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Q.

An inductor of 5H inductance carries a steady current of 2A. How can a 50V self induced emf be made to appear in the inductor? 

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a

t = 0.5 sec

b

t = 0.4 sec

c

t = 0.6 sec

d

t = 0.2 sec

answer is A.

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Detailed Solution

e=Ldidt (or)

dt=Ldie=5×250=15=0.2sec

So, the desired emf can be produced by reducing the given current to zero in 0.2second.

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