Anyone knows the very common glass fuses in electronics etc. But I am uncertaion about the breaking conditions.

Example: A fuse rated as 5A/250V, it is the breaking point. But I have seen such used in devices with a much lower voltage, like 12V or 24V.

The breaking of the fuse is based on a maximum energy generated as heat to melt the internal lead by Ohm'S law, aka P = UxI.

But this does not correlate with the current (A) at different voltages. In the example, the nominal is P = 5A*250V = 1.250, which used with a 24V case, would be (I = P/U) = 1.250/24 = 52A which is way beyond the rated 5A.

In other words: We have different currents at different voltages to generate the same energy, if we are to follow Ohm'S law and that the breaking requires the same energy in any case.

This is a little confusing, because it must be the current that are the dominator and the current is driven by the voltage, so the voltage is certainly a factor...


4 Answers 4


Your calculation gives the power dissipated in the load. A fuse blows by the power dissipated in the fuse - not the load. The fuse resistance will be very low and the power to blow the fuse will be correspondingly low.

  • The current rating is that nominal value which will blow the fuse in a certain time.
  • The voltage rating is the maximum voltage that can be reliably interrupted without forming and sustaining an arc across the burnt-out fuse wire.

enter image description here

Figure 1. GDC fuse time vs current. Source: Cooper Industries.

Note in Figure 1 how the 1 A fuse can sustain about 1.8 A indefinitely (1). On the other hand for a 1 s trip time (2) the fault current would have to be 3 A (3).

How about the resettable fuses (with button) ? Is this the same ? because they are voltage rated too.

They will have the same limitations. Find a datasheet and study it in the light of my Figure 1. If you take my approach you will learn the new words and practice pronouncing them then start to figure out what they mean from the context in the datasheet and further reading. Read, read, read. It's worked well for me.

  • \$\begingroup\$ Transistor:: Thanks very much - How about the resettable fuses (with button) ? Is this the same ? because they are voltage rated too... \$\endgroup\$
    – Gearlos
    Commented Jul 8, 2018 at 14:39
  • 2
    \$\begingroup\$ BTW: I think i made a BIG mistake in the calculation (which agrees with mr. Trans..). I should have used the voltage across the fuse which makes the ampere factor very alike at any voltage when the the resistance is very low (A little embarrasing, but i apparantly needed a kick in the ass to start using my brain ;) \$\endgroup\$
    – Gearlos
    Commented Jul 8, 2018 at 14:45
  • \$\begingroup\$ Every type of fuse should have an isolation distance big enough for the rated voltage when interrupted. Resettable fuses and those for one time use. There are special fuses for medium voltages above 2.4 kV, they are much longer than those for voltages below 0.6 kV. \$\endgroup\$
    – Uwe
    Commented Jul 8, 2018 at 14:48
  • 1
    \$\begingroup\$ @Gearlos we all need a kick in the ass now and then so why not make amends by marking this anwser as formally accepted. Here's how it works and while you are at it review your other questions and, where appropriate do the same. \$\endgroup\$
    – Andy aka
    Commented Jul 8, 2018 at 14:49
  • 1
    \$\begingroup\$ It's a common mistake. Someone else did it a couple of days ago with SSRs that kept failing. He made the same error reckoning that the power was being dissipated in the SSR so if he ran at lower voltage he should be able to run at much higher current. \$\endgroup\$
    – Transistor
    Commented Jul 8, 2018 at 14:49

I guess you are confusing the rated voltage, breaking capacity and the holding current.

Fuses are specified under IEC 60269. This standard defines time/current intervention curves based upon P=RI^2 fuse element dissipation. The higher the power, the lower the intervention time. This is not directly related to voltage.

Voltage ratings on the fuse are related to the maximum voltage (AC voltage, notice that most cartridge fuses are specified in AC only, and DC voltage, if specified, is generally lower) that they are able to break. This is somehow correlated with the breaking capacity, expressed in kA. A small fuse would not be able to open high current sparks inside its body. As the current increases and the open circuit voltage increases, electric arcs are more difficult to open. This consideration is valid for domestic circuit breakers too: your short circuit protection device shall be able to stop the maximum short circuit current at the specified point and be operational at the mains voltage.

One last thing: as obvious there is no such a "ideal" voltage source. Whenever you short two wires together, there is an inherent resistance of the power distribution system, the wires, and the fuse. So, your short circuit current is never "infinite". A fuse has an iherent short circuit current limiting capability (I^2dt) that is the increment of internal resistance with the current, that helps with higher currents.

To put the things together, have a look at the datasheet of a fuse such the following: http://m.littelfuse.com/~/media/electronics/datasheets/fuses/littelfuse_fuse_209_datasheet.pdf.pdf 1A rating: it means (more or less) it will sustain up to 1A RMS currents at indefinite time 350V rated voltage: the maximum open circuit voltage the device is guaranteed to be able to open Interrupt rating: 100A @ 350Vac the breaking capacity of the fuse


The contributers has kindly responded to answer the question to my best understanding of the problem. Thanks all !!

...sometimes the solution is more simple than You think...

In addition I will add a little comment

enter image description here

  • \$\begingroup\$ A superconducting fuse wouldn't get hot and wouldn't blow. \$\endgroup\$
    – Transistor
    Commented Jul 8, 2018 at 15:45
  • 4
    \$\begingroup\$ @Transistor A real superconductor fuse would conduct perfectly until it reached the critical current density Jc and then it would go normal and get hot. \$\endgroup\$ Commented Jul 8, 2018 at 15:53
  • \$\begingroup\$ well - My last sentence about a superconductor resistance was not about a fuse, but merely about Ohm's law that would not rule in such case.. \$\endgroup\$
    – Gearlos
    Commented Jul 8, 2018 at 19:16
  • \$\begingroup\$ @Gearlos Ohm's law still holds for superconductors, its just that by definition R = 0, therefore voltage drop is always zero too, and power dissipated always zero. \$\endgroup\$
    – MikeB
    Commented Nov 21, 2023 at 10:29

Typical small fuses are designed to blow in watt seconds across fuse above a minimum W holding power loss in the fuse and an insulation voltage rating

You can compute this from the curves.


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