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 A
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 calculated 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 $X_L = 2\pi f L$, with $f$ being the frequency and $L$ the inductance. Here, $X_L = 2\pi(400)(0.045)$ Ohms, and thus $I_{max} = \frac{20}{X_L}$, which calculates to approximately 0.18 A.

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