 Hello and welcome to lecture 11. In this lecture, we will look at few more involved theorems. Before I go into the theorem, I will introduce the notion of a graph of a circuit. So, if you have some circuit, each of these could be any elements, we do not make any distinctions between them at this level. A graph of this is basically a picture that contains a node for every node of the circuit and the elements are simply represented as branches. This is basically an abstract representation of a circuit and is useful for discussing some very general properties of a circuit. So, this is the graph of this circuit. Now, let us say we take two different circuits. This is circuit 1 and I have another circuit. It could have any element. You see that both circuits have four nodes and both circuits have elements between corresponding nodes, the same corresponding nodes. In the first circuit, I have an element between 1 and 2, in the second one also between 1 and 2 and so on. It is also clear that both of these circuits have the same graph. So, a graph representation is not useful for discussing particular properties of some circuit, but for general properties of all circuits sharing certain graphs. So, with this short introduction, let us move on. So, first let me take a circuit. In fact, I will take the same kind of circuit I had before. I have all these branches and as I mentioned earlier, this circuit has this particular graph and let me number the branches like this 1, 2, 3, 4, 5, 6 and for each of them, I will show the voltage across the branch V1, V2, V3, V4, V5 and V6. Now, also show the currents with the passive sign convention. That is, I will take the current as entering that terminal of the element whose voltage is defined to be positive. So, for instance V6 is defined like this. So, I will take I6 in that direction, I1 this way, I5, I2, I4 and I3. And first, what I want to look at is basically some of the products of voltage and current over all branches. I introduced this expression in the previous class, but now we will examine it and see what it comes out to and evaluate it exactly. So, first I will do it for this graph, but I will quickly illustrate that it is not true for this particular graph. This will be true for every circuit. So, now, what I will first do is, I will represent every voltage as difference between node voltages with respect to some reference node. So, let me take this as the reference node. If I call these nodes A, B and C. So, clearly I can write, first I will expand these things. It is V1, I1 plus V2, I2 plus V3, I3 plus V4, I4 plus V5, I5 plus V6, I6. And this whole thing equals, you can clearly see from Kirchhoff's voltage law that V1 equals VA minus the reference node voltage which is by definition 0. So, V1 equals VA, V2 equals VA minus VP, V3 equals VP and so on. So, if I complete this for every one of them, I will have VA times I1 plus VA minus VB times I2 plus VB times I3 plus VB minus VC times I4 plus VC times I5 plus VA minus VC times I6. And finally, first I had this form where I summed V and I over all branches. You know that Vk, Ik is the power dissipated in the kth branch at that particular instant of time. Now, we do not worry about whether Vk and Ik are time varying or constant or what kind of elements we have. But at every instant of time, the circuit will obey Kirchhoff's current law and Kirchhoff's voltage law. So, we can take this product, some of these products and decompose them like this and turn it into a form like this. Then, instead of having a summation over all branches, I will now group the node voltages together. So, VA, I have it here, here and here. And how many terms will I have? Three, because there are three branches connected to node A. And what is multiplying VA? It is I1 plus I2 plus I6. And similarly, VB, there are three branches connected to node B. So, I expect three terms and indeed there are three of them, here, here and here. And I will have minus I2 plus I3 plus I4. And finally, VC, again three branches connected to node C. So, I will have one, two and three occurrences of VC. I will have minus I4 plus I5 minus I6. And if I look at each of these terms, what is I1 plus I2 plus I6? It is simply the total current leaving node A. And obviously, by Kirchhoff's current law, this whole thing equals zero. Similarly, minus I2 plus I3 plus I4 is nothing but the total current leaving node B. I2 is entering node B. So, we have minus I2 and I3 and I4 are leaving node B. So, we have plus I3 plus I4. So, this sum also equals zero. And similarly, minus I4 plus I5 plus I6, sorry, minus I6, minus I4 plus I5 minus I6 will be zero because that is the total current leaving node C. So, obviously, the entire product will be equal to zero. So, though I did it for this particular circuit, I can show that it is true for a general circuit, what did I do? I took summation of VK times IK, where K basically runs over all the branches of the circuit. Then I represent each voltage as the difference between two node voltages. This is always possible. It could be that the second node voltage could be zero. That is, if some branch is connected to the reference node, then one of these voltages will simply be the reference node. But obviously, any branch voltage equals the difference between the voltage at one terminal and voltage at the other terminal, where VP and VN are measured with respect to the reference node. Now, if either of these two nodes itself happens to be the reference node, then VP or VN will be equal to zero. But this description holds in any case. So, I can sum over all branches with this modified form. Then what I will do is, I will convert this into a summation over all nodes. So, let me call this J, where J runs over all nodes VJ and that will be multiplying some terms containing currents. So, now the way we define the voltages, if you have a branch K whose voltage is VK, then the current IK goes this way. It goes from the plus terminal where as VK is defined to the minus terminal. We use the passive sign convention. So, whichever node is the positive terminal, that is this VP, it will get multiplied by plus IK. And VN will get multiplied by minus IK. This IK is simply flowing from this node P to this node N. Let me name these nodes P and N. So, if a current is leaving a node, then it will have a positive sign that is we have VP times IK with a positive sign. And if the current is entering a node, then we will have a negative sign. I will have minus VN times IK. So, I will have sum of currents multiplying this and basically this will be nothing but sum of currents leaving the node when I collect all the terms. Because I will have currents leaving this node with a positive sign and there could be some other branch and currents entering this node that will have a negative sign. So, basically if I change this to summing over all the nodes, then the node voltage VJ will be multiplied by something and that something is sum of currents leaving the node, which by Kirchhoff's current law equals 0. So, every node voltage gets multiplied by 0, which means that summation of VK IK over all branches equals 0. And this is nothing but a statement of a conservation of power. And with this we will prove even more interesting results about circuits and finally, also come to reciprocity theorem, which is quite useful when we have circuits with an input and output port that is two port networks. Now, let us say I have this network N and network N hat, which have exactly the same graph. Of course, the elements may not have anything to do with each other and of course, as usual we have not yet used linearity or anything. So, it could be non-linear also. I could for instance in the first in the network N, the branch one could be a voltage source in the other one it could be a resistor or a diode or whatever it is, all that is possible. And I will denote voltages and currents in this by VK and IK and in this by VK hat and IK hat. Clearly, from what we just now said some of VK IK over all branches I will just say over K that means K running through all the branches is 0. And here what would it be? VK hat IK hat would be 0, this is fine. Now, let me show it a little bigger with some boxes for elements this same graph and this is N and N hat. And let us say in the network N hat I have currents I1, let me show that in a different color I1, I2, I3, I4, I5, I6. These are the currents that are flowing in the network and that what exactly there depends on what elements there are in the circuit. But let us not worry about that. What I will do is I will take a current source whose value is the first branch current whose value is I1 and connect it across this branch. So, actually let me not abuse notation these are hatted quantities I1 hat I2 hat and so on up to I6 hat. So, what I will do is I will take I1 hat that is I will look at the network N hat, I will look at whatever current is flowing in branch 1 and make a current source equal to that value and connect that current source across branch 1 of network N. And similarly I do it for all branches. Now, let us say originally the branch voltages and currents in this circuit N were vk and ik that you got by solving for this circuit. Now, after I connect these current sources what will be the solutions? No, I am talking about currents in these black elements voltages. What will be the branch voltages in this circuit in the new circuit? Superposition of those two. So, you are saying the branch voltages will get added up and currents will get added up why? What is superposition? Superposition I mean please understand what is superposition? Superposition is you have a network and you disable some sources in it and you disable some other sources you add up the solutions individually that will be the final solution with all the sources active. But, here that is not what I have done I have changed the network. I originally had the only the black elements here to that I have added these blue current sources. Now, these blue current sources are not arbitrary I took another circuit with the same graph I copied the current values from that into this. So, tell me what will be the branch voltages in this vk vk hat why remains the same why I mean have a current something in parallel things would not remain the same. Now, how would you go about solving for this? First of all let us say I did not have this current sources I had the black network how would I go and solve for this one? What method would I use? Nodal analysis. Nodal analysis now after I add the current sources what happens to the nodal analysis equations? Source vector will change. How will it change? By how much will it change? What is that? No change. No change why? See I have not taken arbitrary current share it is true to each node I have added some current, but what are those currents? The sum of those currents added are also 0 because it comes from another circuit which is valid which follows KCL. For instance, if you think of the original nodal equations at let us say node N1 I will show it only for this you would say these black current I1 plus I2 plus I6 would be 0. Now, what else do you have? Would be equal to the currents flowing into this will yet change to minus I1 hat plus I2 hat plus I6 hat which is also 0. So, if you let us say you think of the methodology of solving this circuit by writing down nodal analysis equations you see that at every node the equation does not change at all because the net current added to every node is 0 because the I have not added a single current I have added all the currents by copying it from some other circuit with the same graph. So, at every node if you compute the sum of the blue current sources in the appropriate direction it will be 0 that means that the nodal analysis equations will remain exactly the same as before. So, what is the what are the branch voltages now? Same as before vk is that fine. Now, I can think of this entire thing as a branch that is possible I mean my two terminal element could consist of my original branch plus this current source in parallel. What will be the first of all and what will be the branch currents through the black branches alone? Same as before Ik. So, now with this new branch which is the parallel combination of the old branch and the current source in parallel with it what is the total current? Some of the two Ik plus Ik hat please follow the reasoning carefully this looks very simple but just follow the steps logically these branches are have currents Ik plus Ik hat. Now, this is a third circuit with the same graph now each branch will be the original branch in the black circuit plus the current source which is copied from the second circuit the hat circuit. But this is also a graph with the I mean this is also a circuit with the same graph and if you apply this to the new circuit what will you get? What did we say for every circuit sum of vk sum of branch products of branch voltages and branch currents equals 0. Now, I can do this to the new combined circuit isn't it? I have to do it for the branch voltage here which is v 1 and the branch current here which is I 1 may be I will call it some I 1 dash or something and v 1 dash. So, what is the result? Sum of v 1 dash I 1 dash will also be 0 and what is sorry sum of vk dash I k dash not v 1 dash I 1 dash and what is vk dash? vk the original circuit original solution to the network n. And what is I k dash? I k plus I k hat equals 0. So, vk I k plus vk I k hat equals 0 vk I k plus vk I k hat equals 0. What next? First term we already know is 0. So, vk I k this was 0. So, we have sum of vk I k hat equals 0. Now, this is I mean please understand the significance of this result and we have not used anything other than to say that the circuit obeys KCL and KBL. You have two circuits completely unrelated except that they have the same graph. So, you take the voltages branch voltages from circuit 1 and corresponding branch currents from circuit 2 form this current times voltage products and sum them that will also be equal to 0. And this is just the consequence of the circuit having obeyed KCL and KBL. This vk I times I k hat that number does not have any meaning for the same I mean in this if you take vk and I k from the same circuit that will be the power dissipated in that branch. Otherwise it has no meaning at all because in fact the elements also could be completely different this here the kth branch could be a voltage source there it is a current source and so on. But this theorem is true is this fine. What we did was we took a we took two circuits which had the same graph then we formed a third circuit which was circuit number 1 plus current sources copied from circuit number 2 and we did that for every branch that is very important otherwise it would not be true. And now you get a third circuit whose branch voltages remain the same branch currents are the sum of currents in the first two circuits. And for that also this sum of branch voltage I mean sum of this branch voltage times branch current products will be 0. From that we get this very interesting result that you take two circuits with the same graph you take the voltage from this current from that from products sum them over all branches that will also be equal to 0. Any questions what do you think of the product sum of V k hat I k? 0. 0 how do you go about proving that? You could take it you could take the currents from the first circuit put it into the added circuit so that is also 0. Now again this is consequence of only kcl and kvl so it can include the elements that we have not considered so far anything that obeys kcl and kvl. It will work it will also work for capacitors and inductors and so on. Now also this network N and N prime need not be two separate circuits it can be the same circuit at time 1 and time 2. So, it is also true that in the same circuit V k V k at T 1 and I k at sum T 2 this will also be 0 you can just think of it as you have two different cases where kcl and kvl are obeyed and of course it is also true that if you have the voltages from the first network at a time T 1 and currents from the second network at a time T 2 this will also be equal to no no V k I am saying I can take the voltages from the first circuit at time T 1 and current from the second circuit at different time that is also true I mean that is yet another case that has the same graph and obeys kcl and kvl that is all. And this result is also true for any I mean we have we are of course used to circuits, but anything where these loss are obeyed that is the sum of flow from a node is 0 the total sum of the total flow from a node will be equal to 0 and also if you go around the loop the sum of branch variables the across variables will be 0. Now, for us the across variable is the voltage and the through variable is the current, but it works for many other things also like fluid flow and things like that. And this extremely interesting result is known as Telegans theorem named after the person who first proved it. Again the way we have gone about proving that is very simple, but you have to be convergent enough with kcl and kvl to follow the logic correctly. Now, this theorem itself is rather too general in fact it is I would say in some ways too general to be useful directly because now we are talking about general properties of networks whereas, in this course our concern are some specific circuits. So, what we will do is we will derive some specific results from this Telegans theorem any questions so far about the proof or statement of Telegans theorem. Now, let me take a network N and this has and this is the this has only resistors this is more I mean less general than before we do not have control sources and we have only resistors in this. And let us say it has 2 pairs of terminals to which we can I mean with only resistors if you have only that of course, all the solutions will be 0 all the currents will be 0 and voltages will be 0. We have to connect sources somewhere and let us say we have 2 pairs of terminals where we can do that. And let us say I have connected V 1 here actually let me say I have connected V A here and V B the sources could be anything it could be 2 current sources 2 voltage sources or 1 voltage and 1 current source. I am just showing the case of 2 voltage sources and you have this currents I and I B. Now, let me first make V A equal to 0 that means I will short circuit this and in the other case I will make V B equal to 0 I will short circuit that one. Now, let me in the first case think of this one as the input and this I A as the result for the output and similarly in this case I think of V A as the input there is only 1 source and I B as the output. And this is a very common view you have some network you connect something here you look at the output there. Now, what do you think is the relationship let us say I connect 1 volt here and I will get some I A if I get if I apply 1 volt here will I be able to tell what I B is. So, please write down the Delegant Delegant standard equations for this and then see what comes out. So, you can think of these 2 circuits and to avoid confusion let me put hats on this it is the same network N, but it is not the same I A and I B that is flowing there we do not know that. So, I have to put hats on these things. So, now I apply a Delegant's theorem to this whole circuit and then see what comes out of that. If I take the voltages from this and the currents from the second circuit and form the products and add them up that will be 0. Similarly, voltages from the second one and the current from the first one that will also be 0. Now, for a consistent polarity I have to multiply V B with minus I B hat because I have to take the direction going in from up to I mean from top to bottom. So, first I will take if I call this network N and N hat the voltages from N and the currents from N hat. So, what is that V B times minus I B hat plus the sum of all voltages inside this resistive network times currents in that hearted currents in the corresponding hearted currents in the resistive network plus the voltage across this, but that is 0. So, that I will ignore this is over k branches inside the resistive network. Next, I will take the voltages from N hat and the currents from N what will I get what is the first term V A hat times minus I A because of that is because of the way we have defined the currents plus summation of V k hat I k over all branches equal 0. So, far we have not used anything about it being a resistive network. So, where will we use it? So, V k would be V k is the branch current times the branch resistance. Similarly, V k hat is I k hat times the branch resistance. So, what will we get out of it? And I will use the resistive network part here and similarly, in the other case V A hat minus I A plus k V k hat I k was 0. And if I use the fact that it is a resistive network V A hat minus I A plus k R k I k hat I k equals 0. So, what do you get from these results? So, clearly this part is common to the two equations. So, this has to be equal to that one I can of course, now remove the minus signs. Now, what is this? V A hat this is the pattern is this fine. So, what is that saying about these cases? So, what is it saying about these two cases? I have a resistive network that is all I know about it. It could have thousands of resistors, but there are two terminals exposed on this side and on that side. I apply one volt here short circuit the other side that is I could apply voltage sources to both sides and then I set one of the voltage sources to 0. Then I measure the short circuit current it will be something. If I apply one volt to the other side measure the short circuit current here what will it be? Exactly the same not merely proportional it will be the same it will of course, be proportional because it is resistive network, but that is not what this is saying right. What is the final result we got I A by V B is I B hat by V A hat is it not? It is a very interesting result if you have resistors alone this is known as reciprocity theorem the network is reciprocal that is if you apply some stimulus here measure the effect there and if you apply stimulus on the other side measure the effect here there will be exactly the same that is the ratio of the response to stimulus will be the same. So, please understand this theorem carefully. We already proved reciprocity theorem when we had voltages exciting from one side or the other side. Now, there could be other cases that is you could have current sources from either side or voltage source from one side and current source from another side. I will not prove reciprocity theorem for all these cases, but you can follow the exact same steps as I did earlier and prove all of these as well. So, just for completeness let me take a resistive network with two volts 1 1 prime and 2 2 prime I have excited this side with V 1 and short circuit at this side I will measure the current I 2 and in another case I have the same network N of course that is very important I excite this side with V 2 hat and short circuit this side measure I 1 hat here and reciprocity theorem says that I 2 by V 1 is exactly equal to I 1 hat divided by V 2 hat. Now, let us look at the other cases again a resistive network N I excite it with the current source I 1 and measure the voltage V 2 here in this case the excitation or the cause is the current and the response or the effect is the voltage I inject I 2 hat from this side and V 1 hat I obtain on the other side and again by reciprocity it turns out that V 2 by I 1 exactly equals I 2 hat by sorry V 1 hat by I 2 hat that is the ratio of response to excitation when port 1 is excited is exactly the same as the ratio of response to excitation when port 2 is excited. Now finally we have the third case where again I have a resistive network N and I excite the first side port 1 with a voltage and measure the voltage that comes out on this side and in the other case I take the exact same network N I excite the second side with a current I 2 hat and measure the current that flows in the first side. Please mind the directions of voltages and currents otherwise you will end up with extra minus signs in different places. So, reciprocity in this case says that V 2 by V 1 which is the response by excitation in the upper circuit exactly equals minus I 1 hat by I 2 hat which is the response by excitation in the lower circuit. You get this minus sign because of the direction of the current chosen. If we had chosen it outwards we would have got a plus sign, but for consistency we have taken all currents as flowing inwards into the resistive port. Now, we have a last case that is left which is trivial that is exciting the left side by a current and right side by a voltage that is simply flipping this picture sideways. So, I am not going to discuss that. Now, I will show all of these things with an extremely simple circuit as an example. My resistive network is just this as three resistors R 1, R 2 and R 3 and let me connect first a voltage source to this side and I will short circuit the second side and in the second case I will connect a voltage source V 2 hat to the right side and short circuit the left side. I will measure the current I 2 in this case and I 1 hat in that case. So, now it is very clear that because of the short circuit this voltage simply appears across R 2. So, the voltage source is connected across R 2. So, the current in R 2 is simply V 1 divided by R 2 also because of the short circuit no current flows in R 3 and is I 2 simply equals minus V 1 by R 2 that is because of the directions chosen V 1 by R 2 is flowing from left to right whereas, I 2 is taken as from right to left. So, I 2 equals minus V 1 by R 2 or I 2 by V 1 which is response by excitation equals minus 1 by R 2. If you now look at the lower circuit again because of the short circuit this voltage V 2 appears across R 2 and therefore the current in R 2 is V 2 hat by R 2 and again because of the short circuit no current flows through R 1 and because of the way we have chosen the direction of I 1 hat I 1 hat will be equal to minus V 2 hat by R 2 which means that I 1 hat by V 2 hat is minus 1 by R 2 minus 1 by R 2. So, this illustrates reciprocity we have already proved it for the general case but it is always a good idea to work out some real examples and convince yourselves that this is indeed the case. Let me excite the circuit with currents instead of voltages. I will excite it with I 1 on the left side and measure the voltage V 2 that is produced on the right side and the other case I will take the exact same network, excite it with a current on the right side and I 2 hat and see the voltage that is produced on the left side. So, now what is the voltage developed here? First of all is I 1 divides into two parts one part goes through R 1 the other one goes through the series combination of R 2 and R 3 and by results on current division you know that the current that flows here is I 1 times R 1 divided by R 1 plus R 2 plus R 3 and V 2 is nothing but this current times the resistance R 3. So, this is equal to V 2 the response by excitation V 2 by 1 will be simply R 1 R 3 by R 1 plus R 2 plus R 3. Now, we can evaluate V 1 hat in exactly the same way this current I 2 hat gets divided into two parts one through R 3 another one through R 1 plus R 2. So, part of the current that flows through R 2 is given by I 2 hat times R 3 by R 3 plus R 1 plus R 2 this is by current division theorem and the voltage that is developed across R 1 is nothing but this current which we have evaluated here times R 1. So, this times R 1 is the voltage V 1 hat. So, the response by excitation V 1 hat by I 2 hat equals R 1 R 3 by R 1 plus R 2 plus R 3 which is exactly the same as in this case. So, we have exactly the same response by excitation in the two cases as we expected and finally, let me illustrate it for a case where you have a voltage on one side and a current on the other side. I am also illustrating this the purpose of illustrating is this is also that when you encounter such a situation you apply reciprocity appropriately. So, let me take V 1 on this side and measure V 2 on the right side and in the other case I will apply a current I 2 hat on this side and measure the current I 1 hat that flows over there. Now, in the upper circuit this V 1 appears across the series combination of R 2 and R 3. We also have R 1, but that is directly across the voltage source V 1. So, it has no effect. So, V 2 is simply V 1 times the resistance divider ratio which is R 3 by R 2 plus R 3 or V 2 by V 1 is simply R 3 by R 2 plus R 3. And in the other case what happens we have a short circuit on this side. So, that means that no current flows through R 1 across that we have a short circuit. So, we just have two resistors R 3 and R 2 in parallel with the current source. I hope all of you are able to see that otherwise you can see that the terminals of the current sources are here and terminals of R 2 are exactly the same and terminals of R 3 are also exactly the same. That is they are connected to the same nodes. Now, the current divides between R 3 and R 2. The current flowing through R 2 in this direction is given by I 2 hat times R 3 by R 3 plus R 2. This is from the current division theorem. Now, all of it flows into this short circuit, but this current I 1 hat is taken in the opposite direction. So, the actual current I 1 hat is minus I 2 hat times R 3 by R 3 plus R 2. Again, if we calculate the response by excitation I 2 hat sorry I 1 hat by I 2 hat equals minus R 3 by R 2 plus R 3. And you can see that these two ratios of response by excitation are related by a factor minus 1 exactly as we expected. It turns out that this reciprocity theorem is very useful in many practical situations. And one such situation is where you have a number of sources in a circuit and you want to calculate the response at a given point, whether it is current or a voltage between two points. Now, instead of calculating the response from multiple sources, what you can do is use reciprocity, interchange the locations of the source and the response and calculate all these responses to a single source which is definitely easier. And by using reciprocity theorem, you can get all of the original responses that you wanted. It turns out this is very useful and it is used in circuit simulators also particularly for noise analysis where you have noises from many parts of the circuit affecting your output. You have already seen the proof of reciprocity in resistive networks using Telegance theorem. Here, what I will show is an alternative proof using basic circuit analysis. So, what I will do is I will consider a three terminal two port. This is port one, port two and this is the common terminal and the important thing is that this is resistive. Now, let us say just for the sake of it, I will drive it with I1 on this side and I2 on that side. The reason to do this is pretty obvious. I would like to use nodal analysis for the analysis of this. Now, we know that nodal analysis is most convenient when you have only resistors and independent current sources. Now, what I will do is this network itself is quite arbitrary. There can be any number of nodes inside. So, I will label this as node N1 and this as node N2 and inside there could be any more number of nodes and I will call them N3 to N capital N. So, there are a total of capital N nodes in the circuit. Now, if you set up the nodal analysis equations for this N1, N2, a number of rows up to N capital N. And remember, this is a purely resistive network and the only independent sources in the circuit are I1 and I2. So, now, I will have some G matrix. This is the conductance matrix times the variable vector V, which is V1, V2, where this is V1 with respect to the common terminal and the voltage here is V2. And after that, you have V3 all the way to V capital N. And G times V, the unknown vector equals the source vector and the source vector consists only of two non-zero elements, I1, which is being injected into node N1 and I2, which is being injected into node N2 and the rest of it is 0. Now, of course, because the network is purely resistive, this G matrix is symmetric. So, let us say I have G11, G12 all the way to G1N here. Now, the first element of the second row, first column and second row will be also G12, G22, etc., G2N. And similarly, here we have G13, G23 and so on. So, this will be symmetric because we do not have any controlled sources in the circuit. The network is purely resistive. Now, for reciprocity, basically I need to find the relationship between let us say V1 and I2 and V2 and I1. So, what I will do is I will try to eliminate all the remaining variables. So, for that, let me subdivide this matrix into four pieces. So, there are four sub matrices here as you can clearly see. You have one and another one and another one here and yet another one there. So, let me write those things as GA and GB, GC and GD. Now, clearly GA is just a 2 by 2 matrix, GB has 2 rows and N minus 2 columns. It is 2 times N minus 2 and GC is the complement of that which is N minus 2 rows and 2 columns and GD has N minus 2 rows and N minus 2 columns. And this times the variable vector V equals the source vector I. Now, because the G matrix is symmetric, we can tell a few things about this. First of all, GA will be a symmetric matrix. GD will also be a symmetric matrix. And also, if you look at GB and GC, GC is nothing but GB transpose. Now, what I will do is this variable vector V, let me write it as V1, V2 and then the vector of the rest of them which I will simply call Vx. This itself is a vector Vx has N minus 2 elements in it and that will be equal to I1, I2 and all 0s here. Now, I will write this as two separate equations. The first part corresponding to these two rows is GA times V1, V2 plus GB, let me denote it like this to make it clear that these are matrices GB times Vx equals I1, I2 and for this part I have GC times V1, V2 plus GD times Vx equals vector of all 0s. So, what I will do is basically eliminate Vx from these two equations, that is all. It is just like eliminating a variable from two linear equations except that we have now vectors and matrices instead of scalar coefficients and scalar variables. But that does not pose any particular complication. So, if you call this 1 and 2, if I solve for Vx in the second one, it is clear that Vx will be GD inverse times minus GC times V1, V2 and I will use the N1. By the way, this means that GD has to be invertible and we will not go into the complication. We will assume that GD is invertible. Then what we have is, if we use this N1, it is pretty straightforward. We will have GA minus GB times GD inverse times GC, this whole thing. Remember this is a 2 by 2 matrix, V1, V2 equals I1, I2. Now from this, it is pretty clear that this 2 by 2 matrix times V1, V2 equals I1, I2. So, if you think of this as a two port, this entire thing is the Y matrix of the two port. Or the inverse of this would be the Z matrix of the two port. Because what is the Y matrix after all? Y times V1, V2 equals I1, I2. It is the matrix that relates the port voltages to port currents. So, now let us examine the Y matrix of this. The Y matrix of our two port is GA minus GB, GD inverse GC. So, first of all we know that GC is GB transpose. So, this is equal to GA minus GB, GD inverse and GB transpose. So, first of all GA itself is symmetric that we knew earlier, we discussed that. And if you look at this part, we have GB, GD inverse and GB transpose. And to check for symmetry, obviously you just take the transpose of this. What will you get? If you have three matrices ABC transpose, you know that basically this is C transpose times B transpose times A transpose. So, if you transpose this what you will get is GB transpose transpose that is corresponding to this one times GD inverse and transpose and finally GB transpose. This comes from here. And clearly this is equal to GB, GD inverse transpose. Remember, GD itself was symmetric, therefore GD inverse is symmetric. So, GD inverse transpose is the same as GD inverse and finally GB transpose. So, the transpose of the matrix is the same as the matrix itself. So, this is also symmetric. This was symmetric. We knew this earlier. And from these two clearly if you subtract one symmetric matrix from another, you will get another symmetric matrix. So, what this shows is that this Y matrix is symmetric, which means Y to 1 is Y12. Therefore, our resistive two port is reciprocal. Reciprocity means that Y12 equals Y21. So, instead of starting from Telegram's theorem and getting these results, we could also do it by defining a two port and we have some two port parameters. We try to calculate the two port parameters from basic circuit analysis, in this case nodal analysis. And you can find that the Y matrix happens to be symmetrical. Of course, if you find any one parameter set to be symmetrical, the rest of them follow the same symmetry. Now, the Z matrix will also be symmetrical. And if you look at G or H parameters, G12 will be minus G21 or H12 will be minus H21 and so on. So, this is an alternative proof of reciprocity in resistive two port networks.