Electrical Machines - I

Prof. Tapas Kumar Bhattacharya

Department of Electrical Engineering
Indian Institute of Technology, Kharagpur

Lecture - 09
Modelling of Practical Transformer - II

(Refer Slide Time: 00:26)

Welcome to the 9th lecture and you recall we were gradually going towards a practical
transformer. And in my last class, this was the slide, where this is a practical transformer.
Here there is no 𝑋𝑚 connected; you must understand that this is a practical transformer in
the laboratory ok; two secondary, two primary, there was nothing like 𝑋𝑚 connected there.

But it was our thought process, which led us to believe that this whatever is happening to
a practical transformer, what is happening with S opened, current drawn from the supply
is finite, magnetizing current 𝐼𝑚 and when S closed, it must be 𝐼𝑚 + 𝐼2′ , because net MMF
in the circuit has to be 𝑁1 𝐼𝑚 , why? Because flux in the core is decided by supply voltage
and frequency, nobody has any say on that.

But if you are trying to disturb the secondary, I mean if you are trying to pass some current
through the secondary, primary cannot be a mere spectator to this happening. It will react
immediately by drawing an extra current 𝐼2′ , and that 𝐼2′ cannot be of any magnitude; it has
 𝐼
to be 𝐼2′ = 𝑎2, so that 𝑁1 𝐼2′ = 𝑁2 𝐼2 , so that net MMF once again is 𝑁1 𝐼𝑚 and this is what we

have discussed last class.

So, this is a reactance, this is an ideal transformer and it correctly models it. So, 𝑉1, 𝐼𝑚 I
have drawn, then I am completing the phasor diagram once again. And suppose, this is 𝑉1
to the ideal transformer applied voltage is 𝑉1, so this is 𝐸1 , this is 𝐸2 and this is 𝑉2; 𝑉2 =
𝐸2 .

So, depending upon the turns ratios your, so this 𝑉1 is nothing but 𝐸1 , they still remains
nothing in between, and your this will be your 𝑉2, which is same as 𝐸2 . And this is suppose
the load power factor decide what is the current here 𝐼2 , then current drawn from the
primary will be 𝐼2′ , 𝐼2 we have got, get 𝐼2 .

And then this 𝐼2′ + 𝐼𝑚 if you add, you will get primary current 𝐼1 , understood. This is how
magnetizing current can be taken into account of a it is somewhat a practical transformer,
still not a full practical transformer. I have neglected, so many other things till now, only
magnetizing current, I have been incorporated and for that an external element 𝑗𝑋𝑚 is to
be selected.

If you like you put 𝑗𝑋𝑚1; side 1 ok, because we have seen parameter value changes from
side this side to that side and so on. Therefore, what will be the equivalent circuit looking
from the primary side, it will be 𝑉1, here is a reactance; no windings this one 𝑗𝑋𝑚 and here
will be the impedance 𝑎2 𝑍2 , you know this will be the equivalent circuit refer to the
primary side 𝑗𝑋𝑚1 .

So, whatever impedance, voltage, this that are there and this current I will show as 𝐼2′
reflected current and this current I will show as magnetizing current and this current is
your 𝐼1 . Mind you I am not drawing, but these are dots that is very important with respect
to this. So, this is the equivalent circuit refer to the primary side.

What will be the equivalent circuit refer to the secondary side, refer to source side; refer
to primary or source side, primary side. And refer to load side or secondary side, it will be
secondary voltage remain secondary voltage; this will be 𝑍2 , secondary things I should not
disturbed, they are already there 𝑍2 , this voltage is 𝑉2, this current is 𝐼2 , I will show it.
 ′
And this fellow the transfer of impedance from this to that side will be 𝑋𝑚1 , what is X m1
dashed

′
𝑋𝑚1
𝑋𝑚1 =
𝑎2

what is this 𝑉2?

𝑉1
𝑉2 =
𝑎

So, this is the equivalent circuit refer to the secondary side. So, you solve this you write
′
𝐼𝑚 reflected current. And then you will what current you will get, I will get 𝐼1′ .

So, you either solve this circuit get everything, because if this somewhat practical
transformer, solve this circuit get 𝐼2′ and then predict what will be 𝐼2 . So, people always
refer to work on a equivalent circuit instead of drawing some coupled coils, then
individually calculated. We can do that, but this is a better way of doing things.

(Refer Slide Time: 07:36)

After I have done this, our next reality which is present in a practical transformer is let us
assume, I will assume winding resistance. Till now, I have neglected winding resistances
𝑟 not 0, that is 𝑟1 is present, 𝑟2 is present that is the what is 𝑟1, this winding; one side, side
𝑙
1 resistance and this winding has also got 𝑟2 , because resistance is after all 𝜌 𝑎.
So, there is some resistivity of this conducting material, some cross sectional area, so many
turns are there, so many length. Of course, the resistance will be pretty small, it is made of
very good material for example, copper. But none the less there will be some finite
resistances of primary side and secondary side.

And it looks like, this is resistance which is of course, distributed can be considered to be
lumped and I can show it like this. Similarly for this coil, this resistance, I will considered
it to be lumped and represent it like this, this is how I can represent.

Now, the question is how this 𝑟1, 𝑟2 should be shown in this transformer. In this
transformer, there was finite magnetizing current; no resistance, no leakage flux, fine and
this is 𝐼𝑚 magnetizing current taking care of by an 𝑋𝑚 .

So, now once again I will add some parameters to this circuit that is 𝑟1, 𝑟2 in appropriate
place that is very important, so that effect of 𝑟1 and 𝑟2 will be addressed by this model.
Question is where should I put 𝑟1, should I put it here, should I put it there, where that is
an interesting question.

This portion is ideal mind you therefore, if there is winding resistance, what it is going to
do. There will be whenever this coil will carry current your practical transformer, there
will be a voltage drop here in the resistance. Applied voltage minus this drop is going to
create flux, not this full voltage is going to produce your magnetizing current, because a
portion of the voltage will be dropped in 𝑟1, got the point.

Therefore, in this circuit I will spoil this circuit, do not mind; what I will do now if I add
something here, our previous thing will also get disturbed. So, what I am telling is this
what I will do now listen carefully, here I will draw the ideal transformer and I will put a
dotted box around it in order to indicate that and then these are the terminals of the ideal
transformer ok.

And then I am telling to this ideal transformer, this winding is not purely this thing. So, so
you are there will be resistances in series, I will connect it here; which is small resistance
nonetheless 𝑟1, it will come in series.

Now, the question is should I put that magnetizing current branch 𝑋𝑚 , before this or after
this, I must put it after this 𝑗𝑋𝑚 . Similarly here of course, there is no magnetizing branch,
𝑟2 is simply comes here. I have shown only one way I am showing not like this resistance;
because these resistances are small, it only indicates that a small resistance in series.
Therefore, you know this is this thing and here is your supply voltage, frequency 𝑓.

So, whatever current it supplies which is decided by it may be the load connected here,
ultimately some current is drawn and when that current flows through the winding, there
will be a voltage drop 𝑟1; and that voltage drop must be subtracted from your supply
voltage and the remaining voltage is responsible for creating flux and giving you the
transformer action, so that is why 𝑟1 should not be shown here, it must be after this.
Therefore, this is 𝑉1, this is ideal transformer, this is 𝑟2 .

Now, the moment 𝑟1 is present I must also distinguish between 𝐸1 and 𝑉1, there will be a
drop here in series. 𝑋𝑚 what do you think; its value will be low or high, its value will be
high you should not choose a magnetic material which requires very large magnetizing
current. See 𝑋𝑚 is what,

𝑉1
𝑋𝑚 ≈
𝐼𝑚

I should not choose a magnetic material which requires very high value of magnetizing
current, then 𝑋𝑚 will be low. Better and better material I use, which is not certainly ideal
its 𝜇𝑟 → ∞, may be 𝜇𝑟 = 5000 quite a large number.

So, 𝐼𝑚 will be small, 𝑉1 is fixed, so 𝑋𝑚 is general high that is why, I have written capital
letter and we so many turns, just to indicate that ok, 𝑟1 is small, small 𝑟2 . Therefore, the
𝑉
magnetizing current which will be flowing here, 𝐼𝑚 is not 𝑋 1 ; 𝑉1 minus this drop divided
𝑚

by 𝑋𝑚 .

And as you know, depending upon the degree of loading the magnitude of the current
drawn from the supply will change. Therefore, drop in this resistance, 𝑉1 minus this drop
is the voltage what is coming here across the ideal transformer. Therefore, the magnitude
of the voltage apply to this ideal transformer is will also we will not remain constant, as
we were thinking in case of ideal transformer apply 𝑉1. I was telling the level of flux 𝜑𝑚𝑎𝑥
𝑉1
is equal to applied voltage in an ideal transformer is equal to 𝜑𝑚𝑎𝑥 = and if applied
√2𝜋𝑓𝑁1

voltage and frequency is constant, 𝜑𝑚𝑎𝑥 gets decided.

But now I come to know ok, the applied voltage to this ideal transformer strictly speaking
will not remain constant because of the presence of 𝑟1, this is constant mind you; 𝑉1 is
constant no doubt, but 𝑉1 minus this drop is what is applied here, what is this drop, this
drop depend on the magnitude of 𝐼1 this your practical transformer is carrying and the
magnitude of 𝐼1 = (𝐼2′ + 𝐼𝑚 ), 𝐼2′ depend on 𝐼2 and depend on 𝑍2 .

So, by as you change 𝑍2 , 𝐼2 is going to change, 𝐼2′ is going to change therefore, 𝐼1 is also
going to change, therefore drop in 𝑟1 is not constant as you change loading. Therefore
applied voltage to this ideal primary winding of this ideal transformer; strictly speaking is
not constant, only consolation is this 𝑟1 is quite small.

Therefore, what people say is this ok, 𝜑𝑚𝑎𝑥 will be approximately constant, because you
have not certainly designed a transformer with high value of 𝑟1 and 𝑟2 , then no one is going
to buy your transformer I mean why, there will be unnecessary power loss in the windings.
Therefore, you must see that very good material is used for example, copper whose
𝑙
resistivity is very low. So, 𝜌 𝑎 the winding resistances are small, so that that way this

assumption that 𝜑𝑚𝑎𝑥 practically remain same from no load to full load is good enough.

Anyway none the less, let us see try to draw the phasor diagram of this one and try to
understand the implication of this. Now, what I will be doing here listen carefully; here I
will start with 𝐸1 and 𝐸2 ok, this voltage and this voltage. Mind you, here is now that 𝑉2
and 𝐸2 will not be same is not, similarly 𝑉1 and 𝐸1 will not be same, because in between
this two sources some 𝐼1 𝑟1 drop here, some 𝐼2 𝑟2 drop here will come, they cannot be same.

But nonetheless, we can do this things. Suppose how to start the phasor diagram drawing,
I will do it like this. Suppose, I draw it will be slightly clumsy, but let it be, but follow my
argument ok; you will draw 𝑉2 first whatever it will be; I draw 𝑉2 arbitrarily, vertically. If
I know 𝑉2, then I can fix up where the 𝐼2 will be is not, because load power factor angle is
known. Suppose, power factor angle of the load is 𝜃2 , 𝑉2 and 𝐼2 , I have drawn with S
closed.

Then I will say, look here your

𝐸2 = 𝑉2 + 𝐼2 𝑟2
follow the logic, the diagrams may be a bit clumsy, but this is what is going to happen.
Suppose, 𝑉2 is known, 𝑉2 and 𝐼2 you draw; then I have to add to 𝑉2 , this 𝐼2 𝑟2 drop. Then I
will add it what is there and 𝐼2 𝑟2 drop will be very small, because 𝑟2 is small, 𝐼2 𝑟2 follow
the logic that is all.

If you do that, then what I am telling you will get your 𝐸2 . The induced voltage in this
ideal transformer, which I will draw by a red line here, this will be your 𝐸2 . It is an ideal
transformer, this is 𝐸2 , then you can confidently draw your 𝐸2 . If number of turns of this
side is higher, then it length of 𝐸1 will be higher than 𝐸2 , 𝐸1 is drawn. So, this will be your
𝐸1 .

And if this is 𝐸1 , this is 𝐸2 , your magnetizing current will be perpendicular to 𝐸1 , 90°

𝐸
lagging. Here is your finite magnetizing current, 𝐼𝑚 is not? I have got 𝐸1 , so 𝑗𝑋1 = 𝐼𝑚 , I
𝑚

get; so, this is the magnetizing current. And it is an ideal transformer with S opened, 𝐼2
was 0 this current was 0 with S closed, 𝐼2′ will appear here, nowhere else it is an ideal
transformer this portion.

So, 𝐼2 is known and 𝐼2′ will be in same phase with that so I will get 𝐼2′ ; whether its length
will be higher than 𝐼2 or not that depends upon the ratios. So, this will be your. .

Student: (Refer Time: 23:33).

It will be less. So, what I will do is I will make these as 𝐼2 , and this is 𝐼2′ , because I have
shown 𝐸1 𝐸1 is higher. So, reflected current must be lower 𝐼2′ , so this is your 𝐼2′ .

Then I will say that this current 𝐼1 = 𝐼2′ + 𝐼𝑚 . So, 𝐼𝑚 is known, 𝐼2′ is known; so I will add
this two, mind you I have shown 𝐼𝑚 slightly higher length it is not so, 𝐼𝑚 is small, anyway
whatever I have drawn. So, this will be your 𝐼1 and if this is 𝐼1 , then I will say your 𝑉1 =
𝐸1 + 𝐼1 𝑟1. So, 𝐸1 is known to this add 𝐼1 𝑟1 parallel to this and from this to this, then
wherever you will end up that will be your 𝑉1.

I think you have got the idea, see life will be will not be so much complicated as we
proceed, but what I am telling this is exactly what, how to draw phasor diagram. Start with
I could start with 𝑉1, but it is better you, because the primary current drawn decided by
load, it is much more easier. So, I will quickly go through this step, so that you can
understand.
Suppose, switch is closed ok, i will clean this and once again redraw. So, I will just I will
(Refer Time: 26:00), but the idea is very important you must keep this in mind. So, I am
redrawing once again very quickly, what I am telling ok, you have close the switch the
circuit is operating, choose 𝑉2, you start with 𝑉2 ok.

(Refer Slide Time: 26:26)

Voltage applied across the load is 𝑉2, then no one can contest me, then the current is

𝑉2
𝐼2 =
𝑍2

𝐼2 suppose lagging power factor load, 𝐼2 will lag here, 𝐼2 will be higher low voltage side;
I assumed I mean, while drawing ok, so 𝐼2 . The moment you know 𝐼2 you can find out 𝐸2 ,
because

𝐸2 = 𝑉2 + 𝐼2 𝑟2

These lengths are small.

And then you get this length to be 𝐸1 , I am simply repeating, because for the first time you
𝑁
are doing this you get 𝐸1 , 𝐸2 sorry. If you get 𝐸2 , this ratio of voltage is 𝑁1 absolutely no
2

𝑉 𝑁 𝑁
doubt; mind you this 𝑉1 may not be 𝑁1 strictly speaking, but this induced voltages are 𝑁1.
2 2 2
So, 𝐸2 is this, then I know 𝐸2 and 𝐸1 are in phase, so I can get 𝐸1 assuming 𝑁1 is greater
than 𝑁2 , it will be this.

Then once I know 𝐼1 , I know the magnetizing current which will be 90° lagging. And this
current is small, let me draw it now correctly 𝐼𝑚 small current; 𝐼𝑚 and once I know 𝐼𝑚 and
𝐼2 is known. So, 𝐼2′ will be this one here will be your 𝐼2′ . So, current drawn from the source
will be 𝐼1 = 𝐼2′ + 𝐼𝑚 , this is your 𝐼1 ; and if this is 𝐼1 , use this one 𝑉1 = 𝐸1 + 𝐼1 𝑟1, whatever
it is, parallel to this 𝐼1 𝑟1 and you get your 𝑉1.

Anyway I am stopping now, but try to understand. So, this will now truly represent, it is
somewhat practical circuit with the finite magnetizing current and winding resistance. In
the next class, we will bring other realities into an ideal transformer to get a somewhat
better picture of the model.

Thank you.