MWBC= more heat or Less heat?

[gar]

090707-2114 EST

I agree with mivey and Larry. In the physical world with a finite resistance and zero potential difference across the resistance this means there is no current flowing.

Because you can create a mathematical analog of the real world where you have for analytical purposes two exactly equal currents that cancel each other does not mean those currents exist in the real world.

If there is no voltage, no current, and a finite resistance, then there is no power dissipation.

.

 

[david luchini]

Where did the potential go?

Where did the potential go?

Larry seems to be saying that the equation x+y=z if valid if x=2 and y=-1 so that z will equal 1, but if x=2 and y=-2, then the equation will no longer be valid because z will equal 0. Kirchoff's current law in this case In=Ia+Ib applies whether the loads are balanced or not.

And what happened to the potential? You say if there is zero potential between two points, then zero current will flow. True enough, but how did the potential magically disappear? Imagine a transformer secondary of 240/120 center tapped. I will measure a potential difference from A to N of 120V and from B to N of 120V. Now run a conductor from A to 1000 watt light bulb, and from the bulb back to the neutral. I will still measure a potential difference between A and N of 120V, and a current of 8.3A will flow through the completed circuit. Next run a conductor from B to a 1000 watt light bulb, and from the bulb back to neutral. I will still measure a potential difference from B ton N of 120V, and a current of 8.3A will flow through the completed circuit.

Now imagine a jumper is installed between the neutral connections at the two light bulbs. I will still measure a potential difference of 120V from A to N, of of 120V from B to N. You are suggesting that this potential no longer exists. This is not true. The potential did not magically disappear. If instead you are suggesting that there is zero potential from the shared neutral connection back to the transformer neutral (so that no current will flow) this is irrelevant. How much potential exists between terminal A and the light bulb along the supply conductor? It is essentially zero, yet current still flows along that conductor.

Also, you suggest that since opening the neutral will have no effect on the current, that my theory is wrong. Your belief is completely false. As I stated earlier, if you open the neutral in this circuit, you have created a completely new circuit. In the original circuit, KCL tells us that In=Ia+Ib (neutral current out of the node equals the sum of the two phase currents into the node) because there are three legs from the common node. Once the neutral is opened, KCL tells us that Ia=-Ib (or that the sum of the two phase currents into the node equals 0.) By having to change the circuit in order to prove your point, you have demonstrated that my theory is correct.

 

[LarryFine]

Larry seems to be saying that the equation x+y=z if valid if x=2 and y=-1 so that z will equal 1, but if x=2 and y=-2, then the equation will no longer be valid because z will equal 0.

What's wrong with z being equal to 0?

Kirchoff's current law in this case In=Ia+Ib applies whether the loads are balanced or not.

What I'm saying is that, in your 3-wire node example, if Ia = Ib, there is nothing left for In. If In = 0a, why must there still be equal-but-opposite currrents to be 0a, when there can simply be no current at all to be 0a?

And what happened to the potential? You say if there is zero potential between two points, then zero current will flow. True enough, but how did the potential magically disappear?

By the voltage drops across A and B being equal.

Imagine a transformer secondary of 240/120 center tapped. I will measure a potential difference from A to N of 120V and from B to N of 120V. Now run a conductor from A to 1000 watt light bulb, and from the bulb back to the neutral. I will still measure a potential difference between A and N of 120V, and a current of 8.3A will flow through the completed circuit. Next run a conductor from B to a 1000 watt light bulb, and from the bulb back to neutral. I will still measure a potential difference from B ton N of 120V, and a current of 8.3A will flow through the completed circuit.

Correct. That gives us two independent 2-wire circuits, with four current-carrying conductors along with their associated I^2R losses. This is what we want to compare a 3-wire MWBC to.

Now imagine a jumper is installed between the neutral connections at the two light bulbs. I will still measure a potential difference of 120V from A to N, of of 120V from B to N. You are suggesting that this potential no longer exists.

No, we're not. We're saying that the voltage between the source N and the load N is non-existent as long as the load currents, and thus their respective voltage drops, are equal.

This is not true. The potential did not magically disappear. If instead you are suggesting that there is zero potential from the shared neutral connection back to the transformer neutral (so that no current will flow) this is irrelevant.

No, that's the core of our point. We're saying that, because the two original neutral conductors are now in parallel, they are effectively one (and can be replaced by one), and that one conductor will carry no current with balanced loads.

How much potential exists between terminal A and the light bulb along the supply conductor? It is essentially zero, yet current still flows along that conductor.

Yes, if we agree that, for the sake of these discussions, voltage drop on conductors is zero.

Even if we consider that, assign a 2v drop over each wire, and say each load sees 116v with the 4-wire twin circuits, each will see 118v with either a 3-wire MWBC or with them in series across 240v with no neutral.

Also, you suggest that since opening the neutral will have no effect on the current, that my theory is wrong. Your belief is completely false. As I stated earlier, if you open the neutral in this circuit, you have created a completely new circuit.

I agree that, physically speaking, removing the neutral creates a new circuit, but, electrically speaking, the two circuits function identically (as long as the loads remain balanced, of course.)

In the original circuit, KCL tells us that In=Ia+Ib (neutral current out of the node equals the sum of the two phase currents into the node) because there are three legs from the common node.

Okay so far.

Once the neutral is opened, KCL tells us that Ia=-Ib (or that the sum of the two phase currents into the node equals 0.) By having to change the circuit in order to prove your point, you have demonstrated that my theory is correct.

I don't agree with that assessment. Earlier, we agreed that Ia + Ib = 0. I also agree that Ia = -Ib.

I don't agree that the two are mutually exclusive. I believe that Ia + Ib = 0 applies whether the node has two or three paths.

The cruxt of our point is that removing the neutral is not a change of the circuit, functionally speaking, and that it creates no change in current flow proves our point.

 

[SAC]

I found a pretty cool circuit simulator that runs in a browser. For those who are interested, I've put up two examples showing the basic principles being discussed here for a mwbc and a swbc. If you hover the mouse over the "scope plots" it will highlight the element that it is showing the information for. It is fully editable and has a link to the author's page for the app with full instructions.

http://ee.mxpark.com/mwbc/mwbc.html
http://ee.mxpark.com/mwbc/swbc.html

 

[LarryFine]

I found a pretty cool circuit simulator that runs in a browser. For those who are interested, I've put up two examples showing the basic principles being discussed here for a mwbc and a swbc. If you hover the mouse over the "scope plots" it will highlight the element that it is showing the information for. It is fully editable and has a link to the author's page for the app with full instructions.

http://ee.mxpark.com/mwbc/mwbc.html
http://ee.mxpark.com/mwbc/swbc.html

Way cool! You can vary the movements, or halt them completely, then hover to read the numbers.

 

[david luchini]

The cruxt (sic) of our point is that removing the neutral is not a change of the circuit

Larry, I'm astounded that you would say that if you have a MWBC with two phases and a shared neutral, that removing the neutral is NOT a change of circuit. Removing the neutral creates a COMPLETELY different circuit.

No, we're not. We're saying that the voltage between the source N and the load N is non-existent as long as the load currents, and thus their respective voltage drops, are equal.

This is wrong, the "voltage between the source N and the load N is non-existant whether the load is balanced, unbalanced, or if you have a single branch circuit (again assuming and ideal circuit where the conductor impedances do not figure into the circuit.) Consider a single branch circuit with 120V from A-N and a 1200W resistive load in the circuit. The measured voltage from A-N is 120V, the measured voltage across the load will be 120V, the measured voltage from A-source to A-load will be 0, and the measured voltage from N-source to N-load will be zero. There is not a potential difference along the neutral which pushes the current along the neutral. In an unbalanced MWBC or a balanced MWBC there is still a measured voltage of 0 from N-source to N-load.

By the voltage drops across A and B being equal.

In your theory, no current flows down the neutral because the potential to push the current has disappeared. In your reasoning, this potential has disappeared "by the voltage drops across A and B being equal." Here is where you go wrong. Consider an unbalanced MWBC. Lets say at the source, we have 120V<0 from A-N and 120V<180 from B-N. On the A phase, we install a 1200W resistive load (12 Ohms) and 10 Amps flows from A, through the load, and returns over the neutral. On the B phase we install a 1000W resistive load (14.4 Ohms) and 8.33 Amps flows from B, through the load and returns over the shared neutral. The shared neutral see a current flow of 1.67 Amps.

Now lets look at the voltage drop across Loads A & B. Per Ohm's laws, the voltage drop across the loads will be V=IR. For load A, the voltage drop is 10Amps x 12 Ohms = a 120 Volt drop. For load B, the voltage drop is 8.33 Amps x 14.4 Ohms = a 120 Volt drop. The voltage drops across A and B ARE EQUAL, yet current flows through the neutral. Under your explanation, equal voltage drops across A & B will REMOVE the potential to return current along the neutral, but here we have current flowing on the neutral even when the voltage drops across A & B are equal.

I don't agree with that assessment. Earlier, we agreed that Ia + Ib = 0. I also agree that Ia = -Ib. I don't agree that the two are mutually exclusive. I believe that Ia + Ib = 0 applies whether the node has two or three paths.

Again, you are going wrong here also. In the 3 wire MWBC, KCL tells us that Ia + Ib + In = 0 (3 wires, 3 paths into the node.) If you remove the neutral to create a 2 wire circuit, KCL tells us that Ia + Ib = 0. The equation Ia + Ib = 0 CANNOT apply if the node has three paths. If the node has three paths, the KCL must have three terms. If the node had five paths, then KCL would have five terms (Ia +Ib + Ic + Id + Ie = 0.)

Lets return to the balanced MWBC with a 1200W load from A-N and a 1200W load from B-N. The voltage from A-N at the source is 120V<0 and from B-N is 120V<180. The current flowing through the load A is 10A<0 and the current flowing through load B is 10A<180. From KCL, the current flowing "out" of the common node along the neutral will be In=Ia + Ib, or In = 10A<0 + 10A<180.

Consider, more fundamentally, that the current flowing in this 60Hz Alternating Current system takes the form of a sine wave which completes a full cycle 60 times in every second. The current flowing through load A will take the form Ia=10*sin(21600*t) where "t" is time. At the beginning of the first cycle (time = 0 seconds), the value of Ia will be zero amps. At 1/4 of the first cycle (or 0.004166 seconds) the value of Ia will be 10 Amps. At 1/2 of the cycle (or 0.008333 seconds) the value of Ia will be 0 Amps, and at 3/4 of the cycle the value of Ia will be -10 Amps. The current flowing through load B will take the form Ib=10*sin(21600*t-180) as the current is 180 degrees out of phase will A. Therefore the current Ib will be -10 Amps at 1/4 of the first cycle, 0 at 1/2 cycle and 10 Amps at 3/4 cycle. The neutral current (by KCL) will then be In = Ia + Ib= 10*sin(21600*t) + 10*sin(21600*t-180). Thus at time zero, the current on In will be 0+0=0, at 1/4 cycle will be 10+(-10)=0, at 1/2 cycle will be 0+0=0, and at 3/4 cycle will be (-10)+10=0. Just to carry it further, at 1.11 seconds after the start of the first cycle the current Ia will be -5.88 Amps and current Ib will be 5.88 Amps, and at 52.32 seconds after the start of the first cycle the current Ia will be 9.51 Amps and the current Ib will be -9.51 Amps.

The neutral current IS NOT zero at all times because the neutral has "been bypassed," or because the neutral is "like an open circuit," or because "the potential to push the current down the neutral has been removed," or because "the voltage drops across A and B or equal. The current in the neutral will be 0 Amps at ALL times, precisely BECAUSE the sum of the two current flowing from Load A & B will equal 0 at all times during each cycle of the AC current.

 

[gar]

090709-1036 EST

david luchini:

I believe for this thread it has to be assumed that any conductor has a non-zero resistance. However, if you want to consider a theoretical zero resistance conductor, and use voltage across that conductor to determine the current in said conductor for a particular circuit, then it is necessary to determine how current varies as resistance approaches zero.

To just use R=0 in the equation
I=E/R
results in the fatal mistake of dividing by zero.

You can only define the value of I when R = 0 if you know how I varies as R approaches 0. Otherwise you must treat I as indeterminate.

.

 

[david luchini]

I believe for this thread it has to be assumed that any conductor has a non-zero resistance. However, if you want to consider a theoretical zero resistance conductor, and use voltage across that conductor to determine the current in said conductor for a particular circuit, then it is necessary to determine how current varies as resistance approaches zero.

To just use R=0 in the equation
I=E/R
results in the fatal mistake of dividing by zero.

Gar, this is not relevant to the discussion. Nowhere did I put forth that R=0 in the equation I=E/R, so I have not made the fatal mistake of dividing by zero. I have assumed that the impedance of the circuit conductor are zero (or small enough to be negligible in the given circuit.) The R in my example unbalanced circuit is 12 ohms in the A-N circuit (I=120V/12 oms= 10Amps) and 14.4 ohms in the B-N circuit (I=120V/14.4 ohms=8.33 Amps)

Where in this example do you see an instance of R=0?

 

[Rick Christopherson]

This is wrong, the "voltage between the source N and the load N is non-existant whether the load is balanced, unbalanced, or if you have a single branch circuit (again assuming and ideal circuit where the conductor impedances do not figure into the circuit.)

I hadn't looked at this thread in a while, but this statement caught my eye and made me read the newest discussion.

While it is true that we normally assume "ideal conductors" in most circuit analysis, at the level of the discussion you are examining, this cannot be properly assumed. Because the wire has some resistance, there must be a voltage drop across the wire for current to flow through the wire. Ohm's Law cannot be violated or it is no longer a "Law". If there is zero volts across the length of a wire, then Ohm's Law mandates that there will be zero amps flowing through the wire. To perform a circuit analysis down at the level you are examining, you need to replace the ideal conductors with a small resistor.

Larry, I'm astounded that you would say that if you have a MWBC with two phases and a shared neutral, that removing the neutral is NOT a change of circuit. Removing the neutral creates a COMPLETELY different circuit.

When there is no current flowing through the neutral, Larry is correct that removing the wire does not alter the circuit. I think I understand why this is confusing you, but I cannot find the quote that pointed this out to me. Oh, here it is:

From KCL, the current flowing "out" of the common node along the neutral will be In=Ia + Ib, or In = 10A<0 + 10A<180.

I think someone else might have already pointed this out, but I believe you might be confusing the mathematical tools we use for solving a circuit with the actual characteristics of the circuit. (Please pardon me if that assumption is wrong.) Even though our mathematical tools tell us that there are two currents flowing in the neutral and that they cancel each other out, in real life, currents do not collide and anylate each other. It is only the mathematical models that suggest this.

Another thing that I noticed is that you listed the two 120 volt sources as being opposite to each other (out of phase). That is perfectly fine to do mathematically, but if you forget that this is true only because we arbitrarily choose the neutral point as a zero reference point, it might cause you to visualize the two currents as also being out of phase. By labeling one of the voltage sources as being 180 degrees, it hides the minus sign.

In other words, in your mind's-eye, you might be visualizing the two currents as being separate and opposite, when in reality, they are both the same current, and flowing in the same direction.

Oh, to give credit where due, it was Gar that already suggested this.

Because you can create a mathematical analog of the real world where you have for analytical purposes two exactly equal currents that cancel each other does not mean those currents exist in the real world.

 

[david luchini]

there must be a voltage drop across the wire for current to flow through the wire

Rick, you truly surprise me. If this statement was true then Ohm's law would not work in an ideal case. Consider a circuit with a 12V battery with terminals + and -. From + we run an ideal conductor (with no impedance) to a 12 ohm resistor (terminal R1.) On the other end of the resistor (terminal R2) we route an ideal conductor to the battery terminal -. According to Ohm's law I=V/R, there will be a 1 Amp current flowing through the circuit. However, if we isolate the conductor from terminals + to R1, it is plain to see that there is no voltage drop from + to R1, yet there is clearly 1 Amp flowing along this conductor. From your statement above, this could not be happening, so Ohm's law must be wrong. Let me correct your statement - it should say "there must be a current flow through a wire for there to be a voltage drop across it."

It makes no difference, in my example circuits, if you consider the conductors to be ideal, or if you include a small resistance for the wire. Ohm's law, Kirchoff's Current Law, and Kirchoff's Voltage law still apply.

When there is no current flowing through the neutral, Larry is correct that removing the wire does not alter the circuit. I think I understand why this is confusing you...

I am not confused on this matter, but perhaps you are. Your statement is wrong! A three wire circuit and a two wire circuit ARE NOT and CANNOT be the same circuit. Just because the current flow is the same in both cases, does not make the circuits the same. Lets say for example, we connect a 1200W (120V) light bulb (or 12 ohms) from A-N and from B-N we connect two 2400W (120V) light bulb (or 6 ohms) in series. Ia will be 10 Amps, and Ib will be 10 Amps, but this is NOT the same circuit as connecting a 1200W light bulb between each phase and neutral despite the fact the the current flow in the two circuits is the same.

Even though our mathematical tools tell us that there are two currents flowing in the neutral and that they cancel each other out, in real life, currents do not collide and anylate (sic) each other.

I have not suggested that the two current are flowing in the neutral and that they are cancelling each other. And I do pardon you for assuming that I am confusing the mathematical tool with the actual circuit, because I do not have that confusion.

Another thing that I noticed is that you listed the two 120 volt sources as being opposite to each other (out of phase). That is perfectly fine to do mathematically, but if you forget that this is true only because we arbitrarily choose the neutral point as a zero reference point, it might cause you to visualize the two currents as also being out of phase. By labeling one of the voltage sources as being 180 degrees, it hides the minus sign.

I'm not quite sure what you mean by this statement, perhaps you can expound upon it. But in reality the A-N and B-N voltage are out of phase from each other. It does not matter how you state it or what method you use to illustrate the point, the fact remains the same. I am open to notating the voltages as Van=120<0 (120V at phase angle 0) while Vbn=120<0. Or if you prefer, I would be just a happy with Van=120+j0 volts and Vbn=-120+j0 volts, or perhaps you favorite method would be Van=120sin(21660*t) and Vbn=120sin(21660*t-180). If you don't like 0 and 180 as the phase angles, we could use 30 and 210. It does not matter. The results are the same.

In other words, in your mind's-eye, you might be visualizing the two currents as being separate and opposite, when in reality, they are both the same current, and flowing in the same direction.

Now I think I see where you are going with that last statement, but again, I don't understand your reasoning. If Ia=10<0 and Ib=10<180, then KCL is correct that In=Ia+Ib or In=10<0 + 10<180, which of course equals 0. Where I think you are confused on this matter is that if In=10<0 + 10<180=0, then of course Ia+Ib=0 becomes Ia=-Ib. Plugging our current values into Ia=-Ib gives us 10<0=-10<180, or 10<0=10<0 which is undoubtedly true. So in reality, Ia and Ib are not both the same current flowing in the same direction, Ia and -Ib are both the same current flowing in the same direction.

Because you can create a mathematical analog of the real world where you have for analytical purposes two exactly equal currents that cancel each other does not mean those currents exist in the real world.

And finally (with respect to Gar,) if the two exactly equal currents that cancel each other out do not exist in the real world, the the whole point and indeed the whole world is moot. If Ia and Ib, exactly equal (and opposite) currents that cancel each other out in the KCL equation DO NOT EXIST in the real world, then our lights would never light up. The currents Ia and Ib DO and MUST exist in the real world, or we would not be able to have this discussion on the internet.

 

[hurk27]

I'm not quite sure what you mean by this statement, perhaps you can expound upon it. But in reality the A-N and B-N voltage are out of phase from each other. It does not matter how you state it or what method you use to illustrate the point, the fact remains the same. I am open to notating the voltages as Van=120<0 (120V at phase angle 0) while Vbn=120<0. Or if you prefer, I would be just a happy with Van=120+j0 volts and Vbn=-120+j0 volts, or perhaps you favorite method would be Van=120sin(21660*t) and Vbn=120sin(21660*t-180). If you don't like 0 and 180 as the phase angles, we could use 30 and 210. It does not matter. The results are the same.

If the source was a two phase generator wired via a common connection between them at opposite phase angles then you equation would work, but the circuit would not be an additive, it would cancel each other and no current would flow, just like placing two battery's together with both positive poles together.

But this is not the circuit in the discussion, the source in the discussion is two windings on a common core transformer connected in series with a tap placed on this common connection we call a neutral.

As I said in post 88 when point A is positive point A-N will be negative and like wise point B-N will be positive and point B will be negative, that is because both windings are in perfect phase with each other and are an additive to each other, to say other wise would not allow you to get 240 volts from point A to point B the neutral current does not cancel out across the neutral but rather takes the path of the source which is the two windings of the transformer, this can be proved by the current across the neutral point between the two windings (A-N to B-N) will be equal to the circuit as a whole at any point in the circuit.

I with Larry and Rick can ya tell:grin:

 

[hurk27]

Oh and yes KCL would work for a polly phase source, because of the phase rotation angle, but in the above discussion we really don't have a set of phases rather a set of legs from a transformer fed from a single phase and neutral of a polly phase generator. mixing the equation of these two circuits will produce the wrong answer's

 

[crossman gary]

David,

It is my opinion that Rick's post is correct. It is also my opinion that you are missing some of the subtleties of the issue.

Assuming zero resistance in a conductor immediately removes us from reality to the theoretical. Sure, it works to get an approximation for most calculations that we do. However, as Mivey previously stated, to accurately model the phenomenon being discussed, you must include the resistance of the conductors. Assuming zero resistance conductors is essentially saying that the loads are connected directly to the source terminals, and that means that a MWBC would be impossible to achieve, which renders the discussion moot.

Ohms Law states that E = I/R for any given component in a circuit. Assuming zero resistance absolutely results in a division by zero which gives us an arbitrary or undefined result and is of no use in this technical discussion.

If you will take the time to draw out and do the calculations of a 120v/240v MWBC with, say 1 ohm in each hot wire and 1 ohm in the neutral, and then loads of say, 50 ohms on one phase to neutral and 75 ohms on the other phase to neutral, you will discover something interesting. The voltage drop on the loads will not be 120v, nor will they be equal. This difference in voltage drop of the two loads is where the voltage across the neutral wire comes from, and is the potential difference to cause the current flow in the neutral.

It is really important for you to work this out if you want to understand what is going on. If you do not care to do the work, perhaps I can find time tomorrow to do it for you. I understand that you are adamant in your argument, and that we will have a hard time convincing you that you are not looking at the subject deeply enough. If you would take the time to work out the problem I gave, the light bulb will go poof in your head and you will think "Oh, now I see what is going on!" And there is no shame in that. I have certainly been wrong on this forum, and when someone else finally made me see the light, it was actually a good feeling to admit the mistake because I learned something in the process.

It is absolutely a fact, that at normal earthly temperatures and using copper or aluminum conductors readily available from the supply house, that to have current flow in the conductor, there must be a difference of potential from one end of the conductor to the other end.

 

[Rick Christopherson]

P.S. I type slow, so I already know that several postings have been made while I was typing this. When I started this posting, it was a direct reply to the previous posting. I know these other postings exist, but I have not read them yet.

there must be a voltage drop across the wire for current to flow through the wire.

Rick, you truly surprise me. If this statement was true then Ohm's law would not work in an ideal case.

Oh man! I am at a compete loss here.

I just looked up your profile and you are listed as not only an E.E, but a P.E. too. We should not be having this caliber of discussion with your credentials. When I wrote my previous message, I mistakenly assumed that your education was not that of an EE. This is not a topic that I should need to explain to a PE, EE. You have the appropriate education, so there is no reason for me to explain how to convert from an ideal circuit to a real circuit. (I mean no offense, but you really caught me off guard with this.)

Let me correct your statement - it should say "there must be a current flow through a wire for there to be a voltage drop across it."

Yes and No. Ohm's law is universal. If no current flows, then there is no voltage differential. Conversly, if there is no voltage differential, then no current flows. You need a force on the electrons to get them to move (aside from random drift).

A three wire circuit and a two wire circuit ARE NOT and CANNOT be the same circuit.

You are correct in saying that they are not the "same" circuit, but they are "equivalent" circuits. Electrically they are the same. Physically they are different. We use many tools in this trade to convert back and forth between equivalent circuits, and this is just one of them.

I'm not quite sure what you mean by this statement, perhaps you can expound upon it. But in reality the A-N and B-N voltage are out of phase from each other. It does not matter how you state it or what method you use to illustrate the point, the fact remains the same. I am open to notating the voltages as Van=120<0 (120V at phase angle 0) while Vbn=120<0. Or if you prefer, I would be just a happy with Van=120+j0 volts and Vbn=-120+j0 volts, or perhaps you favorite method would be Van=120sin(21660*t) and Vbn=120sin(21660*t-180). If you don't like 0 and 180 as the phase angles, we could use 30 and 210. It does not matter. The results are the same.

BINGO, we have a cause!!!
You have been so deeply programmed into this situation that phases A and B are opposite that you have lost sight that they are only opposite with respect to the neutral.

When the voltage between A and N is positive, so is the voltage between N and B (note my polarity). It has become so commonplace with you that you didn't even realize than A-N is pointing in the opposite direction from B-N. There is your missing minus sign, and this is the reason why I oppose people using a-n and b-n without understanding that they have reversed their normal convention without realizing it.

Mathematically, the minus signs cancel out and the analysis works properly, but it still leaves the person performing the analysis thinking that the two systems are in opposition.

If Ia=10<0 and Ib=10<180, then KCL is correct that In=Ia+Ib or In=10<0 + 10<180, which of course equals 0. Where I think you are confused on this matter is that if In=10<0 + 10<180=0, then of course Ia+Ib=0 becomes Ia=-Ib. Plugging our current values into Ia=-Ib gives us 10<0=-10<180, or 10<0=10<0 which is undoubtedly true. So in reality, Ia and Ib are not both the same current flowing in the same direction, Ia and -Ib are both the same current flowing in the same direction.

No, I am not confused on this issue because I have never lost sight of where the minus sign exists. Your answers come out mathematically the same, but visually and mentally, you forget that there is a missing minus sign.

If Ia and Ib, exactly equal (and opposite) currents that cancel each other out in the KCL equation DO NOT EXIST in the real world, then our lights would never light up. The currents Ia and Ib DO and MUST exist in the real world, or we would not be able to have this discussion on the internet.

Only someone that misapplies the mathematical model as being the same as the real-world model would make this error.

 

[SAC]

Give your theory a try...

Give your theory a try...

I'd suggest that anyone can easily "try out" their theories on the circuits that I posted links to. I included both load and conductor resistances in the circuit. To edit the values of the resistors, simply right click (or control click on a Mac) the resistor and choose "edit". Then change the value. You can see exactly how the currents change. Also, the circuits are fully editable. If you right click on the empty space, you can add wires, resistors, etc... If someone comes up with something that they think makes a particular discussion point, choose "export" from the menu and PM me the contents and I'll install the circuit and PM back the link to use to see it. This really shouldn't be a matter of opinion - electricity really only works one way and if you can demonstrate it that is pretty convincing.

http://ee.mxpark.com/mwbc/mwbc.html
http://ee.mxpark.com/mwbc/swbc.html

 

[david luchini]

If the source was a two phase generator wired via a common connection between them at opposite phase angles then you equation would work, but the circuit would not be an additive, it would cancel each other and no current would flow, just like placing two battery's together with both positive poles together.

But this is not the circuit in the discussion, the source in the discussion is two windings on a common core transformer connected in series with a tap placed on this common connection we call a neutral.

Hurk, you are incorrect. My equation works with the common core transformer with a center tap commonly called a neutral. In the case of this transformer, the voltage from one tap to neutral, lets say A-N would be Va=120<0. The voltage from the other tap, B-N is Vb=120<180. It follows then that the voltage from A to B is Vab=Va-Vb, and the voltage from B to A is Vba=Vb-Va.

Therefore, Vab=Va-Vb= 120<0 - 120<180 which equals 240<0.

and, Vba=Vb-Va = 120<180 - 120<0 which equals 240<180.

...to say other wise would not allow you to get 240 volts from point A to point B...

Really, as demonstrated above, I get 240 volts from point A to point B. You are making a mistake in assigning a "postive" and a "negative" to this circuit and assuming that no current can flow because the two "negatives" are together. This is an AC circuit, there is no postive and negative.

 

[LarryFine]

Removing the neutral creates a COMPLETELY different circuit.

Are we talking hardware or theory here?

This is wrong, the "voltage between the source N and the load N is non-existant whether the load is balanced, unbalanced, or if you have a single branch circuit (again assuming and ideal circuit where the conductor impedances do not figure into the circuit.)

No, it's not. I'm talking about the circuit with the neutral connection broken. As long as the loads are balanced, the voltage between source-N and load-N will be zero even with no neutral.

Two 120v bulbs of equal wattage in series across 240v will behave identically with or without a neutral. Every series circuit made of equal elements depends on this property to function, such as X-mas lights.

There is not a potential difference along the neutral which pushes the current along the neutral. In an unbalanced MWBC or a balanced MWBC there is still a measured voltage of 0 from N-source to N-load.

Of course not. In this discussion, we agree that we're treating the conductors as if they have no impedance. That's not what we're talking about. Continue:

Consider an unbalanced MWBC. Lets say at the source, we have 120V<0 from A-N and 120V<180 from B-N. On the A phase, we install a 1200W resistive load (12 Ohms) and 10 Amps flows from A, through the load, and returns over the neutral. On the B phase we install a 1000W resistive load (14.4 Ohms) and 8.33 Amps flows from B, through the load and returns over the shared neutral. The shared neutral see a current flow of 1.67 Amps.

Agreed so far.

Now lets look at the voltage drop across Loads A & B. Per Ohm's laws, the voltage drop across the loads will be V=IR. For load A, the voltage drop is 10Amps x 12 Ohms = a 120 Volt drop. For load B, the voltage drop is 8.33 Amps x 14.4 Ohms = a 120 Volt drop. The voltage drops across A and B ARE EQUAL, yet current flows through the neutral.

Agreed so far again.

Under your explanation, equal voltage drops across A & B will REMOVE the potential to return current along the neutral, but here we have current flowing on the neutral even when the voltage drops across A & B are equal.

That's because you're using imbalanced loads. We weren't.

In the 3 wire MWBC, KCL tells us that Ia + Ib + In = 0 (3 wires, 3 paths into the node.) If you remove the neutral to create a 2 wire circuit, KCL tells us that Ia + Ib = 0. The equation Ia + Ib = 0 CANNOT apply if the node has three paths.

Note that both equations express zero neutral current, which has been our point: One circuit behaves just like the other one: no current between the source neutral and the load neutral.

Of course, the two circuits are physically different. The point is to show why no real current flows in the neutral. Not just no mathematical or theoretical current, but no real current.

If the node has three paths, the KCL must have three terms. If the node had five paths, then KCL would have five terms (Ia +Ib + Ic + Id + Ie = 0.)

Who says we have to use the same KCL equation to show that two different circuits behave the same? That they both conclude with zero neutral current shows why there is none.

Lets return to the balanced MWBC with a 1200W load from A-N and a 1200W load from B-N. The voltage from A-N at the source is 120V<0 and from B-N is 120V<180. The current flowing through the load A is 10A<0 and the current flowing through load B is 10A<180. From KCL, the current flowing "out" of the common node along the neutral will be In=Ia + Ib, or In = 10A<0 + 10A<180.

That's zero, right?

Consider, more fundamentally, that the current flowing in this 60Hz Alternating Current system takes the form of a sine wave which completes a full cycle 60 times in every second. The current flowing through load A will take the form Ia=10*sin(21600*t) where "t" is time. At the beginning of the first cycle (time = 0 seconds), the value of Ia will be zero amps. At 1/4 of the first cycle (or 0.004166 seconds) the value of Ia will be 10 Amps. At 1/2 of the cycle (or 0.008333 seconds) the value of Ia will be 0 Amps, and at 3/4 of the cycle the value of Ia will be -10 Amps. The current flowing through load B will take the form Ib=10*sin(21600*t-180) as the current is 180 degrees out of phase will A. Therefore the current Ib will be -10 Amps at 1/4 of the first cycle, 0 at 1/2 cycle and 10 Amps at 3/4 cycle. The neutral current (by KCL) will then be In = Ia + Ib= 10*sin(21600*t) + 10*sin(21600*t-180). Thus at time zero, the current on In will be 0+0=0, at 1/4 cycle will be 10+(-10)=0, at 1/2 cycle will be 0+0=0, and at 3/4 cycle will be (-10)+10=0. Just to carry it further, at 1.11 seconds after the start of the first cycle the current Ia will be -5.88 Amps and current Ib will be 5.88 Amps, and at 52.32 seconds after the start of the first cycle the current Ia will be 9.51 Amps and the current Ib will be -9.51 Amps.

Correct me if I'm wrong, but doesn't all of that say "zero neutral current?"

The neutral current IS NOT zero at all times because the neutral has "been bypassed," or because the neutral is "like an open circuit," or because "the potential to push the current down the neutral has been removed," or because "the voltage drops across A and B or equal. The current in the neutral will be 0 Amps at ALL times, precisely BECAUSE the sum of the two current flowing from Load A & B will equal 0 at all times during each cycle of the AC current.

So, after all of that, we reach the same conclusions!










So, where do we differ? :-? (Other than my spelling of "crux," one misspell in all of that. :roll

 

[david luchini]

David,

It is my opinion that Rick's post is correct. It is also my opinion that you are missing some of the subtleties of the issue.

Assuming zero resistance in a conductor immediately removes us from reality to the theoretical. Sure, it works to get an approximation for most calculations that we do. However, as Mivey previously stated, to accurately model the phenomenon being discussed, you must include the resistance of the conductors. Assuming zero resistance conductors is essentially saying that the loads are connected directly to the source terminals, and that means that a MWBC would be impossible to achieve, which renders the discussion moot.

Ohms Law states that E = I/R for any given component in a circuit. Assuming zero resistance absolutely results in a division by zero which gives us an arbitrary or undefined result and is of no use in this technical discussion.

If you will take the time to draw out and do the calculations of a 120v/240v MWBC with, say 1 ohm in each hot wire and 1 ohm in the neutral, and then loads of say, 50 ohms on one phase to neutral and 75 ohms on the other phase to neutral, you will discover something interesting. The voltage drop on the loads will not be 120v, nor will they be equal. This difference in voltage drop of the two loads is where the voltage across the neutral wire comes from, and is the potential difference to cause the current flow in the neutral.

It is really important for you to work this out if you want to understand what is going on. If you do not care to do the work, perhaps I can find time tomorrow to do it for you. I understand that you are adamant in your argument, and that we will have a hard time convincing you that you are not looking at the subject deeply enough. If you would take the time to work out the problem I gave, the light bulb will go poof in your head and you will think "Oh, now I see what is going on!" And there is no shame in that. I have certainly been wrong on this forum, and when someone else finally made me see the light, it was actually a good feeling to admit the mistake because I learned something in the process.

It is absolutely a fact, that at normal earthly temperatures and using copper or aluminum conductors readily available from the supply house, that to have current flow in the conductor, there must be a difference of potential from one end of the conductor to the other end.

Crossman Gary, it is not I who am missing the subtleties of the issue, I am highlighting the subtleties of the issue. You must have missed my example of a 12v battery with a 12 ohm resistor with "ideal" circuit conductors connecting the battery to the resistor. In this case the circuit conductors have no resistance, but yet 1 Amp will be flowing through these conductors. How is this possible under your theory? You would be dividing by zero. The answer is you are not dividing by zero, you are dividing by 12. The battery is the potential source in the circuit, the resistance in circuit is the TOTAL resistance of the conductors and the load resistor which is 12 ohms. If it was true that current cannot flow through a conductor with zero resistance because there will be no voltage drop, then Ohm's law is WRONG. Who wants to tell Ohm?

It is NOT relevant whether there is impedance in the circuit conductors or not. Ohm's law does not change, KCL does not change, KVL does not change, and the result does not change.

 

[LarryFine]

If there is zero volts across the length of a wire, then Ohm's Law mandates that there will be zero amps flowing through the wire.

Oops! No, zero volts across the length of a conductor carrying current is proof of zero resistance, not zero current.

To perform a circuit analysis down at the level you are examining, you need to replace the ideal conductors with a small resistor.

Oops again! For this discussion, we can ignore conductor resisnace as long as we apply that to all conductors.

 

[LarryFine]

Really, as demonstrated above, I get 240 volts from point A to point B. You are making a mistake in assigning a "postive" and a "negative" to this circuit and assuming that no current can flow because the two "negatives" are together. This is an AC circuit, there is no postive and negative.

So, remove the AC component from this discussion, and use two 120v batteries with the center tap.