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(A) When the switch is in position $1$ for a long time,the capacitor $C$ charges to the $EMF$ $E$ of the battery. So,$V_C = E$.At $t = 0$,the switch i

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$\frac{E^2R_1}{(R_1+R_2)^2} e^{-2t/(R_1+R_2)C}$

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(A) When the switch is in position $1$ for a long time,the capacitor $C$ charges to the $EMF$ $E$ of the battery. So,$V_C = E$.At $t = 0$,the switch is moved to position $2$. The capacitor now discharges through the series combination of resistors $R_1$ and $R_2$.The total resistance in the circuit is $R_{eq} = R_1 + R_2$.The discharge current $i(t)$ is given by $i(t) = \frac{V_C}{R_{eq}} e^{-t/(R_{eq}C)} = \frac{E}{R_1 + R_2} e^{-t/((R_1 + R_2)C)}$.The thermal power $P_1(t)$ generated in resistance $R_1$ is $P_1(t) = i(t)^2 R_1$.Substituting the expression for $i(t)$:$P_1(t) = \left( \frac{E}{R_1 + R_2} e^{-t/((R_1 + R_2)C)} \right)^2 R_1 = \frac{E^2 R_1}{(R_1 + R_2)^2} e^{-2t/((R_1 + R_2)C)}$.

Switch $S$ of the circuit shown in the figure is in position $1$ for a long time. At instant $t = 0$,it is thrown from position $1$ to $2$. Find the thermal power $P_1(t)$ generated in resistance $R_1$.

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