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Q.

When the current change from + 2A to – 2A in 0.05 second, an e.m.f. of 8 V is induced in a coil. The coefficient of self-induction of the coil is

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a

0.1 H

b

0.2 H

c

0.4 H

d

0.8 H

answer is A.

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Detailed Solution

e=Ldidt⇒8=L×2−−20.05⇒L=0.1H
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