Create the Thevevenin equivalent circuit

In summary, the conversation discusses a problem with calculating voltages in a circuit using node analysis. The speaker has obtained equations and used MATLAB to solve the equations, but the calculated voltages do not match the real voltages. A classmate suggests including a resistor, R11, in the circuit to get the correct results. The speaker also discusses how to calculate the open circuit voltage and the short-circuit current. The conversation ends with a clarification on how to get the correct open circuit voltage by taking into account the voltage drop across R10.
  • #1
David331
31
1
Homework Statement
Replace the circuit A-B with a theveninequivalent.
Relevant Equations
??
Hey, I am having big issues with the following question. I have done node analysis to get the following equations 1-5 and the equation for A i have removed because it was not needed apparently. When solving the equationsystem I get that the voltage in node 1-5 is (using matlab)
V =

-3.259676116091027 (V1)
-1.665272203916370 (V2)
-2.152861289845328 (V3)
-1.231695676964311 (V4)
5.530879269335376 (V5)
But my calculated voltage in the nodes differes from the real voltage. I don't know what I have done wrong, as I have controlled everything 100 times and also done the calculation many times. Could someone please help a bit?

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  • #2
What are the real voltages? Where did they come from? Did the problem statement include the answers, or did the professor or a classmate of yours provide them?

In the last row of the stimulus vector you have E2/R2+Io2 (or so it appears); why do you have R2 in the denominator?
 
  • #3
Thanks for the response! I got the real answers using multisim:
--- Operating Point ---
V(n001): 0.683191 voltage
V(3): -2.31681 voltage
V(2): -2.01304 voltage
V(1): -3.56637 voltage
V(4): -1.333 voltage
V(5): 4.6758 voltage
V(n002): 22.6758 voltage
V(n003): 4.67604 voltage
1632324269419.png


I see that I have written a R2 instead of R10, but I have used R10 when I have made my calculations. Does the calculations and equationsystem look correct though?
 
  • #4
David331 said:
Thanks for the response! I got the real answers using multisim:
--- Operating Point ---
V(n001): 0.683191 voltage
V(3): -2.31681 voltage
V(2): -2.01304 voltage
V(1): -3.56637 voltage
V(4): -1.333 voltage
V(5): 4.6758 voltage
V(n002): 22.6758 voltage
V(n003): 4.67604 voltage
View attachment 289513

I see that I have written a R2 instead of R10, but I have used R10 when I have made my calculations. Does the calculations and equationsystem look correct though?
The circuit you solved with multisim has a 10000k resistor, R11, that your circuit in post #1 does not have. If you include that resistor you will get the same result as multisim:

GMAT.png
 
  • #5
Ohhh I did not know that R11 affected the results, only used it for easier simulation. I am so thankful for the help, I have been stuck on this assignment very long! So now to get the open circuit voltage I have to take U0=VA-VB=VA-0=VA=V5-E2=5,53088-6,00000=-1,3242 voltage. And to get the short-circuit current I take I0=(V5-E2-0)/R10=(5,53088-6,00000)/1800=-0,2606...mA?
 
  • #6
David331 said:
Ohhh I did not know that R11 affected the results, only used it for easier simulation. I am so thankful for the help, I have been stuck on this assignment very long! So now to get the open circuit voltage I have to take U0=VA-VB=VA-0=VA=V5-E2=5,53088-6,00000=-1,3242 voltage. And to get the short-circuit current I take I0=(V5-E2-0)/R10=(5,53088-6,00000)/1800=-0,2606...mA?
You have the short circuit current correct, but for the open circuit voltage you need the voltage at node 5 without R10 and E2 connected from node 5 to ground. It's easy to delete R10 from the 5,5 element of the matrix in post #1 and from the 5,1 element of the corresponding stimulus vector and solve. What do you get if you do that?
 
  • #7
In that case I get
V =

-3.566523917173452 (V1)
-2.013218390804596 (V2)
-2.316892200252596 (V3)
-1.333051549017734 (V4)
4.675368017745513 (V5)
Which is the same as the simulation gave.
In this case I get U0=V5-E2=4.675368...-6.000000=-1.324632 V.(I see that I got this in last also, but computing something completely else...). So this is the correct open circuit voltage? How come we don't need E2 and R10 in our calculations?
 
  • #8
David331 said:
In that case I get
V =

-3.566523917173452 (V1)
-2.013218390804596 (V2)
-2.316892200252596 (V3)
-1.333051549017734 (V4)
4.675368017745513 (V5)
Which is the same as the simulation gave.
In this case I get U0=V5-E2=4.675368...-6.000000=-1.324632 V.(I see that I got this in last also, but computing something completely else...). So this is the correct open circuit voltage? How come we don't need E2 and R10 in our calculations?
These voltages are not exactly the same as the simulation; R11 does have a very small effect.

If the A-B port is open circuited then the right end of E2 should not be grounded. The matrix you used in post #1 has the right end of E2 grounded (A-B shorted). But once you have the voltage at node 5 with R10 and E2 not grounded (A-B open circuited), the voltage at A is just the same as the voltage at node 5 minus the voltage of E2 since there is no voltage drop across R10; R10 carries no current if the right end of E2 is floating.
 
  • #9
Oh now I see, I did not think about that there was no voltage drop across R10, since as you say, no current flows through when it is open circuited. Thanks a lot for the explantions, I have learned much! Just to clarify, the U0 voltage is now correct right?
 
  • #10
David331 said:
Oh now I see, I did not think about that there was no voltage drop across R10, since as you say, no current flows through when it is open circuited. Thanks a lot for the explantions, I have learned much! Just to clarify, the U0 voltage is now correct right?
Provided you don't use the value of V5 from post #3, but rather from the procedure I described in post #6.
 
  • #11
Yes of course, now I used the values computed in #7 (V5=4.675368017745513 V), which indeed differ a bit from the values in the simulation.
 
  • #12
David331 said:
Yes of course, now I used the values computed in #7 (V5=4.675368017745513 V), which indeed differ a bit from the values in the simulation.
Good luck with the rest of your studies.
You might find post #62 here interesting showing use of nodal analysis: https://www.physicsforums.com/goto/post?id=6421623
 
  • Like
Likes David331
  • #13
Thanks and once again thanks for all the help! I will save that link for later use, as I progress throughout my engineering courses, it will definitely be helpful :)
 

1. What is Thevenin's theorem?

Thevenin's theorem is a fundamental concept in electrical engineering that states any linear circuit can be represented by an equivalent circuit with a single voltage source and a single resistor.

2. How do you create Thevenin equivalent circuit?

To create the Thevenin equivalent circuit, you need to follow these steps:

  • 1. Identify the load resistor in the original circuit.
  • 2. Remove the load resistor from the circuit.
  • 3. Calculate the open-circuit voltage (Voc) by using the voltage divider rule.
  • 4. Calculate the equivalent resistance (Req) by shorting all voltage sources and removing all current sources from the original circuit, then calculating the resistance between the load terminals.
  • 5. Draw the Thevenin equivalent circuit with the calculated Voc and Req.

3. Why is Thevenin equivalent circuit useful?

The Thevenin equivalent circuit is useful because it simplifies complex circuits into a single voltage source and a single resistor, making it easier to analyze and design circuits. It also helps in determining the maximum power that can be delivered to a load.

4. What are the assumptions of Thevenin's theorem?

The assumptions of Thevenin's theorem are:

  • 1. The circuit is linear, which means the relationship between voltage and current is constant.
  • 2. The circuit is in steady-state, which means all voltages and currents have reached their final values.
  • 3. The circuit contains only independent sources and resistors.

5. Can Thevenin equivalent circuit be used for non-linear circuits?

No, Thevenin's theorem only applies to linear circuits. Non-linear circuits do not have a constant relationship between voltage and current, so they cannot be simplified into an equivalent circuit with a single voltage source and a single resistor.

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