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Sloving a depentant voltage source problem using node voltage

  1. Feb 24, 2012 #1
    Hi everyone,

    I currently stumped in the process of approaching a dependent voltage source problem using node voltages. Attached is the circuit.

    What I know:

    the Node north of the 60V supply with also be 60V

    there are four nodes, two when supernodes are used


    What I'm confused about:

    with the super node being the 6iψ, KCL states the sums must be zero

    when I try to solve, I end up getting the equation (using v2 as the junction of the dependent, and both 3Ω resistors):

    [(60-v1)/2]+[(V2-V3)/3]+[V2/3]=0
    I am not sure how to calculate the current across the 60Volt source

    :confused:


    In addition, I end up having a recursion problem trying to solve for iψ

    Since V2 is 60+6iψ, and iψ is ([V2-V2]/3), I end up with

    V2=60+6([V2-V3]/3)


    any help would be greatly appreciated

    Joe
     

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  2. jcsd
  3. Feb 25, 2012 #2
    Simply treat A and C as a one big (super) node.
    So we only have one unknown node Vb

    attachment.php?attachmentid=44355&stc=1&d=1330171555.jpg

    Current leaving the node is equal current entering the node.

    For node B

    Vb/24 + (Vb - 60)/2 + (Vb - Vc)/3 = 0 (1)

    And one additional equation

    Vc = Va + 6*Iψ = Va + 6*(Vc - Vb)/3 = 2Vb-Va = 2Vb - 60 (2)

    So we have

    Vb/24 + (Vb - 60)/2 + (Vb - (2Vb - 60))/3 = 0

    http://www.wolframalpha.com/input/?i=+b/24+++(b+-+60)/2+++(b+-+(2b+-+60))/3+=+0

    But we can remove the supernode from the circuit. We you can transform this circuit, legally, and get a new circuit which is equivalent to the old one

    attachment.php?attachmentid=44354&stc=1&d=1330171517.jpg
     

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    • 3.9a.JPG
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    Last edited: Feb 25, 2012
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