A total of n conducting wires of same dimensions but having resistivities 1, 2, 3,...n are connected in series. The equivalent resistivity of the combination is

\frac{n (n + 1)}{2}

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\frac{n + 1}{2}

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\frac{n + 1}{2 n}

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\frac{2 n}{n + 1}

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Solution

The correct option is C $\frac{n (n + 1)}{2}$
$R_{1} = \frac{ρ_{1} L}{A}$

$R_{2} = \frac{ρ_{2} L}{A}$

$R_{n} = \frac{ρ_{n} L}{A}$

$R_{e q} = \sum_{i}^{n} = \frac{ρ_{i} L}{A}$

$\frac{ρ_{e q} L}{A} = \frac{L}{A} [ρ_{1} + ρ_{2} . . . . . + ρ_{n}]$

$ρ_{e q} = [1 + 2 + 3..... + n] = \frac{n (n + 1)}{2}$

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