# A hard resistor network

the question is , to find the net resistance across the points a and b as marked in the figure(attached).

Here I think that the method of Symmetry should work . But I am not sure about it and moreover its not clear to me how to use it.

## The Attempt at a Solution

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SteamKing
Staff Emeritus
Homework Helper
Was it foggy in your room when you took these pix? They're barely legible. :(

phinds
Gold Member
2021 Award
I can hardly see your diagram at all (this is bad form, by the way ... posting illegible diagrams and expecting people to be able to decipher them). Do you know about delta-Y transforms? I seem to recall that there is a trick to the pyramid construct and you are right that it involves symmetry but I don't remember for sure that there is and certainly not how to do it. Delta-Y transforms will give you a brute-force method for getting it.

gneill
Mentor
Here's a depiction of the problem:

To begin with all line segments between nodes represent a 1 Ohm resistors.

By symmetry all points on the center-line (dotted blue vertical line passing through point C) are at the same potential if a potential difference is applied between A and B. So use that fact to recognize new parallel resistance opportunities. Hint: Don't be afraid to cut a 1 Ohm resistor into two 1/2 Ohm resistors in series.

The problem should reduce to something more tractable at least. Still may require an application of Delta-Y or other other circuit analysis methods, but it will be on a trivial circuit.

mooncrater
I studied you post , actually i didn't understand the symmetry method , moreover i didn't know about delta star methods . So i searched about them , studied them , applied them(delta star) in the problem , and in 3 long(+hard) tries I DID IT . Thank you.

gneill
Mentor
@mooncrater : The symmetry method relies on the fact that when separate nodes are at the same potential you can connect them together with a wire or remove components connected to both without changing the circuit behavior in any way.

The simplest case is that of a bridge circuit where the arms of the bridge are balanced and there is no potential across the "bridge".

These circuits have identical characteristics

No current can flow if there is no potential difference, so if there's a component on the bridge it can be removed. If there's no component there you can add what you want, even a shorting wire, and no change will occur to the circuit's operation. In the above diagram the placement of resistor R between the nodes a and b has no effect. Recognizing that two nodes previously thought separate can be treated as one can lead to new parallel component reduction opportunities, or serial reduction opportunities if you can remove a redundant component.

mfb
Mentor
You need at most one delta-star transformation at the very end, everything else can be solved with parallel/series resistor simplifications (after using the symmetry).