I am currently self-studying through MITs 6.002 Spring 2007 course. There are currently no solutions for the assignments afaik. Normally, this isn't an issue, but I am having some trouble with this Thevenin equivalent problem. I was hoping to get a pointer as to why and how my analysis is incorrect.
The question asks us to determine the Thevenin equivalent of the following circuit, where \$\alpha\$ has units of Ohms.
My solution attempt is as follows:
To find the Thevenin resistance \$R_{TH}\$, we treat the current source as an open circuit. Since the dependent voltage source has a voltage given by \$\alpha\$, we know that the resistance through the dependent voltage source is \$\alpha\$. Thus, calculating the resistance from the positive terminal, we see:
$$ R_{TH} = \frac{1}{\frac{1}{R_2} + \frac{1}{R_1 + \alpha}} = \frac{(R_1 + \alpha)R_2}{R_1 + R_2 + \alpha} $$
Next, my intuition is to short the output terminals and determine the current between the terminals as \$I_N\$. We can then multiply by \$R_{TH}\$ to find the Thevenin voltage \$V_{TH}\$.
To do this, I treat the negative port as ground and apply KCL to the current flowing into the "ground node".
I need to determine the branch current through \$R_2\$ first. Considering the node where the dependent voltage source meets the positive output terminal, I believe that the following is true:
$$ i + I_{N} = i_2 $$
where \$i_2\$ is the current flowing through \$R_2\$.
So, putting together these 4 currents, I think we can correctly say the following about the ground node:
$$ -I_0 + i + (i + I_N) + I_N = 0 \\ \implies I_N = \frac{I_0 - 2i}{2} $$
(At this point, I think something is wrong with my analysis...)
Thus,
$$ V_{TH} = I_NR_{TH} = \frac{I_0 - 2i}{2} \frac{(R_1 + \alpha)R_2}{R_1 + R_2 + \alpha} $$
EDIT: I did not realize that a dependent voltage source also has no internal resistance (since it is an ideal), and thus the short circuit is obviously just \$I_N = I_0\$.