Electrical Machines - I

Prof. Tapas Kumar Bhattacharya

Department of Electrical Engineering
Indian Institute of Technology, Kharagpur

Lecture - 07
Transformer with Multiple Coils

A welcome to this Machine - I course. We are discussing about an ideal transformer, and
in my last lecture I told you how to solve a circuit symbol being several ideal transformers
in the circuit because ideal transformer is so simple. But the problems are interesting.
Actually, what impedance you connect, at the end of the across the secondary coil that
values get transferred and we know how to do it.

(Refer Slide Time: 01:12)

Another interesting problem, I will discuss so that you become really conversant with ideal
transformer. For example, consider this is the core first I will draw, this is necessary to
explain what is happening. Suppose you have this transformer with a, so far two coils I
have considered, ok. What we will do is I connect another third coil, another third coil and
perhaps here also I will connect a one switch and load one.

Another switch this is S1 say S2, this is a S3, there is no S1, S1 may be there. So, this is
suppose 𝑍2 and this is suppose 𝑍3 . See, what I am telling I am slightly greedy, that is I will
energize this with a Voltage 𝑉1 frequency 𝑓 and I want to generate a two levels of Voltage
because same flux will be linking this coil to therefore, the induced Voltage here will be
𝑉 𝑉
also 𝑁1 𝑁3 here the Voltage will be, here the Voltage will be 𝑁1 𝑁2 .
1 1

One thing I will tell you in the primary and secondary Voltages are different, currents are
different KVA are same, that is all fine. But also try to understand this one that is suppose
you have a transformer, of turns 𝑁1 and 𝑁2 it is very useful while solving problems and
𝑉 𝑁
other things. What happens is this 𝑉2 = 𝑁2 that is good, but it is also true that
1 1

𝑉1 𝑉2
=
𝑁1 𝑁2

This equation is very useful, at least I find it is tells you that in a several coupled coils what
𝑉 𝑉 𝑉
remains constant is Voltage per turns. What is 𝑁1 ? Voltage per turn is constant 𝑁1 = 𝑁2 it
1 1 2

has to be.

Therefore, what you do to calculate Voltage induced here, you simply calculate Voltage
per turn then multiply with respective turns to get the Voltages here and there. So, you can
have more than one secondary coils. For example, here so, what is Voltage per turn here
𝑉1 𝑉1
if it is 𝑁1 ? that will remain constant. What will be the Voltage available here? 𝑁3 ,
𝑁1 𝑁1

multiply with actual number of turns.

Here also this Voltage will be 𝑉3 say, 𝑉3 will be equal to Voltage per turn that what where
from I get from the supplied Voltage and its number of turns into 𝑁3 ; Voltage per turn
here, what will be this Voltage? It will be

𝑉1
𝑁
𝑁1 2

Therefore, you can have two sources available to you of different Voltage level and you
can supply two different loads that is what I am trying to tell. Let us take a numerical
numbers, so that things become much more easier.
(Refer Slide Time: 05:58)

And this sort of thing I can then draw it like this. Suppose, you have a transformer, I will
draw now simplified diagrams, and you have two separate secondary coils and there is a
common magnetic circuit. This is how I can represent. And suppose the dots are here, this
is dot, this is dot, this is dot. Suppose, I say this number of turns 𝑁1 is equal to say 200 and
here you apply an AC Voltage of 400 V then what I am telling, immediately you know

400
𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑝𝑒𝑟 𝑡𝑢𝑟𝑛 = =2
200

2 V per turn.

Now, suppose I say this one is 𝑁2 is equal to 50 turns and this one I say 𝑁3 is equal to say
25 turns then I will immediate say this Voltage will be Voltage per turn 2 × 50 you will
get here 100 V. And 2 × 25 you will get here 50 V. Are you with me? Then this is the
how. So, Voltage per turn is a very useful concept in transformers that remains same in
several coupled coils.

Therefore, suppose the problem is like this is 200 V, 𝑁1 400 V is the applied Voltage 50
Hz, these are the dots and if the secondary terminals are open circuited. So, I can now
supply two loads, one load demanding 50 V I will supply, another load demanding 100 V
I will supply. The question is that, when you supply loads how the primary currents how
do I calculate? Ok. It is slightly not that easy problem, ok.
Let me a put me in this way suppose you have a switch here and you have connected a
resistive load here of say 10 Ω, and also I have a switch here and here I will connect a
capacitive load−𝑗5Ω. I have planned to connect these two loads.

So, my problem is find out currents in all the windings and this switch is suppose S1 this
which is S2. Calculate I will indicate the currents as 𝐼2 , this is as 𝐼3 , and this is as 𝐼1 .
Calculate 𝐼1 , 𝐼2 and 𝐼3 . Part 1, when only S1 is closed S2 opened. Part 2, only S2 closed
and S1 opened. And finally, both S1 and S2 are closed. I to have this find out. And let this
transformer be ideal. What does it mean? If you close the switch and with both the switches
are open no current is drawn magnetizing current necessary is 0. So, that is the problem.

First part of the problem is pretty simple; this one S1 is close. So, it is like primary
secondary. This fellow although you will get Voltage here, but no current no
complications; so, first part of the problem will be Voltage per turn is this. So, here you
will get

𝑉2 = 100𝑉

100
𝐼2 = = 10𝐴
10

𝐼3 of course will be 0, S2 is open. And what will be 𝐼1 ?

𝐼2 𝐼2 100
𝐼1 = = = = 2.5𝐴
𝑎 200⁄ 4
50

And let us also draw the phasor diagram under this ideal transformer. What is the phasor
diagram? Phasor diagram will be like this. This is your 𝑉1, , 400 V, this will be your 𝑉2
100 V. All the induced Voltage will be in phase. 𝑉2 = 100𝑉 and to a slightly practical I
will this is one-forth, so it will be much higher this length. This is 𝑉1 = 400𝑉

All Voltages are in phase 𝑉1, 𝑉2, 𝐸1 , 𝐸2 , there is no distinction between 𝐸1 , 𝐸2 , 𝐸3 , 𝐸4 .

And what will be 𝐸3 will be there, 𝐸3 is how much? 50 V so, 𝐸3 𝑉3 will be there. 𝑉3 =
50𝑉 these will be the Voltages. Where is your flux? Flux is along this line. Magnetizing
current is also along this line, but that is 0 that is the thing.
Now, if S1 is close there will be 𝐼2 , and this 𝐼2 will be in phase with that of a because it is
resistive load. So, the current 𝐼2 will be I have calculated it 10 A it will be in some another
scale, 𝐼2 will be like this. This 𝐼2 will try to upset them in a balance therefore, primary has
𝐼2
to draw extra current. And that value is 𝐼2 ′ = 𝑎

So, what will be the primary current? And this is the H V side. So, H V side current will
400
be less, ratio is = 4,so it should be divided by 4. So, 𝐼2 = 10𝐴. So, primary current
100

𝐼1 = 2.5 𝐴. 𝐼3 nothing is there, your problem is over. If you wish you can calculate the
equivalent impedance in by the source and you can verify it is equal to 𝑎2 × 10. This part
is very simple.

Second part; in the second part this current 𝐼3 , S1 is opened only S2 is closed. So,
everything remains same. So, I will better not to re-draw here only thing what I will do
part 2, I am solving. So, Voltages will be as it is induced, but the only thing is now 𝐼2 is
not there although 𝑉2 will be there open circuited and 𝐼3 is present. How much is 𝐼3 ?

50
𝐼3 = 10𝐴
5

but this current will lead 𝑉3 this is your 𝑉3 ; is not.

(Refer Slide Time: 18:00)

It will lead 𝑉3 by. 90°.

And what is the magnitude of this current? So, second part 𝐼3 Voltages remain same it will
be suppose I take Voltage on reference

50∠0°
𝐼3 = = 10∠90°
5∠ − 90°

Therefore, 𝐼3 will be here The moment 𝐼3 flows then will be 𝐼3 ′ and that will be equal to
your 𝐼1 and this is LV side this is HV side, so current will be less. What is the ratio of turns
200
here? =8
25

So, your 𝐼1 and they will be in phase why because they have to balance of the mmfs.

10
𝐼1 = 𝐼3′ = ∠90°
8

Please try to follow me.

So, far I was in the previous examples just telling impedance values to tell you about the
rated current. But now I am in a position to go slightly in advance stage that, impedance
may be complex and how to calculate the currents. Had it been RL circuit the phase angle
of the impedance, I could also take into account. But where ever is your secondary current
the reflected current because of that must be there. So, this is the thing.

Now, the last part which is slightly tricky for this what I will do is this. I will copy this, so
that it is easier to talk.
(Refer Slide Time: 20:55)

So, now copy go to the next page, paste it. And this one may be deleted. This part you just
try to see how it is to be done. I am sorry for taking some time, they will be edit it.

Now, listen carefully what I am telling. So, this is the thing now. Now, both the coils
switches are closed what do I do, that is the part 3. Both S2 and S3 are closed. Therefore,
what is going to happen? Thing is that the clue to this problem, once again is that
fundamental thing. What is that fundamental thing? No, matter what is the positions of the
switch either closed or open flux in the core of the transformer cannot change. It has to
remain same. Why? Because KVL is to be satisfied on the primary side either S2 is closed,
S3 is closed or both of them are closed it does not matter this is S3 actually and this is or
S1 S2 I told it whatever it is, both the switches are closed.

Then once again what I will do is this first I will calculating Voltage per turn, Voltage per
turn which is equal 2 which I have already calculated and this Voltage 100 open circuit
Voltage has got. And then I will draw the phasor diagram. Phasor diagram is this one. This
is the 400 V, 𝑉1 400 V; then your 100 V and 50 V. So, this is suppose 100 V 𝑉2. Scale you
forget and this is your 50 V, 𝑉3 = 50𝑉. All are in phase. And suppose this I take as
reference phasor all Voltages are magnitude angle 0 °, 0 °, 0 ° and so on.

Now, you see that this is resistive now both 𝐼2 and 𝐼3 exist.

100∠0°
𝐼2 = = 10𝐴
10∠0°
 50∠0°
𝐼3 = = 10∠90°𝐴
5∠ − 90°

both 𝐼2 and 𝐼3 are present, ok. First, I will draw 𝐼2 and 𝐼3 . So, 𝐼2 will be in phase with 𝑉2
because it is resistive load. Suppose this is 𝐼2 phasor and 𝐼3 will be leading 𝑉2 by 90 °, so
of same magnitude mind you I have taken the numbers such that they become, so it will
be 90°.

Now, the big question is what is 𝐼1 . Two ways we can, I will tell you much simpler method,
but I will go by the basic rule. Basic rule is because these two coils carry current extra
mmf is introduced into the transformer that is namely 𝑁2 𝐼2 in this direction 𝑁3 𝐼3 .
Therefore, these extra current drawn from the supply should be such that we it will nullify
both these mmfs. It has to; because applied Voltage is fixed here KVL is to be satisfied.
Therefore, what I will do? I will calculate 𝐼2 ′ , because of this there will be an 𝐼2 ′ , because
of this there will be an 𝐼3 ′ and then these 𝐼2 ′ and 𝐼3 ′ I will add vectorially phasor by phasor
sum.

So, what is this is 𝐼2 , 𝐼2 is how much?

𝐼2 = 10∠0°

𝐼3 = 10∠90°

On the H V side, I am calculating current. So, 𝐼2 ′ will be lesser current It will be how
𝐼2
much? therefore,
𝑎

Student: (Refer Time: 27:51).

10
𝐼2 ′ = ∠0° = 2.5∠0°A
4

Similarly,

10∠90°
𝐼3 ′ = = 1.25∠90°A
8

So, this is 𝐼2 ′ and this is𝐼3 ′ And then I will say

𝐼1 = 𝐼2 ′ + 𝐼3 ′
This is one complete this problem you must go through very carefully what are the steps
are because there may be situations you from the same source you want to get on the same
core, you connect a several coils like this 2 coils, may be 3 coils another coil is connected,
get different level of Voltages. Of course, in a two winding transformer only two windings
will be present. But if you do this exercise it will further strengthen your understanding of
reflected current. For this there will be reflected current here. This current divided by this
turns ratio. For this current there will be reflected current here. What will be the reflected
current? These actual current divided by turns ratio between these two and so on.

Now, this I will not do, but I will request you to think, ok. These you have found out, 𝐼1
you will find out. Once you find out 𝐼1 therefore, mathematically you will say equivalent
circuit of all these things across the supply will be this 400 V supply and it is delivering
this 𝐼1 therefore, impedance seen by the transformer 𝑍 ′ will be simply this Voltage by this
current, are you getting. 𝐼1 you have to calculate it will have some magnitude and some
angle.

This Voltage divided by this current will then give you the impedance seen by the
transformer. I will request you to think about it that, this is suppose in general 𝑍2 , this is
in general 𝑍3 , are you getting. That ok, supply sees what impedance what should be
equivalent circuit, I have to draw with respect to the supply side. How this R will appear
here, how this Z will appear here that you think in the next class. Of course, I will discuss
about that that is also very interesting.

So, I hope you have understood that if it is ideal transformer magnetizing current is 0 and
everything is so simple. If any coil secondary coil carriers current, primary coil will
immediately draw reflected current corresponding to that winding current what will be that
reflected current. If it is 𝐼2 on the secondary coil it will be 𝐼2 ′ . And we have also seen that
𝐼 𝑁
how to take any general impedance into account. So, 𝐼2 ′ = 𝑎2. What is 𝑎? 𝑎 = 𝑁1 .
2

Similarly, if there is another coil supplying another load wound on the same magnetic core,
if you know 𝐼3 there will be 𝐼3 ′ here, such that the flux remains same. And how to calculate
𝐼3 ′ ? It will be 𝐼3 by turns ratio between the primary coil and this new another coil that is
𝑁1
whatever it is now. So, these things if you take in to account and coupled with this
𝑁3

phasor diagram; the comfort zone is all the Voltages are in safe phase.
So, you have you can take them on reference. And then first calculate the secondary coil
currents, and then calculate reflected currents for each of these components add them up
to get the current which will be draw from the primary side. And then the equivalent
𝑉1
impedance seen by the source can then be calculated easily without because I will do
𝐼1

and say that, these big circuit to the supply is nothing, but this 400 V supply across which
𝑉
you have connected an impedance of 𝐼1 .
1

But what I am asking you to do, in terms of 𝑍2 and 𝑍3 find out that equivalent impedance.
It will be some 𝑎2 𝑍2 this turns ratio square into 𝑍2 , this turns ratio square into 𝑍3 and so
on. You think about it. We will discuss it in the next class.

Thank you.