A 60.0-Hz AC generator with a peak voltage of 110 V drives a series RL circuit with R = 10.0 Ω and L = 10.0 mH. The power factor, cos φ, is
A)−1.00.
B)−0.936.
C)+0.943.
D)+0.936.
E)+1.00.

Question

Quizplus · Accepted Answer

Using the given values, we can find the impedance of the circuit:

$$Z=\sqrt{R^2+(\omega L)^2}=\sqrt{10^2+{(2\pi 	imes 60 	imes 10^{-3})}^2}=10.61\ \Omega$$

The current in the circuit is:

$$I=\frac{V_{peak}}{Z}=\frac{110}{10.61}=10.36\ 	ext{A}$$

The power factor is:

$$\cos\phi=\frac{R}{Z}=\frac{10}{10.61}=0.943$$

Therefore, the best choice is D.

A 600-Hz AC Generator with a Peak Voltage of 110 V