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Mixed Combination of Cells

In the circui...

Question

In the circuit in Figure, if no current flows through the galvanometer when the key $k$ 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_{2}^{1}}{C_{2}^{2}} = \frac{R_{2}}{R_{1}}

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Solution

The correct option is C $\frac{C_{1}}{C_{2}} = \frac{R_{2}}{R_{1}}$
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
$V_{A} - V_{B} = V_{A} - V_{D}$
$I_{1} R_{1} = \frac{q}{C_{1}}$ (i)
Similarly, $V_{B} - V_{C} = V_{D} - V_{C}$
$I_{2} R_{2} = \frac{q}{C_{2}}$ (ii)
Dividing Eq. (i) by Eq. (ii), we get
$\frac{I_{1} R_{1}}{I_{2} R_{2}} = \frac{q / C_{1}}{q / C_{2}} = \frac{C_{2}}{C_{1}} o r \frac{C_{1}}{C_{2}} = \frac{R_{2}}{R_{1}}$

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Which of the following is(are) correct?

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

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