Assuming I put a 60amp fuse to an AWG 8 wire, whats the worst case scenario when a short circuit happens?

Therory about short circuits says the current goes to "infinity" however, shouldnt the fuse blow first before the wiring melting even when the fuse is "bigger" than the maximun current the wire can handle??

  • 2
    \$\begingroup\$ If the fuse is larger gauge (e.g. physically thicker) then the wiring, the wiring is now the fuse. \$\endgroup\$ Jan 10, 2015 at 10:13
  • 1
    \$\begingroup\$ Connor Wolf , 60A fuse is much smaller than 8AWG conductor. \$\endgroup\$
    – sparky Al
    Jan 10, 2015 at 16:01
  • \$\begingroup\$ You have to consider the power source. Fuses have different interrupt ratings. This is especially important in DC applications. If you use a 60 Amp fuse, and the fault current is 1000 amps, the fuse may catastrophically deconstruct, or it may arc-over and continue to conduct. This is especially important in high-power, low-voltage DC applications (e.g., off-grid solar) because the available fault current may be 100's or even 1000's of Amps. The interrupt rating of a fuse is more dependent on fuse physical size. It is mostly independent of the Ampere rating (rating where the fuse will blow). \$\endgroup\$
    – mkeith
    Jan 11, 2015 at 6:52

2 Answers 2


8AWG requires a 50A fuse. It protects it from a heat build up if loaded more than 50A.

From a short circuit it is protected by the preceding transformers impedance, speed of the fuse and max current interrupting capability of the fuse.

That circuit should not be loaded to more than 80% of 50A = 40A.


Worst case? The fuse stays intact and the world comes to an end.

Theory says \$ I= \frac{E}{R} \$

From this Wikipedia article,

Have a look at the fusing currents for 8 AWG copper wire:

enter image description here


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