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Question: On each side of a polygon of n sides a capacitor of capacitance C is placed as shown in figure. Equi...

On each side of a polygon of n sides a capacitor of capacitance C is placed as shown in figure. Equivalent capacitance across A and B is

A

(n−1)Cn\frac{(n-1)C}{n}

B

nCn−1\frac{nC}{n-1}

C

(n-1)C

D

nC

Answer

nCn−1\frac{nC}{n-1}

Explanation

Solution

The circuit between adjacent vertices A and B consists of two parallel branches. One branch is the direct capacitor C connecting A and B. The other branch is the series combination of the remaining (n-1) capacitors forming the rest of the polygon's perimeter. The equivalent capacitance of (n-1) capacitors in series is C/(n−1)C/(n-1). Since these two branches are in parallel, the total equivalent capacitance is the sum of their individual capacitances: C+C/(n−1)=C(1+1/(n−1))=C((n−1+1)/(n−1))=nC/(n−1)C + C/(n-1) = C(1 + 1/(n-1)) = C((n-1+1)/(n-1)) = nC/(n-1).