We usually specify the maximum current that a conductor (such as a fuse) can handle without burning. But doesn't the conductor really fail when a certain amount of energy/heat has been dissipated in the conductor? Then the conductor is at a too high temperature and burns/melts.

Let's say I have a fuse that's rated for 10A. Why is it then that I can operate the fuse continuously at a lower current like 9A without the fuse burning too, but just a bit later?

We also know that power, voltage and current are related by Ohm's law. So if we have a 10A fuse, and it has some arbitrary resistance such as 100 ohms, why don't we instead call it 1kV fuse (10A * 100 ohms), or 10kW fuse (10A * 10A * 100 ohms)? These numbers are completely arbitrary so I know they don't reflect reality but they make my point clear.

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    \$\begingroup\$ " why don't we instead call it 1kV fuse " because you don't know the voltage over the fuse. It depends on the load after the fuse, which normally should have most of the voltage over it. \$\endgroup\$ – Oldfart Aug 31 at 19:37
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    \$\begingroup\$ @Oldfart If we know the resistance of the fuse and maximum current, we do know the maximum voltage, by calculating it using Ohms law. \$\endgroup\$ – S. Rotos Aug 31 at 20:02
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    \$\begingroup\$ They burn at a specific current because that's their purpose. All other units don't matter. You install a fuse in your circuit cause you want to protect it from overcurrent and ideally it has no voltage drop so you don't lose power. Hence putting the resistance in the rating is unnecessary, it should be nearing zero. \$\endgroup\$ – Swedgin Aug 31 at 20:11
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    \$\begingroup\$ @S.Rotos You miss Oldfart's point. You never put a fuse by itself across a voltage source. What is the purpose of a fuse? To limit voltage across itself? No. To limit power it is dissipating in itself? No. To limit current flowing through itself? Suprisingly, it's not this either! A fuse's purpose is to limit the current flowing through the load. You could argue that a fuse's purpose might be to limit the power or voltage across the load, but now all your fuse ratings are dependent on the characteristics of the specific load it is being used with (load power/voltage cannot be seen by the fuse) \$\endgroup\$ – DKNguyen Aug 31 at 20:50
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    \$\begingroup\$ @DKNguyen Ah, I think I understand it now, my misunderstanding was something of a brain-fart. Thank you very much! \$\endgroup\$ – S. Rotos Aug 31 at 21:14

So if we have a 10A fuse, and it has some arbitrary resistance such as 100 ohms, ...

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This typical 10 A fuse has a resistance of 5 mΩ. So your guess was out by a factor of about 20,000. At 10 A the power dissipated is given by \$ P = I^2R = 10^2 \times 5m = 500 \ \text {mW} \$.

RESISTANCE: The resistance of a fuse is usually an insignificant part of the total circuit resistance. Since the resistance of fractional amperage fuses can be several ohms, this fact should be considered when using them in low-voltage circuits. Actual values can be obtained by contacting Littelfuse. Source: Littlefuse Fuseology Application Guide (which is well worth a read).

The reason for higher resistance in fractional ampere fuses is that the fuse wire is about the same length as the 10 A version but would have to be much finer to blow at, for example, 100 mA. A 100 mA fuse may be protecting a circuit normally drawing, say, 50 mA. If the fuse resistance was 1 Ω then there would be a 50 mV drop across it in service.

The required diameter of a fuse wire can be calculated from $$ d = \left( \frac {I_f}{C} \right)^{\frac {2}{3}} $$ where where If is the fusing current in amps, C is Preece’s Coefficient for the particular metal in use. (Source: Ness Engineering.) From this we can see that a 10 A and 0.1 A (a factor of 100) fuses of the same material would result in the 10 A fuse having a wire diameter \$ 100^{\frac {2}{3}} = 21.5\$ times that of the 0.1 A fuse.

... why don't we instead call it 1kV fuse (10 A * 100 ohms), or 10 kW fuse (10 A * 10 A * 100 ohms)?

Because it is an over-current protection device. Fuses already have a voltage rating that means something completely different. See below.

The fuse needs several ratings:

  • The current (which I think is obvious enough).
  • The voltage rating of the fuse. This specifies the maximum voltage it can break reliably without forming and sustaining an internal arc.
  • The time rating - how quickly it will blow.

The Littlefuse article covers all of these in great detail so there is no need to reproduce it here.

  • \$\begingroup\$ Thank you, I think I understand it now. \$\endgroup\$ – S. Rotos Aug 31 at 21:14
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    \$\begingroup\$ Fuses normally also have a "breaking capacity" rating. If the fault current is too high then the fuse may sustain an arc. \$\endgroup\$ – Peter Green Sep 2 at 1:02
  • \$\begingroup\$ From your link : "Preece’s Law can be used to generate an estimate for the approximate dc fusing current for a given wire size and material. The actual fusing current can unfortunately depend on the detailed heat transfer from the wire which can be influenced by the enclosure, conduction of heat through the wire to the terminals on both ends, and other physical conditions. A one-dimensional heat equation or more complicated thermal analysis can therefore be used to better determine the exact fusing current. However, as a quickly determined estimate, Preece’s Law can be valuable." \$\endgroup\$ – Uwe Sep 2 at 14:22
  • \$\begingroup\$ @PeterGreen: Beware the dreaded nail fuse. \$\endgroup\$ – Joshua Sep 2 at 16:41
  1. The conductor fails when it reaches a certain temperature. Because the fuse is in thermal contact with ambient, it can dissipate a certain amount of power before it blows.
  2. Your 10A fuse is designed to blow at 10A (plus or minus some tolerance). So it should run at 9A all day.
    • But that 10A fuse will take a good long time to blow at 10A, and will blow much quicker at 20A, and may misbehave if you shove 100A through it. There's a whole, mostly neglected science to fuses.
    • And if you run that 10A fuse at 9A or 9.8A all day, it'll run hot and slowly degrade.
    • Which all means that if it really matters how fast it blows, or how long it lasts, that you need to talk to the fuse manufacturer.
  3. Fuses are rated in amps because that's what most folks installing fuses care about. The ideal 10A fuse drops no voltage, and neither blows nor degrades at even a nanoamp below 10A, but blows immediately (or after a well-defined time) above that. No ideal fuses exist.
  4. While you're pondering all of this, you may want to dig up some fuse data sheets and look. The good companies (Bussman, Littlefuse, etc.) specify this -- and there are such things as slow-blow fuses that are designed for temporary overload, and fast-blow fuses that are designed to react more quickly than "ordinary" fuses. If the way the fuse needs to respond is non-standard and critical, it can become quite the engineering exercise to design one in.
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    \$\begingroup\$ The UK wiring regulations have charts of the "time to blow" for a given overcurrent on the fuse... \$\endgroup\$ – Solar Mike Aug 31 at 19:49
  • \$\begingroup\$ How does an ideal fuse not dropping voltage make sense? It needs a non zero resistance to generate heat in order to blow, so therefore it should drop voltage too by Ohms law. \$\endgroup\$ – S. Rotos Aug 31 at 20:07
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    \$\begingroup\$ @S.Rotos Yes that are the reasons why ideal components don't exist. \$\endgroup\$ – Christian Aug 31 at 20:33
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    \$\begingroup\$ A 10A fuse will work just fine at 9.8A. If it "runs hot", it's defective. \$\endgroup\$ – Dmitry Grigoryev Sep 2 at 7:13
  • \$\begingroup\$ @DmitryGrigoryev: The ability of a 10A fuse to sustain 9.9 amps would depend its environment. If a 10A fuse is operated for a sustained period in an enclosure that can't dissipate heat, the enclosure may eventually heat up to the point that the fuse fails; using a 15A fuse may reduce the self-heating sufficiently that the fuse never fails. \$\endgroup\$ – supercat Sep 2 at 18:21

Normally, a fuse doesn't know what voltage circuit it is used in - it only knows the current that is flowing through it, so that is the only thing that can cause it to blow.

Fuses also have a voltage rating because, once the fuse blows, it will have the full circuit voltage across it, so it must be designed to safely handle that voltage without arcing.

  • \$\begingroup\$ "fuse doesn't know.." Why not? We know its resistance and the maximum current, so the maximum voltage across it is by Ohms law its resistance times the maximum current. \$\endgroup\$ – S. Rotos Aug 31 at 20:04
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    \$\begingroup\$ @S.Rotos you have calculated the voltage dropped across the fuse, that is not the applied voltage - it could be 10 volts or 200000 volts \$\endgroup\$ – Solar Mike Aug 31 at 20:06
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    \$\begingroup\$ @S.Rotos In one example the fuse is 10A and is 5 milliohms, so by that logic voltage must be 2000V. And yet, that doesn't make sense, does it? The fuse can't know the voltage because the voltage is the same on both sides (until it blows). \$\endgroup\$ – Harper - Reinstate Monica Sep 2 at 0:30

Ask yourself: What is the purpose of a fuse?

  1. Limit the voltage across the fuse? No. this is pointless.
  2. Limit the power a fuse is internally dissipating? No. This is also pointless.
  3. Limit the current going through the fuse? Surprisingly, no! A fuse's job isn't to protect itself from anything. A fuse's job is to protect the load. So suppose you care about the current in the load, then you would only care about the current that the fuse blows at as a secondary concern since the load and fuse currents happen to be the same
  4. Limit the voltage in the load? Arguably, yes but there are issues rating fuses this way which I will discuss below.
  5. Limit the power in the load? Arguably, yes but there are issues rating fuses this way which I will discuss below.
  6. Limit the current in the load? Yes! The fuse's ultimate purpose is to protect the load. I will discuss why current is more valid than either voltage or power in #4 or #5

The load is king. A fuse isn't designed to blow just for its own sake. A fuse is designed to protect the load. You're missing the forest for trees if all you're focused on is when the fuse blows. In the end, I don't care AT ALL what voltage is across the fuse or how much power the fuse is dissipating when it blows. What I care about is the current through the load is when the fuse blows (and by extension the current in the fuse when it blows).

You could argue it's to limit the power AT THE LOAD or the voltage AT THE LOAD, but you can't rate fuses based on the load power or the voltage because those numbers are dependent on the load itself. In other words, that means the fuse cannot be rated such a way without knowing exactly the characteristics of the load it is being used with.

In more rigorous terms, this is because the fuse's position in the circuit does not allow it to observe the power or voltage across the load. It can only observe the current going to the load. Sure, the fuse can observe its own voltage drop or dissipated power from its position in the circuit, but we've already established that isn't relevant to protecting the system.

If you give me a fuse rated using the voltage or watts across it, I have to go through a bunch of needless calculations that take into account the characteristics my load just to figure out if the current the fuse blows at is going to be to protect my load from overcurrent, overvoltage, or overload.


The crucial point to understand is the material which fuse wires are made of. It's plain, simple metal. However, metal has the property of being a cold conductor: If you heat up a wire, it becomes less and less of a conductor and more of a resistor.

Now, if you have a fuse that's operating below its current limit, it turns a tiny bit of electrical energy into heat, which is quickly dissipated, and the wire stays cool. Accordingly, it has a very low resistance, so only a tiny amount of voltage drops at the fuse.

When the current through the fuse rises above the threshold, the fuse wire becomes warmer. This means that its resistance rises, that a larger portion of the voltage drops across the fuse, and as such, that it turns more electricity into heat. The heat in the fuse wire causes more heat to be produced. This is a self-amplifying process, and because there is so much electrical energy available that was simply flowing through the fuse when it was cold, the hot fuse can draw a lot of power from the current even before impacting the voltage at the appliance significantly.

Because of this self-amplifying heating process, the fuse quickly overheats, braking the circuit.


It's true that the fuse conductor heats up in response to the current flowing through it. The wire itself is designed to dissipate that heat by conduction to its environment so the fuse doesn't melt- until the power being dissipated in it exceeds the wire's capacity to conduct away that heat. Then the heat builds to the point where the fuse wire melts. By adding mass to the wire, its thermal time constant is increased which furnishes the ability for it to handle brief surges of overcurrent- resulting in a slo-blo fuse.


But doesn't the conductor really fail when a certain amount of energy/heat has been dissipated in the conductor? Then the conductor is at a too high temperature and burns/melts. [...] Why is it then that I can operate the fuse continuously at a lower current like 9A without the fuse burning too, but just a bit later?

It doesn't matter how much energy has been dissipated in the fuse. What matters is rate at which energy is dissipated into the fuse (that's power -- I2R) compared to the rate at which energy is dissipated out of the fuse via radiated heat and heat conduction.

When energy is going into the fuse faster than it's going out, then the fuse heats up. When the fuse heats up, however, the rate at which energy is dissipated out of the fuse increases. The temperature will increase until the heat power flowing out of the fuse matches the heat power going in (I2R).

So the fuse will quickly reach an equilibrium temperature that is determined by the current. When this temperature is too high, the fuse will blow.

Depending on the fuse material, it could blow when the equilibrium temperature reaches the material melting point, or it could blow by the thermal runaway that @cmaster mentions in his answer. At that point, the increasing temperature in the fuse increases power in faster than it increases power out, and the equilibrium is lost.


Fuses are rated for operating current. A 10A fuse will not blow (or "slowly degrade") at 9A, or even 10A. That it's labelled 10A only means the manufacturer guarantees it will work as expected as long as you don't exceed the rating.

Obviously, that means a 10A fuse will not blow the moment you exceed 10A. In fact, if you look in a datasheet, you'll see that you need something like 20A to blow a 10A fuse at all, and maybe 30+A if you want it to happen reasonably quickly.

Fuses also have voltage drop ratings, indeed, you need both current and voltage to blow up a piece of wire. But since end users typically want a precise current rating, manufacturers don't measure voltage drop precisely and provide only a typical/max value for it. Imagine I tell you that I have a 150 mV / 5 mOhm fuse: do you think that would be enough to protect e.g. a 1kW mains load? You'll have to find out the current rating to tell.


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