In the circuit shown in the figure,initially $K_1$ is closed and $K_2$ is open. What are the charges on each capacitor? Then $K_1$ was opened and $K_2$ was closed (order is important),what will be the charge on each capacitor now? [Given: $C = 1 \,\mu F$,$C_1 = 6C$,$C_2 = 3C$,$C_3 = 3C$,$E = 9 \, V$]

(N/A) Case $1$: $K_1$ is closed and $K_2$ is open.
The capacitors $C_1$ and $C_2$ are in series with the battery $E = 9 \, V$.
The equivalent capacitance is $C_{eq} = \frac{C_1 C_2}{C_1 + C_2} = \frac{(6C)(3C)}{6C + 3C} = \frac{18C^2}{9C} = 2C = 2 \,\mu F$.
The charge on each capacitor is $Q = C_{eq} E = 2 \,\mu F \times 9 \, V = 18 \,\mu C$.
Thus,$Q_1 = 18 \,\mu C$,$Q_2 = 18 \,\mu C$,and $Q_3 = 0 \,\mu C$.
Case $2$: $K_1$ is opened and $K_2$ is closed.
Now,$C_1$ is disconnected from the battery. The capacitor $C_2$ (charged to $18 \,\mu C$) is connected in parallel with $C_3$ (initially uncharged).
The total charge $Q_{total} = 18 \,\mu C$ is shared between $C_2$ and $C_3$.
Since $C_2 = 3C$ and $C_3 = 3C$,the charge is shared equally.
$Q_2' = Q_3' = \frac{Q_{total}}{2} = \frac{18 \,\mu C}{2} = 9 \,\mu C$.
Since $K_1$ is open,$C_1$ remains charged at $18 \,\mu C$.
Final charges: $Q_1 = 18 \,\mu C$,$Q_2 = 9 \,\mu C$,$Q_3 = 9 \,\mu C$.