Equivalent Resistance

Discussion in 'Homework Help' started by zyzxzwzvz, Jan 11, 2014.

  1. zyzxzwzvz

    Thread Starter New Member

    Jan 11, 2014
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    I made up this problem, and I have been thinking about it for a few days, but I do not know if it has a solution. Looking at other posts, this problem seems more mathematically oriented. :) Any help would be appreciated.

    Suppose that we have an unlimited supply of resistors with resistances R1 Ω and R2 Ω. Is it always possible to make a circuit with equivalent resistance R1*R2 Ω using only a finite number of such resistors? (Note that R1 and R2 can be arbitrary real numbers, such as π=3.1415...)

    The problem statement seems so simple, but the problem itself is actually quite elusive. I suspect that it is not possible, since, if R1=sqrt(3) and r2=sqrt(12), we need to make a circuit with equivalent resistance 6. A construction of such a circuit using only series and parallel combinations seems highly unlikely (I have tried in vain), and expressions for more complicated circuits get messy quickly.

    If possible, could someone provide such a construction? If not possible, could someone provide a (rigorous) proof or counterexample explaining why it is not possible?

    Thanks!
     
  2. studiot

    AAC Fanatic!

    Nov 9, 2007
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    Welcome to AAC.

    Looking at your profile I don't see where you are coming from or why this is in homework, where there are special rules. In particuar are you coming from an electrical or a maths background?

    As an intellectual exercise you should note that you can never increase a resistance with parallel combinations. only decrease.

    So to obtain a resistance which is a numerical multiple of R1 we must use series.

    So have you explored the strategy of stacking up (R2+n) R1 resistors in series and then looking for a suitable parallel combination to bring the value back down to the desired value?

    Here is a useful formula for this

    If Rd is the desired value and R3 the cumulative value you have,

    the required shunt to R3 to give Rd is

    \frac{{{R_3}{R_d}}}{{\left( {{R_3} - {R_d}} \right)}}
     
  3. WBahn

    Moderator

    Mar 31, 2012
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    When you say that R1 and R2 can be any arbitrary real numbers, that means that they can be R1=1Ω and R2=2Ω. Since both R1 and R2 can be constructed from two of the other resistor, this effectively reduces to a set of resistors that are all 1Ω.

    With such a set of resistors, you can create any resistance that is rational, but no resistances that are irrational.

    You can, of course, come as close to any irrational value as you choose.
     
  4. zyzxzwzvz

    Thread Starter New Member

    Jan 11, 2014
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    Hello studiot, I was not sure about where to post this, but it seemed more of a homework-type question that a general electronics chat question. I come from more of a mathematical background. I am creating a problem set for a class of mine, and this is one of the problems that I thought of.

    As to the strategy in your response, I have tried that, but it does not work for all choices of R1 and R2 (consider letting R1 and R2 be transcendental numbers). However, even though this strategy does not work in all cases, there may be other strategies that work in all cases, such as making delta circuits and whatnot (basically anything but series and parallel).

    Hello WBahn, I guess that the question statement was unclear. Let me rephrase it this way:

    Suppose that we have an unlimited supply of resistors with resistances R1 Ω and R2 Ω. Is it always possible to make a circuit with equivalent resistance R1*R2 Ω using only a finite number of such resistors, no matter the choice of R1 and R2? (Note that R1 and R2 can be arbitrary real numbers, such as π=3.1415...)

    In your post, you let R1 = 1Ω and R2 = 2Ω, so the goal is to create a circuit with equivalent resistance 1*2 = 2Ω. This can easily be achieved; in fact, if al least one of R1, R2 are rational values, it will always be possible.

    Here is such a construction: Let R1 = A/B and R2 = C, where A/B is in simplest terms, and C is not necessarily rational. We wish to create a circuit with equivalent resistance A*C/B. Take A resistors with resistance C and place them in series. Then take B of these series combinations and put them in parallel. This circuit has a finite number of resistors and it is easy to check that it has the desired resistance.

    But the difficulty of the problem arises when both R1 and R2 are not rational. I am not sure of how to proceed from here.

    Sorry, I probably should have put this in the first post.

    Thanks for the replies so far!
     
  5. studiot

    AAC Fanatic!

    Nov 9, 2007
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    OK so you stack up n R1 resistors with m R2 resistors in series and parallel this with another series combo of p R1 and q R2 resistors so that


    {R_1}{R_2} = \frac{{\left( {n{R_1} + m{R_2}} \right)\left( {p{R_1} + q{R_2}} \right)}}{{\left( {n + p} \right){R_1} + \left( {m + q} \right){R_2}}}


    Will this not do it, trancendental or not, if you can find integers n,m, p, q?
     
  6. Tesla23

    Active Member

    May 10, 2009
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    It is not always possible.

    If you scale the problem by dividing all resistors by R1, then the problem becomes to make the resistance

    1*R2 = R2

    from a finite combination of resistors 1 and R2/R1.

    clearly this is not always possible, for example if R2 is irrational and R2/R1 is rational.
     
    zyzxzwzvz likes this.
  7. studiot

    AAC Fanatic!

    Nov 9, 2007
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    Yes this division is a good thought but, if you replace with an equivalent problem where R3 = R2/R1 and you have a box of 1Ω resistors and a box of R3 resistors then the problem is simply resolved by taking one single R3 resistor and zero 1Ω resistors.

    In terms of my integers, n=p=q=0 and m=1.

    My scheme becomes unwieldy when both R1 and R2 are less than 1Ω so their product is smaller than either.
    It may be worth interchanging the words series and parallel in that case.
     
  8. zyzxzwzvz

    Thread Starter New Member

    Jan 11, 2014
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    I believe that Tesla23's solution works; I think the math was a bit unclear, though.

    Basically, if we are trying to make an equivalent resistance of R1*R2 Ω from some R1 Ω and R2 Ω resistors, we can scale everything down by R1. Then the problem becomes to make an equivalent resistance of R2 Ω from some 1 Ω and R1/R2 Ω resistors. Then we may scale up by R1 at the end.

    As Tesla23 mentioned, this is not always possible. Let's make this more rigorous. Take, for example, R1=sqrt(3) and R2=2*sqrt(3)). Then after scaling, we want to make a circuit with resistance 2*sqrt(3) Ω from some 1 Ω and 2 Ω resistors. However, every circuit can be decomposed into series and parallel combinations (see here: http://tinyurl.com/l8r76pm). This ensures that the equivalent resistance can be written as some complex fraction using only 1 and 2 along with division and addition in its expression. This fraction, then, must be rational. But 2*sqrt(3) is irrational, which is a contradiction.

    I think that studiot's strategy could work if it were more general (in particular, it only covers one case; the parallel resistors need not have the same resistance).
     
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