Thevenin Equivalent Problem

Discussion in 'Homework Help' started by Henry_EE, Nov 12, 2007.

  1. Henry_EE

    Thread Starter New Member

    Nov 12, 2007
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    Hey guys, help me find the Thevenin Equivalent of this. I don't know where to start or what to do. None of the problems I've encountered on Thv Equiv. dealt with dependent sources. My instructor mentioned that the equivalent voltage (Vth) would be zero, since it's a dependent source, but other than that, I'm lost.
     
  2. Ron H

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    Apr 14, 2005
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    There is a clue here.
     
  3. Henry_EE

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    Nov 12, 2007
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    That helps me understand a little more what a dependent source is, but I'm still pretty much lost >_<
     
  4. The Electrician

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    Oct 9, 2007
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    Has your instructor required that you use a particular method to solve this problem, or can you use a method of your own choosing?
     
  5. Henry_EE

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    Nov 12, 2007
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    Any solution/help is appreciated. I really want to understand this.

    Edit:

    Okay, I looked through some more things online, I found out that with dependent sources, the equivalent circuit is just the equivalent resistance of the circuit, Req. So all I have to do is ignore the dependent source completely and just calculate the equivalent resistance of that circuit right?
     
  6. The Electrician

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    I'm not sure what you mean by "equivalent resistance", but you can't ignore the dependent source. What you must do is imagine applying a voltage source of, say, 1 volt to the terminals a and b, and calculating the current from that source. Then divide the voltage of that source (1 volt) by the current. That will be the Thevenin resistance at the terminals a and b. Of course, the Thevenin voltage will be zero.

    I would label the dependent source V1 and the 1 volt source applied to terminals a and b, as V2. The write down the mesh equations. I would use the obvious 3 loops. Label the current in the lower left loop (containing V1, 10 ohms and 60 ohms) as i1; the upper loop current (with 30 ohms, 15 ohms and 10 ohms) as i2; and the lower right loop current as i3. The first equation for the i1 loop would be (with the currents in all 3 loops going clockwise):

    -V1 + 10 i1 - 10 i2 + 60 i1 - 60 i3 = 0

    I'm sure you can write down the next two.

    After you have all three, rearrange them so they are suitable for solution as a linear system using a matrix solver. For example, the first one would be:

    70 i1 - 10 i2 - 60 i3 = V1

    Now, to take care of the dependent source, make the substitution V1 = 30*(i3 -i1).

    Then the first equation becomes:

    100 i1 -10 i2 - 90 i3 = 0

    with the next two equations written just below so that you have a linear system that a matrix solver can solve (or do it yourself by hand!). Solve for i1, i2 and i3 in terms of V2. Then the value V2/i3 will be the resistance seen at the terminals a and b.

    Let us know what final result you get, and if you have any more problems along the way.
     
  7. Henry_EE

    Thread Starter New Member

    Nov 12, 2007
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    Why is it 30*(i3-i1) and not 30*(i1-i3)?
     
  8. The Electrician

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    Look at the schematic you posted. The current Ia is directed upward. If you select the currents i1, i2 and i3 as I described, flowing clockwise, then i3 is directed upward and I1 is directed downward in the 60 ohm resistor. Since i3 is in the same direction as Ia, it is given a positive sign, and i1 is given a minus sign in the expression 30*(i3-i1). That is, Ia is in fact the sum of i3 and -i1.
     
  9. IrVine48

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    Oct 6, 2008
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    sorry to disturb. from the question above , is it the current equal to zero too ???
     
  10. The Electrician

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    Which current are you referring to?
     
  11. hitmen

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    Sep 21, 2008
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    The question can be solved by placing a 1 volt source across AB and then using nodal analysis. Rth = 1/I(ab).

    Letting Vb be the position between the 10 and 15Ω.

    I got Vb = 0.5V
    I x = Vb/ 60 = 1/120
    30Ix = 0.25

    Using KCL at Node A ,

    Iab = 1/30 - 1/40 = 1/120A

    Thus Rth = 120Ω .

    Vth = 0 since there are no independent sources.

    BTW, I am also a student like you so I am not sure of my answer.

    Do verify it however.:)
     
  12. neon9

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    Oct 8, 2008
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    he is sneeky he gives you I as opposed to volts. thevening only deals with volts not I. IS THERE A SOLUTION SURE FIND THE total resistance then apply the current to it and you have the voltage with the voltage then revert back to find the thevenin equivalence. good luck.
     
  13. hgmjr

    Moderator

    Jan 28, 2005
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    Take a look at this material on Thevenin's Theorem in the AAC ebook. There are some good worked examples for you to practice on.

    hgmjr
     
  14. hugoleon

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    Jun 15, 2011
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    i need help with thevenin equivalent
     
  15. hgmjr

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    Jan 28, 2005
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    Open your own thread in the Homework section and provide us with the problem statement the circuit and your effort to solve it.

    hgmjr
     
    hugoleon likes this.
  16. victorhugo289

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    Aug 24, 2010
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    There is a problem with your circuit. I don't understand much of it. Haha.
    What is that rhombus thing and its value? Sorry I haven't seen that before, I don't have that much experience.
    And the LOAD, if it is a resistor, what is its value?
    hmm... I think the LOAD might just be your voltmeter (10M ohm) connected across those points.
    This circuit is easy to solve, but you need to clarify the strange rhombus.
     
  17. victorhugo289

    Member

    Aug 24, 2010
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    I got it.

    It redraws like a bridge circuit, like this:

    [​IMG]

    The Thevenin equivalent is:

    [​IMG]

    Ethevenin and Rthevenin respectively.

    Go ahead, place any load and you'll always get the result.

    EDIT: OK, but I don't know what you mean with 'Dependent source'!!! And what you teacher said makes no sense to me. o.o?

    Ah ah, dependent sources... I know nothing about them yet. :D
    Anyway I hope someone finds this useful.
     
    Last edited: Jun 18, 2011
  18. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    You have the Vth incorrect. The teacher was correct in stating the Vth is zero. But by your own admission, you haven't learned dependent sources yet.
     
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