In the circuit shown below, $C_1 = 6 μF, C_2 = 3 μF$ and battery $B = 40 V$. Initially, switch $A_1$ is opened and the switch $A_2$ is closed. Now switch $A_2$ is opened and $A_1$ is closed. The charge on $C_2$ will be

80 μC

The correct answer is Option (3) → 80 μC

Given:

$C_1 = 6\times10^{-6}\,\text{F}$

$C_2 = 3\times10^{-6}\,\text{F}$

$B = 40\,\text{V}$

Initially $A_1$ open and $A_2$ closed.

Initial charges:

$Q_{1i} = C_1 B = 6\times10^{-6}\times40 = 240\times10^{-6}\,\text{C}$

$Q_{2i} = 0$

After $A_2$ opened and $A_1$ closed, charge conserved:

$Q_{\text{total}} = Q_{1i}+Q_{2i} = 240\times10^{-6}\,\text{C}$

Equivalent capacitance:

$C_{\text{eq}} = C_1 + C_2 = 9\times10^{-6}\,\text{F}$

Final common voltage:

$V_f = \frac{Q_{\text{total}}}{C_{\text{eq}}} = \frac{240\times10^{-6}}{9\times10^{-6}} = \frac{240}{9} = \frac{80}{3}\,\text{V} \approx 26.67\,\text{V}$

Final charge on $C_2$:

$Q_{2f} = C_2 V_f = 3\times10^{-6}\times\frac{80}{3} = 80\times10^{-6}\,\text{C}$

Final Answer: $Q_2 = 80\,\mu\text{C}$