Three capacitors each of capacity $4\,\mu F$ are to be connected in such a way that the effective capacitance is $6\,\mu F$. This can be done by:

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 $\mu F$)

The equivalent capacitance of three capacitors of capacitance $C_1, C_2$ and $C_3$ connected in parallel is $12$ units and their product is $C_1 \cdot C_2 \cdot C_3 = 48$. When the capacitors $C_1$ and $C_2$ are connected in parallel,the equivalent capacitance is $6$ units. Then the capacitances are:

Two capacitors $C_1$ and $C_2$ in a circuit are joined as shown in the figure. The potential of point $A$ is $V_1$ and that of point $B$ is $V_2$. The potential at point $D$ will be

Two capacitors of capacitance $2\ \mu F$ and $3\ \mu F$ are joined in series. The outer plate of the first capacitor is at $1000\ V$ and the outer plate of the second capacitor is earthed (grounded). The potential on the inner plates of each capacitor will be: (in $V$)

The difference between the equivalent capacitances of two identical capacitors connected in parallel and in series is $6 \mu F$. The value of the capacitance of each capacitor is (in $\mu F$)