# Q: Finding the current through the capacitor

Having a hard time trying to find the current through the capacitor I don't know what method to use. simulate this circuit – Schematic created using CircuitLab

This is how far I have gotten:

Using DC analysis:

1V Positive Cycle input: The op amp is turned off during the positive cycle thus: simulate this circuit

Nodal analysis:

$$\ \frac{V_{out}-V_{in}}{4k\Omega}+c\dot V_{out} = 0\$$

$$\ \frac{V_{out}}{4k\Omega}-\frac{V_{in}}{4k\Omega}+c\dot V_{out} = 0\$$

$$\ \frac{V_{out}}{4k\Omega}+c\dot V_{out} = \frac{V_{in}}{4k\Omega}\$$

Laplace Transform:

$$\ \frac{V_{out}(s)}{4k\Omega}+sc V_{out} = \frac{V_{in}(s)}{4k\Omega}\$$

$$\ V_{out}(s)[\frac{1}{4k\Omega}+s470uF]=\frac{V_{in}(s)}{4k\Omega}\$$

$$\ \frac{V_{out}(s)}{V_{in}(s)} = \frac{1}{1.88s+1}\$$

as S->0

Becomes DC gain of 1 $$\ \frac{V_{out}(s)}{V_{in}(s)} = \frac{1}{1.88(0)+1}\$$

$$\ \frac{V_{out}(s)}{V_{in}(s)} =1\$$

Thus Vout = 1V

EDIT: Just to add on to the discussion Positive cycle becomes a RC circuit, to be exact a

$$\ H(s) = \frac{1}{1.88s+1}\$$

You can even predict the output via transfer function

$$\ H(jw) = \frac{1}{\sqrt{(1.88*2*50*\pi)^2+1^2}}\$$

$$\ H(jw) = 0.008465676V\$$

$$\ H(jw) = 8.47mV\$$ which is exact answer.

However looking at the whole thing positive cycle it doesnt match? Should it be the same thing? As we model? During the positive cycle it should have a behavior of a RC circuit?  • What have you figured out so far? Do you know what the voltage across the capacitor looks like? – Elliot Alderson Oct 22 '18 at 18:52
• Yeah, Let me put that up there – Pllsz Oct 22 '18 at 18:52
• Updated it, I just dont understand cause in DC a cap is open so not sure its even possible. I get it cause I used a DC power supply however when I use an AC supply I get the same Vout but cant find how they got the current. – Pllsz Oct 22 '18 at 18:58
• Maybe what i am asking is how to do analysis this circuit using AC? – Pllsz Oct 22 '18 at 19:01
• For AC or DC? In general, the capacitor current is proportional the rate of voltage change across it $I = C \frac{dV}{dt}$. electronics.stackexchange.com/questions/287394/… – G36 Oct 22 '18 at 19:35