Lecture 01: Magnetic Circuit and Transformer

Welcome to this course on Electrical Machines - I, where we will discuss about transformers

and DC machines. These are the two major topics. And we will begin withtransformers.

Transformersas we will see will generally consist of a magnetic core material made of

subtitle, over which there will be at least two coils wound. Andone of the coil will be

energized withAC source of known frequency and in the other coil will getvoltage of same

frequency, but at different levels. So, that is primarily the job of a transformer is,

that is suppose you have a 200 volt 50 hertz supply, you require 400 volt 50 hertz supply,

then I will use a transformer to change the level of the voltage from 200 to 400 volt

keeping the frequency constant. It can be similarly a step down situation

where you have a say 400 volt you want to step it down to 100 volt level then use a

transformer. Transformer is a static device its efficiency is very large. In case of power

transformer efficiency could be as large as 99 percent unlike a rotating machine efficiency

which may be 80 percent, 85 percent very good efficiency because there is no rotating parts

in it. There will be of course, losses which accounts

for that 1 percent or 2 percent loss in power, that is there because of conductors will carry

current it will have a high square or loss. And also core we will see when it is having

a time varying flux then alsothere will be heat loss inside the core of the material.

Anyway, those things we will discuss in detail. But it looks like then we have to deal with

this situations, that issuppose you have a magnetic core of this kind, ok. And it has

got a thickness, like this in 3 dimension I am trying to draw, ok andit

will be like this. So, this is called it it is a made of soft iron and I have drawn it

right now as a solid iron block. We will see what is to be donebecause solid irons are

not used. Butthis is the structure of the iron andover

which there will be coils wound like this, ok. When there is a single coil and there

is a core like this, core material then if this coil carries current it will produce

flux inside the core. So, we start with magnetic circuit, ok. We

just review that because that will be essential in understanding the. ah

. Magnetic circuit

because there is a magnetic material, there will be coils wound over it and the coils

are supposed to carry current therefore, they are going to produce flux in the core. And

we know thatifthis is the direction of the current, suppose DC current first then the

direction of the current will be like this in the coil, I. Suppose we have connected

DC source this is the current. Then what happens, in the core there will be a flux produced

phi. Deduction of the flux will be as you knowyou wrap your fingers around the coils

and thumb will give you the direction of the flux produced.

Of course, if it is a constant DC current whatever flux will be produced inside the

core that too will have constant values. And how to estimate that flux? You know we apply

the ampere circuital law to find out the flux produced in the core. For example, you seeif

if I draw the sectional view of the core, like this, then if you draw the sectional

view the conductors can be shown to be like this, this is the coil sections.

How many coils sections you will see? As is the number of turns of this one. And the direction

of the current as showncan beshown by cross and dot like this this will be cross current,

cross and these will be dot, is it not. Then,the fluxinside the core what do we do

is we take a mean path of the flux which I am showing by dotted lines. This is the flux,

and this red one is the length of the mean path, length of mean path of the flux flux

that length let me call l. Ifthese are N then ampere circuital laws says that integral H

dot dl over a closed path is equal to current enclosed, is it. But in this casethe direction

of H and l are same. Therefore, what will come out to be H into

l that dot product because the direction of H and l as I move they are same, direction

of H, direction of B, direction of phi they are all same and also length is like that,

H dot dl means H into l should be equal to the current enclosed. How much current is

enclosed by this path? N into I, N into I. We know this, so I will not spend much of

this time So, this is NI by l. Its unit is ampere turns

per meter this will be the case. Once I know H then I will calculate B, B will be equal

to mu 0 mu r into H, where mu are is called the relative permeability of the core and

if the relationship between B and H is linear, then mu r is a constant value andso this will

be equal to mu 0 mu r and for H I can write NI by l.

Then thethen I go here, then the flux produced which is in vapor will be equal to B into

the cross sectional area A. What is this A? A is this cross sectional area that is this,

this, this, this, through which flux will be flowing. So, this is A, ok.

So, this cross sectional area is perpendicular to the flux at any sections therefore,flux

in vapor will be B into A, and B already I have calculated, so I substitute that mu 0

mu r NI by l, this is H into area. Now, if you seethis this can be written as NI I am

sorry, if you see this it will be NI and bring all the other things in the denominator which

will become 1 by mu 0 mu r l by A. Why I have written in this fashion is this,

that this is the cause I can identify this, this is the cause and phi is the effect, is

it. I have supplied with mmf and this magnetic circuit returns c way flux phi. And how to

calculate that flux? mmf, this is called mmf, and you this is called reluctance. And that

is why we name magnetic circuit. In case of electrical circuit, I is equal

to EMF by resistance, and if you recall EMF is EMF voltage resistance is rho l by A. So,

there is a striking similarity and that is why it is called reluctance. Reluctancelimits

the value of the flux when you apply an mmf. Although, the the mmf and the flux in electrical

circuit as you know the circuit is like this resistance and this is your EMF say E E by

R. And this is the current, but in magnetic circuit this mmf and this flux wherever it

is flowing they are totally decoupled, I mean it is not that flux is in the conductor.

But, nonethelessthis equation prompts us tosimplify the matter and say that, asso far as calculation

of mmf is concerned you do it like this NI and connect it by a relzctance which is shown

by a curly letter R and here you show the flux. Although, in thisdiagram I show there

as if connected, but they are not in practice, but only prompted by thisrelationship corresponding

in the existing in the electrical circuit, it is better, that is why it is called magnetic

circuit. So, if mmf is known. I will first calculate,

what is the reluctance. Reluctance of the magnetic circuit per depends upon its geometry,

what is the length, what is the cross sectional area, cross sectional area is this one of

this, this is the cross sectional area then you calculate H and I by l, and then multiply

it withmu 0 mu r to get B, then multiply with a and get that one.

Therefore, in a simpleexcitation with DC current this value of the flux whatever you will get,

if you divide the mmf with reluctance that two will be constant and its direction will

be also fixed in this case it is clockwise, ok.

The reverse problem is also very simple. In that case I will say that I want to create

a certain amount of flux what should be the current necessary. So, I will calculate reluctance,

reluctance into flux will give me NI and if I know the number of turns I will simply divide

that NI with m to get the current necessary. Anyway, these areknown stuff andwe have a

magnetic circuit like this, ok. And I will presume that you know about it, so no question

of further telling about it . Now, we will go to next page, next page ah.

l thatkeyboard is useful. Let us go to next page.

Now, I will tell you that what happens if the same magnetic circuit that isthis one,

this point you listen very carefully and this is the most interesting thing, sorry, I am

sorry.Let me try to draw another thing here, that is the core I am drawing you must understand.

Why it is getting circle, anyway hm. So, suppose this is the core of the material

let me draw slowly, ok. Andsuppose you havethe let us consider a single coil same magnetic

circuit with N turns, but this time what I will do is this I will connect it to an AC

source this is AC source of known voltage.

For example, V 1 is equal to root 2 I am sorry will write it like this v 1 t is equal to

root 2 V 1, some say sin omega t or cos omega t I will write it applied voltage to this

coil. Now, if you pass a sinusoidal voltage across the coil, in case of DC circuit magnetic

circuit what I was telling when the current is constant then NI you calculate mmf divide

it by reluctance get the flux. But in AC circuit AC magnetic circuit the coil is connected

across a AC supply voltage of known frequency, rms value V 1 and supply frequency is f.

Then what I am telling, first I will tell this statement the flux in the core gets fixed

I mean no question of once the supply voltage and rms value is known, I mean we we do not

start telling that I will first calculate current, then calculate mmf, thendivide that

mmf with reluctance to get the flux, not like that; it will be the moment you apply a known

rms voltage at certain frequency f where omega equal to 2 phi f and AC voltage across the

coil. The level of flux which will be produced which

will be also time varying it is expected flux gets fixed. Now, what is the reason for that?

Reason for that is it is expected whatever current it draws that will be also sinusoidal

varying, because after all it is some sort of coil or inductance we have connected across

an AC supply, some alternating current it will draw andsince current value and magnitude

is changing with time the this phi 2 will be time dependent, sometimes it will be flowing

to clockwise sometimes in the anticlockwise direction sometimes it will be 0 when the

instantaneous value of the current will be 0 and so on. But the moment an alternating

flux is created inside the core of the coil, inside the core of this magnetic circuit between

these two points they are appears an AC induced voltage.

Suppose the number of turns of this coil is N or say N 1 single coil l 1. Then according

to Faraday if there is a coil ifthere is a time varying flux in the coil

then this coil itself become a seat of EMF. The

value of instantaneous value of which is some minus N 1 d phi d t about that sign it is

not necessarily so important, but what I am telling this is the induced voltage the flux

linking the coil is chaining with time and therefore, there will be induced voltage across

this coil. In this case this flux is created by the current

carried by the coil itself, but it does not matter, Faraday says, if there is a change

of flux rate of change of flux exists linking a coil it is time bearing then between these

two points it will become a seat of EMF. Negative sign is the Langer's law it tells that the

polarity of this induced voltage will be such that it will try to oppose the very cause

for which it is due. For example, this coil can be modeled as here

is your AC supply V 1 t I have applied. Then what I am telling across this coil if I neglect

the resistance of the coil there appears another source here and that induced voltage is E

1, and the polarity of this voltage will be such that if this side is d phi dt is positive

then it will try to oppose the cause, so its polarity will be like this, ok. Or in other

words what I am telling, so these these these thing can be now be modeled like these applied

voltage and there is another induced EMF. In case ofDC magnetic circuit DC 1 will not

be there, there is applied voltage which is constant. Of course, current will be limited

by the resistance of the circuit V by r that is why current gets fixed I. But here what

I am telling the induced voltage in the coil will be same as the applied voltage, only

thing about the polarity its polarity will be such that it will try to oppose the cause

very cause that is the flux it will try to oppose it. What was the reason for flux existing

and increasing? This current it was increasing in the positive direction. So, it will try

to limit reduce that value of the current that was the reason.

Anyway, after this I can say that in this loop k V l equation is to be satisfied. Therefore,

applied rms voltage must be equal tothe rms voltage of V 1 what else, is it. Now, you

see this flux will be some phi max, say let us forget about this cos omega d sin suppose

I start with these one phi max equal to sin omega t.

Why I am assuming this? Because the circuit cannot, but draw sinusoidally varying current

let call phi is equal to phi max sin omega t. If that be the case then the rms value

of the then the induced voltage in the coil E 1 will be equal to N 1 d phi 1 dt, ok.

If you differentiate these this will become equal to N 1 phi max omega sum cos omega t.

This will the value of the induced voltage, is it. So, what will be the rms value of that

voltage? It will be the peak value of the voltage phi max, omega by root 2. For omega

if you substitute 2 phi f, so it will become root 2 phi f phi max into N 1.

See I am not so much bothered about this minus 1, I should not struggle, my intention here

is to calculate what is the rms value of the induced voltage, ok. Differentiate it, peak

value you get divide it by root 2 and that will give you rms voltage.

What I am telling is in the circuit k V l equation must be satisfied or is vanishingly

small. So, applied rms voltage must be equal to the induced rms voltage it cannot be other

than that. Therefore, if the applied rms voltage is V 1 it must be equal to root 2 phi f phi

max into N 1 is it it has to be f nothing other than that.

So, what I have told you that in case of AC E same magnetic circuit which I considered

with DC excitation, if you connect it to an external voltage source which is alternating

in nature sinusoidally and the rms value of that applied voltage is known current whatever

flows has to be sinusoidal therefore, phi created inside the core of this magnetic material

2 will be time dependent and vary sinusoidally. And if this flux vary sinusoidally then Faraday

tells us that across this two points this two points there will be induced voltage.

What is the magnitude of this induced voltage rms value? That magnitude of this rms voltage

is root 2 phi f phi max into N 1. Applied rms voltage is known V 1, induced rms voltage

is known capital E 1, and these two must be same because k V l is to be satisfied in the

primary. If that be the case in this equation what are the things I know, V 1 is known,

supply frequency is known, N 1 is known number of turns, and your phi max is equal to V 1

by root 2 phi f into N 1. This is the crucial thing.

As I told you see mind you in case of alternating magnetic circuit if you apply an AC voltage

the level of flux phi max is decided, decided by whom? By the supply voltage rms value by

the supply frequency and number of turns. So, is it not? Something different from DC

circuit; DC circuit you apply some known current, DC current in the circuit calculate mmf divide

by reluctance, get the flux. But here oh it is somewhat interesting.

The moment you connectan AC voltage V 1 and f I can tell you the flux is sinusoidally

varying. What is the maximum value of that flux? It is fixed it is

this one. Therefore,this phi max how much will be produced inside the core is decided.

I do not have any control over it. In fact, I will go a step ahead I will tell you.

If this core material it has got a permeability of mu or one you replace this core material

with another magnetic material having relative permeability mu r 2 then also phi max is V

1 by these one. So, no matter what is the kind of magnetic material it is good magnetic

material, back bad magnetic material, the moment supply rms voltage and supply frequency

are known phi max in the core is decided. A very crucial point to go ahead with the

concept of transformer. We will continue with that.

Thank you.