Seven capacitors,each of capacity 2,mu F,are | vedclass

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(A) Let the capacity of each capacitor be $C = 2\,\mu F$. We need a total equivalent capacity $C_{eq} = \frac{10}{11}\,\mu F$.Let $n$ capacitors be co

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(A) Let the capacity of each capacitor be $C = 2\,\mu F$. We need a total equivalent capacity $C_{eq} = \frac{10}{11}\,\mu F$.Let $n$ capacitors be connected in parallel and $m$ capacitors be connected in series with this parallel combination.The equivalent capacity of $n$ capacitors in parallel is $C_p = nC = n(2)\,\mu F$.The total equivalent capacity $C_{eq}$ is given by the series combination of $C_p$ and $m$ capacitors in series:$\frac{1}{C_{eq}} = \frac{1}{C_p} + \frac{m}{C} = \frac{1}{2n} + \frac{m}{2} = \frac{1 + nm}{2n}$.Given $C_{eq} = \frac{10}{11}\,\mu F$,we have $\frac{2n}{1 + nm} = \frac{10}{11}$.$22n = 10 + 10nm \implies 11n = 5 + 5nm \implies 5nm = 11n - 5$.Since the total number of capacitors is $n + m = 7$,we have $m = 7 - n$.Substituting $m$: $5n(7 - n) = 11n - 5 \implies 35n - 5n^2 = 11n - 5 \implies 5n^2 - 24n - 5 = 0$.Solving the quadratic equation: $5n^2 - 25n + n - 5 = 0 \implies 5n(n - 5) + 1(n - 5) = 0 \implies (5n + 1)(n - 5) = 0$.Thus,$n = 5$ and $m = 7 - 5 = 2$.This means $5$ capacitors are in parallel and $2$ capacitors are in series with this parallel group. This matches the arrangement in figure $(a)$.

Seven capacitors,each of capacity $2\,\mu F$,are to be connected to have a total capacity of $\frac{10}{11}\,\mu F$. Which of the following figures shows the correct arrangement?

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