A parallel plate capacitor is made by stacking $n$ equally spaced plates connected alternately. If the capacitance between any two plates is $C$, then the resultant capacitance is 

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 

Four plates of the same area of cross-section are joined as shown in the figure. The distance between each plate is $d$. The equivalent capacity across $A$ and $B$ will be where ${\epsilon }_0$ is permittivity of free space.

Four metal plates are arranged as shown. Capacitance between $X$ and $Y(A\to$ Area each plate, $d\to$ distance between the plate)

A capacitor $C_{1}=4 \mu F$ is connected in series with another capacitor $C_{2}=1$ $\mu F$. The combination is connected across d.c. source of $200 \mathrm{V}$. The ratio of potential across $C_2 \text{to }{C}_1$ is

In the circuit diagram shown in the figure, the resultant capacitance between $P \text{and} Q$ is 

Five capacitors;each of capacitance value $C$ are connected as shown in the figure. The ratio of capacitance between $P$ and $R$ and the capacitance between $P$ and $Q$ is 

Equivalent capacitance between $A$ and $B$ in the figure is

When three capacitors of equal capacities are connected in parallel and one of the same capacity is connected in series with its combination. The resultant capacity is $3.75 \mu \mathrm{F}$ . The capacity of each capacitor is