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R = PI Ohms

  1. Jun 23, 2004 #1
    Tommy has just ten resistors, a 1-ohm, a 2-ohm, a 3-ohm and so on up to a 10-ohm.

    He wants to wire them up in a combination that gives a total resistance as near as possible to π ohms, that is to say 3.14159265(etc.) ohms.

    He doesn't have to use all the resistors - if he can get nearer to π ohms by only using (say) eight of the resistors, that's fine.

    What's the best he can do?
     
  2. jcsd
  3. Jun 23, 2004 #2

    NateTG

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    There's a lot of possible arrangements, and they're hard to describe.

    For example: getting [tex]\frac{22}{7}[/tex] is possible by using:

    a 3 ohm and 4 ohm resistor in parrelel to get a [tex]\frac{12}{7}[/tex] Ohm resistor
    and a 2 ohm and 5 ohm resistor in parralel to get a [tex]\frac{10}{7}[/tex] Ohm resistor
    and then putting the two in series. That's a pretty good set-up.
    [tex]\frac{22}{7}=3.\bar{1}\bar{4}\bar{2}\bar{8}\bar{5}\bar{7}[/tex]
    so the error is less than .002.

    That leaves 6 resistors unused.
     
    Last edited: Jun 23, 2004
  4. Jun 23, 2004 #3

    AKG

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    Quick guess, I'm sure you can get closer, but of all the possible combinations with only 2 resistors, putting 5 and 8 in parallel gives:

    3.076923077
     
  5. Jun 23, 2004 #4

    NateTG

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    Here's a slightly better answer:
    5 2 Ohm resistors in parralel for a [tex]\frac{2}{5}[/tex] Ohm resistor
    put it in series with a 10 Ohm resistance for a [tex]\frac{52}{5}[/tex] Ohm resistor
    Put it in parralel with two 9 ohm resistors.
    That makes for a [tex]\frac{468}{149} \approx 3.14093[/tex]

    This is a pretty ugly problem...
     
  6. Jun 24, 2004 #5

    Gokul43201

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    Any reasonable way to make 355/113 ~3.1415929 ?
     
  7. Jun 24, 2004 #6

    NateTG

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    Code tags used for fixed spacing.
    Numbers represent resistors of the appropriate value.

    Code (Text):

       |-1-|
     |-|   |---------|
     | |-2-|         |
    -|               |-2--
     | |-4-| |-2-|   |
     |-|   |-|   |-1-|
       |-5-| |-7-|
     
    Please double check my math.
     
  8. Jun 24, 2004 #7
    You used the two-ohm and one-ohm resistors more than once.
     
  9. Jun 24, 2004 #8

    NateTG

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    Oh. I thought it was any 10 resistors with values 1-10...
     
  10. Jun 24, 2004 #9

    verty

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    Somebody should write a quick proggy to brute force it. I will see if I can do that.
     
  11. Jun 24, 2004 #10

    Gokul43201

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    I doubt you can do better than 22/7 without repeats...
     
  12. Jun 25, 2004 #11
    I am assuming that we are using the proper way of calculating electronic resistance otherwise I cannot do this with as much ease.

    The Bob (2004 ©)
     
  13. Jun 25, 2004 #12
    NateTG. I just want to double check my double checking of your calculations. I believe your resistance would be 3.709302326 etc. Ω. Here are my calculations of your answer:

    1Ω and 2Ω in Parellel = 1/1 + 1/2 = 2/2 + 1/2 = 3/2. Take Reciprocal = 2/3
    4Ω and 5Ω in Parellel = 1/4 + 1/5 = 5/20 + 4/20 = 9/20. Take Reciprocal = 20/9
    2Ω and 7 Ω in Parellel = 1/2 + 1/7 = 7/14 + 2/14 = 9/14. Take Reciprocal 14/9

    Then 20/9 + 14/9 + 1 = 20/9 + 14/9 + 9/9 = 43/9

    Then the parellel resistance of the two sides (e.g. the 2/3 and the 43/9):
    1/(43/9) + 1/(2/3) = 1.709302326 etc.

    1.709302326 + 2 = 3.709302326Ω

    Hope I am right in you being right. :biggrin:

    The Bob (2004 ©)
     
  14. Jun 25, 2004 #13
    Out of intrest ceptimus, had NateTG not got it in post 2#?

    The Bob (2004 ©)
     
  15. Jun 25, 2004 #14

    NateTG

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    I knew there was an error.
    It should have been more like:

    Code (Text):

       |-1-|
     |-|   |-1-------|
     | |-1-|         |
    -|               |-2--
     | |-4-| |-2-|   |
     |-|   |-|   |-1-|
       |-5-| |-7-|
     
    Then it would be:
    [tex]\frac{1}{\frac{3}{2}}+\frac{1}{\frac{43}{9}}=\frac{2}{3}+\frac{9}{43}=\frac{43*2+9*3}{43*3}=\frac{113}{129}[/tex]
    So the resistance is the reciprocal:
    [tex]\frac{129}{113}[/tex]
    Then the total resistance is:
    [tex]\frac{129}{113}+\frac{226}{113}=\frac{355}{113}[/tex]

    Unfortunately the condidition was that only one resistor of each value could be used, so the solution wasn't legal.

    I can replace one of the 2 Ohm resistors with a 6 and a 3 Ohm resistor in parralel. That leaves the 8, 9 and 10 Ohm resistors, but even putting the lot of them in parralel gives a resistance that's too large
     
  16. Jun 25, 2004 #15
    No. You can do much better. Here are my best efforts so far using 3, 4, and 5 resistors. By using 6, 7, 8, 9 or all 10 of the resistors, you can get much closer to PI.

    Code (Text):

    3 resistors: 3.158  (error: 1.6 * 10^-2)
    -+--7-----8--+-
     |           |
     +--4--------+

    4 resistors: 3.14102  (error: 5.67 * 10^-4)
    -+--1--+--3--+-
     |     |     |
     +--5--+-10--+

    5 resistors: 3.141553  (error: 4.01 * 10^-5)
    -+--7-----9--+-
     |           |
     +--1--+--3--+
     |     |  
     +-10--+
     
     
    Last edited: Jun 25, 2004
  17. Jun 25, 2004 #16

    Gokul43201

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    hummm...nice.
     
  18. Jun 25, 2004 #17

    NateTG

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    Brute force or Brain Teaser?

    Is there an elegant solution, or should I just dust off my C skills?
     
  19. Jun 25, 2004 #18
    It's brute force, so far as I know. But optimizing the program so that you're not looking at years of run time, is a nice Brain Teaser.
     
  20. Jun 26, 2004 #19
    Can I ask that we post the resistors like NateTG in post 6# pleaes because I just got confused (at a glance) with what ceptimus put in the last few posts (with the code diagram).

    The Bob (2004 ©)
     
  21. Jul 3, 2004 #20
    How are the programs coming along?

    Here is a 6-resistor version:
    Code (Text):

    6 resistors: 3.14159292  (Error: 2.67 * 10^-7)
    -+-------10--------+-
     |                 |
     +--5--+-----+--2--+
     |     |     |
     +--8--+     |
     |           |
     +--7-----9--+
     
    With 9 resistors, I've found 105768/33667

    I'll keep my 10-resistor solution secret for now.
     
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