# Heating of an non-ohmic conductor

So I know that if you increase the voltage across a wire then the current will increase. But an increase in current leads to a increase in heat production though $P=I^2R$, but as the temperature increase the vibrations of the metal ions increase and so the current is more restricted implying the resistance increases. But as the resistance has increased, the heating has increased again and this will go in a cycle leading to infinite resistance and heat production so where am I going wrong?

## 3 Answers

I am not completely sure but if $R$ increases the current $i$ will decrease and $P=Vi=\frac{V^2}{R}$ will decrease as well because V is constant.

• Re, "V is constant." That would be true if the light bulb* was connected to a constant voltage power supply. In fact, the electical power grid and most modern electronic power supplies do a pretty good job of maintaining constant voltage, but that's by design. There's no physical law that says voltage must be constant in any given circuit. If the bulb were connected to a constant current power supply, then the power would continue to increase until something happened. [*This problem sounds exactly like what happens when you switch on an incandescent light bulb] Commented May 13, 2017 at 18:03

Short answer: Yes, the increase of resistivity with temperature can cause a "thermal runaway" process in which a wire fails quickly by melting from an applied current. This is one of the failure mechanisms of incandescent light bulbs.

Let's consider a small region of the wire, small enough to have an approximately uniform temperature. The volumetric heating in this region is $J^2\rho(T)$, where $J$ is the current density (which is generally a function of many things, including the wire geometry and resistance outside our small region) and $\rho(T)$ is the resistivity, a material property that depends on the temperature of that region.

Mathematically, we can apply heat transfer models to solve for the temperature in our region. For example, considering conduction to the surrounding regions along with our heat generation term, we can write $$k\nabla^2T+J^2\rho=cd\dot T$$ where $k$ is the thermal conductivity, $\nabla^2$ is the spatial second derivative, $c$ is the specific heat, $d$ is the density, and $\dot T$ is the time derivative and use the boundary conditions to solve for the temperature T as a function of time. (For simplicity, I'm considering material properties other than resistivity to be temperature independent.) What this equation says is that of the thermal energy generated by resistive heating, what isn't conducted to the neighboring regions must increase the temperature of our own region.

But more simply, I think you're wondering why, if the resistivity $\rho(T)$ increases with temperature $T$, as it generally does in metals, why the heating doesn't spiral out of control. The answer is that it can; the increasing resistivity can cause a fuse-like effect in which the wire melts and fails. (Of course, this effect can occur with sufficient current if the metal resistivity is temperature independent or even if the resistivity decreases with temperature, but the process is accelerated by a positive temperature coefficient of resistivity.) So there's nothing wrong with your intuition here.

What about the fact that $\rho(T)$ contributes to the total resistance of the circuit and thus influences $J$ in conditions other than constant-current control? Here, we must ask whether the resistance of the heated region dominates over all other resistances. If so, then under constant-voltage conditions, for example, the current decreases concomitantly with the resistance increase, and the configuration may be stable. This occurs in an incandescent light bulb, for example, for which the current changes as the bulb heats up.

If the hot region is relatively small, however, then even a large increase in local resistivity won't change the total load resistance significantly, and thermal runaway can occur even in constant-voltage conditions. This can also occur in incandescent light bulbs; any local damage or imperfection can produce a hot spot that fails, flashing much whiter than the rest of the filament as it undergoes rapid heating and melting.

(I studied this process (Appendix A here, p.61, with plenty of thermal modeling and material property discussions) in the context of thin metal films undergoing resistive heating when surrounded by water. The context was an implantable medical device containing tiny reservoirs that could be opened by an electrical pulse. In this case, the film was adjacent to liquid, so another runaway effect occurred: a vapor bubble next to the film, which cut off easy heat transfer to the liquid and thus accelerated the temperature increase.)

You're going wrong because you're making the fallacy that a monotically increasing function will eventually diverge. This is not true. Sure, it can (as Chemomechanics points out), but it needs not. Unfortunately this hasn't been addressed in neither of the two answers so far.

Details:

Suppose the room temperature is $T_0$ and the resistance of the wire is $R=R_0+R_1T_0$, because $R(T) \approx R_0+R_1T$. Note that I use a linear temperature dependence here because this is accurate for most metals when the change in temperature is small enough (but nothing prevents you from using a quadratic or cubic or more complex dependence, my point will still stand). You turn on a current. This creates a Joule heating proportional to $RI^2=(R_0+R_1T_0)I^2$, which in turn creates a rise in temperature $\Delta T_0$. The resistance of the wire is thus now equal to $R_0+R_1(T_0+\Delta T_0)$. This implies that the Joule heat is now proportional to $R_0+R_1(T_0+\Delta T_0)I^2$. Using the same arguments recursively again and again, one finds the result that:

$T=T_0+\sum_{i=0}^\infty \Delta T_i$.

Your fallacy can be resumed in the false belief that the above sum diverges regardless of $\Delta T_i$. However, from calculus, it can be shown that the series converges if $\Delta T_{i+1} / \Delta T_i <1$ for instance, in which case it would physically correspond to a standard heating of a wire a few degrees above room temperature. It is also possible that the series diverges, which would physically correspond to the melting of the wire, but it is by no mean a necessity.

Addendum for the curious:

The values of $\Delta T_{i+1} / \Delta T_i$ depend on the material's properties (such as the resistance's temperature dependence) and on the current. This is why, in general, using a high current will eventually melt the sample, and also why some materials are heated up to higher temperatures than others, for a same current.