$A$ capacitor of capacity $C_1$ is charged to a potential of $V_0$. After disconnecting it from the battery,it is connected to a capacitor of capacity $C_2$ as shown in the figure. The ratio of the total energy before and after the connection of switch $S$ is:

Two circuits $(a)$ and $(b)$ have charged capacitors with capacitances and charges as shown in the figures. The switches are initially open. On closing the switches, what happens to the charge flow?

$A$ $2\,\mu F$ capacitor is charged to a potential of $10\,V$. Another $4\,\mu F$ capacitor is charged to a potential of $20\,V$. The two capacitors are then connected in a single loop,with the positive plate of one connected to the negative plate of the other. What heat is evolved in the circuit in $\mu J$?

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: $C = 1 \,\mu F$,$C_1 = 6C$,$C_2 = 3C$,$C_3 = 3C$,$E = 9 \, V$]

Two metal spheres of radii $0.01 \ m$ and $0.02 \ m$ are given a charge of $15 \ mC$ and $45 \ mC$ respectively. They are then connected by a wire. The final charge on the first sphere is $\ldots \ldots \ldots \times 10^{-3} \ C$.

In the following figure,$C_1=5 \mu F$,$C_2=C_3=10 \mu F$ and $\varepsilon=20 \text{ V}$. Initially,the switch $S$ is connected to point $A$ until capacitor $C_1$ is fully charged. Afterwards,the switch is thrown to the left side and connected to point $B$. The charge on capacitor $C_3$ after equilibrium is reached will be (in $\mu C$)