thevenin's theorem

Discussion in 'Homework Help' started by NoMore, Oct 9, 2014.

  1. NoMore

    Thread Starter New Member

    Oct 9, 2014
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    [​IMG]

    I can't solve this problem. Please help me.
     
  2. studiot

    AAC Fanatic!

    Nov 9, 2007
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    There are four steps to obtaining the Thevenin equivalent, what have you done so far?
     
  3. NoMore

    Thread Starter New Member

    Oct 9, 2014
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    actually , i don't know how to start . I can solve the independent sources's questions. But there is a dependent current source . I've tried to solve this five or six times and different ways but i couldn't get close to the answer.
     
  4. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Try to use a superposition and find Vth voltage
    http://forum.allaboutcircuits.com/t...rrent-and-voltage-sources.101488/#post-764893

    Ix = 6V/(5 + 3 + 4) + 1.5Ix* 5/12 (current divider rule)
    12Ix = 6 + 7.5Ix
    4.5Ix = 6
    Ix = 6/4.5 = 1.3333A

    So Vth = Ix * 4 ohm = 5.33V

    Next you need to find a Rth, so you need to short A with B and find Isc current.

    Ixsc = 6V/(5 + 3 ) + 1.5Ixsc* 5/8

    Ixsc = 12A

    Rth = Vht/Ixsc = 5.333V/12A = 0.444Ohm
     
    Last edited: Oct 10, 2014
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