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Beginner Circuitry Analysis - Assistance Appreciated =)

  1. Mar 5, 2014 #1
    HrEiw3y.jpg


    I'm given that i1=5.4A and i2=1.5A. I know that I need to apply mesh-current and then nodal-voltage analysis, but I have never dealt with two current sources. It seems like however I set it up, I am missing a piece.

    My first attempt at a KVL equation using mesh analysis incorrectly established that the voltage differences V1 and V2 both had to be zero. Unless there is a voltage difference over a current source, I do not know where to start a correct solution.

    Here was my first set of equations in full, with my currents numbered 1-4, left to right, in the clockwise direction.

    KVL 1) (i1-i2)(40)=0
    KVL 2) (i2-i1)(40)+(i2)(80)+(i2-i3)(80)=0
    KVL 3) (i3-i2)(80)+(i3-i4)(120)=0
    KVL 4) (i4-i3)(120)=0
     
    Last edited: Mar 5, 2014
  2. jcsd
  3. Mar 5, 2014 #2

    donpacino

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    current source #1 is defined as i1
    current source #2 is defined as i2

    you redefined i2 as i4. in the future it would be better practice to call the current from left to right i1,i3,i4,i2...

    now that being said you are making this very difficult to solve. by adding the currents i3 and i4, you are adding extra variables to solve.

    If i were you, I would writ 2 KCls, then use the resulting equations to solve for v1 and v2
     
  4. Mar 5, 2014 #3

    FOIWATER

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    You won't need mesh and nodal, just use nodal
     
  5. Mar 5, 2014 #4
    Am I on the right track now? I took the 'just use nodal-voltage analysis' advice.

    pEFlLfE.jpg

    For Va) [(Va-Vb)/8]+[Va/40]-[5.4]=0
    For Vb) [(Vb-Va)/8]+[Vb/80]+[Vb/120]+[1.5]=0

    Solving that, Va=96V and Vb=72V

    Solving for V1 and V2 in the original problem, I again get V1=96V and V2=72V. This seems right to me. The final circuit satisfies KCL and KVL for all loops and nodes. Can somebody please verify for me?
     
    Last edited: Mar 5, 2014
  6. Mar 5, 2014 #5

    donpacino

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    Your method is correct!

    I did not check the arithmetic.
     
  7. Mar 5, 2014 #6
    Alright, I'll double check the arithmetic because that is certainly not your job :)

    Thank you to everyone.
     
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