Question

The energy stored in a coil of inductance 50 mH carrying a current of 2 A is:

• A. 1 J
• B. 0.1 J
• C. 0.05 J
• D. 0.5 J

Hint:

Use the formula $E = \frac{1}{2} L I^2$ to calculate the energy stored in an inductor. Ensure inductance is in henries and current in amperes.

Answer 1

The Correct Option is B

Energy stored in an inductor: $E = \frac{1}{2} L I^2$.
Given: $L = 50$ mH = $0.05$ H, $I = 2$ A.
$E = \frac{1}{2} \times 0.05 \times (2)^2 = \frac{1}{2} \times 0.05 \times 4 = 0.1$ J.