# How does a fuse blow at its current rating, regardless of voltage?

I know from reading elsewhere that it's safe to use a fuse with a higher voltage rating when replacing one, so long as the current rating and reaction speed is the same.

For example, if a fuse is rated 125V 1A, then a 250V 1A can be used.

Let's say these two example fuses have a resistance of 0.153 and 0.237 ohms, respectively. (Littelfuse 5x20mm fast-action cartridge type.)

Is it therefore correct to say that the 125V 1A fuse in theory should blow at 153 mW and the 250V 1A fuse blows at 237 mW? (Using $P = I^2R$)

• Another question about fuses made me think of the math behind this. Dec 15, 2012 at 7:38
• Russell's answer is the most complete. Just to add a refinement, When the fuse is whole and conducting the voltage it sees is only the voltage drop across it's resistance. So very little. the voltage rating pertains to when the fuse is BLOWN and it now has to support the full voltage across the case without out conducting. Dec 15, 2012 at 17:41
• Because E = I^2R, an equation in which voltage does not appear. Oct 22, 2016 at 9:55
• @EJP I reference P=I²R at the end of this question which is power equal to the square of current times resistance. E typically stands for electromotive force, which is the same as voltage. I'm not sure what your formula is if it's not also power. Oct 22, 2016 at 15:25
• @Jyelton My mistake, I should have used P. It's still the answer to your question. Oct 22, 2016 at 20:03

The current rating of a fuse represents the minimum sustained current the fuse will blow at ... eventually. A 1A fuse will take 1A for a very long time without blowing, and if the fuse can dump some heat into the PCB or has airflow across it, may never blow at 1A.

The critical parameter is the $$\I^2 \cdot t\$$ rating, which gives you an idea of the energy (power and time) needed to blow it. (Remember that fuses are really meant to protect circuits when there are catastrophic failures.)

It's crucially important to match $$\I^2 \cdot t\$$ ratings, since if you replace a fast acting fuse with a slow-blow type, even thought they both say 1A, it's going to take radically different energy levels to actually blow them.

When the fuse is intact, you only have an $$\I \cdot R\$$ voltage drop across it. This drop is going to be nowhere close to the voltage rating of the fuse (else it acts like a big resistor and limits the energy available to your circuit.) Once the fuse blows, the voltage rating comes into play, which represents how much voltage potential the open fuse can withstand without flashing over and re-energizing the compromised load circuit.

• The drop across it (voltage) is IR, not IIR. Aug 15, 2016 at 5:23
• @gbarry: Forget my first reply, I was to quick. He is indeed talking about the voltage drop, not the power loss. I edited the answer.
– Rev
Aug 15, 2016 at 6:05

There are already some excellent answers to this question, but I'd approach the answer slightly differently. Consider the circuit below. Under normal operation (i.e. fuse not blown), Vf is IL*R, where R is the inherent fuse resistance. The current, IL, flows through both fuse and load. The voltage across the load, VL = VB - Vf, where VB >> Vf. The majority of the voltage is dropped by the load, and only a small amount is dropped by the fuse.

As pointed out by others, the power dissipated in the fuse is IL2R. At some level of dissipation, the fuse will open. As the fuse opens, an arc forms that burns away more of the fuse material. During this process, Vf will start out being IL*R (as defined above), but will become VB as IL drops to zero and fuse opens fully. At the end of the clearing event, all of VB appears across Vf and current flow stops completely.

The voltage rating (and AC/DC specification) of the fuse comes into play only after the fuse opens. A fuse with inadequate voltage rating may be unable to quench the resulting arc, leading to rapid failure of the fuse. Similarly, a fuse or breaker rated for use with AC will likely depend on a zero crossing to quench the arc, where DC-rated fuses (especially high voltage DC fuses) are often tightly packed with sand or other arc-quenching material in order to prevent the power dissipated in the arc (theoretically up to VB * IL) from catastrophically destroying the fuse and to ensure that current does not continue to flow via a continuous arc (i.e. the fuse blows yet current continues to flow via plasma between fuse internals).

If the fuse never blows, the voltage rating of the fuse doesn't matter. At the moment it does blow, the current rating ceases to matter and you'll quickly know if you spec'd the appropriate voltage fuse for your application.

• This is a very well-written, clear answer. Thanks. Dec 16, 2012 at 6:02
• Not only to keep the power from catastrophically destroying the fuse--also to stop the current. It would not be inconceivable that a fuse which was was driven with DC and used beyond its rated abilities might sit perfectly happily with an arc across it, perfectly happily dissipating the heat generated by the arc, but not doing much to achieve its design objective of stopping current flow. Dec 16, 2012 at 20:39
• @supercat, thanks for pointing out that omission; you're absolutely right. I've edited my post to be more complete. Dec 17, 2012 at 1:22

The fuse "sees" largely only it's own environment. The fuse wire melts when the net thermal input is enough to cause enough temperature rise to melt the wire or other fusible element.

To get local energy dissipation you need some voltage drop across the fuse.
Power = I^2 x R = V^2/R = V x I
All these are equivalent here.
The frst relates to current carried and fuse resistance.
The 2nd relates to voltage drop across the fuse and fuse resistance.
The 3rd relates to voltage drop across the fuse x current carried.

Net thermal input is energy dissipated - energy radiated per time.

Here is a fuse search engine . specificy parameters (mainly fusing current here) search for fuses. Read resistance value. Some examples here

Two examples:

100 mA: A FRS-R-1/10 600 V 0.1 A Mersen Class RK5 600V Time Delay has about 90 milli-Ohms resistance. V = IR = 0.1 x 0.09 ~= 10 mV !
Power = I^2 x R =~ 1 mW !!!

10 A: A 9F57CAA010 10 A Mersen Oil Cutout Fuse Link has about 10 milli-Ohm resistance.
Voltage drop = IR = 10 x 0.010 = 0.1 V
Power = I^2 R = 10^2 x 0.01 = 1 Watt !

Simple answer is is that moving electrons produce heat independent of voltage. Voltage doesn't matter in this production of heat it is the same regardless of voltage. One amp produces the same amount of heat because of the friction of the electrons bouncing around. Hence one amp of DC is the same amount of heat as one ACamp rms.

• This answer is a little confusing. Voltage does matter in the production of heat, insofar as voltage sets the amount of current flowing through a fixed resistance. Mar 21, 2013 at 12:23

When a fuse blows, it interrupts a (in some cases quite large) current. The fuse does not go instantly from "normal" to "completely blown" - the wire heats up and melts, creating a short break which expands because the wire does not cool down immediately. When the break is small, you can get an arc (especially if the load is inductive), which is extinguished soon after because 1) the instantaneous current reaches zero (since this is AC) and by the time the voltage comes back to peak, the gap is wide enough for an arc.

So, the higher voltage, the wider gap there must be. However, using a higher voltage fuse is not a problem.

Imagine using the small 250V fuse with, say, 10kV - it would arc over the entire fuse.

As for the power at which the fuse blows - it is tiny compared to the power of the system, but it does imply a limit on how low the voltage of the system can be. If the fuse has 0.237ohm resistance and 1A current, then it drops 0.237V, so if your system runs on similar voltage you will have problems.

• I'm familiar with the reason for the voltage rating. On another forum, someone made a comment that, with regard to fuses, only amperage matters, not power/wattage; this sounds folly. Dec 15, 2012 at 8:27
• On further thought, perhaps my question title is misleading. It might be better stated as: "Why does a 1A fuse blow at 1A, regardless of voltage below its rating?" Dec 15, 2012 at 8:35
• The fuse blows at 1A because the wire dissipates enough power (as you know, P=IIR) to heat up and melt. The voltage rating applies only to the effects after the wire melts. Dec 15, 2012 at 8:37
• To expand on rated voltage, see en.wikipedia.org/wiki/Fuse_(electrical)#Rated_voltage Dec 15, 2012 at 12:36
• The low-voltage scenario can sometimes be a problem with self-resetting fuses. The resistance of a typical self-resetting fuse will changes greatly with temperature; the power required to maintain the fuse at its current temperature is about the same when the resistance is ten ohms as when it's ten megs. The fuse may be modeled as a device which absorbs a uniform amount of power. Normally, as the resistance increases, this will increase the amount of voltage dropped across the fuse; for any given load current, this will in turn increase the power and thus temperature and resistance. Dec 16, 2012 at 20:45

1amp DC still produces the same amount of heat as 1 amp rms AC. 1 amp AC regardless of voltage produces the same amount of heat. Don't confuse heat with power consumption which would fall in the realm of voltage drop, such as across a load wether that load be intentional or not such as vd on transmission wires or branch circuits S