North America uses 120 V for most circuits. Power is the product of voltage and current.
At 1 Amp, 120 watts are dissipated by the circuit. About the heat of two incandescent light bulbs.
At 10 Amps, 1200 watts are dissipated by the circuit, about the heat of a space heater.
At 551 Amps, 66,000 watts are dissipated by the circuit. I don’t even have a good comparison. That’s like the power draw of 50 homes all at once.
The higher the gauge, the lower the diameter of the wire. The lower the diameter of the wire, the more of that 66,000 watts is going to be dissipated by the wire itself instead of the load where it is desired. At 22 gauge, basically all of it will be dissipated by the wire, at least for the first fraction of a second before the wire vaporizes in a small explosion.
EDIT: In this scenario, the total resistance of the circuit must be at most 0.22 Ω. Otherwise, the current will not reach 551 A due to Ohm’s Law, V=I×R. This resistance corresponds to a maximum length of 13 feet for copper wire and no load.
I ran this by my brother who’s an electrician and he inferred that might be where the number is coming from, some data on how many amps you can dump into various wire gauges before they simply stop being solids.
Wires have a small resistance which causes a voltage drop over the wire if the current is big enough (U=RI), and therefore it draws power (P=UI) which warms it up. Thinner wires have more resistance.
North America uses 120 V for most circuits. Power is the product of voltage and current.
At 1 Amp, 120 watts are dissipated by the circuit. About the heat of two incandescent light bulbs.
At 10 Amps, 1200 watts are dissipated by the circuit, about the heat of a space heater.
At 551 Amps, 66,000 watts are dissipated by the circuit. I don’t even have a good comparison. That’s like the power draw of 50 homes all at once.
The higher the gauge, the lower the diameter of the wire. The lower the diameter of the wire, the more of that 66,000 watts is going to be dissipated by the wire itself instead of the load where it is desired. At 22 gauge, basically all of it will be dissipated by the wire, at least for the first fraction of a second before the wire vaporizes in a small explosion.
EDIT: In this scenario, the total resistance of the circuit must be at most 0.22 Ω. Otherwise, the current will not reach 551 A due to Ohm’s Law, V=I×R. This resistance corresponds to a maximum length of 13 feet for copper wire and no load.
So the average crypto mining rig?
about $75k of mining rigs actually. 66kW is a lot of heat to dissipate
I ran this by my brother who’s an electrician and he inferred that might be where the number is coming from, some data on how many amps you can dump into various wire gauges before they simply stop being solids.
By dissipated by the wire instead of the load what do you mean?
The wire heats up.
Wires have a small resistance which causes a voltage drop over the wire if the current is big enough (U=RI), and therefore it draws power (P=UI) which warms it up. Thinner wires have more resistance.
The rebel alliance was famously all thin wires