# Current in AC circuit

The problem is to find the current on the capacitor. $I''$ should be correct, but I don't know how to construct the formula for $I'$. I managed to get the value for $I_c$ using Thevenin and Norton equivalents, and they're the same, so that should be correct.

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What is the thingy with just a horizontal line in the top schematic? You really need to properly label things. If it's meant to be a voltage source, then you can ignore R and L since they have no bearing on the voltage accross the voltage source and therefore the voltage accross the capacitor. –  Olin Lathrop May 23 '13 at 19:11
Would electronics.stackexchange.com be a better home for this question? –  Qmechanic May 23 '13 at 19:13
That means an open circuit with which you replace i(t). –  Garfailed May 23 '13 at 19:16
@Qmech: EE.SE would be good, but not without a proper definition of whatever the second thingy from the left is. Without it being fixed, I'd jump on it immediately as not a real question on the grounds it is difficult to tell what is being asked here, and I think it wouldn't take long for 4 others to do the same. –  Olin Lathrop May 23 '13 at 20:58

Studying the problem some, I think what your showing is superposition of a voltage source and a current source. In the first circuit, you've zeroed the current source (making it an open circuit), and in the second circuit, you've zeroed the voltage source (making it a short circuit).

In both cases, you appear to be trying to solve for the current through the capacitor and then summing the two results to find the total.

IF the sources are sinusoidal with angular frequency $\omega$, then this is a very easy problem to solve.

For the first circuit, the capacitor and resistor are in parallel. Find the source current and then use current division to find the capacitor current:

$I'_c = \dfrac{U_s}{(R||-jX_C) + jX_L} \dfrac{R}{R -jX_C}$

For the second circuit, you've used current division correctly.

However, it's not clear that AC analysis is appropriate since the sources are shown as functions of time. Since AC analysis is done in the phasor domain, the sources are not shown as time dependent but, rather, as a complex constant.

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I admit, the picture was wrong, I used the original circuit picture everywhere, even when using constants. I fixed those too. Your solution is somewhat correct, I don't really get why did you use negative signs here and there, but I got the correct value for I' with this: i.imgur.com/8DJaxJY.jpg –  Garfailed May 23 '13 at 20:37
@Garfailed, the negative signs come from this: $\frac{1}{j} = -j$. Also, $X_C = \frac{1}{\omega C}$. Thus: $\frac{1}{j \omega C} = -jX_C$. –  Alfred Centauri May 23 '13 at 20:45
Oh, I see. It makes sense now. –  Garfailed May 23 '13 at 21:03