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https://www.electrical4u.com/temperature-coefficient-of-resistance/

Referring to this article, I'd like to compile two key formulas for the temperature coefficient of resistance.

First is the formula at zero temperature.

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Next is the formula at a specific temperature.

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To put it more simply, the reciprocal of the temperature coefficient of resistance in the second formula is equal to the inferred absolute zero temperature, $ t_o $. So the second formula can be interpreted as, if you want to get the temperature coefficient of resistance at a given temperature that isn't zero you have to add that value to the inferred absolute zero temperature.

I'd like to know, is there a derivation for these formulas? I mean, I totally get how you could define the first formula to just be a reciprocal of that quantity. But with the second, I don't get the intuition behind just adding in the denominator. I mean, it doesn't have to be a derivation but at least some intuition on why it is like that so it would help me remember it other than just straight up memorizing.

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Does this help you? $$ \frac{R_{t_1}}{R_o} = \frac{t_o + t_1}{t_o + 0}, \tag{1} $$ $$ \frac{R_{t_2}}{R_o} = \frac{t_o + t_2}{t_o + 0}, \tag{2} $$ $$ (1) \div (2) \Rightarrow \frac{R_{t_1}}{R_{t_2}} = \frac{t_o + t_1}{t_o + t_2} = \frac{(t_o + t_2) + (t_1 - t_2)}{t_o + t_2} = 1 + \frac{t_1 - t_2}{t_o + t_2} = 1 + \alpha_{t_2} (t_1 - t_2), $$ where \$\alpha_{t_2} = 1 / (t_o + t_2)\$, and $$ R_{t_1} = R_{t_2} + \alpha_{t_2} R_{t_2} (t_1 - t_2). $$

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  • \$\begingroup\$ I do understand the process behind that formula and how it was derived from similar triangles. My question pertains to, what is the origin of the formula used to relate the temperature coefficient and inferred temperature? Just like how your first formula came from similar triangles, is there an origin to the formula of temperature coefficient? \$\endgroup\$ – AndroidV11 Jan 17 at 6:30
  • \$\begingroup\$ If you mean physical models by 'origin', I have no idea. Anyway, \$\alpha_{t_2} = 1 / (t_o + t_2) = 1 / (t_2 + (1 / \alpha_o))\$ gives your second formula by substituting \$t_2\$ with \$t\$ and I have no more idea beyond the formula, sorry. \$\endgroup\$ – chomeyama Jan 17 at 9:17
  • \$\begingroup\$ Not necessarily a physical model, but where it came from. Is it a definition or derived from another formula? I have no trouble accepting that the first formula I mentioned is a definition. Like X is the reciprocal of Y. There are so many concepts just defined like that, example is conductance to resistance. My problem is the second formula I mentioned, like adding to the denominator is quite specific. I intend to know its derivation to understand why it is like that. \$\endgroup\$ – AndroidV11 Jan 17 at 10:19

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