Source equivalence problem

Discussion in 'Homework Help' started by hunterage2000, Dec 15, 2010.

  1. hunterage2000

    Thread Starter Active Member

    May 2, 2010
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    Can someone tell me how I would go about finding the voltage through R5 using the source equivalent method. I still dont really understand this method.
     
  2. Jony130

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    Feb 17, 2009
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    Do you mean Thévenin's theorem ?
     
  3. hunterage2000

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    May 2, 2010
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    Yeah but first reducing the amount of sources by combining the current sources in parallel and voltage sources in series.

    reduce all the sources to one source and one resistance.
     
  4. Jony130

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    Feb 17, 2009
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    So first you need is to disconnect R5 and then find Vth voltage.
    [​IMG]
    To find Vth we could use superposition

    Vth'1
    [​IMG]

    Vth'2
    [​IMG]

    Vth = Vth'1+Vth'2

    Next we replace all voltage sources with short circuits, and all current sources with open circuits.
    And calculate the resistance between terminals A - B

    [​IMG]
     
    Last edited: Dec 15, 2010
  5. hunterage2000

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    May 2, 2010
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    The question says determine the Thévenin equivalent circuit of the above network using the “source equivalent” method. Hence calculate the load voltage vL.

    I assumed that you would reduce the circuit to one source with one resistance.
     
  6. Jony130

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  7. hunterage2000

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    I thought you would have to reduce it first using source equivalence. example below.
     
  8. Georacer

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    Nov 25, 2009
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    They is essentially the same, but I see two ways to solve this.

    You can use the supperposition theorem to find the components of the voltage across the R5, nullifying all the sources but one each time. I trust you know how to do this.

    The other way, which I think is what your tutor wants you to do, is to break the circuit to smaller pieces, merging all the sources, two at a time.

    For example, starting from the left, find the Norton equivalent of R1 and V1. This will give you a resistance and a current source in parallel. That current source will be parallel with I1 and those can be combined.
    Work this way, switching between Norton and Thevenin equivalents, until you reach R5.

    Is that clear?
     
  9. hunterage2000

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    May 2, 2010
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    this is what Im stuck with. It says parallel sources can be combined so from the start can I1 and I2 BE combined to be 6.6m amps?

    if so can you convert all voltages in series into a current source and add them together?
     
  10. Jony130

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    Ok I start for you but you must finish the job.
     
  11. hunterage2000

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    May 2, 2010
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    nice one mate I'll have a look tomorrow. That example I posted was the only thing I got so thats why Im a bit puzzled with the bigger examples.
     
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