The minimum number of capacitors of $2\,\mu F$ capacitance each required to obtain a resultant capacitance of $5\,\mu F$ is

In the circuit shown in the figure,each capacitor has a capacity of $3\,\mu F$. The equivalent capacity between $A$ and $B$ is.......$\mu F$.

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 $\mu C$)

Capacitors of capacities $C_1$ and $C_2$ are connected in series. If the combination is connected to a supply of $V$ volt, then the potential difference across capacitor $C_2$ is:

$A$ series combination of $10$ capacitors,each of value $C_1$,is charged by a source of potential difference $4V$. When another parallel combination of $8$ capacitors,each of value $C_2$,is charged by a source of potential difference $V$,it has the same total energy stored in it as in the first combination. The value of $C_2$ is

Two capacitors,each having a capacitance of $2 \mu F$,are connected in series. What will be the effective capacitance of these capacitors (in $\mu F$)?