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Resistive ladder of infinite height

  1. Mar 5, 2012 #1
    1. The problem statement, all variables and given/known data
    Identical resistors, R, make up the legs and rungs of a resistive ladder of infinite height. Find Req (equivalent resistance) of ladder, measured between points A and B.


    2. Relevant equations



    3. The attempt at a solution

    Here's my thought process:
    First make all the parallel resistors (the rungs) into one equivalent resistor. Since it's infinitely tall, this would be R^infinity/(infinity*R)
    Second, find the total Req by adding the equivalent resistance for the rungs to the sum of the resistors on the legs. This would make Req=R^infinity/(infinity*R)+(infinity*R)

    This makes Req equal to infinity.
    But the answer is Req=(1+sqrt(3))*R

    Is my process for setting this up wrong?
     

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  2. jcsd
  3. Mar 5, 2012 #2

    gneill

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    Staff: Mentor

    It's at best specious :smile: What justification do you have for "making all the parallel resistors (the rungs) into one equivalent resistor"? They are clearly not in parallel since their leads are not all directly connected to each other.

    Also, doing math with infinities is not a straightforward process. Best avoided!

    Instead, consider what would happen to Req if you were to add an additional ladder stage at AB (thus creating a new bottom end, say A' B'). How would the equivalent resistance of the infinite ladder be affected?
     
  4. Mar 6, 2012 #3
    If you added another ladder stage, wouldn't the Req just increase?

    Also, I'm not sure I understand what you meant by :
    So if the rungs were in parallel, there wouldn't be any resistors in the legs (above the first rung)?
     
  5. Mar 6, 2012 #4

    gneill

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    Staff: Mentor

    By how much would you expect the total to increase if you add 1 to infinity?

    The ladder with equivalent resistance Req already has an infinite number of ladder segments. What do you suppose will happen if one more is added?
    Components are in parallel if they share exactly two nodes for their connections. When I look at that ladder I see a lot more than two nodes!
     
  6. Mar 6, 2012 #5
    If you added one more, then the Req would just stay the same, since there's already an infinite amount.
     
  7. Mar 6, 2012 #6

    gneill

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    Yup. So draw the circuit with the current ladder represented as Req, and add another ladder stage. Solve for the 'new' equivalent resistance...
     
  8. Mar 6, 2012 #7
    So I'm still confused how to find the "original" Req.

    To find the "new" Req, would you set it up so the old Req and the resistor in the rung are in parallel? And then add the two R.
     
  9. Mar 6, 2012 #8

    gneill

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    Staff: Mentor

    The original Req is "Req". It's a variable representing the currently unknown value.
    Yup.
     
  10. Mar 6, 2012 #9
    But is there a way to find that variable in terms of just R?
     
  11. Mar 6, 2012 #10

    gneill

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    That's what you're doing by finding the equivalent resistance with one more stage tacked on.
     
  12. Mar 6, 2012 #11
    Ok, I see now, thank you
     
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