Does it matter if the resistor is on top or below a voltage source...

  1. Does it matter if the resistor is on top or below a voltage source after doing a source transformation?

    According to this picture from wikipiedia, the resistor is on top of the voltage source:

    [​IMG]

    But even if the resistor (Z) was on the other side of the voltage source, it would still be equivalent to the current source parallel to Z, right? Because the current-voltage characteristic is still the same?
     
  2. jcsd
  3. Totally the same. Most designers use the lower line as the lowest voltage -V or gnd; as such the resistor is shown on the high side the the voltage source, but the results between A & B are just the same as you determined.
     
  4. phinds

    phinds 8,342
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    Your choice of description ("on top or below") seems to me to be not only non-standard but inaccurate. you are talking about a series circuit and its equivalent parallel circuit. Rather than saying the resistor in the first diagram is "on top", it is better described as "in series with".
     
  5. jim hardy

    jim hardy 4,509
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    it is so difficult to make ourselves slow down and speak unambiguously...
    A question well stated is half answered.

    "on top (above?) or below voltage source" refers to before the transformation. After the transformation it is alongside a current source.

    do you mean: "if i started with the resistor drawn below instead of above the voltage source, that is between V- and B instead of between V+ and A, would the result of my transformation be the same, as shown on the right of my drawing?
     
    Last edited: Apr 6, 2014
  6. davenn

    davenn 3,476
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    incorrect

    in the left diag, the resistor is in series with the load across A and B

    in the right diag, the resistor is in parallel with the load across A and B


    Dave
     
  7. phinds

    phinds 8,342
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    The two circuits are legitimate equivalents so I'm not sure that distinction matters. The load sees identical circuits. The second is the Norton equivalent of the first (or, if you prefer, the first is the Thevenin equivalent of the second)
     
  8. I took the OP at the position that he understood the series Thevenin and parallel Norton equal. I understood the question to be "Does the series component ordering of the resistor and voltage source matter in the Thevenin circuit?" The answer to that question is NO. I took him to understand that the parallel Norton circuit was as stated and he just wanted clarification on the ordering. Hopefully if the OP was asking a different question, he will restate it more clearly.
     
  9. Here is what I originally meant in my question.

    I know from Wikipedia that these two circuits are equivalent:
    1. [​IMG]
    2. [​IMG]

    However, I am unsure if these two circuits are equivalent:
    1. [​IMG]
    2. [​IMG]
     
  10. phinds

    phinds 8,342
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    yes, as far as the outside world (the load) is concerned, they are equivalent, but that does assume you take your ground point as being to the right of the resistor in the series circuit.

    This is more clear if you simply draw the resistor vertically above or below the voltage source.
     
  11. jim hardy

    jim hardy 4,509
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    Nice clarification, Vishera !
     
    Last edited: Apr 7, 2014
  12. davenn

    davenn 3,476
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    I really must be having a brain fade ! :frown:

    that really doesn't make sense to me


    Dave
     
  13. phinds

    phinds 8,342
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    Dave, I don't really know how to respond to that. Do you understand what Thevenin and Norton equivalent circuits are?
     
  14. Dave, intro to Electrical Engineering introduces Thevenin & Norton circuits. Thevenin uses ideal voltage source with 0 (zero) impedance, Norton uses ideal current source with infinite impedance. So when you determine the impedance for a Thevenin you short any voltage source and calculate the impedance. For the Norton you open circuit any current source and calculate the impedance. If you follow this on the diagrams above you will see that between A & B the impedance is the same. The Norton current is the Thevenin voltage divided by the impedance; or the Thevenin voltage is the Norton current * impedance. This is from early introduction to electrical engineering analysis for circuits.
     
  15. davenn

    davenn 3,476
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    Yeah I do

    was just having a "senior moment" Am on the same page now

    mjhilger .... sorry for the disagreement

    cheers
    Dave
     
  16. jim hardy

    jim hardy 4,509
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    the old technician's riddle -

    You are given a sealed black box and told only "it contains an equivalent source".
    The box has two terminals with some voltage between them.
    Without either opening or X-raying the box, but anything else goes
    How do you figure out whether it contains a Thevenin or a Norton ?
     
  17. I know the answer to this. 42! Am I getting warm?
     
    Last edited: Apr 7, 2014
    1 person likes this.
  18. jim hardy

    jim hardy 4,509
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    :biggrin::thumbs:

    42, you say ?

    that's my lotto series - 42, 21, 14, 7, 3, 2
     
  19. phinds

    phinds 8,342
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    HA ! I never heard that one before, but I like it.

    spoiler altert
    Put a very low-ohm, low power resistor across them. A Thevenin equivalent will just cause the resistor to have a very small voltage across it whereas the Norton equivalent will make it explode (or at least burn out very quickly).
     
  20. jim hardy

    jim hardy 4,509
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    mBG was warmer....
     
  21. That's an interesting sequence, but I'm terrible at decoding sequences. I would expect to see a 6.
     
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