Find the Thevenin Equivalent circuit

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Discussion Overview

The discussion revolves around finding the Thevenin equivalent circuit external to a load resistor (RL) in a given electrical circuit. Participants engage in calculations related to resistances, voltages, and currents, while exploring the differences between Thevenin and Norton equivalents. The scope includes theoretical calculations and practical applications of circuit analysis.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant calculates the Thevenin resistance (Rth) as 7Ω and attempts to find the Thevenin voltage (Vth) based on circuit configurations.
  • Another participant questions the identification of a 2Ω resistor in the circuit, indicating a lack of clarity in the initial post.
  • Some participants provide simulation results, indicating a Thevenin current of 1.786A, but express confusion about current values in series and parallel configurations.
  • There is a discussion about the differences between Thevenin and Norton equivalents, with some participants noting that textbooks may mix the two concepts.
  • Several participants attempt to clarify the steps for calculating the Thevenin equivalent, emphasizing the need to find the open circuit voltage (VAB) when RL is removed.
  • One participant calculates V3 as 25V and Isc as 175A, but another challenges the correctness of these calculations based on circuit configurations.
  • There are multiple calculations presented for voltages across different resistors, with participants attempting to reconcile their results with the expected values.
  • Confusion arises regarding the voltage across R2 and the relationship to Vth, with participants debating the correct approach to find these values.
  • One participant expresses uncertainty about the origin of a specific current value (2.173), indicating a need for clarification on its calculation.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the calculations and interpretations of the Thevenin equivalent. There are competing views on the correct approach to finding the Thevenin voltage and resistance, as well as confusion regarding the definitions and applications of Thevenin and Norton theorems.

Contextual Notes

Some calculations depend on assumptions about circuit configurations, and there are unresolved mathematical steps regarding the relationships between different resistors and currents. The discussion reflects varying levels of understanding and interpretations of circuit analysis principles.

Fatima Hasan
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Thread moved from the technical forums, so no Homework Template is shown
Find Thevenins Equivalent circuit external to RL, then find I.
screenshot_17.png


Here's my work :
5Ω & 5Ω series → R(1) = 5+5 = 10 Ω
10Ω & R(1) parallel→ R(2) = (10*10)/(10+10) = 5 Ω
2Ω & R(2) series → R(3) = R Th = 5+2 = 7Ω
screenshot_19.png

V = I R
R = (R Th * RL)/(R Th + RL)
R = 70/17 Ω
V = 5 * (70/17) = 350 / 17 V = 20.6 V

Could somebody check my answer please ?
 

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Fatima Hasan said:
. . .
2Ω & R(2) series → R(3) = R Th = 5+2 = 7Ω
. . .

Hi ! Will you please explain what is 2Ω here?
 
AlexCaledin said:
Hi ! Will you please explain what is 2Ω here?
I forgot to put it ...
screenshot_21.png
 

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These are the currents at RL=0 as calculated by a simulation program;
the "Thevenin current" is 1.786A ( to be then multiplied by 70/17 Ω).

Currents1.PNG


( 357.143m = 0.357143A )
 

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Last edited:
AlexCaledin said:
These are the currents at RL=0 as calculated by a simulation program; the Thevenin current is 1.786A, to be multiplied by 70/17 Ω.

View attachment 221969
The current in a series circuit is everywhere the same , why I2 doesn't equal to I1 ?
 
Why is it series?
5 amperes are rather applied to parallel resistors:
one is the R1 = 5ohm and the other is {R2 + R3-and-R4} = 5 + 20/12 = 40/6 = 6.66666etc ohm;
so in the R2 there is less current.

____________

Thanks to the Electrician, now I know about the "Norton", as opposed to "Thevenin". I have been taught about "The Equivalent Source Theorem", combining both of them.
 
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The Electrician said:
See the section on this page: https://en.wikipedia.org/wiki/Thévenin's_theorem

with the heading "Calculating the Thevenin equivalent".

What do you get if you follow that method?
RNo = RTh = 7 Ω
INo * RNo = VTh
VTh=5*7 = 35 V
VNo = 1.786*70/17 =7.354 V
ITh = VNo / RNo =7.354/7 = 1.051 A
 
  • #10
Fatima Hasan said:
VTh=5*7 = 35 V
That is incorrect.
You need to remove RL and calculate the open circuit voltage VAB.
Fatima Hasan said:
RNo = RTh =
Right.
 
  • #11
cnh1995 said:
You need to remove RL and calculate the open circuit voltage VAB.
VAB = V3 (parallel)
VAB= I3 * R3
= 0.357143 * 10 = 3.57 V
 
  • #12
Fatima Hasan said:
VAB = V3 (parallel)
VAB= I3 * R3
= 0.357143 * 10 = 3.57 V
No. You get I3*R3=3.57V when you 'short circuit' the terminals A and B and VAB=0 in that çase.

Do you know the steps for finding the Thevenin equivalent of a network? It looks like you are mixing Thevenin and Norton without knowing the fundamental differences between them.
 
  • #13
cnh1995 said:
. . . you are mixing Thevenin and Norton . . .

- the thing is, they may be mixed in the textbooks. In our Russian textbooks, there is typically one "Equivalent Source Theorem" mixing Thevenin and Norton straight away, never mentioning any of them, like here:

http://energ2010.ru/Toe/Toe_lekcii_VUZ/Toe_lekcii_belarus/2_11_Teorema_ekvivalentnom_generatore.html

- and in some textbooks that same mixed theorem is called also "Thevenin theorem", as its second name. And of course lots of people - including teachers - from many countries have studied here in Russia so this confusion may well be typical.
 
  • #14
The link you have provided shows in Fig 27 two circuits. The one on the left, designated а) is known on this forum as the Thevenin equivalent. The one on the right, designated б) is known on this forum as the Norton equivalent. They are shown as two different circuits. Are they not given different names in that Russian text?

At any rate, I pointed out the two different circuits and their names used here in posts #7 and #8.
 
  • #15
- they are called equivalent voltage and equivalent current sources - making the point that you can always use this or that, whatever is better for your problem.
 
  • #16
So far Fatima Hasan hasn't calculated either one correctly. Let's hope he (or she) is still with us and tells us which one is needed.
 
  • #17
The Electrician said:
So far Fatima Hasan hasn't calculated either one correctly. Let's hope he (or she) is still with us and tells us which one is needed.
I want to calculate the Thevenin's Equivalent .
 
  • #18
Fatima Hasan said:
I want to calculate the Thevenin's Equivalent .

OK. You have already correctly calculated Rth. Next you need to respond to post #12. You must calculate the voltage across terminals A & B when RL is not present. That will be Vth.
 
  • #19
The Electrician said:
OK. You have already correctly calculated Rth. Next you need to respond to post #12. You must calculate the voltage across terminals A & B when RL is not present. That will be Vth.
R = R1 + R2 = 5 + 5 = 10 Ω
Is = 5 A
Vs = Is * R = 5*10 = 50 V
V3= Vs /(R+R3) * R3
V3 = 50/(10+10) * 10 = 25 V
V3 = VTh = 25 V
Isc = RTh * Voc
=7 *25 = 175 A
 
  • #20
The resistance R should be the resistance seen by the current source, but it's not just R1+R2. There is some current also in R3, so R2+R3 is in parallel with R1; that equivalent resistance is the load on the source.

Also, the next to last thing in your calculations is Isc = RTh * Voc. That can't be right because the units are wrong.
 
  • #21
The Electrician said:
Isc = RTh * Voc
I sc = Voc / RTh
 
  • #22
Fatima Hasan said:
I sc = Voc / RTh

That's good. Now calculate V5 with the R1, R2, R3 combination load.
 
  • #23
The Electrician said:
That's good. Now calculate V5 with the R1, R2, R3 combination load.
screenshot_22.png
 

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  • #24
Fatima Hasan said:
View attachment 222096
You have correctly calculated the voltage V5 across the current source, which is also the voltage across R1. Your last calculation is wrong. You should be calculating the voltage V3 across the resistor R3. You have labeled the left side of that equation V5, but it should be V3, which is also Vth.

If you know the voltage at the left end of R2, which is V5, then what would be the voltage at the right end of R2 considering that R2 and R3 form a voltage divider?
 
  • #25
The Electrician said:
You have correctly calculated the voltage V5 across the current source, which is also the voltage across R1. Your last calculation is wrong. You should be calculating the voltage V3 across the resistor R3. You have labeled the left side of that equation V5, but it should be V3, which is also Vth.

If you know the voltage at the left end of R2, which is V5, then what would be the voltage at the right end of R2 considering that R2 and R3 form a voltage divider?
V2 = R2 /(R1+R2)*Vs
= 5/(10+5) * 18.75
= 6.25 V
 
  • #26
Fatima Hasan said:
V2 = R2 /(R1+R2)*Vs
= 5/(10+5) * 18.75
= 6.25 V
I'm not sure why you would want the voltage across R2 (which we would call V2). Vth is the same as V3 which is the voltage across R3.

If you replace the current source and R1 with an 18.75 volt battery, could you calculate the voltage across R3? If all you had was a circuit with an 18.75 volt battery on the left side, with R2 and R3 forming a voltage divider, what would be the voltage across R3?
 
  • #27
The Electrician said:
I'm not sure why you would want the voltage across R2 (which we would call V2). Vth is the same as V3 which is the voltage across R3.

If you replace the current source and R1 with an 18.75 volt battery, could you calculate the voltage across R3? If all you had was a circuit with an 18.75 volt battery on the left side, with R2 and R3 forming a voltage divider, what would be the voltage across R3?
V3 = 10/15*18.75 = 12.5 V
 
  • #28
Congratulations! This is correct. You now have Rth and Vth. Do you understand all the steps now?
 
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  • #29
The Electrician said:
Congratulations! This is correct. You now have Rth and Vth. Do you understand all the steps now?
The voltage across R2 is 6.25 V
V2= I2 * R2 --> = 5 *2.173 = 10.865
Why I didn't get the same answer ?
 
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  • #30
Fatima Hasan said:
The voltage across R2 is 6.25 V
V2= I2 * R2 --> = 5 *2.173 = 10.865
Why I didn't get the same answer ?

I don't know where that value of 2.173 came from, so I can't explain. You'll have to tell me how you calculated I2.
 

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