Dereck
I had a chance to look at the problem and I will stay with the origional statement. Lets look at your example.
100 watt bulb 144 ohms
2 wire conductor resistance of 2 ohms ea for a total of 4 ohms.
amp flow = 120v/148 ohms = 0.81 amps
watts cable loss = .81 x .81 x 4 = 262 watts
bulb watts = .81 x .81 x 144 = 94.5 watts
Assume that you reduce the wire ohms to 0.5
for a total of 1 ohm.
amp flow = 120/145 ohms = 0.83 amps
watts cable loss = 0.83 x 0.83 x 1 =.69 watts
bulb watts = 0.83 x 0.83 x 144 = 99.2 watts.
Your quote "Bob, I have to disagree with your statement with respect to resistive loads. If you were to use a larger wire with less resistance on a resistive load, you would increase the load current, thereby cost more to operate, not save". is not correct because there is not a direct relationship between the resistance and watts loss. It is a current squared relationship.
