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
![](https://app-content.cdn.examgoal.net/fly/@width/image/jaoe38c1m3o0eigl/ec226b11-0002-4a04-8ce5-45169e33cc35/70e8e250-a636-11ef-a300-b723d4598e73/file-jaoe38c1m3o0eigm.png?format=png)

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

Choose the correct answer: