Thanks Zryn, however, I'm still not clear how that implies the voltage drop on a three-phase load involves the Sqrt(3). After all, there is no return load on a balanced three-phase load, and thus why would one consider line-to-line voltages? It just seems to me that considering any one of the three phase conductors at a given voltage (say 120V) would result in a voltage drop of V=IZ where Z is the impedance of the conductor and I being the motor's draw. I'm not convinced yet :(
Shouldn't voltage Vab be the sum of the two phase voltages Va and Vb, not the difference as Zyrn suggests???
is it really Vab = Va-Vb, that would suggest The difference between A and B is smaller than Va to ground, which isn't the case, because Vab is obviously 1.73 times larger... So the proof works, but it isn't correct, is it?
Don't mean to dredge this up from the past, but I think you've got the vector addition incorrect.
Vab = (-Va) + Vb = Vb - Va
As was stated earlier, Vab is a voltage phasor (or vector) whose tail is at "a" and whose head is at "b", thus meaning phasor from "a" to "b". To traverse this vector path one has to go backwards along Va (hence the minus) and then forwards along Vb, this leads to the minus sign when the re-arrangement is done as above.
The reason for the "Root Three" turning up is the trigonometry involved with the phasors. Your three phasors point from the origin in three directions, separated by 120°. The potential between any two of these phasors is 2V0Cos(60°), which is √3V0. Draw out the vector diagram and work out the distance between the vertices of the triangle made up of the ends of the vectors.