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

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(A) Let $n$ capacitors be connected in parallel,each of $2\,\mu F$,so their equivalent capacitance is $C_p = n 	imes 2\,\mu F = 2n\,\mu F$.Let $m$ su

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$5$ in parallel,$2$ in series

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(A) Let $n$ capacitors be connected in parallel,each of $2\,\mu F$,so their equivalent capacitance is $C_p = n 	imes 2\,\mu F = 2n\,\mu F$.Let $m$ such parallel groups be connected in series,where the total number of capacitors is $N = n 	imes m = 7$.The equivalent capacitance of $m$ such groups in series is given by $C_{eq} = \frac{C_p}{m} = \frac{2n}{m} = \frac{10}{11}$.From this,$\frac{n}{m} = \frac{5}{11}$,which implies $11n = 5m$.Since $n 	imes m = 7$,we test the combinations:If $n=2$ and $m=5$,then $N = 2 	imes 5 = 10$ (Incorrect,we need $7$ capacitors).If we consider a mixed combination where $5$ capacitors are in parallel $(C_p = 5 	imes 2 = 10\,\mu F)$ and this group is in series with $2$ capacitors in parallel $(C_p' = 2 	imes 2 = 4\,\mu F)$,this does not yield $10/11$.However,checking option $D$: $2$ capacitors in parallel give $C_p = 4\,\mu F$. If we have $5$ such groups in series,$C_{eq} = 4/5 = 0.8\,\mu F$.Correct approach: Let $n$ capacitors be in parallel $(2n)$ and $m$ such sets be in series. If $n=5$ and $m=11$ (not possible). Actually,for $n=5$ capacitors in parallel $(10\,\mu F)$ in series with $1$ capacitor $(2\,\mu F)$ is not it. Re-evaluating: $5$ capacitors in parallel $(10\,\mu F)$ in series with $2$ capacitors in series $(1\,\mu F)$: $C_{eq} = (10 	imes 1) / (10 + 1) = 10/11\,\mu F$. Total capacitors = $5 + 2 = 7$. Thus,option $A$ is correct.

Seven capacitors,each of capacitance $2\,\mu F$,are to be connected to obtain a total equivalent capacitance of $10/11\,\mu F$. Which of the following combinations is possible?

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