Question

A 32.0-μF capacitor is connected to a 53.0-Ω resistor and a generator whose rms output is 30.0 V at 60.0 Hz.

(a) Find the rms current in the circuit.
______ A

(b) Find the rms voltage drop across the resistor.
______ V

(c) Find the rms voltage drop across the capacitor.
_______ V

(d) Find the phase angle for the circuit.
The voltage _______ leads ahead of lags behind the current by _____°.

Answer 1

One important point has been overlooked here. You neglected to say HOW they are connected, in series or parallel.
I'll assume they are in series. You will have to adjust things if they are in parallel.

a) Recall the following formula...
Capacitive reactance, XC = 1 / (2π f C)

Reactance is just ac resistance. So we can replace the capacitor with an 'ac resistor' that has a certain value at 60 Hz. Then we just have two resistors in series. Don't forget that Ohm's law works for ac circuits as well as for dc ones.

XC = 1 /( 2π * 60 * 32 * 10^-6 ) = 82.9 Ω

Total circuit impedance, Z, = √(XC^2 + R^2) = √(82.9^2 + 53^2) = 98.38 Ω

Current = Voltage / Z = 30 / 98.38 = 0.305 A

b) Voltage drop across R = I*R = 0.305 * 53 = 16.18 V

c) Voltage drop across capacitor = I * XC = 0.305 * 82.9 = 25.28 V

d) Recall that the phase angle, Ѳ, is given by

cos Ѳ = R / Z

Ѳ = arc cos (R/Z) = arc cos (53 / 82.9) = 50.3 °

phase angle is 50.3° .