How to calculate thevenin's voltage for this circuit?

Question

^{simulate this circuit – Schematic created using CircuitLab}

How to calculate the Thevenin's voltage for this circuit? I am supposed to get 12.5V but I am getting 8.5V.

I tried it using KCL with node at R1. When I did the KCL formula, considering V1 as voltage at R1, I got the equation as 2 - (V1/10k) + ((5 - V1)/10k) = 0. This is the equation i got.. Is it right? Could anyone guide me as to how to solve and get to the right answer?

Answer 1

I am supposed to get 12.5V but I am getting 8.5V. I tried it using KCL and node at R1.

Here's a clue to get you started: -

Answer 2

Since it's all linear superpostion applies and here is a simple method to calcualte the Thevenin equivalent.

Definition: "turning off a source" means to replace a voltage source with short or replace a current source with an open.

1. To get the equivalent impedance, turn off all sources. Calculate the impedance as seen from the terminals. In this case you just have two 10k Ohms in parallel so it's 5 kOhms
2. To get the equivalent voltage, turn on one source at a time. Calculate the terminal voltage for just that sournce. Sum over all sources. The current source alone yields 10 V (2mA through 5kOhms) and the voltage source alone yields 2.5V (5 V over a 10k/10k voltage divider) for a total of 12.5V.

Answer 3

First, I will present a method that uses Mathematica to solve this problem because you already have a very good answer by @Andyaka. 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:

$$ \begin{cases} \text{I}_\text{k}=\text{I}_1+\text{I}_4\\ \\ \text{I}_3=\text{I}_2+\text{I}_4\\ \\ \text{I}_2=\text{I}_3+\text{I}_5\\ \\ \text{I}_1=\text{I}_\text{k}+\text{I}_5 \end{cases}\tag1 $$

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

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

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

In[1]:=FullSimplify[
 Solve[{Ik == I1 + I4, I3 == I2 + I4, I2 == I3 + I5, I1 == Ik + I5, 
   I1 == V1/R1, I2 == (Vi - V1)/R2, I3 == V1/R3}, {I1, I2, I3, I4, I5,
    V1}]]

Out[1]={{I1 -> (R3 (Ik R2 + Vi))/(R2 R3 + R1 (R2 + R3)), 
  I2 -> (-Ik R1 R3 + (R1 + R3) Vi)/(R2 R3 + R1 (R2 + R3)), 
  I3 -> (R1 (Ik R2 + Vi))/(R2 R3 + R1 (R2 + R3)), 
  I4 -> Ik - (R3 (Ik R2 + Vi))/(R2 R3 + R1 (R2 + R3)), 
  I5 -> -Ik + (R3 (Ik R2 + Vi))/(R2 R3 + R1 (R2 + R3)), 
  V1 -> (R1 R3 (Ik R2 + Vi))/(R2 R3 + R1 (R2 + R3))}}

Now, we can find:

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

Where I used the following Mathematica-codes:

In[3]:=FullSimplify[
 Limit[(R1 R3 (Ik R2 + Vi))/(R2 R3 + R1 (R2 + R3)), R3 -> Infinity]]

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

In[4]:=FullSimplify[Limit[(R1 (Ik R2 + Vi))/(R2 R3 + R1 (R2 + R3)), R3 -> 0]]

Out[4]=Ik + Vi/R2

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

Out[4]=(R1 R2)/(R1 + R2)

Using your values we get:

• $$\text{V}_\text{th}=\frac{25}{2}=12.5\space\text{V}\tag6$$
• $$\text{I}_\text{th}=\frac{1}{4000}=0.0025\space\text{A}\tag7$$
• $$\text{R}_\text{th}=5000\space\Omega\tag8$$