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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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Verified Answer
The correct answer is:
$\frac{3}{22}$
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
$$
\therefore \text { Ratio }=\frac{Q^{\prime}}{Q}=\frac{6 \mathrm{CV}}{11 \times 4 C V}=\frac{3}{22}
$$
$$
\frac{1}{C^{\prime}}=\frac{1}{C}+\frac{1}{2 C}+\frac{1}{3 C}
$$
$$
\begin{aligned}
\frac{1}{C^{\prime}} & =\frac{6+3+2}{6 C}=\frac{11}{6 C} \\
\Rightarrow \quad C & =\frac{6 C}{11}
\end{aligned}
$$
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
$$
\therefore \text { Ratio }=\frac{Q^{\prime}}{Q}=\frac{6 \mathrm{CV}}{11 \times 4 C V}=\frac{3}{22}
$$
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