# How to calculate the maximum current a wire can handle?

Is there a formula for calculating the maximum current a wire can conduct without melting? I found one:

I = αd^(3/2) in which I = fusing current, d = wire diameter in inches, and α = a constant depending upon the material, which for copper is 10244.

But if I transform this formula to calculate the minimum diameter for a 1 A current I get a diameter of around 0.5 cm, which seems pretty large to me.

• Any calculation for maximum current needs to consider "time". It's fundamental. Jan 21 at 13:15

That formula must be a bit simplistic, since it doesn't incorporate the environment of that wire, which at least for currents in your range is a dominant factor, as it enables (or not) the transport of heat away from the wire before things go bad.

But: you don't have to do any math. You honestly just stick to a standard, like e.g. given in the many tables for AWG/wire amperage you find online; you then take a table and look up the resistance for your chosen diameter, check whether the rest of what you're building can live with the voltage drop, and if it does, you go with it.

Wires getting any more than warm at 1 A is not something you even want to think about, unless saving one tenth of a cent per device is an explicit constraint you have. So, you’d rather overdimension.

I have tried to do that before but it is dead end. There are no theorical approach for this. You can,

1. Calculate power loss to heat.
2. Calculate heat resistance of insulator.

Now for insulator surface to air. If we modeled static air with no flow, the wire is much hotter than experimental result. To model the wire with air flow is too complicated. The closest things we have is passive air flow around hot cylinder that provide difference answer for each orientation. So we just use experimental/practical approach for it.

If you assume the wire is uninsulated, and in a perfect vacuum such that heat loss is only by thermal radiation;

you should be able to use the Stefan-Boltzmann law of radiation to calculate the rate of heat loss by thermal radiation using the emissivity value of copper, the surface area of a specific length, and at a temperature just below the melting point. Then figure out how much current will disipate that amount of heat and no more.

Remember when calculating the current that the resistance of metals increases when they're heated.

In real world situations of course the value may be higher or lower, being liquid cooled for instance would increase the current capacity but reflective insulation that prevents heat escaping will reduce the current.