First slide

Inductance

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

Easy

A

1 H
B

4 H
C

$10^{- 3} H$
D

$4 \times 10^{- 3} H$

Solution

$e = - L \frac{di}{dt}; But e = 4 V and \frac{di}{dt} = \frac{0 - 1}{10^{- 3}} = - 1 / 10^{- 3}$
$∴ \frac{- 1}{10^{- 3}} (- L) = 4 = L = 4 \times 10^{- 3} H$

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