Hello, I understand why I only need to find Rth in this case. But as you can see in second photo, first, I am confused with outer loop calculation whether it's possible and second, how come I1 = Vx/5. Can someone let me know how I1 = Vx/5 comes out?

 

I am confused with outer loop calculation whether it's possible

Why do you think that it is impossible?

Can someone let me know how I1 = Vx/5 comes out?

Notice that Vx is a voltage drop across 5 ohm's resistor. So Vx = I1 * 5Ω thus I1 = Vx/5Ω

 

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Hello, I understand why I only need to find Rth in this case. But as you can see in second photo, first, I am confused with outer loop calculation whether it's possible and second, how come I1 = Vx/5. Can someone let me know how I1 = Vx/5 comes out?

Hello,

Very simply, if you know the voltage across a resistor then the current obeys Ohms Law which means:
I=V/R

and since V=Vx and R=5, the current must be Vx/5.

 

View attachment 271549View attachment 271548View attachment 271550
Hello, I understand why I only need to find Rth in this case. But as you can see in second photo, first, I am confused with outer loop calculation whether it's possible and second, how come I1 = Vx/5. Can someone let me know how I1 = Vx/5 comes out?

The author of that example is doing you a disservice by playing fast and loose with the units. Also, they are making the analysis needlessly obscure by fixing Vt at a specific voltage and by not eliminating Vx as soon as possible.

Finally, if you do the analysis symbolically, as shown in the attached PDF, a couple of interesting things pop out that are otherwise completely hidden.

What happens to Rth if R2 were chosen to be 1/3 of R1?

What happens to Rth if R2 is chosen to be 1/3 of the sum of the other two resistors?

For what range of values of R2 is Rth negative?

Can you think of any other interesting questions that could be asked about this circuit?

 

Just thought I would take this opportunity to play around with Maple.
Note that equations (1) & (6) are the circuit equations.

 

This problem can be solved in one's head by noticing that the dependent source is equivalent to a -20 ohm resistor.

 

This problem can be solved in one's head by noticing that the dependent source is equivalent to a -20 ohm resistor.

I doubt that that is something the very many people would notice.

But, it is a worthwhile exercise for people to figure out why this is the case and what it is about the circuit that makes it behave this way.

Hint: It's important that the controlling current passes through a device that is in series with the dependent source such that the voltage across the source is dependent only on the current through it (all other parameters that matter being constants).