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(D) From the circuit diagram,capacitors $C_{1}$,$C_{2}$,and $C_{3}$ are in series with each other,and this combination is in parallel with capacitor $

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$\frac{3}{22}$

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(D) From the circuit diagram,capacitors $C_{1}$,$C_{2}$,and $C_{3}$ are in series with each other,and this combination is in parallel with capacitor $C_{4}$.Given: $C_{1} = C$,$C_{2} = 2C$,$C_{3} = 3C$,and $C_{4} = 4C$.The equivalent capacitance of the series combination of $C_{1}$,$C_{2}$,and $C_{3}$ is given by:$\frac{1}{C_{eq}} = \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}$$\Rightarrow C_{eq} = \frac{6}{11}C$The charge on the series combination (which is the same for each capacitor $C_{1}$,$C_{2}$,and $C_{3}$) is:$Q_{series} = C_{eq} V = \frac{6}{11}CV$Since $C_{2}$ is in this series branch,the charge on $C_{2}$ is $Q_{2} = \frac{6}{11}CV$.The charge on capacitor $C_{4}$ (which is in parallel with the battery) is:$Q_{4} = C_{4} V = (4C)V = 4CV$The ratio of the charges on $C_{2}$ and $C_{4}$ is:$\frac{Q_{2}}{Q_{4}} = \frac{\frac{6}{11}CV}{4CV} = \frac{6}{11 	imes 4} = \frac{6}{44} = \frac{3}{22}$

$A$ network of $4$ capacitors is connected to a battery as shown. The ratio of the charges on capacitors $C_{2}$ and $C_{4}$ is

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