MWBC= more heat or Less heat?

[david luchini]

Are we talking hardware or theory here?

Larry, we are talking hardware AND theory here. It is the same. If you remove the neutral from the circuit, you have a different circuit both in the physical world and theoretical world.

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.

Larry, you start out disagreeing with me here, but end up stating the same thing that I did. The is no potential difference from N load to N transformer (with an ideal circuit) regardless of whether the neutral is connected or broken, whether the loads are balanced or not. As you state, the voltage between source-N and load-N will be zero.

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

You are missing the point of that example. It was put forth by someone that there has to be a potential difference from source-n to neutral-n in order for current to flow in the neutral. In that example, there is clearly NOT a potential difference from source-n to neutral-n, but there IS current flowing in the neutral.

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.

Yes, both equations express zero neutral current. That has been my point as well. Of course no "real" current flows in the neutral, just as the mathematical or theoretical current will be zero. The point is that just because one circuit "behaves" like the other does not make them the same circuit. My original point is that the neutral has not been bypassed, it is still an active part of the circuit. Nor is there no current flowing in the neutral because there is "no potential difference along the neutral in a balanced circuit" as has been suggested. It just happens that there is zero current flowing on it. That is not the same as an open circuit at the neutral.

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

Yes, it does. That's what I've said all along. I guess the only place we disagree is in your post #119 where you say that if there is zero voltage difference between two points, a conductor between them will carry no current. My earlier example of a 12 volt battery with a 12 ohm resistor connected by a zero impedance (ideal) conductor shows that there will be no voltage difference between the battery terminal and resistor terminal, but current will obviously be flowing through that conductor.

 

[david luchini]

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

Larry, it seems we find more to agree on.

 

[mivey]

I am picking up at about #130. Sorry about the slow typing but I am not going to throw all this out b/c I'm having fun with text drawn circuits. I'll read on.

Balanced circuit at equilibrium where removing or adding a neutral wire at the equipotential point (the intersection of the loads) does nothing to the voltages and currents:

......-120V+...+1V-...+119V-..10A
...0V|--Ea------Za-----Ra------>---|.0V
.....|.........0.1Ω...11.9Ω........|
.10A.^.............................v.10A
.....|.............................|
.....|.............................|
.....|.............................|
.....|+120V-...-1V+...-119V+..10A..|
...0V|--Eb------Zb-----Rb------<---|.0V
.....|.........0.1Ω...11.9Ω........|
.....|.............................|
..0A.^.............................v.0A
.....|.............................|
.....|.............................|
.....|.............................|
...0V|----------Zn-------------<---|.0V
...............0.1Ω............0A


Unbalanced circuit at equilibrium where removing or adding a neutral wire at the equipotential point (inside the b phase resistor) does nothing to the voltages and currents:

.......-120V+...+1.33V-...+77.78V-...13.3A
....0V|--Ea-------Za--------Ra--------->---|.40.89V
......|..........0.1Ω......5.85Ω...........|
13.3A.^....................................v.13.3A
......|....................................|
......|....................................|
......|....................................|
......|+120V-...-1.33V+..-159.56V+...13.3A.|
....0V|--Eb-------Zb--------Rb---------<---|.40.89V
......|..........0.1Ω......12Ω
......|.......................|.0V
......|.......................|
...0A.^.......................v.0A
......|.......................|
......|.......................|
....0V|-----------Zn------<---|.0V
.................0.1Ω.....0A


Unbalanced circuit (initially at equilibrium) where removing or adding a neutral wire at the intersection of the loads causes the circuit to move away from its natural equilibrium point. Tapping in at the intersection of the two loads, where there is a natural difference in potential across the loads, forces current to travel down the neutral and will consume more power:

......-120V+...+2V-...+117V-..20A
...0V|--Ea------Za-----Ra------>---|.1V
.....|.........0.1Ω...5.85Ω........|
.20A.^.............................v.20A
.....|.............................|
.....|.............................|
.....|.............................|
.....|+120V-...-1V+...-120V+..10A..|
...0V|--Eb------Zb-----Rb------<---|.1V
.....|.........0.1Ω....12Ω.........|
.....|.............................|
.10A.^.............................v.10A
.....|.............................|
.....|.............................|
.....|.........-1V+................|
...0V|----------Zn-------------<---|.1V
...............0.1Ω...........10A

Note that there is a difference between the voltage at the source neutral wire and the load neutral wire because of the conductor impedance (Zn). Also note the difference in potential across the loads.

 

[mivey]

If you are going to get hung up on the conductor impedances, lets take them out:

Balanced ideal circuit at equilibrium where removing or adding a neutral wire at the equipotential point (the intersection of the loads) does nothing to the voltages and currents:

......-120V+..........+120V-..10A
...0V|--Ea-------------Ra------>---|.0V
.....|.................12Ω.........|
.10A.^.............................v.10A
.....|.............................|
.....|.............................|
.....|.............................|
.....|+120V-..........-120V+..10A..|
...0V|--Eb-------------Rb------<---|.0V
.....|.................12Ω.........|
.....|.............................|
..0A.^.............................v.0A
.....|.............................|
.....|.............................|
.....|.............................|
...0V|-------------------------<---|.0V
...............................0A


Unbalanced ideal circuit at equilibrium where removing or adding a neutral wire at the equipotential point (inside the b phase resistor) does nothing to the voltages and currents:

.......-120V+..............+80V-....13.33A
....0V|--Ea-----------------Ra--------->---|.40V
......|.....................6Ω.............|
13.33A^....................................v.13.33A
......|....................................|
......|....................................|
......|....................................|
......|+120V-.............-160V+....13.33A.|
....0V|--Eb-----------------Rb---------<---|.40V
......|....................12Ω
......|.......................|.0V
......|.......................|
...0A.^.......................v.0A
......|.......................|
......|.......................|
....0V|-------------------<---|.0V
..........................0A


Unbalanced ideal circuit (initially at equilibrium) where removing or adding a neutral wire at the intersection of the loads causes the circuit to move away from its natural equilibrium point. Tapping in at the intersection of the two loads, where there is a natural difference in potential across the loads, forces current to travel down the neutral and will consume more power:

......-120V+.........+120V-...20A
...0V|--Ea-------------Ra------>---|.0V
.....|.................6Ω..........|
.20A.^.............................v.20A
.....|.............................|
.....|.............................|
.....|.............................|
.....|+120V-.........-120V+...10A..|
...0V|--Eb-------------Rb------<---|.0V
.....|.................12Ω.........|
.....|.............................|
.10A.^.............................v.10A
.....|.............................|
.....|.............................|
.....|.............................|
...0V|-------------------------<---|.0V
..............................10A

Note that there is no longer a difference between the voltage at the source neutral wire and the load neutral wire because the conductor impedance (Zn) is gone. Also note there is no difference in potential across the loads.

This circuit has equal voltage drops across the load but is not balanced. Removing the neutral will show that the voltages will stabilize such that the voltage drops across the loads are different. This potential difference causes the intersection to have a non-zero potential. Forcing this potential to be zero will force current to flow on the neutral.

And yes, the direction arrows are correct. They are indicating the direction that the holes are moving (opposite of the direction the electrons are moving). The electrons are not butting heads at the neutral point and canceling each other out.

 

[david luchini]

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

We can do that with no problems, the equations would be the same. If we consider two 120V batteries, with a center
tap connecting the two negative terminals, and consider A to be the positive of one battery and B to be the positive
terminal of the second battery, then, Va= Van = +120V. Likewise, Vb=Vbn = +120v.

To determine the voltage from A to B, using the earlier equation, Vab = Va - Vb, or Vab = 120V -120V = 0 volts. This would
appear to be accurate to me. If we instead, connected to batteries together with a center tap that connect the negative
of one battery to the positive of the second battery. Node A will be the positive of the first battery, and B will be the negative
of the second battery. Then Va = Van = +120V and Vb = Vbn = -120V. The voltage Vab for the batteries in this configuration
will therefore be Vab = Va- Vb, or Vab = 120V - (-120V) = 240 Volts.

 

[rattus]

A little thought experiment:

A little thought experiment:

Try this in your mind only because the ideal components are out of stock at Radio Shack:

Imagine a current source driving an ideal rheostat. The voltage across the rheostat will be,

Vr = IxR

Start with 1.0 ohms and 1 amp say, and drop the resistance in 0.1 ohm steps,

Vr = 1V, 0.9V, 0.8V....................0.2V, 0.1V, 0V

Voila, we have 0V and 1 amp, and we didn't divide by zero!

 

[mivey]

It was put forth by someone that there has to be a potential difference from source-n to neutral-n in order for current to flow in the neutral.

There has been discussion of two potentials.

For a normal conductor, there has to be a potential to overcome the wire resistance. Without it, you will not have current flow. See the 1 volt needed in the third diagram of my #143. With the ideal conductor, this goes away (3rd diagram of my post #144)

The other is the difference in potential across the loads when it is at it's natural equilibrium state. See the 40.89 volts in diagram #2 of my post #143 and the 40 volts in diagram #2 of my post #144.

Look at the ideal conductor case. Forcing this 40 volt potential to zero is forcing current down the neutral. Balancing the loads to get this potential to zero reduces the neutral current to zero. Without this unbalanced potential, you will have no neutral current.

Also, in the unbalanced case with no neutral, the voltage sources are supplying 3200 W of power. In the unbalanced case with a neutral, the voltage sources are supplying 3600 W of power. The forced case uses more power.

 

[mivey]

Try this in your mind only because the ideal components are out of stock at Radio Shack:

Imagine a current source driving an ideal rheostat. The voltage across the rheostat will be,

Vr = IxR

Start with 1.0 ohms and 1 amp say, and drop the resistance in 0.1 ohm steps,

Vr = 1V, 0.9V, 0.8V....................0.2V, 0.1V, 0V

Voila, we have 0V and 1 amp, and we didn't divide by zero!

That's because the voltage you are referencing is the rheostat voltage. Now tell me the current source voltage.

The current source is essentially a voltage in series with a resistor. You have just cut one of the resistor leads, called the two lead ends a current source, and stuck your rheostat between the lead ends.

 

[mivey]

That's because the voltage you are referencing is the rheostat voltage. Now tell me the current source voltage.

The current source is essentially a voltage in series with a resistor. You have just cut one of the resistor leads, called the two lead ends a current source, and stuck your rheostat between the lead ends.

FWIW, you keep the current constant by varying the source resistance or the source voltage.

 

[david luchini]

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.)

Rick, no offense (and I haven't look up your profile,) but no need to explain
ideal and real circuits to me. What you need to explain is mistaken beliefs such as "there must be a voltage drop across a wire for current to flow through it." As I have demonstrated, in an ideal circuit, this statement is completely false. I will spell it out again. Let's imagine of 12V battery, its positive terminal will be node A. From A we run and ideal conductor (no impedance) to the terminal of a 12 ohm resistor. We will call this node B. From the other side of the resistor, node C, we will run an ideal conductor to the negative battery terminal, node D. I think we can see that we have a potential source (the battery) which provide a potential of +12 Volts from A-D. From A to B, we have a zero ohm resistance, from B to C we have a 12 ohm resistance, and from C to D we have a zero ohm resistance. Our total load including our load resistor and our circuit conductors is 12 ohms. Per ohm's law, the current flowing in the circuit will be 1 Amp. This current flow through our complete circuit, starting at terminal A, the 1 Amp flows through the A-B conductor, through the load resistor and finally through the C-D conductor to return to the battery.

The voltage drop in this circuit are from zero volts drop from A to B (V=IR = 1 Amp x 0 Ohms,) 12 volts from B to C (V = IR = 1 Amp x 12 Ohms,) and zero volts from C to D (identical to A-B.) Here we see 1 Amp flowing from A-B and 1 Amp flowing from C-D, yet there is not potential difference along either of those. In your theory, there could not be current flow in A-B or C-D because there is not potential difference. It is clear that said theory is not correct.

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).

This is so elemental that I don't know where to begin. Per Ohm's law if there is no current, then there is no voltage drop along the circuit. I would disagree with your statement that if there is no voltage differential then no current flows. I have shown above current from from node A to node B, when there is no voltage differential between A and B. What you mean to say is if there is no "potential source" then there is no current flow. In a non ideal circuit, one where there is an impedance on the circuit conductors and therefore a voltage drop along the conductor, it is NOT the potential difference caused by the voltage drop from A to B that causes the current to travel down the conductor. It is the potential source, in this case the battery, which is the "force" which gets the electrons to move. The electrons will move from A to B in BOTH a "real" circuit with a voltage drop from A-B and in an "ideal" circuit with no voltage drop from A-B.

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.

I am even more astounded now than I was before. I have not been programmed at all that phases A and B are "opposite." And what is your obsession with the "Minus Sign?" (By the way, I am not missing a minus sign.) Please try to follow this:

When the voltage from A to N is "positive," then the voltage from N to B will also be "positive." In this case, the voltage Van=120Volts at phase angle 0 (Van=120<0.) The voltage from N to B is also "positive" therefore, Vnb=120V<0. To obtain the voltage from A to B, we will clearly add the voltage from A-N plus the voltage from N-B (Vab=Van+Vnb.)

In our case, this will give us 120V<0 + 120V<0 which will equal 240V<0. Now let us consider that if Vnb is a +120V<0, then Vbn (voltage from B to N - please note my polarity) will be -120V<0. Vbn is opposite from Vnb, so that if one voltage is positive the other is negative. So Vbn could be listed as -120V<0, or 120<180 (this was a typo in my previous post) or -120+j0 volts (as noted in my previous post) or 120sin(21660*t-180.) And then the voltage from A-B will be (as I posted earlier) Van-Vbn.
This equals 120V<0 - 120V<180 which equals 240V<0.

​

You said earlier that "By labeling one of the voltage sources as being 180 degrees, it hides the minus sign" (again the minus sign obsession) but that is clearly not true. By labeling one of the source as being 180, I am "incorporating" the minus sign. Again, if Vnb = +120V<0, then Vbn = -120V<0 which equals +120V<180. You should try it sometime. If you label the voltages as being at 0 degrees and 180 degrees, then you can stop worrying about the minus sign because it will take care of itself.

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.

Again with the crazy minus sign. The minus sign in irrelevant. My answers come out mathematically the same because I recognize that the voltage from A to N and from B to N are 180 degrees out of phase. With this recognition I could not care less about the minus sign, as it will not affect my equations.

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

Exactly what error are you refering to? It is not me who mistakenly believes that if Ia and Ib cancel out, then Ia and Ib do not exist. According to my equations, Ia and Ib will exist both in the mathematical model and in the real-world model.

 

[rattus]

That's because the voltage you are referencing is the rheostat voltage. Now tell me the current source voltage.

The current source is essentially a voltage in series with a resistor. You have just cut one of the resistor leads, called the two lead ends a current source, and stuck your rheostat between the lead ends.

Mivey, this is an ideal current source--that is all. No mention of the MWBC. This post is an attempt to demonstrate current through zero resistance.

Now, the MWBC does approximate a current source into the neutral wire, so the analogy should hold.

BTW: Pretty good current sources can be made with electronic circuits.

 

[Rick Christopherson]

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

Larry, it seems we find more to agree on.

You guys must have a bigger budget than me, because I cannot afford super conductors operating at near absolute zero degrees.

If you have current flowing through a conductor, you will have a voltage drop, regardless how large the conductor or how low the resistance. The only exception to this is a super conductor, and I do not know how Ohm's Law applies to these, but obviously it still does or it would no longer be called a "Law".

It doesn't matter if you have 1 amp flowing through 22 ga wire or 4/0 wire, you will have a voltage drop commensurate with Ohm's Law. If you do not have a voltage difference from one end of the wire to the other end, there is no force causing the electrons to move.

Since I assume we are not discussing super conductors in this thread, you both are wrong for dismissing the previous comment. I wasn't saying this just because I had nothing better to type, but because it pertained directly to several comments made in this thread.

 

[mivey]

Mivey, this is an ideal current source--that is all. No mention of the MWBC. This post is an attempt to demonstrate current through zero resistance.

Then just call it the current through the lead of whatever device you have because the lead has an insignificant resistance (and thus an insignificant voltage drop)

If you have current flowing through a conductor, you will have a voltage drop, regardless how large the conductor or how low the resistance. The only exception to this is a super conductor, and I do not know how Ohm's Law applies to these, but obviously it still does or it would no longer be called a "Law".

Ohm's law still applies. Apply an ideal voltage source to a super conductor and you will get infinite current. Just take the limit. The internal resistance of a real voltage source would prevent this from happening.

You theoretically could put a current on a superconducting loop and it would flow forever. I don't think we are quite there yet.

 

[david luchini]

You guys must have a bigger budget than me, because I cannot afford super conductors operating at near absolute zero degrees.

If you have current flowing through a conductor, you will have a voltage drop, regardless how large the conductor or how low the resistance. The only exception to this is a super conductor, and I do not know how Ohm's Law applies to these, but obviously it still does or it would no longer be called a "Law".

It doesn't matter if you have 1 amp flowing through 22 ga wire or 4/0 wire, you will have a voltage drop commensurate with Ohm's Law. If you do not have a voltage difference from one end of the wire to the other end, there is no force causing the electrons to move.

Since I assume we are not discussing super conductors in this thread, you both are wrong for dismissing the previous comment. I wasn't saying this just because I had nothing better to type, but because it pertained directly to several comments made in this thread.

Rick, you miss the point completely. It does not matter if you have an ideal circuit or a real world circuit, but as can be see from the ideal "super conductor" circuit, you can have a current flowing through a conductor when there is not voltage drop across that conductor. You are correct that Ohm's law must apply universally even to the ideal case, or it would not be a "law."

So if current CAN flow across a wire where there is no voltage drop, you must consider what is driving that current. The driver of the current is the voltage or potential source. In the examples we have been discussing, the potential source has been the battery connected to the resistor, or the transformer secondary connected to balanced or unbalanced lighting loads. (Technically, the potential source is not the transformer secondary, but the generator at the power plant which connected through the distribution grid, and various substations, to the electrical service in our building where it then feeds the primary of our transformer, but for arguments sake, we'll call the secondary the potential source.)

The potential source and the potential source alone is what DRIVES the current in the circuit. The voltage drop across the conductors and across the load is a RESULT of the current flow, not a cause of the current flow.

You are correct in saying that (in the real world) "It doesn't matter if you have 1 amp flowing through 22 ga wire or 4/0 wire, you will have a voltage drop commensurate with Ohm's Law." However, after that is where you lose the plot. It is NOT the voltage difference from one end of the wire to the other that is the force that moves the current. It is the potential source (transformer or battery) which is moving the current.

Try to think of it in this manner. If I have a 12V battery in one hand, a 12 ohm resistor in the other, and some #12 awg on the table. The battery has potential. It is a power source. The wire and resistor are inert. They have no potential to move current on their own, but the battery has potential just waiting to be used. If I connect a length of wire which has an impedance of 0.5 ohms to each side of the resistor, I have created a new "resistor" of a total of 13 Ohms, (because resistors in series are additive.)

If I then connect my new "resistor" to the battery, a current will flow through both the wires and 12 ohm resistor equal to 0.923 Amps. The resistor and wire are both part of the "load" on the circuit. By KVL, the voltage drop across the load must be equal to the potential provided by the source. The voltage drop over the total load is V=IR, 0.923Amps x 13 Oms = 12 volts. The voltage drop over the total load is CAUSED by the current flowing through the load. It does not CREATE the current.

If you wanted to break the load down into its components, you could do that, but it does not change the functioning of the circuit. Broken down, we would see 0.923 Amps flow throught the first wire, creating a 0.462 Voltage drop, then through the 12 ohm resistor creating a 11.076 Volt drop, and finally through the second wire creating another 0.462 Volt drop.
By KVL again, these individual voltage drops must be equal to the potential from the source, and 0.462+11.076+0.492 equals 12 Volts.

It should be clear, that the potential source is the force which causes the current to move in the circuit. After all, connecting two resistors with a length of wire at each end will not cause a current flow. The voltage drop along each component of the circuits load, is created by the current flowing through the impedance of each component. There does not need to be a voltage drop along a component (as seen in the ideal condition) in order for current to flow through that component, and conversely, the presence of a voltage drop along any of the load components does not create the current in that component.

Sorry if this has been wordy, getting a little tired, but it should be clear that it is NOT the voltage drop that causes current to flow, but ONLY the potential source - the battery or transformer - which causes current flow.

 

[mivey]

Now, the MWBC does approximate a current source into the neutral wire, so the analogy should hold.

Sure. Voltage is the force that makes the electrons move. No additional resistance = no additional voltage force needed.

In a circuit with resistance, the voltage is needed to overcome this resistance and to get the electrons on their merry way. Voltage is like the rocket engine and resistance is like the air friction.

In the parts of the circuit with "no resistance", the electrons keep on moving and no extra force is needed to get through that part of the circuit. Kind of like when you pass through a vaccum.

In a circuit with no resistance, if the electrons are already moving, they would keep on moving, no voltage needed to prod them along and force them to move. Kind of like being in space.

A body in motion will to stay in motion, and a body at rest will stay at rest unless acted on by an unbalanced force.

Without the unbalanced force, the electrons will not start moving on their own down the neutral path.

 

[mivey]

The potential source and the potential source alone is what DRIVES the current in the circuit. The voltage drop across the conductors and across the load is a RESULT of the current flow, not a cause of the current flow.

If you want the current from the rest of the circuit to flow through the wire in question, you must add enough additional potential force to overcome the wire resistance or some of the current is going to flow through a different channel.

If it takes 40 volts to push 10 amps of current through a circuit with superconductor wire, you are going to have to add more voltage force to push the same circuit current through a wire with some resistance.

 

[mivey]

Sorry if this has been wordy, getting a little tired, but it should be clear that it is NOT the voltage drop that causes current to flow, but ONLY the potential source - the battery or transformer - which causes current flow.

Draw the Thevenin circuit for the ends of the wire. Now what is the source? Is it two secondary coils, the primary coil, the substation transformer, the generator...

The voltage across the wire in question is the measure of the potential energy supplied to push current through the wire resistance. Why are you saying identifying the actual source is relevant?

Somewhere in the world a little mouse had to run harder on the wheel so we could push electrons through the wire. Who cares where it came from?

That is like saying the kWh did not do the work but the water over the dam did. OK, but we say the kWh at the load was needed to get the work done not x gallons over the dam.

Again, the potential across the wire is essentially a measure of how much force we apply to push the required number of electrons past the wire resistance.

 

[crossman gary]

Ideal components are simply a tool used to simplify real world problems.

The implications of Ohm's Law break down under certain ideal conditions. For example: Considering a fixed resistance connected to a voltage source - if you increase the voltage, the current increases. With ideal components, there are cases where this relationship is not true.

There are times when the use of ideal components is misleading. The case of the current in the neutral wire is one of those cases. I firmly believe that the current flow in the neutral is a result of the potential difference from one end of the conductor to the other.

 

[crossman gary]

Concerning the previous example of 2 batteries, the negative of one connected to the positive of the other, with two loads of differing resistances connected to form a MWBC. We have three cases to consider:

1. The real life situation where all components of the MWBC have resistance. The wire, the source, and of course the load, all have resistance. >>>> Doing the calculations will result in potential differences which are the cause of all the current flows. The various resistances in the circuit act as a voltage divider... yes, the potential differences are created by the source, but the differences exist in various quantities around the circuit.

2. The theoretical situation where the wires have no resistance, but the source does. >>>>> Doing the calculations still result in a voltage divider circuit and there is a potential difference from one end of the neutral to the other.

3. The theoretical situation where the wires and the batteries are ideal, meaning they have no resistance. In this case, it does indeed appear that current flows through the zero resistance neutral without a potential difference across it. But since this situation cannot exist in reality, why would anyone suggest that it is indeed reality or has any bearing on reality?

For example, if I take one of the ideal batteries and short it with one of the ideal conductors, an infinite amount of electrons will flow. Clearly this is a physical impossibility, because every electron in the universe plus even more would have to be flowing in that circuit. Therefore, this ideal circuit is not universally applicable to all real world circuits.

 

[crossman gary]

It is the potential source, in this case the battery, which is the "force" which gets the electrons to move. The electrons will move from A to B in BOTH a "real" circuit with a voltage drop from A-B and in an "ideal" circuit with no voltage drop from A-B.

The second half of the second sentence is false. Electrons exist and move only in the real world. In your theoretical contrivance of ideal zero impedance conductors and zero impedance infinite sources, there is nothing that moves at all. It is all pencil and paper and math. Show me your theoretical circuit and put a volt meter and amp meter on it and show me that real electrons are moving on the neutral without a potential difference from one end of the neutral to the other. You can't.

Your theoretical circuits are only an approximation of what is actually happening. There are real world MWBC neutrals all over the world with potential differences from one end to the other, with electrons moving because of these differences.

And, not a single real electron ever ever moved in a theoretically perfect circuit in all the history of the earth, and therefore we cannot use the perfect circuit to make blanket statements about what is really happening.