Just to clarify.

[Rick Christopherson]

The idea that the current all of a sudden becomes "out of phase" with the voltage was part of the OP.

As you have already noted, the direction of the current relative to the voltage dictates whether the object is a supply of power or a consumer or power. From what you have said today, the lower half of the power supply (between b and n) is a consumer of power, not a supplier of power.

If the voltage source between (b) and is not consuming power, then the statements you have made are not accurate. Like Rattus, you have lost sight of the minus sign that must be present.

Yes. This entire discussion is about a missing minus sign!

 

[mivey]

Respect for Jim

Respect for Jim

rattus understands that the phase shift happens when crossing the neutral point (the midpoint of the resistor), Jim does not.

Perhaps this needs to be stated differently as it may make it seem like I'm saying Jim doesn't know what he is talking about. On the contrary, I have the utmost respect for Jim. So here is the re-statement:

rattus seems to be applying the concept that the phase shift happens at the neutral point. Jim does not appear to agree with that concept.

Please, if I ruffle any feathers, it is not my intent.

 

[mivey]

minus sign?

minus sign?

As you have already noted, the direction of the current relative to the voltage dictates whether the object is a supply of power or a consumer or power. From what you have said today, the lower half of the power supply (between b and n) is a consumer of power, not a supplier of power.

If the voltage source between (b) and is not consuming power, then the statements you have made are not accurate. Like Rattus, you have lost sight of the minus sign that must be present.

Yes. This entire discussion is about a missing minus sign!

I'm not sure what you mean. If it is about the original single phase circuit drawing, I have stated that the circuit was drawn incorrectly as there could not be any current flow. I belive rattus has stated his intentions many times. He seemed to be fine with the concept of two sources in series feeding through a resistor. He just wants to move the reference point from the corner to in-between the two sources. He may have mis-worded something along the way (I really don't recall) but a reversed sign did not appear to be his intent. The teaser circuit he offered in post #66 makes that clear to me.

Perhaps I need to clarify my position. I am saying that with or without a neutral reference, all of the currents are in phase with their voltages. Without the neutral reference, the corner driving voltage is clearly in phase with the currents.

With the neutral reference, what may not be obvious is that there is a phase (or call it direction if you want) change when you cross the neutral point. Both the current and voltage experience a phase change, not just one or the other.

This is not because the electrons made a u-turn. The electrons are doing the same thing they were doing before they crossed the neutral point. What changed was the observation perspective i.e., the electrons relation to the reference point.

The currents above the neutral are in phase with their voltages and the current below the neutral are in phase with their voltages.

 

[coulter]

...rattus understands that the phase shift happens when crossing the neutral point (the midpoint of the resistor), Jim does not. The north pole was an attempt at explaining that. ...

I don't get this at all. How could the current in a series loop change at all? There is no magic happens to the current phase angle at the point marked N. Once a reference angle was picked, and most of the discussion picked the upper source as phase angle zero, then the current anywhere in the loop is phase angle zero.

Help me out here - where/how can a phase shift occur in a series loop? The current entering a node has a certain magnitude and phase angle. According to Mr Norton, the current leaving the nde has the same magnitude and phase angle. This is true of the N as well as any other point in the loop.

carl

 

[coulter]

...The currents above the neutral are in phase with their voltages and the current below the neutral are in phase with their voltages.

Posts crossed in the air. I get your point. You want to use two different references for zero phase angle. As you said above the N point zero phase angle is one direction, and below the N zero phase angle a different direction. Interesting, the thought would have never occured to me to change the zero phase angle reference depending on where I was looking at the circuit.

As you and rattus have said (paraphrased), "you can do this if you want just add the appropriate minus signs to make the math work out."

Still baffles me why one would want to do this.

carl

 

[coulter]

Let me use a "<" to mean phase angle as in "<0" = phase angle zero

Start with a single source 240 <0 with a single series 24 ohm resistor. anywhere in the loop i = 10<0.

Split the source into two equal series pieces, label each 120<0, don't label the center N. Anywhere in the loop, i = 10<0

Now label the center N. I think the current can still accurately be called i = 10<0 anywhere in the loop.

Now split the 24 ohm resistor into two 12 ohm pieces. Current is still i = 10<0 anywhere in the loop.

Now add a wire from N to the center of the two resistors. Current through the two resistors is still i = 10<0. Current through the two sources is still i = 10<0.

As you know, Norton says the curent entering a node ... So, if you insist on a phase angle change at the N point, then where is the other phase angle change. Oh yeah, it is at the Vbn point. And I would say, Why bother introducing two minus signs to deal with a current that doesn't change?

carl

 

[mivey]

I don't get this at all. How could the current in a series loop change at all? There is no magic happens to the current phase angle at the point marked N. Once a reference angle was picked, and most of the discussion picked the upper source as phase angle zero, then the current anywhere in the loop is phase angle zero.

Help me out here - where/how can a phase shift occur in a series loop? The current entering a node has a certain magnitude and phase angle. According to Mr Norton, the current leaving the nde has the same magnitude and phase angle. This is true of the N as well as any other point in the loop.

carl

It is because of the neutral point. Let me state it by using an example of what we see. If you take a meter and reverse the leads, as far as the meter is concerned, there is a direction change. This change has not happened to the electrons, it happened with our meter.

Take a voltage source across a resistor. Take one lead of a meter and put it on the bottom of the resistor, and put the other lead at the top. Now swap the leads. This observation is what I'm calling the phase shift. Not that the circuit reversed, only that our measurement of that circuit reversed.

Now take one lead and stick it in the middle of the resistor. Take the other lead and put it on the bottom of the resistor. Assume the resistor has an exposed core. Slide the lead up the resistor and through the neutral (mid) point. As you approach the neutral, the voltage will slowly drop until it reaches zero. As you move beyond the neutral, the voltage will start increasing. This is what I'm describing as the voltages being opposite in phase across the neutral. The currents also experience the phase shift across the neutral point.

Because we have defined a reference point, we measure voltages as either positive from it or negative from it. This is by definition what we decided to call a positive and negative voltage. We are using the neutral as a reference.

We also do the same for current. We define a current as positive or negative based on what direction it is going. On one side of the neutral, current is coming towards the neutral. On the other side of the neutral current is going away from the neutral. The same way we would define current as either entering or leaving a node.

The reference point, by definition, determines when the voltages and currents are positive.

Did that help you understand what I'm trying to say, but can't seem to express?

 

[coulter]

It is because of the neutral point. Let me state it by using an example of what we see. If you take a meter and reverse the leads, as far as the meter is concerned, there is a direction change. ...
Did that help you understand what I'm trying to say, but can't seem to express?

I think so. You want to put an ammeter in one part of the circut and measure the current. Then you move the meter to a different part of the same series loop, reverse the leads, and say, "Oh look, the current is a different phase angle."

If that's it, I still baffled as to why one would want to do that.

edited punctuation

carl

 

[mivey]

Posts crossed in the air. I get your point. You want to use two different references for zero phase angle. As you said above the N point zero phase angle is one direction, and below the N zero phase angle a different direction. Interesting, the thought would have never occured to me to change the zero phase angle reference depending on where I was looking at the circuit.

As you and rattus have said (paraphrased), "you can do this if you want just add the appropriate minus signs to make the math work out."

Still baffles me why one would want to do this.

carl

I would not do this for the simple resistor circuit. This is a concept circuit. For the simple resistor circuit, the step analysis is easier if you don't use the neutral as a reference.

What is not clear to me is if not using the neutral would make the analysis of a circuit like I've shown in post #59 any easier.

If it does make it easier, I'm all for it because it brings me value. If it does not make my life easier, then I just see it as a matter of preference.

 

[mivey]

And I would say, Why bother introducing two minus signs to deal with a current that doesn't change?

carl

For the simple circuit, I agree. The neutral does nothing if attached to the equilibrium point. If you were to run the wire to a point that is say 6 ohms on one side and 18 on the other, you will have neutral current. Even in this case, you could use step analysis, which would be simplified by not using the neutral as the reference.

But what if we have RLC compenents in the neutral leg and unbalanced RLC components elsewhere? You are going to have to solve some complicated loops then and would the 240 volt reference make it easier? I'm not sure.

reverse the leads, and say, "Oh look, the current is a different phase angle."

That does sound funny when you put it like that.

I took it to be the assumption was that the decision was already made (upper management?) so how are we going to describe our measurements now.

I don't remember this thread being about the merits of the neutral point (but this bled over from elsewhere so who knows). The disagreement was over whether or not voltages and currents are out of phase across the reference (neutral) point.

 

[coulter]

...What is not clear to me is if not using the neutral would make the analysis of a circuit like I've shown in post #59 any easier...

The a, b, n in post 59 circuit don't give me any useful information. What would help, is if the two sources were each labeled something like 120<0

Defining the voltages at a and b (reference to n) with Van = 120<0 and Vbn = 120<180 would make me go, "Huh?" However that could be because I'm not sure I have ever solved or been asked to solve a single phase two pole problem in my career.

carl

 

[mivey]

What would help, is if the two sources were each labeled something like 120<0

I wanted respondent to pick the reference point based on which analysis would bring me value, like being easier to solve. The assumption up to that point was that the voltages were equal in magnitiude and in phase. I guess I should have put that on the diagram. or just used "V" for both.

 

[coulter]

...But what if we have RLC compenents in the neutral leg and unbalanced RLC components elsewhere? You are going to have to solve some complicated loops then and would the 240 volt reference make it easier? I'm not sure....

Attached a sketch for my partial solution of post 59. All the currents are CW (even the ones with out the arrows

I picked 120<0 for each of the two sources. I did not define the voltages Van or Vbn. Oh I just noticed the 120<0 scribble at the top is just that - a scribble.

So, the first equation is: Starting at the upper left corner

(i1 + i6) j0.012 + (i1 + i6)j0.1 + (i1 - i3)24 + (i1- i4)(6+ j5) + (i1 - i2)(-j2.7) + (i1-i2)(1 + j1.5) - 120<0 = 0

Before you say anything about the -120<0, that's because the current is CW, so the voltage drops are all + to - CW until I get to the source that one goes - to +.

Since the i2 current is also CW, the lower source will also end up as a -120<0.

The a, b, n have no bearing on the equations.

Equation 6 will sum up the voltage drops around the loop ending in ... -120<0 -120<0 for the lower and upper sources. There is no magic phase shift in i6.

PS Please don't make me feel too dumb if I fat fingered the equations.

spelling edit

carl

 

[mivey]

Then you move the meter to a different part of the same series loop, reverse the leads, and say, "Oh look, the current is a different phase angle."

I'm not interpreting what you wrote to be the same as what I meant.

I did not say move the meter and swap the leads. One lead is fixed. I'm just probing all around the circuit with the other lead. I notice that when I probe above the neutral point the voltage reads positive and when I probe below the neutral point, the voltage is negative. I get the same result with my "current probe" that has one "lead" fixed at the neutral point.

[edit: clarity]

 

[coulter]

... I get the same result with my "current probe" that has one "lead" fixed at the neutral point. ...]

I see what you are saying. i'm a post ahead of you - i wait till you catch up

later

carl

 

[mivey]

My goodness. I did not mean for you to tackle that math at this time of day. I just posted the circuit to probe the merits of a 120<0+120<0 series source and comparing the analysis with and without a reference point between them.

If you are going to look for the magic phase shift, pick a simple resistor circuit because the concept is the same.

I'm not going to bet my life on the concept. I just want to ultimately find out if it is a valid way to look at the circuit.

As for the other circuit, I just want to know if ignoring the neutral point saves me any analysis steps when I try to determine the inner currents and voltages. It doesn't even have to save me steps, it can just make it easier. I'm looking for the value of one method vs the other.

 

[winnie]

Now add a wire from N to the center of the two resistors. Current through the two resistors is still i = 10<0. Current through the two sources is still i = 10<0.

As you know, Norton says the curent entering a node ... So, if you insist on a phase angle change at the N point, then where is the other phase angle change. Oh yeah, it is at the Vbn point.

I'd say that you were using two different ways of accounting for current.

1) You look at the current following closed circuit paths. In the case of a single loop, the current must be the same everywhere, and thus in your example i = 10<0 A.

2) You look at the current entering each _node_. Since each node has two paths into it (for a single loop), there are _two_ currents that must sum to zero. In your example circuit, these two currents are 10<0A and 10<180A.

-Jon

 

[jim dungar]

spiraling out of control

spiraling out of control

From the beginning I have said "just because you can doesn't mean you should". I have been trying to point out the issue of choosing voltage and current directions.

First everyone seems to be misapplying Kirchhoff to the single current two node example. Effectively, Kirchhoff says "what goes in must come out" because currents entering must sum to zero. In the circuit of Coming and Going the currents cannot be summed to zero, if they do then no current is leaving. Now if the circuit was Coming1 and the other was Coming2 then per KCL they must sum to zero.

Second, I don't believe I have ever said to pick a corner reference point, nor to not pick a neutral reference point (if I have I apologize). I have just wanted a single current, flowing through two sources in series, to be in the same direction as the driving voltages. This is a concept I learned when first learning about mesh currents. Before we had personal computers and calculators that could solve complex matrices we tried to simplify the problem visually first, by choosing appropriate mesh (loop) currents. My reference is Electric Circuits by Joseph A. Edminister dated 1965: "The positive sign is used if the source drives in the direction of the mesh current, and the negative sign if it drives against the mesh current". This philosophy supports my statement, "In a single 2-wire circuit there is only 1 mesh/loop current to solve for therefore all voltage sources should be identified in the same direction".

 

[Rick Christopherson]

With the neutral reference, what may not be obvious is that there is a phase (or call it direction if you want) change when you cross the neutral point. Both the current and voltage experience a phase change, not just one or the other.

So what you are saying is, "two wrongs make a right". You first violate Kirchoff's Voltage law by requiring subtraction, then you violate Kirchoff's Current Law by saying the current leaving a node in one direction is equal to the current leaving the node in the other direction.

Yes, two wrongs do make a mathematical right, but is it really "Good Engineering" to do so?

Edit Sorry, I type slow. I didn't know that Jim said something similar to this already.

I'm just probing all around the circuit with the other lead. I notice that when I probe above the neutral point the voltage reads positive and when I probe below the neutral point, the voltage is negative.

What you are doing is intermixing nodal analysis with circuit analysis. This can be done as long as you keep your minus signs in check, but if you lose your minus signs, then you have what Rattus is describing.

I get the same result with my "current probe" that has one "lead" fixed at the neutral point.

That's a very interesting current probe you have there. Where can I get one like that?

 

[mivey]

try to see rattus's viewpoint

try to see rattus's viewpoint

Rick, as for the current probe, it is just a shunt. I can give you a link to some websites where you buy varying sizes if you would like one. Putting the words "current probe" in quotes was a shorter description.

First, I am not arguing that a single voltage/current direction is not the best way to analyze the circuit. Actually, for the circuit given, I think it is the best way because, as Jim pointed out, it simplifies the analysis.

But given that rattus (or someone) proposed using the neutral point as a reference, you now have what "scopes out" to be a phase shift.

When rattus was describing V1 & V2 as being "out of phase" across the neutral, that is fine with me, let's help him describe it. Whether we like it or not, let's try to look at it from his standpoint. We are not promoting "violating" any laws, just a way to describe what he wants.

Getting away from the fact that the original simple circuit was not drawn this way, it seems apparant that what rattus wanted was the following: Given that the two voltages shown are identical and in series. It appears rattus wants to call the two voltages across the neutral as "out of phase" across the neutral. This is what you would see on an oscope with the neutral as the common point. Still fine with me, let's help him do that without "violating" any laws.

There seemed to be agreement that we were going to "allow" the voltage across the neutral to be called a "phase shift" for the sake of rattus's analysis. The problem than was that I believe it was alluded to that if you allow this, then the voltages would be "out of phase" with their currents. This absolutely can not be in the resistor circuit. We really would be "violating" some laws.

My thought was that if you have a concept that allows the "voltage crossing the neutral" to be called a "phase shift" then, by the laws of physics, you have to have a reciprocal concept for the current. If the phase of the voltage "shifts" then so must the phase of the current. Using the "current probe" was a way to show this in the physical world.

If you can observe the "phase shift" on an oscope for the voltage, and if you found a way to scope the current, you would HAVE to be able to observe the same "phase shift" on the current. Actually we do have a way to observe the current wave on the oscope, and it is by using a measuring shunt or "current probe".

The concept of a simultaneous phase shift of voltage and current when you cross the neutral allows rattus to continue using the convention he wants and we can not accuse him of "violating" any laws.

The concept of the voltage shift was easy for everyone to see in the physical world because the oscope or meter is the obvious example. The north pole, "coming vs going", etc was my attempt at trying to give a physical world example of the current shift. It would seem I have failed miserably so far.