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AC Circuit with Source Transformation; find Thevenin equivalent

  1. May 29, 2013 #1
    1. The problem statement, all variables and given/known data


    Use source transformation to find the Thevenin equivalent circuit with respect to terminals, a, b.


    [Broken]



    2. Relevant equations

    Voltage Division: (V in)*(R1/R1+R2)


    Thevenin / Norton / source transformation procedures

    RTh = RNo

    VTh = INo*RNo


    Polar Form conversion:

    Z = (R2 + X2)1/2

    θ = arctan(j / R)


    For Complex numbers:

    j*j = -1

    1/j = -j


    3. The attempt at a solution


    I simplified the circuit to this first:


    [Broken]


    For source transformation, you find VTh by removing the load and then finding the drop from A back to the source right?

    And you find RTh by finding the eq resistance from A to B with the voltage source suppressed right?



    So trying to get VTh for the final answer, I think I do this:

    [Broken]

    because I'll want the drop from A right?




    so for VTh:



    (200∠0deg)V* [(50Ω+60Ωj)/(80Ω-40Ωj)]
    in polar form is


    (200∠0deg)V* [ (78.1∠50.1)/(61.4∠26.5) ]

    = (254∠23.6deg)V


    So I think the drop from A is

    (200∠0)V - (254∠23.6deg)V

    -> VTh = (-54 ∠ -23.6 deg)V



    Now for RTh:


    RTh = [ (1/80Ω + 60Ωj) + (-1/100Ωj) ]-1
    so


    = (100Ωj - 80Ω - 60Ω)/(8000Ωj2j + 6000Ω2j2)


    =[ (40Ωj - 80Ω)/(8000Ω2j - 6000Ω2) ] * [ (-8000Ω2j - 6000Ω2)/-8000Ω2j - 6000Ω2) ]


    = (-320000Ω3j2 - 240000Ω3j + 640000Ω3j + 480000Ω3) / (-64000000Ω4j2 + 36000000Ω4)


    = [(80 x 1043 + (40 x 1043]/(100 x 106Ω4


    RTh = 0.008Ω + 0.004Ωj




    If I did anything wrong, I hope I am on the right track at least?? Thank you
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. May 29, 2013 #2

    gneill

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    Staff: Mentor

    The final results that you've obtained are not good.

    Your first simplification is okay, but you can simplify it further by noting that you can form the Thevenin equivalent that combines the rest of the impedances except the capacitor into a single Thevenin impedance. In the figure below, that would be the Thevenin equivalent for everything left of the red arrow.

    attachment.php?attachmentid=59138&stc=1&d=1369865448.gif

    Then you have a voltage divider situation:

    attachment.php?attachmentid=59139&stc=1&d=1369865541.gif

    and finding the Thevenin equivalent of that should be straight forward.
     

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  4. May 31, 2013 #3
    Wow, I sure missed that hard.

    I still may need more help with source transformation itself.

    To find the Thevenin Voltage now for this, you find the voltage across the top impedance (80Ω + 60Ωj) ?

    So with voltage division that would be

    = 200∠0*(80+60j/80-40j)

    => 200∠0*(100∠36.8 / 89.4∠-26.5)

    = (223∠63.3)V ?

    (Which looks strange since the magnitude of this is higher than that of the source)
     
  5. May 31, 2013 #4

    gneill

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    Staff: Mentor

    Which impedance is a-b across?
    While that's the wrong impedance to be taking the voltage across, it is not strange to find voltages in an RLC circuit that are higher than the source voltage. This is due to resonance effects (energy storage and swapping between the L and C components).
     
  6. May 31, 2013 #5
    Okay, that is helpful to know, even though I did wrong impedance.


    So then the Thevenin voltage is:

    200∠0*(0 -100j/80-40j)

    => 200∠0*(100∠ -90 / 89.4 ∠ -26.5 )

    = (2.23∠-63.5)V ?


    By the way, I had for finding polar form of the -100j capacitor arctan(-100/0), can i just take limit instead so you avoid dividing by 0 (because there is no real resistance component)?
     
  7. May 31, 2013 #6

    gneill

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    Staff: Mentor

    The angle looks good, but I think you've lost a factor of 100 on the magnitude.
    You could, but it's probably easier to simply recognize that any vector on the imaginary axis of the complex plane must have an angle of ±##\pi/2##.
     
  8. Jun 1, 2013 #7
    So the Thevenin voltage is

    = (223∠-63.5)V then.


    This still looks really similar for the other impedance too though.


    The Thevenin resistance (impedance) is still basically finding [ (1/80Ω + 60Ωj) + (-1/100Ωj) ]-1 also, right?
     
  9. Jun 1, 2013 #8

    gneill

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    Staff: Mentor

    Yup. A coincidence due to choice of component values.
    Yup.
     
  10. Jun 2, 2013 #9
    I redid that for the Thevenin resistance now and got nearly the exact same answer as before, (except off by a factor of 10 for the final answer.)


    I got

    RTh = 0.08Ω + 0.04Ωj
     
  11. Jun 2, 2013 #10

    gneill

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    Staff: Mentor

    No, your Rth is not right. It should be much larger in magnitude.
     
  12. Jun 2, 2013 #11
    Oh no, I forgot to divide the ^-1 power out so it was actually 1/(0.08Ω + 0.04Ωj) all along. My bad.


    so

    RTh = 12.5Ω + 25Ωj ?
     
  13. Jun 2, 2013 #12

    gneill

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    Staff: Mentor

    No, that's still to small in magnitude.

    You have (80 + j60)Ω in parallel with -j100Ω. Show your work, one step at a time and annotate it with what you're doing for each step.
     
  14. Jun 2, 2013 #13
    Okay. It seems I did make another mistake with forgetting to flip over the fraction from parallel before multiplying conjugate, so I did it again and will type out everything and all the zeroes here too to be sure:


    RTh = [ (1/80Ω + 60Ωj) + (-1/100Ωj) ]-1


    = (100Ωj - 80Ωj - 60Ωj)/(8000Ω2j + 6000Ω2j2) , adding fractions, and disregarding the whole thing to the power of -1 for now.


    = (40Ωj - 80Ω)/(8000Ω2j - 6000Ω2) , simplifying the fraction, 100 - 60 on top, simplifying j*j for denominator,


    Switching them around now because it was to the power -1 becomes

    = (8000Ω2j - 6000Ω2)/(40Ωj - 80Ω)

    Multiplying by opposite complex conjugate now means:

    = (8000Ω2j - 6000Ω2)/(40Ωj - 80Ω) * (-40Ωj - 80Ω)/(-40Ωj - 80Ω)


    = ( -320000Ω3j2 - 640000Ω3j + 240000Ω3j + 480000Ω3)/(-1600Ω2j2 - 3200Ω2j + 3200Ω2j + 6400Ω2)


    Simplifying j*j = -1 means


    = (320000Ω3j + 640000Ω3j + 240000Ω3j + 480000Ω3)/(1600Ω2 - 3200Ω2j + 3200Ω2j + 6400Ω2)


    Now dividing out 100 from everything means


    = (100/100) * (3200Ω3j + 6400Ω3j+ 2400Ω3j + 4800Ω3)/(16Ω2 - 32Ω2j + 32Ω2j + 64Ω2)


    Combining like terms now means:


    = (8000Ω3 + 8800Ω3j)/80Ω2

    which reduces to

    100Ω + 110Ωj
     
  15. Jun 2, 2013 #14

    gneill

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    Staff: Mentor

    The -80Ω term in the numerator should not be imaginary.
    Okay so far.
    There should be no "j" left on the 320000Ω3 term, since j*j = -1 as you stated above.

    You should combine all the like terms at this point to simplify to a complex#/complex# form. You should get:
    $$\frac{(800000 - 400000j)Ω^3}{8000Ω^2}$$
    Then, since the denominator is purely real, divide the numerator terms accordingly.
     
  16. Jun 3, 2013 #15
    I'm doing it on paper too (although in pen) and I still left out the j on the term when I added it, but I changed the sign on the 640000Ω3 for no reason there by mistake on the second line you quoted so it's really


    (320000Ω3 - 640000Ω3j + 240000Ω3j + 480000Ω3)/(1600Ω2 - 3200Ω2j + 3200Ω2j + 6400Ω2)


    Now it simplies to

    (100/100)* (8000Ω3 - 4000Ω3j)/80Ω2


    like you said


    So RTh = 100Ω - 50Ωj


    Thanks again for being patient with me.
     
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