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(B) The circuit consists of two parallel branches connected to a battery of potential $V$.Branch $1$ contains capacitors $C_1, C_2,$ and $C_3$ in seri

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$\frac{3}{22}$

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(B) The circuit consists of two parallel branches connected to a battery of potential $V$.Branch $1$ contains capacitors $C_1, C_2,$ and $C_3$ in series. The equivalent capacitance $C_{eq1}$ of this branch is given by:$\frac{1}{C_{eq1}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} = \frac{1}{C} + \frac{1}{2C} + \frac{1}{3C} = \frac{6+3+2}{6C} = \frac{11}{6C}$Thus,$C_{eq1} = \frac{6C}{11}$.The charge $Q_1$ on each capacitor in this series branch is $Q_1 = C_{eq1} V = \frac{6CV}{11}$.Since $C_1, C_2,$ and $C_3$ are in series,the charge on $C_2$ is $Q_{C2} = Q_1 = \frac{6CV}{11}$.Branch $2$ contains only capacitor $C_4$ connected directly across the battery. The charge on $C_4$ is $Q_{C4} = C_4 V = 4CV$.The ratio of the charges on $C_2$ and $C_4$ is:$\frac{Q_{C2}}{Q_{C4}} = \frac{\frac{6CV}{11}}{4CV} = \frac{6}{11 	imes 4} = \frac{6}{44} = \frac{3}{22}$.

$A$ network of four capacitors with capacitances $C_1 = C$,$C_2 = 2C$,$C_3 = 3C$,and $C_4 = 4C$ is connected to a battery of potential $V$ as shown in the figure. The ratio of the charges on $C_2$ and $C_4$ is:

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