# Determining Thévenin voltage using nodal analysis

Here I want to determine $$\V_{th}\$$ without using mesh analysis and just by using nodal analysis. Now, $$\V_{th}\$$ is the same as the voltage across the $$\6\ \Omega\$$ resistor since no current passes through the outer $$\2\ \Omega\$$ resistor. Now, let us the set the ground reference voltage as $$\0\$$. Now, $$\V-V_x=2V_x\$$ or $$\V=3V_x\$$ due to the dependent voltage source.

So our $$\V_{th}\$$ is now $$\(3V_x-0=3V_x)\$$. Now, if we try to apply KCL at node $$\V_x\$$ (upper side of $$\4\ \Omega\$$ resistor), we see that $$\-5\ \mathrm{A},\frac{V_x}{4},\frac{V_x-3V_x}{2}\$$ currents are exiting the node, but there is one more current we need to take into consideration which is the current flowing across the dependent source.

So I need some insights on how to solve this question using KCL and nodal analysis.

• let us the set the ground reference voltage as 0 <-- a ground reference voltage is not set to be 0 volts; it is 0 volts but, unfortunately you haven't shown that node on your diagram. You then mention $V$ but, I'm not sure what that refers to. Dec 3, 2023 at 9:44
• You have a voltage source (dependent voltage source) in parallel with 2-ohm resistor. And you do not have to include this resistor in your equations. Because we have a supernode. The only equation you need is this $-5A + \frac{V_X}{4\Omega} + \frac{3 V_X}{6\Omega} = 0$ Becouse no mate what Vth will be 3Vx.
– G36
Dec 3, 2023 at 9:55

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{I}_\text{s}&=\text{I}_0+\text{I}_1+\text{I}_3\\ \\ \text{I}_0+\text{I}_3&=\text{I}_2+\text{I}_4\\ \\ 0&=\text{I}_2+\text{I}_4+\text{I}_5\\ \\ \text{I}_1&=\text{I}_\text{s}+\text{I}_5 \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}_2-0}{\displaystyle\text{R}_2}\\ \\ \text{I}_3&=\frac{\displaystyle\text{V}_1-\text{V}_2}{\displaystyle\text{R}_3}\\ \\ \text{I}_4&=\frac{\displaystyle\text{V}_2-\text{V}_3}{\displaystyle\text{R}_4}\\ \\ \text{I}_4&=\frac{\displaystyle\text{V}_3-0}{\displaystyle\text{R}_5} \end{alignat*} \end{cases}\tag2

We also know that $$\\displaystyle\text{V}_2-\text{V}_1=\text{n}\cdot\text{V}_1\$$.

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

In[1]:=Clear["Global*"];
FullSimplify[
Solve[{Is == I0 + I1 + I3, I0 + I3 == I2 + I4, 0 == I2 + I4 + I5,
I1 == Is + I5, I1 == (V1 - 0)/R1, I2 == (V2 - 0)/R2,
I3 == (V1 - V2)/R3, I4 == (V2 - V3)/R4, I4 == (V3 - 0)/R5,
V2 - V1 == n*V1}, {I0, I1, I2, I3, I4, I5, V1, V2, V3}]]

Out[1]={{I0 -> (Is R1 ((1 + n) R2 R3 +
n R2 (R4 + R5) + (1 + n) R3 (R4 + R5)))/(
R3 (R2 (R4 + R5) + (1 + n) R1 (R2 + R4 + R5))),
I1 -> (Is R2 (R4 + R5))/(R2 (R4 + R5) + (1 + n) R1 (R2 + R4 + R5)),
I2 -> (Is (1 + n) R1 (R4 + R5))/(
R2 (R4 + R5) + (1 + n) R1 (R2 + R4 + R5)),
I3 -> -((Is n R1 R2 (R4 + R5))/(
R3 (R2 (R4 + R5) + (1 + n) R1 (R2 + R4 + R5)))),
I4 -> (Is (1 + n) R1 R2)/(R2 (R4 + R5) + (1 + n) R1 (R2 + R4 + R5)),
I5 -> -((Is (1 + n) R1 (R2 + R4 + R5))/(
R2 (R4 + R5) + (1 + n) R1 (R2 + R4 + R5))),
V1 -> (Is R1 R2 (R4 + R5))/(
R2 (R4 + R5) + (1 + n) R1 (R2 + R4 + R5)),
V2 -> (Is (1 + n) R1 R2 (R4 + R5))/(
R2 (R4 + R5) + (1 + n) R1 (R2 + R4 + R5)),
V3 -> (Is (1 + n) R1 R2 R5)/(
R2 (R4 + R5) + (1 + n) R1 (R2 + R4 + R5))}}


Now, we can find:

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

Where I used the following Mathematica codes:

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

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

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

Out[3]=(Is (1 + n) R1 R2)/(R2 R4 + (1 + n) R1 (R2 + R4))

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

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

$$\text{V}_\text{th}=20\space\text{V}\space\wedge\space\text{I}_\text{th}=\frac{10}{3}\approx3.33\space\text{A}\space\wedge\space\text{R}_\text{th}=6\space\Omega\tag6$$

You can keep the sum of the currents of R2 and the controlled source as one. I named it to I2.

Write the ordinary "nodal method" current summing equations for the top ends of resistors R1 and R3. Note that the node voltage at the top end of R3 must be 3Vx because the controlled 2Vx is in series with the unknown node voltage Vx.

The equations are

(I1) + (I2) = (Vx)/(R1) and (I2) + (3Vx)/(R3) = 0

Eliminate unknown I2 and solve Vx. The wanted no-load Voc is 3Vx.

This all can be found in a short form from already given comments.