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

Answer 1

Answer: (B)

Charge on a capacitor is the product of capacitance and potential difference across it.
The charge flowing through $C_4$ is
$q_4=C_4\times V=4CV$
The series combination of $C_1,C_2$ and $C_3$ gives
$\frac{1}{C}=\frac{1}{C}+\frac{1}{2C}+\frac{1}{3C}$
$=\frac{6+3+2}{6C}=\frac{11}{6C}$
$\Rightarrow$ $C=\frac{6C}{11}$
Now, $C$ and $C_4$ form parallel combination giving
$C=C+C_4$ $=\frac{6C}{11}+4C=\frac{50C}{11}$
New charge $q=CV=\frac{50}{11}CV$
Total charge flowing through $C_1,C_2,C_3$ will be
$q=q-q_4=\frac{50}{11}CV-4CV=\frac{6CV}{11}$
Since, $C_1,C_2$ and $C_3$ are in, series combination,
hence charge flowing through these will be same.
Hence, $q_2=q_1=q_3=q=\frac{6CV}{11}$
Thus. $\frac{q_2}{q_4}=\frac{6CV/11}{4CV}=\frac{3}{22}$