SMIN 10 pcs F2AL 250V Glass Fast-Blow Fuse 6x30 mm (2Amp)

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Suppose you have a circuit that runs on 240 V and uses 0.5 A. You (wrongly) protect this circuit with a 1 A, 50 V fuse. When the fuse is intact (not blown) there is no issue, no more than 0.5 A flows through the fuse so it does not blow. Voltage is very much in play. But, it's not the supply voltage or rated voltage of the fuse, it's the Voltage drop across the fuse along the R/Temp curve. Perhaps I'm being pedantic, but a fuse's usefulness really depends on it's non-linear R combined with the fundamental Power transfer law in a series circuit and power is a function of Voltage and Current.

The fuse does blow due to power (heat), but the voltage rating of a fuse is not the voltage drop across the fuse in operation, so isn't used to calculate the power required to blow the fuse. A fuse is a non-linear device and it is designed to take advantage of being in series with a load. In normal operation, it dissipates very little power. But, as the current reaches the fuse's rated current, the power goes up, the heat goes up and the Resistance goes up, which in turn increases the V*I=Power=Heat... and poof goes the fuse's conductor opening the circuit as it is designed to do. To answer the question about a fuse blowing at rated current regardless of applied voltage... yes, but it's because it is a non-linear resistance in series with the load. It is fundamentally wrong to say 'it has nothing to do with the voltage' though. Ohm's Law tells us that without voltage, in this case the voltage drop across the fuse, there is no current (V/R=I). Power is defined as V*I=P. The fuse's conductor has a very non-linear resistance to temperature coefficient, meaning, that as the power (V*I) dissipated goes up, it reaches a point at which R rises rapidly, the VI applied melts the conductor and blows the fuse. The power distribution in this series circuit shows the power dissipated by the fuse is very low when its R is low (in safe operation) and the power redistributes from the load to the fuse when the current reaches a critical point (trip or 'blow' point) in the R to Temp curve. The electric power in watts produced by an electric current I consisting of a charge of Q coulombs every t seconds passing through an electric potential (voltage) difference of V isThen a fault develops in the circuit making more current flow and blowing the fuse. The fuse then opens the circuit and the 240 V develops across the fuse. 240 V across a fuse rated for 50 V! So the fuse might break or arc-over and no longer protect the circuit. This is why the voltage rating is also important but it only becomes important after the fuse has blown. After the fuse has blown the circuit becomes open so a voltage develops across the fuse (usually the supply voltage like mains voltage or the battery voltage). The fuse must be able to withstand that voltage and keep the circuit open. That means that the voltage rating on the fuse must be higher than the voltages used in the circuit you're protecting. So if a fuse is rated for 12V DC and 20 A, this would be equal to 240 watts. If a different voltage is supplied, will this change the current at which the fuse will break? Does the fuse technically 'blow' at 240 watts?

If 6V DC was applied to this example fuse, 240 watts in this condition would be 40 A when the fuse would 'blow'. Am I correct? Or does the fuse always 'blow' at 20 A, regardless of the voltage?" I believe the confusion with voltage 'not mattering' in this discussion, is that, the R of a fuse is NOT constant and non-linear... on purpose. Similar to an incandescent lamp filament, (for a 100W standard household filament, the R when cold is low (5-10 Ohms) and higher when hot (100 Ohms)). The fuse typically has a very low R when cold, but as the Power goes up, Vfuse*I=P), so does the heat.