A network of four capacitors capacity equal $C_1=C, C_2=2 C, C_3=3 C$ and $C_4=$ $4 C$ are conducted to a battery as shown in the figure. The ratio of the change on $C_2$ and $C_4$ is:

• A

  $\frac{4}{7}$
• Correct answer
  B

  $\frac{3}{22}$
• C

  $\frac{7}{4}$
• D

  $\frac{22}{3}$

Solution

Here $C_1, C_2$ and $C_3$ are in series.
$$ \frac{1}{C^{\prime}}=\frac{1}{C}+\frac{1}{2 C}+\frac{1}{3 C} $$ $$ \begin{aligned} \frac{1}{C^{\prime}} & =\frac{6+3+2}{6 C}=\frac{11}{6 C} \\ \Rightarrow \quad C & =\frac{6 C}{11} \end{aligned} $$ All the capacitors in branch number 1 is in series so the charge on each number capacitor is: $$ Q^{\prime}=\frac{6}{11} C V $$ Also charge on capacitor $C_4$ is $Q=4$ V $$ \therefore \text { Ratio }=\frac{Q^{\prime}}{Q}=\frac{6 \mathrm{CV}}{11 \times 4 C V}=\frac{3}{22} $$

Hence, the correct answer is option: 'B'.