Sometimes you get stuck when you are simplifying a resistor network. Some resistor networks cannot be simplified using the usual series and parallel combinations. This situation can often be handled by trying the $\Delta - \text Y$ transformation, or ‘Delta-Wye’ transformation.

Written by Willy McAllister.

Contents

The Delta and Wye names come from the shape of the circuits, which resemble letters. The transformation allows you to replace three resistors in a $\Delta$ configuration by three resistors in a $\text Y$ configuration, and the other way around. Drawing $\Delta$ and $\text Y$ this way emphasizes the 3 terminals. The configurations can be redrawn to square up the resistors. This is called a $\pi - \text T$ configuration, The $\pi - \text T$ style is what you might find in a typical schematic. The transformation equations apply to $\pi - \text T$ as well.

The $\text Y$ configuration is also called a star.

Something to notice: The two configurations have a different number of nodes. $\Delta$ has three nodes while $\text Y$ has four nodes (the one in the center).

Transformations

For the transformation to be valid, the resistance between each pair of terminals must be the same before and after. DIAGRAM 1 $- \,\Delta$ and $\text Y$ configurations with labeled nodes and resistors.

It is possible to write three simultaneous equations to capture this constraint.

Consider terminals $x$ and $y$ on the $\text Y$ side. Assume terminal $z$ isn’t connected to anything, so the current in $R3$ is $0$. 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.)

In the $\Delta$ configuration, the resistance between $x$ and $y$ is $Rc$ in parallel with $Ra +Rb$. On the $\text Y$ side, the resistance between $x$ and $y$ is the series combination $R1+R2$.

We set these equal to each other to get the first of three simultaneous equations,

$R1+R2 = \dfrac{Rc\,(Ra+Rb)}{Rc+(Ra+Rb)}$

We can write two similar expressions for the other two pairs of terminals. Notice the $\Delta$ resistors have letter names, $(Ra$, etc.$)$ and the $\text Y$ resistors have number names, $(R1$, etc.$)$.

After solving the simultaneous equations (not shown), we get the equations to transform either network into the other. Find a full derivation of the transform equations is in this article.

$\Delta \rightarrow \text Y$ transformation

Equations for transforming a $\Delta$ network into a $\text Y$ network,

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

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

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

Transforming from $\Delta$ to $\text Y$ introduces one additional node.

$\text Y \rightarrow\Delta$ transformation

Equations for transforming a $\text Y$ network into a $\Delta$ network,

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

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

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

Transforming from $\text Y$ to $\Delta$ removes one node.

Example 1 - symmetric

Let’s do a nice symmetric example. Assume we have a $\Delta$ circuit with $3\,\Omega$ resistors.

Let’s derive the $\text Y$ equivalent by using the $\Delta \rightarrow \text Y$ equations.

$R1 = \dfrac{Rb\,Rc}{Ra + Rb + Rc} = \dfrac{3 \cdot 3}{3 + 3 + 3} = 1\,\Omega$

$R2 = \dfrac{Ra\,Rc}{Ra + Rb + Rc} = \dfrac{3 \cdot 3}{3 + 3 + 3} = 1\,\Omega$

$R3 = \dfrac{Ra\,Rb}{Ra + Rb + Rc} = \dfrac{3 \cdot 3}{3 + 3 + 3} = 1\,\Omega$ Going in the other direction, from $\text Y \rightarrow\Delta$, looks like this,

$Ra = \dfrac{R1\,R2 + R2\,R3 + R3\,R1}{R1} = \dfrac{1\cdot1 + 1\cdot1 + 1\cdot1}{1} = 3\,\Omega$

$Rb = \dfrac{R1\,R2 + R2\,R3 + R3\,R1}{R2} = \dfrac{1\cdot1 + 1\cdot1 + 1\cdot1}{1} = 3\,\Omega$

$Rc = \dfrac{R1\,R2 + R2\,R3 + R3\,R1}{R3} = \dfrac{1\cdot1 + 1\cdot1 + 1\cdot1}{1} = 3\,\Omega$

This $3:1$ resistor ratio is the classic Delta-Wye test case.

Look like a wizard: If you come across a balanced $\Delta$ or $\text Y$ (all the resistors are the same value) you can instantly do the transformation if you remember this $3:1$ property.

• For the $\Delta \rightarrow \text Y$ direction divide the resistors by $3$.
• For the $\text Y \rightarrow \Delta$ direction multiply the resistors by $3$.

Example 2 - less tidy

Now for an example that’s a little less tidy. Let’s find the equivalent resistance between the top and bottom terminals of this circuit, Try as we might, there are no resistors in series or in parallel. But we are not stuck. First, let’s redraw the schematic to emphasize we have two $\Delta$ connections stacked one on top of the other, Now select one of the $\Delta$’s to convert to a $\text Y$. We will perform a $\Delta \rightarrow \text Y$ transformation on the bottom $\Delta$ (an arbitrary choice). This might break the log jam, opening up other opportunities for simplification.

Very carefully label the resistors and nodes. It is critical to keep the resistor names and node names straight. $Rc$ must connect between nodes $x$ and $y$, and so on for the other resistors. Refer to DIAGRAM 1 above for the labeling convention. When we do the transform on the lower $\Delta$, the three black $\Delta$ resistors will be replaced by three new gray $\text Y$ resistors, like this, Convert the lower $\Delta$ to a $\text Y$. Draw the new circuit.

See if you can perform the transform yourself before peeking ahead at the answer. Be sure to select the right set of transform equations.

show the $\Delta$ to Y transformation

Apply the transformation equations for $\Delta \rightarrow \text Y$,

$R1 = \dfrac{Rb\,Rc}{Ra + Rb + Rc} = \dfrac{5\cdot 3}{4 + 5 + 3} = \dfrac{15}{12}= 1.25\,\Omega$

$R2 = \dfrac{Ra\,Rc}{Ra + Rb + Rc} = \dfrac{4\cdot 3}{4 + 5 + 3} = \dfrac{12}{12}= 1\,\Omega$

$R3 = \dfrac{Ra\,Rb}{Ra + Rb + Rc}= \dfrac{4\cdot 5}{4 + 5 + 3} = \dfrac{20}{12}= 1.66\,\Omega$

To draw the new circuit, substitute the equivalent $\text Y$ network in place of the $\Delta$ resistors. Make sure the $\text Y$ resistor names connect between the proper node names. Refer to DIAGRAM 1 above for the labeling convention.

Voilà! Here’s our circuit with the $\text Y$ resistors in place of the bottom $\Delta$. The circuit now has series and parallel resistors where there were none before. Here’s the circuit redrawn all squared up in a familiar style, We continue simplification using series and parallel resistor combinations, just as we did in the article on Simplifying resistor networks.

On the left branch, $3.125 + 1.25 = 4.375 \,\Omega$
On the right branch, $4 + 1 = 5\,\Omega$ The two parallel resistors combine as,

$4.375 \parallel 5 = \dfrac{4.375 \cdot 5}{4.375 + 5} = 2.33\,\Omega$

And we finish by adding the last two series resistors together,

$R_{equivalent} = 2.33 + 1.66 = 4\,\Omega$ The original five resistors simplify down to a single $4\,\Omega$ resistor.

Simulation model

Here is a simulation model of the transformation we just did. Open the link. The left side of the schematic is the original Example 2 circuit. The right side is the $4\,\Omega$ resistor we derived. Click on DC in the top menu bar to perform a DC analysis.

Notice how the current flowing into the Delta-Wye network is the same as the current flowing into the $4\,\Omega$ resistor. That shows the network and the single resistor are equivalent.

Summary

$\Delta - \text Y$ transformations are another tool in your bag of tricks for simplifying circuits.

Don’t bother to memorize the transformation equations. If the need arises, you can look them up.

For the curious: Find a full derivation of the transform equations is in this article.