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Find current with source transforms

  1. Oct 13, 2016 #1
    1. The problem statement, all variables and given/known data
    Mgf1aK7.png

    2. Relevant equations
    V=IR

    3. The attempt at a solution
    First I combined the -j3 capacitor and 4ohm resistor to get:
    [tex]\frac{32-j48}{25}[/tex].
    Then I combined that with the j4 ohm inductor to get:
    [tex]\frac{32+j52}{25}[/tex].
    Then I combined that with the -j2 ohm capacitor to get:
    [tex]\frac{104-72j}{36+j2}[/tex]
    My circuit now looks something like:
    2B5XjUC.png

    I then transform the Current Source into a voltage source:
    3<30° -> [tex]\frac{3\sqrt{3}}{2} + 1.5i[/tex]
    Z total = [tex]\frac{36-28j}{13}[/tex]
    V=I * Z = [tex]\frac{84+103\sqrt3}{26} + \frac{54-84sqrt3j}{26}[/tex]

    I now have something like this:
    upload_2016-10-12_22-45-51.jpeg
    Have I been doing everything right so far? How do I find I?
     
  2. jcsd
  3. Oct 13, 2016 #2

    gneill

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    Staff: Mentor

    You might want to check that result.

    As you "absorb" components you must modify the source value, too. Effectively you are converting between current and voltage sources as you progress, doing a Norton → Thevenin → Norton shuffle.
     
  4. Oct 13, 2016 #3
    What is wrong with the result? Am I not supposed to combine those two parts?
     
  5. Oct 13, 2016 #4

    gneill

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    Staff: Mentor

    No, combining them is fine. Check your calculated result.
     
  6. Oct 13, 2016 #5
    The two steps in red are where you made your mistake. You must combine the j4 and -j2 in series, and then that equivalent combines in parallel with what you have up to that point.
     
  7. Oct 13, 2016 #6
    upload_2016-10-13_11-54-8.png
    this is what my circuit looks like now, with Z3 = (96-72j)/(36+2j)
    Everything good?
    how do I solve for current I
     
  8. Oct 13, 2016 #7

    gneill

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    Staff: Mentor

    Not good. Your current source has to evolve as you absorb the components. Think of it as taking successive Thevenin or Norton equivalents as you go.

    You started by reducing the -j3 capacitor and 4 Ohm resistor to a single impedance, then took it and the current source as a Norton model. All good so far. Then you wanted to absorb the j4 inductor. To do that you have to convert the Norton circuit into its Thevenin equivalent first. So the Norton impedance becomes the series impedance for the Thevenin model and you also have to convert the current source to a Thevenin voltage by multiplying it by that impedance. Then you can include the j4 inductor. The same thing goes for absorbing the -j2 capacitor. You need to convert your new Thevenin model back to a Norton model (source current and parallel impedance). The original current source will be "buried" behind a set of multipliers that it accumulates during the switching back and forth between model types.

    Your first change of source goes like this:
    upload_2016-10-13_15-37-12.png
     
  9. Oct 13, 2016 #8
    upload_2016-10-13_13-11-10.png

    I = 4.47515-2.2127j
    Z = (4-72j)/(36-48j)

    To get the current, can I use a current divider here or do I need another conversion?
    I ended up getting
    I = 1.32 - 1.44i
     
    Last edited: Oct 13, 2016
  10. Oct 13, 2016 #9

    gneill

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    Staff: Mentor

    Practically speaking you can do it either way. If they are expecting you to use source changes though, you might want to follow through to the end.

    You should check your results using another method when you're done. Perhaps using Nodal analysis.
     
  11. Oct 13, 2016 #10
    I ended up getting
    1.32 - 1.44i Amps
     
  12. Oct 13, 2016 #11

    gneill

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    Staff: Mentor

    Not what I'm seeing. I did the whole series of source changes down to a final Thevenin model in series with the 2 Ohm load, and I checked using nodal analysis to find the voltage at the output node then determined the current through the load from that. The results agreed, and they weren't the same as your result above :frown:
     
  13. Oct 13, 2016 #12
    My Vth is 3-5.2i
    is that off as well?
     
  14. Oct 13, 2016 #13
    Is it right up to here?
    upload_2016-10-13_13-11-10-png.107410.png

    I = 4.47515-2.2127j
    Z = (4-72j)/(36-48j)
     
  15. Oct 13, 2016 #14

    gneill

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    Staff: Mentor

    I don't believe so, no. In that configuration your Z should match my final Thevenin impedance and it doesn't.

    Your Z:
    upload_2016-10-13_17-47-48.png

    Mine:
    upload_2016-10-13_17-48-26.png
     
  16. Oct 13, 2016 #15
    So your Z is 2.769 - 2.154j at the point in my picture?
     
  17. Oct 13, 2016 #16

    gneill

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    Staff: Mentor

    That would be my understanding, yes.

    Perhaps we need to go through the transformations one at a time from the beginning?
     
  18. Oct 13, 2016 #17
    I know where I screwed up.
    Final I should be:
    -0.067 - 1.9i A
     
  19. Oct 13, 2016 #18

    gneill

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    Staff: Mentor

    Yes, that looks good. :approve:
     
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