# Derivation of formula for temperature coefficient of resistance

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.

Next is the formula at a specific temperature.

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.

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).$$
• 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. – chomeyama Jan 17 at 9:17