# Thevenin's theorem

#### MWalden

Joined Apr 26, 2007
24
In The Art of Electronics it says the thevein's thereom is simplifying the circuit and it gives an example Of 30V in serioes with 2 10k resistors and the output taken from one of the 10k resistors like a voltage divider but it says the thevenin's equivalent circuit is a 15 V source in series with one 5k resistor.
Page 11

Why are the resistors in series added like they are in paralell in this example? Did I miss something?

Also there is alot of math in the next few pages dealing with tunnel diodes and zener diodes etc. which I read, but the next day completely forgot. Do you think I should try to practice these problems before I move onto the next or will I get the practice in the areas I need throughout the book? It kinda talks briefly about something and moves on to the next real quick, or atleast so far. I am only on page 19 now. I thought I'd be going over alot of stuff I already knew from other books but I am learning new things.

#### hgmjr

Joined Jan 28, 2005
9,029

#### recca02

Joined Apr 2, 2007
1,214
can u post a dia of the above example?

#### techroomt

Joined May 19, 2004
198
thevenin basically states that any circuit can be viewed between 2 terminals (any 2 points on a circuit) as a voltage and a resistance (inseries).

to solve for thevenin's voltage: calculate the voltage potential across the 2 terminals. your example was a 30vdc suppy and two 10k resistors in series (thevenizing across one of the 10k resistors). the voltage drop of one of the 10k resistors, or Vth = 15vdc

to solve for the thevenin's resistance: replace the original circuit supply (30vdc) with a short, and calculate the total resistence. this is sometimes hard to recognize, but the two 10k resistors are now in parallel. so total resistence or Rth = 15 Kohm

So the equivalent thevenin circuit woul look like a 15vdc supply in series with a 15kohm resistance. by the way, very complex circuiys can be reduced to this type of circuit.

say now, a new load of 7.5kohm is placed across the 2 terminals, you would have a 15vdc supply and a (Rth) 15kohm and (RL)7.5kohm in series, for a total of 22.5kohm resistance. voltage drops of (VRth) 10vdc and (VRL)5vdc respectively and a currect of .67 amps.

substituting new load resistances can be calculated easily without having to perform the dc analysis on the entire complex circuit. just the thevenin circuit.

#### recca02

Joined Apr 2, 2007
1,214
now that i get the ckt dia in my mind (i was too casual last time),
i wud also like to give some explanation.
you are making eq ckt for the load,
so for the load the voltage is going to be the one appearing across the resistor
from where u take the o/p.
for these two points the rest of the ckt appears in parallel hence the two resistors are in parallel when taking the equivalent ckt.
ur confusion will be removed automatically if u try analysing similar ckts.

#### JoeJester

Joined Apr 26, 2005
4,259
Recca,

Attached is figure 1.10 from AOE.

The question was based on exercise 1.9 with the 30 volts in and R1=R2=10k.

#### recca02

Joined Apr 2, 2007
1,214
Thanks Mr Joe,
i did get the idea abt the ckt .
since the question was more focused on why the resistors are taken in paralell,
it happens since it is the equivalent resistance of the circuit as seen by the load,
voltage across r2 is same as voltage across the rest of the circuit across those two points from which o/p is taken.
so the resistor r1 along with the source actually appears in parallel across r2,
it wud have appeared in series for the closed ckt or if o/p was taken across r1 and r2 , but for those two points shown in dia it appears in parallel.