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DEvens

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I think it should be perimeter, but all textbooks show that R = ρ*L/S, where R is the resistance, ρ is specific resistance, L is the resistor length, and S is the cross section.

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NascentOxygen

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I presume roughly proportional to perimeter for frequencies/conditions where skin effect dominates. (I think I've read that skin effect can be seen on massive power transmission cables, such that current density near the core may be, say, half what it is nearer the surface. So a high tensile steel core need not reduce an aluminium cable's resistance appreciably.)I though all metals have this property... Anyway if I take these types of materials, where the flow of electrons is at the skin, and double the width of the wire, will I get half the resistance or a quarter of the resistance? That is, is resistance linearly dependant on the cross section or on the perimeter?

That's because they are dealing with ordinary bulk properties under DC and low frequency conditions.all textbooks show that R = ρ*L/S, where R is the resistance, ρ is specific resistance, L is the resistor length, and S is the cross section.

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nasu

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If the skin depth is small compared with the diameter yes, the resistance is approximately inverse proportional to the diameter (and not diameter squared).

I think it should be perimeter, but all textbooks show that R = ρ*L/S, where R is the resistance, ρ is specific resistance, L is the resistor length, and S is the cross section.

See here for example:

http://chemandy.com/calculators/round-wire-ac-resistance-calculator.htm

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Consider the flow of electrons through a wire analogous to the flow of water through a pipe.

More the cross-sectional area, more the amount of water that can flow in a specific time. Lesser the CSA, lesser the amount of water, i.e., more the resistance.

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