$$R = R_0 * (1 + \alpha (T - T_0))$$
Where $R_0$ and $T_0$ are the reference resistance and temperature.
Is it then safe to say that: $$\rho = \rho _0 * (1 + \alpha (T - T_0))$$
(where $\rho _0$ is the reference resistivity)?
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It only takes a minute to sign up.Sign up to join this community
Yes; under the relationship $R=\rho L/A$, the two expressions are equivalent for a given geometry. Resistivity generally depends on temperature in a complex way, but for small temperature differences, this dependence is approximately linear, as captured by the coefficient $\alpha$. All resistivity values for materials (should) include a reference temperature because of this temperature sensitivity. In this Wikipedia list of material resistivities, for example, the reference temperature is 20°C.
Yes! But no
This expression will work for lower temperatures but gets complex for higher temperatures
Also you should know that it was by luck that you got the right answer since you forgot to consider thermal expansion of length and cross-sectional area
Now if you think about it you might think that expression is wrong it indeed is but certainly gives a good approximation of correct answer
And why is that ?
That's because coefficient of thermal expansion is of the order of 10^-6 Whereas coefficient (alpha) we are talking here is roughly of the order of 10^-3
Hope this helps !