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(C) From the circuit diagram,the capacitors $C_1 = 1C$,$C_2 = 2C$,and $C_3 = 3C$ are connected in series. The equivalent capacitance $C_{eq}$ of this

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$\frac{3}{22}$

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(C) From the circuit diagram,the capacitors $C_1 = 1C$,$C_2 = 2C$,and $C_3 = 3C$ are connected in series. The equivalent capacitance $C_{eq}$ of this series branch 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}$$C_{eq} = \frac{6}{11}C$Since this branch is connected in parallel with the capacitor $C_4 = 4C$ across the battery of voltage $V$,the potential difference across the series branch is $V$.The charge on the series branch (which is the same for $C_1, C_2,$ and $C_3$) is:$Q_{series} = C_{eq} V = \frac{6}{11}CV$The charge on capacitor $C_4$ is:$Q_4 = C_4 V = (4C)V = 4CV$The ratio of the charge on $C_2$ to the charge on $C_4$ is:$\frac{Q_2}{Q_4} = \frac{\frac{6}{11}CV}{4CV} = \frac{6}{11 	imes 4} = \frac{6}{44} = \frac{3}{22}$

The network of four capacitors is connected to a battery as shown in the figure. The ratio of charges on capacitors $C_2$ and $C_4$ is

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