A coil with a self-inductance of 0.75 mH experiences a constant current buildup from zero to 15 A in 0.25 s.What is the induced emf during this interval?
A) 0.045 V
B) 0.47 V
C) 0.030 V
D) 0.019 V

Question

Quizplus · Accepted Answer

The induced emf (ε) in a coil is given by the formula ε = -L * (ΔI/Δt), where L is the self-inductance, ΔI is the change in current, and Δt is the time interval. Substituting the given values: ε = -0.75 mH * (15 A / 0.25 s) = -0.75 * 10^-3 H * 60 A/s = -0.045 V. The negative sign indicates the direction of the induced emf, but the magnitude is 0.045 V.

A Coil with a Self-Inductance of 0