# Thevenins theorem and dependent sources

Discussion in 'Homework Help' started by alexei_kom, Nov 22, 2009.

1. ### alexei_kom Thread Starter New Member

Nov 22, 2009
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Hello all!

As I searched the internet I didn't find the answer to a question that bothers me a lot.
Every article tells that with dependent sources there must be used another method, but no one of them explains WHY.
I can't understand WHY I can't treat the dependent sources like the independent? Why can't they be zeroed like the independent?

Thanks a lot!

P.S. Sorry for my English.

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
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I guess we are talking about superposition.

Suppose a network comprises n independent voltage sources and one dependent voltage source. The current Ix in a resistor Rx in the network will be given by a general relationship

Ix=f1(V1)+f2(V2)+.....+fn(Vn)+fz(V1,V2,v3,....Vn)

where the function fz() is the representation of that part of the current in Rx contributed by the dependent source. The terms f1(), f2(), ...etc are the current contributions from the independent sources.

The current Iy in another network resistor Ry would have a different but similar relationship

Iy=g1(V1)+g2(V2)+.....+gn(Vn)+gz(V1,V2,V3,...Vn)

By superposition one follows the process of sequentially making all (voltage) sources zero but one and finding the individual contribution from that particular (non-zeroed) source. After every source has been considered the results are added algebraically to give the total current.

How would one do that for finding current Ix above?

Suppose all sources other than V1 are made zero. The contribution to Ix would then be

Ix=f1(V1)+fz(V1,0,0,0...0)

Since the dependent component fz() also contains a V1 term, it cannot be zeroed without 'casting' an incorrect value for the particular contribution to Ix from V1.

In principle one could follow a modified superposition process by zeroing all independent (voltage) sources but one and include the dependent source in each successive solution - i.e. solve for the individual current contributions with two sources. But this probably wouldn't be all that useful.

A similar reasoning would apply to dependent and independent current sources - where the latter are considered as an open circuit when not included in a particular step in the superposition process.

3. ### alexei_kom Thread Starter New Member

Nov 22, 2009
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I think I'm starting to understand. Correct me if I'm wrong =).

If we have a circuit with independent sources only, we zero them and can measure the R(th) by applying an ohmmeter. In this case the only voltage is from the ohmmeter and the measurement of the resistence is correct.

But if we have a circuit with independent sources and one depend source, we can't zero the dependent source because according to the superposition it also has a "part" of every independent source. In that case I can't use the ohmmeter to measure the R(th) because dependent source must be presented and that means that in addition to the voltage of the ohmmeter there will be also a voltage from the dependent source. That means that the measurement of the resistence will be wrong and that's the reason why I use a test source when I have a dependent source.

Oct 9, 2007
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Mar 6, 2009
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6. ### alexei_kom Thread Starter New Member

Nov 22, 2009
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I'm sorry, but I don't have a time to read the article. Can you guys tell me if I was right about what I wrote?

Thanks again.

7. ### The Electrician AAC Fanatic!

Oct 9, 2007
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I'm puzzled, Alexei. In your first post you told us:
You said that you searched the internet, but now that I've provided a paper with an answer to your question which is "...detailed as possible...", you tell us that you "don't have a time to read the article."

To paraphrase Euclid, "There is no royal road to learning."

See http://en.wikipedia.org/wiki/Royal_Road about 2/3 of the way down the page, under "Cultural references to the Royal Road".

8. ### alexei_kom Thread Starter New Member

Nov 22, 2009
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The Electrician,
There is no need to be puzzled. It may be (and I'm sure it is) easy for you to read this article because:
A. You know english very well.
B. You are not a novice in circuit analysing.

As for me, I have to sit with a dictionary and translate the text and then try to understand it. I work and study, and couple of free hours is a luxury for me.
What I wanted is a short answer to know if what I wrote was right or not. If you say that the key is in the article, fine. I will read it and thank you. But there is no need to tell that I'm searching for a "royal road to learning" because it's not true.

9. ### The Electrician AAC Fanatic!

Oct 9, 2007
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338
Alexei, you asked "Please let your answer be detailed as possible." You should have known that since you posted your request on a forum where the language is English that you were likely to get a response in English.

If you don't have the time for understanding a "detailed" answer in English, then you shouldn't have asked for such an answer.

The paper I referenced is controversial. Some other people think he is wrong. I think he is correct, but there are certain conditions for using superposition to solve a circuit containing dependent sources, which are discussed in the reference.

A simple answer for you is this: if a circuit contains dependent sources, then use another method instead of superposition, and you won't have any problems.

Your English seemed to me to be good enough to understand the article. I didn't realize that you had so much difficulty with English, and if I have unfairly accused you of seeking a "Royal Road", I apologize. The very fact that you asked the question indicates an above average desire to understand; that's why I was puzzled when you seemed to want an answer without doing the work. Good luck with your studies.

10. ### alexei_kom Thread Starter New Member

Nov 22, 2009
9
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Thank you everybody for help!