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Kirchoff's Tension Law, resistor circuit.

  1. Feb 27, 2012 #1
    1. The problem statement, all variables and given/known data

    Find I1, I2, I3, I6 (intensities at R1, R2, R3 and R6) using Kirchoff's Tension Law.

    R1 = 1/3 Ohm, R2 = 1/4, R3 = 1/3, R4 = 1/2, R5 = 1, R6 = 1/4
    E3 = 32V, E1 = 2V, E2 = 4V.

    http://dl.dropbox.com/u/9301772/Capture.PNG [Broken]


    3. The attempt at a solution

    The schematic has been vastly simplified, and I haven doubts about my simplifications. Anyway, I keep getting the following system, which yields me very weird negative answers.

    If you notice, I1 = I2+I6 (see schematic)

    Left loop: 2 = R1 I2 + R1 I6 + R2 I2 + 4
    Right loop: 4 = (R3+R4) I3 + R5(I3+I6)
    Upper loop: 32 = -(R3+R4) I3 - R2 I2 + R6 I6

    Which yields negative and very high values for I2 and I3.

    Any help would be GREATLY appreciated,
    Wassim H.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Feb 27, 2012 #2

    gneill

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    Staff: Mentor

    Hi Whoracle, welcome to Physics Forums.

    I've taken a quick look at your loop equations and they appear to be okay.

    What do you consider to be "very high values"?
     
    Last edited by a moderator: May 5, 2017
  4. Feb 27, 2012 #3
    I2 = -25.666
    I3 = -19
    I6 = 39

    Pretty sure they're wrong. I tried solving with a software called Solve Elec (just gives me the values, and it gives me more reasonable ones (all positive and under 10A).
     
  5. Feb 27, 2012 #4

    gneill

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    The values match my own calculations for the circuit as drawn and with the component values indicated. I used mesh analysis (a form of loop analysis), and didn't depend upon your loop equations. So, two methods at least give the same result....

    I'm not familiar with Solve Elec.
     
  6. Feb 27, 2012 #5
    I'm starting to have doubts about my simplifications. I'll give you the original circuit in a minute.
    (Iy = 6A, Iz = 12A)

    I used the Norton equivalence to replace current sources with tension sources.

    http://dl.dropbox.com/u/9301772/original.PNG [Broken]
     
    Last edited by a moderator: May 5, 2017
  7. Feb 27, 2012 #6

    gneill

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    Ah. Once you replace a given set of components with a circuit equivalent, even if the resistor values are numerically the same they are no longer the same components. There's no guarantee that the current carried by the R in a Norton equivalent will be the same as the current through the same valued R in its Thevenin equivalent.

    In such a case a way to proceed is to use your simplified circuit to find parameters that are identical in both circuits, such as node voltages or currents through components that are left unchanged. Then use those values to find the missing bits in the original circuit.

    EDIT: Also, I see that your R3 should have ended up in series with the 4V supply that replaces iz, and not in series with R4.
     
    Last edited: Feb 27, 2012
  8. Feb 27, 2012 #7
    I think we're on to something here ! I'll try it and let you know. How did I not see that D: !
     
  9. Feb 27, 2012 #8
    I2 = -30.08
    I3 = 2.97
    I6 = 43.68

    ... I don't get it.
     
  10. Feb 27, 2012 #9

    gneill

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    Without seeing what happened there's not much I can comment on....

    Did you try a node equation approach? It seems that all of the resistances in the circuit have been chosen so that they would make nice round-figure conductances.
     
  11. Feb 27, 2012 #10
    Sorry for the late reply, but I actually found the right values :) I had just miscalculated.

    The next question is finding it using the node method, as you said. Something I'm not very good at it. I'll keep you updated !
     
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