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Find i1 (KCL)

  1. May 27, 2013 #1
    1. The problem statement, all variables and given/known data
    12.jpg

    Find i1

    2. Relevant equations
    KCL
    V=IR


    3. The attempt at a solution
    See picture. What am I doing wrong?

    Also, the current through 7 ohms = 2+i1.
    So for V1 (left loop), can it be written as V1=-5-0.5i1+7(2+i1)?
    Then V1=-5-0.5i1+14+7i1=9+6.5i1
    For V1 (right loop), can it be written as V1=5-10i1?

    Many thanks in advance!
     
    Last edited: May 27, 2013
  2. jcsd
  3. May 27, 2013 #2

    gneill

    User Avatar

    Staff: Mentor

    I don't understand the node equation in your diagram; Why doesn't V1 appear in it? V1 should be the unknown potential at the node that determines the currents in the branches.
    It doesn't look like you've assigned the correct signs to the potential changes with respect to the reference node. i1+2 flowing through the 7Ω resistor in the direction indicated should cause a potential drop, but the 5V source must cause a +5V potential rise. Similarly, the controlled source makes another potential rise of 0.5*i1 on the way from the reference node to V1.
    Nope. Pay attention to the direction with which i1 passes through the 10Ω resistor. What's the resulting polarity of the potential change?
     
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