Question

The back e.m.f. induced in a coil, when current changes from 1 ampere to zero in one millisecond, is 4 volts, the self inductance of the coil is

• A. 1 H
• B. 4 H
• C. ${10}^{-3}$ H
• D. $4\times {10}^{-3}$ H

Answer 1

The correct option is D $4\times {10}^{-3}$ H

Solution:(d) $e=-L\frac{di}{dt}$ but $e=4V$and $\frac{di}{dt}=\frac{0-1}{{10}^{-3}}=\frac{-1}{{10}^{-3}}$
$\therefore \frac{-1}{{10}^{-3}}(-L)=4\Rightarrow$ L = 4 $\times {10}^{-3}$ henry