Physics again. Most detailed answer gets ten points!?

Question

Resistor, R=20 Ω and a capacitor C=200 µF are in a series circuit of alternating current with a voltage of U=120 V and a frequency of ƒ=60 Hz. What is the current that flows through the circuit? What are the voltage drops across the resistor and the capacitor?

Answer 1

So you have real impedance from the resistor:
Z = R

You have imaginary impedance from the capacitor:
Z = -i / (omega)C, where omega = 2 pi f

Use the pythagorean theorem to get the magnitude of the total impedance:

magnitude Z = sqrt (R^2 + 1/ (2 pi f C)^2)

Then magnitude of current = U / Zmag

The magnitude of the voltage drop across the resistor will just be I(mag) R = UR / Zmag.

The magnitude of the voltage drop across the capacitor will be U / (2 pi f C Zmag)

The magnitudes of the elements don't add up to the total magnitude of voltage, U, because they are out of phase.

Answer 2

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Answer 3

Xc = 1/ 2pifC = 1/200*10^-6*6.28*2*60 = 6.63
120/(20+6.63) = 4.5 amperes current flow
4.5*20 = 90 volts drop across R
4.5*6.63 = 30 volts across the capacitor

Answer 4

if you post your e-mail address i'll write it all out for you scan it and e-mail it to you. it'd be too annoying to type it all out. there are several straight forward equations that will solve the problem.