A network of four capacitors with capacitance | vedclass

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(B) The capacitor $C_4$ is connected directly across the battery of voltage $V$. Therefore,the charge on $C_4$ is $q_4 = C_4 	imes V = 4CV$.The capac

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$\frac{3}{22}$

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(B) The capacitor $C_4$ is connected directly across the battery of voltage $V$. Therefore,the charge on $C_4$ is $q_4 = C_4 	imes V = 4CV$.The capacitors $C_1$,$C_2$,and $C_3$ are connected in series with each other,and this series combination is connected in parallel with $C_4$ across the battery.Let the equivalent capacitance of the series combination of $C_1$,$C_2$,and $C_3$ be $C'$.$\frac{1}{C'} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} = \frac{1}{C} + \frac{1}{2C} + \frac{1}{3C} = \frac{6+3+2}{6C} = \frac{11}{6C}$.Thus,$C' = \frac{6C}{11}$.The charge $q'$ flowing through the series combination of $C_1$,$C_2$,and $C_3$ is given by $q' = C' 	imes V = \frac{6C}{11} 	imes V = \frac{6CV}{11}$.Since $C_1$,$C_2$,and $C_3$ are in series,the charge on each of them is the same,which is $q'$.Therefore,the charge on $C_2$ is $q_2 = q' = \frac{6CV}{11}$.The ratio of the charges on $C_2$ and $C_4$ is $\frac{q_2}{q_4} = \frac{6CV/11}{4CV} = \frac{6}{44} = \frac{3}{22}$.

Element	Co-factor
$A$. $-1$	$(1)$ $-2$
$B$. $1$	$(2)$ $32$
$C$. $3$	$(3)$ $4$
$D$. $6$	$(4)$ $6$
	$(5)$ $-6$

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

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