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Exercise about Thevenin equivalent circuit

  1. Sep 14, 2015 #1
    Hi all :)

    I'm trying to figure out the Thevenin equivalent of this circuit:

    Circuito.jpg

    I start by switching off every source and calculating the equivalent resistance:

    RTh = RN = (6 + 14) || 5 = 4 Ω


    Then I proceed with finding the voltage at 2.

    I note that the nodes between 1 and 2 form a supernode:

    Supern.jpg

    So I set up the (super)nodal equation:

    1 = v1/6 + 3 + v2/5

    Then I proceed with mesh analysis. I note than the voltage source and the current source I1 form a supermesh.

    Superm.jpg

    And now I have some problems at setting up the equation :/

    I know that:

    -14 + 14*i1 + 5*i2 - 6(1 - i1) = 0

    To bring this equation as a function of v1 and v2

    I tried to think up about these equivalences, but I probably mistakened since the equation would become v1 = 0

    14i1 = 14 - v2
    5i2 = v2
    6(1 - i1) = v1


    Thanks in advance :)
     
  2. jcsd
  3. Sep 14, 2015 #2
    If you have a problem with a mesh analysis why you do not proceed with the node analysis ? Also you can use a source transformation and convert a current source into a voltage source.
    And your supernode equation is also wrong because V2 - V1 is not equal to 14V.

    EDIT

    The correct super mash look like this:
    Circuitoa.JPG

    So we have
    I1 = 3A
    I2 = 1A

    and for loop I3 we have:

    -14V + 14Ω*I3 + 5Ω*(I3 - I1) + 6Ω*(I3 - I2) = 0

    And please notice that I3 is always in first place in parentheses.

    And the solution is
    http://www.wolframalpha.com/input/?i=-14+++14*x+++5*(x+-+3)+++6*(x+-+1)+=+0

    I3 = 7/5A = 1.4A and the current flowing through 5 ohm resistor is equal to:
    IR1 = I3 - I1 = 1.4A - 3A = -1.6A and this means that V2 voltage is equal to :
    V2 = -1.6A*5Ω = -8V
     
    Last edited: Sep 14, 2015
  4. Sep 15, 2015 #3
    Thanks for the answer, I had mistakened with R1 resistance :)
     
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