A 60.0-Hz ac generator with a peak voltage of 110 V drives a series RL circuit with R = 10.0 $\Omega$ and L = 10.0 mH.The impedance is:
A)3.77 $\Omega$ .
B)9.26 $\Omega$ .
C)10.0 $\Omega$ .
D)10.7 $\Omega$ .
E)13.8 $\Omega$ .

Question

A 60.0-Hz ac generator with a peak voltage of 110 V drives a series RL circuit with R = 10.0  $\Omega$  and L = 10.0 mH.The impedance is:
A)3.77  $\Omega$  .
B)9.26  $\Omega$  .
C)10.0  $\Omega$  .
D)10.7  $\Omega$  .
E)13.8  $\Omega$  .

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 in radians per second, and is given by:

$\omega = 2\pi f$

where $f$ is the frequency in Hertz.

In this problem, the frequency is given as 60.0 Hz, so:

$\omega = 2\pi (60.0	ext{ Hz}) \approx 377	ext{ rad/s}$

Substituting the given values of R, L, and $\omega$ into the impedance equation, we get:

$Z = \sqrt{(10.0	ext{ }\Omega)^2 + (377	ext{ }	ext{rad/s})(10.0	ext{ mH})^2} \approx 10.7	ext{ }\Omega$

Therefore, the answer is D.

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