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