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Proving Thevenin's Theorem 
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#1
Jun604, 11:33 PM

P: 1,778

Hi
We've been taught how to solve circuits (of a resistive nature only, so far) using node voltage/mesh current methods and also by Thevenizing them...but I have not found (so far on the net) a proof of Thevenin's Theorem. Can anyone suggest a resource which mentions it? Cheers Vivek 


#2
Jun704, 12:14 AM

P: 199

My professor told me there wasn't a proof, but I might have misunderstood him.
Edit: Or maybe it was that there is no way to say that a Norton equivalent circuit is identical to a Thevenin equivalent circuit; I can't remember. I can say that my professor and my TA's were very uncomfortable taking a firm position on this subject. 


#3
Jun704, 01:57 PM

P: 1,778

Cheers Vivek 


#4
Jun2504, 05:15 AM

P: 1,778

Proving Thevenin's Theorem
Then there are also the stardelta rearrangement techniques and the symmetry method which work wonders for most of our preundergraduate circuits such as unbalanced bridges, meshes, etc. (for finding network resistance and further simplification). I have found no rigourous mathematical proof of these theorems.
I am not sure but if we were to replace resistors by algebraic symbols or entitities in a pattern, we might be able to associate (and possibly address the problem) such patterns with graph/tree theory. I would like to know if such proofs for stardelta and symmetry methods exist. Otherwise of course, we can prove them by accounting for currents in the (proposed) equivalent arrangements on subjecting them to a common voltage/current source (and thereby justify their equivalence). However, such methods of proof seem heuristic at the moment, at least to me. Any inputs are greatly appreciated... 


#5
Jul404, 11:07 PM

P: 59

The general proof for ANY resistive network is quite cumbersome.
If you want to prove it for RLC networks then just write the two networks (the original and the Thevenin transform) and write their kirchoff voltage and current laws, as seen from one pair of terminals, and show that they are the same Greetings Javier 


#6
Jul504, 07:27 PM

HW Helper
P: 2,327

Kirchoff's laws offer the proof. Just keep in mind that these laws are, themselves, axiomatic to low speed circuit theory, and are not strictly correct.



#7
Jul2304, 07:41 AM

Emeritus
Sci Advisor
PF Gold
P: 6,236

Let us consider a circuit containing components and sources, with two output wires. If we accept that the circuit is linear, meaning that the superposition principle holds (I'll come to that) then we can connect these two output wires to an external ideal voltage source. The current that will flow in that ideal voltage source can be written as: I = (internal source 1) * A + (internal source 2) * B + ... + (internal source n) * C  (external source) * G This is dictated by the superposition principle: there are n internal sources, and one external, so all voltages and currents anywhere in the circuit are a linear combination of these source values (if this is not the case, by definition, the circuit is nonlinear, and indeed, thevenin's theorem doesn't hold). The coefficients do not, of course, depend on the sources (otherwise this is not a true linear combination). Now the n first terms only depend on the internal workings of the circuit (coefficients and internal sources), so we can replace this, as the external source is concerned, by a constant, D. So we have: I = D  V_ext * G Define V_th = D / G and G = 1/R and we have: I = 1/R * (V_th  V_ext) This is of course the equivalent relationship with a thevenin circuit with voltage V_th and resistance R, coupled to our external voltage source V_ext. You could object that connecting an external voltage source is not the most general thing one could do, but this is not true. In ANY application, there will be A potential difference between the two wires ; it is evident that we can then replace the entire network by a voltage source with exactly this potential difference, and our circuit will not notice the difference. The other facet is a bit more cumbersome: one has to show that any complicated circuit, only consisting of resistors and sources, constitutes a linear network. This can be done using Kirchhoff's laws. cheers, Patrick. 


#8
Jul2304, 12:22 PM

P: 284

How do you 'prove' Ohms law  it is a mathematical model of a physical observation
it's proof is that is accurately works given sufficient circumstances. Thevnin circuit series or parallel is a model of an electric source ( mathematically the two are eqivalent) since they yield the same output for open and short load conditions , and in the simple case linearity is assumed. I think of it as a 'black box' you test what it does from the reaction to a load  s/c produces I , o/c produces E , and if I obtain V=E/2 by a load I find that RL = E/I , this gives the 'linear' model where Rs = RL . By the same token you could test for ac. phase and magnitude of s/c I and o/c E with frequency and attempt to model the source as a bunch of components , this can work for very simple circuits but gets too complicated for general networks , especially ones with active components such as transistors , this does not invalidate the measurements it just means that an equivalent model is too complicated. 


#10
Jul2504, 02:39 AM

P: 1,778

The proof for the wyedelta transformation exists and so does one exist for Thevenin's Theorem...strictly they are based on topology and different from the ones that have been discussed here so far.
Cheers Vivek 


#11
Jul2504, 06:52 AM

P: 284

As I understand Topology it is concerned with connective relations that does not pin down( or so it would seem) the values of equivalent components in a thevnin source
which maybe real , imaginary,complex , linear, nonlinear , and dynamically unstable. 


#12
Jul2604, 09:40 AM

P: 34

As someone already said Kirchhoff (almost) IS the proof. The equations form a linear system whose solution must be of the form V = aI + b (note that there is one equation less in the system because the load is not connected). The values a and b can be found by solving the system. One way to do it easily is to put I=0 and "measure" V, and then put V=0 and "measure" I. (for those that might draw the graph of V(I)... a is usually a negative number).



#13
Jul2704, 05:05 AM

P: 1

dear mavrick ,
there's a quite sort of proof of thevivins theorem , but it's very very complicated one that we have taken in our microwave circuit synthesis and design , so if you wish i can post it to your mail . if so send me a mail message at : gamede109@yahoo.com best regards 


#14
Aug104, 02:58 AM

P: 55

Recently this one for delta Y however it requires port variable theory :
For a T network the transfer matrix equation may be found by multiplying transfer matricies : [itex] \[ \left[ \begin{array}{l} 1\,\,\,\,\,\,\,Z_1 \\ 0\,\,\,\,\,\,\,1 \\ \end{array} \right]\left[ \begin{array}{l} \,\,1\,\,\,\,\,\,\,\,\,\,\,0 \\ 1/Z_3 \,\,\,\,\,1 \\ \end{array} \right]\left[ \begin{array}{l} 1\,\,\,\,\,\,\,\,Z_2 \\ 0\,\,\,\,\,\,\,\,1 \\ \end{array} \right] = \left[ \begin{array}{l} 1 + \frac{{Z_1 }}{{Z_3 }}\,\,\,\,\,\,\,\,\,Z_1 + Z_2 + \frac{{Z_1 Z_2 }}{{Z_3 }} \\ \,\,\,\,\,\frac{1}{{Z_3 }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 + \frac{{Z_2 }}{{Z_3 }} \\ \end{array} \right] \] [/itex] And for a pie network (of addmittances) the transfer matrix equation is : [itex] \[ \left[ \begin{array}{l} \,\,1\,\,\,\,\,\,\,\,0 \\ \,Y_1 \,\,\,\,\,\,\,1 \\ \end{array} \right]\left[ \begin{array}{l} 1\,\,\,\,\,\,\,1/Y_3 \\ 0\,\,\,\,\,\,\,\,\,\,\,1 \\ \end{array} \right]\left[ \begin{array}{l} \,\,1\,\,\,\,\,\,\,\,\,0 \\ \,\,Y_2 \,\,\,\,\,\,\,1 \\ \end{array} \right] = \left[ \begin{array}{l} \,\,\,\,\,\,\,\,\,1 + \frac{{Y_2 }}{{Y_3 }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{1}{{Y_3 }} \\ \,\,Y_1 + Y_2 + \frac{{Y_1 Y_2 }}{{Y_3 }}\,\,\,\,\,\,\,\,\,\,\,\,1 + \frac{{Y_1 }}{{Y_3 }} \\ \end{array} \right] \] [/itex] Equate the transfer matrix components and you have the Tpi or DeltaY transforms. Best 


#15
May2909, 02:21 AM

P: 1

I think thevenins theorem can be proved directly from SUPERPOSITION THEOREM which itself can be proved from kirchoffs laws .
To prove that the equivalent resistance of the circuit can be calculated from assuming individual sources V1,V2,V3,V4..... are all 0V and then looking into the circuit from the right hand side: Assume the following circuit : THE MESH V1,...Vn to be simplified is connected to Veq in series . Then the voltage at the node is 0 and so is the current. Now, from superposition theorem the current at the node(0amp) is the sum of currents due to V1,V2...Vn=0V and Veq=0v INDEPENDENTLY. so taking Mesh voltages =0v the current and voltage is equal to the ve of that taking Veq=0. So with mesh voltages as 0V we can calculate V/I looking in from the node . this gives us the mesh resistance or Req. A formal proof would proceed along these lines . Superposition allows us to verify that the equivalent circuit is consistent with the original mesh in terms of Voltage and current at the output terminals. 


#16
May2909, 04:50 PM

P: 758

Proofs of the theorems of Thevenin and Norton, as well as the superposition theorem, can be found in the classic text "Communication Engineering", by W. L. Everitt. This book was published in the 1930s, and can probably be found in your local university library.
Thevenin's original paper is: "Sur un nouveau theoreme d'electiicite dynamique", Comptes Rendues, Vol. 97, 1883, pp.159161. 


#17
Jul2109, 03:23 AM

P: 1

i am first year student in University of Malaya..i would like to ask..where should i get reference on thevenin theorem..and can someone explain to me about thevenin theorem..how to calculate it and so on...?



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