The figure shows an L-R circuit. When the swi | vedclass

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(B) At $t=0 \, s$,the inductor opposes any sudden change in current. Therefore,it acts as an open circuit (infinite resistance) at the instant the swi

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$I_1=\frac{E}{R_1}, I_2=I_3=0$

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(B) At $t=0 \, s$,the inductor opposes any sudden change in current. Therefore,it acts as an open circuit (infinite resistance) at the instant the switch is closed.$1$. The branch containing $R_1$ is purely resistive,so the current $I_1$ flows immediately: $I_1 = \frac{E}{R_1}$.$2$. The branches containing $L_1$ and $L_2$ act as open circuits at $t=0 \, s$ due to the back electromotive force induced in the inductors.$3$. Consequently,no current flows through the branches containing inductors at $t=0 \, s$,meaning $I_2 = 0$ and $I_3 = 0$.Thus,the correct values are $I_1 = \frac{E}{R_1}, I_2 = 0, I_3 = 0$.

The figure shows an $L-R$ circuit. When the switch $S$ is closed,the currents through resistors $R_1, R_2$ and $R_3$ are $I_1, I_2$ and $I_3$ respectively. The values of $I_1, I_2$ and $I_3$ at $t=0 \, s$ are:

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