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(B) In the steady state,no current passes through the capacitors. Let the charge on each capacitor be $q$. Since the current through the galvanometer

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$\frac{C_{1}}{C_{2}}=\frac{R_{2}}{R_{1}}$

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(B) In the steady state,no current passes through the capacitors. Let the charge on each capacitor be $q$. Since the current through the galvanometer is zero,the potential at point $B$ is equal to the potential at point $D$ $(V_{B} = V_{D})$.Since no current flows through the galvanometer,the current $I$ from the source splits into two branches: one through $R_{1}$ and $C_{1}$,and the other through $R_{2}$ and $C_{2}$. Let the current through the left branch be $I_{1}$ and through the right branch be $I_{2}$.For the potential difference across the galvanometer to be zero,the potential drop across $R_{1}$ must equal the potential drop across $C_{1}$ relative to the source,and similarly for the right side.Specifically,$V_{A} - V_{B} = V_{A} - V_{D} \implies I_{1} R_{1} = \frac{q}{C_{1}} \quad (i)$Similarly,for the other side: $V_{B} - V_{C} = V_{D} - V_{C} \implies I_{2} R_{2} = \frac{q}{C_{2}} \quad (ii)$Since the branches are in series with the source,$I_{1} = I_{2} = I$.Dividing equation $(i)$ by equation $(ii)$:$\frac{I_{1} R_{1}}{I_{2} R_{2}} = \frac{q / C_{1}}{q / C_{2}}$$\frac{R_{1}}{R_{2}} = \frac{C_{2}}{C_{1}}$Therefore,$\frac{C_{1}}{C_{2}} = \frac{R_{2}}{R_{1}}$.

In the circuit shown in the figure,if no current flows through the galvanometer when the key $K$ is closed,the bridge is balanced. The balancing condition for the bridge is

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