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