Consider the circuit shown where $C_1=6 \mu \mathrm{F}, C_2=3 \mu \mathrm{F}$ and $V=20 \mathrm{V}$. Capacitor $C_1$ is first charged by closing the switch $S_1$. Switch $S_1$ is then opened, and the charged capacitor is connected to the uncharged capacitor $C_2$ by closing $S_2$.

In the circuit shown in the figure, initially key $K_1$ is closed and key $K_2$ is open. Then, $K_1$ is opened and $K_2$ is closed (order is important). [Take $Q_1^{\text{'}}$ and $Q_2^{\text{'}}$ as charges on $C_1$ and $C_2$ and $V_1$ and $V_2$ as voltage, respectively.] Then,

Two capacitors $C_1$ and $C_2$ are charged to the same potential $V$, but with opposite polarity, as shown in the figure. The switches $S_1$ and $S_2$ are then closed.

Two capacitors of equal capacitance $\left(C_1=C_2\right)$ are as shown in the figure. Initially, while the switch is open (as shown) one of the capacitors is uncharged and the other carries charge $Q_0$. The energy stored in the charged capacitor is $U_0$. Sometime after the switch is closed, the capacitors $C_1$ and $C_2$ carry charges $Q_1$ and $Q_2$, respectively. The energies stored in the capacitors are $U_1$ and $U_2$ respectively. Which of the following expression is correct?