Capacitors with capacitances C_1 = C,C_2 = 2C | vedclass

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(B) The circuit consists of two parallel branches connected to a battery of potential $V$.Branch $A$ contains capacitors $C_1$,$C_2$,and $C_3$ in seri

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$3/22$

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(B) The circuit consists of two parallel branches connected to a battery of potential $V$.Branch $A$ contains capacitors $C_1$,$C_2$,and $C_3$ in series. The equivalent capacitance of this branch is given by:$\frac{1}{C_A} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} = \frac{1}{C} + \frac{1}{2C} + \frac{1}{3C} = \frac{6+3+2}{6C} = \frac{11}{6C}$Thus,$C_A = \frac{6C}{11}$.The charge on branch $A$ is $Q_A = C_A V = \frac{6CV}{11}$. Since $C_1, C_2, C_3$ are in series,the charge on each is the same,so $Q_{C2} = Q_A = \frac{6CV}{11}$.Branch $B$ contains only capacitor $C_4$ in parallel with branch $A$. The potential across $C_4$ is $V$,so the charge on $C_4$ is $Q_{C4} = C_4 V = 4CV$.The ratio of charges on $C_2$ and $C_4$ is:$\frac{Q_{C2}}{Q_{C4}} = \frac{6CV/11}{4CV} = \frac{6}{44} = \frac{3}{22}$.

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

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