Solving Problems with Thevenin: Is My Teacher Wrong?

In summary, the conversation was about a problem with Thevenin's theorem and finding the equivalent resistance and voltage of a circuit. The original poster had some confusion about how to solve the problem and asked for help from others. After discussing different approaches, it was concluded that the correct way to solve the problem was to use the formula R = 1/(1/R1 + 1/R2) to find the equivalent resistance and Vth = Vi * R2/(R1+R2) to find the equivalent voltage.
  • #1
24
0
Have some problems with thevenin (or my teacher is wrong, which i believe :)).


[PLAIN]http://www.fmf.nl/~wim/studie/cir1.png [Broken] [Broken]

example from my book:
Vi=15V, R1=6, R2=R3=3

Voc = Vi R2/(R1+R2) = 5 V (potential divider)
R2//R3=1.5 Ohm
R1+(R2//R3)=7.5 Ohm --> I = 15/7.5 = 2 A
Cause R2 and R3 are equal half of this current will flow to R2 and other half to R3 -->Isc = 1 (how will this change if R2 != R3?)
R = Voc/Isc = 5/1 = 5 Ohm

How i will solve this problem:
Leave R3 first out of the problem
Voc = Vi R2/(R1+R2) = 5 V
R1//R2 = (R1+R2)/(R1R2) = 9/16=2 Ohm
Now i have this:
http://www.fmf.nl/~wim/studie/cir2.png [Broken]
R=(R1//R2)+R3=2+3=5 Ohm

This looks good, same as the book



On my test i got this question:
[PLAIN]http://www.fmf.nl/~wim/studie/cir1.png [Broken] [Broken]

With Vi=10 V and
R1=R2=R3=5 Ohm

Answer here must be, according to teacher: Voc=5V and R=3.75 Ohm

If i do this the same way as the book:
Voc = Vi R2/(R1+R2) = 5 V
R2//R3=2.5 Ohm
R1+(R2//R3)=7.5 Ohm --> I = 10/7.5 = 1.33 A
Cause R2 and R3 are equal half of this current will flow to R2 and other half to R3 -->Isc = 2/3 (how will this change if R2 != R3?)
R = Voc/Isc = 5/(2/3) = 7.5 Ohm (This is 2*3.75 :?)

Doing it my way:
Leave R3 first out of the problem
Voc = Vi R2/(R1+R2) = 5 V
R1//R2 = (R1+R2)/(R1R2) = 10/25=0.4 Ohm
R=(R1//R2)+R3=0.4+5=5.4 Ohm (different than the other answers :?)

What am i doing wrong, What is the good way to do this

edit: fixed url's
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
Hi ottjes,
sorry I don't see what your teacher wants from you. Could you please explain?
 
  • #3
from the second circuit i got 3 different answers, i want to know which one is the right one? and if possible what is wrong about the others
 
  • #4
I can only see the 1st picture, the other link produces a server error.

Do they want you to find the total resistance R of the circuit?
If yes, I should say R = R1 + R2, since no current can go thru R3.
 
  • #5
fixed the url

the meaning of thevenin is that you can big circuits very small, ie a circuit that contains only a resistor and a voltagesource (thevenin), or a circuit with a resistance and a current generator (norton)

example:
calculating voltage across A and B of this circuit is diffucult
http://www.phys.rug.nl/pleit/electronics/O00b1{image0}.gif [Broken]

after applying thevenin and norton a few times you see that the circuit is above is equil to:
http://www.phys.rug.nl/pleit/electronics/O00b1{image4}.gif [Broken]back to my question, anybody knows what i did wrong?
 
Last edited by a moderator:
  • #6


Originally posted by ottjes
R1//R2 = (R1+R2)/(R1R2) = 9/16=2 Ohm

In your original post, this is the first line that looks false.
Because in the 1st picture, R1 and R2 are not parallel.
 
  • #7
they are cause i left R3 out of the circuit there, not the wire there

edit: post 5K
 
  • #8


Ah! Now I think I understand what happens.
I wrote down all the Kirchhoff rules that apply here, and worked it out. I found that you can indeed replace this by just Voc and R. My result was:
Voc = Vi * R2/(R1+R2)
R = (R1||R2) + R3.
I think these are the same as your formulae.

Now your error is probably here:
Originally posted by ottjes
R1//R2 = (R1+R2)/(R1R2) = 10/25=0.4 Ohm
Should read:
R1||R2 = (R1R2)/(R1+R2)=25/10 Ohm = 2.5 Ohm
So, R = 7.5 Ohm.

I think you are correct and 3.75 Ohm is wrong.
 
  • #9
I got the same answer that you did...

First step, disconnect R3 from the circuit.

Vth is the voltage across R2, which would be 5V.

Rth is found by treating the two resistors as if they were in parallel...

Rth = 1/(1/R1+1/R2) = 2.5 ohms.

If you then re-hook up R3, you find total resistance of the circuit

Rtot = Rth + R3 = 7.5 ohms

and IR3 = Vth/Rtot = 2/3 A

EDITED to make it more readable
 
Last edited:
  • #10
Thanx all, glad to know i did understand it ;)

My fault was that i did R=1/R1 + 1/R2 which is (R1+R2)/(R1R2)
But R = 1 / (1/R1 + 1/R2) which is (R1R2)/(R1+R2)
 

1. What is Thevenin's Theorem and how is it used to solve problems?

Thevenin's Theorem is a method used to simplify complex electrical circuits into a single equivalent circuit. It states that any linear circuit can be represented by a voltage source in series with a resistor. This equivalent circuit, known as the Thevenin equivalent circuit, can be used to analyze the behavior of the original circuit.

2. How do I determine the Thevenin voltage and resistance of a circuit?

To determine the Thevenin voltage, the circuit must be simplified down to its Thevenin equivalent circuit. The Thevenin voltage is then equal to the open circuit voltage of the equivalent circuit. To determine the Thevenin resistance, the circuit must be simplified down to its Thevenin equivalent circuit and the internal voltage sources must be removed. The Thevenin resistance is then equal to the total resistance seen from the load terminals.

3. Can Thevenin's Theorem be used for non-linear circuits?

No, Thevenin's Theorem only applies to linear circuits. Non-linear circuits cannot be simplified using this method as the voltage and current relationships are not linear.

4. How accurate is Thevenin's Theorem in predicting circuit behavior?

Thevenin's Theorem is very accurate for predicting the behavior of linear circuits. However, it is an approximation and may not be as accurate for non-linear circuits or circuits with significant parasitic elements.

5. Can Thevenin's Theorem be used to analyze AC circuits?

Yes, Thevenin's Theorem can be applied to both DC and AC circuits. However, the calculations for determining the Thevenin voltage and resistance may differ for AC circuits due to the presence of reactance.

Suggested for: Solving Problems with Thevenin: Is My Teacher Wrong?

Back
Top