# Finding Thevenin Equivalent

• Engineering
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 I0 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:

RTH = 1 / (1 / R2 + 1 / (R1 + α)) = (R1 + α)R2 / (R1 + R2 + α)

Now, for the Thevenin voltage VTH, I thought it might be easier to find the short circuit current IN through the output terminals and then simply multiply by RTH to get VTH.

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:

I0 = I1 + I1 (since the current through the CCVC is I1)

I1 = I2 + IN (can I ignore I2 since the current will completely avoid the resistor and prefer the shorted path?)

If I can ignore the resistor, then I1 = 0 + IN.

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 IN = I1, 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 ITH = I0, 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.

#### Attachments

lewando
Homework Helper
Gold Member
...the resistance of the dependent voltage source is α (since V=IR and the voltage across the dependent voltage source is given by αi.
α is just a scale factor for determining the voltage of the CCVS. It happens to have units of ohms, but it is not the internal resistance of the CCVS. A dependent source is like an independent source--it has no internal resistance.

• DaveE
α is just a scale factor for determining the voltage of the CCVS. It happens to have units of ohms, but it is not the internal resistance of the CCVS. A dependent source is like an independent source--it has no internal resistance.

Ah got it! Thanks very much! So we've effectively shorted the entire circuit, since the current will take the path of least resistance, meaning ##I_N = I_0##? I think this means the current through the CCVC must be ##I_0##? If so, that's super intuitive and makes perfect sense. If not...I guess I'm even more confused ...

That also means that ##R_{TH}## was calculated incorrectly. So we have

$$R_{TH} = \frac{R_1R_2}{R_1 + R_2} \quad\text{and}\quad I_N = I_0$$

Thus, $$V_{TH} = \frac{I_0R_1R_2}{R_1 + R_2}$$

I still feel like I'm missing something, since I haven't used the CCVC at all.

Last edited:
DaveE
Gold Member
I still feel like I'm missing something
Does it make sense to you that α doesn't appear anywhere in your solution? What if Io, R1, and R2 were all set to 1 and α varied from 0 → ∞. Would you expect the circuit to behave differently?

One way to find the Thevenin equivalent is to use the calculations for the open circuit voltage and the short circuit current. Try again with that approach.

lewando
Homework Helper
Gold Member
I could have been clearer, so my apologies. I did not mean zero resistance. I should have said these are ideal active devices and cannot be modeled a resistor. It is true that for a given DC operating point there will be a voltage across and current through a dependent source leading you to think about a resistance. But under different load conditions, or different circuit component values, the dependent source will try to maintain its constant voltage (or current) and yet allow the corresponding current (or voltage) to change as needed.

DaveE