The switch S of the circuit shown in the figu | vedclass

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(C) When the switch $S$ is closed at $t = 0$,the current $i$ in the $RL$ circuit starts from $0$ and grows according to the equation $i(t) = \frac{E}{

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(C) When the switch $S$ is closed at $t = 0$,the current $i$ in the $RL$ circuit starts from $0$ and grows according to the equation $i(t) = \frac{E}{R}(1 - e^{-Rt/L})$.The induced emf $e$ across the inductor $L$ is given by $e = L \frac{di}{dt}$.Differentiating the current equation with respect to time,we get $\frac{di}{dt} = \frac{E}{L} e^{-Rt/L}$.Therefore,$e = L \left( \frac{E}{L} e^{-Rt/L} ight) = E e^{-Rt/L}$.At $t = 0$,$e = E$ (maximum value) and $i = 0$.As $t 	o \infty$,$e 	o 0$ and $i 	o \frac{E}{R}$.Thus,the graph of $e$ versus $t$ is an exponential decay curve starting from $E$ at $t = 0$ and approaching $0$ as $t$ increases,which corresponds to the graph shown in option $C$.

The switch $S$ of the circuit shown in the figure is closed at $t = 0$. If $e$ denotes the induced emf in the inductor $L$ and $i$ denotes the current flowing through the circuit at time $t$,which of the following graphs correctly represents the variation of $e$ with time $t$?

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