It might look like this: http://www.adafruit.com/products/357 - Buy your lighting by the foot. 12V DC may make more sense for homes these days than 115 AC. The incandescent bulb was the "killer app" for a whole power ecosystem. In older houses in places like San Francisco you sometimes find wiring going through the old gas lamp lines.
> 12V DC may make more sense for homes these days than 115 AC
Have you ever sized any 12v wiring? To carry a load of 22amps 20 feet with a voltage drop of <2% with a 12V system requires 4awg wire. That's crazy big and gets very expensive.
22amps might sound like a lot and lighting may be getting more efficient, but you'll still want to run things like vacuum cleaners.
> To carry a load of 22amps 20 feet with a voltage drop of <2% with a 12V system requires 4awg wire.
The wire size needed to carry a load of a specific number of amps does not vary with the voltage between that wire and some other wire somewhere. Perhaps you mean, "To carry a load equivalent to 22 amps on 120 volts, a 12-volt system needs to carry 220 amps, which requires 4AWG wire to lose less than 2% over 20 feet."
> does not vary with the voltage between that wire and some other wire somewhere.
I'm not following what you're saying here. I was just pointing out that voltage drop can be a significant concern with low voltage systems and controlling that can require thick wires.
From my rough calculations, voltage drop at 12V over 20' of 4awg copper @ 20A is ~1.7% and @ 100A is ~8.6%.
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.
