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 ΔY\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 Δ\Delta configuration by three resistors in a Y\text Y configuration, and the other way around.
The ΔY\Delta - \text Y drawing style emphasizes these are 3-terminal configurations. Something to notice is the different number of nodes in the two configurations. Δ\Delta has three nodes, while Y\text Y has four nodes (one extra in the center).
The configurations can be redrawn to square up the resistors. This is called a πT\pi - \text T configuration,
The πT\pi - \text T style is a more conventional drawing you would find in a typical schematic. The transformation equations developed next apply to πT\pi - \text T as well.

ΔY\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 xx and yy (and for the moment assume terminal zz isn't connected to anything, so the current in R3\text R3 is 00 ). In the Δ\Delta configuration, the resistance between xx and yy is RcRc in parallel with Ra+RbRa +Rb .
On the Y\text Y side, the resistance between xx and yy is the series combination R1+R2R1+R2 (again, assume terminal zz isn't connected to anything, so R1\text R1 and R2\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=Rc(Ra+Rb)Rc+(Ra+Rb)R1+R2 = \dfrac{Rc\,(Ra+Rb)}{Rc+(Ra+Rb)}
We can write two similar expressions for the other two pairs of terminals. Notice the Δ\Delta resistors have letter names, (Ra(Ra, etc.)) and the Y\text Y resistors have number names, (R1(R1, etc.)).
After solving the simultaneous equations (not shown), we get the equations to transform either network into the other.

ΔY\Delta \rightarrow \text Y transformation

Equations for transforming a Δ\Delta network into a Y\text Y network:
R1=RbRcRa+Rb+RcR1 = \dfrac{Rb\,Rc}{Ra + Rb + Rc}
R2=RaRcRa+Rb+RcR2 = \dfrac{Ra\,Rc}{Ra + Rb + Rc}
R3=RaRbRa+Rb+RcR3 = \dfrac{Ra\,Rb}{Ra + Rb + Rc}
Transforming from Δ\Delta to Y\text Y introduces one additional node.

YΔ\text Y \rightarrow\Delta transformation

Equations for transforming a Y\text Y network into a Δ\Delta network:
Ra=R1R2+R2R3+R3R1R1Ra = \dfrac{R1\,R2 + R2\,R3 + R3\,R1}{R1}
Rb=R1R2+R2R3+R3R1R2Rb = \dfrac{R1\,R2 + R2\,R3 + R3\,R1}{R2}
Rc=R1R2+R2R3+R3R1R3Rc = \dfrac{R1\,R2 + R2\,R3 + R3\,R1}{R3}
Transforming from Y\text Y to Δ\Delta removes one node.


Let's do a symmetric example. Assume we have a Δ\Delta circuit with 3Ω3\,\Omega resistors. Derive the Y\text Y equivalent by using the ΔY\Delta \rightarrow \text Y equations.
R1=RbRcRa+Rb+Rc=333+3+3=1ΩR1 = \dfrac{Rb\,Rc}{Ra + Rb + Rc} = \dfrac{3 \cdot 3}{3 + 3 + 3} = 1\,\Omega
R2=RaRcRa+Rb+Rc=333+3+3=1ΩR2 = \dfrac{Ra\,Rc}{Ra + Rb + Rc} = \dfrac{3 \cdot 3}{3 + 3 + 3} = 1\,\Omega
R3=RaRbRa+Rb+Rc=333+3+3=1ΩR3 = \dfrac{Ra\,Rb}{Ra + Rb + Rc} = \dfrac{3 \cdot 3}{3 + 3 + 3} = 1\,\Omega
Going in the other direction, from YΔ\text Y \rightarrow\Delta, looks like this,
Ra=R1R2+R2R3+R3R1R1=11+11+111=3ΩRa = \dfrac{R1\,R2 + R2\,R3 + R3\,R1}{R1} = \dfrac{1\cdot1 + 1\cdot1 + 1\cdot1}{1} = 3\,\Omega
Rb=R1R2+R2R3+R3R1R2=11+11+111=3ΩRb = \dfrac{R1\,R2 + R2\,R3 + R3\,R1}{R2} = \dfrac{1\cdot1 + 1\cdot1 + 1\cdot1}{1} = 3\,\Omega
Rc=R1R2+R2R3+R3R1R3=11+11+111=3ΩRc = \dfrac{R1\,R2 + R2\,R3 + R3\,R1}{R3} = \dfrac{1\cdot1 + 1\cdot1 + 1\cdot1}{1} = 3\,\Omega


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 Δ\Delta connections stacked one on the other.
Now select one of the Δ\Delta's to convert to a Y\text Y. We will perform a ΔY\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 Δ\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. RcRc must connect between nodes xx and yy, and so on for the other resistors. Refer to Diagram 1 above for the labeling convention.
When we perform the transform on the lower Δ\Delta, the black Δ\Delta resistors will be replaced by the new gray Y\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 Δ\Delta to a Y\text Y, and draw the complete circuit.
Apply the transformation equations for ΔY\Delta \rightarrow \text Y.
R1=RbRcRa+Rb+Rc=534+5+3=1512=1.25ΩR1 = \dfrac{Rb\,Rc}{Ra + Rb + Rc} = \dfrac{5\cdot 3}{4 + 5 + 3} = \dfrac{15}{12}= 1.25\,\Omega
R2=RaRcRa+Rb+Rc=434+5+3=1212=1ΩR2 = \dfrac{Ra\,Rc}{Ra + Rb + Rc} = \dfrac{4\cdot 3}{4 + 5 + 3} = \dfrac{12}{12}= 1\,\Omega
R3=RaRbRa+Rb+Rc=454+5+3=2012=1.66ΩR3 = \dfrac{Ra\,Rb}{Ra + Rb + Rc}= \dfrac{4\cdot 5}{4 + 5 + 3} = \dfrac{20}{12}= 1.66\,\Omega
Derive an equivalent Y\text Y network by substituting the Δ\Delta resistors. Make sure the Y\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=4.375Ω3.125 + 1.25 = 4.375 \,\Omega
On the right branch, 4+1=5Ω4 + 1 = 5\,\Omega
The two parallel resistors combine as 4.3755=4.37554.375+5=2.33Ω4.375\,||\, 5 = \dfrac{4.375 \cdot 5}{4.375 + 5} = 2.33\,\Omega
And we finish by adding the last two series resistors together,
Requivalent=2.33+1.66=4ΩR_{equivalent} = 2.33 + 1.66 = 4\,\Omega


ΔY\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.
This article is licensed under CC BY-NC-SA 4.0.