A network of four capacitors with capacitance | vedclass

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(C) The capacitor $C_4$ is connected directly across the battery,so the potential difference across it is $V$. The charge on $C_4$ is $Q_4 = C_4 V = (

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$\frac{3}{22}$

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(C) The capacitor $C_4$ is connected directly across the battery,so the potential difference across it is $V$. The charge on $C_4$ is $Q_4 = C_4 V = (4C)V = 4CV$.The capacitors $C_3$,$C_2$,and $C_1$ are in series. Their equivalent capacitance $C_{eq}$ is given by $\frac{1}{C_{eq}} = \frac{1}{C_3} + \frac{1}{C_2} + \frac{1}{C_1} = \frac{1}{3C} + \frac{1}{2C} + \frac{1}{C} = \frac{2+3+6}{6C} = \frac{11}{6C}$.Thus,$C_{eq} = \frac{6C}{11}$.The charge on this series combination is $Q_{series} = C_{eq} V = \frac{6CV}{11}$.Since $C_2$ is part of this series branch,the charge on $C_2$ is $Q_2 = Q_{series} = \frac{6CV}{11}$.The ratio of the charges is $\frac{Q_2}{Q_4} = \frac{6CV/11}{4CV} = \frac{6}{44} = \frac{3}{22}$.

$A$ 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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