# Fuse (I²t) ratings of copper cables

There are plenty of charts/calculators giving the maximal current for given wire section/gauge (they don't always give the same results).

Those ratings are for continuous current. If there is instead a big inrush current, then I expect those wires to act like a fuse, i.e. for inrush currents, the limitation isn't the current but the energy of the inrush (expressed as I²t for a fuse (of given resistance)).

So how can I compute/find the I²t ratings for a copper wire (ideally the one for starting to damage the cable rather than the one to fuse the copper completely).

Why do I ask this? Because I'm quite unsure of my inrush current (currently designing the first prototype, so I can't measure it yet), and I'm hesitating between using a fast or a slow fuse; in case I use the slow one (53 A²s), I would like to be sure it melts before my wires (20 AWG) do.

• What is the power source? Maybe an 11 kV line? Or something DC from an AC/DC convertor at 5 V? Commented Jun 16, 2023 at 18:02
• For the use by our company, a big DC/DC converter (320V in, 24V out). As a standalone product, it can be any DC source in the 20-60V range). So potentially a source with a huge capacitor bank Commented Jun 19, 2023 at 7:44

## 2 Answers

I've seen this many times mentioned on the Wikipedia page about AWG sizes. This looks to be a complex matter (as I could expect) and dated way back in the past, as pointed out by this article (still linked from Wikipedia).

If you go to the AWG page on wikipedia, there's a table halfway down giving fusing currents for pulses of 10 s, 1 s, and 32 ms. You ought to be able to estimate an I2t from these, or at least the two shorter times. Just running my eye down the 1 s and 32 ms columns, there appears to be a ratio of about 5 or 6 between the columns, which would fit perfectly with the conductor being tested having a constant I2t, to within measurement and reporting errors.

You can alternatively calculate the heat capacity of a unit length of wire, and from that the energy needed to melt it, and with the resistance versus temperature come up with a current.