Obtaining a maximum resistance given a set of resistors

[LongApple]

My gut instinct is that putting all the resistors in series will give max resistance, but I am not sure how to give a more rigorous either mathematical or just in words reasoning why. Or maybe I'm wrong! But it seems like the fraction introduced from parallel won't help

In any case, assume all resistors have positive integer resistances with most greater than 1

 

[sophiecentaur]

If you could prove that A+B - 1/(1/A +1/B) >0 then that would do what you want (Difference between series and parallel connection formulae)
Have a go (with algebra - not numbers).

 

[LongApple]

If you could prove that A+B - 1/(1/A +1/B) >0 then that would do what you want (Difference between series and parallel connection formulae)
Have a go (with algebra - not numbers).

I am not entirely convinced because that sounds like just a check for two resistors. How do we know that the fractions in R1 * R1 / (R1+R2) won't be able to stomp R1 + R2 with some values or that with say 3 resistors we would be able to create a frankenstein resistor network with a monster amount of resistors?

I am a bit dramatic.

 

[collinsmark]

Have you considered modeling the combination of identical resistors by using more fundamental principles?

By that I mean, for a simple resistor shaped like a cylinder or a rod, the things which determine its resistance are its resistivity, cross sectional area, and its length.

• If you connect two resistors in parallel, how does that affect the overall, cross sectional area?
• If you connect two resistors in series, how does that affect the overall length?

This approach might not apply well if your resistors are all different values. But for identical resistors, it might be worth consideration.

 

[sophiecentaur]

I am not entirely convinced because that sounds like just a check for two resistors. How do we know that the fractions in R1 * R1 / (R1+R2) won't be able to stomp R1 + R2 with some values or that with say 3 resistors we would be able to create a frankenstein resistor network with a monster amount of resistors?

I am a bit dramatic.

Dramatic -haha!
But no harm in being suspicious.
Many mathematical proofs can be arrived at, validly, by splitting a set into a set of pairs, again and again. If you can identify just one pair of resistors that are connected in parallel then you can increase the effective resistance by putting them in series instead (by the above argument) of parallel and, for the calculation, replace them with R1 + R2. You can repeat this with the next and then the next pair.
Just adding complexity to your network will not suddenly produce a magical result. (We are assuming here that the resistors are all ideal (Ohmic) resistors and that the temperature is kept constant, of course.)

 

[Muti]

Ohm law states that you may need to increase voltage if you added a resistance and again want same current to flow through the "circuit" .

 

[Redbelly98]

I think I have a reasonable proof for the series arrangement. Since this does not look at all like a homework problem, here goes.

Let's suppose we are putting some fixed current through the resistor network. In that case, maximizing the resistance is equivalent to maximizing the voltage across the network.

If there is a resistor arrangement that results in (1) each resistor having the maximum voltage possible and (2) simply adding up all the resistor voltages gives the voltage across the entire network, then that is the maximum voltage (and hence resistance) that can be attained for a network built from the given set of resistors.

By placing the resistors all in series, each resistor gets the full applied current and therefore the maximum voltage possible for that resistor, satisfying condition (1). The series arrangement also satisfies condition (2).

Hope that is clear enough.

 

[LongApple]

I hope I wasn't misleading at the start of the thread

I have no idea if in series is best for maxing anything

 

[sophiecentaur]

Is there actually any point in introducing 'operating volts' into a discussion of Equivalent Resistance? Resistance is a ratio and you can stick with that, surely. All you need to consider is the formulae for series and parallel connection and we always assume that resistors are ohmic (linear) in simple theory.

I have no idea if in series is best for maxing anything

"Maxing"? It's not rocket science to show that the series connection for two resistors always produces a larger equivalent value than the parallel connection. That extends to any complexity of resistor network.

 

[willem2]

Is there actually any point in introducing 'operating volts' into a discussion of Equivalent Resistance? Resistance is a ratio and you can stick with that, surely. All you need to consider is the formulae for series and parallel connection and we always assume that resistors are ohmic (linear) in simple theory.

But not every circuit can be built op with only parallel and series combinations of resistors.

 

[sophiecentaur]

But not every circuit can be built op with only parallel and series combinations of resistors.

You want to introduce the star delta transformation and others? But that doesn't allow for the series parallel change that the OP was asking about. You want other components, too? Again, not included in the original question.
This is all just about a bit of Maths and the inequality is 'unassailable'. (Now there's a challenge)

 

[willem2]

You want to introduce the star delta transformation and others? But that doesn't allow for the series parallel change that the OP was asking about. You want other components, too? Again, not included in the original question.

I got the impression you thought rebelly98's proof was overkill, because you only need to consider circuits with only series and parallel combinations of resistors. I don't see the need to limit the question to only those circuits. The OP had already noticed that having parallel resistances would result in a lower resistance, so the question can't have been only about them.

 

[sophiecentaur]

I got the impression you thought rebelly98's proof was overkill, because you only need to consider circuits with only series and parallel combinations of resistors. I don't see the need to limit the question to only those circuits. The OP had already noticed that having parallel resistances would result in a lower resistance, so the question can't have been only about them.

I think, if you look at the wording of Redbelly's proof, what he is dong is effectively the same thing as deriving the series and parallel connection equivalents - except he is including (or implying) the same steps as the textbook derivations for those formulae. i.e going back to the basics further.

I think that the original question does, in fact, limit the cases to where there are series or parallel connections because there is no way to do the same 'swap' for star or delta unit in the circuit. You would have to leave them untouched. The proof that a parallel / series swap would always increase the equivalent resistance of such a circuit would need to be a fair bit more complicated. I suspect that you'd have to come up with all possible distinct combinations and prove it for each case. This could be like the old Map Colouring Theorem. Otoh, someone here may have a smart solution - it sometimes happens.
Let's wait - but don't hold your breath. :)

 

[Redbelly98]

My proof was meant to be very general -- applying to any combination of resistors, and not limited to series, parallel, or combinations of series & parallel.

It's not clear to me that the OP intended to limit the discussion to series & parallel -- but if so, it still seems reasonable to consider more general networks too.

 

[sophiecentaur]

My proof was meant to be very general -- applying to any combination of resistors, and not limited to series, parallel, or combinations of series & parallel.

It's not clear to me that the OP intended to limit the discussion to series & parallel -- but if so, it still seems reasonable to consider more general networks too.

I am old fashioned enough to rely on Maths to give me a definitive answer and for the Logic of Maths to point me in the right direction. Without some sort of proof / derivation that works with 'Maths', I can't be happy. A proof that comes up with an A>=B type of answer is necessary for me (and a lot of people) to rely on a statement.

As for networks with not just series and parallel components, how would you propose to take a delta of resistors and re-arrange them validly to satisfy the simple Parallel to series swap? I was limiting my answer to something that I can actually demonstrate to be true. If someone can arrive at a simple inequality expression that's the equivalent to the one I was working with then that's fine. Talking in terms of Volts for a given current is actually precisely the same as comparing resistances and doesn't have an easier solution, imo.

 

[Redbelly98]

I am old fashioned enough to rely on Maths to give me a definitive answer and for the Logic of Maths to point me in the right direction. Without some sort of proof / derivation that works with 'Maths', I can't be happy. A proof that comes up with an A>=B type of answer is necessary for me (and a lot of people) to rely on a statement.

This alternative wording of my proof, which includes math, may be more to your liking:

Consider an arrangement of resistors, with a fixed current I₀ passing through the arrangement. Suppose this arrangement is NOT a series connection of all resistors in the given set.

Next, consider some path through the network, and sum up the voltages across each resistor along this path. Since the resistors are not all in series, the path must omit at least one resistor (*** see note below). Moreover, each resistor in the path has at most the applied current going through it. Thus, the sum of the voltages -- and therefore the resistance -- will be less than that of a complete series arrangement.

In mathematical terms:

Note that I_i ≤ I₀ for all resistors, where I_i is the current in the i-th resistor of the non-series arrangement.

Now consider the equivalent resistance,

R_equiv = ΔV / I₀ = ∑(R_i⋅I_i) / I₀
(the summation is calculated using only the resistors along the chosen path)

and compare that to the resistance in a completely series arrangement,

R_series = ΔV_series / I₀ = (∑R_i⋅I₀) / I₀
(this summation is calculated using every resistor, since the only path in the series arrangement is the one that passes through each resistor)

For every term in the sum used to calculate R_equiv, there is a term of equal or greater value in the sum used to calculate R_series. And there will be additional terms in the R_series summation, all positive, so that

R_equiv ≤ R_series

*** Note on paths: While we could make a path pass through every resistor by including closed loops in the path, any loops would have zero voltage by Kirchoff's loop rule and may therefore be removed from consideration.

As for networks with not just series and parallel components, how would you propose to take a delta of resistors and re-arrange them validly to satisfy the simple Parallel to series swap?

I'm not sure why you say we are restricted only to do parallel-to-series swaps.The title of the thread is very general: "Obtaining a maximum resistance given a set of resistors". I take that to mean all possible combinations are under consideration. We only need to compare the delta to a completely series arrangement. It's not necessary that we be able to get there via doing parallel-to-series swaps.

I was limiting my answer to something that I can actually demonstrate to be true. If someone can arrive at a simple inequality expression that's the equivalent to the one I was working with then that's fine. Talking in terms of Volts for a given current is actually precisely the same as comparing resistances and doesn't have an easier solution, imo.

I find it easier to think in terms of comparing voltages -- I guess because it's a more fundamental concept than resistance -- and then dividing by the current in the end to compare the resistances.

 

[sophiecentaur]

Actually, that is totally fine. It was the "path through" bit that got my vote. Cheers.