Homework Help: Capacitors and Circuits

1. Aug 26, 2010

jofree87

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?

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2. Aug 26, 2010

Mindscrape

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.

3. Aug 26, 2010

jofree87

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)?

4. Aug 26, 2010

Mindscrape

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?

5. Aug 26, 2010

jofree87

Here is my attempt at nodal voltage analysis attached below. Can somebody look it over? My answer doesn't seem right.

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6. Aug 26, 2010

Mindscrape

Your node voltage looks great, but then you start doing some funky stuff 3/4 way down the page.

$$V=I=C\frac{dV}{dt}$$

so use separation of variables and integrate both sides to get

$$\frac{1}{C}\int_{t'=0}^{t'=t} dt' = \int_{V(0)}^{V(t)} \frac{dV}{V}$$

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.

7. Aug 26, 2010

jofree87

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

8. Aug 27, 2010

Mindscrape

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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