Help with Thevenin / Norton Equivalent

Discussion in 'Homework Help' started by jboavida, Jul 16, 2008.

  1. jboavida

    Thread Starter Member

    Jul 10, 2008
    23
    0
    Hi All,

    Can you please help me on the following circuit:

    I want to calculate the Thevenin / Norton equivalent.

    What is the influence of the current source on the Thevenin resistence?
    Which is the best way to calculate this circuit? Using KVL on both loops?
    When we calculate the Thevenin voltage equivalente we live out the current source and use the sobreposition theorem?



    Thanks in advance for any help.

    Joaquim
     
  2. jboavida

    Thread Starter Member

    Jul 10, 2008
    23
    0
    From another website I had found that the current source is taked out for the calculation of the Rthev.
    So Rthev will be 7.69 ohm right? (considering ideal voltage sources).

    Knowing that the current in the outer loop is 5A the volatage drop across the 10R resistor will be 50V?

    Sorry, I'm learning this for myself. I just need to confirm my calculations and some guidance.

    Joaquim
     
  3. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    jboavida,

    You are correct. Current sources are considered to have infinite impedance and voltage soureces are considered to have zero impedance.

    No, 10R is in a different loop. I would solve the problem by breaking the circuit into three loops. The west loop is already given as -5A. Solve for the current of the north and south loops using two equations for two unknowns, and you can easily determine the voltage across A-B. Ask again if you need help setting up the equations. Ratch
     
  4. jboavida

    Thread Starter Member

    Jul 10, 2008
    23
    0
    Tks for your help Ratch. I will like help for the equations, but here is my attempt:
    Consigering North I1, South I2 we have (KVL):

    North: -40-50=(10*I1)+(15*I1)+(25*I2)
    South: 50+20=(20*I2)+(25*I1)

    Are these correct?

    Thanks
    Joaquim
     
  5. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    jboavida,

    Assume mathematically that currents are positive for clockwise direction. Then determine all the elements in each loop and the common elements between each loop.

    So:

    50*I1 - 25*I2 = (-40-50-75)

    -25*I1 +45*I2 = 50 + 20

    Easily solved for I1 = -3.4923, I2 = -0.38461

    25 ohms is the common resistance between the two loops. -75 volts is contribution of the 5A source on the 15 ohm resistor. The minus signs indicate that mathematically, the current is in the opposite or CCW direction. Voltage across the A-B points is 20*(I2) = -0.76923.

    Ask if you have any questions.

    Ratch
     
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