 This is the 4th lecture and the topic is network theorems. In particular today we propose to talk about Thevenin and Norton. We started in a very simple manner in the last lecture stating the Thevenin theorem. We want to go today in a more comprehensive manner. Both of these theorems Thevenin and Norton have something to do with equivalence of networks. That is 2 networks N1 and N2, 2 networks N1 and N2 are said to be equivalent if their terminal behaviours are the same. Networks as you know terminal behaviour of a network is characterized by its voltage current relationship. If the voltage current relationships of 2 networks N1 and N2 at the terminals are the same then the networks are said to be equivalent. In particular Thevenin and Norton theorems have to do with 1 port equivalence. Equivalence of 1 port networks that is 1 port means the network contains only 2 terminals. Very simple 1 ports can be thought of like this. Let us say 2 resistances R1 and R2 in series then you know that this 1 port is equivalent to a single resistance of value R1 plus R2 agreed. On the other hand if I have 1 port in which 2 resistances are in parallel then you know that a single resistance is equivalent to this whose value is R1 parallel R2 is given by R1 R2 divided by R1 plus R2. These are examples of very simple equivalences. To take this further we know that if 2 inductors are in series then a single inductor whose value is equal to the sum of the 2 inductors is an equivalent network. If 2 capacitors are in series C1 and C2 then the equivalent network consists of a single capacitor of value C1 C2 over C1 plus C2. Capacitors add when they are connected in parallel alright. Now Thevenin and Norton theorems go beyond this. They say let us consider a network N which not only contains resistors, we are considering basically resistive networks, linear resistive networks along with DC sources. This is our consideration at the present moment. We are not admitting inductors and capacitors. Suppose we have a network N which not only contains resistors but also contains voltage sources and current sources alright. So we are considering a composite 1 port which contains resistors, voltage sources, current sources and these sources could be either independent or dependent. What we mean by dependent we will consider a little later but what we are considering is a 1 port network N which contains resistors and sources. Let us use a single term sources. Sources could be voltage or current, sources could be dependent or independent. The meaning of dependent sources as I said we will illustrate first then we shall go into the explanation. What is important is that you understand the concepts. Thevenin's theorem says that if there is a network N which is linear connected to another network, another network let us say N1 which could be either a linear or non-linear it does not matter. N1 is an arbitrary network which is also in 1 port and capital N is the 1 port which drives N1 that is capital N is the driving network and N1 is the driven network. This is driving and this is driven then as far as the current injected into N1 is concerned in so far as the load or the driven network is concerned the network N of the driving network can be replaced by an equivalent network and this equivalent network consists of a voltage source VT an ideal voltage source in series with a resistance RT. This is Thevenin's theorem. Thevenin's theorem says that if a linear network drives a load this load could be linear or non-linear it could be any combination of resistors and sources but the driving network has to be linear and the driving network in so far as the load is concerned in so far as the current injected into the load or the voltage appearing across the load terminals is concerned the driving network can be replaced by a very simple equivalent network which consists of a series connection of a voltage source VT and a resistance RT. The subscript capital T is for Thevenin is the first letter of Thevenin, T-H-E-V-E-N-I-N we shall we shall extend it to inductive and capacitive networks later but let us understand the concept with regard to with reference to a resisting network. We shall require slightly different techniques for networks containing inductors and capacitors which we shall consider later. Now to be able to understand this a little more in the perspective let us say that under the actual network connection let us say the load current is I L if I L is going then I L must come out also alright this current the current coming out must be I L let us say that the current is I L then Thevenin's theorem says that if this equivalent network is connected to the same network N1 then the current here will also be I L this is what equivalence means, equivalence means that the terminal behaviour is the same that is if this voltage is V L then this voltage shall also be V L this is what equivalence means alright. The interpretation of VT and RT are as follows you have taken the network N which contains resistors voltage sources and current sources and you want to find out it is equivalent you want to find out it is Thevenin equivalent VT plus minus and the Thevenin equivalent resistance RT VT is simply the voltage appearing at the terminal to the one port with the one port open circuited in other words no current the one port N does not drive any current under that condition whatever the voltage is is called VT so it is the VT is the open circuited voltage that is when it is not connected to any resistance the resistance is infinite between the 2 terminals the voltage that appears across this is the VT and RT is the ratio of VT to the current that flows when these 2 terminals are short circuited that is N the terminals are short circuited so this is ISC the short circuit current the ratio of the open circuit voltage to the short circuit current is RT this is one interpretation of RT the other interpretation is that RT is the resistance looking back into N RT is the resistance measured across this one port with the voltage sources short circuited that is voltage sources replaced by a short circuit and current sources current sources replaced by an open circuit that is like this that is you open the current sources and short the voltage sources inside N under that condition whatever resistance is measured across the port is RT this is the interpretation of RT and this is where the concept of dependent and independent sources arise let me make a formal definition a dependent source is 1 now we are talking of dependent and independent sources what I have said about short circuiting a voltage source an open circuiting a current source applies only to independent sources in finding RT dependent sources should not be touched dependent sources should remain intact question is what is a dependent source a dependent source is 1 whose voltage or current a source is either a voltage source or a current source a dependent voltage source is 1 whose voltage depends on some current or voltage at some other point in the circuit all right a dependent source is 1 whose voltage or current depends on some other voltage or current in the circuit that is the simple definition of a source a dependent source for example for example if I have let us say a voltage source VS in parallel with a resistance let us say R1 then we have another resistance R2 then I have let us say current source another resistance suppose this is my circuit where this current is I1 and this current source is let us say beta times I1 agreed beta times I1 then this current source is not an independent current source because the current generated by this beta I1 depends on the current at some other point in the circuit depends on the current through R1 you understand this this voltage source VS for example is an independent voltage source because it generates a voltage VS irrespective of what happens to the rest of the world it does not care whereas this current generator is not so independent its current the current generated by this source depends on another current at some other point in the circuit that is I1 you understand what you mean by dependent and independent sources a dependent source generates a voltage or current irrespective of what happens to the rest of the world a dependent source is not as free as this its current or voltage depends on some other current or voltage it is not necessary that a current source should depend on another current it could depend on another voltage as you can see here I can write I1 as VS by R1 and therefore I could write this as gamma VS where gamma is beta by R1 and therefore this current source could depend on either a voltage or a current similarly and a dependent voltage source could depend on either a voltage or a current and the important point that we are mentioning is that in finding RT the Thevenin equivalent resistance as I said it is the resistance looking back into N the network which you are trying to replace with the following modifications that is if you have a voltage source you short circuit this if you have an independent voltage source you short circuit this which means that physically you remove the source and put a short circuit if you have a current source independent current source you replace this by an open circuit so that what you really do is physically replace physically remove the current source whereas if you have a dependent voltage source dependent sources do not touch they are protected by a shield you do not touch them they remain as they are alright this point has to be borne in mind it is a most important point this is where most of the circuit analysis particularly by non-electrical engineers goes wrong and therefore you must remember this very carefully dependent sources cannot be touched they are barred yes independent current source or that is very simple a transistor output is behaves like an independent current source well we will come across circuits which can be considered as current generators alright electronic circuits basically it is not as simple as a battery yes yes that is independent but suppose it was 3 times V1 where V1 is somewhere else in the circuit then it would have been independent current source is not available in the laboratory in the laboratory what you get are batteries or the the mains current so the mains voltage so that these are all voltage sources fortunately or unfortunately the generators that one makes in practice are voltage generators that one can make current generators initially come across circuits in future when you deal with transistor circuits which behave as current generators we will consider them at a later point of time so the point that I have made is that the that one port a linear one port containing resistors and sources sources of two kinds current and voltage sources of two different characters dependent or independent can be replaced in so far as the driven network is concerned by an equivalent circuit consisting of a simple series connection of a voltage source and the resistance and we have called the voltage source as capital V subscript T and the equivalent resistance is capital R subscript T and capital V T is the open circuit voltage and to determine RT one can proceed in two ways that is one can find out the equivalent resistance by making some changes in the sources in the independent sources that is independent voltage sources have to be short circuited independent current sources have to be open dependent sources not to be touched to be left intact then you measure the equivalent resistance that is RT or the other thing that you can do is forget about touching any of the sources you simply find out the short circuit current then VT by ISC is RT all right now before we take an example to illustrate this I want you to understand the significance of RT in another way N can be connected to N the network which you are trying to replace by a given equivalent can be connected to let us say resistance RL mode now if N is equivalent to VT and in series RT then the current in the load in either a network should be the same if this current is IL then this current should also be equal to IL all right now if IL is 0 under what condition does that happen if RL is infinite that is if the network is open then if RL is infinity then you see the voltage VL let us say would be equal to simply the open circuit voltage which by definition is VT so VL is VT okay if RL is 0 that is if this is short circuited then the current in the original network is ISC and in the Thevenin equivalent what is the current it is VT divided by RT because RL is 0 and therefore RT is simply equal to VT divided by ISC all right this checks with our interpretation of RT RT can be measured in two ways you can either find out the short circuit current or you can find out the equivalent resistance looking into this to illustrate Thevenin's theorem if you have any questions I would like to answer them now before I take the example well the only constraint is it must pass a current it cannot it should not be open circuit that is all the constraint it could be linear it could be nonlinear it could contain resistors inductors capacitors I don't care but since we are considering direct current sources DC sources even if there is an inductor the inductor behaves as a short circuit if there is a capacitor the capacitor gives us open circuit so basically with DC sources under equilibrium condition it behaves as a resistive network we shall extend all these concepts to RLC networks in a at a later point of time but we would like to grasp this concept with resistive networks first any other question the example that we are about to do we have already done in the last lecture when we solved a network containing a voltage source and the current source we find out the currents in all the branches and so on the same example you take and we shall solve by applying Thevenin's theorem and it is not a trivial exercise so I would like you to follow the steps very carefully the circuit is the following we have a 50 volt source 50 volt source along with two resistances 12 ohm and 2 ohms a third resistance 5 ohm and the current generator of 3 amperes here then there is another resistance here of value 10 ohms this is my circuit and to make things a bit more complicated to make it look really complex let us find out the current through these 2 ohms and let us call this current as I now it is a same same exercise that we have taken to illustrate KCL KVL loop analysis and node analysis we want to solve it by applying Thevenin's theorem all right now the first thing we do is to find out the Thevenin equivalent we call these points as let us say A and B to find out Thevenin equivalent we would like to find out VT and RT to find VT we must open this up in other words our new circuit to find VT shall be like this 50 12 then A and B is an open then 5 ohm and you have a 10 ohms here in addition you have a current generator 3 ampere here what is VT VT is the voltage difference between A and B if you call this voltage is VA and this voltage is VB then VT is this difference with this polarity why this polarity because the current was assumed from left to right so this is the polarity if you connect a 2 ampere obviously the current will go from left to right is the point clear we are going to calculate now VT obviously VT as I said is equal to VA minus VB so let us find out what is VA you see that in calculating VA we have a current source here 3 ampere what can this current source do it can only drive a current through the 12 ohm here it cannot flow in any other direction this is the only freedom it is left to it and therefore it will cause a voltage drop across 12 ohm of 36 volts 3 times 12 36 with what polarity plus here and minus here right the polarity will be plus here and minus here and this battery is plus here and minus here so can you tell me what VA is VA is obviously 36 plus plus 50 or minus 50 plus minus plus minus 86 volt okay VA is 86 volt now I come to VB VB is the voltage across this 5 ohm resistance agreed and you see the configuration this can be done by inspection 50 volt is connected to a series combination of 10 ohms and 5 ohms so very simple voltage division takes place and it is 5 divided by 10 plus 5 times 50 this is point here 50 volt appears across a series connection of 10 ohms and 5 ohms remember this is open so the rest of the circuit does not interact with it at all so 10 ohms and 5 ohms what we are trying to find is the drop across 5 ohms and obviously voltage division R2 over R1 plus R2 multiplied by 50 this is fine clear how much is this 50 divided by 3 agreed and therefore and therefore VT our Thevenin equivalent is simply 86 minus 50 divided by 3 which is equal to 3 times 6 18 1 2 5 8 minus 50 divided by 3 is equal to 200 divided by 3 volt I beg your pardon 208 this is VT now we want to find out RT let us find out let us find RT by both ways both interpretations that is by finding the short circuit current and also by measuring the resistance equivalent resistance it turns out that the second approach namely finding out the equivalent resistance by killing sources that is by killing which sources independent sources independent sources that turns out to be easier as you shall see right away our circuit was 50 volt 12 then I beg your pardon I this would be a short circuit now that 2 ohm would be a short circuit you have a 3 ampere source here 3 ampere source then you have a 5 ohm and a 10 ohm here let me draw it again I was feeling lazy I have a 50 12 3 ampere and this place of 2 ohms which was open circuited for finding VT shall now be short circuited this is A this is B and this current is the current that we are interested in finding out ISC all right then I have the 5 ohm resistance and the this resistance is 10 all right I can find out ISC like this suppose I consider this current as I1 this current is I1 then ISC obviously shall be equal to 3 plus I1 by which law have we applied KSE alright we also have to have to know this current let us call this as I2 alright then how many unknowns are there 2 just 2 so let us express everything in terms of ISC and we write 2 loop equations alright 2 loop equations so the first loop is 50 12 5 agreed then I get 50 is equal to 12 I1 plus what is this current I2 plus ISC 5 I2 plus ISC and the other loop equation is this 10 and 12 obviously 10 I2 shall be equal to 12 I1 is that easy to see it is the same difference of potential either you apply KVL or you argue like this you got the same 2 point and therefore 10 I2 should be equal to 12 I1 now I will leave the algebra to you I have done the algebra and I leave the algebra to you I will only give you the final result what you do is you replace I1 by 3 minus ISC and solve from solve for ISC from these 2 equations I don't know a substitution or by Kremers method I don't care the final result is that ISC is equal to let us say I did it somewhere 208 divided by 46 ampere this is the final result my VT was equal to what is equal to how much 208 by 3 volts and therefore my RT shall be equal to what 46 by 3 46 by 3 ohms alright now if you ask me now what did we gain by given is there it didn't look like much simplified one because for ISC I have to solve 2 loop equations anyway isn't it it turns out that the other method instead of calculating ISC the other method is a better choice here that is finding out RT let us see let us see how that solves the problem what we have is 50, 12, A, B this is where we have to measure RT between A and B after killing the sources we will see how to kill them this is 5 this is 3 ampere and there is a 10 ohms here okay in the previous method what I did was I short circuited this and found out ISC the short circuit current now we shall argue like this that RT is the resistance between A and B with independent voltage sources short circuited so this is replaced by short circuit now and independent current sources open circuited so break this up remove this source physically what are you left with you are left with a 12 ohm in series with a parallel combination of 10 and 5 isn't that right what you have is 12 ohms then a 10 ohms here and a 5 ohms here this is the equivalent circuit you are measuring the resistance between these 2 points so RT is almost obtained by inspection it is 12 plus 10 parallel 5 which is 50 divided by 15 which is equal to 12 plus 10 by 3 which is 46 by 3 this is what we had obtained after a bit long calculation in the ISC method the short circuit current method nevertheless the short circuit current has its own importance as we shall see a little later now therefore our Thevenin equivalent circuit you remember our original problem was this 12, 2, 5, 3 ohms and 10 this was 50 volt we were trying to find out I this is my load and what I have found out is that my VT is 208 divided by 3 and RT is 46 by 3 this point is A and this point is B A and B this is my equivalent circuit then I connect the 2 ohm the current through 2 ohms must be the same as in the original circuit so what is the value of I? I is 4 amperes here that is correct and it checks it what we found out by loop analysis or load analysis this is a non-trivial application of Thevenin's theorem and I wanted you to know that even a complicated case like this where there is a voltage source there is a current source and it is a fairly complicated 1, 2, 3, 4 resistors can be solved by simple application of Thevenin's theorem. Now next I shall take another example which contains dependent sources and see how to apply Thevenin's theorem there and this example as we shall see later is a equivalent circuit for a transistor but just look at it the way I am drawing it now forget about what it represents we are interested in solving this problem there is a dependent current source beta I 1 in parallel with a resistance R 2 and this drives a load R L all right what we are interested in is let us say the current in R L if I know I R L if I know the current in R L then I of course know the voltage in R L all right this is what I used to find out I L by application of Thevenin's theorem to that part of the network which is to the left of the dotted line all right I wish to find out the Thevenin equivalent and for that the first thing I do is to find out V T all right I did not show I 1 I 1 is this current I 1 is this current thank you now the obviously beta I 1 is now a dependent source so let us see how we solve this problem the first thing to do is to find out the open circuit voltage let me draw the circuit again this is R 1 and this current is I 1 R 3 beta I 1 and there is a resistance R 2 this voltage is V T open circuit R L is disconnected this is the open circuit now to find out V T V T what are you going to apply there is a voltage source there is a current source and to complicate matters the current source is a dependent one it depends on another current at some part some other part of the circuit the simplest method why you can think of loop analysis mode analysis you can think of branch currents and loop currents and so on the simplest method obviously shall be mode analysis because there is only one voltage which is unknown and this voltage is this mode voltage after all these two are the same modes so all that you have to find out is V T so you write one KCL equation that is sum up all the currents that lead this mode V T obviously these are V T by R 2 this is the current through R 2 plus beta I 1 beta I 1 leaves this mode but what is I 1 I 1 is V S divided by R 1 so beta V S divided by R 1 all right plus V T plus the current through R 3 there are 3 currents that we have to take care of this one this one and this one these are the 3 currents leaving the node and the sum of them should be equal to 0 so the third current the current through R 3 is V T this potential minus V S divided by R 3 so V T minus V S divided by R 3 and this should be equal to 0 which immediately gets you what is V T if I use this symbol 1 by R equal to G capital G then V T can be written in one stroke that V T times G 2 plus G 3 shall be equal to V S I take them to the right hand side G 3 minus beta G 1 is that okay is it all right where I have used G I as equal to 1 over R I G I obviously is conductance what is the unit or Mohs M H O S all right the Americans prefer Mohs because Siemens was a German if Siemens was an American they just said Siemens and you study American text to Smith is an American text book so he writes Mohs minor differences but obviously you can see that V T is given by V S G 3 minus beta G 1 divided by G 2 plus G 3 all that you needed in solving this problem in finding out V T was KCL KVL and common sense which is the strongest tool of an engineer the other point other thing that has to be found out is is RT I shall find out RT by both methods and then you decide which one is simpler let us look at the circuit again what we have to do is V S then we have R 1 R 3 beta I 1 R 2 let us find out RT by direct measurement across these terminals looking back after killing the independent sources dependent sources should not be touched so what I do is I short circuit this that is the only thing that I have to do is not it this is the only independent source and killing it means that I short circuit this I do not touch I do not touch beta I 1 recall that this is I 1 now you see that as a result of shorting this what is I 1 I 1 becomes equal to 0 and therefore beta I 1 becomes equal to 0 which means that the current generator is not there all right so what is RT then it is a parallel combination of R 2 and R 3 because R 3 goes to short circuit R 1 is ineffective so it is simply R 2 R 3 divided by R 2 plus R 3 okay now let us see let us calculate it by the other method that is by calculating the short circuit current our circuit if you recall is V S then R 1 this is I 1 R 3 beta I 1 R 2 and I am going to short circuit this this is the current that I want to find out I SC all right if if this is short circuited if this is short circuited then obviously ineffective R 2 is ineffective no current in R 2 all right and therefore I SC shall consist of 2 components if you apply KCL now here then this current this current minus beta I 1 should be equal to I SC agreed if I apply KCL here this current now what is this current this voltage is V S what is this voltage 0 it is connected to the ground is not this voltage equal to 0 it is short circuited and therefore the drop across R 3 is simply V S therefore the current in R 3 is V S by R 3 minus beta what is I 1 I 1 is still V S divided by R 1 this should be equal to I SC that is I SC is equal to V S G 3 with the same terminology conductance minus beta G 1 this is I SC let us collect our our equations is there any question here this calculation all we did was common sense one application of KCL and the other is common sense nothing else we found out I SC let us let us write our equations again I SC is V S G 3 minus beta G 1 and what was V T V T was V S G 3 minus beta G 1 divided by G 2 plus G 3 and by definition R T shall be the ratio of the 2 what does the ratio give 1 over G 2 plus G 3 which is precisely equal to R 2 R 3 divided by R 2 plus R 3 so I have been able to find out the equivalent thevenin resistance by either method now it is for you to judge which one which method was better in this case which method was yes direct calculation well both of them are equally in the first one it was much easier well then it is a personal preference I do not like green colour some somebody else likes green colour so it is okay to me both of them are equally easy once you see what is going on once you see what is the game is then you can play the game according to the rules of the game alright and hope to win alright so these 2 fairly involved examples should convince you about effectiveness of Thevenin's theorem and along with Thevenin now yes there could be a dependent voltage source in the same example you will have to work it out beta I 1 means beta beta V S divided by R 1 and you will have to work it out alright let us look at this circuit it is not a it is not a bad question it is a it is a very good question what he says is R 1 I 1 R 3 and he says let this be a dependent voltage source beta I 1 instead of a dependent current source then we have an R 2 what did you want to calculate Thevenin resistance before that you also want to calculate V T open circuit voltage the open circuit voltage here obviously shall be equal to beta I 1 alright now if you want to calculate Thevenin resistance R T what you do is you short circuit this short circuit this so R 1 is ineffective R 1 is ineffective and if R 1 is ineffective then what is I 1 I 1 is 0 if I 1 is 0 then beta I 1 is 0 if beta I 1 is 0 what does it mean it means an open circuit or short circuit a voltage source if the voltage is 0 what does it mean no it is a short circuit there can exist a voltage across an open circuit but not across a short circuit so this is a short circuit so R T becomes identically equal to what are you talking of who said 0 who dares to say 0 I appreciate that courage because it is indeed 0 why because between these 2 points there exists a short circuit current is no compulsion to go any other way it will go through the short circuit that is indeed the Thevenin resistance alright okay so the problem becomes much not simplified if it is a controlled voltage source in dependent voltage source instead of dependent current in this situation not necessarily in every situation no in other situations it may be more complicated now along with Thevenin comes the name of Norton Norton was an American working in bell telephone laboratories and he enunciated a theorem which goes by his name Norton's theorem which is precisely the dual of Thevenin's theorem dual that means what Thevenin enunciated in terms of voltage Norton enunciated in terms of the current Thevenin's theorem says that the 1 port N is equivalent to a voltage source in series with a resistance Norton says that N can be replaced by a current source in parallel with a resistance alright current source I N in parallel with a resistance which Norton prefer to call a conductance G N alright and the interpretations of I N and G N you can very easily see that I N is nothing but I SC that is if you short circuit these 2 terminals the current that flows to the short circuit is I SC and you can also see that R N is nothing but what if you measure if you measure the resistance looking from here what resistance would you see 1 upon G N and therefore R N is simply equal to R T is this clear to you the Thevenin resistance the Norton resistance is the same as the Thevenin resistance you recall why is it so you recall that in finding out Thevenin equivalent we found out VOC which is V T we find out found out I SC and we found out R T as V T divided by I SC what we are doing in Norton is simply simply representing it in terms of I SC and an equivalent resistance alright suppose suppose I take I SC which is V T divided by R T in parallel with R T and find the Thevenin equivalent of this with the current source in parallel to resistance what is the open circuit voltage simply V T if this is kept open circuit then this current flows through R T so V T by R T multiplied by R T is V T and if you measure the resistance looking in this direction after killing the source only source it is simply R T so Norton Norton's theorem is simply another way of looking at the Thevenin equivalent alright what was considered as a voltage source by Thevenin is now being converted to current source so both of these problems that we solved also solves for Norton equivalent because we found out R T R T is the same in both in one of them it is a series resistance in Norton equivalent is a parallel resistance the source is I SC which we have to find for Thevenin equivalent also for Thevenin equivalent we found out V T and I SC question may be that if I if I do not find the short circuit current I know the Thevenin equivalent on all that you have to do is to find is to divide the open circuit voltage V T by the Thevenin resistance R T whichever way you find it does not matter and therefore whatever we have done for the two examples in finding out the Thevenin equivalent also solves for the Norton equivalent and this leads us to a more generalization in the case of a generalization in the case of sources that is if we have a source V which is not ideal which has a series resistance let us say R then it is a non-ideal voltage source this is equivalent to a non-ideal current source of current equal to V by R the short circuit current and a parallel resistance equal to R this is called source transformation a big name but a very simple application of Thevenin's theorem similarly if I have a current source I which is non-ideal that is it has a parallel resistance R then this is equivalent to a Thevenin equivalent a voltage source I times R in series with R you will see that this simple transformations the two transformations which go by the big name of source transformation shall be extremely useful in solving network problems without loop analysis without node analysis. In fact Thevenin and Norton applied judiciously can save the labour of network analysis however complicated it may be and we shall see some of the some examples next time is there any question all right the next time.