Question

A galvanometer (G) of $2 \, \Omega$ resistance is connected in the given circuit. The ratio of charge stored in $C_1$ and $C_2$ is:

• A. $\frac{2}{3}$
• B. $\frac{3}{2}$
• C. 1
• D. $\frac{1}{2}$

Answer 1

Answer: (D)

$\frac{1}{2}$

Answer 2

Given: - Capacitance of $C_1 = 4 \, \mu F$ - Capacitance of $C_2 = 6 \, \mu F$ - Voltage across the circuit: $V = 6 \, V$

Step 1: Charge Stored on Capacitors

The charge stored on a capacitor is given by:

$Q = C \times V$

where $C$ is the capacitance and $V$ is the potential difference across the capacitor.

Step 2: Calculating Charge on $C_1$

The charge stored on $C_1$ is:

$Q_1 = C_1 \times V = 4 \times 10^{-6} \, F \times 6 \, V$ $Q_1 = 24 \times 10^{-6} \, C = 24 \, \mu C$

Step 3: Calculating Charge on $C_2$

The charge stored on $C_2$ is:

$Q_2 = C_2 \times V = 6 \times 10^{-6} \, F \times 6 \, V$ $Q_2 = 36 \times 10^{-6} \, C = 36 \, \mu C$

Step 4: Ratio of Charge Stored in $C_1$ and $C_2$

The ratio of the charge stored in $C_1$ to the charge stored in $C_2$ is:

$\text{Ratio} = \frac{Q_1}{Q_2} = \frac{24 \, \mu C}{36 \, \mu C} = \frac{2}{3}$

Conclusion:

The correct ratio of the charge stored in $C_1$ to $C_2$ is $\frac{1}{2}$.