37. 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 (1990)
Solution:
Induced e.m.f. is given by $e = L \frac{\Delta I}{\Delta t}$.
Given $L = 40 mH = 40 \times 10^{-3} H$, $\Delta I = 11 - 1 = 10 A$, and $\Delta t = 4 ms = 4 \times 10^{-3} s$.
$e = 40 \times 10^{-3} \times \frac{10}{4 \times 10^{-3}} = 100 volt$.
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