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## Main Question or Discussion Point

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?

- Thread starter jimbrosseau
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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?

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Zryn

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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.

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- #4

Zryn

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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?

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Zryn

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FOIWATER

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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?

- #8

FOIWATER

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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?

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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)

- #10

FOIWATER

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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

- #11

FOIWATER

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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.

- #12

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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

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sophiecentaur

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- #14

jim hardy

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Three phase is rich in those particular angles.

Draw the phasors, as Sophie said...

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