# How to calculate thevenin's voltage for this circuit?

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?

• This is not a homework solution service. We expect you to show your work and ask a specific question if you get stuck. Oct 30, 2020 at 14:03
• Ok, sorry for it, I will show it in a while Oct 30, 2020 at 14:15
• Using nodal analysis and KCL with node at R1, I got this equation: 2 - (V1/10k) + ((5 - V1)/10k) = 0 Oct 30, 2020 at 14:24
• @SourabhMisal your equation is correct and gives 12.5V. How are you getting the 8.5V? Oct 30, 2020 at 15:33
• @PaulGhobril yes, I realized it, I was doing a silly mistake of taking 2 mA as 2/10k instead of 2/1k and screwed it up Nov 1, 2020 at 4:26

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: -

• Thanks @Andy aka, I understood the method. Could you guide me if I do it using nodal analysis, with node at R1, I get the equation as 2 - (V1/10k) + ((5 - V1)/10k) = 0 which is wrong. How to get the equation right? How to get my analysis right? Oct 30, 2020 at 14:23
• @SourabhMisal I don't do KVL or KCL because I find it to be pointless. If you look at the source conversion I did you will nearly instantly see that the thevenin voltage is 12.5 volts so, you figure out where you went wrong with your equations. Oct 30, 2020 at 14:27
• Ok, and thanks for the guidance though @Andy aka Oct 30, 2020 at 14:31

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.
• This is very useful, thanks @Hilmar Nov 1, 2020 at 4:28

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)


• $$\text{V}_\text{th}=\frac{25}{2}=12.5\space\text{V}\tag6$$
• $$\text{I}_\text{th}=\frac{1}{4000}=0.0025\space\text{A}\tag7$$
• $$\text{R}_\text{th}=5000\space\Omega\tag8$$
• That's certainly correct but feels needlessly complicated. Oct 31, 2020 at 12:33
• @Hilmar For this example I agree, but this method works always even with very (very) complicated circuits. Oct 31, 2020 at 12:36
• It certainly does, but so does superposition and super position is almost always easier to do. Oct 31, 2020 at 14:43
• Thanks, this helps a ton!! @Jan Nov 1, 2020 at 4:28