase or 180 degree shift regard to the referenced. But we don't measure it at the same time but separately. How can we know the sign of ac signal with this measurement? UserIdTAG: 443589 UserNameTAG: ngokhong CreateTimeTAG: 2012-09-16T10:52:02Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL and KCL CommentableIdTAG: 6002x_kvl_and_kcl NumberOfReplyTAG: 1 FirstChildTAG: there is nothing as such as direction of current in AC....we measure the current in terms of it magnitude and angle(|I|angle)....that's how it works in 3 phase ac circuits the currents differ in terms of magnitude as well as angle.....that's what is called as [Phasor][1] [1]: http://en.wikipedia.org/wiki/Phasor check it out FirstChildUserIdTAG: 102529 FirstChildUserNameTAG: laksh FirstChildCreateTimeTAG: 2012-09-16T11:02:36Z IndexTAG: 4273 TitleTAG: How to submit the answer how to post 3R its giving an error message UserIdTAG: 443968 UserNameTAG: shobhitsajwan CreateTimeTAG: 2012-09-16T10:51:44Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: just post only 3....:D FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-09-16T10:54:21Z SecondChildTAG: u should write 3*R or R+R+R. SecondChildUserIdTAG: 369780 SecondChildUserNameTAG: afnan94 SecondChildCreateTimeTAG: 2012-09-16T11:30:54Z SecondChildTAG: thanks SecondChildUserIdTAG: 443968 SecondChildUserNameTAG: shobhitsajwan SecondChildCreateTimeTAG: 2012-09-16T12:41:44Z IndexTAG: 4274 TitleTAG: doubt do the video lectures long last for the entire course?? UserIdTAG: 431594 UserNameTAG: pvssunilkumar CreateTimeTAG: 2012-09-16T10:43:25Z VoteTAG: 0 CoursewareTAG: Week 1 / Course overview CommentableIdTAG: 6002x_S1V1toS1V4 NumberOfReplyTAG: 1 FirstChildTAG: even i have the same doubt.. FirstChildUserIdTAG: 390803 FirstChildUserNameTAG: Spurty FirstChildCreateTimeTAG: 2012-09-16T19:34:37Z IndexTAG: 4275 TitleTAG: a1,a2,b1,b2,c1,c2 ?????????? What do you mean by a1,a2,b1,b2,c1,c2........Is it a genius UserIdTAG: 348501 UserNameTAG: Abdelhamid CreateTimeTAG: 2012-09-16T10:38:13Z VoteTAG: 0 CoursewareTAG: Week 2 / Linear combinations of source strenghts CommentableIdTAG: 6002x_S3E2_Linear_Combinations_of_Source_Strengths NumberOfReplyTAG: 1 FirstChildTAG: قيمة المقاومات RESISTANCES FirstChildUserIdTAG: 156535 FirstChildUserNameTAG: badawiIiIi FirstChildCreateTimeTAG: 2012-09-16T11:43:26Z SecondChildTAG: they maybe constants.. SecondChildUserIdTAG: 342886 SecondChildUserNameTAG: hthimz SecondChildCreateTimeTAG: 2012-09-23T07:31:50Z IndexTAG: 4276 TitleTAG: couldnt parse when Im done solving a problem and I typed the answer in the box ...and the moment i checked it ... the message appeared like couldnt parse " answer i have written in the box " as a formula... what does that mean??? please help me out... UserIdTAG: 140065 UserNameTAG: Madhav92 CreateTimeTAG: 2012-09-16T10:37:03Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 4277 TitleTAG: HW Submission how do i submit my homework? UserIdTAG: 145503 UserNameTAG: chirag3553 CreateTimeTAG: 2012-09-16T10:20:04Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: you have no need to submit just check your answer it will automatically submit.... FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-09-16T10:26:34Z SecondChildTAG: thanx asadbhatti42 SecondChildUserIdTAG: 145503 SecondChildUserNameTAG: chirag3553 SecondChildCreateTimeTAG: 2012-09-16T10:58:15Z IndexTAG: 4278 TitleTAG: Passing Grade? Passing Grade for this Course? UserIdTAG: 349139 UserNameTAG: 1977ROYELMER CreateTimeTAG: 2012-09-16T09:56:41Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: you get a C grade at 60% FirstChildUserIdTAG: 169784 FirstChildUserNameTAG: thewiredbear FirstChildCreateTimeTAG: 2012-09-16T10:34:52Z IndexTAG: 4279 TitleTAG: Help on Video Lessons I simply cannot view the videos. I've used Firefox and IE, but I can't even see the video window. And when I try to play the Youtube links, it says "This video is currently unavailable." Could someone help me please. UserIdTAG: 429543 UserNameTAG: Baktash CreateTimeTAG: 2012-09-16T09:50:17Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Troubleshooting NumberOfReplyTAG: 3 FirstChildTAG: you can also use 'Mozilla Firefox'....:p FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-09-16T10:27:47Z FirstChildTAG: When I started and attempted to use IE, It said there might be a problem. May not be able to use complete program. Suggestion use Chrome. FirstChildUserIdTAG: 359302 FirstChildUserNameTAG: GaryG FirstChildCreateTimeTAG: 2012-09-16T10:16:12Z FirstChildTAG: I use Firefox and I am not having any problem. May be you have to install the latest version of flash payer. FirstChildUserIdTAG: 199839 FirstChildUserNameTAG: vijaykrishnankk FirstChildCreateTimeTAG: 2012-09-16T10:45:03Z SecondChildTAG: Try out Google Chrome and if it still doesn't work update your Flash Player. And remember to keep us updated!! SecondChildUserIdTAG: 317445 SecondChildUserNameTAG: arghya33 SecondChildCreateTimeTAG: 2012-09-16T20:13:20Z SecondChildTAG: you tube is blocked in my region,so i can't see the lectures ,please five me somr alternate soloution to see the videos. SecondChildUserIdTAG: 294230 SecondChildUserNameTAG: AWAISUMT SecondChildCreateTimeTAG: 2012-09-18T19:01:06Z IndexTAG: 4280 TitleTAG: Slight misunderstandability in the last question here... In the last sentence (question) of that last part, "What is the current (in Amperes) that will flow?", at first glance, I understood (MISunderstood) it to mean, the resultant current from the combined battery of cells, off and into the potential load (the rest of the circuit) - however, it seems the question was directed towards the flow from the stronger battery to the lesser one - just in case anyone else falls into the same thought (and assuming I didn't think this through wrongly lol). UserIdTAG: 340856 UserNameTAG: RanmaSaotome CreateTimeTAG: 2012-09-16T09:25:33Z VoteTAG: 0 CoursewareTAG: Week 1 / Battery model CommentableIdTAG: 6002x_S1E9_Battery_Model NumberOfReplyTAG: 0 IndexTAG: 4281 TitleTAG: independent What do you mean by independent KVL/KCL equations?(doesn't have any same variables?) UserIdTAG: 254325 UserNameTAG: bondablack CreateTimeTAG: 2012-09-16T09:20:50Z VoteTAG: 0 CoursewareTAG: Week 1 / Circuit topology CommentableIdTAG: 6002x_S2E1_Circuit_Topology NumberOfReplyTAG: 3 FirstChildTAG: That means that any of the equations cannot be derived from the others. For example, if I have the following equations: 3a+2b=6 and 12a+11b=14, then by adding or subtracting them I get dependent equations, because I derived them from these two. FirstChildUserIdTAG: 370573 FirstChildUserNameTAG: pberkovich FirstChildCreateTimeTAG: 2012-09-16T14:28:14Z FirstChildTAG: For a conceptual answer, see my response to fettah901, above. FirstChildUserIdTAG: 154526 FirstChildUserNameTAG: silentquasar FirstChildCreateTimeTAG: 2012-09-17T02:38:05Z FirstChildTAG: You have d node => you have (d-1) equations of KCL. You have n branch => You have (n-d+1) equations of KVL. FirstChildUserIdTAG: 471408 FirstChildUserNameTAG: SplendorVN FirstChildCreateTimeTAG: 2012-09-21T10:59:27Z IndexTAG: 4282 TitleTAG: Hey guys!! i made discussion or small talk group on fb fb address is "http://www.facebook.com/groups/mitxfallsemester LET'S GET BUSY :) UserIdTAG: 208923 UserNameTAG: Kylekim CreateTimeTAG: 2012-09-16T09:06:58Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: joined....... FirstChildUserIdTAG: 329051 FirstChildUserNameTAG: asadbhatti42 FirstChildCreateTimeTAG: 2012-09-16T10:31:07Z FirstChildTAG: already joined... FirstChildUserIdTAG: 169784 FirstChildUserNameTAG: thewiredbear FirstChildCreateTimeTAG: 2012-09-16T09:49:09Z IndexTAG: 4283 TitleTAG: Is this lab connection correct or not.Please Help me ![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13477862071946008.png UserIdTAG: 396372 UserNameTAG: vsnarendran CreateTimeTAG: 2012-09-16T09:03:58Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: pls help anybody FirstChildUserIdTAG: 396372 FirstChildUserNameTAG: vsnarendran FirstChildCreateTimeTAG: 2012-09-16T12:11:26Z FirstChildTAG: or this pls![enter image description here][1] [1]: https://edxuploads.s3.amazonaws.com/13477864518474439.png FirstChildUserIdTAG: 396372 FirstChildUserNameTAG: vsnarendran FirstChildCreateTimeTAG: 2012-09-16T09:07:44Z SecondChildTAG: it seem to be correct but try it on sandbox and run DC to check problem... SecondChildUserIdTAG: 329051 SecondChildUserNameTAG: asadbhatti42 SecondChildCreateTimeTAG: 2012-09-16T10:32:23Z IndexTAG: 4284 TitleTAG: why swithed to ac for measuring current at node d as u can use the multimeter ther also UserIdTAG: 328894 UserNameTAG: lk001 CreateTimeTAG: 2012-09-16T08:38:02Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL and KCL CommentableIdTAG: 6002x_kvl_and_kcl NumberOfReplyTAG: 1 FirstChildTAG: You can use the multimeter to do that but you muss put the multimeter serial to the resistor. This means that you must break the circuit. The professor has used the current clamp to measure the current without breaking his circuit and I think his clamp can only measure AC current. There are also clamp that can measure DC current but it's very expensive. FirstChildUserIdTAG: 443589 FirstChildUserNameTAG: ngokhong FirstChildCreateTimeTAG: 2012-09-16T10:58:40Z IndexTAG: 4285 TitleTAG: electron speed drift speed of electron is very small in magnitude but current travels from one end of the conductor to other with great speed, i want to know whether time taken for electric field propagation is responsible for this effect/cause? please help UserIdTAG: 249945 UserNameTAG: gourav19 CreateTimeTAG: 2012-09-16T08:31:50Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: I think it's caused by the movement of electrons due to the presence of holes and they are quickly replaced by the adjacent atoms. Like doping transistors in order to control the flow of current. FirstChildUserIdTAG: 209082 FirstChildUserNameTAG: Nikemurton FirstChildCreateTimeTAG: 2012-09-16T08:43:16Z FirstChildTAG: good question,reson behind it is the ability to flow of charge w.r.t time i.e current depends upon its density of charges and rate of scattering FirstChildUserIdTAG: 397298 FirstChildUserNameTAG: ZA90 FirstChildCreateTimeTAG: 2012-09-16T10:00:16Z FirstChildTAG: search for "Drude theory" FirstChildUserIdTAG: 240577 FirstChildUserNameTAG: Alias2 FirstChildCreateTimeTAG: 2012-09-16T10:55:22Z IndexTAG: 4286 TitleTAG: My dyslexia is driving me nuts... But half the time I get the correct answer... however, I keep forgetting to write the polarity, so I get it wrong *ANYWAYS*... I'm about as sharp as a bag of wet mice... too many wrong answers... UserIdTAG: 340856 UserNameTAG: RanmaSaotome CreateTimeTAG: 2012-09-16T08:26:48Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL and KCL exercises CommentableIdTAG: 6002x_kvl_kcl_exercises NumberOfReplyTAG: 2 FirstChildTAG: dont worry its not a big problem.Here goes the solution for this,be cool while solving FirstChildUserIdTAG: 397298 FirstChildUserNameTAG: ZA90 FirstChildCreateTimeTAG: 2012-09-16T10:02:46Z FirstChildTAG: I think I figured it out, though (other than my forgetfulness for inputting the polarity signs) - I've been looking at the loop relationships all wrong, and taking into account more elements than I needed to. Anyone out there who may be having the same problem, keep in mind the "Path of Least Resistance" (as is applied in general, like with water flowing down a hill). FirstChildUserIdTAG: 340856 FirstChildUserNameTAG: RanmaSaotome FirstChildCreateTimeTAG: 2012-09-16T08:46:01Z IndexTAG: 4287 TitleTAG: H1P2 Good day (but not so good for me.) I have: 8 Amps, 1 Ohm, 3 Ohms, 4V. As in the forum. The power dissipated in resistors is 25 Watts (for 1 Ohm) and 27 Watts (for 3 Ohm). I'm sure in it! I checked calculations many time, I've done a model by CircuitLab. Maybe, Ohm was incorrect with his equations? I don't understand - what's wrong? The power savings law also works. Please, help. UserIdTAG: 20141 UserNameTAG: YgReEk CreateTimeTAG: 2012-09-16T08:20:54Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Troubleshooting NumberOfReplyTAG: 1 FirstChildTAG: You can consider three nodes at the voltage source, current source and at the two resistors also. But the current direction is changed to confuse you. Dont divide the current as current divider in this circuit. The real concept is 7A is passing through 1ohms. Therefore power dissipated through 1 ohm is sqr(7)*1 = 49W The current through the 3 ohms is 1 A. therefore the power dissipated is 9W FirstChildUserIdTAG: 396372 FirstChildUserNameTAG: vsnarendran FirstChildCreateTimeTAG: 2012-09-16T08:37:20Z SecondChildTAG: Thanks a lot! I'll try not to make such mistakes. SecondChildUserIdTAG: 20141 SecondChildUserNameTAG: YgReEk SecondChildCreateTimeTAG: 2012-09-16T09:02:51Z IndexTAG: 4288 TitleTAG: H1P1 last question Hey can someone pls explain me the last question from h1p1? I cant seem to understand what exactly is given and what is asked. UserIdTAG: 440331 UserNameTAG: Shaun94 CreateTimeTAG: 2012-09-16T08:13:02Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: someone name theweirdbear has answered the question in the discussions FirstChildUserIdTAG: 202177 FirstChildUserNameTAG: ramanan4561 FirstChildCreateTimeTAG: 2012-09-16T08:22:07Z SecondChildTAG: Where's that? I can't get the total power in smallest-valued resistors combination.....help!! SecondChildUserIdTAG: 268104 SecondChildUserNameTAG: SaulCardenasG SecondChildCreateTimeTAG: 2012-09-16T08:30:15Z FirstChildTAG: 2.67........your welcome FirstChildUserIdTAG: 169784 FirstChildUserNameTAG: thewiredbear FirstChildCreateTimeTAG: 2012-09-16T08:32:12Z SecondChildTAG: Спасибо) SecondChildUserIdTAG: 202275 SecondChildUserNameTAG: Valya SecondChildCreateTimeTAG: 2012-09-16T08:59:10Z SecondChildTAG: And how to calculate? SecondChildUserIdTAG: 202275 SecondChildUserNameTAG: Valya SecondChildCreateTimeTAG: 2012-09-16T08:59:58Z SecondChildTAG: Hi, Thank you for the answer but I would like to know, Is it the total power dissipated over the smallest-valued resistor that can be fabricated by combining these three resistors? The question is very unclear because it says : "Given that each individual resistor can dissipate up to 1 watt of power before burning up, how much total power in watts ( W ) can the smallest-valued composite resistor dissipate before burning up?" "smallest Composite Resistor" meaning the smallest resistor that is composed of some material but they do not specify if they are referring to the question above where all resistors are combined in parallel to form 1 small resistor or if they are referring to one of the given 4ohm resistors. SecondChildUserIdTAG: 256356 SecondChildUserNameTAG: WesleyHanes SecondChildCreateTimeTAG: 2012-09-16T09:11:02Z SecondChildTAG: The Three Resistors Are In Parallel . That Is Two 4 Ohm And One 6 Ohm. To Calculate Power Dissipation . To Find V Given Each Resistor Can Dissipate 1 Watt. Equate Power = V^2/R For The 4 Ohm Resistor. V = x Then V^2/(Rp) Where Rp = Equivalent Resistance In Parallel Then Add Power Dissipated By Each Resistor. SecondChildUserIdTAG: 131789 SecondChildUserNameTAG: Vijayaraghavan SecondChildCreateTimeTAG: 2012-09-16T10:04:26Z FirstChildTAG: Could you tell us how you reach at that result?? Thx FirstChildUserIdTAG: 443442 FirstChildUserNameTAG: Nelutu FirstChildCreateTimeTAG: 2012-09-16T09:04:51Z FirstChildTAG: OK I get it, 2.97W is the total power that would cause all resistors to blow in a parallel circuit. Any answer smaller than this(between 2.67 and 2.81) is correct. each 4 ohm resistor can handle 0.49A but will blow at 0.5A and the 6 ohm resistor can handle 0.39 A but will blow at roughly 0.408 A. add all these currents together and find your voltage and then you are half way there to finding the total power required. Remember you need the power before it blows, and we are working with all resistors in Parallel like the previous question FirstChildUserIdTAG: 256356 FirstChildUserNameTAG: WesleyHanes FirstChildCreateTimeTAG: 2012-09-16T10:01:43Z IndexTAG: 4289 TitleTAG: help me (how can i get certificate?) hey guys. if i didnt send H/W1 or Lab1, do i fail this class? UserIdTAG: 208923 UserNameTAG: Kylekim CreateTimeTAG: 2012-09-16T07:20:10Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: do you want atleast B grade answers for homework 1? and you won't fail FirstChildUserIdTAG: 169784 FirstChildUserNameTAG: thewiredbear FirstChildCreateTimeTAG: 2012-09-16T07:29:54Z SecondChildTAG: I will be happy to help here..... SecondChildUserIdTAG: 169784 SecondChildUserNameTAG: thewiredbear SecondChildCreateTimeTAG: 2012-09-16T07:32:41Z FirstChildTAG: Your least two scored lab and HW will be dropped in the end of the session.And that will be the least one in the end. FirstChildUserIdTAG: 401259 FirstChildUserNameTAG: amangodne FirstChildCreateTimeTAG: 2012-09-16T07:39:27Z FirstChildTAG: Ofcoursly not..:-)you are free to skip any 2 labwork and homework. They are gonna consider best of 10 lab and homeworks out of 12 given. So noworry.. be happy:-) FirstChildUserIdTAG: 364473 FirstChildUserNameTAG: vibgyor FirstChildCreateTimeTAG: 2012-09-16T07:40:50Z SecondChildTAG: :-) thks SecondChildUserIdTAG: 208923 SecondChildUserNameTAG: Kylekim SecondChildCreateTimeTAG: 2012-09-16T07:47:20Z FirstChildTAG: Not at all buddy, you can miss two homeworks and labworks without being degraded, in addition one homework has 1.5 points from the total 100 marks and so does the lab so in case you lose your mark, you will lose only 3 marks, means you can still have an A for the course. FirstChildUserIdTAG: 441835 FirstChildUserNameTAG: KAMBIZAMEEN FirstChildCreateTimeTAG: 2012-09-16T07:52:32Z IndexTAG: 4290 TitleTAG: H1P3 In the 2nd part of the question,we neither have the resistances of the heaters nor the individual currents through them.how are we to find the power dissipated by each component then?? UserIdTAG: 210146 UserNameTAG: Divyakumari CreateTimeTAG: 2012-09-16T07:10:43Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: hold, on i just started the question, will help, soon, and you can maybe rely on me, I got all the above answers correct! FirstChildUserIdTAG: 169784 FirstChildUserNameTAG: thewiredbear FirstChildCreateTimeTAG: 2012-09-16T07:16:19Z SecondChildTAG: for starters they give you the power and voltage of all the heaters... remember the relation. P=V^2/R....solve further, and P=I^2 X R, so this gives the currents or also P= VI SecondChildUserIdTAG: 169784 SecondChildUserNameTAG: thewiredbear SecondChildCreateTimeTAG: 2012-09-16T07:18:56Z SecondChildTAG: i have got the currents right. SecondChildUserIdTAG: 210146 SecondChildUserNameTAG: Divyakumari SecondChildCreateTimeTAG: 2012-09-16T07:19:54Z SecondChildTAG: 35.7500000001....this is a very funny number, maybe it is the answer to your question SecondChildUserIdTAG: 169784 SecondChildUserNameTAG: thewiredbear SecondChildCreateTimeTAG: 2012-09-16T07:21:51Z SecondChildTAG: how did u get that??? SecondChildUserIdTAG: 210146 SecondChildUserNameTAG: Divyakumari SecondChildCreateTimeTAG: 2012-09-16T07:38:43Z SecondChildTAG: i am stuck at the last 3 answers, help me there.... and i will help you SecondChildUserIdTAG: 169784 SecondChildUserNameTAG: thewiredbear SecondChildCreateTimeTAG: 2012-09-16T07:59:30Z SecondChildTAG: ok so total resistance is 18.11 , there fore if H1=H2=H3=H then H =18.11/3= 6.03 ohms. P=I2R , my answer is still wrong . Please can some one help. SecondChildUserIdTAG: 178115 SecondChildUserNameTAG: nehamakhija SecondChildCreateTimeTAG: 2012-09-16T18:10:48Z SecondChildTAG: please help me SecondChildUserIdTAG: 178115 SecondChildUserNameTAG: nehamakhija SecondChildCreateTimeTAG: 2012-09-16T18:11:00Z FirstChildTAG: Well in the first part you have their rated power and voltage from which you can find their resistances, which are equal of course. In the second part you have their resistances so you can apply the voltage division methode to find the voltage drop accross each heater, by having the voltage and resistance you are there to find their power dissipation. FirstChildUserIdTAG: 441835 FirstChildUserNameTAG: KAMBIZAMEEN FirstChildCreateTimeTAG: 2012-09-16T07:39:27Z SecondChildTAG: thank u! it really did help. SecondChildUserIdTAG: 210146 SecondChildUserNameTAG: Divyakumari SecondChildCreateTimeTAG: 2012-09-16T08:05:13Z SecondChildTAG: voltage, is 240, from first part, R is 5.95, then for each, then add the combo of parallel to get R from them as 5.95/2 then rule is VR1/R1+R2.....apply it and my answer is wrong.......please help SecondChildUserIdTAG: 169784 SecondChildUserNameTAG: thewiredbear SecondChildCreateTimeTAG: 2012-09-16T08:18:07Z SecondChildTAG: Why? Req = 240^2 / 850 ? So R = 3* Req? In 2nd part you get v1=160 y v2 = 80 right? Please help !!! SecondChildUserIdTAG: 268104 SecondChildUserNameTAG: SaulCardenasG SecondChildCreateTimeTAG: 2012-09-16T11:04:16Z SecondChildTAG: hahaha thank you a lot man , you are brilliant SecondChildUserIdTAG: 231318 SecondChildUserNameTAG: moutasem SecondChildCreateTimeTAG: 2012-09-17T09:47:39Z FirstChildTAG: I'll try to explain without giving any answers. I did the first part using P = VI so I = P/V The second part, I had trouble because I had no values. I was stuck like you are. However, I realized that with the voltage divider V dropped to 1/3 of the original problem with the parallel and 2/3 in H1. First I found P in the parallel heaters. Because V dropped 1/3, and because V = IR but R is a constant, I realized that Current must have dropped 1/3. If Power = I^2*R and current dropped 1/3, power must have dropped 1/9. The same applies to the power in H1 except that power drops by 4/9. FirstChildUserIdTAG: 139942 FirstChildUserNameTAG: NORRISD FirstChildCreateTimeTAG: 2012-09-16T08:24:43Z SecondChildTAG: Sir, you are....... LEGEN......wait for it.....DARY! SecondChildUserIdTAG: 169784 SecondChildUserNameTAG: thewiredbear SecondChildCreateTimeTAG: 2012-09-16T08:29:41Z SecondChildTAG: @NORRISD but why is the drop 1/3 and 4/3 SecondChildUserIdTAG: 169784 SecondChildUserNameTAG: thewiredbear SecondChildCreateTimeTAG: 2012-09-16T08:30:58Z SecondChildTAG: i still didnt understand SecondChildUserIdTAG: 444708 SecondChildUserNameTAG: jagadeeshv16 SecondChildCreateTimeTAG: 2012-09-16T19:00:51Z SecondChildTAG: me either didn't understand what you mean. kindly help me SecondChildUserIdTAG: 130118 SecondChildUserNameTAG: manis SecondChildCreateTimeTAG: 2012-09-16T19:41:20Z IndexTAG: 4291 TitleTAG: h1p1 Given that each individual resistor can dissipate up to 1 watt of power before burning up, what its means , 1ohm resister dissipate 1 watt , or one resistor dissipate 1 watt , please explain how i calculate it , UserIdTAG: 133891 UserNameTAG: junaidkarim CreateTimeTAG: 2012-09-16T06:48:26Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: It's 1 resistor dissipate 1 watt.first calculate voltage across parallel resistors with the formula v=sqrt(R*P) and then find the equivalent resistance.Atlast calculate power by formula P=V^2*R(equivalent). FirstChildUserIdTAG: 364473 FirstChildUserNameTAG: vibgyor FirstChildCreateTimeTAG: 2012-09-16T07:35:01Z FirstChildTAG: I don't understand well what this question means FirstChildUserIdTAG: 445664 FirstChildUserNameTAG: Cesar_Enriquez FirstChildCreateTimeTAG: 2012-09-16T08:40:10Z IndexTAG: 4292 TitleTAG: math processing error can anybody help me with this.... UserIdTAG: 413613 UserNameTAG: shaliesh CreateTimeTAG: 2012-09-16T06:45:20Z VoteTAG: 0 CoursewareTAG: Week 1 / Battery model CommentableIdTAG: 6002x_S1E9_Battery_Model NumberOfReplyTAG: 1 FirstChildTAG: Specifics? FirstChildUserIdTAG: 340856 FirstChildUserNameTAG: RanmaSaotome FirstChildCreateTimeTAG: 2012-09-16T09:11:06Z SecondChildTAG: don't use IE. try firefox or chrome. SecondChildUserIdTAG: 251793 SecondChildUserNameTAG: chellaton SecondChildCreateTimeTAG: 2012-09-20T14:56:03Z IndexTAG: 4293 TitleTAG: book how to download the given textbook? UserIdTAG: 445270 UserNameTAG: Deepanshu123 CreateTimeTAG: 2012-09-16T06:20:05Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: you cannot FirstChildUserIdTAG: 169784 FirstChildUserNameTAG: thewiredbear FirstChildCreateTimeTAG: 2012-09-16T06:36:31Z IndexTAG: 4294 TitleTAG: piotr and gerry how are you saying that branch(consisting voltage source) can be taken as a single node? I'm confused.. help me.. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-16T06:16:32Z VoteTAG: 0 CoursewareTAG: Week 1 / Nodal analysis with floating voltage CommentableIdTAG: 6002x_FloatingVoltage NumberOfReplyTAG: 1 FirstChildTAG: listen dude ..for example we suppose that we have point A in between voltage source and resistive load we take this point as a Node 1 ..... because total voltage will present at this point .. FirstChildUserIdTAG: 430030 FirstChildUserNameTAG: imali FirstChildCreateTimeTAG: 2012-09-16T08:13:45Z SecondChildTAG: The branch with the voltage source can be considered as a single isolated node and amount of current flowing into it should be equal to amount of current flowing out of it. SecondChildUserIdTAG: 138239 SecondChildUserNameTAG: LaFolle SecondChildCreateTimeTAG: 2012-09-16T11:37:43Z IndexTAG: 4295 TitleTAG: PIECE WISE LINEAR **DIODE IS A PIECE WISE LINEAR.....** because current doesn't vary linearly with the voltage and it is linear only in higher voltage region UserIdTAG: 431034 UserNameTAG: shiva0154 CreateTimeTAG: 2012-09-16T06:05:29Z VoteTAG: 0 CoursewareTAG: Week 2 / Is it linear? CommentableIdTAG: 6002x_is_it_linear NumberOfReplyTAG: 0 IndexTAG: 4296 TitleTAG: independent equations i dont get what independent equations are? UserIdTAG: 81511 UserNameTAG: MOAZZA CreateTimeTAG: 2012-09-16T06:02:32Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL and KCL CommentableIdTAG: 6002x_kvl_and_kcl NumberOfReplyTAG: 4 FirstChildTAG: . FirstChildUserIdTAG: 263241 FirstChildUserNameTAG: yvesauad FirstChildCreateTimeTAG: 2012-09-16T07:20:41Z FirstChildTAG: see the recomended book page 150 and above.... FirstChildUserIdTAG: 168396 FirstChildUserNameTAG: rik2008 FirstChildCreateTimeTAG: 2012-09-16T06:04:31Z FirstChildTAG: when you equate all the equations, you may get 4 equations, but if the top 3 already have all the unknowns, then the 4th equation is REDUNDANT, and is therefore dependant on other 3 equations, you can choose any 1 equation redundant and the others as independent, totally upto you FirstChildUserIdTAG: 169784 FirstChildUserNameTAG: thewiredbear FirstChildCreateTimeTAG: 2012-09-16T06:16:43Z SecondChildTAG: @thewiredbear ~ Thanks 4 ur explanation. SecondChildUserIdTAG: 156225 SecondChildUserNameTAG: sanjana_m SecondChildCreateTimeTAG: 2012-09-23T07:11:33Z FirstChildTAG: if you have 3 equations, sometimes one of those can be reescripted with the other two ones. For instance: a+b=5 2a-84b=245 11a-247b=760 If you multiply the first by 5, the second by 3 and sum both, you´ll get the third. Best Regards. FirstChildUserIdTAG: 263241 FirstChildUserNameTAG: yvesauad FirstChildCreateTimeTAG: 2012-09-16T07:20:23Z IndexTAG: 4297 TitleTAG: please help me out of this.... Given that each individual resistor can dissipate up to 1 watt of power before burning up, how much total power in watts ( W ) can the smallest-valued composite resistor dissipate before burning up? UserIdTAG: 168396 UserNameTAG: rik2008 CreateTimeTAG: 2012-09-16T06:02:22Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: same here,,,,please help FirstChildUserIdTAG: 324463 FirstChildUserNameTAG: Junbacs FirstChildCreateTimeTAG: 2012-09-16T07:49:02Z FirstChildTAG: im lost too FirstChildUserIdTAG: 233221 FirstChildUserNameTAG: JPShukla FirstChildCreateTimeTAG: 2012-09-16T06:19:44Z FirstChildTAG: the least value of resistor is 4ohms, so to get a power of 1W here, you need 2volts, by the relation P=V^2/R.... and then change resistance from 4ohms to the total reisitance that you calculated, i.e., 1.5ohms, and you get you answer as 4/1.5, 2.67.....hope i helped, please help me too, click on my username to see my latest problem FirstChildUserIdTAG: 169784 FirstChildUserNameTAG: thewiredbear FirstChildCreateTimeTAG: 2012-09-16T06:12:31Z SecondChildTAG: @thewiredbear can u gie me method for sloving last three uestion of problem 3? SecondChildUserIdTAG: 227777 SecondChildUserNameTAG: umairhassankhanniazi SecondChildCreateTimeTAG: 2012-09-16T07:23:10Z IndexTAG: 4298 TitleTAG: H2P2 - Help How to find "What is the power (in Watts) that is delivered to this best load resistance?" UserIdTAG: 199839 UserNameTAG: vijaykrishnankk CreateTimeTAG: 2012-09-16T05:59:19Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: Please help.. FirstChildUserIdTAG: 199839 FirstChildUserNameTAG: vijaykrishnankk FirstChildCreateTimeTAG: 2012-09-16T06:00:04Z SecondChildTAG: Anybody out there... Please help me!!!! SecondChildUserIdTAG: 199839 SecondChildUserNameTAG: vijaykrishnankk SecondChildCreateTimeTAG: 2012-09-16T06:56:35Z FirstChildTAG: Look at Maximum Power Transfere Theorem; calculate RL using calculus. Then find out RTH and VTH from Thevenin. Calculate VL on rezistor RL. Then P = VL*VL/RL FirstChildUserIdTAG: 357924 FirstChildUserNameTAG: DanI20 FirstChildCreateTimeTAG: 2012-09-16T07:10:30Z SecondChildTAG: @DanI20 :even i too did the same wt u tel...bt am getn incorrect ans. SecondChildUserIdTAG: 356056 SecondChildUserNameTAG: Ganesh2810 SecondChildCreateTimeTAG: 2012-09-16T08:17:50Z FirstChildTAG: Thanks Danl20. I have used many equations which I know, even the one which you told. Still, it is showing 'incorrect answer'. FirstChildUserIdTAG: 199839 FirstChildUserNameTAG: vijaykrishnankk FirstChildCreateTimeTAG: 2012-09-16T10:27:26Z FirstChildTAG: Any other way to find the power for the given circuit? Or is there any problem with the question? FirstChildUserIdTAG: 199839 FirstChildUserNameTAG: vijaykrishnankk FirstChildCreateTimeTAG: 2012-09-16T10:50:26Z FirstChildTAG: Hey, read the Maximum Power transfer theorem (Rth = RL). Make the thevenin equivalent (i.e, Rth and Vth). With this you have to make i = Vth / (Rth + RL). Now you can use the formula of power -> P = i^2 * RL. FirstChildUserIdTAG: 300169 FirstChildUserNameTAG: castrogfx FirstChildCreateTimeTAG: 2012-09-17T10:47:16Z IndexTAG: 4299 TitleTAG: How do we get the certificates..?? Please any staff member answer... How are we going to get the certificate..?? i mean will it be mailed to us at our addresses (that is we get the hard copy) or we will have just an online copy that we have to print at our place...!! UserIdTAG: 210992 UserNameTAG: neerajnatu CreateTimeTAG: 2012-09-16T05:41:26Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: They will send you an email on successful completion of course and ask you to download your certificate from profile page. This is what they did last time. FirstChildUserIdTAG: 364473 FirstChildUserNameTAG: vibgyor FirstChildCreateTimeTAG: 2012-09-16T08:04:06Z SecondChildTAG: thanks a lot... SecondChildUserIdTAG: 210992 SecondChildUserNameTAG: neerajnatu SecondChildCreateTimeTAG: 2012-09-22T07:02:46Z IndexTAG: 4300 TitleTAG: bug My h1p3 check button is not working, What should i do? UserIdTAG: 145737 UserNameTAG: saad26 CreateTimeTAG: 2012-09-16T05:31:31Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Can you elaborate on what's happening when you click the check button? FirstChildUserIdTAG: 12663 FirstChildUserNameTAG: arjun FirstChildCreateTimeTAG: 2012-09-16T06:39:50Z IndexTAG: 4301 TitleTAG: lab 1 plz somebody help me in doing lab 1.i have done wid all the problems in homework 1 but cant do lab 1.plz plz plz somebody help me!!!!!!!! UserIdTAG: 276808 UserNameTAG: DEBASMITAMAJUMDER CreateTimeTAG: 2012-09-16T05:02:58Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: i couldnt understand pls do ellaborate FirstChildUserIdTAG: 276808 FirstChildUserNameTAG: DEBASMITAMAJUMDER FirstChildCreateTimeTAG: 2012-09-16T05:36:54Z FirstChildTAG: It is quite simple... You have two circuits, one where the OPEN CIRCUIT voltage should be less than 2V and the voltage with a 3ohms resistor connected should be 1.5 ohms. Just solve them together. While solving, remember that the Open Circuit Voltage is nothing but the voltage obtained from the voltage divider equation from the lecture. Hope this helps !!! Regards, Jaychandran FirstChildUserIdTAG: 137686 FirstChildUserNameTAG: Jaychandran FirstChildCreateTimeTAG: 2012-09-16T05:08:06Z FirstChildTAG: Hi DEBASMITAMAJUMDER! Sure, I will like to help you. Were are you lost? You have to read carefully the statement (remember that Students can have different values in their assigments). You have a 6-volt battery (assumed ideal) and a 1.5-volt flashlight bulb, which is known to draw 0.5A when the bulb voltage is 1.5V (see figure below). Design a network of resistors to go between the battery and the bulb to give v**s=1.5V when the bulb is connected**, yet ensures that **vs does not rise above 2V when the bulb is disconnected.** ![enter image description here][1] Ok, two visual hints for you: ![enter image description here][2] They give you TWO CONDITIONS . One is the voltage (green arrow) when the bulb it is connected and the other one it is when the bulb it is not connected (blue arrow). You will have two equations now. So, you have to discover what is inside the box ;)... The last Hint is in the Textbook [Voltage and Current Dividers][3] I hope this can help you. Myriam. [1]: https://www.edx.org/static/content-mit-6002x/images/circuits/Lab1_1.b3102bb33467.png [2]: https://edxuploads.s3.amazonaws.com/13477741287450155.png [3]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/97 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-16T05:50:25Z FirstChildTAG: thank u so much:D FirstChildUserIdTAG: 276808 FirstChildUserNameTAG: DEBASMITAMAJUMDER FirstChildCreateTimeTAG: 2012-09-16T06:21:50Z IndexTAG: 4302 TitleTAG: great problem in signs of values **I have a great problem in the signs of my values (+ve or -ve) what should i do or what should i read or understand and what is the missing part i don't understand that cause these conflicts ?** UserIdTAG: 290966 UserNameTAG: AhmedImam CreateTimeTAG: 2012-09-16T04:02:47Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: Whats about KCL? https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/ You define a sign sense, and considering the current you can say when the Tension at o + or - whith yours convention. FirstChildUserIdTAG: 274475 FirstChildUserNameTAG: Konredus FirstChildCreateTimeTAG: 2012-09-16T04:13:30Z FirstChildTAG: It is best that you note the directions of the arrows of the currents. However the "trick" I use it that, the positive side of the voltage source forces the current to forward. If however the current assigned goes into the voltage source from the positive terminal, we just change the sign to the opposite one. FirstChildUserIdTAG: 209082 FirstChildUserNameTAG: Nikemurton FirstChildCreateTimeTAG: 2012-09-16T05:00:05Z FirstChildTAG: There is something called the "Associated Variables Discipline" that has been clearly described in the lectures. It is very useful in determining the sign of the values while working with the node method. I guess it is the node method that makes the calculation easy, the AVD is just a convention. FirstChildUserIdTAG: 137686 FirstChildUserNameTAG: Jaychandran FirstChildCreateTimeTAG: 2012-09-16T05:11:05Z IndexTAG: 4303 TitleTAG: Basic question after course. My question is: once approved, there's a chance to continue with another course of electronics or directly at MIT? Thanks for giving us the opportunity to study at MIT..>!!! UserIdTAG: 274475 UserNameTAG: Konredus CreateTimeTAG: 2012-09-16T04:00:02Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Feedback NumberOfReplyTAG: 2 FirstChildTAG: You may be able to take a further electronics courses in the future through edX if a course becomes available.(MITx) Some of these courses at edX may have the option to take a proctored exam, which would give you a credit. Apparently one will be proctored from this group of courses. I think you would have to apply via conventional means to "continue with another course of electronics or directly at MIT". This requires a boat-load of money and top-shelf grades however. I am very satisfied with the compromise that is presented to us, the certification is nice. Potentially having the option to take proctored exams is icing on the cake. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-16T04:20:40Z SecondChildTAG: thanks for your reply, but i do not understand what you whit proctored exam mean? or who can help me to study in the USA? SecondChildUserIdTAG: 274475 SecondChildUserNameTAG: Konredus SecondChildCreateTimeTAG: 2012-09-16T04:47:33Z SecondChildTAG: A proctored exam is an exam that is supervised by someone who represents edX (MITx). Essentially the exam is taken in a similar environment as actual students in a real school and is graded by a real person. This proctored exam will give you university credit for the particular class you are tested on. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-16T05:08:57Z FirstChildTAG: Here's a link to a news story about the proctored exams: http://chronicle.com/blogs/wiredcampus/edx-offers-proctored-exams-for-open-online-course/39656 Although in talking to people who have taken proctored exams, they are still done on computers and graded automatically, just like this final will be. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-09-16T05:16:38Z IndexTAG: 4304 TitleTAG: I couldnt do the lap in my android 2.2.1 phone can any one suggest me a way I couldnt do the lab portion in my android 2.2.1 phone can any one suggest me a way UserIdTAG: 145544 UserNameTAG: pandiya CreateTimeTAG: 2012-09-16T01:39:46Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You may have to gain access to a proper computer. Have fun. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-16T01:45:06Z IndexTAG: 4305 TitleTAG: simplifying algebra Hi, i'm not that great at algebra, I got the correct answer, but i don't understand how it equals the given answer, could anyone help me see. For B1: I got: (R2*R3-(R1*R2+R2*R3+R1*R3))/(R1*(R1*R2+R2*R3+R1*R3)) (which was correct) but the given answer is: -(R2+R3)/(R1*R2+R1*R3+R2*R3) could you show me how the algebra works to get the given answer? thank you! UserIdTAG: 155972 UserNameTAG: sreicks CreateTimeTAG: 2012-09-16T00:41:10Z VoteTAG: 0 CoursewareTAG: Week 2 / Linear combinations of source strenghts CommentableIdTAG: 6002x_S3E2_Linear_Combinations_of_Source_Strengths NumberOfReplyTAG: 1 FirstChildTAG: ( R2*R3 - (R1*R2 +R2*R3 +R1*R3)) / (R1 * (R1*R2 + R2*R3 +R1*R3)) = (R2*R3 -R1*R2 -R2*R3 -R1*R3) / (R1 * (R1*R2 + R2*R3 +R1*R3)) = ( -R1*R2 -R1*R3 ) / (R1 * (R1*R2 + R2*R3 +R1*R3)) = -R1( R2+R3) / (R1 * (R1*R2 + R2*R3 +R1*R3)) = -(R2+R3) / (R1*R2 + R2*R3 + R1*R3) that's ur answer. FirstChildUserIdTAG: 119355 FirstChildUserNameTAG: A_KUMAR FirstChildCreateTimeTAG: 2012-09-16T03:00:31Z SecondChildTAG: A_Kumar good solution thanks SecondChildUserIdTAG: 296303 SecondChildUserNameTAG: DEEPatXUniv SecondChildCreateTimeTAG: 2012-09-16T15:38:18Z SecondChildTAG: Thx a lot SecondChildUserIdTAG: 388555 SecondChildUserNameTAG: 4lk4tr43 SecondChildCreateTimeTAG: 2012-09-18T12:03:55Z IndexTAG: 4306 TitleTAG: hi all,about the certification how can i get it? and "honor code certificate from MITx" what does it mean? thanks UserIdTAG: 395257 UserNameTAG: blackguitar CreateTimeTAG: 2012-09-16T00:24:54Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: By completing the course you will gain a "Certificate of Mastery" from edX undersigning MITx. For one of the seven courses offered by edX this fall, you might be able to take a proctored exam for a fee, which I would assume gives you a proper "credit" I would imagine that sometime in the future you would be able to take a proctored exam for all of the courses. Make a note to yourself that the courses may not always be free. Strike while the iron is hot. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-16T00:39:04Z FirstChildTAG: Actually it doesn't say "Certificate of Mastery" all it says is that you successfully completed the 6.002x course within the terms of the honor code. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-09-16T00:55:30Z SecondChildTAG: It does not say "Certificate of Mastery", but that is what it is. Interview with Prof. Agarwal: "I teach the MIT course 6002, and my team and I teaching 6002X online -- it's the same hard course. The grading uses the same standards, and so if several people are able to show mastery of a subject, so be it. They deserve to be recognized with a certificate." Also a quote from the Harvard Crimson: "Until now, edX planned to have its students agree to an honor code before taking exams. Students would receive a Certificate of Mastery upon passing the course. Now, for a currently undisclosed fee, students can choose to visit a test center to complete their final exams and, upon passing, receive a certificate indicating completion of the course with a proctored test, according to an edX press release." SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-16T01:12:47Z IndexTAG: 4307 TitleTAG: some question disappeared how to find some question in my homework1 its really disappeared its unfair for me if i never found those questions i can't answer UserIdTAG: 235727 UserNameTAG: amor CreateTimeTAG: 2012-09-15T23:50:41Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Troubleshooting NumberOfReplyTAG: 3 FirstChildTAG: ![Here is the image for the first part of the question.][1] [1]: https://www.edx.org/static/content-mit-6002x/images/circuits/H1P1_resistor_combinations.8534a09fbd2c.gif FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-16T00:13:34Z SecondChildTAG: The questions for part one are posted down below. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-16T00:15:14Z SecondChildTAG: penny r u there? i want to ask the right keyword SecondChildUserIdTAG: 235727 SecondChildUserNameTAG: amor SecondChildCreateTimeTAG: 2012-09-16T12:00:56Z FirstChildTAG: H1P2: KCL/KVL VS. NODE METHOD This problem explores the difference between solving a circuit using the KCL/KVL method and the node method. The circuit shown below has four elements: two resistors, a current source and a voltage source. The resistance of the resistors and the strengths of the sources are all given. The four branch currents (i1 to i4) and the four branch voltages (v1 to v4) are also defined in the circuit using the associated variables convention. Recall that solving a circuit means solving for these branch variables (currents and voltages). How many nodes are there in the circuit? correct A KCL equation can be written at each of these nodes. How many of these KCL equations are independent? correct How many loops are there in the circuit? correct A KVL equation can be written around each of these loops. How many of the KVL equations are independent? correct How many additional independent equations do we need to solve this circuit? correct We can get these additional equations from the constitutive relations (or element laws) of the four elements. The total number of independent equations needed to solve a circuit using the KCL/KVL method is twice the number of elements in the circuit (eight in this example) - matching the number of unknown branch variables. On the other hand, the node method can require far fewer equations. In this example, if the ground node is appropriately chosen there is only one unknown node voltage. Hence, only one node equation is needed. Using either the KCL/KVL method or the node method, solve this circuit. The value (in Volts) of v1 is: correct The value (in Amperes) of i4 is: correct The power (in watts) coming out of the current source: correct The power (in watts) coming out of the voltage source: correct The power (in watts) dissipated in the 1Ω resistor: incorrect The power (in watts) dissipated in the 3Ω resistor: correct By conservation of power, the net power coming out of the two sources must equal the total power dissipated in the two resistors. If this is not true, then you made a mistake. H1P3: POOR WORKMANSHIP Joe wants to heat his 12'X20' workshop with electric heat. He has hired the HACME electrician company to build the system. They propose to use three 920.0W 240V baseboard heaters to provide a total heating capacity of 2760.0W. (A heater is basically a resistor. This is not quite true, because there is a thermostatic switch incorporated into the heater and because the resistance of a heater varies a bit with its temperature. But we will use a linear resistor as a model of a heater.) In the proposed system the heaters are connected in parallel with the 240V 60Hz AC power line (modeled by a voltage source) as shown in the diagram: Remember (from Exercise S1E3: AC power) that AC power-line voltages and currents are specified as RMS values. So 120V AC heats a given resistance exactly as much as 120V DC would heat that same resistance. How much current is expected to be drawn from the power line by this heating system when all three heaters are on? correct If instead, HACME chose to implement the system with 120V heaters, how much current would have been needed? correct Notice that this would require much heavier and more expensive wire to distribute the power. Back to the original plan with 240V power. Unfortunately, Sparky, who works for HACME, was a little sleepy that day. He accidentally connected the heaters as shown below: As a consequence, Joe found his workshop too cold. H1 was weak; H2 and H3 barely warmed up. What power was being dissipated in H1? correct What power was being dissipated in H2 (or in H3)? correct So the total heating power in Joe's shop was: correct No wonder Joe was cold. FirstChildUserIdTAG: 235727 FirstChildUserNameTAG: amor FirstChildCreateTimeTAG: 2012-09-15T23:54:21Z SecondChildTAG: this is only my homeworks so some questions are disappeared its so unfair for me SecondChildUserIdTAG: 235727 SecondChildUserNameTAG: amor SecondChildCreateTimeTAG: 2012-09-15T23:56:19Z SecondChildTAG: What did you type in the H1P1, if you remember? SecondChildUserIdTAG: 79361 SecondChildUserNameTAG: kimt SecondChildCreateTimeTAG: 2012-09-16T00:03:50Z SecondChildTAG: what you mean by that kimt? u mean H1? or u mean those disappered questions? those disappeared question when i answered it said invalid keyword coz i type R= R1 + R2 + R3 so that i try to figure out the right keyword said to me need to type 3*R right? but those questions are disappeared SecondChildUserIdTAG: 235727 SecondChildUserNameTAG: amor SecondChildCreateTimeTAG: 2012-09-16T00:10:45Z SecondChildTAG: by the wau kimt i have 1 mistake im so comfuse why my answer is wrong thats make me crazy...this is the questions that its so hard for me to get a correct answer The power (in watts) dissipated in the 1Ω resistor i tried to calculate this my answer is 1 but its wrong :( SecondChildUserIdTAG: 235727 SecondChildUserNameTAG: amor SecondChildCreateTimeTAG: 2012-09-16T00:14:42Z FirstChildTAG: Amor, the staff member "kimt" is trying to resolve your problem. Here are my questions for part one of the homework for week one. Note: Some numbers may be different then what you had, so it may not help, but you can try. Also the picture is missing for the first part, but maybe you can at least do the second part. Good Luck. -------------------------------------------------------------------- H1P1: Resistor Combinations This problem investigates how resistors combine. Consider the three resistor networks shown below: What is the equivalent resistance as an algebraic expression (in terms of R) of network A as viewed from its port? What is the equivalent resistance as an algebraic expression (in terms of R) of network B as viewed from its port? What is the equivalent resistance as an algebraic expression (in terms of R) of network as C viewed from its port? You are given three resistors: two 4Ω resistors and one 6Ω resistor. What is the value in Ohms ( Ω ) of the largest-valued resistor that can be fabricated by combining these three resistors? What is the value in Ohms ( Ω ) of the smallest-valued resistor that can be fabricated by combining these three resistors? Given that each individual resistor can dissipate up to 1 watt of power before burning up, how much total power in watts ( W ) can the smallest-valued composite resistor dissipate before burning up? FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-16T00:10:19Z SecondChildTAG: yah this is questions disappeared how to find this question can i answer this? SecondChildUserIdTAG: 235727 SecondChildUserNameTAG: amor SecondChildCreateTimeTAG: 2012-09-16T00:23:35Z SecondChildTAG: Oh, I thought you had answer boxes with no questions, my apologies. Have you tried a different computer or browser, perhaps make sure you are scrolled to the top of the page? SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-16T00:26:54Z SecondChildTAG: so you mean i answer here? i still never found in my homework1 if i answer here only direct answer or need to copy the questions and put and answer never found the answer box and also they can see all my answers? SecondChildUserIdTAG: 235727 SecondChildUserNameTAG: amor SecondChildCreateTimeTAG: 2012-09-16T00:33:11Z SecondChildTAG: What is the equivalent resistance as an algebraic expression (in terms of R) of network A as viewed from its port? R= R1+R2+R3 or 3*R this is my answer question 1 so this is what i need to do penny? SecondChildUserIdTAG: 235727 SecondChildUserNameTAG: amor SecondChildCreateTimeTAG: 2012-09-16T00:41:10Z SecondChildTAG: @amor: Please do not enter answers in this forum. SecondChildUserIdTAG: 79361 SecondChildUserNameTAG: kimt SecondChildCreateTimeTAG: 2012-09-16T00:46:40Z SecondChildTAG: sorry kimt so how to answer for this question kimt? still never found in my homework1 its really sad for me i felt unfair to be honest im really really sad SecondChildUserIdTAG: 235727 SecondChildUserNameTAG: amor SecondChildCreateTimeTAG: 2012-09-16T00:50:26Z SecondChildTAG: Working on it. Please hold. SecondChildUserIdTAG: 79361 SecondChildUserNameTAG: kimt SecondChildCreateTimeTAG: 2012-09-16T00:53:39Z SecondChildTAG: ok kimt i will wait thanx for that SecondChildUserIdTAG: 235727 SecondChildUserNameTAG: amor SecondChildCreateTimeTAG: 2012-09-16T00:59:57Z SecondChildTAG: Hi amor, I sent you an email. Things will be okay :) SecondChildUserIdTAG: 79361 SecondChildUserNameTAG: kimt SecondChildCreateTimeTAG: 2012-09-16T01:21:00Z SecondChildTAG: thanx kimt really appreciated :) SecondChildUserIdTAG: 235727 SecondChildUserNameTAG: amor SecondChildCreateTimeTAG: 2012-09-16T01:48:25Z SecondChildTAG: sent back SecondChildUserIdTAG: 235727 SecondChildUserNameTAG: amor SecondChildCreateTimeTAG: 2012-09-16T01:52:23Z SecondChildTAG: Please check the HW now? SecondChildUserIdTAG: 11 SecondChildUserNameTAG: dormsbee SecondChildCreateTimeTAG: 2012-09-16T04:12:00Z SecondChildTAG: hi dormsbee need to ask coz i dont know the right keyword the question no.3 coz only 1 and 2 they gave the right keyword SecondChildUserIdTAG: 235727 SecondChildUserNameTAG: amor SecondChildCreateTimeTAG: 2012-09-16T12:08:48Z SecondChildTAG: what is the right keyword of this formula Rp= R2(R3+R4)/R2+(R3+R4) or Rt+ R1+ Rp so what is the right keyword ? SecondChildUserIdTAG: 235727 SecondChildUserNameTAG: amor SecondChildCreateTimeTAG: 2012-09-16T12:10:58Z SecondChildTAG: You need to use an "*" for any multiplication input. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-16T16:01:59Z IndexTAG: 4308 TitleTAG: plz I need help H1P1 DISAPPEAR I CAN'T FAIND H1P1 QUESTION UserIdTAG: 156535 UserNameTAG: badawiIiIi CreateTimeTAG: 2012-09-15T23:24:36Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Please put issues such as this in the troubleshooting forum. What is the last thing you typed into H1P1, if any? ![Troubleshooting][1] [1]: https://dl.dropbox.com/u/45575056/troubleshooting.png FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-15T23:47:52Z SecondChildTAG: I am fixing your problem. SecondChildUserIdTAG: 79361 SecondChildUserNameTAG: kimt SecondChildCreateTimeTAG: 2012-09-16T01:55:46Z FirstChildTAG: Hi Badawilili! Can I help you? You have the Link [Here][1] to Homework 1. Do you have difficulties to see the web page? Have you an updated version of a web browser? If it helps, You can Press F5 to refresh your Web page. I hope that you could solve your issue. See you! Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_1/Basic_Circuit_Analysis/ FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-16T01:51:04Z IndexTAG: 4309 TitleTAG: Anyone able to solve Lab2 work It does not seem so straight forward... UserIdTAG: 364851 UserNameTAG: Varindra CreateTimeTAG: 2012-09-15T22:37:45Z VoteTAG: 0 CoursewareTAG: Week 2 / Superposition experiment CommentableIdTAG: 6002x_Superposition_experiment NumberOfReplyTAG: 4 FirstChildTAG: This is a superposition exercise. The equation listed in the lab basically shows both states: Vout=(1/2)*V1 + (1/6)*V2 is equal to: this state + that state. Take them one at a time and then add together. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-09-16T00:11:00Z FirstChildTAG: is easy, you have to combine resistors to have in the output the function (1/2)V1+ (1/6)V2, you only have to think a little to get the correct combination, and how many of resistors you need. FirstChildUserIdTAG: 73837 FirstChildUserNameTAG: ricardo87 FirstChildCreateTimeTAG: 2012-09-15T23:03:20Z SecondChildTAG: I would caution against using words like "easy". What is easy for you may not be so easy for others. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-09-15T23:22:56Z SecondChildTAG: ok sorry, i will bear in mind that,i dont want to hurt feelings, i just want to help, SecondChildUserIdTAG: 73837 SecondChildUserNameTAG: ricardo87 SecondChildCreateTimeTAG: 2012-09-15T23:28:31Z SecondChildTAG: No worries. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-09-15T23:38:45Z SecondChildTAG: Must be include third resistor between R1 and R2 - R3. And one terminal R3 connect to ground. With two resistor R1 and R2 no decision. SecondChildUserIdTAG: 402617 SecondChildUserNameTAG: Vl SecondChildCreateTimeTAG: 2012-09-16T00:09:06Z SecondChildTAG: Use a voltage divider network for both inputs so that it adds up as desired at the output. SecondChildUserIdTAG: 199839 SecondChildUserNameTAG: vijaykrishnankk SecondChildCreateTimeTAG: 2012-09-16T03:59:06Z SecondChildTAG: so you believe that we need 4 resistors SecondChildUserIdTAG: 400832 SecondChildUserNameTAG: panosbr SecondChildCreateTimeTAG: 2012-09-16T18:38:47Z SecondChildTAG: i hav done the lab by using 3 resistors as said by 'VI' SecondChildUserIdTAG: 183169 SecondChildUserNameTAG: JerinS SecondChildCreateTimeTAG: 2012-09-18T10:03:30Z SecondChildTAG: consider the two voltage as two. for v1 fully neglect v2 part(like superposition).then make it to give value of v1/2 by adding appropriate resistors(by voltage divider rule). now go to v2, make it to give v2/6. by solving these two as separate, then make connection between them. observe the output graph. again adjust the resistors for small change in value. remember you will get answer in between 3ohm to 1ohm range including 3 and 1. SecondChildUserIdTAG: 378967 SecondChildUserNameTAG: krish2012 SecondChildCreateTimeTAG: 2012-09-19T18:59:14Z FirstChildTAG: Please note: drawing the wire to the bottom of the ground symbol...not the top of it (inverted "T"). For a long time I could not figure out why I could not get the voltage divider to not work.. FirstChildUserIdTAG: 222527 FirstChildUserNameTAG: sebinanithottam FirstChildCreateTimeTAG: 2012-09-18T22:23:44Z FirstChildTAG: Hi Varindra ! Now that the deadline has passed you can take a look at [here][1] :) [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_S3E2_Linear_Combinations_of_Source_Strengths/threads/506a280d036af91f000000c9 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-02T00:25:55Z IndexTAG: 4310 TitleTAG: lab 2 hey everyone, pls help me with lab2. i hv spent hrs on it but no luck thus far. UserIdTAG: 285945 UserNameTAG: lindalapiso CreateTimeTAG: 2012-09-15T22:36:23Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: where did you stick? FirstChildUserIdTAG: 73837 FirstChildUserNameTAG: ricardo87 FirstChildCreateTimeTAG: 2012-09-15T23:04:58Z FirstChildTAG: Must be include third resistor between R1 and R2 - R3. And one terminal R3 connect to ground another terminal between R1 and R2. With two resistor R1 and R2 no decision FirstChildUserIdTAG: 402617 FirstChildUserNameTAG: Vl FirstChildCreateTimeTAG: 2012-09-16T00:18:18Z SecondChildTAG: thannks SecondChildUserIdTAG: 197539 SecondChildUserNameTAG: irfansultan94 SecondChildCreateTimeTAG: 2012-09-17T07:19:27Z SecondChildTAG: but i cant find out the exact so,ution yet???????????? SecondChildUserIdTAG: 197539 SecondChildUserNameTAG: irfansultan94 SecondChildCreateTimeTAG: 2012-09-17T07:49:26Z SecondChildTAG: how does u decide about the no. of resistors is 3? plz give the concept... SecondChildUserIdTAG: 114864 SecondChildUserNameTAG: Vikaash SecondChildCreateTimeTAG: 2012-09-17T11:46:49Z SecondChildTAG: consider the two voltage as two. for v1 fully neglect v2 part(like superposition).then make it to give value of v1/2 by adding appropriate resistors(by voltage divider rule). now go to v2, make it to give v2/6. by solving these two as separate, then make connection between them. observe the output graph. again adjust the resistors for small change in value. remember you will get answer in between 3ohm to 1ohm range including 3 and 1. SecondChildUserIdTAG: 378967 SecondChildUserNameTAG: krish2012 SecondChildCreateTimeTAG: 2012-09-19T18:59:56Z FirstChildTAG: Hi lindalapiso ! Now that the deadline has passed you can take a look at [here][1] :) [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_S3E2_Linear_Combinations_of_Source_Strengths/threads/506a280d036af91f000000c9 FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-10-02T00:25:10Z IndexTAG: 4311 TitleTAG: Help !!! Hi i have a problem, i was making the homework and at the time of upload the page say me. Invalid input: Could not parse 'R+((R)(R+R))/(R+R+R)' as a formula Please helpme i'm new in this course. thanks UserIdTAG: 251230 UserNameTAG: esle CreateTimeTAG: 2012-09-15T22:22:55Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: If somebody did it HW1P1 please let me know or help me with questions answers. FirstChildUserIdTAG: 243273 FirstChildUserNameTAG: verma1129 FirstChildCreateTimeTAG: 2012-09-16T00:40:26Z FirstChildTAG: Rewrite this R+2*R/3 FirstChildUserIdTAG: 402617 FirstChildUserNameTAG: Vl FirstChildCreateTimeTAG: 2012-09-15T23:46:53Z FirstChildTAG: you have to be carfully with the multiplication sign, alltime you have to use * FirstChildUserIdTAG: 73837 FirstChildUserNameTAG: ricardo87 FirstChildCreateTimeTAG: 2012-09-15T22:27:33Z SecondChildTAG: thanks is corrected SecondChildUserIdTAG: 251230 SecondChildUserNameTAG: esle SecondChildCreateTimeTAG: 2012-09-15T23:47:40Z IndexTAG: 4312 TitleTAG: S3.E1 The answer must be -1.69V. The answer is 6.2V if the voltage source is not rotated. UserIdTAG: 119131 UserNameTAG: DavidNementzik CreateTimeTAG: 2012-09-15T22:02:50Z VoteTAG: 0 CoursewareTAG: Week 2 / Node equation review CommentableIdTAG: 6002x_S3E1_NodeEquationReview NumberOfReplyTAG: 1 FirstChildTAG: I got -1.69 originally as well. To do it properly you have to reverse the measured voltage V2 from -7.2 to 7.2 *because it is rotated*. It is very tricky wording and I'm not positive why they made us do it. FirstChildUserIdTAG: 74863 FirstChildUserNameTAG: Drolyt FirstChildCreateTimeTAG: 2012-09-16T02:45:27Z SecondChildTAG: Well, I will say that after you get burned a few times by the signs, you pay close attention to their orientation. SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-09-18T05:53:55Z IndexTAG: 4313 TitleTAG: getting all the answers to H1P3 WRONG EVEN THEY ALL ARE RIGHT :( ALL THE ANSWERS FOR MY THIRD QUESTIONS ARE RIGHT AND CHECKED AND COUNTER CHECKED THEM BUT IT IS SHOWING THEM CROSSS AND WRONG WHY IS THIS SO? UserIdTAG: 227777 UserNameTAG: umairhassankhanniazi CreateTimeTAG: 2012-09-15T21:48:19Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: First, remove the shift and Caps lock key from your keyboard with a screwdriver. Second, do forum search for H1P3, choose from the many threads that already cover H1P3. It's tricky, but with perseverance you will succeed. Have fun. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-15T22:13:02Z SecondChildTAG: thanx for that pennypacker thats the problem of my friend umair but how he answer now he is not online i guess, its his birthday today maybe he is busy feel so sorry for him coz if he never back to online he can't pass his homework its deadline right SecondChildUserIdTAG: 235727 SecondChildUserNameTAG: amor SecondChildCreateTimeTAG: 2012-09-15T22:22:24Z SecondChildTAG: Happy Birthday, Umairhassankhanniazi! He can miss 2 out of the 10 homeworks without consequence. So as long as he is not having too many Birthdays in the next two months he should be OK. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-15T23:00:07Z SecondChildTAG: thanx all i got it now :) SecondChildUserIdTAG: 227777 SecondChildUserNameTAG: umairhassankhanniazi SecondChildCreateTimeTAG: 2012-09-16T06:31:07Z IndexTAG: 4314 TitleTAG: its unfair for me what happen some of question in my homework 1 are never found so i can't answer? 1st i saw the question but i never found now that i really sure what is those answer so how to get a perfect score? UserIdTAG: 235727 UserNameTAG: amor CreateTimeTAG: 2012-09-15T21:45:40Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Are you saying you can't actually see the questions to homework 1? FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-15T22:32:27Z SecondChildTAG: yah when i answer first i saw the question then when i answered it said invalid keyword so i set aside the time and answer some question but after those i try to answer back to find a right keywork but i never found those question now SecondChildUserIdTAG: 235727 SecondChildUserNameTAG: amor SecondChildCreateTimeTAG: 2012-09-15T22:47:53Z SecondChildTAG: these question never found in my homeworks 1. what is the equivalent restance as an algebraic expression (in terms of R) of network A as viewed from its port 2.network B 3. network C 4. what is the value ohms the largest valued 5 the smallest valued 6. how much the total power in watts these all question never found in my homework so when i try to see my progress i never perfect its 70% coz i never answer all but coz i never found these question SecondChildUserIdTAG: 235727 SecondChildUserNameTAG: amor SecondChildCreateTimeTAG: 2012-09-15T22:53:05Z SecondChildTAG: Best to file a bug report, either that or depending on where you live it may be too late, is it still the 15th there? SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-15T22:56:09Z SecondChildTAG: its 16 here sunday may not too late have few hrs to go i guess but how to do a bug report? SecondChildUserIdTAG: 235727 SecondChildUserNameTAG: amor SecondChildCreateTimeTAG: 2012-09-15T23:06:08Z SecondChildTAG: penny pls just tell my how to file a bug report need to find some of the questions of my home works i have few hrs to go maybe 2 hrs left or not sure but try want to answer those to get a perfect score SecondChildUserIdTAG: 235727 SecondChildUserNameTAG: amor SecondChildCreateTimeTAG: 2012-09-15T23:23:06Z SecondChildTAG: Just start a new thread in the troubleshooting section. In the old forum, you could add the tag [bug] to your thread. I am unsure if this still works. Good luck! SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-15T23:32:53Z SecondChildTAG: Please post in the troubleshooting forum. Can you take a screenshot of the homework page and what you entered before it disappeared? SecondChildUserIdTAG: 79361 SecondChildUserNameTAG: kimt SecondChildCreateTimeTAG: 2012-09-15T23:39:45Z SecondChildTAG: yah it disappeared how to post a troubleshooting forum i tried to find it never know i have few hrs left to go right? i want to answer those disappeared questions SecondChildUserIdTAG: 235727 SecondChildUserNameTAG: amor SecondChildCreateTimeTAG: 2012-09-15T23:48:16Z SecondChildTAG: thanx penny and thanx kimt i found the troubleshooting forum...kimt how many hrs left its sunday here 16 sept so i dont know the exact time deadline of my homeworks SecondChildUserIdTAG: 235727 SecondChildUserNameTAG: amor SecondChildCreateTimeTAG: 2012-09-16T00:00:12Z IndexTAG: 4315 TitleTAG: H1P2 I can't seem to get an answer for this question. "How many of the KVL equations are independent?" I've tried entering in values from 0-10 and they are all incorrect. UserIdTAG: 234060 UserNameTAG: jipjipbee CreateTimeTAG: 2012-09-15T21:27:50Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: "A KVL equation can be written around each of these loops." Hint: So if you can write a KVL equation for each of the loops, how many are independent? FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-15T21:34:36Z SecondChildTAG: That is really weird because the first answer I typed in shows up as correct now. Probably a glitch I guess. SecondChildUserIdTAG: 234060 SecondChildUserNameTAG: jipjipbee SecondChildCreateTimeTAG: 2012-09-15T21:43:34Z SecondChildTAG: 2 SecondChildUserIdTAG: 202249 SecondChildUserNameTAG: ABZ SecondChildCreateTimeTAG: 2012-09-15T22:33:27Z SecondChildTAG: (You should remove your answer ABZ.) Glad you got it. Sometimes you need to wait a few seconds for the green check mark to show up. You were too fast! lol, you might see a little box pop up in the lower left of Firefox that says "Typsetting Math %100", or something similar when you submit your answers. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-15T22:41:37Z SecondChildTAG: to Pennypacker Re:You should remove your answer ABZ Clarify,pls ? SecondChildUserIdTAG: 202249 SecondChildUserNameTAG: ABZ SecondChildCreateTimeTAG: 2012-09-15T22:46:08Z SecondChildTAG: You can't post answers to homework 1 before the 16th. SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-15T22:49:18Z SecondChildTAG: Sorry, in my city is 16/09. SecondChildUserIdTAG: 202249 SecondChildUserNameTAG: ABZ SecondChildCreateTimeTAG: 2012-09-15T22:53:21Z FirstChildTAG: the right answer is two. I actually think that some questions are silly. For instance: it´s obvious that we have 3 loops in that circuit, but it´s also in our face that the little ones forms an indenpendent systems of equations.. Best Rergards FirstChildUserIdTAG: 263241 FirstChildUserNameTAG: yvesauad FirstChildCreateTimeTAG: 2012-09-15T22:55:45Z IndexTAG: 4316 TitleTAG: Is there a limit to how many times I can check my homework? Thank you. Is there a limit to how many times I can check my homework? Thank you. UserIdTAG: 234060 UserNameTAG: jipjipbee CreateTimeTAG: 2012-09-15T21:10:10Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: No. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-15T21:14:37Z FirstChildTAG: HOW DO YOU TURN YOUR HOMEWORK IN. I DO NT SEE A LIMIT AS I CHECKED ALL AND IT ALLOWS YOU TO CHECK AND RE DO. FirstChildUserIdTAG: 233700 FirstChildUserNameTAG: ANDREW77 FirstChildCreateTimeTAG: 2012-09-15T21:19:55Z SecondChildTAG: Please do not shout, we can all hear you. Once your homework is "checked", it is already submitted. :) SecondChildUserIdTAG: 292543 SecondChildUserNameTAG: Pennypacker SecondChildCreateTimeTAG: 2012-09-15T21:25:10Z FirstChildTAG: In other words don't hit "check" unless you are ready to have all your answers scored. That action is irreversible. Doesn't make any difference why the answer is incorrect: typo, whatever. FirstChildUserIdTAG: 241170 FirstChildUserNameTAG: Mauette FirstChildCreateTimeTAG: 2012-09-15T23:23:26Z SecondChildTAG: Although since you there is no limit to how many times you can have it checked, you don't have to worry about when you hit "check". SecondChildUserIdTAG: 21541 SecondChildUserNameTAG: JSChambers SecondChildCreateTimeTAG: 2012-09-16T00:28:27Z IndexTAG: 4317 TitleTAG: Wave cycle && wave form Are these 2 same? UserIdTAG: 138239 UserNameTAG: LaFolle CreateTimeTAG: 2012-09-15T20:51:00Z VoteTAG: 0 CoursewareTAG: Week 1 / AC power CommentableIdTAG: 6002x_S1E3_AC_power NumberOfReplyTAG: 1 FirstChildTAG: Waveform is the shape of the wave. ie: square, sinusoidal etc. A wave cycle could be measured in cycles per second aka Hertz. One Hertz = One cycle per second. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-15T21:12:58Z IndexTAG: 4318 TitleTAG: what can i do ? the homework check does not accept the answer and tells me:Invalid input: Could not parse '3R' as a formula !!! what is wrong ??? UserIdTAG: 284144 UserNameTAG: reem28 CreateTimeTAG: 2012-09-15T20:43:20Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: 3*R,use the symbols that are convention. FirstChildUserIdTAG: 156974 FirstChildUserNameTAG: ManojKumar FirstChildCreateTimeTAG: 2012-09-15T20:45:45Z SecondChildTAG: I`ve plased R+R+R :-) SecondChildUserIdTAG: 202249 SecondChildUserNameTAG: ABZ SecondChildCreateTimeTAG: 2012-09-15T22:35:05Z FirstChildTAG: 3R should be written as 3*R FirstChildUserIdTAG: 57012 FirstChildUserNameTAG: sam747 FirstChildCreateTimeTAG: 2012-09-15T21:38:13Z FirstChildTAG: lOOK AT YOUR SYNTEX FORMAT. I HAD THE SAME PROBLEM, IT IS OICKY ON THE FORM. FirstChildUserIdTAG: 233700 FirstChildUserNameTAG: ANDREW77 FirstChildCreateTimeTAG: 2012-09-15T21:20:55Z IndexTAG: 4319 TitleTAG: lab2 how do we have to perform lab2..plz help.. UserIdTAG: 137918 UserNameTAG: jaisneha CreateTimeTAG: 2012-09-15T19:47:39Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Search function, ;) this has been asked many, many times already. Have fun. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-15T20:54:08Z SecondChildTAG: i guess that i did the lab well but still show me that is wrong, who can give me a hand with this??, i spent so much time on the lab, and mathematically i got the function, (1/2)V1 + (1/6)V2 SecondChildUserIdTAG: 73837 SecondChildUserNameTAG: ricardo87 SecondChildCreateTimeTAG: 2012-09-15T21:46:03Z SecondChildTAG: consider the two voltage as two. for v1 fully neglect v2 part(like superposition).then make it to give value of v1/2 by adding appropriate resistors(by voltage divider rule). now go to v2, make it to give v2/6. by solving these two as separate, then make connection between them. observe the output graph. again adjust the resistors for small change in value. remember you will get answer in between 3ohm to 1ohm range including 3 and 1. SecondChildUserIdTAG: 378967 SecondChildUserNameTAG: krish2012 SecondChildCreateTimeTAG: 2012-09-19T19:00:09Z FirstChildTAG: finally i did it!!! i dont know why, but when you put the value of a resistor in the lab you cant put it with math operations, an example: wrong way 200/3 good way 66.666666 i hope this can help you.... FirstChildUserIdTAG: 73837 FirstChildUserNameTAG: ricardo87 FirstChildCreateTimeTAG: 2012-09-15T22:31:45Z SecondChildTAG: Seriously i wasted 2 days on this ...oh god :( SecondChildUserIdTAG: 185921 SecondChildUserNameTAG: Ravi_Teja SecondChildCreateTimeTAG: 2012-09-17T18:42:37Z SecondChildTAG: Thanks [@ricardo87][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/users/73837 SecondChildUserIdTAG: 185921 SecondChildUserNameTAG: Ravi_Teja SecondChildCreateTimeTAG: 2012-09-17T18:43:28Z IndexTAG: 4320 TitleTAG: week1 homework: H1P3 Hi all, I've the following concern about two questions of the poor workmanship task: question: what power was being dissipated in H1? To calculate it, I first calculated the current flowing through H1 (calculated equivalent resistor of H1 and H2 and H3 together, since voltage is known, from Ohm's low I got current). Then calculated the voltage drop on H1 from voltage divider, since H1 and H2 are equal, the voltage on H1 is half of the one from voltage supply. Than power dissipated on the H1 = calculated current * calculated voltage. However calculating it this way, I can not get the correct value... question: what power was being dissipated in H2 (or in H3)? From the previous calculations, I divide the current by 2 (since H2 and H3 are equal, and from KCL: flowing in current must be equal to the flowing out current). From the previous calculation I also have the voltage drop on H2 and H3 - they are the same since H2 and H3 are in parallel - this is half of the supply voltage. And again P=V*I ... and again wrong result. What is curious, taking this numbers I can calculate correct answer for the last question: total heating power on Joe's shop. (multiplying current flowing through H1 and supplied voltage) Thanks in advance for your feedback! UserIdTAG: 354638 UserNameTAG: reszelaz CreateTimeTAG: 2012-09-15T19:43:36Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: You made an error when you said that since H1 and H2 are equal, H1 has half the voltage across it. You forgot that H2 and H3 are in parallel. FirstChildUserIdTAG: 270011 FirstChildUserNameTAG: SnowmanZA FirstChildCreateTimeTAG: 2012-09-15T19:53:43Z SecondChildTAG: Hi SnowmanZA, Thanks for your reply. Now I got it! To properly calculate to voltage drops across the heaters using the voltage divider: First I had to calculate resistance of parallel heaters H2 and H3 together. Than from voltage divider VH2H3 = V*RH2H3/(RH2H3+RH1). Now my results are correct. Cheers! SecondChildUserIdTAG: 354638 SecondChildUserNameTAG: reszelaz SecondChildCreateTimeTAG: 2012-09-16T10:57:06Z SecondChildTAG: Yes, indeed! Checking out the Voltage Divider theory is enough to solve the rest of this exercise...thank you all for your help! SecondChildUserIdTAG: 314477 SecondChildUserNameTAG: ignos SecondChildCreateTimeTAG: 2012-09-16T15:24:07Z IndexTAG: 4321 TitleTAG: Should I be able to login to the MITx Site with my edX account? There are a few links to the MITx site from within this forum, but when I attempt to click through to them, I am redirected to the 6.002x course front page where it would seem I need to login to get furthur. My edX username/pw do not seem to work there.... Urrr? Why are they independent? UserIdTAG: 204944 UserNameTAG: squirrelerriuqs CreateTimeTAG: 2012-09-15T19:32:04Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I would imagine. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-15T20:07:27Z IndexTAG: 4322 TitleTAG: lab-2 hi everybody i am not just able to solve the problem in lab-2.. i am not getting how is it possible to reduce the sqr wave to half of its magnitude at the output terminals. if possible help me . thnx UserIdTAG: 395966 UserNameTAG: sambo007 CreateTimeTAG: 2012-09-15T19:20:01Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: Think of a volume control in a circuit.(Voltage Divider) Combine half of V1, with one sixth of V2. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-15T20:05:52Z FirstChildTAG: Must be include third resistor between R1 and R2 - R3. And one terminal R3 connect to ground. With two resistor R1 and R2 no decision FirstChildUserIdTAG: 402617 FirstChildUserNameTAG: Vl FirstChildCreateTimeTAG: 2012-09-16T06:10:20Z IndexTAG: 4323 TitleTAG: Hi hi , h have a problem , there is no numbers and i find that : Math Processing Error !! what can i do ?? UserIdTAG: 284144 UserNameTAG: reem28 CreateTimeTAG: 2012-09-15T19:10:09Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: post in details.... math processing errors are those errors whn the said mathematical symbols or digits are not present FirstChildUserIdTAG: 395966 FirstChildUserNameTAG: sambo007 FirstChildCreateTimeTAG: 2012-09-15T19:16:53Z FirstChildTAG: What is the problem on which you get Math Processing Error? Does it happen intermittently (i.e. what happens if you refresh) or all the time? Also, the "Troubleshooting" forum (in the forum dropdown) is the best place for these reports. FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-15T19:20:54Z IndexTAG: 4324 TitleTAG: H1P3: Poor Workmanship Joe wants to heat his 12'X20' workshop with electric heat. He has hired the HACME electrician company to build the system. They propose to use three 1560.0W 240V baseboard heaters to provide a total heating capacity of 4680.0W. (A heater is basically a resistor. This is not quite true, because there is a thermostatic switch incorporated into the heater and because the resistance of a heater varies a bit with its temperature. But we will use a linear resistor as a model of a heater.) In the proposed system the heaters are connected in parallel with the 240V 60Hz AC power line. ![enter image description here][1] The problem is to find power dissipated in resistors when they are conected like below. [1]: https://www.edx.org/static/content-mit-6002x/images/circuits/heaters-bad.39b87d3bf0e6.gif So i have tried to calculate resistances when resistors are connected parallel H = U^2/P = 36.92. Each resistor dissipates the same amount of power and by Kirchoff's laws each of them have the same voltage (in parallel) so resistance should be the same for each resistor. Then modified network in picture calculating power in H1 gives P = I^2 * H1 incorrect answer (I = U/(H+H/2)). Back to beginning (resistors in parallel) there is given total power 4680.0W and total resistance which i took R = H^3/3H = 454.36 . Taking total resistance gives total power P = U^2/R = 126.77 , so why is this happening? :D I also tried forgetting that each resistor dissipates the same power and by using Kirchoff's law (on shown picture) and total power H1H2H3/H1+H2+H3 = U^2/P from first condition but answer still is incorrect. Could somebody help? What is the way of getting correct answer? UserIdTAG: 373763 UserNameTAG: 54321 CreateTimeTAG: 2012-09-15T19:00:31Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: I am going with problem regarding H1P1,While writing the expression OF R...So help me out FirstChildUserIdTAG: 282892 FirstChildUserNameTAG: wiky FirstChildCreateTimeTAG: 2012-09-15T22:36:46Z FirstChildTAG: Find the resistance of each heater. P=V^2/R so 1560 = 240^2/R so 57600/1560 = R Then you can use a simple voltage divider method to work out the voltage across each heater. Then you can easily work out how much power each one dissipates. FirstChildUserIdTAG: 270011 FirstChildUserNameTAG: SnowmanZA FirstChildCreateTimeTAG: 2012-09-15T19:57:47Z FirstChildTAG: First get the total current drawn when all the heaters are on, by simply using I=P/V, then the current drawn by each resistor = 1/3 of total current since all the resistors have the same value. With current drawn for each resistor known, you can calculate the resistance for each resistor using R=V/I, and V=240V since they are all connected in parallel. You can then move to Q3 and calculate the equivalent resistance. This will allow you to calculate the current drawn; since voltage is known (240V). Then Power dissipated = I^2 R...I got the correct answer. For Q4, both H2 and H3 draw 1/2 the current calculated above in Q3. Therefore, power dissipated in H2 is I^2 R, and we know R and I...leading to correct answer. For Q5, sum of power dissipated by all three resistors is calculated by adding power dissipated individually for each resistor, using P=I^2R for each resistor, leading to correct answer FirstChildUserIdTAG: 57012 FirstChildUserNameTAG: sam747 FirstChildCreateTimeTAG: 2012-09-15T21:11:02Z SecondChildTAG: that is rigth, you answer was very helpfull SecondChildUserIdTAG: 281856 SecondChildUserNameTAG: LuisMag SecondChildCreateTimeTAG: 2012-09-15T21:41:12Z SecondChildTAG: does R=36 ohms? SecondChildUserIdTAG: 290014 SecondChildUserNameTAG: Nazdery SecondChildCreateTimeTAG: 2012-09-15T21:54:34Z SecondChildTAG: nvm got it, was making a calculator mistake! thank you SecondChildUserIdTAG: 290014 SecondChildUserNameTAG: Nazdery SecondChildCreateTimeTAG: 2012-09-15T22:00:54Z SecondChildTAG: Thank you so much. You explained it perfectly. I spent almost 2 days trying to figure out the last 3 questions and no other post really made sense to me but yours. SecondChildUserIdTAG: 224302 SecondChildUserNameTAG: tcc626 SecondChildCreateTimeTAG: 2012-09-15T22:32:32Z SecondChildTAG: sam747 has given a good method..I want to add further in Q4 you can use this formula which is much easy at the equivalent circuit,P=V^2/R for H2....It will save time ...Well there should have knowledge of both technique... SecondChildUserIdTAG: 282892 SecondChildUserNameTAG: wiky SecondChildCreateTimeTAG: 2012-09-15T22:35:19Z SecondChildTAG: i realy dont understand it yet :) pld hep SecondChildUserIdTAG: 227777 SecondChildUserNameTAG: umairhassankhanniazi SecondChildCreateTimeTAG: 2012-09-16T06:52:34Z SecondChildTAG: ok so total resistance is 18.11 , there fore if H1=H2=H3=H then H =18.11/3= 6.03 ohms. P=I2R , my answer is still wrong . Please can some one help. SecondChildUserIdTAG: 178115 SecondChildUserNameTAG: nehamakhija SecondChildCreateTimeTAG: 2012-09-16T18:06:26Z IndexTAG: 4325 TitleTAG: Still struggle to get i1... I'm still struggling to understand how to get i1 from this diagram. I really struggled with the last exercise as well. Is there an easier trick to remembering how to get to it? I've tried to use KVL and KCL, but I'm not quite sure where to pull the figures from. Any help in the matter would be greatly appreciated. UserIdTAG: 207204 UserNameTAG: hgustavii CreateTimeTAG: 2012-09-15T18:23:15Z VoteTAG: 0 CoursewareTAG: Week 1 / Series and parallel example CommentableIdTAG: 6002x_S2E4_Series_and_Parallel NumberOfReplyTAG: 3 FirstChildTAG: You have got a full resistance of circuit (from third question), and you have got a single voltage source in the circuit. So, you can just calculate it by i=V/R. But, I still don't understand, why i1 is positive :) I think, it should be negative, because it's going from '-' to '+'... FirstChildUserIdTAG: 322444 FirstChildUserNameTAG: koluch FirstChildCreateTimeTAG: 2012-09-15T20:34:45Z FirstChildTAG: I calculated the equivalent resistance of the entire circuit and then got i1 by using I=V/R, since i1 is current drawn across the entire circuit. FirstChildUserIdTAG: 57012 FirstChildUserNameTAG: sam747 FirstChildCreateTimeTAG: 2012-09-15T21:36:25Z FirstChildTAG: Thanks for your help! Got it now. Now for the voltages... ;) FirstChildUserIdTAG: 207204 FirstChildUserNameTAG: hgustavii FirstChildCreateTimeTAG: 2012-09-16T08:03:54Z IndexTAG: 4326 TitleTAG: practice questions will practice questions within lectures will b evaluated or not ??? UserIdTAG: 72647 UserNameTAG: NEEL11 CreateTimeTAG: 2012-09-15T18:22:27Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: No your homework, labs and test are the only ones that count towards your overall grade which you can see on your progress tab! Hope that helps :D FirstChildUserIdTAG: 238005 FirstChildUserNameTAG: isisbocardo FirstChildCreateTimeTAG: 2012-09-15T20:34:03Z IndexTAG: 4327 TitleTAG: H1P2 Can anyone help me find the value of i4 ?? and also the power of the voltage source and resistors? UserIdTAG: 75849 UserNameTAG: Sachin_Joshi CreateTimeTAG: 2012-09-15T18:13:23Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: i4 = i3 = -(i1+i2) i1 = - current source(given) i2 = V2/R2 = V1/R2 --V1(from part 6) -- R2(given) Power of voltage source P=IV I = i4 V = v4 (given) FirstChildUserIdTAG: 132171 FirstChildUserNameTAG: sallysalem FirstChildCreateTimeTAG: 2012-09-15T18:42:09Z SecondChildTAG: Power of voltage source seems incorrect P must be 4W, but this walue marked as wrong any time SecondChildUserIdTAG: 202249 SecondChildUserNameTAG: ABZ SecondChildCreateTimeTAG: 2012-09-15T22:41:32Z SecondChildTAG: How can I get V1 ? SecondChildUserIdTAG: 410943 SecondChildUserNameTAG: JPalaciosRoman SecondChildCreateTimeTAG: 2012-09-15T23:45:09Z SecondChildTAG: it gives wrong answer for value of i4 SecondChildUserIdTAG: 320074 SecondChildUserNameTAG: salmanumer76 SecondChildCreateTimeTAG: 2012-09-16T05:25:44Z SecondChildTAG: @ABZ Check the sign of current is Positive if move from - to + but in this case current move from + to - so use -i4 SecondChildUserIdTAG: 132171 SecondChildUserNameTAG: sallysalem SecondChildCreateTimeTAG: 2012-09-16T21:31:35Z SecondChildTAG: @ JPalaciosRoman use kcl and kvl v1 = v2 = v3+v4 i1 + i2 + i3 = 0 i1 = -8 (given) i2 = v2/R2 -- v2 = v1 -- R2 (given) i3 = (v2-v4)/R3 --v2 = v1 --v4(given) -- R3(given) Solve i1 + i2 + i3 = 0 using only V1 as unknown -8 + v1 + (v1-4)/3 = 0 So v1 = 7 = v2 SecondChildUserIdTAG: 132171 SecondChildUserNameTAG: sallysalem SecondChildCreateTimeTAG: 2012-09-16T21:37:48Z SecondChildTAG: @salmanumer76 i4 = -(i1+i2) i1 = -8 i2 = v2/R2 = 7/1 i4 = - (-8 + 7) = - (-1) = 1 A SecondChildUserIdTAG: 132171 SecondChildUserNameTAG: sallysalem SecondChildCreateTimeTAG: 2012-09-16T21:39:37Z IndexTAG: 4328 TitleTAG: enrolled on an android! Hi, I enrolled on an Android phone and now can't log in using a PC. Silly me.This means I can't complete the lab 1. When I try to log in using a PC or laptop, I get an error message saying "login or password incorrect". I can't enrol with a new name as there seems to be a cut-off date. Help! UserIdTAG: 394146 UserNameTAG: shp CreateTimeTAG: 2012-09-15T18:01:45Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Change your password. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-15T19:57:32Z SecondChildTAG: Yes..use the forgot password link..I suppose you remember the email you used to sign up..that should be enough to reset your password and get a new one.. SecondChildUserIdTAG: 173147 SecondChildUserNameTAG: cruiser_rahit SecondChildCreateTimeTAG: 2012-09-15T21:04:54Z IndexTAG: 4329 TitleTAG: finding v2 sorry i really need help finding v2... the rest should be a piece of cake... thank you. UserIdTAG: 133084 UserNameTAG: Warrensiggs CreateTimeTAG: 2012-09-15T18:00:33Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL/KCL example CommentableIdTAG: 6002x_S2E3_Using_KVL_KCL_and_VI_constraints NumberOfReplyTAG: 1 FirstChildTAG: KVL and KCL, V1+V2-V=0 .... i1 + I - i2 =0 but i1 = V1/R1 and i2=V2/R2 so you'll get V2=7.78 and V1=-5.78 FirstChildUserIdTAG: 268104 FirstChildUserNameTAG: SaulCardenasG FirstChildCreateTimeTAG: 2012-09-16T05:18:13Z IndexTAG: 4330 TitleTAG: Summary of Maxwell's equations Summary of Maxwell's Equations for those who want it $\epsilon_0\oint E\cdot dA = \sum Q$ $\oint B\cdot ds = \mu_0\int J\cdot dA + \mu_0\epsilon_0\frac{d}{dt}\int E\cdot dA$ $\oint E\cdot ds = -\frac{d}{dt}\int B\cdot dA$ $\oint B\cdot dA = 0$ UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-15T17:53:02Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 4331 TitleTAG: Problems with lab2 I made a mess in the lab2 and i need to reset its workspace , but i dont get how to make it, i tried to click on "check" but nothing happen, somebody could help me? thanks! UserIdTAG: 73837 UserNameTAG: ricardo87 CreateTimeTAG: 2012-09-15T17:47:15Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I'm looking into this for you. It would help me if you could show me your mess (a screenshot). In the future, please use the "Troubleshooting" forums for technical issues like this. FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-15T17:56:02Z SecondChildTAG: Hi, lab 2 grader was updated. Could you please try again? SecondChildUserIdTAG: 79361 SecondChildUserNameTAG: kimt SecondChildCreateTimeTAG: 2012-09-15T19:43:08Z SecondChildTAG: now i could clear the workspace, thanks for the support... SecondChildUserIdTAG: 73837 SecondChildUserNameTAG: ricardo87 SecondChildCreateTimeTAG: 2012-09-15T21:37:09Z SecondChildTAG: consider the two voltage as two. for v1 fully neglect v2 part(like superposition).then make it to give value of v1/2 by adding appropriate resistors(by voltage divider rule). now go to v2, make it to give v2/6. by solving these two as separate, then make connection between them. observe the output graph. again adjust the resistors for small change in value. remember you will get answer in between 3ohm to 1ohm range including 3 and 1. SecondChildUserIdTAG: 378967 SecondChildUserNameTAG: krish2012 SecondChildCreateTimeTAG: 2012-09-19T19:00:22Z IndexTAG: 4332 TitleTAG: home work 1 i know the effective resistance of three resistances in series each of resistance R, is 3R ... but if i type that it says parse error or wrong formula somethin can anyone tell me how to get give the answer UserIdTAG: 250869 UserNameTAG: hp07 CreateTimeTAG: 2012-09-15T17:41:24Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: use a '*' multiplication sign btwn the number and variable FirstChildUserIdTAG: 332941 FirstChildUserNameTAG: SabaSiddiqi FirstChildCreateTimeTAG: 2012-09-15T17:59:08Z FirstChildTAG: you have to type using + in between R FirstChildUserIdTAG: 168396 FirstChildUserNameTAG: rik2008 FirstChildCreateTimeTAG: 2012-09-15T17:44:24Z IndexTAG: 4333 TitleTAG: regarding several answers hello frnds in some of the questions there are more than one answer exists for eg: in ques S3EI: it shows two right answers "5.99" and "6.20" so i m confused which one is right , can anyone help me UserIdTAG: 301618 UserNameTAG: HIMANSHU0192 CreateTimeTAG: 2012-09-15T17:30:47Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: The correct answer is 6.2V, however the exercises will give you some margin in the answer to account for rounding errors in your calculations. FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-15T17:49:17Z IndexTAG: 4334 TitleTAG: independent equations how to identify independent equations both in KVL and KCL??? UserIdTAG: 140065 UserNameTAG: Madhav92 CreateTimeTAG: 2012-09-15T17:19:50Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL and KCL CommentableIdTAG: 6002x_kvl_and_kcl NumberOfReplyTAG: 1 FirstChildTAG: K = (n-1) + m. K-this equation, n - the number of nodes . m - number of simple loops FirstChildUserIdTAG: 202275 FirstChildUserNameTAG: Valya FirstChildCreateTimeTAG: 2012-09-15T19:26:02Z IndexTAG: 4335 TitleTAG: homework 1 problem ![enter image description here![\]\[1\]][1] [1]: https://edxuploads.s3.amazonaws.com/13477293202581103.jpg UserIdTAG: 357993 UserNameTAG: DEEPAK1100 CreateTimeTAG: 2012-09-15T17:15:55Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Troubleshooting NumberOfReplyTAG: 8 FirstChildTAG: use algebric expressions. eg. to add 2 resistors in series.. write R+R and similarly for parallel circuit. FirstChildUserIdTAG: 431224 FirstChildUserNameTAG: erachopra10 FirstChildCreateTimeTAG: 2012-09-15T18:03:06Z FirstChildTAG: what is the last solution to the power problem?? confused!! :( 3 is also not the right answer?? FirstChildUserIdTAG: 127800 FirstChildUserNameTAG: MNB FirstChildCreateTimeTAG: 2012-09-16T02:30:14Z FirstChildTAG: Use a * sign to indicate multiplication, so that instead of writing 3R write 3*R. Also, try to use parenthesis to write the solutions FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-15T17:51:46Z FirstChildTAG: Your pictures are so small I can't see what the problem is. Please explain. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-09-15T17:21:47Z FirstChildTAG: just only the formula of series that is network A network B is a parallel and C is series and parallel FirstChildUserIdTAG: 235727 FirstChildUserNameTAG: amor FirstChildCreateTimeTAG: 2012-09-15T17:39:04Z FirstChildTAG: Hi DEEPAK1100! I saw your images! I suggest you to: 1)Modify the AR for A*R (use *in multiplication). The same goes if A is a fraction like (1/2). 2) Where they ask you the disipation in watts, dont put in the box eg. 5 watts, only put 5 , without units ;) Try with that. If you have any doubt please ask. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-15T18:03:30Z FirstChildTAG: HW1-Q2: I am not sure what it expects in the answers, I know I have the right answer, I have tried several combinations but it does not like it for some reason. Can someone help. Does not like this: ![H1Q2-1][1] or this![H1Q2-2][2] or this ![H1Q2-3][3] What am I doing wrong? Help! [1]: https://edxuploads.s3.amazonaws.com/13477335441343693.jpg [2]: https://edxuploads.s3.amazonaws.com/13477335832667124.jpg [3]: https://edxuploads.s3.amazonaws.com/13477336919548556.jpg FirstChildUserIdTAG: 385224 FirstChildUserNameTAG: shilpiji991 FirstChildCreateTimeTAG: 2012-09-15T18:29:52Z SecondChildTAG: your answers are wrong. SecondChildUserIdTAG: 395953 SecondChildUserNameTAG: dustinge SecondChildCreateTimeTAG: 2012-09-15T18:42:13Z SecondChildTAG: Hi shilpiji991! Something it is wrong with your answers... If you see: ![enter image description here][1] [Ohm][Ohm][Ohm] / [Ohm] =[Ohm]^3/[Ohm]=[Ohm]^2 So, your result it is giving a Ohm^2 Unit as a Resistance and a Resistance Unit is Ohm without the ^2...So, something is wrong. Are you sure that you are applying well the Combination rules of Resistances? Remember that if you have two resistors in series, the total resistence will be the sum of that two resistors, that is to say: Rseries=R1+R2 Remember that if you have two resistors in paralell, the total resistence will be something like this: Rparalell=R1*R2/(R1+R2) I will like to help you. Can I do something? ;) Myriam. [1]: https://edxuploads.s3.amazonaws.com/13477335441343693.jpg SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-16T02:13:26Z FirstChildTAG: can u say me how to submit the homework FirstChildUserIdTAG: 278178 FirstChildUserNameTAG: sandy12992 FirstChildCreateTimeTAG: 2012-09-27T00:30:33Z IndexTAG: 4336 TitleTAG: help Given that each individual resistor can dissipate up to 1 watt of power before burning up, how much total power in watts ( W ) can the smallest-valued composite resistor dissipate before burning up? and guide me for last three questions of heater UserIdTAG: 164689 UserNameTAG: muhammadfaizan CreateTimeTAG: 2012-09-15T17:04:59Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: For the question you are asking and the heater questions you need to understand the concept of electric power. I recommend going through examples 1.13 to 1.16 of the textbook (pages 27-29) to see some similar examples. The equations you need to know are ohms law V = I*R and the ones shown in page 27 of the textbook FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-15T18:03:50Z IndexTAG: 4337 TitleTAG: Question around 6:40 in video You say that output when c = 0 and switch is open should be just straight Vs which makes sense, but I thought Vs was 10V and on the graph you have the output at 5V. Is this a mistake, or am I missing something? And 10/11 is .909, not 0.45, I think the example got screwed up somehow. 5/11 = 0.45 UserIdTAG: 105732 UserNameTAG: nunchucksuka CreateTimeTAG: 2012-09-15T16:29:46Z VoteTAG: 0 CoursewareTAG: Week 3 / Switch Resistor Model of a MOSFET CommentableIdTAG: 6002x_switch_resistor_model NumberOfReplyTAG: 0 IndexTAG: 4338 TitleTAG: h1-p2 hey can u help me through this number of nodes in the given ckt i am pretty much sure the ans:2 but it is indicating wrong can u help me UserIdTAG: 338265 UserNameTAG: Rehan37 CreateTimeTAG: 2012-09-15T16:20:42Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 3 FirstChildTAG: 3 nodes :) FirstChildUserIdTAG: 235727 FirstChildUserNameTAG: amor FirstChildCreateTimeTAG: 2012-09-15T16:22:04Z SecondChildTAG: hw is it possible SecondChildUserIdTAG: 338265 SecondChildUserNameTAG: Rehan37 SecondChildCreateTimeTAG: 2012-09-15T16:22:46Z SecondChildTAG: tell me how it came there r only three branches parallel to each other so there is only 2 nodes possible SecondChildUserIdTAG: 338265 SecondChildUserNameTAG: Rehan37 SecondChildCreateTimeTAG: 2012-09-15T16:27:00Z FirstChildTAG: tell me how it came there r only three branches parallel to each other so there is only 2 nodes possible FirstChildUserIdTAG: 338265 FirstChildUserNameTAG: Rehan37 FirstChildCreateTimeTAG: 2012-09-15T16:27:37Z FirstChildTAG: even i didn't understand that. By the way were you able to solve the rest ??? FirstChildUserIdTAG: 317445 FirstChildUserNameTAG: arghya33 FirstChildCreateTimeTAG: 2012-09-15T17:08:16Z IndexTAG: 4339 TitleTAG: Power Supplied by Current Source Was not able to derive the power supplied by current source. Can somebody help ? UserIdTAG: 374870 UserNameTAG: SandeepTorgal CreateTimeTAG: 2012-09-15T16:20:04Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL/KCL example CommentableIdTAG: 6002x_S2E3_Using_KVL_KCL_and_VI_constraints NumberOfReplyTAG: 2 FirstChildTAG: Sure! We have the equation "P=V*I", so we have to find the current and the voltage passing by the current source. The exercise present the value of the current source, 3.0 amperes, so we just need to find the voltage. Doing the loop (KVC), we know that the current source's voltage + v2 is equals to 0, but both of them are with the current pointing up, so we have to invert the current source's voltage. Thus, we have "v2-CSV=0", "v2=CSV", "CSV=7.77". Getting the voltage, we multiply V*I (7.77*3) and we find the power supplied by the current source: 23.33. I wrote the values approximately, and I am considering that you already find the value of v2 :) FirstChildUserIdTAG: 291362 FirstChildUserNameTAG: Gudson FirstChildCreateTimeTAG: 2012-09-15T17:04:07Z SecondChildTAG: how were you able to find v2...? SecondChildUserIdTAG: 133084 SecondChildUserNameTAG: Warrensiggs SecondChildCreateTimeTAG: 2012-09-15T18:26:20Z SecondChildTAG: @Gudson: In the 2nd paragraph, you explained that we have to invert the current source's voltage...I didn't understand why you are inverting it. I am getting a equation v2+CSV = 0. Please can you elaborate why so? SecondChildUserIdTAG: 31898 SecondChildUserNameTAG: hariharan SecondChildCreateTimeTAG: 2012-09-16T04:10:44Z SecondChildTAG: @hariharan, in a circuit loop, we have a circular movement, like in a clock. In this case, the current source (left) induce an up current, but the v2 (right) too make an up current, so they are incompatible. Inverting the voltage (And, consequently, the current), we will have a real concordance in the circuit, can you understand? If no, remember that when you have a circuit with a voltage source and a resistance, if the negative face of both the elements is turn-down, the resistance's voltage or the voltage source's value will be negative. SecondChildUserIdTAG: 291362 SecondChildUserNameTAG: Gudson SecondChildCreateTimeTAG: 2012-09-16T19:31:04Z FirstChildTAG: Using the formula p=vi and the fact that R2 is in parallel with the current source. FirstChildUserIdTAG: 265711 FirstChildUserNameTAG: Cristianxzaqws FirstChildCreateTimeTAG: 2012-09-15T17:20:22Z IndexTAG: 4340 TitleTAG: Exact time homework is due PLEASE!!!!!!! WHAT TIME ON SUNDAY IS THE HOMEWORK AND LAB DUE? Someone please tell me. UserIdTAG: 145676 UserNameTAG: CathyPK CreateTimeTAG: 2012-09-15T16:17:46Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Midnight of your local time. FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-15T16:22:22Z SecondChildTAG: how about my time? im here in phil its midnight 12:24am here now sept 17 sunday so what the exact time of my lab and homeworks? my sunday night(later)? SecondChildUserIdTAG: 235727 SecondChildUserNameTAG: amor SecondChildCreateTimeTAG: 2012-09-15T16:25:00Z SecondChildTAG: If it is midnight (11:59 PM) at my time, it is 11:59 PM at your time too. Thanks Kimt, I couldn't find the time ANYWHERE. SecondChildUserIdTAG: 145676 SecondChildUserNameTAG: CathyPK SecondChildCreateTimeTAG: 2012-09-15T16:36:52Z SecondChildTAG: It's now posted on the course info. :) SecondChildUserIdTAG: 79361 SecondChildUserNameTAG: kimt SecondChildCreateTimeTAG: 2012-09-15T16:43:21Z SecondChildTAG: kimt how to complete answer of my homewrokd if some questions are gone ? i mean the 1st part question of my homeworks about algebraic expression right i tried to answered it was invalid coz wrong keyword that i used so i set aside for a while and answer some question but when i plan to answer the question i never found so how to find those question coz i want to answer makes me sad for me coz i know the answer but only the keywork i used are wrong but now maybe i know the keywork now coz some student told me but never found the question SecondChildUserIdTAG: 235727 SecondChildUserNameTAG: amor SecondChildCreateTimeTAG: 2012-09-15T16:51:46Z SecondChildTAG: What do you mean that the questions are gone? Could you post a screenshot? SecondChildUserIdTAG: 79361 SecondChildUserNameTAG: kimt SecondChildCreateTimeTAG: 2012-09-15T16:56:08Z SecondChildTAG: in my homework1 never find the 1st question about the equivalent resistance as an algebraic expression network A-C all question about network A-C i never found it i want to answer those question but never found how to answer those question i know those all answer but when i answer it invalid the keyword that i used but now i want to answer the right keyword but never found it the question SecondChildUserIdTAG: 235727 SecondChildUserNameTAG: amor SecondChildCreateTimeTAG: 2012-09-15T17:18:24Z SecondChildTAG: Does this mean that we have all Sunday to finish the assignments/homework due, is that so? The word 'midnight' is truly ambiguous... SecondChildUserIdTAG: 240174 SecondChildUserNameTAG: takaforo SecondChildCreateTimeTAG: 2012-09-15T18:56:46Z SecondChildTAG: this is really unfair for me kimt i can't get a perfect score coz some question never found SecondChildUserIdTAG: 235727 SecondChildUserNameTAG: amor SecondChildCreateTimeTAG: 2012-09-15T22:30:19Z SecondChildTAG: I am having the same problem... I just got back from work and it is midnight in one hour. I was under the impression that I could turn it in Sunday. Is it Midnight as in Sunday 0:00 am or Monday 0:00 am? SecondChildUserIdTAG: 230787 SecondChildUserNameTAG: Gabriel007 SecondChildCreateTimeTAG: 2012-09-16T06:02:07Z IndexTAG: 4341 TitleTAG: Joe would need an AC transformer, right? To make its 120V 60Hz line become 240V *JOE WOULD NEED A TRANSFORMER*, right or wrong? EDIT: I forgot the most important part of the sentence (joe would need a transformer), ouch! Thx U! UserIdTAG: 398594 UserNameTAG: DaveyJC CreateTimeTAG: 2012-09-15T15:45:02Z VoteTAG: 0 CoursewareTAG: Week 1 / Power delivery modeling CommentableIdTAG: 6002x_S2E6_Modeling NumberOfReplyTAG: 2 FirstChildTAG: Can you be a little more elaborate? atleast I cant understand the question... FirstChildUserIdTAG: 257806 FirstChildUserNameTAG: skoda FirstChildCreateTimeTAG: 2012-09-15T15:46:52Z FirstChildTAG: The problem assumes that the power line is 240V 60Hz at the barn. In general though, you are correct that one would need to use a transformer to convert 120 V 60 Hz into 240 V 60 Hz FirstChildUserIdTAG: 321830 FirstChildUserNameTAG: kimth FirstChildCreateTimeTAG: 2012-09-15T16:00:27Z SecondChildTAG: ok so total resistance is 18.11 , there fore if H1=H2=H3=H then H =18.11/3= 6.03 ohms. P=I2R , my answer is still wrong . Please can some one help. SecondChildUserIdTAG: 178115 SecondChildUserNameTAG: nehamakhija SecondChildCreateTimeTAG: 2012-09-16T18:06:08Z IndexTAG: 4342 TitleTAG: INTERCONNECTION PROBLEM IN LAB1 m finding a problem while interconnecting the various connections. i am not able to connect the elements.. UserIdTAG: 431224 UserNameTAG: erachopra10 CreateTimeTAG: 2012-09-15T15:31:52Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: there is a icon like " - " that is the line icon. or you can just doube click the point from where u want start drawing and then click on the ending point. try and let me know. FirstChildUserIdTAG: 154172 FirstChildUserNameTAG: mofassair FirstChildCreateTimeTAG: 2012-09-15T16:34:45Z SecondChildTAG: thanks mofassair. i hv tried this before.. but there is some problem.i hv completed 1 problem although it took a lot of my tym..hope i will b able to fix it soon. SecondChildUserIdTAG: 431224 SecondChildUserNameTAG: erachopra10 SecondChildCreateTimeTAG: 2012-09-15T17:58:09Z FirstChildTAG: Here's [Myrimit's tutorial][1] on how to wire up the circuit simulator [1]: https://www.edx.org/wiki/6.002x/MyrimitCircuitSimulatorTutorial/ FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-15T16:43:50Z SecondChildTAG: thank u SecondChildUserIdTAG: 431224 SecondChildUserNameTAG: erachopra10 SecondChildCreateTimeTAG: 2012-09-15T17:58:16Z IndexTAG: 4343 TitleTAG: finding i1 can someone pls help me find out what i1 is... once found the rest should be a piece of cake... thank you. UserIdTAG: 133084 UserNameTAG: Warrensiggs CreateTimeTAG: 2012-09-15T15:31:46Z VoteTAG: 0 CoursewareTAG: Week 1 / Associated reference directions CommentableIdTAG: 6002x_S2E2_Associated_Reference_Directions NumberOfReplyTAG: 1 FirstChildTAG: as we know ..... i = v/r s0 just divide that you will get the answer .... while if u know the convention about the signs .... you can easily know how negative sign is in the answer .:) FirstChildUserIdTAG: 359074 FirstChildUserNameTAG: AMADMUB FirstChildCreateTimeTAG: 2012-09-15T15:46:36Z SecondChildTAG: u know i did that all along.... all i had to do was add the other 33... like instead of -0.3.. i had to put -0.333... cause i thought 0.3 is not so different from 0.333 you know lol... thank you though SecondChildUserIdTAG: 133084 SecondChildUserNameTAG: Warrensiggs SecondChildCreateTimeTAG: 2012-09-15T16:09:22Z IndexTAG: 4344 TitleTAG: Incorrect question In the problem is stated : "The internal resistances of small batteries are about 0.2Ω, but they vary a bit. Let's assume that R1=0.25 and R2=0.32". I found that really confusing, since I don't understand what does it means "they vary a bit", and also why it is given after later it is not used at all for computing the answers. Even initially I thought the R1 and R2 are without the internal resistance so I added up and my answers were shown as totally wrong... UserIdTAG: 415375 UserNameTAG: ZWX CreateTimeTAG: 2012-09-15T15:07:21Z VoteTAG: 0 CoursewareTAG: Week 1 / Battery model CommentableIdTAG: 6002x_S1E9_Battery_Model NumberOfReplyTAG: 2 FirstChildTAG: They mean the batteries vary between 0.25 - 0.32 ohms, or anywhere in between. Consider the batteries as being a pair of parallel resistors, in the example one has a resistance of 0.25ohms and 0.32ohms for the other. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-15T15:26:14Z FirstChildTAG: I believe they mean that in the real world, internal resistances of batteries vary slightly from battery to battery, due to the chemical nature of the battery. It probably varies by manufacturer, batch, and even temperature. But for the purpose of answering the question, use the values given. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-09-15T20:19:33Z IndexTAG: 4345 TitleTAG: Where is the reference of this section? Which part of the textbook does this section belong to? I want to clear this properly! UserIdTAG: 159427 UserNameTAG: Suyog CreateTimeTAG: 2012-09-15T14:46:55Z VoteTAG: 0 CoursewareTAG: Week 1 / Battery model CommentableIdTAG: 6002x_S1E9_Battery_Model NumberOfReplyTAG: 1 FirstChildTAG: If I'm not mistaken it's in "1.7 Modeling physical elements" FirstChildUserIdTAG: 38538 FirstChildUserNameTAG: Alienista FirstChildCreateTimeTAG: 2012-09-16T03:44:23Z IndexTAG: 4346 TitleTAG: Dropping Homeworks I thought I read somewhere that up 2 to 2 homeworks will be dropped from all of our homeworks, but I can't find that now. Is that true or am I confusing that with another course? Also, I never came across a specific time deadline for the homework submission. I am assuming we have up to midnight tomorrow night to submit it but I just wanted to be sure. Thank you. UserIdTAG: 376061 UserNameTAG: AndrejR CreateTimeTAG: 2012-09-15T14:43:44Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi, the homework should be submitted by midnight of your local time. You are also correct that 2 out of 12 assignments will be dropped from your course grade. Please see the [course syllabus][1] for detailed information. [1]: https://www.edx.org/static/content-mit-6002x/handouts/syllabus.a477535058a1.pdf FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-15T14:53:46Z IndexTAG: 4347 TitleTAG: Progress confusion Hi I have completed homework1 and lab1 with 100% progress in both. If I keep my cursor on Total progress bar it shows: labs = 1.5% of possible 15% homework = 1.5% of possible 15% What does it mean? UserIdTAG: 365309 UserNameTAG: chetnasinghaldas CreateTimeTAG: 2012-09-15T14:27:56Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: The final bar is your grade in the entire course. Homework is worth 15% of the overall course grade. You finished 1 of the 10 required homework assignments, so you have 1.5% score in the course overall. FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-15T14:32:51Z SecondChildTAG: thanks so much for clearing my confusion.. :) SecondChildUserIdTAG: 365309 SecondChildUserNameTAG: chetnasinghaldas SecondChildCreateTimeTAG: 2012-09-15T14:38:50Z SecondChildTAG: No problem. Good luck! :) SecondChildUserIdTAG: 79361 SecondChildUserNameTAG: kimt SecondChildCreateTimeTAG: 2012-09-15T14:54:05Z SecondChildTAG: I AM SORRY I AM INTERFERING BUT I WANT INFORMATION AS SOON AS POSSIBLE AS WE KNOW THAT MITx: 6.002x Circuits and Electronics HAS STARTED FROM 5TH SEPTEMBER. CAN I TAKE COURSE NOW PLEASE?I AM IN GRADE 12. I LIVE IN PAKISTAN. I HAVE STUDIED PHYSICS SINCE CLASS 6TH.I AM REALLY INTERESTED IN IT. I WILL WORK HARD. I WILL WORK FOR THE THINGS WHICH I LEFT BEHIND. KINDLY GUIDE ME PLEASE REGARDS SAFIAN TASEER SecondChildUserIdTAG: 442115 SecondChildUserNameTAG: safiantaseer SecondChildCreateTimeTAG: 2012-09-15T15:25:33Z IndexTAG: 4348 TitleTAG: Confusion about i1 Can anybody explain, please, why i1 is positive? It's going from '-' pole to '+' pole, but we assume, that current is positive, when it's going from '+' to '-'. Where I am wrong? UserIdTAG: 322444 UserNameTAG: koluch CreateTimeTAG: 2012-09-15T14:27:49Z VoteTAG: 0 CoursewareTAG: Week 1 / Series and parallel example CommentableIdTAG: 6002x_S2E4_Series_and_Parallel NumberOfReplyTAG: 0 IndexTAG: 4349 TitleTAG: help with battery model Hello all: I am trying to understand the battery model exercise from class 1, it has been difficult to me to determine the voltage in the open circuit and the current when short circuit is present. I use the circuit simulator but results do not make any sense to me, any hints would be great UserIdTAG: 153308 UserNameTAG: dmiralles CreateTimeTAG: 2012-09-15T13:52:51Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Can you put an image of the circuit which you tried and the result which was didnt make much sense to you.. I would be interested in helping... FirstChildUserIdTAG: 257806 FirstChildUserNameTAG: skoda FirstChildCreateTimeTAG: 2012-09-15T15:57:43Z IndexTAG: 4350 TitleTAG: LAB 1 couldnt understand what to do in lab 1 plz help me out UserIdTAG: 379593 UserNameTAG: varshney123 CreateTimeTAG: 2012-09-15T13:44:18Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Please be more specific FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-15T14:34:44Z SecondChildTAG: in the lab..the first circuit to be done!! I used the potential divider but i it happened to be wrong when i do the circuit!! :S I don't understand how to make it right!! help plz thnx SecondChildUserIdTAG: 151205 SecondChildUserNameTAG: jesher777 SecondChildCreateTimeTAG: 2012-09-15T18:07:35Z IndexTAG: 4351 TitleTAG: LAB 1 HOW TO ROTATE A RESISTOR?? MEANS HOW TO PLACE IT HORIZONTALLY?????? UserIdTAG: 316761 UserNameTAG: gk_goel CreateTimeTAG: 2012-09-15T13:19:42Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 5 FirstChildTAG: Click on the resistor then press "r" on your keyboard. FirstChildUserIdTAG: 94596 FirstChildUserNameTAG: Alkhdour FirstChildCreateTimeTAG: 2012-09-15T15:54:24Z FirstChildTAG: type r FirstChildUserIdTAG: 251262 FirstChildUserNameTAG: HoangNguyen FirstChildCreateTimeTAG: 2012-09-15T15:02:35Z FirstChildTAG: click and press 'R' FirstChildUserIdTAG: 199839 FirstChildUserNameTAG: vijaykrishnankk FirstChildCreateTimeTAG: 2012-09-15T13:24:32Z FirstChildTAG: FIRST UNDO YOUR CAPS LOCK, then review the tutorial on using the circuit simulator. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-15T13:26:36Z SecondChildTAG: Yes, caps lock posting is not very friendly :( SecondChildUserIdTAG: 79361 SecondChildUserNameTAG: kimt SecondChildCreateTimeTAG: 2012-09-15T14:29:01Z FirstChildTAG: i have a problem in lab 1.i couldnt understand what to do? plz help FirstChildUserIdTAG: 276808 FirstChildUserNameTAG: DEBASMITAMAJUMDER FirstChildCreateTimeTAG: 2012-09-15T13:29:38Z IndexTAG: 4352 TitleTAG: Homework II - Help needed Dear friends, as I wont be able to access internet in the coming week, I have to finish homework 2 and lab 2. I finished everything, except one question in homework 2. "What is the power (in Watts) that is delivered to this best load resistance? " Can anybody please help me find the answer. UserIdTAG: 199839 UserNameTAG: vijaykrishnankk CreateTimeTAG: 2012-09-15T13:05:11Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Please take a look at the maximum power transfer theorem.. That should help you answer this question FirstChildUserIdTAG: 257806 FirstChildUserNameTAG: skoda FirstChildCreateTimeTAG: 2012-09-15T13:43:28Z SecondChildTAG: Another approach would be to optimize the power transferred as a function of the load resistance using calculus. SecondChildUserIdTAG: 79361 SecondChildUserNameTAG: kimt SecondChildCreateTimeTAG: 2012-09-15T14:29:45Z IndexTAG: 4353 TitleTAG: H1 last question please answer Joe wants to heat his 12'X20' workshop with electric heat. He has hired the HACME electrician company to build the system. They propose to use three 1520.0W 240V baseboard heaters to provide a total heating capacity of 4560.0W. (A heater is basically a resistor. This is not quite true, because there is a thermostatic switch incorporated into the heater and because the resistance of a heater varies a bit with its temperature. But we will use a linear resistor as a model of a heater.) In the proposed system the heaters are connected in parallel with the 240V 60Hz AC power line (modeled by a voltage source) as shown in the diagram: My question is that....! How much current is expected to be drawn from the power line by this heating system when all three heaters are on? UserIdTAG: 269696 UserNameTAG: hassankazmi CreateTimeTAG: 2012-09-15T13:00:14Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Using the power known, we can find the total current. Since P = VI or P = V^2/R, we can find the value of the resistances(P = known = 1270W). V = 240. Now what's remaining(2nd ckt) is a resistor in series with parallel combination of two similar resistors. You can find the current and hence the power. FirstChildUserIdTAG: 199839 FirstChildUserNameTAG: vijaykrishnankk FirstChildCreateTimeTAG: 2012-09-15T13:09:56Z SecondChildTAG: Where did you get the 1270W from? SecondChildUserIdTAG: 234060 SecondChildUserNameTAG: jipjipbee SecondChildCreateTimeTAG: 2012-09-15T21:14:02Z SecondChildTAG: ok so total resistance is 18.11 , there fore if H1=H2=H3=H then H =18.11/3= 6.03 ohms. P=I2R , my answer is still wrong . Please can some one help. SecondChildUserIdTAG: 178115 SecondChildUserNameTAG: nehamakhija SecondChildCreateTimeTAG: 2012-09-16T18:05:54Z IndexTAG: 4354 TitleTAG: S2E6 Sorry by my error, we use virgule and not point in Brazilian Metrical System. Thanks. UserIdTAG: 260994 UserNameTAG: capvl CreateTimeTAG: 2012-09-15T12:36:36Z VoteTAG: 0 CoursewareTAG: Week 1 / Power delivery modeling CommentableIdTAG: 6002x_S2E6_Modeling NumberOfReplyTAG: 1 FirstChildTAG: Virgula em inglês é "comma" FirstChildUserIdTAG: 38538 FirstChildUserNameTAG: Alienista FirstChildCreateTimeTAG: 2012-09-18T06:12:56Z IndexTAG: 4355 TitleTAG: Hello,I can not complete my Lab-2 of week2! How can we apply the resistance so as to get respective peak values of current as mentioned in the question? UserIdTAG: 232076 UserNameTAG: JOMANUSO CreateTimeTAG: 2012-09-15T12:33:32Z VoteTAG: 0 CoursewareTAG: Week 2 / Superposition CommentableIdTAG: 6002x_superposition_tutorial NumberOfReplyTAG: 1 FirstChildTAG: Its better to use two resistors each for sine and square i/p's(voltage divider network) to get the waveforms. FirstChildUserIdTAG: 199839 FirstChildUserNameTAG: vijaykrishnankk FirstChildCreateTimeTAG: 2012-09-15T13:23:51Z IndexTAG: 4356 TitleTAG: H1P1 last question can someone give me hint to solve this??? UserIdTAG: 316761 UserNameTAG: gk_goel CreateTimeTAG: 2012-09-15T12:25:26Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: the 6ohm resistor can tolerate maximum voltage of 2.4 V but the other two resistors can withstand only 2 V.So the equivalent resistance can tolerate only 2V. this is the clue. FirstChildUserIdTAG: 356056 FirstChildUserNameTAG: Ganesh2810 FirstChildCreateTimeTAG: 2012-09-15T12:32:56Z SecondChildTAG: but 2 V IS ALSO WRONG SecondChildUserIdTAG: 316761 SecondChildUserNameTAG: gk_goel SecondChildCreateTimeTAG: 2012-09-15T12:39:20Z SecondChildTAG: the voltage across the parallel combo of all 3 resistors can't exceed 2V.. so maximum power is (2*2)/Equivalent Resistance=4/1.5=2.67 SecondChildUserIdTAG: 356056 SecondChildUserNameTAG: Ganesh2810 SecondChildCreateTimeTAG: 2012-09-15T12:56:43Z SecondChildTAG: Power, P is inversely proportional to the resistance, R P = V^2 /R So, you see the smaller resistors (i.e 4 ohms) will burn up first when dissipating maximum power (1 watt). Use the above equation to find the voltage, V, where P=1watt and R=4. Then use the value of V to determine Power dissipated by the 6 ohm resistance. So in the end, total power dissipated will be (1 + 1 + 2/3) i.e. 8/3 watts SecondChildUserIdTAG: 152516 SecondChildUserNameTAG: shuvo915 SecondChildCreateTimeTAG: 2012-09-15T13:12:42Z SecondChildTAG: I didn´t understand the question at all. I was really confuse and with your clue I get it. Thanks Ganesh2810. SecondChildUserIdTAG: 433574 SecondChildUserNameTAG: endika86 SecondChildCreateTimeTAG: 2012-09-15T14:53:04Z SecondChildTAG: This question didn't make any sense to me at all. Thanks for the clarification. SecondChildUserIdTAG: 330706 SecondChildUserNameTAG: curtie44 SecondChildCreateTimeTAG: 2012-09-15T17:44:10Z SecondChildTAG: 2.6666 W SecondChildUserIdTAG: 382505 SecondChildUserNameTAG: AhmedGalal2 SecondChildCreateTimeTAG: 2012-09-15T22:21:28Z SecondChildTAG: telling the answer is not the right approach towards a better learning!!:( SecondChildUserIdTAG: 127800 SecondChildUserNameTAG: MNB SecondChildCreateTimeTAG: 2012-09-16T03:56:34Z IndexTAG: 4357 TitleTAG: About -ve current! Looking to figure given, we see current 'i' moving from -ve terminal to +ve terminal, which in case is not possible. So current 'i' should flow in opposite direction. Hence in answer current appears to be -ve. Thank you UserIdTAG: 159427 UserNameTAG: Suyog CreateTimeTAG: 2012-09-15T12:18:51Z VoteTAG: 0 CoursewareTAG: Week 1 / Power CommentableIdTAG: 6002x_S1E2_Power NumberOfReplyTAG: 0 IndexTAG: 4358 TitleTAG: Homework 1 I have a problem in a homework 1 in exercise 1, network B and C. I find the solutions but the space when can we write the equations show me, for wrong question . I now I have right answer in this part. Can you explain me why? UserIdTAG: 422394 UserNameTAG: KreshMech CreateTimeTAG: 2012-09-15T10:37:43Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: I have this problem. Can you help me? FirstChildUserIdTAG: 149194 FirstChildUserNameTAG: andrey_samara FirstChildCreateTimeTAG: 2012-09-15T14:17:50Z FirstChildTAG: Same problem whit me to but i have got only in Problem circuit C. My B is solved FirstChildUserIdTAG: 269696 FirstChildUserNameTAG: hassankazmi FirstChildCreateTimeTAG: 2012-09-15T13:02:40Z FirstChildTAG: maybe your letter cases are not matching, check them FirstChildUserIdTAG: 254607 FirstChildUserNameTAG: werehenry FirstChildCreateTimeTAG: 2012-09-15T10:49:56Z SecondChildTAG: dont simplify the eq resistance to a final value in terms of R....leave it as an algebraic expn itslf. SecondChildUserIdTAG: 356056 SecondChildUserNameTAG: Ganesh2810 SecondChildCreateTimeTAG: 2012-09-15T13:00:09Z FirstChildTAG: i have a problem that the answer to even simple question is wrong although like when i put the value 3R it say that can not be kept please help FirstChildUserIdTAG: 314241 FirstChildUserNameTAG: SUBIGYA FirstChildCreateTimeTAG: 2012-09-15T14:04:34Z SecondChildTAG: When I type answers in the boxes I use notation in the following example: 10*R I think the website requires the use of the * to denote multiplication. I think just sticking a letter next to a number makes the website assume it's a label. SecondChildUserIdTAG: 328138 SecondChildUserNameTAG: MikeSchenck SecondChildCreateTimeTAG: 2012-09-15T14:18:11Z SecondChildTAG: it should be 3*R rather 3R.... :) Like how you write in programming languages... SecondChildUserIdTAG: 257806 SecondChildUserNameTAG: skoda SecondChildCreateTimeTAG: 2012-09-15T14:24:22Z IndexTAG: 4359 TitleTAG: HW1 HELP..... CAN ANY ONE HOW I SOLVE HW1 LAST 3 (HEATER) QUESTION ANSWER.......I COULD NOT FIND ANS....... UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-15T09:59:47Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_Troubleshooting NumberOfReplyTAG: 1 FirstChildTAG: Please do not ask for or give answers to the problems. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-09-15T13:33:03Z IndexTAG: 4360 TitleTAG: alumnos en mexico Hola que tal quisiera contactar alumnos qu esten tomando este curso aqui en mexico,para intercambiar puntos de vista de las tareas o laboratorios,espero respuesta ,gracias. UserIdTAG: 299697 UserNameTAG: ramirito CreateTimeTAG: 2012-09-15T09:43:24Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hola, yo vivo en mexico y tambien estoy tomando el curso FirstChildUserIdTAG: 238005 FirstChildUserNameTAG: isisbocardo FirstChildCreateTimeTAG: 2012-09-15T19:25:46Z SecondChildTAG: hola SecondChildUserIdTAG: 290270 SecondChildUserNameTAG: Diebo SecondChildCreateTimeTAG: 2012-09-15T21:21:15Z SecondChildTAG: Que tal isisbocardo,por donde podremos comunicarnos?,gracias espero tu respuesta,saludos. SecondChildUserIdTAG: 299697 SecondChildUserNameTAG: ramirito SecondChildCreateTimeTAG: 2012-09-16T00:26:49Z SecondChildTAG: Argentina les parece muy lejano? :) Salu2 gente...!!! SecondChildUserIdTAG: 274475 SecondChildUserNameTAG: Konredus SecondChildCreateTimeTAG: 2012-09-16T03:54:50Z SecondChildTAG: disculpa, mi computadora tuvo un problema y me la acaban de regresar, nos podemos comunicar por aqui o por un grupo en fb o por correo, uds dicen. SecondChildUserIdTAG: 238005 SecondChildUserNameTAG: isisbocardo SecondChildCreateTimeTAG: 2012-09-29T06:07:59Z IndexTAG: 4361 TitleTAG: What time do we have to post the homework on sunday? Thanks for your help :)) UserIdTAG: 342966 UserNameTAG: SandraNavarro CreateTimeTAG: 2012-09-15T09:35:38Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: ANYTIME BEFORE OR ON SUNDAY. FirstChildUserIdTAG: 184830 FirstChildUserNameTAG: pt_sunny2k1 FirstChildCreateTimeTAG: 2012-09-15T10:13:33Z SecondChildTAG: which also means on Sunday 22.00 pm possible ? Thanks!! SecondChildUserIdTAG: 342966 SecondChildUserNameTAG: SandraNavarro SecondChildCreateTimeTAG: 2012-09-15T15:00:38Z IndexTAG: 4362 TitleTAG: ANSWER POSTING TIME.......... CAN I POST THE INCOMPLETE ANSWER OF HOME WORK AND LAB BEFORE DEADLINE DATE???????? BCOZ I AM NOT able to solve few problems....plz suggest me someone quickly...... UserIdTAG: 307088 UserNameTAG: PRAKHAR2012 CreateTimeTAG: 2012-09-15T09:35:19Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: YOU CAN.. BUT IT WILL BE BEST TO SOLVE, KEEP ON TRYING. YOU CAN DO THIS. FirstChildUserIdTAG: 184830 FirstChildUserNameTAG: pt_sunny2k1 FirstChildCreateTimeTAG: 2012-09-15T10:14:10Z SecondChildTAG: Me too!!! the i'm getting some troubles with one question in homework and the lab :S!! plz someone help me SecondChildUserIdTAG: 151205 SecondChildUserNameTAG: jesher777 SecondChildCreateTimeTAG: 2012-09-15T14:16:53Z IndexTAG: 4363 TitleTAG: Bug Textbook, page 72, Example 2.14. "In other words, v = 0.5 V." - is wrong. V has to be equal 2V. UserIdTAG: 269960 UserNameTAG: Stiga CreateTimeTAG: 2012-09-15T09:26:08Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: This and other errata can be found here: https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/Book/ FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-15T15:05:53Z IndexTAG: 4364 TitleTAG: H1P1 ayuda ultima pregunta Hola que tal,alguien podria ayudarme a resolver el ultimo reactivo o pregunta ,el de la potencia en la resistencia de menor valor antes de quemarse,no entiendo la pregunta o no encuentro como resolverla,de antemano agradezco su ayuda UserIdTAG: 299697 UserNameTAG: ramirito CreateTimeTAG: 2012-09-15T09:22:32Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: ramirito, en la pregunta te dicen que cada resistencia puede disipar hasta 1 watt de potencia hasta quemarse. Usando la formula p=V²/R con una de las resistencias de 4 ohms puedes determinar el voltaje, ya que tienes p=1W y R=4ohms. Con ese mismo voltaje puedes determinar p para la resistencia de 6ohms. Para calcular p total solo sumas todas las p para las 3 resistencias (1+1+x), espero que eso te ayude. Saludos FirstChildUserIdTAG: 35642 FirstChildUserNameTAG: caled FirstChildCreateTimeTAG: 2012-09-16T00:01:40Z SecondChildTAG: muchas gracias Caled,ya hace un rato encontre la solucion,saludos SecondChildUserIdTAG: 299697 SecondChildUserNameTAG: ramirito SecondChildCreateTimeTAG: 2012-09-16T00:20:39Z IndexTAG: 4365 TitleTAG: How do you find the posts that you are following? Since the discussion board's layout has changed, I can't find any of the posts that I am following. UserIdTAG: 270011 UserNameTAG: SnowmanZA CreateTimeTAG: 2012-09-15T09:10:40Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: This will be coming in a forthcoming update, stay tuned. For now you can at least see if you're following a post by the star at the top right. FirstChildUserIdTAG: 323387 FirstChildUserNameTAG: ibrahimawwal FirstChildCreateTimeTAG: 2012-09-15T09:46:41Z SecondChildTAG: Ok, thank you. SecondChildUserIdTAG: 270011 SecondChildUserNameTAG: SnowmanZA SecondChildCreateTimeTAG: 2012-09-15T19:59:41Z IndexTAG: 4366 TitleTAG: H1P2 Node Count Seems to me that the requested value is the number of branches, not the number of nodes. UserIdTAG: 194450 UserNameTAG: wrbuckley CreateTimeTAG: 2012-09-15T08:22:28Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: agree with you here, I'm confused cause the (correct) answer in my opinion doesn't make any sense ... :-( ... FirstChildUserIdTAG: 157259 FirstChildUserNameTAG: JulianTerzyk FirstChildCreateTimeTAG: 2012-09-15T14:50:10Z SecondChildTAG: The instructors or their assistants should directly address this point. SecondChildUserIdTAG: 194450 SecondChildUserNameTAG: wrbuckley SecondChildCreateTimeTAG: 2012-09-15T15:39:14Z SecondChildTAG: There is one hint, in that three current variables are defined. The justification for same is the issue here. SecondChildUserIdTAG: 194450 SecondChildUserNameTAG: wrbuckley SecondChildCreateTimeTAG: 2012-09-15T15:40:14Z SecondChildTAG: they explain this in Tutorials: Nodal Analysis. It has to do with Floating Voltage Source. SecondChildUserIdTAG: 351903 SecondChildUserNameTAG: Sheybfreak SecondChildCreateTimeTAG: 2012-09-17T01:27:14Z IndexTAG: 4367 TitleTAG: help me!!!! I don't know how to write the answer S1V6 KVL. UserIdTAG: 226017 UserNameTAG: huynhthanhnhan CreateTimeTAG: 2012-09-15T08:14:22Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL and KCL exercises CommentableIdTAG: 6002x_kvl_kcl_exercises NumberOfReplyTAG: 2 FirstChildTAG: if solution is V+v1+v2+v3, just write that. FirstChildUserIdTAG: 138239 FirstChildUserNameTAG: LaFolle FirstChildCreateTimeTAG: 2012-09-15T22:26:57Z SecondChildTAG: the problem is V not permitted in answer SecondChildUserIdTAG: 279053 SecondChildUserNameTAG: MrZaKaRiA SecondChildCreateTimeTAG: 2012-09-15T23:18:31Z FirstChildTAG: the known quantities are V , v1 and v2.So we have to form our answers only using these 3 known quantities.so at 1st, find out the loop containing the unknown quantity and any one or more of these known quantities V, v1 and v2. SO if your answer is v3= V-v1, just write V-v1 in the box specified to input the answers. FirstChildUserIdTAG: 156225 FirstChildUserNameTAG: sanjana_m FirstChildCreateTimeTAG: 2012-09-16T18:39:50Z SecondChildTAG: Thanks a bunch sanjana! SecondChildUserIdTAG: 296914 SecondChildUserNameTAG: Varunika SecondChildCreateTimeTAG: 2012-09-19T15:40:33Z IndexTAG: 4368 TitleTAG: independent equations confused about independent kcl equations &kvl independent equations UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-15T07:36:13Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: I just need some one to help me out here.What are independent equations and how do you know that an equation is independent. FirstChildUserIdTAG: 429851 FirstChildUserNameTAG: ssembajjwe FirstChildCreateTimeTAG: 2012-09-15T08:46:24Z SecondChildTAG: what i understood is that we may be able to write many equations as u know using kvl or kcl or nyother mtd.. but if one equation can be derived from others then that equation is dependent.. eg:- say, a+b=1 --(1); b+d=2 ---(2); d-a=1 ---(3).. here any equation can be derived using other two eqns.. hence independent number of equations will be two.. bcos, using 1 f those we won't be able to derive the latter.. here , any two of the equations can be considered as independent eqns.. SecondChildUserIdTAG: 363980 SecondChildUserNameTAG: AArchchunah SecondChildCreateTimeTAG: 2012-09-15T09:25:20Z SecondChildTAG: Great.Thanks got it. SecondChildUserIdTAG: 429851 SecondChildUserNameTAG: ssembajjwe SecondChildCreateTimeTAG: 2012-09-15T15:22:41Z IndexTAG: 4369 TitleTAG: Lab 2 Help I'm not quite sure what circuit design to use for lab 2. I assume it's a combination of two voltage divider circuits (one for each voltage source), but I'm not really sure how to progress from there. Any help would be appreciated! UserIdTAG: 398786 UserNameTAG: cww501 CreateTimeTAG: 2012-09-15T07:22:17Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: hi, Try the circuit given by them as an example. As you can see it is a series combination. Write Kvl or kcl equation. Most probably you won't find a solution. But it would give a good insight into the problem. Next try parallel in a similar manner. And then a combination of series and parallel. Extra clue- You have to use superposition and/or thevenin theorem also. FirstChildUserIdTAG: 330269 FirstChildUserNameTAG: Jaikr FirstChildCreateTimeTAG: 2012-09-15T07:27:32Z SecondChildTAG: @Jaikr hi, what do you mean by parallel? The resistors in parallel?Or the V1 and V2 in parallel? SecondChildUserIdTAG: 157273 SecondChildUserNameTAG: ongchihang SecondChildCreateTimeTAG: 2012-09-15T08:36:36Z FirstChildTAG: Use one devider in addition to two resistors you have. Two resistor are not enough to solve the task. FirstChildUserIdTAG: 209041 FirstChildUserNameTAG: SmartEngine FirstChildCreateTimeTAG: 2012-09-16T16:20:49Z IndexTAG: 4370 TitleTAG: homework the questions in the homework displays "math processing error" why is it so? UserIdTAG: 341241 UserNameTAG: nyikanet CreateTimeTAG: 2012-09-15T06:58:58Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 4371 TitleTAG: can't the V2 be solved with a voltage divider equation hi, my name is billy, im an electronics engineering student (4th year) from the philippines.. and i was just wondering, because this is driving me nuts, and the V2 be solved with just a voltage divider solution V2= (Vs*R2)/(R1+R2) the answer i got is 1.11 . can anybody clear this out.. thanks UserIdTAG: 277065 UserNameTAG: beleloy808 CreateTimeTAG: 2012-09-15T06:33:25Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL/KCL example CommentableIdTAG: 6002x_S2E3_Using_KVL_KCL_and_VI_constraints NumberOfReplyTAG: 3 FirstChildTAG: hi there, I think here we need to consider the effect of the 3A current supply from the current source. FirstChildUserIdTAG: 201508 FirstChildUserNameTAG: datle FirstChildCreateTimeTAG: 2012-09-15T14:29:38Z FirstChildTAG: I used 3 equations. you need an equation for the divider with the bulb connected. In this equation are three unknown variables: R1, R2, Rtotal you need 2 more equations: how must be the ratio of r1 and r2, if the divider is without the bulb. write down the quation for Rtotal convert these equations and replace the 2 unknown variables. At least you get an equation like these: R2=x*RL FirstChildUserIdTAG: 389333 FirstChildUserNameTAG: juergen FirstChildCreateTimeTAG: 2012-09-15T08:35:51Z SecondChildTAG: hi there, I think here we need to consider the effect of the 3A current supply from the current source. SecondChildUserIdTAG: 201508 SecondChildUserNameTAG: datle SecondChildCreateTimeTAG: 2012-09-15T14:29:04Z SecondChildTAG: oh, sorry for posting the wrong place! SecondChildUserIdTAG: 201508 SecondChildUserNameTAG: datle SecondChildCreateTimeTAG: 2012-09-15T14:29:57Z SecondChildTAG: I don't see a bulb? what are you talking about a bulb for? SecondChildUserIdTAG: 288381 SecondChildUserNameTAG: NAB86 SecondChildCreateTimeTAG: 2012-09-15T14:42:53Z FirstChildTAG: Don't forget the effect of the current source, you can not use the voltage divider formula because the resistors are not in series (not the same current). If you consider Vs alone, you must then consider I alone and use the superposition (you will find 6.66V when using I alone). It has not been seen in this part of the course. FirstChildUserIdTAG: 396446 FirstChildUserNameTAG: RousseauxS FirstChildCreateTimeTAG: 2012-09-15T09:16:10Z SecondChildTAG: oh, i get it.. i thought i could use voltage divider because the current source is in parallel with the R2, and so i assumed that the voltage would be actually the same .. .. but could you please give me a further explanation, :) SecondChildUserIdTAG: 277065 SecondChildUserNameTAG: beleloy808 SecondChildCreateTimeTAG: 2012-09-15T11:57:48Z SecondChildTAG: I think that this is the right explication and solution...the technique of superposition is useful when working with more then one source in a circuit...thanks RousseauxS...:) SecondChildUserIdTAG: 394374 SecondChildUserNameTAG: Eubyse SecondChildCreateTimeTAG: 2012-09-15T16:23:09Z IndexTAG: 4372 TitleTAG: solve how to solve the questions of lab and assignment???? UserIdTAG: 382263 UserNameTAG: Syed_Usama_1993 CreateTimeTAG: 2012-09-15T06:14:21Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 4373 TitleTAG: Section 2.5 More Examples I don't understand hoe we arrive at equation 2.147? Con somebody explain UserIdTAG: 178115 UserNameTAG: nehamakhija CreateTimeTAG: 2012-09-15T06:01:10Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 0 IndexTAG: 4374 TitleTAG: do time has any effect in the matter of power on long term ??? We calculate the power on a resistor by applying p=vi but that is in just 1 second so does that power could make any changes or affecting the efficiency of the resistor on a long term ???? UserIdTAG: 138871 UserNameTAG: corabict CreateTimeTAG: 2012-09-15T05:55:40Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 4 FirstChildTAG: experimenting$$ \begin{equation}\mathbf{A} = \begin{bmatrix}\alpha\phantom{XX}0\newline0\phantom{XX} \beta\end{bmatrix}\end{equation} $$ experimenting FirstChildUserIdTAG: 58798 FirstChildUserNameTAG: Copetin FirstChildCreateTimeTAG: 2012-09-15T06:48:17Z FirstChildTAG: ... \begin{equation}\mathbf{A} = \begin{bmatrix}\alpha 0 \newline0 \beta\end{bmatrix}\end{equation} FirstChildUserIdTAG: 58798 FirstChildUserNameTAG: Copetin FirstChildCreateTimeTAG: 2012-09-15T06:52:56Z FirstChildTAG: ... \begin{equation}\mathbf{A} = \begin{bmatrix}\alpha &\ 0 \newline0 &\ \beta\end{bmatrix}\end{equation} FirstChildUserIdTAG: 58798 FirstChildUserNameTAG: Copetin FirstChildCreateTimeTAG: 2012-09-15T06:56:06Z FirstChildTAG: \begin{equation}\mathbf{A} = \begin{bmatrix}\alpha\phantom{XX}0\newline\\\begin{array}[pos]{spalten} \end{array}0\phantom{XX} \beta\end{bmatrix}\end{equation} $$ experiment \begin{equation}\mathbf{A} = \begin{bmatrix}\alpha &0 \newline0 & \beta\end{bmatrix}\end{equation} FirstChildUserIdTAG: 58798 FirstChildUserNameTAG: Copetin FirstChildCreateTimeTAG: 2012-09-15T06:59:03Z SecondChildTAG: \begin{equation}\mathbf{A} = \begin{bmatrix}\alpha\phantom{XX}0\newline\\\begin{array}[pos]{spalten} \end{array}0\phantom{XX} \beta\end{bmatrix}\end{equation} SecondChildUserIdTAG: 58798 SecondChildUserNameTAG: Copetin SecondChildCreateTimeTAG: 2012-09-15T08:34:50Z SecondChildTAG: oh i couldn't get the message ......!!!! SecondChildUserIdTAG: 138871 SecondChildUserNameTAG: corabict SecondChildCreateTimeTAG: 2012-09-15T14:30:11Z IndexTAG: 4375 TitleTAG: H1P3: Poor Workmanship I figured out all the answer except for the last three problems. All they give us is 240 Vac which equals 239.75 RMS and total power is 3630 Watts. Using ohms law doesn't solve the problem. As a consequence, Joe found his workshop too cold. H1 was weak; H2 and H3 barely warmed up. What power was being dissipated in H1? Any suggestions would be nice. UserIdTAG: 224302 UserNameTAG: tcc626 CreateTimeTAG: 2012-09-15T05:46:44Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Hi tcc626! You can see this [Post][1]. I hope this hints can help you! :) [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_S1E3_AC_power/threads/50536b48624a00230000000e FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-15T06:00:42Z SecondChildTAG: ok so total resistance is 18.11 , there fore if H1=H2=H3=H then H =18.11/3= 6.03 ohms. P=I2R , my answer is still wrong . Please can some one help. SecondChildUserIdTAG: 178115 SecondChildUserNameTAG: nehamakhija SecondChildCreateTimeTAG: 2012-09-16T18:05:43Z IndexTAG: 4376 TitleTAG: Comment on book. It is interesting to me that authors go to much effort in figure 2.20 to show circuits that both violate KVL and are therefore not within the scope of the book (or course) and yet, on the very same page we find in Example 2.8 (figure 2.22) the same circuit, being analysed with KVL. Seems a bit inconsistent to me. UserIdTAG: 194450 UserNameTAG: wrbuckley CreateTimeTAG: 2012-09-15T05:25:26Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: The problem in 2.20 is that every branch voltage is forced with different values and that can not happen. In 2.22, we have "degrees of freedom" by not forcing everything. We can use KVL on the circuit to discover the unknowns. FirstChildUserIdTAG: 396446 FirstChildUserNameTAG: RousseauxS FirstChildCreateTimeTAG: 2012-09-15T09:21:44Z SecondChildTAG: The issue is that in one case the graphic image is of the entire circuit, in the other case it is not. Yet, this point is nowhere explicated; not in the text, nor in other course materials. This is, instead, an assumed point, and it is exactly such assumptions that limit understanding of new concepts by the novice. My point was to clearly demarcate this issue, mostly for the benefit of the instructors/authors of the book; i.e. Dr. Agarwal. SecondChildUserIdTAG: 194450 SecondChildUserNameTAG: wrbuckley SecondChildCreateTimeTAG: 2012-09-15T09:48:02Z SecondChildTAG: By that I mean that the second case is an extracted sub-circuit, so that some of the assumptions (like no where for accumulated charge to go) don't necessarily apply. SecondChildUserIdTAG: 194450 SecondChildUserNameTAG: wrbuckley SecondChildCreateTimeTAG: 2012-09-15T09:49:10Z IndexTAG: 4377 TitleTAG: LAB 1 my potential differences between the resistors are right.The Value Of The Difference is coming as 1.5 and 2v.i have run the DC analysis and the voltages are showing up.i am using the same resistors.yet whenever i check my answers they come as wrong . please help. my submission deadline is 16th UserIdTAG: 386372 UserNameTAG: akashrao1991 CreateTimeTAG: 2012-09-15T04:22:29Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 2 FirstChildTAG: i'm not getting your problem absolutely akash.. give some more details so that anyone can understand your problem.. in the first experiment , the voltage across A&B(across the bulb) should be 1.5 V.. in the second case across A&B when the bulb is removed(open circuit).. will this help you??? FirstChildUserIdTAG: 363980 FirstChildUserNameTAG: AArchchunah FirstChildCreateTimeTAG: 2012-09-15T04:50:19Z FirstChildTAG: I am getting the same problem as akash.... I tried to simulate the circuit on EWB. Yet the o/p come out to be correct but not here.... HELP ME!! FirstChildUserIdTAG: 127800 FirstChildUserNameTAG: MNB FirstChildCreateTimeTAG: 2012-09-15T06:00:24Z IndexTAG: 4378 TitleTAG: lab 2 help Using superposition I found equation R2=2.55R1. Can anyone tell how I can find the value of R1 and R2 from one equation. UserIdTAG: 197641 UserNameTAG: mtanju CreateTimeTAG: 2012-09-15T03:06:29Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 6002x_Fall2012_General NumberOfReplyTAG: 1 FirstChildTAG: Use more resistors, little hint. FirstChildUserIdTAG: 81712 FirstChildUserNameTAG: dzhon FirstChildCreateTimeTAG: 2012-09-15T05:18:42Z IndexTAG: 4379 TitleTAG: Video player help, the youtube player is not working, I can do?. but when you play a video on the official website of youtube there is no problem I CAN DO????' UserIdTAG: 64183 UserNameTAG: jesuslayton CreateTimeTAG: 2012-09-15T02:38:55Z VoteTAG: 0 CoursewareTAG: Week 2 / Superposition CommentableIdTAG: 6002x_superposition NumberOfReplyTAG: 2 FirstChildTAG: The player is working fine now FirstChildUserIdTAG: 133606 FirstChildUserNameTAG: smadi FirstChildCreateTimeTAG: 2012-09-15T09:13:28Z FirstChildTAG: I'm using chrome and when it stops I move the speed and start it again and then put the speed back to 1.0. FirstChildUserIdTAG: 247286 FirstChildUserNameTAG: ronald_borunda FirstChildCreateTimeTAG: 2012-09-17T01:35:19Z IndexTAG: 4380 TitleTAG: Transient Analysis Hi, I have problems with the transient analysis, i have the following values: Params: * Offset value = 1 * Amplitude = 1 * Frequency = 1 I got the following data: * Point A: 1.008 V * Point B: 0.839 V * Ponit C: 0.335 V That results are wrong...the results are A = 1.995, B=1.662, C 0.665...somebody have any idea?? UserIdTAG: 25509 UserNameTAG: laninor CreateTimeTAG: 2012-09-15T02:25:39Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 3 FirstChildTAG: al parecer algo te esta fallando, ponga en la frecuencia: 1000Hz. porque en estos momentos tienes solo 1Hz. FirstChildUserIdTAG: 231685 FirstChildUserNameTAG: ArcangelM FirstChildCreateTimeTAG: 2012-09-15T04:43:08Z FirstChildTAG: Hi, You should put 1k instead of just 1 for frequency FirstChildUserIdTAG: 421380 FirstChildUserNameTAG: ADimitrov FirstChildCreateTimeTAG: 2012-09-15T10:38:23Z SecondChildTAG: agreed. make frequency 1000 instead of 1 SecondChildUserIdTAG: 149420 SecondChildUserNameTAG: Muthukumar SecondChildCreateTimeTAG: 2012-09-20T21:18:26Z FirstChildTAG: u MAYBE NOT USING THE RIGHT TRANSIENT TIME!! FirstChildUserIdTAG: 127800 FirstChildUserNameTAG: MNB FirstChildCreateTimeTAG: 2012-09-15T06:05:18Z IndexTAG: 4381 TitleTAG: AC voltage source if i have a circuit has DC voltage source , say it is 120 volts and a resistance 5 ohms , so that current will be i=v/r=120/5= 24 ampers if i have AC Voltage source 120 and the same resistance 5 ohms , what will be the current ?? UserIdTAG: 392663 UserNameTAG: matto0o CreateTimeTAG: 2012-09-15T02:10:03Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: The current will be the same if the given value of voltage is the RMS value that is root mean square value. FirstChildUserIdTAG: 371240 FirstChildUserNameTAG: Phani55 FirstChildCreateTimeTAG: 2012-09-15T08:40:22Z IndexTAG: 4382 TitleTAG: "Invalid input: Could not parse '3R' as a formula" (help me out) Please someone should tell me what to do i am having problem with homework H1p1, i got the message "Invalid input: Could not parse '3R' as a formula" please help me out. Thank You UserIdTAG: 337904 UserNameTAG: bashykeny CreateTimeTAG: 2012-09-15T01:42:59Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 4 FirstChildTAG: INSERT * (SHIFT 8) BETWEEN 3 7 R i.e 3*R FirstChildUserIdTAG: 950 FirstChildUserNameTAG: Olawale FirstChildCreateTimeTAG: 2012-09-15T05:52:55Z FirstChildTAG: same here help plz FirstChildUserIdTAG: 273622 FirstChildUserNameTAG: hazem91 FirstChildCreateTimeTAG: 2012-09-15T03:40:32Z SecondChildTAG: try "3*R" in place of 3R and see the magic!! SecondChildUserIdTAG: 314624 SecondChildUserNameTAG: Owais001 SecondChildCreateTimeTAG: 2012-09-15T05:10:46Z FirstChildTAG: Bashykeny, it is supposed to be expressed as 3*R and not 3R. Regards, Ugo FirstChildUserIdTAG: 329879 FirstChildUserNameTAG: Ugochukwu FirstChildCreateTimeTAG: 2012-09-15T01:55:20Z FirstChildTAG: Hi bashykeny! You have to put 3*R instead of 3R. Please search in search posts there is a lot of information about this concern ;) FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-15T01:57:54Z SecondChildTAG: YES , IT IS TRUE i faced the same problem -don`t forget to write capital R not r :) SecondChildUserIdTAG: 392663 SecondChildUserNameTAG: matto0o SecondChildCreateTimeTAG: 2012-09-15T02:30:16Z IndexTAG: 4383 TitleTAG: The reverse polarity on V2 The trick of the matter is to realize that the source V2 is reversed, node method takes into account this factor. The formula is: (e - V1) G1 + (e + V2) G2 = 0 UserIdTAG: 402643 UserNameTAG: OzorioNeto CreateTimeTAG: 2012-09-15T01:34:54Z VoteTAG: 0 CoursewareTAG: Week 2 / Node equation review CommentableIdTAG: 6002x_S3E1_NodeEquationReview NumberOfReplyTAG: 1 FirstChildTAG: the true formula is: (e-V1)G1+(e-V2)G2=0 or you can also write (e-V1)/R1+(e-V2)/R2=0 FirstChildUserIdTAG: 358712 FirstChildUserNameTAG: emansabah FirstChildCreateTimeTAG: 2012-09-16T12:17:35Z SecondChildTAG: I´m the same opinion as you, but if calculate I get a value of -2.05798.... This isn´t the right answer I gues. I get e = V1*(R1+R2)/R2 + V2*(R1+R2)/R1.... Did I commit an error? SecondChildUserIdTAG: 417013 SecondChildUserNameTAG: Annaloe SecondChildCreateTimeTAG: 2012-09-18T17:23:21Z IndexTAG: 4384 TitleTAG: lumped parameter lumped parameter UserIdTAG: 393532 UserNameTAG: RAAAHUL CreateTimeTAG: 2012-09-15T01:33:11Z VoteTAG: 0 CoursewareTAG: Week 1 / Course overview CommentableIdTAG: 6002x_S1V1toS1V4 NumberOfReplyTAG: 2 FirstChildTAG: TIME PASS!! FirstChildUserIdTAG: 127800 FirstChildUserNameTAG: MNB FirstChildCreateTimeTAG: 2012-09-15T06:02:26Z FirstChildTAG: Hi RAAAHUL! What did you mean with writing only lumped parameter? ;) Do you need help? Can I help you? FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-15T05:58:39Z IndexTAG: 4385 TitleTAG: I have a problem with home work 1 h1p1 the first 3 questions which are so easy ! , and i`m sure about my solution , everytime i solve it , it tells me that my answer wrong ?? what can i do ?? because of that , when i solve any problem i don`t know whether i`m realy solved it wrong , or there is something wrong with the website! UserIdTAG: 392663 UserNameTAG: matto0o CreateTimeTAG: 2012-09-15T00:54:03Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 4 FirstChildTAG: Hi! Can I help you mattoO0? In what part are you lost? FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-15T01:28:06Z SecondChildTAG: please , it is the equivelent resistance of homework 1 .....network "A,B,C" SecondChildUserIdTAG: 392663 SecondChildUserNameTAG: matto0o SecondChildCreateTimeTAG: 2012-09-15T01:32:43Z FirstChildTAG: may be you forgot to mention it in terms of R...the question says express it in terms of resistance R... FirstChildUserIdTAG: 371466 FirstChildUserNameTAG: abhiab25 FirstChildCreateTimeTAG: 2012-09-15T02:19:36Z SecondChildTAG: i did it , but that error message "Invalid input: Could not parse '1.66666666 R' as a formula" was the response ! SecondChildUserIdTAG: 392663 SecondChildUserNameTAG: matto0o SecondChildCreateTimeTAG: 2012-09-15T02:24:42Z SecondChildTAG: you are supposed to put an asterisk (*) in between multiplications like A*B*C in place of ABC, 9/13*R in place of 9/13R I hope it helps SecondChildUserIdTAG: 314624 SecondChildUserNameTAG: Owais001 SecondChildCreateTimeTAG: 2012-09-15T05:41:03Z FirstChildTAG: I got those questions right, so perhaps you are soling it wrong. If you give more information about what you did, maybe others could give some suggestions. FirstChildUserIdTAG: 310007 FirstChildUserNameTAG: ErinF FirstChildCreateTimeTAG: 2012-09-15T01:10:48Z SecondChildTAG: the soluton of 1st one"equivelent resistance of A is "3" ,2nd is "1/3" , 3rd "5/3" >>>>>>> is it true ?? SecondChildUserIdTAG: 392663 SecondChildUserNameTAG: matto0o SecondChildCreateTimeTAG: 2012-09-15T01:29:36Z SecondChildTAG: no.. you need to include Rs in the answer, for example, if i had 2 resistors em parallel the answer would be R/2. Then, add R in your answers.. SecondChildUserIdTAG: 116471 SecondChildUserNameTAG: haravez SecondChildCreateTimeTAG: 2012-09-15T03:08:45Z FirstChildTAG: Please tell me what exactly is meant by burning up of the composite resistor in the following H1P1 question? *Given that each individual resistor can dissipate up to 1 watt of power before burning up, how much total power in watts ( W ) can the smallest-valued composite resistor dissipate before burning up?* FirstChildUserIdTAG: 314624 FirstChildUserNameTAG: Owais001 FirstChildCreateTimeTAG: 2012-09-15T05:47:59Z SecondChildTAG: every physical resistor has a power rating which indicates the maximum power it can dissipate.... if the power dissipation exceeds this rating the resistor gets physically damaged,i.e burns, on account of excess current flowing through it........the power rating depends on the physical size of the resistor! SecondChildUserIdTAG: 367639 SecondChildUserNameTAG: paikartik SecondChildCreateTimeTAG: 2012-09-15T09:40:43Z SecondChildTAG: i really didnot know how to answer the question please help SecondChildUserIdTAG: 314241 SecondChildUserNameTAG: SUBIGYA SecondChildCreateTimeTAG: 2012-09-16T03:23:22Z IndexTAG: 4386 TitleTAG: isn't the combination of two resistors in parallel shape is 1/R1+1/R2 in this video, when the current source is acting alone. he summed the resistance and the answer was 4/3, and when I sum the resistance I get 3/4. and the method that I use is : 1/R1 + 1/R2 when combining two resistors in a parallel. so who is right? UserIdTAG: 292360 UserNameTAG: Loai CreateTimeTAG: 2012-09-15T00:12:31Z VoteTAG: 0 CoursewareTAG: Week 2 / Superposition CommentableIdTAG: 6002x_superposition_tutorial NumberOfReplyTAG: 2 FirstChildTAG: Close, but no cigar. Should be **1/**(1/R1+1/R2) FirstChildUserIdTAG: 367248 FirstChildUserNameTAG: MacMillan FirstChildCreateTimeTAG: 2012-09-15T00:31:07Z FirstChildTAG: The resistance is the reciprocal of your expression. Think about the units to check the formula. FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-15T00:27:52Z IndexTAG: 4387 TitleTAG: I didn't get that all I know up till now is that when the resistance in a circuit is related to the inputs of v and i in a steady line, then the circuit is a linear circuit. am i write? UserIdTAG: 292360 UserNameTAG: Loai CreateTimeTAG: 2012-09-14T23:07:56Z VoteTAG: 0 CoursewareTAG: Week 2 / Is it linear? CommentableIdTAG: 6002x_is_it_linear NumberOfReplyTAG: 1 FirstChildTAG: the v-i characteristics should pass through zero FirstChildUserIdTAG: 372535 FirstChildUserNameTAG: _Infinity FirstChildCreateTimeTAG: 2012-09-20T19:45:52Z IndexTAG: 4388 TitleTAG: Again about power of current source. There is my initials : V=2 , I=3 , R1=4 , R2=5. Well i solved algebraic system with KVL KCL and Element Relationships. All my answers are correct except the last one. I'm calculating power supplied by the current source as P = I*(-V2). And the answer is negative, while it should be positive. There i made a mistake? I assumed that in a loop with power supply and R2 KVL provides aquation : V2 + Vv = 0 , so Vv = - V2. UserIdTAG: 138620 UserNameTAG: ZatoIchi CreateTimeTAG: 2012-09-14T22:24:49Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL/KCL example CommentableIdTAG: 6002x_S2E3_Using_KVL_KCL_and_VI_constraints NumberOfReplyTAG: 0 IndexTAG: 4389 TitleTAG: Why is the video in B/W? Can any one tell me why is this video in B/W? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-14T21:59:14Z VoteTAG: 0 CoursewareTAG: Week 1 / Three series resistors CommentableIdTAG: 6002x_three_series_resistors NumberOfReplyTAG: 1 FirstChildTAG: accident. But we thought that it was more useful than not, so we included the tutorial anyway. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-09-15T02:38:10Z SecondChildTAG: You are lucky that getting atleast a B/W video. Due to ban on U-tube at our place, we are unable to see theses tutorials at all. SecondChildUserIdTAG: 455950 SecondChildUserNameTAG: Wahabbaluch SecondChildCreateTimeTAG: 2012-10-03T07:23:27Z IndexTAG: 4390 TitleTAG: maybe this would work for those who find it all very tough try watching the whole lectures from the OCW resources at once, and then go to the exercises if you require, the motive here is that you do not loose continuity, I mean you don't lose hope if your are not able to do a exercise, so watch the whole lecture there at once, it is all the same....hope I helped....FYI I am also one of those people( though i am still in school,i.e., did not AP Physics!) UserIdTAG: 169784 UserNameTAG: thewiredbear CreateTimeTAG: 2012-09-14T21:19:46Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: This is a great suggestion and it is easy just to open another window and work on the excercises while the lecture is playing. Unfortunately, this does not always clear up the issues. FirstChildUserIdTAG: 29275 FirstChildUserNameTAG: Mbarsalou FirstChildCreateTimeTAG: 2012-09-14T21:54:52Z SecondChildTAG: I get the sarcasm, and that is not what I meant, go through the whole lecture first and then the exercises, I mean prof. taught the node analysis method at the end, if you saw the whole lecture, you could have solved everything easily, and FYI every video player has a pause button, you could have always used that SecondChildUserIdTAG: 169784 SecondChildUserNameTAG: thewiredbear SecondChildCreateTimeTAG: 2012-09-15T17:40:13Z IndexTAG: 4391 TitleTAG: Proof of number of independent loops using KVL. Hello. Can anyone direct me to a proof of the formula which gives the number of independent loops when applying KVL? I can sort of see why if you have n nodes there are n-1 independent nodes but I am not seeing at all how you derive the number of independent looks from the total number of loops(or even how you derive the total number of loops from the number of branches). UserIdTAG: 371819 UserNameTAG: Lokiie CreateTimeTAG: 2012-09-14T21:15:33Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4392 TitleTAG: Vl question Vl= I*[Rn*Rl/(Rn+Rl)]=1.43*[0.1834/1.934]=0.135606 The correct answer is different Vl=0.13309 Where is mistake? UserIdTAG: 205311 UserNameTAG: Sergtronix CreateTimeTAG: 2012-09-14T20:43:00Z VoteTAG: 0 CoursewareTAG: Week 2 / Norton method example CommentableIdTAG: 6002x_S3E6_Norton_Model NumberOfReplyTAG: 2 FirstChildTAG: Hi ! By shorting the output, you will have a circuit with one current source and two paralel resistances Rp and 2Rs.It will help if you just draw it.The current IN is not I, but IN = I*Rp/(Rp+2Rs), precisely 1.40348. So 1.40348*[0.1834/1.934]=0.13309 .The only mistake is that the Rp value is so hi.If you need further help, i will post a picture. FirstChildUserIdTAG: 240487 FirstChildUserNameTAG: AlexAlexandrescu FirstChildCreateTimeTAG: 2012-09-14T21:30:40Z SecondChildTAG: Hi! Thank you for the answer! I made mistake when wrote the expession. Yes, IN=Vth/Rth=(I*Rp)/(Rp+2Rs)=1.403v. I shouldnt learning at night.. Thanks! SecondChildUserIdTAG: 205311 SecondChildUserNameTAG: Sergtronix SecondChildCreateTimeTAG: 2012-09-15T13:10:40Z SecondChildTAG: I get the same result! Where is the mistake? Bug? SecondChildUserIdTAG: 299251 SecondChildUserNameTAG: TheRedBlackOne SecondChildCreateTimeTAG: 2012-09-17T01:52:17Z FirstChildTAG: To compute VI you should use IN instead of I. Where IN is defined when the circuit between the Rs resistors are short: IN = I(Rp/(Rp+2*Rs)). FirstChildUserIdTAG: 375724 FirstChildUserNameTAG: sergestus FirstChildCreateTimeTAG: 2012-09-15T05:55:44Z IndexTAG: 4393 TitleTAG: ocw.mit.edu sir make the lectures downloadable, i cant access them coz internet speed is an issue in India. if it isn't possible can i use video lectures uploaded on ocw.mit.edu course 6.002 i have downloaded all of them. UserIdTAG: 154418 UserNameTAG: TECHJNKI CreateTimeTAG: 2012-09-14T19:48:31Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: well, I'm not sure if this it's ok but my internet speed is also too slow so I copy all the URLs from the youtube and I paste it in some file.txt then I use a download manager (youtube-dl it's very helpful in linux) FirstChildUserIdTAG: 63232 FirstChildUserNameTAG: VictorMChQ FirstChildCreateTimeTAG: 2012-09-15T06:51:20Z IndexTAG: 4394 TitleTAG: error Invalid input: Could not parse '3r' as a formula.error cming everytym i press check icon. UserIdTAG: 310313 UserNameTAG: ashu91 CreateTimeTAG: 2012-09-14T19:43:45Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: you must use 3*R or R+R+R it Could not parse '3r' as a formula FirstChildUserIdTAG: 282351 FirstChildUserNameTAG: HalfDead FirstChildCreateTimeTAG: 2012-09-14T19:53:48Z FirstChildTAG: It has already be said on the forum, you must use "3*R" with the multiplication "*", not only 3R. FirstChildUserIdTAG: 396446 FirstChildUserNameTAG: RousseauxS FirstChildCreateTimeTAG: 2012-09-14T19:49:20Z IndexTAG: 4395 TitleTAG: Help!! Lecture videos sill don't work! It says "An error occurred. Please try again later." It says the same thing if I go to the video in Youtube. I'm currently using Safari 5.1.7. UserIdTAG: 270160 UserNameTAG: Sethhhh CreateTimeTAG: 2012-09-14T18:15:55Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: It might have something to do with Safari because I haven't tried watching lectures on this computer before, but I can't install anything else because of permission restrictions. FirstChildUserIdTAG: 270160 FirstChildUserNameTAG: Sethhhh FirstChildCreateTimeTAG: 2012-09-14T18:18:24Z SecondChildTAG: Hmmm.. tested it on a non-edx Youtube video and that didn't work either. Must be the computer, then. SecondChildUserIdTAG: 270160 SecondChildUserNameTAG: Sethhhh SecondChildCreateTimeTAG: 2012-09-14T18:21:40Z IndexTAG: 4396 TitleTAG: You caught me This problem is a nice trick :) I did it the incorrect way the first time and had to read the comments to see what went wrong :) UserIdTAG: 252722 UserNameTAG: vaboro CreateTimeTAG: 2012-09-14T18:12:29Z VoteTAG: 0 CoursewareTAG: Week 2 / Node equation review CommentableIdTAG: 6002x_S3E1_NodeEquationReview NumberOfReplyTAG: 0 IndexTAG: 4397 TitleTAG: Puzzled with the sign conventions can any body help. so as to when to follow diagram arrows as sign convention or that the signs with numerical values UserIdTAG: 430650 UserNameTAG: saifhpr1 CreateTimeTAG: 2012-09-14T17:18:40Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL and KCL exercises CommentableIdTAG: 6002x_kvl_kcl_exercises NumberOfReplyTAG: 3 FirstChildTAG: Use the arrow to write down your equations, it is a "guess" sign. If you are wrong, the current will become negative in your equations and that is not a problem ; ). FirstChildUserIdTAG: 396446 FirstChildUserNameTAG: RousseauxS FirstChildCreateTimeTAG: 2012-09-14T18:49:37Z FirstChildTAG: A good way to keep track of it is to think either: all current INTO a node equals zero, or use all current OUT OF a node equals zero. That way if the directions are all mixed up, just use negative signs if it is pointing the wrong direction (for the convention you are using). If you keep everything straight the signs will then work themselves out. For example from S1E8: The bottom node is easy example to use "INTO" convention: i1 + i2 + i6 = 0. The top right node is a good use for "OUT OF" convention with a negative sign for i5: i3 + i6 - i5 = 0. If i5 ends up a negative value (meaning current is flowing in the opposite direction of the arrow) the math will work itself out, as long as you are consistent in the convention. Hope that helps. FirstChildUserIdTAG: 326961 FirstChildUserNameTAG: mrlogue FirstChildCreateTimeTAG: 2012-09-14T20:57:18Z FirstChildTAG: you must use both. the sign gives the value seen on the ammeter, when measured with the assumption that current is flowing in the shown direction. for eg, in the problem, i1 = -0.7 and it is coming out of the node. say, you follow the sign convention that current flowing into the node is positive and that leaving the node is negative. so your KCL equation will be -i1-i4-i5 = 0. substituting i1, -(-0.7)-i4 -i5 = 0. FirstChildUserIdTAG: 364086 FirstChildUserNameTAG: jordan3110 FirstChildCreateTimeTAG: 2012-09-16T01:32:27Z IndexTAG: 4398 TitleTAG: To staff sir i completed last circuit & electronics course 6.002 but due to some reasons i missed the final exam.may i appear only in final exam or i have to do this course again... UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-14T17:17:31Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Unfortunately, no. If you would like to take the course again, you must take the entire course. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-09-14T18:27:10Z IndexTAG: 4399 TitleTAG: S1E8 i am not able to enter the answer in the acceptable format, because simple variable i that i type is not acceptable. can someone help UserIdTAG: 430650 UserNameTAG: saifhpr1 CreateTimeTAG: 2012-09-14T16:57:38Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL and KCL exercises CommentableIdTAG: 6002x_kvl_kcl_exercises NumberOfReplyTAG: 1 FirstChildTAG: The type of variable is i1, i2, i3, etc. How do you type it ? FirstChildUserIdTAG: 396446 FirstChildUserNameTAG: RousseauxS FirstChildCreateTimeTAG: 2012-09-14T19:00:17Z IndexTAG: 4400 TitleTAG: CONFUSION OVER DEL Q/ DEL T Okay now, Current= differentiation of charge with respect to time? (since, change in charge per unit time is current) If this is zero, Does it mean no current? Also, How is delq/delt different from differentiation of q with respect to time. ( q= charge, t= time) UserIdTAG: 254325 UserNameTAG: bondablack CreateTimeTAG: 2012-09-14T16:42:13Z VoteTAG: 0 CoursewareTAG: Week 1 / Lumped Element Abstraction CommentableIdTAG: 6002x_lumped_element_abstraction NumberOfReplyTAG: 1 FirstChildTAG: Current is *defined* to be the rate of change of charge with respect to time: $i(t)=\frac{\partial q}{\partial t}$ However, net charge $q$ itself can be zero, but current can still be non-zero at parts of the system. Imagine a bunch of positive and negative particles. If a new negative particle is introduce every time an old negative particle is removed from the system, then the net charge stays at zero. The net change in charge is also zero, meaning that there is zero net current, but if you look at the part of the system where charge is exiting or the part where charge is entering, then current is non-zero at those parts of the system. Holistically, these currents should all add up to zero, as expected due to Kirchoff's Current Law. FirstChildUserIdTAG: 151427 FirstChildUserNameTAG: RichardZ FirstChildCreateTimeTAG: 2012-09-14T17:26:43Z IndexTAG: 4401 TitleTAG: homework Sir, in first three problems of the homework it says to write the answer as an algebraic expression. I know the answers but i dont know in which way to write it. I tried to write the answer in all the possible ways i could but says "could not parse (my answer) as formula".What does this mean? UserIdTAG: 425976 UserNameTAG: P_V_Rohith CreateTimeTAG: 2012-09-14T16:29:07Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 3 FirstChildTAG: suppose u want to write any number multiplied by a variable, you have to put the "*" sign between them and then it won't show that problem..! :) FirstChildUserIdTAG: 413819 FirstChildUserNameTAG: sharadmirani FirstChildCreateTimeTAG: 2012-09-14T16:32:46Z SecondChildTAG: Thanks SecondChildUserIdTAG: 425976 SecondChildUserNameTAG: P_V_Rohith SecondChildCreateTimeTAG: 2012-09-14T18:08:38Z FirstChildTAG: There is an exercise at the very start of the course on how to input algebraic expressions as your answer. FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-14T17:06:39Z FirstChildTAG: can you plz tell me the ans of homework so i can match my ans with you FirstChildUserIdTAG: 402830 FirstChildUserNameTAG: fahadmushtaq FirstChildCreateTimeTAG: 2012-09-14T17:53:16Z IndexTAG: 4402 TitleTAG: S1E7 I have problem with answers. whether the follow signs of the values of current or use the arrow direction. UserIdTAG: 430650 UserNameTAG: saifhpr1 CreateTimeTAG: 2012-09-14T16:26:02Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL and KCL exercises CommentableIdTAG: 6002x_kvl_kcl_exercises NumberOfReplyTAG: 1 FirstChildTAG: you must use both. the sign gives the value seen on the ammeter, when measured with the assumption that current is flowing in the shown direction. for eg, in the problem, i1 = -0.7 and it is coming out of the node. say, you follow the sign convention that current flowing into the node is positive and that leaving the node is negative. so your KCL equation will be -i1-i4-i5 = 0. substituting i1, -(-0.7)-i4 -i5 = 0. FirstChildUserIdTAG: 364086 FirstChildUserNameTAG: jordan3110 FirstChildCreateTimeTAG: 2012-09-16T01:31:23Z IndexTAG: 4403 TitleTAG: lab submission error Sir my check button in the lab platform doesn't work properly{nothing happen when it press}, but save button is working, could you please help me? my due date is almost near UserIdTAG: 325882 UserNameTAG: charlesandf CreateTimeTAG: 2012-09-14T16:22:47Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4404 TitleTAG: my DC and trans not clic my computer es una prueba para los resultados de serie de resistores para probar DC y TRAN, pronto mi solucion. estos solucion es favorable. adios my email: luisrbm@yahoo.es UserIdTAG: 363298 UserNameTAG: porquistun CreateTimeTAG: 2012-09-14T15:56:02Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: Hola porquistun! No comprendo muy bien lo que has querido decir. Cuál es tu dificultad? Puedo ayudarte en algo? Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-14T16:34:59Z IndexTAG: 4405 TitleTAG: S1E6 hi i have doubt in answers, of part1 and part3 UserIdTAG: 430650 UserNameTAG: saifhpr1 CreateTimeTAG: 2012-09-14T15:53:34Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL and KCL exercises CommentableIdTAG: 6002x_kvl_kcl_exercises NumberOfReplyTAG: 1 FirstChildTAG: Hi, Could you clarify what is your doubt? Some tips that you may find helpful: - Eg. You should input v1 instead of V1 - You can't use v3,v4 or v5 in your answers. - You should write your KVL equations and dispel the unknowns (My primary lenguage is spanish, so I'm not sure if the word "dispel" is the correct word to use) - You shouldn't type your answers including the unknown element you are searching. Eg.v3=V-v1, just type V-v1. FirstChildUserIdTAG: 265027 FirstChildUserNameTAG: edumm1 FirstChildCreateTimeTAG: 2012-09-14T22:36:28Z SecondChildTAG: ¡Gracias!for the fouth tip. How to writte the answer was making me crazy SecondChildUserIdTAG: 301442 SecondChildUserNameTAG: Casama SecondChildCreateTimeTAG: 2012-09-16T16:40:28Z IndexTAG: 4406 TitleTAG: All the best guys I have just finished my HW and lab1 with 100% each thank to the lectures . UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-14T15:43:21Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: can you please help me with H1P1 last question?? got confused many times!!! FirstChildUserIdTAG: 170475 FirstChildUserNameTAG: sandeshacharya FirstChildCreateTimeTAG: 2012-09-14T16:03:51Z FirstChildTAG: kindly help me in last Q of HW 1 plxxxxxxx FirstChildUserIdTAG: 430030 FirstChildUserNameTAG: imali FirstChildCreateTimeTAG: 2012-09-14T16:23:25Z IndexTAG: 4407 TitleTAG: Power of current source? The power of the current source is positive. So it's consuming power rather than supplying power? UserIdTAG: 201508 UserNameTAG: datle CreateTimeTAG: 2012-09-14T15:31:50Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL/KCL example CommentableIdTAG: 6002x_S2E3_Using_KVL_KCL_and_VI_constraints NumberOfReplyTAG: 1 FirstChildTAG: the power of the current source is positive implies that the current source is supplying power to the circuit.... had it been negative then it means that the circuit is forcing the current to oppose the direction of the current source, thus supplying power TO THE source, i.e, the current source would be consuming power in this case..... FirstChildUserIdTAG: 367639 FirstChildUserNameTAG: paikartik FirstChildCreateTimeTAG: 2012-09-14T15:52:12Z SecondChildTAG: thank you! SecondChildUserIdTAG: 201508 SecondChildUserNameTAG: datle SecondChildCreateTimeTAG: 2012-09-15T14:22:28Z SecondChildTAG: the current source would consume power provided if R2 is its internal resistance. SecondChildUserIdTAG: 140065 SecondChildUserNameTAG: Madhav92 SecondChildCreateTimeTAG: 2012-09-15T19:34:34Z IndexTAG: 4408 TitleTAG: Week 1 Tutorials I am not being able to see the tutorials for week 1 please advise thanks my personal email is aimstermill51@aol.com UserIdTAG: 411228 UserNameTAG: aimster CreateTimeTAG: 2012-09-14T15:28:51Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4409 TitleTAG: confirming the answer which i got hi i want to know how many of you got the same answer as mine x1 = .6 x2 = 5.6 y1 = -.225 y2 = 1.34 UserIdTAG: 217045 UserNameTAG: ANURAG1993 CreateTimeTAG: 2012-09-14T14:55:53Z VoteTAG: 0 CoursewareTAG: Week 2 / Superposition example CommentableIdTAG: 6002x_S3E3_superposition NumberOfReplyTAG: 3 FirstChildTAG: y2 is not (x2-V2)/R2. y2 = x2/R1 FirstChildUserIdTAG: 222527 FirstChildUserNameTAG: sebinanithottam FirstChildCreateTimeTAG: 2012-09-16T10:52:09Z FirstChildTAG: Hi, I got x1 = 0.42 V, x2 = 3.94 V, y1 = -0.228 and y2 = 0.563 FirstChildUserIdTAG: 220837 FirstChildUserNameTAG: Calsomus FirstChildCreateTimeTAG: 2012-09-15T00:02:29Z FirstChildTAG: Hey. I got the same answers as you did while using a calculator on the page. Then i used my own. And the answers I got were true (0.42 etc) Try it one more time FirstChildUserIdTAG: 308508 FirstChildUserNameTAG: Alex-Fion FirstChildCreateTimeTAG: 2012-09-17T10:57:06Z SecondChildTAG: The question is to find the current passing through resistor R1(ie..i1) using superposition. So we get y2 as +.56 SecondChildUserIdTAG: 439427 SecondChildUserNameTAG: dany87 SecondChildCreateTimeTAG: 2012-09-18T18:31:26Z IndexTAG: 4410 TitleTAG: current sources &kcl why does using current sources in building circuits voilate kcl ? UserIdTAG: 221617 UserNameTAG: konan CreateTimeTAG: 2012-09-14T14:53:53Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Using current sources does not necessarily violate KCL, it depends how you use them. Although they only violate KCL in theory. If you were to attempt to create such a circuit in practice it would either not work as expected or something would break. FirstChildUserIdTAG: 270011 FirstChildUserNameTAG: SnowmanZA FirstChildCreateTimeTAG: 2012-09-14T17:20:27Z IndexTAG: 4411 TitleTAG: invalid input Why does it say invalid input: v0 not permitted in answer when I type in (e1-v0)/R1 UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-14T13:53:45Z VoteTAG: 0 CoursewareTAG: Week 1 / Node analysis practice, part 2 CommentableIdTAG: 6002x_L2Node1 NumberOfReplyTAG: 2 FirstChildTAG: change the v letter to capital V FirstChildUserIdTAG: 136959 FirstChildUserNameTAG: christerpher FirstChildCreateTimeTAG: 2012-09-14T14:05:45Z SecondChildTAG: Check to make sure that you do not have double space in between the expressions. I keyed in the same thing and was marked right. SecondChildUserIdTAG: 214990 SecondChildUserNameTAG: Shuaibu SecondChildCreateTimeTAG: 2012-09-17T20:16:06Z FirstChildTAG: In fact, you must type V0 (in capital letters). FirstChildUserIdTAG: 396446 FirstChildUserNameTAG: RousseauxS FirstChildCreateTimeTAG: 2012-09-14T14:06:55Z SecondChildTAG: Thanks how foolish of me. SecondChildUserIdTAG: undefined SecondChildUserNameTAG: undefined SecondChildCreateTimeTAG: 2012-09-14T21:02:02Z SecondChildTAG: I figured it out after posting the question. SecondChildUserIdTAG: undefined SecondChildUserNameTAG: undefined SecondChildCreateTimeTAG: 2012-09-14T21:03:12Z IndexTAG: 4412 TitleTAG: I have not been able to watch the lecture sequence I need your attention here as i have not been able to watch the sequence lectures since i joined the class. I am sure of the good status of the internet facility, is there anything i should do. Thanks for your swift response. UserIdTAG: 423384 UserNameTAG: oliveola2015 CreateTimeTAG: 2012-09-14T13:43:52Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: where are you from? Make sure that the network you access isn't blocked.for instance,you can't watch these videos in China unless you use an agent or VPN FirstChildUserIdTAG: 136959 FirstChildUserNameTAG: christerpher FirstChildCreateTimeTAG: 2012-09-14T14:07:56Z IndexTAG: 4413 TitleTAG: LABS sir please anyone tell me from where we can prepare lab work theory????? not tool?? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-14T13:41:35Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4414 TitleTAG: S3E2 I am still struggling with finding b1 & b2. Writting the KCL and KVL equations I am ok with but manipulating them to create coefficients for V1 and V2 is still a struggle. Even substituting in V3 I seem to be making a mess of it. Is there any one who can provide a walk through of the solution? Thanks! UserIdTAG: 244115 UserNameTAG: Pedro1969 CreateTimeTAG: 2012-09-14T13:00:59Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: I assume you found a1 and a2 (or v3). Let's find $i_1$ that is equal to $i_1=\frac{v_3-V1}{R_1}$. For $a_1$, I have $a_1=\frac{1/R1}{1/R1+1/R2+1/R3}$ and for $a_2=\frac{1/R2}{1/R1+1/R2+1/R3}$. (I was too lazy to rearrange the terms ;)). For the current: $\begin{equation} i_1=\frac{v_3-V1}{R_1}\\ =1/R1\cdot[\frac{1/R1}{1/R1+1/R2+1/R3}\cdot V_1+\frac{1/R2}{1/R1+1/R2+1/R3}\cdot V_2-V_1]\\ =1/R1\cdot[(\frac{1/R1}{1/R1+1/R2+1/R3}-1)\cdot V_1+\frac{1/R2}{1/R1+1/R2+1/R3}\cdot V_2] \end{equation}$ This is it, you can rearrange the terms if you want to find the exact same answers but you can type it like this (it is equivalent). FirstChildUserIdTAG: 396446 FirstChildUserNameTAG: RousseauxS FirstChildCreateTimeTAG: 2012-09-14T14:00:52Z IndexTAG: 4415 TitleTAG: S6V4 video 1:32 correct text Norton instead of *node* word time 1:32 "... preferable, to replace the rest of the circuit with a **node** method. Either way is fine. You can do a **node** or a Thevenin, and the two are equivalent." - correct text is "**Norton**" instead of [ node ] UserIdTAG: 331483 UserNameTAG: Doru CreateTimeTAG: 2012-09-14T10:24:21Z VoteTAG: 0 CoursewareTAG: Week 3 / Analyzing Nonlinear Circuits CommentableIdTAG: 6002x_analyzing_nonlinear_circuits NumberOfReplyTAG: 0 IndexTAG: 4416 TitleTAG: lab1 have managed to do lab1, qn1 but am hustling with qn2, someone help me out UserIdTAG: 254607 UserNameTAG: werehenry CreateTimeTAG: 2012-09-14T09:35:58Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: haha, i ve gotten it FirstChildUserIdTAG: 254607 FirstChildUserNameTAG: werehenry FirstChildCreateTimeTAG: 2012-09-14T10:15:06Z FirstChildTAG: actually you have to solve both questions simultaneously. In question 2 there is the limit of 2 volt and bulb is off so just you have a voltage divider circuit. The voltage across the second resistor should be 2 volt and here you will find an equation involving R1 and R2. 1st question will also give an equation solve them simultaneously to find resistor values. Good Luck! FirstChildUserIdTAG: 429821 FirstChildUserNameTAG: prince7ali FirstChildCreateTimeTAG: 2012-09-14T11:06:37Z IndexTAG: 4417 TitleTAG: H1P3 3,4,5 questions In the questions 3,4,5 how to calculate resistances for H1,H2,H3...? If we use individual powers of H1,H2,H3 and 240V supply while calculating resistances using formula P=V^2/R then we get all the resistances H1,H2,H3 equal...but it is not true..isn't it.(since currents are different)...can anyone please help me... UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-14T08:40:42Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: If H1,H2 and H3 have the same resistance, exactly 1/3 of the current will flow through each branch. In other words the current aren't different. You can use P=I^2*R as well. If the resistances are to be the same, the currents will be too. FirstChildUserIdTAG: 236188 FirstChildUserNameTAG: Rince FirstChildCreateTimeTAG: 2012-09-14T09:57:39Z SecondChildTAG: hi...thanks for ur reply...sorry i cant get that...please tell me some what detailed with formulas..please...i am struggling with these 3 questions from 3 days... SecondChildUserIdTAG: undefined SecondChildUserNameTAG: undefined SecondChildCreateTimeTAG: 2012-09-14T14:15:24Z SecondChildTAG: if i in H1 is total current/3 then i will be 4.625A isn't it...and again if i calculate power i got that same individual powers...as given in problem... SecondChildUserIdTAG: undefined SecondChildUserNameTAG: undefined SecondChildCreateTimeTAG: 2012-09-14T14:17:03Z IndexTAG: 4418 TitleTAG: Not able to CHECK my LAB 1 Sir, I have saved my Lab 1 circuit, now when I am pressing CHECK button, there is no response. Also my progress in LAB 1 is showing 0%. What I should do ? UserIdTAG: 341850 UserNameTAG: Juhi CreateTimeTAG: 2012-09-14T07:52:01Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Did you remember to run the DC analysis first? Only after the values show up can you "check" your circuit. FirstChildUserIdTAG: 236188 FirstChildUserNameTAG: Rince FirstChildCreateTimeTAG: 2012-09-14T08:00:39Z SecondChildTAG: Also, the node A must have a valid voltage after running DC analysis for the submission to work. SecondChildUserIdTAG: 79361 SecondChildUserNameTAG: kimt SecondChildCreateTimeTAG: 2012-09-14T10:50:03Z IndexTAG: 4419 TitleTAG: The current source example Hello, the example of a voltage source is a battary, but what is the example of a current source? UserIdTAG: 375724 UserNameTAG: sergestus CreateTimeTAG: 2012-09-14T07:12:05Z VoteTAG: 0 CoursewareTAG: Week 2 / Superposition CommentableIdTAG: 6002x_superposition NumberOfReplyTAG: 3 FirstChildTAG: A current mirror, diode connected transistor, etc. FirstChildUserIdTAG: 330269 FirstChildUserNameTAG: Jaikr FirstChildCreateTimeTAG: 2012-09-14T08:42:34Z FirstChildTAG: In my opinion they are interrelated. http://en.wikipedia.org/wiki/Current_source So far we have used a lumped-circuit current source, which is just an abstraction. FirstChildUserIdTAG: 252722 FirstChildUserNameTAG: vaboro FirstChildCreateTimeTAG: 2012-09-14T20:58:11Z FirstChildTAG: The majority of current sources that I've seen in my job is a bootstrap circuit which is two transistors in series. They call them Darlington pairs. FirstChildUserIdTAG: 247286 FirstChildUserNameTAG: ronald_borunda FirstChildCreateTimeTAG: 2012-09-17T01:44:38Z SecondChildTAG: After several lecture sequences I started to think of a current source as a device that can somehow produce a straight line corresponding to a constant current in its v-i relationship graph. Is this correct? SecondChildUserIdTAG: 252722 SecondChildUserNameTAG: vaboro SecondChildCreateTimeTAG: 2012-10-06T01:06:58Z SecondChildTAG: correct with some non uniformities ofcourse! SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-12-19T12:58:26Z IndexTAG: 4420 TitleTAG: use of tools...... how can i add two or more element by simple wire or line ....pls suggest me it by tools provided. UserIdTAG: 307088 UserNameTAG: PRAKHAR2012 CreateTimeTAG: 2012-09-14T06:52:44Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: Hi PRAKHAR2012! I will copy and paste [my answer of one request of this Post][1].If you have any doubt please ask, I will be pleasured to help you! STEP 1: We will choose a Resistor. Go to that element. And select it with the mousse. Click on that element without releasing and dragg it to the work place. ![enter image description here][2] STEP 2: You will see something like this. The resistor will be appear note in blue, but yes in green. Ok, if you click on the work place you will find that the resistor will change to blue. When it is green is because we are selecting that element, if not not. ![enter image description here][3] STEP 3: Lets try to select another resistor and connect them with a wire. So, repeat the STEP 1 again. And dragg the other resistor in other different place from the first resistor. ![enter image description here][4] STEP 4: Now the difficult part. Lets use the wire ;) ![enter image description here][5] STEP 5: oK! Done. You should see something like this. But wait a minute... I made a mistake, if I want to delet that wire??? How can I do that? Ok, see STEP 6. ![enter image description here][6] STEP 6: Select the wire with the mose and press delete in your keyboard ;). You will delete the wire that you have created before. ![enter image description here][7] Myriam. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/MITx_6_002x_2012_Fall/threads/50474faf685941270000000d [2]: http://s7.postimage.org/xiogau7e3/STEP_1.png [3]: http://s16.postimage.org/z4c1erqg5/STEP2.png [4]: http://s16.postimage.org/vfqxt15t1/step_3.png [5]: http://s13.postimage.org/mri1a7b7r/STEP4.png [6]: http://s9.postimage.org/t0ubwrlxr/STEP5.png [7]: http://s9.postimage.org/et98luymn/step6.png FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-14T12:12:38Z SecondChildTAG: Hi Myriam, I posted your help to a [wiki page][1]. I hope you don't mind! [1]: https://www.edx.org/wiki/6.002x/MyrimitCircuitSimulatorTutorial/ SecondChildUserIdTAG: 79361 SecondChildUserNameTAG: kimt SecondChildCreateTimeTAG: 2012-09-15T16:50:54Z IndexTAG: 4421 TitleTAG: lab 1 i got vs=1.5 for the first circuit and vs=2 for the second circuit, still i am getting that answer is incorrect, anybody got this same problem? UserIdTAG: 141709 UserNameTAG: charbelantonios CreateTimeTAG: 2012-09-14T05:22:11Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Did you use the same value resistors in both circuits? They should both be exactly the same besides the light bulb. FirstChildUserIdTAG: 270011 FirstChildUserNameTAG: SnowmanZA FirstChildCreateTimeTAG: 2012-09-14T05:26:53Z IndexTAG: 4422 TitleTAG: home work 1 H1P1 LAST QUESTION HOW TO SOLVE IT ??? AND THERE IS DEADLINE BUT NO OPTION FOR SUBMISSION BUTTON...?? HOW TO SUBMIT HOMEWORK AFTER ALL I'VE COMPLETED IT UserIdTAG: 387287 UserNameTAG: sailesh_dahal CreateTimeTAG: 2012-09-14T04:39:54Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: just check it..it gets submitted automatically..check out Progress section after checking all answers.. FirstChildUserIdTAG: 80645 FirstChildUserNameTAG: maya_swarnky FirstChildCreateTimeTAG: 2012-09-14T04:48:40Z SecondChildTAG: hey .... have u done last Q of first home work ???? SecondChildUserIdTAG: 430030 SecondChildUserNameTAG: imali SecondChildCreateTimeTAG: 2012-09-14T04:54:23Z FirstChildTAG: mean composite resistor ?? FirstChildUserIdTAG: 430030 FirstChildUserNameTAG: imali FirstChildCreateTimeTAG: 2012-09-14T04:47:56Z IndexTAG: 4423 TitleTAG: voltage across resistor: i could understand how to calculate the voltage across the resistor.Please anyone can help me out ... UserIdTAG: 429168 UserNameTAG: arunprakashavm CreateTimeTAG: 2012-09-14T03:54:38Z VoteTAG: 0 CoursewareTAG: Week 1 / Series and parallel example CommentableIdTAG: 6002x_S2E4_Series_and_Parallel NumberOfReplyTAG: 1 FirstChildTAG: hope you have calculated i1 i.e. the current in the circuit.The current on coming to node where R1, R2 and R3 meet, divides in the ratio of resistances FirstChildUserIdTAG: 201414 FirstChildUserNameTAG: ArcherJeffrey FirstChildCreateTimeTAG: 2012-09-14T06:18:47Z IndexTAG: 4424 TitleTAG: Why Why don’t you use R1, R2 and R3. UserIdTAG: 337902 UserNameTAG: Dan_Pop CreateTimeTAG: 2012-09-14T03:26:40Z VoteTAG: 0 CoursewareTAG: Week 1 / Nodal analysis example CommentableIdTAG: 6002x_NodalAnalysisExample NumberOfReplyTAG: 0 IndexTAG: 4425 TitleTAG: Element law for a current source I think the element law for a current source (Eq. 1.21) should be $i=-I$ by the associated variable convention and the positive terminal assignment shown in Fig. 1.34. In fact, as can be seen on the bottom part of Fig. 1.34, the constant current line has a negative $y$-axis intercept. UserIdTAG: 153760 UserNameTAG: Kavka CreateTimeTAG: 2012-09-14T02:47:57Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4426 TitleTAG: power (in Watts) supplied by the current source What is the power (in Watts) supplied by the current source? How I do this? thankx UserIdTAG: 166869 UserNameTAG: Vasco CreateTimeTAG: 2012-09-14T01:27:58Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL/KCL example CommentableIdTAG: 6002x_S2E3_Using_KVL_KCL_and_VI_constraints NumberOfReplyTAG: 1 FirstChildTAG: you have the current source parallel with a resistor so v2 cross resistor is also cross current source p=vi v =v2 i = i from current source (given ) FirstChildUserIdTAG: 418898 FirstChildUserNameTAG: Mostafa_Elboss FirstChildCreateTimeTAG: 2012-09-14T01:41:30Z SecondChildTAG: um.. what? SecondChildUserIdTAG: undefined SecondChildUserNameTAG: undefined SecondChildCreateTimeTAG: 2012-09-14T03:49:39Z IndexTAG: 4427 TitleTAG: I forgot what EECS stands for... It's kind of silly, I know... However I would suggest EdX to make some sort of "Index or Key words section" for each week or at least for the whole course. I think it would help the non-native speakers a bit, specially since many of us don't use acronyms in our own cultures as often as english speakers. Still, the course is quite good. Your efforts are well appreciated EdX!! UserIdTAG: 256249 UserNameTAG: And90 CreateTimeTAG: 2012-09-13T23:20:45Z VoteTAG: 0 CoursewareTAG: Week 1 / Lumped Element Abstraction CommentableIdTAG: 6002x_lumped_element_abstraction NumberOfReplyTAG: 2 FirstChildTAG: By the way, EECS stands for electrical engineering and computer science. FirstChildUserIdTAG: 151427 FirstChildUserNameTAG: RichardZ FirstChildCreateTimeTAG: 2012-09-13T23:48:57Z FirstChildTAG: We're looking into this. In the mean time, the textbook has an index, and that's a great way to look up key words that you don't understand. FirstChildUserIdTAG: 151427 FirstChildUserNameTAG: RichardZ FirstChildCreateTimeTAG: 2012-09-13T23:35:15Z SecondChildTAG: Good to know. Thank you very much! SecondChildUserIdTAG: 256249 SecondChildUserNameTAG: And90 SecondChildCreateTimeTAG: 2012-09-14T00:24:06Z IndexTAG: 4428 TitleTAG: Wrong problem Battery voltage is 2 Volts , both resisters are 9 Ohms The current i1 will be 0.2222 A. How come current I is 3 A? UserIdTAG: 429518 UserNameTAG: Fang777 CreateTimeTAG: 2012-09-13T22:43:45Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL/KCL example CommentableIdTAG: 6002x_S2E3_Using_KVL_KCL_and_VI_constraints NumberOfReplyTAG: 1 FirstChildTAG: There's a current source as well as a voltage source in the circuit. FirstChildUserIdTAG: 312035 FirstChildUserNameTAG: philhiggins FirstChildCreateTimeTAG: 2012-09-13T23:43:52Z IndexTAG: 4429 TitleTAG: check tutorials i checked week 1 tutorials , nodal analysis section and i applied it , and it worked great UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-13T22:22:47Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL/KCL example CommentableIdTAG: 6002x_S2E3_Using_KVL_KCL_and_VI_constraints NumberOfReplyTAG: 0 IndexTAG: 4430 TitleTAG: [tech] In my lab 1 section I have completed my circuit, but the check button is not working so my Lab 1 assignment is not evaluated, Please help me my lab is due by 16th December, 2012. UserIdTAG: 333877 UserNameTAG: gaurab CreateTimeTAG: 2012-09-13T19:41:29Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: you need the label A for the joint, I dont't know how you can create this Label, because I can choose only a resistor FirstChildUserIdTAG: 389333 FirstChildUserNameTAG: juergen FirstChildCreateTimeTAG: 2012-09-13T20:49:32Z SecondChildTAG: You can reset the problem and get the defaults. SecondChildUserIdTAG: 79361 SecondChildUserNameTAG: kimt SecondChildCreateTimeTAG: 2012-09-13T21:12:33Z FirstChildTAG: my reset button is not on the screen. it is showing only check and save.though the check button is not working but if i click to the save button it tells that hit the check button to grade your answer. but when i click to the check button it is not grading my answer. So my lab is still showing incomplete. FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-14T18:14:13Z IndexTAG: 4431 TitleTAG: H1P2 I can't find the value of V1, neither i4. Could anyone find the answer? I need help. UserIdTAG: 51667 UserNameTAG: Assuncao CreateTimeTAG: 2012-09-13T18:52:38Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: hi For V 1 u have to apply KVL over the boundary of Circuit .... For i 4 u have to aplly mesh analysis ... i have also 1 Q from u ..... ?? FirstChildUserIdTAG: 415376 FirstChildUserNameTAG: imab90 FirstChildCreateTimeTAG: 2012-09-13T18:54:39Z IndexTAG: 4432 TitleTAG: Circuit diagram please Can someone please give me a circuit diagram of the about prob??? Please UserIdTAG: 156974 UserNameTAG: ManojKumar CreateTimeTAG: 2012-09-13T18:41:57Z VoteTAG: 0 CoursewareTAG: Week 1 / Power delivery modeling CommentableIdTAG: 6002x_S2E6_Modeling NumberOfReplyTAG: 1 FirstChildTAG: Sure, I'll bite. I'll also use the circuit simulator :) ![enter image description here][1] [1]: https://dl.dropbox.com/u/45575056/ACline.png FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-13T18:50:48Z SecondChildTAG: Thanks :) I figured out.. :) SecondChildUserIdTAG: 156974 SecondChildUserNameTAG: ManojKumar SecondChildCreateTimeTAG: 2012-09-15T17:59:56Z IndexTAG: 4433 TitleTAG: Link to the Wikipedia reference ? What did he say at the end of video? He addressed a Wikipedia article, kelvin four tuple measurement or something like this? can somebody please provide that link...My English is not that good UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-13T17:55:25Z VoteTAG: 0 CoursewareTAG: Week 1 / Black-Box Resistor Network CommentableIdTAG: 6002x_black_box_resistor_network NumberOfReplyTAG: 2 FirstChildTAG: do u mean the last link regarding AWG??? If so.. don't worry, its some standards. Its not related to this learning.--- Manoj FirstChildUserIdTAG: 156974 FirstChildUserNameTAG: ManojKumar FirstChildCreateTimeTAG: 2012-09-13T19:07:48Z FirstChildTAG: He talks about a "Kelvin four terminal measurement". (Note: turn-on the CC of the video is you don't understand everything ; )) When measuring low value resistances, the risk is to also measure the resistance of the probe used for the measurement (assuming 1ohm (not realistic) for each probe/test lead and a measured resistance of 1ohm, the ohmmeter will give 3ohms). A technique to avoid this problem is the 4-terminal : you use two probes to put some current in the measured resistor and two other probes to measure the voltage across the resistor. No current will flow in these two second probes and you won't measure the test leads. http://en.wikipedia.org/wiki/Four-terminal_sensing FirstChildUserIdTAG: 396446 FirstChildUserNameTAG: RousseauxS FirstChildCreateTimeTAG: 2012-09-13T19:13:30Z SecondChildTAG: its a kelvin double bridge meter used explicitly to measure very low resistances... not required for this course SecondChildUserIdTAG: 368712 SecondChildUserNameTAG: jmen SecondChildCreateTimeTAG: 2012-10-17T14:13:18Z IndexTAG: 4434 TitleTAG: So long - and thanks for all the fish Just wanted to thank you all for taking the effort to organise this course, with all it's materials etc. However, it's quite clear that in the decade since I last touched study I've lost far too much maths to participate. Integrating trigonometric functions is beyond me know. I might be back in a year or two, but there's simply no chance I'll be able to relearn enough maths to keep up, do the day job and keep up with the course. But you, the staff deserve thanks for putting this together. Thanks. UserIdTAG: 413536 UserNameTAG: ThreeTailsCrossing CreateTimeTAG: 2012-09-13T17:18:18Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4435 TitleTAG: H1 i did my homework 1 , anybody knows what now, is there any submit option ,because in my courseware it is still showing homework 1 due 16 september UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-13T17:12:20Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: kindly help me in last Q of HW 1 ...what is value of H1 H2 and total and please do tell me the whole procedure ...i shall be very kind 2 u :) FirstChildUserIdTAG: 415376 FirstChildUserNameTAG: imab90 FirstChildCreateTimeTAG: 2012-09-14T17:26:15Z IndexTAG: 4436 TitleTAG: video runs at 1.5x. Cannot bring it back to normal my video runs at 1.5x speed. I cannot change it to normal ie 1.0 Please help UserIdTAG: 130166 UserNameTAG: sandeepsingh CreateTimeTAG: 2012-09-13T14:54:26Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: This has to do with the hurried transition to the YouTube player after Google changed the external API. Fixing it now :D FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-13T15:21:49Z FirstChildTAG: You can choose any speed you like, in .1 increments. See: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_Fall2012_General/threads/505f2ee8786ce92b00000014 FirstChildUserIdTAG: 4076 FirstChildUserNameTAG: damians FirstChildCreateTimeTAG: 2012-10-02T17:09:55Z IndexTAG: 4437 TitleTAG: Current from the superposition equation There is an superposition equation from the lecture, v= sum(am*Vm) + sum(bn*In) + i*R. And the i*R part is said to be obtained by set the Vm and In to zeros. This is what confuses me. How can we obtain i when there is basically no current flowing through the circuit. I understand when there is no current i=0 but I wonder where that i comes from in the first place? What if i said the superposition equation is v= sum(am*Vm) + sum(bn*In) + c*R, and c is some sort of unknown. So we get c= 1/R* v- (sum(am*Vm) + sum(bn*In)). And i dont see any sort of connection between c and i, so how can c=i as in the correct equation. I hope someone can help me understand more about the problem. UserIdTAG: 146814 UserNameTAG: humink CreateTimeTAG: 2012-09-13T14:22:58Z VoteTAG: 0 CoursewareTAG: Week 2 / Thevenin method CommentableIdTAG: 6002x_thevenin NumberOfReplyTAG: 2 FirstChildTAG: The fact is the last "i" is a forced known current (external excitation) not related to the system. You must distinguish the system itself that you want to simplify and the external world (in our case "i"). You can also see it as a current used to determine Rth. You get your system, you turn off every source (short circuit for voltage source and open circuit for current source) and then apply a current "i" from the output. What you will get is a voltage with a value v=Rth*i; knowing i, you find Rth. FirstChildUserIdTAG: 396446 FirstChildUserNameTAG: RousseauxS FirstChildCreateTimeTAG: 2012-09-13T19:33:11Z FirstChildTAG: i is the current coming from the external network E. So if you are setting all V's and I's of your arbitrary network to zero that has no impact for the current coming in the aribtrary network. You can also say that the external network E is a second arbitrary network and i is the current VTHE/RTHE whereby E means external network. A example: So if the external network exists of a resistance RE, the current i is zero. If the external network exists of a current source IE the current i is IE. I hope I understood it right. If there are some mistakes in my thoughts please correct me. FirstChildUserIdTAG: 230701 FirstChildUserNameTAG: MEng_IIT FirstChildCreateTimeTAG: 2012-09-13T19:35:38Z SecondChildTAG: By doing the exercises I kinda figure out i depends on the external excitation, but I quite doubt ur idea that when the external network have a resistance Re, the current is zero. I think that the current i will be negative, as there is only current flowing out of the Thevenin model. And so v = VTH + i*RTH < VTH. And i can be calculated by i= -VTH/(RTH+Re). SecondChildUserIdTAG: 146814 SecondChildUserNameTAG: humink SecondChildCreateTimeTAG: 2012-09-14T05:25:52Z IndexTAG: 4438 TitleTAG: Which time zone is taken as a basis for the homework? Which time zone is taken as a basis for the homework? I cannot find the answer? Thank you UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-13T14:22:54Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: All dates and times posted are times on the East Coast of the US, i.e. time in Boston. FirstChildUserIdTAG: 151427 FirstChildUserNameTAG: RichardZ FirstChildCreateTimeTAG: 2012-09-13T17:18:35Z SecondChildTAG: UTC/GMT -5 hours SecondChildUserIdTAG: 26837 SecondChildUserNameTAG: carloscorp SecondChildCreateTimeTAG: 2012-09-14T01:11:57Z IndexTAG: 4439 TitleTAG: doubt in lab how to save the lab exercise answers UserIdTAG: 151520 UserNameTAG: venila CreateTimeTAG: 2012-09-13T14:07:07Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4440 TitleTAG: home work exercises i made the homework H1P1 equivalent resistances of networks A&C are wrong. i write for network A is 3R,because three resistances are in series so the equivalent resistance is 3R but it shows wrong. Same way for Network C the equivalent resistance is '5R/3' or '1.666666667R', it show wrong why? UserIdTAG: 276803 UserNameTAG: srinivasaram CreateTimeTAG: 2012-09-13T14:04:11Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 3 FirstChildTAG: Dear srinivasaram, write answer in check book R+R+R FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-13T14:17:14Z FirstChildTAG: Try and make sure you are indicating multiplication (write 3*R instead of just 3R, or 5*R/3 instead of 5R/3). If you type your answers like that they should work. FirstChildUserIdTAG: 390034 FirstChildUserNameTAG: kahlil FirstChildCreateTimeTAG: 2012-09-13T14:09:59Z SecondChildTAG: Is power divide for parallel circuits? SecondChildUserIdTAG: 276803 SecondChildUserNameTAG: srinivasaram SecondChildCreateTimeTAG: 2012-09-13T14:18:30Z SecondChildTAG: Doing homework H1P1. Figure "B", all three of resistance are connected in parallel to them vormula 1 / R +1 / R +1 / R. Says that is not true. What is wrong? SecondChildUserIdTAG: 202275 SecondChildUserNameTAG: Valya SecondChildCreateTimeTAG: 2012-09-13T14:57:14Z SecondChildTAG: Hey Kahlil, thanks! The '*' really makes the difference!!! SecondChildUserIdTAG: 124785 SecondChildUserNameTAG: santosdanillo SecondChildCreateTimeTAG: 2012-09-15T03:44:01Z FirstChildTAG: srinivasram, you have to use correct algebric form.......for 3r, u have to use 3*r. FirstChildUserIdTAG: 114864 FirstChildUserNameTAG: Vikaash FirstChildCreateTimeTAG: 2012-09-15T11:42:08Z IndexTAG: 4441 TitleTAG: Bug video solved As i see it may have been a subtitles problem. Dont worry, i think the subtitles on the right side were not so useful. I prefer when they are judt under the video. The eye is closest , its better to read and see everything. Thanks a lot ! UserIdTAG: 373752 UserNameTAG: Croak CreateTimeTAG: 2012-09-13T13:41:20Z VoteTAG: 0 CoursewareTAG: Week 3 / Static Power of a MOSFET CommentableIdTAG: 6002x_mosfet_static_power NumberOfReplyTAG: 0 IndexTAG: 4442 TitleTAG: methink The current in the circuit is 177uA this pass through 2 resisters R1 and R2, drop across R1 sum up with 5v i.e 1.206+5=6.20V...which is noted as e is my theory correct?? UserIdTAG: 212417 UserNameTAG: saiteja25 CreateTimeTAG: 2012-09-13T13:23:27Z VoteTAG: 0 CoursewareTAG: Week 2 / Node equation review CommentableIdTAG: 6002x_S3E1_NodeEquationReview NumberOfReplyTAG: 1 FirstChildTAG: Yes, that is exactly what is happening. You can also calculate voltage drop at R2 and substract from the 7.5V source (just being careful with the signs) FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-13T13:45:09Z IndexTAG: 4443 TitleTAG: Help If anyone is still having troubles here, just check out https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Week_1/Week_1_Tutorials/ . This circuit is simpler than that. UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-09-13T13:13:32Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL/KCL example CommentableIdTAG: 6002x_S2E3_Using_KVL_KCL_and_VI_constraints NumberOfReplyTAG: 0 IndexTAG: 4444 TitleTAG: how can i calculate Rth ???? how can i get Rth??? UserIdTAG: 206794 UserNameTAG: Abdofarrag CreateTimeTAG: 2012-09-13T12:51:29Z VoteTAG: 0 CoursewareTAG: Week 2 / Thevenin model CommentableIdTAG: 6002x_S3E5_Thevenin_Model NumberOfReplyTAG: 1 FirstChildTAG: Have you watch videos S3V7 to S3V9, and/or read book section 3.6 (starting at page 157)? The method is clearly explained in both. Let me know if you have further doubts FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-13T13:36:57Z IndexTAG: 4445 TitleTAG: Help . . .!! Can't understand the Concept of Bracnch Voltage ; S1E6 KVL i don't understand how to take a branch voltage. and i think that loop and branch are same things. . ?? right or wrong UserIdTAG: 146930 UserNameTAG: fayzan_007 CreateTimeTAG: 2012-09-13T12:43:06Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL and KCL exercises CommentableIdTAG: 6002x_kvl_kcl_exercises NumberOfReplyTAG: 1 FirstChildTAG: A branch is just part of the circuit connecting two different nodes that might be at different voltages. That is why you can say there is a voltage across the branch. For example, a resistor connecting two nodes constitute a branch A loop, however, is a closed path in the circuit, meaning that it always starts and ends at the same node. Since you finish where you begin, the initial and final voltages are always the same (and that is KVL). It derives from here that a loop always contains two or more branches... can you see why? FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-13T13:54:37Z SecondChildTAG: thankz :) SecondChildUserIdTAG: 146930 SecondChildUserNameTAG: fayzan_007 SecondChildCreateTimeTAG: 2012-09-14T18:55:37Z IndexTAG: 4446 TitleTAG: V=IR rule How does the rule comes to play here? and how can we get a mobile version of this video chip to stay fresh on the tutorial UserIdTAG: 422742 UserNameTAG: lawaldebssy CreateTimeTAG: 2012-09-13T12:28:55Z VoteTAG: 0 CoursewareTAG: Week 1 / Combination Rules CommentableIdTAG: 6002x_CombinationRules NumberOfReplyTAG: 0 IndexTAG: 4447 TitleTAG: vldeo offline? how is it possible for me to download the video tutorials for offline reference? anybody have an idea??? UserIdTAG: 387026 UserNameTAG: Adefilaedward CreateTimeTAG: 2012-09-13T12:26:53Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4448 TitleTAG: systematic? How is this "Systematic way"? Looks like brute force to me...but the duality was interesting BTW, What was the brilliant way? UserIdTAG: 91706 UserNameTAG: PooyaM CreateTimeTAG: 2012-09-13T12:10:48Z VoteTAG: 0 CoursewareTAG: Week 1 / Equivalent Resistances CommentableIdTAG: 6002x_equivalent_resistances NumberOfReplyTAG: 0 IndexTAG: 4449 TitleTAG: Homework Problem. In homework one, there is a question: With three resistors in parallel, 4 ohm 4 ohm and 6 ohm. What is the maximum power rating for the composite resistor given that each resistor has a rating of 1W. My answer is 3W, why is this wrong? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-13T12:02:52Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Because 3W on the three resistors in // means 2.12V across each resistor ($U=\sqrt{P\cdot R}$) and then U^2/R=(2.12V)^2/4ohm=1.124W for the 4ohm resistors. It is higher than 1W and can not be correct. Don't forget than we do not have three equals resistors. FirstChildUserIdTAG: 396446 FirstChildUserNameTAG: RousseauxS FirstChildCreateTimeTAG: 2012-09-13T12:29:03Z IndexTAG: 4450 TitleTAG: Error in Textbook I happened to find this error in the textbook on page 72 Example 2.14 : Answer should have been 2V. UserIdTAG: 12819 UserNameTAG: srinib CreateTimeTAG: 2012-09-13T11:49:19Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: You are right, a list of errata is available here : https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/Book/ FirstChildUserIdTAG: 396446 FirstChildUserNameTAG: RousseauxS FirstChildCreateTimeTAG: 2012-09-13T13:14:39Z IndexTAG: 4451 TitleTAG: problem i want to be solved converting from speech to words have problem i want please to solve it it appears when it be on youtube UserIdTAG: 295983 UserNameTAG: qassam CreateTimeTAG: 2012-09-13T10:52:26Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4452 TitleTAG: How to enter algebric expression in H1P1 I am not getting how to write algebraic expression in the first homework set. Even through i known my ans is correct, Please provide some help. UserIdTAG: 314668 UserNameTAG: singhharsh21 CreateTimeTAG: 2012-09-13T10:34:11Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Use * for denoting product. ex 5R is incorrect while 5*R is correct. Also dont hesitate to use (). FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-09-13T10:54:51Z IndexTAG: 4453 TitleTAG: last part of Q 1 of first HW ?? Given that each individual resistor can dissipate up to 1 watt of power before burning up, how much total power in watts ( W ) can the smallest-valued composite resistor dissipate before burning up? any 1 done then kindly let me know thanks UserIdTAG: 430030 UserNameTAG: imali CreateTimeTAG: 2012-09-13T08:23:51Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 3 FirstChildTAG: I'll give you a clue : Assume that the parallel combination has 2V applied across it. 2V for the reason that Power disipated by 4ohm resistor will be 1W. For any higher value the resistor blows up. FirstChildUserIdTAG: 211715 FirstChildUserNameTAG: pitankar FirstChildCreateTimeTAG: 2012-09-13T08:44:19Z SecondChildTAG: @pitanker: that's not undertood. kindly elaborate more....the equivalent power is of the equivalent resistors...i.e.smallest-valued composite resistor not of individual resistors....?? SecondChildUserIdTAG: 107038 SecondChildUserNameTAG: Hassankhan SecondChildCreateTimeTAG: 2012-09-13T10:27:00Z FirstChildTAG: Clue! sketch the diagram. find the least of maximam voltages that can be applioed to different valued resistors, and work with it :-) FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-09-13T11:03:33Z FirstChildTAG: Hi, I can only give you a clue. How do you calculate power? what is common for resistors in parallel? I have almost given you the answer. FirstChildUserIdTAG: 330269 FirstChildUserNameTAG: Jaikr FirstChildCreateTimeTAG: 2012-09-13T08:27:04Z SecondChildTAG: Pt = P1+P2+p3 but this is wrong i suppose?????????? SecondChildUserIdTAG: 430030 SecondChildUserNameTAG: imali SecondChildCreateTimeTAG: 2012-09-13T08:31:13Z IndexTAG: 4454 TitleTAG: Problem The videos are not opening on my iPad.I opened at 13.26 India time. UserIdTAG: 99628 UserNameTAG: Pranjal16 CreateTimeTAG: 2012-09-13T07:56:49Z VoteTAG: 0 CoursewareTAG: Week 2 / Logic Gates CommentableIdTAG: 6002x_logic_gates_tutorial NumberOfReplyTAG: 1 FirstChildTAG: Ipad supports only selective formats :( FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-09-13T10:17:57Z SecondChildTAG: It was opening till yesterday,today it is not opening. SecondChildUserIdTAG: 99628 SecondChildUserNameTAG: Pranjal16 SecondChildCreateTimeTAG: 2012-09-13T11:19:56Z SecondChildTAG: ios 6 issue maybe? SecondChildUserIdTAG: 270284 SecondChildUserNameTAG: nkukushkin SecondChildCreateTimeTAG: 2012-09-20T13:34:39Z IndexTAG: 4455 TitleTAG: Doesn't V0 change? When I have a voltage divider like that, doesn't V0 change with the load applied in my receiver? UserIdTAG: 390558 UserNameTAG: ilhan89bln CreateTimeTAG: 2012-09-13T06:28:34Z VoteTAG: 0 CoursewareTAG: Week 2 / Digital Abstraction CommentableIdTAG: 6002x_digital NumberOfReplyTAG: 1 FirstChildTAG: Hi, I would like to explain using two cases- 1) Low output impedance - This causes very less current to flow to the load and hence the voltage across it will be very less different from the output without load. 2) High output impedance- The load(having Z less than output Z of system) will draw more current and hence the output voltage drops by a considerable amount from the designed value. Please go through thevenin's theorem for o/p Z calculation. FirstChildUserIdTAG: 330269 FirstChildUserNameTAG: Jaikr FirstChildCreateTimeTAG: 2012-09-13T06:53:19Z IndexTAG: 4456 TitleTAG: Page numbers for recommended reading Can you guys please write down the page numbers for the recommended readings? The way it is done is is kind of confusing me and I can't tell what is recommended reading and what is not so important. Cheers, Ilhan UserIdTAG: 390558 UserNameTAG: ilhan89bln CreateTimeTAG: 2012-09-13T06:16:22Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4457 TitleTAG: Have to take a week off but do not want to fall badly behind. What are my options? I'm full time employed, and as a part of my work duty I occasionally have to travel to very remote locations. It means no internet of any kind beside satellite one, that is just too expensive. Next week I have to leave for one of such trips. Trip is three weeks long but only for the last two weeks of trip I'll be in location with no internet. What are my options? UserIdTAG: 167413 UserNameTAG: TeTAn CreateTimeTAG: 2012-09-13T06:04:36Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 3 FirstChildTAG: The handouts are given in two week advance. so try to complete it before then. also only 10 h/w are considered :-) FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-09-13T10:19:49Z FirstChildTAG: The videos will be up indefinitely once released. Also, only 10 out of 12 homework and lab marks will count towards your grade. So you can travel for 2 weeks, watch 4 videos when you get back and as long as you don't miss any more homework you should be fine. FirstChildUserIdTAG: 151427 FirstChildUserNameTAG: RichardZ FirstChildCreateTimeTAG: 2012-09-13T23:46:03Z FirstChildTAG: That is exactly what I'm doing - trying to get ahead with homework. Thank you everybody! FirstChildUserIdTAG: 167413 FirstChildUserNameTAG: TeTAn FirstChildCreateTimeTAG: 2012-09-22T15:14:04Z IndexTAG: 4458 TitleTAG: Bug in LAB 1 While performing the lab 1 questions in circuit sandbox it showed me the required values, but when performing the same circuit in LAB1 it is showing the circuit to be incorrect. any suggestions??? UserIdTAG: 178810 UserNameTAG: sumitbawri CreateTimeTAG: 2012-09-13T04:26:07Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Are you using more than 2 resistors? My first instinct was to use 3 resistors, but it would seem the lab only accepts solutions with only 2 resistors even if the voltage conditions are still satisfied by another configuration. FirstChildUserIdTAG: 248902 FirstChildUserNameTAG: Chanute FirstChildCreateTimeTAG: 2012-09-13T04:46:35Z SecondChildTAG: no i am using 2 resistors SecondChildUserIdTAG: 178810 SecondChildUserNameTAG: sumitbawri SecondChildCreateTimeTAG: 2012-09-13T16:50:39Z IndexTAG: 4459 TitleTAG: Lab 1 Help I tried working Lab 1, but keep hitting dead ends. I'm pretty sure I need to make the second circuit like a Voltage divider, with node A on one end of a resistor, and the start of the second. However, I cannot seem to solve the first circuit. Side note, I already know the solution (I think) because I messed around with possible values for the two resistors before attempting any math. Regardless, I still want to solve the Lab "correctly". In short, I would appreciate a nudge in the right direction. EDIT: The values I thought worked ummm.... didn't. UserIdTAG: 248807 UserNameTAG: 1kingsman CreateTimeTAG: 2012-09-13T04:15:12Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 4 FirstChildTAG: any one please help me in solving lab 1 FirstChildUserIdTAG: 220425 FirstChildUserNameTAG: rishienggju FirstChildCreateTimeTAG: 2012-09-15T20:33:03Z FirstChildTAG: Both of them are supposed to have the same section that you've designed; the only difference between the first and second parts should be the presence or absence of the resistor representing the light bulb. If you set up a voltage divider, you should have two unknowns: the two resistances. The problem gives you two sets of constraints that both need to be satisfied simultaneously. Hopefully that is enough of a hint to get you on the path to solution! FirstChildUserIdTAG: 248902 FirstChildUserNameTAG: Chanute FirstChildCreateTimeTAG: 2012-09-13T04:29:22Z SecondChildTAG: Hehe, close. This much I know from the hint at the bottom of the page. After some solving and fiddling, I got R2=0.857 Ohms. While this gets the second circuit to work, the first one is labeled as wrong. So, I've got to try again. Been at it all day (9 hrs now) soo, I'm really ready to move on. SecondChildUserIdTAG: 248807 SecondChildUserNameTAG: 1kingsman SecondChildCreateTimeTAG: 2012-09-13T04:57:46Z SecondChildTAG: I'm curious - does anyone know why the post I'm replying to has a beige background? SecondChildUserIdTAG: 248902 SecondChildUserNameTAG: Chanute SecondChildCreateTimeTAG: 2012-09-14T00:18:59Z FirstChildTAG: Well, I still cant get the equation to stop having something like $ R_1 \cdot (R_2+R_{bulb}) $ If I try to cancel it out with another equation, or substitute something else, I keep getting more stuff that is awkward to replace. Hmm, should I use conductances? I'll try that. FirstChildUserIdTAG: 248807 FirstChildUserNameTAG: 1kingsman FirstChildCreateTimeTAG: 2012-09-15T01:06:45Z SecondChildTAG: I have read the previous comments and now I have the solution, and the lab done. Like Chanute said: "In the first situation you have the voltage source in series with your chosen R1, in series with the parallel combination of your chosen R2 and the bulb. In the second situation you have the voltage source in series with R1 and R2." My best recommendation to you is to stick to the >>>voltage divider equations on the R2<<< resistor. Your first equation will be a voltage on R2 = 1.5V, interacting with R1,R2 and bulb, your second equation will be a voltage on R2 = 2V, interacting only with R1 and R2. SecondChildUserIdTAG: 265027 SecondChildUserNameTAG: edumm1 SecondChildCreateTimeTAG: 2012-09-15T02:05:27Z FirstChildTAG: Victory!!! If I only replace $ R_{bulb} $ with the given, and then multiply both sides by the denominator, then I can factor out one of the resistors, and solve what's left! In short, my lab is finished 100%, and it's all thanks to your help. Well, about 3 days of headaches, but at least I know that I can do it correctly now. Going to go over my notes one more time, and hopefully, I'll not make the same mistakes next time. Thanks so much Chanute! FirstChildUserIdTAG: 248807 FirstChildUserNameTAG: 1kingsman FirstChildCreateTimeTAG: 2012-09-15T02:02:12Z SecondChildTAG: HORRAY! I'm so happy you got it! Glad to have helped =) Up-votes for you! SecondChildUserIdTAG: 248902 SecondChildUserNameTAG: Chanute SecondChildCreateTimeTAG: 2012-09-16T22:05:13Z IndexTAG: 4460 TitleTAG: PROBLEM H2P1 : what does it really mean "Come up with resistors R1 and R2 such that the divider ratio Vout/Vin is within 10% of the requirement." ???? hlp hlp UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-13T04:14:33Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: The homework gives you specific values for Vin and desired Vout. The question is asking you to take the given values of the resistors (ignoring variance) to design a circuit that produces a Vout value. That value won't be exactly the desired Vout because you are limited to your choices of resistors. The question wants you to design it so that Vout(actual)/Vin is within 10% of Vout(desired)/Vin FirstChildUserIdTAG: 248902 FirstChildUserNameTAG: Chanute FirstChildCreateTimeTAG: 2012-09-13T04:34:03Z IndexTAG: 4461 TitleTAG: Transient Analysis does not display Like many others, I cannot get the transient analysis graph to display either in the 'Circuit Sandbox' or 'Using The Tools' section. I don't think it is a browser issue as I've tried 3 different browsers on 3 different computers, turned off ad-block and pop-up blockers. I've created the circuit correctly and added the probes. I click the 'Trans' button, and enter 5ms for the stop time, but the window doesn't appear. I've tried all the suggestions from other users, different browsers, clearing the cache. I hope this can be resolved before it is required for homework. PS. I started the previous version of the course, so I do know how to use the tool from the first go around. UserIdTAG: 29468 UserNameTAG: randallroman CreateTimeTAG: 2012-09-13T03:07:47Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Above you said "I click the 'Trans' button, and enter 5ms for the stop time", I trust you entered "5m" and not "5ms" in the box. I apologize for going against the grain of your PS, but I just wanted to make sure since you didn't specifically write "5m". FirstChildUserIdTAG: 132828 FirstChildUserNameTAG: rharris FirstChildCreateTimeTAG: 2012-09-20T19:44:54Z IndexTAG: 4462 TitleTAG: lecture viewing problem - video stops after roughly 1 minute I am experiencing same problem seen by many people. I find that exact same problem occurs with: firefox 10.0.7 in windows 7 google chrome browser in windows 7 firefox 14.0.1 running in ubuntu linux 12.04 Seems to point to a problem with the lecture video server. The courseware software is elegantly designed - hope you find bug soon. UserIdTAG: 410846 UserNameTAG: tbrunner CreateTimeTAG: 2012-09-13T02:35:47Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: Please see [this thread][1] for a quick workaround. [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/MITx_6_002x_2012_Fall/threads/5051442739934a2b0000001a FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-13T02:38:05Z FirstChildTAG: I have the same problem on Mac OS X 10.8 with Safari 6.0 and Google Chrome 22. FirstChildUserIdTAG: 166420 FirstChildUserNameTAG: ftt FirstChildCreateTimeTAG: 2012-09-13T02:43:04Z IndexTAG: 4463 TitleTAG: H3P1 help please Need some help with this one, Do I deed V output to get RPuI min. nees help with that part What is the minimum value of the pullup resistor RPuI (in Ohms) for which this inverter can obey the required static discipline? UserIdTAG: 286954 UserNameTAG: ododo CreateTimeTAG: 2012-09-13T01:36:05Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Hi, The pull up resistor also plays a role during pull down operation, as in given a set of transistors in the pull down, based on ron of those you have to decide rpull up. You can assume one of them and design the other. FirstChildUserIdTAG: 330269 FirstChildUserNameTAG: Jaikr FirstChildCreateTimeTAG: 2012-09-13T04:04:16Z IndexTAG: 4464 TitleTAG: how can we edit our personal information in dashboard? While filling in my personal details, i didn't mention my country in the mailing address section. Therefore, I wanted to edit it. I tried doing it in my dashboard but couldn't figure out how to do so. UserIdTAG: 401175 UserNameTAG: Raghav12 CreateTimeTAG: 2012-09-13T00:30:42Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: I don't see any way to do it either. I don't think there is a way to edit that, at least not yet. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-09-13T01:39:07Z IndexTAG: 4465 TitleTAG: H1P3: POOR WORKMANSHIP **how to calculate individual power dissipation?** which tutorial help for that UserIdTAG: 126496 UserNameTAG: rsathish CreateTimeTAG: 2012-09-13T00:28:28Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4466 TitleTAG: By what time on September 16th is the hmwk, lab due? By what time on September 16th is the hmwk, lab due? UserIdTAG: 145676 UserNameTAG: CathyPK CreateTimeTAG: 2012-09-12T22:51:43Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: The system will accept the assignments for as long as it is still September 16th somewhere in the world. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-09-13T01:16:35Z SecondChildTAG: Just to confirm, this means they're effectively due at 10:00 GMT on September 17th? SecondChildUserIdTAG: 138690 SecondChildUserNameTAG: Alyxi SecondChildCreateTimeTAG: 2012-09-15T03:08:17Z SecondChildTAG: Or is UTC-12 being used as the baseline (12:00 GMT Sept 17th) SecondChildUserIdTAG: 138690 SecondChildUserNameTAG: Alyxi SecondChildCreateTimeTAG: 2012-09-15T03:13:01Z SecondChildTAG: means we dont have to worry about what time it is outside our house !!! SecondChildUserIdTAG: 314624 SecondChildUserNameTAG: Owais001 SecondChildCreateTimeTAG: 2012-09-15T05:17:12Z IndexTAG: 4467 TitleTAG: BUG This video suddenly stops without any reason and in order to continue watching the video I have to start it all over again because when I press play nothing happens. I have to take the purple circle of the time line back to the start, I'm getting mad with that. My system is windows 7 ultimate 64 bits (SP1), and my browser is mozilla firefox 15.0.1 HELP! UserIdTAG: 181432 UserNameTAG: enriqueferreralcala CreateTimeTAG: 2012-09-12T22:19:37Z VoteTAG: 0 CoursewareTAG: Week 1 / Nodal analysis example CommentableIdTAG: 6002x_NodalAnalysisExample NumberOfReplyTAG: 3 FirstChildTAG: YouTube made a recent change to their Flash player that broke our video embeds. As a temporary workaround while we figure things out, you can use YouTube's HTML5 player. To do so, please go to YouTube's [Join the HTML5 Trial][1] page and click "Join the HTML5 Trial" at the bottom. Then reload our site in your browser. Thank you for your patience as we work on this. FirstChildUserIdTAG: 11 FirstChildUserNameTAG: dormsbee FirstChildCreateTimeTAG: 2012-09-13T02:29:19Z FirstChildTAG: possible work around for freezing video,….. right click on the video in your browser and choose “Copy video URL”, open a new tab in your browser and paste the URL, press enter key and it should navigate you directly to the You Tube web page... you will lose the caption functionality but should be able to view the video without issues. FirstChildUserIdTAG: 13866 FirstChildUserNameTAG: Halo_55 FirstChildCreateTimeTAG: 2012-09-12T23:05:53Z FirstChildTAG: Thanks for the bug report. We are working on it now. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-09-13T01:19:09Z IndexTAG: 4468 TitleTAG: lab work tutorial please, i want somebody to tell me how to get the tutorial that explain more on the lab work. thank you UserIdTAG: 337904 UserNameTAG: bashykeny CreateTimeTAG: 2012-09-12T21:06:33Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Go in the "Overview" panel on your left (where you find "Week 1", "Week 2", etc.). FirstChildUserIdTAG: 396446 FirstChildUserNameTAG: RousseauxS FirstChildCreateTimeTAG: 2012-09-13T07:57:33Z IndexTAG: 4469 TitleTAG: Help! How am I supposed to know if i had submitted my homework and lab on time. I just completed and there still is a tiny blue reminder of the due date on the left. But my progress tab shows 100 percen in h/w and lab.. UserIdTAG: 148523 UserNameTAG: span993 CreateTimeTAG: 2012-09-12T20:58:57Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: CONGRATULATIONS FOR 100 PERCENT,DON'T WORRY YOUR HOMEWORK HAS BEEN SUBMITTED. FirstChildUserIdTAG: 40209 FirstChildUserNameTAG: sadaf78 FirstChildCreateTimeTAG: 2012-09-12T21:06:37Z SecondChildTAG: Despite your all caps, I like this message. SecondChildUserIdTAG: 79361 SecondChildUserNameTAG: kimt SecondChildCreateTimeTAG: 2012-09-12T21:17:29Z SecondChildTAG: Thanks sadf78 :-) SecondChildUserIdTAG: 148523 SecondChildUserNameTAG: span993 SecondChildCreateTimeTAG: 2012-09-13T07:43:36Z FirstChildTAG: :-) FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-09-13T07:42:40Z IndexTAG: 4470 TitleTAG: w/l in transistor? Please give explanation what is w/l in transistor? UserIdTAG: 72796 UserNameTAG: belos CreateTimeTAG: 2012-09-12T20:37:04Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: This refers to the ratio of width to length of the device. In a MOSFET biased into its active region, it is another parameter on which the drain current depends on. However for the most part when doing problems here we abstract away this parameter and a few others into the single constant "K". FirstChildUserIdTAG: 390034 FirstChildUserNameTAG: kahlil FirstChildCreateTimeTAG: 2012-09-12T20:43:00Z IndexTAG: 4471 TitleTAG: H2P1 Confussion It seems that I can't find the right answer. To get Vmax value, I put 90% value of R1 and 110% value of R2, and for the Vmin value, I put 110% value of R1 and 90% value of R2. But all the possibilities I got are always more than 110% of 24 or less then 90% value of 24. Can somebody help me, please, in understanding the requirement? Thank you UserIdTAG: 422981 UserNameTAG: anugerahf CreateTimeTAG: 2012-09-12T19:29:15Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Try to input your results anyway. I found the best possible answer from given resistors, and even though V extremes weren't within 10%, my answers were marked right. FirstChildUserIdTAG: 267993 FirstChildUserNameTAG: mavlijas FirstChildCreateTimeTAG: 2012-09-12T21:54:53Z SecondChildTAG: I think question need to be stated little bit more clear, as it said here (point h2p1): https://6002x.mitx.mit.edu/wiki/view/CourseErrors SecondChildUserIdTAG: 264920 SecondChildUserNameTAG: Whatsoon SecondChildCreateTimeTAG: 2012-09-16T23:55:14Z IndexTAG: 4472 TitleTAG: Last task Good afternoon. At me the last task didn't turn out. It turns out that the part of tension of V2 will pass to V1, at this V1 should become the consumer, but on what formula to calculate I, R and U? Thanks. UserIdTAG: 233394 UserNameTAG: ishpanec CreateTimeTAG: 2012-09-12T19:21:15Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: undefined NumberOfReplyTAG: 0 IndexTAG: 4473 TitleTAG: Software?? Does anyone know the software he'd used to simulate the characteristics? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-12T18:16:21Z VoteTAG: 0 CoursewareTAG: Week 1 / Lumped Element Abstraction CommentableIdTAG: 6002x_lumped_element_abstraction NumberOfReplyTAG: 2 FirstChildTAG: Hi, If you are talking about the part of the demonstration where Prof. Agarwal is talking about v-i characteristics, and up on the screen the variable resistor is shown to have a straight-line v-i characteristic, and when he varies the slope of the line by varying the resistance (physically by turning a potentiometer, which is a variable resistor used for such things as volume control on a radio): - that does not appear to be a **simulation** program, but a **demonstration** in lab. - it looks like he has a potentiometer hooked up to a computer-based **oscilloscope**. FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-09-14T18:20:39Z FirstChildTAG: In the lab assignment area, there is software for building circuits... HOWEVER, there is a similar (and much, much better, graphically) software phone app (unsure if an iPhone app exist - I assume it does - but I have it on my Amdroid) called, "Every Circuit"... free and pro versions available. It's excellent for making lumped circuit abstractions, seeing if they work, and figuring out KVL and KCL... you oughtta check it out! ^_^ FirstChildUserIdTAG: 340856 FirstChildUserNameTAG: RanmaSaotome FirstChildCreateTimeTAG: 2012-09-16T03:02:41Z SecondChildTAG: Sorry... my spelling abilities before coffee leaves much to be desired lol SecondChildUserIdTAG: 340856 SecondChildUserNameTAG: RanmaSaotome SecondChildCreateTimeTAG: 2012-09-16T03:03:35Z IndexTAG: 4474 TitleTAG: exactly 2.5V Based on the convention anything ABOVE 2.5V will be interpreted as "1" and anything BELOW 2.5V will be interpreted as "0", what if the sender provides EXACTLY 2.5V? Thanks! UserIdTAG: 371000 UserNameTAG: Joseph090892 CreateTimeTAG: 2012-09-12T18:12:49Z VoteTAG: 0 CoursewareTAG: Week 2 / Static Discipline CommentableIdTAG: 6002x_static_discipline NumberOfReplyTAG: 1 FirstChildTAG: Please disregard this question, the answer is provided on the next slide. Sorry! FirstChildUserIdTAG: 371000 FirstChildUserNameTAG: Joseph090892 FirstChildCreateTimeTAG: 2012-09-12T18:15:55Z IndexTAG: 4475 TitleTAG: H1P3 - Watt confusion I'm stuck on this problem. Some of the discussions make note of 3 heaters at 1530w. I'm looking at the problem and seeing 3 heaters at 830w. Another comment has them at 940w. Why the reference to different values.? I calculated answers for 1 and 2 correctly, but can't seem to get 3 4 and 5. UserIdTAG: 151255 UserNameTAG: justinmknott CreateTimeTAG: 2012-09-12T18:01:49Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: H1P3: Poor Workmanship Joe wants to heat his 12'X20' workshop with electric heat. He has hired the HACME electrician company to build the system. They propose to use three 1360.0W 240V baseboard heaters to provide a total heating capacity of 4080.0W. FirstChildUserIdTAG: 270011 FirstChildUserNameTAG: SnowmanZA FirstChildCreateTimeTAG: 2012-09-12T18:49:39Z SecondChildTAG: Joe wants to heat his 12'X20' workshop with electric heat. He has hired the HACME electrician company to build the system. They propose to use three 830.0W 240V baseboard heaters to provide a total heating capacity of 2490.0W. Is what mine says. SecondChildUserIdTAG: 151255 SecondChildUserNameTAG: justinmknott SecondChildCreateTimeTAG: 2012-09-12T19:07:55Z IndexTAG: 4476 TitleTAG: b1 why b1 must be with - ? UserIdTAG: 192114 UserNameTAG: chuba CreateTimeTAG: 2012-09-12T17:38:26Z VoteTAG: 0 CoursewareTAG: Week 2 / Linear combinations of source strenghts CommentableIdTAG: 6002x_S3E2_Linear_Combinations_of_Source_Strengths NumberOfReplyTAG: 1 FirstChildTAG: Because the direction of i1 is opposite to the direction of the current. FirstChildUserIdTAG: 213452 FirstChildUserNameTAG: JoJosida FirstChildCreateTimeTAG: 2012-09-14T07:09:04Z IndexTAG: 4477 TitleTAG: Seeing "[Math Processing Error]" in Homework section (Problem solved after reload Google Chrome) I am seeing "[Math Processing Error]" every now and then in Homework section. For example, in H1P1: "You are given three resistors: two [Math Processing Error] resistors and one [Math Processing Error] resistor. ......" Anyone seeing such problem? I am using Google Chrome in Win7. UserIdTAG: 400207 UserNameTAG: childs CreateTimeTAG: 2012-09-12T16:09:30Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 3 FirstChildTAG: Does this problem persist when refreshing, or occur only occasionally? FirstChildUserIdTAG: 321830 FirstChildUserNameTAG: kimth FirstChildCreateTimeTAG: 2012-09-12T16:41:10Z FirstChildTAG: I am running Windows 7 and IE9 and I see the same thing. FirstChildUserIdTAG: 397595 FirstChildUserNameTAG: mholin FirstChildCreateTimeTAG: 2012-09-12T18:13:10Z SecondChildTAG: i am not facing this problem i am using chrome on Window 7 SecondChildUserIdTAG: 176353 SecondChildUserNameTAG: amJunaid SecondChildCreateTimeTAG: 2012-09-16T14:15:28Z FirstChildTAG: I'm seeing the same thing today. Can't do homework because of it. FirstChildUserIdTAG: 49861 FirstChildUserNameTAG: codymartin FirstChildCreateTimeTAG: 2012-10-03T23:47:42Z IndexTAG: 4478 TitleTAG: node analysis i used node analysis but i got another results ... why ? UserIdTAG: 293070 UserNameTAG: Makary CreateTimeTAG: 2012-09-12T16:08:35Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL/KCL example CommentableIdTAG: 6002x_S2E3_Using_KVL_KCL_and_VI_constraints NumberOfReplyTAG: 2 FirstChildTAG: Are you sure you wrote down all your equations correctly? Nodal analysis is a legitimate way to approach this problem and should give exactly the same answer. FirstChildUserIdTAG: 390034 FirstChildUserNameTAG: kahlil FirstChildCreateTimeTAG: 2012-09-12T16:10:36Z SecondChildTAG: thx alot ... yea it was some calculating mistack SecondChildUserIdTAG: 293070 SecondChildUserNameTAG: Makary SecondChildCreateTimeTAG: 2012-09-12T16:15:02Z FirstChildTAG: do you mean you got different results for the branch variables or did the node voltages come out different? the node voltages depend on the ground reference chosen. if the branch variables come out different recheck your calculations. FirstChildUserIdTAG: 152516 FirstChildUserNameTAG: shuvo915 FirstChildCreateTimeTAG: 2012-09-12T16:16:21Z SecondChildTAG: thx SecondChildUserIdTAG: 293070 SecondChildUserNameTAG: Makary SecondChildCreateTimeTAG: 2012-09-12T16:22:01Z IndexTAG: 4479 TitleTAG: H3P4 - Max Current What is the maximum current (in Amperes) that can go through diode D1? **Isn't it right? (VSmax-V1)/R or (9-3)/6.8e3** Method: When the diode D1 is leading current, short-circuit occurs in D1 and the right R get no current through it. All the current passes through D1. UserIdTAG: 39980 UserNameTAG: MarinoJr CreateTimeTAG: 2012-09-12T15:14:14Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: Hi, You are right about the short circuit. But as you can see there is a voltage source in that branch. If u apply either KVL or KCL, you will get the current through the branch.As you can see, there is potential across other res which you cannot neglect.I think, I have given enough clues. FirstChildUserIdTAG: 330269 FirstChildUserNameTAG: Jaikr FirstChildCreateTimeTAG: 2012-09-12T16:57:36Z SecondChildTAG: I realized my mistake ignoring the current through the right R. I got blind when I thought that all the current of the circuit passed through D1. Actually, the voltage over right R also produces a current through it. Thanks. SecondChildUserIdTAG: 39980 SecondChildUserNameTAG: MarinoJr SecondChildCreateTimeTAG: 2012-09-12T17:22:22Z FirstChildTAG: Hi, how can i get the max negative voltage for v? I manage to get the max positive voltage for v. Please help :) FirstChildUserIdTAG: 157273 FirstChildUserNameTAG: ongchihang FirstChildCreateTimeTAG: 2012-09-19T15:11:36Z SecondChildTAG: Analyze what happens (with the circuit) when you arrive at the negative peak voltage of the source. SecondChildUserIdTAG: 138981 SecondChildUserNameTAG: Pr0bability SecondChildCreateTimeTAG: 2012-09-19T19:31:02Z IndexTAG: 4480 TitleTAG: Current direction across the voltage source The right answer is possible only if in KVL -V + I1*R1 + I2*R2 = 0. Why V<0? Current flows from plus to minus, isn't it? I think V must be whith a plus in KVL. UserIdTAG: 378976 UserNameTAG: Kubasov CreateTimeTAG: 2012-09-12T15:13:20Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL/KCL example CommentableIdTAG: 6002x_S2E3_Using_KVL_KCL_and_VI_constraints NumberOfReplyTAG: 1 FirstChildTAG: When writing KVL loop equations, when moving across a voltage source from its positive terminal to its negative terminal, you lose V volts, so the term has a negative sign. Since the voltage source is a source of power current actually flows out of its plus terminal. FirstChildUserIdTAG: 390034 FirstChildUserNameTAG: kahlil FirstChildCreateTimeTAG: 2012-09-12T15:52:48Z SecondChildTAG: OK, Thank You, I forget one important thing. We usually wrote all losses of voltage in left and all voltage sources in right in my University. Like: I1*R1 + I2*R2 = V. Thank you and sorry for my bad english, i'm from Russia SecondChildUserIdTAG: 378976 SecondChildUserNameTAG: Kubasov SecondChildCreateTimeTAG: 2012-09-12T16:59:14Z IndexTAG: 4481 TitleTAG: problem 1.4 a square wave generator is used as the source. If the square wave signal has a peak-to-peak of 20 V and a zero average value, determine the average power supplied by the source. UserIdTAG: 221617 UserNameTAG: konan CreateTimeTAG: 2012-09-12T12:10:32Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: 0.1 Watt FirstChildUserIdTAG: 299375 FirstChildUserNameTAG: VinayVashisht FirstChildCreateTimeTAG: 2012-09-12T13:35:10Z IndexTAG: 4482 TitleTAG: Don't quite understand the vertical deflection and horizontal deflection part. In the p.33 of the text book, it says that "As can be seen from the circuit, the horizontal deflection will be proportional to vD, and the vertical deflection will be proportional to vR." What is the "vertical deflection" and "horizontal deflection?" and why has it have anything to do with vR and vD? UserIdTAG: 98259 UserNameTAG: rogerloh0 CreateTimeTAG: 2012-09-12T11:59:09Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: I agree that the terminology is quite confusing. I believe that "deflection" is referring as how much an axis changes in the oscilloscope. So the "horizontal deflection" or change in the x-axis is proportional to vD and the "vertical deflection" or change in the y-axis is proportional to vR (or iD). So that by the plot generated in the oscilloscope we get the v-i characteristics of the two terminal device. FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-12T13:17:53Z IndexTAG: 4483 TitleTAG: Mistake? I think there is a mistake in S2V4: Method 1 - KVL, KCL Method @ 2:40 "I5 leaving the node minus ***I4 leaving the node*** ". Clearly I4 is entering the node, and by the convention outlined, a current entering the node is negative. so how can I4 be negative and leaving the node? UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-09-12T11:23:17Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL and KCL CommentableIdTAG: 6002x_kvl_and_kcl NumberOfReplyTAG: 2 FirstChildTAG: He did make a mistake in saying i4 leaving the node. You are correct in saying that it was entering. i4 is not negative and leaving the node; it is negative because it is entering the node. Personally I like the convention of currents_in = currents_out which doesn't worry about signs as long as you put similarly directed currents on the same side (something I find easy.) FirstChildUserIdTAG: 164898 FirstChildUserNameTAG: jbparkes FirstChildCreateTimeTAG: 2012-09-12T13:31:26Z FirstChildTAG: Hello. First, this is long, for which I now apologize, but know of no way around this and still give one with enough information and addressing issues now that will save some time and effort later. I am having big problems with sign convention, especially with reference to any and all circuits. Have looked at a variety of sites, none help. I know there's a difference be able to give and element's properties and this conventional / associated variables discipline a The other difficulty is that one of the videos dealing with what I just mentioned wont play after about 1 minute, for which I made a bug report. I have found annotated slides, but you still lose something. I did not know signs could be determined by whether it is going into or out of a node as I thought that it would make good, logic-based sense in one situation and be irrational in the next. If it is true that the signs can be chosen for nodes based upon their currents entering or leaving a node, please tell me. To make certain we are on the same page, here's the link to the video I found, which is lecture 2(Spring 2007) of Electronics course 6.002: http://www.academicearth.org/lectures/basic-circuit-analysis-method-kvl-and-kcl-mmethod It would help if someone could give the equations for node c, and for d since the professor did not give them. And, please tell me how you determined if the currents were going into or leaving those particular nodes. Thanks FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-13T08:15:23Z IndexTAG: 4484 TitleTAG: Remark I've Very small remark - there is no dimension (it's obvious ampere) for current I. I=3 **A** UserIdTAG: 345464 UserNameTAG: Constantine_ru CreateTimeTAG: 2012-09-12T10:55:28Z VoteTAG: 0 CoursewareTAG: Week 1 / Node method example CommentableIdTAG: 6002x_S2E5_Node_Method NumberOfReplyTAG: 0 IndexTAG: 4485 TitleTAG: number of nodes how many nodes are there 3 or 4....coz i think it shud b 4....bt anant sir talks abt only 3 ..... UserIdTAG: 343262 UserNameTAG: rgdixitt CreateTimeTAG: 2012-09-12T10:50:40Z VoteTAG: 0 CoursewareTAG: Week 2 / Linearity CommentableIdTAG: 6002x_linearity NumberOfReplyTAG: 1 FirstChildTAG: sorry sorry.....it's only 3....my bad FirstChildUserIdTAG: 343262 FirstChildUserNameTAG: rgdixitt FirstChildCreateTimeTAG: 2012-09-12T10:51:48Z IndexTAG: 4486 TitleTAG: Problem: I can't Reset LAB2 Hi I can't Check or Reset LAB2 I delete the V sources and the output by wrong, and now, I can't Check or Reset the LAB I used Chrome and IE UserIdTAG: 171946 UserNameTAG: abohammed CreateTimeTAG: 2012-09-12T10:44:54Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Hi, Is the Reset/Check button completely gone? Can you tell me a bit more about why you can't check or reset the lab? In Chrome, can you try logging out of the website, clearing the cache, and logging back in? Lab 2 should have reset button enabled. FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-12T11:26:13Z SecondChildTAG: only check button and save button are appeared when i press save it works, but when i press check nothing happened i will try clean the cashe SecondChildUserIdTAG: 171946 SecondChildUserNameTAG: abohammed SecondChildCreateTimeTAG: 2012-09-12T11:29:04Z IndexTAG: 4487 TitleTAG: H1P2 in H1P2 the question will be .. How many loops are there in the circuit? ans why don't be two.please tell me any one........... UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-12T09:32:07Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: One big loop plus two inner loops. FirstChildUserIdTAG: 128561 FirstChildUserNameTAG: gusevoy FirstChildCreateTimeTAG: 2012-09-12T09:56:22Z IndexTAG: 4488 TitleTAG: H1P3 #3,4 and 5 can someone help me how to calculate the resistance of each component? i got hard times on solving it without any given value of the resistor.., UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-12T08:55:00Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: With the given power and voltages, you can calculate resistance by using P=V^2/R. FirstChildUserIdTAG: 201414 FirstChildUserNameTAG: ArcherJeffrey FirstChildCreateTimeTAG: 2012-09-12T09:32:23Z SecondChildTAG: yeah given Pt=4020w and Vsource=240 i can compute for Rtotal using that equation., but how about the individual resistances of each resistor? SecondChildUserIdTAG: undefined SecondChildUserNameTAG: undefined SecondChildCreateTimeTAG: 2012-09-12T09:54:50Z FirstChildTAG: you can solve it by applying nodal or mesh analysis.or by star to delta transformation & vice-versa FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-12T10:28:27Z IndexTAG: 4489 TitleTAG: MEASUREMENT OF CURRENT!!!!!!!!!!!!! how can i measure the current??? i mean by this tool or by theory ??????? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-12T08:29:49Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: you measure it by clicking on DC. It will show you the current. FirstChildUserIdTAG: 170371 FirstChildUserNameTAG: Ziad_Yunes FirstChildCreateTimeTAG: 2012-09-12T09:09:35Z IndexTAG: 4490 TitleTAG: Web Access Hi all. I am having issues with submitting the homework and seeing the lab. What internet software should i use? Currently, am accessing the site using microsoft internet explorer. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-12T08:16:50Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: I suggest to try Chrome or Firefox FirstChildUserIdTAG: 143593 FirstChildUserNameTAG: Tsybulkin FirstChildCreateTimeTAG: 2012-09-12T09:21:29Z SecondChildTAG: Thank you guys. I will try it SecondChildUserIdTAG: 313509 SecondChildUserNameTAG: Manu18 SecondChildCreateTimeTAG: 2012-09-16T08:13:51Z FirstChildTAG: I found the IE doesn't work well either. I switched to Firefox and I have no problems now. FirstChildUserIdTAG: 270011 FirstChildUserNameTAG: SnowmanZA FirstChildCreateTimeTAG: 2012-09-12T18:58:54Z IndexTAG: 4491 TitleTAG: Energy Conservation Hi, In text book Page 72,Example 2.14. Isn't the voltage across the resistor v=2V?and not 0.5V. If we use the ohms law formula v=ir for the same problem i = 2mA r = 1k ohm so, v = 2m * 1k v = 2V I think the math in the text is wrong for that particular problem 0.002*v = 0.001*v^2 0.002 = 0.001*v cancelling v on both the sides therefore,v = 2V Please correct me if I am wrong. UserIdTAG: 11538 UserNameTAG: trishul CreateTimeTAG: 2012-09-12T07:13:21Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4492 TitleTAG: sorta right sorta wrong. i started the expression with I1- but the answer ends the expression with -I1. it looked the same to me but just in case, is there a reason that its wrong? UserIdTAG: 345233 UserNameTAG: erbbie CreateTimeTAG: 2012-09-12T07:08:56Z VoteTAG: 0 CoursewareTAG: Week 1 / Node analysis practice, part 3 CommentableIdTAG: 6002x_L2Node2 NumberOfReplyTAG: 1 FirstChildTAG: The reason is that the question asks for the expression that represents the sum of the currents *leaving* the node. Using positive I1 represents the sum of the currents *entering* the node. FirstChildUserIdTAG: 248902 FirstChildUserNameTAG: Chanute FirstChildCreateTimeTAG: 2012-09-12T08:02:14Z IndexTAG: 4493 TitleTAG: H1P2 I was able to work this problem out using the node analysis. My problem is that I got the - and + all over the place. Can someone explain this problem to me? (Change the numbers please or whatever necessary) My problem is really trying to figure out the - vs + thanks! UserIdTAG: 230787 UserNameTAG: Gabriel007 CreateTimeTAG: 2012-09-12T05:03:29Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4494 TitleTAG: Lab 2-got the right form, but not the right values.... So, what I did on lab 2 was use a basic voltage divider for each one-two 1 ohm dividers for the first, a 5 and a 1 for the second. I put the first one on top or below the beginning of the second and grounded it there. My result was I got something in the form of diagram 4.1, but not the right values. However, I did the voltage equation correctly, or so I thought.... Help? I don't believe I'm understanding this as I'm supposed to. UserIdTAG: 344331 UserNameTAG: intellectualwanderer CreateTimeTAG: 2012-09-12T04:51:19Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Hi, You should understand that output voltage of second source will not only be affected by the resistors connected to it, but also by the resistors connected to the first source. Please go through superposition theorem once again. You will be able to identify the problem by yourself, once you understand that. FirstChildUserIdTAG: 330269 FirstChildUserNameTAG: Jaikr FirstChildCreateTimeTAG: 2012-09-12T05:25:13Z IndexTAG: 4495 TitleTAG: Q2 difficulty someone knows how to work out Q2 like our instructor would have loved us to? there must be something important for us to understand by taking integral of instantaneous power over one cycle. my questions are: 1. what is the period of the wave represented (from there we'll get the limits of the integration)? 2. what is instantaneous power? UserIdTAG: 213158 UserNameTAG: xl CreateTimeTAG: 2012-09-12T04:37:38Z VoteTAG: 0 CoursewareTAG: Week 1 / AC power CommentableIdTAG: 6002x_S1E3_AC_power NumberOfReplyTAG: 3 FirstChildTAG: The period `p` is `1/f`, where `f` is the frequency (60 HZ). Using the Wikipedia [definition][1] of RMS, you can find `V_rms` over the interval `0` <= `t` <= `0.1`, which is also known as "the average voltage." This can in turn be used to calculate the average power. [Here][2]'s the calculation on Wolfram. Since `V_rms = 120`, the power, `P`, is `V_rms^2/R = 120^2/110 = 130.9`. [1]: http://en.wikipedia.org/wiki/Root_mean_square#Definition [2]: http://www.wolframalpha.com/input/?i=sqrt%2810*integrate%20%28120*sqrt%282%29*cos%28120*pi*x%29%29%5E2%20from%200%20to%200.1%29 FirstChildUserIdTAG: 194509 FirstChildUserNameTAG: blackcompe FirstChildCreateTimeTAG: 2012-09-12T05:30:02Z FirstChildTAG: I think we just need to understand that the RMS voltage * RMS current is the same as taking the integral of the instantaneous power over one cycle. Instantaneous power is exactly that, it is the power that is flowing at a particular instant in time, which in AC circuits is very different from the average power since the power is changing all the time. FirstChildUserIdTAG: 270011 FirstChildUserNameTAG: SnowmanZA FirstChildCreateTimeTAG: 2012-09-12T19:07:44Z FirstChildTAG: Answer to question number 1 Time Period = 0.015sec Answer to question number 2 instantaneous power is already given in the question = 120⋅√2⋅cos(2π⋅60⋅t) Volts FirstChildUserIdTAG: 118462 FirstChildUserNameTAG: Naino FirstChildCreateTimeTAG: 2012-09-13T06:55:14Z IndexTAG: 4496 TitleTAG: please help me out hey there ,i have done my homework and also checked it ,my answers are right does it mean that i have submitted my homework ? UserIdTAG: 230108 UserNameTAG: u11ee044 CreateTimeTAG: 2012-09-12T03:43:54Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: Yup sir. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-12T03:53:48Z FirstChildTAG: Yes. Check out the 'Progress' tab on the top-level navigation. If you have the vertical bars, then you are good! FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-12T03:54:03Z IndexTAG: 4497 TitleTAG: where I can purchase the book where I can purchase the book? thanks UserIdTAG: 45303 UserNameTAG: brenda CreateTimeTAG: 2012-09-12T03:43:04Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: http://www.amazon.com/Anant-Agarwal/e/B001K8QU2S FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-12T03:54:33Z IndexTAG: 4498 TitleTAG: BUG: Lab 1 Sandbox I'm using the most recent version of Firefox. None of the components are showing up except for the button for resistors. UserIdTAG: 141376 UserNameTAG: krtica CreateTimeTAG: 2012-09-12T02:49:05Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 3 FirstChildTAG: Firefox bug fix for the circuit simulator is coming up shortly. In the meanwhile, please use Google Chrome :) FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-12T03:08:35Z SecondChildTAG: I am using Firefox. Added a resistor circuit that I thought was correct, and when I clicked on "Check", they were incorrect. When I clicked on "Reset", the two circuits appeared properly. SecondChildUserIdTAG: 334855 SecondChildUserNameTAG: JimMonty SecondChildCreateTimeTAG: 2012-09-12T03:53:34Z FirstChildTAG: This is happening with me too. I tried internet explorer and firefox both. I cann't draw cirrcuit. Sandbox is showing me only resistor. can any one help in this regard. FirstChildUserIdTAG: 197641 FirstChildUserNameTAG: mtanju FirstChildCreateTimeTAG: 2012-09-12T03:11:06Z SecondChildTAG: Can you let me know what happens in Chrome? SecondChildUserIdTAG: 79361 SecondChildUserNameTAG: kimt SecondChildCreateTimeTAG: 2012-09-12T03:19:57Z FirstChildTAG: I don't see this schematic That you have sent me. I see all blank page.I could solve my problem if I can see the schematic u sent. I know that except voltage source and ground I just ve to add resistor. FirstChildUserIdTAG: 197641 FirstChildUserNameTAG: mtanju FirstChildCreateTimeTAG: 2012-09-12T04:17:04Z IndexTAG: 4499 TitleTAG: Sign convention(al) foul up? Sign-convention(al) foul-up? Hi, there. I thought I had this sign-convention thing figured out. Thinking that, according to the + going into the object made the sign convention for that object positive? Apparently I got that mixed-up somehow...don't know is I can blame this on my dyslexia or not... Where did, and how did I go wrong? Hey, if you're a TA out there I would greatly appreciate your help and insight into this matter. Thanks UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-12T02:35:37Z VoteTAG: 0 CoursewareTAG: Week 1 / Power CommentableIdTAG: 6002x_S1E2_Power NumberOfReplyTAG: 1 FirstChildTAG: Here's my logic. When you assign branch variables to the element, you assign `i_0` going into the positive terminal (->). When you finally compute `i_0`, you get 1.17 A, but the **actual** current is going into the network (<-) as shown in the model, so `i_0` becomes negative. It's quite confusing at first, but if you work through enough examples in the text, you'll catch on to what's happening with the conventions and signs. FirstChildUserIdTAG: 194509 FirstChildUserNameTAG: blackcompe FirstChildCreateTimeTAG: 2012-09-12T04:55:28Z IndexTAG: 4500 TitleTAG: H3P1 What is the low noise margin (in Volts)? I do not understand when referred to as noise or calculated, Can someone help UserIdTAG: 58618 UserNameTAG: ingeniero13 CreateTimeTAG: 2012-09-12T00:08:56Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: The "noise margins" refer to the boundaries we set with the Static Discipline. There is a week 2 lecture sequence that covers this exact topic and discusses the concepts of low and high noise margins if you need a quick refresher FirstChildUserIdTAG: 390034 FirstChildUserNameTAG: kahlil FirstChildCreateTimeTAG: 2012-09-12T00:26:31Z SecondChildTAG: ok thank, My problem was solved SecondChildUserIdTAG: 58618 SecondChildUserNameTAG: ingeniero13 SecondChildCreateTimeTAG: 2012-09-12T01:00:15Z IndexTAG: 4501 TitleTAG: How do we calculate the current in the last question? Please tell me the calculations required to be done to get the answer of the last question. UserIdTAG: 317445 UserNameTAG: arghya33 CreateTimeTAG: 2012-09-11T22:48:59Z VoteTAG: 0 CoursewareTAG: Week 1 / Battery model CommentableIdTAG: 6002x_S1E9_Battery_Model NumberOfReplyTAG: 2 FirstChildTAG: We want to find the current flowing through the two resistors as the V2 source charges the V1 source. This is found by ohms law- I = V/R = (V2-V1)/(R1 + R2) FirstChildUserIdTAG: 390034 FirstChildUserNameTAG: kahlil FirstChildCreateTimeTAG: 2012-09-12T01:06:32Z SecondChildTAG: But why do we do R1+R2 though they are in parallel & not in series? SecondChildUserIdTAG: 317445 SecondChildUserNameTAG: arghya33 SecondChildCreateTimeTAG: 2012-09-12T12:14:32Z FirstChildTAG: use KVL in loop simply.-V2+IR2+IR1+v1=0 FirstChildUserIdTAG: 73115 FirstChildUserNameTAG: jamsindhi FirstChildCreateTimeTAG: 2012-09-13T21:18:14Z IndexTAG: 4502 TitleTAG: incomplete video there are a problem with the video, stops at 4:19. i can't watch the rest. UserIdTAG: 149549 UserNameTAG: Java_Dido CreateTimeTAG: 2012-09-11T21:44:17Z VoteTAG: 0 CoursewareTAG: Week 1 / Node analysis CommentableIdTAG: 6002x_node_analysis NumberOfReplyTAG: 3 FirstChildTAG: It happened to me too, the only way is going to YouTube and listening the rest... FirstChildUserIdTAG: 265133 FirstChildUserNameTAG: Elienai FirstChildCreateTimeTAG: 2012-09-12T21:18:36Z FirstChildTAG: im having a huge problem with ALL the videos it will only let me watch them for a few seconds then just stops... will not restart at all.. i refresh the page... and again the video will only play for a few seconds then just stop... im at a loss for what to do about this FirstChildUserIdTAG: 309318 FirstChildUserNameTAG: angel220 FirstChildCreateTimeTAG: 2012-09-12T21:38:27Z FirstChildTAG: For me it stops at 3:54 :/ #edit: didn't notice that the video continues later FirstChildUserIdTAG: 119090 FirstChildUserNameTAG: Bezi FirstChildCreateTimeTAG: 2012-09-12T22:38:45Z IndexTAG: 4503 TitleTAG: What is the logic used to decide what the +- signs are for the last video? In the video, he says " I guess the signs are...." Is this really a guess or is there some logic to it? This is causing me to get problems wrong on a continual basis. UserIdTAG: 29275 UserNameTAG: Mbarsalou CreateTimeTAG: 2012-09-11T21:29:51Z VoteTAG: 0 CoursewareTAG: Week 2 / Superposition CommentableIdTAG: 6002x_superposition_tutorial NumberOfReplyTAG: 1 FirstChildTAG: u can think in this way.Voltage is defined as energy transferred per unit charge that flows through the element,here signs have no significance they just mean opp direction.If current is passing in one direction where battery is giving energy to charge,then if its direction gets reversed the energy is transferred from electron to the battery FirstChildUserIdTAG: 397298 FirstChildUserNameTAG: ZA90 FirstChildCreateTimeTAG: 2012-09-11T23:39:08Z SecondChildTAG: This makes sense. However, when looking at some of the video's, the choice of plus or minus sign doesn't seem to follow any sort of particular method. As if they are randomly deciding which direction the current goes. SecondChildUserIdTAG: 29275 SecondChildUserNameTAG: Mbarsalou SecondChildCreateTimeTAG: 2012-09-12T23:22:26Z IndexTAG: 4504 TitleTAG: Problems with approximations I had some problems with the approximations....4*1.4^2=7.84, but 4*1.44^2=8.29, and finally 4*1.444^2=8.34. I realized what happened only when i clicked on "Show Answer" UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-11T20:58:17Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL/KCL example CommentableIdTAG: 6002x_S2E3_Using_KVL_KCL_and_VI_constraints NumberOfReplyTAG: 1 FirstChildTAG: It looks like the system accepts values with 5% tolerance. The problems will disappear, if all calculations are done with 1% error (1.44 for this quiz). Or you can place the math expression instead of the number, for example, (1 + 1) ^ 2 - sqrt(18/2) instead of 1 . FirstChildUserIdTAG: 261503 FirstChildUserNameTAG: EugenyL FirstChildCreateTimeTAG: 2012-09-11T22:50:48Z SecondChildTAG: A little bit more about 1% suggestion. The **absolute error**: eps = (1.4 - 1.4444) is divided by a **value** (1.4444), and we get the **percent error**: delta = -0.03 * 100% = -3%. It is less than 5%. However, we input not the current itself, but the calculated power (P = R * I * I). We collect errors from each factor: delta(P) = delta(R) + delta(I) + delta(I) = 0 + (-3%) + (-3%) = -6%. The error is doubled (due the square), and become worse than 5%. But if we take 1% values, they may be multiplied at least 5 times in order to reach the 5%-limit. I think, it is enough for all our tasks =) ---------- And a simple test for the system to verify the 5% of the input tolerance . i_real = 1.(4) ; I is input value; delta = (i_real - I) / i_real. Native values: | I | delta | 2*delta | result | | 1.4 | 3% | 6% | reject | | 1.44 | 0.3% | 0.6% | accept | Critical values: | I | delta | 2*delta | result | | 1.4078 | 2.54 | 5.07% | reject | | 1.4080 | 2.5% | 5% | accept | As it was expected :) ---------- Truly speaking, the value | I | delta | 2*delta | result | | 1.4079 | 2.53% | 5.06% | accept | is also accepted, because my error-computation also have a little error inside =D And directly computed error is 4.995% instead of 5.06%, so even here all is OK. SecondChildUserIdTAG: 261503 SecondChildUserNameTAG: EugenyL SecondChildCreateTimeTAG: 2012-09-11T23:54:36Z IndexTAG: 4505 TitleTAG: T - period value Integration part and RMS equitation is pretty clear. The only thing I don't understand where did you take that T value is 1/60 from? UserIdTAG: 395639 UserNameTAG: artem8889 CreateTimeTAG: 2012-09-11T20:44:01Z VoteTAG: 0 CoursewareTAG: Week 1 / AC power CommentableIdTAG: 6002x_S1E3_AC_power NumberOfReplyTAG: 2 FirstChildTAG: It's because the frequency of the sine is $f=60$ Hz. The period is given by $T=1/f=1/60$ seconds. FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-11T20:54:53Z FirstChildTAG: Hi,there(again) though I clicked to submit the following, but it did not seem to register. Look at what the graph is telling you. It shows there are 6 cycles in 0.10 seconds. So, there are ten x 0.10 intervals in one second. Meaning there are 60 cycles in one second (6x0.10)x10. Hope that helps. Now to my own problem(s) -me in nowhere...or very near it...folks. FirstChildUserIdTAG: 119440 FirstChildUserNameTAG: WG FirstChildCreateTimeTAG: 2012-09-12T03:23:35Z IndexTAG: 4506 TitleTAG: iv characteristicks.. there is any difference if we call iv characteristicks as vi characteristicks???? UserIdTAG: 80970 UserNameTAG: cruse CreateTimeTAG: 2012-09-11T20:30:20Z VoteTAG: 0 CoursewareTAG: Week 1 / Lumped Element Abstraction CommentableIdTAG: 6002x_lumped_element_abstraction NumberOfReplyTAG: 3 FirstChildTAG: Nope. FirstChildUserIdTAG: 321830 FirstChildUserNameTAG: kimth FirstChildCreateTimeTAG: 2012-09-11T21:03:42Z FirstChildTAG: No, as there is no difference between (3)(2) or (2)(3) FirstChildUserIdTAG: 433312 FirstChildUserNameTAG: FreddyRE FirstChildCreateTimeTAG: 2012-09-13T21:29:53Z FirstChildTAG: As **kimth** and **FreddyRE** already said, you can say "i-v" or "v-i" characteristics; they are equally valid. There is a *convention*, though: If you look at Prof. Agarwal's v-i graphs, voltage **(v)** is almost always plotted along the **x-axis** (horizontal axis), and is usually the independent variable (e.g. the input) and current **(i)** is almost always plotted along the **y-axis** (vertical axis), and is usually the dependent variable (e.g. the output). Hence v=input, i=output; *there must be an input to get an output*, so conventionally it is v-i characteristics. FirstChildUserIdTAG: 292546 FirstChildUserNameTAG: JerseyMark FirstChildCreateTimeTAG: 2012-09-14T18:31:58Z IndexTAG: 4507 TitleTAG: Nodes Why points with minus 6 and zero volts are called "nodes"? As I know "node" is point with 3 or more wires. UserIdTAG: 164611 UserNameTAG: IgorStrelnikov CreateTimeTAG: 2012-09-11T20:00:35Z VoteTAG: 0 CoursewareTAG: Week 1 / Nodal analysis example CommentableIdTAG: 6002x_NodalAnalysisExample NumberOfReplyTAG: 1 FirstChildTAG: When doing KCL equations, for all practical purposes your assumed definition of a node is correct, because only those such points will give you a non-useless equation. But a node can also be thought of as the connection point between any two or more devices, not just three. You might end up a few extra redundant "i1 = i2" equations but there's nothing wrong with that. FirstChildUserIdTAG: 390034 FirstChildUserNameTAG: kahlil FirstChildCreateTimeTAG: 2012-09-12T00:34:48Z IndexTAG: 4508 TitleTAG: H1p3 I need help in h1P3 m not getting the problem clear is there any one to explain problem and tell which formula use for this problem. thanks UserIdTAG: 197422 UserNameTAG: sheryar CreateTimeTAG: 2012-09-11T19:03:12Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Use the total watts/voltage to get total amps, then use volts/amps to get total resistance from that you can get the individual resistance values by using the formula resistance parallel = resistor value/ number of resistors in parallel FirstChildUserIdTAG: 73809 FirstChildUserNameTAG: rhyssouth FirstChildCreateTimeTAG: 2012-09-11T19:35:01Z SecondChildTAG: total resistance calculated and using that current calculation for question 1&2 goes wrong help in this SecondChildUserIdTAG: undefined SecondChildUserNameTAG: undefined SecondChildCreateTimeTAG: 2012-09-14T05:25:44Z IndexTAG: 4509 TitleTAG: Use scientific notation? The first answer of the third point, says that the units is Amps, so I wrote the answer with its zeros, and the second one admit all zeros, I have to write the answer in scientific notation or in normal mode? UserIdTAG: 26082 UserNameTAG: aiaa CreateTimeTAG: 2012-09-11T18:14:35Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: you can write in both scientific or normal mode, 500e-6 or 0.0005, but should be in amps FirstChildUserIdTAG: 133460 FirstChildUserNameTAG: urgossip FirstChildCreateTimeTAG: 2012-09-12T00:16:18Z SecondChildTAG: you can also write : "500u". It is also acceptable. The system they use to check is so cool. SecondChildUserIdTAG: 418182 SecondChildUserNameTAG: euldji2005 SecondChildCreateTimeTAG: 2012-09-12T02:43:13Z IndexTAG: 4510 TitleTAG: Buffering Problem I am facing the problem that video buffers upto some seconds and then stops.On resuming it starts from beginning. UserIdTAG: 370278 UserNameTAG: HassanTouqeer CreateTimeTAG: 2012-09-11T17:45:35Z VoteTAG: 0 CoursewareTAG: Week 1 / Lumped Element Abstraction CommentableIdTAG: 6002x_lumped_element_abstraction NumberOfReplyTAG: 1 FirstChildTAG: Try another browser. I had luck with Opera, but not Firefox. FirstChildUserIdTAG: 129978 FirstChildUserNameTAG: StuRat FirstChildCreateTimeTAG: 2012-09-11T17:51:23Z SecondChildTAG: I try Chrome and Internet explorer and I have the same problem SecondChildUserIdTAG: 250179 SecondChildUserNameTAG: diegoCmC SecondChildCreateTimeTAG: 2012-09-12T23:20:31Z IndexTAG: 4511 TitleTAG: In the real word how do you make del q/del t=0 ? del phi/del t =0? while i understand that this simplifies the calculations, how is this achieved with real world elements? UserIdTAG: 370093 UserNameTAG: DNA03 CreateTimeTAG: 2012-09-11T16:37:35Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: With real world elements, in most cases the approximation still works. So the calculations will agree with the real world. In the real world the $\frac{\partial q}{\partial t}$ and $\frac{\partial \phi}{\partial t}$ are very very tiny so they are equal to zero for all practical purposes. However there are limits to these calculations. In week 1's demo, you can see what happens when Lorenzo increases the voltage across the resistor beyond the limit. You can also see what happens when you do the same with a pickle! :-) The point is that there are limits beyond which our approximations won't work. These limits are reasonable in most cases. Wherever they aren't reasonable, we need to do things the hard way (i.e. Maxwell's equations, maybe even Quantum mechanics). FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-09-11T17:01:17Z SecondChildTAG: Thanks! SecondChildUserIdTAG: 370093 SecondChildUserNameTAG: DNA03 SecondChildCreateTimeTAG: 2012-09-11T18:26:46Z IndexTAG: 4512 TitleTAG: minus I1 Why the I1 is minus when the current goes to e2? S2V9: Method 3 - Node Analysis. Min. 0:43 UserIdTAG: 166869 UserNameTAG: Vasco CreateTimeTAG: 2012-09-11T16:36:33Z VoteTAG: 0 CoursewareTAG: Week 1 / Node analysis CommentableIdTAG: 6002x_node_analysis NumberOfReplyTAG: 2 FirstChildTAG: Because Dr. Agarwal choose that "+" is for current that exit the node. I is going to the node, so I is negative. FirstChildUserIdTAG: 300169 FirstChildUserNameTAG: castrogfx FirstChildCreateTimeTAG: 2012-09-11T17:17:08Z SecondChildTAG: Thanks castrogfx SecondChildUserIdTAG: 166869 SecondChildUserNameTAG: Vasco SecondChildCreateTimeTAG: 2012-09-13T00:45:54Z FirstChildTAG: I am from Europe and in Europe we have a little different way of doing things - that is why I don't uderstand something. When we are analyzing the e1 node, we say that current flows from e1 to e2. Than when analyzing e2 node we suppose it is going from e2 to e1. I don't get it - current can't go from both directions. FirstChildUserIdTAG: 405494 FirstChildUserNameTAG: holycow FirstChildCreateTimeTAG: 2012-09-27T21:31:26Z IndexTAG: 4513 TitleTAG: homework hey guys how does homework and lab work? i know the exercises you could submit as many times as you want and get them wrong and whatever, but get them right at the end. does homework give you multiple chances and stuff just like the exercises? thanks! UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-11T16:05:33Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Yes. You can try getting the answers right with any number of attempts in the homework and labs as well. This will not be the case in the exams however so I would suggest not getting used to it. FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-09-11T16:16:16Z IndexTAG: 4514 TitleTAG: KVL associations > *In this case, I hit the plus sign. So it is plus v1.* I think it's not really a proper way to define a rule for KVL equations so people don't get lost in the signs and don't do mistakes. Because those are just the associations: plus terminal means plus and there's no other meaning. The smarter way is, **plus** means **higher potential** and **minus** - **lower potential**, so when I go from "**-**" to "**+**" I go *"up"* the hill, the potential grows so the voltage change is *positive*. If i go from "**+**" to "**-**" I "*fall down*" so the voltage drops => *minus* sign. It's just the same only it has a meaning, not plain associations :) UserIdTAG: 410033 UserNameTAG: sagitta CreateTimeTAG: 2012-09-11T15:30:34Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL and KCL CommentableIdTAG: 6002x_kvl_and_kcl NumberOfReplyTAG: 2 FirstChildTAG: That's what I use as well! Good to know that other people think that this way of thinking is logical :) FirstChildUserIdTAG: 416659 FirstChildUserNameTAG: Jonas3000 FirstChildCreateTimeTAG: 2012-09-13T09:30:03Z FirstChildTAG: that is a good idea,but you can still think in this way.Voltage is defined as energy transferred per unit charge that flows through the element,here signs have no significance they just mean opp direction.If current is passing in one direction where battery is giving energy to charge,then if its direction gets reversed the energy is transferred from electron to the battery FirstChildUserIdTAG: 397298 FirstChildUserNameTAG: ZA90 FirstChildCreateTimeTAG: 2012-09-11T16:02:59Z IndexTAG: 4515 TitleTAG: h1 p2 how will i calculate power desipated in voltage sourse? is it=(voltage throug sourse*current)? i hve calculated iike this, v value is alrdy given,and i got the current through it,that is i4 which s the answer of other subquestion,but when i am multiplying both its giving a wrong answer.. UserIdTAG: 171170 UserNameTAG: rinutituschakkattil CreateTimeTAG: 2012-09-11T15:20:35Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: mention the problem clearly FirstChildUserIdTAG: 397298 FirstChildUserNameTAG: ZA90 FirstChildCreateTimeTAG: 2012-09-11T16:14:49Z FirstChildTAG: the power in any source is multiplying its current and voltage and see what is the direction of the current always the supply gives power didn't desipate power and its signal is minus FirstChildUserIdTAG: 282351 FirstChildUserNameTAG: HalfDead FirstChildCreateTimeTAG: 2012-09-11T16:00:23Z SecondChildTAG: THANK YOU. SecondChildUserIdTAG: 171170 SecondChildUserNameTAG: rinutituschakkattil SecondChildCreateTimeTAG: 2012-09-12T05:58:24Z IndexTAG: 4516 TitleTAG: BUG or a MISS ! **I dont know if we can call it a bug but there is certainly something which should not be there. When we write answers to questions and check, we may get some answers right and some wrong but the miss here is that because of the "show answer" option, we are immediately provided answers to our wrong questions without letting us think a little more. For example, I didnt put a minus sign in an answer and the show answer option didnt give me a chance to get the right answer and I ,sort of,copy pasted the answer! ** UserIdTAG: 314624 UserNameTAG: Owais001 CreateTimeTAG: 2012-09-11T14:30:33Z VoteTAG: 0 CoursewareTAG: Week 1 / Power CommentableIdTAG: 6002x_S1E2_Power NumberOfReplyTAG: 2 FirstChildTAG: The **Show Answer** option only works for Exercises. These questions are not graded so it doesn't matter if you get it right or not for a grade. The purpose of the exercises is to immediately check if you've understood the lecture sequence up to that point so it doesn't count towards your score. However, I suggest using the **Show Answer** option only when you are unable to figure out what is wrong after a good number of attempts. You still have the **Check** button which tells if you are right or wrong without telling you the correct answer. Use that. For the graded assignments/labs you won't see a **Show Answer** button till the deadline. FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-09-11T14:45:36Z SecondChildTAG: Thanks! but I think I got a 2/2 for it when I saw in the Progress. Thanks anyway! By the way, tell me whether the votes for a comment are going to affect us in any way SecondChildUserIdTAG: 314624 SecondChildUserNameTAG: Owais001 SecondChildCreateTimeTAG: 2012-09-11T17:53:45Z SecondChildTAG: they should have put a warning besides the "show button" to only use if u have tried solving the question and also to check out the discussion to clear your doubts. SecondChildUserIdTAG: 539003 SecondChildUserNameTAG: amitgupta25121993 SecondChildCreateTimeTAG: 2012-10-04T00:39:01Z FirstChildTAG: Yes, show answer does indeed *show* the answer. If you don't want to know the answer just yet, simply checking will let you know that it's wrong... I'm not really understanding what you expected the show answer button to do. FirstChildUserIdTAG: 414313 FirstChildUserNameTAG: staticvoid FirstChildCreateTimeTAG: 2012-09-11T14:46:14Z SecondChildTAG: True that it is up to you when you gonna see the answers but I still feel that it somehow reduces the transparency! Anyway **Ashwith** cleared it. SecondChildUserIdTAG: 314624 SecondChildUserNameTAG: Owais001 SecondChildCreateTimeTAG: 2012-09-11T17:59:48Z IndexTAG: 4517 TitleTAG: S2E3 How to find power supplied by voltage and current source??? UserIdTAG: 114864 UserNameTAG: Vikaash CreateTimeTAG: 2012-09-11T14:17:17Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Remember there are three ways to calculate Power. Each of these formulas use two quantities: Voltage+Current, Voltage+Resistance or Current+Resistance. First pick which of the two quantities you know for a voltage source and which one you can calculate. Use that to decide which formula you use. Next remember that you need to figure out whether the source is generating power of absorbing power (If there is another "stronger" source in the circuit, the weaker source may absorb power just the way a battery charger charges batteries). This will be needed to decide the sign of the power absorbed by the source. FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-09-11T14:34:15Z IndexTAG: 4518 TitleTAG: Dependent equation Which node is dependent and why? UserIdTAG: 357929 UserNameTAG: chips67 CreateTimeTAG: 2012-09-11T13:26:26Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL and KCL CommentableIdTAG: 6002x_kvl_and_kcl NumberOfReplyTAG: 2 FirstChildTAG: There is not a unique "dependent node". You could leave out whichever node you want. This dependancy is due to the structure of the circuit. Understanding why an equation is always dependent needs a background in topology knowledge. FirstChildUserIdTAG: 9806 FirstChildUserNameTAG: AneHCraM FirstChildCreateTimeTAG: 2012-09-11T15:18:07Z FirstChildTAG: Basically, by writing current equations for n-1 nodes, we "cover" all currents in the equation (go ahead and check to see if they're all there), and therefore do no need further equations explaining the relationships between currents. This is proven by the fact that any current coming into any node left from some other node, and so all currents used to calculate the KCL equation for a node are mentioned in the equations of the other nodes. (and no current would flow through a wire leaving and entering the same node because that node has no voltage potential difference.) FirstChildUserIdTAG: 164898 FirstChildUserNameTAG: jbparkes FirstChildCreateTimeTAG: 2012-09-12T13:36:51Z SecondChildTAG: Thanks, the first sentence of your answer cleared my confusion up completely. Seems simple now. SecondChildUserIdTAG: 106250 SecondChildUserNameTAG: craigmarshall SecondChildCreateTimeTAG: 2012-09-16T06:13:47Z SecondChildTAG: Mine too i did not understand what they meant by dependent equations which thankfully you clarified. This is like the Eulerian walk.(the Konisberg problem solved by Euler.) SecondChildUserIdTAG: 396673 SecondChildUserNameTAG: anthonypraveen SecondChildCreateTimeTAG: 2012-09-20T13:53:58Z IndexTAG: 4519 TitleTAG: What exactly mean by algebraic expression?? Friends, I could not understand what exactly algebraic expression mean. I have solved the problems but not able to write the answers in the format software accepts, please help. E.g in first question R+R+R = 3R but, this format is not accepted so please tell me how to enter the answer in acceptable format. UserIdTAG: 351425 UserNameTAG: brain2boom CreateTimeTAG: 2012-09-11T11:51:47Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: just R+R+R FirstChildUserIdTAG: 360479 FirstChildUserNameTAG: MurphyFu FirstChildCreateTimeTAG: 2012-09-11T12:05:32Z FirstChildTAG: Try "3*R". FirstChildUserIdTAG: 144694 FirstChildUserNameTAG: johndoe31415 FirstChildCreateTimeTAG: 2012-09-11T11:52:18Z SecondChildTAG: Thanx dude, it worked. SecondChildUserIdTAG: 351425 SecondChildUserNameTAG: brain2boom SecondChildCreateTimeTAG: 2012-09-11T12:00:35Z IndexTAG: 4520 TitleTAG: node and equation since no.of nodes is 4 ....why don't we hav 3 independent equations ...why only 2...is the third one given by voltage of reference node =0 UserIdTAG: 343262 UserNameTAG: rgdixitt CreateTimeTAG: 2012-09-11T11:05:14Z VoteTAG: 0 CoursewareTAG: Week 1 / Node analysis CommentableIdTAG: 6002x_node_analysis NumberOfReplyTAG: 1 FirstChildTAG: The purpose of Nodal analysis is to figure the voltage at each of the nodes with respect to a ground node (you can choose your ground node). In the circuit which Prof. Agarwal explain in the lecture, we have 4 nodes as you rightly said. Of these, we take 1 node as ground. This leaves us with 3 nodes remaining. Now notice that know the voltage of one of these nodes is $V_0$ so we have only 2 unknown node potentials and hence need only two equations. FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-09-11T15:09:26Z SecondChildTAG: Thanks ashwith. SecondChildUserIdTAG: 166869 SecondChildUserNameTAG: Vasco SecondChildCreateTimeTAG: 2012-09-11T16:25:46Z IndexTAG: 4521 TitleTAG: BUG Lab 1 Sandbox Hi! I am using Google Chrome 21.0.1180.89 m on Windows 7 OS. When I access the Lab1 I cannot see the wires that I draw on the schematic also the terminals of the resistors are sometimes there sometimes not, that is when I try to zoom in and out of the schematic! So for some specific zoom I see the wires that are initially on the schematic, for some other zoom I can see the wires that I blindly drown between my resistors for witch the terminals can be seen only on some other zoom level. VERY ANNOYING!!! Please fix it! Regards, Ionut UserIdTAG: 319221 UserNameTAG: stefan31i CreateTimeTAG: 2012-09-11T10:40:33Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: I have a problem also in Lab 1, I have only Resistor in the components bar, I think the lab needs other components to be completed. FirstChildUserIdTAG: 9375 FirstChildUserNameTAG: mabdallah FirstChildCreateTimeTAG: 2012-09-12T10:00:51Z IndexTAG: 4522 TitleTAG: not able to see the voltage source and bulb in lab 1 assignment i am not able to see the voltage source and bulb in lab 1 assignment can any one help? UserIdTAG: 85797 UserNameTAG: Nilayam CreateTimeTAG: 2012-09-11T09:44:46Z VoteTAG: 0 CoursewareTAG: Week 1 / Power delivery modeling CommentableIdTAG: 6002x_S2E6_Modeling NumberOfReplyTAG: 1 FirstChildTAG: same problem with me. there is only resistor component in lab 1.... :( FirstChildUserIdTAG: 259693 FirstChildUserNameTAG: MehrozKhan FirstChildCreateTimeTAG: 2012-09-11T10:39:51Z IndexTAG: 4523 TitleTAG: H1P3 : when H1 became series i found the solution to "H1P3" , but in case of the load H1 became on series ,i couldn't found total current nor delt v through H1 ...can you help UserIdTAG: 231318 UserNameTAG: moutasem CreateTimeTAG: 2012-09-11T09:19:30Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4524 TitleTAG: e2 in first question Hello. I have some trouble to calculate e2 in first question. Here's my calculations: let's label current throgh R1 as i1, direction from e2 to e1, current through R2 as i2, direction from e2 to e3. Then we have: i1+i2-I=0 where i1=(e2-e1)*R1, i2=(e2-e3)*R2, so finally: (e2-e1)*R1+(e2-e3)*R2-I=0 but this leads to incorrect asnwer. Where I made mistake? edit: got the error UserIdTAG: 324219 UserNameTAG: Dmitry79 CreateTimeTAG: 2012-09-11T09:06:17Z VoteTAG: 0 CoursewareTAG: Week 1 / Node method example CommentableIdTAG: 6002x_S2E5_Node_Method NumberOfReplyTAG: 2 FirstChildTAG: I am confused!Why do we need to take current direction from e2 to e1?And yes If we don't take that direction the answer is not right! please guide me. :) FirstChildUserIdTAG: 281159 FirstChildUserNameTAG: ManasiS FirstChildCreateTimeTAG: 2012-09-11T23:46:09Z FirstChildTAG: it is not * R1, It is /R1 FirstChildUserIdTAG: 162670 FirstChildUserNameTAG: charlesbabyt FirstChildCreateTimeTAG: 2012-09-11T09:42:03Z IndexTAG: 4525 TitleTAG: H3P1 Static Discipline for inverter R_Pul of the inverter shown to the left must be selected. I get the correct answer so that when the input is high the output has the desired voltage. However, when the inverter has a low input, in the SR model, R_OFF will be infinity. Therefore, V_Out will always be V_S. This violates the static discipline, doesn't it? So shouldn't the correct answer be that with the construction to the left, it's not possible to design a inverter that fits the specs? UserIdTAG: 144694 UserNameTAG: johndoe31415 CreateTimeTAG: 2012-09-11T08:13:35Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: As long as VS > VOH, you're good. Low input wants high output from an inverter, so this is exactly the behaviour you're after. You may want to review S4V8. FirstChildUserIdTAG: 248902 FirstChildUserNameTAG: Chanute FirstChildCreateTimeTAG: 2012-09-12T05:47:31Z SecondChildTAG: Yes, you are correct. Thank you so much, I somehow screwed this *completely* up in my notes and got the right answer by pure accident. Now I actually know what I'm doing. Thanks again! SecondChildUserIdTAG: 144694 SecondChildUserNameTAG: johndoe31415 SecondChildCreateTimeTAG: 2012-09-12T11:17:16Z IndexTAG: 4526 TitleTAG: PRACTICE PROBLEM is it necessary to complete the practice problem UserIdTAG: 249945 UserNameTAG: gourav19 CreateTimeTAG: 2012-09-11T07:43:13Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4527 TitleTAG: I'm really very confused about how transforms work Could anybody recommend a book about them?it's hard to understand them UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-11T07:38:09Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4528 TitleTAG: Pakistanis Salaam Every one , I hope to find many Pakistanis In here Good Luck, and Comment to let me know if there are any..... UserIdTAG: 37464 UserNameTAG: DoubleA CreateTimeTAG: 2012-09-11T07:09:27Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4529 TitleTAG: Homework checking problems It seems to be marking my homework wrong when I'm sure it is right. On the very first question of the first week, I tried 3R, 3, and 3.0, but it was marked wrong every time. What's up ? UserIdTAG: 129978 UserNameTAG: StuRat CreateTimeTAG: 2012-09-11T06:49:51Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 3 FirstChildTAG: When answering questions, you need to put * to indicate multiplication. FirstChildUserIdTAG: 312035 FirstChildUserNameTAG: philhiggins FirstChildCreateTimeTAG: 2012-09-11T06:53:47Z FirstChildTAG: try r+r+r FirstChildUserIdTAG: 117028 FirstChildUserNameTAG: arslanbuttar FirstChildCreateTimeTAG: 2012-09-11T14:13:17Z FirstChildTAG: Thanks. So far the following failed: 3,3R,3r,r+r+r. However, I was able to get the following to work: R+R+R, 3*R, 3*r. Apparently, it's not able to figure out that the case on the R doesn't matter, or to recognize standard algebraic notation, like 3R. Instead, it looks like it's just doing an exact character match on what I type in to what answers were listed as correct. Not good. Now let me ask about fractions. It doesn't seem to like R/3, so what's the trick to getting it to accept a fraction ? FirstChildUserIdTAG: 129978 FirstChildUserNameTAG: StuRat FirstChildCreateTimeTAG: 2012-09-11T17:39:53Z SecondChildTAG: Correction: 3*r doesn't work, either. When I tried it, apparently the answer was rejected because it contained a lowercase "r", leaving the former correct answer in it's place. They need to make the error message point out which questions were rejected for having an invalid format ("Couldn't be parsed" or "Not recognized as a formula" errors). SecondChildUserIdTAG: 129978 SecondChildUserNameTAG: StuRat SecondChildCreateTimeTAG: 2012-09-11T17:47:46Z SecondChildTAG: I did get it to take R/3, not sure why it didn't like it the first time. Perhaps it's because I had a lowercase "r" on another question. It looks like I'm going to end up spending far more time getting each answer into a format the checker can recognize than actually finding the answer. Even worse, if it marks a question wrong, I don't know if it's actually wrong, or just not recognized as right because it's not in the format they expected. SecondChildUserIdTAG: 129978 SecondChildUserNameTAG: StuRat SecondChildCreateTimeTAG: 2012-09-11T17:56:33Z SecondChildTAG: Case does matter in math and engineering. There will be plenty of times when you have both V and v in a circuit, and lots of others. SecondChildUserIdTAG: 312035 SecondChildUserNameTAG: philhiggins SecondChildCreateTimeTAG: 2012-09-11T22:38:11Z SecondChildTAG: StuRat, we apologize for the frustration that you've experienced. Case and notation are both very important in science and engineering. "3R" is the written standard algebraic notation, but when you are inputting into a computer program (MATLAB, Excel, Wolfram Alpha, even handheld calculators), then you must explicitly indicate multiplication because "3R" could be interpreted as the name of a variable. We have some videos in week 0 to explain notations and the use of our interface. These videos may be helpful for you. SecondChildUserIdTAG: 151427 SecondChildUserNameTAG: RichardZ SecondChildCreateTimeTAG: 2012-09-13T17:32:06Z IndexTAG: 4530 TitleTAG: trouble viewing lab components I can't see any of the components in the lab unless I zoom out to the point where some of the print is getting hard to read. (Seems to always be four clicks.) Then it suddenly all pops back into view. The only thing I can see when I zoom in closer than that is the circles making up the nodes (and sometimes a few wires). I've tried this in both the sandbox and lab one. It worked perfectly for me last spring. I'm using Chrome 21.0.1180.89(up to date) and Mac 10.6.8. It works on the computers at my school using Safari 5.1.7 and the same OS version. UserIdTAG: 270160 UserNameTAG: Sethhhh CreateTimeTAG: 2012-09-11T05:56:11Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4531 TitleTAG: Charge accumulation at the nodes As I understand the professor here uses the fact that dq/dt = 0 as our assumption, "the rule of the playground", something axiomatic. But as I remember when a teacher explained Kirchhoff's laws to me there actually was kind of a ***proof** that charge can **NOT** accumulate at the nodes* (not about capacitors, just in case of several wires and the node connecting them). Intuitively I imagined it as the charge being *"washed off"* by the current so it cannot be stationary. But I remember there was some simple explanation... Like, assume that we *do* have some stationary charge at the node. Then it will create an electric field... And then? So, **1)** Is it actually possible to prove something like that or is it an axiom we make for the inside of our playground? **2)** If it is possible, then what was the proof? :) UserIdTAG: 410033 UserNameTAG: sagitta CreateTimeTAG: 2012-09-11T05:35:57Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL and KCL CommentableIdTAG: 6002x_kvl_and_kcl NumberOfReplyTAG: 2 FirstChildTAG: This is a good question. We can say that we are simplifying the analysis by saying that there is no charge accumulation in the nodes. This comes from the fact that we consider the wires connecting the elements to be perfect conductors. If you recall from Electromagnetism, there can't charge accumulation inside a perfect conductor (it will just move to the edges). However, in real life we usually use copper wires for our circuits and they have an intrinsic capacitance and hence they store a little bit of charge (i.e. dq/dt is not zero for a brief period of time). Since the effect is very small, we are ignoring it in our analysis. FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-11T14:06:56Z FirstChildTAG: Good question.Look if suppose charge got accumulated at the node then there develops a electrostatic force between the static charge and the moving charge there by push themselves aside and keep moving in the circuit with out any accumulation. FirstChildUserIdTAG: 397298 FirstChildUserNameTAG: ZA90 FirstChildCreateTimeTAG: 2012-09-11T16:10:38Z IndexTAG: 4532 TitleTAG: planar resistor in example I.Io its said that the ratio of resistors didn;t change but the values of resistors it self changed i don't understand why don't they change dimension of microprocessor? UserIdTAG: 221617 UserNameTAG: konan CreateTimeTAG: 2012-09-11T05:00:48Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4533 TitleTAG: just use... superposition! There is a positive voltage at e due to each battery and these add up: 14/31*7.2+17/31*5 UserIdTAG: 253630 UserNameTAG: PhillipAdkins CreateTimeTAG: 2012-09-11T02:37:33Z VoteTAG: 0 CoursewareTAG: Week 2 / Node equation review CommentableIdTAG: 6002x_S3E1_NodeEquationReview NumberOfReplyTAG: 0 IndexTAG: 4534 TitleTAG: H1P1 I am not sure if I am mistyping the information for set "C", but I cannot get anything to be accepted. It is a fairly simple concept of networks, although I can't seem to enter the information correctly. Is anyone else having difficulty with this?????? UserIdTAG: 298538 UserNameTAG: BrianBenedict CreateTimeTAG: 2012-09-11T02:35:25Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Struggled with it for a bit. It was clearer after I went through the Week 1 tutorials especially the one with the ladder network combination. FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-11T03:44:14Z IndexTAG: 4535 TitleTAG: Wording Given that each individual resistor can dissipate up to 1 watt of power before burning up, how much total power in watts ( W ) can the smallest-valued composite resistor dissipate before burning up? Maybe I'm not understanding but the wording could some one explain? UserIdTAG: 130031 UserNameTAG: tass CreateTimeTAG: 2012-09-11T02:14:10Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: the key point is that this applies to the smallest-valued composite register. in other words the combination of all the 3 registers in series or parallel or some combination. In that configuration, when any one burns up (due to power dissipation > 1W) the whole circuit collapses. FirstChildUserIdTAG: 279379 FirstChildUserNameTAG: RajaSrinivasan FirstChildCreateTimeTAG: 2012-09-11T02:53:30Z IndexTAG: 4536 TitleTAG: Timing of lectures Will the lectures continue to be posted in advance (like week 2 was), or will we have to wait for each week from now on? I only ask because I have a sporadic work schedule that fluctuates significantly from week to week. Thanks in advance! UserIdTAG: 248902 UserNameTAG: Chanute CreateTimeTAG: 2012-09-11T01:12:56Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: The lectures will appear week by week so ther is always about a two week window. FirstChildUserIdTAG: 66101 FirstChildUserNameTAG: bluefin FirstChildCreateTimeTAG: 2012-09-11T03:51:26Z SecondChildTAG: Thanks! SecondChildUserIdTAG: 248902 SecondChildUserNameTAG: Chanute SecondChildCreateTimeTAG: 2012-09-13T00:11:01Z IndexTAG: 4537 TitleTAG: Divergence theorem? Did anybody understand what he meant with the thought? What is this related to Stokes's Therorem? UserIdTAG: 265133 UserNameTAG: Elienai CreateTimeTAG: 2012-09-11T00:18:51Z VoteTAG: 0 CoursewareTAG: Week 1 / Circuit topology CommentableIdTAG: 6002x_S2E1_Circuit_Topology NumberOfReplyTAG: 1 FirstChildTAG: The divergence theorem and Stoke's theorem are both very advanced mathematics compared to those encountered in 6.002x. The thought is far outside the curriculum of 6.002x. Don't stress out too much if it doesn't make sense! If you are curious, I can tell you that KCL and KVL are derived from Maxwell's equations. You obtain KCL by taking the divergence of Ampere's law and applying the divergence theorem. You get KVL by applying Stoke's theorem to Faraday's law. For this reason, the two paragraphs above that thought can be understood as descriptors of the divergence theorem and Stoke's theorem respectively. You can get a feel for the maths by watching videos on Khan Academy: http://www.khanacademy.org/math/calculus/multivariable-calculus FirstChildUserIdTAG: 151427 FirstChildUserNameTAG: RichardZ FirstChildCreateTimeTAG: 2012-09-11T01:08:58Z IndexTAG: 4538 TitleTAG: Question How I Could Know The sign of the voltage change ?? Voltage change across R1 (volts, with correct sign): -.5 Why ?? While Voltage change across Vsupply (volts, with correct sign): is 3 ??? UserIdTAG: 247612 UserNameTAG: Radwan CreateTimeTAG: 2012-09-10T23:06:54Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 4 FirstChildTAG: This is because sum of all voltage sources and voltage drops have to be equal to "0" in this case. So when you will start with node A and travel clockwise you will get voltage drop on R1 V = 0.5 on R2 V = 1.5 on R3 V = 1. And because this are voltage drops you are writing it with "-" minus signs. Finally voltage source gives you V = 3 and sum of all value give you 0 at the and. FirstChildUserIdTAG: 409148 FirstChildUserNameTAG: mlody FirstChildCreateTimeTAG: 2012-09-11T09:11:08Z FirstChildTAG: it is because as we go through a resistor in the direction of the current the voltage drops and hence the voltage change (final - initial) is negative . on the other hand the voltage increases as we go from the negative to the positive terminal of the voltage source . hence the voltage change is positive . FirstChildUserIdTAG: 275256 FirstChildUserNameTAG: Riyansh FirstChildCreateTimeTAG: 2012-09-11T13:26:06Z FirstChildTAG: In the question it is mentioned that "Choose the sign of the change for a component using the first terminal you come to in the clockwise traversal as the reference node". Hence, for R1, node A is the reference. So with respect to A, there will be a 0.5 **voltage drop** at B hence the negative sign. Similarly at Vsupply, moving clockwise we will take node D(0V) as the reference. Here the **voltage will increase** by 3V across Vsupply (i.e node A(3V)) and hence positive sign. FirstChildUserIdTAG: 413585 FirstChildUserNameTAG: swapneel91 FirstChildCreateTimeTAG: 2012-09-11T18:39:55Z FirstChildTAG: In addition to what the previous repliers indicated. I'd like to give you one more hint, in the same time, try to draw on a piece of paper what I try to say: - The voltage accros R1 ( I will call it VR1 ) is calculated such that VR1 = VB -> VA, such that VB is the potential of node B, and VA is that of node A (draw an arrow going from node B to Node A). VA can be represented as if you have drawn a large arrow going from the ground to node A, and VB then is as if you draw another one from the ground to node B. - Then, draw a curved arrow that represents the clockwise orientation. What can you notice right now ? As you can notice, both arrows are in opposite orientation to each other. Thus they have opposite signs. In order to understand the physical phenomenon, read the previous posts. They are very handy. FirstChildUserIdTAG: 418182 FirstChildUserNameTAG: euldji2005 FirstChildCreateTimeTAG: 2012-09-12T03:02:15Z IndexTAG: 4539 TitleTAG: Weird..... So on H2P1, I put in my answers and it said they were wrong. OK. But when I look on the progress page, I got credit for them(and also another that I hadn't even answered yet). ? UserIdTAG: 344331 UserNameTAG: intellectualwanderer CreateTimeTAG: 2012-09-10T22:30:56Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: I have the same problem!! FirstChildUserIdTAG: 244841 FirstChildUserNameTAG: eyubero FirstChildCreateTimeTAG: 2012-09-11T16:47:12Z FirstChildTAG: same the case with me. is there any bug FirstChildUserIdTAG: 113109 FirstChildUserNameTAG: Akif FirstChildCreateTimeTAG: 2012-09-12T15:40:37Z IndexTAG: 4540 TitleTAG: S2V2: Demo- KVL, KCL At 2.37, Prof Anant Agrawal says: "What I'll do now is change to an AC voltage so that I can go ahead and measure the current without breaking my circuit." What does he mean by 'without breaking my circuit' ??? UserIdTAG: 15237 UserNameTAG: anubhavsinha CreateTimeTAG: 2012-09-10T22:25:00Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4541 TitleTAG: prerequisites (physics/math) How much of this course is going to fly right over my head if I don't know Physics and don't have a strong background in Math in general, they simply didnt teach Physics when/where I went to school, but I really, really want to learn about Electronics so I'm jumping on every online source I can find atm and this course seems way to good to pass up. - Would I be wasting my time until I study 6.01, 8.02, 18.03 on ocw.edu? - Will this course be repeated again? I know the answer to these questions is most likely yes you will be wasting your time, but please try and give some reasons, dont just answer y/n or dismiss me with a one liner, any fool can find a reason why something might not work. If I just have to drill a handful of formulas into my head... I could nail this and check out 6.01, 8.02, 18.03 in my spare time. I'm sorry if this is a re-post, I did search for 'Prerequisites' and 'Physics', got nothing along these lines and there are 87 pages of threads. Thanks in advance. UserIdTAG: 410771 UserNameTAG: MarkF120 CreateTimeTAG: 2012-09-10T22:24:41Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 3 FirstChildTAG: You can't just nail a handful of formulas in your head, electronics require understanding. You won't know what to do with formulas if you just memorize them or worse, you'll apply them wrong. That said, if you're willing to put in the work I don't see why you wouldn't be able to catch up. It would be hard - from what I've seen the math really picks up later in the course, and you'll be missing some key insights if you haven't studied at least basic physics. Want to do it? Start now. Find somewhere to learn calculus from, pick up a book on electricity and magnetism and have fun. FirstChildUserIdTAG: 267993 FirstChildUserNameTAG: mavlijas FirstChildCreateTimeTAG: 2012-09-10T22:47:14Z SecondChildTAG: I think I'm biting off more then I can chew because I have other courses running atm and after some calculations I think there wont be enough hours in the week for me to do them all justice, I would have to study from 7 in the morning until 12 o clock at night, reading on the bus and at lunch times being careful not to daydream or slow down... I just hope they repeat this course. I'll keep this site bookmarked and check it every now and then, so much info coming available atm I just can't help but try and grab it all before they make it 'pay or go away', I have a ridiculous amount of course lectures downloaded, its like when I was a kid and I got my hands on limewire etc :D Thanks for the reply friend, ^I thought you deserved my thoughts before I signed out, see you again some time. :) SecondChildUserIdTAG: 410771 SecondChildUserNameTAG: MarkF120 SecondChildCreateTimeTAG: 2012-09-11T00:25:58Z FirstChildTAG: I agree with everyone in this thread, it will be tough but it can be done. The best web site you will ever, ever find is the Khan Academy. He explains everything you will ever want to know about math and physics, you will be amazed. Just search on Google and you will find it. As for this course, it was taught in its entirety in 2007 at MIT's OpenCourseWare site. Goto http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/ and look for this course. Good Luck FirstChildUserIdTAG: 132828 FirstChildUserNameTAG: rharris FirstChildCreateTimeTAG: 2012-09-11T06:26:52Z SecondChildTAG: i agree with harris.i have downloaded khan academy tutorial about 2 years ago.its fast and very helpful. SecondChildUserIdTAG: 378967 SecondChildUserNameTAG: krish2012 SecondChildCreateTimeTAG: 2012-09-11T08:11:35Z FirstChildTAG: This course is purely based on circuit analysis.i have studied these in my first year of engineering.don't worry about physics or maths.The book that given for this course covers fundamental things as well.when you face problems with basic things,go online and check its intuitive concept.my suggestion is that don't try to learn everything at a sudden moment,like physics course and maths at same time. when you face problems with that from this course lectures then go for it. when i came to last section of chapter 2,their is a problem that uses mathematical concept of matrix calculation(cramer's rule). But i forgot cramer's rule.then i gone to Wikipedia and searched the article about that.From that i solved that one very easily using cramer's rule.For that i never gone for every bits of matrix calculation,but i referred only cramer's rule. FirstChildUserIdTAG: 378967 FirstChildUserNameTAG: krish2012 FirstChildCreateTimeTAG: 2012-09-11T03:58:19Z IndexTAG: 4542 TitleTAG: proper way of answering can any one tell me proper format of answering in ans box regarding h1p1 in algebraic expression form as my ans r not being excepted. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-10T21:05:45Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: the answer is case sensitive so use R and remember that if you are multiplying to use * There is a tutorial document under the "Course Info" tab FirstChildUserIdTAG: 303762 FirstChildUserNameTAG: leocarrasco FirstChildCreateTimeTAG: 2012-09-10T21:08:36Z FirstChildTAG: A good way to type variable names is to cut-and-paste from the problem statement. The answer boxes will always accept variables which are noted in the problem set. The variables are case-sensitive, which will become important later on in the course. Capital I is not the same as lower case i and so on. Sometimes it's hard to interpret the variable. This is especially true when the variables have subscripts and are in various cases. The solution is to simply cut-and-paste the variable from the problem description. The description will have the cases correct, and you won't have to distinguish ambiguous letters such as I and one, O and zero, v and nu, and so on. The parsing engine can be somewhat quirky. Always use * explicitly everywhere you want multiplication. Variable names can include digits, so the system cannot distinguish between X2 and X*2 unless you explicitly tell it. The parsing engine also sometimes doesn't recognize equivalent forms of the same equation. Sometimes changing X^(-1) with 1/X will cause it to work. Use parentheses liberally, especially in cases where you don't have to. Be very explicit about the order of operations, and don't simplify the equation by rearranging things. Use parentheses especially when doing division. On paper we can write a long bar and everything below the bar is understood to be the divisor. In an equation, a/x+y doesn't imply that "x+y" is the divisor, you need to type a/(x+y) explicitly if that is what you mean. FirstChildUserIdTAG: 10523 FirstChildUserNameTAG: Barrabas FirstChildCreateTimeTAG: 2012-09-10T22:00:41Z IndexTAG: 4543 TitleTAG: Homework from video taped version of 6002x in 2007 I thought I saw we were asked to complete the first 4 or 5 homework problems from the 6002x class given in 2007, but now cannot find where it told me that. Did anybody see this, and if so please let me know where? - Thanks. UserIdTAG: 132828 UserNameTAG: rharris CreateTimeTAG: 2012-09-10T20:23:30Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: 6.002x didn't exist in 2007 that would have been the inhouse MIT 6.002 course. You can find the material on OCW at: http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-electronics-spring-2007/ FirstChildUserIdTAG: 9147 FirstChildUserNameTAG: SueP FirstChildCreateTimeTAG: 2012-09-11T00:11:15Z SecondChildTAG: Oops, I didn't mean to put the "x" on 6.002x. I was referring to the in-house course back in 2007. I thought I saw directions to do some of the in-house problems for 6.002x, but evidently I saw it somewhere else. I just don't want to miss any assignments. Thanks SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-09-11T00:26:06Z FirstChildTAG: Under Week 1 Tutorials, there's mention of 2007 homework problems that are used in the tutorial videos if that's what you're thinking of. --Tom FirstChildUserIdTAG: 184529 FirstChildUserNameTAG: tfors FirstChildCreateTimeTAG: 2012-09-10T22:22:55Z IndexTAG: 4544 TitleTAG: Two questions on HW1 and Lab 1 Firstly, on H1P3, on the last problem that asks for the total power, am I supposed to be doing something different. I just add up the previous two powers for H1 and H2(which were right).... am I just being dumb? That's just one HW problem, so I guess it's no big deal(but I'd still appreciate some help). What I'm really concerned about is the lab. More importantly, on the Lab, I checked the DC analysis, it checked out-1.5 V for the top and 2 for the bottom, yet it was still counted as wrong. Was there an error in grading? I put one resistor to the left of the node, the other beneath it, parallel to the bulb. Was that wrong? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-10T18:52:39Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: In H1P3 it was asked not the power dissapated by the resistors(heaters) but power delivered by source!!!. did u got the answr for h1p1: power delivered by least composite FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-10T18:59:59Z SecondChildTAG: First off, I fixed the lab. Turns out I did have the right answers, I just connected then in a goofy fashion. Yeah, I got that answer. SecondChildUserIdTAG: undefined SecondChildUserNameTAG: undefined SecondChildCreateTimeTAG: 2012-09-10T19:43:57Z IndexTAG: 4545 TitleTAG: Explain Can anyone please explain it to me elaborately ??? UserIdTAG: 336001 UserNameTAG: syd_buet12 CreateTimeTAG: 2012-09-10T18:42:59Z VoteTAG: 0 CoursewareTAG: Week 2 / Linear combinations of source strenghts CommentableIdTAG: 6002x_S3E2_Linear_Combinations_of_Source_Strengths NumberOfReplyTAG: 1 FirstChildTAG: use the node method . the ground node in the bottom and the e node in the top (the two nodes in the intersection between the three branches) and use also the kvl in the three loops you should have three equations . with the node method you should have this equation use it you will found a1 and a2 the equation is : (v3-v1)/R1+(v3-v2)/R2+(V3/R3)=0 in the loops use this equations and replace it in the v3 equation the two equations are: -v1-i1/R1+v3=0 -v2-i2/R2+v3=0 now you should find all the coefficients good luck FirstChildUserIdTAG: 266386 FirstChildUserNameTAG: zakzak200 FirstChildCreateTimeTAG: 2012-09-10T21:58:29Z SecondChildTAG: Using the node method I can easily get (v3-v1)/R1+(v3-v2)/R2+(V3/R3)=0 but beyond that I am finally stumped. I take it a1 and a2 are just some combination of R1 and R2 by rearranging the equation? SecondChildUserIdTAG: 244115 SecondChildUserNameTAG: Pedro1969 SecondChildCreateTimeTAG: 2012-09-11T23:47:14Z IndexTAG: 4546 TitleTAG: S1E9 Help Hi, can anyone helpe me with the last exercise on S1E9: "Now, suppose that the voltages of the two component batteries are not quite the same. For example, suppose that V 2 =1.6 . Then when we hook the two batteries together current will flow and the higher voltage battery will charge the lower voltage one. What is the current (in Amperes) that will flow?"? What technique is used to calculate this? I've been going about and I can't seem to work it out. Thanks! UserIdTAG: 304905 UserNameTAG: SergioSilva CreateTimeTAG: 2012-09-10T18:24:59Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Simple KCL. just draw the schematic... you'll have a voltage source of 1.5 volts and one of 1.6 volts connected by the 2 resistors in series... so the voltage drop over the resistors in series is 0.1 volt... and there you go... FirstChildUserIdTAG: 141235 FirstChildUserNameTAG: ilitzroth FirstChildCreateTimeTAG: 2012-09-10T19:06:24Z IndexTAG: 4547 TitleTAG: the direction i0 Why in the scheme, when due KVL and KCL (time 10:12), then the direction of the current i0 shown that way. As far as I know the current flows from positive to negative. (Sorry for the grammatical errors) UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-10T18:07:51Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL and KCL CommentableIdTAG: 6002x_kvl_and_kcl NumberOfReplyTAG: 1 FirstChildTAG: Current flows whichever way it will. Surely you've seen stuff go up before it comes down? FirstChildUserIdTAG: 141235 FirstChildUserNameTAG: ilitzroth FirstChildCreateTimeTAG: 2012-09-10T19:08:41Z IndexTAG: 4548 TitleTAG: resistor combinations am unable to post the results in the home work page. all the time it is showing invalid input.can any one help me out UserIdTAG: 358302 UserNameTAG: girishgowtham CreateTimeTAG: 2012-09-10T17:53:22Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: You can write + for series resistors and || for parallel resistors. If you have multiplication you have to use *. So 3R is wrong... 3*R and R + R + R are correct. FirstChildUserIdTAG: 300169 FirstChildUserNameTAG: castrogfx FirstChildCreateTimeTAG: 2012-09-10T19:09:38Z IndexTAG: 4549 TitleTAG: S2E1 circuit topology part 3 I did not understand that I count 4 or 5 loops but answer says 7 loops plz tell me how? UserIdTAG: 40209 UserNameTAG: sadaf78 CreateTimeTAG: 2012-09-10T17:36:59Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: Have you considered the loop V-R4-R3-R2? FirstChildUserIdTAG: 66101 FirstChildUserNameTAG: bluefin FirstChildCreateTimeTAG: 2012-09-10T17:41:52Z FirstChildTAG: 1 - V-R4-R5-V 2 - V-R1-R2-V 3 - V-R1-R3-R5-V 4 - V-R2-R3-R4-V 5 - R1-R2-R3-R4-R1 6 - R1-R3-R4-R1 7 - R2-R3-R5-R2 FirstChildUserIdTAG: 258500 FirstChildUserNameTAG: karas FirstChildCreateTimeTAG: 2012-09-10T17:45:27Z IndexTAG: 4550 TitleTAG: Homework 1 The answers are marked as incorrect i know how to do it, but i dont know how to write it well some help... for example the b one is 1/(1/R+1/R+1/R) UserIdTAG: 80392 UserNameTAG: abh015 CreateTimeTAG: 2012-09-10T17:24:39Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: 1/(1/R+1/R+1/R) = 1/(3/R)=R/3 FirstChildUserIdTAG: 370989 FirstChildUserNameTAG: Menendez91 FirstChildCreateTimeTAG: 2012-09-10T17:42:19Z FirstChildTAG: Hey, you can use + for series resistors and || for parallel resistors. So in your case, the correct answer would be R || R || R. FirstChildUserIdTAG: 300169 FirstChildUserNameTAG: castrogfx FirstChildCreateTimeTAG: 2012-09-10T18:51:03Z IndexTAG: 4551 TitleTAG: lab1 help please Can anyone help with lab 1. I got right answers but my R1 and R2 on question 1 is different then on question 2. I have feeling that I get problem wrong. IS R1 and R2 have to be same value on question 1 and question 2? UserIdTAG: 286954 UserNameTAG: ododo CreateTimeTAG: 2012-09-10T16:46:47Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: They have to be same...cannot change them in the two scenarios... dont go by the diagrams given... FirstChildUserIdTAG: 180402 FirstChildUserNameTAG: smeiraj FirstChildCreateTimeTAG: 2012-09-10T17:11:34Z FirstChildTAG: Tips: Look at the hint in the bottom and don't go by the diagrams given. In the textbook 2.3.4.1 explain about voltage dividers. You have the same voltage in parallel resistors. FirstChildUserIdTAG: 300169 FirstChildUserNameTAG: castrogfx FirstChildCreateTimeTAG: 2012-09-10T18:54:43Z SecondChildTAG: Now I got R1 and R2 same in the two scenarios so its right now, but I simply got answer by try and error. It count it right, but I am not sure. Can it steel be wrong? SecondChildUserIdTAG: 286954 SecondChildUserNameTAG: ododo SecondChildCreateTimeTAG: 2012-09-12T01:52:23Z IndexTAG: 4552 TitleTAG: answer marked as wrong even though its correct... In week1 home work the 3 resistors are in series then the answer is (R+R+R)=3R but why it is marking answer as the wrong one..? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-10T16:28:00Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: Try 3*R, maybe? FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-10T16:28:29Z SecondChildTAG: Correct. SecondChildUserIdTAG: 5714 SecondChildUserNameTAG: willingc SecondChildCreateTimeTAG: 2012-09-10T16:34:03Z FirstChildTAG: 3R =/= 3*R try it. FirstChildUserIdTAG: 370989 FirstChildUserNameTAG: Menendez91 FirstChildCreateTimeTAG: 2012-09-10T17:57:55Z IndexTAG: 4553 TitleTAG: S3E2 I solved it Easily by superposition but could not even make correct diagram.Please give figure or node potential at each branch referring r2 node as ground. UserIdTAG: 99628 UserNameTAG: Pranjal16 CreateTimeTAG: 2012-09-10T16:20:23Z VoteTAG: 0 CoursewareTAG: Week 2 / Linear combinations of source strenghts CommentableIdTAG: 6002x_S3E2_Linear_Combinations_of_Source_Strengths NumberOfReplyTAG: 1 FirstChildTAG: If you solved for a1 & a2 by superposition, then you know the node potential. It is v3. The the current to V1 is (v3-V1)/R1. Thus b1 = (a1-1)/R1. Now just replace the a1 from the first answer. FirstChildUserIdTAG: 66101 FirstChildUserNameTAG: bluefin FirstChildCreateTimeTAG: 2012-09-10T17:36:34Z IndexTAG: 4554 TitleTAG: S3E3 I am getting y1 as half of given in S3E3. The answer matches from the expression of same as given in S3E3 .where am I mistaking ? UserIdTAG: 99628 UserNameTAG: Pranjal16 CreateTimeTAG: 2012-09-10T14:33:15Z VoteTAG: 0 CoursewareTAG: Week 2 / Superposition example CommentableIdTAG: 6002x_S3E3_superposition NumberOfReplyTAG: 2 FirstChildTAG: y1 is just the current through R1 when V1 is acting alone, (x1-V1)/R1 = (0.4225-2)/7 = -0.2254 Amps. Trick is to use the network circuit with V2 shorted. FirstChildUserIdTAG: 183947 FirstChildUserNameTAG: gfenton06 FirstChildCreateTimeTAG: 2012-09-11T15:13:18Z FirstChildTAG: It's asking for the current due to V1, not the coefficient you would multiply by V1. So y1=b1*V1, with b1 from the previous exercise. FirstChildUserIdTAG: 208854 FirstChildUserNameTAG: BrynnleeEaton FirstChildCreateTimeTAG: 2012-09-15T04:45:07Z IndexTAG: 4555 TitleTAG: Homework - 1 Hi, could someone please explain to me why question two in homework one marks my answer as incorrect when I put R/3? UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-09-10T13:49:41Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 3 FirstChildTAG: It looks like my problem was, I was using Internet Explorer instead of Google Chrome. That was why I was getting the math processing error. I viewed my homework on Google Chrome and it was fine. However, my answer of R/2 was still correct. Are you using Google Chrome? FirstChildUserIdTAG: 132828 FirstChildUserNameTAG: rharris FirstChildCreateTimeTAG: 2012-09-10T14:46:15Z SecondChildTAG: I'm using Firefox, I hope they fix this problem. SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-09-10T14:54:12Z FirstChildTAG: If you are referring to the question: What is the equivalent resistance as an algebraic expression (in terms of R) of network B as viewed from its port? My answer was R/3, and I got it right. When I opened the homework page this morning, I had many parts of the questions say "Math Processing Error", where math equations were given. I'm assuming then the system is not functioning properly. Stay tuned, I bet the problem will go away when MIT discoveres the problem. FirstChildUserIdTAG: 132828 FirstChildUserNameTAG: rharris FirstChildCreateTimeTAG: 2012-09-10T14:38:41Z SecondChildTAG: I think I know what my problem is, when I entered the answer to the first question I clicked check at the bottom, the computer then thinks I have answered all questions and will not let you modify your answer. So, now, when I put in legitimate answers (i.e. 1.5 for question 5) I get a wrong answer. SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-09-10T14:58:05Z FirstChildTAG: Try with R*1/3 ..worked for me..in firefox.. FirstChildUserIdTAG: 173147 FirstChildUserNameTAG: cruiser_rahit FirstChildCreateTimeTAG: 2012-09-10T19:05:32Z IndexTAG: 4556 TitleTAG: the Powers of ten (10) The powers of 10 are very important in circuit calculations as we can see from the LAB 0 simulation. UserIdTAG: 336872 UserNameTAG: campbewil CreateTimeTAG: 2012-09-10T13:08:46Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 0 IndexTAG: 4557 TitleTAG: week 3 videos available,when? thanks!! UserIdTAG: 342966 UserNameTAG: SandraNavarro CreateTimeTAG: 2012-09-10T12:38:01Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: where are week 3 videos FirstChildUserIdTAG: 342221 FirstChildUserNameTAG: deepkar FirstChildCreateTimeTAG: 2012-09-10T14:20:59Z FirstChildTAG: Up top, click on "Course Info", then on the right-hand side you will see a link labeled "Calender". You will be happy to note that week 3 Homework and Labs are out sometime today, the week 3 video sequences will not be out until next Monday (17th). https://www.edx.org/static/content-mit-6002x/handouts/calendar.86570ce56279.pdf FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-10T14:52:30Z SecondChildTAG: when will homework 3 release SecondChildUserIdTAG: 342221 SecondChildUserNameTAG: deepkar SecondChildCreateTimeTAG: 2012-09-10T15:01:17Z SecondChildTAG: which time is this course following .. SecondChildUserIdTAG: undefined SecondChildUserNameTAG: undefined SecondChildCreateTimeTAG: 2012-09-10T15:20:19Z SecondChildTAG: Ok Thanks, you're right!! But it's weird anyway that HW and Labs3 are posted today and the corresponding videos next week... will see SecondChildUserIdTAG: 342966 SecondChildUserNameTAG: SandraNavarro SecondChildCreateTimeTAG: 2012-09-10T16:11:11Z IndexTAG: 4558 TitleTAG: Incorrect answer Actually the answer must be -1.69 Volts UserIdTAG: 341358 UserNameTAG: akbar13 CreateTimeTAG: 2012-09-10T12:13:43Z VoteTAG: 0 CoursewareTAG: Week 2 / Node equation review CommentableIdTAG: 6002x_S3E1_NodeEquationReview NumberOfReplyTAG: 2 FirstChildTAG: Did you notice that V2 has a negative sign and that it is also inverted with respect to V1? FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-10T13:10:23Z SecondChildTAG: Uh, i thought they write -7.2, cuz its already inverted. So its like double inverted. Tricked me nicely tbh SecondChildUserIdTAG: 329254 SecondChildUserNameTAG: KGabor SecondChildCreateTimeTAG: 2012-09-10T14:53:21Z SecondChildTAG: Can you give us a REAL example of a -7.2 dc power supply? Thank you ! SecondChildUserIdTAG: 240487 SecondChildUserNameTAG: AlexAlexandrescu SecondChildCreateTimeTAG: 2012-09-12T11:54:40Z FirstChildTAG: YES DA CORRECT ANSWER IS -1.69V.GONE MAD WEN IT WAS TELNG THE ANSWER IS WRONG FirstChildUserIdTAG: 351262 FirstChildUserNameTAG: subbucmr FirstChildCreateTimeTAG: 2012-09-12T12:53:04Z IndexTAG: 4559 TitleTAG: Hello Hello, does anyone from Poland or know polish language? ;) UserIdTAG: 408796 UserNameTAG: GrzegorzPietrzak CreateTimeTAG: 2012-09-10T12:12:49Z VoteTAG: 0 CoursewareTAG: Week 1 / Course overview CommentableIdTAG: 6002x_S1V1toS1V4 NumberOfReplyTAG: 2 FirstChildTAG: Me :) FirstChildUserIdTAG: 153454 FirstChildUserNameTAG: Inshiqaq FirstChildCreateTimeTAG: 2012-09-14T14:49:05Z FirstChildTAG: And me (i ja) :) FirstChildUserIdTAG: 41557 FirstChildUserNameTAG: TomaszJerzy FirstChildCreateTimeTAG: 2012-09-16T17:50:06Z IndexTAG: 4560 TitleTAG: How to submit your lab or homework results? I don't see the option to submit lab or homework results. I just have the option of checking. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-10T12:08:48Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: when you check it automatically submitts FirstChildUserIdTAG: 301736 FirstChildUserNameTAG: MOjangole FirstChildCreateTimeTAG: 2012-09-10T12:18:43Z IndexTAG: 4561 TitleTAG: Last question In a last part we are asked to find voltage across load. 1.49 Volts is a voltage across the transmission line. Actually the correct answer is 238.5 Volts. UserIdTAG: 341358 UserNameTAG: akbar13 CreateTimeTAG: 2012-09-10T12:01:43Z VoteTAG: 0 CoursewareTAG: Week 1 / Power delivery modeling CommentableIdTAG: 6002x_S2E6_Modeling NumberOfReplyTAG: 1 FirstChildTAG: No, it is asked to find the drop from the house to the load due to the transmission cable (or voltage across cable) so 1.49V is the correct answer. FirstChildUserIdTAG: 396446 FirstChildUserNameTAG: RousseauxS FirstChildCreateTimeTAG: 2012-09-10T14:48:31Z SecondChildTAG: Actually, I think the drop between the house and the barn should be 0.74V (and then again 0.74V from the barn back to the house, for an overall 1.49V). SecondChildUserIdTAG: 108190 SecondChildUserNameTAG: MickaelFromBrest SecondChildCreateTimeTAG: 2012-09-12T16:22:34Z SecondChildTAG: yes, i thought the same way, yet the answer is 1,49 SecondChildUserIdTAG: 351575 SecondChildUserNameTAG: bigmek SecondChildCreateTimeTAG: 2012-09-16T23:28:59Z SecondChildTAG: Was thinking the same thing. Voltage drop to the barn should be .74V, then another .74 on the way back. SecondChildUserIdTAG: 108926 SecondChildUserNameTAG: kayakMike SecondChildCreateTimeTAG: 2012-09-16T23:53:42Z IndexTAG: 4562 TitleTAG: dq/dt not = 0 Hello. Just curious: can you give an exaple of a material where dq/dt is not 0? I suppose it's about an edge case, e.g. when the resistor burns out: charge accepted is not going anytime back. Or I'm wrong? UserIdTAG: 213170 UserNameTAG: AdamKuch CreateTimeTAG: 2012-09-10T11:18:07Z VoteTAG: 0 CoursewareTAG: Week 1 / Lumped Element Abstraction CommentableIdTAG: 6002x_lumped_element_abstraction NumberOfReplyTAG: 3 FirstChildTAG: A uniform charge density on a surface...were , rate of change of charge with respect to time is constant FirstChildUserIdTAG: 381619 FirstChildUserNameTAG: siddhantmishra007 FirstChildCreateTimeTAG: 2012-09-10T11:22:48Z FirstChildTAG: A typical example of an element with dq/dt different than zero is a capacitor (under certain conditions). All materials have some capacitance within itself, so at least for a brief period of time the dq/dt will be different than zero FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-10T13:34:48Z FirstChildTAG: I was thinking a good example of when dq/dt is no longer 0 might be when you spill water all over your new Laptop computer. The circuits inside will not function as desired if charge *leaks* out of your wires into the surrounding water. FirstChildUserIdTAG: 714237 FirstChildUserNameTAG: PaxPolaris FirstChildCreateTimeTAG: 2012-10-25T06:22:37Z IndexTAG: 4563 TitleTAG: attempts- how many? will the number of attempts to make the correct answer count in progress performance? UserIdTAG: 347644 UserNameTAG: svganesh93 CreateTimeTAG: 2012-09-10T10:59:44Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Looks like it doesn't, at least for homework. I resubmitted answers until I got full credit. I'm assuming this won't be the case for exams, but I'm not sure of course. FirstChildUserIdTAG: 160489 FirstChildUserNameTAG: Spellstealz FirstChildCreateTimeTAG: 2012-09-10T11:38:49Z SecondChildTAG: This is correct. SecondChildUserIdTAG: 79361 SecondChildUserNameTAG: kimt SecondChildCreateTimeTAG: 2012-09-10T13:07:47Z IndexTAG: 4564 TitleTAG: Bug lab1 In lab 1 I cannot submit answer, I get no red cross or green mark. UserIdTAG: 95280 UserNameTAG: Fortyq CreateTimeTAG: 2012-09-10T09:26:43Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 3 FirstChildTAG: it works. my bad. FirstChildUserIdTAG: 95280 FirstChildUserNameTAG: Fortyq FirstChildCreateTimeTAG: 2012-09-10T09:39:46Z FirstChildTAG: Try Mozilla. Often other browsers show this kind of problem. FirstChildUserIdTAG: 97340 FirstChildUserNameTAG: AnkitRana FirstChildCreateTimeTAG: 2012-09-10T09:58:17Z FirstChildTAG: In week two LAB, m facing the same problem.. FirstChildUserIdTAG: 381619 FirstChildUserNameTAG: siddhantmishra007 FirstChildCreateTimeTAG: 2012-09-10T11:23:30Z IndexTAG: 4565 TitleTAG: tech In lab 1 i am not able to add ground... plz help UserIdTAG: 394187 UserNameTAG: AMOL4165654 CreateTimeTAG: 2012-09-10T09:14:38Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Try resetting the lab. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-09-10T10:44:38Z IndexTAG: 4566 TitleTAG: BUG Hi , I tried to watch the training videos for 6.002x circuits and electronics but unfortunately cannot see anything. Have tried to use OS windows vista,XP alongwith internet explorer and firefox but still unlucky....Please could you help to resolve the issue. Thanks. UserIdTAG: 380053 UserNameTAG: SatishPM CreateTimeTAG: 2012-09-10T08:52:56Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: Try with chrome. FirstChildUserIdTAG: 193033 FirstChildUserNameTAG: NIslam FirstChildCreateTimeTAG: 2012-09-10T09:43:57Z FirstChildTAG: There is a known issue that users in countries that block youtube cannot access the videos for the course. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-09-10T10:45:39Z IndexTAG: 4567 TitleTAG: H1P1: RESISTOR COMBINATIONS iN THE PARALLEL AND SERIES COMBINATION OF RESISTANCES I AM ENTERING THE THE VALUES FOR CIRCUITS IS B AND C I M WRITING AS 1/R+1/R+1/R AND R+R+1/R+1/R BUT I AM GETTING MY ANSWER WRONG..PLEASE NEED SOME HELP HERE UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-10T08:35:52Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Your answers are wrong. The formula for parallel resistance is $\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+...+\frac{1}{R_n}$ FirstChildUserIdTAG: 267993 FirstChildUserNameTAG: mavlijas FirstChildCreateTimeTAG: 2012-09-10T08:54:20Z SecondChildTAG: deleted SecondChildUserIdTAG: 207456 SecondChildUserNameTAG: Octomos SecondChildCreateTimeTAG: 2012-09-10T09:52:45Z SecondChildTAG: I know the formula but their are only three resister shown in figure of parallel combination. and what about the other answer SecondChildUserIdTAG: undefined SecondChildUserNameTAG: undefined SecondChildCreateTimeTAG: 2012-09-10T12:39:52Z IndexTAG: 4568 TitleTAG: bug hello to all, who try hard for improvment in presenting courses, Unfortunately the youtube is filtered in iran and seeing or downloading the video lectures is very hard and slow or for someones like me, is impossible :( I'll be so grateful if there would be a link for download the videos from your server ( sure not from youtube ).I'm sure about powerful servers, you access. Best Regards AHRT UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-10T08:18:19Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Search the web for "free vpn service", most work with youtube . FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-10T10:06:06Z IndexTAG: 4569 TitleTAG: EDITING PROFILE INFO I AM UNABLE TO EDIT MY PROFILE INFO PLEASE HELP BECAUSE MY NAME APPEARING ON PROFILE IS WRONG UserIdTAG: 313263 UserNameTAG: MANISHRPV CreateTimeTAG: 2012-09-10T07:45:24Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4570 TitleTAG: Voltage Source Problems (Question 3) I don't understand why the equation isn't e3/R2 + I + (e3-e1)/R1. That part with the voltage source seems to be throwing a lot of people off, and I don't understand the explanations so far. Help, please?! UserIdTAG: 270160 UserNameTAG: Sethhhh CreateTimeTAG: 2012-09-10T07:22:26Z VoteTAG: 0 CoursewareTAG: Week 1 / Node method example CommentableIdTAG: 6002x_S2E5_Node_Method NumberOfReplyTAG: 1 FirstChildTAG: If you follow the current flow it will help: the voltage at e1 is like two batteries are stacked -+-+ assuming e3 is + at e1, you have e3+V and the current flow into R1 (away from node e3) so current away from node e3 through R1 is (e3+V)x1/R1 current flow away from e3 through R2 is e3/R2 lastly current flow I is away from e3 so it is +ve bringing all these together: (e3+V)/R1+e3/R1+I=0 e3=(-I-V/R1)/)12/R1+1/R2) =(-3-5/3)/(1/3+1/5) =-8.75 FirstChildUserIdTAG: 314386 FirstChildUserNameTAG: jid FirstChildCreateTimeTAG: 2012-09-10T21:02:01Z IndexTAG: 4571 TitleTAG: Alternate Reference Textbook The recommended book is too costly to buy in India. Could you please recommend any alternative reference textbook for this course which is available in India for a low price. UserIdTAG: 134209 UserNameTAG: abhaynayak CreateTimeTAG: 2012-09-10T05:33:13Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 3 FirstChildTAG: Use the online textbook, use wikipedia, use allaboutcircuits.com and if you absolutely need a textbook buy "Engineering Circuit Analysis" by Hayt, Kemmerly and Durbin. FirstChildUserIdTAG: 7281 FirstChildUserNameTAG: SubitC FirstChildCreateTimeTAG: 2012-09-10T05:48:17Z FirstChildTAG: In India, Flipkart sells the legal Indian reprint for Rs. 5547. http://www.flipkart.com/foundations-analog-digital-electronic-circuits-1558607358/p/itmdygszbsqmyrhb?pid=9781558607354&ref=f08ce896-983a-4adf-9717-96cffca0d80b FYI, I live in the USA, and I am using a legal Indian reprint that I purchased on eBay for about USD$35, which is about 1/3 the price of the US edition. I assume the price is much lower in India for the seller to make a profit on eBay. International editions are legal for importation for personal use in the USA. FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-09-10T06:05:14Z FirstChildTAG: friend, try to download it at freebookspot.es for free, if it is allowed in your location. Thanks. FirstChildUserIdTAG: 349139 FirstChildUserNameTAG: 1977ROYELMER FirstChildCreateTimeTAG: 2012-09-10T13:51:48Z IndexTAG: 4572 TitleTAG: answers written when i write my answer in box. and chk the answer ... some error occure and write this line "Could not parse '2R/3' as a formula" what i do.plz help me UserIdTAG: 294836 UserNameTAG: kakashi1 CreateTimeTAG: 2012-09-10T04:41:50Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: You may try 2*R/3 FirstChildUserIdTAG: 143593 FirstChildUserNameTAG: Tsybulkin FirstChildCreateTimeTAG: 2012-09-10T04:55:38Z SecondChildTAG: yes..that works.. SecondChildUserIdTAG: 173147 SecondChildUserNameTAG: cruiser_rahit SecondChildCreateTimeTAG: 2012-09-10T19:59:49Z FirstChildTAG: I think you should write the equation not the final result. FirstChildUserIdTAG: 214863 FirstChildUserNameTAG: qahwagi FirstChildCreateTimeTAG: 2012-09-10T04:49:40Z IndexTAG: 4573 TitleTAG: I'm still a little confused with the 3rd constraint of the lumped circuit abstraction I know there's another post about the 3rd constraint in lumped circuit abstraction, but I'm still a little confused, and no one seems to see that post anymore, so I'll just make a new post. How does the signal changes? Does it have anything to do with electromagnetic waves? And what will change if the signal changes? Will the value for the voltage change? UserIdTAG: 98259 UserNameTAG: rogerloh0 CreateTimeTAG: 2012-09-10T03:39:35Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4574 TitleTAG: EECS What does the acronym EECS stand for? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-10T02:22:49Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: google says that this is Electrical Engineering and Computer Science FirstChildUserIdTAG: 394836 FirstChildUserNameTAG: v2g6ch4 FirstChildCreateTimeTAG: 2012-09-10T02:56:43Z FirstChildTAG: Electrical Engineering and Computer Science. As in the EECS department at MIT. FirstChildUserIdTAG: 132828 FirstChildUserNameTAG: rharris FirstChildCreateTimeTAG: 2012-09-10T02:26:24Z IndexTAG: 4575 TitleTAG: ending video screen It would be neat if you could make videos retain the last screen after playback. That way we wouldn't have to hunt down the relevant screen when the lecturer asks a question. UserIdTAG: 267993 UserNameTAG: mavlijas CreateTimeTAG: 2012-09-10T02:09:02Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: No need to hunt down. Just note the time shown on the video when you press [Pause] and go to that place when you restart, FirstChildUserIdTAG: 382646 FirstChildUserNameTAG: newfet12 FirstChildCreateTimeTAG: 2012-09-10T06:46:25Z IndexTAG: 4576 TitleTAG: How Many Students in 6.002x fall 2012? How many of us are there in the fall 2012 6.002x? UserIdTAG: 88550 UserNameTAG: Neil_S_Berry CreateTimeTAG: 2012-09-09T23:28:06Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4577 TitleTAG: What is the voltage (in Volts) v2 across the resistor with resistance R2? S2E3: Using KVL, KCL, and VI constraints - I did pretty good in S2E2, but I can't see how to get to the answer of 7.778 for the first question of S2E3. Should I be putting all equations together that I learned from KCL, KVL Method in lecture that preceded this exercise? Any insight would be appreciated. Thank you, Eric UserIdTAG: 235887 UserNameTAG: ersmith0529 CreateTimeTAG: 2012-09-09T22:50:46Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Hello Eric ! Here's how i worked through this exercise: It is obvious that : v2=R2*i2 => v2=5*i2 **[1]** from KCL for the upper node : -I - i1+ i2 =0 => i2= I+i1 **[2]** Also i1=v1/R1 => i1=v1/4 **[3]** From KVL : V-v1-v2=0 => v1=V-v2 **[4]** so, combining **[2] [3]** and **[4]** we have : i2=I+((V-v2)/4)) then we will substitute the numerical values of I, V and use **[1]** to get: i2=3+((2-(5*i2))/4) and after we clean up the mess we have the value for current i2. Then, from **[1]** we can calculate v2. hope i was helpful :) cheers ! G.G. FirstChildUserIdTAG: 216242 FirstChildUserNameTAG: Maelstrom FirstChildCreateTimeTAG: 2012-09-14T12:32:22Z SecondChildTAG: How do you clean up? You still have i2 on both sides which are unknown quantities SecondChildUserIdTAG: 22791 SecondChildUserNameTAG: Andyj SecondChildCreateTimeTAG: 2012-09-14T17:51:10Z IndexTAG: 4578 TitleTAG: CIRCUIT SANDBOX WHAT IS THIS LAB ABOUT? UserIdTAG: 400892 UserNameTAG: ahmadyousaf CreateTimeTAG: 2012-09-09T22:14:32Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 3 FirstChildTAG: http://en.wikipedia.org/wiki/Sandpit It is a page for you to learn by trying any circuit that you want. You can draw some circuits, simulate them and see how the voltage and current changes. For example, if you have done some electronics before you may remember simple series and parallel circuits. You could remind yourself about them by drawing a voltage source and two 1000ohm resistors. Try the resistors in series and parallel, see which way results in most current flowing. This might help you with one of the questions in week 1. FirstChildUserIdTAG: 397236 FirstChildUserNameTAG: farfield FirstChildCreateTimeTAG: 2012-09-09T22:37:33Z SecondChildTAG: thanks SecondChildUserIdTAG: 400892 SecondChildUserNameTAG: ahmadyousaf SecondChildCreateTimeTAG: 2012-09-09T22:49:55Z FirstChildTAG: it is for practice and for experiments you can build and analyze your own circuits and analyze any circuit as you wish. FirstChildUserIdTAG: 40209 FirstChildUserNameTAG: sadaf78 FirstChildCreateTimeTAG: 2012-09-09T22:40:11Z SecondChildTAG: thanks SecondChildUserIdTAG: 400892 SecondChildUserNameTAG: ahmadyousaf SecondChildCreateTimeTAG: 2012-09-09T22:50:02Z FirstChildTAG: Hi ahmadyousaf! How are you? That is an interesting question! When I started the Prototype Course I wondered the same at first. "The sandbox it is a work place where you can simulate the Circuits that you want. This will be really helpful for you during this course, believe me ;). You can try to create any circuit and analyze it. Try it, it is funny. You can take some solved examples of the Textbook and corroborate the results - voltage , current, etc..." https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Overview/Circuit_Sandbox/ I hope that you are enjoying this Course as much as I did in the Past version of this Course! If you have any doubt you can ask me , I am here to help you. Of course, there are a lot of Students re-taking this Course that can also help you. You can do it! My best wish to you! Myriam. ----- Hola ahmadyousaf! Cómo estás? Es una pregunta interesante! Cuando había comenzado el Curso Prototipo me pregunté lo mismo. "El Sandbox es una especie de Espacio de Trabajo en donde puedes simular los Circuitos que quieras. Créeme que te será de mucha ayuda durante la cursada ;). Aquí, puedes crear cualquier Circuito y analizarlo. Inténtalo, es muy divertido. Si quieres, puedes probar con algún ejercicio que ya está resuelto en el Textbook y corroborar los resultados con los tuyos - tensión, corriente, etc..." https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Overview/Circuit_Sandbox/ Espero que estés disfrutando de este Curso tanto como yo lo he hecho en el Pasado Curso! Si tienes alguna duda puedes preguntarme, estoy aquí para ayudarles. Desde ya, que hay muchos alumnis que están nuevamente cursando este Curso que pueden también ayudarte. Buenos ánimos para tí! Tú puedes hacerlo! Mi mejor deseo, Myriam. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-10T02:32:07Z IndexTAG: 4579 TitleTAG: honor-code certificate wanted to know if the honor-code certificate will have validadação for my career? UserIdTAG: 402643 UserNameTAG: OzorioNeto CreateTimeTAG: 2012-09-09T22:13:17Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: it is indeed validate,but there is another thing which is more precious than this certificate and that is **knowledge and skill** which is indeed a pillar of your career. FirstChildUserIdTAG: 40209 FirstChildUserNameTAG: sadaf78 FirstChildCreateTimeTAG: 2012-09-09T22:44:53Z IndexTAG: 4580 TitleTAG: Accidentally pressed "check" button (S1E9) Accidentally I pressed the "check" button and I had not answered the questions in this section(S1E9). Is there any way to restart this section? UserIdTAG: 248905 UserNameTAG: AJGP CreateTimeTAG: 2012-09-09T21:25:16Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 3 FirstChildTAG: If I'm not mistaken, you can check it as many times as you want. FirstChildUserIdTAG: 405566 FirstChildUserNameTAG: Adrian17 FirstChildCreateTimeTAG: 2012-09-09T21:33:35Z FirstChildTAG: I think you can click check all the times you want. It doesn't affect your work or your scores FirstChildUserIdTAG: 45307 FirstChildUserNameTAG: josejimenez2 FirstChildCreateTimeTAG: 2012-09-09T22:42:52Z FirstChildTAG: You can use the 'Check' button unlimited times on sequence exercises, homeworks, and labs. For homeworks and labs, after the deadline has passed, the 'Check' and 'Save' buttons are replaced with a 'Show Answers' button. On exams, you are limited to three attempts to submit your answers. Also note that the sequence exercises are optional, although highly recommended. They are not used for purposes of your grade, they exist only to help you learn and understand the course material. Only the homeworks, labs, and exams are graded and count toward your course progress, grade, and certificate. FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-09-10T05:20:11Z IndexTAG: 4581 TitleTAG: mysterious K what does 'K' (in denominator of conductivity in S2V11 3:59 1.0x speed) stand for. It is not multiplayer, because is is capital. Also not symbol of a unit, because prof doesn`t use them. What`s the K for? UserIdTAG: 404284 UserNameTAG: tomdrifmach CreateTimeTAG: 2012-09-09T19:19:55Z VoteTAG: 0 CoursewareTAG: Week 1 / Node analysis CommentableIdTAG: 6002x_node_analysis NumberOfReplyTAG: 2 FirstChildTAG: I think it is still for Kilo. (but I see your point, SI says 'k' for 1.000) FirstChildUserIdTAG: 131747 FirstChildUserNameTAG: hasi FirstChildCreateTimeTAG: 2012-09-09T19:33:16Z SecondChildTAG: http://www.unc.edu/~rowlett/units/dictK.html So basically it is just 10^3 when skipping unit symbol SecondChildUserIdTAG: 404284 SecondChildUserNameTAG: tomdrifmach SecondChildCreateTimeTAG: 2012-09-09T20:54:27Z FirstChildTAG: I'm guessing it is supposed to be a 'k' for kOhms. FirstChildUserIdTAG: 183711 FirstChildUserNameTAG: MMatheson FirstChildCreateTimeTAG: 2012-09-09T19:33:47Z IndexTAG: 4582 TitleTAG: Show formula/calculation I like the Check Answer buttons; I try to avoid the Show Answer buttons. I would really like a Show Formula(s) button and a Show Calculation(s) button. Given that I am taking this course 100% online, and that links to the textbook are not very specific, I am finding it cumbersome to find appropriate formulas in the textbook. Thank you. Also ... it would be nice to have these buttons next to each problem, so that answers/etc are not revealed to subsequent problems when trying to solve a prior problem. UserIdTAG: 330688 UserNameTAG: chota300 CreateTimeTAG: 2012-09-09T19:18:54Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4583 TitleTAG: S1V9 Demo volume Despite maximizing the embedded video volume and my pc speaker volume, it is difficult to hear this demo. Please increase the recorded video volume for classroom demos. The non-classroom video volume levels have been fine. Perhaps uploading classroom demos to YouTube ( and adding direct links ) would mitigate volume issues. Thank you. UserIdTAG: 330688 UserNameTAG: chota300 CreateTimeTAG: 2012-09-09T19:07:44Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4584 TitleTAG: S1E5 The third component would have to be a power source right, otherwise how would the circuit run? So therefore, it has to have the sum of the voltages. Which means $1.4 + -0.9 = 0.5$. Am I right? I am really confused about the question, how do you come to an answer? How can you have a negative voltage? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-09T18:54:01Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL and KCL exercises CommentableIdTAG: 6002x_kvl_kcl_exercises NumberOfReplyTAG: 3 FirstChildTAG: the thitd component can be a composed power supply like two battery v3 ( + - .... - + ) FirstChildUserIdTAG: 236136 FirstChildUserNameTAG: Hamid-ch FirstChildCreateTimeTAG: 2012-09-09T21:16:58Z FirstChildTAG: for negative : if we consider a direction of a curent then the mesurment give us a negative sign => this mean that the real direction of the curent is th opposite direction that we have choose (there is no negative curent , it tell us that the direction we consider is wrong ) FirstChildUserIdTAG: 236136 FirstChildUserNameTAG: Hamid-ch FirstChildCreateTimeTAG: 2012-09-09T21:24:31Z FirstChildTAG: Whatever you are thinking is perfectly right, the voltage isnt negative in this case. "Joe was debugging part of an experimental apparatus, probing around with his voltmeter" Joe used his probes in a wrong fashion,exchanging the +ve and -ve potential,so instead of 0.9 we get -0.9 (because he used +ve end of the probe at the -ve terminal of the resistor and vice versa) FirstChildUserIdTAG: 211963 FirstChildUserNameTAG: AmpCET FirstChildCreateTimeTAG: 2012-09-10T06:49:40Z IndexTAG: 4585 TitleTAG: "BUG" I AM UNABLE TO WATCH VIDEOS.I AM FACING THIS PROBLEM SINCE FIVE HOURS, BEFORE EVERY THING WENT OK. I AM USING WINDOWS 7 AND FIREFOX WHEN I PLAY THE VIDEO IT SAYS "THIS VIDEO IS CURRENTLY UNAVAILABLE" UserIdTAG: 40209 UserNameTAG: sadaf78 CreateTimeTAG: 2012-09-09T18:52:40Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4586 TitleTAG: help! im unable to solve how to get the power supplied by voltage source! could someone help me out? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-09T18:31:54Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL/KCL example CommentableIdTAG: 6002x_S2E3_Using_KVL_KCL_and_VI_constraints NumberOfReplyTAG: 2 FirstChildTAG: It is the product of the current through the voltage source by its voltage. Test whether it is positive or negative, because an ideal voltage source is not dissipative. FirstChildUserIdTAG: 361823 FirstChildUserNameTAG: EliasOak FirstChildCreateTimeTAG: 2012-09-09T18:53:29Z FirstChildTAG: you are supposed to use KCL and KVL and then you have to label each component with its own value as an equation, This way you will have enough data to get the result. But to tell you the truth I have used superposition and it's much easier FirstChildUserIdTAG: 45307 FirstChildUserNameTAG: josejimenez2 FirstChildCreateTimeTAG: 2012-09-09T20:30:03Z IndexTAG: 4587 TitleTAG: Clear explanation of Maxwell's equation I can see that the professor is trying to explain things in Maxwell's equations, but one has no idea of them. Upon closer inspection, the derived laws in Maxwell's equation are laws that one has used, for example the law about series and parallel resistance. Can someone please explain, if I need to understand all these difficult things, since magnetic flux is something that I have learnt a lot later, and frankly does not seem relevant without magnets or electromagnetic forces. UserIdTAG: 193033 UserNameTAG: NIslam CreateTimeTAG: 2012-09-09T18:02:57Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL and KCL CommentableIdTAG: 6002x_kvl_and_kcl NumberOfReplyTAG: 2 FirstChildTAG: If you are an enginner, you don´t need to understand this, you have to know how to use it. If you are a physics guy, yes you need it. FirstChildUserIdTAG: 239780 FirstChildUserNameTAG: dprado FirstChildCreateTimeTAG: 2012-09-09T18:11:30Z FirstChildTAG: Any electrical current generates a magnetic field, and any build of current creates a changing magnetic field which is magnetic flux. Magnetism is always around when electricity is around and vice versa. We just don't really care about it for what we are doing. FirstChildUserIdTAG: 164898 FirstChildUserNameTAG: jbparkes FirstChildCreateTimeTAG: 2012-09-12T13:40:55Z IndexTAG: 4588 TitleTAG: HELP I,m not able to watch any of the lecture videos.I'm using windows7 and latest version of chrome.Please help me out UserIdTAG: 367633 UserNameTAG: jasimmunirmalik CreateTimeTAG: 2012-09-09T17:31:14Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4589 TitleTAG: v1 and v2 same sign? and what with KVL? I don't understand why v1 and v2 have the same sign? KVL must be v1 + v2 = 0, right? so they can't have the same sign. Where am I wrong? UserIdTAG: 189407 UserNameTAG: Olstrill CreateTimeTAG: 2012-09-09T16:49:34Z VoteTAG: 0 CoursewareTAG: Week 1 / Associated reference directions CommentableIdTAG: 6002x_S2E2_Associated_Reference_Directions NumberOfReplyTAG: 1 FirstChildTAG: Actually, in this case V1 and v2 have the same sign, because of the reference system used, also, when you do the loop, you have v1-v2=0, or v1=v2; the same happens on the second case, v3-v4=0. You're mistaken in the sequence you do the loop. Just check again and you'll get it right! :) FirstChildUserIdTAG: 83726 FirstChildUserNameTAG: Enriquejgc FirstChildCreateTimeTAG: 2012-09-09T17:43:08Z SecondChildTAG: got it, thank you! SecondChildUserIdTAG: 189407 SecondChildUserNameTAG: Olstrill SecondChildCreateTimeTAG: 2012-09-09T22:17:39Z IndexTAG: 4590 TitleTAG: Invalid input When i put the values with there units,and try to check it ,says invalid input "couldn't interpret. UserIdTAG: 223126 UserNameTAG: gurusubs CreateTimeTAG: 2012-09-09T16:22:17Z VoteTAG: 0 CoursewareTAG: Week 1 / Associated reference directions CommentableIdTAG: 6002x_S2E2_Associated_Reference_Directions NumberOfReplyTAG: 1 FirstChildTAG: The questions usually tells you what units the answer should be in. So there is no need to enter the units in the answer. Only the numerical value. FirstChildUserIdTAG: 303762 FirstChildUserNameTAG: leocarrasco FirstChildCreateTimeTAG: 2012-09-09T16:25:53Z IndexTAG: 4591 TitleTAG: power dissipated by current source i have used the the associated variable convention since using that i took a current i3 opposite to I and power equation becomes i3*v3 which is -ve but the opposite is given as answer clarify plz UserIdTAG: 375082 UserNameTAG: satya1889 CreateTimeTAG: 2012-09-09T15:59:20Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL/KCL example CommentableIdTAG: 6002x_S2E3_Using_KVL_KCL_and_VI_constraints NumberOfReplyTAG: 1 FirstChildTAG: what i mean to say is that how could a source power be positive? FirstChildUserIdTAG: 375082 FirstChildUserNameTAG: satya1889 FirstChildCreateTimeTAG: 2012-09-09T17:45:20Z IndexTAG: 4592 TitleTAG: H1P3 Hi..im not able to solve H1P3 as I'm not able to get the concept behind it.. please help me!! UserIdTAG: 333107 UserNameTAG: Mrugen CreateTimeTAG: 2012-09-09T15:03:28Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 3 FirstChildTAG: It would help if could be a little more specific about what aspect of the problem is confusing you. If you're confused about the talk of heating systems you can ignore that; the problem boils down to a standard circuit with a 240V voltage source and three resistors (the "heaters). The question deals with the concept of power in circuit elements. If you recall, from the lecture sequences, if we have an element with am associated voltage v and current i, the equation is Power = i*v. That knowledge alone should be enough to get you through at least the first half. If you are still confused please feel free to elaborate. FirstChildUserIdTAG: 390034 FirstChildUserNameTAG: kahlil FirstChildCreateTimeTAG: 2012-09-09T15:28:06Z SecondChildTAG: I am confused in Last 3 questions.I am not getting how do I Proceed with individual power.If I go with the given figures.And calculate Rtotal then the power is same.Can you please guide me in this?I think I am going wrong somewhere. SecondChildUserIdTAG: 281159 SecondChildUserNameTAG: ManasiS SecondChildCreateTimeTAG: 2012-09-14T01:00:59Z FirstChildTAG: The idea based on the fact that if you know power dissipation (on the element) and voltage (on the element ) you can calculate the current (through the element) and his resistance. FirstChildUserIdTAG: 277168 FirstChildUserNameTAG: AndrewKiselev FirstChildCreateTimeTAG: 2012-09-09T15:29:15Z FirstChildTAG: The big, big picture lesson from this HW question is that when you look at an Ohm's Law Wheel, like this: http://www.ohmslawcalculator.com/ohms_law_wheel.php, that you understand that each aspect of electricity cannot exist without the others, and that is the reason you can derive the unknown aspects from the known aspects. Quantify the known, solve for the unknown. You can do this homework question with just the Ohms Law Wheel. FirstChildUserIdTAG: 21541 FirstChildUserNameTAG: JSChambers FirstChildCreateTimeTAG: 2012-09-09T16:22:58Z IndexTAG: 4593 TitleTAG: Bug: Math Processing Error Bug: Sample Numeric Problem page shows Math Processing Error before I could enter the answer. Please help UserIdTAG: 300213 UserNameTAG: Artimehta CreateTimeTAG: 2012-09-09T14:57:59Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: I had the same problem with IE. Try to use other browser FirstChildUserIdTAG: 205311 FirstChildUserNameTAG: Sergtronix FirstChildCreateTimeTAG: 2012-09-09T18:45:57Z FirstChildTAG: I answered that question here: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_S2E2_Associated_Reference_Directions/threads/504a0c61afe5cd2300000010#504c0c382a555b1f0000000d (Search is your friend, please use search to find answers before posting duplicate questions) That error is due to how your browser interacts with MathJax. MathJax is used by the server to present mathematical symbols on your screen. Here's how to fix it: 1) Place your cursor on the *[Math Processing Error]* message and right-click. 2) You will see a pull-down MathJax menu. 3) Select Math Settings, Math Renderer, and choose either MathML or SVG. I prefer the way MathML looks, but it drops some symbols when I print, so I have to use SVG. 4) You can also play with the zoom settings on the MathJax menu to adjust the zoom functionality. I hope this helps. FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-09-10T05:35:57Z IndexTAG: 4594 TitleTAG: HELP!!! LAB 1 I mistakenly deleted the node label A and now cant get anything done while clicking on the check botton.. That is, i cant get tha lab1 graded though it is saving but while pressing check button it isn't yielding anything. UserIdTAG: 102370 UserNameTAG: aninda CreateTimeTAG: 2012-09-09T14:37:49Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: I found I had to re-draw it after checking mistakes, maybe "reset" and draw it again. It's easy enough to draw quickly if you remember the values for R. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-09T14:44:22Z SecondChildTAG: The reset button isn't appearing for me.. SecondChildUserIdTAG: 102370 SecondChildUserNameTAG: aninda SecondChildCreateTimeTAG: 2012-09-09T14:51:20Z IndexTAG: 4595 TitleTAG: answers for the home work sir, will the answers to these questions be shown up after the deadline?. if it is, it will be useful, sir. UserIdTAG: 368150 UserNameTAG: vasanthan3886 CreateTimeTAG: 2012-09-09T14:35:34Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4596 TitleTAG: Mesh Analysis I am trying to apply mesh analysis but i cant get my answers right :( UserIdTAG: 336001 UserNameTAG: syd_buet12 CreateTimeTAG: 2012-09-09T11:50:34Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL/KCL example CommentableIdTAG: 6002x_S2E3_Using_KVL_KCL_and_VI_constraints NumberOfReplyTAG: 2 FirstChildTAG: Dont worry now i am getting it :D FirstChildUserIdTAG: 336001 FirstChildUserNameTAG: syd_buet12 FirstChildCreateTimeTAG: 2012-09-09T11:53:30Z SecondChildTAG: Then go around the forums and help others! :D SecondChildUserIdTAG: 79361 SecondChildUserNameTAG: kimt SecondChildCreateTimeTAG: 2012-09-09T12:58:37Z FirstChildTAG: just use kvl and kcl assign arbitrary variables for i1 r1 also for I ull get set of equations with kvl and kcl now give values of r to kvl equations ull get them in i1 and i2 now use kcl equation and solve both ull get i1 and i2 hnce the answer use simply kcl and kvl FirstChildUserIdTAG: 375082 FirstChildUserNameTAG: satya1889 FirstChildCreateTimeTAG: 2012-09-09T16:30:46Z IndexTAG: 4597 TitleTAG: Total Source Voltage Sources are like two batteries +5 V and +7.2 V connected with ends reversed. Net voltage is 2.2 V. Resistance is 6800 + 5600 = 12400 Ohms. Current i= 2.2/12400 amps. e=7.2-(2.2/12400)*5600 = 6.206. Current flow is determined by the source with larger supply voltage. Negative sign may mean the same thing as polarity reversed and it may be redundant. UserIdTAG: 380229 UserNameTAG: kkappagantula CreateTimeTAG: 2012-09-09T11:41:50Z VoteTAG: 0 CoursewareTAG: Week 2 / Node equation review CommentableIdTAG: 6002x_S3E1_NodeEquationReview NumberOfReplyTAG: 1 FirstChildTAG: When I put my two batteries in to photoapparat. The positions of them are reversed to each others. But I see that net voltage is the sum of each. How do you think about it ? FirstChildUserIdTAG: 333731 FirstChildUserNameTAG: Timophei_NhaTrang FirstChildCreateTimeTAG: 2012-09-13T06:50:52Z SecondChildTAG: net voltage=0 =) KVL SecondChildUserIdTAG: 210380 SecondChildUserNameTAG: MurdocRus SecondChildCreateTimeTAG: 2012-09-16T13:31:32Z IndexTAG: 4598 TitleTAG: S1E6 KVL part 3 Write an algebraic expression for the branch voltage v5 I did not understand the answer that how v1+v2-**V**=v5 can any one guide me? UserIdTAG: 40209 UserNameTAG: sadaf78 CreateTimeTAG: 2012-09-09T11:40:42Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Do KVL in upper loop.-v4+v5-v1=0.substitute value of v4.v4 can be found by applying KVL in outer large loop.-v4+v2-v=0 FirstChildUserIdTAG: 55345 FirstChildUserNameTAG: sarmaji FirstChildCreateTimeTAG: 2012-09-09T13:03:56Z SecondChildTAG: THANK YOU SecondChildUserIdTAG: 40209 SecondChildUserNameTAG: sadaf78 SecondChildCreateTimeTAG: 2012-09-09T18:54:17Z IndexTAG: 4599 TitleTAG: Lower Case vs Upper Case Hi, I'm finding this hard to grasp... We have a Voltage source labeled "v" and also "V". What is the difference? if then V=1.5V does v not also equal 1.5V? UserIdTAG: 37348 UserNameTAG: alecjcook CreateTimeTAG: 2012-09-09T09:49:35Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: (assuming this is coming from S1V6) As far as I understand, V refers to a specific property of the voltage source whereas when you're talking about little v it's concerning the v-i relationship more generally. So yes, since for a voltage source v = V, then V=1.5V implies v=1.5V. ...that wasn't very helpful, sorry. Hopefully someone with a better understanding will give a more succinct answer. FirstChildUserIdTAG: 403503 FirstChildUserNameTAG: jmohrmann FirstChildCreateTimeTAG: 2012-09-09T11:34:30Z IndexTAG: 4600 TitleTAG: video unable to play video of S3V3 UserIdTAG: 206368 UserNameTAG: mani48 CreateTimeTAG: 2012-09-09T09:49:28Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4601 TitleTAG: transient analysis when i had clicked the button of the transient analysis it is showing no probes in the diagram error.. UserIdTAG: 130826 UserNameTAG: BILALAMIN90 CreateTimeTAG: 2012-09-09T08:54:23Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 3 FirstChildTAG: same error is with me FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-09T09:19:53Z FirstChildTAG: u have to put probes. on ckt FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-09T15:27:25Z FirstChildTAG: i could not understand transient analysis . pl help me FirstChildUserIdTAG: 181266 FirstChildUserNameTAG: abbasedayadan FirstChildCreateTimeTAG: 2012-09-13T07:57:43Z IndexTAG: 4602 TitleTAG: Found v3 you shall write out 2 calculate: 1/ (e1-e2)/r1-v0=0 2/ e3/r2+i+v0=0 then applied e1=e3+v0 finally got the result v3 UserIdTAG: 265607 UserNameTAG: Harnette CreateTimeTAG: 2012-09-09T08:22:00Z VoteTAG: 0 CoursewareTAG: Week 1 / Node method example CommentableIdTAG: 6002x_S2E5_Node_Method NumberOfReplyTAG: 0 IndexTAG: 4603 TitleTAG: What is the equivalent resistance as an algebraic expression (in terms of R) of network as C viewed from its port? R+(1/(1/R)+1/(1/2R)) how i write the answer because system is not accepting the answer even if i write 5R/3. Please help me. UserIdTAG: 347789 UserNameTAG: engr_zohaib CreateTimeTAG: 2012-09-09T07:58:00Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 3 FirstChildTAG: try 5*R/3 FirstChildUserIdTAG: 230943 FirstChildUserNameTAG: samsung FirstChildCreateTimeTAG: 2012-09-09T08:14:58Z FirstChildTAG: write it as (5/3)*R.... FirstChildUserIdTAG: 334613 FirstChildUserNameTAG: aswinshankar FirstChildCreateTimeTAG: 2012-09-09T08:15:28Z FirstChildTAG: You forget the operator "*", so please enter 5*R/3 instead of 5R/3. FirstChildUserIdTAG: 18430 FirstChildUserNameTAG: ikeli FirstChildCreateTimeTAG: 2012-09-09T08:15:48Z IndexTAG: 4604 TitleTAG: independent kcl or kvl? when it comes to independent kcl or kvl eqns, if one can be derived from others then that wont b an independent one... so does that mean that if i get , v1=v2----(1) v2-v3=4-----(2) v1-v3=4------(3) none of these are independent equations??? as dey can b derived from other 2?? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-09T07:15:36Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: oh.... just now i read dat old post.. got it.. thnx... FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-09T07:18:35Z IndexTAG: 4605 TitleTAG: Figure 3.1b in Textbook In the textbook Sect. 3.2 - The Node Voltage when considering Fig. 3.1b how does one notice seemingly almost by mere observation that the node voltage e_b = 1.5 V ? The simplest way that I see to convince myself of that is by superposition, but that is not introduced until Sect. 3.5 - Superposition. In the meantime, should I just accept this as a given (though not initially stated as such) or am I missing something? UserIdTAG: 188609 UserNameTAG: Laureen CreateTimeTAG: 2012-09-09T06:17:54Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4606 TitleTAG: Q2 HOW can we get the average power dissipated by the resistance. anyone care to explain it ? UserIdTAG: 116433 UserNameTAG: Milkyasd CreateTimeTAG: 2012-09-09T05:58:14Z VoteTAG: 0 CoursewareTAG: Week 1 / AC power CommentableIdTAG: 6002x_S1E3_AC_power NumberOfReplyTAG: 6 FirstChildTAG: Average power = 1/T { ( INTEGRAL FROM 0 TO T ) iv dt } FirstChildUserIdTAG: 209930 FirstChildUserNameTAG: saikiraniitr FirstChildCreateTimeTAG: 2012-09-09T06:14:13Z FirstChildTAG: In period of time dt->0 resistor dissipate power P=(u(t))^2*dt/R, u(t) - value of voltage in time T. if we get periodical signal of u(t), average power is equal to (Q) integral by t t=0, t=T(period) (u(t))^2/R. Then, Q/T - average power dissipation on resistor. Sorry for poor english ) FirstChildUserIdTAG: 394836 FirstChildUserNameTAG: v2g6ch4 FirstChildCreateTimeTAG: 2012-09-09T06:05:12Z FirstChildTAG: first we calculate average voltage in a cycle which is V= V/sqrt2 here V is 120 sqrt2 then you can use the normal formula i.e. P=(V^2)/R FirstChildUserIdTAG: 160677 FirstChildUserNameTAG: ARMACK FirstChildCreateTimeTAG: 2012-09-09T06:15:57Z FirstChildTAG: ![enter image description here][1] [1]: http://s10.postimage.org/ghqxzesgp/image.png Maybe thats help ) FirstChildUserIdTAG: 394836 FirstChildUserNameTAG: v2g6ch4 FirstChildCreateTimeTAG: 2012-09-09T06:21:29Z SecondChildTAG: S- total dissipated energy by time T, S/T - average power dissipation SecondChildUserIdTAG: 394836 SecondChildUserNameTAG: v2g6ch4 SecondChildCreateTimeTAG: 2012-09-09T06:25:55Z FirstChildTAG: average power is same as I^2*R or V^2\R but this is true only for DC for ac the same formulae are used but for AC avg voltage is known as RMS (root mean square) and is calculated as Vrms = 0.7*Vpeak & Vpeak = 1.4Vrms for sine or cosine waves std form is Vpeak*cos(2*pie*freq*t) thus Peak power is (Vpeak)^2\R and Avg power is (0.7*Vpeak)^2\R FirstChildUserIdTAG: 232499 FirstChildUserNameTAG: Asim09 FirstChildCreateTimeTAG: 2012-09-10T17:40:41Z FirstChildTAG: its simple... think like this.. power(t) = V(t)^2/R if V(t) = A.cos(theta) => power(t) = (A^2)/R * (cos(theta))^2 as (A^2)/R is a constant term.. avg value of power will be governed by the avg value of cos(theta)^2.. which is 1/2.. so, power(avg) = costant_part * 1/2..thats it.. FirstChildUserIdTAG: 15237 FirstChildUserNameTAG: anubhavsinha FirstChildCreateTimeTAG: 2012-09-10T20:37:07Z IndexTAG: 4607 TitleTAG: How to write correct answer Hi, how to write correct answer (R*R)/(R+R) or 1/R+1/R for parallel resistors? UserIdTAG: 327723 UserNameTAG: Useris CreateTimeTAG: 2012-09-09T05:36:00Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 5 FirstChildTAG: i think if there are 4 resistances connected in series then we just need to Right 4R. Regards FirstChildUserIdTAG: 347789 FirstChildUserNameTAG: engr_zohaib FirstChildCreateTimeTAG: 2012-09-09T06:33:40Z FirstChildTAG: Remember algebra? When multipling exponent with same base, write base and add exponent. example: X + X = 2X When adding exponent with same base, add how many you have. example: X*X = X^2 Let me know if this helps. FirstChildUserIdTAG: 303762 FirstChildUserNameTAG: leocarrasco FirstChildCreateTimeTAG: 2012-09-09T05:47:46Z FirstChildTAG: if all the resistance values are same you must write it as the value of resistance divided by the number of resistance.......the general formula is r/n where n is the number of resistors FirstChildUserIdTAG: 278577 FirstChildUserNameTAG: dhoni FirstChildCreateTimeTAG: 2012-09-09T06:04:57Z SecondChildTAG: Thank's SecondChildUserIdTAG: 327723 SecondChildUserNameTAG: Useris SecondChildCreateTimeTAG: 2012-09-09T06:07:03Z FirstChildTAG: i think if there are 4 resistances connected in series then we just need to write 4R. Regards FirstChildUserIdTAG: 347789 FirstChildUserNameTAG: engr_zohaib FirstChildCreateTimeTAG: 2012-09-09T06:36:07Z FirstChildTAG: write it as R/2 or 2/R.. FirstChildUserIdTAG: 334613 FirstChildUserNameTAG: aswinshankar FirstChildCreateTimeTAG: 2012-09-09T06:40:04Z SecondChildTAG: how i write this one : R+(1/(1/R)+1/(1/2R)) plz help me. SecondChildUserIdTAG: 347789 SecondChildUserNameTAG: engr_zohaib SecondChildCreateTimeTAG: 2012-09-09T07:59:46Z IndexTAG: 4608 TitleTAG: Textbook Reader I have observed that the textbook reading website quite often gets the next page wrong when using the arrow operators. For instance, I went (in one click of the right arrow) from reading the author's preface (where the chapter overviews are given) to page 186 in Chapter Three. Now, the slowness of delivery is one issue - I need a wireless connection, via microwave, to a tower about 16 miles from my home (I live in a very rural part of Eastern Washington) - but, having problems selecting on a page by page basis will present a strong difficulty in using the text to complete the course. I can image the greater burden on persons not so close to MIT. UserIdTAG: 194450 UserNameTAG: wrbuckley CreateTimeTAG: 2012-09-09T04:54:58Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: A PDF version is available online if you search for the title. Save it to your computer. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-09T14:34:21Z IndexTAG: 4609 TitleTAG: Got everything right except.. Voltage for node C I got 666.504 and the answer was .665. Anyone know why? UserIdTAG: 175876 UserNameTAG: Sunden CreateTimeTAG: 2012-09-09T04:52:41Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 6 FirstChildTAG: You got 666.504 m which is = .666 FirstChildUserIdTAG: 434043 FirstChildUserNameTAG: EslamSameh FirstChildCreateTimeTAG: 2012-09-14T16:52:59Z FirstChildTAG: Make sure that your units are correct. mili-Volts (mV) vs Volts (V) Remember that v=ir (Volts=Amperes*Ohms) so verify that your Amps are not mili-Amps and/or Ohms instead of mili-Ohms. Hope this helps FirstChildUserIdTAG: 303762 FirstChildUserNameTAG: leocarrasco FirstChildCreateTimeTAG: 2012-09-09T05:14:15Z FirstChildTAG: The plot actually displays 666.504m, which indicates it is in milli-volts. The other numbers don't display a letter, which means they're in volts. FirstChildUserIdTAG: 268226 FirstChildUserNameTAG: Mengo7 FirstChildCreateTimeTAG: 2012-09-09T06:09:40Z FirstChildTAG: Sunden if you observe the Transient Analisis graphic you can note the reading says: "m" where "m" means "mili" (1/1000) FirstChildUserIdTAG: 397078 FirstChildUserNameTAG: FJIHERRERA1 FirstChildCreateTimeTAG: 2012-09-09T06:42:07Z FirstChildTAG: It probably says 666.504mV, which is ~0.665V. FirstChildUserIdTAG: 306244 FirstChildUserNameTAG: kevinsysum FirstChildCreateTimeTAG: 2012-09-09T12:01:04Z SecondChildTAG: Hi all, I realise everyone here is jumping on Sunden as he did not type the full unit however... 666.504 /1000 is not 0.665V it's 0.666504 Even rounded to three digits it would be 0.666 so why is listed as .665?? or is it just trying to avoid the number of the devil from appearing?? !!! SecondChildUserIdTAG: 80230 SecondChildUserNameTAG: niod SecondChildCreateTimeTAG: 2012-09-10T09:21:08Z FirstChildTAG: hey even I am having the same problem over here.What am i supposed to do?I have done everything correctly but still I keep getting the wrong answer?Can anyone help me? FirstChildUserIdTAG: 333783 FirstChildUserNameTAG: Sruthi_Suresh FirstChildCreateTimeTAG: 2012-09-14T14:35:00Z IndexTAG: 4610 TitleTAG: Simulator On Tablet I am trying to use the simulator for lab 1 using an ipad2, but I am unable to drag components into the circuit. Has anyone else had luck using the simulator on a tablet? Please share any tricks. Thanks! UserIdTAG: 198899 UserNameTAG: trids64 CreateTimeTAG: 2012-09-09T04:52:23Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Hi, I tried to use the circuit simulator on a tablet during the last offering of 6.002x and found it not possible :( FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-09T12:10:09Z IndexTAG: 4611 TitleTAG: Ohm's law for resistance in parallel? Why is the total resistance the sum and not the product/sum (ohm's law for parallel resistance) of the 2 wires? UserIdTAG: 139810 UserNameTAG: nachumk CreateTimeTAG: 2012-09-09T04:22:20Z VoteTAG: 0 CoursewareTAG: Week 1 / Power delivery modeling CommentableIdTAG: 6002x_S2E6_Modeling NumberOfReplyTAG: 2 FirstChildTAG: ![It's looks like this:][1] [1]: http://s10.postimage.org/sckqr0brt/image.png FirstChildUserIdTAG: 394836 FirstChildUserNameTAG: v2g6ch4 FirstChildCreateTimeTAG: 2012-09-09T05:09:14Z SecondChildTAG: is this what 12 AWG looks like? i am not sure if 0.179ohm should times 2 (in series) or devided (in parallel) by 2. SecondChildUserIdTAG: 176234 SecondChildUserNameTAG: hejinjie SecondChildCreateTimeTAG: 2012-09-10T09:04:47Z FirstChildTAG: The wires are in series with each other with respect to the house. The current must pass through the first wire and out the second, making them in series with each other. Parallel resistances start and end at the same node, so if the barn was ground you would be correct. FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-11T08:50:43Z IndexTAG: 4612 TitleTAG: could not parse my answer is not been recognized by the system no matter how many times i write UserIdTAG: 156713 UserNameTAG: HARITEJAREDDYP CreateTimeTAG: 2012-09-09T03:16:25Z VoteTAG: 0 CoursewareTAG: Week 1 / Node analysis practice, part 3 CommentableIdTAG: 6002x_L2Node2 NumberOfReplyTAG: 1 FirstChildTAG: when u put a answer in do it as 4*R not 4R and so on FirstChildUserIdTAG: 246951 FirstChildUserNameTAG: justin_petty FirstChildCreateTimeTAG: 2012-09-09T03:47:14Z SecondChildTAG: and instead of A+B=C, just type A+B. as told above. SecondChildUserIdTAG: 401175 SecondChildUserNameTAG: Raghav12 SecondChildCreateTimeTAG: 2012-09-14T15:05:14Z SecondChildTAG: I answer the question with (e2-V0)*R4 + (e2-e1)*R3 + (e2*R5) - I1, but the system not accept. SecondChildUserIdTAG: 457034 SecondChildUserNameTAG: Ichihara SecondChildCreateTimeTAG: 2012-09-19T14:01:30Z SecondChildTAG: It should be (e2-V0)/R4... Divided by, not times SecondChildUserIdTAG: 600559 SecondChildUserNameTAG: SergioFP SecondChildCreateTimeTAG: 2012-10-15T17:25:30Z IndexTAG: 4613 TitleTAG: h1p1 network 3 I'm not sure how to put together all of the resistors. I add the 2 series resistors and get 2R the add the 2 parallel together and (R^2)/2. I put the together and get a different answer every time. Sometimes I get (5R)/2, sometimes I get (5R^2)/2R. Then sometimes I get a completely different answer that off the wall. I'm lost. Plz help. P.S. I know that they have to be in equation form, like (5*R)/2 or R/5 UserIdTAG: 246951 UserNameTAG: justin_petty CreateTimeTAG: 2012-09-09T02:55:37Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: I finally got it. If you read this and your having trouble got to http://www.electronics-tutorials.ws/resistor/res_5.html It helped a lot FirstChildUserIdTAG: 246951 FirstChildUserNameTAG: justin_petty FirstChildCreateTimeTAG: 2012-09-09T04:04:10Z IndexTAG: 4614 TitleTAG: R3 positive negative terminal What determines which side of the resistor is positive or negative? Why does R3 get the positive (+) on the left side and the negative (-) on the right side? UserIdTAG: 139810 UserNameTAG: nachumk CreateTimeTAG: 2012-09-09T00:45:21Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL and KCL CommentableIdTAG: 6002x_kvl_and_kcl NumberOfReplyTAG: 1 FirstChildTAG: I believe the assignment of positive and negative sides is arbitrary. I think the key is to make sure that when you assign a variable for current, that current flows into the positive side. So if you reversed the positive and negative values for R3, you would have to show the current as flowing into the right side of R3 instead of the left. Try to work it out with the signs on R3 reversed, and you should get the same results. FirstChildUserIdTAG: 142402 FirstChildUserNameTAG: aaronrod FirstChildCreateTimeTAG: 2012-09-09T01:19:03Z IndexTAG: 4615 TitleTAG: V2 sign confusion The question is about what potential is at the right of R2? V2 is -7.2V from bottom to top. So from top to down it is +7.2V. That's why we should subtract +7.2V from e. Where am I wrong? What statement? UserIdTAG: 208296 UserNameTAG: OZ1 CreateTimeTAG: 2012-09-08T23:57:24Z VoteTAG: 0 CoursewareTAG: Week 2 / Node equation review CommentableIdTAG: 6002x_S3E1_NodeEquationReview NumberOfReplyTAG: 3 FirstChildTAG: Apply node analysis for two leaving currents from e. FirstChildUserIdTAG: 55345 FirstChildUserNameTAG: sarmaji FirstChildCreateTimeTAG: 2012-09-09T01:27:00Z FirstChildTAG: From my point of view V2=-7.2v it's from TOP to Bottom!! (according to - and + signs at the PowerSupply element diagram) and it's means that TOP has +7.2v (from ground). I agree with you that it's very strange. I think that it deal with wide interpretation of possible "voltages" off power supply in this emulation software. But in real live all power supplies are marked by + and - in such manner when + contact is always has more high potential than - contact. FirstChildUserIdTAG: 277168 FirstChildUserNameTAG: AndrewKiselev FirstChildCreateTimeTAG: 2012-09-09T11:06:05Z FirstChildTAG: The problem is where you choose your ground node. If you choose your ground node at -7.2 to be 0v the other side has to be 0--7.2 = 7.2v so your answer is right. FirstChildUserIdTAG: 244352 FirstChildUserNameTAG: foo1226 FirstChildCreateTimeTAG: 2012-09-10T20:01:42Z IndexTAG: 4616 TitleTAG: Reverse In the example with voltages at the end of the video there is a small video error. 1V signify a logical '0' and 3V signify a logical '1' value. UserIdTAG: 304587 UserNameTAG: Vasso CreateTimeTAG: 2012-09-08T22:52:28Z VoteTAG: 0 CoursewareTAG: Week 2 / Static Discipline CommentableIdTAG: 6002x_static_discipline NumberOfReplyTAG: 1 FirstChildTAG: i think it is your wish. If u want u can take 1V as logical '0' and 3V as logical '1' or 3V as logical '0' and 1V as logical '1'. The former is known as 'positive logic' and the latter is 'negative logic'. FirstChildUserIdTAG: 132685 FirstChildUserNameTAG: deepakmurali FirstChildCreateTimeTAG: 2012-09-09T06:21:35Z IndexTAG: 4617 TitleTAG: S2E4 can anybody explain why v4= 0.33 ? UserIdTAG: 281159 UserNameTAG: ManasiS CreateTimeTAG: 2012-09-08T22:28:57Z VoteTAG: 0 CoursewareTAG: Week 1 / Series and parallel example CommentableIdTAG: 6002x_S2E4_Series_and_Parallel NumberOfReplyTAG: 5 FirstChildTAG: i1=V/Requivalent.apply KVL to find i2 FirstChildUserIdTAG: 55345 FirstChildUserNameTAG: sarmaji FirstChildCreateTimeTAG: 2012-09-09T01:32:49Z FirstChildTAG: i1=V/Requivalent.apply KVL to find i2. FirstChildUserIdTAG: 55345 FirstChildUserNameTAG: sarmaji FirstChildCreateTimeTAG: 2012-09-09T01:33:02Z FirstChildTAG: i4=i1-i2.v4=i4*R4 FirstChildUserIdTAG: 55345 FirstChildUserNameTAG: sarmaji FirstChildCreateTimeTAG: 2012-09-09T01:31:21Z FirstChildTAG: This is couse the V in the Rt=r3 +r4 is equal at the voltage in R2 and r3 and r4 are =... FirstChildUserIdTAG: 92624 FirstChildUserNameTAG: Adyprado FirstChildCreateTimeTAG: 2012-09-08T22:53:59Z FirstChildTAG: applying KCL i3= i1-i2 so i3 = 0.333333-0.16666666 = 0.166666666 now v4 = R4(i4) = 2(0.16666666666) = 0.333333333 FirstChildUserIdTAG: 187265 FirstChildUserNameTAG: ferrerfad FirstChildCreateTimeTAG: 2012-09-11T14:54:44Z IndexTAG: 4618 TitleTAG: Circuit Sandbox I try to use FETs, and i guess it doesn't work properly. Is it true, or I just don't understand something? UserIdTAG: 191310 UserNameTAG: Kalashnikov CreateTimeTAG: 2012-09-08T22:24:14Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4619 TitleTAG: Current Negative ? I dont Understand beacuse the current its negative its rare! UserIdTAG: 257168 UserNameTAG: Valenciamc CreateTimeTAG: 2012-09-08T22:01:24Z VoteTAG: 0 CoursewareTAG: Week 1 / Power CommentableIdTAG: 6002x_S1E2_Power NumberOfReplyTAG: 3 FirstChildTAG: because when the polarity of the element is - V + and current go to right (i→) intro in the terminal negative FirstChildUserIdTAG: 318210 FirstChildUserNameTAG: Alvin12 FirstChildCreateTimeTAG: 2012-09-09T06:40:17Z FirstChildTAG: porque cuando la polaridad de el elemento es - v + y la corriente va hacia la derecha entra en el terminal negativo FirstChildUserIdTAG: 318210 FirstChildUserNameTAG: Alvin12 FirstChildCreateTimeTAG: 2012-09-09T06:43:02Z FirstChildTAG: It is really quite simple if under the given set of rules, voltage is traveling from + toward - . then the current would be traveling in the opposite direction also shown in diagram. so if the question is about current and not voltage coming out from the node it will be negative . I think that is how to look at it. FirstChildUserIdTAG: 359302 FirstChildUserNameTAG: GaryG FirstChildCreateTimeTAG: 2012-09-10T23:21:51Z IndexTAG: 4620 TitleTAG: Future Courses Do you have any plan to launch more specialized courses like control systems/signal processing in near future? UserIdTAG: 198567 UserNameTAG: rafee1344 CreateTimeTAG: 2012-09-08T21:39:54Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4621 TitleTAG: H1P2: KCL/KVL VS. NODE METHOD AM NOT ABLE TO FIND OUT THE VALUE OF V1 AND I4 UserIdTAG: 364274 UserNameTAG: akshay_edx CreateTimeTAG: 2012-09-08T20:11:44Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: it's easier to do it by superposition, that way you just get a voltage divider and a current divider FirstChildUserIdTAG: 56070 FirstChildUserNameTAG: estebananaya FirstChildCreateTimeTAG: 2012-09-08T20:42:13Z FirstChildTAG: apply node eqn at the junction where current i2 and i3 leaving...!! FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-08T21:03:30Z SecondChildTAG: how do you aplly node eqn? SecondChildUserIdTAG: 348251 SecondChildUserNameTAG: PaulchenPanther SecondChildCreateTimeTAG: 2012-09-16T21:51:55Z IndexTAG: 4622 TitleTAG: [Correction] No circuit diagram!! in H1P2: KCL/KVL vs. Node Method There seems to be an error or it is stupid me but there is no circuit in the sixth question in homework 1 (H1P2: KCL/KVL vs. Node Method) the line says: Using either the KCL/KVL method or the node method, **solve this circuit** The value (in Volts) of v1 is: (where is the circuit??) or do i need to consider the circuit given above?? Thanks! UserIdTAG: 232499 UserNameTAG: Asim09 CreateTimeTAG: 2012-09-08T17:50:04Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Yes, You should have to use the Circuit above. That is to say this one:![enter image description here][1] Remember that Students can have different values. So, you can have the same Circuit but with different values.This is the Circuit that was given in my Statement. I hope this Can help you :) [1]: https://www.edx.org/static/content-mit-6002x/images/circuits/H1P2.79e52b492b94.png ---- Sí, debes utilizar el Circuito que tienes arriba. Es decir, este: ![enter image description here][1] Recuerda que los Alumnos podemos tener valores diferentes en nuestros Homeworks. Es por ello, que puedes tener el mismo Circuito que yo pero con diferentes valores en los componentes.Espero que esto pueda serte de ayuda :) FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-08T18:23:23Z IndexTAG: 4623 TitleTAG: Onscreen calculator Dear technical team, greetings!! I would be so kind of you, if you can bring in a more advanced onscreen calculator. It makes life much easier!! UserIdTAG: 232499 UserNameTAG: Asim09 CreateTimeTAG: 2012-09-08T17:36:21Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4624 TitleTAG: Saving place How do I come back to location after quitting for the day? UserIdTAG: 137279 UserNameTAG: rogerp CreateTimeTAG: 2012-09-08T16:43:25Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4625 TitleTAG: lab 1 / confusion about voltages It seems like there is some confusions about the voltage at node A and the voltage across the bulb (Vs). Is the voltage at A the voltage across the bulb or the voltage with regard to the ground? If it's with regard to the ground, then when the bulb is removed, the voltage at A will always be 6 volts and Vs across the bulb opening will always be 6 volts also, if nothing replaces the opening. Can someone clarify this? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-08T16:26:51Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: The voltage at node A is in reference to ground. It is possible to have the voltage at A to be something other than 6V if you connect 2 resistors in the right way. I suggest playing around with the possible connections of 2 resistors to the open circuit as given, and see if you can get a potential at A other than 6V. FirstChildUserIdTAG: 310007 FirstChildUserNameTAG: ErinF FirstChildCreateTimeTAG: 2012-09-08T19:57:31Z SecondChildTAG: I figured out, thanks! SecondChildUserIdTAG: undefined SecondChildUserNameTAG: undefined SecondChildCreateTimeTAG: 2012-09-09T12:13:36Z IndexTAG: 4626 TitleTAG: remember me sorry but some one remember me why there is aminus sign in the -dq/dt equation UserIdTAG: 289035 UserNameTAG: Ayman2468 CreateTimeTAG: 2012-09-08T16:25:22Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL and KCL CommentableIdTAG: 6002x_kvl_and_kcl NumberOfReplyTAG: 0 IndexTAG: 4627 TitleTAG: S2E6 given: distance from the barn 113 feeet 12 AWG copper wire have resistance of 1.588 ohm per 1000 feet therefore, 12 Awg copper wire have resis of 1.588 /1000 per feet total distance = 113, thus tot rest = 1.588/1000 * 113 =0.17944 ohm since, circuit requires 2 lengths of the wire, thus total resis=0.17944*2=0.358 2) p=vi=v^2/R, 1000*R=v^2, v=240, R=57.6 3) i=v/Rt= 670.391, p=vi, v=1000/670.391=1.491 UserIdTAG: 132474 UserNameTAG: m_abid CreateTimeTAG: 2012-09-08T15:43:28Z VoteTAG: 0 CoursewareTAG: Week 1 / Power delivery modeling CommentableIdTAG: 6002x_S2E6_Modeling NumberOfReplyTAG: 1 FirstChildTAG: Omg. Is 670.391 supposed to be 670.391A? FirstChildUserIdTAG: 190618 FirstChildUserNameTAG: Kirbabaev FirstChildCreateTimeTAG: 2012-09-11T21:00:02Z SecondChildTAG: 670.391 is the short circuit current in the above equation. Ie.just short out the load in the barn. Yes it is high and in Amp SecondChildUserIdTAG: 264940 SecondChildUserNameTAG: fallinga SecondChildCreateTimeTAG: 2012-09-15T01:47:31Z IndexTAG: 4628 TitleTAG: loops without a volt source does KVL work in loop, ad, db , ba. UserIdTAG: 388273 UserNameTAG: kilmarta CreateTimeTAG: 2012-09-08T15:36:10Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL and KCL CommentableIdTAG: 6002x_kvl_and_kcl NumberOfReplyTAG: 1 FirstChildTAG: yes, KVL works in ANY loops FirstChildUserIdTAG: 394836 FirstChildUserNameTAG: v2g6ch4 FirstChildCreateTimeTAG: 2012-09-08T15:44:54Z IndexTAG: 4629 TitleTAG: Measurements slightly off With mine, I got it mostly right except the sin voltages I got were 2.000, 1.666, and 666.504 rather than the answers listed. It accepted the first two since they round correctly but the third was too far off. The second weird this was the current measurement I entered -500u for amps.. but it didn't accept the u.. only 500e-06. which is a bit of a pain. UserIdTAG: 132733 UserNameTAG: CapAnson CreateTimeTAG: 2012-09-08T15:20:15Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 3 FirstChildTAG: Observe carefully.....at C plot shows 666.504'm' and not 666.504. So, the answer should be 666.504m, where m stands for milli, which should be equal to 0.666504 FirstChildUserIdTAG: 395953 FirstChildUserNameTAG: dustinge FirstChildCreateTimeTAG: 2012-09-08T17:48:07Z FirstChildTAG: -500u is not equal to 500e-06. Notice that the first is negative and the second isn't. FirstChildUserIdTAG: 267993 FirstChildUserNameTAG: mavlijas FirstChildCreateTimeTAG: 2012-09-08T15:38:20Z FirstChildTAG: 5ms = .005 s Watch your units. FirstChildUserIdTAG: 183711 FirstChildUserNameTAG: MMatheson FirstChildCreateTimeTAG: 2012-09-08T23:56:43Z IndexTAG: 4630 TitleTAG: third postulate on lumped matter what is meant by electromagnetic propagation delay ? and why if the filament was enough long or dt was enough small we will have different in & out current ? UserIdTAG: 221617 UserNameTAG: konan CreateTimeTAG: 2012-09-08T14:40:42Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: Electromagnetic propagation delay is the time it takes a signal to travel from the start to the destination. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-08T15:03:05Z FirstChildTAG: Electromagnetic waves have a speed limit (think speed of light through some medium) and so do electrons. Think about sunlight reaching Earth. It takes a significant time for that light to reach us. In this course, we're only working with distances such that the light would reach us practically instantaneously. Think about a simple circuit where you just have a current source that is off connected to itself. Now turn the current source on. If we were observing the current at very very fast intervals (small dt) at some point far down the wire away from the current source, it could take many observations before we actually see any current, so the in current (current from the source) would not equal the out current (current viewed from the point on the wire). FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-08T15:11:23Z IndexTAG: 4631 TitleTAG: parser issues my correct answer (per next page) is not accepted! many variations of this were not evaluated by the page. (did not say wrong/right) (e2-e1)/R3+(e2-V0)/R4+e2/R5-I1=0 UserIdTAG: 279379 UserNameTAG: RajaSrinivasan CreateTimeTAG: 2012-09-08T14:19:22Z VoteTAG: 0 CoursewareTAG: Week 1 / Node analysis practice, part 3 CommentableIdTAG: 6002x_L2Node2 NumberOfReplyTAG: 2 FirstChildTAG: If you leave out the "=0" part, it'll work. FirstChildUserIdTAG: 144694 FirstChildUserNameTAG: johndoe31415 FirstChildCreateTimeTAG: 2012-09-08T14:33:24Z SecondChildTAG: thanks, just could not see it. SecondChildUserIdTAG: 183507 SecondChildUserNameTAG: obiradaniel SecondChildCreateTimeTAG: 2012-09-09T14:04:23Z FirstChildTAG: You should just write your expression without '=0' FirstChildUserIdTAG: 156568 FirstChildUserNameTAG: DO_SL FirstChildCreateTimeTAG: 2012-09-08T20:37:39Z SecondChildTAG: Sorry, I didn't see previous comment. SecondChildUserIdTAG: 156568 SecondChildUserNameTAG: DO_SL SecondChildCreateTimeTAG: 2012-09-08T20:39:00Z SecondChildTAG: Thanks SecondChildUserIdTAG: 254346 SecondChildUserNameTAG: moijes12 SecondChildCreateTimeTAG: 2012-09-12T04:52:15Z IndexTAG: 4632 TitleTAG: The individual state report! I think that we should have an **item which shows participant's process of the course**. For instance, the participant can check their process when they need: **the homework and labs submission** (to ensure their assigns are done or not), **the result of their course at the moment** (to know what mark they got). I think that this is feasible because the web resource needed for this is fairly tiny. UserIdTAG: 378675 UserNameTAG: NhatVu CreateTimeTAG: 2012-09-08T12:23:15Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: It's under the "Progress" Tab, near the top of this and most other pages. FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-08T13:16:39Z IndexTAG: 4633 TitleTAG: If you got a problem People, try first to make DC analysis with DC source and after this change DC source to sin and make transient analysis UserIdTAG: 214584 UserNameTAG: Askeroff CreateTimeTAG: 2012-09-08T12:16:49Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 0 IndexTAG: 4634 TitleTAG: 1 week tutorial videos don't work Why the 1 week tutorials don't work? Actually, i can't watch videos. UserIdTAG: 92315 UserNameTAG: GooCoder CreateTimeTAG: 2012-09-08T12:01:38Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Did you forget adobe flash player? ) FirstChildUserIdTAG: 394836 FirstChildUserNameTAG: v2g6ch4 FirstChildCreateTimeTAG: 2012-09-08T12:03:34Z SecondChildTAG: i have adobe fp. SecondChildUserIdTAG: 92315 SecondChildUserNameTAG: GooCoder SecondChildCreateTimeTAG: 2012-09-10T09:57:38Z IndexTAG: 4635 TitleTAG: Geographical position Hey, it's so boring... Let's talk about globalization ). Where are you from, peoples? ) Vote for theme ) Minuses? Are u hate Russia? xDD UserIdTAG: 394836 UserNameTAG: v2g6ch4 CreateTimeTAG: 2012-09-08T11:42:26Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 4 FirstChildTAG: I'm from Russia. Sakha republic. My city within 10 coldest places of the world :) FirstChildUserIdTAG: 394836 FirstChildUserNameTAG: v2g6ch4 FirstChildCreateTimeTAG: 2012-09-08T11:45:02Z FirstChildTAG: i m from new delhi..india :-) FirstChildUserIdTAG: 126705 FirstChildUserNameTAG: arpitchugh FirstChildCreateTimeTAG: 2012-09-08T11:49:09Z FirstChildTAG: Ontario, Canada FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-08T13:19:02Z FirstChildTAG: I am from IIT Delhi India, FirstChildUserIdTAG: 232499 FirstChildUserNameTAG: Asim09 FirstChildCreateTimeTAG: 2012-09-22T19:45:23Z IndexTAG: 4636 TitleTAG: What a mess I think they want to say -7.2V without considering that V2 is reversed so it's like a +7.2V source. What a mess!! UserIdTAG: 359677 UserNameTAG: guillermb CreateTimeTAG: 2012-09-08T11:24:57Z VoteTAG: 0 CoursewareTAG: Week 2 / Node equation review CommentableIdTAG: 6002x_S3E1_NodeEquationReview NumberOfReplyTAG: 1 FirstChildTAG: No, the question is correct. FirstChildUserIdTAG: 394836 FirstChildUserNameTAG: v2g6ch4 FirstChildCreateTimeTAG: 2012-09-08T11:33:06Z IndexTAG: 4637 TitleTAG: bug Video S2V11 stops at 4:19, but it is not complete. UserIdTAG: 146969 UserNameTAG: Andreas-N CreateTimeTAG: 2012-09-08T10:21:28Z VoteTAG: 0 CoursewareTAG: Week 1 / Node analysis CommentableIdTAG: 6002x_node_analysis NumberOfReplyTAG: 1 FirstChildTAG: Try speed 1.0X FirstChildUserIdTAG: 314386 FirstChildUserNameTAG: jid FirstChildCreateTimeTAG: 2012-09-10T21:20:27Z IndexTAG: 4638 TitleTAG: The sign convention In i4 , it follows as the prof. said "from +ve to -ve" so why it's -ve!!? UserIdTAG: 185715 UserNameTAG: amirengineer CreateTimeTAG: 2012-09-08T08:50:03Z VoteTAG: 0 CoursewareTAG: Week 1 / Associated reference directions CommentableIdTAG: 6002x_S2E2_Associated_Reference_Directions NumberOfReplyTAG: 1 FirstChildTAG: The physical current in this circuit flows in the clockwise direction: from the + end of the battery to the - end. All v's and i's where the direction from + end to - end are in the clockwise direction are positive (i2 and i3). The other ones (i1 and i4 ) are negative. The confusing difference is between the physical current (clockwise) and the symbolic naming of individual branch currents (from + to -). FirstChildUserIdTAG: 379036 FirstChildUserNameTAG: iefrat FirstChildCreateTimeTAG: 2012-09-08T15:22:33Z IndexTAG: 4639 TitleTAG: Can't understand the EDx tutorial! Should I be understanding all what's instructed in the tutorial about the circuit simulation?? I felt lost after he finished the DC analysis part and went into the next part! :( UserIdTAG: 398306 UserNameTAG: Soliman CreateTimeTAG: 2012-09-08T08:16:29Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Same here. FirstChildUserIdTAG: 230787 FirstChildUserNameTAG: Gabriel007 FirstChildCreateTimeTAG: 2012-09-08T14:06:47Z IndexTAG: 4640 TitleTAG: Easy Peazy I got almost Corrected. I think it would be nice if it so like this! UserIdTAG: 232546 UserNameTAG: MJMEENU CreateTimeTAG: 2012-09-08T07:15:55Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL/KCL example CommentableIdTAG: 6002x_S2E3_Using_KVL_KCL_and_VI_constraints NumberOfReplyTAG: 0 IndexTAG: 4641 TitleTAG: Cool Cool i am love with circuits !! UserIdTAG: 289949 UserNameTAG: manolito CreateTimeTAG: 2012-09-08T06:02:39Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 0 IndexTAG: 4642 TitleTAG: *Bug* (Possibly) Transient analysis Transient analysis does not bring up the plot. Watched the tutorials, have done all other types of analysis (even AC analysis in sandbox), however the transient does not work. Some additional info: - Connected probes properly (on different nodes, with and without "node label"); all the probes are set to different colors - tried Chrome (latest - only installed to verify this problem), Firefox, IE - Flash plugin is up to date, checked with adobe website - OS info: Win 7, Ultimate, 64bit, SP1 - all updates are installed. - Hardware: ASUS N53-SN: Core i7 2630QM, 8Gig RAM, Graphic: GT550m, 1GB DDR3 VRAM, all drivers are up to date, and working fine. UserIdTAG: 43909 UserNameTAG: hhomayouni CreateTimeTAG: 2012-09-08T05:55:02Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 2 FirstChildTAG: yeah, i have also tried it is not working,so if you come up with any information please post it..... FirstChildUserIdTAG: 375731 FirstChildUserNameTAG: shubham5690 FirstChildCreateTimeTAG: 2012-09-08T19:53:04Z SecondChildTAG: Will do! SecondChildUserIdTAG: 43909 SecondChildUserNameTAG: hhomayouni SecondChildCreateTimeTAG: 2012-09-08T21:06:18Z FirstChildTAG: I tried adding "edx.org" to exception list of pop-up blocker in firefox, and it is now fixed! It's a bit strange as why 'Tran' is the only component that was affected (while AC analysis worked fine). There is also a possibility that they fixed the bug and it just coincided with me changing this setting !!! Who knows! Anyway, the version my firefox is 15.0.1, went to option -> content (TAB) -> clicked on 'exception' in front of the "block pop-up" and entered "edx.org" (without quotation marks) to the line, and clicked "allow", and then "OK" in the main option box. FirstChildUserIdTAG: 43909 FirstChildUserNameTAG: hhomayouni FirstChildCreateTimeTAG: 2012-09-08T21:26:20Z IndexTAG: 4643 TitleTAG: the easy way in question 2 there is no need to integrate you just put Vavg= Vmax*cos45 UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-08T05:00:07Z VoteTAG: 0 CoursewareTAG: Week 1 / AC power CommentableIdTAG: 6002x_S1E3_AC_power NumberOfReplyTAG: 3 FirstChildTAG: Doing it the integral way, allows you to understand it. FirstChildUserIdTAG: 193033 FirstChildUserNameTAG: NIslam FirstChildCreateTimeTAG: 2012-09-09T15:23:00Z FirstChildTAG: that`s the direct method to get the average voltage FirstChildUserIdTAG: 396071 FirstChildUserNameTAG: hamail FirstChildCreateTimeTAG: 2012-09-09T01:02:14Z FirstChildTAG: but note it`s Vavg= Vmax*cos45 due to the natural o fThe actual voltage. here is "Vmax⋅2√⋅cos(2π⋅60⋅t)" when u integrate it u will get u "Vavg= Vmax*cos45" if the actual is something else it will not work :) FirstChildUserIdTAG: 396071 FirstChildUserNameTAG: hamail FirstChildCreateTimeTAG: 2012-09-09T01:08:05Z IndexTAG: 4644 TitleTAG: cant see the video i have tried different browser (mozilla, google chrome,internet explorer) but cant see the tutorial video. can anybody help or is anybody experiencing same? UserIdTAG: 300989 UserNameTAG: seek_007 CreateTimeTAG: 2012-09-08T04:41:28Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: I use firefox and the videos won't play. Have you solved the problem? FirstChildUserIdTAG: 227139 FirstChildUserNameTAG: qiuyi FirstChildCreateTimeTAG: 2012-09-08T07:55:17Z SecondChildTAG: i have not solved the problem, am searching for the video on you tube SecondChildUserIdTAG: 300989 SecondChildUserNameTAG: seek_007 SecondChildCreateTimeTAG: 2012-09-08T13:40:03Z IndexTAG: 4645 TitleTAG: current flow i think we are using Conventional Current Flow here.. UserIdTAG: 247990 UserNameTAG: kenneth007 CreateTimeTAG: 2012-09-08T04:32:19Z VoteTAG: 0 CoursewareTAG: Week 1 / Associated reference directions CommentableIdTAG: 6002x_S2E2_Associated_Reference_Directions NumberOfReplyTAG: 0 IndexTAG: 4646 TitleTAG: Power source in a loop Is it necessary to have a power source in a loop? UserIdTAG: 373994 UserNameTAG: RoboSobo CreateTimeTAG: 2012-09-08T04:27:55Z VoteTAG: 0 CoursewareTAG: Week 1 / Circuit topology CommentableIdTAG: 6002x_S2E1_Circuit_Topology NumberOfReplyTAG: 2 FirstChildTAG: In fact, not. It is not necessary FirstChildUserIdTAG: 298653 FirstChildUserNameTAG: WilyP FirstChildCreateTimeTAG: 2012-09-08T05:19:02Z FirstChildTAG: Obviously, power source can be used in loop only as it is needed to complete the path of current flow, if it is not connected in loop it acts just as open circuit. FirstChildUserIdTAG: 309803 FirstChildUserNameTAG: PurnenduK FirstChildCreateTimeTAG: 2012-09-08T04:40:09Z SecondChildTAG: alright. SecondChildUserIdTAG: 373994 SecondChildUserNameTAG: RoboSobo SecondChildCreateTimeTAG: 2012-09-08T04:41:52Z IndexTAG: 4647 TitleTAG: kvl law sir i dont no how the voltage value is changing in node b.... UserIdTAG: 361329 UserNameTAG: manjuhasinivenkatachalal CreateTimeTAG: 2012-09-08T04:14:29Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 0 IndexTAG: 4648 TitleTAG: bug Dear Sir , Thanks for your reply But still iam facing the problem with connecting wire ,but I am able to connecting wire when I am using another lap top which as l so using Google chrome as browser my lap top is windows XP and my lap to is Dell latitude E5410 Please me tell us which operating system and browser is suitable for this course ware Thanks MK.Prasanth India UserIdTAG: 156694 UserNameTAG: mkprasanth CreateTimeTAG: 2012-09-08T03:59:08Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 0 IndexTAG: 4649 TitleTAG: Interesting analysis This is very interesting analysis. I like. UserIdTAG: 146995 UserNameTAG: ZYJ CreateTimeTAG: 2012-09-08T02:22:33Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 0 IndexTAG: 4650 TitleTAG: Ohms Law I am a little confused about the e1+6/3 + e1-e2/2 - 3 = 0 I=V/R so, why are we having I/R? UserIdTAG: 230787 UserNameTAG: Gabriel007 CreateTimeTAG: 2012-09-08T01:37:27Z VoteTAG: 0 CoursewareTAG: Week 1 / Nodal analysis example CommentableIdTAG: 6002x_NodalAnalysisExample NumberOfReplyTAG: 2 FirstChildTAG: e is representing voltage, so the equations are in the form I=v/R, where the voltage across the branch is represented by the difference of the voltages at the nodes. FirstChildUserIdTAG: 142402 FirstChildUserNameTAG: aaronrod FirstChildCreateTimeTAG: 2012-09-10T14:24:39Z FirstChildTAG: Its e/R.that is voltage/R FirstChildUserIdTAG: 55345 FirstChildUserNameTAG: sarmaji FirstChildCreateTimeTAG: 2012-09-08T01:57:34Z SecondChildTAG: Thanks! What is e/r? SecondChildUserIdTAG: 230787 SecondChildUserNameTAG: Gabriel007 SecondChildCreateTimeTAG: 2012-09-08T02:13:57Z IndexTAG: 4651 TitleTAG: Why current is negative I think that current is negative, because current going from + to - through resistor, but on the scheme variable 'i' has opposite direction. But, I think, that question is not clear enough... UserIdTAG: 322444 UserNameTAG: koluch CreateTimeTAG: 2012-09-08T01:36:59Z VoteTAG: 0 CoursewareTAG: Week 1 / Power CommentableIdTAG: 6002x_S1E2_Power NumberOfReplyTAG: 5 FirstChildTAG: Mathematical and physical currents are not the same. Direction of the current on the scheme - is only assessment. In your case - current has a minus (-10 A, for example). Physical current flows opposite current direction on the scheme. Sorry for grammar :] FirstChildUserIdTAG: 394836 FirstChildUserNameTAG: v2g6ch4 FirstChildCreateTimeTAG: 2012-09-08T04:08:40Z FirstChildTAG: I agree with you - the question is not clear. In the question it is only asked for value, but not for direction. FirstChildUserIdTAG: 202250 FirstChildUserNameTAG: Zisel FirstChildCreateTimeTAG: 2012-09-08T19:24:56Z FirstChildTAG: True that. The value of I should be positive because that is what is indicated in the diagram and that is what is being asked. FirstChildUserIdTAG: 278301 FirstChildUserNameTAG: prateektaneja FirstChildCreateTimeTAG: 2012-09-08T19:48:19Z FirstChildTAG: The -ve sign is really all about the direction. You are free to choose any direction before solving the circuit as +ve (then you have to stick to this convention until the circuit is solved). If the calculated value of current turns out to be negative, then that simply means the current is flowing in the opposite direction (of what you have chosen as your convention). For eg : Assuming current "i" to be positive if it is leaving a node. now if the value of i turns out to be -10A, then that would simply mean that the current is 10A in magnitude and flows into the node (the -ve sign just tells you that the current is not flowing out from the node, but into it. Just opposite of the convention we choose). FirstChildUserIdTAG: 211715 FirstChildUserNameTAG: pitankar FirstChildCreateTimeTAG: 2012-09-08T05:38:28Z SecondChildTAG: In this case current direction is given. KCL absract of a resistor states Sum of current flowing into a node with current flowing out of the node being neg. SecondChildUserIdTAG: 359302 SecondChildUserNameTAG: GaryG SecondChildCreateTimeTAG: 2012-09-10T23:59:54Z FirstChildTAG: In this case direction of the current is given. FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-10T23:54:48Z IndexTAG: 4652 TitleTAG: Superposition Evidently power is a nonlinear quantity in a general circuit and superposition does not work, Linear values of voltages and currents can be computed using superposition. Then total values for the circuit have to be used for computing power supplied and absorbed by the sources. UserIdTAG: 380229 UserNameTAG: kkappagantula CreateTimeTAG: 2012-09-08T00:35:31Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL/KCL example CommentableIdTAG: 6002x_S2E3_Using_KVL_KCL_and_VI_constraints NumberOfReplyTAG: 0 IndexTAG: 4653 TitleTAG: current flowing out only I am not too sure how to treat a node that has current flowing out only. Can someone explain? - I have read the posts without avail. UserIdTAG: 230787 UserNameTAG: Gabriel007 CreateTimeTAG: 2012-09-07T23:34:13Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL and KCL exercises CommentableIdTAG: 6002x_kvl_kcl_exercises NumberOfReplyTAG: 4 FirstChildTAG: At least one of the currents flowing out will be negative and at least one will be positive so that the sum will be 0 (unless all are 0). FirstChildUserIdTAG: 266912 FirstChildUserNameTAG: pietvo FirstChildCreateTimeTAG: 2012-09-07T23:37:07Z FirstChildTAG: Well Kirchoff's law says that the sum of the currents flowing out of a node is zero, and the sum of the currents flowing in the node is zero too. So to easily write the KCL for a specific node, you may write all the currents in one side of the equation and zero in the other one. Now, if the current is flowing out of the node, then it has a negative sign, and if it is flowing in the node it has a positive sign. Then, you can rearrange the equation as you want. FirstChildUserIdTAG: 365201 FirstChildUserNameTAG: sirajmuhammad FirstChildCreateTimeTAG: 2012-09-07T23:39:47Z FirstChildTAG: Let me share my thought process and see where I am failing... So I can see i4 and i3 will merge and the resulting i2 is a combination of both. So if i4 is negative and i3 is positive (example -2 and 1, then i2 will be - or 2 and -1 then i2 will be +) then i2 might be either. Algebraically I can say that i4+i3=i2 (but I don't know if i2 is negative or positive) First problem asks for i4, so i4 = i2-i3 For the node i1,i2,i6 all of them are coming into the node. so 0=i1+i2+i6 Which is the negative? if i1 is the negative then 0=-i1+i2+i6 the problem asks for i6, so i6=i2-i1 we can only get -(i1+i2) if i6 is the negative one. How do you determine that i6 must be the negative one? FirstChildUserIdTAG: 230787 FirstChildUserNameTAG: Gabriel007 FirstChildCreateTimeTAG: 2012-09-08T00:56:23Z SecondChildTAG: It does not matter what you show as the sign of a current entering or leaving a node. The sum of the physical currents must be zero. If you have several unknowns then you can assign whatever sign you like because you will not know if they are positive or negative. When you have enough independent equations to solve all unknown variables then algebra will show you what the sign is. SecondChildUserIdTAG: 345958 SecondChildUserNameTAG: PWilson123 SecondChildCreateTimeTAG: 2012-09-08T04:41:47Z SecondChildTAG: I was stuck a bit as well until I realised that the sign of the of the variable *i* and the sign of it's value can be different, and *i* can have a negative value. This lets KCL work even if all of the currents are shown to be entering the node on the drawing. So for the first part where you got *i4=i2-i3* consider where you started from. KCL says that all currents flowing into a node must sum to zero. For convenience, we label currents flowing into the node as positive, and those flowing out as negative. So for that center node, that means: *-i2 + i3 +i4 = 0* for that node only. Solve for i4 and what do you get? Use the same process for solving for i6. Don't worry that you are starting out with all currents flowing into the node, because you don't know which are negative values and which are positive. For this example, you have to assume that the known values will allow the circuit to follow KCL, therefore at least one of the values will have a different sign than the others. SecondChildUserIdTAG: 142402 SecondChildUserNameTAG: aaronrod SecondChildCreateTimeTAG: 2012-09-08T14:06:16Z SecondChildTAG: It seems like you misunderstood me, the signs that prefix the currents are NOT the signs of the values! It's just an assumption. For example, if we have a node with three currents flowing IN (or OUT) it, according to Kirchoff's law: i1 + i2 + i3 = 0 **=>** i1 = - i2 - i3 ![enter image description here][1] Now, if i1 is flowing in, i2 is flowing out and i3 is flowing out: i1 - i2 - i3 = 0 ![enter image description here][2] [1]: http://i.imgur.com/DKAtw.jpg [2]: http://i.imgur.com/hkrEp.jpg SecondChildUserIdTAG: 365201 SecondChildUserNameTAG: sirajmuhammad SecondChildCreateTimeTAG: 2012-09-08T16:28:56Z FirstChildTAG: It's simple. The algebraic sum at the node is zero. Then -i4-i1-i5=0 Don't matters the amount of current into the node, is just for to realize the sum. FirstChildUserIdTAG: 80601 FirstChildUserNameTAG: MiltonM FirstChildCreateTimeTAG: 2012-09-12T05:31:12Z IndexTAG: 4654 TitleTAG: current's direction why the direction of current is negative? UserIdTAG: 391927 UserNameTAG: shanuiet CreateTimeTAG: 2012-09-07T21:26:08Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: Because it enters the positive terminal of the supply, while the current direction is from the positive to the negative (outside the supply). FirstChildUserIdTAG: 365201 FirstChildUserNameTAG: sirajmuhammad FirstChildCreateTimeTAG: 2012-09-07T23:31:26Z IndexTAG: 4655 TitleTAG: Cannot re CHECK lab exercises without first pressing RESET Once I press it, the CHECK button disappears in the Lab exercise pages and turns into a RESET button. But then, how do I change my circuit diagram and check it again without clearing everything out first? In comparison, the Homework pages always have the CHECK button there even after I've already checked my solutions. UserIdTAG: 153760 UserNameTAG: Kavka CreateTimeTAG: 2012-09-07T21:14:39Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: its the only way to modify your answer by doing it from the begining FirstChildUserIdTAG: 326507 FirstChildUserNameTAG: mohamed200 FirstChildCreateTimeTAG: 2012-09-08T00:13:19Z IndexTAG: 4656 TitleTAG: e What is e? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-07T19:52:36Z VoteTAG: 0 CoursewareTAG: Week 1 / Nodal analysis example CommentableIdTAG: 6002x_NodalAnalysisExample NumberOfReplyTAG: 3 FirstChildTAG: the potential of a node in reference to the ground FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-07T20:00:04Z FirstChildTAG: I am assuming it is voltage at the node? FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-07T19:54:43Z SecondChildTAG: Correct, e is the node voltage. The wiki provides useful details by course section. See https://www.edx.org/courses/MITx/6.002x/2012_Fall/wiki/6.002x/BasicCircuitAnalysis/ SecondChildUserIdTAG: 5714 SecondChildUserNameTAG: willingc SecondChildCreateTimeTAG: 2012-09-07T20:03:36Z FirstChildTAG: You need to give the context as that has many answers. 1. Its the irrational number 2.71828... 2. Potential at a node voltage 3. 5th letter of the alphabet Probably other I have not thought of. FirstChildUserIdTAG: 15344 FirstChildUserNameTAG: kob FirstChildCreateTimeTAG: 2012-09-07T20:55:57Z SecondChildTAG: Going with (2) here :) SecondChildUserIdTAG: 79361 SecondChildUserNameTAG: kimt SecondChildCreateTimeTAG: 2012-09-08T03:50:13Z IndexTAG: 4657 TitleTAG: S2E6 how to find out load resistance at Joe's barn? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-07T19:33:21Z VoteTAG: 0 CoursewareTAG: Week 1 / Power delivery modeling CommentableIdTAG: 6002x_S2E6_Modeling NumberOfReplyTAG: 2 FirstChildTAG: it's simple we know that P = V^2/R, here power to be delivered is 1000w therefore P = 1000, and V = 240 so there u have it go ahead and calculate R. FirstChildUserIdTAG: 12144 FirstChildUserNameTAG: Sonam FirstChildCreateTimeTAG: 2012-09-08T04:00:40Z SecondChildTAG: why do you use P=V^2/R? wouldn't that be peak power? shouldn't the formula really be P=V^2/(R*2), where 1/2 comes from averaging (as was done in S1E3)? SecondChildUserIdTAG: 410840 SecondChildUserNameTAG: sdomin SecondChildCreateTimeTAG: 2012-09-12T20:44:57Z SecondChildTAG: Yeah this was confusing, but I think the rating (1000W) is nominal too, therefore calculate with nominal current (as it was DC). (resistance doubles) OR they made a mistake... SecondChildUserIdTAG: 324332 SecondChildUserNameTAG: petsol SecondChildCreateTimeTAG: 2012-09-16T19:03:21Z FirstChildTAG: You have the desired voltage and power, with this you will find the load resistance. FirstChildUserIdTAG: 39685 FirstChildUserNameTAG: galeano FirstChildCreateTimeTAG: 2012-09-07T20:14:53Z IndexTAG: 4658 TitleTAG: invalid input what will be the correct method of writing this expression?----------------------- Invalid input: Could not parse '(e2-e1)/R3+(e2-V0)/R4+e2/R5-I1=0' as a formula UserIdTAG: 171378 UserNameTAG: ABHISHEKFROMINDIA CreateTimeTAG: 2012-09-07T18:40:16Z VoteTAG: 0 CoursewareTAG: Week 1 / Node analysis practice, part 3 CommentableIdTAG: 6002x_L2Node2 NumberOfReplyTAG: 2 FirstChildTAG: Drop the '=0' bit. FirstChildUserIdTAG: 267993 FirstChildUserNameTAG: mavlijas FirstChildCreateTimeTAG: 2012-09-07T18:54:27Z SecondChildTAG: After all your answer is incorrect. Because you should multiply R not divide. SecondChildUserIdTAG: 457034 SecondChildUserNameTAG: Ichihara SecondChildCreateTimeTAG: 2012-09-19T13:57:56Z FirstChildTAG: Use an asterisk * to multiply eg x*3 is x times 3 an ^ to mean 'to the power of' eg x^2 is x squared and sqrt(x) to mean square root eg sqrt of x Be sure to use () correctly eg 2/R-1 is not the same as 2/(R-1) FirstChildUserIdTAG: 9147 FirstChildUserNameTAG: SueP FirstChildCreateTimeTAG: 2012-09-07T18:53:55Z IndexTAG: 4659 TitleTAG: Mistakenly deleted the Sources (LAB 2) I mistakenly deleted the Voltage Sources and hit the check button. Now everything is gone , and i cant get the Sources back. Any suggestion ? UserIdTAG: 149844 UserNameTAG: amitaratnayake CreateTimeTAG: 2012-09-07T18:02:16Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: You should be able to "Check" then "Reset" now :) FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-07T18:41:14Z SecondChildTAG: Thanks SecondChildUserIdTAG: 149844 SecondChildUserNameTAG: amitaratnayake SecondChildCreateTimeTAG: 2012-09-07T19:02:59Z IndexTAG: 4660 TitleTAG: Sharing a model from circuit sandbox Is there a way to post the circuit model from my sandbox on this forum? Or somehow link to it? Or attach one to a post? We can post code-blocks and images, why not circuits? I would like some help on the model circuit I'm building but have no way of sharing it. Thx UserIdTAG: 5325 UserNameTAG: vkz CreateTimeTAG: 2012-09-07T17:33:05Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: A possibility is to use the Wiki, which has ability to store circuits FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-07T19:07:28Z FirstChildTAG: There is no way to paste from the sandbox to here. Take a screenshot and use an image editor to trim it. FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-07T18:56:06Z IndexTAG: 4661 TitleTAG: bug am using google chrome and windows 7 os....... not facing any problem with the editor UserIdTAG: 375261 UserNameTAG: vrkotteswaran CreateTimeTAG: 2012-09-07T17:24:21Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4662 TitleTAG: LMD ¿Can you give us an example where it is an advantage or just necesary to not use the lumped matter abstraction? (english is not my native language, sorry if any mistakes) UserIdTAG: 210561 UserNameTAG: Alejovillapar CreateTimeTAG: 2012-09-07T17:02:26Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL and KCL CommentableIdTAG: 6002x_kvl_and_kcl NumberOfReplyTAG: 1 FirstChildTAG: hablas español? por que no entiendo bien tu pregunta FirstChildUserIdTAG: 58618 FirstChildUserNameTAG: ingeniero13 FirstChildCreateTimeTAG: 2012-09-07T17:20:44Z SecondChildTAG: Mi pregunta es si nos pueden dar un ejemplo en el que sea ventajoso o necesario no usar la abstracción de parámetros concentrados (lumped matter abstraction) para ilustrar un caso en el que seria interesante salirse de esta simplificación. SecondChildUserIdTAG: 210561 SecondChildUserNameTAG: Alejovillapar SecondChildCreateTimeTAG: 2012-09-07T18:12:34Z IndexTAG: 4663 TitleTAG: Can't label nodes... I can't seem to label nodes for whatever reason they all remain as ??? no matter what I try. UserIdTAG: 175876 UserNameTAG: Sunden CreateTimeTAG: 2012-09-07T16:51:25Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 2 FirstChildTAG: Can you try double-clicking the wire, rather than the ??? directly? FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-07T17:13:21Z FirstChildTAG: Click on the line of side of label. It shows the box text. Then, change it. FirstChildUserIdTAG: 80601 FirstChildUserNameTAG: MiltonM FirstChildCreateTimeTAG: 2012-09-07T17:32:38Z SecondChildTAG: I've tried clicking on both but I never see a box where I can change text. SecondChildUserIdTAG: 175876 SecondChildUserNameTAG: Sunden SecondChildCreateTimeTAG: 2012-09-07T21:29:16Z IndexTAG: 4664 TitleTAG: question 1 please help me with how to calculate the total resistance of the transmission line?? i am getting incorrect answer.. UserIdTAG: 316353 UserNameTAG: Vivekanand123 CreateTimeTAG: 2012-09-07T16:44:37Z VoteTAG: 0 CoursewareTAG: Week 1 / Power delivery modeling CommentableIdTAG: 6002x_S2E6_Modeling NumberOfReplyTAG: 1 FirstChildTAG: 1000ft -> 1.588Ohm 113*2ft -> x Ohm. So you get x = 226/1000 * 1.588 FirstChildUserIdTAG: 329254 FirstChildUserNameTAG: KGabor FirstChildCreateTimeTAG: 2012-09-07T17:30:23Z IndexTAG: 4665 TitleTAG: LAB 2 show only 3 icons: GND, RESISTOR, TEST POINT I don't understand why I see only 3 icons. I can't put the generators. I use Firefox!!! Help me UserIdTAG: 323209 UserNameTAG: Consogno CreateTimeTAG: 2012-09-07T16:28:33Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Can you try "Check" then "Reset"? FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-07T16:44:18Z SecondChildTAG: I have only CHECK BUTTON but if I press it nothing to do. :-( SecondChildUserIdTAG: 323209 SecondChildUserNameTAG: Consogno SecondChildCreateTimeTAG: 2012-09-07T16:51:08Z IndexTAG: 4666 TitleTAG: 3rd question question3: What is the voltage (in Volts) v2 across the resistor with resistance R2 (keeping e3 as reference)? here how r we getting 8.75 as answer?? this answer is correct since the voltage drops r the same unmindful of the reference nodes.... can anyone please say the steps?? i tried with (e2-e3)/R2 = (8.75-0)/5 which gives me 1.6 which is incorrect. how do i calculate to get the right answer then?? UserIdTAG: 316353 UserNameTAG: Vivekanand123 CreateTimeTAG: 2012-09-07T16:26:38Z VoteTAG: 0 CoursewareTAG: Week 1 / Node method example CommentableIdTAG: 6002x_S2E5_Node_Method NumberOfReplyTAG: 2 FirstChildTAG: e2-e3=8.75 its the right answer why you divide by R2 we don't want to find e2 use this equation : e2/R2+(e2-e1)/R1-I=0 and you will find that e2=8.75 v2=e2-e3=8.75 FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-08T00:29:04Z FirstChildTAG: You tried to calculate (e2-e3)/R2 = 1.6, that's actually the current through the resistor R2. All you needed to do was to determine v2 which is simply e2-e3 = e2 FirstChildUserIdTAG: 67347 FirstChildUserNameTAG: eprime FirstChildCreateTimeTAG: 2012-09-08T22:37:46Z IndexTAG: 4667 TitleTAG: how to find independent kvl and kcl? how to find independent kvl and kcl? UserIdTAG: 118611 UserNameTAG: mitianhari CreateTimeTAG: 2012-09-07T16:14:54Z VoteTAG: 0 CoursewareTAG: Week 1 / Circuit topology CommentableIdTAG: 6002x_S2E1_Circuit_Topology NumberOfReplyTAG: 1 FirstChildTAG: independent kcl = no of nodes -1 independent kvl = total no of loops-no of nodes FirstChildUserIdTAG: 148523 FirstChildUserNameTAG: span993 FirstChildCreateTimeTAG: 2012-09-08T14:06:02Z SecondChildTAG: "no of nodes -1 independent kvl = total no of loops-no of nodes" Is this always true within lumped matter discipline, or only for this specific case? SecondChildUserIdTAG: 142402 SecondChildUserNameTAG: aaronrod SecondChildCreateTimeTAG: 2012-09-09T00:39:32Z IndexTAG: 4668 TitleTAG: Discovery Interesting thing I discovered today. While going through Prof. Agarwal's lecture, run this video ["http://www.youtube.com/watch?v=-hJf4ZffkoI"] with 35-55% volume in the background. The lecture just gets more interesting with this background music. :) UserIdTAG: 152317 UserNameTAG: easyMIT CreateTimeTAG: 2012-09-07T16:11:46Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4669 TitleTAG: awais sir, hoa can i download textbook. I want bthis book so that i can read it with complete subject. UserIdTAG: 72281 UserNameTAG: awais937 CreateTimeTAG: 2012-09-07T14:35:50Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: it's not allowed FirstChildUserIdTAG: 176234 FirstChildUserNameTAG: hejinjie FirstChildCreateTimeTAG: 2012-09-07T14:52:04Z IndexTAG: 4670 TitleTAG: what is meant by lumped matter discipline? Give me a clear idea about the lumped matter discipline. UserIdTAG: 219679 UserNameTAG: kiruthika3 CreateTimeTAG: 2012-09-07T13:29:43Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL and KCL CommentableIdTAG: 6002x_kvl_and_kcl NumberOfReplyTAG: 1 FirstChildTAG: I believe this to be a term unique to MIT. I have never heard it used in university EE courses elsewhere. It is just a term to mean 'lump this stuff together so I can make some usefull, close enough, calculations. FirstChildUserIdTAG: 167370 FirstChildUserNameTAG: alphathreethree FirstChildCreateTimeTAG: 2012-09-08T01:04:01Z IndexTAG: 4671 TitleTAG: edX Timezone Anyone know what timezone edX use? UserIdTAG: 364364 UserNameTAG: agungsantoso CreateTimeTAG: 2012-09-07T13:20:39Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: Gmt FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-07T14:41:21Z FirstChildTAG: We are located in Boston, so the edX people are in EDT (and will move to EST in November). However, our servers are running using UTC which is basically the same thing as GMT. The deadlines for our problem sets are such that they will be accepted as long as there is somewhere in the world where the local time is before the deadline. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-09-07T21:53:45Z IndexTAG: 4672 TitleTAG: current negative why current is negative ? i can't understand, please help. UserIdTAG: 154101 UserNameTAG: srbhjosh1 CreateTimeTAG: 2012-09-07T11:48:47Z VoteTAG: 0 CoursewareTAG: Week 1 / Power CommentableIdTAG: 6002x_S1E2_Power NumberOfReplyTAG: 4 FirstChildTAG: It is just showing direction.if u say 1A flowing upwards,i can say -1A flowing downwards FirstChildUserIdTAG: 55345 FirstChildUserNameTAG: sarmaji FirstChildCreateTimeTAG: 2012-09-07T11:57:29Z FirstChildTAG: I think it is according notation, positive current flow from + to -, negative in the opposite direction from - to +. FirstChildUserIdTAG: 324554 FirstChildUserNameTAG: ekolodzinsky FirstChildCreateTimeTAG: 2012-09-07T12:01:17Z FirstChildTAG: The negative direction just shows that the current's direction you assumed was just **opposite** to what the actual current direction is . If you have assumed the direction to be from node **A** to node **B**, but actually the current is flowing from node **B** to node **A** . FirstChildUserIdTAG: 102529 FirstChildUserNameTAG: laksh FirstChildCreateTimeTAG: 2012-09-07T13:17:44Z FirstChildTAG: As given in the S1E2 Problem, we must first find out the voltage across the resistor labeled 'R' having 8 ohms.However, we must first find out the current flowing towards the resistor 'R' using the derived formula I = sqrt(P/R), where I is current(amp), P is power(watt), and R is resistance(ohms). Therefore, this formula gives us a positive current of 1.17 Amperes flowing through the resistor 'R'. Thus, to find out the voltage we use the voltage formula V = R x I, where 8 ohms times 1.17 amperes gives us 9.38 voltage dissipated across resistor 'R'. Remember now, all of these values for voltage and current are positive because we are defining the direction of current to be moving towards our resistor 'R', which gives us a positive value of 1.17 ampere current entering the resistor 'R', not the Network N. However, if the same positive value of 1.17 ampere current turns to its opposite direction and entered Network N instead, than it will give us a negative value of (-1.17) ampere. Therefore, the value of current i entering Network N from resistor is a negative value: -1.17 ampere. FirstChildUserIdTAG: 181732 FirstChildUserNameTAG: Anthony341994 FirstChildCreateTimeTAG: 2012-09-11T02:50:19Z IndexTAG: 4673 TitleTAG: Calculation of voltage R2 and R4. Hello, could you please help me. I do not fully understand how to calculate voltage across R2 and R4. Can we represent this schema as voltage divider ? UserIdTAG: 324554 UserNameTAG: ekolodzinsky CreateTimeTAG: 2012-09-07T11:38:45Z VoteTAG: 0 CoursewareTAG: Week 1 / Series and parallel example CommentableIdTAG: 6002x_S2E4_Series_and_Parallel NumberOfReplyTAG: 2 FirstChildTAG: Assign i2 as flowing thru R2.and i4,thru R4.use KVL in first loop to find i2.v2=i2*R2.use KCL to find i4.now,v4=i4*R4 FirstChildUserIdTAG: 55345 FirstChildUserNameTAG: sarmaji FirstChildCreateTimeTAG: 2012-09-07T11:48:51Z SecondChildTAG: OK, i got it. thank you. SecondChildUserIdTAG: 324554 SecondChildUserNameTAG: ekolodzinsky SecondChildCreateTimeTAG: 2012-09-07T12:26:16Z SecondChildTAG: Can you please show the steps to find i4? SecondChildUserIdTAG: 29275 SecondChildUserNameTAG: Mbarsalou SecondChildCreateTimeTAG: 2012-09-11T15:59:43Z FirstChildTAG: one can notice that R_2 = R_3+R_4, so the current will be divided equally into those branches. As you have calculated i_1, then half of it goes through R_2 and the other half through R_3 and R_4. The Ohm's law U=I/R FirstChildUserIdTAG: 157610 FirstChildUserNameTAG: mradziwo FirstChildCreateTimeTAG: 2012-09-07T21:46:26Z SecondChildTAG: So this means that i4 is created by dividing the half that is going through R_3 and R_4 in half again? If not, this confused me even more. SecondChildUserIdTAG: 29275 SecondChildUserNameTAG: Mbarsalou SecondChildCreateTimeTAG: 2012-09-12T23:24:50Z IndexTAG: 4674 TitleTAG: H1P1 I need a little push in this one. How can I calculate the power dissipated by a composite resistor. Not looking for the answer just some sugestions:/ UserIdTAG: 23183 UserNameTAG: A1 CreateTimeTAG: 2012-09-07T07:21:48Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: The question seems a bit confusing to me. still working on it FirstChildUserIdTAG: 231554 FirstChildUserNameTAG: MChaudhary FirstChildCreateTimeTAG: 2012-09-07T08:34:29Z FirstChildTAG: voltage drop on resistor * current FirstChildUserIdTAG: 332449 FirstChildUserNameTAG: 4a4ik FirstChildCreateTimeTAG: 2012-09-07T07:37:21Z IndexTAG: 4675 TitleTAG: tran isn't working when i hit 'tran' nothing appears,what to do?? UserIdTAG: 332163 UserNameTAG: rritambhar CreateTimeTAG: 2012-09-07T06:27:38Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 3 FirstChildTAG: I have the same issue using firefox, DC works fine but TRAN does not. I will try it in chrome too. -Amit FirstChildUserIdTAG: 385224 FirstChildUserNameTAG: shilpiji991 FirstChildCreateTimeTAG: 2012-09-08T01:09:08Z FirstChildTAG: have you made sure the probes are connected at the correct places? FirstChildUserIdTAG: 231554 FirstChildUserNameTAG: MChaudhary FirstChildCreateTimeTAG: 2012-09-07T09:15:43Z FirstChildTAG: perhaps it works better using Chrome browser. I know it doesn't work with Internet Explorer FirstChildUserIdTAG: 228637 FirstChildUserNameTAG: DCounsell FirstChildCreateTimeTAG: 2012-09-07T15:58:15Z IndexTAG: 4676 TitleTAG: Node analysis part3 My check button is not working in tis question.i answered it using braces().what is the problem? UserIdTAG: 55345 UserNameTAG: sarmaji CreateTimeTAG: 2012-09-07T05:46:52Z VoteTAG: 0 CoursewareTAG: Week 1 / Node analysis practice, part 3 CommentableIdTAG: 6002x_L2Node2 NumberOfReplyTAG: 1 FirstChildTAG: Sorry4tat.i entered R1 in the place of R4,.that was the mistake FirstChildUserIdTAG: 55345 FirstChildUserNameTAG: sarmaji FirstChildCreateTimeTAG: 2012-09-07T06:27:37Z IndexTAG: 4677 TitleTAG: 6.002x discussion on facebook All interested people are requested to join this group on facebook'GATE,IES & PSU(Preparation With Fun)'............. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-07T05:22:56Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4678 TitleTAG: trans analysis what does it mean to run trans analysis for 5msec UserIdTAG: 198054 UserNameTAG: bi_bat010 CreateTimeTAG: 2012-09-07T05:04:32Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4679 TitleTAG: Section shown as unfinished I finished this group of questions, but for some reason, on the top bar, it still shows it as unfinished, does anybody now why? UserIdTAG: 155972 UserNameTAG: sreicks CreateTimeTAG: 2012-09-07T04:40:38Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL and KCL exercises CommentableIdTAG: 6002x_kvl_kcl_exercises NumberOfReplyTAG: 2 FirstChildTAG: It shows the progress by the correct questions only. If you have incorrect answers, the progress will not go to 100%. FirstChildUserIdTAG: 8803 FirstChildUserNameTAG: SteveM FirstChildCreateTimeTAG: 2012-09-07T06:43:41Z SecondChildTAG: I have the same, but I miscalculated 2 fields. Fair would be imo, if i could not reach 100% if I would have clicked the show solution, but i simply recalculated my answer. :-( SecondChildUserIdTAG: 329254 SecondChildUserNameTAG: KGabor SecondChildCreateTimeTAG: 2012-09-07T11:39:08Z FirstChildTAG: It updates instantly, but like happened to me, I answered all questions unlogged, I don't know it is possible, but I'd done so. But when I updated the page, I realized why it hasn't reached 100%. FirstChildUserIdTAG: 265133 FirstChildUserNameTAG: Elienai FirstChildCreateTimeTAG: 2012-09-09T01:00:22Z IndexTAG: 4680 TitleTAG: offset voltage what is offset voltage UserIdTAG: 181793 UserNameTAG: mitrahul CreateTimeTAG: 2012-09-07T04:37:41Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: offset voltage means the value at which the circuit will have low potential , if you make that 0 then it means you are taking the low voltage as 0 and allowing -ve values . and if you keep offset as +ve then graph shifts above 0 level to the offset voltage , now your reference is offset voltage rather than 0 ( though coordinate 0 lies at same position ) . :) - hope this will help :) - thank you FirstChildUserIdTAG: 362198 FirstChildUserNameTAG: Sampathram FirstChildCreateTimeTAG: 2012-09-07T12:38:26Z SecondChildTAG: I didn't quite get it. Ok, if I set an offset voltage of say, +2V, then does my entire waveform get clamped by +2V? Because, first I set the offset voltage at 0V and hit Tran - that got me all three waveforms asked in the example lab symmetric about y=0. i.e. at 0V, the waveform starts having low potential and negative values are allowed. Now, I set the offset voltage to 1V and hit Tran - that should get me all three waveforms symmetric about y=1 right? i.e. at 1V, the waveform starts having low potential or in other words, y=1 acts as the X-axis right? However, the Tran output shows only the waveform at node A symmetric about y=1, the other two waveforms are not. But all three waveforms have their negative peak at y=0. What does this mean? SecondChildUserIdTAG: 281154 SecondChildUserNameTAG: NoelThomasNicholas SecondChildCreateTimeTAG: 2012-09-07T19:18:18Z IndexTAG: 4681 TitleTAG: Who is from Pakistan and from which Institute. Hey how many of you are from Pakistan and why they are doing this course????? Give your comments about this course, its contents and way of instruction and evaluation. UserIdTAG: 113109 UserNameTAG: Akif CreateTimeTAG: 2012-09-07T03:55:27Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Not many I guess FirstChildUserIdTAG: 37464 FirstChildUserNameTAG: DoubleA FirstChildCreateTimeTAG: 2012-09-13T05:31:24Z IndexTAG: 4682 TitleTAG: Sandbox Transient Analysis. I am making a half wave rectifier in sandbox and doing transient analysis for 5ms but it did not work please check it out if there is any issue. UserIdTAG: 113109 UserNameTAG: Akif CreateTimeTAG: 2012-09-07T03:46:57Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4683 TitleTAG: KCL Why for i5?: Invalid input: Could not parse 'i3+(-i1+-i2)' as a formula UserIdTAG: 369511 UserNameTAG: Year CreateTimeTAG: 2012-09-07T03:27:23Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL and KCL exercises CommentableIdTAG: 6002x_kvl_kcl_exercises NumberOfReplyTAG: 2 FirstChildTAG: Maybe it's the +- bit. Try putting -2 in parentheses, or just writing a - FirstChildUserIdTAG: 267993 FirstChildUserNameTAG: mavlijas FirstChildCreateTimeTAG: 2012-09-07T03:39:59Z FirstChildTAG: You've missed some brackets and that's why programm cannot read it. Try to write: "-i3+((-i1)+(-i2))". FirstChildUserIdTAG: 202880 FirstChildUserNameTAG: Adilzhan FirstChildCreateTimeTAG: 2012-09-07T19:16:30Z IndexTAG: 4684 TitleTAG: homework As a consequence, Joe found his workshop too cold. H1 was weak; H2 and H3 barely warmed up. What power was being dissipated in H1? I don't know what do help me please? UserIdTAG: 58618 UserNameTAG: ingeniero13 CreateTimeTAG: 2012-09-07T02:51:40Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 4 FirstChildTAG: you could get the total resistance of the three heaters then use the formula I=V/R to get the total current and that current is the one passing through H1. After that, use the formula P=I^2*R to get the power being dissipated in H1. sorry if my explanation wasn't clear FirstChildUserIdTAG: 371000 FirstChildUserNameTAG: Joseph090892 FirstChildCreateTimeTAG: 2012-09-07T03:04:24Z SecondChildTAG: i don't understand because, i have only voltage, but the resistance don't have SecondChildUserIdTAG: 58618 SecondChildUserNameTAG: ingeniero13 SecondChildCreateTimeTAG: 2012-09-07T03:26:31Z SecondChildTAG: Thank you! That helped a lot. I kept getting stuck on what current to use! SecondChildUserIdTAG: 208854 SecondChildUserNameTAG: BrynnleeEaton SecondChildCreateTimeTAG: 2012-09-08T05:01:50Z SecondChildTAG: ok so total resistance is 18.11 , there fore if H1=H2=H3=H then H =18.11/3= 6.03 ohms. P=I2R , my answer is still wrong . Please can some one help. SecondChildUserIdTAG: 178115 SecondChildUserNameTAG: nehamakhija SecondChildCreateTimeTAG: 2012-09-16T18:12:45Z FirstChildTAG: excuse me but i don't understand y si alguien habla español seria mas sencillo gracias FirstChildUserIdTAG: 58618 FirstChildUserNameTAG: ingeniero13 FirstChildCreateTimeTAG: 2012-09-07T03:47:50Z SecondChildTAG: Sorry, don't speak Spanish. I'm sure someone will answer later, but for now, [look here][1] [1]: http://2.bp.blogspot.com/_IGPyEGm6pT0/TEr7980y-9I/AAAAAAAAAh4/Via21LTOjgM/s1600/OhmsWattsLaws.png and find resistance from power rating and voltage. SecondChildUserIdTAG: 267993 SecondChildUserNameTAG: mavlijas SecondChildCreateTimeTAG: 2012-09-07T03:54:54Z FirstChildTAG: what happens is this, I know the power that supports each heater is 840W if they were parallel with the source, I would know that voltage drops in each element, but as work with the heater power if I am asked to find another is not given sorry for my poor English FirstChildUserIdTAG: 58618 FirstChildUserNameTAG: ingeniero13 FirstChildCreateTimeTAG: 2012-09-07T04:01:06Z SecondChildTAG: Not sure how you got 840W. Do we have the same numbers? "They propose to use three 1360.0W 240V baseboard heaters" So you have the power and the voltage. From there you can calculate the resistance. The heaters stay the same for the second part of the problem so you can continue using that resistance. SecondChildUserIdTAG: 267993 SecondChildUserNameTAG: mavlijas SecondChildCreateTimeTAG: 2012-09-07T04:12:43Z SecondChildTAG: Power rating is given.therefore,find resistance using R=V^2/P.now,that u got R,find total resistance of circuit.now,I=V/(equivalentR) SecondChildUserIdTAG: 55345 SecondChildUserNameTAG: sarmaji SecondChildCreateTimeTAG: 2012-09-07T04:12:55Z SecondChildTAG: ok so total resistance is 18.11 , there fore if H1=H2=H3=H then H =18.11/3= 6.03 ohms. P=I2R , my answer is still wrong . Please can some one help. SecondChildUserIdTAG: 178115 SecondChildUserNameTAG: nehamakhija SecondChildCreateTimeTAG: 2012-09-16T18:13:00Z FirstChildTAG: I have a problem with H1P3. H2 and H3 are parallel,and then they are in series with H1?First,do I calculate resistance of H2 and H3 parallel and then with H1 series or? How do I calculate power dissipated on H1 and H2(H3)? FirstChildUserIdTAG: 286111 FirstChildUserNameTAG: Bomba_Klc FirstChildCreateTimeTAG: 2012-09-09T10:01:14Z IndexTAG: 4685 TitleTAG: KCL 0 current i2 In KCL 0 why is the current i2 -3.6? I can't figure this one... UserIdTAG: 183711 UserNameTAG: MMatheson CreateTimeTAG: 2012-09-07T02:50:21Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL and KCL exercises CommentableIdTAG: 6002x_kvl_kcl_exercises NumberOfReplyTAG: 2 FirstChildTAG: Nevermind, got it. FirstChildUserIdTAG: 183711 FirstChildUserNameTAG: MMatheson FirstChildCreateTimeTAG: 2012-09-07T03:00:08Z FirstChildTAG: First you have to get the current i5 then in the node where the currents i5, i2 and i1 gets out, the algebraic expression is -i1-i5-i2=0 :) FirstChildUserIdTAG: 253807 FirstChildUserNameTAG: DCan10 FirstChildCreateTimeTAG: 2012-09-07T05:16:51Z SecondChildTAG: you got it and i didn't see it haha SecondChildUserIdTAG: 253807 SecondChildUserNameTAG: DCan10 SecondChildCreateTimeTAG: 2012-09-07T05:18:18Z IndexTAG: 4686 TitleTAG: CAN'T SEE THE COMMENTS I COULDN'T SEE THE RESPONSES TO THE COMMENTS DOES ANY ONE HAVE THE SAME PROBLEM PLEASE UserIdTAG: 129934 UserNameTAG: miguel210 CreateTimeTAG: 2012-09-07T02:15:06Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: I had the same problem using internet explorer, so I changed to google chrome and it worked FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-07T04:43:57Z FirstChildTAG: Are you using Internet Explorer? Please use Google Chrome and all will be okay. FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-07T13:13:06Z IndexTAG: 4687 TitleTAG: Independent Equations Now, I started to have a headache because of the independent and dependent equations !! I just don't get it. How I discover them ? What does independent and dependent really means ?!! I hope some one can explain it in more detail or to be clarified more by the staff. UserIdTAG: 289472 UserNameTAG: OsamaAdel CreateTimeTAG: 2012-09-07T02:12:27Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL and KCL CommentableIdTAG: 6002x_kvl_and_kcl NumberOfReplyTAG: 3 FirstChildTAG: an equation dependent means that this is the sum of other independents equations. In KVL, you got "dependent loop" if it contains subloops, i mean, for example, the big loop contains three sub-loops (one at left and two at right), so, it is a dependent loop. I hope it was helpful. FirstChildUserIdTAG: 268104 FirstChildUserNameTAG: SaulCardenasG FirstChildCreateTimeTAG: 2012-09-11T13:28:40Z FirstChildTAG: a=(1,0,0), b=(0,1,0), c=(0,0,1) they are independents equations(another way: a=x,b=y,c=z) but d=(1,2,-3) is a dependent equation because d = a+2b-3c. If you are looking for KVL independent equations, you have to look for independent loops, for example: Vac, abd , bdc (so you got 3 loops, independent loops) but if you join abd and bdc you got another loop, in this case, dependent loop, in same way, if you join Vac-abd, Vac-bdc, Vac-abd-bdc, so you got 7 KVL equations (or loops) in total (3 independents equations and 4 dependent equations). FirstChildUserIdTAG: 268104 FirstChildUserNameTAG: SaulCardenasG FirstChildCreateTimeTAG: 2012-09-13T03:59:51Z FirstChildTAG: Well, it's about set of independent equations, to be more specific. To solve a set of equations with x unknowns (I hope you understand what I mean, english isn't my native language) you need to have x independent equations. For example: a=0.5b+c, 3a+9=10c, 3b=8c - these are independent equations, none of it is a linear combination of others. With these 3 equations you can easily solve this set. But try to solve it with equations: a=0.5b+c, 3a+9=10c, 6a-1.5b-13c=-9. You can't do that, because third equation is a linear combination of first 2 (just multiply 1st by 3, add to second and group variables) - it means it is dependent. For more info please refer to Wikipedia: linear combination. And about KVL/KCL and dependent/independent equations - you can make many equations with KCL/KVL, but only fixed number of them (depending on number of branches, loops and nodes) will lead you to solving a circuit. I hope I was helpful, regards. FirstChildUserIdTAG: 119090 FirstChildUserNameTAG: Bezi FirstChildCreateTimeTAG: 2012-09-07T14:03:36Z IndexTAG: 4688 TitleTAG: HI problems in HW1 in the box where i have to put the ecation i write 1/R+1/R+1/R and this si a valid ecuation and is the fact is real way to solved but can u tell me please the format to put into>?? thansk a lot.! UserIdTAG: 156060 UserNameTAG: radeon9550 CreateTimeTAG: 2012-09-07T02:08:38Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: if you are trying to solve for the total resistance of parallel resistors, there is something missing with the expression you wrote FirstChildUserIdTAG: 371000 FirstChildUserNameTAG: Joseph090892 FirstChildCreateTimeTAG: 2012-09-07T02:42:18Z IndexTAG: 4689 TitleTAG: negative sign applied Iam not understanding why the negative sign is applied to the current UserIdTAG: 302849 UserNameTAG: coolguy8064 CreateTimeTAG: 2012-09-07T01:17:06Z VoteTAG: 0 CoursewareTAG: Week 1 / Power CommentableIdTAG: 6002x_S1E2_Power NumberOfReplyTAG: 2 FirstChildTAG: it only becomes negative when you assumed the wrong direction for the current flow FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-07T01:30:13Z SecondChildTAG: It's not really thinking of it as flowing in the wrong direction but of thinking of it flowing in both directions. voltage flowing from (pos) to(neg) ,current flowing in the opposite direction at the same time from neg to positive. That is why you can look it in reverse as long as you remember which way you use. In this case I or current direction is stated> SecondChildUserIdTAG: 359302 SecondChildUserNameTAG: GaryG SecondChildCreateTimeTAG: 2012-09-10T23:32:30Z FirstChildTAG: The convention is that current is positive when it flows from a positive terminal into a negative terminal. In this diagram, the current arrow specifies that it is flowing from negative to positive. Therefore, it is flowing opposite to our convention, so a negative sign appears. FirstChildUserIdTAG: 183711 FirstChildUserNameTAG: MMatheson FirstChildCreateTimeTAG: 2012-09-07T01:27:53Z SecondChildTAG: Physically, resistor is not a source. It dissipates power because current is entering the resistor from the network. In other words current is supplied by the network, which means current entering network is negative. Signs are important like credit and debit! SecondChildUserIdTAG: 380229 SecondChildUserNameTAG: kkappagantula SecondChildCreateTimeTAG: 2012-09-08T07:10:48Z IndexTAG: 4690 TitleTAG: Why does the bottom right resistor get used? Why is the resistance of the 10k resistor on the bottom right counted? It seems like the circuit should go out through the negative terminal and come back in through the positive terminal to form the circuit and not need or pass to that resistor. Any thoughts? UserIdTAG: 22443 UserNameTAG: Damionw CreateTimeTAG: 2012-09-07T00:41:47Z VoteTAG: 0 CoursewareTAG: Week 1 / Combination Rules CommentableIdTAG: 6002x_CombinationRules NumberOfReplyTAG: 1 FirstChildTAG: Consider this. Instead of having that 10k resistor there, I replace it with a 10 Ohm resistor. What would happen then? Do you still think current would go out the negative terminal and come back in through the positive? What if it were a 1 Ohm resistor? What if I bridged them together? What I'm trying to illustrate here is the fact that all components participate in the circuit somehow (unless it is removed by an open or short circuit). FirstChildUserIdTAG: 151427 FirstChildUserNameTAG: RichardZ FirstChildCreateTimeTAG: 2012-09-07T00:48:23Z IndexTAG: 4691 TitleTAG: Error in S2V3: METHOD 1 -- KVL, KCL METHOD In S2V3: METHOD 1 -- KVL, KCL METHOD, at time 2:01 at 1.0x speed, the lecturer says: "Then as a third step, you go through the loops one by one and you **write KVL for each of the nodes.**" Should read "write KVL for each of the loops". ps. This is amazing software. UserIdTAG: 180268 UserNameTAG: Syntelic CreateTimeTAG: 2012-09-07T00:41:19Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: yes you got it right.. FirstChildUserIdTAG: 381619 FirstChildUserNameTAG: siddhantmishra007 FirstChildCreateTimeTAG: 2012-09-10T11:26:50Z IndexTAG: 4692 TitleTAG: phiB being zero outside Why do we want phiB to be zero *outside* the elements? Isn't it never zero there when current flows? And, wouldn't changing flux inside the element be what would induces an emf in it? UserIdTAG: 325730 UserNameTAG: gadzin1203 CreateTimeTAG: 2012-09-06T23:02:48Z VoteTAG: 0 CoursewareTAG: Week 1 / Lumped Element Abstraction CommentableIdTAG: 6002x_lumped_element_abstraction NumberOfReplyTAG: 1 FirstChildTAG: Currents produce a magnetic field, which will produce a flux through surfaces connecting the terminals of lumped elements. But Faraday's law involves the derivative of Magnetic flux with respect to time. So as long as the currents aren't changing, the flux isn't changing, and d(phiB)/dt = 0. When currents are changing, such as in AC current, this no longer applies. For low frequencies, however, it can be a good approximation. The way I understand it (someone correct me if I'm wrong), for high frequencies and other systems that involve quickly changing currents we will have to invent an ideal inductor, and we will lump all of the properties of the circuit associated with Faraday's law into that inductor (Just like we separate a real wire into an ideal wire and an ideal resistor) so that we may still use the Lumped Matter Discipline at high frequencies. FirstChildUserIdTAG: 180365 FirstChildUserNameTAG: Opus_723 FirstChildCreateTimeTAG: 2012-09-07T00:42:36Z SecondChildTAG: Do I understand right, Opus_723, that these generalizations we're discussing now are only for resistive elements? That would make it much clearer. SecondChildUserIdTAG: 280220 SecondChildUserNameTAG: pilgrimCycle SecondChildCreateTimeTAG: 2012-09-07T02:40:28Z SecondChildTAG: I think that when you define a lumped element you are considering a box with a pair of terminals. Anything that occurs inside this box (even changes in the magnetic flux) do not matter four our level of abstraction. In the real world may be we can understand this box as the physical space occupied by the element. In this case we will only concern with if the flux change outside this physical space. SecondChildUserIdTAG: 138934 SecondChildUserNameTAG: KarelNN SecondChildCreateTimeTAG: 2012-09-07T13:02:40Z IndexTAG: 4693 TitleTAG: Textbook problem When I click on textbook, I'm taken to where I left off. However, when I click to go to the next page, a page in another part of the book comes up. Is anyone else having that problem? Any solutions? UserIdTAG: 237167 UserNameTAG: Dug CreateTimeTAG: 2012-09-06T22:34:11Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4694 TitleTAG: Calculate Power across a current source Now P = vi, so i have 'i', but not v....so how do i go about ? Thanks UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-06T21:48:24Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL/KCL example CommentableIdTAG: 6002x_S2E3_Using_KVL_KCL_and_VI_constraints NumberOfReplyTAG: 2 FirstChildTAG: The voltage across that source is same as the one accross R2 FirstChildUserIdTAG: 204745 FirstChildUserNameTAG: allwynmendes FirstChildCreateTimeTAG: 2012-09-06T21:53:12Z SecondChildTAG: with what voltage orientation? is top or bottom of current source positive? SecondChildUserIdTAG: undefined SecondChildUserNameTAG: undefined SecondChildCreateTimeTAG: 2012-09-08T00:05:33Z FirstChildTAG: u can use the formula P=I^2*R if your R is given FirstChildUserIdTAG: 371000 FirstChildUserNameTAG: Joseph090892 FirstChildCreateTimeTAG: 2012-09-06T21:55:03Z IndexTAG: 4695 TitleTAG: Power I thought power never was negative... UserIdTAG: 249375 UserNameTAG: MarIgones CreateTimeTAG: 2012-09-06T21:38:53Z VoteTAG: 0 CoursewareTAG: Week 1 / Associated reference directions CommentableIdTAG: 6002x_S2E2_Associated_Reference_Directions NumberOfReplyTAG: 1 FirstChildTAG: "Negative" power depends on your reference- you can think of it as power going "in" as opposed to going "out", or vice-versa. If one is talking about the power "entering" an independent voltage source, a negative number could be interpreted as the voltage source actually "sourcing" power, which makes sense. Or if talking about the power "dissipated" by a resistor, negative power would mean the resistor is (somehow) sourcing power! FirstChildUserIdTAG: 390034 FirstChildUserNameTAG: kahlil FirstChildCreateTimeTAG: 2012-09-06T21:46:55Z SecondChildTAG: One question about what you said and what is in the homework H1P2??? I Realized that almost always the supplied power get a negative sign and the dissipated power get a positive sign. but in the home work both supplied power and dissipated power should be expressed in positive sign to be correct. and just the supplied power by the voltage source is negative because it has just an opposite direction with respect to major current of the circuit which is the current of the current source??? why are all of them should be stated in positive (except voltage source power) and also why does voltage source has a negative power????? SecondChildUserIdTAG: 153707 SecondChildUserNameTAG: masoud_np SecondChildCreateTimeTAG: 2012-09-13T12:04:49Z IndexTAG: 4696 TitleTAG: Vca stands for ? During S2V2 on 1:32 it shows Vca=3V. That stands for Va-Vc according to the measurements on the circuit. However page 65 of textbook, example 2.10 determines the Vga to be -6V. That clearly is Vg-Va. Please advise. UserIdTAG: 206669 UserNameTAG: dimitrios66 CreateTimeTAG: 2012-09-06T21:08:17Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL and KCL CommentableIdTAG: 6002x_kvl_and_kcl NumberOfReplyTAG: 0 IndexTAG: 4697 TitleTAG: Rotate parts I can't rotate the parts, does anyone else have this problem? UserIdTAG: 319221 UserNameTAG: stefan31i CreateTimeTAG: 2012-09-06T19:57:38Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 6 FirstChildTAG: Hi stefan, I am able to rotate the parts by selecting the part and then pressing r. I am using chrome browser and windows xp. Best regards FirstChildUserIdTAG: 83937 FirstChildUserNameTAG: yousif_salman FirstChildCreateTimeTAG: 2012-09-06T20:20:32Z FirstChildTAG: Hey just try another browser! may be that will work.. FirstChildUserIdTAG: 140697 FirstChildUserNameTAG: crocroaz FirstChildCreateTimeTAG: 2012-09-06T21:25:57Z FirstChildTAG: When dragging a component, hit 'R' on the keyboard to rotate. FirstChildUserIdTAG: 43909 FirstChildUserNameTAG: hhomayouni FirstChildCreateTimeTAG: 2012-09-07T03:36:19Z FirstChildTAG: I can confirm (Chromium 20 / Linux) that in some cases pressing R doesn't work. However, I noticed that at the same time shortcuts that usually involve the Ctrl key (like Ctrl+X and Ctrl+V) now work without it, i.e. just pressing X will cut out the selected part. My hypothesis: there's a bug in the schematics tool that in some situations marks the Ctrl key as pressed when in the reality it isn't, thus preventing the R keyboard shortcut from working and altering the other shortcuts. FirstChildUserIdTAG: 99996 FirstChildUserNameTAG: rumith FirstChildCreateTimeTAG: 2012-09-07T10:01:30Z FirstChildTAG: just drag the component to drawing sheet and press R in the kay board it will ratate automatically FirstChildUserIdTAG: 140076 FirstChildUserNameTAG: snarayana FirstChildCreateTimeTAG: 2012-09-06T21:00:19Z FirstChildTAG: R to rotate, which Browser are you using? Make sure it's compatible. FirstChildUserIdTAG: 3856 FirstChildUserNameTAG: g6ejd FirstChildCreateTimeTAG: 2012-09-06T21:12:20Z IndexTAG: 4698 TitleTAG: Ridiculous, power is not entering the source Power NEVER enters a voltage source unless there is a parallel voltage source that has a higher voltage, and therefore the higher voltage source will cause current to flow into the lower voltage source, and it would either burn out the output circuit, or the crowbar circuit would activate. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-06T19:40:21Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 5_Simple_Power NumberOfReplyTAG: 1 FirstChildTAG: Well, if power is not entering, then power is leaving, which can be expressed by saying that the power entering is "negative," right? So, it's just about language convention. FirstChildUserIdTAG: 4463 FirstChildUserNameTAG: pmj FirstChildCreateTimeTAG: 2012-09-07T01:11:34Z SecondChildTAG: Thanks for that. I kept on typing the answer, and I knew it was correct, but I just did not get it. I did not know that power could be expressed as a vector. SecondChildUserIdTAG: 193033 SecondChildUserNameTAG: NIslam SecondChildCreateTimeTAG: 2012-09-07T17:42:31Z SecondChildTAG: As I understand, the power is entering the source because first, it passes trough the resistor, and then it continues until it gets to the source, and since it arrives into the negative pole, the value of it, it's negative SecondChildUserIdTAG: 306985 SecondChildUserNameTAG: morrodisey SecondChildCreateTimeTAG: 2012-09-08T14:52:29Z IndexTAG: 4699 TitleTAG: please help Hey please help me... I am stuck with a question H1P1 last part.... can anybody help me how to solve it? UserIdTAG: 108929 UserNameTAG: namit CreateTimeTAG: 2012-09-06T19:38:42Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4700 TitleTAG: Tran no works In TRAn no works. A window with the next dialog appear: **Trasient analysis, there are no probes in the diagram.** what can i Do? UserIdTAG: 257920 UserNameTAG: Mrkarurosu CreateTimeTAG: 2012-09-06T19:35:04Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 9 FirstChildTAG: Yes my tran does not work too. I insert transient analyses to 5ms and graph does not apear FirstChildUserIdTAG: 370575 FirstChildUserNameTAG: manweb FirstChildCreateTimeTAG: 2012-09-06T19:43:02Z FirstChildTAG: hi, is necessary install test point with the tool "voltage probe" on the points A-B-C FirstChildUserIdTAG: 343291 FirstChildUserNameTAG: LuisEduardoH FirstChildCreateTimeTAG: 2012-09-06T19:50:43Z FirstChildTAG: I also had the issue, but moved the probes to the circuit instead of the extension A-B-C. FirstChildUserIdTAG: 393045 FirstChildUserNameTAG: SrChasJC FirstChildCreateTimeTAG: 2012-09-07T00:38:08Z FirstChildTAG: DRAG THE PROBE FROM RIGHT SIDE AND THEN PLACE IT OVER THE CONCERNED POINT FirstChildUserIdTAG: 133068 FirstChildUserNameTAG: swarn99 FirstChildCreateTimeTAG: 2012-09-07T04:22:10Z FirstChildTAG: Try using Chrome. It doesn't work on FF for me either. FirstChildUserIdTAG: 267993 FirstChildUserNameTAG: mavlijas FirstChildCreateTimeTAG: 2012-09-07T03:46:39Z FirstChildTAG: I'm having the same problem. I've tried putting the probes in the nodes between the resistors, and on the "A, B, C" extensions. I've played around a lot with how values are entered in the voltage source. I'm beginning to wonder if maybe the tool doesn't work with my software. I'm running Windows7 and Firefox browser. It also has a weird feature in which "DC" and "TRAN" do not appear in the toolbar until I hit "CHECK". If I refresh the screen, the entire circuit vanishes. FirstChildUserIdTAG: 283726 FirstChildUserNameTAG: ByronInLawrence FirstChildCreateTimeTAG: 2012-09-07T02:48:02Z SecondChildTAG: correction: I don't get any error message. Simply nothing happens. SecondChildUserIdTAG: 283726 SecondChildUserNameTAG: ByronInLawrence SecondChildCreateTimeTAG: 2012-09-07T02:49:26Z FirstChildTAG: Same as everyone else here, the plot does not appear. Tried connecting the probe to different part of the circuit, nothing. Tried both IE and FireFox, nothing (OS: Win 7 - 64bit). FirstChildUserIdTAG: 43909 FirstChildUserNameTAG: hhomayouni FirstChildCreateTimeTAG: 2012-09-07T03:42:50Z FirstChildTAG: Add probes before running the analysis. FirstChildUserIdTAG: 86810 FirstChildUserNameTAG: Ariel12 FirstChildCreateTimeTAG: 2012-09-07T05:58:40Z FirstChildTAG: Try to put probe on point where resistor begins, not on point with letter A B or C. FirstChildUserIdTAG: 379420 FirstChildUserNameTAG: JasminMNE FirstChildCreateTimeTAG: 2012-09-07T06:31:20Z IndexTAG: 4701 TitleTAG: ans current direction dependent on the kvl loop that is applied UserIdTAG: 386683 UserNameTAG: chethankumar CreateTimeTAG: 2012-09-06T18:58:19Z VoteTAG: 0 CoursewareTAG: Week 1 / Power CommentableIdTAG: 6002x_S1E2_Power NumberOfReplyTAG: 0 IndexTAG: 4702 TitleTAG: Diagram I can't get the diagram right at the beginning. It won't let me delete certain things. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-06T18:43:58Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 4 FirstChildTAG: good afternoon, you can use zoom tool with scroll mouse and delete with backspace key FirstChildUserIdTAG: 343291 FirstChildUserNameTAG: LuisEduardoH FirstChildCreateTimeTAG: 2012-09-06T19:51:56Z FirstChildTAG: You can select any thing on diagram by two ways: 1. Select thing that you want to delete by click-and-move-around (green rectange) and press backspace. 2. Max zoom, click on thing (thing become green) and press backspace. FirstChildUserIdTAG: 258059 FirstChildUserNameTAG: Volant FirstChildCreateTimeTAG: 2012-09-06T21:47:59Z FirstChildTAG: just select the item which u want to delete and then click on cut symbol on the top left of your graph FirstChildUserIdTAG: 181793 FirstChildUserNameTAG: mitrahul FirstChildCreateTimeTAG: 2012-09-07T04:22:11Z FirstChildTAG: You can also move entire circuit up and over by a few squares. By disconnecting from the unwanted items makes it easier to highlight and then delete them. FirstChildUserIdTAG: 359302 FirstChildUserNameTAG: GaryG FirstChildCreateTimeTAG: 2012-09-07T11:33:44Z IndexTAG: 4703 TitleTAG: equation KCL at node e2 I figured out he answer by myself now, but i'm still not sure why it works like that. the convention is all currents leaving the node add up to 0. I am unsure if the current trough R1 is concerned. If I write down (e2-e1)/R1 it works, but why would it not work with -(e2-e1)/R1, because I go trough the negative terminal of R1. I think I illegally mix up the direction of current in R1 somehow. Can anyone try explain a bit deeper? UserIdTAG: 375898 UserNameTAG: madlab CreateTimeTAG: 2012-09-06T18:39:03Z VoteTAG: 0 CoursewareTAG: Week 1 / Node method example CommentableIdTAG: 6002x_S2E5_Node_Method NumberOfReplyTAG: 1 FirstChildTAG: I don't know the answer either, but I think another way of asking the question is this --> Shouldn't the KCL at e2 be i2=i1+3 ? How are we supposed to know that the 3A current source overwhelms the 5V and pushes current into (or against) the 5V source, making i1 the same direction as i2, which is away from the node e2?? FirstChildUserIdTAG: 180600 FirstChildUserNameTAG: Jesster78 FirstChildCreateTimeTAG: 2012-09-08T13:32:42Z IndexTAG: 4704 TitleTAG: Lab average and homework average wat does lab average and homework average depict in progress? anyone having an idea? and does it matter if we try to check our answers and re-correct them ? UserIdTAG: 368712 UserNameTAG: jmen CreateTimeTAG: 2012-09-06T18:18:44Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: They show the average result over your 10 highest marks from the homeworks and the labs. Only the best ten from the 12 will be used. FirstChildUserIdTAG: 80920 FirstChildUserNameTAG: MichaelSturgess FirstChildCreateTimeTAG: 2012-09-06T18:36:43Z FirstChildTAG: It is giving us the score that we are earning for our final evaluation. It will increase when we solve our labs/home works week by week. Try to get 100% , it doesn't matter how many times u try to get it correct. Don't use random try, in most cases, it is in your steps that would have resulted in wrong answer.. All the best. FirstChildUserIdTAG: 823 FirstChildUserNameTAG: naveenatmit FirstChildCreateTimeTAG: 2012-09-06T18:58:06Z IndexTAG: 4705 TitleTAG: Average power formulae let p(t) be instantaneous power p(t)=[v(t)*v(t)]/R. Therfore, avg power= (1/T)(integral of p(t) over time period T). hope this will be of use... UserIdTAG: 148523 UserNameTAG: span993 CreateTimeTAG: 2012-09-06T17:53:35Z VoteTAG: 0 CoursewareTAG: Week 1 / AC power CommentableIdTAG: 6002x_S1E3_AC_power NumberOfReplyTAG: 0 IndexTAG: 4706 TitleTAG: Stuck in Lab 2 Please Someone help me out in solving Lab 2. How to build the mixer with the proportion 1/2 V1 +1/6 V2. Any Idea??? UserIdTAG: 77046 UserNameTAG: Sarwar20 CreateTimeTAG: 2012-09-06T17:50:12Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: I am working on it too. if you insert only two resistors then the output coefficient must sum up to 1, but 1/2 and 1/6 does not some up to 1. so i am confused too. FirstChildUserIdTAG: 131426 FirstChildUserNameTAG: M_Inam_Ul_Haq FirstChildCreateTimeTAG: 2012-09-09T13:14:23Z IndexTAG: 4707 TitleTAG: DC AC TRAN Sometimes I have to refresh the page to see this tools, does anyone knows why? thanks a lot UserIdTAG: 278910 UserNameTAG: Carlos92 CreateTimeTAG: 2012-09-06T17:15:32Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 3 FirstChildTAG: I was facing problem in wiring in Google chrome but when I SWITCHED to Firefox many problems have solved. FirstChildUserIdTAG: 40209 FirstChildUserNameTAG: sadaf78 FirstChildCreateTimeTAG: 2012-09-06T20:58:34Z FirstChildTAG: Hi Carlos92! What web broser are you using? Try with the Last Updated version of Chrome or Firefox. I have not that problem... ---- Hola Carlos92! Qué Navegador estás utilizando? Intenta con la última versión actualizada de Chrome o Firefox. Quizás sea por ello que no puedes visualizar correctamente los componentes... yo no tengo ese problema... FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-06T17:41:24Z FirstChildTAG: Issues like yours could be browser-related; I spent almost 2hrs trying to draw a wire between two connection points in Chrome. I had to switch to Firefox to complete the circuit. FirstChildUserIdTAG: 363999 FirstChildUserNameTAG: uzosky FirstChildCreateTimeTAG: 2012-09-06T18:22:47Z SecondChildTAG: I don't think it is browser-related,because I am also facing same wiring problem but when i switched to Firefox this problem has solved BUT TRAN is not working please tell me what should i do to complete the task SecondChildUserIdTAG: 40209 SecondChildUserNameTAG: sadaf78 SecondChildCreateTimeTAG: 2012-09-06T22:15:48Z IndexTAG: 4708 TitleTAG: Question 5 the answers correct? I think the answers for the Question 5 are not correct. As the transient analysis shows the voltages for the three nodes A, B and C are 1.008V, 0.83V and 0.33V respectively for the sinusoid V supply with 1V amplitude, 1V offset and 1kHz frequency at 1.25ms. UserIdTAG: 161128 UserNameTAG: AvishekC CreateTimeTAG: 2012-09-06T17:10:48Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 5 FirstChildTAG: Your values are half expected. Check your frequency input. FirstChildUserIdTAG: 141987 FirstChildUserNameTAG: lawrence26 FirstChildCreateTimeTAG: 2012-09-06T19:10:31Z FirstChildTAG: thanks you were right! my frequency input was incorrect. it was 1 Hz instead of 1 kHz. Thanks! FirstChildUserIdTAG: 161128 FirstChildUserNameTAG: AvishekC FirstChildCreateTimeTAG: 2012-09-06T22:10:50Z SecondChildTAG: I had the same problem as you, now solved the same way you did SecondChildUserIdTAG: 154373 SecondChildUserNameTAG: JReyes87 SecondChildCreateTimeTAG: 2012-09-07T05:46:49Z FirstChildTAG: You're using DC analysis, I got the same numbers. Click TRAN and set to 1.25ms and then mouse over the right numbers. I got 2V, 1.667V and .666V respectively and it accepted the answer. FirstChildUserIdTAG: 354941 FirstChildUserNameTAG: zeke8402 FirstChildCreateTimeTAG: 2012-09-06T17:17:41Z SecondChildTAG: Actually, you may have been using TRANS analysis but somethings not right in your stop time. Your numbers are correct when stop time is 1ms, not 1.25 SecondChildUserIdTAG: 354941 SecondChildUserNameTAG: zeke8402 SecondChildCreateTimeTAG: 2012-09-06T17:18:54Z SecondChildTAG: I got basically the same numbers, but it won"t let me enter third number as .666V and mine showed more like what looked like mA on third one but same number any ideas? SecondChildUserIdTAG: 359302 SecondChildUserNameTAG: GaryG SecondChildCreateTimeTAG: 2012-09-07T03:43:55Z FirstChildTAG: you set an offset voltage of 1V ? FirstChildUserIdTAG: 332449 FirstChildUserNameTAG: 4a4ik FirstChildCreateTimeTAG: 2012-09-06T19:16:53Z SecondChildTAG: Yeah I did. But my frequency input was not right. Thanks! SecondChildUserIdTAG: 161128 SecondChildUserNameTAG: AvishekC SecondChildCreateTimeTAG: 2012-09-06T22:12:49Z FirstChildTAG: i got thr right answer check ur circuit FirstChildUserIdTAG: 140076 FirstChildUserNameTAG: snarayana FirstChildCreateTimeTAG: 2012-09-06T21:06:08Z SecondChildTAG: Thanks I checked and found out that my frequency input were wrong. SecondChildUserIdTAG: 161128 SecondChildUserNameTAG: AvishekC SecondChildCreateTimeTAG: 2012-09-06T22:13:42Z IndexTAG: 4709 TitleTAG: Done I got every answer correct except one, 500e-6. Even though I got the 500 part correct, I don't get what e-6 means? Is it a logarithmic expression for e^-6? UserIdTAG: 143402 UserNameTAG: Colobus_T CreateTimeTAG: 2012-09-06T16:57:06Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 4 FirstChildTAG: In scientific notation, "e" stands for "10^" so that `500e-6` = $500\times 10^{-6}$ FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-06T17:05:38Z FirstChildTAG: 500e-6 = 500*10^-6 = 0,0005 For example: 2e3 = 2*10^3 = 2000 2e-3 = 2*10^-3 = 0,002 FirstChildUserIdTAG: 256632 FirstChildUserNameTAG: costasilvar FirstChildCreateTimeTAG: 2012-09-06T17:03:04Z FirstChildTAG: In scientific notation "e" refers to EXPONENT of powers of 10, so 500e-6 or 500E-6 is the same as 500 * 10^(-6), which you can also write here as 500u, since "u" is "micro" or 10^(-6). FirstChildUserIdTAG: 4463 FirstChildUserNameTAG: pmj FirstChildCreateTimeTAG: 2012-09-06T17:07:16Z FirstChildTAG: They gave you the scientific notation explanation, but I think what your problem is this... They are asking you to give the answer in amps, The current flowing from source given with the DC analysis is in microamps, so you have to convert that to amps. 500 microamps is equal to .0000005 amps or... 500 x 10^-6 amps or simply... 500e-6 FirstChildUserIdTAG: 354941 FirstChildUserNameTAG: zeke8402 FirstChildCreateTimeTAG: 2012-09-06T17:12:43Z SecondChildTAG: This is not exactly correct. 500 microamps is 0.000500 amps (need to "move" the decimal 6 places to the left) not 0.0000005 which is actually 5e-7. The answer would be either 5e-4 or 500e-6. SecondChildUserIdTAG: 141987 SecondChildUserNameTAG: lawrence26 SecondChildCreateTimeTAG: 2012-09-06T19:15:51Z IndexTAG: 4710 TitleTAG: proper management of technological tools good morning as I would use technological tools, such as the circuit simulator thanks UserIdTAG: 45303 UserNameTAG: brenda CreateTimeTAG: 2012-09-06T16:36:02Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4711 TitleTAG: Value only, unit not included I learned that only the value is necessary, not the unit i.e. 500uA must be 500u to be check by the system. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-06T16:34:42Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 0 IndexTAG: 4712 TitleTAG: Parallel Resistance I cant figure out how the power of composite parallel resistance depends or related to power of individual resistances in parallel? UserIdTAG: 131426 UserNameTAG: M_Inam_Ul_Haq CreateTimeTAG: 2012-09-06T16:26:34Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: As a hint draw the circuit using the 3 resistors in parallel. Next from the information given calculate all the unknown values for the circuit. From these unknowns you should then be able to calculate the power of the composite resistor. FirstChildUserIdTAG: 372866 FirstChildUserNameTAG: jonmark67 FirstChildCreateTimeTAG: 2012-09-06T16:43:51Z SecondChildTAG: One thing that could make the analysis more easy is attribute one value of voltage and use the formulas like P=V^2/I. SecondChildUserIdTAG: 309428 SecondChildUserNameTAG: ThiagoNunes SecondChildCreateTimeTAG: 2012-09-06T16:55:11Z IndexTAG: 4713 TitleTAG: Current I did it all but the only snag is current flowing from source into the circuit.both sign and amplitude of the current ? UserIdTAG: 132475 UserNameTAG: mekazanc CreateTimeTAG: 2012-09-06T15:59:23Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: Nope, only magnitude FirstChildUserIdTAG: 131426 FirstChildUserNameTAG: M_Inam_Ul_Haq FirstChildCreateTimeTAG: 2012-09-06T16:25:01Z IndexTAG: 4714 TitleTAG: recognized values I entered in a value .40V can anyone tell me why that would not be a recognized value or how it would be expressed? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-06T15:32:33Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 4 FirstChildTAG: enter it as 0.4*V FirstChildUserIdTAG: 213700 FirstChildUserNameTAG: bhaveshreddy FirstChildCreateTimeTAG: 2012-09-06T16:34:36Z FirstChildTAG: Are you entering the V? What happens if you enter 0.40? FirstChildUserIdTAG: 9147 FirstChildUserNameTAG: SueP FirstChildCreateTimeTAG: 2012-09-06T15:41:16Z FirstChildTAG: 0.40 might work FirstChildUserIdTAG: 122513 FirstChildUserNameTAG: Mona77 FirstChildCreateTimeTAG: 2012-09-06T16:21:51Z FirstChildTAG: You need to either type "0.4" or "400m". Don't type the "V". FirstChildUserIdTAG: 283726 FirstChildUserNameTAG: ByronInLawrence FirstChildCreateTimeTAG: 2012-09-07T03:00:52Z IndexTAG: 4715 TitleTAG: Textbook pages Sorry for the naive question but I was wondering if there is a way to go from a page in the textbook to another that is not immediately following this one. For example from page 10 to page 23 without passing through all the intermediate pages. Thanks for the answers! UserIdTAG: 147593 UserNameTAG: knack CreateTimeTAG: 2012-09-06T15:31:01Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: add my facebook, i counld send u a copy of the original textbook in pdf format. then u counld do that jumping by using the adobe reader etc. ps: weipanli FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-06T16:30:00Z SecondChildTAG: This could be considered ilegal in some countries.... SecondChildUserIdTAG: undefined SecondChildUserNameTAG: undefined SecondChildCreateTimeTAG: 2012-09-06T16:48:42Z FirstChildTAG: If you like to go to a page, just use this link, and adjust the last number by adding the page number +24. This is for page 10: https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/34 To go to page 23, add 23+24=47, and adjust the last number. https://www.edx.org/courses/MITx/6.002x/2012_Fall/book/0/47 Furthermore, you can use the TOC on the left side. FirstChildUserIdTAG: 297279 FirstChildUserNameTAG: ZlatkoF FirstChildCreateTimeTAG: 2012-09-06T16:30:23Z IndexTAG: 4716 TitleTAG: Transient analysis became tough! From last set of questions,I Had a Doubt. Why does the graph doesn't show the exact value of node voltage at A,B & c when it's pointing to 1.25ms time? UserIdTAG: 232076 UserNameTAG: JOMANUSO CreateTimeTAG: 2012-09-06T15:22:58Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: that´s right! we can see the amplitude but not the voltage, that in my opinion in more useful FirstChildUserIdTAG: 278910 FirstChildUserNameTAG: Carlos92 FirstChildCreateTimeTAG: 2012-09-06T17:12:54Z IndexTAG: 4717 TitleTAG: Anyone missing on this exclusive facebook Group. I guess many of us are still unaware of the exclusive facebook group for the students of EDX 6.002x. Link is here- http://www.facebook.com/groups/483354858342165/ UserIdTAG: 253902 UserNameTAG: sajalok CreateTimeTAG: 2012-09-06T14:29:24Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: Hello Sajal I requested access to your group, however I am concerned you will not know I am signed up for the class, therefore I contacting you through this discussion group. May I please enter the group. Thank you, Richard Harris FirstChildUserIdTAG: 132828 FirstChildUserNameTAG: rharris FirstChildCreateTimeTAG: 2012-09-08T04:10:28Z FirstChildTAG: The search box at the top of the screen is available for anyone to search for facebook references. FirstChildUserIdTAG: 9147 FirstChildUserNameTAG: SueP FirstChildCreateTimeTAG: 2012-09-06T15:40:13Z IndexTAG: 4718 TitleTAG: Calculator Can't find the calculator... What softwares are required? Please help!!! UserIdTAG: 360609 UserNameTAG: mohitkarkhanis7 CreateTimeTAG: 2012-09-06T14:02:54Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: The calculator is now back after a brief absence. FirstChildUserIdTAG: 123717 FirstChildUserNameTAG: cpennington FirstChildCreateTimeTAG: 2012-09-07T23:09:27Z IndexTAG: 4719 TitleTAG: Equation "We know that e1` = e3 + v"... but why??? edit: Ok i got it :P UserIdTAG: 357847 UserNameTAG: PG13 CreateTimeTAG: 2012-09-06T14:00:42Z VoteTAG: 0 CoursewareTAG: Week 1 / Node method example CommentableIdTAG: 6002x_S2E5_Node_Method NumberOfReplyTAG: 1 FirstChildTAG: because the equation of e3 is a function of e1, I and R1, R2: e3/R2 + e1/R1 + I = 0 and as the current flowing through the voltage source is the same as that flowing through the resistor R1 FirstChildUserIdTAG: 51696 FirstChildUserNameTAG: blasco23 FirstChildCreateTimeTAG: 2012-09-07T16:39:29Z SecondChildTAG: This works, but why is the current through voltage source only effected by R1? Why doesn't it supply current to R2? SecondChildUserIdTAG: 248807 SecondChildUserNameTAG: 1kingsman SecondChildCreateTimeTAG: 2012-09-12T03:31:56Z SecondChildTAG: dont think too much .. v = e1-e3 (voltage is the differnce in potentials)... hence e1=V+e3 SecondChildUserIdTAG: 185450 SecondChildUserNameTAG: ajoe SecondChildCreateTimeTAG: 2012-09-17T16:35:10Z IndexTAG: 4720 TitleTAG: NOTE!!! Notice that these two voltage drops add up to the battery supply voltage, 3.00volts---as they should in a series circuit. UserIdTAG: 117661 UserNameTAG: BUNDAY CreateTimeTAG: 2012-09-06T13:41:13Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 0 IndexTAG: 4721 TitleTAG: week 3 When Week 3 is going to upload. UserIdTAG: 908 UserNameTAG: m_umair72001 CreateTimeTAG: 2012-09-06T13:35:55Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: New weeks will go up on Mondays. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-09-06T13:36:52Z IndexTAG: 4722 TitleTAG: Is it the same course? Hi All, I registered for 6002x in FEB this year but didn't really followed it due to a couple of reasons. So, I want to ask that is this course similar to previous one offered earlier this year. Anyone who could answer!! UserIdTAG: 75391 UserNameTAG: muneebullah CreateTimeTAG: 2012-09-06T11:47:29Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: yup this is d sam FirstChildUserIdTAG: 72647 FirstChildUserNameTAG: NEEL11 FirstChildCreateTimeTAG: 2012-09-06T11:52:33Z FirstChildTAG: Muneebullah - I believe it is the same course, but it is at a different url: www.edx.org? I am still confused about the url... I did lab 0 yesterday. But I closed the browser by mistake and I had a terrible time trying to get back to the course. I am still confused. FirstChildUserIdTAG: 145676 FirstChildUserNameTAG: CathyPK FirstChildCreateTimeTAG: 2012-09-06T11:53:39Z IndexTAG: 4723 TitleTAG: Explantion of Answer of Question S1E1.5: SIMPLE POWER. Could anybody please explain the answer of Question S1E1.5: SIMPLE POWER? UserIdTAG: 2956 UserNameTAG: akshayb CreateTimeTAG: 2012-09-06T11:31:38Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: 5_Simple_Power NumberOfReplyTAG: 2 FirstChildTAG: power = v2/R FirstChildUserIdTAG: 151062 FirstChildUserNameTAG: 1232 FirstChildCreateTimeTAG: 2012-09-06T12:04:40Z SecondChildTAG: just remeber P=VI & V=IR hence, P=(V^2)/R or P=(I^2)R SecondChildUserIdTAG: 625118 SecondChildUserNameTAG: alex_003 SecondChildCreateTimeTAG: 2012-10-17T18:00:46Z FirstChildTAG: Your first equation is V=IR ....(1) Your second equation is p=VI....(2) If you change I from eq.(1) in eq.(2) you have p=(V^2)/R or if you change V from eq.(1) in eq.(2) you have p=(I^2)*R FirstChildUserIdTAG: 268104 FirstChildUserNameTAG: SaulCardenasG FirstChildCreateTimeTAG: 2012-09-08T04:28:07Z IndexTAG: 4724 TitleTAG: wires are not avaliable please solve the problem i am unable to wire the components in lab 0 plz tell me what is the main problem. UserIdTAG: 40209 UserNameTAG: sadaf78 CreateTimeTAG: 2012-09-06T11:20:35Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: click on terminal of component and **DRAG** it. FirstChildUserIdTAG: 908 FirstChildUserNameTAG: m_umair72001 FirstChildCreateTimeTAG: 2012-09-06T13:24:55Z IndexTAG: 4725 TitleTAG: sign of current well n/w here is equivalent to a source for the resistor and the term "current entering n/w N from the resistor" is to define the direction of current.As n/w N is acting like a source so current entering its +ve terminal must be with a -ve sign UserIdTAG: 227524 UserNameTAG: sanju36 CreateTimeTAG: 2012-09-06T10:39:19Z VoteTAG: 0 CoursewareTAG: Week 1 / Power CommentableIdTAG: 6002x_S1E2_Power NumberOfReplyTAG: 1 FirstChildTAG: (btw, n/w = nowhere?) Ok, so if there is nowhere a source inside N then we cant have + or - signs at all as no current can flow anywhere and power cannot be consumed if it is not provided - it doesnt exist. No psu => then I=zero, but zeros do not have a sign. Therefore imo, as long as a single + or - sign exits, or a current is assumed to flow, then a source is implicit. To avoid the sign problem elegantly, question should of been "what is the current passing trough the resistor?". This way we dont have to bother with a convention notation that hasnt been made explicit and we dont need to give names to resistors legs. We can answer that question regardless which direction current may be assumed to flow into. I may be wrong, but i think that should of been better way to put the question. FirstChildUserIdTAG: 331483 FirstChildUserNameTAG: Doru FirstChildCreateTimeTAG: 2012-09-06T12:00:04Z SecondChildTAG: agreed SecondChildUserIdTAG: undefined SecondChildUserNameTAG: undefined SecondChildCreateTimeTAG: 2012-09-06T21:53:12Z IndexTAG: 4726 TitleTAG: KVL I have a problem when I introduce my algebraic expressions, when i put voltage I introduce and v, like v4 or v1 but it told me that this is and invalid imput and i don't know what i have to put instead. Because in KCL I put and i when i mean current and it is correct. Anyone can help me please UserIdTAG: 307811 UserNameTAG: AlejandraBlanco CreateTimeTAG: 2012-09-06T10:16:35Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4727 TitleTAG: facing the problem my tool board is not showing the AC. Mozilla/5.0 (Windows NT 6.1; WOW64; rv:14.0) Gecko/20100101 Firefox/14.0.1 UserIdTAG: 358602 UserNameTAG: qpawelvrn CreateTimeTAG: 2012-09-06T10:11:45Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 2 FirstChildTAG: you have to double click at the voltage source and select the type as 'sin' FirstChildUserIdTAG: 350947 FirstChildUserNameTAG: kamath777 FirstChildCreateTimeTAG: 2012-09-06T10:29:47Z FirstChildTAG: One should use Google Chrome for entire course and recommended. FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-06T10:55:20Z IndexTAG: 4728 TitleTAG: S1E5: KVL-0 Please check S1E5: KVL-0 task. I suppose that point between resistors 2 and 3 (let call it point A) can not have the lowest potential (than potentials 1-2 and 2-3 ). In this case current will flow from point A in both sides, so dq/dt !=0 UserIdTAG: 191310 UserNameTAG: Kalashnikov CreateTimeTAG: 2012-09-06T09:45:14Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4729 TitleTAG: No video, no sand box, no lab. Nothing I've tried IE and Chrome, various PC's and can see nothing so far. Any ideas any one? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-06T09:43:54Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: This server or platform I and others are running on is still broken. Lots are assuming it is a browser issue. It is NOT the browser. I have logged into my old MITx course and my old course-ware still shows the simulator working correctly. So, the problem is probably due to how they allocate resources as obviously not all students will be sharing same platform/server. My guess is there are Lots of platforms/servers running identical code but all linked to a central Database. Those of us seeing these bugs are on a server which is flaky or with simulator software corrupted on it in some way (probably needs latest patch or something). Everyone else who are not seeing these problems are almost certainly on platforms which are stable. Not much we can do but keep posting that it is broken as much as you can, until they either let us know what is happening or get it fixed. FirstChildUserIdTAG: 15344 FirstChildUserNameTAG: kob FirstChildCreateTimeTAG: 2012-09-06T10:28:26Z FirstChildTAG: I'm pretty sure you need to install(or reinstall) adobe flash player. http://get.adobe.com/ru/flashplayer/ FirstChildUserIdTAG: 128561 FirstChildUserNameTAG: gusevoy FirstChildCreateTimeTAG: 2012-09-06T10:31:18Z SecondChildTAG: No that is not it, as I run same code on MITx, same machine, same browser, the newest flash software, same identical simulator code, no problem what so ever. It is a resource problem at their end. SecondChildUserIdTAG: 15344 SecondChildUserNameTAG: kob SecondChildCreateTimeTAG: 2012-09-06T10:39:33Z IndexTAG: 4730 TitleTAG: For Downloading Video How can I Download the Video UserIdTAG: 254187 UserNameTAG: SURU13 CreateTimeTAG: 2012-09-06T08:55:52Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: Hi, there are many ways to download then: Using a software: aTubeCatcher, Real Player, and other programs like theese or using web services like KeepVid.com (it's free), just google "download youtube" and you'll find many ways to solve your problem. FirstChildUserIdTAG: 43833 FirstChildUserNameTAG: gmcentenom FirstChildCreateTimeTAG: 2012-09-06T09:01:55Z FirstChildTAG: You can install JDownloader, a freeware program written in java from http://jdownloader.org/ FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-09-06T17:01:39Z IndexTAG: 4731 TitleTAG: video of lecture 1 i want to download the video of the first lecture UserIdTAG: 303524 UserNameTAG: waelnageb CreateTimeTAG: 2012-09-06T08:31:05Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: There are many ways to download them: Using a software: aTubeCatcher, Real Player, and other programs like theese or using web services like KeepVid.com (it's free), just google "download youtube" and you'll find many ways to solve your problem.** FirstChildUserIdTAG: 43833 FirstChildUserNameTAG: gmcentenom FirstChildCreateTimeTAG: 2012-09-06T09:03:35Z IndexTAG: 4732 TitleTAG: H2P1: Not accepting my answer Hi there, I'm new here so please excuse if I'm doing something wrong. I have a problem submitting the resistor divider of H2P1. Since I don't want to post my solution (that is marked as wrong), since this is probably illegal here, I'll just write this: In my calculation, I get a output voltage of 18.57V (16.2V < 18.57V < 19.8V, i.e. within +-10% of the 18V target) and a resistor sum of 18.9k (within 10k-30k) with both resistors chosen from the E12 series. I've even verified it with DC analysis, telling me the same thing (i.e. the output voltage is 18.57V). My answer is not accepted, however. Could somebody please tell me what I'm doing wrong? Thanks, Joe UserIdTAG: 144694 UserNameTAG: johndoe31415 CreateTimeTAG: 2012-09-06T08:20:37Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: "Thevenin resistance as seen from the output terminals". It is not resistor sum, they parallel FirstChildUserIdTAG: 19803 FirstChildUserNameTAG: algarkavy FirstChildCreateTimeTAG: 2012-09-07T20:23:44Z SecondChildTAG: AAAAAAAAAHHHH! Thank you so much! SecondChildUserIdTAG: 144694 SecondChildUserNameTAG: johndoe31415 SecondChildCreateTimeTAG: 2012-09-08T09:42:12Z IndexTAG: 4733 TitleTAG: How to Write Letters A,B and Etc at the End of Each points? With Due Respect If any one have an idea about to tell me How to Write Letters A,B and Etc at the End of Each points? Best Regards Abid Ali UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-06T07:17:00Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: You don't have to write the Letter A, B, and etc. Just make the circuit and you can answer the question without write those letter FirstChildUserIdTAG: 185114 FirstChildUserNameTAG: m_akbar_s FirstChildCreateTimeTAG: 2012-09-06T07:48:28Z FirstChildTAG: After clicking on the node label tool, drag it and drop at the desired position. Then rotate the vertical tool until it points the ??? head outwards. Next, double click on it to bring up its properties for labeling and Label it A and so on. N.B: Ensure the tool is green before clicking for the property to be labeled otherwise you would not be able to edit it. I hope this helps. FirstChildUserIdTAG: 329879 FirstChildUserNameTAG: Ugochukwu FirstChildCreateTimeTAG: 2012-09-06T08:02:16Z IndexTAG: 4734 TitleTAG: Lab0 - error in auto calculation I enetered schematic, press Check and get 'correct' mark on it. But when I press DC button, I get following results: ![dc_wrong_esult][1] [1]: http://s09.radikal.ru/i182/1209/c0/e7bada11453d.png A=3V B=2.25V C=0.00V D=0.00V I=-749uA This is wrong answer, so I calculated it myself and get correct answers, which passed check. What's wrong with the tool or with my way of using it? UserIdTAG: 324219 UserNameTAG: Dmitry79 CreateTimeTAG: 2012-09-06T06:53:06Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 3 FirstChildTAG: mira te has equivocado en la tercera resistencia es 2kohms FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-06T06:57:59Z FirstChildTAG: You've got R3 set to 2 Ohms, change it to 2k and it should work properly. FirstChildUserIdTAG: 373220 FirstChildUserNameTAG: oohall FirstChildCreateTimeTAG: 2012-09-06T06:59:15Z SecondChildTAG: Thank you! It works now. SecondChildUserIdTAG: 324219 SecondChildUserNameTAG: Dmitry79 SecondChildCreateTimeTAG: 2012-09-06T07:06:31Z FirstChildTAG: The tool is accurate. I got I=-500uA, V at node B=2.50V and V at node C=1.00v Voltage exiting Vsupply=3V Try to remove the 3rd probe, then fix the probes - one at ground point and the other at pts B or C for their respective values. FirstChildUserIdTAG: 329879 FirstChildUserNameTAG: Ugochukwu FirstChildCreateTimeTAG: 2012-09-06T08:07:58Z IndexTAG: 4735 TitleTAG: I am getting it wrong I guess there is something wrong with this problem. I am unable to get the answers posted. What I got satisfy the power balance condition posted at the bottom, but my values are quite diiferent from what is posted above. What I got are What is the voltage (in Volts) v2 across the resistor with resistance R2? 6.667volts What is the power (in Watts) dissipated by the resistor with resistance R2? 8.888watts What is the current (in Amperes) i1 through the resistor with resistance R1? -1.167amperes What is the power (in Watts) dissipated by the resistor with resistance R1? 5.444 watts What is the power (in Watts) supplied by the voltage source? -2.333watts What is the power (in Watts) supplied by the current source? 16.667watts Please, if you have an idea of what i did wrong, let me know UserIdTAG: 332113 UserNameTAG: latonbaba CreateTimeTAG: 2012-09-06T06:42:02Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL/KCL example CommentableIdTAG: 6002x_S2E3_Using_KVL_KCL_and_VI_constraints NumberOfReplyTAG: 3 FirstChildTAG: you have to use the method give in lectures !! write first all the ohms laws: voltage source v=V ; v1=r1*i1 ; v2=r2*i2 secondly KVL: V-v1-v2=0 thridly KCL: i1+I-i2=0 combine thos three equation and you will have the result FirstChildUserIdTAG: 151444 FirstChildUserNameTAG: mael FirstChildCreateTimeTAG: 2012-09-06T07:30:11Z FirstChildTAG: jst check it [here][1] [1]: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/MITx_6_002x_2012_Fall/threads/5047a8af6859412700000021 FirstChildUserIdTAG: 82571 FirstChildUserNameTAG: Nidhi3 FirstChildCreateTimeTAG: 2012-09-06T08:01:16Z FirstChildTAG: nodes: i2=i1+I Loop: ---Vo-R1i1-R2i2=0----- ---Vo=R1i1+R2(i1+I)------ ---Vo=R1i1+R2i1+IR2---- ---i1= (Vo-IR2)/(R1+R2)---- --i2=i1+I---- ----What is the power (in Watts) supplied by the current source? ---- ----r2xi2xI---- ----r2xi2-->Voltage, this one x I = Power of Current Source.--- FirstChildUserIdTAG: 143835 FirstChildUserNameTAG: sanpablc FirstChildCreateTimeTAG: 2012-09-07T22:05:52Z IndexTAG: 4736 TitleTAG: Explain 1, the open circuit od a battery always is the same as the baterry, the battery resistence in low so it almost change the voltage. 2,The cortocircuit its the sum of the two I so: 1.5/0.25+1.5+0.32 3, These are a pararel resistance R1||R2 so: 1/Rt = 1/R1 + 1/R2; Rt = R1*R2/(R1+R2) = 0.25*0.35/(0.25+0.35) 4, The another battery shoul reach 0.1V so: 1/(0.32+0.25) UserIdTAG: 299251 UserNameTAG: TheRedBlackOne CreateTimeTAG: 2012-09-06T06:32:06Z VoteTAG: 0 CoursewareTAG: Week 1 / Battery model CommentableIdTAG: 6002x_S1E9_Battery_Model NumberOfReplyTAG: 1 FirstChildTAG: why not we take internal resistance of the battery to calculate 2,3,4 part of the question. FirstChildUserIdTAG: 375307 FirstChildUserNameTAG: imajeet FirstChildCreateTimeTAG: 2012-09-06T13:21:47Z SecondChildTAG: Brother, but according to kvl 1.Must be the negation of the voltage.? 2. this point was not mentioned in the lecture (AM I RIGHT)? 3. Good solution 4. plz elaborate SecondChildUserIdTAG: 128599 SecondChildUserNameTAG: usamabinqasim SecondChildCreateTimeTAG: 2012-09-16T16:50:30Z IndexTAG: 4737 TitleTAG: Solution The RMS Voltage value of this wave is 120V so, 120^2/110 =131 W is the PowerRMS. In this case Power RMS and Power dissipated by the resistor are the same. We can say, the Power RMS is the power makes the resistor gets hot. And it's true that mean value is zero. Guillermo. UserIdTAG: 359677 UserNameTAG: guillermb CreateTimeTAG: 2012-09-06T05:42:45Z VoteTAG: 0 CoursewareTAG: Week 1 / AC power CommentableIdTAG: 6002x_S1E3_AC_power NumberOfReplyTAG: 0 IndexTAG: 4738 TitleTAG: Answer to problem I think that the best form to solve the problem is based on Newton's second law that we demostrate in elementary physics classes (f=m*a=m*(dx/dt)=m*(d^2t/d^2)=dp/dt, this is because this law shows as a force that is applied affects to any body UserIdTAG: 335890 UserNameTAG: neosalvation CreateTimeTAG: 2012-09-06T05:25:43Z VoteTAG: 0 CoursewareTAG: Week 1 / Lumped Element Abstraction CommentableIdTAG: 6002x_lumped_element_abstraction NumberOfReplyTAG: 1 FirstChildTAG: $ f=m*a=m*(dv/dt)=d(m*p)/dt$ FirstChildUserIdTAG: 539003 FirstChildUserNameTAG: amitgupta25121993 FirstChildCreateTimeTAG: 2012-10-04T23:21:13Z IndexTAG: 4739 TitleTAG: Exercise Questions I've successfully completed the exercise after the video lecture, but it shows up as half done in the exercise icon above. Why??? UserIdTAG: 204745 UserNameTAG: allwynmendes CreateTimeTAG: 2012-09-06T05:23:42Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: I'm really sorry guys, but it seems that the problem auto-corrected itself somehow. But weird that for a while it showed up as half done. FirstChildUserIdTAG: 204745 FirstChildUserNameTAG: allwynmendes FirstChildCreateTimeTAG: 2012-09-06T05:28:51Z IndexTAG: 4740 TitleTAG: Loops It took me a while to find the last loop. This course is quite interesting, especially since I have no physics. Thanks for listing the loops here. UserIdTAG: 39123 UserNameTAG: Bdandaraw CreateTimeTAG: 2012-09-06T04:39:26Z VoteTAG: 0 CoursewareTAG: Week 1 / Circuit topology CommentableIdTAG: 6002x_S2E1_Circuit_Topology NumberOfReplyTAG: 0 IndexTAG: 4741 TitleTAG: Happy Well I fell so Happy here and I've finished everething Amazing UserIdTAG: 83519 UserNameTAG: Lidy CreateTimeTAG: 2012-09-06T04:32:03Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4742 TitleTAG: lecture video not playing Android I participated in the pilot course using my galaxy 10.1 tablet running honeycomb 3.2 with no issues. I'm now running ice cream sandwich and can't get the lecture videos to load. Is anyone else having issues watching videos on Android? UserIdTAG: 24887 UserNameTAG: Jeff260z CreateTimeTAG: 2012-09-06T03:24:41Z VoteTAG: 0 CoursewareTAG: Week 1 / Course overview CommentableIdTAG: 6002x_S1V1toS1V4 NumberOfReplyTAG: 3 FirstChildTAG: im using my laptop and im now watching video very well FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-06T09:14:23Z FirstChildTAG: I am using android w ICS and I see the video fine. However, closed captioning on the side of the video only partially works (jumps to top instead of next line) . Bethany FirstChildUserIdTAG: 339698 FirstChildUserNameTAG: BethanyBovard FirstChildCreateTimeTAG: 2012-09-07T11:33:19Z FirstChildTAG: Do you have flash player app installed? FirstChildUserIdTAG: 207456 FirstChildUserNameTAG: Octomos FirstChildCreateTimeTAG: 2012-09-08T10:37:06Z SecondChildTAG: Google removed flash support from Android in August. There are work-a-rounds, but it still boggles the mind why they did it, especially as probably their most used service, Youtube, is flash based SecondChildUserIdTAG: 3007 SecondChildUserNameTAG: badness10000 SecondChildCreateTimeTAG: 2012-09-14T10:35:44Z IndexTAG: 4743 TitleTAG: signs of current I think when the current goes from negative pole to the positive pole then the current is negative and viceversa , because the electron charge is negative and the electrical current is an electrons flux. Sorry by my horrible english. greeting everyone, very interesting iniciative these courses online :D UserIdTAG: 144678 UserNameTAG: jorge07 CreateTimeTAG: 2012-09-06T03:01:17Z VoteTAG: 0 CoursewareTAG: Week 1 / Power CommentableIdTAG: 6002x_S1E2_Power NumberOfReplyTAG: 1 FirstChildTAG: I don't really think you're right cause it seems that the generator is unknown ;let assum that the generator is where the network N is the current direction will be + to - , and if it is contrary the sign will be - such as when we throw sth upwards the sign is -;because it is opposite to gravity force. FirstChildUserIdTAG: 273039 FirstChildUserNameTAG: SOCRATE23 FirstChildCreateTimeTAG: 2012-09-06T14:53:16Z IndexTAG: 4744 TitleTAG: Sequence 1 & 4 exercise numbering I noticed that Sequence 1 exercises are numbered strangely. There's an S1E1.5, but no S1E4. Same story on Sequence 4, there's an S4E2 and S4E3, but no S4E1. Mistake, or intentional? UserIdTAG: 94557 UserNameTAG: g_hopper CreateTimeTAG: 2012-09-06T02:09:39Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: I think S1E1.5 is a intentional way to name the set of excercises between S1E1 and S1E2 FirstChildUserIdTAG: 43833 FirstChildUserNameTAG: gmcentenom FirstChildCreateTimeTAG: 2012-09-06T09:13:20Z IndexTAG: 4745 TitleTAG: Keyboard Shortcuts needed Need an "f" key to zoom/fill. UserIdTAG: 169809 UserNameTAG: sodhi CreateTimeTAG: 2012-09-06T01:47:20Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: Can't agree with it any more FirstChildUserIdTAG: 176234 FirstChildUserNameTAG: hejinjie FirstChildCreateTimeTAG: 2012-09-06T06:34:43Z IndexTAG: 4746 TitleTAG: power supplied by individual sources usando el teorema de superpocicion tambien es posible resolverlo :D UserIdTAG: 247953 UserNameTAG: oescamilla CreateTimeTAG: 2012-09-06T00:26:14Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL/KCL example CommentableIdTAG: 6002x_S2E3_Using_KVL_KCL_and_VI_constraints NumberOfReplyTAG: 1 FirstChildTAG: common language of the course is English.. FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-07T21:40:30Z SecondChildTAG: Y? SecondChildUserIdTAG: 230787 SecondChildUserNameTAG: Gabriel007 SecondChildCreateTimeTAG: 2012-09-08T18:32:58Z IndexTAG: 4747 TitleTAG: Transient Analysis timepoints parameter is missing UserIdTAG: 308246 UserNameTAG: AashishV CreateTimeTAG: 2012-09-05T23:54:05Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4748 TitleTAG: only mistake First, this course is very cool, i will learn a lot, but in the first lab, my only mistake is in the last part: Transient analysis.I did in the worng way, and i dont uderstand. UserIdTAG: 218081 UserNameTAG: Bneia CreateTimeTAG: 2012-09-05T23:25:48Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: Even i am also working towards it... FirstChildUserIdTAG: 224620 FirstChildUserNameTAG: TomVO FirstChildCreateTimeTAG: 2012-09-06T10:18:41Z IndexTAG: 4749 TitleTAG: Unable to complete I signed up for this Circuits and Electronics course, but found out after I signed up that I had to have a Physics course completed, and had not started one yet. So I will be doing OpenCourseWare's Physics 8.02 instead of this course, it seems. UserIdTAG: 248807 UserNameTAG: 1kingsman CreateTimeTAG: 2012-09-05T23:21:30Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 4 FirstChildTAG: They say that, but you don't need one. Its really not a difficult course. Just basic math and common sense will get you through. This is an intro to electronics so don't worry. Have fun. FirstChildUserIdTAG: 15344 FirstChildUserNameTAG: kob FirstChildCreateTimeTAG: 2012-09-05T23:44:41Z SecondChildTAG: Alright, its good to know then. However, I am still going to try to do the Physics course, just at a slower rate. SecondChildUserIdTAG: 248807 SecondChildUserNameTAG: 1kingsman SecondChildCreateTimeTAG: 2012-09-05T23:56:09Z FirstChildTAG: Depending on your science background and your ability to learn "just in time," you may not need to defer 6.002x in favor of OCW 8.02, you may be able to do both simultaneously (or just review the physics material covering electricity & magnetism from 8.02 or another source). I had equivalent physics 40 years ago, but I scrambled with the math "just in time" as I went through the pilot 6.002x course, and it worked out. Of course, your mileage may vary. You'll know very quickly (week 2 or so) if you are in over your head with the physics. For me, the math was the mountain I had to climb in weeks 8-10. FirstChildUserIdTAG: 94557 FirstChildUserNameTAG: g_hopper FirstChildCreateTimeTAG: 2012-09-06T00:18:13Z SecondChildTAG: I'll try my best to do what I can, in regards to learning both at once. As you said, I assume that soon, I will see how much I know/understand. SecondChildUserIdTAG: 248807 SecondChildUserNameTAG: 1kingsman SecondChildCreateTimeTAG: 2012-09-06T00:42:29Z FirstChildTAG: You really don't need much of an understanding of physics, as the topics here are relatively self contained. I would suggest you look up some magnetism videos when we get to inductors, as I was fairly confused when they were brought up and had a hard time reconciling concepts of magnetic fields/flux and electric current. Other than that, not too much physics is needed that can't be picked up on the spot. FirstChildUserIdTAG: 26931 FirstChildUserNameTAG: Csscade FirstChildCreateTimeTAG: 2012-09-06T02:20:08Z FirstChildTAG: Don't give up 1kingsman. Try to follow this Course. You can do it! ;) I will help you. FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-06T00:11:51Z SecondChildTAG: Thanks for the head's up! I will post here if i have too much of a problem. Well, somewhere on these boards. SecondChildUserIdTAG: 248807 SecondChildUserNameTAG: 1kingsman SecondChildCreateTimeTAG: 2012-09-06T08:19:30Z IndexTAG: 4750 TitleTAG: writing system to submit the homework hello, im answering homework 1. i put an equation, then i check, it tells that invalid input: could not parse "answer" as a formula What is the solution??? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-05T23:01:49Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 3 FirstChildTAG: its sensitive to the register so dont forget about big A and the quotes! :) FirstChildUserIdTAG: 341884 FirstChildUserNameTAG: Michaeldumchev FirstChildCreateTimeTAG: 2012-09-05T23:17:43Z FirstChildTAG: I'm sure that all you need to do is add * where there is multiplication :P You know, I have no problem writing, for example, (3*R)/2 or 833+1/3, even more complicated answer like 3750/(240/2) is still valid FirstChildUserIdTAG: 342241 FirstChildUserNameTAG: hieuxlu FirstChildCreateTimeTAG: 2012-09-05T23:23:42Z FirstChildTAG: One answer box wants a number, the other wants a equation. Are you answering the right question? FirstChildUserIdTAG: 88550 FirstChildUserNameTAG: Neil_S_Berry FirstChildCreateTimeTAG: 2012-09-05T23:44:21Z IndexTAG: 4751 TitleTAG: typing i dont know how to type expressions with the keyboard please some one help me UserIdTAG: 326507 UserNameTAG: mohamed200 CreateTimeTAG: 2012-09-05T22:56:05Z VoteTAG: 0 CoursewareTAG: Week 1 / Associated reference directions CommentableIdTAG: 6002x_S2E2_Associated_Reference_Directions NumberOfReplyTAG: 3 FirstChildTAG: All of these answers are numeric, so you only have to enter the numbers that correspond to the answers. FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-09-05T23:05:45Z FirstChildTAG: Take a look at the Overview section, it explains how to enter expressions and numbers in the answers and also how to use the simulator. I think it may help you. FirstChildUserIdTAG: 66241 FirstChildUserNameTAG: Ephexis FirstChildCreateTimeTAG: 2012-09-05T23:21:47Z FirstChildTAG: Hi mohamed200! How are you? ;) I will give you some easy examples of how to write the equation expressions in the Homeworks so that you can try it ;). Example 1): $ [a]^2 $ It is a^2 Example 2): $ \frac{1}{a} $ It is 1/a Example 3). Suppose now that a= jwC $ \left [ \frac{1}{j \omega C} \right ] $ It is 1/(j*w*C) Example 4). Now suppouse that we want the Example 3 expression but like the Example 1): $ \left [ \frac{1}{j \omega C} \right ]^2 $ It is (1/(j*w*C))^2 Example 5). We want to find the square of a. $ \sqrt{a}$ It is (a)^1/2 Or sqrt(a) Remember that Products are separated with asteric like 3*R not 3R. Now it is your turn. You can do it! ;) Also remember that r is different from R . FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-06T00:03:13Z SecondChildTAG: En español es exactamente lo mismo (vean los Ejemplos del 1 al 5). Básicamente hay que tener cuidado en la multiplicación, por ejemplo 3 por R es 3*R y no 3R. También hay que tener cuidado con las expresiones en mayúsculas y en minúscula. No es lo mismo tipear r que R. Ahora es su turno! ;) Espero haberles sido de ayuda, cualquier cosa pregunten estaré encantada de responderles. SecondChildUserIdTAG: 58095 SecondChildUserNameTAG: Myrimit SecondChildCreateTimeTAG: 2012-09-06T00:07:23Z SecondChildTAG: thanx SecondChildUserIdTAG: 326507 SecondChildUserNameTAG: mohamed200 SecondChildCreateTimeTAG: 2012-09-08T00:08:09Z IndexTAG: 4752 TitleTAG: General Whats the difference between peak and max when referring to any sort of quantitative notation be it Voltage, Current and any relevant other ? Thanks UserIdTAG: 204745 UserNameTAG: allwynmendes CreateTimeTAG: 2012-09-05T22:52:07Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4753 TitleTAG: Nice tool Great Lab, I was trying to label the terminals with the letters, but no significant problem. UserIdTAG: 124360 UserNameTAG: nlama CreateTimeTAG: 2012-09-05T22:35:18Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: I don't see any video uploading. FirstChildUserIdTAG: 338980 FirstChildUserNameTAG: my4boys FirstChildCreateTimeTAG: 2012-09-05T22:52:42Z IndexTAG: 4754 TitleTAG: [Lab 0] DC test button doesn't appear When I attempt to run a DC test, sometimes the buttons do not appear, causing me to have to refresh the lab in order to get it to work. However, even refreshing doesn't always fix it, and I do not fully understand why. Running Win 7, Firefox browser. EDIT: updated tags UserIdTAG: 248807 UserNameTAG: 1kingsman CreateTimeTAG: 2012-09-05T21:54:22Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: working in my firefox,after i refreshed the browser FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-05T22:19:18Z FirstChildTAG: Hi 1kingsman! I have no Problems with Chrome. I could simulate Lab0... Try to use the last updated version of Firefox or Chrome or press F5 in order to refresh the web page... ------ Hola 1kingsman!Yo no tuve inconvenientes con Chrome. Pude simular el Lab0... Intenta con la última versión actualizada de Chrome o Firefox, o también puedes probar con la tecla F5 para refrescar la Pantalla... FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-05T23:14:55Z SecondChildTAG: Well, I just pulled open lab 0 again, and refreshing the window with F5 or the refresh button doesn't change anything. Firefox is updated, as is Adobe Flash. I don't have Chrome at the moment, and would rather not install it, as I don't care for it much. SecondChildUserIdTAG: 248807 SecondChildUserNameTAG: 1kingsman SecondChildCreateTimeTAG: 2012-09-05T23:58:54Z SecondChildTAG: Does not matter if your using chrome it won't fix it (I've tried all software combinations and computers, with no joy) as its a problem on one of their flaky servers. Those who are not having the problem have there accounts running from stable servers. All we can do is wait for them to fix or at least acknowledge the problem with part of their system. Please take note this is not a client side (web browser) issue. SecondChildUserIdTAG: 15344 SecondChildUserNameTAG: kob SecondChildCreateTimeTAG: 2012-09-06T11:21:26Z IndexTAG: 4755 TitleTAG: 25% discount code MIT25 not working on ebook Hi all, I am having a problem applying the MIT25 discount when attempting to purchase the textbook online as an ebook when I click the apply button it recalculates but does not apply any discount. The page says "Save 25% on any print or electronic title. Use code MIT25" so this should work right? Anyone else having a problem (or success) with this discount code? Regards Gary UserIdTAG: 46075 UserNameTAG: GaryHope CreateTimeTAG: 2012-09-05T21:24:36Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 3 FirstChildTAG: If you try again or keep trying or clicking backwards and forwards enough times it appears to work. Randomly. Really. But I also experimented with the MIT40 code and could not get that to work. This in Firefox. If you are using Firefox you might try Chrome or Internet Exploder. I suspect Elsevier isn't always well-behaved with FF. FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-05T21:47:47Z FirstChildTAG: I sent a message to the publisher via their website. I'll let you know what I find out (if anything), and I encourage you to send them a message, too. Thanks for pointing this out! FirstChildUserIdTAG: 376895 FirstChildUserNameTAG: alecgg FirstChildCreateTimeTAG: 2012-09-05T23:23:26Z FirstChildTAG: I got an answer back from the publisher, and it appears that the discount is working for the eBook now. FirstChildUserIdTAG: 376895 FirstChildUserNameTAG: alecgg FirstChildCreateTimeTAG: 2012-09-07T04:04:25Z IndexTAG: 4756 TitleTAG: Meaning What does "My own BS" mean? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-05T21:23:46Z VoteTAG: 0 CoursewareTAG: Week 1 / Lumped Element Abstraction CommentableIdTAG: 6002x_lumped_element_abstraction NumberOfReplyTAG: 1 FirstChildTAG: BS is short for "Bovine Supposition" ;) , or something similar. i.e assumptions used in solving a problem or explaining a system. FirstChildUserIdTAG: 4463 FirstChildUserNameTAG: pmj FirstChildCreateTimeTAG: 2012-09-05T21:58:06Z IndexTAG: 4757 TitleTAG: Students From Pakistan Can add me sufiz@hotmail.com (mail and FB) or else send me u email so we can join each other UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-05T21:14:48Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4758 TitleTAG: Brazilian Group / Grupo de Brasileiros Hello everybody! There is a Facebook group composed of people who have participated in the course 6.002xe pessos who are participating now. The grpo is very good to ask questions and exchange information. https://www.facebook.com/groups/222463064518149/312269855537469 Olá pessoal! Existe um grupo no Facebook composto por pessoas que já participaram do curso 6.002x e as pessos que estão participando agora. O grpo é muito bom para tirar dúvidas e trocar informação. https://www.facebook.com/groups/222463064518149/312269855537469 UserIdTAG: 49688 UserNameTAG: baquelo CreateTimeTAG: 2012-09-05T21:07:03Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4759 TitleTAG: Great start I greet you that's great start UserIdTAG: 283354 UserNameTAG: Alsaka CreateTimeTAG: 2012-09-05T20:20:04Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 0 IndexTAG: 4760 TitleTAG: homework hello, im answering homework 1. i put an equation, then i check, it tells that invalid input: could not parse "answer" as a formula UserIdTAG: 374543 UserNameTAG: laudimar CreateTimeTAG: 2012-09-05T20:11:00Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Remember to write answers like you're putting them in a calculator. E.g. 2*x^(3/2) instead of 2x^3/2. FirstChildUserIdTAG: 376895 FirstChildUserNameTAG: alecgg FirstChildCreateTimeTAG: 2012-09-06T01:52:27Z IndexTAG: 4761 TitleTAG: reason why the pwm is 261.8 we have to take note the peak to peak voltage i mean is 130.9w from 0 positive peak! and 130.9w from 0 top negative peak and if u sum the values is exatly 261.8.! UserIdTAG: 156060 UserNameTAG: radeon9550 CreateTimeTAG: 2012-09-05T19:26:00Z VoteTAG: 0 CoursewareTAG: Week 1 / AC power CommentableIdTAG: 6002x_S1E3_AC_power NumberOfReplyTAG: 2 FirstChildTAG: If you calc sqr(120 * sqrt(2)) / 110 you see same value.. Сoincidence? FirstChildUserIdTAG: 199010 FirstChildUserNameTAG: DantaliaN FirstChildCreateTimeTAG: 2012-09-05T19:47:45Z FirstChildTAG: Sorry! but i don't think u are correct. Peak Power means from 0 to positive peak... because peak power = (Voltage Amplitude)X(Current Amplitude) FirstChildUserIdTAG: 226838 FirstChildUserNameTAG: SaSSer FirstChildCreateTimeTAG: 2012-09-06T04:42:36Z IndexTAG: 4762 TitleTAG: lab 0, measure current In lab 0, how is the answer for Current flowing from source into the circuit 500e-6 ? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-05T19:12:27Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 5 FirstChildTAG: 3volts/6kohms = 500microAmps or 500x10-6(-6 in superscript) or it can be written as 500e-6 where e stands for exponential or x10 to the power of a value in this case -6. I hope this helps you. FirstChildUserIdTAG: 372866 FirstChildUserNameTAG: jonmark67 FirstChildCreateTimeTAG: 2012-09-05T19:22:20Z SecondChildTAG: but 10-6 is not same as e-6. both has different values. SecondChildUserIdTAG: 257699 SecondChildUserNameTAG: aashishg11 SecondChildCreateTimeTAG: 2012-09-05T19:47:40Z FirstChildTAG: Did anyone else notice that in the dc analysis it reports the current as minus 500uA? Can anyone explain why it does that? FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-05T22:18:31Z SecondChildTAG: "the source current sign is negative if the current is flowing out of the negative terminal of the source, through the circuit, and into the positive terminal of the source" SecondChildUserIdTAG: 307779 SecondChildUserNameTAG: Izabella SecondChildCreateTimeTAG: 2012-09-05T22:27:31Z SecondChildTAG: Note the direction of the arrow. If it is pointing the opposite direction of current flow, it will give you a negative current. Positive current moves from positive to negative terminals, but the arrow on the DC analysis is reversed. SecondChildUserIdTAG: 174297 SecondChildUserNameTAG: Megan SecondChildCreateTimeTAG: 2012-09-05T22:54:47Z FirstChildTAG: This lab have a error in the "Result of Transient Analysis", I start the Result and never appear FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-06T02:57:30Z FirstChildTAG: It is 500e-6 (500 uA) because R = 6k ohm , V = 3 volts, Now apply ohm's law V = IR ==> I = V/R I = 3/6k = 3/6000 = 0.0005 (Answer) The E-6 means you move the decimal point 6 places to the left. 0.000001 500e-6 = 5e-4 = 0.5e-3 = 0.05e-2 = 0.005e-1 = 0.0005 FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-06T06:01:28Z FirstChildTAG: It is 500e-6 (500 uA) because R = 6k ohm , V = 3 volts, nNow apply ohm's law V = IR ==> I = V/R I = 3/6k = 0.5mA = 500 uA = 500e-6 I hope, you understood now. FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-06T05:53:31Z IndexTAG: 4763 TitleTAG: Invalid input I draw the circuit and evaluated it, but when I am putting the values in the given answer box it shows invalid input.. after correcting units many times it still shows an error like "Invalid input: could not interpret '2.50 volts' as a number". Now what I have to do? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-05T18:49:51Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 6 FirstChildTAG: But I think you can use the suffixes u, k, M ... FirstChildUserIdTAG: 116489 FirstChildUserNameTAG: matheusribeiro FirstChildCreateTimeTAG: 2012-09-05T19:05:21Z SecondChildTAG: i wrote 500u in this question > Blockquote Current flowing from source into the circuit (amps, with correct sign) > Blockquote but it gave me error so i think we should write (500e-6) SecondChildUserIdTAG: 279239 SecondChildUserNameTAG: abdallah-ali SecondChildCreateTimeTAG: 2012-09-05T19:38:33Z FirstChildTAG: Do not write words like volts in edit box. Just number: 2.5 Also do not use suffixes u,k,M and so on. Instead of 500uA write 0.0005(without A) FirstChildUserIdTAG: 258500 FirstChildUserNameTAG: karas FirstChildCreateTimeTAG: 2012-09-05T18:54:07Z FirstChildTAG: also, for really big/small numbers use e notation. Ex, .000006 would be 1e-6 FirstChildUserIdTAG: 364798 FirstChildUserNameTAG: atari1994 FirstChildCreateTimeTAG: 2012-09-05T19:02:24Z FirstChildTAG: you only have to input the numeric value, without the volts. FirstChildUserIdTAG: 83276 FirstChildUserNameTAG: salsero FirstChildCreateTimeTAG: 2012-09-05T20:03:48Z FirstChildTAG: just write (2.50) with the word (volts) FirstChildUserIdTAG: 279239 FirstChildUserNameTAG: abdallah-ali FirstChildCreateTimeTAG: 2012-09-05T19:34:57Z FirstChildTAG: Just write 2.5 Don't write '2.5 volts' FirstChildUserIdTAG: 259238 FirstChildUserNameTAG: omidsadeghi FirstChildCreateTimeTAG: 2012-09-05T19:55:07Z IndexTAG: 4764 TitleTAG: regarding lab how can we increase the value of resistances & voltages UserIdTAG: 350334 UserNameTAG: nikhilpadho CreateTimeTAG: 2012-09-05T18:48:32Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: Double click on the discrete elements to bring up the properties of the elements. You can edit the properties from there. FirstChildUserIdTAG: 85490 FirstChildUserNameTAG: yashsavani FirstChildCreateTimeTAG: 2012-09-05T19:07:28Z FirstChildTAG: just double click on the resistance or voltage source then you will get editing options. FirstChildUserIdTAG: 48151 FirstChildUserNameTAG: santoshhalagatti FirstChildCreateTimeTAG: 2012-09-05T19:07:58Z IndexTAG: 4765 TitleTAG: Exclusive Facebook group for the students attending Edx 6.002x. Hey Hi Guys, Firstly please do accept my well wishes. I have created this group exclusively for the students attending the Edx 6.002x Circuits and Electronics. This group is to share ur doubts, queries, suggestions and whatever u feel like. Lets get together and start exploring. Cheers.:) Below is the link- http://www.facebook.com/groups/483354858342165/ UserIdTAG: 253902 UserNameTAG: sajalok CreateTimeTAG: 2012-09-05T18:04:38Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: With my respect, facebook will drive my attention away. Instead, I find this discussion room focused and has the necessary tools for everything. FirstChildUserIdTAG: 289472 FirstChildUserNameTAG: OsamaAdel FirstChildCreateTimeTAG: 2012-09-05T18:39:24Z FirstChildTAG: Thanks for the initiative FirstChildUserIdTAG: 305965 FirstChildUserNameTAG: jorno FirstChildCreateTimeTAG: 2012-09-05T18:42:00Z IndexTAG: 4766 TitleTAG: Transient analysis in Transient analysis how is the correct answer above 1V and my amplitude is 1 V ?? UserIdTAG: 218038 UserNameTAG: sonic_sie CreateTimeTAG: 2012-09-05T17:48:16Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 3 FirstChildTAG: I can't get Transient alalysis to work..I am sure that i am doing the right configuration of Vsupply [sin(1,1,1k,0)] and for the stop time too.. FirstChildUserIdTAG: 134685 FirstChildUserNameTAG: coreos FirstChildCreateTimeTAG: 2012-09-05T18:03:13Z SecondChildTAG: you have type in the full value of "1000" instead of "1k". Most of this tool does not understand the letter symbols so you always type in the digit representation. SecondChildUserIdTAG: 282092 SecondChildUserNameTAG: richspoon SecondChildCreateTimeTAG: 2012-09-05T18:17:18Z FirstChildTAG: Because we have offset voltage 1V and signal changing from 0 to 2V (without offset -1V to 1V) FirstChildUserIdTAG: 258500 FirstChildUserNameTAG: karas FirstChildCreateTimeTAG: 2012-09-05T18:14:14Z FirstChildTAG: I think because there is an offset of 1V. Try to put the pointer at 0 second and notice the voltage for point A, it's 1V although the source should have begun with 0 voltage and rose up to 1V.(Sorry for the bad English if any mistake observed) FirstChildUserIdTAG: 289472 FirstChildUserNameTAG: OsamaAdel FirstChildCreateTimeTAG: 2012-09-05T18:32:42Z IndexTAG: 4767 TitleTAG: Akshay Baraik Hello. My name is Akshay Baraik. Thanks for the Course. Facebook ID is facebook.com/akshayb4. Anybody from Delhi? All the Best. Thanks. Bye. UserIdTAG: 2956 UserNameTAG: akshayb CreateTimeTAG: 2012-09-05T17:30:24Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4768 TitleTAG: Suggestions for interactive tool 1) Use bit bigger signs of 'wire' and 'ground' buttons on the right side (too small I think) 2) Ability to hold vertical line for analysis tool 3) Button to show/hide hotkeys map at the bottom of interactive tool UserIdTAG: 204680 UserNameTAG: XVilka CreateTimeTAG: 2012-09-05T17:24:49Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 0 IndexTAG: 4769 TitleTAG: need help getting started im new to all of this where do i go to put on the homework and all assignments UserIdTAG: 160242 UserNameTAG: Mlevins35 CreateTimeTAG: 2012-09-05T17:06:40Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: You go to the homework tab from week 1 on the left side of the screen and put all the data in the fields then click check button after completing.If the values are correct you will see check signs on each answer or "X" sign if you are wrong and need to study more to answer that particular check-box. First you should try listening to the courses and instructions. Which are in video format and you will know how to use the platform. Good Luck! FirstChildUserIdTAG: 64973 FirstChildUserNameTAG: GabrielC FirstChildCreateTimeTAG: 2012-09-05T17:15:54Z FirstChildTAG: thanks i am new to this. I am really interested in this course. It said that it started at noon, also that we would interact with the instructors. No one gave any information as far as how and when you need to log on to the class, if anyone knows how this works please let me know FirstChildUserIdTAG: 160242 FirstChildUserNameTAG: Mlevins35 FirstChildCreateTimeTAG: 2012-09-05T22:32:34Z IndexTAG: 4770 TitleTAG: Unable to view the lecture I am unable to view and hear anything in Lecture sequence .Only blank screen is coming. Please help. UserIdTAG: 258007 UserNameTAG: Mathew19 CreateTimeTAG: 2012-09-05T16:51:02Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Plz try changing ur browser. FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-05T16:57:37Z IndexTAG: 4771 TitleTAG: confusing how to find out transient analysis? UserIdTAG: 291529 UserNameTAG: prasanthmandalapu CreateTimeTAG: 2012-09-05T16:38:14Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: Some times the icon/button do not appear. Try to refresh the page and if that do not work get out of that page and try again. FirstChildUserIdTAG: 300827 FirstChildUserNameTAG: PFonseca FirstChildCreateTimeTAG: 2012-09-05T16:49:29Z SecondChildTAG: If you're asking how to use the tool, in order to proceed through the transient analysis, I would recommend viewing the edX tutorial on how to use it for labs. That tutorial has many examples, one which explains the procedure of transient analysis. SecondChildUserIdTAG: 158985 SecondChildUserNameTAG: n1xx SecondChildCreateTimeTAG: 2012-09-05T18:12:59Z IndexTAG: 4772 TitleTAG: Confused about entering equations in the answer tab. Absolutely confused as to how we are supposed to answer in the answer tab. do we write a full equation like R+R+R = 3R or do we just answer it as 3R? UserIdTAG: 360609 UserNameTAG: mohitkarkhanis7 CreateTimeTAG: 2012-09-05T16:37:03Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: Both will work, i.e. "R+R+R" and "3*R" (note the asterisk in the latter) FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-05T16:48:38Z SecondChildTAG: yes, try putting a star in between 3 and R i.e. u shud write 3*R SecondChildUserIdTAG: undefined SecondChildUserNameTAG: undefined SecondChildCreateTimeTAG: 2012-09-05T16:59:09Z FirstChildTAG: R+R+R, or 3*R, or (R+R+R), or ((R+R)+R) or R+(R+R) or R+(2*R) but not 3R FirstChildUserIdTAG: 88550 FirstChildUserNameTAG: Neil_S_Berry FirstChildCreateTimeTAG: 2012-09-06T01:41:55Z IndexTAG: 4773 TitleTAG: Regarding first quizz on week one I put correct answer for quizz one and the site told its wrong,, in the answer of the site is the same answer as mine ,, I am using an iPad UserIdTAG: 279909 UserNameTAG: hussainkhalaf CreateTimeTAG: 2012-09-05T16:33:43Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4774 TitleTAG: how to get the sim soft wear to work. hello i would like to know how to get the simulated circuit to work i tried to connect the componets but nothing happens so can anyone tell me how it works, is there anything i'm doing wrong. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-05T16:25:24Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4775 TitleTAG: New look....... Good look from the prototype course . Hope 2 have a great time in exploring in electronics and love to view Anant sir aha Moments. UserIdTAG: 45699 UserNameTAG: Praveenkrish CreateTimeTAG: 2012-09-05T16:24:24Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4776 TitleTAG: syntax what is the syntax for exponential numbers? and also other numbers? UserIdTAG: 365792 UserNameTAG: pabhijeet CreateTimeTAG: 2012-09-05T16:21:36Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: The syntax for exponents is the caret: ^ So, for example 2^3 is $2^3=8$ FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-05T16:25:55Z IndexTAG: 4777 TitleTAG: how to change value for resistor hi can i have little help ..plz can anyone plz tell me how to change value of the resistors.. for me its all 1 ohm. UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-05T16:14:37Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 4 FirstChildTAG: just double click on the resistance u placed. A box will open and u will be able to change the value.... FirstChildUserIdTAG: 308571 FirstChildUserNameTAG: Sayantani FirstChildCreateTimeTAG: 2012-09-05T16:18:56Z SecondChildTAG: thanks alot .. SecondChildUserIdTAG: 177128 SecondChildUserNameTAG: MARYAAM SecondChildCreateTimeTAG: 2012-09-05T16:22:35Z SecondChildTAG: It worked for resistor but for voltage supply its not help me plz SecondChildUserIdTAG: 383699 SecondChildUserNameTAG: Gogineni SecondChildCreateTimeTAG: 2012-09-05T18:09:15Z FirstChildTAG: Double click the resistor and you will get a dialog box , in it you can change the value of resistors. FirstChildUserIdTAG: 45699 FirstChildUserNameTAG: Praveenkrish FirstChildCreateTimeTAG: 2012-09-05T16:19:06Z SecondChildTAG: thanks SecondChildUserIdTAG: 177128 SecondChildUserNameTAG: MARYAAM SecondChildCreateTimeTAG: 2012-09-05T16:22:47Z FirstChildTAG: take some time to watch the demo before trying out dude. it s all taught there FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-05T16:23:36Z SecondChildTAG: ok thnks i will do SecondChildUserIdTAG: undefined SecondChildUserNameTAG: undefined SecondChildCreateTimeTAG: 2012-09-05T16:41:51Z FirstChildTAG: double click on resistor . A window will be opened. here you can enter Resistance name and its Values as well. Use K for Kilo-Ohms etc Regards FirstChildUserIdTAG: 347789 FirstChildUserNameTAG: engr_zohaib FirstChildCreateTimeTAG: 2012-09-05T16:23:55Z SecondChildTAG: thnx SecondChildUserIdTAG: 177128 SecondChildUserNameTAG: MARYAAM SecondChildCreateTimeTAG: 2012-09-05T16:42:11Z IndexTAG: 4778 TitleTAG: Explanation required can we replace that lumped element with capacitors and inductors? UserIdTAG: 344704 UserNameTAG: roop CreateTimeTAG: 2012-09-05T16:02:18Z VoteTAG: 0 CoursewareTAG: Week 1 / Lumped Element Abstraction CommentableIdTAG: 6002x_lumped_element_abstraction NumberOfReplyTAG: 4 FirstChildTAG: yes, but for capacitors and inductors equations are slightly more complicated, and are not linear but there is also an abstraction FirstChildUserIdTAG: 51696 FirstChildUserNameTAG: blasco23 FirstChildCreateTimeTAG: 2012-09-05T18:32:58Z FirstChildTAG: yes you can, but when you use ohw law, you will use other caracteristic equations like 1/jwc or jml, to capacitors and inductors, but your question will be reply by professor very soon in the next videos. FirstChildUserIdTAG: 250557 FirstChildUserNameTAG: PedroCunha FirstChildCreateTimeTAG: 2012-09-06T08:20:53Z FirstChildTAG: I have also problem with understanding of this syntagm. Is there any other group of words similar to lumped elements,lumped abstraction and lumped matter discipline?...Thanks in advance..:-) FirstChildUserIdTAG: 137527 FirstChildUserNameTAG: drlatech FirstChildCreateTimeTAG: 2012-09-05T16:28:13Z SecondChildTAG: I think that professor means....passive elements, resistors, capacitors and inductors, elements that receive energy. SecondChildUserIdTAG: 250557 SecondChildUserNameTAG: PedroCunha SecondChildCreateTimeTAG: 2012-09-06T08:26:33Z FirstChildTAG: The lumped element (also called lumped parameter, or lumped component) simplifies the description of the behaviour of spatially distributed physical systems into a topology consisting of discrete entities that approximate the behaviour of the distributed system under certain assumptions. It is useful in electrical systems (including electronics), mechanical multibody systems, heat transfer, acoustics. Ha ha! Some1 can now understand. right? FirstChildUserIdTAG: 117661 FirstChildUserNameTAG: BUNDAY FirstChildCreateTimeTAG: 2012-09-07T12:11:56Z IndexTAG: 4779 TitleTAG: Calculating Voltage It's been a while :) So how is this calculated again? 11/V = V/8 --> V=? Tnx! UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-05T15:52:09Z VoteTAG: 0 CoursewareTAG: Week 1 / Power CommentableIdTAG: 6002x_S1E2_Power NumberOfReplyTAG: 3 FirstChildTAG: V*V=88. sqrt 88=V FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-05T16:10:30Z FirstChildTAG: The law you were looking for is P = V^2 / R, so in this case 11W = V*V / 8ohms. This gives you V = sqrt(88). Also take care that its negative, as the current does not flow from + to - as the normal annotation would suggest. Current is then simply calculated from ohms law FirstChildUserIdTAG: 329254 FirstChildUserNameTAG: KGabor FirstChildCreateTimeTAG: 2012-09-05T19:00:48Z SecondChildTAG: I think the correct equation is P=I^2/R, correct me if im wrong SecondChildUserIdTAG: 153198 SecondChildUserNameTAG: alvin_aditya SecondChildCreateTimeTAG: 2012-09-05T23:05:12Z FirstChildTAG: W = (I^2)R 11/8 = I^2 sqrt(11.0/8) = I or W = (V^2)/R 11*8 = V^2 sqrt(11.0 * 8) = V FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-06T06:55:55Z IndexTAG: 4780 TitleTAG: Atlast edx begins, but lectures are the same. I am glad 6.002x began once again. I took this course in the summer at MITx and now I am going to recommend this to my friends. The course went very successfully then and I hope it does now. Well done edX team! :) UserIdTAG: 118131 UserNameTAG: ammubhave CreateTimeTAG: 2012-09-05T15:46:46Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 3 FirstChildTAG: Agreed, this was way too fun to not re-enroll FirstChildUserIdTAG: 64566 FirstChildUserNameTAG: tvl FirstChildCreateTimeTAG: 2012-09-05T16:10:58Z FirstChildTAG: Hello ammubhave, I hope 6.003z is going well FirstChildUserIdTAG: 79361 FirstChildUserNameTAG: kimt FirstChildCreateTimeTAG: 2012-09-05T16:19:59Z FirstChildTAG: Lots easier to, as no need to watch all the lectures so more time to really understand. Good move to do it again. Nice to be here with you again. FirstChildUserIdTAG: 15344 FirstChildUserNameTAG: kob FirstChildCreateTimeTAG: 2012-09-06T02:15:55Z IndexTAG: 4781 TitleTAG: algebraic expression for the current i5 Why in task S1E8: KCL, when we find i5, we used i6? in priveous task we find i4 and we can also use it to wright down algebraic expression for the current i5. UserIdTAG: 192114 UserNameTAG: chuba CreateTimeTAG: 2012-09-05T15:46:04Z VoteTAG: 0 CoursewareTAG: Week 1 / KVL and KCL exercises CommentableIdTAG: 6002x_kvl_kcl_exercises NumberOfReplyTAG: 2 FirstChildTAG: You don't have to use i6. You can also use i4 and i1. FirstChildUserIdTAG: 138981 FirstChildUserNameTAG: Pr0bability FirstChildCreateTimeTAG: 2012-09-05T18:25:45Z FirstChildTAG: look what happens is that first you have to use the kcl on the node where this i4, i2 and i3, i4 and then you clear now using KCl appears on the node where i1, i5 and i4 as the only unknown is i5 FirstChildUserIdTAG: 51696 FirstChildUserNameTAG: blasco23 FirstChildCreateTimeTAG: 2012-09-06T14:45:17Z IndexTAG: 4782 TitleTAG: problem in starting course actualli i am not getting any proper direction to start my course program since this is my first online course UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-05T15:44:36Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: i agree with u FirstChildUserIdTAG: 368292 FirstChildUserNameTAG: nailur FirstChildCreateTimeTAG: 2012-09-05T15:53:18Z FirstChildTAG: Hi. The links at the top of the screen are the most important ones for learning to find your way around. If you go to the top left one (CourseWare) or click this link: https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware You will see the actual course content - this is where you will find the videos, exercises, homeworks and lab work. Click on 'Welcome to 6.002x' and you should see the first video. To see more content you can expand the Week 1 and Week 2. You can work through the videos at your own pace. The deadlines for homework and lab work are given by each link. At the deadline your grade for that homework/lab is frozen, but you have until then to retry as many time as you like. Sometimes you will see some exercises between video segments. They are not graded but are just to help reinforce your learning. E.g. https://www.edx.org/courses/MITx/6.002x/2012_Fall/courseware/Overview/edx_introduction/ has a few sample questions after the first two videos. Also be sure to check the Course Info link at the top regularly, for announcements. These are the places to start, and of course ask questions here if you have problems. Good luck! FirstChildUserIdTAG: 9147 FirstChildUserNameTAG: SueP FirstChildCreateTimeTAG: 2012-09-05T15:56:17Z IndexTAG: 4783 TitleTAG: third question can anyone draw me the schematic diagram for the third question? UserIdTAG: 171378 UserNameTAG: ABHISHEKFROMINDIA CreateTimeTAG: 2012-09-05T15:41:30Z VoteTAG: 0 CoursewareTAG: Week 1 / Battery model CommentableIdTAG: 6002x_S1E9_Battery_Model NumberOfReplyTAG: 5 FirstChildTAG: R=R1*R2/R1+R2 FirstChildUserIdTAG: 192114 FirstChildUserNameTAG: chuba FirstChildCreateTimeTAG: 2012-09-05T15:55:35Z SecondChildTAG: if it was a series network then what will we do ? SecondChildUserIdTAG: undefined SecondChildUserNameTAG: undefined SecondChildCreateTimeTAG: 2012-09-06T01:52:03Z FirstChildTAG: No there schematic, you should use the first and second answer, for find the resistance. FirstChildUserIdTAG: 39685 FirstChildUserNameTAG: galeano FirstChildCreateTimeTAG: 2012-09-05T20:19:58Z FirstChildTAG: it is just Ohm law. V = R*I. The Voltage difference is 0.1. Hence we get: V = (0.1)/(0.25+0.33) FirstChildUserIdTAG: 327015 FirstChildUserNameTAG: Fabi2607 FirstChildCreateTimeTAG: 2012-09-05T16:35:03Z SecondChildTAG: sry wrong post SecondChildUserIdTAG: 327015 SecondChildUserNameTAG: Fabi2607 SecondChildCreateTimeTAG: 2012-09-05T16:35:52Z FirstChildTAG: It's consistes only of one eq.battery and one rezistor. And 3 wires. That's all FirstChildUserIdTAG: 345464 FirstChildUserNameTAG: Constantine_ru FirstChildCreateTimeTAG: 2012-09-11T19:17:07Z FirstChildTAG: The voltages are just in parallel. And their resistances are also in parallel . just apply formula for parallel network resistances. FirstChildUserIdTAG: 211840 FirstChildUserNameTAG: edxian FirstChildCreateTimeTAG: 2012-09-06T06:37:24Z IndexTAG: 4784 TitleTAG: transient Analysis Hello. Good, but in the transient Analysis, the Y voltage axis, don't show the values exactly, so i had to suppose the values according the height of the each color UserIdTAG: 149549 UserNameTAG: Java_Dido CreateTimeTAG: 2012-09-05T15:21:33Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 2 FirstChildTAG: It shows values of both Y (in the top left corner, as I recall) and X axes. Maybe you've got some bug. FirstChildUserIdTAG: 320303 FirstChildUserNameTAG: yoshimi FirstChildCreateTimeTAG: 2012-09-05T15:30:41Z FirstChildTAG: in my case, the values of the nodes are also written in the upper left corner. You don't need to read it from the axis. FirstChildUserIdTAG: 375898 FirstChildUserNameTAG: madlab FirstChildCreateTimeTAG: 2012-09-05T15:52:14Z IndexTAG: 4785 TitleTAG: Lab solutions still visible at the old MITx site It seems that some Lab questions for this course are similar, if not the same, as the ones used in the previous MITx course. Given that the solutions for Lab questions are available at the old site, wouldn't that cause wrong incentives for somebody inclined that way? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-05T15:19:28Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: Not if somebody else is tracking IP addresses. Or I'll put it another way 'not if the mal-inclined somebody's have their own best interests at heart'. Remember what happened to some accounts last time? FirstChildUserIdTAG: 9147 FirstChildUserNameTAG: SueP FirstChildCreateTimeTAG: 2012-09-05T15:33:20Z FirstChildTAG: Also its daft to cheat as the point of this course is to learn. Plus why bother, as there is no recognized qualification at end of it anyway. Let them cheat if they want. Only fools any way if thy do so as only cheating themselves. FirstChildUserIdTAG: 15344 FirstChildUserNameTAG: kob FirstChildCreateTimeTAG: 2012-09-06T02:19:36Z IndexTAG: 4786 TitleTAG: Error in wires why the wires are not working.When i click on the wires the wires are not responding.can anyone help me out in these UserIdTAG: 150120 UserNameTAG: krishna1993 CreateTimeTAG: 2012-09-05T15:05:40Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 4 FirstChildTAG: Which browser are you using? FirstChildUserIdTAG: 30011 FirstChildUserNameTAG: euler FirstChildCreateTimeTAG: 2012-09-05T15:47:59Z FirstChildTAG: I am facing same issue . ... :( FirstChildUserIdTAG: 347789 FirstChildUserNameTAG: engr_zohaib FirstChildCreateTimeTAG: 2012-09-05T15:48:35Z FirstChildTAG: Hi. You must press and hold the left mouse button on the circle, which is at the end of the element (for example, a resistor has two circles), move the mouse( where you want to draw a line), and release the left mouse button. Sorry for my English. FirstChildUserIdTAG: 324289 FirstChildUserNameTAG: Feax FirstChildCreateTimeTAG: 2012-09-05T15:51:06Z FirstChildTAG: What web browser are you using? I had the same problem a few minutes ago. I change my browser to Mozilla and it solved the problem. Please try other web browsers FirstChildUserIdTAG: 132229 FirstChildUserNameTAG: ahlo FirstChildCreateTimeTAG: 2012-09-05T16:06:11Z IndexTAG: 4787 TitleTAG: Wiring Problems i cannot wire the components .if i click the terminals of resistor or voltage source the whole component moves UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-05T14:54:57Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4788 TitleTAG: Rotate. letter R Hello, im Rodrigo Costa From Chile! the **letter "R" rotates** the items in the sand box! UserIdTAG: 222786 UserNameTAG: rocoma CreateTimeTAG: 2012-09-05T14:46:55Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: Thanks FirstChildUserIdTAG: 383637 FirstChildUserNameTAG: IanCull FirstChildCreateTimeTAG: 2012-09-05T17:36:01Z SecondChildTAG: But sometimes it doesnt work well, please check it. Thanks SecondChildUserIdTAG: 359787 SecondChildUserNameTAG: El_Bakan SecondChildCreateTimeTAG: 2012-09-06T05:46:18Z IndexTAG: 4789 TitleTAG: Courseware Problem Hi I can see the video for courseware , plz can any one help me UserIdTAG: 369560 UserNameTAG: Yasseros CreateTimeTAG: 2012-09-05T14:42:27Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 3 FirstChildTAG: What browser are you using? Try using Google Chrome. FirstChildUserIdTAG: 42833 FirstChildUserNameTAG: KKA FirstChildCreateTimeTAG: 2012-09-05T14:44:48Z FirstChildTAG: You said you can see the video, do you want us to help you not see it any more ? FirstChildUserIdTAG: 112757 FirstChildUserNameTAG: MImran FirstChildCreateTimeTAG: 2012-09-05T14:46:59Z FirstChildTAG: I am assuming you cannot see the video. Else it wouldn't be a problem. Try updating Flash Player. I am not gonna advertise any browser here saying this is better than that, but a simple update to flash player could solve your problem, FirstChildUserIdTAG: 152317 FirstChildUserNameTAG: easyMIT FirstChildCreateTimeTAG: 2012-09-05T15:10:48Z IndexTAG: 4790 TitleTAG: Interactive Tool Graph not functioning properly AC,DC not working properly.......... UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-05T14:39:55Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: try reload, it works for me! FirstChildUserIdTAG: 222786 FirstChildUserNameTAG: rocoma FirstChildCreateTimeTAG: 2012-09-05T14:48:12Z IndexTAG: 4791 TitleTAG: Error in ToolBox Itz very exciting to learn in Edx 6.002x. I couldn't change the voltage rating of supply voltage in the Lab0.Can u help me in that????? UserIdTAG: 178752 UserNameTAG: Renjith CreateTimeTAG: 2012-09-05T14:14:46Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: just double click on the voltage supply .you can see the editor and then provide the required voltage ,this is the way hoe it works.. FirstChildUserIdTAG: 372283 FirstChildUserNameTAG: akshayshah101 FirstChildCreateTimeTAG: 2012-09-05T14:18:48Z IndexTAG: 4792 TitleTAG: iPad and the interactive tool that use to enter and analyze circuits. Dear Sirs, How I can drag elements in the interactive tool that use to enter and analyze circuits by iPad? P.S. Photon browser is usefull for elements, but it's not good idea for wire. Unfortunately, my fingers are too thick. UserIdTAG: 151867 UserNameTAG: AndreiStoiakin CreateTimeTAG: 2012-09-05T14:11:05Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: Sorry to answer you outta blue. I am no Sir and I am no iPad app developer, but I think the iPad browsers are having hard time understanding the object type of the circuit elements. Me too facing the same problem. Lack of drag and drop support in iPad browsers might be the problem. iPads understand the drag/pinch to zoom and any touch as scroll or select. The best possible solution with labs and schematic elements would be to use a computer supporting Javascript. FirstChildUserIdTAG: 152317 FirstChildUserNameTAG: easyMIT FirstChildCreateTimeTAG: 2012-09-05T15:03:11Z FirstChildTAG: Had same issue ..... FirstChildUserIdTAG: 68760 FirstChildUserNameTAG: akberbana FirstChildCreateTimeTAG: 2012-09-13T15:33:01Z IndexTAG: 4793 TitleTAG: Exclusive Facebook group for the students attending Edx 6.002x. Hey Hi Guys, Firstly please do accept my well wishes. I have created this group exclusively for the students attending the Edx 6.002x Circuits and Electronics. This group is to share ur doubts, queries, suggestions and whatever u feel like. Lets get together and start exploring. Cheers.:) Below is the link- http://www.facebook.com/groups/483354858342165/ UserIdTAG: 253902 UserNameTAG: sajalok CreateTimeTAG: 2012-09-05T14:11:02Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4794 TitleTAG: I have wrong notes when i run DC Where is mistake? ![enter image description here][1] [1]: http://eom.com.ua/1.jpg UserIdTAG: 379711 UserNameTAG: andery CreateTimeTAG: 2012-09-05T13:46:21Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 2 FirstChildTAG: double check R1, you put there 1к instead of 1k (wrong letter, I guess Cyrillic one) FirstChildUserIdTAG: 320863 FirstChildUserNameTAG: Extrapolator FirstChildCreateTimeTAG: 2012-09-05T13:54:21Z SecondChildTAG: thanks!! SecondChildUserIdTAG: 379711 SecondChildUserNameTAG: andery SecondChildCreateTimeTAG: 2012-09-05T13:55:39Z FirstChildTAG: edit--- sorry I didn't notice that somebody already answered you FirstChildUserIdTAG: 144869 FirstChildUserNameTAG: csberes FirstChildCreateTimeTAG: 2012-09-05T14:10:48Z IndexTAG: 4795 TitleTAG: No Toolbar Not able to see DC button while using the tools tutorial... UserIdTAG: 219706 UserNameTAG: saigiri00 CreateTimeTAG: 2012-09-05T13:44:26Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: Hi, I had the same problem. Using firefox with adblock plus. Disabling adblock for edX.org made the missing buttons appear. Then they disappeared again. I would suggest using the circuit sandbox instead, as this appears to have the tools. Hope that helps! FirstChildUserIdTAG: 265279 FirstChildUserNameTAG: smcl FirstChildCreateTimeTAG: 2012-09-05T14:01:58Z SecondChildTAG: thanks for suggestion and when i did the same prob rectified... SecondChildUserIdTAG: 219706 SecondChildUserNameTAG: saigiri00 SecondChildCreateTimeTAG: 2012-09-06T03:29:50Z FirstChildTAG: I also had this issue. Taking the advice from smcl, I added edx.org to the exception list under Firefox's security section. The DC and TRAN options then appeared. Had to redo my lab. Thankfully, it was just the tutorial. FirstChildUserIdTAG: 266602 FirstChildUserNameTAG: XKaliber FirstChildCreateTimeTAG: 2012-09-06T23:18:34Z IndexTAG: 4796 TitleTAG: Error **Anyone else using safari on a Mac and freezing when clicking OK on transient analysis?** UserIdTAG: 243189 UserNameTAG: NickNy516 CreateTimeTAG: 2012-09-05T13:42:25Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 3 FirstChildTAG: Maybe is because you put the time wrong,my froze when I tried to put 1 minute.. FirstChildUserIdTAG: 252983 FirstChildUserNameTAG: IgorCastelo FirstChildCreateTimeTAG: 2012-09-05T13:59:10Z FirstChildTAG: Try updating your flash player for Safari - I had to do that so Safari would run the interactive question. FirstChildUserIdTAG: 243871 FirstChildUserNameTAG: Vanilly FirstChildCreateTimeTAG: 2012-09-05T14:34:50Z FirstChildTAG: use the Circuit box on the left top and do the analysis..i don't have mac but all component are working in circuit box FirstChildUserIdTAG: 375082 FirstChildUserNameTAG: satya1889 FirstChildCreateTimeTAG: 2012-09-05T13:55:39Z IndexTAG: 4797 TitleTAG: Question Four Does anyone know why the answer to Question-4: "Total resistance between pos and neg terminals (ohms)" is equal to 6000? I would have thought it would be R1 + R2 + R3 = 6r. Many thanks in advance. UserIdTAG: 365551 UserNameTAG: hazel1919 CreateTimeTAG: 2012-09-05T13:39:23Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 4 FirstChildTAG: 6Kiloohms=6000Oohms FirstChildUserIdTAG: 209363 FirstChildUserNameTAG: phaneendraaa FirstChildCreateTimeTAG: 2012-09-05T13:42:25Z FirstChildTAG: Cause the resistance here is in kiloohms. FirstChildUserIdTAG: 381769 FirstChildUserNameTAG: leonepein FirstChildCreateTimeTAG: 2012-09-05T13:43:07Z FirstChildTAG: the values of the resistors are in Kohms. 1Kohm = 1000ohm FirstChildUserIdTAG: 373629 FirstChildUserNameTAG: Gupu25 FirstChildCreateTimeTAG: 2012-09-05T13:43:11Z SecondChildTAG: Probably should have spotted that. Many thanks. SecondChildUserIdTAG: 365551 SecondChildUserNameTAG: hazel1919 SecondChildCreateTimeTAG: 2012-09-05T13:44:18Z FirstChildTAG: The resitance values are R1=1kOhm, R2=3kOhm, R3=2kOhm, where k means kilo, so you have to multiply it x1000 to get value in Ohms. But writing 6k works as well. FirstChildUserIdTAG: 320303 FirstChildUserNameTAG: yoshimi FirstChildCreateTimeTAG: 2012-09-05T13:44:04Z IndexTAG: 4798 TitleTAG: Cantidad de bucles No entiendo, la cantidad de bucles debe ser la parte de circuito cerrado en el ejemplo mostrado hay 3 ? Soy uruguayo y no hablo ingles estoy usando traductores Gracias UserIdTAG: 228270 UserNameTAG: MauriRod CreateTimeTAG: 2012-09-05T13:31:29Z VoteTAG: 0 CoursewareTAG: Week 1 / Circuit topology CommentableIdTAG: 6002x_S2E1_Circuit_Topology NumberOfReplyTAG: 6 FirstChildTAG: Yo soy español y creo que es así: Bucle 1 = a-d-c-a Bucle 2 = a-d-c-b-a Bucle 3 = a-d-b-a Bucle 4 = b-d-c-b Bucle 5 = a-b-d-c-a Bucle 6 = a-d-b-c-a Saludos FirstChildUserIdTAG: 270769 FirstChildUserNameTAG: 2214sanchez FirstChildCreateTimeTAG: 2012-09-07T09:57:50Z FirstChildTAG: tenes 4 loops, los tres interiores y el externo, el que pasa por el generador, R4 y R5. FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-05T13:38:54Z FirstChildTAG: No, en el ejemplo hay 4 loops. FirstChildUserIdTAG: 325634 FirstChildUserNameTAG: Aleksey_Stepanov FirstChildCreateTimeTAG: 2012-09-05T14:34:52Z FirstChildTAG: si no me equivoco 1- S a b c S 2- S a d b c S 3- S a b d c S 4- S a d c S 5- a d b a 6- b d c b 7- a d c b a FirstChildUserIdTAG: 298612 FirstChildUserNameTAG: fabianh FirstChildCreateTimeTAG: 2012-09-05T16:09:38Z FirstChildTAG: compadre lo que pasa es que cual trayectoria cerrada puede considerarse una malla, por eso es que hay 7 por que hay siete posibles trayectorias en el circuito que cierran todas totalmente diferenstes aqui en colomiba creamos un grupo que se llama edx(español) espero que te nos unas FirstChildUserIdTAG: 51696 FirstChildUserNameTAG: blasco23 FirstChildCreateTimeTAG: 2012-09-06T16:56:28Z SecondChildTAG: Donde se encuentra este grupo? SecondChildUserIdTAG: 210561 SecondChildUserNameTAG: Alejovillapar SecondChildCreateTimeTAG: 2012-09-07T17:59:34Z SecondChildTAG: ¿Como puede unirme con ustedes? SecondChildUserIdTAG: 256189 SecondChildUserNameTAG: MarquezMario SecondChildCreateTimeTAG: 2012-09-08T05:00:26Z SecondChildTAG: Como uno a este grupo yo tambien soy de Colombia.... SecondChildUserIdTAG: 344373 SecondChildUserNameTAG: steverodriguez SecondChildCreateTimeTAG: 2012-09-09T17:35:16Z SecondChildTAG: Yo soy de españa y tambien me interesa. Como me uno? SecondChildUserIdTAG: 433574 SecondChildUserNameTAG: endika86 SecondChildCreateTimeTAG: 2012-09-15T16:29:20Z FirstChildTAG: Eso es correcto, tienes que establecer todos los caminos cerrados que llevan hasta la fuente, el total de mayas te va dar 7. FirstChildUserIdTAG: 344940 FirstChildUserNameTAG: elbujo FirstChildCreateTimeTAG: 2012-09-07T01:21:51Z IndexTAG: 4799 TitleTAG: Downvotes I know I'm inviting the inevitable, but I really think downvotes are a bad idea.... Not that I ever get them, of course! :-) But they can lead to flame wars. Positive votes will soon separate the valuable posts from the less valuable without all the negativity. Perhaps there should be a "?" button to allow questioning the accuracy of a post - monitored by the karma gods. (And to use it, you have to write a comment yourself explaining your concerns.) UserIdTAG: 79337 UserNameTAG: AppliedImagination CreateTimeTAG: 2012-09-05T13:14:31Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: I have a feeling that we are going to have student moderators this time. The "Student, Student, Student, Student" on everyone's profile seems to hint at that - as though it's a title. If it does happen, flame wars can be controlled. FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-09-05T13:38:46Z FirstChildTAG: I really think nuclear weapon is a bad idea... But people learned not to use it too often. Of course, not everyone. I think one should use downvotes as well as upvotes, because otherwise you'll not be able to tell a good post from a bad one. What I think should be used rarely (except nukes) is downvoting into negative. it's very rude and should be put only against real brutes. FirstChildUserIdTAG: 332194 FirstChildUserNameTAG: IronAlex FirstChildCreateTimeTAG: 2012-09-05T17:22:33Z SecondChildTAG: Having taken this course during the spring, I can confirm that it really caused no major problems, so if the forum community is anywhere near as helpful as last time, I doubt anyone will have much difficulty with it this session either. It's important to remember that people prone to starting big flame wars rarely do so on topics like inductor-capacitor duality or op-amp function; at least it's hard for me to imagine. SecondChildUserIdTAG: 26931 SecondChildUserNameTAG: Csscade SecondChildCreateTimeTAG: 2012-09-06T00:45:32Z SecondChildTAG: Looks like the new forums don't have down votes AI! :-) SecondChildUserIdTAG: 14386 SecondChildUserNameTAG: ashwith SecondChildCreateTimeTAG: 2012-09-16T18:53:00Z IndexTAG: 4800 TitleTAG: Homework and Labs Am wondering how you submit finished homework because i cannot see anything to do with submit! its just a check button. UserIdTAG: 301736 UserNameTAG: MOjangole CreateTimeTAG: 2012-09-05T13:08:56Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: You just have to press the Check button, If any answers are wrong correct them and press check again, when all the answers are correct you have submitted the homework or Lab... Just simple. FirstChildUserIdTAG: 77046 FirstChildUserNameTAG: Sarwar20 FirstChildCreateTimeTAG: 2012-09-06T18:03:35Z SecondChildTAG: what if some of answers are wrong? SecondChildUserIdTAG: undefined SecondChildUserNameTAG: undefined SecondChildCreateTimeTAG: 2012-09-12T14:48:35Z IndexTAG: 4801 TitleTAG: Current flowing from source into the circuit? I don't understand why "Current flowing from source into the circuit(amps, with correct sign)" equal 500e-6 :( UserIdTAG: 343016 UserNameTAG: MagicPaul CreateTimeTAG: 2012-09-05T13:08:08Z VoteTAG: 0 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 5 FirstChildTAG: Because 500e-6A = 5e-4A = 0.0005A = 0.5mA = 500uA FirstChildUserIdTAG: 191549 FirstChildUserNameTAG: Fodin FirstChildCreateTimeTAG: 2012-09-05T13:17:55Z FirstChildTAG: 500e-6 = 500 * 10 raised to pow -6 i.e. 500 uA FirstChildUserIdTAG: 330881 FirstChildUserNameTAG: mmmishra FirstChildCreateTimeTAG: 2012-09-05T13:16:07Z FirstChildTAG: 500e-6 is the scientific notation for saying [ 500*(10^-6) = 500 / (10^6) = 0.0005 ] Amps that is equal to 500 uA FirstChildUserIdTAG: 105018 FirstChildUserNameTAG: ema_cuo FirstChildCreateTimeTAG: 2012-09-05T13:16:08Z SecondChildTAG: thanks for answering the question. SecondChildUserIdTAG: 204017 SecondChildUserNameTAG: KOMMURI SecondChildCreateTimeTAG: 2012-09-05T13:31:27Z SecondChildTAG: thanks SecondChildUserIdTAG: 209722 SecondChildUserNameTAG: pushkarraj6 SecondChildCreateTimeTAG: 2012-09-05T13:36:54Z SecondChildTAG: the correct answer given by the circuit tool (DC) is 500mA not 500uA -- the green check should be reserved for 500e-3 SecondChildUserIdTAG: 10512 SecondChildUserNameTAG: asicok SecondChildCreateTimeTAG: 2012-09-08T00:34:50Z FirstChildTAG: actually don't know it show error indeed its true answer , they have to provide little information about it how to type answer ? what you say guys FirstChildUserIdTAG: 158348 FirstChildUserNameTAG: mudz FirstChildCreateTimeTAG: 2012-09-05T14:04:12Z FirstChildTAG: what happens is that when you run the simulation circuit gives a direction of the current, this current enters the positive terminal of the source, so this trend is positive, I hope you understand, excuse the bad English FirstChildUserIdTAG: 51696 FirstChildUserNameTAG: blasco23 FirstChildCreateTimeTAG: 2012-09-05T14:38:54Z SecondChildTAG: Sorry to correct your answer blasco23, but according to their notes and I state "And the source current sign is negative if the current is flowing out of the negative terminal of the source, through the circuit, and into the positive terminal of the source." I am also confuse with their answer as why positive instead of negative because the arrow indicates into the positive terminal. can someone please explain? SecondChildUserIdTAG: 132229 SecondChildUserNameTAG: ahlo SecondChildCreateTimeTAG: 2012-09-05T16:25:56Z IndexTAG: 4802 TitleTAG: Anyone from Ethiopia? Hello, My name is Brook am from Addis Ababa. Is there anyone here taking the course who is also from Addis Ababa? UserIdTAG: 65378 UserNameTAG: Brook CreateTimeTAG: 2012-09-05T12:55:30Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4803 TitleTAG: Lab 0: Using the Tools On lab 0, there is not a way to do the voltage testing. It says click on DC but there is not a button called DC like in the video. Or the AC or Trans either. UserIdTAG: 271670 UserNameTAG: moncapitane CreateTimeTAG: 2012-09-05T12:51:29Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: I tried reloading (i.e. logout then login) and tools appeared, but only DC is working FirstChildUserIdTAG: 148389 FirstChildUserNameTAG: chento FirstChildCreateTimeTAG: 2012-09-06T02:42:05Z IndexTAG: 4804 TitleTAG: Adding tags to posts... A minor problem: I wanted to add "discussion" to a post as a tag. It won't let me - "discussion-feedback" floats up as a tip. If I want to write about the discus in the paralympics (a Welshman won!) it won't let me tag it. Also I can't seem to edit tags when I edit a post - I wanted to get rid of my experimental "aaaa" tag, but it isn't shown when I'm editing. I'm tagging this "buglet", "discussionbuglet" and "staff". Actually it let me have buglet and discussionbuglet, but not staff! (Edit: trying the last two again....) UserIdTAG: 79337 UserNameTAG: AppliedImagination CreateTimeTAG: 2012-09-05T12:48:35Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 3 FirstChildTAG: Another bug, I am able to upvote myself. FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-09-05T13:03:21Z SecondChildTAG: Hmmm - I wondered about that but managed to resist the temptation! ;-) Have a vote! SecondChildUserIdTAG: 79337 SecondChildUserNameTAG: AppliedImagination SecondChildCreateTimeTAG: 2012-09-05T13:08:19Z FirstChildTAG: Hmmm - when editing a post you get presented with an empty tag box: you've lost your original. Also, the box seems to only accept ONE tag after that... FirstChildUserIdTAG: 79337 FirstChildUserNameTAG: AppliedImagination FirstChildCreateTimeTAG: 2012-09-05T13:02:54Z FirstChildTAG: I cannot find any documentation or discussion on how to use or add tags to my post and what tags are available. From what I have seen so far, "sandbox" is a tag, as I saw it highlighted in one of my posts when I used the word. When I saw "tag" highlighted, that confirmed it. Hmm, maybe I should put some words in this post, ones that I think could be a tag, and see if they actually are. Here goes: tag, sandbox, staff, Edx. Anyway, does anybody know where I can get more information on tags specifically related to Edx? Thank, you. FirstChildUserIdTAG: 132828 FirstChildUserNameTAG: rharris FirstChildCreateTimeTAG: 2012-09-30T06:25:31Z SecondChildTAG: Well, that didn't seem to work. Maybe I have to highlight the word. Let's see: tag, post, sandbox. What is this "emphasis" button, when I emphasize a word, is that what tags it? Let's see. *tag*, *sandbox*, *staff*. SecondChildUserIdTAG: 132828 SecondChildUserNameTAG: rharris SecondChildCreateTimeTAG: 2012-09-30T06:31:27Z IndexTAG: 4805 TitleTAG: Join 6002x Study group on skype **Join 6002x Study group on skype. ID: m.umair72001** UserIdTAG: 908 UserNameTAG: m_umair72001 CreateTimeTAG: 2012-09-05T12:34:05Z VoteTAG: 0 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 6 FirstChildTAG: آپ پہلے کی طرح اپنا گروہ لے کر یہاں سے فرار ہونا چاہتے ہیں؟ :-) Enjoy! FirstChildUserIdTAG: 112757 FirstChildUserNameTAG: MImran FirstChildCreateTimeTAG: 2012-09-05T14:44:17Z SecondChildTAG: nahin janab, bohat say logon nay kaha k ka ya group nai hona chahiay is liaay manay delete kar dia. SecondChildUserIdTAG: 908 SecondChildUserNameTAG: m_umair72001 SecondChildCreateTimeTAG: 2012-09-06T04:35:33Z FirstChildTAG: Hi, may I participate on this gorup please? I sent you an Skype invite. Thanks, FirstChildUserIdTAG: 221013 FirstChildUserNameTAG: RodrigoAlv FirstChildCreateTimeTAG: 2012-09-05T20:15:32Z FirstChildTAG: bdandaraw just added you as a contact. FirstChildUserIdTAG: 39123 FirstChildUserNameTAG: Bdandaraw FirstChildCreateTimeTAG: 2012-09-06T04:48:50Z FirstChildTAG: hi FirstChildUserIdTAG: 229448 FirstChildUserNameTAG: Shiref FirstChildCreateTimeTAG: 2012-09-06T07:47:47Z SecondChildTAG: can you also add me in the group ? Skype ID = ranierimusella SecondChildUserIdTAG: 200496 SecondChildUserNameTAG: Ranieri SecondChildCreateTimeTAG: 2012-09-06T15:49:42Z FirstChildTAG: can you also add me in the group ? Skype ID = ranierimusella FirstChildUserIdTAG: 200496 FirstChildUserNameTAG: Ranieri FirstChildCreateTimeTAG: 2012-09-06T15:50:21Z FirstChildTAG: I'm sending you a Invite too!:P FirstChildUserIdTAG: 43397 FirstChildUserNameTAG: JesusxFan FirstChildCreateTimeTAG: 2012-09-06T21:52:12Z IndexTAG: 4806 TitleTAG: BUG! report!!! How may I change the voltage source to AC. The tools seems to have a bug. I will do it on Proteus. UserIdTAG: 421471 UserNameTAG: NDJules CreateTimeTAG: 2012-09-13T19:10:42Z VoteTAG: -1 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: Double click on the source and change the "type" to "sin" from the drop down menu. FirstChildUserIdTAG: 416770 FirstChildUserNameTAG: RafaeAshfaq FirstChildCreateTimeTAG: 2012-09-13T19:53:02Z SecondChildTAG: Even I am finding difficulty in changing the Values/Type of voltage source. I suppose everyone else is facing the same problem. Is it? SecondChildUserIdTAG: 163264 SecondChildUserNameTAG: Aamir_edx SecondChildCreateTimeTAG: 2012-09-14T04:12:44Z SecondChildTAG: I also have this problem. Double click on the source not opens the drop down menu. It doesn't cause any action. SecondChildUserIdTAG: 333534 SecondChildUserNameTAG: Anta SecondChildCreateTimeTAG: 2012-09-14T19:59:19Z SecondChildTAG: same here: voltage source is not responding my double click... :/ SecondChildUserIdTAG: 217768 SecondChildUserNameTAG: brunocunha SecondChildCreateTimeTAG: 2012-09-15T02:58:13Z IndexTAG: 4807 TitleTAG: Help! Hi staff...am new i just registered for the course today and i need help so that i can catch up and submit the lab work and the homework before 16th.With a little assistance i will be good to go since i am already a second year student in electrical and electronics engineering inn a university in Kenya.Please assist. UserIdTAG: 431942 UserNameTAG: Neshizzo CreateTimeTAG: 2012-09-13T13:55:30Z VoteTAG: -1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 2 FirstChildTAG: Hi you can solve the lab very easy your target is getting voltage 1.5v in bulb so put resistance in series and change the value or resistance on clicking the resistor you can rotate resistor clicking resistor and press r. FirstChildUserIdTAG: 318906 FirstChildUserNameTAG: eqbal2111 FirstChildCreateTimeTAG: 2012-09-13T14:20:13Z FirstChildTAG: Can I help you? In which Part are you lost? FirstChildUserIdTAG: 58095 FirstChildUserNameTAG: Myrimit FirstChildCreateTimeTAG: 2012-09-13T20:18:04Z IndexTAG: 4808 TitleTAG: Asking About The Exercise What is the question? Where is it? UserIdTAG: undefined UserNameTAG: undefined CreateTimeTAG: 2012-09-13T13:32:46Z VoteTAG: -1 CoursewareTAG: Week 1 / Resistor Voltages CommentableIdTAG: 6002x_resistor_voltages NumberOfReplyTAG: 1 FirstChildTAG: The problem description should be above the video. Is it not showing? FirstChildUserIdTAG: 381746 FirstChildUserNameTAG: jelizon FirstChildCreateTimeTAG: 2012-09-13T13:42:30Z IndexTAG: 4809 TitleTAG: Entering the (t) on this calculator How do I enter the T (t) on this calculator. UserIdTAG: 359302 UserNameTAG: GaryG CreateTimeTAG: 2012-09-11T23:25:23Z VoteTAG: -1 CoursewareTAG: Week 1 / AC power CommentableIdTAG: 6002x_S1E3_AC_power NumberOfReplyTAG: 1 FirstChildTAG: Hi, there...1st, on my dual-responses to another question....oops You can't do that with this basic calculator. To see what you can do with this calculator, click on the, "i." That shows you are really limited to basic functions, like square, sqrt(square root), trig. functions, basic log functions. You have to do the Integration the old-fashioned way, either by hand or by calculator. Will recommend the HP 50G or one of the other companies, like whatever Texas Instrument is offering at the top of their line. You can expect to pay roughly $135 US dollars. Hope that helps. Now, time for me to be confused... FirstChildUserIdTAG: 119440 FirstChildUserNameTAG: WG FirstChildCreateTimeTAG: 2012-09-12T03:33:14Z SecondChildTAG: Thanks, Guess I just bought the wrong one. It has almost everything else but no Time (t). SecondChildUserIdTAG: 359302 SecondChildUserNameTAG: GaryG SecondChildCreateTimeTAG: 2012-09-12T22:02:10Z SecondChildTAG: i have a ti-89 and i get twice the answer :/ SecondChildUserIdTAG: 338871 SecondChildUserNameTAG: jtcochran SecondChildCreateTimeTAG: 2012-09-17T01:53:10Z SecondChildTAG: i take that back...i changed it into radian mode and everything was fine. just a coincidence that it was almost exactly double SecondChildUserIdTAG: 338871 SecondChildUserNameTAG: jtcochran SecondChildCreateTimeTAG: 2012-09-17T02:04:02Z IndexTAG: 4810 TitleTAG: Turn off background music in slides TAs, I find background music during the lectures/slides sequence annoying. Is there any way to turn the music off? Thanks UserIdTAG: 213493 UserNameTAG: StudentZ CreateTimeTAG: 2012-09-10T07:22:58Z VoteTAG: -1 CoursewareTAG: Week 1 / Lumped Element Abstraction CommentableIdTAG: 6002x_lumped_element_abstraction NumberOfReplyTAG: 0 IndexTAG: 4811 TitleTAG: voltage and current have opposite sign The resister dissipates power to water, therefore it becomes power source and the power dissipated obtains negative sign. That's why the current is chosen to have non-conventional direction; and therefore voltage and current have opposite sign. UserIdTAG: 161058 UserNameTAG: van_tho CreateTimeTAG: 2012-09-10T02:04:20Z VoteTAG: -1 CoursewareTAG: Week 1 / Power CommentableIdTAG: 6002x_S1E2_Power NumberOfReplyTAG: 1 FirstChildTAG: But if we measure the current with a ammeter we would see that current goes from the network to the resistor. Or not? FirstChildUserIdTAG: 340568 FirstChildUserNameTAG: witedonkey FirstChildCreateTimeTAG: 2012-09-10T20:03:31Z IndexTAG: 4812 TitleTAG: Is there anywhere on the internet (or elsewhere) that provides a cheat sheet (not a book, video or tutorial) with full examples on translating mathematical calculus calculations into computer code? Is there anywhere on the internet (or elsewhere) that provides a cheat sheet (not a book, video or tutorial) with full examples on translating mathematical calculus calculations into computer code? I am having a very hard time interpreting pure mathematical symbols into a language that is relevant to my area of knowledge (my current profession is web development). Mathematica may have the beginnings of this somewhere, but this is closed source (something I try to avoid at all costs [both moral and financial]) and for some reason is geared towards people with a firm grasp of math and a limited grasp on coding. What I would like to use is [Sage][1], but as far as I can see, it does not solve integral calculations and there is only limited translation of mathematic symbols to python. Is anyone else familiar with sage and knows how to do Maxwell and general integral calculations using it? Thanks! (P.S. There is https://en.wikipedia.org/wiki/Mathematical_symbols which is extremely useful) [1]: http://www.sagemath.org UserIdTAG: 306110 UserNameTAG: AEJ CreateTimeTAG: 2012-09-09T19:48:00Z VoteTAG: -1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 4 FirstChildTAG: Im not realy sure weather this answers your problem. In fact I have quite foggy idea of what you are trying to do xD if you want to do mathematical functions in programing languages, then ive not heard of any librarys. put some more fabulous symboles on paper/site - you might try latex Just do computations for given functions, avoiding mathematica - http://www.scilab.org/ open source. FirstChildUserIdTAG: 404284 FirstChildUserNameTAG: tomdrifmach FirstChildCreateTimeTAG: 2012-09-09T20:52:17Z SecondChildTAG: Thanks, this may be just what I'm looking for! SecondChildUserIdTAG: 306110 SecondChildUserNameTAG: AEJ SecondChildCreateTimeTAG: 2012-09-10T01:56:44Z FirstChildTAG: By the way, by 'cheat sheet', I was not referring to a cheat sheet for the homework, just a table that correlates calculus to something that can actually be calculated FirstChildUserIdTAG: 306110 FirstChildUserNameTAG: AEJ FirstChildCreateTimeTAG: 2012-09-11T02:20:31Z FirstChildTAG: Are either of these what you are looking for? [Rosetta Stone for CAS][1] [Comparison of Computer Algebra Systems][2] [1]: http://web.archive.org/web/20070512135930/wiki.axiom-developer.org/images/axiom--test--1/src/doc/Rosetta.pdf "Rosetta stone for CAS" [2]: http://en.wikipedia.org/wiki/Comparison_of_computer_algebra_systems "Comparison of CAS" FirstChildUserIdTAG: 148542 FirstChildUserNameTAG: planetscape FirstChildCreateTimeTAG: 2012-09-11T12:18:19Z FirstChildTAG: Have you checked out [SciPy][1]? It's a python module for efficient mathematical computation, with a fairly easy programmatic interface, and support for complex numbers, signal processing, logic operations and calculus, among many other features. It depends on numpy, so it also provides fast matrix manipulation. [1]: http://www.scipy.org/ FirstChildUserIdTAG: 185338 FirstChildUserNameTAG: mgaldieri FirstChildCreateTimeTAG: 2012-09-21T22:16:12Z IndexTAG: 4813 TitleTAG: Video Problem I'm not able to watch any of the lecture videos.I'm using windows7 and latest version of chrome.Please help me out UserIdTAG: 98600 UserNameTAG: Himalay CreateTimeTAG: 2012-09-09T16:36:54Z VoteTAG: -1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 5 FirstChildTAG: try firefox FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-09T16:51:37Z FirstChildTAG: May be flash player is required. Download latest from adobe. FirstChildUserIdTAG: 222234 FirstChildUserNameTAG: ivsprasad FirstChildCreateTimeTAG: 2012-09-10T10:56:11Z FirstChildTAG: does it say any error message? maybe it could solve your problem. FirstChildUserIdTAG: 132229 FirstChildUserNameTAG: ahlo FirstChildCreateTimeTAG: 2012-09-10T17:55:30Z FirstChildTAG: For me, on Windows XP, it didn't work on Firefox, but did on the Opera web browser. So, just try every browser until you hopefully find one that works. FirstChildUserIdTAG: 129978 FirstChildUserNameTAG: StuRat FirstChildCreateTimeTAG: 2012-09-11T06:45:24Z FirstChildTAG: You may be in a country that does not allow YouTube? FirstChildUserIdTAG: 292543 FirstChildUserNameTAG: Pennypacker FirstChildCreateTimeTAG: 2012-09-09T17:28:35Z IndexTAG: 4814 TitleTAG: total power total power for Resistors which connected in parallel equal Pt= 1/Pn as Rt= 1/Rn ? UserIdTAG: 399292 UserNameTAG: Qabali CreateTimeTAG: 2012-09-09T14:46:53Z VoteTAG: -1 CoursewareTAG: Week 1 / Circuit topology CommentableIdTAG: 6002x_S2E1_Circuit_Topology NumberOfReplyTAG: 3 FirstChildTAG: I'm not completely clear on what your question is, so please clarify for me if this doesn't answer what you want. When you connect a network of resistors to a voltage source in parallel, the total power dissipated by the system actually increases with the number of resistors you connect, not decreases. You can think of it like this- connecting resistors in parallel always will decrease the net equivalent resistance. Less resistance with a constant voltage source means more current. And thus the total power delivered by the voltage source- given by multiplying i times v- will increase. FirstChildUserIdTAG: 390034 FirstChildUserNameTAG: kahlil FirstChildCreateTimeTAG: 2012-09-09T15:07:47Z FirstChildTAG: No, Pt=P1+P2+P3.. where P1=I1*V ; P2=I2*V .. and I1=V/R1 ; I2=V/R2 .. so P1=V*V/R1 P2=V*V/R2 P3=V*V/R3 . (And V=V1=V2=V3 because " Resistors are connected in parallel") FirstChildUserIdTAG: 277168 FirstChildUserNameTAG: AndrewKiselev FirstChildCreateTimeTAG: 2012-09-09T15:15:13Z FirstChildTAG: nop...Power always added in any scheme whether it is serial or parrell combination FirstChildUserIdTAG: 415376 FirstChildUserNameTAG: imab90 FirstChildCreateTimeTAG: 2012-09-12T15:42:36Z IndexTAG: 4815 TitleTAG: Q2 You can simply do the following: (120*0.707*sqrt(2)*cos(2*pi*60*t))^2/110 UserIdTAG: 239736 UserNameTAG: khatal CreateTimeTAG: 2012-09-09T05:12:09Z VoteTAG: -1 CoursewareTAG: Week 1 / AC power CommentableIdTAG: 6002x_S1E3_AC_power NumberOfReplyTAG: 2 FirstChildTAG: 131*(cos(377*t))^2 ? FirstChildUserIdTAG: 394836 FirstChildUserNameTAG: v2g6ch4 FirstChildCreateTimeTAG: 2012-09-09T05:15:52Z FirstChildTAG: https://www.edx.org/courses/MITx/6.002x/2012_Fall/discussion/forum/6002x_S1E3_AC_power/threads/504c53e607d5d11f00000004 Question 2 done there in video. FirstChildUserIdTAG: 193033 FirstChildUserNameTAG: NIslam FirstChildCreateTimeTAG: 2012-09-09T15:22:28Z IndexTAG: 4816 TitleTAG: Catching Up The negative current on the DC analysis had me stuck for a bit but then I noticed the current is flowing clockwise from the positive end. Also have to remember to include the k for the frequency value before the transient analysis. Using Chrome on Windows 7 and it works perfectly. UserIdTAG: 150589 UserNameTAG: TriciaC CreateTimeTAG: 2012-09-09T04:13:54Z VoteTAG: -1 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 0 IndexTAG: 4817 TitleTAG: KCL and number of the nodes Why there are just three nodes. I can see 4 nodes in this schematic, they are a,b,c,d. and if they are only 3 nodes, so what labels they are? UserIdTAG: 292360 UserNameTAG: Loai CreateTimeTAG: 2012-09-09T02:39:29Z VoteTAG: -1 CoursewareTAG: Week 1 / Circuit topology CommentableIdTAG: 6002x_S2E1_Circuit_Topology NumberOfReplyTAG: 4 FirstChildTAG: coz total number of nodes is no.of branches-total nodes+1 FirstChildUserIdTAG: 351358 FirstChildUserNameTAG: tabi FirstChildCreateTimeTAG: 2012-09-09T03:57:24Z FirstChildTAG: it says 4 nodes FirstChildUserIdTAG: 137545 FirstChildUserNameTAG: engguzman FirstChildCreateTimeTAG: 2012-09-09T02:56:00Z FirstChildTAG: the number of nodes is 4 ; the number of **independent** KCL equations is 3 FirstChildUserIdTAG: 236136 FirstChildUserNameTAG: Hamid-ch FirstChildCreateTimeTAG: 2012-09-09T23:41:15Z FirstChildTAG: There are 4 nodes actually FirstChildUserIdTAG: 416421 FirstChildUserNameTAG: EngineerMaaz FirstChildCreateTimeTAG: 2012-09-11T20:10:32Z IndexTAG: 4818 TitleTAG: Doubt In my lab homework,check button is not responding.Can anyone help me? UserIdTAG: 182669 UserNameTAG: srikmss CreateTimeTAG: 2012-09-08T16:01:28Z VoteTAG: -1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 0 IndexTAG: 4819 TitleTAG: kvl sign i got everything right except for kvl .i got opp signs compared to the ans.. UserIdTAG: 337430 UserNameTAG: aparna_b CreateTimeTAG: 2012-09-08T15:27:47Z VoteTAG: -1 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: First , i too got the same ... but when i read the instructions just above ... i understood what mistake i have done... just see at R1 ...GOING CLOCKWISE...Va=3V ..Vb=2.5V ... change = -0.5 V ..That's it... FirstChildUserIdTAG: 209930 FirstChildUserNameTAG: saikiraniitr FirstChildCreateTimeTAG: 2012-09-08T16:02:37Z SecondChildTAG: ok..i understood now. thanks for the explanation.. SecondChildUserIdTAG: 337430 SecondChildUserNameTAG: aparna_b SecondChildCreateTimeTAG: 2012-09-09T04:16:44Z SecondChildTAG: for correction Vab=Vb-Va, for clockwise (a-->b, and b-->c .... Vbc=Vc-Vb ... Thinks SecondChildUserIdTAG: 851568 SecondChildUserNameTAG: aniss SecondChildCreateTimeTAG: 2012-12-05T23:47:47Z IndexTAG: 4820 TitleTAG: Compatibility issues of the videos in IE8 browser Hi , I am trying to run the video course lectures but they are not showing up in IE8 but are running on youtube.com, access to which does not go well with my company's policy. So, I want to know, is there a prerequisite in terms of hardware/firmware/software to run the lectures while I am logged-in in my edx account, without having to resort to youtube.com UserIdTAG: 259278 UserNameTAG: August04 CreateTimeTAG: 2012-09-07T11:12:47Z VoteTAG: -1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Use Firefox FirstChildUserIdTAG: 142324 FirstChildUserNameTAG: continuity FirstChildCreateTimeTAG: 2012-09-07T13:15:02Z SecondChildTAG: Actually Chrome is the browser of choice. SecondChildUserIdTAG: 79361 SecondChildUserNameTAG: kimt SecondChildCreateTimeTAG: 2012-09-07T13:16:23Z IndexTAG: 4821 TitleTAG: S1E6 KVL. correct answer marked like wrong on the S1E6 KVL. "Write an algebraic expression for the branch voltage v5" i answered: -2*V+2*v1+2*v2 and was marked like a wrong answer but that expression is equivalent to v1+v2-V . Well i guess that i should to simplify the expression, :s UserIdTAG: 144678 UserNameTAG: jorge07 CreateTimeTAG: 2012-09-07T02:53:12Z VoteTAG: -1 CoursewareTAG: Week 1 / KVL and KCL exercises CommentableIdTAG: 6002x_kvl_kcl_exercises NumberOfReplyTAG: 1 FirstChildTAG: Jorge07, "-2*V+2*v1+2*v2 isn't equal to v1+v2-V but it's equal to 2(v1+v2-V). It means that you've answered 2*v5 but not v5". Be careful) FirstChildUserIdTAG: 202880 FirstChildUserNameTAG: Adilzhan FirstChildCreateTimeTAG: 2012-09-07T19:21:26Z IndexTAG: 4822 TitleTAG: user interface it´s really "friendly user" UserIdTAG: 304561 UserNameTAG: willow CreateTimeTAG: 2012-09-06T06:06:11Z VoteTAG: -1 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 0 IndexTAG: 4823 TitleTAG: I think the answer to 5. transient analysis is wrong! I think the answer to 5. transient analysis is wrong! UserIdTAG: 156413 UserNameTAG: isping CreateTimeTAG: 2012-09-06T01:19:59Z VoteTAG: -1 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 4 FirstChildTAG: yeah FirstChildUserIdTAG: 131426 FirstChildUserNameTAG: M_Inam_Ul_Haq FirstChildCreateTimeTAG: 2012-09-06T01:51:42Z SecondChildTAG: How so? Are you running into rounding errors? SecondChildUserIdTAG: 382956 SecondChildUserNameTAG: JustinBautista SecondChildCreateTimeTAG: 2012-09-06T02:13:17Z FirstChildTAG: I have some doubts on this transient analysis. Is there anyone to help me out? or it would be great if you could re-illustrate the way you have proceeded. look forward to hear from you... FirstChildUserIdTAG: 224620 FirstChildUserNameTAG: TomVO FirstChildCreateTimeTAG: 2012-09-06T10:17:03Z FirstChildTAG: I followed the instructions and got the first two parts of #5 correct, but my third value seemed to be off... the reason being that the third value was shown in mV on the graph, while the other two were still in V. I wasn't expecting that, nor was I paying attention lol. FirstChildUserIdTAG: 309877 FirstChildUserNameTAG: ste9 FirstChildCreateTimeTAG: 2012-09-06T03:23:55Z FirstChildTAG: I followed the exercise instructions and I got the right answers, check if you missed something. ;) good luck. FirstChildUserIdTAG: 385692 FirstChildUserNameTAG: tpfslima FirstChildCreateTimeTAG: 2012-09-06T03:33:21Z IndexTAG: 4824 TitleTAG: un finished task if i havent finish my task, and i hit check, can i finish it later? UserIdTAG: 222786 UserNameTAG: rocoma CreateTimeTAG: 2012-09-05T20:01:27Z VoteTAG: -1 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 3 FirstChildTAG: Why you don´t try with lab 0, It cost you nothing. (Just ask yourself: what if...:)) FirstChildUserIdTAG: 164171 FirstChildUserNameTAG: PacoJuarez FirstChildCreateTimeTAG: 2012-09-05T20:22:10Z FirstChildTAG: Yep FirstChildUserIdTAG: 96452 FirstChildUserNameTAG: Lyla FirstChildCreateTimeTAG: 2012-09-05T20:05:07Z FirstChildTAG: Yes! FirstChildUserIdTAG: 59671 FirstChildUserNameTAG: neovitwin FirstChildCreateTimeTAG: 2012-09-05T20:04:38Z IndexTAG: 4825 TitleTAG: Error in changing properties of supply This is irritating me i am trying from 3 hrs anybody could help me,i had changed the resistors values by double clicking but voltage source value and name are not changing UserIdTAG: 383699 UserNameTAG: Gogineni CreateTimeTAG: 2012-09-05T18:07:41Z VoteTAG: -1 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 8 FirstChildTAG: Double-clicking the voltage source in the middle should bring up its properties to edit. Some people are having problems with browsers. Chrome seems to work best. FirstChildUserIdTAG: 282092 FirstChildUserNameTAG: richspoon FirstChildCreateTimeTAG: 2012-09-05T18:15:52Z SecondChildTAG: My head got stuck up with this in chrome wiring is not working in mozilla properties(specifically supply) are not working what to do brother please help me SecondChildUserIdTAG: 383699 SecondChildUserNameTAG: Gogineni SecondChildCreateTimeTAG: 2012-09-05T18:18:36Z FirstChildTAG: I use Mozilla, I could not get it to work on my Windows PC, but it worked fine in my MAC. You might try a different browser. FirstChildUserIdTAG: 351356 FirstChildUserNameTAG: gerrie FirstChildCreateTimeTAG: 2012-09-05T18:17:58Z FirstChildTAG: Hi what browser are you using? FirstChildUserIdTAG: 296419 FirstChildUserNameTAG: oscman FirstChildCreateTimeTAG: 2012-09-05T18:19:49Z SecondChildTAG: i am using opera, try it! SecondChildUserIdTAG: 488779 SecondChildUserNameTAG: nadir22 SecondChildCreateTimeTAG: 2012-10-11T17:13:50Z FirstChildTAG: Im using google chrome and all was fine FirstChildUserIdTAG: 296419 FirstChildUserNameTAG: oscman FirstChildCreateTimeTAG: 2012-09-05T18:21:24Z FirstChildTAG: I'm using Mozilla Firefox on Linux(Ubuntu) and it works fine. I think you may refresh your page(something similar happened to me and the DC button didn't appear) FirstChildUserIdTAG: 289472 FirstChildUserNameTAG: OsamaAdel FirstChildCreateTimeTAG: 2012-09-05T18:37:31Z SecondChildTAG: I'm using firefox in ubuntu too, but the transient analysis doesn't work, even after refreshing. SecondChildUserIdTAG: 347289 SecondChildUserNameTAG: kellrobinson SecondChildCreateTimeTAG: 2012-09-05T20:18:27Z FirstChildTAG: I'm using google chrome and things are weird here... some times it seems that the changes i make are not applied... FirstChildUserIdTAG: 305965 FirstChildUserNameTAG: jorno FirstChildCreateTimeTAG: 2012-09-05T18:38:02Z FirstChildTAG: Gogeneni, does the "Edit Properties" box appear when you double click the voltage source? FirstChildUserIdTAG: 347289 FirstChildUserNameTAG: kellrobinson FirstChildCreateTimeTAG: 2012-09-05T20:10:44Z FirstChildTAG: you should double click the voltage source but if there is nothing umm...I think it's an error from your browser use Google-Chrome or Firefox or Opera and try again :) FirstChildUserIdTAG: 279239 FirstChildUserNameTAG: abdallah-ali FirstChildCreateTimeTAG: 2012-09-05T19:40:57Z IndexTAG: 4826 TitleTAG: Problem with Lab 1 In the Lab 1, I can't see any elements provided but resistor. It's impossible to run a simulation without voltage source and the system also shows an error that I need to put GND on the scheme when I try to check my work. It looks like I can't really do anything with that, maybe it's global mistake? Let me know if you have the same problem or managed to solve it. UserIdTAG: 103874 UserNameTAG: Morill CreateTimeTAG: 2012-09-05T17:53:38Z VoteTAG: -1 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 4 FirstChildTAG: I can't see anything also. FirstChildUserIdTAG: 88163 FirstChildUserNameTAG: cbulbul FirstChildCreateTimeTAG: 2012-09-05T18:28:44Z FirstChildTAG: Hi. I really did not have any problem with the Lab1. You may find the GND on the right side. FirstChildUserIdTAG: 125472 FirstChildUserNameTAG: fabioirmao FirstChildCreateTimeTAG: 2012-09-05T18:00:42Z FirstChildTAG: Could you add voltage source and ground? FirstChildUserIdTAG: 88163 FirstChildUserNameTAG: cbulbul FirstChildCreateTimeTAG: 2012-09-05T18:31:28Z FirstChildTAG: Maybe the browser is the guilty, looks (behaves?) different on Google Chrome than on Firefox. In Firefox do not appears DC and TRAN, but in Chrome they do. I hope they will correct this, or give us some guidance. FirstChildUserIdTAG: 164171 FirstChildUserNameTAG: PacoJuarez FirstChildCreateTimeTAG: 2012-09-05T18:32:41Z SecondChildTAG: I tried all of my browsers but in Chrome I have the same situation and with IE it doesn't work at all. SecondChildUserIdTAG: 103874 SecondChildUserNameTAG: Morill SecondChildCreateTimeTAG: 2012-09-05T19:26:01Z IndexTAG: 4827 TitleTAG: A little typo in "Kirchoff's Voltage Law" In task 4: Kirchoff's Voltage Law. I believe correct is **Kirchhoff's** (double h) circuit laws UserIdTAG: 321074 UserNameTAG: Slashdev CreateTimeTAG: 2012-09-05T13:12:35Z VoteTAG: -1 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 1 FirstChildTAG: True FirstChildUserIdTAG: 105018 FirstChildUserNameTAG: ema_cuo FirstChildCreateTimeTAG: 2012-09-05T13:19:24Z SecondChildTAG: com'n man its not an english grammer class SecondChildUserIdTAG: undefined SecondChildUserNameTAG: undefined SecondChildCreateTimeTAG: 2012-09-05T13:31:41Z IndexTAG: 4828 TitleTAG: Suponer sentidos a las corrientes que no se conocen Primero se debe suponer un sentido para las corrientes en el nodo e2, ya sea que entren o salgan. Se hace la sumatoria de las corrientes que debe ser cero. Para que dé la respuesta tal como se muestra, se suponen que todas las corrientes salen, menos I1 que entra. UserIdTAG: 53004 UserNameTAG: jfdelacruz CreateTimeTAG: 2012-09-10T01:08:38Z VoteTAG: -2 CoursewareTAG: Week 1 / Node analysis practice, part 3 CommentableIdTAG: 6002x_L2Node2 NumberOfReplyTAG: 1 FirstChildTAG: Это было бы логично, но тут для "исходящего" тока использована именно ветвь R5 почему-то. FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-10T12:26:59Z SecondChildTAG: c'est la vie=) SecondChildUserIdTAG: undefined SecondChildUserNameTAG: undefined SecondChildCreateTimeTAG: 2012-09-10T16:41:14Z SecondChildTAG: Спасибо за ответ! -И, Google Translate. SecondChildUserIdTAG: 230787 SecondChildUserNameTAG: Gabriel007 SecondChildCreateTimeTAG: 2012-09-10T17:06:29Z SecondChildTAG: отлично ответил! ) SecondChildUserIdTAG: 341884 SecondChildUserNameTAG: Michaeldumchev SecondChildCreateTimeTAG: 2012-09-11T17:10:00Z IndexTAG: 4829 TitleTAG: Negative current solved, bug possibly found I know there are a lot of similar post regarding the negative current but I guess this is more about a possible bug than the current issue. I also got the current wrong and I understood why when I saw a post saying that the arrow is an **assumption** and further understood that it is the **software analysis' assumption**. The analysis assumes first a counter-clockwise direction of the current and evaluates the current based on that direction, thus the negative current. I was curious enough to try and correct the current sign by reversing the diagram and hoping that the analysis will assume the same counter-clockwise direction, and this is what I got: ![enter image description here][1] It may be that the negative values here are still correct although not conventionally sound (or totally wrong). It's like I am buying a 3 dollar meal so I am "getting" a -$3 from my pocket ;-) [1]: http://www.dudesign.co/public_images/analysis_bug.png UserIdTAG: 136862 UserNameTAG: bongdu CreateTimeTAG: 2012-09-08T22:22:51Z VoteTAG: -2 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 2 FirstChildTAG: No bugs. Your ground is being measured from the high point in potential ( the positive end of the source ). Because voltage is a difference in potential relative to ground, the negative values show that the voltage is dropping from your zero point. The arrow is a bit trickier, it represents the direction of the actual flow of particles ( electrons from - to + ). Because they have a negative charge, any direction they move in will have a negative electric current, giving a positive value to the electric current in the opposite direction. FirstChildUserIdTAG: 212185 FirstChildUserNameTAG: stantonwjones FirstChildCreateTimeTAG: 2012-09-09T03:38:28Z SecondChildTAG: http://en.wikipedia.org/wiki/Voltage SecondChildUserIdTAG: 212185 SecondChildUserNameTAG: stantonwjones SecondChildCreateTimeTAG: 2012-09-09T03:38:53Z SecondChildTAG: Oh yes, Silly. I forgot to move the ground as well on the other side ;-). Thanks for clearing this up for me. SecondChildUserIdTAG: 136862 SecondChildUserNameTAG: bongdu SecondChildCreateTimeTAG: 2012-09-09T21:14:48Z FirstChildTAG: It doesnt matter. FirstChildUserIdTAG: 394836 FirstChildUserNameTAG: v2g6ch4 FirstChildCreateTimeTAG: 2012-09-09T02:57:06Z SecondChildTAG: OK ! In USA, the current arrow has it's real direction ( but a negative sign ). In France we putted the arrow flowing out of the + ok ok...good SecondChildUserIdTAG: 373752 SecondChildUserNameTAG: Croak SecondChildCreateTimeTAG: 2012-09-10T14:00:40Z IndexTAG: 4830 TitleTAG: Lab 0 usage support request Dear Sir , Thanks for your reply But still aim facing the problem with connecting wire ,but I am able to connecting wire when I am using another lap top which as l so using Google chrome as browser I attempted to adjust the settings of my lap top touch pad but still not able to success My lap top is dell Latitude E5410 any way I am approaching to hardware person for this defect When I am adding the value of LAB 0: USING THE TOOLS WHEN I PRESS CHECK BUTTON IAM GETTING THIS MESSAGE “Invalid input: could not interpret '500uA' as a number’’ UserIdTAG: 156694 UserNameTAG: mkprasanth CreateTimeTAG: 2012-09-07T17:22:06Z VoteTAG: -2 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 2 FirstChildTAG: Only put the numbers, without letters. FirstChildUserIdTAG: 80601 FirstChildUserNameTAG: MiltonM FirstChildCreateTimeTAG: 2012-09-07T17:31:12Z FirstChildTAG: Yeah, try '500u'. The unit doesn't need to be mentioned. FirstChildUserIdTAG: undefined FirstChildUserNameTAG: undefined FirstChildCreateTimeTAG: 2012-09-07T18:58:12Z IndexTAG: 4831 TitleTAG: H1P3: Poor Workmanship from which lecture or video (i mean any kind of stuff) i can get enough knowledge to solve this? UserIdTAG: 400892 UserNameTAG: ahmadyousaf CreateTimeTAG: 2012-09-09T22:56:36Z VoteTAG: -3 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: It would be best if you go through chapter 1 of the textbook. It gives you basics of ac power and every other stuff you need to solve that problem!!! FirstChildUserIdTAG: 385662 FirstChildUserNameTAG: droidme FirstChildCreateTimeTAG: 2012-09-10T01:03:40Z SecondChildTAG: thanks SecondChildUserIdTAG: 400892 SecondChildUserNameTAG: ahmadyousaf SecondChildCreateTimeTAG: 2012-09-11T16:53:14Z IndexTAG: 4832 TitleTAG: Calling Hobbie Are you here Hobbie? If you are welcome back Kob UserIdTAG: 15344 UserNameTAG: kob CreateTimeTAG: 2012-09-05T14:55:13Z VoteTAG: -3 CoursewareTAG: undefined CommentableIdTAG: MITx_6_002x_2012_Fall NumberOfReplyTAG: 1 FirstChildTAG: Hey kob! Good to see you again :-) It's Hobie btw :P FirstChildUserIdTAG: 14386 FirstChildUserNameTAG: ashwith FirstChildCreateTimeTAG: 2012-09-05T14:56:42Z SecondChildTAG: Hey great to see a familiar name. Hope you doing well ashwith. This should be more fun this time as no pressure like our first run! SecondChildUserIdTAG: 15344 SecondChildUserNameTAG: kob SecondChildCreateTimeTAG: 2012-09-05T18:38:11Z IndexTAG: 4833 TitleTAG: Kirchhoff Why can I not choose my own sign for the voltage drops? Providing they all sum to zero, the calculation is therefore correct. UserIdTAG: 40228 UserNameTAG: MikeHa CreateTimeTAG: 2012-09-05T14:32:30Z VoteTAG: -3 CoursewareTAG: Overview / Using the tools CommentableIdTAG: 6002x_Lab_0_Using_the_Tools NumberOfReplyTAG: 3 FirstChildTAG: Hi, I am also confused about the first circuit when we did DC analysis. Why its showing -500mA. Why current is orginating from negative terminal and end at Positive terminal. It should be other way around. Please correct me if I am wrong!! FirstChildUserIdTAG: 181368 FirstChildUserNameTAG: Hanumanta FirstChildCreateTimeTAG: 2012-09-05T14:36:44Z SecondChildTAG: Simple answer is: check arrow near -500uA But it's very confusing, yep. SecondChildUserIdTAG: 128561 SecondChildUserNameTAG: gusevoy SecondChildCreateTimeTAG: 2012-09-05T14:48:44Z FirstChildTAG: Because by your definition its a voltage drop! Therefore an increase in voltage would indicate a negative voltage drop and a decrease a positive voltage drop. Its all about the reference. FirstChildUserIdTAG: 80920 FirstChildUserNameTAG: MichaelSturgess FirstChildCreateTimeTAG: 2012-09-05T14:49:39Z FirstChildTAG: V(VOLTAGE)=I(CURRENT)R(RESISTANCE). R1= k ohm, I total= -500uA / -500X10^-6 ,V1=? V1=I total*R1, = [-500*10^-6]*1000 =-0.5V. i hope this would be useful. FirstChildUserIdTAG: 357170 FirstChildUserNameTAG: Rishan FirstChildCreateTimeTAG: 2012-09-05T17:10:35Z SecondChildTAG: Hi, that´s what i thaught.. but i put that answer and it´s was wrong? somebody knows what i´m doing wrong? i don´t understand very good what means to choose the sign?? i´m not a native english, someone guide me what´s my mistake... thank you from the south SecondChildUserIdTAG: 149351 SecondChildUserNameTAG: pamelaf SecondChildCreateTimeTAG: 2012-09-12T02:09:53Z