Question

A network of four capacitors of capacity 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)

Charge on a capacitor is the product of capacitance and potential difference across it.
The charge flowing through $C_{4}$ is
$q_{4}=C_{4} \times V=4 C V$
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} $
$\Rightarrow $ $C=\frac{6 C}{11}$
Now, $C$ and $C_{4}$ form parallel combination giving
$C=C+C_{4}$ $=\frac{6 C}{11}+4 C=\frac{50 C}{11}$
New 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. $\frac{q_{2}}{q_{4}}=\frac{6 C V / 11}{4 C V}=\frac{3}{22}$