The resultant capacitance of the given circuit is

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 $\mu F$)

Two capacitors $A$ and $B$ are connected in series as shown in the figure. When the switch $S$ is closed and both capacitors are fully charged,then:

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 $V$)

An infinite number of identical capacitors,each with a capacitance of $1 \ \mu F$,are connected as shown in the figure. Find the equivalent capacitance between $A$ and $B$ in $\mu F$.

Assertion : If three capacitors of capacitances $C_1 < C_2 < C_3$ are connected in parallel,then their equivalent capacitance $C_P > C_S$,where $C_S$ is the equivalent capacitance in series.
Reason : $\frac{1}{C_P} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$