# Solution Verification: Thevenin Equivalent with Dependent Voltage Source

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\$$.

First, I will present a method that uses Mathematica to solve this problem. I know that this approach is not 'smart' but this method will work all the time, even when the circuit is way complicated than this one. This makes my answer valuable in my opinion.

Well, we are trying to analyze the following circuit: simulate this circuit – Schematic created using CircuitLab

When we use and apply KCL, we can write the following set of equations:

$$\begin{cases} \text{I}_0=\text{I}_1+\text{I}_4\\ \\ \text{I}_4=\text{I}_2+\text{I}_3\\ \\ \text{I}_5=\text{I}_2+\text{I}_3\\ \\ \text{I}_0=\text{I}_1+\text{I}_5 \end{cases}\tag1$$

When we use and apply Ohm's law, we can write the following set of equations:

$$\begin{cases} \text{I}_1=\frac{\text{V}_1}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_2}{\text{R}_2}\\ \\ \text{I}_3=\frac{\text{V}_2}{\text{R}_3} \end{cases}\tag2$$

And we also know that $$\\text{V}_2-\text{V}_1=\text{n}\cdot\text{I}_1\$$.

We can use $$\(2)\$$ and $$\\text{V}_2-\text{V}_1=\text{n}\cdot\text{I}_1\$$ in order to rewrite $$\(1)\$$:

$$\begin{cases} \text{I}_0=\frac{\text{V}_1}{\text{R}_1}+\text{I}_4\\ \\ \text{I}_4=\frac{\text{V}_2}{\text{R}_2}+\frac{\text{V}_2}{\text{R}_3}\\ \\ \text{I}_5=\frac{\text{V}_2}{\text{R}_2}+\frac{\text{V}_2}{\text{R}_3}\\ \\ \text{I}_0=\frac{\text{V}_1}{\text{R}_1}+\text{I}_5\\ \\ \text{V}_2-\text{V}_1=\text{n}\cdot\frac{\text{V}_1}{\text{R}_1} \end{cases}\tag3$$

Now, we can set up a Mathematica-code to solve for all the voltages and currents:

In:=Clear["Global*"];
FullSimplify[
Solve[{I0 == I1 + I4, I4 == I2 + I3, I5 == I2 + I3, I0 == I1 + I5,
I1 == V1/R1, I2 == V2/R2, I3 == V2/R3, V2 - V1 == n*I1}, {I1, I2,
I3, I4, I5, V1, V2}]]

Out={{I1 -> (I0 R2 R3)/((n + R1) R2 + (n + R1 + R2) R3),
I2 -> (I0 (n + R1) R3)/((n + R1) R2 + (n + R1 + R2) R3),
I3 -> (I0 (n + R1) R2)/((n + R1) R2 + (n + R1 + R2) R3),
I4 -> (I0 (n + R1) (R2 + R3))/((n + R1) R2 + (n + R1 + R2) R3),
I5 -> (I0 (n + R1) (R2 + R3))/((n + R1) R2 + (n + R1 + R2) R3),
V1 -> (I0 R1 R2 R3)/((n + R1) R2 + (n + R1 + R2) R3),
V2 -> (I0 (n + R1) R2 R3)/((n + R1) R2 + (n + R1 + R2) R3)}}


Now, we can find:

• $$\\text{V}_\text{th}\$$ we get by finding $$\\text{V}_2\$$ and letting $$\\text{R}_3\to\infty\$$: $$\text{V}_\text{th}=\frac{\text{I}_0\text{R}_2\left(\text{n}+\text{R}_1\right)}{\text{R}_1+\text{R}_2+\text{n}}\tag4$$
• $$\\text{I}_\text{th}\$$ we get by finding $$\\text{I}_3\$$ and letting $$\\text{R}_3\to0\$$: $$\text{I}_\text{th}=\text{I}_0\tag5$$
• $$\\text{R}_\text{th}\$$ we get by finding: $$\text{R}_\text{th}=\frac{\text{V}_\text{th}}{\text{I}_\text{th}}=\frac{\text{R}_2\left(\text{R}_1+\text{n}\right)}{\text{R}_1+\text{R}_2+\text{n}}\tag6$$

Where I used the following Mathematica-codes:

In:=FullSimplify[
Limit[(I0 (n + R1) R2 R3)/((n + R1) R2 + (n + R1 + R2) R3),
R3 -> Infinity]]

Out=(I0 (n + R1) R2)/(n + R1 + R2)

In:=FullSimplify[
Limit[(I0 (n + R1) R2)/((n + R1) R2 + (n + R1 + R2) R3), R3 -> 0]]

Out=I0

In:=FullSimplify[%2/%3]

Out=((n + R1) R2)/(n + R1 + R2)
`
• Wow, thanks for this tip. I did not know Mathematica could solve circuits. Nov 24, 2021 at 17:57
• @clo_jur you're welcome. Mathematica can solve mathematics and when you write the equations for your circuit Mathematica can solve it. Nov 24, 2021 at 17:59
• Am I being dense, or is it the case that $I_{TH}$ must equal $I_0$ because $I_0$ is an independent current source, meaning that it will guarantee $I_0$ throughout the circuit? For some reason, I thought that the current source would still get divided through the circuit nodes that branch. Is this incorrect thinking? Nov 24, 2021 at 17:59
• @clo_jur When you find the short circuit current you will see that the current source will be the only contributor. Nov 24, 2021 at 18:01
• :facepalm: Thanks very much. Not sure why my brain wasn't working on this one. Nov 24, 2021 at 18:03

As you've determined, the Thevenin resistance Rth = R2//(R1 + a) by turning off (open circuit) the independent current source Io.

For the short circuit current Isc, the short circuit at the output terminals would not only by-pass R2 such that i2 = 0, but also connect the dependent voltage source directly across R1. So the voltage across R1 is (R1)i = -ai. It follows that i = 0, and therefore Isc must be equal to Io. Notice that the dependent voltage source is also off (short circuit) since ai = 0.