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 the RL circuit is given by:
Z = √(R^2 + (ωL)^2) where ω = 2πf
Substituting the given values, we get:
Z = √(10^2 + (2π(60)(0.01))^2)
Z ≈ 10.7 Ω

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