Reducing all circuit resistors to only parallel and series?

solour
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Homework Statement


The problem from the textbook is:

Is it possible to connect resistors together in a way that cannot be reduced to some combination of series and parallel combinations?

Homework Equations


V = IR
kirchhoff's current law
kirchhoff's loop law

The Attempt at a Solution


I am completely lost as to where to begin. My intuition tells me that if there is only 1 voltage source, the answer would be no, all resistor combinations can be reduced to a single equivalent resistor(I do not know whether this statement is true or not). However, if the voltage source becomes more than 1 at different locations of the circuit, I do not know how this will effect the answer.
 
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solour said:

Homework Statement


The problem from the textbook is:

Is it possible to connect resistors together in a way that cannot be reduced to some combination of series and parallel combinations?

Homework Equations


V = IR
kirchhoff's current law
kirchhoff's loop law

The Attempt at a Solution


I am completely lost as to where to begin. My intuition tells me that if there is only 1 voltage source, the answer would be no, all resistor combinations can be reduced to a single equivalent resistor(I do not know whether this statement is true or not). However, if the voltage source becomes more than 1 at different locations of the circuit, I do not know how this will effect the answer.
Welcome tpo the PF.

I think that the classic infinite cube of resistors cannot be reduced. What have you found in your Google searches so far?
 
berkeman said:
Welcome tpo the PF.

I think that the classic infinite cube of resistors cannot be reduced. What have you found in your Google searches so far?

Thankyou for the heart warming welcome!

I managed to find the answer for this question on chegg, and it appears to be a yes. Unfortunately I do not have an account and therefore cannot see how they proved it.

I believe the Resistor Cube you suggested also answers the question, though I need to dig in a little deeper to understand how that one works.
 

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