If there are $n$ capacitors,each of capacitance $C$,connected in parallel to a $V$ volt source,then the energy stored is equal to

The combined capacity of the parallel combination of two capacitors is four times their combined capacity when connected in series. This means that:

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 : $\frac{1}{C_P} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$

Five capacitors each of capacity $C$ are connected as shown in the figure. If their resultant capacity is $2 \mu F$,then the capacity of each capacitor is: (in $\mu F$)

The equivalent capacitance between $A$ and $B$ is (in $\mu F$):

Five capacitors,each of capacity $C$,are connected as shown in the figure. The resultant capacity between $A$ and $B$ is $14 \mu F$. The capacity of each capacitor is (in $\mu F$)