Question about thevenin circuit analysis with voltage across voltage source

Question

I have a certain circuit in mind and I want to find its Thevenin equivalent resistor:

I see that there is an dependent voltage source so I add a voltage source of 1V between a and b and make the other voltage source equal to 0 (the independent 9v). when I use nodal analysis on the new circuit, I am not sure how to calculate the current i3. I thought about using Kirchhoff voltage law on the middle loop but for now I am confused. If I have three voltage sources, do I just take a wild guess about which way the current goes when I apply KVL on the middle loop? because if they didn't note the direction of the current on the Vx resistor, I would have said that Vx+2i3-2Vx+1=0, or would that be wrong because I assumed i3's direction to go to the right?

tl;dr

• how would I do a KVL equation on the middle loop
• is there an easier way to solve it using nodal analysis (specifically, to find i3)?

Sorry if the question seems a bit confused, because I'm confused.

Answer 1

This is how i think i would solve the problem:

I don't see it necessary to add a 1V excitation source since there already a 9v in the circuit. The trick when analyzing circuit with dependent sources is to avoid the Short circuiting you would normally do when having a circuit with independent sources alone:

_Things to note that may help

Focus on node A:

define a current from the 9v source direction to node A and also define a current from 6ohm to A

_not Vx =Va

9 - Vx = I1

Vx/6 = I2

and i reckon Vab = 2Vx..

Hope this helps you!

Answer 2

If your objective is to find the Thevenin resistance, then I'm not sure why you're approaching it this way.

If you can find the Thevenin resistance \$R'_{th}\$ of the circuit to the left of the \$1\Omega\$ resistor, the total equivalent resistance is just

$$R_{th} = R'_{th}|| 1\Omega $$

And, I recommend using a current test source which will make finding \$R'_{th}\$ almost trivial.

Answer 3

For this particular question, i think the source conversion technique would be the best.

You convert the independent voltage source and 3 ohm resistor to a current source of 3A with 3 ohm res in parallel.

Next, this 3 ohm res is in parallel with 6 ohm res. Reduce it to a single res of 2 ohm. So now you have a current source of 3A in parallel with 2 ohm res.

At this point again use source conversion. You get a voltage source of 6V in series with a 2 ohm res, combined in series with 2 ohm res and dependent source. The remainder of the network can be easily solved.

A major benefit of this technique is that you have reduced one loop completely.

Hope i am correct.