Three capacitors of capacitance $1.0 \mu F$, $2.0 \mu F$, and $5.0 \mu F$ are connected in series to a $10 \ V$ source. The potential difference across the $2.0 \mu F$ capacitor is

In the figure,three capacitors each of capacitance $6\,pF$ are connected in series. The total capacitance of the combination will be

Obtain the equivalent capacitance of the network in the figure. For a $300 \; V$ supply,determine the charge and voltage across each capacitor.

Three capacitors each of capacitance $1\,\mu F$ are connected in parallel. To this combination,a fourth capacitor of capacitance $1\,\mu F$ is connected in series. The resultant capacitance of the system is.......$\mu F$

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:

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