# Thevenin's Theorem: Finding Rth with Dependent Source

I was working on the problem below (which I was supposed to find Vo using Thevenin's Theorem) and tried to find R_TH but got the wrong answer.

The issue is that there is a dependent current source, so I needed to inject voltage to find R_TH. The goal was to find the current flowing through the voltage source, and then R_TH = 1 V / I, but I got the wrong answer. Does anyone know what I did wrong?

Also, I did a source transformation before I stimulated the circuit with the 1 V, but that should be fine I think.

• If you source transform the left side resistor and the dependent current source, you should get a dependent voltage source and a thevenin resistance. Right? And this, without impairing Ix, either. You can then just move the 6 V source backwards (it's all in series, anyway) to ground. So you have a +6 source, followed by a dependent source, followed by 3k followed by the 1k to ground. And where the loop current is Ix. Not so? I find (6 + 2k * Ix)/4k = Ix. And just solve for Ix as 3 mA. Easy. Commented Oct 16, 2023 at 2:30

The issue was I was doing KVL incorrectly. It should have been 1 - 2Ix +3Ix = 0, which gives me Ix=-1 mA, which is correct.

• Not correct. See here. Commented Oct 16, 2023 at 18:14

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) more complicated than this one. Also, this method will check your work.

Well, we are trying the 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} \begin{alignat*}{1} \text{n}\cdot\text{I}_2&=\text{I}_1+\text{I}_2\\ \\ \text{I}_2&=\text{I}_3+\text{I}_4\\ \\ \text{I}_0&=\text{I}_3+\text{I}_4\\ \\ \text{n}\cdot\text{I}_2&=\text{I}_0+\text{I}_1 \end{alignat*} \end{cases}\tag1

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

\begin{cases} \begin{alignat*}{1} \text{I}_1&=\frac{\displaystyle\text{V}_1-0}{\displaystyle\text{R}_1}\\ \\ \text{I}_2&=\frac{\displaystyle\text{V}_1-\text{V}_2}{\displaystyle\text{R}_2}\\ \\ \text{I}_3&=\frac{\displaystyle\text{V}_3-0}{\displaystyle\text{R}_3}\\ \\ \text{I}_4&=\frac{\displaystyle\text{V}_3-0}{\displaystyle\text{R}_4} \end{alignat*} \end{cases}\tag2

We also know that $$\\displaystyle\text{V}_3-\text{V}_2=\text{V}_\text{i}\$$.

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

In[1]:=Clear["Global*"];
FullSimplify[
Solve[{n*I2 == I1 + I2, I2 == I3 + I4, I0 == I3 + I4,
n*I2 == I0 + I1, I1 == (V1 - 0)/R1, I2 == (V1 - V2)/R2,
I3 == (V3 - 0)/R3, I4 == (V3 - 0)/R4, V3 - V2 == Vi}, {I0, I1, I2,
I3, I4, V1, V2, V3}]]

Out[1]={{I0 -> ((R3 + R4) Vi)/(R3 R4 - (-1 + n) R1 (R3 + R4) + R2 (R3 + R4)),
I1 -> -(((-1 + n) (R3 + R4) Vi)/(-R3 R4 + (-1 + n) R1 (R3 + R4) -
R2 (R3 + R4))),
I2 -> ((R3 + R4) Vi)/(R3 R4 - (-1 + n) R1 (R3 + R4) + R2 (R3 + R4)),
I3 -> (R4 Vi)/(R3 R4 - (-1 + n) R1 (R3 + R4) + R2 (R3 + R4)),
I4 -> (R3 Vi)/(R3 R4 - (-1 + n) R1 (R3 + R4) + R2 (R3 + R4)),
V1 -> -(((-1 + n) R1 (R3 + R4) Vi)/(-R3 R4 + (-1 + n) R1 (R3 + R4) -
R2 (R3 + R4))),
V2 -> -((((-1 + n) R1 - R2) (R3 +
R4) Vi)/(-R3 R4 + (-1 + n) R1 (R3 + R4) - R2 (R3 + R4))),
V3 -> (R3 R4 Vi)/(R3 R4 - (-1 + n) R1 (R3 + R4) + R2 (R3 + R4))}}


Now, we can find:

• $$\\text{V}_\text{th}\$$ we get by finding $$\\text{V}_3\$$ and letting $$\\text{R}_4\to\infty\$$: $$\text{V}_\text{th}=\frac{\displaystyle\text{R}_3\text{V}_\text{i}}{\displaystyle\text{R}_1\left(1-\text{n}\right)+\text{R}_2+\text{R}_3}\tag3$$
• $$\\text{I}_\text{th}\$$ we get by finding $$\\text{I}_4\$$ and letting $$\\text{R}_4\to0\$$: $$\text{I}_\text{th}=\frac{\displaystyle\text{V}_\text{i}}{\displaystyle\text{R}_1\left(1-\text{n}\right)+\text{R}_2}\tag4$$
• $$\\text{R}_\text{th}\$$ we get by finding: $$\text{R}_\text{th}=\frac{\displaystyle\text{V}_\text{th}}{\displaystyle\text{I}_\text{th}}=\frac{\displaystyle\text{R}_3\left(\text{R}_1\left(1-\text{n}\right)+\text{R}_2\right)}{\displaystyle\text{R}_1\left(1-\text{n}\right)+\text{R}_2+\text{R}_3}\tag5$$

Where I used the following Mathematica codes:

In[2]:=FullSimplify[
Limit[(R3 R4 Vi)/(R3 R4 - (-1 + n) R1 (R3 + R4) + R2 (R3 + R4)),
R4 -> Infinity]]

Out[2]=(R3 Vi)/(R1 - n R1 + R2 + R3)

In[3]:=FullSimplify[
Limit[(R3 Vi)/(R3 R4 - (-1 + n) R1 (R3 + R4) + R2 (R3 + R4)),
R4 -> 0]]

Out[3]=Vi/(R1 - n R1 + R2)

In[4]:=FullSimplify[%2/%3]

Out[4]=((R1 - n R1 + R2) R3)/(R1 - n R1 + R2 + R3)
`

$$\text{V}_\text{th}=3\space\text{V}\space\wedge\space\text{I}_\text{th}=\frac{3}{500}=0.006\space\text{A}\space\wedge\space\text{R}_\text{th}=500\space\Omega\tag6$$

The following solution gives details for the approach offered by periblepsis.

First, redraw the circuit giving names to the resistors, voltages, and currents so they can be identified unambiguously. There are two $$\ 1\,\mathrm{k}\Omega \$$ resistors, note, but in general do algebra with symbols, then substitute values and do arithmetic.

Using the Thévenin-equivalent for $$\R_1\$$ and the dependent current source gives this:

It's not clear that the problem asked for the Thévenin resistance, but, to obtain it, set $$\V_\mathrm{s}=0\$$, apply a test voltage $$\V_\mathrm{test}\$$ across the terminals in place of $$\V_\mathrm{o}\$$, find current $$\I_\mathrm{test}\$$, and solve for

\begin{align*} R_\mathrm{thévenin} &= \frac{V_\mathrm{test}}{I_\mathrm{test}}. \end{align*}

The current $$\I_x\$$ is given by the voltage across the series combination of $$\R_1\$$ and $$\R_2\$$ and the sum of the two resistances.

\begin{align*} \frac{ 2 R_1 I_x - V_\mathrm{test}}{R_1 + R_2} &= I_x \\ 2 R_1 I_x - V_\mathrm{test} &= \left( R_1 + R_2 \right) I_x. \end{align*}

Solving this for $$\V_\mathrm{test}\$$ gives \begin{align*} V_\mathrm{test} &= \left( 2 R_1 - R_1 - R_2 \right) I_x\\ &= \left( R_1 - R_2 \right) I_x, \end{align*} from which it is easy to obtain \begin{align*} I_x &= \frac{V_\mathrm{test}}{R_1 - R_2}. \end{align*}

Current $$\I_\mathrm{test}\$$ is the sum of $$\-I_x\$$ and the current flowing down through $$\R_3\$$: \begin{align*} I_\mathrm{test} &= \frac{V_\mathrm{test}}{R_3} - I_x \\ &= \frac{V_\mathrm{test}}{R_3} - \frac{V_\mathrm{test}}{R_1 - R_2}\\ &= V_\mathrm{test} \left( \frac{R_1 - R_2 - R_3}{R_3 \left( R_1 - R_2 \right)} \right) \end{align*}

So we finally can solve for \begin{align*} R_\mathrm{thévenin} &= \frac{V_\mathrm{test}}{I_\mathrm{test}} \\ &= \frac{R_3 \left( R_1 - R_2 \right)}{R_1 - R_2 - R_3}\\ &= \frac{\left( 1\,\mathrm{k}\Omega \right) \left( 1\,\mathrm{k}\Omega - 2\,\mathrm{k}\Omega \right)}{1\,\mathrm{k}\Omega - 2\,\mathrm{k}\Omega - 1\,\mathrm{k}\Omega}\\ &= 500\,\Omega. \end{align*}

Next we solve for the Thévenin voltage, which is the same as voltage $$\V_\mathrm{o}\$$.

\begin{align*} V_\mathrm{o} &= 2 R_1 I_x - I_x \left( R_1 + R_2 \right) + V_\mathrm{s} \\ &= V_\mathrm{s} + \left( R_1 - R_2 \right) I_x \end{align*} But current $$\I_x\$$ also flows through $$\R_3\$$, giving another expression for $$\V_\mathrm{o}\$$: \begin{align*} V_\mathrm{o} &= R_3 I_x \end{align*}

Solving this for $$\I_x\$$ and substituting it into the first equation for $V_\mathrm{o}$ gives \begin{align*} V_\mathrm{o} &= V_\mathrm{s} + \left( R_1 - R_2 \right) \frac{V_\mathrm{o}}{R_3}\\ R_3 V_\mathrm{o} &= R_3 V_\mathrm{s} + \left( R_1 - R_2 \right) V_\mathrm{o}. \end{align*}

Solving this for $$\V_\mathrm{o}\$$ gives \begin{align*} V_\mathrm{o} &= \left( \frac{R_3}{R_3 - R_1 + R_2} \right) V_\mathrm{s}\\ &= \left( \frac{1\,\mathrm{k}\Omega}{1\,\mathrm{k}\Omega - 1\,\mathrm{k}\Omega + 2\,\mathrm{k}\Omega} \right) \left( 6\,\mathrm{V} \right)\\ &= 3\,\mathrm{V}. \end{align*}

So the Thévinin-equivalent circuit is