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(B) polygon with $n$ sides has $n$ capacitors. One capacitor is placed directly between points $A$ and $B$.The remaining $(n - 1)$ capacitors are conn

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$\frac{nC}{n - 1}$

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(B) polygon with $n$ sides has $n$ capacitors. One capacitor is placed directly between points $A$ and $B$.The remaining $(n - 1)$ capacitors are connected in series along the other path between $A$ and $B$.The equivalent capacitance of these $(n - 1)$ capacitors in series is given by:$C_{series} = \frac{C}{n - 1}$Now,this equivalent capacitor is in parallel with the single capacitor $C$ connected directly between $A$ and $B$.The total equivalent capacitance $C_{eq}$ is:$C_{eq} = C + C_{series} = C + \frac{C}{n - 1}$$C_{eq} = \frac{C(n - 1) + C}{n - 1} = \frac{Cn - C + C}{n - 1} = \frac{nC}{n - 1}$

On each side of a polygon of $n$ sides,a capacitor of capacitance $C$ is placed as shown in the figure. The equivalent capacitance across $A$ and $B$ is:

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