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Thevenin/Norton circuit with dependent voltage source

  1. Sep 23, 2012 #1
    1. The problem statement, all variables and given/known data

    This problem has two parts, so I might as well inquire about both.

    https://www.edx.org/static/content-mit-6002x/images/circuits/H4P3%20Thevenin.2b9528ba9f62.png [Broken]

    The figure above shows a circuit with a linear current-controlled-voltage-source and its Thevenin equivalent model as seen from Port A. Given that I0=2A,Z=2Ω,R1=2Ω,R2=4Ω, Determine the Thevenin voltage VTH and the Thevenin resistance RTH.

    https://www.edx.org/static/content-mit-6002x/images/circuits/H4P3%20Norton.cbefdbaf69ef.png [Broken]

    The figure above shows a circuit with a linear voltage-controlled-voltage-source and its Norton equivalent model as seen from Port B. Given that Vo=5V,A=2,R1=1Ω,R2=3Ω,R3=5Ω, determine the Norton current IN and Norton resistance RN.

    For the first, they want the equivalent Thevenin voltage and resistance, for the second, the equivalent Norton current and resistance.


    2. Relevant equations

    KCL, KVL, Thevenin/Norton, all the standard stuff you'd expect.


    3. The attempt at a solution

    OK, let's start with the second one and work from there, as that is the one I got further on.

    Firstly, I used KCL between R1 and R2 and got 7.5 V for the value of u, giving me the dependent voltage source's value.

    I then used KCL at the node above R3 and the dependent voltage source-got -35 volts, which seems weird. Maybe I screwed up on that(I'm checking again but I'm getting the same thing)?

    After that, I should find the resistance-and using both that and the Thevenin voltage, I should be able to find the Norton current, as Thevenin resistance is equal to Norton resistance.

    What I'm stuck on is the dependent voltage source... what exactly should I do about that? I know the independent one becomes a short, should I use a test resistance? I thought that was only for when you had only a dependent source.

    Thanks in advance.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Sep 23, 2012 #2

    gneill

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    Under what conditions did you find that u = 7.5V? Open circuit at Port B? That's not what I see. Can you show your work?

    Also note that the Thevenin and Norton models are closely related, and that you can find the resistance by taking the open-circuit (Thevenin) voltage and dividing it by the short-circuit (Norton) current.
     
  4. Sep 23, 2012 #3
    I used KCL between R1 and R2. I guess I'm not supposed to do that? I just use the standard voltage division to find u-I think I was making this too complicated.

    (Vu - Vo)/R1 = (Vu)/R2.


    Ah yes-I forgot about doing it that way. I'd just do an additional current going out at the second node. Now, I'm getting Vth=-5.

    I'm using KVL to find IN. Got 2.75 which is correct. But that VTH value can't be correct, it gives a negative resistance.
     
    Last edited: Sep 23, 2012
  5. Sep 23, 2012 #4

    gneill

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    That KCL expression doesn't look right to me. The left hand side looks like the current out of the node heading towards the voltage source. Fine. The right hand side looks like the current flowing out of the node towards the node at the junction of R2 and R3. These two current should sum to zero for KCL to hold at the "u node", so one side or the other should be negative. Also, the equation holds only of the output potential at Port B is zero (shorted output). Is that the case?
    If the output is shorted, it will carry the currents from both branches. Should be straightforward.
     
  6. Sep 23, 2012 #5
    Yeah, I'm not using KCL between those two resistors anymore. I just did it at the node above R3 and I got -5, which can't be correct. I'll try it again, but I'm not seeing what I'm doing wrong....


    Vth-Vo/(R1+R2) = Vth-Au/R3
     
  7. Sep 23, 2012 #6

    gneill

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    That doesn't look right; you've got some terms that are voltages and others that are currents.

    I still don't know the conditions under which you're writing the equations. Can you state the condition of Port B, and what method you're employing for each equation you write?
     
  8. Sep 23, 2012 #7

    I'm trying to find the Thevenin voltage, Port B is open. For the method mentioned I'm using KCL.

    I'm assuming the value for Au is the same as it was when I was using KVL to find the Norton current.
     
  9. Sep 23, 2012 #8

    gneill

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    Well, the A will be the same, but the Norton current is found when the output is shorted, so the u will be different. In the open-circuit case the potential across R2 will then be Vu - Vth.

    You might want to apply nodal analysis using two nodes: one for Vu and one for Vth.
     
  10. Sep 23, 2012 #9


    OK-so if I do KCL in the open circuit case I get two KCL equations.

    Vu-Vo/R1 = Vu-Vth/R2

    Vth-Vu/R2 = Vth-Au/R3

    And from there it is just algebra.

    Hold on, I'm getting a ridiculous large number here.
     
    Last edited: Sep 23, 2012
  11. Sep 23, 2012 #10

    gneill

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    You've got sign issues in your node equations. Current directions matter. Try writing the equations as the sum of currents (either entering or leaving, your choice) and set the sum equal to zero. So, for the "u" node, for example, summing as outgoing currents:

    (u - Vo)/R1 + (u - Vth)/R2 = 0

    You'll also need to make a substitution for u. Perhaps using the current through R1 from the above to determine the potential drop from Vo through R1...
     
  12. Sep 23, 2012 #11
    OK. So I'm still getting the wrong answer...


    So, using the equation below, I get Vth - 3Vo = 4u

    Now-the second KCL equation I got was:

    (Vth-u)/3 + (Vth - 2u)/5 = 0

    Leading to Vth = 3u/4

    When I plugged that all in, I ended up getting Vth equal to about 1.26 ohms.

    Just an algebra error, or something more fundamental? I'm going to go through it again to make sure.
     
    Last edited: Sep 23, 2012
  13. Sep 23, 2012 #12

    gneill

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    You should use parentheses to make your equations clear. Your second KCL equation looks as though you're adding voltages and currents.

    For the first equation, I think you've got a sign issue in the terms. Probably an algebra glitch.
     
  14. Sep 23, 2012 #13
    Finally got it(used an online system of equations solver).

    Thank you very much. Later tonight, I'll try the other problem again.
     
  15. Oct 3, 2012 #14
    Can you please elaborate:

    I also got (u-V0)/R1 + (u-Vth)/R2=0
    (Vth-u)/r2+ (Vth-Au)/R3 = 0

    Substituting, I got u = 40/7. and Vth = 55/7

    Next, I was planning to compute Rth as (R1 + R2) || R3 (by shorting Vo voltage source and Au voltage source).

    And calulate IN = Vth/Rth.

    Looks like my Rth is wrong. I am not sure why. I would appreciate if you can point my error.
     
  16. Oct 3, 2012 #15

    gneill

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    When calculating Rth, independent sources may be suppressed but not dependent ones. Instead, place a 'probe' source on the output and determine the voltage and current it injects. In this case a current source placed at the output would be handy for nodal analysis. Find the resulting voltage at the output node. Divide that voltage by the injected current to determine the Thevenin resistance.
     
  17. Oct 3, 2012 #16
    Hi,

    I am sorry! not clear about it. Mathematically, how can I do it? How will the Rth equation change?
     
  18. Oct 3, 2012 #17

    gneill

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    Suppress the independent voltage source. Add a current source to feed current into the output node. Use whatever analysis methods you're comfortable with to find the voltage at the output node. Divide that voltage by the injected current value to yield your Thevenin resistance. Alternatively, use a voltage source on the output and determine the current it injects.
     
  19. Oct 3, 2012 #18
    Hi, I tried with a current source (I) between port A & B. I used node method to calculate U and e (voltage across the new current source) as follows:

    (u-V0)/R1+ (u-e)/R2=0
    and
    (e-u)/R2+(e-Au)/R3 - I = 0

    Then I calculated e from the above pair of equations and the resistance Rth = e/I. However, the solution I got is not correct. Did I formulate it correctly?
     
  20. Oct 3, 2012 #19

    gneill

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    You should suppress the independent source Vo. For a voltage source that means setting it to zero (a short circuit).
     
  21. Oct 3, 2012 #20
    Thanks! It works! Thanks for your patience! For the next step, I need to find norton current. Do I need to fine open circuit voltage Voc and the In = Voc/Rth where Rth is resistance calculated with test source?
     
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