Help with Thevenin's & Norton's theorem

Discussion in 'Homework Help' started by killer6008, Jan 26, 2010.

  1. killer6008

    Thread Starter New Member

    Jan 26, 2010
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    Can someone please help me solve these two problems? Also please explain by writing all the steps if possible so I can do similar problems myself.

    Thank you so much.

    James

    [​IMG]
     
  2. beenthere

    Retired Moderator

    Apr 20, 2004
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    You might have missed the sticky note:
    Otherwise, you're just asking us to do your work.
     
  3. killer6008

    Thread Starter New Member

    Jan 26, 2010
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    Sorry I didn't read that.

    I am not very good with this. First time I am studying it. Been sitting here for 2 hours trying to figure out how to do this. I only have two examples infront of me and these questions are completely different to them.

    For question 1 between A and B the 15 ohm resistor will go to take off I. Also -j40 Vs needs to be removed. I can't really finish the diagram to move onto the next stage.

    Can anyone help me with that please?
     
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    If you are new to this then Q1 is quite a challenge. So I will give you a reasonable "lead".

    Without solving anything, I have attached the steps required to reduce the network to determine ZT across terminals AB. This is not the only approach - just a suggested one.

    If you've not used delta-star transformation you will have to look that up in a book.
     
  5. killer6008

    Thread Starter New Member

    Jan 26, 2010
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    Thanks a lot for that.

    I tried to have a go not sure if I am right or wrong though.

    We calculate Z A*xy and Z xy-B now dont we?

    so for Z A*xy 20*j30/20+j30 than you convert them into polar for division and convert them back to cartesian. i got (13.83 + j9.22) for this.

    and for Z xy-B you do j20.40/40+j20 and same steps with answer ( 8 + j16)

    I doubt this is right but anyone please check this for me ?

    Thank you
     
  6. killer6008

    Thread Starter New Member

    Jan 26, 2010
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    Rest of calculations to show that I am trying.

    Vay= j30/ 20 + j30 * 15<0 = (10.36 + j6.91) V

    Vby = 40 / 40+j20 * 15 < 0 = (6.01 + j12.01) V

    Vab = Vay - Vby = 6.7 < -49.54 = Vth


    Is any of this right? Can someone please confirm for me?
     
  7. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Look at the circuit (with the 15Ω removed) and assume there is a voltage Vx at the mid-point node formed by the junction of the j30Ω, the 20Ω and the j20Ω.

    Using the voltage divider rule the voltage at A or VA will be

    VA=Vx*(-j40)/(20-j40)

    And the voltage at B or VB will be

    VB=Vx*(40)/(40+j20)

    What do you notice about the ratios (-j40)/(20-j40) and (40)/(40+j20) ???

    What can you then conclude about the values of VA and VB and hence what is VAB?

    As to the Zthevenin - how did you do the delta-star transformation?
     
  8. killer6008

    Thread Starter New Member

    Jan 26, 2010
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    Aint the ratios same for both of them. I mean when you divide them you get the same thing for VA and VB. Isn't VAB going to be zero??

    I didn't do the delta-star transformation. I tried to do it with your diagrams but I guess that's wrong then. Can you help me with transformation please?
     
  9. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Yes VAB will be zero.

    Rather than just take my word for it you should prove that to yourself using an alternate method - such as mesh analysis. You don't need Delta-Star transform to do that.

    The Star-Delta (& vice-versa) transform is a well-known circuit analysis operation

    see

    http://en.wikipedia.org/wiki/Y-Δ_transform

    You should attempt to do that part by yourself and come back with some clearer working - even if it is only showing that particular part of the overall analysis. As stated previously, this isn't a homework service.
     
  10. killer6008

    Thread Starter New Member

    Jan 26, 2010
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    Thanks for your reply.

    I am a bit confused here if VAB is zero isn't everything else is going to be zero? because you mutiply and divide in various steps to calculate things.

    I have attached an image of what I tried to do.
     
  11. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Consider the Wheatstone Bridge at balanced condition - you have two nodes at the same potential but there are non-zero voltages elsewhere. This is a similar situation.

    You failed to show any working on your delta-star diagram. You continue to show a singular reluctance to reveal what working processes are going on when you attempt to solve a problem.

    In any case I've given a starting point for you in the attachment.
     
  12. yasinswar

    New Member

    Feb 10, 2010
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    my attempt so far not too sure im on the right track..
     
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