Thevenin Theorem Help (Existence of Dependent Source) X.X

Discussion in 'Homework Help' started by xiahbaby, Apr 14, 2009.

  1. xiahbaby

    Thread Starter New Member

    Feb 4, 2009
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    [​IMG]

    What I did was to short-circuit the 20V, but then I was told I couldn't do that due to the existence of a dependent source.
    Hence how should I proceed from here?:confused::confused:
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    To find Thevenin's equivalent at terminals a & b you need to find

    1. The open circuit voltage Vab = Vth.
    2. The current flowing from a to b when a & b are shorted together.
    3. Hence find Rth = Vab(open circuit)/Iab(short circuit)

    To find the open circuit Vab [Vth], apply Kirchoff's voltage law in the left hand loop to find the current and then Vab = 6*i.

    To find the short circuit current between a & b apply Kirchoff's rules for the two-loop network formed when a and b are shorted.

    If you're not sure how to start with the first part

    from KVL

    20 = 6*i + 2*i + 6*i

    Hence i = ?

    Then Vab(open circuit) = Vth = 6*i

    There's a slightly more "direct" way to get Vab(open circuit) if you can see the following works.

    Remember Vab = 6*i, and thus i=Vab/6

    So using KVL

    20 = 6*(Vab/6) + 2*(Vab/6) +6*(Vab/6)

    Gives Vab = Vth directly after some re-arranging.

    You should try to find Iab (short circuit) and then Rth yourself.
     
  3. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    Shouldn't it be:

    20 = 6*i - 2*i + 6*i

    and also:

    20 = 6*(Vab/6) - 2*(Vab/6) +6*(Vab/6)
     
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  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Sorry - my mistake.

    Thanks "electrician" - well spotted as usual!
     
  5. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    When the day comes that I don't make mistakes like that myself, we'll know that the Millenium has arrived!
     
    Last edited: Apr 14, 2009
    ujnikm likes this.
  6. hgmjr

    Moderator

    Jan 28, 2005
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    Both good hints guys. It will be interesting to see what xiahbaby can do with the information.

    hgmjr
     
  7. jake_up@100

    New Member

    Feb 23, 2009
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    Why consider the Vth equal to the voltage across the 6 ohm resistor when there is the existence of the 10 ohm resistor?

    Given that the 10 ohm resistor is omitted because it is connected to an open circuit, then Voltage at a and b should also be omitted isn't? confusing.

    thanks friends!
     
  8. Nominal

    New Member

    Sep 12, 2009
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    The voltage drop across a resistor is given by V=IR. For the 10 ohm resistor there is no current flowing across it (on one end is an open circuit), so there can be no voltage drop. That means that the voltage on one side of the resistor is the same as on the other side, hence the same as node c.
     
  9. ujnikm

    New Member

    Nov 15, 2012
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    how to get Isc ????????? please help me
     
  10. WBahn

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    Mar 31, 2012
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