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

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

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sophiecentaur

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Have a go (with algebra - not numbers).

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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.

- #4

collinsmark

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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?

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sophiecentaur

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Dramatic -haha!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.

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.)

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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.

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

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sophiecentaur

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

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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.

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sophiecentaur

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

This is all just about a bit of Maths and the inequality is 'unassailable'. (Now there's a challenge)

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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.

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sophiecentaur

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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 text book derivations for those formulae. i.e going back to the basics further.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 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. :)

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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.

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sophiecentaur

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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.

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.

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.

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

Consider an arrangement of resistors, with a fixed current I

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

Now consider the equivalent resistance,

R

(the summation is calculated using only the resistors along the chosen path)

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

R

(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

R

*** 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.

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.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 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.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.

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sophiecentaur

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

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