# How to find Thevenin equivalent of this circuit? Dependent current source

How to find the Thevenin equivalent circuit as seen from terminal a-b.

I found out the value of Zth with ease but I'm not able to find the value of Vth.

I tried using both nodal and mesh analysis and I got an answer of 57.8378 - 2.972j V for Vth, but it is wrong.

KVL equation (4-2j)I1 + (8+4j)I2 + Vth=0 (Considering current flowing in lower loop as I1 and upper loop as I2).

KCL at node 2: -V0/(8+4j) = 5 + 0.2V0.

I don't know how to get the correct answer.

• What is the correct answer, that you know? Commented Nov 1, 2020 at 17:01
• @Jan Yes. The one you posted as the answer is the correct one. 80/37 + (260/37)j Commented Nov 1, 2020 at 17:05

First, I will present a method that uses Mathematica to solve this problem. When I was studying this stuff I used the method all the time (without using Mathematica of course).

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:

$$\begin{cases} 0=\text{I}_\text{k}+\text{I}_1+\text{I}_4\\ \\ \text{I}_2=\text{I}_\text{k}+\text{n}\cdot\left(\text{V}_2-\text{V}_3\right)\\ \\ \text{I}_3=\text{I}_2+\text{I}_4\\ \\ \text{n}\cdot\left(\text{V}_2-\text{V}_3\right)=\text{I}_1+\text{I}_3 \end{cases}\tag1$$

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

$$\begin{cases} \text{I}_1=\frac{\text{V}_2-\text{V}_1}{\text{R}_1}\\ \\ \text{I}_1=\frac{\text{V}_1}{\text{R}_2}\\ \\ \text{I}_3=\frac{\text{V}_3}{\text{R}_3}\\ \\ \text{I}_4=\frac{\text{V}_2-\text{V}_4}{\text{R}_4}\\ \\ \text{I}_4=\frac{\text{V}_4-\text{V}_3}{\text{R}_5} \end{cases}\tag2$$

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

In[1]:=FullSimplify[
Solve[{0 == Ik + I1 + I4, I2 == Ik + n*(V2 - V3), I3 == I2 + I4,
n*(V2 - V3) == I1 + I3, I1 == (V2 - V1)/R1, I1 == V1/R2,
I3 == V3/R3, I4 == (V2 - V4)/R4, I4 == (V4 - V3)/R5}, {I1, I2, I3,
I4, V1, V2, V3, V4}]]

Out[1]={{I1 -> -((Ik (1 + n R3) (R4 + R5))/(
R1 + R2 + R3 + R4 + R5 + n R3 (R4 + R5))),
I2 -> (Ik (R1 + R2 + R3 + R4 + R5 - n R1 (R4 + R5) -
n R2 (R4 + R5)))/(R1 + R2 + R3 + R4 + R5 + n R3 (R4 + R5)),
I3 -> -((Ik (-1 + n (R1 + R2)) (R4 + R5))/(
R1 + R2 + R3 + R4 + R5 + n R3 (R4 + R5))),
I4 -> -((Ik (R1 + R2 + R3))/(
R1 + R2 + R3 + R4 + R5 + n R3 (R4 + R5))),
V1 -> -((Ik R2 (1 + n R3) (R4 + R5))/(
R1 + R2 + R3 + R4 + R5 + n R3 (R4 + R5))),
V2 -> -((Ik (R1 + R2) (1 + n R3) (R4 + R5))/(
R1 + R2 + R3 + R4 + R5 + n R3 (R4 + R5))),
V3 -> -((Ik (-1 + n (R1 + R2)) R3 (R4 + R5))/(
R1 + R2 + R3 + R4 + R5 + n R3 (R4 + R5))),
V4 -> Ik (R4 - ((1 + n R3) (R1 + R2 + R4) (R4 + R5))/(
R1 + R2 + R3 + R4 + R5 + n R3 (R4 + R5)))}}


Now, we can find:

• $$\\text{V}_\text{th}\$$ we get by finding $$\\text{V}_3\$$ and letting $$\\text{R}_3\to\infty\$$: $$\text{V}_\text{th}=\frac{\text{I}_\text{k}\left(\text{R}_4+\text{R}_5\right)\left(1-\text{n}\left(\text{R}_1+\text{R}_2\right)\right)}{\text{n}\left(\text{R}_4+\text{R}_5\right)+1}\tag3$$
• $$\\text{I}_\text{th}\$$ we get by finding $$\\text{I}_3\$$ and letting $$\\text{R}_3\to0\$$: $$\text{I}_\text{th}=\frac{\text{I}_\text{k}\left(\text{R}_4+\text{R}_5\right)\left(1-\text{n}\left(\text{R}_1+\text{R}_2\right)\right)}{\text{R}_1+\text{R}_2+\text{R}_4+\text{R}_5}\tag4$$
• $$\\text{R}_\text{th}\$$ we get by finding: $$\text{R}_\text{th}=\frac{\text{V}_\text{th}}{\text{I}_\text{th}}=\frac{\text{R}_1+\text{R}_2+\text{R}_4+\text{R}_5}{\text{n}\left(\text{R}_4+\text{R}_5\right)+1}\tag5$$

Where I used the following Mathematica-codes:

In[2]:=FullSimplify[
Limit[-((Ik (-1 + n (R1 + R2)) R3 (R4 + R5))/(
R1 + R2 + R3 + R4 + R5 + n R3 (R4 + R5))), R3 -> Infinity]]

Out[2]=-((Ik (-1 + n (R1 + R2)) (R4 + R5))/(1 + n (R4 + R5)))

In[3]:=FullSimplify[
Limit[-((Ik (-1 + n (R1 + R2)) (R4 + R5))/(
R1 + R2 + R3 + R4 + R5 + n R3 (R4 + R5))), R3 -> 0]]

Out[3]=-((Ik (-1 + n (R1 + R2)) (R4 + R5))/(R1 + R2 + R4 + R5))

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

Out[4]=(R1 + R2 + R4 + R5)/(1 + n (R4 + R5))


Now, using your values we get:

• $$\underline{\text{V}}_{\space\text{th}}=\frac{80}{37}+\frac{260}{37}\cdot\text{j}\tag6$$
• $$\underline{\text{I}}_{\space\text{th}}=\frac{10}{37}+\frac{60}{37}\cdot\text{j}\tag7$$
• $$\underline{\text{Z}}_{\space\text{th}}=\frac{164}{37}-\frac{22}{37}\cdot\text{j}\tag8$$

Where $$\\underline{x}\$$ implies that the value is a complex number, so $$\\underline{x}\in\mathbb{C}\$$.

• Thanks for the answer. But I'm not really used to solving circuits using Mathematica. If possible can you please provide an answer using traditional methods? Commented Nov 1, 2020 at 16:56
• @RakshithKrish you're welcome, it is possible to solve this without Mathematica. The only thing I did was using Mathematica to solve the equations, but in the time I was studying this stuff I did this by hand instead of using a software. Commented Nov 1, 2020 at 16:57
• Ok. I actually got the answer just using KCL at node at 2 and 3. By solving both the KCL equations I found out V3 which is the answer. Commented Nov 1, 2020 at 17:37