Need some clarification regarding equivalent circuits?

When trying to find an equivalent circuit using norton or thevenin theorems, how does one decide which theorem to use? Does it even matter? I was just wondering if someone could clear this up for me. Thanks!

6 Answers

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  • Steven
    Lv 7
    1 month ago

    Which ever seems easiest for the circuit you need to analyze. As soon as you get out of school, you can forget these old methods and use computer simulation, aka "SPICE".

  • Anonymous
    1 month ago

    Norton uses current. Thevenin uses voltage. Whatever most convenient or accurate for particular case.

  • Roger
    Lv 7
    1 month ago

    Thevenin is better for mesh equations where you are dealing with voltage rises and drops

    Norton is better for nodal analysis, currents into and out of a node

  • 1 month ago

    The Thevenin Eq. at two terminals has a voltage source = to the open circuit voltage Voc (no load voltage) at those terminals. The Norton equivalent has current source Isc = the current through a short ckt at those two terminals.

    The Thevenin voltage source is in series with a resistance Rth and the Norton current source is in parallel with that same resistance Rth = Rn = Voc/Isc

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  • Anonymous
    1 month ago

    The only difference between Norton and Thevenin is source transformation.

  • 1 month ago

    whichever one works easiest, they are almost identical. Or superposition, which works in a lot of cases. Or mesh equations, or one of the other many methods. 

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