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Determining Rth from a simple circuit

  1. Mar 21, 2015 #1
    1. The problem statement, all variables and given/known data
    Selection_023.png

    2. Relevant equations
    V=IR


    From my textbook:
    If the circuit contains both independent and dependent sources, the open‐circuit terminals are shorted and the short‐circuit current between these terminals is determined. The ratio of the open‐circuit voltage to the short‐circuit current is the resistance Rth.

    3. The attempt at a solution
    I determined that Rth and Vo = 0. Now I have to find Rth. But it's not zero? What? The open circuit volage divided by the short circuit current is Rth. But we know for a fact that the open circuit voltage is 0. Rth = 0 / Ix. But this is wrong?
     
  2. jcsd
  3. Mar 21, 2015 #2

    NascentOxygen

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    Staff: Mentor

    You determined Vo by inspection?

    Have you determined Io when output is short-circuited?

    You can then look at evaluating their quotient.
     
  4. Mar 21, 2015 #3
    If I'm not mistaken, I don't need to determine Io because Rth =Vth/Io = 0/Io = 0

    I determined Vth/Vo by nodal analysis, it's 0 (its correct, green)
     
  5. Mar 21, 2015 #4

    gneill

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    This is one of those cases where you should stick a fixed voltage source on the output and measure the resulting current that it drives into the circuit.

    If you're sharp you should be able to determine almost by inspection what the resulting current will be, and thus the Thevenin resistance.
     
  6. Mar 21, 2015 #5
    Ah yes. That also was in my book, but the theory behind it wasn't explained. Where can I read about why we do this, and why it works?
     
  7. Mar 21, 2015 #6

    NascentOxygen

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    Yes, you are mistaken. A non-zero value of Io is essential for this quotient to be defined.
     
  8. Mar 21, 2015 #7
    I still can't figure it out, using KCL right next to the + Vo terminal (after attaching a 1V source) I get:

    Reference is the bottom node
    V2 (the node in the center of the circuit, intercepted by 2 lines (looks like a +)) is -6 V
    V1 = +1V

    -(v2-v1)/2+v1/4-3=i

    r=1/i=1.333 k ohms

    This is wrong
     
  9. Mar 21, 2015 #8

    gneill

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    Strange, to me it looks like 1.333 k should be correct. 4k || 2k yields 1.333k.
     
  10. Mar 21, 2015 #9
    You need to open up the independent current source and short the independent voltage source. What is your test source then connected across?

    Edit:
    The reasoning behind it is as follows:

    Let's assume you accept Thévenin's theorem, so you can replace any two-terminal linear network as shown here:
    Thevenin_equivalent.png

    The left circuit must then behave exactly like the right circuit, so to find Rth, we could simply turn off all the independent sources in the circuit, which leaves Vth = 0 V (a short, effectively). Your test source is then applied directly across Rth.
     
    Last edited: Mar 21, 2015
  11. Mar 21, 2015 #10

    NascentOxygen

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    If you are allowed multiple guesses, try 2kΩ
     
  12. Mar 21, 2015 #11
    I have infinite guesses. 2k was right. How did you figure?

    In all the problems I've encountered, the book specifically says never to short a independent voltage source or open a independent current source if there are any dependent sources in the circuit. We always have had to work around them (independent is another story though)
     
  13. Mar 22, 2015 #12
    If your book says that, then you're now taking it out of context. I can imagine it says something to that effect when it covers, for instance, source transformations, where it's very important that you don't break the dependency between independent and dependent sources.

    I think if you try to work trough the reasoning I posted, you'll find that the idea behind all this is actually very, very simple. You just have to focus on the Thévenin equivalent circuit, and think about, if you had access to such a simple circuit, what tests could you come up with to determine Vth and Rth? Assume you can turn Vth on and off.

    And I have to emphasize this part: Don't think in terms of the complex circuit - only think in terms of the Thévenin equivalent circuit.
     
  14. Mar 22, 2015 #13

    NascentOxygen

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    The mistake you've made in this exercise is that you have not followed instructions. Even were your numerical results all correct, in a written exam you could expect 0/10 because you have failed to do what was asked.

    Let's review the exercise:

    The instructions indicate that at the conclusion of this you would be able to determine Vo using Thévenin's theorem. But what did you do? Your very first move was to calculate Vo using a method not involving Thévenin!

    EPIC FAIL!

    The second instruction was that you should remove the 4kΩ resistor and calculate RTH of the remaining circuit seen at the output terminals. This instruction could have been stated more clearly. Still, you overlooked it, so calculated for the wrong arrangement.

    The required final step is to picture the Thévenin circuit driving the 4k resistor, and hence determine Vo of the circuit as drawn.

    There is no prohibition re "shorting out" all voltage sources and "open circuiting" all current sources here, because the circuit is drawn to trick you. The dependent source is irrelevant to Vo and RTH. Here, Vo is wholly determined by the pair of independent sources near it. The technique of applying an external source is uncalled for.

    There's a lesson in this: take careful note of instructions, and follow them to the letter!
     
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