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(A) Let $m$ capacitors be connected in parallel and $n$ capacitors be connected in series,such that the total number of capacitors is $m + n = 7$.For

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$5$ in parallel and $2$ in series

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(A) Let $m$ capacitors be connected in parallel and $n$ capacitors be connected in series,such that the total number of capacitors is $m + n = 7$.For $m$ capacitors in parallel,the equivalent capacitance is $C_p = mC$.For $n$ capacitors in series,the equivalent capacitance is $C_s = C/n$.When these two groups are connected in series,the total equivalent capacitance $C_{	ext{net}}$ is given by:$\frac{1}{C_{	ext{net}}} = \frac{1}{C_p} + \frac{1}{C_s} = \frac{1}{mC} + \frac{n}{C} = \frac{1 + mn}{mC}$.Given $C_{	ext{net}} = \frac{10}{11} \mu F$ and $C = 2 \mu F$:$\frac{11}{10} = \frac{1 + mn}{2m} \implies \frac{11}{5} = \frac{1 + mn}{m} = \frac{1}{m} + n$.Testing the options where $m + n = 7$:If $m = 5$ and $n = 2$,then $\frac{1}{5} + 2 = 0.2 + 2 = 2.2 = \frac{11}{5}$.This matches the required value. Thus,$5$ capacitors in parallel and $2$ in series is the correct combination.

Which of the following combinations of $7$ identical capacitors each of $2 \mu F$ gives a capacitance of $\frac{10}{11} \mu F$?

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