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

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(C) Let the capacitance of each capacitor be $C = 2 \mu F$. We need an equivalent capacitance $C_{eq} = \frac{6}{13} \mu F$.If we connect $n$ capacito

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$3$ capacitors in parallel and then $4$ capacitors in series.

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(C) Let the capacitance of each capacitor be $C = 2 \mu F$. We need an equivalent capacitance $C_{eq} = \frac{6}{13} \mu F$.If we connect $n$ capacitors in parallel,their equivalent capacitance is $nC$. If we connect $m$ such parallel groups in series,the total equivalent capacitance $C_{eq}$ is given by $\frac{1}{C_{eq}} = \frac{1}{nC} + \frac{1}{nC} + \dots + \frac{1}{nC} = \frac{m}{nC}$.Given $C = 2 \mu F$,we have $C_{eq} = \frac{nC}{m} = \frac{n 	imes 2}{m} = \frac{6}{13}$.This implies $\frac{n}{m} = \frac{3}{13}$,which does not use all $7$ capacitors. Alternatively,consider a configuration where $3$ capacitors are in parallel $(C_p = 3C = 6 \mu F)$ and this group is in series with $4$ individual capacitors ($C = 2 \mu F$ each).The equivalent capacitance is $\frac{1}{C_{eq}} = \frac{1}{3C} + \frac{1}{C} + \frac{1}{C} + \frac{1}{C} + \frac{1}{C} = \frac{1}{3C} + \frac{4}{C} = \frac{1 + 12}{3C} = \frac{13}{3C}$.Thus,$C_{eq} = \frac{3C}{13} = \frac{3 	imes 2}{13} = \frac{6}{13} \mu F$.Therefore,the correct configuration is $3$ capacitors in parallel and $4$ capacitors in series.

Seven capacitors,each of capacitance $2 \mu F$,are connected in a configuration to obtain an effective capacitance of $\frac{6}{13} \mu F$. The combination that will achieve this is:

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