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(A) Let $n_1$ capacitors be connected in parallel and $n_2$ capacitors be connected in series,such that $n_1 + n_2 = 7$.For $n_1$ capacitors of capaci

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(A) Let $n_1$ capacitors be connected in parallel and $n_2$ capacitors be connected in series,such that $n_1 + n_2 = 7$.For $n_1$ capacitors of capacitance $C$ in parallel,the equivalent capacitance is $C_p = n_1 C$.For $n_2$ capacitors of capacitance $C$ in series,the equivalent capacitance is $C_s = \frac{C}{n_2}$.When these two combinations are connected in series,the total effective capacitance $C_{eff}$ is given by:$\frac{1}{C_{eff}} = \frac{1}{C_p} + \frac{1}{C_s} = \frac{1}{n_1 C} + \frac{n_2}{C} = \frac{1 + n_1 n_2}{n_1 C}$.Given $C = 2 \mu F$ and $C_{eff} = \frac{10}{11} \mu F$,we have:$\frac{11}{10} = \frac{1 + n_1 n_2}{2 n_1} \implies 22 n_1 = 10 + 10 n_1 n_2 \implies 11 n_1 = 5 + 5 n_1 n_2$.Since $n_2 = 7 - n_1$,substitute this into the equation:$11 n_1 = 5 + 5 n_1 (7 - n_1) \implies 11 n_1 = 5 + 35 n_1 - 5 n_1^2$.$5 n_1^2 - 24 n_1 + 5 = 0$.Solving the quadratic equation: $n_1 = \frac{24 \pm \sqrt{576 - 100}}{10} = \frac{24 \pm \sqrt{476}}{10}$. This does not yield an integer.Let's re-examine the circuit: The circuit consists of $n_1$ capacitors in parallel,which are then in series with $n_2$ capacitors in series.If $n_1 = 5$ and $n_2 = 2$,$C_p = 5 	imes 2 = 10 \mu F$ and $C_s = 2 / 2 = 1 \mu F$. $C_{eff} = (10 	imes 1) / (10 + 1) = 10/11 \mu F$. This matches the requirement.

Seven capacitors each of capacitance $2 \mu F$ are to be connected in a configuration to obtain an effective capacitance $\left(\frac{10}{11}\right) \mu F$. The combination is

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