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Thevenin's Equivalent Help

Discussion in 'Electronics Homework Help' started by sjgallagher2, Aug 13, 2013.

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  1. sjgallagher2

    sjgallagher2

    17
    0
    Jan 27, 2013
    For some reason I find immense difficulty in figuring out these circuits. I must be missing something somewhere :p Here's my issue. I have this circuit:
    [​IMG]
    And I need to find the Thevenins equivalent circuit. When I first got the problem I confidently threw some voltage divider equations around and got the thevenin voltage as 0.75V. But that's wrong. It's actually 0.5V. Why? Thats where I need help! If you could explain that to me I would be very thankful.

    More info on what I did to get my answer.
    First I used a voltage divider equation on the 200ohm resistors. That resulted in 1/2(3V) = 1.5V. Then I put that voltage in the next voltage divider equation with the 100ohm resistors, which came out as 1/2(1.5) = 0.75. It seemed so simple, but I went wrong. Where?

    Someone suggested putting the two 100ohm resistors in series and solving from there but I'm not sure that's the right thing to do
     
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,496
    2,837
    Jan 21, 2010
    Homework?

    The trick for Thevinin that I usually employ is to find the open circuit voltage and the short circuit current. The rest then just falls out.

    The open circuit voltage is simple, and can almost be done by inspection.

    First determine the voltage at the junction of the two 200 ohm and the 100 ohm resistors.

    Knowing that you can determine what the voltage between A and B is.

    Those can be done by inspection!

    (you've tried to do this, but you need to realise that the voltage divider at this point is not just two 200 ohm resistors. That arrangement of one of the 200 ohm and the two 100 ohm resistors produces a resistance of...?

    Now determine the short circuit current. Do this by shorting A and B and determining the current at this point.

    This is a little trickier to do by inspection, but if you re-draw the circuit you'll find that the only difficulty is that the numbers aren't quite so nice so the figure doesn't fall out so simply.
     
  3. sjgallagher2

    sjgallagher2

    17
    0
    Jan 27, 2013
    Finally! Someone that could tell me what I was doing wrong, you're exactly right! Everyone else was really unclear about that. Now I get what I was doing wrong. Thanks so much.
     
  4. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,496
    2,837
    Jan 21, 2010
    No problems. Glad to help.
     
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