# Dependent current sources in Thevenin equivalent

In this question to find out ZTH, why have we put a 1 A current source?
Can't we just short circuit the VTH potential and find the current across that to find ZTH?

Source: https://i.sstatic.net/9KVGu.jpg

My solution :https://ibb.co/t87bb3c

The answer:https://ibb.co/1JBSWpq

• Can you show us the circuit? May 14, 2023 at 9:40
• sorry i forgot to put the image. the circuit is the first one May 14, 2023 at 10:27

## 2 Answers

There are different methods to determine the Thèvenin equivalent impedance ZTH.

ZTH can be found using what the theorem explicitly says, i.e. calculating the ratio between the open circuit voltage and the short circuit current at the terminals of the network.

However, you could also find it using the definition of impedance at those two terminals, i.e. deactivating all independent sources in the circuit and calculating the impedance.

This latter method in general amounts to applying a test current (or voltage) source at the terminals and then determining the corresponding voltage (or current) and then calculating the ratio of the two.

The method you follow is your choice and the result you get is the same in all cases. The only difference is that sometimes one approach is much easier to carry out than the others because calculations turn out to be simpler.

It takes some experience in such kind of exercises to find the simpler approach.

• Can you check my solution. I tried solving it but it came out to be incorrect?? May 14, 2023 at 11:07

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.

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{I}_\text{i}&=\text{I}_1+\text{I}_2\\ \\ \text{I}_2&=\text{I}_3+\text{n}\cdot\text{I}_1\\ \\ 0&=\text{I}_0+\text{I}_3+\text{n}\cdot\text{I}_1\\ \\ \text{I}_1&=\text{I}_\text{i}+\text{I}_0 \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{\text{V}_1}{\text{R}_1}\\ \\ \text{I}_2&=\frac{\text{V}_1-\text{V}_2}{\text{R}_2}\\ \\ \text{I}_3&=\frac{\text{V}_2}{\text{R}_3} \end{alignat*} \end{cases}\tag2

Using $$\(2)\$$ we can rewrite $$\(1)\$$ as follows:

\begin{cases} \begin{alignat*}{1} \text{I}_\text{i}&=\frac{\text{V}_1}{\text{R}_1}+\frac{\text{V}_1-\text{V}_2}{\text{R}_2}\\ \\ \frac{\text{V}_1-\text{V}_2}{\text{R}_2}&=\frac{\text{V}_2}{\text{R}_3}+\text{n}\cdot\frac{\text{V}_1}{\text{R}_1}\\ \\ 0&=\text{I}_0+\frac{\text{V}_2}{\text{R}_3}+\text{n}\cdot\frac{\text{V}_1}{\text{R}_1}\\ \\ \frac{\text{V}_1}{\text{R}_1}&=\text{I}_\text{i}+\text{I}_0 \end{alignat*} \end{cases}\tag3

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

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

Out[1]={{I0 -> -((Ii (R1 + n R3))/(R1 + R2 + R3 + n R3)),
I1 -> (Ii (R2 + R3))/(R1 + R2 + R3 + n R3),
I2 -> (Ii (R1 + n R3))/(R1 + R2 + R3 + n R3),
I3 -> (Ii (R1 - n R2))/(R1 + R2 + R3 + n R3),
V1 -> (Ii R1 (R2 + R3))/(R1 + R2 + R3 + n R3),
V2 -> (Ii (R1 - n R2) R3)/(R1 + R2 + R3 + n 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}_\text{i}\left(\text{R}_1-\text{nR}_2\right)}{1+\text{n}}\tag4$$
• $$\\text{I}_\text{th}\$$ we get by finding $$\\text{I}_3\$$ and letting $$\\text{R}_3\to0\$$: $$\text{I}_\text{th}=\frac{\text{I}_\text{i}\left(\text{R}_1-\text{nR}_2\right)}{\text{R}_1+\text{R}_2}\tag5$$
• $$\\text{R}_\text{th}\$$ we get by finding: $$\text{R}_\text{th}=\frac{\text{V}_\text{th}}{\text{I}_\text{th}}=\frac{\text{R}_1+\text{R}_2}{1+\text{n}}\tag6$$

Where I used the following Mathematica codes:

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

Out[2]=(Ii (R1 - n R2))/(1 + n)

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

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

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

Out[4]=(R1 + R2)/(1 + n)
`

So, using your values we get:

• $$\left|\text{V}_\text{th}\right|=\displaystyle\left|\frac{\displaystyle15\left(2-4\text{j}-\frac{1}{2}\cdot\left(4+3\text{j}\right)\right)}{\displaystyle1+\frac{1}{2}}\right|=55\space\text{V}\tag7$$
• $$\left|\text{I}_\text{th}\right|=\displaystyle\left|\frac{\displaystyle15\left(2-4\text{j}-\frac{1}{2}\cdot\left(4+3\text{j}\right)\right)}{\displaystyle2-4\text{j}+4+3\text{j}}\right|=\frac{165}{2 \sqrt{37}}\approx13.5629\space\text{A}\tag8$$
• $$\left|\text{R}_\text{th}\right|=\displaystyle\left|\frac{\displaystyle2-4\text{j}+4+3\text{j}}{\displaystyle1+\frac{1}{2}}\right|=\frac{2 \sqrt{37}}{3}\approx4.05518\space\Omega\tag9$$
• $$\arg\left(\text{V}_\text{th}\right)=\displaystyle\arg\left(\frac{\displaystyle15\left(2-4\text{j}-\frac{1}{2}\cdot\left(4+3\text{j}\right)\right)}{\displaystyle1+\frac{1}{2}}\right)=-\frac{\pi}{2}\approx-1.5708\space\text{rad}\tag{10}$$
• $$\arg\left(\text{I}_\text{th}\right)=\displaystyle\arg\left(\frac{\displaystyle15\left(2-4\text{j}-\frac{1}{2}\cdot\left(4+3\text{j}\right)\right)}{\displaystyle2-4\text{j}+4+3\text{j}}\right)=-\arctan\left(6\right)\approx-1.40565\space\text{rad}\tag{11}$$
• $$\arg\left(\text{R}_\text{th}\right)=\displaystyle\arg\left(\frac{\displaystyle2-4\text{j}+4+3\text{j}}{\displaystyle1+\frac{1}{2}}\right)=-\text{arccot}\left(6\right)\approx-0.165149\space\text{rad}\tag{12}$$