Current carrying conductors

[Jpflex]

No, B = C = 10A, so it would be SQRT(10² + 10² - 10*10) = 10A.

Which is a way of doing the math I mentioned in the middle paragraph of my first post, for the case of 120 degree phase shift.

Cheers, Wayne

Ok got yea

 

[wwhitney]

However I don’t yet get why in the second problem that how removing a load in a third phase will affect the remaining two phases to have a current in the neutral unless it’s as the last post stated that doing so causes the neutral to rise exactly how much amperes was removed from the line and how the third phase cancels the current on the adjacent phase. 120 degrees out ?

There's a variety of ways to explain the neutral cancellation, one would be this:

Say you walk forward 1 ft; then turn exactly 120 degrees to the left and walk forward 1 ft; then turn exactly 120 degrees to the left and walk forward 1 ft. You will end up exactly where you started, having traced the outline of an equilateral triangle.

Which is analogous to saying that if A-N, B-N, and C-N have identical 1 amp loads on them, then the 1 amp from A-N is 120 degrees from the 1 amp from B-N, which is 120 degrees from the 1 amp from C-N. And then when those 3 individual currents add up in the neutral, they exactly cancel.

Cheers, Wayne

 

[Jpflex]

You need to study 3-phase AC, as it is different than simple DC.
Poly-phase systems require some understanding of trigonometry, in particular the wine and cosine functions.

I have studied 3 phase and 120 phase displacement between phases and vectors, a2 + b2 = c2 similar to high school math

 

[Jpflex]

There's a variety of ways to explain the neutral cancellation, one would be this:

Say you walk forward 1 ft; then turn exactly 120 degrees to the left and walk forward 1 ft; then turn exactly 120 degrees to the left and walk forward 1 ft. You will end up exactly where you started, having traced the outline of an equilateral triangle.

Which is analogous to saying that if A-N, B-N, and C-N have identical 1 amp loads on them, then the 1 amp from A-N is 120 degrees from the 1 amp from B-N, which is 120 degrees from the 1 amp from C-N. And then when those 3 individual currents add up in the neutral, they exactly cancel.

Cheers, Wayne

lol good example lol

 

[LarryFine]

However I don’t yet get why in the second problem that how removing a load in a third phase will affect the remaining two phases to have a current in the neutral unless it’s as the last post stated that doing so causes the neutral to rise exactly how much amperes was removed from the line and how the third phase cancels the current on the adjacent phase. 120 degrees out ?

Presuming only L-N loads, removing current from one line has no effect on the other two lines (ignoring things like voltage drop). It only affects the shared-neutral current.

 

[Jpflex]

Charlie B’s response to my question was:

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Understanding when to count a neutral​

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  • Jpflex
  • charlie b
• Start dateToday at 2:12 PM

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Jpflex​

Senior Member​

Today at 2:12 PM
Hi Sir, this is obviously a hard concept to grasp for beginners but perhaps you might have a simple reasoning?

When further studying when to count a neutral as a current carrying conductor it is understandable that for a 3 phase 120/208 Y? volt system supplying balanced single phase loads on each of the 3 phases and derived neutral, the neutral is NOT counted.

Whether correct or not I believe, the pair of phases A and B become a “single phase / single voltage.” This creates a series circuit between two loads and the neutral junction which joins the two loads between the two phases (potentials) and therefore the neutral is not needed for the circuit to work - just as you hook up a 240 volt single phase load with A and B phase without a neutral

However, I would expect the same answer to Not count the neutral in my trivia book for the same 120/208 “Y” volt 3 phase system but now with only balanced loads between phase B and neutral and phase C and neutral with no load on phase A. Instead in this case my book says to count the neutral as current carrying

How can this be since phase B and C creat the same series circuit 208 volts with 2 x the load resistance as the first example?

At this point I would only understand to not count a neutral if I connect an ammeter around a neutral and find no current on a live functioning circuit

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charlie b​

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7 minutes agoNew

However, I would expect the same answer to Not count the neutral in my trivia book for the same 120/208 “Y” volt 3 phase system but now with only balanced loads between phase B and neutral and phase C and neutral with no load on phase A. Instead in this case my book says to count the neutral as current carrying.

Your book is wrong. If B and C have the same load and A has no load, then the neutral will have the same load as B and C. The total heat generated by any combination of loads on the three phases, with the neutral only carrying the unbalanced load among the three phases, will never be higher than that generated from balanced load on the three phases.

Now, please post any further questions on the open forum. It's best to get more than one opinion on technical issues.

 

[LarryFine]

Say you walk forward 1 ft; then turn exactly 120 degrees to the left and walk forward 1 ft; then turn exactly 120 degrees to the left and walk forward 1 ft. You will end up exactly where you started, having traced the outline of an equilateral triangle.

To expand on that, imaging three guys of equal weight standing in a triangle with their left feet on the ground and their right feet pressed together in the center of the triangle, up off the ground.

As long as they apply equal pressure with their right feet, they're as stable as any tripod.

Now imagine one guy walks away.

 

[Jpflex]

According to Charlie my book is giving a wrong answer on when to count a neutral

The second problem removes one load from a phase and says to count the neutral answer on third picture

 

[jim dungar]

I have studied 3 phase and 120 phase displacement between phases and vectors, a2 + b2 = c2 similar to high school math.

Then you know how to add two or three vectors together.

 

[Jpflex]

Then you know how to add two or three vectors together.

It’s been a while I would have to review my book on vector sums but it’s at work 3 hours away. I will have to keep at this

 

[jim dungar]

It’s been a while I would have to review my book on vector sums but it’s at work 3 hours away. I will have to keep at this

You don't need to go deep into the weeds for this general discussion. A simple sketch, using approximate angles, may be useful. For example you know that you connect one vector "arrow" to the end of the vector you are adding it to. This process is continued until you get back to your starting point.

If you do this with 3 equal phase currents, A, B, and C, you will create a triangle. If the phase currents are not equal your diagram will become a 4 sided rhombus, instead of a triangle, because of the added neutral vector.

The formulas you have been given can be used instead of a graphical method.