 So, let me use the same circuit that I was using last time which was this circuit with a number of resistors and a current source. I called this node N0, N1, N2 and N3 and the way I would go about doing nodal analysis is I write I make N0 my reference node, you could pick anyone in this case I have picked N0 and I write down the voltages of each node with respect to N0 as V1, V2, V3 and express all my relationships in terms of V1, V2 and V3. Now, so that means the voltage here is V1, voltage here is V2 and the voltage there is V3 and let me call this R11, R22, R33, R12, R1, R23 and R13. This is I1 and this is I3. Now, in this case what I first do is write KCL at nodes N1, N2 and N3. If I do that and I group together all the variables, I group together the coefficients of each of these variables V1, V2, V3, I get the equations in a certain form and I can express that in a neat matrix form which I am going to write down now. We saw that the diagonal elements of this matrix basically correspond to the total conductance connected to each node. So, the first row corresponds to the KCL at N1, second row to N2 and third row to N3 and I will use the same ordering for the variables also. That is V1, V2 and V3 and the diagonal elements, this element A11 of the matrix corresponds to the total conductance connected to node N1 which is G11, the reciprocal of R11, G12 and G13. Similarly, here this element 22 will consist of G22, G12 and G23. And finally, here in this position we have the total conductance at this node. The off diagonal elements, this entry A12 for instance is the negative of the conductance between node 1 and node 2. So, this is minus G12 and it is a symmetric matrix, this is also minus G12 and here we have the negative of the conductance between node 1 and 3 and here also the same thing. So, this will be minus G13 and finally, this will be these two will be minus G23. And on the right hand side, we will have the current sources and in the first one we will have the current source connected to node 1 in the direction that it pumps current into the node. It is basically the total current coming into node 1 that is plus I1 and nothing is no current source independent current source is connected to node 2. So, this will be 0. Finally, here in the current I3 is pushing current into node N3. So, this will be I3. So, this is what we had and this is what by solving this we get the node voltages and this is what we call nodal. This is what is termed as nodal analysis. Now, the circuit is not solved completely, but once you know all the node voltages, you know all the branch voltages. The branch voltages will be either the node voltage itself for instance for this branch and this branch the branch voltage equals V1. And if you take this branch for instance R12 the branch voltages V1 minus V2. So, that way you can find all the branch voltages and you can also find the branch currents by using branch voltages and the element VI relationships ok. So, this is what is known as nodal analysis ok. And as I said these will be conductances total conductance at the nodes and these the off diagonals once will be negative of the conductance between two nodes ok. So, that is all that is there to it and we get a nice symmetric matrix ok. Now, the only restriction is in this case we have taken a circuit which has resistors and current sources only. We do not have other types of independent sources that is a voltage source or we do not have dependent sources yet we will add them as we go along ok. So, that is a quick review of what we did in the previous class any questions any confusion regarding that and we have this equation in the form the conductance matrix times the node voltage vector equals the vector of independent current sources ok. And this this bar below the letter means that it is a vector and finally, this is the vector of independent sources ok. And to solve for this we have to invert the matrix and multiply by the independent sources ok. Now, right now I will not worry about solving this we will see examples of this later, but there are ways to invert the matrix if it is a 2 by 2 matrix or a 3 by 3 matrix you can do it by hand beyond that you usually do it using a computer ok. But the point here is there is a systematic method of calculating all the node voltages you first set up all the node equations and then there is a procedure whereby you will get all of the node voltages ok. There is a question why why did I call this a matrix and this a vector? If you have a matrix with a single column that is usually called a vector ok a column single column matrix is usually called a vector. So, we have v 1, v 2, v 3 which is you can call it a 3 by 1 matrix, but it is common to call it a vector ok. Now, so like I said I will not worry about solving this right now it is assume that we will be somehow able to solve it. So, what I will do now is to add other kinds of element which we have not added before ok. So, let me take the same circuit and modify it slightly. Now, let me remove this resistor R 2 3 and connect an independent voltage source ok and I will call this v naught ok. So, we want to do a nodal analysis similar analysis as we did before, but with independent voltage sources ok. So, what I did was I had a resistor here I could have put a voltage source anywhere even in parallel with it, but I have just removed that resistor and connected this voltage source ok. So, now my question is can we go ahead exactly as we did before if not what is the difficulty with this ok. The network connection broke and then something happened yeah as some of you recognized the problem is with the problem is with writing KCL here and there where the voltage is connected ok. KCL is still valid the current going here plus the current going there plus the current going there is 0, but the problem is earlier when we had a resistor we could express the current as this conductance times the voltage difference v 2 minus v 3. Now, we do not know what the current is through a voltage source by definition it will support any value of current ok. So, that is the problem with doing nodal analysis when you have independent voltage sources. So, now what do we have to do? So, we have an extra unknown let me call that Ix and I will use the passive sign convention as always I will show the current through the voltage source as going from plus to minus ok that is as we said this Ix is not known ok, but I can still write the KCL at these three equations if I denote this as Ix and that is what I will do ok. So, at n 1 I will write n 1 the equation is exactly the same as before ok. So, I will write it straight away at n 2 what we have are the currents flowing out that is v 2 times g 2 2 v 2 minus v 1 times g 1 2 and minus Ix Ix is flowing in. So, minus Ix is flowing out and similarly at n 3 what do we have? We have minus g 1 3 times v 1 and plus Ix because Ix is flowing out of the node and plus g 1 3 plus g 3 3 times v 3 ok equals I 3 ok. So, we have an extra variable Ix. Now, so we have the first equation is the same as before and the second one and the third one have this extra variable Ix. So, now I totally have four variables v 1, v 2, v 3 and Ix ok and only three equations. So, how do I go about solving this? Is there an equation I am missing? This is a question for the participants I mean I have now four variables and then only three equations to solve for four variables I need four equations right four independent equations. So, where will I get the extra equation from? So, clearly so far I have not described the voltage source itself and that gives me the extra equation. So, this difference v 3 minus v 2 will be equal to v 0 ok. Let me write it with the same order minus v 2 plus v 3 equals v 0. Some of you gave the answer that v 2 minus v 3 is v 0 no it is v 3 minus v 2 equals v 0 with the polarity. So, now I have four equations that is the KCL of the three nodes plus the equation for the voltage source and this is a system of four equations in four unknowns and I will be able to solve this one ok and I will write this in terms of the matrix. So, what I have are for the case of nodal analysis with independent voltage source I have the variables to be the node voltages in general n minus 1 node voltages plus currents through the voltage sources. In my circuit I have a single voltage source but let us say if I have l voltage sources I will have l currents through the l voltage sources. So, the number of variables increases by l ok and the number of equations also increases by l because each voltage source provides a constraint ok. Now, I will write out the equations in matrix form including the extra variable Ix this will be g 1 1 plus g 1 2 plus g 1 3 and minus g 1 2 minus g 1 3 and if you look at the variable vector it will be the node voltage v 1, v 2, v 3 and Ix ok and in the first equation for node n 1 the variable Ix does not appear. So, the coefficient here is 0 ok and in this case I will have minus g 1 2 and here g 2 2 plus g 1 2 ok and here I do not have anything and minus 1 because I have minus Ix this is the KCL equation at node n 2 and at node n 3 I have minus g 1 3 ok nothing for v 2 and g 1 3 plus g 3 3 and I have plus 1 because plus Ix appears in the equation and I have the constraint for the voltage source which is that v 3 minus v 2 that is minus v 2 plus v 3 and these things do not appear equal I 1 0 I 3 these are the currents into node 1 and node 3 and the last equation is minus v 2 plus v 3 equals v 0 ok. So, this is the nodal analysis expressed in matrix form and this has been modified to include an auxiliary variable and that auxiliary variable is the current through an independent voltage source ok and this scheme is known as modified nodal analysis ok. So, whenever you have independent voltage sources you need to have the current through the voltage source as extra variables and then you will be able to again set up the equation systematically and solve for them ok. Any questions about this? So, the modification includes basically extra variable which is the current through the voltage source ok. So, any questions about this? How to deal with how to deal with an independent voltage source in the circuit? When we have an independent voltage source we define an auxiliary current which is the current through the voltage source. So, that gives us an extra variable we get an extra equation from the equation for the voltage source ok. Now, obviously if you have many voltage sources the size of the vector will grow, but I have to emphasize here that the main point of doing this that is being able to write the equation systematically is so that you can solve it even for large circuits ok. It is not intended that you invert this matrix by hand ok it is only through a computer right. I mean for any circuit of any reasonable size you will need a computer to do that, but you have to set up your program so that it looks at the circuit and sets up the equations or this matrix equation correctly ok. And there are routines for inverting the matrix that you can utilize to solve this. Now, this matrix I will still continue to call it conductance matrix, but not all the elements are conductances clearly the ones in the last row and last column they are not conductances ok. Now, on the right side we have the vector of independent sources it is not just independent current sources. Now, we have independent current and voltage sources independent current and voltage sources, but otherwise the structure is exactly the same as before which is the matrix G as I said I still continue to call it G, but not all elements in this are conductances these are all conductances, but these are basically dimensionless ok the last row and last column. G times V that originally was the vector of node voltages, but now it also includes a current and finally equals I and again this is a vector of independent variables it has the current sources as well as the voltage source. Now, I do not want to introduce a new notation so I am going to keep it as it is ok. Now, like I said this is good for analyzing things to the computer because the procedure is very systematic right for every voltage source you simply add a new variable which is the current through the voltage source and also solve for it and you add equations corresponding to each voltage source. Now, when you are solving it by hand you would not want to increase the number of equations like this generally you solve small circuits by hand and then going from a 3 by 3 matrix to a 4 by 4 matrix is tremendously complicated ok when you are doing hand calculations, but as a computer can handle matrices with thousands of rows and columns quite easily ok. So, now for hand analysis the nodal analysis is done with a very slight modification ok. So, let me put down these equations again, so we introduce this extra variable Ix right. Now, what we do is for instance let us say I write up the equations like this and the voltage source is connected between N2 and N3 ok. So, I add up the equations KCL equations for N2 and N3 that is I take this equation here plus that equation there and I add them up to form a new equation and we get something the main point is that this now includes the left hand side of each equation as the total current flowing out of the nodes that is current flowing out of each node in every branch connected to that node ok. So, when I add up these two what it means is it is the sum of all the currents flowing out of N2 and N3 together ok. So, let me do that I will get minus G12 plus G13 times V1 plus G12 plus G22 times V2 plus G13 plus G33 times V3 to be equal to I3. So, the important point is when I add this plus Ix here ok. So, the minus Ix here cancels with plus Ix there ok. Because I have connected I have taken the two nodes at between which the voltage source is connected one of them will have plus Ix the other one will have minus Ix. So, they will cancel ok. So, that is the important part. So, what does it mean now this the left hand side here is the total current flowing out of N2 and N3 ok. And any current that is flowing between N2 and N3 will get cancelled out because it will have plus in one contribution and minus in the other contribution. So, essentially what we are doing is forming what is known as a super node that is a combination of multiple nodes in this case V2 and V3 we always combine the two nodes across which the voltage source is connected. And write KCL in terms of all the currents flowing out of that node ok. For instance we add that current that is flowing out of the super node this current that is flowing out of the super node the current here and the current there and the current there ok. So, we write it as the total current flowing out of the super node to be equal to the total amount of independent current source currents being pumped into that node ok. Now, you can see that the current in the voltage source is not flowing out of this super node or into the super node ok. So, it is not cutting the super node. So, it disappears from the picture. Now, this is a technique to reduce the number of equations and not have this auxiliary variable ok. So, typically when you do hand analysis you would use this and by introducing a super node you will end up with cancelling ix and have only one equation. And we have this extra equation that is we have still lost one equation originally we had three for the three nodes because of the voltage source and combining these two nodes we have a single super node equation. So, we have two this is one and that is the other one ok, but we do have the additional constraint imposed by the voltage source. So, finally we have three equations and three variables ok. And this is one fewer than what we did using modified nodal analysis. So, for hand analysis when you have voltage sources you introduce super nodes and go ahead with the analysis ok. Now, one very important thing is that when I say I combine these two into a super node I am not shorting this to that I still retain this voltage as V2 and this one as V3 ok. So, all the currents must be expressed appropriately that is the current in R12 will be V2 minus V1 times G12, R22 will be V2 times G22 whereas, R33 would be V3 times G33 and R13 would be V3 minus V1 times G13 ok. So, the voltages should not be changed it is not like we are shorting them it is just that we are writing the KCL for the combination of the nodes ok. Any questions about this? Ok there was a question about how do we choose the direction of Ix? Now in principle it does not matter because it will always contribute plus Ix to one side and minus Ix to the other side, but as always use passive sign convention. So, if the voltage source has positive on this side and negative on the other side that is the voltage source symbol is drawn with the voltage is defined with positive on the right side and negative on the left side you choose Ix also to be flowing from right to left ok. And finally you have the voltage source equation. So, then you can again invert the matrix and get the result ok. Now when I say invert the matrix that is how you would set it up and solve on a computer. When you are doing hand analysis obviously you will eliminate variables or eliminate variables one by one sort of by trial and error and so on, but whatever it is if you solve this effectively you have inverted the matrix ok. So, that is what I mean. Now next we take other components which we have so far not included that is some types of controlled sources ok. So, let me again have the same exact same circuit as before I will use this without the independent voltage source we can also include that one if necessary ok. So, this is the resistor R13 right. So, let me take this I have my original circuit and also the matrix that is set up ok. So, this is the original stuff that I had. Now to this what I will do is I will add a voltage controlled current source and let me say I add it here ok and this current source equals some gy times v1 minus v2 ok. So, this is my new circuit. Now what I would like to know is first of all which equations get modified ok which rows get modified. I will set up the question yes Raj please go ahead the question is which of the node equations get modified ok only for node 1, only for node 2, only for node 3 or node 1 and 2 or node 2 and 3 or something else ok. It is clear that this current source is added to node 3 ok. Now if it was between 2 nodes then 2 node voltage node 2 node equations would get modified. In this case this extra current source that I added is added only to node 3. So, only n 3 will get modified ok. Now the next question is how will it get modified ok where will this gy appear is it in column 1, column 2 or column 3, column 1 correspond to coefficients of v2 and column 3 to coefficient of v3 ok. So, which of the columns get modified because of this control source is it column A that is coefficient of v1, column B coefficient of v2 or column C coefficient of v3 or is it something else ok. So, the question is because of this voltage controlled current source which of these columns will get modified is it this one is it this one or is it that one A, B or C ok or is it more than one ok. Many of you answered that it is the third column that will get modified ok, but that is not correct. I think one of you said that it is these two that is coefficients of v1 and v2 that will get modified and that is correct because you do add it to the third row because it belongs to the Kirchhoff's law for node n3, but if you look at this it is gy times v1 minus v2 ok. So, the value of the current is related to the voltages here and those are the ones that will get modified ok because if you look at this current if it was a resistor the current in this would be related to v3 it would be v3 times g33 whereas, this control current source it the current is gy times v1 minus v2 ok it is related to voltages somewhere else. So, it is this basically coefficients of v1 and v2 that will get modified ok. I hope that is clear if there are any questions I will take them ok. So, the point is that VCCS connected to n3. So, n3 equation modified and the voltage controlled current source is controlled by v1 and v2. So, these coefficients get modified ok and how will they be modified? So, first my question is what happens to this coefficient minus g13 originally the coefficient of v1 in this equation is minus g13. How will it get modified? What happens to minus g13? So, the coefficient of this one will get modified and this one also. So, how will this be modified? So, one of you answered this basically the current drawn out of this is gy times v1 minus gy times v2 ok. So, because plus gy times v1 is flowing out of this node because of the controlled current source this minus g13 will be modified to minus g13 plus gy ok. Now similarly, the current drawn out has minus gy times v2 ok. So, that minus gy gets added to this one and this will be minus gy ok. So, now this is how you can handle voltage controlled current sources you do not need any extra variables or equations because this voltage controlled current source simply adds currents to one of the nodes or maybe two nodes ok. In this case the voltage controlled current source is connected between node N3 and the reference node, but it could be connected let us say between node 2 and node 3 also. So, wherever it is connected it will modify those equations and which of the coefficients will get modified that depends on the controlling voltage. In this case the controlling voltage is v1 minus v2. So, the coefficients of v1 and v2 will get modified and the signs you choose appropriately anything on the left hand side is a sum of currents flowing away from the node. So, gy times v1 is flowing away from the node. So, plus gy gets added to the coefficient of v1 minus gy times v2 is flowing away from the node. So, minus gy gets added to the coefficient of v2 ok. Is this clear? Are there any questions? So, one thing you observe is that because of the voltage controlled current source this matrix is not symmetric ok. So, if you had only independent voltage sources and resistors it would be symmetric, but if you have voltage controlled current sources it is not going to be symmetric ok. So, now we know how to handle these extra cases. First we started with only resistors and independent current sources. Then we added independent voltage sources that you can handle either by an auxiliary variable which is easier when you are setting it up for the computer because it is more systematic or you can identify the voltages to which the nodes to which the voltage source is connected as a super node and write a single equation and that is more useful for hand calculation. If you have a voltage controlled current source then you have to modify the matrix it will become asymmetrical, but in general so that also you can do. It will modify the equations at nodes to which the control source is connected and some coefficients will get modified ok. So, next we will take another example with a different kind of control source which is a voltage controlled voltage source ok. So, let me take this circuit again and what I will do is instead of R33 I will have a voltage controlled voltage source ok and this voltage I say is some K times V1 minus V2 ok. So, this control source is K times V1 minus V2. So, what do we do here? First of all just like a voltage source when we had an independent voltage source add an extra variable ok and that is the current through the voltage source I will still call it Ix ok. So, the equations will be exactly the same as before with this Ix substituting for the current that was flowing in the resistor ok. So, if I write down the equations quickly for this one what happens is at node 1 nothing is changed. So, we have G11 plus G12 plus G13 minus G12 and minus G13 ok and at node 2 also nothing is changed. So, we have minus G12 G12 plus G22 plus G23 minus G23. So, this is resistance R23 by the way not R13 and then at N3 we have to change it. So, the equation here would be at N3 would be minus G13 times V1 minus G23 times V2 and here we will have the sum of conductance which is G13 and G23, but also the current flowing out includes Ix ok. So, my variable vector includes Ix ok and it is flowing away. So, I have a plus 1 over here Ix does not appear in the node KCL equations for node 1 and 2. So, I have 0 over there and finally, I have to write the equation for the dependent source ok. Now, in case of the independent source we wrote for instance V3 minus V2 equals V0 and independent source value was on the right hand side. In case of the dependent source V3 equals K times V1 minus V2, but V1 and minus V2 themselves are variables. So, what we do is V3 in this case happens to be K times V1 minus V2, but V3 itself is V1 and V2 themselves are variables. So, we take it over to the left hand side all the variables will be on the left hand side. So, V3 minus KV1 plus KV2 equals 0 ok. So, what do we have minus K plus K and V3 that is plus 1 and 0 and I do not have the room to write the source vector. So, that I will copy over to the next page and write ok. And this equals the total independent current source current flowing into node 1 which is I1 into node 2 which is 0 into node 3 which is I3 and the last one this is simply the equation for the independent sorry the dependent voltage source. So, this is just 0 ok. So, again we have an extra variable Ix and an extra equation for VCVS ok. So, the way to handle the independent voltage source that is voltage controlled voltage source is exactly the same as handling the independent voltage source you define an auxiliary variable and go with it. The only thing is that when you write the equation for the voltage controlled voltage source you group all of the variables to the left hand side ok. So, that is all that is there to it. Any question I think one of you have this question what you do when you have voltage controlled voltage source and this is what you do. Now, the alternative method was to use a super node I am going to go to that but before that if there are any questions about this I will take them ok. It appears there are no questions about this but there are some questions about what we did before with the voltage controlled current source here and I think the question is why this entry is minus G1 3 plus GY ok. So, that comes from writing the KCL for this node let me write it in a different color here. So, the KCL for this node includes the current there plus the current there plus the current there plus the current there equals the current flowing in from the independent current source. So, the current in R1 3 is G1 3 times V3 minus V1 current in R2 3 is G2 3 times V3 minus V2 and the current in R3 3 is G3 3 times V3 and the current in the dependent source is plus GY times V1 minus V2 ok. So, if you look at the coefficient of V1 it is plus GY minus G1 3 ok. So, V1 appears here and here so it is GY minus G1 3 and V2 it is minus GY minus G2 3 ok. So, I hope that is clear ok. So, this is the what is known as modified nodal analysis with voltage controlled voltage source ok. Now, we will try to do the same using the super node now which is the super node that is basically the combination of node N3 and the reference node that is where the voltage source is connected it is a dependent voltage source, but that is where it is. So, we had 3 KCL equations for N1, N2 and N3. Now, which of these equations will get modified because of the super node? This is a question please try to answer that which of these 3 equations which of the KCL equations at these 3 nodes will get modified as a result of making a super node as shown here. So, I hope all of you are able to hear my question and understand it. My question is we have a voltage controlled voltage source and we would like to solve this using super node. So, I have identified this super node that is combining the nodes across which the voltage source is connected ok. So, now which of the node equations will get modified because of this super node? And couple of you answered that it is node N3 that is correct. So, what happens to that equation? What should we write? What is the new equation that we should write? So, now because we this voltage source is connected between this node and the reference node, the super node is the combination of this node and the reference node. Now, you do not write any KCL equation at the reference node. So, if the reference node is part of a super node, the node equation simply goes away ok. Earlier when we had a voltage source between N3 and N2, we combined them into a single equation ok that is the total current flowing out of this combined node. When you have something combined with a super node that equation simply goes away ok because when you combine it with the reference node you normally you do not write any equation for the reference node. So, when any super node when combined with the reference one also goes away ok. So, you have two equations one at N1 and one at N2 which are exactly the same as before. And now you also have an additional equation because of this voltage controlled voltage source. We have this additional constraint that V3 equals K times V1 minus V2. So, as usual we write it with all the variables on the left side which is minus K V1 plus K V2 minus V3 equals 0. And the equations for N1 and N2 remain exactly the same as before that is G11 plus G12 plus G13 times V1 minus G12 V2 minus G13 times V3 equals 0. Similarly, minus G12 V1 plus G12 plus G22 plus G23 times V2 minus G23 V3 equals 0 sorry the first equation it is not 0 it is. So, I hope this part is clear. Now my question to you is what happens to this I3? I3 does not seem to appear in any of the equations why is that? So, if you look at these three equations this one this one and that one. So, we have three equations and I3 does not appear in the equations why? Yes a couple of you answer that it is because it is connected to a super node that is correct, but my question is this super node stuff is some method that we use to analyze circuits ok. Now essentially what this means is that from these equations forget the fact that we use super node we got these equations now I3 does not appear anywhere, but I3 is very much in the circuit. So, if I3 does not appear in any of the equations what it means is the solution does not depend on I3 right. So, why does it not depend on I3? Can you look at the circuit and tell that is my real question ok. So, I got a number of responses some of you said it is not it is because there is no parallel resistance and so on I am not very sure what is meant what was in your mind when you wrote that, but one of you did answer correctly see the point is that this is a current source across a voltage source ok. Now this is a dependent voltage source, but anyway because this is a voltage source the voltage at this node is set by this voltage source ok. So, whatever even if you have current sources here it is not going to be able to change the voltage here ok this voltage source will absorb any current that is put into it. So, because you have this current source across this voltage source it does not appear it is not it does not matter what the value of I3 is the solution will remain the same. So, the correct answer is that it appears across the voltage source ok. So, what we have done is so far to do nodal analysis of circuits and we did it for cases with current sources and resistances. In all cases the equation can be set up as some matrix sometimes some variable vector equal the vector of independent sources ok. So, we can do this in all cases. Now when you have only current sources and resistances this is the cleanest case we have a symmetric g and in all these cases of course the vector is obtained as g inverse times i ok and when we have independent voltage sources. So, we can do modified nodal analysis which is MNA or basically we do modified nodal analysis by introducing extra variables or alternatively we can treat the two nodes to which the voltage source is connected as a super node and go ahead with the analysis. Now modified nodal analysis gives you one more variable and one more equation for every voltage source whereas the super node stuff this will not give you no extra variables ok and finally if you have a voltage controlled current source there are no extra variables or equations but you will have the g matrix will become asymmetric ok and finally when you have a voltage controlled voltage source you handle it in the same way that you did the independent voltage source. Now when you do the independent voltage source what happens is you will have an extra entry in the source matrix on the right hand side ok in the source vector whereas when you have a voltage controlled voltage source there is no independent source that is added whatever is on this side this whole thing should have only independent sources ok. So, that is why you can solve for it correctly that you have this variable equals the inverse of this matrix which is related to the circuit topology times the source vector ok and in case of voltage controlled voltage source there is no extra addition to the source vector and you handle it in exactly the same way as independent voltage source by introducing an extra variable and equation or by using a super node in which case you do not have extra variables. Now this modified nodal analysis you use it when you are doing things with a computer mainly because it is a systematic way of setting up equations and solving it when you do the super node business it is a little more ad hoc but it reduces the number of equations so when you are doing hand analysis you use the super node ok. So, any questions on these things now we have two other types of control sources that is a current control voltage source and a current control current source I will not discuss them in detail here if you are interested you can go to the other course that I am offering this semester at IIT Madras I will give you the URL for that and the details will be there and you can also try to do it by yourselves you add a voltage control sorry current control current source or a current control voltage source and see how the equations turn up if you run into any difficulty please ask me in one of the following classes and I will try to explain ok or you can raise it in the forum as well. So, if you have any questions about anything we did so far please ask so there is a question from Saurav Mahajan asking about why IX is removed now please be more specific are you talking about why IX is not there when we use the super node is that the question ok. So, let me go back to where we had that. So, first of all super node means this entire thing here ok. So, whatever is inside it could have components inside it is the super node and we know from Kirchhoff's current law that all the currents flowing out of a node will sum to 0 but if you have any closed surface all of the currents flowing out of that surface will be 0 ok. So, if you look at this the currents flowing out of this surface will be through the wires that are cutting this red box that I have drawn and that is through R 12, R 22, R 33 this current source I 3 and R 13 ok. The current through the voltage source is completely inside this box and it is not cutting this so it will not appear in the picture ok. Similarly, if you want to think about it from the equations originally we wrote the equations for N 2 and N 3 ok. So, for N 2 you get minus ix and for N 3 you get plus ix and when you write the KCL for the whole super node you are essentially adding up the equations for N 2 and N 3. So, minus ix will always cancel with plus ix ok. Is that fine? Is that the question you wanted to ask? Ok. Now, let me take this particular case so this one I have the equation here and this referred to a circuit with voltage sources and current sources we do not have dependent sources but that is ok. So, I just want to use this to prove certain things about the kind of circuits we have ok I meant this one where I have the independent sources. So, this is the circuit let me copy this over to the end and the circuit itself is not so important so I will make it a little small. So, this is in the form of some matrix G times the variable vector V ok this is the matrix G times the variable vector V equals the source vector i. Like I pointed out because this has both independent voltage and current sources this vector V has voltages and currents and this vector i also has voltages and currents but in order not to change the notation I will continue using this. Now let me expand this out this G times V equals i right. So, obviously the vector V which is the variable we want to solve for as G inverse times i. Now, let me write out i completely V equals G inverse times i 1 0 i 3 V 0 and this can also be written as G inverse times i 1 0 0 0 that is I take only one element here plus G inverse times 0 0 i 3 and 0 in this case I have taken i 3 alone plus this is the inverse of the matrix and again G inverse times 0 0 0 and V 0 ok. So, this is the source vector and these are basically source vectors with one non-zero source at a time ok. So, I can write this vector as i 1 with all 0s and here 0 0 i 3 0 and 0 0 0 and V 0 ok and obviously G inverse times the sum of these three will be G inverse times this plus G inverse times that plus G inverse times that. I hope all of you agree with this expansion ok. So, if there any doubts about this please ask me, but what is the point that I am trying to prove here ok. What does it mean? So, here I have all of them together and here I have all except one to be 0 here i 1 is non-zero and here it is non-zero and here that one is non-zero ok. What is it that what is the point I am trying to make? Ok, I think many of you got the answer what I am trying to prove is the property of superposition of these linear circuits ok. Linear circuits give you this system of linear equations and it follows superposition, but this expansion makes it very clear ok. Now what is this? This is this part is solution with may be it will be clear if I do this ok. So, this is solution with only i 1 being non-zero I will say only i 1 active and this is the solution with only i 3 active and finally this is the solution with only v 0 active ok. And we can activate only one source at a time find the solution and add up the solutions due to all the sources in the circuit. This is the principle of superposition and it is all used very widely to prove certain things about circuits as well as while doing hand analysis. We will continue from this in the next class ok. So, for now if you have any questions about this please ask me and I will clarify them ok. With that we come to the end of today's class. Thanks for attending see you next week.