Conjecture about parallel and series circuits

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Discussion Overview

The discussion revolves around the conjecture regarding the minimum number of resistors, n, required to create a circuit configuration where the equivalent resistance cannot be determined through standard series and parallel reduction methods. The focus is on exploring the cases for n=5 and below, with references to specific circuit configurations.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant conjectures that n=5 is the lowest number of resistors for which the equivalent resistance cannot be determined using traditional methods.
  • Another participant questions the interpretation of "having n resistors" and suggests that the simplest example for n=12 is a resistor cube, asking for a specific example for n=5.
  • A participant identifies the Wheatstone bridge as a configuration for n=5.
  • Another participant outlines a reasoning process for proving that n=4 allows for equivalent resistance calculations through various configurations, suggesting that all combinations lead to solvable cases.
  • This participant also mentions that for n=6 and above, counterexamples can be found, specifically referencing the Wheatstone bridge with additional resistors in series.

Areas of Agreement / Disagreement

Participants express differing views on the conjecture, with some supporting the idea that n=5 is a critical threshold while others provide counterexamples and reasoning for lower values. The discussion remains unresolved regarding the validity of the conjecture.

Contextual Notes

The discussion includes assumptions about circuit configurations and the definitions of equivalent resistance. The reasoning provided does not conclusively prove or disprove the conjecture, and the limitations of the proposed proofs are not fully explored.

bcrowell
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If we have n (possibly unequal) resistors, we can combine them in various ways to produce a device with two terminals. In many cases, the equivalent resistance of this device can be found by repeatedly breaking the circuit down into parallel and series parts. Conjecture: n=5 is the lowest for which a circuit exists such that this strategy cannot be applied.

Can anyone prove this, or provide a counterexample for n<5?
 
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What do you mean "have n resistors". Are you trying to find equivalent resistance between two points? The simplest one I can think of right now is the resistor cube but that is n=12. Can you show me one for n=5?
 
Last edited:
n=5 is the Wheatstone bridge circuit.
 
Shouldn't be that hard to prove. With n=4 you have 4 elements. Connect two together, and you get 2 resistors in series. Now with the 3rd resistor, you can either put it in series again, or connect it in between the first two to make a T.

If you do the first, then you have 3 resistors in series. The forth one must either go at the ends or somewhere in between, making either 4 or 3 or 2 resistors in series (effectively shorting out other resistors). This can be solved using equivalent combination.

If you use the 3rd element to make a T, then the fourth must either be in parallel with two other resistors (three possibilities) or it can go in series and short one resistor out. This can also be solved using equivalent combination.

So yeah, this is a sketch of proof for n=4. With n=3 it's easier (the proof is actually inside the proof for n=4) and n=2 and n=1 are trivial.
For n=6+ you can find a counterexample, such as a Wheatstone bridge with resistors attached in series.
 

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