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### Course: Electrical engineering > Unit 2

Lesson 2: Resistor circuits- Series resistors
- Series resistors
- Parallel resistors (part 1)
- Parallel resistors (part 2)
- Parallel resistors (part 3)
- Parallel resistors
- Parallel conductance
- Series and parallel resistors
- Simplifying resistor networks
- Simplifying resistor networks
- Delta-Wye resistor networks
- Voltage divider
- Voltage divider
- Analyzing a resistor circuit with two batteries

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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 $\mathrm{\Delta}-\text{Y}$

*transformation*, or 'Delta-Wye' transformation.The names $\mathrm{\Delta}$ configuration by three resistors in a $\text{Y}$ configuration, and the other way around.

*Delta*and*Wye*come from the shape of the schematics, which resemble letters. The transformation allows you to replace three resistors in aThe $\mathrm{\Delta}-\text{Y}$ drawing style emphasizes these are 3-terminal configurations. Something to notice is the different number of nodes in the two configurations. $\mathrm{\Delta}$ 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.

$\mathrm{\Delta}-\text{Y}$ 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 $\text{R}3$ is $0$ ). In the $\mathrm{\Delta}$ 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 $\text{R}1$ and $\text{R}2$ 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,

We can write two similar expressions for the other two pairs of terminals. Notice the $\mathrm{\Delta}$ 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.

$\mathrm{\Delta}\to \text{Y}$ transformation

Equations for transforming a $\mathrm{\Delta}$ network into a $\text{Y}$ network:

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

$\text{Y}\to \mathrm{\Delta}$ transformation

Equations for transforming a $\text{Y}$ network into a $\mathrm{\Delta}$ network:

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

## Example

Let's do a symmetric example. Assume we have a $\mathrm{\Delta}$ circuit with $3{\textstyle \phantom{\rule{0.167em}{0ex}}}\mathrm{\Omega}$ resistors. Derive the $\text{Y}$ equivalent by using the $\mathrm{\Delta}\to \text{Y}$ equations.

Going in the other direction, from $\text{Y}\to \mathrm{\Delta}$ , looks like this,

## 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 $\mathrm{\Delta}$ connections stacked one on the other.

Now select one of the $\mathrm{\Delta}$ 's to convert to a $\text{Y}$ . We will perform a $\mathrm{\Delta}\to \text{Y}$ transformation and see if it breaks the logjam, opening up other opportunities for simplification.

We go to work on the bottom $\mathrm{\Delta}$ (an arbitrary choice). $Rc$ must connect between nodes $x$ and $y$ , and so on for the other resistors. Refer to

*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.*Diagram**1*above for the labeling convention.When we perform the transform on the lower $\mathrm{\Delta}$ , the black $\mathrm{\Delta}$ 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**$\mathrm{\Delta}$ to a $\text{Y}$ , 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, $3.125+1.25=4.375{\textstyle \phantom{\rule{0.167em}{0ex}}}\mathrm{\Omega}$

On the right branch, $4+1=5{\textstyle \phantom{\rule{0.167em}{0ex}}}\mathrm{\Omega}$

The two parallel resistors combine as $4.375{\textstyle \phantom{\rule{0.167em}{0ex}}}||{\textstyle \phantom{\rule{0.167em}{0ex}}}5={\displaystyle \frac{4.375\cdot 5}{4.375+5}}=2.33{\textstyle \phantom{\rule{0.167em}{0ex}}}\mathrm{\Omega}$

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

## Summary

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.(16 votes)- 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(7 votes)

- Can you provide the work for solving the 3 simultaneous equations that was not shown?(6 votes)
- 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!(5 votes)

- I want u to ask a full way to derive R1 and Ra plz give a derivation how we get R1 and Ra ?(5 votes)
- 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 @

https://www.khanacademy.org/science/electrical-engineering/ee-circuit-analysis-topic/ee-resistor-circuits/a/ee-delta-wye-resistor-networks?qa_expand_key=kaencrypted_ad2413cde556298b64a98d41d01fd684_cd6a16e8a8a657a21f768c0ff04804160d977a939ac691b3ecd6f7ee4a7f3b908c3a74bcab01337d6533d3b2cd73bf8f6ff245d796c6ba5cd4935e7434b49ff5a41f9fde6e3e3b40568f65d949ebef1c2069c478d967b52cc4b9a4782497244937b838cd8908e0ad3a3a524e87a524e6f816e4f6f1312d1554b5f5586074aab1f7d6c61e4d62263fecf53eb29522e61469af6deb96ff91ae7303c04481bb2de0(11 votes)

- why would we want to go from y-delta(5 votes)
- Hello Professor,

They can be used to simplify the occasional circuit as shown in this video. They really come in handy when working with three-phase system. The transform simplifies the math :)

This link may help: http://www.belden.com/blog/datacenters/3-Phase-Power-Wye-It-Matters.cfm

Regards,

APD(6 votes)

- so if they delta and y are then transformed into a combination circuit.. why have those??(2 votes)
- 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.(5 votes)

- I can derive R1, R2 & R3 easily, but I can't derive Ra at all. I can't figure it out !(2 votes)
- Scan down through the comments here and find a response by learner
*phidot*. There's a derivation in two separate comments.(2 votes)

- A video explanation might made it easier to understand.(2 votes)
- If we used these configurations for inductors instead of resistors, would that create a magnetic field in the shape of the configuration?(1 vote)
- In an inductor most of the magnetic field is concentrated on the inside of the coil. The field is much weaker around the outside of the inductor. If you tuck three straight inductors tight together in a triangle you pretty much get three magnetic fields in a triangle shape. But, if you curve the inductors into a circle the magnetic fields get together to make one more powerful field. See the second to last image of a circular-wound inductor in this article: https://www.khanacademy.org/science/electrical-engineering/ee-circuit-analysis-topic/circuit-elements/a/ee-real-world-circuit-elements(3 votes)

- I didn't understand the derivation of the three simultaneous equations. Can someone please explain them.(1 vote)
- I found this video to be very helpful: https://youtu.be/igvqOyJYAoA(2 votes)

- Do these transformations only work for planar circuits, or would they work for 3D structured circuits as well?(2 votes)
- The Delta-Wye transforms work for deltas and wyes, which are 2D arrangements.

Most 3D circuits are physics puzzles, like this one, http://web.physics.ucsb.edu/~lecturedemonstrations/Composer/Pages/64.42.html

You work these out from first principles or by exploiting symmetries to simplify down to 2D series and/or parallel circuits you do the usual way.(1 vote)