Why does a fuse voltage rating only signify the dielectric strength?

From what I understand, a typical fuse wire will melt once it's current rating is breached by some factor.

However, a fuse also has a voltage rating, which signifies its dielectric strength. From what I've read, this voltage rating means that, "beyond this particular voltage, the fuse may conduct electricity even if the current being drawn is much much higher".

Does this mean that 1A @ 5V and 1A @ 100V builds the same amount of heat in a fuse wire? It shouldn't theoretically since the power rating P=VI is way different, right? To put it in another way, can I use a 1A @ 220V fuse in an environment of 1A @ 5V?

• The dielectric is air - once the fuse has blown. Commented Jan 15, 2021 at 11:18
• So does that mean that 1A @ 5V and 1A @ 100V builds the same amount of heat in a fuse wire? Commented Jan 15, 2021 at 11:25
• I think this answers my question: electronics.stackexchange.com/a/21797/120711 It doesn't depend on voltage at all, since the resistance is super low. Commented Jan 15, 2021 at 11:36
• Ask yourself, where are you measuring the voltage? across the fuse or from one end of the fuse to ground?
– RJR
Commented Jan 15, 2021 at 12:00
• The point is the fuse knows nothing about the voltage of the circuit until the fuse blows (because until then it's just a piece of wire). Once the fuse starts to rupture it learns all about the voltage because arcing allows current to continue flowing. The fuse now needs to separate far enough that arcing is quenched, current stops flowing and the circuit is made safe. Commented Jan 15, 2021 at 13:29

To put it in another way, can I use a 1A @ 220V fuse in an environment of 1A @ 5V?

Yes you can. It's the 1 amp rating that determines at what level of current the fuse initially melts.

However, if you used a 100 volt rated fuse on a 250 VAC application, then it may blow but continue to arc and conduct unsafely.

• So does that mean that 1A @ 5V and 1A @ 100V builds the same amount of heat in a fuse wire? Commented Jan 15, 2021 at 11:25
• It entirely does @judepereira. The voltage isn't across the fuse - the actual volt-drop across the fuse is milli-volts usually but, as the fuse material warms (due to current i.e. $I^2R$) then, at some point the material melts. Commented Jan 15, 2021 at 11:37
• Thanks @andy-aka! I just read electronics.stackexchange.com/a/21797/120711 and that stated the same relation! Commented Jan 15, 2021 at 11:38

1A @ 5V and 1A @ 100V

That’s not what will usually happen. A fuse has relatively low impedance, so it won’t see full circuit voltage until it’s open.

What it will see is I^2*R heat dissipation - where the current is determined by the “short circuit” capacity of whatever circuit is driving the current into the fuse. Of course if you connect a very stiff (low impedance) voltage source across the fuse, it may be able to maintain that voltage on the fuse even when the fuse is closed. All it means is that the fuse will open very quickly since a lot of energy will be dissipated in short time.

And that’s why fuses have interrupting current ratings: the largest current it can break. If you connect a tiny 5x20mm glass fuse to a voltage source that can drive 10kA into that fuse, the presence of the fuse will be largely immaterial. It will explode since the vaporized metal wire will maintain plasma, with some resistance, and thus the current flow will heat up that plsma and make it expand, an arc will form, and the current will keep on flowing.