The equivalent capacitance of the combination shown in the figure is:

In the circuit,two capacitors $C_1$ and $C_2$ are connected as shown in the figure. The potential at point $A$ is $V_1$ and at point $B$ is $V_2$. What is the potential at point $D$?

Two capacitors of $1 \ \mu F$ and $2 \ \mu F$ are connected in series. This combination is charged to a potential difference of $120 \ V$. What is the potential difference across the $1 \ \mu F$ capacitor in volts?

Find the equivalent capacitance between $A$ and $B$ of the following arrangement:

Two capacitors of capacities $2 \mu F$ and $4 \mu F$ are connected in parallel. $A$ third capacitor of $6 \mu F$ capacity is connected in series with this combination. $A$ battery of $12 \text{ V}$ is connected across this combination. The charge on the $2 \mu F$ capacitor is: (in $\mu C$)

Consider the combination of $2$ capacitors $C_{1}$ and $C_{2}$,with $C_{2} > C_{1}$. When connected in parallel,the equivalent capacitance is $\frac{15}{4}$ times the equivalent capacitance of the same capacitors connected in series. Calculate the ratio of the capacitors,$\frac{C_{2}}{C_{1}}$.