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(D) Let $n$ be the number of capacitors in parallel,each of capacitance $C = 2 \mu F$. The equivalent capacitance of this parallel branch is $C_p = nC

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$4$ in parallel and $5$ in series

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(D) Let $n$ be the number of capacitors in parallel,each of capacitance $C = 2 \mu F$. The equivalent capacitance of this parallel branch is $C_p = nC = 2n \mu F$.Let $m$ be the number of such branches connected in series. The total equivalent capacitance $C_{eq}$ is given by $\frac{1}{C_{eq}} = \sum \frac{1}{C_p} = \frac{m}{C_p}$.Thus,$C_{eq} = \frac{C_p}{m} = \frac{2n}{m}$.We are given $C_{eq} = \frac{10}{11} \mu F$ and the total number of capacitors is $n 	imes m = 7$.From $C_{eq} = \frac{2n}{m} = \frac{10}{11}$,we get $\frac{n}{m} = \frac{5}{11}$,which implies $11n = 5m$.Since $n 	imes m = 7$,we test the options. For option $C$,if we have $3$ in parallel $(n=3)$ and $4$ in series $(m=4)$,the total capacitors used is $3 	imes 4 = 12$ (Incorrect).Re-evaluating the combination: If we have $5$ capacitors in parallel $(C_p = 5 	imes 2 = 10 \mu F)$ and these are connected in series with another group,the calculation $\frac{10}{11} \mu F$ is achieved by $5$ capacitors in parallel $(10 \mu F)$ in series with $11$ capacitors (not possible with $7$ total).Given the constraint of $7$ capacitors: If $n=5$ and $m=11$ is not possible,let's check $C_{eq} = \frac{10}{11} \mu F$. The correct combination for $7$ capacitors of $2 \mu F$ is $5$ capacitors in parallel $(10 \mu F)$ in series with $2$ capacitors in parallel $(4 \mu F)$,but that is not the standard series-parallel formula. The correct configuration is $5$ capacitors in parallel $(10 \mu F)$ in series with $1$ capacitor $(2 \mu F)$ resulting in $\frac{10 	imes 2}{10+2} = \frac{20}{12} = 1.66 \mu F$. Given the options provided,the question implies a specific grouping. Based on $\frac{10}{11} \mu F$,the configuration is $5$ capacitors in parallel $(10 \mu F)$ in series with $11$ capacitors,which is impossible with $7$. However,if the question meant $5$ capacitors in parallel $(10 \mu F)$ and $1$ in series,it doesn't match. The closest mathematical fit for $7$ capacitors is $5$ in parallel $(10 \mu F)$ and $2$ in series $(1 \mu F)$,giving $\frac{10}{11} \mu F$.

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

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