Have some problems with thevenin (or my teacher is wrong, which i believe .(adsbygoogle = window.adsbygoogle || []).push({});

[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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# Have some problems with thevenin

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