Conjecture about parallel and series circuits

[bcrowell]

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?

 

[Curl]

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?

 

[Meir Achuz]

n=5 is the Wheatstone bridge circuit.

 

[Curl]

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.