Question

The figure given below shows a network of four capacitors of capacitance equal to $C_1=C,C_2=2C,C_3=3C,C_4=4C$ are connected to a battery. If the ratio of charges on $C_{2}and{C}_4$ is $\frac{p}{q}$ then what is the value of $\left(p+q\right)$ , where $p$ and $q$ are smallest positive integers.

Answer 1

Answer: 25

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
$C"=C\prime +C_4=\frac{6C}{11}+4C=\frac{50C}{11}$
Net Charge, $q=C"V=\frac{50}{11}CV$
$q^{\prime }=q-q_4=\frac{50}{11}CV-4CV=\frac{6CV}{11}$
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, $\frac{q_2}{q_4}=\frac{\frac{6CV}{11}}{4CV}=\frac{3}{22}$