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Homework Help: Capacitors and Circuits

  1. Aug 26, 2010 #1
    Diagram of problem attached below.

    I don't know where to begin with this problem. How can I find i(t) if I am not given v(t)?

    Since i = C dv/dt , dont I need the function v(t) to solve this problem?

    Is it possible to use the function v(t) = v(0) e-t/(RC) , if so, then what would the value of R be?

    Attached Files:

  2. jcsd
  3. Aug 26, 2010 #2
    Be a little more clear. I don't know what V you are talking about. From the voltage source you know that v(t)=3*i(t), and the voltage of the capacitor is something you need to find.
  4. Aug 26, 2010 #3
    By v(t), I meant the voltage cross the capacitor, I'll just call it vc(t). And I need to find ic(t) which is the current through the capacitor.

    I don't think the dependent source and vc(t) are the same since they're not parallel, right?

    So I need to find vc(t) to get ic(t). How do I find vc(t)?
  5. Aug 26, 2010 #4
    You're right, the dependent voltage source and the v_c are not the same voltages.

    Do you know how to do node voltage analysis, or do you know KCL?
  6. Aug 26, 2010 #5
    Here is my attempt at nodal voltage analysis attached below. Can somebody look it over? My answer doesn't seem right.

    Attached Files:

  7. Aug 26, 2010 #6
    Your node voltage looks great, but then you start doing some funky stuff 3/4 way down the page.


    so use separation of variables and integrate both sides to get

    [tex]\frac{1}{C}\int_{t'=0}^{t'=t} dt' = \int_{V(0)}^{V(t)} \frac{dV}{V}[/tex]

    Then you know Vc, and use I=C dVc/dt once again to get I.

    Or.. you can find I with chain rule. I=C dVc/dt = C dV/dI dI/dt. Then if Vc = I, then dVc/dI = 1, so you really get the same thing, I=C dI/dt.

    You were using both I and V in your integrals, and it doesn't quite work that way.
  8. Aug 26, 2010 #7
    I think I am starting to get it now, but where did you get V = I in the first integral?

    Thanks for helping me out btw
  9. Aug 27, 2010 #8
    Oh, I got the Vc=Ic from the node voltage you did (I also did up to this point to check and it was right). When everything is simplified, we get Vc=Ic. Sorry, I intermittently dropped subscripts, but all the Vs and the Is were the Vs and Is of the capacitor.
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