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Kirchoff's Laws Problem - Power

  1. Oct 3, 2011 #1

    KRC

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    1. The problem statement, all variables and given/known data
    At what rate is electrical energy converted to internal energy in the 4.56Ω and the 5.91Ω resistors in the figure?

    2. Relevant equations
    Kirchoff's Laws:
    ƩI(in) = ƩI(out) Junction Rule

    ƩΔV = 0 Loop Rule


    3. The attempt at a solution
    I have set the top branch current to go from right to left = I(3)
    I have set the middle branch current to go from right to left = I(2)
    I have set the bottom branch current to go from left to right = I(1)

    The left middle junction results in the following I(3) + I(2) = I(1)...Junction#1
    Loop #1 = 0 = 6.96V-5.91I(2)-4.56I(3)
    Loop #2 = 0 = -9.56I(1)-5.91I(2)+1.86V

    I get I1=2.7502
    I2=-4.1340
    I3=6.8842

    The answers I get when using the P=I^(2)*R...do not yield results consistent with the online homework

    I have tried this several different ways and the answer according to my homework do not match. Please provide some insight here.

    Thank You.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Oct 3, 2011 #2

    gneill

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    Staff: Mentor

    I think you're not following your own stipulations for current directions. Once you choose them, if you start writing equations you need to follow through.

    When you "pass though" a resistor in the direction of the assumed current flow, you expect a voltage drop. If you pass through against the flow then you get a voltage rise.

    Just a thought: you might find a nodal analysis approach to be easy to apply here.
     
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