help with a thevenin question

Discussion in 'Homework Help' started by shuujin01, Nov 30, 2013.

  1. shuujin01

    Thread Starter New Member

    Nov 30, 2013
    16
    0
    I am trying to find the equivalent thevenin voltage across ab in the attached ac circuit ,
    f=1000Hz
    Ra = -34.22+j30.15 Ω
    Rc = 90.45+ j102.67 Ω
    Z4 = 119 Ω
    Vs = 78 + j270 V
    Vx = 4 + j140 V
    this is what i did just want to make sure if what i am doing is correct, if not please show me the correct way to find Vth.
    Zt=Za+Zc+Z4
    It = (Vs-Vx)/Zt
    Vza = It*Za
    Vzc = It*Zc
    Vz4 = It*Z4

    Eth = Vzc+Vz4-Vx
     
  2. shteii01

    AAC Fanatic!

    Feb 19, 2010
    3,377
    494
    It seems to me:
    Vthevenin = Vs
    Rthevenin = RA
     
  3. shuujin01

    Thread Starter New Member

    Nov 30, 2013
    16
    0
    i already took the load out, and the circuit has two ac voltage sources, i think Vth is voltage across ab which contains Rc, Vx and Z4
     
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    That would be incorrect.
     
  5. shteii01

    AAC Fanatic!

    Feb 19, 2010
    3,377
    494
    Ah. Ok. I thought the load was still in.
     
  6. WBahn

    Moderator

    Mar 31, 2012
    17,715
    4,788
    You are almost correct.

    It's hard to check your work when you don't indicate the polarity of any of your voltages or currents. That is where you need to start -- annotate your diagram so that these are clear. If you do that, you should see your mistake.

    Of course, I'm having to assume things such as the polarity of your two voltage sources. I shouldn't have to be doing that.
     
  7. shuujin01

    Thread Starter New Member

    Nov 30, 2013
    16
    0
    here is the circuit and i indicated the polarity on it, sorry about that
     
  8. WBahn

    Moderator

    Mar 31, 2012
    17,715
    4,788
    Thanks for updating your diagram.

    With those annotations, several of your earlier equations are now wrong.

    Notice that the current It is going through all of the components in the same direction. But the polarities of your Ra, Rc, and Z4 are not all in the same direction, so some of the voltages need to have the opposite polarity.

    You don't indicate what direction It is flowing in and, instead, force the reader to practice either mind reading or reverse engineering. Note that this is NOT the way to impress either a grader or a boss. Doing the reverse engineering, your equation for It implies that It is the current flowing clockwise in the circuit.

    This means that

    Zt=Za+Zc+Z4
    It = (Vs-Vx)/Zt

    Vza = -It*Za
    Vzc = -It*Zc
    Vz4 = It*Z4

    Now look at your circuit and see that you have two possible paths to get from a to b. You can compute the voltage drop along either path and they must yield the same result -- the implies a simple way to check your results.

    Going up the left branch, we have

    Vab = Vs + Vza

    Going up the right branch, we have

    Vab = Vz4 + Vx - Vza

    Are these, in fact, the same? Do the math and see. If you have any problems, post your efforts and we can go from there.
     
  9. shuujin01

    Thread Starter New Member

    Nov 30, 2013
    16
    0
    thank you for your help, i did the math and they are indeed the same, but let me make things more clear, i am attaching the real circuit that i started with. I did a delta to wye transformation that's how i got Ra and Rc and after that i tried solving for Vth so i simplified the circuit. so can u check with me please to see if what i did is correct and thank very much for ur time
     
  10. WBahn

    Moderator

    Mar 31, 2012
    17,715
    4,788
    See if what is correct?

    You haven't given enough information to determine if your transformation was done correctly or not, let alone if your end result is correct or not.
     
  11. shuujin01

    Thread Starter New Member

    Nov 30, 2013
    16
    0
    this is my transformation
    Za =(Z1*Z2)/(Z1+Z2+Z3)
    Zb =(Z2*Z3)/(Z1+Z2+Z3)
    Zc =(Z1*Z3)/(Z1+Z2+Z3)

    so can you tell me how to find the thevenin voltage after i take out the load
     
  12. shteii01

    AAC Fanatic!

    Feb 19, 2010
    3,377
    494
    I got Vthevenin=-99.476+19.687j
    What have you got?
     
  13. WBahn

    Moderator

    Mar 31, 2012
    17,715
    4,788
    May not matter what he got. He has someone that worked the problem and posted an answer and so he can, if he wants to, just turn that in without any further work.

    This particular OP may not do so, be we certainly get a lot that would.
     
  14. WBahn

    Moderator

    Mar 31, 2012
    17,715
    4,788
    The circuit that you set up looks fine.
     
  15. shteii01

    AAC Fanatic!

    Feb 19, 2010
    3,377
    494
    The suspense was killing me.
     
  16. shuujin01

    Thread Starter New Member

    Nov 30, 2013
    16
    0
    ok the thevenin volt would be either
    Eth = Vs+Vza
    Eth = -Vzc+Vz4+Vx
    like you said, right? or if you walk me thru on how to find it. thank you
     
  17. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    This is presumably a point at which one might ask the OP to start again and restate the problem with all component values as given in the original problem formulation. I would be reluctant to trust your transformations without the requisite working shown. Where is node "a" in the original circuit attached in post #9? What are the values of Z1, Z2, Z3 & Z5?

    I'm curious that you can obtain an impedance with a negative real part [see "Rc"] when the circuit offers no clue as to how this has arisen.
     
  18. WBahn

    Moderator

    Mar 31, 2012
    17,715
    4,788
    I already did. See Post #8. See if you can follow where I got everything and then ask specific questions at the points where you can't.
     
  19. shuujin01

    Thread Starter New Member

    Nov 30, 2013
    16
    0
    thank you guys for the hlep, i am just gonna post a new thread with more details on how i started the problem
     
  20. WBahn

    Moderator

    Mar 31, 2012
    17,715
    4,788
    I don't think there's any need to start a new thread dealing with this same problem. It will just cause confusion as people end up posting to one, the other, or both threads. Let's keep it all in one place, okay?
     
Loading...