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Electromagnetic Induction — Easy JEE physics MCQ

**[JEE Mains 2026]** In a series R–L circuit, the voltage of the battery is 12 V. Resistance is 4 Ω and inductance is 2 H. Find the energy stored in the inductor when the current reaches half of its maximum value.
  1. A. 2.25 J
  2. B. 4.5 J
  3. C. 9 J
  4. D. 1.125 J

Solution

Maximum current I₀ = V/R = 12/4 = 3 A. When I = I₀/2 = 1.5 A. Energy = (1/2)LI² = (1/2) × 2 × (1.5)² = 2.25 J.

PHYSICS

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[JEE Mains 2026] In a series R–L circuit, the voltage of the battery is 12 V. Resistance is 4 Ω and inductance is 2 H. Find the energy stored in the inductor when the current reaches half of its maximum value.
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