$A$ capacitor is charged with a battery and the energy stored is $U$. After disconnecting the battery,another capacitor of the same capacity is connected in parallel to the first capacitor. Then the energy stored in each capacitor is

Two capacitors of $100 \mu F$ and $50 \mu F$ are connected in parallel. If the potential difference across $100 \mu F$ is $20 \text{ V}$ and across $50 \mu F$ is $40 \text{ V}$,then the common potential of the parallel combination will be (assuming same polarities of the capacitors are connected together).

Two metal spheres of capacitance $C_1$ and $C_2$ carry some charges. They are put in contact and then separated. The final charges $Q_1$ and $Q_2$ on them will satisfy

Two capacitors of capacitances $3 \mu F$ and $6 \mu F$ are charged to a potential of $12 V$ each. They are now connected to each other,with the positive plate of each joined to the negative plate of the other. The potential difference across each will be $volt$.

Two capacitors having capacitance $C_{1}$ and $C_{2}$ respectively are connected as shown in the figure. Initially,capacitor $C_{1}$ is charged to a potential difference $V$ volt by a battery. The battery is then removed and the charged capacitor $C_{1}$ is now connected to uncharged capacitor $C_{2}$ by closing the switch $S$. The amount of charge on the capacitor $C_{2}$ after equilibrium is

Two capacitors of $1\,\mu F$ and $2\,\mu F$ are charged to $1200\,V$. If they are connected in parallel such that the positive plate of one is connected to the negative plate of the other,what will be the new charge on each capacitor?