Question
The back $e.m.f$. induced in a coil, when current changes from $1$ ampere to zero in one milli second, is $4$ volts, the self inductance of the coil is
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Answer
(d) $e = - L\frac{{di}}{{dt}}$ but $e =\,4\,V$ and $\frac{{di}}{{dt}} = \frac{{0 - 1}}{{{{10}^{ - 3}}}} = - 1/{10^{ - 3}}$
$\therefore \;\frac{{ - 1}}{{{{10}^{ - 3}}}}( - L) = 4 \Rightarrow L = 4 \times {10^{ - 3}}henry$
$\therefore \;\frac{{ - 1}}{{{{10}^{ - 3}}}}( - L) = 4 \Rightarrow L = 4 \times {10^{ - 3}}henry$
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