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(D) From the figure,$C_1 = C$,$C_2 = 2C$,and $C_3 = 3C$.$C_1$ and $C_2$ are connected in series across the battery of voltage $V$.The equivalent capac

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$4$

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(D) From the figure,$C_1 = C$,$C_2 = 2C$,and $C_3 = 3C$.$C_1$ and $C_2$ are connected in series across the battery of voltage $V$.The equivalent capacitance of the series combination is given by $\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{C} + \frac{1}{2C} = \frac{3}{2C}$.Thus,$C_s = \frac{2C}{3}$.The charge on the series combination is $Q_s = C_s V = \frac{2CV}{3}$.Since charge is the same on capacitors in series,the charge on $C_1$ is $Q_1 = Q_s = \frac{2CV}{3}$.Capacitor $C_3$ is connected in parallel to the battery,so the voltage across it is $V$.The charge on $C_3$ is $Q_3 = C_3 V = (3C)V = 3CV$.The ratio of charge on $C_3$ to $C_1$ is $\frac{Q_3}{Q_1} = \frac{3CV}{\frac{2CV}{3}} = \frac{9}{2} = 4.5$.

Three capacitors are connected to a battery as shown in the figure. The ratio of charge on capacitors $C_3$ and $C_1$ is (in $.5$)

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