Thevenin and Norton equivalent circuits, Plz help

Discussion in 'Homework Help' started by banna, Feb 3, 2013.

  1. banna

    Thread Starter New Member

    Feb 3, 2013
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    Draw the Thevenin and Norton equivalent circuits for the circuit given below.

    [​IMG]
     
    Last edited: Feb 3, 2013
  2. antonv

    Member

    Nov 27, 2012
    149
    27
    On the surface the answer is easy: A Thevenin voltage source and a series Thevenin equivalent resistance for the Thevenin circuit and, likewise, a Norton current source with a Norton equivalent parallel resistance for the Norton circuit.

    Question is, do they want you to come up with equations for the sources and resistances? Then you will have to patiently work your way through all the sources and resistances and use superposition to calculate the final equations.
     
  3. WBahn

    Moderator

    Mar 31, 2012
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    There are a number of ways to go about solving it. What methods are fair game for this particular exercise?

    What work have you done in an effort to solve your problem -- and note that it is YOUR problem. We will gladly look over your work and guide you toward finding your mistakes or seeing what the next step should be. We will not just do your work for you.
     
  4. banna

    Thread Starter New Member

    Feb 3, 2013
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    0
    I hv tried my best here is the output....

    For the Thevenin equivalent resistance(Rth) , all the voltage source short and current source open than I got Rth= (((R12||R3)+R4)||R5)+R6. Is it Okay?

    For calculate Thevenin voltage source using superposition (is it the best (understandable) one?).

    1. For voltage sources (U2-U1):
    I found Ref= (R4+R5) ||R3 then voltage on resister R5 by voltage divider rule
    VR5=Ref/ (Ref+R12) *(U2-U1)
    VR5 (1)=R5/(R5+R4) * Vref

    2. For voltage source U5 :
    Ref=R12||R3 then voltage on resister R5 by voltage divider rule
    VR5 =R5/ (Ref+R4+R5) * U5
    VR5(2)=U5- Vref

    3. For current source I3:
    Total parallel Ref= R12||R3|| (R5+R4)
    Current on R5 by current divider rule I R5 = Ref/R5* I3
    Then voltage on resister R5 VR5 (3)=R5*I R5

    4. For current source I4:
    Total parallel Ref= R4||R5
    Current on R5 by current divider rule I R5 = Ref/R5* I4
    Then voltage on resister R5 VR5 (4)=R5*I R5

    So now Vth = VR5 (1)+ VR5 (2)+ VR5 (3)+ VR5(4).
    I can find Norton current by ohm law I=Vth/Rth ..

    I think there are mistakes please help me.
     
  5. WBahn

    Moderator

    Mar 31, 2012
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    Yes. So far so good.

    It's not an unreasonable method to use for this one, though I would have used Nodal Analysis since you only have two non-trivial nodes and one of those is the value you want.

    1. For voltage sources (U2-U1):
    I found Ref= (R4+R5) ||R3 then voltage on resister R5 by voltage divider rule
    VR5=Ref/ (Ref+R12) *(U2-U1)
    VR5 (1)=R5/(R5+R4) * Vref
    [/QUOTE]

    I think you have a typo here. Ref appears to be the total resistance not including R12. That means that your next line yields the voltage on the top-left node, not the node on top of R5. Did you mean to call that Vref and not VR5? If so, then your are still doing good.

    Okay, you are being sloppy with your notation. What is VR5 and what is Vref?

    I think you are getting the concepts right, but you need to clean up your notation a bit.

    This looks correct, but the notation is making it hard to follow. You have R5 and then you have I R5. The space makes it look like these are two different quantities, one of which is the resistance R5. It's admittedly hard with just straight text when you don't have subscripts. A common option is to use an underscore to indicate a subscript, so I_R5. You also need to indicate the polarities of your quantities; is I_R5 the current going up or the current doing down through R5? It's pretty easy to figure out which you mean, but the reader (or grader) should have to figure out what you mean -- it's your job to make it clear what you mean.

    So the current from I4 can either go through R4 and back to the source or it can go through R5 and back to the source (without passing through any other components)?

    Rethink this one.

    You are actually quite close.

    You need to develop the ability to check your own work. So take your results and then make simplifications to the circuit and throw in some easy numbers and see if your results are consistent with them.
     
  6. malazajs

    New Member

    Apr 11, 2013
    2
    0
    help pls wth thevinin equivalent basic clear eg. RLC ccts
     
  7. WBahn

    Moderator

    Mar 31, 2012
    17,724
    4,788
    Try the following:

    1) Start a new thread instead of hijacking someone else's.
    2) Ask specific questions about a specific problem.
    3) Post a schematic or at least a very clear word description of the circuit you are working with.
    4) Post YOUR efforts at a solution so that we have something to start with.
    5) Write in complete and fully spelled out English, at least to the degree that your English skill will permit.
     
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