An inductor with inductance L = 10 mH is connected to a power supply which provides a current I = 5 mA sin $\omega$t.The maximum back EMF is
A)5 mV
B)10 mV
C)25  V
D)50  V
E)0 V

Question

Quizplus · Accepted Answer

The back EMF induced in the inductor is given by
$$
E = -L\frac{di}{dt}
$$
Taking the derivative of the given current gives
$$
\frac{di}{dt} = 5 mA \cos(\omega_0 t) \cdot \omega_0
$$
Substituting this into the first equation gives
$$
E = -10 mH \cdot 5 mA \cos(\omega_0 t) \cdot \omega_0 = -50 \mu V \cos(\omega_0 t)
$$
The maximum value of $\cos(\omega_0 t)$ is 1, so the maximum back EMF is $|-50 \mu V| = 50 \mu V = 50 \cdot 10^{-6} V = 50 \mu F10V$. Therefore, the best choice is D.

An Inductor with Inductance L = 10 MH Is Connected ω\omegaω

An Inductor with Inductance L = 10 MH Is Connected $\omega$