A 80.0 Hz generator with an rms voltage of 150 V is connected in series to a 3.20 k ohm resistor and a 3.50 micro farad capacitor. Find the phase angle between the current and the voltage.

Answer 1

f=80 Hz, V=150V, R=3200 Ω, C=3.5•10⁻⁶ F

ω=2πf =2π•80=160π(rad/s)
ω²=1/T² =1/(2π)²LC =>
L=1/(2π ω)²C= ....
X(L) =ωL= ...
X(C) =1/ωC= ...
tan φ=(X(L) –X(C))/R = ...

Answer 2

Xc = 1/2pi*F*C = 1/(6.28*80*3.5*10^-6) =
569 Ohms

Z=R-jXc = 3200 - j569=3250 Ohms[-10.1o]
The negative impedance angle means the
circuit is capacitive and the current leads the applied voltage by 10.1o.

Tan A = Xc/R = -569/3200 = -0.17772
A = -10.1o