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Basic equivalent resistance problem

  1. Oct 3, 2012 #1
    1. The problem statement, all variables and given/known data
    Find Rab
    https://sphotos-a.xx.fbcdn.net/hphotos-ash4/199915_496321627054024_1130826860_n.jpg [Broken]

    2. Relevant equations

    Req = R1R2 / R1+ R2

    3. The attempt at a solution
    bottom 2Ω resistors in series ---> Req1 = 4Ω

    top right 12 and 4 Ω in parallel ---> Req2 = 3Ω

    this 3Ω and the left top 6Ω are in parallel ---> Req3 = 2Ω

    2 resistors of 4Ω bottom left are in parallel -->Req4 = 2Ω

    Req4 and Req3 are in series --->Req5 = 4Ω

    Req5 and Req1 are in parallel ----> Req6 = 2Ω

    Rab = 3Ω + 2Ω = 5Ω

    please let me know if I have gotten the correct answer.
    Thanks for your time.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Oct 3, 2012 #2

    CWatters

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    I also make it 5 Ohms slightly different way..

    [{(6//12//4) + (4//4)} // {2+2}] + 3

    [{2 + 2} // {4}] + 3

    [4//4] +3

    2 + 3 = 5
     
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