Help with Thévenin and Norton equivalent circuits

Discussion in 'Homework Help' started by phusion9, Jul 10, 2008.

  1. phusion9

    Thread Starter New Member

    Jul 10, 2008
    3
    0
    I'm having some issues with these two exercises about Thevenin and Norton equivalent circuits with dependent voltages.

    1) Determine the equivalent Thevenin Circuit seen by the terminals a-b

    [​IMG]

    So first thing, I apply KVL:

    V1-10*I1+I1*R1+I1*R2=0 \Leftrightarrow I1 = -1 A
    V1 + R2*I1 - Vth = 0 \Leftrightarrow Vth = 10 V
    (Vth = Vab)

    All fine. But now to determine the Rth I would need to know the Isc, the current in short circuit that passes through a and b.

    [​IMG]

    Again I try to apply KVL

    -10 V1 + V1 + R1*I2 + R2*I1=0

    and KCL

    Isc + I1 = I2

    Results in a equation with 2 variables I1 and Isc which I can't solve.

    -10 I1 + V1 + R1*(Isc + I1) + R2*I1 = 0

    What I am missing or doing wrong?

    2) Determine the equivalent of Thevenin and Norton seen by the terminals a-b

    [​IMG]
    I try to apply KVL and KCL.

    I3 = Current that passes right-left through the dependent voltage 5*Ix
    IR2 = Current that passes down-up through the resistance R2

    Ix = I1+IR2+I2

    5*Ix-Ix*R1-IR2*R2=0 \Leftrightarrow Ix = (-R2*I1-R2*I2) / (5-R1-R2) = 5.45 A

    With Ix I determine IR2= -9.55 A

    The Vth or Vab would be IR2*R2 = -38.2 V which is wrong.

    What am I doing wrong?
    How would I then determine the IN?
    Should I try to apply the sobreposition theorem and to the first exercise as well?

    (IN = Current equivalent of Norton, short circuit)

    By the way the solution is:
    1) Vth = 10 V; Rth = 16 ohm
    2) Vth = 48.571 V; IN=8 A; Rth=RN= 3.238 ohm
     
    Last edited: Jul 10, 2008
  2. blurium

    Member

    Jul 9, 2008
    18
    0
    For the first question, why dont you try writing a second KVL for example:
    10*I1=R1*I2
     
  3. phusion9

    Thread Starter New Member

    Jul 10, 2008
    3
    0
    Yea I was missing that simple KVL loop... Thank you
    So the solution for Isc would be:

    V1=R2*I1
    10*I1=R1*I2
    Results in
    I1= 1.25 A
    I2 = 0.625 A
    Isc (down-up) + I2 = I1 \Leftrightarrow Isc=0.625 A

    Vth/Isc = 16 ohm (correct)

    Would you take a look to the second exercise as well?
     
  4. phusion9

    Thread Starter New Member

    Jul 10, 2008
    3
    0
    Ok never mind I got help and finaly managed to solve it.
     
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