The equivalent capacitance between $A$ and $B$ in the given circuit is (in $\mu F$)

Three capacitors each of capacitance $C$ and of breakdown voltage $V$ are joined in series. The capacitance and breakdown voltage of the combination will be

$A$ 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 $\text{J}$)

When three capacitors of equal capacities are connected in parallel and one of the same capacitor is connected in series with this combination,the resultant capacity is $3.75 \mu F$. The capacity of each capacitor is: (in $\mu F$)

Four capacitors,each of capacitance $16 \mu F$,are connected as shown in the figure. The equivalent capacitance between points $A$ and $B$ is: . . . . . . (in $\mu F$).

$A$ capacitor of capacitance $1 \ \mu F$ withstands a maximum voltage of $6 \ kV$,while a capacitor of $2 \ \mu F$ withstands a maximum voltage of $4 \ kV$. What maximum voltage will the system of these two capacitors withstand if they are connected in series? (in $kV$)