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Homework Help: Kirchhoff's law problem

  1. Jan 7, 2010 #1
    1. The problem statement, all variables and given/known data

    Find I1,I2,I3

    2. Relevant equations

    Attached the circuit diagram

    2. The attempt at a solution

    I1 - I2 + I3 = 0

    -120I1 - 60I2 + 0I3 = -1.8

    0I1 -60I2 -20I3 = -1.2

    I1 = 1/150 A

    I2 = 1/60 A

    I3 = 1/100 A

    Is that correct ? The answer is my textbook is different :(
     

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  3. Jan 7, 2010 #2

    tiny-tim

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    Welcome to PF!

    Hi Masterx00! Welcome to PF! :smile:
    Nooo … :redface:
     
  4. Jan 7, 2010 #3
    Thanks :)
    I took the 1st loop clockwise and the second loop anticlockwise, so why the 1st equation is wrong then ?
    Can you please explain further ? :)
     
  5. Jan 7, 2010 #4

    tiny-tim

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    Your 1st equation (I1 - I2 + I3 = 0) doesn't involve loops, it's for a vertex, so it only involves "in" or "out".

    According to the diagram, I1 is in, and both I2 and I3 are out. :smile:
     
  6. Jan 7, 2010 #5
    But, If I made it I1 - I2 - I3 = 0, I1 = 0, I2 = 0.03, I3=-0.03, that sounds correct ?
     
  7. Jan 7, 2010 #6

    tiny-tim

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    No … how did you get that? :confused:
     
  8. Jan 7, 2010 #7
    Kirchhoff's first law says that the current into a juction is equal to the juntion out. So use that on the top middle junction and have another think about that first question (remembering the directions of currents!)

    Could I also ask if the current directions were given or if you chose them?
     
  9. Jan 7, 2010 #8
    By substituting it with the other 2 equations (using the calculator eqn solver):
    I1 - I2 - I3 = 0

    -120I1 - 60I2 + 0I3 = -1.8

    0I1 -60I2 -20I3 = -1.2
     
  10. Jan 7, 2010 #9

    tiny-tim

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    No, I2 and I3 are in opposite directions.
     
  11. Jan 7, 2010 #10
    aha
    Now it gives correct answer (0.02,-0.01,0.03), I though direction of I3 was same like I1, because of the direction of the drawn battery poles :shy:
    Thanks for help tiny-tim :smile:
     
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