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How to make total resistance of 1 ohm?

  1. Nov 30, 2016 #1
    1. The problem statement, all variables and given/known data

    We have 1Ω, 2Ω, 2Ω, 4Ω, 5Ω and 6Ω resistors. We have to make a circuit with all of them and total resistance must be 1 ohm.
    2. Relevant equations
    Parallel: 1/R1+1/R2=1/R
    In series: R1+R2=R

    3. The attempt at a solution
    I have only made a circuit from 4 resistors:
    1 ohm and 2 ohms in series, 6 ohms parallel to them and 2 ohms parallel to them all
     
  2. jcsd
  3. Nov 30, 2016 #2

    BvU

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    Hi Abs, :welcome:

    Unfortunately, that doesn't count as an attempt (more like an arbitrary guess).
    You need a bit more than that; considerations, for example. Like:
    Your last step will be a parallel circuit (because a series will exceed 1 ##\Omega##).
    That parallel circuit will have ##\ge## two sub-circuits, etc. etc.
    Fortunately, I don't have the answer either. :smile: so you must take the first next step

    [edit] With a little trial and error I stumbled on the (or: a ) solution. How far are you ?
     
    Last edited: Nov 30, 2016
  4. Nov 30, 2016 #3
    Oh, I suppose there could be two parallel circuit from 2 ohms and another sub-circuit totally of 2 ohms and so on, I understand how these laws work, but I can't construct such circuit. Just playing on with resistors.
     
  5. Nov 30, 2016 #4

    BvU

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    Yes, one of the considerations is: the 1 ##\Omega## can't be a single parallel branch (total would go under 1 ##\Omega##).

    So the first candidate as a parallel branch would be the 2 ##\Omega##. That reduces the problem to making another 2 ##\Omega## with the remaining 5 resistors (1,1,4,5,6) (since 2 ##\Omega\, \parallel \, 2 \Omega = 1 \Omega ## )

    And so on. You're almost there :rolleyes: !
     
  6. Nov 30, 2016 #5
    And this is where I stop. Is it possible to combine 2 ohms in 2 parallel branches from remaining 5 resistors? Or do we need 3 branches?
     
  7. Nov 30, 2016 #6

    BvU

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    Yes. So in total you end up with three branches: one of 2 ohm.
    The remainder needs to be 2 ohm. Can't be one branch (I think -- why not ?) but two branches is worth investigating
    [edit] turns out to be mistaken. See below .
     
    Last edited: Nov 30, 2016
  8. Nov 30, 2016 #7
    Oh, I've just found an answer. It's (6 ∥ 5+1∥ 4+2) ∥ 2
    Was your solution simpler?
     
  9. Nov 30, 2016 #8

    BvU

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    Brilliant ! Well done.

    My solution was for the wrong problem (1,1,2,4,5,6) o:) instead of (1,2,2,4,5,6). Want to try that too ?
     
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