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Another Thevenin Question

Discussion in 'General Electronics' started by Mrs. Kerchief, Nov 4, 2003.

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  1. There's a self-test at the end of each chapter of my EE textbook, and
    I couldn't figure out--because I didn't understand--how the correct
    response was arrived at in the following question.

    "A certain circuit produces 15 V across its open output terminals, and
    when a 10K load is connected across its output terminals, it produces
    12 V. The Thevenin equivalent for this circuit is...(and the answer
    is 15 V, 2.5K Ohms).

    I didn't know where to begin to figure the answer out, and neither
    does my tutor. My instructor goes way too fast for people taking
    their first electronics course.

    Anyway, thanks for any help.
  2. Ken Taylor

    Ken Taylor Guest

    In an ideal world a voltage source has no series resistance, so no matter
    what load you put on, the voltage at the output is the same. The Thevenin
    circuit assumes a simple series resistance, which then acts as a part of a
    voltage divider at the output. Therefor if you put a 10k load on your 15V
    source and see 12V at the output, you just transpose a voltage divider
    equation to find your unknown resistance.

    ie. Vo = Vs x (Rl/(Rl+Rth))

    where Vo is the Output Voltage, Vs is the Source voltage, Rl is the Load
    Resistance and Rth is the Thevenin equivalent series resistance.

    Okey dokey?


  3. JeffM

    JeffM Guest

    A certain circuit produces 15 V across its open output terminals, and

    Thevenin Analysis
    --- 15V
    Looking back with no load, we read 15V. So Vsource must be 15V.
    (No current draw == no voltage drop across Rs == no voltage divider)

    | | ^
    ----- / |
    --- \10k |
    ----- / 12V
    --- \ |
    | | v
    12V / 10k = 1.2mA

    (15V - 12V) / 1.2mA = 2k5
    (what is dropped across Rs) - (current thru Rs) = Rs
  4. (Mrs. Kerchief curtsies.) Mrs. Kerchief and tutor input 12 volts for
    the output voltage. Thanks, Ken (and also Jeff).
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