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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100 volt

0.4 volt

4.0 volt

440 volt

answer is A.

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Detailed Solution

e=-LdidtGiven that, L = 40 x 10-3 Hdi = 77A - 1A = 10Aand dt = 4x 10-3 s∴e=40×10−3×104×10−3=100V

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