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Q.

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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a

100 volt

b

440 volt

c

0.4 volt

d

4.0 volt

answer is A.

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Detailed Solution

e=-Ldidt
Given that, L = 40 x 10-3 H
di = 77A - 1A = 10A
and dt = 4x 10-3 s
∴e=40×10−3×104×10−3=100V

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