Question

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

• A. $\frac{C_1}{C_2}=\frac{R_1}{R_2}$
• B. $\frac{C_1}{C_2}=\frac{R_2}{R_1}$
• C. $\frac{C_1^2}{C_2^2}=\frac{R_1}{R_2}$
• D. $\frac{C_1^2}{C_2^2}=\frac{R_2}{R_1}$

Answer 1

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 caapcitor be $q$. Since the current through galvanometer is zero.

$\therefore I_1=I_2$


In the balanced Wheatstone bridge, the potential difference between the ends of the galvanometer will be zero.

$\therefore V_A-V_B=V_A-V_D$

$\Rightarrow I_1{R}_1=\frac{q}{C_1}....\left(1\right)$

Similarly, $V_B-V_C=V_D-V_C$

$\Rightarrow I_2{R}_2=\frac{q}{C_2}....\left(2\right)$

Divinding Eq (1) by eq (2) we get,

$\frac{I_1{R}_1}{I_2{R}_2}=\frac{\frac{q}{C_1}}{\frac{q}{C_2}}$

$\Rightarrow \frac{R_1}{R_2}=\frac{C_2}{C_1}$

$\therefore \frac{C_1}{C_2}=\frac{R_2}{R_1}$

Hence, option $\left(b\right)$ is correct.