# Delta-Wye derivations

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

The first derivation is from $\Delta$ to $\text Y$.

The second derivation goes the opposite direction, from $\text Y$ to $\Delta$. The algebra for this one is harder and there are two versions presented. The first version does it strictly by algebraic manipulation. It works, but it seems a bit convoluted. The second version casts the resistors as conductances before diving into the math. The result is pleasingly symmetric.

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

Written by Willy McAllister.

### Contents

There is no need to memorize these transformation equations. If the need arises, you can look them up. No engineer ever needs to produce the following derivations on the spot. They are presented here for your amusement.

## Objective

The resistance between any pair of terminals has to be the same in both the $\Delta$ and $\text Y$ configurations. For example, in $\Delta$, the resistance across the top between terminals $x$ and $y$ is $Rc \parallel (Ra+Rb)$. In $\text Y$, the resistance between terminals $x$ and $y$ is $R1 + R2$.

Given the three resistor values of one configuration we are going to derive the resistor values for the other, going both ways.

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

This derivation was contributed by Khan Academy learner *phidot*. Let’s figure out $R1$ in the $\text Y$ configuration in terms of $\Delta$ resistors $(Ra, Rb, Rc)$. We write a set of three simultaneous equations describing the resistance between each pair of terminals.

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

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

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

The left side is the resistance in the $\text Y$ configuration, the right side is the resistance in the $\Delta$ configuration. $R_{xy}$ stands for the resistance between terminals $x$ and $y$.

The symbol $\parallel$ is shorthand notation for “in parallel with.”

$R_i \parallel R_j = R_i \,R_j / (R_i + R_j)$

When we look at terminals $x$ and $y$, we assume terminal $z$ isn’t connected to anything, so the current in $\text R3$ is $0$.

## Tell me more about the assumption

We can make this assumption because we know resistors are linear devices and we can apply the principle of superposition. (If you haven't studied linearity and superposition yet, please trust me for now that the assumption is a good one.)

Let’s attempt to isolate $R1$ on the left side. We combine the equations with this operation,

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

Fill in all the left sides,

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

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

This verifies the operation isolates $R1$ on the left side. Now do the same operation on the right side,

$R1 = ( \,Rc \parallel (Ra+Rb) + Rb \parallel (Rc+Ra) - Ra \parallel (Rb+Rc)\, ) / 2$

Replace the $\parallel$ symbol with the proper formula for two parallel resistors,

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

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

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

$\qquad \quad ) \,/ 2$

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 search 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 tells us how to compute $\text Y$ resistor $R1$ from the given $\Delta$ resistors. The procedure for finding $R2$ 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 $

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

Going in this direction the algebra is trickier. I found two derivations I admire.

- Start with the $\Delta \rightarrow \text Y$ equastions and use algebra to solve in reverse
- Convert the resistors ($R$) to conductances ($G$) and start from scratch

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

This first derivation is based on a video by Mohiuddin Jewel.

We’re going to start with the three equations from the $\Delta$ to $\text Y$ transformation and solve for the lettered resistors $(Ra, Rb, Rc)$. To review,

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

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

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

These steps may seem a bit goofy, but they work. 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, leaving,

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

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, giving us,

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

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, leaving,

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

Now we work on the denominator. The LCM of the terms in the denominator is $R1\,R2$. Perform the addition by multiplying each term by the appropriate form of $1$,

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

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

Two $R1$ terms and two $R1$ 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 can 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.

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

This second derivation involves treating the resistors as conductances. After changing to conductance, the derivation follows exactly the steps we did in the first derivation in this article, $\Delta$ to $\text Y$. The resistance model and conductance model are *duals* of each other.

Each resistor is replaced with its equivalent conductance, $G = \dfrac{1}{R}.$

The rule for conductances in parallel is the sum of the conductances,

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

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

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

Our strategy is to short circuit the conductance opposite the terminal in question and figure out the equivalent conductance. For example, suppose we are looking at terminal $x$. We short out $Ga$ on the opposite side by connecting a wire between $y$ and $z$. You can think of it as connecting both $y$ and $z$ to ground. Shorting out $Ga$ is the dual of leaving resistor $R1$ open while figuring out $R_{yz}$ up above. Then we compute the conductance between terminal $x$ and ground and call this $G_x$.

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

$G_x = Gb + Gc$

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

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

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

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

Now construct two more equations to describe all three terminals,

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

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

Notice how these three equations are nearly identical to their duals shown at the beginning of the $\Delta$ to $\text Y$ derivation, $R_{xy}, R_{yz}, R_{zx}$.

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

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

The left side becomes,

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

We get a bunch of cancellation,

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

This verifies that the operation isolates $Gb$. Now apply the same operation to the right side,

$Gb = ($

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

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

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

$\qquad ) \,/ 2$

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,

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

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

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

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. You get to come up with the combining operation to apply to the simultaneous equations.

If you want to convert $Gb$ to resistance replace $G$ with $\dfrac{1}{R}$, etc.

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

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,

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

And a bit more 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 $\Delta$. Derive the equations for $Ra$ and $Rc$ with the same process.

## Questions

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. ( :