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

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(B) Let the capacitance of each capacitor be $C = 2\,\mu F$. We need an equivalent capacitance $C_{eq} = \frac{6}{13}\,\mu F$.Consider the configurati

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(B) Let the capacitance of each capacitor be $C = 2\,\mu F$. We need an equivalent capacitance $C_{eq} = \frac{6}{13}\,\mu F$.Consider the configuration in figure $B$. It consists of $3$ capacitors in parallel,which are then connected in series with $4$ capacitors in series.$1$. The equivalent capacitance of $3$ capacitors in parallel is $C_p = 3C = 3 	imes 2 = 6\,\mu F$.$2$. The equivalent capacitance of $4$ capacitors in series is $C_s = \frac{C}{4} = \frac{2}{4} = 0.5\,\mu F$.$3$. Now,$C_p$ and $C_s$ are in series. The total equivalent capacitance is:$\frac{1}{C_{eq}} = \frac{1}{C_p} + \frac{1}{C_s} = \frac{1}{6} + \frac{1}{0.5} = \frac{1}{6} + 2 = \frac{1 + 12}{6} = \frac{13}{6}\,\mu F^{-1}$.Therefore,$C_{eq} = \frac{6}{13}\,\mu F$. This matches the required value.

Seven capacitors,each of capacitance $2\,\mu F$,are to be connected in a configuration to obtain an effective capacitance of $\left( \frac{6}{13} \right)\,\mu F$. Which of the combinations,shown in figures below,will achieve the desired value?

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