1.

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

Answer»

100 VOLT
0.4 volt
4.0 volt
440 volt

Solution :MAGNITUDE of induced e.m.f, `|epsilon|=L(di)/(dt)`
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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