Question 4: Current Electricity

Question

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

Accepted Answer

In the steady state, no current is passing
through capacitor. Let the charge on each
capacitor be q. Since, the current through
galvanometer is zero.
$$	herefore \quad I_1=I_2$$
The potential difference between ends of
galvanometer will be zero.
$$\begin{array}{ll}
	herefore & V_A-V_B=V_A-V_D \
	herefore & I_1 R_1=\frac{q}{C_1} \quad 	ext{..... (i)}
\end{array}$$
$$\begin{aligned}
	ext{Similarly,} \quad V_B-V_C & =V_D-V_C \
I_2 R_2 & =\frac{q}{C_2} \quad 	ext{..... (ii)}
\end{aligned}$$
On dividing Eq. (i) by Eq. (ii), we get
$$\begin{aligned}
\frac{I_1 R_1}{I_2 R_2} & =\frac{q / C_1}{q / C_2}=\frac{C_2}{C_1} \
	herefore \quad \quad \quad \frac{C_1}{C_2} & =\frac{R_2}{R_1}
\end{aligned}$$

Answer

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

Answer

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

Answer

$$\frac{C_1^2}{C_2^2}=\frac{R_1^2}{R_2^2}$$

Answer

$$\frac{C_1^2}{C_2^2}=\frac{R_2}{R_1}$$

Current Electricity

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

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