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Mesh-current method question

  1. Sep 7, 2014 #1
    1. The problem statement, all variables and given/known data
    Use the mesh-current method to find v0.
    Find the power delivered by the dependent source.

    2. Relevant equations
    v1+v2+..vn=0
    i=v/R
    p=iv

    3. The attempt at a solution
    eq #1)-i1*10+i2*20=25V
    eq #2) i2*20+i2*24+6iΔ=0
    eq #3) iΔ*14=21

    eq#3) iΔ=1.5A
    eq#3 sub into eq#2) i2=0.204A
    i2 sub into eq#1) -1.6A

    Power delivered=
    p=iv
    p=0.204A*(6*1.5)V = 1.839 W

    I posted the solution below. I don't understand how they got it. They added up resistances, and also said i1*20 in the first equation, but i2*20 in the second. Also, I'm not sure how to get the polarities when the + and i signs aren't given. For example, the 24Ω resistor.
     

    Attached Files:

  2. jcsd
  3. Sep 8, 2014 #2

    NascentOxygen

    User Avatar

    Staff: Mentor

    You are not applying mesh analysis correctly. The currents denoted as i1 and i2 are loop currents, this means each flows in a closed loop, so it's best that you show them doing so by marking in each loop current as a complete circle even though the sample answer has set a poor example by not doing so.

    When done properly, you can see that the current in the 10Ω resistor is i1, and the current in the 20Ω resistor is the sum of two currents: i1 down and i2 upwards, making it (i1 - i2) total. So the voltage across the 20Ω resistor is going to involve that current, the resistor's total current.

    Make a fresh start. Ignore the calculations shown in the sample answer, for the time being.
     
    Last edited: Sep 8, 2014
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