Two capacitors of capacitance $3 \ \mu F$ and $6 \ \mu F$ are charged to a potential of $12 \ V$. They are then connected by joining their positive plates together and their negative plates together. What will be the potential difference across each capacitor in $V$?

$A$ 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 capacitors $C_1$ and $C_2 = 2C_1$ are connected in a circuit with a switch between them as shown in the figure. Initially,the switch is open and $C_1$ holds charge $Q$. The switch is closed. At steady state,the charge on the capacitors will be:

Two identical capacitors have the same capacitance $C$. One of them is charged to potential $V_1$ and the other to $V_2$. The negative ends of the capacitors are connected together. When the positive ends are also connected,the decrease in energy of the combined system is

$A$ $10\,\mu F$ capacitor and a $20\,\mu F$ capacitor are connected in series across a $200\,V$ supply line. The charged capacitors are then disconnected from the line and reconnected with their positive plates together and negative plates together,and no external voltage is applied. What is the potential difference across each capacitor?

$A$ capacitor is charged with a battery and the energy stored is $U$. After disconnecting the battery,another identical uncharged capacitor is connected in parallel with it. The total energy of the system of capacitors is . . . . . . .