Switch S of the circuit shown in the figure i | vedclass

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(B) When the switch $S$ is closed at $t=0$,the current $i$ in the $RL$ circuit grows according to the equation:$i = \frac{E}{R} (1 - e^{-tR/L})$This i

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(B) When the switch $S$ is closed at $t=0$,the current $i$ in the $RL$ circuit grows according to the equation:$i = \frac{E}{R} (1 - e^{-tR/L})$This is an increasing exponential function starting from $i=0$ at $t=0$ and approaching $i=E/R$ as $t 	o \infty$. This behavior is represented by graph $D$.The induced $emf$ $e$ across the inductor $L$ is given by:$e = L \frac{di}{dt} = L \frac{d}{dt} [\frac{E}{R} (1 - e^{-tR/L})] = L \cdot \frac{E}{R} \cdot \frac{R}{L} e^{-tR/L} = E e^{-tR/L}$At $t=0$,$e=E$,and as $t 	o \infty$,$e 	o 0$. This is a decreasing exponential function,which is represented by graph $C$.Therefore,graph $C$ represents $e$ and graph $D$ represents $i$.

Switch $S$ of the circuit shown in the figure is closed at $t = 0$. If $e$ denotes the induced $emf$ in $L$ and $i$ the current flowing through the circuit at time $t$,which of the following graphs is correct?

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