Seven capacitors,each of capacitance 2 mu F,a | vedclass

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(A) Let $n$ capacitors of $2\ \mu F$ be connected in parallel,and $(7-n)$ capacitors of $2\ \mu F$ be connected in series with this parallel combinati

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(A) Let $n$ capacitors of $2\ \mu F$ be connected in parallel,and $(7-n)$ capacitors of $2\ \mu F$ be connected in series with this parallel combination.The equivalent capacitance of the parallel part is $C_p = n 	imes 2\ \mu F = 2n\ \mu F$.The equivalent capacitance of the $(7-n)$ capacitors in series is $C_s = \frac{2}{7-n}\ \mu F$.The total equivalent capacitance $C_{eq}$ is given by the series combination of $C_p$ and $C_s$:$\frac{1}{C_{eq}} = \frac{1}{C_p} + \frac{1}{C_s} = \frac{1}{2n} + \frac{7-n}{2} = \frac{1 + n(7-n)}{2n} = \frac{1 + 7n - n^2}{2n}$.Given $C_{eq} = \frac{10}{11}\ \mu F$,so $\frac{1}{C_{eq}} = \frac{11}{10}$.Equating the expressions: $\frac{11}{10} = \frac{1 + 7n - n^2}{2n}$.$22n = 10 + 70n - 10n^2 \Rightarrow 10n^2 - 48n + 10 = 0 \Rightarrow 5n^2 - 24n + 5 = 0$.Solving for $n$: $5n^2 - 25n + n + 5 = 0 \Rightarrow 5n(n-5) + 1(n-5) = 0 \Rightarrow (5n+1)(n-5) = 0$.Thus,$n=5$. So,$5$ capacitors are in parallel and $7-5=2$ capacitors are in series with the parallel block. This matches the diagram in option $A$.

Seven capacitors,each of capacitance $2\ \mu F$,are connected in such a way that the total equivalent capacitance is $\frac{10}{11}\ \mu F$. Which of the following circuit diagrams represents this arrangement?

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