An ac generator supplies a peak (not rms) voltage of 150 V at 60.0 Hz. The generator is connected in series with a 35-mH inductor, a 45-µF capacitor and an 85-$\varOmega$ resistor.
-Determine the rms voltage of the generator.
A)220 V
B)110 V
C)82 V
D)75 V
E)53 V

Question

Quizplus · Accepted Answer

First, we need to find the impedance of the circuit:

XL = 2πfL = 2π(60)(0.035) = 13.23 Ω (inductive reactance)
XC = 1/(2πfC) = 1/(2π(60)(0.000045)) = 58.48 Ω (capacitive reactance)

The total impedance is:

Z = R + j(XL - XC) = 85 + j(13.23 - 58.48) = 85 - j45.25 Ω

The peak voltage across the circuit is:

Vp = 150 V

The rms voltage is:

Vrms = Vp/√2 = 150/√2 = 106.07 V

However, the voltage across the circuit is not the same as the voltage across the generator, since there is a voltage drop across the impedance. We need to use voltage division:

Vgen = Vrms(Zgen/Ztotal) = 106.07(85/(85 - j45.25)) = 84.9 V

Therefore, the rms voltage of the generator is 84.9 V, which is closest to answer choice B) 110 V.

An Ac Generator Supplies a Peak (Not Rms) Voltage of 150