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Equivalent Resistance: Series or Parallel Triangles?

  1. Apr 4, 2012 #1
    1. The problem statement, all variables and given/known data
    1jlfr.jpg

    Okay. I thought I knew how to do these type of questions, but here goes. The node at a is attached to the positive terminal of a voltage source and c is attached to the negative. I'm completely bemused as to how to reduce this circuit into a single-resistor equivalent.

    I thought that resistors ab and ae could be reduced into a series resistor, but my study partner says they are in parallel because the current does not go through them one after another but divides at a. However, they can't be reduced to parallel resistors because of the set-up of the junctions (or can they?) I'm actually stumped by how to approach this problem.

    2. Relevant equations

    1/RP=1/R1+1/R2...

    RS=R1+R2...

    R|| = R1R2/R1+R2

    3. The attempt at a solution

    The circuit has already been reduced from a more complex one and I've tried searching through the questions posted in homework for inspiration. Any help would be much appreciated!
     
  2. jcsd
  3. Apr 4, 2012 #2

    gneill

    User Avatar

    Staff: Mentor

    Hi sparkleshine, Welcome to Physics Forums.

    This is one of those cases where there are no resistors that are in parallel or in series, so progress is blocked if you stick to the those two methods.

    Fortunately there are several possible ways to proceed. One way would be to assume some voltage V between terminals a and c as you suggest and then write and solve the KVL loop equations to find the current I that the source drives through the network. Then Req = V/I.

    A second way forward is to employ what is called a Delta-Y transformation on one of the groups of resistors that's laid out in the form of a Delta (Δ). So, look up Delta-Y and Y-Delta transformations.
     
  4. Apr 5, 2012 #3
    Thank you very much! :)
     
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