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Nodal Analysis help

  1. Oct 9, 2016 #1
    The problem statement, all variables and given/known data
    pH7je63.png


    Attempted Solution:

    For my two nodal equations, I got:
    40j + 20 - 12Va*j + 2Vb *j +4Va - 4Vb = 0
    and
    2Vb - 2Va - 6Vb*j - 4Va*j-10-20j

    Are these right? How do I solve this system of equations?
     
  2. jcsd
  3. Oct 10, 2016 #2

    gneill

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    Staff: Mentor

    Hi eehelp150. Welcome to Physics Forums.

    An actual problem statement (words) would have been nice. A circuit diagram alone does not comprise a problem statement. A thread title does not comprise a problem statement. A problem statement should inform the reader what the goal is. A circuit diagram is just a circuit. (Yes we're a bit fussy about format here :smile:).

    For Relevant equations you might have listed KVL, KCL and Ohm's law.

    Regarding your node equations, I don't believe that they are correct; I don't get the same equations, and I can't manipulate mine to match yours.

    Can you show some detail regarding how you arrived at your equations?
     
  4. Oct 10, 2016 #3
    Use nodal analysis to find Va and Vb

    Node at Va:
    -1 + (Va/[4-j2]) + (Va-Vb)/(-j10)=0
    (-1)(-j10) + [(-j10)(Va)]/(4-j2)] + Va - Vb = 0
    j10 + [(-j10)(Va)]/(4-j2)] + Va - Vb = 0
    j10(4-j2) + (-j10)(Va) + (4-j2)(Va - Vb) = 0
    j40 - 20j^2 - j10Va + 4Va - 4Vb - j2Va + j2Vb = 0
    j40 + 20 - j12Va + 4Va - 4Vb + j2Vb = 0
    Node at Vb:
    (Vb-Va)/(-j10) + Vb/(2+j4) + 0.5<-90°
    (Vb-Va)/(-j10) + Vb/(2+j4) + [0 - 0.5j]
    (Vb-Va) + [(-j10)Vb]/(2+j4) + (-10j)(-0.5j) = 0
    (Vb-Va) + [(-j10)Vb]/(2+j4) + 5j^2 = 0
    (Vb-Va) + [(-j10)Vb]/(2+j4) - 5 = 0
    (Vb-Va)(2+j4) + (-j10)Vb - 5(2+j4) = 0
    2Vb - 2Va - j6Vb - j4Va - 10 - 20j
     
  5. Oct 10, 2016 #4

    gneill

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    Staff: Mentor

    Interesting. Your reduction method differs from what I'm used to but it is correct as far as it goes. So my apologies for doubting you! I think though that your method leaves the expressions more complicated than they need to be.

    What I did was normalize the individual terms first using the complex conjugates:

    Node a:
    ##-1 + \frac{Va}{4 - j2} + \frac{Va - Vb}{-j10} = 0##

    ##-1 + \frac{Va(4 + j2)}{20} + \frac{(Va - Vb)(j)}{10} = 0~~~~~~## Clearing the complex denominators

    ##-20 + Va(4 + j2) + (Va - Vb)2j = 0~~~~~~~~## Multiply through by 20 and lose the denominator

    ##-10 + (2 + j2)Va - jVb = 0~~~~~~~~## Collect terms, divide by 2

    Node b:
    ##\frac{Vb - Va}{-j 10} + \frac{Vb}{2 + j4} - \frac{j}{2} = 0##

    ##\frac{(Vb - Va)j}{10} + \frac{Vb(2 - j4)}{20} - \frac{j}{2} = 0##

    ##(Vb - Va)2j + Vb(2 - j4) - j10 = 0~~~~~~~~~## Multiply through by 20 and lose the denominator

    ##Va + (1 + j)Vb + 5 = 0~~~~~~~~~~~~## Collect terms, multiply through by -j/2

    Presumably we should arrive at the same values for Va and Vb when all is said and done.

    The equations can be solved by the usual methods (but using complex arithmetic). Substitution looks like a good choice.
     
  6. Oct 10, 2016 #5
    I ended up getting:
    Va = -1 + 4/j
    Vb = -4/j
    Are these right?
     
  7. Oct 10, 2016 #6

    gneill

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    Staff: Mentor

    Unfortunately those aren't the values I obtained.
     
  8. Oct 10, 2016 #7
    It looks like you're having trouble with your solution to the equations. If you would post your work, we can give you help; if you don't post anything, we can't help.
     
  9. Oct 11, 2016 #8
    I redid it and got:
    Vb = 2-4j
    Va = 2j - 11
     
  10. Oct 11, 2016 #9
    I think your result for Vb is the negative of the correct value.

    For Va, do you really mean that -11 for the real part; could you have been hit with key bounce? I get -2j + 1, again you also have a sign error.

    Perhaps gneill will verify my result.

    Are you doing all the complex arithmetic by hand? Are you just beginning to study complex arithmetic? If your instructor doesn't require you to do the complex arithmetic by hand, I would recommend getting a calculator that can do complex arithmetic, or using a web application, because it's very easy to make mistakes when doing a lot of complex computations, as you have no doubt discovered lately. :biggrin:
     
  11. Oct 11, 2016 #10
    What calculator do you recommend?
    This is how I'm doing it:

    NodeVB: Va + Vb + jVb + 5 = 0
    Solve for Va: Va = -5 - Vb(1+j)
    Sub that into equation of Node Va
    -10 + 2Va + j2Va - jVb = 0
    -10 + [-10 - 2Vb - 2jVb - j10 - j2Vb - j^2*Vb] - jVb
    -10 + [-10 -4jVb - j10] -jVb
    -20 - j10 = 5jVb
    Vb = 2-4j
     
  12. Oct 11, 2016 #11
    The problem is just what I thought, an arithmetic error.
    You are calculating (-20-j10)/(5j)
    I hope you're not changing the sign of 5j when you brought it over to the left side to divide it into -20-j10 are you?

    In other words, did you do (-20-j10)/(-5j) ?

    Your problem can be broken down into 2 parts: -20/5j = 4j and -j10/5j = -2

    Show how you solved for Va.

    A lot of the modern scientific calculators can do complex calculations. The older TI-86, TI-92, HP48, HP50G can do it.
     
  13. Oct 11, 2016 #12
    I mistaked -25j^2 for -25 instead of 25. My new answer for Vb is now 4j-2
    -20-5jVb-j10 = 0
    -20-j10 = 5jVb
    (-20-j10)/(5j) = Vb
    Vb = [(-20-j10)/(5j)] * [(-5j)/(-5j)]
    Vb = (100j+50j^2)/(-25j^2)
    Vb = (100j-50)/(25)
    Vb = 4j-2
     
  14. Oct 11, 2016 #13
    Correct. Now what is your answer for Va?
     
  15. Oct 11, 2016 #14
    I got 1-2j
     
  16. Oct 11, 2016 #15
  17. Oct 11, 2016 #16
    An alternative to solving equations like yours is to use modern mathematical software, such as Maple, Mathematica, Matlab, Mathcad, etc.

    Here's the solution using Mathematica. First the solution for your original equations:

    CPX1.png

    Here is the solution for gneill's initial nodal formulation:

    CPX2.png

    I also get the same result using my HP50G calculator.
     
  18. Oct 11, 2016 #17
    My exams will not allow computers so I'd much rather work them out by hand for the practice. I will definitely look into getting a good calculator though.
     
  19. Oct 11, 2016 #18
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