Question

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

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

Answer 1

Answer: (B)

$\frac{3}{22}$

Equivalent capacitance for three capacitors in series is given by

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

$\frac{C_2{C}_3+C_3{C}_1+C_1{C}_2}{C_1{C}_2{C}_3}$

$C_{eq}=\frac{C\left(2C\right)\left(3C\right)}{C\left(2C\right)+2C\left(3C\right)+3C\left(C\right)}=\frac{6}{11}C$

$Q=C_{eq}V=\frac{6V}{11}$

charge on capacitor $C_4=C_{4}V=4CV$

$\frac{\text{charge on}{C}_2}{\text{charge on}{C}_4}=\frac{\frac{6C}{11}V}{4CV}=\frac{3}{22}$