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The current in self inductance L = 40 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» 100 VOLT GIVEN that, `L=40xx10^(-3)H`, `di=11 A-1A=10 A` and `dt=4XX10^(-3)s` `therefore |epsilon|=40xx10^(-3)XX((10)/(4xx10^(_3)))=100 V` |
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