Question:
The energy stored in a coil of inductance 50 mH carrying a current of 2 A is:
The energy stored in a coil of inductance 50 mH carrying a current of 2 A is:
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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.
Updated On: Jun 3, 2025
- 1 J
- 0.1 J
- 0.05 J
- 0.5 J
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The Correct Option is B
Solution and Explanation
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.
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.
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