I understand that the point of the wye-delta transform is to have the equivalent resistance between every pair of nodes be the same in each form of the circuit.

However, when I see the derivation for the concept, they treat the equivalent resistance between two nodes in the circuit's wye form to be as if no current flows into the other node. My question is, why? If one looks at the attached picture, for example, the equivalent resistance between nodes 1 and 2 is R1 + R3. The resistors are treated as if they are in series. In the real circuit, however, R2 could have drawn current. My instinct is that that would affect the situation and make it differ from treating the circuit as if R2 didn't exist at all and that all the current from R1 flows to R3.

I think my question might come from a lack of understanding of equivalent resistance, as lame as that may be. I am used to thinking of Req in terms of maintaining voltage-current character in a circuit (as in, the same current flows into and out of the equivalent resistor as it did the original element, and the same voltage is across the element), so I don't understand the method of finding equivalent resistance between two points while ignoring the current going into the third resistor. I don't understand how voltage-current character is conserved in that case.

Thanks to anyone who can help me understand this better!


EDIT: The derivation I'm referring to is as follows:

https://i.sstatic.net/wgljk.png (the editor is screwing up and not letting me post the image in here). Note they set the equivalent resistances equal for each configuration. This is intuitive, except when the calculate the equivalent resistance of each branch they do not consider current flowing through the other branch (as in, Rxy = R1 + R2 as if they were in series, except the two are not in series. I don't understand equivalent resistance in this context)

  • \$\begingroup\$ Can you provide a reference for this derivation? Because I haven't seen it done that way, and it doesn't seem right to ignore the branch with R2 \$\endgroup\$
    – user28910
    Commented Jun 1, 2017 at 18:33
  • \$\begingroup\$ I have edited the post. \$\endgroup\$ Commented Jun 1, 2017 at 20:44

2 Answers 2


My previous answer was unhelpful to the OP, so I will attempt a different approach and delete the old answer.

Wye-delta transformations are mathematical and built on the same principle as Thevenin and Norton Equivalents - the superposition theorem for electric circuits.

I won't get into the details of the proof, but if you like it is explained fairly well here. Fundamentally, it states that the response of a bilateral* linear** circuit with multiple sources can be determined by algebraically summing the responses of the circuit to each source, with the remaining sources replaced by characteristic impedances.

*response is independent of current direction
**containing no non-linear elements (capacitors, inductors, etc)

To do this transformation, we apply an ideal current to nodes 1, 2, and 3 (respectively) in each step of the transformation:

1) (I1-I2), -(I1-I2), 0
2) 0, (I2-I3), -(I2-I3)
3) -(I3-I1), 0, (I3-I1)

This is consistent with Kirchoff's Current Law, as I1 + I2 + I3 = 0.

By choosing this approach, we eliminate one of the nodes from each step, allowing us to compute the equivalent resistance across the remaining nodes as we would normal series and parallel resistances. We then linearly sum those results to give a final equivalence.

In short, it is a choice backed by mathematical principles of superposition to make the problem inherently easier to solve.

  • \$\begingroup\$ I think this almost has me there. I just don't understand how to apply superposition here. There are three current sources used (one for each node) in this proof, if I'm correct. Is that right? Now, if one current source is active at a time, with the others closed off, how do we obtain the three sets of current values that we use for superposition? I would think that closing the other two sources off would cause no current to flow to the other nodes. I think I need to develop my understanding more of these theorems because it doesn't feel natural yet. \$\endgroup\$ Commented Jun 1, 2017 at 23:50
  • \$\begingroup\$ The wikipedia page mentions that each problem 1) (I1-I2), -(I1-I2), 0 2) 0, (I2-I3), -(I2-I3) 3) -(I3-I1), 0, (I3-I1) \$\endgroup\$ Commented Jun 2, 2017 at 0:01
  • \$\begingroup\$ If the question involves three current sources, one in each branch, and the listed problems 1) 2) and 3) in your answer are the currents in each branch while each respective current source is active, I believe I fully understand the wye-delta concept by superposition. But I'm not sure if that is the case. \$\endgroup\$ Commented Jun 2, 2017 at 0:07
  • \$\begingroup\$ Yes, those are the currents in the respective branches. I apologize if that was not clear. \$\endgroup\$
    – Chris M.
    Commented Jun 2, 2017 at 0:15

Depending on your intentions you can use the Thévenin's theorem to get an equivalent circuit. With this method you calculate one time the open circuit voltage (in this case voltage drop across R3).

enter image description here

Uout = Uinput * R3/(R1+R3) = 0.5 V

Then you have to calculate the equivalent resistance Ri by short-circuiting the input voltage source.

enter image description here

Ri = (R1*R3/R1+R3) + R2 = 150 Ohm

This delivers you the new circuit. Now the open circuit voltage is the new input voltage of your circuit and Ri (R9 in the simulation) is your equivalent resistor. As an example I added R8 & R10 as a load to the circuit.

enter image description here

Schematic created using CircuitLab.

Or you can transform it by using the wye-delta transformation...

Hopefully this helps.


  • \$\begingroup\$ This is helpful for analyzing the wye's effect on elements connected to it and is understandable. I'm more concerned with the derivation of wye-delta which I put in my original post. This is helpful though. \$\endgroup\$ Commented Jun 1, 2017 at 20:57

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