Obtain Expression For Voltage Across Capacitor

[Captain1024]

Homework Statement

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

Homework Equations

V = IR

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

 

[gneill]

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.

 

[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