# Heating effect of electric current: Will this really happen? [duplicate]

I've read that the electrical resistance of a metal increases with its temperature. So if i pass current through a iron rod then it would heat up a little with time and that should increase the resistance, making it hotter as the current flows which should further increase the resistance and the iron rod would become even hotter. It seems the rod would keep on heating limitlessly at a faster rate than before. Would it really happen ? If not, why? If yes, can you provide an explanation as to why we don't observe this phenomenon in our everyday life?

## marked as duplicate by stafusa, Cosmas Zachos, Kyle Kanos, Yashas, rob♦Jun 13 '18 at 23:20

• It's not really surprising that positive feedback loops exist. Do you think this is some kind of perpetual motion machine? All that's going on is that the rod will absorb energy faster from whatever power source you're using. Energy isn't produced out of nowhere. – knzhou May 8 '18 at 11:48
• If the resistance increases, the current could decrease and the power consumption would decrease if the supply voltage is constant. Then the feedback would be negative. – Bill N Jun 11 '18 at 3:01

In everyday life power supplies produce a constant voltage, not a constant current. The power delivered to heat up the rod is $V^2 \over R$ and as temperature and resistance increase, the current drops and the power falls. Negative feedback.

• Very good point! I had implicitly assumed constant current in my answer, when in reality this is probably the most realistic reason. That said, this negative feedback loop probably still has an equilibrium point at a very high temperature, so a short-circuited piece of metal will still, of course, get very hot. – probably_someone May 8 '18 at 13:29
• @probably_someone, You can observe the phenomenon for yourself if you hook up an ordinary incandescent light bulb to power. The filament of the light bulb is metal, and it is "shorted" across the power supply, and it does indeed get very hot. – Solomon Slow May 8 '18 at 14:32
• @jameslarge I have seen this used to protect delicate components. An incandescent bulb is connected between the component and the power supply. If there is a power surge then more current flows and the bulb's resistance rises, protecting the component. Like a fuse or circuit breaker but doesn't need repairing or resetting afterwards. – RogerJBarlow May 8 '18 at 14:50
• @rogerjbarlow the feedback effect (resistivity rise with temperature) was the basis for one of Hewlett and Packard's first patents for an amplitude-stabilizing circuit in a variable frequency generator. it used a little incandescent light bulb in the feedback control circuit; the bulb temperature and hence the resistance depended on the average power flowing through it in a way that was independent of the frequency range across which the device operated. – niels nielsen May 8 '18 at 16:46
• @rogerjbarlow bulbs are occasionally used for this purpose in various musical devices. Some of the characteristics of loudspeakers in musical instrument amplifiers depend on the changes in resistance of the voice coils as they get hot, and you can simulate some of this with incandescent bulbs. if you want to drive the speaker at less domestically-hostile power levels. – tfb May 9 '18 at 14:25

It definitely happens, up to a point. There are several limiting factors here. The first is the fact that as the metal gets hotter, it radiates more and more heat away. According to the Stefan-Boltzmann law, a perfect absorber with temperature $T$ and surface area $A$ will radiate with a luminosity $L$ determined by

$$L=\sigma A T^4$$

As you can see, the radiated energy grows very quickly with temperature, so it stands to reason that there should be an equilibrium point where the energy per unit time deposited in the metal via resistive heating should be equal to the energy radiated away, at which point the metal will not heat up any further.

Another limiting factor is the fact that the resistivity-temperature relationship does not necessarily hold for extremely high temperatures. This paper goes into the details: http://ns.ihed.ras.ru/ltlp/Literature/LASER_SOLID_INTERACT/KINETIC_COEFFICIENTS/Resistivity/MilchbergPRL2364-1998.pdf

The resistivity of a metal flattens out at extremely high temperatures, as the lattice spacing imposes a limit on the mean free path of an electron.

Also, it should probably be noted that before the metal is likely to hit either of these limits, it will probably hit its melting point, turn to liquid, and cut off the current. This is the operating principle of fuses; too much drawn current will cause a tiny wire to disintegrate, shutting off the power before the wires in the house heat up enough to cause a fire.

That said, all of these limits exist at very high temperatures. There isn't much preventing a piece of metal from getting very hot under too much current. We do see this in everyday life; it's one of the more common causes of house fires.

There are, of course, practical applications where we have to use constant current.

For instance, constant current is needed to maintain a constant torque in a motor. When windings get hot, their resistance increases, which leads to the increase of the IR voltage drop. If we used a constant voltage source, the current would decrease and so would the torque. So we need to increase the voltage to keep the current and the torque unchanged.

I think the reason we don't normally observe thermal runaways in metals is because temperature coefficients of resistance are too low and, as a result, the positive feedback is not strong enough to cause thermal runaway.

Let's consider a simple setup with a constant current flowing through an iron rod.

Ambient temperature, $T_a: 20C$

Iron rod temperature due to the resistive losses $I^2R$, $T_0: 100C$

Say, we increase the resistance of the rod by 1% by stretching it a bit.

The power dissipation, P=$I^2R$, will increase by 1% as well. The temperature rise, $\Delta$T=P$\Theta$, where $\Theta$ is a thermal resistance, would increase by 1% as well. So $T_1-T_a=1.01(T_0-T_a)$ = $1.01\times80$ = 80.8C or $T_1=20+80.8=100.8$C, i.e., the temperature of the rod has changed by $0.8C$.

The corresponding change of the resistance could be calculated from this formula: $\frac {\Delta R}{R_0}$ = $\alpha \Delta T$ = $0.005\times 0.8$=$0.004$ or $0.4$%. Here $R_0$ is the initial resistance of the rod, used as a reference point, and 0.005 is the temperature coefficient for iron.

So, the initial change of the resistance of $1$% led, through the positive feedback mechanism, to the additional change of $0.4$%. Since the gain here is less than 1, we won't have thermal runaway, i.e., the temperature of the rod will quickly converge.

Yes it can happen but you'd require very high current and for long time. Heat = I×I×R R = R(1 + c∆T) now c is generally very small so the change in resistance due to change in temperature is very small. So to make a considerable change in resistance you'll need to give very large currents.

I hadn't seen V.F.'s answer when I wrote mine. His answer is the best so far. Below is my formulation.

Short answer: a monotonous increasing function needs not diverge.

Long answer: In the lab, if you take a conducting wire and send a constant current, it is very easy to just heat a little bit the wire above room temperature. Of course, this steady-state temperature depends on how strong the current is and on the resistance of the wire, which conductor it is and so on. So you certainly can melt the wire with a high current, or bring the wire at very high temperatures above room temperature like we used to do in common house lightning with tungsten based light bulbs, but you can also easily manage to keep the wire's temperature near room temperature. But the point is that while it is true that passing a constant current (or an AC with constant amplitude) increases the temperature and therefore the resistivity of the wire, which in turns intensifies again the Joule heat which increases the temperature and thus the resistivity, and so on, these effects are each time less pronounced. Mathematically what you get is a monotonous increasing function for the temperature. And such functions can, but need not, diverge. Another famous monotonous increasing function that never cross a threshold in Physics is the voltage of a charging capacitor. Its voltage always increases in time, yet it will never cross a threshold and it will remain finite.

Still not convinced? Here's a slightly more rigorous treatment. Let the room temperature be $T$ and the resistance be $R$. At first, you turn the constant current source on. Joule heat is proportional to $R I_0^2=(R_0+R_1T)I_0^2$. This causes a rise in the temperature of the wire (which may or may not be uniform depending on the boundary conditions of the wire, but it doesn't matter.). So its temperature is now $T+\Delta T_1$ and so its resistance is now $R_0+R_1(T+\Delta T_1)$.

So now the Joule heat is $[R_0+R_1(T+\Delta T_1)]I_0^2$. This rises the temperature to $T+\Delta T_1 + \Delta T_2$. Which means the resistance has increased, etc.

Overall you get that the temperature can be written as $T'=T+\Delta T_1+\Delta T_2+ ... =T + \sum_{i=1}^{\infty}\Delta T_i$. And your question boils down to "can such a series converge?". From calculus, it can. For instance it converges if $\Delta T_{i+1}/\Delta T_i<1$. In reality, it depends on the current, $\rho_1$ and possibly $\rho_2$ (indeed, because if you get to temperature much higher than room temperature, the resistance will stop to be linear and then you might want to take into consideration the nonlinear dependence of the resistance with temperature). Thus, it depends on the material and on the current. Therefore, for a given material, there could be a critical current from which the temperature should diverge (i.e. the wire will melt).