Question 674: Electrostatics & Capacitance

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 $\_\_\_\_$

Accepted Answer

The charge flowing through $C_4$ is,
$$ q_4=C_4 	imes V=4 \mathrm{CV} $$
The series combination of $C_1, C_2$ and $C_3$ gives
$$ \begin{aligned} \frac{1}{C^{\prime}} & =\frac{1}{C_0}+\frac{1}{2 C}+\frac{1}{3 C} \ \frac{1}{C^{\prime}} & =\frac{6+3+2}{6 C}=\frac{11}{6 C} \ \Rightarrow \quad C^{\prime} & =\frac{6 C}{11} \end{aligned} $$
Now, $C^{\prime}$ and $C_4$ form parallel combination giving
$$ \begin{aligned} C^{\prime \prime} & =C^{\prime}+C_4 \ & =\frac{6 C}{11}+4 C=\frac{50 C}{11} \end{aligned} $$
Net charge, $q=C^{\prime \prime} V=\frac{50}{11} \mathrm{CV}$
Total charge flowing through $C_1, C_2, C_3$ will be
$$ q^{\prime}=q-q_4=\frac{50}{11} \mathrm{CV}-4 \mathrm{CV}=\frac{6 \mathrm{CV}}{11} $$
Since $C_1, C_2$ and $C_3$ are in series combination. Hence, charge flowing through these will be same i.e. $q_2=q_1=q_3=q^{\prime}=\frac{6 \mathrm{CV}}{11}$
Thus, $\frac{q_2}{q_4}=\frac{6 \mathrm{CV} / 11}{4 \mathrm{CV}}=\frac{3}{22}$.

Answer

$22 / 3$

Answer

$3 / 22$

Answer

$7 / 4$

Answer

$4 / 7$

Electrostatics & Capacitance

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 \_\_\_\_

Choose the correct answer: