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So...I have to find \$v0\$ using exclusively Thévenin/Norton equivalents. I am completely lost since in this case I have a dependent voltage source which relies on \$v0\$ itself.

\$v1,a,R1,R2\$ and \$Rq\$ are known values.

If I understand the procedure correctly, I should obtain \$R_{TH}\$ by "removing" dependent sources and then use Norton equivalents, where I would end with:

\$V_{TH} = R_{TH}*I_{N}\$

enter image description here

Any help/clues are appreciated.

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I already understood my misinterpretation of the question before. vo is the output voltage (measured between A and B) and not the voltage through the resistor. Therefore, the dependent source depends on the voltage between A and B and not on the voltage through the resistor.

That should not cause any trouble because what happens in the open-circuit test is that \$ v_O\$ will be equal to \$ V_{Th}\$ (it means, it's equal to the open-circuit voltage, in this case). Therefore we simply apply KVL and ohm's law:

$$V_{Th} = R_2 * I$$ $$R_1*I + a*V_{Th} + R_2*I - v_1 =0$$

After doing some manipulation

$$Vth=\frac{R_2}{R_1 + (1+a) R_2}*v_1$$

Since the equivalent resistance was correct, I'll reproduce my previous resolution:

Now for the equivalente resistance, I'm going to attach to AB a voltage source equal to \$v_o\$. A current \$i_o\$ will flow to the circuit and my ratio \$v_o/i_o\$ will be the equivalent resistance. Don't forget of course to short-circuit the independent voltage source. Applying KVL and KCL:

$$- v_0 - a v_0 + R_1 * i_1=0$$ $$v_0=R_2*i_2$$ $$i_0=i_1+i_2$$

And therefore

$$v_0=\frac{R_1*i_1}{1+a}$$ $$i_0=i_1+\frac{v_0}{R_2}$$

And after some manipulation:

$$i_1=\frac{v_0*(1+a)}{R_1}$$ $$i_0=\frac{v_0*(1+a)}{R_1}+ \frac{v_0}{R_2}$$ $$i_0=\frac{R_2(1+a)+R_1}{R_1*R_2}*v_0$$ $$R_{eq}=\frac{v_0}{i_0}=\frac{R_1*R_2}{R_2(1+a)+R_1}$$

Ok, now you have your Thèvenin circuit, computing the output voltage must be easy. Sorry for the confusion, I wrongly assumed that the voltage vO was the voltage across the resistor (and therefore the dependent voltage source depended on the voltage across the resistor) which is not the case. Good luck!

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  • \$\begingroup\$ And there we go with the assholes downvoting again. \$\endgroup\$ Mar 21, 2019 at 16:33
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    \$\begingroup\$ I haven't read your answer above. I want to comment about down-voting. People have differing opinions about completed answers. Some of those who don't want completed answers given will down-vote. I feel that while a completed answer may hurt the questioner (who is 'successfully cheating' because of it), it can help hundreds more who need to learn to work such problems. So I feel the value of a completed answer is MORE important than my policing people. Answers help more than they hurt. So don't fret the down-vote. That's their problem, not yours. Stay the course if you feel strongly about it. \$\endgroup\$
    – jonk
    Mar 21, 2019 at 18:19
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    \$\begingroup\$ Over time, you will find a balancing point that mitigates the down-votes. You can find better ways to provide much of the more difficult aspects that leaves an almost-complete answer. But where the OP still needs to take a few further steps they should be able to achieve. This still helps everyone reading the question (because they, too, should be able to handle the remaining aspects with less difficulty) and you will slightly less often trip up over the opinions of some who want to force you into being a cooperating policeman here (their problem, as I said, not yours.) \$\endgroup\$
    – jonk
    Mar 21, 2019 at 18:26

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