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 series with $R_1$ and the battery $E$. In steady state,no current flows through the capacitor branch,so

Vedclass · Accepted Answer

$1.67$

Vedclass · Answer

(A) When $K_1$ is open,the capacitor $C$ is in series with $R_1$ and the battery $E$. In steady state,no current flows through the capacitor branch,so the potential difference across the capacitor is equal to the $EMF$ of the battery,$V_1 = E$. Thus,the charge $q_1 = CE$.When $K_1$ is closed,the circuit forms a potential divider. The capacitor is in parallel with resistor $R_2$. The potential difference across the capacitor is the voltage across $R_2$,which is given by the voltage divider rule: $V_2 = E 	imes \frac{R_2}{R_1 + R_2} = E 	imes \frac{3}{2 + 3} = E 	imes \frac{3}{5} = 0.6E$.Thus,the charge $q_2 = C 	imes V_2 = 0.6CE$.The ratio of charges is $\frac{q_1}{q_2} = \frac{CE}{0.6CE} = \frac{1}{0.6} = 1.666... \approx 1.67$.

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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