Help Applying method to find Thevenin and Norton Equivalents

Discussion in 'Homework Help' started by IronBrain, Mar 3, 2010.

  1. IronBrain

    Thread Starter New Member

    Mar 3, 2010
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    I am to find the Thevin and Norton equivalents of the following circuits, my problem being I am new to these method and any help is appreciated also is there any way to simplify this circuit properly?


    [​IMG]
     
  2. LoganFife

    New Member

    Feb 7, 2010
    13
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    what have you tried? do you know the basic steps you should be taking?

    ... is that thing in the middle a current source?
     
  3. IronBrain

    Thread Starter New Member

    Mar 3, 2010
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    Yes, it is a current source.
    I haven't a good idea of how to approach this problem.
    Is there anything I can simplify in the circuit?
    If, so how would I procede on accurately?
     
  4. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    The symbol as drawn is somewhat ambiguous. The + and - labels makes it appear to be a current-dependent-voltage-source. The arrow inside the diamond makes it appear to be a current-controlled-current-source.

    It makes a significant difference which one it happens to be. Can you clarify the symbol for us?

    hgmjr
     
  5. IronBrain

    Thread Starter New Member

    Mar 3, 2010
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    ....I stated the clarifaction above.
    It is a dependent current controlled current source, the indication of the polarties indicate the direction of the current source, and it is dependent on the current Ix I indicated above.
     
  6. hgmjr

    Moderator

    Jan 28, 2005
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    The direction of the arrow (inside the symbol) is all that is needed to know what direction the current is flowing in. The + and - sign are redundant.

    hgmjr
     
  7. Heavydoody

    Active Member

    Jul 31, 2009
    140
    11
    I assume you are looking in from the right side. Find the open circuit voltage between those two ends. Next, short out those two ends and determine your short circuit current (the current through your imaginary short). The open circuit voltage is the Thevenin voltage. That voltage divided by the short circuit current is the Thevenin resistance. Together, they make up the Thevenin equivalent (the resistance in series with the voltage source). Perform source transformation on that and you wind up with a Norton equivalent (the current source in parallel with the resistance).
     
  8. IronBrain

    Thread Starter New Member

    Mar 3, 2010
    6
    0
    Thanks for the feedback.
    Ill try that out and report back if any problems, so I assume there is nothing I can simplify in the circuit right now?

    Does the dependent source conflict with me applying these methods?
     
  9. Heavydoody

    Active Member

    Jul 31, 2009
    140
    11
    Sorry I disappeared last night. It was a long day. So, I am back from school and I solved this problem.

    First, to answer your last question, no further simplification is necessary. The dependent source only adds a third equation to your two nodal equations (three unknowns, three equations).

    The numbers here seemed a bit awkward, so if anyone else can confirm this, it would be cool. I came up with -5 volts and 11/2 ohms for the Thevenin and -10/11 amps with the 11/2 ohms for the Norton.
     
  10. hgmjr

    Moderator

    Jan 28, 2005
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    214
    I calculated Vth = -5 Volts and Rth = 11/2 Ohms.

    hgmjr
     
  11. Heavydoody

    Active Member

    Jul 31, 2009
    140
    11
    Thanks for the confirmation. The numbers had me wondering.
     
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