# Have some problems with thevenin

ottjes
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

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## Answers and Replies

arcnets
Hi ottjes,
sorry I don't see what your teacher wants from you. Could you please explain?

ottjes
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

arcnets
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.

ottjes
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?

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arcnets

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.

ottjes
they are cause i left R3 out of the circuit there, not the wire there

edit: post 5K

arcnets

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
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.

Staff Emeritus
Gold Member
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

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ottjes
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)