# Why the Square Root of 3 in 3-Phase Voltage Drops?

• jimbrosseau
In summary, Zyrn's attempts to explain to a coworker why the square root of three shows up in a three-phase voltage drop calculation, led to confusion and frustration. Zyrn attempted to explain that the voltage drop on a three-phase load involves the square root of three, but ended up being wrong. The voltages between A and B are not the sum of Va and Vb, but the difference. The voltage drop on a three-phase load is determined by the trigonometry involved with the phasors.
jimbrosseau
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 anyone 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 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.

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...

## 1. What is the significance of the square root of 3 in 3-phase voltage drops?

The square root of 3 in 3-phase voltage drops represents the relationship between line and phase voltages in a 3-phase power system. It is a constant factor of √3 that is used to convert between the two types of voltage.

## 2. How is the square root of 3 derived in 3-phase voltage drops?

The square root of 3 is derived from the trigonometric relationship between the phase and line voltages in a 3-phase system. It is the ratio of the line voltage to the phase voltage in a balanced 3-phase system.

## 3. Why is the square root of 3 important in 3-phase voltage drops?

The square root of 3 is important because it allows for a more efficient distribution of power in 3-phase systems. It reduces the amount of power and equipment needed to transmit the same amount of power as a single-phase system.

## 4. Can the square root of 3 be used in any type of 3-phase system?

Yes, the square root of 3 can be used in any type of 3-phase system, as long as it is a balanced system with equal phase voltages. In an unbalanced system, the square root of 3 may not accurately represent the relationship between line and phase voltages.

## 5. Is the square root of 3 the only constant factor used in 3-phase voltage drops?

No, there are other constant factors used in 3-phase voltage drops, such as the power factor and the impedance of the system. However, the square root of 3 is the most commonly used factor in 3-phase systems.

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