In the given figure,the capacitors $C_1, C_3, C_4, C_5$ have a capacitance of $4\,\mu F$ each. If the capacitor $C_2$ has a capacitance of $10\,\mu F$,then the effective capacitance between $A$ and $B$ will be.....$\mu F$.

$A$ parallel plate capacitor of capacitance $C$ is connected to a battery and is charged to a potential difference $V$. Another capacitor of capacitance $2C$ is connected to another battery and is charged to a potential difference $2V$. The charging batteries are now disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is

Three charged capacitors,$C_1 = 17 \ \mu F$,$C_2 = 34 \ \mu F$,$C_3 = 41 \ \mu F$ and two open switches,$S_1$ and $S_2$ are assembled into a network with initial voltages and polarities,as shown in the figure. The final status of the network is attained when the two switches,$S_1$ and $S_2$ are closed. In the figure,the final charge on capacitor $C_3$ in $mC$ is closest to:

When two identical capacitors are charged individually to different potentials and then connected in parallel,after disconnecting from the source . . . . . .

The diagram shows four capacitors with capacitances and breakdown voltages as mentioned. What should be the maximum value of the external emf source such that no capacitor breaks down? (in $kV$)

If the potential of $A$ is $10\,V$,then the potential of $B$ is