Help with Thévenin and Norton equivalent circuits

Thread Starter


Joined 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:

Thread Starter


Joined 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?