# Would a square piece of copper have a similar maximum current rating as a copper wire?

If I have a 3mm x 3mm x 3mm piece of copper that conducts a DC current, is there a way to determine its maximum safe current?

I looked into here, and found information for copper wires, shown below. From the chart, a diameter of 3.264 mm corresponds to an area of around 8.36 $$\mm^2\$$ (area = $$\ \pi [d/2]^2\$$), while the copper piece has a cross sectional area of 9 $$\mm^2\$$.

Would I be able to take its cross sectional area and assume similar fusing current limit for a square copper piece?

• A cylinder has less cross-sectional area to surface area than a square prism of the same cross-sectional area. So the square prism will run cooler. If the square prism has more area, then it will cooler still. Sep 29, 2020 at 4:33
• The fusing limit is based on longer wires I believe. For a cube, the fusing current will depend on how much heat can be removed at the ends. What I mean is, if the ends can remove lots of heat, then it will need more current to fuse. And if the ends are less thermally conductive than 3mm x 3mm copper square bar, then it will need less current to fuse. So the end effects need to be considered. Sep 29, 2020 at 5:01
• In other words, fusing current tables are for long and skinny wires. Short fat wires may behave differently. Sep 29, 2020 at 5:01
• That chart looks familiar. ;-) Fusing current is when the temperature gets up to the melting point of the metal, which is not what most people would call "maximum safe" current. You could define a maximum temperature that you'd like to see, and then consider heat loss by convection, radiation and conduction to the environment. Resistance of copper as a function of temperature is well known. Radiation and convection are quite nonlinear so it's not an easy problem to solve without simulation. Sep 29, 2020 at 5:24
• The table you have linked, IMO is a current when that piece of metal melts down - fusing current. No way that 9mm2 conductor carry 472A, it would be used approx. up to 50A for a insulated installation wire. Sep 29, 2020 at 7:43

A quick and dirty, approximate way to approach this would by by using an ampacity table rather than a fusing current table. As you observed, your conductor has approximately the same cross-sectional area as 8 AWG wire. Ampacity tables for 8 AWG wire show current limits ranging from 40-50 Amps or so.

Ampacity tables are designed with the goal of allowing you to use wire and be confident that it will not exceed the insulation temperature rating. So they generally assume that the wire is in conduit with other wire. Also, they are for long wires. So they do not strictly apply to the situation described in this question. But it still gives a rough idea.

If it is acceptable in your application for the copper cube to become hot, then it is likely that you could safely exceed 50 Amps. The fusing limit given in the chart in the question is DEFINITELY too much current and could risk actually melting the copper.

If your use case involves intermittent high current, the situation may not be too bad. It takes some time for a piece of copper to heat up. So it can probably withstand the fusing current for 10 ms or more without melting. But it may require a long time to cool off again before the next pulse.

In situations like this, experimentation is what I would recommend to gain confidence. Take reasonable precautions during your experiments to make sure you don't get hurt or burn down a building or something. Keep your distance, use protective gear or barriers, and make sure you can de-energize the device remotely (like with a switch or breaker or whatever). Do not put the device between you and your exit path (in case you need to run away). Keep a fire extinguisher handy.

The resistance of a wire is defined by material, length and cross sectional area,

so apart from temperature, packing, and HF effects at high frequencies, shape doesn't matter.