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Nodal Analysis on Electrical Circuit

  1. Feb 3, 2013 #1
    1. The problem statement, all variables and given/known data
    Nodal.png

    3. The attempt at a solution
    According to the similar problem
    Node X:
    I1=(22-Vx)/66
    I2=(66-Vx)/44
    Il=Vx/264
    I3=I1+I2+Il
    Node Y:
    I4=Vy/240
    I5=6
    I4=I3+I5

    After the calculations I get 1.08 for Vx, which I think is wrong.

    Thanks
     
  2. jcsd
  3. Feb 3, 2013 #2

    gneill

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    Staff: Mentor

    Yeah, Vx is definitely greater than 1.08V.

    Can you show more of your work? How about writing out the node equations?
     
  4. Feb 3, 2013 #3
    If the direction of I2 is that shown on the picture, then shouldn't it be I2=(Vx-66)/44
    and for I3, I3=(Vy-Vx)/24. If these assumptions are correct, then the calculations are easy, but I think the problem is somewhere in the expressions of the currents.

    Thanks for the reply
     
  5. Feb 3, 2013 #4

    gneill

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    Staff: Mentor

    Yes, they look fine.
    With the expressions that you now have for the currents you should be able to write and solve the two node equations (based on KCL at each node).
     
  6. Feb 4, 2013 #5
    Are the equations for the X node:
    I1+I2+Il = I3?
    and for the Y node:
    I4=I3+I5?
    Because I don't understand how the current direction should be expressed by the Kirchoff's law.
    which will made the equations look like these:
    ((22-Vx)/66)+((Vx-66)/44)+(Vx/264)=((Vy-Vx)/24)
    and
    Node Y:
    Vy/240=((Vy-Vx)/24)+6
     
  7. Feb 4, 2013 #6

    gneill

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    Staff: Mentor

    If you choose to assume the current directions as depicted in the diagram, then some are flowing into the X node and some are flowing out. According to KCL the sum of the ones flowing in must balance the sum of those flowing out.

    Regarding the math of current directions, suppose that you had a couple of nodes A and B with potentials Va and Vb as follows:

    attachment.php?attachmentid=55378&stc=1&d=1360016286.gif

    The current flowing from node A into node B would be given by ##Iab = \frac{Va - Vb}{R}## .

    The current flowing from node B into node A would be given by ##Iba = \frac{Vb - Va}{R}## .

    In other words, you specify the assumed current direction by the order in which you take the difference in the potentials.
     

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    Last edited: Feb 4, 2013
  8. Feb 5, 2013 #7
    I2+Il=I1+I3
    I3+I5=I4
    Are these the correct equations for the currents?
     
    Last edited: Feb 5, 2013
  9. Feb 5, 2013 #8

    gneill

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    Staff: Mentor

    Assuming the current directions as depicted in the diagram, the first equation looks okay but the second equation does not look okay. Note that both I3 and I4 are shown flowing OUT of node Y. For I5 (which isn't defined on the circuit diagram) you can just use a constant 6 Amps.
     
  10. Feb 5, 2013 #9
    Is it possible Vx to be 160.3V and Vy to be 276.63V?
    I found plausible values for the currents:
    I1=(22-160.3)/66=-2.095A
    I2=(160.3-66)/44=2.143A
    I3=(276.63-160.3)/24=4.847A
    Il=160.3/264=0.607A
    I4=276.63/240=1.152A
     
  11. Feb 5, 2013 #10

    gneill

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    Staff: Mentor

    Yes, it's possible. In fact it's more than likely :smile:
     
  12. Feb 5, 2013 #11
    Thank you for the help. :)
     
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