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KCl and node voltages, need help

  1. Mar 4, 2015 #1
    1. The problem statement, all variables and given/known data

    2dw183.jpg

    2. Relevant equations


    3. The attempt at a solution

    I'm not getting the correct answers, here are my equations

    Node A:
    80 = ix + i2
    80 = Va/0.143 + (Va-(Vb+10))/0.2..........................1

    Node B:
    20 = iy-i2
    20 = Vb/0.125 - (Va-(Vb+10))/0.2..........................2

    Are these equations correct ?
     
  2. jcsd
  3. Mar 4, 2015 #2

    donpacino

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    Gold Member

    equation 1... i see a sign error
    equation 2....same problem
     
  4. Mar 4, 2015 #3
    I don't see where the sign error is
     
  5. Mar 4, 2015 #4

    donpacino

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    Gold Member

    There is an error with your I2 term. Look at the relationship between VB and the voltage source
     
  6. Mar 5, 2015 #5
    I don't know, Is it -10V ?, I'm confused with the signs :frown:
     
  7. Mar 5, 2015 #6

    NascentOxygen

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    Staff: Mentor

    You drop down from Vb by 10v to get to the 0.2 ohm resistor. Should be ....(Vb-10)....

    If you draw an arrow from (-) to (+) on the battery, which you should do and label it +10V, you can see the drop in potential in going from Vb towards the 0.2 ohm. The bottom of the battery is 10v less than the top.
     
  8. Mar 5, 2015 #7
    I still don't get the correct answers. I get Va = 5.65V and Vb = 7.56V
    These are my equations
    80 = Va/0.143 + (Va-(Vb-10))/0.2

    20 = Vb/0.125 - (Va-(Vb+10))/0.2
     
  9. Mar 5, 2015 #8

    donpacino

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    that is what I got too. either we both made the same mistake, or the 'solution' is incorrect.
     
  10. Mar 5, 2015 #9

    NascentOxygen

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    The answers accompanying the question in post #1 are consistent with the 20A source being directed downwards, so are not appropriate for the circuit as drawn.
     
  11. Mar 6, 2015 #10
    Ok, maybe the solutions are incorrect. I'm a bit confused with the second equation though, why is it Vb+10, because the 0.2 ohm resistor is still connected to the negative end of the battery. Shouldn't it be Vb-10 ?
     
  12. Mar 6, 2015 #11

    NascentOxygen

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    Staff: Mentor

    It should be Vb - 10 for the reason I gave.
     
  13. Mar 6, 2015 #12
    For both the equations ?
     
  14. Mar 6, 2015 #13

    NascentOxygen

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    Staff: Mentor

    Yes, it defines the voltage at one end of that 0.2 Ω resistor.
     
  15. Mar 6, 2015 #14
    Ok thanks guys
     
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