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NEET 1990 (AIPMT) · Self-Inductance & Energy

Self-Inductance & Energy question from NEET 1990 (AIPMT)

Q89NEET 1990 · AIPMTEasy

Self-Inductance & Energy

The current in self inductance L=40 mHL = 40\text{ mH} is to be increased uniformly from 1 amp to 11 amp in 4 milliseconds. The e.m.f. induced in inductor during process is

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Answer: (a)

  1. Induced emf e=L ΔIΔte = L\,\dfrac{\Delta I}{\Delta t}.
  2. e=40×10−3×11−14×10−3=40×10−3×2500e = 40\times10^{-3}\times\dfrac{11-1}{4\times10^{-3}} = 40\times10^{-3}\times 2500
  3. e=100 Ve = 100\text{ V}
  4. So the answer is (a) 100 volt.
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