| An inductor\[(L=100mH)\], a resistor\[(R=100\,\Omega )\] and a battery\[(E=100\,V)\]are initially connected in series as shown in the figure. 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 [AIEEE 2006] |
 |
A) \[1/e\,A\]
B) \[e\text{ }A\]
C) \[0.1A\]
D) \[1\,A\]
Correct Answer: A
Solution :
| [a] This is a combined example of growth and decay of current in an LR circuit. |
 |
| The current through circuit just before shorting the battery, |
| \[{{l}_{0}}=E/R=1A\] |
| (as inductor would be shorted in steady state) |
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| After this, decay of current starts in the circuit according to the equation |
| \[l={{l}_{0}}{{e}^{-t/\tau }}\] |
| where\[\tau =L/R\] |
| \[l=1\times {{e}^{-(1\times {{10}^{-3}})/(100\times {{10}^{-3}}/100)}}=(1/e)A\] |
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