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 $3 \mu \mathrm{F}\text{?}$

Three identical condensers are connected in parallel. The combination is connected in series with one identical condenser. If resultant capacity $3.75 \mu \mathrm{F}\text{,}$ then the capacity of individual condenser is

Two metal spheres of capacitance $C_{1}$ and $C_{2}$ carry some charges. They are put in contact and then separated. The final charges $Q_{1}$ and $Q_{2}$ on them will satisfy

The capacitance between the points $P$ and $Q$ in the following circuit, is

Four capacitors are connected as shown in the figure. The equivalent capacitance between $A$ and is $B$

Between the plates of a parallel plate condenser, a plate of thickness $t_{1}$ and dielectric constant $k_{1}$ is placed. In the rest of the space, there is another plate of thickness $t_{2}$ and dielectric constant $k_{2}$. The potential difference across the condenser will be