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

The Correct Option is D

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 \( \frac{1}{2} \).