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 Homework Statement:
 Find the Thevenin equivalent circuit of the attached circuit. Assume that α has units of Ohms.
 Relevant Equations:

KCL: The sum of currents into and out of a node in a circuit equals 0.
KVL: The sum of voltages in any closed loop is 0.
V = IR
First, I calculate the Thevenin resistance by treating the independent current source I_{0} as an open circuit and noticing that the resistance of the dependent voltage source is α (since V=IR and the voltage across the dependent voltage source is given by αi.
In this way, I have:
R_{TH} = 1 / (1 / R_{2} + 1 / (R_{1} + α)) = (R_{1} + α)R_{2} / (R_{1} + R_{2} + α)
Now, for the Thevenin voltage V_{TH}, I thought it might be easier to find the short circuit current I_{N} through the output terminals and then simply multiply by R_{TH} to get V_{TH}.
This is where my analysis gets confused.
I've attached my redrawn circuit when I short the output terminals. Using this circuit, I think I can say the following using KCL:
I_{0} = I_{1} + I_{1} (since the current through the CCVC is I_{1})
I_{1} = I_{2} + I_{N} (can I ignore I_{2} since the current will completely avoid the resistor and prefer the shorted path?)
If I can ignore the resistor, then I_{1} = 0 + I_{N}.
At this point, I get a bit stuck. The equations don't really seem to tell me anything useful, and I don't think that I_{N} = I_{1}, but I don't know where my confusion lies exactly.
If I write out all the branch currents explicitly and solve it in a very granular fashion, I find that I_{TH} = I_{0}, but I do not know how to derive this result in a more "intuitive" manner.
If someone could point me in the right direction, that would be very helpful. Thanks.
In this way, I have:
R_{TH} = 1 / (1 / R_{2} + 1 / (R_{1} + α)) = (R_{1} + α)R_{2} / (R_{1} + R_{2} + α)
Now, for the Thevenin voltage V_{TH}, I thought it might be easier to find the short circuit current I_{N} through the output terminals and then simply multiply by R_{TH} to get V_{TH}.
This is where my analysis gets confused.
I've attached my redrawn circuit when I short the output terminals. Using this circuit, I think I can say the following using KCL:
I_{0} = I_{1} + I_{1} (since the current through the CCVC is I_{1})
I_{1} = I_{2} + I_{N} (can I ignore I_{2} since the current will completely avoid the resistor and prefer the shorted path?)
If I can ignore the resistor, then I_{1} = 0 + I_{N}.
At this point, I get a bit stuck. The equations don't really seem to tell me anything useful, and I don't think that I_{N} = I_{1}, but I don't know where my confusion lies exactly.
If I write out all the branch currents explicitly and solve it in a very granular fashion, I find that I_{TH} = I_{0}, but I do not know how to derive this result in a more "intuitive" manner.
If someone could point me in the right direction, that would be very helpful. Thanks.