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JEE Main 2022 (26 Jun, Shift 1) · Self-Inductance & Energy

Self-Inductance & Energy question from JEE Main 2022 (26 Jun, Shift 1)

Q63JEE Main 2022 · 26 JUN · SHIFT 1Easy

Self-Inductance & Energy

A 10 Ω\Omega, 20 mH coil carrying constant current is connected to a battery of 20 V through a switch. Now after switch is opened current becomes zero in 100 μ\mus. The average e.m.f. induced in the coil is ____________ V.

Numerical answer type. Work it out, then open the answer.

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Answer: 400

  1. Steady current I=VR=2010=2 AI = \dfrac{V}{R} = \dfrac{20}{10} = 2\text{ A}.
  2. Average emf e=L ΔIΔt=20×10−3×2100×10−6e = L\,\dfrac{\Delta I}{\Delta t} = 20\times10^{-3}\times\dfrac{2}{100\times10^{-6}}
  3. e=20×10−3×2×104=400 Ve = 20\times10^{-3}\times2\times10^{4} = 400\text{ V}
  4. So the average induced emf is 400 V.
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