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Thevenin Equivalent Circuit

  1. Oct 2, 2013 #1
    1. The problem statement, all variables and given/known data

    Above given is the circuit I am supposed to simplify where the 1/5Ω resistor is the load. I'm unsure how to convert this circuit into an equivalent Thevenin Circuit.

    2. Relevant equations

    V = I*R (Ohm's Law)

    3. The attempt at a solution

    Using Microcap and it's Dynamic DC Analysis Tool, I measured the current flow at the load. I managed to break it down to this:

    http://img844.imageshack.us/img844/282/5x8t.png [Broken]

    Where V5 was a 2/3V voltage source. (2A current source in series with the 1/3Ω resistor).

    If this is correct, I would then say that the 2/3V voltage source and the 4A current source were in series and that the V4 voltage source and the 1/6Ω resistor were in series. Is this the correct approach?
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Oct 2, 2013 #2


    User Avatar

    Staff: Mentor

    Remember that a Thevenin equivalent is a single voltage source with a single series resistance (for DC circuits).

    What's the usual algorithm for finding the Thevenin equivalent analytically?
  4. Oct 2, 2013 #3
    Would the first step be setting all those current sources to 0? Thus left with the remaining voltage source and the two resistors?

    http://img820.imageshack.us/img820/2318/q2z4.png [Broken]
    Last edited by a moderator: May 6, 2017
  5. Oct 2, 2013 #4


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    Staff: Mentor

    No. You must suppress ALL sources to find the Thevenin resistance: Remove all current sources and short all voltage sources.

    To find the Thevenin voltage you must find the open-circuit potential across the load (remove the load and find the potential across the open terminals).

    Your course notes or text should cover this!
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