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Find Thevenin equivalent, but off by a small amount

  1. Jul 5, 2013 #1
    1. The problem statement, all variables and given/known data


    [Broken]



    Apply source transformation to find the Thevenin equivalent circuit



    2. Relevant equations

    voltage division, source transformation


    3. The attempt at a solution

    Simplify the far left current source to get

    [Broken]


    Then Vth of this = (50V + 20V)*(100/200)

    = 35V

    and Rth of it = 100||100 = 50Ω

    so now:


    [Broken]


    Then Vth of this = (35V - 10V)*(100/150)

    = 16.6667V

    Rth of it = 50||100 = 33.333Ω


    so finally this:

    [Broken]


    And then the final Vth I got

    = 16.6667*(100/133.3333)

    VTh = 12.5V

    but from the given solution from my professor the answer is really 10V for the Thevenin voltage and she did the whole problem differently by instead making everything Norton equivalent instead (changing voltage sources to current sources and moving the resistors in front to parallel them instead). Shouldn't the answer still be the same though or did I just make a mistake somewhere?

    I got the same RTh though which is

    33.333||100 = 25Ω

    RTh = 25Ω
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jul 5, 2013 #2
    This is not correct: [itex]V_{th}=15V[/itex]
     
    Last edited by a moderator: May 6, 2017
  4. Jul 5, 2013 #3

    gneill

    User Avatar

    Staff: Mentor

    Did you consider the option of first converting all the voltage sources to current sources? Can you see any advantage that might bring?
     
  5. Jul 5, 2013 #4
    So you can't add the voltage sources with each other in same direction here?

    Here it's unique because even with that though I guess you would add 10V to the next simplification, thereby still getting 25V for the next simplification.


    No, but I see it; you do that and then just use KCL and still get Thevenin resistance / voltage. I just can't seem to get the same answer doing it this way though so I think I'm missing something.
     
  6. Jul 5, 2013 #5

    gneill

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    Staff: Mentor

    Everything ends up in parallel. Just gather all the currents into one, gather all the resistors into one. You end up with a simple Norton model which can be converted to its Thevenin equivalent rather trivially.
     
  7. Jul 5, 2013 #6
    What you calculated is a voltage on resistor. To get [itex]V_{th}[/itex] you have to subtract the source voltage :[itex]35V-20V=15V[/itex]. The next simplification gives:

    [itex]V_{th}=(15-10)\cdot\frac{100}{150}+10=\frac{40}{3}V[/itex]
     
    Last edited: Jul 5, 2013
  8. Jul 5, 2013 #7

    NascentOxygen

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    Staff: Mentor

    If in doubt, work it out a couple of ways. If the answers don't match up, determine where your thinking is wrong.

    Output voltage = 50V - the drop across the 100 Ω
    The loop current = 70V / 200Ω, so output voltage = .......

    Good luck with your circuit analysis course! :smile:
     
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