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‘n’ conducting wires of same dimensions but having resistivity 1,2,3,.........n are connected in series the equivalent resistivity of the combination is

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

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

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

$\frac{2 n}{n + 1}$

answer is B.

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Detailed Solution

We know that

$R = \frac{ρ L}{A}$

When they are connected in series

$R_{eq} = R_{1} + R_{2} + \dots \dots \dots R_{3}$

$\Rightarrow \frac{ρ_{e q} (n L)}{A} = \frac{L}{A} (1 + 2 + 3 + \dots . n) \Rightarrow ρ_{e q} = \frac{n + 1}{2}$

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