First slide

Inductance

Question

The magnetic potential energy stored in a certain inductor is $25 m J$ , when the current in the inductor is $60 m A$ . This inductor is of inductance

Easy

A

0.138 H
B

138.88 H
C

1.389 H
D

13.89 H

Solution

Magnetic potential energy stored in an inductor is given by
$\begin{array}{l} U = \frac{1}{2} L i^{2} \Rightarrow 25 \times 10^{- 3} = \frac{1}{2} \times L \times {(60 \times 10^{- 3})}^{2} \\ L = \frac{25 \times 2 \times 10^{6} \times 10^{- 3}}{3600} = \frac{500}{36} = 13.89 H \end{array}$

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