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# 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, minus, start text, Y, end text 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 start text, Y, end text configuration, and the other way around.
The delta, minus, start text, Y, end text 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 start text, Y, end text has four nodes (one extra in the center).
The configurations can be redrawn to square up the resistors. This is called a pi, minus, start text, T, end text configuration,
The pi, minus, start text, T, end text style is a more conventional drawing you would find in a typical schematic. The transformation equations developed next apply to pi, minus, start text, T, end text as well.

## $\Delta - \text Y$delta, minus, start text, Y, end text 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 start text, R, end text, 3 is 0 ). In the delta configuration, the resistance between x and y is R, c in parallel with R, a, plus, R, b .
On the start text, Y, end text side, the resistance between x and y is the series combination R, 1, plus, R, 2 (again, assume terminal z isn't connected to anything, so start text, R, end text, 1 and start text, R, end text, 2 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,
R, 1, plus, R, 2, equals, start fraction, R, c, left parenthesis, R, a, plus, R, b, right parenthesis, divided by, R, c, plus, left parenthesis, R, a, plus, R, b, right parenthesis, end fraction
We can write two similar expressions for the other two pairs of terminals. Notice the delta resistors have letter names, left parenthesis, R, a, etc.right parenthesis and the start text, Y, end text resistors have number names, left parenthesis, R, 1, etc.right parenthesis.
After solving the simultaneous equations (not shown), we get the equations to transform either network into the other.

### $\Delta \rightarrow \text Y$delta, right arrow, start text, Y, end text transformation

Equations for transforming a delta network into a start text, Y, end text network:
R, 1, equals, start fraction, R, b, R, c, divided by, R, a, plus, R, b, plus, R, c, end fraction
R, 2, equals, start fraction, R, a, R, c, divided by, R, a, plus, R, b, plus, R, c, end fraction
R, 3, equals, start fraction, R, a, R, b, divided by, R, a, plus, R, b, plus, R, c, end fraction
Transforming from delta to start text, Y, end text introduces one additional node.

### $\text Y \rightarrow\Delta$start text, Y, end text, right arrow, delta transformation

Equations for transforming a start text, Y, end text network into a delta network:
R, a, equals, start fraction, R, 1, R, 2, plus, R, 2, R, 3, plus, R, 3, R, 1, divided by, R, 1, end fraction
R, b, equals, start fraction, R, 1, R, 2, plus, R, 2, R, 3, plus, R, 3, R, 1, divided by, R, 2, end fraction
R, c, equals, start fraction, R, 1, R, 2, plus, R, 2, R, 3, plus, R, 3, R, 1, divided by, R, 3, end fraction
Transforming from start text, Y, end text to delta removes one node.

## Example

Let's do a symmetric example. Assume we have a delta circuit with 3, \Omega resistors. Derive the start text, Y, end text equivalent by using the delta, right arrow, start text, Y, end text equations.
R, 1, equals, start fraction, R, b, R, c, divided by, R, a, plus, R, b, plus, R, c, end fraction, equals, start fraction, 3, dot, 3, divided by, 3, plus, 3, plus, 3, end fraction, equals, 1, \Omega
R, 2, equals, start fraction, R, a, R, c, divided by, R, a, plus, R, b, plus, R, c, end fraction, equals, start fraction, 3, dot, 3, divided by, 3, plus, 3, plus, 3, end fraction, equals, 1, \Omega
R, 3, equals, start fraction, R, a, R, b, divided by, R, a, plus, R, b, plus, R, c, end fraction, equals, start fraction, 3, dot, 3, divided by, 3, plus, 3, plus, 3, end fraction, equals, 1, \Omega
Going in the other direction, from start text, Y, end text, right arrow, delta, looks like this,
R, a, equals, start fraction, R, 1, R, 2, plus, R, 2, R, 3, plus, R, 3, R, 1, divided by, R, 1, end fraction, equals, start fraction, 1, dot, 1, plus, 1, dot, 1, plus, 1, dot, 1, divided by, 1, end fraction, equals, 3, \Omega
R, b, equals, start fraction, R, 1, R, 2, plus, R, 2, R, 3, plus, R, 3, R, 1, divided by, R, 2, end fraction, equals, start fraction, 1, dot, 1, plus, 1, dot, 1, plus, 1, dot, 1, divided by, 1, end fraction, equals, 3, \Omega
R, c, equals, start fraction, R, 1, R, 2, plus, R, 2, R, 3, plus, R, 3, R, 1, divided by, R, 3, end fraction, equals, start fraction, 1, dot, 1, plus, 1, dot, 1, plus, 1, dot, 1, divided by, 1, end fraction, equals, 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 start text, Y, end text. We will perform a delta, right arrow, start text, Y, end text 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. R, c 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 start text, Y, end text 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 start text, Y, end text, and draw the complete circuit.
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, point, 125, plus, 1, point, 25, equals, 4, point, 375, \Omega
On the right branch, 4, plus, 1, equals, 5, \Omega
The two parallel resistors combine as 4, point, 375, vertical bar, vertical bar, 5, equals, start fraction, 4, point, 375, dot, 5, divided by, 4, point, 375, plus, 5, end fraction, equals, 2, point, 33, \Omega
And we finish by adding the last two series resistors together,
R, start subscript, e, q, u, i, v, a, l, e, n, t, end subscript, equals, 2, point, 33, plus, 1, point, 66, equals, 4, \Omega

## Summary

delta, minus, start text, Y, end text 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.

## Want to join the conversation?

• Why we can assume that terminal z isn't connect to anything?
I mean, in the example, it clearly show that terminal z did connect to something and must have a current flow through.
• Hello Fasteric,

Try to think of each circuit as a black box - assume you don't know what is inside. When you "look" from terminal A to B you see a certain resistance. When you "look" from terminal B to C you see a certain resistance. Finally when you "look" from terminal C to A you again see a resistance. It's important that you "look" at the circuit in isolation. If it was connected to other things it would be difficult to do this thought experiment.

Please give it a try you should be able to convince yourself that the wye and delta look the same. If it helps try calculating the resistances when the delta has three resistors each with a value of 3 Ohm. Then try the wye with three resistors each having a 1 Ohm value.

Regards,

APD
• Can you provide the work for solving the 3 simultaneous equations that was not shown?
• This is awesome. Thank you for explaining. It's very hard to find a straightforward, detailed explanation of the transformations on the web, but you've presented one here. You should do a YouTube video. It would blow people's minds!
• I want u to ask a full way to derive R1 and Ra plz give a derivation how we get R1 and Ra ?
• since,

(resistance between X&Y in Y) = (resistance between X&Y in delta)
(resistance between Y&Z in Y) = (resistance between Y&Z in delta)
(resistance between Z&X in Y) = (resistance between Z&X in delta)

so,
(R1 + R2) = ( Rc || Ra+Rb ) .. [1]
(R2 + R3) = ( Ra || Rb+Rc ) .. [2]
(R3 + R1) = ( Rb || Rc+Ra ) .. [3]

To get R1, we do ([1] + [3] - [2]) / 2, at the left hand side :

( (R1 + R2) + (R3 + R1) - (R2 + R3) ) / 2 = (R1+R2+R3+R1-R2-R3)/2 = (2 x R1)/2 = R1 ..[4]

Do the same for right hand side :

( ( Rc || Ra+Rb ) + ( Rb || Rc+Ra ) - ( Ra || Rb+Rc ) ) / 2
=( (Rc x (Ra+Rb))/(Rc+(Ra+Rb)) + (Rb x (Rc+Ra))/(Rb+(Rc+Ra)) - (Ra x (Rb+Rc))/(Ra+(Rb+Rc)) ) / 2
=( ((Rc x (Ra+Rb)) + (Rb x (Rc+Ra)) - (Ra x (Rb+Rc)) / Ra+Rb+Rc ) / 2
=( ((RbRc+RaRc) + (RaRb+RbRc) - (RaRb+RaRc)) / Ra+Rb+Rc ) / 2
=( (2 x RbRc) / Ra+Rb+Rc ) / 2
=(RbRc)/(Ra+Rb+Rc) .. [5]

Since, left hand side = right hand side, or [4] = [5],

R1 = (Rb x Rc)/(Ra+Rb+Rc)

Done, we got our R1. :D

ref : previous question @
• why would we want to go from y-delta
• so if they delta and y are then transformed into a combination circuit.. why have those??
• One of the common places a Delta resistor configuration shows up is in a "bridge" circuit. If you search for "Whetstone Bridge" on the web you will see deltas in the resistor bridge. This circuit is used to make very sensitive differential voltage measurements.
• I can derive R1, R2 & R3 easily, but I can't derive Ra at all. I can't figure it out !
• Scan down through the comments here and find a response by learner phidot. There's a derivation in two separate comments.
• Hi Professor,

For another alias Phi and T of Delta-Wye. If I flip the circuit upside-down and let the positive wire (i-in) connect to top and negative wire (i-out) connect to downside of the flipped circuit, are these Delta-Wye transformation analysis still valid?