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

 

For the first question, why dont you try writing a second KVL for example:
10*I1=R1*I2

 

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?

 

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