Heating a Material with Negative temperature coefficient of resistance

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?

• please use mathjax Apr 6 at 11:02

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.