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Equivalent Resistance of a Ciruit

  1. Feb 6, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the equivalent resistance of the circuit as shown in the diagram below; where, R1 = 2 Ω, R2 = 1 Ω, R3 = 2 Ω, R4 = 1 Ω, R5 = 4 Ω, R6 = 4 Ω, and R7 = 4 Ω.

    http://i.imgur.com/OsAs2.gif

    2. Relevant equations


    3. The attempt at a solution

    R3 and R4 are in parallel, found R of those 2 then added that to R2 (in series)

    R6 and R5 in series, added those together

    this is where i get lost

    I added R1 + R7 + (R6+R5) in parallel.

    then added that to the 1st R-eq i found - in parallel

    wrong.

    i'm clearly confused about the relationships

    thanks
     
  2. jcsd
  3. Feb 6, 2012 #2

    PeterO

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    Homework Helper

    The R 5&6 combination [in series with each other] is in parallel to the R2,3,4 combination you established. Then that whole combination is in series with R 1&7.
     
  4. Feb 6, 2012 #3

    PeterO

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    Homework Helper

    It sometimes helps to draw the resistor set up in a straigh line, rather than 3 sides of a square.

    This would start with R1 , then divide to two branches, with R5&R6 on the bottom, and R2, along with a parallel R3&R4 on the top, the the branches re-joining to get to R7.
     
    Last edited: Feb 6, 2012
  5. Feb 6, 2012 #4
    hmm, i never thought to look at it that way.
    thanks guys
     
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