Three capacitors $3 \mu F$, $6 \mu F$, and $6 \mu F$ are connected in series to a source of $120 V$. The potential difference, in volts, across the $3 \mu F$ capacitor will be:

Three capacitors $2 \mu F$,$3 \mu F$,and $5 \mu F$ can withstand voltages up to $3 \, V$,$2 \, V$,and $1 \, V$ respectively. Their series combination can withstand a maximum voltage equal to.....$V$.

Two capacitors of capacitances $1 \ \mu F$ and $2 \ \mu F$ can separately withstand potentials of $6 \ kV$ and $4 \ kV$ respectively. The total potential they together can withstand when they are connected in series is (in $kV$)

When three capacitors of equal capacities are connected in parallel and one of the same capacity is connected in series with the combination,the resultant capacity is $4.5 \mu F$. The capacity of each capacitor is: (in $\mu F$)

Four capacitors,each of capacitance $8 \mu F$,are connected as shown in the figure. The equivalent capacitance between points $A$ and $B$ is: (in $\mu F$)

Find the equivalent capacitance of the circuit between points $A$ and $B$.