# Delta-Wye derivations

We derive the $\Delta \leftrightarrow \text Y$ transformation equations.

Find an introduction to the Delta-Wye transformation in this article.

Written by Willy McAllister.

### Contents

Please don’t memorize these transformations. If the need arises, you can look them up. These derivations never appear on a test. They are presented here for your amusement.

## $\Delta$ to $\text Y$ derivation

Our goal is to find a relationship between the three given $\Delta$ resistances $R_{\Delta abc}$ and the three unknown $\text Y$ resistances $R_{\text Y 123}$ so the overall resistance between pairs of terminals is the same. For example, we want $R_{xy}$ to be the same for $\Delta$ as it is for $\text Y$.

Finding $R_{xy}$ and the others is a bit tricky without a surrounding circuit for context. We build a context and apply the principle of superposition.

### Superposition

If you have an arbitrary network of resistors, how do you find its resistance? You might attach a voltage or current source and measure $I$ and $V$. The ratio $V/I$ is the overall resistance $R$ of the network.

Imagine a black box with three resistors inside. The resistors are connected as either a $\Delta$ or $\text Y$ — you don’t know which. To find the overall resistance between terminals of a black box we construct an imaginary experimental setup. Connect three current sources to the black box like this,

We seek a relationship between $R_{\Delta abc}$ and $R_{\text Y 123}$ such that the currents and voltages—and therefore the resistances—are identical for $\Delta$ and $\text Y$. If we find such a relationship then we say the $\Delta$ and $\text Y$ are *equivalent*. *Equivalent* means if we choose just the right resistor values we can’t tell if $\Delta$ or $\text Y$ is inside the box based on observations we make from outside. If that’s the case, then $\Delta$ and $\text Y$ are interchangeable.

If we open the lid of the black box there are the two possible circuits inside,

Let’s work out expressions for ratios of $V / I = R$ for both sides.

When you see multiple power sources connected to linear elements it’s a great time to consider the Principle of Superposition. This experimental setup is basically an excuse to invoke Superposition.

Here is the first of three superposition sub-circuits with two current sources suppressed,

In the $\Delta$ configuration the resistance between terminals $x$ and $y$ is $Rc$ in parallel with the series combination of $Ra$ and $Rb$,

$R_{xy} = Rc \parallel (Ra \text{—} Rb) = \dfrac{Rc\, (Ra + Rb)}{Rc + (Ra + Rb)}$

Symbol $\parallel$ is a shorthand notation for “in parallel with,” $\,R_i \parallel R_j = R_i \,R_j / (R_i + R_j)$ Symbol $\text{—}$ is a shorthand notation for “in series with,” $\,R_i \,\text{—}\, R_j = R_i + R_j$

In the $\text Y$ configuration the resistance between $x$ and $y$ is $R1$ and $R2$ in series,

$R_{xy} = R1 + R2$

We want $R_{xy}$ to be the same in both the $\Delta$ and $\text Y$ configurations, so set them equal,

$R_{xy}: \quad R1 + R2 = Rc \,(Ra + Rb)/(Ra + Rb + Rc)$

Now draw the other two sub-circuits for yourself and find $R_{yz}$ and $R_{zx}$. Each sub-circuit has one active current source and two suppressed sources. You end up with a set of three simultaneous equations for the resistance between each pair of terminals,

$R_{xy}: \quad R1 + R2 = Rc \,(Ra + Rb)/(Ra + Rb + Rc)$

$R_{yz}: \quad R2 + R3 = Ra \,(Rb + Rc)/(Ra + Rb + Rc)$

$R_{zx}: \quad R3 + R1 = Rb \,(Rc + Ra)/(Ra + Rb + Rc)$

### Solve

Let’s start by isolating $R1$. Combine the three equations with this operation,

$(\quad R_{xy}\quad + \quad R_{zx}\quad - \quad R_{yz}\quad)\, / 2$

Verify this operation isolates $R1$ on the left side,

$( \,[R1 + R2] + [R3 + R1] - [R2 + R3] \,) / 2$

$(R1+\,\cancel{R2}\,+\,\cancel{R3}+R1\,-\,\cancel{R2}\,-\,\cancel{R3})/2$

$2R1/2$

$R1\quad \checkmark$

Now do the same operation on the right side,

$R1 = (Rc(Ra+Rb)/(Ra+Rb+Rc) \,+$

$\qquad\quad Rb(Rc+Ra)/(Ra+Rb+Rc) \,-$

$\qquad\quad Ra(Rb+Rc)/(Ra+Rb+Rc)\,)\, / 2$

Notice the denominator is the same in all three terms,

$R1 = \dfrac{Rc(Ra+Rb) + Rb(Rc+Ra) - Ra(Rb+Rc)}{2(Ra+Rb+Rc)}$

Multiply everything out and hunt for cancellations,

$R1 = \dfrac{\cancel{RaRc} \,+ RcRb + RbRc + \,\cancel{RbRa} \,- \,\cancel{RaRb} \,-\, \cancel{RaRc}}{2(Ra+Rb+Rc)}$

$R1 = \dfrac{\cancel{2}\,RbRc}{\cancel{2}( Ra+Rb+Rc )}$

$R1 = \dfrac{Rb\,Rc}{Ra+Rb+Rc}$

That’s it! This expression lets you compute $\text Y$ resistor $R1$ from the given $\Delta$ resistors.

The procedure to find $R2$ and $R3$ is the same, with different subscripts. The specific operations you use on the equations are,

To isolate $R2: (\,[R_{xy}] + [R_{yz}] - [R_{zx}]\,) / 2 $

To isolate $R3: (\,[R_{yz}] + [R_{zx}] - [R_{xy}]\,) / 2 $

The $\Delta$ to $\text Y$ transformation is,

$\quad R1 = \dfrac{Rb\,Rc}{Ra + Rb + Rc}$

$\quad R2 = \dfrac{Ra\,Rc}{Ra + Rb + Rc}$

$\quad R3 = \dfrac{Ra\,Rb}{Ra + Rb + Rc}$

(Notice all the denominators are the same.)

## $\text Y$ to $\Delta$ derivation

The algebra for $\text Y$ to $\Delta$ is harder. I found two derivations I admire,

- Start with the result from the $\Delta \rightarrow \text Y$ transformation and use algebra to solve in reverse. It works, but seems rather convoluted.
- Convert from resistance $R$ to conductance $G$ and solve from scratch. The result is symmetric with the $\Delta$ to $\text Y$ derivation. An elegant use of conductance.

### $\text Y$ to $\Delta$ derivation with just algebra

We start with the three transformation equations from the previous $\Delta$ to $\text Y$ transformation. These equations express $R_{\text Y123}$ in terms of $R_{\Delta abc}$. If the system of equations is a sock, reach into the sock and pull it inside out to find each $\Delta$ resistor $R_{\Delta abc}$ in terms of $R_{\text Y123}$.

To review, the $\Delta$ to $\text Y$ equations are,

$R1 = \dfrac{Rb\,Rc}{Ra + Rb + Rc}$

$R2 = \dfrac{Ra\,Rc}{Ra + Rb + Rc}$

$R3 = \dfrac{Ra\,Rb}{Ra + Rb + Rc}$

We go after $Rc$ first. Divide the $R3$ equation by the $R1$ equation,

$\dfrac{R3}{R1} = \cfrac{\cfrac{Ra\,Rb}{Ra + Rb + Rc}}{\cfrac{Rb\,Rc}{Ra + Rb + Rc}}$

The $Ra + Rb + Rc$ terms are common to the top and bottom, so they cancel,

$\dfrac{R3}{R1} = \dfrac{Ra\,Rb}{Rb\,Rc} = \dfrac{Ra}{Rc}$

or, solving for $Ra$,

$Ra = \dfrac{R3\,Rc}{R1}$

Next, divide the equation for $R3$ by the equation for $R2$. The same kind of cancellation happens,

$\dfrac{R3}{R2} = \dfrac{Ra\,Rb}{Ra\,Rc} = \dfrac{Rb}{Rc}$

or, solving for $Rb$,

$Rb = \dfrac{R3\,Rc}{R2}$

Now we plug in our expressions for $Ra$ and $Rb$ into the equation for $R2$,

$R2 = \cfrac{\cfrac{R3\,Rc}{R1}\,Rc}{\cfrac{R3\,Rc}{R1} + \cfrac{R3\,Rc}{R2} + Rc}$

A bunch of common $Rc$ terms cancel out,

$R2 = \cfrac{\cfrac{R3\,Rc}{R1}}{\dfrac{R3}{R1} + \cfrac{R3}{R2} + 1}$

The least common multiple of the terms in the denominator is $R1\,R2$. Perform the addition in the denominator by multiplying each term by the appropriate form of $1$,

$R2 = \cfrac{\dfrac{R3\,Rc}{R1}}{\cfrac{R3}{R1}\cfrac{R2}{R2} + \cfrac{R3}{R2}\cfrac{R1}{R1} + 1\cfrac{R1}{R1}\cfrac{R2}{R2}}$

leaving us with,

$R2 = \dfrac{\dfrac{R3\,Rc}{R1}}{\cfrac{R3\,R2 + R3\,R1 + R1\,R2}{R1\,R2}}$

Two $R1$ terms and two $R2$ terms cancel,

$\cancel{R2} \rightarrow 1 = \cfrac{\cfrac{R3\,Rc}{\cancel{R1}}}{\cfrac{R3\,R2 + R3\,R1 + R1\,R2}{\cancel{R1}\,\cancel{R2}}}$

$1 = \dfrac{R3\,Rc}{R3\,R2 + R3\,R1 + R1\,R2}$

and now we solve for $Rc$,

$Rc = \dfrac{R3\,R2 + R3\,R1 + R1\,R2}{R3}$

Done! This expression tells us how to compute $\Delta$ resistor $Rc$ from the given $\text Y$ resistors. You can derive $Ra$ and $Rb$ with the same technique using different patterns of dividing equations.

The $\text Y$ to $\Delta$ transformation is,

$\quad Ra = \dfrac{R3\,R2 + R3\,R1 + R1\,R2}{R1}$

$\quad Rb = \dfrac{R3\,R2 + R3\,R1 + R1\,R2}{R2}$

$\quad Rc = \dfrac{R3\,R2 + R3\,R1 + R1\,R2}{R3}$

(Notice all the numerators are the same.)

## yet another derivation

Wikipedia has a similar derivation, with equally mind-numbing algebra.

Let $R_T = Ra+Rb+Rc$.

As a reminder, the $\Delta$ to $\text Y$ equations are,

$R_{1} = {\dfrac {RbRc}{R_T}}$ (1)

$R_{2} = {\dfrac {RaRc}{R_T}}$ (2)

$R_{3} = {\dfrac {RaRb}{R_T}}$ (3)

Multiply pairs of equations,

$R_{1}R_{2}={\dfrac{RaRbRc^{2}}{R_T^{2}}}$ (4)

$R_{1}R_{3}={\dfrac{RaRb^{2}Rc}{R_T^{2}}}$ (5)

$R_{2}R_{3}={\dfrac{Ra^{2}RbRc}{R_T^{2}}}$ (6)

and the sum of these equations is,

$R_{1}R_{2}+R_{1}R_{3}+R_{2}R_{3}={\dfrac {RaRbRc^{2}+RaRb^{2}Rc+Ra^{2}RbRc}{R_T^{2}}}$ (7)

Factor $RaRbRc$ from the right side, which leaves $R_T$ in the numerator, canceling with an $R_T$ in the denominator.

${\begin{aligned} R_{1}R_{2}+R_{1}R_{3}+R_{2}R_{3} &= {}{\dfrac {\left(RaRbRc\right)\left(Ra+Rb+Rc\right)}{R_T^{2}}} \\ &={}{\dfrac {RaRbRc}{R_T}} \end{aligned}}$ (8)

Divide (8) by (1)

${\begin{aligned} \dfrac{R_{1}R_{2}+R_{1}R_{3}+R_{2}R_{3}}{R_{1}} &= {} \dfrac{RaRbRc}{R_T} \dfrac{R_T}{RbRc} \\ &= {}Ra \end{aligned}}$

which is the $\text Y$ to $\Delta$ equation for $Ra$.

Divide (8) by expressions for $R_{2}$ or $R_{3}$ to find the other two equations.

### $\text Y$ to $\Delta$ derivation with conductance

This derivation converts resistance to conductance. The system of equations is the dual of the derivation for $\Delta$ to $\text Y$.

Replace each resistor with its corresponding conductance, $G = \dfrac{1}{R}.$

## review: equivalent conductance

The rule to combine conductances in *parallel* is the sum of the conductances,

$G_{\text{parallel}} = G_i + G_j$

The rule to combine two conductances in *series* is similar to two resistors in parallel,

$G_{\text{series}} = \dfrac{G_i\,G_j}{G_i + G_j}$

If you have $\text N$ conductances in series combine them with,

$\dfrac{1}{G_{series}} = \dfrac{1}{\dfrac{1}{G_i}+\dfrac{1}{G_j}+…+\dfrac{1}{G_\text N}}$

With conductance the form of Ohm’s Law we use is $\text I = \text G \, \text V$, or $\text G = \dfrac{\text I}{\text V}$.

Our goal is to find a relationship between the three $\Delta$ conductances $G_{\Delta abc}$ and the three $\text Y$ conductances $G_{\text Y 123}$ such that the overall conductance between any pair of terminals is the same.

### Superposition

Imagine a black box with three resistors inside. They are connected as either a $\Delta$ or $\text Y$ — you don’t know which. Now build an imaginary experimental setup. Connect three voltage sources like this,

We are trying to find the right relationship between $G_{\Delta abc}$ and $G_{\text Y 123}$ so the given three voltages and the measured three currents are identical. If we find that relationship then the $\Delta$ and $\text Y$ are *equivalent*, which means either configuration could be inside the black box and we can’t tell them apart by with observations made from outside. That means they are interchangeable.

If we open the lid of the black box, here are the two possible circuits,

Let’s develop expressions for ratios of $I / V = G$ for both sides.

When you see multiple power sources connected to linear elements (conductances are linear) it’s a great time to crack open the Principle of Superposition. This experimental setup is a clever way to let us invoke Superposition.

Here is the first of three superposition sub-circuits with two voltage sources suppressed,

Start with terminal $x$. Compute the ratio $I_x / V_x$, which is the conductance between terminal $x$ and ground. Call this $G_x$.

In the $\Delta$ configuration the conductance from $x$ to ground is $Gb$ in parallel with $Gc$,

$G_x = Gb + Gc$

Notice that $G_a$ connects between two ground points. So there is no current flowing in $G_a$.

In the $\text Y$ configuration the conductance from $x$ to ground is $G1$ in series with the parallel combination of $G2$ and $G3$,

$G_x = G1 \,\text{—}\, (G2 \parallel G3) = \dfrac{G1 \, (G2+G3)}{G1+(G2+G3)}$

Symbol $\parallel$ is shorthand for “in parallel with,” $\quad G_i \,\parallel \, G_j = G_i + G_j$

Symbol $\text{—}$ is shorthand for “in series with,” $\quad G_i \,\text{—}\, G_j = G_i \,G_j / (G_i + G_j)$

We want $G_x$ to be the same for $\Delta$ and $\text Y$, so set them equal,

$G_x: \quad Gb + Gc = G1(G2+G3)/(G1+G2+G3)$

Now draw the other two sub-circuits on your own. Construct two more equations,

$G_y: \quad Ga + Gc = G2(G1+G3)/(G1+G2+G3)$

$G_z: \quad Ga + Gb = G3(G1+G2)/(G1+G2+G3)$

This is our system of three equations. Notice they are nearly identical to the equations at the beginning of the $\Delta$ to $\text Y$ derivation, $R_{xy}, R_{yz}, R_{zx}$. The rest of the algebra is the same.

### Solve

Let’s use the three equations to isolate $Gb$. Combine the equations with this operation,

$(\quad[G_x]\quad + \quad [G_z]\quad - \quad [G_y]\quad)\,/2$

The left side becomes,

$([Gb + Gc] + [Ga + Gb] - [Ga + Gc])\,/2$

There’s a bunch of cancellation,

$(Gb+\,\cancel{Gc}\,+\,\cancel{Ga}\,+\,Gb\,-\,\cancel{Ga}\,-\,\cancel{Gc})\,/2$

$\dfrac{2Gb}{2}$

$Gb\quad \checkmark\quad$ This verifies the combining operation isolates $Gb$.

Now apply the same operation to the right side,

$Gb = [\,G1(G2+G3)/(G1+G2+G3) \,+$

$\qquad\quad G3(G1+G2)/(G1+G2+G3) \,-$

$\qquad\quad G2(G1+G3)/(G1+G2+G3) \,]\,/ 2$

Notice the denominator is the same in all three terms,

$Gb = \dfrac{G1(G2+G3) + G3(G1+G2) - G2(G1+G3)}{2(G1+G2+G3)}$

Multiply everything out and search for cancellations,

$Gb = \dfrac{\cancel{G1\,G2} \,+ G1\,G3 + G3\,G1 + \,\cancel{G3\,G2} \,-\, \cancel{G2\,G1} \,- \,\cancel{G2\,G3}}{2(G1+G2+G3)}$

$Gb = \dfrac{\cancel{2}\,G1\,G3}{\cancel{2}(G1+G2+G3)}$

$Gb = \dfrac{G1\,G3}{G1+G2+G3}$

Done! This tells us how to compute $\Delta$ conductance $Gb$ from the given $\text Y$ conductances.

The procedure for finding $Ga$ and $Gc$ is the same.

#### Optional: convert conductance back to resistance

Do you want to convert this conductance form back to resistance? We should end up with the same equation as the all-algebra derivation. Replace every $G$ with $1/R$,

$\dfrac{1}{Rb}=\cfrac{\cfrac{1}{R1}\cfrac{1}{R3}}{\cfrac{1}{R1}+\cfrac{1}{R2}+\cfrac{1}{R3}}$

Go to work on the denominator. The least common multiple of the three fractions is $R1\,R2\,R3$. Multiply each term by an appropriate form of $1$ and add fractions.

$\dfrac{1}{Rb}=\cfrac{\cfrac{1}{R1\,R3}}{\cfrac{R2\,R3}{R1\,R2\,R3}+\cfrac{R1\,R3}{R1\,R2\,R3}+\cfrac{R1\,R2}{R1\,R2\,R3}}$

$\dfrac{1}{Rb}=\cfrac{\cfrac{1}{R1\,R3}}{\cfrac{R2\,R3+R1\,R3+R1\,R2}{R1\,R2\,R3}}$

Now we turn a somersault to bring $R1\,R2\,R3$ to the numerator,

$\dfrac{1}{Rb}=\dfrac{\dfrac{1}{\cancel{R1}\,\cancel{R3}}\,(\cancel{R1}\,R2\,\cancel{R3})}{R2\,R3+R1\,R3+R1\,R2}$

Some cancellation gives us,

$\dfrac{1}{Rb}=\dfrac{R2}{R2\,R3+R1\,R3+R1\,R2}$

One more flip to get the expression for $Rb$ we’ve been looking for,

$Rb=\dfrac{R2\,R3+R1\,R3+R1\,R2}{R2}$

This matches the result from the algebraic derivation we did for $\text Y$ to $\Delta$.

## Summary

As I mentioned in the introduction, please don’t bother to memorize these equations, it’s not worth the effort. These circuits are rare (except in power engineering where they are common). You can always look up the transformation equations.

## Credits

The $\Delta$ to $\text Y$ derivation was contributed by Khan Academy learner *phidot*.

The first $\text Y$ to $\Delta$ derivation is based on a video by Mohiuddin Jewel.

## Questions

Hi Willy, Thank you for your derivation, it does help me a lot to understand these transformations. However, I still have some curiosity (or skepticism) regarding superposition argument that being used here. All derivation methods above developed from the step of equating resistance between nodes (Rxy, Rxz, Ryz) of 2 configurations (Y-delta). How exactly you use superposition principle to justify that? For reference, I already read your article about superposition, linearity, and its application on simple circuit i.e. How to decompose problem with more than one source, switching off voltage/current source, etc. I just can’t make connection between those simple example to this transformation. Would you kindle explain that to me? Thank you in advance. Regards - Rahmat

Rahmat - Good question, and I appreciate your skepticism. I will respond by revising the article with more appreciation of the linearity/superposition reasoning. It will take me another week or so to edit, create circuit diagrams, and publish. The goal is find out the resistor relationships so that from the outside you can’t tell if the network is a Delta or a Wye. The equivalent resistance measured between any two nodes has to be the same. I can measure the resistance by creating an arbitrary surrounding circuit that applies V’s and I’s to the terminals and measure the ratio V/I = R_equiv. I will add three voltage sources connected to each terminal. THAT circuit with multiple sources can be analyzed by the Superposition Method you learned earlier. Hang on a week and I’ll show you what all this means.

Rahmat - I just published a revised version of the article that shows how superposition is used to justify the state of the other terminals, either open or shorted.

Thank you Willy! you sure are dedicated. Now I get a clearer idea on how superposition applied on this transformation. By assuming the initial network as blackbox, we could find any equivalent circuit by equating the V/I values (which are located outside the blackbox). It is somehow similar to Thevenin and Norton Theorem with addition of superposition. There is one thing that nags me a bit. In the initial derivation, you supress the current source while in a later derivation (the one using conductance instead resistance) you supress the voltage instead. I wonder if I could get same results by suppressing voltage source using resistance instead of conductance (I will manage it myself, it will be a good exercise) Back to current article, I think you really put effort into it. Adding circuit for illustration helps a lot. And you even revise the symbol for series element (before, you use + which might get confused with addition operator). Though I almost mistook it for subtraction before I realize that it goes a bit longer. All in all, I’d say thank you for keeping this site floating as it is now.

I’m so glad the revised article was clear. The superposition argument is applied in a nearly identical way in the proof of Thevenin’s Theorem. It took me a fair amount of fiddling to come up with the two external circuits (current sources for one, voltage sources for the other). Suppressing the extra two sources has to give you just the right open or short behavior. If you come up with an alternative I would be interested to know what you did. Thanks for the feedback on the “in series” symbol. The one I picked looks a bit like a minus sign. You may see that change if I find a better symbol. There is no standard “in series with” symbol like the vertical bars for “in parallel with”.

Hey similar typo to the one Hch pointed out, Under the Y to delta derivation using conductance,

$Gb = \dfrac{\cancel{G1\,G2} \,+ G1\,G3 + G3\,G1 + \,\cancel{G3\,G2} \,-\, \cancel{G2\,G1} \,+ \,\cancel{G2\,G3}}{2(G1+G2+G3)}$

the last term should be a minus instead of a plus

it helped me a lot …

Ali - I’m glad this article was helpful. - Willy

Small typo/mistake under “Delta to Y derivation” Multiply everything out and search for cancellations,

$R1 = \dfrac{\cancel{RaRc} \,+ RcRb + RbRc + \,\cancel{RbRa} \,- \,\cancel{RaRb} \,+\, \cancel{RaRc}}{2(Ra+Rb+Rc)}$

last term should be minus instead of plus

Wow. Nice catch. Thanks for reporting the error. It has been fixed.

In “Y to Δ derivation”: why do we use G instead of R?

Whenever you have a resistor circuit you are allowed to treat it as either resistors or conductances, your choice. The reason you would do this is to take advantage of the new combining equations for series and parallel. That’s why I used conductance for one of the derivations. To combine parallel conductance you simply add. That change of perspective from R to G made the steps of the Y-to-Delta derivation look just like the Delta-to-Y algebra. I didn’t come up with this technique myself, but it’s very clever.

We can use 1/R1 instead of G1. The result will be the same. It is just for the convenience (for the learner) of reducing the amount of “1/” (one over + the fraction symbol) on screen. ( :