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Suppose I have a (finite of course!) network of resistors only. Is it always possible to calculate the equivalent resistance between any two points using only parallel, series, Y-Δ and Δ-Y (wye-delta and delta-wye) transforms?

Conversely, is there a resistor network where the star-mesh transform for n>3 is absolutely necessary to find the equivalent resistance between some two nodes?

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Place a resistor at each line of this diagram and apply voltage between any two opposite nodes.

You can't reduce this circuit with the operations you mentioned.

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    \$\begingroup\$ Thank you! Great explanation. That's the kind of answer I was looking for. \$\endgroup\$ – Fibonacci Sep 10 '15 at 4:40
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Given a network of resistors placed on the edges of a cube and connected at the corners, what is the resistance of this network across the body diagonal of the cube?

Take a look at this image

image of resistor cube

Try turning Y into Δ and Δ into Y a couple of times and see if that helps you solve the problem.

The image is from here where you can also find the solution.

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    \$\begingroup\$ I was able to reduce it to a Wheatstone bridge after many Y-Δ and Δ-Y transformations. \$\endgroup\$ – Fibonacci Jul 24 '15 at 2:01
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The basis of Delta-Y transformation is two-port networks. It has been devised with possible connection of impedences in a two-port network. So, it gives us a set of input variables (i,v) and an output one. Since the transformation in itself is independent of nature of connected impedences(should at least follow the lumped parameter abstraction) and how they are connected within the black-box which gives out two-ports, it should work for all sorts of resistors.

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