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The magnetic potential energy stored in a certain inductor is 25 mJ, when the current in the inductor is 60 mA. This inductor is of inductance
- 0.138 H
- 138.88 H
- 13.89 H
- 1.389 H
Correct answer: 0.138 H
Solution
The magnetic potential energy stored in an inductor is given by the formula U = (1/2)LI². Rearranging for L, we get L = 2U/I². Substituting U = 25 × 10⁻³ J and I = 60 × 10⁻³ A, we find L = 2 × (25 × 10⁻³) / (60 × 10⁻³)² = 0.138 H.
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