 Now, then we were looking at some general representations of linear circuits, which can be used more widely ok. So, we were in the middle of sort of generating an equivalent circuit for a complicated circuit ok. So, in the meanwhile if you have any questions about what we did in the previous class or anything before that please go ahead. Any questions about things we were discussing in the previous class? Ok. So, let me get back to what we were doing. What we were looking to do was to take a circuit of this sort which consists of linear elements and linear controlled sources, which are controlled by quantities inside. This is this network N and then it is loaded by current source Ix, which could be a current source or it could be the equivalent of any other load or complicated network that is connected. We know from substitution theorem that if we had some other network here, we could replace it by the current source that it is drawing ok. So, now what we did was this is a linear network. So, superposition applies and we can analyze this in two steps ok, which is what I have shown here. Let me copy this stuff over. So, I will do a two step analysis of this. I have a number of independent sources inside and I also have this independent current source Ix which could be a current source or it could be a substitution for some other element. So, some other load ok. So, first I will deactivate the current source that is I will set Ix equal to 0 and analyze this circuit that is at these terminals 1 1 prime I will find the voltage Vx1 ok. Next I will set all the sources inside all the independent sources inside to 0. So, that is known as nulling the circuit then I will analyze the circuit with Ix alone ok. So, Ix is active in this case. So, clearly by superposition in this case when everything is acting that is when you have the load that is represented by Ix plus when you have all the independent sources inside the circuit you will get a certain value of Vx and this Vx which is the solution you are looking for is nothing but Vx1 plus Vx2 ok. What you calculate from this step by setting this Ix to 0 and then this step by setting all the independent sources in the circuit to 0 and that is nulling the circuit and finding the voltage ok. Is this part clear? I am splitting up the circuit like this. Any questions about splitting up the circuit in this manner that is first I will activate all the current sources inside deactivate the Ix then I will null the circuit and have only Ix active. So, let me just imagine that I have only one source inside ok one independent source outside which is Ix. So, clearly by superposition what is the meaning of superposition? If you have a number of independent sources in a circuit we can and then let us say you want to find a certain quantity let us say this Vx ok at these terminals 1 1 prime then what you can do is analyze the circuit with each independent source I have activated at a time ok. So, if you have a number of independent sources you have set all the independent sources to 0 except one of them you find the value of Vx from that one and next you take the next source and set that one to a nonzero value and everything else to 0 then find the Vx from that. Do this for all independent sources you will get a number of different values of Vx and you add all of them up that will be the value of Vx with all the sources active ok. So, now you do not have to do it one by one let us say you had the five independent sources you could take three of them once and then two of them the other time or four of them once and one of them the other time and so on ok. So, here what I am doing is I will first deactivate this Ix which corresponds to the load which can be substituting for the load and find this voltage and then I will null everything else that is null the circuit and find the value of Vx 2 from this ok. I hope that is clear. Now the question is that is it possible to find the current instead of the voltage yeah it is very much possible ok it is not a problem at all later you will see that ok. First of all let me call this Vx and then there is certain branch and then that is a y or something ok. Now if this circuit is linear then this Vx and Iy any of these quantities can be found from superposition ok that is any branch voltage or any branch current can be found from superposition ok. So, that applies to any quantity that you would like to find. Now let me look at the first part of this that is Vx 1. Now you see that Vx 1 is nothing but the voltage that appears across 1 1 prime when it is open circuited ok it is the open circuit voltage or the voltage that appears across 1 1 prime when it is externally open circuited ok that is no load is connected ok that is the idea ok. So, one part of this is the open circuit voltage. Now remember what we are trying to do is we have a very complicated circuit and only these terminals 1 1 prime are accessible to us and this might be connected to yet another circuit. Now what we would like to do is to have instead of analyzing this complicated circuit n every time we would like an equivalent representation of n that is simple ok. So, that is our goal that is what we are driving at because we are not interested in the internal details of n we have only this 1 1 prime and this will be driving the other circuit. So, we are only interested in what happens to this terminals 1 1 prime and in the other circuit we do not need to analyze all of n every time ok that is the idea. Now that is the first piece of the puzzle the open circuit voltage. Now the next part is with n being null that is all independent sources inside n are set to 0 ok and we have only a single independent source in the circuit. We have a single independent source ix in the circuit ok. So, now we will have this vx2. So, my question is please try to answer this. What is the general form of vx2? What is the general form of vx2? That is we are now talking about a circuit which has only linear components and all the independent sources are set to 0. So, we have a single independent source ix. What I would like you to answer is what will be the general form of vx2 ok in term meaning how would it be related to ix? Yeah I think many of you are able to answer it vx2 will be proportional to ix and if the polarity is given it will be of the form minus ix times some r ok where r is a property of the circuit r is basically the resistance looking into the terminals 1 1 prime ok. Because we have a single independent source here ix and then all the other sources are nulled and all the other components are linear. So, any quantity in this every branch voltage and current will be just proportional to ix and that will apply to vx2 also ok. And with the sign that I have taken ok because the ix is flowing outwards I will say vx2 is minus ix times r ok because the passive sign convention 1 1 prime if ix would be flowing that way then I would so if i were flowing that way I would take v in that polarity in this case i and v are in the opposite polarity to this convention. So, I will say it is minus ix r ok and what is this r? It is the resistance I will say looking into 1 1 prime with all independent sources set to 0 ok. So, that is the resistance of the network m looking into the terminals 1 1 prime ok. And this voltage vx1 is the open circuit voltage I could also call it boc ok or let me call it vth ok the reason for this will become clear. The open circuit voltage is called vth what I do is I simply measure the voltage across 1 1 prime with nothing connected to 1 1 prime ok. And this r I will call as rth again the region will become very clear ok. So, now so this quantity is called what I called vth ok the open circuit voltage and then here this quantity will be minus ix rth where rth is the resistance looking into 1 1 prime. So, the total voltage vx will be equal to vth minus ix rth ok where vth is the open circuit voltage and rth is the resistance looking into 1 1 prime with all the independent sources set to 0 ok. So, this is the voltage that will be present when this network is connected to a load current of ix ok. So, this is vx in the complete circuit now the network is active and the load current is connected to it this ix ok. Now, because vx is in this form it is very easy to see that this can be generated by having a voltage vth and a resistance rth in series with it. So, if I connect a resistance ix the current source ix here you are very easily see that all of this ix will flow through rth. So, the voltage drop across this is ix rth ok and the voltage between these two terminals will be vth minus ix rth ok. So, the voltage here will be exactly the same as that one ok. So, this entire thing here this is terminal 1 1 prime the whatever is inside this red box is equivalent to the original complicated circuit N between the terminals 1 1 prime ok. So, it is only that that is equivalent internal details of this we do not know the internally here we have only a single voltage source and a single resistance. But as far as the terminals 1 1 prime are concerned the voltage across this and the current through this will be exactly same as in the original network N ok. Now, this is regardless of what is connected to it because we did not make any assumptions about ix ok. So, ix is some current which could be a current source or which could be a substitution for some other load. But whatever it is this entire thing which is a set of independent sources and linear elements can be replaced by a voltage source vth in series with the resistance rth. And this particular representation probably many some of you know this already this is known as the Thevenin circuit or the Thevenin equivalent circuit of this part ok. So, now as I said earlier if you are not interested in the internal details of this. So, let us say you design a very complicated circuit with hundreds of resistors and control sources and many independent sources and you want somebody else to use it. But you have only two terminals that are brought out which is what they are going to use. So, in that case they do not need to know all the internal details of this all you have to do is to give the Thevenin equivalent. You give the value of vth and rth because you know the internal details you can calculate these things. But once this is done as far as the terminals 1 1 prime are concerned this very simple representation and whatever stuff you have inside this are completely equivalent ok. And this the idea that any circuit can be reduced to this is known as Thevenin's theorem ok. So, any questions about this what we did was we just use the linearity of the circuit n and proved that it can be represented by a voltage source in series with the resistance ok. Essentially we proved what is known as Thevenin's theorem ok. So, any questions about the proof or the idea of Thevenin's theorem? There is a question on when we cascade two networks is there any chance to change the voltage between the two networks? Actually this question is not clear to me which voltage or what is being referred to ok. So, I have to either draw a figure and email it to me then I will discuss that ok. So, I am not clear about what is meant by cascading of two networks and what is the voltage between the two networks. So, any other questions about this Thevenin's theorem and its proof? Actually the proof is quite simple how we use was linearity ok. And any questions that you have if you want to send something in detail please use the forum discussion forum for given for this course ok on the NPTEL website ok. So, the voltage Vth is open circuit voltage across the terminals and Rth is resistance between the two terminals. So, looking into the two terminals with independent voltage sources nulled ok. It appeared that because of some network problem the connection broke and also the audio and video went away for a while I hope it is solved now. Are you able to hear me if there is no audio then again let me know. So, as I will repeat what I had said I was just saying that Thevenin's theorem says that if you have a two terminal network with independent sources and linear elements inside at the two terminals it can be equally represented by a series combination of a voltage source and a resistance. The voltage source is formed by determining the voltage across the two terminals with nothing connected to them and the resistance is formed by nulling all the independent sources in the network and finding the resistance looking into the two terminals ok. So, as I said we will work out an example and the example circuit I took was this very simple circuit I have a 10 volt source 3 kilo ohm resistance series with 7 kilo ohms and these are the two terminals at which I want to find the Thevenin equivalent. This is very important I mean what is equivalent when you say it is the Thevenin equivalent of a circuit. The Thevenin equivalent applies only to some two terminals specified in this case it is just one and one prime ok. So, now my question is what will be the Thevenin voltage across one one prime in this circuit please try to answer this ok I think all of you are able to immediately recognize that it is this resistive divider 7 kilo ohm divided by 7 plus 3 kilo ohms times 10 volts so it is 7 volts that is the right answer ok. So, again there are some complaints that audio is not ok but it seems that many other people are able to hear so the problem may be with your setup you please try it again ok. So, now the next thing is what is the value of RDH what is the value of RDH looking into these terminals again what you should do is you should take the independent circuit independent voltage source in this circuit make it 0 and find the resistance looking back into the circuit ok. So, I have to make this 0 volts or a short circuit I have to look at the resistance here again the answer is very simple this is a short circuit and you will be able to easily see that the 3 kilo ohm and 7 kilo ohm are in parallel. So, the answer is 2.1 kilo ohm again I got some answers like 2.1 and so on please do not do this always specify the units for any quantity that has dimensions ok. Now, somebody said 7 k parallel 3 k that is correct so I was looking for the numerical value ok. So, what it means is that this entire circuit can be replaced by a 7 volt voltage source in series with a 2.1 kilo ohm resistance ok. Is this part clear make the circuit more complicated I will add a 10 milliamp current source as shown in red ok. So, now we have to find there are now of course according to Thevenin's theorem at 1 1 prime I should have marked the terminals here this is 1 and 1 prime at 1 1 prime I can represent this new circuit also with a single voltage source and a resistance ok. Now, my first question is what will be the value of the resistance in this new circuit what is the value of RTH again as many of you are able to recognize it is exactly the same because when you know the voltage and the current source you short circuit this and you open circuit the current source you are left with exactly what you had before. So, value of RTH does not change at all ok. Now, what will be the value of VTH the Thevenin voltage source what is the value going to be what will be the value of the VTH voltage source VTH with the new with the current source added ok what is the value of VTH. So, we already say there are many ways of doing it you can redo the circuit analysis with the voltage and the current source together you can also use superposition here to calculate VTH as well because after all it is a linear circuit with independent sources and superposition means you have to find the voltage VTH due to 10 volts and the voltage VTH due to the 10 milliamp current source the voltage due to 10 volts have already found previously ok. So, the only thing you have to do is to find the voltage due to the 10 milliamp current source and add it to the voltage due to the 10 volt voltage source ok. So, the question is about the VTH value with this new circuit which has the 10 volt voltage source 10 milliamp current source and the two resistors ok. Now, also given some choices. So, please do the calculation and let me know what the voltage value VTH is or indicate it on the pole or just give me the value ok. I think again many of you are able to do it correctly although one person said 0 volts I am not sure why that is the case. So, let me do it and show it although this also is quite simple to do this with superposition which means that first I will set the 10 milliamp source to 0 which is an open circuit and then I will set the 10 volt source to 0 which is a short circuit ok. Now, this one we have already done before. So, I am simply going to write down the answer. So, the answer here is 7 volts and this other one you should be able to see that this 3 kilo ohm and 7 kilo ohm are in parallel and they are in parallel across this current source ok. So, we have an equivalent picture like this. This is the 10 milliamp current source and this is 3 kilo ohm and 7 kilo ohm in parallel which is 2.1 kilo ohm ok. Now, by the way this is 1 and 1 prime. So, the voltage across this from 1 to 1 prime that is with 1 being positive is minus 10 milliamps times 2.1 kilo ohms because of this polarity of the 10 milliamp current source. So, 10 milliamp is being drawn like this. So, this voltage is minus 10 milliamp times 2.1 kilo ohm which is minus 21 volts ok. So, again if you have got some different answers please look at this and then figure out where you are going wrong. So, the total voltage is the sum of this due to 10 volt source and this due to the 10 milliamp source. So, clearly minus 21 plus 7 is minus 14 volts and this is the correct answer ok. So, with the current source added what we have is minus 21 volts and sorry minus 14 volts and 2.1 kilo ohm in series ok 1, 1 prime. So, what I try to demonstrate is that even if you make the circuit more complicated you can represent it equivalently ok. Now, this will be equivalent circuit regardless of what you connect to 1 and 1 prime ok. This hopefully you understood this point that I connected some I X here and that I X can be either a current source or a representation for any kind of load. So, whatever you connect between 1 and 1 prime that does not matter. This part what we are replacing is this part here that can be represented by this minus 14 volts and 2.1 kilo ohms in series. In fact, what you connect to this can be non-linear. The part that we are converting to thermally equivalent that has to have only linear elements and linear elements and independent voltage and current sources ok. Now, there is a question about the polarity of the current source I think you got the answer right anyway. So, what I am saying is that if you have a current like this and then you have a resistance ok. So, the voltage measured with this polarity will be minus I times R ok. So, if the current was flowing like that the current I is flowing like that the voltage in this polarity will be I R ok. Now, if the current is in the opposite direction which it is if I is like that it will be minus I R that is all I said ok. Is that clear? Ok. So, now just for we will take it just a small step further ok. Let me add another resistance like this which is 1.9 kilo ohms ok. I will take some value and again at 1 1 prime I have to find Vth and Rth ok. Now, please tell me how will Vth and Rth change for this new circuit with this 1.9 kilo ohm resistance what will be the value of Vth? Previously when we did not have this this was the value of Vth ok. So, what is the value of Vth when we have that 1.9 kilo ohms? I think many of you recognize that Vth will be the same because when this is open circuited that is how you find Vth there is 1.9 kilo ohm is floating and no current is flowing through it. So, whatever voltage appears here will be the same as or what appears there ok. So, this is a special case with the way I have connected this 1.9 kilo ohms, but when I open circuit this no current is flowing through that. So, whether you have this 1.9 kilo ohm or not you will get the same voltage ok. What will be the value of Rth? Again this is simple because we have the 3 kilo ohm and 7 kilo ohm in series and this 1.9 kilo ohm like that. So, we know that looking here between these terminals we had 2.1 kilo ohms and this 1.9 kilo ohm is simply in series with that. So, looking this way it is 4 kilo ohm that is 2.1 plus 1.9 ok. So, the equivalent circuit for this including that would be minus 14 volts and 4 kilo ohms ok. So, this is just to illustrate the point that you can make the circuits more complicated, but the representation can remain exactly the same and you can find the representation ok. Now, this circuit is still too simple even if you had 100 resistors and then control sources and so on you can still find the Thevenin equivalent circuit ok. So, any questions so far on any of these calculations or proof of Thevenin's theorem ok. So, that seems pretty clear. There is another theorem which I will not prove you can try to prove it yourself by following another methods and if you get stuck somewhere you can ask me later. So, instead of using a voltage source and a resistance in series ok, any network of the same type, any network with linear and independent sources can be represented by a current source I n and a resistance R n in parallel ok. So, instead of a voltage source Vth and a resistance Rth in series we can represent the same by a current source I n and a resistance R n in parallel ok. So, now it turns out that this I n is now what is known as the short circuit current and the R n is what is known as the Norton resistance ok. So, we have the short circuit current and the Norton resistance. Now, what is the short circuit current? The short circuit current is you take this terminals 1 1 prime short circuit and measure the current ok. So, you short circuit this one and measure the current flowing from 1 to 1 prime that is the short circuit current I n ok. And R n is the same as the resistance looking into the circuit with all the independent sources nulled which will be the same as R dh ok. So, the Norton equivalent the resistance in the Norton equivalent circuit and the Thornian equivalent circuit will be exactly the same because they are nothing but the resistance looking into the circuit N with all the independent sources set to 0. And this I m is the short circuit current that is when you are short circuit 1 1 prime you will have a current I m and this Vth is the open circuit voltage that is if you have an open circuit across 1 1 then the voltage that appears at 1 1 is Vth ok. Now, this I n is of course related to Vth in a very simple way and the very easy way to see that is first of all this is the Norton equivalent circuit of this and this is the Thornian equivalent circuit of the same circuit. So, clearly these two also have to be equivalent to each other and if you look at the open circuit voltage between 1 1 prime what we have is Vth itself because by definition there is the open circuit voltage and here also you see that no current flows through R dh. Now, in this case if you open circuit 1 1 prime all of this I n will flow into R n and the voltage that appears is I n times R n and these two have to be equal to each other ok. So, I n times R n equals Vth or I n times Rth equals Vth and R n and Rth are the same thing ok. So, I hope this is clear as well. So, instead of using a voltage source and resistance in series you can also use a current source and resistance in parallel as an equivalent to any two terminal circuit which has independent sources and linear elements ok. And this idea of using a current source and resistance in parallel is known as Norton's theorem. Just for completeness let me write out the statement R dh and R n of the same and I n is short circuit current between the two terminals when you short circuit it that is the current that flows there ok. So, again I hope this is clear. So, very quickly we will evaluate this what would be the Norton equivalent this has to be represented by a current source I n and the resistance R n in parallel what will be the value of R n in this case what is the value of R n. So, clearly it is the same as R dh we already calculated. So, this is 2.1 kilo ohms and what is the value of I n and also we are able to recognize that it is 10 by 3 milliamps or 3.33 milliamps you can either get it from the previously calculated thermion equivalent circuit that is we had 7 volts and 2.1 kilo ohms and we know that I n times R dh equals V dh from this you can calculate or even in this case looking at the circuit if you short circuit this what happens is that we have this 10 volts and 3 kilo ohms and a circuit like this no current flows through 7 kilo ohms because it has 0 volts across it and the current that flows this way is 10 volts divided by 3 kilo ohms or 3.33 milliamps okay. So, now you can also try it with all these other things added when I added the 10 milliamp etcetera okay you can figure out what happens okay. So, that is the story with Newton's theorem and you can prove it in exactly the same way that we proved the thermion theorem to prove the thermion theorem what we did was we use the current source here and then use superposition in some way. Now instead of a current source you can use a voltage source like any load can be substituted either by a voltage or a current source. So, let us say you substitute with a voltage source here and then you apply superposition and then you can try proving Newton's theorem okay there was a question which I think was regarding this circuit saying if there is yet another current source here 3 milliamp bond happens. So, this is quite simple there is 3 milliamp and 10 milliamp are in parallel it is like having just a 7 milliamp current pointing downwards okay. So, instead of 10 milliamps you have 7 milliamps now the contribution of the current source so this was minus 21 volts. So, instead of that it will be in proportion okay minus 21 volts times 7 by 10 that is the meaning of linearity right. So, you can calculate how much this is this will be minus 14.7 volts I believe. So, instead of adding minus 21 to 7 you have to add minus 14.7 to 7. So, the net voltage will be minus 7.7 volts okay this is fine any questions. So, now we have discussed the 7.7's and Newton's theorems they are very very useful and they are used very frequently for representing circuits of two terminals okay and this happens all the time you design some circuit it is very complicated but it has two terminals available to the outside world and what you have to do is to give an equivalent picture to the user and this will do exactly that you will get with the Thevenin or Norton equivalent circuit which is very simple you will get the equivalence of the circuit at those two terminals okay you will not get an internal picture of what is happening but at those two terminals the behavior of voltage and current will be equivalent to that of the original circuit okay. There is a question on where to use Norton's theorem I guess the question probably is no one to use Thevenin or Norton and that is entirely based on convenience okay sometimes we use Thevenin's theorem and sometimes we use Norton's theorem. In fact there is an interesting thing here that we can discuss briefly. So, now like I said any circuit so there is also another question what is the proper condition for Thevenin and Norton I guess what is meant is that what kind of circuits can be represented by Thevenin and Norton and that I wrote out here so this is exactly the condition any two terminal network with linear elements and independent voltage and current sources if you have a circuit that is like this then it can be represented with either Thevenin or Norton equivalent circuit okay. So, if you have nonlinear elements you cannot do this okay and also one thing which I mentioned last week as well that if you do have dependent sources in this that is okay but it should be dependent only on quantities inside N okay you cannot have them externally controlled. Now like I said any two terminal any circuit with only two terminals available can be equivalently represented at the two terminals using either Thevenin or Norton equivalent circuit okay. So, now let me say this is my circuit okay this is now an extremely simple circuit I have a voltage source okay. Now what is the Thevenin equivalent of this and what is the Norton equivalent of this voltage source please try to reason out and answer this question this is not so much of calculation but reasoning out I have only a voltage source connected to terminals 1 and 1 prime okay. So, what is the Thevenin equivalent of this and what is the Norton equivalent of this and answer for Thevenin equivalent is very clear it is the same as the circuit okay with RTH equal to 0 VTH will be 10 volts and RTH will be 0 okay but what is the Norton equivalent of this circuit yeah again I have got some answers saying either the current will be infinite or it does not exist all these are correct. First of all there is also a question why the RTH of this is 0 and that is quite simple so in this you have a single independent source and that you set to 0 okay. So, you make it 0 volt which is a short circuit and then you are looking at the 1 1 prime so what you see is a 0 ohm resistance okay. So, RTH equal to 0 for this that is pretty clear okay. Now, the Norton equivalent does not exist because IN is VTH by RTH and if RTH is exactly 0 if it is non-zero then you can find it if RTH is very small you can find this IN which will become very large but if RTH is 0 this will tend to infinity okay that is you have an infinite current and across it you will have 0 resistance because RN which is RTH is 0 okay. Now and this infinite current through the 0 resistance can give you some voltage in this case 10 volts but this is really a useless representation okay when you have infinities in the circuit you cannot calculate anything but I am just making the equivalence here. The way to imagine it is you imagine some resistance in series with this voltage source and take the limit as it goes to 0 okay. So, my point here is that while in general any circuit can be represented by either traveling or mountain it is possible that one of them does not exist okay. So, one or the other will surely exist but if you have a pure voltage source between terminals 1 and 1 prime the Norton equivalent does not exist because we cannot short-circuit and find the current or in other words if you short-circuit an infinite current will flow okay. Similarly, if you have a pure current source between the terminals 1 milliamp 1 and 1 prime this again the Thevenin equivalent does not exist and the Norton equivalent is the same as this it is just a current source okay. Now there are a couple of questions first of all if we reverse the case we can get any voltage that is exactly the point here it is indeterminate okay so when you have infinity here it is indeterminate. Now the other question is if you have a voltage source and a current source do you have a Norton's equivalent? If you have a voltage source and a current source then in parallel the result is just a voltage source okay. Just mentioned a few times before if you have a voltage source in parallel with anything okay whether it is a current source or whatever else it is the same as the voltage source itself okay but I have a current source or not it does not matter the only thing that it will change is the current through the voltage source otherwise nothing will change okay. So clearly if there is a voltage source and there is no legitimate Norton equivalent circuit for this one okay. So I hope that part is clear as well. So there can be degenerate cases where one of the other equivalent circuit does not exist but at least one of them will surely exist okay and again somebody asked about the application of this theorem as I said it is widely applicable. The basic point is to abstract out all the details of a very complicated circuit. You have a very complicated circuit with lots of components instead of using all of that in every circuit analysis if you are interested in the behavior only at two terminals you can use either the Thevenin or Norton equivalent circuit okay. Any other questions? So again there is some confusion regarding the load. This equivalence is independent of the load whatever you load you connect to it the behavior at one and one prime will remain the same as in the original circuit okay. So if you want to find the effect of the load on the original circuit if you instead connect the load across the Thevenin equivalent circuit and you will find exactly the same behavior okay. So now that we have discussed Thevenin and Norton theorem let us move on to another very powerful theorem which is also used widely. It is mainly used to get more theorems out of circuit behavior okay. Now let us say we have some circuit. I will show arbitrary branches I do not really care what the branches are okay. There can be any elements and I will number the branches let us say 1, 2, 3, 4, 5, 6 and I will label the voltage of every branch be 6 and then I will also label the currents consistent with the passive same convention I1, I2, I3, I4, I5 and I6 okay. So now basically I have a circuit but I mean this I have shown that 6 branches but any number of branches is okay right. So now let us say we compute the sum of power in all the elements that is sum of vkik over all branches. What do you think this is going to be? Please give me a guess. Actually many people are still asking questions. I will deal with them soon after I deal with this particular thing. That is in this particular case I sum the power in all the branches I1 plus v2I2 plus v3I3 plus v6I6. What is it going to be? It turns out that this number will be 0 okay and this is known as conservation of okay. Now this conservation of power what it means is that if you have a circuit and if some sources are dissipating power and some others will be generating power okay. Whatever power is dissipated has to be generated in some part of the circuit okay. So there is no other external source of power right. So we have only electrical variables here and electrical connections between elements. So whatever power is generated in some elements will be dissipated in other elements or whatever power is dissipated in certain elements has to be generated in some other elements of the same circuit okay. So that is what this statement means that power is conserved in a circuit okay. Now this also can be proved and this can be proved basically using only Kirchhoff's voltage law and Kirchhoff's current law okay. Now this is completely unrelated to what kind of elements we have. It could be linear, non-linear or whatever. But as long as our elements obey KVL and KCL which is basically a fundamental property of the circuit this conservation of power holds good okay. So in fact I would strongly encourage you to try and prove this using only KVL and KCL okay. And I will give you a hint. So let me not name them N12. I will label these nodes N0 which is the reference node. You can consider this to be the reference node. It does not matter which one it is. And NA node NA node ND and node NC okay. So you can try to prove this by making these products VKIK that is V1, I1, V2, I2 etc. And this V1, I1, V2 and so on are the branch voltages okay. Now you try to represent these branch voltages in terms of the node voltages that is VA, VB, VC with respect to the reference node okay. And then you group all the terms containing VA and VB and VC together and then you use Kirchhoff's current law you will be able to prove this okay. So please take it as an exercise and do that. Hopefully you will be able to do it otherwise we will discuss it later. So for this you have to represent branch voltages in terms of node voltages and group the coefficients of each node voltage. And then you will be able to pretty much immediately see the answer okay. So I will stop with this discussion here. I will continue with this in the next lecture. But there were a few questions regarding Ternian equivalent circuit. The first one is what do we do when we have dependent sources? Now dependent sources you find it in exactly the same way as we did with resistors okay. Now with resistors what we did was to use the resistor formula to calculate the resistance but in general if you have any network with two terminals let's say one and one prime. The way to find the resistance looking into one and one prime is to apply a test voltage V test and analyze the circuit in case of Ternian's theorem whatever is inside will be nulled. All the independent sources will be nulled and you find I test. Now because you have a single independent source V test the I test will be proportional to V test. So the Ternian equivalent circuit will be Ternian resistance sorry will be V test by I test okay. Now this is true whether you have dependent sources or not. Now when you have resistors only sometimes you may be able to use the formula for series of parallel combination of resistors. But in general even with resistors this is the method to use okay. So let me take a very simple example again this is a voltage controlled current source whose value is let me call this Vy and the value of this is Vy divided by 2 kilo ohms or basically 0.5 millisiemens times Vy okay. Let us say this is 10 volts. So how would you go about doing this? You have to find the open circuit voltage which of course you do by in the normal way one one prime. You do circuit analysis to find out the voltage across this okay. And you have to also find the Ternian resistance for that you set the 10 volt to 0 so that becomes a short circuit. We have 5 kilo ohms here and we have Vy and this all these things will remain exactly the same okay. You do not change anything in the dependent source. Then you apply this V test and find I test. The ratio of the two will give you RTH looking into this. Now this is a very simple circuit but whatever arbitrary connection of dependent sources you have you can still follow exactly the same procedure okay. I hope this is clear. So procedure is exactly the same. You have to find the resistance between one and one prime. If you are given a black box let us say in the lab and ask to find out the resistance between one and one prime what will you do? You will apply some voltage across one and one prime. Find the current that flows through it take the ratio or you will post a current from one prime to one. Or see the voltage that is developed across it and take the ratio okay. So that is how you do it for any circuit. Now it just happens that if you have only resistors you may be able to take some shortcuts okay. Now the other question was somebody said how do we give the direction? I am not sure what is meant by this. Maybe what is meant is the direction of the current source okay. So when you find the not an equivalent circuit you will short circuit the output from one to one prime. And you measure the current flowing from one to one prime okay. So that gives you the direction right. So if the current is actually flowing from one to one prime then inside you will have a current source like this okay. If in the short circuit current is flowing that way then inside ion will be like that. If it is in opposite direction then ion will also be in the opposite direction okay. So there is no confusion here. Now other question is can source transformation be applied if the network contains non-linear elements? Now like I said before the part that you are transforming has to contain linear elements. So what I mean by this is the following. So I have taken the example of this network N with independent sources and linear elements. And linear elements includes the resistances and controlled sources. The only condition that the controlled sources have to be controlled by something inside N. Some branch voltage or branch current okay. And you have one one prime. Now the part that you are transforming this has to be linear. Now if you have non-linear elements elsewhere in the circuit something like this then that is okay. This is not what you are transforming. So whatever I have shown inside the red box that is linear and that can be represented by this Thevenin or Morton Equivalent Circuit. And the source transformation means that you either represent I mean you go from voltage to current source representation which means that that is also applicable only to linear circuits okay. Any other questions? Okay so if there are no more questions please try to do some of the things that I indicated that is trying to prove the Morton's representation that it is valid for any circuit. That is one thing you can do and you can follow the lines of what we did for Thevenin representation. And you can also try to do the other one that is try to prove conservation of power for any circuit using KCLN KVL okay. Okay there is another question again I am not clear of where this question is. It says across a single element if there are two voltage sources of same magnitude superposition fails and finding the current in the resistor. The question is are there voltage sources in series? Is that what is meant here? Now see if you have two voltage sources of same value in parallel okay. Now this is the same as a single voltage source. So that is one thing. Now if you insist on using superposition you will get a digital rate answer. Because if this is V and this is V and you have to set this to 0 then you have to short circuit it okay. So you will have infinite current here and some voltage across the resistor and so on okay. So this is basically a digital rate condition and you cannot represent things like this okay.