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

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(A) When $K_1$ is open,the capacitor $C$ is in parallel with the battery $E$ because no current flows through $R_1$ in steady state. Thus,the voltage

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$5/3$

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(A) When $K_1$ is open,the capacitor $C$ is in parallel with the battery $E$ because no current flows through $R_1$ in steady state. Thus,the voltage across the capacitor is $V_1 = E$. Therefore,$q_1 = CE$.When $K_1$ is closed,the circuit forms a series combination of $R_1$ and $R_2$ connected to the battery $E$. The capacitor $C$ is in parallel with resistor $R_2$. The voltage across $R_2$ is given by the voltage divider rule: $V_2 = E 	imes \frac{R_2}{R_1 + R_2} = E 	imes \frac{3}{2 + 3} = \frac{3}{5}E$.Thus,the charge on the capacitor is $q_2 = C 	imes V_2 = C 	imes \frac{3}{5}E = \frac{3}{5}CE$.The ratio of the charges is $\frac{q_1}{q_2} = \frac{CE}{\frac{3}{5}CE} = \frac{5}{3}$.

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 $K_1$ 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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