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Simple Thevenin equivalent resistance problem

  1. Oct 4, 2011 #1
    Gentlemen.

    1. The problem statement, all variables and given/known data
    21e5aa8.jpg

    Calculate the Thevenin equivalent resistance.

    2. Relevant equations

    Rseries - Rtot = R1 + R2 + ... + Rn

    Rparallel - Rtot = 1/(R1^-1 + R2^-1 .... + Rn^-1)


    3. The attempt at a solution

    As the picture shows, The second ohm-meter tells me 5.7 Ohms is the correct answer. I also have this answer given to me in the task.

    But I must be inept at calculating resistances, because I read it this way:

    Rth = (R1||R2) + R3 + (R4||R5)

    Rth = 2.045 + 5 + 3.6

    Rth = 10,645

    Obviously wrong. Even if I disregard R3, I get 5.6045 != 5.7.
     
  2. jcsd
  3. Oct 4, 2011 #2

    gneill

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    Staff: Mentor

    R1 is not in parallel with R2. R3 is in the way! In order to be in parallel they'd have to share exactly two nodes.
     
  4. Oct 4, 2011 #3

    phinds

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    I don't understand the left side of your diagram. What are you doing with a multi-meter set to measure ohms and hooked up to a live circuit?
     
  5. Oct 5, 2011 #4
    Thanks! I'm new to this, and unfortunately terrible at recognizing what's parallel and what's serial. I'll review it again today.


    This picture is just something I threw up real quick for illustration. The left side is largely irrelevant. I just didn't know how to draw terminals, so I felt I had to hook it up to the ohmmeter to illustrate.
     
  6. Oct 5, 2011 #5

    gneill

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    So what's your new plan of attack for finding the equivalent resistance?
     
  7. Oct 5, 2011 #6

    phinds

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    OK, understood, but I'd avise you not to ever do that in a real situation unless you don't care if you burn out the ohmmeter.
     
  8. Oct 5, 2011 #7
    I'm trying to find another way of look at the circuit, perhaps redrawing it to make it more intuitive. I'm also gonna try just brute forcing it and then from that make a new drawing and see if I can understand it.
     
  9. Oct 5, 2011 #8

    gneill

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    You may not be able to "redraw away" the inherent difficulty of not having any series or parallel opportunities to exploit for simplification. When brute force falters, try a little finesse :smile:

    Two suggestions:
    1) Investigate what are called Y-Delta and Delta-Y transformations.
    2) Place a voltage source V where your Ohm meter is and use circuit analysis methods to find the current that this voltage would push into the circuit. Then R = V/I.
     
  10. Oct 5, 2011 #9
    214cwgo.jpg

    Alright, so this circuit redrawing is not equivalent to the original, then.

    Yeah we covered Y-Delta and Delta-Y in lectures, I've just never done it myself. I'm looking into it.

    Thanks a lot by the way.
     
  11. Oct 11, 2011 #10
    I left this work for a while to focus on some other papers I had to hand in.

    But I picked it up today, and I worked it out, by R3, R4 and R5 from a PI into a T, and got the 5.7 Ohm correct.

    Thanks for the help lads!
     
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