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Thevenin without removing load resistor

  1. Feb 24, 2009 #1
    I'm having trouble with Thevenin models, specifically this question where the V_oc is not given.

    The problem:
    Assume an unknown circuit provides the specified current through the load resistors as given below:
    100k - 2.3mA
    1.2M - 510uA
    Find a Thevenin model for the unknown circuit.

    All I've been able to find on Thevenin requires me to find V_oc, which I don't know how to do from here. I've tried just using Ohm's Law, but that nets me 230V for the first case, and 612V for the second. Suggestions on how to proceed would be appreciated.
  2. jcsd
  3. Feb 24, 2009 #2

    The Electrician

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    Gold Member

    No one will be able to help you until you post a schematic of your circuit.
  4. Feb 24, 2009 #3
    There is no schematic of the circuit given. The load resistor is connected across two terminals of a "black box", I assume.
  5. Feb 24, 2009 #4


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    Science Advisor

    Actually, this question does not require a schematic. Since this sounds like homework, I won't tell you the answer. However, you can very easily figure out what the question is actually asking for by drawing a generic Thevenin circuit. Actually, it might be easier to use a Norton equivalent in this case.

    In either case, you'll have two equations and two unknowns.

    EDIT: This illustrates one of the principles of the "black box": that you can figure out an equivalent model of the box without actually cracking it open, just by how it drives a load.
    Last edited: Feb 24, 2009
  6. Feb 24, 2009 #5
    I've never actually thought of Thevenin or Norton models' usefulness, until I saw this thread. I've always thought they were to make analysis easier.
  7. Feb 24, 2009 #6
    Hint: remember that the load resistor is not the only resistor in each of the 2 circuits. You have 2 unknowns, and 2 separate cases with part of the circuit resistance given along with the entire source currents. Think KVL.
  8. Feb 24, 2009 #7
    Thanks for the help, guys. I just set the Norton model up as a current divider, and solved from there.
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