In the circuit shown the switch $S$ is closed at $t=0$ and $i_1$ and $i_2$ are the instantaneous cur

Question

In the circuit shown the switch $S$ is closed at $t=0$ and $i_1$ and $i_2$ are the instantaneous currents.

Professor PC Thomas Classes · Answer

$\begin{aligned} & i_1=\frac{E}{R}\left(1-e^{-R t / L}ight) \ & i_2=\frac{E}{R} e^{-t / R C} \ & 	ext { - } i_{1 \max }=\frac{E}{R} 	ext { at } t=\infty 	ext { and } i_{2 \max }=\frac{E}{R} 	ext { at } t=0 \ & i_{1 \max }=i_{2 \max } \ & i_1=i_2 	ext { at } 1-e^{-R t / L}=e^{-R t / L} \ & 	ext { or } t=\frac{L}{R} \ln 2\end{aligned}$