$A$ circuit has a section $AB$ as shown in the figure. If the potential difference between points $A$ and $B$ is $V$ volts,then the potential difference across $C_1$ is

Two capacitors each having a capacitance $2 \times 10^{-6} \ F$ and a breakdown voltage $5000 \ V$,are joined in series. What will be the resultant capacitance and the breakdown voltage of the combination?

Three capacitors of capacitance $3\,\mu F$ are connected in a circuit. Then their maximum and minimum capacitances will be

Two capacitors $C_1 = 3 \mu F$ and $C_2 = 2 \mu F$ are connected in series across a d.c. source of $100 \ V$. The ratio of the potential across $C_2$ to $C_1$ is

Explain the necessity of the combination of capacitors and the ways of their connections.

$A$ parallel combination of two capacitors of capacities $2C$ and $C$ is connected across a $5 \text{ V}$ battery. When they are fully charged,the charges and energies stored in them are $Q_1, Q_2$ and $E_1, E_2$ respectively. Then $\frac{E_1-E_2}{Q_1-Q_2}$ in $\text{J/C}$ is (capacity is in Farad,charge in Coulomb and energy in $\text{J}$)