Norton Theroem Thevenin Equiv.

Discussion in 'Homework Help' started by Hawkeye87, Oct 7, 2008.

  1. Hawkeye87

    Thread Starter Active Member

    Oct 7, 2008
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    I'm having a bit of trouble with this one. It says to find the Thevenin Equivalent and my teacher said to use Norton's theorem to find IL. I wouldn't know what to do with the two voltage sources. I'm not sure if that image got attached.
     
  2. hobbyist

    Distinguished Member

    Aug 10, 2008
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    Hi,
    The way I would do it would be to take out the RL out of the circuit.
    Then algebraically add the two voltage sources. Then using Volt. divide. equation. solve for the voltage drop across points (a) and (b).
    that will be your Vth. seen at the output.
    Now put a short across the 2 volt sources. and calculate the parrallel resitance of the 2 resistors. That will be your Rth.
    Redraw your schematic with a single volt source Vth in series with a resistance Rth.

    Thats the Thevinin equivalent circuit. What that means is that is your open circuit voltage and resistance (no load ).

    I realize your wanting the Norton equivalent so you put a short across points (a) and (b) and then Vth divided by Rth would be your current source. (short circuit current)

    Draw a current source (In.) and put Rth from above calculation, in parralel with the current source, then hook your RL in parallel with this and use Current divider equation to show current through RL.

    Some one please correct me if I gave wrong information about Nortonizing the circuit.

    Thankyou.
     
  3. Hawkeye87

    Thread Starter Active Member

    Oct 7, 2008
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    The voltage sources are flowing in the opposite directions, would i still add them algebraically? So the sum of the two voltages would be 5V flowing up? Or 15V?
     
  4. Hawkeye87

    Thread Starter Active Member

    Oct 7, 2008
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    why are the resistors on parallel? wouldn't they be in series? (after you remove RL?)
     
  5. hitmen

    Active Member

    Sep 21, 2008
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    I know Vth = 8.97V.

    I wish to know if it is series or parallel too. Can someone enlighten?
     
  6. Hawkeye87

    Thread Starter Active Member

    Oct 7, 2008
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    how did you get 8.97? i got something like 1.79. and i think the resistors are in parallel because your looking at them through node A in which the "current" splits into to paths, one going through the bottom resistor and one through the top. Which i could be wrong.
     
    Last edited: Oct 8, 2008
  7. hobbyist

    Distinguished Member

    Aug 10, 2008
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    Yes, I got 1.79 volts across (a) and (b)

    Yes your right about the current paths remember you need to put a short across the voltages to measure the equivalent resistance of a circuit.
     
  8. hgmjr

    Moderator

    Jan 28, 2005
    9,030
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    Greetings hawkeye87,

    I recommend you take a look at the presentation on Thevenin's Theorem in the AAC ebook.

    hgmjr
     
  9. hitmen

    Active Member

    Sep 21, 2008
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    I got 8.97 which is the pd across the 56k. I thibk I am wrong. Can you explain how u got 1.79?

    If you open circuit the ONLY loop, how are you going to get current flow?

    I dont understand.:confused:

    Rth = 100k//56k. am i right?
     
  10. hobbyist

    Distinguished Member

    Aug 10, 2008
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    56
    This is how I got 1.79.v.
     
  11. hitmen

    Active Member

    Sep 21, 2008
    159
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    I got the polarities of the battery wrong. No wonder I got an incorrect answer!
     
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