In the circuit shown in the figure,K_1 is ope | vedclass

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(A) When the key $K_1$ is open,no current flows through the resistor $R_2$. The capacitor $C$ is connected in parallel with the battery $E$ through th

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$5/3$

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(A) When the key $K_1$ is open,no current flows through the resistor $R_2$. The capacitor $C$ is connected in parallel with the battery $E$ through the resistor $R_1$. Since there is no current,there is no potential drop across $R_1$. Therefore,the voltage across the capacitor is $V_C = E$.The charge on the capacitor is $q_1 = CE$.When the key $K_1$ is closed,the circuit reaches a steady state where the capacitor acts as an open circuit. The resistors $R_1 = 2 \ \Omega$ and $R_2 = 3 \ \Omega$ are in series. The voltage across the capacitor $C$ is the voltage across the resistor $R_2$.Using the voltage divider rule,the voltage across $C$ is $V_C' = \frac{R_2}{R_1 + R_2} 	imes E = \frac{3}{2 + 3} 	imes E = \frac{3}{5} E$.The charge on the capacitor is $q_2 = C V_C' = C 	imes \frac{3}{5} E = \frac{3}{5} CE$.Calculating the ratio $q_1/q_2$:$\frac{q_1}{q_2} = \frac{CE}{\frac{3}{5} CE} = \frac{5}{3}$.

In the circuit shown in the figure,$K_1$ is open. The charge on capacitor $C$ in steady state is $q_1$. Now,the key is closed and at steady state,the charge on $C$ is $q_2$. The ratio of charges $q_1/q_2$ is

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