Yes, there is. If your wire is too fat, the short-circuit current won't be enough to trip your breaker. The breaker will pass overrated current, which can heat up the northbound side of your wire. It's not rocket science. And no need to shout misinformation.
I hope you can understand my confusion on this one.
A circuit breaker is rated for let's use 20 amp as an example. The breaker will hold nominally 20 amps. As you draw more than the nominal 20 amps, the breaker will trip. The more above 20 amps you go, the quicker the breaker will trip. Example: 135% above the rating might be a couple minutes, 200% might be a couple seconds, 1000% trips close to instantly.
Your position is that it is possible to draw so much current that the circuit will not trip fast enough to avoid damage on the north (source?) side of the breaker. And this level of of damge indcuing current is the result wiring capable of drawing too much current. Due to it's large size and low resistence.
The question I have: In our example of a 20 amp breaker, (120 VAC) how big of a wire is too big?
Now I am going to do some calcualtions:
Looking a a wire gauge chart
12 gauge wire is ~1.6 ohms/1,000 feet
6 gauge is ~ 0.40 ohms/1,00 feet.
100 foot run of 12 gauge, 120 volts shorted = 375 amps @ 45 K watts
100 foot run of 6 gauge, 120 volts shorted = 1,500 amps @ 180 K watts
The second value, the higher power, is enough to damage the north side of the breaker when the first one would not.
This is where I get my puzzlements. If you have a 25 foot run of 12 gauge shorted out, you would get 1,500 amp @ 180 K watts. Seems to me it is more complicated than "Do not use that wire gauge, it is too big,"