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 peak current in the circuit is
A)0.963 A.
B)10.3 A.
C)11.0 A.
D)11.9 A.
E)29.2 A.

Question

Quizplus · Accepted Answer

The impedance of the circuit can be found using $Z=\sqrt{R^2+(\omega L)^2}$, where $\omega=2\pi f$ is the angular frequency. Thus, $Z=\sqrt{10^2+(2\pi	imes60	imes0.01)^2}\approx10.7 \Omega$.

The peak current can be found using $I_{max}=\frac{V_{max}}{Z}$, where $V_{max}$ is the peak voltage. Thus, $I_{max}=\frac{110}{10.7}\approx10.3	ext{A}$.

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