Just to clarify.

[jim dungar]

Quote:
Originally Posted by jim dungar
If the current is I12 and it is in phase with one source V1n then it must also be in phase with the other source.

I expect resistive current to be in phase with its source voltage. Yours can not be, especially if the source is a center tapped transformer referenced as you have.


Jim,

But Jim, it can be, e.g.,

I12 = (V1n - V2n)/R = 240/R @ 0

Now V2n = 120 @ 180 which is out of phase with I12.

Compute I21 and its phase angle is 180 degrees.

The basis of the question I posed was: Two voltage sources are in series there is nothing connected to the common reference point "n" except the sources. There is a single resistor connected between points 1 and 2.

So are you saying a single source current I12 which is out of phase with your reference V2n, simply needs to be inverted into I21 so that it is back in phase? If yes, then isn't the new source current I21 now out of phase with V1n?

Or are you saying that the current flowing into a source is always out of phase with the source?

There is a single load resistor, so isn't there a single source current either flowing from 1->2 or from 2->1. The fact that you are using phasors should not change basic 2-wire circuit analysis.

Please save the issue of single phase wye circuits for another discussion.

 

[mivey]

time domain

time domain

Jim, from a previous post, you proposed two voltage sources, both referenced to the same ground (actually a center-tapped transformer). A 24 ohm resistor was across the 120@0 source (V1n), and a 12 ohm resistor across the 120@180 source (V2n)and a 48 ohm resistor across both sources.

I worked the circuit with vectors and did not see any issues, but there is a problem. The problem with this method is, when using the phasor representation, you have transformed the wave out of the time domain. Unless you keep the time component (e^jwt), you lose the time relationship between V1 and V2.

I will work this in the time domain and I think it may demonstrate what you were saying.

 

[mivey]

it may demonstrate what you were saying.

What you were saying in the previous post anyway. I'm not sure what ya'll have going on here.

 

[jim dungar]

I will work this in the time domain and I think it may demonstrate what you were saying.

You are correct I have asked about two different circuits, one with an unbalanced 3-wire set of resistors and the other a single 2-wire resistor.

I know there are different ways to represent and solve the circuits. The point I have tried to make is that what ever method is used to describe the voltages should not require an adjustment in order to describe the currents. And, just because I see something doesn't mean it is real.

I have never tried to post graphs or pictures so feel free to do so.

 

[rattus]

The basis of the question I posed was: Two voltage sources are in series there is nothing connected to the common reference point "n" except the sources. There is a single resistor connected between points 1 and 2.

So are you saying a single source current I12 which is out of phase with your reference V2n, simply needs to be inverted into I21 so that it is back in phase? If yes, then isn't the new source current I21 now out of phase with V1n?

Or are you saying that the current flowing into a source is always out of phase with the source?

There is a single load resistor, so isn't there a single source current either flowing from 1->2 or from 2->1. The fact that you are using phasors should not change basic 2-wire circuit analysis.

Please save the issue of single phase wye circuits for another discussion.

Jim,

Let us define the current as I12 (external). Then I12 is in phase with V1n.

But look; the sense of I12 is OUT of V1n and INTO V2n. Since the return current flows into, not out of V2n; it would be 180 degrees out of phase with V2n.

If you reverse the current direction, the opposite situation would obtain.

 

[jim dungar]

Jim,

Let us define the current as I12 (external). Then I12 is in phase with V1n.

But look; the sense of I12 is OUT of V1n and INTO V2n. Since the return current flows into, not out of V2n; it would be 180 degrees out of phase with V2n.

If you reverse the current direction, the opposite situation would obtain.

Rattus, if there is a single external current, then there must also be a single current internal to the sources (well actually in that little tiny connection point between the sources), it can not change as it goes from source V2n to V1n. Draw the single loop current and the two sources. All I can see is that in a 2-wire node analysis what goes in to the node must come out the presence or absence of a voltage source doesn't change anything.

 

[rattus]

Rattus, if there is a single external current, then there must also be a single current internal to the sources (well actually in that little tiny connection point between the sources), it can not change as it goes from source V2n to V1n. Draw the single loop current and the two sources. All I can see is that in a 2-wire node analysis what goes in to the node must come out the presence or absence of a voltage source doesn't change anything.

Yes, the same current will be seen at any point in the circuit. But, we have the option of defining the current as coming out of V1n or out of V2n. In the first case we would have I12 @ 0. In the other case, we would have I21 @ 180. Tricky!

 

[jim dungar]

Yes, the same current will be seen at any point in the circuit. But, we have the option of defining the current as coming out of V1n or out of V2n. In the first case we would have I12 @ 0. In the other case, we would have I21 @ 180. Tricky!

I never new that a single current loop (with voltages in series) required tricks to solve it.

Let me get this staight, there is only one loop current but we will let it have whatever direction we want, whenever we want, so the source voltage references do not look wrong. All I can see is a single current if it is I21 then it is out of direction with your V1n, if it is I12 then it is out of direction with your V2n.

Some one needs to post this loop current.

 

[Rick Christopherson]

I have never tried to post graphs or pictures so feel free to do so.

Jim, does this graphic help you out? If not, just tell me what needs to be modified. (I didn't feel like making a whole new drawing, so this is just photoshoped from the text book scan I took earlier. That's why the names are different.)

 

[jim dungar]

Jim, does this graphic help you out? If not, just tell me what needs to be modified. (I didn't feel like making a whole new drawing, so this is just photoshoped from the text book scan I took earlier. That's why the names are different.)

Yes, this describes the circuit I was thinking. It shows that if I2 is in the same direction as Van it must be opposite Vbn.

 

[rattus]

Jims sketch.

Jims sketch.

I never new that a single current loop (with voltages in series) required tricks to solve it.

Let me get this staight, there is only one loop current but we will let it have whatever direction we want, whenever we want, so the source voltage references do not look wrong. All I can see is a single current if it is I21 then it is out of direction with your V1n, if it is I12 then it is out of direction with your V2n.

Some one needs to post this loop current.

Jim, take a look at the attachment. Give credit to Rick for the diagram.

 

[jim dungar]

Jim, take a look at the attachment. Give credit to Rick for the diagram.

Rattus, that is not the circuit I have described. Please stick to solving the 2-wire load circuit, that Rick kindly posted this morning.

 

[Rick Christopherson]

Jim, take a look at the attachment. Give credit to Rick for the diagram.

Just FYI, that was the diagram I photoshopped to create the one shown above, hence the similarities.

 

[Rick Christopherson]

The basis of the question I posed was: Two voltage sources are in series there is nothing connected to the common reference point "n" except the sources. There is a single resistor connected between points 1 and 2.

Jim, I was wondering, would it help your discussion to look at Power calculations instead of just current?

 

[rattus]

Jim, does this graphic help you out? If not, just tell me what needs to be modified. (I didn't feel like making a whole new drawing, so this is just photoshoped from the text book scan I took earlier. That's why the names are different.)

Jim,

Let "b" be the reference point, then Vab = 230 @ 0, and Iab = 2.3A @ 0. Since Iab returns to the reference point, there is no voltage angle associated with this point, therefore no problem, right?

Now let us make "n" the reference point. Now Van = 120V @ 0, and Vbn = 120V @ 180.

The current, Iab, is still 2.3A @ 0, but now it returns to a poiint where the voltage, Vbn, is 120 @ 180.

The key point is Iab flows TO rather than FROM point b. Therefore, you would expect the 180 degree phase shift.

Now this is much ado about not much for such a simple problem, but for more complicated problems I find it advantageous to define a common reference such as the neutral or ground.

 

[jim dungar]

Jim,

Let "b" be the reference point, then Vab = 230 @ 0, and Iab = 2.3A @ 0. Since Iab returns to the reference point, there is no voltage angle associated with this point, therefore no problem, right?

Now let us make "n" the reference point. Now Van = 120V @ 0, and Vbn = 120V @ 180.

The current, Iab, is still 2.3A @ 0, but now it returns to a poiint where the voltage, Vbn, is 120 @ 180.

The key point is Iab flows TO rather than FROM point b. Therefore, you would expect the 180 degree phase shift.

Now this is much ado about not much for such a simple problem, but for more complicated problems I find it advantageous to define a common reference such as the neutral or ground.

You agree that the current Iab is at 0?, what about the currents Ibn and Ina and don't they have to be identical (i.e. Iab=Ibn=Ian)? But how can the angle of these currents be identical if those of their voltage sources are not, or is it acceptable to have a resistive current "out of phase" with its voltage?

This is not much ado about nothing, discussing current flow is an extension from your discussion of voltage sources.

 

[rattus]

You agree that the current Iab is at 0?, what about the currents Ibn and Ina and don't they have to be identical (i.e. Iab=Ibn=Ian)? But how can the angle of these currents be identical if those of their voltage sources are not, or is it acceptable to have a resistive current "out of phase" with its voltage?

QUOTE]

Yep, it is all one current, and it is in phase with Van and Vnb as you can plainly see. However, if we define the voltage as Vbn, then that voltage will be at 180 degrees to Vnb.

It is a matter of the way we define the second voltage--Vnb or Vbn.

Now if we consider the total voltage across R,

Vab = Van - Vbn = 230V @ 0, and Iab will be in phase with Vab as you would expect.

It is all in the way you look at it.

 

[jim dungar]

It is a matter of the way we define the second voltage--Vnb or Vbn.

It is all in the way you look at it.

Yes, that is what I have been saying. Your voltage definition results in at least one source current having a 180? phase shift from its driving voltage.

If we extend this discussion by adding a center tapped point to my 2-wire resistor so that there is a balanced 3-wire circuit. One of two things must occur when using your voltage definition, either the current flow through source Vbn suddenly changes direction or our standard power equations seem to get messed up.

Given that Iab=Ibn=Ina=I and your use of Vbn = -Vna,
then Pna = IVan and Pbn = I(-Van)
so Pna=-Pbn.

Yes, I realize that this can be continued to |Pna|=|Pbn| but the intermediate results certainly look different than I would expect.

 

[LarryFine]

Jim, you hit on the way I look at it. Sure, from the center tap, A and C resemble opposing phases, but that's only because it's "traditional" to measure everything from the grounded conductor (if there is one).

I've long had a way of looking at things as either fact or opinion, loosely speaking. Facts stand alone without requiring a belief or specific perspective, whereas opinions require a human input to exist.

Insisting on placing the black probe on the neutral is a personal decision. Drounded conductors are, in a way, arbitrarily chosen. Any system will function (GFCI's aside) regardless of which, if any, conductor is grounded.

Center tap aside, a Delta supply is symmetrical. From any line conductor's point of view, the wire to the right is +120 deg., and the one to the left is -120 deg. Grounding one secondary's center tap doesn't change anything else.

Such a Delta can be used as either a 240v 3ph supply or a 120/240v 1ph supply at the same time, but not both to a single piece of equipment. The 3ph load ignores the existence of the neutral, and the 1ph load ignores the high leg.

I like yummy.

 

[rattus]

Yes, that is what I have been saying. Your voltage definition results in at least one source current having a 180? phase shift from its driving voltage.

If we extend this discussion by adding a center tapped point to my 2-wire resistor so that there is a balanced 3-wire circuit. One of two things must occur when using your voltage definition, either the current flow through source Vbn suddenly changes direction or our standard power equations seem to get messed up.

Given that Iab=Ibn=Ina=I and your use of Vbn = -Vna,
then Pna = IVan and Pbn = I(-Van)
so Pna=-Pbn.

Yes, I realize that this can be continued to |Pna|=|Pbn| but the intermediate results certainly look different than I would expect.

It certainly is confusing isn't it? Now, when computing average power, current must be LEAVING the source in question. We do that on paper just by reversing the arrow and shifting the phase by 180 degrees. The average power is positive as it has to be.

All in the way you look at it.