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
    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


    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.


    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

    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


    Jul 9, 2008
    For the first question, why dont you try writing a second KVL for example:
  3. phusion9

    Thread Starter New Member

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

    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
    Ok never mind I got help and finaly managed to solve it.