TCR (Temperature Coefficient of Resistance): $R = R_0[1 + α(T - T₀)]$ Where: $R$ is the resistance at temperature $T$, $R_0$ is the resistance at a reference temperature $T_0$, $\alpha$ is the TCR of the material (in Ω/Ω/°C or 1/°C)

Joule's law: $P = I^2 R$ Where: $P$ is the power dissipated (in watts) $I$ is the current flowing through the resistor (in amperes) R is the resistance of the material (in ohms)

Problem: If a material has $α=-5×10^{-4}$, can that material be heated up using a current? If R decreases, I would increase since I = V/R (assuming power supply V is fixed), and P can either increase or remain the same (the increase in I compensate the decrease in R).

I'm confused which one is it?

  • $\begingroup$ please use mathjax $\endgroup$
    – hyportnex
    Apr 6 at 11:02

1 Answer 1


So, if you have thermal equilibrium, such that the rate of electrical heating is equal to the rate of convective heat loss from the resistor, the heat balance on the resistor reads $$\frac{V^2}{R}=hA(T-T_0)$$or $$\frac{V^2}{R_0}=hA(T-T_0)[1+\alpha(T-T_0)]$$where h is the convective heat transfer coefficient, A is the surface area of the resistor, and $T_0$ is the room temperature. This certainly can have a real positive root for T, depending on the other parameter values.


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