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(A) The energy stored in an inductor is given by $U = \frac{1}{2}Li^2$.The rate at which energy is stored is $P = \frac{dU}{dt} = \frac{d}{dt}(\frac{1

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(A) The energy stored in an inductor is given by $U = \frac{1}{2}Li^2$.The rate at which energy is stored is $P = \frac{dU}{dt} = \frac{d}{dt}(\frac{1}{2}Li^2) = Li(\frac{di}{dt})$.In an $L-R$ circuit,the current $i$ grows as $i = i_0(1 - e^{-Rt/L})$,where $i_0 = \frac{E}{R}$.The rate of change of current is $\frac{di}{dt} = \frac{E}{L}e^{-Rt/L}$.Substituting these,the power $P = L \cdot [i_0(1 - e^{-Rt/L})] \cdot [\frac{E}{L}e^{-Rt/L}] = Ei_0(e^{-Rt/L} - e^{-2Rt/L})$.At $t = 0$,$i = 0$,so $P = 0$.As $t 	o \infty$,$\frac{di}{dt} 	o 0$,so $P 	o 0$.Since the rate is zero at both $t = 0$ and $t = \infty$ and positive in between,the graph must show a peak. Thus,the curve in image $144-$a15 is correct.

In an $L-R$ circuit connected to a battery,the rate at which energy is stored in the inductor is plotted against time during the growth of the current in the circuit. Which of the following best represents the resulting curve?

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