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Question 20238
JEE Physics · Hard

Electromagnetic Induction and Alternating Current: Self-Inductance from Energy MCQ

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Self-Inductance from EnergyHardQuestion 20238

Question

If the energy stored in a 5 H inductor increases from 0 to 1000 J, the final current is:
  1. A
    10 A
  2. B
    20 A
    Correct
  3. C
    √400 A
  4. D
    √200 A

Correct answer

20 A

Explanation

U = ½LI² → I = √(2U/L) = √(2000/5) = √400 = 20 A.

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