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

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

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$5$ in parallel and $2$ in series

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(C) Let $n$ capacitors be connected in parallel,each of capacitance $C = 2 \mu F$. The equivalent capacitance of this parallel group is $C_p = nC = 2n \mu F$.Let $m$ such parallel groups be connected in series. The total number of capacitors is $N = n 	imes m = 7$.The equivalent capacitance of $m$ such groups in series is given by $\frac{1}{C_{eq}} = \frac{1}{C_p} + \frac{1}{C_p} + ... (m 	ext{ times}) = \frac{m}{C_p}$.Thus,$C_{eq} = \frac{C_p}{m} = \frac{2n}{m}$.Given $C_{eq} = \frac{10}{11} \mu F$,we have $\frac{2n}{m} = \frac{10}{11}$,which simplifies to $\frac{n}{m} = \frac{5}{11}$.This implies $11n = 5m$. Since $n 	imes m = 7$,this does not yield integer solutions for $n$ and $m$ directly. Let us re-evaluate the combination: If we have $n$ parallel branches each containing $m$ capacitors in series,the equivalent capacitance is $C_{eq} = \frac{n}{m} C$.Substituting $C = 2 \mu F$ and $C_{eq} = \frac{10}{11} \mu F$,we get $\frac{10}{11} = \frac{n}{m} 	imes 2$,so $\frac{n}{m} = \frac{5}{11}$.Given $n+m$ is not necessarily $7$,but the total capacitors $n 	imes m = 7$ is not correct. The total capacitors $N = n 	imes m = 7$ is not possible for these ratios. Checking option $C$: $5$ in parallel and $2$ in series would mean $n=5, m=2$,$C_{eq} = \frac{5}{2} 	imes 2 = 5 \mu F$. Checking option $D$: $4$ in parallel and $3$ in series would mean $n=4, m=3$,$C_{eq} = \frac{4}{3} 	imes 2 = 2.66 \mu F$. Wait,if we have $n$ capacitors in series and $m$ such rows in parallel,$C_{eq} = \frac{m}{n} C = \frac{m}{n} 	imes 2 = \frac{10}{11} \implies \frac{m}{n} = \frac{5}{11}$. Total capacitors $m 	imes n = 7$ is not possible. Re-reading: If $5$ capacitors are in series and $2$ are in parallel,$C_{eq} = \frac{2}{5} 	imes 2 = 0.8 \mu F$. If $2$ capacitors are in series and $5$ are in parallel,$C_{eq} = \frac{5}{2} 	imes 2 = 5 \mu F$. Actually,for $C_{eq} = \frac{10}{11} \mu F$ with $C=2 \mu F$,we need $\frac{m}{n} = \frac{5}{11}$. This is not possible with $7$ capacitors. However,if the question implies a mixed grouping,the closest match is $5$ in series and $2$ in parallel.

Seven capacitors,each of capacitance $2 \mu F$,are to be connected to obtain an equivalent capacitance of $\left(\frac{10}{11}\right) \mu F$. Which of the following combinations is possible?

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