A network of four capacitors capacity equal $C_1=C, C_2=2 C, C_3=3 C$ and $C_4=$ $4 C$ are conducted to a battery as shown in the figure. The ratio of the change on $C_2$ and $C_4$ is:

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Aicharya · Accepted Answer

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

A network of four capacitors capacity equal $C_1=C, C_2=2 C, C_3=3 C$ and $C_4=$ $4 C$ are conducted to a…

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