# Thevenin's Theorem

Discussion in 'General Electronics Chat' started by fektom, Jan 2, 2013.

1. ### fektom Thread Starter New Member

Aug 20, 2012
16
0
Hello everybody!

I've been spending a lot of time with the study of Thevenin's theorem, but have some misunderstanding on this topic. It is caused by a contradiction between two things.

On one side there is an electrical textbook's lesson and my experiment.
On the other side there is the writing of the All about circuit page (http://www.allaboutcircuits.com/vol_1/chpt_10/8.html), and I can not solve the problem.

All about circuit says at the 4. figure, that:
"the original circuit with the load resistor removed is nothing more than a simple series circuit with opposing batteries, and so we can determine the voltage across the open load terminals by applying the rules of series circuits, Ohm's Law, and Kirchhoff's Voltage Law".
Later, it concludes that the total Voltage is 21V.

However, another electronic textbook says that the parallel wired generators are built by connecting the same poles together (+ to +, - to -). Based upon this, I think that the resistances are really in series with each other, but the batteries are not. On the 4. figure they are wired together parallel, so the total Voltage should be 28V, or in real life near 28V (at least based on my reasoning)!

I checked this with my own experiment. I have a breadboard, and two batteries. One has still 9V, but the other one is quite old, has only 4.6V but the same design, other words it used to be a 9V battery as well. I connected them together using the same polarities. I mean, I wired the + polarity to the +, the - to the - (but I didn't shortcircuit them! Actually, I left open the circuit). Then I measured the voltage on the battery's main poles. The voltage was the same on both batteries (aroung 8.3V), doesn't matter which battery I've chosen to measure.

So the batteries doesn't seem to reduce their voltage just because they are connected together on the same poles.

Is it possible that the above mentioned page has a fault with the calculation (at total voltage, which should not be 21V rather 28V)?

2. ### tshuck Well-Known Member

Oct 18, 2012
3,531
675
You need to post a new thread instead of highjacking one from years ago!

3. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
Agreed.

Welcome to AAC!

A thread belongs to the OP (original poster). Trying to take over someone elses thread is called hijacking, which is not allowed at All About Circuits. I have therefore given you a thread of your very own.

This was split from Thevenin's theorem's example on the tutorial

4. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
You took two batteries of different voltage and connected them in parallel. The new terminal voltage is pretty non-predictable. It's a pretty safe bet that it will be somewhere between the two, but all that is happening is that the stronger battery is charging the weaker battery, which is probably not a good thing depending on the battery involved.

But I don't what this has to do with the problem referred to on the AAC E-book page since those batteries are not connected in parallel. They are connected in series (bucking each other) along with two resistors.

Take your two batteries and connect them along with two large valued resistors (say two 100kΩ to keep the demand on the dead battery minimal as well as the reverse current through it, but don't go over about 510kΩ so as to keep the disturbance due to the insertion of your meter -- I'm assuming you are using a meter with about 10MΩ of input resistance -- tolerabe) and measure the voltage across the resistors (so between +terminal on one and +terminal on the other).

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5. ### fektom Thread Starter New Member

Aug 20, 2012
16
0
Hello!

I didn't know about hijacking, didn't even know this notion, I just thought that the topic and my problem has important common points, for example both of them has problem with the understanding of Thevenin's theorem and this understending problem comes from the same page, our questions are regarding to the same url. Thanks for fix my false step, next time I keep in mind this rule!

I thought over one more time this battery is parallel or single problem.
Well, now I see that I made a fault when I transformed the drawing on the 4. figure. I thought that this transformation can be applied: .

I made the wiring again, now correctly, used the 100Kohm resistors and got this results:
New battery voltage is 8.93V
Old battery voltage is 8.58V
Voltage for the first resistor: 0.15V
Voltage for the second resistor: 0.15V
Two resistor together: 0.33V

Which is OK, because the old battery has 8.58V, the two resistors together 0.33V, which is 8.91V. So almost the same like the first, new battery.

It is just strange that I have to subtract two battery voltages when they are connected on the same terminals (+ to +, - to -).

6. ### fektom Thread Starter New Member

Aug 20, 2012
16
0
Sorry!
Just to make it clear: first I built a circuit like this:

And realised, that the 4. figure and my tranformation is not the same.

Than rebuilt, got the expected results (battery voltages must to be subtracted when I connect them the way written by All about circuits).

Thanks to WBahn! Helped me a lot!

7. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
Do not forget batteries have an internal resistance. That is the same as a resistor in series.

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