Thevenin Theorem lab

Thread Starter

metelskiy

Joined Oct 22, 2010
66
I'm for about to do a lab about Thevenin theorem and i need to do some calculations before actually doing lab (I have just read Thevenin Theorem explanation here on Allaboutcircuit).
Here i have a circuit:
Lab 6.jpg
And they ask to Calculate Thevenin voltage and resistance for the network to the left of points a-b using measured resistance values (R4 is the load). Show all work. First step i know i have to remove the load in that case would be R4 and so i got:
Lab 6_1.png
With given circuit, the voltage between the two points where the load resistor used to be attached needs to be determined. Need help with that.
 

t_n_k

Joined Mar 6, 2009
5,455
With a-b open, R1 & R3 in conjunction with the 12V source form a voltage divider. Vab will be the same as the voltage across R3 - no current is flowing in R2 (as denoted in the first circuit).

You have two R3's in the second rendering.
 
Last edited:

Georacer

Joined Nov 25, 2009
5,182
As for Rth, you need to nullify the source (make it a short circuit) and find the resistance between point A and Ground.
 

Thread Starter

metelskiy

Joined Oct 22, 2010
66
Thanks replies guys, i'm now clear on that circuit.
But i have another one:
28a.png
Where they ask to determine the Thevenin equivalent as seen from terminals A and B. I have found Rth by shorting power source, now to find Vth i need to remove the RL (load resistor) and calculate Va-Vb=Vth but if RL will be removed, wouldn't voltage be the same at points A and B?
 

Thread Starter

metelskiy

Joined Oct 22, 2010
66
28a_Rth.png
I shorted power source and found Rth between A and B by:
Rth=R4+(R2||(R1+R3))= 72Ω

The thing I'm confused about is in calculating Vth, if we remove RL, wouldn't Voltage at A and B be the same? What steps should i approach in order to find Vth? Thanks.
 

JoeJester

Joined Apr 26, 2005
4,390
The thing I'm confused about is in calculating Vth, if we remove RL, wouldn't Voltage at A and B be the same?
The same as what?

from the link I referenced earlier ...

Thevenin's Theorem states that it is possible to simplify any linear circuit, no matter how complex, to an equivalent circuit with just a single voltage source and series resistance connected to a load.
Your Rth is incorrect.

Look at the attached and recalculate. Post your results.
 

Attachments

Last edited:
no they wouldn't be the same voltage because there is still a current in that circuit going through R1 - R2 - R3 thus there would be a potential difference around R2 ( i.e. between a and b).
 

JoeJester

Joined Apr 26, 2005
4,390
Post your "new" original circuit ... with all the values.

How do you know you got the other one correct? Was the answers present in your instructions from the professor?
 

Thread Starter

metelskiy

Joined Oct 22, 2010
66
Here is the original circuit in which i need to find Vth and Rth:
28a.png
I found Rth but shorting power source:
28a_Rth.png
And came out with 72Ω
I need help to find Vth. Thanks.
 
if i'm not mistaken, then Vth is simply the voltage at the top-centre node. As no current flows through R4, then A will be at the same voltage as the top-centre node.

You find this by the voltage division of the R1 and the equivalent resistance of R2&R3.
 

Thread Starter

metelskiy

Joined Oct 22, 2010
66
As no current flows through R4, then A will be at the same voltage as the top-centre node.

You find this by the voltage division of the R1 and the equivalent resistance of R2&R3.
Why wouldn't there be current flowing through R4? I believe in order to find Vth first we remove the RL than find Vth (Va-Vb) so instead of RL now there will be just a connection and R4 would be connected to R2&R3? Just my thoughts, please correct if I'm wrong.
 

JoeJester

Joined Apr 26, 2005
4,390
Why wouldn't there be current flowing through R4? I believe in order to find Vth first we remove the RL than find Vth (Va-Vb) so instead of RL now there will be just a connection and R4 would be connected to R2&R3? Just my thoughts, please correct if I'm wrong.
When you replace the load with a voltmeter ... the same potential appears at points A and B as that which is dropped across R2.

There is no current flow through R3 until a load is connected. Yes, the meter is a load that is of sufficient high impedance that it doesn't affect the voltage reading.

I am close to your 72 ohm Rth calculation ... My figure is higher because I choose to go with three decimal places.

What did you compute for Vth? 1.39V is not the answer.
 

Thread Starter

metelskiy

Joined Oct 22, 2010
66
I used voltage devider formula to find V at junction of R2&R3, That I believe would be my Vth right? (47Ω/117Ω)*2.5V=1V
Vth=1V and Rth=72Ω correct?
 
Top