Question

A network of $4$ capacitors is connected to a battery as shown in the figure. The ratio of charges on $C_2$ and $C_4$ is

• A. $\frac{7}{4}$
  
• B. $\frac{22}{3}$
• C. $\frac{4}{7}$
• D. $\frac{3}{22}$

Answer 1

The correct option is D $\frac{3}{22}$

Potential difference across capacitor $C_4$ is $V$.

So, the charge on $C_4$ be

$Q_4=C_{4}V$

As, $C_4=4C$

$\Rightarrow Q_4=4CV$

Since, $C_1$, $C_2$ and $C_3$ are in series connection so, their equivalent capacitance will be

$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$

$\Rightarrow \frac{1}{C_{eq}}=\frac{1}{C}+\frac{1}{2C}+\frac{1}{3C}$

$\Rightarrow C_{eq}=\frac{6C}{11}$

So, the charge across the $C_2$ will be
$Q_2=C_{eq}V$

$\Rightarrow Q_2=\frac{6CV}{11}$

So, the ratio of the charges on $C_2$ and $C_4$

$\frac{Q_2}{Q_4}=\frac{6CV/11}{4CV}=\frac{3}{22}$

Hence, option (d) is the correct answer.