# Is there well known equation (formula) to calculate maximum practical current which copper wire of fixed diameter can handle

I found many online calculators or tables which can give maximum current for the wire. Like 1 mm diameter can handle up to 10 Amperes, 2 mm diameter wire can handle up to 30 Amperes, etc.

Is there any practical equation (formula) for this which will work in a wide range of currents $$10^{-3}-10^2$$ Amperes (1 milliampere - 100 Amperes) and 0.01-100 mm wire diameter?

Is it linear? If yes, I can make up the formula from all these tables.

I need a simplified case, like a single bare copper cylindrical (round) wire without insulation, in the air with 20-25 Celcius room temperature, etc.

It does not matter for me if the formula will be depending on wire diameter or wire area. I can change it if needed.

UPDATE 1

I tried the answer from Peter Green $$I=15.774 A^{0.6077}$$ and I got this graph

The horizontal axis is wire cross-section (A) in $$mm^2$$ and vertical axis is current in Amperes. Orange is a calculated graph. Blue is from one of the tables from the internet, here for example https://www.powerstream.com/Wire_Size.htm.

As I see the shape is matching, but values are not. For the current less than 10A, the calculated value is higher, for current less than 10A, calculated value is smaller.

If I put it like this $$I=22 A^{0.6077}$$ then it works much better for the current more than 15A, but still not very good for small currents.

Looks like using these tables is the only way :( I will try to get an equation from the table with the assumptions that true equations should be

$$I=y A^{x}$$

I will try to find y and x and post it here if I will be successful.

• Can handle before what? Getting so hot it melts? The insulation melts? (Depends on the insulation...) May 17 '20 at 14:50
• Are you trying to build a heater? May 17 '20 at 14:54
• @JonathanRSwift I need a simplified case, bare wire, no insulation, in normal atmospheric condition, room temperature (20-25 Celsius). "Before what" I would define as "wire works as normal, not too hot to increase resistance dramatically. can work with this current unlimited time without problems like melting, increase resistance, etc". Like in that tables, which can be found online. May 17 '20 at 18:40
• @SolarMike no, not heater. May 17 '20 at 18:43
• The electrical engineers I know use dedicated calculation software that takes, among other things, cable length an insulation into account. There surely is an ICIE or ISO or EN or DIN or VDE code for it, and probably the math is not straightforward.
– mart
May 18 '20 at 20:32

There are tables by various standards authorities which give current ratings for various wire gauges based on their insulation, bundling, free-air, conduit, etc. If you find one of those and plot the data you can generate your own best-fit curve. You might be able to get it down to a quadratic equation for a certain range of wire sizes.

• This is I can do. It is clear. The question is about the existing formula for simiple case. May 17 '20 at 18:42

For whatever reason standards bodies seem to prefer to give a tabulated list of values without giving any rationale to how those figures were obtained.

I once tried to do some first-principles reasoning on this.

Let $$S$$ be the surface area per unit length of the cable, $$R$$ be the resistance per unit length of the cable, $$P$$ be the power that can be safely dissipated per unit length of the cable, $$D$$ be the diameter of the cable and $$A$$ be the cross-sectional area of the cable.

First lets assume that the power can safely be dissipated by a cable is proportional to it's surface area. We also know that the surface area per unit length of a cylinder is proportional to it's diameter.

$$P \propto S \propto D$$

We also know that the cross-sectional area of a cylinder is proportional to it's diameter squared.

$$A \propto D^2$$

And that the resistance of a conductor is inversely proportional to it's cross-sectional area.

$$R \propto \frac{1}{A}$$

And finally the equation for power in terms of current and resistance.

$$P = I^2 R$$

Now lets do some substitution and simplify so we get I in terms of D or A.

$$D \propto I^2 \frac{1}{D^2}$$

$$D^3 \propto I^2$$

$$I \propto D^{1.5} \propto A^{0.75}$$

So how does this reasoning compare to actual tabulated figures in standards? not very well. It seems to overestimate the growth of current carrying capacity with diameter. My best guess is that the assumption that safe power dissipation was proportional to diameter was off.

Nevertheless this sort of reasoning suggests if a curve is to be found then it is likely of the form of a power law. When I brought this up in discussion on a forum another user ran a power law curve fit on some figures from BS7671* and found that

$$I = 15.774 A^{0.6077}$$

was a pretty close fit.

* Specifically the figures for twin and earth cable clipped directly to the surface of a wall.

• So far the best answer. Thanks a lot. I did some calculations and put them as update 1. May 19 '20 at 13:47