OK, let me try to explain it:
Short version: The power dissipation in a resistor is equal to the square of the current times the resistance. This is all the fuse (being a low-ohm resistor) cares about: It's resistance and the current flowing through it. Of course if you know the current flowing through it and and its resistance, you can also calculate the resulting voltage drop across it, but that isn't necessary.
Long version: The fuse is very low-ohm resistor in series with a substantially higher-ohm load. Because it's a series circuit, the current through both resistors is intrinsically the same, and how much current is flowing is is primarily determined by the (relatively high) resistance of the load, where you will then find almost all of the voltage drop. If you do the Volts x Amps (or I^2 * R) calculation for both the load and the fuse, the load is consuming almost all the power and the fuse is consuming almost none (which is why it doesn't blow). Now what happens if we short the load? The current flowing in the circuit will rise until something limits it somehow. Assuming a perfect power supply and wiring, the only non-negligible resistance left in the circuit is the fuse. So that's where all the voltage drop is forced to go. You'll have a whopping current through (and all your voltage drop at) the fuse, which will then be "consuming" some crazy amount of power, which cause it to blow.
Incidentally, in the absence of a fuse (or other current limiting device) the battery becomes the fuse, but batteries typcally "blow" less gracefully than fuses do.![]()
This makes sense. Very good explanation. Thanks


