About source transformation and Thevenin/Norton theorems

Discussion in 'Homework Help' started by kaiosama, Oct 16, 2011.

  1. kaiosama

    Thread Starter New Member

    Dec 6, 2010
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    I was doing general source transformation problems (like in this thread http://forum.allaboutcircuits.com/showthread.php?t=60736). The kind of problem where you have a large circuit and successfully transform voltage sources in series with a resistor to a current source in parallel with the same resistor... and vice-versa... all the way down to a single voltage source in series with a resistor or a current source in parallel with the same resistor. The final circuit would then be equivalent with respect to two terminals a and b at the far right of the circuit. At this point I had not studied Thevenin/Norton.

    I then learned about Thevenin / Norton and began working some problems and then I realized that Thevenin / Norton isn't much more than the previous general transformation problems I was doing previously (am I right?). For example, when I did general transformations as described in the first part and ended up with only a voltage source in series with a resistor, then this is exactly the Thevenin circuit... (same with Norton and the current source parallel to resistor case)... Am I right?

    It seems to me that Thevenin/Norton is the very same general transformation and simplification I was already doing in the first place... Am I right?

    Say for example I begin with a large circuit and simplify it down to a single voltage source in series with a resistor at the left of terminals a and b.... This would be exactly the Thevenin circuit... All Thevenin/Norton do is giving "alternative" methods to find the same equivalent circuit (which I am not even sure it is easier... in this case, what really is the use of Thevenin/Norton?)

    I just want to make sure Im understanding correctly

    Thank you
     
  2. justtrying

    Active Member

    Mar 9, 2011
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    There is a systematic approach to applying Thevenin and Norton (they are similar to a great exetent). If before you just progressively simplified your circuit until it was at its simplest form, these two theorems tell you how to do it the smart way. For both of them you find equivalent voltage and resistance using specific methods. I personally like Thevenin (do not like using current sources) and it is the one that is most commonly used as far as I have seen. Try it, it is good, but you will still have to use superposition or other methods to get Vthevenin.
     
  3. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    In your example, I used both Norton's to reduce your example circuit, then converted the final reduction to the Thevenin's equvalent.

    It was five formulaes on the spreadsheet to arrive at the solution in your example ....
     
  4. kaiosama

    Thread Starter New Member

    Dec 6, 2010
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    Ok, this is what I was understanding. A few questions related:

    When I have a resistor in parallel with a Voltage source, I can remove the resistor without affecting the output at A and B terminals. Same thing with current source in series with resistor. A Spice simulation shows me that it is true.....Can anyone confirm?

    Also, I've read somewhere that If a circuit has only dependent sources, then the thevenin voltage and the norton current is equal to 0. Is it true? Why??? Say a attach a load resistor to terminals A and B, couldn't I find a non-zero V between terminal A and B using nodal or mesh analysis??
     
  5. JoeJester

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    Apr 26, 2005
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    Look at the attached. The operative word in your statement is only dependent sources.
     
  6. kaiosama

    Thread Starter New Member

    Dec 6, 2010
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    Does the CCVS depend on the current through the 21k resistor? I will think about it and come back later with my analysis
     
  7. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    Current Controlled Voltage Source ... and yes, it is dependent on the current through R2, the 1k resistor. That will produce zero volts and zero current at points A and B. If your simulator disagrees, you'll have to post the diagram of your circuit in your simulator. Remember, dependent sources are not self-start.

    And the answer to ...
    ... is yes, that is true. Use the attached circuit and see for yourself ...
     
    Last edited: Oct 17, 2011
  8. kaiosama

    Thread Starter New Member

    Dec 6, 2010
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    Ok, I have found RThevenin = 500 ohms and Vthevenin = 0.
    Can I generalize and say that whenever a circuit has only dependent sources, then VThevenin = 0? Does it still have a thevenin equivalent circuit consisting of the resistor alone? Or Does it mean it simply does not have a thevenin equivalent circuit?
     
  9. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    Yes, you can generalize that. Remember it doesn't have any Norton current either. It does have Rth, but to start the dependent sources, you have to start it with a voltage or current source.
     
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