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

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

The combined capacity of the parallel combination of two capacitors is four times their combined capacity when connected in series. This means that:

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 : $\frac{1}{C_P} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$

Two equal capacitors are first connected in series and then in parallel. The ratio of the equivalent capacities in the two cases will be

Assertion $(A)$: Two condensers of same capacity are connected first in parallel and then in series. The ratio of resultant capacities in the two cases will be $4: 1$.
Reason $(R)$: In parallel,capacity increases and in series,capacity decreases.