First slide

Inductance

Question

An inductor (L = 100 mH) a resistor (R = 100 $Ω$ )and a battery (E = 100 V) are initially connected in series as shown in the Fig. 9.89. After a long time the battery is disconnected after short circuiting the points A and B. The current in the circuit 1 ms after the short circuit is :

Easy

A

1/e A
B

e A
C

0.1 A
D

1 A

Solution

Long time after, current in the circuit
$I_{0} = \frac{E}{R} = 1 A$
On short-circuiting
$I = I_{0} e^{- \frac{R}{L} t} = (1) e^{- \frac{100}{100 x 10^{- 3}} x 10^{- 3}} = (\frac{1}{e}) A$

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