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Question

An inductor (L=100mH), a resistor (R=100Ω) and a battery (E=100V) 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 1ms after the short circuit is:

22449.png

A
1/e A
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B
e A
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C
0.1A
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D
1
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Solution

The correct option is B 1/e A
energy stored in Inductor =12L(ER)2
Io=ER
i=IoetRL
=100100×e103×102100×103
i=e1A
58405_22449_ans.jpg

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