question_answer

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  Solved  Paper-2006

A) \[1/e\,A\]                          

B) \[e\text{ }A\]                   

C)                        \[0.1A\]               

D)        \[1\,A\]

Correct Answer: A

Solution :

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) After this, decay of current starts in the circuit according to the equation \[l={{l}_{0}}{{e}^{-t/\tau }}\] where\[\tau =L/R\] \[l=1\times {{e}^{-(1\times {{10}^{-3}})/(100\times {{10}^{-3}}/100)}}=(1/e)A\]