What is the maximum voltage that a circuit with a 20ohm load and 2.5amp fuse can be connected so that the fuse does not blow?
Can someone answer this with an explanation please?
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"Ohms" law Jesse !
U = I * R -> (2.5 * 20) = 50V
That's for an average value, the fuse might be able to handle a VERY short burst of currents greater than 2.5A
So for your application, under the presumption that the load doesn't change any voltage <= 50V will not blow the fuse.
If you want to get fancy, you include the tolerance in the load (i.e. +/- 5%) So a 20ohm load might have a real value of 21 or 19 ohm, always take the worst case scenario!
** Edit - The load might not be the only thing of interest, you say a "circuit" - So what kind of components are in this circuit ? Can they handle 2.5 A? **
As Sorenp already wrote, the voltage to reach the rated current is calculated by Ohms law and is 50V is this case.
But with fuses you have to be aware of there reaction time.
Fuses are rated with the speed of their actions, a typical 2.5A fuse will only blow at 2.5A, if this current is there for quite a long time. To safely trip the fuse in a short time, a higher current is necessary.
Take for example this diagram from this datasheet:
As you can see, to trip the fuse with a rating of 0.5A in 1s, it takes almost double the current. To trip it in 1ms, it takes 3A already.
This timings are different for the different types of fuses, some are faster, others are slower (but more tolerant to short spikes).