Question

A network of four capacitors of capacity equal to $C_1=C,C_2=2C,$ $C_3=3C$ and $C_4=4C$ are connected to a battery as shown in the figure.

The ratio of the charges on $C_2$ and $C_4$ is

• A. 4/7
• B. 3/22
• C. 7/4
• D. 22/3

Answer 1

Answer: (B)

$C_1,C_2$ and $C_3$ are in series
$\frac{1}{C^{\prime }}=\frac{1}{C}+\frac{1}{2C}+\frac{1}{3C}$
or, $\frac{1}{C^{\prime }}=\frac{6+3+2}{6C}=\frac{11}{6C}$
or, C' = $\frac{6C}{11}$
All the capacitors in branch 1 is in series so the charge on each capacitor is Q' = $\frac{6}{11}$ CV
Also charge on capacitor $C_4$ is Q = 4 CV
$\therefore$ Ratio $=\frac{Q{}^{\prime }}{Q}=\frac{6CV}{11\times 4CV}=\frac{3}{22}$.