$A$ 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 . . . . . . .

$A$ parallel plate capacitor of capacitance $C$ is charged to a potential difference $V$ by connecting it to a battery. Another capacitor of capacitance $2C$ is charged to a potential difference $2V$ by connecting it to another battery. Now,the charging batteries are removed and the capacitors are connected in parallel such that the positive plate of one is connected to the negative plate of the other and the negative plate of the first is connected to the positive plate of the second. Find the final energy of this configuration.

$A$ $10 \mu F$ capacitor is charged by a $100 V$ battery. It is disconnected from the battery and is connected to another uncharged capacitor of capacitance $30 \mu F$. During this process,the electrostatic energy lost by the first capacitor is

Circuits $(a)$ and $(b)$ have charged capacitors with capacitances $C, 2C$ and $3C$ as shown,with an open switch $S$. When the switch is closed,determine the direction of charge flow for each circuit.

$A$ $10 \ \mu F$ capacitor is charged to a potential difference of $50 \ V$ and connected in parallel to another uncharged capacitor. The common potential difference becomes $20 \ V$. The capacitance of the second capacitor is ........ $\mu F$.

$A$ $2\, \mu F$ capacitor $C_{1}$ is first charged to a potential difference of $10\, V$ using a battery. Then the battery is removed and the capacitor is connected to an uncharged capacitor $C_{2}$ of $8\, \mu F$. The charge in $C_{2}$ on equilibrium condition is $\ldots\, \mu C$. (Round off to the Nearest Integer)