 Welcome to lecture 9 of basic electrical circuits, in the previous lecture we looked at mesh analysis, it's an analysis method where you start with Kirchhoff's voltage law around different measures in the circuit, it's a sub case of this loop analysis. So now we see that it's an alternative to nodal analysis and as at the very end of the lecture we were trying to figure out which one would be used nodal analysis or mesh analysis, okay. So before we get there if there are any questions about any of the previous lectures please ask, so we can discuss those things. Any questions about the previous lecture? Okay, in that case let's go ahead with the lecture. At the end of the previous lecture I asked whether we would use nodal analysis or mesh analysis, I mean I said mesh analysis, so in general a loop analysis and the way to decide is that nodal analysis has n-1 KCL equations and loop analysis will have b-n plus 1 KBL equations which you have to solve for a circuit with n nodes and b branches, okay. So what we have to decide is n-1 a larger number or b-n plus 1 a larger number, okay. So to answer this I asked you, first of all the number of branches depends very much on the circuit but what is the maximum number of branches you can have in node circuit if there are branches between every node and every other node that is if there are branches connecting all nodes to all other nodes, how many branches will be there, okay please try to answer this question how many branches will be there, okay there were a couple of answers saying 2n-1 and other one was n times n-1 by 2 and that's the correct answer, the correct answer is that it will be n choose 2 or n times n-1 by 2, okay. So every branch can be connected to n-1 other branches that's why you get this n-1 here and then you have n possibilities for selecting the starting node and also you get divided by 2 because if node a is connected to node b it's the same as connecting node b to node a, okay. Alternatively you can think of while adding branches you have to pick 2 nodes, you have n possible nodes and out of that you pick 2 nodes and you can do that in n choose 2 ways, okay n times n-1 by 2. So this I will simply approximate it by n square by 2 for large n, okay. So this n-1 is approximated by n. So if every node is connected to every other node by branches this number b-n plus 1 will be of the order of n square by 2, okay for large n and the number of KVL equations will be lot more than the number of KCL equations, okay. Now in reality it will be somewhere in the middle it's possible that you have very few branches that is you just have a single loop, okay. For instance you can have n nodes and branches only like that. In this case clearly number of KVL equations will be smaller, alternatively you can have everything connected to everything else, okay something like this, right. In this case you have lot more KVL equations than KCL equations. Now reality will be somewhere in between it's not that all the nodes will be arranged in a single loop or it's not that every node will be connected to every other node using a branch. So the number of KVL equations will be smaller than this but in general it tends to be a little more than the number of KCL equations, okay. That is one reason although this is not a hard and fast rule and it's possible to have a single KVL equation for an n node circuit if you happen to have a single loop. In that case you choose KVL. Now it turns out that this analysis of circuits using a computer that uses nodal analysis. There is another reason for it, okay that is that it's very easy to identify nodes they are just given to you and this is true even for human beings. You look at a circuit you know what the nodes are but a little harder to identify all the loops, okay so you have to first identify a tree then start adding links to the tree to form loops, okay. So because of these reasons usually if you want to do systematic analysis of large circuits you end up using nodal analysis. Is the audio okay, okay it seems fine for some people but maybe for those of you for whom the audio is not good you have to check the setup at your end, okay. So this is just to say that you end up using nodal analysis for systematic analysis of large circuits, okay. So far what we have done is to look at nodal analysis and mesh analysis we can systematically write down equations for the two cases. If you have only current sources and resistors nodal analysis comes out with a nice structure. Similarly if you have only voltage sources and resistors mesh analysis equations have a nice structure but either of these can be used even when you have other sources that is you can use nodal analysis when you have voltage sources and mesh analysis when you have current sources. In case of nodal analysis with voltage sources you either define an auxiliary variable or use super node. Similarly in case of mesh analysis with current sources you use an auxiliary variable for the voltage across the current source or use a super mesh and when you have controlled sources things tend to become more complicated but you should be able to figure out those things as well and if you are trying to solve some specific problem but run into some trouble then please raise it in the class and we will discuss that, okay. So any questions on either nodal analysis or mesh analysis, any questions, okay then let's move on to the next topic in our course which is certain theorems involving circuits. What I showed so far was ways of systematically analyzing large circuits, okay so that's how you use nodal and mesh analysis. Now when you are doing hand analysis of small circuits you tend to use a number of ad hoc ways in fact I will discuss many of those things after I discuss the circuit theorems, okay. So we will start with some very simple things first, okay. First of all these are not called circuit theorems but some transformations that you can use and we have already discussed these but I will briefly touch upon them. If you have a voltage source in parallel with anything it is the same as the voltage source, okay this is just to remind you of some basic results, okay so this is V0 this will also be V0 and similarly if you have a current source in series with anything it is equivalent to a current source, okay. I assume that there will be no questions about this these are really basic results, clearly whatever element you have in series the current coming out here will be I0 so current source in series with anything will be just a current source. Similarly whatever this be across this you have a voltage V0 so whether you analyze the when I say equivalent what it means is if you analyze the circuit with this or without this you will get exactly the same result. The only difference will be that this may draw some extra current from the voltage source, okay. So except for the current in the voltage source having this or not having this there will be no difference. Similarly here except for the voltage across the current source having this or not having this will not make any difference at all, okay. Now let us move on to slightly more sophisticated things. Let us say we have a node with a number of branches and this is the node, okay and this is branch number 1, 2, 3 and 4, okay. So this is just an illustration and let me assume that in my particular circuit in series with these three branches there happen to be voltage sources, okay and all of the same value V0, okay. So all I am saying is there is a node and in my example I have taken four branches and in three of those branches there is a voltage source with the same consistent polarity and the same value V0, okay. Let us say this is the circuit this is just the assumption. Now in general you can have n nodes with n minus 1 branches sorry you can have one node with n branches and n minus 1 of those n branches will have the same valued voltage source, okay. So this is the circuit. So now what I would like answers from you is let me call these nodes n1, n2 and n3, okay. What will be the, how will be the voltages at n1, n2 and n3 be related to each other? My question is there will be some voltage at n1 with respect to the reference node of the circuit and n2 and n3, okay. So how are these going to be related to each other? Please try to answer this. So the question is how are the voltages at n1, n2 and n3 related to each other? So a couple of you have given your answers, they will be the same voltage and that is pretty clear because let us assume that the voltage here is some V1. Now the voltage at n1 will be V1 plus this V0 and here it will be V1 plus V0 again and here also it is V1 plus V0, okay. So they will all be equal to each other. Now it turns out that if two nodes in a circuit have the same voltage, you can connect them without changing the circuit, okay. So if you have two nodes and you know that the voltages are identical, okay. If this is V1, this is also V1 then you can short these two without affecting the circuit, okay. Now this can be used done in most cases but there are some cases where it cannot be done. I will not go into them now but in most general, most regular circuits you can do this, okay. So what does this mean? Because the voltage here, there and there are the same, I can connect all of them together, okay. And in fact, if I had more branches with voltage sources like this, I could connect all of them together, okay. So now if I observe the circuit, I see that this voltage source and that voltage source and that voltage source are in parallel with each other, okay. So these three voltage sources are in parallel with each other. Now if you have voltage sources of equal value in parallel with each other, first of all if there are one equal value, you cannot even connect them in parallel. But if there are equal value and parallel with each other, it is the same as having a single voltage source, okay. So this is the same as doing this with a single voltage source V0 and this branch number 1, 2, 3 and 4, okay. And this node represents the union of the three nodes, basically whatever I have shorted and you have a single voltage source, okay. So what this whole exercise is showing is that if you have a picture like this with voltage source on one of the branches, it is exactly the same as having a picture like this with this voltage source moved to all the other branches, okay. Any questions about this? Okay. So now this theorem is sometimes useful to prove other theorems. I may not use it directly in this course but this is known as pushing the voltage source through a node. Now I proved it by starting with these multiple voltage sources and combining them into this one but it is usually more useful to go the other way around. If you have a voltage source in a single branch, you can push it into all the other branches connected to that node, okay. So typically you end up going in this direction and this is, you can call this as pushing a voltage source through a node, okay. So this is about voltage sources and similarly for current sources, there is something, okay. Now let's say you have a current source I0 in a circuit, okay and it is exactly the same as having two current sources in series of identical value, okay. So this by definition is exactly equal to that one, okay. Now if you, we have formed a new node in the middle and if you take a wire from that node, what will be the current flowing here? What is that current going to be? Whatever be the value of I0, any current source you take it, split it into two sources in series and form a node in the middle and if you connect a wire to that middle node, what will be the current flowing through that wire? Yeah. Let it clear that this Ix will be 0, okay. So now this is useful in the following way. Because the current here is identically going to be 0, you can take this node and connect it to any other node in the circuit, okay, okay. Now this is sometimes useful. So if you have a current source connected between let's say some nodes N1 and N2, you can split that into two current sources and you can connect the middle to anywhere and usually it's convenient to connect it to the reference node, okay. So instead of a single current source between N1 and N2, you will have a current source from N1 to the reference node and from the reference node to N2 and sometimes this is easier to analyze. Again if we, I will point out if there are examples that make use of this but this is some general property that you can use to simplify circuits and visualize certain things about circuits, okay. Any questions about this? Okay. So these are quite simple so we can move forward. Now the next thing is a little more interesting. Let's say I have any circuit N, okay, any circuit or network N and there is some element connected to it. I will take a resistor as an example but this can be any element, okay and it doesn't even have to be linear. So let's say I have a resistance R and this has all kinds of things in it. It can have resistors, current sources and voltage sources and so on, okay. Now the question is, okay, can I take this circuit, remove the resistor and put something else here, okay, some box such that all the voltages and currents in this circuit are exactly the same as voltages and currents in this circuit, okay. Let me assume that in the original circuit there is a certain voltage across the resistor and there is a current through the resistor, okay. So under these conditions, can I replace the resistor with something else so that all the solution to the circuit that is all the branch voltages and branch currents in this circuit will be exactly the same as the ones in this circuit. Is that possible? If so, what will I replace it with? Please try to answer this. So can I replace the resistor with something else without changing any of the branch voltages or branch currents in the circuit N? So please try to answer this, I am not getting any responses. So because of the setup here, I will not be able to hear your questions over audio. So somebody has raised their hand but please use the chat window to ask your question, okay. I have got only one response which says that this could be replaced by a voltage controlled current source. Now I am not sure what is the intent of this answer, perhaps it means that we have seen that a voltage controlled current source can behave as a resistor if the controlling voltage happens to be across the current source, okay. Now this is not my question really, I mean this is fine, this is true but this is a resistor, okay. A voltage controlled current source with the controlling voltage across the, with the controlling voltage across the current source will be exactly the same as a resistor. So my question was can we replace it by any other element and somebody said a voltage source, okay. Now that is possible, we will see how. Now it turns out that a resistor can be replaced with either a voltage source or a current source of specific value, okay. Now please understand that this will not work in general that is you have a circuit N with some particular value of sources inside, okay. So in under those conditions you will have some particular current and particular voltage across this resistor. So in that condition you can replace this with a voltage source or a current source. Now if you change the sources inside N they will not be the same anymore, okay. They will be the same only for a particular value of the sources inside the network, okay. Now let's see how we can prove this. The proof is very simple and it involves no algebra, just some simple logic, okay. Is there a problem with the audio again, okay. So it looks like things are working. So let me copy over this circuit, okay. Now as I said for particular values of some sources inside this could have any number of resistors, voltage sources, current sources, ion control sources and so on and I am showing one particular component here. We could do this with more than one component. Now for this set of values there is a current IR through the resistor and a voltage VR across the resistor, okay. Now let me do this. What I will do is I will connect a current source of value IR in this direction and another current source of exactly the same value IR in the other direction, okay. So I have connected two equal and opposite current sources which means that I have not really connected anything, okay because the current here will be exactly zero, right. All I did was to there is nothing connected here and that nothing I represented as a parallel combination of two equal current sources in opposite direction. The value of each current source I took to be the current flowing in the resistor, okay. Now what I will do is I will just interchange the positions, okay. You see that this resistor, this current source and that current source are in parallel. I will put this current source on the left side and the resistor in the middle, okay. This is IR and across this we have VR and the current through the resistor will be exactly the same as before because nothing has really changed, okay. From my original circuit I have not made any changes. I added a net of zero current here and then here I just simply changed the way I drew the circuit, okay. But the interesting thing now is what will be the current in this part of the wire, this part of the circuit, this wire let me show that in blue. What will be the current in these blue wires, okay. I think all of you can immediately recognize that this current is zero, okay. So now if a wire is carrying zero current there is no point having that wire I may as well cut it off, okay. So this current will be zero so I can cut this off and nothing will change in the circuit, okay. Whether I have an open circuit that is a cut wire or a wire with exactly zero current it is exactly the same thing, okay. So it won't disturb the circuit so I can cut off those things. So what is the bottom line here? My original circuit was like this and after I did all the transformations I described to you and cut off this wire my circuit is like that and all through the steps of logic we saw that nothing really changes in this circuit. So the voltages and currents in this circuit will be exactly the same as the voltages and currents in that circuit, okay. So what this means is that a resistor with a current IR flowing through it can be substituted by a current source of value IR and in the appropriate direction of course IR is flowing downwards here so this current source has to point downwards, okay. And this will be without changing any branch voltages or currents, okay. I hope this is convincing and this particular result is known as the substitution theorem, okay. So this it turns out is quite useful in many applications I mean when I say applications to solve other problems and maybe prove other circuit theorems, okay. So is this convincing any questions about this chain of reasoning and the proof of substitution theorem, okay. So now we can also quickly go through another possible substitution what I will do is initially I mean previously I added zero current and the zero current I represented as IR two current sources of value IR in opposite directions in parallel, okay. Now what I will do is in series with this particular wire I will break this wire, okay and add two voltage sources of equal and opposite value and the value I choose to be VR and what is VR whatever voltage was across this resistor, okay. So now clearly this voltage will be the same as that voltage and essentially this combination two equal and opposite voltage sources in series it is nothing but a short circuit, okay. So nothing has changed in the circuit. Now with respect to some reference voltage let me know let's say let me call this I found a new node here let me call this N1 and N2 how will the voltage at N1 be related to voltage at N2, okay. So maybe I will say the value of VN1 minus VN2 that is the voltage at N1 minus the voltage at N2 what is this value going to be please try and answer this question. My question is how is this voltage related to that voltage, okay. Now I think it must be pretty convincing to you that I have not changed the circuit in any way all I did was to connect zero volts in series with the wire which is like not disturbing the wire at all but that zero volt is represented as two equal and opposite voltages of value VR in series with each other, okay. So there will be at the same voltage again this is pretty clear if you go from here you have a voltage rise of VR and voltage fall of VR so these two will be at exactly the same voltage. Now given that they are at the same voltage what I can do is connect them up like that, okay. So my circuit becomes I will redraw it slightly differently I will have VR across this, okay and you have something hanging from here which is VR and this resistor but you can see again that the current in this wire is definitely going to be zero, okay so that part can be removed from the circuit, okay. So what we have done is that the resistor which was across these two terminals has been substituted by voltage VR where VR is the actual voltage across the resistor in the circuit. So this is another variant of the substitution theorem. If you have a resistor with a voltage VR across it, it can be substituted by an independent voltage source value VR without changing the circuit solution, okay. So this is fine. So this is another variant of substitution theorem. So in summary what it says is that if you have a circuit with a resistor R which has a voltage VR across it and a current IR through it, you can substitute that resistor by a current source of value IR in the appropriate direction or a voltage source of value VR in the appropriate polarity and the circuit solutions will be exactly the same that is the branch voltages and currents here, here and here will be exactly the same, okay and this is known as substitution theorem, okay. Any questions about this? Any questions about the statement of substitution theorem or the proof? Any questions? Okay. So let's move forward. Now one thing I want to point out is we took a resistor and then substituted it. First of all you can do this for multiple resistors in a circuit. It doesn't have to be only one and also it doesn't have to be a resistor. It can be any element, okay. So instead of a resistor we could have any element, okay, which has a current IR through it and a voltage VR through it. I will still call it IR and VR. This may not be a, this need not be a resistor. This can be any two terminal element, okay. Also when you say any two terminal element that itself can be a complicated network with only two terminals exposed, okay. So this could be another network. The only condition is that only two terminals are brought out like this and they are connected and this entire thing can be substituted by, let's say the voltage across these two terminals is VR and the current flowing that way is IR and this entire thing can be substituted by a current source of value IR or a voltage source of value VR, okay. So I started with a resistor but you can work out for yourself that the logic of the proof holds good even if it is non-linear element it can be any element. In fact you can substitute a voltage source by current source and vice versa and also you can substitute a complicated circuit as long as it is only at two terminals, okay. So if you have a complicated network with only two terminals brought out you can substitute the entire network by this voltage source or current source, okay. I hope that part is clear. So there is a question asking about is there any limitation of this theorem? I mean there really is no limitation. The only thing is this theorem by itself is somewhat limited in that it works for specific values of voltage and current. I will take an example after that it will become clear but like I said there is no other limit. You can replace any element with a voltage source or a current source, okay. And when I say element it can be a simple thing like a resistor or some non-linear element like a diode but it can also be a complicated network, okay. So as long as only two terminals are brought out it does not matter what the network is inside it could have thousands of components but at those two terminals it can be replaced by either a current source or voltage source but this current source and voltage source are not arbitrary things they are the actual currents or voltages flowing in the original circuit, okay. So I hope that is clear and that also kind of summarizes the substitution theorem which someone else asked for. And as far as the application of this is concerned in fact we will see shortly we will prove other theorems using this theorem, okay. So under the question what is superposition theorem, superposition we have discussed earlier, okay. We saw that if you we proved it with the nodal analysis but you can do it with any which way you want. The unknown vector in a nodal analysis in the nodal analysis method will come out to be G inverse times the source vector, okay. I call this I and V but we know that this can contain independent current sources and voltage sources, okay. So let's say this independent source vector had two current sources I1, I2 and many zeros and then a voltage source also, okay. So this is the same as what I am writing now. So this is the solution to the node voltages to the unknown vector when all sources are acting together. Now this is the solution when I1 is active and other independent sources are set to zero. Similarly the second one is when I2 is active and other independent sources are set to zero and this finally is when only V1 is active and other independent sources are set to zero. So this is superposition theorem what it says is if you have a number of independent sources in a circuit the solution when all of them are active can be obtained as the solution some of solutions when each one is active and all the others are set to zero, okay. So that is you can activate the sources one by one and set all the other sources to zero. In this case you first do the analysis with I1 only and I2 only and then V1 only and add up all the solutions. And this works for any circuit that is linear, okay and a linear circuit means it has besides these independent sources it has resistors and linear controlled sources, okay. So this is superposition, okay. So I hope that part is clear. Now let's see let's take a numerical example also to illustrate substitution theorem and to get a bit of practice with circuit analysis I will take a very simple circuit let's say it has a single source 16 volts, okay. So now what I would like you to find is are these quantities please don't do it right now in fact you have to do a little bit of working out this circuit is not very difficult but you still have to do that. So please take it as an exercise and do it before next Tuesday and we will discuss this circuit and how to solve it in the lecture, okay. So please calculate the voltage across this I will call that V1 and the current through this resistor I will call that I2 and the current through this resistor I will call it I3, okay. Calculate V1, I2, I3 and you can also calculate V3 here across this, okay so please take this up as an exercise and do the analysis before the next lecture, okay and to illustrate superposition you can also try this in addition to this let me say I have a current source of value 5 milliamps, okay and then you can see you can solve for this one, okay. Again with this calculate the same things, okay so these are the exercises as usual please go through it please go through the circuit step by step and do it while understanding every step, okay the analysis of this is very simple but you should be able to do it with confidence, okay somebody answered saying V1 will be 8 volts that is not correct please do the analysis in detail and then solve for it, okay. Now let's go to something else we will prove another theorem based on substitution theorem and this is something that is very, very widely used, okay and in fact it will use substitution theorem and superposition so I am glad you brought up the topic of superposition, okay. Now let's say we have a circuit N with independent sources and linear elements, okay when I say linear elements it has resistors and controlled sources, okay and it can be any network, right the connections inside we are not imposing any condition so it will have independent sources and linear elements and linear elements means resistors and controlled sources. The only condition I will say is all the controlled sources have controlling quantities inside N, okay it's not that any controlled source is controlled by a voltage that is outside somewhere else, okay if it's a voltage controlled source that voltage will be inside N and similarly if it's a current controlled source that current will be inside N, okay so that's the only condition I will impose, okay. Now let's say that a current I, current source Ix is connected to it it has two terminals I will call them 1, 1 prime that are coming out and I have this current source connected up like this 1, 2, 1 prime, okay is this clear the definition of the circuit I have. Now the voltage here will be of some form, okay now what I will do is I will try to solve it by superposition, superposition means that when I have independent sources I don't take all of them together I can take one by one and I don't have to do it one by one if I have ten of them I don't have to do it ten times I can take five ones and then the remaining five the other time and so on, okay in this case how I will split it is I will do it with only this current source active and everything inside inactive and then I will make this inactive the outside current source Ix inactive and everything inside active, okay so that is also valid so when I say superposition you don't necessarily have to take it one by one so if I have let's say inside the circuit let's say I have five independent sources two current sources and three voltage sources I can do superposition with I1 alone active and all other zero I2 and V1, V2, V3 and then add up the results, okay instead I can also do it like this I can take I1 and I2 and V1 active, okay and V2 and in that case I will set V2 and V3 to zero next I will take V2 and V3 active and set the rest to zero and I add up the two results, okay so this will give me exactly the same answer, okay so just so you don't get confused with this let me show it with an example so please try to solve this so let me take a very simple circuit again so this is the circuit, okay now first of all you can solve for this in a number of ways you can either use nodal analysis or mesh analysis and solve for it so please do that and let me know the voltage across this 6 kilo ohm resistor let me call that V1 please take this up as an exercise and use nodal analysis to find V1, okay so this is an extremely simple example there are only two nodes in fact I suggest that you take this as the reference node there are only two nodes in the circuit and this node has a voltage source connected to the reference node so this node voltage is already known so you only have to write one equation one nodal equation and solve for V1 so please do that please take it as an exercise and let me know the value of this V1 in this circuit I hope the question is clear all I'm asking for is this value of V1 and just do it by nodal analysis by writing the KCL equation at this particular node, okay I've got one response anybody else so I've got three responses all different from each other so hopefully others also have solved it okay so maybe or you can just answer it in the poll, okay I've opened the poll with all the responses that I've got please just click on the poll and let me know your answer okay so this is the result of the poll I mean somebody has voted for all of these choices but the point is only one of these can be correct so I will work it out in detail and you can figure out for yourself where you went wrong this is very important because obviously three of these answers are incorrect now then you can figure out where you went wrong and then fix it for the next time, okay now I said write KCL at this node and from there calculate V1 so the voltage of this node with respect to the reference node is nothing but V1 okay so the current when we write KCL the currents flowing out will be together be equal to 0 so the current flowing in the 4 kilo ohm resistor that is flowing away from this node will be V1 minus 10 volts divided by 4 kilo ohms plus the current flowing in the 6 kilo ohm resistor is V1 by 6 kilo ohms plus the current flowing here it is a current source so we know the current the current flowing in this direction will be minus 1 amperes 1 minus 1 milliampere okay will be equal to 0 right so V1 if I group all of that together I will have V1 times 1 by 4 kilo ohm plus 1 by 6 kilo ohm equals 10 volt divided by 4 kilo ohm okay plus 1 milliampere okay this is what we will have so V1 will be 10 volt by 4 kilo ohm 1 by 1 over 4 kilo ohm plus 1 by 6 kilo ohms plus 1 milliampere divided by 1 by 4 kilo ohm plus 1 by 6 kilo ohms okay so if I take this inside I will get 10 volt 1 divided by 4 kilo ohm this gives me 1 plus 4 by 6 this gives me 2 thirds okay and this one here if I calculate this I will get 6 kilo ohm plus 4 kilo ohms divided by 4 kilo ohm times 6 kilo ohm okay which is basically 10 by 24 kilo ohm in the denominator we have kilo ohm square in the numerator we have kilo ohm so we have 10 by 24 kilo ohm so we have 1 milliamp times 24 kilo ohms divided by 10 okay so you can see that this will be 6 volts that is 10 volts times 3 by 5 plus this will be 1 milliamp times 2.4 kilo ohms 2.4 volts which is equal to 8.4 volts okay so the correct answer is 8.4 volts so I hope it's very clear how to do this so if you do this nodal analysis you will get the final answer in one shot now let me also do it by superposition just to demonstrate it to you okay so what I will do is I will first set this 10 volts independent source to 0 and find V1 as a result of the 1 milliamp current source and then I will set this to 0 the 1 milliamp source to 0 and find this V1 as a result of the 10 volt source and then I add up the values of V1 from this circuit and that circuit and that will give me the same answer as before or it should if superposition is valid. Now this is a linear circuit we have independent sources and except the independent sources we have just resistors which are linear so we can apply superposition okay. Now when I set 10 volt source to 0 what does it mean what should I replace the 10 volt source by when I say set to 0 what does it mean what should I replace it by okay so many of you answered that it is replaced by a short circuit so when I this is a voltage source when I make the voltage source a 0 volt source it is a short circuit okay so here I have to replace it with a short circuit. Now similarly when I set the current source to 0 what should I replace it by so clearly I have to replace it with a 0 current source and a 0 current means no current which is an open circuit okay so there should be no confusion in this sometimes you hear terms like I remove the sources just try not to use that because it will only lead to confusion so you just make that source value equal to 0 so if it is a voltage source it becomes a short circuit if it is a current source it becomes an open circuit okay. So now in the first case what is the value of V1 if you observe the 6 kilo ohm and 4 kilo ohm are in parallel okay it is drawn in a strange way but really this voltage sorry this terminal is common to the two resistors and this also okay so both the resistors are connected between this node and that node so that means they are in parallel okay so what I have really is a 1 milliamp current source going into this 4 kilo ohm and 6 kilo ohm in parallel and we know what happens when you connect resistors in parallel it is equivalent to a single resistor of value 1 by the reciprocal of the resistors in fact when you have only two resistors in parallel you can use the formula R1 R2 by R1 plus R2 so you will get 4 kilo ohm times 6 kilo ohms divided by 4 kilo ohm plus 6 kilo ohm and this is 24 kilo ohms divided by 10 which is 2.4 kilo ohms okay and if 1 milliamp current flows through 2.4 kilo ohms so this milliamp and kilo ohm multiplied will give you volts so you will get 2.4 volts and by the way when you use superposition you make sure that whatever quantity you have to solve for you have put that in a consistent direction everywhere okay V1 is the voltage across 6 kilo ohm resistor with the upper terminal positive it is the same here and in this case V1 will be here across 6 kilo ohm and that is equal to 1 milliamp times 2.4 kilo ohm which is 2.4 volts okay and similarly if I have this particular case okay you can calculate the total current what is the current flowing here what is the current flowing in this loop you have 2 resistors in series 4 kilo ohms and 6 kilo ohms so what is the current flowing here some of you have answered just one always these are quantities with dimensions so please always give the units along with the number and some of you have answered one ampere so please be careful about what you are doing and give me the correct answer the series combination of 4 and 6 kilo ohms is a single resistor of 4 kilo ohms plus 6 kilo ohms which is 10 kilo ohms so we have 10 volts divided by 10 kilo ohms which gives you 1 milliamp here it is not 1 amp here and do not try not to give answers like 1 because that is only going to get confusing so this is 1 milliamp here and this 1 milliamp here flows through both of these so the voltage V1 will be 1 milliamp times 6 kilo ohm which is equal to 6 volts okay so it is very simple and alternatively you can recognize that this is a voltage divider and use the voltage divider formula the voltage across the 6 kilo ohm resistor is nothing but 6 kilo ohm divided by 6 kilo ohm plus 4 kilo ohm times the applied voltage source which is 10 volts which gives you 6 volts either way you get the voltage across the 6 kilo ohm resistor to be 6 volts okay so here we have 2.4 volts and here we have 6 volts okay so the actual total voltage will be the sum of these two which is nothing but 8.4 volts and it is exactly the same as what we have here 8.4 volts okay in fact the way I have written out the expressions this is the contribution from the voltage source and this is the contribution from the current source and they get added together. Now in general if you have multiple inputs that is multiple independent sources okay you have any voltage or any current in the circuit will be a linear combination of all the sources okay for instance let me make these variables just for illustration let me call this VS and let me call this IS if I write Kirchhoff's current law at this node this part will be exactly the same okay let me so instead of 10 volts I will have VS here and instead of minus 1 milliamp here I will have minus IS okay so if I take it to that other equation I will end up getting instead of 10 volts I will have VS and instead of this 1 milliamp I will have IS okay so here I will have VS and there I will have IS the reason I did that is to just show that this V1 will be some linear combination of VS and IS it will be some number times VS plus some other number times IS okay and if you have many many sources it will always be like this it will be in a linear combination form and it is not only V1 you take the voltage or current in any part of the circuit it will be a linear combination of all the independent sources applied to the circuit okay so in this particular case we will get V1 to be 0.6 times VS plus 2.4 kilo ohms times IS okay so if you substitute VS equal to 10 volt and IS equal to 1 milliamp you will get the total to be 8.4 volts okay so hopefully this is clear both how to solve circuits like this is a very simple circuit and also the general idea of superposition okay so we solve the circuit with all the sources in one shot and then we did it with one source by one source setting all the other sources to 0 and we got the same answer as we should have got as we expected from a linear circuit and we also showed here that any voltage or current will appear as a linear combination of all the independent sources okay so that's what superposition is so that's how we solve for it okay so now in this case we had only two sources if we had more than two if we had three sources we could have taken two sources at a time and another source or we could have taken one source at a time three times and so on okay so now let's get back to what we were trying to do we have a network N with many independent sources and linear elements linear elements means registers and linear control sources and the only condition is that all the controlling quantities are inside this network N okay and there are two terminals that are coming out that are visible to the outside world one and one prime and I connect a current source ix from one to one prime okay so to find this voltage across one one prime let me call this vx what I can do is many things I can solve the entire circuit but here I am trying to do a particular I am trying to prove a particular circuit theorem so I will do it in this way I will take Ix alone okay and find vx that is when I say Ix alone all the independent sources inside of set to 0 and I will set Ix to 0 and use only the independent sources inside I activate them and find this vx and then I can add up the two results and find the total value okay so let me just do that let me draw the picture corresponding to that and we can end this lecture and continue from here in the next lecture so first what I will do is so inside the independent sources will be active and I will set Ix to 0 okay so when I set Ix to 0 what happens this is an open circuit okay and in the next case this Ix will be active and this independent sources any number of independent sources inside will all be set to 0 okay and this is known as nulling the circuit so I have nulled the network in that means any independent sources I have set to 0 but Ix is active okay so I will get let's say I call this vx1 when Ix is inactive and this vx2 when Ix is active and then I add up the two values of vx to get the final value okay so this will continue in the next lecture in fact you can take this also as an exercise and find the general form of the solution that you get for vx1 and in particular for vx2 okay what will be the form of vx2 what will it be what will be its dependence on Ix okay and we will take it from there in the next lecture if there are any questions regarding whatever we discussed today or anything else please let me know I will clear those and end the lecture. So there are several questions one is asking is pushing of voltage sources through a node a new theorem I am not sure what it is what is meant by this I mean certainly I didn't invent this it's a well known result okay and relatively obvious result that is sometimes useful and the next thing is about current division now current division is the counter part of voltage division I will show it with two resistors R1 and R2 clearly if you have two parallel resistors R1 R2 the voltage across them will be the same let me call it vx and the current through this will be vx by R1 here it will be vx by R2 and by applying KCL at this node we know that vx by R1 plus vx by R2 equals let me call this is okay so vx will be is divided by 1 by R1 plus 1 by R2 which is is times R1 R2 by R1 plus R2 I could have written down this directly by noticing that R1 and R2 are in parallel so is times R1 parallel R2 is the voltage vx now if I look at any particular current let's say current flowing through R1 that will be equal to vx divided by R1 okay so this current I R1 is vx divided by R1 which is equal to is and R1 cancels here R2 by R1 plus R2 and this is analogous to the voltage divided formula if I have vs the voltage across this is vs R2 by R1 plus R2 okay now it is some ratio of resistors in both cases in case of current division the current through R1 will have R2 in the numerator and in case of voltage division the voltage across R2 will have R2 in the numerator that's all okay now the next question is actually the question is not clear I guess the question is what to do with dependent sources when you apply superposition you do not do anything you certainly do not set dependent sources to 0 dependent sources will have whatever dependence they have okay for instance a voltage controlled voltage source will be dependent on some voltage vx okay and let's say it is some k times vx you don't change anything here you simply do the analysis as usual okay so if this is vx that is k times vx okay and someone asked for the slide showing superposition so it's here all that is saying is the source vector which has many elements can be thought of as sum of one element at a time I1 here I2 here and V1 there and that could have been more things I have not shown it and you see that this basically comes from linearity right so if you have these vectors you can show the test sum of this one and that one and that one and G inverse times this is G inverse times that plus G inverse times that plus G inverse times that and you can interpret each of them as solution when only one of the sources is active okay so I hope that clears up all of those things then if you have any other questions please do feel free to raise them in the forum or in the next lecture okay thanks for attending I will see you on Tuesday.