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Combining several resistors to achieve a specific value?

  1. Jul 30, 2011 #1
    Hi guys,

    I have an interesting and difficult problem as follow:

    Find the smallest number of 12 Ohm resistors (and of course how to connect them) to achieve an equivalent resistor of 7.5 Ohm?

    I have worked on this problem for many hours but cannot find the answer yet.

    Somebody can help me?

    Thanks ^^
     
  2. jcsd
  3. Jul 30, 2011 #2
    try this configuration
    https://docs.google.com/leaf?id=0B2...MDg1Yy00YmUwLWExYTgtZmUyZGVkNWE1Yzlm&hl=en_GB

    here one zig zag line show a resistance.
    How it works.
    combined resistance of first two resistances is 3 and other parllel system is 1.5 it sums to be 7.5
    really it is not so much typical question
     
    Last edited: Jul 30, 2011
  4. Jul 30, 2011 #3
    Hi vkash,

    Thank you for your answer but the value of the resistors is 12 Ohm, not 6 Ohm. Moreover, how can you prove that the combination you suggested has the smallest possible number of resistors?
     
  5. Jul 30, 2011 #4

    PeterO

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    If you add 12 Ohm resistors in Series [1 then 2 then 3 etc], you can get the values 12, 24, 36, 48, 60 etc [not all that useful]

    If you add 12 ohm resistors in parallel [first 2, then 3, then 4 etc You get effective values of 6, 4, 3, 2.4, 2, 12/7 , 1.5, ....

    Can you see how to get 7.5 from the combination of any of those values.

    For example: 7.5 is one eighth of of 60, so if you made 8 equal branches of 60 [5 resistors in series], and connected those 8 branches in parallel you would get 7.5.
    This would also use a lot of resistors, and I am sure it could be done with fewer.
     
  6. Jul 30, 2011 #5
    Hi PeterO,

    Thank you for your answer. I could find a combination satisfying the requirement myself. The difficult part of the problem lies in finding the combination having the smallest possible number of resistors.

    regards,
     
  7. Jul 30, 2011 #6

    PeterO

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    I think you just need "confidence" in your combination. The 8 branch solution I mentioned would take 40 resistors. How many did you use? I can do it a couple of ways with 10.

    It is not as if you are going to apply calculus and set up a max/min problem.
     
  8. Jul 30, 2011 #7
    Oh sorry there are eight resistance in parallel not four.
    there are no mathematical or physical formula to prove this ,that it is smallest combination(as i have read).
    How i solve this.
    required resistance is 7.5 Ohm. It is 6+1.5. and 6 is two parllel resistance and 1.5 is eight parallel resistance. that's it.
    what is it's real answer (in some books answer is written in end of book).
     
  9. Jul 31, 2011 #8
    If it is the answer, then there must be a way to prove it, otherwise the question is meaningless.

    I don't know. This is not an excercise from the books I have.

    You guys have got any other ideas?
     
  10. Jul 31, 2011 #9

    PeterO

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    The alternative to meaningless is obvious.

    How would you go about proving that the minimum number of resistors needed to make a 24 Ohm resistance is 2.
     
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