Welcome to the course on Electrical Machines II, in this course we will primarily focus on three-phase induction motors starting from its basic operations and then single phase induction motors and then synchronous motors. Anyway, before going to three-phase induction motors it is essential to have some knowledge on windings on electrical machines, that we will be doing before I start three-phase induction motors. And before that of course, we will discuss about the basic principles of any rotating machines. Basic operating principles of any rotating machines, we will see that there is a some common underlying principles that govern the operations of all the kinds of standard electrical motors. Be it DC motors, be it three-phase induction motors, be it single-phase or three-phase synchronous motors. So, the study we will center around, the steady-state performance analysis of those machines in somewhat detail. Before of course, I thought that essentially what you will see that when we will study any type of electrical motors, we will try to draw its equivalent circuit and then analyze that circuit to predict the performance of that particular machine. Since the course is on electrical machines II it is presumed that you have undergone a course on transformers and DC machines, all though DC machines are somewhat difficult at its constructional level. But, nonetheless generally the electrical machine course centers around transformer and DC machines. If time permits I will also touch upon some interesting aspects of DC machines as well. Therefore the course begins with electrical machines II and as I told you equivalent circuit is one of them. So, let us not leave a transformer in its totality. Because the equivalent circuit of the transformer you must be remembering, we will start from that; assuming you have undergone a transformer course, but I will try to arrive at the equivalent circuit of a transformer in a from the viewpoint of circuit analysis. And before that, if I want to derive the equivalent circuit of transformers, which you must be knowing, It is drawn like this and this transformer lets assume it has got no core loss. So, I have not drawn parallel resistance. And it’s this is how it is drawn- r1, xl1, this is the what is known as magnetizing inductance jXm and this is a2 r2 and this is a2 xl2 where xl1 and xl2 are called leakage reactance of the transformer; r 1 and r2 are winding resistances of primary and secondary side and a is the ratio of number of turns; N1 by N2. Where, primary side is that side where source has been connected that is here V1, ok. This equivalent circuit must be familiar to you and it has been derived based on some physical considerations of the leakage flux, mutual flux and so on. And this impedances r2 is the actual secondary winding resistance, xl2 is the actual secondary leakage reactance. These are to be multiplied by a2 in order to draw the equivalent circuit referred to what is known as primary side. Similarly, the equivalent circuit can be drawn referred to the secondary side as well, but now as I told you this equivalent circuit we will try to arrive at by purely from the consideration of circuit analysis; after all a transformer is nothing but a collection of two coils which are magnetically coupled. So, this is the transformer core, this is the primary coil say, this is the secondary coil. And, these two coils have got their separate identities with terminals brought out and if you look at this configuration; you will immediately notice that these two coils have mutual coupling between them. Because, if one coil passes current, flux will be created, it is going to link the others and thereby, inducing a voltage in the other coil; the number of turns of the primary the N1 and that of the secondary is N2, it is like this. Now, a coil has got also a self inductance and how to find it out. Suppose you concentrate on one coil; this part I will do rather quickly. Suppose this is a coil and I want to find out its inductance. Now, to find out the inductance of a coil we will start from the fundamental definition of inductance; inductance of a coil is flux linkage with the coil per ampere per 1 ampere. If it is carrying current I1, then divided by I1. Now, what is flux linkage? Flux linkage is if you pass a current I1 through coil 1, it is going to produce a flux like this, is PHI. So, this N1 into PHI divided by I1 is called the inductance of the coil. Now, in a transformer what happens is this, the flux which will be created when a coil carries a current. If I draw this portion in a bigger way, it will be like this, this is the code, this is the turns. There will be some flux which will be confined to the core and that I call PHIm mutual flux and there will be some flux lines which will complete their path through the air gap and this is called leakage flux PHIl. But the definition of inductance says that it is the total flux linkage per unit ampere. So, if I say that I have energized coil 1 by current I1 and total flux created by it, I will write it like this PHIT1, is the total flux. Then this total flux will comprise can, in fact, it constitutes of two pieces: one is the mutual flux PHIm1 plus PHIl1, the leakage flux. So, if it is coil 1, it is carrying I1 current. So, PHIm1 and PHIl1 together is your PHIT1. This is the total flux linkage, this one. Therefore, by definition inductance, self inductance of the coil will be flux linkage with the coils, So, N1 into PHIm1 plus leakage flux PHIl1 divided by I1 and this can be broken up into two terms as N1PHIm1 by I1 plus N1PHIl1 by I1, is it not. So, this is the thing. Now, this part, which is not confined to the core, that is the iron material of the arrangement is called the leakage inductance, this is also an inductance because N PHI by I. So, this is written as N1 PHIm1 by I1 plus this I will write it as Ll1, that is a leakage inductance of the coil. So, this is what this Ll1 means. Therefore, self inductance of a coil if you want to find out, you have to consider the total flux which consists of two parts. That is PHIm1, the flux which will be within the magnetic material and PHIl1 is the flux which crosses or closes its path largely through the air gap, ok. So, this is the leakage flux. So, I think we have understood what self inductance means so, this is page 1. Now, coming back to the transformer we have to we have now got two coils instead of a single coil. So, once again I mean referring to this coil, this is one coil, this is another coil. So, suppose we say that let me re-draw it this is better. This is the core and these are your coils. See, while drawing the coil I have drawn it rather in detail showing its sense of the winding I have not drawn it like this, never draw like that, and better draw in this fashion. And this is the another coil this has got number of turns N2, this has got number of turns N1 and this is the core. Now, let us imagine I have only energized the primary coil with the DC current I1, ok. then as I have told you there will be flux created which will have two components PHIm1 and there will be flux lines linking only the this winding that is called leakage flux PHIl1 and based on that I have defined the in the previous page L1 to be N1 PHIm1 by I1 plus Ll1. You know this is the self inductance of the coil. Now what is the mutual inductance when there are two or more coils sharing a common magnetic circuit, then if any of the coil carries current, there is going to be flux linkage with the other coils. So, in this case is very simple, two coils are there, one coil is carrying current and it is this mutual flux PHIm1 that is going to link the second coil N2, PHIl1 after all cannot link the second coil. And this is therefore, I can define mutual inductance, like this M21 as flux linkage with the second coil, with N2 divided by 1 A flowing through coil 1. Once again, the unit of M will be same as L, Henry that is flux linkage, Weber turns per ampere. So, this is flux linkage with N2 divided by I1, second coil I have not touched, nothing is flowing through the second coil. So, this is called the mutual inductance of the second coil, the second suffix indicates that the first coil is carrying current. So, this will be flux linkage with N2 by I1 and this is; obviously, the flux linkage with N2 because of I1 is nothing, but N2 PHIm1 divided by I1, is it not, this will be the value of M21. So, this is the mutual inductance when coil 1 carries current, coil 2 nothing is flowing and you get M21 and L21, these are the two things you get. Now, let us say I am once again redrawing the circuit and I am telling that this time what we will do is this I will pass some current through the second coil. Let it be constant current DC current I2 and N1 and this coil 1, keep it open, nothing is connected this is N2 and this is N1 and you pass a current like this. If you pass a current in this direction as you can see the flux produced you can show the direction of the current if you draw in this particular way as I told you and decide about the direction of the flux. So, flux will be once again clockwise and it will be like this, isn’t. And the total flux created by secondary coils second coil, I will write it as PHI2T and in the same way can be broken up into two terms. One is the mutual flux PHIm2 and this time I will write 2 because coil 2 is carrying current and it has created the flux. So, PHIm2 is the mutual flux which is confined to the core and there will be a component of flux which will completes its path largely through the air gap and that will be termed as PHIl2 in the same way as we have done in the case of the first coil. So, same explanation goes this is the total flux created by coil two. It is carrying a current I2 and the total flux is nothing, but sum of mutual flux plus the leakage flux. So, this is this therefore, the inductance, self-inductance of the second coil will be once again the flux linkage with the second coil into total flux linkage PHI2T per unit ampere that is I2. Self inductance of this second coil. And this can be after you put these two things here it can be broken up into N2 PHIm2 divided by I2 plus N2 PHIl2 divided by I2. So, the self inductance of the second coil will be this term N2 PHIm2 by I2 and this term I will call leakage inductance of the second coil Ll2. This is how these things comes, now this is this. So, L2 is this one. So, let us go to third page. So, recall that now I have not going to draw, but what results I have got till now is L1 is equal to N1 PHIm1 by I1 plus Ll2 and L2, self inductance of the second coil is N2 PHIm2 by I2 plus Ll2. This is sorry 1, leakage inductance 1. These are the two things. As I told you inductance mutual inductance in this case M21 is nothing, but N2 PHIm1 by I1, similarly I can define the mutual inductance when second coil is carrying current. So, I will write it as M12, this is the flux linkage with the first coil N1 and the mutual flux this time is PHIm2 and divided by per unit ampere flowing in the second coil divided by I2, this will be the thing. Fine. Now, let us concentrate on these two equations, here this two coils mind you they are sharing one common magnetic circuit, ok. So, what will be the flux, we have done magnetic circuit analysis. If a coil is energized how to calculate the flux inside the core? If I assume it is a linear magnetic circuit PHIm1 will be nothing, but mmf divided by reluctance. In this case when coil 1 is energized, mmf is N1 into I1 and divided by the reluctance curly R; reluctance of what? Reluctance of this flux path which completes its path in the code; so, this curly R is the reluctance. Similarly PHIm2 will be nothing but mmf, in this case I am only energizing second coil, in the second row. So, this is N2 I2 by the reluctance. So, this is the thing where reluctance I am assuming it to be constant and assuming linear magnetic circuit. So, I can always calculate the reluctance of the magnetic circuit, you know it is like one by µ0 µl lmean by A (cross section). I am not going into all those details, but I know that flux is mmf by reluctance; mmf is n into I divided by reluctance. So, I have got this; now let me play with these two equations. what I will do is this M21, I will, it is already obtained as N2 PHIm1 by I1 .This is the thing, isn’t? N2 PHIm1 by I 1 and M12 is this one. Now what I will do is this I can write it as N2 by I1 and PHIm1 from this I will plug in this PHIm1 here N2 by I1 is N1 I1 by reluctance, where I1 goes. And it is nothing, but N1 N2 by reluctance. Similarly M12 if you calculate, it is from this it is N1 PHIm2 by I2, but PHIm2 is nothing, but this. So, put it here. So, you will be getting N1 by I2 into PHIm2 is N2 I2 by reluctance, I2 goes and it is equal to N1 N2 by reluctance. So, what is the conclusion M12 is equal to M21. This is a very important thing, there is no reason to distinguish between these two, provided these two coils share the same magnetic circuit. So, and therefore, the mutual inductance can be represented by a single number M, not M12, no suffix I will write. Of course, for self and mutual inductance we have to use L1 and L2 therefore, what we have done till now in this unit as I have told you my target goal will be to derive the equivalent circuit of a transformer. Looking at it as in terms of self and mutual inductances and try to get this equivalent circuit where I have assumed that the transformer is having no core losses, that is hysteresis and eddy current losses. Whether, such an equivalent circuit can be arrived at using the circuit concept and using the concept of inductances; so more in the next unit. Thank you.