Identical dielectric slabs are inserted into two identical capacitors $A$ and $B.$ These capacitors and a battery are connected as shown in the figure. Now, the slab of capacitor $B$ is pulled out with battery remaining connected.

 

Find equivalent capacitance between $X$ and $Y$ if each capacitor is $4 \mathrm{\mu F}$.

If the number of condensers are connected in series, then

In parallel arrangement of capacitors,

i.the potential difference across individual capacitor is same.
ii. the charge is shared by the capacitor in the ratio of the capacitance.
iii. the resultant capacitance is equal to sum of the capacitance of capacitors used.

Four capacitors of equal capacitance have an equivalent capacitance $C_1$ when connected in series and an equivalent capacitance $C_2$ when connected in parallel. The ratio $\frac{C_1}{C_2}$ is

Three capacitors of capacitances $3 \mathrm{\mu F}\text{, }9 \mathrm{\mu F}$ and $18 \mathrm{\mu } \mathrm{F}$ are connected once in series and another time in parallel. The ratio of equivalent capacitances in the two cases $\left(\frac{C_s}{C_p}\right)$ will be

How do you arrange four equal capacitors of $4 \mu \mathrm{F}$ to get effective capacitance $3 \mu \mathrm{F}\text{?}$