A 45-mH inductor is connected to an ac source of emf with a frequency of 400 Hz and a maximum emf of 20 V. The maximum current is:
A) 0
B) 0.18 A
C) 1.1 A
D) 360 A
E) 2300 A

Question

Quizplus · Accepted Answer

The maximum current in an inductor connected to an AC source can be found using the formula $I_{max} = \frac{E_{max}}{X_L}$, where $E_{max}$ is the maximum emf, and $X_L$ is the inductive reactance given by $X_L = 2\pi fL$, with $f$ being the frequency and $L$ the inductance. Here, $X_L = 2\pi(400)(0.045)$ Ohms, and thus $I_{max} = \frac{20}{X_L}$ calculates to approximately 0.18 A.

A 45-MH Inductor Is Connected to an Ac Source of Emf