Thevenin equivalent circuit with independent and dependent sources

Thread Starter

DogLoaf

Joined Oct 10, 2018
1
Hello, I have practice questions on Thevenin theorem. Where I seemed to have gotten stuck. With independent sources it's very simple to find the Thevenin equivalent circuit but this question throws one dependent source thus the steps change.

I know that to solve this, I have to:
1) Find Voc.. Which I did using nodal analysis
2) short circuit terminals a and b and then find In current passing through that short circuit
3) use Rth = Voc/In to find the Thevenin resistor

The problem arises when I short circuit the terminals. I tried using mesh analysis by putting 4 current loops but I ended up with a negative current, thus resulting in a negative Rth which isn't possible apparently.

I don't need anyone to do it for me. I just need some reassurance that i'm going in the right direction. Additionally, I also thought that when I short circuit a and b doesn't that break the whole circuit and render it useless? Anyway, would appreciate any thoughts.
 

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Jony130

Joined Feb 17, 2009
4,969
Can you show us a circuit diagram in a normal format (PNG /JPG)? I do not trust docx files.
 

WBahn

Joined Mar 31, 2012
24,684
Hello, I have practice questions on Thevenin theorem. Where I seemed to have gotten stuck. With independent sources it's very simple to find the Thevenin equivalent circuit but this question throws one dependent source thus the steps change.

I know that to solve this, I have to:
1) Find Voc.. Which I did using nodal analysis
2) short circuit terminals a and b and then find In current passing through that short circuit
3) use Rth = Voc/In to find the Thevenin resistor
That looks correct (provided you adhere to the passive sign convention when assigning the polarities of Voc and Iss).

The problem arises when I short circuit the terminals. I tried using mesh analysis by putting 4 current loops but I ended up with a negative current, thus resulting in a negative Rth which isn't possible apparently.
This is a common misconception. Since you have an active device in the circuit, there is nothing that prevents it from behaving such that the current goes down as the voltage goes up.

I don't need anyone to do it for me. I just need some reassurance that i'm going in the right direction. Additionally, I also thought that when I short circuit a and b doesn't that break the whole circuit and render it useless? Anyway, would appreciate any thoughts.
Why would it break the whole circuit and render it useless to any greater degree than open circuiting the output would?

You don't HAVE to do the analysis open-circuit and short circuit. It's a linear circuit, so the I-V relationship at the terminals is a straight line. You just need to find two points on that line. The usual approach is to find (Voc, 0 A) and (0 V, Iss). But you can place R1 and R2 across the terminals instead and find (V1, I1) and (V2, I2) and go from there. With a real circuit, you would usually do something like this, at least for one of the operating points, since you usually won't get correct data under short-circuit conditions because the circuit will usually be driven non-linear at some point prior to that.
 
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