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Alternating Current question

2018 · Q124
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Alternating Current question

2018 · Q124

NEETPhysicsAlternating CurrentMCQ+4 / −1
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
  1. A
    0.138 H
  2. B
    138.88 H
  3. C
    1.389 H
  4. D
    13.89 H
View written solutionFree

Correct answer: D

From question energy stored in inductor, U = 25 × 10–3 J

Current, I = 60 mA

Also, energy stored in inductor, U = 12LI2{1 \over 2}L{I^2}21​LI2

25 × 10–3 = 12×L×(60×10−3)2{1 \over 2} \times L \times {\left( {60 \times {{10}^{ - 3}}} \right)^2}21​×L×(60×10−3)2

⇒\Rightarrow⇒ L = 25×2×106×10−33600{{25 \times 2 \times {{10}^6} \times {{10}^{ - 3}}} \over {3600}}360025×2×106×10−3​ = 13.89 H

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