Thevenin, Norton Theorems

Discussion in 'Homework Help' started by cheyenne, Dec 9, 2007.

  1. cheyenne

    Thread Starter Active Member

    Dec 6, 2007
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    Hi, can someone please show and explain to me how to find a Norton and Thevenin equivalent circuit of the attached circuit diagram. Please help me understand the process and understand why each step is done. The variable resistor confuses me. Thanks. This is not homework I just need help in understanding these examples.
     
  2. hgmjr

    Moderator

    Jan 28, 2005
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    Did you mean to leave off the voltages for the two DC voltage sources?

    hgmjr
     
  3. cheyenne

    Thread Starter Active Member

    Dec 6, 2007
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    I just need to figure out how to get the equivalent Norton and Thevenin resistances and draw the equivalent Norton and Thevenin circuits. Can you help me with this?
     
  4. hgmjr

    Moderator

    Jan 28, 2005
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    All you should have to do is consider RL to be a specific resistance R and calculate the Thevenin resistance the way you always have. The only diference is that the Thevenin resistance expression will contain the variable R in it.

    hgmjr
     
  5. cheyenne

    Thread Starter Active Member

    Dec 6, 2007
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    Does this mean the equivalent resistance will be 120R? Also, because there are two sources, how is the equivalent source found so the equivalent circuit diagram can be drawn?
     
  6. hgmjr

    Moderator

    Jan 28, 2005
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    For this problem, I imagine the purpose of applying Thevenin's Theorem would be to determine the voltage at the intersection of the 200 ohm, the 300 ohm and RL.

    Thevenin's Theorem is typically used as a tool for reducing the complexity of the circuit to make is easier to calculate the voltage at a particular node in the circuit.

    In your example, Thevenin's Theorem can be applied to either the DC source on the left or the DC source on the right.

    Applying TT to the left side for example, you will end up with a Thevenin's equivalent voltage and a Thevenin's equivalent resistance. To start with you momentarily focus your attention on the left side voltage source, the 200 ohm resistor and RL. You can temporarily pretend that the 300 ohm resistor and the right side DC source are out of the circuit.

    Using what you know about Thevenin's Theorem, what Thevenin's equivalent voltage source and what Thevenin's equivalent resistance would you calculate. Remember, the 300 ohm and the right side DC source are out of the picture for this calculation.

    Assume that the left side voltage source has the voltage V1.

    hgmjr
     
  7. cheyenne

    Thread Starter Active Member

    Dec 6, 2007
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    Could you please tell me what you get for the Thevenin resistance so I can see if my answer is correct. Thanks
     
  8. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    cheyenne,

    It would be better if you told him what you got. Then if your incorrect, you can show him how you obtained that answer and he can point out where the error was.
     
  9. cheyenne

    Thread Starter Active Member

    Dec 6, 2007
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    Well, I got 120 Ohms, but this is just the same as the total resistance. Am I taking the correct approach?
     
  10. hgmjr

    Moderator

    Jan 28, 2005
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    Provided the object of this example is to determine the Thevenin's equivalent resistance and Thevenin's equivalent voltage seen by the resistance labeled RL looking back into the circuit, then the Thevenin's equivalent resistance is 120 ohms as you have calculated.

    Can you take the next step and calculate the Thevenin's equivalent voltage provided both DC sources are 5V?

    Note that the voltage source on the right-hand side has its positive terminal connected to ground.

    hgmjr
     
  11. cheyenne

    Thread Starter Active Member

    Dec 6, 2007
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    My experimental results show my source AD to be 5.04V and source CD to be -3.71V. Using these values I am trying to find the equivalent voltage.

    Attempt:
    I = (8.75V/500 ohms) = 17.5mA

    Voltage drop across 200 ohms resitor:
    V = 17.5mA*200 = 3.5V

    5.04-3.5 = 1.54V is the voltage drop across BD. Is the the equivalent Thevenin Resistance.
     
  12. hgmjr

    Moderator

    Jan 28, 2005
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    I see that you are referencing various voltage drops using the designations AD, CD, and BD. The diagram you posted was not annotated to provide labels to assist me in interpreting your description.

    hgmjr
     
  13. cheyenne

    Thread Starter Active Member

    Dec 6, 2007
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    I have another question regarding Thevenins Theorem. In the "Thevenin Textbook" circuit firstly I short circuited the voltage source placed an open circuit instead of the current source. Using this I got a Thevenin resitance of 1.58k Ohms.

    I assume next I have to use superposition to find the thevenin source.
    So, put open circuit instead of current source. Is the thevenin source the same as the voltage across the 5.6k resistor? If so because it is in parallel with the other resistor does that mean it is 16V as source voltage is the same as all paralle branches.

    Please help me with this problem. Thanks
     
  14. recca02

    Senior Member

    Apr 2, 2007
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    the thevenin voltage is the voltage that will appear across the load which is same as the voltage across 5.6K resistor as they are in parallel ,so yes.
    can u rephrase that? i didnt get it.
     
  15. cheyenne

    Thread Starter Active Member

    Dec 6, 2007
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    Basically I'm trying to find the voltage in the 5.6k resistor. The source is 16V. Is this the same as the voltage through the 5.6k resistor since they are in parallel? Thanks for your help :D
     
  16. recca02

    Senior Member

    Apr 2, 2007
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    remember there will be a voltage drop across the 2.2 k resistance.
    the remaining voltage will appear across the 5.6 K resistor.
    for this remove load resistor and the current source(if u are using superposition).
    it remains a simple one loop circuit from there. solve it to find the voltage across 5.6 k resistor.
     
  17. cheyenne

    Thread Starter Active Member

    Dec 6, 2007
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    Thanks for your help. I think I got it. Using superposition, with just voltage source. I get the voltage drop across the 5.6k resistor to be 11.49V

    Then when only considering current source:
    Because the two resistors are in parallel I used resistive current division to get 2.26mA across 5.6k resistor. Then voltage to be 12.64V.

    12.64-11.49 = 1.15V
    Due to the direction of the voltage source does this mean the voltage become -1.15V. Does everything else seem okay?
     
  18. recca02

    Senior Member

    Apr 2, 2007
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    that seems correct to me.
     
  19. cheyenne

    Thread Starter Active Member

    Dec 6, 2007
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    So it is -1.15V due to polarity of voltage source?
     
  20. recca02

    Senior Member

    Apr 2, 2007
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    -1.15 volts if u consider the direction of current thru it as send by the voltage source and the polarity of voltage drops as such.
    did u mean to ask something else?
     
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