[Music] hello friends so in the previous session we have just started with the interaction and how the course flow of the AC machinery fundamentals that we are going to cover so today is going to be the first lecture session where we are going to cover some simple concepts all right so today let us start with a simple loop in a uniform magnetic field you already know from the DC machines that the best example or the simplest example for a generating alternating voltage is a simple loop of wire through a uniform magnetic field but however in case of AC machines as this is not entirely accurate because here we are assuming the field to be constant right it is a constant field which is existing then you are rotating a the loop of wire through that but in AC machines neither is this value constant nor it is stationary okay so that we will find out when we deal with rotating magnetic fields but however that Vantage here is that yep the equations that we derived about the induced voltage and torque etcetera hold holds good it's just some construction features that we have to add extra so let's start with this particular session we did have a simple equation like so this is the coil we have seen it a lot of times in DC machines alright so if you see it from this angle here this particular coil what you are going to see you are going to see something like this right we are going to see this conductor here a B and you are going to see can't this conductor here which is C D ok and if it is rotating in this particular direction like this the velocity vectors we have already discussed it would be something like this alright so now we have just going little bit behind in time so that we get an angle in between them so that is this particular down so let us take this instant in times where this is a little bit tilted now ok so I have drawn a neat diagram here ok so now let us mark some angles first of all ok so this is the direction of V from north to south so this angle let us call it as theta a B okay and from this particular angle between the and the CD vector this is angle CD so you do not confuse get confused here we will be drawing individual segments all right so this is the construction of the coil right so it has a length L and it has a distance from the center you can call it is not the entire width it is half of the width which is the length R okay so this is basically the construction of the coil and let us start developing an equation for the total EMF e total which is got across this coin so let's start with segment a B so we will start with segment a B and the equation that we are going to use is induced is equal to V cross B remember we used this to find the direction right here we have to add another term dot L through the EMF which is obtained out of field theory for a conductor anyway it is the V cross B dot L right so let us and the direction of L is the direction in which you are moving alright so if you are going from A to B so direction of L will be like this we are coming from C to D the direction of L will be like so it's not a confusing statement here so let's take the first segment a B here so the first segment a-b let us take the first segment a B here like this okay so V we can see that it is having a direction like this this is the VAB right so this is VAB okay and what is the direction of B it is like this B and this angle is Theta a B okay so V cross B okay we if I cross would be like this so what will I get i get a clockwise movement and that means it is a into the paper okay now this particular direction which is into the paper like this okay to get here it is into the paper and L is also having the some same direction right L is also having the same direction so here E is equal to V cross B dot L right so dot means a cos theta component is going to come okay so you know that sine cos if I put 0 and 90 degree this is 0 this is 1 this is 1 and this is 0 so for segment a B the L vector and the V cross B vector are in the same direction see they are the same direction right so this would be equal to this angle will be cos 0 V BL and cos 0 therefore here you will get BL V sine theta a B now the sine theta is coming due to this V cross B all right now let's find the EMF induced for the segment first half of the segment BC so BC you can divide it to two right here so BC you can divide into two so let us find it for this particular segment okay so let us find for the first half of this segment BC okay so here so you can clearly see because it is going into a counter clockwise direction we are rotating it in a counter clockwise direction so V will be in this particular direction so this would be the direction of V and this is the direction of B okay so V cross B will be into the page we cross P will be into the page that I have to have a dot product with the L okay now what is the L direction the L is from like this only at least in the same L is in the direction of the plane of this paper okay so it is from this direction so L has a direction like this and this has a into the page direction that means the angle between these will be 90 degree okay so a 90 degree means the cost component will be 0 that means e is equal to V cross B dot L for the BC segment will be equal to zero that means there is no EMF induced in this segment and the third part is the second half of the segment BC that means this particular component here okay now here you can see that it is going to have a downward direction that it is going to have a downward direction so we will be like this there and B of course will be like this so V cross B will be having a counterclockwise direction and counter clockwise if you move a screw it will come outside right so this is the direction of V cross B that you have to dot with L and L Direction is the same direction legs you are moving from B you see right so L will have the same direction these both are having a 90 degree but this is coming out of the page and this is inside in the plane of the page okay so it's like this so like this is coming towards you this it's coming towards you and this is the direction of L right so this angle is 90 degree so this also because 90 degree means the cost component will be zero so here also a will be equal to V cross B dot L will be equal to zero because the cost 90 equals to zero here okay cost 9208 to take the final segment which is CD okay CD is here that is the most you that is an important segment also so the segment CD let us take the segment CD now okay so what was the velocity direction here sorry yeah so the velocity of CDs in this direction and the bees in this direction okay so if you take week this is Theta CD so V cross B means it will be a counter clockwise direction so V cross B is going to be outward so that you have to dot with the value of L okay and what is the direction of L here so it is the same direction as you are moving so same outward direction so the angle between these two components are 0 cos 0 equal to 1 that means this value V cross B dot L will be equal to b lv sine theta of CD okay and that means let us mark some polarities now okay so this is into the paper right so let us take all that this voltage because it is having at this particular direction into the paper we can call that voltage as EBA like a generator writes coming out power is coming out of here and here we can call it as EDC okay we can call it as EDC that means the total voltage e total will be equal to e VA plus EB C plus E DC okay and this is already you have found the value to be equal to 0 and this is VL v sine theta of a B and this is VL V sine theta of C D okay now we are having two angles here right theta a B and theta C D now if you look at these two angles here so this is V and B and this is Theta a B you can see that you can use a simple trigonometric identity to find that the theta a B is nothing but 180 minus theta C D okay and you know that de sine theta of sine 180 minus theta is nothing but equal to sine theta that means sine theta a B will be nothing but sine theta C that means these two values will be same so e total will be equal to 2b LV sine theta okay so if you draw a graph for this okay what will you get you will get a sine wave okay so this is theta 0 PI by 2 pi 3 PI by 2 and 2 pi so the EMF induced will be having this particular value now there is another way to derive the same expression alright so let us see what is the another way to derive the same expression which will bring us closer to the equation that we like for a summation you know that in a real machine the machine is going to rotate at a speed Omega right that means this theta the angular displacement will be nothing but the speed angular speed multiplied by the time and also you can relate that V will be equal to R Omega and that is why you see the definition of our was from half of the distance all right so V is equal to R Omega and theta is equal to Omega T so e total de induced will be equal to to be L into R Omega into sine Omega T all right now you take the coil here okay this was L length was L okay and this distance was R okay so what is the area of the coil the area of the coil will be equal to 2 into R - R into L right so 2 R into L so you can clearly see here you are having a 2 R and L so I can combine that two to break the area of the coil so this will be equal to B into a ok into Omega into sine omega-t okay so e total will be equal to B into a into Omega into sine omega-t okay now you know that this when the north and south is like this this is the position where you are having the maximum flux which is cutting the coil okay now what is the value of that maximum flux that maximum value of flux will be equal to B into a because here the entire area is getting exposed to the flux density so here will be the area where you are having the maximum flux therefore I can define something as Phi max is equal to B into a that means e induced I'm just replacing e total by induced will be equal to Phi max into Omega into sine Omega T now why this expression is important is because it tells us basically three things okay so first one is that the EMF induced in a machine okay it depends upon the flux inside the machine flux inside the machine which is represented by this Phi max second it will per it depends upon the speed of rotation it depends upon the speed of rotation which is Omega okay and third thing it actually depends upon the construction of the Machine constructional features of the machine which means that number of loops turns conductors etcetera now that has not come into this expression because we use the single loop generator but when we actually derive for the AC machine after sometime maybe after discussing about the armature flux distribution that time we can see what is the actual expression but this expression is good enough to just get into the basic concepts of AC machine so I hope you have understood today's session please excuse me for the voice I have told you a little bit of throat problem is still there so if you liked this video please like share and subscribe my channel and I'll see you in the next video thank you