1. Initial charge on $C_1$

Given: $C_1 = 6\,\mu F, \qquad V_0 = 5\,V$

Initially, capacitor $C_1$ is charged by a $5V$ battery, so its charge is $Q_0 = C_1V_0 = 6\times 5 = 30\,\mu C$

So before connecting $C_2$, capacitor $C_1$ has charge: $Q_0 = 30\,\mu C$

2. Battery is removed and $C_2$ is connected

Now $C_2 = 12\,\mu F$ is connected in parallel with $C_1$ (in place of the battery). When switch $S$ is closed, charge redistributes until both capacitors have the same final potential $V_f$.

Since the battery is removed, total charge is conserved.

Thus, $q_1 + q_2 = 30\,\mu C$

3. Condition at equilibrium

At equilibrium, both capacitors have the same voltage: $V_f = \frac{q_1}{C_1} = \frac{q_2}{C_2}$

So, $\frac{q_1}{6} = \frac{q_2}{12}$

This gives $q_2 = 2q_1$

4. Use charge conservation

From $q_1 + q_2 = 30$ and $q_2 = 2q_1$,

we get $q_1 + 2q_1 = 30$ $3q_1 = 30$ $q_1 = 10\,\mu C$

Then, $q_2 = 20\,\mu C$

5. Check using final voltage

$V_f = \frac{q_1}{C_1} = \frac{10}{6} = \frac{5}{3}V$

Also, $V_f = \frac{q_2}{C_2} = \frac{20}{12} = \frac{5}{3}V$

Both match, so the result is correct.

6. Option check

• A: $q_1 = 10\,\mu C,\ q_2 = 20\,\mu C$ ✅
• B: Total charge $=45\,\mu C$ ❌
• C: Voltage not equal on both capacitors ❌
• D: Total charge $=45\,\mu C$ ❌

Therefore, the correct option is A.