Derivation of formula for temperature coefficient of resistance

Question

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.

Answer 1

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 $$ R_{t_1} = R_{t_2} + \alpha_{t_2} R_{t_2} (t_1 - t_2). $$

Answer 2

See if this derivation helps... And also for intution, we can look into temp coefficient of resistance definition. Based on the definition you have to calculate (R2-R1) /R1 (t2-t1=1 degree C as per definition) as mentioned in the below derivation:

Also based on above highlighted text, i understand that (alpha t) is temperature coefficient when the increment is based on initial temperature t