So far a resistance has been a number on a component. In a plant it is a length of copper, and you can compute it from the metal and the geometry alone.

$R = \dfrac{\rho L}{A}$, read aloud R equals rho L over A, with $\rho$ the Greek letter rho. $R$ is the resistance in ohms; $\rho$ is the material's resistivity in ohm-metres — a property of the substance, 1.68 × 10⁻⁸ Ω·m for copper and about 1.7 times that for aluminium; $L$ is the length of the run in metres; and $A$ is the conductor's cross-sectional area in square metres. Longer run, more resistance. Fatter conductor, less. Both of those are obvious, and that is the point — the relation is just bookkeeping on two obvious facts.

The units are where this lesson bites. Cable is specified in square millimetres and resistivity is quoted in ohm-metres, so the area must be converted before it goes anywhere near the fraction: $1\ \mathrm{mm^2} = 10^{-6}\ \mathrm{m^2}$. Forget it and your answer is off by a factor of a million, in a direction that looks entirely plausible on the page.

Then the second correction, the one that separates a catalogue number from a working one. Metals get MORE resistive as they heat: $R_2 = R_1\left[1 + \alpha (T_2 - T_1)\right]$ — R-two equals R-one, bracket, one plus alpha times T-two minus T-one. The subscripts here mean states, not components: 1 is the reference state where you measured $R_1$, usually the bench at 20 °C, and 2 is the state you care about, the conductor hot in service. $\alpha$ — alpha — is the material's temperature coefficient, 0.00393 per kelvin for copper. Because $T_2 - T_1$ is a DIFFERENCE, kelvin and Celsius give the same number and no conversion is needed. A copper winding that reads 10 Ω cold is genuinely about 12 Ω hot, and a protection setting calculated on the cold figure is a setting calculated on a conductor that does not exist yet.