# Delta-Wye Transformation

#### Rey Arthur Regis

Joined Jul 13, 2015
12
Can someone help me? for a. i got 87.14 ohms

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#### Lool

Joined May 8, 2013
116
I didn't work it out, but doesn't your answer seem too high? I can see several direct paths between points A and x that have much lower resistance than your answer.

Problem A is easier than B, I think, because you have a lot of symmetry and you can simplify the circuit rather quickly.

There might be a trick for problem B too, but looking quickly I don't see it.

#### WBahn

Joined Mar 31, 2012
30,077
Can someone help me? for a. i got 87.14 ohms
You need to show your work. Without it, the most we can tell you is whether your answer is right or wrong. If it's wrong, we can't help you find where it is wrong or what you need to do differently because we can't see what you did period.

The first step is to redraw the circuit with the connections as described.

Then the second step should be to put bounds on the possible resistance values. Without even redrawing the circuit and just picking the most obvious path between Node A and Node x shows that the MAXIMUM resistance between those two nodes, even without making any of the described connections, is 60 Ω. With the connections there is a path that is only 40 Ω.

You also know that the absolute minimum resistance is 10 Ω and can actually bound it at 15 Ω very easily.

So before you begin you know the answer has to be between 15 Ω and 40 Ω.

The first problem can be done by inspection (particularly once it is redrawn) without using delta-wye transforms. But if you want (or are required) to use them, then the following article might help.

#### Rey Arthur Regis

Joined Jul 13, 2015
12
is 87 too high? our instructor gave us a clue that we have to transform it twice to thrice. I'll just have to redraw it like you said and think of other ways.

#### Rey Arthur Regis

Joined Jul 13, 2015
12
You need to show your work. Without it, the most we can tell you is whether your answer is right or wrong. If it's wrong, we can't help you find where it is wrong or what you need to do differently because we can't see what you did period.

The first step is to redraw the circuit with the connections as described.

Then the second step should be to put bounds on the possible resistance values. Without even redrawing the circuit and just picking the most obvious path between Node A and Node x shows that the MAXIMUM resistance between those two nodes, even without making any of the described connections, is 60 Ω. With the connections there is a path that is only 40 Ω.

You also know that the absolute minimum resistance is 10 Ω and can actually bound it at 15 Ω very easily.

So before you begin you know the answer has to be between 15 Ω and 40 Ω.

The first problem can be done by inspection (particularly once it is redrawn) without using delta-wye transforms. But if you want (or are required) to use them, then the following article might help.

i really find it hard to redraw it. any other hints sir?

#### Rey Arthur Regis

Joined Jul 13, 2015
12
i got 28.3333

#### WBahn

Joined Mar 31, 2012
30,077
is 87 too high? our instructor gave us a clue that we have to transform it twice to thrice. I'll just have to redraw it like you said and think of other ways.
Consider just the red path between A and x. What is the resistance of that path? Since any other paths are going to be, at least in part, in parallel with this path, you know that total resistance from A to x has to be less than the resistance of this path.

But now consider the green path, which is valid because A and B are connected. What is the resistance of this path? If it's lower than the red path, it sets a new minimum value.

If we consider the blue path and a yellow path that would go from A over to D and then on to x we can reduce this upper limit even further.

#### WBahn

Joined Mar 31, 2012
30,077
i got 28.3333
Nope.

Look at the post I just made and see what the upper bound is by considering the blue and the yellow (described in words) paths. That bound is less than your result, so we know that it is wrong.

Draw the circuit with the connections made. Since you have several points that are all connected together, use that as your common node along the bottom. Then draw the circuit from x to that common node (which includes A). You may need to sketch it a couple of times to get it clean. But once you have this this diagram you can use a simple symmetry argument to effectively remove one of the resistors and turn it into a simple series-parallel network.

#### Rey Arthur Regis

Joined Jul 13, 2015
12
Nope.

Look at the post I just made and see what the upper bound is by considering the blue and the yellow (described in words) paths. That bound is less than your result, so we know that it is wrong.

Draw the circuit with the connections made. Since you have several points that are all connected together, use that as your common node along the bottom. Then draw the circuit from x to that common node (which includes A). You may need to sketch it a couple of times to get it clean. But once you have this this diagram you can use a simple symmetry argument to effectively remove one of the resistors and turn it into a simple series-parallel network.

i really still don't get it really much. please be patient with me. I have redrawn it but i still get 3 different values which i know is wrong. the circuit is all connected together except for the 10 ohms on the bottom most . am i right?

can you give me a simpler description?

#### WBahn

Joined Mar 31, 2012
30,077

How can I tell you what might be wrong with your redrawn circuit or your work when you won't show either your redrawn circuit or your work?!

#### Rey Arthur Regis

Joined Jul 13, 2015
12

How can I tell you what might be wrong with your redrawn circuit or your work when you won't show either your redrawn circuit or your work?!
here is it. sorry i replied late.

#### WBahn

Joined Mar 31, 2012
30,077
here is it. sorry i replied late.

Here WHAT is? You still haven't shown your redrawn circuit or any work.

#### Rey Arthur Regis

Joined Jul 13, 2015
12
here

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#### WBahn

Joined Mar 31, 2012
30,077
Looks like you got it this time.

FYI: There's really no need to use a 2.5 MB file when a 60 kB file will work just fine:

Are you interested in seeing how to analyze it without doing any transforms at all?

#### Rey Arthur Regis

Joined Jul 13, 2015
12

sure? how?

#### Rey Arthur Regis

Joined Jul 13, 2015
12
I'll take the preliminary exams tomorrow but still it's nice to know other ways.

#### MrAl

Joined Jun 17, 2014
11,496
Hi,

For one thing you can use the concept of a virtual equal node. We might call it an "equanode". It's like a virtual ground only instead of being a virtual zero volts it is some other virtual voltage. Because the second node is equal in voltage to the first node, they can be connected with a short and that reduces the complexity because then some resistances appear in parallel which can then be reduced to one single resistor instead of two.
Note this is for calculation purposes only.

#### Rey Arthur Regis

Joined Jul 13, 2015
12
Hi,

For one thing you can use the concept of a virtual equal node. We might call it an "equanode". It's like a virtual ground only instead of being a virtual zero volts it is some other virtual voltage. Because the second node is equal in voltage to the first node, they can be connected with a short and that reduces the complexity because then some resistances appear in parallel which can then be reduced to one single resistor instead of two.
Note this is for calculation purposes only.
I can't really absorb what you're trying to say. Can you give me a more simpler explanation or a figure sir?

#### MrAl

Joined Jun 17, 2014
11,496
Hi,

Sure. A very simple example should illustrate this concept quite clearly.

Say we have two resistor networks, each set has only two resistors each. The first is made from two 10 ohm resistors in series with a voltage source so it forms a simple voltage divider. The second set is identical and connected to the same voltage source of say 10 volts.
So we have two voltage dividers, both made of two 10 ohm resistors in series, and this means the center node voltage of both networks is have the supply voltage.
So the first network center node voltage is 5 volts, and the second network voltage is 5 volts.
Two identical networks, with both central nodes at 5v, yet they are separate networks.
If we connect an ammeter between the two central nodes, what current do we measure? We measure zero amps, because they are at the same potential.
So what is the total resistance presented to the supply source of 10 volts?
We could calculate the two 10 ohm resistors on the left, which comes out to 20 ohms, then the same on the right, and two 20 ohm resistors in parallel comes out to 10 ohms, so the total resistance is 10 ohms.

But we could also do this another way. Because the two central nodes have EXACTLY the same voltage, we could short them together, noting that nothing about the network currents or voltages changes. Once shorted together, we have two 10 ohm resistors in parallel on the top, and two 10 ohm resistors on the bottom. So calculating the two 10 ohms on top in parallel, we get 5 ohms, and the same for the two bottom resistors. So the two are now equivalent to two 5 ohm resistors in series. This again, comes out to a total of 10 ohms.

So we can look at these two networks as two 20 ohm resistors in parallel, or two 5 ohm resistors in series. The second case is possible because of the concept of equal node voltages, where if two nodes have the same voltage for all time, then can be shorted together and nothing changes about the network.

There are some cautions to observe when using this idea though.
The first is that the voltage must be exactly equal. If there is even a tiny tiny difference, it may make it impossible to short the two. This is why this is mainly a theoretical concept not something you would want to do to a real circuit unless you first investigated the consequences carefully.
The second is that they must be equal for all time. This means that they can not vary for even a microsecond in some cases.

A good example of when this is applicable is when calculating certain things about a simple op amp amplifier circuit, where the 'virtual ground' is considered to be the some voltage as the ground line.
A good example of when this is not applicable is for the same circuit, only trying to calculate the error voltage at the inverting terminal due to the internal gain of the amplifier.
The reason it works for the first analysis and not the second is because for the first we can get away with assuming the inverting terminal is at ground potential, but for the second analysis the result depends highly on a very very tiny difference between the inverting terminal and ground, which can be as little as 10uv. Shorting that to ground would obviously mean we come out with the rwrong result of 0.000000v when the correct result would have been 0.000010v.
For resistor networks in theory though, this idea should always work as long as the voltages are equal.

Another quick example might be a little harder to imagine as being valid.
Again for two voltage divider networks:
First network is 30 ohms in series with 10 ohms, and the second network is 300 ohms in series with 100 ohms, the top resistors are the 30 and 300 which connect to the source voltage supply. Can we short the two central nodes together for analysis?
Since the output of the first network is 1/4 of the source voltage and the second network output is 1/4 of the source voltage, we can short the two network central nodes together (for the purpose of analysis). This places a 30 ohm in parallel with 300 ohm, and a 10 ohm in parallel with a 100 ohm.

This idea can be extended to resistors that do not form the same ratio, but i'd have to look that one up as that is a lot more tricky to implement and i dont remember the details offhand which involves a little formula.

Last edited:

#### WBahn

Joined Mar 31, 2012
30,077

sure? how?
First, let's redraw the circuit with the stated connections:

The editor I use doesn't like two components with the same reference designators, otherwise I would have kept them the same from one diagram to the other. I have combined R3 and R5 to give R17 and, similarly, R5 and R12 to give R27. I have also combined R2 and R4 to give R16. But be sure you satisfy yourself that these are the same circuit when A-B-C-D are all connected (top node in right diagram). The labeled nodes M, N, R, S should help.

Speaking of those nodes, notice the symmetry of the circuit. We can exploit that symmetry and claim that the voltage at Node M will be the same as the voltage at Node N. Similarly, that the voltage at Node R will be the same as the voltage at Node S. As a result, we know that there will be no current flowing in R16, which means that we could simply remove it as it has no effect. Another option is to connect a wire between Node M and Node N, which we also know will have no effect because the two nodes are the same and, hence, even with a wire connected between them no current will flow in this wire. We can do the same with Nodes R and S.

We now have R19 and R28 in parallel, as well as R26 and R33, as well as R32 and R34. We can reduce all of these by inspection.

When we combine R24 and R35 we are left with three 20 Ω resistors in parallel and those are in series with 15 Ω (R39 + R40). So the total resistance from A to X is then

R_AX = 15 Ω + (20/3) Ω = 21.7 Ω

And now that I look at your diagram, it appears that you didn't get it. I thought you answer was 21.67 Ω but I know see that it is 21.17 Ω. So I'll look at your work a bit more closely as I had just spot checked it before.