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MHT CET 2025 (5 May, Shift 2) · Self-Inductance & Energy

Self-Inductance & Energy question from MHT CET 2025 (5 May, Shift 2)

Q30MHT CET 2025 · 5 MAY · SHIFT 2Medium

Self-Inductance & Energy

Two inductors of 80 mH each are joined in parallel. The current passing through the combination is 2.1 A. The energy stored in this combination of inductors is

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Answer: (c)

  1. Parallel: Lp=80×80160=40L_p = \dfrac{80 \times 80}{160} = 40 mH.
  2. U=12LpI2=12×0.04×(2.1)2=0.02×4.41U = \dfrac{1}{2}L_p I^2 = \dfrac{1}{2}\times 0.04 \times (2.1)^2 = 0.02 \times 4.41
  3. U=8.82×10−2U = 8.82\times 10^{-2} J
  4. So the answer is (c).
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