Talk:Y-Δ transform

message: This seems to read more like a how-to than a proper encyclopedic article on the subject. 
----
The following old comments seem no longer applicable. --SlothMcCarty (talk) 05:05, 2 December 2011 (UTC)
thread_title: (No section)

message: Here are a few suggestions:
# The article could give an example of a bridge circuit that can't be broken down into parallel and series parts unless the Y-delta transformation is used. Otherwise it's not clear from the article why the transformation is useful.
# The transformation can be derived from the condition that the resistances of the two circuits are equal when voltages are applied to any two terminals, and the third is left floating. However, it's not trivially obvious that this also makes them completely equivalent in all other cases, e.g., where no terminal is floating. I'm not a big fan of mathematical proofs in WP articles, but the logic could be made more clear without giving the actual proofs.
# Can the Y-delta transformation be used to break *any* network of resistors into series and parallel parts?
--Bcrowell 16:47, 26 September 2005 (UTC)
replies: text: If someone would care to transcribe the Anderson Bridge circuit [], it is an excelent example of where a Delta-star transform helps its analysis.  The bridge allows measurement of inductors and capacitors but requiring only variable resistors in the bridge itself.  The problem with analysis is that the bridge doesn't quite resemble a conventional bridge as the detector is not connected to a corner.  However performing a delta-star transform on the delta formed by P, r and C (in the referenced diagram), converts the bridge into a conventional bridge circuit, although the detector does have an impedance in series with it - but as the detector current is zero at balance, it conveniently disappears. 109.153.242.10 (talk) 16:57, 25 January 2012 (UTC)
thread_title: Suggestions

message: Can someone explain why there is a 30 degree phase shift between delta-wye and wye-delta transformer configurations.  I've check numerous sites, but all the sites say is that there is a phase shift of either +30 or -30 degrees.
If a transformer has DELTA in primary and both STAR & DELTA in secondary so what will be the phase difference in secondary between STAR and DELTA.
:(The comment above me was unsigned.) The difference of the phase shifts of the line-to-line phasor voltages of a delta-wye transformer will depend on the spatial orientation of the windings, which is equivalent to the location of the dots using the dot convention. It also depends on the phase sequence of the system.
:Some D-Y transformers have such a phase difference of 30°, 330° (= -30°), 150°, 210° ( = 150°). --Alej27 (talk) 12:55, 10 January 2021 (UTC)
thread_title:  30 degree phase shift 

message: I’ve tried to clean up the article a bit. Since I did not understand the intent of the statement
:Balanced System:  R_{\Delta} = 3 \times R_y 
I’ve moved it here. Can somebody put it back in context or explain it? —xyzzyn 01:12, 14 January 2007 (UTC)
replies: text: It means that when the three resistors or impedances are exactly the same, the equation you posted above is true and can be used to convert between delta and wye configurations (and vice versa). That equation is derived from the more general ones shown in the article. ---Alej27 (talk) 20:01, 13 January 2021 (UTC)
thread_title:  Balance 

message: I do not agree with BCrowell above
I agree assuming a node is at zero potential is not feasable but that is not done here. The voltages of the nodes are never preserved (note that there is one more node in the graph) only the currents coming in and out of the nodes are preserved.
This seems to imply superimposing the 3 cases
(assume the inputs at nodes N1 ,N2,N3 are current sources and not voltage sources)
1. we have N1
2. we have N2 
3. we have N3
in each of these cases, we have to compute the current leaving each node w.r.t "a zero reference (that we add in the graph)" being equivalent.
so in other words
we would have between N1 and N2
N1-N2=Ib.Rb
N2-N3=Ic.Rc
N3-N1=Ia.Ra
hence between N1 and N2 we have 2 effective circuits
N1-N2=Ib.Rb
and 
N2-N1=IcRc+IaRa
because we assume N3 is non existent (nonzero but non existent) we can assume it is parallel to the 1st and hence Ic=Ia
Likewise for the other 2 cases
So in other words we are superimposing the 3 conditions to find the equivalent (assume the inputs at nodes N1 ,N2,N3 are current sources and not voltage sources)
-Alok 06:11, 20 January 2010 (UTC)
replies: text: When you convert a delta-connected three-phase device to an equivalent wye-connected three-phase device (or vice versa), as far as I know the line (or line-to-line) phasor voltages and line phasor currents are preserved, they don't change in numerical values. --Alej27 (talk) 20:06, 13 January 2021 (UTC)
thread_title:  practical usage 

message: I would like to see the diagram modified slightly in order to make the equations simpler and more elegant in my view. A non-exhaustive perusal of talk entries indicates that this topic is not covered. I would have liked to demonstrate the changes, but I could not figure out how to edit mathematical Wikipedia expressions.
I refer to the left hand (delta) portion of the article's main diagram. Using my representation for subscripted symbols, I would replace Rc, Ra, and Rb by Ra, Rb, and Rc respectively. Doing so will associate the first three letters of the alphabet, abc, with the first three natural numbers, 123 in order. This will correspond to node order numbers. That will remove one source of confusion when trying to follow the details. There will be more symmetry to the transformation equations.  —Preceding unsigned comment added by PEBill (talk • contribs) 23:10, 10 September 2010 (UTC)
replies: text: I agree. Dauto (talk) 15:14, 13 February 2011 (UTC)

text: Done. I would appreciate someone checking my work. --SlothMcCarty (talk) 05:00, 2 December 2011 (UTC)
thread_title:  Improve symmetry of transformation equations 

message: The topic of the article is backwards vs. what you write in the article!
The topic says: "Y-Δ transform", where the first character is pronounced "Star" and the second one "Delta", but the article talks about "Delta Star transform".
Seems like some other languages they've got this already right, but how come it's still backwards here?!
Please fix to avoid more confusion or explain yourself in the article. Thanks!
188.238.22.14 (talk) 19:26, 14 September 2013 (UTC)
thread_title:  Topic is backwards 

message: The purpose of this section is to prove how to convert a delta-connected practical generator (balanced or unbalanced) into an equivalent wye-connected generator. The word practical means the voltage sources have series impedance.
I added the final formulas to the main article (this version).
The following was taught to me during the last course on Electric Circuits (in Spanish). According to the professor, the name of the method was Neutral shift method (in Spanish, El método del desplazamiento del neutro). Unfortunately, I haven't found this method in any classical textbook on circuit theory, so I can't provide references. Because of that, I decided to prove the formulas, so that the reader can be sure this method works and how to derive it.
=== The derivation of the formulas ===
Consider the following network. I will denote phasors with a tilde (~) on top, and impedances with a circumflex (^) on top. Just for clarification on the notation, for phase A, B and C:
* The line phasor currents are respectively \tilde{I}_\text{A}, \tilde{I}_\text{B}, \tilde{I}_\text{C}.
* The phase phasor currents are respectively \tilde{I}_\text{BA}, \tilde{I}_\text{CB}, \tilde{I}_\text{AC}.
* The terminal line (or line-to-line) phasor voltages are respectively \tilde{V}_\text{AB}, \tilde{V}_\text{BC}, \tilde{V}_\text{CA}.
* The terminal phase phasor voltages are equal to the respective terminal line phasor voltages because the generator is in delta.
* The internal line (or line-to-line) phasor voltages are respectively \tilde{V}_{\text{A} ' \text{B}}=\tilde{V}_\text{s1}, \tilde{V}_{\text{B} ' \text{C}}=\tilde{V}_\text{s2}, \tilde{V}_{\text{C} ' \text{A}}=\tilde{V}_\text{s3}.
* The internal phase phasor voltages are equal to the respective terminal line phasor voltages because the generator is in delta.
* The internal series impedances of each armature winding are respectively {\hat Z}_\text{s1}, {\hat Z}_\text{s2}, {\hat Z}_\text{s3}.
First we apply source transformation to transform the practical voltage sources into practical current sources:
{\tilde I}_\text{s1} = \dfrac{{\tilde V}_\text{s1}}{{\hat Z}_\text{s1}}
{\tilde I}_\text{s2} = \dfrac{{\tilde V}_\text{s2}}{{\hat Z}_\text{s2}}
{\tilde I}_\text{s3} = \dfrac{{\tilde V}_\text{s3}}{{\hat Z}_\text{s3}}
This results in the following circuit diagram.
Now we transform the delta-connected impedances into wye-connected impedances. Taking into account the labeling of the impedances in the previous figure and the labeling of the impedances in the following figure, we have:
{\hat Z}_\text{s1Y} = \dfrac{{\hat Z}_\text{s1} \, {\hat Z}_\text{s3}}{{\hat Z}_\text{s1} + {\hat Z}_\text{s2} + {\hat Z}_\text{s3}}     (1)
{\hat Z}_\text{s2Y} = \dfrac{{\hat Z}_\text{s1} \, {\hat Z}_\text{s2}}{{\hat Z}_\text{s1} + {\hat Z}_\text{s2} + {\hat Z}_\text{s3}}     (2)
{\hat Z}_\text{s3Y} = \dfrac{{\hat Z}_\text{s2} \, {\hat Z}_\text{s3}}{{\hat Z}_\text{s1} + {\hat Z}_\text{s2} + {\hat Z}_\text{s3}}     (3)
This results in the following circuit diagram.
Now we apply a technique known as source splitting and source shifting. For example, we can shift or move the current source {\tilde I}_\text{s2} that is located between nodes C and B into the nodes C and N and also into the nodes N and B. In this way, the KCL equation at node C, at node N, and at node B is not altered: at node C there still leaves a current {\tilde I}_\text{s2} (that now enters to node N), at node B there still enters a current {\tilde I}_\text{s2} (that now leaves from node N), and at node N no net current {\tilde I}_\text{s2} enters or leaves (because now the current {\tilde I}_\text{s2} enters and leaves the node N, resulting in no net current). The previous procedure is also applied for the current sources {\tilde I}_\text{s1} and {\tilde I}_\text{s3}. This results in the following circuit diagram.
If you are still in doubt as to whether the circuit of figure 4 is indeed equivalent to the one of figure 3, you can apply KCL at the nodes A, B, C and N of both figures, and you’ll find out they are the same equations, meaning those two circuits are indeed equivalent.
Now we combine the current sources in parallel by applying KCL. Taking into account the direction of the arrow of the current sources in the previous figure and the direction of the arrow of the current source in the following figure, we have:
{\tilde I}_\text{s1T} = {\tilde I}_\text{s1} - {\tilde I}_\text{s3}
{\tilde I}_\text{s2T} = {\tilde I}_\text{s2} - {\tilde I}_\text{s1}
{\tilde I}_\text{s3T} = {\tilde I}_\text{s3} - {\tilde I}_\text{s2}
This results in the following circuit diagram.
Lastly we apply source transformation to transform the practical current sources into practical voltage sources:
{\tilde V}_\text{s1Y} = {\tilde I}_\text{s1T} \, {\hat Z}_\text{s1Y}
{\tilde V}_\text{s2Y} = {\tilde I}_\text{s2T} \, {\hat Z}_\text{s2Y}
{\tilde V}_\text{s3Y} = {\tilde I}_\text{s3T} \, {\hat Z}_\text{s3Y}
Taking into account the direction of the arrow of the current sources in the previous figure, this results in the following circuit diagram.
Now we revert the change of variables we did throughout the derivation, going from the last equations to the first ones. After some math, we get:
{\tilde V}_\text{s1Y} = \left( \dfrac{{\tilde V}_\text{s1}}{{\hat Z}_\text{s1}} - \dfrac{{\tilde V}_\text{s3}}{{\hat Z}_\text{s3}} \right) \, {\hat Z}_\text{s1Y}     (4)
{\tilde V}_\text{s2Y} = \left( \dfrac{{\tilde V}_\text{s2}}{{\hat Z}_\text{s2}} - \dfrac{{\tilde V}_\text{s1}}{{\hat Z}_\text{s1}} \right) \, {\hat Z}_\text{s2Y}     (5)
{\tilde V}_\text{s3Y} = \left( \dfrac{{\tilde V}_\text{s3}}{{\hat Z}_\text{s3}} - \dfrac{{\tilde V}_\text{s2}}{{\hat Z}_\text{s2}} \right) \, {\hat Z}_\text{s3Y}     (6)
In this way, we’ve shown the delta-connected generator of figure 1 is equivalent to the wye-connected generator of figure 6; the relationship between them is given by equations (1) to (6).
—It shall be noted the neutral node of the equivalent network (figure 6) is fictitious or imaginary, it doesn’t exist in real life. The terminal fictitious line-to-neutral phasor voltages are {\tilde V}_\text{AN}, {\tilde V}_\text{BN}, {\tilde V}_\text{CN}. The internal fictitious line-to-neutral phasor voltages are {\tilde V}_{\text{A}  \text{N}} = {\tilde V}_\text{s1Y}, {\tilde V}_{\text{B}  \text{N}} = {\tilde V}_\text{s2Y}, {\tilde V}_{\text{C}  \text{N}} = {\tilde V}_\text{s3Y}.
Notice in figure 6 the new internal nodes of the equivalent/fictitious wye generator are A, B and C, while in figure 1 the original internal nodes of the actual/real delta generator are A', B' and C'.
It shall also be stated those networks are equivalent even if the generator is unbalanced and/or the rest of the circuit/power system is unbalanced, since the expressions that relate these networks don’t depend on the line or phase currents, which indeed depend on whether the system is balanced or not.
Also notice the terminal line (or line-to-line) phasor voltages and the line phasor currents of both networks (figure 1 and 6) are not affected after the transformation.
—If the delta-connected generator of figure 1 is balanced (which means the magnitude of the voltage sources is the same, they’re phase-shifted by 120° between each other, and the impedances are the same), then equations (1) to (6) reduce to the following:
{\hat Z}_\text{sY} = \dfrac{{\hat Z}_\text{s}}{3}     (7)
{\tilde V}_\text{s1Y} = \dfrac{{\tilde V}_\text{s1}}{\sqrt{3} \, \angle \pm 30^\circ}     (8)
{\tilde V}_\text{s2Y} = \dfrac{{\tilde V}_\text{s2}}{\sqrt{3} \, \angle \pm 30^\circ}     (9)
{\tilde V}_\text{s3Y} = \dfrac{{\tilde V}_\text{s3}}{\sqrt{3} \, \angle \pm 30^\circ}     (10)
where for equations (8) to (10) you use the first sign (+) if the phase sequence is positive/abc or the second sign (-) if the phase sequence is negative/acb.
--Alej27 (talk) 18:53, 27 January 2021 (UTC)
thread_title:  Derivation of the formulas for converting a delta to wye practical generator 

message: Consider the following network. I will denote phasors with a tilde (~) on top, and impedances with a circumflex (^) on top. Just for clarification on the notation, for phase A, B and C:
* The line phasor currents are respectively \tilde{I}_\text{A}, \tilde{I}_\text{B}, \tilde{I}_\text{C}.
* The phase phasor currents are respectively \tilde{I}_\text{BA}, \tilde{I}_\text{CB}, \tilde{I}_\text{AC}.
* The terminal line (or line-to-line) phasor voltages are respectively \tilde{V}_\text{AB}, \tilde{V}_\text{BC}, \tilde{V}_\text{CA}.
* The terminal phase phasor voltages are equal to the respective terminal line phasor voltages because the generator is in delta.
* The internal line (or line-to-line) phasor voltages are respectively \tilde{V}_{\text{A} ' \text{B}}=\tilde{V}_\text{s1}, \tilde{V}_{\text{B} ' \text{C}}=\tilde{V}_\text{s2}, \tilde{V}_{\text{C} ' \text{A}}=\tilde{V}_\text{s3}.
* The internal phase phasor voltages are equal to the respective terminal line phasor voltages because the generator is in delta.
* The internal series impedances of each armature winding are respectively {\hat Z}_\text{s1}, {\hat Z}_\text{s2}, {\hat Z}_\text{s3}.
First we apply source transformation to transform the practical voltage sources into practical current sources:
{\tilde I}_\text{s1} = \dfrac{{\tilde V}_\text{s1}}{{\hat Z}_\text{s1}}
{\tilde I}_\text{s2} = \dfrac{{\tilde V}_\text{s2}}{{\hat Z}_\text{s2}}
{\tilde I}_\text{s3} = \dfrac{{\tilde V}_\text{s3}}{{\hat Z}_\text{s3}}
This results in the following circuit diagram.
Now we transform the delta-connected impedances into wye-connected impedances. Taking into account the labeling of the impedances in the previous figure and the labeling of the impedances in the following figure, we have:
{\hat Z}_\text{s1Y} = \dfrac{{\hat Z}_\text{s1} \, {\hat Z}_\text{s3}}{{\hat Z}_\text{s1} + {\hat Z}_\text{s2} + {\hat Z}_\text{s3}}     (1)
{\hat Z}_\text{s2Y} = \dfrac{{\hat Z}_\text{s1} \, {\hat Z}_\text{s2}}{{\hat Z}_\text{s1} + {\hat Z}_\text{s2} + {\hat Z}_\text{s3}}     (2)
{\hat Z}_\text{s3Y} = \dfrac{{\hat Z}_\text{s2} \, {\hat Z}_\text{s3}}{{\hat Z}_\text{s1} + {\hat Z}_\text{s2} + {\hat Z}_\text{s3}}     (3)
This results in the following circuit diagram.
Now we apply a technique known as source splitting and source shifting. For example, we can shift or move the current source {\tilde I}_\text{s2} that is located between nodes C and B into the nodes C and N and also into the nodes N and B. In this way, the KCL equation at node C, at node N, and at node B is not altered: at node C there still leaves a current {\tilde I}_\text{s2} (that now enters to node N), at node B there still enters a current {\tilde I}_\text{s2} (that now leaves from node N), and at node N no net current {\tilde I}_\text{s2} enters or leaves (because now the current {\tilde I}_\text{s2} enters and leaves the node N, resulting in no net current). The previous procedure is also applied for the current sources {\tilde I}_\text{s1} and {\tilde I}_\text{s3}. This results in the following circuit diagram.
If you are still in doubt as to whether the circuit of figure 4 is indeed equivalent to the one of figure 3, you can apply KCL at the nodes A, B, C and N of both figures, and you’ll find out they are the same equations, meaning those two circuits are indeed equivalent.
Now we combine the current sources in parallel by applying KCL. Taking into account the direction of the arrow of the current sources in the previous figure and the direction of the arrow of the current source in the following figure, we have:
{\tilde I}_\text{s1T} = {\tilde I}_\text{s1} - {\tilde I}_\text{s3}
{\tilde I}_\text{s2T} = {\tilde I}_\text{s2} - {\tilde I}_\text{s1}
{\tilde I}_\text{s3T} = {\tilde I}_\text{s3} - {\tilde I}_\text{s2}
This results in the following circuit diagram.
Lastly we apply source transformation to transform the practical current sources into practical voltage sources:
{\tilde V}_\text{s1Y} = {\tilde I}_\text{s1T} \, {\hat Z}_\text{s1Y}
{\tilde V}_\text{s2Y} = {\tilde I}_\text{s2T} \, {\hat Z}_\text{s2Y}
{\tilde V}_\text{s3Y} = {\tilde I}_\text{s3T} \, {\hat Z}_\text{s3Y}
Taking into account the direction of the arrow of the current sources in the previous figure, this results in the following circuit diagram.
Now we revert the change of variables we did throughout the derivation, going from the last equations to the first ones. After some math, we get:
{\tilde V}_\text{s1Y} = \left( \dfrac{{\tilde V}_\text{s1}}{{\hat Z}_\text{s1}} - \dfrac{{\tilde V}_\text{s3}}{{\hat Z}_\text{s3}} \right) \, {\hat Z}_\text{s1Y}     (4)
{\tilde V}_\text{s2Y} = \left( \dfrac{{\tilde V}_\text{s2}}{{\hat Z}_\text{s2}} - \dfrac{{\tilde V}_\text{s1}}{{\hat Z}_\text{s1}} \right) \, {\hat Z}_\text{s2Y}     (5)
{\tilde V}_\text{s3Y} = \left( \dfrac{{\tilde V}_\text{s3}}{{\hat Z}_\text{s3}} - \dfrac{{\tilde V}_\text{s2}}{{\hat Z}_\text{s2}} \right) \, {\hat Z}_\text{s3Y}     (6)
In this way, we’ve shown the delta-connected generator of figure 1 is equivalent to the wye-connected generator of figure 6; the relationship between them is given by equations (1) to (6).
—It shall be noted the neutral node of the equivalent network (figure 6) is fictitious or imaginary, it doesn’t exist in real life. The terminal fictitious line-to-neutral phasor voltages are {\tilde V}_\text{AN}, {\tilde V}_\text{BN}, {\tilde V}_\text{CN}. The internal fictitious line-to-neutral phasor voltages are {\tilde V}_{\text{A}  \text{N}} = {\tilde V}_\text{s1Y}, {\tilde V}_{\text{B}  \text{N}} = {\tilde V}_\text{s2Y}, {\tilde V}_{\text{C}  \text{N}} = {\tilde V}_\text{s3Y}.
Notice in figure 6 the new internal nodes of the equivalent/fictitious wye generator are A, B and C, while in figure 1 the original internal nodes of the actual/real delta generator are A', B' and C'.
It shall also be stated those networks are equivalent even if the generator is unbalanced and/or the rest of the circuit/power system is unbalanced, since the expressions that relate these networks don’t depend on the line or phase currents, which indeed depend on whether the system is balanced or not.
Also notice the terminal line (or line-to-line) phasor voltages and the line phasor currents of both networks (figure 1 and 6) are not affected after the transformation.
—If the delta-connected generator of figure 1 is balanced (which means the magnitude of the voltage sources is the same, they’re phase-shifted by 120° between each other, and the impedances are the same), then equations (1) to (6) reduce to the following:
{\hat Z}_\text{sY} = \dfrac{{\hat Z}_\text{s}}{3}     (7)
{\tilde V}_\text{s1Y} = \dfrac{{\tilde V}_\text{s1}}{\sqrt{3} \, \angle \pm 30^\circ}     (8)
{\tilde V}_\text{s2Y} = \dfrac{{\tilde V}_\text{s2}}{\sqrt{3} \, \angle \pm 30^\circ}     (9)
{\tilde V}_\text{s3Y} = \dfrac{{\tilde V}_\text{s3}}{\sqrt{3} \, \angle \pm 30^\circ}     (10)
where for equations (8) to (10) you use the first sign (+) if the phase sequence is positive/abc or the second sign (-) if the phase sequence is negative/acb.
--Alej27 (talk) 18:53, 27 January 2021 (UTC)
thread_title:  The derivation of the formulas 

message: :The following is a closed discussion of a requested move. Please do not modify it. Subsequent comments should be made in a new section on the talk page. Editors desiring to contest the closing decision should consider a move review after discussing it on the closer's talk page. No further edits should be made to this discussion.
The result of the move request was: not moved. Already primary topic, no compelling reason for disambiguation (closed by non-admin page mover) Polyamorph (talk) 14:29, 7 November 2023 (UTC)
replies: text: Y-Δ transform →  – This article is about a particular application of a mathematical technique explained in YΔ- and ΔY-transformations. It could be considered a subsection of YΔ- and ΔY-transformations, but is currently too long and technical for this.
The current titles "Y-Δ transform" and "YΔ- and ΔY-transformations" are variants of each other (they refer to the same concept), and "Y-Δ transform" should better redirect to YΔ- and ΔY-transformations. 
More generally, any variant of "Y-Δ transform" that is not specific to electrical engineering should redirect to YΔ- and ΔY-transformations.
I therefore suggest to use a title that is specific to electrical engineering.
While eventually more consistent, I suggest to not move to YΔ- and ΔY-transformations (electrical engineering), because the article uses a different notation ("Y-Δ transform" vs. "YΔ-transformation"). 
My suggestion is to move to Y-Δ transform (electrical engineering) for now. MWinter4 (talk) 11:29, 31 October 2023 (UTC)
thread_title:  Requested move 31 October 2023