A current of 2A is increasing at a rate of 4 A/s through a coil of inductance 2H. The energy stored in the inductor per unit time is

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2 J/s

1 J/s

16 J/s

4 J/s

answer is C.

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Detailed Solution

Potential difference across coil is V=Ldidt or V = (2) (4) = 8 VNow energy stored per unit time= power = Vi= (8) (2)= 16 J/s

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