Banging my head against a wall trying to prove to a coworker why the square root of three shows up in a three-phase voltage drop calculation...any ideas?
The voltage at VAB comprises of VA - VB (which are equal in magnitude): VAB = VA - VB = V@0 - V@-120 = V [1@0 - 1@-120] V [1@0 - 1@-120] = V [cos(0) + jsin(0) - {cos(-120) + jsin(-120)}] = V [1 - cos(120) + jsin(120)] V [1 - cos(120) + jsin(120)] = V [1 + 1/2 +j sqrt(3)/2] = V [ 3/2 + j sqrt(3)/2] V [ 3/2 + j sqrt(3)/2] = V [sqrt(3)@30] Therefore the Voltage between A & B is sqrt(3) the magnitude of either A or B and is 30 degrees out of phase.
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 :(
If you have a 'Delta' connection the line and phase voltages are the same, so I'll assume you have a 'Wye' connection. In a 'Wye' connection VA (line) = sqrt(3) VAB (phase), so when you say V = IZ and you are talking about a single conductor, do you mean V/sqrt(3) = IZ?
That's the thing!!! It doesn't matter. You look up any 3-phase voltage drop calculator and they do not discern between Delta or Wye connected loads or sources.
Do you have a link to the specific calculator you are using? They are probably just using certain assumptions which validate the result.
"After all, there is no return load on a balanced three-phase load, and thus why would one consider line-to-line voltages?" on a 3 phase system, phase b is phase a return, phase c is phase b return, and phase a is phase c return. The voltages are different line to line than line to neutral because of the proof than zyrn posted. So there are in fact cases where the line to line voltages are important... i.e. any load connected between two or even three phases?
sorry to bump this old thread but wait... 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? Can someone please set me straight here?
Yes, Zyrn is right. Voltage is potential _difference_, hence the subtraction. If you know phasor notation/diagram (similar to vectors), look here: http://www.electronics-tutorials.ws/accircuits/acp50.gif Where Vab is an phasor with arrow at Va, and tail at Vb. Hence Vab=Va-Vb (normal vector subtraction/addition)
that link does not work for me. Yeah, I drew the circuit, Vab is a vector whos head is at a , tails at b. Current flows from a, returns on b so voltage is dropped across a and b in the opposite direction as Vab?... therefore its the sums of Va and Vb that make up Vab? Still not getting this
is the volt drop on Va opposite with respect to the volt drop on Vb? I mean current only flows in one direction through the two phases, That is why I have trouble seeing how it could be the DIFFERENCE of the voltages Hmm... I think I might get it now. Maybe Va and Vb are opposite.
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. Regards Darryl
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 2V_{0}Cos(60°), which is √3V_{0}. Draw out the vector diagram and work out the distance between the vertices of the triangle made up of the ends of the vectors.
Recall that a 30-60-90 degree triangle has sides with proportion 1-2-√3. Three phase is rich in those particular angles. Draw the phasors, as Sophie said...