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The network of four capacitors is connected to a battery as shown in the figure. The ratio of charges on capacitors $C_2$ and $C_4$ is
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$\frac{3}{22}$
Equivalent capacitance for three capacitors $C_1, C_2$ and $C_3$ in series is given by
$\begin{aligned} & \frac{1}{C_{e q}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}=\frac{C_2 C_3+C_3 C_1+C_1 C_2}{C_1 C_2 C_3} \\ & \Rightarrow C_{e q}=\frac{C_1 C_2 C_3}{C_1 C_2+C_2 C_3+C_3 C_1} \\ & \Rightarrow C_{e q}=\frac{\mathrm{C}(2 \mathrm{C})(3 \mathrm{C})}{(3 \mathrm{C})(2 \mathrm{C})+(2 \mathrm{C})+(3 \mathrm{C}) \mathrm{C}}=\frac{6}{11} \mathrm{C} \\ & \Rightarrow \text { Charge on capacitors }\left(C_1, C_2 \& C_3\right) \text { in series } \\ & =C_{e q} V=\frac{6}{11} C V\end{aligned}$
$=C_{e q} V=\frac{6}{11} C V$
Charge on capacitor $C_4=C_4 V=4 C V$
$\frac{\text { Charge on } C_2}{\text { Charge on } C_4}=\frac{\left(\frac{6 C}{11}\right) V}{(4 C) V} \times \frac{1}{4}=\frac{3}{22}$
$\begin{aligned} & \frac{1}{C_{e q}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}=\frac{C_2 C_3+C_3 C_1+C_1 C_2}{C_1 C_2 C_3} \\ & \Rightarrow C_{e q}=\frac{C_1 C_2 C_3}{C_1 C_2+C_2 C_3+C_3 C_1} \\ & \Rightarrow C_{e q}=\frac{\mathrm{C}(2 \mathrm{C})(3 \mathrm{C})}{(3 \mathrm{C})(2 \mathrm{C})+(2 \mathrm{C})+(3 \mathrm{C}) \mathrm{C}}=\frac{6}{11} \mathrm{C} \\ & \Rightarrow \text { Charge on capacitors }\left(C_1, C_2 \& C_3\right) \text { in series } \\ & =C_{e q} V=\frac{6}{11} C V\end{aligned}$
$=C_{e q} V=\frac{6}{11} C V$
Charge on capacitor $C_4=C_4 V=4 C V$
$\frac{\text { Charge on } C_2}{\text { Charge on } C_4}=\frac{\left(\frac{6 C}{11}\right) V}{(4 C) V} \times \frac{1}{4}=\frac{3}{22}$
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