When three capacitors of equal capacities are connected in parallel and one of the same capacity is connected in series with the combination,the resultant capacity is $4.5 \mu F$. The capacity of each capacitor is: (in $\mu F$)

The equivalent capacitance of the combination shown is

Four metal plates are arranged as shown. Find the equivalent capacitance between $ X $ and $ Y $. ($ A $ = Area of each plate,$ d $ = distance between adjacent plates)

Assertion : If three capacitors of capacitances $C_1 < C_2 < C_3$ are connected in parallel,then their equivalent capacitance $C_P > C_S$,where $C_S$ is the equivalent capacitance in series.
Reason : $\frac{1}{C_P} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$

Find the equivalent capacitance of the circuit between points $A$ and $B$.

$A$ capacitor of capacitance $C_1 = 1\ \mu F$ can withstand a maximum voltage $V_1 = 6\ kV$ and another capacitor of capacitance $C_2 = 3\ \mu F$ can withstand a maximum voltage $V_2 = 4\ kV$. When the two capacitors are connected in series,the combined system can withstand a maximum voltage of......$kV$.