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A current of 1 A through a coil of inductance of 200 mH is increasing at a rate of `0.5As^(-1)`. The energy stored in the inductor per second isA. `0.5Js^(-1)`B. `5.0Js^(-1)`C. `0.1Js^(-1)`D. `2.0Js^(-1)` |
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Answer» Correct Answer - C The energy stored in an inductor is `U=(1)/(2)LI^(2)` The energy stored in the indutor per second is `(dU)/(dt)=LI(dI)/(dt)=200xx10^(-3)Hxx1Axx0.5 As^(-1)=0.1 J s^(-1)` |
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