# Homework Help: Obtain Expression For Voltage Across Capacitor

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1. Dec 18, 2014

### Captain1024

1. The problem statement, all variables and given/known data

A voltage source given by $v_s(t)=25cos(2\pi\times10^3t-30^\circ)$ is connected to a series RC load. If R = 1MOhm and C = 200pF, obtain an expression for $v_c(t)$, the voltage across the capacitor.

Answer known to be: $v_c(t)=15.57cos(2\pi\times10^3t-81.5^\circ)$

2. Relevant equations

V = IR

3. The attempt at a solution

By KVL: $25\angle-30^\circ=RI+\frac{1}{j\omega C}I$

$\Rightarrow\ 25\angle-30^\circ=I(R-\frac{j}{\omega C})$

$\Rightarrow\ I=\frac{25\angle-30^\circ}{R-\frac{j}{\omega C}}$

Plugging in values: $I=1.95\times10^{-5}\angle-68.52^\circ$

Now, my thinking was that Ohm's law works for phasor impedances: $V_c=IZ$, where $Z=\frac{1}{j\omega C}$. However, that yielded an answer with the same magnitude as the current I found above. What am I missing?

Thanks
-Captain1024

2. Dec 18, 2014

### Staff: Mentor

Your current magnitude looks about right, but the phase angle looks off to me. Check your math on that.

Ohm's law does indeed work for phasor values.

3. Dec 18, 2014

### Captain1024

I redid the math. The angle of the current should be positive 8.52 deg. Then, earlier I made the mistake of trusting my calculator to convert rec to polar. I did it by hand with the correct current and got the right answer. Thanks for the guidance.

-Captain1024