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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a
10 mJ
b
37.5 mJ
c
75 mJ
d
Zero
answer is B.
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Detailed Solution
At t = 1 sec, i = 2+(31) = 5A and |e| =Ldidt⇒L=3×10−3So energy U=12Li2=12(3×10−3)×(5)2 = 37.5 mJ.