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Nodal Analysis

  1. Oct 12, 2016 #1
    1. The problem statement, all variables and given/known data
    Find Va(t) using Nodal Analysis:

    2. Relevant equations
    Zc = 1/(jwC)
    Zl = jw*L

    3. The attempt at a solution
    W = 250
    Z36mH = j9 ohm
    Z80mH = j20 ohm
    Z0.25mF = -j16 ohm (is capacitor impedance always negative? It says so in my instructor's notes but I forgot why)

    20cos(250t) -> 20 <0° V
    1.2cos(250t+45°) -> 1.2<45° A

    Are the following Nodal equations correct? Is Node1 simply Va?
    [tex]\frac{Node1-20<0}{8+9j} + \frac{Node1}{-j16}=0[/tex]

    [tex]-1.2<45° + \frac{Node2-4Va}{j20}=0[/tex]
  2. jcsd
  3. Oct 12, 2016 #2
    Node1 and Node2 are just one node. Nodes are not defined by black dots. Those two dots are connected by a wire of perfect conductivity and the voltages at the two dots are the same, so they are really only one node.
  4. Oct 12, 2016 #3
    I assume Va is the voltage across the Capacitor?
    Last edited: Oct 13, 2016
  5. Oct 12, 2016 #4
    That looks OK; now all you have to do is solve for Va.

    By the way, the reason that capacitor impedance is negative is because 1/j = -j
  6. Oct 13, 2016 #5
    Of course. I feel like an idiot now.
  7. Oct 13, 2016 #6
    I ended up getting
    Va = 4528.6 - 911.4i
    Is that correct?
  8. Oct 13, 2016 #7
    No, I don't get that. As a sanity check, for Va to be 4000 volts plus seems a little high to me.

    What happened to your equation in post #3?

    If you'll show your work, we can probably find your error.
  9. Oct 13, 2016 #8
    Don't know what happened to it

    I plugged the equation into MyAlgebra.com and it spit out the answer. Here is the equation again:
    [tex]\frac{Va-20}{8+j9}-\frac{Va}{j16}-\frac{3Va}{j20} = \frac{1.2\sqrt{2}}{2}+ \frac{1.2\sqrt{2}}{2}i[/tex]
  10. Oct 13, 2016 #9


    User Avatar

    Staff: Mentor

    It's not what I get, if that helps :smile:
  11. Oct 13, 2016 #10
    I tried another calculator and got:
    1.89293-12.2816 i
    is that correct?
  12. Oct 13, 2016 #11


    User Avatar

    Staff: Mentor


    Sometimes plugging in all the complex values into a node equation and lugging them around in the reduction is more tedious and error prone than using symbols. Why not assign variables to the branch impedances and the current and solve? That way we can follow your steps.
  13. Oct 13, 2016 #12
    I read the output wrong, it was 1.89293-12.2816 i.
  14. Oct 13, 2016 #13
    I think you may be having a problem with your complex constant j. In Mathematica there is a difference between j9 and j 9.

    Without a space between the j and the 9 (in other words, like this j9), Mathematica treats that as a variable name. With a space, like j 9, that is treated like j*9, the product of j and 9. Also at the very end of your equation you have i instead of j.

    What does MyAlgebra use for the complex constant, j or i? Whichever it is, put the right one in your equation and instead of j9 and j20, etc., put j*9 and j*20 (or possibly i*9 and i*20 if i is the symbol for the complex constant) and run the equation again.
  15. Oct 13, 2016 #14
  16. Oct 13, 2016 #15
    This is correct.
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