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Node-voltage question

  1. Sep 7, 2014 #1
    1. The problem statement, all variables and given/known data
    Use the node-voltage method to calculate the power delivered by the dependent voltage source.

    2. Relevant equations
    i1+i2+..in=0(KCL)
    i=v/R
    p=i*v
    3. The attempt at a solution
    I posted the solution below, but I'm going to post my own workings, based on just the circuit. I'm confused how they got their answer in the solution.

    My workings:

    v0-160/10Ω+i0/100Ω+150i0/50=0

    I can reduce it to say i0*x=v0*x but I don't see how I could solve for either.

    In their solution,they have (v0-150i0)/50 for the last part. Wouldn't it not be v0, since it's a different source?

    Also, I have two unknowns, why did they only use one equation? I'm unsure of when I have to use one or two or more equations for node voltage questions. In this question, I felt like I would need two equations to solve it, but their solution only has one.
     

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  3. Sep 7, 2014 #2

    ehild

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    The equation is wrong. You miss parentheses. You need to have sum of currents, but what is i0/100Ω? In the middle branch, you have iσ current. The dependent voltage source has emf=150 iσ. The current in the branch on the right the current is io, not the same as iσ. io flows through the potential difference V0-150 iσ.


    ehild
     
  4. Sep 7, 2014 #3
    Okay, I understand it better since I last posted. My equation now is the same as the solution,

    (v0-160)/10 + v0/100 + (v0-150iσ)/50=0

    Once I reduce it to iσ=(-v0/100), how do you solve for iσ or v0? I have only one equation and two unknowns.
     
  5. Sep 8, 2014 #4

    ehild

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    Substitute iσ=(-v0/100) into the first equation. One unknown remains: v0. Solve for v0, use the value to get iσ=(-v0/100).

    ehild
     
  6. Sep 8, 2014 #5
    Where does iσ=(-v0/100) come from? Is it because of the middle?
     
  7. Sep 8, 2014 #6

    ehild

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    The downward current on the middle branch is v0/100. The upward current is denoted by iσ.

    ehild
     
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