Physics - Capacitor Question with Solution | TestHub

Here $C_1,C_2$ and $C_3$ are in series.
$$\frac{1}{C^{\prime }}=\frac{1}{C}+\frac{1}{2C}+\frac{1}{3C}$$ $$\begin{array}{rl}\frac{1}{C^{\prime }}& =\frac{6+3+2}{6C}=\frac{11}{6C}\\ \Rightarrow C& =\frac{6C}{11}\end{array}$$ All the capacitors in branch number 1 is in series so the charge on each number capacitor is: $$Q^{\prime }=\frac{6}{11}CV$$ Also charge on capacitor $C_4$ is $Q=4$ V $$\therefore \text{ Ratio }=\frac{Q^{\prime }}{Q}=\frac{6\mathrm{CV}}{11\times 4CV}=\frac{3}{22}$$