Stumped by a Thevenin Problem

[Mrs. Kerchief]

I have to get a better understanding of Thevenin's theorem and have
been unable even with the help of a tutor to come up with the Rth
specified by my textbook for the following circuit:

Vs--50V

R1=22 Ohms || R2=22 Ohms, leading at the first node to R3=47 Ohms,
then to the second node R4=33 Ohms. I've just started to learn how to
Thevenize a circuit, and looking back from the open terminals, the
circuit looks like this:


___22___
_____| |______*______________A
+| |__22___| |
50| |
-| 47
| |
|_______33__________|______________B

I apparently got the right Vth by using the voltage divider formula
(47/91) x 50V. But when I tried to get the "right" (according to the
textbook) Rth of 22.7, there was no way I could get that. Looking
back from the open terminals, I put 11 Ohms (the first two resistors
equivalent resistance) in parallel with 47, then that in parallel with
33. I continually came up with an answer of approximately 7.01, not
22.7.

If anyone could tell me what I did wrong, I would appreciate greatly
hearing from you here on this group. Thanks greatly.

 

[Ken Taylor]

I have to get a better understanding of Thevenin's theorem and have
been unable even with the help of a tutor to come up with the Rth
specified by my textbook for the following circuit:

Vs--50V

R1=22 Ohms || R2=22 Ohms, leading at the first node to R3=47 Ohms,
then to the second node R4=33 Ohms. I've just started to learn how to
Thevenize a circuit, and looking back from the open terminals, the
circuit looks like this:


___22___
_____| |______*______________A
+| |__22___| |
50| |
-| 47
| |
|_______33__________|______________B

I apparently got the right Vth by using the voltage divider formula
(47/91) x 50V. But when I tried to get the "right" (according to the
textbook) Rth of 22.7, there was no way I could get that. Looking
back from the open terminals, I put 11 Ohms (the first two resistors
equivalent resistance) in parallel with 47, then that in parallel with
33. I continually came up with an answer of approximately 7.01, not
22.7.

If anyone could tell me what I did wrong, I would appreciate greatly
hearing from you here on this group. Thanks greatly.

Do you need to parallel up any resistances - look more closely at the
circuit.

Cheers.

Ken

 

[Glenn Hamblin]

You s/b putting the 47 in parallel w/ the series combo of 11(22||22)in
series w/ the 33. ie. 44||33

 

[JeffM]

To do the resistance analysis,
replace each voltace source with its internal resistance (zero ohms)
and each current source with its internal resistance (open circuit).
Then you get 44 || 47.

 

[Mrs. Kerchief]

jeffm wrote

To do the resistance analysis,
replace each voltace source with its internal resistance (zero ohms)
and each current source with its internal resistance (open circuit).
Then you get 44 || 47.

You added the equivalent resistance of the 22 resistors to the 33
resistor and put that in parallel with 47, but I don't understand.
The textbook arrives at Vth=25.8 (which could be arrived at only if
you use RT of 91), so why does RT "change?"

According to the book ("Principles of Electric Circuits, Conventional
Current Version, Floyd), "The Thevenin equivalent resistance is the
total resistance appearing between two terminals in a given circuit
with all sources replaced by their internal resistances." Heck, maybe
"internal resistance" is somehow different from plain "resistance."
I'm a newbie, what do I know.

The author's first schematic explaining Thevenin Resistance is:
________R1_______._________R3_________.A
| |
| |
| |
(Vs replaced by short)
|
| R2 |
| |
| |
|________________._____________________.B

"Rth=R3 + R1 || R2."
-------------------------------------------------------------

In the problem I asked help with, there's no resistor where R3 is in
the above circuit. But if R1 and R2 are to be considered in parallel,
what I don't understand is why R4 in the original problem (at the
antipodes of the circuit) shouldn't also be considered in parallel.

If anyone would be kind enough to offer a little more disquisition on
the logic/procedure for computing Rth, I would be as grateful as I am
for the answers you already provided. Thanks to everybody.

 

[JeffM]

I don't understand is why R4...shouldn't also be considered in parallel

Mrs. Kerchief

The question is "Why should it?".

You're making it harder than it is by making up new rules.

There are only 2 steps needed here to solve for R equiv:
1) Replace each voltage source with a short.
2) Solve for the resistance between the specified points.

 

[Fraze]

Everyone is telling you the right answer, but perhaps you just need to
see the circuit drawn a little different. First off, replace the 50V
supply with a short. Then move R1 and R2 down the line to the left
side.

___________________*______________A
| |
___|___ |
| | |
| | |
22 22 47
| | |
| | |
|______| |
| |

|_______33__________|______________B

Now, I think we can agree that R1 and R2 are in Parallel. Lets go
ahead and make an R equivalent instead of the R1 and R2.

_________________*______________A
| |
| |
| |
| |
11 47
| |
| |
| |
| |

|_______33__________|_____________B

Now lets move R4, the 33-ohm resistor down the line to the left.

_________________*______________A
| |
| |
| |
| |
11 47
| |
| |
33 |
| |

|___________________|____________B

Now, our Req of 11 ohms and R4, 33 ohms are obviously in series, so
lets add them and make another Req of 44 ohms.

_________________*______________A
| |
| |
| |
| |
44 47
| |
| |
| |
| |

|___________________|____________B

From here it should be evident that the new Req and R3(47ohms) are in
parallel. This combination would equal 22.7ohms and there is your
answer.

When deriving Vth the circuit would look a little different.
(R1||R2)+R3+R4=Rt You apparently did this right and came up the
correct Vth of 25.2688V. When the power supply is a short and you are
looking in from terminals A & B, the circuit would look like the above
figure.

Hope this helps.
Fraze

 

[Mrs. Kerchief]

Fraze wrote

Everyone is telling you the right answer, but perhaps you just need to
see the circuit drawn a little different.

Yeah! Now I finally saw it. THanks, Fraze. I never thought I'd
spend so much time on a simple problem.