The current in an inductor is given by i = (2 + 3t) A where t is in second. The self induced emf in it is 9 mV at t=1 second, then the energy stored in the inductor at that instant is

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10 mJ

37.5 mJ

75 mJ

Zero

answer is B.

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Detailed Solution

At t = 1 sec, i = 2+(31) = 5A and |e| =Ldidt⇒L=3×10−3So energy U=12Li2=12(3×10−3)×(5)2 = 37.5 mJ.

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