First slide

Faraday's law of Electromagnetic induction

Question

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

Easy

A

100 volt
B

0.4 volt
C

4.0 volt
D

440 volt

Solution

$e = - L \frac{di}{dt}$
Given that, L = 40 x 10^-3 H
di = 77A - 1A = 10A
and dt = 4x 10^-3 s
$∴ e = 40 \times 10^{- 3} \times (\frac{10}{4 \times 10^{- 3}}) = 100 V$

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