# Delta-Wye resistor networks

The Delta-Wye transformation is an extra technique for transforming certain resistor combinations that cannot be handled by the series and parallel equations. This is also referred to as a Pi - T transformation. Written by Willy McAllister.
Sometimes when you are simplifying a resistor network, you get stuck. 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.
The names Delta and Wye come from the shape of the schematics, 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.
The $\Delta - \text Y$ drawing style emphasizes these are 3-terminal configurations. Something to notice is the different number of nodes in the two configurations. $\Delta$ has three nodes, while $\text Y$ has four nodes (one extra in the center).
The configurations can be redrawn to square up the resistors. This is called a $\pi - \text T$ configuration,
The $\pi - \text T$ style is a more conventional drawing you would find in a typical schematic. The transformation equations developed next apply to $\pi - \text T$ as well.

## $\Delta - \text Y$ transformation

For the transformation to be equivalent, the resistance between each pair of terminals must be the same before and after. It is possible to write three simultaneous equations to capture this constraint.
Consider terminals $x$ and $y$ (and for the moment assume terminal $z$ isn't connected to anything, so the current in $\text R3$ is $0$ ). 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 $R1+R2$ (again, assume terminal $z$ isn't connected to anything, so $\text R1$ and $\text R2$ carry the same current and can be considered in series). 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.

### $\Delta \rightarrow \text Y$ transformation

Equations for transforming a $\Delta$ network into a $\text Y$ network:
$R1 = \dfrac{Rb\,Rc}{Ra + Rb + Rc}$
$R2 = \dfrac{Ra\,Rc}{Ra + Rb + Rc}$
$R3 = \dfrac{Ra\,Rb}{Ra + Rb + Rc}$
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:
$Ra = \dfrac{R1\,R2 + R2\,R3 + R3\,R1}{R1}$
$Rb = \dfrac{R1\,R2 + R2\,R3 + R3\,R1}{R2}$
$Rc = \dfrac{R1\,R2 + R2\,R3 + R3\,R1}{R3}$
Transforming from $\text Y$ to $\Delta$ removes one node.

## Example

Let's do a symmetric example. Assume we have a $\Delta$ circuit with $3\,\Omega$ resistors. 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,
$Ra = \dfrac{R1\,R2 + R2\,R3 + R3\,R1}{R1} = \dfrac{1\cdot1 + 1\cdot1 + 1\cdot1}{1} = 3\,\Omega$
$Rb = \dfrac{R1\,R2 + R2\,R3 + R3\,R1}{R2} = \dfrac{1\cdot1 + 1\cdot1 + 1\cdot1}{1} = 3\,\Omega$
$Rc = \dfrac{R1\,R2 + R2\,R3 + R3\,R1}{R3} = \dfrac{1\cdot1 + 1\cdot1 + 1\cdot1}{1} = 3\,\Omega$

## Example

Now for an example that's a little less tidy. We want to find the equivalent resistance between the top and bottom terminals.
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 the other.
Now select one of the $\Delta$'s to convert to a $\text Y$. We will perform a $\Delta \rightarrow \text Y$ transformation and see if it breaks the logjam, opening up other opportunities for simplification.
We go to work on the bottom $\Delta$ (an arbitrary choice). Very carefully label the resistors and nodes. To get the right answers from the transformation equations, 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 perform the transform on the lower $\Delta$, the black $\Delta$ resistors will be replaced by the new gray $\text Y$ resistors, like this:
Perform the transform yourself before looking at the answer. Check that you select the right set of equations.
Compute three new resistor values to convert the $\Delta$ to a $\text Y$, and draw the complete circuit.
Apply the transformation equations for $\Delta \rightarrow \text Y$.
$R1 = \dfrac{Rb\,Rc}{Ra + Rb + Rc} = \dfrac{5\cdot 3}{4 + 5 + 3} = \dfrac{15}{12}= 1.25\,\Omega$
$R2 = \dfrac{Ra\,Rc}{Ra + Rb + Rc} = \dfrac{4\cdot 3}{4 + 5 + 3} = \dfrac{12}{12}= 1\,\Omega$
$R3 = \dfrac{Ra\,Rb}{Ra + Rb + Rc}= \dfrac{4\cdot 5}{4 + 5 + 3} = \dfrac{20}{12}= 1.66\,\Omega$
Derive an equivalent $\text Y$ network by substituting 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.
And voilà! Check out our circuit. It now has series and parallel resistors where it had none before. Continue simplification with series and parallel combinations until we get down to a single resistor between the terminals. Redraw the schematic again to square up the symbols into a familiar style.
We proceed through the remaining simplification steps just as we did before in the article on Resistor Network Simplification.
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 $4.375\,||\, 5 = \dfrac{4.375 \cdot 5}{4.375 + 5} = 2.33\,\Omega$
And we finish by adding the last two series resistors together,
$R_{equivalent} = 2.33 + 1.66 = 4\,\Omega$

## Summary

$\Delta - \text Y$ transformations are another tool in your bag of tricks for simplifying circuits prior to detailed analysis.
Don't memorize the transformation equations. If the need arises, you can look them up.