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

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

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

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

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

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Solution

The correct option is B
$\frac{C_{1}}{C_{2}} = \frac{R_{2}}{R_{1}}$

In the steady state , no current is passing through the capacitor. Let the charge on each capacitor be $q$ . Since the current through galvanometer is zero.

$I_{1} = I_{2}$

The potential difference between the ends of the galvanometer will be zero. Therefore,

$V_{A} - V_{B} = V_{A} - V_{D}$

$I_{1} R_{1} = \frac{q}{C_{1}}$ ..............(1)

Similarly , $V_{B} - V_{C} = V_{D} - V_{C}$

$I_{2} R_{2} = \frac{q}{C_{2}}$ .............(2)
Divide (1) by (2) we get
$\frac{C_{1}}{C_{2}} = \frac{R_{2}}{R_{1}}$

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