Question

The magnetic potential energy stored in certain inductor is $64 \times 10^{-3}$ J, when the current in the inductor is 80 mA. This inductor is of inductance

• A. 0.20 H
• B. 2.0 H
• C. 20 H
• D. 200 H

Answer 1

Answer: (C)

20 H

Answer 2

The energy stored in an inductor is given by the formula:

$$E = \frac{1}{2} L I^2$$

Rearrange to solve for $L$:

$$L = \frac{2E}{I^2}$$

Substitute the given values $E = 64 \times 10^{-3} \, \text{J}$ and $I = 80 \, \text{mA} = 0.08 \, \text{A}$:

$$L = \frac{2 \times 64 \times 10^{-3}}{(0.08)^2} = \frac{0.128}{0.0064} = 20 \, \text{H}$$