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Nodal analysis, setting up the equations

  1. Sep 14, 2010 #1
    1. The problem statement, all variables and given/known data

    [PLAIN]http://img844.imageshack.us/img844/8508/circuit.png [Broken]

    2. Relevant equations
    KCL set ups


    3. The attempt at a solution
    I am having trouble setting up the KCL equations

    The 30V voltage source on the left with the resistor is messing me up, how do I incorporate that?

    Do I just do 30/15 and assume a 2A current source is going into node A?

    Here are my equations:

    VD = reference node = 0 V

    KCL @ Node B
    (VB-VA)/25 + (VB-VC)/50 - 1 + 4 = 0

    KCL @ Node C
    (VC-VD)/50 + (VC-VB)/50 + 1 = 0

    KCL @ Node A?
    (VA-VB)/25 + (VA-VD)/31.25 - 4 + ?? = 0
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 14, 2010 #2

    ehild

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    Homework Helper
    Gold Member

    You have 30V at the other end of the 15 ohm resistor. So the current through it is (VA-30)/15.

    ehild
     
  4. Sep 14, 2010 #3
    Ok, then the entire equation for voltage at node A is:

    (VA-30)/15 + (VA-VB)/25 + (VA-VD)/31.25 - 4 = 0?

    After I solve for all the voltages at the nodes, how do I find power at the independent voltage a current sources?

    Do I just use voltage drops?
    IE, Power at 4A = (VA-VB)*4
     
  5. Sep 14, 2010 #4

    ehild

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    Homework Helper
    Gold Member

    This is correct.

    Yes, knowing the potential difference U across the terminals of a source and the current I flowing through it, the power is U*I.


    ehild
     
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