What is the Net Resistance in this Complex Resistor Problem?

[Fisherlam]

Hey there! I'm having some real trouble deciphering this complex resistor problem. I have heard of the Kirchhoff voltage and current rules and do know how to use them to solve some problems but I'm not sure how to apply them in this context, or if they are even used to solve this. As seen the the diagram, each block is a resistor with resistance of r, and I'm tasked to find out the net resistance across A and B, the answer given by the book is 2r. I would really appreciate some help in understanding how to solve them and perhaps a universal way to tackle such problems.

THANKS A LOT!

 

[phinds]

Look for obvious series resistors and look for symmetry. If there is symmetry, often two points that do not otherwise seem related can be considered as joined because they are at the same voltage whether joined or not. This can lead to some simplifications.

EDIT: yeah, symmetry is the trick in this one. Once you joint symmetrical nodes, things just start falling out in series and parallel in rather obvious ways.

 

[Fisherlam]

Hmm, thanks for some interesting ideas. In the uploaded picture, I have highlighted the nodes, with same colour nodes have same voltage. I looked up online and found that we can treat the nodes as a parallel circuit, as in the cube resistor problem, and as you mentioned. So we can surely divide the diagram into two along the blue nodes, the red nodes can be represented as a parallel circuit with two branches, each branch with one resistor. However, I am not too sure about what to do after the red nodes, that is the green nodes and the unmarked one. Can I ignore the unmarked node, which i assume to have a different voltage as compared to the green ons, and assume that there are four parallel branches, each with 2r resistance, pouring into the blue middle nodes? In that case, I have ((1/+1/)^-1)+(1/2+1/2+1/2+1/2)^-1)*2 = 2, which coincidentally is also the ans? IS that right? Thanks in advance.

 

[phinds]

The green nodes should be red and the unmarked nodes have to be treated as part of the circuit after all the red nodes are used for simplification.

 

[Fisherlam]

Um I'm sorry but why are the green nodes red?

 

[phinds]

For exactly the same reason the red nodes are red. Same voltage because of symmetry (that is the ones directly up/down from each other)

 

[Fisherlam]

Okay I see but I'm lost how do I continue?

 

[phinds]

Okay I see but I'm lost how do I continue?

Draw wires between equivalent nodes (up/down pairs) and then reduce series / parallel combinations.

 

[Fisherlam]

What do you mean by draw wires?

 

[phinds]

What do you mean by draw wires?

I can't think of any other way to say something that simple. Draw wires.

 

[Fisherlam]

You mean this?

 

[phinds]

FIRST, I'd suggest just drawing wires on the original circuit. I'm not going to try to figure out if your new diagram is exactly that but it looks like the right idea at least.

 

[Fisherlam]

then?

 

[phinds]

Reread post #8

 

[Fisherlam]

Have already done that, i just need confirmation and I am more than sure you have the capability to understand the diagrams I drew and verify.

 

[phinds]

Have already done that, i just need confirmation and I am more than sure you have the capability to understand the diagrams I drew and verify.

Show the wires on the original diagram and label the resistors (R1, R2, etc) and I'll give you another hint.

 

[Fisherlam]

i think it's alright, i get the rough idea. Nice receiving your help, and thanks, anyways just curious, which university did you graduate from? :)

 

[phinds]

i think it's alright, i get the rough idea. Nice receiving your help, and thanks, anyways just curious, which university did you graduate from? :)

Ga. Tech, 1968