Solving Problems with Thevenin: Is My Teacher Wrong?

[ottjes]

Have some problems with thevenin (or my teacher is wrong, which i believe .


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

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
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

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

 

[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

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 back to my question, anybody knows what i did wrong?

 

[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

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.

 

[enigma]

I got the same answer that you did...

First step, disconnect R₃ from the circuit.

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

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

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

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

R_tot = R_th + R₃ = 7.5 ohms

and I_R3 = V_th/R_tot = 2/3 A

EDITED to make it more readable

 

[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)