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PGRE | Resistor Combination Circuit

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  1. Nov 8, 2011 #1
    1. The problem statement, all variables and given/known data
    Hello forum, please see the attached image.

    This problem amounts to finding the equivalent resistance for the circuit and using Ohm's law to solve for the current.


    2. Relevant equations

    V=IR

    3. The attempt at a solution
    What I really need help with is learning how to recognize series and parallel combinations in complicated circuits like this one. It's difficult for me to see any possible combination, series or parallel, in this circuit. It looks to me that if you follow the current down from the top of the resistors, that the first three are definitely parallel, but then those are also in parallel with the others so I really don't see how you're supposed to combine them.
     

    Attached Files:

  2. jcsd
  3. Nov 8, 2011 #2

    gneill

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    Staff: Mentor

    Your confusion is not without good reason: there are no explicit series or parallel resistor opportunities to exploit in the circuit as it is.

    For two components to be in parallel, their leads must share exactly two nodes; you should be able to trace a continuous loop through both components without passing through any other component. No such opportunities exist in the diagram.

    For two components to be in series, one lead from each component must share the same node alone; no other connections are permitted to that node. Again, no such opportunities present themselves in the diagram.

    What you can do, though, is take advantage of the symmetry that exists in the circuit. If two separate nodes have the same potential, then any component connected between those nodes will carry no current and that component can be removed or shorted without affecting the circuit operation! Can you spot any candidate components for removal?
     
    Last edited: Nov 8, 2011
  4. Nov 8, 2011 #3
    I had not thought of that before. So I think I can eliminate 2 resistors using this method which I crossed out in the "Equipotential nodes" image I have attached. I can do this since the potential difference between nodes 1 and 2 is the same as the potential difference between nodes 1 and 3 and 1 and 4 correct? Therefore, nodes 2, 3, and 4 should all be at the same potential allowing me to remove the resistors in between them. Then I end up with the reduced circuit I have drawn in the final attached image. Which is now a combination of two sets of parallel resistors. 1, 2, and 3 can be added in parallel and the same goes with 4, 5, and 6. Which leaves me with 2 resistors in series.

    Thank you so much gneill.
     

    Attached Files:

  5. Nov 8, 2011 #4
    I am not confident about doing this one but I agree with gneill.
    Look at the symmetry, I think a lot can be ignored (discounted) because of the symmetry.
    Half way is half way !!! It does not sound rigorous but that would get me started.
     
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