Doesn't this support my readings then? I'm using 2 phases of a 3 phase system. Even if my 2 phases, A and B, are carrying equal current, the neutral will not carry 0 current because the C phase vector (which I'm not using) of 0 amps needs to be included in the calculation. This makes sense because A and B are 120° and not 180 out of phase.

I would say, yes.
Can't get a straight answer out of these guys. They seem to be back tracking.

I did the math and used an online calculator. Everything I find agrees with your readings.

Who are you gonna believe? Your lying eyes?

If I'm not convinced by tomorrow, I'll connect a couple 300 watt bulbs across 208. Then run a wire from the center tap to neutral and see for myself.

 

you should agree on the definition of neutral. Next to ground, it's probably the most confused. Neutrals do not exist in 2 wire circuits, so the statement
is incorrect. Neutrals, by definition, only carry the imbalanced current.
Grounding has nothing to do with the neutral definition, nor it's calculated currents.

Consider a three-phase wye-connected load that is only connected to the three lines. It has a floating neutral point. No matter how imbalanced the loads are between the phases, their is no neutral current because there is no path for them to flow through. Now ground the neutral (and I mean connect it to a big hunk of metal that is driven many feet into the earth and assuming that the supplying generator/transformer is also grounded) and you will have current in the neutral if there is an imbalance in the loads.

The 'extra' power comes from the fact that you are employing different voltages. These supplies are typically identified as 120/208Y.

Not much of an explanation. After all, the same statement applies to a 120V/240V system.

 

There has to be voltage from center of resistors to neutral.
Current flows!

Yep!

Are you changing your story now? It seems you're taking my side.
There either will be neutral current in a balanced multi-circuit, as I picture it, or there is not.

Don't go back to it must be either/or.

I'm not changing my story, it is you who are changing yours! And that's a good thing. Remember, you started out from a position involving claims such as, "The neutral carries MORE than the unbalance however," and, "There can be no conditions where neutral current is zero and phase currents are more than zero."

You've accepted that these are not guaranteed true statements. That does not mean that the converse of the last one is true, however.

I've been hoping that you would do the math.

You have a two-wire service from a 120V/208V drop. The wiring within the establishment is then supplemented with a neutral grounded at the service entrance (does not describe how it is actually done in every part of the world, of course). Relative to the neutral, the lines are

Vph1 = 120V @ +60°
Vph2 = 120V @ -60°

Of course, the phase reference is arbitrary; what matters is that they are 120° apart.

The line-to-line voltage, Vph1-Vph2, is 208V @ +90°.

Now, let's connect up two equal resistors in series across the two lines.

The voltage at the center point of the resistors is either

Vcp = Vph1 - 104V @ +90° = 120V @ +60° - 104V @ +90°

or

Vcp = Vph1 + 104V @ +90° = 120V @ -60° + 104V @ +90°

Take your pick.

In either case, you end up with

Vcp = 60V @ 0°

So the center of this "balanced" two-phase load is definitely NOT 0V. This means that if you tie this point to the neutral point of the supply, you will get a current flowing in the neutral. That current is going to change the currents flowing in the two resistors -- which are no longer in series -- such that they now each have 120V across them but at the phase angles of the respective phases they are being powered by.

There's nothing magical here. You've just got to run the numbers and do the accounting carefully.

 

There has to be voltage from center of resistors to neutral.
Current flows!

Yep!



Don't go back to it must be either/or.

I'm not changing my story, it is you who are changing yours! And that's a good thing. Remember, you started out from a position involving claims such as, "The neutral carries MORE than the unbalance however," By unbalanced I meant difference. Wrong term maybe.....and, "There can be no conditions where neutral current is zero and phase currents are more than zero." Explained that. We were talking mullti-circuits. That implies neutral connected to center point of the load.

You've accepted that these are not guaranteed true statements. That does not mean that the converse of the last one is true, however.

I've been hoping that you would do the math.

You have a two-wire service from a 120V/208V drop. The wiring within the establishment is then supplemented with a neutral grounded at the service entrance (does not describe how it is actually done in every part of the world, of course). Relative to the neutral, the lines are

Vph1 = 120V @ +60°
Vph2 = 120V @ -60°

Of course, the phase reference is arbitrary; what matters is that they are 120° apart.

The line-to-line voltage, Vph1-Vph2, is 208V @ +90°.

Now, let's connect up two equal resistors in series across the two lines.

The voltage at the center point of the resistors is either

Vcp = Vph1 - 104V @ +90° = 120V @ +60° - 104V @ +90°

or

Vcp = Vph1 + 104V @ +90° = 120V @ -60° + 104V @ +90°

Take your pick.

In either case, you end up with

Vcp = 60V @ 0°

So the center of this "balanced" two-phase load is definitely NOT 0V. This means that if you tie this point to the neutral point of the supply, you will get a current flowing in the neutral. That current is going to change the currents flowing in the two resistors -- which are no longer in series -- such that they now each have 120V across them but at the phase angles of the respective phases they are being powered by.

There's nothing magical here. You've just got to run the numbers and do the accounting carefully.

In electrician terminology, we call the difference in load, of one phase to the other on a muti-circuit, the unbalance.

Wasn't looking for a lesson. In fairness you may have thought it was the homework forum.
I was just backing up the OPs well explained observation that he had a high neutral current in a fairly even loaded muti-circuit.

A simple agreement with his findings would have put it to rest.

It's easy to pick fault with a poor choice of words. Electricians may use different terms than teachers, mathematicians, and engineers.

 

Consider a three-phase wye-connected load that is only connected to the three lines. It has a floating neutral point. No matter how imbalanced the loads are between the phases, their is no neutral current because there is no path for them to flow through. Now ground the neutral (and I mean connect it to a big hunk of metal that is driven many feet into the earth and assuming that the supplying generator/transformer is also grounded) and you will have current in the neutral if there is an imbalance in the loads

sure, and a tree could fall across your transmission lines, but I doubt that would justify redefining your electrical terms. Are you suggesting that a high impedance, discontinuous path be termed 'neutral'?

Not much of an explanation. After all, the same statement applies to a 120V/240V system.

well no actually, there is a substantial difference, however it appears the OP may have that one figured out.

Doesn't this support my readings then? I'm using 2 phases of a 3 phase system. Even if my 2 phases, A and B, are carrying equal current, the neutral will not carry 0 current because the C phase vector (which I'm not using) of 0 amps needs to be included in the calculation. This makes sense because A and B are 120° and not 180 out of phase.

you can see then that you can't add your currents directly, they must be added vectorially.

 

Now ground the neutral (and I mean connect it to a big hunk of metal that is driven many feet into the earth and assuming that the supplying generator/transformer is also grounded) and you will have current in the neutral if there is an imbalance in the loads.

There is some thinking that earth grounding a conductor changes its characteristics somehow?
All it does is reference the particular conductor to earth ground, there are no consequences in the change of any current flow what so ever.
In a star point of a 3ph supply the star point is termed neutral whether it is earth grounded or not.
Nothing otherwise changes.
Max.

 

In a star point of a 3ph supply the star point is termed neutral whether it is earth grounded or not.
Nothing otherwise changes.

and, a wire (white by code) that extends this point through out the system, such that currents from more than one phase are able to flow within, are deemed to be the 'neutral'. The NEC has graciously (or confusingly) allowed the definition to continue with a wire that is beyond the multi phase point, to include a single phase potential, by virtue of being connected to the 'neutral'. The CEC has always maintained a distinction of the later single phase wire, by naming it the 'identified' wire.

 

There has to be voltage from center of resistors to neutral.
Current flows!

Yep!



Don't go back to it must be either/or.

I'm not changing my story, it is you who are changing yours! And that's a good thing. Remember, you started out from a position involving claims such as, "The neutral carries MORE than the unbalance however," and, "There can be no conditions where neutral current is zero and phase currents are more than zero."

You've accepted that these are not guaranteed true statements. That does not mean that the converse of the last one is true, however.

I've been hoping that you would do the math.

You have a two-wire service from a 120V/208V drop. The wiring within the establishment is then supplemented with a neutral grounded at the service entrance (does not describe how it is actually done in every part of the world, of course). Relative to the neutral, the lines are

Vph1 = 120V @ +60°
Vph2 = 120V @ -60°

Of course, the phase reference is arbitrary; what matters is that they are 120° apart.

The line-to-line voltage, Vph1-Vph2, is 208V @ +90°.

Now, let's connect up two equal resistors in series across the two lines.

The voltage at the center point of the resistors is either

Vcp = Vph1 - 104V @ +90° = 120V @ +60° - 104V @ +90°

or

Vcp = Vph1 + 104V @ +90° = 120V @ -60° + 104V @ +90°

Take your pick.

In either case, you end up with

Vcp = 60V @ 0°

So the center of this "balanced" two-phase load is definitely NOT 0V. This means that if you tie this point to the neutral point of the supply, you will get a current flowing in the neutral. That current is going to change the currents flowing in the two resistors -- which are no longer in series -- such that they now each have 120V across them but at the phase angles of the respective phases they are being powered by.

There's nothing magical here. You've just got to run the numbers and do the accounting carefully.

Almost all of this makes sense to me but I'm still a little confused as to why the neutral will never carry more current than the max on either of the 2 phases. So with A and B both carrying 20 amps, C carrying 0, the neutral will carry 20 amps.

I found this online calculator that is pretty cool.

http://www.electrician2.com/calculators/el_calculator_vector.html

 

sure, and a tree could fall across your transmission lines, but I doubt that would justify redefining your electrical terms. Are you suggesting that a high impedance, discontinuous path be termed 'neutral'?

It is my understanding -- with the caveat that terminology varies across the board all over the world -- that the common point of a wye-connect generator set or of a wye-connected load is often referred to as the "neutral point" whether is it grounded or not and whether it is run back to the generator or not. If it is not referred back then it is identified as being a "floating neutral".

I'm not saying I like that terminology and I'm most certainly not saying that it is engraved-in-stone universal. I would prefer something like "common point" to identify the common point of components that are why- or star- connected.