- #1

LearnerErnie

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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

In this way, I have:

R

Now, for the Thevenin voltage V

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

I

If I can ignore the resistor, then I

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

If I write out all the branch currents explicitly and solve it in a very granular fashion, I find that I

If someone could point me in the right direction, that would be very helpful. Thanks.

_{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 α*i*.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.