Resistors have a power rating. This means that they usually can work at a range of voltages as long as it is not so high as to increase the amperage to a high enough level to cause excessive wattage to be dissipated through the resistor to cause it to burn up.

This would lead one to believe that fuses work on this concept as well. Excessive amperage causes excessive wattage which would cause the fuse to melt.

However, how is it that a fuse marked 250 20 amp can be used to provide over current protection at a lower voltage? (20 amp 120 volt is 2400 watt while 20 amp 250 volt is 5000 watt)

 

Fuses are more complex than we think.

In part the melting is governed by Joule heating [check Google for a description of Joule Heating]

The voltage rating has more to do with the issue of the fuse physical structure being able to clear (cool) the arc that forms when the fuse element ruptures.

 

The voltage rating is independent of the current rating because they are different mechanisms.

Say your 250V, 20A fuse has a resistance of 50 milliohms.
At 20 A the power dissipation is (20A)^2 * 0.05 Ohms = 20 W
The current is about the actual fusing where the conductor will melt.
To have 20 W in this fuse you would need Sqrt(20W*0.05 Ohms) = 1V across the fuse.

To have 250V across the fuse the current would need to be 250V/0.05 Ohms = 5000A.
The power dissipation in the fuse would be (250V)^2 / 0.05 Ohms = 1.25 MW !!

The key step here is to understand that the power relationship is between current through the fuse and its resistance, or equivalently voltage across the fuse and its resistance. In a resistive material voltage and current are strictly related through resistance.

The fuse voltage rating is related to more complicated breakdown mechanisms than just overheating and breaking.
For example a fuse can only interrupt a certain voltage - if the voltage is too high the current can continue flowing even if the fuse conductor melted away.