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A current of 1 A through a coil of inductance of 200 mH is increasing at a rate of \mathrm{0.5 \mathrm{~A} \mathrm{~s}^{-1}} . The energy stored in the inductor per second is:

Option: 1

\mathrm{0.5 \mathrm{~J}_{\mathrm{s}}-1}


Option: 2

\mathrm{5.0 \mathrm{~J}_{\mathrm{s}}-1}


Option: 3

\mathrm{0.1 \mathrm{~J}_{\mathrm{s}}-1}


Option: 4

\mathrm{2.0 \mathrm{~J}_{\mathrm{s}}{ }^{-1}}


Answers (1)

best_answer

The energy stored in the inductor is \mathrm{U}=\frac{1}{2} \mathrm{LI}^2

The energy stored in the inductor per second is \frac{\mathrm{dU}}{\mathrm{dt}}=\mathrm{LI} \frac{\mathrm{dI}}{\mathrm{dt}}

\mathrm{=\left(200 \times 10^{-3} \mathrm{H}\right)(1 \mathrm{~A})\left(0.5 \mathrm{As}^{-1}\right)=0.1 \mathrm{Js}^{-1}}

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Pankaj

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