In a book I am following, it is assumed that the temperature resistance relationship is linear.

I would expect, then, that the resistance at a given temperature would be given by the formula

$$ R_{t} = mt + R_{0} $$

Where m is increase in resistance per unit temperature. Therefore by taking another known resistance at temperature A we could calculate this m

$$ m = \frac{R_{A} - R_{0}}{A - 0} $$

However the book (and everywhere else) defines the temperature coefficient as the above gradient divided by the resistance at temperature 0.

$$ \alpha = \frac{R_{A} - R_{0}}{R_{0}(A - 0)} $$

and then the formula for resistance at temperature t as

$$ R_{t} = \alpha R_{0}t + R_{0} \equiv R_{0}(\alpha t + 1) $$

My question is: what is the purpose of dividing by resistance at temperature 0 just to multiply by it again later, and what does this alpha coefficient actually 'mean' (in the same way I understand gradient m 'means' resistance increase by increasing temperature by 1 unit)?

  • \$\begingroup\$ Zero Kelvin implies zero resistance, so no intercept \$\endgroup\$ Oct 13, 2018 at 14:57
  • \$\begingroup\$ I am no expert but am not sure this is always true. In any case, I did not specify Kelvin. This could be Centigrade :) \$\endgroup\$
    – Josh
    Oct 13, 2018 at 15:01
  • \$\begingroup\$ @Josh: For the formula to be true it has to work on any temperature scale, K, C or F. The actual \$ \alpha \$ value will, of course, change for the F scale by \$ \frac {100}{180} \$. \$\endgroup\$
    – Transistor
    Oct 13, 2018 at 15:09

1 Answer 1


If we used your scheme, \$ R_{t} = mt + R_{0} \$ then we would need a unique m for every resistance value. This would be a pain.

A more useful parameter is the temperature coefficient expressed as a percentage or factor. So, given a temperature coefficient of resistance of 0.2%/K and, say, a ΔT of 12°C then I know that a 1k resistor will increase by \$ 1k \cdot 0.002 \cdot 12 \$ and 47 Ω resistor will increase by the same ratio or \$ 47 \cdot 0.002 \cdot 12 \$.

The big advantage is that the one parameter can be applied to all the resistors in a catalogue series.

I would write the formula as

$$ \alpha = \frac {R_{T1} - R_{T0}}{R_{T0} (T_1 - T_0)} $$

where \$ R_{T0} \$ is the reference resistance at temperature \$ T_0 \$, \$ R_{T1} \$ is the resistance at temperature \$ T_1 \$ and \$ \alpha \$ is the temperature coefficient.

  • \$\begingroup\$ This is great, thank you. It was that I didn't understand this coefficient is a property of the material and we then use the reference temperature for the specific resistor we are using made from this material! \$\endgroup\$
    – Josh
    Oct 13, 2018 at 14:48
  • \$\begingroup\$ I don't understand how this is different from a simple slope \$\endgroup\$ Oct 13, 2018 at 14:59
  • \$\begingroup\$ @Scott: For the OP's formula \$ R_t = mt + R_0 \$ my 1k example would give an m of \$ 2 \ \Omega\text {/K} \$ and for a 10k resistor would give \$ 20 \ \Omega\text {/K} \$. The correct formula converts this to a simple ratio independent of the resistor's value. \$\endgroup\$
    – Transistor
    Oct 13, 2018 at 15:05
  • \$\begingroup\$ Put another way: doing the calculation the way it's done means that the temperature coefficient depends on the stuff you're making your resistor out of, not what you do with that stuff. As long as the resistance element makes up the bulk of the resistance, and it doesn't matter how that resistance element is made longer or shorter or fatter or skinnier, then the one number will apply to all the resistors in a series, rather than each one individually. (Note that this may not apply for super-low resistance resistors, where, for instance, the end cap resistance may matter). \$\endgroup\$
    – TimWescott
    Oct 13, 2018 at 15:39

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