All About Circuits Forum Thevenin's theorem's example on the tutorial
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#1
04-06-2007, 11:05 PM
 DrJones New Member Join Date: Apr 2007 Posts: 2
Thevenin's theorem's example on the tutorial

Greetings! I'm not very good at physics, so it was a bless to find this site while looking for tutorials about electronics.

The problem is, I've been reading the whole day the chapter about Thevenin's Theorem, and I still don't get it completely. Could someone help me?

I have no problem coming up with the equivalent resistance of the example, but the equivalent tension drop is a mistery. 11.2V seems to come from 28V minus 16.8V but... isn't there another battery on the circuit? Why doesn't it affect the outcome? Why isn't the equivalent tension drop, say, 7V - 4.2V = 2.8V?

Thank you,
DrJones.
#2
04-07-2007, 10:54 AM
 recca02 Senior Member Join Date: Apr 2007 Location: India-22.35° N 82.68° E. Posts: 1,211

please check the polarity of the voltage drop across the resistors. the voltages add up so it is
7v+4.2v = 11.2v
this is because the direction of current flow from positive terminal of 7V battery to the -ve terminal (since 28v is larger than 7v or u can choose arbitrary direction of current and mark polarity of voltage drop accordingly, depending on which u'll get + or - with the magnitude of current and potential drops) hence the drop actually is 7v + 4.2 volt across resistor.
#3
04-07-2007, 11:12 AM
 antseezee Junior Member Join Date: Sep 2006 Posts: 45

The other battery DOES affect the outcome. It's just a matter of how you view the circuit. When finding Eth, you are looking from points A to B of the load removed in the original circuit. Point A is considered to be on top of the resistor; point B at the bottom of the resistor. This is now an open. So we have a completely series circuit.

Since the two sources are next to each other, we can combine them. The one polarity is opposite of the other, so the resulting Voltage is 21V:

28V - 7V = 21V

This source is on the left where the 28V was. There are two resistors in series, a 4 + 1 = 5 ohms. Therefore, we can find the total current in this series circuit:

21 V / 5 ohms = 4.2A

The whole purpose of combining the sources & resistors was so we could find the total current in the circuit. Now, go back to the original circuit with the split voltage sources, two resistors, and open load. We know this:

4.2A * 4 = 16.8V across R1. The voltage drop is left to right (+ to -). Therefore, tracing the voltages from point A to B, we find:

-16.8V + 28V = Eth
Eth = 11.2V

The 7V source on the right does have an effect on the circuit. It helps determine the total current when you combine it with the other source in series. The mistake you cannot do is trying to do the calculations with the combined 21V source, because this alters the circuit entirely, and will not permit for a correct Eth drop from points A to B.

A little tip in the future. In most Thevenin circuits when finding Eth, I find it best to use an analysis method. Unless you find an easy method to apply (such as combining resistors or sources), do Loop Analysis, Nodal Analysis, Superposition or other methods to find Eth. Usually the circuit looks messy with the load resistor removed, so using another analysis method for finding Eth is VERY useful without making mistakes.
#4
04-07-2007, 11:58 AM
 DrJones New Member Join Date: Apr 2007 Posts: 2

Now I understand it. Thank you very much for the quick replies!

 Tags theorem, thévenin, tutorial

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