Two conducting spheres of radii $R_1$ and $R_2$ are charged with charges $Q_1$ and $Q_2$ respectively. On bringing them in contact,there is:

$A$ capacitor of capacity $C_1$ is charged to potential $V_1$ and then disconnected. An uncharged capacitor of capacity $C_2$ is connected in parallel with $C_1$. The resultant potential $V_2$ 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).

$A$ body of capacity $4\,\mu F$ is charged to $80\,V$ and another body of capacity $6\,\mu F$ is charged to $30\,V$. When they are connected,the energy lost by the $4\,\mu F$ capacitor is.......$mJ$.

Two capacitors $C_1$ and $C_2 = 2C_1$ are connected with a switch $S$ as shown in the figure. Initially,the switch is open and the charge on capacitor $C_1$ is $Q$. Now,when the switch is closed,the final charges on the capacitors are:

$A$ parallel plate capacitor of capacitance $C$ is charged to a potential $V$ by a battery. Another capacitor of capacitance $2C$ is charged to a potential $2V$ by another battery. After removing the batteries,they are connected in parallel such that the positive plate of one is connected to the negative plate of the other. Calculate the final energy of the system.