# Find Thevenin equivalent of given circuit with two voltage sources

I have to find Thevenin equivalent circuit between M and N. So I understand that I have to get rid of R7, and find voltage of Vm and Vn, then Vth = Vm-Vn. But I don't understand how to calculate it. The two opposite voltage sources trip me out.

As for Rth, I calculate it as R8/R7/(R4+(R5/R6)) and get the result 3.75kohms.

I want to understand this problem, so any help would be appreciated.

Edit: So after simplifying the circuit, I calculate I1 = 0.5 mA and I2 = 0, then Vmn = 10 x 0.5 = 5 (V).

Is this correct?

• Simplify the resistances on the right hand side and compute the current in the single loop. Then figure out the voltage difference across M,N. Jan 12, 2022 at 5:54
• I might understand why you removed R7 first, but your calculation even without R7 is incorrect and you still need to add the R7 back to the circuit or you won't have a correct result. Jan 12, 2022 at 6:03
• Is R7 part of the equivalent circuit or a load? Jan 12, 2022 at 6:05
• I assume you already have done these Thevenin equalent circuits previously. Its straight forward. Find Rth first by replacing all Sources to their equalent circuits. Find Voltage across M & N pins for Vth. Jan 12, 2022 at 6:12
• @copper.hat so Rth is 10k/10k/15k = 3.75k yes? Jan 12, 2022 at 6:19

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 complicated than this one. In combination with the other answers, my answer is valuable.

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:

$$\begin{cases} \text{I}_1=\text{I}_2+\text{I}_5\\ \\ \text{I}_3=\text{I}_4+\text{I}_5\\ \\ \text{I}_4=\text{I}_3+\text{I}_6\\ \\ \text{I}_2=\text{I}_1+\text{I}_6 \end{cases}\tag1$$

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

$$\begin{cases} \text{I}_1=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_1}{\text{R}_2}\\ \\ \text{I}_3=\frac{\text{V}_1}{\text{R}_3}\\ \\ \text{I}_4=\frac{\text{V}_3-\text{V}_2}{\text{R}_4}\\ \\ \text{I}_4=\frac{0-\text{V}_3}{\text{R}_5} \end{cases}\tag2$$

Using $$\(2)\$$ and $$\\text{V}_1-\text{V}_2=\text{V}_x\$$ we can rewrite $$\(1)\$$ as follows:

$$\begin{cases} \frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}=\frac{\text{V}_1}{\text{R}_2}+\text{I}_5\\ \\ \frac{\text{V}_1}{\text{R}_3}=\frac{\text{V}_3-\text{V}_2}{\text{R}_4}+\text{I}_5\\ \\ \frac{\text{V}_3-\text{V}_2}{\text{R}_4}=\frac{\text{V}_1}{\text{R}_3}+\text{I}_6\\ \\ \frac{\text{V}_1}{\text{R}_2}=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}+\text{I}_6\\ \\ \frac{\text{V}_1}{\text{R}_3}=\frac{0-\text{V}_3}{\text{R}_5}+\text{I}_5\\ \\ \frac{0-\text{V}_3}{\text{R}_5}=\frac{\text{V}_1}{\text{R}_3}+\text{I}_6\\ \\ \text{V}_1-\text{V}_2=\text{V}_x \end{cases}\tag3$$

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

In[1]:=Clear["Global*"];
FullSimplify[
Solve[{I1 == I2 + I5, I3 == I4 + I5, I4 == I3 + I6, I2 == I1 + I6,
I1 == (Vi - V1)/R1, I2 == V1/R2, I3 == V1/R3, I4 == (V3 - V2)/R4,
I4 == (0 - V3)/R5, V1 - V2 == Vx}, {I1, I2, I3, I4, I5, I6, V1, V2,
V3}]]

Out[1]={{I1 -> (R3 (R4 + R5) Vi + R2 (R3 + R4 + R5) Vi - R2 R3 Vx)/(
R1 R3 (R4 + R5) + R2 R3 (R4 + R5) + R1 R2 (R3 + R4 + R5)),
I2 -> (R3 ((R4 + R5) Vi + R1 Vx))/(
R1 R3 (R4 + R5) + R2 R3 (R4 + R5) + R1 R2 (R3 + R4 + R5)),
I3 -> (R2 ((R4 + R5) Vi + R1 Vx))/(
R1 R3 (R4 + R5) + R2 R3 (R4 + R5) + R1 R2 (R3 + R4 + R5)),
I4 -> (-R2 R3 Vi + R1 R3 Vx + R2 (R1 + R3) Vx)/(
R1 R3 (R4 + R5) + R2 R3 (R4 + R5) + R1 R2 (R3 + R4 + R5)),
I5 -> (R2 (R3 + R4 + R5) Vi - (R1 + R2) R3 Vx)/(
R1 R3 (R4 + R5) + R2 R3 (R4 + R5) + R1 R2 (R3 + R4 + R5)),
I6 -> (-R2 (R3 + R4 + R5) Vi + (R1 + R2) R3 Vx)/(
R1 R3 (R4 + R5) + R2 R3 (R4 + R5) + R1 R2 (R3 + R4 + R5)),
V1 -> (R2 R3 ((R4 + R5) Vi + R1 Vx))/(
R1 R3 (R4 + R5) + R2 R3 (R4 + R5) + R1 R2 (R3 + R4 + R5)),
V2 -> -(((R4 + R5) (-R2 R3 Vi + R1 R3 Vx + R2 (R1 + R3) Vx))/(
R1 R3 (R4 + R5) + R2 R3 (R4 + R5) + R1 R2 (R3 + R4 + R5))),
V3 -> (R2 R3 R5 Vi - (R2 R3 + R1 (R2 + R3)) R5 Vx)/(
R1 R3 (R4 + R5) + R2 R3 (R4 + R5) + R1 R2 (R3 + R4 + R5))}}


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}_2\left(\text{V}_\text{i}\left(\text{R}_4+\text{R}_5\right)+\text{V}_x\text{R}_1\right)}{\text{R}_1\left(\text{R}_2+\text{R}_4+\text{R}_5\right)+\text{R}_2\left(\text{R}_4+\text{R}_5\right)}\tag4$$
• $$\\text{I}_\text{th}\$$ we get by finding $$\\text{I}_3\$$ and letting $$\\text{R}_3\to0\$$: $$\text{I}_\text{th}=\frac{\text{V}_\text{i}}{\text{R}_1}+\frac{\text{V}_x}{\text{R}_4+\text{R}_5}\tag5$$
• $$\\text{R}_\text{th}\$$ we get by finding: $$\text{R}_\text{th}=\frac{\text{V}_\text{th}}{\text{I}_\text{th}}=\frac{\text{R}_1\text{R}_2\left(\text{R}_4+\text{R}_5\right)}{\text{R}_1\left(\text{R}_2+\text{R}_4+\text{R}_5\right)+\text{R}_2\left(\text{R}_4+\text{R}_5\right)}\tag6$$

Where I used the following Mathematica-codes:

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

Out[2]=FullSimplify[
Limit[(R2 R3 ((R4 + R5) Vi + R1 Vx))/(
R1 R3 (R4 + R5) + R2 R3 (R4 + R5) + R1 R2 (R3 + R4 + R5)),
R3 -> Infinity]]

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

Out[3]=Vi/R1 + Vx/(R4 + R5)

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

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

So, using your values we get:

• $$\text{V}_\text{th}=5\space\text{V}\tag7$$
• $$\text{I}_\text{th}=\frac{1}{750}\approx0.00133333\space\text{A}\tag8$$
• $$\text{R}_\text{th}=3750\space\Omega\tag9$$
• Thank you so much for the very well-motivated downvote, I really appreciate it. Jan 13, 2022 at 9:33