Question

A network of four capacitors of capacities equal to $C_{1}=C, C_{2}=2 C, C_{3}=3 C$ and $C_{4}=4 C$ are connected to a battery as shown in the figure. The ratio of the charges on $C_{2}$ and $C_{4}$ is

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

Answer 1

Answer: (B)

$\frac{3}{22}$

Answer 2

The charge flowing through $C_{4}$ is $q_{4}=C_{4} \times V=4\, C V$ The series combination of $C_{1}, C_{2}$ and $C_{3}$ gives $\frac{1}{C'} =\frac{1}{C}+\frac{1}{2 C}+\frac{1}{3 C}$ $=\frac{6+3+2}{6 C}=\frac{11}{6 C}$ $C'=\frac{6 C}{11}$ $\Rightarrow C'=\frac{6 C}{11}$ Now, $C'$ and $C_{4}$ form parallel combination giving $C''=C'+C_{4}$ $=\frac{6 C}{11}+4 C=\frac{50 C}{11}$ Net charge $q=C'' V$ $=\frac{50}{11} C V$ Total charge flowing through $C_{1}, C_{2}, C_{3}$ will be $q'=q-q_{4}$ $=\frac{50}{11} C V-4 C V=\frac{6 C V}{11}$ Since, $C_{1}, C_{2}$ and $C_{3}$ are in series combination hence, charge flowing through these will be same Hence, $q_{2}=q_{1}=q_{3}=q'=\frac{6 C V}{11}$ Thus, $\frac{q_{2}}{q_{4}}=\frac{6 CV / 11}{4 CV }=\frac{3}{22}$