Four capacitors with capacitances C_1 = 1 C,C | vedclass

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(C) From the circuit diagram,$C_4$ is connected directly across the battery of potential $V$. Therefore,the charge on $C_4$ is $Q_4 = C_4 V = (4C)V =

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$\frac{3}{22}$

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(C) From the circuit diagram,$C_4$ is connected directly across the battery of potential $V$. Therefore,the charge on $C_4$ is $Q_4 = C_4 V = (4C)V = 4CV$.The upper branch consists of $C_3$,$C_2$,and $C_1$ in series. The equivalent capacitance $C_{eq}$ of this branch 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 branch is $Q_{branch} = C_{eq} V = \frac{6CV}{11}$.Since $C_2$ is in series in this branch,the charge on $C_2$ is $Q_2 = Q_{branch} = \frac{6CV}{11}$.The ratio of charges is $\frac{Q_2}{Q_4} = \frac{6CV/11}{4CV} = \frac{6}{11 	imes 4} = \frac{6}{44} = \frac{3}{22}$.

Four capacitors with capacitances $C_1 = 1\ C$,$C_2 = 2\ C$,$C_3 = 3\ C$,and $C_4 = 4\ C$ are connected to a battery of potential $V$ as shown in the figure. Find the ratio of charges on $C_2$ and $C_4$.

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