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 impedance is
A) 3.77 Ω.
B) 9.26 Ω.
C) 10.0 Ω.
D) 10.7 Ω.
E) 13.8 Ω.

Question

Quizplus · Accepted Answer

The impedance of a series RL circuit is given by:
$$Z=\sqrt{R^2+(\omega L)^2}$$
where $\omega$ is the angular frequency given by $\omega = 2\pi f$. Plugging in the values:
$$\omega = 2\pi(60.0 	ext{ Hz}) = 377 	ext{ rad/s}$$
$$Z = \sqrt{(10.0 	ext{ }\Omega)^2 + (377 	ext{ }	ext{rad/s} \cdot 0.01 	ext{ H})^2} = 10.7 	ext{ }\Omega$$
Therefore, the correct answer is D.

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