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Circuit (equivalent resistance)

  1. Aug 17, 2015 #1
    1. The problem statement, all variables and given/known data
    Find the equivalent resistance
    Aib5Oqj.jpg
    2. Relevant equations
    Req(series) = R1 + R2...
    Req(parallel)=(1/R1+1/R2...)^-1
    3. The attempt at a solution
    I'm having problems understanding how to simplify this circuit.
    There should be parallel and series "loops" to simplify but I'm having trouble seeing them.
     
    Last edited: Aug 17, 2015
  2. jcsd
  3. Aug 17, 2015 #2
    Mark the junctions where potential is a . Consider a random voltage V , which is the potential in between the 2Ω and 3Ω resistors .

    Now draw a fresh diagram , now marking a , b and V , and connecting resistors between the three appropriately . This should help in simplifying the circuit , making it easy to see the series and parallel connections .

    Hope this helps .
     
  4. Aug 17, 2015 #3
    Updated the image with the resistors numerated in red.

    Do you have to consider potential?
     
  5. Aug 17, 2015 #4
    Yes , the potential is what I mentioned in my previous post .
     
  6. Aug 17, 2015 #5
    What i know is that potential stays the same across parallel sections and is summed up in series.

    How does that help me if i cant distinguish which one is in parallel or in series?

    Would you mind to explain a little further please, I'm kind of new to this.
     
  7. Aug 17, 2015 #6

    CWatters

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    Look at the bottom end of R2...imagine sliding it to the left and around the corner.
     
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