A network of four capacitors with capacitance | vedclass

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(B) From the circuit diagram,capacitors $C_1$,$C_2$,and $C_3$ are connected in series,and this combination is connected in parallel with capacitor $C_

Vedclass · Accepted Answer

$3/22$

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(B) From the circuit diagram,capacitors $C_1$,$C_2$,and $C_3$ are connected in series,and this combination is connected in parallel with capacitor $C_4$ across the battery of potential $V$.$1$. The equivalent capacitance $C_{eq}'$ 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}$$\Rightarrow C_{eq}' = \frac{6}{11}C$$2$. The charge $Q_{series}$ on each of the capacitors $C_1$,$C_2$,and $C_3$ is the same and is given by:$Q_{series} = C_{eq}' V = \left(\frac{6}{11}Cight) V = \frac{6}{11}CV$$3$. The charge $Q_4$ on capacitor $C_4$ is:$Q_4 = C_4 V = (4C) V = 4CV$$4$. The ratio of the charge on $C_2$ to the charge on $C_4$ is:$\frac{Q_2}{Q_4} = \frac{Q_{series}}{Q_4} = \frac{\frac{6}{11}CV}{4CV} = \frac{6}{11 	imes 4} = \frac{6}{44} = \frac{3}{22}$

$A$ network of four capacitors with capacitances $C_1 = C$,$C_2 = 2C$,$C_3 = 3C$,and $C_4 = 4C$ is connected to a battery as shown in the figure. The ratio of the charges on $C_2$ and $C_4$ is

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