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A network of four capacitors of capacity equal to $C_1 = C,$ $C_2 = 2C,$ $C_3 = 3C$ and $C_4 = 4C$ are conducted to a battery as shown in the figure. The ratio of the charges on $C_2$ and $C_4$ is
$4/7$
$3/22$
$7/4$
$22/3$
Solution
Equivalent capacitance for three capacitors $\left(\mathrm{C}_{1}, \mathrm{C}_{2} \& \mathrm{C}_{3}\right)$ in series is given by
$\frac{1}{\mathrm{C}_{\mathrm{eq}}}=\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}}+\frac{1}{\mathrm{C}_{3}}=\frac{\mathrm{C}_{2} \mathrm{C}_{3}+\mathrm{C}_{3} \mathrm{C}_{1}+\mathrm{C}_{1} \mathrm{C}_{2}}{\mathrm{C}_{1} \mathrm{C}_{2} \mathrm{C}_{3}}$
$\Rightarrow$ $C_{\mathrm{eq}}=\frac{\mathrm{C}_{1} \mathrm{C}_{2} \mathrm{C}_{3}}{\mathrm{C}_{1} \mathrm{C}_{2}+\mathrm{C}_{2} \mathrm{C}_{3}+\mathrm{C}_{3} \mathrm{C}_{1}}$
$C_{\mathrm{eq}}=\frac{C(2 C)(3 \mathrm{C})}{C(2 \mathrm{C})+(2 \mathrm{C})(3 \mathrm{C})+(3 \mathrm{C}) \mathrm{C}}=\frac{6}{11}\, \mathrm{C}$
Charge on capacitors $({C_1},{C_2}\& {C_3})$
in series $=\mathrm{C}_{\mathrm{eq}} \mathrm{V}=\frac{6 \mathrm{C}}{11} \mathrm{V}$
Charge on capacitor $C_{4}=C_{4} V=4 C V$
$\frac{\text { Charge on } \mathrm{C}_{2}}{\text { Charge on } \mathrm{C}_{4}}=\frac{\frac{6 \mathrm{C}}{11} \mathrm{V}}{4 \mathrm{CV}}=\frac{6}{11} \times \frac{1}{4}=\frac{3}{22}$
