Gentlemen, you can't fight in here. This is the War Room.
I should know better than to step in the middle of a pissing contest, but sometimes I just do irrational things.
Drop/Supply = % drop
.5 / 240 = .2 % AC line sag for 240Vac 10 Gauge 20amp circuit
1.3 / 120 = 1.1 % AC line sag for 120Vac 10 Gauge 20amp circuit
Hmmmm. P=IE doesn't seem to hold true. Splain that to me, Lucy. I can't wait to see this math. Tell us Nostradamus how you achieved such a feat?
MudPuppy is usually spot on about things electrical, but I think either he's mistaken here, or he hasn't explained well enough for me to understand why the result bstang is seeing is wrong.
For instance, how does P = IE apply to the quote above?
He has the exact same physical branch circuit wires, so presumably the resistance of this wiring will be the same in both cases. In one case they're wired as a 120 V circuit (white tied to the neutral bus in the panel, black to a 20 A circuit breaker) and the other case as a 240V circuit (white tied to the second breaker of a CB pair). Is that right, bstang?
Theory says to expect one quarter the power loss delivering the same amount of power at twice the voltage. This is close to what we're seeing here. This is why power is distributed at high voltages.
Power Loss = I * Vdrop
120V supply:
PL = 5A * 1.3 V = 6.5 W (5 A is presumed from earlier test)
240V supply:
PL = 2.5A * .5V = 1.25 W (current is presumed to be half previous case since it's what would be expected, and E = IR voltage drop is approximately half, through constant R)
It's not a laboratory experiment, and the results aren't exact, but they are certainly in line with the expected values.
What am I missing?
BTW, 'stang: those voltage drops still seem WAY TOO HIGH even though they are much better than they were. 1.3V sag with a 5A load @ 120V (if that's correct) would be 5.2V with a full 20A load. While that's (barely) less than 5%, it should be much better because you're using oversize wiring (#10 on a 20A circuit). How long is the run from panel to outlet?