Why does a 10A fuse blow at 10A and 5V or 10A and 500V?

Something that's always puzzled me:

Wire and fuses can handle a given max current, pretty much regardless of voltage. A 10A fuse will blow if you exceed 10A, if the voltage is 5V or 500V.

Why is that in terms of the physics? The total energy through the fuse is amps times volts.

Intuitively, I would expect a fuse to blow at some power limit. I'd expect a fuse to blow based one the number of watts of power pushed through it. How is it that a skinny little wire in a 10A fuse can handle 5000 watts at 500V, but will blow with only 50 watts at 5 volts.

What blows the fuse is really the power wasted at it. $p(t)=u(t)\cdot i(t)$.

But the voltage here is not the supply voltage of the circuit, but the voltage drop across the fuse, which is determined by the current passed through it. $p(t)=R\cdot i^2(t)$. R is constant, so it really depends only on the current.

In typical setting, the resistance of the fuse will be negligible compared to resistance of the load, so the current running through it will be determined solely by the load resistance (or impedance in general).

In case of short circuit, there will be no load impedance and current will be limited only by the fuse's filament. In such case, supply with higher voltage will make it blow faster. ($p(t)=\frac{u^2(t)}{R}$)

Both voltage and current are equally important. There is another thing that is important as voltage and current which is time.

A fuse is essentially a resistor but with very low resistance. Depending on the type of material of the fuse it has a specific melting point. If it reaches the melting point it will blow. However, to reach the melting point temperature, we need heat. Heat is energy not power. The product of voltage and current is power not energy. Power is work done or energy used in 1 s. We can calculate Heat using the equation H = VIt where the resulting heat is in joule.

The most important thing here is that the voltage V is voltage drop of the fuse. A fuse should be connected in series with the load resistance. So what we have now is voltage divider circuit. However the load resistance is many times greater than the resistance of the fuse. So voltage drop of the fuse is very low. Therefore the 500V you are referring to is not the voltage drop of the fuse. However if you force the fuse with that much voltage by creating short circuit then it will definitely blow.

Sometimes a fuse can take some time to blow because the voltage and the current provided may not be enough to produce the necessary heat. So it will store the heat until the melting point is reached and finally melt. So you are probably thinking that no matter what is the voltage and current is, if we wait long enough the fuse will blow. No. when the fuse produces heat the environment absorb the heat but the rate of production and absorption might not be the same. It is only then when the rate of production of heat is greater than the rate of absorption the temperature increases and finally reaches the melting point.

Here's the short version: Fuses are very small-valued resistors which blow when the resistance dissipates enough power to raise the filament's temperature to where it will melt. From basic electronics, power dissipated is the square of current times resistance, or $P = I^2 R$. There's no voltage involved in this equation, so voltage has no influence on the fuse's rating.

If you'd like to insist that voltage has to be involved somehow since $P = I V$, note that the only voltage involved is the voltage drop across the resistance. What would that voltage drop be? From Ohm's law, $V = IR$. In other words, current through a resistance causes the voltage drop. Substitute that into $P = IV$ and you're back to $P = I^2 R$.

What really matters when you try to push current through a wire is the current, not the voltage. The wire doesn't care what voltage it happens to be sitting at. Only the insulation might care, as too high a voltage across the insulation will result in a breakdown and then an arc.

The current carrying capacity of a wire is determined by the resistance of the wire and how much heat is generated as a result. The resistance of the wire determines the magnitude of the voltage drop across it. Let's say you have 100 feet if wire with a resistance of 0.01 ohms per foot. The total resistance is 1 ohm. If I put 1 amp through that, it will generate a 1 volt differential between the ends of the wire and the wire will dissipate 1 amp * 1 volt = 1 watt of heat. A fuse is just a wire sized in such a way that it will generate enough heat to melt when a specific amount of current passes through it. With proper cooling, a wire can carry an essentially unlimited amount of current. Wire current ratings are generally designed so that the wire can only produce a certain temperature rise at the rated current, under specific environmental conditions. Naturally this will be affected by ambient temperature, insulation (both thermal and electrical), airflow, etc. Bigger wires have a lower resistance and as a result can carry more current with the same power dissipation.

This voltage drop is why long distance power transmission uses very high voltages. Since the power lost in the wires is proportional only to the current, the voltage is stepped up in transformers as much as is feasible to reduce line losses. The end result is that the wires carry a tiny fraction of the current that they would have to if the voltage was not stepped up. Our electric grid is AC for this reason: it's very easy to build a transformer and step the voltage up or down. It's not so easy for DC.

• Your initial sentence may confuse many, as current is the result of a voltage potential. – whatsisname Sep 8 '14 at 3:44

Calculating power loss can get tricky. If you apply 500v across a 10amp fuse it will blow. If you apply 5v its still gonna blow because there is not enough resistance to limit the current below 10amp. However if u limit the current by connecting a load in series with the fuse the voltage across the fuse is going drop.

There is good way to understand this. Take for example a large pipe connected to a water pump. The water pump cannot pump enough water to fill the pipe so the pressure is very low. But if u replace the large pipe with a narrow one the pressure increases and may rupture the pipe. But if you could limit the water flow by having a narrow outlet the pipe is going to survive. The total pressure of the pump is the same as before but the narrow opening is talking most of that pressure so the pressure in the pipe is very low.

It is my understanding, though, that the voltage rating in a fuse is the maximum voltage at which the fuse woul clear the fault without arcing. Which guarantees that occurrent would cease, therefore the fuse would do its job.

• Clear the fault without arcing and without violent self-destruction. – mkeith Mar 14 at 3:45