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When ‘n’ wires which are identical are connected in series, the effective resistance exceeds that when they are in parallel by X/Y $Ω$ . Then the resistance of each wire is

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$\frac{xn}{y (n^{2} - 1)}$

$\frac{yn}{x (n^{2} - 1)}$

$\frac{xn}{y (n - 1)}$

$\frac{yn}{x (n - 1)}$

answer is A.

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Detailed Solution

$R_{s} = R_{p} + (\frac{X}{Y})$

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