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Resistive circuit - ladder problem

  1. Sep 1, 2014 #1

    Rectifier

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    Hey!
    1. The problem
    Figure shows a very long ladder where every part has a resistanse of R. What is the resistance between A and B if the ladder can be approximated to be semi-infinite (the ladder has a beginning but no end).


    2.The figure
    IOobaQp.png


    3. The attempt
    The first three resistances have a resistance of 3R the next step add 3R in paralell to R. The next step after that provides 3 more resistances in paralell with on of the resistances in step two.

    Step 1: ##3R## [##2R+\frac {3R}{3}##]
    Step 2: ##\frac{3R \cdot R}{R+3R}+2R=2R+\frac{3R}{4}##
    Step 3: ##2R+\frac{3R}{5}##
    Step 4: ##2R+\frac{3R}{6}##
    Step 5: ##2R+\frac{3R}{7}##

    There is a pattern here. Namely
    ##Rab=2R+\frac {3R}{2+n}## for n> 0 and n being integer.

    The answere is 2.73 but i get 2.000... something for high values on n.

    Please help me.
     
    Last edited: Sep 1, 2014
  2. jcsd
  3. Sep 1, 2014 #2

    phinds

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    You might want to recalculate step 3
     
  4. Sep 1, 2014 #3

    NascentOxygen

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    There was one of these questions just recently. The technique for solving it involves drawing a vertical line and saying the resistance to the right of that line equals the resistance seen from terminals A and B. Let's call it M. You can redraw the circuit so it looks much simpler now, and solve for M.
     
  5. Sep 2, 2014 #4

    Rectifier

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    Is this right for step 3?
    ## \frac{41R}{15}##
     
  6. Sep 2, 2014 #5

    Rectifier

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    I am not that I have heared of this technique before. I will try to use it anyway. At what "step" should I draw the line?

    Btw, is there a name for it so I can read more about it?
     
  7. Sep 2, 2014 #6

    NascentOxygen

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    Draw it where the network remaining to its right will look exactly like the network you started with.

    I can't think of a good name for it right now.
     
  8. Sep 2, 2014 #7

    ehild

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    We think the way that the resultant resistance X does not change if one unit of the ladder is added or removed from the infinite number of units. See picture of the ladder and the equivalent circuit. Solve for x.

    ehild
     

    Attached Files:

  9. Sep 2, 2014 #8

    Rectifier

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    Thank you for the reply.

    Okay. Lets say I draw a vertical line at step 2 "above" the resistor that is connected in parallell with the rest of the crcuit. Now I am left with 2 identical circuits.

    How do I continue now?
     
  10. Sep 2, 2014 #9

    BvU

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    The technique is called: realizing you have the same situation if you cut off the leftmost three resistors. So apparently the whole thing has the same resistance as 2R + R//(the whole thing).

    // meaning: parallel to in my sloppy notation

    (If you are as lazy as I am, you want to draw the line at the first step where it makes sense, but not so far as to cause more work than absolutely necessary).

    I don't have a good word for the approach: assume you know the answer and work backwards isn't good enough. Someone else ?
     
  11. Sep 2, 2014 #10

    Rectifier

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    Thank you for your replies I will read and comment back in some minutes.

    EDIT:
    I will have a lecture now and I will ask the professor if he has time. Sorry that I am being dumb on this one.
     
    Last edited: Sep 2, 2014
  12. Sep 2, 2014 #11

    NascentOxygen

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    Let's say the total resistance of the circuit on the right is equivalent to a resistance M.

    You can also say that when you add 3 resistors to that resistor M you make a circuit which again has an equivalent resistance M.

    Solve for M. :smile:

    Perhaps this approach could be called a recursive technique! :approve:
     
  13. Sep 2, 2014 #12

    BvU

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    Re technique: at some point I learned to deal with geometric series, e.g.$$S =1+{1\over2}+{1\over 4} + {1\over 8} + \dots$$and the way to tackle them is to distinguish that $${1\over 2}S = {1\over2}+{1\over 4} + {1\over 8} + \dots$$so that $$S = 1+ {S\over 2}$$see the analogy ?
     
  14. Sep 2, 2014 #13

    phinds

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    No, it's not what I get. How about you show your work (in detail, step by step, and DRAW the exact circuit you are analyzing in that step)
     
  15. Sep 2, 2014 #14

    ehild

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    Yes, it is right.
    What is the next?

    Evaluating the results, you have a sequence 3R, 2.75R, 2.733R ... They appear to tend somewhere.:biggrin: The differences get smaller and smaller. At the end, the difference between the last two elements of the sequence is so small that they can be considered the same.
    If you have the nth term, the n+1th is Rn+1=2R+RRn/(R+Rn), and at the end you can consider Rn+1=Rn. What do you get for Rn?

    ehild
     
  16. Sep 2, 2014 #15

    phinds

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    Damn. I guess *I* must have made a mistake. Perhaps you can show me where I went wrong.

    OOPS. NEVER MIND. I just didn't expand it the way you did. We agree.

    https://www.physicsforums.com/newreply.php?do=newreply&p=4839495 [Broken]
     

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    Last edited by a moderator: May 6, 2017
  17. Sep 2, 2014 #16

    ehild

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    I am glad to hear. :smile:

    ehild
     
  18. Sep 2, 2014 #17

    Rectifier

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    This is what my professor told me too. But I dont understand why you get the same resistance when you add stuff :,(.
     
  19. Sep 2, 2014 #18

    Rectifier

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    Thanks for the comment.

    Why not? :eek:
     
  20. Sep 2, 2014 #19

    Rectifier

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    Here it is again :eek:
    Cant get it to add up like that in my head.
     
  21. Sep 2, 2014 #20

    Rectifier

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    I read the thread more careful and found the answer to my question above in a thread above.

    "At the end, the difference between the last two elements of the sequence is so small that they can be considered the same."

    ##R_{ab}=R+R//R_{ab}+R \\ R_{ab}=2R+\frac{R \cdot R_{ab}}{R+R_{ab}} \\ R_{ab}(R+R_{ab})=2R(R+R_{ab})+R \cdot R_{ab} \\ R_{ab} \cdot R+R_{ab}^2=2R^2+2R_{ab}\cdot R+R \cdot R_{ab} \\ R_{ab}^2=2R^2+2R_{ab}\cdot R \\ 2R^2+2R_{ab}\cdot R-R_{ab}^2=0 \\ ##

    Solve for ##R_{ab}## where ##R_{ab}>0##

    ##R_{ab}^2-2 R \cdot R_{ab}-2R^2=0 \\ R_{ab}= R \pm \sqrt{R^2+2R^2} \\ R_{ab}= R \pm R\sqrt{3}##

    ##R_{ab}= (1 + \sqrt{3})R ## since ## R_{ab}= (1 - \sqrt{3})R ## is undefiened.

    Yey! :D It is the right answer.
     
    Last edited: Sep 2, 2014
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