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In this article, the author talks about 3 phase and the magic of a missing neutral wire. And there's the following paragraph:

In a 3-phase system the voltage between any two phases is 3 times higher than the voltage of an individual phase by a factor of 1.73 (square root of 3 to be exact). If your X-N (and Y-N and Z-N) voltage is 120V (common in the US), the X-Y (and Y-Z and Z-X) voltages (a.k.a. “cross-phase” voltages) will be 120V * 1.73 = 208V.

When he says the X-N voltage is 120V, with N being the reference voltage, is the reference voltage 0V or does it have some value?

From the diagram (in the link,) I realise that the neutral wire is grounded, but is this the same as earth grounding? Doesn't the neutral wire go back to the transformer so as to provide a closed circuit?

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    \$\begingroup\$ I think you're not understanding something here. There is no absolute value of voltage, just voltage differences. \$\endgroup\$ – Hearth May 5 at 12:43
  • \$\begingroup\$ No I did understand that part. I'm saying the 120V voltage difference between X-N can be either 121-1 or 120-0. Which one is it? \$\endgroup\$ – noorav May 5 at 12:45
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    \$\begingroup\$ The fact that you're asking that shows that you don't understand. \$\endgroup\$ – Hearth May 5 at 12:46
  • \$\begingroup\$ I'm sorry, but where is my understanding flawed? This is how I've been learning about voltage difference for so long. Could you please correct me \$\endgroup\$ – noorav May 5 at 12:47
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    \$\begingroup\$ Well, you're asking whether there's a voltage on the neutral wire. A voltage relative to what? \$\endgroup\$ – Hearth May 5 at 12:49
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The neutral wire is defined to be 0V. And each phase individually has a voltage relative to neutral of 115V if you are in the US, and 230V if you are in most of Europe.

But because the three phases are 120 deg out of phase compared to each other this results in a voltage difference of 115V * sqrt(3) = 200V (or 230V * sqrt(3) = 400V) between any two phases.

Depending on the type of mains distribution system you've got in the country you live (I don't know how it is in the US) the neutral wire might be connected to earth at the power company. Keep in mind however, that even though the neutral wire might be connected to earth, this doesn't mean that it is actually at earth potential, because a return current might be flowing in the neutral wire, and due to the resistance in the wire this will cause a voltage at your end of the neutral wire.

It is also important to keep in mind that in many countries the mains plugs we use can actually be turned 180 deg putting line where neutral was supposed to be, meaning that with these kinds of systems you can never be sure if your neutral is actually neutral or line (BE CAREFUL!)

So to sum it up: The neutral wire is defined to be 0V. But will typically be at some voltage potential (relative to earth).

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    \$\begingroup\$ Sometimes, earth is defined to be zero volts instead of neutral. Though they're nominally the same, they can differ. This can be important, and I feel like it ought to be mentioned at least. \$\endgroup\$ – Hearth May 5 at 16:21
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The original article you've quoted is garbled.

... the voltage between any two phases is 3 times higher than the voltage of an individual phase by a factor of 1.73 (square root of 3 to be exact).

It can't be three times higher and \$ \sqrt 3\$ times higher simultaneously. The correct value is \$ \sqrt 3\$ times higher.

enter image description here

Figure 1. The diagram in question. Source: PacketPower.

When he says the X-N voltage is 120 V, with N being the reference voltage, is the reference voltage 0 V or does it have some value?

The reference is the wye (star) point where the three windings share a common node. In this diagram it is neutralised by connecting to Earth. (That means that this conductor should not see any significant voltage on it with respect to earth.) There will be 120 V between each of the phases and the wye point whether it is earthed or not.

From the diagram (in the link,) I realise that the neutral wire is grounded, but is this the same as earth grounding? Doesn't the neutral wire go back to the transformer so as to provide a closed circuit?

It depends on local regulations. If we're dealing with a transformer secondary here then typically the primary will be delta powered so there may be no neutral on the incoming supply. The solution is to ground the wye to the building's earth bonding in which case it would be the same as earth grounding. In the case of a ship, for example, the "earth" would be the ship's hull.

enter image description here

Figure 2. A delta-wye (delta-star) transformer connection. Source: Gamatronic.

Here we can see that there is no neutral connection on the incoming supply. Using a wye configuration on the secondary allows us to create one for internal use. The wye point can be left floating or can be grounded.

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The neutral wire is grounded, so if you literally took a multi-meter and tried to read voltage between the neutral wire and ground, you will read 0 voltage. Please note, this is for a healthy working balanced system.

So yes, consider the neutral wire 0V if everything is in working order.

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  • \$\begingroup\$ That depends completely on the type of distribution system is used in the country you live in. In most of Europe the neutral is indeed connected to earth at the power company, but this is not always the case/ not the case in all countries.. \$\endgroup\$ – Vinzent May 5 at 13:02
  • \$\begingroup\$ Do you have an example where someone is running power through the neutral wire? Would be interesting to see this setup. \$\endgroup\$ – Busta May 5 at 13:05
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    \$\begingroup\$ You can read this wiki page about earthing systems: en.wikipedia.org/wiki/Earthing_system, It describes the different types of earthing systems, such as TN, TT, IN and IT.. \$\endgroup\$ – Vinzent May 5 at 13:10
  • \$\begingroup\$ Thanks for the link. So I will default that my answer works in traditional WYE (Star) / Delta systems that use a 4-wire 3-phase traditional set-up where the neutral is tied to the ground. Hence, if you tried to read voltage between neutral and ground...theoretically you should get 0volts but your answer explained why you may see some voltage.. The original question gave the assumption that the neutral wire is grounded. \$\endgroup\$ – Busta May 5 at 13:18
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"Here we can see that there is no neutral connection on the incoming supply."

So what provides the return path for flow of current?

This was asked as a comment, but to answer it I need far more space than a comment allows.

With a simple two wire AC connection, the outgoing current on the hot wire is has the exact same absolute magnitude as the return current in the neutral wire, but they have opposite signs. In addition to this, the current is actually a sine wave.

Now, lets take three copies of this, but adjust the phase of two of the circuits so that all three are 120 degrees out of phase with respect to each other.

Next, lets replace the three neutral wires with one big wire three times the capacity. Up to this point, everything is working just fine.

However, lets now take a close look at the current flowing in the neutral wire. It'll be the sum of the three individual component flows, i.e. the sum of three sine waves, each 120 degrees out of phase with respect to the others.

If you either draw it out, or sum it up mathematically, those three sine waves always add up to zero, at any point in time. This in turn means that there's actually no current flowing in that huge return wire. So why not just remove it, and let the three hot wires carry on doing what they're doing.

In practice, this works well enough that we can transmit power over large distances without an explicit return wire. Due to imbalances in the loads, there will be some return current, but it's generally small enough that using Earth as a return is sufficient.

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  • \$\begingroup\$ Actually, you don't even need the earth as a return. A delta-star transformer (what is found in most cases where voltage is stepped down for final use) converts line-neutral loads to line-line loads. Any imbalance in loading simply shows up as an imbalance of phase currents, not also a neutral current. \$\endgroup\$ – Someone Somewhere May 6 at 0:10
  • \$\begingroup\$ Star systems typically use a neutral conductor capable of carrying the same current as the phase conductors (or slightly more due to harmonics). Going smaller is not advisable as it is entirely possible for one phase to be very heavily loaded while the others are light. \$\endgroup\$ – Someone Somewhere May 6 at 0:12

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