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MHT CET 2026 (17 Apr, Shift 1) · Self-Inductance & Energy

Self-Inductance & Energy question from MHT CET 2026 (17 Apr, Shift 1)

Q19MHT CET 2026 · 17 APR · SHIFT 1Medium

Self-Inductance & Energy

Two coils have self inductance L1L_1 and L2L_2. The current through them is increasing at constant rate. If the power dissipated in both the coils is same, then the ratio of energy stored in the coil having inductance L1L_1 to that in L2L_2 is

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Answer: (d)

  1. P=LIdIdtP = L I \dfrac{dI}{dt}; equal PP and equal rate give L1I1=L2I2L_1 I_1 = L_2 I_2, so I1I2=L2L1\dfrac{I_1}{I_2} = \dfrac{L_2}{L_1}.
  2. U1U2=12L1I1212L2I22=I1I2=L2L1\dfrac{U_1}{U_2} = \dfrac{\tfrac12 L_1 I_1^2}{\tfrac12 L_2 I_2^2} = \dfrac{I_1}{I_2} = \dfrac{L_2}{L_1}
  3. So the answer is (d) L2L1\dfrac{L_2}{L_1}.
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