Which of the following combination of 7 identical capacitors each of $2 \mu \mathrm{F}$ gives a capacitance…
Which of the following combination of 7 identical capacitors each of $2 \mu \mathrm{F}$ gives a capacitance of $\frac{10}{11} \mu \mathrm{F}$ ?
5 in parallel and 2 in series
4 in parallel and 3 in series
3 in parallel and 4 in series
2 in parallel and 5 in series
Solution
For $\mathrm{n}$ identical capacitors connected in series, the equivalent capacitance is, $\mathrm{C}_{\mathrm{s}}=\frac{\mathrm{C}}{\mathrm{n}}$
Similarly, for $\mathrm{m}$ identical capacitors connected parallel to each other, the equivalent capacitance is, $\mathrm{C}_{\mathrm{p}}=\mathrm{mC}$
Assuming the two combinations are connected in series, the net capacitance,
$\begin{aligned} \frac{1}{\mathrm{C}_{\text {net }}} & =\frac{1}{\mathrm{mC}}+\frac{\mathrm{n}}{\mathrm{C}}=\frac{11}{10} \mu \mathrm{F} \quad \ldots\left(\because \mathrm{C}_{\text {net }}=\frac{10}{11} \mu \mathrm{F}\right) \\ \therefore \quad \text { for } \mathrm{C} & =2 \mu \mathrm{F}, \\ \frac{11 \times 2}{10} & =\frac{1}{\mathrm{~m}}+\mathrm{n}\end{aligned}$
$\therefore \quad \frac{1}{\mathrm{~m}}+\mathrm{n}=\frac{11}{5}...(i)$
Substituting the values for $\mathrm{m}$ and $\mathrm{n}$ in equation (i) from each option, the correct answer can be found to be (A).
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