Question

Figure given below shows a network of four capacitors of capacitance equal to $C_1=C$, $C_2=2C$, $C_3=3C$, $C_4=4C$ connected to a battery. The ratio of charges on $C_2$ and $C_4$ is :

• A. $7/4$
• B. $7/22$
• C. $3/22$
• D. $4/7$

Answer 1

The correct option is C $3/22$

The charge flowing through $C_4$ is
$q_4=C_4\times V=4C_V$
The series combination of $C_1$, $C_2$ and $C_3$
$\frac{1}{C^{\prime }}=\frac{1}{C}+\frac{1}{2C}+\frac{1}{3C}$
$\frac{1}{C^{\prime }}=\frac{6+3+2}{6C}=\frac{11}{6C}\Rightarrow C^{\prime }=\frac{6C}{11}$
Now, $C^{\prime }$ and $C_4$ form parallel combination giving
$C^{\prime \prime }=C^{\prime }+C_4=\frac{6C}{11}+4C=\frac{50C}{11}$
Net charge, $q=C^{\prime \prime }V=\frac{50}{11}CV$
Total charge flowing through $C_1$, $C_2$ and $C_3$ will be
$q^{\prime }=q-q_4=\frac{50}{11}CV-4CV=\frac{6CV}{11}$
Since, $C_1$, $C_2$, $C_3$ are in series combination. Hence, charge flowing through these will be same.
Hence, $q_2=q_1=q_3=q^{\prime }=\frac{6CV}{11}$
Thus, $\frac{q_2}{q_4}=\frac{6CV/11}{4CV}=\frac{3}{22}$