> The resistance to current flow could be as low as 0.001Ω.
I wondered what the real value was.
The bolt is fat. The sheetmetal is thin but wide. Seems to me the bottleneck is where the bolt meets the sheetmetal.
Say 1/4" bolt in 1/16" panel. Say it makes perfect contact all around the circumference. (That won't happen in DIY hole-drilling, but partial contact under the wider head will give a similar result.) Circumference of the bolt is 0.25"*pi= 0.8". Panel is 0.0625" thick. 0.8*0.0625"= 0.050 square inches.
The total length of the path through the panel may average 10 inches. But it will also be ~10" wide. Right at the bolt, the circumference expands from 0.8" at the bolt to 2.4" at 1/4" away (3/4" diameter), to 4" at another 1/4" (5/4" diameter), etc. Most of the resistance is in about the first bolt-diameter. If we wanna avoid radial math and integration, we can pretend we have a conductor with cross-section 0.25"*0.0625", and 0.25" long.
The 0.25"*0.0625"= 0.050 sq.in. cross-section turns out to be the same as a US #2 wire (0.258" diameter, car-battery cable). Say the panel is Aluminum. I don't have an Al wire table handy, but Al is about twice copper the same size. #2 Copper is 0.159Ω per 1,000 feet. So Al would be 0.32Ω per 1.000 feet. We want to know the ohms of 0.0208 feet (0.25") of #2 Al wire. 0.02/1,000= 0.000,02, and 0.000,02 times 0.32Ω/1,000' = 0.000,0067Ω.
The bolt has two ends, double that number.
Whoops: I didn't notice that the cross-section of the panel is around the same as the bolt. Most of the panel resistance is in the first fraction of an inch, but the bolt is 1.75" long. A brass bolt is better than aluminum, a steel bolt is worse. Oy, round up to 4 inches of 1/4" diameter copper. (0.16Ω/1000')/3,000= 0.000,05Ω.
Assuming the transformer wants to run at 0.1 Volts per turn, 0.1V/0.000,05Ω= 1,875 Amps.
If that could actually happen: at 150 Amps, #2 Al wire runs as hot as common insulation can stand. So 1,800 Amps would be far-far-far too hot for any nearby insulation. (I suspect the bolt would glow and melt.)
But at 120V primary, 0.1 volts per turn, we have a 1,200:1 step-down ratio from line primary to the bolt+case shorted turn. So the primary current, neglecting primary resistance, is 1,875/1200 or about 1.5 Amps. Not enough to burn the house, not enough to pop a 2 Amp fuse.
The accidental shorted turn appears at the 120V side as 120V/1.5A= 80 ohms. We already estimated that a 120V 25VA 20% reg transformer has about 144 ohms effective resistance. I neglected to note that half of that is primary, half is reflected secondary, and the normal secondary is overwhelmed by the accidental bolt+case secondary. So the line sees 72 ohms primary loss, 80 ohms load in bolt+case. Current is about 120/(72+80)= 0.79 Amps. Power is 120V*0.79A= 95 Watts. This is divided between primary copper and bolt+case in 72:80 proportion.
About 45 Watts of heat appears in the primary, about 50 Watts of heat appears in bolt and case. The 25VA 20% transformer is sized for about 5 or 6 Watts of internal dissipation. Now it has 45 Watts of heat inside it, and another 50 Watts of heat in its bolt and the case near it. It cooks. Yet the 95 Watts of power sucked from the wall is no problem for house-fuse, house wiring, or even a 1A line-fuse.