Yes, you're right. I was wrong. Thanks for explaining. My calculations are below. Hopefully you have enlightened me; please let me know if I'm still confused.
4awg wire is ¼Ω per thousand feet (6dB of resistance below 10AWG), or 0.005Ω for 20 feet. 20 amps × 0.005Ω = 0.1 volts. So the voltage drop is 0.1 volts, which is 0.8% of 12V or 0.08% of 120V. If it's 20 feet each way, it's 1.7% or 0.17%.
The low-voltage gets you double: you need more amps to carry the same power (which is what I thought you were saying) and you also need thicker wire to carry the same current.
So, to take a worst-case example, if you have a 900W microwave oven that can tolerate a power supply 10% lower than nominal, and you're running it on 120V 20 feet from the breaker box, you can deal with 6 volts of drop on each side of the romex, at which point you need 900/(120×.9) = 8⅓A. 6V/8⅓A = 0.72Ω, which means your wire can have up to 36Ω per thousand feet, which would be 25AWG, which is about half a millimeter across and commonly used for Ethernet. Not safe, because you're dissipating more than a watt per foot, which might set something on fire, but the microwave will still run. (For safety you'd probably use 12AWG instead, which is 2mm across, if not 10AWG.)
If instead you're running a different 900W microwave off 12V, you need 900W/(12V×0.9), which is 83 amps. But you can't afford more than 0.6V drop on either side of the rails, and 0.6V/83A = 7.2mΩ, so you can't afford more than 0.36Ω per thousand feet, which is 5AWG, which is a copper bar 4.6mm in diameter, thicker than any wire in a normal person's house. Again, that's not to be safe — that's to get the microwave to work at all. You're still dissipating over a watt per foot of cable. To be safe, you need 2AWG, which is 6.5mm in diameter, and a pain in the ass to bend if you don't get stranded wire. 20 feet of copper 2AWG triple-stranded wire costs US$150, which is more than the microwave, and weighs 15 pounds.
4awg wire is ¼Ω per thousand feet (6dB of resistance below 10AWG), or 0.005Ω for 20 feet. 20 amps × 0.005Ω = 0.1 volts. So the voltage drop is 0.1 volts, which is 0.8% of 12V or 0.08% of 120V. If it's 20 feet each way, it's 1.7% or 0.17%.
The low-voltage gets you double: you need more amps to carry the same power (which is what I thought you were saying) and you also need thicker wire to carry the same current.
So, to take a worst-case example, if you have a 900W microwave oven that can tolerate a power supply 10% lower than nominal, and you're running it on 120V 20 feet from the breaker box, you can deal with 6 volts of drop on each side of the romex, at which point you need 900/(120×.9) = 8⅓A. 6V/8⅓A = 0.72Ω, which means your wire can have up to 36Ω per thousand feet, which would be 25AWG, which is about half a millimeter across and commonly used for Ethernet. Not safe, because you're dissipating more than a watt per foot, which might set something on fire, but the microwave will still run. (For safety you'd probably use 12AWG instead, which is 2mm across, if not 10AWG.)
If instead you're running a different 900W microwave off 12V, you need 900W/(12V×0.9), which is 83 amps. But you can't afford more than 0.6V drop on either side of the rails, and 0.6V/83A = 7.2mΩ, so you can't afford more than 0.36Ω per thousand feet, which is 5AWG, which is a copper bar 4.6mm in diameter, thicker than any wire in a normal person's house. Again, that's not to be safe — that's to get the microwave to work at all. You're still dissipating over a watt per foot of cable. To be safe, you need 2AWG, which is 6.5mm in diameter, and a pain in the ass to bend if you don't get stranded wire. 20 feet of copper 2AWG triple-stranded wire costs US$150, which is more than the microwave, and weighs 15 pounds.