A network of 4 capacitors is connected to a battery as shown. The ratio of the charges on capacitors…
A network of 4 capacitors is connected to a battery as shown. The ratio of the charges on capacitors $\mathrm{C}_{2}$ and $\mathrm{C}_{4}$ is
- $\frac{3}{13}$
- $\frac{3}{19}$
- $\frac{3}{17}$
- $\frac{3}{22}$
Solution
Equivalent capacitance for three capacitors $\mathrm{C}_{1}, \mathrm{C}_{2}$ and $\left.\mathrm{C}_{3}\right)$ in series is given by
$\begin{array}{l}
\frac{1}{\mathrm{C}_{\mathrm{eq}}}=\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}}+\frac{1}{\mathrm{C}_{3}}=\frac{\mathrm{C}_{2} \mathrm{C}_{3}+\mathrm{C}_{3} \mathrm{C}_{1}+\mathrm{C}_{\mathrm{C}} \mathrm{C}_{2}}{\mathrm{C}_{\mathrm{C}} \mathrm{C} \mathrm{C}_{3}} \\
\Rightarrow \mathrm{C}_{\mathrm{eq}}=\frac{\mathrm{C}_{1} \mathrm{C}_{2} \mathrm{C}_{3}}{\mathrm{C}_{1} \mathrm{C}_{2}+\mathrm{C}_{2} \mathrm{C}_{3}+\mathrm{C}_{3} \mathrm{C}_{1}} \\
\Rightarrow \mathrm{C}_{\mathrm{eq}}=\frac{\mathrm{C}(2 \mathrm{C})(3 \mathrm{C})}{\mathrm{C}(2 \mathrm{C})+(2 \mathrm{C})(3 \mathrm{C})+(3 \mathrm{C}) \mathrm{C}}=\frac{6}{11} \mathrm{C}
\end{array}$
$\Rightarrow$ Charge on capacitors $\left(\mathrm{C}_{1}, \mathrm{C}_{2} \& \mathrm{C}_{3}\right)$ in series
$=\mathrm{C}_{\mathrm{eq}} \mathrm{V}=\frac{6 \mathrm{C}}{11} \mathrm{v}$
Charge on capacitor $\mathrm{C}_{4}=\mathrm{C}_{4} \mathrm{~V}=4 \mathrm{CV}$
$\frac{\text { Charge } \mathrm{onC}_{2}}{\text { Charge } \mathrm{onC}_{4}}=\frac{\frac{6 \mathrm{C}}{11} \mathrm{~V}}{4 \mathrm{CV}}=\frac{6}{11} \times \frac{1}{4}=\frac{3}{22}$
.
Asked in: MHT CET 2020 (14 Oct Shift 1)
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