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 inductive reactance (X_L) is calculated using the formula X_L = 2πfL, where f is the frequency and L is the inductance. Substituting the given values, X_L = 2π(400)(0.045) = 113.1 ohms. The maximum current (I_max) is then given by I_max = V_max / X_L = 20 V / 113.1 ohms ≈ 0.18 A.

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